[
  {
    "id": 30001,
    "subject": "Mathematics (Multi-modal)",
    "question": "We call a prime number $p$ cute if there exists a prime $q$ so that both $pq - 2$ and $pq + 2$ are also primes. We call $p$ wonderful if both $p$ and $p+2$ are cute primes. Find all wonderful numbers.",
    "options": [],
    "answer": "3, 5",
    "solution": "Let $p$ be a cute prime. The numbers $pq - 2$, $pq$ and $pq + 2$ give all the possible remainders modulo $3$, so one of them must be divisible by $3$. If $3 \\mid pq - 2$, then $pq - 2 = 3$ and $pq = 5$, which is impossible, as $p$, $q$ are primes. Similarly if $3 \\mid pq + 2$, then $pq + 2 = 3$ and $pq = 1$, which is also impossible. So $3 \\mid pq$, which means that either $p = 3$ or $q = 3$. Given that in the case $p = 3$ we can have $q = 3$, we have shown that $p$ is cute iff both $3p - 2$ and $3p + 2$ are primes.\n\nTherefore $p$ is wonderful if $p$, $p + 2$, $3p - 2$, $3p + 2$, $3p + 4$ and $3p + 8$ are all primes. We know that the numbers $3p - 2$, $3p$, $3p + 2$, $3p + 4$ and $3p + 6$ give all of the possible remainders modulo $5$, so one of those numbers is divisible by $5$. If it's $3p - 2$, $3p + 2$ or $3p + 4$, then due to primality it must be equal to $5$, but then $p$ is not prime. If it's $3p$ or $3p + 6$, then either $5 \\mid p$ or $5 \\mid p + 2$ respectively. Thus $p = 3$ or $p = 5$. We can verify that both work.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30002,
    "subject": "Mathematics (Multi-modal)",
    "question": "A function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfies for all real $x$ and $y$ the equation $2f(x)(f(y))^2 + y^2f(-x|y|) = f(xy^2)$. Find all possible values of $f(1)$.",
    "options": [],
    "answer": "-1, 0, 1",
    "solution": "Denote $f(1) = a$ and $f(-1) = b$. Choosing $x = y = 1$ and $x = y = -1$ yields the system of equations\n$$\n\\begin{cases} 2a^3 + b = a \\\\ 2b^3 + a = b \\end{cases}\n$$\nAdding the equations and cancelling equal terms, we get $2b^3 = -2a^3$. Therefore $b = -a$. Substituting this into the first equation yields $2a^3 - 2a = 0$. So $f(1)$ is one of the numbers $-1$, $0$, $1$. These values can be obtained by the functions $f(x) = -x|x|$, $f(x) = 0$ and $f(x) = x|x|$ respectively, which satisfy the equations.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be the orthocenter of an acute triangle $ABC$. Let $M$ be the midpoint of $BC$. Let $A'$ and $H'$ be the reflections of the points $A$ and $H$ across the point $M$. Prove that the points $B$ and $C$ and the reflections of $A'$ over the lines $BH'$ and $CH'$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $O$ be the circumcenter of $ABC$ and $G$ the antipode of $A$ (Fig. 11). Then $BG \\perp AB$ and $CG \\perp AC$. As $CH \\perp AB$ and $BH \\perp AC$, this means that $CG \\parallel BH$ and $BG \\parallel CH$. So $BHCG$ is a parallelogram. Therefore $M$, the midpoint of $BC$, is also the midpoint of $HG$. Consequently $H' = G$. Hence $BH' \\perp AB$ and $CH' \\perp AC$. Also, by the choice of $A'$, $ABA'C$ is a parallelogram, meaning $A'B \\parallel AC$ and $A'C \\parallel AB$. Thus $A'B \\perp CH'$ and $A'C \\perp BH'$, which means that the reflections of $A'$ over the lines $BH'$ and $CH'$ lie on lines $A'C$ and $A'B$ respectively; denote them by $X$ and $Y$ (Fig. 12). Then $CX \\parallel AB$ and $BX = BA' = AC$, analogously also $BY \\parallel AC$ and $CY = CA' = AB$. Therefore $ABCX$ and $ABCY$ are isosceles trapezoids. Isosceles trapezoids are cyclic quadrilaterals, thus both $X$ and $Y$ must lie on the circumcircle of $ABC$. The desired claim follows.\n\n![](attached_image_2.png)\nLike in the previous solution, notice that $ABA'C$ is a parallelogram. By symmetry with respect to $M$ notice that triangles $ABC$ and $A'CB$ are congruent and that $H'$ is the orthocenter of $A'CB$.\nLet $X$ and $Y$ be the reflections of $A'$ over $BH'$ and $CH'$ respectively (Fig. 13). Then $A'X \\perp BH'$ and $A'Y \\perp CH'$ and since $H'$ is the orthocenter of $A'CB$, we also have $A'C \\perp BH'$ and $A'B \\perp CH'$. So $X$ and $Y$ lie on the lines $A'C$ and $A'B$ respectively.\nThe choice of $X$ and $Y$ implies that the triangles $A'BX$ and $A'CY$ are isosceles. Therefore $\\angle A'XB = \\angle A'YC = \\angle BA'C$. The desired claim follows.\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n > 2$ be positive integers. In each cell of an $m \\times n$ grid there is a lamp, which can either be turned on or off. In one switch, we can change the state of five lamps, which are placed in cells forming a cross (see the diagram). Initially all the lamps are turned off. How many possible arrangements of lamps can be formed with these switches? Arrangements obtained from each other by rotations and reflections are considered different.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "2^{(m-2)(n-2)}",
    "solution": "Answer: $2^{(m-2)(n-2)}$.\n\nFor convenience, assume that the switch for any cross is located at the central cell of that cross. Then switches are located in all cells not on the edge of the grid. There are $(m-2)(n-2)$ such cells.\n\nThe final state of each lamp is determined by whether there have been an even or odd number of switches affecting it, meaning we can cancel out all double presses of the same switch. So any possible arrangement can be achieved by using each switch at most once. There are $2^{(m-2)(n-2)}$ possibilities for the choice of switches to use.\n\nConsider any two combinations differing by at least one switch used. Let $L$ be the leftmost square, whose switch is used in one combination, but not in the other (if there are several options, choose at random). Then the left neighbour $K$ of $L$ will have a different end state for these arrangements. So different combinations yield different lamp arrangements. Therefore there are exactly $2^{(m-2)(n-2)}$ possible arrangements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mother wants to divide a cake of triangular shape between three kids. She makes a straight cut from one vertex to the midpoint of the opposite side and then another straight cut from another vertex to the midpoint of the opposite side. She gives the piece of quadrilateral shape to Anna, the triangular piece opposite to it to Berta and the remaining two triangular pieces to Clara. Who gets the largest part of cake?",
    "options": [],
    "answer": "All receive equal amounts.",
    "solution": "*Answer:* all kids get the same amount.\n\nLet $S$ be the area of the initial triangle, $S_A$ be the area of the quadrilateral piece, $S_B$ be the area of Berta's triangle, and $S_1$ and $S_2$ be the areas of the remaining two triangles (Fig. 14). Then $S_1 + S_B = S_2 + S_B = \\frac{1}{2}S$ (a common altitude while the ratio of the corresponding bases being $\\frac{1}{2}$) and $S_B = 2S_1$ (for similar reasons). Thus $S_1 = S_2 = \\frac{1}{6}S$ and $S_B = \\frac{1}{3}S$, whence also $S_1 + S_2 = \\frac{1}{3}S$ and $S_A = \\frac{1}{3}S$.\n\n![](attached_image_1.png)\nSuppose that mother makes one more cut from the third vertex to the midpoint of the opposite side. As all medians of a triangle meet in one point, the new cut divides Anna's and Berta's pieces into two parts while not touching Clara's pieces. So every child gets exactly two pieces. We show that medians of a triangle divide the triangle into six parts of equal area; this implies that all kids get the same amount of cake. Let the triangle be $ABC$, its medians be $AD, BE$ and $CF$, and the centroid be $G$ (Fig. 15). The length of the side $BD$ of the triangle $BGD$ is $\\frac{1}{2}$ of the length of the side $BC$ of the triangle $ABC$, the length of the corresponding altitude in the triangle $BGD$ is $\\frac{1}{3}$ of the length of the corresponding altitude in the triangle $ABC$ (since $|AD| = 3|GD|$, the perpendicular drawn from the point $A$ to the line $BC$ is 3 times longer than the perpendicular drawn from the point $G$ to the same line). Thus the area of the triangle $BGD$ equals $\\frac{1}{6}$ of the area of the triangle $ABC$. The same holds for other pieces. Consequently, all pieces have the same area.\n\n![](attached_image_2.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of integers $(x, y)$ that satisfy the equation $y^4 = x(2x^2 + y)^3$.",
    "options": [],
    "answer": "(x, y) = (z^3(z-1)/2, z^6(z-1)/2) for all integers z",
    "solution": "*Answer:* $(\\frac{z^3(z-1)}{2}, \\frac{z^6(z-1)}{2})$ where $z$ is arbitrary integer.\n\nSolution:\nIf $x = 0$ then $y = 0$. We now assume that $x \\neq 0$.\nLet $d = \\gcd(x, y) > 0$ and $x = da, y = db$. Dividing the sides of the equation by $d^4$, we get $b^4 = a(2a^2d + b)^3$. Thus $a \\mid b^4$. Since $a$ and $b$ are coprime, the only possibility is $|a| = 1$ and the equation reduces to\n$$\nb^4 = \\pm (2d + b)^3.\n$$\nAs the same integer is simultaneously a cube and a fourth power, it is a twelfth power of some integer, yielding $b = z^3$ for some integer $z$. Substituting $b = z^3$ into the equation and taking cube root gives $z^4 = \\pm(2d + z^3)$ or, equivalently, $\\pm z^4 = 2d + z^3$. Since $-z^4 \\le -|z^3| \\le z^3 < 2d + z^3$, the minus sign is not possible, therefore $d = \\frac{z^4-z^3}{2} = \\frac{z^3(z-1)}{2}$. We conclude that $x = \\frac{z^3(z-1)}{2}$ and $y = \\frac{z^6(z-1)}{2}$, where $z$ is an arbitrary integer different from $0$ and $1$. In this case $d > 0$ and the solution satisfies the original equation. If $z = 0$ or $z = 1$, we get the initial solution $(0, 0)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least possible sum of 2021 terms of the sequence $a_1, a_2, a_3, \\dots$, where $a_1 = 0$, $a_2 = a_3 = 1$ and $a_{i+j} > a_i + a_j$ for every $i, j \\ge 2$.",
    "options": [],
    "answer": "2040200",
    "solution": "**Answer:** $2 \\cdot 1010^2$.\n\nWe show that the least sum arises in the case of the sequence $0, 1, 1, 3, 3, 5, 5, \\dots$ ($a_{2i} = a_{2i+1} = 2i - 1$ for every $i \\ge 1$). Firstly, we show that this sequence meets the conditions of the problem. Indeed, if $j$ and $k$ are of the same parity then $a_{j+k} = j + k - 1 > (j - 1) + (k - 1) \\ge a_j + a_k$. If $j$ and $k$ have different parities then letting, w.l.o.g., $j$ be even, we obtain $a_{j+k} = j + k - 2 > (j - 1) + (k - 2) = a_j + a_k$.\n\nWe prove now that if $a_1, a_2, a_3, \\dots$ is an arbitrary sequence of integers satisfying the conditions of the problems then $a_{2n} \\ge 2n - 1$ if $a_{2n+1} \\ge 2n - 1$ for any positive integer $n$. We proceed by induction on $n$. The base cases $a_2 \\ge 1$ and $a_3 \\ge 1$ hold. If $a_{2n} \\ge 2n - 1$ then, from the conditions of the problem,\n$$\na_{2n+2} \\ge a_{2n} + a_2 + 1 \\ge 2n - 1 + 1 + 1 = 2(n + 1) - 1,\n$$\nwhich proves the induction step for even indices. The same holds for odd indices: If $a_{2n+1} \\ge 2n - 1$ then\n$$\na_{2n+3} \\ge a_{2n+1} + a_2 + 1 \\ge 2n - 1 + 1 + 1 = 2(n + 1) - 1.\n$$\nThus the least sum of 2021 terms is $2(1 + 3 + \\dots + 2019)$, i.e., $2 \\cdot 1010^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ satisfies $AB = AC$. Medians $AD$ and $BE$ intersect at $G$. Let $P$ be the midpoint of the line segment $GE$.\n\na. Prove that if $GP = GD$ then the quadrilateral $CEPD$ is cyclic.\n\nb. Does it hold that if the quadrilateral $CEPD$ is cyclic then $GP = GD$?",
    "options": [],
    "answer": "Yes; GP = GD if and only if quadrilateral CEPD is cyclic.",
    "solution": "Let $BD = DC = x$ and $GP = PE = y$. Then $BG = 2 \\cdot 2y = 4y$, $BP = 4y + y = 5y$ and $BE = 4y + 2y = 6y$. Thus $GP = GD$ if and only if $y^2 = BG^2 - BD^2 = 16y^2 - x^2$ or, equivalently, $x^2 = 15y^2$. The quadrilateral $CEPD$ is cyclic if and only if $BP \\cdot BE = BD \\cdot BC$, i.e., $5y \\cdot 6y = x \\cdot 2x$. The latter simplifies to $x^2 = 15y^2$, too. Hence $GP = GD$ if and only if the quadrilateral $CEPD$ is cyclic, which solves both parts of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Little Juku writes all integers from $1$ to $n$ on a blackboard, but as he does not know the digit $4$ yet, he skips all numbers that contain $4$. Juku's sister Mari erases two numbers on the blackboard and writes the absolute value of the difference of these numbers on the blackboard. Then Mari again erases two numbers on the blackboard and writes the absolute value of their difference on the blackboard, etc. Can it happen after a finite number of such steps that there are all integers from $1$ to $n$ that contain the digit $4$ and only these on the blackboard, each one exactly once, if\n\na. $n = 2021$;\n\nb. $n = 10000$?",
    "options": [],
    "answer": "(a) No; (b) No",
    "solution": "*Answer:* (a) No; (b) No.\n\na. The difference and the sum of two integers have equal parity, likewise are an integer and its absolute value of equal parity. Thus Mari's every step keeps the parity of the sum of all numbers on the blackboard unchanged. Among $1, 2, \\ldots, 2021$, there are $1011$ odd numbers. As $1011$ is odd itself, the sum of all numbers $1, 2, \\ldots, 2021$ is odd. This implies that the sum of all integers from $1$ to $2021$ that contain $4$ and the sum of integers from $1$ to $2021$ that do not contain $4$ have different parities. Hence the set of all natural numbers from $1$ to $2021$ that contain $4$ can never appear on the blackboard.\n\nb. Let there be $a$ positive integers from $1$ to $10000$ that do not contain $4$. Then there are $10000 - a$ positive integers from $1$ to $10000$ that contain $4$. As each Mari's step decreases the number of numbers on the blackboard by $1$, she should make $a - (10000 - a)$ or, equivalently, $2a - 10000$ steps to reach the desired state. As only one new number can appear at each step, introducing $10000 - a$ new numbers assumes $2a - 10000 \\ge 10000 - a$ which is equivalent to $a \\ge \\frac{20000}{3}$. But there are only $9^4$ numbers among the first $10000$ positive integers that do not contain $4$; since $9^4 = 6561 < \\frac{20000}{3}$, achieving the desired state is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $n$ such that across all representations $n = ab$, the digits $0, 1, \\dots, 9$ are all present as the final digit of $a^b$ at least once?",
    "options": [],
    "answer": "No",
    "solution": "**Answer:** No.\n\nIf $n$ is odd, both $a$ and $b$ will always be odd, but then $a^b$ will never end with a $0$. If $n$ is even, then $a$ or $b$ will be even, so $a^b$ will be even or a perfect square. In either case, it will never end in a $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the value of\n$$\n\\frac{1 \\cdot 3}{3 \\cdot 5} + \\frac{2 \\cdot 4}{5 \\cdot 7} + \\frac{3 \\cdot 5}{7 \\cdot 9} + \\frac{4 \\cdot 6}{9 \\cdot 11} + \\dots + \\frac{1009 \\cdot 1011}{2019 \\cdot 2021}\n$$",
    "options": [],
    "answer": "509545/2021",
    "solution": "**Answer:** $\\frac{505 \\cdot 1009}{2021} = \\frac{509545}{2021} = 252 + \\frac{253}{2021}$.\n\nThe sum consists of $1009$ terms, where the $i$-th term is of the form $\\frac{i(i+2)}{(2i+1)(2i+3)}$. Let $s$ be the desired sum. Notice that\n$$\n\\frac{i(i+2)}{(2i+1)(2i+3)} = \\frac{1}{4} - \\frac{3}{4} \\cdot \\frac{1}{(2i+1)(2i+3)} = \\frac{1}{4} - \\frac{3}{8} \\cdot \\left( \\frac{1}{2i+1} - \\frac{1}{2i+3} \\right)\n$$\nTherefore\n$$\n\\begin{aligned}\ns &= 1009 \\cdot \\frac{1}{4} - \\frac{3}{8} \\cdot \\left( \\left( \\frac{1}{3} - \\frac{1}{5} \\right) + \\left( \\frac{1}{5} - \\frac{1}{7} \\right) + \\dots + \\left( \\frac{1}{2019} - \\frac{1}{2021} \\right) \\right) \\\\\n&= \\frac{1009}{4} - \\frac{3}{8} \\cdot \\left( \\frac{1}{3} - \\frac{1}{2021} \\right) = \\frac{1009}{4} - \\frac{3}{8} \\cdot \\frac{2021-3}{3 \\cdot 2021} = \\frac{1009}{4} - \\frac{2018}{8 \\cdot 2021} \\\\\n&= \\frac{1009}{4} - \\frac{1009}{4 \\cdot 2021} = \\frac{1009}{4} \\cdot \\left( 1 - \\frac{1}{2021} \\right) = \\frac{1009}{4} \\cdot \\frac{2020}{2021} = \\frac{505 \\cdot 1009}{2021} = \\frac{509545}{2021}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be a circle with center $O$ and diameter $AB$. A circle with center $B$ intersects $\\omega$ at $C$ and $AB$ at $D$. The line $CD$ intersects $\\omega$ at the point $E$ ($E \\neq C$). The intersection of lines $OE$ and $BC$ is $F$.\n\na. Prove that the triangle $OBF$ is isosceles.\n\nb. Find the ratio $\\frac{FB}{BD}$, given that $D$ is the midpoint of $OB$.",
    "options": [],
    "answer": "4",
    "solution": "(a) Let $\\angle BCD = \\alpha$, then the equal radii $BC = BD$ give us $\\angle BDC = \\alpha$ (Fig. 2) and $\\angle CBD = 180^\\circ - 2\\alpha$. Then $\\angle OBF = \\angle CBD = 180^\\circ - 2\\alpha$. But on the other hand $\\angle BOE = 2\\angle BCE = 2\\angle BCD = 2\\alpha$ and therefore $\\angle BOF = 180^\\circ - 2\\alpha$. Since $\\angle OBF = \\angle BOF$, the triangle $OBF$ is isosceles.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n(b) Due to equal radii the triangle $CBO$ is isosceles (Fig. 3). Furthermore, it is similar to $OBF$ as they share the angle at $B$. We have $OB = 2BD = 2BC$, so by the similarity also $FB = 2OB$. Therefore $FB = 4BD$ and $\\frac{FB}{BD} = 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30013,
    "subject": "Mathematics (Multi-modal)",
    "question": "A field is a $2020 \\times 2021$ grid with a positive integer written into each cell, such that no number repeats in any row or column. An onion consists of $4$ consecutive cells in a row or column, whose numbers add up to exactly $4 \\cdot 2021$. Find the largest possible number of onions on the field.",
    "options": [],
    "answer": "1009 * 4041",
    "solution": "*Answer:* $1009 \\cdot 4041$.\n\nIn a column of length $2020$ there are $2017$ potential onions. However there cannot be two consecutive onions, as they share $3$ cells, meaning their fourth numbers would have to be equal. So a column can have at most $1009$ onions. Analogously a row of length $2021$ can also have at most $1009$ onions. As there are $2020 + 2021 = 4041$ rows and columns in total, there can be at most $1009 \\cdot 4041$ onions.\n\nWe will show there exists a field with $1009 \\cdot 4041$ onions. We will first construct a $2020 \\times 2020$ grid, choosing the first two rows to be\n\n1, -1,  2, -2,  3, -3, ..., 1009, -1009, 1010, -1010,\n-1,   1, -2,   2, -3,   3, ..., -1009,   1009, -1010,   1010.\n\nThe next two rows will be shifted the cells to the left:\n\n2, -2,  3, -3, ..., 1009, -1009, 1010, -1010,   1, -1,\n-2,   2, -3,   3, ..., -1009,   1009, -1010,  1010, -1,   1.\n\nSimilarly we construct the other rows. Finally we add the rightmost column consisting of the numbers\n$1011, -1011, 1012, -1012, \\dots, 2019, -2019, 2020, -2020$.\n\nWe obtain the desired field by adding $2 \\cdot 2021$ to all negative numbers in the grid.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle where $AB = AC$. Let $E$ be the foot of its altitude from the vertex $B$. Given that either $BEA$ or $BEC$ is an isosceles triangle, find all possibilities of the size of the angle at the vertex $A$ of the triangle $ABC$.",
    "options": [],
    "answer": "45°, 90°, 135°",
    "solution": "*Answer:* $45^\\circ$, $90^\\circ$, $135^\\circ$.\n\nLet $\\alpha = \\angle BAC$. Both triangles $BEA$ and $BEC$ have right angle at vertex $E$. Hence these triangles can be isosceles only if their other angles have size $45^\\circ$.\n\nSuppose that $BEC$ is isosceles (see figure below).\nThen $\\angle BCE = 45^\\circ$. Since $\\angle BCE = \\angle BCA = \\frac{180^\\circ - \\alpha}{2}$, we have $\\alpha = 180^\\circ - 2 \\cdot 45^\\circ = 90^\\circ$.\n\nSuppose now that $BEA$ is isosceles. Then $\\angle BAE = 45^\\circ$.\nIf $E$ lies on the line segment $AC$ (see figure below) then $\\angle BAE = \\angle BAC = \\alpha$, implying $\\alpha = 45^\\circ$.\nIf $E$ lies outside the line segment $AC$ (see figure below) then $\\angle BAE = 180^\\circ - \\angle BAC = 180^\\circ - \\alpha$, implying $\\alpha = 180^\\circ - 45^\\circ = 135^\\circ$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\nLet $\\alpha = \\angle BAC$. Then $\\angle ABC = \\angle ACB = \\frac{180^\\circ - \\alpha}{2}$ and $\\angle EBC = 90^\\circ - \\frac{180^\\circ - \\alpha}{2} = \\frac{\\alpha}{2}$.\nBoth triangles $BEA$ and $BEC$ have right angle at vertex $E$. Hence these triangles can be isosceles only if their other two angles have equal size.\n\nSuppose that $BEC$ is isosceles. Then $\\frac{180^\\circ - \\alpha}{2} = \\frac{\\alpha}{2}$, implying $\\alpha = 90^\\circ$.\n\nSuppose now that $BEA$ is isosceles. If the triangle $ABC$ is acute, we must have $\\angle BAE = \\alpha$ and $\\angle ABE = \\frac{180^\\circ - \\alpha}{2} - \\frac{\\alpha}{2} = 90^\\circ - \\alpha$. Thus $\\alpha = 90^\\circ - \\alpha$, implying $\\alpha = 45^\\circ$.\nIf the triangle $ABC$ is obtuse then $\\angle BAE = 180^\\circ - \\alpha$ and $\\angle ABE = \\frac{\\alpha}{2} - \\frac{180^\\circ - \\alpha}{2} = \\alpha - 90^\\circ$. Hence $180^\\circ - \\alpha = \\alpha - 90^\\circ$, implying $\\alpha = 135^\\circ$.\nThe triangle $ABC$ is not right as, otherwise, the triangle $BEA$ would have two right angles.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist integers $x$ and $y$ such that:\n\na. $x^2 + (x + 1)^2 + (x + 2)^2 = y^2$?\n\nb. $x^2 + (x + 1)^2 + (x + 2)^2 + (x + 3)^2 = y^2$?",
    "options": [],
    "answer": "(a) No; (b) No.",
    "solution": "Answer: (a) No; (b) No.\n\na. The integers $x$, $x+1$, $x+2$ are congruent to $0$, $1$, $2$ modulo $3$ in some order. The squares of these integers are congruent to $0$, $1$, $1$ modulo $3$, respectively. Hence the l.h.s. of the equation is congruent to $2$ while the r.h.s. is congruent to $0$ or $1$ modulo $3$. Thus the equality cannot hold.\n\nb. Among the integers $x$, $x+1$, $x+2$, $x+3$, there are two even numbers and two odd numbers. The squares of even numbers are divisible by $4$ while the squares of odd numbers are congruent to $1$ modulo $4$. Thus the l.h.s. of the equation is congruent to $2$ while the r.h.s. is congruent to $0$ or $1$ modulo $4$. Hence the equality cannot hold.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ which satisfy the inequality $f(x) + f(x+y) \\le f(xy) + f(y)$ for all real numbers $x, y$.",
    "options": [],
    "answer": "All constant functions: f(x) = c for all real x, where c is any real constant.",
    "solution": "**Answer:** All constant functions $f(x) = c$ where $c$ is arbitrary real number.\n\nDenote the given inequality by $V(x,y)$. Then $V(x,0)$ together with simplification gives\n$$\nf(x) \\le f(0) \\qquad (3)\n$$\nfor every real number $x$. On the other hand, adding $V(x,y)$ and $V(y,x)$ gives $f(x+y) \\le f(xy)$, where taking $y = -x$ leads to $f(0) \\le f(-x^2)$. Along with (3) this implies that\n$$\nf(z) = f(0) \\qquad (4)\n$$\nfor any non-positive real number $z$. Now $V(z, -1)$ with non-positive $z$, simplified by (4), gives $f(0) \\le f(-z)$. The latter along with (3) implies $f(x) = f(0)$ for all positive real numbers $x$.\nThus $f(x) = f(0)$ for every real number $x$, i.e., $f$ is a constant function. All constant functions clearly satisfy the conditions of the problem.\nDenote the given inequality by $V(x,y)$.\nFirstly, note that $V(x, 1)$ along with simplification leads to $f(x+1) \\le f(1)$. As $x+1$ takes all real values, the function $f$ obtains its maximum value at 1. Secondly, note that $V(1, y)$ leads to $f(1) + f(y+1) \\le 2f(y)$. Along with the inequality $f(y) \\le f(1)$ obtained above, this implies $f(y+1) \\le f(y)$\nfor all real numbers $y$. By applying the latter inequality to both $y = 0$ and $y = 1$ and taking into account that $f(1)$ is the maximum value of $f$, one gets $f(1) = f(0) = f(-1)$.\nThirdly, note that $V(-1, y)$ gives $f(-1) + f(y - 1) \\le f(-y) + f(y)$. As $f(y) \\le f(y - 1)$ by the above, the inequality $f(-1) \\le f(-y)$ must hold for every real number $y$. Since $-y$ obtains all real values, the function $f$ obtains its minimum value at $-1$. As $f(1) = f(-1)$, the maximum and minimum value coincide which means that $f$ is a constant function.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a fixed prime number. Juku and Miku play the following game. One of the players chooses a natural number $a$ such that $a > 1$ and $a$ is not divisible by $p$, his opponent chooses any natural number $n$ such that $n > 1$. Miku wins if the natural number written as $n$ ones in the positional numeral system with radix $a$ is divisible by $p$, otherwise Juku wins. Which player has a winning strategy if:\n\na. Juku chooses the number $a$, tells it to Miku and then Miku chooses the number $n$;\n\nb. Juku chooses the number $n$, tells it to Miku and then Miku chooses the number $a$?",
    "options": [],
    "answer": "a) Miku; b) Juku",
    "solution": "*Answer*: (a) Miku; (b) Juku.\n\nThe positional representation with radix $a$ consisting of $n$ ones denotes the sum $a^{n-1} + a^{n-2} + \\dots + a + 1$ which equals $\\frac{a^n-1}{a-1}$.\n\na. Let $a \\equiv 1 \\pmod{p}$. If $p > 2$ or $a \\equiv 1 \\pmod{4}$ then, by the lifting-the-exponent lemma, the exponent of the prime $p$ in the canonical representation of $\\frac{a^n-1}{a-1}$ equals that in the canonical representation of $n$. Thus to win, Miku may choose any number $n$ that is divisible by $p$. If $p = 2$ and $a \\equiv -1 \\pmod{4}$ then the exponent of the prime 2 in the canonical representation of $a-1$ is 1, while that in the canonical representation of $a^2-1$ is larger as $a^2 \\equiv 1 \\pmod{4}$. Thus to win, Miku may choose $n = 2$.\n\nLet now $a \\not\\equiv 1 \\pmod{p}$. By Fermat's little theorem, $a^{p-1} \\equiv 1 \\pmod{p}$. Hence $a^n \\equiv 1 \\pmod{p}$ whenever $n$ is a multiple of $p-1$. Then the numerator of the fraction $\\frac{a^n-1}{a-1}$ is divisible by $p$ while the denominator is not, whence the value of the fraction is divisible by $p$. Consequently, Miku can win by choosing any multiple of $p-1$ greater than 1 as $n$.\n\nb. We show that Juku wins by choosing $n = 2p-1$. Let $a$ be the number chosen by Miku.\n\nLet $a \\equiv 1 \\pmod{p}$. If $p > 2$ or $a \\equiv 1 \\pmod{4}$ then, by the lifting-the-exponent lemma, the exponent of $p$ in the canonical representation of $\\frac{a^n-1}{a-1}$ equals that in the canonical representation of $n$. As $p$ does not divide $2p-1$, it does not divide $\\frac{a^n-1}{a-1}$ either. If $p = 2$ and $a \\equiv -1 \\pmod{4}$ then $a^n \\equiv -1 \\pmod{4}$, whence 2 does not divide $a^n-1$. Therefore 2 does not divide $\\frac{a^n-1}{a-1}$.\n\nLet now $a \\not\\equiv 1 \\pmod{p}$. By Fermat's little theorem, $a^{p-1} \\equiv 1 \\pmod{p}$ and $a^p \\equiv a \\pmod{p}$, giving $a^n = a^{2p-1} \\equiv a \\pmod{p}$. Thus $a^n \\not\\equiv 1 \\pmod{p}$. As $p$ does not divide the numerator of the fraction $\\frac{a^n-1}{a-1}$, it cannot divide the value of the fraction either.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30018,
    "subject": "Mathematics (Multi-modal)",
    "question": "(a) Is it true that, for arbitrary integer $n$ greater than $1$ and distinct positive integers $i$ and $j$ not greater than $n$, the set of any $n$ consecutive integers contains distinct numbers $i'$ and $j'$ whose product $i'j'$ is divisible by the product $ij$?\n\n(b) Is it true that, for arbitrary integer $n$ greater than $2$ and distinct positive integers $i, j, k$ not greater than $n$, the set of any $n$ consecutive integers contains distinct numbers $i', j', k'$ whose product $i'j'k'$ is divisible by the product $ijk$?",
    "options": [],
    "answer": "(a) Yes; (b) No",
    "solution": "*Answer:* (a) Yes; (b) No.\n\n(a) Consider $n$ consecutive integers $k+1, k+2, \\dots, k+n$. As $i \\le n$, there must be a multiple of $i$ among them; denote it $i'$. As $j \\le n$, there must also be a multiple $x$ of $j$ among them. If $x \\ne i'$ then one may choose $j' = x$; then the product $i'j'$ is divisible by the product $ij$. Suppose in the rest that $x = i'$; then $i'$ as a common multiple of $i$ and $j$ is divisible by $\\text{lcm}(i, j)$.\nNote that if $\\text{gcd}(i, j) > \\frac{n}{2}$ then $\\text{gcd}(i, j)$ could not have two distinct multiples $i$ and $j$ among $1, 2, \\dots, n$. Thus $\\text{gcd}(i, j) \\le \\frac{n}{2}$, which in turn implies that $\\text{gcd}(i, j)$ must have two distinct multiples among $k+1, k+2, \\dots, k+n$. At least one of them differs from $i'$; let that be $j'$. Then the product $i'j'$ is divisible by the product $\\text{lcm}(i, j)\\ \\text{gcd}(i, j)$ which equals $ij$.\n\n(b) Let $n = 143$ and $i = 77 = 7 \\cdot 11$, $j = 91 = 7 \\cdot 13$, and $k = 143 = 11 \\cdot 13$; then $ijk = 7^2 \\cdot 11^2 \\cdot 13^2$. We show that among $143$ consecutive integers $p-71, p-70, \\dots, p+70, p+71$, where $p = 6006 = 2 \\cdot 3 \\cdot 7 \\cdot 11 \\cdot 13$, no three distinct numbers can have a product divisible by $7^2 \\cdot 11^2 \\cdot 13^2$. For that, note that the largest number less than $p$ that is divisible by at least two numbers among $7, 11$ and $13$ is $p-77$, and similarly, the least number greater than $p$ that is divisible by at least two numbers among $7, 11$ and $13$ is $p+77$. Both lie outside the region under consideration. Consequently, at most one prime among $7, 11$ and $13$ can belong to the canonical representation of any of the numbers $p-71, p-70, \\dots, p-1, p+1, \\dots, p+70, p+71$. Now choose any three numbers among $p-71, p-70, \\dots, p+70, p+71$.\n* Let $p$ be among these three numbers. The primes $7, 11$ and $13$ have exponent $1$ in its canonical representation. As shown above, only one of these three primes can occur in the canonical representation of either of the other two chosen numbers. Thus the product of the chosen three numbers cannot be divisible by $7^2 \\cdot 11^2 \\cdot 13^2$.\n* Let all these three numbers differ from $p$. As shown above, each of these numbers can be divisible by at most one prime among $7, 11$ and $13$. Hence, for their product to be divisible by $7^2 \\cdot 11^2 \\cdot 13^2$, one of the numbers should be divisible by $13^2$. But $\\frac{p}{13} = 6 \\cdot 77 \\equiv -6 \\pmod{13}$ which implies $p \\equiv -6 \\cdot 13 \\pmod{13^2}$. Thus the nearest to $p$ numbers divisible by $13^2$ are $p-7 \\cdot 13$ and $p+6 \\cdot 13$ which lie outside the region under consideration. Consequently, the product of the chosen three number cannot be divisible by $7^2 \\cdot 11^2 \\cdot 13^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30019,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any non-negative integer $i$, denote by $d_i$ the first digit of the number $2^i$. Let $n$ be a positive integer. Prove that there exists a non-zero digit that occurs in the tuple $(d_0, d_1, \\dots, d_{n-1})$ less than $\\frac{n}{17}$ times.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The claim obviously holds for $n=1$, whence assume in the following that $n \\ge 2$. Let $k$ be the minimal number of occurrences of a non-zero digit in the tuple $(d_0, d_1, \\dots, d_{n-1})$. Then the total number of occurrences of digits $5$, $6$, $7$, $8$, $9$ is at least $5k$. As each digit $5$, $6$, $7$, $8$, $9$ that is not the last digit of the tuple is followed by a digit $1$, and also $d_0 = 1$, the number of occurrences of digit $1$ is at least $5k$. Each digit $1$ that is not the last digit of the tuple is followed by either $2$ or $3$. Thus if the last digit is not $1$ then the total number of occurrences of $2$ and $3$ is at least $5k$. But if the last digit of the tuple is $1$ then the first $1$ of the tuple was previously not counted, whence the total number of occurrences of digits $2$ and $3$ is at least $5k$ in this case, too. Therefore, the number of occurrences of the only digit not counted yet, the digit $4$, is at most $n-15k$. As each occurrence of $8$ or $9$ follows a digit $4$, the total number of occurrences of $8$ and $9$ is also at most $n-15k$. Hence $n-15k \\ge 2k$, implying $k \\le \\frac{n}{17}$.\n\nTo prove that $k < \\frac{n}{17}$, suppose that $k = \\frac{n}{17}$, i.e., $n = 17k$. This implies that, in the argument above, every inequality must hold as an equality, i.e., each of the digits $5$, $6$, $7$, $8$, $9$ occurs exactly $k$ times and the digit $1$ occurs exactly $5k$ times. As $d_3 = d_{13} = d_{23} = 8$, the digit $8$ occurs more than once in the tuple $(d_0, d_1, \\dots, d_{17-1})$ and more than twice in the tuple $(d_0, d_1, \\dots, d_{2 \\cdot 17-1})$. Hence $k \\ge 3$.\n\nWe show that each segment consisting of exactly $17$ consecutive terms of the tuple contains at least $5$ occurrences of the digit $1$. Indeed, the last term of the tuple $(2^i, 2^{i+1}, \\dots, 2^{i+16})$ is exactly $65536$ times the first term, whence the last term contains at least $4$ more digits than the first term. As the first power of $2$ containing a certain number of digits definitely starts with $1$, the tuple $(2^{i+1}, \\dots, 2^{i+16})$ contains at least $4$ terms starting with $1$. If also $2^i$ starts with $1$ then there are at least $5$ such terms altogether; but if $2^i$ starts with a larger digit then the last term contains at least $5$ more digits than $2^i$, implying that there are still $5$ terms that start with $1$.\n\nIt remains to notice that the tuple $(d_0, d_1, \\dots, d_{3 \\cdot 17-1})$ contains the digit $1$ at least $16$ times because $d_0 = 1$ and $2^{3 \\cdot 17-1} = 2^{50} = (2^{1024})^5 > (10^3)^5 = 10^{15}$, implying that the number $2^{3 \\cdot 17-1}$ has at least $16$ digits. As every segment $(d_{17i}, d_{17i+1}, \\dots, d_{17(i+1)-1})$ contains the digit $1$ at least $5$ times, the digit $1$ occurs in the tuple $(d_0, d_1, \\dots, d_{17k-1})$ more than $5k$ times. The contradiction shows that $k < \\frac{n}{17}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(a,b,c)$ of integers which satisfy the equation\n$$ (a-b)^3(a+b)^2 = c^2 + 2(a-b) + 1. $$",
    "options": [],
    "answer": "(0,1,0), (-1,0,0)",
    "solution": "Answer: $(0,1,0), (-1,0,0)$.\n\nSubstituting $x = a - b$, $y = a + b$ we obtain the equation\n$$\nx^3 y^2 = c^2 + 2x + 1,\n$$\nwhich suffices to be solved in integers such that $x$ and $y$ have equal parity.\nWe shall consider the equivalent equation\n$$\nx(x^2 y^2 - 2) = c^2 + 1.\n$$\nIf both $x$ and $y$ are even then $x^2 y^2 - 2$ is even, whence the l.h.s. is divisible by 4. Thus $c^2 \\equiv -1 \\pmod 4$, but this is impossible since $-1$ is not a quadratic residue modulo 4.\nLet now both $x$ and $y$ be odd; then $x^2 y^2$ is a positive odd number. Consider\ntwo cases:\n* If $x^2 y^2 = 1$ then $x^2 = 1$ and $y^2 = 1$. Since $x^2 y^2 - 2 = -1 < 0$ whereas the product $x(x^2 y^2 - 2)$ equals the positive number $c^2 + 1$, we must have $x = -1$. The possibilities $x = -1, y = 1$ and $x = -1, y = -1$ give $a = 0, b = 1$ and $a = -1, b = 0$, respectively. In both cases $c = 0$.\n* If $x^2 y^2 > 1$ then $x^2 y^2 - 2 > 0$. Squares of odd numbers give remainder 1 upon division by 4, whence $x^2 y^2 - 2 \\equiv 1 \\cdot 1 - 2 = -1 \\pmod 4$. Consequently there exists a prime divisor $p$ of $x^2 y^2 - 2$ such that $p \\equiv -1 \\pmod 4$. But then $c^2 \\equiv -1 \\pmod p$, which is impossible since $-1$ is not a quadratic residue modulo $p$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30021,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an unopened pack, there are red, green and blue candies that weigh $2$ g, $5$ g and $25$ g respectively. There are the same number of candies of each color. After Mary eats some candies from that pack, the remaining candies weigh exactly $787$ g in total. Find the least number of candies that Mary could have eaten.",
    "options": [],
    "answer": "4",
    "solution": "Let there be $n$ candies of each color in the unopened pack. Since one red, one green and one blue candy weigh $32$ grams in total, the total weight of all candies in the unopened pack is $32n$ grams. Therefore Mary eats $32n - 787$ or $32(n - 25) + 13$ grams of candies.\n\nNote that if $n = 27$, then Mary could have eaten $32(n - 25) + 13 = 77$ grams of candy by eating $3$ blue and $1$ red candy, for a total of $4$ candies. We will show that this is the minimum possible number of candies by considering all possible cases:\n\n* If $n = 25$, then $32(n - 25) + 13 = 13$ grams can only be achieved by eating $1$ green and $4$ red candies, which is more than $4$ candies.\n* If $n = 26$, then Mary ate $32(n - 25) + 13 = 45$ grams of candy. If Mary ate a blue candy, then the rest of the lighter candies had to weigh $45 - 25 = 20$ grams. For that, at least $4$ lighter candies are needed, since up to $3$ lighter candies weighs at most $3 \\cdot 5 = 15$ grams. If Mary did not eat a blue candy, then she had to eat at least $9$ lighter candies, since up to $8$ lighter candies weighs at most $8 \\cdot 5 = 40$ grams. Either way, the amount of candies needed is greater than $4$.\n* If $n = 27$, then Mary ate $32(n - 25) + 13 = 77$ grams of candy. If Mary ate $0$, $1$ or $2$ blue candies, then the rest of the lighter candies had to weigh $77$, $52$ or $27$ grams respectively. In the last case, at least $6$ lighter candies are needed, since up to $5$ lighter candies weighs at most $5 \\cdot 5 = 25$ grams; in the first and second case even more lighter candies are obviously needed. Hence more than $4$ candies would have been eaten.\n* If $n = 28$, then Mary ate $32(n - 25) + 13 = 109$ grams of candy, so she needed to eat more than $4$ candies, since any $4$ candies weigh at most $4 \\cdot 25 = 100$ grams. If $n > 28$, then the total weight of candies eaten is even greater and hence more than $4$ candies had to be eaten.\n\nSince in all cases at least $4$ candies had to be eaten, $4$ is the least number of candies that Mary could have eaten.\nSince one red, one green and one blue candy weigh $32$ grams in total and there are the same amount of each color candies in the pack, the total weight of the candies is a multiple of $32$.\n\nSuppose Mary ate $k$ candies. Note that $k = 4$ is possible, since if there were $27$ candies of each color in the original pack, the total weight of the candies was $864$ grams and Mary could have eaten three blue candies and one red candy with a combined weight of $77$ grams, leaving $787$ grams of candy in the pack.\n\nLet us show that $k$ cannot be less than $4$.\n\n* If $k = 1$, then Mary would have eaten either one red, one green or one blue candy. In that case, the total weight of candies in the unopened back would have been $789$, $792$ or $812$ grams. Since none of these are divisible by $32$, they cannot be the original weight of the pack, so $k = 1$ is not possible.\n* If $k = 2$, then Mary would have eaten either two red, two green, two blue, one red and one green, one green and one blue or one red and one blue candy. Then the original total weight of the candies would have been $791$, $797$, $837$, $794$, $817$ or $814$ grams respectively. Since none of these are divisible by $32$, $k = 2$ is not possible either.\n* If $k = 3$, then Mary would have eaten either three red, three green, three blue, one red and two green, one red and two blue, one green and two red, one green and two blue, one blue and two red, one blue and two green, or one red, one green and one blue candy. Then the original total weight of the candies would have been $793$, $802$, $862$, $799$, $839$, $796$, $842$, $816$, $822$ or $819$ grams respectively. Since none of these is divisible by $32$, $k = 3$ is not possible either.\n\nHence the least number of candies eaten is $4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30022,
    "subject": "Mathematics (Multi-modal)",
    "question": "An acute-angled triangle $ABC$ with $AC > AB$ is given. The perpendicular bisector of side $BC$ intersects the lines $AC$ and $AB$ at points $D$ and $E$, respectively. The circle with diameter $DE$ intersects the lines $AC$ and $AB$ at points $K$ and $L$, respectively ($K \\neq D, L \\neq E$). Let $M$ be the midpoint of side $BC$. Prove that the points $K, L$, and $M$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the points $E, K, L$, and $D$ are concyclic (Fig. 28), it follows that $\\angle CKL = \\angle DKL = \\angle DEL = \\angle MEB$. From the problem conditions, $\\angle CME = 90^\\circ$, and by Thales' theorem, $\\angle CKE = \\angle DKE = 90^\\circ$. Therefore, the points $C, M, K, E$ are also concyclic. Consequently, $\\angle CKM = \\angle CEM$. Since point $E$ lies on the perpendicular bisector of side $BC$, the triangles $BEM$ and $CEM$ are congruent, thus $\\angle MEB = \\angle CEM$. From the previous result, $\\angle CKL = \\angle CKM$. Since the points $L$ and $M$ lie on the same side of line $CK$, the points $K, L$, and $M$ are collinear.\n\n![](attached_image_1.png)\nFig. 28",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30023,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 8 distinct points marked on a circle. Juku wants to draw as many triangles as possible in such a way that all vertices of each triangle he draws are at the marked points, and no two of these triangles share a side. Find the largest number of triangles that can be drawn under these conditions.",
    "options": [],
    "answer": "8",
    "solution": "Assume w.l.o.g. that the points marked on the circle are equally spaced and number the marked points counterclockwise with natural numbers $0$, $1$, $\\ldots$, $7$. Consider a triangle with vertices marked at points $0$, $1$, $3$ and its $7$ copies obtained by rotating the original triangle counterclockwise by $\\frac{1}{8}$, $\\frac{2}{8}$, $\\ldots$, $\\frac{7}{8}$ of a full turn around the center of the circle (illustrated in Fig. 29 with different colors). These $8$ triangles do not share any sides because all sides of the original triangle have different lengths, and each rotation of a side with a specific length results in different segments. Therefore, it is possible to draw $8$ triangles under the given conditions.\n\n![](attached_image_1.png)\nFig. 29\n\nOn the other hand, note that from each marked point, at most $7$ segments can be drawn to the remaining marked points. Each triangle uses either $0$ or $2$ of these segments. Thus, each marked point can be the endpoint of at most $6$ different triangle sides in total. Since we count each side twice (once at each endpoint), there can be at most $\\frac{8 \\cdot 6}{2}$, or $24$ different triangle sides. Since each triangle has $3$ sides, there can be at most $\\frac{24}{3}$, or $8$ triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30024,
    "subject": "Mathematics (Multi-modal)",
    "question": "The internal angle bisectors at vertices $B$ and $C$ of triangle $ABC$ intersect the circumcircle of triangle $ABC$ at $E$ and $F$, respectively. Given that $BE = CF \\neq 0$, may we be sure that triangle $ABC$ is isosceles?",
    "options": [],
    "answer": "No",
    "solution": "Let $\\angle ABC = 30^\\circ$ and $\\angle BCA = 90^\\circ$. Let $O$ be the circumcenter of triangle $ABC$; it is also the midpoint of the hypotenuse $AB$. We have $\\angle OAC = \\angle BAC = 60^\\circ$, and since $AO = CO$, it follows that $\\angle OCA = 60^\\circ$ (Fig. 30). Therefore, $\\angle FCO = 60^\\circ - \\frac{90^\\circ}{2} = 15^\\circ$. Let $C'$ be the reflection of vertex $C$ over the point $O$ (Fig. 31); then $\\angle FCC' = 15^\\circ$ as well. On the other hand, $\\angle ABE = \\frac{30}{2}^\\circ = 15^\\circ$. In conclusion, we see that the arcs $AE$ and $FC'$ subtend equal inscribed angles on the circumcircle of triangle $ABC$, hence the corresponding arcs are equal. Since $AB$ and $CC'$ are diameters, the remaining arcs $BE$ and $CF$ are also equal. Therefore, the corresponding chords $BE = CF$ are equal. Thus, the equality $BE = CF$ can hold even in a non-isosceles triangle.\n\n![](attached_image_1.png)\nFig. 30\n![](attached_image_2.png)\nFig. 31\nLet $\\angle CAB = 60^\\circ$ and $AC < AB$, and let $O$ be the circumcenter of triangle $ABC$ (Fig. 32). Then $\\angle BOC = 2 \\cdot 60^\\circ = 120^\\circ$. Thus, $\\angle COE + \\angle EOA + \\angle AOF + \\angle FOB = 360^\\circ - 120^\\circ = 240^\\circ$. Since $\\angle COE = \\angle EOA$ and $\\angle AOF = \\angle FOB$, we have $\\angle EOA + \\angle AOF = \\frac{240^\\circ}{2} = 120^\\circ$. Consequently, $\\angle EOF = \\angle BOC$, from which it follows that\n$$\n\\angle COF = \\angle COE + \\angle EOF = \\angle BOC + \\angle COE = \\angle BOE.\n$$\nIn conclusion, we see that the central angles subtended by the shorter arcs $BE$ and $CF$ of the circumcircle of triangle $ABC$ are equal, hence the corresponding arcs are equal. Therefore, the corresponding chords $BE$ and $CF$ are equal. Thus, the equality $BE = CF$ can hold in a non-isosceles triangle.\n\n![](attached_image_3.png)\nFig. 32",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a positive integer $n$ such that\n$$\n1950^n + 1934^n = 2024^n?\n$$",
    "options": [],
    "answer": "No",
    "solution": "Answer: No.\n\nThe numbers $1950$, $1934$, and $2024$ give remainders $4$, $2$, and $1$, respectively, when divided by $7$. Raising $4$ to powers $n = 1, 2, 3, 4, 5, 6, \\dots$ results in remainders $4, 2, 1, 4, 2, 1, \\dots$, and raising $2$ to the same powers results in remainders $2, 4, 1, 2, 4, 1, \\dots$. Thus, the remainders of the left side of the given equation for powers $n = 1, 2, 3, 4, 5, 6, \\dots$ are $6, 6, 2, 6, 6, 2, \\dots$. The remainder of the right side is $1$ for every $n$. The contradiction shows that there is no positive integer $n$ that satisfies the given equation.\n\nClearly $n = 1$ is not suitable, because $1950 + 1934 > 2024$. It is also easy to verify that\n\n$$\n1950^2 + 1934^2 = 3802500 + 3740356 = 7542856 > 4096576 = 2024^2.\n$$\n\nFor $n \\ge 3$, there are no integer solutions to the equation $a^n + b^n = c^n$ by Fermat's Last Theorem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist an integer $a$ such that\n$$\n\\frac{1}{\\sqrt{2024}} < \\frac{1}{\\sqrt{a+1}} + \\frac{1}{\\sqrt{a+2}} + \\dots + \\frac{1}{\\sqrt{a+2023}} < \\frac{1}{\\sqrt{2023}}?\n$$",
    "options": [],
    "answer": "a = 2023^3",
    "solution": "For each $i = 1, 2, \\dots, 2023$, we have\n$$\n\\frac{1}{\\sqrt{2023^3 + i}} < \\frac{1}{\\sqrt{2023^3}} = \\frac{1}{2023\\sqrt{2023}},\n$$\nthus\n$$\n\\frac{1}{\\sqrt{2023^3+1}} + \\frac{1}{\\sqrt{2023^3+2}} + \\dots + \\frac{1}{\\sqrt{2023^3+2023}} < 2023 \\cdot \\frac{1}{2023\\sqrt{2023}} = \\frac{1}{\\sqrt{2023}}.\n$$\n\nSimilarly, by estimating all fractions from below, we see that\n$$\n\\begin{aligned}\n& \\frac{1}{\\sqrt{2023^3+1}} + \\frac{1}{\\sqrt{2023^3+2}} + \\dots + \\frac{1}{\\sqrt{2023^3+2023}} \\\\\n& > 2023 \\cdot \\frac{1}{\\sqrt{2023^3+2023^2}} = \\frac{2023}{2023\\sqrt{2024}} = \\frac{1}{\\sqrt{2024}}\n\\end{aligned}\n$$\nTherefore, $a = 2023^3$ is suitable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circle $\\omega$ with center $O$ and circle $\\alpha$ with center $A$ intersect at two distinct points $C$ and $D$, whereas $\\angle OCA = 90^\\circ$. A point $E$ is chosen on circle $\\omega$ inside circle $\\alpha$. Let $F$ be the reflection of point $E$ over the point $O$, and $G$ the intersection of line $CE$ with circle $\\alpha$ ($G \\neq C$). Prove that points $G$, $D$, and $F$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The line $OC$ perpendicular to the radius $AC$ of the circle $\\alpha$ is tangent to this circle. Hence we get $\\angle GDC = 180^\\circ - \\angle OCG = 180^\\circ - \\angle OCE$. From the equality of inscribed angles, we get $\\angle CDF = \\angle CEF = \\angle CEO$. Since $OC = OE$, we finally get $\\angle CEO = \\angle OCE$. In conclusion,\n$$\n\\angle GDF = \\angle GDC + \\angle CDF = 180^\\circ - \\angle OCE + \\angle CEO = 180^\\circ.\n$$\n\nTherefore, points $G$, $D$, and $F$ are collinear.\n\nSolution 2:\n\nWe express the value of angle $GDF$ as the sum of the values of angles $GDA$, $ADO$, and $ODF$ (Fig. 35). Let $\\angle DCE = \\gamma$.\nFirstly, note that $AD = AG$. Using this and the relationship between central and inscribed angles in circle $\\alpha$, we get\n$$\n\\begin{aligned}\n\\angle GDA &= \\frac{180^\\circ - \\angle GAD}{2} = 90^\\circ - \\frac{\\angle GAD}{2} = 90^\\circ - \\angle GCD \\\\\n&= 90^\\circ - \\angle ECD = 90^\\circ - \\gamma.\n\\end{aligned}\n$$\n\nBy symmetry, $\\angle ADO = \\angle ACO = 90^\\circ$.\nNext, note that $OD = OF$. Using this and the equality of inscribed angles,\n![](attached_image_1.png)\n\n$$\n\\angle GDF = \\angle GDA + \\angle ADO + \\angle ODF = (90^\\circ - \\gamma) + 90^\\circ + \\gamma = 180^\\circ.\n$$\n\nTherefore, points $G$, $D$, and $F$ are collinear.\n\nSolution 3:\n\nWe express the value of angle $GDF$ as the sum of the values of angles $GDA$, $ADE$, and $EDF$ (Fig. 36). Let $\\angle DCE = \\gamma$.\nFirstly, note that $AD = AG$. Using this and the relationship between central and inscribed angles in circle $\\alpha$, we get\n$$\n\\begin{aligned} \\angle GDA &= \\frac{180^\\circ - \\angle GAD}{2} = 90^\\circ - \\frac{\\angle GAD}{2} = 90^\\circ - \\angle GCD \\\\ &= 90^\\circ - \\angle ECD = 90^\\circ - \\gamma. \\end{aligned}\n$$\nBy symmetry, $\\angle ADO = \\angle ACO = 90^\\circ$. The line $AD$, which is perpendicular to the radius $OD$ of the circle $\\omega$, is tangent to circle $\\omega$. Therefore, $\\angle ADE = \\angle ECD = \\gamma$. Since $EF$ is a diameter of circle $\\omega$, $\\angle EDF = 90^\\circ$. In conclusion, $\\angle GDF = \\angle GDA + \\angle ADE + \\angle EDF = (90^\\circ - \\gamma) + \\gamma + 90^\\circ = 180^\\circ$. Therefore, points $G$, $D$, and $F$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Pieces of cardboard of dimensions $1 \\times 4$ are placed on a $10 \\times 10$ grid in such a way that each piece covers exactly 4 adjacent unit squares (either horizontally or vertically) and no two pieces touch each other side-to-side, edge-to-edge, or corner-to-corner. Find the largest possible number of cardboard pieces.",
    "options": [],
    "answer": "12",
    "solution": "Let the pieces of cardboard be placed on the grid as required. Since each piece covers exactly 4 unit squares and no two pieces touch, there is at least a 1-unit wide space between every two pieces. Therefore, if we draw a half-unit wide \"no-go zone\" around each piece, the areas covered by the pieces and their no-go zones will have dimensions of $2 \\times 5$ and will not overlap. Since these areas extend over the edges of the grid by at most half a unit, all these areas can fit within a $11 \\times 11$ square. Therefore, no more than $\\lfloor \\frac{11 \\cdot 11}{2 \\cdot 5} \\rfloor = 12$ pieces can be placed on the grid.\nThis limit case is achievable, as shown in Fig. 37.\n\n![](attached_image_1.png)\nFig. 37\nDivide the $10 \\times 10$ grid into squares of dimensions $2 \\times 2$ (Fig. 38).\n\nNote that at most 2 unit squares can be covered in any $2 \\times 2$ square because otherwise at least two different pieces must be used which would then touch each other. Since there are 25 squares of $2 \\times 2$, the pieces can cover at most $25 \\cdot 2 = 50$ unit squares, and thus there can be at most $\\lfloor (50/2) \\rfloor = 12$ pieces. The fact that 12 pieces are possible is shown in Fig. 37.\n\n![](attached_image_2.png)\nFig. 38",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Kadi has a square pasture with a side length of $2$, where a goat lives. Since there is no fence around the pasture, Kadi fears that the goat might run away at any moment. Instead of getting a fence, Kadi decided that it would be cheaper to place scarecrows around the pasture, which the goat is afraid of. It is known that the goat will not dare to be at a distance of $1$ or less from any scarecrow. Find the minimum number of scarecrows that Kadi needs to confine the goat to a finite area that entirely contains the area within the boundaries of the pasture.",
    "options": [],
    "answer": "7",
    "solution": "Let $P$ be the center of the square pasture and let some scarecrow be at point $O$. Since the area bounded by the scarecrows must entirely contain the interior of the square and the points of the square are at least distance $1$ from the center of the square, we have $PO \\ge 1+1=2$ with the equality holding only if the midpoint of $PO$ coincides with the midpoint of one of the sides of the square. Consider first the case where $PO = 2$. Let $A$ and $B$ be points on the unit circle centered at $O$ such that $PA$ and $PB$ are tangents to this circle, and let $C$ be the midpoint of segment $PO$ (Fig. 39). Since the tangents $PA$ and $PB$ are perpendicular to the radii $OA$ and $OB$ at the points of tangency, the points $P, A, O, B$ lie on a circle with diameter $PO$. Thus, $OA = OB = OC = PC = AC = BC = 1$, and since triangles $OAC$ and $OBC$ are equilateral, $\\angle AOB = 60^\\circ + 60^\\circ = 120^\\circ$ and $\\angle APB = 180^\\circ - 120^\\circ = 60^\\circ$.\n\n![](attached_image_1.png)\nFig. 39\n![](attached_image_2.png)\nFig. 40\n\nHence, a scarecrow covers an angle of $60^\\circ$ of the horizon as viewed from the center of the pasture. If $PO > 2$, the covered angle is obviously smaller. Since the scarecrows must cover the entire $360^\\circ$ horizon as seen from $P$, at least $6$ scarecrows are needed. However, covering this area with exactly $6$ scarecrows would only be possible if each scarecrow covers exactly $60^\\circ$. This would require all scarecrows to be at a distance of $2$ from the center of the pasture. This, in turn, would be possible only if the midpoint of each segment connecting the scarecrow to the center of the square is at the midpoint of one of the sides of the square. Since the square has only $4$ sides, it is impossible to place $6$ scarecrows this way. Therefore, at least $7$ scarecrows are needed.\n\nWe will show that $7$ scarecrows are sufficient. Let the coordinate origin be the center of the pasture and let the coordinate axes be parallel to the sides of the pasture. The scarecrows can be placed, for example, at the points $O_1 = (2,1)$, $O_2 = (2-\\sqrt{3}, 2)$, $O_3 = (-1.6, 1.8)$, $O_4 = (-2,0)$, $O_5 = (-1.6, -1.8)$, $O_6 = (2-\\sqrt{3}, -2)$ and $O_7 = (2,-1)$ (Fig. 40). We will prove this.\n\nFirst, it is clear that the areas affected by the scarecrows at points $O_1, O_2, O_4, O_6$, and $O_7$ touch the square but do not overlap with it. Since the distance of point $O_3$ from the upper left corner of the square is $1$, and it is even further from the other points of the square, the influence area of the scarecrow at point $O_3$ does not overlap with the pasture. The same holds for $O_5$.\n\nSecondly, we show that the scarecrows cover the entire $360^\\circ$ horizon as seen from the pasture. For this, we show that the length of each side of the polygon $O_1O_2...O_7$ is at most $2$. Clearly, $|O_7O_1| = 2$ and direct inspection shows that $O_1O_2 = O_6O_7 = 2$. It remains to see that\n$$\n\\begin{aligned}\n(O_2O_3)^2 &= (2 - \\sqrt{3} + 1.6)^2 + (2 - 1.8)^2 = (3.6 - \\sqrt{3})^2 + 0.2^2 \\\\\n&< (3.6 - \\sqrt{2.89})^2 + 0.2^2 = (3.6 - 1.7)^2 + 0.2^2 \\\\\n&= 1.9^2 + 0.2^2 = 3.61 + 0.04 < 4, \\\\[1em]\n(O_3O_4)^2 &= (-1.6 + 2)^2 + 1.8^2 = 0.4^2 + 1.8^2 = 0.16 + 3.24 = 3.4 < 4.\n\\end{aligned}\n$$\nThus $O_2O_3 < 2$ and $O_3O_4 < 2$; by symmetry $O_4O_5 < 2$ and $O_5O_6 < 2$. Therefore, $7$ scarecrows are sufficient to confine the goat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer divisible by $4$. Prove that\n$$\\sin^2\\left(1 \\cdot \\frac{360^\\circ}{n}\\right) + \\sin^2\\left(2 \\cdot \\frac{360^\\circ}{n}\\right) + \\sin^2\\left(3 \\cdot \\frac{360^\\circ}{n}\\right) + \\dots + \\sin^2\\left(n \\cdot \\frac{360^\\circ}{n}\\right) = \\frac{n}{2}.$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n = 4k$. For each $i = 1, 2, \\dots, n$, denote $\\alpha_i = i \\cdot \\frac{360^\\circ}{n}$ and $x_i = \\sin^2 \\alpha_i$. We pair the terms as $(x_1, x_{k+1}), (x_2, x_{k+2}), \\dots, (x_k, x_{2k})$ and $(x_{2k+1}, x_{3k+1}), (x_{2k+2}, x_{3k+2}), \\dots, (x_{3k}, x_{4k})$. Since\n$$\n\\alpha_{i+k} = (i + k) \\cdot \\frac{360^\\circ}{n} = i \\cdot \\frac{360^\\circ}{n} + k \\cdot \\frac{360^\\circ}{n} = \\alpha_i + 90^\\circ,\n$$\nwe have\n$$\n\\sin^2 \\alpha_{i+k} = \\sin^2 (\\alpha_i + 90^\\circ) = \\sin^2 (90^\\circ - (-\\alpha_i)) = \\cos^2 (-\\alpha_i) = \\cos^2 \\alpha_i.\n$$\nTherefore $x_i + x_{i+k} = \\sin^2 \\alpha_i + \\cos^2 \\alpha_i = 1$. Thus, the sum of the numbers of each pair is $1$. Since we have $\\frac{n}{2}$ pairs, the sum of all numbers is $\\frac{n}{2}$.\nLet $n = 4k$. Note that\n$$\n\\sin^2 \\left( i \\cdot \\frac{360^\\circ}{n} \\right) = \\frac{1 - \\cos \\left( 2i \\cdot \\frac{360^\\circ}{n} \\right)}{2} = \\frac{1}{2} - \\frac{1}{2} \\cos \\left( 2i \\cdot \\frac{360^\\circ}{n} \\right).\n$$\nFor each $i = 1, 2, \\dots, n$, denote $\\alpha_i = i \\cdot \\frac{360^\\circ}{n}$ and $y_i = \\cos 2\\alpha_i$; then the equation to be proven takes the form $\\frac{1}{2}n - \\frac{1}{2}(y_1 + y_2 + \\dots + y_n) = \\frac{1}{2}n$, which is equivalent to $y_1 + y_2 + \\dots + y_n = 0$. We are going to prove this equality in the rest. We pair the terms as $(y_1, y_{k+1}), (y_2, y_{k+2}), \\dots, (y_k, y_{2k})$ and $(y_{2k+1}, y_{3k+1}), (y_{2k+2}, y_{3k+2}), \\dots, (y_{3k}, y_{4k})$ and prove the angle equality $\\alpha_{i+k} = \\alpha_i + 90^\\circ$ as in Solution 1. Next, we see that\n$$\n\\cos 2(\\alpha_i + 90^\\circ) = \\cos (2\\alpha_i + 180^\\circ) = -\\cos 2\\alpha_i,\n$$\nfrom which it follows that $y_i + y_{i+k} = \\cos 2\\alpha_i - \\cos 2\\alpha_i = 0$. Since the sum of the numbers of each pair is $0$, the sum of all numbers is also $0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30031,
    "subject": "Mathematics (Multi-modal)",
    "question": "In Kükametsa School of Science, the number of secondary school participants in the \"Kangaroo\" competition was exactly one fourth greater than in the previous year. This year, the number of participants from the 11th and 12th grade was respectively 3 and 4 less than the number of participants from the 10th and 11th grade in the previous year, but the number of participants from the 10th grade this year was twice as high as the number of participants from the 12th grade last year. In the previous year, the average number of participants per grade level in Kükametsa School of Science was more than 60, but less than 65. Find all possibilities how many participants from the 10th grade could have participated in the \"Kangaroo\" competition this year.",
    "options": [],
    "answer": "106, 108, 110",
    "solution": "Let $e$ be the number of secondary school participants in the \"Kangaroo\" competition in the previous year. This year, the number of participants was greater by $\\frac{1}{4}e$. Hence $e$ is divisible by 4. Since the total number of participants in the 11th and 12th grade this year was $3+4=7$ less than the number of participants in the 10th and 11th grade last year, but the number of participants from the 10th grade this year was twice the number of participants from the 12th grade last year, the total number of secondary school participants increased by a number that is 7 less than the number of participants from the 12th grade last year. Hence the number of participants from 12th grade last year was 7 more than the number of additional participants this year, or $\\frac{1}{4}e+7$. The number of participants from the 10th grade this year was twice that, or $\\frac{1}{2}e+14$.\n\nThe average amount of participants per grade level last year is $\\frac{1}{3}e$. Since $e$ is divisible by 4, the number $\\frac{1}{3}e$ can be represented as a common fraction with a numerator that is a multiple of 4 and the denominator 3. Considering all such common fractions between 60 and 65, we get three options $\\frac{184}{3}$, $\\frac{188}{3}$ and $\\frac{192}{3}$. So $e$ can be either 184, 188 or 192 and $\\frac{1}{2}e+14$ is either 106, 108 or 110 respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30032,
    "subject": "Mathematics (Multi-modal)",
    "question": "The altitudes of an acute-angled triangle $ABC$ intersect at point $H$. The tangent at point $A$ to the circumcircle of triangle $AHB$ intersects the line $CH$ at point $K$. The tangent at point $A$ to the circumcircle of triangle $AHC$ intersects the line $BH$ at point $L$. Prove that the points $B$, $C$, $K$, $L$ lie on the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, and $\\angle BCA = \\gamma$. By tangency, $\\angle KAH = \\angle ABH = 90^\\circ - \\alpha$ (Fig. 41). Since $\\angle ACH = 90^\\circ - \\alpha$ as well, triangles $AHC$ and $KHA$ are similar. Consequently, the corresponding third angles are equal, i.e., $\\angle AKH = \\angle CAH = 90^\\circ - \\gamma$. Similarly, we get $\\angle LAH = 90^\\circ - \\alpha$ and $\\angle ALH = 90^\\circ - \\beta$. Thus, $\\angle KAL = \\angle KAH + \\angle LAH = 180^\\circ - 2\\alpha$, with $AH$ bisecting the angle $KAL$.\n\nOn the other hand,\n$$\n\\begin{align*} \n\\angle KHL &= \\angle CHB = 180^\\circ - \\angle BCH - \\angle CBH \\\\ \n&= 180^\\circ - (90^\\circ - \\gamma) - (90^\\circ - \\beta) \\\\ \n&= \\beta + \\gamma = 180^\\circ - \\alpha. \n\\end{align*}\n$$\nNow note (Fig. 42) that for the incenter $I$ of triangle $AKL$,\n$$\n\\begin{align*} \n\\angle KIL &= 180^\\circ - \\angle IKL - \\angle ILK = 180^\\circ - \\frac{\\angle AKL}{2} - \\frac{\\angle ALK}{2} \\\\ \n&= 180^\\circ - \\frac{\\angle AKL + \\angle ALK}{2} \\\\ \n&= 180^\\circ - \\frac{180^\\circ - \\angle KAL}{2} = 180^\\circ - \\frac{180^\\circ - (180^\\circ - 2\\alpha)}{2} = 180^\\circ - \\alpha, \n\\end{align*}\n$$\nso $\\angle KIL = \\angle KHL$. Since points $H$ and $I$ lie on the same side of line $KL$, points $K$, $H$, $I$, and $L$ lie on the same circle. Given that both $H$ and $I$ lie on the angle bisector $AH$ of angle $KAL$, which intersects the chord $KL$, it follows that $H = I$. Consequently, $\\angle LKH = \\angle AKH = 90^\\circ - \\gamma$ and $\\angle KLH = \\angle ALH = 90^\\circ - \\beta$. In addition, we have $\\angle HLK = 90^\\circ - \\beta = \\angle HCB$, from which it follows that points $B$, $C$, $K$, $L$ lie on the same circle.\nAs in Solution 1, we note that triangles $AHC$ and $KHA$ are similar, hence $\\frac{HA}{HK} = \\frac{HC}{HA}$, or $HA^2 = HK \\cdot HC$. Similarly, triangles $AHB$ and $LHA$ are similar, giving $HA^2 = HL \\cdot HB$. Therefore, $HB \\cdot HL = HC \\cdot HK$, from which it follows that points $B$, $C$, $K$, $L$ lie on the same circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30033,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is given. There are $n$ computers in a network, numbered with natural numbers $1, 2, \\dots, n$. The computers are connected with one-way communication lines such that information can be sent from any computer to all other computers either directly or through other computers in the network. Initially, each computer only knows its own number. When a certain procedure is initiated on any computer, the computer outputs all the numbers of computers it knows and communicates them to all the computers it has direct lines to. The computer operator initiates this procedure on any selected computers in any chosen sequence (possibly multiple times on the same computer).\nProve that the minimum total number of procedure initiations, for which it is possible for all computers to have output the numbers of all computers at least once, does not depend on the network of lines, but only on the number $n$, and find this number of initiations.",
    "options": [],
    "answer": "2n - 1",
    "solution": "We will show that $2n - 1$ initiations are necessary regardless of the network of lines. Consider an arbitrary sequence of initiations, at the end of which all computers have output the numbers of all computers. Let $A$ be the computer with the latest time of the initiation of the first procedure. Since the procedure has been initiated at least once in all other computers by that time, the order number of this initiation is at least $n$. Therefore, after the first $n - 1$ initiations, no computer has output the number of computer $A$. To ensure that each computer outputs the number of computer $A$ at least once, $n$ more initiations are needed. Thus, at least $2n - 1$ initiations are necessary.\n\nNow we will show that $2n - 1$ initiations are sufficient regardless of the network of lines. Choose an arbitrary computer $A$ and construct a sequence of the remaining computers $B_1, B_2, \\dots, B_{n-1}$ without repetition, such that each computer in this sequence can receive information directly from some preceding computer in this sequence or from computer $A$. Such a sequence of length $n - 1$ indeed exists, because if at any step adding the next computer was not possible, then from none of the already included computers in the sequence nor from computer $A$ could information be sent directly to any of the remaining computers – which contradicts the condition of the problem that information can move between these computers via the network. Similarly, construct a sequence of computers $C_1, C_2, \\dots, C_{n-1}$ without repetition, such that each computer in this sequence can send information directly to some preceding computer in this sequence or to computer $A$. By symmetry considerations, such a sequence also exists. It turns out that a suitable\nsequence of procedure initiations of length $2n - 1$ is\n$C_{n-1}, C_{n-2}, \\dots, C_1, A, B_1, B_2, \\dots, B_{n-1}$.\n\nWe will prove this statement; for simplicity, denote $C_0 = A$. According to the construction of the sequence $C_1, C_2, \\dots, C_{n-1}$, each computer $C_i$ where $0 < i$, sends its number to at least one computer $C_{i'}$ where $0 \\le i' < i$. If $0 < i'$, then computer $C_{i'}$ in turn sends the number of computer $C_i$ to at least one computer $C_{i''}$ where $0 \\le i'' < i'$. Continuing in this way, the number of computer $C_i$ reaches computer $C_j$ for successively smaller indices $j$. Since there are $n-1$ computers besides $A$, the number of computer $C_i$ must reach computer $A$ in at most $n-1$ steps. Since $C_i$ was chosen arbitrarily, the numbers of all other computers reach computer $A$ within the first $n-1$ initiations of the procedure. It remains to note that during each subsequent initiation of the procedure, all computer numbers are output and transmitted. This is evidently done by computer $A$, and when all computers $B_1, B_2, \\dots, B_{i-1}$ have done this, then by the construction of the sequence $B_1, B_2, \\dots, B_{n-1}$, computer $B_i$ also does this. Consequently, after $2n-1$ initiations of the procedure, all computers have output the numbers of all computers at least once.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a function $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfies for all real numbers $x, y$ the equation\n$$\nf(x) + f(f(y)) + f(f(f(x))) = x + y?\n$$",
    "options": [],
    "answer": "No, there is no such function.",
    "solution": "Taking $y = f(x)$ in the given equation, we get\n$$\nf(x) + f(f(f(x))) + f(f(f(x))) = x + f(x),\n$$\nfrom which, by expressing $f(f(f(x)))$, we find\n$$\nf(f(f(x))) = \\frac{x}{2}. \\qquad (7)\n$$\nTaking $y = 0$ in the given equation, we get $f(x) + f(f(0)) + f(f(f(x))) = x + 0$; using the equality (7) here, we get after simplification\n$$\nf(x) = \\frac{x}{2} - f(f(0)). \\qquad (8)\n$$\nSubstituting $x = f(f(0))$ into equation (8) and applying equation (7) on the left hand side, we get\n$$\n\\frac{0}{2} = \\frac{f(f(0))}{2} - f(f(0)), \\qquad (9)\n$$\nwhich after simplification gives $f(f(0)) = 0$. Consequently, equation (8) takes the form $f(x) = \\frac{x}{2}$. But in this case, $f(f(f(x))) = \\frac{x}{8}$, whereas from equation (7) $f(f(f(x))) = \\frac{x}{2}$. Since in general $\\frac{x}{2} \\ne \\frac{x}{8}$, there are no functions that satisfy the given functional equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30035,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$, the internal angle bisectors at vertices $B$ and $C$ intersect at point $I$ and intersect the sides $CA$ and $AB$ at points $E$ and $F$, respectively. Let $M$ and $N$ be the midpoints of segments $BI$ and $CI$, respectively. The line $FM$ intersects the external angle bisector at vertex $B$ of triangle $ABC$ at point $K$, and the line $EN$ intersects the external angle bisector at vertex $C$ of triangle $ABC$ at point $L$. Prove that the points $K, B, C$, and $L$ lie on the same circle.\n\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $\\angle KBE = 90^\\circ$ as $BE$ and $BK$ are the internal and external angle bisectors at the same vertex (Fig. 43). We will now show $\\angle BKC = 90^\\circ$. Let $X$ be a point on line $AB$ such that $CX \\parallel BE$; then the external angle bisector at vertex $B$ of triangle $ABC$ is also perpendicular to line $CX$ (Fig. 44). Denote $\\angle ABC = \\beta$. Then $\\angle BCX = \\angle CBE = \\frac{\\beta}{2}$ and $\\angle XBC = 180^\\circ - \\angle ABC = 180^\\circ - \\beta$. Therefore, $\\angle BXC = 180^\\circ - (\\angle BCX + \\angle XBC) = \\frac{\\beta}{2}$, so the triangle $XBC$ is isosceles with apex at $B$. Thus, the altitude from vertex $B$ of triangle $XBC$ bisects the base $CX$. As previously established, this altitude lies on the external angle bisector at vertex $B$ of triangle $ABC$.\nLet $k = \\frac{|FX|}{|FB|} = \\frac{|FC|}{|FI|}$. A homothety with center $F$ and ratio $k$ sends points $B$ and $I$ to points $X$ and $C$, respectively, so it sends the midpoint $M$ of segment $BI$ to the midpoint of segment $CX$. Therefore, the line $FM$ passes through the midpoint of segment $CX$. Consequently, the intersection point $K$ of line $FM$ and the external angle bisector at vertex $B$ of triangle $ABC$ lies at the midpoint of segment $CX$, and $\\angle BKC = 90^\\circ$, as we wanted to show. Similarly, we can show that $\\angle BLC = 90^\\circ$. Therefore, points $K$ and $L$ lie on the circle with diameter $BC$. The statement of the problem follows.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30036,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number is called a *palindrome* if reversing its digits results in the exact same number.\nGiven an integer $n > 1$, how many $n$-digit natural numbers are there such that when added to the number obtained by reversing the order of its digits, the result is a palindrome?",
    "options": [],
    "answer": "If n is even: 36 * 55^{(n-2)/2} + 8 * 9^{(n-2)/2}. If n is odd: 36 * 55^{(n-3)/2} * 5 + 8 * 9^{(n-3)/2}.",
    "solution": "Let $\\overline{a_1a_2...a_n}$ be an $n$-digit number such that the sum of the number with the number obtained by reversing its digits is a palindrome. We consider two cases separately.\n\n* If the sum is an $n$-digit number, then obviously $a_1 + a_n < 10$. We show that no carry occurs when adding the numbers $\\overline{a_1a_2...a_n}$ and $\\overline{a_n a_{n-1}...a_1}$. Suppose, to the contrary, that a carry occurs. Let the result of the addition be the number $\\overline{b_1b_2...b_n}$, with the first carry occurring in the $i$-th position from the end. Then $a_n + a_1 = b_n$, $a_{n-1} + a_2 = b_{n-1}$, $\\dots$, $a_{n+1-(i-1)} + a_{i-1} = b_{n+1-(i-1)}$ and $a_{n+1-i} + a_i \\ge 10$. Due to the last inequality, a carry also occurs in the $(n+1-i)$-th position from the end. Therefore, $b_{i-1} = a_{i-1} + a_{n+1-(i-1)} + 1 = b_{n+1-(i-1)} + 1$. But for $\\overline{b_1b_2...b_n}$ to be a palindrome, the equality $b_{i-1} = b_{n+1-(i-1)}$ must hold, which is a contradiction.\n\n* If the sum is a $(n+1)$-digit number, its first digit must be 1; let the remaining digits of the sum be $\\overline{b_1b_2...b_n}$. We will show that for each $i=1, \\dots, n$, either $a_{n+1-i} = a_i = 0$ or $a_{n+1-i} + a_i = 11$. The statement is obviously true for $i=1$, since $b_n=1$, but $a_n+a_1 \\ne 1$ due to $a_n>0$, $a_1>0$. Now let $i>1$ and assume that the statement holds for $i-1$. Consider the case $a_{n+1-(i-1)} + a_{i-1} = 11$. Depending on whether or not there is a carry in the $(n+1-i)$-th position from the end, $b_{i-1}=2$ or $b_{i-1}=1$. Accordingly, $b_{n-(i-1)}=2$ or $b_{n-(i-1)}=1$, that is, $b_{n+1-i}=2$ or $b_{n+1-i}=1$. Since there is a carry in the $(i-1)$-th position from the end, the last digit of $a_{n+1-i}+a_i$ must be 1 or 0, respectively.\nSimilarly, in the case $a_{n+1-(i-1)} = a_{i-1} = 0$, depending on whether or not there is a carry in the $(n+1-i)$-th position from the end, $b_{i-1}=1$ or $b_{i-1}=0$. Accordingly, $b_{n-(i-1)}=1$ or $b_{n-(i-1)}=0$, that is, $b_{n+1-i}=1$ or $b_{n+1-i}=0$. Since there is no carry in the $(i-1)$-th position from the end, the last digit of $a_{n+1-i}+a_i$ is 1 or 0, respectively.\nIn conclusion, we have shown that if there is a carry in the $(n+1-i)$-th position from the end, the last digit of $a_{n+1-i}+a_i$ is 1, and if there is no carry in the $(n+1-i)$-th position from the end, the last digit of $a_{n+1-i}+a_i$ is 0. In the first case, $a_{n+1-i}+a_i = 11$, and in the second case, $a_{n+1-i} = a_i = 0$. This completes the induction step and proves the desired statement.\n\nAll described numbers trivially satisfy the problem's condition. Let's count them. If $n$ is even, the number of $n$-digit numbers for which no carry occurs when adding the number and the number obtained by reversing its digits is $36 \\cdot 55^{\\frac{n-2}{2}}$, because the number of pairs of digits $(a, b)$ whose sum is less than 10 is $8+7+\\dots+1$ or 36, when zero is not allowed (as in the first and last positions), and $10+9+\\dots+1$ or 55, when zero is allowed (as in the other positions). The number of $n$-digit numbers for which the sum of the first and last digits is 11 and the sum of the digits equidistant from the middle is either 0 or 11 is $8 \\cdot 9^{\\frac{n-2}{2}}$. Similarly, if $n$ is odd, the number of $n$-digit numbers for which no carry occurs when adding the number and the number obtained by reversing its digits is $36 \\cdot 55^{\\frac{n-3}{2}} \\cdot 5$. The number of $n$-digit numbers for which the sum of the first and last digits is 11 and the sum of the digits equidistant from the middle is either 0 or 11 is $8 \\cdot 9^{\\frac{n-3}{2}}$. Thus, the total number of such numbers is $36 \\cdot 55^{\\frac{n-2}{2}} + 8 \\cdot 9^{\\frac{n-2}{2}}$ if $n$ is even and $36 \\cdot 55^{\\frac{n-3}{2}} \\cdot 5 + 8 \\cdot 9^{\\frac{n-3}{2}}$ if $n$ is odd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which it is possible to write all positive integers from $1$ to $n$ in some order on a circle, each exactly once, such that every number is coprime with the number that is two positions away from it.",
    "options": [],
    "answer": "All positive integers n with n ≡ 0, 1, or 3 mod 4 (i.e., n not ≡ 2 mod 4).",
    "solution": "Let $n$ be odd. Write the number $1$ at some point on the circle and, after every $\\frac{2}{n}$ full turns, write the next number by size (Fig. 45 shows the case for $n = 11$). In this way, all the numbers from $1$ to $n$ can be written in different points. Choosing any number on the circle and the number two positions away from it, we have either consecutive integers or the numbers $n$ and $1$. In both cases, the chosen numbers are coprime.\n\nIf $n$ is divisible by $4$, let $n = 2m$, where $m$ is an even number. Like described for odd $n$, write the numbers from $1$ to $m$ on the circle and in the remaining points, write the numbers from $m + 1$ to $n$ in order (Fig. 46 shows the case for $n = 12$). Choosing any number on the circle and the number two positions away from it, we have either consecutive integers or the numbers $m$ and $1$ or the numbers $n$ and $m + 1$. In the first two cases, the chosen numbers are coprime. If the numbers $n$ and $m + 1$ had a common divisor $d$, then $d$ would also be a divisor of $n - (m + 1)$, that is, $m - 1$, and also a divisor of $(m + 1) - (m - 1)$, that is, $2$. But $d \\neq 2$ because $m$ is even, so $m + 1$ is odd and has no factor $2$. Thus, the only possibility is $d = 1$, meaning $\\gcd(n, m + 1) = 1$. Therefore, the chosen numbers are coprime in all cases.\n\n![](attached_image_1.png)\nFig. 45\n![](attached_image_2.png)\nFig. 46\n\nNow let $n$ be an even number that is not divisible by $4$, i.e., $n = 2m$, where $m$ is an odd number. Suppose that the positive integers from $1$ to $n$ are arranged in some order on a circle. This divides these numbers into two groups, each containing $m$ numbers: the first group contains the number $1$ and all those that can be reached by repeatedly jumping to the number two positions away, while the second group contains all the remaining numbers. Among all the written numbers, there are $m$ even numbers, and at least half of them must belong to one group; since $m$ is odd, more than half of the even numbers must belong to one group. This means that there must be two even numbers that are consecutive within the group, i.e., one is two positions away from the other on the circle. These two numbers have a common factor $2$. Therefore, it is not possible to arrange the numbers from $1$ to $n$ on the circle such that every number is coprime with the number two positions away from it.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30038,
    "subject": "Mathematics (Multi-modal)",
    "question": "All integers from $1$ to $12$ are written on the edges of a cube so that every edge has exactly one integer. Two robot ants stand in the same vertex of the cube and wish to arrive at the vertex that is furthest away from them. Each ant picks a path consisting of exactly three edges of the cube. They multiply the integers on their path. The product of the integers on the path of the first ant is divisible by $100$, but not by $200$. The sum of the digits of the product of the integers on the path of the second ant is $2$ and the integer $1$ is not on its path. Find all possible products that either ant can get.",
    "options": [],
    "answer": "First ant: 300; Second ant: 110",
    "solution": "Since $100$ is divisible by $5^2$, but none of the numbers on the edges of the cube are divisible by $5^2$, the path of the first ant has to contain two numbers that are divisible by $5$. The only two such numbers are $5$ and $10$. Their product is divisible by $2$, but not by $2^2$. Since $100 = 2^2 \\cdot 5^2$, the third number on the path of the first ant has to be divisible by $2$ but not by $2^2$. The numbers on the edges of the cube that are divisible by $2$, but not by $2^2$, are $2$, $6$ and $10$. Since $10$ has already been used, the third number on the path of the first ant has to be either $2$ or $6$.\n\nFor the sum of the digits of the product of the second ant to be $2$, it has to consist of either one two or two ones, in addition to any number of zeroes. Since the product of the three largest numbers on the edges of the cube is $12 \\cdot 11 \\cdot 10 = 1320$, the possible products for the second ant are $2$, $11$, $20$, $101$, $110$, $200$, $1001$, $1010$ and $1100$. Of these $2$, $11$ and $101$ are prime numbers and cannot be the product of three integers that are larger than $1$. The number $20$ can also be eliminated, since the only way to write it as a product of three integers larger than one is $2 \\cdot 2 \\cdot 5$, which includes the number $2$ twice. The numbers $1001$ and $1010$ can be eliminated, since $1001 = 7 \\cdot 11 \\cdot 13$ and there is no $13$ on the edges of the cube, and $1010 = 2 \\cdot 5 \\cdot 101$ and there is no $101$ on the edges of the cube. This leaves $110$, $200$ and $1100$.\n\nNow note that $200$ and $1100$ are both divisible by $5^2$. To get these products, there have to be two numbers on the path of the second ant that are divisible by $5$. The only such numbers are $5$ and $10$. Hence we can eliminate $1100$, since to get this product the third number would have to be $22$, which is not on the cube. To get the product $200$, the numbers should be $5$, $10$ and $4$. To get the product $110$ the numbers should be $2$, $5$ and $11$.\n\nHowever, the paths of the two ants can have either $0$, $1$ or $3$ edges in common, but not exactly $2$ edges. Hence the numbers on the path of the second ant cannot be $5$, $10$ and $4$, and so the numbers on the path of the second ant have to be $2$, $5$ and $11$. This leaves the first ant with the numbers $5$, $6$ and $10$, which give a product of $300$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30039,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Does there exist a positive integer $n$ such that the eight last digits of the number $n^2 + 1$ are the same as in the number $2n$, but the ninth digit from the end of these two numbers are different?\n\nb. Does there exist a positive integer $n$ such that the nine last digits of the number $n^2 + 1$ are the same as in the number $2n$, but the tenth digit from the end of these two numbers are different?",
    "options": [],
    "answer": "a) Yes; for example n = 100010001. b) No; such an integer does not exist.",
    "solution": "The condition that the last $k$ digits of two numbers are the same is fulfilled if and only if the difference of these two numbers ends with exactly $k$ zeroes. Note that $n^2 + 1 - 2n = (n-1)^2$.\n\na. Let $n = 100010001$, then the number $n-1$ ends with 4 zeroes and the fifth digit from the end is 1. Hence the number $(n-1)^2$ ends with 8 zeroes and the ninth digit from the end is 1. Hence this $n$ fits.\n\nb. If a number ends with exactly $k$ zeroes, then the square of this number ends with exactly $2k$ zeroes. Hence $(n-1)^2$ cannot end with exactly 9 zeroes, since 9 is odd, and so there are no such integers $n$ that would fulfill the condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30040,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the base $BC$ of the isosceles triangle $ABC$ points $D$ and $E$ are chosen such that $BD = EC = 7$ cm. A line perpendicular to side $AB$ and passing through point $A$ intersects base $BC$ at point $F$. An altitude of triangle $ACE$ is $EG$, a median of triangle $AEF$ is $AH$. It is known that $\\angle DAF = \\frac{1}{2}\\angle BAC$. Find the length of the segment $GH$.",
    "options": [],
    "answer": "3.5 cm",
    "solution": "Since triangle $BAC$ is isosceles (Fig. 25), its base angles are equal; denote $\\angle ABC = \\angle BCA = \\beta$. Since the sum of the interior angles of triangle $BAC$ is $180^\\circ$, we have $\\angle BAC = 180^\\circ - 2\\beta$. Then $\\angle DAF = \\frac{180^\\circ - 2\\beta}{2} = 90^\\circ - \\beta$ because $\\angle DAF = \\frac{1}{2}\\angle BAC$. Since $FA$ is perpendicular to line $BA$,\n$$\n\\angle BAD = \\angle BAF - \\angle DAF = 90^\\circ - (90^\\circ - \\beta) = \\beta.\n$$\nThus triangle $ADB$ is isosceles, because angles $DBA$ and $BAD$ are equal.\nConsequently $AD = BD = 7$ cm.\n![](attached_image_1.png)\nFig. 25\nAs points $D$ and $E$ are symmetric w.r.t. the perpendicular bisector of $BC$ and so are points $B$ and $C$, we also have $AE = CE = 7$ cm. Hence the triangle $AEC$ is isosceles, too. From triangle $BAF$ we now get\n$$\n\\angle AFB = \\angle AFD = 180^\\circ - 90^\\circ - \\beta = 90^\\circ - \\beta.\n$$\nThus triangle $DAF$ is isosceles, because angles $DAF$ and $DFA$ are equal. Hence $DF = 7$ cm. Therefore $DF = EC$.\nAs the altitude driven from the apex of an isosceles triangle bisects the base, $G$ must be the midpoint of $AC$. Since $H$ is a midpoint of $EF$ and $DF = EC$ (Figures 26 and 27 depict two possible cases), $H$ is also a midpoint of $DC$. In conclusion $HG$ is a midsegment of triangle $ACD$ parallel to side $AD$. So $HG = \\frac{1}{2}AD = 3.5$ cm.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku and Miku play the following game on a grid of dimensions $n \\times m$: In the beginning, all unit squares are white. Each player on their turn paints one white unit square either red or blue of their choice, but no two unit squares with a common side or a common vertex can be painted the same color. The players take turns and Juku starts. A player who cannot make the allowed move has lost. Is it possible for Juku to win the game regardless of how Miku plays if:\n\na) $n = 2023$ and $m = 2023$;\n\nb) $n = 2023$ and $m = 2024$;\n\nc) $n = 2024$ and $m = 2024$?",
    "options": [],
    "answer": "a) Yes; b) No; c) No",
    "solution": "If $n$ and $m$ are even, then there is a middle square on the grid. Let Juku paint the middle square any color on the first move. From now on, each of Juku's moves should mirror Miku's last move relative to the center of the grid. If before Miku's move the position is symmetrical with respect to the center of the grid, then Juku can certainly respond symmetrically, and before Miku's next move the position is again symmetrical with respect to the center of the grid. Therefore Juku always wins.\n\nIf $n$ or $m$ is even, then Miku can mirror Juku's last move with respect to the center of the grid in each of his moves, but with the opposite color. Then, before each move by Juku the unit squares symmetrical to the center are of the opposite color. So if Juku can make a move, Miku can respond symmetrically to the center. Consequently with this strategy Miku wins. It follows that on $2023 \\times 2023$ grid Juku can win for any moves of Miku, on $2023 \\times 2024$ and $2024 \\times 2024$ grids it is not always possible.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Do there exist positive integers $x$ and $y$ such that\n$$\n11x^5 + 33y = 13y^5 + 31x + 2024?\n$$",
    "options": [],
    "answer": "No",
    "solution": "The given equation is equivalent to\n$$\n(11x^5 - 31x) + (33y - 13y^5) = 2024.\n$$\nThe fifth power of any positive integer ends with the same digit as the number itself. Therefore, $11x^5$ ends with the same digit as $11x$, which in turn ends with the same digit as $x$. Since $31x$ also ends with the same digit, the difference $11x^5 - 31x$ ends with zero. Similarly, the difference $33y - 13y^5$ ends with zero. However, the sum of numbers ending in zero cannot end in four. Therefore, there are no integers that satisfy the given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest positive integer $n$ for which there exist two distinct pairs of positive integers $(x, y)$ such that $(x^2 - 1)(y^2 - 1) = n$ and $x \\le y$.",
    "options": [],
    "answer": "360",
    "solution": "Notice that for $n = 360$, the pairs $x = 2, y = 11$ and $x = 4, y = 5$ satisfy the condition.\n\nWe will show that for smaller numbers $n$, there do not exist two distinct suitable pairs $(x, y)$. If $x = 1$, then $(x^2 - 1)(y^2 - 1)$ is not positive. Thus, we can assume that $1 < x \\le y$. Now, as $x$ or $y$ increases, $(x^2 - 1)(y^2 - 1)$ also increases. Let's examine the cases.\n\n* If $x = 2$, then $y = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, \\dots$ give us respectively $(x^2 - 1)(y^2 - 1) = 9, 24, 45, 72, 105, 144, 189, 240, 297, 360, \\dots$.\n* If $x = 3$, then $y = 3, 4, 5, 6, 7, \\dots$ give similarly $(x^2 - 1)(y^2 - 1) = 64, 120, 192, 280, 384, \\dots$.\n* If $x = 4$, then $y = 4, 5, \\dots$ give $(x^2 - 1)(y^2 - 1) = 225, 360, \\dots$.\n\nNo positive number smaller than 360 appeared repeatedly. If we continue the inspection for $x = 5, 6, \\dots$, even the first case $y = x$ would give $(x^2 - 1)(y^2 - 1) > 360$, since $x = 4$ gives 360 or a larger number for the same $y$. In conclusion, 360 is the smallest number with the required property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of natural numbers $(x, y, z)$ satisfying the system of equations\n$$\n\\begin{cases} x + y - z = 23, \\\\ x^2 + y^2 - z^2 = 23. \\end{cases}\n$$",
    "options": [],
    "answer": "The solutions are (24, 276, 277), (276, 24, 277), (34, 46, 57), and (46, 34, 57).",
    "solution": "Substituting $z = x + y - 23$ from the first equation into the second yields $x^2 + y^2 - (x + y - 23)^2 = 23$ which simplifies to\n$xy - 23x - 23y = -23 \\cdot 12.$\nAdding $23 \\cdot 23$ to both sides and factoring yields\n$$\n(x - 23)(y - 23) = 23 \\cdot 11.\n$$\nAs $23$ and $11$ are primes, the only factors on the right hand side are $1$, $11$, $23$ and $11 \\cdot 23$. Thus $x = 23 + 1 = 24$, $x = 23 + 11 = 34$, $x = 23 + 23 = 46$ or $x = 12 \\cdot 23 = 276$; The corresponding values of $y$ are $276$, $46$, $34$, $24$ and the values of $z$ are $277$, $57$, $57$, $277$. The negative factors of $23 \\cdot 11$ don't yield solutions, as $z$ would be negative.\nSubstituting $z = x + y - 23$ from the first equation into the second and simplifying like in Solution 1 yields\n$$\nxy = 23x + 23y - 23 \\cdot 12. \\tag{1}\n$$\nThe right hand side of (1) is divisible by $23$, so the left side must be as well.\nThus $x$ or $y$ is divisible by $23$. Let $x = 23k$ (in the other case we may swap $x$ and $y$ due to symmetry). Substituting it into (1) and dividing the sides by $23$ yields $ky = 23k + y - 12$, from which $k = \\frac{y-12}{y-23} = 1 + \\frac{11}{y-23}$. For $k$ to be a positive integer, there are two possibilities:\n* if $y - 23 = 11$, then $k - 1 = 1$, which yields $y = 34$, $x = 46$, $z = 57$;\n* if $y - 23 = 1$, then $k - 1 = 11$, which yields $y = 24$, $x = 276$, $z = 277$.\nIn addition to those, we get the solutions with $x$ and $y$ swapped.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30045,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $X$ is chosen on the median $AD$ of triangle $ABC$. The circumcircle of triangle $ABX$ intersects the median $BE$ of triangle $ABC$ at point $Y \\neq B$. The circumcircle of triangle $EXY$ intersects the line $DE$ at point $K \\neq E$. Prove that the location of $K$ does not depend on $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will use directed angles (Fig. 3 and 4 depict both possible configurations). As $DE$ is a midline in $ABC$, we have $DE \\parallel AB$. Thus\n$$\n\\angle DEY = \\angle DEB = \\angle ABE = \\angle ABY = \\angle AXY = \\angle DXY.\n$$\nTherefore points $D$, $E$, $X$, $Y$ are concyclic. This means that the circumcircle of $EXY$ intersects $DE$ at $D$. Thus $K = D$, i.e. the midpoint of $BC$, regardless of the choice of $X$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku and Miku are playing the following game. In the beginning, there is a positive integer on the board. Each turn, a player subtracts from the number on the board a non-zero digit that appears in his or his opponent's ID code, and replaces the number on the board with the result. Players take turns, Juku starts. The player whose move ends up in a negative number on the board loses. Prove that, among any 10 consecutive positive integers, there is a number $n$ such that, if initially the number $n$ is on the board, then Juku can win the game regardless of his opponent's counterplay.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a, a+1, \\dots, a+9$ be 10 arbitrary consecutive positive integers. If there exists a number $n$ among $a, a+1, \\dots, a+8$ for which Juku has a winning strategy, then we are done. We will now assume that if any of the numbers $a, a+1, \\dots, a+8$ is on the board, then the active player loses if his opponent plays perfectly. Then, if $n = a+9$, Juku can start by subtracting any non-zero digit, after which the number on the board is one of $a, a+1, \\dots, a+8$, putting Miku in a losing position, which means that Juku will win. There must be a non-zero digit in either code, as both codes cannot be strings of zeroes, as they must be distinct and have the same length.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30047,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the classroom there is a clock, whose minute and hour hands move with a constant angular velocity. The minute hand works correctly, but the hour hand moves at half the angular velocity of the minute hand. At 10:00, the clock shows the correct time. When will the clock show the correct time for the next time?",
    "options": [],
    "answer": "12:24",
    "solution": "Imagine a correctly working clock next to the broken clock. The hour hand of the working clock moves 12 times slower than its minute hand, which means that its hour hand moves 6 times slower than the hour hand of the broken clock. As the minute hand of the broken clock is always at the correct position, the broken clock next shows the correct time when its hour hand has moved one more full circle compared to the hour hand of the correct clock, meaning that their indicated times differ by 12 hours.\n\nDenote the number of hours passed by this point by $x$. Based on the previous information, we have the equation $6x = x + 12$ or $5x = 12$, which yields $x = 2.4$. Thus the broken clock shows the correct time after 2 hours and $0.4 \\cdot 60 = 24$ minutes or at 12:24.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Non-negative integers $a$, $b$, $q$, $r$ are all less than $5$ and satisfy conditions $q < a$ and $r < b \\le a$. Dividing $qb + r$ by $5$ gives a remainder of $a$. Can we be certain that dividing $a$ by $b$ gives a remainder of $r$?",
    "options": [],
    "answer": "No",
    "solution": "*Solution 1:* The numbers $a = b = 3$ and $q = r = 2$ satisfy the conditions of the problem, but dividing $3$ by $3$ gives a remainder of $0$, not $2$.\n\n\n*Solution 2:* The numbers $a = b = 4$, $q = 2$ and $r = 1$ satisfy the conditions of the problem, but dividing $4$ by $4$ gives a remainder of $0$, not $1$.\n\n\n*Solution 3:* The numbers $a = 4$, $b = 3$, $q = 3$ and $r = 0$ satisfy the conditions of the problem, but dividing $4$ by $3$ gives a remainder of $1$, not $0$.\n\n\n*Solution 4:* The numbers $a = b = 4$, $q = 3$ and $r = 2$ satisfy the conditions of the problem, but dividing $4$ by $4$ gives a remainder of $0$, not $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which you can replace a digit $4$ with the digits $22$ and obtain a number divisible by $n$.",
    "options": [],
    "answer": "All n of the form 45*10^t or 2475*10^t for integer t >= 0.",
    "solution": "Let $n$ satisfy the condition and $n'$ be the number obtained by the replacement of digits. Denote by $x$ the number formed by the digits to the left of the replaced digit and denote by $y$ the number formed by the digits to the right of the replaced digit. Let $k$ be the number of digits in $y$. Then $n = x \\cdot 10^{k+1} + 4 \\cdot 10^k + y$ and $n' = x \\cdot 10^{k+2} + 22 \\cdot 10^k + y$. As $n'$ is divisible by $n$, then so must also be $n' - n$ and $10n - n'$, which yields\n$$\nx \\cdot 10^{k+1} + 4 \\cdot 10^k + y \\mid 9x \\cdot 10^{k+1} + 18 \\cdot 10^k, \\qquad (2)\n$$\n$$\nx \\cdot 10^{k+1} + 4 \\cdot 10^k + y \\mid 18 \\cdot 10^k + 9y. \\qquad (3)\n$$\nWe will consider the following cases.\n\n* If $x = 0$, then by (2), $t (4 \\cdot 10^k + y) = 18 \\cdot 10^k$ for some integer $t$. As\n$5 (4 \\cdot 10^k + y) \\ge 5 \\cdot 4 \\cdot 10^k > 18 \\cdot 10^k > 3 \\cdot 5 \\cdot 10^k > 3 (4 \\cdot 10^k + y)$,\nthe only option is $t = 4$. The equation $4 (4 \\cdot 10^k + y) = 18 \\cdot 10^k$ yields\n$y = \\frac{2 \\cdot 10^k}{4} = 5 \\cdot 10^{k-1} < 10^k$. Altogether $n = 45 \\cdot 10^{k-1}$.\n\n* If $x = 1$, then by (2), $t (14 \\cdot 10^k + y) = 108 \\cdot 10^k$ for some integer $t$. As\n$8 (14 \\cdot 10^k + y) \\ge 8 \\cdot 14 \\cdot 10^k > 108 \\cdot 10^k > 6 \\cdot 15 \\cdot 10^k > 6 (14 \\cdot 10^k + y)$,\nthe only option is $t = 7$, but the equation $7 (14 \\cdot 10^k + y) = 108 \\cdot 10^k$ has no integer solutions, because $7 \\nmid 108 \\cdot 10^k$.\n\n* If $x = 2$, then by (2), $t (24 \\cdot 10^k + y) = 198 \\cdot 10^k$ for some integer $t$. As\n$9 (24 \\cdot 10^k + y) \\ge 9 \\cdot 24 \\cdot 10^k > 198 \\cdot 10^k > 7 \\cdot 25 \\cdot 10^k > 7 (24 \\cdot 10^k + y)$,\nthe only option is $t = 8$. The equation $8 (24 \\cdot 10^k + y) = 198 \\cdot 10^k$ yields $y = \\frac{6 \\cdot 10^k}{8} = 75 \\cdot 10^{k-2} < 10^k$. Altogether $n = 2475 \\cdot 10^{k-2}$.\n\n* If $x \\ge 3$, then by (3) we have $x \\cdot 10^{k+1} + 4 \\cdot 10^k + y \\le 18 \\cdot 10^k + 9y$, which yields $x \\cdot 10^{k+1} \\le 14 \\cdot 10^k + 8y$. On the other hand $x \\cdot 10^{k+1} \\ge 30 \\cdot 10^k > 22 \\cdot 10^k > 14 \\cdot 10^k + 8y$. The equations contradict each other, so no such $n$ can exist.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $2023^{10} > 10 \\cdot (1^9 + 2^9 + 3^9 + \\dots + 2022^9)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Multiplying out $(k+1)^{10}$ yields among others the terms $k^{10}$ and $10k^9$. All of the other terms are positive for a positive $k$. Thus for all positive integers $k$ we have $(k+1)^{10} > k^{10} + 10k^9$. Combining this for $k = 2022, 2021, \\dots, 2, 1$ yields\n$$\n\\begin{aligned}\n2023^{10} &> 2022^{10} + 10 \\cdot 2022^9 \\\\\n&> 2021^{10} + 10 \\cdot 2021^9 + 10 \\cdot 2022^9 \\\\\n&> \\dots \\\\\n&> 1^{10} + 10 \\cdot 1^9 + 10 \\cdot 2^9 + \\dots + 10 \\cdot 2021^9 + 10 \\cdot 2022^9 \\\\\n&> 10 \\cdot (1^9 + 2^9 + 3^9 + \\dots + 2022^9).\n\\end{aligned}\n$$\nWe will prove that for all integers $a, n$ greater than 1,\n$$\na^n > n \\cdot (1^{n-1} + 2^{n-1} + 3^{n-1} + \\dots + (a-1)^{n-1}); \\quad (4)\n$$\nwhere taking $a = 2023$, $n = 10$ yields the desired result.\nWe will prove this via mathematical induction on the variable $a$. For $a = 2$, the inequality $2^n > n \\cdot 1^{n-1} = n$ is well known. For the induction step we will show that if (4) holds, then so does\n$$\n(a+1)^n > n \\cdot (1^{n-1} + 2^{n-1} + 3^{n-1} + \\dots + (a-1)^{n-1} + a^{n-1}).\n$$\nWe will show that the left hand side increases by at least as much as the right hand side or $(a+1)^n - a^n \\ge n \\cdot a^{n-1}$. Using the formula for the difference of powers together with $a+1 > a$ yields\n$$\n\\begin{aligned} (a+1)^n - a^n &= ((a+1)-a)((a+1)^{n-1} + (a+1)^{n-2}a + \\dots + (a+1)a^{n-2} + a^{n-1}) \\\\ &> 1 \\cdot (a^{n-1} + a^{n-1} + \\dots + a^{n-1} + a^{n-1}) = n \\cdot a^{n-1}, \\end{aligned}\n$$\nwhich proves the induction step and thus finishes the proof.\nAs the function $y = x^9$ is increasing in $[1; 2023]$, we have\n$$\n1^9 + 2^9 + 3^9 + \\dots + 2022^9 < \\int_1^{2023} x^9 dx.\n$$\nAs the integral of $y = x^9$ is $y = \\frac{1}{10}x^{10}$, the Newton-Leibniz formula gives\n$$\n\\int_{1}^{2023} x^{9} dx = \\frac{1}{10} \\cdot 2023^{10} - \\frac{1}{10} \\cdot 1^{10} < \\frac{1}{10} \\cdot 2023^{10}.\n$$\nAltogether, $10 \\cdot (1^9 + 2^9 + 3^9 + \\dots + 2022^9) < 2023^{10}$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30051,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ with $AB < AC$, the altitudes $BE$ and $CF$ intersect at $H$. The tangent to the circumcircle of $ABC$ at $A$ intersects the circumcircle of $AEF$ at $T \\neq A$. The circumcircles of $TBE$ and $TCF$ intersect at $K \\neq T$. Prove that $\\angle KHB = \\angle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From $\\angle HEA = 90^\\circ = 180^\\circ - 90^\\circ = 180^\\circ - \\angle HFA$ (Fig. 5) we deduce that $AEHF$ is cyclic. By definition, $T$ lies on this circle as well, so\n$$\n\\angle THE = 180^\\circ - \\angle TAE = 180^\\circ - \\angle TAC = \\angle ABC,\n$$\nwhere the final equality is due to the tangent-chord theorem.\n\nLet $K'$ be the intersection of the lines $TH$ and $BC$ (Fig. 6). It's sufficient to show that $K' = K$ or that $K'$ lies on the circumcircles of $TBE$ and $TCF$, as we have $\\angle K'HB = \\angle THE = \\angle ABC$. To show this we notice that\n$$\n\\angle ETK' = \\angle ETH = \\angle EAH = \\angle CAH = 90^\\circ - \\angle ACB = \\angle EBC = \\angle EBK',\n$$\n![](attached_image_1.png)\nFig. 5\n![](attached_image_2.png)\nFig. 6\n\nwhich shows that the points $T$, $B$, $K'$ and $E$ are concyclic. Analogously we show that $T$, $C$, $K'$ and $F$ are concyclic. Thus $K' = K$, as desired.\nAnalogously to Solution 1 we show that $\\angle THE = \\angle ABC$.\n\nSince $\\angle BEC = \\angle BFC = 90^\\circ$, we see that $BCEF$ is cyclic. Thus the lines $TK$, $BE$ and $CF$ are the radical axes of the circumcircles of $BCEF$, $TCF$ and $TBE$. As these intersect at a point, we have that $H$ lies on the line $TK$. This yields $\\angle KHB = \\angle THE = \\angle ABC$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mari and Jüri play the following game on an infinite grid: They take turns with Mari starting, Mari colours one uncolored square red in each of her turns, Jüri colours one uncolored square blue in each of his turns. If the centers of any four squares of the same colour form the corners of some square, then this player wins. Does either player have a winning strategy? If so, what is the least number of turns to guarantee a victory?",
    "options": [],
    "answer": "Mari has a winning strategy; the least number of her turns needed to guarantee victory is 5.",
    "solution": "We will first prove that Mari can win in 5 turns. We will denote squares in the grid by the coordinates of their centers, taking Mari's first square to be $(0,0)$. W.l.o.g., let Jüri's first turn be $(1,0)$, allowing him to also color squares with fractional coordinates in future turns (Fig. 7). Mari will color $(2,0)$ on her second move (Fig. 8). W.l.o.g., assume that the $y$-coordinate of Jüri's second move is nonpositive.\nThe only option for him to color three vertices of some square together with $(1,0)$ and $(0,2)$ is by choosing $(-1,-1)$ (Fig. 9). The only option for him to color three vertices of some square together with $(1,0)$ and $(2,2)$ is by\n![](attached_image_1.png)\nFig. 8\n![](attached_image_2.png)\nFig. 9\n![](attached_image_3.png)\nFig. 10\n---\nchoosing $(3, -1)$ (Fig. 10). Thus we consider three cases.\n\n* If Jüri colors $(-1, -1)$ or $(3, -1)$ on his second turn, let it be $(-1, -1)$ without loss of generality (due to symmetry). Then Mari will color $(0, 2)$ (Fig. 11), after which Jüri is forced to color $(2, 2)$ to avoid immediate defeat. Then Mari can color $(-2, 0)$ (Fig. 12), threatening to win by $(-2, 2)$ or $(0, -2)$. Jüri cannot block both squares at once or win himself, so Mari will win on the next move.\n\n* If Jüri colors $(-2, 0)$, $(0, -2)$, $(2, -2)$ or $(4, 0)$, then assume without loss of generality that the $x$-coordinate of this move is greater than $1$. Mari will color $(0, 2)$ again, after which Jüri is forced to color $(2, 2)$ to avoid immediate defeat (the two possible positions are shown in Figures 13 and 14). As $(-2, 2)$, $(-2, 0)$ and $(0, -2)$ are all uncolored, Mari can win exactly like in the previous case.\n\n* If Jüri colors any other square on his second turn, then Mari can color $(0, 2)$ and win just like in the other cases.\n\n![](attached_image_4.png)\n![](attached_image_5.png)\n![](attached_image_6.png)\n![](attached_image_7.png)\nFig. 11\nFig. 12\nFig. 13\nFig. 14\n\nWe finally note that Mari cannot win in 4 moves, as after her 3rd move there exists at most 1 square which would yield an immediate victory for her, so Jüri can just block it.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a positive integer $n$ supereven, if its largest odd factor $d$ is less than $\\frac{n}{d}$. How many positive integers less than 1000 are supereven?",
    "options": [],
    "answer": "46",
    "solution": "Let $n$ be supereven and $d$ its greatest odd divisor. If $\\frac{n}{d}$ were divisible by some odd $p > 1$, then $pd > d$ would also be a factor of $n$, contradiction. Thus $\\frac{n}{d}$ is a power of 2. Hence supereven numbers are exactly those that can be expressed as a product of a power of two and an odd number less than that power of two.\nWe will find the supereven numbers less than 1000 by their largest odd factors $d$.\n\n* If $d = 1$, then the power of 2 can be one of 2, 4, 8, 16, 32, 64, 128, 256, 512. We obtain 9 supereven numbers.\n\n* If $d = 3$, then the power of 2 can be one of 4, 8, 16, 32, 64, 128, 256. We obtain 7 supereven numbers.\n\n* If $d = 5$ or $d = 7$, then the power of 2 can be one of 8, 16, 32, 64, 128. We obtain $2 \\cdot 5 = 10$ supereven numbers.\n\n* If $d$ is one of the numbers 9, 11, 13, 15, then the power of 2 can be one of 16, 32, 64. We obtain $4 \\cdot 3 = 12$ supereven numbers.\n\n* If $d$ is one of the numbers 17, 19, 21, 23, 25, 27, 29, 31, then the power of 2 can be only 32. We obtain 8 supereven numbers.\n\nThus there are $9 + 7 + 10 + 12 + 8 = 46$ supereven numbers less than 1000.\nLike in Solution 1, we will show that supereven numbers are exactly those that can be expressed as a product of a power of two and an odd number less than that power of 2. We will call this power of 2 the *even part* of a supereven integer. As every other integer is odd, there are $\\frac{2^k}{2} = 2^{k-1}$ odd positive integers less than $2^k$. The exponents $k = 1, 2, 3, 4, 5$ thus yield 1, 2, 4, 8, 16 supereven integers with an even part of $2^k$. Of those, the greatest is $2^5 \\cdot (2^5 - 1) = 992 < 1000$, so all of them are less than 1000. For $k = 6, 7, 8, 9$, the greatest odd factor keeping the product under 1000 is 15, 7, 3, 1 respectively, yielding $8 + 4 + 2 + 1 = 15$ more supereven numbers less than 1000. Thus the total number of supereven integers less than 1000 is $1 + 2 + 4 + 8 + 16 + 15 = 46$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two rectangles are drawn on a piece of paper. The length and width of one rectangle are both $5$ cm greater than the corresponding measures of the other rectangle. The area of the larger rectangle is $1$ dm$^2$ greater than the area of the smaller rectangle. Find the perimeter of the smaller rectangle.",
    "options": [],
    "answer": "30 cm",
    "solution": "Let the sidelengths of the smaller rectangle (in cm) be $a$ and $b$. Then the sidelengths of the larger rectangle are $a+5$ and $b+5$. The area of the smaller rectangle is $ab$, whereas the area of the larger rectangle is $(a+5)(b+5) = ab + 5a + 5b + 25$, which by the information given is equal to $ab + 100$. Thus $5a + 5b + 25 = 100$, which yields $a+b = 15$. Thus the perimeter of the smaller rectangle (in cm) is $2(a+b) = 30$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Anu and Bert each play the following game with ChatGPT. Anu's game starts with the number $2023!$ on the board, Bert's game with $2024!$ on the board. Each move consists of two parts. First the active player divides the number on the board by one of its composite factors $d$. Then their opponent must do one of the following actions:\n(1) multiply the number on the board by any factor $d'$ of $d$ satisfying $1 < d' < d$;\n(2) multiply the number on the board by $7$;\n(3) divide the number on the board by $10$ and multiply the result by $2023$ (this option cannot be chosen if the result is not an integer).\nOn the next move, players interchange their roles. The player who cannot make the required action loses the game. ChatGPT starts both games. Prove that ChatGPT can win at least one of the games.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As $2024! = 2023! \\cdot 2024$ and $2024 = 2^3 \\cdot 11^1 \\cdot 23^1$, the sum of the exponents of all primes in the prime factorisation of the number on the board is odd in one of the games and even in the other. We will show that ChatGPT can win the game where the sum is even.\n\nOn any of its moves, ChatGPT chooses $d$ to be a product of exactly two (not necessarily distinct) primes dividing the number on the board. Then, no matter what its opponent does, the sum of exponents will increase by $1$. This is obvious for the first two choices; for the third choice we notice that $10 = 2 \\cdot 5$ has two prime factors and $2023 = 7 \\cdot 17^2$ has three prime factors. Thus the move as a whole decreases the sum of exponents by $1$.\n\nWhatever the opponent chooses as $d$, ChatGPT chooses the first option with\n\nd'$ = \\frac{d}{p}$, where $p$ is any prime factor of $d$. Thus the move as a whole decreases the sum of exponents by $1$.\n\nContinuing with this strategy, ChatGPT ensures that at the beginning of each of its turns the sum of exponents is even. As the sum of exponents decreases by one during every move, there will eventually be a situation where ChatGPT's opponent will start its move with the sum of exponents being $1$, i.e. with a prime on the board. Thus the player cannot make their move, meaning that ChatGPT wins.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\setminus \\{1\\} \\to \\mathbb{R} \\setminus \\{0\\}$ which satisfy the equations $f(0) = 1$ and $f(f(xy)) = 1 - \\frac{1}{y f(f(x))}$ for any real numbers $x$ and $y$ such that $xy \\neq 0,\\ xy \\neq 1$ and $x \\neq 1$.",
    "options": [],
    "answer": "f(x) = 1/(1 - x)",
    "solution": "Note that if $x \\neq 0$ and $x \\neq 1$ then $f(x) \\neq 0$ and $f(x) \\neq 1$. Indeed, difference from 0 is given in the problem and, if we supposed $f(x) = 1$, taking $y = 1$ would yield $f(f(xy)) = f(f(x)) = f(1)$, meaning that $f(f(xy))$ were undefined and could not satisfy the equation given in the problem. Hence one can apply $f$ infinitely to any real number $x \\notin \\{0, 1\\}$.\n\nPlugging $x = 2, y = \\frac{z}{2}$, where $z \\notin \\{0, 1\\}$, into the original equation yields\n$$\nf(f(z)) = 1 - \\frac{2}{z f(f(f(2)))}.\n$$\nWe see that $f(f(z))$ obtains all real values except 1 and $1 - \\frac{2}{f(f(f(2)))}$, when $z$ obtains all real values except 0 and 1. As $f(f(z))$ cannot obtain the value 0, the only option is $1 - \\frac{2}{f(f(f(2)))} = 0$, i.e., $f(f(z))$ obtains all real values except 0 and 1.\n\nNow substituting $x = z$ and $y = 1$, where still $z \\notin \\{0, 1\\}$, into the original equation yields\n$$\nf(f(z)) = 1 - \\frac{1}{f(f(f(z)))}.\n$$\nDenoting $f(f(z)) = x$, we obtain $x = 1 - \\frac{1}{f(x)}$, i.e.,\n$$\nf(x) = \\frac{1}{1-x} \\qquad (5)\n$$\nThis holds for any real number $x$ except 0 and 1. By the condition of the problem, $f(0) = 1$, which implies the equality (5) also in the case $x = 0$.\n\nAn easy check shows that $f(x) = \\frac{1}{1-x}$ satisfies all conditions of the problem.\nPlugging in $y = \\frac{c}{x}$ for $c \\notin \\{0, 1\\}$ gives $f(f(c)) = 1 - \\frac{x}{c f(f(x))}$.\n\nOn the other hand, $y = 1$ and $x = c$ gives $f(f(c)) = 1 - \\frac{1}{f(f(f(c)))}$, from which we conclude that $\\frac{x}{f(f(f(x)))}$ is constant for all $x \\neq 0, x \\neq 1$. Thus $f(f(f(x))) = a x$ for some constant $a$. Now, after substituting $y = 1$ into the original equation, applying $f(f(f(x))) = a x$ gives\n$$\nf(f(x)) = 1 - \\frac{1}{a x} \\qquad (6)\n$$\nLike in Solution 1, note that $f(x) \\neq 0$, $f(x) \\neq 1$ whenever $x \\neq 0, x \\neq 1$. Thus we can substitute $f(x)$ for $x$ in (6), yielding\n$$\na x = f(f(f(x))) = 1 - \\frac{1}{a f(x)},\n$$\nwhich in turn implies $f(x) = \\frac{1}{a} \\cdot \\frac{1}{1-a x}$. Thus\n$$\n\\begin{aligned}\nf(f(x)) &= \\frac{1}{a} \\cdot \\frac{1}{1-a \\cdot \\frac{1}{a} \\cdot \\frac{1}{1-a x}} = \\frac{1}{a} \\cdot \\left(1 - \\frac{1}{a x}\\right), \\\\\nf(f(f(x))) &= \\frac{1}{a} \\cdot \\left(1 - \\frac{1}{a \\cdot \\frac{1}{a} \\cdot \\frac{1}{1-a x}}\\right) = x.\n\\end{aligned}\n$$\nConsequently, $a = 1$ and $f(x) = \\frac{1}{1-x}$ whenever $x \\neq 0, x \\neq 1$. The condition $f(0) = 1$ implies that $f(x) = \\frac{1}{1-x}$ holds for $x = 0$, too. An easy check shows that $f(x) = \\frac{1}{1-x}$ satisfies all conditions of the problem.\nLike in Solution 1, we show that if $x \\neq 0, x \\neq 1$ then $f$ can be applied to $x$ infinitely many times.\n\nWe note that if $x \\notin \\{0,1\\}$ then $f(f(f(x))) = x$. Indeed, if we suppose the contrary for any $x \\notin \\{0,1\\}$, plugging $y = \\frac{1}{f(f(f(x)))}$ into the given equation yields $f\\left(f\\left(\\frac{x}{f(f(f(x)))}\\right)\\right) = 0$. This violates the condition of the problem.\n\nHence the original equation reduces to $f(f(xy)) = 1 - \\frac{1}{xy}$ whenever $xy \\neq 0, xy \\neq 1, x \\neq 1$. After applying $f$ to both sides of this equation, using $f(f(f(x))) = x$ yields $xy = f\\left(1 - \\frac{1}{xy}\\right)$. Denoting $z = 1 - \\frac{1}{xy}$ gives $f(z) = \\frac{1}{1-z}$ for all real numbers $z$ except 0 and 1, because all such real numbers can be represented in the form $1 - \\frac{1}{xy}$. The condition $f(0) = 1$ implies that $f(x) = \\frac{1}{1-x}$ holds for $x = 0$, too. An easy check shows that $f(x) = \\frac{1}{1-x}$ satisfies all conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $E$ and $F$ are chosen respectively on the sides $CA$ and $AB$ of triangle $ABC$. Lines $BE$ and $CF$ intersect at $P$. Let $Q$ be a point such that $PBQC$ is a parallelogram and $R$ a point such that $AERF$ is a parallelogram. Prove that $PR \\parallel AQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be a point such that $PESF$ is a parallelogram (Fig. 15). We first show that $PR \\parallel AS$. For this note that $\\angle PER = \\angle SFA$ since $PE \\parallel FS$ and $ER \\parallel AF$. Additionally $PE = FS$ and $ER = AF$, thus triangles $PER$ and $SFA$ are congruent. From this we deduce that $PR \\parallel AS$.\n\nIn order to prove the problem statement it now suffices to show that $A$, $S$ and $Q$ are collinear. Let $X$ be the point of intersection of $ES$ and $AF$ and $Y$ be the point of intersection of $FS$ and $AE$ (Fig. 16).\n\nFirst we show that $XY \\parallel BC$. Denote $\\frac{AE}{AC} = \\kappa$ and $\\frac{AF}{AB} = \\lambda$. As $EX \\parallel CF$, we have $\\frac{AX}{AF} = \\frac{AE}{AC} = \\kappa$. Thus $\\frac{AX}{AB} = \\frac{AX}{AF} \\cdot \\frac{AF}{AB} = \\kappa \\cdot \\lambda$. Analogously, we get $\\frac{AY}{AC} = \\kappa \\cdot \\lambda$. Consequently, $\\frac{AX}{AB} = \\frac{AY}{AC}$, which implies $XY \\parallel BC$.\n\nThe fact just proven implies $\\frac{XY}{BC} = \\frac{AX}{AB}$. It suffices to notice that the triangles $BCQ$ and $XYS$ are similar as their respective sides are parallel. Thus $\\overrightarrow{XS} = \\overrightarrow{XY}$ which implies $\\overrightarrow{XS} = \\frac{AX}{AB}$. Since $XS \\parallel BQ$, this implies that the points $A$, $S$ and $Q$ are collinear.\n\n![](attached_image_1.png)\nFig. 15\n![](attached_image_2.png)\nFig. 16\nDenote $\\vec{AC} = \\vec{e}_1$ and $\\vec{AB} = \\vec{e}_2$. Take $\\kappa, \\lambda$ such that $\\vec{AE} = \\kappa \\vec{e}_1$ and $\\vec{AF} = \\lambda \\vec{e}_2$. Then\n$$\n\\begin{aligned}\n\\vec{AR} &= \\vec{AE} + \\vec{AF} = \\kappa \\vec{e}_1 + \\lambda \\vec{e}_2, \\\\\n\\vec{EB} &= \\vec{AB} - \\vec{AE} = \\vec{e}_2 - \\kappa \\vec{e}_1, \\\\\n\\vec{FC} &= \\vec{AC} - \\vec{AF} = \\vec{e}_1 - \\lambda \\vec{e}_2.\n\\end{aligned}\n$$\nTake also $\\alpha, \\beta$ such that $\\vec{EP} = \\alpha \\vec{EB}$ and $\\vec{FP} = \\beta \\vec{FC}$. Then\n$$\n\\begin{aligned}\n\\vec{AP} &= \\vec{AF} + \\beta \\vec{FC} = \\lambda \\vec{e}_2 + \\beta(\\vec{e}_1 - \\lambda \\vec{e}_2) = \\beta \\vec{e}_1 + (1 - \\beta)\\lambda \\vec{e}_2, \\\\\n\\vec{AP} &= \\vec{AE} + \\alpha \\vec{EB} = \\kappa \\vec{e}_1 + \\alpha(\\vec{e}_2 - \\kappa \\vec{e}_1) = (1 - \\alpha)\\kappa \\vec{e}_1 + \\alpha \\vec{e}_2.\n\\end{aligned}\n$$\nConsequently, $\\beta \\vec{e}_1 + (1 - \\beta) \\lambda \\vec{e}_2 = (1 - \\alpha) \\kappa \\vec{e}_1 + \\alpha \\vec{e}_2$. As vectors $\\vec{e}_1$ and $\\vec{e}_2$ are not parallel, the equality can hold only if the respective coefficients are equal, i.e., $\\beta = (1 - \\alpha)\\kappa$ and $(1 - \\beta)\\lambda = \\alpha$. This implies $\\vec{AP} = \\beta \\vec{e}_1 + \\alpha \\vec{e}_2$. Now we obtain\n$$\n\\begin{aligned}\n\\vec{PR} &= \\vec{AR} - \\vec{AP} = (\\kappa \\vec{e}_1 + \\lambda \\vec{e}_2) - (\\beta \\vec{e}_1 + \\alpha \\vec{e}_2) = (\\kappa - \\beta) \\vec{e}_1 + (\\lambda - \\alpha) \\vec{e}_2, \\\\\n\\vec{AQ} &= \\vec{AC} + \\vec{CQ} = \\vec{AC} + \\vec{PB} = \\vec{AC} + \\vec{AB} - \\vec{AP} \\\\\n&= \\vec{e}_1 + \\vec{e}_2 - (\\beta \\vec{e}_1 + \\alpha \\vec{e}_2) = (1 - \\beta) \\vec{e}_1 + (1 - \\alpha) \\vec{e}_2.\n\\end{aligned}\n$$\nThe lines $PR$ and $AQ$ being parallel is equivalent to the vectors $\\vec{PR}$ and $\\vec{AQ}$ being parallel, which in turn is equivalent to their coefficients being proportional. Hence it suffices to show that $\\frac{\\kappa - \\beta}{1 - \\beta} = \\frac{\\lambda - \\alpha}{1 - \\alpha}$, or equivalently,\n$$\n(\\kappa - \\beta)(1 - \\alpha) = (\\lambda - \\alpha)(1 - \\beta).\n$$\nAs $\\kappa(1 - \\alpha) = \\beta$ and $\\lambda(1 - \\beta) = \\alpha$, we have\n$$\n\\begin{aligned}\n(\\kappa - \\beta)(1 - \\alpha) &= \\kappa(1 - \\alpha) - \\beta(1 - \\alpha) = \\beta - \\beta + \\alpha\\beta = \\alpha\\beta, \\\\\n(\\lambda - \\alpha)(1 - \\beta) &= \\lambda(1 - \\beta) - \\alpha(1 - \\beta) = \\alpha - \\alpha + \\alpha\\beta = \\alpha\\beta.\n\\end{aligned}\n$$\nHence the desired equality holds and the proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least number of buttons that can be placed on the squares of a $5 \\times 5$ grid so that no two buttons are on the same square or on squares with a common side (buttons may be on squares with a common vertex) and no buttons can be added to the grid under the same conditions.",
    "options": [],
    "answer": "7",
    "solution": "We say that a button *covers* a square if the button lies on either the square itself or one of its neighbors. Consider the $2 \\times 2$ corner areas and the central cross consisting of 5 squares (colored with green and red respectively in Fig. 17).\n\nEvery button covering a corner has to be located in the corresponding $2 \\times 2$ corner area, thus there is at least one button in each of them. Such a button covers exactly 3 squares in its area, meaning there exists a square in each area covered by other buttons.\n\nIf there is a button in the central square, then it covers no squares in the corner areas. One button can cover the missing squares of only 2 corner areas at once. Thus we would need at least 3 other buttons in addition to the 4 buttons located in the corner areas, meaning a total of at least 7 buttons.\n\nIf there is no button in the central square, then each button can cover at most 2 squares in the central cross, meaning at least 3 buttons needed to cover it. However, no such button can cover a corner square, meaning 4 other buttons to cover them for a total of at least 7 buttons.\n\n![](attached_image_1.png)\n\nOne possibility for the covering of the $5 \\times 5$ square with 7 buttons is shown in Fig. 18.\nDefine covering like in Solution 1. Assume for contradiction that we have covered the grid with 6 buttons. Divide the $5 \\times 5$ grid into an *edge zone* and a *central zone* (green and red respectively in Fig. 19).\n\nA button placed in the edge zone always covers exactly 3 consecutive edge zone squares. If it is in a corner square, then it covers no central zone squares, otherwise it covers exactly 1 central zone square. To cover the corner squares, we need at least 4 buttons in the edge zone, meaning at most 2 buttons in the central zone. The buttons located in the edge zone cover 12 edge zone squares, but there are 16 squares in total.\n\n![](attached_image_2.png)\nFig. 19\n![](attached_image_3.png)\nFig. 20\n\nA button placed in the central zone covers at most 2 edge zone squares, but 2 is only possible if there is a single corner square between them (Fig. 20). Thus a central square button covers at most 1 edge zone square that is not already covered by the edge zone buttons. Thus, in order to cover all squares of the edge zone, we would need at least 5 buttons in it. These cover at most 15 squares in the edge zone and 5 squares in the central zone. Thus the final button has to cover at least 1 edge zone square and 4 central zone squares. This is only possible if the covered edge zone square is the middle square of a side (Fig. 21). But then the only way to place 5 buttons in the edge zone to cover 15 of its squares is to place 2 of them in the corners. But buttons placed in the corners do not cover squares in the central region, which leaves the center uncovered (Fig. 22). The contradiction shows that we need at least 7 buttons.\n\n![](attached_image_4.png)\nFig. 21\n\n![](attached_image_5.png)\nFig. 22\n\nOne possibility for the covering of the $5 \\times 5$ square with 7 buttons is shown in Fig. 18.\nDefine covering like in Solution 1. To cover the corner squares, we need at least 4 buttons located either in corner squares or in squares adjacent to them.\n\nIn a corner square, a button covers 3 squares. On the edge, but not in the corner, a button covers 4 squares and elsewhere it covers 5 squares. Consider two cases.\n\n* If there are $k > 1$ buttons in corner squares, then the 4 buttons covering the corners cover at most $3k + 4(4-k) = 16-k$ squares, meaning\n\n$25 - (16 - k) = 9 + k$ squares remain uncovered by them. As $9 + k > 10$, this requires at least 3 more buttons for a total of at least 7.\n\n* If there is at most 1 button in a corner square, then there exists a pair of empty opposite corners (w.l.o.g., these are the upper left and lower right corner, with the button covering the upper left corner in the left-most column, as in Fig. 23). Then to cover the red square in Fig. 23, another button needs to be located on it or in the central square of the top row. Repeating the same argument for the other empty corner shows that there must be 2 buttons in both red regions in Fig. 24, alongside at least 1 button in both green regions. Thus we have a total of 6 buttons on the edge of the grid, meaning another button is necessary to cover the central square. Thus we need at least 7 buttons in total.\n\n![](attached_image_6.png)\nFig. 23\n![](attached_image_7.png)\nFig. 24\n\nOne possibility for the covering of the $5 \\times 5$ square with 7 buttons is shown in Fig. 18.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and $d$ one of its positive divisors. Integers $1$ to $n$ are written in columns of length $d$ (the first column contains the numbers $1$ to $d$, starting from the top, the second column contains the numbers $d+1$ to $2d$ etc.). Then, one finds the greatest common divisor of each row and finally the least common multiple of the greatest common divisors. Prove that if $d < n$, then the final result is equal to $d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The first two numbers of the $i$-th row are $i$ and $i+d$. Denote the greatest common divisor of this row by $a$; since $a$ divides both $i$ and $i+d$, it must also divide their difference $d$. Thus all the greatest common divisors divide $d$, meaning their least common multiple is at most $d$. But as the final row is $d, 2d, \\dots, n$ with the greatest common divisor $d$, the least common multiple of the greatest common divisors is exactly $d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the greatest possible value of the expression\n$$\n(((a * b) * c) * d) * e,\n$$\nif each star is replaced with one of the operations $+$, $-$, $\\cdot$ and the numbers $a, b, c, d, e$ are $-2, -1, 0, 1, 2$ in some order. Different stars can correspond to different operations.",
    "options": [],
    "answer": "8",
    "solution": "We will show that a value greater than $8$ is impossible. For this, let us assume the opposite. In this case, we can simplify the expression by removing the $0$ and the operation to the left of it. The remaining operation consists of the numbers $-2, -1, 1, 2$ and three operations and has a value of at least $9$. We will consider two cases.\n\n* If the rightmost number of the simplified expression is $2$ or $-2$, then the absolute value of the subexpression to the left of it needs to be at least $5$. Then, if the second number from the right is also $2$ or $-2$, then the absolute value of the subexpression to the left of it needs to be at least $3$, which is impossible using just the numbers $-1$ and $1$. Similarly, if the second number from the right is $-1$ or $1$, then the absolute value of the subexpression to the left of it needs to be at least $4$, which is impossible using two numbers with absolute values $1$ and $2$.\n\n* If the rightmost number of the simplified expression is $1$ or $-1$, then the absolute value of the subexpression to the left of it needs to be at least $8$. Then, if the second number from the right is also $1$ or $-1$, then the absolute value of the subexpression to the left of it needs to be at least $7$, which is impossible using just the numbers $-2$ and $2$. Similarly, if the second number from the right is $-2$ or $2$, then the absolute value of the subexpression to the left of it needs to be at least $4$, which is impossible using two numbers with absolute values $1$ and $2$.\n\nWe obtained a contradiction in all cases, which means that values greater than $8$ are impossible.\n\nOn the other hand, if $a = -1$, $b = 1$, $c = -2$, $d = 2$, $e = 0$, then $(((a - b) \\cdot c) \\cdot d) + e = 8$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 30061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $D \\notin \\{B, C\\}$ is chosen on segment $BC$ and point $A$ is chosen on the plane such that it does not lie on the line $BC$. Points $P$ and $Q$ are chosen such that $ABPD$ and $ACQD$ are parallelograms (with the vertices in this order). The circumcircles of triangles $BDP$ and $CDQ$ intersect in $R \\neq D$. Prove that the points $P, Q, R$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A reflection across the centre of the segment $BD$ sends the triangle $BDP$ into the triangle $DBA$. Thus the circumcircle of $BDP$ goes to the circumcircle of $DBA$. Analogously we see that a reflection across the centre of $CD$ sends the circumcircle of $CDQ$ into the circumcircle of $DCA$.\n\n![](attached_image_1.png)\n\nThe reflection of a circle across the midpoint of one of its chords is equivalent to a reflection across the line corresponding to the chord. Thus the reflections of the circumcircles of $BDP$ and $CDQ$ across $BC$ are the circumcircles of $ABD$ and $ACD$, respectively. The first two circles intersect in points $A$ and $D$, whereas the latter two circles intersect in $R$ and $D$. Thus $A$ and $R$ are reflections of each other across $BC$. Therefore the points $P$, $Q$, $R$ are all located on the opposite side of $BC$ compared to $A$, but at the same distance from $BC$ as $A$. Thus $P$, $Q$, $R$ lie on one line parallel to $BC$.\nWe will use directed angles. Angle chasing yields\n$$\\angle QRD = \\angle QCD = \\angle ADC = \\angle ADB = \\angle PBD = \\angle PRD.$$ \nThe equality $\\angle QRD = \\angle PRD$ means that $P, R, Q$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Today, on September 23, 2023, twins Mari and Jüri received a total of $5000$ candies for their $10$th birthdays. Starting from this day, their mother allows them both to take candies once per day, such that the amount of candies taken by any child on any day is less than their age in full years (on their birthday, they already use their new age). Neither child can resist taking at least one candy every day. Jüri allows Mari to take candies first on every day. The children agreed that whoever takes the last candy, has to buy new candies. Which child can ensure that they do not have to buy new candies, no matter how the other child takes their candies?",
    "options": [],
    "answer": "Mari",
    "solution": "We will show that Mari can avoid taking the last candy. To achieve this, she will take $8$ candies today, and on every following day, she will take candies so that along with Jüri's candies from the previous day, the total is the age of the children on the previous day. Since $2024$ is a leap year, they will be $10$ years old for $366$ days; thus on the $11$th birthday of the children, there will be $5000 - 8 - 366 \\cdot 10 = 1332$ candies remaining. After $121$ more days, there will be $1332 - 121 \\cdot 11 = 1$ candy remaining after Mari has taken her candies for the day. Thus Jüri has no choice but to take the last candy.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least number of buttons that can be placed on the squares of a $4 \\times 4$ grid so that no two buttons are on the same square or on squares with a common side (buttons may be on squares with a common vertex) and no buttons can be added to the grid under the same conditions.",
    "options": [],
    "answer": "4",
    "solution": "If there are 3 or fewer buttons placed, then it is always possible to add more, as each button blocks up to 5 squares: the square it is in and its 4 neighbors. Thus 3 buttons cover up to $3 \\cdot 5 = 15$ of the $4 \\cdot 4 = 16$ squares of the grid, meaning that a button can be placed in at least one of the squares. It remains to observe that by placing 4 buttons as shown in Fig. 2, no more buttons can be added.\n\n![](attached_image_1.png)\nWe say that a button covers a square if the button lies on either the square itself or one of its neighbors. Observe that one button can cover at most one of the 4 corner squares of the grid, meaning that at least 4 buttons are needed to cover the grid. To show that 4 buttons are sufficient, we can use the same placement as in Solution 1.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Juku multiplied four consecutive positive integers and divided the result by four. Find all the possibilities for the digit in which this quotient can end.",
    "options": [],
    "answer": "0 and 6",
    "solution": "Among 4 consecutive integers there always exist two even numbers, one of which is also divisible by 4. Thus the quotient is even, excluding all odd final digits.\nIf the final digit of the quotient upon division by 4 is 0, 2, 4, 6 or 8, then the\nfinal digit of the dividend is 0, 8, 6, 4 or 2, respectively.\n\n* If the four consecutive integers have final digits 1, 2, 3 and 4, then the final digit of their product is 4.\n* If the four consecutive integers have final digits 6, 7, 8 and 9, then the final digit of their product is 4.\n* In all other cases one of the four integers is divisible by 5. Since the product is even, its final digit can only be 0.\n\nSo the product can have a final digit of 4 or 0. Upon division by 4, the final digit of the quotient is 6 or 0, respectively.\nAmong 4 consecutive integers there always exist two even numbers, one of which is also divisible by 4. Thus the product is divisible by 8. We consider two cases.\n\n* If one of the factors is divisible by 5, then the product is also divisible by 5, and thus also by 40. Therefore the quotient upon division by 4 is divisible by 10 and thus ends with a 0.\n\n* If none of the factors are divisible by 5, then they are of the form $5n + 1, 5n + 2, 5n + 3$ and $5n + 4$. The expression $(5n + 1)(5n + 2)(5n + 3)(5n + 4)$ can be written as $5m + 1 \\cdot 2 \\cdot 3 \\cdot 4$ for an integer $m$. As the product of the four numbers and the term $1 \\cdot 2 \\cdot 3 \\cdot 4$ are divisible by 8, then so must also be $5m$ and thus also $m$. Therefore the product is of the form $40k + 24$, which yields $10k + 6$ upon division by 4. This number ends with a 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a number *interesting* if it can be represented as the sum of squares of three distinct non-negative integers. For example, the number $5$ is interesting, because $5 = 0^2 + 1^2 + 2^2$. Call a number *special* if it is not interesting, but can be represented as the product of two distinct interesting numbers.\n\na. Find one special number.\n\nb. Prove that there are infinitely many special numbers.",
    "options": [],
    "answer": "One example is 175. There are infinitely many special numbers.",
    "solution": "The factorisation $(2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$, where one of the numbers $k$ and $k+1$ is always even, shows that the square of any odd number gives a remainder of $1$ upon division by $8$. The square of an even number not divisible by $4$ gives a remainder of $4$ and the square of an even number divisible by $4$ gives a remainder of $0$. Thus the only possible remainders of squares modulo $8$ are $0$, $1$ and $4$.\n\nTherefore the sum of squares of three odd numbers gives a remainder of $3$ modulo $8$. The sum of squares of one number divisible by $4$, one number divisible by $2$, but not by $4$, and one odd number gives a remainder of $5$. Thus there are infinitely many interesting numbers giving remainders $3$ and $5$ modulo $8$, and therefore also infinitely many products of distinct interesting numbers with a remainder of $3 \\cdot 5 \\equiv 7 \\pmod{8}$. These numbers are indeed special, as three numbers with remainders of $0$, $1$ or $4$ cannot add up to a remainder of $7$ modulo $8$.\n\nThe smallest example is obtained by choosing the interesting numbers $1^2 + 3^2 + 5^2$ and $0^2 + 1^2 + 2^2$, whose product is $35 \\cdot 5 = 175$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Acute triangle $ABC$ with $AB < AC$ has circumcircle $\\omega$. Let $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, $AB$ correspondingly. Ray $AD$ intersects $\\omega$ at point $K$ ($K \\neq A$) and the circumcircle of triangle $AEF$ at point $M$ ($M \\neq A$). On circle $\\omega$ a point $T$ such that $AT \\parallel BC$ is chosen. The circumcircles of triangles $KDT$ and $AEF$ intersect at point $L$ on the shorter arc $EF$. Prove that circle $\\omega$ and the circumcircle of triangle $KLM$ are tangent to each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of $\\omega$ and $Q$ be the intersection point of lines $DO$ and $AT$. Since $ABCT$ is a trapezoid and a cyclic quadrilateral, it is an isosceles trapezoid and the points $A$ and $T$ are symmetric with respect\n![](attached_image_1.png)\nFig. 47\n![](attached_image_2.png)\nFig. 48\n\nto line $DO$ (Fig. 47). We have $\\angle DKT = \\angle AKT = \\frac{1}{2} \\angle AOT = \\angle QOT = 180^\\circ - \\angle DOT$, whereas $K$ and $O$ are on different sides of line $DT$, so point $O$ is on the circumcircle of triangle $KDT$.\n\nSince $O$ is the point of intersection of the perpendicular bisectors of triangle $ABC$, we have $\\angle OEA = \\angle OFA = 90^\\circ$ (Fig. 48). Thus points $E$ and $F$ are on the circle with diameter $OA$ and $O$ is on the circumcircle of triangle $AEF$. Since triangle $ABC$ is acute, triangle $AEF$ is also acute (because $EF \\parallel BC$), hence its center is inside the triangle. So point $O$ is on the shorter arc $EF$. Thus the circumcircles of triangles $KDT$ and $AEF$ intersect at $O$ on the shorter arc $EF$, so $O = L$.\n\nSince $AL$ is a diameter of the circumcircle of triangle $AEF$ and point $M$ is on the circle, $ML$ and $AK$ are perpendicular (Fig. 49). Therefore $KL$ is a diameter of the circumcircle of triangle $KLM$. Since $KL$ is also a radius of circle $\\omega$, the line through $K$ perpendicular to line $KL$ is tangent to both circles. Consequently circle $\\omega$ and the circumcircle of triangle $KLM$ are tangent to each other.\n\n![](attached_image_3.png)\nFig. 49",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $k$ be positive integers. On the board $k$ different positive integers not exceeding $n$ are written. Prove that the equation\n$$\nx + y = z + w\n$$\nhas at least $\\frac{k^4}{2n-1}$ solutions $(x, y, z, w)$, where $x, y, z, w$ are some (not necessarily different) numbers written on the board.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For positive integer $s$ denote by $a_s$ the number of pairs $(x, y)$, where $x, y$ are numbers on the board and $x + y = s$. Since the sum of two numbers on the board cannot be less than $2$ or greater than $2n$, the number of solutions of equation $x + y = z + w$ is $a_2^2 + \\dots + a_{2n}^2$. Applying the AM-QM inequality to $a_2, \\dots, a_{2n}$ gives\n$$\na_2^2 + \\dots + a_{2n}^2 \\ge \\frac{(a_2 + \\dots + a_{2n})^2}{2n - 1}.\n$$\nSince each pair $(x, y)$, where $x, y$ are numbers on the board, is counted exactly once in the sum $a_2 + \\dots + a_{2n}$, the sum is equal to the number of pairs $k^2$. Hence\n$$\n\\frac{(a_2 + \\dots + a_{2n})^2}{2n - 1} = \\frac{k^4}{2n - 1},\n$$\nwhich proves the claim.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30068,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many integers $k$ are there for which $1 \\le k \\le n$ and the sum $1 + 2 + \\dots + k$ is divisible by $n$, if\n\na) $n = 2023^{2024}$;\nb) $n = 2024^{2023}$?",
    "options": [],
    "answer": "a) 4; b) 3",
    "solution": "Note that $1+2+\\dots+k = \\frac{k(k+1)}{2}$.\n\na) Since $n$ is odd, the number $\\frac{k(k+1)}{2}$ is divisible by $n$ if and only if $k(k+1)$ is divisible by $n$.\n\nSince $2023 = 7 \\cdot 17^2$, the number $n$ can be expressed as $p_1^{\\alpha_1} p_2^{\\alpha_2}$, where $p_1 = 7$, $p_2 = 17$ and $\\alpha_1, \\alpha_2$ are positive integers. The number $k(k+1)$ is divisible by $n$ iff $k(k+1)$ is divisible by $p_1^{\\alpha_1}$ and $p_2^{\\alpha_2}$. Since $k$ and $k+1$ have no common factor, their product is divisible by $p_i^{\\alpha_i}$ iff one of the factors is divisible by this number. If $k$ is divisible by $p_i^{\\alpha_i}$ then $k \\equiv 0 \\pmod{p_i^{\\alpha_i}}$, if $k+1$ is divisible by $p_i^{\\alpha_i}$ then $k \\equiv -1 \\pmod{p_i^{\\alpha_i}}$.\n\nBy Chinese Remainder Theorem the congruence system\n$$\n\\begin{cases} k \\equiv d_1 \\pmod{p_1^{\\alpha_1}} \\\\ k \\equiv d_2 \\pmod{p_2^{\\alpha_2}} \\end{cases}\n$$\nhas exactly one solution in the interval $1 \\le k \\le p_1^{\\alpha_1} \\cdot p_2^{\\alpha_2} = n$ for any integers $d_1$ and $d_2$. The solution $k$ satisfies the conditions of the problem iff $d_1, d_2 \\in \\{0, -1\\}$. There are 4 ways to choose such a vector of the right hand sides, each of them gives a different solution $k$.\n\nb) The number $\\frac{k(k+1)}{2}$ is divisible by $n$ iff $k(k+1)$ is divisible by $2n$.\n\nSince $2024 = 2^3 \\cdot 11 \\cdot 23$, the number $n$ can be expressed as $2^{\\beta_0} q_1^{\\beta_1} q_2^{\\beta_2}$, where $q_1 = 11$, $q_2 = 23$ and $\\beta_0, \\beta_1, \\beta_2$ are positive integers. The number $k(k+1)$ is divisible by $2n$ iff $k(k+1)$ is divisible by $2^{\\beta_0+1}$, $q_1^{\\beta_1}$ and $q_2^{\\beta_2}$. Since $k$ and $k+1$ have no common factor, their product is divisible by each of the powers of the primes iff one of the factors is divisible by this number. So $k(k+1)$ is divisible by $2^{\\beta_0+1}$ iff $k \\equiv 0 \\pmod{2^{\\beta_0+1}}$ or $k \\equiv -1 \\pmod{2^{\\beta_0+1}}$, and by $q_i^{\\beta_i}$ ($i \\in \\{1, 2\\}$) iff $k \\equiv 0 \\pmod{q_i^{\\beta_i}}$ or $k \\equiv -1 \\pmod{q_i^{\\beta_i}}$.\n\nBy Chinese Remainder Theorem the congruence system\n$$\n\\left\\{\n\\begin{array}{ll}\nk \\equiv d_0 & (\\text{mod } 2^{\\beta_0}) \\\\\nk \\equiv d_1 & (\\text{mod } q_1^{\\beta_1}) \\\\\nk \\equiv d_2 & (\\text{mod } q_2^{\\beta_2})\n\\end{array}\n\\right.\n$$\nhas exactly one solution in the interval $1 \\le k \\le 2^{\\beta_0} \\cdot q_1^{\\beta_1} \\cdot q_2^{\\beta_2} = n$ for any integers $d_0, d_1$ and $d_2$. The solution $k$ satisfies the conditions of the problem iff $d_0, d_1, d_2 \\in \\{0, -1\\}$ and $k \\equiv d_0 \\pmod{2^{\\beta_0+1}}$. Note that if $d_0 = d_1 = d_2 = 0$ then $k = n \\ne 0 \\pmod{2^{\\beta_0+1}}$, and if $d_0 = d_1 = d_2 = -1$ then $k = n - 1 \\ne -1 \\pmod{2^{\\beta_0+1}}$. Otherwise let $d'_i = -1 - d_i$ and define $k'$ as the solution of the congruence system\n$$\n\\left\\{\n\\begin{array}{ll}\nk' \\equiv d'_0 & (\\text{mod } 2^{\\beta_0}) \\\\\nk' \\equiv d'_1 & (\\text{mod } q_1^{\\beta_1}) \\\\\nk' \\equiv d'_2 & (\\text{mod } q_2^{\\beta_2})\n\\end{array}\n\\right.\n$$\nin the interval $1 \\le k' \\le n$. Note that $k + k' \\equiv -1 \\pmod{n}$; since $1 \\le k, k' \\le n-2$, we have $-1 < k + k' < 2n-1$, hence $k + k' = n-1$. Also note that $k \\equiv d_0 \\pmod{2^{\\beta_0}}$ means that exactly one of claims $k \\equiv d_0 \\pmod{2^{\\beta_0+1}}$ and $k \\equiv n+d_0 \\pmod{2^{\\beta_0+1}}$ is true, the same holds for $k'$. It follows that exactly one of claims $k \\equiv d_0 \\pmod{2^{\\beta_0+1}}$ and $k' \\equiv d'_0 \\pmod{2^{\\beta_0+1}}$ is true. Indeed, if both were true then $n-1 = k+k' \\equiv d_0+d'_0 = -1 \\pmod{2^{\\beta_0+1}}$, if neither of them were valid, then $n-1 = k+k' \\equiv 2n+d_0+d'_0 = 2n-1 \\pmod{2^{\\beta_0+1}}$ – a contradiction in both cases. Consequently there are $\\frac{8-2}{2} = 3$ integers satisfying the conditions.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30069,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $(a_n)$ satisfies $a_1 = 2$, $a_2 = 3$, $a_3 = 5$ and $a_n = a_{n-1}^2$ for any $n \\ge 4$. A sequence $(b_n)$ satisfies $b_1 = 2$, $b_2 = 3$, $b_3 = 5$ and $b_n = b_{n-1} \\cdot b_{n-2} \\cdot b_{n-3}^2$ for any $n \\ge 4$. A sequence $(c_n)$ satisfies $c_1 = 2$, $c_2 = 3$, $c_3 = 5$ and $c_n = c_1 \\cdot c_2 \\cdots c_{n-1}$ for any $n \\ge 4$. Order the numbers $a_{1000}$, $b_{1000}$ and $c_{1000}$ by size.",
    "options": [],
    "answer": "a_{1000} < c_{1000} < b_{1000}",
    "solution": "$$\nc_n = (c_1 \\cdots c_{n-2}) \\cdot c_{n-1} = c_{n-1}^2.\n$$\nSince $a_4 = 5^2 = 25$ and $c_4 = 2 \\cdot 3 \\cdot 5 = 30$, we have $a_n < c_n$ for any $n \\ge 4$.\nNext note that $c_1 = b_1 = 2$, $c_2 = b_2 = 3$, $c_3 = b_3 = 5$ and\n$$\nc_4 = 30 < 60 = b_4, \\quad c_5 = 900 < 2700 = b_5, \\quad c_6 = 810000 < 4050000 = b_6.\n$$\nNow prove by induction that $c_n < b_n$ for any $n \\ge 7$. Assume that the statement holds for every $m$ with $4 \\le m < n$. As $c_{n-3} = c_{n-4} \\cdot c_{n-5} \\cdot \\dots \\cdot c_1$, we get $c_n = c_{n-1}c_{n-2}c_{n-3}c_{n-4}\\dots c_1 = c_{n-1}c_{n-2}c_{n-3}^2 < b_{n-1}b_{n-2}b_{n-3}^2 = b_n$, which we wanted to prove.\nHence $a_n < c_n < b_n$ for any $n \\ge 4$.\nAs in Solution 1 show that $a_n < c_n$ for any $n \\ge 4$. To show that $c_n < b_n$ for any $n \\ge 4$ we prove by induction that $b_n \\ge c_n c_{n-3}$; since $c_{n-3} > 1$ for $n \\ge 4$, it follows $b_n > c_n$ for any $n \\ge 4$. For simplicity define $c_0 = c_{-1} = c_{-2} = 1$; then the inequality $b_n \\ge c_n c_{n-3}$ holds for $n = 1, 2, 3$ as equality and for $n \\ge 4$ we have $c_n = c_{n-1} c_{n-2} \\dots c_{-2}$. So assuming the hypothesis holds for $n-1, n-2$ and $n-3$, we get for $n \\ge 4$\n$$\n\\begin{aligned}\nb_n &= b_{n-1}b_{n-2}b_{n-3}^2 \\ge c_{n-1}c_{n-4}c_{n-2}c_{n-5}c_{n-3}c_{n-6}c_{n-3}c_{n-6} \\\\\n&= (c_{n-1} \\dots c_{n-6}) c_{n-3}c_{n-6}.\n\\end{aligned}\n$$\nIf $n \\le 8$ or $n \\ge 10$ then $c_{n-6} \\ge c_{n-7} \\dots c_{-2}$, hence\n$$\nc_{n-1} \\dots c_{n-6} \\cdot c_{n-6} \\ge c_{n-1} \\dots c_{-2} = c_n.\n$$\nSo in these cases $b_n \\ge c_n \\cdot c_{n-3}$. It remains to observe that\n$$\n\\begin{aligned}\nb_9 &= b_8b_7b_6^2 = b_7b_6b_5^2 \\cdot b_7b_6^2 = b_7^2b_6^2b_5^2 \\cdot b_6 \\\\\n&\\ge (c_7c_4)^2(c_6c_3)^2(c_5c_2)^2 \\cdot c_6c_3 = (c_7c_6c_5c_4c_3c_2)^2 \\cdot c_6 \\cdot 5 \\\\\n&> (c_7c_6c_5c_4c_3c_2)^2 \\cdot c_6 \\cdot 4 = (c_7c_6c_5c_4c_3c_2c_1)^2 \\cdot c_6 = c_8^2 \\cdot c_6 = c_9 \\cdot c_6.\n\\end{aligned}\n$$\nConsequently the inequality holds in all cases.\nBy mathematical induction it is easy to see that $a_n = 5^{2^{n-3}}$ for any $n \\ge 3$. Indeed it is true for $n = 3$ and if $n > 3$ and the claim holds for $n-1$ then $a_n = a_{n-1}^2 = (5^{2^{n-4}})^2 = 5^{2^{n-4} \\cdot 2} = 5^{2^{n-3}}$ so it holds for $n$.\nAlso $c_n = 2^{2^{n-4}} 3^{2^{n-4}} 5^{2^{n-4}}$ for any $n \\ge 4$. Indeed it is true for $n = 4$ because $c_4 = 2 \\cdot 3 \\cdot 5 = 2^1 3^1 5^1 = 2^{2^0} 3^{2^0} 5^{2^0}$, and if $n > 4$ and the claim holds for $n-1$ then\n$$\n\\begin{aligned}\nc_n &= c_{n-1} \\cdot c_{n-2} \\dots c_1 = c_{n-1}^2 \\\\\n&= (2^{2^{n-5}} 3^{2^{n-5}} 5^{2^{n-5}})^2 = (2^{2^{n-5}})^2 (3^{2^{n-5}})^2 (5^{2^{n-5}})^2 \\\\\n&= 2^{2^{n-5} \\cdot 2} 3^{2^{n-5} \\cdot 2} 5^{2^{n-5} \\cdot 2} = 2^{2^{n-4}} 3^{2^{n-4}} 5^{2^{n-4}}\n\\end{aligned}\n$$\nso the claim holds for $n$. Hence for any $n \\ge 4$ we have\n$$\nc_n = 6^{2^{n-4}} 5^{2^{n-4}} > 5^{2^{n-4}} 5^{2^{n-4}} = 5^{2^{n-4}+2^{n-4}} = 5^{2^{n-3}} = a_n.\n$$\nConsider the sequence $b_n$. Clearly $b_n$ is a product of powers of primes 2, 3 and 5. From the definition of $(b_n)$ for all $n \\ge 4$ the powers of $p = 2, 3, 5$ satisfy $v_p(b_n) = v_p(b_{n-1}) + v_p(b_{n-2}) + 2v_p(b_{n-3})$. Calculate the powers of 2, 3 and 5 in the canonical representation of the first 9 terms:\n\n| n | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |\n|---|---|---|---|---|---|---|---|---|---|\n| $v_2(b_n)$ | 1 | 0 | 0 | 2 | 2 | 4 | 10 | 18 | 36 |\n| $v_3(b_n)$ | 0 | 1 | 0 | 1 | 3 | 4 | 9 | 19 | 36 |\n| $v_5(b_n)$ | 0 | 0 | 1 | 1 | 2 | 5 | 9 | 18 | 37 |\n\nWe see that $v_p(b_n) > 2^{n-4}$ for $n = 7, 8, 9$ and $p = 2, 3, 5$. But if $v_p(b_{n-1}) > 2^{n-5}$, $v_p(b_{n-2}) > 2^{n-6}$ and $v_p(b_{n-3}) > 2^{n-7}$ for some $n \\ge 4$, then\n$$\n\\begin{aligned}\nv_p(b_n) &= v_p(b_{n-1}) + v_p(b_{n-2}) + 2v_p(b_{n-3}) \\\\\n&> 2^{n-5} + 2^{n-6} + 2 \\cdot 2^{n-7} \\\\\n&= 2^{n-5} + 2^{n-6} + 2^{n-6} = 2^{n-5} + 2^{n-5} = 2^{n-4}.\n\\end{aligned}\n$$\nSo by mathematical induction $v_p(b_n) > 2^{n-4}$ for all $n \\ge 7$ and $p = 2, 3, 5$. Consequently $c_n < b_n$ for all $n \\ge 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30070,
    "subject": "Mathematics (Multi-modal)",
    "question": "The number $\\sqrt{2}$ is on the computer screen. In one step, Juku can multiply the number currently on the screen by any positive rational number or add any natural number to it. Is there a positive integer $n$ such that every number that Juku can get after a finite number of steps can be obtained by at most $n$ steps? If so, find the least suitable $n$.",
    "options": [],
    "answer": "3",
    "solution": "Any number that can be obtained after a finite number of steps is of the form $r\\sqrt{2} + s$, where $r$ is positive and $s$ is a nonnegative rational number. Indeed, the initial number $\\sqrt{2}$ is of the form $1\\sqrt{2} + 0$, and if we multiply a number of the form $r\\sqrt{2} + s$ by a positive rational number $t$ or add a natural number $k$ to it, the result is $rt\\sqrt{2} + st$ or $r\\sqrt{2} + (s + k)$ correspondingly, which is of the same form.\n\nShow that any number of the form $r\\sqrt{2} + s$, where $r$ is positive and $s$ is a nonnegative rational number, can be obtained by at most 3 steps. If $s > 0$ then multiply $\\sqrt{2}$ by a positive rational number $\\frac{r}{s}$, add 1 to the result, and then multiply the result by the positive rational number $s$. If $s = 0$ then it is enough to multiply $\\sqrt{2}$ by the positive rational number $r$.\n\nFinally, we show that 2 steps are not always enough. First, make sure that in the representation $r\\sqrt{2} + s$ of a number, $r$ and $s$ are uniquely determined. Indeed, if $r\\sqrt{2} + s = r'\\sqrt{2} + s'$ then $(r - r')\\sqrt{2} = s' - s$. If $r - r' \\ne 0$ then $\\sqrt{2} = \\frac{s' - s}{r - r'}$, which is not possible, because $\\frac{s' - s}{r - r'}$ is a rational number, but $\\sqrt{2}$ is an irrational number. Hence $r - r' = 0$ or $r' = r$; from $(r - r')\\sqrt{2} = s' - s$ also $s' = s$.\n\nConsider the number $r\\sqrt{2} + s$ where $r = 1$ and $s$ is an arbitrary fractional number. We cannot get this number by multiplications only, because then $s$ remains zero, or with additions only, because then $s$ would be an integer. If we multiply by a rational number $t$ and add a natural number $k$ in either order then the result is either $t\\sqrt{2} + k$ or $t(\\sqrt{2} + k)$. In both cases we must have $t = r = 1$. But multiplication by 1 does not change the result, and we already saw that by addition only it is not possible to get the result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Fix a natural number $n$. A function $f : \\{0, 1, \\dots, n\\} \\to \\{0, 1, \\dots, n\\}$ is called *regular* if $f(0) = 0$ and $f(i) \\in \\{i-1, f(i-1), f(f(i-1)), \\dots\\}$ for every $i = 1, \\dots, n$. If, for instance, $n = 3$, then the function $f(0) = f(1) = 0, f(2) = f(3) = 1$ is regular, but the function $f(0) = f(1) = f(2) = 0, f(3) = 1$ is not (in the latter case, $f(3)$ violates the regularity condition). Juku chooses a regular function $f$ and tells Miku for each number $k = 0, 1, \\dots, n$, how many different arguments $i$ are there such that $f(i) = k$. Can Miku always determine based on this information which regular function Juku had chosen?",
    "options": [],
    "answer": "Yes",
    "solution": "Firstly, we prove by induction that if $f$ is a regular function then $f(i) < i$ for each positive argument $i$. For that, assume that the claim holds for all smaller arguments. We have $f(i) \\in \\{i-1, f(i-1), f(f(i-1)), \\dots\\}$ by regularity. By the induction hypothesis and the assumption $f(0) = 0$, we have $i-1 \\ge f(i-1) \\ge f(f(i-1)) \\ge \\dots$. Hence $f(i) < i$ indeed.\n\nNow we prove by induction on $n$ that the information given to Miku by Juku uniquely determines the regular function. If $n = 0$ then only one regular function exists whence the claim holds trivially. Assume in the rest that $n > 0$ and that the claim holds for all smaller numbers. As $f(1) < 1$ by the lemma proven in the beginning, we must have $f(1) = 0$.\n\nLet $k$ be the largest number in $\\{0, 1, \\dots, n\\}$ such that $f(k) = 0$. Then $k \\ge 1$ since $f(1) = 0$. By the lemma proven in the beginning, $0 < i \\le k$ always implies $f(i) < i$ and hence also $f(i) < k$. On the other hand, $k < i \\le n$ implies $f(i) \\ge k$. Indeed, assume that $f(j) \\ge k$ for each $j = k+1, \\dots, i-1$; let exactly $s$ initial members of the sequence $i-1, f(i-1), f(f(i-1)), \\dots$ be larger than or equal to $k$. Then the $s$th member must be $k$ because otherwise the $s$ initial members would all be in $\\{k+1, \\dots, i-1\\}$, implying that the next member would be larger than or equal to $k$, too. Thus the next member is $f(k)$ which equals 0 and all following members are 0, too. But as $f(i) \\ne 0$, we must have $f(i) \\ge k$. We can conclude that the number of arguments $i$ such that $f(i) \\ge k$ is exactly $n-k$.\n\nNow if $0 < l < k$ then similarly to what we did before we can prove that $0 < i \\le l$ always implies $f(i) < l$. But among the arguments $i$ such that $l < i \\le n$, there is $k$ for which $f(k) = 0$. Thus the number of arguments $i$ such that $f(i) \\ge l$ is less than $n-l$. Consequently, $k$ is the least positive integer for which the number of arguments $i$ such that $f(i) \\ge k$ is $n-k$. According to this property, Miku can determine $k$.\n\nRestricting $f$ to $\\{0, 1, \\dots, k-1\\}$, all conditions of regularity hold. Hence, by the induction hypothesis, Miku can determine $f(1), \\dots, f(k-1)$. Note that if we forget about arguments $1, \\dots, k$ and decrease the other argument-value pairs of $f$ by exactly $k$, the conditions of regularity again hold. Hence, by the induction hypothesis, Miku can also determine $f(k+1), \\dots, f(n)$. Consequently, Miku can determine the values of $f$ at all its arguments.\nWe start like in Solution 1 by showing that if $f$ is regular then $f(i) < i$ for every positive argument $i$.\n\nNow we prove the following lemma: Whenever $0 < k \\le i \\le n$, either $f(i) \\ge k$ or $f(i) \\le f(k)$. To this end, fix $k$ and proceed by induction on $i$. The base case $i = k$ is trivial. Assume now that $i > k$ and, for every $i'$ such that $k \\le i' < i$, either $f(i') \\ge k$ or $f(i') \\le f(k)$. By regularity, $f(i)$ is a term in the sequence $i-1, f(i-1), f(f(i-1)), \\dots$. This sequence decreases until it reaches zero. Let the sequence contain exactly $s$ terms larger than or equal to $k$. As all terms are less than $i$, the induction hypothesis implies that the term number $s+1$ is at least $k$ (which is impossible by the choice of $s$) or at most $f(k)$. Then the following terms are also at most $f(k)$. Thus $f(i)$ must be either at least $k$ or at most $f(k)$.\n\nMiku can determine Juku's regular function as follows. Let Miku put the numbers of repetitions of values in the order of decreasing value (i.e., the number of repetitions of $n$ first, then the number of repetitions of $n-1$, etc.). Every time a number of repetitions of some value $k$ equals a positive number $a$, let Miku define the value of $f$ on the least $a$ arguments greater than $k$ where the value is not defined so far to be $k$. We show that this results in the only possible regular function with the given numbers of repetitions of values. Suppose the contrary. Let $k$ be the largest value of the function for which a divergence appears. We know that all arguments on which the function obtains the value $k$ must be larger than $k$. Hence there exist $i'$ and $i$ such that $k < i' < i$, $f(i) = k$ and $f(i') \\neq k$. Since all values greater than $k$ are already assigned by the algorithm and they are correct by the choice of $k$, the only option is $f(i') < k$. But then the lemma proven above implies either $f(i) \\ge i' > k$ or $f(i) \\le f(i') < k$, contradiction. Consequently, no divergence between the actual function and the one constructed by the algorithm can arise. Hence Miku can always determine Juku's function.\nLike in Solution 1, we show that if $f$ is regular then $f(i) < i$ for every positive argument $i$. Like in Solution 2, we show that $0 < k \\le i \\le n$ always implies either $f(i) \\ge k$ or $f(i) \\le f(k)$.\n\nMiku can determine Juku's regular function as follows. Let Miku start by defining $f(0) = 0, f(1) = 0$. These are clearly the only options. Whenever $f(0), f(1), \\dots, f(i-1)$ have been fixed, let Miku define $f(i)$ to be the largest number among $0, 1, \\dots, i-1$ whose presumed number of occurrences is not achieved yet. As $f(i) < i$, such number must exist. We show now that this definition is the only possible. Suppose the contrary. Let the first divergence arise at the definition of $f(i)$. Let the algorithm assign the value $k$ to it; then actually $f(i) < k < i$, because the algorithm finds the largest number less than $i$ that can be used. But since the value $k$ is not yet saturated, there must exist $i'$ such that $i' > i$ and $f(i') = k$. By the lemma proven at the beginning of the solution, either $f(i') \\ge i > k$ or $f(i') \\le f(i) < k$, contradiction. Hence Miku can always determine Juku's function.\n\nNow we prove by induction on $n$ that the information given to Miku by Juku uniquely determines the regular function. For $n=0$, there exists only one regular function, whence the claim holds. Assume in the following that $n > 0$ and the claim holds for $n - 1$. Let $a_0, a_1, \\dots, a_n$ be the numbers of repetitions of values of $f$. By the property proven at the beginning of the solution, $n$ cannot occur among the values of $f$, hence $a_n = 0$. Let $l$ be the least positive integer such that $a_l = 0$. As $a_{l-1} > 0$, we have $f(l) = l - 1$ by the above. Define $b_0, b_1, \\dots, b_{n-1}$ by\n$$\nb_i = \\begin{cases} a_i, & \\text{if } i < l - 1, \\\\ a_i - 1, & \\text{if } i = l - 1, \\\\ a_{i+1}, & \\text{if } i \\ge l. \\end{cases}\n$$\nNote that $b_0, b_1, \\dots, b_{n-1}$ are the numbers of repetitions of values of function $g$ defined by\n$$\ng(i) = \\begin{cases} f(i), & \\text{if } i < l, \\\\ f(i+1), & \\text{if } f(i+1) < l \\le i, \\\\ f(i+1) - 1, & \\text{if } l < f(i+1). \\end{cases}\n$$\nIn other words, function $g$ is obtained from $f$ by removing the argument-value pair $(l, l-1)$ (this removes the only occurrence of $l$ in these pairs) and decreasing the numbers $l+1, \\dots, n$ by 1 in all other pairs. It is easy to check that regularity of $f$ implies regularity of $g$. By the induction hypothesis, $g$ is the only regular function with numbers of repetitions of values are $b_0, b_1, \\dots, b_{n-1}$. But $f$ can be defined in terms of $g$:\n$$\nf(i) = \\begin{cases} g(i), & \\text{if } i < l, \\\\ l-1, & \\text{if } i = l, \\\\ g(i-1), & \\text{if } g(i-1) < l < i, \\\\ g(i-1)+1, & \\text{if } l \\le g(i-1). \\end{cases}\n$$\nIn other words, add the argument-value pair $(l, l-1)$ and increase all numbers $l, \\dots, n-1$ by 1 in other pairs. This completes the solution.\nLike in Solution 1, we show that if $f$ is regular then $f(i) < i$ for every positive argument $i$. Next we show that $f(i) = i-1$ whenever $f$ is regular and $i-1$ occurs among the values of $f$. Indeed, let $k$ be the least positive integer such that $f(k) = i-1$; by the above, $k \\ge i$. By regularity, $i-1$ occurs in the sequence $k-1, f(k-1), f(f(k-1)), \\dots$. If $i-1 = k-1$ then $f(i) = i-1$, otherwise $i-1$ can be represented in the form $f(j)$ where $j \\le k-1 < k$, which contradicts the choice of $k$.\n\nNow we prove by induction on $n$ that the information given to Miku by Juku uniquely determines the regular function. For $n=0$, there exists only one regular function, whence the claim holds. Assume in the following that $n > 0$ and the claim holds for $n - 1$. Let $a_0, a_1, \\dots, a_n$ be the numbers of repetitions of values of $f$. By the property proven at the beginning of the solution, $n$ cannot occur among the values of $f$, hence $a_n = 0$. Let $l$ be the least positive integer such that $a_l = 0$. As $a_{l-1} > 0$, we have $f(l) = l - 1$ by the above. Define $b_0, b_1, \\dots, b_{n-1}$ by\n$$\nb_i = \\begin{cases} a_i, & \\text{if } i < l - 1, \\\\ a_i - 1, & \\text{if } i = l - 1, \\\\ a_{i+1}, & \\text{if } i \\ge l. \\end{cases}\n$$\nNote that $b_0, b_1, \\dots, b_{n-1}$ are the numbers of repetitions of values of function $g$ defined by\n$$\ng(i) = \\begin{cases} f(i), & \\text{if } i < l, \\\\ f(i+1), & \\text{if } f(i+1) < l \\le i, \\\\ f(i+1) - 1, & \\text{if } l < f(i+1). \\end{cases}\n$$\nIn other words, function $g$ is obtained from $f$ by removing the argument-value pair $(l, l-1)$ (this removes the only occurrence of $l$ in these pairs) and decreasing the numbers $l+1, \\dots, n$ by 1 in all other pairs. It is easy to check that regularity of $f$ implies regularity of $g$. By the induction hypothesis, $g$ is the only regular function with numbers of repetitions of values are $b_0, b_1, \\dots, b_{n-1}$. But $f$ can be defined in terms of $g$:\n$$\nf(i) = \\begin{cases} g(i), & \\text{if } i < l, \\\\ l-1, & \\text{if } i = l, \\\\ g(i-1), & \\text{if } g(i-1) < l < i, \\\\ g(i-1)+1, & \\text{if } l \\le g(i-1). \\end{cases}\n$$\nIn other words, add the argument-value pair $(l, l-1)$ and increase all numbers $l, \\dots, n-1$ by 1 in other pairs. This completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30072,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangular grid is divided by two perpendicular straight lines into four smaller rectangles with integral side lengths. It is possible to remove one among these four rectangles in such a way that the remaining figure can be exactly covered by rectangles of size $2 \\times 3$ and $3 \\times 2$. Prove that it is possible to exactly cover one among these four smaller rectangles by rectangles of size $2 \\times 3$ and $3 \\times 2$. (By *exact covering* we mean covering without gaps, overlaps and overflows.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Call a figure *coverable* if it can be exactly covered by rectangles of size $2 \\times 3$ and $3 \\times 2$. Let the original rectangle be of size $(a+d) \\times (b+c)$ and let the figure remaining after cutting out the upper left corner of size $a \\times b$ be *coverable* (Fig. 52). This would be impossible if $c=1$ or $d=1$, whence we can assume in the rest that $c \\ge 2$ and $d \\ge 2$. Coverability implies that $6 \\mid ac+bd+cd$. Also one can notice that a rectangle of integral side lengths $n \\times m$ where $n \\ge 2$ and $m \\ge 2$ is *coverable* if and only if $6 \\mid nm$. Indeed, rectangles of size $2k \\times 3l$ and $3l \\times 2k$ are obviously *coverable*,\n![](attached_image_1.png)\nFig. 52\n\nwhereas rectangles of size $6k \\times (6l \\pm 1)$ can be divided into rectangles of size $6k \\times 3$ ja $6k \\times 2s$, both of which are coverable by the above; the symmetric case is analogous.\nSince $2 \\mid ac + bd + cd$, at least one of $a, b, c$ and $d$ must be even. We show next that at least one of $a, b, c$ and $d$ must be divisible by 3. For that, suppose the contrary, i.e., none of $a, b, c$ and $d$ being divisible by 3. If either $3 \\mid a + d$ or $3 \\mid b + c$ then $3 \\mid bd$ or $3 \\mid ac$, respectively, because $3 \\mid ac + bd + cd$. Since 3 is prime, we obtain a contradiction. Hence $3 \\nmid a + d$ and $3 \\nmid b + c$. Thus either $a \\equiv d \\equiv 1 \\pmod 3$ or $a \\equiv d \\equiv 2 \\pmod 3$, as well as either $b \\equiv c \\equiv 1 \\pmod 3$ or $b \\equiv c \\equiv 2 \\pmod 3$. If either $a \\equiv b \\equiv c \\equiv d \\equiv 1 \\pmod 3$ or $a \\equiv b \\equiv c \\equiv d \\equiv 2 \\pmod 3$ then color the squares of the figure by the rising diagonals with 3 colors. If either $a \\equiv d \\equiv 1 \\pmod 3$ and $b \\equiv c \\equiv 2 \\pmod 3$ or $a \\equiv d \\equiv 2 \\pmod 3$ and $b \\equiv c \\equiv 1 \\pmod 3$ then color the squares of the figure by the falling diagonals with 3 colors. The area to be covered, except for the neighborhood of the meeting point of the cutting lines of the original rectangle that can be of four different shape (in Figures 53–56, this area is marked by bullets and framed with bold line), can be cut into strips of size $3k \\times 1$ and $1 \\times 3l$ containing an equal amount of unit squares of each color. Hence the area that should be coverable by assumption contains different number of unit squares of different color, which is impossible as every $2 \\times 3$ and $3 \\times 2$ rectangle contains the same number of unit squares of each color. As we got a contradiction in every case, we have proven that at least one of $a, b, c$ and $d$ is divisible by 3.\n![](attached_image_2.png)\nFig. 53\n![](attached_image_3.png)\nFig. 54\n![](attached_image_4.png)\nFig. 55\n![](attached_image_5.png)\nFig. 56\nIt remains to show that one of the small rectangles is coverable whenever one of $a, b, c$ and $d$ is divisible by 2 and also one of $a, b, c$ and $d$ is divisible by 3. The following table shows for each case which of the small rectangles is coverable:\n\n|         | 3 \\mid a | 3 \\mid b | 3 \\mid c | 3 \\mid d |\n|---------|----------|----------|----------|----------|\n| 2 \\mid a| a \\times c| a \\times b| a \\times c| a \\times c|\n| 2 \\mid b| a \\times b| d \\times b| d \\times b| d \\times b|\n| 2 \\mid c| a \\times c| d \\times b| d \\times c| d \\times c|\n| 2 \\mid d| a \\times c| d \\times b| d \\times c| d \\times c|\n\nOne can show this by the following case study:\n* If one of $a$ and $d$ is divisible by one of 2 and 3 and one of $b$ and $c$ is divisible by the other of 2 and 3 then the corresponding small rectangle is coverable. (Fills all squares outside the two long diagonals of the table.)\n* If $6 \\mid a$ or $6 \\mid d$ then $a \\times c$ or $d \\times c$ is coverable, respectively, since $c \\ge 2$ and the other side length is long enough by divisibility. Similarly, if $6 \\mid b$ or $6 \\mid c$ then $d \\times b$ or $d \\times c$ is coverable, respectively. (Fills all squares of the main diagonal of the table.)\n* If $a$ is divisible by one of 2 and 3 and $d$ is divisible by the other of 2 and 3 then, by $6 \\mid ac+bd+cd$, we must have $6 \\mid ac$. As $a \\ge 2$ (by divisibility) and $c \\ge 2$, the rectangle $a \\times c$ is coverable. Similarly, if $b$ is divisible by one of 2 and 3 and $c$ is divisible by the other of 2 and 3 then $d \\times b$ is coverable. (Fills all squares of the secondary diagonal of the table.)\nThe central claim that neither $a \\equiv d \\not\\equiv 0 \\pmod 3$ nor $b \\equiv c \\not\\equiv 0 \\pmod 3$ can hold can be proven as follows. Color the grid by rows with three colors. Note that each rectangle of size $2 \\times 3$ or $3 \\times 2$ contains either exactly 2 unit squares of every color or 3 unit squares of each of some two colors and 0 squares of the third color. In both cases, the numbers of covered unit squares of all colors are congruent modulo 3. Thus if the figure obtained after cutting out the rectangle of size $a \\times b$ were coverable, the numbers of unit squares of all colors within this area should be congruent modulo 3. We can show that in no cases this is so. As the numbers of unit squares of all colors in 3 consecutive rows of equal length are equal, it suffices to consider the area in the middle that remains after repeatedly removing horizontal strips of width 3 from both top and bottom.\n* If $b \\equiv c \\equiv 1 \\pmod 3$ (Figures 57 and 58) then the numbers of unit squares of the three colors in the remaining area are $a+d$, $d$ and 0. These numbers are incongruent modulo 3, regardless of the remainders of $a$ and $d$.\n* If $b \\equiv c \\equiv 2 \\pmod 3$ (Figures 59 and 60) then the numbers of unit squares of the three colors in the remaining area are $a+2d$, $a+d$ and $d$. These numbers are also incongruent modulo 3, regardless of the remainders of $a$ and $d$.\n![](attached_image_6.png)\nFig. 57\n![](attached_image_7.png)\nFig. 58\n![](attached_image_8.png)\nFig. 59\n![](attached_image_9.png)\nFig. 60",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30073,
    "subject": "Mathematics (Multi-modal)",
    "question": "The product $abc$ of positive integers $a$, $b$, and $c$ is divisible by $3$, and the equations $a = \\frac{b^2}{2} = \\frac{c}{4}$ hold. Find the smallest possible sum of the numbers $a$, $b$, and $c$ under these conditions.",
    "options": [],
    "answer": "96",
    "solution": "The given equations are equivalent to $2a = b^2 = \\frac{c}{2}$. So $b^2$ is divisible by $2$, therefore $b$ is also divisible by $2$. Since the product $abc$ is divisible by $3$, one of the numbers $a$, $b$ and $c$ must be divisible by $3$. If $a$ is divisible by $3$, then $b^2$ is also divisible by $3$, therefore $b$ must be divisible by $3$. If $c$ is divisible by $3$, then $b^2$ must also be divisible by $3$. So in all cases $b$ is divisible by $3$. Therefore $b$ is divisible by both $2$ and $3$, so $b$ must be at least $2 \\cdot 3 = 6$. If $b = 6$, then $a = \\frac{b^2}{2} = 18$ and $c = 2b^2 = 72$ and $a + b + c = 96$. If $b$ is larger, then $b^2$ and also $a$ and $c$ would be larger, hence the sum found is the smallest possible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30074,
    "subject": "Mathematics (Multi-modal)",
    "question": "A mother has 7 apples, 6 pears, and 5 oranges. She wants to divide them among 2 children so that each gets the same number of fruits. In how many different ways can this be done?\n\n*Remark:* We consider the distributions of fruit to be different if a child receives a different number of some types of fruit.",
    "options": [],
    "answer": "36",
    "solution": "According to the conditions, each child must receive 9 fruits. It suffices to find how many possibilities there are to give the first child 9 fruits, because the second child receives all the remaining fruits.\nThe first child can be given 0 to 6 pears and 0 to 5 oranges. Disregarding the condition that he must receive 9 fruit in total, there are a total of $7 \\cdot 6$, or 42, possibilities for giving pears and oranges. The possibilities where the first child receives more than 9 of pears and oranges alone, and also those where the first child gets less than 2 of pears and oranges are not suitable. The possibilities where the total number of pears and oranges is more than 9 are 3 (6 + 4, 5 + 5 and 6 + 5), while the possibilities where the total number of pears and oranges is less than 2 are also 3 (1 + 0, 0 + 1 and 0 + 0). Thus, there are $42 - 3 - 3 = 36$ suitable possibilities.\nAgain it suffices to find how many possibilities there are to give the first child 9 fruits.\nIf the first child gets more apples than the second child, then the first child gets 4 apples and 5 more fruits. Mark with 5 circles the fruits – apples, followed by pears, and finally oranges – and with 2 dashes the places where one kind of fruit changes to another. Then all possible choices of 5 fruits are represented as a word consisting of 7 characters, and the choice consists in determining the positions where the dashes are located. We get $\\frac{7 \\cdot 6}{2} = 21$ possibilities. But the possibilities where before the first dash there are more than 3 circles are not suitable, because we have only 3 apples left. In this case the dashes are either on fifth and sixth, fifth and seventh or sixth and seventh position. Hence $21 - 3 = 18$ possibilities remain.\nThe possibilities where the first child gets fewer apples than the second child are symmetrical, so there are the same number of them. So there are a total of $18 \\cdot 2 = 36$ different ways to distribute the fruits.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a natural number *twistable* if it does not contain digits $3$, $4$, $7$ and its last digit is not zero. The *twisting* of a twistable number is the number obtained after the following two steps:\n* Reverse the order of digits of the given number;\n* Twist each digit: $0$, $1$ and $8$ remain unchanged, $2$ and $5$ are turned into each other, $6$ and $9$ are turned into each other.\n\nFor instance, the twisting of the number $68012$ is $51089$ and the twisting of the number $69$ is $69$.\n\nFind all integers that can be represented as the ratio of a twistable positive integer $n$ and its twisting $k$.",
    "options": [],
    "answer": "1",
    "solution": "Since the numbers $n$ and $k$ are positive integers of the same length, the quotient $\\frac{n}{k}$ must be a single-digit positive number, because multiplying by a multi-digit number increases the number of digits. The quotient $1$ is obviously possible (e.g. $\\frac{69}{69} = 1$). We show that no other quotient is possible.\n\n* If the first digit of the number $k$ is $1$, then the last digit of the number $n$ is $1$. The quotient $\\frac{n}{k}$ cannot be $2$, $4$, $5$, $6$, or $8$, because the multiples of these numbers cannot end with the digit $1$. If the quotient $\\frac{n}{k}$ were $3$ or $7$, then the last digit of $k$ should be $7$ or $3$, respectively. However, these digits cannot occur in a twisting. If the quotient $\\frac{n}{k}$ were $9$, then the last digit of the number $k$ should be $9$ and the first digit of the number $n$ should therefore be $6$. However, since $n = 9k$, the number $n$ can only start with the digit $9$. Therefore, the only possibility is $\\frac{n}{k} = 1$.\n\n* If the first digit of the number $k$ is $2$, then the last digit of the number $n$ is $5$. The quotient $\\frac{n}{k}$ cannot be $5$, $6$, $7$, $8$, or $9$, because in these cases the number $n$ would have more digits than the number $k$. The quotient $\\frac{n}{k}$ cannot be $2$ or $4$, because the multiples of these numbers cannot end with the digit $5$. If the quotient $\\frac{n}{k}$ were $3$, then the last digit of the number $k$ should also be $5$ and the first digit of the number $n$ should therefore be $2$. However, since $n = 3k$, the number $n$ can only start with the digits $6$, $7$, and $8$. So again, the only possibility is $\\frac{n}{k} = 1$.\n\n* The first digit of the number $k$ cannot be $3$ or $4$, because these numbers do not occur in a twisting.\n\n* If the first digit of the number $k$ is $5$, $6$, $7$, $8$, or $9$, then $\\frac{n}{k} \\ge 2$ is not possible, because then there would be more digits in the number $n$ than in the number $k$. Thus, the only possibility is $\\frac{n}{k} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the value of the expression\n$$\nx^4 - 2x^3 - 88x^2 + 90x + 2025\n$$\nis positive regardless of the value of $x$ on the real line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using equalities $2025 = 45^2$ and $90 = 2 \\cdot 45$ we get\n$$\n\\begin{aligned}\nx^4 - 2x^3 - 88x^2 + 90x + 2025 &= x^4 - 2x^3 - 89x^2 + (x + 45)^2 \\\\\n&= (x^2)^2 - 2x^2(x + 45) + (x + 45)^2 + x^2 \\\\\n&= (x^2 - x - 45)^2 + x^2.\n\\end{aligned}\n$$\nBut $(x^2 - x - 45)^2 + x^2 \\ge 0$. Equality is impossible because it could occur only if $x = 0$ and also $x^2 - x - 45 = 0$. Thus indeed $(x^2 - x - 45)^2 + x^2 > 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 30077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point on the side $BC$ of an acute triangle $ABC$ and let $E$ be a point on the line segment $AD$. Let $F$ and $G$ be the feet of the altitudes drawn from the vertex $D$ in triangles $ABD$ and $ACD$, respectively. The line $BE$ intersects the circumcircle of the triangle $DEG$ at point $H \\neq E$. Prove that points $B, F, G$ and $H$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle AFD = \\angle AGD = 90^\\circ$, the quadrilateral $AFDG$ is cyclic (Fig. 28). Using inscribed angles in the circumcircle of the quadrilateral $AFDG$, we get\n$$\n\\angle BFG = 180^\\circ - \\angle AFG = 180^\\circ - \\angle ADG = 180^\\circ - \\angle EDG.\n$$\n\n![](attached_image_1.png)\nFig. 28\n\n![](attached_image_2.png)\nFig. 29\n\n![](attached_image_3.png)\nFig. 31\n\n$\\angle BHG = \\angle EHG = \\angle EDG$. Thus $\\angle BFG = 180^\\circ - \\angle BHG$, so the points $B, F, G$ and $H$ are concyclic.\n\n* If the order of the points is $D, E, H, G$ (Fig. 30), then opposite angles of the cyclic quadrilateral $DEHG$ give $\\angle BHG = \\angle EHG = 180^\\circ - \\angle EDG$. So $\\angle BFG = \\angle BHG$, hence $B, F, H$ and $G$ are concyclic.\n\n* If the order of the points is $D, H, E, G$ (Fig. 31), then we similarly get $\\angle BHG = 180^\\circ - \\angle EHG = 180^\\circ - \\angle EDG$. So again $\\angle BFG = \\angle BHG$, hence $B, F, H$ and $G$ are concyclic.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Ats and Pets take turns to write representations of the number $15$ as the sum of three distinct single-digit positive integers. On every move, each player must write a sum that has exactly one common addend with the previous sum, no common addends with the second previous sum, and less than three common addends with any sum written earlier. Ats starts and can choose the first sum arbitrarily. The player who cannot write a sum loses. Which player can win regardless of his opponent's play?",
    "options": [],
    "answer": "Pets (the second player)",
    "solution": "There are $8$ possible sums:\n$$\n\\begin{array}{ccc}\n1+5+9, & 2+6+7, & 3+4+8, \\\\\n1+6+8, & 2+4+9, & 3+5+7, \\\\\n2+5+8, & 4+5+6.\n\\end{array}\n$$\nNote that no two of the three sums in the first row have any common addends, and the same applies to the three sums in the second row. Moreover, each sum in the first two rows has one common addend with every sum outside its own row, including the sums in the third row. Thus, Pets can choose one of the first two rows (one from which Ats did not choose a sum in the first move) and write any sum from it on his turn. This guarantees a common addend with the sum that Ats has written on his turn. Since Ats cannot write a sum from the same row from which Pets has chosen his sum on his second move, Pets can choose any unused sum from that row on his second move, because it has a common addend with the sum written by Ats and no common addends with the sum last written by Pets. Analogously, Pets can make his third move if Ats has not already lost by then.\n\nWhen Pets has made his third move, there are only two sums left unused. This means that if Ats were to make another move, one of the sums in the third row should have been written on the board either on this move or on an earlier move. According to Pets' strategy, Ats had to do this. However, the sums in the third row have one common addend with every other sum, so according to the rules they cannot have either the second previous sum or the sum after the next sum. This contradicts the assumption that the game can continue after Pets' third move. So in fact, Ats has no more suitable sums left after Pets' third move and has lost.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30079,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Every side and diagonal of a regular $2025$-gon is coloured either red or blue. Can it happen that the same number of red and blue line segments meet at each vertex?\n\nb. The same question if only the diagonals are coloured.",
    "options": [],
    "answer": "a: Yes. b: No.",
    "solution": "a. Each side and diagonal connects vertices that are $1$ to $1012$ side lengths apart as you move along the polygon. Exactly $2$ sides or diagonals of a given length meet at each vertex. We color the sides and diagonals that have an odd number of sides between their endpoints red and all other diagonals blue. Since there are an equal number of odd and even lengths, there are also an equal number of $1012$ red and $1012$ blue segments meeting at each vertex.\n\nb. Suppose we can color the diagonals as required. Then each vertex has $1011$ red (and $1011$ blue) diagonals. Adding up the numbers of red diagonals that meet at each vertex, we get $1011 \\cdot 2025$, which is an odd number. On the other hand, each diagonal has been counted exactly $2$ times (once at each of its endpoints), so the sum should be even. The contradiction shows that the situation is not possible.\nConsider a graph whose vertices are the vertices of a given polygon and whose edges are its colored sides and diagonals.\n\na. We note that each vertex has $2024$ adjacent edges, which is an even number. Hence there is an Euler cycle in this graph. In total, there are $\\frac{2025 \\cdot 2024}{2} = 2025 \\cdot 1012$ edges in this graph, which is also an even number. We can color the edges in the resulting Euler cycle alternately red and blue. Then there will be one red and one blue edge at each vertex when we pass through that vertex. Since we eventually pass through all the edges and return to the starting vertex, the numbers of blue and red edges at each vertex will be equal.\n\nb. Now there are $\\frac{2025 \\cdot 2022}{2} = 2025 \\cdot 1011$ edges in total, which is odd. If the same number of red and blue line segments meet at each vertex then the total numbers of red and blue segments must be the same. So the total number of edges should be even, a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Anna, Berta and Carol make fruit drinks from syrup. Anna makes $a$ litres of drink by mixing water and syrup in the proportion of $a : 1$. Berta makes $b$ litres of drink by mixing water and syrup in the proportion of $b : 1$. Carol makes $c$ litres of drink by mixing water and syrup in the proportion of $c : 1$. (It is not known if $a$, $b$ and $c$ are integers.) They make $6$ litres of drink in total. Prove that they use at most $2$ litres of syrup.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The percentage of syrup in Anna's drink is $\\frac{1}{a+1}$, thus it contains $\\frac{a}{a+1}$ litres of syrup. Similarly, Berta uses $\\frac{b}{b+1}$ litres and Carol uses $\\frac{c}{c+1}$ litres of syrup. Hence we have to prove that $a + b + c = 6$ implies\n$$\n\\frac{a}{a+1} + \\frac{b}{b+1} + \\frac{c}{c+1} \\le 2.\n$$\nAs $\\frac{a}{a+1} = 1 - \\frac{1}{a+1}$ and similarly $\\frac{b}{b+1} = 1 - \\frac{1}{b+1}$ and $\\frac{c}{c+1} = 1 - \\frac{1}{c+1}$, this is equivalent to the inequality\n$$\n\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1} \\ge 1. \\quad (12)\n$$\nMultiplying both sides by $3$ leads to the equivalent inequality\n$$\n\\frac{3}{a+1} + \\frac{3}{b+1} + \\frac{3}{c+1} \\ge 3.\n$$\nAdding $3$ to both sides and applying $3 = \\frac{6+3}{3} = \\frac{a+b+c+3}{3} = \\frac{a+1}{3} + \\frac{b+1}{3} + \\frac{c+1}{3}$ in the left hand side, we obtain the equivalent inequality\n$$\n\\frac{3}{a+1} + \\frac{a+1}{3} + \\frac{3}{b+1} + \\frac{b+1}{3} + \\frac{3}{c+1} + \\frac{c+1}{3} \\ge 6. \\quad (13)\n$$\nThe sum of every positive real number and its reciprocal is at least $2$. Hence the inequality (13) holds for every $a$, $b$ and $c$.\nAs in Solution 1, we reduce the problem to the inequality (12). After converting the fractions to a common denominator, removing parentheses and collecting similar terms, it suffices to show that\n$$\n\\frac{ab + bc + ca + 2(a + b + c) + 3}{abc + ab + bc + ca + a + b + c + 1} \\ge 1,\n$$\nor equivalently,\n$$\nab + bc + ca + 2(a + b + c) + 3 \\ge abc + ab + bc + ca + a + b + c + 1.\n$$\nApplying the assumption $a+b+c=6$ and collecting similar terms reduces this inequality to $8 \\ge abc$. By AM-GM, $2 = \\frac{6}{3} = \\frac{a+b+c}{3} \\ge \\sqrt[3]{abc}$. Hence $8 = 2^3 \\ge abc$, completing the proof.\nAs in Solution 1, we reduce the problem to the inequality (12). The latter is equivalent to the inequality\n$$\n\\frac{\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1}}{3} \\ge \\frac{1}{3},\n$$\nBy AM-HM,\n$$\n\\frac{3}{\\frac{1}{a+1} + \\frac{1}{b+1} + \\frac{1}{c+1}} \\le \\frac{(a+1) + (b+1) + (c+1)}{3} = \\frac{a+b+c+3}{3}.\n$$\nAs $a+b+c=6$, we obtain $\\frac{a+b+c+3}{3} = 3$, completing the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30081,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incentre of a triangle $ABC$ is $I$. Points $D$ and $E$ on the sides $AB$ and $AC$, respectively, satisfy $DI \\perp BI$ and $EI \\perp CI$. Prove that the line $DE$ is tangent to the incircle of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X$ and $Y$ be the reflections of points $D$ and $E$, respectively, across the point $I$ (Fig. 32). Then $\\angle XBI = \\angle IBD = \\angle IBA = \\angle CBI$, implying that $X$ lies on the line $BC$. As $\\angle DIB = 90^\\circ$, points $D$, $I$ and $X$ lie on a line, i.e., $X$ is the point of intersection of lines $ID$ and $BC$. Analogously, we see that $Y$ is the point of intersection of lines $IE$ and $BC$. Hence $\\angle EID = \\angle YIX$, which along with the equalities $IX = ID$ and $IY = IE$ shows that triangles $DEI$ and $XYI$ are equal.\n\nThus also the altitudes drawn from the vertex $I$ in triangles $DEI$ and $XYI$ are equal. The altitude drawn from the vertex $I$ of the triangle $XYI$ equals the inradius of the triangle $ABC$. Hence the same holds for the triangle $DEI$, i.e., the incircle of the triangle $ABC$ passes through the foot of the altitude drawn from the vertex $I$ of the triangle $DEI$. The line $DE$ is perpendicular to this altitude which is the inradius of the triangle $ABC$; consequently, the line $DE$ is tangent to the incircle of the triangle $ABC$.\n\n![](attached_image_1.png)\nFig. 32\nLet $U$ and $V$ be the reflections of points $B$ and $C$, respectively, across the point $I$ (Fig. 33). Then $IU = IB$ and $IV = IC$, implying that $BC \\parallel UV$. As the line $BC$ is tangent to the incircle of the triangle $ABC$ whose centre $I$ is the centre of reflection, the line $UV$ is also tangent to the incircle of the triangle $ABC$ by symmetry.\n\nNow let the line $UV$ intersect the side $AB$ and $AC$ at points $D'$ and $E'$, respectively (Fig. 34). Then $\\angle D'UB = \\angle CBU = \\angle CBI = \\angle IBA = \\angle UBD'$, implying $D'B = D'U$. As $D'I$ is the median drawn from the vertex angle of the isosceles triangle $D'BU$, it must also be its altitude. This implies $D'I \\perp BI$ yielding $D' = D$.\n\nAnalogously, we get $E' = E$. Hence $DE = D'E' = UV$. Altogether, we have shown that the line $DE$ is tangent to the incircle of the triangle $ABC$.\n\n![](attached_image_2.png)\nFig. 33\n![](attached_image_3.png)\nFig. 34\nDenote $\\angle BAI = \\angle IAC = \\alpha$, as well as $\\angle CBI = \\angle IBA = \\beta$ and $\\angle ACI = \\angle ICB = \\gamma$. Let $K$ be the circumcentre of the triangle $DEI$ (Fig. 35). By construction, $\\alpha + \\beta + \\gamma = \\frac{180^\\circ}{2} = 90^\\circ$.\n\nFrom the triangle $BCI$, we get\n$$\n\\angle BIC = 180^\\circ - (\\beta + \\gamma) = 180^\\circ - (90^\\circ - \\alpha) = 90^\\circ + \\alpha,\n$$\nhence $\\angle EID = 360^\\circ - 90^\\circ - 90^\\circ - (90^\\circ + \\alpha) = 90^\\circ - \\alpha$. From the circum-circle of the triangle $DEI$, we now get\n$$\n\\angle EKD = 2(90^\\circ - \\alpha) = 180^\\circ - 2\\alpha = 180^\\circ - \\angle DAE.\n$$\nThus the quadrilateral $ADKE$ is cyclic. Its chords $DK$ and $EK$ are equal, implying that the corresponding inscribed angles are also equal, i.e., $\\angle DAK = \\angle KAE$. Hence $K$ lies on the bisector of the angle $BAC$, i.e., on the line $AI$.\n\n![](attached_image_4.png)\nFig. 35\n![](attached_image_5.png)\nFig. 36\n\nFrom the triangle $ABI$, we get\n$$\n\\angle AIB = 180^\\circ - (\\alpha + \\beta) = 180^\\circ - (90^\\circ - \\gamma) = 90^\\circ + \\gamma,\n$$\nhence $\\angle KID = \\angle AID = (90^\\circ + \\gamma) - 90^\\circ = \\gamma$. Therefore also $\\angle IDK = \\gamma$ as $DK = IK$. On the other hand, $\\angle KDE = \\frac{180^\\circ - (180^\\circ - 2\\alpha)}{2} = \\alpha$, implying $\\angle IDE = \\gamma + \\alpha = 90^\\circ - \\beta$. Since also $\\angle BDI = 90^\\circ - \\beta$, the line $DI$ is the external bisector of the angle $ADE$, whereas $I$ is the point of intersection of this external bisector and the internal bisector of $DAE$. This means that $I$ is the excentre of the triangle $ADE$. As the incentre of the triangle $ABC$ is $I$ and the incircle of $ABC$ is tangent to the prolongation of the side $AD$ of the triangle $ADE$, these circles must coincide. Hence $DE$ is tangent to the incircle of the triangle $ABC$.\nLet $r$ be the inradius of the triangle $ABC$. We show that the point $D$ lies at distance $2r$ from the side $BC$.\n\nTo this end, let $D'$ be the projection of the point $D$ on the side $BC$ (Fig. 36) and let $\\angle CBI = \\angle IBA = \\beta$. Then $BI = \\frac{r}{\\sin\\beta}$, $DB = \\frac{BI}{\\cos\\beta} = \\frac{r}{\\sin\\beta\\cos\\beta}$ and $DD' = DB \\sin 2\\beta = \\frac{r}{\\sin\\beta\\cos\\beta} \\cdot 2\\sin\\beta\\cos\\beta = 2r$.\n\nAnalogously, we can show that the point $E$ lies at distance $2r$ from the side $BC$. Thus the line $DE$ is parallel to the side $BC$ and at distance $2r$ from it. As the line $BC$ is tangent to the incircle of the triangle $ABC$, also the line $DE$ is tangent to this circle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be any natural number. Find the least natural number $k$ such that it is possible to write a natural number from $1$ through $k$ into every cell of an $n \\times n$ table in such a way that the sum of every two cells with a common side differs from all other such sums. The numbers in different cells do not have to be distinct.",
    "options": [],
    "answer": "n^2 - n + 1",
    "solution": "The least and the largest among the sums of two natural numbers from $1$ through $k$ are $1+1=2$ and $k+k=2k$, respectively. Thus there can be at most $2k-1$ distinct sums. The number of distinct pairs of cells with a common side is $n-1$ in every row and column. As the total number of rows and columns is $2n$, the number of such pairs is $2n(n-1) = 2n^2-2n$. Hence $2k-1 \\ge 2n^2-2n$ which implies\n$$\nk \\ge \\frac{2n^2 - 2n + 1}{2} = n^2 - n + \\frac{1}{2}.\n$$\nAs $k$ is an integer, we have $k \\ge n^2 - n + 1$.\n\nNow we show that $k = n^2 - n + 1$ is achievable. To this end, partition the $n \\times n$ table into $2n-1$ diagonals (directed from top right to bottom left; see Fig. 37). We fill the diagonals starting from the top left corner with consecutive integers $1, 2, \\dots$, but whenever switching from a diagonal with an odd number to the next diagonal with an even number, we repeat the number just written. This repetition happens $n-1$ times. Hence the last\n\n![](attached_image_1.png)\nFig. 37\n![](attached_image_2.png)\nFig. 38\nnumber written into the bottom right cell is $n^2 - n + 1$ as desired. Figure 38 depicts the situation for $n = 4, k = 13$.\nWe show that the sum of every two cells with a common side is unique. Each sum under consideration is obtained by adding numbers in cells of two consecutive diagonals. All sums of cells of the same two diagonals are obviously distinct, because when moving from the top right end to the bottom left end, the sum strictly increases at every step. When considering distinct pairs of consecutive diagonals, the sums must also be distinct. Indeed, let one pair under consideration contain diagonals No $a$ and $a+1$ and let the other pair contain diagonals No $b$ and $b+1$, where $a < b$. Then numbers in the first diagonal of the first pair do not exceed numbers in the first diagonal of the second pair, whereas numbers in the second diagonal of the first pair do not exceed numbers in the second diagonal of the second pair. In either case, equality can hold only if $b = a + 1$. But the equality cannot hold for both cases simultaneously, because $a$ and $a+1$ have distinct parities, implying that, by construction, when switching either from the diagonal No $a$ to the diagonal No $a+1$ or from the diagonal No $b$ to the diagonal No $b+1$, the last number is not repeated.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30083,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $E$ is chosen on the side $AB$ of a rectangle $ABCD$ ($E \\neq A, E \\neq B$). The line segments $BD$ and $CE$ intersect at point $F$. Among the triangles $ADE$, $DEF$, $DCF$, $BCF$ and $BEF$, there are exactly two pairs of triangles with equal area (the order of components in a pair is not taken into account). Find the ratio of the lengths of the line segments $EB$ and $AB$.",
    "options": [],
    "answer": "sqrt(2)/2",
    "solution": "Denote the area of a figure $K$ by $S_K$. As\n$$\nS_{DEF} + S_{DCF} = S_{CDE} = \\frac{1}{2} S_{ABCD} = S_{BCD} = S_{BCF} + S_{DCF},\n$$\nwe have $S_{DEF} = S_{BCF}$ (Fig. 39). Hence ($DEF$, $BCF$) is one pair of triangles with equal area regardless of the choice of point $E$.\n\nIn order to have exactly two such pairs, none of the remaining three triangles can have the same area as triangles *DEF* and *BCF*. Thus the second pair must come from among triangles *ADE*, *BEF* and *DCF*. On the other hand,\n$$\nS_{DEF} + S_{DCF} = S_{CDE} = \\frac{1}{2} S_{ABCD} = S_{ABD} = S_{ADE} + S_{DEF} + S_{BEF},\n$$\nimplying $S_{DCF} = S_{ADE} + S_{BEF}$. Hence each of triangles *ADE* and *BEF* has an area smaller than that of the triangle *DCF*. Thus the second pair of triangles with equal area is (*ADE*, *BEF*). Taking into account the equality $S_{ADE} + S_{BEF} = S_{DCF}$, we obtain $S_{ADE} = S_{BEF} = \\frac{1}{2} S_{DCF}$.\n\nAs $\\angle EBF = \\angle CDF$ and $\\angle FEB = \\angle FCD$, triangles *BEF* and *DCF* are similar. Hence $\\frac{S_{BEF}}{S_{DCF}} = \\left(\\frac{EB}{CD}\\right)^2 = \\left(\\frac{EB}{AB}\\right)^2$, implying\n$$\n\\frac{EB}{AB} = \\sqrt{\\frac{S_{BEF}}{S_{DCF}}} = \\sqrt{\\frac{\\frac{1}{2} S_{DCF}}{S_{DCF}}} = \\sqrt{\\frac{1}{2}} = \\frac{\\sqrt{2}}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30084,
    "subject": "Mathematics (Multi-modal)",
    "question": "The letters $M$, $A$, $T$, $E$, $I$ and $K$ are assigned the numbers $1$, $2$, $3$, $4$, $5$ and $6$ in a certain order so that different letters correspond to different numbers. The sum of the numbers corresponding to the letters of the word $MATEMAATIK$ (taking into account repetitions) is $42$ and the sum of the numbers corresponding to the letters of the word $KEEMIK$ (taking into account repetitions) is $13$. Find the sum of the numbers corresponding to the letters of the word $IT$.",
    "options": [],
    "answer": "8",
    "solution": "In the words $MATEMAATIK$, $KEEMIK$ and $IT$, all the given letters occur exactly $3$ times. Since the sum of the numbers corresponding to the letters $M$, $A$, $T$, $E$, $I$ and $K$ is $1+2+3+4+5+6=21$, the sum of the numbers corresponding to the letters in these three words is $3 \\cdot 21 = 63$. Consequently, the sum of the numbers corresponding to the letters in the word $IT$ is $63 - 42 - 13 = 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the least positive integer $n$ such that:\n\na. both $n$ and $n+1$ are divisible by the squares of two distinct prime numbers;\n\nb. both $n$ and $n+3$ are divisible by the squares of two distinct prime numbers.",
    "options": [],
    "answer": "a: 675; b: 2025",
    "solution": "Both the pair $(n, n + 1)$ and $(n, n + 3)$ must contain one odd number. This odd number must be divisible by the square of one of the following numbers:\n$$\n15 = 3 \\cdot 5,\\ 21 = 3 \\cdot 7,\\ 33 = 3 \\cdot 11,\\ 35 = 5 \\cdot 7,\\ 39 = 3 \\cdot 13,\\ 45 = 3^2 \\cdot 5, \\dots\n$$\nTheir squares are $225$, $441$, $1089$, $1225$, $1521$, $2025$, .... Thus either $n$ or $n + 1$ or $n + 3$ must be a multiple of one of these squares.\n\na. We cannot have $n + 1 = 225$ or $n = 225$, as neither $224 = 2^5 \\cdot 7$ nor $226 = 2 \\cdot 113$ is divisible by two prime squares. Similarly, we cannot have $n + 1 = 441$ or $n = 441$, as $440 = 2^3 \\cdot 5 \\cdot 11$ and $442 = 2 \\cdot 13 \\cdot 17$. We also cannot have $n + 1 = 3 \\cdot 225 = 675$, as $674 = 2 \\cdot 337$. But $n = 675$ works, as $676 = 2^2 \\cdot 13^2$. Thus the desired $n$ is $675$.\n\nb. In the solution to part (a) we saw that both $675$ and $676$ are divisible by two prime squares. Therefore, so must also the numbers $3 \\cdot 675 = 2025$ and $3 \\cdot 676 = 2028$. Thus $n = 2025$ has the desired properties. It remains to show that there are no smaller such numbers.\nThe only option for $n$ or $n + 3$ not divisible by $2^2$ or $3^2$ is $1225$, but we cannot have either $n + 3 = 1225$ or $n = 1225$, as $1222 = 2 \\cdot 13 \\cdot 47$ and $1228 = 2^2 \\cdot 307$. We may now assume that one of $n$ and $n + 3$ is divisible by $2^2$ and the other by $3^2$, as they cannot be divisible by the same prime square.\n\nThe odd number out of $n$ and $n + 3$ is either one of the odd multiples of $225$, which are $225$, $675$, $1125$ and $1575$, one of the odd multiples of $441$, which are $441$ and $1323$, or one of $1089$ and $1521$. For each of them, only one of the numbers differing from it by $3$ is divisible by $2^2$, so it remains to check that these numbers do not contain any other prime factors squared:\n$$\n\\begin{aligned}\n225 &\\rightarrow 228 = 2^2 \\cdot 3 \\cdot 19, \\\\\n675 &\\rightarrow 672 = 2^2 \\cdot 3 \\cdot 56, \\\\\n1125 &\\rightarrow 1128 = 2^2 \\cdot 3 \\cdot 94, \\\\\n1575 &\\rightarrow 1572 = 2^2 \\cdot 3 \\cdot 131, \\\\\n441 &\\rightarrow 444 = 2^2 \\cdot 3 \\cdot 37, \\\\\n1323 &\\rightarrow 1320 = 2^2 \\cdot 3 \\cdot 110, \\\\\n1089 &\\rightarrow 1092 = 2^2 \\cdot 3 \\cdot 91, \\\\\n1521 &\\rightarrow 1524 = 2^2 \\cdot 3 \\cdot 127.\n\\end{aligned}",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30086,
    "subject": "Mathematics (Multi-modal)",
    "question": "The coefficients of the polynomial\n$$\nP(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0\n$$\nare real numbers such that $a_i = a_{n-i}$ for every $i = 0, 1, \\dots, n$, and $a_n \\neq 0$. Let $x_1, x_2, \\dots, x_k$ be all the real roots of the polynomial $P(x)$ without repetitions.\n\na. Prove that\n$$\n|x_1| + |x_2| + \\dots + |x_k| \\ge k.\n$$\n\nb. Is the inequality definitely strict in the case $k > 1$?",
    "options": [],
    "answer": "a. The sum of the absolute values of the distinct real roots is at least the number of such roots. b. No; the inequality need not be strict. For example, (x−1)²(x+1) has real roots 1 and −1 with sum of absolute values equal to 2.",
    "solution": "a. Let $c$ be any root of $P(x)$. As $P(0) = a_0 = a_n \\neq 0$, we must have $c \\neq 0$. Notice that\n$$\n\\begin{aligned}\nP\\left(\\frac{1}{c}\\right) &= a_n \\left(\\frac{1}{c}\\right)^n + a_{n-1} \\left(\\frac{1}{c}\\right)^{n-1} + \\dots + a_1 \\left(\\frac{1}{c}\\right) + a_0 \\\\\n&= \\left(\\frac{1}{c}\\right)^n (a_0 c^n + a_1 c^{n-1} + \\dots + a_{n-1} c + a_n) \\\\\n&= \\left(\\frac{1}{c}\\right)^n (a_n c^n + a_{n-1} c^{n-1} + \\dots + a_1 c + a_0) \\\\\n&= \\left(\\frac{1}{c}\\right)^n P(c) = 0,\n\\end{aligned}\n$$\nwhich shows that $\\frac{1}{c}$ is also a root of $P(x)$. Thus the roots can be divided into inverse pairs.\nFor each root $c$ we have $|c| + \\left|\\frac{1}{c}\\right| \\ge 2$. The only roots paired with itself are 1 and -1, both of which have an absolute value of 1. If there are $l$ such roots, then\n$$\n|x_1| + |x_2| + \\dots + |x_k| \\ge \\frac{k-l}{2} \\cdot 2 + l \\cdot 1 = k - l + l = k.\n$$\n\nb. The polynomial $P(x) = x^3 - x^2 - x + 1 = (x-1)^2(x+1)$ satisfies the conditions of the problem and its real roots are $x_1 = 1$ and $x_2 = -1$. Thus $k = 2 > 1$, but $|x_1| + |x_2| = 2 = k$, so the inequality is non-strict.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $\\omega_1$ and $\\omega_2$ touch at point $K$. The line through the centres of the circles intersects the circle $\\omega_1$ once more at point $A$. A line through the point $A$ intersects the circle $\\omega_1$ once more at point $B$ and the circle $\\omega_2$ at points $C$ and $D$, where the points $A, B, C, D$ lie on the line in the order. Given that the line segments $AB$, $BC$ and $CD$ have equal lengths, find the ratio of the radii of the circles $\\omega_1$ and $\\omega_2$.",
    "options": [],
    "answer": "1/3",
    "solution": "Let the radii of $\\omega_1$ and $\\omega_2$ be $r_1$ and $r_2$ respectively. Let $E$ be the second intersection of $AK$ and $\\omega_2$ (Fig. 40). As $AK$ and $KE$ are diameters of $\\omega_1$ and $\\omega_2$ respectively, the angles $ABK$ and $KDE$ must be right angles. In triangle $AKC$, the segment $KB$ is both a median and an altitude, thus $KA = KC$ and $\\angle KAC = \\angle ACK$. As $CDEK$ is cyclic, we have\n$$\n\\angle EAD = \\angle KAC = \\angle ACK = \\angle DEA.\n$$\nThus $ADE$ is isosceles, also $KBA$ and $KDE$ are similar. Therefore\n$$\n\\frac{r_1}{r_2} = \\frac{AK}{KE} = \\frac{AB}{DE} = \\frac{AB}{AD} = \\frac{1}{3}.\n$$\n\n![](attached_image_1.png)\nFig. 40\nLet the radii of $\\omega_1$ and $\\omega_2$ be $r_1$ and $r_2$ respectively. Let $P$ be the midpoint of $CD$ and $Q$ the center of $\\omega_2$ (Fig. 41). Then $PQ$ is perpendicular to $CD$, as it connects the midpoint of the chord $CD$ of $\\omega_2$ and the center of $\\omega_2$. Thus $\\angle APQ = \\angle CPQ = 90^\\circ$. As $AK$ is a diameter of $\\omega_1$, we have $\\angle ABK = 90^\\circ$. Therefore $BK \\parallel PQ$. Together with $AB = BC = CD$, we get $\\frac{AQ}{AK} = \\frac{AP}{AB} = 2.5$ or $2AQ = 5AK$. From here $2(2r_1 + r_2) = 5 \\cdot 2r_1$, from which $\\frac{r_1}{r_2} = \\frac{1}{3}$.\n\n![](attached_image_2.png)\nFig. 41",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which all $10^n$ non-negative integers consisting of $n$ digits can be ordered in such a way that all the following conditions are met:\n(1) the first number consists of zeros only;\n(2) every two numbers that are consecutive in this order differ at exactly one position and the digits at this position differ by exactly 1;\n(3) the last number consists of nines only.\n\n*Remark:* In this problem, we allow numbers to begin with zero(s).",
    "options": [],
    "answer": "All odd positive integers n",
    "solution": "Assume that for some $n$, a suitable ordering exists. From condition 2 we see that the sums of digits of any two consecutive numbers differ by exactly 1, and thus have opposite parities. As the first number has an even sum of digits, the $10^n$-th number must have an odd sum of digits. As the latter sum is $9n$, this means that the conditions can only be satisfied for odd $n$.\n\nNow let $n$ be odd. We prove by induction that a suitable ordering exists. For $n = 1$, the natural ordering works. Now let $n > 1$ and assume that an ordering exists for $n - 2$. We first order all numbers starting with 00, using the existing ordering for the other digits. Then we order the numbers starting with 01 in the reverse existing ordering for the other digits, then the numbers starting with 02 in the original order, and so on. During this process, we change the first two digits in the following order:\n$$\n\\begin{array}{r@{\\quad}l}\n00 & \\rightarrow 01 \\rightarrow \\dots \\rightarrow 09 \\rightarrow 19 \\rightarrow 18 \\rightarrow \\dots \\rightarrow 10 \\rightarrow \\\\\n& \\rightarrow 20 \\rightarrow 21 \\rightarrow \\dots \\rightarrow 29 \\rightarrow 39 \\rightarrow 38 \\rightarrow \\dots \\rightarrow 30 \\rightarrow \\\\\n& \\rightarrow 40 \\rightarrow 41 \\rightarrow \\dots \\rightarrow 49 \\rightarrow 59 \\rightarrow 58 \\rightarrow \\dots \\rightarrow 50 \\rightarrow \\\\\n& \\rightarrow 60 \\rightarrow 61 \\rightarrow \\dots \\rightarrow 69 \\rightarrow 79 \\rightarrow 78 \\rightarrow \\dots \\rightarrow 70 \\rightarrow \\\\\n& \\rightarrow 80 \\rightarrow 81 \\rightarrow \\dots \\rightarrow 89.\n\\end{array}\n$$\n\nThe last number written so far ends with zeros. It remains to order the numbers starting with a 9. We divide them into groups based on their last $n-2$ digits. We order these groups by the ordering existing for numbers with $n-2$ digits. Within the first group, we order the numbers in the descending order, within the second group in the ascending order etc. There are an even number of groups, so the final group is in the ascending order. Thus the created ordering will satisfy the conditions of the problem.\nWe exclude the even $n$ like in Solution 1.\nLike in Solution 1, we find an ordering for odd $n$ by induction. We again divide the numbers into groups based on their last $n-2$ digits and order the groups by the existing ordering. There are again an even number of groups.\n\nIn odd-numbered groups, we order the numbers in the group based on their first two digits, using the ordering in Fig. 42, starting from 00 and ending with 54. In even-numbered groups, we use the same ordering backwards, except for the final group, where we use the ordering in Fig. 43, starting with 54 and ending with 99.\n\n![](attached_image_1.png)\n\nFig. 42\n![](attached_image_2.png)\nFig. 43",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every night, Juku listens to exactly $14$ songs from a playlist containing exactly $100$ songs. Every time a song ends, the next song is chosen from among all $100$ songs with equal probability (the same song may also repeat). Prove that, on more than half of all nights, Juku listens some song more than once.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that the probability of listening to the same song multiple times is greater than $0.5$, which is equivalent to the probability of listening to $14$ distinct songs being less than $0.5$. The latter probability can be expressed as\n$$\n1 \\cdot 0.99 \\cdot 0.98 \\cdot 0.97 \\cdot 0.96 \\cdot 0.95 \\cdot 0.94 \\cdot 0.93 \\cdot 0.92 \\cdot 0.91 \\cdot 0.9 \\cdot 0.89 \\cdot 0.88 \\cdot 0.87.\n$$\n\nTo prove the bound, notice that\n$$\n0.93 \\cdot 0.87 = (0.9 + 0.03)(0.9 - 0.03) = 0.9^2 - 0.03^2 < 0.9^2,\n$$\n$$\n0.92 \\cdot 0.88 = (0.9 + 0.02)(0.9 - 0.02) = 0.9^2 - 0.02^2 < 0.9^2,\n$$\n$$\n0.91 \\cdot 0.89 = (0.9 + 0.01)(0.9 - 0.01) = 0.9^2 - 0.01^2 < 0.9^2.\n$$\nThus the product above is less than $0.9^7$. We will now show that $0.9^7 < 0.5$. Using the property $0.9 \\cdot \\overline{0.ab} = \\overline{0.ab} - \\overline{0.0ab} \\le \\overline{0.ab} - \\overline{0.0a}$ repeatedly, we get\n$$\n\\begin{aligned}\n0.9^7 &= 0.9 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\\\\n&= 0.81 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\\\\n&\\le 0.73 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\\\\n&\\le 0.66 \\cdot 0.9 \\cdot 0.9 \\cdot 0.9 \\\\\n&\\le 0.6 \\cdot 0.9 \\cdot 0.9 \\\\\n&= 0.54 \\cdot 0.9 \\\\\n&\\le 0.49 < 0.5.\n\\end{aligned}\n$$\n\nThis finishes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Integers are assigned to variables $x$, $y$ and $z$ to satisfy the equation\n$$\nx^3 + y^3 + z^3 - 3xyz = 2025.\n$$\nFind all possible values of the sum $x + y + z$.",
    "options": [],
    "answer": "3, 9, 27, 75, 225, 675",
    "solution": "The given equation is equivalent to the equation\n$$\nx^3 + y^3 + z^3 = 2025 + 3xyz.\n$$\nAs $3 \\mid 2025$ and $3 \\mid 3xyz$, also $3 \\mid x^3 + y^3 + z^3$. Thus $3 \\mid x + y + z$ as an integer and its cube are congruent modulo 3. Let $x + y + z = 3m$.\n\nW.l.o.g., let $x \\le y \\le z$ and $x = m-a$, $y = m+b$, $z = m+c$, where $a \\ge 0$, $c \\ge 0$ and $b = a-c$. Substituting for $x$, $y$, $z$ in the given equation and simplifying leads to\n$$\n3m(a^2 + b^2 + c^2 + ab + ac - bc) - a^3 + b^3 + c^3 + 3abc = 2025.\n$$\nSubstituting also $b = a-c$ now gives $3m(3a^2 - 3ac + 3c^2) = 2025$ or, equivalently,\n$$\nm(a^2 - ac + c^2) = 225.\n$$\nAs $a^2 - ac + c^2 = (a-c)^2 + ac \\ge 0$, the number $m$ is a positive divisor of 225. The positive divisors of 225 are 1, 3, 5, 9, 15, 25, 45, 75, 225.\n\nIf $a = 0$ then $mc^2 = 225$; we obtain 1, 9, 25, 225 as possible values of $m$, giving 3, 27, 75, 675 as the corresponding values of $x + y + z$. If $a = 2c$ then $mc^2 = 75$; now $m$ can be either 3 or 75, giving 9 and 225 as the corresponding values of $x + y + z$. In the remaining cases, $5 \\mid m$ but $25 \\nmid m$; then also $5 \\mid a^2 - ac + c^2$ but $25 \\nmid a^2 - ac + c^2$. Thus $5 \\mid (a^2 - ac + c^2)(a+c) = a^3 + c^3$. Cubing integers that are incongruent modulo 5 produces results incongruent modulo 5. As $a^3 = -c^3 \\equiv (-c)^3 \\pmod 5$, we must have $a \\equiv -c \\pmod 5$. Hence $0 \\equiv a^2 - ac + c^2 \\equiv (-c)^2 - (-c)c + c^2 = 3c^2 \\pmod 5$, implying that $5 \\nmid c$. But then also $5 \\mid a$ and $25 \\mid a^2 - ac + c^2$. The contradiction shows that there are no other solutions.\nIt is easy to see that\n$$\nx^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx). \\quad (15)\n$$\nAs $3 \\mid 2025$, either $3 \\mid x + y + z$ or $3 \\mid x^2 + y^2 + z^2 - xy - yz - zx$. On the other hand, the representation\n$$\nx^2 + y^2 + z^2 - xy - yz - zx = (x + y + z)^2 - 3(xy + yz + zx)\n$$\nshows that $3 \\mid x^2 + y^2 + z^2 - xy - yz - zx$ if and only if $3 \\mid x + y + z$. Hence $3 \\mid x + y + z$ anyway.\n\nLet $x + y + z = 3m$. W.l.o.g., let $x \\le y \\le z$ and $x = m-a$, $y = m+b$, $z = m+c$, where $a \\ge 0$, $c \\ge 0$ and $a = b+c$. Using the equality (15) along with the equality\n$$\nx^2 + y^2 + z^2 - xy - yz - zx = \\frac{1}{2}((x-y)^2 + (y-z)^2 + (z-x)^2),\n$$\nas well as the equations about $x$, $y$ and $z$, the initial equation can be converted to\n$$\n3m((a+b)^2 + (b-c)^2 + (a+c)^2) = 4050.\n$$\nSubstituting also $a = b + c$ gives $3m(6b^2 + 6bc + 6c^2) = 4050$ which reduces to\n$$\nm(b^2 + bc + c^2) = 225.\n$$\nNote that $b^2 + bc + c^2 = \\frac{1}{2}((b+c)^2 + b^2 + c^2) \\ge 0$; hence $m$ is a positive divisor of 225. The positive divisors of 225 are 1, 3, 5, 9, 15, 25, 45, 75, 225. If $b=0$ then $mc^2 = 225$; we obtain 1, 9, 25, 225 as possible values of $m$. If $b=c$ then $mc^2 = 75$; now $m$ can be either 3 or 75. It remains to check the possibilities $m=5$, $m=15$ and $m=45$. In these cases $b^2 + bc + c^2 = d$ where $d$ is 45, 15 or 5, respectively. As $b^2 + bc + c^2 = d$ is equivalent to $3b^2 + (b+2c)^2 = 4d$, the number $4d - 3b^2$ is a perfect square. If $d=5$ then $20 - 3b^2$ must be a perfect square which is impossible. If $d=15$ then we have $3 \\mid 4d - 3b^2$, hence $60 - 3b^2 = (3k)^2$, implying $20 - 3k^2 = b^2$ for some integer $k$ which is impossible. If $d=45$ then $3 \\mid 4d - 3b^2$ again, hence $180 - 3b^2 = (3k)^2$, implying that $60 - 3k^2 = b^2$ for some integer $k$ which is impossible.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f$ from the set of all non-negative real numbers to the set of all real numbers such that $f(1) = 1$ and\n$$\n(f(x+y))^2 \\le f(x^2 - 2xy + y^2)\n$$\nfor all real numbers $x$ and $y$ that satisfy the inequality $x + y \\ge 0$.",
    "options": [],
    "answer": "f(x) = 1 for all non-negative real x",
    "solution": "The given inequality can be rewritten as\n$$\n(f(x+y))^2 \\le f((x-y)^2).\n$$\n\nTake $y = 1-x$. Then $1 = 1^2 = (f(1))^2 \\le f((2x-1)^2)$. As $(2x-1)^2$ obtains all non-negative real values, we can conclude that $1 \\le f(z)$ for every non-negative real number $z$.\nNow take $x \\ge -\\frac{1}{2}$ and $y = 1+x$. Then $(f(2x+1))^2 \\le f(1) = 1$. This implies $f(2x+1) \\le 1$ because all values of $f$ are non-negative by the above. As $2x+1$ obtains all non-negative real values, we can deduce that $f(z) \\le 1$ for every non-negative real number $z$.\n\nConsequently, $f(z) = 1$ for every non-negative real number $z$. This function satisfies all conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30092,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$ with $AB < AC$, points $D$, $E$ and $F$ are the feet of the altitudes drawn from vertices $A$, $B$ and $C$, respectively. Let the orthocenter of $ABC$ be $H$ and the midpoint of the side $BC$ be $M$. Point $K$ on the prolongation of the line segment $EM$ beyond $M$ and point $L$ on the line segment $FM$ satisfy $MK = ML = MD$. Prove that points $K$, $L$ and $H$ lie on a line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly, we prove that $BDLF$ is an isosceles trapezium (Fig. 44). As $\\angle CEB = 90^\\circ = \\angle CFB$, points $B$, $C$, $E$ and $F$ lie on a circle with diameter $BC$. Hence $M$ is the center of this circle. Consequently, $MB = MF$, implying\n\n![](attached_image_1.png)\nFig. 44\n\n$DB = MB - MD = MF - ML = LF$. Hence $DL \\parallel BF$ and the desired claim follows.\n\nConsequently, points $B$, $D$, $L$ and $F$ also lie on a circle (Fig. 45). Since $\\angle BDH = 90^\\circ = \\angle HFB$, we know that $BH$ is a diameter of this circle. Thus $H$ lies on this circle, too. Now\n$$\n\\begin{aligned}\n\\angle MLK &= \\frac{\\angle 180^\\circ - \\angle KML}{2} = \\frac{\\angle LME}{2} = \\frac{\\angle FME}{2} \\\\\n&= \\angle FBE = \\angle FBH = \\angle FLH.\n\\end{aligned}\n$$\n\nBut $\\angle MLK = \\angle FLH$ implies that points $K$, $L$, and $H$ lie on a line.\nAs in Solution 1, we prove that $BDLF$ is an isosceles trapezium whose circumcircle passes through $H$. By interchanging the roles of $B$ and $C$, the roles of $E$ and $F$, and the roles of $K$ and $L$, we analogously obtain that $CKDE$ is an isosceles trapezium whose circumcircle passes through $H$ (Fig. 46). Now $\\angle LHB = \\angle LFB = \\angle FBD = \\angle ABC$, but, on the other hand,\n$$\n\\angle KHB = 180^\\circ - \\angle EHK = \\angle KCE = \\angle CED.\n$$\nAs $\\angle BDA = 90^\\circ = \\angle BEA$, the quadrilateral $ABDE$ is cyclic, implying that $\\angle CED = \\angle ABC$. Consequently, $\\angle LHB = \\angle KHB$, implying that points $K$, $L$ and $H$ lie on a line.\nAs in Solution 1, we prove that points $B$, $D$, $L$, $F$ and $H$ lie on a circle with diameter $BH$. By interchanging the roles of $B$ and $C$, the roles of $E$ and $F$, and the roles of $K$ and $L$, we analogously obtain that points $C$, $D$, $K$, $E$ and $H$ lie on a circle with diameter $CH$. As $HL \\perp BL$ and $HK \\perp CK$ by Thales' theorem (Fig. 47), it suffices to prove that $BL \\parallel CK$. To this end, note that\n$$\n\\begin{aligned}\n\\angle LBM &= \\angle LBD = \\angle LFD = \\angle MFD, \\\\\n\\angle KCM &= \\angle KCD = \\angle KED = \\angle MED.\n\\end{aligned}\n$$\n\n![](attached_image_2.png)\nFig. 46\n![](attached_image_3.png)\nFig. 47\n\nIt is known that the midpoints of sides of a triangle and the feet of altitudes of the triangle lie on a circle (so-called nine-point circle). Hence $M$, $D$, $E$ and $F$ are concyclic, implying that $\\angle MFD = \\angle MED$. Consequently, $\\angle LBM = \\angle KCM$, proving the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30093,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a plane, 5 points are chosen arbitrarily. Find the largest possible number of distinct right triangles with all vertices in the chosen points.",
    "options": [],
    "answer": "8",
    "solution": "A square $ABCD$ and its centre $E$ determine 8 distinct right triangles: $ABC$, $BCD$, $CDA$, $DAB$, $AEB$, $BEC$, $CED$, $DEA$.\n\n![](attached_image_1.png)\nFig. 48\n\nWe show that more than 8 right triangles is impossible. Firstly, note that among any 5 points, one can choose 4 points that are vertices of a rectangle in at most one way. Indeed, suppose that this can be done in two different ways. Then these two quadruples of points have 3 points in common. But 3 vertices of a rectangle (even parallelogram) uniquely determine the last vertex.\n\nSecondly, note that a quadrangle whose every three vertices form a right triangle is a rectangle. Indeed, these four triangles must have right angle at distinct vertices, otherwise three points would lie on a line. Now if $A$, $B$, $C$ are the vertices of a right triangle with right angle at $B$ then the remaining vertex $D$ must be located in such a way that some triangle could have right triangle at $A$ and some triangle could have right angle at $C$. Hence $D$ must lie on a line passing through $A$ and perpendicular to either $AB$ or $AC$, and also on a line passing through $C$ and perpendicular to either $AC$ or $BC$ (in Fig. 49, these lines are drawn using dashes of distinct colours).\n\n![](attached_image_2.png)\nFig. 49\n\nAs lines perpendicular to $AC$ do not meet, $D$ must lie on the line passing through $A$ perpendicular to $AB$ or on the line passing through $C$ perpendicular to $BC$. If $D$ lies on both these lines, $ABCD$ is a rectangle. Hence consider the case with $D$ lying on exactly one of these lines. W.l.o.g., let $D$ lie on the line passing through $A$ perpendicular to $AB$ and on the line passing through $C$ perpendicular to $AC$. But then $BCD$ is not a right triangle. Hence the only possibility for $D$ is such that $ABCD$ is a rectangle.\n\nLet now 5 points be chosen on a plane. Let us count right triangles with vertices in the chosen points by quadrangles, among the vertices of which the vertices of the triangle are. As there are 5 possibilities to extract 4 points out of the given 5 points and, by the above, at most one of the obtained quadruples enable 4 right triangles and the others consequently enable at most 3 right triangles, we can have at most $4 + 4 \\cdot 3 = 16$ right triangles. But each of these right triangles is counted twice as the fourth vertex can be either of the two remaining points. Hence there are at most 8 distinct right triangles.\nAs in Solution 1, we show that it is possible to find 8 right triangles.\n\nNow we prove that more than 8 right triangles is impossible. Suppose the opposite, i.e., there can be 9 right triangles. As there are $C_5^3 = 10$ triples of chosen points in total, at most one triple does not determine a right triangle. Let $A$ and $B$ be chosen points at maximal distance from each other (if there are several pairs of points with this property, take any of them). By choice, $AB$ can only be the hypotenuse of a right triangle. Hence at least two of the remaining three chosen points must lie on the circle with diameter $AB$; let these points be $C$ and $D$. At least one of $ACD$ and $BCD$ must be a right triangle. As all vertices of this triangle lie on the circle with diameter $AB$, two vertices of this triangle must be the endpoints of another diameter of this circle. These points can only be $C$ and $D$. Therefore also $CD$ is a line segment of maximal length and can only be the hypotenuse of a right triangle.\n\nIf the last chosen point $E$ also lies on the circle with diameter $AB$ then $ACE$ and $BCE$ are not right triangles because no pair of vertices can be the endpoints of a diameter of this circle. But if $E$ does not lie on this circle then $AEB$ and $CED$ are not right angles, implying that $ABE$ and $CDE$ cannot be right triangles. The contradiction shows that there cannot be 9 right triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a geometric progression, among the members of which there are\n\na. $3$, $45$ and $2025$;\n\nb. $3$, $\\frac{45}{\\sqrt{5}}$ and $2025$?",
    "options": [],
    "answer": "a: no; b: yes",
    "solution": "a.\nLet the common ratio of the progression be $q$. W.l.o.g., assume that $q > 1$. One can also assume that the first term of the progression is $3$. Then $45 = 3 \\cdot q^k$ and $2025 = 3 \\cdot q^l$, where $k$ and $l$ are integers. This implies $q^k = 15$ and $q^l = 675$. Therefore $q^{kl} = 15^l$, as well as $q^{kl} = 675^k$. Since $15 = 3 \\cdot 5$ and $675 = 3^3 \\cdot 5^2$, the equality $15^l = 675^k$ implies $3^l \\cdot 5^l = 3^{3k} \\cdot 5^{2k}$ which simplifies to $5^{l-2k} = 3^{3k-1}$. As $k$ and $l$ are integers, this is possible only if $l-2k = 3k-l = 0$. But then $k = (l-2k) + (3k-l) = 0$ which is obviously false.\n\nb.\nIf $q = \\frac{15}{\\sqrt{3}}$ then the next term after $3$ is $\\frac{45}{\\sqrt{3}}$ and the term after the next term is $\\frac{3 \\cdot 15^3}{5} = 3^4 \\cdot 5^2 = 2025$. Hence there exists a suitable geometric progression.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30095,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of the quadrilateral *ABDE* meet at *C*. The segments *AB* and *CE* are of equal length 8 cm, and the segments *AE* and *CD* are also of equal length. The perimeter of the triangle *CDE* is 35 cm. Given that $\\angle BAC = \\angle AEC$, find the perimeter of the pentagon *ABCDE*.",
    "options": [],
    "answer": "54 cm",
    "solution": "Using the condition of the problem, we get\n$$\n\\angle BAE = \\angle BAC + \\angle CAE = \\angle AEC + \\angle CAE\n$$\n(Fig. 22). From the triangle $ACE$ we get $\\angle AEC + \\angle CAE = \\angle DCE$. Thus $\\angle BAE = \\angle DCE$. At the same time, $AE = CD$ and $AB = CE$. Consequently, the triangles $AEB$ and $CDE$ are equal, hence the perimeter of the triangle $AEB$ is $35$ cm.\n\n![](attached_image_1.png)\nFig. 22\n\nNow we get\n$$\n\\begin{aligned}\nEA+AB+BC+CD+DE &= (EA+AB+BE-CE) + (CD+DE+EC-CE) \\\\\n&= (EA+AB+BE) + (CD+DE+EC) - 2CE.\n\\end{aligned}\n$$\nSince $EA + AB + BE = CD + DE + EC = 35 \\text{ cm}$ and $CE = 8 \\text{ cm}$, the perimeter of the pentagon $ABCDE$ is $2 \\cdot 35 \\text{ cm} - 2 \\cdot 8 \\text{ cm} = 54 \\text{ cm}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30096,
    "subject": "Mathematics (Multi-modal)",
    "question": "In how many ways can one choose 5 numbers from the list\n$\\frac{1}{8}, \\frac{1}{7}, \\frac{1}{6}, \\frac{1}{5}, \\frac{1}{4}, \\frac{1}{3}, \\frac{1}{2}, 1, 2, 3, 4, 5, 6, 7, 8$\nso that the product of the chosen numbers is 1?\n\n*Remark:* Two choices are considered different if one choice contains a number that the other choice does not. The order of numbers is not important.",
    "options": [],
    "answer": "41",
    "solution": "Let's first count the choices that include the number $1$. If some three of the remaining numbers were $2$, $3$, and $4$, then the conditions of the problem would be violated, because the denominators of all the fractions in the list are smaller than $2 \\cdot 3 \\cdot 4 = 24$, and the product of the selected 5 numbers could not be $1$. If instead of $2$, $3$, and $4$ some other 3 integers are chosen, their product will be even larger, so the conditions of the problem cannot be met. The situation is similar if 3 fractions are chosen next to the number $1$. Thus, in addition to the number $1$, the choice must include 2 integers and 2 fractions, with the product of the integers equal to the product of the denominators of the fractions. There are exactly 21 ways to choose 2 integers in addition to the number $1$, and the resulting products are in the following table:\n\n|   | 3 | 4 | 5 | 6 | 7 | 8 |\n|---|---|---|---|---|---|---|\n| 2 | 6 | 8 | 10 | 12 | 14 | 16 |\n| 3 |   | 12 | 15 | 18 | 21 | 24 |\n| 4 |   |    | 20 | 24 | 28 | 32 |\n| 5 |   |    |    | 30 | 35 | 40 |\n| 6 |   |    |    |    | 42 | 48 |\n| 7 |   |    |    |    |    | 56 |\n\nWe see that the numbers $12$ and $24$ appear in the table 2 times, the remaining numbers are unique. Thus, we get 21 options where the selected integers are the same as the denominators of the selected fractions, and in addition 4 options where the selected integers and the denominators of the selected fractions are not the same, but their product is either $12$ or $24$. In total, there are 25 options with the number $1$.\n\nNow we count the options that do not have the number $1$. We have to choose either 3 integers and 2 fractions or 3 fractions and 2 integers. Since, due to symmetry, there are the same number of options of both types, we count the options involving 3 integers and 2 fractions. If $2$ were not included in the selection, the product of the selected integers should be at least $3 \\cdot 4 \\cdot 5 = 60$, but according to the multiplication table, the product of the denominators of any 2 fractions is less than $60$, which is why the condition of the problem cannot be met. Therefore, $2$ must be included in the selection. Similarly, we see that either $3$ or $4$ must also be included in the selection. We get the possibilities\n$$\n2 \\cdot 3 \\cdot 4 = 24, \\quad 2 \\cdot 3 \\cdot 5 = 30, \\quad 2 \\cdot 3 \\cdot 6 = 36, \\quad 2 \\cdot 3 \\cdot 7 = 42, \\quad 2 \\cdot 3 \\cdot 8 = 48, \\\\\n2 \\cdot 4 \\cdot 5 = 40, \\quad 2 \\cdot 4 \\cdot 6 = 48, \\quad 2 \\cdot 4 \\cdot 7 = 56, \\quad 2 \\cdot 4 \\cdot 8 = 64.\n$$\nOf these 9 products, only $36$ and $64$ do not appear in the multiplication table above, the product $24$ appears there 2 times and the remaining 6 products 1 time. So we get $2 \\cdot 0 + 1 \\cdot 2 + 6 \\cdot 1 = 8$ possibilities. There are the same number of possibilities with 3 fractions and 2 integers. So there are $2 \\cdot 8 = 16$ possibilities without the number $1$.\nHence the total number of possibilities is $25 + 16 = 41$.\nThe list of numbers contains 7 pairs of reciprocals in the form $(\\frac{1}{a}, a)$. At most 2 such pairs can occur among the 5 numbers chosen.\nIf there are two pairs of reciprocals in the choice, then the last number must be $1$. To choose 2 pairs from the 7 existing pairs of reciprocals, there are $\\frac{7 \\cdot 6}{2} = 21$ possibilities.\nIf there is one pair of reciprocals in the choice, then the product of the remaining 3 numbers must be $1$. If the integers $a$ and $b$ are among these 3 numbers, then the last number must be $\\frac{1}{ab}$, so the integer $ab$ is also in the original list. Similarly if the fractions $\\frac{1}{a}$ and $\\frac{1}{b}$ are among these 3 numbers, then $a$, $b$ and $ab$ must be in the original list. The only such numbers are $2$, $3$, $6$ and $2$, $4$, $8$, each of which gives 2 choices for the 3 numbers from the original list. Then there are $7 - 3 = 4$ options for the pair of reciprocal numbers. So in this case there are $2 \\cdot 2 \\cdot 4 = 16$ possibilities.\nIf there are no pairs of reciprocals in the selection, then each integer can appear in the selection at most once, either by itself or in the denominator of the reciprocal. The prime numbers $5$ and $7$ cannot appear in either role. The prime number $3$ only appears as a factor in the numbers $3$ and $6$; if both of them were omitted, there would not be enough numbers left in the selection, so they must both be in the selection, either as $\\frac{1}{3}$ and $\\frac{1}{6}$ or as $\\frac{2}{9}$ and $\\frac{2}{3}$. The product of these numbers is $\\frac{1}{2}$ or $2$, respectively. The remaining numbers are $1$, $2$, $4$, and $8$, from which 3 must be chosen. If one does not choose $1$, then it is not possible to get the number $2$ or $\\frac{1}{2}$ by using each remaining number or its reciprocal as a factor only once. Thus, the number $1$ must be in the selection. From the remaining numbers and their reciprocals, the number $2$ or $\\frac{1}{2}$ can be composed in 4 ways: $\\frac{1}{2} \\cdot 4$, $\\frac{1}{4} \\cdot 8$, $2 \\cdot \\frac{1}{4}$, and $4 \\cdot \\frac{1}{8}$. So the total number of choices is $21 + 16 + 4 = 41$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Inside a regular hexagon $ABCDEF$, equal rectangles $ABZY$, $CDXZ$, and $EFYX$ are drawn. How much of the area of the hexagon $ABCDEF$ do these rectangles cover?\n![](attached_image_1.png)",
    "options": [],
    "answer": "2/3",
    "solution": "Each interior angle of a regular hexagon has a size of $120^\\circ$. Thus, a regular hexagon can be divided into $6$ equilateral triangles with side lengths equal to the hexagon itself (Fig. 23). Denoting the area of such a triangle as $S$, the area of the hexagon $ABCDEF$ is therefore $6S$.\n\nThe opposite sides of a rectangle are of equal length. Hence $XY = EF$, $YZ = AB$ and $XZ = CD$, so $XYZ$ is also an equilateral triangle with the same side length. Since the rectangles $ABZY$, $CDXZ$ and $EFYX$ are equal, the triangles $AYF$, $BZC$ and $DXE$ are isosceles. Their bases are the sides of the hexagon, the base angle is $120^\\circ - 90^\\circ = 30^\\circ$ and the vertex angle is $180^\\circ - 2 \\cdot 30^\\circ = 120^\\circ$. Thus, it is possible to form one equilateral triangle from these three triangles, whose side length is equal to the side length of the hexagon (Fig. 24).\n\nIn total, the area not covered by the rectangles is $2S$. Therefore, the total area of the rectangles is $6S - 2S = 4S$, which is $\\frac{2}{3}$ of the area of the hexagon.\n\n![](attached_image_2.png)\nFig. 23\n![](attached_image_3.png)\nFig. 24",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers whose last digit is not zero and deleting the first digit of which gives a number exactly 25 times smaller.",
    "options": [],
    "answer": "625, 3125, 9375",
    "solution": "Let $n$ be the desired number, and let $m$ be the number obtained by deleting the first digit. Since the last digit of the product $m \\cdot 25$ is not 0, it can only be 5. Since $5 \\cdot 25$ has three digits, it cannot be the number $n$ itself,\nso the number $m$ must have more digits. Consider a digit $d$ in both $n$ and $m$. Then in the product $d$ is obtained from multiplying all lower-order digits by 25 and adding the last digit of the number $d \\cdot 5$ to the corresponding digit in the product. If $d$ is even then the corresponding digit in the product of lower order digits by 25 is $d$, i.e. an even number, while in the case of odd $d$, it is $d - 5$ or $d + 5$, i.e. again an even number.\nSince the tens digit of $5 \\cdot 25$ is 2, the tens digit of $n$ and $m$ must be either 2 or 7. If it is 2 and $m$ does not have more digits, then $n = 25 \\cdot 25 = 625$ which satisfies the conditions of the problem. If there are more digits in the number $m$, then the hundreds digit of the numbers $m$ and $n$ must be 6 or 1, because the hundreds digit of the intermediate result of multiplying lower-order numbers is 6. Now $n$ cannot be $625 \\cdot 25$ because the product has 5 digits. Since the thousands digit 5 of this product is odd, it is not possible to add more digits to the number $m$. If the hundreds digit is 1 then we get the possibility $m = 125$ and $n = 3125$, which satisfies the conditions of the problem, and since the thousands digit in this number is odd, it is not possible to add more digits to the number $m$ here either.\nIf the tens digit of $m$ and $n$ is 7, then $n$ cannot be $75 \\cdot 25$ because the product has 4 digits. Since the hundreds digit of $75 \\cdot 25$ is 8, the hundreds digit of $m$ can be 8 or 3. Again $n$ cannot be $875 \\cdot 25$ because the product has 5 digits, and since the thousands digit 1 of the product is odd, it is not possible to add more digits to the number $m$. If the hundreds digit of $m$ is 3 we get $m = 375$ and $n = 9375$, which satisfies the conditions of the problem. Since the thousands digit of the product is odd, it is not possible to add more digits to the number $m$ here either.\nThus only 625, 3125 and 9375 satisfy the conditions of the problem.\nLet $n$ be the desired number. Let the first digit of this number be $a$ and the number formed from the remaining digits be $m$, with the number $m$ having $k$ digits. Then $n = 10^k a + m = 25m$, implying $10^k a = 24m$. Thus, the number $10^k a = 2^k 5^k a$ is divisible by the number 24. Consequently, the number $2^k 5^k a$ must be divisible by the numbers $2^3$ and $3$. Since the number $2^k 5^k$ is not divisible by 3, the number $a$ must be divisible by 3, which leaves the possibilities $a = 3, a = 6$ and $a = 9$.\n* If $a = 3$, then $2^k 5^k$ must be divisible by $2^3$. Therefore $k \\ge 3$ and we get $m = \\frac{10^k a}{2^4} = \\frac{3000 \\cdot 10^{k-3}}{2^4} = 125 \\cdot 10^{k-3}$. Since last digit of $m$ is not zero, we have $m = 125$, whence $n = 3125$.\n* If $a = 6$, then $2^k 5^k$ must be divisible by $2^2$. Therefore $k \\ge 2$ and we get $m = \\frac{10^k a}{2^4} = \\frac{600 \\cdot 10^{k-2}}{2^4} = 25 \\cdot 10^{k-2}$. Since the last digit is not zero, we have $m = 25$, whence $n = 625$.\n* If $a = 9$, then $2^k 5^k$ must be divisible by $2^3$. Therefore $k \\ge 3$ and we get $m = \\frac{10^k a}{2^4} = \\frac{9000 \\cdot 10^{k-3}}{2^4} = 375 \\cdot 10^{k-3}$. Since the last digit of $m$ is not zero, we have $m = 375$, whence $n = 9375$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $P$ a point inside it. Let $A'$, $B'$, $C'$ be the reflections of $A$, $B$, $C$ from point $P$, respectively. Find the ratio of the areas of the hexagon $AB'CA'BC'$ and the triangle $ABC$.",
    "options": [],
    "answer": "2",
    "solution": "Denote the area of the figure $K$ by $S_K$. Note (Fig. 25) that\n$$\nS_{AB'CA'BC'} = S_{APB'} + S_{B'PC} + S_{CPA'} + S_{A'PB} + S_{BPC'} + S_{C'PA}.\n$$\nFurthermore, $S_{APB'} = S_{APB}$, because triangles $APB'$ and $APB$ have the same altitude drawn from vertex $A$ and the corresponding bases are equal. Similarly we get $S_{B'PC} = S_{BPC}$, $S_{CPA'} = S_{CPA}$, $S_{A'PB} = S_{APB}$, $S_{BPC'} = S_{BPC}$ and $S_{C'PA} = S_{CPA}$. Hence\n$$\nS_{AB'CA'BC'} = 2S_{APB} + 2S_{BPC} + 2S_{CPA} = 2(S_{APB} + S_{BPC} + S_{CPA}) = 2S_{ABC}.\n$$\n\n![](attached_image_1.png)\nFig. 25",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30100,
    "subject": "Mathematics (Multi-modal)",
    "question": "The price of the old model of smartwatch A differs from the price of the old model of smartwatch B by $p\\%$ ($0 < p < 100$). The new model of watch A is $q\\%$ more expensive than the old model of watch A, and the new model of watch B is $q\\%$ cheaper than the old model of watch B ($0 < q < 100$). The price of the new model of watch A differs from the price of the new model of watch B by $p\\%$.\n\na. Is the new model of watch B more expensive, cheaper, or just as expensive as the old model of watch A?\n\nb. Is the new model of watch A more expensive, cheaper, or just as expensive as the old model of watch B?",
    "options": [],
    "answer": "a. Just as expensive.\nb. Cheaper.",
    "solution": "Since the new model of watch A is more expensive than the old model of watch A and the new model of watch B is cheaper than the old model of watch B, but the difference in price is still $p\\%$, the old model of watch A must be $p\\%$ cheaper than the old model of watch B, and the new model of watch A must be $p\\%$ more expensive than the new model of watch B.\n\nLet the price of the old model of watch B be $x$. Then the price of the new model of watch B is $(1 - \\frac{q}{100})x$. The old model of watch A then costs $(1 - \\frac{p}{100})x$, while the new model of watch A costs $(1 + \\frac{q}{100})(1 - \\frac{p}{100})x$. Since from the previous paragraph the price of the new model of watch A is $(1 + \\frac{p}{100})(1 - \\frac{q}{100})x$, we have\n$$\n\\left(1 + \\frac{q}{100}\\right) \\left(1 - \\frac{p}{100}\\right) x = \\left(1 + \\frac{p}{100}\\right) \\left(1 - \\frac{q}{100}\\right) x.\n$$\nAfter cancelling $x$ from both sides, opening the parentheses and simplifying, we get $p = q$.\n\na. Since $p = q$, the prices of the new model of watch B and of the old model of watch A are both $(1 - \\frac{p}{100})x$.\n\nb. Since $p = q$, the price of the new model of watch A is $(1 - \\frac{p^2}{10000})x$ which is less than the price of the old model of watch B.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30101,
    "subject": "Mathematics (Multi-modal)",
    "question": "The midpoint of the hypotenuse $AB$ of a right-angled triangle $ABC$ is $M$. A point $D$ lies on the side $BC$ such that the circumcircle of the triangle $ACD$ intersects the line $DM$ at some point $K$ between the points $D$ and $M$. Let $L$ be the reflection of the point $K$ from the point $M$. The circumcircles of the triangles $ACD$ and $LBC$ intersect at the point $N$, $N \\neq C$. Find the size of the angle $KNL$.",
    "options": [],
    "answer": "90°",
    "solution": "The right angle $DCA$ subtends the chord $AD$ of the circumcircle of triangle $ACD$ (Fig. 26), therefore $AD$ is a diameter of this circle. Since point $K$ lies on the same circle, $\\angle AKD = 90^\\circ$. From supplementary angles we get $\\angle MKA = 180^\\circ - \\angle AKD = 90^\\circ$.\n\nFrom the conditions of the problem $AM = MB$ and $KM = ML$ and from the vertex angles we get $\\angle AMK = \\angle BML$. Thus, the triangles $AMK$ and $BML$ are equal. Consequently $\\angle MLB = \\angle MKA = 90^\\circ$.\n\nNext, we show that $\\angle NLK = \\angle NKA$. It suffices to show the equality $\\angle NLB = \\angle NKD$ since $\\angle NLK = \\angle NLB - \\angle KLB = \\angle NLB - 90^\\circ$ and\n\n![](attached_image_1.png)\n\n$\\angle NKA = \\angle NKD - \\angle AKD = \\angle NKD - 90^\\circ$. Furthermore, since $N, L, B$ and $C$ lie on the same circle in this order, we have $\\angle NLB = 180^\\circ - \\angle BCN$, while the points $N, K, D$ and $C$ lying on one circle in this order implies $\\angle NKD = 180^\\circ - \\angle DCN = 180^\\circ - \\angle BCN$. So $\\angle NLK = \\angle NKA$ as desired. Hence $\\angle NLK + \\angle LKN = \\angle NLK + \\angle LKA - \\angle NKA = \\angle LKA = 90^\\circ$, which implies $\\angle KNL = 180^\\circ - 90^\\circ = 90^\\circ$.\nAs in Solution 1 we show that $AD$ is a diameter of the circumcircle of triangle $ACD$ and $\\angle MLB = 90^\\circ$. Since point $N$ lies on the circumcircle of triangle $ACD$ we have $\\angle AND = 90^\\circ$.\n\nLet $T$ be the intersection of the lines $LD$ and $AC$ (Fig. 27). From supplementary angles we get $\\angle TCB = 180^\\circ - \\angle BCA = 90^\\circ$. Since $\\angle TLB = 90^\\circ$, points $C$ and $L$ lie on the circle with the diameter $TB$, whence $T$ lies on the circumcircle of triangle $LBC$.\n\nWe show that $\\angle DNL = \\angle ANK$. Since the points $C, N, L, T$ lie on the same circle in this order, we have $\\angle CNL + \\angle LTC = 180^\\circ$. Hence\n$$\n\\angle DNL = \\angle CNL - \\angle CND = 180^\\circ - \\angle LTC - \\angle CND \\\\ = 180^\\circ - \\angle DTA - \\angle TAD.\n$$\nFrom triangle $TAD$ we get $180^\\circ - \\angle DTA - \\angle TAD = \\angle ADT$ and from supplementary angles $\\angle ADT = 180^\\circ - \\angle KDA$. Since $A, N, K, D$ lie on the same circle in this order, we get $180^\\circ - \\angle KDA = \\angle ANK$. So indeed $\\angle DNL = \\angle ANK$.\n\nFinally notice that $\\angle ANK = \\angle AND + \\angle DNK = 90^\\circ + \\angle DNK$, hence $\\angle DNL = 90^\\circ + \\angle DNK$. Consequently $\\angle KNL = \\angle DNL - \\angle DNK = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a positive integer $n$ interesting if both the sum of digits of $n$ and the sum of digits of $n+1$ are perfect squares, whereas $n$ and $n+1$ have the same number of digits. Find all positive integers $k$ for which there exists an interesting $k$-digit number.",
    "options": [],
    "answer": "all integers k ≥ 10",
    "solution": "For any positive integer $a$, let $s(a)$ denote the sum of digits of $a$. Clearly $s(n + 1) = s(n) + 1$ unless the last digit of $n + 1$ is zero. The only two consecutive integers that are both perfect squares are $0$ and $1$, but $s(n) = 0$ is impossible for a positive $n$. The contradiction shows that the last digit of $n + 1$ must be $0$ and the last digit of $n$ must be $9$.\n\nLet $n$ be an interesting positive integer that ends with exactly $m$ digits $9$. Clearly $s(n) - s(n + 1) = 9m - 1$. Note that a perfect square is congruent to $0$, $1$, $4$ or $7$ modulo $9$. As $s(n) - s(n + 1) = 9m - 1 \\equiv 8 \\pmod 9$, we must have $s(n) \\equiv 0 \\pmod 9$ and $s(n + 1) \\equiv 1 \\pmod 9$. But $s(n + 1) = 1$ is impossible, because this would require $n + 1$ to be a power of $10$, in the case of which $n$ and $n + 1$ would not consist of the same number of digits. Thus $s(n + 1) \\ge 8^2 = 64$, as $8^2$ is the next smallest perfect square congruent to $1$ modulo $9$. But as $s(n) > s(n + 1)$, we have $s(n) \\ge 9^2 = 81$. This implies that $n$ must contain at least $9$ digits. Containing exactly $9$ digits would be possible only if $n$ consisted entirely of nines, but in such case $n + 1$ would contain one more digit. Hence $n$ must contain at least $10$ digits.\n\nConsider now the $10$-digit number $n = 788888899$. In this case $m = 2$, $s(n) = 81 = 9^2$ and $s(n + 1) = 81 - 9 \\cdot 2 + 1 = 64 = 8^2$, thus $n$ is interesting. Inserting zeros between $7$ and $8$ does not influence $s(n)$ and $s(n + 1)$. So we can obtain an $m$-digit interesting number for every $m \\ge 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30103,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $m$ is called *usual* if the square of every prime divisor of $m$ is less than $m$.\n\na. Prove that there are infinitely many positive integers $n$ such that both $n$ and $n+1$ are usual.\n\nb. Is there a positive integer $n$ such that $n$, $n+1$ and $n+2$ are all usual?",
    "options": [],
    "answer": "a) Infinitely many such n exist. For example, if a and a+1 are both composite, then n = a^2 − 1 works. b) Yes. For instance, n = 440 gives 440, 441, 442 all usual (also n = 9800 gives 9800, 9801, 9802).",
    "solution": "Let $a$ be a composite number such that $a+1$ is composite, too. Then $a^2$ is usual since each of its prime divisors is a prime divisor of $a$ and therefore less than $a$. We show that $a^2-1$ is usual, too. To this end, observe that $a^2-1 = (a-1)(a+1)$. As $a+1$ is composite, all prime divisors of $a+1$ are less than $a+1$. But as the difference of $a+1$ and its arbitrary prime divisor is also divisible by this divisor, the prime divisors of $a+1$ cannot be larger than $a-1$. Thus no prime divisor of $a^2-1$ is larger than $a-1$, meaning that the squares of prime divisors of $a^2-1$ do not exceed $(a-1)^2$. As $(a^2-1) - (a-1)^2 = (a-1)((a+1) - (a-1)) = 2(a-1) > 0$, squares of prime divisors of $a^2-1$ are less than $a^2-1$, implying that $a^2-1$ is usual. Consequently, $n = a^2-1$ is a suitable example.\nAs there exist arbitrarily long sequences of consecutive composite numbers, one can choose two consecutive composite numbers in infinitely many ways. Hence there exist infinitely many integers with the desired property. It remains to notice that $a = 21$ implies $a^2 + 1 = 442 = 2 \\cdot 13 \\cdot 17$, whereas $2^2 < 13^2 < 17^2 = 289 < 442$. Hence (440, 441, 442) is a triple of consecutive positive integers, all of which are usual.\nConsider the so-called Pell's equation $x^2 - 2y^2 = 1$. It has the trivial solution $(x_0, y_0) = (1, 0)$, and whenever $(x_{k-1}, y_{k-1})$ is a solution, $(x_k, y_k) = (3x_{k-1} + 4y_{k-1}, 2x_{k-1} + 3y_{k-1})$ is a solution, too.\nAs $x_0 < x_1 < x_2 < \\dots$ and $y_0 < y_1 < y_2 < \\dots$, all these solutions are distinct. Note that $3 \\mid y_0$ and, whenever $3 \\mid y_{2k}$, also $3 \\mid 3x_{2k} + 4y_{2k} = x_{2k+1}$, and whenever $3 \\mid x_{2k+1}$, also $3 \\mid 2x_{2k+1} + 3y_{2k+1} = y_{2k+2}$. Hence $3 \\mid x_{2k+1}$ for every natural number $k$.\nLet $k$ be any positive integer. We show that $n = 2y_{2k+1}^2$ meets the conditions. To this end, let $p$ be any prime divisor of $n$. If $p > 2$ then $p \\mid y_{2k+1}^2$, implying $p \\mid y_{2k+1}$ which gives $p \\le y_{2k+1}$. But if $p = 2$ then $p = y_1 < y_{2k+1}$ still. Thus $p^2 \\le y_{2k+1}^2 < 2y_{2k+1}^2 = n$ in any case. Let $p$ now be a prime divisor of $n+1 = x_{2k+1}^2$. Then also $p \\mid x_{2k+1}$. We previously showed that $3 \\mid x_{2k+1}$. As $3 = x_1 < x_{2k+1}$, the number $x_{2k+1}$ is composite, implying that $p < x_{2k+1}$. Thus $p^2 < x_{2k+1}^2 = n+1$. Consequently, $n$ and $n+1$ are both usual. This solves part (a) of the problem since the parameter $k$ can be chosen in infinitely many ways which all produce different solutions.\nTo solve part (b), it suffices to note that if $k = 1$ then $n = 2 \\cdot 70^2 = 9800$, $n+1 = 99^2 = 9801$ and $n+2 = 9802 = 2 \\cdot 13^2 \\cdot 29$. Moreover, observe that $2^2 < 13^2 < 29^2 = 841 < 9802$. Hence 9800, 9801 and 9802 are three consecutive positive integers which are all usual.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ that satisfy\n$$\nf(y^2f(x) - f(xy)) = f(y^2) + 2(x^2 - f(x))(f(y) - 1) + 1\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x^2 + 1",
    "solution": "Substituting $y = 1$ into the given equation, we obtain\n$$\nf(0) = f(1) + 2(x^2 - f(x))(f(1) - 1) + 1\n$$\nwhich is equivalent to\n$$\n2f(x)(f(1) - 1) = f(1) - f(0) + 2x^2(f(1) - 1). \\quad (3)\n$$\n\nIf $f(1) = 1$ then (3) implies $0 = 1 - f(0) + 1$, or equivalently, $f(0) = 2$.\nSubstituting now $y = 0$ into the original equation and applying $f(0) = 2$ gives us\n$$\nf(-2) = 2 + 2(x^2 - f(x)) + 1,\n$$\nor equivalently, $f(x) = x^2 + c$ where $c = \\frac{3-f(-2)}{2}$. In the other case, $f(x)$ is expressed in the same form with $c = \\frac{f(1)-f(0)+1}{2(f(1)-1)}$.\nWe show that the function $f(x) = x^2 + c$ satisfies the original equation if and only if $c = 1$. Substituting $f(x) = x^2 + c$ into the original equation and simplifying leads to\n$$\n(c^2 - 1) y^4 + 2c(1-c)y^2 + (3c^2 - 2c - 1) = 0.\n$$\nThis equality must hold for every real number $y$. For that, all coefficients in the left hand side must be zeros. From the leading term, we get $c^2 - 1 = 0$, implying $c = 1$ or $c = -1$. From the quadratic term, we get $2c(1-c) = 0$, implying $c = 0$ or $c = 1$. Altogether, only $c = 1$ works. It makes the constant term also zero. Hence $f(x) = x^2 + 1$ is the only function that satisfies the given equation.\nSubstituting $y = 0$ into the original equation gives\n$$\nf(-f(0)) = f(0) + 2(x^2 - f(x))(f(0) - 1) + 1,\n$$\nor equivalently,\n$$\n2f(x)(f(0)-1) = f(0) - f(-f(0)) + 1 + 2x^2(f(0)-1). \\quad (4)\n$$\nThis implies that either $f(0) = 1$ or $f(x) = x^2 + \\frac{f(0)-f(-f(0))+1}{2(f(0)-1)}$ for any real number $x$.\n\nConsider the case $f(0) = 1$. Substituting $x = 0$ into the original equation leads to\n$$\nf(y^2 - 1) = f(y^2) - 2(f(y) - 1) + 1,\n$$\nwhich must hold for any real number $y$. Substituting $-y$ for $y$ gives\n$$\nf(y^2 - 1) = f(y^2) - 2(f(-y) - 1) + 1,\n$$\nwhich must also hold for any real number $y$. Hence $f(y) = f(-y)$ for any real number $y$, i.e., $f$ is an even function. Eliminating the two terms with $f(0) - 1$ in (4) gives $0 = 1 - f(-1) + 1$, or equivalently, $f(-1) = 2$. Substituting now $y = -1$ into the original equation gives\n$$\nf(f(x) - f(-x)) = f(1) + 2(x^2 - f(x))(f(-1) - 1) + 1. \\quad (5)\n$$\nAs $f$ is an even function and $f(1) = f(-1) = 2$, the equation (5) simplifies to $1 = 2 + 2(x^2 - f(x)) + 1$ which is equivalent to $f(x) = x^2 + 1$.\n\nHence $f(x) = x^2 + c$ for any real number $x$, where $c$ is some constant. We proceed like in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre of an acute triangle $ABC$. Points $D$ and $E$ are chosen on the side $BC$ such that $AD$ is an altitude of the triangle $ABC$ and $AE$ bisects the angle $CAD$. Bisectors of the triangle $AOB$ meet at point $J$. Prove that the triangle $JBE$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1:\n\nLet $\\angle BCA = \\gamma$ and let $M$ be the point of intersection of lines $OJ$ and $AB$ (Figures 10 and 11).\nAs $OA = OB$ and $OM$ bisects the angle $AOB$, we have $OM \\perp AB$. We also have $\\angle AOM = \\angle AOB = \\angle ACB = \\gamma$, because $O$ lies inside the triangle $ABC$. As $\\angle AMO = 90^\\circ = \\angle ADC$, triangles $AMO$ and $ADC$ are similar. Hence also $\\angle MAO = \\angle DAC = 90^\\circ - \\gamma$, implying\n$$\n\\angle MAJ = \\angle JAO = \\angle DAE = \\angle EAC = \\frac{1}{2} (90^\\circ - \\gamma). \\quad (6)\n$$\nHence triangles $AJO$ and $AEC$ are similar, too. This implies the similarity of triangles $AJE$ and $AOC$ (spiral similarity). As $OA = OC$, the latter similarity implies $JA = JE$. By symmetry in the isosceles triangle $AOB$, we obtain $JA = JB$. Consequently, $JB = JE$. Hence the triangle $JBE$ is isosceles.\n\n![](attached_image_1.png)\nFig. 10\n![](attached_image_2.png)\nFig. 11\nSolution 2:\n\nLet $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$ and $\\angle BCA = \\gamma$. We start by proving the equalities (6) as in Solution 1. Next, let lines $OJ$ and $BC$ meet in $F$, whereas let lines $AJ$ and $BC$ meet in $X$ (Figures 12 and 13 depict situations that differ by the order of points $E$ and $F$). We show that points $A, J, E, F$ are concyclic. From equalities (6) we get\n$$\n\\begin{aligned} \n\\angle JAE &= \\angle BAC - \\angle BAJ - \\angle EAC \\\\ \n&= \\alpha - \\frac{1}{2}(90^\\circ - \\gamma) - \\frac{1}{2}(90^\\circ - \\gamma) \\\\ \n&= \\alpha - (90^\\circ - \\gamma) = \\alpha + \\gamma - 90^\\circ = 180^\\circ - \\beta - 90^\\circ = 90^\\circ - \\beta. \n\\end{aligned}\n$$\nAs $OA = OB$, the line $OJ$ is the perpendicular bisector of the line segment $AB$. Thus\n$$\n\\angle XFJ = \\angle BFJ = 90^\\circ - \\angle ABF = 90^\\circ - \\beta.\n$$\n\n![](attached_image_3.png)\nFig. 12\n![](attached_image_4.png)\nFig. 13\nAltogether, we have $\\angle XAE = \\angle JAE = 90^\\circ - \\beta = \\angle XFJ$, implying that $A, J, E, F$ are concyclic.\n\nTherefore $\\angle JEA = \\angle JFA = \\angle JFB = \\angle JFX = \\angle XAE = \\angle JAE$, implying $JA = JE$. But $JA = JB$ as $J$ lies on the perpendicular bisector of the line segment $AB$. Consequently, $JB = JE$. Hence the triangle $JBE$ is isosceles.\nSolution 3:\n\nWe use the known fact that lines drawn from a vertex of a triangle to its orthocentre and its circumcentre are isogonals which implies that the angles between these lines and the respective sides are equal.\nAs $AD$ and $AO$ are the altitudes and the circumradius of the triangle $ABC$, both drawn from vertex $A$, we have $\\angle BAO = \\angle CAD$. But $AE$ bisects the angle $CAD$ and $J$ is the point of intersection of angle bisectors of the triangle $ABO$, hence $\\angle BAJ = \\angle CAD = \\angle EAD$. This shows that rays $AJ$ and $AD$ are isogonals in the triangle $ABE$. As $AD$ is also an altitude of the triangle $ABE$, the line $AJ$ must pass through the circumcentre of the triangle $ABE$. But the circumcentre of the triangle $ABE$ must also lie on the perpendicular bisector of the side $AB$. As $OA = OB$, the bisector of the angle $AOB$ coincides with the perpendicular bisector of the side $AB$. Hence the circumcentre of the triangle $ABE$ is the point $J$ of intersection of lines $AJ$ and $OJ$. Consequently, $JB = JE$, implying that the triangle $JBE$ is isosceles.\nSolution 4:\n\nAs $OA = OB$, the bisector of the angle $AOB$ coincides with the perpendicular bisector of the side $AB$. Hence the point $J$ lies on the perpendicular bisector of the side $AB$, implying that points $A$ and $B$ lie on some circle $\\omega$ with centre $J$.\nWe show that $\\angle AJB = 2\\angle AEB$. Indeed,\n$$\n\\begin{aligned}\n\\angle AEB &= \\angle AED = 90^\\circ - \\angle EAD \\\\\n&= 90^\\circ - \\frac{\\angle CAD}{2} = 90^\\circ - \\frac{90^\\circ - \\angle ACD}{2} = 90^\\circ + \\frac{\\angle ACD}{2}.\n\\end{aligned}\n$$\nAs $JA = JB$ implies $\\angle JAB = \\angle JBA$, we obtain\n$$\n\\begin{aligned}\n\\angle AJB &= 180^\\circ - 2\\angle JAB = 180^\\circ - \\angle OAB \\\\\n&= 180^\\circ - \\frac{180^\\circ - \\angle AOB}{2} = 90^\\circ + \\frac{\\angle AOB}{2} = 90^\\circ + \\angle ACB = 2\\angle AEB.\n\\end{aligned}\n$$\n\nHence $\\omega$ is the circumcircle of the triangle $ABE$. Consequently, $JB = JE$, implying that the triangle $JBE$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30106,
    "subject": "Mathematics (Multi-modal)",
    "question": "A gardener Andres wants to plant one currant bush to each cell of his garden of shape $24 \\times 2024$. He wants to plant as many blackcurrant bushes as possible under the following conditions: There must be at least one redcurrant and at least one whitecurrant bush, and for any cell with a blackcurrant bush, cells that have a common side with it must contain equally many redcurrant and whitecurrant bushes (maybe also 0 of both). Find the largest number of blackcurrant bushes Andres can plant.",
    "options": [],
    "answer": "46632",
    "solution": "Firstly, we show that $46632$ blackcurrant bushes is possible. Let's alternate redcurrant and whitecurrant bushes on the falling diagonal starting from the top left corner; when we reach the bottom edge of the garden, skip one column and proceed similarly along a rising diagonal starting from the next column, and so on. Into all remaining cells, we put blackcurrant bushes (Figure 14 depicts the beginning of the garden).\nThis way, blocks consisting of $24$ columns, each containing $23$ blackcurrant bushes and $1$ non-blackcurrant bush, alternate with singleton columns that contain blackcurrants only. As $2024 = 80 \\cdot 25 + 24$, the last diagonal containing non-blackcurrant bushes ends in the bottom right corner of the garden (no column of blackcurrants follows). Each cell next to a diagonal of non-blackcurrant bushes (including the utmost cell of each column of blackcurrants) has exactly one red and one white neighbour. Other blackcurrant cells have only black neighbours. There are $2024 \\cdot 24 - 81 \\cdot 24 = 46632$ blackcurrant bushes in total.\n\n![](attached_image_1.png)\nFig. 14\n![](attached_image_2.png)\nFig. 15\n\nSecondly, we show that there cannot be more blackcurrant bushes. To this end, consider an arbitrary situation that satisfies the conditions. Let the garden have $24$ rows and $2024$ columns.\nIf there is a column containing blackcurrants solely, there must definitely exist two consecutive columns, one of which contains non-blackcurrant bushes and the other does not. Then in the latter column, the blackcurrant bush next to a non-blackcurrant bush of the former column must have another non-black neighbour on the other side (Fig. 15). Thus there cannot be two consecutive columns with blackcurrant bushes only, nor can the first or the last column contain blackcurrants only.\nConsider now a block of $m$ consecutive columns, each containing at least one non-black cell. Assume that this block cannot be extended to obtain a bigger block with the same property. We show that if $m < 24$ then this block must contain at least $m + 1$ non-black cells in total. Consider two cases:\n* If there is a row that contains no non-black cells within the block then there must be two consecutive rows, one of which contains non-black cells within the block and the other does not. Like above, we see that in the latter row, the black cell next to a non-black cell of the former row must have another non-black neighbour on the other side. Thus the column of this cell must contain two non-black cells. As every column of the block contains at least one non-black cell, the block must contain at least $m + 1$ non-black cells in total.\n* If every row contains non-black cells within the block then the block contains at least $24$ non-black cells in total. As $m < 24$, this implies that the block contains at least $m + 1$ non-black cells.\nLet's add one more column of blackcurrants to the beginning of the garden; by the above, this does not give rise to two consecutive columns of blackcurrants. (We do not apply the condition of the problem to the extra column, it is introduced for convenience only.) Divide the garden into blocks, each of which begins with a column of blackcurrants and ends with the last column before the next column of blackcurrants (or with the last column of the garden if no columns of blackcurrants follow). By the definition of blocks, each block of $m$ columns contains at least $m - 1$ non-black cells; but if $m < 25$ then the block must contain at least $m$ non-black cells. Since $\\frac{2025}{25} = 81$, the number of blocks with $1$ less non-black cells than columns can be at most $81$. Thus the garden contains at least $2025 - 81$ non-black cells and at most $2024 \\cdot 24 - (2025 - 81) = 46632$ black cells.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is there a positive integer $n$ such that $88$ divides $2^n + n^3$?",
    "options": [],
    "answer": "n = 10",
    "solution": "Taking $n = 10$ gives $2^n + n^3 = 1024 + 1000 = 2024 = 88 \\cdot 23$, so $88$ divides $2^{10} + 10^3$.\n\nWe consider divisibility by $11$ and by $8$ separately. By Fermat's little theorem, we have $2^{10} \\equiv 1 \\pmod{11}$, whereas $10^3 \\equiv (-1)^3 = -1 \\pmod{11}$. In summary $2^{10} + 10^3 \\equiv 1 - 1 = 0 \\pmod{11}$, meaning that $11$ divides $2^{10} + 10^3$. As $8$ divides both $2^{10}$ and $10^3$, it divides the sum $2^{10} + 10^3$. Thus $88$ divides $2^{10} + 10^3$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. In an $n \\times n$ grid there are three invisible monsters: one in the upper right corner square and one in each of its neighboring squares. In a $2 \\times 2$ area in the opposite corner of the grid some frogs are placed. A square may contain more than one frog.\n\nThe frogs and the monsters take turns making the following moves. On the frogs' turn, each frog makes a knight move, jumping either two squares up and one to the right or one square up and two to the right. On the monsters' turn, they may make up to a total of 3 knight moves, jumping either two squares down and one to the left or one square down and two to the left. The frogs start. If a frog and a monster ever land on the same square, the monster will eat the frog.\n\nFind the least number of frogs needed to ensure that at least one frog could make it to a spot where both possible jumps would take it off the grid.",
    "options": [],
    "answer": "1",
    "solution": "We color the square in 3 colors by descending diagonals (in Fig. 16 the colors are denoted by A, B, C and X, Y, Z, which are still the same colors, but their order depends on the value of $n$). Notice that both frogs and monsters can only jump on squares of the same color as the one they started from. As the monsters cover only two colors of squares, but a $2 \\times 2$ area contains squares of all three colors, it is sufficient to place one frog on a square of the color not covered by the monsters. In this case, no matter the moves, the frog will be safe.\n\n![](attached_image_1.png)\nFig. 16",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of positive integers, such that\n$$3 \\cdot x! + 4 \\cdot y! = 5 \\cdot z!.$$",
    "options": [],
    "answer": "(2, 1, 2) and (2, 3, 3)",
    "solution": "**Solution 1:** If $x \\le y$, then $4 \\cdot y! < 3 \\cdot x! + 4 \\cdot y! \\le 7 \\cdot y!$, so $\\frac{4}{5} \\cdot y! < z! < \\frac{7}{5} \\cdot y!$. Two distinct factorials differ by a factor of at least 2, so $z < y$ would yield $z! \\le \\frac{1}{2} \\cdot y! < \\frac{4}{5} \\cdot y! < z!$, contradiction. Analogously $z > y$ would yield $z! \\ge 2 \\cdot y! > \\frac{7}{5} \\cdot y! > z!$, contradiction. Therefore the only option is $z = y$. Then the given equation simplifies to $3 \\cdot x! = y!$. Here $y > x$ and the product of $x + 1, x + 2, \\dots, y$ must be 3. This can only happen when $y = x + 1 = 3$, giving the solution $(x, y, z) = (2, 3, 3)$.\nIf $x > y$, then $3 \\cdot x! < 3 \\cdot x! + 4 \\cdot y! < 7 \\cdot x!$, so $\\frac{3}{5} \\cdot x! < z! < \\frac{7}{5} \\cdot x!$. Analogously to the previous case we get $z = x$. Then the given equation simplifies to $4 \\cdot y! = 2 \\cdot x!$ or $2 \\cdot y! = x!$. Analogously to the previous case, this is only possible when $x = y + 1 = 2$, giving the solution $(x, y, z) = (2, 1, 2)$.\n\n\n**Solution 2:** Notice that $x \\le z$ and $y \\le z$, because if either $x \\ge z + 1$ or $y \\ge z + 1$, then\n$$\n5 \\cdot z! = 3 \\cdot x! + 4 \\cdot y! \\ge 3 \\cdot \\max(x!, y!) \\ge 3 \\cdot (z+1)!,\n$$\nwhere dividing by $z!$ gives $5 \\ge 3(z+1)$, from which $z \\le \\frac{2}{3} < 1$, contradiction.\n\n* If $x = z$, then like in Solution 1, we only get the solution $(2, 1, 2)$.\n* If $y = z$, then like in Solution 1, we only get the solution $(2, 3, 3)$.\n* If $x \\le z - 1$ and $y \\le z - 1$, then\n$$\n5 \\cdot z! = 3 \\cdot x! + 4 \\cdot y! \\le 3 \\cdot (z-1)! + 4 \\cdot (z-1)! = 7 \\cdot (z-1)!,\n$$\nwhere dividing by $(z-1)!$ gives $5z \\le 7$. This leaves only the option $z = 1$, which is impossible by the assumptions $x \\le z - 1$ and $y \\le z - 1$.\n\n\n**Solution 3:** We divide the sides $3x! + 4y! = 5z!$ by the smallest factorial present in the equation. This leaves an equation $3a + 4b = 5c$, where $a, b, c$ are positive integers, of which at least one is equal to 1.\nIf $c = 1$, then $3a + 4b = 5$, giving no solutions. If $a = b = 1$, then $7 = 5c$, also giving no solutions.\nIf $a = 1, b > 1, c > 1$ and $3 + 4b = 5c$, then $y > x$ and $z > x$, meaning that both $b$ and $c$ are divisible by $x + 1$. Then 3 is also divisible by $x + 1$, which gives $x = 2$. Then $z = 3$, or else the right hand side of the equation is divisible by 4, whereas the left hand side isn't. Then also $y = 3$.\nIf $b = 1, a > 1, c > 1$ and $3a + 4 = 5c$, then $x > y$ and $z > y$, meaning that both $a$ and $c$ are divisible by $y + 1$. Then 4 is also divisible by $y + 1$, which gives $y = 1$ or $y = 3$. If $y = 1$, then $z = 2$, or else the right hand side of the equation is divisible by 3, whereas the left hand side isn't. Then also $x = 2$. But if $y = 3$, then the left hand side is never divisible by 5, so there are no solutions.\nThus the only suitable triples are $(2, 1, 2)$ and $(2, 3, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mama snail and her child want to visit a neighbour who lives at distance $75$ cm. Every hour, they have planned to use $45$ minutes to move and $15$ minutes to rest. On the $n$-th hour, they move $\\frac{1}{n^2+1}$ metres forward, but instead of resting, the child pulls them backwards by $\\frac{1}{n+1}$ of this hour's distance. Will they ever reach the neighbour, and if so, when?",
    "options": [],
    "answer": "They will never reach the neighbour.",
    "solution": "Combining both parts of the $n$-th hour, the total distance travelled forward is $\\frac{1}{n^2+1} - \\frac{1}{n+1} \\cdot \\frac{1}{n^2+1} = \\frac{n}{(n+1)(n^2+1)}$ metres. On the first hour, this means $\\frac{1}{4}$ metres. Notice that $\\frac{n}{(n+1)(n^2+1)} < \\frac{n}{(n+1)n} = \\frac{1}{n} - \\frac{1}{n+1}$. Thus by the end of the $m$-th hour, where $m \\ge 2$, they have travelled less than $\\frac{1}{4} + (\\frac{1}{2} - \\frac{1}{3}) + (\\frac{1}{3} - \\frac{1}{4}) + \\dots + (\\frac{1}{m} - \\frac{1}{m+1}) = \\frac{3}{4} - \\frac{1}{m+1}$ metres, which is less than $75$ centimetres. Thus they will never reach the neighbour by the end of the hour.\n\nWe will show that they will also not reach the neighbour between the two parts of an hour. Indeed, $45$ minutes after starting they have gone $\\frac{1}{2}$ metres and $1$ hour and $45$ minutes after starting $\\frac{1}{4} + \\frac{1}{5}$ metres, both less than $75$ centimetres. But if $m \\ge 2$, then $m$ hours and $45$ minutes after starting they have travelled less than $\\frac{3}{4} - \\frac{1}{m+1} + \\frac{1}{(m+1)^2+1}$ metres, which is still less than $75$ centimetres, as $\\frac{1}{(m+1)^2+1} < \\frac{1}{m+1}$.\n\nThus they will never reach the neighbour.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30111,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$, the extension of the altitude $AD$ over $D$ intersects the circumcircle at $E$. The midpoint of $CE$ is $F$. The circumcircles of $ABC$ and $DEF$ intersect at $G \\neq E$. The foot of the altitude drawn from $A$ to $FG$ is $P$. Prove that $DA = DP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that $\\angle BGF = 90^\\circ$ (Fig. 17). For this we notice that $F$ as the midpoint of the hypothenuse in $CDE$ is also its circumcenter, so $FC = FD$. Together with $BEGC$ being cyclic, we get\n$$\n\\angle EGB = \\angle ECB = \\angle FCD = \\angle FDC.\n$$\nThus\n$$\n\\begin{align*}\n\\angle BGF &= \\angle EGF - \\angle EGB \\\\\n&= 180^\\circ - \\angle EDF - \\angle FDC = 180^\\circ - \\angle EDC = 180^\\circ - 90^\\circ = 90^\\circ.\n\\end{align*}\n$$\n\n![](attached_image_1.png)\nFig. 17\n![](attached_image_2.png)\nFig. 18\n\nLet $A'$ and $K$ be the intersections of $GF$ with $AD$ and with the circumcircle of $ABC$, respectively (Fig. 18). As $\\angle BGK = \\angle BGF = 90^\\circ$, we have that $BK$ is a diameter of the circumcircle of $ABC$. Then also $\\angle BCK = 90^\\circ$, from which $CK \\parallel AE$. Now $AKCE$ as a cyclic quadrilateral with two parallel sides must be an isosceles trapezium. Combining this with $FD = FE$, we get $\\angle EDF = \\angle DEF = \\angle AEC = \\angle EAK$, so $DF \\parallel AK$. From $FE = FC$ and the parallel lines we get equal triangles $FCK$ and $FEA'$, from which $FA' = FK$. Thus $DF$ is a midline of $AKA'$, from which $D$ is the midpoint of $AA'$. In conclusion $D$ is the circumcenter of the right-angled triangle $APA'$, which finishes the proof.\nLet $A'$ be the reflection of $A$ over $BC$ and let $BK$ be a diameter of the circumcircle of $ABC$.\nFirst, we will show that $F$ lies on $A'K$ (Fig. 19). As $KC \\perp BC$ and $AD \\perp BC$, we have $KC \\parallel EA'$. Also\n$$\n\\begin{aligned}\n\\angle AEK &= \\angle ACK = 90^\\circ - \\angle ACB, \\\\\n\\angle AA'C &= \\angle DA'C = \\angle DAC = 90^\\circ - \\angle ACB.\n\\end{aligned}\n$$\nThus $\\angle AEK = \\angle AA'C$, from which $EK \\parallel A'C$. Therefore $A'CKE$ is a parallelogram. As the intersection of the diagonals of a parallelogram divides both of its diagonals in half and $F$ is the midpoint of $CE$, it must also be the midpoint of $A'K$, so $A', K, F$ are indeed collinear.\n\n![](attached_image_3.png)\nFig. 19\n\nNow we show that $G$ also lies on $A'K$. For this, let $G'$ be the second intersection of $A'K$ and the circumcircle of $ABC$ (Fig. 20). Notice that $DF$ is a midline of $A'AK$, so $DF \\parallel AK$. Therefore\n$$\n\\angle EDF = \\angle EAK = 180^\\circ - \\angle EG'K = 180^\\circ - \\angle EG'F,\n$$\nwhich shows that $G'$ lies on the circumcircle of $DEF$, meaning that $G' = G$. We have shown that $A'$ lies on $GF$, which also contains $P$. Thus $APA'$ is a right triangle, whose circumcenter is the midpoint $D$ of $AA'$. Therefore $DA = DP$, as desired.\nLike in Solution 1, we show that $FC = FD$ and $BG \\perp GF$. Since also $AP \\perp GF$ (Fig. 21), we have $AP \\parallel BG$. From this\n$$\n\\angle GAP = \\angle AGB = \\angle ACB = \\angle ACD,\n$$\nso the right triangles $GAP$ and $ACD$ are similar. Therefore\n$$\n\\frac{GA}{AC} = \\frac{PG}{DA}. \\qquad (7)\n$$\n\n![](attached_image_4.png)\nFig. 21\n\nOn the other hand, we have the equalities\n$$\n\\angle GDF = \\angle GEF = \\angle GEC = \\angle GAC,\n$$\n$$\n\\angle FGD = \\angle FED = \\angle CEA = \\angle CGA,\n$$\nfrom which it follows that the triangles $GDF$ and $GAC$ are similar. Thus\n$$\n\\frac{GA}{AC} = \\frac{GD}{DF}. \\qquad (8)\n$$\n\nFrom (7) and (8) we get $\\frac{PG}{DA} = \\frac{GD}{DF}$ or\n$$\n\\frac{GD}{PG} = \\frac{DF}{DA}. \\qquad (9)\n$$\nNotice also that there must exist a spiral similarity at $G$ between the similar triangles $GDF$ and $GAC$. Hence $AGD$ and $CGF$ are also similar, implying\n$$\n\\frac{GC}{GA} = \\frac{FC}{DA}. \\qquad (10)\n$$\nAs $DF = FC$, the right hand sides of (9) and (10) are equal. So the left hand sides must be equal as well, i.e., $\\frac{GD}{PG} = \\frac{GC}{GA}$. As $\\angle PGD = \\angle FGD = \\angle CGA$, triangles $PGD$ and $AGC$ must also be similar. Therefore\n$$\n\\frac{GA}{AC} = \\frac{PG}{DP}. \\qquad (11)\n$$\nCombining (7) and (11) gives $DA = DP$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a rectangle. The bisector of the angle $CAD$ meets the side $CD$ at point $L$. Let $M$ be the midpoint of the line segment $AL$. The line $DM$ meets lines $AC$ and $AB$ at points $E$ and $F$, respectively. Given that line segments $AE$ and $AF$ are equal, prove that $ABCD$ is a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle DAL = \\angle LAC = \\alpha$. Then $\\angle FAE = 90^\\circ - 2\\alpha$ (Fig. 1).\n\nAs $AE = AF$, we obtain $\\angle AFE = \\frac{180^\\circ - (90^\\circ - 2\\alpha)}{2} = 45^\\circ + \\alpha$.\n\nBut as $M$ bisects the hypotenuse $AL$ of the right triangle $ALD$, it follows that $M$ is the circumcentre of $ALD$, implying $MA = MD$. Hence $\\angle MDA = \\angle MAD = \\alpha$, implying $\\angle AFE = \\angle AFD = 90^\\circ - \\alpha$.\n\nWe obtain $45^\\circ + \\alpha = 90^\\circ - \\alpha$ which implies $\\angle CAD = 2\\alpha = 45^\\circ$. Hence also $\\angle ACD = 45^\\circ$, implying $AD = CD$. Consequently, $ABCD$ is a square.\n\n![](attached_image_1.png)\nFig. 1\nAs $AE = AF$, the triangle $AEF$ is isosceles with vertex angle at $A$ (Fig. 1). Denote $\\angle AEF = \\angle AFE = \\beta$.\n\nNote that triangles $MAF$ and $MLD$ are equal because $\\angle FMA = \\angle DML$, $\\angle MAF = \\angle MLD$ and $MA = ML$. Thus $AF = DL$ which shows that $ADLF$ is a rectangle. As the diagonals of a rectangle are equal and bisect each other, we have $MA = MF$, implying that the triangle $MAF$ is isosceles with vertex angle at $M$. Hence $\\angle MAF = \\angle MFA = \\beta$.\n\nAs $\\angle EAF = 180^\\circ - 2\\beta$, we have $\\angle MAE = \\beta - (180^\\circ - 2\\beta) = 3\\beta - 180^\\circ$. Hence also $\\angle MAD = 3\\beta - 180^\\circ$, because $AM$ bisects the angle $DAE$. We obtain the equation $\\beta + (3\\beta - 180^\\circ) = 90^\\circ$ which gives $\\beta = 67.5^\\circ$. Thus $\\angle BAC = 180^\\circ - 2 \\cdot 67.5^\\circ = 45^\\circ$, implying that the diagonal of the rectangle $ABCD$ bisects its angle. Consequently, $ABCD$ is a square.\n\n![](attached_image_1.png)\nFig. 1",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30113,
    "subject": "Mathematics (Multi-modal)",
    "question": "A TV show airs every 28 days. This century there was a year when the show aired on both January 1 and January 29. In how many years will the show air twice in January again?",
    "options": [],
    "answer": "21",
    "solution": "In a common year, there are $365 = 13 \\cdot 28 + 1$ days. Thus, after a common year, the dates of the airings shift 1 day earlier. After a leap year, the airings shift 2 days earlier. In every 4 years, there are 3 common years and 1 leap year, which means a total shift of $3 \\cdot 1 + 2 = 5$ days. Thus in $5 \\cdot 4 = 20$ years the total shift is $5 \\cdot 5 = 25$ days.\nThus, if in year $a$ the show airs on January 29, then in year $a + 20$ it will air on January 4, and in the intermediate years it will air between these dates. In these years, there cannot be a second airing in January. However, in year $a + 21$, the first airing will be on January 2 or January 3, which also means a second airing in January. Thus 21 years is the desired answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Digits $A$, $B$, $C$ are given distinct values from $1$ to $9$ to make the value of the expression $2024 \\cdot AB \\cdot CC \\cdot BA$ a perfect square. How many distinct values can the expression $A+B+C$ obtain?",
    "options": [],
    "answer": "6",
    "solution": "Denote the value of the given expression by $k$. We know that $2024 = 2^3 \\cdot 11 \\cdot 23$ and $CC = C \\cdot 11$. In a perfect square, all prime exponents are even. Thus one of the numbers $AB$, $BA$, $CC$ must be divisible by $23$. This cannot be $CC$, so let it be $AB$ without loss of generality (the other option is symmetrical). The two-digit multiples of $23$ are $23$, $46$, $69$ and $92$.\n\n* If $AB = 23$, then $BA = 32 = 2^5$, so $k = 2^8 \\cdot 11^2 \\cdot 23^2 \\cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C = 1$, $C = 4$ or $C = 9$. The sum $A+B+C$ is $6$, $9$ or $14$ respectively.\n\n* If $AB = 46$, then $BA = 64 = 2^6$, so $k = 2^{10} \\cdot 11^2 \\cdot 23^2 \\cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C = 1$ or $C = 9$ (the digit $4$ is already in use). The sum $A+B+C$ is $11$ or $19$ respectively.\n\n* If $AB = 69$, then $BA = 96 = 2^5 \\cdot 3$, so $k = 2^8 \\cdot 3^2 \\cdot 11^2 \\cdot 23^2 \\cdot C$. Then $k$ is a perfect square iff $C$ is a perfect square, meaning $C = 1$ or $C = 4$ (the digit $9$ is already in use). The sum $A+B+C$ is $16$ or $19$ respectively.\n\n* If $AB = 92$, then $BA = 29$, so $k = 2^5 \\cdot 11^2 \\cdot 23^2 \\cdot 29 \\cdot C$. This cannot be a perfect square, as $C$ cannot be divisible by $29$.\n\nThe sum $A+B+C$ can thus obtain $6$ different values ($19$ was present in two different cases).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 30115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve the system of equations\n$$\n\\begin{cases}\nx + y = z, \\\\\nx^2 + y^2 = 4z, \\\\\nx^3 + y^3 = 18z.\n\\end{cases}\n$$",
    "options": [],
    "answer": "(x, y, z) = (0, 0, 0) or z = 6 with {x, y} = {3 + sqrt(3), 3 − sqrt(3)}",
    "solution": "**Solution 1:** If $x \\neq 0$ or $y \\neq 0$, then from the second equation $z > 0$. Thus $z = 0$ can only hold if $x = y = 0$. The triple $(x, y, z) = (0, 0, 0)$ satisfies all equations. Now assume $z \\neq 0$.\nSquaring the first equation yields $x^2 + 2xy + y^2 = z^2$. Subtracting the second equation yields\n$$\n2xy = z^2 - 4z. \\quad (1)\n$$\nCubing the first equation yields $x^3 + 3x^2y + 3xy^2 + y^3 = z^3$. Subtracting the third equation and factoring yields\n$$\n3xy(x + y) = z(z^2 - 18).\n$$\nUsing $x+y=z \\neq 0$, this can be reduced to\n$$\n3xy = z^2 - 18. \\quad (2)\n$$\nExpressing $xy$ from both (1) and (2) yields $3(z^2 - 4z) = 2(z^2 - 18)$. This simplifies to $z^2 - 12z + 36 = 0$ or $(z - 6)^2 = 0$, from which $z = 6$.\n\nNow equation (1) yields $2xy = 36 - 24 = 12$, from which $xy = 6$. On the other hand $x + y = z = 6$. Combining by Viète's formulas yields the quadratic equation $x^2 - 6x + 6 = 0$, from which $x = 3 \\pm \\sqrt{3}$ and respectively $y = 3 \\mp \\sqrt{3}$. The triples $(x,y,z) = (3+\\sqrt{3},3-\\sqrt{3},6)$ and $(x,y,z) = (3-\\sqrt{3},3+\\sqrt{3},6)$ satisfy all three equations.\n**Solution 2:** Like in Solution 1, we show that $z = 0$ yields only the solution $(x,y,z) = (0,0,0)$. We also similarly deduce (1). We then use the identity $x^3 + y^3 = (x+y)(x^2-xy+y^2)$. Substituting $x+y, x^2+y^2$ and $x^3+y^3$ from the given equations and $xy$ from (1) yields\n$$\n18z = z \\left( 4z - \\frac{z^2 - 4z}{2} \\right).\n$$\nDividing both sides by $z \\neq 0$ and simplifying yields $z^2 - 12z + 36 = 0$ or $(z-6)^2 = 0$, from which $z = 6$. We proceed like in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 30116,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an isosceles triangle $ABC$ with $AB = AC$, the bisector of the angle $BAC$ intersects $BC$ at $D$. The bisector of the angle $ABC$ intersects the perpendicular bisector of $AD$ at $E$. Prove that the bisector of the angle $ACB$ is perpendicular to $DE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "As the $A$-bisector is also the altitude, we have $BD \\perp AD$. Let $I$ be the incenter of $ABC$ and $E' \\neq B$ the intersection of $BI$ and the circumcircle of $ABD$ (Fig. 2). As the angles $\\angle ABE' = \\angle DBE'$ correspond to the arcs $E'A$ and $E'D$ of this circle, we have $E'A = E'D$. Thus $E'$ lies on the perpendicular bisector of $AD$, meaning $E' = E$. Hence $ABDE$ is a cyclic quadrilateral (Fig. 3).\nIt remains to verify that $\\angle EDI + \\angle CID = 90^\\circ$, which would yield $CI \\perp DE$, as desired. To show this, let $\\angle ACB = \\angle ABC = \\gamma$. Then\n$$\n\\angle EDI = \\angle EDA = \\angle EBA = \\frac{\\gamma}{2},\n$$\nwhereas\n$$\n\\angle CID = 180^\\circ - \\angle CDI - \\angle DCI = 180^\\circ - 90^\\circ - \\frac{\\gamma}{2} = 90^\\circ - \\frac{\\gamma}{2}.\n$$\nThus $\\angle EDI + \\angle CID = 90^\\circ$, finishing the proof.\nAs the $A$-bisector is also the altitude, we have $BD \\perp AD$. Thus the perpendicular bisector of $AD$ is parallel to $BC$, therefore containing the midline of $ABC$ and bisecting $AB$ and $AC$. Let $L$ be the midpoint of $AB$ and $I$ the incenter of $ABC$ (Fig. 4). Then $\\angle LEB = \\angle CBE = \\angle LBE$, meaning that $LBE$ is isosceles and $LE = LB = LA$. Thus $L$ is the circumcenter of $ABE$ and $AB$ is a diameter of the circle. Now as $ADB = 90^{\\circ}$, the point $D$ also lies on this circle, meaning $ABDE$ is cyclic. We proceed like in Solution 1.\nAs the $A$-bisector is also the altitude, we have $BD \\perp AD$. Thus the perpendicular bisector of $AD$ is parallel to $BC$, therefore containing the midline of $ABC$ and bisecting $AB$ and $AC$. Let $L$ and $M$ be the midpoints of $AB$ and $AC$, respectively. Then $\\angle LEB = \\angle CBE = \\angle LBE$, meaning that $LBE$ is isosceles and $LE = LB = LA$. Also $MC = MA = LB = LE$, and as $LM$ is the midline of $ABC$, we have $LM = \\frac{1}{2}BC = CD$.\nLet $N$ be the intersection of $DE$ and $AC$; depending on the orientation of points (Figures 5 and 6) we have\n$$\nCN = CM \\pm MN, \\\\\nCD = LM = LE \\pm EM = CM \\pm ME.\n$$\nTriangles $NCD$ and $NME$ are similar by parallel sides, giving $\\frac{CD}{ME} = \\frac{CN}{MN}$. Substituting in the previous equation, we get $\\frac{CM \\pm ME}{ME} = \\frac{CM \\pm MN}{MN}$, from which $\\frac{CM}{ME} \\pm 1 = \\frac{CM}{MN} \\pm 1$, which in turn implies $\\frac{CM}{ME} = \\frac{CM}{MN}$. Therefore $ME = MN$, from which $CD = CN$. The claim of the problem follows, as the $C$-bisector in the isosceles triangle $CDN$ and the side $DN$ are perpendicular.\n\n![](attached_image_1.png)\nFig. 5\n![](attached_image_2.png)\nFig. 6\nLet $I$ be the incenter of $ABC$. Clearly $D$ is the point of tangency of the incircle and side $BC$. Let $N$ be the tangency point with side $AC$ (Fig. 7), then $CD = CN$. In the isosceles triangle $CDN$, the line $CI$ is both the angle bisector and altitude, so $DN$ and $CI$ are perpendicular. It remains to show that $E$ lies on $DN$.\nBy the Iran lemma, the angle bisector $BI$ and the extensions of the incircle chord $DN$ and the midline corresponding to $BC$ intersect at one point. Of those, the first and the third intersect at $E$, so $E$ must lie on $DN$, as desired.\n\n![](attached_image_3.png)\nFig. 7",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30117,
    "subject": "Mathematics (Multi-modal)",
    "question": "The website of a company has multiple pages, which may point to each other. A list containing each of these pages exactly once is called *natural*, if whenever one page points to another, it is written before the other page.\nThe webmaster of the company has a natural list of pages. Then she reorders the pages, writing down first all pages which are not pointed to by any other page, then all pages which the first page of the original list points to, then all pages which the second page of the original list points to and which are not listed yet etc.\n\na. It is known that each page is pointed to by at most one other page. Can we be certain that the new list is natural?\n\nb. It is not known whether each page is pointed to by at most one other page. Can we be certain that at least one possible new list is natural?",
    "options": [],
    "answer": "a. Yes. b. No.",
    "solution": "We write $x \\to y$ whenever the page $x$ points to the page $y$.\n\na. Consider any two pages $a$ and $b$ with $a \\to b$. By the assumption, $b$ is not pointed to by any other page. Thus $b$ is located in the new list in the group of pages pointed to by page $a$. We consider two cases.\n* If $a$ is not pointed to by any page, then it is in the first group, and thus also before $b$.\n* If $a$ is pointed to by a page $c$, then $a$ is in the group of pages pointed to by $c$. But as the initial list is natural, $c$ must be before $a$ in it. Thus in the new list, $a$ must be before $b$.\n\nIn either case, $a$ is before $b$ in the new list, making it natural, as desired.\n\n![](attached_image_1.png)\nFig. 8\n\nb. Let there be 4 pages $a, b, c, d$ with $a \\to b$, $b \\to c$, $c \\to d$ and $a \\to d$ (Fig. 8). Then the list $a, b, c, d$ is natural. The new list can then be either $a, b, d, c$ or $a, d, b, c$. But neither of them is natural, as the last page $c$ points to the page $d$ before it.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(m, n)$ of natural numbers that satisfy $m - n = 96$ and $\\text{lcm}(m, n) = 2024$.",
    "options": [],
    "answer": "(184, 88)",
    "solution": "As $\\text{lcm}(m,n) = 2024 = 8 \\cdot 253$ and $8 = 2^3$, at least one of the numbers $m$ and $n$ is divisible by $8$. Since $8 \\mid 96 = m-n$, the other one must also be divisible by $8$. Both $m$ and $n$ are divisors of $2024$. All divisors of $2024$ that are divisible by $8$ are $8$, $88$, $184$ and $2024$. The only two of these with difference $96$ are $184$ and $88$. A straightforward check shows that $\\text{lcm}(184,88) = \\text{lcm}(8 \\cdot 23, 8 \\cdot 11) = 8 \\cdot 23 \\cdot 11 = 2024$ indeed.\nAs $\\text{lcm}(m,n) = 2024$, both $m$ and $n$ are divisors of $2024$. All divisors of $2024$ are\n1, 2, 4, 8, 11, 22, 23, 44, 46, 88, 92, 184, 253, 506, 1012, 2024.\nAs $m-n=96$, we must have $m > 96$. This observation cuts out all cases except $m = 184, m = 253, m = 506, m = 1012$ and $m = 2024$. As differences between $253$, $506$, $1012$ and $2024$ are larger than $96$, it suffices to check the options $m = 184$ and $m = 253$ which imply $n = 88$ and $n = 157$, respectively. As $157 \\nmid 2024$, the only solution can be $(m,n) = (184, 88)$. An easy check shows that $\\text{lcm}(184,88) = \\text{lcm}(8 \\cdot 23, 8 \\cdot 11) = 8 \\cdot 23 \\cdot 11 = 2024$ indeed.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30119,
    "subject": "Mathematics (Multi-modal)",
    "question": "According to a message sent by extraterrestrial creatures who are millions of years ahead of us in development, the height of the highest two places of their planet, measured from the sea level, is $h$, whereas the lowest point on mainland has height $l$ (where $h \\ge 0 \\ge l$). The radius of the planet (i.e., the distance of the sea level from the centre of the planet) is $r$. Express the largest enabled by these conditions distance between two points on this planet, one of which can be visible from the other one.",
    "options": [],
    "answer": "2√((2r + h + l)(h − l))",
    "solution": "![](attached_image_1.png)\nFig. 9\n\nIf the distance is maximal, the line connecting these points must be a tangent of the planet, otherwise one could increase the distance by pushing the points along the surface of the planet farther away. Let the centre be $O$; let the two points under consideration be $P_1$ and $P_2$ with the tangent point $P_3$ between them (Fig. 9). Denote $OP_i = d_i$ for $i = 1, 2, 3$. As the tangent line is perpendicular to the radius drawn to the tangent point, $OP_1P_3$ and $OP_2P_3$ are right triangles with hypothenuses $OP_1$ and $OP_2$, respectively. Hence\n$$\nP_1P_2 = P_1P_3 + P_2P_3 = \\sqrt{d_1^2 - d_3^2} + \\sqrt{d_2^2 - d_3^2}.\n$$\nThe value of this expression is maximal if $d_1$ and $d_2$ are as large as possible and $d_3$ is as small as possible, i.e., $d_1 = d_2 = r+h$ and $d_3 = r+l$. Substituting these values gives the desired distance $2\\sqrt{(r+h)^2 - (r+l)^2}$, or equivalently, $2\\sqrt{(2r+h+l)(h-l)}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be any natural number. A real number is written into every vertex of a regular $n$-gon in such a way that numbers in any two neighbouring vertices differ by at most $1$. Find the least non-negative real number $C$ such that, regardless of the choice of the numbers in the vertices, there exist two neighbouring vertices in which numbers differ by at most $C$.",
    "options": [],
    "answer": "C = 1 if n is even; C = (n−1)/(n+1) if n is odd",
    "solution": "Denote the vertices of the polygon as $V_1, \\dots, V_n$ and let the real numbers in these vertices be $v_1, v_2, \\dots, v_n$, respectively. Moreover, denote $V_{n+1} = V_1$ and $v_{n+1} = v_1$. Define the value of the side $V_iV_{i+1}$ of the polygon to be $s_i = v_{i+1} - v_i$.\n\nObserve that $v_1, \\dots, v_n$ satisfy the conditions of the problem if and only if the corresponding differences $s_1, \\dots, s_n$ satisfy $\\max\\{|s_1|, \\dots, |s_n|\\} \\le 1$ and $s_1 + \\dots + s_n = 0$. Therefore the problem can be reformulated as finding the least non-negative real number $C$ such that, for arbitrary real numbers $s_1, \\dots, s_n$, assumptions $\\max\\{|s_1|, \\dots, |s_n|\\} \\le 1$ and $s_1 + \\dots + s_n = 0$ would imply $\\min\\{|s_1|, \\dots, |s_n|\\} \\le C$.\n\nClearly $\\min\\{|s_1|, \\dots, |s_n|\\} \\le 1$ for any choice of $s_1, \\dots, s_n$ satisfying the assumptions. If $n=2k$ for some $k \\in \\mathbb{N}$ then choosing $s_1 = \\dots = s_k = 1$ and $s_{k+1} = \\dots = s_{2k} = -1$ would imply $\\min\\{|s_1|, \\dots, |s_n|\\} = 1$. Hence $C=1$.\n\nLet now be $n=2k+1$ for some $k \\in \\mathbb{N}$. Taking $s_1 = s_2 = \\dots = s_{k+1} = \\frac{k}{k+1}$ and $s_{k+2} = \\dots = s_{2k+1} = -1$ establishes $\\min\\{|s_1|, \\dots, |s_n|\\} = \\frac{k}{k+1}$. We show that $\\min\\{|s_1|, \\dots, |s_n|\\} \\le \\frac{k}{k+1}$ whenever $s_1, \\dots, s_n$ satisfy the assumptions. To this end, suppose the contrary, i.e. $\\min\\{|s_1|, \\dots, |s_n|\\} > \\frac{k}{k+1}$. Reorder $s_1, \\dots, s_n$ as $d_1, \\dots, d_n$ so that $d_1 \\ge d_2 \\ge \\dots \\ge d_n$. Note that there exists $l \\in \\{1, \\dots, n\\}$ such that $d_l > \\frac{k}{k+1}$ and $d_{l+1} < -\\frac{k}{k+1}$.\n\nIf $l \\ge k+1$ then\n$$\n0 = s_1 + \\dots + s_n = d_1 + \\dots + d_n \\ge l \\cdot d_l + (n-l)d_n > l \\cdot \\frac{k}{k+1} - (n-l) \\ge (k+1) \\cdot \\frac{k}{k+1} - k = k - k = 0,\n$$\ncontradiction. In the case $l < k+1$, the proof is analogous (one can consider the opposite differences $-d_n, -d_{n-1}, \\dots, -d_1$). Hence $C = \\frac{k}{k+1} = \\frac{\\lfloor \\frac{n}{2} \\rfloor}{\\lfloor \\frac{n}{2} \\rfloor + 1}$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Kati writes the numbers\n$$\n2^0, 2^1, 2^2, \\dots, 2^{100}, 3^0, 3^1, 3^2, \\dots, 3^{100}, 6^0, 6^1, 6^2, \\dots, 6^{100}\n$$\non the board. In each step, she performs one of the following operations:\n(1) She can pick two numbers and replace them with their greatest common divisor and least common multiple; or\n(2) She can pick two numbers, one of which is divisible by the other, and replace them with some two numbers whose greatest common divisor and least common multiple would be the two picked numbers.\nFind the least and the greatest sum of all numbers on the board that can be achieved via finitely many steps.",
    "options": [],
    "answer": "Greatest sum: 101 + 2*(6^101 - 1)/5. Least sum: 2^102 - 3*2^51 + 2*3^50 + 3^101.",
    "solution": "At all points throughout the process, every number on the board can be written as $2^\\alpha 3^\\beta$ for some $0 \\le \\alpha, \\beta \\le 100$. Moreover, the list of exponents of both $2$ and $3$ does not change throughout the process. Indeed, in a step of the first kind, two numbers $2^{\\alpha_1} 3^{\\beta_1}$ and $2^{\\alpha_2} 3^{\\beta_2}$ are turned into $2^{\\min(\\alpha_1, \\alpha_2)} 3^{\\min(\\beta_1, \\beta_2)}$ and $2^{\\max(\\alpha_1, \\alpha_2)} 3^{\\max(\\beta_1, \\beta_2)}$, preserving both the exponents $\\alpha_1, \\alpha_2$ of $2$ and the exponents $\\beta_1, \\beta_2$ of $3$. A step of the second kind is just the reverse of a step of the first kind, so the exponents are preserved again.\nThus the lists of both the exponents of $2$ and $3$ are\n$$\n\\underbrace{0, \\dots, 0}_{103 \\text{ copies}}; 1, 1, 2, 2, \\dots, 100, 100.\n$$\n\nBy the rearrangement inequality, we obtain that the sum of the numbers on the board is the greatest when they are\n$$\n\\underbrace{1, \\dots, 1}_{103 \\text{ copies}}; 6, 6, 6^2, 6^2, \\dots, 6^{100}, 6^{100}.\n$$\nThe sum of these numbers equals $101 + 2 \\cdot \\frac{6^{101}-1}{5}$. These numbers can be achieved by performing a step of the first kind on the numbers $2^\\gamma$ and $3^\\gamma$ for every $\\gamma = 1, \\dots, 100$.\n\nBy the rearrangement inequality, we obtain that the sum of the numbers on the board is the least when the lists of the exponents of $2$ and $3$ are sorted in opposite directions. This means that the numbers would be\n$$\n\\begin{array}{ccccccccccc}\n2^{100} & 2^{100} & 2^{99} & 2^{99} & 2^{98} & 2^{98} & \\dots & 2^{51} & 2^{51} & 2^{50} \\\\\n2^{50} & 2^{49} & 2^{49} \\cdot 3 & 2^{48} \\cdot 3 & 2^{48} \\cdot 3^2 & 2^{47} \\cdot 3^2 & \\dots & 2 \\cdot 3^{49} & 3^{49} & 3^{50} \\\\\n3^{50} & 3^{51} & 3^{51} & 3^{52} & 3^{52} & 3^{53} & \\dots & 3^{99} & 3^{100} & 3^{100}.\n\\end{array}\n$$\nUsing the formula for the sum of a geometric series for each row, separately for numbers at odd positions and even positions, yields:\n* for the first row, $2^{101} - 2^{50}$ and $2^{101} - 2^{51}$;\n* for the second row, $3^{51} - 2^{51}$ and $3^{50} - 2^{50}$;\n* for the third row, $\\frac{1}{2}(3^{101} - 3^{50})$ and $\\frac{1}{2}(3^{101} - 3^{51})$.\n\nThus the sum of all the numbers is $2^{102} - 3 \\cdot 2^{51} + 2 \\cdot 3^{50} + 3^{101}$. This is indeed achievable. For this, we will pair the first desired number with the last one, the second first number with the second last one, etc. The least common multiples of the pairs are $6^{100}, 6^{100}, \\dots, 6^{26}, 6^{26}, 6^{25}$ and the greatest common divisors are $\\underbrace{1, 1, \\dots, 1}_{103 \\text{ copies}}; 6, 6, 6^2, 6^2, \\dots, 6^{24}, 6^{24}$. Finally, $6^{25}$ is left over. This means that we can use steps of the second kind to achieve the desired numbers from the situation with the maximal sum.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30122,
    "subject": "Mathematics (Multi-modal)",
    "question": "The angle bisectors of an acute triangle $ABC$ meet at point $I$. The line $AI$ meets the circumcircle of the triangle $ABC$ at point $D$ ($D \\neq A$) and the side $BC$ at point $E$. The line $BI$ meets the circumcircle of the triangle $CDI$ at point $K$ whereas the line $CI$ meets the circumcircle of the triangle $BDI$ at point $L$ ($K \\neq I, L \\neq I$).\n\na. Prove that the line $DI$ is tangent to the circumcircle of the triangle $IKL$.\n\nb. Prove that points $A, K, L, E$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha = \\angle CAI = \\angle IAB$, $\\beta = \\angle ABI = \\angle IBC$, $\\gamma = \\angle BCI = \\angle ICA$. Then $\\alpha + \\beta + \\gamma = 90^\\circ$ and $\\angle CBD = \\angle CAD = \\alpha = \\angle DAB = \\angle DCB$, yielding\n$$\n\\angle KBD = \\angle IBD = \\alpha + \\beta = 90^\\circ - \\gamma, \\\\\n\\angle DCL = \\angle DCI = \\alpha + \\gamma = 90^\\circ - \\beta.\n$$\nDepending on the location of the point $K$ (Figures 50 and 51), we have either $\\angle DKB = \\angle DKI = \\angle DCI$ or $\\angle DKB = 180^\\circ - \\angle IKD = \\angle DCI$; in each case $\\angle DKB = 90^\\circ - \\beta$. Analogously, we obtain $\\angle CLD = 90^\\circ - \\gamma$. Hence\n$$\n\\angle BDK = 180^\\circ - (90^\\circ - \\gamma) - (90^\\circ - \\beta) = \\beta + \\gamma = 90^\\circ - \\alpha, \\\\\n\\angle LDC = 180^\\circ - (90^\\circ - \\beta) - (90^\\circ - \\gamma) = \\beta + \\gamma = 90^\\circ - \\alpha.\n$$\nOn the other hand, we have $\\angle BDC = 180^\\circ - 2\\alpha = 2(90^\\circ - \\alpha)$, meaning that both $DK$ and $DL$ bisect the angle $BDC$. Consequently, points $D, K$ and $L$ lie on a line. As the bisector of the vertex angle of the isosceles triangle $BCD$ is also the perpendicular bisector of the line segment $BC$, symmetry yields $\\angle DCK = \\angle KBD = 90^\\circ - \\gamma$ and $\\angle LBD = \\angle DCL = 90^\\circ - \\beta$.\n\na. If the point $K$ lies between points $C$ and $I$ then\n$$\n\\angle DKI = \\angle DKB = 90^\\circ - \\beta = \\angle LBD = \\angle LID.\n$$\nHence the line $DI$ is tangent to the circumcircle of the triangle $IKL$ (as points $D, K, L$ lie on a line). If the point $K$ lies between points $I$ and $D$ then\n$$\n\\angle ILD = \\angle CLD = 90^\\circ - \\gamma = \\angle DCK = \\angle DIK.\n$$\nAnalogously to the previous case, the line $DI$ must be tangent to the circumcircle of the triangle $IKL$.\n\nb. Since $\\angle DKB = 90^\\circ - \\beta = \\angle LBD$ and points $D, K, L$ lie on a line, the line $DB$ is tangent to the circumcircle of the triangle $BKL$. On the other hand, we have $\\angle DAB = \\alpha = \\angle CBD = \\angle EBD$, implying that $DB$ is also tangent to the circumcircle of the triangle $BEA$. Using the power of the point $D$ w.r.t. to these circles, we get $DB^2 = DK \\cdot DL$ and $DB^2 = DA \\cdot DE$, respectively. Altogether, we obtain $DA \\cdot DE = DK \\cdot DL$. Hence points $A, K, L, E$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30123,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ with $AB < BC$ and an obtuse angle at vertex $B$ is given. The incircle of the triangle $ABC$ with incentre $I$ touches the sides $BC$, $CA$ and $AB$ at points $D$, $E$ and $F$, respectively. The line $AI$ intersects the side $BC$ at point $K$. The ray $IB$ intersects the circumcircle of the triangle $KIF$ at point $P$ ($P \\neq I$), whereas the ray $IC$ intersects the circumcircle of the triangle $KIE$ at point $Q$ ($Q \\neq I$). Prove that the line $PQ$ bisects the line segment $DK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFig. 52\nSolution:\nLet $\\angle BAC = 2\\alpha$, $\\angle CBA = 2\\beta$, $\\angle ACB = 2\\gamma$; then $\\alpha + \\beta + \\gamma = 90^\\circ$. By the two tangents theorem, we have $AE = AF$ (Fig. 52), and because $\\angle FAK = \\alpha = \\angle KAE$, the triangles $AEK$ and $AFK$ are equal. Thus also $KE = KF$.\n\nOn the other hand, $\\angle EIA = 90^\\circ - \\alpha = \\angle AIF$, implying that $\\angle KIE = 180^\\circ - \\angle EIA = 180^\\circ - \\angle AIF = \\angle FIK$. This shows that the angles inscribed on equal chords $KE$ and $KF$ of the circumcircles of the triangles $KIE$ and $KIF$ are equal. Consequently, these circles have equal radii and angles inscribed on equal chords of these circles are always equal.\n\nFrom the triangle $AIC$, we obtain $\\angle KIQ = \\angle KIC = \\alpha + \\gamma = 90^\\circ - \\beta$ and $\\angle FIP = \\angle FIB = 90^\\circ - \\beta$. Thus angles inscribed on the chords $KQ$ and $PF$ of the circumcircles of the triangles $KIE$ and $KIF$ are equal, which implies $KQ = PF$. By the two tangents theorem, we also have $BF = BD$. As $\\angle FBP = 180^\\circ - \\beta = \\angle PBD$, the triangles $FBP$ and $DBP$ are equal. Hence also $PF = PD$. Altogether, we obtain $KQ = PD$.\n\nAnalogously, we get $PK = DQ$. Hence the quadrilateral $PDQK$ is a parallelogram and its diagonals $PQ$ and $DK$ bisect each other.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 30124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABCD$ be a convex quadrilateral with $\\angle DAB = \\angle BCD = 90^{\\circ}$ and $\\angle ABC > \\angle CDA$. Let $Q$ and $R$ be points on the segments $BC$ and $CD$, respectively, such that the line $QR$ intersects lines $AB$ and $AD$ at points $P$ and $S$, respectively. It is given that $PQ = RS$. Let the midpoint of $BD$ be $M$ and the midpoint of $QR$ be $N$. Prove that $M, N, A$ and $C$ lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that $N$ is also the midpoint of $PS$. From right-angled triangles $PAS$ and $CQR$ we obtain $\\angle ANP = 2 \\angle ASP$, $\\angle CNQ = 2 \\angle CRQ$, hence\n$$\n\\angle ANC = \\angle ANP + \\angle CNQ = 2(\\angle ASP + \\angle CRQ) = 2(\\angle RSD + \\angle DRS) = 2 \\angle ADC.\n$$\nSimilarly, using right-angled triangles $BAD$ and $BCD$, we obtain $\\angle AMC = 2 \\angle ADC$.\nThus $\\angle AMC = \\angle ANC$, and the required statement follows.\n\n![](attached_image_1.png)\nSolution:\n\nIn this proof we show that we have $\\angle NCM = \\angle NAM$ instead. From right-angled triangles $BCD$ and $QCR$ we get $\\angle DRS = \\angle CRQ = \\angle RCN$ and $\\angle BDC = \\angle DCM$. Hence $\\angle NCM = \\angle DCM - \\angle RCN$. From right-angled triangle $APS$ we get $\\angle PSA = \\angle SAN$. From right-angled triangle $BAD$ we have $\\angle MAD = \\angle BDA$. Moreover, $\\angle BDA = \\angle DRS + \\angle RSD - \\angle RDB$.\nTherefore $\\angle NAM = \\angle NAS - \\angle MAD = \\angle CDB - \\angle DRS = \\angle NCM$, and the required statement follows.\nSolution:\n\nAs $N$ is also the midpoint of $PS$, we can shrink triangle $APS$ to a triangle $A_0 QR$ (where $P$ is sent to $Q$ and $S$ is sent to $R$). Then $A_0, Q, R$ and $C$ lie on a circle with center $N$. According to the shrinking, the line $A_0R$ is parallel to the line $AD$. Therefore $\\angle CNA = \\angle CNA_0 = 2 \\angle CRA_0 = 2 \\angle CDA = \\angle CMA$. The required statement follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest positive integer $k$ for which there exist a colouring of the positive integers $\\mathbb{Z}_{>0}$ with $k$ colours and a function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ with the following two properties:\n(i) For all positive integers $m, n$ of the same colour, $f(m+n)=f(m)+f(n)$.\n(ii) There are positive integers $m, n$ such that $f(m+n) \\neq f(m)+f(n)$.\n\nIn a colouring of $\\mathbb{Z}_{>0}$ with $k$ colours, every integer is coloured in exactly one of the $k$ colours. In both (i) and (ii) the positive integers $m, n$ are not necessarily different.",
    "options": [],
    "answer": "k = 3",
    "solution": "Solution:\nThe answer is $k=3$.\n\nFirst we show that there is such a function and coloring for $k=3$. Consider $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ given by $f(n)=n$ for all $n \\equiv 1$ or $2$ modulo $3$, and $f(n)=2 n$ for $n \\equiv 0$ modulo $3$. Moreover, give a positive integer $n$ the $i$-th color if $n \\equiv i$ (mod $3$).\n\nBy construction we have $f(1+2)=6 \\neq 3=f(1)+f(2)$ and hence $f$ has property (ii).\n\nNow let $n, m$ be positive integers with the same color $i$. If $i=0$, then $n+m$ has color $0$, so $f(n+m)=2(n+m)=2 n+2 m=f(n)+f(m)$. If $i=1$, then $n+m$ has color $2$, so $f(n+m)=n+m=f(n)+f(m)$. Finally, if $i=2$, then $n+m$ has color $1$, so $f(n+m)=n+m=f(n)+f(m)$. Therefore $f$ also satisfies condition (i).\n\nNext we show that there is no such function and coloring for $k=2$.\n\nConsider any coloring of $\\mathbb{Z}_{>0}$ with $2$ colors and any function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Then there exist positive integers $m$ and $n$ such that $f(m+n) \\neq f(m)+f(n)$. Choose $m$ and $n$ such that their sum is minimal among all such $m, n$ and define $a=m+n$. Then in particular for every $b<a$ we have $f(b)=b f(1)$ and $f(a) \\neq a f(1)$.\n\nIf $a$ is even, then condition (i) for $m=n=\\frac{a}{2}$ implies $f(a)=f\\left(\\frac{a}{2}\\right)+f\\left(\\frac{a}{2}\\right)=f(1) a$, a contradiction. Hence $a$ is odd. We will prove two lemmas.\n\nLemma 1. Any odd integer $b<a$ has a different color than $a$.\n\nProof. Suppose that $b<a$ is an odd integer, and that $a$ and $b$ have the same color. Then on the one hand, $f(a+b)=f(a)+b f(1)$. On the other hand, we also have $f(a+b)=f\\left(\\frac{a+b}{2}\\right)+f\\left(\\frac{a+b}{2}\\right)=(a+b) f(1)$, as $\\frac{a+b}{2}$ is a positive integer smaller than $a$. Hence $f(a)=f(a+b)-b f(1)=(a+b) f(1)-b f(1)=a f(1)$, which is again a contradiction. Therefore all odd integers smaller than $a$ have a color different from that of $a$.\n\nLemma 2. Any even integer $b<a$ has the same color as $a$\n\nProof. Suppose $b<a$ is an even integer, and that $a$ and $b$ have different colors. Then $a-b$ is an odd integer smaller than $a$, so it has the same color as $b$. Thus $f(a)=f(a-b)+f(b)=(a-b) f(1)+b f(1)=a f(1)$, a contradiction. Hence all even integers smaller than $a$ have the same color as $a$.\n\nSuppose now $a+1$ has the same color as $a$. As $a>1$, we have $\\frac{a+1}{2}<a$ and therefore $f(a+1)=2 f\\left(\\frac{a+1}{2}\\right)=(a+1) f(1)$. As $a-1$ is an even integer smaller than $a$, we have by Lemma 2 that $a-1$ also has the same color as $a$. Hence $2 f(a)=f(2 a)=f(a+1)+f(a-1)=(a+1) f(1)+(a-1) f(1)=2 a f(1)$, which implies that $f(a)=a f(1)$, a contradiction. So $a$ and $a+1$ have different colors.\n\nSince $a-2$ is an odd integer smaller than $a$, by Lemma 1 it has a color different from that of $a$, so $a-2$ and $a+1$ have the same color. Also, we have seen by Lemma 2 that $a-1$ and $a$ have the same color. So $f(a)+f(a-1)=f(2 a-1)=f(a+1)+f(a-2)=(a+1) f(1)+(a-2) f(1)=(2 a-1) f(1)$, from which it follows that $f(a)=(2 a-1) f(1)-f(a-1)=(2 a-1) f(1)-(a-1) f(1)=a f(1)$, which contradicts our choice of $a$ and finishes the proof.\nSolution:\nWe prove that $k \\leq 3$ just as in first solution.\n\nNext we show that there is no such function and coloring for $k=2$.\n\nConsider any coloring of $\\mathbb{Z}_{>0}$ with $2$ colors and any function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). We first notice with $m=n$ that $f(2 n)=2 f(n)$.\n\nLemma 3. For every $n \\in \\mathbb{Z}_{>0}, f(3 n)=3 f(n)$ holds.\n\nProof. Define $c=f(n), d=f(3 n)$. Then we have the relations\n$$\nf(2 n)=2 c, \\quad f(4 n)=4 c, \\quad f(6 n)=2 d\n$$\n- If $n$ and $2 n$ have the same color, then $f(3 n)=f(n)+f(2 n)=3 c=3 f(n)$.\n- If $n$ and $3 n$ have the same color, then $4 c=f(4 n)=f(n)+f(3 n)=c+f(3 n)$, so $f(3 n)=3 f(n)$.\n- If $2 n$ and $4 n$ have the same color, then $2 d=f(6 n)=f(2 n)+f(4 n)=2 c+4 c=6 c$, so $f(3 n)=d=3 c$.\n- Otherwise $n$ and $4 n$ have the same color, and $2 n$ and $3 n$ both have the opposite color to $n$. Therefore we compute $5 c=f(n)+f(4 n)=f(5 n)=f(2 n)+f(3 n)=2 c+f(3 n)$ so $f(3 n)=3 f(n)$.\n\nConsequently, for $k=2$ we necessarily have $f(3 n)=3 f(n)$.\n\nNow let $a$ be the smallest integer such that $f(a) \\neq a f(1)$. In particular $a$ is odd and $a>3$. Consider the three integers $a, \\frac{a-3}{2}, \\frac{a+3}{2}$. By pigeonhole principle two of them have the same color.\n\n- If $\\frac{a-3}{2}$ and $\\frac{a+3}{2}$ have the same color, then $f(a)=\\frac{a-3}{2} f(1)+\\frac{a+3}{2} f(1)=a f(1)$.\n- If $a$ and $\\frac{a-3}{2}$ have the same color, then $3 \\frac{a-1}{2} f(1)=3 f\\left(\\frac{a-1}{2}\\right)=f\\left(\\frac{3 a-3}{2}\\right)=f(a)+f\\left(\\frac{a-3}{2}\\right)=f(a)+\\frac{a-3}{2} f(1)$, so $f(a)=a f(1)$.\n- If $a$ and $\\frac{a+3}{2}$ have the same color, then $3 \\frac{a+1}{2} f(1)=3 f\\left(\\frac{a+1}{2}\\right)=f\\left(\\frac{3 a+3}{2}\\right)=f(a)+f\\left(\\frac{a+3}{2}\\right)=f(a)+\\frac{a+3}{2} f(1)$, so $f(a)=a f(1)$.\n\nIn the three cases we find a contradiction with $f(a) \\neq a f(1)$, so it finishes the proof.\nSolution:\nAs before we prove that $k \\leq 3$ and for any such function and colouring we have $f(2 n)=2 f(n)$.\n\nNow we show that there is no such function and coloring for $k=2$.\n\nConsider any coloring of $\\mathbb{Z}_{>0}$ with $2$ colors and any function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ satisfying conditions (i) and (ii). Say the two colors are white (W) and black (B). Pick $m, n$ any two integers such that $f(m+n)=f(m)+f(n)$. Without loss of generality we may assume that $m+n, m$ are black and $n$ is white.\n\nLemma 4. For all $l \\in \\mathbb{Z}_{>0}$ and every $x$ whose color is black, we have $x+l m$ is black and $f(x+l m)=f(x)+l f(m)$.\n\nProof. We proceed by induction. It is clearly true for $l=0$. If $x+l m$ is black and satisfies $f(x+l m)=f(x)+l f(m)$, then $f(x+(l+1) m)=f(x+l m)+f(m)=f(x)+(l+1) f(m)$ and $f(x+(l+1) m+n)=f(x+l m)+f(m+n)=f(x)+l f(m)+f(m+n) \\neq f(x)+(l+1) f(m)+f(n)=f(x+(l+1) m)+f(n)$, so $x+(l+1) m$ is not the same color of $n$, therefore $x+(l+1) m$ is black. This completes the induction.\n\nIn particular we then must have that $2^{l} n$ is white for every $l$, because otherwise since $2^{l} m$ is black we would have $2^{l} f(m+n)=f\\left(2^{l} m+2^{l} n\\right)=f\\left(2^{l} m\\right)+f\\left(2^{l} n\\right)=2^{l}(f(m)+f(n))$, and consequently $f(m+n)=f(m)+f(n)$.\n\nLemma 5. For every $l \\geq 1, 2^{l} m+2^{l-1} n$ is black.\n\nProof. On the one hand we have $2^{l} f(m+n)=f\\left(2^{l} m+2^{l} n\\right)=f\\left(2^{l-1}(2 m+n)+2^{l-1} n\\right)$. On the other hand we have\n$2^{l} f(m+n)=2^{l-1} \\cdot 2 f(m+n) \\neq 2^{l-1}(f(m+n)+f(m)+f(n))=2^{l-1}(f(2 m+n)+f(n))=f\\left(2^{l} m+2^{l-1} n\\right)+f\\left(2^{l-1} n\\right)$.\nTherefore $2^{l} m+2^{l-1} n$ and $2^{l-1} n$ have different color, which means $2^{l} m+2^{l-1} n$ is black.\n\nCombining the two lemmas give $j m+2^{l-1} n$ is black for all $j \\geq 2^{l}$ and every $l \\geq 1$.\n\nNow write $m=2^{l-1} m'$ with $m'$ odd. Let $t$ be a number such that $\\frac{2^{t}-1}{m'}$ is an integer and $j=\\frac{2^{t}-1}{m'} n \\geq 2^{l}$, i.e. $t$ is some multiple of $\\phi\\left(m'\\right)$. Then we must have that $j m+2^{l-1} n$ is black, but by definition $j m+2^{l-1} n=\\left(2^{t}-1\\right) 2^{l-1} n+2^{l-1} n=2^{t+l-1} n$ is white. This is a contradiction, so $k=2$ is impossible.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are $2017$ lines in a plane such that no $3$ of them go through the same point. Turbo the snail can slide along the lines in the following fashion: she initially moves on one of the lines and continues moving on a given line until she reaches an intersection of $2$ lines. At the intersection, she follows her journey on the other line turning left or right, alternating the direction she chooses at each intersection point she passes. Can it happen that she slides through a line segment for a second time in her journey but in the opposite direction as she did for the first time?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nWe show that this is not possible.\nThe lines divide the plane into disjoint regions. We claim that there exists an alternating $2$-coloring of these regions, that is each region can be colored in black or white, such that if two regions share a line segment, they have a different color. We show this inductively.\nIf there are no lines, this is obvious. Consider now an arrangement of $n$ lines in the plane and an alternating $2$-coloring of the regions. If we add a line $g$, we can simply switch the colors of all regions in one of the half planes divided by $g$ from white to black and vice versa. Any line segment not in $g$ will still be between two regions of different color. Any line segment in $g$ cuts a region determined by the $n$ lines in two, and since we switched colors on one side of $g$ this segment will also lie between two regions of different color.\nNow without loss of generality we may assume, that Turbo starts on a line segment with a white region on her left and a black one on her right. At any intersection, if she turns right, she will keep the black tile to her right. If she turns left, she will keep the white tile to her left. Thus wherever she goes, she will always have a white tile on her left and a black tile on her right. As a consequence, she can cross every line segment only in the direction where she has a white tile to her left, and never the opposite direction where she would have a black tile to the left.\nSolution:\n\nSuppose the assumption is true.\nLet's label each segment in the snail's path with $\\mathbf{L}$ or $\\mathbf{R}$ depending on the direction that Turbo chose at the start point of this segment (one segment can have several labels if it has been visited several times).\nConsider the first segment that has been visited twice in different directions, name this segment $s_{1}$. Assume without loss of generality that it is labeled with $\\mathbf{L}$. Then next segment must be labeled with $\\mathbf{R}$, name this one $s_{2}$.\nLet's look at the label which $s_{1}$ can get on the second visit. If it gets $\\mathbf{L}$ then the previous segment in the path must be $s_{2}$. But in this case $s_{1}$ is not the first segment that has been visited twice in different directions because $s_{2}$ has been visited earlier. So the second label of $s_{1}$ must be $\\mathbf{R}$, and Turbo must have come from the opposite side of $s_{2}$.\n\n![](attached_image_1.png)\n\nSince Turbo alters directions at each point, labels in her path also alter. And because two labels of $s_{1}$ are different, the number of visited segments between these two visits must be even.\nNow let's make the following observation: each segment in the path corresponds to exactly one line, and its previous and next segments are on opposite sides of this line.\n\n![](attached_image_2.png)\n\nAgain consider the path between two visits of $s_{1}$.\nEach line intersecting this path must be crossed an even number of times because Turbo has to return to the initial side of each line. Therefore, an even number of segments of Turbo's path are contained on each of these lines. But the line containing $s_{1}$ must be crossed an odd number times. Since each crossing corresponds to exactly one segment in the path, the number of segments must be odd.\nHere we get the contradiction. Therefore, the assumption is false.\nSolution:\n\nSuppose that the snail always slides slightly to the right of the line segments on her path. When turning to the right, she does not cross any line, whereas when turning to the left, she crosses exactly two lines. This means that at any time of her journey, she has crossed an even number of lines.\nAssuming that at some point she slides along a segment for the second time, but in the opposite direction, we argue that she needs to cross an odd number of lines. Let $\\ell$ be the line on which the revisit happens. In order to get to the other side of $\\ell$, the snail has to cross $\\ell$ an odd number of times. To visit the same segment of $\\ell$, she must cross every other line an even number of times.\nSolution:\n\nLet us color in red all intersection points of the given lines and let us choose one of two possible directions on each segment (draw an arrow on each segment). Consider a red point $R$ where two given lines $a$ and $b$ meet, and the four segments $a_{1}, a_{2}, b_{1}, b_{2}$ with endpoint $R$ (so that $a_{i} \\subset a, b_{j} \\subset b$). $R$ is called a saddle if on $a_{1}, a_{2}$ the arrows go out of $R$ while on $b_{1}, b_{2}$ the arrows enter $R$, or vice versa, on $b_{1}, b_{2}$ the arrows go out of $R$ while on $a_{1}, a_{2}$ the arrows enter $R$. The set of arrows (chosen on all segments) is said to be good if all red points are saddles. It is sufficient to prove that there exists a good set of arrows. Indeed, if initially Turbo is moving along (or opposite) the arrow, then this condition holds after she turns at a red point.\nThe given lines cut the plane into regions. Further we need the following property of the good set of arrows (this property directly follows from the definition): the boundary of any bounded region is a directed cycle of arrows; the boundary of any unbounded region is a directed chain of arrows.\nWe construct a good set of arrows by induction on $n$ with trivial base $n=1$. Now erase one of $n$ given lines and assume we have a good set of arrows for remaining $n-1$ lines. Now restore the $n$-th line $\\ell$, assume that $\\ell$ is horizontal. Denote by $A_{1}, \\ldots, A_{n-1}$ all new red points on $\\ell$ from the left to the right. Each of $A_{i}$ belongs to some old segment $m_{i}$ of the line $\\ell_{i}$. Let us call $A_{i}$ ascending if the arrow on $m_{i}$ goes up, and descending if the arrow on $m_{i}$ goes down. Consider the region containing the segment $A_{i}A_{i+1}$. By the property, $A_{i}$ and $A_{i+1}$ can not be both ascending or both descending. Thus we can choose arrows on all pieces of $\\ell$ so that each arrow goes from a descending to an ascending vertex.\nEach of points $A_{i}$ cuts $m_{i}$ into two new pieces; the direction of new pieces supposed to be the same as on $m_{i}$. Now simultaneously change the direction of arrows on all pieces below the line $\\ell$. It is easy to see that $A_{1}, \\ldots, A_{n-1}$ become saddles, while the other red points remain saddles. This completes the induction step.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n \\geq 1$ be an integer and let $t_{1}<t_{2}<\\ldots<t_{n}$ be positive integers. In a group of $t_{n}+1$ people, some games of chess are played. Two people can play each other at most once. Prove that it is possible for the following conditions to hold at the same time:\n\ni) The number of games played by each person is one of $t_{1}, t_{2}, \\ldots, t_{n}$,\n\nii) For every $i$ with $1 \\leq i \\leq n$, there is someone who has played exactly $t_{i}$ games of chess.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $\\mathcal{T}=\\{t_{1}, \\ldots, t_{n}\\}$. The proof proceeds by induction on $n=|\\mathcal{T}|$. If $n=1$ and $\\mathcal{T}=\\{t\\}$, choose a group of $t+1$ people and let every pair of two persons play against each other. Then every person has played $t$ games and the conditions of the problem are satisfied.\n\nIn the inductive step, suppose that $\\mathcal{T}$ has $n \\geq 2$ elements $t_{1}<t_{2}<\\cdots<t_{n}$. Consider the set\n$$\n\\mathcal{T}'=\\{t_{n}-t_{n-1}, t_{n}-t_{n-2}, \\ldots, t_{n}-t_{1}\\} .\n$$\nBy the inductive hypothesis, there exists a group $G'$ of $t_{n}-t_{1}+1$ people that satisfies the conditions of the problem for $T'$.\n\nNext construct a group $G''$ of $t_{n}+1$ people by adding $t_{1}$ people who do not know any of the other $t_{n}-t_{1}+1$ people in $G'$. Finally, construct a group $G$ by complementing the knowledge relation in $G''$ : two persons play against each other in $G$ if and only if they do not play against each other in $G''$.\n\nBy construction $t \\in \\mathcal{T}$ if and only if there exists a person in $G''$ that played against exactly $t_{n}-t$ other people (if $t=t_{n}$, choose one of the $t_{1}$ people added to $G'$). That person knows $t_{n}-(t_{n}-t)=t$ other students in $G$, completing the proof.\nSolution:\n\nLet $\\mathcal{T}=\\{t_{1}, \\ldots, t_{n}\\}$. The proof proceeds by induction on $n=|\\mathcal{T}|$. If $n=1$ and $\\mathcal{T}=\\{t\\}$, we choose a group of $t+1$ people such that everyone plays with everyone else. If $n=2$ and $\\mathcal{T}=\\{t_{1}, t_{2}\\}$ with $t_{1}<t_{2}$, divide the $t_{2}+1$ people into groups $A$ resp. $B$ of size $t_{1}$ resp. $t_{2}-t_{1}+1$ such that everyone from group $A$ played with everyone else whereas people from group $B$ only played with the people from group $A$. Then the people from group $A$ resp. $B$ played with exactly $t_{2}$ resp. $t_{1}$ other people.\n\nIn the inductive step, suppose that $T$ has $n>2$ elements $t_{1}<\\ldots<t_{n}$. Consider the set\n$$\n\\mathcal{T}'=\\left(\\mathcal{T} \\setminus\\{t_{1}, t_{n}\\}\\right)-t_{1}=\\{t_{n-1}-t_{1}, t_{n-2}-t_{1}, \\ldots, t_{2}-t_{1}\\}\n$$\nBy the induction hypothesis there exists a group $C$ of $t_{n-1}-t_{1}+1$ people that satisfies the conditions of the problem for $T'$. Next add groups $D$ resp. $E$ of $t_{1}$ resp. $t_{n}-t_{n-1}$ people such that people from group $D$ played with everyone else whereas people from group $E$ only played with the people from group $D$. Then the people from group $C$ played with $t$ other people if and only if they played with $t-t_{1}$ many people among $C$, i.e. if and only if $t \\in\\{t_{2}, \\ldots, t_{n-1}\\}=\\mathcal{T} \\setminus\\{t_{1}, t_{n}\\}$. People from group $D$ resp. $E$ played with $t_{n}$ resp $t_{1}$ people, which completes the proof.\nSolution:\n\nThe proof proceeds by induction on $|t_{n}|$. If $t_{n}=1$ we have $n=1$ and we can consider two persons that play against each other. Then every player has played 1 game and the conditions of the problem are satisfied.\n\nIf $t_{n}>1$ we distinguish the two cases $t_{1}>1$ and $t_{1}=1$.\n\nIf $t_{1}>1$ there exists, by the induction hypothesis, a group $A$ of size $t_{n}$ that satisfies the conditions of the problem for $t_{1}'=t_{1}-1, \\ldots, t_{n}'=t_{n}-1$. Now add a new person to $A$ and let him/her play against everyone from $A$. The new group will be of size $t_{n}+1$ and there exists a person which has played $t$ games if and only if there exists a person that has played $t-1$ games within $A$, i.e. if and only if $t \\in\\{t_{1}, \\ldots, t_{n}\\}$. Hence the conditions of the problem are satisfied.\n\nIf $t_{1}=1$ there exists, by the induction hypothesis, a group $B$ of size $t_{n-1}$ that satisfies the conditions of the problem for $t_{2}-1, \\ldots, t_{n-1}-1$. Now add a new person $P$ and let him/her play with everyone from group $B$ and a group $C$ of size $t_{n}-t_{n-1}>0$ and let them play with $P$. The new group will be of size $t_{n-1}+1+(t_{n}-t_{n-1})+1=t_{n}+1$. Since person $P$ has played against everyone he will have played $t_{n}$ games. The people in $C$ will have played $1=t_{1}$ games. There exists a person in $B$ that has played $t$ games if and only if there exist a person in $B$ that has played $t-1$ games within $B$, i.e. if and only if $t \\in\\{t_{2}, \\ldots, t_{n-1}\\}$. Hence the conditions of the problem are satisfied.\nSolution:\n\nWe generalize the construction for $\\mathcal{T}=\\{1, \\ldots, n\\}$\n\nConstruction\n\nTake sets of people $A_{1}, \\ldots, A_{n}$. Let all people of $A_{i}$ play chess with all people in $A_{j}$ with $j \\geq n-i+1$\n\n![](attached_image_1.png)\n\nNow the number of games played by anyone in $A_{i}$ is\n$\\left(\\sum_{j \\geq n-i+1}|A_{j}|\\right)$ or $\\left(\\sum_{j \\geq n-i+1}|A_{j}|\\right)-1$ if $i \\geq n-i+1$.\nNow if we start with one person in each $A_{i}$ and two people in $A_{\\left\\lceil\\frac{n}{2}\\right\\rceil}$. The number of played games for anyone in $A_{i}$ is equal to $i$. In particular this is a construction for $\\mathcal{T}=\\{1, \\ldots, n\\}$\nNow to get to numbers of general sets $\\mathcal{T}$ of size $n$ we can change the sizes of $A_{i}$ but keep the construction.\n\nVariant 1\n\nObservation 1 Adding a person to a set $A_{i}$ increases the number of games played in $A_{j}$ for $j \\geq n-i+1$, by exactly one.\nStart with the construction above and then add $t_{1}-1$ people to group $A_{n}$, making the new set of games played equal to $\\{t_{1}, t_{1}+1, \\ldots, n+t_{1}-1\\}$. Then add $t_{2}-t_{1}-1$ to $A_{n-1}$ to get set of games played to $\\{t_{1}, t_{2}, t_{2}+1, \\ldots, n+t_{2}-2\\}$ and repeat until we get to the set $\\mathcal{T}$ adding a total of $\\sum_{j=1}^{n} t_{j}-t_{j-1}-1=t_{n}-n$ people (let $t_{0}=0$ ), so we get $t_{n}+1$ people in the end.\nClearly we can start by adding vertices to $A_{1}$ or any other set instead of $A_{n}$ first and obtain an equivalent construction with the same number of people.\n\nVariant 2\n\nIt is also possible to calculate the necessary sizes of $A_{i}$'s all at once. We have by construction the number of games played in $A_{1}$ is less than the number of games played in $A_{2}$ etc. So we have that in the end we want the games played in $A_{i}$ to be exactly $t_{i}$.\nSo $(t_{1}, t_{2}, t_{3}, \\ldots, t_{n}) \\stackrel{!}{=}( |A_{n}|, |A_{n}|+|A_{n-1}|, \\ldots, (\\sum_{j=2}^{n}|A_{j}|)-1, (\\sum_{j=1}^{n}|A_{j}|)-1 )$.\nThis gives us by induction that $|A_{n}|= t_{1}, |A_{n-1}|= t_{2}-t_{1}, \\ldots, |A_{1}|= t_{n}-t_{n-1}$ and a quick calculation shows that the sum of all sets is exactly $1+\\sum_{j=1}^{n}(t_{j}-t_{j-1})=t_{n}+1$. (where the $+1$ comes from the set $A_{\\left\\lceil\\frac{n}{2}\\right\\rceil}$ and $t_{0}=0$.)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n \\geq 2$ be an integer. An $n$-tuple $(a_{1}, a_{2}, \\ldots, a_{n})$ of positive integers is expensive if there exists a positive integer $k$ such that\n$$\n(a_{1}+a_{2})(a_{2}+a_{3}) \\cdots (a_{n-1}+a_{n})(a_{n}+a_{1})=2^{2k-1}.\n$$\na) Find all positive integers $n \\geq 2$ for which there exists an expensive $n$-tuple.\nb) Prove that for every positive integer $m$ there exists an integer $n \\geq 2$ such that $m$ belongs to an expensive $n$-tuple.\n\nThere are exactly $n$ factors in the product on the left hand side.",
    "options": [],
    "answer": "a) Expensive tuples exist if and only if n is odd.\nb) For every odd positive integer m there exists some n ≥ 2 such that m appears in an expensive n-tuple.",
    "solution": "Solution:\n\na) Notice that for odd integers $n>2$, the tuple $(1,1, \\ldots, 1)$ is expensive. We will prove that there are no expensive $n$-tuples for even $n$.\n\nLemma 0.1. If an expensive $n$-tuple exists for some $n \\geq 4$, then also an expensive $n-2$-tuple.\n\nProof. In what follows all indices are considered modulo $n$. Let $(a_{1}, a_{2}, \\ldots, a_{n})$ be an expensive $n$-tuple and $a_{t}$ the largest element of the tuple. We have the inequalities\n$$\n\\begin{aligned}\na_{t-1}+a_{t} & \\leq 2 a_{t}<2(a_{t}+a_{t+1}) \\\\\na_{t}+a_{t+1} & \\leq 2 a_{t}<2(a_{t-1}+a_{t}) .\n\\end{aligned}\n$$\nSince both $a_{t-1}+a_{t}$ and $a_{t}+a_{t+1}$ are powers of 2 (they are divisors of a power of 2), we deduce from (1) and (2)\n$$\na_{t-1}+a_{t}=a_{t}+a_{t+1}=2^{r}\n$$\nfor some positive integer $r$, and in particular $a_{t-1}=a_{t+1}$.\n\nConsider now the $n-2$-tuple $(b_{1}, \\ldots, b_{n-2})$ obtained by removing $a_{t}$ and $a_{t+1}$ from $(a_{1}, a_{2}, \\ldots, a_{n})$. By what we just said we have\n$$\n\\prod_{i=1}^{n-2}(b_{i}+b_{i+1})=\\frac{\\prod_{i=1}^{n}(a_{i}+a_{i+1})}{(a_{t-1}+a_{t})(a_{t}+a_{t+1})}=2^{2(k-r)-1}\n$$\nand hence $(b_{1}, \\ldots, b_{n-2})$ is again expensive.\n\nFrom the lemma we now conclude that if there exists an expensive $n$-tuple for some even $n$, then also an expensive 2-tuple, i.e.\n$$\n(a_{1}+a_{2})^{2}=2^{2k-1}\n$$\nfor some positive integers $a_{1}, a_{2}$, which is impossible since the right hand side is not a square.\n\nb) We prove this by induction. In a) we saw that 1 belongs to an expensive $n$-tuple. Assume now that all odd positive integers less than $2^{k}$ belong to an expensive $n$-tuple, for some $k \\geq 1$. Hence for any odd $r<2^{k}$ there is an integer $n$ and an expensive $n$-tuple $(a_{1}, \\ldots, r, \\ldots, a_{n})$. We notice that then also $(a_{1}, \\ldots, r, 2^{k+1}-r, r, \\ldots, a_{n})$ is expensive. Since $2^{k+1}-r$ can take all odd values between $2^{k}$ and $2^{k+1}$ the induction step is complete.\nSolution:\n\na) For odd $n$ the tuple $(1,1, \\ldots, 1)$ is a solution.\n\nNow consider $n$ even. Since the product $\\prod(a_{i}+a_{i+1})$ is a power of two, every factor needs to be a power of two. We are going to prove that for all tuples $(a_{1}, \\ldots, a_{n})$ such that $a_{i}+a_{i+1}$ is always a power of two, it is the case that $\\prod(a_{i}+a_{i+1})$ is equal to an even power of two. We are going to prove this with strong induction on $\\sum a_{i}$.\n\nWhen all $a_{i}$ are equal to one this is certainly the case. Since $a_{i}+a_{i+1}>1$ it is even and we conclude that the $a_{i}$ are either all odd or all even. In the case they are all even, then consider the tuple $(b_{1}, \\ldots, b_{n})$ with $b_{i}=a_{i} / 2$. This tuple clearly satisfies the hypothesis as well and we have $\\sum b_{i}<\\sum a_{i}$. Furthermore we have $\\prod(a_{i}+a_{i+1})=2^{n} \\prod(b_{i}+b_{i+1})$ and since $n$ is even we are done in this case.\n\nNow all $a_{i}$ are odd. Suppose none of them are one, then consider the tuple $(b_{1}, \\ldots, b_{n})$ with $b_{i}=(a_{i}+(-1)^{i}) / 2$. Since all $a_{i}$ are odd and strictly larger than one, the $b_{i}$ are positive integers and satisfy $b_{i}+b_{i+1}=(a_{i}+a_{i+1}) / 2$, a power of two. Since $\\sum b_{i}<\\sum a_{i}$ and $\\prod(a_{i}+a_{i+1})=2^{n} \\prod(b_{i}+b_{i+1})$ we are done in this case again.\n\nNow there is at least one $a_{i}$ being one. We may assume $i=1$, because the condition is cyclic. Moreover we may also assume that $a_{2}>1$ since not all of the $a_{i}$ are equal to one. Let now $k$ be the smallest index larger than one such that $a_{k}$ is equal to one. We are not excluding the case $k=n+1$, yet. Now for $i=1, \\ldots, k-1$ we have $a_{i}+a_{i+1}>2$ and thus divisible by four. By induction it easily follows that $a_{i} \\equiv(-1)^{i+1} \\bmod 4$ for $i=1, \\ldots, k-1$. In particular, since $a_{k}=1$ we find that $k$ is odd and at least three.\n\nNow consider the tuple $(b_{1}, \\ldots, b_{n})$ with $b_{i}=(a_{i}-(-1)^{i}) / 2$ for $i=1, \\ldots, k$ and $b_{i}=a_{i}$ otherwise. This is again a tuple that satisfies the hypothesis, since $b_{1}=a_{1}=1=b_{k}=a_{k}$. Moreover $b_{2}<a_{2}$ and thus $\\sum b_{i}<\\sum a_{i}$. Finally we have $\\prod(a_{i}+a_{i+1})=2^{k-1} \\prod(b_{i}+b_{i+1})$ and since $k$ is odd we conclude the proof.\n\nb) We use some of the ideas from a). Consider the operators $T_{\\pm}(n)=2n\\pm 1$. We claim that for every odd integer $m$ there is an integer $r$ and signs $\\epsilon_{i} \\in\\{+,-\\}$ for $i=1, \\ldots, r$ such that $T_{\\epsilon_{r}} \\circ \\ldots \\circ T_{\\epsilon_{1}}(1)=m$. This is certainly true for $m=1$ and for $m>1$ we find that $m=T_{-}((m+1)/2)$ if $m \\equiv 1 \\bmod 4$ and $m=T_{+}((m-1)/2)$ if $m \\equiv 3 \\bmod 4$. Note that both $(m+1)/2$ and $(m-1)/2$ are odd integers in their respective cases and $(m-1)/2 \\leq (m+1)/2 < m$ for $m>1$. Therefore iterating the procedure will eventually terminate in one.\n\nFor the construction it is most convenient to set $n=2l+1$ and label the tuple $(a_{-l}, a_{-l+1}, \\ldots, a_{l})$. For $m=1$ we have the expensive tuple $(1,1,1)$. For $m>1$ we will define operators $T_{\\pm}$ on expensive tuples with the condition $a_{-l}=a_{l}=1$ that give rise to a new expensive tuple $(b_{-l'}, \\ldots, b_{l'})$ with $b_{-l'}=b_{l'}=1$ and $b_{0}=T_{\\pm} a_{0}$. It is then clear that $T_{\\epsilon_{r}} \\circ \\cdots \\circ T_{\\epsilon_{1}}((1,1,1))$ is an expensive tuple containing $m$.\n\nWe define $T_{\\pm}$ as follows: set $l'=l+1$ and $b_{-l'}=b_{l'}=1$ and $b_{i}=T_{\\pm(-1)^{i}}(a_{i})$ for $i=-l, \\ldots, l$. Here we identify $+$ with $+1$ and $-$ with $-1$. We are left to prove that the new tuple is indeed expensive. If $\\pm(-1)^{l}=-1$, then $\\prod(b_{i}+b_{i+1})=4 \\cdot 2^{2l} \\prod(a_{i}+a_{i+1})$, and if $\\pm(-1)^{l}=+1$, then $\\prod(b_{i}+b_{i+1})=4 \\cdot 2^{2l+2} \\prod(a_{i}+a_{i+1})$. In both cases we end up with an expensive tuple again.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A B C$ be a triangle with circumcentre $O$. The points $D, E$ and $F$ lie in the interiors of the sides $B C, C A$ and $A B$ respectively, such that $D E$ is perpendicular to $C O$ and $D F$ is perpendicular to $B O$. (By interior we mean, for example, that the point $D$ lies on the line $B C$ and $D$ is between $B$ and $C$ on that line.)\nLet $K$ be the circumcentre of triangle $A F E$. Prove that the lines $D K$ and $B C$ are perpendicular.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $\\ell_{C}$ be the tangent at $C$ to the circumcircle of $\\triangle A B C$. As $C O \\perp \\ell_{C}$, the lines $D E$ and $\\ell_{C}$ are parallel. Now we find that\n$$\n\\angle C D E=\\angle\\left(B C, \\ell_{C}\\right)=\\angle B A C,\n$$\nhence the quadrilateral $B D E A$ is cyclic. Analogously, we find that the quadrilateral $C D F A$ is cyclic. As we now have $\\angle C D E=\\angle A=\\angle F D B$, we conclude that the line $B C$ is the external angle bisector of $\\angle E D F$. Furthermore, $\\angle E D F=180^{\\circ}-2 \\angle A$. Since $K$ is the circumcentre of $\\triangle A E F, \\angle F K E=2 \\angle F A E=2 \\angle A$. So $\\angle F K E+\\angle E D F=180^{\\circ}$, hence $K$ lies on the circumcircle of $\\triangle D E F$. As $|K E|=|K F|$, we have that $K$ is the midpoint of the arc $E F$ of this circumcircle. It is well known that this point lies on the internal angle bisector of $\\angle E D F$. We conclude that $D K$ is the internal angle bisector of $\\angle E D F$. Together with the fact that $B C$ is the external angle bisector of $\\angle E D F$, this yields that $D K \\perp B C$, as desired.\nSolution:\n\nAs in the previous solution, we show that the quadrilaterals $B D E A$ and $C D F A$ are both cyclic. Denote by $M$ and $L$ respectively the circumcentres of these quadrilaterals. We will show that the quadrilateral $K L O M$ is a parallelogram. The lines $K L$ and $M O$ are the perpendicular bisectors of the line segments $A F$ and $A B$, respectively. Hence both $K L$ and $M O$ are perpendicular to $A B$, which yields $K L \\| M O$. In the same way we can show that the lines $K M$ and $L O$ are both perpendicular to $A C$ and hence parallel as well. We conclude that $K L O M$ is indeed a parallelogram. Now, let $K^{\\prime}, L^{\\prime}, O^{\\prime}$ and $M^{\\prime}$ be the respective projections of $K, L, O$ and $M$ to $B C$. We have to show that $K^{\\prime}=D$. As $L$ lies on the perpendicular bisector of $C D$, we have that $L^{\\prime}$ is the midpoint of $C D$. Similarly, $M^{\\prime}$ is the midpoint of $B D$ and $O^{\\prime}$ is the midpoint of $B C$. Now we are going to use directed lengths. Since $K L O M$ is a parallelogram, $M^{\\prime} K^{\\prime}=O^{\\prime} L^{\\prime}$. As\n$$\nO^{\\prime} L^{\\prime}=O^{\\prime} C-L^{\\prime} C=\\frac{1}{2} \\cdot(B C-D C)=\\frac{1}{2} \\cdot B D=M^{\\prime} D\n$$\nwe find that $M^{\\prime} K^{\\prime}=M^{\\prime} D$, hence $K^{\\prime}=D$, as desired.\nSolution:\n\nDenote by $\\ell_{A}, \\ell_{B}$ and $\\ell_{C}$ the tangents at $A, B$ and $C$ to the circumcircle of $\\triangle A B C$. Let $A^{\\prime}$ be the point of intersection of $\\ell_{B}$ and $\\ell_{C}$ and define $B^{\\prime}$ and $C^{\\prime}$ analogously. As in the first solution, we find that $D E \\| \\ell_{C}$ and $D F \\| \\ell_{B}$. Now, let $Q$ be the point of intersection of $D E$ and $\\ell_{A}$ and let $R$ be the point of intersection of $D F$ and $\\ell_{A}$. We easily find $\\triangle A Q E \\sim \\triangle A B^{\\prime} C$. As $\\left|B^{\\prime} A\\right|=\\left|B^{\\prime} C\\right|$, we must have $|Q A|=|Q E|$, hence $\\triangle A Q E$ is isosceles. Therefore the perpendicular bisector of $A E$ is the internal angle bisector of $\\angle E Q A=\\angle D Q R$. Analogously, the perpendicular bisector of $A F$ is the internal angle bisector of $\\angle D R Q$. We conclude that $K$ is the incentre of $\\triangle D Q R$, thus $D K$ is the angle bisector of $\\angle Q D R$. Because the sides of the triangles $\\triangle Q D R$ and $\\triangle B^{\\prime} A^{\\prime} C^{\\prime}$ are pairwise parallel, the angle bisector $D K$ of $\\angle Q D R$ is parallel to the angle bisector of $\\angle B^{\\prime} A^{\\prime} C^{\\prime}$. Finally, as the angle bisector of $\\angle B^{\\prime} A^{\\prime} C^{\\prime}$ is easily seen to be perpendicular to $B C$ (as it is the perpendicular bisector of this segment), we find that $D K \\perp B C$, as desired.\nSolution:\n\nThis is a simplified variant of Solution 1. $\\angle C O B=2 \\angle A$ (angle at centre of circle $A B C$ ) and $O B=O C$ so $\\angle O B C=\\angle B C O=90^{\\circ}-\\angle A$. Likewise $\\angle E K F=2 \\angle A$ and $\\angle K F E=\\angle F E K=90^{\\circ}-\\angle A$. Now because $D E \\perp C O, \\angle E D C=90^{\\circ}-\\angle D C O=90^{\\circ}-\\angle B C O=\\angle A$ and similarly $\\angle B D F=\\angle A$, so $\\angle F D E=180^{\\circ}-2 \\angle A$. So quadrilateral $K F D E$ is cyclic (opposite angles), so (same segment) $\\angle K D E=\\angle K F E=90^{\\circ}-\\angle A$, so $\\angle K D C=90^{\\circ}$ and $D K$ is perpendicular to $B C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a positive integer. Find the greatest possible integer $m$, in terms of $n$, with the following property: a table with $m$ rows and $n$ columns can be filled with real numbers in such a manner that for any two different rows $[a_{1}, a_{2}, \\ldots, a_{n}]$ and $[b_{1}, b_{2}, \\ldots, b_{n}]$ the following holds:\n$$\n\\max \\left(\\left|a_{1}-b_{1}\\right|,\\left|a_{2}-b_{2}\\right|, \\ldots,\\left|a_{n}-b_{n}\\right|\\right)=1\n$$",
    "options": [],
    "answer": "2^n",
    "solution": "Solution:\nThe largest possible $m$ is equal to $2^{n}$.\n\nIn order to see that the value $2^{n}$ can be indeed achieved, consider all binary vectors of length $n$ as rows of the table. We now proceed with proving that this is the maximum value.\n\nLet $[a_{k}^{i}]$ be a feasible table, where $i=1, \\ldots, m$ and $k=1, \\ldots, n$. Let us define undirected graphs $G_{1}, G_{2}, \\ldots, G_{n}$, each with vertex set $\\{1,2, \\ldots, m\\}$, where $ij \\in E(G_{k})$ if and only if $|a_{k}^{i}-a_{k}^{j}|=1$ (by $E(G_{k})$ we denote the edge set of the graph $G_{k}$). Observe the following two properties.\n\n(1) Each graph $G_{k}$ is bipartite. Indeed, if it contained a cycle of odd length, then the sum of $\\pm 1$ along this cycle would need to be equal to $0$, which contradicts the length of the cycle being odd.\n\n(2) For every $i \\neq j$, $ij \\in E(G_{k})$ for some $k$. This follows directly from the problem statement.\n\nFor every graph $G_{k}$ fix some bipartition $(A_{k}, B_{k})$ of $\\{1,2, \\ldots, m\\}$, i.e., a partition of $\\{1,2, \\ldots, m\\}$ into two disjoint sets $A_{k}, B_{k}$ such that the edges of $G_{k}$ traverse only between $A_{k}$ and $B_{k}$. If $m>2^{n}$, then there are two distinct indices $i, j$ such that they belong to exactly the same parts $A_{k}, B_{k}$, that is, $i \\in A_{k}$ if and only if $j \\in A_{k}$ for all $k=1,2, \\ldots, n$. However, this means that the edge $ij$ cannot be present in any of the graphs $G_{1}, G_{2}, \\ldots, G_{n}$, which contradicts (2). Therefore, $m \\leq 2^{n}$.\n\n\nSolution 2:\nIn any table with the given property, the least and greatest values in a column cannot differ by more than $1$. Thus, if each value that is neither least nor greatest in its column is changed to be equal to either the least or the greatest value in its column (arbitrarily), this does not affect any $|a_{i}-b_{i}|=1$, nor does it increase any difference above $1$, so the table still has that given property. But after such a change, for any choice of what the least and greatest values in each column are, there are only two possible choices for each entry in the table (either the least or the greatest value in its column); that is, only $2^{n}$ possible distinct rows, and the given property implies that all rows must be distinct. As in the previous solution, we see that this number can be achieved.\n\n\nSolution 3:\nWe prove by induction on $n$ that $m \\leq 2^{n}$.\n\nFirst suppose $n=1$. If real numbers $x$ and $y$ have $|x-y|=1$ then $\\lfloor x\\rfloor$ and $\\lfloor y\\rfloor$ have opposite parities and hence it is impossible to find three real numbers with all differences $1$. Thus $m \\leq 2$.\n\nSuppose instead $n>1$. Let $a$ be the smallest number appearing in the first column of the table; then every entry in the first column of the table lies in the interval $[a, a+1]$. Let $A$ be the collection of rows with first entry $a$ and $B$ be the collection of rows with first entry in $(a, a+1]$. No two rows in $A$ differ by $1$ in their first entries, so if we list the rows in $A$ and delete their first entries we obtain a table satisfying the conditions of the problem with $n$ replaced by $n-1$; thus, by the induction hypothesis, there are at most $2^{n-1}$ rows in $A$. Similarly, there are at most $2^{n-1}$ rows in $B$. Hence $m \\leq 2^{n-1}+2^{n-1}=2^{n}$. As before, this number can be achieved.\n\n\nSolution 4:\nConsider the rows of the table as points of $\\mathbb{R}^{n}$. As the values in each column differ by at most $1$, these points must lie in some $n$-dimensional unit cube $C$. Consider the unit cubes centred on each of the $m$ points. The conditions of the problem imply that the interiors of these unit cubes are pairwise disjoint. But now $C$ has volume $1$, and each of these cubes intersects $C$ in volume at least $2^{-n}$: indeed, if the unit cube centred on a point of $C$ is divided into $2^{n}$ cubes of equal size then one of these cubes must lie entirely within $C$. Hence $m \\leq 2^{n}$. As before, this number can be achieved.\n\n\nSolution 5:\nAgain consider the rows of the table as points of $\\mathbb{R}^{n}$. The conditions of the problem imply that these points must all lie in some $n$-dimensional unit cube $C$, but no two of the points lie in any smaller cube. Thus if $C$ is divided into $2^{n}$ equally-sized subcubes, each of these subcubes contains at most one row of the table, giving $m \\leq 2^{n}$. As before, this number can be achieved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA set $A$ of integers is called sum-full if $A \\subseteq A+A$, i.e. each element $a \\in A$ is the sum of some pair of (not necessarily different) elements $b, c \\in A$. A set $A$ of integers is said to be zero-sum-free if $0$ is the only integer that cannot be expressed as the sum of the elements of a finite nonempty subset of $A$.\n\nDoes there exist a sum-full zero-sum-free set of integers?",
    "options": [],
    "answer": "Yes; for example A = {F_{2n} : n ≥ 1} ∪ {−F_{2n+1} : n ≥ 1}, where F_k are Fibonacci numbers.",
    "solution": "Solution:\nThe set $A=\\{F_{2 n}: n=1,2, \\ldots\\} \\cup\\{-F_{2 n+1}: n=1,2, \\ldots\\}$, where $F_{k}$ is the $k^{\\text{th}}$ Fibonacci number ($F_{1}=1$, $F_{2}=1$, $F_{k+2}=F_{k+1}+F_{k}$ for $k \\geq 1$) qualifies for an example. We then have $F_{2 n}=F_{2 n+2}+(-F_{2 n+1})$ and $-F_{2 n+1}=(-F_{2 n+3})+F_{2 n+2}$ for all $n \\geq 1$, so $A$ is sum-full (and even with unique representations). On the other hand, we can never have\n$$\n0=\\sum_{i=1}^{s} F_{2 n_{i}}-\\sum_{j=1}^{t} F_{2 n_{j}+1}\n$$\nowing to the fact that Zeckendorf representations are known to be unique.\n\nIt remains to be shown that all nonzero values can be represented as sums of distinct numbers $1,-2,3,-5,8,-13,21, \\ldots$. This may be done using a greedy algorithm: when representing $n$, the number largest in magnitude that is used is the element $m= \\pm F_{k}$ of $A$ that is closest to $0$ subject to having the same sign as $n$ and $|m| \\geq |n|$. That this algorithm terminates without using any member of $A$ twice is a straightforward induction on $k$; the base case is $k=2$ ($m=1$) and the induction hypothesis is that for all $n$ for which the above algorithm starts with $\\pm F_{\\ell}$ with $\\ell \\leq k$, it terminates without having used any member of $A$ twice and without having used any $\\pm F_{j}$ with $j>\\ell$.\n\nRemark (James Aaronson and Adam P Goucher). Let $n$ be a positive integer, and write $u=2^{n}$; we claim that the set\n$$\n\\{1,2,4, \\ldots, 2^{n-1},-u, u+1,-(2 u+1), 3 u+2,-(5 u+3), 8 u+5, \\ldots\\}\n$$\nis a sum-full zero-sum-free set. The proof is similar to that used for the standard examples.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30132,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers $p$ and $q$ are prime and satisfy\n$$\n\\frac{p}{p+1}+\\frac{q+1}{q}=\\frac{2 n}{n+2}\n$$\nfor some positive integer $n$. Find all possible values of $q-p$.",
    "options": [],
    "answer": "{2, 3, 5}",
    "solution": "Solution:\n\nRearranging the equation, $2 q n(p+1) = (n+2)(2 p q + p + q + 1)$. The left hand side is even, so either $n+2$ or $p+q+1$ is even, so either $p=2$ or $q=2$ since $p$ and $q$ are prime, or $n$ is even.\n\nIf $p=2$, $6 q n = (n+2)(5 q + 3)$, so $(q-3)(n-10) = 36$. Considering the divisors of $36$ for which $q$ is prime, we find the possible solutions $(p, q, n)$ in this case are $(2,5,28)$ and $(2,7,19)$ (both of which satisfy the equation).\n\nIf $q=2$, $4 n(p+1) = (n+2)(5 p + 3)$, so $n = p n + 10 p + 6$, a contradiction since $n < p n$, so there is no solution with $q=2$.\n\nFinally, suppose that $n=2 k$ is even. We may suppose also that $p$ and $q$ are odd primes. The equation becomes $2 k q(p+1) = (k+1)(2 p q + p + q + 1)$. The left hand side is even and $2 p q + p + q + 1$ is odd, so $k+1$ is even, so $k = 2 \\ell + 1$ is odd. We now have\n$$\nq(p+1)(2 \\ell + 1) = (\\ell + 1)(2 p q + p + q + 1)\n$$\nor equivalently\n$$\n\\ell q(p+1) = (\\ell + 1)(p q + p + 1)\n$$\nNote that $q \\mid p q + p + 1$ if and only if $q \\mid p+1$. Furthermore, because $(p, p+1) = 1$ and $q$ is prime, $(p+1, p q + p + 1) = (p+1, p q) = (p+1, q) > 1$ if and only if $q \\mid p+1$.\n\nSince $(\\ell, \\ell+1) = 1$, we see that, if $q \\nmid p+1$, then $\\ell = p q + p + 1$ and $\\ell + 1 = q(p+1)$, so $q = p+2$ (and $(p, p+2, 2(2 p^{2} + 6 p + 3))$ satisfies the original equation). In the contrary case, suppose $p+1 = r q$, so $\\ell(p+1) = (\\ell+1)(p+r)$, a contradiction since $\\ell < \\ell+1$ and $p+1 \\leq p+r$.\n\nThus the possible values of $q-p$ are $2, 3$ and $5$.\n\n\nSolution 2:\n\nSubtracting $2$ and multiplying by $-1$, the condition is equivalent to\n$$\n\\frac{1}{p+1} - \\frac{1}{q} = \\frac{4}{n+2}\n$$\nThus $q > p+1$. Rearranging,\n$$\nq-p-1 = \\frac{4(p+1)q}{n+2}\n$$\nThe expression on the right is a positive integer, and $q$ must cancel into $n+2$ else $q$ would divide $p+1 < q$. Let $(n+2)/q = u$ a positive integer.\n\nSo\n$$\nq-p-1 = \\frac{4(p+1)}{u}\n$$\n$$\nu q - u(p+1) = 4(p+1)\n$$\nso $p+1$ divides $u q$. However, $q$ is prime and $p+1 < q$, therefore $p+1$ divides $u$. Let $v$ be the integer $u/(p+1)$. Now\n$$\nq-p = 1 + \\frac{4}{v} \\in \\{2,3,5\\}\n$$\nAll three cases can occur, where $(p, q, n)$ is $(3,5,78)$, $(2,5,28)$ or $(2,7,19)$. Note that all pairs of twin primes $q = p+2$ yield solutions $(p, p+2, 2(2 p^{2} + 6 p + 3))$.\n\n\nSolution 3:\n\nSubtract $2$ from both sides to get\n$$\n\\frac{1}{p+1} - \\frac{1}{q} = \\frac{4}{n+2}\n$$\nFrom this, since $n$ is positive, we have that $q > p+1$. Therefore $q$ and $p+1$ are coprime, since $q$ is prime.\n\nGroup the terms on the LHS to get\n$$\n\\frac{q-p-1}{q(p+1)} = \\frac{4}{n+2}\n$$\nNow $(q, q-p-1) = (q, p+1) = 1$ and $(p+1, q-p-1) = (p+1, q) = 1$ so the fraction on the left is in lowest terms. Therefore the numerator must divide the numerator on the right, which is $4$. Since $q-p-1$ is positive, it must be $1, 2$ or $4$, so that $q-p$ must be $2, 3$ or $5$. All of these can be attained, by $(p, q, n) = (3,5,78)$, $(2,5,28)$ and $(2,7,19)$ respectively.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are infinitely many people registered on the social network Mugbook. Some pairs of (different) users are registered as friends, but each person has only finitely many friends. Every user has at least one friend. (Friendship is symmetric; that is, if $A$ is a friend of $B$, then $B$ is a friend of $A$.)\n\nEach person is required to designate one of their friends as their best friend. If $A$ designates $B$ as her best friend, then (unfortunately) it does not follow that $B$ necessarily designates $A$ as her best friend. Someone designated as a best friend is called a 1-best friend. More generally, if $n>1$ is a positive integer, then a user is an $n$-best friend provided that they have been designated the best friend of someone who is an $(n-1)$-best friend. Someone who is a $k$-best friend for every positive integer $k$ is called popular.\n\na. Prove that every popular person is the best friend of a popular person.\n\nb. Show that if people can have infinitely many friends, then it is possible that a popular person is not the best friend of a popular person.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFor any person $A$, let $f^{0}(x)=x$, let $f(A)$ be $A$'s best friend, and define $f^{k+1}(A)=f\\left(f^{k}(A)\\right)$, so any person who is a $k$-best friend is $f^{k}(A)$ for some person $A$; clearly a $k$-best friend is also an $\\ell$-best friend for all $\\ell<k$. Let $X$ be a popular person. For each positive integer $k$, let $x_{k}$ be a person with $f^{k}\\left(x_{k}\\right)=X$. Because $X$ only has finitely many friends, infinitely many of the $f^{k-1}\\left(x_{k}\\right)$ (all of whom designated $X$ as best friend) must be the same person, who must be popular.\n\nIf people can have infinitely many friends, consider people $X_{i}$ for positive integers $i$ and $P_{i, j}$ for $i<j$ positive integers. $X_{i}$ designates $X_{i+1}$ as her best friend; $P_{i, i}$ designates $X_{1}$ as her best friend; $P_{i, j}$ designates $P_{i+1, j}$ as her best friend if $i<j$. Then all $X_{i}$ are popular, but $X_{1}$ is not the best friend of a popular person.\nSolution:\n\nFor any set $S$ of people, let $f^{-1}(S)$ be the set of people who designated someone in $S$ as their best friend. Since each person has only finitely many friends, if $S$ is finite then $f^{-1}(S)$ is finite.\n\nLet $X$ be a popular person and put $V_{0}=\\{X\\}$ and $V_{k}=f^{-1}\\left(V_{k-1}\\right)$. All $V_{i}$ are finite and (since $X$ is popular) nonempty.\n\nIf any two sets $V_{i}, V_{j}$, with $0 \\leq i<j$ are not disjoint, define $f^{i}(x)$ for positive integers $i$ as in Solution 1. It follows $\\emptyset \\neq f^{i}\\left(V_{i} \\cap V_{j}\\right) \\subseteq f^{i}\\left(V_{i}\\right) \\cap f^{i}\\left(V_{j}\\right) \\subseteq V_{0} \\cap V_{j-i}$, thus $X \\in V_{j-i}$. But this means that $f^{j-i}(X)=X$, therefore $f^{n(j-i)}(X)=X$. Furthermore, if $Y=f^{j-i-1}(X)$, then $f(Y)=X$ and $f^{n(j-i)}(Y)=Y$, so $X$ is the best friend of $Y$, who is popular.\n\nIf all sets $V_{n}$ are disjoint, by König's infinity lemma there exists an infinite sequence of (distinct) $x_{i}, i \\geq 0$, with $x_{i} \\in V_{i}$ and $x_{i}=f\\left(x_{i+1}\\right)$ for all $i$. Now $x_{1}$ is popular and her best friend is $x_{0}=X$.\n\nIf people can have infinitely many friends, proceed as in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30134,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be an acute-angled triangle with circumcircle $\\Gamma$ and orthocentre $H$. Let $K$ be a point of $\\Gamma$ on the other side of $BC$ from $A$. Let $L$ be the reflection of $K$ in the line $AB$, and let $M$ be the reflection of $K$ in the line $BC$. Let $E$ be the second point of intersection of $\\Gamma$ with the circumcircle of triangle $BLM$. Show that the lines $KH$, $EM$ and $BC$ are concurrent. (The orthocentre of a triangle is the point on all three of its altitudes.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince the quadrilateral $B M E L$ is cyclic, we have $\\angle B E M = \\angle B L M$. By construction, $|B K| = |B L| = |B M|$, and so (using directed angles)\n$$\n\\begin{aligned}\n\\angle B L M & = 90^\\circ - \\frac{1}{2} \\angle M B L = 90^\\circ - \\left(180^\\circ - \\frac{1}{2} \\angle L B K - \\frac{1}{2} \\angle K B M\\right) \\\\\n& = \\left(\\frac{1}{2} \\angle L B K + \\frac{1}{2} \\angle K B M\\right) - 90^\\circ = \\left(180^\\circ - \\angle B\\right) - 90^\\circ = 90^\\circ - B.\n\\end{aligned}\n$$\nWe see also that $\\angle B E M = \\angle B A H$, and so the point $N$ of intersection of $E M$ and $A H$ lies on $\\Gamma$.\nLet $X$ be the point of intersection of $K H$ and $B C$, and let $N'$ be the point of intersection of $M X$ and $A H$. Since $B C$ bisects the segment $K M$ by construction, the triangle $K X M$ is isosceles; as $A H \\parallel M K$, $H X N'$ is isosceles. Since $A H \\perp B C$, $N'$ is the reflection of $H$ in the line $B C$. It is well known that this reflection lies on $\\Gamma$, and so $N' = N$. Thus $E, M, N$ and $M, X, N'$ all lie on the same line $M N$; that is, $E M$ passes through $X$.\n\n![](attached_image_1.png)\nSolution:\nWe work with directed angles. Let $H K$ meet $B C$ at $X$. Let $M X$ meet $A H$ at $H_{A}$ on $\\Gamma$ (where $H_{A}$ is the reflection of $H$ in $B C$). Define $E'$ to be where $H_{A} M$ meets $\\Gamma$ (again). Our task is to show that $\\angle M E' B = \\angle M L B$.\nObserve that\n$$\n\\begin{aligned}\n\\angle M E' B & = \\angle H_{A} A B & \\text{ (angles in same segment) } \\\\\n& = B^{c} &\n\\end{aligned}\n$$\nNow\n$$\n\\begin{aligned}\n\\angle M L B & = \\angle H L B \\\\\n& = \\angle B K H_{C} \\\\\n& = \\angle B C H_{C} \\\\\n& = B^{c} .\n\\end{aligned}\n$$\n(Simson line, doubled)\n(reflecting in the line $A B$)\n(angles in the same segment)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA word is a finite sequence of letters from some alphabet. A word is repetitive if it is a concatenation of at least two identical subwords (for example, $a b a b a b$ and $a b c a b c$ are repetitive, but $a b a b a$ and $a a b b$ are not). Prove that if a word has the property that swapping any two adjacent letters makes the word repetitive, then all its letters are identical. (Note that one may swap two adjacent identical letters, leaving a word unchanged.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIn this and the subsequent solutions we refer to a word with all letters identical as constant.\nLet us consider a nonconstant word $W$, of length $|W|=w$, and reach a contradiction. Since the word $W$ must contain two distinct adjacent letters, be it $W=A a b B$ with $a \\neq b$, we may assume $B=c C$ to be non-empty, and so $W=A a b c C$. By the proper transpositions we get the repetitive words $W' = A b a c C = P^{w / p}$, of a period $P$ of length $p \\mid w, 1 < p < w$, and $W'' = A a c b C = Q^{w / q}$, of a period $Q$ of length $q \\mid w, 1 < q < w$. However, if a word $UV$ is repetitive, then the word $VU$ is also repetitive, of a same period length; therefore we can work in the sequel with the repetitive words $W_0' = C A b a c$, of a period $P'$ of length $p$, and $W_0'' = C A a c b$, of a period $Q'$ of length $q$. The main idea now is that the common prefix of two repetitive words cannot be too long.\nNow, if a word $a_1 a_2 \\ldots a_w = T^{w / t}$ is repetitive, of a period $T$ of length $t \\mid w, 1 \\leq t < w$, then the word (and any subword of it) is $t$-periodic, i.e. $a_k = a_{k+t}$, for all $1 \\leq k \\leq w-t$. Therefore the word $C A$ is both $p$-periodic and $q$-periodic.\nWe now use the following classical result:\nWilf-Fine Theorem. Let $p, q$ be positive integers, and let $N$ be a word of length $n$, which is both $p$-periodic and $q$-periodic. If $n \\geq p+q-\\operatorname{gcd}(p, q)$ then the word $N$ is $\\operatorname{gcd}(p, q)$-periodic (but this need not be the case if instead $n \\leq p+q-\\operatorname{gcd}(p, q)-1$).\nBy this we need $|C A| \\leq p+q-\\operatorname{gcd}(p, q)-1 \\leq p+q-2$, hence $w \\leq p+q+1$, otherwise $W_0'$ and $W_0''$ would be identical, absurd. Since $p \\mid w$ and $1 < p < w$, we have $2p \\leq w \\leq p+q+1$, and so $p \\leq q+1$; similarly we have $q \\leq p+1$.\nIf $p=q$, then $|C A| \\leq p+p-\\operatorname{gcd}(p, p)-1 = p-1$, so $2p \\leq w \\leq p+2$, implying $p \\leq 2$. But the three-letter suffix $a c b$ is not periodic (not even for $c=a$ or $c=b$), thus must be contained in $Q'$, forcing $q \\geq 3$, contradiction.\nIf $p \\neq q$, then $\\max(p, q) = \\min(p, q) + 1$, so $3 \\min(p, q) \\leq w \\leq 2 \\min(p, q) + 2$, hence $\\min(p, q) \\leq 2$, forcing $\\min(p, q) = 2$ and $\\max(p, q) = 3$; by an above observation, we may even say $q = 3$ and $p = 2$, leading to $c = b$. It follows $6 = 3 \\min(p, q) \\leq w \\leq 2 \\min(p, q) + 2 = 6$, forcing $w = 6$. This leads to $C A = a b a = a b b$, contradiction.\nSolution:\nWe will take over from the solution above, just before invoking the Wilf-Fine Theorem, by replacing it with a weaker lemma, also built upon a seminal result of combinatorics on words.\nLemma. Let $p, q$ be positive integers, and let $N$ be a word of length $n$, which is both $p$-periodic and $q$-periodic. If $n \\geq p+q$ then the word $N$ is $\\operatorname{gcd}(p, q)$-periodic.\nProof. Let us first prove that two not-null words $U, V$ commute, i.e. $UV = VU$, if and only if there exists a word $W$ with $|W| = \\operatorname{gcd}(|U|, |V|)$, such that $U = W^{|U| / |W|}, V = W^{|V| / |W|}$. The \"if\" part being trivial, we will prove the \"only if\" part, by strong induction on $|U| + |V|$. Indeed, for the base step $|U| + |V| = 2$ we have $|U| = |V| = 1$, and so clearly we can take $W = U = V$. Now, for $|U| + |V| > 2$, if $|U| = |V|$ it follows $U = V$, and so we can again take $W = U = V$. If not, assume without loss of generality $|U| < |V|$; then $V = U V'$, so $U U V' = U V' U$, whence $U V' = V' U$. Since $|V'| < |V|$, it follows $2 \\leq |U| + |V'| < |U| + |V|$, so by the induction hypothesis there exists a suitable word $W$ such that $U = W^{|U| / |W|}, V' = W^{|V'| / |W|}$, so $V = U V' = W^{|U| / |W|} W^{|V'| / |W|} = W^{(|U| + |V'|) / |W|} = W^{|V| / |W|}$.\nNow, assuming without loss of generality $p \\leq q, q = k p + r$, we have $N = Q P S$, with $|Q| = q, |P| = p$. If $r = 0$ all is clear; otherwise it follows we can write $P = U V, Q = V (U V)^k$, with $|V| = r$, whence $U V = V U$, implying $P Q = Q P$, and so by the above result there will exist a word $W$ of length $\\operatorname{gcd}(p, q)$ such that $P = W^{p / \\operatorname{gcd}(p, q)}$, $Q = W^{q / \\operatorname{gcd}(p, q)}$, therefore $N$ is $\\operatorname{gcd}(p, q)$-periodic.\nBy this we need $|C A| \\leq p+q-1$, hence $w \\leq p+q+2$, otherwise by the previous lemma $W_0'$ and $W_0''$ would be identical, absurd. Since $p \\mid w$ and $1 < p < w$, we have $2p \\leq w \\leq p+q+2$, and so $p \\leq q+2$; similarly we have $q \\leq p+2$. That implies $\\max(p, q) \\leq \\min(p, q) + 2$. Now, from $k \\max(p, q) = w \\leq p+q+2 \\leq 2 \\max(p, q) + 2$ we will have $(k-2) \\max(p, q) \\leq 2$; but $\\max(p, q) \\leq 2$ is impossible, since the three-letter suffix $a c b$ is not periodic (not even for $c=a$ or $c=b$), thus must be contained in $Q'$, forcing $q \\geq 3$. Therefore $k = 2$, and so $w = 2 \\max(p, q)$.\nIf $\\max(p, q) = \\min(p, q)$, then $w = 2p = 2q$, for a quick contradiction.\nIf $\\max(p, q) = \\min(p, q) + 1$, it follows $3 \\min(p, q) \\leq w = 2 \\max(p, q) = 2 \\min(p, q) + 2$, hence $\\min(p, q) \\leq 2$, forcing $\\min(p, q) = 2$ and $\\max(p, q) = 3$; by an above observation, we may even say $q = 3$ and $p = 2$, leading to $c = b$. It follows $w = 2 \\max(p, q) = 6$, leading to $C A = a b a = a b b$, contradiction.\nIf $\\max(p, q) = \\min(p, q) + 2$, it follows $3 \\min(p, q) \\leq w = 2 \\max(p, q) = 2 \\min(p, q) + 4$, hence $\\min(p, q) \\leq 4$. From $\\min(p, q) \\mid w = 2 \\max(p, q)$ then follows either $\\min(p, q) = 2$ and $\\max(p, q) = 4$, thus $w = 8$, clearly contradictory, or else $\\min(p, q) = 4$ and $\\max(p, q) = 6$, thus $w = 12$, which also leads to contradiction, by just a little deeper analysis.\nSolution:\nWe define the distance between two words of the same length to be the number of positions in which those two words have different letters. Any two words related by a transposition have distance 0 or 2; any two words related by a sequence of two transpositions have distance $0, 2, 3$ or $4$.\nSay the period of a repetitive word is the least $k$ such that the word is the concatenation of two or more identical subwords of length $k$. We use the following lemma on distances between repetitive words.\nLemma. Consider a pair of distinct, nonconstant repetitive words with periods $g a$ and $g b$, where $(a, b) = 1$ and $a, b > 1$, the first word is made up of $k b$ repetitions of the subword of length $g a$ and the second word is made up of $k a$ repetitions of the subword of length $g b$. These two words have distance at least $\\max(k a, k b)$.\nProof. We may assume $k = 1$, since the distance between the words is $k$ times the distance between their initial subwords of length $g a b$. Without loss of generality suppose $b > a$.\nFor each positive integer $m$, look at the subsequence in each word of letters in positions congruent to $m$ $(\\bmod g)$. Those subsequences (of length $a b$) have periods dividing $a$ and $b$ respectively. If they are equal, then they are constant (since each letter is equal to those $a$ and $b$ before and after it, mod $a b$, and $(a, b) = 1$). Because $a > 1$, there is some $m$ for which the first subsequence is not constant, and so is unequal to the second subsequence. Restrict attention to those subsequences.\nWe now have two distinct repetitive words, one (nonconstant) made up of $b$ repetitions of a subword of length $a$ and one made up of $a$ repetitions of a subword of length $b$. Looking at the first of those words, for any $1 \\leq t \\leq b$ consider the letters in positions $t, t+b, \\ldots, t+(a-1) b$. These letters cover every position (mod $a$); since the first word is not constant, the letters are not all equal, but the letters in the corresponding positions in the second word are all equal. At least one of these letters in the first word must change to make them all equal to those in the corresponding positions in the second word; repeating for each $t$, at least $b$ letters must change, so the words have distance at least $b$.\nIn the original problem, consider all the words (which we suppose to be repetitive) obtained by a transposition of two adjacent letters from the original nonconstant word; say that word has length $n$. Suppose those words include two distinct words with periods $n / a$ and $n / b$; those words have distance at most $4$. If $a > 4$ or $b > 4$, we have a contradiction unless $a \\mid b$ or $b \\mid a$. If $a > 4$ is the greatest number of repetitions in any of the words ($n / a$ is the smallest period), then unless all the numbers of repetitions divide each other there must be words with $2$ or $4$ repetitions, words with $3$ repetitions and all larger numbers of repetitions must divide each other and be divisible by $6$.\nWe now divide into three cases: all the numbers of repetitions may divide either other; or there may be words with (multiples of) $2, 3$ and $6$ repetitions; or all words may have at most $4$ repetitions, with at least one word having $3$ repetitions and at least one having $2$ or $4$ repetitions.\n\nCase 1. Suppose all the numbers of repetitions divide each other. Let $k$ be the least number of repetitions. Consider the word as being divided into $k$ blocks, each of $\\ell$ letters; any transposition of two adjacent letters leaves those blocks identical. If any two adjacent letters within a block are the same, then this means all the blocks are already identical; since the word is not constant, the letters in the first block are not all identical, so there are two distinct adjacent letters in the first block, and transposing them leaves it distinct from the other blocks, a contradiction. Otherwise, all pairs of adjacent letters within each block are distinct; transposing any adjacent pair within the first block leaves it identical to the second block. If the first block has more than two letters, this is impossible since transposing the first two letters has a different result from transposing the second two. So the blocks all have length $2$; similarly, there are just two blocks, the arrangement is $a b b a$ but transposing the adjacent letters $b b$ does not leave the word repetitive.\n\nCase 2. Suppose some word resulting from a transposition is made of (a multiple of) $6$ repetitions, some of $3$ repetitions and some of $2$ repetitions (or $4$ repetitions, counted as $2$). Consider it as a sequence of $6$ blocks, each of length $\\ell$. If the six blocks are already identical, then as the word is not constant, there are some two distinct adjacent letters within the first block; transposing them leaves a result where the blocks form a pattern $B A A A A A$, which cannot have two, three or six repetitions. So the six blocks are not already identical. If a transposition within a block results in them being identical, the blocks form a pattern (without loss of generality) $B A A A A A, A B A A A A$ or $A A B A A A$. In any of these cases, apply the same transposition (that converts between $A$ and $B$) to an $A$ block adjacent to the $B$ block, and the result cannot have two, three or six repetitions. Finally, consider the case where some transposition between two adjacent blocks results in all six blocks being identical. The patterns are $B C A A A A, A B C A A A$ and $A A B C A A$ (and considering the letters at the start and end of each block shows $B \\neq C$). In all cases, transposing two adjacent distinct letters within an $A$ block produces a result that cannot have two, three or six repetitions.\n\nCase 3. In the remaining case, all words have at most $4$ repetitions, at least one has $3$ repetitions and at least one has $2$ or $4$ repetitions. For the purposes of this case we will think of $4$-repetition words as being $2$-repetition words. The number of each letter is a multiple of $6$, so $n \\geq 12$; consider the word as made of six blocks of length $\\ell$.\nIf the word is already repetitive with $2$ repetitions, pattern $A B C A B C$, any transposition between two distinct letters leaves it no longer repetitive with two repetitions, so it must instead have three repetitions after the transposition. If $A B$ is not all one letter, transposing two adjacent letters within $A B$ implies that $C A = B C$, so $A = B = C$, the word has pattern $A A A A A A$ but transposing within the initial $A A$ means it no longer has $3$ repetitions. This implies that $A B$ is all one letter, but similarly $B C$ must also be all one letter and so the word is constant, a contradiction.\nIf the word is already repetitive with $3$ repetitions, it has pattern $A B A B A B$ and any transposition leaves it no longer having $3$ repetitions, so having $2$ repetitions instead. $A B A$ is not made all of one letter (since the word is not constant) and any transposition between two adjacent distinct letters therein turns it into $B A B$; such a transposition affects at most two of the blocks, so $A = B$, the word has pattern $A A A A A A$ and transposing two adjacent distinct letters within the first half cannot leave it with two repetitions.\nSo the word is not already repetitive, and so no two adjacent letters are the same; all transpositions give distinct strings. Consider transpositions of adjacent letters within the first four letters; three different words result, of which at most one is periodic with two repetitions (it must be made of two copies of the second half of the word) and at most one is periodic with three repetitions, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe side $BC$ of the triangle $ABC$ is extended beyond $C$ to $D$ so that $CD = BC$. The side $CA$ is extended beyond $A$ to $E$ so that $AE = 2CA$.\nProve that if $AD = BE$, then the triangle $ABC$ is right-angled.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDefine $F$ so that $ABFD$ is a parallelogram. Then $E, A, C, F$ are collinear (as diagonals of a parallelogram bisect each other) and $BF = AD = BE$. Further, $A$ is the midpoint of $EF$, since $AF = 2AC$, and thus $AB$ is an altitude of the isosceles triangle $EBF$ with apex $B$. Therefore $AB \\perp AC$.\n\n![](attached_image_1.png)\n\nA Variant. Let $P$ be the midpoint of $[AE]$, so that $AP = AB$ because $AE = 2AB$. Let $Q$ be the midpoint of $[AB]$. Then $PQ = \\frac{1}{2} BE = \\frac{1}{2} AD = CQ$. Hence $PA$ is a median of the isosceles triangle $CPQ$. In other words, $PA \\perp AB$, which completes the proof.\n\n![](attached_image_2.png)\nSolution:\n\nNotice that $A$ is the centroid of triangle $BDE$, since $C$ is the midpoint of $[BD]$ and $AE = 2CA$. Let $M$ be the midpoint of $[BE]$. Then $M, A, D$ lie on a line, and further, $AM = \\frac{1}{2} AD = \\frac{1}{2} BE$. This implies that $\\angle EAB = 90^{\\circ}$.\n\n![](attached_image_3.png)\nSolution:\n\nLet $P$ be the midpoint $[AE]$. Since $C$ is the midpoint of $[BD]$, and, moreover, $AC = EP$, we have\n$$\n[ACD] = [ABC] = [EBP]\n$$\nBut $AD = BE$, and, as mentioned previously, $AC = EP$, so this implies that\n$$\n\\angle BEP = \\angle CAD \\text{ or } \\angle BEP = 180^{\\circ} - \\angle CAD.\n$$\nBut $\\angle BEP < \\angle BAC$ and $\\angle BAC + \\angle CAD = \\angle BAD < 180^{\\circ}$, so we must be in the first case, i.e. $\\angle BEP = \\angle CAD$. It follows that triangles $BEP$ and $DAC$ are congruent, and thus $\\angle BPA = \\angle ACB$. But $AP = AC$, so $BA$ is a median of the isosceles triangle $BCP$. Thus $AB \\perp PC$, completing the proof.\n\n![](attached_image_4.png)\nSolution:\n\nWrite $\\beta = \\angle ECB$, and let $x = AC$, $y = BC = CD$, $z = BE = AD$. Notice that $EC = 3x$. Then, using the cosine theorem,\n$$\n\\begin{array}{ll}\nz^{2} = x^{2} + y^{2} + 2xy \\cos \\beta & \\text{ in triangle } ACD; \\\\\nz^{2} = 9x^{2} + y^{2} - 6xy \\cos \\beta & \\text{ in triangle } BCE.\n\\end{array}\n$$\nHence $4z^{2} = 12x^{2} + 4y^{2}$ or $z^{2} - y^{2} = 3x^{2}$. Let $H$ be the foot of the perpendicular through $B$ to $AC$, and write $h = BH$. Then\n$$\ny^{2} - h^{2} = CH^{2}, \\quad z^{2} - h^{2} = EH^{2}\n$$\nHence $z^{2} - y^{2} = EH^{2} - CH^{2}$. Substituting from the above,\n$$\nEH^{2} - CH^{2} = 3x^{2} = EA^{2} - CA^{2}.\n$$\nThus $H = A$, and hence the triangle $ABC$ is right-angled at $A$.\nSolution:\n\nWriting $a = BC$, $b = CA$, $c = AB$, we have\n$$\n\\left.\\begin{array}{rlrl}\na^{2} & = b^{2} + c^{2} - 2bc \\cos \\angle A \\\\\nc^{2} & = a^{2} + b^{2} - 2ab \\cos \\angle C\n\\end{array}\\right\\} \\quad \\begin{array}{ll}\n\\text{ in triangle } ABC; \\\\\nEB^{2} & = 4b^{2} + c^{2} + 4bc \\cos \\angle A \\\\\nAD^{2} & = a^{2} + b^{2} + 2ab \\cos \\angle C\n\\end{array} \\quad \\begin{aligned}\n& \\text{ in triangle } AEB; \\\\\n& \\text{ in triangle } ACD.\n\\end{aligned}\n$$\nThus\n$$\n\\begin{aligned}\n6b^{2} + 3c^{2} - 2a^{2} & = 4b^{2} + c^{2} + 4bc \\cos \\angle A = EB^{2} = AD^{2} \\\\\n& = a^{2} + b^{2} + 2ab \\cos \\angle C = 2a^{2} + 2b^{2} - c^{2}\n\\end{aligned}\n$$\nwhich gives $a^{2} = b^{2} + c^{2}$. Therefore $\\angle BAC$ is a right angle by the converse of the theorem of Pythagoras.\nSolution:\n\nLet $\\overrightarrow{AC} = \\vec{x}$ and $\\overrightarrow{AB} = \\vec{y}$. Now $\\overrightarrow{AD} = 2\\vec{x} - \\vec{y}$ and $\\overrightarrow{EB} = 2\\vec{x} + \\vec{y}$. Then\n$$\n\\overrightarrow{BE} \\cdot \\overrightarrow{BE} = \\overrightarrow{AD} \\cdot \\overrightarrow{AD} \\quad \\Longleftrightarrow \\quad (2\\vec{x} + \\vec{y})^{2} = (2\\vec{x} - \\vec{y})^{2} \\quad \\Longleftrightarrow \\quad \\vec{x} \\cdot \\vec{y} = 0\n$$\nand thus $AC \\perp AB$, whence triangle $ABC$ is right-angled at $A$.\nSolution:\n\nLet $a, b, c, d, e$ denote the complex co-ordinates of the points $A, B, C, D, E$ and take the unit circle to be the circumcircle of $ABC$. We have\n$$\nd = b + 2(c - b) = 2c - b \\quad \\text{ and } \\quad e = c + 3(a - c) = 3a - 2c.\n$$\nThus $b - e = (d - a) + 2(b - a)$, and hence\n$$\n\\begin{aligned}\nBE = AD & \\Longleftrightarrow (b - e)(\\overline{b - e}) = (d - a)(\\overline{d - a}) \\\\\n& \\Longleftrightarrow 2(d - a)(\\overline{b - a}) + 2(\\overline{d - a})(b - a) + 4(b - a)(\\overline{b - a}) = 0 \\\\\n& \\Longleftrightarrow 2(d - a)(a - b) + 2(\\overline{d - a})(b - a)ab + 4(b - a)(a - b) = 0 \\\\\n& \\Longleftrightarrow (d - a) - (\\overline{d - a})ab + 2(b - a) = 0 \\\\\n& \\Longleftrightarrow 2c - b - a - 2\\bar{c}ab + a + b + 2(b - a) = 0 \\\\\n& \\Longleftrightarrow c^{2} - ab + bc - ac = 0 \\\\\n& \\Longleftrightarrow (b + c)(c - a) = 0,\n\\end{aligned}\n$$\nimplying $c = -b$ and that triangle $ABC$ is right-angled at $A$.\nSolution:\n\nWe use areal co-ordinates with reference to the triangle $ABC$. Recall that if $\\left(x_{1}, y_{1}, z_{1}\\right)$ and $\\left(x_{2}, y_{2}, z_{2}\\right)$ are points in the plane, then the square of the distance between these two points is $-a^{2}vw - b^{2}wu - c^{2}uv$, where $(u, v, w) = \\left(x_{1} - x_{2}, y_{1} - y_{2}, z_{1} - z_{2}\\right)$.\nIn our case $A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1)$, so $E = (3, 0, 2)$ and, introducing point $F$ as in the first solution, $F = (-1, 0, 2)$. Then\n$$\nBE^{2} = AD^{2} \\quad \\Longleftrightarrow \\quad -2a^{2} + 6b^{2} + 3c^{2} = 2a^{2} + 2b^{2} - c^{2},\n$$\nand thus $a^{2} = b^{2} + c^{2}$. Therefore $\\angle BAC$ is a right angle by the converse of the theorem of Pythagoras.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all integers $m$ for which the $m \\times m$ square can be dissected into five rectangles, the side lengths of which are the integers $1,2,3, \\ldots, 10$ in some order.",
    "options": [],
    "answer": "11 and 13",
    "solution": "Solution:\nThe solution naturally divides into three different parts: we first obtain some bounds on $m$. We then describe the structure of possible dissections, and finally, we deal with the few remaining cases.\n\nIn the first part of the solution, we get rid of the cases with $m \\leqslant 10$ or $m \\geqslant 14$. Let $\\ell_{1}, \\ldots, \\ell_{5}$ and $w_{1}, \\ldots, w_{5}$ be the lengths and widths of the five rectangles. Then the rearrangement inequality yields the lower bound\n$$\n\\begin{aligned}\n\\ell_{1} w_{1} & +\\ell_{2} w_{2}+\\ell_{3} w_{3}+\\ell_{4} w_{4}+\\ell_{5} w_{5} \\\\\n& =\\frac{1}{2}\\left(\\ell_{1} w_{1}+\\ell_{2} w_{2}+\\ell_{3} w_{3}+\\ell_{4} w_{4}+\\ell_{5} w_{5}+w_{1} \\ell_{1}+w_{2} \\ell_{2}+w_{3} \\ell_{3}+w_{4} \\ell_{4}+w_{5} \\ell_{5}\\right) \\\\\n& \\geqslant \\frac{1}{2}(1 \\cdot 10+2 \\cdot 9+3 \\cdot 8+\\cdots+8 \\cdot 3+9 \\cdot 2+10 \\cdot 1)=110\n\\end{aligned}\n$$\nand the upper bound\n$$\n\\begin{aligned}\n\\ell_{1} w_{1} & +\\ell_{2} w_{2}+\\ell_{3} w_{3}+\\ell_{4} w_{4}+\\ell_{5} w_{5} \\\\\n& =\\frac{1}{2}\\left(\\ell_{1} w_{1}+\\ell_{2} w_{2}+\\ell_{3} w_{3}+\\ell_{4} w_{4}+\\ell_{5} w_{5}+w_{1} \\ell_{1}+w_{2} \\ell_{2}+w_{3} \\ell_{3}+w_{4} \\ell_{4}+w_{5} \\ell_{5}\\right) \\\\\n& \\leqslant \\frac{1}{2}(1 \\cdot 1+2 \\cdot 2+3 \\cdot 3+\\cdots+8 \\cdot 8+9 \\cdot 9+10 \\cdot 10)=192.5\n\\end{aligned}\n$$\nAs the area of the square is sandwiched between $110$ and $192.5$, the only possible candidates for $m$ are $11,12$, and $13$.\n\nIn the second part of the solution, we show that a dissection of the square into five rectangles must consist of a single inner rectangle and four outer rectangles that each cover one of the four corners of the square. Indeed, if one of the sides of the square had three rectangles adjacent to it, removing these three rectangles would leave a polygon with eight vertices, which is clearly not the union of two rectangles. Moreover, since $m>10$, each side of the square has at least two adjacent rectangles. Hence each side of the square has precisely two adjacent rectangles, and thus the only way of partitioning the square into five rectangles is to have a single inner rectangle and four outer rectangles each covering one of the four corners of the square, as claimed.\n\nLet us now show that a square of size $12 \\times 12$ cannot be dissected in the desired way. Let $R_{1}, R_{2}, R_{3}$ and $R_{4}$ be the outer rectangles (in clockwise orientation along the boundary of the square). If an outer rectangle has a side of length $s$, then some adjacent outer rectangle must have a side of length $12-s$. Therefore, neither of $s=1$ or $s=6$ can be sidelengths of an outer rectangle, so the inner rectangle must have dimensions $1 \\times 6$. One of the outer rectangles (say $R_{1}$) must have dimensions $10 \\times x$, and an adjacent rectangle (say $R_{2}$) must thus have dimensions $2 \\times y$. Rectangle $R_{3}$ then has dimensions $(12-y) \\times z$, and rectangle $R_{4}$ has dimensions $(12-z) \\times (12-x)$. Note that exactly one of the three numbers $x, y, z$ is even (and equals $4$ or $8$), while the other two numbers are odd. Now, the total area of all five rectangles is\n$$\n144=6+10x+2y+(12-y)z+(12-z)(12-x),\n$$\nwhich simplifies to $(y-x)(z-2)=6$. As exactly one of the three numbers $x, y, z$ is even, the factors $y-x$ and $z-2$ are either both even or both odd, so their product cannot equal $6$, and thus there is no solution with $m=12$.\n\nFinally, we handle the cases $m=11$ and $m=13$, which indeed are solutions. The corresponding rectangle sets are $10 \\times 5, 1 \\times 9, 8 \\times 2, 7 \\times 4$ and $3 \\times 6$ for $m=11$, and $10 \\times 5, 9 \\times 8, 4 \\times 6, 3 \\times 7$ and $1 \\times 2$ for $m=13$. These sets can be found by trial and error. The corresponding partitions are shown in the figure below.\n\n![](attached_image_1.png)\n\nA Variant for Obtaining Bounds. We first exclude the cases $m \\leqslant 9$ by the observation that one of the small rectangles has a side of length $10$ and must fit into the square; hence $m \\geqslant 10$.\nTo exclude the cases $m \\geqslant 14$, we work via the perimeter: as every rectangle has at most two sides on the boundary of the $m \\times m$ square, the perimeter $4m$ of the square is bounded by $1+2+3+4+5+6+7+8+9+10=55$; hence $m \\leqslant 13$.\n\nWe are left to deal with the case $m=10$: clearly, the rectangle with side length $10$ must have one of its sides of length $10$ along the boundary of the square. The remaining rectangle $R$ of dimensions $10 \\times s$, say, would have to be divided into four rectangles with different sidelengths strictly less than $10$. If there were at least two rectangles adjacent to one of the sides of length $s$ of $R$, removing these two rectangles from $R$ would leave a polygon with at least six vertices (since the sidelengths of the rectangles partitioning $R$ are strictly less than $10$). It is clearly impossible to partition such a polygon into no more than two rectangles with different sidelengths. Hence, given a side of length $s$ of $R$, there is only one rectangle adjacent to that side, so the rectangles adjacent to the sides of length $s$ of $R$ would have to have the same length $s$, a contradiction.\n\nA Variant for Dealing with $\\boldsymbol{m}=12$. As in the previous solution, we show that the inner rectangle must have dimensions $1 \\times 6$. Since the area of the square and the area of the inner rectangle are even, the areas of the four outer rectangles must sum to an even number. Now the four sides of the square are divided into segments of lengths $2$ and $10$, $3$ and $9$, $4$ and $8$, and $5$ and $7$. Hence the sides with adjacent segments of lengths $3$ and $9$, and $5$ and $7$ must be opposite sides of the square (otherwise, exactly one of the outer rectangles would have odd area). However, the difference of two rectangle side lengths on opposite sides of the square must be $1$ or $6$ (in order to accommodate the inner rectangle). This is not the case, so there is no solution with $m=12$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer.\n\na. Prove that there exists a set $S$ of $6 n$ pairwise different positive integers, such that the least common multiple of any two elements of $S$ is no larger than $32 n^{2}$.\n\nb. Prove that every set $T$ of $6 n$ pairwise different positive integers contains two elements the least common multiple of which is larger than $9 n^{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nLet the set $A$ consist of the $4 n$ integers $1, 2, \\ldots, 4 n$ and let the set $B$ consist of the $2 n$ even integers $4 n + 2, 4 n + 4, \\ldots, 8 n$. We claim that the $6 n$-element set $S = A \\cup B$ has the desired property.\n\nIndeed, the least common multiple of two (even) elements of $B$ is no larger than $8 n \\cdot (8 n / 2) = 32 n^{2}$, and the least common multiple of some element of $A$ and some element of $A \\cup B$ is at most their product, which is at most $4 n \\cdot 8 n = 32 n^{2}$.\n\nb.\nWe prove the following lemma: \"If a set $U$ contains $m + 1$ integers, where $m \\geqslant 2$, that are all not less than $m$, then some two of its elements have least common multiple strictly larger than $m^{2}$.\"\n\nLet the elements of $U$ be $u_{1} > u_{2} > \\cdots > u_{m+1} \\geqslant m$. Note that $1 / u_{1} \\leqslant 1 / u_{i} \\leqslant 1 / m$ for $1 \\leqslant i \\leqslant m+1$. We partition the interval $[1 / u_{1} ; 1 / m]$ into $m$ subintervals of equal length. By the pigeonhole principle, there exist indices $i, j$ with $1 \\leqslant i < j \\leqslant m+1$ such that $1 / u_{i}$ and $1 / u_{j}$ belong to the same subinterval. Hence\n$$\n0 < \\frac{1}{u_{j}} - \\frac{1}{u_{i}} \\leqslant \\frac{1}{m} \\left( \\frac{1}{m} - \\frac{1}{u_{1}} \\right) < \\frac{1}{m^{2}}.\n$$\nNow $1 / u_{j} - 1 / u_{i}$ is a positive fraction with denominator $\\operatorname{lcm}(u_{i}, u_{j})$. The above thus yields the lower bound $\\operatorname{lcm}(u_{i}, u_{j}) > m^{2}$, completing the proof of the lemma.\n\nApplying the lemma with $m = 3 n$ to the $3 n + 1$ largest elements of $T$, which are all not less than $3 n$, we arrive at the desired statement.\n\n\nA Variant.\n\nAlternatively, for part (b), we prove the following lemma: \"If a set $U$ contains $m \\geqslant 2$ integers that all are greater than $m$, then some two of its elements have least common multiple strictly larger than $m^{2}$.\"\n\nLet $u_{1} > u_{2} > \\cdots > u_{m}$ be the elements of $U$. Since $u_{m} > m = m^{2} / m$, there exists a smallest index $k$ such that $u_{k} > m^{2} / k$. If $k = 1$, then $u_{1} > m^{2}$, and the least common multiple of $u_{1}$ and $u_{2}$ is strictly larger than $m^{2}$. So let us suppose $k > 1$ from now on, so that we have $u_{k} > m^{2} / k$ and $u_{k-1} \\leqslant m^{2} / (k-1)$. The greatest common divisor $d$ of $u_{k-1}$ and $u_{k}$ satisfies\n$$\nd \\leqslant u_{k-1} - u_{k} < \\frac{m^{2}}{k-1} - \\frac{m^{2}}{k} = \\frac{m^{2}}{(k-1)k}\n$$\nThis implies $m^{2} / (d k) > k-1$ and $u_{k} / d > k-1$, and hence $u_{k} / d \\geqslant k$. But then the least common multiple of $u_{k-1}$ and $u_{k}$ equals\n$$\n\\frac{u_{k-1} u_{k}}{d} \\geqslant u_{k} \\cdot \\frac{u_{k}}{d} > \\frac{m^{2}}{k} \\cdot k = m^{2}\n$$\nand the proof of the lemma is complete.\n\nIf we remove the $3 n$ smallest elements from set $T$ and apply the lemma with $m = 3 n$ to the remaining elements, we arrive at the desired statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $a$ and $b$ for which there are three consecutive integers at which the polynomial\n$$\nP(n)=\\frac{n^{5}+a}{b}\n$$\ntakes integer values.",
    "options": [],
    "answer": "b = 1 with any positive integer a; or b = 11 with a congruent to plus or minus 1 modulo 11",
    "solution": "Solution:\nDenote the three consecutive integers by $x-1, x$, and $x+1$, so that\n$$\n(x-1)^{5}+a \\equiv 0 \\quad(\\bmod b), \\quad x^{5}+a \\equiv 0 \\quad(\\bmod b), \\quad(x+1)^{5}+a \\equiv 0 \\quad(\\bmod b) .\n$$\nBy computing the differences of the equations in (1) we get\n$$\n\\begin{aligned}\nA & :=(x+1)^{5}-(x-1)^{5}=10 x^{4}+20 x^{2}+2 \\equiv 0 \\quad(\\bmod b) \\\\\nB & :=(x+1)^{5}-x^{5}=5 x^{4}+10 x^{3}+10 x^{2}+5 x+1 \\equiv 0 \\quad(\\bmod b)\n\\end{aligned}\n$$\nAdding the first and third equation in (1) and subtracting twice the second equation yields\n$$\nC:=(x+1)^{5}+(x-1)^{5}-2 x^{5}=20 x^{3}+10 x \\equiv 0 \\quad(\\bmod b) .\n$$\nNext, (2) and (4) together yield\n$$\nD:=4 x A-\\left(2 x^{2}+3\\right) C=-22 x \\equiv 0(\\bmod b) .\n$$\nFinally we combine (3) and (5) to derive\n$$\n22 B+\\left(5 x^{3}+10 x^{2}+10 x+5\\right) D=22 \\equiv 0(\\bmod b) .\n$$\nAs the positive integer $b$ divides 22 , we are left with the four cases $b=1, b=2, b=11$ and $b=22$.\nIf $b$ is even (i.e. $b=2$ or $b=22$ ), then we get a contradiction from (3), because the integer $B=2\\left(5 x^{3}+5 x^{2}\\right)+5\\left(x^{4}+x\\right)+1$ is odd, and hence not divisible by any even integer.\nFor $b=1$, it is trivial to see that a polynomial of the form $P(n)=n^{5}+a$, with $a$ any positive integer, has the desired property.\nFor $b=11$, we note that\n$$\n\\begin{aligned}\n& n \\equiv 0,1,2,3,4,5,6,7,8,9,10 \\quad(\\bmod 11) \\\\\n& \\Longrightarrow \\quad n^{5} \\equiv 0,1,-1,1,1,1,-1,-1,-1,1,-1 \\quad(\\bmod 11)\n\\end{aligned}\n$$\nHence a polynomial of the form $P(n)=\\left(n^{5}+a\\right) / 11$ has the desired property if and only if $a \\equiv \\pm 1(\\bmod 11)$. This completes the proof.\n\nA Variant. We start by following the first solution up to equation (4). We note that $b=1$ is a trivial solution, and assume from now on that $b \\geqslant 2$. As $(x-1)^{5}+a$ and $x^{5}+a$ have different parity, $b$ must be odd. As $B$ in (3) is a multiple of $b$, we conclude that (i) $b$ is not divisible by 5 and that (ii) $b$ and $x$ are relatively prime. As $C=10 x\\left(2 x^{2}+1\\right)$ in (4) is divisible by $b$, we altogether derive\n$$\nE:=2 x^{2}+1 \\equiv 0(\\bmod b)\n$$\nTogether with (2) this implies that\n$$\n5 E^{2}+10 E-2 A=11 \\equiv 0 \\quad(\\bmod b)\n$$\nHence $b=11$ is the only remaining candidate, and it is handled as in the first solution.\nSolution:\nLet $p$ be a prime such that $p$ divides $b$. For some integer $x$, we have\n$$\n(x-1)^{5} \\equiv x^{5} \\equiv(x+1)^{5} \\quad(\\bmod p)\n$$\nNow, there is a primitive root $g$ modulo $p$, so there exist $u, v, w$ such that\n$$\nx-1 \\equiv g^{u} \\quad(\\bmod p), \\quad x \\equiv g^{v} \\quad(\\bmod p), \\quad x+1 \\equiv g^{w} \\quad(\\bmod p)\n$$\nThe condition of the problem is thus\n$$\ng^{5 u} \\equiv g^{5 v} \\equiv g^{5 w} \\quad(\\bmod p) \\quad \\Longrightarrow \\quad 5 u \\equiv 5 v \\equiv 5 w \\quad(\\bmod p-1)\n$$\nIf $p \\not \\equiv 1(\\bmod 5)$, then 5 is invertible modulo $p-1$ and thus $u \\equiv v \\equiv w(\\bmod p-1)$, i.e. $x-1 \\equiv x \\equiv x+1(\\bmod p)$. This is a contradiction. Hence $p \\equiv 1(\\bmod 5)$ and thus $u \\equiv v \\equiv w\\left(\\bmod \\frac{p-1}{5}\\right)$. Thus, from (6), there exist integers $k, \\ell$ such that\n$$\n\\left.\\begin{array}{rl}\nx-1 & \\equiv g^{v+k \\frac{p-1}{5}} \\equiv x t^{k} \\quad(\\bmod p) \\\\\nx+1 & \\equiv g^{v+\\ell \\frac{p-1}{5}} \\\\\n\\equiv x t^{\\ell} & (\\bmod p)\n\\end{array}\\right\\} \\quad \\text { where } t=g^{\\frac{p-1}{5}}\n$$\nLet $r=t^{k}$ and $s=t^{\\ell}$. In particular, the above yields $r, s \\not \\equiv 1(\\bmod p)$, and thus\n$$\nx \\equiv-(r-1)^{-1} \\equiv(s-1)^{-1} \\quad(\\bmod p)\n$$\nIt follows that\n$$\n(r-1)^{-1}+(s-1)^{-1} \\equiv 0 \\quad(\\bmod p) \\quad \\Longrightarrow \\quad r+s \\equiv 2 \\quad(\\bmod p)\n$$\nNow $t^{5} \\equiv 1(\\bmod p)$, so $r$ and $s$ must be congruent, modulo $p$, to some of the non-trivial fifth roots of unity $t, t^{2}, t^{3}, t^{4}$. Observe that, for any pair of these non-trivial roots of unity,\neither one is the other's inverse, or one is the other's square. In the first case, we have $r+r^{-1} \\equiv 2(\\bmod p)$, implying $r \\equiv 1(\\bmod p)$, a contradiction. Hence\n$$\nr+r^{2} \\equiv 2 \\quad(\\bmod p) \\quad \\Longrightarrow \\quad(r-1)(r+2) \\equiv 0 \\quad(\\bmod p),\n$$\nor\n$$\ns^{2}+s \\equiv 2 \\quad(\\bmod p) \\quad \\Longrightarrow \\quad(s-1)(s+2) \\equiv 0 \\quad(\\bmod p)\n$$\nThus, since $r, s \\not \\equiv 1(\\bmod p)$, we have $r \\equiv-2(\\bmod p)$ or $s \\equiv-2(\\bmod p)$, and thus $1 \\equiv r^{5} \\equiv-32(\\bmod p)$ or an analogous equation obtained from $s$. Hence $p \\mid 33$. Since $p \\equiv 1(\\bmod 5)$, it follows that $p=11$, i.e. $b$ is a power of 11 .\nExamining the fifth powers modulo 11 , we see that $b=11$ is indeed a solution with $a \\equiv \\pm 1(\\bmod 11)$ and, correspondingly, $x \\equiv \\pm 4(\\bmod 11)$. Now suppose, for the sake of contradiction, that $11^{2}$ divides $b$. Then, for some integer $m$, we must have\n$$\n(x-1, x, x+1) \\equiv \\pm(3+11 m, 4+11 m, 5+11 m) \\quad(\\bmod 121)\n$$\nand thus, substituting into the condition of the problem,\n$$\n\\begin{aligned}\n3^{5}+55 \\cdot 3^{4} m & \\equiv 4^{5}+55 \\cdot 4^{4} m \\equiv 5^{5}+55 \\cdot 5^{4} m \\quad(\\bmod 121) \\\\\n& \\Longrightarrow \\quad 1-22 m \\equiv 56+44 m \\equiv-21+11 m \\quad(\\bmod 121)\n\\end{aligned}\n$$\nHence $33 m \\equiv 22(\\bmod 121)$ and $33 m \\equiv 44(\\bmod 121)$, so $22 \\equiv 0(\\bmod 121)$, a contradiction. It follows that $b \\mid 11$.\nFinally, we conclude that the positive integers satisfying the original condition are $b=11$, with $a \\equiv \\pm 1(\\bmod 11)$, and $b=1$, for any positive integer $a$.\nSolution:\nDenote the three consecutive integers by $x-1, x$, and $x+1$ as in Solution 1. By computing the differences in (1), we find\n$$\n\\begin{aligned}\n& F:=(x+1)^{5}-x^{5}=5 x^{4}+10 x^{3}+10 x^{2}+5 x+1 \\equiv 0 \\quad(\\bmod b), \\\\\n& G:=x^{5}-(x-1)^{5}=5 x^{4}-10 x^{3}+10 x^{2}-5 x+1 \\equiv 0 \\quad(\\bmod b) .\n\\end{aligned}\n$$\nBy determining the polynomial greatest divisor of $F(x)$ and $G(x)$ using the Euclidean algorithm, we find that\n$$\np(x) F(x)+q(x) G(x)=22\n$$\nwhere\n$$\n\\begin{aligned}\n& p(x)=-15 x^{3}+30 x^{2}-28 x+11 \\\\\n& q(x)=15 x^{3}+30 x^{2}+28 x+11\n\\end{aligned}\n$$\nSince $b \\mid F(x)$ and $b \\mid G(x)$, it follows from (7) that $b \\mid 22$. We now finish off the problem as in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 30140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $\\Omega$ be the circumcircle of the triangle $A B C$. The circle $\\omega$ is tangent to the sides $A C$ and $B C$, and it is internally tangent to $\\Omega$ at the point $P$. A line parallel to $A B$ and intersecting the interior of triangle $A B C$ is tangent to $\\omega$ at $Q$.\nProve that $\\angle A C P=\\angle Q C B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAssume that $\\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively and let $P E, P F, P Q$ meet $\\Omega$ at $K, L, M$, respectively. Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$, and consider the homothety $\\mathscr{H}$ that maps $\\omega$ onto $\\Omega$. Now $K$ is the image of $E$ under $\\mathscr{H}$, and $E I \\perp A C$. Hence $O K \\perp A C$, and thus $K$ is the midpoint of the arc $C A$. Similarly, $L$ is the midpoint of the arc $B C$ and $M$ is the midpoint of the arc $B A$. It follows that arcs $L M$ and $C K$ are equal, because\n$$\n\\begin{aligned}\n\\overparen{B M}=\\overparen{M A} & \\Longrightarrow \\overparen{B L}+\\overparen{L M}=\\overparen{M K}+\\overparen{K A} \\Longrightarrow \\overparen{L C}+\\overparen{L M}=\\overparen{M K}+\\overparen{C K} \\\\\n& \\Longrightarrow 2 \\overparen{L M}+\\overparen{M C}=\\overparen{M C}+2 \\overparen{C K}\n\\end{aligned}\n$$\nThus arcs $F Q$ and $D E$ of $\\omega$ are equal, too, where $D$ is the intersection of $C P$ with $\\omega$. Since $C E$ and $C F$ are tangents to $\\omega$, this implies that $\\angle D E C=\\angle C F Q$. Further, $C E=C F$, and thus triangles $C E D$ and $C F Q$ are congruent. In particular, $\\angle E C D=\\angle Q C F$, as required.\n\n![](attached_image_1.png)\n\nA Variant. As above, we show that $\\operatorname{arcs} F Q$ and $D E$ of $\\omega$ are equal, which implies that $D E F Q$ is an isoceles trapezoid, and so we have $\\angle F E D=\\angle Q F E$. Together with $|F Q|=|D E|$, this implies that, since $E$ and $F$ are images of each other under reflection in the angle bisector $C I$ of $\\angle C$, so are the segments $[E Q]$ and $[F D]$, and, in particular, $D$ and $Q$. In turn, this yields $\\angle E C D=\\angle Q C F$, as required.\n\nRemark. Let $J$ denote the incentre of $A B C$. By Sawayama's theorem, $J$ is the midpoint of $[E F]$, i.e. $P J$ is a median of $P F E$. Since $C$ is the intersection of the tangents $A C$ and $B C$ to the circumcircle of $P F E$ at $E$ and $F$, respectively, $P C$ is a symmedian of $P F E$. Thus $\\angle C P E=\\angle F P J$. But, since the arcs $F Q$ and $D E$ of $\\omega$ are equal, $\\angle C P E=\\angle F P Q$. This shows that $J$ lies on the line $P Q$.\n\nAnother Variant. We show that $\\operatorname{arcs} Q E$ and $F D$ are equal, and then finish as in the main solution. Let $B P$ meet $\\omega$ again at $Z$. Consider the homothety $\\mathscr{H}$ that maps $\\omega$ onto $\\Omega$. Under $\\mathscr{H}, D \\mapsto C$ and $Z \\mapsto B$, so $D Z \\parallel C B$. (This also follows by considering the common tangent to $\\omega$ and $\\Omega$, and tangential angles.) Now, by power of a point,\n$$\nB F^{2}=B Z \\cdot B P, \\quad C F^{2}=C D \\cdot C P\n$$\nNow $D Z \\parallel C B$ implies $B Z / B P=C D / C P$, and so, dividing the two previous equations by each other, and taking square roots, $B F / C F=B P / C P$. Hence $P F$ bissects angle $\\angle B P C$. Now let $\\angle B P F=\\angle F P C=\\beta$. By tangential angles, it follows that $\\angle C F D=\\beta$. Further, $\\angle B A C=\\angle B P C=2 \\beta$. Let the tangent to $\\omega$ through $Q$ and parallel to $A B$ meet $A C$ at $X$. Then $\\angle Q X C=2 \\beta$, so, since $X Q=X E$ by tangency, $\\angle Q E X=\\beta$. By tangential angles, it follows that $\\operatorname{arcs} F D$ and $Q E$ are equal, as claimed.\n\n![](attached_image_2.png)\nSolution:\n\nLet $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. Observe that $C I$ is the angle bisector of angle $\\angle C$, because $\\omega$ is tangent to $A C$ and $B C$. Consider the homothety $\\mathscr{H}$ that maps $\\omega$ onto $\\Omega$. Let $M$ be the image of $Q$ under $\\mathscr{H}$. By construction, $I Q \\perp A B$, so $O M \\perp A B$. Thus the diameter $O M$ of $\\Omega$ passes through the midpoint of the $\\operatorname{arc} A B$ of $\\Omega$, which also lies on the angle bisector $C I$. This implies that $\\angle I C M=90^{\\circ}$. We next show that $P, I, Q, C$ lie on a circle. Notice that\n$$\n\\begin{aligned}\n\\angle P Q I & =90^{\\circ}-\\frac{1}{2} \\angle Q I P=90^{\\circ}-\\frac{1}{2} \\angle M O P=90^{\\circ}-\\left(180^{\\circ}-\\angle P C M\\right) \\\\\n& =(\\angle P C I+\\angle I C M)-90^{\\circ}=\\angle P C I\n\\end{aligned}\n$$\nHence $P, I, Q, C$ lie on a circle. But $P I=I Q$, so $C I$ is the angle bisector of $\\angle P C Q$. Since $C I$ is also the angle bisector of angle $\\angle C$, it follows that $\\angle A C P=\\angle Q C B$, as required.\n\n![](attached_image_3.png)\n\nA Variant. We show that $P I Q C$ is cyclic by chasing angles. Define $\\alpha=\\angle B A C$, $\\beta=\\angle C B A$ and $\\gamma=\\angle A C P$. For convenience, we consider the configuration where $A$ and $P$ lie one the same side of the angle bisector $C I$ of $\\angle C$. In this configuration,\n$$\n\\angle P C I=\\frac{1}{2} \\angle A C B-\\angle A C P=90^{\\circ}-\\frac{1}{2} \\alpha-\\frac{1}{2} \\beta-\\gamma .\n$$\nNow notice that $\\angle P B A=\\angle A C P=\\gamma$, and therefore $\\angle C A P=180^{\\circ}-\\beta-\\gamma$, whence $\\angle P A B=180^{\\circ}-\\alpha-\\beta-\\gamma$. Further, $P O$ is a diameter of $\\Omega$, and therefore $\\angle A P O=90^{\\circ}-\\gamma$. Let $A B$ and $P O$ intersect at $T$. Then\n$$\n\\angle B T O=180^{\\circ}-\\angle P A B-\\angle A P O=\\alpha+\\beta+2 \\gamma-90^{\\circ}\n$$\nBut $Q I \\perp A B$ by construction, and thus\n$$\n\\begin{aligned}\n& \\angle O I Q=90^{\\circ}-\\angle B T O=180^{\\circ}-\\alpha-\\beta-2 \\gamma \\\\\n& \\Longrightarrow \\quad \\angle Q I P=180^{\\circ}-\\angle O I Q=\\alpha+\\beta+2 \\gamma \\\\\n& \\Longrightarrow \\quad \\angle P Q I=90^{\\circ}-\\frac{1}{2} \\alpha-\\frac{1}{2} \\beta-\\gamma .\n\\end{aligned}\n$$\nHence $\\angle I C Q=\\angle P Q I$, and thus $P I Q C$ is cyclic. Since $P I=Q I$, it follows that $C I$ is the angle bisector of $\\angle P C Q$, which completes the proof.\nSolution:\n\nLet $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. Let $D$ be the second point of intersection of $C P$ with $\\omega$, and let $\\ell$ denote the tangent to $\\omega$ at $D$, which meets $A C$ at $S$. Hence $I D \\perp \\ell$. By construction, $P, I, O$ lie one a line, and hence the isosceles triangles $P I D$ and $P O C$ are similar. In particular, it follows that $O C \\perp \\ell$, so $C$ is the midpoint of the arc of $\\Omega$ defined by the points of intersection of $\\ell$ with $\\Omega$. It is easy to see that this implies that\n$$\n\\angle D S C=\\angle A B C\n$$\nUnder reflection in the angle bisector $C I$ of $\\angle C, \\ell$ is thus mapped to a tangent to $\\omega$ parallel to $A B$ and intersecting the interior of $A B C$, since $\\omega$ is mapped to itself under this reflection. In particular, $D$ is mapped to $Q$, and thus $\\angle Q C B=\\angle A C D$, as required.\n\n![](attached_image_4.png)\n\nRemark. Conceptually, this solution is similar to Solution 1, but here, we proceed more directly via the reflectional symmetry. Therefore, this solution links Solution 1 to Solution 4, in which we use an inversion.\nSolution:\n\nLet the tangent to $\\omega$ at $Q$ meet $A C$ and $B C$ at $X$ and $Y$, respectively. Then $A C / X C=B C / Y C$, and thus there is a radius $r$ such that $r^{2}=A C \\cdot Y C=B C \\cdot X C$. Let $\\Gamma$ denote the circle with centre $C$ and radius $r$, and consider the inversion $\\mathscr{I}$ in the circle $\\Gamma$. Under $\\mathscr{I}$,\n$A \\longmapsto A^{\\prime}$, the point on the ray $C A$ satisfying $C A^{\\prime}=C Y$;\n$B \\longmapsto B^{\\prime}$, the point on the ray $C B$ satisfying $C B^{\\prime}=C X$;\n$\\Omega \\longmapsto$ the line $A^{\\prime} B^{\\prime}$;\n$\\omega \\longmapsto \\omega^{\\prime}$, the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$;\n$P \\longmapsto P^{\\prime}$, the point where $\\omega^{\\prime}$ touches $A^{\\prime} B^{\\prime}$;\nIn particular, $\\omega^{\\prime}$, the image of $\\omega$, is a circle tangent to $A C, B C$ and $A^{\\prime} B^{\\prime}$, so it is either the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$, or the incircle of $C A^{\\prime} B^{\\prime}$. Let $\\omega$ be tangent to $B C$ at $F$, and let $F^{\\prime}$ be the image of $F$ under $\\mathscr{I}$. Then $C F \\cdot C F^{\\prime}=B C \\cdot X C$. Now $C F<B C$, so $C F^{\\prime}>C X=C B^{\\prime}$. Hence $\\omega^{\\prime}$ cannot be the incircle, so $\\omega^{\\prime}$ is indeed the excircle of $C A^{\\prime} B^{\\prime}$ opposite $C$.\nNow note that $\\omega$ is the excircle of $C X Y$ opposite $C$. The reflection about the angle bisector of $\\angle C$ maps $X$ to $B^{\\prime}, Y$ to $A^{\\prime}$. It thus maps the triangle $C X Y$ to $C B^{\\prime} A^{\\prime}, \\omega$ to $\\omega^{\\prime}$ and, finally, $Q$ to $P^{\\prime}$. It follows that $\\angle A C P=\\angle A C P^{\\prime}=\\angle Q C B$, as required.\n\n![](attached_image_5.png)\nSolution:\n\nLet $r$ be the radius such that $r^{2}=A C \\cdot B C$. Let $\\mathscr{J}$ denote the composition of the inversion $\\mathscr{I}$ in the circle of centre $C$ and radius $r$, followed by the reflection in the angle bisector of $\\angle C$. Under $\\mathscr{J}$,\n$A \\longmapsto B, B \\mapsto A ;$\n$\\Omega \\longmapsto$ the line $A B$;\n$\\omega \\longmapsto \\omega^{\\prime}$, the excircle of $A B C$ opposite the vertex $C$;\n$P \\longmapsto Q^{\\prime}$, the point where $\\omega^{\\prime}$ touches $A B$;\nIn particular, note that the image $\\omega^{\\prime}$ of $\\omega$ under $\\mathscr{J}$ is a circle tangent to $A C, B C$ and $A B$, so it is either the incircle of $A B C$, or the excircle opposite vertex $C$. Observe that $r \\geqslant \\min \\{A C, B C\\}$, so the image of the points of tangency of $\\omega$ must lie outside $A B C$, and thus $\\omega^{\\prime}$ cannot be the incircle. Thus $\\omega^{\\prime}$ is the excircle opposite vertex $C$ as claimed. Further, the point of tangency $P$ is mapped to $Q^{\\prime}$.\nNow, since the line $C P$ is mapped to itself under the inversion $\\mathscr{I}$, and mapped onto $C Q^{\\prime}$ under $\\mathscr{J}, C P$ and $C Q^{\\prime}$ are images of each other under reflection in the angle bisector of $\\angle C$. But $C, Q, Q^{\\prime}$ lie on a line for there is a homothety with centre $C$ that maps $\\omega$ onto the excircle $\\omega^{\\prime}$. This completes the proof.\n\n![](attached_image_6.png)\nSolution:\n\nAssume that $\\omega$ is tangent to $A C$ and $B C$ at $E$ and $F$, respectively. Assume that $C P$ meets $\\omega$ at $D$. Let $I$ and $O$ denote the respective centres of $\\omega$ and $\\Omega$. To set up a solution in the complex plane, we take the circle $\\omega$ as the unit circle centered at the origin of the complex plane, and let $P O$ be the real axis with $o>0$, where we use the convention that lowercase letters denote complex coordinates of corresponding points in the plane denoted by uppercase letters.\nNow, a point $Z$ on the circle $\\Omega$ satisfies\n$$\n|z-o|^{2}=(o+1)^{2} \\quad \\Longleftrightarrow \\quad z z^{*}-o\\left(z+z^{*}\\right)-2 o-1=0\n$$\nThe triangle $A B C$ is defined by the points $E$ and $F$ on $\\omega$, the intersection $C$ of the corresponding tangents lying on $\\Omega$. Thus $c=2 e f /(e+f)$, and further\n$$\n|c-o|^{2}=(o+1)^{2} \\quad \\Longleftrightarrow \\quad c c^{*}-o\\left(c+c^{*}\\right)-2 o-1=0,\n$$\nand this is the equality defining $o$. The points $A$ and $B$ are the second intersection points of $\\Omega$ with the tangents to $\\omega$ at $E$ and $F$ respectively. A point $Z$ on the tangent through $E$ is given by $z=2 e-e^{2} z^{*}$, and thus $A$ and $C$ satisfy\n$$\n\\begin{aligned}\n\\left(2 e-e^{2} z^{*}\\right) z^{*} & -o\\left(2 e-e^{2} z^{*}+z^{*}\\right)-2 o-1=0 \\\\\n& \\Longleftrightarrow-e^{2} z^{* 2}+\\left(2 e+o e^{2}-o\\right) z^{*}-(2 e o+2 o+1)=0 \\\\\n& \\Longleftrightarrow z^{* 2}-\\left(2 e^{*}+o-o e^{* 2}\\right) z^{*}+\\left(2 e^{*} o+2 o e^{* 2}+e^{* 2}\\right)=0\n\\end{aligned}\n$$\nsince $|e|=1$. Thus\n$$\na^{*}+c^{*}=2 e^{*}+o-o e^{* 2} \\quad \\Longrightarrow \\quad a^{*}=\\frac{2 e^{*} f}{e+f}+o\\left(1-e^{* 2}\\right)\n$$\nand similarly\n$$\nb^{*}=\\frac{2 f^{*} e}{f+e}+o\\left(1-f^{* 2}\\right)\n$$\nThen\n$$\n\\begin{aligned}\nb^{*}-a^{*} & =\\frac{2\\left(e f^{*}-e^{*} f\\right)}{e+f}+o\\left(e^{* 2}-f^{* 2}\\right) \\\\\n& =\\frac{2 e f\\left(f^{* 2}-e^{* 2}\\right)}{e+f}+o\\left(e^{* 2}-f^{* 2}\\right) \\\\\n& =\\left(f^{* 2}-e^{* 2}\\right)\\left(\\frac{2 e f}{e+f}-o\\right)=\\left(f^{* 2}-e^{* 2}\\right)(c-o) .\n\\end{aligned}\n$$\nNow let $Z$ be a point on the tangent to $\\omega$ parallel to $A B$ passing through $Q$. Then\n$$\nz=2 q-q^{2} z^{*} \\quad \\Longleftrightarrow \\quad z-q=q-q^{2} z^{*}=-q^{2}\\left(z^{*}-q^{*}\\right)\n$$\nfor $|q|=1$, and thus\n$$\n\\frac{b-a}{b^{*}-a^{*}}=\\frac{z-q}{z^{*}-q^{*}}=\\frac{-q^{2}\\left(z^{*}-q^{*}\\right)}{z^{*}-q^{*}}=-q^{2} .\n$$\nIt follows that\n$$\n\\begin{aligned}\nq^{2} & =-\\frac{b-a}{b^{*}-a^{*}}=-\\frac{\\left(f^{2}-e^{2}\\right)\\left(c^{*}-o\\right)}{\\left(f^{* 2}-e^{* 2}\\right)(c-o)}=e^{2} f^{2} \\frac{c^{*}-o}{c-o} \\\\\n& =e^{2} f^{2} \\frac{\\left(c^{*}-o\\right)^{2}}{|c-o|^{2}}=e^{2} f^{2} \\frac{\\left(c^{*}-o\\right)^{2}}{(1+o)^{2}}\n\\end{aligned}\n$$\nwhere we have used (1). In particular,\n$$\nq=e f \\frac{c^{*}-o}{1+o}\n$$\nwhere the choice of sign is to be justified a posteriori. Further, the point $D$ satisfies\n$$\n-d p=\\frac{d-p}{d^{*}-p^{*}}=\\frac{c-p}{c^{*}-p^{*}} \\quad \\Longrightarrow \\quad d=-\\frac{c-p}{c^{*} p-1}=\\frac{c+1}{c^{*}+1},\n$$\nusing $p=-1$ to obtain the final equality.\nNow, it suffices to show that (i) $D Q \\parallel E F \\perp C I$ and (ii) the midpoint of $[D Q]$ is on $C I$. The desired equality then follows by symmetry with respect to the angle bisector of the angle $\\angle A C B$. Notice that (i) is equivalent with\n$$\n\\frac{d-q}{d^{*}-q^{*}}=\\frac{e-f}{e^{*}-f^{*}} \\quad \\Longleftrightarrow \\quad d q=e f\n$$\nfor $[D Q]$ and $[E F]$ are chords of $\\omega$. But\n$$\n\\begin{aligned}\nd q=e f & \\Longleftrightarrow \\frac{c+1}{c^{*}+1} e f \\frac{c^{*}-o}{1+o}=e f \\quad \\Longleftrightarrow \\quad(c+1)\\left(c^{*}-o\\right)=\\left(c^{*}+1\\right)(1+o) \\\\\n& \\Longleftrightarrow c c^{*}-o\\left(c+c^{*}\\right)-2 o-1=0 .\n\\end{aligned}\n$$\nThe last equality is precisely the defining relation for $o,(1)$. This proves (i). Further, the midpoint of $[D Q]$ is $\\frac{1}{2}(d+q)$, so it remains to check that\n$$\nd q=\\frac{d+q}{d^{*}+q^{*}}=\\frac{c}{c^{*}}=e f\n$$\nwhere the first equality expresses that $[D Q]$ is a chord of $\\omega$ (obviously) containing its midpoint, the second equality expresses the alignment of the midpoint of $[D Q], C$ and $I$, and the third equality follows from the expression for $c$. But we have just shown that $d q=e f$. This proves (ii), justifies the choice of sign for $q$ a posteriori, and thus completes the solution of the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSnow White and the Seven Dwarves are living in their house in the forest. On each of 16 consecutive days, some of the dwarves worked in the diamond mine while the remaining dwarves collected berries in the forest. No dwarf performed both types of work on the same day. On any two different (not necessarily consecutive) days, at least three dwarves each performed both types of work. Further, on the first day, all seven dwarves worked in the diamond mine.\n\nProve that, on one of these 16 days, all seven dwarves were collecting berries.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe define $V$ as the set of all 128 vectors of length 7 with entries in $\\{0,1\\}$. Every such vector encodes the work schedule of a single day: if the $i$-th entry is 0 then the $i$-th dwarf works in the mine, and if this entry is 1 then the $i$-th dwarf collects berries. The 16 working days correspond to 16 vectors $d_{1}, \\ldots, d_{16}$ in $V$, which we will call day-vectors. The condition imposed on any pair of distinct days means that any two distinct day-vectors $d_{i}$ and $d_{j}$ differ in at least three positions.\n\nWe say that a vector $x \\in V$ covers some vector $y \\in V$, if $x$ and $y$ differ in at most one position; note that every vector in $V$ covers exactly eight vectors. For each of the 16 day-vectors $d_{i}$ we define $B_{i} \\subset V$ as the set of the eight vectors that are covered by $d_{i}$. As, for $i \\neq j$, the day-vectors $d_{i}$ and $d_{j}$ differ in at least three positions, their corresponding sets $B_{i}$ and $B_{j}$ are disjoint. As the sets $B_{1}, \\ldots, B_{16}$ together contain $16 \\cdot 8=128=|V|$ distinct elements, they form a partition of $V$; in other words, every vector in $V$ is covered by precisely one day-vector.\n\nThe weight of a vector $v \\in V$ is defined as the number of 1-entries in $v$. For $k=0,1, \\ldots, 7$, the set $V$ contains $\\binom{7}{k}$ vectors of weight $k$. Let us analyse the 16 day-vectors $d_{1}, \\ldots, d_{16}$ by their weights, and let us discuss how the vectors in $V$ are covered by them.\n\n1. As all seven dwarves work in the diamond mine on the first day, the first day-vector is $d_{1}=(0000000)$. This day-vector covers all vectors in $V$ with weight 0 or 1.\n\n2. No day-vector can have weight 2, as otherwise it would differ from $d_{1}$ in at most two positions. Hence each of the $\\binom{7}{2}=21$ vectors of weight 2 must be covered by some day-vector of weight 3. As every vector of weight 3 covers three vectors of weight 2, exactly $21 / 3=7$ day-vectors have weight 3.\n\n3. How are the $\\binom{7}{3}=35$ vectors of weight 3 covered by the day-vectors? Seven of them are day-vectors, and the remaining 28 ones must be covered by day-vectors of weight 4. As every vector of weight 4 covers four vectors of weight 3, exactly $28 / 4=7$ day-vectors have weight 4.\n\nTo summarize, one day-vector has weight 0, seven have weight 3, and seven have weight 4. None of these 15 day-vectors covers any vector of weight 6 or 7, so that the eight heavyweight vectors in $V$ must be covered by the only remaining day-vector; and this remaining vector must be $(1111111)$. On the day corresponding to $(1111111)$ all seven dwarves are collecting berries, and that is what we wanted to show.\nSolution:\n\nIf a dwarf $X$ performs the same type of work on three days $D_{1}, D_{2}, D_{3}$, then we say that this triple of days is monotonous for $X$. We claim that the following configuration cannot occur: There are three dwarves $X_{1}, X_{2}, X_{3}$ and three days $D_{1}, D_{2}, D_{3}$, such that the triple $(D_{1}, D_{2}, D_{3})$ is monotonous for each of the dwarves $X_{1}, X_{2}, X_{3}$.\n\n(Proof: Suppose that such a configuration occurs. Then among the remaining dwarves there exist three dwarves $Y_{1}, Y_{2}, Y_{3}$ that performed both types of work on day $D_{1}$ and on day $D_{2}$; without loss of generality these three dwarves worked in the mine on day $D_{1}$ and collected berries on day $D_{2}$. On day $D_{3}$, two of $Y_{1}, Y_{2}, Y_{3}$ performed the same type of work, and without loss of generality $Y_{1}$ and $Y_{2}$ worked in the mine. But then on days $D_{1}$ and $D_{3}$, each of the five dwarves $X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}$ performed only one type of work; this is in contradiction with the problem statement.)\n\nNext we consider some fixed triple $X_{1}, X_{2}, X_{3}$ of dwarves. There are eight possible working schedules for $X_{1}, X_{2}, X_{3}$ (like mine-mine-mine, mine-mine-berries, mine-berries-mine, etc). As the above forbidden configuration does not occur, each of these eight working schedules must occur on exactly two of the sixteen days. In particular this implies that every dwarf worked exactly eight times in the mine and exactly eight times in the forest.\n\nFor $0 \\leqslant k \\leqslant 7$ we denote by $d(k)$ the number of days on which exactly $k$ dwarves were collecting berries. Since on the first day all seven dwarves were in the mine, on each of the remaining days at least three dwarves collected berries. This yields $d(0)=1$ and $d(1)=d(2)=0$. We assume, for the sake of contradiction, that $d(7)=0$ and hence\n$$\nd(3)+d(4)+d(5)+d(6)=15\n$$\nAs every dwarf collected berries exactly eight times, we get that, further,\n$$\n3 d(3)+4 d(4)+5 d(5)+6 d(6)=7 \\cdot 8=56\n$$\nNext, let us count the number $q$ of quadruples $(X_{1}, X_{2}, X_{3}, D)$ for which $X_{1}, X_{2}, X_{3}$ are three pairwise distinct dwarves that all collected berries on day $D$. As there are $7 \\cdot 6 \\cdot 5=210$ triples of pairwise distinct dwarves, and as every working schedule for three fixed dwarves occurs on exactly two days, we get $q=420$. As every day on which $k$ dwarves collect berries contributes $k(k-1)(k-2)$ such quadruples, we also have\n$$\n3 \\cdot 2 \\cdot 1 \\cdot d(3)+4 \\cdot 3 \\cdot 2 \\cdot d(4)+5 \\cdot 4 \\cdot 3 \\cdot d(5)+6 \\cdot 5 \\cdot 4 \\cdot d(6)=q=420\n$$\nwhich simplifies to\n$$\nd(3)+4 d(4)+10 d(5)+20 d(6)=70\n$$\nFinally, we count the number $r$ of quadruples $(X_{1}, X_{2}, X_{3}, D)$ for which $X_{1}, X_{2}, X_{3}$ are three pairwise distinct dwarves that all worked in the mine on day $D$. Similarly as above we see that $r=420$ and that\n$$\n7 \\cdot 6 \\cdot 5 \\cdot d(0)+4 \\cdot 3 \\cdot 2 \\cdot d(3)+3 \\cdot 2 \\cdot 1 \\cdot d(4)=r=420\n$$\nwhich simplifies to\n$$\n4 d(3)+d(4)=35\n$$\nMultiplying (1) by $-40$, multiplying (2) by $10$, multiplying (3) by $-1$, multiplying (4) by $4$, and then adding up the four resulting equations yields $5 d(3)=30$ and hence $d(3)=6$. Then (4) yields $d(4)=11$. As $d(3)+d(4)=17$, the total number of days cannot be 16. We have reached the desired contradiction.\n\nA Variant. We follow the second solution up to equation (3). Multiplying (1) by $8$, multiplying (2) by $-3$, and adding the two resulting equations to (3) yields\n$$\n3 d(5)+10 d(6)=22\n$$\nAs $d(5)$ and $d(6)$ are positive integers, (5) implies $0 \\leqslant d(6) \\leqslant 2$. Only the case $d(6)=1$ yields an integral value $d(5)=4$. The equations (1) and (2) then yield $d(3)=10$ and $d(4)=0$.\n\nNow let us look at the $d(3)=10$ special days on which exactly three dwarves were collecting berries. One of the dwarves collected berries on at least five special days (if every dwarf collected berries on at most four special days, this would allow at most $7 \\cdot 4 / 3<10$ special days); we call this dwarf $X$. On at least two out of these five special days, some dwarf $Y$ must have collected berries together with $X$. Then these two days contradict the problem statement. We have reached the desired contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine all real constants $t$ such that whenever $a, b, c$ are the lengths of the sides of a triangle, then so are $a^{2}+b c t$, $b^{2}+c a t$, $c^{2}+a b t$.",
    "options": [],
    "answer": "t in [2/3, 2]",
    "solution": "Solution:\n\nIf $t < 2/3$, take a triangle with sides $c = b = 1$ and $a = 2 - \\epsilon$. Then\n$$\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t = 3 t - 2 + \\epsilon (4 - 2 t - \\epsilon) \\leq 0\n$$\nfor small positive $\\epsilon$; for instance, for any $0 < \\epsilon < (2 - 3 t)/(4 - 2 t)$.\n\nOn the other hand, if $t > 2$, then take a triangle with sides $b = c = 1$ and $a = \\epsilon$. Then\n$$\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t = 2 - t + \\epsilon (2 t - \\epsilon) \\leq 0\n$$\nfor small positive $\\epsilon$; for instance, for any $0 < \\epsilon < (t - 2)/(2 t)$.\n\nNow assume that $2/3 \\leq t \\leq 2$ and $b + c > a$. Then using $(b + c)^{2} \\geq 4 b c$ we obtain\n$$\n\\begin{aligned}\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t &= (b + c)^{2} + a t (b + c) - (2 + t) b c - a^{2} \\\\\n&\\geq (b + c)^{2} + a t (b + c) - \\frac{1}{4}(2 + t)(b + c)^{2} - a^{2} \\\\\n&\\geq \\frac{1}{4}(2 - t)(b + c)^{2} + a t (b + c) - a^{2}\n\\end{aligned}\n$$\nAs $2 - t \\geq 0$ and $t > 0$, this last expression is an increasing function of $b + c$, and hence using $b + c > a$ we obtain\n$$\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t > \\frac{1}{4}(2 - t) a^{2} + t a^{2} - a^{2} = \\frac{3}{4}\\left(t - \\frac{2}{3}\\right) a^{2} \\geq 0\n$$\nas $t \\geq 2/3$. The other two inequalities follow by symmetry.\nSolution:\n\nAfter showing that $t$ must be in the interval $[2/3, 2]$ as in Solution 1, we let $x = (c + a - b)/2$, $y = (a + b - c)/2$ and $z = (b + c - a)/2$ so that $a = x + y$, $b = y + z$, $c = z + x$. Then we have:\n$$\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t = (x^{2} + y^{2} - z^{2} + x y + x z + y z) t + 2(z^{2} + x z + y z - x y)\n$$\nSince this linear function of $t$ is positive both at $t = 2/3$ where\n$$\n\\frac{2}{3}(x^{2} + y^{2} - z^{2} + x y + x z + y z) + 2(z^{2} + x z + y z - x y) = \\frac{2}{3}((x - y)^{2} + 4(x + y)z + 2 z^{2}) > 0\n$$\nand at $t = 2$ where\n$$\n2(x^{2} + y^{2} - z^{2} + x y + x z + y z) + 2(z^{2} + x z + y z - x y) = 2(x^{2} + y^{2}) + 4(x + y)z > 0,\n$$\nit is positive on the entire interval $[2/3, 2]$.\nSolution:\n\nAfter the point in Solution 2 where we obtain\n$$\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t = (x^{2} + y^{2} - z^{2} + x y + x z + y z) t + 2(z^{2} + x z + y z - x y)\n$$\nwe observe that the right hand side can be rewritten as\n$$\n(2 - t) z^{2} + (x - y)^{2} t + (3 t - 2) x y + z(x + y)(2 + t)\n$$\nAs the first three terms are non-negative and the last term is positive, the result follows.\nSolution:\n\nFirst we show that $t$ must be in the interval $[2/3, 2]$ as in Solution 1. Then:\n\nCase 1: If $a \\geq b, c$, then $a b + a c - b c > 0$, $2(b^{2} + c^{2}) \\geq (b + c)^{2} > a^{2}$ and $t \\geq 2/3$ implies:\n$$\n\\begin{aligned}\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t &= b^{2} + c^{2} - a^{2} + (a b + a c - b c) t \\\\\n&\\geq (b^{2} + c^{2} - a^{2}) + \\frac{2}{3}(a b + a c - b c) \\\\\n&\\geq \\frac{1}{3}(3 b^{2} + 3 c^{2} - 3 a^{2} + 2 a b + 2 a c - 2 b c) \\\\\n&\\geq \\frac{1}{3}\\left[(2 b^{2} + 2 c^{2} - a^{2}) + (b - c)^{2} + 2 a(b + c - a)\\right] \\\\\n&> 0\n\\end{aligned}\n$$\n\nCase 2: If $b \\geq a, c$, then $b^{2} + c^{2} - a^{2} > 0$. If also $a b + a c - b c \\geq 0$, then $b^{2} + c a t + c^{2} + a b t - a^{2} - b c t > 0$. If, on the other hand, $a b + a c - b c \\leq 0$, then since $t \\leq 2$, we have:\n$$\n\\begin{aligned}\nb^{2} + c a t + c^{2} + a b t - a^{2} - b c t &\\geq b^{2} + c^{2} - a^{2} + 2(a b + a c - b c) \\\\\n&\\geq (b - c)^{2} + a(b + c - a) + a(b + c) \\\\\n&> 0\n\\end{aligned}\n$$\nBy symmetry, we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $D$ and $E$ be two points on the sides $AB$ and $AC$, respectively, of a triangle $ABC$, such that $DB = BC = CE$, and let $F$ be the point of intersection of the lines $CD$ and $BE$. Prove that the incenter $I$ of the triangle $ABC$, the orthocenter $H$ of the triangle $DEF$ and the midpoint $M$ of the $\\operatorname{arc} BAC$ of the circumcircle of the triangle $ABC$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAs $DB = BC = CE$ we have $BI \\perp CD$ and $CI \\perp BE$. Hence $I$ is orthocenter of triangle $BFC$. Let $K$ be the point of intersection of the lines $BI$ and $CD$, and let $L$ be the point of intersection of the lines $CI$ and $BE$. Then we have the power relation $IB \\cdot IK = IC \\cdot IL$. Let $U$ and $V$ be the feet of the perpendiculars from $D$ to $EF$ and $E$ to $DF$, respectively. Now we have the power relation $DH \\cdot HU = EH \\cdot HV$.\n\n![](attached_image_1.png)\n\nLet $\\omega_1$ and $\\omega_2$ be the circles with diameters $BD$ and $CE$, respectively. From the power relations above we conclude that $IH$ is the radical axis of the circles $\\omega_1$ and $\\omega_2$.\nLet $O_1$ and $O_2$ be centers of $\\omega_1$ and $\\omega_2$, respectively. Then $MB = MC$, $BO_1 = CO_2$ and $\\angle MBO_1 = \\angle MCO_2$, and the triangles $MBO_1$ and $MCO_2$ are congruent. Hence $MO_1 = MO_2$. Since radii of $\\omega_1$ and $\\omega_2$ are equal, this implies that $M$ lies on the radical axis of $\\omega_1$ and $\\omega_2$ and $M, I, H$ are collinear.\n\n\nLet the points $K, L, U, V$ be as in Solution 1. Let $P$ be the point of intersection of $DU$ and $EI$, and let $Q$ be the point of intersection of $EV$ and $DI$.\nSince $DB = BC = CE$, the points $CI$ and $BI$ are perpendicular to $BE$ and $CD$, respectively. Hence the lines $BI$ and $EV$ are parallel and $\\angle IEB = \\angle IBE = \\angle UEH$. Similarly, the lines $CI$ and $DU$ are parallel and $\\angle IDC = \\angle ICD = \\angle VDH$. Since $\\angle UEH = \\angle VDH$, the points $D, Q, F, P, E$ are concyclic. Hence $IP \\cdot IE = IQ \\cdot ID$.\nLet $R$ be the second point intersection of the circumcircle of triangle $HEP$ and the line $HI$. As $IH \\cdot IR = IP \\cdot IE = IQ \\cdot ID$, the points $D, Q, H, R$ are also concyclic. We have $\\angle DQH = \\angle EPH = \\angle DFE = \\angle BFC = 180^\\circ - \\angle BIC = 90^\\circ - \\angle BAC / 2$. Now using the concyclicity of $D, Q, H, R$, and $E, P, H, R$ we obtain $\\angle DRH = \\angle ERH = 180^\\circ - (90^\\circ - \\angle BAC / 2) = 90^\\circ + \\angle BAC / 2$. Hence $R$ is inside the triangle $DEH$ and $\\angle DRE = 360^\\circ - \\angle DRH - \\angle ERH = 180^\\circ - \\angle BAC$ and it follows that the points $A, D, R, E$ are concyclic.\n![](attached_image_2.png)\nAs $MB = MC$, $BD = CE$, $\\angle MBD = \\angle MCE$, the triangles $MBD$ and $MCE$ are congruent and $\\angle MDA = \\angle MEA$. Hence the points $M, D, E, A$ are concyclic. Therefore the points $M, D, R, E, A$ are concyclic. Now we have $\\angle MRE = 180^\\circ - \\angle MAE = 180^\\circ - (90^\\circ + \\angle BAC / 2) = 90^\\circ - \\angle BAC / 2$ and since $\\angle ERH = 90^\\circ + \\angle BAC / 2$, we conclude that the points $I, H, R, M$ are collinear.\n\n\nSuppose that we have a coordinate system and $(b_x, b_y), (c_x, c_y), (d_x, d_y), (e_x, e_y)$ are the coordinates of the points $B, C, D, E$, respectively. From $\\overrightarrow{BI} \\cdot \\overrightarrow{CD} = 0$, $\\overrightarrow{CI} \\cdot \\overrightarrow{BE} = 0$, $\\overrightarrow{EH} \\cdot \\overrightarrow{CD} = 0$, $\\overrightarrow{DH} \\cdot \\overrightarrow{BE} = 0$ we obtain $\\overrightarrow{IH} \\cdot (\\vec{B} - \\vec{C} - \\vec{E} + \\vec{D}) = 0$. Hence the slope of the line $IH$ is $(c_x + e_x - b_x - d_x)/(b_y + d_y - c_y - e_y)$.\nAssume that the $x$-axis lies along the line $BC$, and let $\\alpha = \\angle BAC$, $\\beta = \\angle ABC$, $\\theta = \\angle ACB$. Since $DB = BC = CE$, we have $c_x - b_x = BC$, $e_x - d_x = BC - BC \\cos \\beta - BC \\cos \\theta$, $b_y = c_y = 0$, $d_y - e_y = BC \\sin \\beta - BC \\sin \\theta$. Therefore the slope of $IH$ is $(2 - \\cos \\beta - \\cos \\theta)/(\\sin \\beta - \\sin \\theta)$.\nNow we will show that the slope of the line $MI$ is the same. Let $r$ and $R$ be the inradius and circumradius of the triangle $ABC$, respectively. As $\\angle BMC = \\angle BAC = \\alpha$ and $BM = MC$, we have\n$$\nm_y - i_y = \\frac{BC}{2} \\cot \\left(\\frac{\\alpha}{2}\\right) - r \\text{ and } m_x - i_x = \\frac{AC - AB}{2}\n$$\nwhere $(m_x, m_y)$ and $(i_x, i_y)$ are the coordinates of $M$ and $I$, respectively. Therefore the slope of $MI$ is $(BC \\cot (\\alpha / 2) - 2r)/(AC - AB)$.\nNow the equality of these slopes follows using\n$$\n\\frac{BC}{\\sin \\alpha} = \\frac{AC}{\\sin \\beta} = \\frac{AB}{\\sin \\theta} = 2R\n$$\nhence\n$$\nBC \\cot \\left(\\frac{\\alpha}{2}\\right) = 4R \\cos^2 \\left(\\frac{\\alpha}{2}\\right) = 2R(1 + \\cos \\alpha)\n$$\nand\n$$\n\\frac{r}{R} = \\cos \\alpha + \\cos \\beta + \\cos \\theta - 1\n$$\nas\n$$\n\\frac{BC \\cot (\\alpha / 2) - 2r}{AC - AB} = \\frac{2R(1 + \\cos \\alpha) - 2r}{2R(\\sin \\beta - \\sin \\theta)} = \\frac{2 - \\cos \\beta - \\cos \\theta}{\\sin \\beta - \\sin \\theta}\n$$\ngiving the collinearity of the points $I, H, M$.\n\n\nLet the bisectors $BI$ and $CI$ meet the circumcircle of $ABC$ again at $P$ and $Q$, respectively. Let the altitude of $DEF$ belonging to $D$ meet $BI$ at $R$ and the one belonging to $E$ meet $CI$ at $S$.\nSince $BI$ is angle bisector of the isosceles triangle $CBD$, $BI$ and $CD$ are perpendicular. Since $EH$ and $DF$ are also perpendicular, $HS$ and $RI$ are parallel. Similarly, $HR$ and $SI$ are parallel, and hence $HSIR$ is a parallelogram.\nOn the other hand, as $M$ is the midpoint of the $\\operatorname{arc} BAC$, we have $\\angle MPI = \\angle MPB = \\angle MQC = \\angle MQI$, and $\\angle PIQ = (\\widehat{PA} + \\widehat{CB} + \\widehat{AQ}) / 2 = (\\widehat{PC} + \\widehat{CB} + \\widehat{BQ}) / 2 = \\angle PMQ$. Therefore $MPIQ$ is a parallelogram.\nSince $CI$ is angle bisector of the isosceles triangle $BCE$, the triangle $BSE$ is also isosceles. Hence $\\angle FBS = \\angle EBS = \\angle SEB = \\angle HEF = \\angle HDF = \\angle RDF = \\angle FCS$ and $B, S, F, C$ are concyclic. Similarly, $B, F, R, C$ are concyclic. Therefore $B, S, R, C$ are concyclic. As $B, Q, P, C$ are also concyclic, $SR$ and $QP$ are parallel.\nNow it follows that $HSIR$ and $MQIP$ are homothetic parallelograms, and therefore $M, H, I$ are collinear.\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\\omega(n)=k$ and $d(n)$ does not divide $d\\left(a^{2}+b^{2}\\right)$ for any positive integers $a, b$ satisfying $a+b=n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe will show that any number of the form $n=2^{p-1} m$ where $m$ is a positive integer that has exactly $k-1$ prime factors all of which are greater than $3$ and $p$ is a prime number such that $(5 / 4)^{(p-1) / 2}>m$ satisfies the given condition.\n\nSuppose that $a$ and $b$ are positive integers such that $a+b=n$ and $d(n) \\mid d\\left(a^{2}+b^{2}\\right)$. Then $p \\mid d\\left(a^{2}+b^{2}\\right)$. Hence $a^{2}+b^{2}=q^{c p-1} r$ where $q$ is a prime, $c$ is a positive integer and $r$ is a positive integer not divisible by $q$. If $q \\geq 5$, then\n$$\n2^{2 p-2} m^{2}=n^{2}=(a+b)^{2}>a^{2}+b^{2}=q^{c p-1} r \\geq q^{p-1} \\geq 5^{p-1}\n$$\ngives a contradiction. So $q$ is $2$ or $3$.\n\nIf $q=3$, then $a^{2}+b^{2}$ is divisible by $3$ and this implies that both $a$ and $b$ are divisible by $3$. This means $n=a+b$ is divisible by $3$, a contradiction. Hence $q=2$.\n\nNow we have $a+b=2^{p-1} m$ and $a^{2}+b^{2}=2^{c p-1} r$. If the highest powers of $2$ dividing $a$ and $b$ are different, then $a+b=2^{p-1} m$ implies that the smaller one must be $2^{p-1}$ and this makes $2^{2 p-2}$ the highest power of $2$ dividing $a^{2}+b^{2}=2^{c p-1} r$, or equivalently, $c p-1=2 p-2$, which is not possible. Therefore $a=2^{t} a_{0}$ and $b=2^{t} b_{0}$ for some positive integer $t<p-1$ and odd integers $a_{0}$ and $b_{0}$. Then $a_{0}^{2}+b_{0}^{2}=2^{c p-1-2 t} r$. The left side of this equality is congruent to $2$ modulo $4$, therefore $c p-1-2 t$ must be $1$. But then $t<p-1$ gives $(c / 2) p=t+1<p$, which is not possible either.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine all integers $n \\geq 2$ for which there exist integers $x_{1}, x_{2}, \\ldots, x_{n-1}$ satisfying the condition that if $0 < i < n$, $0 < j < n$, $i \\neq j$ and $n$ divides $2i + j$, then $x_{i} < x_{j}$.",
    "options": [],
    "answer": "All integers of the form 2^k with k ≥ 1 and 3·2^k with k ≥ 0.",
    "solution": "Solution:\n\nSuppose that $n$ has one of these forms. For an integer $i$, let $x_{i}$ be the largest integer such that $2^{x_{i}}$ divides $i$. Now assume that $0 < i < n$, $0 < j < n$, $i \\neq j$, $n$ divides $2i + j$ and $x_{i} \\geq x_{j}$. Then the highest power of $2$ dividing $2i + j$ is $2^{x_{j}}$ and therefore $k \\leq x_{j}$ and $2^{k} \\leq j$. Since $0 < j < n$, this is possible only if $n = 3 \\cdot 2^{k}$ and either $j = 2^{k}$ or $j = 2^{k+1}$. In the first case, $i \\neq j$ and $x_{i} \\geq x_{j}$ imply $i = 2^{k+1}$ leading to the contradiction $3 \\cdot 2^{k} = n \\mid 2i + j = 5 \\cdot 2^{k}$. The second case is not possible as $i \\neq j$ and $x_{i} \\geq x_{j}$ now imply $i \\geq 2^{k+2} > n$.\n\nNow suppose that $n$ does not have one of these forms and $x_{1}, x_{2}, \\ldots, x_{n-1}$ satisfying the given condition exist. For any positive integer $m$, let $a_{m}$ be the remainder of the division of $(-2)^{m}$ by $n$. Then none of $a_{m}$ is $0$ as $n$ is not a power of $2$. Also $a_{m} \\neq a_{m+1}$ for any $m \\geq 1$ as $a_{m} = a_{m+1}$ would lead to $n$ dividing $3 \\cdot 2^{m}$. Moreover $n$ divides $2a_{m} + a_{m+1}$. Hence we must have $x_{a_{1}} < x_{a_{2}} < x_{a_{3}} < \\ldots$ which is not possible as $a_{m}$'s can take on only finitely many values.\n\nLet $E = \\{n / 3, n / 2, 2n / 3\\} \\cap \\{1, 2, \\ldots, n-1\\}$, $D = \\{1, 2, \\ldots, n-1\\} \\setminus E$, and let $f: D \\rightarrow \\{1, 2, \\ldots, n-1\\}$ be the function sending $i$ in $D$ to the unique $f(i)$ in $\\{1, 2, \\ldots, n-1\\}$ such that $f(i) \\equiv -2i \\pmod{n}$.\nThen the condition of the problem is that $x_{i} < x_{f(i)}$ for each $i$ in $D$. Since $D$ is a finite set, the integers $x_{1}, x_{2}, \\ldots, x_{n-1}$ exist if and only if for each $i$ in $D$ there exists a positive integer $k(i)$ such that $f^{k(i)}(i)$ belongs to $E$. This can be seen as follows:\n- If $f^{k}(i)$ does not belong to $E$ for any $k > 0$ for some $i$, then there exists $k_{2} > k_{1} > 0$ such that $f^{k_{1}}(i) = f^{k_{2}}(i)$, leading to the contradiction $x_{f^{k_{1}}(i)} < x_{f^{k_{2}}(i)} = x_{f^{k_{1}}(i)}$.\n- On the other hand, if such $k(i)$ exists for each $i$ in $D$, and if $k_{0}(i)$ denotes the smallest such, then the condition of the problem is satisfied by letting $x_{i} = -k_{0}(i)$ for $i$ in $D$, and $x_{i} = 0$ for $i$ in $E$.\nIn other words, the integers $x_{1}, x_{2}, \\ldots, x_{n-1}$ exist if and only if for each $i$ in $D$ there exists a positive integer $k(i)$ such that $(-2)^{k(i)} i \\equiv n / 3, n / 2$ or $2n / 3 \\pmod{n}$. For $i = 1$, this implies that $n = 2^{k}$ with $k \\geq 1$ or $n = 3 \\cdot 2^{k}$ with $k \\geq 0$. On the other hand, if $n$ has one of these forms, letting $k(i) = k$ does the trick for all $i$ in $D$.\n\nSuppose that $x_{1}, x_{2}, \\ldots, x_{k-1}$ satisfy the condition of the problem for $n = k$. Let $y_{2i} = x_{i}$ for $1 \\leq i \\leq k-1$ and choose $y_{2i-1}$ for $1 \\leq i \\leq k$ to be less than $\\min \\{x_{1}, x_{2}, \\ldots, x_{k-1}\\}$. Now suppose that for $n = 2k$ we have $0 < i < n$, $0 < j < n$, $i \\neq j$, $n$ divides $2i + j$. Then $j$ is even. If $i$ is also even, then $0 < i/2 < k$, $0 < j/2 < k$ and $k$ divides $2(i/2) + (j/2)$; hence $y_{i} = x_{i/2} < x_{j/2} = y_{j}$. On the other hand, if $i$ is odd, then $y_{i} < \\min \\{x_{1}, x_{2}, \\ldots, x_{k-1}\\} \\leq x_{j/2} = y_{j}$. Therefore, $y_{1}, y_{2}, \\ldots, y_{2k-1}$ satisfy the condition of the problem for $n = 2k$.\nSince the condition is vacuous for $n = 2$ and $n = 3$, it follows that $x_{1}, x_{2}, \\ldots, x_{n-1}$ satisfying the condition exist for all $n = 2^{k}$ with $k \\geq 1$ and $n = 3 \\cdot 2^{k}$ with $k \\geq 0$.\nNow suppose that $x_{1}, x_{2}, \\ldots, x_{n-1}$ satisfying the condition of the problem exist for $n = 2^{k} m$ where $k$ is a nonnegative integer and $m > 3$ is an odd number. Let $b_{0} = 2^{k}$ and let $b_{i+1}$ be the remainder of the division of $(-2) b_{i}$ by $n$ for $i \\geq 0$. No terms of this sequence is $0$ and no two consecutive terms are both equal to $b_{1}$ as $m > 3$. On the other hand, as $(-2)^{\\phi(m)} \\equiv 1 \\pmod{m}$, we have $b_{\\phi(m)} \\equiv (-2)^{\\phi(m)} 2^{k} \\equiv 2^{k} \\equiv b_{0} \\pmod{n}$, and hence $b_{\\phi(m)} = b_{0}$. Since $2b_{i} + b_{i+1}$ is divisible by $n$ for all $i \\geq 0$, we have $x_{b_{0}} < x_{b_{1}} < \\cdots < x_{b_{\\phi(m)}} = x_{b_{0}}$, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer. We have $n$ boxes where each box contains a nonnegative number of pebbles. In each move we are allowed to take two pebbles from a box we choose, throw away one of the pebbles and put the other pebble in another box we choose. An initial configuration of pebbles is called solvable if it is possible to reach a configuration with no empty box, in a finite (possibly zero) number of moves. Determine all initial configurations of pebbles which are not solvable, but become solvable when an additional pebble is added to a box, no matter which box is chosen.",
    "options": [],
    "answer": "Exactly those configurations in which every box contains an even number of pebbles and the total number of pebbles is 2n − 2.",
    "solution": "Solution:\n\nNumber the boxes from $1$ through $n$ and denote a configuration by $x = (x_{1}, x_{2}, \\ldots, x_{n})$ where $x_{i}$ is the number of pebbles in the $i$th box. Let\n$$\nD(x) = \\sum_{i=1}^{n} \\left\\lfloor \\frac{x_{i} - 1}{2} \\right\\rfloor\n$$\nfor a configuration $x$. We can rewrite this in the form\n$$\nD(x) = \\frac{1}{2} N(x) - n + \\frac{1}{2} O(x)\n$$\nwhere $N(x)$ is the total number of pebbles and $O(x)$ is the number of boxes with an odd number of pebbles for the configuration $x$.\n\nNote that a move either leaves $D$ the same (if it is made into a box containing an even number of pebbles) or decreases it by $1$ (if it is made into a box with an odd number of pebbles). As $D$ is nonnegative for any configuration which does not have any empty boxes, it is also nonnegative for any solvable configuration. On the other hand, if a configuration has nonnegative $D$, then making $m_{i} = \\left\\lfloor (x_{i} - 1) / 2 \\right\\rfloor$ moves from the $i$th box into $m_{i}$ empty boxes for each $i$ with $m_{i} > 0$ fills all boxes as $D(x) \\geq 0$ means $\\sum_{m_{i} > 0} m_{i} \\geq$ (number of empty boxes).\n\nAs $N(x)$ and $O(x)$ have the same parity, a configuration $x$ is solvable exactly when $O(x) \\geq 2n - N(x)$, and unsolvable exactly when $O(x) \\leq 2n - 2 - N(x)$. In particular, any configuration with $2n - 1$ pebbles is solvable, and a configuration with $2n - 2$ pebbles is unsolvable if and only if all boxes contain even numbers of pebbles.\n\nSuppose that $x'$ is obtained from $x$ by adding a pebble in some box. Then $O(x') = O(x) + 1$ or $O(x') = O(x) - 1$. If $x$ is unsolvable and $x'$ is solvable, then we must have $O(x) \\leq 2n - 2 - N(x)$ and $O(x') \\geq 2n - N(x') = 2n - 1 - N(x)$, and hence $O(x') = O(x) + 1$. That is, the pebble must be added to a box with an even number of pebbles. This can be the case irrespective of where the pebble is added only if all boxes contain even numbers of pebbles, and $0 = O(x) \\leq 2n - 2 - N(x)$ and $1 = O(x') \\geq 2n - 1 - N(x)$; that is, $N(x) = 2n - 2$.\nSolution:\n\nLet $x$ be a configuration and $\\tilde{x}$ be another configuration obtained from $x$ by removing two pebbles from a box and depositing them in another box.\n\nClaim 1: $\\tilde{x}$ is solvable if and only if $x$ is solvable.\n\nLet us call two configurations equivalent if they have the same total number of pebbles and parities of the number of pebbles in the corresponding boxes are the same. (It does not matter whether we consider this equivalence for a fixed ordering of the boxes or up to permutation.) From Claim 1 it follows that two equivalent configurations are both solvable or both unsolvable. In particular, any configuration with $2n - 1$ or more pebbles is solvable, because it is equivalent to a configuration with no empty boxes.\n\nLet us call a configuration with all boxes containing two or fewer pebbles scant. Every unsolvable configuration is equivalent to a scant configuration.\n\nClaim 2: A scant configuration is solvable if and only if it contains no empty boxes.\n\nBy Claim 1 and Claim 2, addition of a pebble to a scant unsolvable configuration makes it solvable if and only if the configuration has exactly one empty box and the pebble is added to the empty box or to a box containing two pebbles. Hence, the addition of a pebble makes an unsolvable scant configuration into a solvable configuration irrespective of where it is added if and only if all boxes have even numbers of pebbles and exactly one of them is empty. Therefore, the addition of a pebble makes an unsolvable configuration into a solvable one irrespective of where the pebble is added if and only if the configuration has $2n - 2$ pebbles and all boxes have even numbers of pebbles.\n\nProof of Claim 1: Suppose that the two pebbles were moved from box $B$ in $x$ to box $\\tilde{B}$ in $\\tilde{x}$, and $x$ is solvable. Then we perform exactly the same sequence of moves for $\\tilde{x}$ as we did for $x$ except that instead of the first move that is made out of $B$ we make a move from $\\tilde{B}$ (into the same box), and if there was no move from $B$, then at the end we make a move from $\\tilde{B}$ to $B$ in case $B$ is now empty.\n\nProof of Claim 2: Any move from a scant configuration either leaves the number of empty boxes the same and the resulting configuration is also scant (if it is made into an empty box), or increases the number of empty boxes by one (if it is made into a nonempty box). In the second case, if the move was made into a box containing one pebble, then the resulting configuration is still scant. On the other hand, if it is made into a box containing two pebbles, then the resulting configuration is equivalent to the scant configuration which has one pebble in the box the move was made into and exactly the same number of pebbles in all other boxes as the original configuration. Therefore, any sequence of moves from a scant configuration results in a configuration with more or the same number of empty boxes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDetermine all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying the condition\n$$\nf\\left(y^{2}+2 x f(y)+f(x)^{2}\\right)=(y+f(x))(x+f(y))\n$$\nfor all real numbers $x$ and $y$.",
    "options": [],
    "answer": "f(x) = x; f(x) = -x; f(x) = 1/2 - x",
    "solution": "Solution:\nIt can be easily checked that the functions $f(x)=x$, $f(x)=-x$ and $f(x)=\\frac{1}{2}-x$ satisfy the given condition. We will show that these are the only functions doing so. Let $y=-f(x)$ in the original equation to obtain\n$$\nf\\left(2 f(x)^{2}+2 x f(-f(x))\\right)=0\n$$\nfor all $x$. In particular, $0$ is a value of $f$. Suppose that $u$ and $v$ are such that $f(u)=0=f(v)$. Plugging $x=u$ or $v$ and $y=u$ or $v$ in the original equations we get $f\\left(u^{2}\\right)=u^{2}$, $f\\left(u^{2}\\right)=uv$, $f\\left(v^{2}\\right)=uv$ and $f\\left(v^{2}\\right)=v^{2}$. We conclude that $u^{2}=uv=v^{2}$ and hence $u=v$. So there is exactly one $a$ mapped to $0$, and\n$$\nf(x)^{2}+x f(-f(x))=\\frac{a}{2}\n$$\nfor all $x$.\nSuppose that $f\\left(x_{1}\\right)=f\\left(x_{2}\\right) \\neq 0$ for some $x_{1}$ and $x_{2}$. Using (*) we obtain $x_{1} f\\left(-f\\left(x_{1}\\right)\\right)=x_{2} f\\left(-f\\left(x_{2}\\right)\\right)=x_{2} f\\left(-f\\left(x_{1}\\right)\\right)$ and hence either $x_{1}=x_{2}$ or $f\\left(x_{1}\\right)=f\\left(x_{2}\\right)=-a$. In the second case, letting $x=a$ and $y=x_{1}$ in the original equation we get $f\\left(x_{1}^{2}-2 a^{2}\\right)=0$, hence $x_{1}^{2}-2 a^{2}=a$. Similarly, $x_{2}^{2}-2 a^{2}=a$, and it follows that $x_{1}=x_{2}$ or $x_{1}=-x_{2}$ in this case.\nUsing the symmetry of the original equation we have\n$$\nf\\left(f(x)^{2}+y^{2}+2 x f(y)\\right)=(x+f(y))(y+f(x))=f\\left(f(y)^{2}+x^{2}+2 y f(x)\\right)\n$$\nfor all $x$ and $y$. Suppose $f(x)^{2}+y^{2}+2 x f(y) \\neq f(y)^{2}+x^{2}+2 y f(x)$ for some $x$ and $y$. Then by the observations above, $(x+f(y))(y+f(x)) \\neq 0$ and $f(x)^{2}+y^{2}+2 x f(y)=-\\left(f(y)^{2}+x^{2}+2 y f(x)\\right)$. But these conditions are contradictory as the second one can be rewritten as $(f(x)+y)^{2}+(f(y)+x)^{2}=0$.\nTherefore from $(**)$ now it follows that\n$$\nf(x)^{2}+y^{2}+2 x f(y)=f(y)^{2}+x^{2}+2 y f(x)\n$$\nfor all $x$ and $y$. In particular, letting $y=0$ we obtain $f(x)^{2}=(f(0)-x)^{2}$ for all $x$. Let $f(x)=s(x)(f(0)-x)$ where $s: \\mathbf{R} \\rightarrow \\{1,-1\\}$. Plugging this in $(***)$ gives\n$$\nx(y s(y)+f(0)(1-s(y)))=y(x s(x)+f(0)(1-s(x)))\n$$\nfor all $x$ and $y$. So $s(x)+f(0)(1-s(x))/x$ must be constant for $x \\neq 0$.\nIf $f(0)=0$ it follows that $s(x)$ is constant for $x \\neq 0$, and therefore either $f(x)=x$ for all $x$ or $f(x)=-x$ for all $x$. Suppose that $f(0) \\neq 0$. If $s(x)$ is $-1$ for all $x \\neq 0$, then $-1+2 f(0)/x$ must be constant for all $x \\neq 0$, which is not possible. On the other hand, if there exist nonzero $x$ and $y$ such that $s(x)=-1$ and $s(y)=1$, then $-1+2 f(0)/x=1$. That is, there can be only one such $x$, that $x$ is $f(0)$, and hence $f(x)=f(0)-x$ for all $x$. Putting this back in the original equation gives $2 f(0)^{2}=f(0)$ and hence $f(0)=1/2$. We are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe $n$ contestants of an EGMO are named $C_{1}, \\ldots, C_{n}$. After the competition they queue in front of the restaurant according to the following rules.\n- The Jury chooses the initial order of the contestants in the queue.\n- Every minute, the Jury chooses an integer $i$ with $1 \\leq i \\leq n$.\n- If contestant $C_{i}$ has at least $i$ other contestants in front of her, she pays one euro to the Jury and moves forward in the queue by exactly $i$ positions.\n- If contestant $C_{i}$ has fewer than $i$ other contestants in front of her, the restaurant opens and the process ends.\n\na. Prove that the process cannot continue indefinitely, regardless of the Jury's choices.\n\nb. Determine for every $n$ the maximum number of euros that the Jury can collect by cunningly choosing the initial order and the sequence of moves.\n\n(Hungary)",
    "options": [],
    "answer": "2^n - n - 1",
    "solution": "Solution:\n\nThe maximal number of euros is $2^{n}-n-1$.\n\nTo begin with, we show that it is possible for the Jury to collect this number of euros. We argue by induction. Let us assume that the Jury can collect $M_{n}$ euros in a configuration with $n$ contestants. Then we show that the Jury can collect at least $2 M_{n}+n$ moves in a configuration with $n+1$ contestants. Indeed, let us begin with all the contestants lined up in reverse order. In the first $M_{n}$ moves the Jury keeps $C_{n+1}$ in first position and reverses the order of the remaining contestants, then in the next $n$ moves all contestants $C_{1}, \\ldots, C_{n}$ (in this order) jump over $C_{n+1}$ and end up in the first $n$ positions of the line in reverse order, and finally in the last $M_{n}$ moves the Jury rearranges the first $n$ positions.\n\nSince $M_{1}=0$ and $M_{n+1} \\geq 2 M_{n}+n$, an easy induction shows that $M_{n} \\geq 2^{n}-n-1$.\n\nLet us show now that at most $2^{n}-n-1$ moves are possible. To this end, let us identify a line of contestants with a permutation $\\sigma$ of $\\{1, \\ldots, n\\}$. To each permutation we associate the set of reverse pairs\n$$\nR(\\sigma):=\\{(i, j): 1 \\leq i<j \\leq n \\text{ and } \\sigma(i)>\\sigma(j)\\}\n$$\nand the nonnegative integer\n$$\nW(\\sigma):=\\sum_{(i, j) \\in R(\\sigma)} 2^{i}\n$$\nwhich we call the total weight of the permutation. We claim that the total weight decreases after any move of the contestants. Indeed, let us assume that $C_{i}$ moves forward in the queue, let $\\sigma$ be the permutation before the move, and let $\\sigma'$ denote the permutation after the move. Since $C_{i}$ jumps over exactly $i$ contestants, necessarily she jumps over at least one contestant $C_{j}$ with index $j>i$. This means that the pair $(i, j)$ is reverse with respect to $\\sigma$ but not with respect to $\\sigma'$, and this yields a reduction of $2^{i}$ in the total weight. On the other hand, the move by $C_{i}$ can create new reverse pairs of the form $(k, i)$ with $k<i$, but their total contribution is at most\n$$\n2^{0}+2^{1}+\\ldots+2^{i-1}=2^{i}-1\n$$\nIn conclusion, when passing from $\\sigma$ to $\\sigma'$, at least one term $2^{i}$ disappears from the computation of the total weight, and the sum of all the new terms that might have been created is at most $2^{i}-1$. This shows that $W\\left(\\sigma'\\right) \\leq W(\\sigma)-1$.\n\nWe conclude by observing that the maximum possible value of $W(\\sigma)$ is realized when all pairs are reverse, in which case\n$$\nW(\\sigma)=\\sum_{i=1}^{n}(i-1) 2^{i}=2^{n}-n-1.\n$$\nThis proves that the number of moves is less than or equal to $2^{n}-n-1$, and in particular it is finite.\nSolution:\n\nAs in the previous solution, the fundamental observation is again that, when a contestant $C_{i}$ moves forward, necessarily she has to jump over at least one contestant $C_{j}$ with $j>i$.\n\nLet us show now that the process ends after a finite number of moves. Let us assume that this is not the case. Then at least one contestant moves infinitely many times. Let $i_{0}$ be the largest index such that $C_{i_{0}}$ moves infinitely many times. Then necessarily $C_{i_{0}}$ jumps infinitely many times over some fixed $C_{j_{0}}$ with $j_{0}>i_{0}$. On the other hand, we know that $C_{j_{0}}$ makes only a finite number of moves, and therefore she can precede $C_{i_{0}}$ in the line only a finite number of times, which is absurd.\n\nIn order to estimate from above the maximal number of moves, we show that the contestant $C_{i}$ can make at most $2^{n-i}-1$ moves. Indeed, let us argue by \"backward extended induction\". To begin with, we observe that the estimate is trivially true for $C_{n}$ because she has no legal move.\n\nLet us assume now that the estimate has been proved for $C_{i}, C_{i+1}, \\ldots, C_{n}$, and let us prove it for $C_{i-1}$. When $C_{i-1}$ moves, at least one contestant $C_{j}$ with $j>i-1$ must precede her in the line. The initial configuration can provide at most $n-i$ contestants with larger index in front of $C_{i-1}$, which means at most $n-i$ moves for $C_{i-1}$. All other moves are possible only if some contestant in the range $C_{i}, C_{i+1}, \\ldots, C_{n}$ jumps over $C_{i-1}$ during her moves. As a consequence, the total number of moves of $C_{i-1}$ is at most\n$$\nn-i+\\sum_{k=i}^{n}\\left(2^{n-k}-1\\right)=2^{n-i+1}-1\n$$\nSumming over all indices we obtain that\n$$\n\\sum_{i=1}^{n}\\left(2^{n-i}-1\\right)=2^{n}-n-1\n$$\nwhich gives an estimate for the total number of moves.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle such that $\\angle CAB > \\angle ABC$, and let $I$ be its incentre. Let $D$ be the point on segment $BC$ such that $\\angle CAD = \\angle ABC$. Let $\\omega$ be the circle tangent to $AC$ at $A$ and passing through $I$. Let $X$ be the second point of intersection of $\\omega$ and the circumcircle of $ABC$. Prove that the angle bisectors of $\\angle DAB$ and $\\angle CXB$ intersect at a point on line $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $S$ be the intersection point of $BC$ and the angle bisector of $\\angle BAD$, and let $T$ be the intersection point of $BC$ and the angle bisector of $\\angle BXC$. We will prove that both quadruples $A, I, B, S$ and $A, I, B, T$ are concyclic, which yields $S = T$.\n\nFirstly denote by $M$ the middle of $\\operatorname{arc} AB$ of the circumcenter of $ABC$ which does not contain $C$. Consider the circle centered at $M$ passing through $A, I$ and $B$ (it is well-known that $MA = MI = MB$); let it intersect $BC$ at $B$ and $S'$. Since $\\angle BAC > \\angle CBA$ it is easy to check that $S'$ lies on side $BC$. Denoting the angles in $ABC$ by $\\alpha, \\beta, \\gamma$ we get\n$$\n\\angle BAD = \\angle BAC - \\angle DAC = \\alpha - \\beta.\n$$\nMoreover since $\\angle MBC = \\angle MBA + \\angle ABC = \\frac{\\gamma}{2} + \\beta$, then\n$$\n\\angle BMS' = 180^\\circ - 2\\angle MBC = 180^\\circ - \\gamma - 2\\beta = \\alpha - \\beta\n$$\nIt follows that $\\angle BAS' = 2\\angle BMS' = 2\\angle BAD$ which gives us $S = S'$.\n\n![](attached_image_1.png)\n\nSecondly let $N$ be the middle of arc $BC$ of the circumcenter of $ABC$ which does not contain $A$. From $\\angle BAC > \\angle CBA$ we conclude that $X$ lies on the $\\operatorname{arc} AB$ of circumcircle of $ABC$ not containing $C$. Obviously both $AI$ and $XT$ are passing through $N$. Since $\\angle NBT = \\frac{\\alpha}{2} = \\angle BXN$ we obtain $\\triangle NBT \\sim \\triangle NXB$, therefore\n$$\nNT \\cdot NX = NB^2 = NI^2\n$$\nIt follows that $\\triangle NTI \\sim \\triangle NIX$. Keeping in mind that $\\angle NBC = \\angle NAC = \\angle IXA$ we get\n$$\n\\angle TIN = \\angle IXN = \\angle NXA - \\angle IXA = \\angle NBA - \\angle NBC = \\angle TBA.\n$$\nIt means that $A, I, B, T$ are concyclic which ends the proof.\nSolution:\n\nLet $\\angle BAC = \\alpha$, $\\angle ABC = \\beta$, $\\angle BCA = \\gamma$, $\\angle ACX = \\phi$. Denote by $W_1$ and $W_2$ the intersections of segment $BC$ with the angle bisectors of $\\angle BXC$ and $\\angle BAD$ respectively. Then $BW_1 / W_1C = BX / XC$ and $BW_2 / W_2D = BA / AD$. We shall show that $BW_1 = BW_2$.\n\nSince $\\angle DAC = \\angle CBA$, triangles $ADC$ and $BAC$ are similar and therefore\n$$\n\\frac{DC}{AC} = \\frac{AC}{BC}\n$$\nBy the Law of sines\n$$\n\\frac{BW_2}{W_2D} = \\frac{BA}{AD} = \\frac{BC}{AC} = \\frac{\\sin \\alpha}{\\sin \\beta}.\n$$\nConsequently\n$$\n\\begin{gathered}\n\\frac{BD}{BW_2} = \\frac{W_2D}{BW_2} + 1 = \\frac{\\sin \\beta}{\\sin \\alpha} + 1, \\\\\n\\frac{BC}{BW_2} = \\frac{BC}{BD} \\cdot \\frac{BD}{BW_2} = \\frac{1}{1 - DC/BC} \\cdot \\frac{BD}{BW_2} = \\frac{1}{1 - AC^2 / BC^2} \\cdot \\frac{BD}{BW_2} = \\\\\n\\frac{\\sin^2 \\alpha}{\\sin^2 \\alpha - \\sin^2 \\beta} \\cdot \\frac{\\sin \\beta + \\sin \\alpha}{\\sin \\alpha} = \\frac{\\sin \\alpha}{\\sin \\alpha - \\sin \\beta}.\n\\end{gathered}\n$$\nNote that $AXBC$ is cyclic and so $\\angle BXC = \\angle BAC = \\alpha$. Hence, $\\angle XBC = 180^\\circ - \\angle BXC - \\angle BCX = 180^\\circ - \\alpha - \\phi$. By the Law of sines for the triangle $BXC$, we have\n$$\n\\begin{gathered}\n\\frac{BC}{W_1B} = \\frac{W_1C}{W_1B} + 1 = \\frac{CX}{BX} + 1 = \\frac{\\sin \\angle CBX}{\\sin \\phi} + 1 = \\\\\n\\frac{\\sin (\\alpha + \\phi)}{\\sin \\phi} + 1 = \\sin \\alpha \\cot \\phi + \\cos \\alpha + 1.\n\\end{gathered}\n$$\nSo, it's enough to prove that\n$$\n\\frac{\\sin \\alpha}{\\sin \\alpha - \\sin \\beta} = \\sin \\alpha \\cot \\phi + \\cos \\alpha\n$$\nSince $AC$ is tangent to the circle $AIX$, we have $\\angle AXI = \\angle IAC = \\alpha / 2$. Moreover $\\angle XAI = \\angle XAB + \\angle BAI = \\phi + \\alpha / 2$ and $\\angle XIA = 180^\\circ - \\angle XAI - \\angle AXI = 180^\\circ - \\alpha - \\phi$. Applying the Law of sines again $XAC, XAI, IAC$ we obtain\n$$\n\\begin{gathered}\n\\frac{AX}{\\sin (\\alpha + \\phi)} = \\frac{AI}{\\sin \\alpha / 2}, \\\\\n\\frac{AX}{\\sin (\\gamma - \\phi)} = \\frac{AC}{\\sin \\angle AXC} = \\frac{AC}{\\sin \\beta}, \\\\\n\\frac{AI}{\\sin \\gamma / 2} = \\frac{AC}{\\sin (\\alpha / 2 + \\gamma / 2)}.\n\\end{gathered}\n$$\nCombining the last three equalities we end up with\n$$\n\\begin{gathered}\n\\frac{\\sin (\\gamma - \\phi)}{\\sin (\\alpha + \\phi)} = \\frac{AI}{AC} \\cdot \\frac{\\sin \\beta}{\\sin \\alpha / 2} = \\frac{\\sin \\beta}{\\sin \\alpha / 2} \\cdot \\frac{\\sin \\gamma / 2}{\\sin (\\alpha / 2 + \\gamma / 2)} \\\\\n\\frac{\\sin (\\gamma - \\phi)}{\\sin (\\alpha + \\phi)} = \\frac{\\sin \\gamma \\cot \\phi - \\cos \\gamma}{\\sin \\alpha \\cot \\phi + \\cos \\alpha} = \\frac{2 \\sin \\beta / 2 \\sin \\gamma / 2}{\\sin \\alpha / 2}\n\\end{gathered}\n$$\n$$\n\\frac{\\sin \\alpha \\sin \\gamma \\cot \\phi - \\sin \\alpha \\cos \\gamma}{\\sin \\gamma \\sin \\alpha \\cot \\phi + \\sin \\gamma \\cos \\alpha} = \\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2}{\\cos \\gamma / 2}\n$$\nSubtracting 1 from both sides yields\n$$\n\\begin{gathered}\n\\frac{-\\sin \\alpha \\cos \\gamma - \\sin \\gamma \\cos \\alpha}{\\sin \\gamma \\sin \\alpha \\cot \\phi + \\sin \\gamma \\cos \\alpha} = \\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2}{\\cos \\gamma / 2} - 1 = \\\\\n\\frac{2 \\sin \\beta / 2 \\cos \\alpha / 2 - \\sin (\\alpha / 2 + \\beta / 2)}{\\cos \\gamma / 2} = \\frac{\\sin \\beta / 2 \\cos \\alpha / 2 - \\sin \\alpha / 2 \\cos \\beta / 2}{\\cos \\gamma / 2}, \\\\\n\\frac{-\\sin (\\alpha + \\gamma)}{\\sin \\gamma \\sin \\alpha \\cot \\phi + \\sin \\gamma \\cos \\alpha} = \\frac{\\sin (\\beta / 2 - \\alpha / 2)}{\\cos \\gamma / 2}, \\\\\n\\frac{-\\sin \\beta}{\\sin \\alpha \\cot \\phi + \\cos \\alpha} = 2 \\sin \\gamma / 2 \\sin (\\beta / 2 - \\alpha / 2) = \\\\\n2 \\cos (\\beta / 2 + \\alpha / 2) \\sin (\\beta / 2 - \\alpha / 2) = \\sin \\beta - \\sin \\alpha\n\\end{gathered}\n$$\nand the result follows. We are left to note that none of the denominators can vanish.\nSolution:\n\nWe first note that\n$$\n\\angle BAD = \\angle BAC - \\angle DAC = \\angle A - \\angle B.\n$$\nLet $CX$ and $AD$ meet at $K$. Then $\\angle CXA = \\angle ABC = \\angle KAC$. Also, we have $\\angle IXA = \\angle A/2$, since $\\omega$ is tangent to $AC$ at $A$. Therefore,\n$$\n\\angle DAI = |\\angle B - \\angle A/2| = |\\angle KXA - \\angle IXA| = \\angle KXI,\n$$\n(the absolute value depends on whether $\\angle B \\geq \\angle A/2$ or not) which means that $XKIA$ is cyclic, i.e. $K$ lies also on $\\omega$.\nLet $IK$ meet $BC$ at $E$. (If $\\angle B = \\angle A/2$, then $IK$ degenerates to the tangent line to $\\omega$ at $I$.) Note that $BEIA$ is cyclic, because\n$$\n\\angle EIA = 180^\\circ - \\angle KXA = 180^\\circ - \\angle ABE.\n$$\nWe have $\\angle EKA = 180^\\circ - \\angle AXI = 180^\\circ - \\angle A/2$ and $\\angle AEI = \\angle ABI = \\angle B/2$. Hence\n$$\n\\begin{aligned}\n\\angle EAK & = 180^\\circ - \\angle EKA - \\angle AEI \\\\\n& = 180^\\circ - (180^\\circ - \\angle A/2) - \\angle B/2 \\\\\n& = (\\angle A - \\angle B)/2 \\\\\n& = \\angle BAD/2.\n\\end{aligned}\n$$\nThis means that $AE$ is the angle bisector of $\\angle BAD$. Next, let $M$ be the point of intersection of $AE$ and $BI$. Then\n$$\n\\angle EMI = 180^\\circ - \\angle B/2 - \\angle BAD/2 = 180^\\circ - \\angle A/2\n$$\nand so, its supplement is\n$$\n\\angle AMI = \\angle A/2 = \\angle AXI\n$$\nso $X, M, K, I, A$ all lie on $\\omega$. Next, we have\n$$\n\\begin{aligned}\n\\angle XMA & = \\angle XKA \\\\\n& = 180^\\circ - \\angle ADC - \\angle XCB \\\\\n& = 180^\\circ - \\angle A - \\angle XCB \\\\\n& = \\angle B + \\angle XCA \\\\\n& = \\angle B + \\angle XBA \\\\\n& = \\angle XBE,\n\\end{aligned}\n$$\nand so $X, B, E, M$ are concyclic. Hence\n$$\n\\begin{aligned}\n\\angle EXC & = \\angle EXM + \\angle MXC \\\\\n& = \\angle MBE + \\angle MAK \\\\\n& = \\angle B/2 + \\angle BAD/2 \\\\\n& = \\angle A/2 \\\\\n& = \\angle BXC/2\n\\end{aligned}\n$$\nThis means that $XE$ is the angle bisector of $\\angle BXC$ and so we are done!\nSolution:\n\nIt is $\\angle ABD = \\angle DAC$, and so $\\overline{AC}$ is tangent to the circumcircle of $\\triangle BAD$ at $A$. Hence $CA^2 = CD \\cdot CB$.\n\n![](attached_image_2.png)\n\nTriangle $ABC$ is similar to triangle $CAD$, because $\\angle C$ is a common angle and $\\angle CAD = \\angle ABC$, and so $\\angle ADC = \\angle BAC = 2\\varphi$.\nLet $Q$ be the point of intersection of $AD$ and $CX$. Since $\\angle BXC = \\angle BAC = 2\\varphi$, it follows that $BDQX$ is cyclic. Therefore, $CD \\cdot CB = CQ \\cdot CX = CA^2$ which implies that $Q$ lies on $\\omega$.\nNext let $P$ be the point of intersection of $AD$ with the circumcircle of triangle $ABC$. Then $\\angle PBC = \\angle PAC = \\angle ABC = \\angle APC$ yielding $CA = CP$. So, let $T$ be on the side $BC$ such that $CT = CA = CP$. Then\n$$\n\\angle TAD = \\angle TAC - \\angle DAC = \\left(90^\\circ - \\frac{\\angle C}{2}\\right) - \\angle B = \\frac{\\angle A - \\angle B}{2} = \\frac{\\angle BAD}{2}\n$$\nthat is, line $AT$ is the angle bisector of $\\angle BAD$. We want to show that $XT$ is the angle bisector of $\\angle BXC$. To this end, it suffices to show that $\\angle TXC = \\varphi$.\nIt is $CT^2 = CA^2 = CQ \\cdot CX$, and so $CT$ is tangent to the circumcircle of $XTQ$ at $T$. Since $\\angle TXQ = \\angle QTC$ and $\\angle QDC = 2\\varphi$, it suffices to show that $\\angle TQD = \\varphi$, or, in other words, that $I, Q$, and $T$ are collinear.\nLet $T'$ be the point of intersection of $IQ$ and $BC$. Then $\\triangle AIC$ is congruent to $\\triangle T'IC$, since they share $CI$ as a common side, $\\angle ACI = \\angle T'CI$, and\n$$\n\\angle IT'D = 2\\varphi - \\angle T'QD = 2\\varphi - \\angle IQA = 2\\varphi - \\angle IXA = \\varphi = \\angle IAC.\n$$\nTherefore, $CT' = CA = CT$, which means that $T$ coincides with $T'$ and completes the proof.\nSolution:\n\nLet $G$ be the point of intersection of $AD$ and $CX$. Since the quadrilateral $AXBC$ is cyclic, it is $\\angle AXC = \\angle ABC$.\n\n![](attached_image_3.png)\n\nLet the line $AD$ meet $\\omega$ at $K$. Then it is $\\angle AXK = \\angle CAD = \\angle ABC$, because the angle that is formed by a chord and a tangent to the circle at an endpoint of the chord equals the inscribed angle to that chord. Therefore, $\\angle AXK = \\angle AXC = \\angle AXG$. This means that the point $G$ coincides with the point $K$ and so $G$ belongs to the circle $\\omega$.\nLet $E$ be the point of intersection of the angle bisector of $\\angle DAB$ with $BC$. It suffices to show that\n$$\n\\frac{CE}{BE} = \\frac{XC}{XB}.\n$$\nLet $F$ be the second point of intersection of $\\omega$ with $AB$. Then we have $\\angle IAF = \\frac{\\angle CAB}{2} = \\angle IXF$, where $I$ is the incenter of $ABC$, because $\\angle IAF$ and $\\angle IXF$ are inscribed in the same arc of $\\omega$. Thus $\\triangle AIF$ is isosceles with $AI = IF$. Since $I$ is the incenter of $ABC$, we have $AF = 2(s-a)$, where $s = (a+b+c)/2$ is the semiperimeter of $ABC$. Also, it is $CE = AC = b$ because in triangle $ACE$, we have\n$$\n\\begin{aligned}\n\\angle AEC & = \\angle ABC + \\angle BAE \\\\\n& = \\angle ABC + \\frac{\\angle BAD}{2} \\\\\n& = \\angle ABC + \\frac{\\angle BAC - \\angle ABC}{2} \\\\\n& = 90^\\circ - \\frac{\\angle ACE}{2}\n\\end{aligned}\n$$\nand so $\\angle CAE = 180^\\circ - \\angle AEC - \\angle ACE = 90^\\circ - \\frac{\\angle ACE}{2} = \\angle AEC$. Hence\n$$\nBF = BA - AF = c - 2(s-a) = a - b = CB - CE = BE.\n$$\nMoreover, triangle $CAX$ is similar to triangle $BFX$, because $\\angle ACX = \\angle FBX$ and\n$$\n\\angle XFB = \\angle XAF + \\angle AXF = \\angle XAF + \\angle CAF = \\angle CAX\n$$\nTherefore\n$$\n\\frac{CE}{BE} = \\frac{AC}{BF} = \\frac{XC}{XB},\n$$\nas desired. The proof is complete.\nSolution:\n\nLet $\\omega$ denote the circle through $A$ and $I$ tangent to $AC$. Let $Y$ be the second point of intersection of the circle $\\omega$ with the line $AD$. Let $L$ be the intersection of $BC$ with the angle bisector of $\\angle BAD$. We will prove $\\angle LXC = \\frac{1}{2} \\angle BAC = \\frac{1}{2} \\angle BXC$.\n\nWe will refer to the angles of $ABC$ as $\\angle A, \\angle B, \\angle C$. Thus $\\angle BAD = \\angle A - \\angle B$.\nOn the circumcircle of $ABC$, we have $\\angle AXC = \\angle ABC = \\angle CAD$, and since $AC$ is tangent to $\\omega$, we have $\\angle CAD = \\angle CAY = \\angle AXY$. Hence $C, X, Y$ are collinear.\nAlso note that $CAL$ is isosceles with $\\angle CAL = \\angle CLA = \\frac{1}{2}(\\angle BAD) + \\angle ABC = \\frac{1}{2}(\\angle A + \\angle B)$ hence $AC = CL$. Moreover, $CI$ is angle bisector to $\\angle ACL$ so it's the symmetry axis for the triangle, hence $\\angle ILC = \\angle IAC = \\frac{1}{2} \\angle A$ and $\\angle ALI = \\angle LIA = \\frac{1}{2} \\angle B$. Since $AC$ is tangent to $\\omega$, we have $\\angle AYI = \\angle IAC = \\frac{1}{2} \\angle A = \\angle LAY + \\angle ALI$. Hence $L, Y, I$ are collinear.\nSince $AC$ is tangent to $\\omega$, we have $\\triangle CAY \\sim \\triangle CXA$ hence $CA^2 = CX \\cdot CY$. However we proved $AC = CL$ hence $CL^2 = CX \\cdot CY$. Hence $\\triangle CLY \\sim \\triangle CXL$ and hence $\\angle CXL = \\angle CLY = \\angle CAI = \\frac{1}{2} \\angle A$.\nSolution:\n\nLet $M$ be the midpoint of the $\\operatorname{arc} BC$. Let $\\omega$ denote the circle through $A$ and $I$ tangent to $AC$. Let $N$ be the second point of intersection of $\\omega$ with $AB$ and $L$ the intersection of $BC$ with the angle bisector of $\\angle BAD$. We know $\\frac{DL}{LB} = \\frac{AD}{AB}$ and want to prove $\\frac{XB}{XC} = \\frac{LB}{LC}$.\n\nFirst note that $CAL$ is isosceles with $\\angle CAL = \\angle CLA = \\frac{1}{2}(\\angle BAD) + \\angle ABC$ hence $AC = CL$ and $\\frac{LB}{LC} = \\frac{LB}{AC}$.\nNow we calculate $\\frac{XB}{XC}$:\nComparing angles on the circles $\\omega$ and the circumcircle of $ABC$ we get $\\triangle XIN \\sim \\triangle XMB$ and hence also $\\triangle XIM \\sim \\triangle XNB$ (having equal angles at $X$ and proportional adjoint sides). Hence $\\frac{XB}{XM} = \\frac{NB}{IM}$.\nAlso comparing angles on the circles $\\omega$ and the circumcircle of $ABC$ and using the tangent $AC$ we get $\\triangle XAI \\sim \\triangle XCM$ and hence also $\\triangle XAC \\sim \\triangle XIM$. Hence $\\frac{XC}{XM} = \\frac{AC}{IM}$.\nComparing the last two equations we get $\\frac{XB}{XC} = \\frac{NB}{AC}$. Comparing with $\\frac{LB}{LC} = \\frac{LB}{AC}$, it remains to prove $NB = LB$.\n\n![](attached_image_4.png)\n\nWe prove $\\triangle INB \\equiv \\triangle ILB$ as follows:\nFirst, we note that $I$ is the circumcentre of $\\triangle ALN$. Indeed, $CI$ is angle bisector in the isosceles triangle $ACL$ so it's perpendicular bisector for $AL$. As well, $\\triangle IAN$ is isosceles with $\\angle INA = \\angle CAI = \\angle IAB$ hence $I$ is also on the perpendicular bisector of $AN$.\nHence $IN = IL$ and also $\\angle NIL = 2\\angle NAL = \\angle A - \\angle B = 2\\angle NIB$ (the last angle is calculated using that the exterior angle of $\\triangle NIB$ is $\\angle INA = \\angle A/2$. Hence $\\angle NIB = \\angle LIB$ and $\\triangle INB \\equiv \\triangle ILB$ by SAS.\nSolution:\n\nLet $M, N$ be the midpoints of arcs $BC, BA$ of the circumcircle $ABC$, respectively. Let $Y$ be the second intersection of $AD$ and circle $ABC$. Let $E$ be the incenter of triangle $ABY$ and note that $E$ lies on the angle bisectors of the triangle, which are the lines $YN$ (immediate), $BC$ (since $\\angle CBY = \\angle CAY = \\angle CAD = \\angle ABC$) and the angle bisector of $\\angle DAB$; so the question reduces to showing that $E$ is also on $XM$, which is the angle bisector of $\\angle CXB$.\n\nWe claim that the three lines $CX, ADY, IE$ are concurrent at a point $D'$. We will complete the proof using this fact, and the proof will appear at the end (and see the solution by HEL5 for an alternative proof of this fact).\nTo show that $XEM$ are collinear, we construct a projective transformation which projects $M$ to $X$ through center $E$. We produce it as a composition of three other projections. Let $O$ be the intersection of lines $AD'DY$ and $CIN$. Projecting the points $YNC M$ on the circle $ABC$ through the (concyclic) point $A$ to the line $CN$ yields the points $ONCI$. Projecting these points through $E$ to the line $AY$ yields $OYDD'$ (here we use the facts that $D'$ lies on $IE$ and $AY$). Projecting these points to the circle $ABC$ through $C$ yields $NYBX$ (here we use the fact that $D'$ lies on $CX$). Composing, we observe that we found a projection of the circle $ABC$ to itself sending $YNC M$ to $NYBX$. Since the projection of the circle through $E$ also sends $YNC$ to $NYB$, and three points determine a projective transformation, the projection through $E$ also sends $M$ to $X$, as claimed.\n\n![](attached_image_5.png)\n\nLet $B', D'$ be the intersections of $AB, AD$ with the circle $AXI$, respectively. We wish to show that this $D'$ is the concurrency point defined above, i.e. that $CD'X$ and $ID'E$ are collinear. Additionally, we will show that $I$ is the circumcenter of $AB'E$.\nConsider the inversion with center $C$ and radius $CA$. The circles $AXI$ and $ABD$ are tangent to $CA$ at $A$ (the former by definition, the latter since $\\angle CAD = \\angle ABC$), so they are preserved under the inversion. In particular, the inversion transposes $D$ and $B$ and preserves $A$, so sends the circle $CAB$ to the line $AD$. Thus $X$, which is the second intersection of circles $ABC$ and $AXI$, is sent by the inversion to the second intersection of $AD$ and circle $AXI$, which is $D'$. In particular $CD'X$ are collinear.\nIn the circle $AIB'$, $AI$ is the angle bisector of $B'A$ and the tangent at $A$, so $I$ is the midpoint of the arc $AB'$, and in particular $AI = IB'$. By angle chasing, we find that $ACE$ is an isosceles triangle:\n$\\angle CAE = \\angle CAD + \\angle DAE = \\angle ABC + \\angle EAB = \\angle ABE + \\angle EAB = \\angle AEB = \\angle AEC$,\nthus the angle bisector $CI$ is the perpendicular bisector of $AE$ and $AI = IE$. Thus $I$ is the circumcenter of $AB'E$.\nWe can now show that $ID'E$ are collinear by angle chasing:\n$$\n\\angle EIB' = 2\\angle EAB' = 2\\angle EAB = \\angle DAB = \\angle D'AB' = \\angle D'IB'.\n$$\nSolution:\n\nLet $W$ be the midpoint of arc $BC$, let $D'$ be the second intersection point of $AD$ and the circle $ABC$. Let $P$ be the intersection of the angle bisector $XW$ of $\\angle CXB$ with $BC$; we wish to prove that $AP$ is the angle bisector of $DAB$. Denote $\\alpha = \\frac{\\angle CAB}{2}$, $\\beta = \\angle ABC$.\nLet $M$ be the intersection of $AD$ and $XC$. Angle chasing finds:\n$$\n\\begin{aligned}\n\\angle MXI & = \\angle AXI - \\angle AXM = \\angle CAI - \\angle AXC = \\alpha - \\beta \\\\\n& = \\angle CAI - \\angle CAD = \\angle DAI = \\angle MAI\n\\end{aligned}\n$$\nAnd in particular $M$ is on $\\omega$. By angle chasing we find\n$$\n\\angle XIA = \\angle IXA + \\angle XAI = \\angle ICA + \\angle XAI = \\angle XAC = \\angle XBC = \\angle XBP\n$$\nand $\\angle PXB = \\alpha = \\angle CAI = \\angle AXI$, and it follows that $\\triangle XIA \\sim \\triangle XBP$. Let $S$ be the second intersection point of the cirumcircles of $XIA$ and $XBP$. Then by the spiral map lemma (or by the equivalent angle chasing) it follows that $ISB$ and $ASP$ are collinear.\nLet $L$ be the second intersection of $\\omega$ and $AB$. We want to prove that $ASP$ is the angle bisector of $DAB = \\angle MAL$, i.e. that $S$ is the midpoint of the $\\operatorname{arc} ML$ of $\\omega$. And this follows easily from chasing angular arc lengths in $\\omega$:\n$$\n\\begin{aligned}\n& \\overparen{AI} = \\angle CAI = \\alpha \\\\\n& \\widehat{IL} = \\angle IAL = \\alpha \\\\\n& \\widehat{MI} = \\angle MXI = \\alpha - \\beta \\\\\n& \\widehat{AI} - \\widehat{SL} = \\angle ABI = \\frac{\\beta}{2}\n\\end{aligned}\n$$\nAnd thus\n$$\n\\widehat{ML} = \\widehat{MI} + \\widehat{IL} = 2\\alpha - \\beta = 2\\left(\\widehat{AI} - \\frac{\\beta}{2}\\right) = 2\\widehat{SL}\n$$\nSolution:\n\nLet $E$ be the intersection of the bisector of $\\angle BAD$ and $BC$, and $N$ be the middle point of arc $BC$ of the circumcircle of $ABC$. Then it suffices to show that $E$ is on line $XN$.\nWe consider the inversion at $A$. Let $P^*$ be the image of a point denoted by $P$. Then $A, B^*, C^*, E^*$ are concyclic, $X^*, B^*, C^*$ are colinear, and $X^* I^*$ and $AC^*$ are parallel. Now it suffices to show that $A, X^*, E^*, N^*$ are concyclic. Let $Y$ be the intersection of $B^*C^*$ and $AE^*$. Then, by the power of a point, we get\n$$\n\\begin{aligned}\nA, X^*, E^*, N^* \\text{ are concyclic } &\\Longleftrightarrow YX^* \\cdot YN^* = YA \\cdot YE^* \\\\\n&\\Longleftrightarrow YX^* \\cdot YN^* = YB^* \\cdot YC^* \\\\\n&\\left(A, B^*, C^*, E^* \\text{ are concyclic }\\right)\n\\end{aligned}\n$$\nHere, by the property of inversion, we have\n$$\n\\angle AI^*B^* = \\angle ABI = \\frac{1}{2} \\angle ABC = \\frac{1}{2} \\angle C^*AD^*.\n$$\n![](attached_image_6.png)\nDefine $Q, R$ as described in the figure, and we get by simple angle chasing\n$$\n\\angle QAI^* = \\angle QI^*A, \\quad \\angle RAI^* = \\angle B^*I^*A\n$$\nEspecially, $B^*R$ and $AI^*$ are parallel, so that we have\n$$\n\\frac{YB^*}{YN^*} = \\frac{YR}{YA} = \\frac{YX^*}{YC^*}\n$$\nand the proof is completed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all lists $\\left(x_{1}, x_{2}, \\ldots, x_{2020}\\right)$ of non-negative real numbers such that the following three conditions are all satisfied:\n(i) $x_{1} \\leq x_{2} \\leq \\ldots \\leq x_{2020}$;\n(ii) $x_{2020} \\leq x_{1}+1$;\n(iii) there is a permutation $\\left(y_{1}, y_{2}, \\ldots, y_{2020}\\right)$ of $\\left(x_{1}, x_{2}, \\ldots, x_{2020}\\right)$ such that\n$$\n\\sum_{i=1}^{2020}\\left(\\left(x_{i}+1\\right)\\left(y_{i}+1\\right)\\right)^{2}=8 \\sum_{i=1}^{2020} x_{i}^{3}\n$$\nA permutation of a list is a list of the same length, with the same entries, but the entries are allowed to be in any order. For example, $(2,1,2)$ is a permutation of $(1,2,2)$, and they are both permutations of $(2,2,1)$. Note that any list is a permutation of itself.",
    "options": [],
    "answer": "Either the list with the first 1010 entries equal to 0 and the last 1010 entries equal to 1, or the list with the first 1010 entries equal to 1 and the last 1010 entries equal to 2.",
    "solution": "Solution:\nWe first prove the inequality\n$$\n((x+1)(y+1))^{2} \\geq 4\\left(x^{3}+y^{3}\\right)\n$$\nfor real numbers $x, y \\geq 0$ satisfying $|x-y| \\leq 1$, with equality if and only if $\\{x, y\\}=\\{0,1\\}$ or $\\{x, y\\}=\\{1,2\\}$.\nIndeed,\n$$\n\\begin{aligned}\n4\\left(x^{3}+y^{3}\\right) & =4(x+y)\\left(x^{2}-x y+y^{2}\\right) \\\\\n& \\leq\\left((x+y)+\\left(x^{2}-x y+y^{2}\\right)\\right)^{2} \\\\\n& =\\left(x y+x+y+(x-y)^{2}\\right)^{2} \\\\\n& \\leq(x y+x+y+1)^{2} \\\\\n& =((x+1)(y+1))^{2},\n\\end{aligned}\n$$\nwhere the first inequality follows by applying the AM-GM inequality on $x+y$ and $x^{2}-x y+y^{2}$ (which are clearly nonnegative). Equality holds in the first inequality precisely if $x+y=x^{2}-x y+y^{2}$ and in the second one if and only if $|x-y|=1$. Combining these equalities we have $x+y=(x-y)^{2}+x y=1+x y$ or $(x-1)(y-1)=0$, which yields the solutions $\\{x, y\\}=\\{0,1\\}$ or $\\{x, y\\}=\\{1,2\\}$.\n\nNow, let $\\left(x_{1}, x_{2}, \\ldots, x_{2020}\\right)$ be any sequence satisfying conditions (i) and (ii) and let $\\left(y_{1}, y_{2}, \\ldots, y_{2020}\\right)$ be any permutation of $\\left(x_{1}, x_{2}, \\ldots, x_{2020}\\right)$. As $0 \\leq \\min \\left(x_{i}, y_{i}\\right) \\leq \\max \\left(x_{i}, y_{i}\\right) \\leq \\min \\left(x_{i}, y_{i}\\right)+1$, we can apply inequality (1) to the pair $\\left(x_{i}, y_{i}\\right)$ and sum over all $1 \\leq i \\leq 2020$ to conclude that\n$$\n\\sum_{i=1}^{2020}\\left(\\left(x_{i}+1\\right)\\left(y_{i}+1\\right)\\right)^{2} \\geq 4 \\sum_{i=1}^{2020}\\left(x_{i}^{3}+y_{i}^{3}\\right)=8 \\sum_{i=1}^{2020} x_{i}^{3}\n$$\nTherefore, in order to satisfy condition (iii), every inequality must be an equality. Hence, for every $1 \\leq i \\leq 2020$ we must have $\\{x_{i}, y_{i}\\}=\\{0,1\\}$ or $\\{x_{i}, y_{i}\\}=\\{1,2\\}$. By condition (ii), we see that either $\\{x_{i}, y_{i}\\}=\\{0,1\\}$ for all $i$ or $\\{x_{i}, y_{i}\\}=\\{1,2\\}$ for all $i$.\n\nIf $\\{x_{i}, y_{i}\\}=\\{0,1\\}$ for every $1 \\leq i \\leq 2020$, this implies that the sequences $\\left(x_{1}, x_{2}, \\ldots, x_{2020}\\right)$ and $\\left(y_{1}, y_{2}, \\ldots, y_{2020}\\right)$ together have 2020 zeroes and 2020 ones. As $\\left(y_{1}, y_{2}, \\ldots, y_{2020}\\right)$ is a permutation of $\\left(x_{1}, x_{2}, \\ldots, x_{2020}\\right)$ this implies that $\\left(x_{1}, x_{2}, \\ldots, x_{2020}\\right)=(0,0, \\ldots, 0,1,1, \\ldots, 1)$ with 1010 zeroes and 1010 ones. Conversely, note that this sequence satisfies conditions (i), (ii), and (iii) (in (iii), we take $\\left(y_{1}, y_{2}, \\ldots, y_{2020}\\right)=\\left(x_{2020}, x_{2019}, \\ldots, x_{1}\\right)$), showing that this sequence indeed works. The same reasoning holds for the case that $\\{x_{i}, y_{i}\\}=\\{1,2\\}$ for all $i$.\n\nTherefore, the only solutions are:\n$$(\\underbrace{0,0, \\ldots, 0}_{1010}, \\underbrace{1,1, \\ldots, 1}_{1010})$$\nand\n$$(\\underbrace{1,1, \\ldots, 1}_{1010}, \\underbrace{2,2, \\ldots, 2}_{1010})$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $A B C D E F$ be a convex hexagon such that $\\angle A=\\angle C=\\angle E$ and $\\angle B=\\angle D=\\angle F$ and the (interior) angle bisectors of $\\angle A, \\angle C$, and $\\angle E$ are concurrent.\nProve that the (interior) angle bisectors of $\\angle B, \\angle D$, and $\\angle F$ must also be concurrent.\n\nNote that $\\angle A=\\angle F A B$. The other interior angles of the hexagon are similarly described.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenote the angle bisector of $A$ by $a$ and similarly for the other bisectors. Thus, given that $a, c, e$ have a common point $M$, we need to prove that $b, d, f$ are concurrent. We write $\\angle(x, y)$ for the value of the directed angle between the lines $x$ and $y$, i.e. the angle of the counterclockwise rotation from $x$ to $y$ (defined $\\bmod 180^\\circ$).\n\nSince the sum of the angles of a convex hexagon is $720^\\circ$, from the angle conditions we get that the sum of any two consecutive angles is equal to $240^\\circ$. In particular, it now follows that $\\angle(b, a)=\\angle(c, b)=\\angle(d, c)=\\angle(e, d)=\\angle(f, e)=\\angle(a, f)=60^\\circ$ (assuming the hexagon is clockwise oriented).\n\nLet $X=AB \\cap CD$, $Y=CD \\cap EF$ and $Z=EF \\cap AB$. Similarly, let $P=BC \\cap DE$, $Q=DE \\cap FA$ and $R=FA \\cap BC$. From $\\angle B+\\angle C=240^\\circ$ it follows that $\\angle(ZX, XY)=\\angle(BX, XC)=60^\\circ$. Similarly we have $\\angle(XY, YZ)=\\angle(YZ, ZX)=60^\\circ$, so triangle $XYZ$ (and similarly triangle $PQR$) is equilateral. We see that the hexagon $ABCDEF$ is obtained by intersecting the two equilateral triangles $XYZ$ and $PQR$.\n\nWe have $\\angle(AM, MC)=\\angle(a, c)=\\angle(a, b)+\\angle(b, c)=60^\\circ$, and since\n$$\n\\angle(AM, MC)=\\angle(AX, XC)=\\angle(AR, RC)=60^\\circ\n$$\n$A, C, M, X, R$ are concyclic. Because $M$ lies on the bisector of angle $\\angle XAR$, we must have $MR=MX$, so triangle $MRX$ is isosceles. Moreover, we have $\\angle(MR, MX)=\\angle(CR, CX)=\\angle(BC, CD)$, which is angle $C$ of the hexagon. We now see that the triangles $MRX, MPY$ and $MQZ$ are isosceles and similar. This implies that there is a rotation centered at $M$ that sends $X, Y$ and $Z$ to $R, P$ and $Q$ respectively. In particular, the equilateral triangles $XYZ$ and $PQR$ are congruent.\n\nIt follows that there also exists a rotation sending $X, Y, Z$ to $P, Q, R$ respectively. Define $N$ as the center of this rotation. Triangles $NXZ$ and $NPR$ are congruent and equally oriented, hence $N$ is equidistant from $XZ$ and $PR$ and lies on the inner bisector $b$ of $\\angle B$ (we know $N$ lies on the inner, not the outer bisector because the rotation centered at $N$ is clockwise). In the same way we can show that $N$ is on $d$ and on $f$, so $b, d, f$ are concurrent at $N$.\n\nRemark. The key observation (a rotation centered at $M$ sends $\\triangle XYZ$ to $\\triangle RQP$) can be established in slightly different ways. E.g., since $A, M, R, X$ are concyclic and $A, M, Q, Z$ are concyclic, $M$ is the Miquel point of the lines $XZ, RQ, XR, ZQ$, hence it is the center of similitude $s$ sending $\\overrightarrow{XZ}$ to $\\overrightarrow{RQ}$. Repeating the same argument for the other pairs of vectors, we obtain that $s$ sends $\\triangle XYZ$ to $\\triangle RQP$. Moreover, $s$ is a rotation, since $M$ is equidistant from $XZ$ and $RQ$.\n\nRemark. The reverse argument can be derived in a different way, e.g., defining $N$ as the common point of the circles $BXP D, D YQ F, FZR B$, and showing that $\\triangle NXZ=\\triangle NPR$, etc.\n\n![](attached_image_1.png)\nSolution:\n\nAs in Solution A, we prove that the hexagon $ABCDEF$ is the intersection of the equilateral triangles $PQR$ and $XYZ$.\n\nLet $d(S, AB)$ denote the signed distance from the point $S$ to the line $AB$, where the negative sign is taken if $AB$ separates $S$ and the hexagon. We define similarly the other distances ($d(S, BC)$, etc). Since $M \\in a$, we have $d(M, ZX)=d(M, QR)$. In the same way, we have $d(M, XY)=d(M, RP)$ and $d(M, YZ)=d(M, PQ)$. Therefore $d(M, ZX)+d(M, XY)+d(M, YZ)=d(M, QR)+d(M, RP)+d(M, PQ)$.\n\nWe now use the following well-known lemma (which can be easily proved using areas) to deduce that triangles $PQR$ and $XYZ$ are congruent.\n\nLemma. The sum of the signed distances from any point to the sidelines of an equilateral triangle (where the signs are taken such that all distances are positive inside the triangle) is constant and equals the length of the altitude.\n\nFor $N=b \\cap d$ we now find $d(N, ZX)=d(N, RP)$ and $d(N, XY)=d(N, PQ)$. Using again the lemma for the point $N$, we get $d(N, ZX)+d(N, XY)+d(N, YZ)=d(N, QR)+d(N, RP)+d(N, PQ)$. Therefore $d(N, YZ)=d(N, QR)$, thus $N \\in f$.\n\nRemark. Instead of using the lemma, it is possible to use some equivalent observation in terms of signed areas.\nSolution:\n\nWe use the same notations as in Solution A. We will show that $a, c$ and $e$ are concurrent if and only if\n$$\nAB+CD+EF=BC+DE+FA,\n$$\nwhich clearly implies the problem statement by symmetry.\n\nLet $\\vec{a}$ be the vector of unit length parallel to $a$ directed from $A$ towards the interior of the hexagon. We define analogously $\\vec{b}$, etc. The angle conditions imply that opposite bisectors of the hexagon are parallel, so we have $\\vec{a}\\parallel\\vec{d}$, $\\vec{b}\\parallel \\vec{e}$ and $\\vec{c} \\parallel \\vec{f}$. Moreover, as in the previous solutions, we know that $\\vec{a}, \\vec{c}$ and $\\vec{e}$ make angles of $120^\\circ$ with each other. Let $M_A=c \\cap e, M_C=e \\cap a$ and $M_E=a \\cap c$. Then $M_A, M_C, M_E$ form an equilateral triangle with side length denoted by $s$. Note that the case $s=0$ is equivalent to $a, c$ and $e$ being concurrent.\n\nProjecting $\\overrightarrow{M_EA}+\\overrightarrow{AB}=\\overrightarrow{M_EB}=\\overrightarrow{M_EC}+\\overrightarrow{CB}$ onto $\\vec{e}=-\\vec{b}$, we obtain\n$$\n\\overrightarrow{AB} \\cdot \\vec{b}-\\overrightarrow{CB} \\cdot \\vec{b}=\\overrightarrow{M_EC} \\cdot \\vec{b}-\\overrightarrow{M_EA} \\cdot \\vec{b}=\\overrightarrow{M_EA} \\cdot \\vec{e}-\\overrightarrow{M_EC} \\cdot \\vec{e}\n$$\nWriting $\\varphi=\\frac{1}{2} \\angle B=\\frac{1}{2} \\angle D=\\frac{1}{2} \\angle F$, we know that $\\overrightarrow{AB} \\cdot \\vec{b}=-AB \\cdot \\cos (\\varphi)$, and similarly $\\overrightarrow{CB} \\cdot \\vec{b}=-CB \\cdot \\cos (\\varphi)$. Because $M_EA$ and $M_EC$ intersect $e$ at $120^\\circ$ angles, we have $\\overrightarrow{M_EA} \\cdot \\vec{e}=\\frac{1}{2} M_EA$ and $\\overrightarrow{M_EC} \\cdot \\vec{e}=\\frac{1}{2} M_EC$. We conclude that\n$$\n2 \\cos (\\varphi)(AB-CB)=M_EC-M_EA\n$$\nAdding the analogous equalities $2 \\cos (\\varphi)(CD-ED)=M_AE-M_AC$ and $2 \\cos (\\varphi)(EF-AF)=M_CA-M_CE$, we obtain\n$$\n2 \\cos (\\varphi)(AB+CD+EF-CB-ED-AF)=M_EC-M_EA+M_AE-M_AC+M_CA-M_CE.\n$$\nBecause $M_A, M_C$ and $M_E$ form an equilateral triangle with side length $s$, we have $M_EC-M_AC=\\pm s, M_CA-M_EA=\\pm s$, and $M_AE-M_CE=\\pm s$. Therefore, the right hand side $M_EC-M_EA+M_AE-M_AC+M_CA-M_CE$ equals $\\pm s \\pm s \\pm s$, which (irrespective of the choices of the $\\pm$-signs) is $0$ if and only if $s=0$. Because $\\cos (\\varphi) \\neq 0$, we conclude that\n$$\nAB+CD+EF=CB+ED+AF \\Longleftrightarrow s=0 \\Longleftrightarrow a, c, e \\text{ concurrent, }\n$$\nas desired.\n\nRemark. Equalities used in the solution could appear in different forms, in particular, in terms of signed lengths.\n\nRemark. Similar solutions could be obtained by projecting onto the line perpendicular to $b$ instead of $b$.\nSolution:\n\nWe use the same notations as in previous solutions and the fact that $a \\parallel d$, $b \\parallel e$ and $c \\parallel f$ make angles of $120^{\\circ}$. Also, we may assume that $E$ and $C$ are not symmetric in $a$ (if they are, the entire figure is symmetric and the conclusion is immediate).\n\nWe consider two mappings: the first one $s: a \\rightarrow BC \\rightarrow d$ sending $A' \\mapsto B' \\mapsto S$ is defined such that $A'B' \\parallel AB$ and $B'S \\parallel b$, and the second one $t: a \\rightarrow EF \\rightarrow d$ sending $A' \\mapsto F' \\mapsto T$ is defined such that $A'F' \\parallel AF$ and $F'T \\parallel f$. Both maps are affine linear since they are compositions of affine transformations. We will prove that they coincide by finding two distinct points $A', A'' \\in a$ for which $s(A')=t(A')$ and $s(A'')=t(A'')$. Then we will obtain that $s(A)=t(A)$, which by construction implies that the bisectors of $\\angle B, \\angle D$ and $\\angle F$ are concurrent.\n\nWe will choose $A'$ to be the reflection of $C$ in $e$ and $A''$ to be the reflection of $E$ in $c$. They are distinct since otherwise $C$ and $E$ would be symmetric in $a$. Applying the above maps $a \\rightarrow BC$ and $a \\rightarrow EF$ to $A'$, we get points $B'$ and $F'$ such that $A'B' C D E F'$ satisfies the problem statement. However, this hexagon is symmetric in $e$, hence the bisectors of $\\angle B', \\angle D, \\angle F'$ are concurrent and $s(A')=t(A')$. The same reasoning yields $s(A'')=t(A'')$, which finishes the solution.\n\nRemark. This solution is based on the fact that two specific affine linear maps coincide. Here it was proved by exhibiting two points where they coincide. One could prove it in another way, exhibiting one such point and proving that the 'slopes' are equal.\n\nRemark. There are similar solutions where claims and proofs could be presented in more 'elementary' terms. For example, an elementary reformulation of the 'slopes' being equal is: if $b'$ passes through $B'$ parallel to $b$, and $f'$ passes through $F'$ parallel to $f$, then the line through $b \\cap f$ and $b' \\cap f'$ is parallel to $a$ (which is parallel to $d$).\nSolution:\n\nWe use the same notations as in previous solutions.\n\nSince the sum of the angles of a convex hexagon is $720^{\\circ}$, from the angle conditions we get $\\angle B+\\angle C=720^{\\circ} / 3=240^{\\circ}$. From $\\angle B+\\angle C=240^{\\circ}$ it follows that the angle between $c$ and $b$ equals $60^{\\circ}$. The same is analogously true for other pairs of bisectors of neighboring angles.\n\nConsider the points $O_a \\in a, O_c \\in c, O_e \\in e$, each at the same distance $d'$ from $M$, where $d'>\\max \\{MA, MC, ME\\}$, and such that the rays $AO_a, CO_c, EO_e$ point out of the hexagon. By construction, $O_a$ and $O_c$ are symmetrical in $e$, hence $O_aO_c \\perp b$. Similarly, $O_cO_e \\perp d, O_eO_a \\perp f$. Thus it suffices to prove that perpendiculars from $B, D, F$ to the sidelines of $\\triangle O_aO_cO_e$ are concurrent. By a well-known criterion, this condition is equivalent to equality\n$$\nO_aB^2-O_cB^2+O_cD^2-O_eD^2+O_eF^2-O_aF^2=0\n$$\nTo prove $(* )$ consider a circle $\\omega_a$ centered at $O_a$ and tangent to $AB$ and $AF$ and define circles $\\omega_c$ and $\\omega_e$ in the same way. Rewrite $O_aB^2$ as $r_a^2+B_aB^2$, where $r_a$ is the radius of $\\omega_a$, and $B_a$ is the touch point of $\\omega_a$ with $AB$. Using similar notation for the other tangent points, transform $(* )$ into\n$$\nB_aB^2-B_cB^2+D_cD^2-D_eD^2+F_eF^2-F_aF^2=0.\n$$\nFurthermore, $\\angle O_cO_aB_a=\\angle MO_aB_a+\\angle O_cO_aM=(90^{\\circ}-\\varphi)+30^{\\circ}=120^{\\circ}-\\varphi$, where $\\varphi=\\frac{1}{2} \\angle A$. (Note that $\\varphi>30^{\\circ}$, since $ABCDEF$ is convex.) By analogous arguments, $\\angle O_aO_cB_c=\\angle O_eO_cD_c=\\angle O_cO_eD_e=\\angle O_aO_eF_e=\\angle O_eO_aF_a=120^{\\circ}-\\varphi$. It follows that rays $O_aB_a$ and $O_cB_c$ (being symmetrical in $e$) intersect at $U_e \\in e$ forming an isosceles triangle $\\triangle O_aU_eO_c$. Similarly define $\\triangle O_cU_aO_e$ and $\\triangle O_eU_cO_a$. These triangles are congruent (equal bases and corresponding angles). Therefore we have $O_aU_c=U_cO_e=O_eU_a=U_aO_c=O_cU_e=U_eO_a$. Moreover, we also have $B_aU_e=O_aU_e-r_a=O_aU_c-r_a=F_aU_c=x$, and thus similarly $D_cU_a=B_cU_e=y$, $F_eU_c=D_eU_a=z$.\n\nNow from quadrilateral $BB_aU_eB_c$ with two opposite right angles $B_aB^2-B_cB^2=B_cU_e^2-B_aU_e^2=y^2-x^2$. Similarly $D_cD^2-D_eD^2=D_eU_a^2-D_cU_a^2=z^2-y^2$ and $F_eF^2-F_aF^2=F_aU_c^2-F_eU_c^2=x^2-z^2$. Finally, we substitute this into $(**)$, and the claim is proved.\n\nRemark. Circles $\\omega_a, \\omega_c$ and $\\omega_e$ could be helpful in some other solutions. In particular, the movement of $A$ along $a$ in Solution D is equivalent to varying $r_a$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA permutation of the integers $1,2, \\ldots, m$ is called fresh if there exists no positive integer $k < m$ such that the first $k$ numbers in the permutation are $1,2, \\ldots, k$ in some order. Let $f_{m}$ be the number of fresh permutations of the integers $1,2, \\ldots, m$.\n\nProve that $f_{n} \\geq n \\cdot f_{n-1}$ for all $n \\geq 3$.\n\nFor example, if $m=4$, then the permutation $(3,1,4,2)$ is fresh, whereas the permutation $(2,3,1,4)$ is not.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $\\sigma = (\\sigma_{1}, \\ldots, \\sigma_{n-1})$ be a fresh permutation of the integers $1,2, \\ldots, n-1$. We claim that for any $1 \\leq i \\leq n-1$ the permutation\n$$\n\\sigma^{(i)} = (\\sigma_{1}, \\ldots, \\sigma_{i-1}, n, \\sigma_{i}, \\ldots, \\sigma_{n-1})\n$$\nis a fresh permutation of the integers $1,2, \\ldots, n$. Indeed, let $1 \\leq k \\leq n-1$. If $k \\geq i$ then we have $n \\in \\{\\sigma_{1}^{(i)}, \\ldots, \\sigma_{k}^{(i)}\\}$, but $n \\notin \\{1,2, \\ldots, k\\}$. And if $k < i$ we have $k < n-1$, and $\\{\\sigma_{1}^{(i)}, \\ldots, \\sigma_{k}^{(i)}\\} = \\{\\sigma_{1}, \\ldots, \\sigma_{k}\\} \\neq \\{1,2, \\ldots, k\\}$, since $\\sigma$ is fresh. Moreover, it is easy to see that when we apply this construction to all fresh permutations of $1,2, \\ldots, n-1$, we obtain $(n-1) \\cdot f_{n-1}$ distinct fresh permutations of $1,2, \\ldots, n$.\n\nNote that a fresh permutation of $1,2, \\ldots, n-1$ cannot end in $n-1$, and hence none of the previously constructed fresh permutations of $1,2, \\ldots, n$ will end in $n-1$ either. Therefore, we will complete the proof by finding $f_{n-1}$ fresh permutations of $1,2, \\ldots, n$ that end in $n-1$. To do this, let $\\sigma = (\\sigma_{1}, \\ldots, \\sigma_{n-1})$ be a fresh permutation of $1,2, \\ldots, n-1$ and let $j$ be such that $\\sigma_{j} = n-1$. Define\n$$\n\\sigma' = (\\sigma_{1}, \\ldots, \\sigma_{j-1}, n, \\sigma_{j+1}, \\ldots, \\sigma_{n-1}, n-1)\n$$\nthen clearly $\\sigma'$ is a permutation of $1,2, \\ldots, n$ that ends in $n-1$. We show that $\\sigma'$ is fresh, so let $1 \\leq k \\leq n-1$. If $k \\geq j$ then $n \\in \\{\\sigma_{1}', \\ldots, \\sigma_{k}'\\}$ but $n \\notin \\{1,2, \\ldots, k\\}$; if $k < j$, then $k < n-1$ and $\\{\\sigma_{1}', \\ldots, \\sigma_{k}'\\} = \\{\\sigma_{1}, \\ldots, \\sigma_{k}\\} \\neq \\{1,2, \\ldots, k\\}$, since $\\sigma$ is fresh. So we have constructed $f_{n-1}$ additional fresh permutations of $1,2, \\ldots, n$ (which again are all different), and the total number $f_{n}$ of fresh permutations of $1,2, \\ldots, n$ must at least be $(n-1) f_{n-1} + f_{n-1} = n f_{n-1}$, as required.\n\nComment. A similar way to construct $n$ fresh permutations of $1,2, \\ldots, n$ for each fresh permutation $\\sigma$ of $1,2, \\ldots, n-1$ is as follows: increase all entries of $\\sigma$ by 1, and then add the number 1 anywhere; all these permutations are fresh, except the ones where 1 is added at the front, which can be made fresh by swapping the 1 and the 2. It is again straightforward to check that we obtain $n f_{n-1}$ permutations that are fresh and distinct, although a little extra care is needed to account for the fact that we increased the entries of our original permutation.\nSolution:\n\nAssuming $n \\geq 3$, we construct $f_{n-1} \\cdot n$ different fresh permutations.\n\nConsider a fresh permutation of the $n-1$ numbers $1,3,4, \\ldots, n$ (the number 2 has been removed), by which we mean a permutation $(x_{1}, \\ldots, x_{n-1})$ such that $x_{1} \\neq 1$ and $\\{x_{1}, \\ldots, x_{k}\\} \\neq \\{1,3, \\ldots, k+1\\}$ for all $k$ with $2 \\leq k \\leq n-2$. There are exactly $f_{n-1}$ such permutations.\n\nBy inserting 2 anywhere in such a permutation, i.e. before or after all entries, or between two entries $x_{i}, x_{i+1}$, we generate the following list of $n$ distinct permutations of $1,2, \\ldots, n$, which we claim are all fresh:\n$$\n(2, x_{1}, \\ldots, x_{n-1}), \\quad \\ldots, \\quad (x_{1}, \\ldots, x_{i-1}, 2, x_{i}, \\ldots, x_{n-1}), \\quad \\ldots, \\quad (x_{1}, \\ldots, x_{n-1}, 2)\n$$\nIn order to verify freshness, suppose that some permutation above is not fresh, that is, for some $1 \\leq k \\leq n-1$, the first $k$ elements are $1, \\ldots, k$. If $k=1$ then the first element is 1; but the first element is either 2 or $x_{1}$ and $x_{1} \\neq 2$, so this is not possible. If $k \\geq 2$ then the first $k$ elements must contain 2 and $x_{1}, \\ldots, x_{k-1}$, so $\\{x_{1}, \\ldots, x_{k-1}\\} = \\{1,3, \\ldots, k\\}$, in contradiction to the fact that $(x_{1}, \\ldots, x_{n-1})$ is fresh.\n\nWe have thus constructed $f_{n-1} \\cdot n$ different fresh permutations, so $f_{n} \\geq f_{n-1} \\cdot n$.\n\nComment. Note that the construction from solution B can also be performed by placing $n-1$ instead of 2, or indeed any $k$ between 2 and $n-1$ (in which case it will work for $n \\geq k+1$).\n\nMoreover, similar arguments can be made while choosing to work with non-fresh permutations rather than fresh permutations. Indeed, in order to show that $n! - f_{n} \\leq n((n-1)! - f_{n-1})$ for all $n \\geq 3$, which is equivalent to the problem statement, one can argue as follows. Given a non-fresh permutation $(\\sigma_{1}, \\ldots, \\sigma_{n})$ of $1,2, \\ldots, n$, erasing $n-1$ from it and changing $n$ to $n-1$ yields a permutation which cannot be fresh. Moreover, each non-fresh permutation of $1,2, \\ldots, n-1$ can be obtained in this way from at most $n$ different non-fresh permutations of $1,2, \\ldots, n$. This type of argument would essentially mirror the one carried out in Solution B.\nSolution:\n\nBy considering all permutations of $1,2, \\ldots, n$ and considering the smallest $k$ for which the first $k$ numbers are $1,2, \\ldots, k$ in some order, one can deduce the recursion\n$$\n\\sum_{k=1}^{n} (n-k)! \\cdot f_{k} = n!\n$$\nwhich holds for any $n \\geq 1$.\n\nTherefore, if $n \\geq 3$ we have\n$$\n\\begin{aligned}\n0 &= n! - (n+1)(n-1)! + (n-1)(n-2)! \\\\\n&= \\sum_{k=1}^{n} (n-k)! \\cdot f_{k} - (n+1) \\sum_{k=1}^{n-1} (n-1-k)! \\cdot f_{k} + (n-1) \\sum_{k=1}^{n-2} (n-2-k)! \\cdot f_{k} \\\\\n&= f_{n} - n f_{n-1} + \\sum_{k=1}^{n-2} (n-2-k)! \\cdot ((n-k)(n-k-1) - (n+1)(n-k-1) + (n-1)) \\cdot f_{k} \\\\\n&= f_{n} - n f_{n-1} + \\sum_{k=1}^{n-2} (n-2-k)! \\cdot k(k+2-n) \\cdot f_{k} \\leq f_{n} - n f_{n-1}\n\\end{aligned}\n$$\nwhich rewrites as $f_{n} \\geq n f_{n-1}$.\nSolution:\n\nWe first show that for any $n \\geq 2$ we have\n$$\nf_{n} = \\sum_{k=1}^{n-1} k (n-1-k)! \\cdot f_{k}\n$$\nTo deduce this relation, we imagine obtaining permutations of $1,2, \\ldots, n$ by inserting $n$ into a permutation of the numbers $1,2, \\ldots, n-1$. Specifically, let $(\\sigma_{1}, \\ldots, \\sigma_{n-1})$ be any permutation of $1,2, \\ldots, n-1$, and let $1 \\leq k \\leq n-1$ be minimal with $\\{\\sigma_{1}, \\ldots, \\sigma_{k}\\} = \\{1,2, \\ldots, k\\}$. Note that there are $(n-1-k)! \\cdot f_{k}$ such permutations. Furthermore, inserting $n$ in this permutation will give a fresh permutation if and only if $n$ is inserted before $\\sigma_{k}$, so there are $k$ options to do so. Consequently, if $n \\geq 3$,\n$$\n\\begin{aligned}\nf_{n} &= \\sum_{k=1}^{n-1} k (n-1-k)! \\cdot f_{k} \\\\\n&= (n-1) \\cdot f_{n-1} + \\sum_{k=1}^{n-2} k (n-1-k)! \\cdot f_{k} \\\\\n&\\geq (n-1) \\cdot f_{n-1} + \\sum_{k=1}^{n-2} k (n-2-k)! \\cdot f_{k} \\\\\n&= (n-1) \\cdot f_{n-1} + f_{n-1} = n \\cdot f_{n-1}\n\\end{aligned}\n$$\ncompleting the proof.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider the triangle $ABC$ with $\\angle BCA > 90^{\\circ}$. The circumcircle $\\Gamma$ of $ABC$ has radius $R$. There is a point $P$ in the interior of the line segment $AB$ such that $PB = PC$ and the length of $PA$ is $R$. The perpendicular bisector of $PB$ intersects $\\Gamma$ at the points $D$ and $E$.\n\nProve that $P$ is the incentre of triangle $CDE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nA. The angle bisector of $\\angle ECD$ intersects the circumcircle of $CDE$ (which is $\\Gamma$) at the midpoint $M$ of arc $DBE$. It is well-known that the incentre is the intersection of the angle bisector segment $CM$ and the circle with centre at $M$ and passing through $D, E$. We will verify this property for $P$.\n\n![](attached_image_1.png)\n\nBy the conditions we have $AP = OA = OB = OC = OD = OE = R$. Both lines $OM$ and $APB$ are perpendicular to $ED$, therefore $AP \\parallel OM$; in the quadrilateral $AOMP$ we have $AP = OA = AM = R$ and $AP \\parallel OM$, so $AOMP$ is a rhombus and its fourth side is $PM = R$.\n\nIn the convex quadrilateral $OMBP$ we have $OM \\parallel PB$, so $OMBP$ is a symmetric trapezoid; the perpendicular bisector of its bases $AO$ and $PB$ coincide. From this symmetry we obtain $MD = OD = R$ and $ME = OE = R$. (Note that the triangles $OEM$ and $OMD$ are equilateral.) We already have $MP = MD = ME = R$, so $P$ indeed lies on the circle with center $M$ and passing through $D, E$. (Notice that this circle is the reflection of $\\Gamma$ about $DE$.)\n\nFrom $PB = PC$ and $OB = OC$ we know that $B$ and $C$ are symmetrical about $OP$; from the rhombus $AOMP$ we find that $A$ and $M$ are also symmetrical about $OP$. By reflecting the collinear points $B, P, A$ (with $P$ lying in the middle) we get that $C, P, M$ are collinear (and $P$ is in the middle). Hence, $P$ lies on the line segments $CM$.\n\n\nB. Let $X$ be the second intersection of $CP$ with $\\Gamma$. Using the power of the point $P$ in the circle $\\Gamma$ and the fact that $PB = PC$, we find that $PX = PA = R$. The quadrilateral $AOXP$ has four sides of equal length, so it is a rhombus and in particular $OX$ is parallel to $AP$. This proves that $OXBP$ is a trapezoid, and because the diagonals $PX$ and $OB$ have equal length, this is even an isosceles trapezoid. Because of that, $DE$ is not only the perpendicular bisector of $PB$, but also of $OX$.\n\n![](attached_image_2.png)\n\nIn particular we have $XD = XP = XO = XE = R$, which proves that $X$ is the middle of the arc $DE$ and $P$ belongs to the circle with center $X$ going through $D$ and $E$. These properties, together with the fact that $C, P, X$ are collinear, determine uniquely the incenter of $CDE$.\n\n\nC. Let $Y$ be the circumcenter of triangle $BPC$. Then from $YB = YP$ it follows that $Y$ lies on $DE$ (we assume $D$ lies in between $Y$ and $E$), and from $YB = YC$ it follows that $Y$ lies on $OP$, where $O$ is the center of $\\Gamma$.\n\nFrom $\\angle AOC = 2\\angle ABC = \\angle APC$ (because $\\angle PBC = \\angle PCB$) we deduce that $AOPC$ is a cyclic quadrilateral, and from $AP = R$ it follows that $AOPC$ is an isosceles trapezoid. We now find that $\\angle YCP = \\angle YPC = 180^{\\circ} - \\angle OPC = 180^{\\circ} - \\angle ACP$, so $Y$ lies on $AC$.\n\n![](attached_image_3.png)\n\nPower of a point gives $YO \\cdot YP = YC \\cdot YA = YD \\cdot YE$, so $D, P, O$ and $E$ are concyclic. It follows that $2\\angle DAE = \\angle DOE = \\angle DPE = \\angle DBE = 180^{\\circ} - \\angle DAE$, so $\\angle DAE = 60^{\\circ}$. We can now finish the proof by angle chasing.\n\nFrom $AB \\perp DE$ we have $\\angle AOD + \\angle BOE = 180^{\\circ}$ and from $\\angle DOE = 2\\angle DAE = 120^{\\circ}$ it follows that $\\angle BOD + \\angle BOE = 120^{\\circ}$. It follows that $\\angle AOD - \\angle BOD = 180^{\\circ} - 120^{\\circ} = 60^{\\circ}$. Let $\\angle OAB = \\angle OBA = 2\\beta$; then $\\angle AOD + \\angle BOD = \\angle AOB = 180^{\\circ} - 4\\beta$. Together with $\\angle AOD - \\angle BOD = 60^{\\circ}$, this yields $\\angle AOD = 120^{\\circ} - 2\\beta$ and $\\angle BOD = 60^{\\circ} - 2\\beta$. We now find $\\angle AED = \\frac{1}{2} \\angle AOD = 60^{\\circ} - \\beta$, which together with $\\angle DAE = 60^{\\circ}$ yields $\\angle ADE = 60^{\\circ} + \\beta$. From the isosceles trapezoid $AOPC$ we have $\\angle CDA = \\angle CBA = \\frac{1}{2} \\angle CPA = \\frac{1}{2} \\angle PAO = \\beta$, so $\\angle CDE = \\angle CDA + \\angle ADE = \\beta + 60^{\\circ} + \\beta = 60^{\\circ} + 2\\beta$.\n\nFrom $\\angle BOD = 60^{\\circ} - 2\\beta$ we deduce that $\\angle BED = 30^{\\circ} - \\beta$; together with $\\angle DBE = 120^{\\circ}$ this yields $\\angle EDB = 30^{\\circ} + \\beta$. We now see that $\\angle PDE = \\angle BDE = 30^{\\circ} + \\beta = \\frac{1}{2} \\angle CDE$, so $P$ is on the angle bisector of $\\angle CDE$. Similarly, $P$ lies on the angle bisector of $\\angle CED$, so $P$ is the incenter of $CDE$.\n\n\nD. We draw the lines $DP$ and $EP$ and let $D'$, resp. $E'$, be the second intersection point with $\\Gamma$.\n\n![](attached_image_4.png)\n\nThe triangles $APD'$ and $DPB$ are similar, and the triangles $APE'$ and $EPB$ are also similar, hence they are all isosceles and it follows that $E', O, P, D'$ lie on a circle with center $A$. In particular $AOD'$ and $AOE'$ are equilateral triangles. Angle chasing gives\n$$\n\\begin{gathered}\n\\angle CDP = \\angle CDD' = \\frac{1}{2} \\angle COD' = \\frac{1}{2}(60^{\\circ} + \\angle COA) \\\\\n\\angle EDP = \\angle EDD' = \\angle EE'D' = \\angle PE'D' = \\frac{1}{2} \\angle PAD' = \\frac{1}{2}(60^{\\circ} + \\angle PAO)\n\\end{gathered}\n$$\nSimilarly we prove $\\angle CEP = \\frac{1}{2}(60^{\\circ} - \\angle COA)$ and $\\angle DEP = \\frac{1}{2}(60^{\\circ} - \\angle PAO)$ so if we can prove that $\\angle COA = \\angle PAO$, we will have proven that $P$ belongs to the angle bisector of $\\angle CED$ and to the angle bisector of $\\angle CDE$, which is enough to prove that $P$ is the incenter of the triangle $CDE$.\n\nLet $\\beta = \\angle ABC = \\angle PCB$. We have $\\angle APC = 2\\beta$ and $\\angle AOC = 2\\beta$, so $AOPC$ is an inscribed quadrilateral. Moreover, since the diagonals $AP$ and $CO$ have equal length, this is actually an isosceles trapezoid, and hence $\\angle PAO = \\angle CPA = \\angle COA$ which concludes the proof.\n\n\nE. Without loss of generality we assume that $D$ and $C$ are in the same half-plane regarding line $AB$.\n\nSince $PC = BP$ and $ABCD$ is inscribed quadrilateral we have $\\angle PCB = \\angle CBP = \\angle CEA = \\alpha$. As in the other solutions, $AOPC$ is an isosceles trapezoid and $2\\alpha = \\angle CPA = \\angle PCO$.\n\n![](attached_image_5.png)\n\nLet $K$ be intersection of $EO$ and $\\Gamma$. Then $\\angle KDE = 90^{\\circ}$, $AB \\parallel DK$ and $KDBA$ is isosceles trapezoid. We obtain $DP = BD = AK$, which implies that $DPAK$ is a parallelogram and hence $DK = PA = R = OD = OK$. We see that $DOK$ is an equilateral triangle. Then $\\angle ECD = \\angle EKD = 60^{\\circ}$.\n\nFurther we prove that $PC$ bisects $\\angle ECD$ using $\\angle DCB = \\angle DKB = \\angle KDA$ (from isosceles trapezoid $DKAB$) and that $\\angle KEC = \\angle OEC = \\angle OCE$ (from isosceles triangle $OCE$):\n$$\n\\angle DCP = \\angle DCB + \\angle BCP = \\angle KDA + \\alpha = \\angle KEC + \\angle CEA + \\alpha = \\angle OCE + 2\\alpha = \\angle PCE\n$$\nFurther by $\\angle DCP = \\angle PCE = \\frac{1}{2} \\angle ECD = 30^{\\circ}$ distances between $P$ and sides $\\triangle CDE$ are $\\frac{1}{2} PC = \\frac{1}{2} PB$ (as ratio between cathetus and hypotenuse in right triangle with angles $60^{\\circ}$ and $30^{\\circ}$). So, we have found incentre.\n\n\nF. Let $F, G, H$ be the projections of $P$ on the sides $DE, DC$ resp. $CE$. If $P$ is indeed the incenter, then the three line segments $PF, PG, PH$ have the same length. This means that the problem is equivalent to proving that $PG = PH = PF = \\frac{1}{2} PB = \\frac{1}{2} PC$ and thus trigonometry in the right-angled triangles $CPG$ and $CPH$ tells us that it is enough to prove that $\\angle DCP = \\angle ECP = 30^{\\circ}$.\n\n![](attached_image_6.png)\n\nWe introduce the point $X$ as the second intersection of the line $CP$ with $\\Gamma$. Because $O$ is the center of $\\Gamma$ we can reduce the problem to proving that $\\angle XOD = \\angle XOE = 60^{\\circ}$, or equivalently that $XOD$ and $XOE$ are equilateral triangles. This last condition is equivalent to $X$ being the reflection of $O$ on the line $DE$. Following the chain of equivalences, we see therefore that in order to solve the problem it is enough to prove that $X$ is the reflection of $O$ on $DE$. We prove this property as in Solution B, using the fact that $OXBP$ is an isosceles trapezoid.\n\n\nG. Assume $AB$ is parallel to the horizontal axis, and that $\\Gamma$ is the unit circle. Write $f(\\theta)$ for the point $(\\cos(\\theta), \\sin(\\theta))$ on $\\Gamma$. As in Solution C, assume that $\\angle AOB = 180^{\\circ} - 4\\beta$; then we can take $B = f(2\\beta)$ and $A = f(180^{\\circ} - 2\\beta)$. As in Solution C, we observe that $AOPC$ is an isosceles trapezoid, which we use to deduce that $\\angle ABC = \\frac{1}{2} \\angle APC = \\frac{1}{2} \\angle OAB = \\beta$. We now know that $C = f(180^{\\circ} - 4\\beta)$.\n\nThe point $P$ lies on $AB$ with $AP = R = 1$, so $P = (\\cos(180^{\\circ} - 2\\beta) + 1, \\sin(2\\beta)) = (1 - \\cos(2\\beta), \\sin(2\\beta))$. The midpoint of $BP$ therefore has coordinates $\\left(\\frac{1}{2}, \\sin(2\\beta)\\right)$, so $D$ and $E$ have $x$-coordinate $\\frac{1}{2}$. Without loss of generality, we take $D = f(60^{\\circ})$ and $E = f(-60^{\\circ})$.\n\nWe have now obtained coordinates for all points in the problem, with one free parameter $(\\beta)$. To show that $P$ is the incenter of $CDE$, we will show that $P$ lies on the bisector of $\\angle CDE$; analogously, one can show that $P$ lies on the bisector of angle $CED$. The bisector of angle $CDE$ passes through the midpoint $M$ of the arc $CE$ not containing $DE$; because $C = f(180^{\\circ} - 4\\beta)$ and $E = f(-60^{\\circ})$, we have $M = f(240^{\\circ} - 2\\beta)$.\n\nIt remains to show that $P = (1 - \\cos(2\\beta), \\sin(2\\beta))$ lies on the line connecting the points $D = (\\cos(60^{\\circ}), \\sin(60^{\\circ}))$ and $M = (\\cos(240^{\\circ} - 2\\beta), \\sin(240^{\\circ} - 2\\beta)) = (-\\cos(60^{\\circ} - 2\\beta), -\\sin(60^{\\circ} - 2\\beta))$. The equation for the line $DM$ is\n$$\n\\left(Y + \\sin(60^{\\circ} - 2\\beta)\\right)\\left(\\cos(60^{\\circ}) + \\cos(60^{\\circ} - 2\\beta)\\right) = \\left(\\sin(60^{\\circ}) + \\sin(60^{\\circ} - 2\\beta)\\right)\\left(X + \\cos(60^{\\circ} - 2\\beta)\\right)\n$$\nwhich, using the fact that $\\cos(60^{\\circ}) + \\cos(60^{\\circ} - 2\\beta) = 2\\cos(\\beta)\\cos(60^{\\circ} - \\beta)$ and $\\sin(60^{\\circ}) + \\sin(60^{\\circ} - 2\\beta) = 2\\cos(\\beta)\\sin(60^{\\circ} - \\beta)$, simplifies to\n$$\n\\left(Y + \\sin(60^{\\circ} - 2\\beta)\\right)\\cos(60^{\\circ} - \\beta) = \\left(X + \\cos(60^{\\circ} - 2\\beta)\\right)\\sin(60^{\\circ} - \\beta)\n$$\nBecause $\\cos(60^{\\circ} - 2\\beta)\\sin(60^{\\circ} - \\beta) - \\sin(60^{\\circ} - 2\\beta)\\cos(60^{\\circ} - \\beta) = \\sin(\\beta)$, this equation further simplifies to\n$$\nY\\cos(60^{\\circ} - \\beta) - X\\sin(60^{\\circ} - \\beta) = \\sin(\\beta).\n$$\nPlugging in the coordinates of $P$, i.e., $X = 1 - \\cos(2\\beta)$ and $Y = \\sin(2\\beta)$, shows that $P$ is on this line: for this choice of $X$ and $Y$, the left hand side equals $\\sin(60^{\\circ} + \\beta) - \\sin(60^{\\circ} - \\beta) = 2\\cos(60^{\\circ})\\sin(\\beta)$, which is indeed equal to $\\sin(\\beta)$. So $P$ lies on the bisector $DM$ of $\\angle CDE$, as desired.\n\n\nH. Let $\\Gamma$ be the complex unit circle and let $AB$ be parallel with the real line and $0 < \\varphi = \\arg b < \\frac{\\pi}{2}$. Then\n$$\n|b| = 1, \\quad a = -\\bar{b}, \\quad p = a + 1 = 1 - \\bar{b}\n$$\nFrom $\\operatorname{Re} d = \\operatorname{Re} e = \\operatorname{Re} \\frac{p + b}{2} = \\frac{1}{2}$ we get that $d = \\frac{1}{2} + \\frac{\\sqrt{3}}{2} i$ and $e = \\frac{1}{2} - \\frac{\\sqrt{3}}{2} i$ are conjugate 6th roots of unity; $d^3 = e^3 = -1, d + e = 1, d^2 = -e, e^2 = -d$ etc.\n\nPoint $C$ is the reflection of $B$ in line $OP$. From $\\arg p = \\arg(1 - \\bar{b}) = \\frac{1}{2}(\\pi - \\varphi)$, we can get $\\arg c = 2\\arg p - \\arg b = \\pi - 2\\varphi$, so $c = -\\bar{b}^2$.\n\nNow we can verify that $EP$ bisects $\\angle CED$. This happens if and only if $(p - e)^2(\\bar{c} - \\bar{e})(\\bar{d} - \\bar{e})$ is real. Since $\\bar{d} - \\bar{e} = -\\sqrt{3} i$, this is equivalent with $\\operatorname{Re}[(p - e)^2(\\bar{c} - \\bar{e})] = 0$. Here\n$$\n\\begin{aligned}\n(p - e)^2(\\bar{c} - \\bar{e}) & = (1 - \\bar{b} - e)^2(-b^2 - d) = (d - \\bar{b})^2(-b^2 - d) \\\\\n& = -|b|^4 + 2d|b|^2 b - d^2 b^2 - d\\bar{b}^2 + 2d^2\\bar{b} - d^3 \\\\\n& = -1 - 2db + \\bar{d}b^2 - d\\bar{b}^2 - 2\\overline{db} + 1 \\\\\n& = -2(db - \\overline{db}) + (\\bar{d}b^2 - d\\bar{b}^2),\n\\end{aligned}\n$$\nwhose real part is zero. It can be proved similarly that $DP$ bisects $\\angle EDC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $m > 1$ be an integer. A sequence $a_{1}, a_{2}, a_{3}, \\ldots$ is defined by $a_{1} = a_{2} = 1$, $a_{3} = 4$, and for all $n \\geq 4$,\n$$\na_{n} = m\\left(a_{n-1} + a_{n-2}\\right) - a_{n-3}.\n$$\nDetermine all integers $m$ such that every term of the sequence is a square.",
    "options": [],
    "answer": "m = 2 or m = 10",
    "solution": "Solution:\nThe only such $m$ are $m = 2$ and $m = 10$.\n\nConsider an integer $m > 1$ for which the sequence defined in the problem statement contains only perfect squares. We shall first show that $m - 1$ is a power of $3$.\n\nSuppose that $m - 1$ is even. Then $a_{4} = 5m - 1$ should be divisible by $4$ and hence $m \\equiv 1 \\pmod{4}$. But then $a_{5} = 5m^{2} + 3m - 1 \\equiv 3 \\pmod{4}$ cannot be a square, a contradiction. Therefore $m - 1$ is odd.\n\nSuppose that an odd prime $p \\neq 3$ divides $m - 1$. Note that $a_{n} - a_{n-1} \\equiv a_{n-2} - a_{n-3} \\pmod{p}$. It follows that modulo $p$ the sequence takes the form $1, 1, 4, 4, 7, 7, 10, 10, \\ldots$; indeed, a simple induction shows that $a_{2k} \\equiv a_{2k-1} \\equiv 3k - 2 \\pmod{p}$ for $k \\geq 1$. Since $\\gcd(p, 3) = 1$ we get that the sequence $a_{n} \\pmod{p}$ contains all the residues modulo $p$, a contradiction since only $(p + 1)/2$ residues modulo $p$ are squares. This shows that $m - 1$ is a power of $3$.\n\nLet $h, k$ be integers such that $m = 3^{k} + 1$ and $a_{4} = h^{2}$. We then have $5 \\cdot 3^{k} = (h - 2)(h + 2)$. Since $\\gcd(h - 2, h + 2) = 1$, it follows that $h - 2$ equals either $1, 3^{k}$ or $5$, and $h + 2$ equals either $5 \\cdot 3^{k}, 5$ or $3^{k}$, respectively. In the first two cases we get $k = 0$ and in the last case we get $k = 2$. This implies that either $m = 2$ or $m = 10$.\n\nWe now show the converse. Suppose that $m = 2$ or $m = 10$. Let $t = 1$ or $t = 3$ so that $m = t^{2} + 1$. Let $b_{1}, b_{2}, b_{3}, \\ldots$ be a sequence of integers defined by $b_{1} = 1, b_{2} = 1, b_{3} = 2$, and\n$$\nb_{n} = t b_{n-1} + b_{n-2}, \\quad \\text{for all } n \\geq 4\n$$\nClearly, $a_{n} = b_{n}^{2}$ for $n = 1, 2, 3$. Note that if $m = 2$ then $a_{4} = 9$ and $b_{4} = 3$, and if $m = 10$ then $a_{4} = 49$ and $b_{4} = 7$. In both the cases we have $a_{4} = b_{4}^{2}$.\n\nIf $n \\geq 5$ then we have\n$$\nb_{n}^{2} + b_{n-3}^{2} = \\left(t b_{n-1} + b_{n-2}\\right)^{2} + \\left(b_{n-1} - t b_{n-2}\\right)^{2} = \\left(t^{2} + 1\\right)\\left(b_{n-1}^{2} + b_{n-2}^{2}\\right) = m\\left(b_{n-1}^{2} + b_{n-2}^{2}\\right).\n$$\nTherefore, it follows by induction that $a_{n} = b_{n}^{2}$ for all $n \\geq 1$. This completes the solution.\nSolution:\nWe present an alternate proof that $m = 2$ and $m = 10$ are the only possible values of $m$ with the required property.\n\nNote that\n$$\n\\begin{aligned}\n& a_{4} = 5m - 1 \\\\\n& a_{5} = 5m^{2} + 3m - 1 \\\\\n& a_{6} = 5m^{3} + 8m^{2} - 2m - 4\n\\end{aligned}\n$$\nSince $a_{4}$ and $a_{6}$ are squares, so is $a_{4} a_{6}$. We have\n$$\n4 a_{4} a_{6} = 100 m^{4} + 140 m^{3} - 72 m^{2} - 72 m + 16\n$$\nNotice that\n$$\n\\begin{aligned}\n& \\left(10 m^{2} + 7m - 7\\right)^{2} = 100 m^{4} + 140 m^{3} - 91 m^{2} - 98 m + 49 < 4 a_{4} a_{6} \\\\\n& \\left(10 m^{2} + 7m - 5\\right)^{2} = 100 m^{4} + 140 m^{3} - 51 m^{2} - 70 m + 25 > 4 a_{4} a_{6}\n\\end{aligned}\n$$\nso we must have\n$$\n4 a_{4} a_{6} = \\left(10 m^{2} + 7m - 6\\right)^{2} = 100 m^{4} + 140 m^{3} - 71 m^{2} - 84 m + 36\n$$\nThis implies that $m^{2} - 12m + 20 = 0$, so $m = 2$ or $m = 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle with an obtuse angle at $A$. Let $E$ and $F$ be the intersections of the external bisector of angle $A$ with the altitudes of $ABC$ through $B$ and $C$ respectively. Let $M$ and $N$ be the points on the segments $EC$ and $FB$ respectively such that $\\angle EMA = \\angle BCA$ and $\\angle ANF = \\angle ABC$. Prove that the points $E, F, N, M$ lie on a circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe first solution is based on the main Lemma. We present this Lemma with two different proofs.\nLemma: Let $ABC$ be an acute triangle with $AB = BC$. Let $P$ be any point on $AC$. Line passing through $P$ perpendicular to $AB$, intersects ray $BC$ in point $T$. If the line $AT$ intersects the circumscribed circle of the triangle $ABC$ the second time at point $K$, then $\\angle AKP = \\angle ABP$.\n\n![](attached_image_1.png)\n\nProof 1:\nLet $H$ be the orthocenter of the triangle $ABP$. Then\n$$\n\\angle BHP = 180^\\circ - \\angle BAC = 180^\\circ - \\angle BCP\n$$\nSo $BHPC$ is cyclic. Then we get\n$$\nTK \\cdot TA = TC \\cdot TB = TP \\cdot TH\n$$\nSo, $AHPK$ is also cyclic. But then\n$$\n\\angle AKP = 180^\\circ - \\angle AHP = \\angle ABP\n$$\n\nProof 2:\nConsider the symmetric points $B'$ and $C'$ of $B$ and $C$, respectively, with respect to the line $PT$. It is clear that\n$$\nTC' \\cdot TB' = TC \\cdot TB = TK \\cdot TA\n$$\nSo $B'C'KA$ is cyclic. Also, because of the symmetry we have\n$$\n\\angle PC'B' = \\angle PCB = \\angle PAB\n$$\nSo $B'C'PA$ is also cyclic. Therefore, the points $B', C', K, P$ and $A$ all lie on the common circle. Because of this fact and because of the symmetry again we have\n$$\n\\angle PKA = \\angle PB'A = \\angle PBA.\n$$\n\nSo, lemma is proved and now return to the problem.\n\n![](attached_image_2.png)\n\nLet $H$ be intersection point of the altitudes at $B$ and $C$. Denote by $M'$ and $N'$ the intersection points of the circumcircle of the triangle $HEF$ with the segments $EC$ and $FB$, respectively. We are going to show that $M = M'$ and $N = N'$ and it will prove the points $E, F, N, M$ lie on a common circle.\nOf course, $A$ is an orthocenter of the triangle $BCH$. Therefore $\\angle BHA = \\angle BCA$, $\\angle CHA = \\angle CBA$ and $\\angle HBA = \\angle HCA$. Thus\n$$\n\\angle HEF = \\angle HBA + \\angle EAB = \\angle HCA + \\angle FAC = \\angle HFE\n$$\nSo, the triangle $HEF$ is isosceles, $HE = HF$.\nBy using lemma, we get\n$$\n\\angle AM'E = \\angle AHE = \\angle ACB\n$$\nand\n$$\n\\angle AN'F = \\angle AHF = \\angle ABC.\n$$\n\n![](attached_image_3.png)\n\nTherefore $M = M'$ and $N = N'$ and we are done.\nSolution:\nLet $X, Y$ be projections of $B$ on $AC$, and $C$ on $AB$, respectively. Let $\\omega$ be circumcircle of $BXYC$. Let $Z$ be intersection of $EC$ and $\\omega$ and $D$ be projection of $E$ on $BA$.\n$$\n\\angle MAC = \\angle AME - \\angle MCA = \\angle XCB - \\angle XCE = \\angle ZCB = \\angle ZXB\n$$\nSince $BXYC$ is cyclic $\\angle ACY = \\angle XBA$, and since $DEXA$ is cyclic\n$$\n\\angle EXD = \\angle EAD = \\angle FAC\n$$\nTherefore, we get that the quadrangles $BZXD$ and $CMAF$ are similar. Hence $\\angle FMC = \\angle DZB$. Since $ZEDB$ is cyclic,\n$$\n\\angle DZB = \\angle DEB = \\angle XAB.\n$$\nThus $\\angle FMC = \\angle XAB$. Similarly, $\\angle ENB = \\angle YAC$. We get that $\\angle FMC = \\angle ENB$ and it implies that the points $E, F, N, M$ lie on a circle.\n\n![](attached_image_4.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be an acute-angled triangle in which $BC < AB$ and $BC < CA$. Let point $P$ lie on segment $AB$ and point $Q$ lie on segment $AC$ such that $P \\neq B$, $Q \\neq C$ and $BQ = BC = CP$. Let $T$ be the circumcentre of triangle $APQ$, $H$ the orthocentre of triangle $ABC$, and $S$ the point of intersection of the lines $BQ$ and $CP$. Prove that $T$, $H$ and $S$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nWe show that $T$ and $H$ are both on the angle bisector $\\ell$ of $\\angle BSC$.\n\nWe first prove that $H \\in \\ell$. The altitude $CH$ in triangle $ABC$ is also the altitude in isosceles triangle $PBC$ with $CP = CB$. Therefore, $CH$ is also the angle bisector of $\\angle PCB$ and hence also of $\\angle SCB$. Analogously, $BH$ is the angle bisector of $\\angle SBC$. We conclude that $H$, as the intersection of two angle bisectors of $\\triangle BSC$, is also on the third angle bisector, which is $\\ell$.\n\nWe now prove that $T \\in \\ell$.\n\nVariant 1. In the isosceles triangles $\\triangle BCP$ and $\\triangle CBQ$ we see that $\\angle BCP = 180^\\circ - 2\\angle B$ and $\\angle CBQ = 180^\\circ - 2\\angle C$. This yields $\\angle PSQ = \\angle BSC = 180^\\circ - (180^\\circ - 2\\angle B) - (180^\\circ - 2\\angle C) = 180^\\circ - 2\\angle A$. Furthermore, $\\angle PTQ = 2\\angle PAQ = 2\\angle A$ ($T$ being circumcentre of $\\triangle APQ$). Now $\\angle PTQ + \\angle PSQ = 180^\\circ$, so $PTQS$ is a cyclic quadrilateral. From $PT = TQ$ we then obtain that $\\angle PST = \\angle PQT = \\angle QPT = \\angle QST$, so $T$ is on the angle bisector $\\angle PSQ$, which is also $\\ell$.\n\nWe conclude that $T$, $S$ and $H$ are collinear.\n\nVariant 2. Let $R$ be the second intersection of $BQ$ and $\\odot APQ$.\n\n$APRQ$ is a cyclic quadrilateral, so $\\angle PRS = \\angle A$, $\\angle APR = \\angle BQC = \\angle C$. On the other hand $\\angle BPC = \\angle B$. Therefore $\\angle RPS = 180^\\circ - \\angle B - \\angle C = \\angle A$. Hence, the triangle $PRS$ is isosceles with $SP = SR$; then $\\ell$ is the perpendicular bisector of the chord $PR$ in the circle that passes through $T$.\n\nNote that if $BQ$ is tangent to $\\odot APQ$, then $CP$ is also tangent to $\\odot APQ$ and triangle $ABC$ is isosceles, so $T$, $S$ and $H$ lie on the altitude from $A$.\n![](attached_image_2.png)\n\nRemark. The fact that $CP$ is also tangent to $\\odot APQ$ could be shown with $PTQS$ being a cyclic quadrilateral like in the first variant of the solution. Otherwise we can consider $R$ the second intersection of $CP$ and $\\odot APQ$ and prove that triangle $QRS$ is isosceles.\n![](attached_image_3.png)\n![](attached_image_4.png)\nSolution:\n![](attached_image_5.png)\nIn the same way as in the previous solution, we see that $\\angle PSQ = 180^\\circ - 2\\angle A$, so $\\angle CSQ = 2\\angle A$. From the cyclic quadrilateral $AEHD$ (with $E$ and $D$ feet of the altitudes $CH$ and $BH$) we see that $\\angle DHC = \\angle DAE = \\angle A$. Since $BH$ is the perpendicular bisector of $CQ$, we have $\\angle DHQ = \\angle A$ as well, so $\\angle CHQ = 2\\angle A$. From $\\angle CHQ = 2\\angle A = \\angle CSQ$, we see $CHSQ$ is a cyclic quadrilateral. This means $\\angle QHS = \\angle QCS$.\n\nSince triangles $PTQ$ and $CHQ$ are both isosceles with apex $2\\angle A$, we get $\\triangle PTQ \\sim \\triangle CHQ$. We see that one can be obtained from the other by a spiral similarity centered at $Q$, so we also obtain $\\triangle QTH \\sim \\triangle QPC$. This means that $\\angle QHT = \\angle QCP$. Combining this with $\\angle QHS = \\angle QCS$, we see that $\\angle QHT = \\angle QCP = \\angle QCS = \\angle QHS$. So $\\angle QHT = \\angle QHS$, which means that $T$, $S$ and $H$ are collinear.\nSolution:\n![](attached_image_6.png)\nLet us draw a parallel $f$ to $BC$ through $A$. Let $B' = BQ \\cap f$, $C' = CP \\cap f$. Then $AQB' \\sim CQB$ and $APC' \\sim BPC$, therefore both $QAB'$ and $APC'$ will be isosceles.\n\nAlso, $BCS \\sim B'C'S$ with respect to the similarity with center $S$, therefore if we take the image of the line $BH$ (which is the perpendicular bisector of the segment $CQ$) through this transformation, it will go through the point $B'$, and be perpendicular to $AQ$ (as the image of $CQ$ is parallel to $CQ$). As $AQB'$ is isosceles, this line is the perpendicular bisector of the segment $AQ$. This means it goes through $T$, the circumcenter of $APQ$. Similarly on the other side the image of $CH$ also goes through $T$. This means that the image of $H$ with respect to the similarity through $S$ will be $T$, so $T$, $S$, $H$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe are given an acute triangle $A B C$. Let $D$ be the point on its circumcircle such that $A D$ is a diameter. Suppose that points $K$ and $L$ lie on segments $A B$ and $A C$, respectively, and that $D K$ and $D L$ are tangent to circle $A K L$.\nShow that line $K L$ passes through the orthocentre of $A B C$.\nThe altitudes of a triangle meet at its orthocentre.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFigure 1: Diagram to solution 1\n\nLet $M$ be the midpoint of $K L$. We will prove that $M$ is the orthocentre of $A B C$. Since $D K$ and $D L$ are tangent to the same circle, $|D K|=|D L|$ and hence $D M \\perp K L$. The theorem of Thales in circle $A B C$ also gives $D B \\perp B A$ and $D C \\perp C A$. The right angles then give that quadrilaterals $B D M K$ and $D M L C$ are cyclic.\nIf $\\angle B A C=\\alpha$, then clearly $\\angle D K M=\\angle M L D=\\alpha$ by angle in the alternate segment of circle $A K L$, and so $\\angle M D K=\\angle L D M=\\frac{\\pi}{2}-\\alpha$, which thanks to cyclic quadrilaterals gives $\\angle M B K=\\angle L C M=\\frac{\\pi}{2}-\\alpha$. From this, we have $B M \\perp A C$ and $C M \\perp A B$, and so $M$ indeed is the orthocentre of $A B C$.\n## Preliminaries\nLet $A B C$ be a triangle with circumcircle $\\Gamma$. Let $X$ be a point in the plane. The Simson line (Wallace-Simson line) is defined via the following theorem. Drop perpendiculars from $X$ to each of the three side lines of $A B C$. The feet of these perpendiculars are collinear (on the Simson line of $X$) if and only if $X$ lies on $\\Gamma$. The Simson line of $X$ in the circumcircle bisects the line segment $X H$ where $H$ is the orthocentre of triangle $A B C$. See Figure 2\n\n![](attached_image_2.png)\nFigure 2: The Wallace-Simson configuration\n\nWhen $X$ is on $\\Gamma$, we can enlarge from $X$ with scale factor 2 (a homothety) to take the Simson line to the doubled Simson line which passes through the orthocentre $H$ and contains the reflections of $X$ in each of the three sides of $A B C$.\n\n## Solution of the problem\n![](attached_image_3.png)\nFigure 3: Three circles do the work\n\nLet $\\Gamma$ be the circle $A B C$, $\\Sigma$ be the circle $A K L$ with centre $O$, and $\\Omega$ be the circle on diameter $O D$ so $K$ and $L$ are on this circle by converse of Thales. Let $\\Omega$ and $\\Gamma$ meet at $D$ and $F$. By Thales in both circles, $\\angle A F D$ and $\\angle O F D$ are both right angles so $A O F$ is a line. Let $A F$ meet $\\Sigma$ again at $T$ so $A T$ (containing $O$) is a diameter of this circle and by Thales, $T L \\perp A C$.\n\nLet $G$ (on $\\Sigma$) be the reflection of $K$ in $A F$. Now $A T$ is the internal angle bisector of $\\angle G A K$ so, by an upmarket use of angles in the same segment (of $\\Sigma$), $T L$ is the internal angle bisector of $\\angle G L K$. Thus the line $G L$ is the reflection of the line $K L$ in $T L$, and so also the reflection of $K L$ in the line $A C$ (internal and external angle bisectors).\n\nOur next project is to show that $L G F$ are collinear. Well $\\angle F L K=\\angle F O K$ (angles in the same segment of $\\Omega$) and $\\angle G L K=\\angle G A K$ (angles in the same segment of $\\Sigma$) $=2 \\angle O A K$ ($A K G$ is isosceles with apex $A)=\\angle T O K$ (since $O A K$ is isosceles with apex $O$, and this is an external angle at $O$). The point $T$ lies in the interior of the line segment $F O$ so $\\angle T O K=\\angle F O K$. Therefore $\\angle F L K=\\angle G L K$ so $L G F$ is a line.\n\nNow from the second paragraph, $F$ is on the reflection of $K L$ in $A C$. By symmetry, $F$ is also on the reflection of $K L$ in $A B$. Therefore the reflections of $F$ in $A B$ and $A C$ are both on $K L$ which must therefore be the doubled Wallace-Simson line of $F$. Therefore the orthocentre of $A B C$ lies on $K L$.\nLet $H$ be the orthocentre of triangle $A B C$ and $\\Sigma$ the circumcircle of $A K L$ with centre $O$. Let $\\Omega$ be the circle with diameter $O D$, which contains $K$ and $L$ by Thales, and let $\\Gamma$ be the circumcircle of $A B C$ containing $D$. Denote the second intersection of $\\Omega$ and $\\Gamma$ by $F$. Since $O D$ and $A D$ are diameters of $\\Omega$ and $\\Gamma$ we have $\\angle O F D=\\frac{\\pi}{2}=\\angle A F D$, so the points $A, O, F$ are collinear. Let $M$ and $N$ be the second intersections of $C H$ and $B H$ with $\\Gamma$, respectively. It is well-known that $M$ and $N$ are the reflections of $H$ in $A B$ and $A C$, respectively (because $\\angle N C A=\\angle N B A=\\angle A C M=\\angle A B M$). By collinearity of $A, O, F$ and the angles in $\\Gamma$ we have\n$$\n\\angle N F O=\\angle N F A=\\angle N B A=\\frac{\\pi}{2}-\\angle B A C=\\frac{\\pi}{2}-\\angle K A L\n$$\nSince $D L$ is tangent to $\\Sigma$ we obtain\n$$\n\\angle N F O=\\frac{\\pi}{2}-\\angle K L D=\\angle L D O,\n$$\nwhere the last equality follows from the fact that $O D$ is bisector of $\\angle L D K$ since $L D$ and $K D$ are tangent to $\\Sigma$. Furthermore, $\\angle L D O=\\angle L F O$ since these are angles in $\\Omega$. Hence, $\\angle N F O=\\angle L F O$, which implies that points $N, L, F$ are collinear. Similarly points $M, K, F$ are collinear. Since $N$ and $M$ are reflections of $H$ in $A C$ and $A B$ we have\n$$\n\\angle L H N=\\angle H N L=\\angle B N F=\\angle B M F=\\angle B M K=\\angle K H B.\n$$\nHence,\n$$\n\\angle L H K=\\angle L H N+\\angle N H K=\\angle K H B+\\angle N H K=\\pi\n$$\nand the points $L, H, K$ are collinear.\nAs in Solution 3 let $M$ and $N$ be the reflections of the orthocentre in $A B$ and $A C$. Let $\\angle B A C=\\alpha$. Then $\\angle N D M=\\pi-\\angle M A N=\\pi-2 \\alpha$.\nLet $M K$ and $N L$ intersect at $F$. See Figure 3.\n\nClaim. $\\angle N F M=\\pi-2 \\alpha$, so $F$ lies on the circumcircle.\n\nProof. Since $K D$ and $L D$ are tangents to circle $A K L$, we have $|D K|=|D L|$ and $\\angle D K L=\\angle K L D=\\alpha$, so $\\angle L D K=\\pi-2 \\alpha$.\nBy definition of $M, N$ and $D, \\angle M N D=\\angle A N D-\\angle A N M=\\frac{\\pi}{2}-\\left(\\frac{\\pi}{2}-\\alpha\\right)=\\alpha$ and analogously $\\angle D M N=\\alpha$. Hence $|D M|=|D N|$.\nFrom $\\angle N D M=\\angle L D K=\\pi-2 \\alpha$ it follows that $\\angle L D N=\\angle K D M$. Since $|D K|=|D L|$ and $|D M|=|D N|$, triangles $M D K$ and $N D L$ are related by a rotation about $D$ through angle $\\pi-2 \\alpha$, and hence the angle between $M K$ and $N L$ is $\\pi-2 \\alpha$, which proved the claim.\n\nWe now finish as in Solution 3:\n$$\n\\begin{gathered}\n\\angle M H K=\\angle K M H=\\angle F M C=\\angle F A C \\\\\n\\angle L H N=\\angle H N L=\\angle B N F=\\angle B A F\n\\end{gathered}\n$$\nAs $\\angle B A F+\\angle F A C=\\alpha$, we have $\\angle L H K=\\alpha+\\angle N H M=\\alpha+\\pi-\\alpha=\\pi$, so $H$ lies on $K L$.\nSince $A D$ is a diameter, it is well known that $D B H C$ is a parallelogram (indeed, both $B D$ and $C H$ are perpendicular to $A B$, hence parallel, and similarly for $D C \\parallel B H$). Let $B'$, $C'$ be the reflections of $D$ in lines $A K B$ and $A L C$, respectively; since $A B D$ and $A C D$ are right angles, these are also the factor-2 homotheties of $B$ and $C$ with respect to $D$, hence $H$ is the midpoint of $B' C'$. We will prove that $B' K C' L$ is a parallelogram: it will then follow that the midpoint of $B' C'$, which is $H$, is also the midpoint of $K L$, and in particular is on the line, as we wanted to show.\n\nWe will prove $B' K C' L$ is a parallelogram by showing that $B' K$ and $C' L$ are the same length and direction. Indeed, for lengths we have $K B'=K D=L D=L C'$, where the first and last equalities arise from the reflections defining $B'$ and $C'$, and the middle one is equality of tangents. For directions, let $\\alpha, \\beta, \\gamma$ denote the angles of triangle $A K L$. Immediate angle chasing in the circle $A K L$, and the properties of the reflections, yield\n$$\n\\begin{aligned}\n\\angle C' L C &=\\angle C L D=\\angle A K L=\\beta \\\\\n\\angle B K B' &=\\angle D K B=\\angle K L A=\\gamma \\\\\n& \\angle L D K=2 \\alpha-\\pi\n\\end{aligned}\n$$\nand therefore in directed angles $(\\bmod 2 \\pi)$ we have\n$\\angle\\left(C' L, B' K\\right)=\\angle C' L C+\\angle C L D+\\angle L D K+\\angle D K B+\\angle B K B'=2 \\alpha+2 \\beta+2 \\gamma-\\pi=\\pi$\nand hence $C' L$ and $B' K$ are parallel and in opposite directions, i.e. $C' L$ and $K B'$ are in the same direction, as claimed.\nThere are a number of \"phantom point\" arguments which define $K'$ and $L'$ in terms of angles and then deduce that these points are actually $K$ and $L$.\n\nNote: In these solutions it is necessary to show that $K$ and $L$ are uniquely determined by the conditions of the problem. One example of doing this is the following:\nTo prove uniqueness of $K$ and $L$, let us consider that there exist two other points $K'$ and $L'$ that satisfy the same properties ($K'$ on $A B$ and $L'$ on $A C$ such that $D K'$ and $D L'$ are tangent to the circle $A K' L'$).\nThen, we have that $D K=D L$ and $D K'=D L'$. We also have that $\\angle K D L=\\angle K' D L'=\\pi-2 \\angle A$. Hence, we deduce $\\angle K D K'=\\angle K D L-\\angle K' D L=\\angle K' D L'-\\angle K' D L=\\angle L D L'$ Thus we have that $\\triangle K D K' \\equiv \\triangle L D L'$, so we deduce $\\angle D K A=\\angle D K K'=\\angle D L L'=\\pi-\\angle A L D$. This implies that $A K D L$ is concyclic, which is clearly a contradiction since $\\angle K A L+\\angle K D L=\\pi-\\angle B A C$.\nWe will use the usual complex number notation, where we will use a capital letter (like $Z$) to denote the point associated to a complex number (like $z$). Consider $\\triangle A K L$ on the unit circle. So, we have $a \\cdot \\bar{a}=k \\cdot \\bar{k}=l \\cdot \\bar{l}=1$. As point $D$ is the intersection of the tangents to the unit circle at $K$ and $L$, we have that\n$$\nd=\\frac{2 k l}{k+l} \\text{ and } \\bar{d}=\\frac{2}{k+l}\n$$\nDefining $B$ as the foot of the perpendicular from $D$ on the line $A K$, and $C$ as the foot of the perpendicular from $D$ on the line $A L$, we have the formulas:\n$$\nb=\\frac{1}{2}\\left(d+\\frac{(a-k) \\bar{d}+\\bar{a} k-a \\bar{k}}{\\bar{a}-\\bar{k}}\\right)\n$$\n$$\nc=\\frac{1}{2}\\left(d+\\frac{(a-l) \\bar{d}+\\bar{a} l-a \\bar{l}}{\\bar{a}-\\bar{l}}\\right)\n$$\nSimplyfing these formulas, we get:\n$$\n\\begin{gathered}\nb=\\frac{1}{2}\\left(d+\\frac{(a-k) \\frac{2}{k+l}+\\frac{k}{a}-\\frac{a}{k}}{\\frac{1}{a}-\\frac{1}{k}}\\right)=\\frac{1}{2}\\left(d+\\frac{\\frac{2(a-k)}{k+l}+\\frac{k^2-a^2}{a k}}{\\frac{k-a}{a k}}\\right) \\\\\nb=\\frac{1}{2}\\left(\\frac{2 k l}{k+l}-\\frac{2 a k}{k+l}+(a+k)\\right)=\\frac{k(l-a)}{k+l}+\\frac{1}{2}(k+a) \\\\\nc=\\frac{1}{2}\\left(d+\\frac{(a-l) \\frac{2}{k+l}+\\frac{l}{a}-\\frac{a}{l}}{\\frac{1}{a}-\\frac{1}{l}}\\right)=\\frac{1}{2}\\left(d+\\frac{\\frac{2(a-l)}{k+l}+\\frac{l^2-a^2}{a l}}{\\frac{l-a}{a l}}\\right) \\\\\nc=\\frac{1}{2}\\left(\\frac{2 k l}{k+l}-\\frac{2 a l}{k+l}+(a+l)\\right)=\\frac{l(k-a)}{k+l}+\\frac{1}{2}(l+a)\n\\end{gathered}\n$$\nLet $O$ be the circumcenter of triangle $\\triangle A B C$. As $A D$ is the diameter of this circle, we have that:\n$$\no=\\frac{a+d}{2}\n$$\nDefining $H$ as the orthocentre of the $\\triangle A B C$, we get that:\n$$\n\\begin{gathered}\nh=a+b+c-2 \\cdot o=a+\\left(\\frac{k(l-a)}{k+l}+\\frac{1}{2}(k+a)\\right)+\\left(\\frac{l(k-a)}{k+l}+\\frac{1}{2}(l+a)\\right)-(a+d) \\\\\nh=a+\\frac{2 k l}{k+l}-\\frac{a(k+l)}{k+l}+\\frac{1}{2} k+\\frac{1}{2} l+a-\\left(a+\\frac{2 k l}{k+l}\\right) \\\\\nh=\\frac{1}{2}(k+l)\n\\end{gathered}\n$$\nHence, we conclude that $H$ is the midpoint of $K L$, so $H, K, L$ are collinear.\nLet us employ the barycentric coordinates. Set $A(1,0,0), K(0,1,0), L(0,0,1)$.\nThe tangent at $K$ of $(A K L)$ is $a^{2} z+c^{2} x=0$, and the tangent at $L$ of $(A K L)$ is $a^{2} y+b^{2} x=0$. Their intersection is\n$$\nD\\left(-a^{2}: b^{2}: c^{2}\\right)\n$$\nSince $B \\in A K$, we can let $B(1-t, t, 0)$. Solving for $\\overrightarrow{A B} \\cdot \\overrightarrow{B D}=0$ gives\n$$\nt=\\frac{3 b^{2}+c^{2}-a^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)} \\Longrightarrow B=\\left(\\frac{-a^{2}-b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, \\frac{-a^{2}+3 b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, 0\\right)\n$$\nLikewise, $C$ has the coordinate\n$$\nC=\\left(\\frac{-a^{2}+b^{2}-c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}, 0, \\frac{-a^{2}+b^{2}+3 c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)\n$$\nThe altitude from $B$ for triangle $A B C$ is\n$$\n-b^{2}\\left(x-z-\\frac{-a^{2}-b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)+\\left(c^{2}-a^{2}\\right)\\left(y-\\frac{-a^{2}+3 b^{2}+c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)=0\n$$\nAlso the altitude from $C$ for triangle $A B C$ is\n$$\n-c^{2}\\left(x-y-\\frac{-a^{2}+b^{2}-c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)+\\left(a^{2}-b^{2}\\right)\\left(z-\\frac{-a^{2}+b^{2}+3 c^{2}}{2\\left(b^{2}+c^{2}-a^{2}\\right)}\\right)=0\n$$\nThe intersection of these two altitudes, which is the orthocenter of triangle $A B C$, has the barycentric coordinate\n$$\nH=(0,1/2,1/2)\n$$\nwhich is the midpoint of the segment $K L$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $k$ be a positive integer. Lexi has a dictionary $\\mathcal{D}$ consisting of some $k$-letter strings containing only the letters $A$ and $B$. Lexi would like to write either the letter $A$ or the letter $B$ in each cell of a $k \\times k$ grid so that each column contains a string from $\\mathcal{D}$ when read from top-to-bottom and each row contains a string from $\\mathcal{D}$ when read from left-to-right.\nWhat is the smallest integer $m$ such that if $\\mathcal{D}$ contains at least $m$ different strings, then Lexi can fill her grid in this manner, no matter what strings are in $\\mathcal{D}$ ?",
    "options": [],
    "answer": "2^{k-1}",
    "solution": "Solution:\nWe claim the minimum value of $m$ is $2^{k-1}$.\n\nFirstly, we provide a set $\\mathcal{S}$ of size $2^{k-1}-1$ for which Lexi cannot fill her grid. Consider the set of all length-$k$ strings containing only $A$'s and $B$'s which end with a $B$, and remove the string consisting of $k$ $B$'s. Clearly there are $2$ independent choices for each of the first $k-1$ letters and $1$ for the last letter, and since exactly one string is excluded, there must be exactly $2^{k-1}-1$ strings in this set.\n\nSuppose Lexi tries to fill her grid. For each row to have a valid string, it must end in a $B$. But then the right column would necessarily contain $k$ $B$'s, and not be in our set. Thus, Lexi cannot fill her grid with our set, and we must have $m \\geqslant 2^{k-1}$.\n\nNow, consider any set $\\mathcal{S}$ with at least $2^{k-1}$ strings. Clearly, if $\\mathcal{S}$ contained either the uniform string with $k$ $A$'s or the string with $k$ $B$'s, then Lexi could fill her grid with all of the relevant letters and each row and column would contain that string.\n\nConsider the case where $\\mathcal{S}$ contains neither of those strings. Among all $2^k$ possible length-$k$ strings with $A$'s and $B$'s, each has a complement which corresponds to the string with $B$'s in every position where the first string had $A$'s and vice-versa. Clearly, the string with all $A$'s is paired with the string with all $B$'s. We may assume that we do not take the two uniform strings and thus applying the pigeonhole principle to the remaining set of strings, we must have two strings which are complementary.\n\nLet this pair of strings be $\\ell, \\ell'$ in $\\mathcal{S}$ in some order. Define the set of indices $\\mathcal{J}$ corresponding to the $A$'s in $\\ell$ and thus the $B$'s in $\\ell'$, and all other indices (not in $\\mathcal{J}$) correspond to $B$'s in $\\ell$ (and thus $A$'s in $\\ell'$). Then, we claim that if Lexi puts an $A$ in the cell in row $r$, column $c$ if $r, c \\in \\mathcal{J}$ or $r, c \\notin \\mathcal{J}$, and a $B$ otherwise, each row and column contains a string in $\\mathcal{S}$.\n\nWe illustrate this with a simple example: If $k=6$ and we have that $AAABAB$ and $BBBAB A$ are both in the dictionary, then Lexi could fill the table as follows:\n\n| A | A | A | B | A | B |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| A | A | A | B | A | B |\n| A | A | A | B | A | B |\n| B | B | B | A | B | A |\n| A | A | A | B | A | B |\n| B | B | B | A | B | A |\n\nSuppose we are looking at row $i$ or column $i$ for $i \\in \\mathcal{J}$. Then by construction the string in this row/column contains $A$'s at indices $k$ with $k \\in \\mathcal{J}$ and $B$'s elsewhere, and thus is precisely $\\ell$. Suppose instead we are looking at row $i$ or column $i$ for $i \\notin \\mathcal{J}$. Then again by construction the string in this row/column contains $A$'s at indices $k$ with $k \\notin \\mathcal{J}$ and $B$'s elsewhere, and thus is precisely $\\ell'$. So each row and column indeed contains a string in $\\mathcal{S}$.\n\nThus, for any $\\mathcal{S}$ with $|\\mathcal{S}| \\geqslant 2^{k-1}$, Lexi can definitely fill the grid appropriately. Since we know $m \\geqslant 2^{k-1}$, $2^{k-1}$ is the minimum possible value of $m$ as claimed.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe are given a positive integer $s \\geqslant 2$. For each positive integer $k$, we define its twist $k^{\\prime}$ as follows: write $k$ as $a s+b$, where $a, b$ are non-negative integers and $b<s$, then $k^{\\prime}=b s+a$. For the positive integer $n$, consider the infinite sequence $d_{1}, d_{2}, \\ldots$ where $d_{1}=n$ and $d_{i+1}$ is the twist of $d_{i}$ for each positive integer $i$.\nProve that this sequence contains 1 if and only if the remainder when $n$ is divided by $s^{2}-1$ is either 1 or $s$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFirst, we consider the difference $k-k^{\\prime \\prime}$. If $k=a s+b$ as in the problem statement, then $k^{\\prime}=b s+a$. We write $a=l s+m$ with $m, l$ non-negative numbers and $m \\leq s-1$. This gives $k^{\\prime \\prime}=m s+(b+l)$ and hence $k-k^{\\prime \\prime}=(a-m) s-l=l\\left(s^{2}-1\\right)$.\nWe conclude\nFact 1.1. $k \\geq k^{\\prime \\prime}$ for every every $k \\geq 1$\nFact 1.2. $s^{2}-1$ divides the difference $k-k^{\\prime \\prime}$.\nFact 1.2 implies that the sequences $d_{1}, d_{3}, d_{5}, \\ldots$ and $d_{2}, d_{4}, d_{6}, \\ldots$ are constant modulo $s^{2}-1$. Moreover, Fact 1.1 says that the sequences are (weakly) decreasing and hence eventually constant. In other words, the sequence $d_{1}, d_{2}, d_{3}, \\ldots$ is 2-periodic modulo $s^{2}-1$ (from the start) and is eventually 2-periodic.\n\nNow, assume that some term in the sequence is equal to 1. The next term is equal to $1^{\\prime}=s$ and since the sequence is 2-periodic from the start modulo $s^{2}-1$, we conclude that $d_{1}$ is either equal to 1 or $s$ modulo $s^{2}-1$. This proves the first implication.\n\nTo prove the other direction, assume that $d_{1}$ is congruent to 1 or $s$ modulo $s^{2}-1$. We need the observation that once one of the sequences $d_{1}, d_{3}, d_{5}, \\ldots$ or $d_{2}, d_{4}, d_{6}, \\ldots$ stabilises, then their value is less than $s^{2}$. This is implied by the following fact.\nFact 1.3. If $k=k^{\\prime \\prime}$, then $k=k^{\\prime \\prime}<s^{2}$.\nProof. We use the expression for $k-k^{\\prime \\prime}$ found before. If $k=k^{\\prime \\prime}$, then $l=0$, and so $k^{\\prime \\prime}=m s+b$. Both $m$ and $b$ are reminders after division by $s$, so they are both $\\leq s-1$. This gives $k^{\\prime \\prime} \\leq(s-1) s+(s-1)<s^{2}$.\n\nUsing Fact 1.2, it follows that the sequence $d_{1}, d_{3}, d_{5}, \\ldots$ is constant to 1 or $s$ modulo $s^{2}-1$ and stabilises to 1 or $s$ by Fact 1.3. Since $s^{\\prime}=1$, we conclude that the sequence contains a 1.\nSolution:\n\nWe make a number of initial observations. Let $k$ be a positive integer.\n\nFact 2.1. If $k \\geq s^{2}$, then $k^{\\prime}<k$.\nProof. Write $k=a s+b$, as in the problem statement. If $k \\geq s^{2}$, then $a \\geq s$ because $b<s$. So, $k^{\\prime}=b s+a \\leq(s-1) s+a \\leq a s \\leq a s+b=k$. Moreover, we cannot have equality since that would imply $s-1=b=0$.\n\nFact 2.2. If $k \\leq s^{2}-1$, then $k^{\\prime} \\leq s^{2}-1$ and $k^{\\prime \\prime}=k$.\nProof. Write $k=a s+b$, as in the problem statement. If $k<s^{2}$, then it must hold $1 \\leq a, b<s$, hence $k^{\\prime}=b s+a<s^{2}$ and $k^{\\prime \\prime}=(b s+a)^{\\prime}=a s+b=k$.\n\nFact 2.3. We have $k^{\\prime} \\equiv s k\\left(\\bmod s^{2}-1\\right)$ (or equivalently $k \\equiv s k^{\\prime}\\left(\\bmod s^{2}-1\\right)$).\nProof. We write $k=a s+b$, as in the problem statement. Now,\n$$\ns k-k^{\\prime}=s(a s+b)-(b s+a)=a\\left(s^{2}-1\\right) \\equiv 0 \\quad\\left(\\bmod s^{2}-1\\right),\n$$\nas desired.\n\nCombining Facts 2.1 and 2.2, we find that the sequence $d_{1}, d_{2}, d_{3} \\ldots$ is eventually periodic with period 2, starting at the first value less than $s^{2}$. From Fact 2.3, it follows that\n$$\nk^{\\prime \\prime} \\equiv s k^{\\prime} \\equiv s^{2} k \\equiv k \\quad\\left(\\bmod s^{2}-1\\right)\n$$\nand hence the sequence is periodic modulo $s^{2}-1$ from the start with period 2.\n\nNow, if the sequence contains 1, the sequence eventually alternates between 1 and $s$ since the twist of 1 is $s$ and vice versa. Using periodicity modulo $s^{2}-1$, we must have $n \\equiv 1, s$ $\\left(\\bmod s^{2}-1\\right)$. Conversely, if $n \\equiv 1, s\\left(\\bmod s^{2}-1\\right)$ then the eventual period must contain at least one value congruent to either 1 or $s$ modulo $s^{2}-1$. Since these values must be less than $s^{2}$, this implies that the sequence eventually alternates between 1 and $s$, showing that it contains a 1.\nSolution:\n\nWe give an alternate proof of the direct implication: if the sequence contains a 1, then the first term is 1 or $s$ modulo $s^{2}-1$. We prove the following fact, which is a combination of Facts 2.1 and 2.3.\n\nFact 3.1. For all $k \\geq s^{2}$, we have $\\left(k-s^{2}+1\\right)^{\\prime} \\in\\left\\{k^{\\prime}, k^{\\prime}-s^{2}+1\\right\\}$.\nProof. We write $k=a s+b$, as in the problem statement. Since $k \\geq s^{2}$, we have $a \\geq s$. If $b<s-1$, then\n$$\n\\left(k-s^{2}+1\\right)^{\\prime}=((a-s) s+(b+1))^{\\prime}=(b+1) s+(a-s)=b s+a=k^{\\prime}\n$$\nOn the other hand, if $b=s-1$, then\n$$\n\\left(k-s^{2}+1\\right)^{\\prime}=((a-s+1) s+0)^{\\prime}=0 s+(a-s+1)=a-s+1=k^{\\prime}-s^{2}+1\n$$\nNow assume $n \\geq s^{2}$ and the sequence $d_{1}, d_{2}, \\ldots$ contains a 1. Denote by $e_{1}, e_{2}, \\ldots$ the sequence constructed as in the problem statement, but with initial value $e_{1}=n-s^{2}+1$. Using the above fact, we deduce that $e_{i} \\equiv d_{i}\\left(\\bmod s^{2}-1\\right)$ and $e_{i} \\leq d_{i}$ for all $i \\geq 1$ by induction on $i$. Hence, the sequence $e_{1}, e_{2}, \\ldots$ also contains a 1.\n\nSince the conclusion we are trying to reach only depends on the residue of $d_{1}$ modulo $s^{2}-1$, we conclude that without loss of generality we can assume $n<s^{2}$.\n\nUsing Fact 2.2, it now follows that the sequence $d_{1}, d_{2}, \\ldots$ is periodic with period two. Since 1 and $s$ are twists of each other, it follows that if this sequence contains a 1, it must be alternating between 1 and $s$. Hence, $d_{1} \\equiv 1, s\\left(\\bmod s^{2}-1\\right)$ as desired.\n\nFor the other direction we can make a similar argument, observing that the second of the two cases in the proof of Fact 3.1 can only apply to finitely many terms of the sequence $d_{1}, d_{2}, d_{3}, \\ldots$, allowing us to also go the other way.\nSolution:\n\nFirst assume that $d_{k}=1$ for some $k$. Let $k$ be the smallest such index. If $k=1$ then $n=1$, so we may assume $k \\geqslant 2$.\nThen $d_{k-1}=a s+b$ for some non-negative integers $a, b$ satisfying $b<s$ and $b s+a=1$. The only solution is $b=0, a=1$, so $d_{k-1}=s$. So, if $k=2$, then $n=s$, so we may assume $k \\geqslant 3$.\nThen there exist non-negative integers $c, d$ satisfying $d_{k-2}=c s+d, d<s$ and $d s+c=s$. We have two solutions: $d=0, c=s$ and $d=1, c=0$. However, in the second case we get $d_{k-2}=1$, which contradicts the minimality of $k$. Hence, $d_{k-2}=s^{2}$. If $k=3$, then $n=d_{1}=s^{2}$, which gives remainder 1 when divided by $s^{2}-1$.\n\nAssume now that $k \\geqslant 4$. We will show that for each $m \\in\\{3,4, \\ldots, k-1\\}$ there exist $b_{1}, b_{2}, \\ldots, b_{m-2} \\in\\{0,1, \\ldots, s-1\\}$ such that\n$$\nd_{k-m}=s^{m}-\\sum_{i=1}^{m-2} b_{i}\\left(s^{m-i}-s^{m-i-2}\\right) .\n$$\nWe will prove this equality by induction on $m$. If $m=3$, then $d_{k-3}=a_{1} s+b_{1}$ for some non-negative integers $a_{1}, b_{1}$ satisfying $b_{1}<s$ and $b_{1} s+a_{1}=d_{k-2}=s^{2}$. Then $a_{1}=s^{2}-b_{1} s$, so $d_{k-3}=s^{3}-b_{1}\\left(s^{2}-1\\right)$, which proves (9) for $m=3$.\n\nAssume that (9) holds for some $m$ and consider $d_{k-(m+1)}$. There exist non-negative integers $a_{m-1}, b_{m-1}$ such that $d_{k-(m+1)}=a_{m-1} s+b_{m-1}, b_{m-1}<s$ and $d_{k-m}=b_{m-1} s+a_{m-1}$. Using the inductive assumption we get\n$$\na_{m-1}=d_{k-m}-b_{m-1} s=s^{m}-\\sum_{i=1}^{m-2} b_{i}\\left(s^{m-i}-s^{m-i-2}\\right)-b_{m-1} s,\n$$\ntherefore\n$$\n\\begin{aligned}\nd_{k-(m+1)} & =a_{m-1} s+b_{m-1}=s^{m+1}-\\sum_{i=1}^{m-2} b_{i}\\left(s^{m-i}-s^{m-i-2}\\right) s-b_{m-1} s^{2}+b_{m-1} \\\\\n& =s^{m+1}-\\sum_{i=1}^{m-1} b_{i}\\left(s^{m+1-i}-s^{m-i-1}\\right)\n\\end{aligned}\n$$\nwhich completes the proof of (9). In particular, for $m=k-1$ we get\n$$\nd_{1}=s^{k-1}-\\sum_{i=1}^{k-3} b_{i}\\left(s^{k-i-1}-s^{k-i-3}\\right) .\n$$\nThe above sum is clearly divisible by $s^{2}-1$, and it is clear that the remainder of $s^{k-1}$ when divided by $s^{2}-1$ is 1 when $k$ is odd, and $s$ when $k$ is even. It follows that the remainder when $n=d_{1}$ is divided by $s^{2}-1$ is either 1 or $s$.\n\nTo prove the other implication, assume that $n$ gives remainder 1 or $s$ when divided by $s^{2}-1$. If $n \\in\\left\\{1, s, s^{2}\\right\\}$, then one of the numbers $d_{1}, d_{2}$ and $d_{3}$ is 1. We therefore assume that $n>s^{2}$. Since the reminder when a power of $s$ is divided by $s^{2}-1$ is either 1 or $s$, there exists a positive integer $m$ such that $s^{m}-n$ is non-negative and divisible by $s^{2}-1$. By our assumption $m \\geqslant 3$. We also take the smallest such $m$, so that $n>s^{m-2}$. The quotient $\\frac{s^{m}-n}{s^{2}-1}$ is therefore smaller than $s^{m-2}$, so there exist $b_{1}, \\ldots, b_{m-2} \\in\\{0,1, \\ldots, s-1\\}$ such that $\\frac{s^{m}-n}{s^{2}-1}=\\sum_{i=1}^{m-2} b_{i} s^{i-1}$. It follows that\n$$\nn=s^{m}-\\sum_{i=1}^{m-2} b_{i}\\left(s^{i+1}-s^{i-1}\\right) .\n$$\nWe now show that\n$$\nd_{j}=s^{m+1-j}-\\sum_{i=1}^{m-1-j} b_{i}\\left(s^{i+1}-s^{i-1}\\right)\n$$\nfor $j=1,2, \\ldots, m-2$ by induction on $j$. For $j=1$ this follows from $d_{1}=n$. Assume now that (10) holds for some $j<m-2$. Then\n$$\nd_{j}=\\left(s^{m-j}-\\sum_{i=2}^{m-1-j} b_{i}\\left(s^{i}-s^{i-2}\\right)-b_{1} s\\right) s+b_{1}\n$$\nAs $d_{j}$ is positive and $b_{1} \\in\\{0,1, \\ldots, s-1\\}$, the expression $s^{m-j}-\\sum_{i=2}^{m-1-j} b_{i}\\left(s^{i}-s^{i-2}\\right)-b_{1} s$ has to be non-negative, so we can compute the twist of $d_{j}$ as\n$$\nd_{j+1}=b_{1} s+s^{m-j}-\\sum_{i=2}^{m-1-j} b_{i}\\left(s^{i}-s^{i-2}\\right)-b_{1} s=s^{m-j}-\\sum_{i=1}^{m-2-j} b_{i}\\left(s^{i+1}-s^{i-1}\\right),\n$$\nwhich finishes the induction.\n\nNow we use (10) for $j=m-2$ and get $d_{m-2}=s^{3}-b_{1}\\left(s^{2}-1\\right)=(s^{2}-b_{1} s)+b_{1}$. Then $d_{m-1}=b_{1} s+s^{2}-b_{1} s=s^{2}=s \\cdot s+0, d_{m}=0 \\cdot s+s=s=1 \\cdot s+0$ and $d_{m+1}=0 \\cdot s+1=1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo different integers $u$ and $v$ are written on a board. We perform a sequence of steps. At each step we do one of the following two operations:\n\n(i) If $a$ and $b$ are different integers on the board, then we can write $a+b$ on the board, if it is not already there.\n\n(ii) If $a, b$ and $c$ are three different integers on the board, and if an integer $x$ satisfies $a x^{2}+b x+c=0$, then we can write $x$ on the board, if it is not already there.\n\nDetermine all pairs of starting numbers $(u, v)$ from which any integer can eventually be written on the board after a finite sequence of steps.",
    "options": [],
    "answer": "All distinct integer pairs (u, v) with u and v nonzero, at least one of u or v positive, and not equal to the pair {−1, 1}.",
    "solution": "Solution:\n\nWe will show that the answer are the integer pairs $(u, v)$ such that $u \\neq 0$, $v \\neq 0$, $\\{u, v\\} \\neq \\{-1,1\\}$ and $u>0$ or $v>0$.\n\nIf $u=0$ or $v=0$, then (i) will never yield a new number and we cannot use (ii) with only two numbers. Hence, if $u=0$ or $v=0$, we cannot reach every possible $y$. From now on, assume $u \\neq 0$ and $v \\neq 0$.\n\nIf both numbers $u, v$ were negative, we will show that there can only be negative numbers on the board. With negative numbers $a, b$, operation (i) will only yield a negative number. The same holds for operation (ii), because for a non-negative $x$ and negative $a, b, c$, we have $a x^{2}+b x+c<0$. Hence, if both $u<0$ and $v<0$, then we cannot reach every possible $y$. From now on, assume that at least one of $u, v$ is positive. Without loss of generality take $u<v$, and so $v>0$.\n\nAfter one step, we can have the numbers $u, v, u+v$, which are mutually distinct due to $u, v$ being nonzero. Notice that the equation $u x^{2}+(u+v) x+v=0$ has a root $-1$, and so we can have $-1$ on the board.\n\nWe now check two cases: case $v=1$, and case $v>1$.\n\nIf $v=1$, then $u<0$. Further split the case of $v=1$ based on whether $u=-1$ or $u<-1$.\n\nIf $v=1$ and $u=-1$, we can only additionally write number $0$ on the board using operation (i); and no additional numbers using (ii) because setting $\\{a, b, c\\}$ to $\\{-1,0,1\\}$ in any order only has solutions for $x$ in $\\{-1,0,1\\}$. Hence, if $\\{u, v\\}=\\{-1,1\\}$, then we cannot reach every possible $y$.\n\nIf $v=1$ and $u<-1$, we can use operation (i) on numbers $u,-1$ (and then repeat choosing the obtained result and $-1$) to get any negative number smaller than $u$, and operation (i) on numbers $(u, 1)$ (and then repeat choosing the obtained result and $1$) to get any negative number larger than $u$, as well as $0$. Then, we set $(a, b, c)=(0,1,-2)$ and apply operation (ii) to additionally get number $2$. Applying (i) on $(2,1)$ (and then repeat choosing the obtained result and $1$), we can get all the remaining integers too.\n\nFrom now on, assume $v>1$. Recall that we can make $u+v$ and $-1$.\n\nWe will now apply operation (i). First, $(v,-1)$ gives $v-1$. Next, $(v, v-1)$ gives $2 v-1$. Since $v>1$, we know $v \\neq 2 v-1$, so we can apply operation (i) on $(v, 2 v-1)$ to get $3 v-1$, and then continue adding $(v, k v-1)$ to get $(k+1) v-1$ for any positive $k$. Since $v>1$, we can get an arbitrarily large integer by repeating this.\n\nIf $a$ is any positive number on the board, applying (i) to $(a,-1)$ gives $a-1$. By repeating this, we have that we can get all numbers smaller than $a$ and larger than or equal to $-1$. Together with previously having found a way to get an arbitrarily large integer, we have that we can get any integer $l \\geq -1$ on the board.\n\nNow, we set $(a, b, c)=(0,1,2)$ and apply operation (ii) to additionally get number $-2$. Then we can repeat operation (i) on $(-1,-2)$ (and afterwards on $-1$ and the obtained result) to get any negative number.\n\nTherefore, if $u \\neq 0, v \\neq 0, \\{u, v\\} \\neq \\{-1,1\\}$ and $u>0$ or $v>0$, we can write every integer on the board.\nSolution:\n\nIf $u=0$ or $v=0$ then we can only get $\\{u, v\\}$.\n\nProof. If $u=0$ or $v=0$, then (i) will never yield a new number and we cannot use (ii) with only two numbers.\n\nIf $\\max (u, v)<0$, we cannot get non-negative numbers.\n\nProof. For $a, b, c<0$ :\n(i) cannot generate a non-negative number as $a+b<0$.\n(ii) cannot generate a non-negative number as for $x \\geq 0$ : $a x^{2}+b x+c \\leq c<0$.\n\nIf $u+v \\neq 0, u, v \\neq 0$ and $\\max (u, v)>0$, we can get every number.\n\nProof. $u \\neq v \\rightarrow u+v$ can be written. $u \\neq 0$, so $u+v \\neq v \\rightarrow u+2 v$ can be written. $v \\neq 0$, so $u+2 v \\neq u$, meaning that $2 u+2 v$ can be written. If for $n>1, n(u+v)$ can be written then $(n+1)(u+v)=n(u+v)+(u+v)$ can also be written, because $u+v \\neq 0$, so $n(u+v) \\neq(u+v)$. Therefore, by induction for all $n>0$ the number $n(u+v)$ can be written.\n\nTaking $n=2,3, \\ldots$ and $(u+v)(x+1)(x+n)=(u+v) x^{2}+(u+v)(n+1) x+(u+v) n=0$ gives $x=-1$ and $x=-n \\rightarrow$ for all $n>0$ we can get $-n$ (we can get all the negative numbers). Additionally, we can get $u+(-u)=0$.\n\nFor $n \\geq 1$ we can take $0 \\cdot x^{2}+u x+(-u n)=0$ as $u \\neq 0 \\rightarrow u,-n u, 0$ are all distinct, therefore we can get $n$. Thus, we can get all the numbers.\n\nIf $u+v=0$ and $\\max (u, v)>1$ we can get all the numbers.\n\nProof. We can get $0=u+v$ as $u \\neq v$. Take $0 \\cdot x^{2}+u x-u=0$ so we can get $1$ written on the board. Take $(u', v')=(1, \\max (u, v))$ and then use the result from claim 3. As $0<u'=1<\\max (u, v)=v'$ and $u'+v'>0$ we can get all the numbers.\n\nRemaining case: If $v=1$ and $u=-1$, we can only additionally write number $0$ on the board using operation (i); and no additional numbers using (ii) because setting $\\{a, b, c\\}$ to $\\{-1,0,1\\}$ in any order only has solutions for $x$ in $\\{-1,0,1\\}$. Hence, if $\\{u, v\\}=\\{-1,1\\}$, then we cannot reach every possible $y$.\nSolution:\n\nWe show none of the initial number can be $0$, as in Solution 1. Then, we split into three cases: initial numbers having different signs, both being positive, and both being negative.\n\nCase 1. Suppose we have two numbers $u, v$ with different signs such that $\\operatorname{gcd}(u, v)=k$ for some $k \\in \\mathbb{Z}^{+}$. Without loss of generality assume that $u>0$ and $v<0$.\n\nCase 1.1. $v<-k$ :\nWe can generate all numbers $y$ such that $k \\mid y$ and $y \\leq u$.\n\nProof. Define $u'=\\frac{u}{k}, v'=\\frac{v}{k}$. Note that by definition, $\\operatorname{gcd}(u', v')=1$. If starting with numbers $u', v'$, we can make a sequence of applications of only the (i) rule to write a number $y=p \\cdot u'+q \\cdot v'$ on the board, then we can apply the same sequence of moves to $u, v$ to write $k \\cdot y=p \\cdot u+q \\cdot v$ on the board.\n\nTherefore, we will prove instead that if $v'<-1$, we can write all numbers $y<u'$ on the board, which is equivalent to the lemma.\n\nWe attempt to write numbers $u'+q \\cdot v'$ for $q \\in \\mathbb{Z}^{+}$ on the board by repeatedly adding $v'$ to $u'$. This process can only ever halt if we reach a point where $u'+q \\cdot v'=v'$. That cannot occur, as it would imply $u'=(1-q) \\cdot v'$. Taking into account that $u', v' \\neq 0$, that implies $\\operatorname{gcd}(u', v')=|v'|$. That is a contradiction, as $|v'|>1$.\n\nTherefore, we can write all numbers of the form $u'+q \\cdot v'$ on the board for $q \\in \\mathbb{Z}^{+}$. We will use these numbers to construct an arbitrary integer $y$.\n\nIf we want to write a number $y<u'$ on the board, and we already have a number $n<y$ on the board such that $y \\equiv n \\pmod{u'}$, then we can construct $y$ by repeatedly adding $u'$ to $n$ until we reach $y$, skipping all numbers that are already on the board. As $y<u'$, none of the numbers we attempt to add $u'$ to will be equal to $u'$.\n\nSuppose we fix any number $y<u'$. As $\\operatorname{gcd}(u', v')=1$, $q \\cdot v'$ takes all residues modulo $u'$ as $q$ runs through the positive integers. Therefore, we will always be able to find a number $n$ of the form $u'+q \\cdot v'$ such that $y \\equiv n \\pmod{u'}$. We can generate an arbitrarily small $n'$ by taking $n'=n+l \\cdot u' \\cdot v'=u'+(q+l \\cdot u') \\cdot v'$ for large enough $l$, making both $n'<y$ and $n' \\equiv y \\pmod{u'}$ true.\n\nTherefore, we can write all numbers $y<u'$ on the board. Thus, starting from $u$ and $v$, we can get all numbers $y \\leq u$ such that $k \\mid y$.\n\nThe numbers $k, 0$ and all negative multiples of $k$ are a subset of the integers $y \\leq u$ such that $k \\mid y$. Therefore, we have all of those numbers on the board.\n\nWe can now get an arbitrary nonzero number by applying (ii) to the polynomial $k \\cdot (x-n)(x+n)=k \\cdot x^{2}+0 \\cdot x-n^{2} \\cdot k$. The coefficients of this polynomial are distinct for all integers $n \\neq 0$, and they are from the set $\\{k, 0\\} \\cup \\{q \\cdot k \\mid q<0\\}$, which we have on the board. Therefore the rule application is valid.\n\nAs this works for all integers $n \\in \\mathbb{Z} \\setminus \\{0\\}$, and $0$ is already on the board, we have proven that we can write all integers on the board.\n\nCase 1.2. $v=-k$ :\n\nCase 1.2.1. $k \\neq 1$.\nIf $k \\neq 1$, we can generate $-1$ from the polynomial $u \\cdot x^{2}+(u+v) x+v$. We can now repeatedly add $-1$ to $v=-k$ until we reach $-2k$ on the board. Now, we can appeal to Case 1.1.\n\nCase 1.2.2. $k=1$.\nTo restate the conditions of this sub-case, $v=-k=-1$, and $u$ is an arbitrary positive number.\n\nCase 1.2.2.1. $u=1$.\nIf $u=1$, we are only ever able to construct the numbers $-1,0,1$, no matter how we apply the rules.\n\nCase 1.2.2.2. $u \\neq 1$.\nWe can subtract $1$ from $u$ until we reach $0$. This procedure also generates $1$. We now add $1$ to $u$ until we get all positive numbers.\n\nNow, we find an arbitrary polynomial with different positive coefficients that have a negative root smaller than $-1$. An example is $(x+3)^{2}$.\n\nWe will now keep adding $-1$ to $-3$ to get all the negative numbers $y \\leq -3$. To get $-2$, we can add $-3$ to $1$, generating all integers.\n\nCase 2. Suppose that both of $u, v$ are positive.\n\nWe can now use any method from the previous solutions to generate a negative number, and then appeal to Case 1.\n\nCase 3. Suppose $u<0, v<0$.\n\nThen there can only be negative numbers on the board. With negative numbers $a, b$, operation (i) will only yield a negative number. The same holds for operation (ii), because for a non-negative $x$ and negative $a, b, c$, we have $a x^{2}+b x+c<0$. Hence, if both $u<0$ and $v<0$, then we cannot reach every possible $y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $AC > AB$, and denote its circumcircle by $\\Omega$ and incentre by $I$. Let its incircle meet sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. Let $X$ and $Y$ be two points on minor arcs $\\widehat{DF}$ and $\\widehat{DE}$ of the incircle, respectively, such that $\\angle BXD = \\angle DYC$. Let line $XY$ meet line $BC$ at $K$. Let $T$ be the point on $\\Omega$ such that $KT$ is tangent to $\\Omega$ and $T$ is on the same side of line $BC$ as $A$. Prove that lines $TD$ and $AI$ meet on $\\Omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBy the alternate segment theorem we have that:\n$$\n180^\\circ = \\angle DCY + \\angle CYD + \\angle YDC = \\angle DCY + \\angle DXB + \\angle YXD = \\angle DCY + \\angle YXB\n$$\nTherefore opposite angle of $BXYC$ are supplementary and so $CYXB$ cyclic.\nOne can apply power of a point at $K$ :\n$$\nKT^2 = KB \\cdot KC = KX \\cdot KY = KD^2 \\Longrightarrow KT = KD.\n$$\n![](attached_image_1.png)\nFigure 1: The proposer's solution using a new point $Q$\n(Alternatively you can sidestep power of a point by observing that $K$ is the radical centre of the incircle $DEF$, the circumcircle $\\Gamma$ and the circle $CYXB$ and so $KT^2 = KD^2$.)\nNow let $AI$ meet $\\Omega$ at $M$, the midpoint of $\\widehat{BC}$ not containing $A$. Let the tangent at $M$ meet $KT$ at $Q$. Observe that $QM \\parallel KD$ so $\\angle TKD = \\angle TQM$ and also $KT = KD$, $QT = QM$ hence $\\triangle TKD \\sim \\triangle TQM$. As $T, K, Q$ are collinear, this means that $T, D, M$ are collinear so $TD$ and $AI$ meet at $M$ which lies on $\\Omega$.\nSolution:\n\nThe role of $X$ and $Y$ in this problem is secondary. Draw any circle through $B$ and $C$ which meets the incircle at $X$ and $Y$ and you determine the same point $K$. This is because $K$ is the radical centre of the three circles in play. Therefore $K$ is the intersection of $BC$ with the radical axis of the circumcircle $\\Omega$ and the incircle $DEF$, and so is independent of the choice of circle through $B$ and $C$ which gives rise to $X$ and $Y$. In the problem as posed, this is disguised, with angle properties of $X$ and $Y$ giving rise to the circle $BXYC$.\nYou can now work with the simplified diagram shown in Figure 2.\nSomehow we must use a characterization of $M$ in order to finish. In the proposer's solution, we used the tangent to the circumcircle $\\Omega$ at $M$ being parallel to $BC$. In this alternative, we use the fact that the internal angle bisector of angle $\\angle CTB$ meets $\\Omega$ again at $M$.\nTriangle $TKD$ is isosceles with apex $K$ so\n$$\n\\angle KDT = \\angle DTK = \\angle DTB + \\angle BTK.\n$$\n![](attached_image_2.png)\nFigure 2: Illustration of an angle chase\nBy the alternate segment theorem, $\\angle BTK = \\angle BCT = \\angle DCT$. Now angle $\\angle KDT$ is an exterior angle of triangle $DCT$ so $\\angle CTD = \\angle DTB$.\nTherefore the line $TD$ is the internal angle bisector of angle $\\angle CTB$ and so must pass through $M$, the midpoint of the arc $\\widehat{BC}$ of $\\Omega$ which does not contain $T$.\nSolution:\n\nThe centre of direct enlargement from the incircle to the circumcircle gives another way to finish the proof. This enlargement carries $D$ to $M$ since the tangent lines to their associated circles are parallel. The centre of enlargement therefore lies on the line $MD$. Let $MD$ meet the incircle again at $U$ and the circumcircle again at $V$. Draw the tangent to the incircle at $U$ to meet $BC$ at $W$ so triangle $WDU$ is isosceles with apex $W$ and has equal base angles $\\angle WDU$ and $\\angle DUW$. The enlargement carries the line $WU$ to the tangent line to the circumcircle at $V$ which meets $BC$ at $S$. Enlargements carry lines to parallel lines, so $\\angle DVK' = \\angle DUW = \\angle WDU = \\angle K'DV$. Therefore triangle $K'DV$ is isosceles with apex $K'$.\n![](attached_image_3.png)\nFigure 3: The three points $K$ on the radical axis.\n![](attached_image_4.png)\nFigure 4: The enlargement of $W$ gives $K'$.\nThis identifies $S$ as the intersection of the radical axis of the two circles with $BC$, so $K' = K$ and $V = T$ and the proof is complete.\nSolution:\n\nLet $\\phi$ be the inversion with center $K$ and radius $KD$. Note that this inversion maps incircle of $ABC$ to itself and $K, X, Y$ are collinear, hence $\\phi(X) = Y$. Also $\\phi(D) = D$, so $\\phi$ maps circle $XBD$ to circle $YDB'$, where $B' := \\phi(B)$ is the point on $BC$ different from $B$ such that $\\angle BXD = \\angle DYB'$, hence $\\phi(B) = C$. From $\\phi(B) = C$ we get that $KD = KT$ as $KD^2 = KB \\cdot KC = KT^2$ since $KD$ is the radius of inversion.\nThe rest of the solutions is the same as in the other solutions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe call a positive integer $n$ peculiar if, for any positive divisor $d$ of $n$, the integer $d(d+1)$ divides $n(n+1)$. Prove that for any four different peculiar positive integers $A, B, C$ and $D$, the following holds:\n$$\n\\operatorname{gcd}(A, B, C, D)=1\n$$\nHere $\\operatorname{gcd}(A, B, C, D)$ is the largest positive integer that divides all of $A, B, C$ and $D$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nObserve that $n=1$ is peculiar and that every prime is peculiar. Consider $\\frac{n}{d}$, where $d>1$ is a divisor of $n$. Then $\\frac{n}{d}\\left(\\frac{n}{d}+1\\right)$ divides $n(n+1)$, equivalent to $n+d$ dividing $d^{2}(n+1)$. Since $n \\equiv -d \\pmod{n+d}$, we obtain that $n+d$ divides $d^{3}-d^{2}$, hence $n+d \\leq d^{3}-d^{2}$ and in particular $n<d^{3}$. Now taking $d$ to be the smallest prime divisor of $n$, we deduce that $n$ cannot have more than two prime factors, counting multiplicities. Hence from now on we work with $n=pq$, where $p$ and $q$ are primes. In fact, $p \\neq q$, as otherwise $p+1 \\mid p^{2}-p$, i.e. $p+1 \\mid 2$, contradiction.\n\nLet us first prove that $n$ is square-free. Assume that $n=p^{k} h$, where $\\operatorname{gcd}(p, h)=1$. Now $p^{k-1} h\\left(p^{k-1} h+1\\right) \\mid p^{k} h\\left(p^{k} h+1\\right)$, which is equivalent to $\\left(p^{k-1} h+1\\right) \\mid p\\left(p^{k} h+1\\right)$. Since $\\operatorname{gcd}\\left(p, p^{k-1} h+1\\right)=1$, we must have\n$$\n\\left(p^{k-1} h+1\\right) \\mid \\left(p^{k} h+1\\right)=p^{k} h+p-p+1\n$$\nso $\\left(p^{k-1} h+1\\right) \\mid (p-1)$. If $k \\geq 2$, this is a contradiction. Hence, $n$ is square-free.\n\nLet us now show that $n$ has at most 2 different prime factors. Write $n=p_{1} p_{2} \\cdots p_{k}$. Now we must have\n$$\np_{2} \\cdots p_{k}\\left(p_{2} \\cdots p_{k}+1\\right) \\mid p_{1} \\cdots p_{k}\\left(p_{1} \\cdots p_{k}+1\\right)\n$$\nwhich is equivalent to\n$$\n\\left(p_{2} \\cdots p_{k}+1\\right) \\mid p_{1}\\left(p_{1} \\cdots p_{k}+1\\right)\n$$\nNow\n$$\np_{1}\\left(p_{1} \\cdots p_{k}+1\\right)-p_{1}^{2}\\left(p_{2} \\cdots p_{k}+1\\right)=p_{1}-p_{1}^{2}\n$$\nHence,\n$$\n\\left(p_{2} \\cdots p_{k}+1\\right) \\mid p_{1}^{2}-p_{1}\n$$\nPick $p_{1}$ to be the smallest prime factor of $n$. Then $p_{2} \\cdots p_{k}>p_{1}^{2}$ if $k>2$, which is a contradiction.\n\nThe only relations of concern are $p(p+1) \\mid n(n+1)$ and $q(q+1) \\mid n(n+1)$. The former yields $p(p+1) \\mid p q(p q+1)$, i.e. $p+1 \\mid q(p q+1)$, that is, $p+1 \\mid q(q-1)$; analogously $q+1 \\mid p(p-1)$. Without loss of generality assume $p>q$. If $q$ does not divide $p+1$, then since $p+1$ and $q$ are relatively prime, we obtain from $p+1 \\mid q(q-1)$ that $p+1$ divides $q-1$ and hence $p<p+1 \\leq q-1<q$, contradiction. Hence $q$ divides $p+1$. Also, $p$ does not divide $q+1$ due to $p>q$ (unless $p=q+1$, i.e. $p=3, q=2$, but then $p+1 \\nmid q(q-1)$ ) and now from $q+1 \\mid p(p-1)$ we deduce that $q+1$ divides $p-1$.\n\nWrite $p+1=m q$ for some positive integer $m$. Then $q+1$ divides $p-1=m q-2= m(q+1)-m-2$, so $q+1$ divides $m+2$. On the other hand, $p+1 \\mid q(q-1)$ implies that $m$ divides $q-1$. If the ratio of the latter is greater than 1, then $m \\leq \\frac{q-1}{2}$, but $q+1$ dividing $m+2$ from above insists on $q \\leq m+1$, so $m \\leq \\frac{q-1}{2} \\leq \\frac{m}{2}$, contradiction. Hence $m=q-1$ and we have proven the following: if $n>1$ is composite and peculiar, then $n=pq$, where $p>q$ are primes and $p=q^{2}-q-1$.\n\nHence from now on we work with $n=ab$, with $a<b$ primes. The only relations of concern are $a(a+1) \\mid n(n+1)$ and $b(b+1) \\mid n(n+1)$. The former yields $a(a+1) \\mid ab(ab+1)$, i.e. $a+1 \\mid b(ab+1)$, that is, $a+1 \\mid b(b-1)$; analogously $b+1 \\mid a(a-1)$. If $a$ does not divide $b+1$, then since $b+1$ and $a$ are relatively prime, we obtain from $b+1 \\mid a(a-1)$ that $b+1$ divides $a-1$ and hence $b<b+1 \\leq a-1<a$, contradiction. Hence $a$ divides $b+1$. Writing $b+1=ac$, from $b+1 \\mid a(a-1)$ we have that $c$ divides $a-1$.\n\nNow from $b+1=ac$ with $c \\mid a-1$ and $a+1 \\mid b(b-1)$ we derive that $b=a^{2}-a-1$ in a different (and longer) manner, without using the primality of $b$ until the final step. Note that since $b=ac-1 \\equiv -c-2 \\pmod{a+1}$, we have $a+1 \\mid (c+1)(c+2)$. Since $c \\mid a-1$, we may write $a-1=cd$, and note that $cd \\equiv -2 \\pmod{a+1}$. We then have\n$$\n0 \\equiv -d(c+1)(c+2) = -c^{2}d-3cd-2d \\equiv 2(c-d)+6 \\quad (\\bmod a+1)\n$$\nSince $1 \\leq c, d \\leq a-1$, we have $|2(c-d)+6| \\leq 2(a-2)+6=2a+2$. Furthermore, we have $c \\geq 2$ (otherwise $b=ac-1<a$, contradiction), and so $d \\leq \\frac{a-1}{2}$, thus $2(c-d)+6 \\geq 11-a>-(a+1)$. From these inequalities and equivalence (1), we obtain $2(c-d)+6 \\in\\{0, a+1,2(a+1)\\}$. It is immediate to verify that these three possible values correspond to the following three cases, respectively:\n(i) $d=c+3$ and $a=c^{2}+3c+1$;\n(ii) $d=2, c=\\frac{a-1}{2}, a=2c+1$; and\n(iii) $d=1, c=a-1$.\nSubstituting these values in $b=ac-1$, we obtain in case (i), $b=c^{3}+3c^{2}+c-1=(c+1)(c^{2}+2c-1)$, and in case (ii), $b=2c^{2}+c-1=(c+1)(2c-1)$. In both cases these decompositions contradict the primality of $b$, as $c \\geq 2$. Thus we are left only with case (iii), where $b=a^{2}-a+1$, as claimed.\n\nIn conclusion, a prime $p$ can only divide the peculiar numbers $p$, $p(p^{2}-p-1)$ (if $p^{2}-p-1$ is prime) and $pq$ where $p=q^{2}-q-1$ (if $q$ is prime). In particular, the greatest common divisor of more than three peculiar numbers must be $1$. (Note that three peculiar numbers may indeed have a common divisor, e.g. $5, 15=3 \\cdot 5$ and $95=5 \\cdot 19$ are all peculiar.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor a sequence $a_{1} < a_{2} < \\cdots < a_{n}$ of integers, a pair $(a_{i}, a_{j})$ with $1 \\leq i < j \\leq n$ is called interesting if there exists a pair $(a_{k}, a_{l})$ of integers with $1 \\leq k < l \\leq n$ such that\n$$\n\\frac{a_{l} - a_{k}}{a_{j} - a_{i}} = 2\n$$\nFor each $n \\geq 3$, find the largest possible number of interesting pairs in a sequence of length $n$.",
    "options": [],
    "answer": "(n-1)(n-2)/2 + 1",
    "solution": "Solution:\n\nAnswer. $\\frac{1}{2}(n-1)(n-2)+1$.\n\nConsider the numbers $a_{i} = 2^{i}$ for $2 \\leq i \\leq n$ and $a_{1} = 0$, and choose any pair $(i, j)$ with $1 \\leq i < j \\leq n$.\nIf $i = 1$, the pair $(1, j)$ is interesting for all $2 \\leq j \\leq n-1$, as $\\frac{a_{j+1} - a_{1}}{a_{j} - a_{1}} = \\frac{2^{j+1}}{2^{j}} = 2$.\nIf $i \\geq 2$, the pair $(i, j)$ is also interesting for all $j$ such that $i+1 \\leq j \\leq n-1$, since $\\frac{a_{j+1} - a_{i+1}}{a_{j} - a_{i}} = \\frac{2^{j+1} - 2^{i+1}}{2^{j} - 2^{i}} = 2$.\nIn addition, the pair $(n-1, n)$ is interesting, as $\\frac{a_{n} - a_{0}}{a_{n} - a_{n-1}} = 2$.\nHence, all pairs in which $j \\leq n-1$, and additionally the pair $(n-1, n)$, are interesting, giving $\\frac{1}{2}(n-1)(n-2)+1$ pairs in total.\n\nNow, let's show that there can't be more interesting pairs. To this end we show that at least $\\frac{1}{2} n(n-1) - \\left(\\frac{1}{2}(n-1)(n-2)+1\\right) = n-2$ pairs are not interesting. Clearly, the pair $(1, n)$ is not interesting.\nLet's make the following observation: if a pair $(i, j)$ is interesting and it's not $(1, n)$, then $a_{j} - a_{i}$ can't exceed $\\frac{1}{2}(a_{n} - a_{1})$. (Since, if $(i, j)$ is interesting then $a_{l} - a_{k} = 2(a_{j} - a_{i})$ for some pair $(k, l)$. If then $2(a_{j} - a_{i}) > a_{n} - a_{1}$ we would have $a_{l} - a_{k} > a_{n} - a_{1}$, which is not possible.)\nFinally, for any $2 \\leq i \\leq n-1$ consider pairs $(1, i)$ and $(i, n)$. If both of them are interesting, then, following the observation above, the only possibility is $a_{i} - a_{1} = a_{n} - a_{i} = \\frac{a_{n} - a_{1}}{2}$. Clearly, this is possible for at most one $i$, so for all other $n-3$ values of $i$, at least one of pairs $(1, i)$ and $(i, n)$ is not interesting. Combining with the pair $(1, n)$, we get at least $n-2$ not interesting pairs in total, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathbb{N}$ denote the set of positive integers. Find all functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that the following assertions are true for every pair of positive integers $(x, y)$:\n- $x$ and $f(x)$ have the same number of positive divisors.\n- If $x$ does not divide $y$ and $y$ does not divide $x$, then\n$$\n\\operatorname{gcd}(f(x), f(y)) > f(\\operatorname{gcd}(x, y))\n$$\nHere $\\operatorname{gcd}(m, n)$ is the largest positive integer that divides both $m$ and $n$.",
    "options": [],
    "answer": "All functions of the form f(n) = q^{d(n)-1} for an arbitrary fixed prime q.",
    "solution": "Solution:\nLet $f$ be a function satisfying these properties. From the first one, it follows that $f(1)=1$ and for each prime number $p$, $f(p)=q_p$ for some prime number $q_p$. Assume $p \\neq 2$, the pair $(2, p)$ in property 2 gives\n$$\n\\operatorname{gcd}\\left(q_2, q_p\\right) = \\operatorname{gcd}(f(2), f(p)) > f(\\operatorname{gcd}(2, p)) = f(1) = 1\n$$\nso $q_p = q_2$ for all $p$. Set $q = q_2$.\nDenote by $d(n)$ the number of divisors of a positive integer $n$.\nWe will prove the following claim by induction on $k \\geq 0$, the number of distinct prime divisors of $n$: For a positive integer $n$, $f(n)$ is a power of $q$.\nThe claim, together with the first property, imply that $f(n) = q^{d(n)-1}$.\nFor $k=0$ the claim is already proved.\nFor $k=1$, we need to prove that $f\\left(p^m\\right)$ is a power of $q$ for all positive $m$. The case $m=1$ was already proved. Assume now the result being proved for all positive integers $\\leq m$. As $d\\left(p^t\\right) = t+1$, we then know $f\\left(p^t\\right) = q^t$ for all $t \\leq m$. Now we consider $t = m+1$.\nLet $r \\neq p$ be a prime. Plug the pair $\\left(p^{m-1} r, p^m\\right)$ into the second property:\n$$\n\\operatorname{gcd}\\left(f\\left(p^{m-1} r\\right), q^m\\right) = \\operatorname{gcd}\\left(f\\left(p^{m-1} r\\right), f\\left(p^m\\right)\\right) > f\\left(p^{m-1}\\right) = q^{m-1}\n$$\nThis implies $q^m$ divides $f\\left(p^{m-1} r\\right)$. Since $f\\left(p^{m-1} r\\right)$ has $2m$ divisors, and $v_q\\left(f\\left(p^{m-1} r\\right)\\right) \\geq m$, it follows that $f\\left(p^{m-1} r\\right)$ does not have prime divisors other than $q$, since it would then have at least $2 \\cdot \\left(v_q\\left(f\\left(p^{m-1} r\\right)\\right)+1\\right) > 2m$ divisors. Thus, $f\\left(p^{m-1} r\\right)$ is a power of $q$. And since it has $2m$ divisors, we must have $f\\left(p^{m-1} r\\right) = q^{2m-1}$.\nNow, plug the pair $\\left(p^{m+1}, p^{m-1} r\\right)$ into the second property. We have\n$$\n\\operatorname{gcd}\\left(f\\left(p^{m+1}\\right), q^{2m-1}\\right) = \\operatorname{gcd}\\left(f\\left(p^{m+1}\\right), f\\left(p^{m-1} r\\right)\\right) > f\\left(p^{m-1}\\right) = q^{m-1}.\n$$\nThen $f\\left(p^{m+1}\\right)$ is divisible by $q^m$. If $f\\left(p^{m+1}\\right)$ had a prime factor other than $q$, it would have at least $2(m+1)$ divisors, but it has $m+2$ divisors and $2(m+1) > m+2$ since $m \\geq 1$. Hence, $f\\left(p^{m+1}\\right)$ must also be a power of $q$. By induction, the proof is finished for $k=1$.\nSuppose the claim is true for all integers $n$ with at most $k$ distinct prime factors, for some $k \\geq 1$. In order to prove the claim for integers with $k+1$ distinct prime factors, it suffices\nto show that for every positive integer $N$ with $k$ distinct prime factors, every positive integer $m$ and every prime $p$ which does not divide $N$, the number $f\\left(N p^m\\right)$ is a power of $q$. We will prove this by induction on $m$. The case $m=0$ follows from the previous induction hypothesis. Now suppose it has already been proved for a certain $m \\geq 0$.\nLet $s$ be a prime divisor of $N$. Consider the numbers $\\left(N p^{m+1}, N s p^m\\right)$. By the second property,\n$$\n\\operatorname{gcd}\\left(f\\left(N p^{m+1}\\right), q^{d\\left(N s p^m\\right)-1}\\right) = \\operatorname{gcd}\\left(f\\left(N p^{m+1}\\right), f\\left(N s p^m\\right)\\right) > f\\left(N p^m\\right) = q^{d\\left(N p^m\\right)-1}\n$$\nThen $q^{d\\left(N p^m\\right)}$ divides $f\\left(N p^{m+1}\\right)$. If $f\\left(N p^{m+1}\\right)$ has a prime factor other than $q$, it would have at least $2\\left(d\\left(N p^m\\right)+1\\right) = 2((m+1)d(N)+1) = (2m+2)d(N)+2$ divisors, but it has only $(m+2)d(N)$ divisors. Hence, $f\\left(N p^{m+1}\\right)$ must also be a power of $q$.\nFinally, let us check that the function $f(n) = q^{d(n)-1}$ does satisfy the properties of the problem for any prime number $q$. The first property obviously holds. To check the second one, note that\n$$\n\\operatorname{gcd}(f(x), f(y)) = q^{\\min (d(x), d(y))} > q^{d(\\operatorname{gcd}(x, y))} = f(\\operatorname{gcd}(x, y))\n$$\nwhenever $x \\nmid y$ and $y \\nmid x$.\nSolution:\nFirstly, one can prove that for any prime $p$ and any $m \\geq 0$, $f\\left(p^m\\right) = q^m$, in the same way as in the first solution.\nWe will prove the claim this time by induction on $k = d(n)$. We notice again that the claim implies that $f(n) = q^{d(n)-1}$.\nThe cases $k=1,2,3$ are then already proved. Now suppose that for some $k \\geq 3$ we have proved the claim for all positive integers $n$ with at most $k$ divisors.\nConsider a positive integer $n$ with $k+1$ divisors. If $n$ is a power of a prime, then it is already proved that $f(n)$ is a power of $q$. Suppose $n$ is not a power of a prime. Consider distinct prime numbers $p_1$ and $p_2$ which divide $n$, such that $v_{p_1}(n) \\leq v_{p_2}(n)$. Consider the number $\\frac{n p_2}{p_1}$: its number of divisors is $\\frac{v_{p_1}(n)(v_{p_2}(n)+2)}{(v_{p_1}(n)+1)(v_{p_2}(n)+1)} d(n)$, which is strictly less than the number of divisors of $n$. Thus, by induction hypothesis, $f\\left(\\frac{n p_2}{p_1}\\right)$ is a power of $q$.\nBy applying the second property to the pair $\\left(n, \\frac{n p_2}{p_1}\\right)$, we have\n$$\n\\operatorname{gcd}\\left(f(n), q^{d\\left(\\frac{n p_2}{p_1}\\right)-1}\\right) = \\operatorname{gcd}\\left(f(n), f\\left(\\frac{n p_2}{p_1}\\right)\\right) > f\\left(\\frac{n}{p_1}\\right) = q^{d\\left(\\frac{n}{p_1}\\right)-1}\n$$\nThis implies that $q^{d\\left(\\frac{n}{p_1}\\right)}$ divides $f(n)$. Again, if other primes than $q$ divided $f(n)$, it would have at least $2\\left(d\\left(\\frac{n}{p_1}\\right)+1\\right)$ divisors. But $2\\left(d\\left(\\frac{n}{p_1}\\right)+1\\right) = 2\\left(\\frac{v_{p_1}(n)}{v_{p_1}(n)+1} d(n)+1\\right) > d(n)$. So $f(n)$ is a power of $q$ and the Claim is proved.\nWe finally check as in the first solution that $f(n) = q^{d(n)-1}$ satisfies the 2 properties for all prime $q$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $d$ for which there exists a degree $d$ polynomial $P$ with real coefficients such that there are at most $d$ different values among $P(0), P(1), \\ldots, P\\left(d^{2}-d\\right)$.",
    "options": [],
    "answer": "{1, 2, 3}",
    "solution": "Solution:\nWe claim that such polynomials exist if and only if $d \\leq 3$. The following examples show that such polynomials do exist for $d \\leq 3$:\n$$\n\\begin{array}{lll}\nd=1: & d^{2}-d=0, & P_{1}(x)=x, \\\\\n\\end{array} \\begin{array}{ll} \n& P(0)=0 ;\n\\end{array}\n$$\nWe can make more examples by adding constants.\n\nNow we will show that there are no examples of degree greater than 3.\nFrom now on we assume (without loss of generality) that the leading coefficient of our polynomial $P$ is positive and that all values $P(i)$ are positive (by adding a constant if necessary) for integers $i$ in the range $0 \\leq i \\leq d^{2}-d+1$.\n\nAssume (for contradiction) that $P$ is a polynomial of degree $d \\geq 4$ that satisfies the conditions of the problem and let $P(0), \\ldots, P\\left(d^{2}-d\\right)$ take values among $p_{1}<\\cdots<p_{d}$. For $i=1, \\ldots, d$, let $n_{i} \\geq 0$ be the number of appearances of $p_{i}$ among $P(0), \\ldots, P\\left(d^{2}-d\\right)$.\nBy definition $n_{1}+\\cdots+n_{d}=d^{2}-d+1$. Since $P$ has degree $d$, $n_{i} \\leq d$. The key observation is now the following.\n\nClaim 1\nMake the convention that $n_{0}=n_{d+1}=0$. If $n_{i}=d$ for some $i$ in the range $1 \\leq i \\leq d$, then $n_{i \\pm 1} \\leq d-2$.\n\nProof. Up to scaling and hence without loss of generality, $P$ has leading coefficient $+1$. Since $n_{i}=d$, there exist non-negative integers $a_{1, i}<\\cdots<a_{d, i} \\leq d^{2}-d$ such that\n$$\nP(X)=\\left(X-a_{1, i}\\right) \\cdots\\left(X-a_{d, i}\\right)+p_{i}.\n$$\nBy construction, each of the $d-1$ intervals $I_{j}=\\left[a_{j, i}, a_{j+1, i}\\right]$ contains at least one local extremum of $P$, so contains exactly one such extremum because $P$, having degree $d$, has at most $d-1$ such extrema. Suppose that $i \\leq d-1$ and that $P(m)=p_{i+1}>p_{i}$ for some $m \\in\\{0, \\ldots, d^{2}-d\\}$. Since $P$ has positive leading coefficient,\n$$\nm \\in\\left(a_{d, i}, \\infty\\right) \\cup\\left(a_{d-2, i}, a_{d-1, i}\\right) \\cup \\cdots \\cup\\left(a_{1, i}, a_{2, i}\\right)\n$$\nif $d$ is odd or\n$$\nm \\in\\left(a_{d, i}, \\infty\\right) \\cup\\left(a_{d-2, i}, a_{d-1, i}\\right) \\cup \\cdots \\cup\\left(-\\infty, a_{1, i}\\right)\n$$\nif $d$ is even.\n\nSuppose that $a_{j, i}<m<a_{j+1, i}$, for some $j \\in\\{1, \\ldots, d-1\\}$. If $a_{j, i}+1<m<a_{j+1, i}-1$, then, because $I_{j}$ contains exactly one local extremum (which is a maximum),\nEither $p_{i+1}=P(m)>P\\left(a_{j, i}+1\\right)$ or $p_{i+1}=P(m)>P\\left(a_{j+1, i}-1\\right)$. Since $P\\left(a_{j, i}+1\\right)>P\\left(a_{j, i}\\right)=p_{i}$ and $P\\left(a_{j, i}-1\\right)>P\\left(a_{j+1, i}\\right)=p_{i}$, this contradicts the requirement that $P\\left(a_{j, i}+1\\right), P\\left(a_{j+1, i}-1\\right) \\in\\{p_{1}, \\ldots, p_{d}\\}$. Hence $m=a_{j, i}+1$ or $m=a_{j, i}-1$. Similarly, if $m>a_{d, i}$, then $m=a_{d, i}+1$, but if $m<a_{1, i}$ (which may arise when $d$ is even), then $m=a_{1, i}-1$. This shows that $m$ belongs to this list:\n$$\na_{d, i}+1, a_{d-1, i}-1, \\ldots, a_{2, i}+(-1)^{d}, a_{1, i}-(-1)^{d}\n$$\nThis list contains at most $d$ different integers. It follows in particular that, if $n_{i+1}>d-2$, then either\n$$\nP\\left(a_{d, i}+1\\right)=p_{i+1}=P\\left(a_{d-1, i}-1\\right)\n$$\nor\n$$\nP\\left(a_{2, i}+(-1)^{d}\\right)=p_{i+1}=P\\left(a_{1, i}-(-1)^{d}\\right)\n$$\nwith, additionally, $a_{2, i}+(-1)^{d} \\neq a_{1, i}-(-1)^{d}$.\n\nWe have\n$$\n\\left|P\\left(a_{1, i} \\pm 1\\right)-p_{i}\\right|=1 \\cdot\\left|a_{1, i} \\pm 1-a_{2, i}\\right| \\cdot \\prod_{j=3}^{d}\\left|a_{1, i} \\pm 1-a_{j, i}\\right|\n$$\nand\n$$\n\\left|P\\left(a_{2, i} \\mp 1\\right)-p_{i}\\right|=\\left|a_{2, i} \\mp 1-a_{1, i}\\right| \\cdot 1 \\cdot \\prod_{j=3}^{d}\\left|a_{2, i} \\mp 1-a_{j, i}\\right|.\n$$\nAs $a_{1, i}<a_{2, i}<a_{3, i}<\\ldots<a_{d, i}$ we have $\\left|a_{1, i} \\pm 1-a_{j, i}\\right| \\geq\\left|a_{2, i} \\mp 1-a_{j, i}\\right|$ with equality possible only if $a_{1, i}+1=a_{2, i}-1$. We also have $\\left|a_{1, i} \\pm 1-a_{2, i}\\right|=\\left|a_{2, i} \\mp 1-a_{1, i}\\right|$, which can be zero only if $a_{1, i}+1=a_{2, i}-1$. We conclude that $\\left|P\\left(a_{1, i} \\pm 1\\right)-p_{i}\\right|>\\left|P\\left(a_{2, i} \\mp 1\\right)-p_{i}\\right|$ or $a_{1, i}+1=a_{2, i}-1$.\n\nLooking at the other end of the list of $\\left(a_{j, i}\\right)$ as $j$ varies, we have\n$$\n\\left|P\\left(a_{d-1, i}-1\\right)-p_{i}\\right|=1 \\cdot\\left|a_{d-1, i}-1-a_{d, i}\\right| \\cdot \\prod_{j=1}^{d-2}\\left|a_{d-1, i}-1-a_{j, i}\\right|\n$$\nand\n$$\n\\left|P\\left(a_{d, i}+1\\right)-p_{i}\\right|=\\left|a_{d, i}+1-a_{d-1, i}\\right| \\cdot 1 \\cdot \\prod_{j=1}^{d-2}\\left|a_{d, i}+1-a_{j, i}\\right|\n$$\nIn these two formulas the shared factor outside the product is at least 2 and so is not 0. Now look at the factors behind the product symbols. As $a_{1, i}<a_{2, i}<a_{3, i}<\\ldots<a_{d, i}$, for $j \\leq d-2$ we have $\\left|a_{d, i}+1-a_{j, i}\\right|>\\left|a_{d-1, i}-1-a_{j, i}\\right|$. We conclude that $\\left|P\\left(a_{d, i}+1\\right)-p_{i}\\right|>\\left|P\\left(a_{d-1, i}-1\\right)-p_{i}\\right|$. Claim 1 is proved.\n\nFor each $i \\in\\{1, \\ldots, d-1\\}$, there are three possibilities:\n- $n_{i}, n_{i+1} \\leq d-1$\n- $n_{i}=d$ and $n_{i+1} \\leq d-2$\n- $n_{i+1}=d$ and $n_{i} \\leq d-2$.\nIn all three cases, $n_{i}+n_{i+1} \\leq 2(d-1)$. If $n$ is even, this leads to the contradiction\n$$\nd^{2}-d+1=\\left(n_{1}+n_{2}\\right)+\\cdots+\\left(n_{d-1}+n_{d}\\right) \\leq(d / 2)[2(d-1)]=d^{2}-d\n$$\nThis is an important staging point in the argument because we have eliminated the possibility of a polynomial of even degree $d$ satisfying the conditions of the problem if $d \\geq 4$.\n\nFrom now on we assume that $d \\geq 5$ is odd, and\n$$\nd^{2}-d+1=\\left(n_{1}+n_{2}\\right)+\\cdots+\\left(n_{d-2}+n_{d-1}\\right)+n_{d} \\leq[(d-1) / 2][2(d-1)]+d=d^{2}-d+1.\n$$\nEquality must therefore hold throughout. Since the sum can also be grouped as\n$$\nn_{1}+\\left(n_{2}+n_{3}\\right)+\\cdots+\\left(n_{d-1}+n_{d}\\right)\n$$\nthis requires $n_{1}=n_{d}=d, n_{i}+n_{i+1}=2(d-1)$ for $i=1,2, \\ldots, d-1$ i.e. $n_{i}=d$ for odd $i$ and $n_{i}=d-2$ for even $i$.\n\nWe are interested in the degree of $P$ being $d \\geq 5$ and odd, and showing that no polynomial $P$ satisfying the conditions of the problem can exist. There are $d-1 \\geq 4$ extremal points which alternate between local maxima and minima (in that order) as you read from left to right (we normalize so that $P$ is monic). For any $p_{i}$ with $i$ odd, the line $y=p_{i}$ (with $i$ odd) crosses the graph of $P$ in $d$ places with $x$-coordinates in the real closed interval $[0, d^{2}-d]$ at points $(z, p_{i})$ so each $z$ must be an integer. Suppose that $J$ is a real interval on the $x$-axis ending at adjacent local extrema. The function defined by $P$ is monotonic on each $J$. The line $y=p_{i}$ ($i$ odd) meets the graph at most once on $J$. Therefore it meets the graph of $P$ exactly once in the interior of each $J$ (there are $d-2$ such intervals) and at the only two possible places outside the union of these intervals.\n\nNow consider $p_{j}$ when $j$ is even (so $n_{j}=d-2$). These $d-2$ intervals $J$ afford $d-2$ real values at which $P$ will take $p_{j}$ as a value where $j$ is fixed and even. The question is, are the corresponding arguments integers? The proof of Claim 1 tells us that in the middle of the run $\\{0,1, \\ldots, d^{2}-d+1\\}$ all is well: the polynomial is assuming the value $p_{j}$ at an integer where the polynomial assumes the values $p_{j-1}$ and $p_{j+1}$ at adjacent integers in some order. The problem is at the ends of the run where $|P(a_{1, i}+1)-p_{i}|>|P(a_{2, i}-1)-p_{i}|$ and $|P(a_{d, i}+1)-p_{i}|>|P(a_{d-1, i}-1)-p_{i}|$. When $j$ is even, two of the roots of $P(x)-p_{j}$ are not integers, and we now know approximately where this trouble is happening (at the ends).\n\nAt this point we could finish if $d \\geq 7$, because the run of regular behaviour in the middle is sufficiently long that we could obtain a contradiction. However we have to work a little harder to include the case $d=5$. We now show that the run of regular behaviour is slightly longer than we have currently established. We do this using Claim 2.\n\nClaim 2. If $d$ is odd, and $n_{i}=d, n_{i \\pm 1}=d-2$ for some $i$, then $P$ attains $p_{i \\pm 1}$ precisely at the $d-2$ integers\n$$\na_{2, i}-1, a_{3, i}+1, \\cdots, a_{d-2, i}-1, a_{d-1, i}+1\n$$\nProof Suppose (for contradiction) $a_{1, i}+1 \\neq a_{2, i}-1$ and $P\\left(a_{1, i}+1\\right)=p_{i+1}$. Now $a_{1, i}<a_{2, i}$ so either $a_{1, i}+1<a_{2, i}-1$ or $a_{1, i}=a_{2, i}-1$. In the latter case, $P\\left(a_{1, i}+1\\right)=P\\left(a_{2, i}\\right)=p_{i}$, a contradiction. In the former case, the proof of Claim 1 shows that\n$$\n|P(a_{2, i}-1)-p_{i}|<|P(a_{1, i}+1)-p_{i}|=|p_{i+1}-p_{i}|=p_{i+1}-p_{i}\n$$\nso $P(a_{2, i}-1)<p_{i+1}$. The polynomial is decreasing on the interval $(a_{2, d}, a_{2,1})$ so $p_{i}<P(a_{2, i}-1)<p_{i+1}$ which is absurd because $P(a_{2, i}-1)=p_{j}$ for some $j$. Therefore $P(a_{2, i}-1)=p_{i+1}$ for all odd $i$. A similar argument shows that $P(a_{d-1, i}+1)=p_{i+1}$ so Claim 2 is established.\n\nNow we have a sequence of alternating falling then rising then falling etc. full runs starting at $(a_{2, d}, p_{d})$ and ending at $(a_{d-1,1}, p_{1})$ so the initial run of $3d+1$ terms of this run of values is\n$$\np_{d}, p_{d-1}, \\cdots p_{1}, p_{1}, p_{2}, \\ldots, p_{d}, p_{d}, p_{d-1}, \\cdots p_{1}, p_{1}\n$$\nwhich starts at $(a_{2, d}, p_{d})$ and ends at $(a_{4,1}+1, p_{1})$ which is fine because $4 \\leq d-1$.\n\nThere are now various ways we can finish.\n\n(a) Consider the run of length $2d$ consecutive values\n$$\np_{d}, p_{d-1}, \\cdots p_{1}, p_{1}, p_{2}, \\ldots, p_{d}\n$$\nThe first $d+1$ points determine $P(x)$. The last $d+1$ values also determine $P$ but the values are in the reverse order, so $P(X)=P(c-X)$ for some constant $c$. However, the coefficients of $X^{d}$ have opposite signs ($d$ is odd) so this is absurd.\n\n(b) The idea in (a) can be expressed in terms of Lagrange interpolation to obtain essentially the same contradiction. Construct $P$ in two ways using Lagrange interpolation on both the first $d+1$ and the last $d+1$ points. The symmetry in the data forces the graph of $P$ to have a vertical axis of symmetry. This is absurd because the degree of $P$ is odd.\n\n(c) The initial fragment length $3d+1$ mentioned above at $(*)$ includes two identical runs of values of $P$ (in the same order) of length $d+1$. The polynomial $P$ is determined by each of them and so $P(X)=P(X+c)$ for some constant $c$ and so the polynomial defines a bounded function which is absurd.\n\nTherefore, the only possible values are $d=1,2,3$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 30166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFor a positive integer $N$, let $c_{1} < c_{2} < \\dots < c_{m}$ be all the positive integers smaller than $N$ that are coprime to $N$. Find all $N \\geq 3$ such that\n$$\ngcd (N, c_{i} + c_{i + 1}) \\neq 1$$\nfor all $1 \\leq i \\leq m - 1$.\nHere $\\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\\gcd(a, b) = 1$.",
    "options": [],
    "answer": "All even integers and all powers of 3.",
    "solution": "Solution:\nThe answer is all even integers and all powers of $3$. First we show that these work.\n\n- When $N$ is even, all $c_{i}$ are odd, and so $2 \\mid \\gcd (N, c_{i} + c_{i + 1})$ for every $i$.\n\n- When $N$ is a power of $3$, the $c_{i}$ are exactly the numbers in the range $1, 2, \\ldots, N - 1$ that are not divisible by $3$. So, the sequence $c_{1}, \\ldots, c_{m}$ alternates between numbers congruent to $1$ (mod $3$) and $2$ (mod $3$). Thus $3 \\mid \\gcd (N, c_{i} + c_{i + 1})$ for every $i$.\n\nNow we show that no other positive integer works. For the sake of contradiction, consider an odd candidate $N$ that is not a power of $3$, and suppose it satisfies the problem condition. Then, since $c_{1} = 1$ and $c_{2} = 2$, we have $\\gcd (N, 1 + 2) \\neq 1$, so $3 \\mid N$. Thus, we can write $N$ as $3^{k} M$, where $k \\geq 1$ and $\\gcd (M, 6) = 1$. We have $M \\geq 5$ (as $M \\neq 1$ by assumption) and $M + 2 < N$.\n\nWe now split into two cases based on $M$ modulo $3$.\n\nCase 1. $M \\equiv 1$ (mod $3$).\nAs $3 \\mid M - 1$ and $M \\mid M$, neither $M - 1$ nor $M$ are in the sequence $\\{c_{i}\\}$. As $M$ is odd and $M \\equiv 1$ (mod $3$), $M - 2$ and $M + 1$ are congruent to $1$ (mod $3$), are coprime to $M$, and are smaller than $N$. Thus, both $M - 2$ and $M + 1$ are in the sequence, and they are consecutive terms. However, this means $\\gcd (N, 2M - 1) \\neq 1$. This gives a contradiction, as both $3$ and $M$ are coprime to $2M - 1$.\n\nCase 2. $M \\equiv 2$ (mod $3$).\nThis case is similar to Case 1. Neither $M$ nor $M + 1$ are in the sequence, but $M - 1$ and $M + 2$ are. We obtain a similar contradiction from $\\gcd (N, 2M + 1) \\neq 1$.\nSolution:\nWe give an alternative way to show that any odd $N$ has to be the power of $3$. Suppose for contradiction that $N$ has at least $2$ distinct prime factors. Choose integers $a, b > 1$ such that\n$$ab \\mid N$$\n- $a$ and $b$ are coprime;\n- Every prime divisor of $N$ divides $ab$.\n\nWe use the Chinese Remainder Theorem, to find an integer $n \\in [1, ab]$ such that\n$$\n\\left\\{ \\begin{array}{ll}\nn \\equiv 1 \\pmod{a}, \\\\\nn \\equiv 2 \\pmod{b}.\n\\end{array} \\right.\n$$\nWe claim that there is an index $i$ such that $c_{i} = n - 3$ and $c_{i + 1} = n$. Indeed, we note the following:\n\n- $n$ is consecutive to a multiple of $a$, so $\\gcd(a, n) = 1$. In a similar way we get that $\\gcd(b, n - 3) = 1$.\n- $n$ is $2$ away from a multiple of $b$, therefore as $2 \\nmid b$, $\\gcd(b, n) = 1$. In a similar way we get that $\\gcd(a, n - 3) = 1$.\n- $n \\leq ab \\leq N$ and $n - 3 \\geq (a + 1) - 3 > 0$ as $a > 2$.\n\nThis implies that $n$ and $n - 3$ are coprime with $N$, whereas neither $n - 2$ nor $n - 1$ are coprime with $N$, which proves our claim. Finally, we claim that $c_{i} + c_{i + 1} = (n - 3) + n = 2n - 3$ is coprime with $N$. Indeed we have $2n - 3 \\equiv -1 \\pmod{a}$ and $2n - 3 \\equiv 1 \\pmod{b}$. This implies that $N$ is a prime power.\n\nWhen $N$ is odd, since $c_{1} = 1$ and $c_{2} = 2$, we have $\\gcd(N, 1 + 2) \\neq 1$, so $3 \\mid N$, which implies that $N$ is necessarily a power of $3$. This concludes our proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn infinite increasing sequence $a_{1} < a_{2} < a_{3} < \\dots$ of positive integers is called central if for every positive integer $n$, the arithmetic mean of the first $a_{n}$ terms of the sequence is equal to $a_{n}$.  \nShow that there exists an infinite sequence $b_{1}, b_{2}, b_{3}, \\ldots$ of positive integers such that for every central sequence $a_{1}, a_{2}, a_{3}, \\ldots$, there are infinitely many positive integers $n$ with $a_{n} = b_{n}$.",
    "options": [],
    "answer": "b_n = 2n - 1 for all n",
    "solution": "Solution:\n\nWe claim that the sequence $b_{1}, b_{2}, b_{3}, \\ldots$ defined by $b_{i} = 2i - 1$ has this property.  \nLet $d_{i} = a_{i} - b_{i} = a_{i} - 2i + 1$. The condition $a_{i} < a_{i + 1}$ now becomes $d_{i} + 2i - 1 < d_{i + 1} + 2i + 1$, which can be rewritten as $d_{i + 1} \\geqslant d_{i} - 1$. Thus, if $d_{i + 1} < d_{i}$, then $d_{i + 1}$ must be equal to $d_{i} - 1$. This implies in particular that if $d_{i_{0}} \\geqslant 0$ but $d_{i_{1}} \\leqslant 0$ for some indices $i_{1} > i_{0}$, there must be some intermediate index $i_{0} \\leqslant i \\leqslant i_{1}$ with $d_{i} = 0$.\n\nBecause the average of the first $a_{n}$ terms of the sequence is equal to $a_{n}$, we know for all $n$ that\n$$\n\\sum_{i = 1}^{a_{n}} d_{i} = \\sum_{i = 1}^{a_{n}} (a_{i} - 2i + 1) = \\sum_{i = 1}^{a_{n}} a_{i} - \\sum_{i = 1}^{a_{n}} (2i - 1) = a_{n}^{2} - a_{n}^{2} = 0.\n$$\nBecause the sequence $(a_{n})$ is increasing, this implies that the sequence $(d_{i})$ contains infinitely many non-negative $(d_{i} \\geqslant 0)$ and infinitely many non-positive $(d_{i} \\leqslant 0)$ terms. In particular, we can find arbitrarily large indices $i_{0} \\leqslant i_{1}$ such that $d_{i_{0}} \\geqslant 0$ and $d_{i_{1}} \\leqslant 0$. By our earlier observation, it follows that there are infinitely many $i$ such that $d_{i} = 0$, as desired.\nSolution:\n\nWe give an alternative proof that the sequence $b_{i} = 2i - 1$ works. This proof is by contradiction, so we assume that there are only finitely many $a_{i}$ such that $a_{i} = 2i - 1$.\n\nLet $S(n) = \\sum_{i = 1}^{n} a_{i}$. We have $S(a_{n}) = a_{n}^{2}$ and $S(a_{n + 1}) = a_{n + 1}^{2}$. If $a_{n + 1} = a_{n} + 1$, then it follows that\n$$\nS(a_{n + 1}) - S(a_{n}) = a_{n + 1}^{2} - a_{n}^{2} = a_{n + 1}^{2} - (a_{n + 1} - 1)^{2} = 2a_{n + 1} - 1.\n$$\nOn the other hand, if $a_{n + 1} = a_{n} + 1$, then $S(a_{n + 1}) - S(a_{n})$ is $a_{a_{n + 1}}$, so it follows that $a_{a_{n + 1}} = 2a_{n + 1} - 1$. By assumption, this can only happen finitely many times, so for all sufficiently large $n$ we must have $a_{n + 1} \\geqslant a_{n} + 2$.\n\nFor large enough $n$, we now know that $a_{n} > 2n - 1$ implies $a_{n + 1} > (2n - 1) + 2 = 2(n + 1) - 1$. This means that there are two cases possible:\n\n(A) For all sufficiently large $n$ (say $n \\geqslant N_{A}$) we have $a_{n} > 2n - 1$\n\n(B) For all sufficiently large $n$ (say $n \\geqslant N_{B}$) we have $a_{n} < 2n - 1$\n\nIn case (A), we know for $m > N_{A}$ that\n$$\nS(m) = S(N_{A}) + \\sum_{i = N_{A} + 1}^{m} a_{i} \\geqslant S(N_{A}) + \\sum_{i = N_{A} + 1}^{m} 2i = S(N_{A}) + m(m + 1) - N_{A}(N_{A} + 1)\n$$\n$$\n\\qquad = m^{2} + m + S(N_{A}) - N_{A}(N_{A} + 1).\n$$\nFor $m$ large enough (e.g. $m > N_{A}(N_{A} + 1)$), this expression is always larger than $m^{2}$, contradicting $S(a_{n}) = a_{n}^{2}$ for all $n$.\n\nSimilarly, in case (B), we similarly know for $m > N_{B}$ that\n$$\nS(m) = S(N_{B}) + \\sum_{i = N_{B} + 1}^{m} a_{i} \\leqslant S(N_{B}) + \\sum_{i = N_{B} + 1}^{m} 2(i - 1) = S(N_{B}) + m(m - 1) - N_{B}(N_{B} - 1)\n$$\n$$\n\\qquad = m^{2} - m + S(N_{B}) - N_{B}(N_{B} - 1).\n$$\nFor $m$ large enough (e.g. $m > S(N_{B})$), this expression is always smaller than $m^{2}$, again contradicting $S(a_{n}) = a_{n}^{2}$ for all $n$.\nSolution:\n\nWe claim that the sequence $b_{1}, b_{2}, b_{3}, \\ldots$ defined by $b_{i} = 2i - 1$ has this property.\n\nLemma. If there are no terms $a_{j}$ such that $a_{j} - a_{j - 1} = 1$, then $a_{j} = a_{j - 1} + 2$ for all $j$.\n\nProof. Let $c$ be such that $a_{d} = c$ for some $d$. Now\n$$\na_{1} + a_{2} + \\dots + a_{c} = c^{2}.\n$$\nEquality holds for $a_{i} = 2i - 1$ for $1 \\leq i \\leq c$, so if any difference between two consecutive terms is greater, the left-hand side of the equation is greater than $c^{2}$, a contradiction. $\\square$\n\nLemma. If both $d$ and $d + 1$ are terms of the sequence, i.e. $a_{c} = d$ and $a_{c + 1} = d + 1$ for some $c$, then $a_{d + 1} = 2d + 1 = b_{d + 1}$.\n\nProof. We have $a_{1} + a_{2} + \\dots + a_{d} = d^{2}$ and $a_{1} + a_{2} + \\dots + a_{d + 1} = (d + 1)^{2}$. Hence $a_{d + 1} = (d + 1)^{2} - d^{2} = 2d + 1$. $\\square$\n\nFrom the observations above, we see that we are done if there are infinitely many gaps of size 1. The only remaining case is one with finitely many gaps of size 1. This will be the subject of the following lemma.\n\nLemma. If there are only finitely many indices $j$ such that $a_{j + 1} - a_{j} = 1$, then there is an index $n_{0}$ such that for all $k > n_{0}$, we have $a_{k} = 2k - 1$.\n\nProof. Let $r$ and $s$ be indices such that for all the $j$ satisfying $a_{j + 1} - a_{j} = 1$, we have $j < r$, $s$. Furthermore, assume $s > r$ and that there are $i_{1}$ and $i_{2}$ such that $a_{i_{1}} = r$ and $a_{i_{2}} = s$. The first goal is to show that $a_{s} \\geq 2s - 1$. If $a_{r} \\geq 2r - 1$, this is clearly the case. Assume now $a_{r} < 2r - 1$. Now $a_{r} \\geq 2r - 1 - m$, where $m$ is the number of indices $j$ with $a_{j + 1} - a_{j} = 1$. Denote $a_{r + 1} = 2r + 1 - m + \\theta_{1}$, $a_{r + 2} = 2r + 3 - m + \\theta_{2}$, etc. Remember that $a_{r + j + 1} - a_{r + j} \\geq 2$ always. Now $0 \\leq \\theta_{1} \\leq \\theta_{2} \\leq \\dots$. Furthermore, write $s = r + h$. Now\n$$\n(r + h)^{2} - r^{2} = a_{r + 1} + a_{r + 2} + \\dots + a_{r + h} = \\sum_{j = 1}^{h} 2r - 1 + 2j - m + \\theta_{j}.\n$$\nFrom this we deduce\n$$\n2r h + h^{2} = 2r h - h - m h + h(h + 1) + \\sum_{j = 1}^{h} \\theta_{j}.\n$$\nSo we obtain $\\sum_{j = 1}^{h} \\theta_{j} = m h$. Since the sequence $\\theta_{j}$ is increasing, we have $\\theta_{h} \\geq m$. Hence, $a_{s} = a_{r + h} \\geq 2r - 1 - m + 2h + m = 2r + 2h - 1 = 2s - 1$.\n\nNow $a_{s}$ is exactly the desired shape. If for any $t > s$, we have $a_{t} - a_{t - 1} > 2$, then\n$$\na_{s} + a_{s + 1} + \\dots + a_{t} > t^{2} - s^{2},\n$$\nagain a contradiction.\nSolution:\n\nNote that $a_{1} = 1$ because if it is not the case, then $a_{1}^{2} = a_{1} + \\dots + a_{a_{1}} > a_{1} + a_{1} + \\dots + a_{1} = a_{1}^{2}$.\n\nAssume by contradiction that there are only finitely many indices $k$ such that $a_{k} = 2k - 1$. Set $i$ to be the largest integer such that $a_{i} = 2i - 1$ (which must exist as $a_{1} = 1$). Assume that there exists $j \\geqslant i$ such that $a_{j + 1} - a_{j} = 1$. Then $2a_{j + 1} - 1 = a_{j + 1}^{2} - a_{j}^{2} = a_{a_{j + 1}}$ and since $a_{k} \\geqslant k$ for all $k$, we have $a_{j + 1} \\geqslant j + 1 > i$, which contradicts the definition of $i$. Thus for all $j \\geqslant i$, we have $a_{j + 1} \\geqslant a_{j} + 2$, which implies by induction that $a_{j} \\geqslant 2j - 1$ for $j \\geqslant i$, and even $a_{j} \\geq 2j$ if $j > i$.\n\nThere are two ways to finish the solution from here.\n\n## First way to finish the solution\nFor all $n$ such that $a_{n} \\geqslant i$, we have\n$$\na_{n + 1}^{2} - a_{n}^{2} = a_{a_{n + 1}} + a_{a_{n + 1} - 1} + \\dots + a_{a_{n + 1}} \\geqslant 2a_{n + 1} + 2(a_{n + 1} - 1) + \\dots + 2(a_{n} + 1)\n$$\n$$\n\\qquad = (a_{n + 1} - a_{n})(a_{n + 1} + a_{n} + 1)\n$$\n$$\n\\qquad > a_{n + 1}^{2} - a_{n}^{2}.\n$$\nThis gives a contradiction.\n\n## Second way to finish the solution\nFor all $n$ such that $a_{n} \\geqslant i$, we introduce $x_{n} = a_{n + 1} - a_{n}$. We have\n$$\nx_{n}^{2} + 2x_{n} a_{n} = a_{n + 1}^{2} - a_{n}^{2} = a_{a_{n + 1}} + a_{a_{n + 1} - 1} + \\dots + a_{a_{n} + 1} \\geqslant \\sum_{j = 1}^{x_{n}} (a_{a_{n}} + 2j) \\geq x_{n} a_{a_{n}} + x_{n}(x_{n} + 1).\n$$\nBy simplifying, we get $a_{a_{n}} \\leq 2a_{n} - 1$, which gives a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30168,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute triangle. Points $B$, $D$, $E$, and $C$ lie on a line in this order and satisfy $BD = DE = EC$. Let $M$ and $N$ be the midpoints of $AD$ and $AE$, respectively. Let $H$ be the orthocentre of triangle $ADE$. Let $P$ and $Q$ be points on lines $BM$ and $CN$, respectively, such that $D$, $H$, $M$, and $P$ are concyclic and $E$, $H$, $N$, and $Q$ are concyclic. Prove that $P$, $Q$, $N$, and $M$ are concyclic.\n\nThe orthocentre of a triangle is the point of intersection of its altitudes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDenote by $B'$ and $C'$ the reflections of $B$ and $C$ in $M$ and $N$, respectively. Points $C'$, $A$, $B'$ are clearly collinear and $DEB'A$ is a parallelogram. Since $EH \\perp AD$, we have $EH \\perp EB'$. Also $HA \\perp AB'$, so points $H, E, B', A$ are concyclic. This gives\n\n$$\n\\angle C'QH = \\angle NQH = \\angle NEH = \\angle AEH = \\angle AB'H = \\angle C'B'H,\n$$\nand so points $C'$, $B'$, $Q$, $H$ are concyclic. Analogously points $C'$, $B'$, $P$, $H$ are concyclic, and so all points $B'$, $C'$, $P$, $Q$, $H$ are. Now we have\n\n$$\n\\angle NMB' = \\angle AB'M = \\angle C'B'P = \\angle C'QP = \\angle NQP,\n$$\nwhich proves that $P, Q, N, M$ are also concyclic.\n\n![](attached_image_1.png)\n\n\nIntroduce points $B'$, $C'$ as above. Also define $A' = B'E \\cap C'D$, so that $EA DA'$ is a parallelogram and $ADE$ is the medial triangle of $A'B'C'$. It that follows that the orthocentre $H$ of $ADE$ is the circumcentre of $A'B'C'$, and in particular\n\n$$\n\\angle C'B'H = 90^{\\circ} - \\angle B'A'C' = 90^{\\circ} - \\angle DAE = \\angle AEH = \\angle NEH = \\angle NQH = \\angle C'QH.\n$$\nSo again we have that $C'$, $B'$, $Q$, $H$ are concyclic and conclude as in Solution 1.\nSolution:\n\nLet $X$ be the second intersection of $(DHM)$ and $(EHN)$ and let $O'$ be the circumcentre of $(AMN)$. Note that $\\angle MXN = \\angle MDH + \\angle NEH = 180^{\\circ} - 2\\angle DAE$ and since $\\angle MO'N = 2\\angle DAE$ we have that $X, M, O', N$ is cyclic and since $\\angle MXH = \\angle NXH$ it means that $HX$ is the angle bisector of $\\angle MXN$ but since $O'M = O'N$ it means that $H, X, O'$ are collinear. Let $BM$ and $CN$ intersect at $T$ and let $K$ and $L$ be the midpoints of $MN$ and $BC$. Note that $L$ is also the midpoint of $DE$. Since $MN$ is parallel to $BC$ it means that $T$, $K$, and $L$ are collinear, but since $A$, $K$, and $L$ are collinear we get that $A$, $T$, $K$, and $L$ are collinear. Now, $\\frac{TL}{TK} = \\frac{BC}{MN} = 6$. Since $KL = KA$ it means that $\\frac{AT}{TK} = \\frac{TL - 2TK}{TK} = 4$ so by the lemma below, $T$ lies on $HO'$. Since $HO'$ is the radical axis of $(DHM)$ and $(EHN)$ we finish the problem using the Radical Axes Theorem $(TM \\cdot TP = TN \\cdot TQ)$.\n\nLemma. Let $A'$ be the reflection of $A$ around the orthocentre $H$ of $\\triangle ABC$ and $O$ and $M$ be the circumcentre of $\\triangle ABC$ and the midpoint of $BC$, respectively. Let $T$ be the intersection of $A'O$ and $AM$. Then $\\frac{AT}{TM} = 4$.\n\nProof. Since $OM \\parallel AA'$ we have $\\frac{AT}{TM} = \\frac{AA'}{OM} = \\frac{2AH}{OM} = \\frac{4OM}{OM} = 4$. We used here that $AH = 2OM$. $\\square$\n\n![](attached_image_2.png)\n\nAs in solution 2, we will prove that $O'$ is both on line $HT$ and the radical axis of the circles, hence $T$ is on the radical axis, from which we conclude the required concyclicity. We present alternative proofs of both facts, discovered by contestants.\n\nLet $M_{1}$, $N_{1}$ be the midpoints of $AM$, $AN$, respectively, so that $AM_{1}:M_{1}D = AN_{1}:N_{1}E = 1:3$. It is easy to verify (e.g. by applying the theorems of Ceva or Menelaus, or by computing in barycentric coordinates as in Solution 3) that $T$ lies on $EM_{1}$ and $DN_{1}$. Note that $M_{1}N_{1} \\parallel DE$, and also $M_{1}O' \\parallel HE$ as both are perpendicular to $AMD$, and similarly $N_{1}O' \\parallel HD$. It follows that $DEH$ and $N_{1}M_{1}O'$ are homothetically similar triangles, and the center of their (negative) homothety is $T = DN_{1} \\cap EM_{1}$. Therefore $T$ also lies on $HO'$, as claimed. (We also have that the homothety is by factor $\\frac{M_{1}N_{1}}{ED} = -\\frac{1}{4}$.)\n\nNow, let $M_{2}$, $N_{2}$ be the second intersection points of $O'M$, $O'N$ with the circumcircles of $HMD$, $HNE$, respectively. To prove that $O'$ is on the radical axis, it suffices to show that $O'M \\cdot O'M_{2} = O'N \\cdot O'N_{2}$. But by definition of $O'$ we have $O'M = O'N$, so we must show $O'M_{2} = O'N_{2}$, which is equivalent to $M_{2}N_{2} \\parallel MN$. Angle chasing in circle $MM_{2}DH$ gives\n\n$$\n\\angle OM_{2}H = \\angle MM_{2}H = \\angle MDH = \\angle ADH = 90^{\\circ} - \\angle EAD = 90^{\\circ} - \\angle NAM = \\angle O'MN,\n$$\nfrom which it follows that $M_{2}H \\parallel MN$. Similarly, we have $N_{2}H \\parallel MN$, and the two facts together imply that $M_{2}, H, N_{2}$ are collinear and the line through them is parallel to $MN$, as claimed. $\\square$\n\n![](attached_image_3.png)\nSolution:\n\nWe compute using linear combinations with respect to $ADE$. We have $B = 2D - E$, $C = 2E - D$, $M = \\frac{A + D}{2}$, and $N = \\frac{A + E}{2}$, from which we immediately obtain that the intersection $T = BM \\cap CN$ is $T = \\frac{3A + D + E}{5} = \\frac{6M - B}{5} = \\frac{6N - C}{5}$. As in solution 2, we reduce to showing that $T$ is on the radical axis of $HDM$ and $HEN$, whence $TM \\cdot TP = TN \\cdot TQ$ proves the concyclicity of $P, Q, N, M$.\n\nSynthetic finish, similar to Solution 2. Let $O$ be the circumcentre of $ADE$ and $O' = \\frac{A + O}{2}$ be the circumcentre of $AMN$. As in Solution 2, we have that $O'$ is on the desired radical axis, so it is enough to show $T \\in O'H$. Let $G = \\frac{A + D + E}{3}$ be the barycentre of $ADE$. By properties of the Euler line, we also have $G = \\frac{H + 2O}{3}$. Now using our known identities we find\n\n$$\nT = \\frac{3A + D + E}{5} = \\frac{2A + 3G}{5} = \\frac{2A + H + 2O}{5} = \\frac{H + 4O'}{5}\n$$\nand in particular $T \\in HO'$, as we wanted to show.\n\nComputational finish. Let $f(T)$ be the power difference at $T$ w.r.t. $DHM$ and $HEN$. We compute $f(A)$ and $f(L)$ where $L = \\frac{D + E}{2}$. Since $T = \\frac{3A + 2L}{5}$, it is enough to show that $3P(A) + 2P(L) = 0$. In the following all lengths are directed. We compute trigonometrically: Let $\\alpha, \\beta, \\gamma$ be the angles of $ADE$ and assume the diameter of its circumcircle is $1$. We have\n\n$$\nf(A) = \\frac{AD^{2} - AE^{2}}{2} = \\frac{\\sin^{2}(\\gamma) - \\sin^{2}(\\beta)}{2}.\n$$\nTo compute $P(L)$, let $D', E'$ be the second intersection points of $HMD, HNE$ with $DE$, and let $M', N', F$ be the feet of the perpendiculars from $H$ to $AD, AE, DE$, respectively. Note that $DM' = \\sin \\alpha \\cos \\beta$, thus\n\n$$\nM'M = DM - DM' = \\frac{\\sin(\\alpha + \\beta)}{2} - \\sin \\alpha \\cos \\beta = \\frac{\\sin(\\beta - \\alpha)}{2}.\n$$\nWe also have $HM' = \\cos \\alpha \\cos \\beta$, $HF = \\cos \\beta \\cos \\gamma$, and $HM'M \\sim HFD'$, therefore\n\n$$\nFD' = \\frac{HF}{HM'} M'M = \\frac{\\cos \\gamma \\sin(\\beta - \\alpha)}{\\cos \\alpha}\n$$\nand similarly\n$$\nE'F = \\frac{\\cos \\beta \\sin(\\gamma - \\alpha)}{\\cos \\alpha}.\n$$\nWe also have the standard $FD = \\sin \\gamma \\cos \\beta$ and $EF = \\sin \\beta \\cos \\gamma$. We can finally compute\n\n$$\nf(L) = LD \\cdot LD' - LE \\cdot LE' = \\frac{\\sin \\alpha}{2} (LD' + LE') = \\frac{\\sin \\alpha}{2} (FD' + FE' - FD - FE)\n$$\n$$\n\\qquad = \\frac{\\sin \\alpha}{4 \\cos \\alpha} (\\cos \\gamma \\sin (\\beta - \\alpha) - \\cos \\beta \\sin (\\gamma - \\alpha) - 2 \\cos \\alpha (\\sin \\gamma \\cos \\beta - \\sin \\beta \\cos \\gamma))\n$$\n$$\n\\qquad = \\frac{3 \\sin \\alpha \\sin(\\beta - \\gamma)}{4} = \\frac{3}{4} (\\sin (\\beta + \\gamma) \\sin (\\beta - \\gamma)) = \\frac{3}{8} (\\cos (2\\gamma) - \\cos (2\\beta))\n$$\n$$\n\\qquad = \\frac{3}{4} (\\sin^{2}(\\beta) - \\sin^{2}(\\gamma)) = -\\frac{3}{2} f(A).\\quad \\square\n$$\n\n![](attached_image_4.png)\nSolution:\n\nLet $T = BM \\cap CN$, let $F$ be the foot of the altitude from $A$ to $BC$, let $O$ be the circumcentre of $(ADE)$, let $D' \\neq D$ and $E' \\neq E$ be the second intersections of $(DHMP)$ and $(EHNQ)$ with line $BC$ and let $U$ and $V$ be the antipodes of $D$ and $E$ on $(DHMP)$ and $(EHNQ)$, respectively.\n\nWe begin with a bit of barycentric coordinates. Set barycentric coordinates in $\\triangle ABC$, set so that $A = (1,0,0)$, $B = (0,1,0)$, and $C = (0,0,1)$. The definitions of $D$ and $E$ give $D = (0,2/3,1/3)$ and $E = (0,1/3,2/3)$, whence $M = (1/2,1/3,1/6)$ and $N = (1/2,1/6,1/3)$. This means that line $BM$ is given by $(1/2:y:1/6)$ while line $CN$ is given by $(1/2:1/6:z)$. So their intersection $T$ is $(1/2:1/6:1/6) = (3:1:1)$, giving $T = \\frac{3A + B + C}{5} = \\frac{3A + D + E}{5}$.\n\nOur next tool is the linearity of the power of a point. Let $f:\\mathbb{R}^{2} \\to \\mathbb{R}$ be defined by\n\n$$\nf(Z) = \\mathrm{Pow}_{(DHMP)}(Z) - \\mathrm{Pow}_{(EHNQ)}(Z).\n$$\nIt suffices to show that $f(T) = 0$; from there, the required concyclicity will follow from $TM \\cdot TP = TN \\cdot TQ$. The key observation is that $f$ is a linear function. In particular, $f(T) = \\frac{3f(A) + f(D) + f(E)}{5}$. So, we need only show that $3f(A) + f(D) + f(E) = 0$. Pick an arbitrary one-dimensional coordinate system on the line $BC$ and let $\\tau$ be the map from a point on $BC$ to its coordinate. We compute\n\n$$\nf(A) = AM \\cdot AD - AN \\cdot AE = \\frac{AD^2 - AE^2}{2} = \\frac{FD^2 - FE^2}{2}\n$$\n$$\n\\qquad = (\\tau(E) - \\tau(D))\\left(\\tau(F) - \\tau\\left(\\frac{D + E}{2}\\right)\\right),\n$$\n$$\nf(D) = -DE \\cdot DE' = (\\tau(E) - \\tau(D))(\\tau(D) - \\tau(E')),\\quad f(E) = ED \\cdot ED' = (\\tau(E) - \\tau(D))(\\tau(E) - \\tau(D')).\n$$\nRearranging, it suffices to show that $3\\tau(F) = \\tau(D') + \\tau(E') + \\tau((D + E)/2)$. This can be rewritten as $3F = D' + E' + (D + E)/2$. By projecting down to line $BC$, it suffices to show that the displacement vector $H + 2A - (O + U + V)$ is perpendicular to line $BC$.\n\nWe do this using complex numbers. Let $(ADE)$ be the unit circle with $A = a$, $D = d$, and $E = e$, so that $O = 0$ and $H = h = a + d + e$. Note that $U$ satisfies $UM \\perp AD$ and $UH \\perp DH \\perp AE$. Translating these conditions into equations, we have $u = a d\\overline{u}$ and $u + a e\\overline{u} = h + a e\\overline{h}$. Rearranging gives\n\n$$\nv := H + 2A - (O + U + V) = h + 2a - \\frac{(d + e)h + 2a d e\\overline{h}}{d + e} = 2\\left(a - \\frac{a d e\\overline{h}}{d + e}\\right) = -\\frac{2de}{d + e}.\n$$\nThis displacement vector $v$ satisfies $v = de\\overline{v}$ and so it is orthogonal to line $DE$, as desired.\nSolution:\n\nThis solution uses almost exclusively complex numbers. As in other solutions, we reduce to showing that $H$, $S := (DHM) \\cap (EHN)$, and $T := BM \\cap CN$ are collinear; this is all of the synthetic information we shall use. (If one computes $T = \\frac{3A + D + E}{5}$ using means other than complex numbers, the solution becomes much shorter.)\n\nWe use complex numbers with $A = a$, $D = d$, and $E = e$ on the unit circle. Note that $H = a + d + e$, $M = \\frac{a + d}{2}$, and $B = 2d - e$. We will make great use of the \"arbitrary line intersection formula,\" which says that the intersection between lines $WX$ and $YZ$ can be written explicitly as\n\n$$\n\\frac{(\\overline{u} x - w\\overline{x})(y - z) - (w - x)(\\overline{y} z - y\\overline{z})}{(\\overline{w} - \\overline{x})(y - z) - (w - x)(\\overline{y} - \\overline{z})}.\n$$\n\nWe first use this to find $T = t$. We compute\n\n$$\nb - m = (2d - e) - \\frac{a + d}{2} = \\frac{3d - a - 2e}{2}\n$$\n$$\n\\overline{b} - m = \\frac{3a e - d e - 2a d}{2a d e}\n$$\n$$\n\\overline{b} m - b\\overline{m} = \\frac{2e - d}{de} \\cdot \\frac{a + d}{2} - (2d - e) \\cdot \\frac{a + d}{2ad} = \\frac{(a + d)(2a e + e^{2} - a d - 2d e)}{2ad e}\n$$\n\nIf $\\mathcal{E}$ is some expression, we use the notation $\\mathcal{E} - \\{\\sim\\}$ to denote $\\mathcal{E}$ minus the expression formed by swapping $d$ and $e$ in $\\mathcal{E}$. We now have\n\n$$\nt = \\frac{(\\overline{b}m - b\\overline{m})(c - n) - \\{\\sim\\}}{(\\overline{b} - \\overline{m})(c - n) - \\{\\sim\\}}\n$$\n$$\n\\quad = \\frac{(a + d)(2a e + e^{2} - a d - 2d e)(3e - a - 2d) - \\{\\sim\\}}{(3a e - d e - 2a d)(3e - a - 2d) - \\{\\sim\\}}\n$$\n$$\n\\quad = \\frac{[a^{3}(d - 2e) + a^{2}(3d^{2} - 7d e + 5e^{2}) + a(2d^{3} - d^{2}e - 3d e^{2} + 3e^{3}) + d e(4d^{2} - 8d e + 3e^{2})] - \\{\\sim\\}}{[a^{2}(2d - 3e) + a(4d^{2} - 11d e + 9e^{2}) + d e(2d - 3e)] - \\{\\sim\\}}\n$$\n$$\n\\quad = \\frac{a^{3}(3(d - e)) - a^{2}(2(d^{2} - e^{2})) + a(-(d^{3} - e^{3}) + 2d e(d - e)) + d e(d^{2} - e^{2})}{a^{2}(5(d - e)) - a(5(d^{2} - e^{2})) + d e(5(d - e))}\n$$\n$$\n\\quad = \\frac{3a^{3} - 2a^{2}(d + e) - a(d^{2} - d e + e^{2}) + d e(d + e)}{5(a^{2} - a(d + e) + d e)}\n$$\n$$\n\\quad = \\frac{(a - d)(a - e)(3a + d + e)}{5(a - d)(a - e)} = \\frac{3a + d + e}{5}.\n$$\n(The factorization in the last line can be motivated by noting that the expression, while cubic in $a$, is only quadratic in $d$. When written out as a polynomial in $d$, each coefficient is divisible by $a - e$; by symmetry, the numerator is divisible by $a - d$ as well, and the factorization follows.)\n\nComputing $S = s$ is slightly harder, as it is the intersection of two circles rather than of two lines. We get around this by noting that $\\{h, d, m, s\\}$ are concyclic if and only if $\\{0, h - d, h - m, h - s\\}$ are concyclic, which happens if and only if $\\{\\frac{1}{h - d}, \\frac{1}{h - m}, \\frac{1}{h - s}\\}$ are collinear. (One can see this by inversion, or just by writing out the cross-ratio in the special case when one of the points is zero.) Thus $\\frac{1}{h - s}$ is the intersection of the line through $w := \\frac{1}{h - d}$ and $x := \\frac{1}{h - m}$ and the line through $y := \\frac{1}{h - e}$ and $z := \\frac{1}{h - n}$. We compute\n\n$$\nw - x = \\frac{1}{a + e} - \\frac{1}{\\frac{a + d}{2} + e} = -\\frac{a - d}{(a + e)(a + d + 2e)}\n$$\n$$\n\\overline{w - x} = \\frac{a e^{2}(a - d)}{(a + e)(2ad + a e + d e)}\n$$\n$$\n\\overline{w} x - w\\overline{x} = \\frac{a e}{a + e} \\cdot \\frac{2}{a + d + 2e} - \\frac{1}{a + e} \\cdot \\frac{2ad e}{2ad + a e + d e}\n$$\n$$\n\\qquad = 2a e \\frac{(2ad + a e + d e) - d(a + d + 2e)}{(a + e)(a + d + 2e)(2ad + a e + d e)}\n$$\n$$\n\\qquad = \\frac{2a e(a - d)(d + e)}{(a + e)(a + d + 2e)(2ad + a e + d e)}.\n$$\n\nUsing the line intersection formula, we have\n\n$$\n\\frac{1}{h - s} = \\frac{(\\overline{w} x - w\\overline{x})(y - z) - \\{\\sim\\}}{(\\overline{w} - \\overline{x})(y - z) - \\{\\sim\\}}\n$$\n$$\n= 2(d + e) \\frac{[e(ad + 2ae + de)] - \\{\\sim\\}}{[e^{2}(a + d + 2e)(ad + 2ae + de)] - \\{\\sim\\}}\n$$\n$$\n= 2(d + e) \\frac{[a(de + 2e^{2}) + de^{2}] - \\{\\sim\\}}{[a^{2}(de^{2} + 2e^{3}) + a(d^{2}e^{2} + 5de^{3} + 4e^{4}) + de(de^{2} + 2e^{3})] - \\{\\sim\\}}\n$$\n$$\n= \\frac{2(d + e)(a(2(e^{2} - d^{2})) + de(e - d))}{(a^{2} + de)(2(e^{3} - d^{3}) + de(e - d)) + a(4(e^{4} - d^{4}) + 5de(e^{2} - e^{2}))}\n$$\n$$\n= \\frac{2(d + e)(2a(d + e) + de)}{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}.\n$$\n\nSince $h - t = \\frac{2a + 4d + 4e}{5}$, this gives us\n\n$$\n\\frac{h - s}{h - t} = \\frac{5}{4} \\cdot \\frac{(a^{2} + de)(2d^{2} + 3de + 2e^{2}) + a(d + e)(4d^{2} + 5de + 4e^{2})}{(d + e)(2ad + 2ae + de)(a + 2d + 2e)}.\n$$\n\nIt is easy to see that this is real by using the symmetry of the expressions (both the numerator and denominator satisfy $\\mathcal{E} = a^{2}d^{3}e^{3}\\overline{\\mathcal{E}}$). We conclude that $H$, $S$, and $T$ are collinear, as desired.\nSolution:\n\nAs usual, we reduce to proving $T = (3A + D + E)/5$ is on the radical axis and compute; this time in Cartesian coordinates.\n\nLet $A(0,h)$, $D(b,0)$, $E(c,0)$ be coordinates for $ADE$. Then $H(0, -\\frac{bc}{h})$, $M(\\frac{b}{2}, \\frac{h}{2})$, $N(\\frac{c}{2}, \\frac{h}{2})$ and $T(\\frac{b + c}{5}, \\frac{3h}{5})$. We compute $O_{D}(x_{D}, y_{D})$ the circumcentre of $DMH$ and obtain $O_{E}$ by symmetry. We then have to verify that $O_{D}O_{E} \\perp HT$, which can be done by comparing slopes.\n\nThe centre $O_{D}$ can be given by the intersection of perpendicular bisectors of $DM$ and $DH$. This gives the following system of equations on $x_{D}, y_{D}$:\n\n$$\nh \\cdot x_{D} + c \\cdot y_{D} = \\frac{b h}{2} - \\frac{b c^{2}}{2 h}\n$$\n$$\n-b \\cdot x_{D} + h \\cdot y_{D} = -\\frac{3 b^{2}}{4} + \\frac{h^{2}}{4}\n$$\nSolving the system gives\n\n$$\n(h^{2} + b c)(x_{D} - x_{E}) = \\frac{b - c}{4} (5b c + 3h^{2})\n$$\n$$\n(h^{2} + b c)(y_{D} - y_{E}) = -\\frac{b - c}{4} h(b + c)\n$$\nSo $\\frac{y_{D} - y_{E}}{x_{D} - x_{E}} = -\\frac{h(b + c)}{5b c + 3h^{2}}$.\n\nThe other slope is more immediate:\n\n$$\n\\frac{y_{T} - y_{H}}{x_{T} - x_{H}} = \\frac{3h / 5 + b c / h}{(b + c) / 5} = \\frac{5b c + 3h^{2}}{h(b + c)} = -\\frac{x_{D} - x_{E}}{y_{D} - y_{E}}\n$$\nso indeed the two slopes correspond to perpendicular lines.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute triangle with incentre $I$ and $AB \\neq AC$. Let lines $BI$ and $CI$ intersect the circumcircle of $ABC$ at $P \\neq B$ and $Q \\neq C$, respectively. Consider points $R$ and $S$ such that $AQRB$ and $ACSP$ are parallelograms (with $AQ \\parallel RB$, $AB \\parallel QR$, $AC \\parallel SP$, and $AP \\parallel CS$). Let $T$ be the point of intersection of lines $RB$ and $SC$. Prove that points $R$, $S$, $T$, and $I$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe will prove that $\\triangle BIR \\sim \\triangle CIS$, since the statement then follows from $\\angle TRI = \\angle BRI = \\angle CSI = \\angle TSI$.\n\n![](attached_image_1.png)\n\nStep 1. Let us prove $\\angle RBI = \\angle SCI$. We will use directed angles:\n\n$$(BR,BI) = (BR,AB) + (AB,BI) = (AQ,AB) + (BI,BC) = (CQ,BC) + (BI,BC) = (CI,CB) + (BI,BC),$$\n\nwhich is symmetric in $B,C$. Therefore, analogously we would obtain the same expression for $(CS,CI)$.\n\nStep 2. Let us prove $BR / BI = CS / CI$. Clearly $BR = AQ$ and $CS = AP$. Angle chasing gives $\\angle ICB = \\angle QCB = \\angle APQ$, and similarly $\\angle PQA = \\angle CBI$, and so $\\triangle IBC \\sim \\triangle AQP$, from which the desired $AQ / BI = AP / CI$ follows. This finishes the solution.\nSolution:\n\nWe use complex numbers, with $(ABC)$ as the unit circle. Set $D = d$, $P = p$, $Q = q$, so that $A = a = - \\frac{pq}{d}$, $B = b = - \\frac{dq}{p}$, and $C = c = - \\frac{dp}{q}$. Write $z \\sim w$ if $z / w$ is a nonzero real number. We observe that\n\n$$\\frac{R - T}{S - T} \\sim \\frac{R - B}{S - C} = \\frac{Q - A}{P - A} = \\frac{q + \\frac{pq}{d}}{p + \\frac{pq}{d}} = \\frac{(d + p)q}{(d + q)p}.$$ \n\nSo, it suffices to show that $\\frac{I - R}{I - S} \\sim \\frac{(d + p)q}{(d + q)p}$. Indeed,\n\n$$I - R = (d + p + q) - (Q + B - A) = d + p + \\frac{dq}{p} - \\frac{pq}{d} = (d + p)\\left(1 + \\frac{(d - p)q}{dp}\\right) = \\frac{(d + p)(dp + dq - pq)}{dp},$$\n\nso\n\n$$\\frac{I - R}{I - S} = \\frac{\\frac{d + p}{dp}}{\\frac{d + q}{dq}} = \\frac{(d + p)q}{(d + q)p}.$$\nSolution:\n\nIn the following, all segment notations denote vectors.\n\nAs mentioned above, we find $\\triangle AQP \\sim \\triangle IBC$, and by definitions of the parallelograms we have $BR = AQ$ and $CS = AP$ as well as $\\angle RTS = \\angle QAP$, so it suffices to show $\\angle RIS = \\angle QAP$. From the similarity $\\triangle AQP \\sim \\triangle IBC$, we have a spiral map $\\lambda$ such that $IB = \\lambda AQ$ and $IC = \\lambda AP$. It follows that $IR = IB + BR = (\\lambda + 1)AQ$ and $IS = IC + CS = (\\lambda + 1)AP$. Because $\\lambda + 1$ is also a spiral map, we have $\\triangle IRS \\sim \\triangle AQP$ and in particular $\\angle RIS = \\angle QAP$, as we wanted to show.\nSolution:\n\nLet $E$, $F$, $G$ be the midpoints of $AI$, $BQ$, $CP$. As in Solution 1, angle chase shows that $\\triangle AQP \\sim \\triangle IBC$.\n\nNote that by the Mean Geometry Theorem we have that $\\frac{1}{2}AQP + \\frac{1}{2}IBC = EFG$ is similar to $\\triangle IBC$. Homothety with center $A$ and scale-factor $2$ maps $EFG$ to $IRS$. Hence $\\angle RIS = \\angle FEG = \\angle QAP = \\angle BTC = \\angle RTS$, so $R,T,I,S$ are concyclic.\n\nRemark. As shown above, $E$ lies on $QP$ and $AI \\perp PQ$. One can prove that $\\angle FEG = \\angle BIC$ in another way. Let $J$ be the midpoint of $PQ$. Then $\\angle BIC = \\angle FJG$ by midlines and $\\angle FJG = \\angle FEG$ by the lemma below applied in $BCPQ$.\n\nLemma. Let $ABCD$ is a cyclic quadrilateral and $E$ is the intersection of its diagonals. Then the midpoints of $AB$, $BC$, $CD$ and the foot of the perpendicular from $E$ to $BC$ are concyclic.\n\n![](attached_image_2.png)\nSolution:\n\nLet $O$ be the circumcenter of $(ABC)$. Let $M$, $N$, and $L$ be the midpoints of $OD$, $PC$, and $QB$ respectively.\n\nClaim 1. $\\triangle OPQ$ and $\\triangle DCB$ are directly similar.\n\nProof. Clearly $DB = DC$ and $OQ = OP$. Also note that $\\angle QOP = 2\\angle QDP = 2\\angle QDA + 2\\angle PDA = \\angle BDA + \\angle CDA = \\angle BDC$. So the two triangles are directly similar by SAS.\n\nClaim 2. $ML = MN$ and $\\angle LMN = 180^\\circ - \\angle BAC$\n\nProof. Note that since $\\triangle OQP \\sim \\triangle DBC$ by the Mean Geometry Theorem, we have that the average of the two triangles is also similar to them, therefore $\\triangle MLN \\sim \\triangle DBC \\Rightarrow ML = MN$ and $\\angle LMN = \\angle BDC = 180^\\circ - \\angle BAC$\n\nLet $K$ be the reflection of $A$ over $M$\n\nClaim 3. $K$ is the circumcenter of $\\triangle RTS$\n\nProof. Note that since $AQRB$ and $APSC$ are parallelograms we have that $A - L - R$ are collinear and that $A - N - S$ are collinear. The homothety centered at $A$ with scale-factor $2$ maps $\\triangle LMN$ to $\\triangle RKS$, therefore $KR = KS$ and $\\angle RKS = \\angle LMN = \\angle BDC = 2(180^\\circ - \\angle RTS)$ (and $K$ and $T$ are in opposite sides of $RS$), implying that $K$ is the circumcenter of $\\triangle RTS$\n\nClaim 4. $KT = KI$\n\nProof. Note that $AOKD$ is a parallelogram. Let $BT$ intersect the $(ABC)$ again at point $G$. Since $\\angle ABG = \\angle ABT = \\angle QAB = \\angle QCA \\Rightarrow AQ = AG$ and also $OQ = OG$ hence $AO \\perp QG$. Then by Reim's theorem we have that $QG \\parallel TI$ and also that $AO \\parallel DK$, so $DK \\perp TI$. Since $DI = DT$, it means that $KD$ is the perpendicular bisector of $TI$, therefore $KT = KI$.\n\nThis means that $RTIS$ is cyclic with center $K$.\n\n![](attached_image_3.png)\nSolution:\n\nAs shown above, we have that $BTIC$ is cyclic. Let $D$ and $E$ be the second intersections of $AC$ and $AB$ with this circle, respectively. Since the center of this circle lies on $AI$ (by symmetry about $AI$), we have that $AB = AD$ and $AC = AE$, therefore $BE = CD$. Note that since $C - I - Q$ and $A - B - E$ are collinear, by Reim's theorem we have that $AQ \\parallel EI$ and since $AQ \\parallel BT$, we have that $BT \\parallel EI$. Similarly, we get $CT \\parallel DI$. Let $F$ and $G$ be the intersections of $ID$ and $IE$ with $PS$ and $QR$, respectively. Clearly, $RGEB$ and $FSCD$ are parallelograms. Since $RGEB$ is parallelogram and $BEIT$ is isosceles trapezoid, we have that $RGIT$ is isosceles trapezoid. Similarly, $SFIT$ is isosceles trapezoid. Hence, both of them are cyclic. Note also that $QR = AB = AD = PF$ and $QG = AE = AC = PS$. Since $QR$ and $PS$ are tangents to the circumcircle of $\\triangle ABC$ we have that $R$ and $F$ are symmetric (reflections) about the perpendicular bisector of $PQ$. Similarly, $G$ and $S$ are symmetric about the perpendicular bisector of $PQ$. This gives us that $QP \\parallel RF \\parallel GS$ and that $RFSG$ is an isosceles trapezoid, hence a cyclic quadrilateral with $\\angle RGS = 180^\\circ - \\angle GQP = 180^\\circ - \\angle QAP = 180^\\circ - \\angle RTS \\Rightarrow R, G, S, T$ are concyclic. Combining all the facts about the cyclic quadrilaterals we proved above, we have that $R, G, S, F, I, T$ are concyclic. Therefore $R, T, I, S$ lie on a circle.\n\n![](attached_image_4.png)\nSolution:\n\nLet $E$ be the $A$-excenter of $\\triangle ABC$. Let the midpoints of $AQ, QB, CP, PA$ be the points $F, G, H, J$, respectively. Both $PD$ and $EC$ are perpendicular to $CI$, hence $PD \\parallel CE$.\n\nSince $PA = PC$ we have that $AJHC$ is an isosceles trapezoid so it is cyclic. Let $K$ be the second intersection of $(AJHC)$ and $AI$. Then $\\angle ADP = \\angle ACP = \\angle ACH = \\angle AKH \\Rightarrow DP \\parallel KH$. So $KH$ is a line passing through the midpoint of the side $CP$ of trapezoid $DPCE$ and parallel to the bases, hence $K$ is the midpoint of $DE$. Similarly, we show that the circle $(AFGB)$ passes through the midpoint of $DE$. Homothety centered at $A$ with scale-factor $2$ maps $(AJH)$ to $(APS)$, $(AFG)$ to $(AQR)$, and line $AI$ to line $AI$. This means that the circles $(AQR)$ and $(APS)$ intersect on $AI$, call it point $L$.\n\n![](attached_image_5.png)\n\nNow, $\\angle ILR = 180^\\circ - \\angle AQR = \\angle QAB = \\angle QCB = 180^\\circ - \\angle ITB = 180^\\circ - \\angle ITR$, therefore $R,L,I,T$ are concyclic. Similarly, we get that $S,L,T,I$ are concyclic. Combining these, it means that $R$ and $S$ belong to the circle $(LIT)$. The conclusion follows.\n\n![](attached_image_6.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n > 1$ be an integer. In a configuration of an $n \\times n$ board, each of the $n^{2}$ cells contains an arrow, either pointing up, down, left, or right. Given a starting configuration, Turbo the snail starts in one of the cells of the board and travels from cell to cell. In each move, Turbo moves one square unit in the direction indicated by the arrow in her cell (possibly leaving the board). After each move, the arrows in all of the cells rotate $90^{\\circ}$ counterclockwise. We call a cell good if, starting from that cell, Turbo visits each cell of the board exactly once, without leaving the board, and returns to her initial cell at the end. Determine, in terms of $n$, the maximum number of good cells over all possible starting configurations.",
    "options": [],
    "answer": "n^2/4 if n is even; 0 if n is odd",
    "solution": "Solution:\n\nWe will show that the maximum number of good cells over all possible starting configurations is\n$$\n\\frac{n^{2}}{4} \\quad \\text{if } n \\text{ is even and}\n$$\n$$\n0 \\quad \\text{if } n \\text{ is odd.}\n$$\n\n## Odd $n$\n\nFirst, we will prove that there are no good cells if $n$ is an odd number.\nFor Turbo to reach her goal, she must return to her initial cell after visiting every cell exactly once. Consider the chessboard coloring of the board. Without loss of generality, we assume that Turbo starts in a black cell. Since, at every step, Turbo moves to a cell of a different color; she will be in a white cell after $n^{2} \\equiv 1 \\mod 2$ moves. Thus, it is impossible for Turbo to come back to her initial black cell on her $n^{2}$-th move, which is a contradiction. Thus there are no good cells.\n\n## Lower bound for even $n$\n\nWe will now construct a starting configuration with $\\frac{n^{2}}{4}$ good cells for even $n$.\nLet $(i,j)$ denote the cell in row $i$ and column $j$. Consider the following cycle\n$$\n\\begin{array}{r l}\n& (1,1)\\rightarrow (1,2)\\rightarrow (1,3)\\rightarrow \\ldots \\rightarrow (1,n)\\\\\n& \\qquad \\rightarrow (2,n)\\rightarrow (2,n - 1)\\rightarrow \\ldots \\rightarrow (2,2)\\\\\n& \\qquad \\ldots\\\\\n& \\qquad \\rightarrow (2i - 1,2)\\rightarrow (2i - 1,3)\\rightarrow \\ldots \\rightarrow (2i - 1,n)\\\\\n& \\qquad \\rightarrow (2i,n)\\rightarrow (2i,n - 1)\\rightarrow \\ldots \\rightarrow (2i,2)\\\\\n& \\qquad \\ldots\\\\\n& \\qquad \\rightarrow (n,n)\\rightarrow (n,n - 1)\\rightarrow \\ldots \\rightarrow (n,2)\\\\\n& \\qquad \\rightarrow (n,1)\\rightarrow (n - 1,1)\\rightarrow \\ldots \\rightarrow (2,1)\\rightarrow (1,1).\n\\end{array} \\quad (1,1)\n$$\n\n![](attached_image_1.png)\n\nNote that the cycle returns to the initial cell after visiting every cell exactly once. To prove that $(1,1)$ is good, we need to find a starting configuration such that Turbo traverses this cycle.\nLet $c_{i}$ be the $(i - 1)$-th cell on the cycle: so we have $c_{0} = (1,1)$, $c_{2} = (1,2)$, ..., $c_{n^{2} - 1} = (2,1)$. For every $i$, we draw an arrow in cell $c_{i}$ pointing towards cell $c_{i + 1}$ (or pointing towards $c_{0}$ if $i = n^{2} - 1$) and then rotate this arrow $i$ times $90^{\\circ}$ in the clockwise direction. After $i$ moves, the arrow in $c_{i}$ will have rotated $i$ times $90^{\\circ}$ counterclockwise and be in the same direction as on the path defined above. Thus, Turbo will traverse the cycle $c_{0}, c_{1}, c_{2}, \\ldots , c_{n^{2} - 1}, c_{0}$ and $(1,1)$ is good.\nEvery four moves, all arrows point in the same direction as in the beginning. Moreover, the board will return to its initial configuration after traversing the full cycle, since $n^{2}$, the length of the cycle, is divisible by 4. Therefore Turbo can also start at any $c_{i}$ with $4 \\mid i$ and follow the same route. Hence the cells $c_{0}, c_{4}, c_{8}, \\ldots , c_{n^{2} - 4}$ are good and there are $\\frac{n^{2}}{4}$ of such cells.\n\n## Upper bound for even $n$\n\nWe will prove that for even $n$ and any start configuration there are at most $\\frac{n^{2}}{4}$ good cells.\nLet $a_{0}$ be a good cell. Let $a_{0},a_{1},a_{2},\\ldots ,a_{n^{2} - 1},a_{n^{2}} = a_{0}$ be the sequence of cells that Turbo visits when she starts at $a_{0}$. Now suppose there is another good cell $b_{0}$ and let $b_{0},b_{1},b_{2},\\ldots ,b_{n^{2} - 1},b_{n^{2}} = b_{0}$ be the sequence of cells that Turbo visits when she starts at $b_{0}$.\nNote that, since $4\\mid n^{2}$, the arrows are back to their initial configuration after $n^{2}$ steps. Thus, if Turbo keeps walking after returning to her initial cell, she would just traverse the same cycle over and over again.\nConsider the upper left corner of the board. With standard row and column numbering, the corner cell is $(1,1)$. This cell has only two neighbours, so both the $a$-route and the $b$-route must have cells $(2,1),(1,1),(1,2)$ in that order or $(1,2),(1,1),(2,1)$ in that order. Without loss of generality, $a_{i - 1} = (2,1)$, $a_{i} = (1,1)$ and $a_{i + 1} = (1,2)$ for some $i$. Let $j$ be such that $b_{j} = (1,1)$. If $b_{j - 1} = (2,1) = a_{i - 1}$ then the arrow in cell $(2,1)$ must be pointed in the same direction after $i - 1$ steps and after $j - 1$ steps, so $i\\equiv j \\pmod{4}$. But then the arrow in cell $b_{j} = (1,1) = a_{i}$ must also be pointed in the same direction after $i$ and after $j$ steps, so Turbo moves to $b_{j + 1} = a_{i + 1}$ in both cases, and again finds the arrow pointed in the same direction in both cases. Continuing, we find that the $b$-route is actually identical to $a_{4t},a_{4t + 1},\\ldots ,a_{n^{2}} = a_{0},a_{1},\\ldots ,a_{4t - 1},a_{4t}$ for some $t$, as any other starting point would have the arrows in the wrong direction initially.\n\nNow suppose instead that $b_{j + 1} = (2,1) = a_{i - 1}$. Considering the $a$-route, the arrows in the upper left corner after $i - 1$ steps must be like this:\n\n![](attached_image_2.png)\n\nConsidering the $b$-route instead, the arrows after $j - 1$ steps must be like this:\n\n![](attached_image_3.png)\n\nFrom the arrows in cell $(1,1)$ we see that $i\\equiv j + 1 \\pmod{4}$. However, for the cells $(2,1)$ and $(1,2)$ this gives a contradiction.\nWe conclude that the only possible good cells are $a_{4t}$ for $t = 0,1,\\ldots ,\\frac{n^{2}}{4} - 1$, which gives at most $\\frac{n^{2}}{4}$ good cells.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn each cell of a $2025 \\times 2025$ board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to $1$, and the sum of the numbers in each column is equal to $1$. Define $r_{i}$ to be the largest value in row $i$, and let $R = r_{1} + r_{2} + \\dots + r_{2025}$. Similarly, define $c_{i}$ to be the largest value in column $i$, and let $C = c_{1} + c_{2} + \\dots + c_{2025}$. What is the largest possible value of $\\frac{R}{C}$?\n\nWhat is the largest possible value of $\\frac{R}{C}$?",
    "options": [],
    "answer": "2025/89",
    "solution": "Solution:\n\nAnswer: $\\frac{2025}{89}$.\nIn general, if the table is $m^{2} \\times m^{2}$, the answer is $\\frac{m^{2}}{2m - 1}$.\n\nThe example is as follows: label rows and columns from $1$ to $m^{2}$, from top to bottom and left to right. For the first $m$ columns, write $\\frac{1}{m}$ in all squares whose coordinates have the same residue modulo $m$ and place $0$ everywhere else. For the remaining $m^{2} - m$ columns, place $\\frac{1}{m^{2}}$ everywhere. Then $R = m^{2} \\cdot \\frac{1}{m} = m$, and $C = m \\cdot \\frac{1}{m} + (m^{2} - m) \\cdot \\frac{1}{m^{2}} = 2 - \\frac{1}{m}$. So the ratio is as claimed.\n\n| 1/2 | 0   | 1/4 | 1/4 |\n|-----|-----|-----|-----|\n| 0   | 1/2 | 1/4 | 1/4 |\n| 1/2 | 0   | 1/4 | 1/4 |\n| 0   | 1/2 | 1/4 | 1/4 |\n\nIn particular, when $n := m^{2} = 2025$, we get $\\frac{2025}{89}$. Now we need to show that $\\frac{R}{C} \\leq \\frac{n}{2\\sqrt{n} - 1}$.\n\nFor each row, select one cell having the largest value appearing in said row and colour it red. Then, without loss of generality, we may rearrange the columns such that the red cells appear in the first $k$ columns from the left, and each column contains at least one red cell, for some $k \\leq n$.\n\nFor the $j^{\\mathrm{th}}$ column, for all $1 \\leq j \\leq k$, let $p_{j}$ and $n_{j}$ denote the sum and number of red cells in it, respectively. We observe that $c_{j}$, the biggest number in the $j^{\\mathrm{th}}$ column, is at least $\\frac{p_{j}}{n_{j}}$, for all $1 \\leq j \\leq k$. For all other columns, the largest value they contain is at least $\\frac{1}{n}$, as their sum is $1$. Thus, $C \\geq \\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}$.\n\nWe can also observe that $R = p_{1} + p_{2} + \\dots + p_{k}$.\n\nTherefore we have to show that:\n$$\np_{1} + p_{2} + \\dots +p_{k} \\leq \\frac{n}{2\\sqrt{n} - 1} \\cdot \\left(\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}\\right). \\quad (*)\n$$\nBy construction, $n_{1} + n_{2} + \\dots + n_{k} = n$, and, as the numbers in every column are nonnegative, we see that $p_{j} \\leq 1$ for every $j$. Also, since each number in a red cell is at least $\\frac{1}{n}$, we also have $p_{j} \\geq \\frac{n_{j}}{n}$.\n\nSince our inequality is linear in each $p_{j}$, it suffices to prove it when each variable equals one of its two critical values. By relabeling, we may assume that $p_{j} = \\frac{n_{j}}{n}$ for $1 \\leq j \\leq t$, and $p_{j} = 1$ for $t + 1 \\leq j \\leq k$, for an integer $0 \\leq t \\leq k$.\n\nFirst, if $t = k$, we observe that $p_{1} + p_{2} + \\dots +p_{k} = \\frac{n_{1}}{n} + \\frac{n_{2}}{n} + \\dots + \\frac{n_{k}}{n} = 1$, and that $\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\dots + \\frac{p_{k}}{n_{k}} = \\frac{k}{n}$, so the inequality becomes $1 \\leq \\frac{n}{2\\sqrt{n} - 1}$, which is true.\n\nFrom now on we may assume that $t < k$. We need to show that:\n$$\n\\frac{n_{1} + \\dots + n_{t}}{n} +k - t \\leq \\frac{n}{2\\sqrt{n} - 1} \\cdot \\left(\\frac{t}{n} + \\frac{1}{n_{t + 1}} + \\dots + \\frac{1}{n_{k}} + \\frac{n - k}{n}\\right).\n$$\nBy Cauchy-Schwarz inequality we have that:\n$$\n\\frac{1}{n_{t + 1}} + \\dots + \\frac{1}{n_{k}} \\geq \\frac{(k - t)^{2}}{n_{t + 1} + \\dots + n_{k}} = \\frac{(k - t)^{2}}{n - (n_{1} + \\dots + n_{t})}. \\quad (CS)\n$$\nLet $n_{1} + \\dots +n_{t} = n\\cdot q$, where $0\\leq q< 1$. Thus, it is now enough to show that:\n$$\nq + k - t\\leq \\frac{n}{2\\sqrt{n} - 1}\\left(\\frac{t}{n} +\\frac{(k - t)^{2}}{n - nq} +\\frac{n - k}{n}\\right).\n$$\nLet $k - t = \\ell \\geq 1$. The inequality becomes:\n$$\nq + \\ell \\leq \\frac{1}{2\\sqrt{n} - 1}\\cdot \\left(n - \\ell +\\frac{\\ell^{2}}{1 - q}\\right).\n$$\nRearranging this we get:\n$$\nn + q + \\frac{\\ell^{2}}{1 - q}\\geq 2(q + \\ell)\\sqrt{n}.\n$$\nIf $q = 0$ the inequality is trivially true by $AM - GM$. Suppose now that $0< q< 1$. Then, by Cauchy-Schwarz, we have:\n$$\nq + \\frac{\\ell^{2}}{1 - q} = \\frac{q^{2}}{q} +\\frac{\\ell^{2}}{1 - q}\\geq (q + \\ell)^{2}.\n$$\nIt is therefore enough to show that $n + (q + \\ell)^{2}\\geq 2(q + \\ell)\\sqrt{n}$, which is true by $AM - GM$, completing the proof.\n\n\nSolution 1'. We prove the main inequality $(\\ast)$ in a slightly different manner. Instead of the strong lower bound $p_{j}\\geq \\frac{n_{j}}{n}$, we use the weaker, simpler and more immediate lower bound $p_{j}\\geq 0$ (thus proving the inequality in a larger regime).\nAs in Solution 1, suppose $p_{j} = 0$ for $1\\leq j\\leq t$ and $p_{j} = 1$ for $t + 1\\leq j\\leq k$, with $\\ell = k - t$. We also denote by $m = n_{t + 1} + \\dots +n_{k}$ and note that $m\\leq n - t$, since $n_{j}\\geq 1$ for each $i\\leq t$. We need to prove that:\n$$\n\\ell \\leq \\frac{n}{2\\sqrt{n} - 1}\\cdot \\left(\\frac{1}{n_{t + 1}} +\\dots +\\frac{1}{n_{k}} +\\frac{n - k}{n}\\right).\n$$\nRearranging and using the same Cauchy-Schwarz (CS) as in Solution 1, we see it suffices to show that:\n$$\n(2\\sqrt{n} -1)\\ell \\leq n - k + \\frac{n\\ell^{2}}{m},\n$$\nor equivalently, that:\n$$\n2\\sqrt{n}\\ell \\leq n - t + \\frac{n\\ell^{2}}{m}.\n$$\nBut since $n - t\\geq m$ this immediately follows from $2\\sqrt{n}\\ell \\leq m + \\frac{n\\ell^{2}}{m}$, which is a simple application of $AM - GM$.\n\n\nSolution 1\". This is an alternative way of getting the upper bound on $\\frac{R}{C}$ from\n$$\n\\frac{R}{C}\\leq \\frac{p_{1} + p_{2} + \\ldots + p_{k}}{\\frac{p_{1}}{n_{1}} + \\frac{p_{2}}{n_{2}} + \\ldots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}}.\n$$\nUsing the fact that $\\sum_{j = 1}^{k}n_{j} = n$, we can rewrite the above right hand side as follows:\n$$\n\\frac{\\sum_{j = 1}^{k}p_{j}}{\\sum_{j = 1}^{k}\\left(\\frac{p_{j}}{n_{j}} +\\frac{n_{j} - 1}{n}\\right)}.\n$$\nWe notice that this is a quotient of affine functions in the $p_{j}$'s, for which the denominator does not vanish over the set defined by $0 \\leq p_{j} \\leq 1$. Therefore the maximum of this function is attained when a certain number of $p_{j}$'s are $1$ and the others are $0$. Without loss of generality we may assume that the first $t$ are equal to $1$ and the other $k - t$ are $0$ for some $0 \\leq t \\leq k$. Then one has that the previous expression is at most\n$$\n\\frac{t}{\\sum_{1 \\leq j \\leq t} \\left(\\frac{1}{n_{j}} + \\frac{n_{j} - 1}{n}\\right) + \\sum_{t < j \\leq k} \\frac{n_{j} - 1}{n}}.\n$$\nWe now lower bound the denominator by observing that the second sum is non negative, while each term of the first sum can be bounded by $AM - GM$ as follows:\n$$\n\\frac{1}{n_{j}} + \\frac{n_{j} - 1}{n} \\geq \\frac{2}{\\sqrt{n}} - \\frac{1}{n}.\n$$\nWe therefore have\n$$\n\\frac{R}{C} \\leq \\max_{1 \\leq t \\leq k} \\frac{t}{\\sum_{1 \\leq j \\leq t} \\left(\\frac{2}{\\sqrt{n}} - \\frac{1}{n}\\right)} = \\frac{n}{2\\sqrt{n} - 1},\n$$\nwhich finishes the proof.\n\n\nSolution $1^{\\prime\\prime}$. We follow the same notation as above. First, we apply Cauchy-Schwarz as follows :\n$$\n\\left(\\sum_{i = 1}^{k} \\frac{p_{i}}{n_{i}}\\right) \\left(\\sum_{i = 1}^{k} p_{i} n_{i}\\right) \\geq \\left(\\sum_{i = 1}^{k} p_{i}\\right)^{2} = R^{2}.\n$$\nWe now write $z_{i} = 1 - p_{i}$ for all $1 \\leq i \\leq k$, and observe that all $z_{i}$ are positive. Moreover, we have that $\\sum_{i = 1}^{k} z_{i} = k - R$, and $\\sum_{i = 1}^{k} p_{i} n_{i} = n - \\sum_{i = 1}^{k} p_{i} z_{i}$. Thus, from our last inequality we get\n$$\n\\frac{p_{1}}{n_{1}} + \\dots + \\frac{p_{k}}{n_{k}} \\geq \\frac{R^{2}}{n - \\sum_{i = 1}^{k} n_{i} z_{i}}.\n$$\nAs before we have $C \\geq \\frac{p_{1}}{n_{1}} + \\dots + \\frac{p_{k}}{n_{k}} + \\frac{n - k}{n}$, and so\n$$\nC \\geq \\frac{R^{2}}{n - \\sum_{i = 1}^{k} n_{i} z_{i}} + \\frac{n - k}{n} = \\frac{R^{2}}{n - \\sum_{i = 1}^{k} n_{i} z_{i}} + \\frac{n - R - \\sum_{i = 1}^{k} z_{i}}{n}.\n$$\nPutting everything together we get\n$$\n\\frac{R}{C} \\leq \\frac{R}{\\frac{R^{2}}{n - \\sum_{i = 1}^{k} n_{i} z_{i}} + \\frac{n - R - \\sum_{i = 1}^{k} z_{i}}{n}} = \\frac{n}{\\frac{R n}{n - \\sum_{i = 1}^{k} n_{i} z_{i}} + \\frac{n - \\sum_{i = 1}^{k} z_{i}}{R} - 1}.\n$$\nApplying $AM - GM$ to the denominator we get $\\frac{R}{C} \\leq \\frac{n}{2 \\sqrt{n \\frac{n - \\sum_{i = 1}^{k} z_{i}}{n - \\sum_{i = 1}^{k} n_{i} z_{i}} - 1}}$, which finishes the proof by noting that $\\frac{n - \\sum_{i = 1}^{k} z_{i}}{n - \\sum_{i = 1}^{k} n_{i} z_{i}} \\geq 1$.\n\n\nSolution 2. This is an alternative approach that goes via an intermediary quantity in order to establish to the upper bound on $R / C$.\nLet $x_{i j}$ be the entry in row $i$ and column $j$. Let $n = 2025$. The key idea is to analyze the expression:\n$$\nT:= \\sum_{i,j}x_{i j}\\max \\left(x_{i j},\\frac{1}{n}\\right).\n$$\nOn one hand, since $c_{j}\\geq 1 / n$, we have $x_{i j}\\max (x_{i j},\\frac{1}{n})\\leq x_{i j}c_{j}$ for every $(i,j)$. So\n$$\nT\\leq \\sum_{j}\\sum_{i}x_{i j}c_{j} = \\sum_{j}c_{j} = C.\n$$\nOn the other hand, let $j_{i}$ be one of the indices for which $r_{i} = x_{i j_{i}}$. We therefore have:\n$$\nT=\\sum_{i}\\left(x_{i j_{i}}\\max \\left(x_{i j_{i}},\\frac{1}{n}\\right)+\\sum_{j\\neq j_{i}}x_{i j}\\max \\left(x_{i j},\\frac{1}{n}\\right)\\right)\n$$\n$$\n\\geq\\sum_{i}\\left(r_{i}^{2}+\\frac{1}{n}\\sum_{j\\neq j_{i}}x_{i j}\\right)\n$$\n$$\n=\\sum_{i}\\left(r_{i}^{2}-\\frac{1}{n}r_{i}+\\frac{1}{n}\\right)\\geq\\sum_{i}\\left(\\frac{2}{\\sqrt{n}}r_{i}-\\frac{1}{n}r_{i}\\right)=\\left(\\frac{2}{\\sqrt{n}}-\\frac{1}{n}\\right)R.\n$$\nThis gives the claimed result.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer tous les entiers $a$ et $b$ tels que $(a+1)(b-1)=a^{2} b^{2}$.",
    "options": [],
    "answer": "(0,1) and (-1,0)",
    "solution": "Solution:\nSoit $a, b$ des entiers tels que $(a+1)(b-1)=a^{2} b^{2}$.\nPuisque d'une part $a$ et $a+1$ sont premiers entre eux, et d'autre part $b$ et $b-1$ sont premiers entre eux, c'est donc que $b-1= \\pm a^{2}$ et que $a+1= \\pm b^{2}$, où les signes sont les mêmes dans les deux relations.\n\n- Si $b-1=a^{2}$ et $a+1=b^{2}$ :\nAlors, $a=b^{2}-1=(b-1)(b+1)=a^{2}\\left(a^{2}+2\\right)$.\nSi $a=0$, on obtient $b=1$, et on vérifie bien que $(a, b)=(0,1)$ est bien une solution du problème.\nSi $a \\neq 0$, après division par $a$, il vient donc $a\\left(a^{2}+2\\right)=1$. Comme $a$ et $a^{2}+2$ sont des entiers, avec $a^{2}+2>2$, cette dernière égalité ne peut être satisfaite.\n\n- Si $b-1=-a^{2}$ et $a+1=-b^{2}$ :\nAlors, $-a-1=b^{2}$ et $-b+1=a^{2}$. Ce sont les équations obtenues au cas précédent, où $(a, b)$ a été remplacé par $(-b,-a)$. Ainsi, la seule solution est $(a, b)=(-1,0)$.\n\nFinalement, les solutions du problème sont $(a, b)=(0,1)$ et $(a, b)=(-1,0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier tel que le quotient de $2^{n}$ par $n$ est une puissance de $2$. Montrer que $n$ est une puissance de $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPar hypothèse, $2^{n} / n$ est une puissance de $2$ (ici, on parle bien de quotient, et non pas de quotient dans la division euclidienne). Il existe donc un entier $k \\geqslant 0$ tel que $2^{n} = 2^{k} n$. Ainsi, $n = 2^{n - k}$, de sorte que $n$ est bien une puissance de $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les triplets $ (a, b, c) $ de réels strictement positifs tels que\n$$\n\\left\\{\\begin{array}{l}\na \\sqrt{b} - c = a \\\\\nb \\sqrt{c} - a = b \\\\\nc \\sqrt{a} - b = c\n\\end{array}\\right.\n$$",
    "options": [],
    "answer": "(4, 4, 4)",
    "solution": "Solution:\nOn remarque que $a = b = c = 4$ est solution. On va montrer que c'est en fait la seule.\n\nSi deux des trois nombres sont égaux à $4$ (disons $a$ et $b$), on vérifie facilement que le troisième aussi, car $4 \\sqrt{4} - c = 4$, donc $c = 4$.\n\nSi un des trois nombres vaut $4$ (disons $a$), alors $c \\sqrt{a} - b = c$ devient $2c - b = c$, donc $b = c$. De plus, on $a \\sqrt{b} - b = 4$, qui se réécrit $(\\sqrt{b} - 2)^2 = 0$, donc $b = 4$ et $c = 4$.\n\nIl reste à montrer qu'il est impossible que les trois nombres soient différents de $4$.\n\nSi c'est le cas, alors soit au moins deux des nombres sont strictement supérieurs à $4$, soit au moins deux sont strictement inférieurs à $4$. On traite les deux cas séparément.\n\nSi au moins deux des nombres sont $> 4$, mettons que $c$ soit le plus petit des trois nombres. Alors $a, b > 4$, donc $a = a \\sqrt{b} - c > 2a - c$, donc $c > a > 4$, ce qui contredit la minimalité de $c$.\n\nDe même, si au moins deux des nombres sont $< 4$, supposons que $c$ est le plus grand. Alors $a, b < 4$, donc $a = a \\sqrt{b} - c < 2a - c$, donc $c < a$, ce qui contredit la maximalité de $c$.\n\nLa seule solution est donc bien $(4, 4, 4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProuver qu'il n'existe qu'un nombre fini de nombres premiers s'écrivant sous la forme $n^{3}+2 n+3$ avec $n \\in \\mathbb{N}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRemarquons que $n^{3}-n = n(n-1)(n+1)$ est un produit de trois entiers consécutifs. Puisque parmi trois entiers consécutifs il y a toujours un multiple de $3$, on obtient que $n^{3}-n$ est divisible par $3$, c'est-à-dire $n^{3} \\equiv n \\pmod{3}$. (On peut le voir aussi en utilisant le petit théorème de Fermat.)\n\nOn a donc $n^{3}+2 n+3 \\equiv n+2 n+3 \\equiv 3 n+3 \\equiv 0 \\pmod{3}$, donc $n^{3}+2 n+3$ est toujours divisible par $3$.\n\nD'autre part, il ne peut être égal à $3$ que pour un nombre fini de valeurs de $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn polyèdre a 6 sommets et 12 arêtes. Montrer que chaque face est un triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $F$ le nombre de faces, $A$ le nombre d'arêtes et $S$ le nombre de sommets. La formule d'Euler donne $F - A + S = 2$, donc $F = A - S + 2 = 8$.\n\nNotons $x_{i}$ ($i = 1, 2, \\ldots, 8$) le nombre d'arêtes de la face $i$. On a $x_{1} + \\cdots + x_{8} = 2A = 24$ car chaque arête appartient à exactement deux faces. On en déduit que\n$$\n\\frac{x_{1} + \\cdots + x_{8}}{8} = 3\n$$\nautrement dit le nombre moyen d'arêtes par face est $3$. Or, chaque face possède au moins $3$ arêtes, donc finalement chaque face est un triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\omega$ et $\\Omega$ deux cercles concentriques (c'est-à-dire qu'ils ont le même centre) de sorte que le cercle $\\omega$ soit à l'intérieur du cercle $\\Omega$. Soient $X$ et $Y$ deux points sur le cercle $\\omega$. On note $P$ et $Q$ les points d'intersection respectifs du cercle $\\Omega$ avec les tangentes en $X$ et $Y$ à $\\omega$, de telle sorte que $P$ et $Q$ soient du même côté par rapport à la droite $(XY)$. Montrer que les points $X, Y, P$ et $Q$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nLe quadrilatère $XYQP$ semble être un trapèze isocèle sur la figure. On commence par le démontrer.\n\nSoit $d$ la médiatrice de $[XY]$. Puisque $O$ est le centre du cercle $\\omega$, $O$ appartient à $d$. Ainsi, les cercles centrés en $O$, et en particulier $\\omega$ et $\\Omega$, sont envoyés sur eux-mêmes par la symétrie d'axe $d$. On note $s$ cette symétrie.\n\nOn a $s(X) = Y$, $s(\\omega) = \\omega$ et $s(\\Omega) = \\Omega$, donc $s$ envoie la tangente à $\\omega$ en $X$ sur la tangente à $s(\\omega) = \\omega$ en $s(X) = Y$. Le point d'intersection $P$ de $\\Omega$ avec la tangente à $\\omega$ en $X$ est envoyé sur le point d'intersection de $s(\\Omega)$ avec la tangente à $s(\\omega)$ en $s(X)$.\n\nIl s'agit du point d'intersection de $\\Omega$ avec la tangente à $\\omega$ en $Y$, à savoir le point $Q$. Autrement dit, $P$ et $Q$ sont symétriques par rapport à la médiatrice de $[XY]$. Les segments $[XY]$ et $[PQ]$ sont perpendiculaires à $d$ donc parallèles entre eux et les segments $[PX]$ et $[QY]$ sont symétriques par rapport à $d$. Ceci suffit à conclure que le quadrilatère $PQYX$ est un trapèze isocèle.\n\nConcluons qu'il s'agit d'un quadrilatère cyclique. Par angles correspondants, $\\widehat{QPX} = 180^\\circ - \\widehat{PXY}$ et puisque $s$ conserve les angles, $\\widehat{YQP} = \\widehat{QPX}$. On déduit que\n$$\n\\widehat{YQP} = \\widehat{QPX} = 180^\\circ - \\widehat{PXY}\n$$\ndonc les points $P, Q, Y$ et $X$ sont bien cocycliques d'après la réciproque du théorème de l'angle inscrit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $p$ tels que $p$, $p+2$ et $p+4$ soient tous les 3 premiers.\n\nUn nombre premier est un entier $\\geqslant 2$ qui n'est divisible que par 1 et lui-même.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nTout d'abord, on peut s'attendre à ce qu'il n'y en ait que très peu. On souhaite obtenir des informations sur ces nombres premiers.\n\nNotons qu'un nombre premier divisible par $3$ est a fortiori égal à $3$.\n\nOn considère 3 cas en fonction du reste de la division de $p$ par $3$ (modulo $3$) :\n\n- $p = 3k$ : $p$ est divisible par $3$ donc $p = 3$. $\\{3, 5, 7\\}$ est bien un triplet de nombres premiers.\n\n- $p = 3k + 1$ : $p + 2$ est divisible par $3$ et premier donc $p = 1$ (impossible).\n\n- $p = 3k + 2$ : $p + 4$ est divisible par $3$ et premier donc $p = -1$ (impossible).\n\nD'où $\\{p\\} = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un carré de côté $1$. À l'intérieur du carré, on trace les arcs de cercles de centres $A$, $B$, $C$, $D$ et de rayon $1$. Déterminer l'aire de chaque portion délimitée à l'intérieur du carré.",
    "options": [],
    "answer": "a = 1 - sqrt(3) + pi/3; b = -1 + sqrt(3)/2 + pi/12; c = 1 - sqrt(3)/4 - pi/6",
    "solution": "Solution:\n\nConsidérons les aires $a$, $b$, $c$ des portions indiquées sur la figure.\n\n![](attached_image_1.png)\n\nComme l'aire du carré est égale à $1$, on a\n$$\na + 4b + 4c = 1.\n$$\nComme l'aire du quart de cercle $ABC$ est égale à $\\pi / 4$, on a\n$$\na + 3b + 2c = \\frac{\\pi}{4}.\n$$\nConsidérons maintenant le secteur angulaire $OMN$ de rayon $1$ et d'angle $\\pi / 3$. Il a pour aire $\\pi / 6$. D'autre part, le triangle $OMN$ a pour aire $\\frac{\\sqrt{3}}{4}$, donc l'aire de la \"lune\" est égale à $\\frac{\\pi}{6} - \\frac{\\sqrt{3}}{4}$. On en déduit que\n$$\na + 2b + c = 2\\left(\\frac{\\pi}{6} - \\frac{\\sqrt{3}}{4}\\right) + \\frac{\\sqrt{3}}{4} = \\frac{\\pi}{3} - \\frac{\\sqrt{3}}{4}.\n$$\nEn soustrayant les équations (1) et (2) et les équations (1) et (3), on en déduit\n$$\n\\begin{aligned}\nb + 2c &= 1 - \\frac{\\pi}{4} \\\\\n2b + 3c &= 1 + \\frac{\\sqrt{3}}{4} - \\frac{\\pi}{3}\n\\end{aligned}\n$$\nce qui donne facilement\n$$\n\\begin{aligned}\nb &= -1 + \\frac{\\sqrt{3}}{2} + \\frac{\\pi}{12} \\\\\nc &= 1 - \\frac{\\sqrt{3}}{4} - \\frac{\\pi}{6} \\\\\na &= 1 - \\sqrt{3} + \\frac{\\pi}{3}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $x, y$ des réels strictement positifs. Montrer que:\n$$\nx + \\frac{y^{2}}{x} \\geqslant 2y\n$$\net trouver les cas d'égalité.",
    "options": [],
    "answer": "Equality holds if and only if x equals y.",
    "solution": "Solution:\nL'inégalité de la moyenne donne\n$$\nx + \\frac{y^{2}}{x} \\geqslant 2 \\sqrt{x \\cdot \\frac{y^{2}}{x}} = 2y\n$$\nSupposons qu'on a égalité, alors $x = \\frac{y^{2}}{x}$ donc $x^{2} = y^{2}$ donc $x = y$. Réciproquement si $x = y$, $x + \\frac{y^{2}}{x} = y + \\frac{y^{2}}{y} = 2y$ on a bien égalité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère et $M, N, P, Q$ les milieux respectifs de $[AB]$, $[BC]$, $[CD]$, $[DA]$.\nMontrer que le quadrilatère $MNPQ$ est un parallélogramme.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPuisque les points $M$ et $N$ sont les milieux respectifs des côtés $[AB]$ et $[BC]$, les droites $(MN)$ et $(AC)$ sont parallèles.\n\nPuisque les points $P$ et $Q$ sont les milieux respectifs des côtés $[CD]$ et $[DA]$, les droites $(PQ)$ et $(AC)$ sont parallèles.\n\nDonc les droites $(PQ)$ et $(MN)$ sont parallèles.\n\nOn obtient de la même façon que les droites $(NP)$ et $(MQ)$ sont parallèles.\n\nDonc les côtés opposés deux à deux du quadrilatère $MNPQ$ sont parallèles, ce qui en fait un parallélogramme.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $a$, $b$ et $c$ des réels tels que\n$$\n|a-b| \\geq |c|,\\ |b-c| \\geq |a| \\text{ et } |c-a| \\geq |b|.\n$$\nProuver que l'un des trois nombres $a$, $b$ et $c$ est la somme des deux autres.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLa première inégalité s'écrit $(a-b)^2 \\geq c^2$, ou encore $(a-b+c)(a-b-c) \\geq 0$. De même, les deux autres inégalités conduisent à $(b-c+a)(b-c-a) \\geq 0$ et $(c-a+b)(c-a-b) \\geq 0$.\nEn multipliant membre à membre ces trois inégalités, il vient\n$$\n(a-b+c)^2(-a+b+c)^2(a+b-c)^2 \\leq 0\n$$\nOr, un carré étant toujours positif, c'est donc que l'un des facteurs est nul, ce qui conclut.\nSolution:\nLes variables $a$, $b$ et $c$ jouent des rôles symétriques. En outre, le problème est invariant si on change simultanément les signes de $a$, $b$ et $c$. Ainsi, il y a au moins deux variables parmi $a$, $b$ et $c$ qui ont le même signe ; on peut donc supposer, sans perte de généralité, que l'on est dans un des deux cas $a \\geq b \\geq c \\geq 0$ ou $a \\geq b \\geq 0 \\geq c$.\nMontrons que la relation $b=a+c$ est vraie dans les deux cas, ce qui conclura le problème.\n1. Si $a \\geq b \\geq c \\geq 0$, alors $b \\geq b-c=|b-c| \\geq |a|=a$; chaque inégalité est donc une égalité, d'où la relation $b=b-c=|b-c|=|a|=a$.\n2. Si $a \\geq b \\geq 0 \\geq c$, alors $b-c=|b-c| \\geq |a|=a$ et $a-b=|a-b| \\geq |c|=-c$, d'où la relation $a+c \\geq b \\geq a+c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes cases d'une grille à 10 lignes et 10 colonnes sont coloriées en blanc et en noir. Un coloriage de ces cases est dit homogène s'il contient un carré $3 \\times 3$ monochrome, et inhomogène sinon. Montrer qu'il existe plus de coloriages inhomogènes que de coloriages homogènes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIl y a $2^{100}$ coloriages possibles. Si $c$ est un carré $3 \\times 3$, le nombre de coloriages tels que $c$ soit monochrome est égal à $2 \\times 2^{91} = 2^{92}$. Or, il y a 64 carrés $3 \\times 3$, donc il y a au plus $64 \\times 2^{92} = 2^{98}$ coloriages homogènes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un trapèze isocèle avec $(AD)//(BC).\n\na. Montrer que $ABCD$ admet un cercle circonscrit.\n\nb. On note $O$ le centre du cercle circonscrit de $ABCD$ et $E$ l'intersection des diagonales. Montrer que $O$ est sur le cercle circonscrit de $AEB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. $ABCD$ est un trapèze isocèle, ainsi $\\widehat{ABC} = \\widehat{DCB}$. On sait également que $(AD)//(CB)$, donc $\\widehat{DCB} = 180 - \\widehat{CDA}$. Avec ces deux égalités on trouve que $\\widehat{ABC} + \\widehat{CDA} = 180$, ce qui montre bien que les points $A$, $B$, $C$ et $D$ sont sur le même cercle.\n\nb. On va noter $\\alpha$ l'angle $\\widehat{ACB}$ et on va essayer de calculer différents angles afin d'arriver à la conclusion demandée. On trouve par symétrie que $\\widehat{DBC} = \\alpha$ également. Ainsi, dans le triangle $EBC$, sachant que la somme des angles vaut $180$, on trouve $\\widehat{AEB} = 180 - \\widehat{BEC} = \\widehat{EBC} + \\widehat{ECB} = 2\\alpha$. Avec le théorème de l'angle au centre, sachant que l'angle inscrit $\\widehat{ACB} = \\alpha$, on a $\\widehat{AOB} = 2\\alpha$. Mais alors, $\\widehat{AOB} = 2\\alpha = \\widehat{AEB}$ ce qui montre bien que les points $A$, $E$, $O$ et $B$ sont cocycliques.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a$, $b$, $c$ des réels strictement positifs. Montrer que\n$$\n\\frac{a}{b c}+\\frac{b}{a c}+\\frac{c}{a b} \\geqslant \\frac{2}{a}+\\frac{2}{b}-\\frac{2}{c} .\n$$\nAttention, il y a bien un \"moins\" dans le membre de droite!",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEn multipliant les deux membres par $a b c$, l'inégalité à montrer se réécrit\n$$\na^{2}+b^{2}+c^{2} \\geqslant 2 b c+2 a c-2 a b\n$$\nEn passant $2 a b$ de l'autre côté, on peut factoriser. L'inégalité à montrer devient\n$$\n(a+b)^{2}+c^{2} \\geqslant 2(a+b) c\n$$\nEn passant tout à gauche, l'inégalité est finalement équivalente à\n$$\n(a+b-c)^{2} \\geqslant 0\n$$\nqui est bien toujours vraie. De plus, on a égalité si et seulement si $c=a+b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les triplets $\\{p, q, r\\}$ de nombres premiers tels que les 3 différences\n$$\n|p-q|,\\ |q-r|,\\ |r-p|\n$$\nsoient également des nombres premiers.",
    "options": [],
    "answer": "{2, 5, 7}",
    "solution": "Solution:\n\nNotons que les trois nombres doivent être deux à deux distincts puisque $0$ n'est pas un nombre premier. On peut donc supposer, quitte à échanger l'ordre des variables, que $p > q > r$. Un nombre premier est impair ou égal à $2$.\n\nOn suppose que $p$, $q$ et $r$ sont tous impairs. Alors $p-q$, $q-r$ et $r-p$ sont pairs. Comme leur valeur absolue est première, ces nombres valent tous $2$. Ainsi les entiers $p$, $p+2$ et $p+4$ sont premiers. Si $p$ est divisible par $3$, alors $p=3$ et $q=5$ et $r=7$.\n\nCependant, le triplet $(3,5,7)$ n'est pas solution du problème : $7-3=4$ n'est pas premier.\n\nSi $p$ n'est pas divisible par $3$, alors $p$ est de la forme $3k+1$ ou $3k+2$. Le premier cas implique que $p+2$ soit divisible par $3$ donc $p+2=3$ mais $p=1$ n'est pas un nombre premier. Le deuxième cas implique que $p+4$ soit divisible par $3$, mais $p+4=3$ ne donne pas de solution strictement positive.\n\nOn suppose que $r=2$. Alors $p$ et $q$ sont impaires et $p-q$ est pair et premier donc égal à $2$. Il vient que $q+2$, $q$ et $q-2$ sont tous les trois des nombres premiers. D'après le cas précédent, cela implique que $q-2=3$ donc $p=7$. Réciproquement, le triplet $(p, q, r) = (2, 5, 7)$ et ses permutations sont donc bien solutions au problème.\n\nLes seuls triplets solutions sont donc $(2, 5, 7)$ ainsi que ses permutations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $d_{1}$, $d_{2}$, $d_{3}$ des droites concourantes et $A, A^{\\prime}$ des points sur la droite $d_{1}$, $B, B^{\\prime}$ des points sur la droite $d_{2}$, $C, C^{\\prime}$ des points sur la droite $d_{3}$ tels que les droites $(AB)$ et $(A^{\\prime}B^{\\prime})$ sont parallèles et les droites $(BC)$ et $(B^{\\prime}C^{\\prime})$ sont parallèles.\nMontrer que les droites $(AC)$ et $(A^{\\prime}C^{\\prime})$ sont parallèles.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $S$ le point de concours des trois droites. Puisque les droites $(AB)$ et $(A^{\\prime}B^{\\prime})$ sont parallèles, d'après le théorème de Thalès,\n$$\n\\frac{AS}{A^{\\prime}S} = \\frac{BS}{B^{\\prime}S}\n$$\nDe même on trouve\n$$\n\\frac{BS}{B^{\\prime}S} = \\frac{CS}{C^{\\prime}S}\n$$\nOn déduit que\n$$\n\\frac{AS}{A^{\\prime}S} = \\frac{CS}{C^{\\prime}S}\n$$\nce qui signifie, d'après le théorème de Thalès, que les droites $(AC)$ et $(A^{\\prime}C^{\\prime})$ sont parallèles.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $L$, $M$ et $N$ les milieux respectifs des côtés $[BC]$, $[CA]$ et $[AB]$. On suppose que $\\widehat{ANC} = \\widehat{ALB}$. Montrer que $\\widehat{CAL} = \\widehat{ABM}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn essaie dans un premier temps d'interpréter la condition de l'énoncé $\\widehat{ANC} = \\widehat{ALB}$. Pour cela on commence par introduire $G$ le centre de gravité du triangle $ABC$ (on rappelle que c'est aussi l'intersection des trois médianes $(AL)$, $(BM)$ et $(CN)$). La condition de l'énoncé revient à dire que le quadrilatère $NGBL$ est cyclique.\n\nOn a ainsi $\\widehat{CAL} = \\widehat{ALN}$ par angles alternes-internes. Puis comme le quadrilatère $NGBL$ est cyclique on a $\\widehat{CAL} = \\widehat{ALN} = \\widehat{GLN} = \\widehat{GBN} = \\widehat{ABM}$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $x, y$ des réels. Montrer que:\n$$\n|x|+|y| \\leqslant |x-y|+|x+y|.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSi $x$ et $y$ sont positifs, $|x|+|y|=x+y=|x+y|$.\n\nSi $x$ et $y$ sont négatifs $|x|+|y|=-x-y=|x+y|$.\n\nSi $x$ est positif et $y$ négatif $|x|+|y|=x-y=|x-y|$.\n\nSi $x$ est négatif et $y$ positif $|x|+|y|=y-x=|x-y|$.\n\nDans tous les cas par positivité de la valeur absolue $|x|+|y| \\leqslant |x-y|+|x+y|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle isocèle et obtus en $A$. Soit $\\Gamma$ le cercle de centre $B$ passant par $A$, et $\\Omega$ le cercle de centre $C$ passant par $A$. Soit $D$ le point d'intersection du cercle $\\Gamma$ avec le segment $[BC]$, $E$ le deuxième point d'intersection de la droite $(AD)$ avec le cercle $\\Omega$, et $F$ le point d'intersection de la droite $(BC)$ avec le cercle $\\Omega$ qui n'est pas sur le segment $[BC]$.\nMontrer que le triangle $DFE$ est isocèle en $F$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $X$ le point d'intersection du cercle $\\Omega$ avec le segment $[BC]$. Les points $B$ et $C$ sont symétriques par rapport à la médiatrice du segment $[BC]$ donc les cercles $\\Gamma$ et $\\Omega$ le sont aussi. Il vient que $D$ et $X$ sont symétriques par rapport à la médiatrice du segment $[BC]$. Le triangle $DAX$ est donc isocèle en $A$ et $\\widehat{ADX} = \\widehat{AXD}$.\n\nLes points $F, E, X$ et $A$ sont cocycliques donc $\\widehat{FEA} = \\widehat{FXA}$. On déduit\n$$\n\\widehat{FED} = \\widehat{FEA} = \\widehat{FXA} = \\widehat{DXA} = \\widehat{XDA} = \\widehat{FDE}\n$$\nCe qui donne bien que le triangle $FDE$ est isocèle en $F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les nombres entiers $z \\in \\mathbb{Z}$ tels que\n$$\n2^{z}+2=r^{2}\n$$\noù $r \\in \\mathbb{Q}$ est un nombre rationnel.\nUn nombre rationnel est un nombre qui s'écrit sous la forme $\\frac{\\mathrm{a}}{\\mathrm{b}}$ avec $\\mathrm{a}, \\mathrm{b}$ des entiers et $\\mathrm{b} \\neq 0$.",
    "options": [],
    "answer": "{-2, 1}",
    "solution": "Solution:\n\nNotons que si $(z, r)$ est un couple solution, $(z,-r)$ est également un couple solution. On peut donc supposer pour la suite que $r \\geqslant 0$. Etant donné que $2^{z}+2>0$, on a même $r>0$. On pose $r=\\frac{a}{b}$, avec $a$ et $b$ des entiers strictement positifs et premiers entre eux.\n\nSi $z \\geqslant 0$, alors $2^{z}+2$ est entier donc $r$ est entier et $b=1$. On doit désormais résoudre l'équation $2^{z}+2=a^{2}$ dans les entiers positifs. Si $z \\geqslant 2$, alors l'équation vue modulo 4 donne $a^{2} \\equiv 2 \\bmod 4$ ce qui n'a pas de solution puisqu'un carré est toujours congru à 0 ou à 1 modulo 4. On déduit que $z=0$ ou $z=1$. Dans le premier cas, on obtient $3=a^{2}$ qui n'admet pas de solution entière. Dans le deuxième cas on trouve $4=a^{2}$ soit $a=2$. Réciproquement, les couples $(1,-2)$ et $(1,2)$ satisfont bien l'équation.\n\nSi $z<0$, alors on pose $z'=-z$, avec $z'>0$. L'équation devient $\\frac{1}{2^{z'}}+2=\\frac{a^{2}}{b^{2}}$. En supprimant les dénominateurs on obtient\n$$\nb^{2}\\left(1+2^{z'+1}\\right)=a^{2} \\cdot 2^{z'}\n$$\nPuisque $1+2^{z'+1}$ est premier avec $2^{z'}$, par le lemme de Gauss on obtient que $1+2^{z'+1}$ divise $a^{2}$. Comme $a$ et $b$ sont premiers entre eux, par le lemme de Gauss on obtient aussi que $a^{2}$ divise $1+2^{z'+1}$. On déduit que $a^{2}=2^{z'+1}+1$, que l'on réécrit $2^{z'+1}=(a-1)(a+1)$. On déduit que $a+1$ et $a-1$ sont tous les deux des puissances de 2 dont la différence vaut 2. On déduit donc que $a+1=4$ et $a-1=2$ soit $a=3$ et $z'=2$. Ainsi $b=2$. Réciproquement, les couples $\\left(-2,-\\frac{3}{2}\\right)$ et $\\left(-2, \\frac{3}{2}\\right)$ sont bien solutions de l'équation. Les solutions sont donc $\\left\\{\\left(-2,-\\frac{3}{2}\\right),\\left(-2, \\frac{3}{2}\\right),(1,-2),(1,2)\\right\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer tous les triplets d'entiers $(a, b, n)$ strictement positifs vérifiant:\n$$\na! + b! = 2^{n}\n$$",
    "options": [],
    "answer": "[(1,1,1), (2,2,2), (2,3,3), (3,2,3)]",
    "solution": "Solution:\nLes factorielles ayant beaucoup de facteurs impairs en commun, on se dit directement qu'obtenir une puissance de 2 va être très contraignant.\nSupposons que $a, b \\geqslant 3$, $3$ divise donc la somme des factorielles et donc $2^{n}$ : c'est absurde. C'est à dire que l'un des deux est dans $\\{1,2\\}$. Par symétrie, on ne traite que deux cas :\n\n1. $a=1$ : encore deux petits cas de figure :\n\n- $b=1$: $n=1$\n- $b \\geqslant 2$ : les deux membres n'ont pas la même parité, c'est impossible\n\n2. $a=2$: $b \\geqslant 2$ (comme vu au-dessus), quelques cas :\n\n- $b=2,3$: $n=2,3$\n- $b \\geqslant 4$ : il y a un problème modulo 4, impossible\n\nD'où : les seules solutions sont dans $\\{(1,1,1), (2,2,2), (2,3,3), (3,2,3)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle acutangle. Soient $P$ et $Q$ deux points sur le segment $[BC]$. On note respectivement $O_1$, $O_2$, $O_3$ et $O_4$ les centres des cercles circonscrits des triangles $ABP$, $ABQ$, $ACP$ et $ACQ$. Montrer que les points $O_1$, $O_2$, $O_3$ et $O_4$ sont cocycliques si et seulement si $\\widehat{BAP} = \\widehat{QAC}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nLa figure comporte beaucoup d'objets, on va donc essayer de la simplifier un peu. On remarque donc dans un premier temps qu'il y a beaucoup d'axes radicaux (les droites $(AB)$, $(AP)$, $(AQ)$ et $(AC)$ sur la figure) et donc beaucoup d'angles droits avec les centres des cercles. On a les perpendicularités suivantes :\n$$\n(O_1O_2) \\perp (AB),\\ (O_1O_3) \\perp (AP),\\ (O_2O_4) \\perp (AQ),\\ (O_3O_4) \\perp (AC)\n$$\nEn particulier, on veut démontrer une relation de cocyclicité et il faut pour cela se préoccuper d'angles entre droites (et on n'a pas besoin de plus d'informations). En particulier :\n$$\n(O_1O_3, O_1O_2) = (O_1O_3, AP) + (AP, AB) + (AB, O_1O_2) = -\\frac{\\pi}{2} + (AP, AB) + \\frac{\\pi}{2} = (AP, AB)\n$$\nDe la même manière,\n$$\n(O_4O_3, O_4O_2) = (AC, AQ)\n$$\nAinsi,\n$$\n\\begin{array}{cl}\n(O_1O_2, O_1O_3) = (O_2O_4, O_3O_4) & (O_1O_2O_3O_4 \\text{ cyclique}) \\\\\n\\mathbb{\\Downarrow} & \\\\\n(AB, AP) = (AQ, AC), & (\\widehat{BAP} = \\widehat{QAC})\n\\end{array}\n$$\nCe qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les polynômes $P$ à coefficients réels tels que pour tous réels $x, y$,\n$$\nx P(x)+y P(y) \\geqslant 2 P(x y)\n$$",
    "options": [],
    "answer": "P(x) = c x with c ≥ 0",
    "solution": "Solution:\nSi $P(X) = c X$ avec $c \\geqslant 0$, $x P(x) + y P(y) = c(x^{2} + y^{2}) \\geqslant 2 c x y = 2 P(x y)$ par inégalité de la moyenne, les polynômes de la forme $c X$ avec $c \\geqslant 0$ conviennent, montrons que ce sont les seuls.\n\nSupposons $P$ non constant. En évaluant en $(x, 0)$ l'inégalité, on obtient $x P(x) \\geqslant 2 P(0)$. En regardant la limite en $+\\infty$, on obtient que le coefficient dominant de $P$ est forcément strictement positif (car $P$ ne peut tendre vers $-\\infty$).\n\nEn évaluant en $(x, x)$, on obtient $x P(x) \\geqslant P(x^{2})$. Le polynôme $X P(X)$ est de degré $\\deg(P) + 1$, $P(X^{2})$ est de degré $2 \\deg(P)$. Si $2 \\deg(P) > \\deg(P) + 1$, alors $P(X^{2}) - X P(X)$ est un polynôme de degré $2 \\deg(P)$ de coefficient dominant strictement positif, donc il est strictement positif pour $x$ assez grand, contradiction. En particulier $2 \\deg(P) \\leqslant \\deg(P) + 1$ donc $P$ est un polynôme de degré au plus $1$ et de coefficient dominant strictement positif.\n\nMaintenant posons $P(X) = a X + b$ avec $a \\geqslant 0$. En évaluant l'inégalité en $(x, x)$ on obtient $a x^{2} + b x \\geqslant a x^{2} + b$ donc $b x \\geqslant b$. Pour $x = 2$, on obtient $b \\geqslant 0$, pour $x = 0$ on obtient $b \\leqslant 0$ donc $b = 0$. Ainsi $P(X) = a X$ avec $a \\geqslant 0$ ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $\\Gamma$ son cercle circonscrit et $\\Omega$ un autre cercle passant par les points $A$ et $B$. La droite $(AC)$ coupe le cercle $\\Omega$ en un point $D$ et la tangente au cercle $\\Gamma$ en $B$ coupe $\\Omega$ en un point $E$.\n\nMontrer que les droites $(BC)$ et $(DE)$ sont parallèles.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLa droite $(BE)$ est tangente au cercle $\\Omega$ en $B$, donc par le théorème de l'angle tangent, $\\widehat{EBA} = \\widehat{BCA}$.\n\nLes points $D$, $A$, $B$ et $E$ sont cocycliques donc $\\widehat{EDA} = 180^\\circ - \\widehat{EBA}$.\n\nOn déduit que $\\widehat{EDA} = 180^\\circ - \\widehat{BCA}$, ce qui donne bien que les droites $(BC)$ et $(DE)$ sont parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. Soit $t$ la tangente en $A$ au cercle circonscrit du triangle $ABC$. On note $A'$ le point de $t$ de telle sorte que la droite $(BC)$ coupe le segment $[AA']$ en son milieu. On note $C'$ le symétrique de $C$ par rapport à $t$. Montrer que les points $B$, $A$, $C'$ et $A'$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn note $X$ le point d'intersection des droites $(AA')$ et $(BC)$. On note de plus $C^*$ le symétrique de $C$ par rapport au point $X$. On va montrer que les points $B$, $A$, $C'$, et $A'$ sont cocycliques sur un cercle qui contient de plus le point $C^*$. On commence par utiliser les longueurs et la puissance du point depuis $X$.\n\n$$\n\\begin{aligned}\nXA \\cdot XA' & = -XA^2 \\\\\n& = -XB \\cdot XC \\\\\n& = XB \\cdot XC^* .\n\\end{aligned}\n$$\n\nCe qui montre bien que les points $B$, $A$, $C^*$ et $A'$ sont cocycliques. Il ne reste plus qu'à placer le point $C'$ sur le cercle. Pour cela on remarque que $\\widehat{AC'A'} = \\widehat{ACA'}$ par symétrie puis que $\\widehat{AC^*A'} = \\widehat{ACA'}$ comme le quadrilatère $AC A' C^*$ est un parallélogramme. Cela conclut donc que $\\widehat{AC'A'} = \\widehat{AC^*A'}$. Et enfin que les points $B$, $A$, $C^*$, $C'$ et $A'$ sont cocycliques.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A$, $B$, $C$ et $D$ quatre points sur un cercle dans cet ordre. Soit $U$ le point d'intersection des droites $(AB)$ et $(CD)$, et $V$ le point d'intersection des droites $(BC)$ et $(DA)$. Soit $K$ le point d'intersection de la bissectrice issue de $U$ dans le triangle $AUC$ et de la bissectrice issue de $V$ dans le triangle $AVC$. Soit $L$ le point d'intersection de la médiatrice du segment $[KU]$ et de la médiatrice du segment $[KV]$.\n\nMontrer que les points $U$, $V$ et $L$ sont alignés.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLe point $L$ est le centre du cercle circonscrit au triangle $UKV$. Pour montrer que les points $U$, $L$ et $V$ sont alignés, il suffit de montrer que $\\widehat{ULV} = 180^{\\circ}$. Par le théorème de l'angle au centre, $\\widehat{UKV} = \\frac{1}{2} \\widehat{ULV}$. Il suffit donc de montrer que $\\widehat{UKV} = 90^{\\circ}$.\n\nLa somme des angles du triangle $UKV$ vaut $180^{\\circ}$, donc il suffit de montrer que $\\widehat{KUV} + \\widehat{KVU} = 90^{\\circ}$. D'une part $\\widehat{KUV} = \\widehat{CUV} - \\widehat{KUC} = \\widehat{CUV} - \\frac{1}{2} \\widehat{BUC}$. Or $\\widehat{BUC} = 180^{\\circ} - \\widehat{UBC} - \\widehat{BCU} = \\widehat{ADC} - \\widehat{UCB}$. On déduit\n$$\n\\widehat{KUV} = \\widehat{CUV} - \\frac{1}{2} \\widehat{ADC} + \\frac{1}{2} \\widehat{UCB}\n$$\nDe la même manière, on obtient\n$$\n\\widehat{KVU} = \\widehat{CVU} - \\frac{1}{2} \\widehat{CBA} + \\frac{1}{2} \\widehat{VCD}\n$$\nOr $\\widehat{CBA} + \\widehat{ADC} = 180^{\\circ}$ et\n$$\n\\widehat{UCB} = \\widehat{VCD} = \\widehat{UCV} = 180^{\\circ} - \\widehat{CUV} - \\widehat{CVU}\n$$\nFinalement\n$$\n\\begin{aligned}\n& \\widehat{KUV} + \\widehat{KVU} = \\widehat{CUV} - \\frac{1}{2} \\widehat{ADC} + \\frac{1}{2} \\widehat{UCB} + \\widehat{CVU} - \\frac{1}{2} \\widehat{CBA} + \\frac{1}{2} \\widehat{VCD} \\\\\n& = \\widehat{CUV} + \\widehat{CVU} - \\frac{1}{2} \\cdot 180^{\\circ} + 2 \\cdot \\frac{1}{2} \\cdot (180^{\\circ} - \\widehat{CUV} - \\widehat{CVU}) = 90^{\\circ}\n\\end{aligned}\n$$\nce qui donne le résultat voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer tous les triplets d'entiers $(x, y, z)$ vérifiant la propriété suivante :\n$$\n\\operatorname{pgcd}(x, y, z)<\\operatorname{pgcd}(x+y, y+z, z+x)\n$$",
    "options": [],
    "answer": "All triples that are positive integer multiples of a triple of odd integers; equivalently, those for which dividing each entry by their greatest common divisor yields three odd integers.",
    "solution": "Solution:\nNotons que puisque $\\operatorname{pgcd}(x, y, z)$ divise chacun des $x, y, z$, il divise également $\\operatorname{pgcd}(x+y, y+z, z+x)$.\nOn remarque que si le triplet $(x, y, z)$ est solution, alors les triplets $(k x, k y, k z)$ sont solutions pour tout $k \\in \\mathbb{N}^{*}$. On peut donc supposer, quitte à diviser chaque variable par $\\operatorname{pgcd}(x, y, z)$, que les entiers $x, y, z$ sont premiers entre eux dans leur ensemble. En particulier ils ne sont pas tous pairs.\n\nSoit $d=\\operatorname{pgcd}(x+y, y+z, z+x)$. Alors $d$ divise $(x+y)+(x+z)-(y+z)=2 x$ et de même $d$ divise $2 y$ et $2 z$. Donc $d$ divise $\\operatorname{pgcd}(2 x, 2 y, 2 z)=2 \\operatorname{pgcd}(x, y, z)=2$. On déduit que $d=1$ ou $d=2$. Mais comme $d>1$, $d=2$ donc les entiers $x, y, z$ sont tous de même parité. Comme ils ne sont pas tous pairs, ils sont tous impairs.\n\nRéciproquement, si $x, y$ et $z$ sont tous les trois impairs, étant donné que $\\operatorname{pgcd}(x, y, z)$ et $2$ sont premiers entre eux et divisent $\\operatorname{pgcd}(x+y, y+z, z+x)$, on a bien\n$$\n\\operatorname{pgcd}(x, y, z)<2 \\operatorname{pgcd}(x, y, z) \\leqslant \\operatorname{pgcd}(x+y, y+z, z+x)\n$$\nLes triplets solutions sont donc les triplets $\\{(k x, k y, k z),\\ k \\in \\mathbb{N}^{*},\\ x, y, z$ impairs $\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver toutes les fonctions $f$ de $\\mathbb{R}$ dans $\\mathbb{R}$ telles que pour tous réels $x, y$,\n$$\nf(x f(x)+f(y))=f(x)^{2}+y\n$$",
    "options": [],
    "answer": "f(x) = x for all real x, and f(x) = -x for all real x",
    "solution": "Solution:\n\nEn évaluant l'égalité en $(0, y)$, $f(f(y))=f(0)^{2}+y$. En particulier $f \\circ f$ est bijective, donc $f$ est injective et surjective donc bijective. Soit $a$ tel que $f(a)=0$. En évaluant en $(a, a)$, $f(0)=f(a f(a)+f(a))=f(a)^{2}+a=a$ donc $f(0)=a$. En particulier on obtient que $f(f(0))=0$. Comme pour tout $y, f(f(y))=f(0)^{2}+y$, en évaluant en $y=0$, on obtient $f(0)^{2}=0$ donc $f(0)=0$. En particulier, pour tout $y, f(f(y))=y$.\n\nÉvaluons désormais l'égalité en $(x, 0)$, on obtient $f(x f(x))=f(x)^{2}$. En évaluant l'égalité précédente en $f(x)$ on obtient $f(f(x) f(f(x)))=f(f(x))^{2}$ donc $f(x f(x))=x^{2}$. En particulier $x^{2}=f(x)^{2}$ pour tout $x$ réel, donc pour tout réel $x, f(x)= \\pm x$.\n\nMontrons désormais que pour tout $x, f(x)=x$ ou pour tout $x, f(x)=-x$. Supposons qu'il existe $a$ et $b$ non nuls tels que $f(a)=a$ et $f(b)=-b$. En évaluant l'égalité initiale en $(a, b)$, on obtient que $f\\left(a^{2}-b\\right)=a^{2}+b$, donc $a^{2}+b$ vaut $a^{2}-b$ ou $b-a^{2}$, ainsi $a=0$ ou $b=0$ ce qui est absurde. On en déduit donc que pour tout $x, f(x)=x$ ou pour tout $x, f(x)=-x$.\n\nRéciproquement si $f(x)=x$ pour tout réel $x$, $f(x f(x)+f(y))=f\\left(x^{2}+y\\right)=x^{2}+y=f(x)^{2}+y$ donc $f$ est solution. Si $f(x)=-x$ pour tout réel $x$, $f(x f(x)+f(y))=f\\left(-x^{2}-y\\right)=x^{2}+y=f(x)^{2}+y$ donc $f$ est solution. Les solutions sont donc la fonction identité et son opposé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExiste-t-il des entiers $a$ et $b$ pour lesquels : $a^{5} b+3$ et $a b^{5}+3$ sont tous deux des cubes parfaits?\n\nUn cube parfait est un entier $n$ pour lequel il existe $m \\in \\mathbb{Z}$ tel que : $n = m^{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous allons montrer qu'il n'y a pas d'entiers $a$ et $b$ satisfaisant les conditions du problème.\n\nSoient $a$ et $b$ des entiers tels que $a^{5} b+3$ et $a b^{5}+3$ sont des cubes parfaits. Soit $m$ et $n$ des entiers tels que $a^{5} b+3=m^{3}$ et $a b^{5}+3=n^{3}$.\n\nSupposons que $3 \\mid n$. Alors $27 \\mid n^{3}$. De plus $3 \\mid a b^{5}$ donc $3 \\mid a$ ou $3 \\mid b^{5}$. Si $3 \\mid a$, alors $3^{5} \\mid a^{5}$ donc $3 \\mid a^{5} b+3=m^{3}$. Mais alors $27 \\mid m^{3}$ et $27 \\mid a^{5}$ donc $27 \\mid m^{3}-a^{5} b=3$ ce qui est absurde. Si $3 \\mid b$, alors $3^{5} \\mid b^{5}$ et en particulier $27 \\mid n^{3}-a b^{5}=3$ ce qui est aussi absurde.\n\nOn déduit que 3 ne divise pas $n$ et de même 3 ne divise pas $m$. Si $3 \\mid a$, alors $3 \\mid m$, ce qui est exclu. On déduit donc que 3 ne divise pas non plus $a$ et $b$.\n\nNotons que\n$$\nm^{3}-n^{3}=a^{5} b-a b^{5}=a b(a-b)(a+b)\\left(a^{2}+b^{2}\\right)\n$$\nComme $a$ et $b$ ne sont pas divisibles par $3$, $a^{2} \\equiv b^{2} \\equiv 1 \\bmod 3$ donc $3 \\mid a^{2}-b^{2}=(a-b)(a+b)$ et donc $3 \\mid m^{3}-n^{3}$.\n\nD'après le théorème de Fermat, $m^{3} \\equiv m \\bmod 3$ et $n^{3} \\equiv n \\bmod 3$. On déduit que $m \\equiv m^{3} \\equiv n^{3} \\equiv n \\bmod 3$. Il vient que\n$$\nm^{2}+m n+n^{2} \\equiv 3 m^{2}=0 \\quad \\bmod 3\n$$\nOn déduit que 3 divise $m-n$ et 3 divise $m^{2}+m n+n^{2}$ donc $9 \\mid (m-n)\\left(m^{2}+m n+n^{2}\\right)=m^{3}-n^{3}$. Il vient que $9 \\mid a b(a-b)(a+b)\\left(a^{2}+b^{2}\\right)$.\n\nMais 3 ne divise pas $a$, $b$ et $a^{2}+b^{2} \\equiv 2 \\bmod 3$ donc 3 ne divise pas non plus $a^{2}+b^{2}$ et donc 9 non plus. On déduit que $9 \\mid a^{2}-b^{2}$, soit $a^{2} \\equiv b^{2} \\bmod 9$.\n\nAinsi\n$$\nm^{3}=a^{5} b+3 \\equiv a^{3} \\cdot a^{2} \\cdot b+3 \\equiv a^{3} b^{2} \\cdot b+3 \\equiv (a b)^{3}+3 \\quad \\bmod 9\n$$\nOr les cubes modulo 9 prennent uniquement les valeurs $-1, 0$ et $1$. On ne peut donc avoir $m^{3}-(a b)^{3} \\equiv 3$ mod 9.\n\nLa condition de l'énoncé ne peut donc pas être satisfaite, comme annoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $BCDE$ un carré et soit $O$ son centre. Soit $A$ un point situé à l'extérieur du carré $BCDE$ tel que le triangle $ABC$ est rectangle en $A$. Montrer que le point $O$ appartient à la bissectrice de l'angle $\\widehat{BAC}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLes diagonales du carré $BCDE$ se coupent perpendiculairement en $O$ donc $\\widehat{BOC} = 90^{\\circ} = \\widehat{BAC}$ donc les points $A$, $B$, $O$ et $C$ sont cocycliques. Comme $O$ est sur la médiatrice du segment $[BC]$, $O$ est le pôle Sud de $A$ dans le triangle $ABC$ donc $O$ est sur la bissectrice de l'angle $\\widehat{BAC}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle non isocèle en $A$ et $I$ le centre de son cercle inscrit. On note $D$ le point de la droite $(BC)$ tel que $\\widehat{DIA} = 90^{\\circ}$. On note $E$ le pied de la hauteur issue de $I$ dans le triangle $ADI$. Montrer que le point $E$ est sur le cercle circonscrit du triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn introduit $\\Omega$ le cercle circonscrit du triangle $ABC$ ainsi que $\\omega$ le cercle antarctique de centre $S$, le milieu de l'arc $BC$ ne contenant pas $A$. $\\omega$ passe ainsi par $B$, $C$ et $I$. On note de plus $\\gamma$ le cercle de diamètre $[AI]$, ce cercle recoupe $\\Omega$ en $A$ et $H$.\n\nRegardons les axes radicaux des cercles $\\omega$, $\\Omega$ et $\\gamma$. Comme les cercles $\\gamma$ et $\\Omega$ sont tangents (leurs centres se trouvent sur la bissectrice depuis $A$ dans le triangle $ABC$), leur axe radical est la droite perpendiculaire à $(AC)$ passant par $I$, soit la droite $(DI)$. Les deux autres axes radicaux sont $(BC)$ et $(AH)$. Comme les axes radicaux sont concourants, on a que la droite $(AH)$ passe par $D$, donc les points $A$, $H$ et $D$ sont alignés. Le cercle $\\gamma$ a pour diamètre $[AI]$ donc on a $\\widehat{AHI} = 90^{\\circ}$, ce qui montre que $H$ est le pied de la hauteur issue de $I$ dans le triangle $DAI$, soit $H = E$. Mais par définition $H$ est sur $\\Omega$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\Omega$ et $\\Gamma$ deux cercles sécants. On note $A$ une de leurs intersections. Soit $d$ une droite quelconque passant par le point $A$. On note $P$ et $Q$ les intersections respectives de la droite $d$ avec les cercles $\\Omega$ et $\\Gamma$ différentes de $A$.\nMontrer qu'il existe un point indépendant de la droite $d$ choisie et qui appartient toujours à la médiatrice du segment $[PQ]$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPar symétrie, on peut supposer que le rayon du cercle $\\Omega$ est supérieur au rayon du cercle $\\Gamma$.\nSoit $O_1$ le centre du cercle $\\Omega$ et $O_2$ le centre du cercle $\\Gamma$. Soit $B$ le point tel que le quadrilatère $AO_1BO_2$ soit un parallélogramme. On va montrer que le point $B$ appartient à la médiatrice du segment $[PQ]$. Comme le point $B$ est indépendant du choix de la droite $d$, ceci montrera bien que les médiatrices des segments $[PQ]$ passent par un point fixe lorsque la droite $d$ varie.\n\nPour montrer que $BP = BQ$, on va montrer que les triangles $PO_1B$ et $BO_2Q$ sont isométriques. On sait déjà que $O_1P = O_1A = O_2B$ et $O_1B = O_2A = O_2Q$. Il reste donc à montrer que $\\widehat{BO_1P} = \\widehat{BO_2Q}$.\n\nD'une part\n$$\n\\widehat{PO_1B} = 360^\\circ - \\widehat{PO_1A} - \\widehat{AO_1B} = 360^\\circ - (180^\\circ - 2\\widehat{O_1AP}) - \\widehat{AO_1B} = 180^\\circ + 2\\widehat{O_1AP} - \\widehat{AO_1B}\n$$\nD'autre part\n$$\n\\widehat{BO_2Q} = \\widehat{BO_2A} + \\widehat{AO_2Q} = \\widehat{BO_2A} + (180^\\circ - 2\\widehat{O_2AQ}) = 180^\\circ + \\widehat{BO_2A} - 2\\widehat{O_2AQ}\n$$\nMais $\\widehat{O_1AP} + \\widehat{O_1AO_2} + \\widehat{O_2AQ} = 180^\\circ$ donc $2\\widehat{O_1AP} + 2\\widehat{O_1AO_2} + 2\\widehat{O_2AQ} = 360^\\circ$ et $\\widehat{O_1AO_2} = 180^\\circ - \\widehat{AO_1B} = 180^\\circ - \\widehat{BO_2A}$ donc\n$$\n2\\widehat{O_1AP} + 180^\\circ - \\widehat{AO_1B} + 180^\\circ - \\widehat{BO_2A} + 2\\widehat{O_2AQ} = 360^\\circ\n$$\ndonc\n$$\n2\\widehat{O_1AP} - \\widehat{AO_1B} = \\widehat{BO_2A} - 2\\widehat{O_2AQ}\n$$\nOn trouve bien $\\widehat{PO_1B} = \\widehat{BO_2Q}$ et le point $B$ est bien sur la médiatrice de $[PQ]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\Omega$ son cercle circonscrit. On note $X$ le point d'intersection des tangentes à $\\Omega$ en $B$ et $C$. On note $\\varphi$ l'angle $\\widehat{B A X}$ et $\\mu$ l'angle $\\widehat{X A C}$. On note $Y$ le point de la droite $(A X)$ tel que $\\widehat{A C Y}=\\varphi$. Montrer que $\\widehat{A B Y}=\\mu$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nOn redéfinit les points de l'énoncé de la manière qui nous arrange. On note $\\omega_{1}$ (resp. $\\omega_{2}$) le cercle passant par $A, B$ (resp. $A$ et $C$) et tangent en $A$ à $(A C)$ (resp. $(A B)$). On note $Y'$ l'intersection des cercles $\\omega_{1}$ et $\\omega_{2}$ autre que $A$. Pour conclure il suffit de montrer que les points $Y'$, $A$ et $X$ sont alignés. En effet, on a par théorème de l'angle tangent que $\\widehat{BAY}=\\widehat{A C Y'}$ ainsi que $\\widehat{C A Y'}=\\widehat{A B Y'}$.\n\nRegardons maintenant $\\mathfrak{J}$, l'inversion de centre $A$ composée avec une symétrie d'axe la bissectrice de l'angle $\\widehat{B A C}$ qui échange $B$ et $C$ (il s'agit de ce que l'on appelle une involution projective). Les cercles $\\omega_{1}$ et $\\omega_{2}$ sont alors envoyés respectivement sur les droites parallèles à $(A B)$ et $(A C)$ passant par $C$ et $B$. Le point $Y'$ est donc envoyé sur le point $A'$, le symétrique de $A$ par rapport au milieu de $[B C]$ par l'involution $\\mathcal{J}$, la droite $(A Y')$ est échangée avec la médiane $(A A')$ dans le triangle $A B C$. Comme la droite $(A X)$ est la symédiane on a donc que la droite $(A Y')$ et la droite $(A X)$ sont la même droite ce qui conclut.\nSolution:\n\n![](attached_image_2.png)\nOn pose $\\alpha=\\widehat{B A C}$, $\\beta=\\widehat{A B C}$ et $\\gamma=\\widehat{B C A}$. On va utiliser un autre outil classique lorsque qu'il y a des symédianes en jeu, la chasse aux sinus. On remarque dans un premier temps que d'après le théorème de l'angle tangent :\n$$\n\\widehat{B C X}=\\widehat{C B X}=\\widehat{B A C}=\\alpha\n$$\nLe triangle $B C X$ est donc isocèle en $X$. On a de plus $\\widehat{A B X}=\\alpha+\\beta=180^{\\circ}-\\gamma$. De même, on a $\\widehat{A C X}=180^{\\circ}-\\beta$. En appliquant la loi des sinus dans le triangle $A B X$, puis dans le triangle $A C X$, on trouve\n$$\n\\frac{\\sin (\\gamma)}{\\sin (\\varphi)}=\\frac{\\sin \\left(180^{\\circ}-\\gamma\\right)}{\\sin (\\varphi)}=\\frac{A X}{B X}=\\frac{A X}{C X}=\\frac{\\sin (\\beta)}{\\sin (\\mu)}\n$$\nEn appliquant la loi des sinus cette fois-ci dans le triangle $A B Y$ puis dans le triangle $A B C$, on a\n$$\n\\frac{Y C}{Y A}=\\frac{\\sin (\\mu)}{\\sin (\\varphi)}=\\frac{\\sin (\\beta)}{\\sin (\\gamma)}=\\frac{A C}{B C}\n$$\nCela montre que les triangles $B Y A$ et $A Y C$ sont semblables. En effet, ils ont un angle $\\varphi$ en commun et un rapport en commun\n$$\n\\frac{B A}{Y A}=\\frac{A C}{C Y}\n$$\nCette relation de similitude implique que $\\widehat{A B Y}=\\widehat{Y A C}$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $p$ un nombre premier impair, $h < p$ un entier, $e \\in \\{1,2\\}$.\nOn pose $n = h \\cdot p^{e} + 1$ et on suppose que :\n$$\n\\left\\{\\begin{array}{r}\nn \\mid 2^{n-1} - 1 \\\\\nn \\nmid 2^{h} - 1\n\\end{array}\\right.\n$$\n\nMontrer que $n$ est premier.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $\\omega$ l'ordre de $2$ modulo $n$. Par hypothèse, $\\omega$ divise $n-1$ mais $\\omega$ ne divise pas $h = \\frac{n-1}{p^{e}}$, donc $p$ divise $\\omega$.\n\n$\\omega$ divise $\\varphi(n)$ par le théorème d'Euler donc $p$ divise $\\varphi(n)$. Évidemment, $p$ ne divise pas $n$ donc il existe $q$ premier qui divise $n$ avec $q \\equiv 1 \\pmod{p}$ (d'après la formule donnant $\\varphi$).\n\nOn écrit $n = q^{a} N$ où $N$ est premier avec $q$. Modulo $p$ on a $1 = 1^{a} \\cdot N$ donc $N \\equiv 1 \\pmod{p}$. On va montrer que $n = q$.\n\nDès lors, si $n \\neq q$, soit $a \\geqslant 2$ soit $N \\geqslant p+1$ et donc $n \\geqslant (p+1)^{2} > p^{2} + 1 > p h + 1$ donc $e = 2$ et le raisonnement montre que soit $a = 1$, soit $N = 1$ et dans tous les cas $a \\leqslant 2$ et $N < p^{2}$.\n\nTraitons deux cas :\n\n- Si $a = 2$, $n \\equiv q^{2} \\equiv (1 + k p)^{2} \\equiv 1 + 2 k p \\pmod{p}$ où $q = 1 + k p$. Or $n \\equiv 1 \\pmod{p^{2}}$ donc $p$ divise $2k$ ; $p$ divise $k$ et donc $q \\geqslant p^{2} + 1$ et $n = q^{2} > p^{3} + 1 > p^{2} h + 1 = n$, contradiction.\n\n- Si $a = 1$, $q = 1 + k p$, $N = 1 + l p$ avec $k, l > 0$ (car $q \\equiv N \\equiv 1 \\pmod{p}$). On a obligatoirement $k, l < p$ car sinon $n = q N \\geqslant (1 + p^{2})(1 + p) > p^{3} + 1 > p^{2} h + 1 = n$. De plus $1 + (k + l) p \\equiv n \\equiv 1 \\pmod{p^{2}}$ donc $k + l = p$. Un parmi $k$ et $l$ est donc impair, ce qui oblige $q$ ou $N$ pair, donc dans tous les cas $n$ est pair, ce qui contredit $n \\mid 2^{n-1} - 1$.\n\nDans tous les cas $n = q$ est premier.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $H$ son orthocentre et $M$ le milieu du segment $[BC]$. Soit $d$ une droite passant par le point $M$. On suppose que $d$ coupe le cercle de diamètre $[AH]$ en $P$ et $Q$. Montrer que l'orthocentre du triangle $APQ$ est sur le cercle circonscrit du triangle $ABC$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $N$ le milieu de $[PQ]$ et $K$ l'orthocentre du triangle $APQ$. Si on appelle $K'$ le symétrique de $K$ par rapport au milieu $N$ du segment $[PQ]$, alors $[PQ]$ et $[KK']$ ont le même milieu $N$ donc $PKQK'$ est un parallélogramme donc $\\widehat{K'PA} = \\widehat{K'PQ} + \\widehat{QPA'} = \\widehat{PQK} + \\widehat{QPA'} = 90^{\\circ}$ (car $(PA)$ et $(QK)$ sont perpendiculaires). De même, $\\widehat{K'QA} = 90^{\\circ}$ donc $K'$ est le point diamétralement opposé à $A$ sur le cercle circonscrit à $APQ$, autrement dit $K' = H$ : le symétrique de $K$ par rapport au milieu $N$ du segment $[PQ]$ est $H$.\n\nSoit $X$ le symétrique de $H$ par rapport à $M$ et $Y$ la seconde intersection de la droite $(MH)$ avec le cercle circonscrit au triangle $ABC$. Par un raisonnement analogue à celui mené précédemment, on sait que $X$ est le point diamétralement opposé à $A$ sur le cercle circonscrit à $ABC$ donc $\\widehat{HYA} = \\widehat{XYA} = 90^{\\circ}$ donc $Y$ est sur le cercle de diamètre $[AH]$. Puisque $X$ est le point diamétralement opposé à $A$ sur le cercle circonscrit à $ABC$, il suffit de montrer que $\\widehat{XKA} = 90^{\\circ}$ pour finir l'exercice.\n\nOn décompose l'angle en trois morceaux : $\\widehat{XKA} = 360^{\\circ} - \\widehat{XKH} - \\widehat{HKQ} - \\widehat{QKA}$\n\nLes points $M$ et $N$ sont les milieux respectifs des segments $[HX]$ et $[HK]$ donc les droites $(MN)$ et $(KX)$ sont parallèles. On déduit $\\widehat{XKH} = \\widehat{MNH} = \\widehat{PNH}$.\n\nLes segments $[PQ]$ et $[KH]$ se coupent en leur milieu donc le quadrilatère $PHQK$ est un parallélogramme. $D'$ où $\\widehat{HKQ} = \\widehat{KHP} = \\widehat{NHP}$.\n\nLe point $Q$ est l'orthocentre du triangle $KPA$ donc $\\widehat{QKA} = 180^{\\circ} - \\widehat{QPA}$.\n\nOn déduit\n$$\n\\widehat{XKA} = 360^{\\circ} - \\widehat{PNH} - \\widehat{NHP} - (180^{\\circ} - \\widehat{QPA}) = 180^{\\circ} - \\widehat{PNH} - \\widehat{NHP} + \\widehat{QPA} = \\widehat{NPH} + \\widehat{QPA} = \\widehat{APH} = 90^{\\circ}\n$$\ncar $P$ est sur le cercle de diamètre $[AH]$. Ainsi, $\\widehat{XKA} = 90^{\\circ}$ donc $K$ est bien sur le cercle circonscrit à $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe tel que $AC = BD$ et tel que les côtés $AB$ et $CD$ ne sont pas parallèles. Soit $P$ le point d'intersection des diagonales $(AC)$ et $(BD)$. Soient $E$ et $F$ des points respectivement sur les segments $[BP]$ et $[AP]$ tels que $PC = PE$ et $PD = PF$. Montrer que le cercle circonscrit au triangle formé par les droites $(AB)$, $(CD)$ et $(EF)$ est tangent au cercle circonscrit au triangle $ABP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn note dans un premier temps $X$ l'intersection des droites $(CD)$ et $(FE)$, $Y$ l'intersection des droites $(AB)$ et $(CD)$ ainsi que $Z$ l'intersection des droites $(AB)$ et $(FE)$.\n\nLa première idée de cet exercice est de considérer un point de Miquel (pour savoir ce qu'est un point de Miquel on fait référence au polycopié de T. Budzinski, Transformations Géométriques, Théorème 4.14), que l'on notera $M$ et qui intervient naturellement dans la configuration. En effet, les triangles plats $AFC$ et $BED$ sont isométriques. Ainsi si on considère la similitude directe qui envoie le segment $[AF]$ sur le segment $[EB]$ il s'agit d'une rotation (comme $AF / EB = 1$) et cette rotation envoie donc également le point $C$ sur le point $D$. Comme annoncé, on note $M$ le centre de la rotation (le point de Miquel). $M$ est donc le point de Miquel des quadrilatères complets suivants : $ABFE$, $ABCD$ et $FECD$. En principe l'ordre des points importe comme il y a 3 classes d'ordre pour 4 mêmes points qui donnent 3 points de Miquel différents ; pour les distinguer on adopte ici les notations du Théorème 4.14 cité précédemment.\n\nLe point $M$ est donc sur les cercles circonscrits de plein de triangles et en particulier sur le cercle circonscrit des $ABP$, $ZAF$ et $BDY$, c'est donc un bon candidat pour être notre point de tangence. Dans la suite on va donc montrer que $M$ est également sur le cercle circonscrit au triangle $XYZ$ et que c'est le point de tangence entre les cercles circonscrits des triangles $XYZ$ et $ABP$.\n\nOn remarque maintenant que le quadrilatère $DCEF$ est un trapèze isocèle avec $(CE) // (DF)$. $M$ étant le point d'intersection des cercles circonscrits à $PFE$ et $PCD$ autre que $P$, on en déduit que $M$ est sur l'axe de symétrie du trapèze $FDCE$. $M$ est donc sur la bissectrice extérieure issue de $P$ dans les triangles $PFE$ et $PCD$, en particulier $M$ est le pôle Nord de ces triangles. Par les propriétés classiques du point de Miquel, $M$ est également le centre de la similitude directe qui envoie $F$ sur $A$ et $E$ sur $B$. Donc $MFE \\sim MAB$, en particulier $MA = MB$ et $M$ est alors le pôle Nord dans le triangle $ABP$.\n\nPour conclure il suffit de montrer que $M$ est le pôle Nord dans le triangle $XYZ$. On montre dans un premier temps que cette assertion implique bien la fin de l'exercice.\n\nSi $M$ est le pôle Nord du triangle $XYZ$, alors il est bien sur le cercle circonscrit de ce triangle. Soit $t$ (resp. $t'$) la tangente au cercle circonscrit de $ABP$ (resp. $XYZ$) en $M$. On voudrait montrer que $t$ et $t'$ sont la même droite. Comme $M$ est le pôle Nord, on a $t \\parallel (AB)$. Mais de la même manière $t' \\parallel (ZY)$. Ce qui montre bien que $t \\parallel t'$ et donc qu'il s'agit de la même droite.\n\nPour conclure on va montrer que $M$ est sur la médiatrice du segment $[ZY]$. Comme $M$ est sur la bissectrice extérieure issue de $X$ (car $M$ est sur la médiatrice commune de $(DF)$ et $(EC)$ et les droites $(EF)$ et $(DC)$ se coupent en $X$), cela conclurait bien. Il suffit donc de montrer que $(MZ, MA) = (MB, MY)$. Or on a plein de cocyclicités qui apparaissent avec le point de Miquel et le trapèze $DFEC$.\n\n$$\n\\begin{aligned}\n(MZ, MA) &= (FZ, FA) \\\\\n&= (FE, FC) \\\\\n&= (DE, DC) \\\\\n&= (MB, MY)\n\\end{aligned}\n$$\n\n($ZAFM$ cyclique)\n\n(angles opposés)\n\n($FECD$ cyclique)\n\n($MBYD$ cyclique).\n\nCe qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $m, n \\geqslant 2$ des entiers vérifiant la propriété suivante:\n$$\na^{n} \\equiv 1 \\quad(\\bmod m) \\quad a=1, \\ldots, n\n$$\nProuver que $m$ est un nombre premier et que $n = m - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par supposer que $m = p$ est un nombre premier. On doit donc montrer $n = p - 1$. Si $n \\geqslant p$, on a $p^{n} \\equiv 1 (\\bmod p)$, une évidente contradiction. Donc $n < p$.\n\nConsidérons $T = X^{n} - 1$ le polynôme de $\\mathbb{Z} / p \\mathbb{Z}[X]$. Il a pour racines par hypothèse $1, \\ldots, n$, qui sont au nombre de $n$ et distinctes, et, puisque $\\mathbb{Z} / p \\mathbb{Z}$ est intègre, et que $T$ est de degré $n$, ce sont les seules. Donc $T = (X-1) \\cdot \\ldots \\cdot (X-n)$. En regardant le coefficient en $X^{n-1}$ de $T$, on voit que $0 \\equiv 1 + \\ldots + n (\\bmod p)$ par les formules de Viète. Ainsi, $p \\mid \\frac{n(n+1)}{2}$, donc (puisque $n < p$) on a nécessairement $n = p - 1$.\n\nRevenons au cas général. Soit $p$ un diviseur premier de $m$. On a $n = p - 1$ puisque l'hypothèse de l'énoncé reste vraie pour $m = p$. Dès lors, si $q$ est un autre diviseur premier de $m$, $q - 1 = n = p - 1$ et $q = p$. Il reste à voir que $m$ ne peut pas être une puissance de $p$ autre que $p$. Il suffit clairement de le voir pour $m = p^{2}$. Or $(p-1)^{p-1} \\equiv (-1)^{p-1} + p \\times (-1)^{p-2} (p^{p}) (\\bmod p^{2})$ par le binôme de Newton. Puisque $p-1$ est pair, et que $p^{2} \\nmid -p(p-1)$, on a bien $(p-1)^{p-1} \\not\\equiv 1 (\\bmod p^{2})$. Ceci achève la démonstration.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 30209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $1 < t < 2$ un nombre réel. Montrer que pour tout entier $d$ suffisamment grand, il existe un polynôme $P(X) = \\sum_{k=0}^{d} a_{k} X^{k}$, avec $a_{d} = 1$ et $a_{0}, \\ldots, a_{d-1} \\in \\{1, -1\\}$, tel que :\n$$\n|P(t) - 2019| \\leqslant 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoient $d \\geqslant 2019$ un entier et $a_{1} < \\ldots < a_{k}$ les $d$-uplets de la forme $(\\pm 1, \\ldots, \\pm 1)$, ordonnés par ordre lexicographique (en partant de la droite, c'est-à-dire que $a_{i} > a_{j}$ si et seulement si le premier terme non nul de $a_{i} - a_{j}$ en partant de la droite est $> 0$), avec $k = 2^{d}$.\n\nPour passer de $a_{i}$ à $a_{i+1}$, on regarde le premier $-1$ en partant de la droite dans $a_{i}$, on le change par un $1$, et on remplace tous les $1$ à sa droite par des $-1$. Par exemple si $d = 3$, les triplets sont dans l'ordre $a_{1} = (-1, -1, -1)$, $a_{2} = (-1, -1, 1)$, $a_{3} = (-1, 1, -1)$, $a_{4} = (-1, 1, 1)$, $a_{5} = (1, -1, -1)$, $a_{6} = (1, -1, 1)$, $a_{7} = (1, 1, -1)$ et $a_{8} = (1, 1, 1)$.\n\nOn associe à $a_{i} = (a_{i1}, \\ldots, a_{id})$ le polynôme $P_{i}(X) = \\sum_{j=1}^{d} a_{ij} X^{j-1}$, de telle sorte que $M := -P_{1}(t) = P_{k}(t) = 1 + \\ldots + t^{d-1} > d$.\n\nOr $|P_{i+1}(t) - P_{i}(t)| = 2|t^{l} - (1 + \\ldots + t^{l-1})|$ pour un certain $l \\geqslant 0$ (qui dépend de $i$) pour tout $i$.\n\nMontrons $t^{l} - (1 + \\ldots + t^{l-1}) \\leqslant 1$ : cela équivaut à $t^{l+1} - t^{l} - (t^{l} - 1) \\leqslant t - 1$ après multiplication par $t-1$, soit après réarrangement $t^{l}(2-t) + t \\geqslant 2$, ce qui est vrai car $t^{l} \\geqslant 1$.\n\nDès lors, $P_{i+1}(t)$ est plus grand que $P_{i}(t)$ d'au plus $2$, ce qui montre que, puisque $2019 - t^{d} \\in [-M ; M]$, il existe $i$ tel que $|P_{i}(t) - (2019 - t^{d})| \\leqslant 1$, ce qui conclut en posant $P = X^{d} + P_{i}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $H$ son orthocentre. On note $\\ell_{1}$ et $\\ell_{2}$ deux droites passant par $H$ et perpendiculaires entre elles. On note $X_{1}, Y_{1}$ et $Z_{1}$ les points d'intersection de $\\ell_{1}$ avec les droites $(BC)$, $(CA)$ et $(AB)$ respectivement. On note également $X_{2}, Y_{2}$ et $Z_{2}$ les points d'intersection de $\\ell_{2}$ avec les droites $(BC)$, $(CA)$ et $(AB)$ respectivement. Montrer que les milieux des segments $[X_{1}X_{2}], [Y_{1}Y_{2}]$ et $[Z_{1}Z_{2}]$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nL'approche présentée dans ce corrigé est longue, mais purement élémentaire et les nombreuses étapes doivent toutes paraître naturelles.\n\nPremière étape : les milieux. L'introduction d'un milieu dans un exercice est toujours une épine dans notre pied car cette condition est difficile à utiliser avec notre outil favori : la chasse aux angles. On peut donc se demander comment retirer cette épine ; ici, une solution doit paraître raisonnable. Du fait de l'angle droit $\\widehat{X_{1}HX_{2}}$, le milieu de $[X_{1}X_{2}]$ est naturellement le milieu du diamètre du cercle circonscrit au triangle $HX_{1}X_{2}$, donc le centre de ce cercle, qui sera maintenant nommé $\\Gamma_{A}$.\n\nSur notre figure, on trace donc trois cercles $\\Gamma_{A}$, $\\Gamma_{B}$ et $\\Gamma_{C}$, où $\\Gamma_{B}$ et $\\Gamma_{C}$ sont définis de manière similaire. On peut ainsi arrêter de penser aux milieux comme des milieux et on y pense comme des centres de cercles.\n\nDeuxième étape : Effacer les centres. Manipuler des centres de cercles est souvent plus agréable que manipuler des milieux de segments, mais ce n'est pas encore tout à fait satisfaisant. En traçant nos trois nouveaux cercles, on constate qu'ils ont l'air de se couper à nouveau en un autre point que $H$. Devant une telle observation, une première interrogation est de se demander ce qu'une telle information impliquerait par symétrie. Par exemple, sur ma figure, peut-être que $B$ semble se trouver sur $\\Gamma_{A}$, mais si c'était toujours le cas, par symétrie, $C$ devrait aussi se situer sur $\\Gamma_{A}$, et sur ma figure ce n'est pas du tout le cas, ma conjecture est donc à éliminer.\n\nIci pas de tel argument, la concourance des trois cercles n'entraîne rien par symétrie. Mieux, regardons maintenant ce qui découlerait de la concourance de ces trois cercles en un point $P$. Les trois centres seraient sur la médiatrice de $[HP]$, corde commune aux trois cercles, ce qui conclurait ! Dans l'autre sens, si les trois centres sont alignés, alors le symétrique de $H$ par rapport à la droite des trois centres est sur chaque cercle, ils se recoupent donc bien. À la place de l'alignement des trois centres, on peut montrer la concourance des trois cercles, une propriété a priori bien plus agréable !\n\nInterlude : Comment savoir si on est allé dans la bonne direction ? C'est généralement une question difficile mais ici on peut répondre par l'affirmative car on peut effacer des morceaux de la figure (ici les trois milieux) et la simplifier. On a tracé le triangle, les deux droites et les 6 points d'intersection, préfère-t-on tracer trois milieux et montrer leur alignement ou trois cercles et montrer leur concourance ? La deuxième option.\n\nMême si les deux options semblent similaires en complexité (ce qui n'est pas la difficulté), un problème de géométrie olympique maximise sa difficulté par rapport à sa complexité donc un énoncé équivalent et de complexité comparable sera presque toujours plus facile, sinon cet énoncé aurait été posé à la place. On est maintenant convaincu d'aller dans la bonne direction, et on n'aura pas peur de s'éloigner encore plus du problème de base. C'est souvent une bonne chose, tant qu'on garde l'équivalence avec le problème de base bien sûr.\n\nTroisième étape : Les propriétés du point $P$. Après une chasse aux angles naïve infructueuse, on cherche des propriétés des cercles $\\Gamma$ ou du point $P$. Deux possibilités : on constate que $P$ est sur le cercle circonscrit à $ABC$, $\\Omega$, parce qu'on a tracé ce cercle. En effet, dans un tel exercice centré sur un triangle $ABC$, on trace toujours son cercle circonscrit sur au moins une figure, si ce n'est sur toutes. Une autre manière d'effectuer cette conjecture présuppose une certaine habitude avec le point de Miquel. Le point d'intersection de $\\Gamma_{B}$ et de $\\Gamma_{C}$ est le point de Miquel de $\\ell_{1}, \\ell_{2}, (AB), (AC)$. Dire qu'il est sur $\\Gamma_{A}$ revient à dire que c'est un point de Miquel du pentalatère $\\ell_{1}, \\ell_{2}, (AB), (AC), (BC)$ (tout cercle circonscrit au trilatère formé par 3 de ces droites passe par $P$).\n\nUne bonne manière de se représenter les points de Miquel de pentalatère (qui n'existent pas pour n'importe quel pentalatère) est de se rappeler de la droite de Simson (ou Steiner). Un point est sur le cercle circonscrit à un trilatère (ou triangle) si et seulement si ses projetés (orthogonaux) sur les trois côtés sont alignés, donc un point est le point de Miquel de 4 droites si et seulement ses 4 projetés sont alignés, et de 5 si et seulement si ses 5 projetés sont alignés. Avec cette vision des choses, en supposant $P$ sur les trois cercles $\\Gamma$, les projetés de $P$ sur les côtés sont tous sur la droite reliant les deux projetés de $P$ sur $\\ell_{1}$ et sur $\\ell_{2}$, ils sont donc alignés et $P$ est sur $\\Omega$.\n\nPourquoi avoir choisi $\\Omega$ alors que $P$ est sur 10 cercles, littéralement ? C'est le seul qui ne fait intervenir que des points déjà introduits et surtout qui conserve la symétrie entre $A, B$ et $C$. Pour montrer la concourance de $\\Gamma_{A}, \\Gamma_{B}$ et $\\Gamma_{C}$, on peut montrer que $\\Gamma_{B}$ et $\\Gamma_{C}$ se coupent sur $\\Omega$, en un point qui sera alors aussi sur $\\Gamma_{A}$.\n\nInterlude : va-t-on dans la bonne direction ? Que préfère-t-on, montrer la concourance de $\\Gamma_{A}, \\Gamma_{B}$ et $\\Gamma_{C}$, ou bien celle de $\\Gamma_{B}, \\Gamma_{C}$ et $\\Omega$ ? Avantage du premier : cela conserve la symétrie. Avantage du deuxième : c'est plus loin de l'énoncé sans être plus complexe. Le deuxième gagne.\n\nQuatrième étape : Il nous reste une hypothèse : $\\ell_{1}$ et $\\ell_{2}$ se coupent en l'orthocentre. J'avais oublié. Corrigeons ça. Comment utiliser l'hypothèse d'orthocentre ? On ne souhaite surtout pas tracer des hauteurs qui ne passent par rien ici. Par contre, le cercle circonscrit est déjà tracé et important, et on sait que de nombreux symétriques de $H$ s'y trouvent. Lesquels choisir ? Par rapport aux côtés ou aux milieux des côtés ? Cette fois, il suffit de se poser la question pour avoir la réponse : les symétriques par rapport aux côtés sont aussi sur les cercles $\\Gamma$, car ceci sont stables par la symétrie axiale ! Ils sont donc beaucoup plus intéressants : c'est clairement la bonne manière d'utiliser l'hypothèse de milieu. Nommons les $A', B'$ et $C'$.\n\nConclusion : On a reformulé tous les éléments de l'énoncé en des éléments équivalents ($H$ est bien le seul point dont les trois symétriques sont sur le cercle circonscrit du triangle $ABC$) et utilisables pour faire une chasse aux angles. On a conservé la symétrie entre $B$ et $C$, essayons de la conserver encore pour la chasse aux angles (ce n'est pas toujours possible). On définit $P$ comme le point d'intersection autre que $H$ de $\\Gamma_{B}$ et $\\Gamma_{C}$. On veut montrer qu'il se trouve sur $\\Omega$. Plutôt que de montrer qu'il est cocyclique avec $A, B, C$, montrons qu'il est cocyclique avec $AB'C'$. En effet, on ne sait rien de la direction de la droite $PB$ ou $PC$, alors que $PB'$ et $PC'$ sont des cordes de $\\Gamma_{B}$ et $\\Gamma_{C}$. De plus, l'angle $\\widehat{B'AC'}$ est connu (c'est-à-dire qu'on peut l'exprimer avec des opérations simples en fonction des angles du triangle $ABC$).\n\nOn veut donc trouver la valeur connue de $\\widehat{B'PC'}$. Pour exploiter la définition de $P$, on éclate cet angle en $\\widehat{B'PH}$ et $\\widehat{HPC'}$. On cherche donc une relation simple entre ces deux angles. Ce sont des angles inscrits, qui interceptent un arc de deux fois la longueur de $HY_{1}$ d'un côté de $HZ_{1}$ de l'autre. On cherche donc une relation entre l'angle $\\widehat{HY_{2}Y_{1}}$ et $\\widehat{Z_{1}Z_{2}H}$, mais ces deux angles sont entre la même droite $\\ell_{2}$ et les côtés du triangle. On a donc une relation, qui, en remontant tout, donne une relation simple entre $\\widehat{B'AC'}$ et $\\widehat{B'PC'}$, qui ne peut être que la somme des deux valant $180^{\\circ}$, ce qui terminerait. On est maintenant convaincu que cette petite chasse aux angles va terminer ce problème, il suffit de l'écrire pour finir la preuve de l'exercice.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $I$ le centre de son cercle inscrit. La perpendiculaire à la droite $(AI)$ passant par le point $I$ coupe la droite $(AB)$ en un point $D$ et la droite $(AC)$ en un point $E$. On suppose qu'il existe deux points $F$ et $G$ sur le segment $[BC]$ tels que $BA = BF$ et $CA = CG$. Soit $T$ le point d'intersection des cercles circonscrits aux triangles $ADF$ et $AEG$.\n\nMontrer que le centre du cercle circonscrit au triangle $AIT$ se trouve sur la droite $(BC)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoient $M$ et $N$ les milieux respectifs des segments $[AD]$ et $[AE]$ et soit $X$ le point d'intersection de la droite $(MN)$ (qui est aussi la médiatrice du segment $[AI]$) avec la droite $(BC)$. Soient $O_1$ et $O_2$ les centres respectifs des cercles circonscrits aux triangles $ADF$ et $AEG$. On souhaite montrer que le point $X$ est le centre du cercle circonscrit à $AIT$ donc qu'il appartient à la médiatrice du segment $[AT]$, c'est-à-dire à la droite $(O_1O_2)$.\n\nSoient $P, Q$ et $R$ les points de contact respectifs du cercle inscrit au triangle $ABC$ avec les côtés $[BC], [AC]$ et $[AB]$. Le point $O_1$ est sur la médiatrice du segment $[AD]$ donc la droite $(O_1M)$ est perpendiculaire à la droite $(AB)$. Les droites $(O_1M)$ et $(IR)$ sont donc parallèles. D'après le théorème de Thalès, on déduit que\n$$\n\\frac{O_1I}{O_1B} = \\frac{MR}{MB}.\n$$\nOn a de même\n$$\n\\frac{O_2C}{O_2I} = \\frac{NC}{NQ}.\n$$\nDe plus, le triangle $ABF$ est isocèle en $B$ donc la médiatrice de $[AF]$ est la bissectrice de $\\widehat{ABF}$ donc $O_1$ est sur $(BI)$ et de même $O_2$ est sur $(CI)$.\n\nD'après le théorème de Ménélaus appliqué aux points $M, N, X$ dans le triangle $ABC$ :\n$$\n\\frac{XB}{XC} \\cdot \\frac{NC}{NA} \\cdot \\frac{MA}{MB} = 1\n$$\net en l'appliquant au triangle $BIC$, l'alignement des points $O_1, O_2$ et $X$ équivaut à\n$$\n\\frac{O_1I}{O_1B} \\cdot \\frac{XB}{XC} \\cdot \\frac{O_2C}{O_2I} = 1.\n$$\nOn souhaite donc montrer cette égalité.\n\nOn sait déjà\n$$\n\\frac{XB}{XC} = \\frac{NA}{NC} \\cdot \\frac{MB}{MA}, \\quad \\frac{O_1I}{O_1B} = \\frac{MR}{MB}, \\quad \\frac{O_2C}{O_2I} = \\frac{NC}{NQ}\n$$\ndonc\n$$\n\\frac{O_1I}{O_1B} \\cdot \\frac{XB}{XC} \\cdot \\frac{O_2C}{O_2I} = \\frac{MR}{MB} \\cdot \\frac{NA}{NC} \\cdot \\frac{MB}{MA} \\cdot \\frac{NC}{NQ} = \\frac{MR \\cdot NA}{MA \\cdot NQ} = 1\n$$\ncar $MA = \\frac{1}{2} DA = \\frac{1}{2} EA = NA$ et $AR = AQ$ donc $MR = NQ$. Ainsi, $O_1, O_2, X$ sont alignés.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $S$ un ensemble non vide d'entiers strictement positifs vérifiant la propriété suivante : Pour tous entiers $a, b \\in S$, l'entier $ab+1$ appartient aussi à $S$.\n\nMontrer que l'ensemble des nombres premiers ne divisant aucun des éléments de $S$ est fini.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $p$ un nombre premier et soit $a_{1}, a_{2}, \\ldots, a_{k}$ les restes possibles des éléments de $S$ modulo $p$. On suppose que $0$ n'appartient pas à $R=\\{a_{1}, \\ldots, a_{k}\\}$, c'est-à-dire que $p$ ne divise aucun élément de $S$.\n\nOn sait que pour tout $i, j$, $a_{i} a_{j}+1 \\in R$. Notons que si $j$ et $l$ sont distincts, alors $a_{i} a_{j}+1$ et $a_{i} a_{l}+1$ sont distincts. En effet\n$$\na_{i} a_{j}+1-\\left(a_{i} a_{l}+1\\right) \\equiv a_{i}\\left(a_{j}-a_{l}\\right) \\not\\equiv 0 \\pmod{p}\n$$\nAinsi pour $i$ fixé, $R=\\{a_{1}, \\ldots, a_{k}\\}=\\{a_{i} a_{1}+1, \\ldots, a_{i} a_{k}+1\\}$. Il vient que\n$$\na_{1}+\\ldots+a_{k} \\equiv a_{i} a_{1}+1+\\ldots+a_{i} a_{k}+1 \\equiv a_{i}\\left(a_{1}+\\ldots+a_{k}\\right)+k \\pmod{p}\n$$\nSi $a_{1}+\\ldots+a_{k} \\equiv 0 \\pmod{p}$, alors $k \\equiv 0 \\pmod{p}$ donc $k=p$ et $0 \\in R$ ce qui est contraire à l'hypothèse. On déduit que $a_{1}+\\ldots+a_{k}$ est inversible modulo $p$ et\n$$\na_{i} \\equiv \\frac{k}{a_{1}+\\ldots+a_{k}} \\pmod{p}\n$$\nCe terme ne dépend pas de $i$. Donc les $a_{i}$ sont égaux et $R$ est un singleton noté $\\{a\\}$. Ainsi $a^{2}+1 \\equiv a \\pmod{p}$ donc $p \\mid a^{2}+1-a$ et $p \\leqslant n^{2}-n+1$ pour tout $n$ dans $S$. Si $n_{0}$ est le plus petit élément de $S$, alors en particulier $p \\leqslant n_{0}^{2}-n_{0}+1$.\n\nEn conclusion, si $p$ ne divise aucun élément de $S$, $p$ est borné par $n_{0}^{2}-n_{0}+1$. Ainsi l'ensemble des nombres premiers ne divisant aucun élément de $S$ est fini.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $P$, $Q$ deux polynômes à coefficients réels, non constants et premiers entre eux. Montrer qu'il existe au plus trois réels $\\lambda$ tels que:\n$$\nP+\\lambda Q=R^{2}\n$$\noù $R \\in \\mathbb{R}[X]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn procède par l'absurde en supposant l'existence de quatre réels distincts $\\lambda_{1}, \\ldots, \\lambda_{4}$ et quatre polynômes $R_{1}, \\ldots, R_{4}$ tels que $P+\\lambda_{i} Q=R_{i}^{2}$ pour $i=1,2,3,4$.\n\nOn a $P'+\\lambda_{i} Q'=2 R_{i} R_{i}'$ et donc $R_{i} \\mid Q'\\left(P+\\lambda_{i} Q\\right)-Q\\left(P'+\\lambda_{i} Q'\\right)=P Q'-P' Q$.\n\nRemarquons que si $T \\mid R_{i}, R_{j}$ alors $T \\mid R_{i}^{2}-R_{j}^{2}=\\left(\\lambda_{i}-\\lambda_{j}\\right) Q$ puis $T \\mid R_{i}^{2}-\\lambda_{i} Q=P$ et donc $T$ est constant, car $P$, $Q$ sont premiers entre eux.\n\nDès lors les $R_{i}$ sont deux à deux premiers entre eux et ainsi $R_{1} R_{2} R_{3} R_{4} \\mid P Q'-P' Q$.\n\nSoient $A, B$ des polynômes non nuls ; le degré de $A+\\lambda B$, si $\\lambda \\in \\mathbb{R}$, est égal à $\\max \\{\\operatorname{deg}(A), \\operatorname{deg}(B)\\}$, sauf si $\\operatorname{deg}(A)=\\operatorname{deg}(B)$ et $\\lambda=-\\frac{a}{b}$ où $a, b$ sont les coefficients dominants de $A$ et $B$ respectivement, auquel cas il est plus petit.\n\nSi $A=P$ et $B=Q$, cela montre que, soit $\\operatorname{deg} R_{i}=\\frac{1}{2} \\max \\{\\operatorname{deg}(P), \\operatorname{deg}(Q)\\}$ pour tout $i$, et donc que $\\operatorname{deg}\\left(R_{1} R_{2} R_{3} R_{4}\\right)=2 \\max \\operatorname{deg}\\{P, Q\\}>\\operatorname{deg}(P)+\\operatorname{deg}(Q)-1 \\geqslant \\operatorname{deg}\\left(P Q'-P' Q\\right)$, soit il y a un $R_{i}$, tel que $\\operatorname{deg}\\left(R_{i}\\right)<\\operatorname{deg}\\left(R_{j}\\right)$ pour $j \\neq i ;$ alors pour $A=P+\\lambda_{i} Q=: P_{1}$ et $B=Q$ on voit que $\\operatorname{deg}\\left(R_{i}\\right)=\\frac{1}{2} \\operatorname{deg}\\left(P_{1}\\right)$ et $\\operatorname{deg} R_{j}=\\frac{1}{2} \\max \\left\\{\\operatorname{deg}\\left(P_{1}\\right), \\operatorname{deg}(Q)\\right\\}$ et $\\operatorname{donc} \\operatorname{deg}\\left(R_{1} R_{2} R_{3} R_{4}\\right)=\\frac{3}{2} \\max \\left\\{\\operatorname{deg}\\left(P_{1}\\right), \\operatorname{deg}(Q)\\right\\}+\\frac{1}{2} \\operatorname{deg}\\left(P_{1}\\right)>$\n$\\operatorname{deg}(Q)+\\operatorname{deg}\\left(P_{1}\\right)-1 \\geqslant \\operatorname{deg}\\left(P_{1} Q'-P_{1}' Q\\right)=\\operatorname{deg}\\left(PQ'-P' Q\\right)$.\n\nPar un argument sur les degrés, on a donc nécessairement $P Q'-P' Q=0$ et ainsi $P \\mid P' Q$. Or $P$ et $Q$ sont premiers entre eux donc $P \\mid P'$, ce qui est une évidente contradiction à $P$ non constant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\omega$ son cercle inscrit dont les points de tangence sur $[BC]$, $[CA]$ et $[AB]$ sont respectivement $D$, $E$ et $F$. On note $\\Omega$ le cercle tangent à $\\omega$ et passant par $B$ et $C$ et $T$ le point de tangence de ces deux cercles. On note $X$ et $Y$ les milieux des segments $[DE]$ et $[DF]$. On note $U$ et $V$ les seconds points d'intersection des droites $(TX)$ et $(TY)$ avec $\\Omega$. On note également $Z$ et $W$ les seconds points d'intersection des droites $(AB)$ et $(AC)$ avec $\\Omega$. Montrer que $(ZU)$, $(WV)$, $(XY)$ et la bissectrice de l'angle $\\widehat{BAC}$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPendant toute la preuve on va utiliser les résultats du lemme qui suit. On commence d'abord par une notation.\nSoit $ABC$ un triangle, on note $\\Omega$ le cercle circonscrit à $ABC$, on note également $\\omega$ le cercle tangent à $(AB)$, $(AC)$ et intérieurement à $\\Omega$, on dit que ce cercle est le cercle mixtilinéaire (intérieur) du triangle $ABC$ depuis $A$.\n\nLemme 3. Soit $ABC$ un triangle, on note $I$ le centre de son cercle inscrit, on note de plus $\\Omega$ le cercle circonscrit à $ABC$ ainsi que $\\omega$ le cercle mixtilinéaire depuis $A$. On note $T$ (resp. $D$ et $F$) le point de tangence de $\\omega$ avec $\\Omega$ (resp. $(AC)$ et $(AB)$) ainsi que $N$ le pôle nord (le milieu de l'arc $BC$ qui contient $A$). Sous ces hypothèses et notations on sait alors que $I$ est le milieu du segment $[DF]$ et que les points $T$, $I$ et $N$ sont alignés.\n\nDémonstration.\n\n![](attached_image_1.png)\n\nNotons $E$ et $W$ les milieux respectifs des arcs $AC$ et $AB$. D'après le lemme du bocal on sait que les points $W$, $D$ et $T$ sont alignés ainsi que les points $T$, $F$ et $E$. On sait de plus d'après le théorème du pôle sud que les points $B$, $I$ et $E$ sont alignés ainsi que les points $C$, $I$ et $W$. On peut alors appliquer le théorème de Pascal à l'hexagone $BETWCA$ pour conclure que les points $D$, $I$ et $F$ sont alignés. Comme l'axe $(AI)$ est un axe de symétrie qui échange $(AB)$ avec $(AC)$ et $\\omega$ avec lui-même on peut en déduire que le point $I$ est bien le milieu du segment $[EF]$.\n\nPour la deuxième partie du lemme, on note $\\beta$ et $\\gamma$ les angles en $B$ et $C$ respectivement dans le triangle $ABC$. Dans le triangle $TDF$ on remarque que la droite $(TI)$ est une médiane d'après la première partie du lemme. De plus, par des propriétés classiques de la construction de la symédiane (voir par exemple le cours d'Alexander Semenov au stage de Valbonne 2022 ou bien celui de Thomas Budzinski à Montpellier 2016) la droite $(TA)$ est la symédiane. Ainsi, $\\widehat{ITC} = \\widehat{ITF} + \\widehat{FTC} = \\widehat{ATD} + \\widehat{FTC} = \\gamma/2 + \\beta/2 = \\widehat{BTI}$ où la dernière égalité découle par symétrie. Cela conclut la deuxième partie du lemme.\n\n![](attached_image_2.png)\n\nOn peut maintenant passer à la preuve de l'exercice.\n\nEn se plaçant dans le triangle $ZBC$ on remarque que $\\omega$ est le cercle mixtilinéaire depuis $B$. Ainsi, d'après le Lemme 3, le point $Y$ est le centre du cercle inscrit de ce triangle et de plus, la droite $(TY)$ passe par le pôle Nord issu de $B$ dans $ZBC$, ce point est le point $V$ de l'énoncé. De la même manière le point $X$ est le centre du cercle inscrit du triangle $WBC$ et $U$ est le milieu de l'arc $BW$ qui contient $C$. Cela montre en particulier que la droite $(WW)$ bissecte extérieurement l'angle $\\widehat{AWZ}$, de la même manière la droite $(ZU)$ est la bissectrice extérieure de l'angle $\\widehat{WZA}$. Ainsi il est maintenant clair que dans le triangle $AWZ$ les droites $(WV)$, $(ZU)$ et la bissectrice issue de $A$ sont concourantes en un point $P$.\n\nOn veut également démontrer que la droite $(YZ)$ passe par ce point $P$. On veut pour cela appliquer le théorème de Pascal à l'hexagone $TUZSWV$, où $S$ dénote le milieu de l'arc $BC$ sur le cercle $\\Omega$. Mais la droite $(ZY)$ est bien la bissectrice de l'angle $\\widehat{BZC}$ comme $Y$ est le centre du cercle inscrit au triangle $ZBC$. De la même manière la droite $(WX)$ passe bien par le point $S$, on peut alors conclure en appliquant le théorème de Pascal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $n \\geqslant 1$ pour lesquels la fonction :\n$$\nx \\longmapsto x^{x}\n$$\nprenne toutes les valeurs possibles modulo $n$ lorsque $x$ parcourt $\\llbracket 0, n-1 \\rrbracket$.\nOn dit que c'est une surjection dans $\\mathbb{Z} / n \\mathbb{Z}$.",
    "options": [],
    "answer": "Exactly those n that are squarefree and whose distinct prime factors p and q satisfy that no prime p divides q minus one; equivalently, n is a product of distinct primes with the property that for any two distinct prime factors p_i and p_j, p_i does not divide p_j − 1. (This includes n = 1.)",
    "solution": "Solution:\n\nNotons déjà que si $n$ vérifie la propriété, alors tout diviseur $d$ de $n$ la vérifie également. Posons $f$ l'application qui à $x$ dans $\\mathbb{N}^{*}$ associe $x^{x}$ dans $\\mathbb{Z} / n \\mathbb{Z}$.\n\nPour $n = p^{2}$ avec $p$ premier, notons que $p$ n'est pas dans l'image de $f$. En effet, si $f(x) \\equiv p \\pmod{p^{2}}$, $p$ divise $x^{x}$ donc $p$ divise $x$. En particulier, comme $x \\geqslant p \\geqslant 2$, $p^{2}$ divise $x^{x} = f(x)$, contradiction. En particulier, si $n$ vérifie la propriété, dans sa décomposition en facteurs premiers toutes les valuations $p$-adiques valent au plus 1.\n\nPour $n = p$ premier, donnons-nous $y \\in \\mathbb{N}$. On cherche $x$ entier positif tel que $x^{x} \\equiv y \\pmod{p}$. Pour cela, on utilise le petit théorème de Fermat : si $p-1$ divise $x-1$, $x^{x} \\equiv x^{x-1} \\times x \\equiv x \\pmod{p}$. En particulier, il suffit de prendre $x$ strictement positif tel que $x \\equiv 1 \\pmod{p-1}$ et $x \\equiv y \\pmod{p}$. Ceci est possible par le théorème des restes chinois, car $p$ et $p-1$ sont premiers entre eux, donc $f$ est surjective.\n\nIl reste à traiter le cas où $n$ est un produit de facteurs premiers deux à deux distincts : supposons $n = p_{1} \\ldots p_{k}$ avec les $p_{i}$ premiers et deux à deux distincts. On peut réessayer d'utiliser l'argument précédent : soit $y \\in \\mathbb{N}$, on cherche $x$ entier strictement positif tel que $x^{x} = y \\pmod{n}$, ceci est équivalent par théorème des restes chinois à $x^{x} = y \\pmod{p_{i}}$ pour tout $i$ entre 1 et $k$. Il suffit de prendre $x$ tel que $x \\equiv 1 \\pmod{p_{i}-1}$ et $x \\equiv y \\pmod{p_{i}}$ pour tout $i$, c'est-à-dire $x \\equiv 1 \\pmod{\\prod_{i=1}^{k}(p_{i}-1)}$ et $x \\equiv y \\pmod{p_{i}}$ pour tout $i$. Notons que les $p_{i}$ sont deux à deux premiers entre eux, mais à priori on ne peut pas affirmer que $p_{i}$ est premier avec $\\prod_{i=1}^{k}(p_{i}-1)$ pour tout $i$.\n\nSi pour tout $(i, j)$ entre 1 et $k$, $p_{i}$ ne divise pas $p_{j}-1$, dans ce cas $p_{i}$ est premier avec $\\prod_{i=1}^{k}(p_{i}-1)$ pour tout $i$, donc par théorème des restes chinois on peut trouver $x$ tel que $x \\equiv 1 \\pmod{\\prod_{i=1}^{k}(p_{i}-1)}$ et $x \\equiv y \\pmod{p_{i}}$ pour tout $i$, ainsi $x^{x} \\equiv y \\pmod{n}$.\n\nSupposons qu'il existe $(i, j)$ tel que $p_{i}$ divise $p_{j}-1$. Comme $i \\neq j$, il suffit dans ce cas de vérifier que $f$ n'est pas surjective modulo $m$ avec $m = p_{i} p_{j}$. Supposons que $f$ est surjective modulo $m$. Soit $a \\in \\mathbb{N}$, $y$ tel que $y \\equiv 0 \\pmod{p_{i}}$ et $y \\equiv a \\pmod{p_{j}}$ par théorème des restes chinois. Supposons qu'il existe $x$ tel que $x^{x} \\equiv y \\pmod{p_{i} p_{j}}$. Dans ce cas $x^{x} \\equiv 0 \\pmod{p_{i}}$ donc $p_{i}$ divise $x$, posons alors $x = k p_{i}$. Dans ce cas, $x^{x} \\equiv (x^{k})^{p_{i}} \\equiv a \\pmod{p_{j}}$. On obtient, en élevant à la puissance $\\frac{p_{j}-1}{p_{i}}$, $a^{\\frac{p_{j}-1}{p_{i}}} = (x^{k})^{p_{j}-1} = 1 \\pmod{p_{j}}$. En particulier, si $a$ est une racine primitive modulo $p_{j}$, comme $a$ est d'ordre exactement $p_{j}-1$, on obtient une contradiction, l'application $f$ n'est pas surjective.\n\nAinsi l'application est surjective si et seulement si $n = p_{1} \\ldots p_{k}$ avec les $p_{i}$ premiers et deux à deux distincts et pour tout $(i, j)$ entre 1 et $k$, $p_{i}$ ne divise pas $p_{j}-1$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 30216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, soit $O$ le centre de son cercle circonscrit. Soit $I$ le centre du cercle inscrit au triangle $ABC$ et $D$ le point de tangence de ce cercle avec le segment $[AC]$. Les droites $(OI)$ et $(AB)$ se coupent en un point $P$. Soit $M$ le milieu de l'arc $AC$ ne contenant pas $B$ et $N$ le milieu de l'arc $BC$ contenant $A$.\n\nMontrer que les droites $(MD)$ et $(NP)$ se coupent sur le cercle circonscrit à $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $X$ le point d'intersection de la droite $(MD)$ avec le cercle circonscrit au triangle $ABC$ et soit $P'$ le point d'intersection de la droite $(XN)$ avec le segment $[AB]$. Soit $S$ le point d'intersection de la bissectrice de l'angle $\\widehat{BAC}$ avec le cercle circonscrit au triangle $ABC$. On sait que les points $N$, $O$ et $S$ sont alignés.\n\nSoit $E$ et $F$ les points de contact respectifs du cercle inscrit au triangle $ABC$ avec les segments $[AB]$ et $[BC]$.\n\nLe point $M$ est le milieu de l'arc $CA$, il est donc sur la bissectrice de l'angle $\\widehat{AXC}$. Le point $D$ est le pied de cette bissectrice. D'après le théorème de la bissectrice, $\\frac{AX}{XC} = \\frac{DA}{DC} = \\frac{AE}{CF}$. Puisque $\\widehat{EAX} = \\widehat{BAX} = \\widehat{BCX} = \\widehat{FCX}$, il vient que les triangles $AEX$ et $CFX$ sont semblables donc le point $X$ est le centre de la similitude qui envoie les points $E$ et $F$ sur les points $A$ et $C$, il est donc le point d'intersection des cercles circonscrits aux triangles $ABC$ et $EFB$.\n\nPuisque $[BI]$ est un diamètre du cercle circonscrit au triangle $EBF$, on déduit $\\widehat{IXB} = 90^\\circ$. Soit $Y$ le point diamétralement opposé au point $B$ dans le cercle circonscrit au triangle $ABC$. Alors $\\widehat{YXB} = 90 = \\widehat{IXB}$. Les points $X$, $I$ et $Y$ sont donc alignés.\n\nD'après le théorème de Pascal appliqué à l'hexagone $SABYXN$, les points $O = (BY) \\cap (SN)$, $I = (YX) \\cap (AS)$ et $P' = (AB) \\cap (XN)$ sont alignés. Le point $P'$ correspond donc au point d'intersection des droites $(OI)$ et $(AB)$ donc $P = P'$ ce qui donne bien que les droites $(MD)$ et $(NP)$ se coupent sur le cercle circonscrit au triangle $ABC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProuver que tout ensemble de 90 nombres choisis dans $\\{1,2, \\cdots, 100\\}$ en contient 10 qui forment une progression arithmétique.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOn dispose les entiers de 1 à 100 dans un tableau carré $10 \\times 10$ comme suit :\n\n| 1  | 2  | 3  | 4  | 5  | 6  | 7  | 8  | 9  | 10  |\n|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|:---:|\n| 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20  |\n| 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30  |\n| 31 | 32 | 33 | 34 | 35 | 36 | 37 | 38 | 39 | 40  |\n| 41 | 42 | 43 | 44 | 45 | 46 | 47 | 48 | 49 | 50  |\n| 51 | 52 | 53 | 54 | 55 | 56 | 57 | 58 | 59 | 60  |\n| 61 | 62 | 63 | 64 | 65 | 66 | 67 | 68 | 69 | 70  |\n| 71 | 72 | 73 | 74 | 75 | 76 | 77 | 78 | 79 | 80  |\n| 81 | 82 | 83 | 84 | 85 | 86 | 87 | 88 | 89 | 90  |\n| 91 | 92 | 93 | 94 | 95 | 96 | 97 | 98 | 99 | 100 |\n\nSi l'on veut éliminer 10 nombres de sorte qu'avec les 90 restants on ne puisse former de suite arithmétique de longueur 10, il faut déjà éliminer au moins un nombre dans chacune des 10 lignes, et chacune des 10 colonnes. Comme il y a justement 10 lignes, et 10 colonnes, c'est donc que l'on doit éliminer exactement un nombre par ligne et par colonne.\n\nDe plus, si pour $1 \\leqslant i \\leqslant 9$, le nombre éliminé dans la ligne $i$ appartient à la colonne $k$, alors le nombre éliminé dans la ligne $i+1$ doit appartenir à une colonne $l$, avec $l<k$, sans quoi on aurait 10 entiers consécutifs parmi les 90 restants. Cela implique que l'on doit éliminer $10,19,28,37,46,55,64,73,82$ et $91$.\n\nMais alors, il reste $11,20,29,38,47,56,65,74,83$ et $92$, qui forment une suite arithmétique de raison $9$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les couples d'entiers $(n, p)$ strictement positifs où $p$ est un nombre premier et tels que $n+p$ soit une puissance de $n$.\n\nUne puissance de $n$ est de la forme $n^k$ pour $k$ entier naturel.",
    "options": [],
    "answer": "(2, 2)",
    "solution": "Solution:\n\nSoit $(p, n)$ un éventuel couple solution. On dispose d'un entier naturel $k$ tel que\n$$\nn + p = n^k\n$$\nTout d'abord, si $k=0$, alors $n+p=1$. Or $p$ étant premier, $p>1$ et $n \\geqslant 0$ donc $n+p=1<p \\leqslant p+n$ ce qui est absurde. On a donc $k \\geqslant 1$.\n\nOn peut alors réécrire l'équation comme ceci :\n$$\np = n(n^{k-1} - 1)\n$$\nOn obtient que $n$ divise $p$. Comme $p$ est premier, il faut donc que $n=1$ ou $n=p$.\n\nSi $n=1$, alors $1+p=1$ ce qui est absurde puisque $p \\neq 0$.\n\nOn a donc $n=p$. Alors $2p = p^k$ donc $2$ divise $p$. Ainsi, $p=2$ et l'équation devient $2+2=2^k$ donc $k=2$. Réciproquement, le couple $(2,2)$ vérifie bien que $2+2=2^2$ et donc que $2+2$ est une puissance de $2$.\n\nLe seul couple solution est donc $(2,2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver toutes les paires $(m,n)$ d'entiers strictement positifs telles que\n$$\nm! + n! = m^n\n$$\n\nRemarque. Pour tout entier $n$, on rappelle que $n!$ désigne l'entier $1 \\times 2 \\times 3 \\times \\cdots \\times n$.",
    "options": [],
    "answer": "(2,2) and (2,3)",
    "solution": "Solution:\n\nOn vérifie facilement que les paires $(2,2)$ et $(2,3)$ sont solutions. On va montrer que ce sont les seules.\n\nOn commence par traiter le cas où $m$ ou $n$ vaut $1$ ou $2$.\n\nSi $m=1$, alors $m^n=1$ mais $m!+n!\\geqslant 2$, donc il n'y a pas de solution.\n\nSi $m=2$, l'équation devient $2 + n! = 2^n$.\nSi $n \\geqslant 4$, alors le membre de gauche est congru à $2$ modulo $4$, et le membre de droite à $0$, donc $n$ vaut $1$, $2$ ou $3$, et $n=1$ ne marche pas.\n\nSi $n=1$, l'équation devient $m! + 1 = m$, mais $m$ divise $m!$ et $m$, donc $m$ divise $1$ et $m=1$, mais $(1,1)$ n'est pas solution.\n\nEnfin, si $n=2$, on obtient $m! + 2 = m^2$. Mais $m$ divise $m!$ et $m^2$ donc $m$ divise $2$, donc $m$ vaut $1$ ou $2$ et on retombe sur des cas déjà traités.\n\nOn traite maintenant le cas $2 < m \\leqslant n$. Dans ce cas $m!$ divise $n!$ donc $m!$ divise $m^n$. En particulier, $m-1$ divise $m!$ donc aussi $m^n$, mais $m-1$ est premier avec $m$, donc aussi avec $m^n$. Par conséquent, $m-1$ ne peut valoir que $1$ donc $m=2$, contradiction.\n\nEnfin, on traite le cas $2 < n < m$. Dans ce cas $n!$ divise $m!$, donc on peut écrire $n!\\left(1+\\frac{m!}{n!}\\right) = m^n$. En particulier, $1+\\frac{m!}{n!}$ divise $m^n$. Mais $\\frac{m!}{n!} = (n+1)(n+2) \\ldots m$ est divisible par $m$, donc $1+\\frac{m!}{n!}$ est premier avec $m$, donc avec $m^n$. On a donc $1+\\frac{m!}{n!}=1$, ce qui est impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRésoudre $x^{4}-6 x^{2}+1=7 \\times 2^{y}$ pour $x$ et $y$ entiers.",
    "options": [],
    "answer": "(3, 2), (-3, 2)",
    "solution": "Solution:\n\nL'équation se réécrit $\\left(x^{2}-3\\right)^{2}=7 \\times 2^{y}+8$.\n\nSi $y \\geq 3$, alors le côté droit s'écrit $8\\left(7 \\times 2^{y-3}+1\\right)$. Le côté gauche étant un carré, sa valuation 2-adique est paire. Ainsi, $7 \\times 2^{y-3}+1$ doit être pair, donc $y=3$. Dans ce cas $x^{2}-3=8$ donc $x^{2}=11$, impossible. Donc $y \\leq 2$.\n\nSi $y=2$, on trouve $x= \\pm 3$.\n\nPour $y<0$, ainsi que pour $y=0$ ou $y=1$ il n'y a pas de solution.\n\nLes seuls couples de solutions sont donc $(x, y)=(3,2)$ et $(-3,2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle rectangle en $B$. Soit $M$ le point d'intersection de la médiane issue de $B$ avec la droite $(AC)$, et $(d)$ la perpendiculaire à la droite $(BC)$ passant par le point $M$. Soit $U$ le milieu du segment $[AB]$, $V$ le milieu du segment $[AM]$, $I$ le point d'intersection de la droite $(UV)$ avec la droite $(d)$, et $J$ le point d'intersection de la droite $(UV)$ avec la droite $(BC)$.\n\nMontrer que $AC = IJ$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLes droites $d$ et $(AB)$ sont perpendiculaires à la droite $(BC)$ donc elles sont parallèles. On obtient par le théorème de Thalès que $$\frac{VI}{VU} = \frac{AV}{VM} = 1$$ donc $V$ est le milieu du segment $[UI]$.\n\nLe quadrilatère $AIMU$ a ses diagonales qui se coupent en leur milieu, c'est donc un parallélogramme. Les points $U$ et $M$ sont les milieux respectifs des segments $[AB]$ et $[AC]$ donc les droites $(UM)$ et $(BC)$ sont parallèles. Donc la droite $(UM)$ est perpendiculaire au segment $[AB]$ et le quadrilatère $AIMU$ est un rectangle.\n\nSoit $l$ la droite perpendiculaire au segment $[BC]$ passant par $V$. Cette droite est la médiatrice du segment $[UM]$ et du segment $[AI]$ donc le point $U$ est le symétrique du point $M$ par rapport à la droite $l$ et le point $A$ est le symétrique du point $I$ par rapport à la droite $l$ donc la droite $(AM)$ est la symétrique de la droite $(UI)$ par rapport à la droite $l$.\n\nPuisque la droite $(BC)$ est sa propre symétrique par rapport à la droite $l$, le point d'intersection de la droite $(BC)$ avec la droite $(UI)$ est le symétrique du point d'intersection de la droite $(AM)$ avec la droite $(BC)$ donc les points $C$ et $J$ sont symétriques par rapport à la droite $l$. La symétrie conserve les longueurs, donc $AC = IJ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les nombres irrationnels $x$ pour lesquels les deux nombres $x^{2}+x$ et $x^{3}+2 x^{2}$ sont des entiers.",
    "options": [],
    "answer": "(-1 - sqrt(5))/2 and (-1 + sqrt(5))/2",
    "solution": "Solution:\n\nSoit $x$ un nombre irrationnel pour lequel les deux nombres $x^{2}+x$ et $x^{3}+2 x^{2}$ sont des entiers.\nOn pose $x^{2}+x=a$ et $x^{3}+2 x^{2}=b$, où $a$ et $b$ sont des entiers. Alors $b-a x=x^{2}=a-x$, et ainsi $x(a-1)=b-a$. Si $a-1 \\neq 0$, on aurait $x=\\frac{b-a}{a-1}$, et $x$ serait rationnel, en contradiction avec l'énoncé.\nPar suite, on a $a=1$ et donc $b=a=1$. Mais alors $x^{2}+x-1=0$ et $x^{3}+2 x^{2}-1=0$. On note que, si $x^{2}+x-1=0$, alors $x^{3}=-x^{2}+x$ et $x=-x^{2}+1$, d'où $x^{3}+2 x^{2}-1=0$.\nAinsi, les solutions du problème sont les solutions irrationnelles de l'équation du second degré $x^{2}+x-1=0$, c'est-à-dire les nombres $x_{1}=\\frac{-1-\\sqrt{5}}{2}$ et $x_{2}=\\frac{-1+\\sqrt{5}}{2}$. Puisque $\\sqrt{5}$ n'est pas rationnel, ces deux nombres sont effectivement des solutions du problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers strictement positifs $m$ et $n$ tels que\n$$\n3 \\cdot 2^{m} + 1 = n^{2}.\n$$",
    "options": [],
    "answer": "m=3, n=5; m=4, n=7",
    "solution": "Solution:\n\nRemarquons tout d'abord que la condition se réécrit $3 \\cdot 2^{m} = (n-1)(n+1)$.\n\n$\\triangleright$ Si $n$ est pair, les entiers $(n+1)$ et $(n-1)$ sont premiers entre eux donc égaux à $2^{m}$ et $3$ ou à $3 \\cdot 2^{m}$ et $1$. Or, ils sont tous les deux impairs (et $m > 0$) : contradiction.\n\n$\\triangleright$ Si $n$ est impair, le pgcd de $(n+1)$ et $(n-1)$ est le pgcd de $(n+1)$ et $n+1-(n-1)=2$ donc $2$. Ainsi, ils sont égaux à $6$ et $2^{m-1}$, ou bien à $2$ et $3 \\cdot 2^{m-1}$.\n\n- Si $n-1=6$, alors $n=7$ et $2^{m-1}=n+1=8$ d'où $m=4$.\n- Si $n+1=6$, alors $n=5$ et $2^{m-1}=n-1=4$ d'où $m=3$.\n- Si $n-1=2$ alors $n=3$ et $3 \\cdot 2^{m}=n^{2}-1=8$, ce qui est impossible car $3$ ne divise pas $8$.\n- Si $n+1=2$ alors $n=1$ et $3 \\cdot 2^{m}=n^{2}-1=0$, ce qui est impossible.\n\nEn conclusion, l'équation n'est vérifiée que pour les couples $(m, n) \\in \\{(4,7), (3,5)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSept élèves d'une classe comparent leurs notes dans 12 épreuves, et remarquent qu'il n'existe pas deux élèves ayant des notes identiques dans chacune des 12 épreuves.\n\nMontrer que l'on peut choisir 6 épreuves telles que deux élèves quelconques aient des notes différentes à au moins l'une de ces épreuves.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $m$ le plus petit entier tel que l'on peut choisir $m$ épreuves $E_{1}, \\ldots, E_{m}$ de sorte que deux élèves quelconques aient des notes différentes à au moins l'une de ces $m$ épreuves. On doit montrer que $m \\leqslant 6$.\n\nComme $m$ est minimal, pour tout $i=1, \\ldots, m$ il existe une paire $A_{i}$ d'élèves ayant la même note à toutes les épreuves sauf $E_{i}$.\n\nOn forme un graphe dont les sommets sont les élèves, et dont les arêtes sont $A_{1}, \\ldots, A_{m}$. Supposons que le graphe possède un cycle $\\left(A_{i_{1}}, \\ldots, A_{i_{r}}\\right)$. Notons $a$ et $b$ les extrémités de $A_{i_{1}}$. Par hypothèse, $a$ et $b$ obtiennent des notes différentes à l'épreuve $i_{1}$. Cependant, le chemin $\\left(A_{i_{2}}, \\ldots, A_{i_{r}}\\right)$ montre que $a$ et $b$ obtiennent des notes identiques à l'épreuve $i_{1}$, ce qui est contradictoire.\n\nPar conséquent, le graphe est un arbre. Comme un arbre a strictement plus de sommets que d'arêtes, on en déduit que $m < 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans un cercle $\\mathcal{C}$ de centre $O$, on trace une corde $[CD]$. Soit $\\Gamma$ le cercle de diamètre $[CD]$ et $O'$ son centre. La médiatrice de $[CD]$ coupe le cercle $\\mathcal{C}$ en deux points $A$ et $B$ tels que $A$ est extérieur à $\\Gamma$. Notons $T$ et $T'$ les points de contact des tangentes à $\\Gamma$ passant par $A$. Soit $F$ le milieu de $[TT']$. Montrer que $O'$ est le milieu de $[BF]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nComme les triangles $AFT$ et $ATO'$ sont semblables, on a $\\frac{AF}{AT} = \\frac{AT}{AO'}$. Or, $AT^2 = AO'^2 - O'T^2 = AO'^2 - O'C^2$ donc $AF = \\frac{AO'^2 - O'C^2}{AO'}$.\n\nComme les triangles $ACB$ et $AO'C$ sont semblables, on a $\\frac{AB}{AC} = \\frac{AC}{AO'}$. Or, $AC^2 = AO'^2 + O'C^2$ donc $AB = \\frac{AO'^2 + O'C^2}{AO'}$.\n\nEn additionnant les égalités précédemment trouvées, on obtient $\\frac{AF + AB}{2} = AO'$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer que si $a, b, c$ sont des nombres réels positifs vérifiant $a+b+c=1$ alors\n$$\n\\frac{7+2 b}{1+a}+\\frac{7+2 c}{1+b}+\\frac{7+2 a}{1+c} \\geqslant \\frac{69}{4}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComme $7+2 b=5+2(1+b)$, on écrit le membre de gauche sous la forme\n$$\n5\\left(\\frac{1}{1+a}+\\frac{1}{1+b}+\\frac{1}{1+c}\\right)+2\\left(\\frac{1+b}{1+a}+\\frac{1+c}{1+b}+\\frac{1+a}{1+c}\\right)\n$$\nEn utilisant l'inégalité $\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z} \\geqslant \\frac{9}{x+y+z}$, on minore le premier terme par $\\frac{45}{4}$.\nEn utilisant l'inégalité $x+y+z \\geqslant 3 \\sqrt[3]{x y z}$, on minore le second terme par $6$. L'assertion à démontrer en découle.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle avec $\\widehat{BAC}=60^{\\circ}$. Soient $D$ et $E$ les pieds des bissectrices issues respectivement des sommets $B$ et $C$. Soit $I$ le point d'intersection des droites $(BD)$ et $(CE)$. Montrer que les points $A$, $I$, $D$ et $E$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn va faire une chasse aux angles pour démontrer que $\\widehat{IDA}+\\widehat{IEA}=180^{\\circ}$. Notons $x=\\widehat{ACE}=\\widehat{ECB}$ ainsi que $y=\\widehat{ABD}=\\widehat{DBC}$. Comme la somme des angles d'un triangle vaut $180^{\\circ}$ on a ainsi\n$$\n\\begin{aligned}\n180^{\\circ} & =\\widehat{BAC}+\\widehat{ABC}+\\widehat{ACB} \\\\\n& =60^{\\circ}+2y+2x\n\\end{aligned}\n$$\nCe qui implique que $x+y=60^{\\circ}$. On peut donc exprimer tous les angles de la figure en fonction de $x$ et en particulier les deux angles qui nous intéressent, $\\widehat{IDA}$ et $\\widehat{IEA}$. On a $\\widehat{IDA}=\\widehat{BDA}=180^{\\circ}-60^{\\circ}-y$ ainsi que $\\widehat{IEA}=\\widehat{CEA}=180^{\\circ}-60^{\\circ}-x$. Donc la somme vaut\n$$\n\\widehat{IDA}+\\widehat{IEA}=180^{\\circ}-60^{\\circ}-y+180^{\\circ}-60^{\\circ}-x=180^{\\circ}+60^{\\circ}-x-y=180^{\\circ}\n$$\nCe qui est l'égalité voulue.\nSolution:\n\n![](attached_image_2.png)\n\nUne fois que l'on a obtenu que $x+y=60^{\\circ}$, on peut conclure d'une autre façon, en démontrant que $\\widehat{DAE}+\\widehat{DIE}=180^{\\circ}$. En effet, dans le triangle $BCI$, on a $\\widehat{BIC}=180^{\\circ}-x-y=120^{\\circ}$, de sorte que $\\widehat{DIE}+\\widehat{DAE}=\\widehat{BIC}+\\widehat{BAC}=120^{\\circ}+60^{\\circ}=180^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $x \\geqslant 0$ un réel. Montrer que :\n$$\n1 + x^{2} + x^{6} + x^{8} \\geqslant 4 x^{4}\n$$\net trouver les cas d'égalité.",
    "options": [],
    "answer": "x = 1",
    "solution": "Solution:\nOn utilise l'inégalité arithmético-géométrique, dans le cas $n = 4$. On obtient\n$$\n1 + x^{2} + x^{6} + x^{8} \\geqslant 4 x^{\\frac{2+6+8}{4}} = 4 x^{4}.\n$$\nSupposons qu'on a égalité. D'après le cas d'égalité, $x^{2} = 1$, donc $x = 1$. Réciproquement, si $x = 1$, on a $1 + x^{2} + x^{6} + x^{8} = 4 = 4 x^{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\omega_{1}$ et $\\omega_{2}$ deux cercles de centre $O_{1}$ et $O_{2}$, on suppose qu'ils se coupent en $A$ et $B$. Le cercle passant par les points $O_{1}$, $O_{2}$ et $B$ coupe le cercle $\\omega_{1}$ en $C$. Montrer que les points $C$, $A$ et $O_{2}$ sont alignés.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour résoudre cet exercice, on va plutôt introduire dans un premier temps $C'$ la deuxième intersection de la droite $(A O_{2})$ avec le cercle $\\omega_{1}$ puis on va montrer que $C = C'$.\n\nOn peut décomposer l'angle plat $\\widehat{O_{2}AC'}$ en deux angles : $\\widehat{C' AO_{1}}$ et $\\widehat{O_{1} AO_{2}}$ dont la somme vaut donc $180$ degrés. D'une part, en faisant une symétrie axiale d'axe $(O_{1} O_{2})$, alors les points $A$ et $B$ sont échangés, mais alors l'angle $\\widehat{O_{1} AO_{2}}$ vaut l'angle $\\widehat{O_{1} BO_{2}}$. D'autre part, $O_{1}A = O_{1}C'$ comme $O_{1}$ est le centre du cercle $\\omega_{1}$, ainsi le triangle $O_{1} A C'$ est isocèle en $O_{1}$, et on trouve que $\\widehat{O_{2} C' O_{1}} = \\widehat{AC' O_{1}} = \\widehat{C' AO_{1}} = 180 - \\widehat{O_{1} AO_{2}} = 180 - \\widehat{O_{1} BO_{2}}$, donc en particulier, $\\widehat{O_{2} C' O_{1}} = 180 - \\widehat{O_{2} BO_{1}}$, ce qui démontre bien que les points $O_{1}$, $B$, $O_{2}$ et $C'$ sont cocycliques.\n\nAinsi, si on reprend la définition du point $C$, on obtient que $O_{1}$, $O_{2}$, $B$ et $C$ sont cocycliques et que $C'$ est sur $\\omega_{1}$, donc que $C' = C$, mais comme les points $A$, $O_{2}$ et $C' = C$ sont alignés on a entre autre que $A$, $O_{2}$ et $C$ sont alignés.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. Soient $a$, $b$ et $c$ trois nombres réels tels que\n$$\n|a| \\geqslant |a+b|,\\ |b| \\geqslant |b+c| \\text{ et } |c| \\geqslant |c+a|.\n$$\nMontrer que $a = b = c = 0$.\n\n2. Soient $a$, $b$, $c$ et $d$ quatre nombres réels tels que\n$$\n|a| \\geqslant |a+b|,\\ |b| \\geqslant |b+c|,\\ |c| \\geqslant |c+d| \\text{ et } |d| \\geqslant |d+a|\n$$\nA-t-on forcément $a = b = c = d = 0$ ?",
    "options": [],
    "answer": "Part 1: a = b = c = 0. Part 2: No; for example, a = c = 1 and b = d = −1 satisfy the conditions.",
    "solution": "Solution:\n\n1. Parmi les trois nombres, il y en a au moins deux de même signe. On peut supposer que ce sont $a$ et $b$ et qu'ils sont positifs, les autres cas se traitant de la même manière. On a alors\n$$\na = |a| \\geqslant |a+b| = a+b \\geqslant a\n$$\ndonc on a égalité partout, donc $b=0$. Comme $|b| \\geqslant |b+c|$, on a donc $0 \\geqslant |c|$, donc $c=0$. Finalement, comme $|c| \\geqslant |c+a|$, on a $0 \\geqslant |a|$ donc $a=0$ donc les trois nombres sont nuls.\n\n2. Non. Par exemple, on peut prendre $a = c = 1$ et $b = d = -1$. On a alors\n$$\na+b = b+c = c+d = d+a = 0\n$$\ndonc la condition est vérifiée, bien que les nombres choisis soient non nuls.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn pose 23 allumettes sur une table et 2 joueurs jouent à un jeu : chacun, à son tour, retire entre 1 et 4 allumettes (inclus). Celui qui prend la dernière gagne. Existe-t-il une stratégie gagnante pour l'un des deux?\n\nUne stratégie gagnante est une manière de jouer qui permet à l'un des deux de gagner peu importe comment son adversaire joue.",
    "options": [],
    "answer": "Yes — the first player. Take three first, then always take enough matches so that your move plus the opponent’s previous move totals five.",
    "solution": "Solution:\n\nLe $1^{\text{er}}$ joueur possède une stratégie gagnante. La voici :\n\n1. $J_1$ prend 3 allumettes : il en reste un multiple de 5.\n\n2. $J_2$ en retire $r \\in \\{1,2,3,4\\}$.\n\n3. $J_1$ en enlève $5 - r$ : il en reste un multiple de 5.\n\nEt ainsi de suite. Donc à chaque fois que $J_2$ joue, il y a un nombre d'allumettes divisible par 5. À chaque fois que $J_1$ joue, en revanche, il y en a un nombre non-divisible par 5 et donc en particulier non nul ! Ce qui assure à $J_1$ de gagner.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un trapèze tel que $(AB)$ et $(CD)$ sont parallèles. On suppose que $AD < CD$ et que $ABCD$ est inscrit dans un cercle $\\Gamma$. Soit $P$ sur $\\Gamma$ tel que $(DP)$ est parallèle à $(AC)$. La tangente à $\\Gamma$ en $D$ recoupe $(AB)$ en $E$, et les cordes $[BP]$ et $[CD]$ s'intersectent en $Q$. Montrer que $EQ = AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn sait que $(CQ)$ est parallèle à $(AE)$, donc l'énoncé revient à montrer que $ACQE$ est un parallélogramme, donc il suffit de montrer que $(EQ)$ et $(AC)$ sont parallèles. Le quadrilatère $BQDE$ est un trapèze, avec $(BE)$ et $(DQ)$ parallèles. On va d'abord montrer que c'est un trapèze isocèle. En effet, on a\n$$\n\\widehat{EDQ} = \\widehat{EDC} = 180^\\circ - \\widehat{DAC}\n$$\nd'après le cas limite du théorème de l'angle inscrit. D'autre part, on a\n$$\n\\widehat{DQB} = 180^\\circ - \\widehat{BDC} - \\widehat{DBP} = 180^\\circ - \\widehat{BAC} - \\widehat{DAP}\n$$\nOr, les quadrilatères $ABCD$ et $ACPD$ sont des trapèzes inscriptibles, donc isocèles, donc $PC = AD = BC$, donc $C$ est le milieu de l'arc entre $B$ et $P$. On a donc $\\widehat{BAC} = \\widehat{CAP}$, donc\n$$\n\\widehat{DQB} = 180^\\circ - \\widehat{CAP} - \\widehat{DAP} = 180^\\circ - \\widehat{DAC} = \\widehat{EDQ}\n$$\ndonc le trapèze $EBQD$ est bien isocèle, et a un axe de symétrie. On en déduit\n$$\n\\widehat{DQE} = \\widehat{QDB} = \\widehat{CDB} = \\widehat{DCA},\n$$\nen utilisant à la fin le fait que $ABCD$ a un axe de symétrie. Cela montre que $(QE)$ et $(AC)$ sont parallèles, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes diagonales du quadrilatère convexe $ABCD$ sont perpendiculaires et se rencontrent en $O$. La perpendiculaire à $(AB)$ passant par $O$ rencontre $(AB)$ en $M$ et $(CD)$ en $M'$. La perpendiculaire à $(BC)$ passant par $O$ rencontre $(BC)$ en $N$ et $(DA)$ en $N'$. La perpendiculaire à $(CD)$ passant par $O$ rencontre $(CD)$ en $P$ et $(AB)$ en $P'$. La perpendiculaire à $(DA)$ passant par $O$ rencontre $(DA)$ en $Q$ et $(BC)$ en $Q'$. \n\nProuver que les points $M, N, P, Q, M', N', P', Q'$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nCommençons par prouver que le quadrilatère $MNPQ$ est inscriptible :\nOn note que le quadrilatère $BMON$ a deux angles droits en $M$ et en $N$, donc il est inscrit dans le cercle de diamètre $[OB]$. De même, les quadrilatères $CNOP$, $DPOQ$ et $AQOM$ sont inscrits respectivement dans les cercles de diamètres $[OC]$, $[OD]$ et $[OA]$. On en déduit que $\\widehat{OMN}=\\widehat{OBN}$, que $\\widehat{OMQ}=\\widehat{OAQ}$, que $\\widehat{OPQ}=\\widehat{ODQ}$ et que $\\widehat{OPN}=\\widehat{OCN}$.\n\n$$\n\\begin{aligned}\n& \\text{Il vient alors} \\\\\n& \\begin{aligned}\n\\widehat{QMN} & = \\widehat{QMO} + \\widehat{OMN} \\\\\n& = \\widehat{QAO} + \\widehat{OBN} \\\\\n& = \\widehat{DAC} + \\widehat{DBC}\n\\end{aligned}\n\\end{aligned}\n$$\n\nEt d'autre part\n\n$$\n\\begin{aligned}\n\\widehat{QPN} & = \\widehat{QPO} + \\widehat{OPN} \\\\\n& = \\widehat{QDO} + \\widehat{OCN} \\\\\n& = \\widehat{ADB} + \\widehat{ACB}\n\\end{aligned}\n$$\n\nOr, puisque $[AC]$ et $[DB]$ sont perpendiculaires, on a $\\widehat{DAC} + \\widehat{ADB} = 90^{\\circ}$, et de même $\\widehat{DBC} + \\widehat{ACB} = 90^{\\circ}$. Ainsi $\\widehat{QMN} = 180^{\\circ} - \\widehat{QPN}$. Cela assure que $MNPQ$ est inscriptible.\n\nOn note $\\Gamma$ le cercle circonscrit à $MNPQ$. Par symétrie des rôles, pour conclure, il suffit de prouver que $Q'$ est sur $\\Gamma$.\n\nSupposons que $N \\in [CQ']$, les autres cas se traitant de façons similaires.\nNous avons $\\widehat{NQ'Q} = \\widehat{NQ'O} = 180^{\\circ} - \\widehat{Q'CO} - \\widehat{COQ'}$.\nMais $\\widehat{Q'CO} = \\widehat{NCO} = \\widehat{NPO}$, puisque $NCPO$ inscriptible.\nD'autre part, comme les côtés sont perpendiculaires deux à deux, on a $\\widehat{COQ'} = \\widehat{ODQ} = \\widehat{OPQ}$, la dernière égalité découlant du fait que $OPDQ$ est inscriptible.\nAinsi, on a $\\widehat{NQ'Q} = 180^{\\circ} - \\widehat{NPO} - \\widehat{OPQ} = 180^{\\circ} - \\widehat{NPQ}$, ce qui assure que $Q'$ appartient bien au cercle circonscrit à $NPQ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nJules et Jim, deux colocataires, reçoivent 10 autres paires de colocataires chez eux. Jules fait un petit sondage au cours de la soirée et interroge les 21 autres personnes. Parmi elles, aucune n'a serré la main de son colocataire, et il n'y en a pas deux qui ont serré le même nombre de mains. Combien de mains a serré Jim?",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nAu cours de la soirée, parmi les 21, chaque personne serre entre 0 et 20 mains. Comme elles serrent toutes un nombre différent de mains, pour tout $n$ compris entre $0$ et $20$ il existe une et une seule personne $A_{n}$ parmi les 21 qui a serré exactement $n$ mains.\n\nComme $A_{20}$ a serré 20 mains, $A_{20}$ a serré la main à tout le monde sauf lui-même et son colocataire. Or, $A_{0}$ n'a serré la main à personne, donc $A_{0}$ est le colocataire de $A_{20}$.\n\nComme $A_{19}$ a serré 19 mains, $A_{19}$ a serré la main à tout le monde sauf lui-même, son colocataire et $A_{0}$. Or, $A_{1}$ a serré la main à $A_{20}$ et à personne d'autre, donc n'a pas serré la main à $A_{19}$. On en déduit que $A_{1}$ est le colocataire de $A_{19}$.\n\nEn continuant le raisonnement, on montre que $A_{n}$ est le colocataire de $A_{20-n}$ pour tout $n = 0, 1, \\ldots, 9$, donc $A_{10}$ est le colocataire de Jules. Finalement, Jules a serré 10 mains.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, et $M$ le milieu de $[BC]$. On note $I_{b}$ et $I_{c}$ les centres des cercles inscrits à $AMB$ et $AMC$. Montrer que le second point d'intersection des cercles circonscrits aux triangles $ABI_{b}$ et $ACI_{c}$ se situe sur la droite $(AM)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $T$ l'intersection entre le cercle de diamètre $[BC]$ et la demi-droite $[MA)$. Il suffit de montrer que $T$ appartient au cercle $ABI_{b}$ (par symétrie des rôles de $B$ et $C$, ceci montrera qu'il appartient également au cercle $ACI_{c}$).\n\nComme $\\widehat{BTC}$ est droit, on a $\\widehat{ATB} = 90^{\\circ} + \\frac{1}{2} \\widehat{AMB} = \\widehat{AI_{b}B}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $x$ un réel strictement positif tel que $x^{5}-x^{3}+x \\geq 3$. Prouver que $x^{6} \\geq 5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLa clé du problème est la factorisation $x^{6}+1=\\left(x^{2}+1\\right)\\left(x^{4}-x^{2}+1\\right)$, valable pour tout réel $x$.\nSoit $x>0$ tel que $x^{5}-x^{3}+x \\geq 3$. D'après l'identité ci-dessus, on a alors\n$$\nx^{6}+1=\\left(x^{2}+1\\right)\\left(x^{4}-x^{2}+1\\right)=\\frac{x^{2}+1}{x}\\left(x^{5}-x^{3}+x\\right)=\\left(x+\\frac{1}{x}\\right)\\left(x^{5}-x^{3}+x\\right)\n$$\net donc $x^{6}+1 \\geq 3\\left(x+\\frac{1}{x}\\right)$.\nOr, d'après l'inégalité arithmético-géométrique (IAG), on a $x+\\frac{1}{x} \\geq 2$, d'où $x^{6}+1 \\geq 6$, et ainsi $x^{6} \\geq 5$, comme souhaité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a, b, c$ des nombres réels. Montrer que:\n$$\n2 a^{2}+20 b^{2}+5 c^{2}+8 a b-4 b c-4 a c \\geqslant 0\n$$\net trouver les cas d'égalité.",
    "options": [],
    "answer": "Equality holds only for a = b = c = 0.",
    "solution": "Solution:\n\nL'inégalité est équivalente à $4 b c+4 a c-8 a b \\leqslant 2 a^{2}+20 b^{2}+5 c^{2}$. Or on sait que $4 a c=2 \\times a \\times(2 c) \\leqslant a^{2}+(2 c)^{2}=a^{2}+4 c^{2}$ et que $4 b c=2 \\times c(2 b) \\leqslant 4 b^{2}+c^{2}$. On a également $-8 a b=2 \\times(-a) \\times(4 b) \\leqslant a^{2}+16 b^{2}$. En sommant toutes ces inégalités, on obtient $4 b c+4 a c-8 a b \\leqslant 2 a^{2}+20 b^{2}+5 c^{2}$.\n\nSupposons qu'on a égalité, dans ce cas on a égalité dans les trois inégalités utilisées : on a donc $a=2 c$, $c=2 b$ donc $a=4 b$ et $-a=4 b$. En particulier $-a=a$ donc $a=0$ donc $b$ et $c$ sont également nuls. Réciproquement, si $a=b=c=0$, on a bien égalité.\n\nOn aurait pu également remarquer que\n$$\n2 a^{2}+20 b^{2}+5 c^{2}+8 a b-4 b c-4 a c=(a+4 b)^{2}+(a-2 c)^{2}+(2 b-c)^{2} \\geqslant 0\n$$\net trouver de la même façon le cas d'égalité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver le plus petit entier positif qui ne s'écrit pas sous la forme $\\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$ pour $a, b, c, d \\in \\mathbb{N}$.",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nSoit $E$ l'ensemble des entiers strictement positifs s'écrivant sous cette forme. Commençons par remarquer que\n$$\n\\frac{2^{a}-2^{b}}{2^{c}-2^{d}}=2^{b-d} \\frac{2^{a-b}-1}{2^{c-d}-1}\n$$\nAinsi, si $x>0$ s'écrit $x=\\frac{2^{a}-2^{b}}{2^{c}-2^{d}}$, alors $b-d=v_{2}(x)$ et si on appelle $y$ l'entier impair tel que $x=2^{v_{2}(x)} y$, alors $y$ s'écrit sous la forme $\\frac{2^{n}-1}{2^{m}-1}$. En particulier, l'entier que nous cherchons est nécessairement impair. D'autre part, on peut tout de suite exclure les entiers de la forme $2^{n}-1$, atteints avec $m=1$.\n\nIl est classique que si $2^{m}-1$ divise $2^{n}-1$, alors $m$ divise $n$. Ainsi, en écrivant $n=m d$, $y$ est de la forme\n$$\ny=\\frac{2^{m d}-1}{2^{m}-1}=2^{(d-1) m}+2^{(d-2) m}+\\ldots+2^{m}+1\n$$\nAutrement dit, en notant $u=\\underbrace{0 \\ldots 0}_{m-1 \\text{ zéros }} 1$, l'écriture en base 2 de $y$ est de la forme $\\overline{1 u \\ldots u}^{2}$, avec $u$ apparaissant $d-1$ fois. Les entiers impairs appartenant à $E$ sont exactement ceux ayant une telle écriture binaire pour $m$ et $d$ bien choisis. Les écritures binaires des premiers entiers impairs qui ne sont pas de la forme $2^{n}-1$ sont\n$$\n5=\\overline{101}^{2},\\ 9=\\overline{1001}^{2},\\ 11=\\overline{1011}^{2}\n$$\nD'après le critère ci-dessus, le plus petit entier cherché est $11$.\n\nRemarque : cet exercice peut aussi se faire en trouvant des manières d'écrire les entiers $1$ à $10$ sous cette forme, puis en montrant que $11$ ne s'écrit pas sous cette forme : en effet, vu que $11$ est impair, on peut supposer $b=d=1$. Ensuite\n$$\n\\left(2^{a}-1\\right)=11\\left(2^{c}-1\\right)\n$$\nse réécrit\n$$\n10=2^{c} \\times 11-2^{a}\n$$\nPuisque $10=2 \\times 5$, nous avons $a=1$ ou $c=1$. Si $a=1$, on trouve $2^{c} \\times 11=12$, qui n'a pas de solution, et si $c=1$, on trouve $2^{a}=12$, qui n'a pas de solution non plus. Donc $11$ est l'entier cherché.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont $O$ est le centre du cercle circonscrit. On note $\\ell$ une droite perpendiculaire à la droite $(AO)$. La droite $(\\ell)$ intersecte les côtés $(AB)$ et $(AC)$ en les points $D$ et $E$. Montrer que les points $B$, $C$, $E$ et $D$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPosons $X$ et $Y$ les projetés orthogonaux de $O$ sur $(AC)$ et $(DE)$ respectivement. Notons que $X$ est le milieu du segment $[AC]$ et aussi le pied de la bissectrice de l'angle $\\widehat{COA}$ dans le triangle $AOC$ isocèle en $O$.\n\nComme $\\widehat{EXO} = 90^{\\circ} = \\widehat{EYO}$, on en déduit que $E$, $X$, $O$ et $Y$ sont cocycliques par le théorème de l'angle inscrit. Par le théorème de l'angle au centre, on a alors :\n$$\n\\widehat{CBA} = \\widehat{COA} / 2 = \\widehat{XOA} = 180^{\\circ} - \\widehat{YEX} = 180^{\\circ} - \\widehat{CED}\n$$\nOn en déduit la cocyclicité des points $B$, $C$, $D$ et $E$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $p$ un nombre premier. Trouver tous les entiers $n \\geqslant 1$ tels que pour tout entier $a \\geqslant 1$, si $a^{n}-1$ est divisible par $p$, alors $a^{n}-1$ est aussi divisible par $p^{2}$.",
    "options": [],
    "answer": "All positive integers n that are multiples of p",
    "solution": "Solution:\n\nMontrons que les entiers $n$ qui conviennent sont exactement les multiples de $p$. Pour cela, nous allons d'abord prouver que lorsque $n=p m$ pour un entier $m \\geqslant 1$, pour tout entier $a \\geqslant 1$, si $a^{n}-1$ est divisible par $p$, alors $a^{n}-1$ est aussi divisible par $p^{2}$. Puis, réciproquement, nous allons prouver que si $p$ ne divise pas $n$, alors il existe un entier $a \\geqslant 1$ tel que $a^{n}-1$ est divisible par $p$ mais tel que $a^{n}-1$ n'est pas divisible par $p^{2}$.\n\n$\\triangleright$ Si $n=p m$ pour un entier $m \\geqslant 1$, alors $a^{n}=a^{p m} \\equiv a^{m} \\pmod{p}$ d'après le petit théorème de Fermat pour tout entier $a$. Supposons dorénavant que l'entier $a$ est tel que $a^{n}-1$ est divisible par $p$. D'après la ligne ci-dessus, $a^{m}-1$ est divisible par $p$. Par ailleurs, la factorisation suivante est vérifiée:\n$$\na^{n}-1=a^{p m}-1=\\left(a^{m}-1\\right)\\left(1+a^{m}+\\cdots+a^{(p-1) m}\\right)\n$$\nComme $p$ divise $a^{m}-1$, $a^{m} \\equiv 1 \\pmod{p}$ donc\n$$\n1+a^{m}+\\cdots+a^{(p-1) m} \\equiv 1+1+\\cdots+1 \\equiv p \\equiv 0 \\pmod{p}\n$$\nEn conclusion, $a^{n}-1$ est le produit de deux entiers divisibles par $p$ donc est divisible par $p^{2}$.\n\n$\\triangleright$ Si $p$ ne divise pas $n$, alors avec l'entier $a=p+1$, on a que $a^{n}-1 \\equiv 1^{n}-1 \\equiv 0 \\pmod{p}$ donc $p$ divise $a^{n}-1$. Or\n$$\n1+a+\\cdots+a^{(n-1)} \\equiv 1+1+\\cdots+1^{n-1} \\equiv n \\not \\equiv 0 \\pmod{p}\n$$\nAinsi, $p$ ne divise pas $1+a+\\cdots+a^{(n-1)}$, et donc $p^{2}$ ne divise pas $p\\left(1+a+\\cdots+a^{(n-1)}\\right)=(a-1)\\left(1+a+\\cdots+a^{(n-1)}\\right)=a^{n}-1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle isocèle en $A$. Soient $M$ et $N$ deux points de $[BC]$. Les droites $[AM]$ et $[AN]$ recoupent le cercle circonscrit à $ABC$ en $P$ et $Q$.\n\n1) Montrer que $M, N, P, Q$ sont cocycliques.\n\n2) Soient $R_{1}$ et $R_{2}$ les rayons des cercles circonscrits à $BMP$ et $CMP$. Calculer $R_{1}+R_{2}$ en fonction des longueurs $AB$ et $BC$.",
    "options": [],
    "answer": "R1 + R2 = (AB × BC) / (2 √(AB^2 − (BC/2)^2))",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\n1) $\\widehat{PMN}=\\widehat{PMC}=\\pi-\\widehat{MCP}-\\widehat{CPM}=\\pi-\\widehat{BCP}-\\widehat{CPA}=\\pi-\\widehat{BCP}-\\widehat{CBA}$, et $\\widehat{NQP}=\\widehat{AQP}=\\widehat{ACP}=\\widehat{ACB}+\\widehat{BCP}=\\widehat{CBA}+\\widehat{BCP}$,\ndonc $\\widehat{PMN}=\\pi-\\widehat{NQP}$, ce qui prouve la cocyclicité de $M, N, P, Q$.\n\n2) D'après la loi des sinus, on a\n$$\n2\\left(R_{1}+R_{2}\\right)=\\frac{BM}{\\sin \\widehat{MPB}}+\\frac{CM}{\\sin \\widehat{CPM}}\n$$\nOr, $\\widehat{MPB}=\\widehat{APB}=\\widehat{ACB}$ et de même $\\widehat{CPM}=\\widehat{ACB}$, donc\n$$\n2\\left(R_{1}+R_{2}\\right)=\\frac{BM+CM}{\\sin \\widehat{ACB}}=\\frac{BC}{\\sin \\widehat{ACB}}.\n$$\nD'autre part, soit $H$ le projeté orthogonal de $A$ sur $BC$. On a $\\sin \\widehat{ACB}=\\frac{AH}{AB}=\\frac{\\sqrt{AB^{2}-(BC / 2)^{2}}}{AB}$, donc\n$$\nR_{1}+R_{2}=\\frac{AB \\times BC}{2 \\sqrt{AB^{2}-(BC / 2)^{2}}}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver toutes les fonctions $f$ de $\\mathbb{R}$ dans $\\mathbb{R}$ telles que, pour tous réels $x$ et $y$, on ait\n$$\nf(x f(y))+x=f(x) f(y+1)\n$$",
    "options": [],
    "answer": "f(x)=x and f(x)=-x",
    "solution": "Solution:\nEn prenant $x=0$, on obtient $f(0)=f(0) f(y+1)$ pour tout $y$, donc soit $f(0)=0$, soit $f(y+1)=1$ pour tout $y$. Dans le second cas, $f$ est constante égale à $1$, mais alors l'équation devient\n$$\n1+x=1 \\times 1\n$$\npour tout $x$, ce qui est impossible. On a donc $f(0)=0$.\n\nEn prenant $y=0$, on obtient alors $x=f(x) f(1)$ pour tout $x$. En particulier, en prenant $x=1$, on obtient $1=f(1)^2$ donc $f(1)$ vaut $1$ ou $-1$.\n\nSi $f(1)=1$, alors $x=f(x)$ donc $f$ est l'identité, qui est bien solution.\n\nSi $f(1)=-1$, on obtient $x=-f(x)$ pour tout $x$, donc $f(x)=-x$ pour tout $x$, et on vérifie que cette fonction aussi est bien solution du problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $H$ son orthocentre. Le cercle circonscrit de $ABH$, de centre $O$, recoupe la droite $(BC)$ en $D$, la droite $(DH)$ intersecte la droite $(AC)$ en $P$. Montrer que le centre du cercle circonscrit de $APD$ est sur le cercle circonscrit de $ODB$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUne construction soigneuse nous permet de conjecturer que le centre du cercle $APD$ est également sur la droite $(AB)$, c'est donc tout naturellement que l'on introduit $O_1$, le point de la droite $(AB)$ qui est à égale distance de $A$ et de $D$, $c'$ est alors un bon candidat pour être le centre du cercle $APD$.\n\nÉtape 1. $O_1$ est le centre du cercle de $APD$.\n\nOn introduit $H_B$ le pied de la hauteur issue de $B$ dans le triangle $ABC$, on pose alors $x = \\widehat{H_BPH}$. Comme le triangle $HH_BP$ est droit en $H_B$, on a $\\widehat{H_BHP} = 90 - \\widehat{H_BPH} = 90 - x$. Mais d'autre part, $x = \\widehat{H_BBP} = \\widehat{DHB}$, ce dernier angle vaut l'angle $\\widehat{DAB} = x$ par cocyclicité des points $A, H, B$ et $D$.\n\nOn note $M$ le milieu du segment $[AD]$, alors $M$ et $O_1$ sont sur la médiatrice du segment $[AD]$, donc $(MO_1) \\perp (AD)$, et ainsi $\\widehat{MO_1A} = 90 - \\widehat{MAO_1}$.\n\nOr $O_1A = OD$, donc $\\widehat{AO_1D} = 2x = 2\\widehat{APD}$. On a maintenant d'une part que $\\widehat{AO_1D} = 2\\widehat{APD}$ mais d'autre part que $OA = O_1D$, on voudrait en déduire que $O_1$ est bien le centre du cercle circonscrit de $APD$. Mais par angle au centre c'est bien le cas.\n\nÉtape 2. $O_1$ est sur le cercle $ODB$.\n\nOn remarque que les points $O, M$ et $O_1$ sont sur la (même) médiatrice de $[AC]$, ils sont donc alignés. On trouve ainsi l'angle $\\widehat{OO_1D} = x$. Mais d'autre part, $\\widehat{DOB} = 2\\widehat{DAB} = 2(90 - x)$. Comme le triangle $ODB$ est isocèle en $O$, on peut calculer l'angle $\\widehat{OBD}$ qui vaut ainsi $x$. Mais alors, $\\widehat{OO_1D} = x = \\widehat{OBD}$, ce qui conclut l'étape 2.\n\nOn sait que $O_1$ est le centre du cercle $APD$ et sur le cercle $ODB$, donc le centre du cercle de $APD$ est bien sur le cercle $ODB$ ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les triplets d'entiers strictement positifs $(a, b, c)$ tels que $6^{a} = 1 + 2^{b} + 3^{c}$.",
    "options": [],
    "answer": "[(1,1,1), (2,3,3), (2,5,1)]",
    "solution": "Solution:\n\nRemarquons tout d'abord que $3$ divise $2^{b} + 1$ donc $b$ est impair (car $2^{b} + 1 \\equiv (-1)^{b} + 1 \\pmod{3}$).\n\n$\\triangleright$ Si $b = 1$, l'équation se réécrit $1 + 3^{c-1} = 2 \\cdot 6^{a-1}$ en divisant par $3$.\n\n- Si $a > 1$, alors $3$ divise $1 + 3^{c-1}$ ce qui est impossible.\n- Sinon, $a = 1$ et par conséquent $c = 1$.\n\n$\\triangleright$ Si $b \\geqslant 3$, alors $a \\geqslant 2$ (car $6 < 1 + 2^{3}$). En réduisant modulo $4$, on obtient la condition nécessaire $1 + 3^{c} \\equiv 0 \\pmod{4}$, et donc $c$ est impair. En réduisant alors modulo $8$, on obtient la nouvelle condition $1 + 2^{b} + 3^{c} \\equiv 1 + 0 + 3 \\pmod{8} \\equiv 4 \\pmod{8}$ donc $6^{a} \\equiv 4 \\pmod{8}$ ce qui entraîne immédiatement $a = 2$ (en effet, $6^{a} \\equiv 0 \\pmod{8}$ pour $a \\geqslant 3$).\n\nL'équation initiale devient alors $2^{b} + 3^{c} = 35$ et une étude exhaustive donne $b = 3$ et $c = 3$ ou $b = 5$ et $c = 1$.\n\nEn conclusion, les triplets solutions sont $\\{(1,1,1), (2,3,3), (2,5,1)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les triplets de nombres premiers $(p, q, r)$ tels que $(p+1)(q+2)(r+3)=4 p q r$.",
    "options": [],
    "answer": "(2,3,5), (7,5,2), (5,3,3)",
    "solution": "Solution:\n\nSi $p=2$ alors on a $3(q+2)(r+3)=8 q r$ ce qui implique $q=3$ ou $r=3$. On a alors la solution $(2,3,5)$.\n\nSi $q=2$ et $p, r>2$ alors on a $(p+1)(r+3)=2 p r$. Puisque $p$ et $r$ sont impairs, le membre de gauche est divisible par 4, alors que le membre de droite est divisible par 2, mais pas par 4, contradiction. Donc $p$ ou $r$ vaut 2 mais aucune solution valide n'en découle.\n\nSi $r=2$ alors $5(p+1)(q+2)=8 p q$ et donc $p$ ou $q$ vaut 5. On obtient la solution $(7,5,2)$.\n\nSinon, $(p+1)$ est pair et $(r+3)$ est pair. Donc on se ramène à\n$$\n\\left(\\frac{p+1}{2}\\right)(q+2)\\left(\\frac{r+3}{2}\\right)=p q r.\n$$\nNous avons donc écrit $p q r$ comme un produit de trois facteurs strictement plus grands que 1. On en déduit que $\\left(\\frac{p+1}{2}, q+2, \\frac{r+3}{2}\\right)$ est une permutation de $(p, q, r)$.\n\nClairement, $p \\neq \\frac{p+1}{2}$ puisque $p>\\frac{p+1}{2}$.\n\nSi $p=\\frac{r+3}{2}$, alors puisque $q \\neq q+2$, nous avons nécessairement $q=\\frac{p+1}{2}=\\frac{r+5}{4}$. Alors $r=q+2=\\frac{r+13}{4}$, et donc $r$ n'est pas entier.\n\nSi $p=q+2$, alors il y a deux cas à considérer :\n\n- $r=\\frac{p+1}{2}=\\frac{q+3}{2}$ et $q=\\frac{r+3}{2}=\\frac{q+9}{4}$. Donc $q=3$ et on obtient la solution $(5,3,3)$.\n\n- $r=\\frac{r+3}{2}$ et $q=\\frac{p+1}{2}=\\frac{q+3}{2}$, ce qui donne $r=q=3$. On obtient alors cette même solution $(5,3,3)$.\n\nFinalement, les seules solutions sont $(2,3,5)$, $(7,5,2)$ et $(5,3,3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn grand carré de côté $n$ est découpé en $n^{2}$ petits carrés de côté $1$. On veut colorier en rouge ou bleu chacun des $(n+1)^{2}$ sommets des petits carrés de telle manière que chacun des petits carrés a exactement $2$ sommets rouges. Combien y a-t-il de coloriages possibles?",
    "options": [],
    "answer": "2^{n+2} - 2",
    "solution": "Solution:\n\nLa réponse est $2^{n+2} - 2$. Pour le montrer, on commence par colorier les $n+1$ sommets les plus hauts. Il y a $2^{n+1}$ manières de le faire.\n\nSi il existe deux sommets consécutifs de même couleur sur la rangée supérieure ($2^{n+1} - 2$ coloriages de la rangée supérieure), alors ils fixent les couleurs sur la rangée juste en-dessous, et ainsi de suite, ce qui donne $2^{n+1} - 2$ coloriages.\n\nSi les couleurs sur la rangée supérieure sont alternées (2 coloriages de la rangée supérieure), alors on est obligé d'alterner les couleurs sur la rangée d'en-dessous, et sur les suivantes... On peut cependant, à chaque ligne, choisir la couleur par laquelle on commence, ce qui donne $2 \\times 2^{n}$ coloriages, donc $2^{n+2} - 2$ au total.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nÀ l'intérieur du cercle $\\Gamma$ se trouvent trois cercles $\\gamma_{1}$, $\\gamma_{2}$ et $\\gamma_{3}$, et tangents à $\\Gamma$ respectivement en $A$, $B$ et $C$, tous distincts. Les cercles $\\gamma_{2}$ et $\\gamma_{3}$ ont un point commun $K$ qui appartient à $[BC]$, les cercles $\\gamma_{3}$ et $\\gamma_{1}$ ont un point commun $L$ qui appartient à $[CA]$, et les cercles $\\gamma_{1}$ et $\\gamma_{2}$ ont un point commun $M$ qui appartient à $[AB]$.\n\nProuver que le centre de $\\Gamma$ appartient à $\\gamma_{1}$, $\\gamma_{2}$ et $\\gamma_{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSoit $X$ un point de la tangente commune à $\\Gamma$ et $\\gamma_{1}$, et situé du côté opposé à $C$ par rapport à la droite $(AB)$. Alors $\\widehat{ALM}=\\widehat{XAM}=\\widehat{XAB}=\\widehat{ACB}$, ce qui assure que $(ML)$ et $(BC)$ sont parallèles. De même, on a $(KM)$ et $(AC)$ parallèles, ainsi que $(KL)$ et $(AB)$ parallèles.\n\nPosons $x=\\frac{AM}{AB}$. D'après le théorème de Thalès, on a alors :\n$$\n\\frac{BK}{BC}=\\frac{BM}{BA}=1-x \\text{ et } \\frac{CL}{CA}=\\frac{CK}{CB}=1-(1-x)=x\n$$\nAinsi $x=\\frac{AM}{AB}=\\frac{AL}{AC}=1-x$, d'où $x=\\frac{1}{2}$.\n\nLes points $M$, $K$, $L$ sont donc les milieux respectifs des segments $[AB]$, $[BC]$, $[CA]$, et les triangles $AML$, $MBK$, $LKC$ sont tous semblables à $ABC$ dans un rapport $\\frac{1}{2}$. Si on note $r_{1}$, $r_{2}$, $r_{3}$ et $R$ les rayons respectifs des cercles $\\gamma_{1}$, $\\gamma_{2}$, $\\gamma_{3}$ et $\\Gamma$, on en déduit que $r_{1}=r_{2}=r_{3}=\\frac{1}{2}R$.\n\nD'autre part, puisque $\\gamma_{1}$ et $\\Gamma$ sont tangents en $A$ les diamètres issus de $A$ dans $\\gamma_{1}$ et $\\Gamma$ sont portés par la même droite. Ainsi, le centre de $\\Gamma$ appartient à $\\gamma_{1}$, et on prouve de même qu'il appartient également à $\\gamma_{2}$ et $\\gamma_{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer tous les réels $t$ pour lesquels le polynôme\n$$\nP(x)=x^{3}+3 t x^{2}+(4 t-1) x+t\n$$\npossède deux racines réelles dont la différence est égale à 1.",
    "options": [],
    "answer": "10/9, 6/5, 0",
    "solution": "Solution:\nSi $t$ est un réel fixé, on pose $P_{t}(x)=x^{3}+3 t x^{2}+(4 t-1) x+t$.\nLa première chose à remarquer est que, pour tout $t$, on a $P_{t}(-1)=0$. Par suite, -1 est une racine du polynôme $P_{t}$, ce qui assure que $P_{t}(x)$ est factorisable par $x+1$. On trouve ainsi que\n$$\nP_{t}(x)=(x+1)\\left(x^{2}+(3 t-1) x+t\\right)\n$$\npour tout réels $x$ et $t$.\nSoit $t$ un réel pour lequel $P_{t}$ possède deux racines réelles dont la différence est égale à 1, disons $a$ et $a+1$.\n- Si $a=-1$ : il faut et il suffit que 0 soit également une racine de $P_{t}$. Comme $P_{t}(0)=t$, cela signifie que $t=0$.\n- Si $a+1=-1$, il faut et il suffit que -2 soit également une racine de $P_{t}$. Comme $P_{t}(-2)=5 t-6$, cela signifie que $t=\\frac{6}{5}$.\n- Si -1 n'est ni $a$ ni $a+1$, c'est donc que $a$ et $a+1$ sont les racines de $x^{2}+(3 t-1) x+t$. On a donc $a^{2}+(3 t-1) a+t=0$ et $(a+1)^{2}+(3 t-1)(a+1)+t=0$. Par soustraction, il vient $2 a+1+3 t-1=0$, soit donc $a=-\\frac{3 t}{2}$. En reportant dans la première équation ci-dessus, il vient alors $-9 t^{2}+10 t=0$, soit $t=0$ ou $t=\\frac{10}{9}$.\nRéciproquement, on a déjà vu que $t=0$ convenait et, pour $t=\\frac{10}{9}$, les racines de $P_{t}$ sont $-1,-\\frac{2}{3}$ et $-\\frac{5}{3}$, ce qui prouve que $t=\\frac{10}{9}$ est bien une solution du problème.\nFinalement, les réels $t$ cherchés sont $\\frac{10}{9}, \\frac{6}{5}$ et $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver toutes les fonctions $f$ de $\\mathbb{R}$ dans $\\mathbb{R}$ telles que pour tous réels $x, y$,\n$$\nf(x) f(y)=f(x y)+x+y\n$$",
    "options": [],
    "answer": "f(x) = 1 + x for all real x",
    "solution": "Solution:\n\nEn évaluant en $y=0$ on obtient $f(x) f(0)=f(0)+x$ pour tout réel $x$.\n\nPour $x=1$, on obtient que $f(0) \\neq 0$, sinon on aurait $1=0$.\n\nEn évaluant en $x=0$ on obtient $f(0)^2=f(0)$ donc $f(0)=1$.\n\nDonc $f(x)=1+x$ pour tout $x$ réel.\n\nRéciproquement, soit $f$ la fonction telle que $f(x)=1+x$ pour tout $x$ réel et soit $(x, y) \\in \\mathbb{R}^2$, alors $f(x) f(y)=(1+x)(1+y)=1+xy+x+y=f(xy)+x+y$ donc $f$ vérifie bien l'équation.\n\nL'unique solution de l'équation est donc la fonction $f$ telle que $f(x)=1+x$ pour tout $x$ réel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier. On suppose qu'il existe exactement 2005 couples $(x, y)$ d'entiers naturels tels que $\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{n}$. Montrer que $n$ est le carré d'un entier.\n\nN.B. Si $x \\neq y$ alors $(x, y) \\neq (y, x)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn remarque que $x > n$ et $y > n$.\nL'équation s'écrit $x y = n(x + y)$, ou encore $n^{2} = (x - n)(y - n)$. On en déduit qu'il y a exactement 2005 couples d'entiers naturels $(u, v)$ tels que $n^{2} = u v$.\nSi $n^{2} = u v$ alors $u$ est un diviseur de $n^{2}$. Réciproquement, si $u$ est un diviseur de $n^{2}$ alors l'équation $n^{2} = u v$ détermine uniquement $v$. On en déduit que $n^{2}$ admet exactement 2005 diviseurs.\n\nÉcrivons $n = p_{1}^{\\alpha_{1}} \\cdots p_{r}^{\\alpha_{r}}$ avec $p_{i}$ premiers distincts et $\\alpha_{i}$ entiers. On a donc $n^{2} = p_{1}^{2 \\alpha_{1}} \\cdots p_{r}^{2 \\alpha_{r}}$. On sait que le nombre de diviseurs de $n^{2}$ est égal à $\\left(2 \\alpha_{1} + 1\\right) \\cdots \\left(2 \\alpha_{r} + 1\\right)$. Or, la décomposition en facteurs premiers de 2005 est $5 \\times 401$, donc soit $r = 1$ et $2 \\alpha_{1} + 1 = 2005$, soit $r = 2$ et $2 \\alpha_{1} + 1 = 5$, $2 \\alpha_{2} + 1 = 401$. Dans tous les cas, les $\\alpha_{i}$ sont pairs donc $n$ est un carré parfait.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\omega_{1}$ et $\\omega_{2}$ deux cercles disjoints. Une tangente commune extérieure à ces deux cercles coupe $\\omega_{1}$ en $A$ et $\\omega_{2}$ en $B$. La deuxième tangente commune extérieure à ces deux cercles coupe $\\omega_{1}$ en $D$ et $\\omega_{2}$ en $C$. La droite $(BD)$ coupe $\\omega_{1}$ en le point $P$ autre que $D$ et $\\omega_{2}$ en le point $Q$ autre que $B$. Montrer que $BQ = DP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nUne symétrie axiale par rapport à la droite joignant les centres des deux cercles échange $A$ et $D$ ainsi que $B$ et $C$. Donc, $AB = DC$. On pense alors à appliquer la puissance d'un point, on a alors :\n$$\nDQ \\cdot DB = DC^{2} = AB^{2} = BP \\cdot BD\n$$\nou la première égalité vient de la puissance du point depuis $D$ par rapport à $\\omega_{2}$ et la troisième égalité vient de la puissance du point depuis $B$ par rapport au cercle $\\omega_{1}$. On peut réécrire cette identité de la manière suivante :\n$$\n(BP - DQ) BD = 0\n$$\nCela implique que $BP = DQ$ et donc que $DP = BQ$, d'où la conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle isocèle et obtus en $A$. Soit $\\Gamma$ le cercle de centre $B$ passant par $A$, et $\\Omega$ le cercle de centre $C$ passant par $A$. Soit $D$ le point d'intersection du cercle $\\Gamma$ avec le segment $[BC]$, $E$ le deuxième point d'intersection de la droite $(AD)$ avec le cercle $\\Omega$, et $F$ le point d'intersection de la droite $(BC)$ avec le cercle $\\Omega$ qui n'est pas sur le segment $[BC]$.\nMontrer que le triangle $DFE$ est isocèle en $F$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $X$ le point d'intersection du cercle $\\Omega$ avec le segment $[BC]$. Les points $B$ et $C$ sont symétriques par rapport à la médiatrice du segment $[BC]$ donc les cercles $\\Gamma$ et $\\Omega$ le sont aussi. Il vient que $D$ et $X$ sont symétriques par rapport à la médiatrice du segment $[BC]$. Le triangle $DAX$ est donc isocèle en $A$ et $\\widehat{ADX} = \\widehat{AXD}$.\n\nLes points $F, E, X$ et $A$ sont cocycliques donc $\\widehat{FEA} = \\widehat{FXA}$. On déduit\n$$\n\\widehat{FED} = \\widehat{FEA} = \\widehat{FXA} = \\widehat{DXA} = \\widehat{XDA} = \\widehat{FDE}\n$$\nCe qui donne bien que le triangle $FDE$ est isocèle en $F$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\left(a_{n}\\right)_{n \\geqslant 1}$ une suite croissante d'entiers strictement positifs et $k$ un entier strictement positif. Supposons que pour un certain $r \\geqslant 1$, on $\\frac{r}{a_{r}}=k+1$. Montrer qu'il existe un entier $s \\geqslant 1$ tel que $\\frac{s}{a_{s}}=k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $v_{n}=n-k a_{n}$ pour tout $n \\geqslant 1$. Le problème revient à montrer qu'il existe $s$ tel que $v_{s}=0$.\n\nOn observe que $v_{1}=1-k a_{1} \\leqslant 0$, que $v_{r}=r-k a_{r}=a_{r}>0$ et qu'enfin $v_{n+1}= (n+1)-k a_{n+1}=n-k a_{n+1}+1 \\leqslant n-k a_{n}+1=v_{n}+1$, car la suite $\\left(a_{n}\\right)_{n}$ est croissante.\n\nIntuitivement, on a donc une suite d'entiers qui part d'un nombre négatif, atteint un nombre strictement positif, et ne peut augmenter que d'au plus 1 à chaque fois : elle doit atteindre 0. Prouvons-le : soit $s$ le plus grand entier inférieur à $r$ tel que $v_{s} \\leqslant 0$ (il existe car $v_{1} \\leqslant 0$). Alors, comme $v_{r}>0$, on a $s<r$, donc $v_{s+1}>0$. On a donc $v_{s+1} \\geqslant 1$, donc $0 \\geqslant v_{s} \\geqslant v_{s+1}-1 \\geqslant 0$, donc $v_{s}=0$, de sorte que $\\frac{s}{a_{s}}=k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $H$ le pied de la hauteur issue de $A$, $L$ le pied de la bissectrice issue de $B$ et $M$ le milieu de $[AB]$. On suppose que dans le triangle $HLM$ on a les deux propriétés : $(AH)$ est une hauteur et $(BL)$ est une bissectrice. Montrer que $(CM)$ est une médiane.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn prend le triangle $ABC$, on va démontrer qu'il est équilatéral, en utilisant les deux propriétés, la troisième découlera.\nOn a $(AH) \\perp (ML)$ et $(AH) \\perp (BC)$, donc $(BC) // (ML)$. On trouve ainsi que $L$ est le milieu de $[AC]$. Cependant, $\\frac{AL}{LC} = \\frac{AB}{BC}$ ce qui montre que $AB = BC$, et $(BL)$ est une bissectrice, une hauteur et une médiane.\n\nIl faut maintenant que $(BL)$ bissecte $\\widehat{MLH}$, ou encore $(AC)$ comme bissectrice extérieure, donc par chasse aux angles $\\beta = \\widehat{HLC} = \\widehat{MLA} = \\alpha$. Donc $\\alpha = \\beta$ et $ABC$ est équilatéral, donc en particulier $(CM)$ est la médiane dans $MHL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\Gamma$ un cercle, $P$ un point à l'extérieur du cercle. Les tangentes au cercle $\\Gamma$ passant par le point $P$ sont tangentes au cercle $\\Gamma$ en $A$ et $B$. Soit $M$ est le milieu du segment $[BP]$ et $C$ le point d'intersection de la droite $ (AM) $ et du cercle $\\Gamma$. Soit $D$ la deuxième intersection de la droite $ (PC) $ et du cercle $\\Gamma$.\n\nMontrer que les droites $(AD)$ et $(BP)$ sont parallèles.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEn utilisant la puissance du point $M$ par rapport au cercle $\\Gamma$, $MB^2 = MC \\cdot MA$. Puisque $M$ est le milieu du segment $[BP]$, $MP^2 = MB^2$ donc $MP^2 = MC \\cdot MA$. On déduit de la réciproque de la puissance d'un point par rapport à un cercle que la droite $(PM)$ est tangente au cercle circonscrit au triangle $PAC$. On obtient du théorème de l'angle tangent que $\\widehat{MPC} = \\widehat{PAC}$. Or la droite $(PA)$ est tangente au cercle $\\Gamma$ donc à nouveau par le théorème de l'angle tangent, $\\widehat{PAC} = \\widehat{ADC}$. En résumé :\n$$\n\\widehat{BPD} = \\widehat{MPC} = \\widehat{PAC} = \\widehat{ADC} = \\widehat{ADP}\n$$\ndonc les droites $(AD)$ et $(BP)$ sont parallèles.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les triplets de réels $\\left(a, b, c\\right)$ vérifiant le système d'égalités :\n$$\n\\left\\{\n\\begin{array}{l}\na\\left(b^{2}+c\\right)=c(c+a b) \\\\\nb\\left(c^{2}+a\\right)=a(a+b c) \\\\\nc\\left(a^{2}+b\\right)=b(b+a c)\n\\end{array}\n\\right.\n$$",
    "options": [],
    "answer": "All real triples with a = b = c (any real value).",
    "solution": "Solution:\n\nLes égalités se réécrivent $a b(b-c)=c(c-a)$, $b c(c-a)=a(a-b)$ et $c a(a-b)=b(b-c)$. En multipliant ces trois égalités, on obtient $(a b c)^{2}(c-a)(b-c)(a-b)=a b c(c-a)(b-c)(a-b)$. En particulier deux cas se présentent :\n\n- Soit $a b c=0$. Les égalités étant cycliques, supposons $a=0$. Par la première égalité $c^{2}=0$ donc $c=0$. Par la troisième égalité $b^{2}=0$ donc $b=0$.\n\n- Soit deux des éléments parmi $a, b, c$ sont égaux, on suppose $a=b$. Dans ce cas $b c(c-a)=0$. Si $b=0$ ou $c=0$ on retombe dans le cas précédent, sinon $a=b=c$.\n\n- Soit $a b c=1$, dans ce cas $a, b, c$ sont non nuls. Les égalités se réécrivent : $b-c=c^{2}(c-a)$, $c-a=a^{2}(a-b)$ et $a-b=b^{2}(b-c)$. En particulier $a-b, b-c$ et $c-a$ sont de même signe, mais de somme nulle et on a donc nécessairement $a-b=b-c=c-a=0$ i.e. $a=b=c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les couples de nombres premiers $\\left(p, q\\right)$ tels que :\n$$\np^{2}\\left(p^{3}-1\\right)=q(q+1)\n$$",
    "options": [],
    "answer": "no solution",
    "solution": "Solution:\nSi $p = q$, alors $p^{4} - p = p + 1$. Or $p^{4} = p \\cdot p^{3} \\geqslant 8p > 2p + 1$ pour tout nombre premier $p$ car $p \\geqslant 2$. On a donc $p \\neq q$ et donc $p$ et $q$ sont premiers entre eux.\n\n$p^{2}$ est premier avec $q$ donc d'après le lemme de Gauss, $p^{2} \\mid q + 1$. On dispose donc de $k \\in \\mathbb{N}^{*}$ tel que $q + 1 = k p^{2}$.\n\nComme $q$ est premier avec $p^{2}$, $q \\mid p^{3} - 1 = (p - 1)(p^{2} + p + 1)$. Donc $q \\mid p - 1$ ou $q \\mid p^{2} + p + 1$.\n\nSi $q \\mid p - 1$, alors\n$$\np < k p^{2} - 1 = q \\leqslant p - 1\n$$\nce qui est absurde.\n\nOn a donc $q \\mid p^{2} + p + 1$ donc $k p^{2} - 1 = q \\leqslant p^{2} + p + 1$ soit $(k - 1) p^{2} \\leqslant p + 2$. Si $k = 1$, alors $q = p^{2} - 1 = (p - 1)(p + 1)$ qui n'est premier que si $p = 2$.\n\nOn a alors $q = 3$ mais $2^{2}(2^{3} - 1) = 28 \\neq 12 = 3(3 + 1)$.\n\nOn en déduit que $k \\geqslant 2$ et $p > 2$, donc $p + 2 < p^{2} \\leqslant (k - 1) p^{2} \\leqslant p + 2$ ce qui est également absurde.\n\nFinalement il n'y a pas de solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n, m \\geqslant 1$ des entiers, avec $m$ impair.\nProuver que $2^{m}-1$ et $2^{n}+1$ sont premiers entre eux.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPar l'absurde : supposons qu'il existe un nombre premier $p$ qui divise à la fois $2^{m}-1$ et $2^{n}+1$.\nAlors, on a $2^{m} \\equiv 1 \\bmod [p]$ et $2^{n} \\equiv -1 \\bmod [p]$,\nd'où $2^{2n} \\equiv 1 \\bmod [p]$.\nSoit $\\omega$ l'ordre de $2$ modulo $p$. On sait qu'alors $\\omega$ divise $m$ et $2n$. Mais, comme $m$ est impair, $\\omega$ est donc impair, ce qui assure qu'en fait $\\omega$ divise $n$.\nOn a alors $2^{n} \\equiv 1 \\bmod [p]$, et toujours $2^{n} \\equiv -1 \\bmod [p]$.\nPar suite, $p$ divise $2$, et donc $p=2$.\nMais, on a $m \\geqslant 1$, d'où $2^{m}-1$ est impair et n'est donc pas divisible par $2$. Contradiction.\nFinalement, $2^{m}-1$ et $2^{n}+1$ sont premiers entre eux.\n\nMontrons d'abord par récurrence sur $a+b$ que le pgcd de $2^{a}-1$ et $2^{b}-1$ est $2^{\\operatorname{pgcd}(a, b)}-1$. Si $a=0$ ou $b=0$ c'est clair. Supposons la propriété vraie pour tous les couples ($a'$, $b'$) tels que $a'+b'<a+b$. On peut supposer $a \\geqslant b>0$. On a\n$$\n\\begin{aligned}\n\\operatorname{pgcd}\\left(2^{a}-1,2^{b}-1\\right) & =\\operatorname{pgcd}\\left(2^{b}-1,2^{a}-1-\\left(2^{b}-1\\right)\\right) \\\\\n& =\\operatorname{pgcd}\\left(2^{b}-1,2^{b}\\left(2^{a-b}-1\\right)\\right) \\\\\n& =\\operatorname{pgcd}\\left(2^{b}-1,2^{a-b}-1\\right) \\quad \\operatorname{car}~2^{b}-1~\\text{et}~2^{b}~\\text{premiers entre eux} \\\\\n& =2^{\\operatorname{pgcd}(b, a-b)}-1 \\quad \\text{par hypothèse de récurrence} \\\\\n& =2^{\\operatorname{pgcd}(a, b)}-1\n\\end{aligned}\n$$\nce qui démontre notre assertion.\nRevenons à l'exercice. Notons $d' = \\operatorname{pgcd}\\left(2^{m}-1,2^{n}+1\\right)$. Soit $d=\\operatorname{pgcd}(m, n)$, alors ce qui précède montre que $2^{d}-1=\\operatorname{pgcd}\\left(2^{m}-1,2^{2n}-1\\right)$. Or, $2^{2n}-1$ est un multiple de $2^{n}+1$, donc $d'$ divise $2^{d}-1$. Comme $2^{d}-1$ divise $2^{n}-1$, $d'$ divise aussi $\\left(2^{n}+1\\right)-\\left(2^{n}-1\\right)=2$. Or, $2^{m}-1$ est impair donc $d'$ est impair. Il vient $d'=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans un club d'échecs de 20 personnes, 14 parties sont jouées et chaque joueur a joué au moins une partie. Montrer qu'il y a 6 parties auxquelles 12 joueurs différents ont participé.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $r$ le plus grand entier tel qu'il existe un ensemble $M$ de $r$ parties faisant intervenir $2r$ joueurs différents.\n\nNotons $J$ l'ensemble des $2r$ personnes qui ont joué les parties de $M$. Par maximalité de $M$, si $a$ et $b$ sont deux joueurs n'appartenant pas à $J$, alors $a$ et $b$ n'ont pas joué entre eux. Par conséquent, ils ont joué une partie contre l'un des $2r$ joueurs de $M$. On en déduit que le nombre total de parties est au moins égal à $r + (20 - 2r) = 20 - r$. Or, par hypothèse le nombre de parties est $14$, donc $14 \\geqslant 20 - r$, ce qui entraîne $r \\geqslant 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers $n \\geqslant 2$ vérifiant la propriété suivante : pour tous entiers $a_{1}, a_{2}, \\ldots, a_{n}$ dont la somme n'est pas divisible par $n$, il existe un indice $i$ tel qu'aucun des nombres\n$$\na_{i}, a_{i}+a_{i+1}, \\ldots, a_{i}+\\cdots+a_{i+n-1}\n$$\nn'est divisible par $n$ (pour $i>n$, on pose $a_{i}=a_{i-n}$ ).",
    "options": [],
    "answer": "All prime numbers",
    "solution": "Solution:\n\nCe sont exactement les nombres premiers!\n\nEn effet, si $n=ab$, on peut prendre $a_{1}=0$ et $a_{2}=\\cdots=a_{n}=a$. La somme des $a_{i}$ vaut $a(n-1)$ donc n'est pas divisible par $n$. Cependant, soit $1 \\leqslant i \\leqslant n$. Si $i+b-1 \\leqslant n$, alors le nombre $a_{i}+\\cdots+a_{i+b-1}=ab=n$ est divisible par $n$. Si $i+b-1>n$, alors le nombre $a_{i}+\\cdots+a_{i+b}=ab=n$ est divisible par $n$.\n\nRéciproquement, supposons $n$ premier, et soient $a_{1}, \\ldots, a_{n}$ des entiers dont la somme n'est pas divisible par $n$. Si $n$ ne vérifie pas la propriété, alors pour tout indice $i$, il existe $j(i)$ avec $i+1 \\leqslant j(i) \\leqslant i+n$ tel que\n$$\na_{i}+a_{i+1}+\\cdots+a_{j(i)-1}\n$$\nest divisible par $n$. De plus, comme la somme des $a_{i}$ n'est pas divisible par $n$, on ne peut pas avoir $j(i)=i+n$, donc $i+1 \\leqslant j(i) \\leqslant i+n-1$. On définit alors par récurrence une suite d'indices $(i_{n})$ par $i_{1}=1$ et $i_{n+1}=j(i_{n})$. On sait que pour tout $k$, l'entier\n$$\na_{i_{k}}+\\cdots+a_{i_{k+1}-1}\n$$\nest divisible par $n$ donc, en sommant, pour tous indices $k<\\ell$, l'entier\n$$\na_{i_{k}}+\\cdots+a_{i_{\\ell}-1}\n$$\nest divisible par $n$. Par le principe des tiroirs, il existe $1 \\leqslant k<\\ell \\leqslant n+1$ tels que $i_{k} \\equiv i_{\\ell} \\pmod{n}$. Le nombre de termes de la somme (1) vaut alors $i_{\\ell}-i_{k}$ donc est divisible par $n$, donc chacun des $a_{i}$ apparaît exactement $\\frac{i_{\\ell}-i_{k}}{n}$ fois. De plus, on sait que $i_{j+1}-i_{j} \\leqslant n-1$ pour tout $j$ et que $\\ell-k \\leqslant n$, donc $i_{\\ell}-i_{k} \\leqslant n(n-1)$. La somme (1) vaut donc\n$$\n\\frac{i_{\\ell}-i_{k}}{n} \\times \\sum_{i=1}^{n} a_{i} .\n$$\nMais $\\frac{i_{\\ell}-i_{k}}{n} \\leqslant n-1$ donc ne peut pas être divisible par $n$, et la somme des $a_{i}$ ne l'est pas non plus. Comme $n$ est premier, la somme (1) n'est pas divisible par $n$, d'où la contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers strictement positifs $n$ tels que $2^{n-1} n+1$ soit un carré parfait.",
    "options": [],
    "answer": "n = 5",
    "solution": "Solution:\n\nOn veut résoudre $2^{n-1} n+1=m^{2}$, c'est-à-dire $2^{n-1} n=(m-1)(m+1)$. Puisque $n=1$ n'est pas solution, on a $n \\geq 2$, et donc $m$ est nécessairement impair, et $m-1$ et $m+1$ sont pairs (en particulier $n \\geq 3$ ). On pose $k=\\frac{m-1}{2}$. Il suffit alors de résoudre $2^{n-3} n=k(k+1)$. Parmi les entiers $k$ et $k+1$ exactement un est pair, et donc ils sont de la forme $2^{n-3} d$ et $\\frac{n}{d}$ avec $d$ un diviseur de $n$. Or un diviseur de $n$ ne peut pas être à distance 1 d'un entier supérieur à $2^{n-3}$ si $n$ est trop grand. Plus précisément, on a $2^{n-3} d \\geq 2^{n-3} \\geq n+2$ si $n \\geq 6$, donc on a nécessairement $n \\leq 5$. Pour $n=5$, on trouve $2^{4} \\times 5+1=9^{2}$, donc 5 est solution. On vérifie que $2,3,4$ ne sont pas solutions. Donc $n=5$ est la seule solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ l'entier $4 \\times 201420142014 \\ldots 2014$ (où $2014$ est écrit $117819$ fois). Montrer que $2014^{3}$ divise $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn peut écrire\n$$\n4 \\times 201420142014 \\ldots 2014 = 4 \\cdot 2014 \\cdot \\frac{10^{4 \\cdot 117819}-1}{10^{4}-1}\n$$\nComme $2014 = 2 \\cdot 19 \\cdot 53$ et $10^{4}-1 = 3^{2} \\cdot 11 \\cdot 101$ sont premiers entre eux, il suffit de montrer que $19^{2}$ et $53^{2}$ divisent $10^{4 \\cdot 117819}-1$. D'après le théorème d'Euler,\n$$\n10^{\\Phi\\left(19^{2}\\right)} \\equiv 1 \\quad (\\bmod 19^{2}) \\quad \\text{et} \\quad 10^{\\Phi\\left(53^{2}\\right)} \\equiv 1 \\quad (\\bmod 53^{2}),\n$$\noù $\\phi$ désigne la fonction indicatrice d'Euler. Il suffit donc de vérifier que $\\phi\\left(19^{2}\\right)$ et $\\phi\\left(53^{2}\\right)$ divisent tous les deux $4 \\cdot 117819$. Ceci découle du fait que $\\phi\\left(19^{2}\\right) = 19^{2}-19 = 2 \\cdot 3^{2} \\cdot 19$, $\\phi\\left(53^{2}\\right) = 53^{2}-53 = 2^{2} \\cdot 13 \\cdot 53$ et $4 \\cdot 117819 = 2^{2} \\cdot 3^{2} \\cdot 13 \\cdot 19 \\cdot 53$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère cyclique. Un cercle passant par $A$ et $B$ coupe $[AC]$ et $[BD]$ en $E$ et $F$. Les droites $(AF)$ et $(BE)$ coupent $[BC]$ et $[AD]$ en $P$ et $Q$ respectivement. Montrer que $(PQ)$ est parallèle à $(CD)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n$\\widehat{AQB} = \\widehat{AQE} = \\pi - \\widehat{QEA} - \\widehat{EAQ} = \\widehat{AEB} - \\widehat{CAD}$ et de même, $\\widehat{APB} = \\widehat{AFB} - \\widehat{CBD}$. Or, $\\widehat{AEB} = \\widehat{AFB}$ et $\\widehat{CAD} = \\widehat{CBD}$, donc $\\widehat{AQB} = \\widehat{APB}$. On en déduit que $A, B, P, Q$ sont cocycliques. Il vient : $(AD, PQ) = (QA, QP) = (BA, BP) = (BA, BC) = (DA, DC)$, donc $(PQ)$ et $(DC)$ sont parallèles.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle acutangle avec $AB < AC < BC$ et soit $\\Omega$ son cercle circonscrit. Soient $D$ et $E$ les points diamètralement opposés respectivement aux points $B$ et $C$ dans le cercle $\\Omega$. Le cercle de centre $A$ et de rayon $AE$ intersecte $[AC]$ en $K$. Le cercle de centre $A$ et de rayon $AD$ coupe $(BA)$ en $L$ (de telle sorte que $A$ se trouve entre $B$ et $L$). Montrer que les droites $(EK)$ et $(DL)$ se coupent sur le cercle $\\Omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $X$ le second point d'intersection de la droite $(EK)$ avec le cercle $\\Omega$. La première chose à comprendre est que les points $K$ et $L$ sont très peu importants et qu'ils ne sont là que pour créer des triangles rectangles isocèles.\n\nAinsi, on sait que $[CE]$ est un diamètre du cercle $\\Omega$ et on obtient donc que $\\widehat{EAK} = 90^\\circ$. Le triangle $EAK$ est isocèle en $A$ car $AK = AE$ d'après les hypothèses de l'énoncé, en combinant cela avec l'information que $\\widehat{EAK} = 90^\\circ$ cela montre que $\\widehat{KEA} = 45^\\circ$.\n\nDe la même manière le triangle $DAL$ est rectangle isocèle en $A$ et donc\n$$\n\\widehat{XDA} = 180^\\circ - \\widehat{AEX} = 180^\\circ - 45^\\circ = 180^\\circ - \\widehat{LDA}\n$$\nCe qui montre bien que les points $X$, $D$ et $L$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver toutes les applications $f: \\mathbb{R} \\longrightarrow \\mathbb{R}$ telles que pour tous $x, y$ dans $\\mathbb{R}$,\n$$\nf(x+y)^2 - f\\left(2x^2\\right) = f(y+x) f(y-x) + 2x f(y)\n$$",
    "options": [],
    "answer": "f(x) = 0 for all x, or f(x) = x for all x",
    "solution": "Solution:\nEn posant $x = y = 0$, on obtient que $f(0)^2 - f(0) = f(0)^2$, donc $f(0) = 0$.\n\nEn prenant $y = 0$, on obtient que\n$$\nf(x)^2 = f\\left(2x^2\\right) + f(x) f(-x)\n$$\ndonc $f^2$ est paire, i.e. $f(x)^2 = f(-x)^2$ pour tout $x$.\n\nEn remplaçant dans l'équation $x$ par $-x$, on obtient\n$$\nf(y-x)^2 - f\\left(2x^2\\right) = f(y+x) f(y-x) - 2x f(y)\n$$\nEn soustrayant membre à membre cette égalité à celle de départ, il s'ensuit\n$$\nf(y+x)^2 - f(y-x)^2 = 4x f(y)\n$$\nEn échangeant $x$ et $y$ on trouve\n$$\nf(y+x)^2 - f(x-y)^2 = 4y f(x)\n$$\nComme $f^2$ est paire, on a toujours $f(y-x)^2 = f(x-y)^2$ donc pour tous réels $x, y$, $x f(y) = y f(x)$.\n\nEn particulier, en prenant $x = 1$, on obtient $f(y) = y f(1)$ pour tout $y$, donc $f$ est de la forme $x \\rightarrow a x$, avec $a \\in \\mathbb{R}$.\n\nEn prenant $x = 1$ et $y = 0$, l'équation de départ devient $a^2 - 2a = -a^2$, donc $a = a^2$, donc $a = 0$ ou $a = 1$.\n\nAinsi, $f$ est la fonction nulle ou l'identité.\n\nRéciproquement, on vérifie sans difficulté que ces fonctions sont des solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $\\omega_{1}$ et $\\omega_{2}$ deux cercles qui se coupent en $P$ et $Q$. On suppose que la tangente commune à $\\omega_{1}$ et $\\omega_{2}$, la plus proche de $P$, touche les cercles $\\omega_{1}$ et $\\omega_{2}$ en $A$ et $B$ respectivement. Les tangentes en $P$ aux cercles $\\omega_{1}$ et $\\omega_{2}$ recoupent $\\omega_{2}$ et $\\omega_{1}$ en $D$ et $C$ respectivement. On pose $H$ et $K$ sur les segments $[A C]$ et $[B D]$ de telle sorte que $A H = A P$ et $B K = B P$. Montrer que les points $A, B, K, Q$ et $H$ sont sur le même cercle.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDémonstration. On conjecture sur une bonne figure que les points $H, P$ et $B$ sont alignés, de même pour les points $K, P$ et $A$.\n\nLemme. Les points $H, P$ et $B$ sont alignés, ainsi que les points $K, P$ et $A$.\n\nOn note $x = \\widehat{PAB}$, $y = \\widehat{ABP}$ et $z = \\widehat{CPD}$. Comme la droite $(PC)$ est tangente à $\\omega_{2}$, on sait que l'angle que la tangente forme avec la droite $(PB)$ est également $x$, de la même manière l'angle entre $(PC)$ et $(AP)$ est $y$ (il s'agit de choisir les bonnes orientations). $z$ est l'angle entre les deux tangentes. Dans ce cas $\\widehat{APB} = x + y + z$ et au total les angles du triangle $ABP$ sont $x, y$ et $x + y + z$, cela montre que $2x + 2y + z = 180$.\n\nPar angle à la tangente, l'angle $\\widehat{CAP} = z$, mais $AH = AP$ donc le triangle $AHP$ est isocèle en $A$, et $\\widehat{AHP} = \\widehat{APH} = \\frac{180 - z}{2} = \\frac{2x + 2y}{2} = x + y$, alors l'angle $\\widehat{HPB} = \\widehat{HPA} + \\widehat{APB} = x + y + x + z + y = 2x + 2y + z = 180$, ce qui montre bien que les points $H, P$ et $B$ sont alignés. Le raisonnement symétrique montre que $K, P$ et $A$ sont alignés.\n\nOn a trouvé dans la preuve du lemme que $\\widehat{AHB} = \\widehat{AHP} = x + y$, mais alors par symétrie $\\widehat{BKA} = \\widehat{BKP} = x + y$, et par angle à la tangente, $\\widehat{PQA} = x$ et $\\widehat{PQB} = y$, donc $\\widehat{AQB} = x + y$, on a ainsi $\\widehat{AHB} = \\widehat{BKA} = \\widehat{BQA}$ ce qui montre bien que les points $A, H, Q, K$ et $B$ sont cocycliques.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn désigne par $K$ la valeur maximale de\n$$\n\\left|x_{1}-x_{2}\\right| \\cdot\\left|x_{1}-x_{3}\\right| \\cdot\\left|x_{1}-x_{4}\\right| \\cdot\\left|x_{2}-x_{3}\\right| \\cdot\\left|x_{2}-x_{4}\\right| \\cdot\\left|x_{3}-x_{4}\\right|\n$$\nou $x_{1}, x_{2}, x_{3}, x_{4} \\in [0 ; 1]$.\n\na. Prouver que $\\frac{4}{243} < K < \\frac{1}{27}$.\n\nb. Déterminer $K$.",
    "options": [],
    "answer": "sqrt(5)/125",
    "solution": "Solution:\n\na. Soit $x_{1}, x_{2}, x_{3}, x_{4} \\in [0 ; 1]$.\nOn pose\n$$\nf\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right) = \\left|x_{1}-x_{2}\\right| \\cdot \\left|x_{1}-x_{3}\\right| \\cdot \\left|x_{1}-x_{4}\\right| \\cdot \\left|x_{2}-x_{3}\\right| \\cdot \\left|x_{2}-x_{4}\\right| \\cdot \\left|x_{3}-x_{4}\\right|.\n$$\nSi deux des nombres $x_{1}, x_{2}, x_{3}, x_{4}$ sont égaux, on a $f\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right) = 0$, qui n'est clairement pas la valeur maximale de $f$. Sans perte de généralité, on peut donc supposer que $x_{1} > x_{2} > x_{3} > x_{4}$. D'après l'inégalité arithmético-géométrique, on a\n$$\n\\sqrt[3]{\\left(x_{1}-x_{2}\\right)\\left(x_{2}-x_{3}\\right)\\left(x_{3}-x_{4}\\right)} \\leqslant \\frac{\\left(x_{1}-x_{2}\\right)+\\left(x_{2}-x_{3}\\right)+\\left(x_{3}-x_{4}\\right)}{3} = \\frac{x_{1}-x_{4}}{3} \\leqslant \\frac{1}{3}\n$$\net donc $\\left|x_{1}-x_{2}\\right| \\cdot \\left|x_{2}-x_{3}\\right| \\cdot \\left|x_{3}-x_{4}\\right| \\leqslant \\frac{1}{27}$.\nD'autre part, on a $\\left|x_{1}-x_{3}\\right| < 1$, $\\left|x_{1}-x_{4}\\right| \\leqslant 1$ et $\\left|x_{2}-x_{4}\\right| < 1$, d'où $f\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right) < \\frac{1}{27}$.\nEt ainsi, on a $K < \\frac{1}{27}$.\nD'autre part, on vérifie facilement que $f\\left(1, \\frac{3}{4}, \\frac{1}{4}, 0\\right) = \\frac{9}{512} > \\frac{4}{243}$, d'où $K > \\frac{4}{243}$.\n\nb. Sans perte de généralité, on suppose toujours que $x_{1} > x_{2} > x_{3} > x_{4}$. Mais, il est facile de constater que $f\\left(x_{1}, x_{2}, x_{3}, x_{4}\\right) \\leqslant f\\left(1, x_{2}, x_{3}, 0\\right)$. Ainsi, en posant $x_{2} = y$ et $x_{3} = z$, on cherche la valeur maximale de\n$$\ng(y, z) = (1-y)(1-z)(y-z) y z\n$$\nsous la contrainte $0 < z < y < 1$.\nSoit $s$ un réel.\nOn rappelle que si $a, b$ sont deux réels tels que $a + b = s$ alors $ab = \\frac{1}{4} s^{2} - \\left(a - \\frac{s}{2}\\right)^{2} \\leqslant \\frac{1}{4} s^{2}$, avec égalité si et seulement si $a = b = \\frac{1}{2} s$.\nSoit alors $t \\in ]0,1[$, et considérons les couples $(y, z)$ pour lesquels $y - z = t$.\nDans ces conditions, les valeurs $(1-y) + z = 1-t$ et $(1-z) + y = 1+t$ sont fixées. D'après le rappel, le produit $(1-y)z$ est donc maximal pour $y + z = 1$, et il en est de même pour le produit $(1-z)y$. Compte-tenu de $y - z = t$, cela correspond à $y = \\frac{1+t}{2}$ et $z = \\frac{1-t}{2}$ et, dans ces conditions, on a $g(y, z) = \\frac{1}{16} t (1-t^{2})^{2}$.\nIl s'agit donc de chercher la valeur maximale de $P(t) = \\frac{1}{16} t (1-t^{2})^{2}$, lorsque $t \\in ]0,1[$.\nIl est facile de vérifier que, pour tout $t \\in ]0,1[$, on a $P'(t) = \\frac{1}{16} (1-t^{2})(1-5t^{2})$. Cela assure que la valeur maximale cherchée est $P\\left(\\frac{1}{\\sqrt{5}}\\right) = \\frac{\\sqrt{5}}{125}$.\nAinsi, on a $K = \\frac{\\sqrt{5}}{125}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A$ un point extérieur à un cercle $\\Gamma$. On mène deux tangentes $[A T]$ et $[A T']$ issues de $A$. Soient $M$ et $M'$ les milieux de $[A T]$ et $[A T']$. Soit $P$ un point de $[M M']$. Notons $[U V]$ la corde de $\\Gamma$ telle que $(P U)$ et $(P V)$ soient tangentes à $\\Gamma$. La droite $(U V)$ coupe $[M M']$ en $Q$. Montrer que le triangle $P A Q$ est rectangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nComme $M A = M T$, $M$ appartient à l'axe radical de $A$ et $\\Gamma$ ($A$ étant considéré comme un cercle de rayon nul). De même, $M'$ appartient à cet axe radical, donc $P$ et $Q$ également. On en déduit que $P A = P U = P V$ et que $Q A^2 = Q U \\cdot Q V$.\n\nNotons $H$ le projeté orthogonal de $P$ sur $(U V)$. On a\n$$\n\\begin{aligned}\nP A^2 + Q A^2 & = P U^2 + Q U \\cdot Q V = P H^2 + H U^2 + (Q H - H U)(Q H + H U) \\\\\n& = P H^2 + H U^2 + Q H^2 - H U^2 = P H^2 + Q H^2 = P Q^2\n\\end{aligned}\n$$\ndonc $P A Q$ est rectangle en $A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle avec $\\widehat{ABC} = \\widehat{ACB} = 40^{\\circ}$. La bissectrice issue du sommet $B$ coupe la droite $(AC)$ au point $D$. Montrer que $BD + DA = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPour résoudre ce genre d'exercice, une bonne idée est souvent d'essayer de reporter les longueurs qui nous intéressent à des endroits où les calculs seront plus faciles à faire. C'est pour cela que l'on introduit $X$ le point du segment $[BC]$ de telle sorte que $BD = BX$. On veut alors démontrer que $XC = AD$ pour conclure.\n\nUne chasse aux angles rapide montre dans un premier temps que $\\widehat{BXD} = 80^{\\circ}$. En effet, on sait par construction du point $X$ que le triangle $BDX$ est isocèle en $B$, on peut donc calculer $\\widehat{BXD} = \\left(180^{\\circ} - 20^{\\circ}\\right) / 2 = 80^{\\circ}$. Cela donne ensuite que $\\widehat{CDX} = \\widehat{DXB} - \\widehat{DCX} = 80^{\\circ} - 40^{\\circ} = 40^{\\circ} = \\widehat{DCX}$ et donc que le triangle $DXC$ est isocèle en $X$. Cela montre que $DX = XC$. Résoudre l'exercice revient donc à montrer que $AD = DX$.\n\nEn regardant très fort le quadrilatère $BADX$ on remarque que la droite $(BD)$ est la bissectrice de l'angle $\\widehat{ABX}$ et comme on veut démontrer que $AD = DX$ on pense naturellement au théorème du pôle sud. Il suffit donc de démontrer que les points $B, A, D$ et $X$ sont cocycliques pour conclure, il s'agit alors d'une deuxième petite chasse aux angles.\n\nEn effet, $\\widehat{BAD} + \\widehat{BXD} = 100^{\\circ} + 80^{\\circ} = 180^{\\circ}$ ce qui montre bien que le quadrilatère $ADXB$ est cyclique, puis d'après le théorème du pôle Sud dans le triangle $BAX$ que $AD = DX$. Cela conclut ainsi la preuve de l'exercice.\nSolution:\n\n![](attached_image_2.png)\n\nOn présente une deuxième solution, correspondant à une deuxième façon de reporter les longueurs à étudier.\n\nSoit $E$ le point de la demi-droite $[BD)$ tel que $BE = BC$. On cherche à montrer que $ED = DA$, de sorte que $BD + DA = BD + DE = BE = BC$. Le triangle $EBC$ est isocèle en $B$, de sorte que $180^{\\circ} = \\widehat{EBC} + \\widehat{ECB} + \\widehat{CEB} = 20^{\\circ} + 2\\widehat{CEB}$. Ainsi, $\\widehat{CEB} = 80^{\\circ}$ et $\\widehat{ECD} = \\widehat{ECB} - \\widehat{DCB} = 80^{\\circ} - 40^{\\circ} = 40^{\\circ}$. Le point $D$ est donc le pied de la bissectrice issue du sommet $C$ dans le triangle $ECB$.\n\nSoit désormais $F$ le point d'intersection des droites $(EC)$ et $(AB)$. L'idée derrière l'introduction de ce point est que l'on connait déjà deux bissectrices de ce triangle : la droite $(BD)$ est la bissectrice de l'angle $\\widehat{ABC}$ et la droite $(CD)$ est la bissectrice de l'angle $\\widehat{FCB}$. Ainsi, le point $D$ est le centre du cercle inscrit au triangle $FCB$. Il est donc sur la bissectrice de l'angle $\\widehat{BFC}$.\n\nOr on a $\\widehat{DEF} = 180^{\\circ} - \\widehat{DEC} = 100^{\\circ}$ et $\\widehat{DAF} = 180^{\\circ} - \\widehat{DAB} = 80^{\\circ}$, de sorte que $\\widehat{DEF} + \\widehat{DAF} = 180^{\\circ}$ et les points $F, E, D$ et $A$ sont cocycliques. Le point $D$ est donc sur le cercle circonscrit au triangle $FEA$ et sur la bissectrice de l'angle $\\widehat{EFA}$, il s'agit donc du pôle Sud du point $F$. À ce titre, $D$ est sur la médiatrice du segment $[AE]$ et $DE = DA$ comme annoncé.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les couples d'entiers strictement positifs $\\left(m, n\\right)$ pour lesquels :\n$$\n1+2^{n}+3^{n}+4^{n}=10^{m}\n$$",
    "options": [],
    "answer": "(m, n) = (1, 1) and (2, 3)",
    "solution": "Solution:\nCette équation est valable pour tous $n$, $m$, elle est donc valable en la passant modulo un entier $k$, c'est-à-dire en ne considérant que les restes de la division par rapport à $k$.\nOn commence par la regarder modulo $3$ :\n$$\n1+(-1)^{n}+0+1^{n} \\equiv 1^{m} \\pmod{3} \\text{ donc } (-1)^{n} \\equiv -1\n$$\nDonc $n$ est impair : soit $k \\in \\mathbb{N}$ tel que $n=2k+1$.\nOn suppose à présent que $n, m \\geqslant 3$ et on regarde modulo $8$ :\n$$\n1+0+3 \\cdot 3^{2k}+0 \\equiv 0 \\pmod{8} \\text{ ie } 1+3 \\cdot 1 \\equiv 0\n$$\nCe qui est absurde.\nOn en déduit que l'un des deux est dans $\\{1,2\\}$.\nIl suffit alors de traiter les cas $n=1$, $n=2$, $m=1$ et $m=2$.\n\nPour $n=1$ on trouve $1+2+3+4=10$, $(1,1)$ est solution.\nPour $n=2$, $1+4+9+16=30$ n'est pas une puissance de $10$ (car $30$ est divisible par $3$).\nComme l'application qui à $n$ associe $1+2^{n}+3^{n}+4^{n}$ est strictement croissante à $m$ fixé on a au plus une solution.\nComme $(1,1)$ et $(3,2)$ sont solutions (car $1+2^{3}+3^{3}+4^{3}=1+8+27+64=100=10^{2}$), ce sont donc les seules solutions avec $m=1$ ou $2$.\nL'ensemble des solutions est donc $\\{(1,1),(2,3)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $m$ et $n$ deux entiers tels que $0 \\leqslant m \\leqslant 2n$. Prouver que le nombre entier\n$$\n2^{2n+2} + 2^{m+2} + 1\n$$\nest un carré parfait si, et seulement si, $m = n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRemarquons que si $m = n$, alors $2^{2n+2} + 2^{m+2} + 1 = \\left(2^{m+1} + 1\\right)^2$ est bien un carré parfait.\n\nRéciproquement, supposons que $2^{2n+2} + 2^{m+2} + 1$ est un carré parfait. Comme il est strictement plus grand que $2^{2n+2} = \\left(2^{n+1}\\right)^2$, il est supérieur ou égal au carré suivant, c'est-à-dire\n$$\n2^{2n+2} + 2^{m+2} + 1 \\geqslant \\left(2^{n+1} + 1\\right)^2.\n$$\nEn développant, il vient $2^{m+2} \\geqslant 2^{n+2}$, donc $m \\geqslant n$. Supposons par l'absurde que $m > n$ soit $2^m > 2^{2n-m}$.\n\nPar hypothèse, la quantité $2^{2n+2} + 2^{m+2} + 1$ est le carré d'un entier impair $2k+1$ strictement plus grand que $1$, d'où\n$$\n2^{2n+2} + 2^{m+2} + 1 = (2k+1)^2\n$$\npour un certain entier $k \\geqslant 1$, ce qui se réécrit sous la forme\n$$\nk(k+1) = 2^{2n} + 2^m = 2^m (2^{2n-m} + 1).\n$$\nOr les entiers $k$ et $k+1$ sont premiers entre eux. On en déduit que :\n- soit $k = 2^m p$ et $p(k+1) = 2^{2n-m} + 1$ ce qui entraîne $(k+1)p < k/p + 1$ : contradiction.\n- soit $k+1 = 2^m p$ et $pk = 2^{2n-m} + 1$ ce qui entraîne $kp < \\frac{k+1}{p} + 1$ donc $kp \\leqslant \\frac{k+1}{p}$, ou encore $k(p^2 - 1) \\leqslant 1$, ce qui impose $p = 1$ puisque $k \\geqslant 1$. Alors, $2^{2n-m} = 2(2^{m-1} - 1)$ donc $2n - m = 1$ et $m = 2$ : contradiction.\n\nAinsi, si $2^{2n+2} + 2^{m+2} + 1 = \\left(2^{m+1} + 1\\right)^2$ est un carré parfait, alors $m = n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A$, $B$, $C$ et $D$ quatre points sur un cercle dans cet ordre. Soit $U$ le point d'intersection des droites $(AB)$ et $(CD)$, et $V$ le point d'intersection des droites $(BC)$ et $(DA)$. Soit $K$ le point d'intersection de la bissectrice issue de $U$ dans le triangle $AUC$ et de la bissectrice issue de $V$ dans le triangle $AVC$. Soit $L$ le point d'intersection de la médiatrice du segment $[KU]$ et de la médiatrice du segment $[KV]$.\n\nMontrer que les points $U$, $V$ et $L$ sont alignés.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLe point $L$ est le centre du cercle circonscrit au triangle $UKV$. Pour montrer que les points $U$, $L$ et $V$ sont alignés, il suffit de montrer que $\\widehat{ULV} = 180^{\\circ}$. Par le théorème de l'angle au centre, $\\widehat{UKV} = \\frac{1}{2} \\widehat{ULV}$. Il suffit donc de montrer que $\\widehat{UKV} = 90^{\\circ}$.\n\nLa somme des angles du triangle $UKV$ vaut $180^{\\circ}$, donc il suffit de montrer que $\\widehat{KUV} + \\widehat{KVU} = 90^{\\circ}$. D'une part $\\widehat{KUV} = \\widehat{CUV} - \\widehat{KUC} = \\widehat{CUV} - \\frac{1}{2} \\widehat{BUC}$. Or $\\widehat{BUC} = 180^{\\circ} - \\widehat{UBC} - \\widehat{BCU} = \\widehat{ADC} - \\widehat{UCB}$. On déduit\n$$\n\\widehat{KUV} = \\widehat{CUV} - \\frac{1}{2} \\widehat{ADC} + \\frac{1}{2} \\widehat{UCB}\n$$\nDe la même manière, on obtient\n$$\n\\widehat{KVU} = \\widehat{CVU} - \\frac{1}{2} \\widehat{CBA} + \\frac{1}{2} \\widehat{VCD}\n$$\nOr $\\widehat{CBA} + \\widehat{ADC} = 180^{\\circ}$ et\n$$\n\\widehat{UCB} = \\widehat{VCD} = \\widehat{UCV} = 180^{\\circ} - \\widehat{CUV} - \\widehat{CVU}\n$$\nFinalement\n$$\n\\begin{aligned}\n& \\widehat{KUV} + \\widehat{KVU} = \\widehat{CUV} - \\frac{1}{2} \\widehat{ADC} + \\frac{1}{2} \\widehat{UCB} + \\widehat{CVU} - \\frac{1}{2} \\widehat{CBA} + \\frac{1}{2} \\widehat{VCD} \\\\\n& = \\widehat{CUV} + \\widehat{CVU} - \\frac{1}{2} \\cdot 180^{\\circ} + 2 \\cdot \\frac{1}{2} \\left(180^{\\circ} - \\widehat{CUV} - \\widehat{CVU}\\right) = 90^{\\circ}\n\\end{aligned}\n$$\nce qui donne le résultat voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $x > 1$ et $y$ des entiers vérifiant $2 x^{2} - 1 = y^{15}$. Montrer que $x$ est divisible par $5$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nL'entier $y$ est clairement impair et strictement plus grand que $1$. On factorise l'équation sous la forme\n$$\nx^{2} = \\left(\\frac{y^{5} + 1}{2}\\right)\\left(y^{10} - y^{5} + 1\\right)\n$$\nRemarquons que\n$$\ny^{10} - y^{5} + 1 \\equiv 3 \\pmod{y^{5} + 1}\n$$\net que donc $\\operatorname{pgcd}(y^{5} + 1, y^{10} - y^{5} + 1)$ est égal à $1$ ou à $3$. S'il valait $1$, alors $y^{10} - y^{5} + 1$ serait un carré. Or pour $y > 0$ nous avons\n$$\n\\left(y^{5} - 1\\right)^{2} = y^{10} - 2y^{5} + 1 < y^{10} - y^{5} + 1 < y^{10} = \\left(y^{5}\\right)^{2},\n$$\nc'est-à-dire que $y^{10} - y^{5} + 1$ est strictement compris entre deux carrés consécutifs, et ne peut pas être lui-même un carré. Donc $\\operatorname{pgcd}(y^{5} + 1, y^{10} - y^{5} + 1) = 3$, de sorte qu'il existe des entiers $a$ et $b$ tels que\n$$\ny^{5} + 1 = 6a^{2} \\quad \\text{et} \\quad y^{10} - y^{5} + 1 = 3b^{2}.\n$$\nOn peut factoriser $(y + 1)\\left(y^{4} - y^{3} + y^{2} - y + 1\\right) = 6a^{2}$. Puisque $y^{5} \\equiv -1 \\pmod{3}$, on a nécessairement $y \\equiv -1 \\pmod{3}$, donc $y + 1$ est divisible par $6$. De même que plus haut, on a\n$$\ny^{4} - y^{3} + y^{2} - y + 1 \\equiv 5 \\pmod{y + 1}\n$$\net donc $\\operatorname{pgcd}(y + 1, y^{4} - y^{3} + y^{2} - y + 1)$ est égal à $1$ ou à $5$. S'il vaut $5$, alors $a$ est divisible par $5$ et donc $x$ aussi, et nous avons terminé. Supposons donc qu'il vaut $1$. Alors $y^{4} - y^{3} + y^{2} - y + 1$ est un carré. Dans ce cas, $4\\left(y^{4} - y^{3} + y^{2} - y + 1\\right)$ est aussi un carré, ce qui est impossible, car pour $y > 1$, on a\n$$\n\\left(2y^{2} - y\\right)^{2} = 4y^{4} - 4y^{3} + y^{2} < 4\\left(y^{4} - y^{3} + y^{2} - y + 1\\right) < 4y^{4} - 4y^{3} + 5y^{2} - 2y + 1 = \\left(2y^{2} - y + 1\\right)^{2}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $D$ est le milieu de l'arc $BC$ du cercle $ABC$ ne contenant pas $A$, $Z$ est l'unique point sur la bissectrice extérieure de $\\widehat{BAC}$ tel que $ZA = ZC$. Montrer que le cercle $ADZ$ passe par le milieu du côté $[AB]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $\\alpha = \\widehat{BAC}$, on pose $M_B$ le milieu de $[AB]$ et $M_C$ le milieu de $[AC]$. On note $X$ l'intersection des droites $(M_BM_C)$ et $(DC)$.\n\nAlors comme $(M_BM_C) \\parallel BC$, on trouve $\\widehat{M_BXC} = \\widehat{BCD} = \\widehat{BAD} = \\frac{\\alpha}{2}$, l'avant-dernière égalité découle du fait que les points $A, B, C$ et $D$ sont cocycliques et que $D$ est le pôle sud de $A$ dans le triangle $ABC$. On calcule maintenant que $\\widehat{CAZ} = 90 - \\frac{\\alpha}{2}$, $Z$ et $M_B$ sont sur la médiatrice de $[AC]$ donc $(ZM_B) \\perp (AC)$, et on trouve alors $\\widehat{AZM_B} = \\widehat{CZ M} = \\frac{\\alpha}{2}$, mais alors $\\widehat{CZ M} = \\widehat{CB M}$, ce qui montre que les points $Z, M_B, C$ et $X$ sont cocycliques.\n\nComme $Z, M_B, C$ et $X$ sont cocycliques et $\\widehat{Z M_B C} = 90$ on doit avoir $\\widehat{ZCX} = 90$. On a de plus, $\\widehat{DAZ} = 90$, ce qui montre que les points $A, Z, X$ et $D$ sont cocycliques.\n\nOn a également, $\\widehat{M_BAD} = \\frac{\\alpha}{2} = \\widehat{M_BXD}$ ce qui montre que les points $A, M_B, D$ et $X$ sont cocycliques. Mais dans ce cas : $A, M_B, D, X$ et $Z$ sont cocycliques. Donc en particulier, le cercle $ADZ$ passe par le milieu du segment $[AB]$.\n\n![](attached_image_1.png)\nSolution:\n\nOn va présenter ici une méthode un peu plus générale que la méthode précédente, il s'agit de composer trois similitudes et d'en trouver certaines propriétés.\n\nOn pose $S_{Z, \\mu, r}$ la similitude directe de centre $Z$, d'angle $\\mu$ (orienté) et de rapport $r$.\n\nAlors, $S_{M_B, 180, 1}$ fixe $M_B$, échange $A$ et $B$, elle échange également $D$ avec un point appelé $D'$. On trouve par chasse aux angles immédiate que $\\widehat{AZC} = \\alpha$.\n\n$S_{Z, \\alpha, 1}$ fixe $Z$, envoie alors $A$ sur $C$.\n\nOn trouve par chasse aux angles que $\\widehat{CDB} = 180 - \\alpha$. Donc $S_{D, 180-\\alpha, 1}$ fixe $D$ et envoie $C$ sur $B$.\n\nOn effectue alors $T = S_{M_B, 180, 1} + S_{Z, \\alpha, 1} + S_{D, 180-\\alpha, 1}$, cette transformation est également une similitude directe de rapport $1 \\times 1 \\times 1$ et d'angle $180 + 180 - \\alpha + \\alpha = 360$, le résultat est donc une translation, mais $T$ envoie $B$ sur $B$, une translation qui a un point fixe est l'identité donc $T$ est en fait l'identité.\n\nMais la transformation $T$ envoie $D$ sur $D'$, puis $D'$ sur $D''$ et $D''$ sur $D''' = D$, $D''$ est envoyé sur $D''' = D$ par la similitude de centre $D$ donc $D'' = D''' = D$, ainsi la similitude de centre $Z$ envoie $D'$ sur $D$, donc en particulier $ZD' = ZD$ et $\\widehat{D'ZD} = \\alpha$. Dans le triangle $A' D Z$, le point $M_B$ est le pied de la hauteur, bissectrice, etc..., donc en particulier, $\\widehat{M_B ZC} = \\frac{\\alpha}{2} = \\widehat{M_BAD}$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCaractériser les entiers $n \\geq 2$ tels que pour tout entier $a$ on ait $a^{n+1} \\equiv a \\pmod{n}$.",
    "options": [],
    "answer": "2, 6, 42, 1806",
    "solution": "Solution:\n\nVoici les $n$ vérifiant cette propriété : $2$, $2 \\cdot 3$, $2 \\cdot 3 \\cdot 7$, $2 \\cdot 3 \\cdot 7 \\cdot 43$.\n\nPour prouver que c'est exhaustif, on procède de la façon suivante : on commence par remarquer que $n$ n'a pas de facteur carré. En effet, si $p^{2}$ divise $n$, alors $p^{n+1}-p$ est divisible par $p^{2}$, ce qui n'est pas possible. Ainsi, $n$ est forcément un produit $p_{1} \\ldots p_{k}$ de nombres premiers distincts. Par conséquent, par le lemme chinois, la condition de l'énoncé est équivalente à $a^{n+1} \\equiv a \\pmod{p}$ pour tout entier $a$ et tout $p \\in \\{p_{1}, \\ldots, p_{k}\\}$. En choisissant $a$ d'ordre $p-1$ modulo $p$, on obtient que cela est équivalent à ce que $p-1$ divise $n$ pour tout $p \\in \\{p_{1}, \\ldots, p_{k}\\}$.\n\nEn résumé : $n$ est produit de premiers distincts $p_{1}, \\ldots, p_{k}$, et $p_{i}-1$ divise $n$ pour tout $i$. On en déduit que pour tout $i$, $p_{i}-1$ est sans facteur carré et $p_{i}-1 = q_{1} \\ldots q_{m}$ où les $q_{j}$ sont des nombres premiers pris parmi les $p_{r}$, $r \\neq i$.\n\nQuitte à re-numéroter les $p_{i}$, on peut supposer que $p_{1} < p_{2} < \\ldots < p_{k}$. D'après la condition ci-dessus, nous avons nécessairement $p_{1} = 2$. Si $k = 1$, cela nous donne l'entier $n = 2$. Si $k > 1$, alors $p_{2} - 1$ est nécessairement égal à $2$, donc $p_{2} = 3$. Si $k = 2$, cela nous donne la solution $n = 6$. Si $k > 2$, on continue en disant que $p_{3} - 1$ ne peut être égal à $2$ ou $3$, donc il est égal à $p_{1} p_{2} = 6$, d'où $p_{3} = 7$. Si $k = 3$, cela donne la solution $n = 2 \\cdot 3 \\cdot 7 = 42$. Si $k > 3$, on voit que de même la seule valeur possible pour $p_{4}$ est $2 \\times 3 \\times 7 + 1 = 43$. Pour $k = 4$, cela donne la solution $n = 2 \\cdot 3 \\cdot 7 \\cdot 43$.\n\nEnfin, supposons que $k > 4$. Alors $p_{5} - 1$ doit être pair et strictement supérieur à $43$, donc les seules valeurs possibles sont $2 \\cdot 43$, $2 \\cdot 3 \\cdot 43$, $2 \\cdot 7 \\cdot 43$, $2 \\cdot 3 \\cdot 7 \\cdot 43$. On vérifie qu'aucune de ces possibilités ne fournit un $p_{5}$ premier. Ainsi on a nécessairement $k \\leq 4$ et les solutions que nous avons trouvées sont les seules.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\omega$ son cercle inscrit. On note $P, Q, R$ les points de contact de $\\omega$ avec $(BC), (CA)$ et $(AB)$. Un cercle passant par $B$ et $C$ est tangent en $X$ à $\\omega$, un cercle passant par $C$ et $A$ est tangent en $Y$ à $\\omega$ et un cercle passant par $A$ et $B$ est tangent en $Z$ à $\\omega$. Montrer que les droites $(PX), (QY)$ et $(RZ)$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRappelons d'abord quelques résultats sur la polarité que nous allons utiliser. Soit $\\mathcal{C}$ un cercle de centre $O$ et de rayon $R$. Si $P$ est un point distinct de $O$, la polaire de $P$ par rapport à $\\mathcal{C}$ est la droite formée des points $M$ tels que $\\overrightarrow{OP} \\cdot \\overrightarrow{OM} = R^{2}$. Le pôle par rapport à $\\mathcal{C}$ d'une droite $\\mathcal{D}$ ne passant pas par $O$ est l'unique point $P$ tel que pour tout $M$ appartenant à $\\mathcal{D}$ on a $\\overrightarrow{OP} \\cdot \\overrightarrow{OM} = R^{2}$. À partir de cette définition, il est facile de montrer les propriétés suivantes:\n(i) si $P$ est un point extérieur à $\\mathcal{C}$ et $(PT), (PT')$ sont les deux tangentes à $\\mathcal{C}$, alors la polaire de $P$ est la droite $(TT')$ ;\n(ii) trois points donnés sont alignés si et seulement si leurs polaires sont concourantes.\n(Remarque : pour la propriété (ii), la droite sur laquelle se trouvent les trois points est la polaire du point de rencontre des trois droites.)\n\n![](attached_image_1.png)\n\nRevenons à l'exercice. Notons $\\Gamma_{A}$ le cercle passant par $B, C$ et $X$. On définit de même les cercles $\\Gamma_{B}$ et $\\Gamma_{C}$. Soit $A'$ le point de rencontre entre la tangente commune à $\\omega$ et $\\Gamma_{A}$, et la droite $(BC)$. On définit de même les points $B'$ et $C'$.\n\nD'après la propriété (ii), la polaire de $A'$ par rapport à $\\omega$ est la droite $(PX)$. Il suffit donc d'après la propriété (i) de montrer que $A', B'$ et $C'$ sont alignés.\n\nSoit $\\Gamma$ le cercle circonscrit à $ABC$. Comme $A'$ appartient à l'axe radical $(BC)$ de $\\Gamma$ et $\\Gamma_{A}$, on a $\\mathcal{P}_{\\Gamma}(A') = \\mathcal{P}_{\\Gamma_{A}}(A') = A'X^{2} = \\mathcal{P}_{\\omega}(A')$. On en déduit que $A'$ appartient à l'axe radical de $\\Gamma$ et $\\omega$. Il en va de même pour $B'$ et $C'$, ce qui prouve que $A', B'$ et $C'$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a$, $b$, $c$ des réels strictement positifs tels que : $ab + bc + ca = abc$. Montrer que :\n$$\n\\frac{1}{a^{2}+b^{2}}+\\frac{1}{b^{2}+c^{2}}+\\frac{1}{c^{2}+a^{2}} \\leqslant \\frac{1}{6},\n$$\net trouver les cas d'égalité.",
    "options": [],
    "answer": "The sum is at most 1/6, with equality when a = b = c = 3.",
    "solution": "Solution:\n\nPosons $u = \\frac{1}{a}$, $v = \\frac{1}{b}$, $w = \\frac{1}{c}$. En divisant par $abc$, la condition se réécrit $u + v + w = 1$.\n\nL'inégalité voulue se réécrit\n$$\n\\frac{u^{2} v^{2}}{u^{2}+v^{2}} + \\frac{u^{2} w^{2}}{u^{2}+w^{2}} + \\frac{v^{2} w^{2}}{v^{2}+w^{2}} \\leqslant \\frac{1}{6}.\n$$\n\nOn utilise l'inégalité arithmético-géométrique sur les dénominateurs : $u^{2} + v^{2} \\geqslant 2uv$ et similairement sur les deux autres. On obtient\n$$\n\\frac{u^{2} v^{2}}{u^{2}+v^{2}} + \\frac{u^{2} w^{2}}{u^{2}+w^{2}} + \\frac{v^{2} w^{2}}{v^{2}+w^{2}} \\leqslant \\frac{1}{2}(uv + vw + uw).\n$$\nOr\n$$\n1 = (u+v+w)^{2} = u^{2} + v^{2} + w^{2} + 2(uv + vw + wu) \\geqslant 3(uv + vw + uw)\n$$\npar le lemme du tourniquet. En particulier $uv + vw + uw \\leqslant \\frac{1}{3}$ donc\n$$\n\\frac{u^{2} v^{2}}{u^{2}+v^{2}} + \\frac{u^{2} w^{2}}{u^{2}+w^{2}} + \\frac{v^{2} w^{2}}{v^{2}+w^{2}} \\leqslant \\frac{1}{6}.\n$$\n\nPour les cas d'égalité, on a égalité dans le lemme du tourniquet, donc $u = v = w$ donc $a = b = c$. En réinjectant dans la condition, $3a^{2} = a^{3}$ donc $a = b = c = 3$.\n\nRéciproquement, si $a = b = c = 3$, on a\n$$\n\\frac{1}{a^{2}+b^{2}} + \\frac{1}{b^{2}+c^{2}} + \\frac{1}{c^{2}+a^{2}} = \\frac{3}{2 \\times 9} = \\frac{1}{6}\n$$\non a bien égalité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$n$ joueurs participent à un tournoi d'échecs. Chaque joueur fait exactement une partie avec chacun des autres joueurs. Une victoire rapporte 1 point, un match nul un demipoint et une défaite aucun point. Une partie est dite anormale si le gagnant de cette partie obtient un score au tournoi strictement plus faible que le perdant.\n\na) Montrer que la proportion de parties anormales est inférieure ou égale à $75\\%$.\n\nb) Est-il possible qu'elle soit supérieure ou égale à $70\\%$ ?",
    "options": [],
    "answer": "a) At most 75%.\nb) Yes; it can be at least 70% (for example, a construction with a large enough number of players exceeds 70%).",
    "solution": "Solution:\n\na. Soit $m = [n / 2]$. On classe les joueurs suivant leur score (parmi les joueurs ayant le même score, on les classe arbitrairement). On dira qu'un joueur classé parmi les $m$ meilleurs est fort; sinon, on dira qu'il est faible.\n\nSoit $x$ le nombre de parties normales entre les joueurs forts et les joueurs faibles. Les joueurs forts ont marqué au total $m(m-1)/2$ points entre eux, et au plus $x$ lors de parties contre les joueurs faibles.\n\nNotons $S_{1}$ le score total des joueurs forts et $S_{2}$ le score total des joueurs faibles. On a donc $S_{1} \\leq m(m-1)/2 + x$ et $S_{1} + S_{2} = n(n-1)/2$.\n\nLe score moyen des joueurs forts, $\\frac{S_{1}}{m}$, est supérieur au score moyen de tous les joueurs: $\\frac{S_{1}}{m} > \\frac{n(n-1)/2}{n}$, i.e. $S_{1} > m(n-1)/2$. Par conséquent,\n$$\nx \\geq S_{1} - m(m-1)/2 > \\frac{m(n-m)}{2} \\geq \\frac{n(n-1)}{8}\n$$\nComme il y a au total $n(n-1)/2$ parties, la proportion de parties normales est supérieure à $1/4$.\n\nb. Considérons un tournoi constitué de $2k+1$ groupes de $k$ joueurs (on a donc $n = k(2k+1)$). On suppose que les joueurs du groupe $i$ font match nul entre eux, perdent contre les joueurs des groupes $i+1, \\ldots, i+k$ et gagnent tous les autres matchs (ici, $i$ est pris modulo $k$). Alors tous les joueurs ont le même score.\n\nOn va modifier les parties des joueurs du $(k+1)$-ième groupe afin que les scores soient différents. Pour tout $i$, on remplace $ik$ victoires de joueurs du $(k+1)$-ième groupe contre des joueurs du $(k+1-i)$-ème groupe par des matchs nuls, de sorte que chaque joueur du $(k+1)$-ième groupe voie son score décroître de $i/2$ tandis que chaque joueur du $(k+1-i)$-ème groupe voie son score augmenter de $i/2$.\n\nDe même, on remplace $ik$ défaites du $(k+1)$-ième groupe contre les $(k+1+i)$-ème groupe par des matchs nuls.\n\nAlors les premières places dans le tournoi sont prises par les joueurs du premier groupe, puis du deuxième, etc.\n\nLes joueurs des groupes $i \\leq k$ ont perdu $k^{3} - ik$ parties, les joueurs des groupes $i > k+1$ perdu $(2k+1-i)k^{2}$ parties anormales, et les joueurs du $(k+1)$-ième groupe ont perdu $k^{3} - k\\left(\\frac{1}{2}k(k+1)\\right)$ parties anormales. Tous calculs faits, on obtient en tout\n$$\n\\frac{3}{2}k^{4} - \\frac{1}{2}k^{3} - k^{2}\n$$\nparties anormales.\n\nIl y a au total $k(2k+1)(k(2k+1)-1)/2$ parties. On vérifie que pour $k=100$, la proportion de parties anormales dépasse $70\\%$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $p \\geqslant 3$ un nombre premier. Pour $k \\in \\mathbb{N}$ vérifiant $1 \\leqslant k \\leqslant p-1$, le nombre de diviseurs de $kp+1$ qui sont compris strictement entre $k$ et $p$ est noté $a_k$.\nQue vaut $a_1+a_2+\\ldots+a_{p-1}$ ?",
    "options": [],
    "answer": "p - 2",
    "solution": "Solution:\n\nLa réponse est $p-2$.\n\nNous allons montrer que chacun de $\\{2, \\ldots, p-1\\}$ contribue exactement une fois au comptage représenté par $a_1+a_2+\\ldots+a_{p-1}$.\n\nSoit $2 \\leqslant m \\leqslant p-1$ un entier, on se propose de montrer deux choses :\n\n1. $m$ est compté au plus une fois\n2. $m$ est bien compté\n\n$m$ est compté au plus une fois :\n\nPar l'absurde, on suppose qu'il existe $1 \\leqslant i < j \\leqslant p-1$ deux entiers vérifiant :\n- $m \\mid ip+1$ et $m > i$\n- $m \\mid jp+1$ et $m > j$\n\nOn a donc $m \\mid p(j-i)$ or $\\gcd(m, p) = 1$ donc $m \\mid j-i$. Cependant, c'est impossible car $0 < j-i < j < m$ : $m$ est compté au plus une fois.\n\n$m$ est bien compté :\n\nOn considère les $m-1$ entiers $\\{p+1, 2p+1, \\ldots, (m-1)p+1\\}$.\n\nExactement comme précédemment, on voit qu'ils sont deux à deux distincts modulo $m$. De plus, aucun d'entre eux n'est $\\equiv 1 \\pmod{m}$ car si on retire les $+1$ aucun entier n'est divisible par $m$.\n\nD'où $m$ est compté une et une fois et cela conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer qu'il existe une infinité de nombres entiers strictement positifs $a$ tels que $a^{2}$ divise $2^{a}+3^{a}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRemarquons que $a=1$ convient puis construisons par récurrence une suite $(u_{n})$ strictement croissante d'entiers impairs vérifiant la propriété.\n\nTout d'abord, on vérifie que $a=1$ et $a=5$ conviennent. Posons ainsi $u_{0}=1$ et $u_{1}=5$.\n\nConsidérons un entier $n \\geqslant 1$ et supposons la suite définie jusqu'au rang $n$. Par définition de $u_{n}$, il existe un entier $q$ tel que $q u_{n}^{2}=2^{u_{n}}+3^{u_{n}}$. Comme $u_{n} \\geqslant u_{1}=5$, on a $q>1$.\n\nSoit $p \\geqslant 3$ un facteur premier (impair) de $q$ et vérifions que $u_{n+1}=p u_{n}$ vérifie la propriété requise. Montrons que\n$$\np \\quad \\text{divise} \\quad 3^{u_{n}(p-1)}-2^{u_{n}} 3^{u_{n}(p-2)}+\\cdots+2^{u_{n}(p-1)}.\n$$\nCeci permet de conclure, car alors, étant donné que $p u_{n}^{2}$ divise $2^{u_{n}}+3^{u_{n}}$, il en découle que $p \\cdot p u_{n}^{2}$ divise\nce qui achève la démonstration.\n\nPour établir (1), on remarque que $p$ divise $2^{u_{n}}+3^{u_{n}}$ (car $p$ divise $q$), de sorte qu'on a la congruence $2^{u_{n}} \\equiv -3^{u_{n}} \\pmod{p}$. Ainsi,\n$$\n\\begin{aligned}\n3^{u_{n}(p-1)}-2^{u_{n}} 3^{u_{n}(p-2)}+\\cdots+2^{u_{n}(p-1)} &\\equiv 3^{u_{n}(p-1)}+3^{u_{n}(p-1)}+\\cdots+3^{u_{n}(p-1)} \\quad (\\bmod p) \\\\\n&\\equiv p \\cdot 3^{u_{n}(p-1)} \\quad (\\bmod p) \\\\\n&\\equiv 0 \\quad (\\bmod p) .\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. On note $D$ et $E$ les milieux respectifs des côtés $AB$ et $AC$. Soit $O$ le centre du cercle circonscrit du triangle $ABC$. On note $K$ le point d'intersection des droites $(OE)$ et $(BC)$. On note $L$ le deuxième point d'intersection de la droite $(OD)$ avec le cercle circonscrit au triangle $OKB$. On note $F$ la projection orthogonale de $A$ sur la droite $(KL)$. Montrer que les points $D$, $E$ et $F$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nEn traçant la figure, on s'aperçoit que le point $A$ appartient également au cercle passant par $O$, $K$ et $B$. On commence par démontrer ce fait.\n\nD'une part, d'après le théorème de Thalès, les droites $(DE)$ et $(BC)$ sont parallèles, on a l'égalité des angles alternes-internes :\n$$\n\\widehat{OKB} = \\widehat{OED}.\n$$\nD'autre part, puisque $\\widehat{ODA} = 90^\\circ = 180^\\circ - \\widehat{OEA}$, les points $O$, $E$, $A$ et $D$ sont cocycliques, de sorte qu'avec le théorème de l'angle inscrit on obtient\n$$\n\\widehat{OAB} = \\widehat{OAD} = \\widehat{OED} = \\widehat{OKB}.\n$$\nCe qui prouve bien que le point $A$ appartient au cercle passant par $O$, $K$ et $B$. Notons que $O$ est le milieu de l'arc $AB$ dans le cercle circonscrit au triangle $AKB$. Le point $O$ est donc le pôle Sud du point $K$ dans le triangle $ABK$. En particulier, $\\widehat{OKB} = \\widehat{AKO}$.\n\nPuisque $\\widehat{AFK} = 90^\\circ = 180^\\circ - \\widehat{AEK}$, les points $A$, $F$, $K$ et $E$ sont cocycliques. On a donc avec le théorème de l'angle inscrit :\n$$\n\\widehat{AFE} = \\widehat{AKE} = \\widehat{AKO}.\n$$\nD'autre part, puisque $\\widehat{AFL} = 90^\\circ = \\widehat{ADL}$, les points $A$, $D$, $F$ et $L$ sont cocycliques. On a alors avec le théorème de l'angle inscrit dans le cercle passant par $A$, $D$, $F$ et $L$ puis dans celui passant par $A$, $L$, $K$ et $O$ :\n$$\n\\widehat{AFD} = \\widehat{ALD} = \\widehat{ALO} = \\widehat{AKO}\n$$\nOn déduit que $\\widehat{AFD} = \\widehat{AKO} = \\widehat{AFE}$. Les points $F$, $D$ et $E$ sont donc alignés, comme voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSi $n>0$ est un entier, on désigne par $d(n)$ le nombre de diviseurs strictement positifs de $n$.\n\na) Existe-t-il une suite $\\left(a_{i}\\right)_{i \\geqslant 1}$ strictement croissante d'entiers strictement positifs tels que, pour tout $i$ suffisamment grand, le nombre $a_{i}$ soit divisible par exactement $d(i)-1$ termes de la suite (y compris lui-même)?\n\nb) Existe-t-il une suite $\\left(a_{i}\\right)_{i \\geqslant 1}$ strictement croissante d'entiers strictement positifs tels que, pour tout $i$ suffisamment grand, le nombre $a_{i}$ soit divisible par exactement $d(i)+1$ termes de la suite (y compris lui-même)?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) La réponse est oui.\nIl suffit de trouver une suite $\\left(a_{i}\\right)$ qui vérifie les deux conditions suivantes :\n- le nombre $a_{1}$ ne divise aucun autre terme de la suite,\n- le nombre $i$ divise $j$ si et seulement si $a_{i}$ divise $a_{j}$, pour tous $i, j \\geqslant 2$.\nOr, il est bien connu que, pour tous entiers $i, j \\geqslant 1$, on a $\\operatorname{pgcd}\\left(2^{i}-1,2^{j}-1\\right)=2^{d}-1$, où $d=\\operatorname{pgcd}(i, j)$. Par conséquent, la suite définie par $a_{1}=2$ et $a_{i}=2^{i}-1$ pour $i>1$ convient.\n\nb) La réponse est également oui.\nEn fait, de façon plus générale, on va prouver que si $f: \\mathbb{N}^{*} \\longrightarrow \\mathbb{N}^{*}$ est une fonction et qu'il existe un entier $N>0$ tel que $f(n) \\leqslant n$ pour tout $n \\geqslant N$, alors il existe une suite strictement croissante $\\left(a_{i}\\right)_{i \\geqslant 1}$ d'entiers strictement positifs telle que, pour tout $i$ suffisamment grand, le nombre $a_{i}$ est divisible par exactement $f(i)$ termes de la suite.\nNotons tout de suite que cela répond à la fois au a) et au b) puisque $d(n)<n$ pour tout $n \\geqslant 3$ (le nombre $n-1$ ne divise pas $n$ pour $n \\geqslant 3$ ).\nDans les conditions ci-dessus donc, on commence par choisir une suite strictement croissante $a_{1}, a_{2}, \\cdots, a_{M}$ de $M$ nombres premiers. Puis, par récurrence sur $k>M$, si $a_{1}, a_{2}, \\cdots, a_{k-1}$ sont définis avec $a_{1}<a_{2}<\\cdots<a_{k-1}$, on choisit un nombre premier $p_{k}$ tel que $p_{k}>a_{k-1}$ et on pose $a_{k}=a_{1} a_{2} \\cdots a_{f(k)-1} p_{k}$.\nOn a alors clairement $a_{k-1}<a_{k}$.\nDe plus, pour tout $i>M$, le nombre $a_{i}$ est divisible que par $a_{1}, a_{2}, \\cdots, a_{f(i)-1}$ et par lui-même. Par contre, $a_{i}$ n'est divisible par aucun autre terme $a_{j}$, avec $j>f(i)-1$ et $j \\neq i$, puisqu'un tel $a_{j}$ contient un facteur premier $p_{j}$ qui ne divise aucun des $a_{k}$ avec $k \\leqslant f(i)-1$ et tel que $p_{j} \\neq p_{i}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeux cercles de centres respectifs $B$ et $C$ et de rayons différents sont tangents extérieurement en un point $A$. Soit $t$ une tangente commune aux deux cercles ne contenant pas le point $A$. La perpendiculaire à la droite $t$ passant par le point $A$ coupe la médiatrice du segment $[BC]$ en un point $F$. Montrer que $BC = 2AF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoient $X$ et $Y$ les points de tangence de $t$ avec les cercles de centre $B$ et $C$ respectivement. Soit $O$ le point d'intersection de la droite $t$ avec la tangente commune aux deux cercles en $A$. Les points $X$ et $A$ sont symétriques par rapport à la droite $(OB)$ et les points $Y$ et $A$ sont symétriques par rapport à la droite $(OC)$. On déduit $\\widehat{BOA} = \\frac{1}{2} \\widehat{XOA}$ et $\\widehat{COA} = \\frac{1}{2} \\widehat{YOA}$. On en déduit :\n$$\n\\widehat{BOC} = \\widehat{BOA} + \\widehat{AOC} = \\frac{1}{2} \\widehat{XOA} + \\frac{1}{2} \\widehat{AOY} = \\frac{1}{2} \\widehat{XOY} = 90^{\\circ}\n$$\nSoit $M$ le milieu de $[BC]$. Le triangle $BOC$ est rectangle en $O$ donc $M$ est le centre du cercle circonscrit à ce triangle. Ceci donne $MO = MB = \\frac{1}{2} BC$. Comme les droites $(OA)$ et $(MF)$ sont perpendiculaires à la droite $(BC)$, elles sont donc parallèles. Enfin,\n$$\n\\widehat{MOY} = \\widehat{MOC} + \\widehat{COY} = \\widehat{MCO} + \\widehat{AOC} = 90^{\\circ}\n$$\ndonc les droites $(OM)$ et $(AF)$ sont perpendiculaires à la droite $t$ donc elles sont parallèles. Donc le quadrilatère $OMFA$ est un parallélogramme et $BC = 2OM = 2AF$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle isocèle en $A$, $\\Gamma$ est un cercle tangent à $(AC)$ en $C$ à l'extérieur du triangle $ABC$. On note $\\omega$ le cercle passant par $A$ et $B$ et tangent intérieurement à $\\Gamma$ en $D$. $E$ est la seconde intersection de $(AD)$ et de $\\Gamma$, montrer que $(BE)$ est tangente à $\\Gamma$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $I_{A}$ l'inversion de centre $A$ et de rayon $r = AB = AC$, ainsi, $B$ et $C$ sont fixes par $I_{A}$, la puissance de $A$ par rapport au cercle $\\omega$ est exactement le rayon au carré donc le cercle $\\omega$ est fixe par inversion. Cela montre que $D$ et $E$ sont échangés dans $I_{A}$. Mais alors le cercle $\\Gamma$ passant par $A, B$ et $D$ va être envoyé sur une droite passant par $B$ et $E$ et tangente au cercle $\\omega$, cela conclut.\n\n\nPreuve 2:\n\nOn construit $B'$, le deuxième point de $\\Gamma$, tel que $AB = AB' = AC$. Alors, on veut démontrer que $l'$, intersection, $E'$ des droites $(BB')$ et $(AD)$ est sur $\\omega$.\n\nOn a $\\widehat{BDA} = x$ et $\\widehat{ABB'} = x$. Par isocélité de $ABB'$ en $A$, on trouve $\\widehat{AB'B} = x$. Donc les triangles $ABD$ et $AEB$ sont semblables, on trouve alors $AB^2 = AE \\times AD$. Donc $E'$ est sur $\\omega$, ce qui montre que $E' = E$. L'homothétie de centre $D$ qui envoie le cercle $\\omega$ sur le cercle $\\Gamma$ envoie ainsi le point $E$ sur le point $A$ et donc la tangente en $E$ à $\\omega$ sur la tangente en $A$ à $\\Gamma$. Il se trouve que comme $AB = AB'$, la tangente en $A$ à $\\Gamma$ est parallèle à la droite $(B')$, mais la tangente en $E$ à $\\omega$ doit être parallèle à la tangente en $A$ à $\\Gamma$. Cela montre que la tangente en $E$ à $\\omega$ et $(BB')$ ne sont en fait qu'une seule et même droite et cela conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les polynômes à coefficients réels $P$ tels que le polynôme\n$$\n(X+1) P(X-1)-(X-1) P(X)\n$$\nsoit constant.",
    "options": [],
    "answer": "All solutions are P(X) = a X(X+1) + c with real a, c; for these, (X+1)P(X-1) − (X-1)P(X) equals the constant 2c.",
    "solution": "Solution:\nOn présente deux solutions : une qui utilise une factorisation polynomiale, et une autre plus calculatoire.\n\nCommençons par la solution par factorisation. Supposons que $(X+1) P(X-1)-(X-1) P(X)$ est constant égal à $2k$. En prenant $X=-1$, on obtient $2P(-1)=2k$. En prenant $X=1$, on obtient $2P(0)=2k$. On a donc $P(0)=P(-1)=k$, donc $0$ et $-1$ sont des racines de $P-k$, donc $P-k$ se factorise par $X$ et $X+1$. Par conséquent, il existe un polynôme $Q$ tel que\n$$\nP(X)=X(X+1) Q(X)+k\n$$\nLa condition se réécrit alors de la manière suivante : le polynôme\n$$\n(X-1) X(X+1)(Q(X-1)-Q(X))+2k\n$$\nest constant égal à $2k$, donc\n$$\n(X-1) X(X+1)(Q(X-1)-Q(X))=0\n$$\nOn a donc $Q(n)=Q(n-1)$ pour tout entier $n \\geqslant 2$, donc $Q$ prend une infinité de fois la valeur $Q(2)$, donc $Q-Q(2)$ a une infinité de racines, donc $Q$ est constant. Finalement, $P$ est donc de la forme\n$$\nP(X)=c X(X+1)+k\n$$\nou $c$ et $k$ sont deux nombres réels. Réciproquement, si $P$ est de cette forme, on vérifie facilement que le polynôme donné est constant, égal à $2k$.\n\nPassons à la solution calculatoire. On note $d$ le degré de $P$, et on écrit\n$$\nP(X)=a_{d} X^{d}+a_{d-1} X^{d-1}+\\ldots\n$$\nav ec $a_{d} \\neq 0$. Le polynôme $(X+1) P(X-1)-(X-1) P(X)$ est au plus de degré $d+1$, et on va calculer ses coefficients de degrés $d+1$ et $d$. On a\n$$\n\\begin{aligned}\nP(X-1) & =a_{d}(X-1)^{d}+a_{d-1}(X-1)^{d-1}+ [\\text{Polynôme de degré au plus } d-2] \\\\\n& =a_{d} X^{d}-d a_{d} X^{d-1}+a_{d-1} X^{d-1}+ [\\text{Polynôme de degré au plus } d-2]\n\\end{aligned}\n$$\nen développant $(X-1)^{d}$ et $(X-1)^{d-1}$ avec le binôme de Newton. On en déduit $(X+1) P(X-1)=a_{d} X^{d+1}-d a_{d} X^{d}+a_{d-1} X^{d}+a_{d} X^{d}+[$ Polynôme de degré au plus $d-2]$ et, similairement,\n$$\n(X-1) P(X)=a_{d} X^{d+1}+a_{d-1} X^{d}-a_{d} X^{d}+ [\\text{Polynôme de degré au plus } d-2]\n$$\nDans $(X+1) P(X-1)-(X-1) P(X)$, le coefficient devant $X^{d+1}$ vaut donc $a_{d}-a_{d}=0$, et le coefficient devant $X^{d}$ vaut $(2-d) a_{d}$, avec $a_{d}$. Si $d>0$, comme le polynôme doit être constant, ce coefficient doit être nul. Comme $a_{d}>0$ par définition, on doit donc avoir $d=2$, donc $P$ est de degré $0$ ou $2$, donc de la forme\n$$\nP(X)=a X^{2}+b X+c\n$$\nEn remplaçant $P$ par cette expression, on obtient que le polynôme suivant doit être constant :\n$$\n(b-a) X+(a-b+2c)\n$$\nCe dernier polynôme est constant si et seulement si $b-a=0$, soit $a=b$, c'est-à-dire si $P$ est de la forme\n$$\nP(X)=a X(X+1)+c\n$$\nav ec $a, c \\in \\mathbb{R}$, et on retrouve la même solution que par la première méthode.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle acutangle non isocèle en $A$. Soit $M$ le milieu du segment $[BC]$, $H$ l'orthocentre du triangle $ABC$, $O_{1}$ le milieu du segment $[AH]$ et $O_{2}$ le centre du cercle circonscrit au triangle $CBH$. Montrer que le quadrilatère $O_{1}AMO_{2}$ est un parallélogramme.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $H_{A}$ le pied de la hauteur issue du sommet $A$, $H_{B}$ le pied de la hauteur issue du sommet $B$ et $O$ le centre du cercle circonscrit au triangle $ABC$.\n\nOn remarque déjà que $O_{2}$ est sur la médiatrice de $[BC]$ donc les droites $(MO_{2})$ et $(AH)$ sont parallèles car elles sont perpendiculaires à $(BC)$. Il suffit donc de montrer que $MO_{2} = AO_{1}$.\n\nSoit $X$ le symétrique du point $H$ par rapport au point $M$. Alors $M$ est le milieu de $[BC]$ et $[XH]$ donc $BHCX$ est un parallélogramme et $\\widehat{XBA} = \\widehat{XBC} + \\widehat{CBA} = \\widehat{HCB} + \\widehat{CBA} = 90^{\\circ}$ (car le triangle $BCH_{C}$ est rectangle en $H_{C}$) et de même $\\widehat{XCA} = 90^{\\circ}$ donc $X$ est le point diamétralement opposé à $A$ sur le cercle circonscrit à $ABC$.\n\nLa symétrie de centre $M$ envoie $B$ sur $C$, $C$ sur $B$ et $H$ sur $X$ donc elle envoie le cercle circonscrit à $BCH$ sur le cercle circonscrit à $BCX$ donc elle envoie $O_{2}$ sur $O$. En particulier, $MO_{2} = MO$.\n\nComme les points $O$ et $M$ sont les milieux respectifs des segments $[XA]$ et $[XH]$, d'après le théorème de Thalès,\n$$\n\\frac{OM}{AH} = \\frac{XM}{XH} = \\frac{1}{2}\n$$\ndonc $AH = 2 OM$ donc\n$$\nMO_{2} = MO = \\frac{1}{2} AH = AO_{1}\n$$\nce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k \\geq 3$ un entier. On définit la suite $\\left(a_{n}\\right)_{n \\geq k}$ par $a_{k}=2k$, et\n$$\na_{n}= \\begin{cases}a_{n-1}+1 & \\text{ si } \\operatorname{pgcd}\\left(a_{n-1}, n\\right)=1 \\\\ 2n & \\text{ sinon. }\\end{cases}\n$$\n\nMontrer que la suite $\\left(a_{n+1}-a_{n}\\right)_{n \\geq k}$ a une infinité de termes qui sont des nombres premiers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPartons d'un entier $n$ tel que $a_{n}=2n$. Montrons par récurrence que si $p$ est le plus petit facteur premier de $n-1$, alors pour tout $i \\in\\{0, \\ldots, p-2\\}, a_{n+i}=2n+i$. En effet, c'est vrai pour $i=0$, et si c'est vrai pour un certain $i<p-2$, alors\n$$\n\\operatorname{pgcd}(n+i+1,2n+i)=\\operatorname{pgcd}(n+i+1, n-1)=\\operatorname{pgcd}(i+2, n-1)=1\n$$\ncar $i+2<p$, et donc $i+2$ est premier avec $n-1$ par définition de $p$, ce qui conclut la récurrence. De même, $\\operatorname{pgcd}(n+p-1,2n+p-2)=\\operatorname{pgcd}(p, n-1)=p \\neq 1$, et donc $a_{n+p-1}=2(n+p-1)$. En particulier,\n$$\na_{n+p-1}-a_{n+p-2}=2n+2p-2-(2n+p-2)=p\n$$\nest premier.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont les angles sont aigus, $\\omega$ son cercle circonscrit et $O$ le centre de $\\omega$. La perpendiculaire issue de $A$ par rapport à $(BC)$ intersecte $(BC)$ en $D$ et $\\omega$ en $E$. Soit $F$ un point sur le segment $[AE]$ tel que $2 \\cdot FD = AE$. Soit $\\ell$ la perpendiculaire à $(OF)$ passant par $F$. Montrer que la droite $\\ell$, la droite $(BC)$ et la tangente à $\\omega$ en $E$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn désire montrer que trois droites sont concourantes. Une stratégie serait donc de considérer le point d'intersection de deux de ces droites et de montrer que ce point appartient à la troisième droite, c'est la stratégie que nous allons adopter ici.\n\nDans le cas de notre exercice, il était possible de choisir n'importe quelle paire de droites parmi les droites $(\\ell)$, $(BC)$ et la tangente à $\\omega$ en $E$ et aboutir.\n\nSoit $X$ le point d'intersection de la tangente à $\\omega$ en $E$ avec la droite $(BC)$. Nous allons montrer que $X$ est sur $\\ell$ en montrant que les droites $(FX)$ et $(FO)$ sont perpendiculaires. Et pour montrer qu'elles sont bien perpendiculaires, on va utiliser la réciproque du théorème de Pythagore en montrant que $XO^2 = OF^2 + FX^2$. On notera dans la suite $R$ le rayon de $\\omega$.\n\nD'une part, puisque $(XE)$ est tangente au cercle $\\omega$, on a $XO^2 = XE^2 + OE^2 = XE^2 + R^2$. D'après le théorème de Pythagore dans le triangle $XDE$ rectangle en $D$, $XE^2 = XD^2 + DE^2$. On en déduit que\n$$\nXO^2 = XD^2 + DE^2 + R^2\n$$\nD'autre part, d'après le théorème de Pythagore dans le triangle $FDX$ rectangle en $D$, $XF^2 = XD^2 + FD^2$. Pour calculer $OF^2$, on utilise la formule de la puissance du point $F$ par rapport au cercle $\\omega$. En effet, cette formule nous dit que la puissance du point $F$ par rapport au cercle $\\omega$ vaut $OF^2 - R^2$. Comme d'autre part elle vaut $-FA \\times FE$, on déduit que $OF^2 = R^2 - FA \\times FE$. On a alors,\n$$\nXF^2 + OF^2 = (XD^2 + FD^2) + (R^2 - FA \\times FE)\n$$\nPour montrer que $XO^2 = XF^2 + OF^2$, il suffit donc de montrer que $DE^2 = FD^2 - FA \\times FE$, ou encore $DE^2 + FA \\times FE = FD^2$. Il s'agit alors de faire apparaître l'hypothèse de longueur sur le point $F$.\n\nNotons $H$ l'orthocentre du triangle $ABC$. On sait que les points $H$ et $E$ sont symétriques par rapport à la droite $(BC)$ (résultat classique de géométrie du triangle que l'on pourra retrouver dans le polycopié de géométrie pour débutant de la POFM, exercice 4.8). On déduit que $DE = DH$. Mais alors,\n$$\nFH = FD - DH = \\frac{1}{2}(AE - HE) = \\frac{1}{2} AH.\n$$\nEn particulier, $F$ est le milieu du segment $[AH]$. Il ne reste plus qu'à calculer $DE^2 + FA \\times FE$ de la manière suivante\n$$\n\\begin{aligned}\nDE^2 + FA \\times FE &= DE^2 + FA \\times (FH + \\underbrace{DH + DE}_{=2DE}) \\\\\n&= DE^2 + FA^2 + 2DE \\times FA \\\\\n&= (FA + DE)^2 \\\\\n&= (FH + DH)^2 = FD^2.\n\\end{aligned}\n$$\nCe qui conclut bien que $XO^2 = XF^2 + FO^2$ et termine la preuve du problème.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. On désigne par $D$ le pied de la bissectrice de $\\widehat{BAC}$, et par $E$ le pied de la hauteur issue de $A$. La médiatrice de $[AD]$ rencontre les demi-cercles de diamètres respectifs $[AB]$ et $[AC]$ construits extérieurement à $ABC$, en $X$ et $Y$.\n\nProuver que les points $X, Y, D, E$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nOn désigne par $M, N$ et $P$ les milieux respectifs de $[AB],[AC]$ et $[AD]$. D'après le théorème des milieux, on a $N, P, M$ alignés et, compte-tenu de ce que $(AP)=(AD)$ est la bissectrice de $\\widehat{BAC}$, on a\n$$\n\\frac{MP}{NP}=\\frac{BD}{CD}=\\frac{AM}{AN}=\\frac{AB}{AC}=\\frac{MX}{NY}.\n$$\nEn particulier, il vient $\\frac{MP}{MX}=\\frac{NP}{NY}$. En utilisant la loi des sinus dans $MPX$ et $NPY$, et puisque $\\widehat{NPY}=\\widehat{MPX}$, c'est donc que $\\sin \\widehat{PXM}=\\sin \\widehat{PYN}$. (1)\n\nOr, on a clairement $\\widehat{PXM}<\\widehat{BXA}=90^\\circ$ et $\\widehat{PYN}<\\widehat{AYC}=90^\\circ$, donc (1) donne $\\widehat{PXM}=\\widehat{PYN}$. Comme $\\widehat{MPX}=\\widehat{NPY}$, c'est donc que les triangles $PXM$ et $PYN$ sont semblables, et que $\\widehat{PMX}=\\widehat{PNY}$. (2)\n\nSans perte de généralité, on peut supposer qu'entre $M$ et $N$, c'est $N$ qui est du même côté que $A$ par rapport à $(XY)$.\nAinsi (2) conduit à $\\widehat{PMA}+\\widehat{AMX}=\\widehat{PNC}+\\widehat{CNY}$,\nd'où $\\widehat{CBA}+2\\widehat{ABX}=180^\\circ-\\widehat{BCA}+180^\\circ-2\\widehat{ACY}$\net donc\n$$\n\\begin{aligned}\n\\widehat{ABX}+\\widehat{ACY} & =180^\\circ-\\widehat{ABC}-\\widehat{BCA}+90^\\circ-\\widehat{ABX}+90^\\circ-\\widehat{ACY} \\\\\n& =\\widehat{BAC}+\\widehat{XAB}+\\widehat{CAY} \\\\\n& =\\widehat{XAY} .\n\\end{aligned}\n$$\nMais, $\\widehat{AEB}=\\widehat{AEC}=90^\\circ$ donc $E$ appartient aux cercles de diamètres $[AB]$ et $[AC]$, soit donc aux cercles circonscrits aux triangles $ABX$ et $ACY$. Par suite, on a $\\widehat{ABX}=\\widehat{AEX}$ et $\\widehat{ACY}=\\widehat{AEY}$. (4)\n\nDe (3) et (4), on déduit que\n$$\n\\widehat{XEY}=\\widehat{XEA}+\\widehat{AEY}=\\widehat{ABX}+\\widehat{ACY}=\\widehat{XAY}\n$$\nD'autre part, puisque $(XY)$ est la médiatrice de $[AD]$, on a $\\widehat{XAY}=\\widehat{XDY}$, et finalement $\\widehat{XEY}=\\widehat{XDY}$, ce qui conclut.\nSolution:\n\nOn applique une inversion $i$ de pôle $A$. Rappelons que les cercles ne passant pas par $A$ sont transformés en des cercles ne passant pas par $A$, que les droites passant par $A$ sont conservées et que les cercles passant par $A$ sont transformés en des droites perpendiculaires à la droite reliant $A$ et le centre du cercle. D'autre part, toute inversion conserve les angles entre deux droites ou cercles sécants.\n\nOn en déduit que $i$ transforme le cercle de diamètre $[AB]$ en la perpendiculaire à $\\left(AB'\\right)$ passant par $B'$, transforme la droite $(BC)$ en le cercle $\\left(AB'C'\\right)$ de sorte que $D'$ est le milieu de l'arc $B'C'$ ne contenant pas $A$ et $E'$ est le point diamétralement opposé à $A$. De plus, $i$ envoie la médiatrice de $[AD]$ sur le cercle de centre $D'$ passant par $A$. Enfin, comme $E$ sur le cercle de diamètre $[AB]$, les points $E', B, X'$ sont alignés et de même $E', C, Y'$ sont alignés.\n\nSupposons que les angles en $B$ et en $C$ du triangle $ABC$ sont aigus (le raisonnement est analogue dans le cas contraire). Alors $E'$ appartient à l'arc $B'C'$ ne contenant pas $A$. Quitte à échanger les rôles de $B$ et de $C$, on peut supposer que $E'$ appartient à l'arc $C'D'$ comme sur la figure.\n![](attached_image_2.png)\n\nLe fait que $E$ ne se trouve pas sur le même demi-cercle de diamètre $[AB]$ que $X$ implique que $B'$ se trouve entre $E'$ et $X'$. De même, $C'$ se trouve entre $E'$ et $Y'$.\n\nRemarquons d'abord que $\\widehat{Y'E'D'}$ et $\\widehat{X'E'D'}$ sont supplémentaires puisque $\\widehat{Y'E'D'}=\\widehat{C'E'D'}=\\pi-\\widehat{C'AD'}=\\pi-\\widehat{D'AB'}=\\pi-\\widehat{D'E'B'}=\\pi-\\widehat{D'E'X'}$.\n\nAppliquons maintenant le théorème de Ceva trigonométrique pour le point $D'$ appartenant au triangle $EX'Y'$.\n$$\n\\frac{\\sin \\left(\\overrightarrow{E'D'}, \\overrightarrow{E'X'}\\right)}{\\sin \\left(\\overrightarrow{E'D'}, \\overrightarrow{E'Y'}\\right)} \\times \\frac{\\sin \\left(\\overrightarrow{X'D'}, \\overrightarrow{X'Y'}\\right)}{\\sin \\left(\\overrightarrow{X'D'}, \\overrightarrow{X'E'}\\right)} \\times \\frac{\\sin \\left(\\overrightarrow{Y'D'}, \\overrightarrow{Y'E'}\\right)}{\\sin \\left(\\overrightarrow{Y'D'}, \\overrightarrow{Y'X'}\\right)}=-1 .\n$$\nComme $\\left(\\overrightarrow{E'D'}, \\overrightarrow{E'X'}\\right)=\\left(\\overrightarrow{E'D'}, \\overrightarrow{E'B'}\\right)=\\left(\\overrightarrow{AD'}, \\overrightarrow{AB'}\\right)$ et $\\left(\\overrightarrow{E'D'}, \\overrightarrow{E'Y'}\\right)=\\left(\\overrightarrow{E'D'}, \\overrightarrow{E'C'}\\right)=\\left(\\overrightarrow{AD'}, \\overrightarrow{AC'}\\right)+\\pi$, la première fraction est égale à 1.\n\nDe plus, comme $D'X'Y'$ est isocèle en $D'$, on a $\\left(\\overrightarrow{X'D'}, \\overrightarrow{X'Y'}\\right)=-\\left(\\overrightarrow{Y'D'}, \\overrightarrow{Y'X'}\\right)$. En reportant dans (1), on en conclut que\n$$\n\\sin \\left(\\overrightarrow{X'D'}, \\overrightarrow{X'E'}\\right)=\\sin \\left(\\overrightarrow{Y'D'}, \\overrightarrow{Y'E'}\\right)\n$$\nce qui prouve que les angles $\\widehat{D'X'E'}$ et $\\widehat{D'Y'E'}$ sont égaux ou supplémentaires.\n\nS'ils étaient supplémentaires, on aurait $\\widehat{X'E'D'}+\\widehat{D'X'E'}+\\widehat{Y'E'D'}+\\widehat{D'Y'E'}=2\\pi$, donc la somme des angles des triangles $D'X'E'$ et $D'Y'E'$ dépasserait $2\\pi$, ce qui est impossible. Par conséquent, $\\widehat{D'X'E'}$ et $\\widehat{D'Y'E'}$ sont égaux. D'après $(2)$, ceci montre que $\\left(\\overrightarrow{X'D'}, \\overrightarrow{X'E'}\\right)=\\left(\\overrightarrow{Y'D'}, \\overrightarrow{Y'E'}\\right)$, d'où la cocyclicité de $D', E', X', Y'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a$ et $b$ deux entiers tels que $a + b$ n'est pas divisible par $3$. Montrer que l'on ne peut pas colorier les entiers relatifs en trois couleurs de sorte que pour tout entier $n$, les trois nombres $n$, $n+a$ et $n+b$ soient de couleurs différentes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons par l'absurde qu'il existe un tel coloriage. On notera $x \\sim y$ si $x$ et $y$ sont de la même couleur.\n\nPour tout $n$, les entiers $(n+a)+b$ et $n+a$ ne sont pas de la même couleur, et de même $n+a+b$ n'est pas de la même couleur que $n+b$, donc $n+a+b \\sim n$. Il vient immédiatement $0 \\sim p(a+b)$ pour tout entier $p$, et en particulier $0 \\sim a(a+b)$.\n\nD'autre part, $n+a$, $n+2a$ et $n+a+b$ sont de couleurs différentes, donc $n$, $n+a$ et $n+2a$ sont de couleurs différentes. En appliquant ce qui précède à $n+a$, on en déduit que $n+3a$ est de même couleur que $n$, donc pour tout entier $j$, le nombre $ja$ est de la même couleur que $ra$ où $r$ est le reste de la division euclidienne de $j$ par $3$.\n\nEn appliquant ceci à $j=a+b$, on en déduit que $0 \\sim ra$ où $r$ est le reste de la division euclidienne de $a+b$ par $3$, ce qui est impossible puisque $0$, $a$ et $2a$ sont de couleurs différentes et $r \\in \\{1,2\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a_{1}, \\ldots, a_{2019}$ des entiers positifs. Montrer qu'il y a équivalence entre:\n\n(i) il existe un réel $x$ tel que pour tout $i \\in \\{1, \\ldots, 2019\\}$, on a : $a_{i}=\\lfloor i x\\rfloor$\n\n(ii) pour tous $i, j \\in \\{1, \\ldots, 2019\\}$ vérifiant $i+j \\leqslant 2019$, on a : $a_{i}+a_{j} \\leqslant a_{i+j} \\leqslant a_{i}+a_{j}+1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nProuvons tout d'abord le sens direct : s'il existe $x$ réel tel que pour tout $i$ entre $1$ et $2019$, $a_{i}=\\lfloor i x\\rfloor$, et si on se donne $i, j$ entre $1$ et $2019$ tels que $i+j \\leqslant 2019$, alors $(i+j)x = i x + j x \\geqslant \\lfloor i x\\rfloor + \\lfloor j x\\rfloor = a_{i} + a_{j}$ et $(i+j)x = i x + j x < \\lfloor i x\\rfloor + 1 + \\lfloor j x\\rfloor + 1 = a_{i} + a_{j} + 2$, donc $a_{i} + a_{j} \\leqslant a_{i+j} < a_{i} + a_{j} + 2$. Comme les $a_{i}$ sont entiers, on a donc $a_{i} + a_{j} \\leqslant a_{i+j} \\leqslant a_{i} + a_{j} + 1$.\n\nRéciproquement, supposons que pour tout $i, j$ entre $1$ et $2019$ tels que $i+j \\leqslant 2019$, $a_{i} + a_{j} \\leqslant a_{i+j} \\leqslant a_{i} + a_{j} + 1$. Trouver $x$ tel que pour tout $i$ entre $1$ et $2019$, $a_{i} = \\lfloor i x\\rfloor$ revient à trouver $x$ tel que pour tout $i$ entre $1$ et $2019$, $a_{i} \\leqslant i x < a_{i} + 1$, c'est-à-dire tel que pour tout $i$, $\\frac{a_{i}}{i} \\leqslant x < \\frac{a_{i} + 1}{i}$.\n\nIl suffit donc de montrer que le maximum des $\\frac{a_{i}}{i}$ pour $i$ entre $1$ et $2019$ est strictement inférieur au minimum des $\\frac{a_{j} + 1}{j}$ pour $j$ entre $1$ et $2019$, ou plus simplement que $\\frac{a_{i}}{i} < \\frac{a_{j} + 1}{j}$ pour tout $i, j$ entre $1$ et $2019$.\n\nMontrons cela par récurrence forte sur $m = \\max(i, j)$. Si $m = 1$, $i = j = 1$ donc l'inégalité est évidente.\n\nPour l'hérédité, supposons l'inégalité vraie pour tout $i, j$ entre $1$ et $m$. On veut montrer l'inégalité $\\frac{a_{i}}{i} < \\frac{a_{j} + 1}{j}$, où $1 \\leqslant i, j \\leqslant m + 1$. Il y a trois cas à traiter :\n\n- Si $i = j$ l'inégalité est évidente.\n\n- Si $i < j$, en appliquant l'hypothèse de récurrence à $(i, j-i)$ : $a_{j} \\geqslant a_{i} + a_{j-i} > a_{i} + (j-i) \\frac{a_{i}}{i} - 1 = a_{i} \\times \\frac{j}{i} - 1$ donc $\\frac{a_{j} + 1}{j} > \\frac{a_{i}}{i}$.\n\n- Si $i > j$, en appliquant l'hypothèse de récurrence à $(i-j, j)$ : $a_{i} \\leqslant a_{j} + a_{i-j} + 1 < a_{j} + \\frac{(i-j)(a_{j} + 1)}{j} = a_{j} \\frac{i}{j} + \\frac{i-j}{j} < (a_{j} + 1) \\frac{i}{j}$ donc $\\frac{a_{i}}{i} < \\frac{a_{j} + 1}{j}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers $n \\geqslant 1$ tels qu'il existe une permutation $(a_{1}, a_{2}, \\ldots, a_{n})$ de $(1,2, \\ldots, n)$ vérifiant la condition suivante :\n$$\nk \\mid a_{1}+a_{2}+\\ldots+a_{k}\n$$\npour tout $k \\in\\{1,2, \\ldots, n\\}$.",
    "options": [],
    "answer": "n = 1 and n = 3",
    "solution": "Solution:\n\nOn commence par regarder ce qu'il se passe pour $n=1,2,3$ : $n=1,3$ sont solutions mais $n=2$ ne l'est pas. Soit $n>3$ vérifiant la propriété de l'énoncé.\nOn a $n \\mid a_{1}+a_{2}+\\ldots+a_{n}=1+2+\\ldots+n=\\frac{n(n+1)}{2}$ et donc $\\frac{n+1}{2} \\in \\mathbb{Z}$ : $n$ est impair.\nEnsuite $n-1 \\mid a_{1}+a_{2}+\\ldots+a_{n-1}=\\frac{n(n+1)}{2}-a_{n}$.\nEt $n-1 \\mid \\frac{(n-1)(n+1)}{2}$ donc $n-1 \\mid \\frac{n+1}{2}-a_{n}$.\nCependant, il y a un \"problème de taille\" : $-(n-1)<\\frac{n+1}{2}-a_{n}<n-1$\nce qui n'est pas absurde, mais implique que $a_{n}=\\frac{n+1}{2}$.\nOn poursuit (pas de souci, $n>3$ ) :\n$n-2 \\mid \\frac{n(n+1)}{2}-\\frac{n+1}{2}-a_{n-1}=\\frac{(n-1)(n+1)}{2}-a_{n-1}$.\nEt $n-2 \\mid \\frac{(n-2)(n+1)}{2}$ donc $n-2 \\mid \\frac{n+1}{2}-a_{n-1}$.\nComme précédemment, $-(n-2)<\\frac{n+1}{2}-a_{n-1}<n-2$ et donc $a_{n}=a_{n-1}=\\frac{n+1}{2}$ : c'est absurde.\n\nLes seules solutions sont donc bien 1 et 3 .",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ADE$ un triangle isocèle en $A$. Soit le cercle $\\omega$, tangent aux droites $(AD)$ et $(AE)$ en $D$ et $E$. Soit $B$ et $C$ deux points au delà de $D$ et $E$ sur les droites $(AD)$ et $(AE)$. On suppose que $BC > BD + CE$. Finalement, soit $F$ et $G$ deux points sur le segment $[BC]$ de telle sorte que $BD = BF$ et $CG = CE$, les droites $(DG)$ et $(FE)$ se coupent en $K$, la tangente à $\\omega$ parallèle à $(BC)$ et entre $A$ et $(BC)$ touche $\\omega$ en $L$. Montrer que le centre du cercle inscrit de $ABC$ est sur $(KL)$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $B'$, $C'$ deux points sur les demi-droites $[AD]$ et $[AC]$ au delà de $D$ et $E$ de telle sorte que $(B'C') \\parallel (BC)$ et que $(B'C')$ soit tangente à $\\omega$ en $L'$\n\nLemme. $D, F$ et $L'$ sont alignés, de même pour $E, G$ et $L'$.\n\nOn a $(BF) \\parallel (B'L')$ et $DB = BF$, ainsi que $DB' = B'L'$, donc $DBF \\sim DB'L'$ par Thalès. On obtient que $D, F$ et $L'$ sont alignés, le raisonnement symétrique montre que $E, G$ et $L'$ sont alignés.\n\nOn remarque que $\\omega$ est le cercle inscrit de $AB'C'$. La tangente en $L$ et la tangente en $L'$ sont deux droites parallèles, donc le segment $[LL']$ est un diamètre de $\\omega$, ainsi $\\widehat{L'EL} = 90^\\circ$ et $\\widehat{L'DL} = 90^\\circ$.\n\nComme $I$ est le centre du cercle inscrit de $ABC$, $I$ est l'intersection des bissectrices des angles $\\widehat{ABC}$, $\\widehat{BCA}$ et $\\widehat{CAB}$ ou encore, par isocélité des triangles $DBF$, $GCE$ et $EAD$, que $I$ est l'intersection des médiatrices de $[DF]$, $[GE]$ et $[ED]$. Le fait que ces médiatrices soient concourantes nous amène à dire que le quadrilatère $DEFG$ est cyclique et de centre du cercle circonscrit $I$.\n\n![](attached_image_2.png)\n\nOn va isoler maintenant le cercle $DEFG$, le point $I$, le point $L$ et le point $K$, en sachant que $I$ est le centre du cercle $DEFG$, que $K$ est l'intersection des diagonales et que $L$ est tel que $\\widehat{L'EL} = 90^\\circ$ et $\\widehat{L'DL} = 90^\\circ$.\n\nOn intersecte maintenant la droite $(EL)$ avec le cercle $EDFG$ en $G'$, et la droite $(DL)$ également avec le cercle $EDG$ en $F'$. En sachant que $\\widehat{L'EL} = 90^\\circ$ et $\\widehat{L'DL} = 90^\\circ$, on trouve que $G'$ (resp. $D'$) est le point diamétralement opposé à $G$ (resp. $D$) dans le cercle $EDFG$, donc $I$ est l'intersection des droites $GG'$ et $DD'$.\n\nOn numérote les points $G'$, $G$, $D$, $F'$, $F$ et $E$ comme 1, 2, 3, 4, 5 et 6 et on applique Pascal : comme les 6 points sont sur le même cercle, alors $(G'G) \\cap (F'F)$, $(GD) \\cap (FE)$ et $(DF') \\cap (EG')$ (c'est-à-dire $(12) \\cap (45)$, $(23) \\cap (56)$ et $(34) \\cap (61)$) sont alignés, ce qui démontre bien que les points $K$, $I$ et $L$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $x_{1}, \\ldots, x_{n}$ des réels quelconques. Montrer que\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n}\\left|x_{i}+x_{j}\\right| \\geqslant n \\sum_{i=1}^{n}\\left|x_{i}\\right| .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nParmi les $n$ nombres, on suppose que $r$ sont positifs (ou nuls), et $s=n-r$ sont strictement négatifs. On note $\\left(a_{i}\\right)_{1 \\leqslant i \\leqslant r}$ la liste des nombres positifs (ou nuls), et $\\left(-b_{j}\\right)_{1 \\leqslant j \\leqslant s}$ la liste des nombres strictement négatifs, de sorte que tous les $a_{i}$ et les $b_{j}$ sont positifs.\nOn pose\n$$\nA=\\sum_{i=1}^{r} a_{i}, \\quad B=\\sum_{j=1}^{s} b_{j}\n$$\nOn note $M_{1}$ et $M_{2}$ le membre de gauche et le membre de droite dans l'énoncé respectivement. Quitte à changer tous les signes, on peut supposer $A \\geqslant B$.\nOn a alors\n$$\nM_{1}=2 r A+2 s B+2 \\sum_{1 \\leqslant i \\leqslant r} \\sum_{1 \\leqslant j \\leqslant s}\\left|a_{i}-b_{j}\\right|,\n$$\nou $2 r A$ est la somme des termes $\\left|x_{i}+x_{j}\\right|$ avec $x_{i}, x_{j} \\geqslant 0$ et de même $2 s B$ est la somme des termes $\\left|x_{i}+x_{j}\\right|$ avec $x_{i}, x_{j}<0$ et $\\sum \\sum\\left|a_{i}-b_{j}\\right|$ est la somme des termes $\\left|x_{i}+x_{j}\\right|$ où $x_{i}$ et $x_{j}$ sont de signes différents. De plus, on a $M_{2}=(r+s)(A+B)$. Montrer que $M_{1} \\geqslant M_{2}$ est donc équivalent à montrer\n$$\n2 \\sum_{i=1}^{r} \\sum_{j=1}^{s}\\left|a_{i}-b_{j}\\right| \\geqslant(s-r)(A-B)\n$$\nSi $s \\leqslant r$, le résultat est évident car le membre de gauche est positif et celui de droite est négatif. On suppose donc $s>r$. On a alors\n$$\n\\begin{aligned}\n2 \\sum_{i=1}^{r} \\sum_{j=1}^{s}\\left|a_{i}-b_{j}\\right| & \\geqslant \\sum_{i=1}^{r} \\sum_{j=1}^{s}\\left|a_{i}-b_{j}\\right| \\\\\n& \\geqslant \\sum_{i=1}^{r} \\sum_{j=1}^{s} a_{i}-b_{j} \\\\\n& =s A-r B \\\\\n& =(s-r) A+r A-r B \\\\\n& \\geqslant(s-r) A \\\\\n& \\geqslant(s-r)(A-B)\n\\end{aligned}\n$$\ncar $s>r$ et $A \\geqslant B$. On a donc montré ce qu'on voulait.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle isocèle en $A$, $D$ le pied de la hauteur issue de $A$ et $M$ un point intérieur à $ADC$ tel que $\\widehat{AMB}$ est obtus et $\\widehat{DBM} + \\widehat{DAM} = \\widehat{MCB}$. Les droites $(CM)$ et $(AD)$ se coupent en $P$, et les droites $(BM)$ et $(AD)$ se coupent en $Q$. Soit $S$ un point de $[AB]$ et $R$ un point de $[AM)$ qui n'est pas sur $[AM]$ tel que $\\widehat{SQB} = \\widehat{DPC}$ et $\\widehat{MRQ} = 2\\widehat{QAM}$. Montrer que $QRS$ est isocèle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $\\alpha = \\widehat{BAD}$, $\\beta = \\widehat{CBM}$, $\\gamma = \\widehat{MCB}$ et $\\theta = \\widehat{DAM}$. Par hypothèse, on a $\\beta + \\theta = \\gamma$, $\\widehat{SQB} = \\widehat{DPC} = \\frac{\\pi}{2} - \\gamma$ et $\\widehat{MRQ} = 2\\theta$.\n\nOn a $\\widehat{QSA} = \\pi - \\widehat{BSQ} = \\widehat{QBS} + \\widehat{SQB} = \\left(\\frac{\\pi}{2} - \\alpha - \\beta\\right) + \\left(\\frac{\\pi}{2} - \\gamma\\right) = \\pi - (\\alpha + \\beta + \\gamma)$.\n\nD'après la loi des sinus pour les triangles $QAS$ et $QAR$, on a\n$$\n\\frac{QS}{QR} = \\frac{QS}{QA} \\times \\frac{QA}{QR} = \\frac{\\sin \\alpha}{\\sin (\\alpha + \\beta + \\gamma)} \\frac{\\sin 2\\theta}{\\sin \\theta} = \\frac{2 \\sin \\alpha \\cos \\theta}{\\sin (\\alpha + \\beta + \\gamma)}\n$$\nOn doit montrer que ce rapport est égal à $1$.\n\nD'après le théorème de Ceva trigonométrique appliqué au point $M$ dans le triangle $ABC$, on a\n$$\n\\frac{\\sin (\\alpha + \\theta)}{\\sin (\\alpha - \\theta)} \\times \\frac{\\sin \\beta}{\\sin \\left(\\frac{\\pi}{2} - \\alpha - \\beta\\right)} \\times \\frac{\\sin \\left(\\frac{\\pi}{2} - \\alpha - \\gamma\\right)}{\\sin \\gamma} = 1\n$$\ndonc $\\sin (\\alpha + \\theta) \\sin \\beta \\cos (\\alpha + \\gamma) = \\sin (\\alpha - \\theta) \\cos (\\alpha + \\beta) \\sin \\gamma$.\n\nEn utilisant l'identité $2 \\sin x \\cos y = \\sin (x + y) + \\sin (x - y)$, on en déduit\n$$\n\\sin (\\alpha + \\theta)[\\sin (\\alpha + \\beta + \\gamma) - \\sin (\\alpha + \\gamma - \\beta)] = \\sin (\\alpha - \\theta)[\\sin (\\alpha + \\beta + \\gamma) - \\sin (\\alpha + \\beta - \\gamma)]\n$$\nCompte tenu de $\\gamma - \\beta = \\theta$,\n$$\n\\sin (\\alpha + \\theta)[\\sin (\\alpha + \\beta + \\gamma) - \\sin (\\alpha + \\theta)] = \\sin (\\alpha - \\theta)[\\sin (\\alpha + \\beta + \\gamma) - \\sin (\\alpha - \\theta)]\n$$\nou encore $(\\sin (\\alpha + \\theta) - \\sin (\\alpha - \\theta))[\\sin (\\alpha + \\beta + \\gamma) - (\\sin (\\alpha + \\theta) + \\sin (\\alpha - \\theta))]$.\n\nIl vient $\\sin (\\alpha + \\beta + \\gamma) = \\sin (\\alpha + \\theta) + \\sin (\\alpha - \\theta) = 2 \\sin \\alpha \\cos \\theta$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle équilatéral. Soit $X$ un point de la droite $(BC)$ différent de $B$ et $C$. Soient $Y$ et $Z$ deux points sur les droites $(AB)$ et $(AC)$ de telle sorte que les deux droites $(BZ)$ et $(CY)$ sont parallèles à la droite $(AX)$. La droite $(XY)$ intersecte la droite $(AC)$ en $M$ et la droite $(XZ)$ intersecte la droite $(AB)$ en $N$. Montrer que la droite $(MN)$ est tangente au cercle inscrit de $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par énoncer deux lemmes qui vont être utiles dans la suite. Si l'utilité du premier lemme sautera aux yeux du lecteur, l'intérêt du deuxième lemme paraîtra sans doute moins clair.\n\nLemme 1. Soit $ABC$ un triangle et $\\omega$ le cercle $A$-exinscrit de ce triangle, on note $J$ le centre de $\\omega$ et $\\alpha$ l'angle en $A$. Alors, $\\widehat{BJC} = 90 - \\alpha/2$, de plus il existe une forme d'unicité. Soit $X$ un point de la droite $(AB)$, si on effectue une rotation de la droite $(JX)$ autour de $J$ par un angle $90 - \\alpha/2$ et que l'on note par $Y$ l'intersection de cette droite avec la droite $(AB)$, alors la droite $(YX)$ est tangente à $\\omega$.\n\nRemarque 2. Le lecteur attentif remarquera que la rotation d'angle $90 - \\alpha/2$ depuis $J$ n'est pas unique, il faut faire attention au sens dans lequel on effectue la rotation (en fait il faut travailler avec des angles orientés).\n\n![](attached_image_1.png)\n\n## Démonstration.\nOn remarque dans un premier temps que l'unicité découle de la première partie du lemme. En effet, le point $Y$ est défini de manière unique et vérifie alors bien la condition $\\widehat{XJY} = 90 - \\alpha/2$. Il ne nous reste plus qu'à démontrer la première partie du lemme. On note $\\beta$ et $\\gamma$ les angles en $B$ et $C$. Alors, $(BJ)$ et $(CJ)$ sont des bissectrices extérieures donc\n$$\n\\widehat{JBC} = (180 - \\beta)/2 = 90 - \\beta/2\n$$\net\n$$\n\\widehat{JCB} = 90 - \\gamma/2\n$$\nComme les angles dans le triangle $BCJ$ ont pour somme $180$, il faut donc que $\\widehat{BJC} = 90 - \\alpha/2$ comme annoncé.\n\nLemme 2. Soit $ABCD$ un trapèze isocèle avec $(AB)$ parallèle à $(CD)$. On note $X$ le point d'intersection des diagonales $(AC)$ et $(BD)$ ainsi que $Y$ le point d'intersection des droites $(AD)$ et $(BC)$. La droite $(XY)$ intersecte la droite $(CD)$ au point $Z$. Le point $Z$ est alors le milieu du segment $[DC]$.\n\nRemarque 3. Il est possible que les droites $(AD)$ et $(BC)$ soient parallèles ; dans ce cas on considère le point $Y$ \"à l'infini\" (c'est en principe une notion bien définie mais on ne le fait pas proprement ici) dans la même direction que la droite $(AD)$ et donc de la droite $(BC)$ par hypothèses. Dans ce cas une droite qui passe par $Y$ est juste une droite qui est parallèle à $(AD)$ (et donc à $(BC)$). Moralement, passer par un point à l'infini c'est forcer une direction.\n\n![](attached_image_2.png)\n\n## Démonstration.\nIl existe une preuve naturelle de ce lemme en utilisant des outils de la géométrie projective mais on va donner ici une preuve élémentaire. Soit $W$ l'intersection de la droite $(XY)$ avec la droite $(AB)$, on peut alors écrire les identités suivantes qui découlent du théorème de Thalès.\nComme les trois droites $(AC)$, $(BD)$ et $(WZ)$ sont concourantes en $Y$ et les droites $(AB)$ et $(CD)$ sont parallèles, on a l'égalité\n$$\n\\frac{AW}{WB} = \\frac{CZ}{ZD}\n$$\nou l'on a pris soin des orientations de longueurs (ici les deux côtés de l'équation sont négatifs). De la même manière, les trois droites $(AD)$, $(BC)$ et $(WZ)$ sont concourantes en $Z$ et les droites $(AB)$ et $(CD)$ sont parallèles, on a donc l'égalité\n$$\n\\frac{BW}{WA} = \\frac{CZ}{ZD}\n$$\nEn combinant ces deux égalités on obtient\n$$\n\\left(\\frac{CZ}{ZD}\\right)^2 = 1\n$$\nEt donc, en prenant soin des signes\n$$\n\\frac{CZ}{ZD} = -1\n$$\nCe qui conclut la preuve du lemme.\n\nRevenons désormais au problème.\n\n![](attached_image_3.png)\n\nOn note $D$, $E$ et $F$ les points de tangence du cercle inscrit dans le triangle $ABC$. On va montrer un résultat a priori plus fort. Si on note $x = \\widehat{CAX}$ on va démontrer que $\\widehat{MIE} = x$. Cela conclurait alors car par symétrie on aurait $\\widehat{MIF} = \\widehat{BAX} = 60 - x$ et donc $\\widehat{MIN} = 120 - (60 - x) - x = 60$ ce qui conclurait d'après le Lemme 1.\n\nSoit $T$ le point d'intersection des droites $(AX)$ et $(BE)$, l'égalité $\\widehat{MIE} = \\widehat{EAT}$ est équivalente à la propriété que les points $A, M, I$ et $T$ sont cocycliques. Une chasse aux angles immédiate montre que cette propriété est encore équivalente à ce que les droites $(MT)$ et $(AB)$ soient parallèles (par exemple on peut utiliser des parallèles/antiparallèles par rapport aux droites $(EB)$ et $(EA)$). On note de plus $P$ l'intersection des droites $(MB)$ et $(AX)$.\n\nOn applique alors le Lemme 2 dans un premier temps au trapèze $AXCY$ pour montrer que le point $P$ est le milieu du segment $[AX]$. Ainsi, d'après la droite des milieux on a $F, P$ et $E$ alignés. On remarque alors que dans le quadrilatère $AMTB$ avec $P$ l'intersection des droites $(AT)$ et $(MB)$ ainsi que $E$ l'intersection des droites $(AM)$ et $(BT)$, la droite $(EP)$ passe par le milieu du segment $[AB]$, il suit que l'on peut alors appliquer la réciproque du Lemme 2 (il y a une unicité laissée au lecteur) pour montrer que le quadrilatère $AMTB$ est bien un trapèze avec $(AB) \\parallel (MT)$ ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a, b, c, d$ des réels strictement positifs tels que $a+b+c+d=1$. Montrer que :\n$$\n\\frac{a^{4}}{a^{3}+a^{2} b+a b^{2}+b^{3}}+\\frac{b^{4}}{b^{3}+b^{2} c+b c^{2}+c^{3}}+\\frac{c^{4}}{c^{3}+c^{2} d+c d^{2}+d^{3}}+\\frac{d^{4}}{d^{3}+d^{2} a+d a^{2}+a^{3}} \\geqslant \\frac{1}{4}\n$$\net déterminer les cas d'égalité.",
    "options": [],
    "answer": "Lower bound is 1/4, with equality if and only if a = b = c = d = 1/4.",
    "solution": "Solution:\n\nDans cette solution, on va utiliser à plusieurs reprises l'inégalité $2\\left(x^{2}+y^{2}\\right) \\geqslant (x+y)^{2}$, valable pour tous réels $x, y$ avec égalité si et seulement si $x=y$.\n\nDéjà notons que $a^{3}+a^{2} b+a b^{2}+b^{3}=\\left(a^{2}+b^{2}\\right)(a+b)$. En particulier l'inégalité se réécrit\n$$\n\\sum_{\\text{cycl.}} \\frac{a^{4}}{\\left(a^{2}+b^{2}\\right)(a+b)} \\geqslant \\frac{1}{4}.\n$$\nNotons $S$ la somme de gauche dans l'inégalité précédente. Chaque dénominateur des termes de $S$ est symétrique en les deux variables qu'il contient, on cherche donc une relation entre $S$ et $\\sum_{\\text{cycl.}} \\frac{b^{4}}{\\left(a^{2}+b^{2}\\right)(a+b)}$.\n\nCalculons la différence des deux termes :\n$$\nS-\\sum_{\\text{cycl.}} \\frac{b^{4}}{\\left(a^{2}+b^{2}\\right)(a+b)}=\\sum_{\\text{cycl.}} \\frac{a^{4}-b^{4}}{\\left(a^{2}+b^{2}\\right)(a+b)}=\\sum_{\\text{cycl.}} \\frac{a^{2}-b^{2}}{(a+b)}=\\sum_{\\text{cycl.}} a-b=0\n$$\nEn particulier $S=\\sum_{\\text{cycl.}} \\frac{b^{4}}{\\left(a^{2}+b^{2}\\right)(a+b)}$ donc $2S=\\sum_{\\text{cycl.}} \\frac{a^{4}+b^{4}}{\\left(a^{2}+b^{2}\\right)(a+b)}$.\n\nPar l'inégalité précédente, $a^{4}+b^{4} \\geqslant \\frac{\\left(a^{2}+b^{2}\\right)^{2}}{2}$ donc en sommant\n$$\n2S \\geqslant \\sum_{\\text{cycl.}} \\frac{\\left(a^{2}+b^{2}\\right)^{2}}{2\\left(a^{2}+b^{2}\\right)(a+b)} = \\sum_{\\text{cycl.}} \\frac{a^{2}+b^{2}}{2(a+b)}.\n$$\nOr, $a^{2}+b^{2} \\geqslant \\frac{(a+b)^{2}}{2}$, donc en sommant\n$$\n2S \\geqslant \\sum_{\\text{cycl.}} \\frac{(a+b)^{2}}{4(a+b)} = \\sum_{\\text{cycl.}} \\frac{a+b}{4} = \\frac{2(a+b+c+d)}{4} = \\frac{1}{2}.\n$$\nDonc $S \\geqslant \\frac{1}{4}$.\n\nSupposons qu'on a égalité. Dans ce cas on a les égalités cycliques $a=b$, soit $a=b=c=d$. Par la condition sur la somme $4a=1$ donc $a=\\frac{1}{4}$.\n\nRéciproquement, si $a=b=c=d=\\frac{1}{4}$, alors $a+b+c+d=1$ et\n$$\nS=\\sum_{\\text{cycl.}} \\frac{a^{4}}{a^{3}+a^{2} b+a b^{2}+b^{3}}=\\sum_{\\text{cycl.}} \\frac{a^{4}}{4 a^{3}}=\\sum_{\\text{cycl.}} \\frac{a}{4}=4 \\cdot \\frac{a}{4}=\\frac{1}{4}\n$$\non a bien égalité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExercice 9. Dans tout l'exercice, on s'intéresse à des tableaux $5 \\times 5$, dont les lignes (resp. les colonnes) sont notées $L_{0}, L_{1}, L_{2}, L_{3}, L_{4}$ du bas vers le haut (resp. $C_{0}, C_{1}, C_{2}, C_{3}, C_{4}$, de gauche à droite).\nDans chacune des 25 cases d'un tel tableau, on écrit un nombre. Le nombre qui est dans la case située sur la ligne $L_{i}$ et la colonne $C_{j}$ est noté $f(i, j)$.\nUn tableau est dit cohérent lorsque, pour tous $i$ et $j$ :\nLe nombre $f(i, j)$ est le nombre de $i$ écrits dans les cases de la ligne $L_{j}$.\nDéterminer tous les tableaux cohérents de taille $5 \\times 5$.",
    "options": [],
    "answer": "[[[1, 0, 0, 0, 0], [2, 1, 0, 0, 0], [0, 1, 0, 1, 0], [1, 3, 2, 1, 1], [1, 0, 3, 3, 4]], [[2, 1, 0, 0, 0], [1, 0, 0, 0, 0], [0, 0, 0, 0, 1], [1, 4, 1, 1, 1], [1, 0, 4, 4, 3]]]",
    "solution": "Solution:\n\nOn se donne un éventuel tableau cohérent. Commençons par dégager quelques unes de ses propriétés.\n\na) Pour tous $i, j$, on a $f(i, j) \\in \\{0,1,2,3,4,5\\}$.\nEn effet, dans un tableau cohérent, chaque $f(i, j)$ représente le nombre de fois où $i$ apparaît sur $L_{j}$, donc $f(i, j)$ est un entier positif ou nul, et il ne peut dépasser 5 car c'est le nombre total de cases de $L_{j}$.\n\nb) On a $f(0,0) \\geqslant 1$.\nPar l'absurde : Si $f(0,0)=0$ alors, il n'y a aucun 0 sur $L_{0}$. Mais, $f(0,0)$ est lui-même sur $L_{0}$. Contradiction.\nAinsi, $f(0,0) \\geqslant 1$.\n\nPour tout $i$ (resp. $j$), on note $x_{i}$ (resp. $y_{j}$ ) la somme des nombres inscrits dans la colonne $C_{i}$ (resp. la ligne $L_{j}$ ).\n\nc) On a $x_{i} \\leqslant 5$ pour tout $i$.\nSoit $i$ fixé. Pour tout $j$, on note $n_{j}$ le nombre de $j$ dans $L_{i}$. Alors :\n$x_{i}=f(0, i)+f(1, i)+f(2, i)+f(3, i)+f(4, i)=n_{0}+n_{1}+n_{2}+n_{3}+n_{4} \\leqslant$ nombre de cases de $L_{i}$ $=5$.\nAinsi, on a $x_{i} \\leqslant 5$, pour tout $i$.\n\nd) Pour tous $i$ et $j$, on a $f(i, j) \\leqslant 4$.\nPar l'absurde : supposons qu'il existe $i$ et $j$ tels que $f(i, j) \\geqslant 5$.\nD'après a), c'est donc que $f(i, j)=5$. Cela signifie que la ligne $L_{j}$ ne contient que des $i$. En particulier, on a $f(j, j)=i$.\nSi $i>0$, la ligne $L_{j}$ doit donc contenir au moins un $j$. Comme elle ne contient que des $i$, c'est que $i=j$ et que $5=f(i, j)=f(j, j)=i$, en contradiction avec $i \\leqslant 4$.\nAinsi, on a $i=0$ et il n'y a que des 0 sur la ligne $L_{j}$. Puisque $x_{j} \\leqslant 5$, il n'y a que des 0 sur $C_{j}$ sauf $f(0, j)=5$. En particulier, chaque ligne $L_{k}$ contient au moins un 0 pour $k>0$. Cela assure que $f(0, k) \\neq 0$ pour tout $k>0$. Mais, on a vu que $f(0,0)>0$ donc il doit y avoir au moins un 0 sur $L_{0}$, alors qu'on vient de prouver qu'aucun des nombres sur $L_{0}$ n'est égal à 0. Contradiction.\nFinalement, on a $f(i, j) \\leqslant 4$, pour tous $i$ et $j$.\n\ne) On a $x_{i}=5$ pour tout $i$.\nMaintenant que l'on sait qu'il n'y a pas de 5 dans le tableau, on reprend le calcul et raisonnement du c), mais il n'y a plus d'inégalité...\nAinsi, on a $x_{i}=5$ pour tout $i$.\n\nf) On a $y_{1}+2 y_{2}+3 y_{3}+4 y_{4}+5 y_{5}=25$.\nIl suffit de remarquer que, pour tout $i$, le nombre $y_{i}$ est le nombre de $i$ qui apparaissent dans le tableau. Donc $0 y_{0}+y_{1}+2 y_{2}+3 y_{3}+4 y_{4}+5 y_{5}$ est la somme de tous les nombres du tableau, et on a $y_{1}+2 y_{2}+3 y_{3}+4 y_{4}+5 y_{5}=x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=25$.\n\ng) On a $f(0,1)=0$ et $f(0, i) \\geqslant 1$ pour tout $i \\geqslant 2$.\nPar l'absurde : Supposons qu'il existe $i \\geqslant 3$ tel que $f(0, i)=0$.\nAlors, il n'y a aucun 0 sur $L_{i}$, ce qui entraine que chaque ligne contient au moins un $i$. Alors, on a $y_{j} \\geqslant i \\geqslant 3$.\nD'après f), il vient alors $25 \\geqslant(0+1+2+3+4) \\times 3=30$. Contradiction.\nAinsi, on a $f(0,3)>0$ et $f(0,4)>0$.\nPar l'absurde: Supposons que $f(0,2)=0$.\nComme ci-dessus, cela signifie que chaque ligne contient au moins un 2. Mais alors, chacun des nombres $0,1,3,4$ (2 aussi, mais il va avoir un traitement particulier) apparaît au moins deux fois dans le tableau, d'où $y_{i} \\geqslant 2$ pour $i=0,1,3,4$.\nD'autre part, il y a 5 cases sur $L_{2}$, aucune ne contenant 0 et au moins une contenant un 2, d'où $y_{2} \\geqslant 6$. En utilisant f), il vient\n$25 \\geqslant(0+1+3+4) \\times 2+2 \\times 6=28$. Contradiction.\nPar suite, on a $f(0,2)>0$.\nOr, on a vu que $f(0,0)>0$ et donc qu'il y a au moins un 0 sur $L_{0}$. Comme $f(0, i)>0$ pour tout $i \\neq 1$, c'est que $f(0,1)=0$.\nFinalement, on a $f(0,1)=0$ et $f(0, i) \\geqslant 1$ pour tout $i \\geqslant 2$.\nIl n'y a donc qu'un seul nombre non nul sur $L_{0}$, d'où $f(0,0)=1$.\n\nh) On a $y_{1} \\geqslant 8$.\nPuisque $f(0,1)=0$, il n'y a aucun 0 sur $L_{1}$. En particulier, $f(1,1) \\geqslant 1$ donc il y a au moins un 1 sur $L_{1}$.\n- Il ne peut y avoir que des 1 sur $L_{1}$, car alors $f(1,1)=1$ et $f(1,1)=5$, ce qui est impossible.\n- S'il y a quatre 1, on a $f(1,1)=4$ et tous les autres sont des 1, d'où $y_{1}=8$.\n- S'il y a trois 1, alors $f(1,1)=3$ et un autre des nombres est supérieur ou égal à 2, donc $y_{1} \\geqslant 3+1+1+1+2=8$.\n- S'il y a deux 1, le même raisonnement conduit à $y_{1} \\geqslant 2+1+1+2+2=8$.\n- S'il n'y a qu'un seul 1, alors c'est $f(1,1)=1$ et les autres sont supérieurs ou égaux à 2, donc $y_{1} \\geqslant 1+4 \\times 2=9$.\nAinsi, on a $y_{1} \\geqslant 8$.\n\ni) On a $f(4,2)=f(4,3)=f(4,4)=0$, et donc que $f(0,4) \\geqslant 3$.\nIl n'y a aucun 0 dans $L_{1}$ donc chaque ligne contient au moins un 1.\nPar l'absurde: supposons qu'il y ait un 4 sur $L_{2}, L_{3}$ ou $L_{4}$.\n- S'il y a un 4 sur $L_{4}$, comme il y a aussi au moins un 1, on a $y_{4} \\geqslant 5$. Compte-tenu de $y_{1} \\geqslant 8$, la formule du f) conduit à $25 \\geqslant 28$, ce qui est impossible. Il n'y a donc pas de 4 sur $L_{4}$.\n- S'il y a un 4 sur $L_{3}$, comme il y a aussi au moins un 1, on a $y_{3} \\geqslant 5$. Compte-tenu de $y_{1} \\geqslant 8$ et $y_{4} \\geqslant 1$, la formule du f) conduit à $25 \\geqslant 27$, ce qui est encore impossible. Il n'y a donc pas de 4 sur $L_{3}$.\n- S'il y a un 4 sur $L_{2}$ : dans chaque colonne, sauf $C_{1}$, les lignes $L_{0}$ et $L_{1}$ ont toutes les deux un nombre non nul. Or, la somme des termes d'une colonne vaut 5 d'après e), donc le 4 de $L_{2}$ ne peut être que dans $C_{1}$, c'est-à-dire $f(2,1)=4$. Il y a donc quatre 2 dans $L_{1}$, d'où $y_{1} \\geqslant 9$. Comme on a $y_{2} \\geqslant 5, y_{3} \\geqslant 1$ et $y_{4} \\geqslant 1$, la formule du f) conduit à $25 \\geqslant 26$, ce qui est encore impossible. Il n'y a donc pas de 4 sur $L_{2}$.\nFinalement, il n'y a aucun 4 dans les cases des lignes $L_{2}, L_{3}$ et $L_{4}$.\nCela prouve que $f(4,2)=f(4,3)=f(4,4)=0$, et donc que $f(0,4) \\geqslant 3$.\n\nj) On a $f(3,2)=f(3,3)=f(3,4)=0$, et donc que $f(0,3) \\geqslant 3$.\nPar l'absurde : supposons qu'il y ait un 3 dans les lignes $L_{2}, L_{3}$ ou $L_{4}$.\n- S'il y a un 3 sur $L_{4}$, compte-tenu des 1 sur chaque ligne et de h), on a $y_{4} \\geqslant 4$ et $y_{1} \\geqslant 8$. Comme au i), la formule du f) conduit alors à une contradiction. Il n'y a donc pas de 3 sur $L_{4}$.\n- S'il y a un 3 sur $L_{3}$, compte-tenu des 1 sur chaque ligne et de h), on a cette fois $y_{4} \\geqslant 1, y_{3} \\geqslant 4, y_{2} \\geqslant 1$ et $y_{1} \\geqslant 8$. Comme au i), la formule du f) conduit alors à une contradiction. Il n'y a donc pas de 3 sur $L_{3}$.\n- S'il y a un 3 sur $L_{2}$ : notons que l'on a $f(0,4) \\geqslant 3$ et $f(1,4) \\geqslant 1$. Puisque $x_{4}=5$, il est impossible que $f(2,4)=3$.\nDe même pour $C_{3}$, comme $f(0,3) \\geqslant 2$ et $f(1,3) \\geqslant 1$, on ne peut avoir $f(2,3)=3$.\nSi $f(2,2)=3$ alors $y_{2} \\geqslant 7$. Avec $y_{4} \\geqslant 1, y_{3} \\geqslant 1$ et $y_{1} \\geqslant 8$, on obtient le même type de contradiction que ci-dessus par utilisation de c).\nSi $f(2,0)=3$ alors, puisque $f(0,0)=1$ et $f(0,1)=0$, la ligne $L_{0}$ est déterminée et, en particulier, on a $f(0,4)=2$, en contradiction avec les trois 0 que l'on a déjà identifiés sur $L_{4}$.\nReste à étudier le cas $f(2,1)=3$.\nOn sait que $y_{4} \\geqslant 1, y_{3} \\geqslant 1, y_{2} \\geqslant 4$ et $y_{1} \\geqslant 8$. En l'état, le f) n'apporte pas de contradiction, mais permet de voir que si $y_{4} \\geqslant 2$ ou $y_{3} \\geqslant 2$, on en obtiendrait une. Donc, on doit avoir $y_{4}=y_{3}=1$, ce qui permet de remplir $L_{3}$ et donc $C_{3}$, et de savoir que $f(0,3)=f(0,4)=4$ ainsi que $f(1,4)=f(1,3)=1$. Comme $x_{4}=x_{3}=5$, les colonnes $C_{3}$ et $C_{4}$ sont complètes. Les trois 2 sur $L_{1}$ permettent de compléter $L_{1}$ et donc $C_{1}$. Ces mêmes 2 sur $L_{1}$ permettent aussi de remplir les trous sur $L_{0}$ et $L_{2}$ et de finir de remplir le tableau. On obtient\n\n| 1 | 0 | 0 | 0 | 0 |\n| :--- | :--- | :--- | :--- | :--- |\n| 0 | 0 | 1 | 0 | 0 |\n| 1 | 3 | 1 | 0 | 0 |\n| 2 | 2 | 2 | 1 | 1 |\n| 1 | 0 | 1 | 4 | 4 |\n\nEn particulier, on a $f(2,0)=1$ et pas de 2 sur $L_{0}$. Contradiction.\nFinalement, il n'y a aucun 3 dans les cases des lignes $L_{2}, L_{3}$ et $L_{4}$.\nCela prouve que $f(3,2)=f(3,3)=f(3,4)=0$, et donc que $f(0,3) \\geqslant 3$.\n\nk) On trouve tous les tableaux cohérents.\nD'après ce qui précède, on part d'un tableau de la forme\n\n| . | . | 0 | 0 | 0 |\n| :--- | :--- | :--- | :--- | :--- |\n| . | . | 0 | 0 | 0 |\n| . | . | . | . | . |\n| . | . | . | . | . |\n| 1 | 0 | . | a | b |\n\navec $a, b \\geqslant 3$, et aucun terme nul sur $L_{1}$. Comme $a, b \\in \\{3,4\\}$ et que la somme par colonne doit toujours donner 5, on tombe assez rapidement sur les deux seuls tableaux cohérents, à savoir :\n\n| 1 | 0 | 0 | 0 | 0 |\n| :--- | :--- | :--- | :--- | :--- |\n| 2 | 1 | 0 | 0 | 0 |\n| 0 | 1 | 0 | 1 | 0 |\n| 1 | 3 | 2 | 1 | 1 |\n| 1 | 0 | 3 | 3 | 4 |\n\n| 2 | 1 | 0 | 0 | 0 |\n| :--- | :--- | :--- | :--- | :--- |\n| 1 | 0 | 0 | 0 | 0 |\n| 0 | 0 | 0 | 0 | 1 |\n| 1 | 4 | 1 | 1 | 1 |\n| 1 | 0 | 4 | 4 | 3 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère une rangée de cases numérotées $0,1, \\ldots, k$ de gauche à droite où, pour chaque $i \\geqslant 1$, la case numéro $i$ contient $x_{i}$ jetons. Il n'y a initialement aucun jeton sur la case numéro $0$. À tour de rôle, Alice et Bob jouent alors selon les règles suivantes :\n- Bob choisit un ensemble $S$ de jetons, pas forcément tous sur la même case.\n- Alice peut ensuite soit éliminer tous les jetons qui ne sont pas dans $S$ mais alors déplacer chaque jeton de $S$ de la case qu'il occupe à la case voisine à sa gauche (un tel jeton passe donc d'une case numéro $i$ à la case numéro $i-1$), soit éliminer tous les jetons qui sont dans $S$ mais alors déplacer chaque jeton qui n'est pas dans $S$ de la case qu'il occupe à la case voisine à sa gauche.\n\nBob gagne la partie s'il arrive à amener un jeton sur la case numéro $0$, et Alice gagne si elle arrive à éliminer tous les jetons.\n\n1) Prouver qu'Alice possède une stratégie gagnante si $\\sum_{i=1}^{k} 2^{-i} x_{i}<1$.\n\n2) Est-il vrai que si $\\sum_{i=1}^{k} 2^{-i} x_{i} \\geqslant 1$ alors Bob possède une stratégie gagnante?",
    "options": [],
    "answer": "Alice has a winning strategy when the halving-weighted sum is less than one. Yes: when the halving-weighted sum is at least one, Bob has a winning strategy.",
    "solution": "Solution:\n\nDans tout ce qui suit, on peut clairement supposer que Bob ne choisit jamais $S=\\emptyset$ ou $S$ l'ensemble de tous les jetons non encore éliminés, sans quoi Alice gagne immédiatement. Du coup, le nombre de jetons diminue strictement après chaque fois qu'Alice joue, ce qui assure que le jeu se termine toujours en un nombre fini de tours, soit par élimination de tous les jetons, soit parce que l'un d'eux aura atteint la case numéro $0$. Ainsi, il n'y aura pas de partie nulle.\nDe plus, s'agissant d'un jeu à information parfaite, il est bien connu que l'un des deux joueurs a alors une stratégie gagnante. Reste à savoir lequel...\n\n1)\n- Approche probabiliste : Fixons une stratégie pour Bob. Alice va alors jouer de façon aléatoire. Plus précisément, à chaque tour, après que Bob ait choisi son ensemble $S$ de jetons, Alice lance une pièce équilibrée. Si elle obtient Face, elle élimine les jetons de $S$ (et déplace donc ceux qui ne sont pas dans $S$ vers la gauche) et, si elle obtient Pile, elle fait le contraire.\nPour chaque jeton $j$, on note $X_{j}$ la variable aléatoire égale à $1$ si le jeton $j$ est sur la case numéro $0$ à la fin de la partie, et égale à $0$ sinon.\nSoit enfin $X=\\sum_{j} X_{j}$, la somme portant sur l'ensemble des jetons de départ.\nOn remarque que $X$ représente le nombre total de jetons qui arrivent sur la case numéro $0$ à la fin du jeu, et donc qu'Alice gagne si et seulement si $X<1$.\nSoit $j$ un des jetons de départ. À chaque tour, Bob peut choisir $S$ de sorte que, si $j$ n'a pas encore été éliminé, on ait $j \\in S$ ou $j \\notin S$ mais, quoi qu'il en soit, la probabilité que $j$ soit alors déplacé vers la gauche est égale à $\\frac{1}{2}$. Si, au début du jeu, le jeton $j$ se trouve sur la case numéro $i$, il arrive sur la case numéro $0$ à la fin du jeu si et seulement si les $i$ premiers lancers de pièces ont tous conduit à des déplacements vers la gauche, ce qui arrive donc avec une probabilité $\\frac{1}{2^{i}}$. Par suite, on a $E\\left(X_{j}\\right)=\\frac{1}{2^{i}}$ pour tout jeton $j$.\nPar linéarité de l'espérance, on a donc\n$$\nE(X)=\\sum_{j} E\\left(X_{j}\\right)=\\sum_{i=1}^{k} 2^{-i} x_{i}<1\n$$\nCela assure que l'événement $[X<1]$ se réalise avec une probabilité non nulle. Ainsi, quelle que soit la stratégie de Bob, Alice a une probabilité non nulle de gagner, ce qui prouve que Bob n'a pas de stratégie gagnante. D'après notre remarque initiale, c'est donc Alice qui en possède une.\n\n- Approche déterministe : Plaçons-nous à un instant donné au cours de la partie, juste avant que ce soit à Bob de jouer, et considérons la configuration $C$ obtenue. Pour chaque $i$, on note $y_{i}$ le nombre de jetons qui sont alors sur la case numéro $i$, et on définit le poids de $C$ par $W(C)=\\sum_{i=1}^{k} y_{i} 2^{-i}$.\nPour faire le lien avec l'approche probabiliste, on peut noter que $W(C)=E(Y)$, où $Y$ est le nombre de jetons qui vont arriver sur la case numéro $0$ si Alice joue le reste de la partie, depuis la configuration $C$, de la façon aléatoire ci-dessus.\nLa stratégie d'Alice va consister alors à toujours choisir des configurations afin de minimiser les poids $W$ successifs :\nSoit donc $C$ une configuration pour laquelle le jeu ne soit pas déjà terminé, de poids $W=W(C)$, et supposons que Bob choisisse l'ensemble $S$ de jetons.\nOn note $W^{+}$ (resp. $W^{-}$) le poids de la configuration obtenue si Alice déplace les jetons de $S$ vers la gauche (resp. élimine les jetons de $S$).\nPour chaque jeton $j$, la contribution de $j$ est nulle ($j$ a été éliminé) dans l'une des sommes $W^{+}$ et $W^{-}$, et vaut le double de ce qu'elle est dans $W$ pour l'autre. Ceci étant vrai pour chaque jeton, on a donc $W=\\frac{1}{2}\\left(W^{+}+W^{-}\\right)$.\nEn particulier, Alice peut alors choisir une configuration $C'$ pour laquelle $W\\left(C'\\right) \\leqslant W(C)$. Or, puisqu'on débute le jeu avec une configuration de poids strictement inférieur à $1$, cela assure que, tout au long du jeu, Alice peut imposer des configurations de poids strictement inférieurs à $1$. Ceci empêche clairement Bob de gagner puisqu'un jeton sur la case numéro $0$ donne, à lui seul, un poids égal à $1$.\n\n2)\nSupposons que $\\sum_{i=1}^{k} 2^{-i} x_{i} \\geqslant 1$.\nOn va prouver que Bob possède une stratégie gagnante. Compte-tenu de notre remarque préliminaire, il suffit de prouver qu'Alice n'a pas de stratégie gagnante.\nSoit $C$ une configuration pour laquelle le jeu n'est pas encore terminé. On note $J$ l'ensemble des jetons de $C$.\nSi $A \\subset J$, on note $C_{A}$ la configuration obtenue à partir de $C$ en ne gardant que les jetons qui sont dans $A$ (sans les déplacer vers la gauche).\n\nLemme. Si $C$ est une configuration telle que $W(C) \\geqslant 1$, il existe une partie $S$ de $J$ telle que\n$$\nW\\left(C_{S}\\right) \\geqslant \\frac{1}{2} \\text{ et } W\\left(C_{\\bar{S}}\\right) \\geqslant \\frac{1}{2} \\text{.}\n$$\nPreuve du lemme.\nTout d'abord, on remarque que si une configuration $C$ vérifie $W(C) \\geqslant 1$ alors il existe $i$ tel que $y_{i} \\geqslant 2$. En effet, dans le cas contraire, on aurait\n$$\nW(C) \\leqslant \\sum_{i=1}^{k} 2^{-i}=1-\\frac{1}{2^{k}}<1\n$$\nOn raisonne par récurrence sur le nombre $n$ de jetons de la configuration :\n- Si la configuration considérée ne possède que deux jetons alors, d'après ci-dessus, ils sont tous les deux sur la même case. De plus, si l'on veut que $W(C) \\geqslant 1$, il est facile de vérifier qu'ils sont en fait sur la case numéro $1$. Ainsi, la partie $S$ formée par l'un des deux jetons convient.\n- Soit $n \\geqslant 3$ et supposons la conclusion assurée pour toute configuration de poids supérieur ou égal à $1$, et contenant $n-1$ jetons. Soit alors $C$ une configuration telle que $W(C) \\geqslant 1$ et à $n$ jetons.\nSi $y_{1} \\geqslant 2$, une partie $S$ formée uniquement d'un des jetons qui sont sur la case numéro $1$ convient. Sinon, soit $i \\geqslant 2$ tel que $y_{i} \\geqslant 2$ et soit $A$ et $B$ deux jetons qui sont sur la case numéro $i$. On considère alors la configuration $C'$ identique à $C$ sauf pour $A$ et $B$ qui sont remplacés par un seul jeton $X$, placé sur la case numéro $i-1$. On a clairement $W\\left(C'\\right)=W(C) \\geqslant 1$, et $C'$ ne contient que $n-1$ jetons. D'après l'hypothèse de récurrence, il existe donc une partie $S'$ de l'ensemble des jetons de $C'$ telle que $W\\left(C_{S'}'\\right) \\geqslant \\frac{1}{2}$ et $W\\left(C_{\\overline{S'}}'\\right) \\geqslant \\frac{1}{2}$.\nIl est alors facile de vérifier que si $S$ est la partie formée des jetons de $S'$, et en remplaçant $X$ par $A$ et $B$, on a bien $W\\left(C_{S}\\right) \\geqslant \\frac{1}{2}$ et $W\\left(C_{\\bar{S}}\\right) \\geqslant \\frac{1}{2}$, ce qui achève la récurrence et la preuve.\n\nRevenons au jeu. Une fois que Bob a choisi un ensemble $S$ de jetons, quel que soit le choix d'Alice, celui-ci conduira à une configuration dont le poids sera le double de celui de la configuration formée uniquement par les jetons qui ne seront pas éliminés. Le lemme assure donc qu'à partir d'une configuration de poids supérieur ou égal à $1$, Bob peut choisir $S$ de sorte qu'après le choix d'Alice, on obtienne une configuration de poids supérieur ou égal à $1$. Puisque la configuration initiale est supposée de poids supérieur ou égal à $1$, Bob peut donc imposer que toute la partie se déroule sur de telles configurations. Cela empêche clairement Alice de gagner (elle cherche à obtenir une configuration de poids nul), d'où la conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $t$ un entier naturel non-nul. Montrer qu'il existe un entier $n>1$ premier avec $t$ tel que pour tout entier $k \\geq 1$, l'entier $n^{k}+t$ ne soit pas une puissance (c'est-à-dire qu'il ne soit pas de la forme $m^{r}$ avec $m \\geq 1$ et $r \\geq 2$ ).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour que $n$ soit premier avec $t$, on va le chercher sous la forme $1+t s$ où $s$ est entier. On aura alors $n^{k}+t \\equiv 1+t\\pmod{s}$. En particulier, si $s$ est divisible par $(t+1)$, alors $n^{k}+t$ l'est également.\n\nOn va d'abord traiter le cas où $t+1$ n'est pas une puissance. Dans ce cas, il suffirait de s'assurer qu'on peut choisir $s$ de telle sorte que $n^{k}+t$ soit divisible par $t+1$, et que le quotient soit premier avec $t+1$. Pour cela, prenons $s=(t+1)^{2}$. Alors $n=1+t(t+1)^{2}$, et\n$$\nn^{k}+t=\\underbrace{\\sum_{i=1}^{k}\\binom{k}{i} t^{i}(t+1)^{2 i}}_{\\text{termes divisibles par }(t+1)^{2}}+1+t=(t+1)(a(t+1)+1)\n$$\npour un certain entier $a$, donc on a gagné.\n\nSupposons maintenant que $t+1$ soit une puissance : $t+1=m^{r}$ avec $m$ qui n'est pas une puissance. Si on garde le même $n$ que ci-dessus, on voit que si $n^{k}+t=b^{d}$ est une puissance (avec $d \\geq 2$ ), alors $t+1$ est nécessairement une puissance $d$-ième, et donc $d$ divise $r$. Ainsi, nous avons une borne sur les $d$ tels que $n^{k}+t$ est puissance $d$-ième pour un certain $k$. On constate alors qu'en remplaçant $n$ par sa puissance $r$-ième, c'est-à-dire en posant $n=n_{0}^{r}$ où $n_{0}=1+t(t+1)^{2}$ (ce qui ne change pas le fait que $t+1$ divise $n^{k}+t$, que le quotient soit premier avec $t+1$, et que donc $n^{k}+t=b^{d}$ implique $d \\mid r$ comme ci-dessus), on arrive à écrire $t$ comme une différence de deux puissances $d$-ièmes :\n$$\nt=b^{d}-\\left(n_{0}^{k e}\\right)^{d}\n$$\noù $e$ est l'entier naturel tel que $r=d e$. Cela n'est pas possible car $n_{0}$ est grand par rapport à $t$. Plus précisément, nous avons:\n$$\nt=b^{d}-\\left(n_{0}^{k e}\\right)^{d}=\\left(b-n_{0}^{k e}\\right)\\left(b^{d-1}+b^{d-2} n_{0}^{k e}+\\ldots+b n_{0}^{k e(d-2)}+n_{0}^{k e(d-1)}\\right) \\geq n_{0}>t\n$$\ncontradiction. Donc pour tout $k$ et pour tout $d, n^{k}+t$ n'est pas une puissance $d$-ième et nous avons terminé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers strictement positifs $a$ tels que l'entier\n$$\n1-8 \\cdot 3^{a}+2^{a+2}\\left(2^{a}-1\\right)\n$$\nsoit un carré parfait.",
    "options": [],
    "answer": "3 and 5",
    "solution": "Solution:\n\nLa quantité (4) est égale à $\\left(2^{a+1}-1\\right)^{2}-2^{3} \\cdot 3^{a}$. On remarque que cette quantité est un carré pour $a=3$ (elle vaut $9=3^{2}$) et $a=5$ (elle vaut $2025=45^{2}$) mais pas pour $a \\in\\{1,2,4,6,7\\}$. On suppose dorénavant $a \\geqslant 8$.\n\nSupposons que la quantité (4) soit un carré, c'est-à-dire, puisqu'elle est impaire, qu'il existe un entier $k$ tel que\n$$\n\\left(2^{a+1}-1\\right)^{2}-2^{3} \\cdot 3^{a}=(2k+1)^{2}\n$$\nAinsi\n$$\n\\left(2^{a+1}-1\\right)^{2}-(2k+1)^{2}=2^{3} \\cdot 3^{a}\n$$\nsoit encore $\\left(2^{a}-k-1\\right)\\left(2^{a}+k\\right)=2 \\cdot 3^{a}$. Il existe alors un entier $b \\geqslant 0$ tel que l'une des deux situations suivantes se produit (en discutant selon la parité de $k$):\n\n- on a $2^{a}+k=2 \\cdot 3^{b}$ et $2^{a}-k-1=3^{a-b}$ donc $2k+1=2 \\cdot 3^{b}-3^{a-b}$ (donc $b \\geqslant a-b$);\n- on a $2^{a}+k=3^{b}$ et $2^{a}-k-1=2 \\cdot 3^{a-b}$ donc $2k+1=3^{b}-2 \\cdot 3^{a-b}$ (donc $b \\geqslant a-b$).\n\nAinsi, $3^{a-b}$ divise $2k+1$, et donc $3^{2(a-b)}$ divise $(2k+1)^{2}+2^{3} \\cdot 3^{a}$. En utilisant (5), il vient $3^{a-b} \\mid 2^{a+1}-1$.\n\nDans la première situation, en additionnant les deux égalités on a $2 \\cdot 3^{b}=2^{a+1}-3^{a-b}-1<2^{a+1}$ donc $3^{b}<2^{a+1}$. On montre de manière analogue que cette inégalité est valable aussi dans la deuxième situation.\n\nRemarquons que $2^{9} \\leqslant 3^{9-3}$ donc, pour tout $a \\geqslant 8$,\n$$\n2^{a+1}=2^{9} \\cdot 2^{a-8} \\leqslant 2^{9} \\cdot 3^{a-8} \\leqslant 3^{6} \\cdot 3^{a-8}=3^{a-2}\n$$\nEn combinant les deux inégalités, $3^{b}<3^{a-2}$ donc $a-b \\geqslant 3$.\n\nComme $3^{a-b}$ divise $2^{a+1}-1$ et $a-b \\geqslant 3$, on a $2^{a+1} \\equiv 1\\ (\\bmod\\ 27)$ d'où 18 divise $a+1$ (en remarquant que le plus petit entier $k>0$ tel que $2^{k} \\equiv 1\\ (\\bmod\\ 27)$ est 18). Fort de cette condition, réduisons la quantité de l'énoncé modulo 19 en utilisant le théorème d'Euler ($x^{18}=x^{\\varphi(19)} \\equiv 1\\ (\\bmod\\ 19)$ pour tout entier $x$ non divisible par 19) et la relation $3 \\cdot 13 \\equiv 1\\ (\\bmod\\ 19)$ :\n$$\n\\left(2^{a+1}-1\\right)^{2}-2^{3} \\cdot 3^{a} \\equiv (1-1)^{2}-2^{3} \\cdot 13 \\quad (\\bmod\\ 19) \\equiv 10 \\quad (\\bmod\\ 19)\n$$\nOn conclut en remarquant alors que $(2k+1)^{18} \\equiv 10^{9}\\ (\\bmod\\ 19)=18\\ (\\bmod\\ 19)$ ce qui contredit le théorème d'Euler. Il n'y a donc pas de solution $a \\geqslant 9$.\n\nEn conclusion, les seules solutions sont 3 et 5.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $\\Gamma$ son cercle circonscrit et $\\omega$ le cercle de même centre que $\\Gamma$ et tangent à la droite $(BC)$. Les tangentes au cercle $\\omega$ passant par $A$ coupent $(BC)$ en un point $X$ du côté de $B$ et en un point $Y$ du côté de $C$. La tangente au cercle $\\Gamma$ en $B$ et la parallèle à la droite $(AC)$ passant par $X$ se coupent en un point $S$ et la tangente au cercle $\\Gamma$ en $C$ et la parallèle à la droite $(AB)$ passant par $Y$ se coupent en un point $T$.\n\nMontrer que la droite $(ST)$ est tangente au cercle $\\Gamma$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAfin d'éviter d'avoir à séparer les différents cas, en fonction de la position du point $Y$ par rapport au segment $[BC]$, nous allons utiliser les angles de droite orientés : $(AB, CD)$ désigne l'angle (relatif) dont il faut tourner la droite $(AB)$ pour qu'elle soit parallèle à la droite $(CD)$.\n\nNous allons montrer que la droite $(TA)$ est tangente au cercle circonscrit au triangle $ABC$ et de même pour la droite $(SA)$, ce qui donnera bien que la droite $(ST)$ est tangente à ce cercle.\n\nPour montrer que la droite $(AT)$ est tangente au cercle $\\Gamma$, il suffit de montrer que $(AC, AT) = (BC, BA)$. Les droites $(TY)$ et $(AB)$ sont parallèles donc $(BC, BA) = (BC, YT) = (YC, YT)$, donc on est ramené à montrer que $(AC, AT) = (YC, YT)$, autrement dit que les points $T, A, Y$ et $C$ sont cocycliques.\n\nPour cela, on peut montrer que $(YT, YA) = (CT, CA)$. En effet, $(YT, YA) = (AB, AY)$ (par parallélisme de $(AB)$ et $(YT)$) et $(CT, CA) = (BC, BA)$ donc il ne reste plus qu'à montrer que $(AB, AY) = (BC, BA)$, autrement dit que le triangle $AYB$ est isocèle en $Y$.\n\nSoit $O$ le centre du cercle $\\Gamma$, qui est aussi le centre du cercle $\\omega$, et soit $M$ et $N$ les points de contact respectifs du cercle $\\omega$ avec les segments $[YA]$ et $[YB]$. Alors les triangles $OMA$ et $ONB$ sont rectangles, $OA = OB$ et $OM = ON$ donc $MA = \\sqrt{OA^2 - OM^2} = \\sqrt{OB^2 - ON^2} = NB$ et puisque $YM = YN$, on déduit $YA = YB$ et le triangle $AYB$ est isocèle en $Y$ comme voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $a \\in ] 0 ; 1[$ et $n>0$ un entier. On note $f_{n}$ la fonction définie sur $\\mathbb{R}$ par $f_{n}(x)=x+\\frac{x^{2}}{n}$, pour tout réel $x$. Prouver que\n$$\n\\frac{a(1-a) n^{2}+2 a^{2} n+a^{3}}{(1-a)^{2} n^{2}+a(2-a) n+a^{2}}<\\underbrace{\\left(f_{n} \\circ f_{n} \\circ \\cdots \\circ f_{n}\\right)}_{n}(a)<\\frac{a n+a^{2}}{(1-a) n+a} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPour tout entier $k \\geq 0$, posons $a_{k}=\\underbrace{\\left(f_{n} \\circ f_{n} \\circ \\cdots \\circ f_{n}\\right)}_{k}(a)$. En particulier, $a_{0}=a$.\nOn remarque que, pour tout $k \\geq 0$, on a $a_{k+1}=a_{k}+\\frac{a_{k}^{2}}{n}>a_{k}$, donc la suite $\\left(a_{k}\\right)$ est strictement croissante. En particulier, $a_{k} \\geq a>0$ pour tout $k \\geq 0$.\nDe plus, pour tout $k \\geq 0$, on a $a_{k+1}=\\frac{a_{k}\\left(a_{k}+n\\right)}{n}$ et donc\n$$\n\\frac{1}{a_{k+1}}=\\frac{1}{a_{k}}-\\frac{1}{a_{k}+n} .\n$$\nAinsi, après sommation de ces égalités pour $k=0, \\ldots, n-1$, on a\n$$\n\\frac{1}{a_{n}}=\\frac{1}{a}-\\sum_{k=0}^{n-1} \\frac{1}{a_{k}+n}\n$$\nPuisque $\\left(a_{k}\\right)$ est strictement croissante et à valeurs positives, on a\n$$\n\\frac{n}{a_{n}+n}<\\sum_{k=0}^{n-1} \\frac{1}{a_{k}+n}<\\frac{n}{a+n}\n$$\net ainsi\n$$\n\\frac{1}{a}-\\frac{n}{a+n}<\\frac{1}{a_{n}}<\\frac{1}{a}-\\frac{n}{a_{n}+n} .\n$$\nL'inégalité de gauche est équivalente à $a_{n}<\\frac{a n+a^{2}}{(1-a) n+a}$, ce qui est l'inégalité de droite de l'encadrement demandé.\nReportons maintenant cette dernière inégalité dans l'inégalité de droite de (1). Il vient\n$$\n\\frac{1}{a_{n}}<\\frac{1}{a}-\\frac{n}{\\frac{a n+a^{2}}{(1-a) n+a}+n}\n$$\nou encore\n$$\n\\frac{a(1-a) n^{2}+2 a^{2} n+a^{3}}{(1-a)^{2} n^{2}+a(2-a) n+a^{2}}<a_{n}\n$$\nce qui est l'inégalité de gauche de l'encadrement demandé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\mathbb{R}_{+}^{*}$ l'ensemble des réels strictement positifs. Trouver toutes les fonctions $f: \\mathbb{R}_{+}^{*} \\longrightarrow \\mathbb{R}_{+}^{*}$ telles que pour tous $x, y>0$, on ait\n$$\nf\\left(\\frac{f(x)}{y f(x)+1}\\right)=\\frac{x}{x f(y)+1}\n$$",
    "options": [],
    "answer": "f(x) = x for all x > 0",
    "solution": "Solution:\n\nOn remarque tout d'abord que pour tout $z>0$, l'application $x \\in \\mathbb{R}^{+*} \\longmapsto \\frac{x}{x z+1} \\in ] 0 ; z^{-1}[$ est une bijection strictement croissante.\n\nMaintenant, si $f(x)=f\\left(x'\\right)$, on trouve que\n$$\n\\frac{x}{x f(1)+1}=f\\left(\\frac{f(x)}{1 \\times f(x)+1}\\right)=f\\left(\\frac{f\\left(x'\\right)}{f\\left(x'\\right) \\times 1+1}\\right)=\\frac{x'}{x' f(1)+1}\n$$\ndonc $x=x'$ et $f$ est injective.\n\nD'autre part, fixons $y>0$. Chaque $z$ compris entre 0 et $f(y)^{-1}$ (strictement) s'écrit comme un $\\frac{x}{x f(y)+1}$ pour un certain $x$ strictement positif, tout $0<z<f(y)^{-1}$ est atteint par $f$. En particulier, si $0<z<f(y)^{-1}$, on peut écrire $z=f\\left(y'\\right)$. Mais alors $f$ atteint tout point entre 0 et $f\\left(y'\\right)^{-1}=z^{-1}$. En faisant tendre $z$ vers 0 (ce qui est permis, la seule contrainte étant $0<z<f(y)^{-1}$ ), on en déduit que $f$ est surjective, donc $f$ est bijective.\n\nSoit $g$ la bijection réciproque de $f$, i.e. la seule fonction telle que $f(g(x))=g(f(x))=x$ pour tout $x>0$. On vérifie que $g$ vérifie la même équation fonctionnelle que $f$ : en appliquant l'équation fonctionnelle de départ à $g(x)$ et $g(y)$, on obtient\n$$\nf\\left(\\frac{x}{x g(y)+1}\\right)=\\frac{g(x)}{y g(x)+1}\n$$\nEn appliquant $g$ des deux côtés, on obtient que $g$ vérifie la même équation que $f$.\n\nSoient maintenant $x, x'>0$ tels que $x'<f(x)$. Alors il existe un $y>0$ tel que $x'=\\frac{f(x)}{y f(x)+1}$ (il suffit de prendre $y=\\frac{1}{x'}-\\frac{1}{f(x)}$ ), donc\n$$\nf\\left(x'\\right)=f\\left(\\frac{f(x)}{y f(x)+1}\\right)=\\frac{x}{x f(y)+1}<x\n$$\nDonc si $x'<f(x)$, alors $f\\left(x'\\right)<x$. En particulier, si $x<f(x)$, alors $f(x)<x$, c'est absurde. Donc pour tout réel $x>0, f(x) \\leqslant x$.\n\nOr, on a vu que $g$ était également solution de l'équation, donc pour tout réel $x>0$, on a $g(x) \\leqslant x$. Notamment, si $x>0, x=g(f(x)) \\leqslant f(x)$. On a donc forcément $f(x)=x$, donc $f$ est l'identité, et on vérifie facilement que l'identité est bien solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans le plan on se donne 2011 points deux à deux distincts colorés soit en bleu, soit en rouge.\n\na. On suppose que pour tout point bleu le disque de centre ce point et de rayon 1 contienne exactement deux points rouges. Quel est le plus grand nombre possible de points bleus?\n\nb. On suppose que pour tout point bleu le cercle de centre ce point et de rayon 1 contienne exactement deux points rouges. Quel est le plus grand nombre possible de points bleus?",
    "options": [],
    "answer": "a) 2009; b) 1966",
    "solution": "Solution:\n\na. S'il existe un point bleu, il doit exister au moins deux points rouges donc il ne peut y avoir plus de 2009 points bleus.\n\nRéciproquement, si l'on considère 2009 disques de rayon 1 ayant un intérieur commun à tous qui soit non vide, il suffit de marquer deux points rouges dans cette partie commune et de marquer en bleu les centres de ces 2009 disques pour obtenir une configuration à 2009 points bleus qui vérifie les conditions de l'énoncé.\n\nLe maximum cherché est donc 2009.\n\nb. Considérons une configuration de $r$ points rouges et $b$ points bleus vérifiant les conditions de l'énoncé, avec $r+b=2011$. Comme ci-dessus, on a $r \\geq 2$.\n\nSi $R_{1}$ et $R_{2}$ sont deux points rouges distincts, on note $n(R_{1}, R_{2})$ le nombre de points bleus qui sont centres de cercles de rayon 1 passant par $R_{1}$ et $R_{2}$.\n\nPuisque tout cercle de rayon 1 et centré en un point bleu passe par exactement deux points rouges, cela assure que:\n- tout point bleu est compté au moins une fois dans un certain $n(R_{1}, R_{2})$,\n- si $\\{R_{1}, R_{2}\\}$ et $\\{R_{1}', R_{2}'\\}$ sont deux paires distinctes de points rouges, alors les points bleus comptés dans $n(R_{1}, R_{2})$ sont deux à deux distincts des points bleus comptés dans $n(R_{1}', R_{2}')$.\n\nAinsi, chaque point bleu est compté une et une seule fois dans un certain $n(R_{1}, R_{2})$. En sommant sur les paires de points rouges, il vient\n$$\n\\sum_{\\{R_{1}, R_{2}\\}} n(R_{1}, R_{2}) = b\n$$\nD'autre part, pour toute paire de points du plan, il n'existe qu'au plus deux cercles de rayon 1 qui passent par ces deux points, on a donc $n(R_{1}, R_{2}) \\leq 2$ pour toute paire $R_{1}, R_{2}$ de points rouges. Comme il y a $r(r-1)/2$ paires de points rouges, on a alors :\n$$\nb = \\sum_{\\{R_{1}, R_{2}\\}} n(R_{1}, R_{2}) \\leq \\sum_{\\{R_{1}, R_{2}\\}} 2 = r(r-1),\n$$\net donc $r(r-1) \\geq 2011 - r$, ou encore $r^{2} \\geq 2011$. On en déduit facilement que $r \\geq 45$, et donc que $b \\leq 1966$.\n\nRéciproquement, considérons 45 points rouges, tous situés sur un même segment de longueur 1. Deux quelconques de ces points rouges appartiennent alors toujours à deux cercles de rayon 1. On trace seulement 1966 de ces cercles (on pourrait en tracer $1980 = 45 \\times 44$), et on marque en bleu les centres de ces cercles. Il est facile de vérifier que l'on obtient ainsi une configuration à 2011 points colorés qui vérifie les conditions de l'énoncé.\n\nLe maximum cherché est donc 1966.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver toutes les fonctions $f$ de $\\mathbb{R}$ dans $\\mathbb{R}$ telles que $f(a b)=f(a+b)$ pour tous nombres irrationnels $a$ et $b$.",
    "options": [],
    "answer": "All constant functions",
    "solution": "Solution:\nIl est clair que les fonctions constantes conviennent. Montrons que ce sont les seules. Soit $u$ un nombre irrationnel positif. Alors $\\sqrt{u}$ est irrationnel. En prenant $(a, b) = (\\sqrt{u}, -\\sqrt{u})$, il vient $f(-u) = f(0)$. En prenant $(a, b) = (-\\sqrt{u}, -\\sqrt{u})$, il vient $f(u) = f(-2\\sqrt{u}) = f(0)$ car $-2\\sqrt{u}$ est un nombre irrationnel négatif. Ainsi, on a $f(u) = f(0)$ pour $u \\notin \\mathbb{Q}$.\n\nMaintenant soit $r \\in \\mathbb{Q}$ non nul, et écrivons $r = \\sqrt{2} + (r - \\sqrt{2})$. Comme $\\sqrt{2}$ est irrationnel, il est clair que $\\sqrt{2}, r - \\sqrt{2}, \\sqrt{2}(r - \\sqrt{2})$ sont tous irrationnels. Alors :\n$$\nf(r) = f(\\sqrt{2}(r - \\sqrt{2})) = f(0)\n$$\nd'après ce qui précède. Donc $f$ est constante.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère inscrit dans un cercle de centre $O$. On note $\\Delta_{1}$ et $\\Delta_{2}$ les images de la droite $(AB)$ par les symétries dont les axes sont respectivement les bissectrices intérieures de $\\widehat{CAD}$ et $\\widehat{CBD}$. Soit $P$ l'intersection de $\\Delta_{1}$ et $\\Delta_{2}$.\n\nProuver que $(OP)$ et $(CD)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn traite uniquement le cas où $O$ est à l'intérieur de $ABCD$ (les autres cas étant similaires). D'après les propriétés de l'angle inscrit, les deux bissectrices considérées s'intersectent en un point du cercle noté $S$, qui est le milieu de l'arc $CD$. En particulier, $(OS)$ et $(CD)$ sont perpendiculaires. Il suffit donc de montrer que $P$, $O$, $S$ sont alignés.\n\n![](attached_image_1.png)\n\nUne chasse aux angles (voir figure) fournit $\\widehat{APB} + \\widehat{AOB} = 180^{\\circ}$. Ainsi, $A$, $P$, $B$, $O$ sont cocycliques. Alors $\\widehat{POB} = \\widehat{PAB} = 180^{\\circ} - 2\\alpha - 2\\beta$. Comme $A$, $B$, $C$, $D$ sont cocycliques, on a $\\widehat{BOC} = 2\\widehat{BAC} = 2\\beta$ et $\\widehat{COS} = 2\\widehat{SBC} = 2\\alpha$. Ainsi, $\\widehat{POB} + \\widehat{BOC} + \\widehat{COS} = 180^{\\circ}$. Les points $P$, $O$, $S$ sont donc alignés, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a, b, c, d$ des entiers naturels tels que $0<|a d-b c|<\\min (c, d)$.\nProuver que pour tous entiers $x, y>1$ premiers entre eux, le nombre $x^{a}+y^{b}$ n'est pas divisible par $x^{c}+y^{d}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPar l'absurde : on suppose que les entiers $x, y>1$ sont premiers entre eux et que $s=x^{c}+y^{d}$ divise $x^{a}+y^{b}$. On a alors:\n$$\nx^{c}=-y^{d} \\quad \\bmod s \\quad \\text { et } \\quad x^{a}=-y^{b} \\quad \\bmod s .\n$$\nD'où\n$$\nx^{a d}=(-1)^{d} y^{b d} \\bmod s \\quad \\text { et } \\quad x^{b c}=(-1)^{c} y^{b d} \\bmod s .\n$$\nAinsi $x^{a d}=(-1)^{b-d} x^{b c} \\bmod s$.\nOr, puisque $c>0$ et que $x$ et $y$ sont premiers entre eux, on a clairement $x$ et $s$ premiers entre eux. On peut donc diviser par $x^{\\min (a d, b c)}$ dans la congruence ci-dessus, et il vient $x^{|a d-b c|}=(-1)^{b-d} \\bmod s$. De même, on a $y^{|a d-b c|}=(-1)^{a-c} \\bmod s$.\nOn en déduit que :\n$$\nx^{|a d-b c|}+y^{|a d-b c|} \\text { ou } x^{|a d-b c|}-y^{|a d-b c|} \\text { est divisible par } s .\n$$\nMais on a $x^{|a d-b c|}-y^{|a d-b c|} \\neq 0$ car $x$ et $y$ sont premiers entre eux, supérieurs à 1 , et $|a d-b c|>0$. De plus $|a d-b c|<\\min (c, d)$ donc on a les inégalités\n$$\n0<\\left|x^{|a d-b c|}-y^{|a d-b c|}\\right|<x^{|a d-b c|}+y^{|a d-b c|}<x^{c}+y^{d}=s,\n$$\nce qui contredit (1).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $n$ et $k$ deux entiers strictement positifs. On considère une assemblée de $k$ personnes telle que, pour tout groupe de $n$ personnes, il y en ait une $(n+1)$-ième qui les connaisse toutes (si $A$ connaît $B$ alors $B$ connaît $A$ ).\n\n1) Si $k=2 n+1$, prouver qu'il existe une personne qui connaît toutes les autres.\n\n2) Si $k=2 n+2$, donner un exemple d'une telle assemblée dans laquelle personne ne connaît tous les autres.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1) On commence par construire, par récurrence, un groupe de $n+1$ personnes qui se connaissent deux à deux : il est clair que l'on peut trouver deux personnes qui se connaissent. Supposons que pour $p \\in \\{2, \\ldots, n\\}$ fixé, on ait réussi à trouver un groupe de $p$ personnes qui se connaissent deux à deux. En complétant ce groupe par $n-p$ personnes quelconques, on forme un groupe de $n$ personnes dont on sait qu'il en existe une $(n+1)^{ième}$ qui les connaît toutes. En ajoutant cette personne à notre groupe de $p$ personnes, on forme ainsi un groupe de $p+1$ personnes qui se connaissent deux à deux.\n\nOn considère donc un groupe $G$ de $n+1$ personnes qui se connaissent deux à deux. Puisque $k=2 n+1$, il reste donc $n$ personnes qui forment un groupe $G'$ disjoint du précédent. Pour ce groupe $G'$, on sait qu'il existe une personne appartenant nécessairement à $G$ qui en connaît tous les membres. Cette personne connaît alors tout le monde.\n\n2) On divise les personnes en $n+1$ paires disjointes, et on suppose que chaque personne connaît toutes les autres sauf celle qui est dans la même paire qu'elle. Ainsi, personne ne connaît tout le monde.\n\nSoit $G$ un groupe de $n$ personnes de cette assemblée. Puisqu'il y a $n+1$ paires, c'est donc qu'il existe une paire, disons $\\{A, B\\}$, dont aucun des deux membres n'est dans $G$. Par suite, $A$ connaît tous les membres de $G$, et les conditions de l'énoncé sont satisfaites.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $x_{1}, x_{2}, \\ldots, x_{5}$ des réels tels que\n$$\n\\left|x_{2}-x_{1}\\right|=2\\left|x_{3}-x_{2}\\right|=3\\left|x_{4}-x_{3}\\right|=4\\left|x_{5}-x_{4}\\right|=5\\left|x_{1}-x_{5}\\right| .\n$$\nMontrer que ces cinq réels sont égaux.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNotons $\\alpha=\\left|x_{2}-x_{1}\\right|$ et montrons par l'absurde que $\\alpha=0$. D'après l'énoncé, il existe des signes $\\varepsilon_{1}, \\ldots, \\varepsilon_{5}$ tels que\n$$\nx_{2}-x_{1}=\\varepsilon_{1} \\alpha, \\quad x_{3}-x_{2}=\\varepsilon_{2} \\frac{\\alpha}{2}, \\quad x_{4}-x_{3}=\\varepsilon_{3} \\frac{\\alpha}{3}, \\quad x_{5}-x_{4}=\\varepsilon_{4} \\frac{\\alpha}{4}, \\quad x_{1}-x_{5}=\\varepsilon_{5} \\frac{\\alpha}{5} .\n$$\nAlors, par télescopage,\n$$\n0=x_{1}-x_{5}+x_{5}-x_{4}+x_{4}-x_{3}+x_{3}-x_{2}+x_{2}-x_{1}=\\varepsilon_{1} \\alpha+\\varepsilon_{2} \\frac{\\alpha}{2}+\\varepsilon_{3} \\frac{\\alpha}{3}+\\varepsilon_{4} \\frac{\\alpha}{4}+\\varepsilon_{5} \\frac{\\alpha}{5} .\n$$\nSi $\\alpha \\neq 0$, il existe donc des signes $\\varepsilon_{1}^{\\prime}, \\ldots, \\varepsilon_{4}^{\\prime}$ tels que\n$$\n\\frac{1}{5}=\\varepsilon_{1}^{\\prime}+\\varepsilon_{2}^{\\prime} \\frac{1}{2}+\\varepsilon_{3}^{\\prime} \\frac{1}{3}+\\varepsilon_{4}^{\\prime} \\frac{1}{4}\n$$\ndonc un entier $n$ tel que $\\frac{1}{5}=\\frac{n}{12}$ ce qui est impossible car $5$ ne divise pas $12$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPeut-on numéroter les arêtes d'un cube de 1 à 12 en sorte que la somme des nombres sur les arêtes entrant dans un sommet du cube soit la même pour tous les sommets?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nNous allons prouver par l'absurde que la réponse à la question est négative.\nSupposons qu'une telle numérotation existe et notons $k$ la somme dans chaque sommet. Sur chaque arête, écrivons son numéro deux fois : une fois à un bout de l'arête et l'autre fois à l'autre bout. Maintenant, calculons la somme des numéros ainsi écrits de deux manières différentes.\n\nPremière manière : on additionne les numéros arête par arête. On trouve alors $2(1+2+\\cdots+12)$, car chaque numéro de 1 à 12 est écrit deux fois.\n\nDeuxième manière : on additionne les numéros sommet par sommet. Par hypothèse, la somme vaut $k$ dans chaque sommet. Au total, on trouve donc $8k$.\n\nConclusion:\n$$\n8k = 2(1+2+\\ldots+12) = 12 \\times 13 = 4 \\times 39\n$$\nd'où $k = 39 / 2$, ce qui est impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k \\geqslant 2$ un entier.\n\n1) Soit $n > k$ un entier.\nExiste-t-il $n$ entiers strictement positifs tels que $k$ quelconques ne sont jamais premiers entre eux dans leur ensemble, mais $k+1$ quelconques sont toujours premiers entre eux dans leur ensemble?\n\n2) Existe-t-il une suite infinie d'entiers strictement positifs vérifiant les deux conditions ci-dessus?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1) Soit $A$ l'ensemble des parties de $\\{1, \\ldots, n\\}$ de cardinal $k$. Comme l'ensemble des nombres premiers est infini, il existe une famille $\\left(p_{I}\\right)_{I \\in A}$ de nombres premiers deux à deux distincts indexée par $A$. Pour tout $i$, notons $A(i)$ l'ensemble des parties $I$ de $\\{1, \\ldots, n\\}$ de cardinal $k$ telles que $i \\in I$. On définit\n$$\na_{i} = \\prod_{I \\in A(i)} p_{I}\n$$\nSi $J$ est une partie de $\\{1, \\ldots, n\\}$, alors les $a_{j}$ ($j \\in J$) ne sont pas premiers entre eux dans leur ensemble\n$\\Longleftrightarrow$ ils ont un facteur premier commun\n$\\Longleftrightarrow$ il existe $I \\in A$ tel que $p_{I} \\mid a_{j}$ pour tout $j \\in J$\n$\\Longleftrightarrow$ il existe $I \\in A$ tel que $j \\in I$ pour tout $j \\in J$\n$\\Longleftrightarrow$ il existe $I \\in A$ tel que $J \\subset I$\n$\\Longleftrightarrow \\operatorname{Card}(J) \\leqslant k$.\nCeci montre que la famille $\\left(a_{i}\\right)_{1 \\leqslant i \\leqslant n}$ convient.\n\n2) Supposons par l'absurde qu'il y a une famille $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ d'entiers strictement positifs telle que $k$ quelconques ne sont jamais premiers entre eux dans leur ensemble.\nNotons $p_{1}, \\ldots, p_{m}$ les diviseurs premiers de $a_{0}$. Comme $k \\geqslant 2$, pour tout $n \\in \\mathbb{N}$, $a_{n}$ possède un diviseur parmi $p_{1}, \\ldots, p_{m}$. Notons $p_{i(n)}$ l'un d'eux. Comme la suite $\\left(p_{i(n)}\\right)_{n \\in \\mathbb{N}}$ ne prend qu'un nombre fini de valeurs, au moins une de ses valeurs (que nous noterons $p_{\\ell}$) est atteinte une infinité de fois. Il existe donc $i_{1} < \\cdots < i_{k+1}$ tels que $p_{\\ell}$ divise $a_{i_{1}}, \\ldots, a_{i_{k+1}}$, ce qui montre que $a_{i_{1}}, \\ldots, a_{i_{k+1}}$ ne sont pas premiers entre eux dans leur ensemble.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ est un trapèze dans lequel les côtés $AD$ et $BC$ sont parallèles, $K$ est un point du côté $AB$ et $L$ un point du côté $CD$. Montrer que si les angles $\\widehat{BAL}$ et $\\widehat{CDK}$ sont égaux alors les angles $\\widehat{BLA}$ et $\\overline{CKD}$ le sont aussi.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn suppose les angles $\\widehat{BAL}$ et $\\widehat{CDK}$ ont la même valeur. Nous allons montrer que les angles $KBL$ et $KCL$ sont égaux (ceci impliquera que les triangles $KDC$ et $BAL$ ont deux angles identiques, et donc que $\\widehat{BLA}=\\widehat{KCL}$).\n\n![](attached_image_1.png)\n\nJe vais d'abord montrer un point amusant sur les trapèzes : si on place des points $K$ et $L$ sur les côtés $AB$ et $CD$ tels que $A, D, K, L$ soient cocycliques, alors les points $B, C, K, L$ sont aussi cocycliques. Il suffit d'utiliser la propriété qu'un quadrilatère est inscrit dans un cercle ssi la somme des angles opposés vaut $180^{\\circ}$, et aussi que comme $AD$ et $BC$ sont parallèles, $\\widehat{A}+\\widehat{B}=\\widehat{C}+\\widehat{D}=180^{\\circ}$.\n\nL'hypothèse $\\widehat{BAL}=\\widehat{CDK}$ revient à dire que $A, D, K, L$ sont cocycliques par le théorème des angles inscrits, donc par le résultat précédent $B, C, K, L$ sont aussi cocycliques et les angles bleus sont égaux.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes sept dixièmes de la surface de la Terre sont couverts par l'océan. Montrer qu'il existe un diamètre de la Terre dont les deux extrémités baignent dans l'océan.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $S$ la surface couverte par l'océan et $S'$ son symétrique par rapport au centre de la Terre. Si les surfaces $S$ et $S'$ étaient disjointes, elles occuperaient ensemble les $14/10$ de la surface de la Terre, ce qui est impossible, car $14/10 > 1$. Donc $S$ et $S'$ ont au moins un point d'intersection, qu'on appellera $P$. Soit $P'$ le symétrique de $P$ par rapport au centre de la Terre. On a donc $P \\in S$ et $P \\in S'$, autrement dit $P \\in S$ et $P' \\in S$. Par conséquent, les deux bouts du diamètre $P P'$ baignent dans l'océan.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\omega$ un cercle qui passe par $A$ et $B$ et recoupe les côtés $[BC]$ et $[AC]$ respectivement en $D$ et $E$. Les points $K$ et $L$ sont respectivement les centres du cercle inscrit dans $DAC$ et du cercle inscrit dans $BEC$. Soit $N$ l'intersection des droites $(EL)$ et $(DK)$.\n\nProuver que le triangle $KNL$ est isocèle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSolution 1. On va montrer l'égalité entre les angles de droites ($KL, KN$) et $(LN, KL)$. Notons d'abord que $C, K, L$ sont alignés (sur la bissectrice intérieure en $C$ du triangle $ABC$).\n$$\n\\begin{aligned}\n(KL, KN) & = (CK, CD) + (DC, DN) = \\frac{1}{2}((\\overrightarrow{CA}, \\overrightarrow{CB}) + (\\overrightarrow{DC}, \\overrightarrow{DA})) \\\\\n& = \\frac{1}{2}[(\\overrightarrow{CA}, \\overrightarrow{CB}) + \\Pi + (\\overrightarrow{DB}, \\overrightarrow{DA})]\n\\end{aligned}\n$$\noù $\\Pi$ est l'angle plat, et\n$$\n\\begin{aligned}\n(LN, KL) & = (EL, EC) + (EC, CK) = \\frac{1}{2}((\\overrightarrow{EB}, \\overrightarrow{EC}) + (\\overrightarrow{CE}, \\overrightarrow{CB})) \\\\\n& = \\frac{1}{2}[(\\overrightarrow{EB}, \\overrightarrow{EA}) + \\Pi + (\\overrightarrow{CA}, \\overrightarrow{CB})]\n\\end{aligned}\n$$\nOr, les angles de vecteurs $(\\overrightarrow{DB}, \\overrightarrow{DA})$ et $(\\overrightarrow{EB}, \\overrightarrow{EA})$ sont égaux puisque $E$ et $D$ sont sur le même arc délimité par $A$ et $B$, ce qui conclut.\n\n\nSolution 2. Comme $A, B, D, E$ sont cocycliques, on a les égalités d'angles orientés de droites\n$$\n\\begin{aligned}\n& (DA, DC) = (DA, DB) = (EA, EB) = (EC, EB) \\\\\n& (AD, AC) = (AD, AE) = (BD, BE) = (BC, BE)\n\\end{aligned}\n$$\ndonc les angles orientés de $CAD$ et $CBE$ sont deux à deux opposés, par conséquent les triangles $CAD$ et $CBE$ sont indirectement semblables. Soit $f$ la similitude indirecte qui envoie $C, A, D$ sur $C, B, E$ respectivement. Alors $f$ envoie le centre du cercle inscrit de $CAD$ sur celui de $CBE$ : autrement dit, $f(K) = L$.\nOn a $(LK, LN) = (LK, EL) = (CL, EL)$ car $C, K, L$ sont alignés (sur la bissectrice intérieure en $C$ du triangle $ABC$). De plus, $f$ envoie $C, D, K$ sur $C, E, L$ respectivement, donc $(CL, EL) = -(CK, DK) = (KD, KC) = (KN, KL)$. Finalement, $(LK, LN) = (KN, KL)$ donc $LKN$ est isocèle en $N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABC$ est un triangle dont tous les angles sont aigus. Soit respectivement $H$ le pied de la hauteur de ce triangle issue de $C$ et $K$ le milieu du côté $AC$. On suppose que $BK = CH$ et que les angles $\\widehat{KBC}$ et $\\widehat{HCB}$ sont égaux. Montrer que le triangle $ABC$ est équilatéral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVoici la figure :\n![](attached_image_1.png)\n\nMaintenant considérons la médiatrice de $[BC]$ (dessinée en pointillés sur la figure) et la symétrie par rapport à cet axe. Il est évident que par cette symétrie $B$ devient $C$ et vice-versa. Appelons $A'$ l'image de $A$ par la symétrie. Les hypothèses que tous les angles sont aigus, $BK = CH$ et $\\widehat{KBC} = \\widehat{HCB}$ impliquent que $K$ est l'image de $H$ par la symétrie. Mais comme $A$ est l'intersection de $BH$ et $CK$, $A'$ est l'intersection de l'image de $BH$ (c-à-d $CK$) et de l'image de $CK$ (c-à-d $BH$), donc $A' = A$, $A$ est sur la médiatrice de $BC$ et le triangle est isocèle en $A$.\n\nMais nous n'avons pas fini, comme $CH \\mapsto BK$ et $BA \\mapsto CA$, l'angle droit est conservé, et $BK$ est à la fois une hauteur et une médiane. Donc le triangle est aussi isocèle en $B$, il est donc équilatéral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $p$ un nombre premier. Trouver tous les entiers $a, b, c \\geq 1$ tels que:\n$$\na^{p}+b^{p}=p^{c}\n$$",
    "options": [],
    "answer": "All solutions are:\n- For p = 2: a = 2^u, b = 2^u, c = 2u + 1 for any integer u ≥ 0.\n- For p = 3: (a, b, c) = (2*3^u, 3^u, 2 + 3u) or (3^u, 2*3^u, 2 + 3u) for any integer u ≥ 0.",
    "solution": "Solution:\n(i) $p=2$ et $(a, b, c)=\\left(2^{u}, 2^{u}, 2u+1\\right)$ pour un entier $u \\geq 0$\n(ii) $p=3$ et $(a, b, c)=\\left(2 \\cdot 3^{u}, 3^{u}, 2+3u\\right)$ ou $(a, b, c)=\\left(3^{u}, 2 \\cdot 3^{u}, 2+3u\\right)$ pour un entier $u \\geq 0$.\n\nPour un entier $n \\geq 1$ et un nombre premier $p$ on note $v_{p}(n)$ le plus grand entier $k \\geq 0$ tel que $p^{k}$ divise $n$. On montre d'abord trois lemmes.\n\nLemme 1\nSoit $p$ un nombre premier. Alors, pour $1 \\leq j \\leq p-1, \\binom{p}{j}$ est divisible par $p$.\nDémonstration: On a $j!(p-j)!\\binom{p}{j}=p!$. Comme $1 \\leq j<p$ et $1 \\leq p-j<p, j$ et $p-j$ sont premiers avec $p$ de sorte que $p$ divise $\\binom{p}{j}$.\n\nLemme 2\nSoit $p$ un nombre premier impair et soient $a, j$ des entiers strictement positifs avec $2 \\leq j \\leq p$. On pose $u=v_{p}(a)$ et on considère un entier $k>u$. Alors:\n$$\nv_{p}\\left(\\binom{p}{j} p^{jk} a^{p-j}\\right) \\geq k+u(p-1)+2\n$$\nDémonstration: D'après le lemme 1, pour $2 \\leq j \\leq p-1$, on a $v_{p}\\left(\\binom{p}{j} p^{jk} a^{p-j}\\right) \\geq 1+jk+u(p-j)$. Il suffit donc de prouver que $1+jk+u(p-j) \\geq k+u(p-1)+2$, ce qui est équivalent au fait que $(j-1)(k-u) \\geq 1$, ce qui est clairement vérifié. Pour $j=p$, on a $v_{p}\\left(\\binom{p}{j} p^{jk} a^{p-j}\\right)=pk$, qui est supérieur ou égal à $k+u(p-1)+2$ car l'inégalité $pk \\geq k+u(p-1)+2$ est équivalente à $(p-1)(k-u) \\geq 2$, qui est clairement vraie.\n\nLemme 3\nOn a $(2n)^{\\frac{1}{n-1}}<2$ pour tout entier $n \\geq 5$.\nDémonstration: On raisonne par récurrence sur $n$. Pour $n=5$, on a clairement $10<2^{4}$. On suppose que $2n<2^{n-1}$. Alors $2(n+1)=2n+2 \\leq 2^{n-1}+1<2^{n}$ car $2^{n}-2^{n-1}=2^{n-1}>1$. Ceci clôt la preuve.\n\nRésolvons maintenant l'exercice. On considère en un premier temps le cas $p=2$. Si $c<2$, la seule solution est $a=b=c=1$. Si $c \\geq 2$ et $a^{2}+b^{2}=2^{c}$, en raisonnant modulo 4, on voit que $a$ et $b$ doivent être pairs. Soit $u=\\min \\left(v_{2}(a), v_{2}(b)\\right)$. Alors $\\left(a / 2^{u}\\right)^{2}+\\left(b / 2^{u}\\right)^{2}=2^{c-2u}$. Si $c-2u \\geq 2$, alors $a / 2^{u}$ et $b / 2^{u}$ sont des entiers pairs, ce qui contredit la définition de $u$. Donc $c-2u=1$ et $a=b=2^{u}$. Pour conclure, on vérifie aisément que pour tout entier $u \\geq 0$ :\n$$\n\\left(2^{u}\\right)^{2}+\\left(2^{u}\\right)^{2}=2^{2u+1} .\n$$\n\nOn suppose maintenant que $p>2$. Soient $a, b, c \\geq 1$ des entiers tels que $a^{p}+b^{p}=p^{c}$. Sans perte de généralité, supposons $a \\geq b$. Comme $a+b$ divise $a^{p}+b^{p}$, il existe un entier $k \\geq 1$ tel que :\n$$\na+b=p^{k}\n$$\nSoit $u=v_{p}(a)$. En particulier, (5) implique que $a<p^{k}$ de sorte que $u<k$. Alors, en utilisant le fait que $p$ est impair\n$$\na^{p}+b^{p}=a^{p}+\\left(p^{k}-a\\right)^{p}=\\sum_{j=2}^{p}(-1)^{j+1} u_{j}+u_{1}, \\quad \\text{ où } \\quad u_{j}=\\binom{p}{j} p^{jk} a^{p-j}\n$$\nD'après le Lemme 2, on a $v_{p}\\left(u_{j}\\right) \\geq k+u(p-1)+2$ pour $2 \\leq j \\leq p$. De plus, comme $v_{p}\\left(u_{1}\\right)=1+k+u(p-1)$, il en découle que $v_{p}\\left(a^{p}+b^{p}\\right)=1+k+u(p-1)$. Comme $a^{p}+b^{p}$ est une puissance de $p$, on conclut que:\n$$\na^{p}+b^{p}=p^{1+k+u(p-1)}=p(a+b) p^{u(p-1)},\n$$\noù l'on a utilisé le fait que $a+b=p^{k}$ pour la dernière égalité. Puisque $a \\geq b$, ceci impose $a^{p} \\leq 2p a p^{u(p-1)}$. Ainsi :\n$$\na \\leq (2p)^{\\frac{1}{p-1}} p^{u}\n$$\nMais par définition $u=v_{p}(a)$, de sorte que $a \\geq p^{u}$. De plus, $a>p^{u}$ implique $a \\geq 2p^{u}$. D'après le Lemme 3, comme $(2p)^{\\frac{1}{p-1}}<2$ pour $p \\geq 5$, on en déduit que $p \\geq 5$ implique $a=p^{u}$, et que $p=3$ implique $a=3^{u}$ ou $a=2 \\cdot 3^{u}$ (le cas $p=2$ est exclu car $p$ est impair).\n\nOn étudie d'abord le cas $a=p^{u}$. On a alors:\n$$\nb^{p}=p^{up}\\left(p^{c-up}-1\\right),\n$$\nde sorte que $v_{p}(b)=p^{u}$. Comme $a \\geq b$ et $a=p^{u}$, ceci implique $a=b$, et donc 2 divise $p$, ce qui est exclu car $p$ est impair.\n\nOn a donc prouvé que ($p$ est impair) :\n$$\na^{p}+b^{p}=p^{c} \\text{ et } a \\geq b \\quad \\Longrightarrow \\quad p=3, \\quad a=2 \\cdot 3^{u} .\n$$\nDans la suite on suppose donc que $p=3$ et $a=2 \\cdot 3^{u}$. Alors $b^{3}=3^{3u}\\left(3^{c-3u}-8\\right)$. Donc $3^{u}$ divise $b$. Par suite $3^{u} \\leq b \\leq a=2 \\cdot 3^{u}$. Si $b=a$, alors 2 divise $p^{c}$, ce qui contredit le fait que $p$ est impair. Donc $b=3^{u}$, ce qui implique $\\left(3^{u}\\right)^{3}+\\left(2 \\cdot 3^{u}\\right)^{3}=3^{3u+2}=3^{c}$. Finalement, $c=3u+2$ et on vérifie aisément que pour tout entier $u \\geq 0$ :\n$$\n\\left(2 \\cdot 3^{u}\\right)^{3}+\\left(3^{u}\\right)^{3}=3^{2+3u} .\n$$\nAinsi $p=3, a=2 \\cdot 3^{u}, b=3^{u}, c=2+3u$ pour un certain entier $u \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProuver que, pour tous réels $a, b, c > 0$ et tout réel $t \\geqslant 0$, on a\n$$\n\\frac{a}{b+t c}+\\frac{b}{c+t a}+\\frac{c}{a+t b} \\geqslant \\frac{3}{1+t} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPosons $x = b + t c$, $y = c + t a$, $z = a + t b$. On vérifie que\n$$\n\\begin{aligned}\n(1 + t^3) a & = z - t x + t^2 y \\\\\n(1 + t^3) b & = x - t y + t^2 z \\\\\n(1 + t^3) c & = y - t z + t^2 x\n\\end{aligned}\n$$\ndonc, d'après l'inégalité arithmético-géométrique,\n$$\n\\begin{aligned}\n(1 + t^3)\\left(\\frac{a}{b + t c} + \\frac{b}{c + t a} + \\frac{c}{a + t b}\\right) & = \\frac{z - t x + t^2 y}{x} + \\frac{x - t y + t^2 z}{y} + \\frac{y - t z + t^2 x}{z} \\\\\n& = \\left(\\frac{z}{x} + \\frac{x}{y} + \\frac{y}{z}\\right) + t^2\\left(\\frac{y}{x} + \\frac{z}{y} + \\frac{x}{z}\\right) - 3 t \\\\\n& \\geqslant 3 \\sqrt[3]{\\frac{z}{x} \\frac{x}{y} \\frac{y}{z}} + 3 t^2 \\sqrt[3]{\\frac{y}{x} \\frac{z}{y} \\frac{x}{z}} - 3 t \\\\\n& = 3(1 - t + t^2) = 3 \\frac{1 + t^3}{1 + t}\n\\end{aligned}\n$$\nd'où la conclusion.\n\n\nSolution 2:\n\nPuisque l'inégalité est homogène en $a, b, c$, on peut supposer que $a + b + c = 1$. Comme le membre de gauche est égal à $a f(x) + b f(y) + c f(z)$ où $x = b + t c$, $y = c + t a$, $z = a + t b$ et $f(x)$ est la fonction convexe $x \\mapsto \\frac{1}{x}$, d'après l'inégalité de Jensen le membre de gauche est supérieur ou égal à\n$$\nf(a x + b y + t z) = \\frac{1}{a b + t a c + b c + t b a + c a + t a b} = \\frac{1}{(1 + t)(a b + b c + c a)},\n$$\ndonc il suffit de prouver que $a b + b c + c a \\leqslant \\frac{1}{3}$.\nOr, $1 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(a b + b c + c a)$ donc il suffit de montrer que $a^2 + b^2 + c^2 \\geqslant \\frac{1}{3}$, ce qui découle de l'inégalité entre la moyenne arithmétique et la moyenne quadratique de $a, b, c$.\n\n\nSolution 3:\n\nD'après l'inégalité de Cauchy-Schwarz,\n$$\n\\left(\\frac{a^2}{a b + t c a} + \\frac{b^2}{b c + t a b} + \\frac{c^2}{c a + t b c}\\right) \\cdot ((a b + t c a) + (b c + t a b) + (c a + t b c)) \\geqslant (a + b + c)^2\n$$\nce qui s'écrit\n$$\n\\left(\\frac{a}{b + t c} + \\frac{b}{c + t a} + \\frac{c}{a + t b}\\right) \\cdot (1 + t)(a b + b c + c a) \\geqslant a^2 + b^2 + c^2 + 2(a b + b c + c a).\n$$\nOr, on a $a^2 + b^2 + c^2 \\geqslant a b + b c + c a$ d'après l'inégalité de réordonnement (ou d'après celle de Cauchy-Schwarz), donc\n$$\n\\left(\\frac{a}{b + t c} + \\frac{b}{c + t a} + \\frac{c}{a + t b}\\right) \\cdot (1 + t)(a b + b c + c a) \\geqslant 3(a b + b c + c a)\n$$\nL'inégalité demandée en découle après division par $(1 + t)(a b + b c + c a)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver cent entiers positifs distincts $n_{1}, \\ldots, n_{100}$ tels que\n$$\n1=\\frac{1}{n_{1}}+\\frac{1}{n_{2}}+\\cdots+\\frac{1}{n_{100}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNous montrerons par récurrence la propriété suivante :\nPour tout entier $k \\geqslant 3$ on peut trouver $k$ entiers positifs $n_{1}<n_{2}<\\cdots<n_{k}$ tels que\n$$\n1=\\frac{1}{n_{1}}+\\frac{1}{n_{2}}+\\cdots+\\frac{1}{n_{k}}\n$$\net que $n_{k}$ soit pair.\n\nPour $k=3$, il suffit de prendre $n_{1}=2, n_{2}=3, n_{3}=6$. On a bien\n$$\n1=\\frac{1}{2}+\\frac{1}{3}+\\frac{1}{6}\n$$\net 6 est bien pair.\n\nPour passer de $k$ à $k+1$, nous allons garder tels quels les nombres $n_{1}, \\ldots, n_{k-1}$ et remplacer $n_{k}$ par $n_{k}'=3 n_{k} / 2$ (qui est entier car $n_{k}$ est pair) et $n_{k+1}'=3 n_{k}$ (qui est pair car $n_{k}$ l'est). Les nombres\n$$\nn_{1}, \\ldots, n_{k-1}, \\frac{3 n_{k}}{2}, 3 n_{k}\n$$\nsont toujours rangés dans l'ordre croissant; le nombre le plus grand est toujours pair; et la somme des inverses n'a pas changé car\n$$\n\\frac{1}{n_{k}'}+\\frac{1}{n_{k+1}'}=\\frac{2}{3 n_{k}}+\\frac{1}{3 n_{k}}=\\frac{3}{3 n_{k}}=\\frac{1}{n_{k}}\n$$\nAinsi, si la propriété est vraie pour $k$, elle est aussi vraie pour $k+1$.\n\nMaintenant, en prenant $k=100$, nous obtenons l'affirmation demandée par l'exercice.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k > 1$ un entier. Une fonction $f: \\mathbb{N}^* \\rightarrow \\mathbb{N}^*$ est dite $k$-tastrophique lorsque pour tout entier $n > 0$, on a $f_k(n) = n^k$ où $f_k$ est la $k$-ième itérée de $f$ :\n$$\nf_k(n) = \\underbrace{f \\circ f \\circ \\cdots \\circ f}_{k \\text{ fois }}(n)\n$$\nPour quels $k$ existe-t-il une fonction $k$-tastrophique?",
    "options": [],
    "answer": "for all integers k at least two",
    "solution": "Solution:\n\nOn va prouver que, pour tout entier $k \\geq 1$, il existe une fonction $k$-tastrophique.\n\nClairement, la fonction $f: n \\longmapsto n$ est 1-tastrophique. On suppose donc que $k \\geq 2$.\n\nOn procède maintenant de la façon suivante. On pose $f(1) = 1$, et si $n$ est le plus petit entier pour lequel $f(n)$ n'est pas encore défini, alors :\n- si $n = a^k$ pour un certain entier $a \\geq 2$, on pose :\n$$\nf(n) = f(a)^k\n$$\n- si $n$ n'est pas la puissance $k$-ième d'un entier, on note $n_1 = n, n_2, \\ldots, n_k$ les $k$ plus petits entiers (dans l'ordre) qui ne sont pas des puissances $k$-ième d'entiers et pour lesquels $f$ n'est pas définie, et on pose :\n$$\nf\\left(n_1\\right) = n_2,\\ f\\left(n_2\\right) = n_3,\\ \\ldots,\\ f\\left(n_{k-1}\\right) = n_k,\\ \\text{ et } f\\left(n_k\\right) = n_1^k = n^k\n$$\nIl est clair que l'on définit ainsi une fonction $f: \\mathbb{N} \\longrightarrow \\mathbb{N}$. Il reste à vérifier qu'elle est $k$-tastrophique. Soit $a \\geq 2$ un entier. On commence par noter que, d'après (1), on a :\n$$\nf_2\\left(a^k\\right) = f\\left(f\\left(a^k\\right)\\right) = f\\left(f(a)^k\\right) = [f(f(a))]^k = \\left[f_2(a)\\right]^k\n$$\net, par une récurrence immédiate, on a\n$$\nf_p\\left(a^k\\right) = \\left[f_p(a)\\right]^k \\text{ pour tout entier } p \\geq 1\n$$\nOn prouve alors que $f$ est $k$-tastrophique par récurrence (forte) : On a évidemment $f_k(1) = 1 = 1^k$. Supposons que $n \\geq 2$ soit un entier et que $f_k(m) = m^k$ pour tout entier naturel $m \\leq n-1$.\n\n- Si $n = a^k$ pour un certain entier $a \\geq 2$. Alors $a \\leq n-1$ et on a\n$$\n\\begin{aligned}\nf_k(n) & = f_k\\left(a^k\\right) = \\left[f_k(a)\\right]^k \\quad \\text{ d'après (4) } \\\\\n& = \\left(a^k\\right)^k \\quad \\text{ d'après l'hypothèse de récurrence } \\\\\n& = n^k\n\\end{aligned}\n$$\n\n- Si $n$ n'est pas la puissance $k$-ième d'un entier, on sait qu'il existe des entiers $n_1, n_2, \\ldots, n_k$ pour lesquels on a (3) et avec $n = n_i$ pour un certain $i \\in \\{1, \\ldots, k\\}$. Alors, d'après (3) et (4), il vient\n$$\n\\begin{aligned}\nf_k(n) & = f_k\\left(n_i\\right) = f_{k+i-1}\\left(n_1\\right) = f_{i-1}\\left(f_k\\left(n_1\\right)\\right) \\\\\n& = f_{i-1}\\left(f\\left(n_k\\right)\\right) = f_{i-1}\\left(n_1^k\\right) = \\left[f_{i-1}\\left(n_1\\right)\\right]^k = n_i^k = n^k\n\\end{aligned}\n$$\n\nAinsi, dans tous les cas, on a $f_k(n) = n^k$. Ce qui achève la récurrence, et la démonstration.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer tous les entiers naturels $m, n$ et tous les nombres premiers $p$ tels que\n$$\nm\\left(4 m^{2}+m+12\\right)=3\\left(p^{n}-1\\right)\n$$",
    "options": [],
    "answer": "m=0, n=0, p any prime; and m=12, n=4, p=7",
    "solution": "Solution:\nOn récrit l'équation sous la forme\n$$\n3 p^{n}=4 m^{3}+m^{2}+12 m+3=(4 m+1)\\left(m^{2}+3\\right)\n$$\nOn en déduit qu'il existe $u, v, a, b$ entiers naturels tels que $4 m+1=3^{u} p^{a}$ et $m^{2}+3=3^{v} p^{b}$ avec $u+v=1$ et $a+b=n$.\nSi $m=0$, l'équation devient $3 p^{n}=3$, donc $n=0$ et tout nombre premier convient.\nSupposons désormais $m \\geqslant 1$. Alors $4 m+1>3$ et $m^{2}+3>3$, donc $a \\geqslant 1$ et $b \\geqslant 1$.\nOn a\n$$\n\\begin{aligned}\n16 \\times 3^{v} p^{b} & =16\\left(m^{2}+3\\right)=(4 m)^{2}+48=((4 m+1)-1)^{2}+48 \\\\\n& =3^{2 u} p^{2 a}-2 \\times 3^{u} p^{a}+49\n\\end{aligned}\n$$\nComme $p$ divise le membre de gauche, il divise le membre de droite, donc $p \\mid 49$, ce qui donne $p=7$. On remplace alors 49 par $p^{2}$ :\n$$\n16 \\times 3^{v} p^{b}=3^{2 u} p^{2 a}-2 \\times 3^{u} p^{a}+p^{2}\n$$\nSi $a=1$ alors $4 m+1$ vaut 7 ou 21. Comme $m$ est un entier, on a $4 m+1=21$, ou encore $m=5$, ce qui entraîne $3^{v} p^{b}=m^{2}+3=28$. Impossible. Donc $a \\geqslant 2$. On en déduit que $3^{2 u} p^{2 a}-2 \\times 3^{u} p^{a}+p^{2}$ est divisible par $p^{2}$, donc $16 \\times 3^{v} p^{b}$ est divisible par $p^{2}$, d'où $b \\geqslant 2$.\nSi $a \\geqslant 3$, alors $16 \\times 3^{v} p^{b}=3^{2 u} p^{2 a}-2 \\times 3^{u} p^{a}+p^{2}$ est congru à $p^{2}$ modulo $p^{3}$, donc n'est pas divisible par $p^{3}$, par conséquent $b=2$. En divisant par $p^{2}$, on obtient que $16 \\times 3^{v} \\equiv 1[p]$. Pour $v=0$ cela donne $16 \\equiv 1[7]$ et pour $v=1$ cela donne $48 \\equiv 1[7]$. Impossible.\nDonc on a nécessairement $a=2$ et $4 m+1=3^{u} \\times 49$. Si $u=0$ alors $m=12$ et $n=4$. Si $u=1$ alors $4 m+1=147$, ce qui est impossible.\nConclusion: les seules solutions sont\n- $m=0, n=0, p$ premier quelconque, et\n- $m=12, n=4, p=7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRuby a effectué une série de mouvements avec son Rubik's cube. (Par exemple, elle peut tourner la face du haut dans le sens des aiguilles d'une montre, puis la face de fond de 180 degrés, puis la face de droite dans le sens contraire des aiguilles d'une montre. Ou n'importe quelle autre série de rotations de faces.) Ensuite elle répète inlassablement la même série de mouvements. Montrer qu'au bout d'un certain nombre de répétitions elle retrouvera la configuration de départ.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $x_{0}$ la configuration de départ et $x_{n}$ la configuration après $n$ répétitions de la série de Ruby. Comme le nombre total de configurations d'un Rubik's cube est fini, à un moment une configuration va se répéter : $x_{n}=x_{m}$ avec $n<m$. Demandons maintenant à Ruby d'inverser sa série de mouvements, autrement dit d'exécuter les rotations dans l'ordre inverse et dans le sens inverse par rapport à sa série de mouvements initiale. En appliquant cette série inversée à la configuration $x_{n}$ nous obtenons $x_{n-1}$. De même, en l'appliquant à $x_{m}$ nous obtenons $x_{m-1}$. Comme $x_{n}=x_{m}$, on en déduit que $x_{n-1}=x_{m-1}$. En continuant à appliquer la série inversée, nous allons arriver au bout de $n$ répétitions à l'égalité $x_{0}=x_{m-n}$.\n\nNous avons donc prouvé que la configuration de départ réapparaîtra au bout $m-n$ répétitions de la série de Ruby.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les polynômes $X^{n}+a_{1} X^{n-1}+\\cdots+a_{n-1} X+a_{n}$, non constants et à coefficients entiers, dont les racines sont exactement les nombres $a_{1}, \\ldots, a_{n-1}, a_{n}$ (avec multiplicité).",
    "options": [],
    "answer": "All solutions are precisely the polynomials of the form X^m, for m ≥ 1, and X^m · (X^2 + X − 2), and X^m · (X^3 + X^2 − X − 1), for m ≥ 0.",
    "solution": "Solution:\n\nSoit $P(X) = X^n + a_1 X^{n-1} + \\cdots + a_{n-1} X + a_n$ un polynôme à coefficients entiers.\nAlors $P(X)$ est une solution du problème ssi\n$$\nP(X)=\\prod_{i=1}^{n}\\left(X-a_{i}\\right)\n$$\nSupposons tout d'abord que $a_{n}=0$. Soit alors $i$ minimal tel que $a_{n-i} \\neq 0$. On a donc $P(X)=X^{i}\\left(X^{n-i}+a_{1} X^{n-i-1}+\\cdots+a_{n-i-1}\\right)$ et ainsi (5) est vérifiée si et seulement si $Q(X)=X^{n-i}+a_{1} X^{n-i-1}+\\cdots+a_{n-i-1}$ est constant ou est lui-même une solution du problème. Le premier cas correspond à $P(X)=X^{n}$, qui convient bien.\nOn supposera donc dans ce qui suit que $P(X)$ est une solution avec $a_{n} \\neq 0$. Dans ces conditions, on en déduit tout de suite que l'on ne peut avoir $X+a_{1}=X-a_{1}$ et donc qu'il n'y a pas de polynôme de degré 1 qui soit solution. On suppose donc que $n \\geq 2$. De $P(0)=a_{n}=(-1)^{n} \\prod_{i=1}^{n} a_{i}$, on déduit alors que\n$$\n(-1)^{n} \\prod_{i=1}^{n-1} a_{i}=1\n$$\nEt, puisque les $a_{i}$ sont des entiers, on a $a_{i}=1$ ou $a_{i}=-1$ pour tout $i \\leq n-1$. On note $k$ le nombre d'entre eux qui valent -1 (et donc $0 \\leq k \\leq n-1$ ). On peut noter que (6) montre alors que $k$ et $n$ ont la même parité. Les relations entre coefficients et racines conduisent à\n$$\n-a_{1}=\\sum_{i=1}^{n} a_{i}=a_{n}+(n-1-k)-k\n$$\nd'où\n$$\na_{n}=2 k-n \\quad \\text { ou } \\quad a_{n}=2 k-n+2\n$$\nPour $n=2$, puisque $a_{2} \\neq 0$ et que $k$ doit être pair et ne pas dépasser 1 , la seule possibilité est $k=0, a_{1}=1, a_{2}=-2$, et donc $P(X)=X^{2}+X-2=(X-1)(X+2)$, qui est bien solution.\nOn suppose maintenant que $n \\geq 3$. Alors\n$$\na_{2}=\\sum_{1 \\leq i<j \\leq n} a_{i} a_{j}=a_{n} \\sum_{i=1}^{n-1} a_{i}+\\frac{k(k-1)}{2}+\\frac{(n-1-k)(n-2-k)}{2}-k(n-1-k)\n$$\net donc\n$$\n2 a_{2}=2 a_{n}(n-1-2 k)+n^{2}+4 k^{2}-4 k n-3 n+4 k+2\n$$\nCompte-tenu de (7), deux cas se présentent:\n- Si $a_{n}=2 k-n$ : De (8), on tire $(n-2 k)^{2}+n=2-2 a_{2}$, avec $a_{2}=1$ ou $a_{2}=-1$. Donc $a_{2}=-1$ et, puisque $n-2 k=-a_{n} \\neq 0$, la seule possibilité est $n=3$ et $(n-2 k)^{2}=1$. De $k \\leq n-1$ et de même parité que $n$, on déduit que $k=1$, puis que $a_{1}=1$ et $a_{3}=-1$, d'où $P(X)=X^{3}+X^{2}-X-X=(X+1)^{2}(X-1)$, qui est bien solution.\n- Si $a_{n}=2 k-n+2$ : Comme ci-dessus, on obtient cette fois\n$$\n(n-2 k-2)^{2}+n=2-2 a_{2}\n$$\net donc $a_{2}=-1, n=3, k=1, a_{1}=1, a_{3}=1$, d'où $P(X)=X^{3}+X^{2}-X+1 \\neq (X-1)^{2}(X+1)$, qui n'est pas solution.\nEn définitive, les solutions du problème sont les polynômes $P(X)=X^{n}$ pour $n \\geq 1$, $P(X)=X^{n}\\left(X^{2}+X-2\\right)$ ou $P(X)=X^{n}\\left(X^{3}+X^{2}-X-X\\right)$ pour $n \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $O$ un point intérieur à ce triangle. $D$ et $E$ sont respectivement les pieds des perpendiculaires abaissées de $O$ sur les droites $(BC)$ et $(AC)$ et $F$ est le milieu du segment $AB$. Montrer que si $DF = EF$ alors les angles $\\widehat{OBD}$ et $\\widehat{OAE}$ sont égaux.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans cet exercice il y a une hypothèse superflue, à savoir que le point $O$ est à l'intérieur du triangle $ABC$. Nous allons résoudre l'exercice sans cette hypothèse.\n\nCommençons par tracer un angle droit $AEO$ et un segment $AB$ et plaçons $F$ au milieu du segment $AB$. Notre but maintenant est de trouver un point $D$ tel que l'angle $BDO$ soit droit et que l'on ait $FD = FE$.\n\n![](attached_image_1.png)\n\nComme la figure ci-dessus l'indique, il existe toujours une façon évidente de placer le point $D$ : il suffit de tracer la droite parallèle à $AE$ passant par $B$ et de prendre le point d'intersection de cette droite avec la droite $EO$. L'angle $BDO$ est alors droit par construction et on a bien $FD = FE$, car $F$ est équidistant des deux droites parallèles. Une telle position du point $D$ ne peut pas apparaître dans l'exercice (et cela va nous servir par la suite) : en effet, l'exercice parle d'un triangle $ABC$, or ici les droites $AE$ et $BD$ ne se coupent pas et il n'y a donc pas de point $C$.\n\nRajoutons à notre figure le milieu $M$ du segment $AO$ et le milieu $N$ du segment $BO$.\n\n![](attached_image_2.png)\n\nOn a alors trois segments égaux représentés en gras sur la figure. En effet, $ME = MO$, car le triangle $AEO$ est rectangle en $E$; et $MO = FN$, car $FNOM$ est un parallélogramme, les lignes des milieux étant parallèles aux côtés du triangle $AOB$.\n\nRevenons maintenant à l'exercice.\n\n![](attached_image_3.png)\n\nNous venons de montrer que $ME = FN$ et on a de même $ND = FM$. Comme on a de plus $FD = FE$ par hypothèse, on conclut que les triangles $NDF$ et $MFE$ sont égaux. Il est important de distinguer deux cas : celui où ces triangles sont directement isométriques, comme sur la figure, et celui où ils sont indirectement isométriques (distinction qui n'a pas été faite correctement dans la solution initialement proposée).\n\nNous allons utiliser les angles orientés.\n\n1er cas : triangles directement isométriques. Soit\n$$\n\\alpha = (MF, MO) = (NO, NF), \\quad \\beta = (ME, MF) = (NF, ND)\n$$\nDans le triangle rectangle $AEO$ on a\n$$\n(AE, AO) = \\frac{1}{2}(ME, MO) \\bmod 180^\\circ\n$$\n(L'égalité n'est vraie que modulo $180^\\circ$ à cause de la division par 2.) On trouve donc, modulo $180^\\circ$,\n$$\n(AE, AO) = \\frac{1}{2}(ME, MO) = \\frac{1}{2}[(ME, MF) + (MF, MO)] = \\frac{1}{2}(\\alpha + \\beta).\n$$\nDe la même manière\n$$\n(BO, BD) = \\frac{1}{2}(NO, ND) = \\frac{1}{2}[(NO, NF) + (NF, ND)] = \\frac{1}{2}(\\alpha + \\beta).\n$$\nOn a donc $(AE, AO) = (BO, BD) \\bmod 180^\\circ$. Or les deux angles font partie de triangles rectangles et sont donc aigus, ce qui implique qu'ils sont géométriquement égaux.\n\n2ème cas : triangles indirectement isométriques. Les calculs sont exactement les mêmes que dans le cas précédent, sauf que cette fois-ci on a $(NF, ND) = -\\beta$. On trouve donc\n$$\n(AE, AO) = \\frac{1}{2}(\\alpha + \\beta), \\quad (BO, BD) = \\frac{1}{2}(\\alpha - \\beta)\n$$\nmodulo $180^\\circ$. Ainsi ces angles ne sont pas égaux entre eux. Avons-nous trouvé une erreur dans l'exercice ? Non : en fait nous allons prouver que dans ce cas les droites $AE$ et $BD$ sont parallèles et nous sommes donc dans la configuration interdite mentionnée au début de la solution.\n\nPour cela il suffit de prouver que $(AE, AB) + (BA, BD) = 0 \\bmod 180^\\circ$. On a\n$$\n(AE, AB) + (BA, BD) = (AE, AO) + (AO, AB) + (BA, BO) + (BO, BD)\n$$\nNous savons déjà que, modulo $180^\\circ$,\n$$\n(AE, AO) + (BO, BD) = \\frac{1}{2}(\\alpha + \\beta) + \\frac{1}{2}(\\alpha - \\beta) = \\alpha\n$$\nDe plus, dans le triangle $AOB$, on a\n$$\n(AO, AB) + (BA, BO) = (OA, OB) \\bmod 180^\\circ.\n$$\nCe dernier angle vaut $180^\\circ - \\alpha = -\\alpha \\bmod 180^\\circ$. On trouve donc bien que $(AE, AB) + (BA, BD) = \\alpha - \\alpha = 0 \\bmod 180^\\circ$. Par conséquent $AE$ et $BD$ sont parallèles. Le cas de triangles indirectement isométriques ne peut donc pas apparaître dans un triangle $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe tel que les droites $(AD)$ et $(BC)$ ne sont pas parallèles. On suppose que les cercles de diamètre $[AB]$ et $[CD]$ se coupent en deux points $E$ et $F$ situés à l'intérieur de $ABCD$. Soit $\\Gamma_{E}$ le cercle qui passe par les projetés orthogonaux de $E$ sur $(AB)$, $(BC)$ et $(CD)$, et $\\Gamma_{F}$ le cercle qui passe par les projetés orthogonaux de $F$ sur $(CD)$, $(DA)$ et $(AB)$.\nProuver que le milieu de $[EF]$ appartient à la droite passant par les deux points d'intersection de $\\Gamma_{E}$ et $\\Gamma_{F}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSoient $P, Q, R, S$ les projections de $E$ sur les droites $(DA)$, $(AB)$, $(BC)$, $(CD)$. Les points $P$ et $Q$ appartiennent au cercle de diamètre $[AE]$ de sorte que $\\widehat{QPE} = \\widehat{QAE}$. De même, $\\widehat{QRE} = \\widehat{QBE}$. Ainsi :\n$$\n\\widehat{QPE} + \\widehat{QRE} = \\widehat{QAE} + \\widehat{QBE} = 90^\\circ\n$$\nDe même, on montre que $\\widehat{QPS} + \\widehat{QRS} = 90^\\circ + 90^\\circ = 180^\\circ$. Ainsi, les points $P, Q, R, S$ appartiennent à $\\Gamma_{E}$. De même, les projetés orthogonaux de $F$ sur les côtés de $ABCD$ sont sur $\\Gamma_{F}$.\n\nSoit $K$ le point d'intersection de $(AD)$ et $(BC)$. Sans perte de généralité, on suppose que $A \\in [DK]$. On voit facilement que $\\widehat{CKD} < 90^\\circ$ (sinon le cercle de diamètre $[CD]$ couvre entièrement $ABCD$). Les droites $(EP)$ et $(BC)$ s'intersectent donc en un point $P'$, et les droites $(ER)$ et $(AD)$ en un point $R'$. Montrons que $P'$ et $R'$ sont sur $\\Gamma_{E}$. Par cocyclicité des points $R, E, Q, B$ :\n$$\n\\widehat{QRK} = \\widehat{QEB} = 90^\\circ - \\widehat{QBE} = \\widehat{QAE} = \\widehat{QPE}.\n$$\nDonc $\\widehat{QRK} = \\widehat{QPP'}$, ce qui implique que $P' \\in \\Gamma_{E}$. De même, $R' \\in \\Gamma_{E}$.\n\nDe même, soient $M$ et $N$ les projections respectives de $F$ sur $(AD)$ et $(BC)$, et soient $M'$ le point d'intersection de $(FM)$ et $(BC)$ et $N'$ le point d'intersection de $(FN)$ et $(AD)$. En utilisant les mêmes arguments que précédemment, on voit que $M'$ et $N'$ sont sur $\\Gamma_{F}$.\n\nSoient finalement $U$ l'intersection de $(NN')$ avec $(PP')$ et $V$ l'intersection de $(MM')$ avec $(RR')$. Comme les angles en $N$ et $P$ sont droits, $N, N', P, P'$ sont cocycliques. La puissance de $U$ par rapport au cercle passant par ces points vaut donc :\n$$\nUN \\cdot UN' = UP \\cdot UP'\n$$\nAinsi, $U$ appartient à l'axe radical de $\\Gamma_{E}$ et $\\Gamma_{F}$. De même $V$ appartient à cet axe radical. Or $EUFV$ est un parallélogramme, ce qui implique que $(UV)$ coupe $[EF]$ en son milieu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient un entier $n$ et des réels $0 < u_{1} < u_{2} < \\ldots < u_{n}$ tels que\n$$\nu_{1} + u_{2} + \\ldots + u_{n} = \\frac{1}{u_{1}^{2}} + \\frac{1}{u_{2}^{2}} + \\ldots + \\frac{1}{u_{n}^{2}}$$\nMontrer que, pour tout entier $k$ inférieur ou égal à $n$, il existe $k$ réels parmi $u_{1}, u_{2}, \\ldots, u_{n}$ dont la somme est supérieure ou égale à $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$\\triangleright$ Commençons par montrer le résultat pour $k = n$. Notons $a = \\frac{u_{1} + \\ldots + u_{n}}{n}$ et $g = (u_{1} \\ldots u_{n})^{\\frac{1}{n}}$ les moyennes arithmétique et géométrique de $u_{1}, \\ldots, u_{n}$. Alors, l'inégalité classique entre moyenne arithmétique-géométrique-harmonique donne\n$$\na = \\frac{\\frac{1}{u_{1}^{2}} + \\ldots + \\frac{1}{u_{n}^{2}}}{n} \\geqslant \\frac{1}{g^{2}} \\geqslant \\frac{1}{a^{2}}\n$$\nAinsi, $a \\geqslant 1$ et, donc, $u_{1} + \\ldots + u_{n} \\geqslant n$.\n\n$\\triangleright$ Soit $k < n$. Supposons que $u_{1} + \\ldots + u_{k} < k, u_{2} + \\ldots + u_{k+1} < k, \\ldots, u_{n} + \\ldots + u_{k-1} < k$. Alors, en additionnant toutes ces inégalités, on obtient\n$$\nn(u_{1} + \\ldots + u_{n}) < n k\n$$\nce qui contredit le point précédent. Ainsi, l'une des sommes $u_{1} + \\ldots + u_{k}, u_{2} + \\ldots + u_{k+1}, \\ldots, u_{n} + \\ldots + u_{k-1}$ est supérieure ou égale à $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrois sauterelles se trouvent aux points $(0,0)$, $(0,1)$ et $(1,0)$ d'une feuille quadrillée. Chaque minute une sauterelle saute sur un autre point de la grille d'une telle façon que son saut soit parallèle à la droite passant par les deux autres sauterelles. Est-il possible qu'au bout d'un certain temps les sauterelles se retrouvent aux points $(-1,0)$, $(1,0)$ et $(0,1)$ ?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nSi $AB$ et $CD$ sont deux droites parallèles, alors l'aire du triangle $ABC$ est égale à celle du triangle $ABD$, car ces deux triangles ont la même base $[AB]$ et la même hauteur. Il découle de cette remarque que l'aire du triangle formé par les trois sauterelles ne change pas lors de leurs sauts. Or l'aire du triangle de départ vaut $1/2$, tandis que celle du triangle qu'on souhaite obtenir à la fin vaut $1$. Il est donc impossible de passer de l'un à l'autre.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans le triangle $ABC$ les bissectrices des angles en $A$ et $C$ rencontrent le cercle circonscrit au triangle $ABC$ respectivement en $A_{0}$ et $C_{0}$. La droite passant par $I$, centre du cercle inscrit dans le triangle $ABC$, et parallèle à la droite $(AC)$ rencontre la droite $(A_{0}C_{0})$ en $P$.\n\nMontrer que la droite $(PB)$ est tangente au cercle circonscrit au triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$\\underline{\\text{Sur la figure on va rajouter } B_{0} \\text{ le point d'intersection de la bissectrice de } \\widehat{ABC} \\text{ avec le cercle :}}$\n![](attached_image_1.png)\nAvant de commencer, je veux démontrer un petit résultat sur les arcs. Regardez la figure de droite (je vais noter $\\widehat{UV}$ l'arc délimité par $U$ et $V$ en radians):\n$$\n\\widehat{UYV}=\\widehat{YVW}+\\widehat{YXW}=\\frac{1}{2} \\widehat{WX}+\\frac{1}{2} \\widehat{UV}.\n$$\nRevenons à nos moutons. Grâce à la formule précédente, on peut calculer :\n$$\n\\widehat{BIC_{0}}=\\frac{1}{2} \\widehat{BC_{0}}+\\frac{1}{2} \\widehat{B_{0}C}=\\frac{1}{2} \\widehat{C_{0}A}+\\frac{1}{2} \\widehat{AB_{0}}=\\widehat{C_{0}BI}.\n$$\nDonc le triangle $BIC_{0}$ est isocèle en $C_{0}$. On prouve de même que $A_{0}BI$ est isocèle en $A_{0}$, donc $(A_{0}C_{0})$ est la médiatrice du segment $BI$.\n\nPar symétrie, $\\widehat{A_{0}BP}=\\widehat{A_{0}IP}$, et comme $(IP)$ et $(AC)$ sont parallèles, $\\widehat{A_{0}IP}=\\widehat{A_{0}AC}=\\widehat{BAA_{0}}$. Comme $\\widehat{A_{0}BP}=\\widehat{BAA_{0}}$, on se retrouve avec le cas extrême du théorème de l'angle inscrit, et par la réciproque du théorème $(PB)$ est tangente au cercle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nChacun des 400 députés d'un parlement a giflé exactement un autre député. Montrer qu'on peut créer une commission parlementaire de 134 députés telle qu'aucun membre de la commission n'ait giflé aucun autre membre.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAppelons deux députés ennemis si l'un d'eux a giflé l'autre. Nous allons résoudre par récurrence un exercice plus général :\nDans un parlement d'au moins $3n-2$ députés, chaque député a giflé 0 ou 1 autre député. Il est alors toujours possible de créer une commission parlementaire de $n$ députés qui ne contienne aucun couple ennemi.\n\nPour $n=1$ la propriété est évidente : on peut toujours former une commission à un député car un député n'est jamais son propre ennemi.\n\nMontrons maintenant comment passer de $n$ à $n+1$. Considérons un parlement avec au moins\n$$\n3(n+1)-2=3n+1\n$$\ndéputés. Comme le nombre total de gifles est inférieur ou égal au nombre de députés, il existe, par le principe des tiroirs, un député qui a été giflé au plus une fois et qui a donc au plus deux ennemis. Mettons ce député dans la commission d'office et écartons son ou ses deux ennemis. Il nous reste au moins $3n-2$ députés parmi lesquels il n'y a aucun ennemi du député sélectionné. Par hypothèse de récurrence, on peut compléter la commission en choisissant encore $n$ députés parmi ceux qui restent sans qu'il y ait aucun couple ennemi parmi eux.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn digicode s'ouvre dès qu'on fait l'unique combinaison correcte de 4 chiffres (qui peut éventuellement contenir des répétitions). Par exemple, si l'on tape la suite des chiffres 000125 le digicode s'ouvrira si le code est soit 0001, soit 0012, soit 0125. Petit Pierre ne connaît pas le code. Combien de chiffres au minimum doit-il taper pour ouvrir le digicode à coup sûr?",
    "options": [],
    "answer": "10003",
    "solution": "Solution:\n\nIl est clair qu'il faut au moins $10003$ chiffres, car il faut essayer toutes les $10000$ combinaisons. Montrons que ce nombre est aussi suffisant.\n\nDessinons $1000$ points correspondant à toutes les suites de $3$ chiffres. Dessinons une flèche entre deux points si les deux derniers chiffres de la première combinaison coïncident avec les deux premiers chiffres de la deuxième combinaison (et vont dans le même ordre). Par exemple, il y aura une flèche allant du point $200$ vers le point $009$ et une autre allant du point $201$ vers le point $017$ ; entre les points $303$ et $030$ il y aura deux flèches allant dans les deux sens ; sur le point $777$ il y aura une flèche en forme de boucle qui revient vers son point de départ. Ainsi, chaque flèche correspond à un unique code possible.\n\nOn voit facilement qu'il y a $10$ flèches entrantes et $10$ flèches sortantes dans chaque point. En effet, pour obtenir une flèche sortante, par exemple, il faut choisir un chiffre de $0$ à $9$ à rajouter aux derniers chiffres du numéro du point, ce qui fait exactement $10$ choix.\n\nIl est facile à voir également que le graphe formé par les points et les flèches est connexe. En effet, on passe d'un point $abc$ au point $def$ en trois flèches $abc \\rightarrow bcd \\rightarrow cde \\rightarrow def$.\n\nLe théorème d'Euler bien connu (voir, par exemple, \"Graphe Eulérien\" dans wikipédia) assure alors qu'on peut parcourir toutes les flèches du graphe sans passer deux fois par la même flèche. Ce parcours donnera la suite de $10003$ chiffres recherchée : il faudra taper les $3$ chiffres du point de départ et ensuite juste le dernier chiffre de chaque point du parcours. Le fait qu'on passe une et une seule fois par chaque flèche du graphe signifie exactement que chaque code possible de $4$ chiffres sera essayé une et une seule fois.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $x$, $y$ et $z$ des réels strictement positifs tels que\n$$\nx + y + z \\geqslant \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\n$$\nMontrer que\n$$\n\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} \\geqslant \\frac{1}{x y} + \\frac{1}{y z} + \\frac{1}{z x}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEn mettant au même dénominateur, l'inégalité recherchée se réécrit\n$$\nx^{2} z + y^{2} x + z^{2} y \\geqslant x + y + z\n$$\nMontrons celle-ci en utilisant successivement l'hypothèse et l'inégalité de Cauchy-Schwarz :\n$$\n\\begin{aligned}\n& x + y + z \\leqslant \\frac{(x + y + z)^{2}}{\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}} \\\\\n& \\leqslant \\frac{\\left(x \\sqrt{z} \\cdot \\frac{1}{\\sqrt{z}} + y \\sqrt{x} \\cdot \\frac{1}{\\sqrt{x}} + z \\sqrt{y} \\cdot \\frac{1}{\\sqrt{y}}\\right)^{2}}{\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}} \\\\\n& \\leqslant x^{2} z + y^{2} x + z^{2} y . \\\\\n& \\text{\\ or\\ }\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dispose de 500 emplacements à billes numérotés de 1 à 500 de gauche à droite. Des billes numérotées de 1 à 500 sont posées dans ces emplacements dans le désordre : leurs numéros ne coïncident pas nécessairement avec les numéros des emplacements. (Dans chaque emplacement il y a exactement une bille.) On dit que l'ordre des billes est plutôt croissant si la plus longue sous-suite décroissante des numéros des billes qu'on peut extraire de cet ordre ne dépasse pas 5 éléments. Montrer qu'il existe au plus $5^{1000}$ ordres plutôt croissants.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nMontrons d'abord que si les billes sont disposées dans un ordre plutôt croissant alors on peut les colorier en 5 couleurs de telle sorte que la suite des billes de chaque couleur soit croissante. Pour cela nous allons ordonner les 5 couleurs dans un certain ordre et colorier les billes une par une de gauche à droite en utilisant à chaque fois la première couleur disponible, c'est à dire la première couleur telle que la précédente bille de cette couleur ait un numéro plus petit que celle que nous sommes en train de colorier. Montrons qu'avec cette méthode nous arriverons en effet à colorier toutes les billes. Supposons par l'absurde que nous sommes arrivés à une bille pour laquelle aucune des 5 couleurs n'est disponible. En particulier cela veut dire qu'on a déjà utilisé la couleur numéro 5 pour colorier une bille dont le numéro était plus grand. Si pour cette bille-là nous avions choisi la couleur numéro 5, c'est que, d'après notre règle, la couleur numéro 4 n'était pas disponible. Autrement dit il y avait une bille encore plus tôt dans la suite qui avait un numéro encore plus grand et qu'on avait coloriée en couleur 4. En remontant de la même manière jusqu'à la couleur numéro 1 on découvrira une suite de 6 billes dont les numéros vont dans l'ordre décroissant. Ceci contredit l'hypothèse que notre ordre était plutôt croissant.\n\nConsidérons maintenant un ordre plutôt croissant et colorions les billes selon la règle ci-dessus. Colorions également chaque emplacement de la même couleur que la bille qu'il contient (il suffira de mettre beaucoup de peinture sur chaque bille, de sorte qu'elle coule aussi dans l'emplacement). Il est facile à voir que l'ordre des billes se rétablit d'une manière inambiguë si l'on connaît la couleur de toutes les billes et de tous les emplacements. En effet, les billes d'une couleur donnée vont dans les emplacements de cette même couleur et y sont rangées dans l'ordre croissant.\n\nOr il y a $5^{500}$ coloriages possibles des billes et $5^{500}$ coloriages possibles des emplacements en 5 couleurs. Donc le nombre d'ordres plutôt croissants est au plus $5^{1000}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un nombre à 9 chiffres est intéressant si chaque chiffre de 1 à 9 y apparaît une unique fois, que les chiffres de 1 à 5 y apparaissent dans l'ordre mais pas les chiffres de 1 à 6, par exemple 189236457.\n\nCombien y a-t-il de nombres intéressants?",
    "options": [],
    "answer": "2520",
    "solution": "Solution:\n\nPour construire un nombre intéressant, on peut placer d'abord les chiffres de 1 à 5, puis intercaler le 6 quelque part, puis le 7, le 8 et le 9. L'ordre des chiffres de 1 à 5 est imposé. Ensuite, le 6 peut être placé n'importe où sauf après le 5 car les chiffres de 1 à 6 ne sont pas dans l'ordre. Il y a donc 5 manières de concerning le 6. Ensuite, on peut placer le 7 n'importe où, donc il y a 7 manières de completely le placer, et de même 8 manières de placer le 8 et 9 pour le 9.\n\nIl y a donc $5 \\times 7 \\times 8 \\times 9 = 2520$ nombres intéressants.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $A, B, C, D$ quatre points sur un même cercle. Notons $A'$, $C'$ les projetés orthogonaux de $A$ et $C$ sur $(BD)$, et $B'$, $D'$ les projetés orthogonaux de $B$ et $D$ sur $(AC)$. Montrer que $A'$, $B'$, $C'$, $D'$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nComme $A, B, A', B'$ sont cocycliques (sur le cercle de diamètre $[AB]$), les angles orientés de droites $(BA', BA)$ et $(B'A', B'A)$ sont égaux, donc $(BD, BA) = (B'A', B'D')$. De même, $(CD, CA) = (C'A', C'D')$. Or, $(CD, CA) = (BD, BA)$ puisque $A, B, C, D$ sont cocycliques, donc $(B'A', B'D') = (C'A', C'D')$, ce qui prouve que $A', B', C', D'$ sont cocycliques.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30335,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nExiste-t-il des réels $a$, $b$, $c$, $d > 0$ et $e$, $f$, $g$, $h < 0$ vérifiant simultanément\n$$\na e + b c > 0,\\quad e f + c g > 0,\\quad f d + g h > 0 \\text{ et } d a + h b > 0 ?\n$$",
    "options": [],
    "answer": "No, such real numbers do not exist.",
    "solution": "Solution:\nNon, il n'en existe pas. Par l'absurde : supposons qu'il existe de tels réels. On commence par réécrire les inégalités, mais avec uniquement des termes positifs. On a donc\n$$\nb c > a(-e) \\text{ et } (-e)(-f) > c(-g) \\text{ et } (-g)(-h) > (-f)d \\text{ et } d a > (-h) b\n$$\nSi l'on multiplie toutes ces inégalités membres à membres (et, comme tout est positif, il n'y a aucun danger), il vient $a b c d e f g h > a b c b d e f g h$, d'où la contradiction cherchée.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle rectangle isocèle en $B$. Soit $M$ un point de l'arc $AC$ du cercle de centre $B$ passant par $A$ et $C$, $H$ son projeté orthogonal sur $(AB)$. On note $I$ le centre du cercle inscrit à $BHM$ et $J$ le centre du cercle exinscrit dans l'angle $B$ ($J$ est donc l'intersection de la bissectrice intérieure en $B$ avec les bissectrices extérieures en $H$ et $M$). Montrer que $MIAJ$ est un carré.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn observe d'abord que $(MI) \\perp (MJ)$ puisque dans un triangle, les bissectrices intérieure et extérieure en un point sont perpendiculaires.\n\nD'autre part, $\\widehat{MIJ} = 180^\\circ - \\widehat{BIM} = \\widehat{IBM} + \\widehat{BMI} = \\frac{1}{2}(\\widehat{HBM} + \\widehat{BMH}) = 45^\\circ$.\n\nComme $MIJ$ est rectangle en $M$, on en déduit qu'il est rectangle isocèle en $M$.\n\nDe plus, $(IJ)$ est la bissectrice de $\\widehat{MBH} = \\widehat{MBA}$, et $MBA$ est isocèle en $B$, donc $(IJ)$ est la médiatrice de $[AM]$.\n\nOn en déduit que $AIJ$ est le symétrique du triangle rectangle isocèle $MIJ$ par rapport à la droite $(IJ)$, donc $MIAJ$ est un carré.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn appelle diviseur propre d'un entier $n$ un diviseur positif de $n$ qui est différent de $1$ et de $n$.\nExiste-t-il un entier $n$ dont le produit des diviseurs propres est égal à $2013$ ?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLa décomposition en facteurs premiers de $2013$ est $3 \\times 11 \\times 61$. Ainsi, si le produit des diviseurs stricts d'un entier $n$ vaut $2013$, cela implique que $n$ est lui-même divisible par $3$, $11$ et $61$. Or, dans ce cas, il compte parmi ses diviseurs stricts au moins les nombres suivants :\n$$\n3, 11, 61, 33, 671\n$$\ndont le produit surpasse $2013$ (il vaut $2013 \\times 33 \\times 671 = 44573859$). Ainsi, un tel entier $n$ n'existe pas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProuver que l'équation\n$$\na^{3}+b^{5}+c^{7}+d^{11}=e^{13}\n$$\nadmet une infinité de solutions en entiers strictement positifs.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIl convient de remarquer que si $(a, b, c, d, e)$ est une solution alors, pour tout entier $k>0$, le quintuplet\n$$\n\\left(a k^{5 \\times 7 \\times 11 \\times 13}, b k^{3 \\times 7 \\times 11 \\times 13}, c k^{3 \\times 5 \\times 11 \\times 13}, d k^{3 \\times 5 \\times 7 \\times 13}, e k^{3 \\times 5 \\times 7 \\times 11}\\right)\n$$\nen est une aussi.\n\nPour conclure, il suffit donc de trouver une solution. On peut penser à en trouver une telle que\n$$\na^{3}=b^{5}=c^{7}=d^{11}=\\frac{1}{4} e^{13}\n$$\nou $e=2^{t}$ pour un certain $t$ à déterminer. Pour cela, $t$ doit être choisi de sorte que $13 t-2$ soit multiple commun de $3,5,7,11$. En cherchant un peu, on voit que l'on peut prendre $13 t-2=3 \\times 5 \\times 7 \\times 11$, ou encore $t=89$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $n \\geqslant 5$ un entier. Soient $a_{1}, \\ldots, a_{n}$ des entiers dans $\\{1, \\ldots, 2n\\}$ deux à deux distincts. Montrer qu'il existe des indices $i, j \\in \\{1, \\ldots, n\\}$ avec $i \\neq j$ tels que\n$$\n\\operatorname{PPCM}\\left(a_{i}, a_{j}\\right) \\leqslant 6\\left(E\\left(\\frac{n}{2}\\right)+1\\right)\n$$\noù $E(x)$ désigne la partie entière du nombre $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSupposons, pour commencer, qu'il existe un indice $i$ tel que $a_{i} \\leqslant n$. S'il existe $j$ tel que $a_{j}=2 a_{i}$, on a alors\n$$\n\\operatorname{PPCM}\\left(a_{i}, a_{j}\\right)=a_{j} \\leqslant 2 n \\leqslant 6 \\cdot\\left(E\\left(\\frac{n}{2}\\right)+1\\right)\n$$\net on a trouvé un couple $(i, j)$ convenable. Si, au contraire, l'entier $2 a_{i}$ n'apparaît pas parmi les $a_{j}$, définissons les entiers $b_{1}, \\ldots, b_{n}$ en posant $b_{i}=2 a_{i}$ et $b_{j}=a_{j}$ pour $j \\neq i$. Les $b_{j}$ sont encore deux à deux distincts et compris entre $1$ et $2n$. De plus, on vérifie immédiatement que $\\operatorname{PPCM}\\left(b_{i}, b_{j}\\right) \\geqslant \\operatorname{PPCM}\\left(a_{i}, a_{j}\\right)$. Ainsi, il suffit de démontrer la propriété de l'énoncé pour la suite des $b_{j}$.\n\nEn appliquant à nouveau le raisonnement précédent — éventuellement plusieurs fois — on en vient à supposer que tous les $a_{j}$ sont strictement supérieurs à $n$. Autrement dit, l'ensemble des $a_{j}$ n'est autre que $\\{n+1, n+2, \\ldots, 2n\\}$. Posons $k=E\\left(\\frac{n}{2}\\right)+1$. Comme on a supposé $n \\geqslant 5$, il est facile de vérifier que l'un des $a_{j}$ vaut $2k$ et un autre vaut $3k$. On conclut en remarquant que le PPCM de ces deux nombres est égal à $6k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a, b, c$ des réels tels que $-1 \\leqslant a x^{2}+b x+c \\leqslant 1$ pour $x=-1, x=0$ et $x=1$. Prouver que\n$$\n-\\frac{5}{4} \\leqslant a x^{2}+b x+c \\leqslant \\frac{5}{4} \\text{ pour tout réel } x \\in[-1,1]\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPosons $P(x)=a x^{2}+b x+c$. Alors $P(-1)=a-b+c$, $P(0)=c$ et $P(1)=a+b+c$. Et, d'après l'énoncé, on a $|P(-1)| \\leqslant 1$, $|P(0)| \\leqslant 1$ et $|P(1)| \\leqslant 1$.\n\nOr, pour tout réel $x$, on vérifie directement que\n$$\nP(x)=\\frac{x(x+1)}{2} P(1)-\\frac{x(1-x)}{2} P(-1)+\\left(1-x^{2}\\right) P(0)\n$$\n\n- Soit $x \\in[0 ; 1]$. D'après (1) et l'inégalité triangulaire, il vient\n$$\n\\begin{aligned}\n|P(x)| & \\leqslant \\frac{x(x+1)}{2}|P(1)|+\\frac{x(1-x)}{2}|P(-1)|+\\left(1-x^{2}\\right)|P(0)| \\\\\n& \\leqslant \\frac{x(x+1)}{2}+\\frac{x(1-x)}{2}+\\left(1-x^{2}\\right) \\\\\n& =-x^{2}+x+1 \\\\\n& =\\frac{5}{4}-\\left(x-\\frac{1}{2}\\right)^{2}\n\\end{aligned}\n$$\nEt ainsi $|P(x)| \\leqslant \\frac{5}{4}$.\n\n- Soit $x \\in[-1 ; 0]$. D'après (1) et l'inégalité triangulaire, il vient cette fois\n$$\n\\begin{aligned}\n|P(x)| & \\leqslant -\\frac{x(x+1)}{2}|P(1)|-\\frac{x(1-x)}{2}|P(-1)|+\\left(1-x^{2}\\right)|P(0)| \\\\\n& \\leqslant -\\frac{x(x+1)}{2}-\\frac{x(1-x)}{2}+\\left(1-x^{2}\\right) \\\\\n& =-x^{2}-x+1 \\\\\n& =\\frac{5}{4}-\\left(x+\\frac{1}{2}\\right)^{2}\n\\end{aligned}\n$$\nEt ainsi $|P(x)| \\leqslant \\frac{5}{4}$, à nouveau.\n\nFinalement, pour tout $x \\in[-1 ; 1]$, on a $|P(x)| \\leqslant \\frac{5}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30341,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSur la diagonale $[BD]$ d'un carré $ABCD$ on a choisi un point $E$. Soient $O_{1}$ et $O_{2}$ les centres des cercles circonscrits aux triangles $ABE$ et $ADE$ respectivement. Montrer que $AO_{1}EO_{2}$ est un carré.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$\\widehat{AO_{1}E} = 2\\widehat{ABE} = 90^{\\circ}$ et $O_{1}A = O_{1}E$ donc $AO_{1}E$ est un triangle isocèle rectangle en $O_{1}$. De même, $AO_{2}E$ est un triangle isocèle rectangle en $O_{2}$.\n\nComme $O_{1}$ et $O_{2}$ ne sont pas dans le même demi-plan délimité par $(AE)$, on en déduit que $AO_{1}EO_{2}$ est un carré.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ une suite croissante et non constante d'entiers strictement positifs tels que $a_{n}$ divise $n^{2}$ pour tout $n \\geqslant 1$. Prouver que l'une des affirmations suivantes est vraie:\n\na) Il existe un entier $n_{1}>0$ tel que $a_{n}=n$ pour tout $n \\geqslant n_{1}$.\n\nb) Il existe un entier $n_{2}>0$ tel que $a_{n}=n^{2}$ pour tout $n \\geqslant n_{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, puisque pour tout entier $n$, on a $a_{n} \\in \\mathbb{N}^{*}$, et que la suite $\\left(a_{n}\\right)$ est croissante et non constante, il existe un entier $n_{0}$ tel que $a_{n} \\geqslant 2$ pour tout $n \\geqslant n_{0}$. Par conséquent, pour tout nombre premier $p>n_{0}$, on a $a_{p}=p$ ou $a_{p}=p^{2}$.\n\n- Supposons que, pour tout entier $n \\geqslant 1$, on ait $a_{n} \\leqslant n$. Soit $p>n_{0}$ un nombre premier. Prouvons par récurrence que $a_{n}=n$ pour tout $n \\geqslant p$. En effet, d'après ci-dessus, on a $a_{p}=p$. De plus, si $a_{n}=n$ pour un certain $n \\geqslant p \\geqslant 2$, alors $n=a_{n} \\leqslant a_{n+1} \\leqslant n+1$, d'où $a_{n+1}=n$ ou $a_{n+1}=n+1$. Mais, $a_{n+1}$ divise $(n+1)^{2}$, tandis que $n$ ne divise pas $(n+1)^{2}$. Ainsi, on a $a_{n+1}=n+1$, ce qui achève la récurrence.\n\nIl suffit alors de choisir $n_{1}=p$ pour conclure dans ce cas.\n\n- Supposons qu'il existe un entier $m \\geqslant 1$ tel que $a_{m}>m$. S'agissant d'entiers, c'est donc que $a_{m} \\geqslant m+1$. Mais, $m+1 \\geqslant 2$ et $m+1$ est premier avec $m^{2}$, donc $m+1$ ne peut diviser $m^{2}$, et on a alors $a_{m} \\geqslant m+2$. On en déduit que $a_{m+1} \\geqslant a_{m}>m+1$. Une récurrence immédiate assure donc que $a_{n}>n$ pour tout $n \\geqslant m$.\n\nEn particulier, compte-tenu de notre remarque initiale, si $p \\geqslant m$ est un nombre premier, on a $a_{p}=p^{2}$. Soit donc $p>m$ un nombre premier impair. Nous allons prouver par récurrence que $a_{n}=n^{2}$ pour tout $n \\geqslant p$. Nous venons de voir que c'est vrai pour $n=p$. Supposons que $a_{n}=n^{2}$ pour un certain entier $n \\geqslant p$. Alors $n \\geqslant 3$, d'où $a_{n+1} \\geqslant a_{n}=n^{2}>\\frac{1}{2}(n+1)^{2}$. Or, $a_{n+1}$ divise $(n+1)^{2}$, donc $a_{n+1}=(n+1)^{2}$. Cela achève la récurrence.\n\nIl suffit donc de choisir $n_{2}=p$ pour conclure dans ce cas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe. Soit $M$ l'intersection entre les bissectrices intérieures des angles $B$ et $C$, et $N$ l'intersection entre les bissectrices intérieures des angles $A$ et $D$. Montrer que les droites $AB$, $CD$ et $MN$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nQuitte à échanger les rôles de $A$ et de $B$ et les rôles de $C$ et de $D$, on peut supposer que $A$ et $D$ se trouvent sur les segments $[EB]$ et $[EC]$ respectivement.\n\n![](attached_image_1.png)\n\nNotons $E$ le point d'intersection entre $(AB)$ et $(CD)$. Alors $M$ est l'intersection des bissectrices intérieures des angles $B$ et $C$ du triangle $EBC$, donc $M$ est le centre du cercle inscrit à $EBC$. En particulier il se trouve sur la bissectrice intérieure en $E$.\n\nDe même, $N$ est le point d'intersection des bissectrices extérieures en $A$ et $D$ du triangle $EAD$, donc $N$ est le centre du cercle exinscrit dans l'angle $E$ de $EAD$. En particulier, il se trouve sur la bissectrice intérieure en $E$. Ceci prouve que $E, M, N$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans un pays, il y a $n$ villes. Deux villes quelconques sont toujours reliées soit par une autoroute, soit par une ligne de train. Montrer qu'un des deux moyens de transport permet de relier n'importe quelle ville à n'importe quelle autre.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn raisonne par récurrence sur $n$ : pour $n=1$ ou $n=2$, le résultat est immédiat. Supposons qu'on l'a montré au rang $n$, et considérons $n+1$ villes :\nOn isole une des villes, appelons-la par exemple Paris. Alors un des deux moyens de transports permet de relier entre elles toutes les villes sauf Paris. Supposons que c'est la voiture :\n- si Paris est relié à une ville par une autoroute, par exemple Lyon, on peut en voiture aller de Paris à Lyon, puis à n'importe quelle autre ville d'après l'hypothèse de récurrence.\n- sinon, Paris est relié à toutes les villes par une ligne de train, donc on peut aller de n'importe quelle ville à n'importe quelle ville en train en passant par Paris.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nChaque nombre rationnel strictement positif est colorié soit en rouge, soit en noir, de telle sorte que:\n- les nombres $x$ et $x+1$ sont de couleurs différentes ;\n- les nombres $x$ et $\\frac{1}{x}$ sont de la même couleur;\n- le nombre $1$ est colorié en rouge.\nQuelle est la couleur de $\\frac{2012}{2013}$ ?\n(On ne demande pas de démontrer l'existence d'un tel coloriage.)",
    "options": [],
    "answer": "red",
    "solution": "Solution:\n\nOn remarque que $\\frac{2012}{2013}$ s'écrit:\n$$\n\\frac{2012}{2013}=\\frac{1}{1+\\frac{1}{2012}}\n$$\nAinsi, d'après les règles de l'énoncé, il a la couleur inverse de $2012$. Or, ce dernier s'obtient en ajoutant $2011$ fois $1$ à $1$ ; il a donc la couleur inverse de $1$ (puisqu'on inverse $2011$ fois la couleur), c'est-à-dire noir. Ainsi $\\frac{2012}{2013}$ est rouge.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProuver que, pour tous entiers strictement positifs $a$, $b$, $c$, $d$, on a\n$$\n(2a-1)(2b-1)(2c-1)(2d-1) \\geqslant 2abcd-1\n$$\net déterminer les cas d'égalité.",
    "options": [],
    "answer": "Equality holds if and only if at least three of a, b, c, d are equal to 1.",
    "solution": "Solution:\nSi $x$ et $y$ sont des entiers strictement positifs, on a $x \\geqslant 1$ et $y \\geqslant 1$, et ainsi $(x-1)(y-1) \\geqslant 0$. Cela conduit à $xy - x - y + 1 \\geqslant 0$, et donc à $(2x-1)(2y-1) \\geqslant 2xy-1$. Notons que l'égalité a lieu si et seulement si $x=1$ ou $y=1$.\n\nSi $a$, $b$, $c$, $d$ sont des entiers strictement positifs, on a donc $(2a-1)(2b-1) \\geqslant 2ab-1$ et $(2c-1)(2d-1) \\geqslant 2cd-1$, mais aussi $(2ab-1)(2cd-1) \\geqslant 2abcd-1$. Ainsi :\n$$\n(2a-1)(2b-1)(2c-1)(2d-1) \\geqslant (2ab-1)(2cd-1) \\geqslant 2abcd-1\n$$\nD'après ci-dessus, l'égalité a lieu si et seulement si trois des nombres $a$, $b$, $c$, $d$ sont égaux à $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProuver que, pour tout réel $a \\geqslant 0$, on a\n$$\na^{3}+2 \\geqslant a^{2}+2 \\sqrt{a}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPour tout réel $a \\geqslant 0$, on a\n$$\n\\begin{aligned}\na^{3}-a^{2}-2 \\sqrt{a}+2 & =a^{2}(a-1)-2(\\sqrt{a}-1) \\\\\n& =a^{2}(\\sqrt{a}-1)(\\sqrt{a}+1)-2(\\sqrt{a}-1) \\\\\n& =(\\sqrt{a}-1)\\left(a^{2}(\\sqrt{a}+1)-2\\right) \\quad (1)\n\\end{aligned}\n$$\nOr:\n- Si $a \\geqslant 1$ alors $\\sqrt{a} \\geqslant 1$ et $a^{2} \\geqslant 1$. Ainsi, on a $\\sqrt{a}-1 \\geqslant 0$ et $a^{2}(\\sqrt{a}+1) \\geqslant 2$. Par suite, chacun des facteurs de (1) est positif, ce qui assure que le produit est positif.\n- Si $a \\leqslant 1$ alors $\\sqrt{a} \\leqslant 1$ et $a^{2} \\leqslant 1$. Ainsi, on a $\\sqrt{a}-1 \\leqslant 0$ et $a^{2}(\\sqrt{a}+1) \\leqslant 2$. Par suite, chacun des deux facteurs de (1) est négatif, et le produit est donc encore positif.\nFinalement, pour tout réel $a \\geqslant 0$, on a $a^{3}-a^{2}-2 \\sqrt{a}+2 \\geqslant 0$ ou encore $a^{3}+2 \\geqslant a^{2}+2 \\sqrt{a}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn quadrilatère $ABCD$ est inscrit dans un cercle. Ses diagonales se coupent au point $K$. Le cercle passant par $A$, $B$, $K$ croise les droites $(BC)$ et $(AD)$ aux points $M$ et $N$ respectivement. Montrer que $KM = KN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nOn a $(NM, NK) = (BM, BK) = (BC, BD) = (AC, AD) = (AK, AN) = (MK, MN)$ donc $MKN$ est isocèle en $K$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2014 scientifiques participent à un congrès, chaque scientifique étant soit un mathématicien, soit un physicien. Bien sûr, les physiciens mentent toujours et les mathématiciens disent toujours la vérité, sauf quand ils se trompent. Lors du dîner final, tous sont assis en rond autour d'une table, et chacun prétend se trouver entre un mathématicien et un physicien. Il se trouve qu'exactement un mathématicien distrait s'est trompé. Combien y a-t-il de physiciens au congrès?",
    "options": [],
    "answer": "671",
    "solution": "Solution:\n\nSi 2 physiciens sont assis côte à côte, comme ils ont menti, ils ne sont pas entre un physicien et un mathématicien, donc chacun est entre deux physiciens, et ainsi de suite, donc le congrès n'accueille que des physiciens, ce qui est absurde car il y a au moins un mathématicien (celui qui s'est trompé...). Chaque physicien est donc entre deux mathématiciens.\n\nD'autre part, chaque mathématicien est entre un mathématicien et un physicien, sauf celui qui s'est trompé. Appelons-le Thomas : si Thomas est entre deux physiciens, alors on a partout ailleurs une alternance de deux mathématiciens, un physicien... Mais alors, si il y a $k$ paires de mathématiciens côte à côte, il y a $2k+1$ mathématiciens et $k+1$ physiciens, donc $3k+2$ scientifiques, ce qui est absurde car $3k+2 \\neq 2014$.\n\nThomas est donc entre deux mathématiciens, donc on a trois mathématiciens côté à côte, et les autres par paires. Si il y a $k$ paires, alors on a $2k+3$ mathématiciens et $k+1$ physiciens donc $3k+4=2014$ donc $k=670$ et il y a 671 physiciens.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère inscriptible, $L = (AC) \\cap (BD)$, $J$ et $K$ les pieds des perpendiculaires à $(AD)$ et $(BC)$ passant par $L$ et $I$ le milieu de $[C, D]$. Montrer que $IJ = IK$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIl est facile de voir en utilisant le théorème de l'angle inscrit que $BLC$ et $ALD$ sont semblables, et que $CKL$ et $DJL$ sont semblables.\n\nNotons $M$ et $N$ les milieux respectifs de $[LD]$ et $[LC]$. Alors $L N I M$ est un parallélogramme. On en déduit que $NL = IM$. Or, $CKL$ est un triangle rectangle en $K$ donc $NL = NK$, d'où $NK = IM$. De même, $MJ = IN$.\n\nComme $MJL$ est isocèle en $M$, on a $\\widehat{JML} = 180^{\\circ} - 2 \\widehat{MLJ} = 180^{\\circ} - 2\\left(90^{\\circ} - \\widehat{LDJ}\\right) = 2 \\widehat{BDA}$, et de même $\\widehat{KNL} = 2 \\widehat{BCA}$. Or, $\\widehat{BDA} = \\widehat{BCA}$ donc $\\widehat{JML} = \\widehat{KNL}$. On en déduit que $\\widehat{IMJ} = 360^{\\circ} - \\widehat{JML} - \\widehat{LMI} = 360^{\\circ} - \\widehat{KNL} - \\widehat{LNI} = \\widehat{KNI}$ (l'avant-dernière égalité découlant du fait que $LMIN$ est un parallélogramme).\n\nDes égalités $\\widehat{IMJ} = \\widehat{KNI}$, $NK = IM$ et $MJ = IN$, on déduit que les triangles $IMJ$ et $KNI$ sont isométriques, et en particulier que $IJ = IK$.\nSolution:\n\n$$\n\\begin{aligned}\nIJ^{2} - IK^{2} & = \\overrightarrow{IJ} \\cdot \\overrightarrow{IJ} - \\overrightarrow{IK} \\cdot \\overrightarrow{IK} = (\\overrightarrow{IJ} + \\overrightarrow{IK}) \\cdot (\\overrightarrow{IJ} - \\overrightarrow{IK}) \\\\\n& = (\\overrightarrow{ID} + \\overrightarrow{DJ} + \\overrightarrow{IC} + \\overrightarrow{CK}) \\cdot \\overrightarrow{KJ} = (\\overrightarrow{DJ} + \\overrightarrow{CK}) \\cdot (\\overrightarrow{LJ} - \\overrightarrow{LK}) \\\\\n& = \\overrightarrow{CK} \\cdot \\overrightarrow{LJ} - \\overrightarrow{DJ} \\cdot \\overrightarrow{LK} \\quad \\text{car } (DJ) \\perp (LJ) \\text{ et } (CK) \\perp (LK) \\\\\n& = CK \\cdot LJ \\cos (\\overrightarrow{CK}, \\overrightarrow{LJ}) - DJ \\cdot LK \\cos (\\overrightarrow{DJ}, \\overrightarrow{LK})\n\\end{aligned}\n$$\n\nOr, $\\overrightarrow{LK}$ fait un angle orienté de $+90^{\\circ}$ par rapport à $\\overrightarrow{CK}$, et $\\overrightarrow{DJ}$ fait un angle orienté de $+90^{\\circ}$ par rapport à $\\overrightarrow{LJ}$, donc les deux cosinus sont égaux. Par conséquent, il suffit de montrer que $CK \\cdot LJ = DJ \\cdot LK$ : ceci découle du fait que\n$$\n\\frac{CK}{LK} = \\cot \\widehat{KCL} = \\cot \\widehat{BCA} = \\cot \\widehat{BDA} = \\cot \\widehat{LDJ} = \\frac{DJ}{LJ}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProuver que pour tous réels strictement positifs $a, b, c$ tels que $a b c=1$, on a\n$$\n\\frac{1}{1+a^{2}+(b+1)^{2}}+\\frac{1}{1+b^{2}+(c+1)^{2}}+\\frac{1}{1+c^{2}+(a+1)^{2}} \\leqslant \\frac{1}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSoit $a, b, c$ des réels strictement positifs.\nCompte tenu de l'inégalité bien connue $x^{2}+y^{2} \\geqslant 2 x y$ valable pour tous réels $x, y$, on a\n$$\n\\frac{1}{1+a^{2}+(b+1)^{2}}=\\frac{1}{2+a^{2}+b^{2}+2 b} \\leqslant \\frac{1}{2(1+a b+b)}\n$$\nDe même, et puisque $a b c=1$, on a\n$$\n\\frac{1}{1+b^{2}+(c+1)^{2}} \\leqslant \\frac{1}{2(1+b c+c)}=\\frac{a b}{2\\left(a b+a b^{2} c+a b c\\right)}=\\frac{a b}{2(1+a b+b)}\n$$\net\n$$\n\\frac{1}{1+c^{2}+(a+1)^{2}} \\leqslant \\frac{1}{2(1+a c+a)}=\\frac{b}{2(b+a b c+a b)}=\\frac{b}{2(1+a b+b)}\n$$\nen sommant membre à membre, il vient\n$$\n\\begin{aligned}\n& \\frac{1}{1+a^{2}+(b+1)^{2}}+\\frac{1}{1+b^{2}+(c+1)^{2}}+\\frac{1}{1+c^{2}+(a+1)^{2}} \\\\\n& \\leqslant \\frac{1}{2}\\left(\\frac{1}{1+a b+b}+\\frac{a b}{1+a b+b}+\\frac{b}{1+a b+b}\\right)=\\frac{1}{2}\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLe point $D$ appartient au côté $[A C]$ du triangle équilatéral $A B C$. On note $F$ le projeté orthogonal de $D$ sur $(B C)$, puis $K$ le projeté orthogonal de $F$ sur $(A B)$, et enfin $E$ le projeté orthogonal de $K$ sur $(C A)$. Soit $L$ le milieu de $[BC]$, et $P$ l'intersection des droites $(K E)$ et $(F D)$.\n\nProuver que la droite $(BP)$ passe par le milieu de $[AL]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPremière solution. Comme les côtés de $PKF$ sont perpendiculaires aux côtés de $ABC$, le triangle $PKF$ a les mêmes angles intérieurs que $A B C$ et est donc équilatéral, et donc $P F = K F$.\n\nNotons $M$ l'intersection de $(BP)$ et $(A L)$. Comme les triangles $BLM$ et $BFP$ sont semblables, on a\n$$\n\\frac{M L}{B L} = \\frac{P F}{B F} = \\frac{K F}{B F} = \\sin 60^{\\circ} = \\frac{\\sqrt{3}}{2}\n$$\nD'autre part, $\\frac{A L}{B L} = \\tan 60^{\\circ} = \\sqrt{3}$, donc $A L = 2 M L$, ce qui prouve que $M$ est le milieu de $[AL]$.\n\n\nDeuxième solution. Posons que les côtés de $A B C$ soient de longueur $2$, et que $A D = 2 a$ avec $a \\in [0,1]$. On en déduit que $DC = 2 - 2 a$, $FC = DC \\cos(60^{\\circ}) = 1 - a$, et $BF = 1 + a$. Par suite, on a\n$$\nBK = BF \\cos(60^{\\circ}) = \\frac{1 + a}{2} \\quad \\text{et} \\quad KF = BF \\sin(60^{\\circ}) = \\frac{\\sqrt{3}}{2}(1 + a)\n$$\nPuisque les côtés de $PKF$ sont perpendiculaires aux côtés de $A B C$, le triangle $PKF$ a les mêmes angles intérieurs que $A B C$ et est donc équilatéral. On a alors $P F = K F = \\frac{\\sqrt{3}}{2}(1 + a)$.\n\nLe point $L$ étant le milieu de $[B C]$ avec $A B C$ équilatéral, les droites $(A L)$ et $(B C)$ sont donc perpendiculaires, ce qui assure que $(M L)$ et $(P F)$ sont parallèles, où $M$ est l'intersection de $(B P)$ et $(A L)$. D'après le théorème de Thalès, il vient alors\n$$\n\\frac{M L}{P F} = \\frac{B L}{B F} = \\frac{1}{1 + a}\n$$\nd'où $M L = \\frac{\\sqrt{3}}{2} = \\frac{1}{2} A L$, ce qui conclut.\n\n\nTroisième solution. Comme ci-dessus, on remarque que $PKF$ est équilatéral, et on note $M$ l'intersection de $(BP)$ et $(AL)$. Soit $Q$ l'intersection de $(FP)$ et $(AB)$. On a $\\widehat{PKQ} = \\widehat{PQK} = 30^{\\circ}$, donc $PQK$ est isocèle en $P$ et on a $P Q = P K$. Mais, $P K = K F$ puisque $PKF$ est équilatéral, donc $P Q = K F$ ce qui assure que $P$ est le milieu de $[QF]$. Ayant leurs côtés communs ou parallèles, les triangles $BPQ$ et $BMA$ d'une part, et les triangles $BPF$ et $BML$ d'autre part, sont semblables. Cela conduit à\n$$\n\\frac{P Q}{A M} = \\frac{P M}{B M} \\quad \\text{et} \\quad \\frac{P F}{M L} = \\frac{P M}{B M}\n$$\nd'où $\\frac{A M}{M L} = \\frac{P Q}{P F} = 1$, et $M$ est bien le milieu de $[A L]$.\n\n\nQuatrième solution. Soit $X$ le milieu de $[A B]$ et $Y$ tel que $A L C Y$ soit un rectangle. Alors $X C = A L = C Y$ et $\\widehat{Y C X} = 60^{\\circ}$, donc $X C Y$ est équilatéral. Deux de ses côtés étant perpendiculaires à deux des côtés du triangle équilatéral $A B C$, il en est de même de son troisième côté. Soit $M$ l'intersection de $(AL)$ et $(BY)$. Puisque $(A L)$ et $(C Y)$ sont parallèles et que $B C = 2 B L$, on a $A L = C Y = 2 M L$, d'où $M$ est le milieu de $[AL]$. Comme ci-dessus, on remarque que $KFP$ est équilatéral car ses côtés sont perpendiculaires à ceux de $A B C$. Or, il en est de même de $X Y C$, donc les triangles $KFP$ et $X C Y$ ont leurs côtés parallèles. Puisque les points $B, F, C$ d'une part, et $B, K, X$ d'autre part sont alignés, il existe donc une homothétie de centre $B$ qui envoie $X C Y$ sur $KFP$. En particulier, le point $P$ appartient à la droite $(BY)$ qui passe par le milieu de $[AL]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLa suite de Fibonacci est construite ainsi : chaque terme est la somme des deux précédents. Ses premiers termes valent:\n$$\n1,1,2,3,5,8,13,21,34,55,89, \\ldots\n$$\nMontrer que la suite de Fibonacci contient un multiple de 1000.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $F_{n}$ le $n$-ième terme de la suite de Fibonacci ($F_{1}=1$, $F_{2}=1$, $F_{3}=2$, etc.) et notons $u_{n}$ le reste de la division euclidienne de $F_{n}$ par $1000$ ($u_{n}$ est donc le nombre formé par les trois derniers chiffres de $F_{n}$). Comme il n'y a qu'un nombre fini de couples $(x, y)$ d'entiers compris entre $0$ et $999$ (il y en a exactement $1000^{2}=1000000$), il existe des entiers $n$ et $N$ tels que $N>0$ et\n$$\nu_{n}=u_{n+N} \\quad ; \\quad u_{n+1}=u_{n+1+N}$$\nOr, $u_{n-1}$ (resp. $u_{n-1+N}$) s'obtient à partir de $u_{n}$ et $u_{n+1}$ (resp. $u_{n+N}$ et $u_{n+1+N}$) comme le reste de la division euclidienne de $u_{n+1}-u_{n}$ (resp. $u_{n+1+N}-u_{n+N}$) par $1000$. Ainsi, des égalités (1), on déduit que $u_{n-1}=u_{n-1+N}$. En continuant ainsi, on obtient $u_{N+1}=u_{1}=1$ et $u_{N+2}=u_{2}=1$, d'où il résulte enfin $u_{N}=0$. L'entier $F_{N}$ est donc multiple de $1000$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer que tout polyèdre a deux faces qui ont le même nombre de sommets.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $n$ le nombre de faces du polyèdre, et soit $F$ une face fixée : les arêtes de $F$ séparent $F$ de faces toutes différentes, donc le nombre d'arêtes de $F$ est inférieur ou égal au nombre de faces autres que $F$. De plus, $F$ a au moins 3 arêtes. Chaque face a autant d'arêtes que de sommet, donc toute face a un nombre de sommets compris entre 3 et $n-1$, soit $n-3$ possibilités. D'après le principe des tiroirs, il existe donc deux faces qui ont le même nombre de sommets.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans le plan, on considère l'ensemble $S$ des points de coordonnées $(x, y)$, où $x, y \\in \\{1,2, \\ldots, 2013\\}$. Deux points de $S$ sont dits voisins s'ils sont à une distance $1$ l'un de l'autre. À chaque seconde, une mouche et des araignées se déplacent sur les points de $S$ de la façon suivante : tout d'abord, la mouche soit ne bouge pas, soit va sur un point voisin de celui sur lequel elle se trouve. Puis, chaque araignée soit ne bouge pas, soit va sur un point voisin de celui sur lequel elle se trouve. Plusieurs araignées peuvent se trouver simultanément sur un même point, et la mouche ainsi que les araignées connaissent les positions respectives des unes et des autres.\n\na) Déterminer le plus petit entier $k$ pour lequel $k$ araignées pourront toujours finir par attraper la mouche en un temps fini, et ce quelles que soient les positions initiales de la mouche et des araignées sur les points de $S$.\n\nb) Répondre à la même question si l'on suppose cette fois que l'on est dans l'espace usuel et que $S$ est l'ensemble des points de coordonnées $(x, y, z)$ où $x, y, z \\in \\{1,2, \\ldots, 2013\\}$.",
    "options": [],
    "answer": "a) 2; b) 2",
    "solution": "Solution:\n\nUne seule araignée ne peut attraper la mouche : en effet, la mouche peut attendre jusqu'à ce que l'araignée soit sur un point voisin de celui sur lequel elle se trouve. À partir de ce moment, elle se déplace à chaque fois sur un point diagonalement opposé à celui sur lequel est l'araignée si celle-ci est sur un point voisin, et sinon attend où elle est.\n\nOn va prouver maintenant que, dans le plan comme dans l'espace, deux araignées suffisent pour attraper la mouche.\n\nPour cela, on note $M$ la mouche, et on désigne par $P$ et $Q$ les deux araignées. Pour chaque animal $A$, on note respectivement $A_{x}, A_{y}$ et $A_{z}$ son abscisse, son ordonnée et sa cote.\n\na)\n\nLes araignées $P$ et $Q$ se placent tout d'abord sur le point de coordonnées $(1,1)$. Puisque $M$ ne peut s'enfuir indéfiniment vers la droite, $P$ peut alors en un temps fini, et en ne se déplaçant qu'horizontalement, se trouver en un point tel que $P_{x}=M_{x}$ et $P_{y}=1$. De même, $Q$ se place en un point tel que $Q_{x}=1$ et $Q_{y}=M_{y}$ (il est clair que si $P$, par exemple, atteint ce premier objectif avant $Q$, elle peut suivre les déplacements de $M$ en conservant $P_{x}=M_{x}$ et $P_{y}=1$ et attendre le positionnement adéquat de $Q$).\n\nÀ partir de ces positions, atteintes au temps $t_{0}$ :\n- si $M$ se déplace horizontalement (c'est-à-dire de sorte que $M_{x}$ change) alors $P$ fait de même afin de conserver $P_{x}=M_{x}$.\n- sinon, $P$ se déplace verticalement, en allant vers $M$.\n\nL'araignée $Q$ procède de la même façon, mais en échangeant les rôles des abscisses et ordonnées.\n\nEn utilisant cette stratégie et à partir de $t_{0}$, la quantité $|P_{y}-M_{y}|+|Q_{x}-M_{x}|$ n'augmente jamais. Mais, cette quantité ne reste constante que lors d'un déplacement de $M$ vers la droite ou vers le haut, ce qui ne peut arriver qu'un nombre fini de fois. Par suite, après un temps fini, au moins l'un des deux termes de la somme va être égal à $0$, signifiant ainsi la perte de la mouche.\n\nb)\n\nLes deux araignées se placent tout d'abord sur le point $(1,1,1)$. Puis, en restant dans le plan $\\pi$ d'équation $z=1$ et par utilisation de la stratégie du a), l'une d'elles va attraper l'ombre de $M$ dans $\\pi$ (i.e. la projection orthogonale de $M$ sur $\\pi$). Sans perte de généralité, on peut supposer qu'il s'agit de $P$, et on a donc $P_{x}=M_{x}$ et $P_{y}=M_{y}$. Une fois ceci réalisé, $P$ calque ses déplacements sur ceux de $M$ de façon à conserver ces deux égalités. Toutefois, si $M$ se déplace parallèlement à l'axe des cotes ou reste immobile, alors $P$ se déplace parallèlement à l'axe des cotes, mais en allant vers $M$ (et on note qu'on a toujours $P_{z} \\leqslant M_{z}$). Puisque $M_{z}$ ne peut augmenter indéfiniment, $M$ ne peut échapper à $P$ qu'en ne faisant qu'un nombre fini de déplacements parallèlement à l'axe des cotes et en ne restant immobile qu'un nombre fini de fois. Par conséquent, à partir d'un instant $t_{1}$, la mouche $M$ ne se déplacera plus que dans un même plan $\\pi'$ d'équation $z=$ constante, et sans rester immobile.\n\nL'araignée $Q$ vient alors se placer dans le plan $\\pi'$. Quitte à rester immobile, elle peut rendre la quantité $f=|Q_{x}-M_{x}|+|Q_{y}-M_{y}|$ paire. À partir de cet instant, noté $t_{2}$, $Q$ suit la procédure suivante :\n- Si $|Q_{x}-M_{x}|>|Q_{y}-M_{y}|$ alors $Q$ se déplace parallèlement à l'axe des abscisses, de sorte que $|Q_{x}-M_{x}|$ n'augmente pas ($Q$ se déplace horizontalement, le long d'une droite $d$, en allant vers la projection de $M$ sur la droite $d$).\n- Sinon, $Q$ se déplace parallèlement à l'axe des ordonnées, dans des conditions analogues.\n\nPour se fixer les idées, on peut supposer que pour tout $t \\in \\mathbb{N}$, la mouche se déplace à l'instant $t+\\frac{1}{3}$ et l'araignée à l'instant $t+\\frac{2}{3}$. Puisque ni $M$ ni $Q$ ne reste immobile, le mouvement de $M$ fait augmenter ou diminuer $f$ d'une unité et le mouvement de $Q$ fait diminuer $f$ d'une unité, donc pour tout $t \\in \\mathbb{N}$, $f(t)$ est pair (en particulier, $Q$ ne se placera jamais sur un point voisin de $M$) et de plus $f(t+1) \\leqslant f(t)$.\n\nSoit $g=\\max\\left(|Q_{x}-M_{x}|,|Q_{y}-M_{y}|\\right)$. On observe que le mouvement de $Q$ ne peut que faire décroître $g$. De plus,\n- Si le mouvement de $M$ à l'instant $t+\\frac{1}{3}$ fait décroître $g$, alors $g(t+1) \\leqslant g(t)$.\n- Supposons que le mouvement de $M$ à l'instant $t+\\frac{1}{3}$ fait croître $g$. Si par exemple $M$ s'est déplacée dans la direction des $x$, nécessairement $|Q_{x}(t)-M_{x}(t)| \\geqslant |Q_{y}(t)-M_{y}(t)|$ et $M$ s'est éloignée de $Q$, donc après le mouvement de $M$ on a $|Q_{x}-M_{x}|>|Q_{y}-M_{y}|$. Par conséquent l'araignée se déplace dans le même sens que la mouche et on a $g(t+1)=g(t)$.\n\nDans tous les cas, on voit que $g(t+1) \\leqslant g(t)$ pour tout $t \\in \\mathbb{N}$, donc il existe $t_{2} \\in \\mathbb{N}$ tel que $g(t)=g(t_{2})$ et $f(t)=f(t_{2})$ pour tout $t \\geqslant t_{2}$. Supposons par l'absurde que $g(t_{2})>0$.\n\nQuitte à changer l'orientation des axes, on peut supposer que $Q_{x}(t_{2}) \\leqslant M_{x}(t_{2})$ et $Q_{y}(t_{2}) \\leqslant M_{y}(t_{2})$.\n- Premier cas: $Q_{y}(t_{2})=M_{y}(t_{2})$. Comme $f(t_{2})$ est pair, on a $M_{x}(t_{2}) \\geqslant Q_{x}(t_{2})+2$. Si entre les instants $t_{2}$ et $t_{2}+1$ la mouche se déplace dans la direction des $y$, alors l'araignée se déplace vers la droite, et $g(t_{2}+1)=g(t_{2})-1$. Impossible. Si la mouche se déplace vers la gauche, alors son déplacement fait diminuer $f$ d'une unité, donc $f(t_{2}+1)=f(t_{2})-2$. Impossible. On en déduit que la mouche se déplace vers la droite, et du coup l'araignée également. Par une récurrence immédiate, on obtient que $M_{x}(t_{2}+k)=M_{x}(t_{2})+k$ pour tout $k \\in \\mathbb{N}$, ce qui est impossible.\n- Deuxième cas: $Q_{x}(t_{2})=M_{x}(t_{2})$. Analogue.\n- Troisième cas: $Q_{x}(t_{2})<M_{x}(t_{2})$ et $Q_{y}(t_{2})<M_{y}(t_{2})$. Comme le déplacement de $Q$ fait diminuer strictement la quantité $f$, le déplacement de $M$ doit faire augmenter $f$ d'une unité. Par conséquent, entre les instants $t_{2}$ et $t_{2}+1$, la mouche doit se déplacer vers la droite ou vers le haut. On en déduit facilement que $Q_{x}(t_{2}+1)<M_{x}(t_{2}+1)$ et $Q_{y}(t_{2}+1)<M_{y}(t_{2}+1)$. Par une récurrence immédiate, la suite $\\left(M_{x}(t_{2}+k)+M_{y}(t_{2}+k)\\right)_{k \\in \\mathbb{N}}$ est strictement croissante, ce qui est impossible puisqu'elle est bornée et à valeurs entières.\n\nDans tous les cas on a abouti à une contradiction, ce qui prouve que $g(t_{2})=0$ et que l'araignée a rattrapé la mouche à l'instant $t_{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. On note $O$ le centre de son cercle circonscrit. Soient $D$, $E$, $F$ des points situés sur $[BC]$, $[CA]$ et $[AB]$ respectivement. On suppose que $FB = FD$ et $ED = EC$. Le cercle de centre $F$ et de rayon $FB$ et le cercle de centre $E$ et de rayon $EC$ se recoupent en $G$. Montrer que $A$, $F$, $O$, $E$, $G$ sont cocycliques.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLes triangles $BFD$ et $CED$ étant isocèles, on a $(DE, DF) = (DE, DC) + (DB, DF) = (CD, CE) + (BF, BD) = (BC, AC) + (AB, BC) = (AB, AC)$.\n\nComme $EG = ED$ et $FG = FD$, les points $E$ et $F$ appartiennent à la médiatrice de $[GD]$ donc le quadrilatère $FGE D$ est symétrique par rapport à $(FE)$. On en déduit que $(DE, DF) = (GF, GE)$ donc $(GF, GE) = (AB, AC) = (AF, AE)$. Il en résulte que $F$, $A$, $G$, $E$ sont cocycliques.\n\nOn a $2(\\overrightarrow{GB}, \\overrightarrow{GD}) = (\\overrightarrow{FB}, \\overrightarrow{FD}) = \\Pi - 2(\\overrightarrow{BD}, \\overrightarrow{BF})$ où $\\Pi$ est l'angle plat (cette dernière égalité provenant du fait que $FBD$ est isocèle en $F$). On montre de même que $2(\\overrightarrow{GD}, \\overrightarrow{GC}) = \\Pi - 2(\\overrightarrow{CE}, \\overrightarrow{CD})$. En additionnant ces deux égalités, on obtient\n$$\n\\begin{aligned}\n2(\\overrightarrow{GB}, \\overrightarrow{GC}) & = -2(\\overrightarrow{BD}, \\overrightarrow{BF}) - 2(\\overrightarrow{CE}, \\overrightarrow{CD}) \\\\\n& = -2(\\overrightarrow{BC}, \\overrightarrow{BA}) - 2(\\overrightarrow{CA}, \\overrightarrow{CB}) - 2(\\overrightarrow{CB}, \\overrightarrow{BC}) \\\\\n& = -2(\\overrightarrow{CA}, \\overrightarrow{BA}) = 2(\\overrightarrow{AB}, \\overrightarrow{AC})\n\\end{aligned}\n$$\nDonc $A$, $B$, $C$, $G$ sont cocycliques.\n\nComme $FG = FB$ et $OB = OG$, les points $G$ et $B$ sont symétriques par rapport à $(OF)$ donc $2(\\overrightarrow{OG}, \\overrightarrow{OF}) = (\\overrightarrow{OG}, \\overrightarrow{OB})$. De même, $2(\\overrightarrow{OE}, \\overrightarrow{OG}) = (\\overrightarrow{OC}, \\overrightarrow{OG})$. En additionnant ces deux égalités, il vient $2(\\overrightarrow{OE}, \\overrightarrow{OF}) = (\\overrightarrow{OC}, \\overrightarrow{OB}) = 2(\\overrightarrow{AC}, \\overrightarrow{AB}) = 2(\\overrightarrow{EA}, \\overrightarrow{AF})$, donc $A$, $O$, $E$, $F$ sont cocycliques.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProuver que si $n$ est un entier strictement positif, l'expression\n$$\n\\frac{\\sqrt{n+\\sqrt{0}}+\\sqrt{n+\\sqrt{1}}+\\sqrt{n+\\sqrt{2}}+\\cdots+\\sqrt{n+\\sqrt{n^{2}-1}}+\\sqrt{n+\\sqrt{n^{2}}}}{\\sqrt{n-\\sqrt{0}}+\\sqrt{n-\\sqrt{1}}+\\sqrt{n-\\sqrt{2}}+\\cdots+\\sqrt{n-\\sqrt{n^{2}-1}}+\\sqrt{n-\\sqrt{n^{2}}}}\n$$\nest indépendante de $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEn calculant le carré de chacun des deux membres, on déduit que, pour tous réels $a, b$ tels que $0 \\leqslant b \\leqslant a$, on a\n$$\n\\sqrt{a+\\sqrt{a^{2}-b^{2}}}=\\sqrt{\\frac{a+b}{2}}+\\sqrt{\\frac{a-b}{2}}\n$$\nEn particulier, pour tous entiers naturels $n$ et $m$, avec $m \\leqslant n^{2}$, on a\n$$\n\\sqrt{n+\\sqrt{m}}=\\sqrt{\\frac{n+\\sqrt{n^{2}-m}}{2}}+\\sqrt{\\frac{n-\\sqrt{n^{2}-m}}{2}}\n$$\nSoit $n>0$ un entier. On a donc\n$$\n\\sum_{m=0}^{n^{2}} \\sqrt{n+\\sqrt{m}}=\\sum_{m=0}^{n^{2}} \\sqrt{\\frac{n+\\sqrt{n^{2}-m}}{2}}+\\sum_{m=0}^{n^{2}} \\sqrt{\\frac{n-\\sqrt{n^{2}-m}}{2}}\n$$\nD'où, après réindexation :\n$$\n\\begin{aligned}\n\\sum_{m=0}^{n^{2}} \\sqrt{n+\\sqrt{m}} & =\\sum_{m=0}^{n^{2}} \\sqrt{\\frac{n+\\sqrt{m}}{2}}+\\sum_{m=0}^{n^{2}} \\sqrt{\\frac{n-\\sqrt{m}}{2}} \\\\\n& =\\frac{1}{\\sqrt{2}} \\sum_{m=0}^{n^{2}} \\sqrt{n+\\sqrt{m}}+\\frac{1}{\\sqrt{2}} \\sum_{m=0}^{n^{2}} \\sqrt{n-\\sqrt{m}}\n\\end{aligned}\n$$\nAinsi\n$$\n\\left(1-\\frac{1}{\\sqrt{2}}\\right) \\sum_{m=0}^{n^{2}} \\sqrt{n+\\sqrt{m}}=\\frac{1}{\\sqrt{2}} \\sum_{m=0}^{n^{2}} \\sqrt{n-\\sqrt{m}}\n$$\net donc\n$$\n\\frac{\\sum_{m=0}^{n^{2}} \\sqrt{n+\\sqrt{m}}}{\\sum_{m=0}^{n^{2}} \\sqrt{n-\\sqrt{m}}}=1+\\sqrt{2}\n$$\nqui est bien une valeur indépendante de $n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 30358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont tous les angles sont aigus, et dont l'angle $\\widehat{B}$ est de 60 degrés ($=\\pi/3$ radian). Les hauteurs $[AD]$ et $[CE]$ se coupent au point $H$. Prouver que le centre du cercle circonscrit au triangle $ABC$ est situé sur la bissectrice commune des angles $\\widehat{AHE}$ et $\\widehat{CHD}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPuisque $B, E, H, D$ sont sur le cercle de diamètre $[BH]$, on a $\\widehat{DHE} = 180^{\\circ} - \\widehat{EBD} = 120^{\\circ}$, donc $\\widehat{AHC} = 120^{\\circ}$. Comme $\\widehat{AOC} = 2\\widehat{ABC} = 120^{\\circ}$, on en déduit que $A, C, O, H$ sont cocycliques.\n\nPar conséquent, $\\widehat{CHO} = \\widehat{CAO} = \\frac{1}{2}(\\widehat{CAO} + \\widehat{OCA}) = \\frac{1}{2}\\left(180^{\\circ} - \\widehat{AOC}\\right) = 90^{\\circ} - \\widehat{ABC} = 30^{\\circ}$.\n\nDe plus, comme $(CH) \\perp (AB)$ et $(AH) \\perp (BC)$, on a $\\widehat{CHD} = \\widehat{ABC} = 60^{\\circ}$, donc $O$ est sur la bissectrice de $\\widehat{CHD}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMontrer que l'équation\n$$\nx(x+2)=y(y+1)\n$$\nn'a pas de solution en nombres entiers strictement positifs.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSi le couple $(x, y)$ est solution, on a $x(x+1)<x(x+2)=y(y+1)$ et donc $x<y$. Mais de même, en supposant toujours $(x, y)$ solution, on peut écrire $(x+1)(x+2)>x(x+2)=y(y+1)$, d'où on déduit $x+1>y$. Ainsi, une solution $(x, y)$ hypothétique devrait donc vérifier $x<y<x+1$, ce qui ne peut se produire. On en déduit que l'équation proposée n'a pas de solution en nombres entiers strictement positifs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe. Pour tout point $M$ à l'intérieur ou sur le bord de $ABCD$, on pose\n$$\nf(M) = MA + MB + MC + MD.\n$$\nProuver que, pour tout $M$, on a\n$$\nf(M) \\leqslant \\max(f(A), f(B), f(C), f(D))\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLemme. Si le point $K$ est sur les bords ou à l'intérieur du triangle $XYZ$, on a\n$$\nXY + XZ \\geqslant KY + KZ.\n$$\nPreuve du lemme. Si $K = Y$, le résultat est immédiat d'après l'inégalité triangulaire. Sinon, on note $P$ l'intersection de $(YK)$ avec le côté $[XZ]$. D'après l'inégalité triangulaire, on a alors\n$$\nXY + XZ = XY + XP + PZ \\geqslant YP + PZ = YK + KP + PZ \\geqslant YK + KZ.\n$$\n\nRevenons à l'exercice, et considérons un point $M$ à l'intérieur ou sur le bord du quadrilatère convexe $ABCD$. On va commencer par prouver que l'on peut trouver un point $P$ du bord de $ABCD$ tel que $f(M) \\leqslant f(P)$. Les diagonales $[AC]$ et $[BD]$ se coupent en $O$ et, sans perte de généralité, on peut supposer que $M$ est dans ou sur les bords du triangle $ADO$. Les droites $(BM)$ et $(CM)$ recoupent alors le côté $[AD]$ respectivement en $X$ et $Y$. Soit $P$ un point de $[XY]$. D'après le lemme, puisque $M$ est dans ou sur les bords du triangle $BDP$, on a $PB + PD \\geqslant MB + MD$. Et, puisque $M$ est dans ou sur les bords du triangle $CAP$, on a $PA + PC \\geqslant MA + MC$. En sommant, il vient $f(M) \\leqslant f(P)$, avec $P \\in [AD]$ comme désiré.\n\nToujours avec le point $P \\in [AD]$ ci-dessus, on montre maintenant que $f(P) \\leqslant f(A)$ ou $f(P) \\leqslant f(D)$, ce qui terminera la démonstration. En effet, puisque $P \\in [AD]$, on a $PA + PD = AD$. On veut maximiser $PB + PC$. Pour cela, on construit le symétrique $B'$ de $B$ par rapport à $(AD)$. On a donc $PB + PC = PB' + PC$. De plus, puisque $ABCD$ est convexe, selon que la droite $(B'C)$ rencontre $(AD)$ en un point de $[AD]$ ou non, le point $P$ est dans ou sur les bords de $AB'C$ ou de $DB'C$. D'après le lemme, on a alors $PB' + PC \\leqslant AB' + AC = AB + AC$ ou $PB' + PC \\leqslant DB' + DC = DB + DC$. Ainsi, on a\n$$\nf(P) = PA + PD + PB + PC \\leqslant AD + AB + AC = f(A)\n$$\nou\n$$\nf(P) = PA + PD + PB + PC \\leqslant AD + DB + DC = f(D),\n$$\ncomme annoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA l'extérieur du triangle $A B C$ on construit les deux points $X$ et $Y$ vérifiant :\n- le triangle $A X B$ est isocèle de base $[A B]$;\n- le triangle $B Y C$ est isocèle de base $[B C]$;\n$-\\widehat{A X B}+\\widehat{B Y C}=180^{\\circ}$.\nSoit $Z$ le milieu de $[A C]$. Montrer que les droites $(X Z)$ et $(Y Z)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $D$ et $E$ les milieux de $[B C]$ et $[A B]$.\n\n![](attached_image_1.png)\n\nOn a\n- $(X E) \\perp (D Z)$ car $(D Z) // (A B)$\n- $(E Z) \\perp (D Y)$ car $(E Z) // (B C)$\n- $\\frac{X E}{D Z}=\\frac{X E}{E A}=\\frac{C D}{D Y}=\\frac{E Z}{D Y}$ (la deuxième égalité provient du fait que les triangles rectangles $X E A$ et $C D Y$ sont semblables)\n\ndonc $X E Z$ et $Z D Y$ sont semblables. Comme $(X E) \\perp (D Z)$, on en déduit que $(X Z) \\perp (Z Y)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$\\mathbb{N}^{*}$ désigne l'ensemble des entiers supérieurs ou égaux à un. Trouver toutes les applications $f: \\mathbb{N}^{*} \\rightarrow \\mathbb{N}^{*}$ vérifiant :\n\ni) quel que soit $n$ appartenant à $\\mathbb{N}^{*}$, $f(n+f(n))=f(n)$\n\nii) $f(2013)=1$.",
    "options": [],
    "answer": "The unique solution is the constant function f(n) = 1 for all positive integers n.",
    "solution": "Solution:\n\nSoit $f$ une solution éventuelle.\nSoit $a \\in \\mathbb{N}^{*}$ tel que $f(a)=1$.\nAlors $f(a+1)=f(a+f(a))=f(a)=1$.\nPuisque $f(2013)=1$, on en déduit par récurrence que $f(n)=1$ pour tout entier $n \\geqslant 2013$.\n\nD'autre part, supposons que $a \\geqslant 2$ soit un entier tel que $f(n)=1$ pour tout $n \\geqslant a$.\nAlors, $a-1 \\geqslant 1$ et $f(a-1) \\geqslant 1$, d'où $a-1+f(a-1) \\geqslant a$ et donc d'après le premier paragraphe $f(a-1)=f(a-1+f(a-1))=1$.\nCela assure que $f(n)=1$ pour tout $n \\geqslant a-1$.\n\nPuisque l'on a vu que $f(n)=1$ pour tout $n \\geqslant 2013$, on en déduit par récurrence descendante que $f(n)=1$ pour tout $n \\geqslant 1$.\n\nRéciproquement, il est clair que la fonction constante $f: n \\longmapsto 1$ est une solution du problème.\n\nFinalement, la seule solution est la fonction constante $f: n \\longmapsto 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $(a_{n})$ une suite définie par $a_{1}, a_{2} \\in [0,100]$ et\n$$\na_{n+1} = a_{n} + \\frac{a_{n-1}}{n^{2}-1} \\quad \\text{ pour tout entier } n \\geqslant 2\n$$\n\nExiste-t-il un entier $n$ tel que $a_{n} > 2013$ ?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nLa réponse est non.\n\nPlus précisément, montrons par récurrence que l'on a $a_{n} \\leqslant 400$, pour tout $n \\geqslant 0$.\n\nL'inégalité est vraie pour $n=1$ et $n=2$, d'après l'énoncé.\n\nSupposons qu'elle soit vraie pour tout $k \\leqslant n$ pour un certain entier $n \\geqslant 2$.\n\nPour tout $k \\in \\{2, \\cdots, n\\}$, on a $a_{k+1} = a_{k} + \\frac{a_{k-1}}{k^{2}-1}$. En sommant, membre à membre, ces relations et après simplification des termes communs, il vient :\n$$\n\\begin{aligned}\na_{n+1} & = a_{2} + \\sum_{k=2}^{n} \\frac{a_{k-1}}{k^{2}-1} \\\\\n& \\leqslant 100 + \\sum_{k=2}^{n} \\frac{400}{k^{2}-1}, \\text{ d'après l'hypothèse de récurrence et l'énoncé } \\\\\n& = 100 + 200 \\sum_{k=2}^{n} \\left( \\frac{1}{k-1} - \\frac{1}{k+1} \\right) \\\\\n& = 100 + 200 \\left( 1 + \\frac{1}{2} - \\frac{1}{n} - \\frac{1}{n+1} \\right) \\text{ après simplification par dominos } \\\\\n& = 400 - \\frac{200}{n} - \\frac{200}{n+1}\n\\end{aligned}\n$$\net donc $a_{n+1} \\leqslant 400$, ce qui achève la récurrence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSur un échiquier $5$ sur $5$, on a placé sur des cases différentes $k$ cavaliers, de telle manière que chacun peut en prendre exactement $2$ autres.\n\nQuelle est la plus grande valeur possible de $k$?",
    "options": [],
    "answer": "16",
    "solution": "Solution:\n\nLa configuration suivante donne un exemple avec $16$ cavaliers :\n\n![](attached_image_1.png)\n\nOn montre maintenant que c'est optimal : quitte à échanger les couleurs, on peut supposer le centre noir. Notons $k$ le nombre de cavaliers sur une case noire, $l$ le nombre de cavaliers sur une case blanche et $N$ le nombre de manières de choisir deux cavaliers en prise : sur deux tels cavaliers, un est sur une case noire donc il y a $k$ manières de le choisir, puis $2$ manières de choisir un cavalier sur une case blanche qu'il peut prendre, donc $N = 2k$, mais le même raisonnement donne $N = 2l$ donc $k = l$.\n\nSi il y a un cavalier au centre de l'échiquier, il a accès à $8$ cases blanches donc au moins $6$ cases blanches sont inoccupées donc $l \\leqslant 12 - 6 = 6$ donc $k + l \\leqslant 12$. On peut donc supposer le centre vide.\n\nSi il y a un cavalier sur une case blanche qui touche le centre, il a accès à $6$ cases, donc $4$ d'entre elles sont inoccupées. Le centre l'est aussi, donc $k \\leqslant 13 - 5 = 8$ et $k + l \\leqslant 16$, ce qu'on veut montrer.\n\nSi enfin le centre et ses voisins immédiats sont inoccupés, alors les coins le sont aussi car ils ne peuvent avoir de cavaliers en prise, ce qui ne laisse que les $16$ cases de la figure ci-dessus.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $n$ et $m$ deux entiers strictement positifs. Montrer que $5^{m}+5^{n}$ s'écrit comme une somme de deux carrés si et seulement si $n$ et $m$ ont même parité.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSupposons pour commencer que $n$ et $m$ n'ont pas même parité. On étudie alors la quantité $5^{n}+5^{m}$ modulo $8$. Par récurrence, on démontre facilement que $5^{k}$ est congru à $1$ (resp. $5$) modulo $8$ si $k$ est pair (resp. impair). Ainsi, d'après notre hypothèse, on a toujours $5^{n}+5^{m} \\equiv 6 \\pmod{8}$. Or, par ailleurs, un carré est toujours congru à $0$, $1$ ou $4$ modulo $8$. Comme il n'y a aucune manière d'écrire $6$ comme somme de deux de ces trois nombres, on en déduit que $5^{n}+5^{m}$ ne peut s'écrire comme somme de deux carrés.\n\nSupposons maintenant que $n$ et $m$ ont même parité. Quitte à échanger $n$ et $m$, on peut supposer en outre que $m \\geqslant n$. Ainsi, il existe un entier $k$ positif ou nul tel que $m=n+2k$. On a alors $5^{m}+5^{n}=5^{n}\\left(5^{2k}+1\\right)$. Comme, manifestement, $5=2^{2}+1^{2}$ et $5^{2k}+1$ s'écrivent comme sommes de deux carrés, il suffit pour conclure de montrer que, si $x$ et $y$ s'écrivent comme somme de deux carrés, alors il en va de même du produit $xy$. Ceci résulte directement de la formule :\n$$\n\\left(a^{2}+b^{2}\\right)\\left(c^{2}+d^{2}\\right)=(ad-bc)^{2}+(ac+bd)^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn considère une grille carrée de $4 \\times 4$ cases.\nDeux cases distinctes ayant un côté commun sont dites voisines. Initialement, toutes les cases sont rouges. Par la suite, une case pourra éventuellement changer de couleur et être soit rouge soit bleue. Effectuer une opération sur la case $c$ signifie que l'on change simultanément la couleur de $c$ ainsi que celles de toutes ses voisines (les cases rouges deviennent bleues et les cases bleues deviennent rouges).\nDonner toutes les valeurs de $n$ pour lesquelles il existe un groupe de $n$ cases distinctes telles qu'après avoir effectué une opération sur chacune de ces $n$ cases, on obtienne une grille entièrement bleue.",
    "options": [],
    "answer": "4, 6, 8, 10, 12",
    "solution": "Solution:\nRegroupons les cases de la grille en quatre tuiles, contenant chacune une case marquée, comme indiqué dans la figure suivante :\n![](attached_image_1.png)\nIl est facile de vérifier qu'une opération quelconque change la couleur d'une et une seule de ces cases marquées et qu'une opération effectuée dans une tuile ne modifie pas la couleur des cases marquées des autres tuiles. Si l'on veut rendre la grille entièrement bleue, il faut donc que, dans chaque tuile, le nombre d'opérations effectuées soit impair. La somme de quatre nombres impairs étant paire, cela signifie que le nombre total $n$ d'opérations doit être pair. De plus, puisque dans chaque tuile il y a quatre cases et que les opérations doivent être effectuées sur des cases distinctes, c'est que dans chaque tuile on a fait une ou trois opérations. Par suite, on a $4 \\leqslant n \\leqslant 12$.\nLes seules valeurs de $n$ possibles sont donc $n=4,6,8,10,12$. Il reste à prouver que ces cinq valeurs sont bien des solutions. Pour cela, on pourra se reporter à la figure ci-dessus et ses quatre cases marquées et aux figures ci-dessous (les points indiquent les cases sur lesquelles on effectue une opération).\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n1) Soient $a$ et $b$ deux nombres réels tels que $a^{n}+b^{n}$ est un entier pour $n=1,2,3,4$. Montrer que $a^{n}+b^{n}$ est un entier pour tout $n \\in \\mathbb{N}^{*}$.\n\n2) Est-il vrai que si $a$ et $b$ sont deux nombres réels tels que $a^{n}+b^{n}$ est un entier pour $n=1,2,3$ alors $a^{n}+b^{n}$ est un entier pour tout $n \\in \\mathbb{N}^{*}$ ?",
    "options": [],
    "answer": "Part 1: True; the sums are integers for all positive exponents. Part 2: False; for example, taking 1 plus and minus one over the square root of two gives integer sums for the first three exponents but not for the fourth.",
    "solution": "Solution:\n1) Notons $s=a+b$ et $p=a b$. Soit $S_{n}=a^{n}+b^{n}$. Par hypothèse, $S_{n}$ est un entier pour $1 \\leqslant n \\leqslant 4$. Comme pour tout $n \\geqslant 1$ on a\n$$\nS_{n+1}=a^{n+1}+b^{n+1}=(a+b)\\left(a^{n}+b^{n}\\right)-a b\\left(a^{n-1}+b^{n-1}\\right)=S_{1} S_{n}-p S_{n-1},\n$$\nsi on montre que $p$ est un entier alors par récurrence immédiate il s'ensuivra que $S_{n}$ est un entier pour tout $n$.\nOn a\n$$\nS_{2}=(a+b)^{2}-2 a b=s^{2}-2 p\n$$\ndonc $2 p=S_{1}^{2}-S_{2}$ est un entier. On a\n$$\nS_{4}=\\left(a^{2}+b^{2}\\right)^{2}-2(a b)^{2}=S_{2}^{2}-2 p^{2}\n$$\ndonc $2 p^{2}=S_{2}^{2}-S_{4}$ est un entier. On en déduit que $(2 p)^{2}=2\\left(2 p^{2}\\right)$ est un entier pair. Comme $2 p$ est entier, il est pair donc $p$ est bien entier.\n\n2) On a\n$$\nS_{3}=(a+b)\\left(a^{2}+b^{2}\\right)-a b^{2}-a^{2} b=S_{1} S_{2}-p s\n$$\ndonc si $p$ est un demi-entier (c'est-à-dire que $p-\\frac{1}{2}$ est un entier) et si $s$ est un entier pair, alors $S_{1}, S_{2}, S_{3}$ sont des entiers alors que $S_{4}$ est un demi-entier. Par exemple, si $a=1+\\frac{1}{\\sqrt{2}}$ et $b=1-\\frac{1}{\\sqrt{2}}$ alors $s=2$, $p=\\frac{1}{2}, S_{1}=2, S_{2}=3, S_{3}=5, S_{4}=17 / 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer la plus grande valeur possible et la plus petite valeur possible de\n$$\n\\sqrt{4-a^{2}}+\\sqrt{4-b^{2}}+\\sqrt{4-c^{2}}\n$$\nlorsque $a, b, c$ sont des réels strictement positifs vérifiant $a^{2}+b^{2}+c^{2}=6$.",
    "options": [],
    "answer": "maximum = 3√2, minimum = 2+√2",
    "solution": "Solution:\n\nTout d'abord, on note que si l'on veut que l'expression ait un sens, il faut $a, b, c \\in [0; 2]$.\n\nD'après l'inégalité de Cauchy-Schwarz, on a\n$$\n\\left(\\sqrt{4-a^{2}}+\\sqrt{4-b^{2}}+\\sqrt{4-c^{2}}\\right)^{2} \\leqslant 3\\left(4-a^{2}+4-b^{2}+4-c^{2}\\right)=18,\n$$\nc'est-à-dire\n$$\n\\sqrt{4-a^{2}}+\\sqrt{4-b^{2}}+\\sqrt{4-c^{2}} \\leqslant 3 \\sqrt{2},\n$$\navec égalité pour $a=b=c=\\sqrt{2}$.\n\nAinsi, la plus grande valeur possible est $3 \\sqrt{2}$.\n\nCherchons maintenant la valeur minimale :\n\nSans perte de généralité, on peut supposer que $a \\leqslant b \\leqslant c$.\nDe $a^{2}+b^{2}+c^{2}=6$, on déduit alors que $3 a^{2} \\leqslant 6$, soit donc $0 \\leqslant a^{2} \\leqslant 2$.\n\nD'autre part, si $x, y \\geqslant 0$, on a clairement $\\sqrt{x}+\\sqrt{y} \\leqslant \\sqrt{x+y}$, avec égalité si et seulement si $x=0$ ou $y=0$.\n\nIl vient alors\n$$\n\\sqrt{4-a^{2}}+\\sqrt{4-b^{2}}+\\sqrt{4-c^{2}} \\geqslant \\sqrt{4-a^{2}}+\\sqrt{8-b^{2}-c^{2}}=\\sqrt{4-a^{2}}+\\sqrt{2+a^{2}},\n$$\net, puisque $0 \\leqslant 4-c^{2} \\leqslant 4-b^{2}$, égalité a lieu si et seulement si $4-c^{2}=0$, c'est-à-dire $c=2$.\n\nIl reste donc à trouver le minimum de l'expression $f(x)=\\sqrt{4-x}+\\sqrt{2+x}$, lorsque $x \\in[0 ; 2]$. Or, puisque tout est positif, cela revient à trouver le minimum de $(f(x))^{2}=6+2 \\sqrt{(4-x)(2+x)}$, sous les mêmes conditions.\n\nComme $(4-x)(2+x)=-x^{2}+2 x+8=9-(x-1)^{2}$, la valeur minimale de $\\sqrt{(4-x)(2+x)}$ est $\\sqrt{8}$, avec égalité pour $x=0$.\n\nAinsi, on a $(f(x))^{2} \\geqslant 6+2 \\sqrt{8}=(2+\\sqrt{2})^{2}$, ou encore $f(x) \\geqslant 2+\\sqrt{2}$, avec égalité pour $x=0$.\n\nPar suite, on a $\\sqrt{4-a^{2}}+\\sqrt{4-b^{2}}+\\sqrt{4-c^{2}} \\geqslant 2+\\sqrt{2}$, avec égalité en particulier pour $(a, b, c)=(0, \\sqrt{2}, 2)$.\n\nCela assure que la plus petite valeur cherchée est $2+\\sqrt{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver le plus grand entier $n \\geqslant 3$, vérifiant:\n\"pour tout entier $k \\in \\{2,3, \\cdots, n\\}$ si $k$ et $n$ sont premiers entre eux alors $k$ est un nombre premier.\"",
    "options": [],
    "answer": "30",
    "solution": "Solution:\n\nOn remarque d'abord que $n=30$ vérifie la propriété. En effet, si $k>1$ est premier avec $n$, alors il est premier avec $2,3,5$. Si de plus $k$ n'est pas premier, alors il admet une factorisation non triviale $k=lm$ avec $\\ell, m>1$. Comme $\\ell$, $m$ sont premiers avec $n$, ils sont premiers avec $2,3,5$, donc $\\ell$, $m \\geqslant 7$, ce qui entraîne que $k \\geqslant 7 \\times 7=49>n$.\n\nRéciproquement, montrons que si $n$ vérifie la propriété alors $n \\leqslant 30$. Supposons par l'absurde que $n>30$. Soit $p$ le plus petit entier premier ne divisant pas $n$. Comme $p^{2}$ n'est pas premier mais est premier avec $n$, on a $p^{2}>n>30>5^{2}$ donc $p \\geqslant 7$. En particulier, $2,3$ et $5$ divisent $n$ donc $30$ divise $n$. Comme $n>30$, on en déduit que $p^{2}>n \\geqslant 60>7^{2}$ donc $p>7$, par conséquent $p \\geqslant 11$.\n\nNotons $p_{1}<p_{2}<p_{3}<\\cdots$ la liste des nombres premiers ($p_{1}=2$, $p_{2}=3$, etc.). Ce qui précède montre que $p=p_{k+1}$ où $k \\geqslant 4$. De plus, $p_{1}, \\ldots, p_{k}$ divisent $n$ donc $p_{k+1}^{2}>n \\geqslant p_{1} \\cdots p_{k}$, ce qui contredit l'inégalité de Bonse.\n\nRemarque: si on ne connaît pas l'inégalité de Bonse, on la retrouve facilement à partir du postulat de Bertrand qui dit que $p_{j+1}<2 p_{j}$ pour tout $j$. En effet,\n$$\np_{k+1}^{2}<4 p_{k}^{2}<8 p_{k-1} p_{k}<2 \\times 3 \\times 5 \\times p_{k-1} p_{k} \\leqslant p_{1} p_{2} \\cdots p_{k}\n$$\nsi $k \\geqslant 5$. De plus, si $k=4$ on vérifie directement que $p_{k+1}^{2}=121<210=2 \\times 3 \\times 5 \\times 7=p_{1} \\cdots p_{k}$.\n\nRemarque : il existe une démonstration élémentaire de l'inégalité de Bonse n'utilisant pas le postulat de Bertrand. On vérifie d'abord à la main que si $4 \\leqslant n \\leqslant 7$ alors $p_{1} p_{2} \\cdots p_{n}>p_{n+1}^{2}$.\n\nSupposons par l'absurde qu'il existe $n \\geqslant 8$ tel que $p_{1} p_{2} \\cdots p_{n} \\leqslant p_{n+1}^{2}$. Soit $m=\\left[\\frac{n}{2}\\right]$. On a\n$$\n\\left(p_{1} p_{2} \\cdots p_{m}\\right)^{2}<p_{1} p_{2} \\cdots p_{n} \\leqslant p_{n+1}^{2}\n$$\ndonc $p_{1} p_{2} \\cdots p_{m}<p_{n+1}$.\n\nConsidérons les entiers $N_{j}=j p_{1} p_{2} \\cdots p_{m-1}-1$ ($1 \\leqslant j \\leqslant p_{m}$). Pour tout $j$, on a $N_{j}<p_{1} p_{2} \\cdots p_{m}<p_{n+1}$ et $N_{j}$ est premier avec $p_{1}, p_{2}, \\ldots, p_{m-1}$. Donc si $q_{j}$ est le plus petit entier premier divisant $N_{j}$, on a $p_{m} \\leqslant q_{j} \\leqslant p_{n}$.\n\nLes $q_{j}$ sont distincts car si $j<\\ell$ et $q_{j}=q_{\\ell}$, alors $q_{j}$ divise $N_{\\ell}-N_{j}=(\\ell-j) p_{1} p_{2} \\cdots p_{m}$, donc $q_{j}$ divise $\\ell-j$, ce qui est impossible puisque $1 \\leqslant \\ell-j<p_{m} \\leqslant q_{j}$.\n\nPar conséquent, il y a au moins $p_{m}$ nombres premiers distincts compris entre $p_{m}$ et $p_{n}$ : autrement dit, $p_{m} \\leqslant n-m+1$. Or, $n \\leqslant 2 m+1$ donc $p_{m} \\leqslant m+2$.\n\nComme $p_{m} \\geqslant 2 m-1$ pour tout $m \\geqslant 1$, cela entraîne $2 m-1 \\leqslant m+2$, donc $m \\leqslant 3$, et donc $n \\leqslant 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCDEF$ un hexagone régulier et $M \\in [A, C], N \\in [C, E]$. On suppose que $\\frac{AM}{AC}$ et $\\frac{CN}{CE}$ sont égaux à un nombre $r > 0$, et que $B, M, N$ sont colinéaires. Déterminer la valeur de $r$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "1/sqrt(3)",
    "solution": "Solution:\n\nOn peut supposer que l'hexagone est inscrit dans un cercle de rayon $1$. On se place dans un repère tel que les coordonnées de $A, B, C$ sont respectivement $A = (0, 1)$, $B = \\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$ et $C = \\left(\\frac{\\sqrt{3}}{2}, -\\frac{1}{2}\\right)$.\n\nComme $CE = \\sqrt{3}$, on a $CN = r \\sqrt{3}$ donc $\\overrightarrow{CN} = (-r \\sqrt{3}, 0)$. On en déduit que\n$$\nN = \\left(\\frac{\\sqrt{3}}{2} - r \\sqrt{3}, -\\frac{1}{2}\\right)\n$$\nComme $\\overrightarrow{AC} = \\left(\\frac{\\sqrt{3}}{2}, -\\frac{3}{2}\\right)$, on a $\\overrightarrow{AM} = \\left(\\frac{\\sqrt{3}}{2} r, -\\frac{3r}{2}\\right)$ donc\n$$\nM = \\left(\\frac{\\sqrt{3}}{2} r, 1 - \\frac{3r}{2}\\right)\n$$\nOn calcule aisément que\n$$\n\\overrightarrow{BM} = \\left(\\frac{\\sqrt{3}}{2}(r-1), \\frac{1}{2} - \\frac{3r}{2}\\right), \\quad \\overrightarrow{BN} = (-r \\sqrt{3}, -1)\n$$\nLe fait que $B, M, N$ sont alignés signifie que ces deux vecteurs sont colinéaires, ce qui équivaut à\n$$\n\\frac{\\sqrt{3}}{2}(1 - r) + r \\sqrt{3}\\left(\\frac{1}{2} - \\frac{3r}{2}\\right) = 0\n$$\nce qui se simplifie en $r = \\frac{1}{\\sqrt{3}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\in \\mathbb{N}^{*}$. Montrer qu'il est possible de partitionner $\\{1,2, \\ldots, n\\}$ en deux sous-ensembles $A$ et $B$ tels que la somme des éléments de $A$ soit égale au produit des éléments de $B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nA priori, l'ensemble $B$ doit être bien plus petit que $A$. On va donc chercher un $B$ avec un petit nombre d'éléments. En testant des petites valeurs de $n$, on peut penser à chercher $B$ sous la forme $\\{1, a, b\\}$ avec $a, b \\neq 1$ et $a \\neq b$.\nOn a alors $\\prod_{x \\in B} x = a b$, et :\n$$\n\\begin{aligned}\n\\sum_{x \\in \\mathcal{A}} & = \\sum_{x=1}^{n} x - 1 - a - b \\\\\n& = \\frac{n(n+1)}{2} - a - b - 1\n\\end{aligned}\n$$\nOn veut donc $\\frac{n(n+1)}{2} = a b + a + b + 1 = (a+1)(b+1)$, donc il suffit d'écrire $\\frac{n(n+1)}{2}$ comme un produit. Or, si $n$ est pair, on peut prendre $a+1 = \\frac{n}{2}$ et $b+1 = n+1$, soit $a = \\frac{n}{2} - 1$ et $b = n$. Si $n$ est impair, on prend $a = \\frac{n-1}{2}$ et $b = n-1$. Dans les deux cas, l'hypothèse $n \\geqslant 5$ permet de vérifier facilement que $1$, $a$ et $b$ sont deux à deux distincts.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLe sommet $B$ d'un angle $\\widehat{A B C}$ se trouve à l'extérieur d'un cercle $\\omega$ tandis que les demi-droites $[B A)$ et $[B C)$ le traversent. Soit $K$ un point d'intersection du cercle $\\omega$ avec $[B A)$. La perpendiculaire à la bissectrice de l'angle $\\widehat{A B C}$ passant par $K$ recoupe le cercle au point $P$, et la droite $(B C)$ au point $M$. Montrer que le segment $[P M]$ est deux fois plus long que la distance entre le centre de $\\omega$ et la bissectrice de $\\widehat{A B C}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nSoit $O$ le centre du cercle $\\omega$. Notons $\\Delta$ la bissectrice de $\\widehat{A B C}$, $\\Delta'$ la parallèle à $\\Delta$ passant par $O$ et $N$ le symétrique de $O$ par rapport à $\\Delta$.\nLa symétrie $s_{\\Delta'}$ par rapport à $\\Delta'$ envoie $O$ sur $O$ et $P$ sur $K$ (car $(P K) \\perp \\Delta'$ et $\\Delta'$ passe par $O$).\nLa symétrie $s_{\\Delta}$ par rapport à $\\Delta$ envoie $O$ sur $N$ et $K$ sur $M$.\nPar conséquent, la composée $s_{\\Delta} \\circ s_{\\Delta'}$ envoie $O$ sur $N$ et $P$ sur $M$. Or, $s_{\\Delta} \\circ s_{\\Delta'}$ est une translation, donc $M N O P$ est un parallélogramme. Il s'ensuit que $P M = O N = 2 d(O, \\Delta)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nExiste-t-il des nombres rationnels positifs ou nuls $x, y$ et $z$ tels que:\n$$\nx^{5}+2 y^{5}+5 z^{5}=11\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNous allons montrer qu'il n'existe pas de tels rationnels $x, y, z$. On raisonne par l'absurde en supposant qu'il en existe. Soit $d$ le plus petit dénominateur commun de $x, y$ et $z$. On peut alors écrire $x=\\frac{a}{d}$, $y=\\frac{b}{d}$ et $z=\\frac{c}{d}$ pour certains entiers $a, b$ et $c$. L'équation que l'on cherche à résoudre devient alors :\n$$\na^{5}+2 b^{5}+5 c^{5}=11 d^{5}\n$$\nÉtudions cette équation modulo $11$. Une recherche exhaustive (ou l'utilisation du petit théorème de Fermat) montre qu'une puissance $5$-ième est congrue à $0, 1$ ou $-1$ modulo $11$. On en déduit que la congruence $a^{5}+2 b^{5}+5 c^{5} \\equiv 0 \\pmod{11}$ implique que $a, b$ et $c$ sont tous les trois multiples de $11$. Ainsi, $a^{5}+2 b^{5}+5 c^{5}$ est divisible par $11^{5}$, d'où on déduit que $d$ est lui aussi divisible par $11$. Les fractions $\\frac{a}{d}, \\frac{b}{d}$ et $\\frac{c}{d}$ peuvent donc, toutes les trois, être simplifiées par $11$. Ceci contredit la minimalité de $d$ et termine la démonstration.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier $n \\geqslant 1$, on pose $S_{n} = x_{1} + x_{2} + \\cdots + x_{n}$, où $x_{k} = \\frac{k(k+1)}{2}$ pour tout $k \\geqslant 1$. Prouver que, pour tout $n \\geqslant 10$, il existe un entier $a_{n}$ tel que $S_{n-1} < a_{n}^{2} < S_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour tout entier $n \\geqslant 1$, on a\n$$\nS_{n} = \\sum_{k=1}^{n} \\frac{k(k+1)}{2} = \\frac{1}{2} \\sum_{k=1}^{n} k^{2} + \\frac{1}{2} \\sum_{k=1}^{n} k = \\frac{n(n+1)(2n+1)}{12} + \\frac{n(n+1)}{4} = \\frac{n(n+1)(n+2)}{6}\n$$\nIl suffit donc de prouver que, pour $n \\geqslant 10$, on a $\\sqrt{S_{n}} - \\sqrt{S_{n-1}} > 1$, car alors l'intervalle $] S_{n-1}, S_{n} [$ contiendra forcément un entier.\n\nOr, d'après le calcul ci-dessus, l'inégalité $\\sqrt{S_{n}} - \\sqrt{S_{n-1}} > 1$ est équivalente à\n$$\n\\sqrt{\\frac{n(n+1)(n+2)}{6}} - \\sqrt{\\frac{n(n+1)(n-1)}{6}} > 1\n$$\nou encore\n$$\n\\sqrt{\\frac{n(n+1)}{6}} (\\sqrt{n+2} - \\sqrt{n-1}) > 1\n$$\nPuisque\n$$\n\\sqrt{n+2} - \\sqrt{n-1} = \\frac{3}{\\sqrt{n+2} + \\sqrt{n-1}}\n$$\non veut prouver que\n$$\n\\sqrt{3 n(n+1)} > \\sqrt{2(n+2)} + \\sqrt{2(n-1)}\n$$\nPour cela, il suffit de montrer que $\\sqrt{3 n(n+1)} > 2 \\sqrt{2(n+2)}$. Cette dernière inégalité s'écrit aussi $3 n^{2} - 5 n - 16 > 0$. Il est facile de vérifier qu'elle est satisfaite pour $n \\geqslant 10$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn se donne $n$ points du plan, tels que trois quelconques d'entre eux ne sont jamais alignés. Chacun est colorié en rouge ou en bleu. On suppose qu'il y a exactement un point bleu à l'intérieur de chaque triangle dont les sommets sont rouges, et un point rouge à l'intérieur de chaque triangle dont les sommets sont bleus.\nQuelle est la plus grande valeur possible de $n$?",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nOn montre d'abord que les points bleus forment un polygone convexe : si ce n'était pas le cas, on pourrait trouver un point bleu à l'intérieur d'un polygone bleu, donc en découpant ce polygone en triangles, on pourrait trouver un point bleu à l'intérieur d'un triangle bleu. Le grand triangle bleu doit contenir un unique point rouge, mais on peut le découper en trois triangles qui contiennent chacun un unique point rouge, ce qui est absurde.\n\nLes points bleus forment donc un polygone convexe, de même que les rouges. De plus, il y a au maximum 2 points rouges à l'intérieur du polygone bleu, car s'il y en avait 3 ils formeraient un triangle sans point bleu à l'intérieur.\n\nEnfin, si $k$ est le nombre de points bleus, on peut découper le polygone bleu en $k-2$ triangles, qui chacun doivent contenir un point rouge, donc $k-2 \\leqslant 2$ et $k \\leqslant 4$. De même, il y a au plus 4 points rouges donc $n \\leqslant 8$. On peut alors vérifier que la configuration à 8 points suivante convient :\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPartant d'un triplet d'entiers relatifs $(x, y, z)$, une opération consiste à ajouter à l'un de ces trois entiers un multiple de l'un des deux autres (ce multiple peut être positif ou négatif). Prouver que si $a, b, c$ sont des entiers premiers entre eux dans leur ensemble, on peut passer du triplet $(a, b, c)$ au triplet $(1,0,0)$ en au plus cinq opérations.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSi $b$ et $c$ sont nuls alors, puisque $\\operatorname{pgcd}(a, b, c)=1$, on doit avoir $a=1$ ou $a=-1$. Dans le premier cas, il n'y a rien à faire. Dans le second, par trois opérations successives, on peut passer de $(-1,0,0)$ à $(-1,1,0)$, puis à $(1,1,0)$ et enfin à $(1,0,0)$.\n\nOn peut donc supposer que $b$ ou $c$ est non nul.\n\nL'objectif est d'atteindre un triplet de la forme $(1, s, t)$ en au plus trois étapes, puisqu'avec deux étapes supplémentaires, il est alors immédiat d'obtenir $(1,0,0)$.\n\nOn suppose que $c \\neq 0$, le cas $b \\neq 0$ se traitant de façon analogue.\n\nAfin d'atteindre l'objectif ci-dessus, il suffit de trouver un entier $n$ tel que $c$ et $b+na$ soient premiers entre eux : en effet, supposons pour le moment qu'un tel entier $n$ ait été déterminé et voyons comment la conclusion va en découler.\n\nOn pose $b' = b + n a$ et, avec une première opération, on peut passer de $(a, b, c)$ à $(a, b', c)$. Ensuite, d'après le théorème de Bézout, on sait qu'il existe deux entiers $u$ et $v$ tels que $u b' + v c = 1$. Après multiplication par $a-1$, il existe donc deux entiers $x$ et $y$ tels que $x b' + y c = a-1$. On utilise alors une seconde opération pour passer de $(a, b', c)$ à $(a - x b', b', c)$, puis une troisième pour passer de $(a - x b', b', c)$ à $(a - x b' - y c, b', c) = (1, b', c)$.\n\nPour conclure, il ne reste donc plus qu'à prouver qu'un tel entier $n$ existe.\n\nSoit $E$ l'ensemble des diviseurs premiers de $c$ qui divisent également au moins un nombre de la forme $b + k a$, où $k \\in \\mathbb{Z}$.\n\nSi $E = \\emptyset$ alors tout entier $n$ est tel que $\\operatorname{pgcd}(c, b + n a) = 1$.\n\nSinon, on note tout d'abord que $E$ est fini car il ne contient que des diviseurs premiers de $c$, avec $c \\neq 0$. Ensuite, pour chaque $p \\in E$, il existe un entier $k_p$ tel que $b + k_p a = 0 \\bmod p$. On remarque qu'un tel nombre $p$ ne divise pas $a$ car sinon $p$ diviserait aussi $b$, en contradiction avec $\\operatorname{pgcd}(a, b, c) = 1$.\n\nD'après le théorème chinois, il existe un entier $n$ tel que $n = k_p + 1 \\bmod p$ pour tout $p \\in E$. Prouvons que cet entier $n$ convient :\n\n- Si $p$ divise $c$ et $p \\notin E$ alors $p$ ne divise aucun nombre de la forme $b + k a$ et donc, en particulier, ne divise pas $b + n a$.\n- Si $p \\in E$ alors $b + n a = b + k_p a + a = a \\neq 0 \\bmod p$.\n\nAinsi, aucun diviseur premier de $c$ ne divise $b + n a$, ce qui assure que $c$ et $b + n a$ sont bien premiers entre eux.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère un ensemble fini $E$ de garçons et de filles.\n\nUne partie $G$ de garçons de $E$ est dite populaire si chaque fille de $E$ connaît au moins un des garçons de $G$. De même, une partie $F$ de filles est dite populaire si chaque garçon de $E$ connaît au moins une des filles de $F$. On suppose que si $a$ connaît $b$ alors $b$ connaît $a$.\n\nProuver que le nombre de parties populaires formées par les garçons a la même parité que le nombre de parties populaires formées par les filles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPremière solution. On dira qu'un ensemble $X$ de garçons est étranger à un groupe $Y$ de filles si aucun garçon de $X$ n'est connu d'une des filles de $Y$. De même, un ensemble $Y$ de filles sera dit étranger à un groupe $X$ de garçons si aucune fille de $Y$ n'est connue d'un des garçons de $X$. Puisque la relation \"se connaître\" est supposée symétrique, on en déduit que $X$ est étranger à $Y$ si et seulement si $Y$ est étranger à $X$. Cela ouvre la porte à un double-comptage :\n\nOn note $N$ le nombre de couples $(X, Y)$ de groupes $X$ de garçons et $Y$ de filles mutuellement étrangers. Soit $X$ un groupe de garçons. On note $Y_{X}$ le plus grand groupe de filles qui soit étranger à $X$. Clairement, tout groupe $Y$ de filles qui est étranger à $X$ est contenu dans $Y_{X}$ et, réciproquement, tout sous-ensemble de $Y_{X}$ est étranger à $X$. Par suite, le nombre de groupes $Y$ qui sont étrangers à $X$ est $2^{|Y_{X}|}$ (où $|A|$ désigne le nombre d'éléments de l'ensemble $A$). On en déduit que\n$$\nN = \\sum_{X} 2^{|Y_{X}|}\n$$\nOr, pour tout groupe $X$ de garçons, on a $X$ populaire si et seulement si $Y_{X} = \\emptyset$, c'est-à-dire $2^{|Y_{X}|} = 1$. Ainsi, le nombre de groupes populaires de garçons est congru à $N$ modulo $2$.\n\nUn raisonnement analogue montre que le nombre de groupes populaires de filles est lui aussi congru à $N$ modulo $2$, ce qui conclut.\n\n\nDeuxième solution. Soit $G$ l'ensemble des garçons et $F$ celui des filles. Nous allons prouver le résultat par récurrence sur $n = |G|$. On note tout d'abord que si $F$ ou $G$ est vide, la conclusion est immédiate (il n'existe aucun groupe populaire). Cela assure le résultat pour $n = 0, 1$ et on considère maintenant un groupe de $n + 1 \\geqslant 2$ personnes dans lequel il y a au moins une fille et au moins un garçon.\n\nSoit $g$ un des garçons. On pose $G' = G - \\{g\\}$ et $F'$ l'ensemble des filles qui ne connaissent pas $g$. Alors :\n- un groupe populaire de garçons dans $G' \\cup F$ l'est encore dans $G \\cup F$,\n- un groupe populaire de garçons dans $G \\cup F$ qui ne l'est plus dans $G' \\cup F$ est la réunion d'un groupe populaire de garçons dans $G' \\cup F'$ et de $\\{g\\}$.\n\nIl en découle que le nombre $N(G, F)$ de groupes populaires de garçons dans $G \\cup F$ est la somme du nombre $N(G', F)$ de groupes populaires de garçons dans $G' \\cup F$ et du nombre $N(G', F')$ de groupes populaires de garçons dans $G' \\cup F'$, soit donc $N(G, F) = N(G', F) + N(G', F')$.\n\nD'autre part :\n- un groupe populaire de filles dans $G \\cup F'$ est encore dans $G' \\cup F$,\n- un groupe populaire de filles dans $G' \\cup F$ qui ne l'est plus dans $G \\cup F$ est un groupe populaire de filles dans $G' \\cup F'$.\n\nPar conséquent, le nombre $M(G, F)$ de groupes populaires de filles dans $G \\cup F$ est la différence entre le nombre $M(G', F)$ de groupes populaires de filles dans $G' \\cup F$ et le nombre $M(G', F')$ de groupes populaires de filles dans $G' \\cup F'$, soit donc $M(G, F) = M(G', F) - M(G', F')$.\n\nOr, d'après l'hypothèse de récurrence, on a $N(G', F) \\equiv M(G', F) \\pmod{2}$ et $N(G', F') \\equiv M(G', F') \\equiv -M(G', F') \\pmod{2}$, d'où $N(G, F) \\equiv M(G, F) \\pmod{2}$, ce qui achève la récurrence.\n\n\nTroisième solution. On note toujours respectivement $F$ et $G$ l'ensemble des filles et celui des garçons. Pour chaque fille $f$ et chaque garçon $g$, on pose $h(f, g) = 0$ si $f$ et $g$ se connaissent, et $h(f, g) = 1$ sinon. Alors, un ensemble $X$ de garçons est populaire si et seulement si\n$$\n\\prod_{f \\in F} \\left(1 - \\prod_{g \\in X} h(f, g)\\right) = 1.\n$$\nPar suite, le nombre $N$ de groupes populaires de garçons est\n$$\n\\begin{aligned}\nN &= \\sum_{X \\subseteq G} \\prod_{f \\in F} \\left(1 - \\prod_{g \\in X} h(f, g)\\right) \\\\\n  &= \\sum_{X \\subseteq G} \\prod_{f \\in F} \\left(1 + \\prod_{g \\in X} h(f, g)\\right) \\pmod{2} \\\\\n  &= \\sum_{X \\subseteq G} \\sum_{Y \\subseteq F} \\prod_{g \\in X} \\prod_{f \\in Y} h(f, g) \\pmod{2}.\n\\end{aligned}\n$$\nOn a une égalité similaire pour le nombre $M$ de groupes populaires de filles, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProuver qu'il existe une unique manière de colorier chaque nombre rationnel strictement positif soit en rouge, soit en bleu, de sorte que :\n- les nombres $x$ et $x+1$ sont de couleurs différentes ;\n- les nombres $x$ et $\\frac{1}{x}$ sont de la même couleur;\n- le nombre $1$ est colorié en rouge.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nMontrons tout d'abord qu'un tel coloriage est nécessairement unique. Pour cela, nous allons démontrer par récurrence sur $\\max (a, b)$ que la couleur de la fraction $\\frac{a}{b}$ (supposée écrite sous forme irréductible) est entièrement déterminée. La conclusion est vraie lorsque $\\max (a, b)=1$ puisque cela impose $a=b=1$ et que l'on sait, par hypothèse, que la couleur de $1$ est rouge. Considérons maintenant une fraction irréductible $\\frac{a}{b}$ et supposons que l'on ait déjà démontré que la couleur de toute fraction irréductible $\\frac{a'}{b'}$ avec $\\max (a', b')<\\max (a, b)$ soit déterminée. Quitte à remplacer $\\frac{a}{b}$ par son inverse $\\frac{b}{a}$ (ce qui ne modifie ni $\\max (a, b)$, ni la couleur de la fraction), on peut supposer que $a>b$. Par hypothèse, $\\frac{a}{b}$ a la couleur inverse de $\\frac{a}{b}-1=\\frac{a-b}{b}$. De plus, $\\max (a-b, b)<\\max (a, b)=a$. Ainsi, la fraction $\\frac{a-b}{b}$ relève de l'hypothèse de récurrence; sa couleur est donc entièrement déterminée et, par suite, celle de $\\frac{a}{b}$ l'est également.\n\nL'existence se démontre de la même façon : en reprenant les arguments précédents, on construit le coloriage par récurrence sur $\\max (a, b)$ en vérifiant que chaque identité entre couleurs est utilisée une et une seule fois lorsqu'on construit le coloriage avec cette méthode.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLa ville de Gotham City est un grand rectangle, découpé en pâtés de maisons rectangulaires plus petits par des rues parallèles aux bords de la ville. On suppose qu'il y a au moins une rue, et qu'aucune rue ne traverse la ville de part en part.\n\nMontrer qu'il existe un pâté de maison qui ne touche pas le bord de la ville.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nBatman explore la ville selon le procédé suivant : il part du bord de la ville (du point rouge sur la figure) et suit jusqu'au bout une rue (l'Avenue du Joker par exemple) qui s'enfonce à l'intérieur. Comme l'Avenue de Joker ne traverse pas la ville de part en part, elle débouche sur une autre (appelons-la Boulevard Double-Face). Mais le Boulevard Double-Face ne traverse pas non plus Gotham City de part en part, donc va déboucher sur une autre rue soit à gauche, soit à droite, soit éventuellement des deux côtés. Si le Boulevard débouche sur une nouvelle rue à gauche, Batman part vers la gauche. Sinon, il part vers la droite. Il continue ainsi...\n\nIl est alors clair que Batman ne sortira jamais de Gotham. Mais le nombre de rues est fini, donc il finira forcément par passer deux fois par un même point. On suppose par exemple que le premier point où il passe deux fois de suite est la Place Ra's Al Ghul (en bleu sur la figure) : entre ses deux passages, il a fait le tour d'une certaine zone, sans toucher le bord de la ville. Tout pâté de maison dans cette zone convient, comme celui colorié en vert.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $P$ et $Q$ deux polynômes à coefficients réels, de degrés $n \\geqslant 0$. On suppose que le coefficient de $x^{n}$ de chacun de ces deux polynômes est égal à $1$ et que, pour tout réel $x$, on a $P(P(x)) = Q(Q(x))$.\n\nProuver que $P = Q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLe résultat est évident si $n = 0$. Dans ce qui suit, on suppose donc que $n \\geqslant 1$.\n\nPar l'absurde : supposons que le polynôme $R = P - Q$ ne soit pas le polynôme nul.\n\nSoit $k$ le degré de $R$.\n\nPuisque $P$ et $Q$ sont tous deux de degré $n$, et de même coefficient dominant, on a $k \\in \\{0, \\cdots, n-1\\}$. De plus, on a\n$$\nP(P(x)) - Q(Q(x)) = [Q(P(x)) - Q(Q(x))] + R(P(x))\n$$\n\nPosons $Q(x) = x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{1} x + a_{0}$.\n\nOn remarque qu'alors\n$$\nQ(P(x)) - Q(Q(x)) = [P^{n}(x) - Q^{n}(x)] + a_{n-1}[P^{n-1}(x) - Q^{n-1}(x)] + \\cdots + a_{1}[P(x) - Q(x)]\n$$\net chacun des termes de cette somme autre que $[P^{n}(x) - Q^{n}(x)]$ est de degré au plus $n(n-1)$. D'autre part, on a\n$$\n[P^{n}(x) - Q^{n}(x)] = R(x)[P^{n-1}(x) + P^{n-2}(x) Q(x) + \\cdots + Q^{n-1}(x)]\n$$\nce qui assure que $[P^{n}(x) - Q^{n}(x)]$ est de degré $n(n-1) + k$ et de coefficient dominant égal à $n$ (on rappelle que $P$ et $Q$ sont tous les deux de coefficient dominant égal à $1$).\n\n- On suppose que $k > 0$.\n\nAlors, le polynôme $Q(P(x)) - Q(Q(x))$ est de degré $n(n-1) + k$.\n\nD'autre part, le degré de $R(P(x))$ est $k n$, et on a $k n \\leqslant n(n-1) < n(n-1) + k$.\n\nDe (1), on déduit que $P(P(x)) - Q(Q(x))$ est de degré $n(n-1) + k$, et donc non nul, en contradiction avec l'énoncé.\n\n- Il reste à étudier le cas où $k = 0$, c'est-à-dire lorsque $R$ est constant.\n\nNotons $c$ cette constante. Notre hypothèse initiale assure que $c \\neq 0$.\n\nDe $P(P(x)) = Q(Q(x))$, il vient $Q(Q(x) + c) = Q(Q(x)) - c$.\n\nPuisque $Q$ n'est pas constant, c'est donc que l'égalité $Q(y + c) = Q(y) - c$ est vraie pour une infinité de réels $y$. S'agissant de polynômes, c'est donc qu'elle est vraie pour tout réel $y$.\n\nUne récurrence immédiate conduit alors à $Q(j c) = a_{0} - j c$ pour tout entier $j \\geqslant 0$.\n\nComme ci-dessus, l'égalité $Q(x) = -x + a_{0}$ étant vraie pour une infinité de valeurs (puisque $c \\neq 0$), elle est donc vraie pour tout $x$. Cela contredit que $Q$ est de coefficient dominant égal à $1$.\n\nAinsi, dans tous les cas, on a obtenu une contradiction. Cela assure que $R$ est bien le polynôme nul, et achève la démonstration.\nSolution:\n\nLe résultat est facile à montrer si $n = 0$ ou $n = 1$. Dans ce qui suit, on suppose donc que $n \\geqslant 2$. Notons $R = P - Q$ et supposons par l'absurde qu'il est non nul. Quitte à permuter les rôles de $P$ et de $Q$, on peut supposer que le coefficient dominant de $R$ est strictement positif.\n\nLemme 1. Pour tout polynôme $R$ ayant un coefficient dominant strictement positif, il existe un réel $a$ tel que pour tout $x \\geqslant a$ on a $R(x) > 0$.\n\nEn effet, $R(x)$ peut s'écrire sous la forme $c x^{m}\\left(1 + \\frac{c_{1}}{x} + \\cdots + \\frac{c_{m}}{x^{m}}\\right)$ avec $c > 0$. Le terme entre parenthèses tend vers $1$ lorsque $x \\rightarrow +\\infty$, donc est strictement positif pour $x$ assez grand.\n\nLemme 2. Pour tout polynôme $P$ de degré $\\geqslant 1$ ayant un coefficient dominant strictement positif, il existe $a$ tel que $P$ est strictement croissant sur $[a, +\\infty[$.\n\nEn effet, le polynôme dérivé $P'$ vérifie les conditions du lemme 1 donc $P'(x)$ est strictement positif pour $x$ assez grand.\n\nRevenons à l'exercice. D'après les deux lemmes précédents, il existe $a$ tel que sur $[a, +\\infty[$, $P$ et $Q$ sont strictement croissants, et $R$ et $R \\circ Q$ sont strictement positifs.\n\nOn a alors pour tout $x \\geqslant a$ :\n$$\n\\begin{aligned}\nP(P(x)) &> P(Q(x)) \\quad \\text{puisque} \\; P(x) > Q(x) \\\\\n&= Q(Q(x)) + R(Q(x)) \\\\\n&> Q(Q(x))\n\\end{aligned}\n$$\nce qui contredit $P(P(x)) = Q(Q(x))$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\omega_{1}$ et $\\omega_{2}$ deux cercles sécants en deux points $X$ et $Y$. Un cercle $\\omega$ est tangent intérieurement à $\\omega_{1}$ et $\\omega_{2}$ en $P$ et $Q$ respectivement. Le segment $[X, Y]$ coupe $\\omega$ en $M$ et $N$. Les demi-droites $[P, M)$ et $[P, N)$ coupent $\\omega_{1}$ en $A$ et $D$; les demi-droites $[Q, M)$ et $[Q, N)$ coupent $\\omega_{2}$ en $B$ et $C$. Montrer que $A B=C D$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $\\ell$ la tangente en $M$ à $\\omega$ et $\\ell_{1}$ la tangente en $A$ à $\\omega_{1}$. Comme $\\omega$ est tangent en $P$ à $\\omega_{1}$, il existe une homothétie $h$ de centre $P$ qui envoie $\\omega$ sur $\\omega_{1}$. Or, $P, M, A$ sont alignés donc $h(M)=A$. On en déduit que $h(\\ell)=\\ell_{1}$, et donc $\\ell \\parallel \\ell_{1}$.\n\nD'autre part, comme $M$ est sur l'axe radical de $\\omega_{1}$ et $\\omega_{2}$, on a $\\overline{M A} \\cdot \\overline{M P}= \\overline{M B} \\cdot \\overline{M Q}$ donc $A, B, P, Q$ sont cocycliques. On en déduit que\n$$\n(A B, A P)=(Q B, Q P)=(Q M, Q P)=(\\ell, M P)=\\left(\\ell_{1}, A P\\right)\n$$\nd'où $(A B)=\\ell_{1}$. Autrement dit, $(A B)$ est tangente en $A$ à $\\omega_{1}$. Mais de même, $(A B)$ est tangente en $B$ à $\\omega_{2}$, donc $(A B)$ est une tangente commune à $\\omega_{1}$ et $\\omega_{2}$. De même, $(C D)$ est une tangente commune à $\\omega_{1}$ et $\\omega_{2}$, donc $A B=C D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. Soient $D \\in [AC]$ et $E \\in [AB]$ tels que $BE = CD$. Soit $P$ le point d'intersection de $(BD)$ et $(CE)$. Les cercles circonscrits à $BEP$ et $CDP$ se recoupent en $Q$. Soient $K$ et $L$ les milieux respectifs de $[BE]$ et $[CD]$. Soit $R$ l'intersection entre la perpendiculaire à $(QK)$ passant par $K$ et la perpendiculaire à $(QL)$ passant par $L$. Montrer que\n\n1) $R$ se trouve sur le cercle circonscrit à $ABC$;\n2) $Q$ se trouve sur la bissectrice de l'angle $\\widehat{BAC}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRappelons d'abord la méthode de construction du centre d'une similitude directe qui envoie deux points distincts $A$ et $B$ sur deux points distincts $A'$ et $B'$ respectivement.\n\n![](attached_image_1.png)\n\nSupposons que $A \\neq A'$, $B \\neq B'$, et que les droites $(AA')$ et $(BB')$ ne sont pas parallèles. Soit $M$ le point d'intersection de $(AA')$ et de $(BB')$. Alors les cercles $MAB$ et $MA'B'$ se recoupent en un point $N$. Comme $(NA, NB) = (MA, MB) = (MA', MB') = (NA', NB')$ et $(AB, AN) = (MB, MN) = (MB', MN) = (A'B', A'N)$, les triangles $NAB$ et $NA'B'$ sont directement semblables, donc la similitude de centre $N$ et de rapport $NA'/NA$ envoie $A$ sur $A'$ et $B$ sur $B'$. On vérifie que le résultat reste vrai si $A = A'$ ou $B = B'$.\n\nSi $(AA')$ et $(BB')$ sont parallèles, alors $N$ est le point d'intersection de $(AB)$ et de $(A'B')$.\n\nRevenons à l'exercice. Montrons d'abord la deuxième assertion.\n\n![](attached_image_2.png)\n\nD'après le préliminaire, $Q$ est le centre de la similitude directe qui envoie $B$ sur $D$ et $E$ sur $C$. Comme $BE = CD$, cette similitude est une rotation que l'on notera $r$. Comme $r$ envoie $(AB)$ sur $(AC)$, $Q$ est équidistant de $(AB)$ et de $(AC)$ donc $Q$ est sur une bissectrice de l'angle $\\widehat{BAC}$. Vérifions que c'est la bissectrice intérieure.\n\nPar un argument de continuité, il suffit de se restreindre au cas où $E$ et $D$ sont strictement à l'intérieur des segments $[AB]$ et $[AC]$. Dans ce cas, $A$, $B$ et $E$ étant alignés donc non-cocycliques, $Q \\neq A$.\n\nDe plus, lorsque $ABC$ est isocèle en $A$ et que $D$ et $E$ tendent vers $A$, les cercles circonscrits à $BPE$ et $CPD$ tendent vers les cercles $\\Gamma_1$ passant par $B$ et tangent à $(AC)$ en $A$ et $\\Gamma_2$ passant par $C$ et tangent à $(AB)$ en $A$. Or, le cercle $\\Gamma_1$ et le triangle sont du même côté de la droite $(AC)$, le cercle $\\Gamma_2$ et le triangle sont du même côté de $(AB)$, donc $Q$ tend vers un point à l'intérieur de l'angle $\\widehat{BAC}$ ou de celui opposé par le sommet. Ce sera donc un point de la bissectrice intérieure (différent de $A$ puisque $(AB) \\neq (AC)$).\n\nEn faisant varier les points $A$, $D$ et $E$, le point $Q$ reste sur la réunion des deux bissectrices et ne passe pas par $A$, donc il doit par continuité rester sur la bissectrice intérieure.\n\nMontrons maintenant la première assertion. Comme $r$ envoie $(QK)$ sur $(QL)$ et $(AK)$ sur $(AL)$, les angles de droites $(QK, QL)$ et $(AK, AL)$ sont égaux donc $A, K, Q, L$ sont cocycliques.\n\nComme les triangles $KQR$ et $LQR$ sont rectangles en $K$ et $L$, le cercle de diamètre $[QR]$ passe par $K$ et $L$, donc le cercle circonscrit à $QKL$ rencontre $R$, ce qui montre que $A, K, Q, L, R$ sont cocycliques.\n\nSoit $r'$ la similitude directe qui envoie $K$ sur $B$ et $L$ sur $C$. D'après le préliminaire, le centre de $r'$ est le second point d'intersection du cercle $ABC$ avec le cercle $AKL$, c'est-à-dire $R$. Comme $r'$ envoie les droites $(RK)$ et $(RL)$ sur $(RB)$ et $(RC)$ respectivement, on en déduit que $(RB, RC) = (RK, RL) = (AK, AL) = (AB, AC)$ où la deuxième égalité provient de la cocyclicité de $A, K, L, R$. Ceci prouve que $A, B, C, R$ sont cocycliques.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLe cercle $\\omega$ passe par les sommets $B$ et $C$ du triangle $A B C$ et coupe les côtés $[A B]$ et $[A C]$ aux points $D$ et $E$ respectivement. Les segments $[C D]$ et $[B E]$ se coupent en un point $O$. Soient $M$ et $N$ les centres des cercles inscrits dans les triangles $A D E$ et $O D E$ respectivement. Montrer que le milieu de l'arc $D E$ (le plus petit des deux arcs) du cercle $\\omega$ se trouve sur la droite ( $M N$ ).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $\\theta=(D E, D F)=(E F, E D)$. On a $2(D F, D M)+2 \\theta=2(D E, D M)=(D E, D A)=\\theta+(D F, D A)=\\theta+(D F, D B)=\\theta+(E F, E B)=2 \\theta+(E D, E O)=\\theta+2(E D, E N)$, donc $\\widehat{F D M}=\\widehat{D E N}$. De même, $\\widehat{M E F}=\\widehat{N D E}$. Notons $\\beta$ et $\\gamma$ ces angles.\n\n\n\nSoient $P$ et $Q$ les seconds points d'intersection des droites $(D F)$ et $(E F)$ avec le cercle circonscrit au triangle DME. On a\n\n\n\nComme $D F=D E$, on a $(P Q) // (D E)$. Avec (4), cela implique que les triangles $D E N$ et $Q M P$ sont homothétiques. De plus, $F$ est le centre de l'homothétie (puisque c'est le point d'intersection des droites $(Q E)$ et $(P D)$ ). Donc la droite $(M N)$ traverse $F$.\n\n\nAutre solution. Si on ne pense pas à introduire les points $P$ et $Q$, on peut terminer la démonstration moyennant des calculs trigonométriques un peu pénibles.\n\nNotons $\\varphi$ l'angle $(\\overrightarrow{F M}, \\overrightarrow{E D})$. D'après la loi des sinus dans les triangles $M D F$ et $M E F$, on a\n$$\n\\begin{aligned}\n\\frac{M F}{\\sin \\beta} & =\\frac{D F}{\\sin (\\beta+\\varphi+\\theta)} \\\\\n\\frac{M F}{\\sin \\gamma} & =\\frac{E F}{\\sin (\\varphi-\\gamma-\\theta)}\n\\end{aligned}\n$$\nComme $D F=E F$, en divisant ces deux égalités on obtient\n$$\n\\frac{\\sin \\gamma}{\\sin \\beta}=\\frac{\\sin (\\varphi-\\gamma-\\theta)}{\\sin (\\beta+\\varphi+\\theta)}\n$$\nEn utilisant la formule $\\sin (x+y)=\\sin x \\cos y+\\cos x \\sin y$, on en déduit\n$$\n\\frac{\\sin \\gamma}{\\sin \\beta}=\\frac{\\sin (\\varphi) \\cos (\\gamma+\\theta)-\\cos \\varphi \\sin (\\gamma+\\theta)}{\\sin (\\varphi) \\cos (\\beta+\\theta)+\\cos \\varphi \\sin (\\beta+\\theta)}\n$$\nOn chasse les dénominateurs et on rassemble les termes en $\\sin \\varphi$ et en $\\cos \\varphi$ :\n$$\n\\cos \\varphi(\\sin \\gamma \\sin (\\beta+\\theta)+\\sin \\beta \\sin (\\gamma+\\theta)))=\\sin \\varphi(-\\sin \\gamma \\cos (\\beta+\\theta)+\\sin \\beta \\cos (\\gamma+\\theta))\n$$\nComme\n$$\n\\begin{aligned}\n- & \\sin \\gamma \\cos (\\beta+\\theta)+\\sin \\beta \\cos (\\gamma+\\theta) \\\\\n& =-\\sin \\gamma(\\cos \\beta \\cos \\theta-\\sin \\beta \\sin \\theta)+\\sin \\beta(\\cos \\gamma \\cos \\theta-\\sin \\gamma \\sin \\theta) \\\\\n& =\\cos \\theta \\sin (\\beta-\\gamma),\n\\end{aligned}\n$$\non obtient\n$$\n\\tan \\varphi=\\frac{\\sin \\gamma \\sin (\\beta+\\theta)+\\sin \\beta \\sin (\\gamma+\\theta)}{\\cos \\theta \\sin (\\beta-\\gamma)}\n$$\nSoit $\\varphi^{\\prime}=(F N, D E)$. Le calcul de $\\tan \\varphi^{\\prime}$ est identique, en remplaçant $(\\beta, \\gamma, \\theta)$ par $(\\beta+\\theta, \\gamma+\\theta,-\\theta)$. Or, l'expression de $\\tan \\varphi$ est invariante par cette transformation, donc $\\tan \\varphi^{\\prime}=\\tan \\varphi$, d'où $F, M, N$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier $k \\geqslant 2$, on note $\\mathrm{P}(k)$ le plus grand diviseur premier de $k$.\nProuver qu'il existe une infinité d'entiers $n$ tels que\n$$\n\\mathrm{P}(n)<\\mathrm{P}(n+1)<\\mathrm{P}(n+2)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $p$ un nombre premier impair. Pour tout entier $k \\geqslant 1$, on note $n_{k}=p^{2^{k}}-1$. On a évidemment $\\mathrm{P}\\left(n_{k}+1\\right)=p$. D'autre part, pour tout entier $i$ tel que $0<i<k$, on a\n$$\np^{2^{k}}=\\left(p^{2^{i}}\\right)^{2^{k-i}}=(-1)^{2^{k-i}}=1 \\quad \\bmod p^{2^{i}}+1,\n$$\nd'où $n_{k}+2=2 \\bmod \\left[n_{i}+2\\right]$.\nLes entiers $n_{k}+2$ et $n_{i}+2$ étant pairs, il vient alors $\\operatorname{pgcd}\\left(n_{k}+2, n_{i}+2\\right)=2$. Or $n_{k}+2$ n'est pas une puissance de 2, car $p^{2^{k}}=1 \\bmod 4$ pour $k \\geqslant 1$ et donc $n_{k}=2 \\bmod 4$. Puisque $n_{k}+2>1$, il existe donc un nombre premier impair $q_{k}$ qui divise $n_{k}+2$ et aucun des $n_{i}+2$ pour $i<k$. Cela assure qu'il existe une infinité de nombres premiers impairs qui divisent chacun au moins un des $n_{k}+2$, et donc que la suite $\\left(\\mathrm{P}\\left(n_{k}+2\\right)\\right)_{k \\geqslant 1}$ n'est pas bornée. On peut donc considérer $k$ minimal tel que $P\\left(n_{k}+2\\right)>p$.\n\nPour conclure, il reste à prouver que, pour ce choix de $k$, on a $P\\left(n_{k}\\right)<p$. Or, on a\n$$\nn_{k}=\\left(p^{2^{k-1}}+1\\right)\\left(p^{2^{k-1}}-1\\right)=\\left(n_{k-1}+2\\right) n_{k-1},\n$$\nd'où, par une récurrence,\n$$\nn_{k}=(p-1) \\prod_{i=1}^{k-1}\\left(n_{i}+2\\right)\n$$\nPuisque $n_{k}+1$ est divisible par $p$, ce ne peut être le cas pour $n_{k}$, et la minimalité de $k$ assure qu'aucun des facteurs $n_{i}+2$ n'est divisible par un nombre premier supérieur à $p$. Ainsi, $n_{k}$ n'a pas de diviseur premier supérieur ou égal à $p$, et donc $\\mathrm{P}\\left(n_{k}\\right)<p$, comme souhaité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les entiers naturels $k>0$ tels que l'équation en $x$ et $y$ :\n$$\nx(x+k)=y(y+1)\n$$\nait une solution en entiers strictement positifs.",
    "options": [],
    "answer": "k = 1 and all integers k ≥ 4",
    "solution": "Solution:\nL'équation de l'énoncé s'écrit encore $\\left(x+\\frac{k}{2}\\right)^{2}=\\left(y+\\frac{1}{2}\\right)^{2}+\\frac{k^{2}-1}{4}$, soit, en factorisant:\n$$\n\\left(x-y+\\frac{k-1}{2}\\right) \\cdot\\left(x+y+\\frac{k+1}{2}\\right)=\\frac{k^{2}-1}{4} .\n$$\nDistinguons deux cas selon la parité de $k$.\n\nSi $k$ est impair, on écrit $k=2 a+1$ et l'équation précédente devient $(x-y+a)(x+y+a+1)=a(a+1)$. En écrivant que le premier facteur vaut 1 et le second $a(a+1)$, on obtient $x=\\frac{a(a-1)}{2}$ et $y=x+(a-1)=\\frac{(a-1)(a+2)}{2}$. Étant donné que le produit de deux entiers consécutifs est toujours pair, les valeurs que l'on vient d'obtenir forment une solution dès que $a>1$. Pour $a=1$ (i.e. $k=3$ ), au contraire, il n'y a pas de solution, car l'égalité $(x-y+1)(x+y+2)=2$ ne peut être satisfaite étant donné que le deuxième facteur est toujours $>2$ lorsque $x$ et $y$ sont strictement positifs. Pour $a=0$, enfin, on a $k=1$ et, clairement, tous les couples $(x, y)$ conviennent.\n\nOn raisonne de manière analogue dans le cas où $k$ est pair. On pose $k=2 a$ et l'équation (2) devient $(2 x-2 y+2 a-1)(2 x+2 y+2 a+1)=4 a^{2}-1$. Comme précédemment, en demandant que le premier facteur vaille 1, on obtient un système en $x$ et $y$ dont les solutions sont $x=a(a-1)$ et $y=(a+1)(a-1)$. Cette solution est acceptable dès que $a>1$. Pour $a=1$, l'équation à résoudre devient $(2 x-2 y+1)(2 x+2 y+3)=3$ et elle n'a pas de solution avec $x, y>0$ étant donné que cette dernière condition implique que le deuxième facteur est $>3$.\n\nEn résumé, les entiers $k$ convenables sont $k=1$ et tous les entiers $k \\geqslant 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn cercle de centre $O$ est inscrit dans un quadrilatère $A B C D$ dont les côtés ne sont pas parallèles. Montrer que le point $O$ coïncide avec le point d'intersection des lignes médianes du quadrilatère si et seulement si $O A \\cdot O C=O B \\cdot O D$. (Une ligne médiane du quadrilatère est une droite reliant les milieux des côtés opposés.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoient $I$ et $J$ les milieux des côtés $[A B]$ et $[C D]$ respectivement. Les droites $(A B)$ et $(C D)$ se coupent au point $P$. Sans perte de généralité, on peut supposer que $P$ est le point d'intersection des demi-droites $[A B)$ et $[C D)$.\n\n![](attached_image_1.png)\n\n1) Supposons que $O$ est le point d'intersection des lignes médianes du quadrilatère $A B C D$. Alors $O$ est le milieu du segment $[I J]$, parce que les milieux des côtés d'un quadrilatère sont les sommets d'un parallélogramme. Puis, $[P O]$ est la bissectrice et la médiane issus de $P$ du triangle $I P J$, donc $I P J$ est isocèle en $P$ et $\\widehat{B I O}=\\widehat{C J O}$. En outre,\n$$\n\\widehat{I B O}+\\widehat{J C O}=\\frac{1}{2}(\\widehat{A B C}+\\widehat{B C D})=\\frac{1}{2}(\\pi+\\widehat{B P C})=\\pi-\\widehat{P I J}=\\widehat{I B O}+\\widehat{I O B}\n$$\nAlors $\\widehat{J C O}=\\widehat{I O B}$ et les triangles $O I B$ et $C J O$ sont semblables. On obtient que $\\frac{O B}{O C}=\\frac{I B}{J O}$. De même, on a $\\frac{O A}{O D}=\\frac{I A}{J O}$. Comme $I B=I A$, on obtient $O A \\cdot O C=O B \\cdot O D$.\n\n2) On suppose maintenant que $O A \\cdot O C=O B \\cdot O D$. On note que\n$$\n\\begin{aligned}\n\\widehat{A O B}+\\widehat{C O D} & = (\\pi-\\widehat{O A B}-\\widehat{O B A}) + (\\pi-\\widehat{O C D}-\\widehat{O D C}) \\\\\n& = \\pi-\\frac{1}{2}(\\widehat{D A B}+\\widehat{A B C}+\\widehat{B C D}+\\widehat{C D A})=\\pi\n\\end{aligned}\n$$\nPar conséquent, si on construit les parallélogrammes $O A K B$ et $C O D L$ à partir des triangles $O A B$ et $O C D$, ces parallélogrammes seront semblables, parce que $\\frac{O A}{A K}=\\frac{O A}{O B}=\\frac{D O}{O C}=\\frac{D O}{D L}$. En particulier, les triangles $O I B$ et $C J O$ sont également semblables, parce qu'ils se correspondent mutuellement dans les parallélogrammes semblables $O A K B$ et $C O D L$. Il vient\n$$\n\\widehat{I O B}=\\widehat{O C J}=\\widehat{O C B}, \\quad \\widehat{C O J}=\\widehat{I B O}=\\widehat{O B C}\n$$\nOn obtient que\n$$\n\\widehat{I O B}+\\widehat{B O C}+\\widehat{C O J}=\\widehat{O C B}+\\widehat{B O C}+\\widehat{O B C}=\\pi,\n$$\ndonc $O$ se trouve sur $(I J)$. De même, $O$ se trouve sur l'autre ligne médiane du quadrilatère $A B C D$.\nSolution:\n\nOn peut résoudre l'exercice par une méthode analytique utilisant les nombres complexes. Les calculs sont longs mais pas impraticables.\n\nNotons $U, V, W, X$ les points de contact des droites $(A B),(B C),(C D)$ et $(D A)$ avec le cercle. On peut supposer que le cercle est de centre 0 et de rayon 1. Notons $a, b, c, d, u, v, w, x$ les affixes des points $A, B, C, D, U, V, W, X$.\n\n![](attached_image_2.png)\n\nUn point d'affixe $z$ appartient à $(A B)$ si et seulement si $z-u$ est perpendiculaire à $u$, ce qui s'écrit $0=\\bar{u}(z-u)+u(\\bar{z}-\\bar{u})$, ou encore $\\bar{u} z+u \\bar{z}=2$. Comme $\\bar{u}=1 / u$, ceci s'écrit encore $z+u^{2} \\bar{z}=2 u$. De même, l'équation de la droite $(A D)$ est $z+x^{2} \\bar{z}=2 x$. En combinant ces deux équations, on trouve que\n$$\na=\\frac{2 u x}{u+x}.\n$$\nDe même, on a $b=\\frac{2 u v}{u+v}$, $c=\\frac{2 v w}{v+w}$ et $d=\\frac{2 w x}{w+x}$.\n\nOn calcule que $4 O A^{-2}=|u+x|^{2}=2+u \\bar{x}+\\bar{u} x$. On a des expressions analogues pour $4 O B^{-2}$, etc. Par conséquent,\n$$\n\\begin{aligned}\nO A \\cdot O C & =O B \\cdot O D \\\\\n& \\Longleftrightarrow \\quad(2+u \\bar{x}+\\bar{u} x)(2+v \\bar{w}+\\bar{v} w)=(2+u \\bar{v}+\\bar{u} v)(2+w \\bar{x}+\\bar{w} x) \\\\\n& \\Longleftrightarrow \\quad u x(2+u \\bar{x}+\\bar{u} x) v w(2+v \\bar{w}+\\bar{v} w)=u v(2+u \\bar{v}+\\bar{u} v) w x(2+w \\bar{x}+\\bar{w} x) \\\\\n& \\Longleftrightarrow \\quad\\left(2 u x+u^{2}+x^{2}\\right)\\left(2 v w+v^{2}+w^{2}\\right)=\\left(2 u v+u^{2}+v^{2}\\right)\\left(2 w x+x^{2}+w^{2}\\right) \\\\\n& \\Longleftrightarrow \\quad 2 u v^{2} x+2 u w^{2} x+2 u^{2} v w+u^{2} v^{2}+2 v w x^{2}+w^{2} x^{2} \\\\\n& =2 u v x^{2}+2 u v w^{2}+2 u^{2} w x+u^{2} x^{2}+2 v^{2} w x+v^{2} w^{2} \\\\\n& \\Longleftrightarrow 2(u-w) v^{2} x-2(u-w) u w x+2(u-w) u v w \\\\\n& \\quad-2(u-w) v x^{2}+(u-w)(u+w) v^{2}-(u-w)(u+w) x^{2}=0 \\\\\n& \\Longleftrightarrow 2 v^{2} x-2 u w x+2 u v w-2 v x^{2}+u v^{2}+v^{2} w-u x^{2}-w x^{2}=0 \\\\\n& \\Longleftrightarrow \\quad 2 v x(v-x)+(v-x) w(v+x)+2 u w(v-x)+u(v+x)(v-x)=0 \\\\\n& \\Longleftrightarrow 2 v x+v w+v x+2 u w+u v+u x=0 \\\\\n& \\Longleftrightarrow 2(u w+x v)+(x+v)(u+w)=0 .\n\\end{aligned}\n$$\n\nD'autre part, $O$ appartient à la droite passant par les milieux de $[A D]$ et de $[B C]$ si et seulement si $0, \\frac{a+d}{2}$ et $\\frac{b+c}{2}$ sont alignés, ce qui se traduit par le fait que $\\frac{a+d}{b+c}$ est réel, c'est-à-dire\n$$\n\\frac{\\frac{u x}{u+x}+\\frac{x w}{x+w}}{\\frac{u v}{u+v}+\\frac{v w}{v+w}}=\\frac{\\frac{1}{u+x}+\\frac{1}{w+x}}{\\frac{1}{u+v}+\\frac{1}{v+w}}\n$$\nOn multiplie les numérateurs de chaque membre par $(u+x)(x+w)$ et les dénominateurs par $(u+v)(v+w)$, ce qui donne successivement\n$$\n\\begin{aligned}\n& \\frac{x[u(x+w)+w(u+x)]}{v[u(v+w)+w(u+v)]}=\\frac{2 x+(u+w)}{2 v+(u+w)} \\\\\n& \\quad \\Longleftrightarrow \\quad x[2 u w+u(u+w)][2 v+(u+w)]=v[2 u w+v(u+w)][2 x+(u+w)] \\\\\n& \\Longleftrightarrow \\quad 2 x u w(u+w)+2 x^{2} v(u+w)+x^{2}(u+w)^{2} \\\\\n& \\quad=2 v u w(u+w)+2 x v^{2}(u+w)+v^{2}(u+w)^{2} \\\\\n& \\quad \\Longleftrightarrow \\quad 2(x-v) u w(u+w)+2(x-v) x v(u+w)+(x-v)(x+v)(u+w)^{2}=0\n\\end{aligned}\n$$\nComme les côtés opposés ne sont pas parallèles, on a $u+w \\neq 0$. En simplifiant par $(x-v)(u+w)$, on obtient la condition $2 u w+2 x v+(x+v)(u+w)=0$ qui est la même que la précédente.\n\nOn en conclut que $O A \\cdot O C=O B \\cdot O D$ si et seulement si $O$ appartient à la droite passant par les milieux de $[A D]$ et de $[B C]$. De même, on montre que $O A \\cdot O C=O B \\cdot O D$ si et seulement si $O$ appartient à la droite passant par les milieux de $[A B]$ et de $[C D]$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $E$ un ensemble de cardinal $n$ et $\\mathscr{F}$ un ensemble de parties de $E$ avec $|\\mathscr{F}| = 2^{n-1}$ tel que pour tous $A$, $B$ et $C$ dans $\\mathscr{F}$, $A \\cap B \\cap C$ est non vide.\n\nMontrer que l'intersection de tous les éléments de $\\mathscr{F}$ est non vide.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSi $A \\subset E$, on notera $A^{c}$ son complémentaire : l'ensemble des parties de $E$ est partitionné en $2^{n-1}$ doubletons de la forme $\\{A, A^{c}\\}$ où $A \\subset E$. Or, il est impossible que les deux éléments d'un doubleton soient dans $\\mathscr{F}$, car leur intersection est vide, donc $\\mathscr{F}$ contient exactement un élément de chaque doubleton.\n\nSoient donc $A$ et $B$ dans $\\mathscr{F}$ : on va montrer que $A \\cap B \\in \\mathscr{F}$. D'après ce qu'on vient de dire, il suffit de montrer $(A \\cap B)^{c} \\notin \\mathscr{F}$. Or, ceci est vrai car $A \\cap B \\cap (A \\cap B)^{c}$ est vide.\n\nPar récurrence sur $k$, on en déduit que pour tous $A_{1}, A_{2}, \\ldots, A_{k}$ de $\\mathscr{F}$, $A_{1} \\cap A_{2} \\cap \\ldots \\cap A_{k} \\in \\mathscr{F}$. En effet, c'est vrai pour $k=2$ et si on l'a montré pour $k$, alors :\n$$\nA_{1} \\cap A_{2} \\cap \\ldots \\cap A_{k} = \\left(A_{1} \\cap A_{2} \\cap \\ldots \\cap A_{k-1}\\right) \\cap A_{k}\n$$\nest l'intersection de deux éléments de $\\mathscr{F}$, donc est dans $\\mathscr{F}$.\n\nComme $\\mathscr{F}$ est fini, l'intersection de tous les éléments de $\\mathscr{F}$ est dans $\\mathscr{F}$. Or, l'ensemble vide ne peut pas être dans $\\mathscr{F}$ car son intersection avec lui-même est vide, donc l'intersection de tous les éléments de $\\mathscr{F}$ n'est pas vide.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $n>0$ un entier et $x_{1}, \\cdots, x_{n}$ des réels strictement positifs. Prouver que :\n$$\n\\begin{gathered}\n\\max _{x_{1}>0, \\cdots, x_{n}>0} \\min \\left(x_{1}, \\frac{1}{x_{1}}+x_{2}, \\cdots, \\frac{1}{x_{n-1}}+x_{n}, \\frac{1}{x_{n}}\\right)= \\\\\n\\min _{x_{1}>0, \\cdots, x_{n}>0} \\max \\left(x_{1}, \\frac{1}{x_{1}}+x_{2}, \\cdots, \\frac{1}{x_{n-1}}+x_{n}, \\frac{1}{x_{n}}\\right)=2 \\cos \\left(\\frac{\\pi}{n+2}\\right) .\n\\end{gathered}\n$$",
    "options": [],
    "answer": "2 cos(π/(n+2))",
    "solution": "Solution:\nSoit $U$ l'ensemble des $n$-uplets de réels strictement positifs. Pour $x=\\left(x_{1}, \\cdots, x_{n}\\right) \\in U$, on pose\n$$\n\\begin{aligned}\n& \\quad m(x)=\\min \\left(x_{1}, \\frac{1}{x_{1}}+x_{2}, \\cdots, \\frac{1}{x_{n-1}}+x_{n}, \\frac{1}{x_{n}}\\right) \\\\\n& \\text{ et } M(x)=\\max \\left(x_{1}, \\frac{1}{x_{1}}+x_{2}, \\cdots, \\frac{1}{x_{n-1}}+x_{n}, \\frac{1}{x_{n}}\\right)\n\\end{aligned}\n$$\nNotre stratégie va consister à prouver qu'il existe $a \\in U$ tel que $m(a)=M(a)$ et que, pour tout $x \\in U$, on a $m(x) \\leqslant m(a)$ et $M(a) \\leqslant M(x)$.\nSoit $a=\\left(a_{1}, \\cdots, a_{n}\\right) \\in U$.\nLa condition $m(a)=M(a)$ s'écrit\n$$\na_{1}=\\frac{1}{a_{1}}+a_{2}=\\cdots=\\frac{1}{a_{n-1}}+a_{n}=\\frac{1}{a_{n}} .\n$$\nMais, admettons pour le moment que l'on ait déjà trouvé $a \\in U$ tel que $m(a)=M(a)$.\nPar l'absurde : On suppose qu'il existe $x \\in U$ tel que $m(x)>m(a)$.\nOn prouve alors par récurrence sur $k$ que $x_{k}>a_{k}$ pour tout $k \\in\\{1, \\cdots, n\\}$ :\nDéjà, on a $x_{1} \\geqslant m(x)>m(a)=a_{1}$.\nD'autre part, si $x_{k}>a_{k}$ pour un certain $k \\in\\{1, \\cdots, n-1\\}$, alors\n$$\n\\frac{1}{x_{k}}+x_{k+1} \\geqslant m(x)>m(a)=\\frac{1}{a_{k}}+a_{k+1}\n$$\nOr, d'après l'hypothèse de récurrence, on a $\\frac{1}{x_{k}}<\\frac{1}{a_{k}}$, d'où $x_{k+1}>a_{k+1}$, ce qui achève la récurrence.\nEn particulier, on a donc $x_{n}>a_{n}$.\nMais, $\\frac{1}{x_{n}} \\geqslant m(x)>m(a)=\\frac{1}{a_{n}}$, d'où $x_{n}<a_{n}$. Contradiction.\nAinsi, pour tout $x \\in U$, on a $m(x) \\leqslant m(a)$.\nOn démontre de même que $M(a) \\leqslant M(x)$, pour tout $x \\in U$.\nDans ces conditions, on a $\\max _{x \\in U}\\{m(x)\\}=m(a)=M(a)=\\min _{x \\in U}\\{M(x)\\}$, comme désiré.\nPour conclure, il ne reste donc plus qu'à trouver $a \\in U$ vérifiant (1).\nMontrons comment trouver un tel $a$ sans trop s'aider de l'énoncé :\nSupposons qu'un tel $a$ existe. On note $\\alpha>0$ la valeur commune dans (1).\nOn vérifie sans difficulté qu'alors, pour tout $k$, on a $a_{k}=\\frac{b_{k}}{b_{k-1}}$, où $b_{0}=1, b_{1}=\\alpha$, et $b_{j}=\\alpha b_{j-1}-b_{j-2}$ pour $j \\geqslant 2$. (2)\nComme $\\alpha=\\frac{1}{a_{n}}$, on doit avoir $b_{n-1}=\\alpha b_{n}$, ce qui signifie que $b_{n+1}=0$.\nRevenons sur $\\alpha$. En fait, on a même $\\alpha<2$ :\nEn effet, supposons que $\\alpha \\geqslant 2$. Alors, $a_{1}=\\alpha \\geqslant 2$ et, par une récurrence sans difficulté, on déduit que $a_{k}=\\alpha-\\frac{1}{a_{k-1}} \\geqslant 1+\\frac{1}{k}$.\nEn particulier, on a $a_{n} \\geqslant 1+\\frac{1}{n}>1$ et $a_{n}=\\frac{1}{\\alpha}<1$, contradiction.\nOn peut donc poser $\\alpha=2 \\cos (t)$ où $t \\in] 0, \\frac{\\pi}{2}[$.\nCompte-tenu de la formule bien connue\n$$\n2 \\cos (a) \\sin (b)=\\sin (b+a)+\\sin (b-a)\n$$\nune autre récurrence sans difficulté à partir de (2) conduit alors à\n$$\nb_{k}=\\frac{\\sin ((k+1) t)}{\\sin (t)} \\text{ pour tout } k \\geqslant 0\n$$\nLa condition $b_{n+1}=0$ impose alors $t=\\frac{\\pi}{n+2}$. Ainsi, on a $\\alpha=2 \\cos \\left(\\frac{\\pi}{n+2}\\right)$ et $a=\\left(a_{1}, \\cdots, a_{n}\\right)$ où $a_{k}=\\frac{\\sin \\left(\\frac{(k+1) \\pi}{n+2}\\right)}{\\sin \\left(\\frac{k \\pi}{n+2}\\right)}$.\nRéciproquement, on vérifie aisément que, dans ces conditions, la chaîne d'égalités (1) est vraie.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn note $I$ le centre du cercle inscrit $\\gamma$ dans le triangle $ABC$ et $O$ le centre de son cercle circonscrit. Soit $\\omega_{A}$ le cercle qui passe par $B$ et $C$ et qui est tangent à $\\gamma$. Les cercles $\\omega_{B}$ et $\\omega_{C}$ sont définis de manière analogue. On note $A'$ l'autre point commun de $\\omega_{B}$ et $\\omega_{C}$. Les points $B'$ et $C'$ sont définis de façon semblable.\n\nProuver que les droites $\\left(AA'\\right), \\left(BB'\\right)$ et $\\left(CC'\\right)$ sont concourantes en un point $K$, qui est aligné avec $O$ et $I$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn note respectivement $A_{1}, B_{1}, C_{1}$ les points de contact de $\\gamma$ avec les côtés $[BC], [CA], [AB]$. Soit $X_{A}$ le point de commun des cercles $\\gamma$ et $\\omega_{A}$.\n\nLe cercle $\\omega_{A}$ est l'image de $\\gamma$ par une homothétie de centre $X_{A}$, qui envoie $A_{1}$ en un point $M_{A}$ de $\\omega_{A}$ en lequel la tangente à $\\omega_{A}$ est parallèle à $(BC)$. Par conséquent, ce point $M_{A}$ est le milieu de l'arc $BC$ de $\\omega_{A}$ qui ne contient pas $X_{A}$, ce qui assure que\n$$\n\\widehat{M_{A} X_{A} B} = \\widehat{M_{A} X_{A} C} = \\widehat{M_{A} BC} = \\widehat{M_{A} B A_{1}}.\n$$\nComme, on a évidemment $\\widehat{A_{1} M_{A} B} = B\\widehat{M_{A}}_{A}$, c'est que les triangles $M_{A} B A_{1}$ et $M_{A} X_{A} B$ sont semblables. Il vient alors\n$$\n\\frac{M_{A} B}{M_{A} X_{A}} = \\frac{M_{A} A_{1}}{M_{A} B},\n$$\nou encore $M_{A} B^{2} = M_{A} A_{1} \\cdot M_{A} X_{A}$.\n\nCette dernière relation signifie que $M_{A}$ appartient à l'axe radical $\\Delta_{B}$ de $B$ et $\\gamma$. De même, $M_{A}$ appartient à l'axe radical $\\Delta_{C}$ de $C$ et $\\gamma$. Notons que cet axe radical $\\Delta_{C}$ est perpendiculaire à la droite $(CI)$. Or, puisque $A_{1}$ et $B_{1}$ sont les points de contact avec $\\gamma$ des deux tangentes à $\\gamma$ issues de $C$, les droites $\\left(A_{1} B_{1}\\right)$ et $(CI)$ sont perpendiculaires, d'où l'on peut affirmer que $\\Delta_{C}$ est parallèle à $\\left(A_{1} B_{1}\\right)$. De même, les droites $\\Delta_{B}$ et $(A_{1} C_{1})$ sont parallèles.\n\nOn définit les points $X_{B}, X_{C}, M_{B}, M_{C}$ et la droite $\\Delta_{A}$ de façon analogue à ci-dessus, et on prouve de même que $M_{B}$ appartient à $\\Delta_{A}$ et $\\Delta_{C}$, que $M_{C}$ appartient à $\\Delta_{A}$ et $\\Delta_{B}$, et que les droites $\\Delta_{A}$ et $\\left(B_{1} C_{1}\\right)$ sont parallèles. On remarque alors que $\\Delta_{A}, \\Delta_{B}$ et $\\Delta_{C}$ portent les côtés du triangle $M_{A} M_{B} M_{C}$, et les parallélismes signalés ci-dessus assurent que le triangle $M_{A} M_{B} M_{C}$ est l'image de $A_{1} B_{1} C_{1}$ par une homothétie $\\Phi$, dont on note $K$ le centre et\n$$\nr = \\frac{M_{A} K}{A_{1} K} = \\frac{M_{B} K}{B_{1} K} = \\frac{M_{C} K}{C_{1} K}\n$$\nle rapport. En particulier, les droites $\\left(M_{A} A_{1}\\right), \\left(M_{B} B_{1}\\right)$ et $\\left(M_{C} C_{1}\\right)$ sont concourantes en $K$.\n\nSi l'on évalue la puissance de $K$ par rapport à $\\gamma$, il vient $KA_{1} \\cdot KX_{A} = KB_{1} \\cdot KX_{B}$. Après multiplication des deux côtés par $r$, on obtient $M_{A} K \\cdot K X_{A} = M_{B} K \\cdot K X_{B}$. Cela signifie que $K$ appartient à l'axe radical $\\left(CC'\\right)$ des cercles $\\omega_{A}$ et $\\omega_{B}$. On prouve de même que $K$ appartient à $\\left(AA'\\right)$ et à $\\left(BB'\\right)$.\n\nSoit $O'$ l'image de $I$ par $\\Phi$. Le point $O'$ appartient donc à la droite passant par $M_{A}$ et parallèle à $\\left(A_{1} I\\right)$ (et donc perpendiculaire à $(BC)$). Comme $M$ est le milieu d'un arc $BC$ de $\\omega_{A}$, cette droite est donc la médiatrice de $[BC]$. De même, le point $O'$ appartient à la médiatrice de $[AC]$, et finalement c'est que $O' = O$. Par suite, les points $I, O, K$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $A, B, C, D, E$ des points dans cet ordre sur un demi-cercle de rayon $1$. Démontrer que\n$$\nA B^{2}+B C^{2}+C D^{2}+D E^{2}+A B \\cdot B C \\cdot C D+B C \\cdot C D \\cdot D E \\leqslant 4\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nEn remplaçant $E$ par le point diamétralement opposé à $A$, la quantité $D E$ augmente et les autres longueurs de la formule ne changent pas, donc on se ramène au cas où $[A E]$ est un diamètre.\nD'après la formule d'Al-Kashi,\n$$\n\\begin{aligned}\nA B^{2}+B C^{2} & =A C^{2}+2 A B \\cdot B C \\cos \\widehat{A B C} \\\\\nC D^{2}+D E^{2} & =C E^{2}+2 C D \\cdot D E \\cos \\widehat{C D E} \\\\\nA C^{2}+C E^{2} & =A E^{2}\n\\end{aligned}\n$$\nEn additionnant ces trois égalités, compte tenu de $A E^{2}=4$, on obtient\n$$\nA B^{2}+B C^{2}+C D^{2}+D E^{2}=4+2 A B \\cdot B C \\cos \\widehat{A B C}+2 C D \\cdot D E \\cos \\widehat{C D E},\n$$\ndonc on se ramène à montrer que\n$$\n\\begin{gathered}\n2 A B \\cdot B C \\cos \\widehat{A B C}+2 C D \\cdot D E \\cos \\widehat{C D E}+A B \\cdot B C \\cdot C D+B C \\cdot C D \\cdot D E \\leqslant 0 . \\\\\n\\text{Or, } C D \\leqslant C E=A E \\cos \\widehat{A E C}=2 \\cos \\left(180^{\\circ}-\\widehat{A B C}\\right)=-2 \\cos \\widehat{A B C} \\text{, donc } \\\\\nA B \\cdot B C \\cdot C D+2 A B \\cdot B C \\cos \\widehat{A B C} \\leqslant 0\n\\end{gathered}\n$$\net de même\n$$\nB C \\cdot C D \\cdot D E+2 C D \\cdot D E \\cos \\widehat{C D E} \\leqslant 0\n$$\nOn conclut en additionnant ces deux dernières inégalités.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $ABCD$ un quadrilatère inscriptible. On note $K$ le point d'intersection des diagonales. Soient $M$ et $N$ les milieux de $[A, C]$ et $[B, D]$. Les cercles circonscrits à $ADM$ et $BCM$ se recoupent en un point $L$. Montrer que $K, L, M, N$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nNotons $E$ le centre radical des trois cercles $ABCD$, $ADM$, $BCM$. Les triplets $(E, A, D)$, $(E, L, M)$ et $(E, B, C)$ sont alignés.\n\nSoit $F$ le point où les cercles $ADK$ et $BCK$ se recoupent. On a $(FA, FD) = (KA, KD) = (KC, KB) = (FC, FB)$ et $(AD, AF) = (KD, KF) = (KB, KF) = (CB, CF)$, donc les triangles $FAD$ et $FCB$ sont semblables. Soit $s$ la similitude de centre $F$ qui envoie $A$ sur $D$ et $C$ sur $B$. Alors $s(M) = N$, donc l'angle de $s$ est à la fois égal à $(FM, FN)$ et à $(AC, DB) = (KM, KN)$. Par conséquent, $F, K, M, N$ sont cocycliques.\n\n$E$ est l'intersection de $(AD)$ et de $(BC)$, donc il est le centre radical des cercles $ABCD$, $ADK$ et $BCK$. Par conséquent, $E, F, K$ sont alignés. En utilisant la puissance d'un point par rapport à un cercle, on a\n$$\n\\overline{EK} \\cdot \\overline{EF} = \\overline{EA} \\cdot \\overline{ED} = \\overline{EB} \\cdot \\overline{EC} = \\overline{EM} \\cdot \\overline{EL},\n$$\ndonc $M, L, K, F$ sont cocycliques.\n\nFinalement, $L$ et $N$ sont sur le cercle $MKF$ donc $K, L, M, N$ sont cocycliques.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les nombres réels $t$ tels que si $a, b, c$ sont les longueurs des côtés d'un triangle, alors il en est de même pour $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$.",
    "options": [],
    "answer": "[2/3, 2]",
    "solution": "Solution:\n\nPar symétrie, nous pouvons ordonner $0<a \\leq b \\leq c$. De plus, les conditions de l'énoncé étant homogènes en $a, b, c$, nous pouvons fixer $a=1$. La seule inégalité triangulaire non triviale à exploiter sera donc $c<1+b$. Comme nous allons le voir, le fait d'ordonner $a, b, c$ permet de rendre deux des trois inégalités à vérifier assez claires.\nNous devons comparer $A=1+b c t, B=b^{2}+c t$ et $C=c^{2}+b t$. Tout d'abord,\n$$\n\\begin{aligned}\nB+C-A & =b^{2}+c^{2}+(b+c) t-b c t-1 \\\\\n& =\\left(b^{2}-2 b c+c^{2}\\right)+2 b c-t b c+(b+c) t-1 \\\\\n& \\geq(b+c)^{2}+(2-t) b c+\\frac{4}{3}-1 \\\\\n& >0 .\n\\end{aligned}\n$$\nD'autre part,\n$$\n\\begin{aligned}\nC+A-B & =c^{2}+b t+1+b c t-b^{2}-c t \\\\\n& =(c-b)(c+b-t)+1+b c t \\\\\n& >(c-b)(2-t) \\geq 0\n\\end{aligned}\n$$\n(En fait, on a même $C>B$.) Moralement, comme $c$ est le plus grand des trois réels, et que $t$ lui-même n'est pas très grand, $C$ est le terme dominant, et il n'est donc pas étonnant que ce soient les deux inégalités où il faut minorer $C$ qui soient les plus faciles à démontrer. Passons à la démonstration de $C<A+B$. Pour cela, il est commode de se débarrasser des inégalités qui lient $b$ et $c$ en posant $c=b+\\epsilon$ avec $0 \\leq \\epsilon<1$. Il nous reste alors à avoir la stricte positivité de\n$$\nQ(b)=t b^{2}+\\epsilon(t-2) b+\\left(1+\\epsilon t-\\epsilon^{2}\\right),\n$$\npour $\\frac{2}{3} \\leq t \\leq 2$ et $0 \\leq \\epsilon<1$. On reconnaît un polynôme du second degré en $b$, dont on calcule le discriminant :\n$$\n\\Delta(t, \\epsilon)=\\left(t^{2}+4\\right) \\epsilon^{2}-4 t^{2} \\epsilon-4 t\n$$\nCe dernier peut lui-même être vu comme un polynôme du second degré en $\\epsilon$. Il suffit de vérifier qu'il est strictement négatif pour $\\epsilon \\in[0,1[$ quel que soit $t \\in[2 / 3,2]$. Puisque son coefficient dominant est strictement positif, il suffit pour cela de vérifier qu'il est strictement négatif en 0 et négatif ou nul en 1 . Or $\\Delta(t, 0)=-4 t<0$ si $2 / 3 \\leq t \\leq 1$, et $\\Delta(t, 1)=-3 t^{2}-4 t+4$. On obtient ainsi un dernier polynôme du second degré, en $t$, dont il faut vérifier qu'il est négatif ou nul entre les valeurs de $t$ qui nous intéressent. Nous avons $\\Delta(2 / 3,1)=0$, donc $2 / 3$ est racine de ce polynôme. D'autre part, il est strictement positif en 0 et son coefficient dominant est négatif, donc l'autre racine est négative, et $\\Delta(t, 1)$ est strictement négatif sur $] 2 / 3,+\\infty[$. Finalement, $\\Delta(t, \\epsilon)$ est donc bien strictement négatif pour $\\epsilon \\in[0,1[$ quel que soit $t \\in[2 / 3,2]$, et donc $Q(b)$ est de signe constant sur $\\mathbf{R}$ pour ces valeurs d'$\\epsilon$ et de $t$, nécessairement positif du fait de son coefficient dominant.\nSolution:\n\nSupposons $2 / 3 \\leq t \\leq 2$ et $b+c>a$. En utilisant $(b+c)^{2} \\geq 4 b c$ on obtient\n$$\n\\begin{aligned}\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t & =(b+c)^{2}+a t(b+c)-(2+t) b c-a^{2} \\\\\n& \\geq(b+c)^{2}+a t(b+c)-\\frac{1}{4}(2+t)(b+c)^{2}-a^{2} \\\\\n& \\geq \\frac{1}{4}(2-t)(b+c)^{2}+a t(b+c)-a^{2}\n\\end{aligned}\n$$\nPuisque $2-t \\geq 0$ et $t>0$, cette dernière expression est croissante en $b+c$, et en utilisant $b+c>a$ on obtient\n$$\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t>\\frac{1}{4}(2-t) a^{2}+t a^{2}-a^{2}=\\frac{3}{4}\\left(t-\\frac{2}{3}\\right) a^{2} \\geq 0\n$$\npuisque $t \\geq 2 / 3$. Les deux autres inégalités s'obtiennent par symétrie.\nSolution:\n\nOn pose $x=(c+a-b) / 2, y=(a+b-c) / 2$ et $z=(b+c-a) / 2$ de sorte que $a=x+y, b=y+z, c=z+x$. Ceci rend les expressions un peu plus compliquées à écrire, mais a l'avantage de remplacer les réels $a, b, c$ sur lesquels nous avons des contraintes pas très faciles à utiliser par les réels $x, y, z$ qui ne sont soumis à aucune contrainte (à part à celle d'être strictement positifs). Alors :\n$$\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t=\\left(x^{2}+y^{2}-z^{2}+x y+x z+y z\\right) t+2\\left(z^{2}+x z+y z-x y\\right)\n$$\nCette expression est une fonction affine de $t$, strictement positive aussi bien en $t=2 / 3$ où\n$$\n\\frac{2}{3}\\left((x-y)^{2}+4(x+y) z+2 z^{2}\\right)>0\n$$\nqu'en $t=2$ où\n$$\n2\\left(x^{2}+y^{2}\\right)+4(x+y) z>0.\n$$\nElle est donc positive sur l'intervalle $[2/3,2]$ tout entier.\n\nVariante\n\nAprès avoir obtenu l'expression (1), on observe qu'elle peut être réécrite sous la forme\n$$\n(2-t) z^{2}+(x-y)^{2} t+(3 t-2) x y+z(x+y)(2+t)\n$$\nLes trois premiers termes sont positifs ou nuls et le dernier est strictement positif, ce qui conclut.\nSolution:\n\nOn distingue deux cas:\n\nPremier cas: Si $a \\geq b, c$, alors $a b+a c-b c>0,2\\left(b^{2}+c^{2}\\right) \\geq(b+c)^{2}>a^{2}$ et $t \\geq 2 / 3$ impliquent:\n$$\n\\begin{aligned}\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t & =b^{2}+c^{2}-a^{2}+(a b+a c-b c) t \\\\\n& \\geq\\left(b^{2}+c^{2}-a^{2}\\right)+\\frac{2}{3}(a b+a c-b c) \\\\\n& \\geq \\frac{1}{3}\\left(3 b^{2}+3 c^{2}-3 a^{2}+2 a b+2 a c-2 b c\\right) \\\\\n& \\geq \\frac{1}{3}\\left[\\left(2 b^{2}+2 c^{2}-a^{2}\\right)+(b-c)^{2}+2 a(b+c-a)\\right] \\\\\n& >0\n\\end{aligned}\n$$\n\nDeuxième cas: Si $b \\geq a, c$, alors $b^{2}+c^{2}-a^{2}>0$. Si on a également $a b+a c-b c \\geq 0$, alors $b^{2}+c a t+c^{2}+a b t-a^{2}-b c t>0$. Si au contraire $a b+a c-b c<0$, alors avec $t \\leq 2$, nous avons :\n$$\n\\begin{aligned}\nb^{2}+c a t+c^{2}+a b t-a^{2}-b c t & \\geq b^{2}+c^{2}-a^{2}+2(a b+a c-b c) \\\\\n& \\geq(b-c)^{2}+a(b+c-a)+a(b+c) \\\\\n& >0\n\\end{aligned}\n$$\nPar symétrie, nous avons fini.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30393,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $D$ et $E$ des points appartenant respectivement aux intérieurs des côtés $[A B]$ et $[A C]$ d'un triangle $A B C$, tels que $D B = B C = C E$. Soient $F$ le point d'intersection des droites $C D$ et $B E$, $I$ le centre du cercle inscrit au triangle $A B C$, $H$ l'orthocentre du triangle $D E F$ et $M$ le milieu de l'arc $B A C$ du cercle circonscrit au triangle $A B C$. Montrer que $I, H$ et $M$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPuisque $D B = B C = C E$, nous avons $(B I) \\perp (C D)$ et $(C I) \\perp (B E)$. Ainsi, $I$ est l'orthocentre du triangle $B F C$. Soit $K$ le point d'intersection des droites $(B I)$ et $(C D)$, et $L$ le point d'intersection des droites $(C I)$ et $(B E)$. Les triangles $I L B$ et $I K C$ sont semblables, donc nous avons la relation $I B \\cdot I K = I C \\cdot I L$. Soient $U$ et $V$ les projetés orthogonaux respectifs de $D$ sur $E F$ et de $E$ sur $D F$. De même, les triangles $D H V$ et $E H U$ étant semblables, on a $D H \\cdot H U = E H \\cdot H V$.\n\n![](attached_image_1.png)\n\nSoient $\\omega_{1}$ et $\\omega_{2}$ les cercles de diamètres respectifs $B D$ et $C E$. En voyant les relations ci-dessus comme des égalités entre les puissances des points $I$ et $H$ par rapport à ces deux cercles, nous pouvons conclure que la droite $(I H)$ est l'axe radical des cercles $\\omega_{1}$ et $\\omega_{2}$.\n\nReste à montrer que $M$ appartient aussi à l'axe radical de $\\omega_{1}$ et $\\omega_{2}$. Puisque ces cercles ont même rayon, il suffit pour cela de prouver que $M$ est à égale distance de leurs centres. Soient donc $O_{1}$ et $O_{2}$ les centres de $\\omega_{1}$ et $\\omega_{2}$, respectivement. Puisque $M$ est le milieu de l'arc $B A C$, nous avons $M B = M C$. D'autre part, $B O_{1} = C O_{2}$ et $\\widehat{M B O_{1}} = \\widehat{M B A} = \\widehat{M C A} = \\widehat{M C O_{2}}$, donc les triangles $M B O_{1}$ et $M C O_{2}$ sont isométriques. Nous avons donc bien $M O_{1} = M O_{2}$, ce qui conclut.\nSolution:\n\nOn définit les points $K, L, U, V$ comme dans la solution 1. Soit $P$ le point d'intersection de $(D U)$ et $(E I)$, et soit $Q$ le point d'intersection de $(E V)$ et $(D I)$.\n\nPuisque $D B = B C = C E$, les droites $(C I)$ et $(B I)$ sont perpendiculaires respectivement à $(B E)$ et $(C D)$. Ainsi, les droites $(B I)$ et $(E V)$ sont parallèles et, le triangle $B I E$ étant isocèle en $I$, $\\widehat{I E B} = \\widehat{I B E} = \\widehat{U E H}$. De même, les droites $(C I)$ et $(D U)$ sont parallèles et $\\widehat{I D C} = \\widehat{I C D} = \\widehat{V D H}$. Puisque $\\widehat{U E H} = \\widehat{V D H}$, les points $D, Q, F, P, E$ sont cocycliques. On tire de là l'égalité des puissances $I P \\cdot I E = I Q \\cdot I D$.\n\nSoit $R$ le deuxième point d'intersection centre du cercle circonscrit au triangle $H E P$ et de la droite $(H I)$. Puisque $I H \\cdot I R = I P \\cdot I E = I Q \\cdot I D$, les points $D, Q, H, R$ sont également cocycliques, de même que les points $E, P, H, R$. Nous avons $\\widehat{D Q H} = \\widehat{E P H} = \\widehat{D F E} = \\widehat{B F C} = 180^{\\circ} - \\widehat{B I C} = 90^{\\circ} - \\widehat{B A C} / 2$. En combinant ce calcul avec la cocyclicité de $D, Q, H, R$, et de $E, P, H, R$, on obtient $\\widehat{D R H} = \\widehat{E R H} = 180^{\\circ} - (90^{\\circ} - \\widehat{B A C} / 2) = 90^{\\circ} + \\widehat{B A C} / 2$. Ainsi, $\\widehat{D R H} + \\widehat{E R H} > 180^{\\circ}$, et donc $R$ est à l'intérieur du triangle $D E H$. Par conséquent, $\\widehat{D R E} = 360^{\\circ} - \\widehat{D R H} - \\widehat{E R H} = 180^{\\circ} - \\widehat{B A C}$ ce qui implique que les points $A, D, R, E$ sont cocycliques.\n\n![](attached_image_2.png)\n\nPuisque $M B = M C$, $B D = C E$ et $\\widehat{M B D} = \\widehat{M C E}$, les triangles $M B D$ et $M C E$ sont isométriques et $\\widehat{M D A} = \\widehat{M E A}$. Ainsi, les points $M, D, E, A$ sont cocycliques. On en conclut que les points $M, D, R, E, A$ sont cocycliques. On a alors $\\widehat{M A E} = \\widehat{B A C} + \\widehat{M A B} = \\widehat{B A C} + \\widehat{M C B} = \\widehat{B A C} + (90^{\\circ} - \\angle B A C / 2)$, d'où $\\widehat{M R E} = 180^{\\circ} - \\widehat{M A E} = 90^{\\circ} - \\angle B A C / 2$. On avait vu plus haut que $\\widehat{E R H} = 90^{\\circ} + \\widehat{B A C} / 2$, donc les points $I, H, R, M$ sont colinéaires.\nSolution:\n\nCette solution est une solution calculatoire : il s'agit de déterminer les pentes des droites $(I H)$ et $(I M)$ dans un certain repère, et de montrer qu'elles sont égales. Supposons que nous avons un repère dans lequel les coordonnées des points $B, C, D, E$ sont données respectivement par $\\left(b_{x}, b_{y}\\right), \\left(c_{x}, c_{y}\\right), \\left(d_{x}, d_{y}\\right), \\left(e_{x}, e_{y}\\right)$. Après avoir remarqué comme dans les solutions précédentes que $(B E) \\perp (C I)$ et $(C D) \\perp (B I)$, on a $\\overrightarrow{I H} = \\overrightarrow{I C} + \\overrightarrow{C D} + \\overrightarrow{D H}$, d'où $\\overrightarrow{I H} \\cdot \\overrightarrow{B E} = \\overrightarrow{C D} \\cdot \\overrightarrow{D E}$. De même, nous avons $\\overrightarrow{I H} = \\overrightarrow{I B} + \\overrightarrow{B E} + \\overrightarrow{E H}$, d'où $\\overrightarrow{I H} \\cdot \\overrightarrow{C D} = \\overrightarrow{C D} \\cdot \\overrightarrow{B E}$. On obtient donc $\\overrightarrow{I H} \\cdot (\\overrightarrow{B E} + \\overrightarrow{D C}) = 0$. On en conclut que la pente de la droite $(I H)$ est $\\left(c_{x} + e_{x} - b_{x} - d_{x}\\right) / \\left(b_{y} + d_{y} - c_{y} - e_{y}\\right)$.\n\nSupposons maintenant que l'axe des abscisses a pour vecteur directeur $\\overrightarrow{B C}$ et coïncide avec la droite $(B C)$, et posons $\\alpha = \\widehat{B A C}, \\beta = \\widehat{A B C}, \\gamma = \\widehat{A C B}$. Puisque $D B = B C = C E$, nous avons $c_{x} - b_{x} = B C, e_{x} - d_{x} = B C - B C \\cos \\beta - B C \\cos \\gamma$, $b_{y} = c_{y} = 0, d_{y} - e_{y} = B C \\sin \\beta - B C \\sin \\gamma$. Ainsi, par la formule obtenue plus haut, la pente de $(I H)$ est $(2 - \\cos \\beta - \\cos \\gamma) / (\\sin \\beta - \\sin \\gamma)$.\n\nMontrons maintenant que la pente de $(M I)$ est la même. Soient $r$ et $R$ respectivement les rayons des cercles inscrit et circonscrit au triangle $A B C$, et $\\left(m_{x}, m_{y}\\right)$ et $\\left(i_{x}, i_{y}\\right)$ les coordonnées de $M$ et de $I$ dans le repère choisi. Nous avons $\\widehat{B M C} = B A C = \\alpha$ et $B M = M C$ (ceci voulant dire que le projeté de $M$ sur la droite $(B C)$ est le milieu du segment $[B C]$), et donc\n$$\nm_{y} - i_{y} = \\frac{B C}{2 \\tan \\left(\\frac{\\alpha}{2}\\right)} - r\n$$\nD'autre part, en posant $A B = u + v, B C = v + w, C A = w + u$, on a $i_{x} = b_{x} + v = c_{x} - w$, donc\n$$\ni_{x} = \\frac{1}{2} \\left(b_{x} + c_{x} + v - w\\right) = m_{x} + \\frac{A C - A B}{2}\n$$\nd'où $m_{x} - i_{x} = \\frac{A C - A B}{2}$. Par conséquent, la pente de la droite $(M I)$ est $\\left(\\frac{B C}{\\tan (\\alpha / 2)} - 2 r\\right) / (A C - A B)$.\n\nLa loi des sinus s'écrit\n$$\n\\frac{B C}{\\sin \\alpha} = \\frac{A C}{\\sin \\beta} = \\frac{A B}{\\sin \\gamma} = 2 R\n$$\nd'où\n$$\n\\frac{B C}{\\tan \\left(\\frac{\\alpha}{2}\\right)} = 4 R \\cos^{2} \\left(\\frac{\\alpha}{2}\\right) = 2 R (1 + \\cos \\alpha)\n$$\nEn utilisant l'identité\n$$\n\\frac{r}{R} = \\cos \\alpha + \\cos \\beta + \\cos \\gamma - 1\n$$\n(voir démonstration plus bas) on obtient l'égalité des pentes\n$$\n\\frac{\\frac{B C}{\\tan (\\alpha / 2)} - 2 r}{A C - A B} = \\frac{2 R (1 + \\cos \\alpha) - 2 r}{2 R (\\sin \\beta - \\sin \\gamma)} = \\frac{2 - \\cos \\beta - \\cos \\gamma}{\\sin \\beta - \\sin \\gamma}\n$$\nqui donne la colinéarité des points $I, H, M$.\n\nDémonstration de l'identité (2) : Commençons par démontrer une autre identité utile :\n$$\n\\frac{r}{R} = 4 \\sin \\frac{\\alpha}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}\n$$\nPour cela, posons encore une fois $A B = u + v, B C = v + w, C A = w + u$. Puisque $v$ est la distance à $B$ du point de contact du cercle inscrit avec le segment $[B C]$, nous avons $r = v \\tan \\frac{\\beta}{2}$, et de même, $r = w \\tan \\frac{\\gamma}{2}$. Ainsi,\n$$\nB C = v + w = r \\left(\\frac{1}{\\tan \\frac{\\beta}{2}} + \\frac{1}{\\tan \\frac{\\gamma}{2}}\\right) = r \\left(\\frac{\\cos \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2} + \\sin \\frac{\\beta}{2} \\cos \\frac{\\gamma}{2}}{\\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}}\\right),\n$$\net donc, en utilisant le fait que $\\alpha + \\beta + \\gamma = \\pi$ et le fait que $\\sin \\left(\\frac{\\pi}{2} - x\\right) = \\cos x$, on a\n$$\nr = \\frac{B C \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2}}{\\cos \\frac{\\alpha}{2}}\n$$\nD'autre part, nous avons $B C = 2 R \\sin \\alpha = 4 R \\sin \\frac{\\alpha}{2} \\cos \\frac{\\alpha}{2}$, d'où le résultat.\n\nReste à en déduire l'identité (2). Rappelons pour cela l'identité trigonométrique $2 \\sin x \\sin y = \\cos (x - y) - \\cos (x + y)$, qui nous permet d'écrire :\n$$\n\\begin{aligned}\n4 \\sin \\frac{\\alpha}{2} \\sin \\frac{\\beta}{2} \\sin \\frac{\\gamma}{2} & = 2 \\sin \\frac{\\gamma}{2} \\left(\\cos \\left(\\frac{\\alpha - \\beta}{2}\\right) - \\cos \\left(\\frac{\\alpha + \\beta}{2}\\right)\\right) \\\\\n& = 2 \\sin \\frac{\\gamma}{2} \\left(\\cos \\left(\\frac{\\alpha - \\beta}{2}\\right) - \\sin \\frac{\\gamma}{2}\\right) \\\\\n& = 2 \\cos \\left(\\frac{\\alpha + \\beta}{2}\\right) \\cos \\left(\\frac{\\alpha - \\beta}{2}\\right) - 2 \\sin^{2} \\left(\\frac{\\gamma}{2}\\right) \\\\\n& = \\cos \\alpha + \\cos \\beta + \\cos \\gamma - 1,\n\\end{aligned}\n$$\noù on a utilisé les identités $\\cos \\alpha + \\cos \\beta = 2 \\cos \\left(\\frac{\\alpha + \\beta}{2}\\right) \\cos \\left(\\frac{\\alpha - \\beta}{2}\\right)$ et $\\cos \\gamma = 1 - 2 \\sin^{2} \\frac{\\gamma}{2}$ pour conclure.\nSolution:\n\nProlongeons les bissectrices $(B I)$ et $(C I)$ de sorte qu'elles intersectent le cercle circonscrit à $A B C$ respectivement en $P$ et en $Q$. Appelons de plus $R$ et $S$ les points d'intersection respectifs de la hauteur de $D E F$ issue de $D$ avec $(B I)$ et de celle issue de $E$ avec $(C I)$.\n\nDe même que dans les solutions précédentes, nous avons que $(B I)$ et $(C D)$ sont perpendiculaires. Puisque $(E H)$ et $(D F)$ sont également perpendiculaires, $(H S)$ et $(R I)$ sont parallèles. De même, $(H R)$ et $(S I)$ sont parallèles, et par conséquent, $H S I R$ est un parallélogramme.\n\nD'autre part, puisque $M$ est le milieu de l'arc $B A C$, nous avons\n$$\n\\widehat{M P I} = \\widehat{M P B} = \\widehat{M C B} = \\widehat{M B C} = \\widehat{M Q C} = \\widehat{M Q I}\n$$\naussi bien que\n$$\n\\widehat{P I Q} = \\widehat{B I C} = 180^{\\circ} - \\widehat{P B C} - \\widehat{Q C B} = 180^{\\circ} - \\frac{1}{2}(\\widehat{A B C} + \\widehat{A C B}) = 90^{\\circ} + \\widehat{A B C}\n$$\net\n$$\n\\begin{aligned}\n\\widehat{P M Q} & = \\widehat{P M C} + \\widehat{C M B} + \\widehat{Q P B} \\\\\n& = \\widehat{C A B} + \\widehat{P B C} + \\widehat{Q C B} \\\\\n& = \\widehat{C A B} + \\frac{1}{2}(\\widehat{A B C} + \\widehat{A C B}) \\\\\n& = 90^{\\circ} + \\widehat{A B C} \\\\\n& = \\widehat{P I Q} .\n\\end{aligned}\n$$\nAinsi, $M P I Q$ est un parallélogramme.\n\nPuisque $C I$ est la médiatrice du segment $[B E]$, le triangle $B S E$ est isocèle, d'où $\\widehat{F B S} = \\widehat{E B S} = \\widehat{S E B} = \\widehat{H E F} = \\widehat{H D F} = \\widehat{R D F} = \\widehat{F C S}$, et donc $B, S, F, C$ sont cocycliques. De même, $B, F, R, C$ sont cocycliques. Il s'ensuit que $B, S, R, C$ sont cocycliques. Puisque $B, Q, P, C$ sont également cocycliques, $(S R)$ et $(Q P)$ sont parallèles.\n\nOn en conclut que $H S I R$ et $M Q I P$ sont des parallélogrammes homothétiques, et que par conséquent $M, H, I$ sont colinéaires.\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier strictement positif. Trouver le plus petit entier $k$ ayant la propriété suivante : pour tous réels $a_{1}, \\ldots, a_{d}$ vérifiant $a_{1}+a_{2}+\\cdots+a_{d}=n$ et $0 \\leqslant a_{i} \\leqslant 1$ pour $i=1,2, \\ldots, d$, il est possible de regrouper les nombres en $k$ paquets (éventuellement vides) de sorte que la somme des nombres de chaque paquet soit $\\leqslant 1$.",
    "options": [],
    "answer": "2n - 1",
    "solution": "Solution:\n\n$\\triangleright$ Les réels $a_{1}=\\ldots=a_{2 n-1}=\\frac{n}{2 n-1}$ appartiennent à l'intervalle $\\left.] \\frac{1}{2}, 1\\right]$ et leur somme vaut $n$. Comme chaque réel est strictement plus grand que $\\frac{1}{2}$, chaque paquet contient au plus un de ces $2 n-1$ réels (car la somme de deux d'entre eux est strictement supérieure à 1 ) : il faut donc au moins $2 n-1$ paquets.\n\n$\\triangleright$ Supposons que l'on ait partitionné notre ensemble de réels $a_{1}, \\ldots, a_{d}$ en $k \\geq 2 n$ paquets et notons $\\alpha_{1} \\leq \\ldots \\leq \\alpha_{k}$ les sommes associées à ces paquets classées par ordre croissant. Alors, en regroupant ces nombres deux par deux, on obtient\n$$\n\\begin{aligned}\nn & =\\alpha_{1}+\\alpha_{2}+\\ldots+\\alpha_{k} \\\\\n& =\\left(\\alpha_{1}+\\alpha_{2}\\right)+\\left(\\alpha_{3}+\\alpha_{4}\\right)+\\ldots \\\\\n& \\geq\\left\\lfloor\\frac{k}{2}\\right\\rfloor\\left(\\alpha_{1}+\\alpha_{2}\\right) \\geq n\\left(\\alpha_{1}+\\alpha_{2}\\right)\n\\end{aligned}\n$$\nAinsi, $\\alpha_{1}+\\alpha_{2} \\leq 1$ et on peut regrouper les deux paquets correspondants en un nouveau paquet de somme totale inférieure ou égale à 1 et on est ramené à $k-1$ paquets.\n\n$\\triangleright$ En conclusion, il est nécessaire d'avoir $2 n-1$ paquets et, si l'on a partitionné en plus de $2 n$ paquets, on peut procéder à un regroupement et faire diminuer le nombre de paquets nécessaires. La valeur recherchée est donc $2 n-1$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCombien existe-t-il de couples d'entiers strictement positifs $(a, b)$ tels que\n$$\n\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{2014} ?\n$$",
    "options": [],
    "answer": "27",
    "solution": "Solution:\n\nSoient $a, b$ deux entiers strictement positifs tels que $\\frac{1}{a} + \\frac{1}{b} = \\frac{1}{2014}$. On a en particulier $a, b > 2014$. On peut donc multiplier l'équation par $ab$ : on cherche en fait les entiers $a, b > 2014$ tels que $ab - 2014a - 2014b = 0$ ou encore, de manière équivalente, tels que $(a - 2014)(b - 2014) = 2014^{2}$. On en déduit que le nombre recherché est le nombre de couples d'entiers strictement positifs $(u, v)$ tels que $uv = 2014^{2}$, autrement dit le nombre de diviseurs positifs de $2014^{2}$. Comme $2014^{2} = 2^{2} \\cdot 19^{2} \\cdot 53^{2}$, $2014^{2}$ possède $(2+1) \\cdot (2+1) \\cdot (2+1) = 27$ diviseurs positifs.\n\nLa réponse est donc $27$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère un échiquier classique $8 \\times 8$. Un zigzag est un chemin sur les cases blanches, qui part d'une case (quelconque) de la ligne du bas, et monte de ligne en ligne jusqu'à atteindre celle du haut (sur n'importe quelle case) : à chaque étape, on monte d'une case en diagonale. Combien y a-t-il de zigzags?",
    "options": [],
    "answer": "296",
    "solution": "Solution:\n\nOn écrit un 1 dans chacune des quatre cases blanches du bas. Puis dans la deuxième ligne, on écrit la somme des cases blanches qui y mènent : c'est le nombre de manières d'atteindre ces cases. On recommence dans la troisième ligne, etc. jusqu'à la huitième ligne : sur chacune des quatre cases, on a le nombre de zigzags qui arrivent à cette case. Le nombre total de zigzags est donc $69+103+35+89=296$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $C$ et $C'$ deux cercles de centres $O$ et $O'$, extérieurs l'un à l'autre. Une tangente commune extérieure coupe les deux tangentes communes intérieures aux points $M$ et $N$.\n\nMontrer que $(OM)$ est perpendiculaire à $(O'M)$ et que $(ON)$ est perpendiculaire à $(O'N)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $(T)$ la tangente commune extérieure de l'énoncé. Soit $(T')$ la tangente commune intérieure passant par $M$. Les droites $(T)$ et $(T')$ sont donc les deux tangentes à $C$ issues de $M$.\n\nComme ces deux tangentes sont symétriques par rapport à $(OM)$, la droite $(OM)$ est une bissectrice de $(T)$ et $(T')$. De même, $(O'M)$ est une bissectrice de $(T)$ et $(T')$.\n\nComme les bissectrices de deux droites sont perpendiculaires, $(OM)$ et $(O'M)$ sont perpendiculaires ou confondues.\n\nSi elles étaient confondues, $O$, $O'$, $M$ seraient alignés, et les deux tangentes à $C$ passant par $M$ seraient extérieures, ce qui n'est pas le cas.\n\nPar conséquent, $(OM) \\perp (O'M)$. On montre de même que $(ON) \\perp (O'N)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCalculer\n$$\n\\sqrt{1+\\frac{1}{1^{2}}+\\frac{1}{2^{2}}}+\\sqrt{1+\\frac{1}{2^{2}}+\\frac{1}{3^{2}}}+\\cdots+\\sqrt{1+\\frac{1}{2014^{2}}+\\frac{1}{2015^{2}}}\n$$",
    "options": [],
    "answer": "2015 - 1/2015",
    "solution": "Solution:\n\nOn réduit au même dénominateur\n$$\n\\begin{aligned}\n1+\\frac{1}{n^{2}}+\\frac{1}{(n+1)^{2}} & =\\frac{n^{2}(n+1)^{2}+(n+1)^{2}+n^{2}}{n^{2}(n+1)^{2}} \\\\\n& =\\frac{n^{4}+n^{2}(2 n+1)+n^{2}+(n+1)^{2}}{n^{2}(n+1)^{2}} \\\\\n& =\\frac{n^{4}+2 n^{2}(n+1)+(n+1)^{2}}{n^{2}(n+1)^{2}} \\\\\n& =\\frac{\\left(n^{2}+(n+1)\\right)^{2}}{n^{2}(n+1)^{2}}\n\\end{aligned}\n$$\nOn en déduit que la somme recherchée vaut\n$$\n\\begin{aligned}\n\\sum_{n=1}^{2014} \\frac{n^{2}+n+1}{n(n+1)} & =\\sum_{n=1}^{2014} 1+\\frac{1}{n(n+1)} \\\\\n& =2014+\\sum_{n=1}^{2014} \\frac{1}{n(n+1)} \\\\\n& =2014+\\sum_{n=1}^{2014} \\frac{1}{n}-\\frac{1}{n+1} \\\\\n& =2014+\\left(1-\\frac{1}{2}+\\frac{1}{2}-\\frac{1}{3}+\\cdots+\\frac{1}{2014}-\\frac{1}{2015}\\right)\n\\end{aligned}\n$$\nOn reconnaît une somme télescopique. Les termes se simplifient deux à deux, sauf 1 et $\\frac{1}{2015}$, donc la somme vaut\n$$\n2014+1-\\frac{1}{2015}=2015-\\frac{1}{2015}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les polynômes $P$ à coefficients entiers pour lesquels l'ensemble $P(\\mathbb{N})$ contient une suite géométrique infinie de raison $a$ avec $a \\notin\\{-1,0,1\\}$ et de premier terme non nul.",
    "options": [],
    "answer": "All such polynomials are exactly P(x) = w (v x + u)^n with integers n > 0, v ≠ 0, and w ≠ 0.",
    "solution": "Solution:\n\nTout d'abord, on note que si $P$ est un polynôme ayant les propriétés de l'énoncé, alors $-P$ les possède aussi (il suffit de changer le premier terme de la suite géométrique en son opposée). On peut donc supposer que le coefficient dominant de $P$ est strictement positif.\n\nSoit $\\left(u_{k}\\right)=\\left(u_{0} \\cdot a^{k}\\right)$ une suite géométrique infinie telle que décrite ci-dessus. Alors, quitte à la remplacer par la suite $\\left(u_{2 k}\\right)=\\left(u_{0} \\cdot\\left(a^{2}\\right)^{k}\\right)$, on peut supposer que $a>0$.\n\nDe plus, notons que $a=\\frac{u_{1}}{u_{0}}$ est rationnel. Si $a$ n'est pas entier, soit $p$ un facteur premier tel que la valuation $p$-adique de $a$ soit strictement négative c'est-à-dire que $v_{p}(a) \\leq -1$. En posant $k=v_{p}\\left(u_{0}\\right)$, on constate alors que $v_{p}\\left(u_{k+1}\\right) \\leq -1$, ce qui contredit le fait que $u_{k+1}$ soit entier, donc que $u_{k+1} \\in P(\\mathbb{N})$. Ainsi, on sait que $a$ est un entier tel que $a>1$.\n\nDéveloppons alors le polynôme $P$ sous la forme $P(x)=p_{n} x^{n}+p_{n-1} x^{n-1}+\\cdots+p_{1} x+p_{0}$. Pour tout entier $b$, on a\n$$\n\\begin{aligned}\nP(a x+b) & =a^{n} p_{n} x^{n}+\\left(n a^{n-1} p_{n} b+a^{n-1} p_{n-1}\\right) x^{n-1}+\\cdots \\\\ \n\\text{et} \\\\\na^{n} P(x) & =a^{n} p_{n} x^{n}+a^{n} p_{n-1} x^{n-1}+\\cdots+a^{n} p_{0} .\n\\end{aligned}\n$$\nLe polynôme $P(a x+b)-a^{n} P(x)$ est un polynôme de degré au plus $n-1$ et dont le coefficient de degré $n-1$ est $a^{n-1}\\left(n p_{n} b+p_{n-1}(1-a)\\right)$. Notons que $P$ est non constant, donc que $n p_{n}>0$ : ce coefficient a donc une expression affine strictement croissante en $b$. Ainsi, il existe un entier $b$ tel que\n$$\nn p_{n}(b+1)+p_{n-1}(1-a)>0>n p_{n}(b-1)+p_{n-1}(1-a) .\n$$\nPar conséquent, il existe même un entier $N \\geq 0$ tel que\n- $P(a x+b-1)<a^{n} P(x)<P(a x+b+1)$ pour tout $x \\geq N$, et\n- $P$ est strictement croissante sur $[N,+\\infty[$.\n\nEn outre, soit $M$ un entier tel que $M \\geq N$ et $a M+b-1 \\geq N$ : quitte à considérer une suite $\\left(u_{k+\\ell}\\right)=\\left(u_{\\ell} \\cdot a^{k}\\right)$ au lieu de la suite $\\left(u_{k}\\right)$, on peut supposer que $\\left|u_{0}\\right|>\\sum_{i=0}^{M}|P(i)|$. Pour tout entier $k \\geq 0$, il existe donc un entier $v_{k}>M$ tel que $P\\left(v_{k}\\right)=u_{k}=u_{0} \\cdot a^{k}$.\n\nEn particulier, notons que\n$$\nP\\left(a v_{k}+b-1\\right)<a^{n} P\\left(v_{k}\\right)=P\\left(v_{k+n}\\right)<P\\left(a v_{k}+b+1\\right),\n$$\ndonc que $v_{k+n}=a v_{k}+b$ et que $a^{n} P\\left(v_{k}\\right)=P\\left(a v_{k}+b\\right)$. Ainsi, chaque entier $v_{k}$ est une racine du polynôme $a^{n} P(X)-P(a X+b)$, qui est donc nul. On a donc l'égalité de polynômes $a^{n} P(X)=P(a X+b)$.\n\nPuisque $P$ est non constant, soit $r$ une racine de $P$. On pose alors\n$$\n\\delta=\\frac{b}{1-a} \\text{ et } \\rho_{k}=a^{k}(r-\\delta)+\\delta\n$$\nAlors $\\rho_{0}=r$ et $\\rho_{k+1}=a\\left(\\rho_{k}-\\delta\\right)+\\delta=a \\rho_{k}+b$. Une récurrence immédiate montre donc que tous les termes $\\rho_{k}$ sont des racines de $P$.\n\nOr, si $r \\neq \\delta$, les termes $\\rho_{k}$ sont deux à deux distincts, et $P$ se retrouve être nul, ce qui est impossible. donc $r=\\delta$ est nécessairement la seule racine de $P$, de sorte que l'on peut écrire $P(X)=p_{n}(X-\\delta)^{n}$. Puisque $P(X)=p_{n} X^{n}-n p_{n} \\delta X^{n-1}+\\cdots+(-1)^{n} p_{n} \\delta^{n}$ est à coefficients entiers, alors $n p_{n} \\delta$ est entier, donc $\\delta$ est rationnel, et peut s'écrire sous forme irréductible $\\delta=\\frac{u}{v}$. Puis $p_{n} \\delta^{n}$ est entier, donc $v^{n}$ divise $p_{n}$, et l'on peut factoriser $p_{n}$ sous la forme $p_{n}=w v^{n}$. On peut alors réécrire $P(X)=w(v X-u)^{n}$, avec $n, u, v$ et $w$ entiers et $n \\geq 1, v \\neq 0, w \\neq 0$.\n\nRéciproquement, supposons que $P(X)=w(v X-u)^{n}$ avec $n, u, v$ et $w$ entiers et $n \\geq 1$, $v \\neq 0, w \\neq 0$.\n- Si $u=0$, alors $P\\left(2^{k}\\right)=v^{n} \\cdot w \\cdot 2^{k n}$ décrit bien une suite géométrique de raison $2^{n} \\notin\\{-1,0,1\\}$ et de premier terme $P(1)=v^{n} \\cdot w$ non nul.\n- Si $u \\cdot v>0$, on pose $x_{k}=(3 u v-1)^{2 k+1}+1$ et $y_{k}=\\frac{u}{v} x_{k}$. Alors $x_{k} \\equiv 0(\\bmod v)$, de sorte que $y_{k} \\in \\mathbb{N}$. En outre, $P\\left(y_{k}\\right)=w \\cdot u^{n} \\cdot(3 u v-1)^{(2 k+1) n}$ décrit bien une suite géométrique de raison $(3 u v-1)^{2 n} \\notin\\{-1,0,1\\}$ et de premier terme $P\\left(3 u^{2}\\right)=w \\cdot u^{n} \\cdot(3 u v-1)^{n}$ non nul.\n- Si $u \\cdot v<0$, on pose $x_{k}=1-(3 u v+1)^{2 k}$ et $y_{k}=\\frac{u}{v} x_{k}$. Alors $x_{k} \\equiv 0(\\bmod v)$, de sorte que $y_{k} \\in \\mathbb{N}$. En outre, $P\\left(y_{k}\\right)=w \\cdot(-u)^{n} \\cdot(3 u v+1)^{(2 k) n}$ décrit bien une suite géométrique de raison $(3 u v+1)^{2 n} \\notin\\{-1,0,1\\}$ et de premier terme $P(0)=w \\cdot(-u)^{n}$ non nul.\n\nLes polynômes recherchés sont donc exactement ceux de la forme $P(x)=w(v x+u)^{n}$, avec $n, u, v$ et $w$ des entiers tels que $n>0, v \\neq 0$ et $w \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\in \\mathbb{N}^{*}$. On appelle $S$ l'ensemble des suites de $2 n$ chiffres comportant $n$ zéros et autant de uns. Deux suites de $S$ sont voisines lorsqu'il suffit de changer la position d'un chiffre de l'une pour obtenir l'autre. Par exemple, 11100010 et 10110010 sont voisines puisqu'en décalant le premier 0 de la deuxième suite de deux \"pas\" vers la droite, on obtient la première suite. Soit $T$ un sous-ensemble dense de cardinal minimal. Montrer que $\\frac{1}{n^{2}+1}|S| \\leq|T| \\leq \\frac{1}{n+1}|S|$.\n\nRemarque : on note $|A|$ le cardinal de $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn traite séparément les inégalités.\n\n- Montrons que $\\frac{1}{2 n^{2}+1}|S| \\leq|T|$. Soit $s$ un élément de $S$. Soit $s_{i, j}$ la suite que l'on peut obtenir à partir de $s$, en déplaçant le $i$-ème chiffre de $s$ pour le mettre en position $j$. Si $s_{i, j} \\neq s$, alors $j \\neq i$. En outre, sans perte de généralité, on suppose que le $i$-ème chiffre de $s$ est un 0 ; si $j>i$ et si le $j$-ème chiffre de $s$ est un 0, alors $s_{i, j}=s_{i, j-1}$; si $j<i$ et si le $j$-ème chiffre de $s$ est un 0, alors $s_{i, j}=s_{i, j+1}$. Ainsi, toute suite $s_{i, j}$ distincte de $s$ peut s'écrire sous la forme $s_{i, k}$, telle que le $k$-ème chiffre de $s$ soit un 1 : à $i$ fixé, il existe au plus $n$ telles suites. Puisque $i$ peut prendre $2 n$ valeurs, la suite $s$ a donc, en effet, au plus $2 n^{2}$ voisins différents d'elle-même. Ainsi, il faut au moins $\\frac{1}{2 n^{2}+1}|S|$ éléments pour faire un ensemble dense.\n\n- Montrons que $|T| \\leq \\frac{1}{n+1}|S|$. Considérons les 0 comme des poteaux fixes, délimitant $n+1$ intervalles de 1 (on compte les extérieurs), certains éventuellement vides. Si $s \\in S$, soit $r_{k}$ le nombre de 1 dans le $k$-ème intervalle en partant de la gauche. On introduit $N(s):=r_{1}+2 r_{2}+\\cdots+n r_{n}$ et $R(s)$ la congruence de $N(s)$ modulo $n+1$. Si on déplace un 0 du $i$-ème intervalle (il y a bien un intervalle non vide) vers le $j$-ème, on obtient $s^{\\prime}$ telle que $N\\left(s^{\\prime}\\right)=N(s)+j-i$. En faisant varier $j$ entre 1 et $n+1$, on obtient toutes les congruences possibles pour $R\\left(s^{\\prime}\\right)$. Si, pour tout $0 \\leq k \\leq n, S_{k}$ est l'ensemble des $s \\in S$ telles que $R(s)=k$, toute suite de $S_{k}$ a une voisine dans $S_{k^{\\prime}}$ pour tout $k^{\\prime} \\neq k$. Comme \"être voisine\" est une relation symétrique (si $s$ voisine de $s^{\\prime}$, alors $s^{\\prime}$ est voisine de $s$), toute suite $s$ qui n'est pas dans $S_{k}$ a une voisine dans $S_{k}$, donc $S_{k}$ est dense. Or les $S_{k}$ forment une partition de $S$ en $n+1$ sous-ensembles, donc il existe un $k_{0}$ tel que $\\left|S_{k_{0}}\\right| \\leq \\frac{1}{n+1}|S|$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier strictement positif $x$, on note $S(x)$ la somme des chiffres de son écriture décimale.\nSoit $k>0$ un entier. On définit la suite $\\left(x_{n}\\right)$ par $x_{1}=1$ et $x_{n+1}=S\\left(k x_{n}\\right)$ pour tout $n>0$.\nProuver que $x_{n}<27 \\sqrt{k}$, pour tout $n>0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLemme. Pour tout entier $s \\geqslant 0$, on a $10^{s} \\geqslant (s+1)^{3}$.\nPreuve du lemme. On raisonne par récurrence sur $s$.\nPour $s=0$, on a $10^{0}=1=(0+1)^{3}$.\nSupposons l'inégalité vraie pour la valeur $s$. Alors :\n$$\n10^{s+1} = 10 \\times 10^{s} \\geqslant 10(s+1)^{3}\n$$\nd'après l'hypothèse de récurrence\n$$\n= 10 s^{3} + 30 s^{2} + 30 s + 10\n$$\nOr,\n$$\n10 s^{3} + 30 s^{2} + 30 s + 10 \\geqslant s^{3} + 6 s^{2} + 12 s + 8 = (s+2)^{3}\n$$\nce qui achève la preuve.\n\nSi $x>0$ est un entier dont l'écriture décimale utilise $s+1$ chiffres, on a $S(x) \\leqslant 9(s+1)$ et $10^{s} \\leqslant x < 10^{s+1}$. Or, d'après le lemme, on a $s+1 \\leqslant 10^{\\frac{s}{3}}$, donc $s+1 \\leqslant x^{\\frac{1}{3}}$ et ainsi $S(x) \\leqslant 9 x^{\\frac{1}{3}}$.\n\nMontrons maintenant par récurrence que $x_{n}<27 \\sqrt{k}$ pour tout $n$. L'assertion est claire pour $n=1$. Supposons $x_{n}<27 \\sqrt{k}$, alors\n$$\nx_{n+1}=S\\left(k x_{n}\\right) \\leqslant 9\\left(k x_{n}\\right)^{\\frac{1}{3}} < 9(27 k \\sqrt{k})^{\\frac{1}{3}} = 9 \\times 3 \\times \\sqrt{k} = 27 \\sqrt{k}\n$$\nd'où le résultat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a$, $b$, $c$ trois nombres réels strictement positifs tels que $a + b + c = 9$. Montrer que\n$$\n\\frac{a^{3} + b^{3}}{a b + 9} + \\frac{b^{3} + c^{3}}{b c + 9} + \\frac{c^{3} + a^{3}}{c a + 9} \\geqslant 9\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn cherche une minoration de la forme $\\frac{a^{3} + b^{3}}{a b + 9} \\geqslant u(a + b) + v$, c'est-à-dire $f(a) \\geqslant 0$ où $f(a) = a^{3} + b^{3} - (u(a + b) + v)(a b + 9)$.\n\nEn sommant les trois inégalités, on en déduirait\n$$\n\\frac{a^{3} + b^{3}}{a b + 9} + \\frac{b^{3} + c^{3}}{b c + 9} + \\frac{c^{3} + a^{3}}{c a + 9} \\geqslant 2u(a + b + c) + 3v = 3(6u + v),\n$$\ndonc il suffit que $6u + v = 3$.\n\nD'autre part, si une telle minoration existe, comme l'inégalité devient une égalité lorsque $a = b = c = 3$, on doit avoir $f(3) = 0$ si $b = 3$. Comme de plus $f(a) \\geqslant 0$ pour tout $a$, on doit nécessairement avoir $f'(3) = 0$ pour $b = 3$.\n\nCeci impose la condition $0 = f'(3) = 27 - 18u - (6u + v) \\times 3 = 18(1 - u)$, donc $u = 1$ et $v = -3$.\n\nMontrons donc $\\frac{a^{3} + b^{3}}{a b + 9} \\geqslant (a + b) - 3$. Comme $a^{3} + b^{3} \\geqslant \\frac{(a + b)^{3}}{4}$ et $a b + 9 \\leqslant \\left(\\frac{a + b}{2}\\right)^{2} + 9$, il suffit de montrer que\n$$\n\\frac{s^{3}}{s^{2} + 36} \\geqslant s - 3\n$$\noù $s = a + b$. On réduit au même dénominateur; l'inégalité se simplifie en $3s^{2} - 36s + 108 \\geqslant 0$, ou encore en $3(s - 6)^{2} \\geqslant 0$, ce qui est vrai.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn prend 1008 entiers distincts compris (au sens large) entre 1 et 2014.\n\na. Montrer qu'il existe trois entiers $a, b, c$ tels que le pgcd de $a$ et $b$ divise $c$ (le pgcd ou Plus Grand Diviseur Commun de deux entiers est le plus grand entier naturel qui les divise tous les deux).\n\nb. Montrer qu'il existe trois entiers $a, b, c$ tels que le pgcd de $a$ et $b$ ne divise pas $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nGroupons les entiers de 1 à 2014 en paires d'entiers consécutifs : $(1, 2), (3, 4),\\ldots, (2013, 2014)$. Il y en a 1007. Par le principe des tiroirs, si on choisit 1008 entiers, il y en aura deux d'une même paire, $a$ et $b$, avec $b=a+1$. Leur pgcd vaut 1 (puisqu'il divise leur différence), donc divise n'importe lequel des 1006 autres nombres, que l'on peut prendre pour $c$.\n\nb.\nIl y a 1007 entiers pairs prenables. Donc parmi 1008 nombres, il y a forcément un impair. S'il y a deux nombres pairs au moins parmi les 1008, notons-les $a$ et $b$. Soit $c$ un nombre impair parmi les 1008. Le $\\operatorname{pgcd}$ de $a$ et $b$ est multiple de 2, puisque 2 divise à la fois $a$ et $b$. (en fait, le $\\operatorname{pgcd}$ de $a$ et $b$ vaut 2 fois le $\\operatorname{pgcd}$ de $a / 2$ et $b / 2$). Donc s'il divise $c$, alors 2 divise $c$, ce qui est absurde.\n\nS'il y a moins de 2 nombres pairs parmi les 1008, ça veut dire qu'on a pris les 1007 impairs, ainsi qu'un nombre pair, noté $c$. On peut trouver des triplets d'au moins deux manières.\n\n- Solution courte : on dispose du triplet $(3,9,1)$, et il est clair que $\\operatorname{pgcd}(3,9)$ ne divise pas 1.\n\n- Solution moins courte : si $c$ est divisible par tous les pgcd de couples d'entiers impairs entre 1 et 2013, ça va lui faire beaucoup de diviseurs premiers : il sera divisible par le pgcd de 3 et 9, le pgcd de 5 et 15, celui de 7 et 21, celui de 11 et 33, etc. Donc il est divisible par $3,5,7,11$ déjà (et par 2, puisqu'il est pair). Le plus petit multiple commun de ces nombres est leur produit (puisqu'ils n'ont deux-à-deux aucun facteur commun). Il vaut $2 \\times 3 \\times 5 \\times 7 \\times 11=2310$ donc $c>2014$, ce qui n'est pas possible. Donc il existe deux nombres impairs $a$ et $b$ tels que leur pgcd ne divise pas $c$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k \\geqslant 1$ un entier. À chaque client, un opérateur téléphonique propose $k$ numéros pour lesquels la communication est gratuite (si une personne $A$ choisit le numéro de $B$, alors les appels de $A$ vers $B$ et de $B$ vers $A$ sont gratuits). On considère un groupe de $n$ personnes.\n\n1) Si $n \\geqslant 2k+2$, montrer qu'il existe deux personnes qui ne pourront pas communiquer gratuitement.\n\n2) Si $n=2k+1$, montrer que les $n$ personnes peuvent faire en sorte que toute personne peut communiquer gratuitement avec n'importe quelle autre.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn dira pour simplifier qu'une personne $A$ choisit une personne $B$ si le numéro de $B$ fait partie des numéros choisis par $A$ vers lesquels la communication est gratuite.\n\n1)\nComme chaque personne peut choisir au plus $k$ autres personnes, il y a au plus $kn$ paires de personnes qui peuvent communiquer gratuitement. Or le nombre total de paires de personnes est $\\dfrac{n(n-1)}{2}$, et on a\n$$\n\\frac{n(n-1)}{2} \\geqslant \\frac{n(2k+1)}{2} > kn\n$$\nAinsi, le nombre total de paires de personnes est strictement supérieur au nombre de paires de personnes qui peuvent communiquer gratuitement. Il existe donc forcément deux personnes qui ne pourront pas communiquer gratuitement.\n\n2)\nPlaçons les $2k+1$ personnes sur un cercle. Chaque personne choisit alors de communiquer gratuitement avec les $k$ personnes qui sont situées juste après elle. On vérifie aisément que cette configuration permet à toute personne de communiquer gratuitement avec n'importe quelle autre : par construction, chaque personne peut communiquer avec les $k$ personnes situées juste après elle ainsi qu'avec les $k$ personnes situées juste avant elles, autrement dit tout le monde.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSur une droite se trouvent 400 points bleus et 200 points verts. Montrer que l'on peut trouver un segment qui contient exactement 200 points bleus et 100 points verts.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $A_{1}, \\ldots, A_{600}$ les points de la droite, alignés dans cet ordre. Il s'agit de montrer qu'il existe un entier $k \\in \\{0, \\ldots, 300\\}$ tel que, parmi les points $A_{k+1}, \\ldots, A_{k+300}$, il y en a exactement 100 verts (il y en aura alors forcément 200 bleus).\n\nIntroduisons la fonction $f : \\{0, \\ldots, 300\\} \\longrightarrow \\mathbb{N}$ qui à un entier $k$ associe le nombre de points verts parmi $A_{k+1}, \\ldots, A_{k+300}$. On veut montrer que $f$ prend la valeur 100. Pour cela, remarquons deux choses :\n\n1. Tout d'abord, la somme $f(0) + f(300)$ compte le nombre total de points verts, et elle vaut donc 200. Ainsi, l'un des deux termes de la somme est supérieur ou égal à 100 tandis que l'autre ne dépasse pas 100. Quitte à changer le sens de parcours des points sur la droite, on peut supposer que $f(0) \\leqslant 100 \\leqslant f(300)$.\n\n2. Intéressons-nous maintenant à la différence $f(k) - f(k-1)$, où $k \\geqslant 1$ :\n- si $A_{k+300}$ et $A_{k}$ sont de même couleur, elle vaut 0 ;\n- si $A_{k+300}$ est vert et $A_{k}$ est bleu, elle vaut 1 ;\n- si $A_{k+300}$ est bleu et $A_{k}$ est vert, elle vaut $-1$.\n\nEn particulier, on a $f(k) - f(k-1) \\leqslant 1$ pour tout $k \\in \\{1, \\ldots, 300\\}$.\n\nConsidérons maintenant le plus petit entier $p$ tel que $f(p) \\geqslant 100$ : cet entier existe bien, puisque $f(300) \\geqslant 100$. Si $p = 0$, il n'y a plus rien à faire : en effet, dans ce cas, on a $f(0) \\geqslant 100$ et $f(0) \\leqslant 100$, d'où $f(0) = 100$. Si $p > 0$, la minimalité de $p$ assure que $f(p-1) \\leqslant 99$ et donc, d'après ci-dessus, on a alors $100 \\leqslant f(p) \\leqslant f(p-1) + 1 \\leqslant 100$, et ainsi $f(p) = 100$.\n\nFinalement, dans tous les cas, la valeur 100 est bien atteinte par la fonction $f$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $n \\geqslant 1$ tels que $2^{n}+12^{n}+2014^{n}$ soit un carré parfait.",
    "options": [],
    "answer": "no positive integers n",
    "solution": "Solution:\n\nRegardons l'expression modulo $3$ : $2^{n}+12^{n}+2014^{n} \\equiv (-1)^{n}+1 \\pmod{3}$. Comme un carré n'est jamais congru à $2$ modulo $3$, on en déduit que $n$ est impair.\n\nRegardons ensuite l'expression modulo $7$ :\n$$\n2^{n}+12^{n}+2014^{n} \\equiv 2^{n}+(-2)^{n}+5^{n} \\equiv 5^{n} \\pmod{7}\n$$\ncar $n$ est impair. Lorsque $n$ est impair, $5^{n}$ ne peut être congru qu'à $3$, $5$ ou $6$ modulo $7$. Or un carré est congru à $0$, $1$, $2$ ou $4$ modulo $7$. Il n'existe donc pas d'entiers $n \\geqslant 1$ tels que $2^{n}+12^{n}+2014^{n}$ soit un carré parfait.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $p$ et $q$ deux nombres premiers supérieurs ou égaux à $7$. Soit $x=\\frac{p^{2012}+q^{2016}}{120}$. Calculer $x-[x]$, où $[x]$ désigne la partie entière de $x$, c'est-à-dire le plus grand entier inférieur ou égal à $x$.",
    "options": [],
    "answer": "1/60",
    "solution": "Solution:\n\nSi $a$ est un entier non divisible par $2$, $3$ et $5$, on vérifie facilement que $a^{4}$ est congru à $1$ modulo $3$, $5$ et $8$. Autrement dit, $a^{4}-1$ est divisible par $3$, $5$ et $8$, donc par $3 \\times 5 \\times 8 = 120$. On en déduit que $p^{4}$ et $q^{4}$ sont congrus à $1$ modulo $120$, donc $p^{2012}+q^{2016} \\equiv 1+1 \\equiv 2 \\pmod{120}$.\n\nOn en déduit que $x-[x]=\\frac{2}{120}=\\frac{1}{60}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer qu'il existe une infinité d'entiers positifs $n$ tels que le plus grand facteur premier de $n^{4}+n^{2}+1$ soit égal au plus grand facteur premier de $(n+1)^{4}+(n+1)^{2}+1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour tout $n \\geq 1$, on note $p_{n}$ le plus grand diviseur premier de $n^{4}+n^{2}+1$, et $q_{n}$ le plus grand diviseur premier de $n^{2}+n+1$.\nOn a donc $p_{n}=q_{n^{2}}$.\n\nDe plus, de l'identité\n$$\nn^{4}+n^{2}+1=\\left(n^{2}+n+1\\right)\\left(n^{2}-n+1\\right)=\\left(n^{2}+n+1\\right)\\left((n-1)^{2}+(n-1)+1\\right)\n$$\nil découle que $p_{n}=\\max \\left(q_{n}, q_{n-1}\\right)$ pour tout $n \\geq 2$.\n\nD'autre part, le nombre $n^{2}-n+1=n(n-1)+1$ est impair, donc\n$$\n\\operatorname{pgcd}\\left(n^{2}+n+1, n^{2}-n+1\\right)=\\operatorname{pgcd}\\left(2 n, n^{2}-n+1\\right)=\\operatorname{pgcd}\\left(n, n^{2}-n+1\\right)=1\n$$\nAinsi, on a $q_{n} \\neq q_{n-1}$.\n\nPour conclure, il suffit alors de prouver que l'ensemble\n$$\nS=\\left\\{n \\geq 2 \\mid q_{n}>q_{n-1} \\text{ et } q_{n}>q_{n+1}\\right\\}\n$$\nest infini, puisque pour tout $n \\in S$, on a\n$$\np_{n}=\\max \\left(q_{n}, q_{n-1}\\right)=q_{n}=\\max \\left(q_{n}, q_{n+1}\\right)=p_{n+1} .\n$$\n\nPar l'absurde: supposons que $S$ soit fini.\nPuisque $q_{2}=q_{4}=7<13=q_{3}$, on peut affirmer que $S$ n'est pas vide. Puisqu'il est fini, il possède alors un plus grand élément, disons $m$.\nLa suite $\\left(q_{n}\\right)_{n \\geq m}$ ne peut être strictement décroissante puisqu'elle est formée d'entiers positifs. Il existe donc $k \\geq m$ tel que $q_{k}<q_{k+1}$ (on rappelle que $q_{k} \\neq q_{k+1}$ ). Mais, il est également impossible d'avoir $\\left(q_{n}\\right)_{n \\geq k}$ strictement croissante car $q_{(k+1)^{2}}=p_{k+1}=\\max \\left(q_{k}, q_{k+1}\\right)=q_{k+1}$.\nNotons alors $t$ le plus petit entier supérieur ou égal à $k+1$ tel que $q_{t}>q_{t+1}$. La minimalité de $t$ assure que l'on a $q_{t-1}<q_{t}$, et donc que $t \\in S$.\nPourtant, on a $t \\geq k+1>k \\geq m$, ce qui contredit la maximalité de $m$.\nPar suite, on a $S$ infini, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCombien y a-t-il de nombres à six chiffres qui ont quatre chiffres pairs, deux chiffres impairs et qui sont multiples de $5$ ?\n\nNote: un nombre ne commence pas par un $0$.",
    "options": [],
    "answer": "40625",
    "solution": "Solution:\n\nUn tel nombre termine par $0$ ou $5$. Nous allons compter séparément les nombres qui se terminent par $0$ et ceux qui se terminent par $5$.\n\nNombres dont le dernier chiffre est $0$ :\n- Si le premier chiffre est pair, nous avons $4$ possibilités pour le choisir ($2,4,6,8$, mais pas $0$). Il reste à choisir les $4$ chiffres du milieu. On a $6$ manières de positionner les impairs :\n$$\n(p, p, i, i),\\ (p, i, p, i),\\ (p, i, i, p),\\ (i, p, i, p),\\ (i, i, p, p),\\ (i, p, p, i)\n$$\n(où $p$ représente un chiffre pair et $i$ un chiffre impair). Ensuite, il y a $5$ possibilités pour le choix de chaque chiffre. Donc $6 \\times 4 \\times 5^{4}$ possibilités.\n\n- Si le premier chiffre est impair, nous avons $5$ possibilités pour le choisir. Il reste $4$ manières de positionner l'impair du milieu. Et de nouveau $5$ choix pour chaque chiffre du milieu. Donc $4 \\times 5^{5}$ nombres possibles.\n\nNombres dont le dernier chiffre est $5$ :\n- Si le premier chiffre est pair, nous avons $4$ possibilités pour le choisir. Il y a $4$ possibilités pour la place de l'impair du milieu, $5$ choix pour chacun des chiffres du milieu, donc $16 \\times 5^{4}$ possibilités.\n\n- Si le premier chiffre est impair il y a $5$ possibilités pour le choisir. Il reste alors simplement à déterminer les pairs du milieu : $5$ choix à chaque fois, donc $5^{5}$ choix en tout.\n\nAu total, il y a $24 \\times 5^{4} + 4 \\times 5^{5} + 16 \\times 5^{4} + 5^{5} = 13 \\times 5^{5} = 40625$ possibilités.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $n \\geqslant 1$ un entier. Déterminer tous les entiers $p \\geqslant 1$ pour lesquels il existe des entiers strictement positifs $x_{1}<x_{2}<\\cdots<x_{n}$ tels que\n$$\n\\frac{1}{x_{1}}+\\frac{2}{x_{2}}+\\cdots+\\frac{n}{x_{n}}=p\n$$",
    "options": [],
    "answer": "All integers p with 1 ≤ p ≤ n",
    "solution": "Solution:\nNous allons montrer que l'ensemble des solutions est l'ensemble $\\{1,2, \\ldots, n\\}$.\n\nSoient $p \\geqslant 1$ un entier et des entiers strictement positifs $x_{1}<x_{2}<\\cdots<x_{n}$ tels que\n$$\n\\frac{1}{x_{1}}+\\frac{2}{x_{2}}+\\cdots+\\frac{n}{x_{n}}=p\n$$\nOn remarque tout d'abord que $p \\leqslant n$. En effet, si $i$ et $j$ sont deux entiers tels que $i>j$, alors $i \\geqslant j+1$. On en déduit que $x_{i} \\geqslant i$ pour tout entier $1 \\leqslant i \\leqslant n$. Ainsi\n$$\n\\frac{1}{x_{1}}+\\frac{2}{x_{2}}+\\cdots+\\frac{n}{x_{n}} \\leqslant \\frac{1}{1}+\\frac{2}{2}+\\cdots+\\frac{n}{n} \\leqslant n .\n$$\nDonc $p \\leqslant n$.\n\nRéciproquement, soit $1 \\leqslant p \\leqslant n$ et montrons qu'il existe des entiers strictement positifs $x_{1}<x_{2}<\\cdots<x_{n}$ tels que\n$$\n\\frac{1}{x_{1}}+\\frac{2}{x_{2}}+\\cdots+\\frac{n}{x_{n}}=p .\n$$\nPour cela, on choisit les $x_{i}$ de sorte que parmi les différents quotients $i / x_{i}$ ($1 \\leqslant i \\leqslant n$) on ait $p-1$ quotients égaux à $1$ et les $n-p+1$ autres égaux à $1 /(n-p+1)$. Plus précisément, on choisit:\n$x_{1}=1,\\ x_{2}=2,\\ \\ldots,\\ x_{p-1}=p-1,\\ x_{p}=p(n-p+1),\\ x_{p+1}=(p+1)(n-p+1),\\ \\ldots,\\ x_{n}=n(n-p+1)$.\nOn vérifie aisément que $x_{1}<x_{2}<\\cdots<x_{n}$ et que\n$$\n\\frac{1}{x_{1}}+\\frac{2}{x_{2}}+\\cdots+\\frac{n}{x_{n}}=p-1+(n-p+1) \\cdot \\frac{1}{n-p+1}=p .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les couples d'entiers relatifs $(x, y)$ tels que $x^{3}+y^{3}=(x+y)^{2}$.",
    "options": [],
    "answer": "All integer pairs of the form (t, -t) for any integer t, together with (2,2), (1,2), (2,1), (1,0), and (0,1).",
    "solution": "Solution:\n\nSoit $(x, y)$ un couple solution. Comme $x^{3}+y^{3}=(x+y)(x^{2}-x y+y^{2})$, on a donc soit $x+y=0$, soit $x^{2}-x y+y^{2}=x+y$.\n\nDans le deuxième cas, on remarque qu'alors nécessairement\n$$(x-1)^{2}+(y-1)^{2}+(x-y)^{2}=2.$$\nParmi les trois entiers $x-1$, $y-1$ et $x-y$, deux sont donc égaux à $1$ ou $-1$ et le troisième est nul. On vérifie que les seules possibilités sont $(2,2)$, $(0,0)$, $(1,2)$, $(1,0)$, $(2,1)$ et $(0,1)$.\n\nAinsi, les solutions sont les couples $(2,2)$, $(1,2)$, $(1,0)$, $(2,1)$, $(0,1)$ et $(x,-x)$ où $x$ est un entier relatif.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. On dessine des triangles équilatéraux $ABE$ et $ACF$ à l'extérieur de $ABC$. Soit $G$ le centre de gravité de $ABE$ et $K$ le milieu de $[EF]$. Déterminer les angles du triangle $KCG$.",
    "options": [],
    "answer": "The triangle KCG is right-angled at K, with angle at G equal to 60° and angle at C equal to 30°.",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nMontrons que $GKC$ est rectangle en $K$ avec $\\widehat{CGK}=60^\\circ$. Pour cela, considérons le symétrique $G'$ de $G$ par rapport à $K$ : il suffit de montrer que $GCG'$ est équilatéral.\n\nComme $EGFG'$ est un parallélogramme, on a $G'F=EG$, et comme $G$ est le centre du cercle circonscrit à $ABE$, on a $GA=GE$, donc $GA=G'F$.\n\nD'autre part, $AC=FC$ car $AFC$ est équilatéral.\n\nEnfin, $\\widehat{GAC}=30^\\circ+\\widehat{BAC}$ et $(\\overrightarrow{FG}, \\overrightarrow{FC})=(\\overrightarrow{GE}, \\overrightarrow{FC})=(\\overrightarrow{GE}, \\overrightarrow{AG})+(\\overrightarrow{AG}, \\overrightarrow{AC})+(\\overrightarrow{AC}, \\overrightarrow{FC})=60^\\circ+(\\overrightarrow{AG}, \\overrightarrow{AC})-60^\\circ=(\\overrightarrow{AG}, \\overrightarrow{AC})$, donc $GAC$ et $G'FC$ sont isométriques.\n\nOr, $F$ est obtenu à partir de $A$ en effectuant une rotation de $60^\\circ$ autour de $C$, donc cette rotation envoie le triangle $GAC$ sur $G'FC$. En particulier, $GC=G'C$ et $\\widehat{GCG'}=60^\\circ$, ce qui conclut.\n\n\nAutre méthode. Si on connaît les nombres complexes, il est facile de résoudre l'exercice par une méthode analytique systématique.\n\nOn prend un repère tel que $A$ soit l'origine. On nomme $b, c, e, f, g, k$ les affixes de $B, C, E, F, G, K$. On a alors $e=e^{-i \\pi / 3} b$, $f=e^{i \\pi / 3} c$, $k=(e+f)/2$, $g=(b+e)/3$. Il est alors facile de vérifier que $c-k=i \\sqrt{3}(g-k)$, ce qui montre que $(CK) \\perp (GK)$ et que $CK=GK \\times \\sqrt{3}$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a, b, c, d > 0$ des réels tels que $a b c d = 1$.\nProuver que\n$$\n\\frac{1}{a+b+2} + \\frac{1}{b+c+2} + \\frac{1}{c+d+2} + \\frac{1}{d+a+2} \\leqslant 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIl est bien connu que, si $x, y > 0$ sont des réels, on a $x + y \\geqslant 2 \\sqrt{x y}$.\n\nPar suite, $\\frac{1}{a+b+2} \\leqslant \\frac{1}{2(\\sqrt{a b} + 1)}$ et $\\frac{1}{c+d+2} \\leqslant \\frac{1}{2(\\sqrt{c d} + 1)}$, d'où\n$$\n\\begin{aligned}\n\\frac{1}{a+b+2} + \\frac{1}{c+d+2} & \\leqslant \\frac{1}{2(\\sqrt{a b} + 1)} + \\frac{1}{2(\\sqrt{c d} + 1)} = \\frac{1}{2(\\sqrt{a b} + 1)} + \\frac{1}{2\\left(\\frac{1}{\\sqrt{a b}} + 1\\right)} \\\\\n& = \\frac{1}{2(\\sqrt{a b} + 1)} + \\frac{\\sqrt{a b}}{2(\\sqrt{a b} + 1)} = \\frac{1}{2}\n\\end{aligned}\n$$\nOn prouve de même que $\\frac{1}{b+c+2} + \\frac{1}{d+a+2} \\leqslant \\frac{1}{2}$.\n\nEn sommant les deux inégalités précédentes, on déduit immédiatement la conclusion désirée.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn note $\\mathbb{N}^*$ l'ensemble des entiers naturels strictement positifs. Trouver toutes les fonctions $f: \\mathbb{N}^* \\rightarrow \\mathbb{N}^*$ telles que\n$$\nm^2+f(n) \\mid m f(m)+n\n$$\npour tous entiers strictement positifs $m$ et $n$.",
    "options": [],
    "answer": "f(n) = n for all positive integers n",
    "solution": "Solution:\n\nSoit $f$ une solution éventuelle du problème.\nOn note $\\left( * \\right)$ la condition \"$m^2+f(n)$ divise $m f(m)+n$\".\n\nEn choisissant $m=n=2$ dans $\\left(*)\\right$, il vient que $4+f(2)$ divise $2 f(2)+2$. Or, on a $2 f(2)+2 < 2(f(2)+4)$, et il faut donc que $f(2)+4=2 f(2)+2$, d'où $f(2)=2$.\n\nEn choisissant maintenant $m=2$, la condition $\\left(*)\\right$ assure que, pour tout $n \\geq 1$, le nombre $4+f(n)$ divise $4+n$, ce qui conduit à $f(n) \\leq n$, pour tout $n \\geq 1$. (i)\n\nD'autre part, en choisissant $m=n$ dans $\\left(*)\\right$, on déduit cette fois que, pour tout $n \\geq 1$, le nombre $n^2+f(n)$ divise $n f(n)+n$, d'où $n^2+f(n) \\leq n f(n)+n$. Ainsi, on a $(n-1)(f(n)-n) \\geq 0$, ce qui conduit à $f(n) \\geq n$, pour tout $n \\geq 2$. Puisqu'on a clairement $f(1) \\geq 1$, c'est donc que $f(n) \\geq n$ pour tout $n \\geq 1$. (ii)\n\nDe (i) et (ii), on déduit que $f(n)=n$ pour tout $n \\geq 1$.\n\nRéciproquement, il est évident que $f: n \\longmapsto n$ est bien une solution du problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les couples d'entiers positifs $(m, n)$ tels que $1+(m+n) m$ divise $(m+n)(n+1)-1$.",
    "options": [],
    "answer": "(m, n) = (1, n) for any positive integer n",
    "solution": "Solution:\n\nSoit $(m, n)$ un couple solution. Alors $1+(m+n) m$ divise $(m+n)(n+1)-1+1+(m+n) m = (m+n)(m+n+1)$. Or, $1+(m+n) m$ est premier avec $m+n$. Ainsi, $1+(m+n) m$ divise $m+n+1$. Donc $m^{2}+m n+1 \\leqslant m+n+1$. Donc $m=0$ ou $m=1$.\n\nRéciproquement, on vérifie que les couples $(0, n)$ et $(1, n)$, où $n \\geqslant 0$ est un entier, conviennent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont les angles sont aigus, et tel que $AB \\neq AC$. On note $D$ le pied de la bissectrice de $\\widehat{BAC}$. Le point $E$ (resp. $F$) désigne le pied de la hauteur issue de $B$ (resp. de $C$). Le cercle circonscrit au triangle $DBF$ rencontre le cercle circonscrit au triangle $DCE$ en un point $M$ autre que $D$.\n\nProuver que $ME = MF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nLemme 1. Les cercles $AEF$, $BDF$ et $CDE$ sont concourants en $M$.\nCela découle immédiatement du théorème de Miquel mais rappelons tout de même la démonstration:\n$$(MF, ME) = (MF, MD) + (MD, ME) = (BF, BD) + (CD, CE) = (AB, CD) + (CD, AC) = (AB, AC)$$\ndonc $A, E, F, M$ sont cocycliques.\n\nLemme 2. Les triangles $AEF$ et $ABC$ sont (indirectement) semblables.\nLes triangles $AFC$ et $AEB$ sont rectangles en $F$ et en $E$, et leurs angles en $A$ sont égaux, donc ils sont semblables. On en déduit que $\\frac{AE}{AF} = \\frac{AB}{AC}$. Ceci s'écrit encore $\\frac{AE}{AB} = \\frac{AF}{AC}$. Comme $\\widehat{EAF} = \\widehat{BAC}$, les triangles $EAF$ et $BAC$ sont semblables. Il est clair que la similitude est indirecte puisque les angles orientés $(\\overrightarrow{AB}, \\overrightarrow{AC})$ et $(\\overrightarrow{AE}, \\overrightarrow{AF})$ sont opposés.\n\nRevenons à l'exercice. Soit $M'$ le point d'intersection autre que $A$ entre la bissectrice de $\\widehat{BAC}$ et le cercle $AEF$. De même, on définit $M''$ comme le point d'intersection autre que $A$ entre la bissectrice de $\\widehat{BAC}$ et le cercle $ABC$. Montrons que $M' = M$.\n\n$$\n\\begin{aligned}\n(M'F, M'D) & = (M'F, M'A) \\quad \\text{car } A, M, D \\text{ alignés} \\\\\n& = (EF, EA) \\quad \\text{car } A, E, F, M \\text{ cocycliques} \\\\\n& = (BA, BC) \\quad \\text{car } AEF \\text{ et } ABC \\text{ indirectement semblables} \\\\\n& = (BF, BF).\n\\end{aligned}\n$$\n\nOn en déduit que $M'$ appartient au cercle $BDF$. De même, il appartient au cercle $CDE$. Par conséquent, il est l'intersection des trois cercles $BDF$, $CDE$ et $AEF$ : ceci prouve que $M' = M$.\n\nEnfin, comme les arcs $MF$ et $ME$ sont égaux (puisque $(AM)$ est la bissectrice de $\\widehat{FAE}$), on a $MF = ME$.\nSolution:\n\nLemme 1. Deux cordes $[BB']$ et $[CC']$ d'un cercle se coupent en un point extérieur $A$. Alors les triangles $ABC$ et $AC'B'$ sont indirectement semblables.\n\n![](attached_image_2.png)\n\nOn a $(B'C', B'A) = (B'C', B'B) = (CC', CB) = (CA, CB)$ et de même $(C'A, C'B') = (BC, BA)$.\n\nLemme 2. $\\frac{DB}{DC} = \\frac{AB}{AC}$.\nD'après la loi des sinus, on a $\\frac{AB}{DB} = \\frac{\\sin \\widehat{ADB}}{\\sin \\frac{\\widehat{BAC}}{2}} = \\frac{\\sin \\widehat{CDA}}{\\sin \\frac{\\widehat{BAC}}{2}} = \\frac{AC}{DC}$.\n\nRevenons à l'exercice. Comme $AFC$ et $AEB$ sont rectangles en $F$ et $E$ et ont le même angle en $A$, ils sont semblables. On en déduit $\\frac{AF}{AE} = \\frac{AC}{AB}$, ce qui s'écrit encore $AF \\cdot AB = AE \\cdot AC$. Autrement dit, $A$ a la même puissance par rapport aux deux cercles $BFD$ et $CED$. Par conséquent, $A$ se trouve sur l'axe radical $(MD)$.\n\nD'autre part, le lemme 1 entraîne que $AFM$ et $ADB$ sont semblables, donc $\\frac{BD}{BA} = \\frac{MF}{MA}$. De même, $\\frac{CD}{CA} = \\frac{ME}{MA}$. D'après le lemme 2, on a $\\frac{BD}{BA} = \\frac{CD}{CA}$ donc $MF = ME$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient deux cercles extérieurs $\\mathcal{C}$ et $\\mathcal{C}^{\\prime}$ de centres $O$ et $O^{\\prime}$. On mène deux rayons parallèles de même sens $[O M]$ et $[O^{\\prime} M^{\\prime}]$, et deux autres rayons parallèles de même sens $[O P]$ et $[O^{\\prime} P^{\\prime}]$. La droite $(M M^{\\prime})$ recoupe $\\mathcal{C}^{\\prime}$ en $N$ et la droite $(P P^{\\prime})$ recoupe $\\mathcal{C}^{\\prime}$ en $Q$. Montrer que $M, N, P, Q$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nLes triangles $OMP$ et $O^{\\prime} M^{\\prime} P^{\\prime}$ sont isocèles en $O$ et $O^{\\prime}$ respectivement. De plus, les angles $(\\overrightarrow{OM}, \\overrightarrow{OP})$ et $(\\overrightarrow{O^{\\prime} M^{\\prime}}, \\overrightarrow{O^{\\prime} P^{\\prime}})$ sont égaux, donc ces deux triangles sont directement semblables. Comme de plus $(OM)$ et $(O^{\\prime} M^{\\prime})$ sont parallèles, $(MP)$ et $(M^{\\prime} P^{\\prime})$ le sont.\n\nOn en déduit $(MP, MN) = (M^{\\prime} P^{\\prime}, M^{\\prime} N) = (QP^{\\prime}, QN) = (QP, QN)$, donc $MNPQ$ est inscriptible.\n\nAutre solution. Supposons que les cercles sont de rayons différents. Soit $\\Omega$ le centre de l'homothétie $h$ de rapport $\\lambda > 0$ qui transforme le premier cercle en le deuxième. Le point $h(M)$ vérifie la même condition de l'énoncé que le point $M^{\\prime}$, donc $h(M) = M^{\\prime}$. De même, $h(P) = P^{\\prime}$. On a $\\Omega N \\cdot \\Omega M^{\\prime} = \\Omega Q \\cdot \\Omega P^{\\prime}$ (puissance d'un point par rapport à un cercle), donc $\\lambda \\Omega N \\cdot \\Omega M = \\lambda Q \\cdot \\Omega P$. En simplifiant par $\\lambda$ et en utilisant encore la puissance d'un point par rapport à un cercle, on en déduit que $M, N, P, Q$ sont cocycliques.\n\nSi les cercles sont de même rayon, alors il existe une translation qui envoie $M$ et $P$ sur $M^{\\prime}$ et $P^{\\prime}$ respectivement, donc comme ci-dessus on a $(MP, MN) = (M^{\\prime} P^{\\prime}, M^{\\prime} N) = (QP^{\\prime}, QN) = (QP, QN)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier strictement positif. On considère $2n+1$ entiers distincts compris au sens large entre $-2n+1$ et $2n-1$. Montrer qu'on peut en choisir 3 dont la somme soit nulle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn raisonne par récurrence sur $n$.\n\nInitialisation : Pour $n=1$, on prend $2n+1=3$ entiers distincts parmi $-1, 0, 1$, donc ces trois entiers. Leur somme est bien nulle.\n\nHérédité : Supposons l'énoncé vrai au rang $n$ : si on prend $2n+1$ entiers distincts compris au sens large entre $-2n+1$ et $2n-1$, alors on peut en choisir 3 dont la somme soit nulle. Maintenant, on veut montrer que parmi $2n+3$ entiers distincts entre $-2n-1$ et $2n+1$, il y en a toujours 3 de somme nulle.\n\nSi parmi ces $2n+3$ entiers, il y en a $2n+1$ entre $-2n+1$ et $2n-1$, alors c'est gagné par hypothèse de récurrence. On peut donc supposer que 3 au moins de ces entiers sont parmi $\\{-2n-1, -2n, 2n, 2n+1\\}$. Par symétrie entre les positifs et les négatifs, on peut supposer qu'on dispose de $2n$ et $2n+1$, ainsi que de $-2n-1$ ou $-2n$, donc on peut supposer qu'on ne dispose pas de $0$, puisque $0 + 2n + (-2n) = 0$ et $0 + (2n+1) + (-2n-1) = 0$ et on aurait alors un triplet de somme nulle.\n\nMaintenant, considérons les $n$ paires\n$$\n\\{-k, -2n-1+k\\}, \\quad 1 \\leq k \\leq n\n$$\nsi l'une d'elles est incluse dans notre sélection de $2n+3$ entiers, alors on a encore trois nombres de somme nulle : $-k + (-2n-1 + k) + (2n+1) = 0$ et on a terminé. Supposons donc que notre sélection ne contienne qu'au plus un élément de ces $n$ paires, donc il reste au moins $n+1$ nombres à choisir parmi $(\\{1, \\ldots, 2n-1\\}$ et $-2n-1$). On pose $A = \\{1, \\ldots, 2n-1\\}$.\n\nOn distingue 2 cas.\n\n- Si on a pris $-2n-1$, on doit prendre au moins $n$ éléments de $A$. On regarde cette fois les $n$ paires\n$$\n\\{k, 2n+1-k\\}, \\quad 1 \\leq k \\leq n\n$$\nSi l'une d'elles est complète, on a un triplet à somme nulle avec $-2n-1$. On a supposé au début que $2n$ était pris, donc si on prend $1$, $\\{1, 2n\\}$ est complète. Sinon on prend $n$ nombres dans les $n-1$ paires restantes, donc par principe des tiroirs, on a une paire complète.\n\n- Si on n'a pas pris $-2n-1$, on doit prendre au moins $n$ éléments de $A$. Et on a pris $-2n$ d'après le début du raisonnement. Et $A$ peut se découper en $n-1$ paires\n$$\n\\{k, 2n-k\\}, \\quad 1 \\leq k \\leq n-1\n$$\nDe nouveau, par le principe des tiroirs on doit prendre une paire complète, donnant un triplet à somme nulle avec $-2n$.\n\nCeci clôt la récurrence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30419,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier.\nOn considère un ensemble de $4n+5$ points du plan, trois jamais alignés, et chacun coloré soit en rouge soit en bleu.\nProuver qu'il existe $n$ triangles dont les sommets sont tous d'une même couleur (la même pour tous les triangles), et dont les intérieurs respectifs sont deux à deux disjoints et ne contiennent aucun point coloré.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLemme. Soit $n \\geqslant 3$ et $m \\geqslant 1$ des entiers.\nOn considère un $n$-gone convexe $P$ et $m$ de ses points intérieurs, trois des $n+m$ points jamais alignés. Alors, on peut trianguler $P$ en $n+2m-2$ triangles dont les sommets sont tous parmi les $n+m$ points, aucun de ces triangles ne contenant un des $m$ points en son intérieur.\n\nPreuve du lemme. On procède par récurrence sur $m$.\n\nPour $m=1$, il suffit de joindre le point intérieur à chacun des sommets, ce qui donne une triangulation adéquate de $P$ en $k$ triangles.\n\nSupposons que l'affirmation soit vraie pour la valeur $m$ et donnons-nous $m+1$ points intérieurs à $P$. Soit $I$ un de ces points intérieurs. À l'aide des $m$ autres points intérieurs, d'après l'hypothèse de récurrence, on peut trianguler $P$ de la façon souhaitée en $k+2m-2$ triangles. Comme $I$ est intérieur à $P$, il est donc intérieur à l'un de ces triangles. En joignant $I$ à chacun des trois sommets de ce triangle, on obtient ainsi une triangulation adéquate en $k+2m-2+2=k+2(m+1)-2$ triangles, ce qui achève la preuve.\n\nRevenons à l'exercice. Notons $r$ (resp. $b$) le nombre de points rouges (resp. bleus). On a donc $r+b=4n+5$.\n\nSans perte de généralité, on peut supposer que $r \\geqslant b$ et donc $r \\geqslant 2n+3$.\n\nSoit $P$ l'enveloppe convexe des points rouges, et on suppose que $P$ est un $k$-gone. En particulier, il n'existe aucun point rouge en dehors de $P$, donc il y a $m$ points rouges intérieurs à $P$, avec $k+m=r \\geqslant 2n+3$. (1)\n\nOn note $w$ le nombre de points bleus intérieurs à $P$.\n\nD'après le lemme, en ne considérant que les points rouges, on peut obtenir une triangulation de $P$ en $k+2m-2$ triangles à sommets rouges et ne contenant intérieurement aucun point rouge. Si au moins $n$ de ces triangles ne contiennent aucun point bleu, c'est fini.\n\nSans quoi, au plus $n-1$ des triangles à sommets rouges ne contiennent pas de point bleu. Cela assure qu'au moins $k+2m-2-(n-1)=k+2m-(n+1)$ triangles à sommets rouges contiennent chacun au moins un point bleu et, ces triangles étant d'intérieurs deux à deux disjoints, on a donc $w \\geqslant k+2m-(n+1)$.\n\nSoit $Q$ l'enveloppe convexe de ces $w$ points bleus intérieurs à $P$. Comme ci-dessus, on peut trianguler $Q$ en au moins $w-2$ triangles ne contenant aucun point bleu en leurs intérieurs.\n\nPar l'absurde : supposons qu'au plus $n-1$ de ces triangles bleus ne contiennent pas de point rouge.\n\nAlors, au moins $w-2-(n-1)$ triangles bleus contiennent au moins un point rouge et on a $m \\geqslant w-(n+1)$. Par suite, on a $m \\geqslant k+2m-(n+1)-(n+1)$, d'où $k+m \\leqslant 2n+2$, en contradiction avec (1).\n\nAinsi, au moins $n$ de ces triangles à sommets bleus ne contiennent aucun point rouge, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère 6 points du plan, disposés de telle sorte que le triangle formé par trois quelconques d'entre eux ait trois côtés de longueurs distinctes. Montrer qu'il existe un triangle dont le plus court côté est également le côté le plus long d'un autre triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn colorie en rouge le plus court côté de chaque triangle. Il suffit de montrer qu'on a un triangle rouge. Appelons $P_{1}, \\ldots, P_{6}$ les points. Parmi les cinq segments $P_{1} P_{2}, P_{1} P_{3}, P_{1} P_{4}, P_{1} P_{5}, P_{1} P_{6}$ ayant $P_{1}$ pour extrémité, il y en a au moins 3 rouges ou bien au moins 3 non coloriés.\n\n- Supposons qu'il y en ait 3 rouges, disons $P_{1} P_{2}, P_{1} P_{3}$ et $P_{1} P_{4}$. Le triangle $P_{2} P_{3} P_{4}$ a au moins un côté rouge, disons $P_{2} P_{3}$, donc le triangle $P_{1} P_{2} P_{3}$ est rouge.\n\n- Supposons qu'il y en ait 3 non coloriées, disons $P_{1} P_{2}, P_{1} P_{3}$ et $P_{1} P_{4}$. Chacun des trois triangles $P_{1} P_{2} P_{3}$, $P_{1} P_{3} P_{4}, P_{1} P_{4} P_{2}$ a au moins un côté rouge, et ce côté n'est pas l'un de ceux ayant $P_{1}$ pour extrémité, donc $P_{2} P_{3}, P_{3} P_{4}, P_{4} P_{2}$ sont rouges. Ainsi, le triangle $P_{2} P_{3} P_{4}$ est rouge.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $n > 0$ un entier, et $a, b, c$ des entiers strictement positifs tels que\n$$\n(a + b c)(b + a c) = 19^{n}\n$$\n\nProuver que $n$ est pair.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nProcédons par l'absurde, et supposons qu'il existe un triplet $(a, b, c)$ tel que $(a + b c)(b + a c) = 19^{n}$, où $n$ est un entier impair. Sans perte de généralité, on peut supposer que $a \\geqslant b$ et que la somme $a + b + c$ est minimale.\n\nPuisque $19$ est premier, on sait que $b + a c \\geqslant 2$ est une puissance de $19$, donc que $b + a c \\equiv 0 \\pmod{19}$. Si $19$ divise $a$, alors $19$ divise $b$ également : alors le triplet $(\\alpha, \\beta, c) = \\left(\\frac{a}{19}, \\frac{b}{19}, c\\right)$ est tel que $(\\alpha + \\beta c)(\\beta + \\alpha c) = 19^{n-2}$, de sorte que la somme $a + b + c$ n'est pas minimale. On sait donc que $\\gcd(a, 19) = 1$.\n\nEn outre, si $c = 1$, alors $(a + b c)(b + a c) = (a + b)^{2}$ est un carré, donc $n$ est nécessairement pair. De même, si $a = b$, alors $(a + b c)(b + a c) = (a + a c)^{2}$ est un carré, et $n$ est pair. On sait donc que $c > 1$ et que $a > b$.\n\nAlors $b + a c = (a + b c) + (a - b)(c - 1) > a + b c$, de sorte qu'il existe deux entiers naturels $u$ et $v$ tels que $a + b c = 19^{u}$ et $b + a c = 19^{u + v}$. En particulier,\n$$\na(c - 1)(c + 1) = a c^{2} - a = (b + a c) c - (a + b c) = 19^{u}(19^{v} c - 1).\n$$\nOn sait donc que $19^{u}$ divise $(c - 1)(c + 1)$ et, puisque $\\gcd(19^{u}, c - 1, c + 1) = \\gcd(19^{u}, c - 1, 2) = 1$, on en déduit que $19^{u}$ divise soit $c - 1$ soit $c + 1$.\n\nOr, on sait que $19^{u} = a + b c > b + b c \\geqslant c + 1 > c - 1 > 0$. Il est donc impossible que $19^{u}$ divise $c - 1$ ou $c + 1$, de sorte que notre supposition initiale était fausse, et donc que $n$ est nécessairement pair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'une suite $\\left(u_{n}\\right)_{n \\geqslant 1}$ est Sicilienne si $u_{1}$ est un entier strictement positif, et si pour tout $n$,\n$$\nu_{n+1}= \\begin{cases}u_{n} / 2 & \\text{ si } u_{n} \\text{ est pair, et } \\\\ u_{n}+\\left[\\sqrt{u_{n}}\\right] & \\text{ si } u_{n} \\text{ est impair. }\\end{cases}$$\nExiste-t-il une suite Sicilienne $\\left(u_{n}\\right)_{n \\geqslant 1}$ telle que $u_{n}>1$ pour tout $n$ ?\n(N.B. $[x]$ désigne la partie entière de $x$. Par exemple, $[2,71828]=2$.)",
    "options": [],
    "answer": "No; every such sequence contains a term equal to one.",
    "solution": "Solution:\n\nCommençons par une simple remarque. Soit $\\mathrm{f}: \\mathbb{N}^{*} \\rightarrow \\mathbb{N}^{*}$ la fonction qui, à tout entier naturel non nul $n$, associe l'entier $n / 2$ si $n$ est pair, et $n+[\\sqrt{n}]$ si $n$ est impair. Il est clair que l'image $f(n)$ est bien un entier naturel non nul.\n\nNous allons maintenant montrer que la réponse à la question de l'énoncé est négative : toute suite Sicilienne $\\left(u_{n}\\right)_{n \\geqslant 1}$ contient un terme $u_{n}=1$.\n\nSoit $\\left(u_{n}\\right)_{n \\geqslant 1}$ une suite Sicilienne : on note que $u_{n+1}=f\\left(u_{n}\\right)$ pour tout entier $n \\geqslant 1$. Soit alors $E=\\left\\{u_{n} \\mid n \\geqslant 1\\right\\}$ l'ensemble des valeurs prises par les termes de cette suite. $E$ est un ensemble d'entiers non vide, de sorte qu'il admet un minimum $M$, tel que $M \\geqslant 1$ : il existe alors un entier $n \\geqslant 1$ tel que $M=u_{n}$. Notre but est de montrer que $M=1$. Nous allons donc procéder par l'absurde, et supposer que $M \\geqslant 2$.\n\nSi $M$ est pair, alors $u_{n+1}=f\\left(u_{n}\\right)=f(M)=M / 2<M$ et, puisque $u_{n+1} \\in E$, cela contredit la minimalité de $M$.\n\nSi $M$ est impair, soit $k$ l'entier naturel non nul tel que $k^{2} \\leqslant M<(k+1)^{2}$. Si $k$ est impair, alors $u_{n+1}=f(M)=M+k$, donc $M+k$ est pair, et $u_{n+2}=f\\left(u_{n+1}\\right)=f(M+k)=\\frac{M+k}{2}$. Or, $M>1$, donc $M-k \\geqslant M-\\sqrt{M}=\\sqrt{M}(\\sqrt{M}-1)>0$, de sorte que $u_{n+2}<M$ : puisque $u_{n+2} \\in E$, cela contredit la minimalité de $M$.\n\nAinsi, on sait que $M$ est impair et que $k$ est pair, donc que $k^{2}+1 \\leqslant M<(k+1)^{2}$.\n\n- Si $k^{2}+1 \\leqslant M \\leqslant k^{2}+k$, alors $k^{2} \\leqslant f(M)=M+k<(k+1)^{2}$, puis $(k+1)^{2} \\leqslant f^{2}(M)= M+2 k<(k+2)^{2}$, donc $f^{3}(M)=M+3 k+1$ est pair, et enfin $f^{4}(M)=\\frac{M+3 k+1}{2}$.\n\n- Si $k^{2}+k+1 \\leqslant M<(k+1)^{2}$, alors $(k+1)^{2} \\leqslant f(M)=M+k<(k+2)^{2}$, donc $f^{2}(M)= M+2 k+1$ est pair, et enfin $f^{3}(M)=\\frac{M+2 k+1}{2}$.\n\nIl s'ensuit que $M=\\min E \\leqslant \\min \\left\\{f^{3}(M), f^{4}(M)\\right\\} \\leqslant \\frac{M+3 k+1}{2}$, donc que $M \\leqslant 3 k+1$.\n\nEn particulier, $k^{2} \\leqslant 3 k+1$, ce qui signifie que $k^{2}-3 k-1=\\left(k-\\frac{3}{2}\\right)^{2}-\\frac{13}{4} \\leqslant 0$, donc que $k \\leqslant \\frac{3}{2}+\\sqrt{\\frac{13}{4}}<2+\\sqrt{4}=4$. Puisque $k$ est pair et $M$ impair, on en déduit que $k=2$ et que $M \\in\\{5,7\\}$. On conclut donc l'exercice en montrant que $\\mathbf{f}^{5}(5)=3<5$ et que $\\mathrm{f}^{4}(7)=3<7$, ce qui permet bien d'établir que nulle valeur $M \\geqslant 2$ n'est possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier strictement positif et $x_{1}, \\ldots, x_{n}$ des réels strictement positifs. Montrer qu'il existe des nombres $a_{1}, \\ldots, a_{n} \\in \\{-1,1\\}$ tels que :\n$$\na_{1} x_{1}^{2}+a_{2} x_{2}^{2}+\\cdots+a_{n} x_{n}^{2} \\geqslant\\left(a_{1} x_{1}+a_{2} x_{2}+\\cdots+a_{n} x_{n}\\right)^{2} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par ordonner les réels $x_{k}$ de sorte à ce que $x_{1} \\geq x_{2} \\geq \\ldots \\geq x_{n}>0$.\nMontrons par récurrence forte qu'une solution est fournie par la suite de coefficients définie par $a_{k}=1$ si $k$ est impair et $a_{k}=-1$ si $k$ est pair.\n\n$\\triangleright$ Si $n=1$, $a_{1} x_{1}^{2}=\\left(a_{1} x_{1}\\right)^{2}$ donc l'inégalité est vérifiée.\n\n$\\triangleright$ Si $n=2$, $a_{1} x_{1}^{2}+a_{2} x_{2}^{2}=x_{1}^{2}-x_{2}^{2} \\geq x_{1}^{2}-2 x_{1} x_{2}+x_{2}^{2}=\\left(x_{1}-x_{2}\\right)^{2}$ car $x_{1} \\geq x_{2}$; l'inégalité est encore vérifiée.\n\n$\\triangleright$ Soit $n \\geq 3$. Supposons que le résultat est établi jusqu'au rang $n-1$ et montrons l'inégalité au rang $n$.\nLa différence des deux membres est alors\n$$\na_{1} x_{1}^{2}+\\ldots+a_{n} x_{n}^{2}-\\left(a_{1} x_{1}+\\ldots+a_{n} x_{n}\\right)^{2}=\\alpha x_{1}+\\beta\n$$\nav ec\n$$\n\\alpha=-2\\left(a_{2} x_{2}+\\ldots+a_{n} x_{n}\\right)=2\\left(\\left(x_{2}-x_{3}\\right)+\\left(x_{4}-x_{5}\\right)+\\ldots\\right) \\leq 0\n$$\ncar $x_{2} \\geq x_{3}, x_{4} \\geq x_{5}, \\ldots$ Par conséquent, en utilisant la croissance de la fonction $x \\mapsto \\alpha x+\\beta$ et $x_{1} \\geq x_{2}$,\n$$\n\\begin{aligned}\na_{1} x_{1}^{2}+\\ldots+a_{n} x_{n}^{2}-\\left(a_{1} x_{1}+\\ldots+a_{n} x_{n}\\right)^{2} & \\geq \\alpha x_{2}+\\beta \\\\\n& \\geq a_{1} x_{2}^{2}+a_{2} x_{2}^{2} \\ldots+a_{n} x_{n}^{2}-\\left(a_{1} x_{2}+a_{2} x_{2}+\\ldots+a_{n} x_{n}\\right)^{2} \\\\\n& \\geq a_{3} x_{1}^{3}+\\ldots+a_{n} x_{n}^{2}-\\left(a_{3} x_{3}+\\ldots+a_{n} x_{n}\\right)^{2}\n\\end{aligned}\n$$\ncar $a_{1}=-a_{2}$ et donc $a_{1} x_{1}^{2}+a_{2} x_{2}^{2}=a_{1} x_{1}+a_{2} x_{2}=0$. Le dernier terme est positif d'après l'hypothèse de récurrence avec les réels $x_{3}, \\ldots, x_{n}$ ce qui termine la preuve de l'inégalité au rang $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer que si la somme de tous les diviseurs positifs d'un entier $n \\geqslant 1$ est une puissance de deux, alors le nombre de diviseurs positifs de $n$ est une puissance de deux.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDécomposons $n$ en produit de facteurs premiers : $n=\\prod_{i=1}^{k} p_{i}^{\\alpha_{i}}$. On voit aisément que la somme des diviseurs positifs de $n$, notée $\\sigma(n)$, vaut\n$$\n\\sigma(n)=\\prod_{i=1}^{k}\\left(1+p_{i}+p_{i}^{2}+\\cdots+p_{i}^{\\alpha_{i}}\\right)\n$$\nAinsi, si $1 \\leqslant i \\leqslant k$, $1+p_{i}+p_{i}^{2}+\\cdots+p_{i}^{\\alpha_{i}}$ est une puissance de deux. Ceci implique clairement que $p_{i} \\neq 2$. Supposons ensuite par l'absurde que $\\alpha_{i}+1$ n'est pas une puissance de deux et choisissons $q$ un nombre premier impair divisant $\\alpha_{i}+1$. Alors\n$$\n1+p_{i}+p_{i}^{2}+\\cdots+p_{i}^{\\alpha_{i}}=\\left(1+p_{i}+p_{i}^{2}+\\cdots+p_{i}^{q-1}\\right) \\cdot\\left(1+p_{i}^{q}+p_{i}^{2 q}+\\cdots+p_{i}^{\\alpha_{i}+1-q}\\right)\n$$\nEn effet, cette égalité découle du fait que $\\left(p_{i}^{\\alpha_{i}+1}-1\\right) /(p-1)=\\left(p_{i}^{q}-1\\right) /(p-1) \\cdot\\left(p_{i}^{\\alpha_{i}+1}-1\\right) /\\left(p^{q}-1\\right)$. Ceci est absurde, car $1+p_{i}+p_{i}^{2}+\\cdots+p_{i}^{q-1}$ est impair. Ainsi, $\\alpha_{i}+1$ est une puissance de deux pour tout $1 \\leqslant i \\leqslant k$. Ceci conclut, car le nombre de diviseurs (positifs) de $n$ est $\\prod_{i=1}^{k}\\left(\\alpha_{i}+1\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30425,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle non isocèle inscrit dans un cercle $\\Gamma$ de rayon $R$. Le cercle passant par $A$ et tangent en $C$ à $[BC]$ recoupe le cercle passant par $B$ et tangent en $C$ à $[AC]$ au point $D$.\n\na) Montrer que $CD \\leqslant R$.\n\nb) Montrer que lorsque $C$ se déplace sur $\\Gamma$, la droite $(CD)$ passe par un point fixe.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\na) On observe d'abord que $(AD, AC) = (CD, CB)$ et $(BC, BD) = (CA, CD)$, donc $DAC$ et $DCB$ sont directement semblables. On en déduit que $D^2 = DA \\cdot DB$.\n\nDe plus, en notant $\\gamma = (\\overrightarrow{CA}, \\overrightarrow{CB})$, on a $(\\overrightarrow{DC}, \\overrightarrow{DA}) = \\pi - \\gamma$ et de même $(\\overrightarrow{DB}, \\overrightarrow{DC}) = \\pi - \\gamma$, donc $(\\overrightarrow{DA}, \\overrightarrow{DB}) = 2\\gamma$. Il vient\n$$\n\\begin{aligned}\nA^2 & = DA^2 + DB^2 - 2 DA \\cdot DB \\cos 2\\gamma \\\\\n& = DA^2 + DB^2 - 2 DA \\cdot DB (1 - 2 \\sin^2 \\gamma) \\\\\n& = (DA - DB)^2 + 4 DA \\cdot DB \\sin^2 \\gamma \\geqslant 4 DA \\cdot DB \\sin^2 \\gamma = (2 DC \\sin \\gamma)^2.\n\\end{aligned}\n$$\nOr, $2R = \\dfrac{AB}{\\sin \\gamma}$, donc $R \\geqslant CD$.\n\nb) On va montrer que $(CD)$ passe par le point de rencontre des tangentes en $A$ et $B$ au cercle circonscrit, que nous appellerons $S$. Il est connu que $(CS)$ est la symédiane du triangle passant par $C$, donc il suffit de montrer que $(CD)$ est une symédiane.\n\nNotons $M$ le milieu de $[AB]$. Comme les aires de $ACM$ et $BCM$ sont égales, on a $\\frac{1}{2} CA \\times CM \\times \\sin \\widehat{ACM} = \\frac{1}{2} CB \\times CM \\times \\sin \\widehat{MCB}$, donc $CA \\sin \\widehat{ACM} = CB \\sin \\widehat{MCB}$.\n\nOn en déduit que si $\\Delta$ est la symédiane et $\\theta$ est l'angle entre $(CA)$ et $\\Delta$, alors $\\frac{\\sin \\theta}{\\sin (\\gamma - \\theta)} = \\frac{\\sin \\widehat{MCB}}{\\sin \\widehat{ACM}} = \\frac{CA}{CB}$. Il suffit donc de montrer que $\\frac{\\sin \\widehat{ACD}}{\\sin \\widehat{DCB}} = \\frac{CA}{CB}$.\n\n$$\n\\begin{aligned}\n\\frac{\\sin \\widehat{ACD}}{\\sin \\widehat{DCB}} & = \\frac{\\sin \\widehat{ACD}}{AD} \\times \\frac{DB}{\\sin \\widehat{DCB}} \\times \\frac{DA}{DB} \\\\\n& = \\frac{\\sin \\widehat{CDA}}{AC} \\times \\frac{BC}{\\sin \\widehat{BDC}} \\times \\frac{DA}{DB}\n\\end{aligned}\n$$\n\nOr, $\\widehat{CDA} = \\widehat{BDC} = \\pi - \\gamma$ et $\\frac{BC}{AC} = \\frac{DC}{DA} = \\frac{DB}{DC}$, donc\n$$\n\\frac{\\sin \\widehat{ACD}}{\\sin \\widehat{DCB}} = \\frac{BC}{AC} \\times \\frac{DA}{DC} \\times \\frac{DC}{DB} = \\frac{DC}{DB} = \\frac{CA}{CB}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les couples d'entiers positifs ou nuls $(x, y)$ pour lesquels $x^{2}+y^{2}$ divise à la fois $x^{3}+y$ et $x+y^{3}$.",
    "options": [],
    "answer": "[(0,0), (1,0), (0,1), (1,1)]",
    "solution": "Solution:\n\nTout d'abord, remarquons que les couples $(x, y) \\in \\{(0,0),(1,0),(0,1),(1,1)\\}$ sont solutions.\n\nOn se place maintenant dans le cas où $(x, y)$ est une solution éventuelle autre que celles-ci. De plus, $x$ et $y$ jouant des rôles symétriques, on suppose ici que $x \\leqslant y$, donc que $y \\geqslant 2$. Si $x=0$, le problème revient à trouver $y$ tel que $y^{2}$ divise à la fois $y$ et $y^{3}$, ce qui est incompatible avec la contrainte $y \\geqslant 2$. Ainsi, $y \\geqslant 2$ et $x \\geqslant 1$, donc $x y-1 \\geqslant 1$.\n\nSoit alors $d$ le PGCD de $x$ et $y$ : on note $x=d X$ et $y=d Y$, avec $X$ et $Y$ premiers entre eux : notons que $d>0$ et que $X^{2}+Y^{2}>0$. Il s'ensuit que $d\\left(X^{2}+Y^{2}\\right)$ divise à la fois $d^{2} X^{3}+Y$ et $X+d^{2} Y^{3}$. En particulier, $d$ divise à la fois $X$ et $Y$, donc $d=1$. Par conséquent, $\\operatorname{PGCD}\\left(x^{2}+y^{2}, x\\right)=\\operatorname{PGCD}\\left(y^{2}, x\\right)=1$.\n\nEn outre, $x^{2}+y^{2}$ divise $y\\left(x^{3}+y\\right)-\\left(x^{2}+y^{2}\\right)=x^{2}(x y-1)$, de sorte que, d'après le théorème de Gauss, $x^{2}+y^{2}$ divise $x y-1$. Puisque $x y-1 \\neq 0$, on en déduit que $x^{2}+y^{2} \\leqslant x y-1$. Or, $x^{2}+y^{2}=(x y-1)+\\left(x y+1+(x-y)^{2}\\right)>x y-1$. On a ainsi montré que les quatre solutions sus-mentionnées sont les seules solutions au problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $n$ et $p$ des entiers $\\geqslant 1$. Dans une assemblée de $n$ personnes, deux personnes quelconques ont au plus $p$ connaissances communes ; bien sûr, si $A$ connaît $B$, alors $B$ connaît $A$. Montrer que le nombre de paires non ordonnées $\\{A, B\\}$ de personnes qui se connaissent est inférieur ou égal à $\\sqrt{\\mathrm{p}n^{3}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $A_{1}, \\cdots, A_{n}$ les personnes et considérons le graphe simple et non orienté dont les sommets sont les $A_{i}$, et où deux personnes $A_{i}$ et $A_{j}$ sont reliées par une arête si et seulement si elles se connaissent.\n\nPour tout $i$, notons $d_{i}$ le degré de $A_{i}$ (c'est-à-dire le nombre de personnes que connaît $A_{i}$). Il s'agit de prouver que le nombre d'arêtes $a$ de ce graphe vérifie $a \\leqslant \\sqrt{\\mathrm{p}n^{3}}$.\n\nIl est bien connu que $2a = \\sum_{i=1}^{n} d_{i}$.\n\nSoit $M$ le nombre de paires d'arêtes $\\{XY, YZ\\}$ ayant un sommet commun.\n\nPour tout $i$, le sommet $A_{i}$ joue le rôle de $Y$ autant de fois que l'on peut choisir deux sommets adjacents à $A_{i}$. Ainsi,\n$$\nM = \\sum_{i=1}^{n} \\binom{d_{i}}{2}\n$$\nD'autre part, d'après l'énoncé, deux sommets quelconques ont toujours au plus $p$ sommets adjacents communs. Ainsi,\n$$\nM \\leqslant p \\binom{n}{2}.\n$$\nDe (1) et (2), on déduit que\n$$\n\\sum_{i=1}^{n} \\binom{d_{i}}{2} \\leqslant p \\binom{n}{2}\n$$\nOr, la fonction $x \\longmapsto \\frac{1}{2} x(x-1)$ est convexe sur $\\mathbb{R}$ donc, d'après l'inégalité de Jensen, on a\n$$\n\\sum_{i=1}^{n} \\binom{d_{i}}{2} \\geqslant n \\binom{\\frac{1}{n} \\sum_{i=1}^{n} d_{i}}{2} = \\frac{n}{2} \\left( \\frac{2a}{n} \\right) \\left( \\frac{2a}{n} - 1 \\right)\n$$\nDe (3) et (4), on obtient que $2a(2a-n) \\leqslant p n(n-1) \\leqslant p n^{3}$, ou encore $4a^{2} - 2n a - p n^{3} \\leqslant 0$.\n\nSi $a > \\sqrt{p n^{3}}$, alors $0 > 4a^{2} - 2n a - a^{2} = (3a - 2n)a \\geqslant (3 \\sqrt{p n} - 2n)a \\geqslant n a \\geqslant 0$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans un cirque, il y a plusieurs clowns. Chacun utilise au moins 5 couleurs, parmi 12 possibles, pour se peindre. Une même couleur est utilisée par au plus 20 clowns. Deux clowns n'ont jamais exactement les mêmes couleurs. Combien y a-t-il de clowns au maximum?",
    "options": [],
    "answer": "48",
    "solution": "Solution:\n\nSoit $n$ le nombre de clowns et $N$ le nombre de paires $(L, c)$ où $L$ est un clown et $c$ une couleur qu'il porte. D'une part, chaque clown ayant au moins 5 couleurs, $N \\geq 5 \\times n$. D'autre part, chacune des 12 couleurs peut apparaître dans au plus 20 paires. Donc $N \\leq 20 \\times 12$. On en déduit que $5 n \\leq 240$, donc $n \\leq 48$.\n\nMaintenant, on considère les quintuplets : $A=\\left(c_{1}, c_{2}, c_{3}, c_{4}, c_{5}\\right), B=\\left(c_{1}, c_{2}, c_{3}, c_{4}, c_{6}\\right), C=\\left(c_{1}, c_{2}, c_{3}, c_{4}, c_{7}\\right)$, $D=\\left(c_{1}, c_{2}, c_{3}, c_{4}, c_{8}\\right)$. On imagine que les 12 couleurs sont disposées dans l'ordre sur un cercle (comme les nombres sur une pendule par exemple), et on note $A_{i}$ le quintuplet obtenu en décalant les couleurs de $A$ de $i$ rangs dans le sens croissant $\\left(A_{0}=A, A_{1}=\\left(c_{1+1}, c_{2+1}, c_{3+1}, c_{4+1}, c_{5+1}\\right)=\\left(c_{2}, c_{3}, c_{4}, c_{5}, c_{6}\\right)\\right.$, $A_{9}=\\left(c_{10}, c_{11}, c_{12}, c_{1}, c_{2}\\right.$, etc. $)$, et on fait de même pour $B, C, D$. Concrètement, chaque tel quintuplet de couleurs consiste donc de quatre couleurs consécutives sur le cercle, puis fait un \"saut\" pour le numéro de la dernière couleur qui est de taille 1 pour les $A_{i}$, de 2 pour les $B_{i}$, de 3 pour les $C_{i}$ et de 4 pour les $D_{i}$.\n\nConsidérons l'ensemble des $K_{i}$ avec $K=A, B, C$ ou $D$ et $0 \\leq i \\leq 11$. Cela donne 48 paquets de 5 couleurs. Et on vérifie aisément que deux paquets ne peuvent être identiques : si pour certains $i$ et $j$ et certains $K$ et $K'$ on avait $K_{i}=K_{j}'$, alors $K=K'$ pour avoir le même \"saut\" entre le \"paquet\" des quatre couleurs consécutives et la cinquième (1 pour $A$, 2 pour $B$, 3 pour $C$, 4 pour $D$), donc $K_{i}=K_{j}$, et on voit bien que $i=j$, pour que les deux paquets de quatre couleurs consécutives soient les mêmes.\n\nEnfin, chaque couleur est utilisée exactement 20 fois. En effet, on peut remarquer que toutes les couleurs jouent ici le même rôle (puisqu'on a \"décalé\" cycliquement les ensembles $A, B, C, D$), donc chacune sera représentée autant que les autres. On a au total $5 \\times 48=240$ utilisations de couleur, donc chacune des 12 couleurs est utilisée $240 / 12=20$ fois. Pour s'en convaincre plus concrètement, on peut fixer une couleur : elle apparaît une et une seule fois comme \"première\" couleur d'un paquet de type $A_{i}$, puis comme \"deuxième\", etc., donc 5 paquets $A_{i}$ utilisent cette couleur, et de même pour $B, C$ et $D$. Elle est donc utilisée 20 fois. Conclusion : on peut bien avoir $n=48$ clowns, en donnant à chaque clown un des 48 paquets de couleurs qu'on vient de construire.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les nombres premiers $p$ et $q$ tels que $p$ divise $5^{q}+1$ et $q$ divise $5^{p}+1$.",
    "options": [],
    "answer": "(2,2), (3,3), (13,2), (2,13), (3,7), (7,3)",
    "solution": "Solution:\n\nNotons $\\alpha$ l'ordre de $5$ modulo $q$ et $\\beta$ l'ordre de $5$ modulo $p$. Comme $5^{2p} \\equiv 1 \\bmod q$, on en déduit que $\\alpha$ divise $2p$. Donc $\\alpha = 1, 2, p$ ou $2p$.\n\nSi $\\alpha = 1$, alors $5 \\equiv 1 \\bmod q$, donc $q = 2$. Si $\\alpha = 2$, alors $25 \\equiv 1 \\bmod q$, donc $q = 2$ ou $q = 3$.\n\nSi $q = 2$, alors $p$ divise $5^{2} + 1$, donc $p = 2$ ou $p = 13$. Réciproquement, $(2,2)$ et $(13,2)$ conviennent.\n\nSi $q = 3$, alors $p$ divise $5^{3} + 1$, donc $p = 2, 3$ ou $7$. Réciproquement, on vérifie que $(3,3), (7,3)$ conviennent, mais que $(2,3)$ ne convient pas.\n\nDe même, on a $\\beta = 1, 2, q$ ou $2q$. Si $\\beta = 1$ ou $2$, on obtient similairement les solutions $(2,13)$ et $(3,7)$.\n\nSupposons maintenant que $\\alpha = p$ ou $2p$ et que $\\beta = q$ ou $2q$. D'après le petit théorème de Fermat, $5^{p-1} \\equiv 1 \\bmod p$ et $5^{q-1} \\equiv 1 \\bmod q$. Donc $\\beta$ divise $p-1$ et $\\alpha$ divise $q-1$. Donc $q$ divise $p-1$ et $p$ divise $q-1$. Donc $q \\leqslant p-1$ et $p \\leqslant q-1$. Ainsi $q \\leqslant q-2$, ce qui est absurde.\n\nLes solutions sont donc $(2,2), (3,3), (13,2), (2,13), (3,7)$ et $(7,3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $P$ un polynôme à coefficients entiers, de degré $n$, avec $n \\leq 10$. On suppose que $|P(10)-P(0)|<1000$ et que, pour tout $k \\in\\{1, \\ldots, 10\\}$, il existe un entier $m$ tel que $P(m)=k$. Montrer que, pour tout entier $k$ il existe un entier $m$ tel que $P(m)=k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour $i \\in\\{1, \\ldots, 10\\}$, on désigne par $c_{i}$ un entier tel que $P\\left(c_{i}\\right)=i$. Pour $i \\in\\{1, \\ldots, 9\\}$, puisque $P$ est à coefficients entiers, l'entier $c_{i+1}-c_{i}$ divise $P\\left(c_{i+1}\\right)-P\\left(c_{i}\\right)=i+1-i=1$, d'où $c_{i+1}-c_{i}= \\pm 1$. De plus, pour $i, j \\in\\{1, \\ldots, 10\\}$ distincts, on a $P\\left(c_{i}\\right)=i \\neq j=P\\left(c_{j}\\right)$, donc $c_{i} \\neq c_{j}$. Ainsi, $c_{1}, c_{2}, \\ldots, c_{10}$ sont, dans cet ordre, dix entiers consécutifs.\n\nQuitte à étudier le polynôme $P(10-X)$ plutôt que le polynôme $P(X)$ lui-même, on peut supposer que $c_{1}<c_{2}<\\cdots<c_{10}$ : on a donc $c_{i}=c_{1}+i-1$. Soit alors les polynômes $Q(X)=X+1-c_{1}$ et $R(X)=P(X)-Q(X)$. Le polynôme $Q$ est affine, donc de degré 1, et $R$ est de degré au plus 10.\n\nEn outre, pour $i=1,2, \\ldots, 10$, on a $R\\left(c_{i}\\right)=P\\left(c_{i}\\right)-Q\\left(c_{i}\\right)=i-i=0$, donc $c_{1}, \\ldots, c_{10}$ sont dix racines distinctes entières du polynôme $R$, à coefficients entiers. On peut donc factoriser $R$ sous la forme $R(X)=S(X) T(X)$, où $S(X)=\\prod_{i=1}^{10}\\left(X-c_{i}\\right)$ et où $T$ est un polynôme à coefficients entiers. Remarquons que $T$ est de degré au plus 0, ce qui signifie que $T$ est constant : il existe un entier $a$ tel que $R=a S$.\n\nSi $a=0$, on a bien $P=Q$, donc $P$ est affine de coefficient $\\pm 1$ et $P(\\mathbb{Z})=\\mathbb{Z}$ : on suppose désormais que $a \\neq 0$. Notons alors que $S(n) \\equiv 0\\ (\\bmod\\ 10!)$ pour tout entier $n$. En effet, soit $k \\geq 10$ un entier. Alors\n$$\nS\\left(c_{1}+k\\right)=\\prod_{i=0}^{9}(k-i)=10!\\binom{k}{10} \\equiv 0 \\quad(\\bmod 10!)\n$$\nLa relation $S\\left(c_{1}+k\\right) \\equiv 0$ est donc vraie dans $\\mathbb{Z} / 10!\\mathbb{Z}$, ce qui signifie bien que $S(n) \\equiv 0$ $(\\bmod 10!)$ pour tout entier $n$.\n\nEn particulier, il s'ensuit que $R(n) \\equiv 0(\\bmod 10!)$ pour tout entier $n$. Or,\n$$\n|R(10)-R(0)| \\leq|P(10)-P(0)|+|Q(10)-Q(0)| \\leq 1010<10!\n$$\nAinsi, $R(10)=R(0)$, donc $S(10)=S(0)$. Le polynôme $S$ est strictement monotone sur $]-\\infty, c_{1}]$ et sur $[c_{10},+\\infty[$, donc on ne peut avoir ni $10 \\leq c_{1}$ ni $0 \\geq c_{10}$. Ainsi, l'un des entiers 0 et 10 doit être une racine de $S$, mais alors l'autre aussi, ce qui est impossible car les racines de $S$ sont mutuellement distantes d'au plus 9.\n\nLe cas $a \\neq 0$ est donc impossible, ce qui conclut le problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\omega$ le cercle circonscrit à un triangle $A B C$. On désigne par $M$ et $N$ les milieux des côtés $[A B]$ et $[A C]$ respectivement, et par $T$ le milieu de l'arc $B C$ de $\\omega$ ne contenant pas $A$. Les cercles circonscrits aux triangles $A M T$ et $A N T$ rencontrent les médiatrices de $[A C]$ et $[A B]$ aux points $X$ et $Y$ respectivement; on suppose que $X$ et $Y$ sont à l'intérieur du triangle $A B C$. Les droites $(M N)$ et $(X Y)$ se coupent en $K$. Montrer que $K A=K T$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $O$ le centre du cercle $\\omega$. On a $O=(M Y) \\cap(N X)$. Soit $\\ell$ la médiatrice de $[A T]$. Elle passe par $O$.\n\nNotons $s$ la symétrie par rapport à $\\ell$. Comme $(A T)$ est la bissectrice de $\\widehat{B A C}$, la droite $s(A B)$ est parallèle à $(A C)$. Comme $(O M) \\perp(A B)$ et $(O N) \\perp(A C)$, la droite $s(O M)$ est parallèle à $(O N)$. Or, elle passe par $O$, donc $s(O M)=(O N)$.\n\nDe plus, le cercle circonscrit $\\gamma$ à $A M T$ est symétrique par rapport à $\\ell$, c'est-à-dire $s(\\gamma)=\\gamma$, donc $s(M)$ appartient à la fois à $\\gamma$ et à $s(O M)=(O N)$. Nécessairement, $s(M)=X$. De même, $s(N)=Y$. On en déduit que $s(M N)=(X Y)$. Le point commun $K$ de $(M N)$ et $(X Y)$ appartient ainsi à la droite $\\ell$, ce qui implique que $K A=K T$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nExiste-t-il des entiers strictement positifs $a$ et $b$ tels que $a^{n}+n^{b}$ et $b^{n}+n^{a}$ soient premiers entre eux pour tout entier $n \\geqslant 0$ ?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPar l'absurde : supposons qu'il existe de tels entiers $a$ et $b$.\nPar symétrie, on peut supposer que $a \\geqslant b$.\nSi $d=\\operatorname{pgcd}(a, b)$ alors, pour $n=d$, il est clair que $d$ divise $a^{d}+d^{b}$ et $b^{d}+d^{a}$. Or ces deux nombres étant censés être premiers entre eux, c'est que $d=1$.\nSoit $N=a^{a}+b^{b}$. On a $N>1$ donc il existe un nombre premier $p$ qui divise $N$. De plus, $p$ ne peut diviser $a$ sans quoi il devrait diviser également $b$, en contradiction avec $d=1$. Ainsi, $a$ est inversible modulo $p$ et il existe un entier $k \\in\\{0, \\cdots, p-1\\}$ tel que $k a=b \\bmod [p]$.\nOn choisit alors $n=k+p(a-b-k+p-1)$. D'après le petit théorème de Fermat, il vient :\n$$\na^{n}+n^{b} \\equiv a^{k} \\cdot a^{a-b-k+p-1}+k^{b} \\equiv a^{a-b}+k^{b} \\bmod [p]\n$$\n$d'$ où $a^{b}\\left(a^{n}+n^{b}\\right) \\equiv a^{a}+b^{b} \\equiv 0 \\bmod [p]$,\net, puisque $p$ ne divise pas $a$, on a alors $a^{n}+n^{b} \\equiv 0 \\bmod [p]$.\nOn prouve de même, on a $b^{n}+n^{a} \\equiv 0 \\bmod [p]$.\nMais, alors $p$ divise à la fois $a^{n}+n^{b}$ et $b^{n}+n^{a}$, ce qui contredit qu'ils soient premiers entre eux.\nFinalement, il n'existe pas de tels entiers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn répartit des poids de respectivement $1~\\mathrm{g}, 2~\\mathrm{g}, \\ldots, 200~\\mathrm{g}$ sur les deux plateaux d'une balance de sorte que chaque plateau contienne 100 poids.\nProuver que l'on peut échanger 50 poids d'un plateau avec 50 poids de l'autre plateau pour que la balance devienne équilibrée.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans cette solution, on dira que deux poids sont complémentaires s'ils pèsent 201 grammes à eux deux.\n\nConsidérons alors les 100 poids situés sur le plateau de gauche. On peut les classer en trois catégories:\n\n1. les poids $a_{1}, \\ldots, a_{n}$ de 100 grammes ou moins, dont le complémentaire est également sur le plateau de gauche;\n2. leurs complémentaires, qui sont les poids $A_{1}, \\ldots, A_{n}$ de 101 grammes ou plus, dont le complémentaire est également sur le plateau de gauche (avec $A_{k}=201-a_{k}$);\n3. enfin, les poids $b_{1}, \\ldots, b_{\\ell}$ dont le complémentaire est sur le plateau de droite.\n\nDe même, on peut classer les 100 poids du plateau de droite en trois catégories :\n\n1. les poids $c_{1}, \\ldots, c_{m}$ de 100 grammes ou moins, dont le complémentaire est également sur le plateau de droite;\n2. leurs complémentaires, qui sont les poids $C_{1}, \\ldots, C_{m}$ de 101 grammes ou plus, dont le complémentaire est également sur le plateau de gauche (avec $C_{k}=201-c_{k}$);\n3. enfin, les poids $B_{1}, \\ldots, B_{\\ell}$ qui sont les complémentaires des poids $b_{1}, \\ldots, b_{\\ell}$ (avec $B_{k}=201-b_{k}$); ce sont les poids du plateau de droite dont le complémentaire est sur le plateau de gauche.\n\nNotons que $100=2n+\\ell=2m+\\ell$, de sorte que $m=n$. Deux possibilités s'offrent alors à nous:\n\n4. si $m=n \\geqslant 25$, on conserve sur le plateau de gauche les 50 poids $a_{1}, \\ldots, a_{25}$ et $A_{1}, \\ldots, A_{25}$, et on y transfère les 50 poids $c_{1}, \\ldots, c_{25}$ et $C_{1}, \\ldots, C_{25}$; tous les autres poids sont mis sur le plateau de droite;\n\n5. si $m=n<25$, cela signifie que $\\ell>50$; dans ce cas, on conserve sur le plateau de gauche les 50 poids $b_{1}, \\ldots, b_{50}$ et on y transfère les 50 poids $B_{1}, \\ldots, B_{50}$; tous les autres poids sont mis sur le plateau de droite.\n\nDans les deux cas, on a transféré 50 poids du plateau de gauche vers le plateau de droite, et réciproquement. En outre, chacun des deux plateaux est maintenant formé de 50 paires de poids complémentaires, de sorte que chacune pèse $50 \\times 201$ grammes, et donc qu'elles ont bien le même poids.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère $n^{2}+2n+1$ points dans un carré de côté $n$. Montrer que trois d'entre eux sont les sommets d'un triangle (éventuellement dégénéré) d'aire au plus $1/2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn raisonne sur l'enveloppe convexe des $(n+1)^{2}$ points : si elle contient beaucoup de points, il y aura 3 sommets voisins de cette enveloppe assez proches pour faire un triangle d'aire petite. Si elle n'en contient pas beaucoup, on la triangule et on regarde les nombreux points dans les petits triangles pour en former un suffisamment petit.\n\nConcrètement, voici une démonstration : Soit $E$ l'enveloppe convexe de nos points. On vérifie aisément que son périmètre est inférieur à celui du carré, à savoir $4n$. Si $E$ a au moins $4n$ sommets, donc autant de côtés, il existe deux côtés consécutifs $c_{1}$ et $c_{2}$ vérifiant $c_{1}+c_{2} \\leq 2$ donc $c_{1}c_{2} \\leq 1$ par inégalité arithméticogéométrique. Si $h$ est la hauteur relative à $c_{1}$ dans le triangle formé par les extrémités de $c_{1}$ et $c_{2}$, $h \\leq c_{2}$, donc l'aire du triangle est bien d'au plus $1/2$.\n\nSi $E$ a $s \\leq 4n-1$ sommets, on la triangule en $s-2$ triangles. Maintenant, prenons un point restant et relions-le aux sommets du triangle dans lequel il se trouve (ou des triangles s'il est sur le côté commun de deux triangles de la triangulation). On prend un autre point restant et on le relie de même aux sommets des triangles de la nouvelle triangulation dans lesquels il se trouve. On continue ainsi jusqu'à ce que tous les points soient intégrés à la triangulation. Ainsi, on a obtenu ce qu'on appelle une triangulation de notre ensemble de points. On avait $s-2$ triangles initialement, on en a rajouté au moins 2 à chaque étape (voire 3), donc on a au moins $s-2+2\\left((n+1)^{2}-s\\right) \\geq 2n^{2}+4n-1-s \\geq 2n^{2}$ triangles disjoints, donc l'un d'eux est d'aire au plus $1/2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30435,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver toutes les fonctions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ telles que $2 m n + m f(m) + n f(n)$ est un carré parfait pour tous entiers positifs $m$ et $n$.",
    "options": [],
    "answer": "f(n) = n for all n",
    "solution": "Solution:\n\nPour simplifier, posons $F(m, n) = 2 m n + m f(m) + n f(n)$. Alors $F(m, 0) = m f(m)$ est un carré pour tout $m \\geqslant 1$.\n\nOn peut donc écrire $f(p) = p a^{2}$ pour un nombre premier $p$. Supposons que $a \\geqslant 2$. Alors $(a p)^{2} + 2 p + f(1) = F(p, 1) > (a p^{2})$ est un carré, donc $(a p)^{2} + 2 p + f(1) \\geqslant (a p + 1)^{2}$ et $2 p + f(1) \\geqslant 2 a p + 1 \\geqslant 4 p + 1$, de sorte que $p \\leqslant (f(1) - 1) / 2$. Ainsi $f(p) = p$ pour tous nombres premiers $p$ suffisamment grands.\n\nSoit maintenant $k \\geqslant 1$ et $p$ un nombre premier suffisamment grand pour que $f(p) = p$, $p > k f(k)$ et $p > k^{2} + 1$. Alors\n$$\n\\begin{gathered}\n(k + p - 1)^{2} = \\left(k^{2} + 1\\right) + p^{2} - 2k - 2p + 2 k p < p^{2} - p - 2k + 2 k p < F(p, k) \\text{ et } \\\\\nF(p, k) = p^{2} + 2 k p + k f(k) < p^{2} + 2 k p + p \\leqslant (p + k + 1)^{2} .\n\\end{gathered}\n$$\nComme $F(p, k)$ est un carré, ceci impose que $k f(k) + p^{2} + 2 k p = (k + p)^{2}$ et donc que $f(k) = k$.\n\nRéciproquement, on vérifie que la fonction identité convient.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes Xantiens sont les habitants, en nombre éventuellement infini, de la planète Xanta. Vis-à-vis d'eux-mêmes et de leurs semblables, les Xantiens sont capables de ressentir deux types d'émotions, qu'ils appellent amour et respect. Il a été observé que:\n- Chaque Xantien aime un et un seul Xantien, et respecte un et un seul Xantien.\n- Si $A$ aime $B$, alors tout Xantien qui respecte $A$ aime également $B$.\n- Si $A$ respecte $B$, alors tout Xantien qui aime $A$ respecte également $B$.\n- Chaque Xantien est aimé d'au moins un Xantien.\nEst-il vrai que chaque Xantien respecte le Xantien qu'il aime?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour chaque Xantien $x$, désignons respectivement par $f(x)$ et $g(x)$ les Xantiens aimés et respectés par $x$.\nLa première condition assure que l'on a bien défini des fonctions $f$ et $g$ de l'ensemble $X$ des Xantiens sur lui-même.\nIl s'agit de savoir si, pour tout $x$, on a $f(x)=g(x)$.\nEn fait, nous allons même prouver que, pour tout $x$, on a $f(x)=g(x)=x$.\nSoit $x$ un Xantien.\nIl respecte $g(x)$ qui, de son côté, aime $f(g(x))$.\nLa seconde condition assure donc que $f(g(x))=f(x)$ pour tout $x$. (1)\nDe même, la troisième condition conduit à $g(f(x))=g(x)$ pour tout $x$. (2)\nEnfin, la quatrième condition signifie que $f$ est surjective.\nSoit $x$ un Xantien. Il existe donc un Xantien $y$ tel que $f(y)=x$.\nOr, on a $f(g(f(y)))=f(g(y))$ d'après (2)\net ainsi, d'après (1), il vient $f(f(y))=f(y)$, ou encore $f(x)=x$.\nDe (1), on déduit alors immédiatement que $g(x)=x$.\nFinalement, pour tout $x$, on a bien $f(x)=g(x)=x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les triplets d'entiers $(a, b, c)$ tels que $a \\neq 0$ et\n$$\n2 a^{4} + 2 a^{2} b^{2} + b^{4} = c^{2}\n$$",
    "options": [],
    "answer": "No such integer triples with nonzero first entry exist.",
    "solution": "Solution:\nSupposons qu'il existe un tel triplet : soit alors $a > 0$ le plus petit entier tel que $(a, b, c)$ soit solution. Si $d = \\operatorname{pgcd}(a, b)$, en divisant l'égalité par $d^{4}$ on voit que $\\left(a / d, b / d, c / d^{2}\\right)$ est également solution. On a donc $d = 1$.\n\nRegardons l'expression $2 a^{4} + 2 a^{2} b^{2} + b^{4}$ modulo $8$, et utilisons le fait qu'un carré est congru à $0, 1$ ou $4$ modulo $8$ : si $a$ est impair, alors $2 a^{4} + 2 a^{2} b^{2} + b^{4}$ est congru à $2$ ou à $5$ modulo $8$ ; $a$ est donc forcément pair, et $b$ est impair. Réécrivons l'égalité de l'énoncé en\n$$\n\\left(a^{2}\\right)^{2} + \\left(a^{2} + b^{2}\\right)^{2} = c^{2}\n$$\nLe triplet $\\left(a^{2}, a^{2} + b^{2}, c\\right)$ est donc pythagoricien. Comme $\\operatorname{pgcd}(a, b) = 1$, il existe des entiers $u, v$ premiers entre eux tels que $a^{2} = 2 u v$, $a^{2} + b^{2} = u^{2} - v^{2}$ et $c = u^{2} + v^{2}$.\n\nEn regardant modulo $4$, on obtient $1 \\equiv a^{2} + b^{2} \\equiv u^{2} - v^{2}$. Puisqu'un carré est congru à $0$ ou $1$ modulo $4$, on en déduit que $u$ est impair et que $v$ est pair. Puisque $a^{2} = 2 u v$ et que $u$ et $v$ sont premiers entre eux, on peut écrire $u = m^{2}$ et $v = 2 n^{2}$ avec $m, n$ premiers entre eux et $m$ impair. De $b^{2} = u^{2} - v^{2} - 2 u v$ il vient $2 v^{2} = (u - v + b)(u - v - b)$. Donc\n$$\n2 n^{4} = \\left(\\frac{m^{2} - 2 n^{2} + b}{2}\\right)\\left(\\frac{m^{2} - 2 n^{2} - b}{2}\\right).\n$$\nPosons $x = \\left(m^{2} - 2 n^{2} + b\\right) / 2$ et $y = \\left(m^{2} - 2 n^{2} - b\\right) / 2$ : ce sont des entiers car $m^{2}$ et $b$ sont impairs. En outre, $0 < a^{2} + b^{2} = (u - v)(u + v)$, et $u + v$ est positif, donc $u - v$ l'est également, et a fortiori $x = (u - v + b) / 2 \\geqslant 0$, puis $y \\geqslant 0$ également.\n\nD'autre part, on obtient $x - y = b$ et $x y = 2 n^{4}$. Comme $\\operatorname{pgcd}(b, n) = 1$, $x$ et $y$ sont forcément premiers entre eux. Puisque $x y = 2 n^{4}$, on peut écrire $x$ et $y$ sous la forme $x = 2 k^{4}$ et $y = l^{4}$, ou bien $x = l^{4}$ et $y = 2 k^{4}$, avec $\\operatorname{pgcd}(k, l) = 1$ et $l$ impair.\n\nAinsi, $2 k^{4} + l^{4} = x + y = m^{2} - 2 n^{2}$, et donc\n$$\nm^{2} = 2 k^{4} + l^{4} + 2 n^{2} = 2 k^{4} + 2 k^{2} l^{2} + l^{4}\n$$\nOr, $2 klm = 2 mn = a > 0$, donc $lm \\geqslant 1$ et $0 < k < a$, ce qui contredit la minimalité de $a$.\n\nIl n'existe donc pas de triplet $(a, b, c)$ solution lorsque $a \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer toutes les fonctions $f: \\mathbb{R}_{+}^{*} \\longmapsto \\mathbb{R}_{+}^{*}$ telles que\n$$\nf\\left(\\frac{y}{f(x+1)}\\right)+f\\left(\\frac{x+1}{x f(y)}\\right)=f(y)\n$$\npour tous $x, y \\in \\mathbb{R}_{+}^{*}$.",
    "options": [],
    "answer": "f(x) = 1/x for all x > 0",
    "solution": "Solution:\nEn cherchant un peu parmi les fonctions usuelles, on constate que la fonction $f: x \\longmapsto \\frac{1}{x}$ est une solution du problème. Nous allons maintenant prouver que c'est la seule.\nSoit $f$ une solution du problème. Si $y>0$ est tel que $y f(y)>1$, posons $x=\\frac{1}{y f(y)-1}$. La relation de l'énoncé donne\n$$\nf\\left(\\frac{y}{f(x+1)}\\right)+f(y)=f(y)\n$$\nce qui est impossible car $f$ est à valeurs strictement positives. Ainsi, $y f(y) \\leq 1$.\nPour tous $x, y \\in \\mathbb{R}_{+}^{*}$, on a donc\n$$\nf(y)=f\\left(\\frac{y}{f(x+1)}\\right)+f\\left(\\frac{x+1}{x f(y)}\\right) \\leq \\frac{f(x+1)}{y}+\\frac{x f(y)}{x+1},\n$$\ndonc $y f(y) \\leq(x+1) f(x+1)$. En particulier, pour tous les réels $a, b>1$, on a bien $a f(a) \\leq b f(b) \\leq a f(a)$ : il existe donc un réel $c \\in] 0,1]$ tel que $x f(x)=c$ pour tout $x>1$.\nSi $y>1$, alors $y>c$ donc $\\frac{y(x+1)}{c}>x+1>1$ et $\\frac{(x+1) y}{x c}>\\frac{y}{c}>1$, et\n$$\n\\begin{aligned}\nc & =y \\cdot f(y)=y \\cdot f\\left(\\frac{y}{f(x+1)}\\right)+y \\cdot f\\left(\\frac{x+1}{x f(y)}\\right) \\\\\n& =y \\cdot f\\left(\\frac{(x+1) \\cdot y}{c}\\right)+y \\cdot f\\left(\\frac{(x+1) \\cdot y}{c \\cdot x}\\right) \\\\\n& =\\frac{c^{2} \\cdot y}{(x+1) \\cdot y}+\\frac{c^{2} \\cdot x \\cdot y}{(x+1) \\cdot y}=c^{2},\n\\end{aligned}\n$$\ndonc $c=1$. On a donc $f(x)=\\frac{1}{x}$ pour tout $x>1$.\nOn va maintenant prouver, par récurrence sur $k \\in \\mathbb{N}$, que $f(x)=\\frac{1}{x}$ pour tout $x>2^{-k}$. Le résultat est déjà connu pour $k=0$. En outre, si on suppose le résultat connu pour $k \\geq 0$, posons $x=1$. Si $y>2^{-1-k}$, alors $2 y>2^{-k}$ et $\\frac{2}{f(y)} \\geq 2 y>2^{-k}$. L'énoncé indique donc que\n$$\nf(y)=f(2 y)+f\\left(\\frac{2}{f(y)}\\right)=\\frac{1}{2 y}+\\frac{f(y)}{2}\n$$\nc'est-à-dire $f(y)=\\frac{1}{y}$. Le résultat est donc vrai aussi pour $k+1$.\nCeci montre que $f: x \\longmapsto \\frac{1}{x}$ est bien l'unique solution du problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dispose de $n$ jetons portant chacun un numéro entier (qui peut être négatif). Si on trouve parmi eux deux jetons portant le même numéro $m$, on les enlève et on met à leur place un jeton portant le numéro $m-1$, et un autre portant le numéro $m+1$. Montrer qu'au bout d'un nombre fini de tels changements, tous les jetons porteront des numéros distincts.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par montrer le lemme suivant:\nSoit $m_{0}$ le plus petit des numéros des jetons au début. On note $m_{1}^{(k)}, \\ldots, m_{j_{k}}^{(k)}$ les numéros inférieurs ou égaux à $m_{0}$ qui sont portés par des jetons après $k$ changements, comptés avec multiplicité et rangés par ordre décroissant de sorte que $m_{j_{k}}^{(k)} \\leq \\ldots \\leq m_{1}^{(k)} \\leq m_{0}^{(k)}=m_{0}$. Alors pour tout $i \\in\\left\\{0, \\ldots, j_{k}-1\\right\\}$, on a $m_{i+1}^{(k)}-m_{i}^{(k)} \\leq 2$.\n\nDémonstration : on raisonne par récurrence sur $k$. Le cas $k=0$ est clair.\n\nSupposons que le résultat soit vérifié pour un certain $k$. Au $(k+1)$-ème changement, si nous remplaçons deux jetons portant un numéro strictement plus grand que $m_{0}$, aucun des $m_{i}^{(k)}$ ne change, donc la propriété est toujours vérifiée. Supposons maintenant que nous remplaçons deux jetons portant un numéro $m_{i}^{(k)}=m_{i+1}^{(k)}=m$. Alors la suite $\\left(m_{i}^{(k+1)}\\right)$ s'obtient à partir de la suite $\\left(m_{i}^{(k)}\\right)$ en remplaçant deux occurrences de $m$ par une occurrence de $m-1$ et une occurrence de $m+1$. La différence entre ces dernières est exactement $2$, donc même s'il reste encore des termes égaux à $m$, leur distance aux termes autour d'eux sera inférieure ou égale à $1$. Celle entre $m-1$ et le terme suivant est inférieure strictement à celle entre $m$ et ce même terme suivant, donc inférieure à $2$ par hypothèse de récurrence. Pour ce qui est de $m+1$, nous avons deux cas à distinguer. Soit $m=m_{0}$ et alors $m+1$ est strictement supérieur à $m_{0}$ et ne nous intéresse plus. Soit $m<m_{0}$, et alors $m+1$ a une distance au terme précédent inférieure à la distance entre $m$ et ce même terme suivant, donc inférieure à $2$ par hypothèse de récurrence.\n\nConséquence du lemme : tous les numéros portés par des jetons sont supérieurs ou égaux à $m_{0}-2n$ à tout moment. Quitte à ajouter $2n-m_{0}+1$ à tous les numéros au départ, ce qui ne change pas l'exercice, on peut donc supposer qu'ils sont tous strictement positifs à tout moment.\n\nOn note alors $P_{k}$ le produit des numéros de tous les jetons après $k$ changements. D'après ce qu'on vient de montrer, $P_{k}$ est un entier strictement positif pour tout $k$. Supposons que le $(k+1)$-ème changement consiste à remplacer deux jetons portant le numéro $m$ par un jeton portant le numéro $m-1$ et un jeton portant le numéro $m+1$. Alors $P_{k}$ s'écrit $m^{2}c$ avec $c>0$, et $P_{k+1}$ s'écrit\n$$\nP_{k+1}=(m-1)(m+1)c=\\left(m^{2}-1\\right)c<m^{2}c=P_{k}\n$$\nAinsi, la suite des $\\left(P_{k}\\right)$ est une suite strictement décroissante d'entiers strictement positifs, elle est donc finie, c'est-à-dire qu'on arrive nécessairement à un moment où aucun changement n'est possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers strictement positifs $a$ et $b$ tels que $4a+1$ et $4b-1$ soient premiers entre eux, et tels que $a+b$ divise $16ab+1$.",
    "options": [],
    "answer": "b = 3a − 1 or a = 3b + 1",
    "solution": "Solution:\n\nNotons (C) la condition de l'énoncé. Les entiers $4a+1$ et $4b-1$ sont premiers entre eux si et seulement si $4a+1$ est premier avec $(4a+1)+(4b-1)=4(a+b)$. Or, $4a+1$ est impair donc il est automatiquement premier avec $4$. On en déduit que $4a+1$ et $4b-1$ sont premiers entre eux si et seulement si $4a+1$ est premier avec $a+b$.\n\nD'autre part, $16ab+1=16a(a+b)+(1-16a^2)=16a(a+b)-(4a-1)(1+4a)$, donc $a+b$ divise $16ab+1$ si et seulement si $a+b$ divise $(4a-1)(1+4a)$. Or, si $a+b$ est premier avec $1+4a$ alors ceci équivaut à $a+b \\mid 4a-1$.\n\nOn en déduit que (C) équivaut à ce que $\\operatorname{pgcd}(a+b, 4a+1)=1$ et $a+b \\mid 4a-1$.\n\nD'autre part, montrons que si $a+b \\mid 4a-1$ alors on a automatiquement $\\operatorname{pgcd}(a+b, 4a+1)=1$. En effet, si un nombre premier $p$ divise $a+b$ et $4a+1$, alors il divise $4a-1$ et $4a+1$, donc il divise $(4a+1)-(4a-1)=2$. Il vient $p=2$, puis $2 \\mid 4a-1$. Impossible.\n\nFinalement, (C) équivaut à $a+b \\mid 4a-1$.\n\nOr, $a+b>a>\\frac{4a-1}{4}$, donc $a+b \\mid 4a-1$ ne peut se produire que si $a+b=4a-1$ ou $a+b=\\frac{4a-1}{2}$ ou $a+b=\\frac{4a-1}{3}$.\n\nLe deuxième cas ne peut pas se produire car $4a-1$ est impair.\n\n(C) équivaut donc à $a+b=4a-1$ ou $a+b=\\frac{4a-1}{3}$, c'est-à-dire $b=3a-1$ ou $a=3b+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver le nombre de suites $\\left(u_{n}\\right)_{n \\geqslant 1}$ d'entiers relatifs telles que $u_{n} \\neq -1$ pour tout entier $n \\geqslant 1$ et telles que\n$$\nu_{n+2}=\\frac{2014+u_{n}}{1+u_{n+1}}$$\npour tout entier $n \\geqslant 1$.",
    "options": [],
    "answer": "14",
    "solution": "Solution:\n\nLa relation de l'énoncé impose que $\\left(u_{n+1}-u_{n-1}\\right)\\left(u_{n}+1\\right)=u_{n}-u_{n-2}$ pour $n \\geqslant 3$. Par récurrence, il vient\n$$\n\\mathbf{u}_{3}-\\mathbf{u}_{1}=\\prod_{i=3}^{n}\\left(u_{i}+1\\right)\\left(u_{n+1}-u_{n-1}\\right)\n$$\nSupposons par l'absurde que $u_{3} \\neq u_{1}$. Ainsi $u_{n+1} \\neq u_{n-1}$ pour tout $n \\geqslant 2$. En faisant croître $n$, on voit qu'il ne peut pas y avoir une infinité de termes $u_{i}$ tels que $u_{i} \\neq 0,-2$. Ainsi, $u_{i}=0$ ou $-2$ à partir d'un certain rang (car $\\mathfrak{u}_{3}-\\mathfrak{u}_{1}$ a un nombre fini de diviseurs). On vérifie aisément à partir de la relation de récurrence de l'énoncé que ceci n'est pas possible. Donc $u_{3}=u_{1}$, et par (1) on en déduit que $u_{n+2}=u_{n}$ pour tout entier $n \\geqslant 1$.\n\nSous cette hypothèse, la relation à vérifier devient $\\left(1+u_{n+1}\\right) u_{n}=2014+u_{n}$, ou encore $u_{n} u_{n+1}=2014$, ce qui est équivalent à $u_{1} u_{2}=2014$. Ainsi, le nombre de suites $\\left(u_{n}\\right)_{n \\geqslant 1}$ qui conviennent est le nombre de paires d'entiers $(a, b)$ telles que $a b=2014$, avec $a \\neq -1$ et $b \\neq -1$ : c'est exactement le nombre de diviseurs (positifs ou négatifs) de $2014$, moins $2$. Comme $2014=2 \\cdot 19 \\cdot 53$, celui-ci a $2 \\cdot (1+1) \\cdot (1+1) \\cdot (1+1)=16$ diviseurs (positifs ou négatifs), donc $14$ suites conviennent.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère un quadrillage formé de tous les petits carrés de côté $1$ entièrement inclus dans le disque donné par l'inéquation $x^{2}+y^{2} \\leq 2014^{2}$ dans le plan muni d'un repère orthonormé $Oxy$. À chaque étape du jeu, chaque carré contient le nombre $1$ ou le nombre $-1$. On suppose que chaque case du quadrillage contient un $1$ au début. Un tour consiste à choisir une ligne, une colonne, ou une diagonale (c'est-à-dire un ensemble de cases maximal tel que les centres de ces dernières soient alignés sur une droite formant un angle de $45$ degrés avec l'un des axes de coordonnées) du quadrillage, et à changer les signes du contenu de toutes les cases de celle-ci. Peut-on à la fin avoir un $-1$ dans une case, et des $1$ dans toutes les autres?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nOn raisonne par l'absurde en supposant qu'on peut arriver à avoir un seul $-1$. On appelle $C_{1}$ le petit carré qui contiendra ce $-1$. Par symétrie de la figure, on peut supposer que le centre de ce petit carré a pour coordonnées $\\left(a-\\frac{1}{2}, b-\\frac{1}{2}\\right)$ avec $a \\geq b \\geq 1$.\n\n1ère étape : On montre que le petit carré de centre $\\left(a-\\frac{3}{2}, b+\\frac{1}{2}\\right)$ appartient également à la figure. Pour cela, il suffit de montrer que le coin supérieur droit de ce petit carré appartient à la figure, c'est-à-dire que $(a-1)^{2}+(b+1)^{2} \\leq 2014^{2}$. On distingue deux cas :\n\n- Si $a>b$ alors $(a-1)^{2}+(b+1)^{2} \\leq a^{2}+b^{2} \\leq 2014^{2}$.\n- Si $a=b$, alors l'inégalité $a^{2}+b^{2}=2 a^{2} \\leq 2014^{2}$ ne peut être une égalité car la valuation $2$-adique du côté gauche est impaire alors que celle du côté droit est paire. Nous avons donc $2 a^{2}+2 \\leq 2014^{2}$, d'où $(a-1)^{2}+(a+1)^{2}=2 a^{2}+2 \\leq 2014^{2}$.\n\nOn appelle $C_{2}$ ce petit carré de centre $\\left(a-\\frac{3}{2}, b+\\frac{1}{2}\\right)$.\n\n2ème étape : Par des considérations de symétrie et de convexité, les petits carrés $C_{1}, \\ldots, C_{8}$ suivants appartiennent tous à la figure :\n\n![](attached_image_1.png)\n\nOr chaque ligne, chaque colonne et chaque diagonale contient soit aucun, soit deux de ces huit petits carrés. Ainsi, la parité du nombre de $-1$ parmi ces carrés ne change pas, contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle tel que $\\widehat{A}=40^{\\circ}$ et $\\widehat{B}=60^{\\circ}$. Soient $D$ et $E$ des points de $[AC]$ et $[AB]$ tels que $\\widehat{CBD}=40^{\\circ}$ et $\\widehat{ECB}=70^{\\circ}$. On note $F$ l'intersection de $(BD)$ et $(CE)$. Montrer que $(AF) \\perp (BC)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn a $\\widehat{BFC}=180^{\\circ}-\\widehat{CBF}-\\widehat{FCB}=180^{\\circ}-40^{\\circ}-70^{\\circ}=70^{\\circ}=\\widehat{FCB}$ donc $BCF$ est isocèle en $B$. Il vient $BC=BF$.\n\nSoit $M$ le point d'intersection entre la bissectrice de $\\widehat{CBF}$ et $(AC)$. On a $\\widehat{BMA}=40^{\\circ}=\\widehat{BAM}$ donc $MA=MB$.\n\nD'autre part, $\\widehat{BMC}=180^{\\circ}-\\widehat{CBM}-\\widehat{MCB}=180^{\\circ}-20^{\\circ}-80^{\\circ}=80^{\\circ}=\\widehat{MCB}$ donc $BC=BM$.\n\nOn déduit de ce qui précède que $MA=BF$.\n\nSoit $B' \\in (BC)$ tel que $AB'$ est équilatéral. Soit $\\Delta$ la bissectrice de $\\widehat{B'BA}$. Comme $BF=BM$ et $\\widehat{FBA}=\\widehat{CBM}$, les points $M$ et $F$ sont symétriques par rapport à $\\Delta$. Or, $A$ et $B'$ sont aussi symétriques par rapport à $\\Delta$, donc $MA=FB'$.\n\nFinalement, $FB'=FB$, donc $F$ appartient à la médiatrice de $[BB']$, qui est aussi la hauteur $(AF)$ du triangle $ABB'$.\n\n\nAutre solution. Notons $H$ le projeté de $F$ sur $[BC]$ et $H'$ le projeté de $A$ sur $[BC]$. Il s'agit de montrer que $H=H'$. Pour cela, il suffit de voir que $BH=BH'$.\n\nOn a $B'=AB \\cos 60$ et $BH=BF \\cos 40=BC \\cos 40$. Comme $\\frac{AB}{BC}=\\frac{\\sin 80}{\\sin 40}$, il suffit de montrer que $\\sin 80 \\cos 60=\\sin 40 \\cos 40$, c'est-à-dire (compte tenu de $\\cos 60=\\frac{1}{2}$) que $\\sin 80=2 \\sin 40 \\cos 40$, ce qui est vrai d'après la formule générale $\\sin 2x=2 \\sin x \\cos x$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMontrer que si $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{n} \\geqslant 0$ et $a_{1}+a_{2}+\\cdots+a_{n}=1$ alors\n$$\na_{1}^{2}+3 a_{2}^{2}+5 a_{3}^{2}+\\cdots+(2 n-1) a_{n}^{2} \\leqslant 1 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nAfin de faire apparaître les termes $a_{i}^{2}$, il est tentant de commencer par élever l'expression $a_{1}+a_{2}+\\cdots+a_{n}=1$ au carré, puis de tout développer en espérant très fort qu'une simplification appraîtra. En outre, on voit bien qu'il faudra utiliser le caractère ordonné des termes $a_{i}$ (c'est-à-dire que $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{n} \\geqslant 0$ ), puisque l'inégalité à démontrer est évidemment fausse si on prend $a_{1}=\\ldots=a_{n-1}=0$ et $a_{n}=1$. On obtient alors directement\n$$\n\\begin{aligned}\n1 & =\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2}=\\sum_{i=1}^{n} a_{i}^{2}+2 \\sum_{i=1}^{n} \\sum_{j=i+1}^{n} a_{i} a_{j} \\\\\n& \\geq \\sum_{i=1}^{n} a_{i}^{2}+2 \\sum_{i=1}^{n} \\sum_{j=i+1}^{n} a_{j}^{2}=a_{1}^{2}+3 a_{2}^{2}+5 a_{3}^{2}+\\cdots+(2 n-1) a_{n}^{2} .\n\\end{aligned}\n$$\n\n\nSolution alternative\nSoit $\\Omega$ l'ensemble des $n$-uplets $\\left(a_{1}, \\ldots, a_{n}\\right)$ de réels tels que $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{n} \\geqslant 0$ et $a_{1}+a_{2}+\\cdots+a_{n}=1$. Considérons la fonction\n$$\nf: \\quad \\begin{aligned}\n& \\Omega \\mapsto \\mathbb{R} \\\\\n& \\left(a_{1}, \\ldots, a_{n}\\right) \\mapsto a_{1}^{2}+3 a_{2}^{2}+5 a_{3}^{2}+\\cdots+(2 n-1) a_{n}^{2}\n\\end{aligned}\n$$\nLa fonction $f$ est convexe car somme de fonctions convexes, donc elle prend son maximum en un point extrémal de $\\Omega$, c'est-à-dire en un point $p \\in \\Omega$ tel que, pour tous $a, b \\in \\Omega$ et pour tout $\\lambda \\in] 0,1[$, si $p=\\lambda a+(1-\\lambda) b$, alors $a=b=p$. On montre alors aisément que les points extrémaux de $\\Omega$ sont nécessairement de la forme $a_{1}=\\ldots=a_{k}=\\frac{1}{k}, a_{k+1}=\\ldots=a_{n}=0$, avec $1 \\leq k \\leq n$. Et alors $f\\left(a_{1}, \\ldots, a_{n}\\right)=\\sum_{i=1}^{k} \\frac{2 i+1}{k^{2}}=\\frac{k(k-1)+k}{k^{2}}=1$.\nOn en déduit donc que $f\\left(a_{1}, \\ldots, a_{n}\\right) \\leq 1$ dès lors que $\\left(a_{1}, \\ldots, a_{n}\\right) \\in \\Omega$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer le plus grand nombre d'entiers que l'on peut extraire de l'ensemble $\\{1,2, \\ldots, 2014\\}$ de sorte que la différence de deux quelconques de ces entiers soit différente de 17.",
    "options": [],
    "answer": "1011",
    "solution": "Solution:\nCet exercice découle en fait d'une utilisation astucieuse du principe des tiroirs. Soit $E$ une partie de $\\{1,2, \\ldots, 2014\\}$ ne contenant que des entiers dont la différence n'est jamais égale à 17.\n\nTout d'abord, pour tout entier $a$, on note $S_{a}$ l'ensemble $\\{a, a+17\\}$, et $T_{a}$ l'ensemble $\\{a, a+1, \\ldots, a+33\\}$ : alors on sait que $E \\cap S_{a}$ contient au plus un élément, quel que soit l'entier $a$ considéré : autrement dit, $|E \\cap S_{a}| \\leq 1$. Puisque $2014 = 34 \\times 59 + 8$, on sait que\n$$\n\\{1,2, \\ldots, 2014\\} \\subseteq \\bigcup_{i=0}^{58} T_{34i+1} \\cup \\bigcup_{a=2009}^{2014} S_{a} = \\bigcup_{i=0}^{58}\\left(\\bigcup_{a=34i+1}^{34i+17} S_{a}\\right) \\cup \\bigcup_{a=2009}^{2014} S_{a}\n$$\nde sorte que\n$$\n|E| = |E \\cap \\{1,2, \\ldots, 2014\\}| \\leq \\sum_{i=0}^{58}\\left(\\sum_{a=34i+1}^{34i+17} |E \\cap S_{a}|\\right) + \\sum_{a=2009}^{2014} |E \\cap S_{a}| = 59 \\times 17 + 8 = 1011\n$$\nD'autre part, en choisissant exactement $E = \\bigcup_{i=0}^{58}\\left(\\bigcup_{a=34i+1}^{34i+17} \\{a\\}\\right) \\cup \\bigcup_{a=2009}^{2014} \\{a\\}$, on obtient bien une partie de $\\{1,2, \\ldots, 2014\\}$ ne contenant que des entiers dont la différence n'est jamais égale à 17, et dont le cardinal est précisément $|E| = 1011$.\n\nAinsi, le nombre recherché était bien 1011.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $P$ un point à l'extérieur d'un cercle $\\mathcal{C}$. Les tangentes à $\\mathcal{C}$ passant par $P$ touchent $\\mathcal{C}$ en $A$ et $B$. Une droite passant par $P$ intersecte $\\mathcal{C}$ aux points $Q$ et $R$. Soit $S$ un point de $\\mathcal{C}$ tel que $(B S) \\parallel (Q R)$. Montrer que $(S A)$ passe par le milieu de $[Q R]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLa puissance du point $P$ par rapport au cercle $\\mathcal{C}$ est égale à $P B^{2} = P Q \\cdot P R$. Ainsi, on a égalité des rapports de longueurs $\\frac{P R}{P B} = \\frac{P B}{P Q}$, ce qui montre que les triangles $P R B$ et $P B Q$ sont indirectement semblables. En particulier, on en déduit que $\\frac{P B}{P R} = \\frac{B Q}{B R}$. Puis, $A$ et $B$ jouant des rôles symétriques, on a également $\\frac{P A}{P R} = \\frac{A Q}{A R}$.\n\nSoit alors $O$ le centre du cercle $\\mathcal{C}$ ainsi que $N_{1}$ le milieu de $[Q R]$ et $N_{2}$ le milieu de $[B S]$. Les cordes $Q R$ et $B S$ étant parallèles, on sait que $(N_{1} O) = (N_{2} O)$ est la médiatrice commune à $[Q R]$ et à $[B S]$, c'est-à-dire un axe de symétrie du trapèze $Q S B R$. Il s'ensuit que $\\frac{B Q}{B R} = \\frac{S R}{S Q}$. On en déduit donc que\n$$\n\\frac{R S}{Q S} \\cdot \\frac{R A}{Q A} = \\frac{B Q}{B R} \\cdot \\frac{P R}{P A} = \\frac{P B}{P R} \\cdot \\frac{P R}{P A} = \\frac{P B}{P A} = 1\n$$\nc'est-à-dire que $A R \\cdot R S = A Q \\cdot Q S$.\n\nNotons ensuite que les points $A, R, S$ et $Q$ sont cocycliques, de sorte que $\\sin (\\widehat{A R S}) = \\sin (\\widehat{A Q S})$, donc que\n$$\n2 \\mathcal{A}_{A R S} = (A R \\cdot R S) \\sin (\\widehat{A R S}) = (A Q \\cdot Q S) \\sin (\\widehat{A Q S}) = 2 \\mathcal{A}_{A Q S}\n$$\nEnfin, soit $M$ le point d'intersection de $(A S)$ et $(Q R)$. On sait que $2 \\mathcal{A}_{A R S} = (A S \\cdot M R) \\sin (\\widehat{A M S})$ et $2 \\mathcal{A}_{A Q S} = (A S \\cdot M Q) \\sin (\\widehat{S M Q})$, d'où l'on déduit que $(A S \\cdot M R) \\sin (\\widehat{A M S}) = (A S \\cdot M Q) \\sin (\\widehat{S M Q})$. Puisque $\\widehat{A M R} = \\widehat{S M Q}$, cela montre en fait que $M R = M Q$, c'est-à-dire $M$ est le milieu de $[Q R]$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les nombres réels $x, y, z$ satisfaisant le système d'équations suivant :\n\n$$\n\\begin{cases}\nx = \\sqrt{2y + 3} \\\\\ny = \\sqrt{2z + 3} \\\\\nz = \\sqrt{2x + 3}\n\\end{cases}\n$$",
    "options": [],
    "answer": "x = y = z = 3",
    "solution": "Solution:\n\nIl est évident que les nombres $x, y, z$ doivent être strictement positifs. Les deux premières équations donnent $x^{2} = 2y + 3$ et $y^{2} = 2z + 3$. En les soustrayant, on obtient $x^{2} - y^{2} = 2(y - z)$. On en déduit que si $x \\leqslant y$ alors $y \\leqslant z$, et de même si $y \\leqslant z$ alors $z \\leqslant x$. Donc si $x \\leqslant y$, on a $x \\leqslant y \\leqslant z \\leqslant x$, ce qui impose que $x = y = z$.\n\nOn montre de même que si $x \\geqslant y$ alors $x = y = z$. Donc dans tous les cas, $x, y, z$ sont égaux, et leur valeur commune satisfait l'équation $x^{2} = 2x + 3$, qui s'écrit encore $(x - 3)(x + 1) = 0$. Or, $x$ est strictement positif, donc nécessairement $x = y = z = 3$.\n\nRéciproquement, on vérifie immédiatement que $x = y = z = 3$ est bien solution du système.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $ABCD$ un quadrilatère convexe (c'est-à-dire que ses diagonales sont à l'intérieur de $ABCD$), et $P$ l'intersection de ses diagonales $[AC]$ et $[BD]$. On note $O_1, O_2, O_3$ et $O_4$ les centres des cercles circonscrits à $ABP$, $BCP$, $CDP$ et $DAP$.\nMontrer que $O_1O_2O_3O_4$ est un parallélogramme.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n$O_1$ et $O_2$ sont sur la médiatrice de $[PB]$, donc $O_1O_2$ est la médiatrice de $[PB]$. De même, $(O_3O_4)$ est la médiatrice de $[PD]$, donc $(O_1O_2)$ et $(O_3O_4)$ sont toutes deux perpendiculaires à $(BD)$, donc elles sont parallèles. De même, $(O_2O_3)$ et $(O_4O_1)$ sont toutes deux perpendiculaires à $(AC)$, donc elles sont parallèles.\n\n$O_1O_2O_3O_4$ a ses côtés opposés parallèles deux à deux, donc c'est un parallélogramme.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a_{1}, a_{2}, \\ldots, a_{2 n}$ des réels tels que $a_{1}+a_{2}+\\cdots+a_{2 n}=0$.\nProuver qu'il existe au moins $2 n-1$ couples $\\left(a_{i}, a_{j}\\right)$ avec $i<j$ tels que $a_{i}+a_{j} \\geqslant 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSans perte de généralité, on peut supposer que $a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{2 n}$. On distingue deux cas :\n\n- Si $a_{n}+a_{2 n-1} \\geq 0$ alors on a $a_{i}+a_{2 n-1} \\geq 0$ pour $i=n, \\cdots, 2 n-2$, et $a_{i}+a_{2 n} \\geq 0$ pour $i=n \\cdots, 2 n-1$. Cela fournit bien $2 n-1$ sommes positives ou nulles.\n\n- Si $a_{n}+a_{2 n-1}<0$ alors\n$$\na_{1}+\\cdots+a_{n-1}+a_{n+1}+\\cdots a_{2 n-2}+a_{2 n}>0\n$$\nD'autre part, on a $0>a_{n}+a_{2 n-1} \\geq a_{n-1}+a_{2 n-2} \\geq \\cdots \\geq a_{2}+a_{n+1}$, donc\n$$\na_{2}+a_{3}+\\cdots+a_{n-1}+a_{n+1}+\\cdots+a_{2 n-2}<0\n$$\nDe (1) et (2), on déduit que $a_{1}+a_{2 n} \\geq 0$, ce qui assure que $a_{i}+a_{2 n} \\geq 0$ pour $i=1, \\cdots 2 n-1$.\n\n\nAutre solution. Notons $b_{1} \\leqslant b_{2} \\leqslant \\cdots \\leqslant b_{\\ell}$ les entiers positifs ou nuls parmi $a_{1}, \\ldots, a_{2 n}$. Premier cas: $\\ell>n$. Alors il y a au moins $\\frac{\\ell(\\ell-1)}{2} \\geq \\frac{n(n+1)}{2}$ couples $\\left(a_{i}, a_{j}\\right)$ avec $i<j, a_{i} \\geqslant 0$ et $a_{j} \\geqslant 0$. Or, $\\frac{n(n+1)}{2}-(2 n-1)=\\frac{n^{2}-3 n+2}{2}=\\frac{(n-1)(n-2)}{2} \\geqslant 0$, donc il y a au moins $2 n-1$ couples $\\left(a_{i}, a_{j}\\right)$ avec $i<j$ tels que $a_{i}+a_{j} \\geqslant 0$.\n\nDeuxième cas : $\\ell \\leqslant n$. Notons $c_{1} \\leqslant \\cdots \\leqslant c_{\\ell}$ les plus petits entiers parmi $a_{1}, \\ldots, a_{2 n}$. Comme $\\ell \\leqslant n$, on a $c_{\\ell}<0$. De plus, $\\sum_{i=1}^{2 n} a_{i}$ est égal à la somme de $\\sum_{i=1}^{\\ell}\\left(b_{i}+c_{i}\\right)$ et de termes négatifs, donc $\\sum_{i=1}^{\\ell}\\left(b_{i}+c_{i}\\right) \\geqslant 0$. Or, $b_{\\ell}+c_{\\ell} \\geqslant b_{i}+c_{i}$ pour tout $i$, donc $b_{\\ell}+c_{\\ell} \\geqslant 0$.\nOn forme ainsi déjà $2 n-\\ell$ couples $\\left(a_{i}, a_{j}\\right)$ en prenant $a_{j}=b_{\\ell}$ et $a_{i}$ autre que $c_{1}, \\ldots, c_{\\ell-1}, b_{\\ell}$.\nDe plus, pour tout $k=1, \\ldots, \\ell-1$, on a $\\sum_{i=1}^{\\ell}\\left(b_{i}+c_{i+k}\\right) \\geqslant 0$ (où par convention $c_{\\ell+1}=c_{1}$, $c_{\\ell+2}=c_{2}$, etc.), donc pour tout $k$ il existe $i$ tel que $b_{i}+c_{i+k} \\geqslant 0$. Ceci fournit encore $\\ell-1$ couples supplémentaires.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans un pays, se trouvent 100 villes. Chacune de ces villes est reliée à exactement trois autres villes par des routes directes dans les deux sens. Prouver qu'il existe une ville $A$ à partir de laquelle on peut aller de ville en ville et revenir en $A$, sans jamais passer deux fois par une même route, et en utilisant un nombre total de routes qui n'est pas divisible par 3 (il n'est pas demandé que toutes les villes du pays soient visitées au cours de ce voyage).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPuisqu'il n'y a qu'un nombre fini de villes, on peut considérer un chemin $C$ de longueur maximale. Soit $v_{0}$ une des villes extrémités de $C$ et parcourons $C$ en partant de $v_{0}$ en numérotant les villes au fur et à mesure. La maximalité de $C$ assure que les trois villes reliées à $v_{0}$ par une route sont dans $C$. Il s'agit de $v_{1}$, $v_{i}$ et $v_{j}$ avec $1 < i < j$. On a ainsi identifié trois cycles :\n$$\n\\begin{gathered}\nv_{0}, v_{1}, \\cdots, v_{i}, v_{0} \\text{, de longueur } i+1, \\\\\nv_{0}, v_{1}, \\cdots, v_{j}, v_{0} \\text{, de longueur } j+1 \\\\\nv_{0}, v_{i}, v_{i+1}, \\cdots, v_{j}, v_{0}, \\text{ de longueur } j-i+2\n\\end{gathered}\n$$\nSi $i+1$ ou $j+1$ n'est pas divisible par $3$, l'un des deux premiers cycles convient. Sinon, c'est que $i = j = -1 \\bmod 3$, d'où $j-i+2 = 2 \\bmod 3$ et le troisième cycle permet de conclure.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$. On place une pièce sur chaque case d'un échiquier $n \\times n$. Un mouvement consiste à déplacer chaque pièce sur une case qui touche la case de départ par un coin exactement (plusieurs pièces peuvent se retrouver sur la même case).\nQuel est le plus petit entier $k$ tel qu'il est possible qu'après un certain nombre de mouvements, seules $k$ cases contiennent au moins une pièce?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nOn note $A$, $B$, $C$ et $D$ les quatre cases qui forment le carré $2 \\times 2$ en haut à droite :\n![](attached_image_1.png)\nChaque pièce peut être amenée sur une des cases $A$, $B$, $C$ et $D$, puis osciller à chaque mouvement entre $A$ et $D$ ou entre $B$ et $C$. Il est donc possible qu'après un certain nombre de mouvements, toutes les pièces soient sur une des quatre cases $A$, $B$, $C$ et $D$.\n\nIl reste à montrer qu'on ne peut pas faire mieux que $4$. On considère le coloriage suivant de l'échiquier (ici pour $n=4$) :\n![](attached_image_2.png)\nComme $n \\geqslant 2$, il y a au début au moins une pièce sur une case de chaque couleur. Mais à chaque mouvement, les pièces qui se trouvent sur des cases bleues passent sur des cases rouges, celles sur des cases rouges passent sur des bleues, celles sur des jaunes passent sur des vertes et celles sur des vertes passent sur des jaunes. À chaque étape, il y a donc des pièces sur au moins une case de chaque couleur, donc il y a toujours au moins $4$ cases qui contiennent au moins une pièce.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30452,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A$ un point extérieur à un cercle $\\mathscr{C}$ de centre $O$. Un point $P$ se déplace sur $\\mathscr{C}$. Soit $M$ le point d'intersection entre $(A P)$ et la bissectrice de $\\widehat{P O A}$. Montrer que $M$ se déplace sur un cercle que l'on décrira.",
    "options": [],
    "answer": "Let d = OA and R be the radius of the given circle, and set k = d / (d + R). The locus of M is the circle with center O′ on the line AO such that AO′ = k · AO = d^2 / (d + R), and radius kR = dR / (d + R). Equivalently, M lies on the circle centered at O′ with radius kR, which also passes through O.",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $d = O A$ et $R$ le rayon du cercle.\n\nD'après le théorème de la bissectrice, on a $M P / M A = O P / O A$, donc\n$$\n\\frac{M A}{A P} = \\frac{M A}{M A + M P} = \\frac{1}{1 + \\frac{M P}{M A}} = \\frac{1}{1 + \\frac{O P}{O A}} = \\frac{O A}{O A + O P} = \\frac{d}{d + R}.\n$$\nNotons $k$ cette valeur.\n\nSoit $O'$ le point de $[A O]$ tel que $A O' = k A O$.\n\nD'après le théorème de Thalès, les droites $(O' M)$ et $(O P)$ sont parallèles, donc\n$$\n\\frac{O' M}{O P} = \\frac{A M}{A P} = k,\n$$\ndonc $O' M = k R$.\n\nCeci prouve que $M$ décrit le cercle de centre $O'$ et de rayon $k R$.\n\nRemarquons que $A O' + k R = k(d + R) = d$, donc ce cercle passe par $O$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30453,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Prouver que, pour tous réels strictement positifs $a, b, k$ tels que $a < b$, on a\n$$\n\\frac{a}{b} < \\frac{a + k}{b + k}\n$$\n\nb) Prouver que\n$$\n\\frac{1}{100} + \\frac{4}{101} + \\frac{7}{102} + \\frac{10}{103} + \\cdots + \\frac{148}{149} > 25\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) On a $a k < b k$, donc $a b + a k < a b + b k$. Ceci s'écrit $a(b + k) < b(a + k)$, ou encore $\\frac{a}{b} < \\frac{a + k}{b + k}$.\n\nb) Soit $A = \\frac{1}{100} + \\frac{4}{101} + \\frac{7}{102} + \\frac{10}{103} + \\cdots + \\frac{148}{149}$. En appliquant ce qui précède, on en déduit\n$$\nA > \\frac{1}{100} + \\frac{3}{100} + \\frac{5}{100} + \\frac{7}{100} + \\cdots + \\frac{99}{100} = \\frac{50 + 2(0 + 1 + 2 + 3 + \\cdots + 49)}{100}\n$$\nIl suffit donc de vérifier que ce dernier terme vaut 25, soit en calculant à la main le numérateur, soit en utilisant la formule\n$$\n1 + 2 + 3 + \\cdots + n = \\frac{n(n + 1)}{2}\n$$\nqui donne\n$$\n\\frac{50 + 2(0 + 1 + 2 + 3 + \\cdots + 49)}{100} = \\frac{50 + 49 \\times 50}{100} = 25\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30454,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers naturels $a$ pour lesquels il existe des nombres premiers $p$, $q$, $r$, pas forcément distincts, tels que\n$$\na = \\frac{p+q}{r} + \\frac{q+r}{p} + \\frac{r+p}{q}\n$$",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nTout d'abord, si $p = q = r$, alors on obtient $a = 6$, qui est bien une solution.\n\nMontrons qu'il s'agit de la seule solution, et supposons qu'il existe une solution $a = \\frac{p+q}{r} + \\frac{q+r}{p} + \\frac{r+p}{q}$ avec $p, q, r$ premiers et non tous égaux.\n\n- Si deux des nombres $p, q, r$ sont égaux, disons $p = q \\neq r$, alors:\n$$\na = 2\\left(\\frac{p}{r} + \\frac{r}{p}\\right) + 2 = \\frac{2(p^2 + r^2)}{pr} + 2\n$$\nAinsi, $\\frac{2(p^2 + r^2)}{pr}$ est un entier. On en déduit que le nombre premier $p$ divise $2 r^2$ tout en étant premier avec $r$, et donc que $p = 2$. Le raisonnement s'adapte mot pour mot pour prouver que $r = 2$, ce qui contredit que $r \\neq p$. Il n'y a donc pas de solution dans ce cas.\n\n- Si $p, q, r$ sont deux à deux distincts alors, sans perte de généralité, on peut supposer que $p > q > r$ (on note qu'alors $p$ et $q$ sont impairs), et alors\n$$\napqr = pq(p+q) + qr(q+r) + rp(r+p)\n$$\nAinsi, $p$ divise $qr(q+r)$. Puisque $p$ est premier avec $qr$, il divise donc $q + r$. Mais $0 < q + r < 2p$ donc $p = q + r$. Cela implique que $r$ est pair, et donc que $r = 2$. On a alors $p = q + 2$ et $a = (q + 1) + 1 + \\frac{q + 4}{q} = q + 3 + \\frac{4}{q}$. En particulier, $\\frac{4}{q}$ doit être un entier, ce qui est impossible pour $q > 1$ impair. Il n'y a ainsi pas de solution dans ce cas, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\Gamma$ un cercle de centre $O$, et $(d)$ une droite qui n'intersecte pas $\\Gamma$. On appelle $P$ le projeté orthogonal de $O$ sur $(d)$. Soit $Q$ un point variable sur la droite $(d)$, et $\\left(t_{1}\\right)$ et $\\left(t_{2}\\right)$ les tangentes à $\\Gamma$ passant par $Q$. On note $A$ et $B$ les projetés orthogonaux de $P$ sur $\\left(t_{1}\\right)$ et $\\left(t_{2}\\right)$. Montrer que le point d'intersection de $(AB)$ et $(OP)$ reste fixe quand $Q$ varie.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $X$ l'intersection de $(AB)$ et $(OP)$, $S$ et $T$ les points où $\\left(t_{1}\\right)$ et $\\left(t_{2}\\right)$ touchent $\\Gamma$, et $Y$ le point d'intersection de $(ST)$ et $(OP)$. Une première remarque qu'on peut faire est que $O, P, Q, S$ et $T$ sont cocycliques sur le cercle de diamètre $[OQ]$. Une bonne figure suggère que $Y$ ne dépend pas de $Q$. Et en effet, on a :\n\n$$\n\\widehat{OSY} = \\widehat{OST} = \\widehat{OQT} = \\widehat{OQS} = \\widehat{OPS}\n$$\n\ndonc les triangles $OSY$ et $OPS$ sont indirectement semblables donc $\\frac{OY}{OS} = \\frac{OS}{OP}$, où $OY = \\frac{r^2}{OP}$, où $r$ est le rayon de $\\Gamma$, donc $OY$ ne dépend pas de $Q$ et $Y$ est fixe (les amateurs de géométrie projective auront reconnu le pôle de $(d)$ par rapport à $\\Gamma$).\n\nSoit maintenant $Z$ le projeté orthogonal de $P$ sur $(ST)$ : comme $P$ est sur le cercle circonscrit à $PST$, la droite de Simson nous garantit que $Z \\in (AB)$. Faisons maintenant un peu de chasse aux angles :\n\n$$\n\\widehat{PZX} = \\widehat{PZA} = \\widehat{PSA} = \\widehat{PSQ} = \\widehat{POQ} = \\widehat{OPZ}\n$$\n\noù on a utilisé successivement que $P, Z, A$ et $S$ sont cocycliques (sur le cercle de diamètre $[PS]$), que $O, P, Q, S$ sont cocycliques et enfin que $(OQ) \\parallel (PZ)$. On a donc que $XPZ$ est isocèle en $X$, donc $X$ est sur la médiatrice de $[PZ]$. Or, le centre du cercle circonscrit à $PYZ$ est sur cette médiatrice, et est sur $(PY)$ car $PYZ$ est rectangle en $Z$.\n\n$X$ est donc le centre du cercle circonscrit à $PYZ$, donc est le milieu de $[PY]$ car $PYZ$ est rectangle en $Z$. Comme $P$ et $Y$ ne dépendent pas de $Q$, $X$ non plus.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30456,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle acutangle tel que $AC > AB$, $D$ sur $BC$ tel que $AD = AB$ et $D$ différent de $B$. Soit $\\Gamma$ le cercle circonscrit à $ABC$, $\\Delta$ la tangente à $\\Gamma$ en $C$, $E$ l'intersection de $(AD)$ et $\\Delta$.\n\nMontrer que $CD^{2} = AD \\cdot DE - BD \\cdot DC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn a $\\widehat{EDC} = \\widehat{ADB} = \\widehat{CBA}$ et $\\widehat{DCE} = \\widehat{BAC}$ donc $ABC$ et $CDE$ sont semblables. On en déduit que $\\frac{AB}{CD} = \\frac{BC}{DE}$, donc\n$$\nAD \\cdot DE - BD \\cdot DC = AB \\cdot DE - BD \\cdot DC = BC \\cdot CD - BD \\cdot CD = (BC - BD) CD = CD^{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 3$ un nombre entier et $a_{1}, a_{2}, \\ldots, a_{n}$ des nombres réels.\n\na. On suppose que $a_{i}<\\max \\left(a_{i-1}, a_{i+1}\\right)$ pour tout $i \\in\\{2,3, \\ldots, n-1\\}$. Montrer que $a_{i}<\\max \\left(a_{1}, a_{n}\\right)$ pour tout $i \\in\\{2,3, \\ldots, n-1\\}$.\n\nb. On suppose que $a_{i} \\leqslant \\max \\left(a_{i-1}, a_{i+1}\\right)$ pour tout $i \\in\\{2,3, \\ldots, n-1\\}$. Est-il vrai que $a_{i} \\leqslant \\max \\left(a_{1}, a_{n}\\right)$ pour tout $i \\in\\{2,3, \\ldots, n-1\\}$ ?\n\nN.B. Si $x$ et $y$ sont deux nombres réels, on note $\\max (x, y)$ le plus grand des deux.",
    "options": [],
    "answer": "Part a: The inequality holds; all interior terms are less than the larger of the two endpoints (the maximum occurs at an endpoint). Part b: No in general; it is true for three terms but false for four or more terms (for example, set both endpoints to one and all interior terms to two).",
    "solution": "Solution:\n\na. Soit $i \\in\\{1,2, \\ldots, n\\}$ un entier tel que $a_{i}=\\max \\left(a_{1}, \\ldots, a_{n}\\right)$. L'énoncé indique que $i \\notin\\{2,3, \\ldots, n-1\\}$. Cela montre à la fois que $i \\in\\{1, n\\}$, donc que $a_{i}=\\max \\left(a_{1}, a_{n}\\right)$, et que $a_{k}<a_{i}$ si $k \\in\\{2,3, \\ldots, n-1\\}$, ce qui était l'inégalité demandée.\n\nb. Le résultat est clairement vrai pour $n=3$. Montrons qu'il n'est pas toujours vrai pour $n \\geqslant 4$ en exhibant un contre-exemple : prenons $a_{1}=a_{n}=1$ et $a_{i}=2$. La consigne de l'énoncé est respectée mais $a_{i}>\\max \\left(a_{1}, a_{n}\\right)$ pour tout $2 \\leqslant i \\leqslant n-1$.\nSolution:\n\nAutre solution pour (a). L'idée est de regarder où la suite \"remonte\" pour la première fois. Pour cela, nous allons distinguer deux cas :\n\nPremier cas : il existe un plus petit entier $1 \\leqslant i_{0} \\leqslant n-1$ tel que $a_{i_{0}} \\leqslant a_{i_{0}+1}$. Ainsi, $a_{1}>\\cdots>a_{i_{0}}$ et $a_{i_{0}} \\leqslant a_{i_{0}+1}$. En particulier, si $i_{0}=n-1$, on remarque qu'on a bien le résultat voulu (car alors $a_{n-1}<a_{1}$ et $a_{n-1} \\leqslant a_{n}$). Supposons maintenant que $i_{0}<n-1$ et montrons que $a_{j}<a_{j+1}$ pour tout entier $i_{0}+1 \\leqslant j \\leqslant n-1$ par récurrence sur $j$.\n\nInitialisation : on a $a_{i_{0}+1}<\\max \\left(a_{i_{0}}, a_{i_{0}+2}\\right)$. Comme $a_{i_{0}} \\leqslant a_{i_{0}+1}$, on doit avoir $\\max \\left(a_{i_{0}}, a_{i_{0}+2}\\right)=a_{i_{0}+2}$, ce qui implique que $a_{i_{0}+1}<a_{i_{0}+2}$.\n\nHérédité : soit $i_{0}+1 \\leqslant j<n-1$ un entier tel que $a_{j}<a_{j+1}$. Montrons que $a_{j+1}<a_{j+2}$. Comme $a_{j+1}<\\max \\left(a_{j}, a_{j+2}\\right)$ et $a_{j}<a_{j+1}$, on doit avoir $\\max \\left(a_{j}, a_{j+2}\\right)=a_{j+2}$, ce qui implique que $a_{j+1}<a_{j+2}$.\n\nOn a donc :\n$$\na_{1}>\\cdots>a_{i_{0}} \\leqslant a_{i_{0}+1}<a_{i_{0}+2}<a_{i_{0}+3}<\\cdots<a_{n}\n$$\nce qui implique que $a_{i}<\\max \\left(a_{1}, a_{n}\\right)$ pour tout $i \\in\\{2,3, \\ldots, n-1\\}$.\n\nDeuxième cas : il n'existe pas d'entier $1 \\leqslant i_{0} \\leqslant n-1$ tel que $a_{i_{0}} \\leqslant a_{i_{0}+1}$. Alors $a_{1}>a_{2}>\\cdots>a_{n}$ et on a bien le résultat voulu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A$ un ensemble de 13 entiers entre 1 et 37. Montrer qu'il existe quatre nombres deux à deux distincts dans $A$ tels que la somme de deux d'entre eux est égale à la somme des deux autres.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$A$ contient 13 entiers, donc il y a $\\frac{13 \\times 12}{2} = 78$ manières de choisir deux nombres de $A$ différents (13 manières de choisir le premier, 12 manières de choisir le deuxième et on divise par 2 car on a compté deux fois chaque ensemble de 2 nombres).\n\nOr, la somme de deux nombres de $A$ vaut toujours au moins $1+2=3$ et au plus $36+37=73$, donc elle peut prendre 71 valeurs différentes.\n\nD'après le principe des tiroirs, il existe donc $a \\neq d$ et $b \\neq c$ dans $A$ tels que $a + d = b + c$, et l'ensemble $\\{a, d\\}$ est différent de $\\{b, c\\}$.\n\nPour conclure, il suffit de s'assurer que $a, b, c$ et $d$ sont deux à deux distincts.\n\nMais si par exemple $a = b$, alors comme $a + d = b + c$, on doit avoir $d = c$ donc $\\{a, d\\} = \\{b, c\\}$, ce qui est faux. On a donc bien trouvé 4 nombres qui vérifient la propriété voulue.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAu club théâtre d'un lycée, on a formé 14 groupes de 4 élèves afin de travailler les scènes d'une pièce. Deux groupes différents ont toujours un et un seul élève en commun.\na) Prouver qu'il existe un élève qui appartient à au moins 5 groupes.\nb) Chaque élève du club théâtre est membre d'au moins un groupe. Combien y a-t-il d'élèves dans ce club?",
    "options": [],
    "answer": "43",
    "solution": "Solution:\n\na) Soit $G$ un des groupes. Chacun des 13 autres groupes a un élève en commun avec $G$. Or $13 = 3 \\times 4 + 1$ donc, d'après le principe des tiroirs, un des quatre élèves de $G$, disons $x$, appartient à au moins 4 des treize autres groupes. Avec $G$, cela signifie que $x$ appartient à au moins 5 groupes.\n\nb) Prouvons que $x$ appartient à tous les groupes. On note $G = G_1, G_2, G_3, G_4, G_5$ des groupes auxquels appartient $x$, puis on procède par l'absurde : supposons qu'il existe un groupe $G'$ tel que $x \\notin G'$.\n\nPuisque $G'$ a un élève en commun avec chacun des groupes $G_1, G_2, G_3, G_4, G_5$ et, comme $G'$ ne contient que quatre élèves, c'est que l'un des membres de $G'$, disons $y$, appartient à deux de ces cinq groupes, disons $G_i$ et $G_j$, avec $1 \\leqslant i < j \\leqslant 5$. Clairement, on a $x \\neq y$, car $x \\notin G'$ et $y \\in G'$. Mais alors les groupes $G_i$ et $G_j$ ont $x$ et $y$ en commun, en contradiction avec l'hypothèse de l'énoncé.\n\nAinsi, $x$ appartient à tous les groupes. En enlevant $x$ de chacun de ces 14 groupes, on obtient alors 14 groupes deux à deux disjoints de 3 élèves, soit $14 \\times 3 = 42$ élèves. Il faut encore y ajouter $x$ et, puisque chaque membre du club appartient à un des groupes, les membres sont alors bien tous comptés une et une seule fois.\n\nIl y a donc exactement 43 membres dans le club.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $n$ et $k$ des entiers strictement positifs. Il y a $n k$ objets (de même taille) et $k$ boîtes qui peuvent contenir chacune $n$ objets. Chaque objet est colorié en une couleur parmi $k$ couleurs possibles. Montrer qu'il est possible de ranger les objets dans les boîtes de sorte que chaque boîte contienne des objets d'au plus 2 couleurs différentes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn le montre par récurrence sur $k$. Si $k=1$ c'est évident. Supposons la propriété vraie pour tous les entiers $<k$.\n\nSi l'une des couleurs $c$ est prise exactement $n$ fois, on range tous les objets dont la couleur est $c$ dans la dernière boîte. L'hypothèse de récurrence permet de ranger les $n(k-1)$ objets restants dans les $k-1$ premières boîtes de sorte que chaque boîte contienne des objets d'au plus 2 couleurs différentes.\n\nSupposons donc qu'aucune couleur ne soit prise exactement $n$ fois. Si toutes les couleurs sont prises $\\leqslant n-1$ fois, alors il y a au plus $(n-1)k$ objets au total, ce qui contredit l'hypothèse. Donc l'une des couleurs $c_1$ est prise $\\geqslant n+1$ fois. De même, l'une des couleurs $c_2$ est prise $\\leqslant n-1$ fois.\n\nOn place alors tous les objets de couleur $c_2$ dans la dernière boîte. On complète cette boîte avec des objets de couleur $c_1$. L'hypothèse de récurrence permet de ranger les $n(k-1)$ objets restants dans les $k-1$ premières boîtes de sorte que chaque boîte contienne des objets d'au plus 2 couleurs différentes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont tous les angles sont aigus, et $H$ son orthocentre. Les bissectrices de $\\widehat{ABH}$ et $\\widehat{ACH}$ se coupent en un point $I$. Montrer que $I$ est aligné avec les milieux de $[BC]$ et de $[AH]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $D, E, F$ les pieds des hauteurs, $O$ le centre du cercle circonscrit, $M$ le milieu de $[BC]$ et $N$ le milieu de $[AH]$. On a $2 \\overrightarrow{MN} = 2 \\overrightarrow{ON} - 2 \\overrightarrow{OM} = \\overrightarrow{OA} + \\overrightarrow{OH} - \\overrightarrow{OB} - \\overrightarrow{OC}$.\n\nOr, $\\overrightarrow{OH} = \\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC}$ donc $\\overrightarrow{MN} = \\overrightarrow{OA}$. Par conséquent, il suffit de montrer que $\\left(MI\\right)$ est parallèle à $\\left(OA\\right)$.\n\nComme $BCF$ est un triangle rectangle en $F$, le point $F$ est situé sur le cercle de centre $M$ passant par $B$ et $C$. Il en va de même pour le point $E$. La bissectrice de $\\widehat{ABH}$ passe par le milieu de l'arc $EF$, et de même pour la bissectrice de $\\widehat{ACH}$, donc $I$ est le milieu de l'arc $EF$. En notant $C'$ le milieu de $[AB]$, on en déduit les égalités d'angles de droites $(MI, MB) = 2(CI, CB) = (CE, CB) + (CF, CB) = (CA, CB) + (CF, AB) + (AB, CB)$.\n\nD'autre part, $(OA, MB) = (OA, OC') + (OC', AB) + (AB, MB) = (CA, CB) + (CF, AB) + (AB, CB)$ car $(OC')$ et $(CF)$ sont parallèles.\n\nPar conséquent, $(OA, MB) = (MI, MB)$, ce qui prouve que $(OA)$ et $(MI)$ sont parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les polynômes $P$ à coefficients dans $\\mathbb{Z}$ tels que, pour tous $p$ premier et $u$ et $v$ dans $\\mathbb{Z}$ tels que $p \\mid u v-1$ on ait : $p \\mid P(u) P(v)-1$.",
    "options": [],
    "answer": "P(X) = ± X^n for some nonnegative integer n",
    "solution": "Solution:\n\nSoit $f$ un tel polynôme, et $n \\geqslant 0$ son degré. Clairement $f$ n'est pas le polynôme nul.\n\nPosons $g(X) = X^{n} f\\left(\\frac{1}{X}\\right)$. Alors $g$ est un polynôme à coefficients entiers.\n\nSoit $x$ un entier non nul, et $p$ un nombre premier avec $p > \\max \\left(|x|, |f(x) g(x) - x^{n}|\\right)$. Alors $x$ est inversible modulo $p$, et il existe $y > 0$ tel que $x y = 1 \\bmod p$. On vérifie facilement que $x^{n} f(y) = g(x) \\bmod p$.\n\nD'autre part, on sait que $f(x) f(y) = 1 \\bmod p$, d'où $x^{n} = x^{n} f(x) f(y) = f(x) g(x) \\bmod p$. Ainsi $p$ divise $f(x) g(x) - x^{n}$ avec $p > |f(x) g(x) - x^{n}|$, d'où $f(x) g(x) = x^{n}$, pour tout $x \\neq 0$ (et donc aussi pour $x = 0$ comme égalité entre deux polynômes).\n\nPuisque $f$ est de degré $n$, on en déduit que $g$ est constant et que $f(X) = a X^{n}$, où $a$ est un entier. En reportant dans $f(x) g(x) = x^{n}$, il vient immédiatement $a = \\pm 1$, et donc $f(X) = \\pm X^{n}$.\n\nRéciproquement, tout polynôme de la forme $f(X) = \\pm X^{n}$ est bien une solution du problème puisque, pour tous entiers $u$ et $v$ et tout nombre premier $p$, si $u v = 1 \\bmod p$ alors $u^{n} v^{n} = 1 \\bmod p$.\n\nFinalement, les polynômes cherchés sont ceux de la forme $f(X) = \\pm X^{n}$, avec $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. $H$ son orthocentre et $P$, $Q$ et $R$ les pieds des hauteurs issues de $A$, $B$ et $C$.\nMontrer que $H$ est le centre du cercle inscrit à $PQR$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn sait que les points $A$, $B$, $P$ et $Q$ sont cocycliques sur le cercle de diamètre $[AB]$, donc :\n\n$$\n\\widehat{HPQ} = \\widehat{APQ} = \\widehat{ABQ} = 90^{\\circ} - \\widehat{BAQ} = 90^{\\circ} - \\widehat{BAC}\n$$\n\nDe même, $A$, $C$, $P$ et $R$ sont cocycliques sur le cercle de diamètre $[AC]$ donc :\n\n$$\n\\widehat{HPR} = \\widehat{APR} = \\widehat{ACR} = 90^{\\circ} - \\widehat{CAR} = 90^{\\circ} - \\widehat{BAC}\n$$\n\nOn a donc $\\widehat{HPQ} = \\widehat{HPR}$ donc $(PH)$ est la bissectrice de $\\widehat{QPR}$. On montre de même que $(QH)$ et $(RH)$ sont les bissectrices de $\\widehat{PQR}$ et $\\widehat{PRQ}$, donc $H$ est le centre du cercle inscrit à $PQR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un entier strictement positif $n$ est amusant si pour tout diviseur strictement positif $d$ de $n$, l'entier $d+2$ est premier. Déterminer tous les entiers amusants dont le nombre de diviseurs est maximum.",
    "options": [],
    "answer": "135",
    "solution": "Solution:\n\nSoit $n$ un entier amusant et $p$ un diviseur premier de $n$. Alors $p$ est impair, puisque $p+2$ est premier.\n\nSupposons que $p \\geqslant 5$. Alors $p+2$ est premier, et $p+2>3$, donc $p+2$ n'est pas divisible par $3$. On en déduit que $p$ n'est pas congru à $1$ modulo $3$. Il est clair que $p$ n'est pas congru à $0$ modulo $3$, donc $p \\equiv 2 \\pmod{3}$.\n\nSupposons que $p^{2} \\mid n$. Alors $p^{2} \\equiv 1 \\pmod{3}$, donc $p^{2}+2 \\equiv 0 \\pmod{3}$. De plus, $p^{2}+2$ est premier, donc $p^{2}+2=3$, ce qui est impossible.\n\nOn déduit de ce qui précède que si $p \\geqslant 5$ est un diviseur premier de $n$, alors $p^{2}$ ne divise pas $n$.\n\nDe même, si $p_{1}$ et $p_{2}$ sont des diviseurs premiers distincts $\\geqslant 5$ de $n$, alors $p_{1} p_{2}$ divise $n$, donc $p_{1} p_{2}+2$ est divisible par $3$, ce qui est impossible.\n\nPar conséquent, tout entier amusant $n$ est de la forme $n=3^{k}$ ou bien $3^{k} p$ où $p$ est premier.\n\nOn vérifie que $3+2, 3^{2}+2, 3^{3}+2$ et $3^{4}+2$ sont premiers mais $3^{5}+2$ ne l'est pas, donc $k \\leqslant 4$. Les entiers amusants de la forme $3^{k}$ ont donc au plus $5$ diviseurs.\n\nSi $n=3^{4} \\times 5=405$, alors $n+2$ n'est pas premier donc $n$ n'est pas amusant.\n\nSi $n=3^{k} p$ avec $p \\geqslant 7$ premier, et $k \\geqslant 3$, alors les nombres $p, 3p, 9p, 27p$ sont des diviseurs de $n$ congrus à $p, 3p, 4p, 2p$ modulo $5$. Aucun n'est congru à $0$ modulo $5$, et les congruences de ces quatre nombres modulo $5$ sont deux à deux distinctes, donc à l'ordre près valent $1,2,3,4$. On en déduit qu'il existe un diviseur $d \\geqslant 7$ de $n$ tel que $d \\equiv 3 \\pmod{5}$. On a donc $d+2>5$ et $5 \\mid d$, ce qui entraîne que $d$ n'est pas premier.\n\nOn déduit de ce qui précède que les entiers amusants de la forme $3^{k} p$ vérifient $k \\leqslant 2$ ou bien $(k \\leqslant 3$ et $p=5$), donc ont au plus $8$ diviseurs, avec égalité seulement lorsque $k=3$ et $p=5$. Réciproquement, on vérifie que $3^{3} \\times 5$ est amusant, donc l'unique entier amusant ayant un nombre maximal de diviseurs est $3^{3} \\times 5=135$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCDEF$ un hexagone ayant tous ses angles égaux à $120^{\\circ}$. Montrer que $AB + BC = DE + EF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn prolonge les côtés $[AB]$, $[CD]$ et $[EF]$, et on appelle $X$ l'intersection de $(AB)$ et $(CD)$, $Y$ celle de $(CD)$ et $(EF)$, et $Z$ celle de $(EF)$ et $(AB)$. Alors $\\widehat{XBC} = 180^{\\circ} - \\widehat{ABC} = 60^{\\circ}$ et de même pour $\\widehat{XCB}$, donc:\n\n$$\n\\widehat{ZXY} = \\widehat{BXC} = 180^{\\circ} - \\widehat{XBC} - \\widehat{XCB} = 180^{\\circ} - 60^{\\circ} - 60^{\\circ} = 60^{\\circ}\n$$\n\net de même pour les angles du triangle $XYZ$ : on a représenté les angles de $120^{\\circ}$ en vert et ceux de $60^{\\circ}$ en rouge sur la figure.\n\nLe triangle $XBC$ est donc équilatéral, donc $AB + BC = AB + BX = AX$ et de même $DE + EF = FY$. Or, $XYZ$ est équilatéral donc $ZX = ZY$ et $ZAF$ est équilatéral donc $ZA = ZF$, d'où $AX = ZX - ZA = ZY - ZF = FY$, d'où le résultat.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe dont les diagonales ne sont pas perpendiculaires et tel que les droites $(AB)$ et $(CD)$ ne sont pas parallèles. On note $O$ le point d'intersection de $[AC]$ et $[BD]$. Soit $H_1$ et $H_2$ les orthocentres respectifs des triangles $AOB$ et $COD$. On désigne par $M$ et $N$ les milieux respectifs de $[AB]$ et $[CD]$.\n\nProuver que les droites $(H_1 H_2)$ et $(MN)$ sont parallèles si et seulement si $AC = BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $A'$, $B'$ les pieds des hauteurs issues respectivement de $A$ et $B$ dans $AOB$. Soit $C'$, $D'$ les pieds des hauteurs issues respectivement de $C$ et $D$ dans $COD$.\n\nLes points $A'$ et $D'$ appartiennent donc au cercle $\\Gamma_1$ de diamètre $[AD]$, et les points $B'$ et $C'$ appartiennent au cercle $\\Gamma_2$ de diamètre $[BC]$. Les points $A'$ et $B'$ appartiennent donc au cercle $\\Gamma_3$ de diamètre $[AB]$. En utilisant la puissance de $H_1$ par rapport à $\\Gamma_3$, il vient $H_1A \\cdot H_1A' = H_1B \\cdot H_1B'$.\n\nOn en déduit que $H_1$ a la même puissance par rapport à $\\Gamma_1$ et $\\Gamma_2$, donc que $H_1$ appartient à l'axe radical de ces deux cercles, que l'on note $\\Delta$. De même, on a $H_2 \\in \\Delta$. Or, l'axe radical de deux cercles étant perpendiculaire à la droite qui relie leur centres, on a $(H_1 H_2) \\perp (PQ)$, où $P$ et $Q$ sont les milieux respectifs de $[AD]$ et $[BC]$.\n\nAinsi, $(H_1 H_2) // (MN)$ si et seulement si $(MN) \\perp (PQ)$. Or, d'après le théorème des milieux, il est facile de vérifier que $PMQN$ est un parallélogramme et que $MP = \\frac{1}{2} BD$ et $MQ = \\frac{1}{2} AC$. Par suite, $(MN) \\perp (PQ)$ si et seulement si $PMQN$ est un losange, ce qui équivaut à $MP = MQ$, c'est-à-dire $AC = BD$.\nSolution:\n\nLes points considérés étant définis uniquement par des intersections de droite et de relations d'orthogonalité (mais ne faisant intervenir ni égalités d'angles ni intersections de cercles ou de cercle et droite), on peut tenter une approche analytique, qui sera pénible mais humainement faisable.\n\nPlaçons-nous dans un repère orthonormé de centre $O$ et tel que les points $A, B, C, D, O$ aient pour coordonnées respectives\n$$\n\\begin{pmatrix}-a \\\\ 0\\end{pmatrix},\\begin{pmatrix}-1 \\\\ -b\\end{pmatrix},\\begin{pmatrix}c \\\\ 0\\end{pmatrix},\\begin{pmatrix}d \\\\ bd\\end{pmatrix},\\begin{pmatrix}0 \\\\ 0\\end{pmatrix}\n$$\nAlors les points $H_1, H_2, M, N$ ont pour coordonnées respectives\n$$\n\\begin{pmatrix}-1 \\\\ y_1\\end{pmatrix}, \\quad \\begin{pmatrix}d \\\\ y_2\\end{pmatrix}, \\quad \\begin{pmatrix}-\\frac{a+1}{2} \\\\ -\\frac{b}{2}\\end{pmatrix}, \\quad \\begin{pmatrix}\\frac{c+d}{2} \\\\ \\frac{bd}{2}\\end{pmatrix}\n$$\nav ec $\\overrightarrow{AH_1} \\cdot \\overrightarrow{OB} = -(a-1) - y_1 b = 0$ et $\\overrightarrow{CH_2} \\cdot \\overrightarrow{OD} = (d-c)d + y_2 bd = 0$, c'est-à-dire\n$$\ny_1 = \\frac{1-a}{b} \\text{ et } y_2 = \\frac{c-d}{b}\n$$\nLes droites $(H_1 H_2)$ et $(MN)$ sont parallèles si et seulement si elles ont même pente, c'est-à-dire si l'égalité\n$$\n\\frac{y_2 - y_1}{d+1} = \\frac{bd + b}{c + d + a + 1}\n$$\nest vérifiée. On calcule donc que\n$$\n\\begin{aligned}\n&\\Leftrightarrow b(d+1)^2 = (c+d+a+1)(y_2 - y_1) = \\frac{1}{b}(c+d+a+1)(c-d+a-1) \\\\\n&\\Leftrightarrow b^2(d+1)^2 = (c+a)^2 - (d+1)^2 \\\\\n&\\Leftrightarrow BD^2 = (d+1)^2(b^2 + 1) = (c+a)^2 = AC^2\n\\end{aligned}\n$$\nce qui permet de conclure.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPar un point $A$ d'un cercle de centre $O$, on mène une tangente à ce cercle, et on prend deux points $B$ et $C$ sur cette tangente, avec $C$ entre $A$ et $B$. De $B$ et $C$, on mène $(B D)$ et $(C E)$ tangentes au cercle. Démontrer que $\\widehat{B O C}=\\widehat{D A E}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $\\alpha=\\widehat{B D E}$ et $\\beta=\\widehat{C E A}$. D'après le théorème de l'angle inscrit, on a $\\alpha=\\widehat{D A E}$.\n\nComme $(C A)$ et $(C E)$ sont les deux tangentes menées à partir de $C$, on a $\\widehat{E A C}=\\widehat{C E A}=\\beta$.\n\n$\\widehat{B O C}=\\widehat{B O A}-\\widehat{C O A}$. Or, $(B O) \\perp (A D)$ et $(O A) \\perp (A C)$, donc $\\widehat{B O A}=\\widehat{D A C}=\\alpha+\\beta$, et de même $\\widehat{C O A}=\\beta$, donc $\\widehat{B O C}=\\alpha=\\widehat{D A E}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier $n \\in \\mathbb{N}^{*}$, on note $v_{3}(n)$ la valuation 3-adique de $n$, c'est-à-dire le plus grand entier $k$ tel que $n$ est divisible par $3^{k}$. On pose $u_{1}=2$ et $u_{n}=4 v_{3}(n)+2-\\frac{2}{u_{n-1}}$ pour tout $n \\geqslant 2$ (si tant est que $u_{n-1}$ soit défini et non nul).\n\nMontrer que, pour tout nombre rationnel strictement positif $q$, il existe un et un seul entier $n \\geqslant 1$ tel que $u_{n}=q$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, notons que $u_{1}=2, u_{2}=1, u_{3}=3, u_{4}=\\frac{3}{2}, u_{5}=\\frac{2}{3}$ et $u_{6}=3$. On prouve par récurrence que, pour tout entier $n \\geq 2$, on a $0<u_{n}$ et\n$$\n0<u_{3 n-1}<1<u_{3 n-2}<2<u_{3 n}=2+u_{n}:\n$$\nc'est déjà vrai pour $n=2$.\nPuisque $u_{3 n}=u_{n}+2$, on montre successivement que $u_{3 n+1}, u_{3 n+2}$ et $u_{3 n+3}$ sont bien définis, avec\n$$\n\\begin{aligned}\n& u_{3 n+1}=2-\\frac{2}{u_{3 n}}=1+\\frac{u_{n}}{2+u_{n}}, \\text{ donc } 1<u_{3 n+1}<2 ; \\\\\n& u_{3 n+2}=2-\\frac{2}{u_{3 n+1}}=1+\\frac{u_{n}}{1+u_{n}}, \\text{ donc } 0<u_{3 n+2}<1 ; \\\\\n& u_{3 n+3}=4 v_{3}(3(n+1))+2-\\frac{2}{u_{3 n+2}}=4 v_{3}(n)+4-\\frac{2}{u_{n}}=2+u_{n+1} .\n\\end{aligned}\n$$\nCeci conclut la récurrence.\n\nMaintenant, considérons la fonction $\\varphi:\\{x \\in \\mathbb{Q}: 0<x$ et $x \\notin\\{1,2\\}\\} \\mapsto\\{x \\in \\mathbb{Q}: 0<x\\}$ telle que\n$$\n\\begin{aligned}\n\\varphi: & x \\mapsto \\frac{x}{1-x} \\text{ si } 0<x<1 ; \\\\\n& x \\mapsto 2 \\frac{x-1}{2-x} \\text{ si } 1<x<2 ; \\\\\n& x \\mapsto x-2 \\text{ si } 2<x\n\\end{aligned}\n$$\nEn outre, pour toute fraction irréductible $\\frac{p}{q}$, avec $p \\geq 0$ et $q>0$, on pose $\\left\\|\\frac{p}{q}\\right\\|=p+q$. Alors :\n- $\\left\\|\\varphi\\left(\\frac{p}{q}\\right)\\right\\|=\\left\\|\\frac{p}{q-p}\\right\\|=p<p+q=\\left\\|\\frac{p}{q}\\right\\|$ si $0<p<q$ ；\n- $\\left\\|\\varphi\\left(1+\\frac{p}{q}\\right)\\right\\|=\\left\\|2 \\frac{p}{q-p}\\right\\| \\leq 2 q<(p+q)+q=\\left\\|1+\\frac{p}{q}\\right\\|$ si $0<p<q$ ；\n- $\\left\\|\\varphi\\left(2+\\frac{p}{q}\\right)\\right\\|=\\left\\|\\frac{p}{q}\\right\\|=p+q<(p+2 q)+q=\\left\\|2+\\frac{p}{q}\\right\\|$ si $0<p$ et $0<q$.\n\nDans tous les cas, si $x$ est un rationnel strictement positif tel que $x \\notin\\{1,2\\}$, on a $\\|\\varphi(x)\\|<\\|x\\|$. Or, pour tout rationnel strictement positif $x$ et pour tout entier $n \\geq 1$ :\n- si $0<x<1$, alors $u_{n}=\\varphi(x) \\Leftrightarrow u_{3 n+2}=x ;$\n- si $1<x<2$, alors $u_{n}=\\varphi(x) \\Leftrightarrow u_{3 n+1}=x$;\n- si $2<x$, alors $u_{n}=\\varphi(x) \\Leftrightarrow u_{3 n}=x$.\n\nUne récurrence sur $\\|x\\|$ montre donc immédiatement que, pour tout rationnel strictement positif $x$, il existe un entier unique entier $n \\geq 1$ tel que $u_{n}=x$, ce qui conclut l'exercice.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $n$ un entier strictement positif tel que $n(n+2015)$ est le carré d'un entier.\na) Prouver que $n$ n'est pas un nombre premier.\nb) Donner un exemple d'un tel entier $n$.",
    "options": [],
    "answer": "n is not prime; an example is n = 1612.",
    "solution": "Solution:\na) Supposons que $n$ est premier et qu'il existe un entier $m$ vérifiant $n(n+2015)=m^{2}$. Alors $n$ divise $m^{2}$, donc $n$ divise $m$. On peut donc écrire $m=n r$. Il vient $n(n+2015)=n^{2} r^{2}$, puis $n+2015=n r^{2}$. Par conséquent, $2015=n r^{2}-n=n\\left(r^{2}-1\\right)$ est divisible par $n$.\nOr, $2015=5 \\times 13 \\times 31$, donc $n$ est l'un des entiers $5$, $13$, $31$.\nSi $n=5$ alors $r^{2}-1=13 \\times 31=403$, ce qui est impossible car $404$ n'est pas un carré parfait. De même, $5 \\times 31+1=156$ et $5 \\times 13+1=66$ ne sont pas des carrés parfaits, ce qui exclut les cas $n=13$ et $n=31$.\n\nb) On cherche $n$ et $m$ entiers tels que\n$$\n(2 m)^{2}=4 n(n+2015)=(2 n)^{2}+2 \\times(2 n) \\times 2015=(2 n+2015)^{2}-2015^{2}\n$$\nCeci équivaut à $2015^{2}=(2 n+2015)^{2}-(2 m)^{2}=(2 n+2015+2 m)(2 n+2015-2 m)$.\nSoient $a=2015 \\times 5$ et $b=\\frac{2015}{5}$. Il suffit donc que\n$$\n2 n+2015+2 m=a \\text{ et } 2 n+2015-2 m=b\n$$\nEn additionnant ces égalités, on trouve que $4 n+4030=a+b=403 \\times 25+403$, donc $\\underline{n=1612}$, et en soustrayant on obtient $4 m=a-b=403 \\times 24$ donc $m=403 \\times 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle non isocèle en $A$ et $\\Gamma$ son cercle circonscrit. Soit $N$ le milieu de l'arc de $\\Gamma$ entre $B$ et $C$ qui contient $A$. Soient $I$ le centre du cercle inscrit à $ABC$, $J$ le centre du cercle $A$-exinscrit à $ABC$ et $K$ le point d'intersection de $BC$ avec la bissectrice extérieure de $\\angle BAC$.\n\nMontrer que $(JN) \\perp (IK)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $J_b$ et $J_c$ les centres des cercles exinscrits dans les angles $\\widehat{B}$ et $\\widehat{C}$. Alors $I$ est l'orthocentre de $J_b J_c$. En utilisant le fait que $B$ et $C$ sont sur le cercle de diamètre $J_b J_c$, et que $A, B, C, N$ sont cocycliques, on a $\\overline{KJ_b} \\cdot \\overline{KJ_c} = \\overline{KB} \\cdot \\overline{KC} = \\overline{KA} \\cdot \\overline{KN}$, donc\n$$\n\\begin{aligned}\n\\left(\\overline{KA} + \\overline{A J_b}\\right)\\left(\\overline{KA} + \\overline{A J_c}\\right) & = \\overline{KA} \\cdot (\\overline{KA} + \\overline{AN}) \\\\\n\\overline{KA}\\left(\\overline{A J_b} + \\overline{A J_c} + \\overline{NA}\\right) & = -\\overline{A J_b} \\cdot \\overline{A J_c}\n\\end{aligned}\n$$\nOr, $\\overline{A J_b} + \\overline{A J_c} + \\overline{NA} = \\overline{N J_b} + \\overline{A J_c} = \\overline{J_c N} + \\overline{A J_c} = \\overline{A N}$, donc $\\overline{KA} = -\\frac{\\overline{A J_b} \\cdot \\overline{A J_c}}{\\overline{A N}}$.\n\nOr, $I$ étant l'orthocentre de $J_b J_c$, on a $\\frac{IA}{A J_b} = \\frac{A J_c}{A J}$. Il vient $\\frac{IA}{KA} = \\frac{A N}{A J}$, donc les triangles rectangles $KAI$ et $JAN$ sont semblables. Comme $(KA) \\perp (JA)$, il vient $(KI) \\perp (JN)$.\n\n\nAutre solution. Il suffit de montrer que $I$ est l'orthocentre de $KJN$. Comme $(AJ) \\perp (KN)$, il suffit de voir que $(NI) \\perp (KJ)$.\n\nSoit $\\omega$ le cercle de diamètre $[IJ]$. Il contient les points $B$ et $C$. Soit $L$ le second point d'intersection de $(KJ)$ avec $\\omega$. On a $\\overline{KL} \\cdot \\overline{KJ} = \\overline{KB} \\cdot \\overline{KC} = \\overline{KA} \\cdot \\overline{KN}$ donc $A, N, L, J$ sont cocycliques.\n\nOr, $(JA) \\perp (AN)$, donc $(JL) \\perp (LN)$.\n\nDe plus, comme $\\omega$ a pour diamètre $[IJ]$, on a $(JL) \\perp (IL)$, donc $N, I, L$ sont alignés. Il vient $(NI) \\perp (JL)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $0 \\leqslant a, b, c, d, e \\leqslant 1$ des nombres réels. Montrer que\n$$\n(1+a+b+c+d+e)^2 \\geqslant 4\\left(a^2+b^2+c^2+d^2+e^2\\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nL'idée est d'utiliser l'inégalité $(x+y)^2 \\geqslant 4 x y$ pour $x=1$ et une valeur particulière de $y$.\n\nPlus précisément, pour tous $x, y \\geqslant 0$, on a $(x-y)^2 \\geqslant 0$, et donc $(x+y)^2 \\geqslant 4 x y$ (cette inégalité est aussi l'inégalité arithmético-géométrique à deux variables). En évaluant cette inégalité avec $x=1$ et $y=a+b+c+d+e$, on obtient:\n$$\n(1+a+b+c+d+e)^2 \\geqslant 4(a+b+c+d+e)\n$$\nOr pour tout nombre réel $x$ tel que $0 \\leqslant x \\leqslant 1$, on a $x \\geqslant x^2$. Donc\n$$\n4(a+b+c+d+e) \\geqslant 4\\left(a^2+b^2+c^2+d^2+e^2\\right)\n$$\nce qui montre finalement que $(1+a+b+c+d+e)^2 \\geqslant 4\\left(a^2+b^2+c^2+d^2+e^2\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes sept nains ont des tailles deux à deux distinctes. Ils se rendent à la mine en colonne dans un certain ordre, de telle manière que le nain en tête est plus grand que le deuxième, qui est plus petit que le troisième, qui est plus grand que le quatrième et ainsi de suite...\n\nCombien y a-t-il de telles manières d'arranger les nains?",
    "options": [],
    "answer": "272",
    "solution": "Solution:\n\nOn note $u_{3}$ le nombre de manières d'arranger 3 nains de manière à ce que le premier est plus grand que le second, qui est plus grand que le troisième, $u_{5}$ le nombre de manières similaires d'arranger 5 nains et $u_{7}$ la solution de l'exercice.\n\nAvec 3 nains, on est obligé de mettre le nain le plus petit au milieu et on peut ensuite placer comme on veut les deux autres.\n\nAvec 5 nains, le nain le plus petit peut être en deuxième ou quatrième position :\n\n- S'il est en position 2, n'importe quel nain peut être en premier, donc on a 4 choix pour le premier nain. Il reste à ordonner les trois derniers : il y a $u_{3}$ manières de le faire, donc on obtient $4 u_{3}$ arrangements.\n\n- S'il est en position 4, on a de même $4 u_{3}$ arrangements (en choisissant d'abord le cinquième).\n\nOn a donc $u_{5} = 4 u_{3} + 4 u_{3} = 8 \\times 2 = 16$.\n\nAvec 7 nains, le nain le plus petit peut être en position 2, 4 ou 6 :\n\n- S'il est en position 2, n'importe quel nain peut être en premier (soit 6 choix), puis il y a $u_{5}$ manières d'ordonner les cinq derniers, soit $6 u_{5}$ ordres possibles.\n\n- S'il est en position 6, n'importe quel nain peut être en dernier (soit 6 choix), puis il y a $u_{5}$ manières d'ordonner les cinq premiers, soit $6 u_{5}$ ordres possibles.\n\n- S'il est en position 4, on peut d'abord trier les nains qui seront devant lui de ceux qui seront derrière. Il y a $\\binom{6}{3} = 20$ manières de faire ce tri. Puis il y a $u_{3}$ manières d'ordonner les trois nains de devant et $u_{3}$ manières d'ordonner les trois de derrière, soit $20 u_{3}^{2}$ ordres possibles.\n\nOn a donc $u_{7} = 6 u_{5} + 6 u_{5} + 20 u_{3}^{2} = 6 \\times 16 + 6 \\times 16 + 20 \\times 2^{2} = 272$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2015 droites deux à deux distinctes sont tracées dans le plan. On suppose qu'elles délimitent moins de 8000 régions (les régions peuvent être non bornées). Montrer que le nombre de régions est égal à 2016, 4030, 6042 ou 6043, et donner un exemple de configuration dans chaque cas.",
    "options": [],
    "answer": "2016, 4030, 6042, 6043",
    "solution": "Solution:\n\nNotons $n = 2015$ le nombre de droites, $f$ le nombre de régions, $p$ le nombre maximal de droites parallèles et $q$ le nombre maximal de droites concourantes.\n\nMontrons d'abord que $f \\geqslant (p+1)(n-p+1)$. En effet, plaçons d'abord les $p$ droites parallèles. Elles forment $p+1$ régions. Il reste ensuite $n-p$ droites à placer, et chacune de ces droites crée au moins $p+1$ régions supplémentaires.\n\nDe même, montrons que $f \\geqslant q(n-q+2)$. En effet, on place d'abord $q$ droites concourantes. Chaque nouvelle droite crée au moins $q-1$ points d'intersection, donc au moins $q$ nouvelles régions. On en déduit que $f \\geqslant (2q) + (n-q)q = q(n-q+2)$.\n\nSupposons que $3 \\leqslant p \\leqslant n-3$. Comme l'expression $g(p) = (p+1)(n-p+1)$ est égale à $(n+2)^2/4 - (p-n/2)^2$, la fonction $g$ croît entre $0$ et $n/2$ et décroît entre $n/2$ et $n$, donc $g(p) \\geqslant g(3) = 4(n-2) > 8000$. Impossible. On en déduit que $p \\in \\{1,2, n-2, n-1, n\\}$ et de même $q \\in \\{1,2,3, n-1, n\\}$.\n\nSi $p = n$ alors toutes les droites sont parallèles, et on a $f = n+1 = 2016$.\n\nSi $p = n-1$ alors toutes les droites sauf une sont parallèles, et on a $f = 2n = 4030$.\n\nSi $p = n-2$ et les deux droites restantes sont parallèles, ou bien sécantes sur l'une des $n-2$ premières droites, alors $f = 3(n-1) = 6042$.\n\nSi $p = n-2$ et les deux droites restantes sont sécantes ailleurs que sur les $n-2$ premières droites, alors $f = 3n-2 = 6043$.\n\nSi $q = n$ alors $f = 2n$.\n\nSi $q = n-1$ alors $f = 3n-3$ ou $3n-2$ selon que la dernière droite est parallèle ou non à l'une des premières droites.\n\nIl reste à traiter le cas où $p \\leqslant 2$ et $q \\leqslant 3$. Notons $D_1, \\ldots, D_{2015}$ les droites. À chaque fois que l'on trace une droite $D_i$ ($i > 1000$), celle-ci coupe au moins 999 des droites $D_1, \\ldots, D_{1000}$, et chacun de ces points d'intersection est commun à au plus deux de ces droites, donc on obtient au moins 500 points d'intersection entre $D_i$ et l'une des 1000 premières droites. Par conséquent, $D_i$ crée au moins 501 régions supplémentaires. On en déduit que $f \\geqslant 1015 \\times 501 > 8000$.\n\nExemples de configurations :\n\n- $f = 2016$ : toutes les droites sont parallèles.\n- $f = 4030$ : toutes les droites sauf une sont parallèles.\n- $f = 6042$ : $n-2$ droites parallèles, et les deux restantes sont parallèles entre elles ou sécantes sur l'une des $n-2$ premières.\n- $f = 6043$ : $n-2$ droites parallèles, et les deux restantes sont sécantes ailleurs que sur les $n-2$ premières droites.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn veut colorier les parties à trois éléments de $\\{1,2,3,4,5,6,7\\}$, de sorte que si deux de ces parties n'ont pas d'élément en commun alors elles soient de couleurs différentes. Quel est le nombre minimum de couleurs pour réaliser cet objectif?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nConsidérons la suite de parties $\\{1,2,3\\},\\{4,5,6\\},\\{1,2,7\\},\\{3,4,6\\},\\{1,5,7\\},\\{2,3,6\\},\\{4,5,7\\}$, $\\{1,2,3\\}$.\nChaque partie doit avoir une couleur différente de la suivante, donc déjà il y a au moins deux couleurs. S'il n'y avait qu'exactement deux couleurs, alors les couleurs devraient alterner, ce qui est impossible car la dernière partie est la même que la première et devrait être de couleur opposée.\n\nRéciproquement, montrons que trois couleurs suffisent:\n- On colorie en bleu les parties qui contiennent au moins deux éléments parmi 1, 2, 3.\n- On colorie en vert les parties non coloriées en bleu et qui contiennent au moins deux éléments parmi 4, 5, 6.\n- On colorie en rouge les parties non coloriées en bleu ou en vert.\n\nIl est évident que deux parties bleues ont un élément en commun parmi 1, 2, 3 ; de même, deux parties vertes ont un élément en commun parmi $4,5,6$. Enfin, toute partie rouge contient trois éléments, dont contient nécessairement l'élément 7 : deux parties rouges ont donc également un élément en commun.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A B C D$ un quadrilatère convexe. On note $P, Q, R$ et $S$ les milieux des côtés $[A B],[B C],[C D]$ et $[D A]$. Montrer que l'aire de $PQRS$ est égale à la moitié de l'aire de $A B C D$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $X$ l'intersection des diagonales de $A B C D$. $[P Q]$ coupe $[B D]$ en $Y$, et $[Q R]$ coupe $[A C]$ en $Z$. D'après le théorème de la droite des milieux, $(P Q)$ est parallèle à $(A C)$, donc $(Q Z)$ est parallèle à $(C X)$ et, de nouveau d'après le théorème de la droite des milieux, $Z$ est le milieu de $[B X]$, donc l'aire du triangle $X Q Z$ est la moitié de l'aire du triangle $X Q B$. De même, l'aire du triangle $X Q Y$ est la moitié de celle du triangle $X Q C$.\n\nLe quadrilatère $PQRS$ couvre donc la moitié du triangle $X B C$. De la même manière, il couvre la moitié des triangles $X C D$, $X D A$ et $X A B$, donc son aire est la moitié de celle de $A B C D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver toutes les fonctions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ telles que $f(f(x)+3y)=12x+f(f(y)-x)$ pour tous $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All functions of the form f(x) = a + 3x for any real constant a.",
    "solution": "Solution:\n\nSoit $f$ une fonction satisfaisant à la condition de l'énoncé. L'idée est de montrer que $f$ est injective en utilisant le fait qu'elle est surjective.\n\nMontrons d'abord que $f$ est surjective. En prenant $y=-f(x)/3$ dans l'équation fonctionnelle, on obtient $f(f(y)-x)=f(0)-12x$. En faisant varier $x$, on voit que le membre de droite de cette dernière égalité peut prendre pour valeur n'importe quel nombre réel, et donc le membre de gauche aussi, ce qui implique que $f$ est surjective.\n\nMontrons que $f$ est injective. Soient $a, b \\in \\mathbb{R}$ tels que $f(a)=f(b)$. Nous allons montrer que $a=b$. En prenant $y=a$ et $y=b$ dans l'équation fonctionnelle, on obtient pour tout $x \\in \\mathbb{R}$\n$$\nf(f(x)+3a)=12x+f(f(a)-x)=12x+f(f(b)-x)=f(f(x)+3b)\n$$\noù on a utilisé le fait que $f(a)=f(b)$ pour la deuxième égalité. Ceci implique que pour tout $y \\in \\mathbb{R}$,\n$$\nf(y)=f(y+3(b-a))\n$$\nEn effet, si on fixe $y \\in \\mathbb{R}$, par surjectivité il existe $x$ tel que $f(x)=y-3a$ et en injectant dans (1), ceci donne (2). En particulier, $f(0)=f(3(b-a))$. Alors\n$$\nf(f(0))=f(f(3(b-a))+3 \\cdot 0)=12 \\cdot 3(b-a)+f(f(0)-3(b-a))=12 \\cdot 3(b-a)+f(f(0))\n$$\noù pour la première égalité on utilise le fait que $f(0)=f(3(b-a))$, pour la deuxième on utilise l'équation fonctionnelle de l'énoncé (avec $x=3(b-a)$ et $y=0$), pour la troisième égalité on utilise (2) (avec $y=f(0)-3(b-a)$). Donc $3(b-a)=0$ et $a=b$.\n\nPour conclure, en prenant $x=0$ dans l'équation de départ et en utilisant l'injectivité de $f$, on obtient\n$$\nf(y)=f(0)+3y\n$$\npour tout réel $y$.\n\nRéciproquement, on vérifie que les fonctions de la forme $f(x)=a+3x$ pour tout $x \\in \\mathbb{R}$ avec $a \\in \\mathbb{R}$ vérifient l'équation fonctionnelle de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. Soit $P$ appartenant au cercle circonscrit. On sait que les projetés de $P$ sur $(BC)$, $(CA)$ et $(AB)$ sont alignés sur la droite dite de Simson. On suppose que cette droite passe par le point diamétralement opposé à $P$. Montrer qu'elle passe également par le centre de gravité de $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $P'$ le point diamétralement opposé à $P$. Soit $\\Delta$ la droite de Simson. Soit $h$ l'homothétie de centre $P$ et de rapport $2$. Alors $h(\\Delta)$ est la droite de Steiner. On sait qu'elle passe par l'orthocentre $H$, donc $\\Delta$ passe par le milieu de $[PH]$. On en déduit que $\\Delta$ est la médiane de $PP'H$ issue de $P'$.\n\nPar ailleurs, $(HO)$ est la médiane de $PP'H$ issue de $H$. Comme $\\overrightarrow{HG}=\\frac{2}{3} \\overrightarrow{HO}$, le point $G$ est le centre de gravité de $PHP'$, donc $\\Delta$ passe par $G$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30478,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$. On a un nombre fini de bouteilles, chacune contenant une quantité d'eau inférieure à 1 litre, telles que la quantité totale d'eau est de $\\frac{n}{2}$ litres. On dispose également de $n$ seaux vides.\n\nMontrer qu'il est possible de vider chaque bouteille dans un seau (on peut vider plusieurs bouteilles dans le même seau, mais pas vider une bouteille en partie dans un seau et en partie dans un autre) de manière à avoir au plus 1 litre d'eau dans chaque seau.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn raisonne par récurrence sur $n$ :\n\nPour $n = 1$, il suffit de tout vider dans l'unique seau.\n\nSupposons le résultat vrai au rang $n$ : si on arrive à vider dans un seau un certain nombre de bouteilles représentant au total au moins $\\frac{1}{2}$ litre et au plus 1 litre, alors il restera $\\frac{n-1}{2}$ litres à vider dans les $n-1$ seaux restants, et on pourra appliquer l'hypothèse de récurrence pour conclure.\n\nS'il existe une bouteille contenant au moins $\\frac{1}{2}$ litre, on peut donc la vider dans le premier seau. Sinon, toutes les bouteilles contiennent moins de $\\frac{1}{2}$ litre. On les vide donc une par une dans le premier seau, et on s'arrête dès que le contenu du seau dépasse $\\frac{1}{2}$ litre : le seau contient alors plus de $\\frac{1}{2}$ litre. De plus, juste avant de vider la dernière bouteille, il contenait moins de $\\frac{1}{2}$ litre, et on a ajouté une bouteille, donc moins de $\\frac{1}{2}$ litre. Le seau contient donc moins de 1 litre, et on peut conclure par hypothèse de récurrence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les polynômes $P$ et $Q$ à coefficients entiers tels que, si l'on définit la suite $\\left(x_{n}\\right)$ par $x_{0}=2015$, $x_{2 n+1}=P\\left(x_{2 n}\\right)$ et $x_{2 n+2}=Q\\left(x_{2 n+1}\\right)$ pour tout $n \\geqslant 0$, alors tout entier $m>0$ divise au moins un terme non nul de la suite.",
    "options": [],
    "answer": "(P(X), Q(X)) = (ε X + b, ε X + d) with ε ∈ {+1, −1} and integers b, d, such that either b + ε d divides 2015 or d + ε b divides ε·2015 + b.",
    "solution": "Solution:\n\nSoit $P$ et $Q$ deux tels polynômes.\nOn dira qu'une suite $\\left(y_{n}\\right)$ d'entiers possède la propriété D si tout entier $m>0$ divise au moins un terme non nul de cette suite. On dira qu'un polynôme $T$ à coefficients entiers possède la propriété D s'il existe un entier $a$ tel que la suite $\\left(x_{n}\\right)$ définie par $x_{0}=a$ et $x_{n+1}=T\\left(x_{n}\\right)$ possède la propriété D.\n\nNous débutons maintenant par trois lemmes.\n\nLemme 1. Soit $(x_{n})$ une suite d'entiers et soit $k$ un entier naturel non nul. Alors $(x_{n})$ possède la propriété D si et seulement si l'une des $k$ suites $\\left(x_{k n}\\right),\\left(x_{k n+1}\\right), \\ldots,\\left(x_{k n+k-1}\\right)$ possède la propriété D.\n\nPreuve du lemme 1. Supposons d'abord qu'aucune des $k$ suites $\\left(x_{k n}\\right),\\left(x_{k n+1}\\right), \\ldots,\\left(x_{k n+k-1}\\right)$ ne possède la propriété $D$. Alors, pour chaque entier $\\ell \\in\\{0,1, \\ldots, k-1\\}$, il existe un entier $M_{\\ell}$ qui ne divise aucun terme non nul de la suite $(x_{k n+\\ell})$. En particulier, l'entier $\\prod_{\\ell=0}^{k-1} M_{\\ell}$ ne peut donc diviser aucun terme non nul de la suite $\\left(x_{n}\\right)$, qui ne peut posséder la propriété D non plus.\n\nRéciproquement, si une suite $\\left(x_{k n+\\ell}\\right)$ a la propriété $D$, alors clairement $\\left(x_{n}\\right)$ a la propriété $D$ aussi.\n\nLemme 2. Soit $T$ un polynôme à coefficients entiers, de degré $d$ et de coefficient dominant $\\rho$. Si $T$ possède la propriété D, alors $d=1$ et $|\\rho|<4$.\n\nPreuve du lemme 2. Si $T$ est constant, la suite $(x_{n})$ ne prend qu'un nombre fini de valeurs distinctes, ce qui lui interdit d'avoir la propriété D. Procédons maintenant par l'absurde, et supposons que $d \\geqslant 2$ ou que $d=1$ et $|\\rho| \\geqslant 4$. Il existe alors un réel $c>0$ tel que\n$$\n|T(x)|>3|x| \\text{ pour tout } x \\text{ tel que }|x|>c. \\tag{1}\n$$\nOn note qu'il n'existe qu'un nombre fini d'entiers dans $[-c; c]$ et que chacun d'eux ne peut être atteint qu'un nombre fini de fois par $T$. Ainsi, il existe un entier $N \\geqslant 0$ tel que $|x_{N}|>c$. Une récurrence immédiate à l'aide de (1) montre qu'alors $|x_{i}|>c$ pour tout $i \\geqslant N$.\n\nDe plus, sans perte de généralité, on peut choisir $N$ minimal ayant cette propriété. On a donc\n$$\n|x_{i}| \\leqslant c \\text{ pour tout } i<N, \\text{ et }|x_{i}|>c \\text{ pour tout } i \\geqslant N. \\tag{2}\n$$\nEn particulier, de (1) et (2), on déduit que $|x_{N+1}|>\\max \\left(c,|x_{0}|, \\cdots,|x_{N}|\\right)$.\n\nPosons $m=|x_{N+1}-x_{N}|$, qui est bien un entier strictement positif. Alors:\n$$\nm \\geqslant|x_{N+1}|-|x_{N}|>2|x_{N}|>\\max (|x_{0}|, \\cdots,|x_{N}|),\n$$\nce qui assure $m$ ne divise aucun $x_{i}$ non nul lorsque $i \\leqslant N$. Mais alors $m$ ne divise pas $x_{N+1}$ non plus.\n\nD'autre part, puisque $T$ est à coefficients entiers, $x_{n+1}-x_{n}=T\\left(x_{n}\\right)-T\\left(x_{n-1}\\right)$ est divisible par $x_{n}-x_{n-1}$ pour tout $n \\geqslant 1$. En particulier, pour tout $n \\geqslant N$, on a $x_{n+1}-x_{n}$ divisible par $m$, et donc\n$$\nx_{n+1}-x_{N+1}=(x_{n+1}-x_{n})+(x_{n}-x_{n-1})+\\cdots+(x_{N+2}-x_{N+1})\n$$\nest lui aussi divisible par $m$. Mais, $m$ ne divisant pas $x_{N+1}$, il ne divise donc pas non plus $x_{n}$ pour tout $n \\geqslant N+1$. Finalement, $m$ ne divise aucun terme non nul de la suite, en contradiction avec l'hypothèse que $\\left(x_{n}\\right)$ possède la propriété $D$.\n\nLemme 3. Soit $T(X)=\\rho X+\\theta$ un polynôme à coefficients entiers. Si $T$ possède la propriété $D$, alors $\\rho=1$.\n\nPreuve du lemme 3. Si $T$ possède la propriété D, soit $x_{0}$ un entier tel que la suite $(x_{n})$ définie par $x_{n+1}=T\\left(x_{n}\\right)$ ait la propriété $D$. D'après le lemme 1, l'une des suites $\\left(x_{2 n}\\right)$ et $\\left(x_{2 n+1}\\right)$ a la propriété $D$, et satisfait bien $x_{n+2}=T\\left(T\\left(x_{n}\\right)\\right)$. Le polynôme $T(T(X))$ a donc également la propriété D.\n\nOr, $T(T(X))$ est de coefficient dominant $\\rho^{2}$. Le lemme 2 montre donc que $\\rho^{2}<4$, donc que $\\rho \\in\\{-1,0,1\\}$. Puisque $\\rho$ est un coefficient dominant de $T$, on a nécessairement $\\rho \\neq 0$. De plus, si $\\rho=-1$, alors $T(T(X))=X$ n'a pas la propriété $D$. Il s'ensuit que $\\rho=1$.\n\nRevenons maintenant à l'exercice.\n\nOn pose $H(X)=P(Q(X))$ et $K(X)=Q(P(X))$.\nD'après le lemme 1, l'une des deux suites $(x_{2k})$ et $\\left(x_{2k+1}\\right)$ possède la propriété D, donc l'un des polynômes $H$ et $K$ possède la propriété D, car $x_{2k+2}=H\\left(x_{2k}\\right)$ et $x_{2k+3}=K\\left(x_{2k+1}\\right)$.\n\nPuisque $\\deg H=(\\deg P) \\cdot(\\deg Q)=\\deg K$, le lemme 2 indique que $\\deg H=\\deg P=\\deg Q=\\deg K=1$. Notons alors $P(X)=a X+b$ et $Q(X)=c X+d$. On a alors $H(X)=a c X+a d+b$ et $K(X)=a c X+b c+d$, donc le lemme 3 indique que $a=c= \\pm 1$.\n\nFinalement, notons que $x_{2 n}=2015+(a d+b) n$ et que $x_{2 n+1}=(a 2015+d)+(b c+d) n$. Or, une suite arithmétique $\\left(y_{n}\\right)$ de raison $r$ a la propriété $D$ si et seulement si $r$ divise $y_{0}$ :\n- si $y_{0}=q r$, et si $m \\in \\mathbb{N}^{*}$, alors $q^{2} m-q \\geqslant 0$ et $m$ divise $y_{q^{2} m-q}=q^{2} m r$;\n- si $r$ ne divise pas $y_{0}$, alors $r$ ne divise aucun entier $y_{n}$.\n\nOr, le lemme 1 indique que $\\left(x_{n}\\right)$ a la propriété $D$ si et seulement si $\\left(x_{2 n}\\right)$ ou $\\left(x_{2 n+1}\\right)$ a la propriété $D$. Dans le premier cas, cela signifie que $a d+b$ divise $x_{0}=2005$. Dans le second cas, cela signifie que $b c+d$ divise $x_{1}=2005 c+d$. Or, notons que $|a d+b|=|b c+d|$.\n\nLes paires des polynômes $(P(X), Q(X))$ recherchées sont donc les paires $(P(X), Q(X))=(\\varepsilon X+b, \\varepsilon X+d)$ telles que $\\varepsilon= \\pm 1$ et $b+\\varepsilon d$ divise $2005$ ou $2005+\\varepsilon d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle non isocèle. Soit $\\omega$ le cercle inscrit et $I$ son centre. On note $M, N, P$ les points de contact de $\\omega$ avec les côtés $[BC], [CA], [AB]$. Soit $J$ le point d'intersection entre $(MN)$ et $(IC)$. La droite $(PJ)$ recoupe $\\omega$ en $K$. Montrer que\n\na) $CKIP$ est cyclique;\n\nb) $(CI)$ est la bissectrice de $\\widehat{PCK}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\na) Comme $(IN) \\perp (NC)$ et $(IM) \\perp (MC)$, les points $M$ et $N$ sont situés sur le cercle de diamètre $[IC]$, donc $I, M, C, N$ sont cocycliques.\n\nD'après la puissance d'un point par rapport à ce cercle, on a $JI \\cdot JC = JM \\cdot JN$. D'autre part, en utilisant la puissance par rapport au cercle inscrit, on a $JM \\cdot JN = JK \\cdot JP$, donc finalement $JI \\cdot JC = JK \\cdot JP$, ce qui entraîne que $P, I, K, C$ sont cocycliques.\n\nb) D'après la question a), on a $\\widehat{ICP} = \\widehat{IKP}$ et $\\widehat{KCI} = \\widehat{KPI}$. Or, $IPK$ est isocèle en $I$, donc $\\widehat{IKP} = \\widehat{KPI}$, ce qui entraîne que $\\widehat{ICP} = \\widehat{KCI}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $S$ un ensemble d'entiers strictement positifs tel que\n$$\n\\lfloor\\sqrt{x}\\rfloor=\\lfloor\\sqrt{y}\\rfloor \\text{ pour tous } x, y \\in S\n$$\nProuver que si $x, y, z, t \\in S$ avec $(x, y) \\neq(z, t)$ et $(x, y) \\neq(t, z)$, alors $x y \\neq z t$.\n( $\\lfloor.\\rfloor$ désigne la partie entière.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSupposons tout d'abord qu'il existe des entiers $x_{1}, x_{2}, x_{3}, x_{4}$ dans $S$ tels que $x_{1} x_{2} \\leqslant x_{3} x_{4}$ et $x_{1}+x_{2}>x_{3}+x_{4}$. Puisqu'il s'agit d'entiers, on a donc $x_{1}+x_{2}-x_{3}-x_{4} \\geqslant 1$. Soit $n=\\left\\lfloor\\sqrt{x_{1}}\\right\\rfloor$. Par définition de $S$, pour $i=1,2,3,4$, il existe donc un entier $w_{i}$ tel que $0 \\leqslant w_{i} \\leqslant 2 n$ et $x_{i}=n^{2}+w_{i}$.\nLes inégalités $x_{1} x_{2} \\leqslant x_{3} x_{4}$ et $x_{1}+x_{2}>x_{3}+x_{4}$ deviennent respectivement\n$$\nw_{1}+w_{2}-w_{3}-w_{4} \\geqslant 1 \\quad \\text{ et } \\quad\\left(w_{1}+w_{2}-w_{3}+w_{4}\\right) n^{2} \\leqslant w_{3} w_{4}-w_{1} w_{2}\n$$\nEn particulier, on doit avoir $w_{3}>0$, et donc\n$$\n\\begin{aligned}\nn^{2} & \\leqslant\\left(w_{1}+w_{2}-w_{3}+w_{4}\\right) n^{2} \\leqslant w_{3} w_{4}-w_{1} w_{2} \\\\\n& <w_{3}\\left(w_{1}+w_{2}-w_{3}\\right)-w_{1} w_{2}=\\left(w_{1}-w_{3}\\right)\\left(w_{3}-w_{2}\\right) \\\\\n& <\\frac{\\left(\\left(w_{1}-w_{3}\\right)+\\left(w_{3}-w_{2}\\right)\\right)^{2}}{4}=\\frac{\\left(w_{1}-w_{2}\\right)^{2}}{4} \\leqslant n^{2}\n\\end{aligned}\n$$\nd'où la contradiction.\n\nSoit alors $x_{1}, x_{2}, x_{3}, x_{4}$ dans $S$ tels que $x_{1} x_{2}=x_{3} x_{4}=a$. D'après le lemme ci-dessus, on a donc $x_{1}+x_{2}= x_{3}+x_{4}=b$. Les nombres $x_{1}$ et $x_{2}$ d'une part, $x_{3}$ et $x_{4}$ d'autre part, sont alors les racines du polynôme $X^{2}-a X+b$. On a alors $x_{1}=x_{3}$ et $x_{2}=x_{4}$, ou $x_{1}=x_{4}$ et $x_{2}=x_{3}$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn note $S$ l'ensemble des entiers de $1$ à $2016$. Combien y a-t-il de manières de partitionner $S$ en deux sous-ensembles $A$ et $B$ de telle manière que ni $A$ ni $B$ ne contient deux entiers dont la somme est une puissance de $2$ ?",
    "options": [],
    "answer": "2048",
    "solution": "Solution:\n\nLa réponse est $2^{11} = 2048$. Plus précisément, on va montrer qu'il est possible de répartir comme on veut les nombres $1, 2, 4, 8, \\ldots, 1024 = 2^{10}$ comme on veut entre $A$ et $B$ et qu'une fois ces nombres placés il existe une unique manière de répartir les autres. On peut alors conclure car il y a $2^{11} = 2048$ manières de répartir ces $11$ nombres entre $A$ et $B$.\n\nLe fait que la répartition de $1, 2, 4, 8, \\ldots, 1024$ impose la position de $k$ se montre par récurrence forte sur $k$ : la position de $1$ et $2$ a déjà été fixée, et $3$ doit être dans l'ensemble qui ne contient pas $1$ car $3 + 1 = 2^{2}$. Supposons maintenant que toutes les positions de nombres de $1$ à $k-1$ soient fixées : si $k$ est une puissance de $2$, alors la position de $k$ est fixée par hypothèse. Sinon, il existe $a$ tel que $2^{a-1} < k < 2^{a}$. Mais alors $0 < 2^{a} - k < k$ donc $2^{a} - k$ est dans $A$ ou $B$, et sa position a déjà été fixée. $k$ est donc nécessairement dans l'autre ensemble. Il existe donc au plus une partition qui convient une fois fixée la répartition des puissances de $2$.\n\nIl reste à montrer qu'une telle répartition existe bien. Mais la preuve précédente fournit un algorithme qui permet de répartir les entiers qui ne sont pas des puissances de $2$ entre $A$ et $B$ de telle manière que pour tous $k$ et $a$ tel que $2^{a-1} < k < 2^{a}$, les nombres $k$ et $2^{a} - k$ ne sont pas dans le même ensemble. Or, si deux nombres $x \\neq y$ ont pour somme une puissance de $2$ notée $2^{a}$, on peut supposer $x > y$. On a alors $x > \\frac{x + y}{2} = 2^{a-1}$, donc $2^{a-1} < x < 2^{a}$ et $y = 2^{a} - x$, donc $x$ et $y$ ne sont pas dans le même ensemble, donc il existe bien toujours une répartition qui convient.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle rectangle en $C$. La bissectrice de $\\widehat{BAC}$ coupe $[BC]$ en $P$, et celle de $\\widehat{ABC}$ coupe $[AC]$ en $Q$. Soient $M$ et $N$ sur $[AB]$ tels que $(MP)$ et $(NQ)$ soient perpendiculaires à $(AB)$. Combien vaut l'angle $\\widehat{MCN}$ ?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "45°",
    "solution": "Solution:\n\nComme les angles $\\widehat{AMP}$ et $\\widehat{ACP}$ sont droits, les points $A, C, P$ et $M$ sont cocycliques sur le cercle de diamètre $[AP]$. De même, les points $B, C, Q$ et $N$ sont cocycliques sur le cercle de diamètre $[BQ]$. On peut donc faire une chasse aux angles :\n\n$$\n\\begin{aligned}\n\\widehat{MCN} & = 90^{\\circ} - \\widehat{MCP} - \\widehat{NCQ} \\\\\n& = 90^{\\circ} - \\widehat{MAP} - \\widehat{NBQ} \\\\\n& = 90^{\\circ} - \\frac{1}{2} \\widehat{ABC} - \\frac{1}{2} \\widehat{BAC} \\\\\n& = \\frac{1}{2}\\left(180^{\\circ} - \\widehat{ABC} - \\widehat{BAC}\\right) \\\\\n& = \\frac{1}{2} \\widehat{ACB} \\\\\n& = 45^{\\circ}\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. On suppose que la médiane $(BM)$ et la bissectrice $(CD)$ se coupent en un point $J$ tel que $JB = JC$. Soit $H$ le pied de la hauteur issue de $A$. Montrer que $JM = JH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $\\gamma = \\widehat{ACB}$. On a $MA = MC = MH$, donc $\\widehat{CHM} = \\gamma$.\n\nOn a aussi $\\widehat{CHM} = 180^\\circ - \\widehat{MHB} = \\widehat{HBM} + \\widehat{BMH}$.\n\nDe plus, $\\widehat{HBM} = \\widehat{DCB} = \\gamma / 2$, donc $\\widehat{BMH} = \\widehat{HMB} = \\gamma / 2$.\n\nOn en déduit que $BH = HM = MC$. Les triangles $BHJ$ et $CMJ$ vérifient $BJ = CJ$, $\\widehat{HBJ} = \\widehat{MCJ}$ et $BH = CM$ donc sont isométriques. Par conséquent, $JH = HM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $a$, $b$, $c > 0$. Montrer que\n$$\n\\sqrt{\\frac{a}{b+c}} + \\sqrt{\\frac{b}{c+a}} + \\sqrt{\\frac{c}{a+b}} > 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nL'idée est d'utiliser l'inégalité arithmético-géométrique pour minorer $\\frac{1}{\\sqrt{x+y}}$. Plus précisément, on a, par inégalité arithmético-géométrique :\n$$\n\\sqrt{\\frac{a}{b+c}} = \\frac{a}{\\sqrt{a(b+c)}} \\geqslant \\frac{a}{(a+b+c)/2} = \\frac{2a}{a+b+c}\n$$\nOn procède de même pour montrer que $\\sqrt{\\frac{b}{c+a}} \\geqslant \\frac{2b}{a+b+c}$ et que $\\sqrt{\\frac{c}{a+b}} \\geqslant \\frac{2c}{a+b+c}$. On en déduit que\n$$\n\\sqrt{\\frac{a}{b+c}} + \\sqrt{\\frac{b}{c+a}} + \\sqrt{\\frac{c}{a+b}} \\geqslant 2\n$$\nIl y a égalité seulement si et seulement si on est dans le cas d'égalité de chacune des trois inégalités arithmético-géométrique utilisées, soit : $a = b + c$, $b = a + c$ et $c = a + b$, ce qui est impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 5$ un entier, et $\\{a_{1}, a_{2}, \\cdots, a_{n}\\}=\\{1,2, \\cdots, n\\}$. Prouver que les nombres $a_{1}, a_{1}+a_{2}, a_{1}+a_{2}+a_{3}, \\cdots, a_{1}+a_{2}+\\cdots+a_{n}$ donnent au moins $\\lfloor\\sqrt{n}\\rfloor+1$ restes distincts modulo $n$.\n\n($\\lfloor.\\rfloor$ désigne la partie entière.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour $i=1, \\cdots, n$, on pose $b_{i}=a_{1}+\\cdots+a_{i}$. Procédons par l'absurde et supposons que les nombres $b_{1}, \\cdots, b_{n}$ ne donnent qu'au plus $\\sqrt{n}$ restes distincts modulo $n$.\n\nNotons que, pour $i=1, \\cdots, n-1$, on a $a_{i+1}=b_{i+1}-b_{i}$, et seul celui des $a_{i}$ qui vaut $n$ peut donc conduire à $b_{i+1}-b_{i} \\equiv 0 \\pmod{n}$. Les nombres $b_{i+1}-b_{i}$ ne peuvent donc prendre, modulo $n$, qu'au plus $\\sqrt{n}(\\sqrt{n}-1)$ valeurs non nulles distinctes. En tenant compte du $a_{i}$ qui est égal à $n$ et de $a_{n}$ (qui n'est pas de la forme $b_{i+1}-b_{i}$), et puisque $n \\geqslant 5$, les $a_{i}$ prennent donc au plus\n$$\nn-\\sqrt{n}+2 \\leqslant n-\\sqrt{5}+2<n\n$$\nvaleurs distinctes modulo $n$. Or, les $a_{i}$ forment une permutation de $\\{1, \\cdots, n\\}$ donc ils représentent tous les restes modulo $n$ : il y a là une contradiction.\n\nLes nombres $b_{1}, \\cdots, b_{n}$ donnent donc au moins $\\lfloor\\sqrt{n}\\rfloor+1$ restes distincts modulo $n$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 30487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a_{1}, a_{2}, \\ldots, a_{n}$ des entiers strictement positifs. Pour tout $k=1,2, \\ldots, n$, on note\n$$\nm_{k}=\\max _{1 \\leq \\ell \\leq k} \\frac{a_{k-\\ell+1}+a_{k-\\ell+2}+\\cdots+a_{k}}{\\ell} .\n$$\nMontrer que pour tout $\\alpha>0$, le nombre d'entiers $k$ tel que $m_{k}>\\alpha$ est strictement plus petit que $\\frac{a_{1}+a_{2}+\\cdots+a_{n}}{\\alpha}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $k_{1}$ le plus grand entier $k$ tel que $m_{k}>\\alpha$. Il existe $\\ell_{1}$ tel que $a_{k_{1}-\\ell_{1}+1}+\\cdots+a_{k_{1}}>\\ell_{1} \\alpha$.\nSoit $k_{2}$ le plus grand entier $\\leqslant k_{1}-\\ell_{1}$ tel que $m_{k}>\\alpha$. Il existe $\\ell_{2}$ tel que $a_{k_{2}-\\ell_{2}+1}+\\cdots+a_{k_{2}}>\\ell_{2} \\alpha$. ...\nOn construit ainsi une suite finie $k_{1}>k_{2}>\\cdots>k_{r}$ telle que tout entier $k$ vérifiant $m_{k}>\\alpha$ se trouve dans l'un des intervalles $I_{j}=\\llbracket k_{j}-\\ell_{j}+1, k_{j} \\rrbracket$. De plus, ces intervalles sont deux à deux disjoints.\nPar conséquent, le nombre d'entiers $k$ tels que $m_{k}>\\alpha$ est inférieur ou égal à la somme des longueurs des intervalles, à savoir $\\ell_{1}+\\cdots+\\ell_{r}$.\nOr, $\\sum_{j=1}^{r} \\ell_{j}<\\frac{1}{\\alpha} \\sum_{j}\\left(a_{k_{j}-\\ell_{j}+1}+\\cdots+a_{k_{j}}\\right) \\leqslant \\frac{1}{\\alpha}\\left(a_{1}+\\cdots+a_{n}\\right)$. CQFD.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle ayant ses trois angles aigus, et $P$ le pied de la hauteur issue de $A$. On note $I_{1}$ et $I_{2}$ les centres de cercles inscrits à $ABP$ et $ACP$. Le cercle inscrit à $ABC$ touche $[BC]$ en $D$. Combien valent les angles du triangle $I_{1}I_{2}D$ ?\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "90°, 45°, 45°",
    "solution": "Solution:\n\nOn note $E$ et $F$ les points où les cercles inscrits à $ABP$ et $ACP$ touchent $[BC]$, et $G$ et $H$ les points où ils touchent $[AP]$ : $PEI_{1}H$ est un carré car les angles en $P$, $E$ et $H$ sont droits, et $I_{1}E = I_{1}H$, donc $EP = EI_{1}$, et de même $FP = FI_{2}$. On a donc :\n$$\nFD = CD - CF = \\frac{CA + CB - AB}{2} - \\frac{CA + CP - AP}{2} = \\frac{CB - CP + AP - AB}{2} = \\frac{BP + AP - AB}{2} = PE = EI_{1}\n$$\net de même $ED = FI_{2}$. Comme de plus $\\widehat{I_{2}FD}$ et $\\widehat{I_{1}ED}$ sont droits, les triangles $I_{1}ED$ et $I_{2}FD$ sont isométriques donc $DI_{1} = DI_{2}$. De plus on a :\n\n$$\n\\widehat{I_{1}DI_{2}} = 180^{\\circ} - \\widehat{I_{1}DE} - \\widehat{I_{2}DF} = 180^{\\circ} - \\widehat{I_{1}DE} - \\widehat{DI_{1}E} = \\widehat{DEI_{1}} = 90^{\\circ}\n$$\n\nLe triangle est donc isocèle rectangle en $D$, donc ses angles valent $90^{\\circ}$, $45^{\\circ}$ et $45^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSur chaque sommet d'un $n$-gone régulier on place un signe $+$ ou $-$. À chaque étape, on a le droit de changer les signes de trois sommets consécutifs du $n$-gone.\n\nQuels sont les $n$ pour lesquels, quelle que soit la configuration de départ, on peut obtenir en un nombre fini d'étapes des $+$ sur tous les sommets ?",
    "options": [],
    "answer": "All positive integers n that are not divisible by 3",
    "solution": "Solution:\n\nOn va montrer que ce sont exactement les $n$ qui ne sont pas divisibles par $3$.\n\nSi $n$ est divisible par $3$, on note $A_{1}, \\ldots, A_{n}$ les sommets du polygone, et on note $P$ le nombre de $k$ qui ne sont pas divisibles par $3$ et tels qu'il y a un signe $-$ sur $A_{k}$. On montre qu'à chaque étape la parité de $P$ ne change pas : à chaque étape, on change les signes de $3$ sommets consécutifs $A_{k}, A_{k-1}$ et $A_{k+1}$ (avec la convention $A_{0}=A_{n}$ et $A_{n+1}=A_{1}$). Exactement un de ces nombres $k-1, k$ et $k+1$ est divisible par $3$, donc on change exactement les signes de $2$ sommets non divisibles par $3$. $P$ peut donc varier de $-2$ (si deux $-$ deviennent des $+$), de $0$ (si un $-$ devient un $+$ et un $+$ devient un $-$) ou de $2$ (si deux $+$ deviennent deux $-$), donc sa parité ne change jamais. En particulier, si $P=1$ au départ, il est impossible d'obtenir $P=0$ donc il est impossible de n'avoir que des $+$.\n\nSi $n$ n'est pas divisible par $3$, on montre qu'il est possible de modifier exactement un signe. On peut alors facilement conclure car il suffit de modifier un par un les signes $-$ :\n\n- Si $n$ est de la forme $3k-1$, on peut modifier les signes de $A_{1}, A_{2}, A_{3}$ puis de $A_{4}, A_{5}, A_{6}$ et ainsi de suite jusqu'à $A_{3k-2}, A_{3k-1}, A_{1}$, ce qui modifie le signe de tout le monde sauf $A_{1}$ (qui a été touché deux fois). On peut ensuite modifier le signe de tout le monde sauf $A_{2}$, puis modifier les signes de $A_{1}, A_{2}, A_{3}$. Cela revient à ne changer que le signe de $A_{3}$.\n\n- Si $n$ est de la forme $3k+1$, on peut modifier les signes de $A_{1}, A_{2}, A_{3}$ puis de $A_{4}, A_{5}, A_{6}$ et ainsi de suite jusqu'à $A_{3k-2}, A_{3k-1}, A_{3k}$, ce qui modifie le signe de tout le monde sauf $A_{3k+1}$ et on conclut comme ci-dessus.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère un échiquier $2015 \\times 2015$, auquel on a retiré le carré $2 \\times 2$ en bas à droite. Montrer qu'on peut le paver par des pièces composées de 3 cases en forme de L (les pièces peuvent être pivotées).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComme l'échiquier peut être décomposé en trois rectangles $2010 \\times 2015$, $5 \\times 2004$ et $5 \\times 11$, et comme $2010$ et $2004$ sont divisibles par $6$, il suffit de montrer que\n\n1) Pour tout $n \\geqslant 2$ et tout entier $m$, un échiquier $6 m \\times n$ peut être pavé par des pièces en forme de $L$;\n\n2) L'échiquier $5 \\times 11$ privé du carré $2 \\times 2$ en bas à droite peut être pavé par des pièces en forme de $L$.\n\nPour le point 1), comme un tel échiquier est réunion d'échiquiers $6 \\times n$, on se ramène à traiter le cas $m=1$.\n\nIl est clair que l'on peut paver un échiquier $3 \\times 2$, donc aussi des échiquiers $6 \\times 2$ et $6 \\times 3$. On peut donc paver tout échiquier de taille $6 \\times 2k$ ou $6 \\times (2k+3)$. Or, tout entier $n \\geqslant 2$ s'écrit sous la forme $2k$ s'il est pair, et $2k+3$ s'il est impair, ce qui conclut.\n\nLe point 2) se démontre en construisant un exemple de pavage:\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, et $D, E, F$ les pieds des hauteurs de $A, B, C$ respectivement. On définit aussi $H$ l'orthocentre de $ABC$, $O$ le centre de son cercle circonscrit, et $X$ le point de la droite $(EF)$ qui vérifie $XA = XD$. Montrer que les droites $(AX)$ et $(OH)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn sait que l'axe radical de deux cercles est une droite perpendiculaire à la droite passant par les centres des deux cercles.\n\nLa condition $XA = XD$ signifie que $X$ est sur la médiatrice de $[AD]$, qui n'est autre que la droite des milieux $\\left(B', C'\\right)$ (avec $B'$ et $C'$ les milieux de $[AC]$ et de $[AB]$). On sait aussi que les points $A, E, F, H$ sont cocycliques d'après la réciproque du théorème de l'angle inscrit. De plus, $[AH]$ étant un diamètre du cercle passant par ces quatre points, le milieu $H'$ de $[AH]$ est le centre de ce cercle.\n\nLe centre du cercle passant par $A, B', C'$ est le milieu $O'$ de $[AO]$ (par propriété de l'homothétie de centre $A$ et de rapport $1/2$). On sait que $\\left(O'H'\\right)$ et $(OH)$ sont parallèles. Le problème revient donc à montrer que $(AX)$ et $\\left(O'H'\\right)$ sont perpendiculaires. Nous allons montrer ici que $(AX)$ est en fait l'axe radical des deux cercles évoqués précédemment, ce qui terminera l'exercice d'après la propriété évoquée au début de la preuve.\n\n$A$ étant sur les deux cercles, est bien sur leur axe radical. On sait de plus que $B', C', E, F$ sont cocycliques sur le cercle d'Euler du triangle $ABC$. Ainsi on a l'égalité de longueurs : $XC' \\times XB' = XE \\times XF$, ce qui signifie que $X$ est bien sur l'axe radical de nos deux cercles.\n\nAinsi on a bien $(AX)$ et $(OH)$ qui sont perpendiculaires.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$. On trace un circuit sur un échiquier $n \\times n$ qui passe exactement une fois par chaque case et revient à son point de départ (deux cases consécutives sur le circuit doivent avoir un côté commun).\n\nMontrer qu'il existe deux cases voisines sur l'échiquier telles que, si on \"coupe\" le circuit en ces deux cases, les deux chemins obtenus ont pour longueur au moins le quart de la longueur du circuit de départ.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRemarquons tout d'abord que si l'échiquier est pavé en noir et blanc de la manière habituelle, le circuit doit alterner des cases noires et blanches et revenir à sa case de départ donc sa longueur doit être paire donc $n^{2}$ est pair. On écrit donc $n=2m$.\n\n![](attached_image_1.png)\n\nOn considère l'ensemble des coins des carrés qui se trouvent à l'intérieur du circuit (ce sont les points bleus sur la figure), et on relie deux de ces points quand ils sont à distance 1 l'un de l'autre (ce sont les arêtes bleues sur la figure). Le graphe obtenu est un arbre : il est connexe car le circuit l'est, et il n'a pas de cycle car un cycle soit créerait un deuxième circuit à l'intérieur du premier, soit enfermerait le centre d'une case qui ne serait donc pas dans le circuit rouge. De plus, dans cet arbre, chaque sommet a au plus 4 voisins.\n\nLe lemme suivant permet de faire le lien entre la longueur du chemin rouge et le nombre de sommets de l'arbre bleu :\n\nLemme (Lemme 1). Soit $T$ un arbre dont les sommets sont sur les coins d'un quadrillage, et $C$ un circuit passant par les centres des cases et faisant le tour de l'arbre. On note $|C|$ la longueur de ce circuit. Alors $|C|=2|T|+2$.\n\nPreuve du lemme 1. Chaque sommet de l'arbre est le centre d'un carré de côté 1. On calcule de deux manières différentes la somme $S$ des périmètres de ces carrés. D'une part on a $S=4|T|$. Mais d'autre part, pour calculer $S$, on prend en compte une fois chaque arête de $C$, et deux fois chaque arête qui sépare deux sommets de l'arbre. Comme l'arbre a $|T|-1$ arêtes, on a donc $S=|C|+2(|T|-1)$. On en déduit le lemme.\n\nOn sait que $|C|=n^{2}=4m^{2}$. On en déduit $|T|=\\frac{1}{2}(|C|-2)=2m^{2}-1$. De plus, couper le circuit en deux cases voisines revient essentiellement à couper une arête de l'arbre. On va donc montrer le lemme suivant :\n\nLemme (Lemme 2). On note $|T|$ le nombre de sommets d'un arbre. Soit $T$ un arbre ayant au moins deux sommets et dont tous les sommets sont de degré au plus 4. Il existe une arête $e$ de $T$ telle qu'en coupant l'arête $e$, l'arbre $T$ est coupé en deux arbres $T_{1}$ et $T_{2}$ tels que $|T_{1}| \\geqslant \\frac{1}{4}(|T|-1)$ et $|T_{2}| \\geqslant \\frac{1}{4}(|T|-1)$.\n\nPreuve du lemme 2. On raisonne par l'absurde. On fixe $x_{0}$ un sommet de $T$ : supprimer $x_{0}$ divise $T$ en $T_{1}, T_{2}, T_{3}$ et $T_{4}$ (si $x_{0}$ n'a pas quatre voisins, certains de ces arbres ont un nombre de sommets nul). D'après le principe des tiroirs, il existe $i$ tel que $|T_{i}| \\geqslant \\frac{1}{4}(|T|-1)$. On suppose que c'est $T_{1}$, et on note $x_{1}$ le sommet de $T_{1}$ qui est relié à $x_{0}$. Mais si $T$ ne vérifie pas le lemme, alors $|T_{1}| > |T| - \\frac{1}{4}(|T|-1) = \\frac{3}{4}(|T|-1)+1$.\n\nOn peut maintenant effectuer le même raisonnement avec $x_{1}$ : il a au plus trois voisins différents de $x_{0}$ qui correspondent à trois arbres $T_{1}', T_{2}', T_{3}'$, et le nombre total de sommets de ces trois arbres vaut au moins $\\frac{3}{4}(|T|-1)$. Par le principe des tiroirs il existe $i$ tel que $|T_{i}'| \\geqslant \\frac{1}{4}(|T|-1)$ donc, comme $T$ ne vérifie pas le lemme, $|T_{1}'| > \\frac{3}{4}(|T|-1)+1$. On note $x_{2}$ le sommet par lequel $T_{1}'$ est relié à $x_{1}$. Comme on n'a considéré que les voisins de $x_{1}$ qui sont différents de $x_{0}$, on a $x_{2} \\neq x_{0}$. On peut alors itérer cet argument. Comme $T$ ne vérifie pas le lemme, le procédé ne s'arrête jamais et on construit une suite $(x_{n})$ de sommets telle que $x_{n+2} \\neq x_{n}$ pour tout $n$. Comme $T$ est un arbre, $(x_{n})$ ne passe jamais deux fois par le même sommet donc $T$ est infini, ce qui est absurde.\n\nD'après le lemme 2, il existe donc une arête $e$ de l'arbre (en vert sur la figure) telle que, si on coupe $e$, on obtient $T_{1}$ et $T_{2}$ tels que $|T_{1}|, |T_{2}| \\geqslant \\frac{1}{4}(|T|-1) = \\frac{m^{2}-1}{2}$. Si on note $C_{1}$ et $C_{2}$ les circuits qui font le tour de $T_{1}$ et de $T_{2}$, on a donc d'après le lemme 1 les inégalités $|C_{1}|, |C_{2}| \\geqslant m^{2}+1$. Mais les deux parties du circuit obtenues en le coupant ont pour longueurs $|C_{1}|-1$ et $|C_{2}|-1$ (on retire l'arête orange) qui valent au moins $m^{2}$, d'où le résultat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ un nombre entier et soient $x_{1}, \\ldots, x_{n}$ des nombres réels positifs tels que $x_{1} x_{2} \\cdots x_{n}=1$. Montrer que\n$$\n\\frac{1}{n-1+x_{1}}+\\frac{1}{n-1+x_{2}}+\\cdots+\\frac{1}{n-1+x_{n}} \\leqslant 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEn réduisant au même dénominateur, il s'agit de montrer que\n$$\n\\sum_{k=1}^{n} \\prod_{j \\neq k}\\left(x_{j}+n-1\\right) \\leqslant \\prod_{k=1}^{n}\\left(x_{k}+n-1\\right)\n$$\nOn note\n$$\n\\sigma_{k}=\\sum_{1 \\leqslant i_{1}<\\cdots<i_{k} \\leqslant n} x_{i_{1}} x_{i_{2}} \\cdots x_{i_{k}}\n$$\nle $k$-ème polynôme élémentaire (de degré $k$ donc) en $x_{1}, \\cdots, x_{n}$. Alors, en développant les deux côtés de (5), on voit que le coefficient de $\\sigma_{k}$ est $(n-1)^{n-k}$ dans celui de droite pour $1 \\leqslant k \\leqslant n$, et $(n-k)(n-1)^{n-k-1}$ dans celui de gauche pour $1 \\leqslant k \\leqslant n-1$. Le coefficient constant vaut $(n-1)^{n}$ à droite et $n(n-1)^{n-1}$ à gauche. Autrement dit,\n$$\n\\prod_{k=1}^{n}\\left(x_{j}+n-1\\right)=\\sum_{k=1}^{n}(n-1)^{n-k} \\sigma_{k}+(n-1)^{n}\n$$\net\n$$\n\\sum_{k=1}^{n} \\prod_{j \\neq k}\\left(x_{j}+n-1\\right)=\\sum_{k=1}^{n-1}(n-k)(n-1)^{n-k-1} \\sigma_{k}+n(n-1)^{n-1}\n$$\nEn notant $D=\\prod_{k=1}^{n}\\left(x_{k}+n-1\\right)-\\sum_{k=1}^{n} \\prod_{j \\neq k}\\left(x_{j}+n-1\\right)$, on a donc\n$$\nD=x_{1} x_{2} \\cdots x_{n}+\\sum_{k=1}^{n-1}(k-1)(n-1)^{n-k-1} \\sigma_{k}-(n-1)^{n-1}\n$$\nOr $\\sigma_{k} \\geqslant\\binom{ n}{k}$ par l'inégalité arithmético-géométrique (car $x_{1} x_{2} \\cdots x_{n}=1$), de sorte que :\n$$\nD \\geqslant \\sum_{k=1}^{n}(k-1)(n-1)^{-k}\\binom{n}{k}-(n-1)^{n-1}\n$$\nou encore, en divisant les deux côtés par $(n-1)^{n-1}$,\n$$\n\\frac{D}{(n-1)^{n-1}} \\geqslant \\sum_{k=1}^{n}(k-1)(n-1)^{-k}\\binom{n}{k}-1=\\sum_{k=1}^{n} k(n-1)^{-k}\\binom{n}{k}-\\sum_{k=1}^{n}(n-1)^{-k}\\binom{n}{k}-1 .\n$$\nOr\n$$\n\\sum_{k=1}^{n} k\\binom{n}{k}(n-1)^{-k}=\\frac{n}{n-1} \\sum_{k=1}^{n}\\binom{n-1}{k-1} \\cdot\\left(\\frac{1}{n-1}\\right)^{k-1}=\\frac{n}{n-1} \\times\\left(\\frac{n}{n-1}\\right)^{n-1}=\\left(\\frac{n}{n-1}\\right)^{n}\n$$\net\n$$\n\\sum_{k=1}^{n}(n-1)^{-k}\\binom{n}{k}=\\sum_{k=0}^{n}(n-1)^{-k}\\binom{n}{k}-1=\\left(\\frac{n}{n-1}\\right)^{n}-1 .\n$$\nOn en déduit que $D \\geqslant 0$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a$, $b$, $c$, $n$ des entiers, avec $n \\geq 2$. Soit $p$ un nombre premier qui divise $a^{2}+a b+b^{2}$ et $a^{n}+b^{n}+c^{n}$, mais qui ne divise pas $a+b+c$.\nProuver que $n$ et $p-1$ ne sont pas premiers entre eux.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSi $p$ divise $a$ et $b$, alors $p$ divise $c^{n}$ donc $p$ divise $c$, ce qui contredit la dernière assertion. Donc, quitte à échanger $a$ et $b$, on suppose que $p$ ne divise pas $b$. Quitte à multiplier $a, b, c$ par un inverse de $b$ modulo $p$, on peut supposer que $b=1$ et donc\n$$\np\\left|a^{2}+a+1, \\quad p\\right| a^{n}+1+c^{n}, \\quad p \\nmid a+1+c .\n$$\nComme $a^{2}+a+1$ est impair, $p$ l'est aussi.\nComme $p$ divise \\left($a^{2}+a+1\\right)(a-1)=a^{3}-1$, l'ordre de $a$ modulo $p$ est 1 ou 3.\n\nPremier cas: $a \\equiv 1\\ (\\bmod\\ p)$. Alors $p=3$, donc $c^{n} \\equiv-1-a^{n} \\equiv 1\\ (\\bmod\\ 3)$.\nEn particulier, $c$ est inversible modulo 3, donc $c \\equiv \\pm 1\\ (\\bmod\\ 3)$. Comme $p \\nmid a+1+c$, on a nécessairement $c \\equiv-1\\ (\\bmod\\ 3)$. Enfin, comme $c^{n} \\equiv 1\\ (\\bmod\\ 3)$, l'entier $n$ est pair donc n'est pas premier avec $p-1$.\n\nDeuxième cas: l'ordre de $a$ modulo $p$ est 3. Comme par ailleurs l'ordre de $a$ modulo $p$ divise $p-1$ en vertu du petit théorème de Fermat, on a $3 \\mid p-1$.\nSupposons que $n$ est premier avec $p-1$. Alors $x \\mapsto x^{n}$ est une bijection de $\\mathbb{Z} / p \\mathbb{Z}$ sur lui-même :\n- si $n \\equiv 1\\ (\\bmod\\ 3)$ alors $c^{n} \\equiv-a^{n}-1 \\equiv-a-1 \\equiv a^{2} \\equiv\\left(a^{2}\\right)^{n}\\ (\\bmod\\ p)$ donc par bijectivité de $x \\mapsto x^{n}$ dans $\\mathbb{Z} / p \\mathbb{Z}$ on a $c \\equiv a^{2} \\equiv-a-1\\ (\\bmod\\ p):$ contradiction!\n- si $n \\equiv 2\\ (\\bmod\\ 3)$ alors $c^{n} \\equiv-a^{n}-1 \\equiv-a^{2}-1 \\equiv a \\equiv\\left(a^{2}\\right)^{n}\\ (\\bmod\\ p)$ donc par bijectivité de $x \\mapsto x^{n}$ dans $\\mathbb{Z} / p \\mathbb{Z}$ on a $c \\equiv a^{2} \\equiv-a-1\\ (\\bmod\\ p):$ contradiction!\nOn en déduit que $n \\equiv 0\\ (\\bmod\\ 3)$, donc 3 est un diviseur commun de $n$ et de $p-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30495,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a_{1}, a_{2}, a_{3}$ des entiers strictement positifs. Pour tout entier $n \\geqslant 3$, on pose\n$$\na_{n+1}=\\operatorname{ppcm}\\left(a_{n}, a_{n-1}\\right)-\\operatorname{ppcm}\\left(a_{n-1}, a_{n-2}\\right)\n$$\nétant entendu que l'on a $\\operatorname{ppcm}(0, x)=0$ pour tout entier $x$.\n\nProuver qu'il existe un entier naturel $k$ tel que $k \\leqslant a_{3}+4$ et $a_{k} \\leqslant 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSi $x$ et $y$ sont des entiers, on note $x \\vee y=\\operatorname{ppcm}(x, y)$ et $x \\wedge y=\\operatorname{pgcd}(x, y)$, avec $\\operatorname{pgcd}(0, x)=x$ pour tout $x$. On rappelle qu'alors $x y=(x \\vee y)(x \\wedge y)$.\n\nPour tout entier $n \\geqslant 3$, $a_{n-1}$ divise $a_{n} \\vee a_{n-1}$ et $a_{n-1} \\vee a_{n-2}$, donc $a_{n-1}$ divise $a_{n+1}$ également. Mais alors, pour tout $n \\geqslant 4$, le nombre $a_{n-2}$ divise $a_{n}$, donc divise $a_{n} \\vee a_{n-1}$ et $a_{n-1} \\vee a_{n-2}$, et enfin divise $a_{n+1}$. Le nombre $a_{n+1}$ est donc divisible par $a_{n-1} \\vee a_{n-2}$.\n\nOn pose maintenant $c_{4}=\\frac{a_{5}}{a_{2} \\vee a_{3}}$. Montrons alors l'égalité ($E_{n}$) suivante pour tout entier $n \\geqslant 4$ tel que $a_{4}, \\ldots, a_{n}>0$ :\n$$\n\\left(a_{n-1} \\vee a_{n-2}\\right)\\left(4+c_{4}-n\\right)=a_{n+1}\n$$\nTout d'abord, par construction, l'égalité ($E_{4}$) est vraie. Supposons maintenant que l'égalité ($E_{n-1}$) est vraie, et déduisons-en que, si $a_{4}, \\ldots, a_{n}>0$, alors ($E_{n}$) est vraie aussi. Puisque $a_{n-3}$ divise $a_{n-1}$, on observe que\n$$\n\\begin{aligned}\n\\left(a_{n-1} \\wedge a_{n}\\right)\\left(a_{n-1} \\vee a_{n-2}\\right) & =\\left(a_{n-1} \\wedge\\left(a_{n-1} \\vee a_{n-2}-a_{n-2} \\vee a_{n-3}\\right)\\right) \\cdot\\left(a_{n-1} \\vee a_{n-2} \\vee a_{n-3}\\right) \\\\\n& =\\left(a_{n-1} \\wedge\\left(a_{n-2} \\vee a_{n-3}\\right)\\right) \\cdot\\left(a_{n-1} \\vee\\left(a_{n-2} \\vee a_{n-3}\\right)\\right) \\\\\n& =a_{n-1}\\left(a_{n-2} \\vee a_{n-3}\\right) .\n\\end{aligned}\n$$\nOr, puisque\n$$\n\\begin{aligned}\n\\left(E_{n}\\right) & \\Leftrightarrow \\frac{a_{n+1}}{a_{n-1} \\vee a_{n-2}}=4+c_{4}-n=\\frac{a_{n}}{a_{n-2} \\vee a_{n-3}}-1 \\\\\n& \\Leftrightarrow \\frac{a_{n} \\vee a_{n-1}}{a_{n-1} \\vee a_{n-2}}=\\frac{a_{n}}{a_{n-1} \\vee a_{n-2}}+1=\\frac{a_{n}}{a_{n-2} \\vee a_{n-3}} \\\\\n& \\Leftrightarrow\\left(a_{n-1} \\vee a_{n-2}\\right) a_{n}=\\left(a_{n-2} \\vee a_{n-3}\\right)\\left(a_{n} \\vee a_{n-1}\\right) \\\\\n& \\Leftrightarrow\\left(a_{n-1} \\vee a_{n-2}\\right) a_{n}\\left(a_{n} \\wedge a_{n-1}\\right)=\\left(a_{n-2} \\vee a_{n-3}\\right) a_{n} a_{n-1},\n\\end{aligned}\n$$\non en déduit que ($E_{n}$) est bien vraie.\n\nIl s'ensuit qu'il existe un entier naturel $k$, minimal, tel que $k \\leqslant c_{4}+5$ et $a_{k} \\leqslant 0$. Si $k \\leqslant 5$, alors on a bien $k \\leqslant a_{3}+4$. Sinon, alors $0<a_{5}<a_{3} \\vee a_{4}$ et $0<a_{4}<a_{2} \\vee a_{3}$, donc\n$$\nc_{4}=\\frac{a_{5}}{a_{2} \\vee a_{3}}<\\frac{a_{3} \\vee a_{4}}{a_{2} \\vee a_{3}} \\leqslant \\frac{a_{3} a_{4}}{a_{2} \\vee a_{3}}<a_{3},\n$$\ndonc $k \\leqslant c_{4}+5 \\leqslant a_{3}+4$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeux cercles $\\Gamma_{1}$ et $\\Gamma_{2}$ de centres $O_{1}$ et $O_{2}$ se coupent en $P$ et $Q$. Une droite passant par $O_{1}$ coupe $\\Gamma_{2}$ en $A$ et $B$, et une droite passant par $O_{2}$ coupe $\\Gamma_{1}$ en $C$ et $D$. Montrer que s'il existe un cercle passant par $A, B, C$ et $D$, alors le centre de ce cercle est sur $(PQ)$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $\\Gamma$ le cercle passant par $A, B, C$ et $D$. D'après le théorème des axes radicaux, $(AB)$, $(CD)$ et $(PQ)$ sont concourantes en un point qu'on appelle $X$. De plus, $(AB)$ est perpendiculaire à $\\left(O_{1}O_{2}\\right)$, donc est la hauteur issue de $O_{1}$ dans $O_{1}O_{2}O$. De même, $(CD)$ est la hauteur issue de $O_{2}$, donc $X$ est l'orthocentre de $O_{1}O_{2}O$. Par conséquent, $(OX)$ est perpendiculaire à $\\left(O_{1}O_{2}\\right)$, mais on sait déjà que la perpendiculaire à $\\left(O_{1}O_{2}\\right)$ passant par $X$ est $(PQ)$, donc les droites $(OX)$ et $(PQ)$ sont confondues, et $O \\in (PQ)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $ABC$ un triangle, et $\\Gamma$ son cercle circonscrit. Soit $M$ le milieu de l'arc $BC$ ne contenant pas $A$. Un cercle $\\mathscr{C}$ est tangent à $[AB), [AC)$ en $D$ et $E$ respectivement, et tangent intérieurement à $\\Gamma$ en $F$. Montrer que $(DE), (BC)$ et $(FM)$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nSoit $I$ le centre du cercle inscrit dans $ABC$. La droite $(EF)$ recoupe $\\Gamma$ en un point $H$. Soit $J$ le point d'intersection de $(BC)$ avec $(FM)$.\n\nIl est facile de voir que $H$ est le milieu de l'arc $AC$ ne contenant pas $B$ : en effet, l'homothétie de centre $F$ qui envoie $\\mathscr{C}$ sur $\\Gamma$ envoie $(AC)$ sur la tangente en $H$ à $\\Gamma$ ; celle-ci est parallèle à $(AC)$, ce qui entraîne que $H$ est le milieu de l'arc, et par conséquent $B, I, H$ sont alignés.\n\nEn appliquant le théorème de Pascal à l'hexagone $AMFHBC$, on obtient que $I, E, J$ sont alignés. De même, $D, I, J$ sont alignés. Ainsi, $(DE), (BC)$ et $(FM)$ se rencontrent en $J$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn note $S(k)$ la somme des chiffres d'un nombre entier $k$. On dit qu'un entier $a$ est d'ordre $n$ s'il existe une suite d'entiers $a_{0}, a_{1}, \\ldots, a_{n}$ tels que $a_{n}=a$ et $a_{i+1}=a_{i}-S\\left(a_{i}\\right)$ pour tout $i=0,1, \\ldots, n-1$. Montrer que pour tout entier $n \\geqslant 1$ il existe un entier $b$ qui soit d'ordre $n$ mais qui ne soit pas d'ordre $n+1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn pose $F(x)=x-S(x)$, et on considère la suite définie par $u_{0}=1$ et $u_{i+1}=10^{u_{i}}$ pour $i \\geqslant 0$. Soit $n \\geqslant 1$ fixé, et considérons un entier $N>u_{n+1}$. Nous allons montrer que $X=F^{(n)}\\left(10^{N}-5 N\\right)$ (on itère $n$ fois $F$) est d'ordre $n$ mais pas $n+1$. Pour cela, il suffit de montrer que $X$ n'est pas d'ordre $n+1$.\n\nOn remarque pour commencer que $F(a)-F(b)$ est toujours divisible par $9$, et que si $a$ et $b$ ne différent que sur leurs $k$ derniers chiffres, alors $|a-b| \\leqslant 10^{k}$.\n\nÉtablissons maintenant une sorte de réciproque à cette remarque. Soient $a, b$ deux entiers et écrivons $a=\\sum_{i \\geqslant 0} c_{i} 10^{i}$ et $b=\\sum_{i \\geqslant 0} c_{i}' 10^{i}$. On a\n$$\nF(a)-F(b)=\\sum_{i \\geqslant 0}\\left(c_{i}-c_{i}'\\right)\\left(10^{i}-1\\right).\n$$\nSi $j$ est le plus grand indice tel que $c_{j}>c_{j}'$, alors\n$$\nF(a)-F(b) \\geqslant\\left(10^{j}-1\\right)+\\sum_{i=0}^{j-1}\\left(c_{i}-c_{i}'\\right)\\left(10^{i}-1\\right) \\geqslant\\left(10^{j}-1\\right)-9 \\sum_{i=0}^{j-1}\\left(10^{i}-1\\right)=9 j\n$$\nOn en déduit que si $|F(a)-F(b)|=9 k$, alors $j \\leqslant k$, et donc si $|F(a)-F(b)|=9 k$, alors $a$ et $b$ ne peuvent différer que sur leurs $k+1$ derniers chiffres. (6)\n\nMontrons maintenant le lemme suivant:\n\n**Lemme.** Si $F^{(n)}(a)=F^{(n)}(b)$ alors $a$ et $b$ ne peuvent différer que sur leurs $u_{n}$ derniers chiffres.\n\nRevenons maintenant à la preuve du lemme en raisonnant par récurrence sur $n$. L'initialisation pour $n=1$ provient de (6) appliqué avec $k=0$. Pour l'hérédité, soit $n \\geqslant 1$ tel que le lemme est vrai jusqu'au rang $n$. Soient $a, b$ deux entiers tels que $F^{(n+1)}(a)=F^{(n+1)}(b)$. Alors $F^{(n)}(F(a))=F^{(n)}(F(b))$, et donc $|F(a)-F(b)| \\leqslant 10^{u_{n}}$ par hypothèse de récurrence. Comme $|F(a)-F(b)|$ est divisible par $9$, écrivons $|F(a)-F(b)|=9 k$ (avec $9 k \\leqslant 10^{u_{n}}$). Alors d'après (6), $a$ et $b$ ne peuvent différer que sur leurs $k+1$ derniers chiffres. Comme $k+1 \\leqslant 10^{u_{n}} / 9+1 \\leqslant 10^{u_{n}}$, $a$ et $b$ ne peuvent différer que sur leurs $10^{u_{n}}=u_{n+1}$ derniers chiffres, ce qui conclut.\n\nRevenons à l'exercice. Raisonnons par l'absurde en supposant l'existence d'un entier $y$ tel que $X=F^{(n+1)}(y)=F^{(n)}\\left(10^{N}-5 N\\right)$. Alors\n$$\nF^{(n)}\\left(10^{N}-5 N\\right)=F^{(n)}(F(y))\n$$\net donc d'après le lemme\n$$\n\\left|10^{N}-5 N-F(y)\\right| \\leqslant 10^{u_{n}} \\leqslant N.\n$$\nDonc\n$$\n10^{N}-4 N \\leqslant F(y) \\leqslant 10^{N}-6 N\n$$\nOr on constate que si $y \\geqslant 10^{N}$, alors $F(y) \\geqslant 10^{N}-1$ (en effet, on vérifie aisément que $F$ est croissante et que $F\\left(10^{N}\\right)=10^{N}-1$). De plus, si $y \\leqslant 10^{N}-1$, alors $F(y) \\leqslant 10^{N}-9 N-1$; en effet si $y=\\sum_{i=0}^{N-1} c_{i} 10^{i}$, alors\n$$\nF(y)=\\sum_{i=0}^{N-1} c_{i}\\left(10^{i}-1\\right) \\leqslant \\sum_{i=0}^{N-1} 9\\left(10^{i}-1\\right)=10^{N}-9 N-1\n$$\nCeci contredit (7), et $X$ vérifie bien la condition cherchée.\nSolution:\n\nAutre solution, un peu plus courte : Comme ci-dessus, on pose $F(x)=x-S(x)$. Il vient immédiatement que $F(a+1)=F(a)$ si $a \\not \\equiv 9\\ (\\bmod\\ 10)$ et que $F(a+1)>F(a)$ si $a \\equiv 9\\ (\\bmod\\ 10)$. Par conséquent, tout entier $k$ a au plus $10$ antécédents par $F$, et a fortiori il a au plus $10^{n}$ antécédents par $F^{(n)}$, où $F^{(n)}$ est la $n^{\\text{ème}}$ itérée de $F$. Il s'ensuit que $F^{(n)}\\left(k+10^{n}\\right)>F^{(n)}(k)$ pour tout entier $k \\geqslant 1$ (en effet, si $F^{(n)}\\left(k+10^{n}\\right)=F^{(n)}(k)$ pour un certain entier $k \\geqslant 1$, on aurait $F^{(n)}(k)=F^{(n)}(i)$ pour tout $k \\leqslant i \\leqslant k+10^{n}$, de sorte que l'entier $F^{(n)}(k)$ aurait au moins $10^{n}+1$ antécédents qui sont $k, k+1, k+2, \\ldots, k+10^{n}$.\n\nMaintenant, soit $\\ell$ un entier tel que $9 \\ell \\geqslant 2 \\cdot 10^{n}$. On observe que $F\\left(10^{\\ell}\\right)=10^{\\ell}-1$ et que $F\\left(10^{\\ell}-1\\right)=10^{\\ell}-1-9 \\ell$, puis on pose $a_{k}=F^{(k)}\\left(10^{\\ell}-10^{n}-1\\right)$ et $b_{k}=F^{(k)}\\left(10^{\\ell}-1\\right)$, pour $k \\geqslant 0$. Puisque $b_{0} \\geqslant a_{0}+10^{n}$, on en déduit que $b_{n}>a_{n}$. De même, puisque $a_{0} \\geqslant b_{1}+10^{n}$, on en déduit que $a_{n}>b_{n+1}$. Cela montre que $b_{0}>i>b_{1}$ pour tout entier $i$ tel que $a_{n}=F^{(n)}(i)$.\n\nOr, on a $F(k) \\geqslant b_{0}$ pour tout $k \\geqslant 10^{\\ell}$ et $F(k) \\leqslant b_{1}$ pour tout $k \\leqslant 10^{\\ell}-1$, ce qui montre que les éléments de l'ensemble $\\{b_{1}+1, \\ldots, b_{0}-1\\}$ sont tous d'ordre $0$ mais pas d'ordre $1$. Il s'ensuit que $a_{n}$ est d'ordre $n$ mais pas d'ordre $n+1$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\in \\mathbb{N}^{*}$. On place des rectangles $1 \\times n$ et $n \\times 1$ sur un échiquier $2 n \\times 2 n$, de telle manière que deux rectangles ne s'intersectent jamais et que chaque rectangle recouvre exactement $n$ cases. Un ensemble de tels rectangles est dit saturé s'il est impossible d'ajouter un rectangle sans intersecter un rectangle déjà en place.\nTrouver le plus petit $k$ tel qu'il existe un ensemble saturé de $k$ rectangles.",
    "options": [],
    "answer": "k = 4 if n = 1; k = 6 if n = 2; k = 2n + 1 if n ≥ 3",
    "solution": "Solution:\n\nCommençons par le cas $n \\geqslant 3$ : la réponse est $k=2 n+1$. L'ensemble suivant (dans le cas $n=5$, la construction se généralisant facilement) convient :\n\n![](attached_image_1.png)\n\nIl reste à montrer qu'avec $2 n$ rectangles il est toujours possible d'en ajouter un qui ne les intersecte pas. Si les $2 n$ rectangles sont orientés dans le même sens, par exemple tous horizontaux, alors il doit y en avoir exactement un sur chaque ligne (sinon il y aurait une ligne vide), et il ne peut pas être tout au bout de la ligne (s'il était tout à gauche, on pourrait ajouter un rectangle à sa droite). Par conséquent, la première et la dernière colonne sont vides, et il est possible d'ajouter un rectangle vertical.\n\nS'il y a des rectangles dans les deux sens, on peut supposer qu'il y a au plus $n$ rectangles verticaux. Supposons que la configuration soit saturée. Soit $1 \\leqslant k \\leqslant n$ tel que la colonne $k$ contient un rectangle vertical noté $r$. En décalant $r$ d'une case vers la gauche, on doit intersecter un rectangle déjà en place, noté $r^{\\prime}$ (sinon on pourrait ajouter un rectangle). Cependant, si $r^{\\prime}$ était horizontal, il devrait intersecter $r$ car il n'y a pas assez de place pour un rectangle horizontal à gauche de $r$. Le rectangle $r^{\\prime}$ est donc vertical, donc la colonne $k-1$ contient un rectangle vertical et de même pour les colonnes $k-2, \\ldots, 2,1$.\n\nDe même, si $n+1 \\leqslant k \\leqslant 2n$ est tel que la colonne $k$ contient un rectangle vertical, alors toutes les colonnes à sa droite contiennent un rectangle vertical. L'ensemble des colonnes ne contenant pas de rectangle vertical est donc un ensemble d'au moins $n$ colonnes consécutives. Il y a donc un rectangle $n \\times 2n$ qui ne rencontre aucun des rectangles verticaux. Si la configuration est saturée, il faut donc $2n$ rectangles horizontaux pour \"bloquer\" chaque ligne dans ce rectangle $n \\times 2n$. Comme on a supposé qu'il y avait au moins un rectangle vertical, il y a donc au moins $2n+1$ rectangles, ce qui est absurde.\n\nPassons aux cas $n=1$ et $n=2$ : le cas $n=1$ est trivial et la réponse est $4$ (il faut tout couvrir, sinon on pourrait ajouter un carré $1$). Pour $n=2$, il n'y a pas d'ensemble saturé de $5$ rectangles.\n\nPour le montrer, on peut soit effectuer une disjonction de cas un peu fastidieuse, soit adapter la preuve précédente : si tous les rectangles ont la même direction, ce n'est pas possible (une seule ligne contient $2$ rectangles donc un seul rectangle peut intersecter la première colonne donc il reste la place pour un domino vertical dans la première colonne). Sinon, on peut supposer qu'il y a au plus $2$ dominos verticaux. La preuve pour $n \\geqslant 3$ montre qu'il y a deux colonnes consécutives libres, donc au moins $4$ dominos horizontaux, donc un seul vertical. De plus, les colonnes sans domino vertical sont consécutives donc on peut supposer que le domino vertical est dans la colonne $1$. Il doit être au milieu de la colonne pour ne pas laisser une place, et il est alors facile de voir qu'il est impossible de \"boucher tous les trous\" avec $4$ dominos horizontaux.\n\nEnfin, la configuration suivante marche avec $k=6$ :\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $D, E, F$ les points de tangence du cercle inscrit sur les côtés $(BC)$, $(AC)$ et $(AB)$, $M$ le milieu de $[BC]$, $I$ le centre du cercle inscrit de $ABC$. $G$ et $H$ sont définis comme les symétriques de $E$ et $F$ par rapport à $I$. On note $Q$ l'intersection entre les droites $(BC)$ et $(GH)$. Montrer que les droites $(IQ)$ et $(IM)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSoit $\\omega$ le cercle inscrit de $ABC$.\nOn notera pour tout point $X$ du plan $p(X)$ la polaire de $X$ par rapport à $\\omega$. On sait que pour tout point $X$ du plan, $p(X)$ est une droite perpendiculaire à $(IX)$. Nous allons donc montrer que $p(Q)$ et $(IM)$ sont parallèles pour en déduire le résultat.\nOn sait que $p(D) = (BC)$, et que $Q \\in (BC)$ donc $Q \\in p(D)$. Ainsi par propriété des polaires (par rapport à un cercle fixé), on en déduit que $D \\in p(Q)$. De plus par construction de la polaire, on sait que $A'$ le point d'intersection des tangentes à $\\omega$ en $G$ et $H$ est aussi sur la polaire de $Q$. Ainsi on a : $p(Q) = (DA')$. Par symétrie de la construction, on sait que $A$ et $A'$ sont symétriques par rapport à $I$.\nSoit maintenant $D'$ le symétrique de $D$ par rapport à $I$ (qui est donc sur $\\omega$) et $D''$ le point de $(BC)$ tel que $M$ soit le milieu de $[DD'']$. Une homothétie envoyant une droite sur une droite qui lui est parallèle, on sait que les droites $(DA')$ et $(D'A)$ sont parallèles, ainsi que les droites $(IM)$ et $(D'D'')$. Or il est connu que les points $A, D', D''$ sont alignés (se montre par une homothétie de centre $A$).\nAinsi les droites $(DA')$ et $(IM)$ sont parallèles, et $(DA')$ et $(QI)$ étant perpendiculaires, on en déduit que $(IQ)$ et $(IM)$ sont bien perpendiculaires.\n\n\nAutre solution. Notons $\\alpha, \\beta$ et $\\gamma$ les angles de $DEF$ et $R$ le rayon du cercle circonscrit à $DEF$. On a $BD = R \\tan \\beta$, $CD = R \\tan \\gamma$ donc\n$$\n2DM = BC - 2BD = R(\\tan \\gamma - \\tan \\beta) = R \\frac{\\sin \\gamma \\cos \\beta - \\sin \\beta \\cos \\gamma}{\\cos \\beta \\cos \\gamma} = R \\frac{\\sin (\\gamma - \\beta)}{\\cos \\beta \\cos \\gamma}.\n$$\nD'autre part, la loi des sinus dans $QDH$ donne $DQ = DH \\frac{\\cos \\gamma}{\\sin (\\beta - \\gamma)} = 2R \\frac{\\cos \\beta \\cos \\gamma}{\\sin (\\beta - \\gamma)}$ donc $QD \\cdot DM = R^2 = DI^2$. Or, $[ID]$ est la hauteur de $IQM$ issue de $I$, donc $IQM$ est rectangle en $I$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn ensemble $E$ fini et non vide de réels strictement positifs est dit puissant lorsque, pour tous $a, b \\in E$ distincts, l'un au moins des nombres $a^{b}$ et $b^{a}$ appartient aussi à $E$. Déterminer le nombre maximal d'éléments que peut contenir un ensemble puissant.",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nTout d'abord, l'ensemble\n$$\n\\left\\{1, \\frac{1}{2}, \\frac{1}{4}, \\frac{1}{16}\\right\\}\n$$\nest un ensemble puissant qui contient quatre éléments. Soit $S$ un ensemble puissant, et $n$ son cardinal : nous allons montrer que $n \\leqslant 4$.\n\nDémontrons d'abord le lemme suivant : il n'existe pas d'éléments $a$ et $b$ de $S$ tels que $a<1<b$. En effet, si l'on avait $a<1<b$, alors $a^{b}<a<1<b^{a}<b$; en prenant $a=\\min S$ et $b=\\min \\{s \\in S: 1<s\\}$, on s'assurerait donc que ni $a^{b}$ ni $b^{a}$ n'appartient à $S$, contredisant le fait que $S$ est puissant.\n\nD'autre part, si $S$ est un ensemble puissant, alors $S \\cup\\{1\\}$ l'est aussi. Notre recherche d'un ensemble puissant de cardinal maximal nous amène donc à considérer deux cas.\n\n1. Si $S=\\left\\{a_{1}, a_{2}, \\cdots, a_{n}\\right\\}$, avec $n \\geqslant 5$ et $1=a_{1}<a_{2}<\\cdots<a_{n}$, alors on observe que\n$$\na_{n}<a_{n}^{a_{2}}<a_{n}^{a_{3}}<\\ldots<a_{n}^{a_{n-1}}\n$$\ndonc que $S$ contient les éléments\n$$\na_{2}<a_{2}^{a_{n}}<a_{3}^{a_{n}}<\\ldots<a_{n-1}^{a_{n}} .\n$$\nCela montre que $a_{i+1}=a_{i}^{a_{n}}$ pour tout $i \\in\\{2, \\ldots, n-1\\}$. On remarque cependant que\n$$\na_{2}<a_{2}^{a_{3}}<a_{2}^{a_{n}}=a_{3}<a_{3}^{a_{2}}<a_{3}^{a_{n}}=a_{4}\n$$\nce qui montre que ni $a_{2}^{a_{3}}$ ni $a_{3}^{a_{2}}$ n'appartient à $S$, qui ne peut donc pas être puissant.\n\n2. Si $S=\\left\\{a_{1}, a_{2}, \\cdots, a_{n}\\right\\}$, avec $n \\geqslant 5$ et $a_{1}<a_{2}<\\cdots<a_{n}=1$, alors on observe que\n$$\na_{n-1}<a_{n-1}^{a_{n-2}}<a_{n-1}^{a_{n-3}}<\\ldots<a_{n-1}^{a_{1}}<a_{n}\n$$\ndonc que $S$ contient les éléments\n$$\na_{1}<a_{1}^{a_{n-1}}<a_{2}^{a_{n-1}}<\\ldots<a_{n-2}^{a_{n-1}}<a_{n}=1\n$$\nCela montre que $a_{i+1}=a_{i}^{a_{n-1}}$ pour tout $i \\in\\{1, \\ldots, n-2\\}$. De même, on observe que\n$$\na_{n-1}=a_{n-2}^{a_{n-1}}<a_{n-2}^{a_{n-3}}<\\ldots<a_{n-2}^{a_{1}}<a_{n}=1\n$$\ndonc que $S$ contient les éléments\n$$\na_{2}=a_{1}^{a_{n-1}}<a_{1}^{a_{n-2}}<a_{2}^{a_{n-2}}<\\ldots<a_{n-3}^{a_{n-2}}<a_{n}=1 .\n$$\nCela montre aussi que $a_{i+2}=a_{i}^{a_{n-2}}=a_{i}^{a_{n-1}^{2}}$ pour tout $i \\in\\{1, \\ldots, n-3\\}$.\n\nOn en déduit que $a_{n-1}=a_{n-2}^{a_{n-1}}=a_{n-1}^{2 a_{n-1}}$ puis que $1=2 a_{n-1}$, donc que $a_{n-i}=2^{-2^{i}}$ pour $i \\in\\{1, \\ldots, n-1\\}$. Cependant, puisque $n \\geqslant 5$, alors $a_{n-4}^{a_{n-3}}=2^{-2^{-4}}$ et $a_{n-3}^{a_{n-4}}=2^{-2^{-13}}$, donc $S$ ne contient ni $a_{n-4}^{a_{n-3}}$ ni $a_{n-3}^{a_{n-4}}$ et ne peut pas être puissant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe. On suppose que les cercles inscrits aux triangles $ABC$, $BCD$, $CDA$ et $DAB$ ont un point commun. Montrer que $ABCD$ est un losange.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $\\omega_A$ le cercle inscrit à $DAB$ et ainsi de suite : les cercles $\\omega_B$ et $\\omega_D$ sont situés de part et d'autre de $(AC)$, donc ils ne peuvent s'intersecter que sur $(BC)$. De même, $\\omega_A$ et $\\omega_C$ ne peuvent s'intersecter que sur $(BD)$, donc l'intersection des 4 cercles ne peut être que l'intersection des diagonales, qu'on note $X$.\n\nDe plus, notons $R$ et $S$ les points de tangence du cercle inscrit à $ABC$ avec $[AB]$ et $[BC]$, et $T$ et $U$ les points de contact du cercle inscrit à $CDA$ avec $[CD]$ et $[DA]$. On a :\n\n$$\nAB + CD = AR + BR + CT + DT = AX + BS + CX + DU = AU + BS + CS + DU = AD + BC\n$$\n\ndonc $ABCD$ est circonscriptible : il existe un cercle à l'intérieur de $ABCD$ tangent à tous les côtés, qu'on note $\\omega$.\n\nLe centre de l'homothétie négative qui envoie $\\omega_A$ sur $\\omega_C$ est $X$, donc est sur $(AC)$. De plus, $A$ est le centre de l'homothétie positive qui envoie $\\omega_A$ sur $\\omega$, et $C$ est le centre de l'homothétie positive qui envoie $\\omega$ sur $\\omega_C$, donc le centre de l'homothétie positive qui envoie $\\omega_A$ sur $\\omega_C$ est aussi sur $(AC)$, donc les centres de $\\omega_A$ et $\\omega_C$ sont sur $(AC)$, donc $(AC)$ est la bissectrice de $\\widehat{BCD}$ et $\\widehat{DAB}$, donc le quadrilatère est symétrique par rapport à $(AC)$ d'où $AB = AD$ et $CB = CD$. On a de même $BA = BC$ et $DA = DC$, donc $ABCD$ est un losange.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $n \\geqslant 2$ et $a_{1}, \\ldots, a_{n}, b_{1}, \\ldots, b_{n}$ des réels strictement positifs. Les $a_{i}$ sont supposés deux à deux distincts, de même que les $b_{i}$. On suppose qu'il est possible de séparer les $a_{i}$ en deux sous-ensembles de même somme, et de même pour les $b_{i}$.\n\nMontrer qu'il existe un $2n$-gone simple ayant exactement $n$ côtés horizontaux et $n$ côtés verticaux, dont les longueurs des côtés horizontaux sont exactement les $a_{i}$ et dont les longueurs des côtés verticaux sont exactement les $b_{i}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nQuitte à changer les indices et éventuellement à échanger les rôles de $a$ et $b$, on peut supposer qu'il existe $\\ell \\leqslant m \\leqslant \\frac{n}{2}$ tels que $a_{1}+\\cdots+a_{\\ell}=a_{\\ell+1}+\\cdots+a_{n}=S_{a}$ et $b_{1}+\\cdots+b_{m}=b_{m+1}+\\cdots+b_{n}$, que $b_{1} \\geqslant \\cdots \\geqslant b_{m}$ et que $b_{m+1} \\leqslant \\cdots \\leqslant b_{n}$. Quand on fait le tour de notre polygone dans le sens horaire, les $a_{i}$ avec $i \\leqslant \\ell$ correspondront aux longueurs des côtés où on va vers la droite, les $a_{i}$ avec $i \\geqslant \\ell+1$ à ceux où on va vers la gauche, les $b_{i}$ avec $i \\leqslant m$ à ceux où on va vers le haut et les $b_{i}$ avec $i \\geqslant m+1$ à ceux où on va vers le bas. Il reste à placer ces côtés correctement les uns par rapport aux autres, de manière à ce qu'ils ne s'intersectent pas.\n\nOn note $b_{i}^{\\prime}=b_{i}$ pour $i \\leqslant m$ et $b_{i}^{\\prime}=-b_{i}$ pour $i \\geqslant m+1$, de manière à avoir $\\sum_{i=1}^{n} b_{i}^{\\prime}=0$ (ce changement de notation permettra de mesurer la variation d'ordonnée quand on passe d'un sommet du polygone à un autre). Les $b_{i}^{\\prime}$ sont alors rangés dans l'ordre décroissant.\n\nLemme. Il existe un sous-ensemble $I \\subset \\llbracket 2, n-1 \\rrbracket$ de cardinal $\\ell-1$ tel que $b_{1}^{\\prime}+\\sum_{i \\in I} b_{i}^{\\prime}>0$ et $b_{n}^{\\prime}+\\sum_{i \\in I} b_{i}^{\\prime}<0$.\n\nPreuve du lemme. L'énoncé du lemme revient à trouver $I$ de cardinal $\\ell-1$ avec $\\sum_{i \\in I} b_{i}^{\\prime} \\in ]-b_{1}^{\\prime},-b_{n}^{\\prime}[$. Soient $I_{\\max }=\\{2, \\ldots, \\ell\\}$ et $I_{\\min }=\\{n-\\ell+1, \\ldots, n-1\\}$. Il est clair que $\\sum_{i \\in I_{\\max }} b_{i}^{\\prime}>-b_{1}^{\\prime}$ (si on a $n$ nombres de somme nulle, la somme des $\\ell$ plus grands est positive) et que $\\sum_{i \\in I_{\\min }} b_{i}^{\\prime}<-b_{n}^{\\prime}$.\n\nSi on remplace un par un les éléments de $I_{\\min }$ par ceux de $I_{\\max }$, la somme augmente à chaque étape d'un nombre de la forme $b_{i}^{\\prime}-b_{j}^{\\prime}<b_{1}^{\\prime}-b_{n}^{\\prime}$ (l'inégalité est stricte car les $b_{i}$ sont deux à deux distincts). On passe donc d'un nombre inférieur à $-b_{n}^{\\prime}$ à un nombre supérieur à $-b_{1}^{\\prime}$ en augmentant à chaque étape d'un nombre strictement inférieur à $b_{1}^{\\prime}-b_{n}^{\\prime}$ donc à un moment la somme se trouvera strictement entre $-b_{1}^{\\prime}$ et $-b_{n}^{\\prime}$.\n\nOn veut construire notre polygone de la manière suivante : $b_{1}$ sera la longueur du côté vertical le plus à gauche (on note $A$ le sommet en bas et $B$ le sommet en haut de ce côté) et $b_{n}$ celle du côté vertical le plus à droite (on nomme $C$ le sommet en haut et $D$ le sommet en bas de ce côté), et l'écart horizontal entre ces deux côtés vaut $S_{a}$. Pour passer de $B$ à $C$, on utilise les côtés horizontaux de longueur $a_{i}$ avec $1 \\leqslant i \\leqslant \\ell$ entre lesquels on intercale les $\\ell-1$ côtés de longueurs $b_{i}$ avec $i \\in I$, en commençant par ceux tels que $i \\leqslant m$. On obtient ainsi une ligne brisée (en bleu) qui relie le haut de $b_{1}$ au haut de $b_{n}$ en alternant les déplacements vers la droite et vers le haut, puis les déplacements vers la droite et vers le bas.\n\nPuis, pour relier $D$ à $A$, on utilise les $n-\\ell$ côtés horizontaux de longueur $a_{i}$ avec $\\ell+1 \\leqslant i \\leqslant n$, entre lesquels on intercale les $b_{i}$ restants, cette fois en commençant par les $i \\geqslant m+1$ pour descendre avant de remonter (cela correspond à la ligne verte sur la figure).\n\n![](attached_image_1.png)\n\nLe polygone ainsi construit a bien les longueurs de côtés voulues. Il reste à vérifier que ces côtés ne s'intersectent pas. Les deux côtés rouges n'en coupent aucun autre car par construction, ils sont respectivement plus à gauche et plus à droite que tous les autres. De plus, deux côtés bleus ne peuvent pas s'intersecter car il y en a toujours un plus à gauche que l'autre, et de même pour deux côtés verts.\n\nIl suffit donc de vérifier qu'un côté bleu ne peut pas intersecter un sommet vert. On va montrer qu'en fait, la ligne bleue est entièrement située au-dessus de la ligne verte. Comme la ligne bleue monte puis redescend, son ordonnée est partout supérieure à $\\min \\{y_{B}, y_{C}\\}$. De même, l'ordonnée sur la ligne verte est partout inférieure à $\\max \\{y_{A}, y_{D}\\}$. Il suffit donc de montrer que $\\min \\{y_{B}, y_{C}\\}>\\max \\{y_{A}, y_{D}\\}$.\n\nComme il est évident que $y_{B}>y_{A}$ et $y_{C}>y_{D}$, il suffit de montrer $y_{C}>y_{A}$ et $y_{B}>y_{D}$. Or, $y_{C}-y_{A}=b_{1}+\\sum_{i \\in I} b_{i}^{\\prime}$, qui est strictement positive d'après le lemme. De même, $y_{D}-y_{B}=\\sum_{i \\in I} b_{i}^{\\prime}-b_{n}^{\\prime}$ qui est strictement négative d'après le lemme. On peut donc tracer une ligne horizontale (en orange sur la figure) telle que la ligne bleue soit strictement au-dessus et la verte strictement en-dessous.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30504,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle non-isocèle en $A$. On note $D, E, F$ les points de tangence du cercle inscrit sur les côtés $(BC)$, $(AC)$ et $(AB)$, $I$ le centre du cercle inscrit de $ABC$. Soient $P$ et $Q$ les intersections de la droite $(EF)$ avec le cercle $\\Omega$ circonscrit à $ABC$. Soient enfin $O_{1}$ et $O_{2}$ les centres des cercles circonscrits à $AIB$ et $AIC$. Montrer que le centre du cercle circonscrit à $DPQ$ se situe sur la droite $\\left(O_{1}O_{2}\\right)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn peut déjà remarquer que $O_{1}$ et $O_{2}$ sont les pôles sud de $C$ et de $B$ respectivement dans le triangle $ABC$. La droite $\\left(O_{1}O_{2}\\right)$ est la médiatrice de $[AI]$. Or les droites $(AI)$ et $(EF)$ sont perpendiculaires. On en déduit alors que les droites $(PQ)$ et $\\left(O_{1}O_{2}\\right)$ sont parallèles. Comme $[PQ]$ et $\\left[O_{1}O_{2}\\right]$ sont aussi des cordes de $\\Omega$, ces segments ont la même médiatrice. En particulier l'intersection entre la médiatrice de $[PQ]$ et $\\left(O_{1}O_{2}\\right)$ est le milieu de $\\left[O_{1}O_{2}\\right]$, disons $T$.\n\nAinsi si l'énoncé est vrai, alors le centre du cercle circonscrit à $DPQ$ doit être $T$.\n\nDans cette situation, on contrôle alors assez mal les deux autres médiatrices du triangle $PQD$. On peut alors chercher un quatrième point appartenant au cercle. En particulier, définir un point comme seconde intersection entre un cercle et une droite peut souvent s'avérer utile.\n\nIci on peut s'intéresser à la seconde intersection de notre cercle avec la droite $(BC)$ : c'est en fait le milieu de $[BC]$. En effet, si on définit $X$ comme l'intersection entre les droites $(PQ)$ et $(BC)$, alors on sait que $XP \\times XQ = XB \\times XC$ et, comme les droites $(AD), (BE), (CF)$ sont concourantes (en le point de Gergonne du triangle $ABC$), la division $(X, D, B, C)$ est harmonique et donc on peut en déduire d'après la relation de Mac Laurin que : $XB \\times XC = XD \\times XM$. On trouve donc $XP \\times XQ = XD \\times XM$, ce qui implique que les points $P, Q, D, M$ sont cocycliques. Il suffit maintenant de montrer que la médiatrice de $[DM]$ passe par $T$. Pour cela, nous allons montrer que le projeté orthogonal de $T$ sur $(BC)$ est le milieu de $[DM]$.\n\nOn associe à présent à chaque point $Z$ du plan le vecteur $\\overrightarrow{OZ}$ qu'il forme avec une origine $O$ quelconque du plan (fixée). On notera $Z$ ce vecteur pour ne pas alourdir les notations.\n\nSoit $N_{A}$ le pôle nord de $A$ dans $ABC$. Par définition, on sait que : $T = \\frac{O_{1} + O_{2}}{2}$. De plus, il est connu que $O_{1}$ est le milieu de $[II_{C}]$ (centre du cercle exinscrit de $C$) et de même $O_{2}$ est le milieu de $[II_{B}]$. Ainsi on trouve : $T = \\frac{I + \\frac{I_{B} + I_{C}}{2}}{2}$. Or on sait aussi que $N_{A}$ est le milieu de $[I_{B}I_{C}]$ et donc on trouve que $T$ est le milieu de $[IN_{A}]$. Le projeté orthogonal de $T$ sur $(BC)$ est le milieu des projetés orthogonaux de $I$ et $N_{A}$ sur $(BC)$, à savoir $D$ et $M$ respectivement.\n\nAinsi on en déduit que le centre du cercle circonscrit à $DPQ$ est $T$, qui est bien sur la droite $\\left(O_{1}O_{2}\\right)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeux cercles $\\omega_{1}$ et $\\omega_{2}$ sont tangents en $S$, avec $\\omega_{1}$ à l'intérieur de $\\omega_{2}$. On note $O$ le centre de $\\omega_{1}$. Une corde $[AB]$ de $\\omega_{2}$ est tangente à $\\omega_{1}$ en $T$. Montrer que $(AO)$, la perpendiculaire à $(AB)$ passant par $B$ et la perpendiculaire à $(ST)$ passant par $S$ sont concourantes.\n\n![](attached_image_1.png)\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $(Bx)$ la perpendiculaire à $(AB)$ passant par $B$, et $(Sy)$ la perpendiculaire à $(ST)$ passant par $S$. Pour obtenir le résultat grâce au théorème de Ceva trigonométrique dans le triangle $ABS$, on doit montrer :\n\n$$\n\\frac{\\sin \\widehat{BAO}}{\\sin \\widehat{SAO}} \\cdot \\frac{\\sin \\widehat{ASy}}{\\sin \\widehat{BSy}} \\cdot \\frac{\\sin \\widehat{SBx}}{\\sin \\widehat{ABx}}=1\n$$\n\nOr, on sait que $\\sin ABx=1$. De plus, l'homothétie de centre $S$ qui envoie $\\omega_{1}$ sur $\\omega_{2}$ envoie $T$ sur le milieu de l'arc $\\widehat{AB}$, donc $(ST)$ est la bissectrice intérieure de $\\widehat{ASB}$ et $(Sy)$ est sa bissectrice extérieure, $d'$, où $\\sin \\widehat{ASy}=\\sin \\widehat{BSy}$. Il reste donc à montrer :\n\n$$\n\\frac{\\sin \\widehat{BAO}}{\\sin \\widehat{SAO}} \\cdot \\sin \\widehat{SBx}=1\n$$\n\nOr, $\\sin \\widehat{BAO}=\\frac{OT}{AO}=\\frac{OS}{AO}=\\frac{\\sin \\widehat{OAS}}{\\sin \\widehat{ASO}}$, donc il ne reste plus qu'à montrer $\\widehat{ASO}=\\widehat{SBx}$. Si on note $O'$ le centre de $\\omega_{2}$, alors :\n\n$$\n\\widehat{ASO}=\\widehat{ASO'}=\\frac{1}{2}\\left(\\pi-\\widehat{AO'S}\\right)=\\frac{\\pi}{2}-\\widehat{SBA}=\\widehat{SBx}\n$$\n\nd'où le résultat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProuver que, pour tout entier $n \\geq 2$, on a :\n$$\n\\sum_{k=2}^{n} \\frac{1}{\\sqrt[k]{(2 k)!}} \\geq \\frac{n-1}{2 n+2}\n$$\nN.B. Si $a>0$, on note $\\sqrt[k]{a}$ l'unique nombre réel $b>0$ tel que $b^{k}=a$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn raisonne par récurrence sur $n \\geq 2$.\n\nPour $n=2$, on a bien $\\frac{1}{\\sqrt{24}} > \\frac{1}{6}$.\n\nSupposons que l'inégalité désirée soit vraie pour la valeur $n-1 \\geq 2$. Pour la valeur $n$, le membre de droite augmente de\n$$\n\\frac{n-1}{2 n+2} - \\frac{n-2}{2 n} = \\frac{1}{n(n+1)}.\n$$\nD'après l'hypothèse de récurrence, il suffit donc de prouver que\n$$\n\\frac{1}{\\sqrt[n]{(2 n)!}} \\geq \\frac{1}{n(n+1)}.\n$$\nOr, pour $k=1,2, \\cdots, n$, on a $(n-k)(n-k+1) \\geq 0$, donc $0 \\leq k(2 n-k+1) \\leq n(n+1)$.\n\nEn multipliant ces inégalités membre à membre, il vient $(2 n)! \\leq (n(n+1))^{n}$, d'où la conclusion.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geq 5$ un entier, et $E_{1}, E_{2}, \\ldots, E_{2 n-1}$ des parties distinctes à deux éléments de $\\{1,2, \\ldots, n\\}$.\nProuver que, parmi les $2 n-1$ parties $E_{i}$, on peut en choisir $n$ de sorte que la réunion de ces $n$ parties ne contienne pas plus de $\\frac{2}{3} n+1$ éléments.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn va prouver par récurrence sur $k \\leq \\frac{2 n-1}{3}$ que l'on peut toujours éliminer $3 k$ des $2 n-1$ parties de sorte que la réunion des $2 n-1-3 k$ restantes ne contienne pas plus de $n-k$ éléments.\n\nLe cas $k=0$ est immédiat.\n\nSoit maintenant $1 \\leq k \\leq \\frac{2 n-1}{3}$. On suppose que l'on a éliminé $3(k-1)$ des $E_{i}$ de sorte que la réunion $U_{k-1}$ des $2 n-1-3(k-1)$ autres ne contienne pas plus de $n-k+1$ éléments.\n\nChacun des $E_{i}$ contient deux éléments. De $2(2 n-1-3(k-1))<4(n-k+1)$, on déduit alors qu'il existe un élément $x$ de $U_{k-1}$ qui n'appartient qu'à au plus trois de $E_{i}$ qui forment $U_{k-1}$. Ainsi, en éliminant encore trois des $E_{i}$ non déjà éliminés, dont tous ceux qui contiennent $x$, cela assure que la réunion des $n-3 k$ restants ne contient pas plus de $n-k$ éléments.\n\nCela achève la récurrence.\n\nPour $k=\\left\\lfloor\\frac{n-1}{3}\\right\\rfloor$, on a $n-\\left\\lfloor\\frac{n-1}{3}\\right\\rfloor \\leq n-\\frac{n-1}{3}=\\frac{2}{3} n+1$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30508,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle non rectangle tel que $AB < AC$. On note $H$ le projeté de $A$ sur $(BC)$, et $E, F$ les projetés respectifs de $H$ sur $(AB)$ et $(AC)$. La droite $(EF)$ coupe $(BC)$ au point $D$. On considère le demi-cercle de diamètre $[CD]$ situé dans le même demi-plan délimité par $(CD)$ que $A$. Soit $K$ le point de ce demi-cercle qui se projette sur $B$. Montrer que $(DK)$ est tangente au cercle $KEF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nComme $AEHF$ sont cocycliques (sur le cercle de diamètre $[AH]$), on a $(EF, EA) = (HF, HA) = (HF, CA) + (CA, CB) + (CB, HA) = (CA, CB) = (CF, CB)$, donc $(EF, EB) = (CF, CB)$. Par conséquent, $B, E, F, C$ sont cocycliques.\n\nD'après la puissance d'un point par rapport à un cercle, on a $DE \\cdot DF = DB \\cdot DC$.\n\nD'autre part, comme $DKC$ est rectangle en $K$, les triangles $DKB$ et $DCK$ sont semblables, ce qui implique $\\frac{DK}{DB} = \\frac{DC}{DK}$, ou encore $DK^{2} = DB \\cdot DC$.\n\nIl vient $DK^{2} = DE \\cdot DF$, et donc d'après la puissance de $D$ par rapport au cercle $KEF$, on en déduit que $(DK)$ est tangente en $K$ au cercle $KEF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. On note $P$ le symétrique de $B$ par rapport à $(AC)$ et $Q$ le symétrique de $C$ par rapport à $(AB)$.\nSoit $T$ l'intersection entre $(PQ)$ et la tangente en $A$ au cercle circonscrit à $(APQ)$.\nMontrer que le symétrique de $T$ par rapport à $A$ appartient à $(BC)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSoient $B'$ et $C'$ les symétriques de $B$ et $C$ par rapport à $A$. Comme le triangle $AB'P$ est isocèle en $A$, on a $\\widehat{AB'P} = \\frac{1}{2}\\left(180^{\\circ} - \\widehat{B'AP}\\right) = \\frac{1}{2} \\widehat{PAB} = \\widehat{BAC}$, donc $(B'P) \\parallel (AC)$.\n\nDe même, $(C'Q) \\parallel (AB)$, donc les droites $(B'P)$ et $(C'Q)$ sont sécantes. Notons $A'$ leur point d'intersection.\n\nLes triangles $A'B'C'$ et $ACB$ sont semblables car leurs côtés sont deux à deux parallèles.\n\nDe plus, $AB'P$ et $AC'Q$ sont semblables car $\\widehat{AB'P} = \\widehat{AC'Q}$ et car ces deux triangles sont isocèles en $A$.\n\nMontrons maintenant l'assertion demandée. Par symétrie par rapport à $A$, il revient au même de montrer que $T, B', C'$ sont alignés. En appliquant le théorème de Ménélaüs dans $A'PQ$, cela équivaut à\n$$\n\\frac{B'P}{B'A'} \\times \\frac{C'A'}{C'Q} \\times \\frac{TQ}{TP} = 1 \\quad (E)\n$$\nComme $TAQ$ et $TPA$ sont semblables, on a $\\frac{TA}{TP} = \\frac{TQ}{TA} = \\frac{AQ}{AP}$, donc $\\frac{TQ}{TP} = \\frac{AQ^2}{AP^2} = \\frac{AC^2}{AB^2}$.\n\nComme $A'B'C'$ et $ACB$ sont semblables, on a $\\frac{C'A'}{B'A'} = \\frac{AB}{AC}$.\n\nComme $AB'P$ et $AC'Q$ sont semblables, on a $\\frac{B'P}{C'Q} = \\frac{AP}{AQ} = \\frac{AB}{AC}$.\n\nOn en déduit que $(E)$ est vraie.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $m \\geqslant 1$ et $n \\geqslant 1$ tels que $\\frac{5^{m}+2^{n+1}}{5^{m}-2^{n+1}}$ soit le carré d'un entier.",
    "options": [],
    "answer": "m = 1, n = 1",
    "solution": "Solution:\n\nLa démonstration qui suit est valable pour $m, n \\in \\mathbb{N}$.\n\nDéjà, $5^{m}-2^{n+1}$ doit diviser $5^{m}+2^{n+1}$, donc divise $5^{m}+2^{n+1}-\\left(5^{m}-2^{n+1}\\right)=2^{n+2}$, par conséquent c'est une puissance de 2. Or, $5^{m}-2^{n+1}$ est impair, donc $5^{m}-2^{n+1}=1$.\n\nÉcrivons $5^{m}+2^{n+1}=a^{2}$. On a donc $(a-1)(a+1)=a^{2}-1=5^{m}+2^{n+1}-5^{m}+2^{n+1}=2^{n+2}$, donc $a-1$ et $a+1$ sont des puissances de 2.\n\nÉcrivons $a-1=2^{c}$ et $a+1=2^{d}$ avec $c+d=n+2$. Alors $c<d$ donc $a-1=2^{c}$ divise $2^{d}-2^{c}=(a+1)-(a-1)=2$, donc $a-1=1$ ou $a-1=2$.\n\nSi $a=2$ alors $2^{d}=a+1=3$, ce qui est impossible.\n\nOn en déduit que $a=3, c=1, d=2$ et $n+2=3$, ce qui donne $n=1$ et $5^{m}=1+2^{n+1}=5$, puis $m=1$.\n\nFinalement, $m=n=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère 7 îles $A_{1}, \\ldots, A_{7}$. On est autorisé à construire des ponts, soit entre une île $A_{i}$ et l'île suivante $A_{i+1}$ (pour $i \\in \\{1,2, \\ldots, 6\\}$ ), soit entre une île $A_{i}$ et la dernière $A_{7}$ (pour $i \\in \\{1,2, \\ldots, 6\\}$ ). De combien de manières peut-on réaliser ces constructions avec le moins de ponts possibles de sorte que l'on puisse se rendre d'une île vers n'importe quelle autre ?\n\nExemple pour 3 îles au lieu de 7 : les trois constructions possibles utilisant deux ponts sont\n1) un pont entre $A_{1}$ et $A_{2}$, et un pont entre $A_{1}$ et $A_{3}$\n2) un pont entre $A_{1}$ et $A_{2}$, et un pont entre $A_{2}$ et $A_{3}$\n3) un pont entre $A_{1}$ et $A_{3}$, et un pont entre $A_{2}$ et $A_{3}$.",
    "options": [],
    "answer": "144",
    "solution": "Solution:\n\nOn dira qu'une configuration est bonne si elle satisfait les conditions de l'énoncé.\nNotons $a_{n}$ le nombre de bonnes configurations avec $n$ îles. On a $a_{1}=a_{2}=1$ et $a_{3}=3$.\n\nPartant d'une bonne configuration avec $n$ îles telle que $A_{n-1}$ et $A_{n}$ ne soient pas reliées, alors $A_{n-1}$ et $A_{n-2}$ sont nécessairement reliées, donc si on supprime l'île $A_{n-1}$ on obtient une bonne configuration avec $n-1$ îles. Réciproquement, toute bonne configuration avec $n-1$ îles provient d'une et une seule bonne configuration avec $n$ îles dont les deux dernières ne sont pas reliées : il suffit en effet d'intercaler une île entre les deux dernières et de mettre un pont entre celle-ci et la $n$ - 2-ième.\nOn en déduit qu'il y a $a_{n-1}$ bonnes configurations avec $n$ îles telles que $A_{n-1}$ et $A_{n}$ ne soient pas reliées.\n\nD'autre part, partant d'une bonne configuration avec des ponts entre $A_{n}$ et $A_{n-1}, A_{n-1}$ et $A_{n-2}, \\ldots$, $A_{n-k+1}$ et $A_{n-k}$ ($k \\geqslant 1$) mais pas entre $A_{n-k}$ et $A_{n_{k}-1}$, si on supprime les îles $A_{n-1}, \\ldots, A_{n-k}$ alors on obtient une bonne configuration avec $n-k$ îles. On en déduit comme ci-dessus qu'il y a $a_{n-1}+a_{n-2}+\\cdots+a_{1}$ bonnes configurations comportant un pont entre $A_{n}$ et $A_{n-1}$.\n\nOn voit donc que $a_{n}=2 a_{n-1}+a_{n-2}+\\cdots+a_{1}$.\nOn en déduit de proche en proche que $a_{4}=8, a_{5}=21, a_{6}=55$ et $a_{7}=144$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30512,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers $a>0$ pour lesquels il existe des entiers strictement positifs $n, s, m_{1}, \\cdots, m_{n}, k_{1}, \\cdots, k_{s}$ tels que\n$$\n\\left(a^{m_{1}}-1\\right) \\cdots\\left(a^{m_{n}}-1\\right)=\\left(a^{k_{1}}+1\\right) \\cdots\\left(a^{k_{s}}+1\\right)\n$$",
    "options": [],
    "answer": "2 and 3",
    "solution": "Solution:\n\nOn va prouver que les entiers $a$ cherchés sont $a=2$ et $a=3$.\n\nTout d'abord, on constate que $2^{2}-1=2+1$ et que $(3-1)(3-1)=3+1$, ce qui assure que $a=2$ et $a=3$ sont effectivement des solutions du problème.\n\nRéciproquement, soit $a, n, s, m_{1}, \\cdots, m_{n}, k_{1}, \\cdots, k_{s}$ des entiers strictement positifs tels que\n$$\n\\left(a^{m_{1}}-1\\right) \\cdots\\left(a^{m_{n}}-1\\right)=\\left(a^{k_{1}}+1\\right) \\cdots\\left(a^{k_{s}}+1\\right)\n$$\nClairement, on a $a \\neq 1$, puisque si $a=1$ le membre de droite vaut 0 mais pas celui de gauche.\n\nPar l'absurde: supposons que $a>3$.\nOn pose $A=\\left(a^{m_{1}}-1\\right) \\cdots\\left(a^{m_{n}}-1\\right)=\\left(a^{k_{1}}+1\\right) \\cdots\\left(a^{k_{s}}+1\\right)$.\n\nLemme 1. Chacun des nombres $m_{1}, \\cdots, m_{n}$ et $a-1$ est une puissance de 2.\n\nPreuve du lemme 1. Soit $i \\in\\{1, \\cdots, n\\}$ et $\\gamma$ un diviseur impair de $m_{i}$ (éventuellement $\\gamma=1$ ), disons $m_{i}=\\gamma \\times b$. Soit $p$ un diviseur premier de $a^{\\gamma}-1$.\n\nPuisque $a^{\\gamma} \\equiv 1\\ (\\bmod\\ p)$, on a $a^{m_{i}} \\equiv\\left(a^{\\gamma}\\right)^{b} \\equiv 1\\ (\\bmod\\ p)$.\nAinsi $A$ est divisible par $p$, ce qui assure qu'il existe $j$ tel que $a^{k_{j}}+1$ soit divisible par $p$. Comme ci-dessus, on a alors $\\left(a^{\\gamma}\\right)^{k_{j}}-1 \\equiv 0\\ (\\bmod\\ p)$ et, puisque $\\gamma$ est impair, on a aussi $\\left(a^{k_{j}}\\right)^{\\gamma} \\equiv-1\\ (\\bmod\\ p)$.\n\nPar suite $p$ divise $\\left(\\left(a^{k_{j}}\\right)^{\\gamma}+1\\right)-\\left(\\left(a^{\\gamma}\\right)^{k_{j}}-1\\right)=2$, d'où $p=2$.\n\nAinsi, le seul diviseur premier de $a^{\\gamma}-1$ est $p=2$, ce qui prouve que $a^{\\gamma}-1$ est une puissance de 2. Il existe donc un entier $c$ tel que\n$$\n2^{c}=a^{\\gamma}-1=(a-1)\\left(a^{\\gamma-1}+\\cdots+a+1\\right) .\n$$\nEn particulier, $a-1$ est donc une puissance de 2 et $a$ est impair. Mais, puisque $\\gamma$ est impair, $a^{\\gamma-1}+\\cdots+a+1$ est impair et doit aussi être une puissance de 2 . Cela implique que $\\gamma=1$ et donc que le seul diviseur impair de $m_{i}$ est 1 , ce qui conclut la preuve du lemme 1.\n\nD'après le lemme, on a $a-1=2^{c}$ et $a-1>2$, donc $a-1$ est divisible par 4. Pour tout entier $k \\geq 0$, on a alors $a^{k}+1 \\equiv 2\\ (\\bmod\\ 4)$.\n\nPour tout $i$, on pose $a_{i}=\\frac{1}{2}\\left(a^{2^{i}}+1\\right)$. On sait donc que $a_{i}$ est un entier impair.\n\nOn vérifie facilement par récurrence que, pour tout entier $d \\geq 0$ :\n$$\na^{2^{d}}-1=(a-1)(a+1)\\left(a^{2}+1\\right) \\cdots\\left(a^{2^{d-1}}+1\\right)=2^{c} \\cdot 2^{d} \\cdot a_{0} a_{1} \\cdots a_{d-1}\n$$\n\nLemme 2. Les nombres $a_{0}, a_{1}, a_{2}, \\cdots$ sont deux à deux premiers entre eux.\n\nPreuve du lemme 2. Soit $0 \\leq i<j$. On note que $a_{i}$ et $a_{j}$ sont impairs donc leur pgcd $d$ est impair. D'après (1), le nombre $a^{2^{j}}-1$ est divisible par $a_{i}$. Puisque $a^{2^{j}}+1=2 a_{j}$, on en déduit que tout diviseur commun à $a_{i}$ et $a_{j}$ divise aussi $\\left(a^{2^{j}}+1\\right)-\\left(a^{2^{j}}-1\\right)=2$. Ainsi $d$ est impair et divise 2, donc $d=1$. On a donc $a_{i}$ et $a_{j}$ premiers entre eux, et cela achève la preuve du lemme 2.\n\nD'après le lemme 1, on sait que chaque $m_{i}$ est une puissance de 2 . En utilisant (1) pour chaque $m_{i}$, on déduit que $A$ peut s'écrire sous la forme\n$$\nA=2^{N}\\left(a_{0}\\right)^{N_{0}} \\cdots\\left(a_{q}\\right)^{N_{q}}\\tag{2}\n$$\nav ec $N>N_{0}+\\cdots N_{q}$.\n\nPuisque chaque $a^{k_{j}}+1$ est congru à 2 modulo 4 , on doit donc avoir $s=N$. Or, pour tout $j$, si l'on pose $k_{j}=2^{r} t$ avec $t$ impair, alors $a^{k_{j}}+1=\\left(a^{2^{r}}\\right)^{t}+1 \\equiv(-1)^{t}+1 \\equiv 0\\ (\\bmod\\ 2)^{r}+1$. Ainsi $a^{k_{j}}+1$ est divisible par $a_{r}$.\n\nAinsi, chacun des nombres $a^{k_{j}}+1$ est divisible par un des $a_{i}$. Puisque $s>N_{0}+\\cdots N_{q}$, il existe $i$ pour lequel $a_{i}$ divise plus de $N_{i}$ des $a^{k_{j}}+1$. Par suite, $A$ est divisible par $\\left(a_{i}\\right)^{N_{i}+1}$, ce qui est impossible d'après (2), puisque $a_{i}$ est impair et premier avec chacun des autres $a_{j}$ d'après le lemme 2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geq 1$ un entier. Un groupe de $2n$ personnes se réunit. Chacune de ces personnes possède au moins $n$ amies dans ce groupe (en particulier, si $A$ est amie avec $B$ alors $B$ est amie avec $A$, et on n'est pas ami avec soi-même). Prouver que l'on peut disposer ces $2n$ personnes autour d'une table ronde de sorte que chacune soit entre deux de ses amies.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsidérons une disposition arbitraire des invités :\nAB.... $A$.\nou $A, B$ désignent des personnes différentes, mais les deux $A$ représentent la même personne (on est autour d'une table ronde).\nDans ce qui suit, on considérera la disposition comme une suite de personnes, ordonnée de gauche à droite.\nSi deux voisins ne sont pas amis, on dira que c'est une tension.\nS'il n'y a aucune tension, c'est fini.\nS'il existe une tension, par symétrie circulaire, on peut toujours supposer que $A$ et $B$ ne sont pas amis.\nComme $B$ a au moins $n$ amis, plus lui-même, on en déduit que $B$ a au plus $n-1$ \"ennemis\".\nSupposons que dans la disposition ci-dessus et après $B$, on ne trouve jamais deux voisins $A^{\\prime}$ et $B^{\\prime}$ (dans cet ordre) qui sont amies respectivement de $A$ et de $B$.\nAlors, comme une telle disposition n'arrive pas non plus avant $B$, on en déduit qu'elle n'apparaît jamais, et donc qu'à droite de chaque ami de $A$ doit se trouver un ennemi de $B$. Donc, le nombre d'ennemis de $B$ est au moins égal au nombre d'amis de $A$, ce qui implique $n-1 \\geq n$. Contradiction.\n\nPar suite, on peut trouver deux voisins $A^{\\prime}$ et $B^{\\prime}$ (dans cet ordre) qui sont amies respectivement de $A$ et de $B$.\nLa disposition est alors de la forme :\n$$\nA B \\ldots A^{\\prime} B^{\\prime} \\ldots A\n$$\nDans ces conditions, on peut éliminer la tension entre $A$ et $B$ en renversant l'ordre des personnes entre $B$ et $A^{\\prime}$ :\n$$\nA\\left(B \\ldots A^{\\prime}\\right) B^{\\prime} \\ldots A \\rightarrow A\\left(A^{\\prime} \\ldots B\\right) B^{\\prime} \\ldots A\n$$\nCela fournit une nouvelle disposition mais avec au moins une tension de moins que dans la précédente, puisque clairement la modification ne crée pas de nouvelle tension (par contre, elle peut éliminer une tension éventuelle entre $A^{\\prime}$ et $B^{\\prime}$ ).\nEn répétant cette procédure autant que nécessaire (un nombre fini de fois en tout cas, puisque le nombre initial de tensions est fini), on fait disparaître toutes les tensions, et l'objectif est atteint.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer toutes les fonctions $f$ de $\\mathbb{R}$ dans $\\mathbb{R}$ telles que, pour tous $x$ et $y$ réels, on ait l'égalité\n$$\nf(x+y)=f(x-y)+f(f(1-x y))\n$$",
    "options": [],
    "answer": "f(x)=0 for all real x",
    "solution": "Solution:\n\nPour $y=0$ on trouve que $f(x)=f(x)+f(f(1))$ donc $f(f(1))=0$.\n\nPour $x=0$ on trouve $f(y)=f(-y)+f(f(1))$ donc $f(y)=f(-y)$ pour tout $y$.\n\nPour $y=1$ on obtient $f(x+1)=f(x-1)+f(f(1-x))$. En remplaçant $x$ par $-x+2$, il vient $f(-x+3)=f(-x+1)+f(f(x-1))$.\n\nOr, $f(x-1)=f(1-x)$ donc $f(f(1-x))=f(f(x-1))$. Par conséquent, $f(x+1)=f(x-1)+f(-x+3)-f(-x+1)=f(x-3)$ puisque $f(-t)=f(t)$ pour tout $t$.\n\nOn en déduit, en remplaçant $x$ par $x+1$, que $f(x+2)=f(x-2)$ pour tout $x$.\n\nPrenons $y=2$ dans l'équation fonctionnelle. On a $f(x+2)=f(x-2)+f(f(1-2 x))$, donc $f(f(1-2 x))=0$ pour tout $x$. En remplaçant $x$ par $(1-t)/2$, on trouve que $f(f(t))=0$ pour tout $t$.\n\nOn revient à l'équation fonctionnelle : $f(x+y)=f(x-y)+f(f(1-x y))$, donc $f(x+y)=f(x-y)$ pour tous $x$ et $y$. On prend $x=y=t/2$ : il vient $f(t)=f(0)$, donc $f$ est constante. Comme $f(f(1))=0$, cette constante est nulle, et finalement $f(x)=0$ pour tout $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Trouver tous les entiers $m \\geqslant 1$ et $n \\geqslant 1$ tels que $\\frac{5^{m}+2^{n+1}}{5^{m}-2^{n+1}}$ soit le carré d'un entier.\n\nb) Plus généralement, trouver tous les entiers $m \\geqslant 1$ et $n \\geqslant 1$, ainsi que les nombres premiers $p$, tels que $\\frac{5^{m}+2^{n} p}{5^{m}-2^{n} p}$ soit le carré d'un entier.",
    "options": [],
    "answer": "a) The only solution is (m, n) = (1, 1).\n\nb) The solutions are exactly (m, n, p) = (1, 1, 2), (2, 2, 5), and (2, 3, 3).",
    "solution": "Solution:\na) Voir exercice 2 ci-dessus.\n\nb) Supposons que $p=5$. Alors $\\frac{5^{m-1}+2^{n}}{5^{m-1}-2^{n}}$ est le carré d'un entier. D'après la partie a) (qui marchait aussi dans le cas d'entiers $\\geqslant 0$), on a $m=n=2$.\n\nSupposons enfin $p \\neq 2$ et $p \\neq 5$. Soit $d$ le PGCD de $5^{m}+2^{n} p$ et de $5^{m}-2^{n} p$. Alors $d$ divise $5^{m}+2^{n} p+5^{m}-2^{n} p=2 \\times 5^{m}$. Comme $d$ est le PGCD de deux nombres impairs, il est impair, donc il divise $5^{m}$.\n\nDe plus, $d$ divise $5^{m}+2^{n} p-\\left(5^{m}-2^{n} p\\right)=2^{n+1} p$ et il est impair, donc il divise $p$. Comme $p$ et $5$ sont premiers entre eux, on en déduit que $d=1$ et que $5^{m}-2^{n} p=1$; de plus, $5^{m}+2^{n} p=a^{2}$ pour un certain entier $a$.\n\nOn a donc $(a-1)(a+1)=a^{2}-1=2^{n+1} p$.\n\nLe PGCD de $a-1$ et de $a+1$ divise $(a+1)-(a-1)=2$, donc il vaut $1$ ou $2$. De plus, $a-1$ et $a+1$ sont de même parité, et leur produit est pair, donc leur PGCD vaut $2$. On en déduit que $(a-1=2,\\ a+1=2^{n} p)$ ou $(a-1=2 p,\\ a+1=2^{n})$ ou $(a-1=2^{n},\\ a+1=2 p)$.\n\nDans le premier cas on aurait $a=3$, donc $4=a+1=2^{n} p$, ce qui est impossible.\n\nDans le deuxième cas, on a $p=2^{n-1}-1$. Comme $p$ est premier, $n=1$ et $n=2$ ne conviennent pas donc $n \\geqslant 3$. Par conséquent, $5^{m}=2^{n} p+1 \\equiv 1\\ [8]$. Comme $5^{2 \\ell+1}=5 \\times 25^{\\ell} \\equiv 5 \\times 1^{\\ell}=5\\ [8]$, l'entier $m$ est nécessairement pair. Posons $m=2 \\ell$, alors $\\left(5^{\\ell}-1\\right)\\left(5^{\\ell}+1\\right)=2^{n} p$. L'un des facteurs est égal à $2 p$ et l'autre à $2^{n-1}$. Comme leur différence est égale à $2$, on a $\\pm 2=2 p-2^{n-1}=2 p-(p+1)=p-1$, donc $p=3$, $n=3$ et $m=2$.\n\nDans le troisième cas, on a $p=2^{n-1}+1$. On ne peut pas avoir $n=1$, sinon $p=2$. Si $n=2$ alors $p=3$ et $5^{m}=1+2^{n} p=13$. Impossible. Donc $n \\geqslant 3$. Le même raisonnement que plus haut conduit à $m=2 \\ell$ avec $\\pm 2=2 p-2^{n-1}=2 p-(p-1)=p+1$, ce qui contredit $p \\geqslant 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30516,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $I$ le centre du cercle inscrit à un triangle $ABC$. Soit $D$ le point diamétralement opposé à $A$ sur le cercle circonscrit. On suppose que le point $E$ de la demi-droite $[BA)$ et le point $F$ de la demi-droite $[CA)$ satisfont la condition\n$$\nBE = CF = \\frac{AB + BC + CA}{2}.\n$$\nMontrer que $(EF) \\perp (DI)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $a, b, c$ les longueurs des côtés, $r$ le rayon du cercle inscrit et $p = (a + b + c)/2$. Comme $[AD]$ est un diamètre, les angles $\\widehat{ABD}$ et $\\widehat{ACD}$ sont droits donc\n$$\n\\begin{aligned}\nDE^2 - DF^2 &= (DB^2 + BE^2) - (DC^2 + CF^2) = DB^2 - DC^2 \\\\\n&= (AD^2 - DC^2) - (AD^2 - DB^2) = AC^2 - AB^2 \\\\\n&= b^2 - c^2.\n\\end{aligned}\n$$\nD'autre part, si $C'$ est le point de contact du cercle inscrit avec $[AB]$, on sait que $BC' = p - b$, donc $C'E = BE - BC' = b$. Il vient $IE^2 = (IC')^2 + (C'E)^2 = r^2 + b^2$, et de même $IF^2 = r^2 + c^2$, donc\n$$\nDE^2 - DF^2 = IE^2 - IF^2\n$$\nCeci signifie que les points $D$ et $I$ ont la même différence de puissances par rapport à $E$ et $F$ (considérés comme des cercles de rayon nul), donc $(DI) \\perp (EF)$.\n\nPrécision : expliquons pourquoi si $C$ et $C'$ sont deux cercles de centres $O$ et $O'$, et si $P_C(A) - P_{C'}(A) = P_C(B) - P_{C'}(B)$ alors $(AB) \\perp (OO')$. Notons $R$ et $R'$ les rayons des cercles et $C, D$ les projetés respectifs de $A$ et $B$ sur $[OO']$.\nOn a $P_C(A) - P_{C'}(A) = OA^2 - O'A^2 - (R^2 - (R')^2) = OC^2 - O'C^2 - (R^2 - (R')^2)$, donc $OC^2 - O'C^2 = OD^2 - O'D^2$. Ceci s'écrit encore $(\\overline{OC} - \\overline{O'C})(\\overline{OC} + \\overline{O'C}) = (\\overline{OD} - \\overline{O'D})(\\overline{OD} + \\overline{O'D})$, ou encore $2\\overline{O'O} \\cdot \\overline{MC} = 2\\overline{O'O} \\cdot \\overline{MD}$ où $M$ est le milieu de $[O'O]$. Par conséquent, $C = D$, et donc $(AB) = (AC) \\perp (OO')$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRésoudre en nombres réels le système d'équations\n$$\n\\begin{gathered}\nx_{1}\\left(x_{1}-1\\right)=x_{2}-1 \\\\\nx_{2}\\left(x_{2}-1\\right)=x_{3}-1 \\\\\n\\cdots \\\\\nx_{2016}\\left(x_{2016}-1\\right)=x_{2017}-1 \\\\\nx_{2017}\\left(x_{2017}-1\\right)=x_{1}-1 .\n\\end{gathered}\n$$",
    "options": [],
    "answer": "x1 = x2 = ... = x2017 = 1",
    "solution": "Solution:\n\nPosons par convention $x_{2018}=x_{1}$. Pour tout $i=1, \\ldots, 2017$, on a\n$$\nx_{i+1}-x_{i}=x_{i}\\left(x_{i}-1\\right)+1-x_{i}=\\left(x_{i}-1\\right)^{2} \\geqslant 0,\n$$\ndonc $x_{1}=x_{2018} \\geqslant x_{2017} \\geqslant \\cdots \\geqslant x_{1}$. On en déduit que les $x_{i}$ sont tous égaux, donc\n$$\n0=x_{i+1}-x_{i}=\\left(x_{i}-1\\right)^{2}=0.\n$$\nPar conséquent, $x_{i}=1$ pour tout $i$. Réciproquement, il est clair que $x_{i}=1$ est solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\Gamma_{1}$ et $\\Gamma_{2}$ deux cercles se coupant en $A$ et $B$ distincts. Notons $O$ le centre de $\\Gamma_{1}$. Soit $C$ un point de $\\Gamma_{1}$, soit $D, E$ les intersections respectives de $(AC)$ et de $(BC)$ avec $\\Gamma_{2}$.\n\nMontrer que $(OC)$ et $(DE)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn procède par chasse aux angles.\nOn présente ici une solution qui se veut très (trop) rigoureuse, il n'y avait pas besoin d'être aussi pointilleux ni de traiter tous ces cas pour recevoir la note maximale à cet exercice.\nComme on remarque que la figure change beaucoup selon que le point $C$ est du même côté de la droite $(AB)$ que $O$ ou non, on a le choix entre distinguer des cas dans notre chasse aux angles en faisant attention aux problèmes d'orientation ou travailler avec des angles orientés de droites. La deuxième méthode est souvent préférable, tant qu'on n'a pas à diviser des angles par deux (le problème est que, si on regarde nos angles orientés de droites modulo $180$ degrés et qu'on divise par $2$, on se retrouve à regarder des angles orientés de droite modulo $90$ degrés, donc on perd de l'information). Ici, on va panacher les deux méthodes.\n\nOn calcule modulo $180$ degrés :\n$$(OC, DE) = (OC, CD) + (CD, DE) = (OC, CD) + (AB, BE) = (OC, CA) + (CB, AB).$$\nOr $2(OC, CA) + (OA, OC) = 0$ et $2(CB, AB) = (OC, OA)$.\nDonc $2(OC, DE) = 0$ modulo $180$ degrés, donc $(OC, DE)$ est un multiple de $90$ degrés (donc $0$ ou $90$ degrés modulo $180$), donc $(OC)$ et $(DE)$ sont parallèles ou perpendiculaires.\n\nEt là, on est embêtés pour exclure le cas parallèle. On peut s'en sortir par disjonction de cas :\n- Si $A, C$ et le point diamétralement opposé à $A$ sur $\\Gamma_{1}$ sont dans cet ordre quand on lit dans le sens trigonométrique (anti-horaire) sur le cercle $\\Gamma_{1}$, alors $(OC, CA) = \\widehat{OCA}$.\n- Sinon, $(OC, CA) = 180^\\circ - \\widehat{OCA}$.\n- Si $A, B, C$ se lisent dans cet ordre sur le cercle $\\Gamma_{1}$ dans le sens trigonométrique, alors $(CB, AB) = \\widehat{CBA}$.\n- Sinon, $(CB, AB) = 180^\\circ - \\widehat{CBA}$.\n- On a dans tous les cas $\\widehat{COA} = 180^\\circ - 2\\widehat{OCA}$.\n- Mais $\\widehat{COA} = 2\\widehat{CBA}$ si et seulement si $O$ et $B$ sont du même côté de la droite $(AC)$.\n- En revanche, si $O$ et $B$ ne sont pas du même côté de la droite $(AC)$, $\\widehat{COA} = 360^\\circ - 2\\widehat{CBA}$.\n\nLes angles géométriques ayant des mesures entre $0$ et $180$ degrés.\nSi on traite tous les cas présentés ci-dessus, en combinant les égalités, on arrive à montrer que les droites $(OC)$ et $(DE)$ sont perpendiculaires.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30519,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère un échiquier $3 \\times 3$. Au début, on écrit le chiffre $0$ dans chacune des $9$ cases. Ensuite, à chaque étape, on effectue l'opération suivante : on choisit deux cases ayant un côté commun, puis on rajoute $1$ au nombre écrit dans ces deux cases, ou bien on retranche $1$ au nombre écrit dans ces deux cases. Est-il possible d'obtenir une configuration où le nombre $2017$ est écrit dans les $9$ cases?",
    "options": [],
    "answer": "Not possible",
    "solution": "Solution:\n\nCe n'est pas possible. On colorie l'échiquier $3 \\times 3$ en noir et blanc de manière usuelle (de sorte qu'une case noire n'ait que des cases blanches comme voisins et qu'une case blanche n'ait que des cases noires comme voisins). On vérifie que la somme des nombres sur les cases noires est toujours égale à la somme des nombres sur les cases blanches. En effet, c'est le cas au début, et si la somme des nombres sur les cases noires est égale à la somme des nombres sur les cases blanches, alors c'est toujours le cas après avoir effectué l'opération de l'énoncé.\n\nLorsque le nombre $2017$ est écrit dans les $9$ cases, ces deux sommes ne sont pas égales, d'où le résultat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les quadruplets $ (p, q, r, n) $ d'entiers strictement positifs vérifiant les trois conditions suivantes :\n- $p$ et $q$ sont premiers,\n- $p+q$ n'est pas divisible par $3$,\n- $p+q = r (p-q)^n$.",
    "options": [],
    "answer": "(2,3,5,2n) for all positive integers n; (3,2,5,n) for all positive integers n; (3,5,2,2); (5,3,1,3); (5,3,2,2); (5,3,4,1)",
    "solution": "Solution:\n\nOn suppose d'abord que $p \\neq 3$ et $q \\neq 3$, et on considère $p$ et $q$ modulo $3$. Puisque $p+q \\not\\equiv 0 \\pmod{3}$, on n'a pas $p \\equiv 1 \\pmod{3}$ et $q \\equiv 2 \\pmod{3}$, ni $p \\equiv 2 \\pmod{3}$ et $q \\equiv 1 \\pmod{3}$. On est donc assuré que $p \\equiv q \\equiv 1 \\pmod{3}$ ou bien que $p \\equiv q \\equiv 2 \\pmod{3}$. Dans ces deux cas, $3$ divise $p-q$ et donc divise $p+q = r (p-q)^n$ puisque $n \\geq 1$. Ceci est absurde et invalide donc notre supposition, ce qui montre que $p=3$ ou $q=3$.\n\nD'autre part, puisque $n \\geq 0$, on sait que $p-q$ divise $p+q$, donc $p \\neq q$, de sorte que $p$ et $q$ sont premiers entre eux. Puisque $p-q$ divise $(p+q)+(p-q)=2p$ et $(p+q)-(p-q)=2q$, on sait donc que $p-q$ divise $\\operatorname{PGCD}(2p, 2q)=2\\operatorname{PGCD}(p, q)=2$. Il s'ensuit que $p-q \\in \\{-2, -1, 1, 2\\}$. On en déduit donc que $(p, q) \\in \\{(2,3), (3,2), (3,5), (5,3)\\}$; on traite ces quatre cas un à un :\n\n- si $(p, q)=(2,3)$, l'équation devient $5 = (-1)^n r$; ses solutions sont $r=5$ et $n \\in 2\\mathbb{N}^*$;\n\n- si $(p, q)=(3,2)$, l'équation devient $5 = 1^n r$; ses solutions sont $r=5$ et $n \\in \\mathbb{N}^*$;\n\n- si $(p, q)=(3,5)$, l'équation devient $8 = (-2)^n r$; ses solutions sont $(r, n) = (2,2)$;\n\n- si $(p, q)=(5,3)$, l'équation devient $8 = 2^n r$; ses solutions sont $(r, n) = (1,3), (2,2)$ et $(4,1)$.\n\nEn conclusion, les solutions sont : $(2,3,5,2n)_{n \\in \\mathbb{N}^*}$, $(3,2,5,n)_{n \\in \\mathbb{N}^*}$, $(3,5,2,2)$, $(5,3,1,3)$, $(5,3,2,2)$ et $(5,3,4,1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUne cellule peut se diviser en 42 ou en 44 petites cellules. Combien de divisions faut-il pour obtenir, à partir d'une cellule, exactement 2017 cellules?",
    "options": [],
    "answer": "48",
    "solution": "Solution:\n\nUne division augmente de $41$ ou $43$ le nombre de cellules. Soit $a$ le nombre de divisions en $42$ cellules, et $b$ le nombre de divisions en $44$ cellules. On veut trouver $a+b$, sachant que\n$$\n2017 = 1 + 41a + 43b = 1 + 41(a+b) + 2b = 1 + 43(a+b) - 2b\n$$\nOn voit que $41(a+b) \\leq 2016$, donc $a+b < 50$.\nDe même, $43(a+b) \\geq 2016$, donc $a+b > 46$.\nOr $2017 = 1 + 41 \\times 49 + 7 = 1 + 41 \\times 48 + 48 = 1 + 41 \\times 47 + 89$. Comme $2b$ est pair, la seule solution est $a+b = 48$ (avec $2b = 48$ donc $a = b = 24$ ).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn définit une suite ainsi:\n$$\n\\left\\{\\begin{array}{l}\nu_{0}=15,\\ u_{1}=57 \\\\\nu_{n}=u_{n-1}+u_{n-2}\\ \\text{ pour tout } n \\geq 2\n\\end{array}\\right.\n$$\nTrouver le plus grand entier $k$ tel que $3^{k} \\mid u_{2017}$.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nSoit $k$ l'entier cherché. Les premiers termes de la suite sont $15$, $57$, $72$, $129$, $201$, $330$, $541$, ... apparemment tous divisibles par $3$. En effet, $u_{0}$ et $u_{1}$ sont multiples de $3$, donc $u_{2}=u_{0}+u_{1}$ aussi. De même, $u_{3}$ est divisible par $3$. Plus généralement, si $u_{n-1}$ et $u_{n-2}$ sont multiples de $3$, alors $u_{n}$ aussi. On prouve ainsi de proche en proche (par récurrence, en fait) que pour tout entier $n$, $u_{n}$ est divisible par $3$, donc $3 \\mid u_{2017}$ et $k \\geq 1$.\n\nRaisonnons maintenant modulo $9$ pour voir si $k \\geq 2$. On trouve :\n$u_{0} \\equiv 6\\ (\\bmod\\ 9)$\n$u_{1} \\equiv 3\\ (\\bmod\\ 9)$\n$u_{2} \\equiv 0\\ (\\bmod\\ 9)$\n$u_{3} \\equiv 3\\ (\\bmod\\ 9)$\n$u_{4} \\equiv 3\\ (\\bmod\\ 9)$\n$u_{5} \\equiv 6\\ (\\bmod\\ 9)$\n$u_{6} \\equiv 0\\ (\\bmod\\ 9)$\n$u_{7} \\equiv 6\\ (\\bmod\\ 9)$\n$u_{8} \\equiv 6\\ (\\bmod\\ 9)$\n$u_{9} \\equiv 3\\ (\\bmod\\ 9)$\nOn constate que $u_{8} \\equiv u_{0}\\ (\\bmod\\ 9)$ et $u_{9} \\equiv u_{1}\\ (\\bmod\\ 9)$, donc $u_{10} \\equiv u_{2}\\ (\\bmod\\ 9)$, car les modulos s'additionnent. Plus généralement, on établit comme précédemment que $u_{n+8} \\equiv u_{n}\\ (\\bmod\\ 9)$ pour tout $n \\in \\mathbb{N}$. Ainsi, la suite $\\left(u_{n}\\ (\\bmod\\ 9)\\right)$ est périodique de période $8$. Or $2017=252 \\times 8+1$, donc $u_{2017} \\equiv u_{1} \\equiv 3\\ (\\bmod\\ 9)$. Donc $k<2$. Conclusion : $k=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver les couples d'entiers $(x, y) \\in \\mathbb{Z}$ solutions de l'équation $y^{2}=x^{5}-4$.",
    "options": [],
    "answer": "No integer solutions.",
    "solution": "Solution:\n\nPar le petit théorème de Fermat, pour tout entier $x$ premier avec $11$, $x^{10} \\equiv 1[11]$. Donc $11$ divise $x^{10}-1=\\left(x^{5}-1\\right)\\left(x^{5}+1\\right)$. Par le lemme de Gauss, $11$ divise $x^{5}-1$ ou $x^{5}+1$. Donc pour tout entier $x$, $x^{5}-4 \\equiv -5[11]$ ou $x^{5}-4 \\equiv -3[11]$ ou, dans le cas où $11$ divise $x$, $x^{5}-4 \\equiv -4[11]$.\n\nPour le terme de gauche, on calcule les résidus quadratiques modulo $11$ : $0^{2} \\equiv 0[11], 1^{2}=(-1)^{2} \\equiv 1[11], 2^{2}=(-2)^{2} \\equiv 4[11], 3^{2}=(-3)^{2} \\equiv -2[11]$, $4^{2}=(-4)^{2} \\equiv 5[11]$, et $5^{2}=(-5)^{2} \\equiv 3[11]$.\n\nAinsi $y^{2} \\not \\equiv x^{5}-4[11]$, donc l'équation n'a pas de solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans la région reculée de Torturie, le vizir dispose en cercle $n$ condamnés, numérotés dans l'ordre entre 1 et $n$. Implacablement, il envoie au bourreau un condamné sur deux : les condamnés $2,4, \\ldots$ et ainsi de suite en tournant autour du cercle, en sautant une personne entre deux supplices, jusqu'au moment où il n'en reste plus qu'un. Calculer en fonction de $n$ le numéro du dernier condamné restant.",
    "options": [],
    "answer": "If n = 2^a + b with 0 ≤ b < 2^a, then the last remaining position is 2b + 1.",
    "solution": "Solution:\n\nOn note $f(n)$ le numéro du survivant. Traitons d'abord le cas où $n$ est une puissance de $2$. Montrons par récurrence que dans ce cas, $f(n)=1$.\n\nSi $n=1$, c'est bien le cas. Si $n=2^{a}$, avec $a \\geq 1$, le vizir élimine d'abord tous les numéros pairs. Il reste alors $\\frac{n}{2}=2^{a-1}$ condamnés, et le prochain à être envoyé au bourreau est le numéro $3$, qui est la deuxième personne restante. Par hypothèse de récurrence, le survivant sera la première personne restante, c'est-à-dire le numéro $1$. Ainsi $f\\left(2^{a}\\right)=1$ pour tout entier $a$.\n\nTraitons maintenant le cas général. On écrit $n=2^{a}+b$, où $a$ et $b$ sont des entiers tels que $0 \\leq b<2^{a}$. Le vizir envoie d'abord au bourreau les numéros $2,4, \\ldots, 2b$. Il reste alors $2^{a}$ condamnés. D'après le cas précédent, le survivant est celui situé juste avant la prochaine victime, à savoir le numéro $2b+1$, car $2b+1 \\leq n$. Ainsi $f\\left(2^{a}+b\\right)=2b+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn place quatre points dans le plan, trois jamais alignés, et on les relie tous deux à deux. On colorie chacun des six segments obtenus soit en bleu soit en rouge. Montrer qu'il existe deux triangles différents coloriés de la même façon. Par exemple, dans l'exemple suivant (où l'on a remplacé rouge par épais et bleu par pointillé), $ABD$ et $ACD$ sont coloriés de la même façon.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIl y a quatre façons différentes de colorier un triangle (avec zéro, un, deux ou trois segments rouges) et il y a quatre triangles dessinés. Donc, soit deux triangles sont coloriés de la même façon, soit tous les triangles sont différents. Montrons que le deuxième cas n'est pas possible. En effet, deux triangles différents partagent toujours deux sommets en commun, donc un côté en commun. Or, si tous les triangles sont différents, il y a un triangle entièrement bleu, et un autre entièrement rouge. Leur côté commun serait alors à la fois rouge et bleu, impossible. Finalement, on a bien deux triangles coloriés de la même façon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30526,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer la valeur minimale de\n$$\n\\left\\lfloor\\frac{a+b+c}{d}\\right\\rfloor+\\left\\lfloor\\frac{b+c+d}{a}\\right\\rfloor+\\left\\lfloor\\frac{c+d+a}{b}\\right\\rfloor+\\left\\lfloor\\frac{d+a+b}{c}\\right\\rfloor\n$$\nlorsque $a, b, c, d$ décrivent $\\mathbb{N}^*$.",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nLe minimum cherché est $9$.\nEn effet, si $a, b, c, d$ sont des entiers naturels non nuls, on pose\n$$\nf(a, b, c, d)=\\left\\lfloor\\frac{a+b+c}{d}\\right\\rfloor+\\left\\lfloor\\frac{b+c+d}{a}\\right\\rfloor+\\left\\lfloor\\frac{c+d+a}{b}\\right\\rfloor+\\left\\lfloor\\frac{d+a+b}{c}\\right\\rfloor.\n$$\nOn rappelle que, pour tout réel $x$, on a $\\lfloor x\\rfloor>x-1$. Ainsi, pour tous entiers $a, b, c, d>0$, on a\n$$\n\\begin{aligned}\nf(a, b, c, d)+4 & >\\frac{a+b+c}{d}+\\frac{b+c+d}{a}+\\frac{c+d+a}{b}+\\frac{d+a+b}{c} \\\\\n& >\\left(\\frac{a}{b}+\\frac{b}{a}\\right)+\\left(\\frac{a}{c}+\\frac{c}{a}\\right)+\\left(\\frac{a}{d}+\\frac{d}{a}\\right)+\\left(\\frac{b}{c}+\\frac{c}{b}\\right)+\\left(\\frac{b}{d}+\\frac{d}{b}\\right)+\\left(\\frac{c}{d}+\\frac{d}{c}\\right) \\\\\n& >6 \\times 2 \\text{ car } t+\\frac{1}{t} \\geqslant 2 \\text{ pour tout réel } t>0.\n\\end{aligned}\n$$\nAinsi, $f(a, b, c, d)>8$. Or, $f(a, b, c, d)$ étant clairement un entier, c'est donc que $f(a, b, c, d) \\geqslant 9$. Il reste à vérifier que $f(5,5,5,4)=9$ pour conclure.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30527,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ un nombre entier. On dispose les $n^{2}$ nombres entiers $1,2, \\ldots, n^{2}$ dans les $n^{2}$ cases d'un échiquier $n \\times n$ comme suit : la première ligne contient les nombres $1,2, \\ldots, n$ (de gauche à droite), la deuxième ligne contient les nombres $n+1, n+2, \\ldots, 2 n$ et ainsi de suite. À chaque étape, on peut choisir deux cases ayant un côté commun, et ajouter ou soustraire un même entier aux deux nombres dans ces deux cases.\n\nTrouver toutes les valeurs de $n$ pour lesquelles il est possible d'obtenir la configuration où il n'y a que des 0, et dans ce cas trouver le nombre minimal d'étapes nécessaires pour y parvenir.",
    "options": [],
    "answer": "n must be even; the minimal number of steps is 3n^2/4.",
    "solution": "Solution:\n\nTout d'abord, on vérifie que la parité de la somme des nombres écrits ne change pas lorsqu'on effectue une étape. La somme des $n^{2}$ premiers entiers vaut $n^{2}(n^{2}+1)/2$, qui est paire si et seulement si $n$ est pair. Comme on veut obtenir la configuration où il n'y a que des 0 (dont la somme des nombres est paire), on voit que $n$ doit être pair.\n\nRéciproquement, si $n$ est pair, montrons qu'il est possible d'atteindre la configuration où il n'y a que des 0, et qu'il faut au moins $3 n^{2} / 4$ étapes pour y arriver. Supposons qu'on ait réussi à atteindre cette configuration en un certain nombre d'étapes. Considérons alors le graphe $G$ dont les sommets sont les cases et deux sommets sont adjacents si on a modifié leur contenu à une même étape. Soit $C$ une composante connexe de $G$. Alors $C$ possède au moins deux sommets (car tous les nombres sont non nuls au départ). De plus, $C$ ne peut pas avoir exactement deux sommets (car tous les nombres sont différents au départ). Supposons que $C$ a trois sommets. Dans ce cas, deux de ces sommets ont degré 1 et un a degré 2 (à cause de la géométrie de l'échiquier), et le nombre inscrit au départ dans la case correspondant à celle de degré 2 doit être la somme des nombres correspondant à celles de degré 1. On vérifie aisément que ce n'est possible. Ainsi, chaque composante connexe a au moins 4 sommets, ce qui montre que $G$ a au plus $\\frac{n^{2}}{4}$ composantes connexes.\n\nNotons $C_{1}, C_{2}, \\ldots, C_{k}$ les composantes connexes de $G$ ainsi que $s_{i}$ et $a_{i}$ respectivement le nombre de sommets et d'arêtes de $C_{i}$. Comme $a_{i} \\geqslant s_{i}-1$ pour tout $1 \\leqslant i \\leqslant k$, on en déduit que le nombre total d'arêtes $A$ de $G$ vérifie\n$$\nA \\geqslant \\sum_{i=1}^{k}\\left(s_{i}-1\\right)=\\left(\\sum_{i=1}^{k} s_{i}\\right)-k=n^{2}-k \\geqslant \\frac{3 n^{2}}{4}\n$$\nce qui montre qu'il faut au moins $3 n^{2} / 4$ étapes.\n\nLorsque $n$ est pair, en subdivisant l'échiquier en petits carrés $2 \\times 2$ (chacun contiendra les nombres $i, i+1, i+n, i+n+1$) on vérifie aisément qu'il est possible d'obtenir des 0 dans chacun de ces petits carrés en utilisant 3 opérations par petit carré (on enlève $i$ aux deux de gauche, on enlève $i+1$ aux deux de gauche puis on enlève $n$ aux deux du bas).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30528,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn possède une liste infinie de cases, les cases étant numérotées par $1,2, \\ldots$. Au départ, toutes les cases contiennent le nombre $1$. À chaque étape, on choisit un nombre $a \\in \\mathbb{N}^*$ tel que :\n- soit toutes les cases numérotées par un multiple de $a$ contiennent $1$ auquel cas on remplace ces $1$ par des $0$,\n- soit toutes les cases numérotées par un multiple de $a$ contiennent $0$ auquel cas on remplace ces $0$ par des $1$.\n\nMontrer que, pour tout $n \\geqslant 2$, on peut faire en sorte que toutes les cases de numéro au plus $n$ contiennent un $0$ sauf la première case qui contient un $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous allons d'abord reformuler le problème. On se place dans $\\mathbb{N}^*$. Si $A$, $B$ sont deux ensembles (dans $\\mathbb{N}^*$), on note\n- $A \\oplus B := A \\cup B$ si $A \\cap B = \\emptyset$, sinon l'opération est interdite, et\n- $A \\ominus B := A \\setminus B$ si $B \\subset A$, sinon l'opération est interdite.\nOn ajoute la convention que $\\oplus$ et $\\ominus$ sont associatives à gauche, c'est-à-dire que les opérations sont effectuées de la gauche vers la droite. Par exemple, $\\{1\\} \\oplus \\{2\\} \\ominus \\{1,2\\} = (\\{1\\} \\oplus \\{2\\}) \\ominus \\{1,2\\} = \\emptyset$, par contre $\\{1\\} \\oplus \\{1,2\\} \\ominus \\{1\\}$ est interdit.\n\nLe problème se reformule donc par : montrer que, pour $n \\in \\mathbb{N}^*$, il existe une formule commençant par $\\mathbb{N}^*$, contenant que des ensembles de multiples d'entiers et utilisant les opérations $\\oplus$ et $\\ominus$ aboutissant à un ensemble $E$ tel que $E \\cap \\{1, \\ldots, n\\} = \\{1\\}$. Pour montrer cela, nous allons utiliser un lemme, que nous démontrerons plus tard, et qui n'est rien d'autre qu'une variante améliorée du principe d'inclusion-exclusion.\n\nLemme : Si $m \\in \\mathbb{N}^*$ et $A_1, \\ldots, A_m$ sont des ensembles, alors il existe une formule utilisant les opérations $\\oplus$ et $\\ominus$ et constituée uniquement d'intersections quelconques des ensembles $A_1, \\ldots, A_m$ dont la valeur est $A_1 \\cup \\cdots \\cup A_m$.\n\nPar exemple, dans le cas $m=2$, $A_1 \\cup A_2 = A_1 \\ominus (A_1 \\cap A_2) \\oplus A_2$.\n\nMaintenant, notons $p_1, p_2, \\ldots$ les nombres premiers dans l'ordre. Notons $A_i$ l'ensemble des multiples (strictement positifs) de $p_i$, pour tout $i \\in \\mathbb{N}^*$. Prenons $n \\geqslant 2$ et soit $m$ le nombre de nombres premiers inférieurs ou égaux à $n$. D'après le lemme, il existe une formule dont la valeur est $A_1 \\cup \\cdots \\cup A_m$. Or cette union est l'ensemble des nombres qui sont multiples d'au moins un nombre parmi $p_1, p_2, \\ldots, p_m$. Il contient donc tous les nombres inférieurs ou égaux à $n$ sauf $1$. En remplaçant les $\\oplus$ par des $\\ominus$ et vice-versa dans la formule et en ajoutant $\\mathbb{N}^* \\ominus$ devant, nous obtenons donc une formule autorisée dont la valeur vérifie ce que nous voulons. Il reste à ajouter pour conclure que, si $1 \\leqslant l \\leqslant m$ et si $1 \\leqslant i_1 < i_2 < \\cdots < i_l \\leqslant m$, alors $A_{i_1} \\cap A_{i_2} \\cap \\cdots \\cap A_{i_l}$ n'est rien d'autre que l'ensemble des multiples de $p_{i_1} p_{i_2} \\ldots p_{i_l}$. Donc la formule que l'on a obtenue n'est constituée que d'ensembles de multiples d'entier et donc convient.\n\nMontrons maintenant le lemme par récurrence sur $m$. Le cas $m=1$ est trivial. De plus, nous avons traité le cas $m=2$, pour l'exemple, ci-dessus. Soit $m \\geqslant 2$ et supposons le théorème vrai pour $m$ ensembles. Soient $A_1, \\ldots, A_{m+1}$ $m+1$ ensembles. On a\n$$\n\\begin{aligned}\n& A_1 \\cup \\cdots \\cup A_{m+1} = A_1 \\cup \\cdots \\cup A_m \\ominus \\left( (A_1 \\cup \\cdots \\cup A_m) \\cap A_{m+1} \\right) \\oplus A_{m+1} \\\\\n& A_1 \\cup \\cdots \\cup A_{m+1} = A_1 \\cup \\cdots \\cup A_m \\ominus \\left( (A_1 \\cap A_{m+1}) \\cup \\cdots \\cup (A_m \\cap A_{m+1}) \\right) \\oplus A_{m+1}\n\\end{aligned}\n$$\nPar hypothèse de récurrence, il existe une formule pour le deuxième terme constituée uniquement d'intersections quelconques de $A_1 \\cap A_{m+1}, \\ldots, A_m \\cap A_{m+1}$; donc uniquement d'intersections quelconques de $A_1, \\ldots, A_{m+1}$. Maintenant, on peut insérer la formule du deuxième terme dans la grosse formule en remplaçant, dans la formule du deuxième terme, les $\\oplus$ par des $\\ominus$ et vice-versa. En remplaçant directement le premier terme par la formule obtenue par l'hypothèse de récurrence, on obtient la formule voulue.\n\nOn conclut la preuve du lemme par récurrence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers naturels $n$ pour lesquels $2^{n} + 12^{n} + 2011^{n}$ est un carré parfait.",
    "options": [],
    "answer": "1",
    "solution": "Solution:\n\nPosons $u_{n} = 2^{n} + 12^{n} + 2011^{n}$. On regarde modulo $3$ : $2 \\equiv -1 \\pmod{3}$, $12 \\equiv 0 \\pmod{3}$ et $2011 \\equiv 1 \\pmod{3}$, donc\n$$\nu_{n} \\equiv (-1)^{n} + 0^{n} + 1^{n} \\equiv (-1)^{n} + 1 \\pmod{3}$$\nSi $n$ est pair, $u_{n} \\equiv 2 \\pmod{3}$. Or aucun carré ne peut être congru à $2$ modulo $3$ : en effet, si $x \\equiv 0 \\pmod{3}$, $x^{2} \\equiv 0 \\pmod{3}$, si $x \\equiv 1 \\pmod{3}$, $x^{2} \\equiv 1 \\pmod{3}$, et si $x \\equiv 2 \\pmod{3}$, $x^{2} \\equiv 1 \\pmod{3}$. \n\nSi $n$ est impair, on remarque que $u_{1} = 2 + 12 + 2011 = 2025 = 45^{2}$ donc $n = 1$ convient.\n\nSi $n \\geq 3$, $2^{n} \\equiv 12^{n} \\equiv 0 \\pmod{4}$. Et $2011 \\equiv 3 \\pmod{4}$. En écrivant $n = 2k + 1$, $k \\in \\mathbb{N}$, on a\n$$\nu_{n} \\equiv 3^{2k+1} \\equiv (3^{2})^{k} \\times 3 \\equiv 9^{k} \\times 3 \\equiv 1^{k} \\times 3 \\equiv 3 \\pmod{4}.$$\nOr un carré n'est jamais congru à $3$ modulo $4$, si on fait un tableau de congruence :\n\n| $x \\pmod{4}$ | $x^{2} \\pmod{4}$ |\n| :---: | :---: |\n| 0 | 0 |\n| 1 | 1 |\n| 2 | 0 |\n| 3 | 1 |\n\nDonc $u_{n}$ ne peut pas être un carré si $n \\geq 3$ est impair.\n\nConclusion : $n = 1$ est la seule solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. $H$ son orthocentre et $P$, $Q$ et $R$ les pieds des hauteurs issues de $A$, $B$ et $C$.\nMontrer que $H$ est le centre du cercle inscrit à $PQR$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn sait que les points $A$, $B$, $P$ et $Q$ sont cocycliques sur le cercle de diamètre $[AB]$, donc :\n$$\n\\widehat{HPQ} = \\widehat{APQ} = \\widehat{ABQ} = 90 - \\widehat{BAQ} = 90 - \\widehat{BAC}\n$$\nDe même, $A$, $C$, $P$ et $R$ sont cocycliques sur le cercle de diamètre $[AC]$ donc :\n$$\n\\widehat{HPR} = \\widehat{APR} = \\widehat{ACR} = 90 - \\widehat{CAR} = 90 - \\widehat{BAC}\n$$\nOn a donc $\\widehat{HPQ} = \\widehat{HPR}$ donc $(PH)$ est la bissectrice de $\\widehat{QPR}$. On montre de même que $(QH)$ et $(RH)$ sont les bissectrices de $\\widehat{PQR}$ et $\\widehat{PRQ}$, donc $H$ est le centre du cercle inscrit à $PQR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn a $n$ lampes, chacune avec un interrupteur. Chaque lampe fonctionne normalement, appuyer sur son interrupteur l'éteint si elle est allumée, et l'allume si elle est éteinte. Toutes les minutes on touche aux interrupteurs.\n\nÀ la première minute : on appuie sur tous les interrupteurs $(1,2,3, \\ldots, n)$.\n\nÀ la deuxième minute : on appuie sur tous les interrupteurs dont la lampe a un numéro pair $(2,4,6, \\ldots)$.\n\nÀ la troisième minute : on appuie sur tous les interrupteurs dont la lampe a un numéro divisible par $3\\ (3,6,9, \\ldots)$.\n\nEt ainsi de suite, à la $k$-ième minute, on appuie sur les interrupteurs avec un numéro divisible par $k$.\n\nQuelles sont les lampes qui sont allumées après $n$ minutes sachant qu'au début toutes les lampes sont éteintes ?",
    "options": [],
    "answer": "Exactly those lamps whose numbers are perfect squares",
    "solution": "Solution:\n\nSoit $i \\in \\{1, \\ldots, n\\}$ le numéro d'une des lampes. On appuie sur l'interrupteur $i$ à la $k$-ième minute si et seulement si $k$ est un diviseur de $i$, donc on a appuyé $d(i)$ fois sur l'interrupteur $i$, avec $d(i)$ le nombre de diviseurs de $i$.\n\nDécomposons $i$ en facteurs premiers : $i = p_{1}^{\\alpha_{1}} \\times \\cdots \\times p_{r}^{\\alpha_{r}}$, avec $p_{1}, \\ldots, p_{r}$ des nombres premiers distincts et $\\alpha_{1}, \\ldots, \\alpha_{r}$ des entiers.\n\nLes diviseurs de $i$ sont les entiers de la forme $p_{1}^{\\beta_{1}} \\times \\cdots \\times p_{r}^{\\beta_{r}}$, avec $\\beta_{1} \\in \\{0,1, \\ldots, \\alpha_{1}\\}, \\ldots, \\beta_{r} \\in \\{0,1, \\ldots, \\alpha_{r}\\}$, il y en a donc $d(i) = (\\alpha_{1}+1) \\ldots (\\alpha_{r}+1)$ (nombre de choix de $\\beta_{1}, \\ldots, \\beta_{r}$).\n\nPour que la lampe $i$ soit allumée à la fin, il faut et il suffit qu'on ait appuyé un nombre impair de fois sur l'interrupteur, donc que $d(i)$ soit impair, donc que tous les $\\alpha_{j},\\ j \\in \\{1, \\ldots, r\\}$, soient pairs, c'est-à-dire que $i$ soit un carré parfait.\n\nAinsi, les lampes allumées après $n$ minutes sont celles dont le numéro est un carré parfait.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit un quadrilatère $ABCD$ convexe. On se donne $E, F$ deux points tels que $E, B, C, F$ soient alignés dans cet ordre. On suppose de plus que $\\widehat{BAE} = \\widehat{CDF}$ et $\\widehat{EAF} = \\widehat{FDE}$. Montrer que $\\widehat{FAC} = \\widehat{EDB}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nC'est encore une chasse aux angles. Comme $\\widehat{FAE} = \\widehat{FDE}$ et $A, D$ sont du même côté de $(EF)$, les points $A, D, E, F$ sont cocycliques.\n\nDe plus, pour montrer que $\\widehat{FAC} = \\widehat{EDB}$, il suffit de montrer que $\\widehat{CAB} = \\widehat{FAB} - \\widehat{FAC}$ est égal à $\\widehat{CDB} = \\widehat{CDE} - \\widehat{EDB}$, autrement dit que $A, B, C, D$ sont cocycliques. Montrons pour cela que $\\widehat{DAB} + \\widehat{DCB} = 180^\\circ$. On a :\n$$\n\\widehat{DAB} = \\widehat{DAF} + \\widehat{FAB} = \\widehat{DEF} + \\widehat{CDE} = 180^\\circ - \\widehat{ECD}\n$$\nce qui prouve la cocyclicité de $A, B, C, D$ et donc le résultat voulu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $m$, $n \\geq 0$ tels que $n^{3}-3 n^{2}+n+2=5^{m}$.",
    "options": [],
    "answer": "(m, n) = (0, 1) and (1, 3)",
    "solution": "Solution:\n\nOn peut factoriser $n^{3}-3 n^{2}+n+2=(n-2)(n^{2}-n-1)$. Pour que l'équation soit vraie, il faut que les deux facteurs soient, au signe près, des puissances de $5$.\n\nSi $n-2$ est négatif, comme $n$ est positif, pour que $-(n-2)$ soit une puissance de $5$, il faut prendre $n=1$, et dans ce cas $n^{3}-3 n^{2}+n+2=1$, donc $m=0$, $n=1$ est solution.\n\nSinon, $(n^{2}-n-1)-(n-2)=n^{2}-2 n+1=(n-1)^{2}>0$, donc $n^{2}-n-1$ est une puissance de $5$ plus grande que $n-2$, donc $n-2$ divise $n^{2}-n-1=(n-2)(n+1)+1$, donc $n-2$ divise $1$, donc $n=3$, et dans ce cas $n^{3}-3 n^{2}+n+2=5$, donc $m=1$, $n=3$ est solution.\n\nAinsi les seules solutions sont $m=0, n=1$ et $m=1, n=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExiste-t-il un sous-ensemble infini $A$ de $\\mathbb{N}$ qui vérifie la propriété suivante : toute somme finie d'éléments distincts de $A$ n'est jamais une puissance d'un entier (c'est-à-dire un entier de la forme $a^{b}$ avec $a$ et $b$ entiers supérieurs ou égaux à 2) ?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn cherche à construire un tel ensemble $A=\\{a_{0}, a_{1}, \\ldots\\}$, avec $(a_{0}, a_{1}, \\ldots)$ une suite croissante d'entiers. On choisit comme premier élément $a_{0}=0$.\n\nSoit $n \\in \\mathbb{N}$. On suppose qu'on a déjà trouvé $a_{0}<\\cdots<a_{n}$ tels que l'ensemble $B$ des sommes d'éléments distincts ne contienne pas de puissance d'un entier. On veut choisir $a_{n+1}>a_{n}$ tel que l'ensemble des sommes d'éléments distincts de $\\{a_{0}, \\ldots, a_{n+1}\\}$ ne contienne pas de puissance d'un entier. Il suffit qu'aucun des $a_{n+1}+b$, avec $b \\in B$, ne soit une puissance d'un entier.\n\nSoit $N \\in \\mathbb{N}$. Majorons la proportion $p(N)$ de puissances d'entiers dans $\\{a_{n}+1, \\ldots, N\\}$. Il y a moins de $\\sqrt{N}$ carrés, moins de $\\sqrt[3]{N}$ cubes, et ainsi de suite jusqu'aux racines $k$-ièmes, avec $k=\\lfloor\\log_{2}(N)\\rfloor$. On peut s'arrêter à $k$ car pour tout $b>k$, $2^{b}>N$ donc il n'y a pas de puissance $b$-ième dans $\\{a_{n}+1, \\ldots, N\\}$.\n\nDonc\n$$\np(N) \\leq \\frac{\\sqrt{N}+\\sqrt[3]{N}+\\cdots+\\sqrt[k]{N}}{N-a_{n}} \\leq \\frac{\\sqrt{N} \\log_{2}(N)}{N-a_{n}} \\underset{N \\rightarrow+\\infty}{\\longrightarrow} 0.\n$$\n\nAinsi, pour $N$ assez grand, on peut trouver $1+\\max B$ nombres consécutifs dans $\\{a_{n}+1, \\ldots, N\\}$ qui ne sont pas des puissances d'entiers. On choisit pour $a_{n+1}$ le plus petit d'entre eux. Alors aucun des $a_{n+1}+b$, avec $b \\in B$, n'est une puissance d'un entier.\n\nFinalement, on a bien construit un ensemble $A$ satisfaisant la propriété demandée.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $S$ l'ensemble des nombres à deux chiffres qui ne contiennent pas le chiffre $0$. Deux nombres de $S$ sont dits amis si leurs plus grands chiffres sont égaux, et si la différence entre leurs plus petits chiffres est égale à $1$. Par exemple, $68$ et $85$ sont amis, $78$ et $88$ sont amis, mais $58$ et $75$ ne sont pas amis.\n\nDéterminer le plus grand entier $m$ tel qu'il existe une partie $T$ de $S$ possédant $m$ éléments, telle que deux éléments quelconques de $T$ ne soient pas amis.",
    "options": [],
    "answer": "45",
    "solution": "Solution:\n\nRéponse : $45$. On peut prendre pour $T$ l'ensemble des nombres dont le plus petit chiffre est impair.\n\nRéciproquement, si $x=\\overline{a b}$ avec $1 \\leqslant b < a \\leqslant 9$ alors $x$ et $x+1$ sont amis. Si $x=\\overline{a b}$ avec $2 \\leqslant a < b \\leqslant 9$ et $a$ pair, alors $x$ et $x+10$ sont amis. On a ainsi trouvé $36$ paires d'amis disjointes. Par conséquent, parmi les $72$ nombres $\\geqslant 21$, $T$ ne peut contenir qu'au plus $36$ nombres, donc $|T| \\leqslant 9+36=45$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30536,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a$ et $b$ des entiers strictement positifs. Prouver que si\n$$\na+\\frac{b}{a}-\\frac{1}{b}\n$$\nest un entier alors c'est un carré.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEn multipliant par $a$, on obtient que $a^{2}+b-\\frac{a}{b}$ est un entier, donc $k=\\frac{a}{b}$ est un entier.\n\nDe même, en multipliant par $b$ on voit que $\\frac{b^{2}}{a}$ est un entier. Comme $\\frac{b^{2}}{a}=\\frac{b}{k}$, on a $b=k c$ où $c$ est un entier.\n\nComme $\\frac{b}{a}-\\frac{1}{b}=\\frac{1}{k}-\\frac{1}{k c}$ est un entier, en multipliant par $k$ on voit que $\\frac{1}{c}$ est un entier, donc $c=1$, $b=k$ et $a=k^{2}$.\n\nFinalement, $a+\\frac{b}{a}-\\frac{1}{b}=k^{2}$ est un carré.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient deux cercles $\\omega_{1}$, $\\omega_{2}$ tangents l'un à l'autre en un point $T$, tels que $\\omega_{1}$ soit à l'intérieur de $\\omega_{2}$. Soient $M$ et $N$ deux points distincts sur $\\omega_{1}$, différents de $T$. Soient $[AB]$ et $[CD]$ deux cordes du cercle $\\omega_{2}$ passant respectivement par $M$ et $N$. On suppose que les segments $[BD]$, $[AC]$, et $[MN]$ s'intersectent en un point $K$.\n\nMontrer que $(TK)$ est la bissectrice de l'angle $\\widehat{MTN}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $E$ et $F$ les points d'intersection respectifs de $(TM)$ et $(TN)$ avec $\\omega_{2}$, autres que $T$. Puisque $\\omega_{1}$ et $\\omega_{2}$ sont tangents en $T$, l'homothétie de centre $T$ et qui envoie $\\omega_{1}$ sur $\\omega_{2}$ envoie également $M$ sur $E$ et $N$ sur $F$. On en déduit que\n$$\n\\frac{\\mathrm{TM}}{\\mathrm{TN}}=\\frac{\\mathrm{ME}}{\\mathrm{NF}}\n$$\nEn outre, la puissance des points $M$ et $N$ par rapport à $\\omega_{2}$ vaut respectivement $AM \\cdot MB=TM \\cdot ME$ et $CN \\cdot ND=TN \\cdot NF$. Cela montre que\n$$\n\\frac{\\mathrm{TM}^2}{\\mathrm{TN}^2}=\\frac{\\mathrm{TM} \\cdot \\mathrm{ME}}{\\mathrm{TN} \\cdot \\mathrm{NF}}=\\frac{AM \\cdot MB}{CN \\cdot ND}\n$$\nD'autre part, dans les triangles $AMK$ et $DNK$, la loi des sinus indique que\n$$\nAM \\cdot \\sin (\\widehat{MAK})=MK \\cdot \\sin (\\widehat{AKM}) \\text{ et que } DN \\cdot \\sin (\\widehat{NDK})=NK \\cdot \\sin (\\widehat{DKN}) .\n$$\nPuisque $A, B, C$ et $D$ sont cocycliques, on a également $\\sin (\\widehat{BAC})=\\sin (\\widehat{BDC})$, de sorte que\n$$\n\\sin (\\widehat{MAK})=\\sin (\\widehat{BAC})=\\sin (\\widehat{BDC})=\\sin (\\widehat{KDN})\n$$\nIl s'ensuit que $\\frac{AM}{DN}=\\frac{MK \\cdot \\sin (\\widehat{AKM})}{NK \\cdot \\sin (\\widehat{DKN})}$ et, de même, que $\\frac{BM}{CN}=\\frac{MK \\cdot \\sin (\\widehat{BKM})}{NK \\cdot \\sin (\\widehat{CKN})}$. Puisque les angles $\\widehat{BKM}$ et $\\widehat{DKN}$ d'une part, et $\\widehat{AKM}$ et $\\widehat{CKM}$ d'autre part, sont opposés par le sommet, donc sont égaux deux à deux, on en déduit que\n$$\n\\frac{AM \\cdot MB}{DN \\cdot NC}=\\frac{MK^2}{NK^2}\n$$\nOn en conclut que $TM \\cdot NK=TN \\cdot MK$.\n\nLa loi des sinus dans les triangles $TKM$ et $TKN$ indique en outre que\n$$\n\\mathrm{TM} \\cdot \\sin (\\widehat{KTM})=\\mathrm{KM} \\cdot \\sin (\\widehat{MKT}) \\text{ et que } \\mathrm{TN} \\cdot \\sin (\\widehat{KTN})=\\mathrm{KN} \\cdot \\sin (\\widehat{NKT})\n$$\nPuisque $\\widehat{MKT}$ et $\\widehat{NKT}$ sont supplémentaires, ils ont même sinus, de sorte que\n$$\n\\frac{\\sin (\\widehat{KTM})}{\\sin (\\widehat{KTN})}=\\frac{\\mathrm{KM} \\cdot \\sin (\\widehat{MKT}) \\cdot \\mathrm{TN}}{\\mathrm{TM} \\cdot \\mathrm{KN} \\cdot \\sin (\\widehat{NKT})}=1\n$$\nce qui prouve que $(TK)$ est la bissectrice de $\\widehat{MTN}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle acutangle. On note $O$ le centre de son cercle circonscrit. On considère le cercle $\\Gamma_{B}$ passant par $A$ et $B$ et qui est tangent à $(AC)$, ainsi que le cercle $\\Gamma_{C}$ passant par $A$ et $C$ qui est tangent à $(AB)$. Une droite $\\Delta$ passant par $A$ recoupe $\\Gamma_{B}$ en $X$ et $\\Gamma_{C}$ en $Y$. Montrer que $OX = OY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $Z$ le point d'intersection du cercle circonscrit de $ABC$ avec $\\Delta$. En faisant une chasse aux angles, on trouve que $\\widehat{XBA} = \\widehat{ZAC} = \\widehat{ZBC}$. On en déduit que $\\widehat{XBZ} = \\widehat{ABC}$. Or, comme $\\widehat{BCA} = \\widehat{BZA}$, on en déduit que les triangles $ABC$ et $XBZ$ sont semblables. Donc\n$$\n\\frac{XZ}{AC} = \\frac{BZ}{BC}.\n$$\nOn montre de même que les triangles $CAY$ et $CBZ$ sont semblables, de sorte que\n$$\n\\frac{AY}{BZ} = \\frac{AC}{BC}\n$$\nOn en déduit que $AY = XZ$, puis que $AX = ZY$. Ceci implique que les médiatrices des segments $[XY]$ et $[AZ]$ sont les mêmes. Donc $O$ appartient à la médiatrice de $[XY]$. Donc $OX = OY$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30539,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $m$ et $n$ des entiers naturels non nuls. Prouver que\n$$\n\\frac{(m+n)!}{(m+n)^{m+n}} < \\frac{m!}{m^{m}} \\frac{n!}{n^{n}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nD'après la formule du binôme, on a\n$$\n(m+n)^{m+n} = \\sum_{k=0}^{m+n} \\binom{m+n}{k} m^{k} n^{m+n-k}\n$$\nLe membre de droite étant une somme de $m+n+1$ nombres strictement positifs, elle est strictement plus grande que n'importe lequel de ces nombres. En particulier, pour $k=m$, on a donc\n$$\n(m+n)^{m+n} > \\binom{m+n}{m} m^{m} n^{n}\n$$\nPuisque $\\binom{m+n}{m} = \\frac{(m+n)!}{m! n!}$, cela conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nPour quel $n \\geqslant 1$ peut-on remplir un échiquier $n \\times n$ avec des pièces de la forme\n![](attached_image_1.png)\nsans qu'elles se recouvrent? (Les rotations sont autorisées).",
    "options": [],
    "answer": "Exactly when n is divisible by 4",
    "solution": "Solution:\nSupposons que nous avons réussi à recouvrir un échiquier $n \\times n$. Une pièce recouvre 4 cases. Donc clairement 4 divise $n^{2}$ donc $n$ est pair. Maintenant, chaque pièce recouvre soit trois cases noires et une case blanche de l'échiquier, soit l'inverse. Or, comme $n$ est pair, il y a autant de cases noires que de cases blanches. Il faut donc autant de pièces recouvrant trois cases noires que de pièces recouvrant trois cases blanches. Par conséquent, il y a un nombre pair de pièces, d'où 8 divise $n^{2}$. Donc, si l'on parvient à couvrir un échiquier $n \\times n$, alors 4 divise $n$.\n\nRéciproquement, si 4 divise $n$, on peut facilement recouvrir l'échiquier avec des blocs $4 \\times 4$.\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer que pour tout entier $n \\geq 1$, il existe $m$ un multiple de $n$ tel que la somme des chiffres de $m$ fasse $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLa suite des puissances de $10$, donc $1, 10, 100, \\cdots$ contient une infinité de termes, qui peuvent prendre un nombre fini de résidus modulo $n$. Il existe donc un résidu $k$ tel qu'une infinité de puissances de $10$ soient congrues à $k$ modulo $n$. Soient donc $0 \\leqslant a_{1} < a_{2} < \\cdots < a_{n}$ des entiers tels que\n$$\n10^{a_{i}} \\equiv k \\quad (\\bmod n)\n$$\npour tout $i \\in \\{1,2, \\cdots, n\\}$. On pose $m = 10^{a_{n}} + 10^{a_{n-1}} + \\cdots + 10^{a_{2}} + 10^{a_{1}}$. $m$ s'écrit avec $n$ chiffres \"1\" et un certain nombre de zéros, donc sa somme des chiffres vaut $n$.\n\nEt $m \\equiv k + \\cdots + k \\equiv k n \\equiv 0 \\pmod{n}$, donc $m$ est divisible par $n$, donc cet entier satisfait aux conditions de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère 2017 droites du plan, qui se rencontrent deux à deux en des points distincts. On appelle $E$ l'ensemble de ces points d'intersection.\nOn veut attribuer une couleur à chacun des points de $E$ de sorte que deux quelconques de ces points qui appartiennent à une même droite et dont le segment qui les relient ne contient aucun autre point de $E$, soient de couleurs différentes.\nCombien faut-il au minimum de couleurs pour pouvoir réaliser une telle coloration?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nLe minimum $m$ cherché est $m=3$ et, avec ce qui suit, il sera assez évident que le résultat reste vrai pour $n \\geq 3$ droites.\n\nTout d'abord, on note que, dans la configuration obtenue, il y a au moins une région non subdivisée qui est un triangle. En effet, trois droites non concourantes et deux jamais parallèles forment un triangle. Or, toute droite qui traverse un triangle, le partage en deux polygones dont au moins un est un triangle et donc, en partant de trois des droites données et en \"ajoutant\" une par une les 2014 autres, on est assuré de l'existence d'une telle région triangulaire dans la configuration finale. Les trois sommets d'un tel triangle doivent être de couleurs distinctes, ce qui implique que $m \\geq 3$.\n\nPour conclure, nous allons construire une coloration adéquate à trois couleurs. On commence par remarquer que, puisqu'il n'y a qu'un nombre fini de points d'intersection, on peut choisir un repère orthogonal dans lequel ces points ont des abscisses deux à deux distinctes. On numérote alors les points $M_{1}, M_{2} \\ldots, M_{k}$ selon les abscisses croissantes (et où $k=\\frac{2017 \\times 2016}{2}$ ). Pour tout $i$, le point $M_{i}$ possède au plus quatre voisins et, s'il appartient à un segment qui joint deux de ses voisins, un seul de ces deux voisins a une abscisse inférieure à celle de $M_{i}$. Cela assure que, pour tout $i \\leq k$, parmi les voisins de $M_{i}$, il y en a au plus deux qui ont des indices inférieurs à $i$. On peut alors colorier les $M_{i}$ dans l'ordre de la numérotation selon la procédure suivante : on colorie $M_{1}$ en vert et $M_{2}$ en rouge. Et, pour tout $i$ tel que $2 \\leq i \\leq k-1$, si l'on suppose que les points $M_{1}, M_{2} \\ldots, M_{i}$ ont été colorés chacun soit en vert, soit en rouge, soit en bleu de sorte que deux points voisins ne soient pas (encore) de la même couleur alors, d'après la remarque précédente, au plus deux voisins de $M_{i+1}$ ont déjà été colorés, ce qui laisse une couleur libre pour $M_{i+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30543,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère 5 nombres entiers positifs. En les ajoutant deux à deux de toutes les façons possibles, on génère 10 entiers. Montrer que ces 10 entiers ne peuvent pas être 10 entiers consécutifs.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn suppose que les 10 entiers sont consécutifs, et on note $n$ le plus petit. Leur somme est $n + (n+1) + \\cdots + (n+9) = 10n + 45$. Mais c'est aussi la somme des 5 nombres de départ, comptés 4 fois chacun, donc c'est un multiple de 4. C'est impossible car pour tout entier $n$, $10n + 45$ est impair. On aboutit à une contradiction, donc les 10 entiers ne peuvent pas être consécutifs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30544,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les entiers strictement positifs $a, b$ et $c$ tels que $\\operatorname{PGCD}(a, b, c)=1$,\n$$\na\\mid (b-c)^2, \\quad b\\mid (a-c)^2, \\quad c \\mid (a-b)^2\n$$\net tels qu'il est possible de construire un triangle non aplati dont les côtés ont pour longueurs $a$, $b$ et $c$.",
    "options": [],
    "answer": "a = b = c = 1",
    "solution": "Solution:\nTout d'abord, vérifions que $\\operatorname{PGCD}(a, b)=1$. Raisonnons par l'absurde en supposant que $p$ est un nombre premier qui divise à la fois $a$ et $b$. Alors $p$ divise $a$, qui lui même divise\n$$\n(b-c)^2 = b(b-2c) + c^2\n$$\ndonc $p$ divise $c^2$ et donc $p$ divise $c$. Ceci contredit le fait que $\\operatorname{PGCD}(a, b, c)=1$.\n\nSupposons tout d'abord que deux des nombres sont égaux, par exemple $a=b$ par symétrie. Comme $a$ et $b$ sont premiers entre eux, on $a = b = 1$. La seule possibilité pour construire un triangle non-aplati est de prendre $c=1$.\n\nSupposons maintenant que $a, b, c$ sont deux à deux différents. Raisonnons par l'absurde, en supposant qu'il est possible de construire un triangle non aplati dont les côtés ont pour longueur $a$, $b$ et $c$. Par symétrie, supposons que $c > b > a$.\n\nMontrons que $a$ divise $(a+b-c)^2$ et que $b$ divise $(a+b-c)^2$. On $a(a+b-c)^2 = a^2 + 2a(b-c) + (b-c)^2$, donc $a$ divise $(a+b-c)^2$. De même $b$ divise $(a+b-c)^2$.\n\nComme $a$ et $b$ sont premiers entre eux, on en déduit que $ab$ divise $(a+b-c)^2$. Or $0 < a+b-c < a$, donc\n$$\nab > a^2 > (a+b-c)^2\n$$\nce qui est absurde.\n\nAinsi, la seule solution est $a = b = c = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans les cases d'un tableau rectangulaire à $n$ lignes et $m$ colonnes sont écrits des nombres réels. On suppose que pour toute ligne ou colonne, la somme des nombres écrits sur cette ligne ou colonne est un entier. Montrer qu'il est possible de remplacer chaque réel $x$ par l'entier $\\lfloor x\\rfloor$ ou $\\lceil x\\rceil$ de sorte que les sommes de chaque colonne et de chaque ligne demeurent inchangées.\n\nRemarque : On rappelle que si $x$ est un nombre réel, $\\lfloor x\\rfloor$ est l'unique entier tel que $\\lfloor x\\rfloor \\leqslant x<\\lfloor x\\rfloor+1$, et $\\lceil x\\rceil$ est l'unique entier tel que $\\lceil x\\rceil-1<x \\leqslant\\lceil x\\rceil$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nQuitte à soustraire sa partie entière à chaque entrée du tableau, on peut supposer que chaque nombre est compris entre $0$ et $1$. On dit qu'une case est « fractionnaire » si son nombre n'est pas entier. Remarquons que, si $C$ est une case fractionnaire, alors il existe nécessairement une autre case fractionnaire sur la même ligne que $C$, et il existe aussi une autre case fractionnaire sur la même colonne que $C$.\n\nChoisissons donc une case $C_{0}$ fractionnaire, puis, pour tout $n \\in \\mathbb{N}$ :\n- si $n$ est pair, on choisit une case $C_{n+1}$ fractionnaire sur la même ligne que $C_{n}$;\n- si $n$ est impair, on choisit une case $C_{n+1}$ fractionnaire sur la même colonne que $C_{n}$.\n\nPar principe des tiroirs, il existe deux entiers $k<\\ell$ tels que $C_{k}=C_{\\ell}$. Sans perte de généralité, on suppose que $\\ell$ est aussi petit que possible, parmi tous les choix de cases $C_{0}, C_{1}, \\ldots$ et d'entiers $k<\\ell$ que l'on aura pu faire.\n\nOn a donc $k=0$, et l'on dispose ainsi d'un « cycle » $C_{0}, \\ldots, C_{\\ell-1}, C_{\\ell}$ de cases fractionnaires deux à deux distinctes (sauf $C_{0}$ et $C_{\\ell}$) et telles que les segments $[C_{i}, C_{i+1}]$ sont alternativement horizontaux et verticaux. Ce cycle étant de longueur $\\ell$ minimale, on sait que chaque ligne et chaque colonne contient soit $0$, soit $2$ cases du cycle, et que ces cases sont consécutives dans le cycle.\n\nPour tout $i<\\ell-1$, on note $x_{i}$ le nombre écrit sur la case $C_{i}$, et l'on pose $y_{i}=x_{i}$ si $i$ est pair, et $y_{i}=1-x_{i}$ si $i$ est impair. Puis l'on remplace $x_{i}$ par le réel\n$$\nx_{i}'=x_{i}+(-1)^{i} \\varepsilon\n$$\nou $\\varepsilon=\\min \\{y_{i} \\mid i<\\ell\\}$. Par construction de $\\varepsilon$, on a donc $0 \\leqslant x_{i}' \\leqslant 1$ pour tout $i<\\ell$, avec $x_{i}'=0$ ou $x_{i}'=1$ pour au moins un entier $i$.\n\nCe procédé n'a modifié la somme d'aucune ligne ni colonne, et a permis de réduire le nombre de cases fractionnaires, tout en affectant à chaque case modifiée une valeur $x_{i}'$ telle que\n$$\n\\{\\lfloor x_{i}'\\rfloor,\\lceil x_{i}'\\rceil\\} \\subseteq \\{0,1\\} = \\{\\lfloor x_{i}\\rfloor,\\lceil x_{i}\\rceil\\}.\n$$\nOn peut donc réitérer ce procédé jusqu'à éliminer toutes les cases fractionnaires, ce qui conclut le problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $P(x) = x^{4} - x^{3} - 3 x^{2} - x + 1$. Montrer qu'il existe une infinité d'entiers $n$ tels que $P\\left(3^{n}\\right)$ ne soit pas premier.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn observe que $3^{2} \\equiv -1 \\pmod{5}$ et $3^{4} \\equiv 1 \\pmod{5}$. Soit $n \\geqslant 1$. Soit $x = 3^{4n+1}$, alors $x = \\left(3^{4}\\right)^{n} \\times 3 \\equiv 3 \\pmod{5}$, donc $P(x) \\equiv 3^{4} - 3^{3} - 3^{2} - 3 + 1 \\equiv 1 + 3 + 3 - 3 + 1 \\equiv 0 \\pmod{5}$.\n\nD'autre part, $P(x) > x^{4} - x^{3} - 3 x^{3} - x^{3} = x^{3}(x-5) > x-5 > 3^{4n} - 5 > 5$, donc $P\\left(3^{4n+1}\\right)$ n'est pas premier.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes points $D$ et $E$ divisent le côté $[A B]$ d'un triangle équilatéral en trois parties égales, de telle manière que $D$ est situé entre $A$ et $E$. Le point $F$ est situé sur $[B C]$ de sorte que $C F = A D$. Calculer la somme des angles $\\widehat{C D F} + \\widehat{C E F}$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "30°",
    "solution": "Solution:\n\nOn a $B F = B D$ et $\\widehat{D B F} = 60^{\\circ}$, donc le triangle $D B F$ est équilatéral. On a donc $\\widehat{B A C} = \\widehat{B D F} = 60^{\\circ}$ donc $(D F) \\parallel (A C)$, donc $\\widehat{C D F} = \\widehat{A C D}$.\n\nD'autre part, $\\widehat{A C D} = \\widehat{B C E}$ par symétrie, donc $\\widehat{C D F} = \\widehat{B C E} = \\widehat{F C E}$. La somme qui nous intéresse vaut donc $\\widehat{F C E} + \\widehat{C E F} = \\widehat{B F E}$ d'après la somme des angles du triangle $F C E$.\n\nComme $(F E)$ est une médiane du triangle équilatéral $D B F$, c'est aussi une bissectrice, donc $\\widehat{B F E} = \\frac{1}{2} \\widehat{B F D} = 30^{\\circ}$ et notre somme est égale à $30^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer que $n! = 1 \\times 2 \\times \\cdots \\times n$ est divisible par $2^{n-1}$ si et seulement si $n$ est une puissance de $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn définit, pour tout nombre premier $p$, la valuation $p$-adique d'un entier $n$ comme le plus grand entier, noté $v_{p}(n)$, tel que $p^{v_{p}(n)}$ divise $n$. La valuation $p$-adique d'un produit d'entiers est la somme de leurs valuations $p$-adiques. Pour $x$ un nombre réel, on note $\\lfloor x\\rfloor$ la partie entière de $x$, c'est-à-dire le plus grand entier inférieur à $x$.\n\nOn observe que $\\{1, \\ldots, n\\}$ contient $\\left\\lfloor\\frac{n}{p}\\right\\rfloor$ multiples de $p$, $\\left\\lfloor\\frac{n}{p^{2}}\\right\\rfloor$ multiples de $p^{2}$, etc. On en déduit la formule de Legendre :\n$$\nv_{p}(n!) = \\left\\lfloor\\frac{n}{p}\\right\\rfloor + \\left\\lfloor\\frac{n}{p^{2}}\\right\\rfloor + \\ldots\n$$\nPour cet exercice, on prend $p=2$, et on définit $k$ l'entier tel que $2^{k} \\leq n < 2^{k+1}$. Alors\n$$\nv_{2}(n!) = \\left\\lfloor\\frac{n}{2}\\right\\rfloor + \\left\\lfloor\\frac{n}{4}\\right\\rfloor + \\cdots + \\left\\lfloor\\frac{n}{2^{k}}\\right\\rfloor \\leq \\frac{n}{2} + \\frac{n}{4} + \\cdots + \\frac{n}{2^{k}} = n\\left(1 - \\frac{1}{2^{k}}\\right) \\leq n-1.\n$$\nPour que $n!$ soit divisible par $2^{n-1}$, c'est-à-dire pour que $v_{2}(n!) \\geq n-1$, il faut être dans les cas d'égalité. En particulier, il faut que $n\\left(1 - \\frac{1}{2^{k}}\\right) = n-1$, donc que $n = 2^{k}$.\n\nRéciproquement, si $n = 2^{k}$, alors on est dans les cas d'égalité, donc $2^{n-1}$ divise $n!$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $[AB]$ le diamètre d'un demi-cercle sur lequel on prend deux points $C$ et $D$. Soit $S$ l'intersection de $(AC)$ et $(BD)$ et $T$ le pied de la perpendiculaire à $[AB]$ issue de $S$.\n\nMontrer que $(ST)$ est la bissectrice de l'angle $\\widehat{CTD}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn s'en sort par chasse aux angles. Les points $A, D, S, T$ sont cocycliques (sur le cercle de diamètre $[AS]$) et de même les points $B, C, S, T$. Donc $\\widehat{STD} = \\widehat{SAD}$, et $\\widehat{STC} = \\widehat{SBC}$. Comme $\\widehat{SAD} = \\widehat{CAD} = \\widehat{CBD} = \\widehat{SBC}$ par cocyclicité des points $A, B, C, D$, on a bien $\\widehat{STC} = \\widehat{STD}$.\n\nOn peut aussi résoudre cet exercice avec de la géométrie projective. En effet, on remarque $(ST)$ est la bissectrice de $\\widehat{CTD}$ si et seulement si $(AB)$, $(ST)$, $(TD)$, $(TC)$ sont harmoniques (grâce au lemme 2/4). Soit $F$ l'intersection de $(BD)$ et de $(TC)$. Il suffit de montrer que $B, S, D, F$ sont harmoniques, donc il suffit de montrer que $B, A, (CD) \\cap (AB), T$ sont harmoniques. Or $(CD) \\cap (AB)$ est par construction sur la polaire de $S$ donc la polaire de $(CD) \\cap (AB)$ passe par $S$. De plus, elle est perpendiculaire à $(AB)$, donc $c'$ est $(ST)$. C'est pourquoi on a bien $(CD) \\cap (AB), T, A, B$ harmoniques, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a_{0} < a_{1} < a_{2} < \\ldots$ une suite infinie d'entiers strictement positifs. Prouver qu'il existe un unique entier $n \\geqslant 1$ tel que\n$$\na_{n} < \\frac{a_{0} + a_{1} + a_{2} + \\cdots + a_{n}}{n} \\leqslant a_{n+1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour $n \\geqslant 1$, on pose $d_{n} = a_{0} + a_{1} + \\cdots + a_{n} - n a_{n}$.\nClairement, l'inégalité de gauche de l'encadrement souhaité est équivalente à $d_{n} > 0$.\nD'autre part, on a $d_{n+1} = a_{0} + \\cdots + a_{n} - n a_{n+1}$, donc l'inégalité de droite est équivalente à $d_{n+1} \\leqslant 0$. Il s'agit donc de prouver qu'il existe un unique $n \\geqslant 1$ tel que $d_{n+1} \\leqslant 0 < d_{n}$, ce qui nous pousse naturellement à étudier les variations de la suite $(d_{i})$.\nPour tout $n \\geqslant 1$, on a $d_{n+1} - d_{n} = n(a_{n} - a_{n+1}) < 0$ car la suite $(a_{i})$ est strictement croissante. De plus, on a $d_{1} = a_{0} > 0$.\nAinsi, la suite $(d_{i})$ est une suite strictement décroissante d'entiers, ce qui assure qu'elle ne prendra des valeurs négatives ou nulles qu'à partir d'un certain rang, et de premier terme strictement positif. Il existe donc bien un unique indice $n$ pour lequel $d_{n+1} \\leqslant 0 < d_{n}$, à savoir le plus grand entier tel que $d_{n} > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30551,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $A_{1}, A_{2}, \\ldots, A_{m}$ des sous-ensembles distincts de $\\{1,2, \\ldots, n\\}$ tels que pour tout $1 \\leqslant i, j \\leqslant n$ on a $A_{i} \\cap A_{j} \\neq \\emptyset$. Montrer $m \\leqslant 2^{n-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsidérons toutes les paires d'ensembles $\\{A, A^{c}\\}$ pour tout $A \\subset \\{1,2, \\ldots, n\\}$. Puisqu'il y a $2^{n-1}$ telles paires, si $m > 2^{n-1}$ on aurait $1 \\leqslant i < j \\leqslant m$ tels que $A_{i}, A_{j} \\in \\{A, A^{c}\\}$ pour un $A \\subset \\{1,2, \\ldots, n\\}$. Donc $A_{i} \\cap A_{j} = \\emptyset$ ce qui contredit notre hypothèse.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle rectangle en $C$. Soient $D$ le pied de la hauteur issue de $C$, et $Z$ le point de $[AB]$ tel que $AC = AZ$. La bissectrice de $\\widehat{BAC}$ coupe $(CB)$ et $(CZ)$ en $X$ et $Y$ respectivement. Montrer que les quatre points $B$, $X$, $Y$, $D$ sont sur un même cercle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\n$CAZ$ est isocèle en $A$ donc $(AY)$, qui est la bissectrice de $\\widehat{CAZ}$, est perpendiculaire à $(CZ)$. Donc $ADYC$ est cyclique.\n\nOn a alors $\\widehat{DYX} = 180^{\\circ} - \\widehat{AYD} = 180^{\\circ} - \\widehat{ACD} = 180^{\\circ} - \\widehat{XBD}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30553,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a$, $b$ et $c$ trois nombres strictement positifs tels que $a + b + c + abc = 4$. Montrer que\n$$\n\\left(1 + \\frac{a}{b} + ca\\right)\\left(1 + \\frac{b}{c} + ab\\right)\\left(1 + \\frac{c}{a} + bc\\right) \\geqslant 27\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEn utilisant l'inégalité arithmético-géométrique, on trouve que\n$$\n4 = a + b + c + abc \\geqslant 4\\left((abc)^2\\right)^{\\frac{1}{4}}\n$$\net donc $abc \\leqslant 1$. Ensuite, en utilisant encore l'inégalité arithmético-géométrique,\n$$\n\\left(1 + \\frac{a}{b} + ca\\right)\\left(1 + \\frac{b}{c} + ab\\right)\\left(1 + \\frac{c}{a} + bc\\right) = \\frac{(4-a)(4-b)(4-c)}{abc} = 15 + 4\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right)\n$$\net en utilisant encore l'inégalité arithmético-géométrique on en déduit que\n$$\n\\left(1 + \\frac{a}{b} + ca\\right)\\left(1 + \\frac{b}{c} + ab\\right)\\left(1 + \\frac{c}{a} + bc\\right) \\geqslant 15 + 12\\left(\\frac{1}{abc}\\right)^{1/3} \\geqslant 27\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont tous les angles sont aigus.\nLes hauteurs $[AA_{1}]$, $[BB_{1}]$ et $[CC_{1}]$ se coupent au point $H$. Soit $A_{2}$ le symétrique de $A$ par rapport à $[B_{1}C_{1}]$, et soit $O$ le centre du cercle circonscrit à $ABC$.\n\na) Prouver que les points $O, A_{2}, B_{1}, C$ sont cocycliques.\n\nb) Prouver que $O, H, A_{1}, A_{2}$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\na) $A, H, B_{1}, C_{1}$ sont cocycliques sur le cercle de diamètre $[AH]$, donc $\\widehat{AB_{1}C_{1}} = \\widehat{AHC_{1}} = 90^{\\circ} - \\widehat{C_{1}AH}$. Or, $(AA_{2}) \\perp (B_{1}C_{1})$ donc $\\widehat{A_{2}AB_{1}} = \\widehat{C_{1}AH}$, i.e. les demi-droites $[AH)$ et $[AA_{2})$ sont symétriques par rapport à la bissectrice de $\\widehat{BAC}$. On en déduit que $A_{2} \\in [AO)$. Comme $AO = CO$, on a $\\widehat{ACO} = \\widehat{OAC} = \\widehat{AA_{2}B_{1}}$, donc $O, A_{2}, B_{1}, C$ sont cocycliques.\n\nb) En utilisant la puissance de $A$ par rapport aux cercles $O, A_{2}, B_{1}, C$ et $H, A_{1}, C, B_{1}$, on a $AB_{1} \\cdot AC = AO \\cdot AA_{2}$ et $AB_{1} \\cdot AC = AH \\cdot AA_{1}$. Par conséquent, $AO \\cdot AA_{2} = AH \\cdot AA_{1}$. D'après la puissance d'un point par rapport à un cercle, on en déduit que $O, A_{2}, H, A_{1}$ sont cocycliques.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un entier naturel $d$ est sympathique si, pour tout couple d'entiers $(x, y)$,\n$$\nd\\left|(x+y)^{5}-x^{5}-y^{5} \\Longleftrightarrow d\\right|(x+y)^{7}-x^{7}-y^{7}\n$$\n\nMontrer qu'il existe une infinité de nombres sympathiques. Est-ce-que 2017, 2018 sont sympathiques ?",
    "options": [],
    "answer": "There are infinitely many: every prime distinct from five and seven is friendly (and any product of coprime friendly numbers is also friendly). Both 2017 and 2018 are friendly.",
    "solution": "Solution:\n\nOn remarque que :\n$$\n\\begin{aligned}\n& (x+y)^{5}-x^{5}-y^{5}=5 x y(x+y)\\left(x^{2}+x y+y^{2}\\right) \\\\\n& (x+y)^{7}-x^{7}-y^{7}=7 x y(x+y)\\left(x^{2}+x y+y^{2}\\right)^{2}\n\\end{aligned}\n$$\nAinsi, si on prend $p$ un nombre premier différent de $5$ et de $7$ :\n- si $p$ divise $(x+y)^{5}-x^{5}-y^{5}$, comme $p$ et $5$ sont premiers entre eux, d'après le lemme de Gauß, $p$ divise aussi $x y(x+y)\\left(x^{2}+x y+y^{2}\\right)$, a fortiori $p$ divise $(x+y)^{7}-x^{7}-y^{7}$.\n- si $p$ divise $(x+y)^{7}-x^{7}-y^{7}$, comme $p$ et $7$ sont premiers entre eux, $p$ divise également $x y(x+y)\\left(x^{2}+x y+y^{2}\\right)^{2}$. Si $p$ divise $x^{2}+x y+y^{2}$, alors $p$ divise $(x+y)^{5}-x^{5}-y^{5}$ (car c'est bien un multiple de $x^{2}+x y+y^{2}$ ). Sinon, comme $p$ premier, cela signifie que $p$ et $x^{2}+x y+y^{2}$ sont premiers entre eux, donc d'après le lemme de Gauß, $p$ divise $x y(x+y)$, d'où l'on déduit que $p$ divise $(x+y)^{5}-x^{5}-y^{5}$.\nDonc un tel nombre $p$ est sympathique.\nComme il y a une infinité de nombres premiers différents de $5$ et de $7$, il y a bien une infinité de nombres sympathiques.\n\nEn particulier, le nombre $2017$ est premier, différent de $5$ et de $7$, donc sympathique.\n\nDe plus, si $a$ et $b$ sont deux entiers sympathiques premiers entre eux, comme on a en général, pour tout entier $n$,\n$$\na b \\text{ divise } n \\Leftrightarrow a \\text{ divise } n \\text{ et } b \\text{ divise } n \\text{, }\n$$\n$a b$ est aussi sympathique. Donc $2018=2 \\cdot 1009$ est sympathique.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDeux cercles $\\mathcal{C}$ et $\\mathcal{C}^{\\prime}$ de centres $O$ et $O^{\\prime}$ sont tangents extérieurement en $B$. Une tangente commune extérieure touche $\\mathcal{C}$ en $M$ et $\\mathcal{C}^{\\prime}$ en $N$. La tangente commune à $\\mathcal{C}$ et $\\mathcal{C}^{\\prime}$ en $B$ coupe $(M N)$ en $A$. On note $C$ l'intersection de $(O A)$ et $(B M)$, et $D$ l'intersection de $\\left(O^{\\prime} A\\right)$ et $(B N)$.\nMontrer que $(C D)$ est parallèle à $(M N)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComme $(A M)$ et $(A B)$ sont tangents au cercle de centre $O$, on a $A M = A B$. On a aussi $O M = O B$, donc $(O A)$ est la médiatrice de $[M B]$, et donc $C$ est le milieu de $[M B]$.\n\nDe même, $D$ est le milieu de $[B N]$, donc $(C D)$ est parallèle à $(M N)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30557,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle tel que $AB < AC$, $H$ son orthocentre, $\\Gamma$ son cercle circonscrit, $d$ la tangente à $\\Gamma$ en $A$. On considère le cercle de centre $B$ passant par $A$. Il coupe $d$ en $D$ et $(AC)$ en $E$.\n\nMontrer que $D$, $E$, $H$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn remarque tout d'abord que la hauteur issue de $B$ dans $ABC$ est la médiatrice de $[AE]$. Soit en effet $L$ son point d'intersection avec $(AC)$. Par la puissance d'un point, $CE \\cdot CA = CB^2 - AB^2 = CL^2 - AL^2 = CA \\cdot (CL - AL)$ donc $LA = LE$.\n\nDonc $HAE$ est isocèle en $H$. Donc, par chasse aux angles,\n$$\n\\widehat{HEA} = \\widehat{HAC} = 90^\\circ - \\widehat{ACB} = 90^\\circ - \\widehat{DAB} = \\frac{1}{2} \\widehat{ABD} = \\widehat{DEA}.\n$$\n\nDonc $D$, $E$, $H$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30558,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers $n \\geqslant 3$ pour lesquels il existe $n$ polynômes $P_{1}, \\ldots, P_{n}$ à coefficients réels et tels que, pour tous $i, j, k$ distincts : $P_{i}+P_{j}$ n'a aucune racine réelle et $P_{i}+P_{j}+P_{k}$ admet au moins une racine réelle.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nProuvons que seul $n=3$ a la propriété désirée.\n\nTout d'abord, en posant $P_{1}(x)=3x^{2}+3$ et $P_{2}(x)=P_{3}(x)=-x^{2}-2$, on vérifie aisément que les polynômes $P_{1}(x)+P_{2}(x)=P_{1}(x)+P_{3}(x)=2x^{2}+1$ et $P_{2}(x)+P_{3}(x)=-2x^{2}-4$ n'ont aucune racine réelle, tandis que $P_{1}(x)+P_{2}(x)+P_{3}(x)=x^{2}-1$ admet $1$ et $-1$ comme racines. Ainsi, $n=3$ est une solution du problème.\n\nRemarquons maintenant que si $n \\geqslant 4$ polynômes $P_{1}, \\ldots, P_{n}$ vérifient les conditions requises, alors $P_{1}, P_{2}, P_{3}$ et $P_{4}$ vérifient déjà ces conditions. Pour conclure, il suffit donc de prouver que $n=4$ n'est pas une solution.\n\nLa clé est alors de noter que si $P$ et $Q$ sont des polynômes tels que $P+Q$ n'a pas de racine réelle, alors $P+Q$ garde un signe constant sur $\\mathbb{R}$.\n\nOn procède donc par l'absurde : supposons que $P_{1}, P_{2}, P_{3}$ et $P_{4}$ soient des polynômes tels que, pour tous $i, j, k$ distincts, $P_{i}+P_{j}$ n'a aucune racine réelle et $P_{i}+P_{j}+P_{k}$ admet au moins une racine réelle.\n\nTraçons alors le graphe complet dont les sommets sont les polynômes $P_{1}, P_{2}, P_{3}, P_{4}$ et colorons l'arête $P_{i}P_{j}$ en rouge si $P_{i}+P_{j}$ est strictement positif, et en vert si $P_{i}+P_{j}$ est strictement négatif. Chaque arête est nécessairement colorée. De plus, ce graphe ne peut contenir de triangles dont les trois arêtes sont de la même couleur, ce sans quoi la somme des trois polynômes qui sont les sommets de ce triangle n'aurait pas de racine réelle.\n\nQuitte à changer chaque $P_{i}$ en $-P_{i}$, on peut supposer que deux des trois arêtes issues de $P_{1}$ sont rouges et, quitte à renuméroter, on peut supposer que $P_{1}P_{2}$ et $P_{1}P_{4}$ sont rouges. Pour éviter le triangle rouge $P_{1}P_{2}P_{4}$, c'est donc que $P_{2}P_{4}$ est verte. Puisque $P_{2}P_{3}P_{4}$ ne peut avoir ses trois arêtes vertes, c'est donc que $P_{2}P_{3}$ ou $P_{4}P_{3}$ est rouge. Par symétrie des rôles, on peut supposer que $P_{2}P_{3}$ est rouge. Mais alors $P_{1}P_{3}$ est verte.\n\nAinsi, pour tout réel $x$, on a simultanément\n$$\n\\begin{aligned}\n& P_{1}(x)+P_{4}(x)>0 \\text{ et } P_{2}(x)+P_{3}(x)>0, \\text{ donc } P_{1}(x)+P_{2}(x)+P_{3}(x)+P_{4}(x)>0, \\text{ et } \\\\\n& P_{1}(x)+P_{3}(x)<0 \\text{ et } P_{2}(x)+P_{4}(x)<0, \\text{ donc } P_{1}(x)+P_{2}(x)+P_{3}(x)+P_{4}(x)<0,\n\\end{aligned}\n$$\nce qui est absurde. Ainsi, $n=4$ n'est pas une solution, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nCombien de tableaux $3 \\times 3$ peut-on construire en les remplissant avec les nombres de 1 à 3 tels qu'il n'y ait pas deux fois le même nombre dans une ligne ni dans une colonne. Et de tableaux $4 \\times 4$ avec les nombres de 1 à 4 ?",
    "options": [],
    "answer": "3x3: 12; 4x4: 576",
    "solution": "Solution:\nNous présentons ci-dessous une preuve complète de l'exercice. Il n'était pas nécessaire d'être aussi précis pour avoir tous les points mais il est bon, lorsque que l'on compte des objets, d'avoir une idée de la preuve pour démontrer que nous n'avons rien oublié ni compté en double.\nNous avons six façons de remplir la première rangée (trois façons de placer le $1$ puis deux de placer le $2$ puis une seule possibilité pour placer le $3$). Ensuite, il nous reste deux façons de remplir la colonne contenant déjà un $1$ (ce qui fait douze en tout). On se retrouve avec, par exemple, le tableau\n$$\n\\left[\\begin{array}{lll}\n2 & 1 & 3 \\\\\n& 2 & A \\\\\n& 3 &\n\\end{array}\\right] .\n$$\nOn se rend compte qu'il n'y a qu'une seule façon de remplir la case $A$, c'est-à-dire celle qui est dans la rangée contenant un $2$ et dans la colonne contenant un $3$. Puis on complète le tableau de la seule manière possible.\nOn a donc douze façons de remplir un tableau $3 \\times 3$ comme dans l'énoncé.\n\nPour les tableaux $4 \\times 4$, c'est à peu près la même idée. Nous avons vingt-quatre façons de remplir la première rangée et six façons de remplir la colonne contenant un $1$. Nous obtenons un tableau ressemblant à\n$$\n\\left[\\begin{array}{cccc}\n2 & 1 & 4 & 3 \\\\\n& 4 & & \\\\\nA_{3,2} & 3 & & \\\\\n& 2 & &\n\\end{array}\\right] .\n$$\nMaintenant, notons $A_{k, l}$, $k, l \\in\\{2,3,4\\}$, la case dans la rangée contenant $k$ et la colonne contenant $l$. Nous voyons qu'il y a deux possibilités pour remplir $A_{3,2}$ :\n- si l'on remplit $A_{3,2}$ avec $1$, nous n'aurons plus qu'une seule possibilité pour $A_{4,2}$ puis, dans l'ordre, $A_{2,2}$, $A_{3,4}$, $A_{3,3}$, $A_{4,4}$, $A_{4,3}$, $A_{2,4}$ et $A_{2,3}$. On obtient l'unique tableau\n$$\n\\left[\\begin{array}{llll}\n2 & 1 & 4 & 3 \\\\\n3 & 4 & 1 & 2 \\\\\n1 & 3 & 2 & 4 \\\\\n4 & 2 & 3 & 1\n\\end{array}\\right]\n$$\n- si l'on remplit $A_{3,2}$ avec $4$ nous avons de nouveau deux possibilités:\n* on remplit $A_{4,2}$ avec $1$ puis on remplit les autres cases de la seule façon possible.\n$$\n\\left[\\begin{array}{llll}\n2 & 1 & 4 & 3 \\\\\n1 & 4 & 3 & 2 \\\\\n4 & 3 & 2 & 1 \\\\\n3 & 2 & 1 & 4\n\\end{array}\\right] .\n$$\n* on remplit $A_{4,2}$ avec $3$ puis on remplit $A_{2,2}$, $A_{2,4}$ et $A_{2,3}$ de la seule façon possible. Il reste deux façons de remplir le tableau obtenu.\n$$\n\\left[\\begin{array}{llll}\n2 & 1 & 4 & 3 \\\\\n3 & 4 & & \\\\\n4 & 3 & & \\\\\n1 & 2 & 3 & 4\n\\end{array}\\right] .\n$$\nRemarque que toutes les affirmations sont toujours vraies, pas seulement dans l'exemple. Donc, quelque soit la façon de remplir la première rangée et la colonne contenant un $1$, il reste toujours quatre façons de remplir le tableau. Il y a donc en tout $24 \\cdot 6 \\cdot 4=576$ façons de remplir un tableau $4 \\times 4$ selon l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a_{0}, a_{1}, \\ldots, a_{99}, b_{0}, b_{1}, \\ldots, b_{99}$ des réels strictement positifs.\nPour $k=0,1, \\ldots, 198$, on pose $S_{k} = \\sum_{i=0}^{198} a_{i} b_{k-i}$, avec $a_{j} = 0$ et $b_{j} = 0$ si $j < 0$ ou $j > 99$.\nEst-il possible que les nombres $S_{0}, S_{1}, \\ldots, S_{198}$ soient tous égaux?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nNon. Raisonnons par l'absurde. En effet, comme $a_{0} b_{0} = a_{0} b_{99} + \\cdots + a_{99} b_{0}$ où les $\\cdots$ représentent des termes strictement positifs, on a $a_{0} b_{0} > a_{0} b_{99}$ donc $b_{0} > b_{99}$, et de même $a_{0} > a_{99}$. Par conséquent, $S_{0} = a_{0} b_{0} > a_{99} b_{99} = S_{198}$, ce qui est contradictoire.\n\n\nAutre démonstration qui ne nécessite pas l'hypothèse de positivité (mais seulement le fait que $a_{99} \\neq 0$ et $b_{99} \\neq 0$) :\nsupposons que $S_{0} = S_{1} = \\ldots = S_{198} = \\alpha$.\nNotons que $\\alpha \\neq 0$.\nOn pose $P(x) = a_{0} + a_{1} x + \\ldots + a_{99} x^{99}$ et $Q(x) = b_{0} + b_{1} x + \\ldots + b_{99} x^{99}$. Comme $P(x) Q(x) = S_{0} + S_{1} x + \\ldots + S_{198} x^{198} = \\alpha (1 + x + \\ldots + x^{198})$\n$= \\alpha \\frac{1 - x^{199}}{1 - x}$ pour $x \\neq 1$, on constate que $P(x) Q(x) \\neq 0$ pour tout $x \\neq 1$. D'autre part, on a $\\alpha \\neq 0$ puisque $P$ et $Q$ sont des polynômes non nuls, donc $P(1) Q(1) = 199 \\alpha \\neq 0$.\nPar conséquent, $P(x) Q(x) \\neq 0$ pour tout réel $x$, et donc $P$ n'admet pas de racine réelle. Ceci contredit le fait que $P$ est de degré impair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $m, n$ des entiers positifs tels que $\\operatorname{pgcd}(m, n)=1$, où $a \\wedge b$ désigne le plus grand diviseur commun de $a$ et $b$. Quelle(s) valeur(s) peut prendre\n$$\n\\left(2^{m}-2^{n} \\wedge 2^{m^{2}+m n+n^{2}}-1\\right) ?\n$$",
    "options": [],
    "answer": "1 and 7",
    "solution": "Solution:\nOn utilise la propriété suivante : si $m \\geq 1$ et $a, b \\in \\mathbb{N}$, alors $\\left(m^{a}-1\\right) \\wedge\\left(m^{b}-1\\right)= m^{a \\wedge b}-1$. On se ramène ainsi à étudier $\\left(m^{2}+m n+n^{2}\\right) \\wedge(m-n)$. Si $d \\mid m^{2}+m n+n^{2}$ et $d \\mid m-n$, alors $d \\mid m^{2}+m n+n^{2}-(m-n)^{2}=3 m n$. Donc $d \\mid 3 m n-3 n(m-n)=3 n^{2}$ et $d \\mid 3 m n+3 m(m-n)=3 m^{2}$, donc $d \\mid\\left(3 m^{2}, 3 n^{2}\\right)$, donc $d \\mid 3$. Ainsi, $\\left(m^{2}+m n+n^{2}\\right) \\wedge(m-n) \\mid 3$, et ne peut prendre qu'au plus deux valeurs, 1 et 3. Vérifions qu'elles sont réalisées.\nSi $m=2$ et $n=1$, $\\left(m^{2}+m n+n^{2}\\right) \\wedge(m-n)=1$ et\n$$\n\\left(2^{m}-2^{n}\\right) \\wedge\\left(2^{m^{2}+m n+n^{2}}-1\\right)=2^{\\left(m^{2}+m n+n^{2}\\right) \\wedge(m-n)}-1=1\n$$\nSi $m=n=1$, $\\left(m^{2}+m n+n^{2}\\right) \\wedge(m-n)=3$ et\n$$\n\\left(2^{m}-2^{n}\\right) \\wedge\\left(2^{m^{2}+m n+n^{2}}-1\\right)=2^{\\left(m^{2}+m n+n^{2}\\right) \\wedge(m-n)}-1=7.\n$$\nDonc les valeurs possibles sont 1 et 7.\n\nPreuve de la propriété : on peut supposer $a \\geq b$. On écrit $a=b q+r$ la division euclidienne de $a$ par $b$. Soit $d:=\\left(m^{a}-1\\right) \\wedge\\left(m^{b}-1\\right)$. $d \\mid\\left(m^{b}-1\\right)+\\left(m^{2 b}-m^{b}\\right)+\\left(m^{3 b}-m^{2 b}\\right)+\\cdots+\\left(m^{q b}-m^{(q-1) b}\\right)$, soit $d \\mid m^{q b}-1$. Donc $d \\mid m^{a}-m^{q b}$. Or $m^{a}-m^{q b}=m^{q b}\\left(m^{r}-1\\right)$. Comme $d$ est un diviseur de $m^{b}-1$ qui est premier avec $m$, alors $d$ est premier avec $m^{q b}$. D'après le lemme de Gauss, $d \\mid m^{r}-1$. Ainsi, $d \\mid\\left(m^{b}-1\\right) \\wedge\\left(m^{r}-1\\right)$. On procède de même en faisant la division euclidienne de $b$ par $r$ et ainsi de suite. On reproduit ainsi l'algorithme d'Euclide. Ce dernier termine sur $a \\wedge b$, donc on obtient à la fin que $d$ divise $m^{a \\wedge b}-1$. Pour conclure, il reste à prouver que $m^{a \\wedge b}-1$ est un diviseur de $m^{b}-1$ et de $m^{a}-1$. Or, si $i=k j$ pour des entiers $i, j, k$, $m^{j}-1$ divise $m^{i}-1$, car :\n$$\nm^{i}-1=\\left(m^{j}\\right)^{k}-1=\\left(m^{j}-1\\right)\\left(\\left(m^{j}\\right)^{k-1}+\\left(m^{j}\\right)^{k-2}+\\cdots+m^{j}+1\\right)\n$$\nCeci permet de conclure.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30562,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $ABC$ un triangle, $\\Gamma$ son cercle circonscrit. Soit $\\omega_{A}$ le cercle inscrit intérieurement à $(AB),(AC)$ et à $\\Gamma$. On note $T_{A}$ le point de tangence de $\\Gamma$ avec $\\omega_{A}$. On définit de même $T_{B}$ et $T_{C}$.\nMontrer que $\\left(AT_{A}\\right),\\left(BT_{B}\\right)$ et $\\left(CT_{C}\\right)$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nQuand on voit beaucoup de cercles tangents, on doit penser aux homothéties qui les échangent. En l'occurrence, on a trois homothéties positives, de centre respectifs $T_{A}, T_{B}, T_{C}$ et qui envoient respectivement $\\omega_{A}, \\omega_{B}$ et $\\omega_{C}$ sur $\\Gamma$.\n\nDe plus, $\\omega_{A}$ étant tangent à $(AB)$ et $(AC)$, on a aussi une homothétie positive de centre $A$ qui envoie $\\omega_{A}$ sur le cercle $\\omega$ inscrit à $ABC$. De même pour $B$ et $C$.\n\nOr la composée de deux homothéties positives de centres $O_{1}$ et $O_{2}$ est une homothétie positive dont le centre est sur la droite $(O_{1}O_{2})$. On a donc construit trois homothéties positives qui envoient toutes $\\Gamma$ sur $\\omega$, et leurs centres respectifs appartiennent à $(AT_{A}),(BT_{B})$ et $(CT_{C})$.\n\nComme il existe au plus une homothétie positive envoyant un cercle donné sur un autre cercle, ces trois homothéties sont en fait une seule et même homothétie, de centre $Z$. On a donc $Z \\in (AT_{A}), Z \\in (BT_{B})$ et $Z \\in (CT_{C})$.\n\nLes droites $(AT_{A}),(BT_{B})$ et $(CT_{C})$ sont donc concourantes.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30563,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont les trois angles sont aigus et $\\Gamma$ son cercle circonscrit. La tangente à $\\Gamma$ en $A$ recoupe $(BC)$ en $P$. On note $M$ le milieu de $[AP]$. La droite $(BM)$ recoupe $\\Gamma$ en $R$ et la droite $(PR)$ recoupe $\\Gamma$ en $S$.\n\nMontrer que $(AP)$ et $(CS)$ sont parallèles.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEn écrivant la puissance de $M$ par rapport à $\\Gamma$ puis le fait que $M$ est le milieu de $[AP]$, on obtient\n$$\nMR \\times MB = MA^{2} = MP^{2}\n$$\nLes triangles $MRP$ et $MPB$ sont donc indirectement semblables, donc $\\widehat{RPM} = \\widehat{PBM}$. On a donc\n$$\n\\begin{aligned}\n\\widehat{CSP} &= \\widehat{CSR} \\\\\n&= \\widehat{CBR} \\\\\n&= \\widehat{PBM} \\\\\n&= \\widehat{RPM} \\\\\n&= \\widehat{SPA},\n\\end{aligned}\n$$\noù $\\widehat{CSR} = \\widehat{CBR}$ par le théorème de l'angle inscrit. Par angles alternes-internes, les droites $(CS)$ et $(AP)$ sont donc parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30564,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a \\in [0 ; 1]$. On définit la suite $\\left(x_{n}\\right)$ par\n$x_{0} = a$ et $x_{n+1} = 1 - \\left|1 - 2 x_{n}\\right|$, pour tout $n \\geq 0$.\n\nProuver que la suite $\\left(x_{n}\\right)$ est périodique à partir d'un certain rang si et seulement si $a$ est un nombre rationnel.\n\n(Note: on dit que $\\left(x_{n}\\right)$ est périodique à partir d'un certain rang s'il existe des entiers $T > 0$ et $n \\geqslant 0$ tels que $x_{k+T} = x_{k}$ pour tout $k \\geqslant n$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note tout d'abord que si $x_{n} \\in [0 ; 1]$ alors $-1 \\leq 1 - 2 x_{n} \\leq 1$, d'où $0 \\leq x_{n+1} \\leq 1$. Puisque $x_{0} \\in [0 ; 1]$, on déduit ainsi par récurrence que $x_{n} \\in [0 ; 1]$ pour tout $n \\geq 0$.\n\n- Supposons que $a$ soit un nombre rationnel. Une récurrence immédiate assure alors que, pour tout $n \\geq 0$, $x_{n}$ est un nombre rationnel.\n\nPour tout $n \\geq 0$, on pose $x_{n} = \\frac{p_{n}}{q_{n}}$, avec $p_{n}$ et $q_{n}$ entiers positifs et premiers entre eux. Alors $x_{n+1} = \\frac{q_{n} - \\left|q_{n} - 2 p_{n}\\right|}{q_{n}}$, d'où $q_{n+1} \\leq q_{n}$.\n\nLa suite $\\left(q_{n}\\right)$, suite décroissante d'entiers naturels, est alors stationnaire à partir d'un certain rang. Il existe donc des entiers $q > 0$ et $N \\geq 0$ tels que $q_{n} = q$ pour tout $n \\geq N$.\n\nPour tout $n \\geq N$, le nombre $x_{n}$ est donc un nombre rationnel de $[0 ; 1]$ dont le dénominateur vaut $q$. Or, il n'y a qu'un nombre fini de tels rationnels et il existe donc $n_{0} \\geq N$ et $k > 0$ tels que $x_{n_{0} + k} = x_{n_{0}}$. La définition par récurrence de $\\left(x_{n}\\right)$ assure alors que $\\left(x_{n}\\right)_{n \\geq n_{0}}$ est périodique de période $k$.\n\n- Supposons maintenant qu'il existe des entiers $k > 0$ et $N \\geq 0$ tels que $x_{n+k} = x_{n}$ pour tout $n \\geq N$.\n\nPour $n \\geq 0$, on a $x_{n+1} = 2 x_{n}$ ou $x_{n+1} = 2 - 2 x_{n}$ selon que $1 - 2 x_{n}$ est positif ou négatif. On peut donc poser $x_{n+1} = a + 2 b x_{n}$, où $a$ est un entier et $b = \\pm 1$. On prouve alors facilement par récurrence que, pour tout $i \\geq 0$, il existe un entier $a_{i}$ tel que $x_{N+i} = a_{i} + 2^{i} b_{i} x_{N}$ avec $b_{i} = \\pm 1$.\n\nEn particulier, on a $x_{N} = x_{N+k} = a_{k} + 2^{k} b_{k} x_{N}$. Puisque $2^{k} b_{k} = \\pm 2^{k} \\neq 1$, cette équation du premier degré en $x_{N}$ admet une unique solution $x_{N} = \\frac{a_{k}}{1 - 2^{k} b_{k}}$. Ainsi, $x_{N}$ est un nombre rationnel. Or, il est facile de vérifier que, pour $m \\geq 0$, si $x_{m+1}$ est rationnel alors $x_{m}$ est rationnel donc, par récurrence descendante, on déduit que $a = x_{0}$ est rationnel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPeut-on arranger les nombres $1,1,2,2, \\ldots, 50,50$ tels que, pour tout $1 \\leqslant k \\leqslant 50$, entre les deux nombres $k$ il y ait exactement $k$ éléments.",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nLa réponse est non. Supposons, par l'absurde, qu'on le peut. Pour tout $1 \\leqslant k \\leqslant 50$, soit $a_{k}$ un entier tel que $a_{k}$ et $a_{k}+k+1$ sont les positions des deux $k$. Évidemment, on a\n$$\n\\sum_{k=1}^{50}\\left(a_{k}+a_{k}+k+1\\right)=1+2+\\ldots+100\n$$\nEn regardant l'égalité précédente modulos 2, on doit avoir $51 \\times 26-1 \\equiv 50 \\times 101\\ (\\bmod\\ 2)$ ce qui est faux.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30566,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Déterminer toutes les fonctions $f: \\mathbb{R} \\longrightarrow \\mathbb{Z}$ telles que\n$$\nf(f(y)-x)^{2}+f(x)^{2}+f(y)^{2}=f(y)(1+2 f(f(y)))\n$$\npour tous réels $x$ et $y$.\nb) Déterminer toutes les fonctions $f: \\mathbb{R} \\longrightarrow \\mathbb{R}$ telles que\n$$\nf(f(y)-x)^{2}+f(x)^{2}+f(y)^{2}=f(y)(1+2 f(f(y)))\n$$\npour tous réels $x$ et $y$.",
    "options": [],
    "answer": "For both parts (a) and (b), the only solutions are the constant functions f(x) = 0 for all real x and f(x) = 1 for all real x.",
    "solution": "Solution:\na) Soit $f$ une éventuelle solution du problème. Notons $P(x, y)$ l'égalité de l'énoncé pour les valeurs $x$ et $y$. On pose $c=f(0)$.\nPour tout $x$, de $P(0, x)$ on déduit que $f(f(x))^{2}+c^{2}+f(x)^{2}=f(x)+2 f(f(x)) f(x)$, d'où $f(x)=c^{2}+(f(f(x))-f(x))^{2}$. Il s'ensuit que $f(x) \\geqslant c^{2} \\geqslant 0$.\n\nPour $x=0$, cela donne $c=c^{2}+(f(c)-c)^{2}$, d'où $c(1-c)=(f(c)-c)^{2} \\geqslant 0$. Puisque $c$ est un entier, il vient $c=0$ ou $c=1$. Dans les deux cas, cela montre que $f(c)-c=0$. Ainsi, on sait que $f(c)=c$.\n\nEnfin, pour tout $x$, de $P(x, 0)$ on déduit que\n$$\nf(c-x)^{2}+f(x)^{2}+c^{2}=c(1+2 c)\n$$\n- Si $c=0$, l'égalité ci-dessus se réécrit comme $f(-x)^{2}+f(x)^{2}=0$, donc $f(x)=0$.\n- Si $c=1$, elle se réécrit comme $f(1-x)^{2}+f(x)^{2}=2$. Or, $f(1-x)$ et $f(x)$ sont entiers naturels, donc $f(x)=f(1-x)=1$.\n\nRéciproquement, les fonctions $f: x \\longmapsto 0$ et $f: x \\longmapsto 1$ sont clairement solutions du problème.\n\nb) Soit $f$ une éventuelle solution du problème. Notons $P(x, y)$ l'égalité de l'énoncé pour les valeurs $x$ et $y$. On pose $c=f(0), d=f(c)$ et $\\mathcal{F}=\\{f(x) \\mid x \\in \\mathbb{R}\\}$.\nPour tout $x \\in \\mathbb{R}$, de $P(0, x)$ on déduit que $f(f(x))^{2}+c^{2}+f(x)^{2}=f(x)+2 f(f(x)) f(x)$, d'où $f(x)=c^{2}+(f(f(x))-f(x))^{2}$. Par conséquent, pour tout $x \\in \\mathcal{F}$, on a $x \\geqslant c^{2} \\geqslant 0$ et $f(x)=x \\pm \\sqrt{x-c^{2}}$.\n\nIl s'ensuit que $c \\geqslant c^{2}$, c'est-à-dire que $c(1-c) \\geqslant 0$, donc que $0 \\leqslant c \\leqslant 1$. On procède alors par l'absurde, et on suppose que $0<c<1$.\n\nPar inégalité arithmético-géométrique, il vient $0<c(1-c) \\leqslant\\left(\\frac{c+1-c}{2}\\right)^{2}=\\frac{1}{4}$. D'autre part, de manière générale, pour tout réel $x \\in] 0 ; 1[$, on sait que $x<\\sqrt{x}$.\n\nOr, on a prouvé ci-dessus que $d=c+\\epsilon \\sqrt{c-c^{2}}$, avec $\\epsilon= \\pm 1$. Si $\\epsilon=-1$, alors $d-c^{2}=\\left(c-c^{2}\\right)-\\sqrt{c-c^{2}}<0$, ce qui est impossible. Il vient donc $\\epsilon=1$, et $d=c+\\sqrt{c-c^{2}}$. Puis on trouve $f(d)=d+\\varepsilon \\sqrt{d-c^{2}}$, avec $\\varepsilon= \\pm 1$. Observons que\n$$\n0<c-c^{2} \\leqslant d-c^{2} \\leqslant\\left(c-c^{2}\\right)+\\sqrt{c-c^{2}} \\leqslant \\frac{1}{4}+\\frac{1}{2}=\\frac{3}{4}<1 .\n$$\nPar conséquent, si $\\varepsilon=-1$, alors $f(d)-c^{2}=\\left(d-c^{2}\\right)-\\sqrt{d-c^{2}}<0$, ce qui est encore une fois impossible. Il vient donc $\\varepsilon=1$, et $f(d)=d+\\sqrt{d-c^{2}}$.\n\nEnfin, $P(d, 0)$ indique que\n$$\n\\begin{aligned}\n0 & =f(c-d)^{2}+f(d)^{2}+c^{2}-c(1+2 d)=f(c-d)^{2}+\\left(d+\\sqrt{d-c^{2}}\\right)^{2}+c^{2}-c(1+2 d) \\\\\n& \\geqslant c^{4}+d^{2}+d-c^{2}+c^{2}-c(1+2 d)=c^{4}+(d-c)^{2}+(d-c)-c^{2} \\\\\n& \\geqslant c^{4}+c-c^{2}+\\sqrt{c-c^{2}}-c^{2} \\geqslant c^{4}+c-c^{2}+c-c^{2}-c^{2}=c(c-1)^{2}(c+2)\n\\end{aligned}\n$$\nce qui est impossible.\n\nIl s'ensuit que $c=0$ ou $c=1$, donc que $c=c^{2}$ et que $f(c)=c \\pm \\sqrt{c-c^{2}}=c$. Par conséquent, pour tout $x, P(x, 0)$ indique que\n$$\nf(c-x)^{2}+f(x)^{2}+c^{2}=c(1+2 c)\n$$\n- Si $c=0$, l'égalité ci-dessus se réécrit comme $f(-x)^{2}+f(x)^{2}=0$, donc $f(x)=0$.\n- Si $c=1$, elle se réécrit comme $f(1-x)^{2}+f(x)^{2}=2$. Or, on a vu plus haut que $f(1-x) \\geqslant c^{2}=1$ et que $f(x) \\geqslant c^{2}=1$, ce qui montre que $f(x)=f(1-x)=1$.\n\nRéciproquement, les fonctions $f: x \\longmapsto 0$ et $f: x \\longmapsto 1$ sont clairement solutions du problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $O$ le centre d'un polygone régulier à 18 côtés de sommets $A_{1}, \\ldots, A_{18}$. Soit $B$ le point de $[OA_{1}]$ tel que $\\widehat{BA A_{2} O}=20^{\\circ}$ et $C$ le point de $[OA_{2}]$ tel que $\\widehat{CA_{1} O}=10^{\\circ}$. Montrer que $BCA_{2} A_{3}$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn prolonge déjà les différentes droites existantes : $A_{1}, O, A_{10}$ sont alignés, $A_{2}, O, A_{11}$ aussi.\n\nOn applique ensuite le théorème de l'angle inscrit : tout angle au centre qui intercepte un segment de longueur $A_{1}A_{2}$ vaut $20^{\\circ}$, tout angle inscrit qui intercepte un tel segment vaut $10^{\\circ}$. Donc $A_{1}, C, A_{9}$ alignés, $A_{2}, B, A_{13}$ alignés.\n\nPar symétrie axiale d'axe $(OA_{2})$, il vient $A_{3}, C, A_{13}$ alignés.\n\nSi on parvient à montrer que $A_{3}, B, A_{15}$ alignés, alors $\\widehat{CA_{3} B}=\\widehat{A_{13} A_{3} A_{15}}=20^{\\circ}=\\widehat{CA_{2} B}$, ce qui établit la cocyclicité demandée dans cet exercice.\n\nMontrons maintenant qu'on a effectivement $A_{3}, B, A_{15}$ alignés. Par symétrie axiale d'axe $(OA_{1})$, cela équivaut à montrer que $A_{17}, B, A_{5}$ sont alignés.\n\nOr, on remarque que $(A_{17}A_{5})$ est la médiatrice du segment $[OA_{2}]$ : en effet, le triangle $OA_{2}A_{5}$ est équilatéral (comme $OA_{2}=OA_{5}$ et $\\widehat{A_{2}OA_{5}}=60^{\\circ}$) donc $A_{5}$ est sur la médiatrice de $[OA_{2}]$, de même pour $A_{17}$.\n\nComme $BO_{2}$ est isocèle en $B$, $B$ est aussi sur la médiatrice de $[OA_{2}]$, ce qui établit l'alignement voulu et termine la solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30568,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les triplets d'entiers naturels $(x, y, z)$ tels que :\n$$\nx^{2}+y^{2}=3 \\cdot 2016^{z}+77.\n$$",
    "options": [],
    "answer": "{(4,8,0),(8,4,0),(35,70,1),(70,35,1),(14,77,1),(77,14,1)}",
    "solution": "Solution:\nSoit un triplet solution $(x, y, z)$. Que peut-on en dire de façon nécessaire ?\n\nSi $z=0$, on a $x^{2}+y^{2}=80$. On fait une table de congruence modulo $4$ : on montre ainsi que, si une somme de deux carrés est congrue modulo $4$, c'est que les deux carrés en question sont tous les deux pairs. On peut donc poser $x=2 x_{0}, y=2 y_{0}$ avec $x_{0}, y_{0}$ entiers et $x_{0}^{2}+y_{0}^{2}=20$. Par le même argument modulo $4$, on arrive $x=4 x_{1}, y=4 y_{1}$ avec $x_{1}, y_{1}$ entiers et $x_{1}^{2}+y_{1}^{2}=5$. Les seuls couples vérifiant cela sont $(x_{1}, y_{1})=(1,2)$ ou $(2,1)$.\n\nSi $z>0$, comme $7 \\mid 2016$, on a modulo $7$ $x^{2}+y^{2}=0$, ce qui donne, en écrivant la table des carrés modulo $7$, $x$ et $y$ congrus modulo $7$. On peut donc poser $x=7 x_{1}$ et $y=7 y_{1}$ avec $x_{1}, y_{1}$ entiers. Donc $49 \\mid x^{2}+y^{2}$, c'est-à-dire $49 \\mid 3 \\cdot 2016^{z}+77$ donc $z=1$. Il vient alors $x_{1}^{2}+y_{1}^{2}=125$. On teste les cas restants un par un pour voir ceux qui marchent à la main : $(x_{1}, y_{1})=(5,10),(2,11),(10,5),(11,2)$.\n\nAinsi, tous les triplets solutions sont dans l'ensemble :\n$$\n\\{(4,8,0),(8,4,0),(35,70,1),(70,35,1),(14,77,1),(77,14,1)\\}\n$$\net on vérifie que, réciproquement, tous ces triplets sont bien des solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProuver qu'il existe un entier $n>0$ tel que parmi les 2016 chiffres de droite dans l'écriture décimale de $2^{n}$, il y a au moins 1008 chiffres 9.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn peut légitimement se demander quand trouver des 9 à la droite de l'écriture décimale de $2^{n}$. On peut penser que c'est quand la puissance de 2 est légèrement inférieure à une puissance de 10. On va donc chercher des nombres de la forme $2^{n}+1$ qui sont divisibles par 5 selon une puissance élevée. Pour un tel nombre, en multipliant par la puissance de 2 adéquate, on obtiendra un nombre de la forme $2^{k}\\left(2^{n}+1\\right)=2^{n+k}+2^{k}$ divisible par une grande puissance de 10, et il suffira de retrancher le $2^{k}$ en question pour obtenir le type de puissance de 2 cherché.\n\nLemme. Pour tout entier $k \\geq 1$, le nombre $a_{k}=2^{2 \\cdot 5^{k-1}}+1$ est divisible par $5^{k}$.\n\nPreuve du lemme. On raisonne par récurrence sur $k$ :\n- Pour $k=1$, on a $a_{1}=5$ qui est bien divisible par 5.\n- Supposons le résultat établi pour un certain $k \\geq 1$.\nPosons $a=4^{5^{k-1}}=a_{k}-1$. On note qu'alors $a=-1 \\bmod 5$.\nOn a alors $a_{k+1}=4^{5^{k}}+1=a^{5}+1=(a+1)\\left(a^{4}-a^{3}+a^{2}-a+1\\right)$.\nD'après l'hypothèse de récurrence, on a $a+1=0 \\bmod 5^{k-1}$.\nD'autre part, puisque $a=-1 \\bmod 5$, on a $a^{4}-a^{3}+a^{2}-a+1=0 \\bmod 5$.\nAinsi, $a_{k+1}$ est divisible par $5^{k}$, ce qui achève la récurrence.\n\nComme prévu, on en déduit que, pour tout $k \\geq 1$, le nombre $2^{2 k}\\left(2^{2 \\cdot 5^{2 k-1}}+1\\right)=2^{2 k+2 \\cdot 5^{2 k-1}}+2^{2 k}$ est divisible par $10^{2 k}$. Son écriture décimale se termine donc par au moins $2 k$ chiffres 0.\n\nOr, pour $k \\geq 1$, on a $2^{2 k}<10^{k}$. Ainsi, l'écriture décimale de $2^{2 k}$ n'utilise pas plus de $k$ chiffres. Par suite, parmi les $2 k$ chiffres de droite de l'écriture décimale de $2^{2 k+2 \\cdot 5^{2 k-1}}$, il y a au moins $k$ chiffres 9.\n\nLe résultat demandé correspond à $k=1008$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30570,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $(I)$ le centre de son cercle inscrit. La droite $(AI)$ recoupe $[BC]$ en $D$. La médiatrice de $[AD]$ recoupe $(BI)$ en $M$ et $(CI)$ en $N$.\n\nMontrer que $A$, $M$, $N$ et $I$ sont cocycliques.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par rappeler le théorème du Pôle Sud :\n\nLemme 1 (Théorème du Pôle Sud). Soit $ABC$ un triangle, $\\Gamma$ son cercle circonscrit et $I$ le centre de son cercle inscrit. Soit $S$ le second point d'intersection de la bissectrice de $\\widehat{BAC}$ avec $\\Gamma$. Alors $SB = SC$. Le point $S$ est appelé Pôle Sud de $A$ dans le triangle $ABC$.\n\nPreuve du lemme. Le théorème de l'angle inscrit dans $\\Gamma$ donne $\\widehat{SBC} = \\widehat{SAC}$ et $\\widehat{SCB} = \\widehat{SAB}$. Or, comme $S$ est sur la bissectrice de $\\widehat{BAC}$ on a $\\widehat{SAC} = \\widehat{SAB}$, d'où $\\widehat{SBC} = \\widehat{SCB}$, donc $SCB$ est isocèle en $S$ et $SB = SC$.\n\nRevenons à notre exercice. Notons que le pôle Sud de $A$ dans un triangle $ABC$ est sur la bissectrice de $\\widehat{BAC}$ et sur la médiatrice de $[BC]$, donc il peut être défini comme l'intersection de ces deux droites. On commence par montrer que $A$, $B$, $D$ et $M$ sont cocycliques. Le point $M$ est sur la médiatrice de $[AD]$ et sur la bissectrice de $\\widehat{ABD}$. C'est donc le pôle Sud de $B$ dans $ABD$, donc il est sur le cercle circonscrit à $ABD$. De même, les points $A$, $C$, $D$ et $N$ sont cocycliques. On peut maintenant faire une chasse aux angles :\n\n$$\n\\widehat{AMI} = \\widehat{AMB} = \\widehat{ADB} = 180^\\circ - \\widehat{ADC} = 180^\\circ - \\widehat{ANC} = 180^\\circ - \\widehat{ANI}\n$$\n\nLes points $A$, $I$, $M$ et $N$ sont donc cocycliques.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1) Pierre répartit les entiers $1,2, \\ldots, 2012$ en deux groupes disjoints dont les sommes respectives des éléments sont égales.\nSans même regarder la répartition choisie par Pierre, Clara affirme alors que l'on peut éliminer deux nombres de chaque groupe de sorte que, dans chaque groupe, les sommes respectives des éléments restants soient égales.\nProuver que Clara a raison.\n\n2) Pierre répartit les entiers $1,2, \\ldots, 20$ en deux groupes disjoints dont les sommes respectives des éléments sont égales.\nSans même regarder la répartition choisie par Pierre, Clara affirme alors que l'on peut éliminer deux nombres de chaque groupe de sorte que, dans chaque groupe, les sommes respectives des éléments restants soient égales.\nProuver que, cette fois, Clara aurait peut-être mieux fait de regarder avant de parler.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn colorie les éléments du groupe $G_{1}$ en rouge, et ceux de $G_{2}$ en bleu. Sans perte de généralité, on peut supposer que $1$ est rouge. Lorsque deux entiers consécutifs ne sont pas dans le même groupe, on dit que l'on a une alternance.\n\n- Si l'on a au moins quatre alternances, on peut trouver la configuration ...rb...br...rb...br..., où les pointillés indiquent toujours que l'on reste sur la même couleur, sauf peut-être pour le dernier paquet. Dans la situation décrite, la conclusion est assurée en éliminant les entiers responsables des première et quatrième alternances (on ne peut utiliser les deux premières alternances car $c'$ est peut-être le même nombre bleu qui est concerné).\n\n- Si l'on a exactement trois alternances ...rb...br...rb... avec au moins deux bleus successifs dans le premier paquet de bleus, on peut effacer les nombres impliqués dans les deux premières alternances. De même, s'il y a au moins deux rouges successifs dans le second paquet de rouges, on peut effacer les nombres impliqués dans les seconde et troisième alternances.\n\nSi l'on n'est pas dans un de ces deux cas, c'est que l'on a ...rbrb...., où le premier groupe de pointillés ne désigne que des rouges, et le second groupe de pointillés ne désigne que des bleus. On appelle $k$ le nombre de rouges dans ce premier paquet. La somme des nombres rouges est alors $\\frac{k(k+1)}{2} + (k+2)$, alors que la somme de tous les entiers de $1$ à $2012$ vaut $2012 \\times 2013 / 2$. Puisque la somme des nombres dans chaque groupe est la même, on doit avoir $\\frac{k(k+1)}{2} + (k+2) = 2012 \\times 2012 / 4$, autrement dit, $k(k+3) + 4 = 2012 \\times 2013 / 2$. Or, cette équation n'a pas de solution modulo $3$, ce qui assure que cette configuration n'est finalement pas possible.\n\n- Si l'on a exactement deux alternances, on est forcément dans la configuration ....rb...br...., où les pointillés indiquent qu'il n'y a pas de changements de couleurs. S'il y a au moins deux bleus, on peut effacer les nombres impliqués dans les deux alternances. Sinon, c'est qu'il n'y a qu'un bleu, mais alors il doit être égal à $1012539$, ce qui est impossible.\n\n- S'il n'y a qu'une alternance, c'est que l'on est dans la configuration ....rb... Si, comme ci-dessus, on note $k$ le nombre de rouges, on doit avoir $\\frac{k(k+1)}{2} = 1012539$, soit $k(k+1) = 2025078$ qui n'a pas non plus de solution entière (on peut le voir modulo $10$).\n\nFinalement, dans tous les cas, on peut éliminer deux nombres de chaque groupe de sorte que, dans chaque groupe, les sommes respectives des éléments restants soient égales.\n\n2)\n\nPar exemple, pour $n=20$, on colorie en rouge les entiers de $1$ à $14$, et les autres en bleu. Les sommes des deux groupes sont égales, mais si l'on élimine deux rouges, la somme des rouges diminuera d'au plus $14+13=27$, alors que si on élimine deux bleus, la somme des bleus diminuera d'au moins $15+16=31$. Il est alors impossible qu'après élimination, les deux groupes aient encore des sommes égales.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère un quadrilatère convexe $ABCD$ dont les diagonales sont perpendiculaires et se coupent en $E$. Soit $P \\in [AD]$ un point tel que $P \\neq A$ et $PE = EC$. On suppose que le cercle circonscrit du triangle $BCD$ recoupe $[AD]$ en un point $Q$ différent de $A$. On note $R$ le deuxième point d'intersection de $[AC]$ avec le cercle passant par $A$ et tangent à $(EP)$ en $P$. On suppose que les points $B, R, Q$ sont alignés. Montrer que $\\widehat{BCD} = 90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIntroduisons $F$, le point d'intersection du cercle circonscrit du triangle $ARD$ et du segment $[DE]$. La puissance de $E$ par rapport au cercle passant par $A, R, F, D$ est $ER \\cdot EA = EF \\cdot ED$. Par ailleurs, la puissance de $E$ par rapport au cercle passant par $A, R, P$ est\n$$\nER \\cdot EA = EP^{2} = EC^{2}.\n$$\nOn en déduit que $ED \\cdot EF = EC^{2}$, ce qui montre que le cercle passant par $F, C, D$ est tangent à $(CE)$.\n\nOn en déduit que $\\widehat{ECF} = \\widehat{EDC}$. Par ailleurs, comme $A, R, F, D$ sont cocycliques, on a $\\widehat{ADE} = \\widehat{ERF}$. On en déduit que $\\widehat{QDB} + \\widehat{BDC} = \\widehat{FRC} + \\widehat{RCF}$, ce qui implique que $\\widehat{RBC} = \\widehat{RFC}$.\n\nComme $(BF)$ et $(CR)$ sont perpendiculaires, ceci implique que $F$ est le symétrique de $B$ par rapport à la droite $(AC)$. Il en découle que $\\widehat{BCR} = \\widehat{RCF} = \\widehat{BDC}$. Comme $\\widehat{BEC} = 90^{\\circ}$, on a $\\widehat{ECD} = 90^{\\circ} - \\widehat{BCE}$ et on conclut que $\\widehat{BCD} = 90^{\\circ}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\na) Soit $x_{1} \\geqslant x_{2}>0$ et $y_{1} \\geqslant y_{2}>0$ des réels tels que\n$$\nx_{1} \\geqslant y_{1} \\text{ et } x_{1} x_{2} \\geqslant y_{1} y_{2}\n$$\nProuver que\n$$\nx_{1}+x_{2} \\geqslant y_{1}+y_{2}\n$$\nb) Soit $x_{1} \\geqslant x_{2} \\geqslant \\ldots \\geqslant x_{n}>0$ et $y_{1} \\geqslant y_{2} \\geqslant \\ldots \\geqslant y_{n}>0$ des réels tels que\n$$\nx_{1} x_{2} \\cdots x_{i} \\geqslant y_{1} y_{2} \\cdots y_{i} \\text{ pour } i=1, \\ldots, n\n$$\nProuver que\n$$\nx_{1}+x_{2}+\\cdots+x_{n} \\geqslant y_{1}+y_{2}+\\cdots+y_{n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\na) Procédons par l'absurde: supposons que $y_{1}+y_{2}>x_{1}+x_{2}$. Alors $y_{2}>x_{1}+x_{2}-y_{1}$ et on a\n$$\n0 \\geqslant y_{1} y_{2}-x_{1} x_{2}>y_{1}\\left(x_{1}+x_{2}-y_{1}\\right)-x_{1} x_{2}=\\left(x_{1}-y_{1}\\right)\\left(y_{1}-x_{2}\\right)\n$$\nPuisque $x_{1} \\geqslant y_{1}$, alors $y_{1}<x_{2}$. Mais alors $x_{1}+x_{2} \\geqslant 2 y_{1} \\geqslant y_{1}+y_{2}$, ce qui contredit notre hypothèse. On a donc bien $y_{1}+y_{2} \\leqslant x_{1}+x_{2}$.\n\nb) Procédons par récurrence forte sur $n$. Tout d'abord, le résultat est évident pour $n=1$, et on vient de le prouver pour $n=2$. On suppose donc $n \\geqslant 3$.\nSoit $\\lambda=\\min _{1 \\leqslant k<n} \\frac{x_{1}+\\ldots+x_{k}}{y_{1}+\\ldots+y_{k}}$, et soit $\\ell$ un entier tel que $1 \\leqslant \\ell<n$ et tel que\n$$\nx_{1}+\\ldots+x_{\\ell}=\\lambda\\left(y_{1}+\\ldots+y_{\\ell}\\right)\n$$\nOn pose alors $y_{1}^{\\prime}=\\lambda y_{1}, y_{n}^{\\prime}=\\frac{y_{n}}{\\lambda}$ et $y_{k}^{\\prime}=y_{k}$ pour $k=2, \\ldots, n-1$. Par construction, on a bien $y_{1}^{\\prime} \\geqslant y_{2} \\geqslant \\ldots \\geqslant y_{n}^{\\prime}$ et, pour $k=1, \\ldots, n$, on a également $x_{1} \\ldots x_{k} \\geqslant y_{1}^{\\prime} \\ldots y_{k}^{\\prime}$, avec égalité quand $k=\\ell<n$. Par hypothèse de récurrence, on sait que $x_{1}+\\ldots+x_{\\ell} \\geqslant y_{1}^{\\prime}+\\ldots+y_{\\ell}^{\\prime}$.\nEn outre, remarquons que $y_{1}^{\\prime} \\geqslant y_{n}^{\\prime}, y_{1} \\geqslant y_{n}, y_{1}^{\\prime} \\geqslant y_{1}$ et $y_{1}^{\\prime} y_{n}^{\\prime}=y_{1} y_{n}$. En appliquant l'hypothèse de récurrence (pour $n=2$ ), on obtient donc $y_{1}^{\\prime}+y_{n}^{\\prime} \\geqslant y_{1}+y_{n}$, et même $y_{1}^{\\prime}+y_{2}^{\\prime}+\\ldots+y_{n}^{\\prime} \\geqslant y_{1}+y_{2}+\\ldots+y_{n}$.\nEnfin, posons $\\bar{x}_{i}=x_{\\ell+i}$ et $\\bar{y}_{i}=y_{\\ell+1}^{\\prime}$, pour $i=1, \\ldots, n-\\ell$. Alors $\\bar{x}_{1} \\geqslant \\ldots \\geqslant \\bar{x}_{n-\\ell}>0$ et $\\bar{y}_{1} \\geqslant \\ldots \\geqslant \\bar{y}_{n-\\ell}>0$. En outre, pour $i=1, \\ldots, n-\\ell$, on a\n$$\n\\left(x_{1} \\ldots x_{\\ell}\\right)\\left(\\bar{x}_{1} \\ldots \\bar{x}_{i}\\right)=x_{1} \\ldots x_{\\ell+i} \\geqslant y_{1}^{\\prime} \\ldots y_{\\ell+i}^{\\prime}=\\left(y_{1}^{\\prime} \\ldots y_{\\ell}^{\\prime}\\right)\\left(\\bar{y}_{1} \\ldots \\bar{y}_{i}\\right)\n$$\nPuisque $x_{1} \\ldots x_{\\ell}=y_{1}^{\\prime} \\ldots y_{\\ell}^{\\prime}>0$, cela montre que $\\bar{x}_{1} \\ldots \\bar{x}_{i} \\geqslant \\bar{y}_{1} \\ldots \\bar{y}_{i}$. On a $n-\\ell<n$, et l'hypothèse de récurrence montre donc que $\\bar{x}_{1}+\\ldots+\\bar{x}_{n-\\ell} \\geqslant \\bar{y}_{1}+\\ldots+\\bar{y}_{n-\\ell}$.\nMises bout à bout, ces inégalités montrent que\n$$\n\\begin{aligned}\nx_{1}+\\ldots+x_{n} & =\\left(x_{1}+\\ldots+x_{\\ell}\\right)+\\left(\\bar{x}_{1}+\\ldots+\\bar{x}_{n-\\ell}\\right) \\\\\n& \\geqslant y_{1}^{\\prime}+\\ldots+y_{\\ell}^{\\prime}+\\bar{y}_{1}+\\ldots+\\bar{y}_{n-\\ell}=y_{1}^{\\prime}+\\ldots+y_{n}^{\\prime} \\\\\n& \\geqslant y_{1}+\\ldots+y_{n} .\n\\end{aligned}\n$$\nSolution:\nOn pose\n$$\nx_{n+1}=\\min \\left(x_{n}, y_{n}\\right) \\frac{y_{1} \\cdots y_{n}}{x_{1} \\cdots x_{n}} \\text{ et } y_{n+1}=\\min \\left(x_{n}, y_{n}\\right)\n$$\nOn a alors $0<x_{n+1} \\leqslant x_{n}, 0<y_{n+1} \\leqslant y_{n}, x_{n+1} \\leqslant y_{n+1}$ et $x_{1} \\cdots x_{n} x_{n+1}=y_{1} \\cdots y_{n} y_{n+1}$. Tout étant homogène, quitte à multiplier les $x_{i}$ et les $y_{i}$ par une même constante, on peut supposer que $x_{i}, y_{i}>1$ pour tout $i$. On pose ensuite $\\alpha_{i}=\\ln \\left(x_{i}\\right)$ et $\\beta_{i}=\\ln \\left(y_{i}\\right)$, pour tout i. On a alors:\n$-\\alpha_{1} \\geqslant \\ldots \\geqslant \\alpha_{n+1}>0$ et $\\beta_{1} \\geqslant \\ldots \\geqslant \\beta_{n+1}>0 ;$\n$-\\beta_{1}+\\cdots+\\beta_{i} \\leqslant \\alpha_{1}+\\cdots+\\alpha_{i}$, pour $i=1, \\ldots, n$;\n$-\\beta_{1}+\\cdots +\\beta_{n+1}=\\alpha_{1}+\\cdots+\\alpha_{n+1}$.\nOn peut donc utiliser l'inégalité de Muirhead, qui assure que, pour tous réels $x_{1}, \\ldots, x_{n+1}$ strictement positifs, on a\n$$\n\\sum_{\\text{sym.}} x_{1}^{\\beta_{1}} \\cdots x_{n+1}^{\\beta_{n+1}} \\leqslant \\sum_{\\text{sym.}} x_{1}^{\\alpha_{1}} \\cdots x_{n+1}^{\\alpha_{n+1}} .\n$$\nEn particulier, si $x_{1}=e$ et $x_{i}=1$ pour $i \\geqslant 2$, il vient $\\sum_{i=1}^{n+1} y_{i} \\leqslant \\sum_{i=1}^{n+1} x_{i}$. Puisque $x_{n+1} \\leqslant y_{n+1}$, il s'ensuit en fait $\\sum_{i=1}^{n} y_{i} \\leqslant \\sum_{i=1}^{n} x_{i}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30574,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $n \\geqslant 2$ un entier et $X$ un ensemble à $n$ éléments. Montrer que le nombre de fonctions $f: X \\rightarrow X$ telles que $f \\circ f$ soit une fonction constante est égale à\n$$\nn \\sum_{i=1}^{n-1}\\binom{n-1}{i} i^{n-i-1}\n$$",
    "options": [],
    "answer": "n \\sum_{i=1}^{n-1} \\binom{n-1}{i} i^{n-i-1}",
    "solution": "Solution:\n\nSoit $f: X \\rightarrow X$ telle que $f \\circ f = c$ pour un $c \\in X$. D'abord on a $f(c) = c$ parce que\n$$\nf \\circ f(c) = c \\Rightarrow f \\circ f \\circ f(c) = f(c),\n$$\net selon la définition $f \\circ f \\circ f(c) = f \\circ f(f(c)) = c$, ce qui nous donne $f(f(c)) = c$ donc $f(c) = c$. Posons l'ensemble\n$$\nA = \\{ x \\in X \\mid f(x) = c \\}\n$$\nOn a montré qu'il contient $c$. Selon la définition, on a $f(x) \\in A \\setminus \\{c\\}$ pour tout $x \\in A^c$ où $A^c$ est le complémentaire de $A$.\nDonc, pour déterminer toutes les fonctions $f: X \\rightarrow X$ telles que $f \\circ f$ est constante, on doit d'abord choisir un $c \\in X$ ($n$ manières de faire), puis un $A \\subset X$ contenant $c$ ($\\binom{n-1}{i}$ manières de faire si $|A| = i+1$), et ensuite choisir les valeurs $f(x)$ pour chaque $x \\in A^c$ dans $A \\setminus \\{c\\}$ ($i^{n-i-1}$ manières de faire). Alors le nombre de telles fonctions sera\n$$\nn \\sum_{i=1}^{n-1}\\binom{n-1}{i} i^{n-i-1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30575,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont l'orthocentre $H$ est distinct des sommets ainsi que du centre du cercle circonscrit $O$. On désigne par $M, N, P$ les centres des cercles circonscrits aux triangles $HBC$, $HCA$ et $HAB$.\n\nMontrer que les droites $(AM)$, $(BN)$, $(CP)$ et $(OH)$ sont concourantes.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $H'$ le symétrique de $H$ par rapport à $(BC)$. On sait que $H'$ est sur le cercle circonscrit à $ABC$ (il est facile de vérifier $\\widehat{B H' C} = \\widehat{B H C} = 180^{\\circ} - \\widehat{BAC}$). Le centre du cercle circonscrit à $H'BC$ est donc $O$. Par symétrie par rapport à $(BC)$, le centre du cercle circonscrit à $BHC$ est donc le symétrique de $O$ par rapport à $(BC)$, donc $M$ est le symétrique de $O$ par rapport à $(BC)$.\n\nOn va maintenant montrer que $AHMO$ est un parallélogramme. Si c'est bien vrai, alors $[AM]$ et $[OH]$ se coupent en leur milieu, donc le milieu de $[OH]$ est sur $(AM)$ et, par le même raisonnement, il est aussi sur $(BN)$ et $(CP)$.\n\nSi on note $D$ le milieu de $[BC]$, alors $\\overrightarrow{OD} = \\frac{1}{2}(\\overrightarrow{OB} + \\overrightarrow{OC}) = \\frac{1}{2}(\\overrightarrow{OH} - \\overrightarrow{OA}) = \\frac{1}{2} \\overrightarrow{AH}$, donc $\\overrightarrow{OM} = \\overrightarrow{AH}$.\n\nEn utilisant la relation $\\overrightarrow{OA} + \\overrightarrow{OB} + \\overrightarrow{OC} = \\overrightarrow{OH}$, qui est équivalente à $\\overrightarrow{OH} = 3 \\overrightarrow{OG}$ (droite d'Euler).\n\nOn donne également une autre manière, moins rapide mais nécessitant moins de connaissances, de montrer que $AHMO$ est un parallélogramme. Les droites $(AH)$ et $(OM)$ sont parallèles, donc il suffit de montrer que $AH = OM$. Pour cela, on peut utiliser la trigonométrie. D'une part, en notant $R$ le rayon du cercle circonscrit à $ABC$ et $\\alpha, \\beta, \\gamma$ ses angles, on a\n$$\nOM = 2OD = 2OB \\sin \\widehat{OBC} = 2R \\sin(90^{\\circ} - \\alpha) = 2R \\cos \\alpha\n$$\nD'autre part, en notant $P$ le pied de la hauteur issue de $C$, on a\n$$\nAH = \\frac{AP}{\\cos \\widehat{PAH}} = \\frac{AP}{\\sin \\beta} = \\frac{AC \\sin \\widehat{ACP}}{\\sin \\beta} = 2R \\sin \\widehat{ACP} = 2R \\sin(90^{\\circ} - \\alpha) = 2R \\cos \\alpha\n$$\nen utilisant la loi des sinus.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $p$ un nombre premier, $m$ un entier naturel. Trouver le plus petit entier $d$ tel qu'il existe un polynôme unitaire $Q$ de degré $d$ à coefficients entiers tel que, pour tout entier $n$, $p^{m} \\mid Q(n)$.",
    "options": [],
    "answer": "d is the smallest integer k such that p^m divides k! (equivalently, the least k with v_p(k!) ≥ m).",
    "solution": "Solution:\n\nMontrons que $d$ est le plus petit entier tel que $p^{m} \\mid d!$. Soit $k$ le plus petit entier tel que $p^{m} \\mid k!$.\n\nTout d'abord, montrons que $k \\geq d$. On considère le polynôme $Q(x) = X(X+1) \\cdots (X+k-1)$ : il est unitaire, de degré $k$, à coefficients entiers. Qui plus est, pour tout $n \\in \\mathbb{N}$, $Q(n) = k! \\binom{n}{k}$ (c'est une propriété des coefficients binomiaux), d'où $p^{m} \\mid k! \\mid Q(n)$. Donc $k$ est un entier pour lequel on dispose d'un polynôme $Q$ unitaire à coefficients entiers de degré $k$ tel que pour tout $n \\in \\mathbb{N}$, $p^{m} \\mid Q(n)$. Donc $k \\geq d$.\n\nReste à montrer le sens réciproque. Autrement dit, soit $Q$ un polynôme unitaire à coefficients entiers de degré $l$ tel que pour tout entier $n$, $p^{m} \\mid Q(n)$. Montrons que $k \\leq l$. Il suffit de montrer que $p^{m} \\mid l!$ (car, par définition, $k$ est le plus petit entier vérifiant cela).\n\nOn introduit la famille des polynômes $P_{i} := X(X+1) \\cdots (X+i-1)$ pour $i \\in \\{0, \\ldots, l\\}$ ($P_{0} := 1$). En effectuant les divisions euclidiennes successives de $Q$ par $P_{l}, P_{l-1}, \\ldots, P_{0}$, on montre que $Q$ s'écrit :\n$$\nQ = \\sum_{i=0}^{l} c_{i} P_{i}\n$$\noù $c_{i} \\in \\mathbb{Z}$ pour tout $i$.\n\nOn montre alors par récurrence sur $i \\in \\{0, \\ldots, l\\}$ la propriété suivante :\n$$\np^{m} \\mid c_{i} P_{i}(-i) \\implies p^{m} \\mid c_{i} P_{i}(n) \\text{ pour tout } n \\in \\mathbb{Z}\n$$\n- pour $i=0$, le seul des $P_{i}$ qui ne s'annule pas en zéro est $P_{0}$, donc $Q(0) = c_{0}$, d'où $p^{m} \\mid c_{0} = c_{0} P_{0}(n)$ pour tout $n \\in \\mathbb{Z}$.\n- soit $i < l$. Supposons que la propriété est vraie pour tous les éléments de $\\{0, \\ldots, i\\}$ et montrons-la pour $i+1$. On constate que si $j > i+1$, $(X+i+1) \\mid P_{j}$ donc $P_{j}(-i-1) = 0$. Ainsi,\n$$\n\\underbrace{Q(-i-1)}_{p^{m} \\mid} = \\underbrace{\\sum_{j=0}^{i} c_{j} P_{j}(-i-1)}_{p^{m} \\mid} + c_{i+1} P_{i+1}(-i-1)\n$$\npar hypothèse de récurrence et par définition de $Q$. Donc $p^{m} \\mid c_{i+1} P_{i+1}(-i-1)$, et comme $P_{i+1}(-i-1) = (i+1)!$ divise tout produit de $i+1$ entiers consécutifs (on peut encore le voir grâce aux coefficients binomiaux), il divise $P_{i+1}(n)$ pour tout $n \\in \\mathbb{Z}$.\n\nAinsi, on obtient en particulier $p^{m} \\mid P_{l}(n)$ pour tout $n \\in \\mathbb{N}$, donc $p^{m} \\mid l!$. Donc $k \\leq l$, d'où $k \\leq d$. Par double encadrement, on a donc bien montré que $k = d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle tel que $AB \\neq AC$. Soit $E$ tel que $AE = BE$ et $(BE)$ perpendiculaire à $(BC)$ et soit $F$ tel que $AF = CF$ et $(CF)$ perpendiculaire à $(BC)$. Soit $D$ le point de $(BC)$ tel que $(AD)$ soit tangente au cercle circonscrit à $ABC$ en $A$.\n\nMontrer que les points $D$, $E$, $F$ sont colinéaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSupposons (sans restreindre la généralité) que $AB < AC$.\n\nSoit $O$ le centre du cercle circonscrit à $ABC$ et $G$ le point de la droite $(AD)$ tel que $(AB)$ soit parallèle à $(GC)$.\n\nMontrons tout d'abord que les angles $\\widehat{BAD}$ et $\\widehat{BCA}$ sont de même mesure $\\gamma$, et que les angles $\\widehat{BOA}$ et $\\widehat{CFA}$ sont tous les deux de mesure $2\\gamma$. Le seul point qui ne découle pas immédiatement du théorème de l'angle inscrit est l'égalité $\\widehat{BOA} = \\widehat{CFA}$.\n\nOr $DBA$ et $DAC$ sont semblables, donc il existe une similitude $s$ de centre $D$ qui envoie $B$ sur $A$ et $A$ sur $C$. Comme $s$ conserve les rapports de longueur, elle envoie la médiatrice de $[AB]$ sur la médiatrice de $[BC]$. Elle envoie aussi la perpendiculaire à $(DA)$ passant par $A$ (c'est-à-dire $(AO)$, puisque $(AD)$ est tangente en $A$ au cercle de centre $O$ passant par $A$) sur la perpendiculaire à $(DC)$ passant par $C$ (c'est-à-dire $(CF)$). Donc $O$ est envoyé sur $F$. Comme $s$ conserve les mesures d'angles géométriques (attention pour ceux qui travaillent en angles orientés, c'est une similitude indirecte!), $\\widehat{BOA} = \\widehat{AFC}$.\n\nOn en déduit que $F$ est le centre du cercle circonscrit à $AGC$. En particulier, $F$ est sur la médiatrice de $[GC]$.\n\nOn considère maintenant l'homothétie $h$ de centre $D$ qui envoie $B$ sur $C$. Comme l'image de toute droite par $h$ est une droite parallèle, $(AB)$ est envoyée sur $(GC)$, et $(DA)$ est fixée donc $A$ est envoyé sur $G$. Comme $h$ conserve les rapports de longueurs, la médiatrice de $[AB]$ est envoyée sur la médiatrice de $[GC]$. De plus, $(EB)$ est envoyée sur $(CF)$ grâce au parallélisme, donc $E$ est envoyé sur $F$.\n\nDonc $D$ (en tant que centre de l'homothétie), $E$ et $F$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30578,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les entiers strictement positifs $a, b$ et $c$ tels que $(a^{3}+b)(b^{3}+a)=2^{c}$.",
    "options": [],
    "answer": "(a,b,c) = (1,1,2), (3,5,12), (5,3,12)",
    "solution": "Solution:\nSoit $(a, b, c)$ un triplet solution.\n\nSi $a=b$, on a $a^{2}(a^{2}+1)^{2}=2^{c}$, ce qui implique $a=b=1$ et $c=2$.\n\nSi $a \\neq b$, supposons par symétrie que $a>b$. On voit aisément que $a$ et $b$ sont impairs et premiers entre eux. Il existe deux entiers $x, y \\geqslant 1$ tels que $2^{x}=a^{3}+b$ et $2^{y}=b^{3}+a$. Par croissance stricte de la fonction $f(x)=x^{3}-x$ sur $[1, \\infty[$, on $a^{3}+b>b^{3}+a$, de sorte que $x>y$.\n\nSi $b=1$, on a $2^{x}=(2^{y}-1)^{3}+1$, et donc $2^{x-y}=3-3 \\cdot 2^{y}+2^{2y}$, ce qui est absurde.\n\nSi $b>1$, comme $b^{3}+a$ divise $(b^{3})^{3}+a^{3}=b^{9}+a^{3}$, $b^{3}+a$ divise $b^{9}+a^{3}-(a^{3}+b)=b^{9}-b$.\n\nComme $a$ et $b$ sont premiers entre eux, on en déduit que $b^{3}+a$ divise\n$$\nb^{8}-1=(b^{4}+1)(b^{2}+1)(b^{2}-1)\n$$\nComme $b^{4}+1$ et $b^{2}+1$ sont congrus à 2 modulo 4, on en déduit que $2^{y-2}$ divise $b^{2}-1$, ce qui implique que $b^{2}-1 \\geqslant 2^{y-2}$. Or $2^{y}>b^{3}$. Ceci implique que $4(b^{2}-1)>b^{3}$, ce qui force $b=3$. Alors $2^{y-2}$ divise 8, donc $y=3,4$ ou 5. On vérifie aisément que cela implique $(a, b)=(5,3)$.\n\nRéciproquement, on vérifie que les seules solutions sont $(a, b, c)=(1,1,2), (a, b, c)=(3,5,12)$ et $(a, b, c)=(5,3,12)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30579,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn constructeur de jouets crée au moins un jouet par jour. Il n'est pas capable de créer plus de 725 jouets par an. Pour tout $n \\in \\mathbb{N}$, montrer qu'il existe des jours consécutifs pendant lesquels il a créé exactement $n$ jouets.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $x_{i} \\in \\mathbb{N}$ le nombre de jouets qu'il a créés pendant le $i$-ème jour. Si\n$$\nx_{1} + x_{2} + \\ldots + x_{n} < 2n\n$$\nétudions les nombres suivants :\n$$\n1 \\leqslant x_{1} < x_{1} + x_{2} < \\ldots < x_{1} + x_{2} + \\ldots + x_{n} < 2n.\n$$\nOn a deux cas. Soit ils représentent tous les résidus modulo $n$, soit il existe $1 \\leqslant i < j \\leqslant n$ tels que\n$$\nx_{1} + x_{2} + \\ldots + x_{i} \\equiv x_{1} + x_{2} + \\ldots + x_{j} \\pmod{n}\n$$\nCe dernier cas nous donne\n$$\nn \\mid x_{i+1} + \\ldots + x_{j}\n$$\npuis\n$$\nx_{i+1} + \\ldots + x_{j} = n\n$$\nDans le premier cas, il existe $1 \\leqslant j \\leqslant n$ tel que\n$$\nn \\mid x_{1} + x_{2} + \\ldots + x_{j}\n$$\ndonc\n$$\nx_{1} + x_{2} + \\ldots + x_{j} = n\n$$\nDans les deux cas, le problème est résolu. Étudions maintenant ce qui se passe si, pour tous $n$ jours consécutifs, le nombre total de jouets est plus grand que $2n$. Alors\n$$\n2n \\times 365 \\leqslant \\sum_{i=1}^{365n} x_{i} \\leqslant 725n\n$$\non aboutit à une contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30580,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle.\n\nPour un point $P$ de $(BC)$ donné, on note $E(P)$ et $F(P)$ les deuxièmes points d'intersection des droites $(AB)$ et $(AC)$ avec le cercle de diamètre $[AP]$. Soit $T(P)$ l'intersection des tangentes à ce cercle en $E(P)$ et $F(P)$.\n\nMontrer que quand $P$ varie sur $(BC)$, le lieu géométrique de $T(P)$ est une droite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\n**Étape 1 : une première remarque importante.**\n\nPosons tout d'abord $H$ le pied de la hauteur issue de $A$ dans $ABC$. Soient deux points $P$ et $P'$ de la droite $(BC)$. On note $E = E(P)$, $F = F(P)$, $T = T(P)$, $E' = E(P')$, $F' = F(P')$, $T' = T(P')$. On introduit également $O$ (respectivement $O'$) le centre du cercle de diamètre $[AP]$ (respectivement $[AP']$).\n\nRemarquons que les cercles circonscrits à $EAF$ et à $E'A F'$ se coupent en $A$ et $H$. En particulier, $H$ est le point de Miquel du quadrilatère de côtés $(EE')$, $(E'F')$, $(F'F)$, $(FE)$, autrement dit il existe une unique similitude directe $s$ de centre $H$ qui envoie $E$ sur $E'$ et $F$ sur $F'$.\n\nD'autre part, on a $\\widehat{EOF} = 2\\widehat{EHF} = 360^\\circ - 2\\widehat{EAF} = 360^\\circ - \\widehat{BAC}$, ce qui ne dépend pas de $P$. Donc $\\widehat{EOF} = \\widehat{E'O'F'}$, donc la similitude directe $s$ envoie $O$ sur $O'$.\n\nOr, le point $T$ est entièrement défini par la donnée de $E, O, F$ (via des opérations de type tracer un cercle, prendre une tangente..., qui sont conservées par les similitudes). Donc $s$ envoie aussi $T$ sur $T'$.\n\nÀ partir de là, le plus dur est fait.\n\n**Étape 2 : un jeu technique avec les similitudes pour conclure.**\n\nSoit $P_0$ un point de $(BC)$. On note $\\sigma$ la similitude directe de centre $H$ qui envoie $E(P_0)$ sur $T(P_0)$. Soit $P$ un point de $(BC)$. On note $\\sigma_P$ la similitude directe (de centre $H$) qui envoie $E(P_0)$ sur $E(P)$. D'après ce qui précède, elle envoie aussi $T(P_0)$ sur $T(P)$.\n\nComme les deux similitudes ont le même centre, $\\sigma_P \\circ \\sigma = \\sigma \\circ \\sigma_P$. En particulier, $\\sigma(E(P)) = \\sigma \\circ \\sigma_P(E(P_0)) = \\sigma_P(T(P_0)) = T(P)$.\n\nDonc le lieu de $T(P)$ est l'image par $\\sigma$ (qui est une similitude directe) du lieu de $E(P)$ (qui est une droite). C'est donc une droite, comme voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30581,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les polynômes $P$ à coefficients entiers tels que $P(P(n)+n)$ est un nombre premier pour une infinité d'entiers $n$.",
    "options": [],
    "answer": "All constant polynomials equal to a prime, and all polynomials P(x) = −2x + b with b an odd integer.",
    "solution": "Solution:\n\nSoit $P$ un polynôme solution qui n'est pas constant. On peut écrire $P(P(X)+X) = P(X) Q(X)$, avec $Q$ un polynôme à coefficients entiers. En effet, si $P(X) = \\sum_{i=0}^{k} a_{i} X^{i}$, avec $a_{i} \\in \\mathbb{Z}$, on a\n$$\nP(P(X)+X) = \\sum_{i=0}^{k} a_{i}(P(X)+X)^{i} = a_{0} + \\sum_{i=1}^{k} a_{i}\\left(P(X) Q_{i}(X) + X^{i}\\right) = P(X) + P(X) \\sum_{i=1}^{k} a_{i} Q_{i}(X) = P(X) Q(X),\n$$\noù les $Q_{i}$ sont des polynômes à coefficients entiers, et donc $Q$ également.\n\nComme $P(X)+1$ et $P(X)-1$ ont un nombre fini de racines et que $P(n+P(n))$ est premier pour une infinité de nombres entiers $n$, on en déduit que $Q(n)=1$ ou $Q(n)=-1$ pour une infinité d'entiers $n$, ce qui implique que $Q$ est constant. En notant $k$ le degré de $P$, on a $\\deg(P(P(X)+X)) = k^{2}$, ce qui implique $k = k^{2}$ et donc $k = 1$.\n\nSi $P(X) = a X + b$ avec $a \\neq 0$, on a\n$$\nP(P(n)+n) = (a+1)(a n + b).\n$$\nComme $a \\neq 0$, pour que ce nombre soit premier pour une infinité d'entiers $n$, on doit avoir $|a+1| = 1$, ce qui implique que $a = -2$. Pour que $2 n - b$ soit un nombre premier pour une infinité de nombres entiers $n$, il faut et il suffit que $b$ soit impair.\n\nAinsi, les solutions sont les polynômes constants égaux à un nombre premier et les polynômes $P$ de la forme $P(X) = -2 X + b$ avec $b$ un entier impair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont les trois angles sont aigus. Soit $D \\in [BC]$ tel que $\\widehat{BAC} = \\widehat{ADB}$. Soit $H$ le pied de la hauteur issue de $B$ dans $ABC$. La perpendiculaire à $(BC)$ passant par $H$ coupe $(AD)$ en $K$. On suppose que $K$ est à l'intérieur du triangle $ABC$. Soit $M$ le milieu de $[AC]$. Montrer que $MH = MK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons que $\\widehat{ABH} = 90^{\\circ} - \\widehat{BAC} = 90^{\\circ} - \\widehat{ADB} = \\widehat{AKH}$. Les points $A, B, K, H$ sont donc cocycliques.\n\nSoit $T$ le pied de la hauteur issue de $A$ dans $ABC$. Alors $\\widehat{BTA} = \\widehat{BHA} = 90^{\\circ}$ donc $A, B, H$ et $T$ sont cocycliques et le cercle qui passe par ces points passe aussi par $K$. On en déduit $\\widehat{ATK} = 180^{\\circ} - \\widehat{AHK} = \\widehat{KHC}$.\n\nOn va maintenant montrer que $M, K$ et $T$ sont alignés. On a $(AT) \\parallel (HK)$ car ces deux droites sont perpendiculaires à $(BC)$, donc $\\widehat{ATK} = \\widehat{KHC} = \\widehat{TAC}$. Mais le triangle $ATC$ est rectangle en $T$ donc $MA = MC = MT$ et $\\widehat{TAM} = \\widehat{ATM}$. On en déduit $\\widehat{ATM} = \\widehat{ATK}$ donc $T, K$ et $M$ sont alignés.\n\nEnfin, $MH = MK$ parce que $AT \\parallel HK$ et $MA = MT$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSi $n \\in \\mathbb{N}^{*}$, on note $d(n)$ son nombre de diviseurs. Quels sont les entiers strictement positifs tels que $d(n)^{3}=4 n$ ?",
    "options": [],
    "answer": "2, 128, 2000",
    "solution": "Solution:\n\nSoit $f(n):=\\frac{d(n)^{3}}{n}$, l'énoncé revient à trouver les antécédents de $4$ par $f$. On remarque que $f(a b)=f(a) f(b)$ si $a$ et $b$ sont premiers entre eux. Ceci nous incite à décomposer $n$ en produit de facteurs premiers :\n\n$$\nn=p_{1}^{a_{1}} \\times p_{2}^{a_{2}} \\times \\cdots \\times p_{l}^{a_{l}},\n$$\noù les $p_{i}, 1 \\leqslant i \\leqslant l$, sont premiers et $a_{i} \\geqslant 1$. On a donc $f(n)=\\prod_{i=1}^{l} f\\left(p_{i}^{a_{i}}\\right)$.\n\nSoit $n \\in \\mathbb{N}^{*}$ vérifiant l'énoncé, $n^{3}=2 \\times(d(n) / 2)^{3}$ est le double d'un cube parfait ( $d(n)$ est nécessairement pair). On en déduit que $v_{p}(n) \\equiv 0\\ (\\bmod\\ 3)$ pour $p$ nombre premier impair, et $v_{2}(n) \\equiv 1\\ (\\bmod\\ 3)$. Ici, $v_{p}(n)$ désigne la valuation $p$-adique de $n$, soit le plus grand entier $k$ tel que $p^{k} \\mid n$, donc $v_{p_{i}}(n)=a_{i}$ avec nos notations. On peut écrire\n\n$$\nf(n)=\\frac{\\left(\\left(a_{1}+1\\right)\\left(a_{2}+1\\right) \\cdots\\left(a_{l}+1\\right)\\right)^{3}}{p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{l}^{a_{l}}}\n$$\n\nD'après la remarque précédente, le numérateur n'est pas divisible par $3$, donc aucun des $p_{i}$ ne peut valoir $3$. Et $n$ est le produit de puissances de cubes de nombres premiers impairs que multiplie le double d'une puissance de $2^{3}$.\n\nOn montre aisément que si $m$ est puissance du cube d'un nombre premier strictement plus grand que $3$, alors $f(m) \\geqslant \\frac{64}{125}$ avec égalité si et seulement si $m=5^{3}$. On prouve à cette fin que pour tout premier $p \\geqslant 5$, $k \\rightarrow f\\left(p^{3 k}\\right)$ est strictement décroissante, et que pour tout $k \\geqslant 1$, $p \\rightarrow f\\left(p^{3 k}\\right)$ l'est aussi. Pour avoir $f(n)=4$, il faut donc utiliser des puissances de $2$. On a $f(2)=f\\left(2^{7}\\right)=4$, $f\\left(2^{4}\\right)=\\frac{125}{16}$ et $f\\left(2^{3 k+1}\\right)<1$ pour $k>2$.\n\nAinsi, $2$ et $128$ sont solution. Et pour en avoir une autre, il faut trouver $m$ un produit de puissances de cubes de nombres premiers impairs tel que $f(m)=\\frac{64}{125}$. D'après l'étude précédente, seul $m=5^{3}$ convient. Donc il y a $3$ entiers $n$ solution au problème : $n=2$, $n=2^{7}$ et $n=2^{4} \\times 5^{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn polynôme est dit « unitaire » si le coefficient de son monôme de plus haut degré vaut $1$. Soient $a < b$ des réels, et soit $P$ un polynôme unitaire non constant tel que $\\max_{x \\in [a, b]} |P(x)| < 2$.\n\nProuver qu'il existe un polynôme $Q$, unitaire et non constant, tel que $\\max_{x \\in [a, b]} |Q(x)| < \\frac{1}{2017}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSi $Q$ désigne un polynôme, on pose\n$$\n\\|Q\\| = \\frac{1}{2} \\max_{x \\in [a, b]} |Q(x)| \\text{ et } T(Q)(x) = Q(x)^2 - 2\\|Q\\|^2\n$$\nIl est clair que si $Q$ est un polynôme unitaire non constant, alors $T(Q)$ l'est également. De plus, on a $\\|T(Q)\\| = \\|Q\\|^2$. Si l'on note $T^n$ la $n$-ième itérée de $T$, alors une récurrence immédiate assure que pour tout entier $n \\geqslant 0$, $T^n(Q)$ est un polynôme non constant dont le coefficient du monôme de plus haut degré vaut $1$ et que $\\left\\|T^n(Q)\\right\\| = \\|Q\\|^{2^n}$.\n\nSoit donc $P$ un polynôme unitaire non constant, tel que $\\max_{x \\in [a, b]} |P(x)| < 2$. Puisque $0 < \\|P\\| < 1$, on a $\\lim_{n \\to +\\infty} \\left\\|T^n(P)\\right\\| = 0$. En particulier, il existe $n \\geqslant 0$ suffisamment grand pour que $\\left\\|T^n(P)\\right\\| < \\frac{1}{2 \\times 2017}$, et le polynôme $Q = T^n(P)$ répond alors à la question.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $A_{1}, A_{2}, \\ldots, A_{n}, B_{1}, B_{2}, \\ldots, B_{n}$ des ensembles de $k$ éléments. Montrer que si $n < 2^{k-1}$ alors il existe des éléments $a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n}$ tels que:\n$$\na_{i} \\in A_{i},\\ b_{j} \\in B_{j},\\ a_{i} \\neq b_{j}\n$$\npour tout $1 \\leqslant i, j \\leqslant n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $X$ l'union de tous les $A_{i}$ et $B_{j}$ pour tous $1 \\leqslant i, j \\leqslant n$ et soit $m=|X|$. On veut montrer qu'il y a un coloriage des éléments de $X$ en rouge et bleu telle que\n- Aucun des $A_{1}, A_{2}, \\ldots, A_{n}$ ne soit totalement rouge,\n- Aucun des $B_{1}, B_{2}, \\ldots, B_{n}$ ne soit totalement bleu.\n\nPour chaque $A_{i}$, le nombre de coloriages tels que $A_{i}$ soit tout en rouge est égal à $2^{m-k}$. Donc, puisque $2 n \\left(2^{m-k}\\right)<2^{m}$, on aura un coloriage avec la propriété précédente. Avec un tel coloriage, on choisit chaque $a_{i}$ parmi les éléments bleus de $A_{i}$ et chaque $b_{j}$ parmi les éléments rouges de $B_{j}$. Ce qui convient.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30586,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $O$ le centre de son cercle circonscrit. Soit $P$ un point sur $(AO)$ et $D, E, F$ les projections orthogonales de $P$ sur $(AB)$, $(BC)$ et $(CA)$. Soit $X$ et $Y$ les intersections des cercles circonscrits à $DEF$ et à $BCP$.\n\nMontrer que $\\widehat{BAX} = \\widehat{YAC}$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nÉtape 1 : Montrons que $(DF)$ est parallèle à $(BC)$.\n\nSoit $P_0$ le point diamétralement opposé à $A$ sur le cercle circonscrit à $ABC$. $(P_0C)$ est perpendiculaire à $(AC)$ donc parallèle à $(PF)$ et $(P_0B)$ est de même parallèle à $(PD)$.\n\nDonc, d'après le théorème de Thalès, $\\frac{AD}{AB} = \\frac{AP}{AP_0} = \\frac{AF}{AC}$, d'où $(DF)$ parallèle à $(BC)$.\n\nÉtape 2 : Montrons que $H_A$, le pied de la hauteur issue de $A$, appartient à $(DEF)$.\n\n![](attached_image_2.png)\n\nOn pose $G = (AH_AA) \\cap (DF)$, $J$ la projection orthogonale de $P$ (celle de $E$ aussi) sur $(DF)$ et $K$ la projection (orthogonale) de $P$ sur $(AH_A)$.\n\n$PKGJ$ est un rectangle, et $P, K, D, F$ sont cocycliques sur le cercle de diamètre $[PA]$. Donc $PKFD$ est un trapèze isocèle, donc $(PJ)$ et $(KG)$ sont symétriques par rapport à la médiatrice de $[DF]$.\n\nComme $EJGH$ est un rectangle et $(H_AG)$ et $(EJ)$ sont symétriques par rapport à la médiatrice de $(EF)$, $D, H_A, E, F$ sont cocycliques.\n\nÉtape 3 : pour conclure.\n\nL'étape 1 permet de montrer qu'il existe une involution de centre $A$ qui échange $B$ et $F$ d'une part, $C$ et $D$ d'autre part. Pour conclure, il suffit de montrer que cette involution échange $X$ et $Y$, c'est-à-dire qu'elle envoie le cercle circonscrit à $BCP$ sur le cercle circonscrit à $DEF$.\n\nLa seule difficulté est de montrer que le point sur lequel l'involution envoie $P$ appartient au cercle circonscrit à $DEF$.\n\nOr $P$ est un point du cercle circonscrit à $ADF$ tel que $(AP)$ et le cercle circonscrit à $ADF$ sont perpendiculaires en $P$. Donc son image $I$ est un point de la droite $(BC)$ tel que $(BC)$ et $(AI)$ sont perpendiculaires en $I$. Donc $I = H_A$, et l'étape 2 permet de conclure.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABCD$ un quadrilatère inscrit dans un cercle. On note $\\Delta_{1}$ la droite parallèle à $(BC)$ passant par $A$ et $\\Delta_{2}$ la droite parallèle à $(AD)$ passant par $B$. On note $E$ le point d'intersection de $\\Delta_{1}$ avec $(CD)$ et $F$ le point d'intersection de $\\Delta_{2}$ avec $(CD)$. La droite perpendiculaire à $\\Delta_{1}$ passant par $A$ coupe $(BC)$ en $P$, et la droite perpendiculaire à $\\Delta_{2}$ passant par $B$ coupe $(AD)$ en $Q$. On note $\\Gamma_{1}$ le cercle circonscrit du triangle $ADE$ et $\\Gamma_{2}$ le cercle circonscrit du triangle $BFC$. Montrer que $\\Gamma_{1}$ et $\\Gamma_{2}$ sont tangents si et seulement si $(DP)$ et $(CQ)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFIGURE 1 - Preuve de la première implication\n\nOn commence par quelques remarques préliminaires. On a $\\widehat{EDA}=180^{\\circ}-\\widehat{ADC}=\\widehat{ABC}=180^{\\circ}-\\widehat{EAB}$, donc $(AB)$ et $\\Gamma_{1}$ sont tangents. De même, $\\Gamma_{2}$ et $(AB)$ sont tangents.\n\nEnsuite, comme $(AP) \\perp (BC)$ et $(AD) \\perp (BQ)$, on a $(PD) \\perp (CQ)$ si et seulement si les triangles $APD$ et $BCQ$ sont semblables, c'est-à-dire si et seulement si $\\frac{AP}{AD}=\\frac{BC}{CQ}$.\n\nOr les triangles $ADE$ et $BFC$ sont semblables, donc $\\frac{AD}{AE}=\\frac{BF}{BC}$. On en déduit que $\\frac{AP}{AD}=\\frac{BC}{CQ}$ si et seulement si $\\frac{AP}{AE}=\\frac{BF}{BQ}$ si et seulement si $PAE$ et $FBQ$ sont semblables (car $\\widehat{PAE}=\\widehat{FBQ}=90^{\\circ}$).\n\nSupposons maintenant que $\\Gamma_{1}$ et $\\Gamma_{2}$ sont tangents en $R$. On voit facilement que $\\widehat{ARB}=90^{\\circ}$, ce qui implique que les points $A, Q, P, B, R$ appartiennent tous au cercle de diamètre $[AB]$. On en déduit que $\\widehat{ARP}=180^{\\circ}-\\widehat{ABP}=180^{\\circ}-\\widehat{ABC}=180^{\\circ}-\\widehat{ADE}=180^{\\circ}-\\widehat{ARE}$. Donc $E, R, P$ sont alignés. On montre de même que $F, R, Q$ sont alignés. Ainsi, $\\widehat{PEA}=\\widehat{REA}=\\widehat{RAB}=\\widehat{RQB}=\\widehat{FQB}$. On en déduit que $PEA$ et $FQB$ sont semblables.\n\nSupposons maintenant que $PEA$ et $FQB$ sont semblables. Soit $S$ le point d'intersection de $(PE)$ et $(QF)$. Nous allons montrer que $\\Gamma_{1}$ et $\\Gamma_{2}$ sont tangents en $S$. On a l'égalité des angles\n$$\n\\widehat{SQB}=\\widehat{FQB}=\\widehat{PEA}=\\widehat{EPC}=180^{\\circ}-\\widehat{SPB}\n$$\nOn en déduit que $S, Q, B, P$ sont cocycliques. Ainsi, $A, Q, P, B, S$ sont cocycliques. Il en découle que\n$$\n\\widehat{ASE}=180^{\\circ}-\\widehat{ASP}=\\widehat{ABP}=\\widehat{ABC}=180^{\\circ}-\\widehat{ADC}=\\widehat{ADE}\n$$\nDonc $S \\in \\Gamma_{1}$. On montre de même que $S \\in \\Gamma_{2}$.\n\nPour conclure, on écrit\n$$\n\\widehat{BSA}=90^{\\circ}=\\widehat{SAB}+\\widehat{ABS}=\\widehat{SDA}+\\widehat{BCS}\n$$\n\n![](attached_image_2.png)\nFigure 2 - Preuve de la deuxième implication.\n\nCeci entraîne aisément que la tangente à $\\Gamma_{1}$ passant par $S$ est tangente à $\\Gamma_{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nProuver qu'il existe une infinité d'entiers $n$ tels que $2^{2^{n}+1}+1$ est divisible par $n$, mais $2^{n}+1$ ne l'est pas.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour $m \\geqslant 1$, posons $a_{m}=2^{3^{m}}+1$. On a\n$$\na_{m}=\\left(2^{3^{m-1}}+1\\right)\\left(\\left(2^{3^{m-1}}\\right)^{2}-2^{3^{m-1}}+1\\right)=a_{m-1}\\left(\\left(2^{3^{m-1}}\\right)^{2}-2^{3^{m-1}}+1\\right)\n$$\nNotons que pour $a \\in \\mathbb{N}^{*}$, si $p \\mid a+1$ et $p \\mid a^{2}-a+1$, alors $p \\mid 2 a-1$ car $2 a-1=a(a+1)-\\left(a^{2}-a+1\\right)$. Et $p \\mid 2 a+2$, donc $p \\mid 3$. Donc $\\left(a_{m-1} \\wedge a_{m}\\right) \\mid 3$ pour tout $m \\in \\mathbb{N} *$. Donc $a_{m}$ possède un facteur premier impair (car $a_{m}$ impair) autre que 3 qui ne divise pas $a_{m-1}$, sauf éventuellement si $a_{m}$ est une puissance de 3, auquel cas $a_{m+1}$ possède un facteur premier que n'a pas $a_{m}$.\n\nDonc pour une infinité d'entiers $m \\in \\mathbb{N}^{*}, a_{m}$ possède un facteur premier $p_{m}>3$ qui ne divise pas $a_{m-1}$. Posons $b_{m}=3^{m-1} p_{m}$. D'après le petit théorème de Fermat,\n$$\n2^{b_{m}}+1 \\equiv 2^{3^{m-1}}+1 \\equiv a_{m-1} \\quad(\\bmod p)_{m}\n$$\nDonc $p_{m} \\nmid 2^{b_{m}}+1$, donc $b_{m} \\nmid 2^{b_{m}}+1$.\n\nComme $p_{m}$ et 3 sont premiers entre eux, il suffit de montrer que $p_{m} \\mid 2^{2^{b m}+1}+1$ et que $3^{m_{1}} \\mid 2^{2^{b_{m}}+1}+1$. Cela se fait assez directement avec LTE mais on peut peut-être trouver une solution alternative. Recherche en cours.\n\nAinsi, une infinité d'entiers $b_{m}$ satisfont la condition de l'énoncé.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle d'orthocentre $H$. Soient $\\left(d_{1}\\right)$ et $\\left(d_{2}\\right)$ deux droites perpendiculaires se coupant en $H$. Soit $A_{1}$ (respectivement $B_{1}, C_{1}$) l'intersection de $\\left(d_{1}\\right)$ avec $(BC)$ (respectivement $(CA), (AB)$). Soit $A_{2}$ (respectivement $B_{2}, C_{2}$) l'intersection de $\\left(d_{2}\\right)$ avec $(BC)$ (respectivement $(CA), (AB)$). Soient $A_{3}, B_{3}, C_{3}$ des points dans le plan tels que $A_{1}A_{2}A_{3}, B_{1}B_{2}B_{3}, C_{1}C_{2}C_{3}$ soient des triangles directement semblables.\n\nMontrer que $A_{3}, B_{3}, C_{3}$ sont alignés.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous donnons une preuve avec des similitudes directes. Il existe d'autres preuves, un peu plus courtes, qui utilisent des nombres complexes.\n\nRappelons tout d'abord le résultat suivant :\n\n**Lemme 2.** Soit $ABC$ un triangle, $\\Gamma$ son cercle circonscrit et $H$ son orthocentre. Alors les symétriques $H_{A}, H_{B}$ et $H_{C}$ de $H$ par rapport à $(BC), (CA)$ et $(AB)$ appartiennent à $\\Gamma$.\n\nPreuve. Soit $H_{A}'$ l'intersection de $AH$ et $\\Gamma$. Par le théorème de l'angle inscrit, on a :\n$$\n\\begin{aligned}\n\\widehat{H_{A}'BC} & = \\widehat{H_{A}'AC} \\\\\n& = 90 - \\widehat{BCA} \\\\\n& = \\widehat{HBC}\n\\end{aligned}\n$$\nAinsi, $\\left(H_{A}'B\\right)$ et $(HB)$ sont symétriques par rapport à $(BC)$. On en déduit que $H_{A}' = H_{A} \\in \\Gamma$. De même $H_{B}, H_{C} \\in \\Gamma$.\n\nReprenons les notations de l'exercice et introduisons $H_{A}, H_{B}$ et $H_{C}$ comme dans le lemme précédent.\n\n**Lemme 3.** Soit $S$ l'intersection de $\\left(A_{1}H_{A}\\right)$ et de $\\left(B_{1}H_{B}\\right)$. Alors $S \\in \\Gamma$.\n\nPreuve. On note $\\mathcal{S}_{yz}$ la symétrie d'axe $(YZ)$. Alors\n$$\n\\begin{aligned}\n\\mathcal{S}_{AC}\\left(\\mathcal{S}_{BC}\\left(H_{A}A_{1}\\right)\\right) & = \\mathcal{S}_{AC}\\left(A_{1}H\\right) \\text{ car } H_{A} \\text{ et } H \\text{ sont symétriques par rapport à } (BC) \\\\\n& = \\mathcal{S}_{AC}\\left(B_{1}H\\right) \\\\\n& = \\left(H_{B}B_{1}\\right) \\text{ car } H \\text{ et } H_{B} \\text{ sont symétriques par rapport à } (AC).\n\\end{aligned}\n$$\nNotons $\\left(XY, ZW\\right)$ l'angle de droite formé par les droites $XY$ et $ZW$. De la proposition 1, on déduit que $\\left(H_{A}A_{1}\\right)$ et $\\left(H_{B}B_{1}\\right)$ sont images l'une de l'autre par la rotation d'angle $2 \\cdot (CB, CA)$ de centre $C$ : $\\left(SH_{A}, SH_{B}\\right) = 2 \\cdot (CB, CA)$. Ainsi,\n$$\n\\begin{aligned}\n\\left(SH_{A}, SH_{B}\\right) & \\equiv 2 \\cdot (CB, CA) \\\\\n& \\equiv (CB, CA) + (CB, CH) + (CH, CA) \\\\\n& \\equiv (CB, CA) + \\left(CH_{A}, CB\\right) + \\frac{\\pi}{2} - (AC, AB) \\\\\n& \\equiv \\left(CH_{A}, CA\\right) + (BA, BH) \\\\\n& \\equiv \\left(BH_{A}, BA\\right) + \\left(BA, BH_{B}\\right) \\\\\n& \\equiv \\left(BH_{A}, BH_{B}\\right) [\\pi]\n\\end{aligned}\n$$\nAinsi, $S \\in \\Gamma$.\n\n**Lemme 4.** Les cercles de diamètre $\\left[A_{1}A_{2}\\right], \\left[B_{1}B_{2}\\right], \\left[C_{1}C_{2}\\right]$ sont concourants en un point $M$ appartenant à $\\Gamma$.\n\nPreuve. Soit $M'$ le point d'intersection des cercles de diamètre $\\left[A_{1}A_{2}\\right]$ (passant donc par $H$ et $H_{A}$) et $\\left[B_{1}B_{2}\\right]$ (passant donc par $H$ et $H_{B}$) différent de $H$. On a :\n$$\n\\begin{aligned}\n\\left(M'H_{A}, M'H_{B}\\right) & \\equiv \\left(M'H_{A}, M'H\\right) + \\left(M'H, M'H_{B}\\right) \\\\\n& \\equiv \\left(A_{1}H_{A}, A_{1}H\\right) + \\left(B_{1}H, B_{1}H_{B}\\right) \\\\\n& \\equiv 2 \\cdot \\left(A_{1}B, A_{1}H\\right) + 2 \\cdot \\left(B_{1}H, B_{1}A\\right) \\\\\n& \\equiv 2 \\cdot \\left(\\frac{\\pi}{2} + \\left(HH_{A}, HA_{1}\\right)\\right) + 2 \\cdot \\left(HB_{2}, HH_{B}\\right) \\\\\n& \\equiv 2 \\cdot \\left(HH_{A}, HA_{1}\\right) + 2 \\cdot \\frac{\\pi}{2} + 2 \\cdot \\left(HB_{2}, HH_{B}\\right) \\\\\n& \\equiv 2 \\cdot \\left(HH_{A}, HH_{B}\\right) \\\\\n& \\equiv 2 \\cdot (CB, CA) \\\\\n& \\equiv \\left(SH_{A}, SH_{B}\\right) [\\pi].\n\\end{aligned}\n$$\nAinsi, $M' \\in \\Gamma$ par le théorème de l'angle inscrit. De la même façon, on montre que l'intersection des cercles de diamètre $\\left[B_{1}B_{2}\\right]$ et $\\left[C_{1}C_{2}\\right]$ est un point $M'' \\in \\Gamma$. Comme l'intersection du cercle de diamètre $\\left[B_{1}B_{2}\\right]$ et de $\\Gamma$ est unique, on en déduit que $M' = M'' = M$, c'est-à-dire que les quatre cercles sont concourants en $M$.\n\nNous pouvons maintenant terminer l'exercice :\n$$\n\\begin{aligned}\n\\left(C_{1}M, C_{1}C_{2}\\right) & \\equiv \\left(HM, HC_{2}\\right) \\text{ car } M, C_{1}, H, C_{2} \\text{ sont cocycliques } \\\\\n& \\equiv \\left(HM, HB_{2}\\right) \\\\\n& \\equiv \\left(B_{1}M, B_{1}B_{2}\\right) [\\pi] \\text{ car } M, B_{1}, H, B_{2} \\text{ cocycliques. }\n\\end{aligned}\n$$\nDe la même façon, on prouve que $\\left(B_{1}M, B_{1}B_{2}\\right) \\equiv \\left(C_{1}M, C_{1}C_{2}\\right)$. Donc, comme $\\left(MA_{1}, MA_{2}\\right) \\equiv \\left(MB_{1}, MB_{2}\\right) \\equiv \\left(MC_{1}, MC_{2}\\right) \\equiv \\frac{\\pi}{2} \\ [\\pi]$, on obtient que $MA_{1}A_{2}, MB_{1}B_{2}$ et $MC_{1}C_{2}$ sont directement semblables, ce que l'on notera désormais $MA_{1}A_{2} \\sim MB_{1}B_{2} \\sim MC_{1}C_{2}$. En outre, $A_{1}A_{2}A_{3} \\sim B_{1}B_{2}B_{3} \\sim C_{1}C_{2}C_{3}$ par hypothèse. Donc $MA_{1}A_{2}A_{3} \\sim MB_{1}B_{2}B_{3} \\sim MC_{1}C_{2}C_{3}$.\n\nL'utilisation de similitudes directes permet alors de conclure directement : une similitude directe $\\mathcal{S}$ de centre $M$ envoie $A_{1}$ sur $A_{3}$, et également $B_{1}$ sur $B_{3}$ et $C_{1}$ sur $C_{3}$. Comme $A_{1}, B_{1}$ et $C_{1}$ sont alignés, on en déduit que $A_{3}, B_{3}$ et $C_{3}$ le sont également.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na) Prouver qu'il existe des entiers $a$, $b$, $c$ tels que $(a, b, c) \\neq (0,0,0)$ et $|a|,|b|,|c|<10^{6}$ pour lesquels\n$$\n|a+b \\sqrt{2}+c \\sqrt{3}|<10^{-11}\n$$\n\nb) Soit $a$, $b$, $c$ des entiers tels que $(a, b, c) \\neq (0,0,0)$ et $|a|,|b|,|c|<10^{6}$. Prouver que\n$$\n|a+b \\sqrt{2}+c \\sqrt{3}|>10^{-21}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na) Soit $E$ l'ensemble des $10^{18}$ nombres de la forme $a+b \\sqrt{2}+c \\sqrt{3}$, avec $a, b, c$ entiers naturels et $a, b, c<10^{6}$. On pose $d=(1+\\sqrt{2}+\\sqrt{3}) 10^{6}$.\nPour tout $x \\in E$, on a $0 \\leqslant x<d$. De plus $0<d<(1+2+3) \\times 10^{6}<10^{7}-10^{-11}$. On divise alors l'intervalle $[0; d[$ en $10^{18}-1$ intervalles\n$$\n\\left[\\frac{(k-1) d}{10^{18}-1} ; \\frac{kd}{10^{18}-1}[,\\right.\n$$\npour $k=1, \\ldots, 10^{18}-1$. D'après le principe des tiroirs, au moins deux des $10^{18}$ éléments de $E$, disons $x$ et $y$, appartiennent à un même de ces intervalles. Leur différence $z=x-y$ est bien de la forme $z=a+b \\sqrt{2}+c \\sqrt{3}$ avec $a, b, c$ entiers et $|a|,|b|,|c|<10^{6}$. De plus, $|z|=|x-y|<\\frac{d}{10^{18}-1}<10^{-11}$.\n\nb) Soit $a, b, c$ des entiers tels que $(a, b, c) \\neq (0,0,0)$ et $|a|,|b|,|c|<10^{6}$. Pour $\\epsilon= \\pm 1$ et $\\varepsilon= \\pm 1$, on pose\n$$\nF_{\\epsilon, \\varepsilon}=a+\\epsilon b \\sqrt{2}+\\varepsilon c \\sqrt{3} .\n$$\nComme ci-dessus, on a $|F_{\\epsilon, \\varepsilon}|<10^{7}$.\nOn vérifie facilement que\n$$\n\\begin{aligned}\nF_{1,1} F_{1,-1} F_{-1,1} F_{-1,-1} & =\\left((a+b \\sqrt{2})^{2}-3 c^{2}\\right)\\left((a-b \\sqrt{2})^{2}-3 c^{2}\\right) \\\\\n& =\\left(a^{2}+2 b^{2}-3 c^{2}+2 a b \\sqrt{2}\\right)\\left(a^{2}+2 b^{2}-3 c^{2}-2 a b \\sqrt{2}\\right) \\\\\n& =\\left(a^{2}+2 b^{2}-3 c^{2}\\right)^{2}-2(2 a b)^{2}\n\\end{aligned}\n$$\nest un entier.\nAfin de montrer qu'il est non nul, on procède par l'absurde, et on suppose que\n$$\nF_{1,1} F_{1,-1} F_{-1,1} F_{-1,-1}=0\n$$\nIl s'ensuit que $x^{2}=2 y^{2}$, avec $x=a^{2}+2 b^{2}-3 c^{2}$ et $y=2 a b$. Si $x$ et $y$ sont non nuls, en considérant la valuation 2-adique, il vient $2 v_{2}(x)=1+2 v_{2}(y)$. C'est absurde, ce qui montre que $x=y=0$. Puis :\n— si $a=0$, et puisque $x=0$, on a $2 b^{2}=3 c^{2}$, et la valuation 2-adique montre de nouveau que $b=c=0$;\n- si $b=0$, et puisque $x=0$, on a $a^{2}=3 c^{2}$, et la valuation 3-adique montre aussi que $a=c=0$.\nPar conséquent, si $(a, b, c) \\neq 0$, l'entier $F_{1,1} F_{1,-1} F_{-1,1} F_{-1,-1}$ est non nul.\nAinsi, $|F_{1,1} F_{1,-1} F_{-1,1} F_{-1,-1}| \\geqslant 1$, donc $|F_{1,1}| \\geqslant \\frac{1}{|F_{1,-1} F_{-1,1} F_{-1,-1}|}>10^{-21}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier positif. Montrer que dans un ensemble $A$ de $2^{n}$ nombres strictement positifs, on peut choisir un sous-ensemble $B$ de taille $n+1$ tel que la somme de deux nombres différents dans $B$ ne soit jamais dans $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoient $n$ et $A$ comme dans l'énoncé. Nous allons montrer, par récurrence sur $m \\leqslant n+1$, que l'algorithme glouton consistant à toujours prendre le plus grand nombre qui ne pose pas de problème convient.\n\nPosons $A_{0}=A$, et $B_{0}=B$. Supposons que, pour $m<n+1$, on a construit $A_{m}$ et $B_{m}$ des sous-ensembles de $A$ tels que :\n\n1. la somme de deux éléments différents de $B_{m}$ ne soit jamais dans $A$,\n2. la somme d'un élément de $A_{m}$ et d'un élément de $B_{m}$ ne soit jamais dans $A$,\n3. $A_{m}$ et $B_{m}$ sont disjoints,\n4. $\\operatorname{card}\\left(A_{m}\\right) \\geqslant 2^{n}-2^{m}+1$ et\n5. $\\operatorname{card}\\left(B_{m}\\right)=m$.\n\nDans ce cas, $A_{m}$ est non vide car $m \\leqslant n$ donc $2^{n}-2^{m}+1 \\geqslant 1$. Soit $b_{m+1}$ le plus grand élément de $A_{m}$. On pose $B_{m+1}=B_{m} \\cup\\left\\{b_{m+1}\\right\\}$. Par l'hypothèse 3 pour l'ordre $m$, on obtient 5 pour l'ordre $m+1$. De même, les hypothèses 1 et 2 nous donne 1 pour l'ordre supérieur. Maintenant, on pose\n\n$$\nA_{m+1}=\\left\\{a \\in A_{m} \\mid a \\neq b_{m+1} \\text{ et } a+b_{m+1} \\notin A\\right\\} .\n$$\n\nPar définition, et par l'hypothèse 2, $A_{m+1}$ vérifie 2 et 3. Le seul point moins évident est 4. Nous voulons montrer que $\\operatorname{card}\\left(A_{m}\\right)-\\operatorname{card}\\left(A_{m+1}\\right) \\leqslant 2^{m}$. Or,\n\n$$\n\\begin{aligned}\nA_{\\mathfrak{m}} \\backslash A_{\\mathfrak{m}+1} & =\\left\\{\\mathbf{b}_{\\mathfrak{m}+1}\\right\\} \\cup\\left\\{a \\in A_{\\mathfrak{m}} \\mid a+b_{\\mathfrak{m}+1} \\in A\\right\\} \\\\\n& \\subset\\left\\{b_{\\mathfrak{m}+1}\\right\\} \\cup\\left\\{a \\in A \\mid a>b_{\\mathfrak{m}+1}\\right\\} \\\\\n& \\subset\\left\\{\\mathbf{b}_{\\mathfrak{m}+1}\\right\\} \\cup\\left(A \\backslash A_{\\mathfrak{m}}\\right) .\n\\end{aligned}\n$$\n\nOr ce dernier ensemble est de taille au plus $1+\\left(2^{n}-\\left(2^{n}-2^{m}+1\\right)\\right)=2^{m}$. Donc, $\\operatorname{card}\\left(A_{m}\\right)-\\operatorname{card}\\left(A_{m+1}\\right) \\leqslant 2^{m}$, puis on obtient 4 à l'ordre $m+1$. Ce qui conclut la récurrence.\n\nMaintenant, on applique la récurrence jusqu'à $m=n+1$ et on obtient un ensemble $B_{n+1}$ de taille $n+1$ qui vérifie l'hypothèse 1, donc qui est bien solution de l'exercice.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30592,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn note $P(n)$ le plus grand diviseur premier de $n$. Montrer qu'il existe une infinité d'entiers $n$ tels que $P(n-1)<P(n)<P(n+1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $p$ un nombre premier impair. On cherche un entier $n$ de la forme $p^{2^{k}}$, avec $k$ entier, tel que $P(n-1)<P(n)<P(n+1)$.\n\nSoit $k>l \\geq 0$, on a\n$$\n\\frac{p^{2^{k}}+1}{2}=\\frac{p^{2^{l}}+1}{2}\\left(p^{2^{k}-2^{l}}-p^{2^{k}-2 \\cdot 2^{l}}+\\cdots-p^{2 \\cdot 2^{l}}+p^{2^{l}}-1\\right)+1.\n$$\nDonc les $\\frac{p^{2^{k}}+1}{2}$, $k \\in \\mathbb{N}$, sont deux à deux premiers entre eux. Cela implique que les $P\\left(\\frac{p^{2^{k}}+1}{2}\\right)=P\\left(p^{2^{k}}+1\\right)$ sont des entiers distincts. En particulier, on peut trouver $k$ tel que $P\\left(p^{2^{k}}+1\\right) \\geq p=P\\left(p^{2^{k}}\\right)$. Soit $k_{p}$ le plus petit $k \\geq 0$ vérifiant cette propriété. On choisit $n=p^{2^{k_{p}}}$.\n\nOn voit que $P(n+1) \\geq P(n)$, or $n$ et $n+1$ sont premiers entre eux, donc $P(n+1)>P(n)$. Il reste à montrer que $P(n-1)<P(n)$. On écrit\n$$\nn-1=p^{2^{k_{p}}}-1=\\left(p^{2^{k_{p}-1}}+1\\right)\\left(p^{2^{k_{p}-2}}+1\\right) \\ldots\\left(p^{2}+1\\right)(p+1)(p-1)\n$$\ndonc\n$$\nP(n-1)=\\max \\left(P\\left(p^{2^{k_{p}-1}}+1\\right), \\ldots, P(p+1), P(p-1)\\right)<p=P(n)\n$$\npar minimalité de $k_{p}$ et parce que $P(p-1) \\leq p-1<p$.\n\nFinalement, pour tout nombre premier impair $p$, on peut trouver $n$ une puissance de $p$ telle que $P(n-1)<P(n)<P(n+1)$. Comme il y a une infinité de nombres premiers impairs, et comme les puissances de nombres premiers distincts sont distinctes, il existe une infinité de tels entiers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans les carrés suivants, on s'autorise à remplacer tous les 0 par des 1 et réciproquement sur toute une ligne ou toute une colonne ou toute une diagonale. Dans chaque cas, peut-on n'obtenir que des 0 ?\n$$\n\\left(\\begin{array}{llll}\n0 & 1 & 1 & 0 \\\\\n1 & 1 & 0 & 1 \\\\\n0 & 0 & 1 & 1 \\\\\n0 & 0 & 1 & 1\n\\end{array}\\right) \\quad\\left(\\begin{array}{llll}\n0 & 1 & 0 & 0 \\\\\n1 & 1 & 0 & 1 \\\\\n0 & 0 & 0 & 1 \\\\\n1 & 0 & 1 & 1\n\\end{array}\\right) \\quad\\left(\\begin{array}{llll}\n0 & 0 & 0 & 0 \\\\\n1 & 1 & 0 & 0 \\\\\n0 & 1 & 0 & 1 \\\\\n1 & 0 & 0 & 1\n\\end{array}\\right)\n$$",
    "options": [],
    "answer": "First grid: no. Second grid: no. Third grid: yes.",
    "solution": "Solution:\n\nDans tous les cas, on va dire que l'on agit sur une ligne, une colonne ou une diagonale si on y remplace les 0 par des 1 et réciproquement.\n\nDans le $1^{\\text{er}}$ cas, on remarque que l'on a un nombre impair de 1 : il y en a 9 dans la grille. Or, lors d'une action, la parité du nombre de 1 ne change pas, puisque le nombre de 1 augmente de $+4,+2,0,-2$ ou $-4$. Par conséquent, on aura toujours un nombre impair de 1. Il est donc impossible de n'obtenir que des 0.\n\nDans le $2^{\\text{nd}}$ cas, on a initialement un nombre pair de 1. Cela dit, parmi les 4 cases du carré $2 \\times 2$ situé en haut à gauche, on a trois 1. Or, nulle action ne change la parité du nombre de 1 dans ce carré, car chaque action change la valeur de 0 ou 2 cases du carré. On aura donc toujours un nombre impair de 1 dans ce carré, et il est impossible de n'obtenir que des 0.\n\nDans le $3^{\\text{ème}}$ cas, on peut obtenir une grille formée uniquement de 0, par exemple en agissant successivement sur la $1^{\\text{ère}}$ colonne, puis la $3^{\\text{ème}}$ ligne et enfin la diagonale « nord-ouest / sud-est » :\n$$\n\\left(\\begin{array}{llll}\n0 & 0 & 0 & 0 \\\\\n1 & 1 & 0 & 0 \\\\\n0 & 1 & 0 & 1 \\\\\n1 & 0 & 0 & 1\n\\end{array}\\right) \\longrightarrow \\left(\\begin{array}{llll}\n1 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0 \\\\\n1 & 1 & 0 & 1 \\\\\n0 & 0 & 0 & 1\n\\end{array}\\right) \\longrightarrow \\left(\\begin{array}{llll}\n1 & 0 & 0 & 0 \\\\\n0 & 1 & 0 & 0 \\\\\n0 & 0 & 1 & 0 \\\\\n0 & 0 & 0 & 1\n\\end{array}\\right) \\longrightarrow \\left(\\begin{array}{llll}\n0 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 0 \\\\\n0 & 0 & 0 & 0\n\\end{array}\\right).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geq 3$ un entier et considérons $n$ droites en position générale (c'est-à-dire que trois droites ne sont jamais concourantes et deux droites jamais parallèles). Combien de triangles sont formés par ces droites?\n\nN.B. Par exemple, dans la figure ci-dessous, il y a 4 triangles.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "C(n,3) = n(n-1)(n-2)/6",
    "solution": "Solution:\n\nSi on numérote de 1 à $n$ les $n$ droites, on remarque que chaque triangle est codé par la donnée d'un sous-ensemble de $\\{1,2, \\ldots, n\\}$ à trois éléments. Il y a donc $\\binom{n}{3} = \\frac{n(n-1)(n-2)}{6}$ triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier naturel $n$, on note $S(n)$ la somme des chiffres de l'écriture décimale de $n$. On dit qu'un entier naturel $n$ est joli si $S(n)=S\\left(n^{2}\\right)$. Déterminer toutes les valeurs possibles de $S(n)$ pour les entiers jolis $n$.",
    "options": [],
    "answer": "All nonnegative integers congruent to 0 or 1 modulo 9",
    "solution": "Solution:\n\nModulo $9$, $S(n)$ vérifie l'équation $x^{2}=x$. On vérifie en testant $x=0,1, \\ldots, 8$ que les seules solutions (modulo $9$) sont $0$ et $1$, donc $S(n)$ est nécessairement congru à $0$ ou $1$ modulo $9$.\n\nRéciproquement, on vérifie que si $n=10^{k}-1=99 \\cdots 9$ ($k$ fois) alors $n^{2}=99 \\cdots 9800 \\cdots 01$ (avec $k-1$ chiffres $9$ et $k-1$ chiffres $0$), donc $S(n)=S\\left(n^{2}\\right)=9k$.\n\nDe même, si $n=2 \\times 10^{k}-1=199 \\cdots 9$ alors $n^{2}=399 \\cdots 9600 \\cdots 01$ donc $S(n)=S\\left(n^{2}\\right)=9k+1$.\n\nOn en conclut que les valeurs possibles de $S(n)$ sont les entiers naturels congrus à $0$ ou à $1$ modulo $9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30596,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver le nombre de solutions de $n^{2} m^{6} = 180 t + 2$ pour $n$, $m$ et $t$ des entiers positifs.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nOn considère l'expression modulo $4$. La gauche est un carré et est donc congrue à $0$ ou $1$ modulo $4$. $180$ est congru à $0$ modulo $4$ donc $180 t + 2$ est congru à $0 \\cdot t + 2 = 2$ modulo $4$. L'équation n'a pas de solutions modulo $4$, elle n'en a donc pas non plus dans $\\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30597,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes entiers $1, 2, \\ldots, 2018$ sont écrits au tableau. On effectue alors 2017 opérations comme suit : choisir deux nombres $a$ et $b$, les effacer, et écrire $a + b + 2ab$ à la place. À la fin, il ne reste qu'un seul entier sur le tableau.\nQuelles sont les valeurs que son chiffre des unités peut prendre?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nPuisqu'il y a 1009 nombres impairs entre 1 et 2018, on sait que la somme des entiers initialement inscrits au tableau est impaire. En outre, l'un de ces entiers est congru à $2 \\pmod{5}$. Une récurrence immédiate permet alors de montrer que, après chaque opération, la somme des entiers écrits sur le tableau reste impaire, et l'un de ces entiers est congru à $2 \\pmod{5}$. Par conséquent, le dernier entier écrit sur le tableau est congru à $1 \\pmod{2}$ et à $2 \\pmod{5}$, c'est-à-dire à $7 \\pmod{10}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30598,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn pays comprend $2018 n+1$ villes, où $n$ est un entier naturel non nul. Certaines paires de villes sont reliées par des lignes directes de chemin de fer, de sorte qu'il y ait au plus une ligne entre deux villes ; chaque ligne va dans les deux sens. La distance entre deux villes $A$ et $B$ est alors le nombre minimal de lignes à prendre pour aller de $A$ à $B$.\n\nTrouver l'ensemble des entiers $n$ pour lesquels il possible de construire un réseau ferré respectant le critère suivant:\n\nPour toute ville $C$ et pour tout $i \\in \\{1,2, \\ldots, 2018\\}$, il y a exactement $n$ villes à distance $i$ de $C$.",
    "options": [],
    "answer": "All even positive integers n",
    "solution": "Solution:\n\nDans toute la suite, on va bien sûr réinterpréter l'énoncé en termes de graphes, et on va montrer que les entiers $n$ recherchés sont les entiers pairs.\n\nSupposons d'abord que l'on dispose d'un entier $n$ et d'un graphe respectant le critère de l'énoncé. Alors tout sommet est de degré $n$. La somme des degrés des sommets du graphe vaut donc $(2018 n+1) n$. Or, cette somme est toujours paire. On en déduit que\n$$\nn \\equiv (2018 n+1) n \\equiv 0 \\quad (\\bmod 2)\n$$\nc'est-à-dire que $n$ est pair.\n\nRéciproquement, si $n$ est pair, on forme d'abord un cycle de $2018 n+1$ sommets. Puis on rajoute des arêtes auxiliaires entre deux sommets $u$ et $v$ dès lors que la distance entre $u$ et $v$ (dans notre cycle sans arêtes auxiliaires) est comprise entre $2$ et $n / 2$. Il est alors aisé de vérifier que ce graphe (avec arêtes auxiliaires) respecte le critère de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer la valeur maximale de $\\sqrt{x}+\\sqrt{2 y+2}+\\sqrt{3 z+6}$ lorsque $x, y, z$ sont des nombres réels strictement positifs vérifiant $x+y+z=3$.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nD'après l'inégalité de Cauchy-Schwarz, on a\n$$\n\\sqrt{x} \\sqrt{1}+\\sqrt{y+1} \\sqrt{2}+\\sqrt{z+2} \\sqrt{3} \\leqslant \\sqrt{x+y+1+z+2} \\sqrt{1+2+3}=6\n$$\nL'égalité est atteinte lorsque $x=y=z=1$. Donc la valeur maximale est $6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les entiers $x, y \\geqslant 1$ tels que\n$$\n\\frac{1}{x}+\\frac{1}{y}=\\frac{1}{2017}\n$$",
    "options": [],
    "answer": "Solutions: (4034, 4034), (4070306, 2018), (2018, 4070306).",
    "solution": "Solution:\nL'équation équivaut à $2017(x+y)=x y$, donc $2017 \\mid x y$. Comme $2017$ est premier, $2017 \\mid x$ ou $2017 \\mid y$. Sans perte de généralité, $x$ et $y$ jouant le même rôle, on peut supposer que $2017 \\mid x$. Soit $x' \\geqslant 1$ tel que $x=2017 x'$. On a donc $2017 x'+y=x' y$, soit\n$$\n2017 x'=y(x'-1)\n$$\nDonc $2017 \\mid y$ ou $2017 \\mid x'-1$, comme précédemment.\n\nSi $2017 \\mid y$, on écrit $y=2017 y'$, et donc $x'+y'=x' y'$. On voit qu'il n'y a pas de solution si $x'=1$ ou $y'=1$. Si $x'=2$, on trouve $y'=2$ et réciproquement. Si $x', y' \\geqslant 3$, $x' y' \\geqslant \\max (3 x', 3 y') > x'+y'$, donc il n'y a pas de solution. On trouve ainsi une unique solution : $x'=y'=2$, soit $x=y=4034$. On vérifie que cela satisfait bien l'équation de départ.\n\nSi $2017 \\mid x'-1$, on écrit $x'=2017 k+1$ avec $k \\in \\mathbb{N}$. On obtient alors $2017 k+1=y k$, soit $(y-2017) k=1$. Ceci implique que $k=1$ et donc que $y-2017=1$. Ainsi, $y=2018$ et $x=2017 \\times 2018$. On vérifie que cela satisfait l'équation initiale.\n\nConclusion : il y a trois solutions, à savoir les couples d'entiers $(4034,4034)$, $(2017 \\times 2018,2018)$ et $(2018,2017 \\times 2018)$. En effet, $x$ et $y$ jouant le même rôle, le deuxième cas que nous avons traité donne deux solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle acutangle, et soit $P$ un point situé à l'intérieur du triangle $ABC$. Soit $D$ le milieu du segment $[PC]$, $E$ le point d'intersection des droites $(AP)$ et $(BC)$, et $Q$ le point d'intersection des droites $(BP)$ et $(DE)$. Montrer que, si les angles $\\widehat{PAC}$ et $\\widehat{PCB}$ sont égaux, alors $\\sin (\\widehat{BCQ}) = \\sin (\\widehat{BAP})$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $R$ le point d'intersection des droites $(CQ)$ et $(AP)$. Il suffit de prouver l'égalité des angles de droites $(CB, CQ) = (AB, AR)$. Pour ce faire, il suffit de prouver que les points $B, C, A, R$ sont cocycliques, c'est-à-dire que $(BR, BC) = (AR, RC)$. Comme $(AR, AC) = (CP, CB)$, il suffit de prouver que $(BR, BC) = (CP, CB)$, donc que les droites $(BR)$ et $(PC)$ sont parallèles.\n\nPour ce dernier point, on peut utiliser que si $X = (CP) \\cap (BR)$ alors $(X, D, P, C)$ est une division harmonique. On peut également se passer de géométrie projective comme suit :\n\nOn utilise le théorème de Ceva dans le triangle $PCE$, puisque les droites $(PB), (DE)$ et $(RC)$ sont concourantes en $Q$, avec $R \\in (PE)$ et $B \\in (CE)$. On en déduit que\n$$\n1 = \\frac{CB}{EB} \\cdot \\frac{PD}{CD} \\cdot \\frac{RE}{RP} = \\frac{CB}{EB} \\cdot \\frac{RE}{RP}\n$$\nc'est-à-dire\n$$\n\\frac{RE}{EB} = \\frac{RP}{CB}\n$$\nEn appliquant la réciproque du théorème de Thalès aux triangles $BRE$ et $CPE$, on déduit de cette dernière égalité que les droites $(BR)$ et $(PC)$ sont bien parallèles, ce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30602,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver la somme des $n$ tels que $n^{2}+8 n+44$ soit un carré parfait.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nSoit $f$ la fonction étudiée, une idée dans ce genre d'exercice est de coincer $f(n)$ entre 2 carrés d'entiers consécutifs $k^{2}<f(n)<(k+1)^{2}$ et de dire par l'absurde que si $f(n)=i^{2}$ alors $k<i<k+1$ ce qui est impossible car on a des entiers.\n\nIci les calculs peuvent s'abréger en remarquant que $f$ a même parité que $n+4$, ainsi si $(n+4)^{2}=n^{2}+8 n+16<f(n)<n^{2}+12 n+36=(n+6)^{2}$ ce qui est vrai dès que $n>2$, $f(n)$ n'est pas un carré. Pour les petits cas $2$ est solution et $1$ non.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30603,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. On note $L, M, N$ les milieux de $[BC]$, $[CA]$ et $[AB]$. Notons $(d)$ la tangente en $A$ au cercle circonscrit à $ABC$. La droite $(LM)$ coupe $(d)$ en $P$, et la droite $(LN)$ coupe $(d)$ en $Q$. Montrer que $(CP)$ et $(BQ)$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nOn a $(AQ, AB) = (CA, CB) = (LQ, LB)$ donc $A, Q, B, L$ sont cocycliques, et de même $A, P, C, M$ sont cocycliques. On a donc $(CP, BC) = (CP, CL) = (AP, AL) = (AQ, AL) = (BQ, BL) = (BQ, BC)$, donc $(CP)$ et $(BQ)$ sont parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn dispose, dans le plan, 16 points deux à deux distincts, que l'on note $A_{i, j}$ pour $i, j \\in \\{1,2,3,4\\}$. Ces points vérifient les relations d'alignement et de cocyclicité suivantes :\n\n$\\triangleright$ pour tout $i \\in \\{1,2,3,4\\}$, les points $A_{i, 1}, A_{i, 2}, A_{i, 3}$ et $A_{i, 4}$ sont alignés;\n\n$\\triangleright$ pour tout $j \\in \\{1,2,3,4\\}$, les points $A_{1, j}, A_{2, j}, A_{3, j}$ et $A_{4, j}$ sont alignés;\n\n$\\triangleright$ les quadrilatères $A_{1,1} A_{1,2} A_{2,2} A_{2,1}, A_{2,1} A_{2,2} A_{3,2} A_{3,1}, A_{3,1} A_{3,2} A_{4,2} A_{4,1}, A_{1,2} A_{1,3} A_{2,3} A_{2,2}$, $A_{1,3} A_{1,4} A_{2,4} A_{2,3}, A_{1,1} A_{2,2} A_{2,3} A_{1,4}$ et $A_{1,1} A_{2,2} A_{3,2} A_{4,1}$ sont cycliques.\n\nMontrer que le quadrilatère $A_{4,1} A_{3,2} A_{3,3} A_{4,4}$ est lui aussi cyclique.\n\nNote: on dit qu'un quadrilatère est cyclique si ses quatre sommets sont sur un même cercle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\n\nCi-dessous, on notera $\\Delta_{i}$ la droite $\\left(A_{i, 1} A_{i, 2}\\right)$ et $\\Delta^{j}$ la droite $\\left(A_{1, j} A_{2, j}\\right)$. Une figure et une chasse aux angles nous montrent que les droites $\\Delta_{1}$ et $\\Delta_{3}$ sont parallèles, puisque ( $\\Delta_{1}, \\Delta^{1}$ ) $\\equiv$ $\\left(\\Delta^{2}, \\Delta_{2}\\right) \\equiv\\left(\\Delta_{3}, \\Delta^{1}\\right)\\left(\\bmod 180^{\\circ}\\right)$. De même, on sait que $\\Delta_{2} / / \\Delta_{4}$, que $\\Delta^{1} / / \\Delta^{3}$ et que $\\Delta^{2} / / \\Delta^{4}$.\n\nCe faisant, et sous réserve que le quadrilatère $A_{1,1} A_{1,2} A_{2,2} A_{2,1}$ formé par les droites $\\Delta_{1}, \\Delta^{1}$, $\\Delta_{2}$ et $\\Delta^{2}$ soit cyclique, il suffit de définir le point $A_{i, j}$ comme le point d'intersection de $\\Delta_{i}$ et $\\Delta^{j}$ pour être assuré de nos relations d'alginement, et pour être certain que nos cinq premiers quadrilatères sont cycliques. En particulier, toujours par chasse aux angles, chaque quadrilatère $A_{i, j} A_{i+1, j} A_{i+1, j+1} A_{i, j+1}$ est alors cyclique.\n\nDans toute la suite, on supposera ces relations d'alignement et de cocyclicité acquises, et on s'intéresse maintenant au quadrilatère $A_{1,1} A_{2,2} A_{2,3} A_{1,4}$. Vu le rôle symétrique qu'il semble jouer avec les quadrilatères $A_{1,1} A_{2,2} A_{3,2} A_{4,1}$ d'une part et $A_{1,1} A_{2,2} A_{3,2} A_{4,1}$ d'autre part, il serait bon de trouver un critère simple qui soit équivalent avec son caractère cyclique.\n\nOr, il ne met en jeu que les droites $\\Delta_{1}$ et $\\Delta_{2}$ d'une part, et les quatre droites $\\Delta^{j}$ d'autre part, les paires $\\left(\\Delta^{1}, \\Delta^{2}\\right)$ et $\\left(\\Delta^{3}, \\Delta^{4}\\right)$ jouant des rôles symétriques. On en vient alors à considérer les points d'intersection $X=\\Delta_{1} \\cap \\Delta_{2}, Y=\\Delta^{1} \\cap \\Delta^{2}$ et $Z=\\Delta^{3} \\cap \\Delta^{4}$, dont la figure semble montrer qu'ils sont alignés. Cela suggère le lemme suivant.\n\nLemme: Le quadrilatère $A_{1,1} A_{2,2} A_{2,3} A_{1,4}$ est cyclique si et seulement si $X$, $Y$ et $Z$ sont alignés.\n\nDémonstration : Supposons d'abord $\\mathrm{X}, \\mathrm{Y}$ et Z alignés. Le théorème de Thalès indique alors que les quadrilatères $A_{1,1} A_{1,2} A_{2,2} A_{2,1}$ et $A_{1,3} A_{2,3} A_{2,4} A_{3,4}$ sont homothétiques l'un de l'autre, donc que $\\left(A_{2,1} A_{1,2}\\right) / /\\left(A_{2,3} A_{1,4}\\right)$. On en déduit que ( $\\left.A_{1,1} A_{1,4}, A_{1,1} A_{2,2}\\right) \\equiv\\left(A_{1,1} A_{1,2}, A_{1,1} A_{2,2}\\right) \\equiv$ $\\left(A_{2,1} A_{1,2}, A_{2,1} A_{2,2}\\right) \\equiv\\left(A_{2,3} A_{1,4}, A_{2,3} A_{2,2}\\right)\\left(\\bmod 180^{\\circ}\\right)$, ce qui signifie que $A_{1,1} A_{1,4} A_{2,3} A_{2,2}$ est bien cyclique.\n\nRéciproquement, une fois fixées les droites $\\Delta_{1}, \\Delta_{2}, \\Delta^{1}, \\Delta^{2}$ et $\\Delta^{3}$, le point $A_{4,1}$ puis la droite $\\Delta^{4}$ sont définis de manière unique, donc $Z$ se retrouve nécessairement aligné avec $X$ et $Y$.\n\nOn en profite aussi pour poser $T=\\Delta_{3} \\cap \\Delta_{4}$. Ici, puisque $A_{1,1} A_{2,2} A_{2,3} A_{1,4}$ et $A_{1,1} A_{2,2} A_{3,2} A_{4,1}$ sont cycliques, le lemme montre l'alignement des points $X, Y$ et $Z$, et $X, Y$ et $T$ d'autre part. Ainsi, les points $T, Y$ et $Z$ sont alignés, et le lemme montre cette fois-ci que le quadrilatère $A_{4,1} A_{3,2} A_{3,3} A_{4,4}$ est cyclique.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30605,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn fixe un entier naturel $n \\geqslant 2$. Déterminer tous les nombres réels $x \\geqslant -1$ tels que pour tous nombres réels $a_{1}, \\ldots, a_{n} \\geqslant 1$ on ait\n$$\n\\frac{a_{1}+x}{2} \\times \\cdots \\times \\frac{a_{n}+x}{2} \\leqslant \\frac{a_{1} a_{2} \\cdots a_{n}+x}{2} .\n$$",
    "options": [],
    "answer": "[-1, 1]",
    "solution": "Solution:\n\nEn prenant $a_{i}=1$ pour tout $i$, on obtient la condition $\\left(\\frac{1+x}{2}\\right)^{n} \\leqslant \\frac{1+x}{2}$. Notons $y=\\frac{1+x}{2}$. Par hypothèse, $y \\geqslant 0$. Comme $y^{n} \\leqslant y$, on a $y^{n-1} \\leqslant 1$ donc $y \\leqslant 1$, ce qui implique $x \\leqslant 1$.\n\nRéciproquement, montrons que tout $x \\in[-1,1]$ convient. On raisonne par récurrence sur $n$. Pour $n=1$ l'inégalité est évidente. Supposons qu'elle soit vraie au rang $n-1$. Le membre de gauche de l'inégalité est alors inférieur ou égal à $\\frac{a+x}{2} \\times \\frac{b+x}{2}$, où $a=a_{1} a_{2} \\cdots a_{n-1}$ et $b=a_{n}$. Il suffit alors de montrer que $\\left(\\frac{a+x}{2}\\right)\\left(\\frac{b+x}{2}\\right) \\leqslant\\left(\\frac{a b+x}{2}\\right)$. En développant, l'inégalité devient $(a+b-2) x+x^{2} \\leqslant a b$. Comme $a+b-2 \\geqslant 0$, et comme $x \\leqslant 1$ et $x^{2} \\leqslant 1$, le membre de gauche de l'inégalité est majoré par $(a+b-2)+1$, donc il suffit de montrer que $(a+b-2)+1 \\leqslant a b$, ou encore que $0 \\leqslant a b-a-b+1$. Or, on a $a b-a-b+1=(a-1)(b-1)$. Comme $a \\geqslant 1$ et $b \\geqslant 1$, l'inégalité $(a-1)(b-1) \\geqslant 0$ est bien vérifiée.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn place les entiers de $1$ à $9$ dans chacune des cases d'une grille $3 \\times 3$. Pour $i=1$, $2$ et $3$, on note $\\ell_{i}$ le plus grand entier présent dans la $i^{\\text{ème}}$ ligne et $c_{i}$ le plus petit entier présent dans la $i^{\\text{ème}}$ colonne.\n\nCombien existe-t-il de grilles telles que $\\min \\left\\{\\ell_{1}, \\ell_{2}, \\ell_{3}\\right\\}=\\max \\left\\{c_{1}, c_{2}, c_{3}\\right\\}=4$ ?",
    "options": [],
    "answer": "25920",
    "solution": "Solution:\n\nSoit $G$ une grille quelconque, et soit $i$ et $j$ les entiers tels que l'entier $4$ se trouve sur la $i^{\\text{ème}}$ ligne et la $j^{\\text{ème}}$ colonne. On dit que la grille $G$ est bonne si $\\ell_{i}=c_{j}=4$.\n\nTout d'abord, $\\min\\left\\{\\ell_{1}, \\ell_{2}, \\ell_{3}\\right\\}=\\max \\left\\{c_{1}, c_{2}, c_{3}\\right\\}=4$, alors $4=\\ell_{i}=c_{j}$, donc $G$ est une bonne grille. Réciproquement, si $G$ est une bonne grille, alors, pour tout entier $j'$, la colonne $j'$ comporte un des entiers de la ligne $i$. Cet entier vaut donc au plus $4$, ce qui montre que $c_{j'} \\leqslant 4$. De même, pour tout entier $i'$, on a $\\ell_{i'} \\geqslant 4$. On a donc bien $\\min \\left\\{\\ell_{1}, \\ell_{2}, \\ell_{3}\\right\\}=\\max \\left\\{c_{1}, c_{2}, c_{3}\\right\\}=4$.\n\nPar conséquent, on cherche en fait à compter combien il existe de bonnes grilles. Pour construire une bonne grille, il suffit de choisir successivement :\n\n$\\triangleright$ les entiers $i$ et $j$, c'est-à-dire l'emplacement du $4$ : on a $9$ choix possibles ;\n\n$\\triangleright$ chacun des deux autres entiers qui se trouvent sur la ligne $i$. Comme $\\ell_{i}=4$, ces deux entiers doivent être plus petits que $4$, donc être choisis dans $\\{1,2,3\\}$. On a $3$ manières de choisir le premier (disons le plus à gauche), puis $2$ manières de choisir le second ;\n\n$\\triangleright$ chacun des deux autres entiers qui se trouvent sur la colonne $j$. Comme $c_{j}=4$, les deux autres entiers doivent être plus grands que $4$, donc être choisis dans $\\{5,6,7,8,9\\}$. On a $5$ manières de choisir le premier, puis $4$ manières de choisir le second ;\n\n$\\triangleright$ chacun des quatre autres entiers qui restent à placer : on a $4!$ manières de procéder.\n\nIl existe donc $9 \\times 3 \\times 2 \\times 5 \\times 4 \\times 4! = 25920$ grilles telles que $\\min \\left\\{\\ell_{1}, \\ell_{2}, \\ell_{3}\\right\\}=\\max \\left\\{c_{1}, c_{2}, c_{3}\\right\\}=4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30607,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAdalbert et Babette jouent aux dominos sur une grille rectangulaire de 2 cases de hauteur et 2018 cases de largeur. Adalbert commence en posant un domino de taille $1 \\times 2$ en position horizontale, de façon à ce qu'il recouvre exactement deux cases de la grille. Puis Babette joue de même un domino $1 \\times 2$ en position verticale, et ainsi de suite. Le premier qui ne peut plus jouer sans chevaucher un domino déjà posé a perdu. Montrer qu'Adalbert a une stratégie gagnante.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDivisons la grille en 1009 blocs consécutifs de taille $2 \\times 2$. Adalbert commence par poser un domino dans un de ces blocs. Il est alors assuré de pouvoir le compléter plus tard par un autre domino. Babette doit alors jouer dans un autre bloc (et elle ne peut pas jouer à cheval entre deux blocs, car ses dominos sont verticaux). Adalbert joue dans un troisième bloc, etc. Lorsqu'il joue pour la 505-ème fois, il entame un 505-ème bloc, ce qui lui garantit la possibilité de jouer encore 505 fois. Il pourra donc poser 1010 dominos au total. Comme on peut poser au plus 2018 dominos au total, Babette pourra en poser au plus 1008, elle sera donc bloquée avant Adalbert, qui l'emporte.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30608,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ un entier naturel. On se donne $2 n$ boules. Sur chacune de ces boules est écrit un nombre. On suppose que, à chaque fois que l'on regroupe les boules en $n$ paires, deux de ces paires ont la même somme.\n\na. Montrer que quatre de ces boules comportent le même nombre.\n\nb. Montrer que le nombre de valeurs distinctes inscrites sur les boules est $\\leqslant n-1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. On note $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{2 n}$ les valeurs prises par les $2 n$ boules dans l'ordre décroissant. On les apparie ainsi : $(a_{1}, a_{2}), (a_{3}, a_{4}), \\ldots$ Comme il existe $i<j$ impairs tels que $a_{i}+a_{i+1}=a_{j}+a_{j+1}$, et comme $a_{i} \\geqslant a_{i+1} \\geqslant a_{j} \\geqslant a_{j+1}$, on a $a_{i}=a_{i+1}=a_{j}=a_{j+1}$.\n\nb. Supposons qu'il y ait au moins $n$ valeurs distinctes, que l'on ordonne par ordre décroissant $b_{1}>b_{2}>\\cdots>b_{n}$. On ordonne les autres boules dans l'ordre décroissant $c_{1} \\geqslant \\cdots \\geqslant c_{n}$. On forme les $n$ paires $(b_{i}, c_{i})$, et on constate que si $i<j$ alors $b_{i}+c_{i}>b_{j}+c_{j}$, ce qui contredit le fait que deux de ces sommes doivent être égales.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $n \\geqslant 3$ tels que, si $a_{1}, \\ldots, a_{n}$ sont des réels strictement positifs tels que $\\max \\left(a_{1}, \\ldots, a_{n}\\right) \\leqslant n \\cdot \\min \\left(a_{1}, \\ldots, a_{n}\\right)$, alors il existe nécessairement trois de ces réels qui sont les longueurs des côtés d'un triangle acutangle, c'est-à-dire d'un triangle dont les trois angles sont strictement aigus.",
    "options": [],
    "answer": "all integers n ≥ 13",
    "solution": "Solution:\n\nSans perte de généralité, on suppose que $1=a_{1} \\leqslant a_{2} \\leqslant \\ldots \\leqslant a_{n}$. On note alors que trois réels $a_{i}, a_{j}$ et $a_{k}$, avec $i<j<k$, sont les longueurs des côtés d'un triangle acutangle si et seulement si $a_{i}^{2}+a_{j}^{2}>a_{k}^{2}$ (on peut le voir par exemple avec le théorème d'AlKashi). Si tel est le cas, alors $a_{k-2}, a_{k-1}$ et $a_{k}$ sont eux aussi les côtés d'un triangle acutangle. Par conséquent, appelons obtusangle une suite de réels $1=a_{1} \\leqslant a_{2} \\leqslant \\ldots \\leqslant a_{n}$ telle que $a_{k}^{2} \\geqslant a_{k-2}^{2}+a_{k-1}^{2}$ pour tout $k \\in\\{3, \\ldots, n\\}$ : le problème revient à chercher les entiers $n \\geqslant 3$ tels que, si $a_{1}, \\ldots, a_{n}$ est une suite obtusangle, alors $a_{n}>n$.\n\nOr, soit $F_{n}$ le $n^{\\text{ème}}$ nombre de Fibonacci, défini par $F_{1}=F_{2}=1$ et $F_{n+2}=F_{n+1}+F_{n}$ pour tout $n \\geqslant 1$. Dans une suite obtusangle $a_{1}, \\ldots, a_{n}$, on a nécessairement $1 \\leqslant a_{1} \\leqslant a_{2}$, c'est-à-dire $F_{1} \\leqslant a_{1}^{2}$ et $F_{2} \\leqslant a_{2}^{2}$. Une récurrence immédiate sur $n$ montre alors que $a_{n}^{2} \\geqslant F_{n}$. Réciproquement, on remarque bien que la suite $a_{1}, \\ldots, a_{n}$ définie par $a_{k}=\\sqrt{F_{k}}$ est obtusangle. Par conséquent, les entiers $n \\geqslant 3$ recherchés sont ceux tels que $\\sqrt{F_{n}}>n$, c'est-à-dire $F_{n}>n^{2}$.\n\nLes premiers termes de la suite de Fibonacci sont $F_{1}=F_{2}=1, F_{3}=2, F_{4}=3, F_{5}=5, F_{6}=8$, $F_{7}=13, F_{8}=21, F_{9}=34, F_{10}=55, F_{11}=89, F_{12}=144, F_{13}=233$ et $F_{14}=377$. On constate donc que $F_{n} \\leqslant n^{2}$ pour tout $n \\leqslant 12$, et que $F_{n}>n^{2}$ pour $n=13$ et $n=14$. En outre, pour tout entier $n \\geqslant 13$ tel que $F_{n}>n^{2}$ et $F_{n+1}>(n+1)^{2}$, on observe que\n$$\nF_{n+2}=F_{n}+F_{n+1}>n^{2}+(n+1)^{2} \\geqslant 5 n+(n+1)^{2} \\geqslant 2 n+3+(n+1)^{2}=(n+2)^{2}\n$$\nUne récurrence immédiate montre donc que les entiers recherchés sont les entiers $n \\geqslant 13$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver toutes les valeurs possibles pour $f(2018)$ où $f$ est une fonction de $\\mathbb{Q}$ dans $\\mathbb{R}_{+}$ vérifiant les trois conditions suivantes:\n- $f(2)=1 / 2$ ;\n- pour tout rationnel $x$, si $f(x) \\leqslant 1$, alors $f(x+1) \\leqslant 1$;\n- $f(x y)=f(x) f(y)$ pour tous rationnels $x$ et $y$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\nTout d'abord, l'égalité $1 / 2=f(2)=f(1 \\times 2)=f(1) f(2)=f(1) / 2$ montre que $f(1)=1$, et donc que $f(n) \\leqslant 1$ pour tout entier naturel non nul $n$.\n\nNotons de plus que pour tous $x, y$ avec $y \\neq 0$ on a $f(y) \\neq 0$, et $f(x / y)=f(x) / f(y)$ puisque $f(x / y) f(y)=f((x / y) y)=f(x)$. En particulier, $f(1 / 2)=2$. Enfin, pour tout rationnel $x$, on a $f(x)^{2}=f\\left(x^{2}\\right)=f(-x)^{2}$, donc $f$ est paire.\n\nSoit alors $k$ un entier naturel impair, de la forme $k=2 n+1$ : on a nécessairement $f(k) \\leqslant 1$. On sait également que $f(-k / 2)>1$, puisque $1 / 2=-k / 2+(n+1)$, donc $f(-2 / k)=1 / f(-k / 2) \\leqslant 1$, ce qui implique que $f\\left(\\frac{k-2}{k}\\right)=f\\left(-\\frac{2}{k}+1\\right) \\leqslant 1$, par conséquent $f(k-2) \\leqslant f(k)$.\n\nCeci permet de montrer par récurrence que pour tout $k \\geqslant 1$ impair on a $f(k) \\geqslant 1$, et donc $f(k)=1$.\n\nOn déduit de tout ceci que $f(2018)=f(2) f(1009)=1 / 2$.\n\nRéciproquement, il reste à vérifier qu'une fonction $f$ telle que décrite dans l'énoncé existe bien. Il suffit pour ce faire de constater que la fonction $f$ définie par\n$$\nf: p / q \\mapsto \\begin{cases}0 & \\text{ si } p=0 \\\\ 2^{v_{2}(q)-v_{2}(p)} & \\text{ si } p \\neq 0\\end{cases}\n$$\noù $v_{2}(n)$ désigne la valuation 2-adique de l'entier $n$, satisfait bien les conditions de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un entier $k$ est olympique s'il existe quatre entiers $a, b, c$ et $d$, tous premiers avec $k$, tels que $k$ divise $a^{4}+b^{4}+c^{4}+d^{4}$. Soit $n$ un entier quelconque.\n\nMontrer que $n^{2}-2$ est olympique.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $p_{1}^{\\alpha_{1}} \\cdots p_{\\ell}^{\\alpha_{\\ell}}$ la décomposition de $n^{2}-2$ en produits de facteurs premiers. Grâce au théorème Chinois, il nous suffit en fait de montrer que chacun des entiers $p_{i}^{\\alpha_{i}}$ est olympique.\n\nSupposons tout d'abord que $p_{i}=2$. Puisque $n^{2}-2 \\equiv\\{2,3\\} (\\bmod 4)$, on sait que $\\alpha_{i} \\leqslant 1$. Or, il est clair que $2$ est olympique, puisque $1+1+1+1 \\equiv 0 (\\bmod 2)$.\n\nOn considère maintenant un diviseur premier $p$ impair de $n^{2}-2$. Puisque $2$ n'est pas carré modulo $3$ ou $5$, on sait en fait que $p \\geqslant 7$. Par ailleurs, si $p^{\\alpha}$ est olympique, alors $p$ l'est clairement lui aussi. Réciproquement, si $p$ est olympique (et impair), le lemme de Hensel montre que les puissances $4^{\\text{èmes}}$ modulo $p^{k}$ sont exactement les entiers dont les résidus (modulo $p$) sont des puissances $4^{\\text{èmes}}$ modulo $p$, et il s'ensuit que $p^{\\alpha}$ est également olympique.\n\nOn se ramène donc à montrer que tout nombre premier $p \\geqslant 7$ tel que $2$ soit un carré modulo $p$ est olympique. On va procéder par disjonction de cas, selon la valeur de $p$ modulo $4$ ou $8$. Cependant, on rappelle d'abord le petit théorème de Fermat, qui stipule que l'ordre $\\omega_{p}(x)$ d'un élément inversible modulo $p$ doit diviser $p-1$. On rappelle également que $\\mathbb{Z} / p \\mathbb{Z}$ contient un élément d'ordre $p-1$, comme l'indique le Théorème 17 du cours d'Arithmétique de la POFM sur les propriétés de $\\mathbb{Z} / \\mathrm{n} \\mathbb{Z}$. On en déduit le lemme suivant.\n\nLemme : Les puissances $k^{\\text{èmes}}$ sont les éléments $x$ dont l'ordre $\\omega_{p}(x)$ divise $(p-1)/\\delta$, où $\\delta=\\operatorname{PGCD}(k, p-1)$.\n\nDémonstration : Soit $z$ un élément d'ordre $p-1$. Les puissances $k^{\\text{èmes}}$ sont les éléments de la forme $z^{a k+b(p-1)}$, avec $a$ et $b$ entiers, c'est-à-dire les éléments de la forme $z^{c \\delta}$ avec $c$ entier. L'ordre de ces éléments modulo $p$ divise clairement $(p-1)/\\delta$.\n\nRéciproquement, $\\omega_{p}(x)$ divise $(p-1)/\\delta$ si et seulement si $x$ est une des racines du polynôme $P(X)=1-X^{(p-1)/\\delta}$. Or, les éléments de la forme $z^{c \\delta}$ constituent déjà $(p-1)/\\delta$ racines de $P(X)$. Par conséquent, les puissances $^{k\\text{èmes}}$ sont exactement les racines de $P(X)$, ce qui conclut la preuve du lemme.\n\nOn procède enfin à la disjonction de cas annoncée ci-dessus.\n\n1. Si $p \\equiv 1 (\\bmod 8)$, et puisque $-1$ est d'ordre $2$, c'est une puissance $4^{\\text{ème}}$. Comme $0 \\equiv 1+1+(-1)+(-1) (\\bmod p)$, l'entier $p$ est olympique.\n\n2. Si $p \\equiv 5 (\\bmod 8)$, alors $-1$ est un carré. Soit $i$ et $-i$ ses racines carrées. Elles sont d'ordre $4$, donc ce ne sont pas des carrés. Or, $(1+i)^{2} \\equiv 2i (\\bmod p)$. Puisque l'ensemble des carrés est clos par produit et par inverse (donc par division), $2$ n'est donc pas un carré, ce qui signifiait en fait que $p$ ne divisait pas $n^{2}-2$. Ce cas s'avère donc impossible.\n\n3. Si $p \\equiv 3 (\\bmod 4)$, alors être une puissance $4^{\\text{ème}}$ revient en fait à être un carré. Il suffit donc de montrer que $0$ est somme de quatre carrés non nuls modulo $p$.\n\nOn va d'abord montrer que c'est une somme de $\\ell$ carrés non nuls, avec $2 \\leqslant \\ell \\leqslant 4$. En effet, soit $Q$ l'ensemble des carrés. Notons que $|Q|=(p+1)/2$, puisque tout carré non nul a exactement deux racines carrées modulo $p$. Ainsi, pour tout $k \\in \\mathbb{Z} / p \\mathbb{Z}$, et par principe des tiroirs, l'ensemble $(k-Q) \\cap Q$ est non vide, ce qui signifie que $k$ est somme de deux carrés. Si $k \\neq 0$, l'un de ces carrés est même non nul. Mais alors, en écrivant $0 \\equiv k+(-k) (\\bmod p)$, on peut bien écrire $0$ comme une somme de quatre carrés dont au moins deux sont non nuls.\n\nPuis, en vertu de l'identité $k^{2} \\equiv (3k/5)^{2} + (4k/5)^{2} (\\bmod p)$, tout carré non nul est en fait somme de deux carrés non nuls. Par conséquent, $0$ est bien somme de quatre carrés non nuls, ce qui signifie que $p$ est olympique.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $p$ et $n$ des entiers strictement positifs, avec $p$ premier, et $n \\geqslant p$. On suppose que $1+n p$ est un carré parfait. Montrer que $n+1$ est une somme de $p$ carrés parfaits non nuls (non nécessairement distincts).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $k$ tel que $1+n p = k^{2}$, alors $n p = (k-1)(k+1)$.\n\nPremier cas : $p$ divise $k-1$. Alors on peut écrire $k-1 = p \\ell$, donc $k = p \\ell + 1$. En reportant dans l'égalité, on obtient que\n$$\nn+1 = p \\ell^{2} + 2 \\ell + 1 = (p-1) \\ell^{2} + (\\ell+1)^{2}.\n$$\n\nDeuxième cas : $p$ divise $k+1$. On obtient avec la même méthode\n$$\nn+1 = p \\ell^{2} - 2 \\ell + 1 = (p-1) \\ell^{2} + (\\ell-1)^{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn entier $n$ est parfait si la somme de ses diviseurs est $2n$. Soit $n$ un entier parfait et $p$ son plus petit diviseur premier. Montrer que l'exposant de $p$ dans la décomposition en produit de puissances de nombres premiers de $n$ est pair.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $n = p_1^{a_1} \\cdots p_k^{a_k}$ avec les $p_i$ premiers distincts. La somme des diviseurs de $n$ est $\\sigma(n) = \\prod_{i=1}^{k} \\left(1 + \\cdots + p_i^{a_i}\\right)$.\n\nSupposons par l'absurde que $a_1$ soit impair, alors\n$$\n1 + \\cdots + p_1^{a_1} = (1 + p_1) + p_1^2 (1 + p_1) + \\cdots + p_1^{a_1-1}(1 + p_1).\n$$\nAlors $p_1 + 1$ divise $1 + \\cdots + p_1^{a_1}$, donc aussi $2n = \\prod_{i=1}^{k} \\left(1 + \\cdots + p_i^{a_i}\\right)$.\n\nSi $p_1 + 1$ n'est pas premier : il a un plus grand diviseur premier $q < p_1$ qui vérifie $q \\mid 2n$ ; alors comme $q$ est premier avec $n$, $q = 2$ ; mais alors $q^2 \\mid p_1 + 1 \\mid 2n$ et donc $q \\mid n$, contradiction.\n\nDonc $p_1 + 1$ est premier et alors nécessairement $p_1 = 2$ et $p_1 + 1 = p_2 = 3$. Alors $n, \\frac{n}{2}, \\frac{n}{3}$ et $\\frac{n}{6}$ sont entiers et divisent $n$. Comme $n > 6$, ils sont différents de $1$. Ainsi\n$$\n2n = \\sigma(n) \\geq n + \\frac{n}{2} + \\frac{n}{3} + \\frac{n}{6} + 1 = 2n + 1,\n$$\ncontradiction. Donc $a_1$ est pair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30614,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit que deux permutations $a_{1}, \\ldots, a_{4035}$ et $b_{1}, \\ldots, b_{4035}$ des entiers $1, \\ldots, 4035$ s'intersectent s'il existe un entier $k \\leqslant 4035$ tel que $a_{k}=b_{k}$. On dit qu'un ensemble $E$ de permutations est inévitable si chaque permutation des entiers $1, \\ldots, 4035$ intersecte une permutation appartenant à E.\n\na) Montrer qu'il existe un ensemble inévitable contenant 2018 permutations.\n\nb) Existe-t-il un ensemble inévitable contenant 2017 permutations?",
    "options": [],
    "answer": "Yes: there exists an unavoidable set of size 2018 (e.g., the 2018 cyclic shifts of the first 2018 positions while fixing the rest). No: there is no unavoidable set of size 2017.",
    "solution": "Solution:\n\na) Pour tout $i \\in \\{1, \\ldots, 2018\\}$, on note $\\sigma^{(i)}$ la permutation\n$$\ni+1, i+2, \\ldots, 2018, 1, 2, \\ldots, i, 2019, 2020, \\ldots, 4035\n$$\nEn d'autres termes, $\\sigma^{(i)}$ est la permutation telle que:\n$\\triangleright$ pour tout $j \\leqslant 2018$, $\\sigma_{j}^{(i)}$ est le seul élément de $\\{1, \\ldots, 2018\\}$ tel que $\\sigma_{j}^{(i)} \\equiv i+j \\pmod{2018}$;\n$\\triangleright$ pour tout $j \\geqslant 2019$, $\\sigma_{j}^{(i)}=j$.\nAlors l'ensemble $E=\\{\\sigma^{(1)}, \\ldots, \\sigma^{(2018)}\\}$ est inévitable. En effet, soit $b_{1}, \\ldots, b_{4035}$ une permutation. Parmi les entiers $b_{1}, \\ldots, b_{2018}$, il y en a au moins un qui est inférieur ou égal à 2018, puisque l'ensemble $\\{2019, \\ldots, 4035\\}$ ne compte que 2017 éléments. Soit $i$ un entier tel que $i \\leqslant 2018$ et $b_{i} \\leqslant 2018$, et soit $j$ l'unique élément de $\\{1, \\ldots, 2018\\}$ tel que $j \\equiv b_{i}-i \\pmod{2018}$. Alors $b_{i} \\equiv i+j \\equiv \\sigma_{i}^{(j)} \\pmod{2018}$, et puisque $1 \\leqslant b_{i}, \\sigma_{i}^{(j)} \\leqslant 2018$, cela montre que $b_{i}=\\sigma_{i}^{(j)}$, donc que $b$ intersecte la permutation $\\sigma^{(j)}$.\n\nb) Soit $E$ un ensemble contenant 2017 permutations. Pour toute permutation $b_{1}, \\ldots, b_{4035}$, on appelle collision de $b$ un entier $i$ tel qu'il existe une permutation $a$ dans $E$ pour laquelle $a_{i}=b_{i}$, et on appelle poids de $b$ le nombre de collisions de $b$. On considère alors une permutation $b$ de poids minimum : on va montrer que ce poids est nul.\n\nEn effet, supposons ce poids non nul. Soit alors $i$ une collision de $b$. L'ensemble $\\bar{X}=\\{x \\in \\{1, \\ldots, 4035\\} \\mid \\exists a \\in E$ tel que $a_{x}=b_{i}\\}$ est de cardinal au plus $|E|=2017$, donc l'ensemble $X=\\{x \\in \\{1, \\ldots, 4035\\} \\mid \\forall a \\in E, a_{x} \\neq b_{i}\\}$ est de cardinal au moins $4035-2017=2018$.\n\nOn considère alors l'ensemble $Y=\\{b_{x} \\mid x \\in X\\}$, de cardinal au moins 2018 lui aussi. Toujours d'après le principe des tiroirs, il existe un entier $y \\in Y$ tel que $a_{i} \\neq y$ pour toute permutation $a$ dans $E$. Soit $x \\in X$ tel que $y=b_{x}$.\n\nOn considère alors la permutation $b'$ obtenue à partir de $b$ en échangeant les valeurs de $b_{i}$ et de $b_{x}$. Par construction, ni $i$ ni $x$ ne sont des collisions de $b'$. Les autres collisions de $b'$ sont toutes des collisions de $b$. Par conséquent, $b'$ est de poids strictement inférieur au poids de $b$. Ceci contredit la définition de $b$, ce qui invalide notre supposition et montre donc que nul ensemble contenant 2017 permutations n'est inévitable.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30615,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe d'aire $S$. On note $a = AB$, $b = BC$, $c = CD$ et $d = DA$. Pour toute permutation $x, y, z, t$ de $a, b, c, d$, montrer que\n$$\nS \\leqslant \\frac{1}{2}(xy + zt)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSi $x$ et $y$ sont adjacents, sans perte de généralité on se ramène à montrer que $S \\leqslant \\frac{1}{2}(ab + cd)$. Cela découle de $S_{ABC} = \\frac{1}{2} AB \\cdot BC \\cdot \\sin \\widehat{ABC} \\leqslant \\frac{1}{2} ab$, et de même $S_{CDA} \\leqslant \\frac{1}{2} cd$.\n\nSi $x$ et $y$ sont des côtés opposés, on doit montrer que $S \\leqslant \\frac{1}{2}(ac + bd)$. Soit $A'$ le symétrique de $A$ par rapport à la médiatrice de $[BD]$. Alors $BA'D$ est isométrique à $DAB$. On applique ce qui précède à $A'BCD$, ce qui donne $S = S_{BAD} + S_{BCD} = S_{BA'D} + S_{BCD} = S_{A'BCD} \\leqslant \\frac{1}{2}(A'B \\cdot BC + CD \\cdot DA') = \\frac{1}{2}(ac + bd)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30616,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier. Pour tout sous-ensemble non vide $A$ de $\\{1,2, \\ldots, n\\}$, on note $P(A)$ le produit de tous les éléments de $A$. Par exemple, pour $A=\\{2,4,7\\}$, on a $P(A)=56$. Déterminer la somme des $\\frac{1}{P(A)}$ lorsque $A$ parcourt tous les sous-ensembles non vides de $\\{1,2, \\ldots, n\\}$.",
    "options": [],
    "answer": "n",
    "solution": "Solution:\n\nOn a l'identité $\\left(1+x_{1}\\right)\\left(1+x_{2}\\right) \\cdots\\left(1+x_{n}\\right)=1+\\sum x_{i}+\\sum_{i<j} x_{i} x_{j}+\\cdots$. En prenant $x_{i}=\\frac{1}{a_{i}}$, on obtient\n\n$$\n\\sum_{k \\geqslant 0} \\sum_{a_{1}<\\cdots<a_{k}} \\frac{1}{a_{1} a_{2} \\cdots a_{k}}=\\left(1+\\frac{1}{1}\\right) \\cdots\\left(1+\\frac{1}{n}\\right)=\\frac{2}{1} \\times \\frac{3}{2} \\times \\frac{4}{3} \\cdots \\times \\frac{n+1}{n}=n+1\n$$\n\nOn a donc $1+\\sum_{A} \\frac{1}{P(A)}=n+1$ donc $\\sum_{A} \\frac{1}{P(A)}=n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier strictement positif. Montrer qu'il existe $n$ entiers 2 à 2 distincts $r_{1}, \\ldots, r_{n}$ tels que chaque $r_{i}$ divise $r_{1}+\\cdots+r_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLa solution s'inspire des fractions égyptiennes. Une fraction égyptienne est un $n$-uplet d'entiers distincts $a_{1}, \\ldots, a_{n}$ vérifiant $1 / a_{1} + \\cdots + 1 / a_{n} = 1$, par exemple $1 / 2 + 1 / 3 + 1 / 6 = 1$.\n\nAlors en posant $r_{i} := \\prod_{j \\neq i} a_{j} = \\frac{a_{1} \\cdots a_{n}}{a_{i}} \\in \\mathbb{Z}$, ils seront distincts et leur somme fera $a_{1} \\cdots a_{n}$, que chaque $r_{i}$ divise.\n\nOn montre par récurrence qu'une fraction égyptienne existe pour chaque $n \\geqslant 3$. L'exemple fait l'initialisation.\n\nHérédité : supposons prouvé pour $n$, prouvons pour $n+1$. On écrit les coefficients triés dans l'ordre croissant, puis on écrit $a_{n}' = 1 + a_{n}$ et $a_{n+1}' = a_{n}(1 + a_{n})$. On remarque $1 / a_{n}' + 1 / a_{n+1}' = 1 / a_{n}$. En posant $a_{i}' = a_{i}$ pour $i \\in \\{1, \\ldots, n-1\\}$, on dispose d'une fraction égyptienne de longueur $n+1$, ce qui conclut la récurrence (depuis l'exemple on a $1 / 7 + 1 / 42 = 1 / 6$ et cela donne : $1 / 2 + 1 / 3 + 1 / 7 + 1 / 42 = 1$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEst-il possible de ranger 601 disques de rayon $1$ dans une boîte rectangulaire de taille $4 \\times 600$ sans que deux disques ne se chevauchent (ils peuvent toutefois se toucher)?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nLa réponse est oui! Pour le montrer, on considère la disposition en triangle sur la figure ci-dessous.\nCalculons $d$. On note $(x_{A}, y_{A})$ les coordonnées du centre du disque $A$, et ainsi de suite. On a donc directement\n$$\nx_{A}=1, \\quad x_{B}=2, \\quad x_{C}=3, \\quad \\text{ et } \\quad y_{A}=1, \\quad y_{B}=?, \\quad y_{C}=1\n$$\nPour déterminer $y_{B}$, on écrit que $(x_{A}-x_{B})^{2}+(y_{A}-y_{B})^{2}=4$, d'où $y_{B}=1+\\sqrt{3}$. On détermine ensuite la position du centre du disque $D$. On sait que $y_{D}=3$, puis on doit avoir $(x_{D}-x_{B})^{2}+(y_{D}-y_{B})^{2}=4$, d'où $x_{D}=2+\\sqrt{4 \\sqrt{3}-3}$. On obtient que les coordonnées du centre de $E$ sont\n$$\nx_{E}=x_{D}+1=3+\\sqrt{4 \\sqrt{3}-3} \\quad \\text{ et } \\quad y_{E}=4-y_{B}=3-\\sqrt{3}\n$$\n![](attached_image_1.png)\nOn vérifie que les disques $E$ et $C$ sont tangents, car $(x_{E}-x_{C})^{2}+(y_{E}-y_{C})^{2}=4$. Donc notre figure est valide (pas de recouvrements). Au final, on trouve $d=x_{E}+1=4+\\sqrt{4 \\sqrt{3}-3}$.\nOn peut alors recommencer notre disposition à partir de l'abscisse $d$. Autrement dit, on peut mettre $6$ disques dans un intervalle $d$, et donc $6k$ disques dans un intervalle $dk$. En prenant $k=100$, on trouve qu'on peut mettre $600$ disques dans un intervalle de taille $100~d \\approx 596.4$. Il reste donc plein de place pour mettre le dernier disque dans le coin en bas à droite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier naturel impair, et soit $a_{1}, \\ldots, a_{n}$ des entiers naturels non nuls. On note $A$ le produit des entiers $a_{i}$, et $d$ leur plus grand diviseur commun.\nMontrer que\n$$\n\\operatorname{PGCD}\\left(a_{1}^{n}+A, a_{2}^{n}+A, \\ldots, a_{n}^{n}+A\\right) \\leqslant 2 d^{n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour tout $i$, on pose $b_{i}=a_{i} / d$, de sorte que $\\operatorname{PGCD}\\left(b_{1}, \\ldots, b_{n}\\right)=1$, et on pose également $B=b_{1} \\times \\ldots \\times b_{n}$ et $\\Delta=\\operatorname{PGCD}\\left(b_{1}^{n}+B, b_{2}^{n}+B, \\ldots, b_{n}^{n}+B\\right)$. Alors $\\operatorname{PGCD}\\left(a_{1}^{n}+A, a_{2}^{n}+A, \\ldots, a_{n}^{n}+A\\right)=d^{n} \\times \\Delta$, et il reste à démontrer que $\\Delta \\leqslant 2$.\n\nOn considère alors un éventuel facteur premier $p$ de $\\Delta$. Si $p$ divise un des entiers $b_{i}$, alors il divise $B$ également, donc il divise chaque entier $b_{j}$, ce qui est absurde. Par conséquent, $p$ ne divise aucun des entiers $b_{i}$, et ne divise pas $B$ non plus. Cela montre que $\\Delta$ est premier avec $B$.\n\nOr, puisque $b_{1}^{n} \\equiv b_{2}^{n} \\equiv \\ldots \\equiv b_{n}^{n}=-B\\ (\\bmod \\Delta)$, alors $B^{n} \\equiv b_{1}^{n} \\times \\ldots \\times b_{n}^{n} \\equiv(-1)^{n} B^{n} \\equiv-B^{n}$ $(\\bmod \\Delta)$. Par conséquent, $\\Delta$ divise $2 B^{n}$. On en déduit que $\\Delta$ divise $2$, ce qui conclut.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30620,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les nombres réels $a$ tels qu'il existe une suite infinie de nombres réels strictement positifs $x_{0}, x_{1}, x_{2}, x_{3}, \\ldots$ vérifiant pour tout $n$ l'égalité\n$$\nx_{n+2}=\\sqrt{a x_{n+1}-x_{n}} .\n$$",
    "options": [],
    "answer": "a > 1",
    "solution": "Solution:\n\nSi $a>1$, on peut prendre la suite constante égale à $a-1$.\n\nSupposons $a \\leqslant 1$. Comme la racine carrée est bien définie, on a nécessairement $a x_{n+1}-x_{n}>0$, donc déjà $a \\geqslant 0$, et de plus $x_{n+1} \\geqslant a x_{n+1}>x_{n}$. Par conséquent, la suite est strictement croissante.\n\nComme $x_{n+1} \\leqslant x_{n+2} \\leqslant \\sqrt{x_{n+1}-x_{n}}<\\sqrt{x_{n+1}}$, on en déduit que $x_{n}<1$ pour tout $n$. De plus, $x_{n+1}^{2} \\leqslant x_{n+1}-x_{n}$ donc $x_{n+1} \\geqslant x_{n+1}^{2}+x_{n} \\geqslant x_{n} x_{0}+x_{n}=\\left(1+x_{0}\\right) x_{n}$, ce qui montre que $x_{n} \\geqslant \\left(1+x_{0}\\right)^{n} x_{0}$ pour tout $n$. Ceci contredit que $x_{n} \\leqslant 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30621,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier positif. Montrer qu'il existe un entier positif $m$ tel que $n! = \\varphi(m)$, où $\\varphi$ est la fonction indicatrice d'Euler. (On rappelle que si $m = p_{1}^{\\alpha_{1}} \\cdots p_{k}^{\\alpha_{k}}$ avec $p_{1}, \\ldots, p_{k}$ des nombres premiers deux à deux distincts, $\\varphi(m) = m \\left(1 - \\frac{1}{p_{1}}\\right) \\cdots \\left(1 - \\frac{1}{p_{k}}\\right)$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn procède par construction en partant des plus grands nombres premiers au plus petit.\nL'idée est de contraindre la valeur de $\\varphi(m)$. La construction peut paraître un peu lourde mais il ne faut pas se laisser impressionner et comprendre l'idée qui est derrière.\n\nSoit $p_{1}, p_{2}, \\ldots$ les nombres premiers, $p_{k} \\leqslant n < p_{k+1}$.\nOn montre par récurrence descendante que pour tout $i$ il existe $m_{i}$ tel que\n- $m_{i}$ est un produit de $p_{j}$ avec $k \\geqslant j \\geqslant i$\n- Pour tout $j \\geqslant i$, $p_{j}$ a le même exposant dans $m_{i}$ et $n!$\n- $\\varphi(m_{i}) \\mid n!$\n\nL'initialisation se fait pour $i = k+1$ avec $m_{k+1} = 1$.\n\nPour l'hérédité, on multiplie $m_{i+1}$ par $p_{i}$ jusqu'à avoir autant de $p_{i}$ dans $\\varphi(m_{i})$ que dans $n!$ pour obtenir $m_{i}$, on respecte ainsi les deux premières hypothèses sur $m_{i}$.\n\nVérifions que la dernière est aussi vérifiée : pour $j \\geqslant i$ on sait que l'on n'a pas de problème, le risque est une trop grosse valuation pour un $p_{l}$ où $l$ petit. Or, vu la construction, si $p_{l}^{\\alpha} \\mid \\varphi(m_{i})$, $p_{l}^{\\alpha} \\mid (p_{i}-1)(p_{i+1}-1) \\cdots (p_{k}-1) \\mid n!$, on est donc assuré que la récurrence marche.\n\nAvec $m = m_{1}$, on obtient une solution à l'exercice.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère un tableau de taille $2018 \\times 2018$ dont chaque case contient un entier naturel non nul. Noémie modifie ces entiers à sa guise, en appliquant les opérations suivantes :\n$\\triangleright$ choisir une ligne puis multiplier par 2 tous les entiers contenus dans cette ligne;\n$\\triangleright$ choisir une colonne puis soustraire 1 à tous les entiers contenus dans cette colonne.\nMontrer que, en appliquant ces opérations, Noémie peut se débrouiller pour obtenir un tableau dont chaque case contient l'entier 0.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn va dire qu'une colonne est positive si elle ne contient que des entiers naturels non nuls, et est nulle si elle ne contient que des 0. Remarquons que, chaque fois que Noémie double les entiers contenus sur une ligne, les colonnes positives restent positives et les colonnes nulles restent nulles. On dit enfin qu'un entier $n$ est une valeur charnière si les colonnes 1 à $n$ sont nulles et les colonnes $n+1$ à 2018 sont positives. Notons qu'une grille n'a pas nécessairement de valeur charnière; ce sera néanmoins le cas de toutes les grilles que nous considérerons.\n\nInitialement, 0 est une valeur charnière. On va montrer que, partant d'une grille ayant une valeur charnière $n$, Noémie peut appliquer les opérations de l'énoncé pour obtenir une grille ayant $n+1$ pour valeur charnière. Pour ce faire, on appelle poids de la $(n+1)^{\\text{ème}}$ colonne le plus grand entier qu'elle contient. Si ce poids est strictement supérieur à 1, Noémie le fait baisser comme suit :\n$\\triangleright$ tout d'abord, elle double les entiers de chacune des lignes dont le $(n+1)$ème entier vaut 1 (et vaudra donc 2 après doublement);\n$\\triangleright$ puis elle retranche 1 à tous les entiers de la $(n+1)^{\\text{ème}}$ colonne.\n\nCe faisant, la $(n+1)^{\\text{ème}}$ colonne a bien vu son poids diminuer de 1, et $n$ est toujours la valeur charnière de la grille. Enfin, si la $(n+1)^{\\text{ème}}$ colonne a un poids de 1, alors Noémie retranche 1 à tous les entiers de cette colonne, et en fait une colonne nulle.\n\nElle obtient ainsi une grille dont $n+1$ est la valeur charnière. En réitérant ce processus, elle finit par aboutir à une grille dont 2018 est la valeur charnière, ce qui signifie que la grille ne contient que l'entier 0.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30623,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $q$ un nombre réel. Margaret a écrit 10 nombres réels, deux à deux distincts, sur une ligne. Puis elle ajoute trois lignes comme suit :\n\n$\\triangleright$ sur la $2^{\\text{nde}}$ ligne, elle écrit tous les nombres de la forme $a-b$, où $a$ et $b$ sont deux réels (non nécessairement distincts) de la $1^{\\text{ère}}$ ligne;\n\n$\\triangleright$ sur la $3^{\\text{ème}}$ ligne, elle écrit tous les nombres de la forme $q a b$, où $a$ et $b$ sont deux réels (non nécessairement distincts) de la $2^{\\text{nde}}$ ligne;\n\n$\\triangleright$ sur la $4^{\\text{ème}}$ ligne, elle écrit tous les nombres de la forme $a^{2}+b^{2}-c^{2}-d^{2}$, où $a, b, c$ et $d$ sont des réels (non nécessairement distincts) de la $2^{\\text{nde}}$ ligne.\n\nTrouver tous les réels $q$ tels que, quels que soient les 10 nombres écrits sur la $1^{\\text{ère}}$ ligne, chaque nombre de la $3^{\\text{ème}}$ ligne soit également sur la $4^{\\text{ème}}$ ligne.",
    "options": [],
    "answer": "-2, 0, 2",
    "solution": "Solution:\n\nOn va dire qu'un réel $q$ est bon s'il a la propriété demandée dans l'énoncé. Tout d'abord, il est clair que, si $q$ est bon, alors $-q$ l'est aussi, et réciproquement. D'autre part, $q=0$ est manifestement bon. On cherche donc les bons réels $q>0$, s'il y en a.\n\nSoit $\\lambda$ un grand nombre réel et soit $\\varepsilon=1/\\lambda$. On suppose que Margaret a écrit dix nombres $x_{1}, \\ldots, x_{10}$ tels que $0 \\leqslant x_{1}<\\ldots<x_{8} \\leqslant \\varepsilon, x_{9}=1$ et $x_{10}=\\lambda$. Alors la $2^{\\text{nde}}$ ligne contient les nombres $\\pm \\lambda, \\pm(\\lambda-1)$ et $\\pm 1$, ainsi que des éléments de l'ensemble $\\Omega_{2}=[0,1] \\cup [\\lambda-\\varepsilon, \\lambda]$ et leurs opposés. En particulier, Margaret écrit donc les réels $q \\lambda^{2}$ et $q \\lambda$ sur la $3^{\\text{ème}}$ ligne.\n\nOn considère alors un nombre de la $4^{\\text{ème}}$ ligne, de la forme $z=a^{2}+b^{2}-c^{2}-d^{2}$, et tel que $z \\geqslant 0$. Chacun des réels $a^{2}, b^{2}, c^{2}, d^{2}$ appartient à l'ensemble $\\{0, \\lambda^{2}\\} + \\{-2\\lambda, 0\\} + [0,2]$, donc $z$ appartient à l'ensemble $\\Omega_{4}=\\{0, \\lambda^{2}, 2\\lambda^{2}\\} + \\{-4\\lambda, -2\\lambda, 0, 2\\lambda, 4\\lambda\\} + [-4,4]$.\n\nSi $q$ est bon, alors $q \\lambda^{2}$ et $q \\lambda$ appartiennent à $\\Omega_{4}$, donc en faisant tendre $\\lambda$ vers $+\\infty$ on en déduit respectivement que $q \\in \\{1,2\\}$ et que $q \\in \\{2,4\\}$. Cela montre que seul $q=2$ est susceptible d'être bon.\n\nRéciproquement, si $q=2$, tout nombre de la $3^{\\text{ème}}$ est de la forme $2(a-b)(c-d)$ avec $a, b, c, d$ des réels de la $1^{\\text{ère}}$ ligne. Puisque $2(a-b)(c-d)=(a-d)^{2}+(b-c)^{2}-(a-c)^{2}-(b-d)^{2}$, ce nombre figure également sur la $4^{\\text{ème}}$ ligne. Par conséquent, les bons réels sont $-2, 0$ et $2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans un pays se trouvent 64 villes. Il existe des routes qui relient certaines paires de villes, mais Bob ne sait pas lesquelles (les routes ne se croisent pas mais peuvent éventuellement passer les unes au-dessus des autres ; de plus, les routes peuvent être parcourues dans les deux sens). Alice, elle, connaît parfaitement le réseau. Bob peut choisir n'importe quelle paire de villes et demander à Alice si elles sont reliées directement par une route. Le but de Bob est de déterminer s'il est possible de voyager de n'importe quelle ville à n'importe quelle autre (en empruntant éventuellement plusieurs routes successives). Montrer qu'il n'existe aucun algorithme (c'est-à-dire aucune stratégie en fonction des réponses successives d'Alice) lui permettant de le savoir à coup sûr en moins de 2016 questions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn remarque que $2016 = 64 \\times 63 / 2 = \\binom{64}{2}$, donc il y a exactement 2016 paires de villes. Le réseau routier est un graphe non orienté ayant 64 sommets. Chaque paire de villes correspond à une question, et on veut savoir si en moins de 2016 questions Bob peut déterminer si le graphe est connexe.\n\nSupposons par l'absurde que Bob possède un algorithme en moins de 2016 questions : par conséquent, pour tout graphe ayant 64 sommets, si Bob suit son algorithme en posant une suite de questions $Q_{1}, Q_{2}, \\ldots$ et s'arrête à une question $Q_{n}$ dès qu'il peut conclure à coup sûr si le graphe est connexe ou non, alors $n \\leqslant 2015$ (l'entier $n$ dépend du graphe). Remarquons que la réponse à $Q_{n}$ est nécessairement \"oui\" si le graphe est connexe et \"non\" si le graphe est non connexe, sinon Bob aurait déjà pu conclure à la question $Q_{n-1}$.\n\nOn part du graphe $G_{0}$ ne possédant aucune arête. Bob, en utilisant son algorithme, pose une suite de questions $Q_{1}, Q_{2}, \\ldots, Q_{n_{0}}$ à Alice.\n\nLa question $Q_{n_{0}}$ correspond à une paire de villes. Soit $G_{1}$ le graphe consistant à rajouter une arête entre ces deux villes. En suivant son algorithme, Bob pose à nouveau les questions $Q_{1}, \\ldots, Q_{n_{0}}$, et d'autres questions $Q_{n_{0}+1}, \\ldots, Q_{n_{1}}$. On a bien $n_{1} > n_{0}$ car, pour le graphe $G_{1}$, la réponse à $Q_{n_{0}}$ est \"oui\", et la réponse à $Q_{n_{1}}$ est \"non\" puisque le graphe $G_{1}$ n'est pas connexe.\n\nOn continue la procédure, et on construit ainsi une suite de graphes $G_{0}, G_{1}, G_{2}, \\ldots$, chaque graphe étant construit à partir du précédent en rajoutant une arête, jusqu'à obtenir un graphe connexe $G_{k}$. Comme $n_{k-1} > n_{k-2} > \\cdots > n_{0} \\geqslant 1$, on a $n_{k-1} \\geqslant k$ donc $k \\leqslant 2015$.\n\nOr, $G_{k}$ possède $k$ arêtes, donc il existe une paire de villes $X$ et $Y$ qui ne sont pas reliées directement. En revanche, comme $G_{k}$ est connexe on peut aller de $X$ vers $Y$ en suivant un chemin consistant en des arêtes de $G_{k}$. Soit $j$ le plus grand indice tel que l'une des arêtes $A$ de ce chemin appartient à $G_{j}$ mais pas à $G_{j-1}$. Soit $G'$ le graphe obtenu en enlevant l'arête $A$ au graphe $G_{k}$ et en ajoutant l'arête $XY$, alors la réponse aux questions $Q_{1}, \\ldots, Q_{n_{j-1}}$ est la même pour $G'$ que pour $G_{j-1}$. Or, Bob a pu conclure à coup sûr que $G_{j-1}$ n'est pas connexe, donc $G'$ ne l'est pas non plus.\n\nMais ceci est contradictoire car, $G_{k}$ étant connexe, $G'$ l'est aussi.",
    "topic": "Discrete Mathematics",
    "subtopic": "Algorithms"
  },
  {
    "id": 30625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 3$ un entier, montrer qu'il existe deux entiers $x$ et $y$ tels que $7 x^{2} + y^{2} = 2^{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEncore une solution par récurrence!\n\nInitialisation: $x_{3} = y_{3} = 1$.\n\nHérédité : On suppose prouvé par récurrence pour $n$, alors\n\n$X = \\dfrac{x_{n} + y_{n}}{2}$, $Y = \\dfrac{|7 x_{n} - y_{n}|}{2}$\n\nou bien\n\n$X = \\dfrac{|x_{n} - y_{n}|}{2}$, $Y = \\dfrac{7 x_{n} + y_{n}}{2}$\n\nLes deux couples vérifient $7 X^{2} + Y^{2} = 2^{n+1}$.\n\nLe premier couple est solution si $x_{n}$ et $y_{n}$ ont même congruence modulo $4$ (alors $X$ et $Y$ sont entiers), sinon le second convient, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer que, parmi $2048$ entiers, on peut toujours en trouver $1024$ dont la somme est divisible par $1024$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn va montrer par récurrence sur $n \\geqslant 0$ la propriété $P(n)$ selon laquelle, parmi $2^{n+1}$ entiers, on peut toujours en trouver $2^{n}$ dont la somme est divisible par $2^{n}$. Tout d'abord, pour $n=0$, le résultat est évident, puisque tout entier est divisible par $2^{0}=1$. La propriété $P(0)$ est donc vraie.\n\nOn considère maintenant un entier $n \\geqslant 0$ tel que $P(n)$ est vraie, et on va montrer $P(n+1)$. Considérons un ensemble de $2^{n+2}$ entiers quelconques, que l'on coupe arbitrairement en deux tas de $2^{n+1}$ entiers chacun. Grâce à la propriété $P(n)$, on extrait du premier tas un ensemble $E_{1}$ de $2^{n}$ entiers dont la somme est divisible par $2^{n}$. De même, on extrait du second tas un ensemble $E_{2}$ de $2^{n}$ entiers dont la somme est divisible par $2^{n}$. Enfin, parmi les $2^{n+1}$ entiers n'appartenant ni à $E_{1}$ ni à $E_{2}$, on extrait de nouveau un ensemble $E_{3}$ de $2^{n}$ entiers dont la somme est divisible par $2^{n}$.\n\nLes trois ensembles $E_{1}, E_{2}$ et $E_{3}$ sont disjoints, et ont des sommes congrues à $0$ ou à $2^{n}$ modulo $2^{n+1}$. Deux de ces sommes sont donc égales modulo $2^{n+1}$. L'union des ensembles associés est donc un ensemble de $2^{n+1}$ entiers dont la somme est divisible par $2^{n+1}$, ce qui montre $P(n+1)$ et conclut l'exercice.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les entiers strictement positifs $p, q$ tels que\n$$\np 2^{q} = q 2^{p}.\n$$",
    "options": [],
    "answer": "All pairs with equal entries (p, p) for any positive integer p, together with (2, 1) and (1, 2).",
    "solution": "Solution:\nPremier cas : si $p = q$, alors l'égalité est vraie.\n\nSecond cas : si $p \\neq q$, on peut supposer sans perte de généralité que $p > q$, le cas $q > p$ se traitant de même. On remarque que tout diviseur impair de $p$ est un diviseur impair de $q$, et réciproquement. Ainsi, si on écrit $p = a 2^{b}$ et $q = c 2^{d}$ avec $a, c$ impairs et $b, d \\in \\mathbb{N}$ (et il est toujours possible de faire ainsi d'après le théorème fondamental de l'arithmétique), alors $a = c$. Le rapport $p / q$ est donc une puissance de 2 (différente de 1 car $p \\neq q$). Soit $e \\in \\mathbb{N}^{*}$ tel que $p = 2^{e} q$. On a alors\n$$\n2^{e+q} = 2^{2^{e} q}\n$$\ndonc $e + q = 2^{e} q$ en identifiant les exposants, soit $e = q (2^{e} - 1)$. Or, une récurrence rapide montre que $2^{n} - 1 > n$ pour tout $n \\geqslant 2$ : c'est en effet vrai pour $n = 2$, et si $2^{n} > 1 + n$ pour un certain entier $n$, alors $2^{n+1} = 2 \\times 2^{n} > 2(1 + n) > 2n + 2 \\geqslant (n+1) + 1$. Donc si $e > 2$, $q (2^{e} - 1) > q e \\geqslant e$, contradiction. Donc $e = 1$, donc $1 = q (2^{1} - 1)$, soit $q = 1$. On en déduit que la seule solution est $p = 2$ et $q = 1$.\n\nConclusion : les solutions du problème sont les couples $(p, p)$ pour $p \\in \\mathbb{N}^{*}$ ainsi que les couples $(2, 1)$ et $(1, 2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $p \\geqslant 5$ un nombre premier. Montrer qu'il existe un entier $n$ tel que pour tout $x \\in \\{n-1, n, n+1\\}$, $p^{2} \\nmid x^{p-1}-1$ et $p \\nmid x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLemme : entre $0$ et $p^{2}$ il y a exactement $p-1$ éléments dont la puissance $(p-1)$-ième est congrue à $1$ modulo $p^{2}$.\n\nPreuve : On prend $\\omega$ une racine primitive modulo $p^{2}$ (pour rappel $\\omega$ une racine primitive modulo $m$ signifie que les puissances de $\\omega$ correspondent à tous les inversibles modulo $m$, ce qui existe dès que $m$ est $1,2,4$ ou une puissance de premier impair ou $2$ fois une puissance de premier impair, lire les polycopiés pour plus de détail), les éléments d'ordre divisant $p-1$, sont alors ceux dont le logarithme discret (log vérifie que $\\log (x)<\\varphi(m)$ et $\\omega^{(\\log (x))}=x\\ (\\bmod\\ m)$) est divisible par $p$, ce qui donne bien $p-1$ possibilités.\n\nOn utilise un principe des tiroirs : les chaussettes sont les $x$ entre $0$ et $p^{2}$ d'ordre divisant $p-1$, il y en a $p-1$ par le lemme, les $p$ tiroirs sont les intervalles $k p+1 ; (k+1) p$, il y donc un tiroir vide d'où l'on peut tirer le $n$ voulu en prenant par exemple $kp+2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $G$ un graphe orienté infini, dont tout sommet est de degré fini. On suppose que, pour chaque sommet $s$, le degré entrant de $s$ est strictement inférieur au degré sortant de $s$. Soit $v$ un sommet de $G$. Pour tout entier $n \\geqslant 1$, on note $V_{n}$ le nombre de sommets que l'on peut atteindre à partir de $v$ en passant par au plus $n$ arêtes, $v$ compris.\n\nQuelle est la plus petite valeur possible de $V_{n}$ ?",
    "options": [],
    "answer": "floor((n+2)^2 / 4)",
    "solution": "Solution:\n\nPour tout entier $n \\geqslant 0$, on note $E_{n}$ l'ensemble des sommets atteints en passant par $n$ arêtes de $G$ ou moins, et $F_{n}$ l'ensemble $E_{n} \\backslash E_{n-1}$, c'est-à-dire l'ensemble des sommets atteints en passant par $n$ arêtes de $G$ au minimum. En outre, on note $a_{n}$ le nombre d'arêtes $(x, y)$ telles que $x, y \\in E_{n}$, et $b_{n}$ le nombre d'arêtes $(x, y)$ telles que $x \\in E_{n}$ et $y \\notin E_{n}$. Remarquons qu'il s'agit exactement des arêtes $(x, y)$ telles que $x \\in F_{n}$ et $y \\in F_{n+1}$.\n\nPar construction, on sait que $a_{n}$ est inférieur ou égal à la somme des degrés entrants des sommets de $E_{n}$, tandis que $a_{n}+b_{n}$ est égal à la somme des degrés sortants des sommets de $E_{n}$. Par conséquent, on a $a_{n}+b_{n} \\geqslant a_{n}+\\left|E_{n}\\right|$, donc $b_{n} \\geqslant\\left|E_{n}\\right|$. Le principe des tiroirs indique alors que l'un des sommets de $F_{n}$ au moins est de degré sortant $b_{n} /\\left|F_{n}\\right|$ ou plus, donc que $\\left|F_{n+1}\\right| \\geqslant b_{n} /\\left|F_{n}\\right| \\geqslant\\left|E_{n}\\right| /\\left|F_{n}\\right|$. Cela montre en particulier que\n$$\n\\left|E_{n+2}\\right|=\\left|E_{n+1}\\right|+\\left|F_{n+2}\\right| \\geqslant\\left|E_{n+1}\\right|+\\left|E_{n+1}\\right| /\\left|F_{n+1}\\right|=\\left|E_{n}\\right|+\\left|F_{n+1}\\right|+\\left(\\left|E_{n}\\right|+\\left|F_{n+1}\\right|\\right) /\\left|F_{n+1}\\right| .\n$$\nL'inégalité arithmético-géométrique montre que $\\left|F_{n+1}\\right|+\\left|E_{n}\\right| /\\left|F_{n+1}\\right| \\geqslant 2 \\sqrt{\\left|E_{n}\\right|}$, d'où l'on déduit en fait que $\\left|E_{n+2}\\right| \\geqslant\\left|E_{n}\\right|+2 \\sqrt{\\left|E_{n}\\right|}+1$.\n\nOn remarque alors que $E_{0}=F_{0}=\\{v\\}$, donc que $\\left|E_{0}\\right|=\\left|F_{0}\\right|=1$, et donc que $\\left|E_{1}\\right|=\\left|E_{0}\\right|+\\left|F_{1}\\right| \\geqslant 1+\\left|E_{0}\\right| /\\left|F_{0}\\right|=2$. On montre alors par récurrence sur $n$ que $\\left|E_{2 n}\\right| \\geqslant(n+1)^{2}$ et que $\\left|E_{2 n+1}\\right| \\geqslant (n+1)(n+2)$. Tout d'abord, c'est vrai pour $n=0$. Par ailleurs, si ces deux égalités sont vraies pour un entier $n \\geqslant 0$, alors:\n$$\n\\begin{aligned}\n& \\triangleright\\left|E_{2 n+2}\\right| \\geqslant\\left|E_{2 n}\\right|+2 \\sqrt{\\left|E_{2 n}\\right|}+1 \\geqslant(n+1)^{2}+2(n+1)+1=(n+2)^{2} ; \\\\\n& \\triangleright\\left|E_{2 n+3}\\right| \\geqslant\\left|E_{2 n+1}\\right|+2 \\sqrt{\\left|E_{2 n+1}\\right|}+1>(n+1)(n+2)+2(n+1)+1=(n+2)(n+3)-1, \\\\\n& \\quad \\text{ donc }\\left|E_{2 n+3}\\right| \\geqslant(n+2)(n+3) .\n\\end{aligned}\n$$\nCeci conclut notre récurrence.\n\nRéciproquement, supposons que les sommets du graphe $G$ sont les paires $(a, b)$ telles que $0 \\leqslant 2 b \\leqslant a$, et dont les arêtes sont les paires $\\left((a, b),\\left(a+1, b^{\\prime}\\right)\\right)$. Alors chaque sommet $(a, b)$ est de degré entrant $\\lceil a / 2\\rceil$ et de degré sortant $\\lceil a / 2\\rceil+1$. Par ailleurs, si $v$ est le sommet $(0,0)$, alors $E_{n}$ est l'ensemble des sommets $(a, b)$ tels que $a \\leqslant n$, ce qui montre que $\\left|E_{2 n}\\right|=\\sum_{k=0}^{n-1} 2(k+1)+(n+1)=(n+1)^{2}$ et que $\\left|E_{2 n+1}\\right|=\\left|E_{2 n}\\right|+(n+1)=(n+1)(n+2)$.\n\nPar conséquent, on peut affirmer que la plus petite valeur possible de $V_{n}=\\left|E_{n}\\right|$ est $(n+2)^{2} / 4$ si $n$ est pair, ou bien $(n+1)(n+3) / 4$ si $n$ est impair, c'est-à-dire $\\left\\lfloor(n+2)^{2} / 4\\right\\rfloor$ dans tous les cas.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 30630,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn définit une suite $\\left(a_{n}\\right)_{n \\geqslant 0}$ ainsi : on choisit $a_{0}, a_{1} \\in \\mathbb{N}^{*}$, et pour tout entier $n \\geqslant 0$, on pose\n$$\na_{n+2}=\\left\\lfloor\\frac{2 a_{n}}{a_{n+1}}\\right\\rfloor+\\left\\lfloor\\frac{2 a_{n+1}}{a_{n}}\\right\\rfloor,\n$$\noù $\\lfloor x\\rfloor$ désigne le plus grand entier qui ne dépasse pas $x$. Prouver qu'il existe $m \\in \\mathbb{N}$ tel que $a_{m}=4$ et $a_{m+1} \\in\\{3,4\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour tous $a, b \\in \\mathbb{N}^{*}$, on note $g(a, b):=\\left\\lfloor\\frac{2 a}{b}\\right\\rfloor+\\left\\lfloor\\frac{2 b}{a}\\right\\rfloor$. Il est bien connu que $\\left(\\frac{\\sqrt{a}}{\\sqrt{b}}-\\frac{\\sqrt{b}}{\\sqrt{a}}\\right)^{2}=\\frac{a}{b}+\\frac{b}{a}-2$, or un carré est toujours positif, donc $\\frac{a}{b}+\\frac{b}{a} \\geqslant 2$, avec égalité si et seulement si $\\frac{\\sqrt{a}}{\\sqrt{b}}=\\frac{\\sqrt{b}}{\\sqrt{a}}$, soit $a=b$. Et pour tout réel $x$, on a $a\\lfloor x\\rfloor>x-1$. Donc $g(a, b)>2\\left(\\frac{a}{b}+\\frac{b}{a}\\right)-2 \\geqslant 2$. Or $g(a, b)$ est entier, donc $g(a, b) \\geqslant 3$. Ainsi, $a_{n} \\geqslant 3$ pour tout $n \\geqslant 2$ (ce qui permet au passage de justifier la bonne définition de la suite $\\left(a_{n}\\right)$ : elle ne s'annule jamais, donc les termes au dénominateur dans l'énoncé ne posent pas de problème).\n\nPour $n \\geqslant 2$, posons $u_{n}=\\max \\left(a_{n}, a_{n+1}\\right)$. On remarque que pour $a \\geqslant b \\geqslant 3$, $g(a, b) \\leqslant \\frac{2 \\max (a, b)}{3}+2$, donc si $\\max (a, b)>6$, on a $g(a, b)<a$. Donc, si $u_{n}>6$, $a_{n+2}<u_{n}$. Si de plus $a_{n+1}<u_{n}$, alors $u_{n+1}<u_{n}$. Sinon, $u_{n+1}=u_{n}$, et $a_{n+3}<u_{n+1}$ comme précédemment, donc $u_{n+2}<u_{n+1}$. Ainsi, si $u_{n}>6$, $u_{n+1}<u_{n}$ ou $u_{n+2}<u_{n}$. Or $\\left(u_{n}\\right)$ est une suite à valeurs entières, donc $\\min \\left(u_{n+1}, u_{n+2}\\right) \\leqslant u_{n}-1$. En continuant de cette façon, on arrive nécessairement après un nombre fini d'étapes sur un entier $l$ tel que $u_{l} \\leqslant 6$. On a donc $a_{l}, a_{l+1} \\in\\{3,4,5,6\\}$. On prouve aisément par récurrence sur $n \\geqslant l$ que $3 \\leqslant a_{n} \\leqslant 6$, en utilisant le fait que $g(a, b) \\leqslant \\frac{2 \\max (a, b)}{3}+2$. Ainsi, la suite $\\left(a_{n}\\right)$ prendra toujours ses valeurs dans $\\{3,4,5,6\\}$. Et, on constate facilement que si $a \\in\\{3,4,5,6\\}$, $g(a, 4)=g(4, a) \\in\\{3,4\\}$. Il suffit ainsi de montrer que $a_{m}=4$, pour un certain $m \\geqslant l$. Il reste donc à étudier le cas où $a_{l}, a_{l+1} \\in\\{3,5,6\\}$. On vérifie par une petite étude de cas que $g(a, b) \\leqslant 5$ pour tout $a, b \\in\\{3,4,5,6\\}$, donc $a_{l+2}, a_{l+3} \\in\\{3,4,5\\}$. Si $a_{l+2}=4$ ou $a_{l+3}=4$, c'est gagné. Sinon, on vérifie que pour $a, b \\in\\{3,5\\}$, $g(a, b)=4$, ce qui nous permet de conclure.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPatricia a placé 2018 points dans le plan, de sorte que les distances entre 2 points quelconques soient deux à deux distinctes. Elle colorie alors chacun de ses 2018 points, en faisant attention à ce que ; pour chaque point $P$, les points $Q$ et $R$ placés le plus près et le plus loin de $P$ sont de la même couleur que $P$.\nCombien de couleurs, au plus, Patricia a-t-elle pu utiliser?",
    "options": [],
    "answer": "504",
    "solution": "Solution:\n\nOn va montrer que Patricia a pu utiliser un maximum de 504 couleurs. Tout d'abord, dès qu'une couleur sert à colorier un sommet $v$, elle sert aussi à colorier les sommets le plus proche et le plus éloigné de $v$, c'est-à-dire 3 sommets au moins.\n\nSupposons maintenant que deux couleurs, au moins, n'ont servi à colorier que 3 sommets chacune. On appelle ces sommets respectivement $A_{1}, A_{2}, A_{3}$ et $B_{1}, B_{2}, B_{3}$. Sans perte de généralité, on suppose que $A_{1}A_{2} < A_{2}A_{3} < A_{3}A_{1}$ et que $B_{1}B_{2} < B_{2}B_{3} < B_{3}B_{1}$. Alors $A_{2}$ est le sommet le plus proche de $A_{3}$ et $B_{3}$ est le sommet le plus éloigné de $B_{2}$, donc $A_{2}A_{3} < B_{2}A_{3} < B_{2}B_{3}$. On montrerait de même que $B_{2}B_{3} < A_{2}A_{3}$, aboutissant à une contradiction. Notre supposition était donc fausse, et au plus une couleur n'a servi à colorier que 3 sommets.\n\nSi Patricia a utilisé $k$ couleurs, elle a donc colorié au moins $4k-1$ sommets. Cela montre que $4k-1 \\leqslant 2018$, donc que\n$$\nk \\leqslant \\left\\lfloor \\frac{2018+1}{4} \\right\\rfloor = 504.\n$$\n\nRéciproquement, voici une figure d'une construction possible pour 14 au lieu de 2018. Nous détaillons ci-dessous la construction pour 2018.\n\n![](attached_image_1.png)\n\nPatricia construit un 1008-gone régulier. Elle numérote ses sommets dans le sens des aiguilles d'une montre, de $A_{1}$ à $A_{1008}$, puis elle les bouge très légèrement, de manière à ce que les distances entre points $A_{i}$ soient deux à deux distinctes : de telles perturbations peuvent être choisies de manière à être arbitrairement petites. Ensuite, à proximité immédiate de chaque point $A_{i}$, et toujours en faisant en sorte que les distances entre points soient deux à deux distinctes, elle place un point $B_{i}$. Enfin, elle place à proximité de $A_{1}$ et de $A_{2}$ deux points $C_{1}$ et $C_{2}$.\n\nAlors, pour chaque sommet $X_{i}$, où $X \\in \\{A, B, C\\}$, le sommet le plus proche est l'un des autres sommets $X_{i}$, et le sommet le plus éloigné est l'un des sommets $X_{504+i}$ (si $1 \\leqslant i \\leqslant 504$) ou $X_{i-504}$ (si $505 \\leqslant i \\leqslant 1008$). Patricia utilise donc des couleurs numérotées de 1 à 504, et peint de la couleur $k$ les sommets $X_{k}$ et $X_{k+504}$. La première phrase du paragraphe montre que ce coloriage respecte bien les conditions de l'énoncé, et Patricia a donc bien pu utiliser 504 couleurs, ce qui conclut l'exercice.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $P$ un polynôme à coefficients rationnels de degré supérieur ou égal à $2$, et $\\left(q_{n}\\right)_{n \\in \\mathbb{N}}$ une suite de rationnels tels que pour tout $n \\geqslant 0$, $q_{n} = P\\left(q_{n+1}\\right)$. Montrer que la suite $q_{n}$ est périodique à partir d'un certain rang.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $u$ entier tel que $u_{1}$ soit entier. Soit $\\tilde{P}(X) = u P\\left(\\frac{1}{u} X\\right)$. Soit $v$ entier tel que $v \\tilde{P}$ soit à coefficients entiers. Soit $a$ le coefficient dominant de $v \\tilde{P}$. On pose $Q(X) = a \\tilde{P}\\left(\\frac{1}{a} X\\right)$, $m = a^{\\operatorname{deg}(P)-2} v$ et pour tout $n \\in \\mathbb{N}$, $r_{n} = u a q_{n}$.\n\nNous avons $m Q(X)$ unitaire à coefficients entiers et pour tout $n$, $r_{n} = Q\\left(r_{n+1}\\right)$. $r_{0}$ est entier, or comme pour tout $n$, $r_{n+1}$ est racine rationnelle du polynôme $m Q(X) - m r_{n}$ unitaire à coefficients entiers, par récurrence, $r_{n}$ est entier pour tout $n$.\n\nComme $\\operatorname{deg}(Q) > 1$, $|Q(x) / x|$ tend vers $+\\infty$ quand $|x|$ tend vers $+\\infty$. Donc il existe $M > |r_{0}|$ tel que pour tout $x$, $|Q(x)| \\leqslant M \\Rightarrow |x| \\leqslant M$. Par récurrence immédiate, $|r_{n}| \\leqslant M$ pour tout $n$. La suite $r$ prend donc ses valeurs dans un ensemble fini. Soit $p$ minimal tel que $\\{r_{0}, \\ldots, r_{p-1}\\} = \\{r_{0}, r_{1}, \\ldots\\}$. Montrons par récurrence sur $n$ que $r_{n+p} = r_{n}$ pour tout $n$. Par hypothèse, $r_{n+p} \\in \\{r_{n}, \\ldots, r_{n+p-1}\\} = \\{r_{0}, \\ldots, r_{p-1}\\}$. Donc $r_{n+p} = r_{n+k}$ avec $k \\in \\{0, \\ldots, p-1\\}$. Alors si $k > 0$, $r_{p-1} = Q^{\\circ(n+1)}\\left(r_{n+p}\\right) = Q^{\\circ(n+1)}\\left(r_{n+k}\\right) = r_{k-1}$ et il y a contradiction avec la minimalité de $p$. Donc $k = 0$ et $r_{n} = r_{n+p}$. Ce qui conclut la récurrence.\n\nNous venons de montrer que la suite est périodique de période $p$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver toutes les fonctions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ telles qu'il existe un réel $A$ vérifiant $A>f(x)$ pour tout $x \\in \\mathbb{R}$, et telles que pour tous réels $x, y$, on ait :\n$$\nf(x f(y)) + y f(x) = x f(y) + f(x y)\n$$",
    "options": [],
    "answer": "Two functions: (1) f(x) = 0 for all x; (2) f(x) = 2x for x < 0 and f(x) = 0 for x ≥ 0.",
    "solution": "Solution:\n\nLa fonction nulle est clairement solution. Supposons désormais que $f$ n'est pas nulle sur tout $\\mathbb{R}$.\n\nEn posant $x=0$, on a $y f(0) = f(0)$ pour tout réel $y$ donc $f(0) = 0$.\n\nEn posant $y=1$, on a $f(x f(1)) = x f(1)$ pour tout réel $x$, donc si $f(1) \\neq 0$, on contredit l'énoncé en choisissant $x = \\frac{A}{f(1)}$. Donc $f(1) = 0$.\n\nEn posant $x=1$, on a $f(f(y)) = 2 f(y)$.\n\nSupposons qu'il existe $z$ tel que $f(z) > 0$. En composant $z$ $n$ fois par $f$, pour $n \\geqslant 0$, on trouve $f(\\ldots f(z) \\ldots) = 2^{n} f(z)$. Pour $n$ assez grand, $2^{n} f(z) > A$, contradiction. Donc pour tout $x \\in \\mathbb{R}$, $f(x) \\leqslant 0$.\n\nSi $f(y) = 0$ pour un certain $y \\neq 0$, alors $y f(x) = f(x y)$ pour tout réel $x$. Si $x$ est tel que $f(x) \\neq 0$, $f(x y) \\neq 0$. Donc $f(x)$ et $f(x y)$ sont du même signe d'après le paragraphe précédent, donc $y > 0$. Il en résulte que $f$ est strictement négative sur $\\mathbb{R}_{-}^{*}$.\n\nPrenons $x \\neq 0$ et $y = \\frac{1}{x}$ dans l'équation initiale. On obtient\n$$\nf\\left(x f\\left(\\frac{1}{x}\\right)\\right) + \\frac{f(x)}{x} = x f\\left(\\frac{1}{x}\\right).\n$$\nEn échangeant $x$ et $\\frac{1}{x}$, on obtient\n$$\nf\\left(\\frac{f(x)}{x}\\right) + x f\\left(\\frac{1}{x}\\right) = \\frac{f(x)}{x}\n$$\nEn additionnant ces deux égalités, on aboutit à\n$$\nf\\left(\\frac{f(x)}{x}\\right) + f\\left(x f\\left(\\frac{1}{x}\\right)\\right) = 0\n$$\nor $f$ est à valeurs négatives, donc $f\\left(\\frac{f(x)}{x}\\right) = f\\left(x f\\left(\\frac{1}{x}\\right)\\right) = 0$ pour tout $x \\neq 0$.\n\nSi $x > 0$, $\\frac{f(x)}{x} \\neq 0$, donc $\\frac{f(x)}{x} = 0$, soit $f(x) = 0$ (on rappelle que $f$ est strictement négative sur $\\mathbb{R}_{-}^{*}$).\n\nEn prenant $y > 0$ dans l'équation initiale, on a $y f(x) = f(x y)$ pour tout $x \\in \\mathbb{R}$. En particulier, pour $x = -1$, on obtient $f(-y) = y f(-1)$ et $f$ est linéaire sur $\\mathbb{R}_{-}$. Donc pour $y < 0$, comme $f(y) < 0$, on a $f(f(y)) = -f(-1) f(y)$. Avec $x = 1$ dans l'équation initiale, on obtient $f(f(y)) = 2 f(y)$, donc $f(-1) = 2$.\n\nAinsi, $f(x) = 2x$ si $x < 0$ et $0$ sinon. On vérifie rapidement qu'elle satisfait bien l'équation de départ. On a ainsi deux fonctions solution du problème, avec la fonction nulle.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour $m$ entier positif, on note $d(m)$ le nombre de diviseurs positifs de $m$ (1 et $m$ compris). Soit $k$ un entier strictement positif. Montrer qu'il existe une infinité d'entiers positifs $n$ tels que $n$ ait exactement $k$ diviseurs premiers distincts et tel que pour tout $a, b$ entiers strictement positifs avec $n=a+b$, $d(n)$ ne divise pas $d\\left(a^{2}+b^{2}\\right)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nProuvons que chaque entier de la forme $n = m 2^{p-1}$ avec $p$ premier impair, possédant $k-1$ facteurs premiers strictement plus grands que $3$ et vérifiant $(5 / 4)^{(p-1) / 2} > m$ est solution.\n\nSi $a + b = n$ et $d(n) \\mid d\\left(a^{2} + b^{2}\\right)$ alors $p \\mid d\\left(a^{2} + b^{2}\\right)$. Donc il existe $q$ premier tel que $q$ ait un exposant $c p - 1$ dans $a^{2} + b^{2}$. Si $q \\geqslant 5$, $a^{2} + b^{2} \\geqslant 5^{q-1} > n^{2}$, or $a^{2} + b^{2} = n^{2} - 2 a b \\leqslant n^{2}$. Donc $q = 2$ ou $3$.\n\nSi $q = 3$, en regardant $a^{2} + b^{2}$ modulo $3$, $3$ divise $a$ et $b$ donc $n$, contradiction avec l'hypothèse.\n\nSi $q = 2$, si $a$ et $b$ ont des valuations $2$-adiques distinctes, la plus petite est $p-1$, celle de la somme des carrés $2p-2$, absurde.\n\nS'ils ont même valuation $2$-adique, $a = 2^{t} a_{0}$, $b = 2^{t} b_{0}$ avec $a_{0}$ et $b_{0}$ impairs, $a_{0}^{2} + b_{0}^{2} = x 2^{cp-1-2t}$, la gauche fait $1$ modulo $4$, donc $cp-1-2t = 1$ or $t < p-1$, donc $c = 1$ est la seule solution, on a un problème de parité absurde, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un ensemble $B$ d'entiers est un intervalle d'entiers s'il existe des entiers $i \\leqslant j$ tels que $B=\\{i, i+1, \\ldots, j\\}$, et on note $\\mathcal{J}$ l'ensemble des intervalles d'entiers. Par ailleurs, si $A=\\{a_1, \\ldots, a_k\\}$ est un ensemble d'entiers, avec $a_1<\\ldots<a_k$, on pose\n$$\nf(A)=\\max_{1 \\leqslant i<k}(a_{i+1}-a_i) \\text{ et } g(A)=\\max_{B \\subseteq A, B \\in \\mathcal{J}} |B|.\n$$\nOn rappelle que la notation $|B|$ désigne le nombre d'éléments de l'ensemble $B$. Si $A$ est vide ou est un singleton, on pose également $f(A)=0$. Enfin, pour tout entier $n \\geqslant 1$, on définit\n$$\nF(n)=\\sum_{A \\subseteq\\{1, \\ldots, n\\}} f(A) \\text{ et } G(n)=\\sum_{A \\subseteq\\{1, \\ldots, n\\}} g(A).\n$$\nMontrer qu'il existe un entier naturel $m$ tel que $F(n)>G(n)$ pour tout entier $n \\geqslant m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans toute cette solution, on notera $E_n$ l'ensemble des parties de $\\{1, \\ldots, n\\}$ et $S_n$ l'ensemble des parties $A$ de $\\{1, \\ldots, n\\}$ telles que $\\{1, n\\} \\subseteq A$. Si $A$ est un ensemble et $k$ un entier, on notera également $A+k$ l'ensemble $\\{x+k \\mid x \\in A\\}$.\n\nPour un ensemble $A \\in E_n$ donné, $f(A)$ est la plus grande différence entre des éléments successifs de $A$, tandis que $g(A)+1$ est la plus grande différence entre des éléments successifs de $\\{0, \\ldots, n+1\\} \\setminus A$, ou encore de $\\{1, \\ldots, n+2\\} \\setminus (A+1)$.\n\n![](attached_image_1.png)\n\nOn a donc $g(A)+1=f(\\{1, \\ldots, n+2\\} \\setminus (A+1))$. Or, quand $A$ parcourt l'ensemble $E_n$, $\\{1, \\ldots, n+2\\} \\setminus (A+1)$ parcourt $S_{n+2}$. On a donc\n\n$$\nG(n)=\\sum_{A \\in S_{n+2}} (f(A)-1)\n$$\n\nOn souhaite alors comparer $F(n)$ à la somme $\\sum_{A \\in S_{n+2}} f(A)$. Pour ce faire, on introduit la bijection $\\varphi: E_n \\to S_{n+2}$ définie comme suit : $\\varphi(\\emptyset)=\\{1, \\ldots, n+2\\}$ et, si $A$ est une partie non vide de $\\{1, \\ldots, n\\}$, de minimum $a$ et de maximum $b$, alors\n$$\n\\varphi(A)=\\{1, \\ldots, a\\} \\cup \\{x+1 \\mid x \\in A,\\ a<x<b\\} \\cup \\{b+2, \\ldots, n+2\\}.\n$$\n\n![](attached_image_2.png)\n\nNotons que, à partir de $\\varphi(A)$, on peut calculer $a$ et $b$, donc retrouver $A$ lui-même, de sorte que $\\varphi$ est injective. Puisque $E_n$ et $S_{n+2}$ sont tous deux de cardinal $2^n$, $\\varphi$ est donc bien une bijection.\n\nEn outre, remarquons que, pour tout $A \\in E_n$ tel que $|A| \\geqslant 2$, de minimum $a$ et de maximum $b$, on a $f(\\varphi(A)) \\leqslant f(A)+1$, avec égalité si et seulement si $a+f(A)$ est le deuxième plus petit élément de $A$, ou $b-f(A)$ est le deuxième plus grand élément de $A$. D'autre part, si $|A| \\leqslant 1$, alors $f(A)=0$ et $f(\\varphi(A)) \\leqslant 2$.\n\nOn dit alors qu'un ensemble $A \\in E_n$ est bon si $|A| \\geqslant 2$ et si $f(\\varphi(A)) \\leqslant f(A)$, et note $e_n$ le nombre de bons ensembles de $E_n$. Alors\n\n$$\n\\begin{aligned}\n\\sum_{A \\in S_{n+2}} f(A) & =\\sum_{A \\in E_n} f(\\varphi(A)) \\\\\n& \\leqslant \\sum_{A \\in E_n, |A| \\geqslant 2,\\ A\\ \\text{bon}} f(A) + \\sum_{A \\in E_n, |A| \\geqslant 2,\\ A\\ \\text{pas bon}} (f(A)+1) + \\sum_{A \\in E_n, |A| \\leqslant 1} 2 \\\\\n& \\leqslant \\sum_{A \\in E_n, |A| \\geqslant 2} (f(A)+1) - e_n + 2(n+1) \\\\\n& = \\sum_{A \\in E_n} (f(A)+1) - e_n + n+1\n\\end{aligned}\n$$\n\nOr, si $n \\geqslant 5$, tout ensemble $A$ de la forme $\\{1,2, n-1, n\\} \\cup (X+3)$, où $X$ est un élément de $E_{n-5}$, est nécessairement bon, puisque $f(A) \\geqslant 2$. Il existe donc au moins $2^{n-5}$ bons ensembles dans $E_n$, ce qui montre que $e_n \\geqslant 2^{n-5}$. En particulier, on a $e_{10} \\geqslant 2^5=32>10+1$, et une récurrence immédiate montre que $e_n>n+1$ pour tout $n \\geqslant 10$. Il s'ensuit que, si $n \\geqslant 10$, alors\n\n$$\n\\begin{aligned}\nG(n) & = \\sum_{A \\in S_{n+2}} f(A) - |S_{n+2}| \\\\\n& \\leqslant \\sum_{A \\in E_n} (f(A)+1) - e_n + n+1 - |S_{n+2}| \\\\\n& = F(n) + |E_n| - e_n + n+1 - |S_{n+2}| \\\\\n& = F(n) - e_n + n+1 < F(n)\n\\end{aligned}\n$$\n\nAinsi, il existe un entier naturel $m$ (par exemple $m=10$) tel que $F(n)>G(n)$ pour tout entier $n \\geqslant m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30636,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $\\Gamma_{1}$ et $\\Gamma_{2}$ deux cercles s'intersectant en $M$ et $N$. Soit $A$ et $B$ les points de contact de ces deux cercles avec leur tangente extérieure commune la plus proche de $M$. Soient $C$ et $D$ les symétriques respectifs de $A$ et $B$ par rapport à $M$, et $E$ et $F$ les points d'intersection (autres que $M$) du cercle circonscrit à $MCD$ avec $\\Gamma_{1}$ et $\\Gamma_{2}$ respectivement. Montrer que les cercles circonscrits à $MEF$ et $NEF$ ont le même rayon.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nL'énoncé ainsi formulé peut sembler curieux, mais on remarque qu'il suffit de prouver que $\\widehat{EMF}+\\widehat{ENF}=180^\\circ$ pour obtenir le résultat voulu.\n\nSoit $L$ l'intersection de $(MN)$ et $(AB)$. $L$ est sur l'axe radical de $\\Gamma_{1}$ et $\\Gamma_{2}$, donc sa puissance par rapport à ces deux cercles est la même, soit $LA^{2}=LB^{2}$, donc $L$ est le milieu de $[AB]$.\n\nPar construction de $C$ et $D$, $ABCD$ est un parallélogramme de centre $M$, donc $(AD)//(BC)//(MN)$.\n\nProuvons que $B$, $C$, $F$ sont alignés : par théorème de l'angle inscrit, $\\widehat{CFM}=\\widehat{CDM}$. Or $\\widehat{CDM}=\\widehat{DBA}=\\widehat{MBA}$ et $\\widehat{MBA}=\\widehat{BFM}$ par théorème de l'angle inscrit à la tangente. Donc $B$, $C$, $F$ sont alignés, et il en va de même pour $A$, $D$, $E$ par symétrie.\n\nAinsi, $AMNF$ et $BMNE$ sont des trapèzes dont les quatre sommets sont cocycliques, le théorème de l'angle inscrit montre qu'ils sont isocèles. Donc $\\widehat{MNE}=\\widehat{AMN}$ et $\\widehat{MNP}=\\widehat{NMB}$, ce qui implique que $\\widehat{ENF}=\\widehat{AMB}=180^\\circ-\\widehat{BMC}$. Il reste donc à établir que $\\widehat{BMC}=\\widehat{EMF}$.\n\nOn prouve même que $CMB$ et $EMF$ sont semblables (ce qui permet de conclure) : comme $C$, $D$, $E$, $F$, $M$ sont cocycliques, $\\widehat{MCB}=180^\\circ-\\widehat{MCF}=\\widehat{MEF}$ et $\\widehat{MBC}=\\widehat{MDA}=180^\\circ-\\widehat{MDE}=\\widehat{MFE}$, donc $CMB$ et $EMF$ sont bien semblables.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30637,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCarine et Cyril jouent au jeu suivant. Tout d'abord, Cyril choisit un entier $n \\geqslant 1$. Puis, dans chacune des cases d'une grille $3 \\times 3$ (apparentée à un jeu de morpion), il écrit un entier.\n\nVient ensuite le tour de Carine. Elle peut, autant de fois qu'elle le souhaite, effectuer l'opération suivante : elle choisit une case $\\mathbf{c}$, puis augmente de 1 la valeur de $\\mathbf{c}$ et de ses voisines (c'est-à-dire des cases qui partagent un côté avec $c$). Carine gagne la partie si elle réussit à faire en sorte que les 9 entiers soient tous égaux modulo $n$.\n\nQuel est le joueur qui dispose d'une stratégie gagnante?",
    "options": [],
    "answer": "Carine",
    "solution": "Solution:\n\nOn va montrer que Carine a une stratégie gagnante. Tout d'abord, adoptons quelques notations. On note $(i, j)$ la case située en ligne $i$ et en colonne $j$, et on note $k_{i, j}$ l'entier écrit sur cette case. Sans perte de généralité, on suppose que nos entiers sont écrits modulo $n$.\n\nOn va également noter $a_{i, j}$ une action de Carine consistant à choisir la case $(i, j)$, donc à augmenter de 1 les entiers $k_{\\hat{i}, \\hat{j}}$ tels que $|i-\\hat{i}|+|j-\\hat{j}| \\leqslant 1$. Au vu de la condition de victoire de Carine, on suppose aussi qu'elle dispose d'une opération $a_{\\infty}$ qui consiste à ôter 1 à toutes les cases.\n\nAlors, en effectuant les actions $a_{1,1}, a_{3,1}, a_{2,3}$ et $a_{\\infty}$, Carine se débrouille pour augmenter l'entier $k_{2,1}$ de 1 sans changer les autres. De même, elle peut augmenter isolément chacun des entiers $k_{2,3}, k_{1,2}$ et $k_{3,2}$. On appelle $b_{2,1}, b_{2,3}, b_{1,2}$ et $b_{3,2}$ les \"actions\" correspondantes.\n\nPuis, en réalisant l'action $a_{1,1}$ et en réalisant $n-1$ fois les actions $b_{2,1}$ et $b_{1,2}$, Carine a augmenté l'entier $k_{1,1}$ sans changer les autres. De même, elle peut augmenter isolément les entiers $k_{1,3}, k_{3,1}$ et $k_{3,3}$.\n\nIl lui suffit donc de modifier les valeurs de chacune des cases autres que $(2,2)$ pour que tous les entiers deviennent égaux, et alors elle aura gagné.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 30638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $S$ un ensemble d'entiers. On dit que $S$ est insommable si, pour tous les entiers $x$ et $y$ appartenant à $S$, la somme $x+y$ n'appartient pas à $S$. Pour tout entier $n \\geqslant 1$, on note $s_{n}$ le nombre de sous-ensembles de $\\{1,2, \\ldots, 2 n\\}$ qui sont insommables.\n\nDémontrer que $s_{n} \\geqslant 2^{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $S$ une partie de $\\{n+1, \\ldots, 2 n\\}$. Une telle partie est insommable, puisque, pour tous les entiers $x, y$ et $z$ appartenant à $S$, on a $x+y \\geqslant 2(n+1)>2 n \\geqslant z$. Or, il existe $2^{n}$ parties de $\\{n+1, \\ldots, 2 n\\}$. On en déduit bien que $s_{n} \\geqslant 2^{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $x$ et $y$ deux entiers tels que $5x + 6y$ et $6x + 5y$ soient des carrés parfaits. Montrer que $x$ et $y$ sont tous deux divisibles par $11$.\n\nNote : on dit qu'un entier $n$ est un carré parfait si c'est le carré d'un entier.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $a$ et $b$ deux entiers tels que $5x + 6y = a^{2}$ et $6x + 5y = b^{2}$. On note que $a^{2} + b^{2} = 11(x + y)$ est divisible par $11$. Or, modulo $11$, les carrés sont $0, 1, 3, 4, 5$ et $9$ : ainsi, la somme de deux carrés est nulle (mod $11$) si et seulement si les deux carrés en question sont nuls (mod $11$) eux aussi.\n\nDans notre cas, cela signifie que $a$ et $b$ sont divisibles par $11$. Il existe donc des entiers $A$ et $B$ tels que $a = 11A$ et $b = 11B$. Mais alors\n$$\n11x = (6 \\times 6 - 5 \\times 5)x = 6(b^{2} - 5y^{2}) - 5(a^{2} - 6y) = 6b^{2} - 5a^{2} = 11^{2}(6B^{2} - 5A^{2})\n$$\nde sorte que $x = 11(6B^{2} - 5A^{2})$ est bien divisible par $11$. De même, on montre que $y = 11(6A^{2} - 5B^{2})$ est lui aussi divisible par $11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a$ et $b$ deux réels tels que $a b \\geqslant a^{3}+b^{3}$.\nDémontrer que $a+b \\leqslant 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPuisque $(a+b)^{2} = (a-b)^{2} + 4 a b \\geqslant 4 a b$, on en déduit que\n$$\n(a+b)^{3} = a^{3} + b^{3} + 3(a+b) a b \\leqslant (3a + 3b + 1) a b \\leqslant \\frac{(3a + 3b + 1)(a+b)^{2}}{4}\n$$\nSi $a+b > 1$, on peut diviser notre inégalité par $(a+b)^{2}/4$, ce qui signifie que $4(a+b) \\leqslant 3a + 3b + 1$, donc que $a+b \\leqslant 1$ tout de même.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30641,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAu début, les 9 cases d'un échiquier $3 \\times 3$ contiennent chacune un $0$. À chaque étape, Pedro choisit deux cases partageant un côté, et ajoute soit $1$ aux deux cases, soit $-1$ aux deux cases. Montrer qu'il est impossible d'atteindre en un nombre fini de coups la situation où toutes les cases sont remplies par un $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nL'idée ici est de faire apparaître un invariant $I$.\n\nColorions l'échiquier naturellement en noir et blanc de telle sorte qu'il y a $4$ cases noires et $5$ blanches. Soit donc $I$ la somme des cases noires moins la somme des cases blanches. Au début, $I = 0$, et dans une hypothétique situation où toutes les cases sont remplies par un $2$, $I = -2$.\n\nOr $I$ ne varie pas ; en effet, à chaque étape on choisit deux cases partageant un côté commun, donc de couleur différente, et on leur ajoute le même nombre. Il en résulte que $I$ n'est pas modifié lors de l'accomplissement d'une étape, d'où le résultat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30642,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont tous les angles sont aigus, avec $AB \\neq AC$. Soit $\\Gamma$ le cercle circonscrit à $ABC$, et $D$ le milieu de l'arc $\\widehat{BC}$ ne contenant pas $A$. Soit $E$ et $F$ des points appartenant respectivement aux segments $[AB]$ et $[AC]$, de sorte que $AE = AF$. Soit $P$ le point d'intersection, autre que $A$, entre $\\Gamma$ et le cercle circonscrit à $AEF$.\nSoit $G$ et $H$ les points d'intersection respectifs, autres que $P$, entre $\\Gamma$ et les droites $(PE)$ et $(PF)$. Enfin, soit $J$ le point d'intersection entre les droites $(AB)$ et $(DG)$, et soit $K$ le point d'intersection entre les droites $(AC)$ et $(DH)$.\nDémontrer que le milieu du segment $[BC]$ appartient à la droite $(JK)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par tracer une figure, et notons $X$ le milieu de $[BC]$, que l'on espère être également le point d'intersection des droites $(BC)$ et $(JK)$. Sans perte de généralité, on suppose également que $AB > AC$.\n\n![](attached_image_1.png)\n\nUne première remarque que l'on peut formuler est que, puisque les arcs $\\widehat{BD}$ et $\\widehat{DC}$ ont même mesure, les angles $\\widehat{BAD}$ et $\\widehat{DAC}$ sont en fait égaux. Cela signifie que la droite $(AD)$ est la bissectrice de l'angle $\\widehat{BAC}$; plus précisément, le point $D$ est en fait le pôle Sud de $A$ dans le triangle $ABC$. On s'empresse donc de dessiner cette bissectrice, d'autant plus que l'égalité $AE = AF$ revient alors à dire que les points $E$ et $F$ sont en fait symétriques par rapport à la droite $(AD)$.\n\nUne seconde remarque frappante est que l'angle $\\widehat{AJD}$ semble droit. Après avoir vérifié qu'il l'était bien sur une deuxième figure, on s'empresse donc de démontrer ce premier résultat. Pour ce faire, on entame donc une chasse aux angles de droites :\n$$\n(AJ, DJ) = (AJ, PG) + (PG, DJ) = (AE, PE) + (PG, DG) = (AE, PE) + (PA, AD)\n$$\nPour aller plus loin, on voudrait remplacer l'angle $(AE, PE)$ par un angle faisant intervenir la droite $(PA)$. On utilise alors le cas limite du théorème de l'angle au centre, comme suit.\n\nSoit $\\mathcal{C}$ le cercle circonscrit à $AEF$ et $\\Delta$ sa tangente en $A$. Alors on sait que $(AE, PE) = (\\Delta, PA)$, et donc que $(AJ, DJ) = (\\Delta, AD)$. Or, la droite $(AD)$ est un axe de symétrie de $\\mathcal{C}$, et elle est donc perpendiculaire à $\\Delta$. Comme souhaité, on a donc $(AJ, DJ) = 90^\\circ$.\n\nMais alors $J$ et $K$ sont les projetés orthogonaux de $D$ sur les droites $(AB)$ et $(AC)$. Puisque $D$ appartient à la bissectrice de $\\widehat{BAC}$, on a donc $DJ = DK$ et $AJ = AK$. Par ailleurs, comme $A$ est le milieu de l'arc $\\overparen{BC}$, on sait que $BD = CD$. Les triangles $BJD$ et $CKD$ étant respectivement rectangles en $J$ et en $K$, ils sont donc isométriques.\n\nPuisque $AC = |x \\pm y|$, $AB = |x \\pm y|$ et $AB > AC$, c'est donc que $AB = x + y$ et $AC = |x - y|$. Autrement dit, on a $\\overrightarrow{AC} = (1 - y/x) \\overrightarrow{AK}$ et $\\overrightarrow{AB} = (1 + y/x) \\overrightarrow{AJ}$. Mais alors\n$$\n\\overrightarrow{AX} = (\\overrightarrow{AB} + \\overrightarrow{AC}) / 2 = (1 - y/x) \\overrightarrow{AK} / 2 + (1 + y/x) \\overrightarrow{AJ} / 2 = \\overrightarrow{AX} + \\overrightarrow{XK} + (1 + y/x) \\overrightarrow{KJ} / 2\n$$\nce qui signifie que $\\overrightarrow{KX} = (1 + y/x) \\overrightarrow{KJ} / 2$, et donc que $X$ se trouve bien sur la droite $(JK)$, comme on devait le démontrer.\n\n\n\\underline{\\text { Solution alternative } n^{\\circ} 1}\n\nOn propose une autre façon de terminer le problème une fois que l'on a obtenu que les angles $\\widehat{AJD}$ et $\\widehat{AKD}$ sont droits.\n\nComme $(AJ, DJ) = 90^\\circ = (DK, AK)$, les points $K, A, J$ et $D$ sont cocycliques. Soit alors $X'$ le point d'intersection des droites $(BC)$ et $(JK)$. Puisque\n$$\n\\left(DK, KX'\\right) = (DK, KJ) = (DA, AJ) = (DA, AB) = (DC, CB) = \\left(DC, CX'\\right)\n$$\nles points $D, C, K$ et $X'$ sont eux aussi cocycliques.\n\nMais alors $\\left(CX', X'D\\right) = (CK, KD) = 90^\\circ$, donc $X'$ est en fait le projeté orthogonal de $D$ sur la droite $(BC)$. Comme $D$ se trouve sur la médiatrice du segment $[BC]$, le point $X'$ est donc le milieu de ce segment, ce qui conclut.\n\n\n\\underline{\\text { Solution alternative } n^{\\circ} 2}\n\nOn propose une troisième façon de terminer le problème : il s'agit là d'une variante de la solution alternative $n^{\\circ} 1$.\n\nPuisque $BD = CD$, le point $X$ est également le projeté orthogonal de $D$ sur la droite $(BC)$. Mais alors, puisque $D$ appartient au cercle circonscrit à $ABC$, les points $J, K$ et $X$ sont en fait alignés sur la droite de Simson de $D$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un trapèze tel que $(AB)$ soit parallèle à $(CD)$. Soit $P$ un point de $[AC]$ et $Q$ un point de $[BD]$ tels que $\\widehat{APD} = \\widehat{BQC}$.\n\nDémontrer que $\\widehat{AQD} = \\widehat{BPC}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nL'énoncé nous donne plein d'égalités d'angles : exploitons-les ! Tout d'abord, on remarque que\n$$\n\\widehat{DPC} = 180^{\\circ} - \\widehat{APD} = 180^{\\circ} - \\widehat{CQB} = \\widehat{DQC},\n$$\nce qui signifie que les points $C$, $D$, $P$ et $Q$ sont cocycliques.\n\nOn en déduit que\n$$\n\\widehat{BQP} = 180^{\\circ} - \\widehat{PQD} = 180^{\\circ} - \\widehat{PCD} = 180^{\\circ} - \\widehat{PAB}\n$$\nce qui signifie là encore que les points $A$, $B$, $P$ et $Q$ sont cocycliques.\n\nOn en conclut que\n$$\n\\begin{aligned}\n\\widehat{AQD} & = \\widehat{AQP} + \\widehat{PQD} = \\widehat{ABP} + \\widehat{PCD} = (\\widehat{ABC} - \\widehat{PBC}) + (\\widehat{BCD} - \\widehat{BCP}) \\\\\n& = (\\widehat{ABC} + \\widehat{BCD}) - (\\widehat{PBC} + \\widehat{BCP}) = 180^{\\circ} - \\left(180^{\\circ} - \\widehat{CPB}\\right) = \\widehat{CPB}\n\\end{aligned}\n$$\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30644,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $m$, $n$, $k$ trois entiers positifs tels que $m^{2} + n = k^{2} + k$. Montrer que $m \\leqslant n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn écrit $(2k+1)^{2} = 4(k^{2} + k) + 1 = 4(m^{2} + n) + 1 = (2m)^{2} + 4n + 1$.\n\nOr si $n < m$, on peut écrire\n$$\n(2m)^{2} < (2m)^{2} + 4n + 1 < 4m^{2} + 4m + 1 = (2m + 1)^{2}\n$$\nce qui est impossible puisque $(2k + 1)^{2} = (2m)^{2} + 4n + 1$ est un carré et ne peut donc pas être entre deux carrés consécutifs. D'où $n \\geqslant m$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver les fonctions $f: \\mathbb{R} \\mapsto \\mathbb{R}$ telles que, pour tous les réels $x$ et $y$, on ait:\n$$\nf\\left(x^{2}+x+f(y)\\right)=y+f(x)+f(x)^{2}\n$$",
    "options": [],
    "answer": "f(x) = x",
    "solution": "Solution:\nTout d'abord, il est clair que la fonction $f: x \\mapsto x$ est une solution. On va montrer que c'est la seule.\n\nDans la suite, on notera $E_{x, y}$ l'équation $f\\left(x^{2}+x+f(y)\\right)=y+f(x)+f(x)^{2}$. En premier lieu, $E_{0, y}$ indique que $f(f(y))=y+f(0)+f(0)^{2}$, ce qui montre que $f$ est à la fois injective et surjective, donc bijective.\n\nD'autre part, pour tous les réels $x$ et $t$, notons que $x^{2}+x=t^{2}+t$ si et seulement si $0= x^{2}-t^{2}+x-t=(x-t)(x+t+1)$, c'est-à-dire si $x=t$ ou $t=-1-x$. En particulier, en posant $t=-1-x$, alors $E_{x, y}$ et $E_{t, y}$ indiquent que $f(x)+x^{2}=f(t)+t^{2}$. Par conséquent, si $x \\neq -1/2$, et puisque l'on a alors $t \\neq x$ et que $f$ est injective, on a également $f(t)=-1-f(x)$ et $f(x) \\neq -1/2$. On en déduit notamment que $f(-1/2)=-1/2$.\n\nOn vient de montrer que $-1/2$ est un point fixe de $f$. Or, si $y$ et $x$ sont des points fixes de $f$, l'équation $E_{x, y}$ montre que $x^{2}+x+y$ en est un également. De même, si $x$ et $x^{2}+x+y$ sont des points fixes de $f$, et puisque $f$ est bijective, alors $y$ en est un également. Ici, en choisissant $x=-1/2$, on constate donc que, dès lors que $y$ est un point fixe de $f$, $y-1/4$ et $y+1/4$ en sont aussi. En particulier, $0$ et $1/4$ sont des points fixes.\n\nDans ces conditions, $E_{x, 0}$ et $E_{0, y}$ montrent respectivement que $f\\left(x^{2}+x\\right)=f(x)+f(x)^{2}$ et que $f(f(y))=y$. Puis, si $t$ est un réel positif, soit $x$ un réel tel que $x^{2}+x=t-1/4$. Alors $E_{x, 1/4}$ indique que\n$$\nf(t)=f\\left(x^{2}+x+f(1/4)\\right)=1/4+f(x)+f(x)^{2} \\geqslant 0\n$$\nPar ailleurs, si $z$ est un réel tel que $z^{2}+z=t$, alors $E_{z, f(y)}$ indique que\n$$\nf(t+y)=f(y)+f(z)+f(z)^{2}=f(y)+f\\left(z^{2}+z\\right)=f(y)+f(t)\n$$\nOn en déduit à la fois que $f$ est croissante et que $f$ est en fait additive.\n\nLa fonction $f$ est donc linéaire, c'est-à-dire de la forme $f: x \\mapsto \\lambda x$. On conclut en observant que, puisque $-1/2$ est un point fixe de $f$, cela impose $\\lambda=1$, donc que $f$ est bien la fonction identité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30646,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $m$ et $n$ deux entiers naturels. Démontrer que $n!\\neq m^{2}+2019$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nProcédons par l'absurde, et supposons que l'on dispose de deux entiers naturels $m$ et $n$ tels que $n! = m^{2} + 2019$. Tout d'abord, on sait que $n! \\geqslant 2019 > 6! = 720$, donc que $n \\geqslant 7$. Par conséquent, $n!$ est divisible par $7! = 2^{4} \\times 3^{2} \\times 5 \\times 7$, et l'on a $0 \\equiv m^{2} + 2019$ $ (\\bmod\\ 7!) $.\n\nEn vertu de la décomposition en produit de facteurs premiers énoncée ci-dessus et du théorème Chinois, cela signifie que $-2019$ est un carré modulo $k$ pour tout entier $k$, parmi $2^{4}$, $3^{2}$, $5$ et $7$.\n\nOn pourrait démontrer que cela ne constitue pas une contradiction pour $k=5$ et $k=7$. En revanche, $-2019$ n'est pas un carré modulo $9$ (ni modulo $16$). En effet, si $m^{2} \\equiv -2019 \\equiv -3$ $ (\\bmod\\ 9) $, alors $3$ divise $m$, donc $9$ divise $m^{2}$, qui devrait donc être congru à la fois à $0$ et à $-3$ modulo $9$.\n\nAinsi, dès lors que $n \\geqslant 7$, on a en fait $n! \\equiv 0 \\not\\equiv m^{2} - 3 \\equiv m^{2} + 2019 (\\bmod\\ 9)$, ce qui montre bien que nulle paire d'entiers naturels ($m, n$) ne peut satisfaire l'équation $n! = m^{2} + 2019$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $a, b, c$ des nombres réels positifs ou nuls tels que $a+b+c=1$.\nDémontrer que\n$$\n\\frac{5+2 b+c^{2}}{1+a}+\\frac{5+2 c+a^{2}}{1+b}+\\frac{5+2 a+b^{2}}{1+c} \\geqslant 13\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSoit $S$ la somme\n$$\n\\frac{5+2 b+c^{2}}{1+a}+\\frac{5+2 c+a^{2}}{1+b}+\\frac{5+2 a+b^{2}}{1+c}\n$$\nNotons également $x_{i}$ le $i^{\\text{ème}}$ plus petit élément de l'ensemble $\\{a, b, c\\}$. L'inégalité du réordonnement indique que\n$$\n\\begin{aligned}\n& \\frac{b}{1+a}+\\frac{c}{1+b}+\\frac{a}{1+c} \\geqslant \\frac{x_{1}}{1+x_{1}}+\\frac{x_{2}}{1+x_{2}}+\\frac{x_{3}}{1+x_{3}} \\text{ et } \\\\\n& \\frac{c^{2}}{1+a}+\\frac{a^{2}}{1+b}+\\frac{b^{2}}{1+c} \\geqslant \\frac{x_{1}^{2}}{1+x_{1}}+\\frac{x_{2}^{2}}{1+x_{2}}+\\frac{x_{3}^{2}}{1+x_{3}},\n\\end{aligned}\n$$\nde sorte que\n$$\n\\begin{aligned}\nS & \\geqslant \\frac{5+2 x_{1}+x_{1}^{2}}{1+x_{1}}+\\frac{5+2 x_{2}+x_{2}^{2}}{1+x_{2}}+\\frac{5+2 x_{3}+x_{3}^{2}}{1+x_{3}} \\\\\n& \\geqslant 4\\left(\\frac{1}{1+x_{1}}+\\frac{1}{1+x_{2}}+\\frac{1}{1+x_{3}}\\right)+\\left(1+x_{1}\\right)+\\left(1+x_{2}\\right)+\\left(1+x_{3}\\right) \\\\\n& \\geqslant 4+4\\left(\\frac{1}{1+x_{1}}+\\frac{1}{1+x_{2}}+\\frac{1}{1+x_{3}}\\right)\n\\end{aligned}\n$$\nLa fonction $\\mathrm{f}: \\mathrm{x} \\mapsto 1 /(1+x)$ étant convexe, on sait en outre que\n$$\n\\frac{1}{1+x_{1}}+\\frac{1}{1+x_{2}}+\\frac{1}{1+x_{3}}=f\\left(x_{1}\\right)+\\mathbf{f}\\left(x_{2}\\right)+\\mathbf{f}\\left(x_{3}\\right) \\geqslant 3 \\mathbf{f}\\left(\\frac{x_{1}+x_{2}+x_{3}}{3}\\right)=3 \\mathbf{f}(1 / 3)=9 / 4\n$$\nOn en déduit que $S \\geqslant 4+9=13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30648,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\Gamma$ un cercle de centre $O$ et de rayon $r$ et $\\ell$ une droite qui ne coupe pas $\\Gamma$. On note $E$ le point d'intersection entre $\\ell$ et la droite perpendiculaire à $\\ell$ passant par $O$.\nSoit $M$ un point de $\\ell$ différent de $E$. Les tangentes au cercle $\\Gamma$ et passant par $M$ touchent $\\Gamma$ en $A$ et $B$. Enfin, soit $H$ le point d'intersection des droites $AB$ et $OE$.\nMontrer que $OH = \\dfrac{r^2}{OE}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, les triangles $OAM$, $OBM$ et $OEM$ sont respectivement rectangles en $A$, $B$ et $E$, de sorte que les points $A$, $O$, $B$, $E$, $M$ appartiennent tous à un même cercle de diamètre $[OM]$.\n\nOr, d'après la loi des sinus dans les triangles $OBH$ et $OEA$, on sait que\n$$\n\\frac{OH \\times OE}{r^{2}} = \\frac{OH}{OB} \\times \\frac{OE}{OA} = \\frac{\\sin (\\widehat{OBA})}{\\sin (\\widehat{OHB})} \\times \\frac{\\sin (\\widehat{OAE})}{\\sin (\\widehat{OEA})}\n$$\n\nLes points $O$, $B$, $E$ et $A$ étant cocycliques, on sait que $\\widehat{OBA} = \\widehat{OEA}$. D'autre part, puisque les droites $(AH)$ et $(HO)$ sont respectivement perpendiculaires à $(OM)$ et $(ME)$, on sait également que\n$$\n\\widehat{OHB} = 180^{\\circ} - \\widehat{AHO} = 180^{\\circ} - \\widehat{OME} = 180^{\\circ} - \\widehat{OAE}\n$$\n\nCela montre que $\\sin (\\widehat{OBA}) = \\sin (\\widehat{OEA})$ et que $\\sin (\\widehat{OHB}) = \\sin (\\widehat{OAE})$, ce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30649,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nChaque case d'un tableau de taille $8 \\times 8$ peut être coloriée en blanc ou en noir, de telle sorte que chaque rectangle de taille $2 \\times 3$ ou $3 \\times 2$ contient deux cases noires ayant un côté en commun. Déterminer le nombre minimal de cases noires qu'il peut y avoir sur le tableau.",
    "options": [],
    "answer": "24",
    "solution": "Solution:\n\nComme d'habitude dans ce genre de problèmes, on commence par rechercher des bornes supérieures (en exhibant une configuration) et inférieures (par des arguments de double comptage et/ou de pavage) sur le nombre minimal de cases noires que doit contenir le tableau.\n\nPar exemple, la configuration ci-dessous (à gauche) nous assure que 28 cases noires sont suffisantes; en effet, tout rectangle vertical (de taille $3 \\times 2$) contient deux cases sur l'une des lignes 3 ou 6, et tout rectangle horizontal (de taille $2 \\times 3$) contient deux cases sur l'une des colonnes 3 ou 6. D'autre part, chacun des 10 rectangles du pavage partiel ci-dessous (à droite) doit contenir au moins 2 cases noires, de sorte que 20 cases noires sont nécessaires.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nEn tâtonnant, on finit par trouver une nouvelle configuration satisfaisant les conditions de l'énoncé, mais qui ne contient que 24 cases noires. Il s'agit de la configuration ci-dessous.\n\n![](attached_image_3.png)\n\nD'autre part, utiliser un simple pavage par des rectangles semble un peu trop brutal pour nous indiquer la meilleure borne inférieure possible sur le nombre de cases noires nécessaires. On étudie un rectangle quelconque donné :\n\n![](attached_image_4.png)\n\nOn constate ici que, parmi les 3 cases hachurées, au moins une doit être une case noire, ce sans quoi notre rectangle ne contiendra jamais deux cases noires adjacentes. En particulier, la case en ligne 1, colonne 2 a nécessairement une voisine noire au sein de ce rectangle. De même, la case en ligne 2, colonne 2, a aussi une voisine noire au sein de notre rectangle.\n\nForts de ce constat, on s'intéresse, pour chaque case $c$ de l'échiquier, au nombre minimal de voisines noires que possède la case $c$ :\n\n$\\triangleright$ si $c$ est un des quatre coins, a priori, elle pourrait n'avoir aucune voisine noire (c'est d'ailleurs le cas dans notre configuration précédente);\n\n$\\triangleright$ si $c$ est une des $2 \\times 6=12$ cases situées sur le côté haut ou bas, elle appartient à la colonne centrale d'un rectangle horizontal, donc elle a au moins une voisine noire;\n\n$\\triangleright$ de même, si $c$ est une des $2 \\times 6=12$ cases situées sur le côté gauche ou droit, elle appartient à la ligne centrale d'un rectangle vertical, donc elle a au moins une voisine noire;\n\n$\\triangleright$ enfin, si $c$ est une des $6 \\times 6=36$ cases situées au centre de l'échiquier, elle a déjà une voisine noire au minimum (disons $d$); mais alors $c$ appartient également à la colonne (respectivement, à la ligne) d'un autre rectangle horizontal (resp., vertical) ne contenant pas $d$, donc $c$ a une deuxième voisine noire.\n\nEn conclusion, on dispose en fait d'au moins $12+12+36 \\times 2=96$ paires $(c, d)$ de cases adjacentes telles que $d$ est une case noire. Or, chaque case noire a au plus quatre voisines, donc appartient à au plus quatre telles paires. On a donc au moins $96 / 4=24$ cases noires sur notre échiquier, ce qui rejoint la borne supérieure trouvée précédemment, et conclut donc.\nSolution:\n\nVoici une autre manière de démontrer que 24 cases noires sont nécessaires une fois que l'on sait que chaque case située au centre de l'échiquier a au moins deux voisines noires. En effet, cela signifie que chaque occurrence du motif représenté ci-dessous doit contenir au moins deux cases noires.\n\n![](attached_image_5.png)\n\nOn peut alors placer de tels motifs à trois ou à quatre cases, sans recouvrement, dans notre échiquier $8 \\times 8$, par exemple en procédant comme suit : dans le dessin ci-dessous, on relie les centres des cases appartenant à un même motif.\n\n![](attached_image_6.png)\n\nOn a donc réussi à placer dans l'échiquier 12 exemplaires de notre motif à quatre cases, ou bien 8 exemplaires du motif à quatre cases et 8 exemplaires du motif à trois cases, ne se chevauchant pas. Dans chaque cas, ces motifs contiennent bien un total de 24 cases noires au minimum, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn coloriage de $\\mathbb{Q}$ consiste à colorier tout nombre rationnel soit en rouge, soit en bleu. On dit qu'un coloriage de $\\mathbb{Q}$ est harmonieux si, pour tous les rationnels $x$ et $y$ d'une même couleur, le rationnel $x+y$ est encore de la même couleur.\n\nTrouver tous les coloriages de $\\mathbb{Q}$ qui sont harmonieux.",
    "options": [],
    "answer": "Exactly these harmonious colorings: (1) all rationals have the same color; (2) nonnegative rationals have one color and negative rationals have the other; (3) strictly positive rationals have one color and nonpositive rationals have the other.",
    "solution": "Solution:\n\nOn va montrer que les coloriages recherchés sont les suivants :\n\n$\\triangleright$ on colorie tous les rationnels avec la même couleur;\n\n$\\triangleright$ on colorie tous les rationnels positifs ou nuls avec une même couleur, et tous les rationnels strictement négatifs avec l'autre couleur;\n\n$\\triangleright$ on colorie tous les rationnels strictement positifs avec une même couleur, et tous les rationnels négatifs ou nuls avec l'autre couleur.\n\nTout d'abord, il est clair que chacun de ces coloriages est harmonieux. Montrons que ce sont les seuls.\n\nDans la suite, on supposera que l'on dispose d'un coloriage harmonieux fixé. On note $B$ l'ensemble des rationnels coloriés en bleu et $R$ l'ensemble des rationnels coloriés en rouge.\n\nAvec ces notations, si $x$ est un rationnel quelconque appartenant à un ensemble $E \\in\\{B, R\\}$, on va d'abord montrer que pour tout nombre rationnel $p / q$ strictement positif, $p x / q$ appartient aussi à $E$. En effet, si $x / q \\notin E$, alors une récurrence immédiate montre que $x=q x / q \\notin E$, ce qui n'est pas possible. On en déduit que $x / q \\in E$, puis que $p x / q \\in E$, toujours par récurrence immédiate.\n\nAinsi, tous les rationnels strictement positifs sont de la même couleur que $1$, et tous les rationnels strictement négatifs sont de la même couleur que $-1$. Vu notre objectif, il reste à montrer que, si $1$ et $-1$ sont de la même couleur, alors $0$ est de cette couleur là également : ceci découle précisément du fait que $0=1+(-1)$ et du caractère harmonieux de notre coloriage.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'une paire d'entiers $(a, b)$ est chypriote si $a \\geqslant b \\geqslant 2$, si $a$ et $b$ sont premiers entre eux, et si $a+b$ divise $a^{b}+b^{a}$.\n\nDémontrer qu'il existe une infinité de paires chypriotes distinctes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPosons $k = a - b$. Puisque $a$ et $b$ sont premiers entre eux, on sait que $a \\neq b$, donc que $k \\geqslant 1$. Ainsi,\n$$\na^{b} + b^{a} \\equiv a^{b} + (-a)^{a} \\equiv a^{b}\\left(1 + (-1)^{a} a^{k}\\right) \\pmod{a+b}\n$$\nOr, $a$ est premier avec $b$ donc avec $a+b$ aussi.\nD'après le théorème de Gauss, on souhaite donc que $a+b = 2a - k$ divise $1 + (-1)^{a} a^{k}$, ou encore $a^{k} + (-1)^{a}$.\n\nFaute de mieux, étudions maintenant les petites valeurs de $k$. Choisir $k=1$ reviendrait à souhaiter que $a+b$ divise $a \\pm 1$, ce qui est impossible puisque $a+b > a+1 > a-1$. On regarde donc $k=2$.\n\nDans ce cas, on souhaite que $2a-2 = 2(a-1)$ divise $a^{2} + (-1)^{a}$. Or, notons que $a^{2} + (-1)^{a} \\equiv 1 + (-1)^{a} \\pmod{a-1}$. Par conséquent, et puisque $a \\geqslant 2$, il nous faut nécessairement choisir $a$ impair. Dans ce cas, il reste à faire en sorte que $2(a-1)$ divise $a^{2} - 1 = (a-1)(a+1)$, c'est-à-dire que $2$ divise $a-1$. Mais c'est justement le cas, puisque $a$ est impair.\n\nEn conclusion, on a bien montré que toutes les paires $(2k+1, 2k-1)$, où $k \\geqslant 2$, étaient chypriotes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $P_{1}, P_{2}, \\ldots, P_{2019}$ des polynômes non constants à coefficients réels tels que\n$$\nP_{1}\\left(P_{2}(x)\\right)=P_{2}\\left(P_{3}(x)\\right)=\\ldots=P_{2019}\\left(P_{1}(x)\\right)\n$$\npour tout réel $x$. Démontrer que $P_{1}=P_{2}=\\ldots=P_{2019}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous allons en fait démontrer le résultat plus général suivant : si $P_{1}, \\ldots, P_{2 n+1}$ sont des polynômes non constants à coefficients réels et tels que $P_{1} \\circ P_{2}= P_{2} \\circ P_{3}=\\ldots=P_{2 n+1} \\circ P_{1}$, alors $P_{1}=\\ldots=P_{2 n+1}$. Par commodité, dans la suite, on pose $P_{2 n+1+k}=P_{k}$ pour tout $k \\geqslant 1$.\n\nTout d'abord, on note $d_{k}$ le degré de $P_{k}$. Puisque $P_{k} \\circ P_{k+1}$ est un polynôme de degré $d_{k} d_{k+1}$, c'est-à-dire que $d_{k} d_{k+1}=d_{k+1} d_{k+2}$, et donc $d_{k}=d_{k+2}$, pour tout $k$. On en déduit que tous les polynômes $P_{k}$ ont degré $d_{1}$, et on pose simplement $d=d_{1}$.\n\nEnsuite, soit $p_{k}$ le coefficient dominant de $P_{k}$, c'est-à-dire son coefficient de degré $d$, et posons $q_{k}=p_{k} / p_{k+1}$. Le coefficient dominant de $P_{k} \\circ P_{k+1}$ est égal à $p_{k} p_{k+1}^{d}$. On en déduit que $p_{k} p_{k+1}^{d}= p_{k+1} p_{k+2}^{d}$, c'est-à-dire que $q_{k}=q_{k+1}^{-d}$. Mais alors\n$$\n1=\\prod_{i=k+1}^{k+2 n+1} q_{i}=q_{k}^{s}\n$$\nou l'on a posé\n$$\ns=\\sum_{i=0}^{2 n}(-d)^{i} \\equiv 1 \\quad(\\bmod 2) .\n$$\nPuisque $s$ est un entier impair, c'est donc que $q_{k}=1$. Cela signifie donc que tous les polynômes $P_{k}$ sont de coefficient dominant $p_{1}$, et on pose simplement $p=p_{1}$.\n\nSupposons maintenant que $d=1$. Dans ce cas, on peut écrire chaque polynôme $P_{k}$ sous la forme $P_{k}(X)=p X+b_{k}$, et alors $P_{k}\\left(P_{k+1}(X)\\right)=p^{2} X+p b_{k+1}+b_{k}$. On pose alors $c_{k}=b_{k+1}-b_{k}$, et puisque l'énoncé indique entre autres que $p b_{k+1}+b_{k}=p b_{k+2}+b_{k+1}$, cela signifie en fait que $c_{k}=-p c_{k+1}$. Là encore, on en déduit que $c_{k}=(-p)^{2 n+1} c_{k}$, de sorte que $c_{k}=0$ pour tout $k$, ou bien que $p=-1$; mais, dans ce second cas, on a tout de même $c_{k}=c_{k+1}$ pour tout $k$, de sorte que $0=\\sum_{i=1}^{2 n+1} c_{i}=(2 n+1) c_{k}$ et que l'on a $c_{k}=0$ de toute façon. Ainsi, quand $d=1$, les polynômes $P_{k}$ sont bien égaux deux à deux.\n\nSupposons enfin que $d \\geqslant 2$. On note $p_{k, \\ell}$ le coefficient de $P_{k}$ de degré $\\ell$, et on va prouver par récurrence sur $d-\\ell$ que $p_{k, \\ell}=p_{1, \\ell}$. Tout d'abord, pour $\\ell=d$, on a bien $p_{k, d}=p=p_{1, d}$. On suppose donc maintenant que $p_{k, m}=p_{1, m}$ pour tout entier $m \\geqslant \\ell+1$, et on étudie le coefficient de degré $d^{2}-d+\\ell$ du polynôme $P_{k} \\circ P_{k+1}$. En écrivant $A \\approx B$ dès lors que deux polynômes $A$ et $B$ ont même coefficient de degré $d^{2}-d+\\ell$, on constate que\n$$\n\\begin{aligned}\nP_{k}\\left(P_{k+1}(X)\\right) & =\\sum_{i=0}^{d} p_{k, i}\\left(\\sum_{j=0}^{d} p_{k+1, j} X^{j}\\right)^{i} \\\\\n& \\approx p\\left(\\sum_{j=0}^{d} p_{k+1, j} X^{j}\\right)^{d}+p_{k, d-1}\\left(\\sum_{j=0}^{d} p_{k+1, j} X^{j}\\right)^{d-1} \\\\\n& \\approx p\\left(p_{k+1, \\ell} X^{\\ell}+\\sum_{j=\\ell+1}^{d} p_{1, j} X^{j}\\right)^{d}+p_{k, d-1}\\left(p X^{d}\\right)^{d-1} \\\\\n& \\approx d p_{k+1, \\ell} p^{d} X^{d^{2}-d+\\ell}+p\\left(\\sum_{j=\\ell+1}^{d} p_{1, j} X^{j}\\right)^{d}+p_{k, d-1} p^{d-1} X^{d^{2}-d}\n\\end{aligned}\n$$\nPour $\\ell \\geqslant 1$, on en déduit que $d p_{k+1, \\ell} p^{d}=d p_{1, \\ell} p^{d}$, ce qui signifie bien que $p_{k+1, \\ell}=p_{1, \\ell}$. Puis, pour $\\ell=0$, on en déduit aussi que $d p_{k+1,0} p^{d}+p_{k, d-1}=d p_{1,0} p^{d}+p_{2 n+1, d-1}$. Or, puisque $d \\geqslant 2$, quand on en arrive à étudier le cas $\\ell=0$, on sait déjà que $p_{k, d-1}=p_{2 n+1, d-1}$. On en déduit donc comme prévu que $p_{k+1,0}=p_{1,0}$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, et soit $M$ le pied de la médiane issue de $A$. Soit également $\\ell_{\\mathrm{b}}$ la bissectrice de $\\widehat{AMB}$ et $\\ell_{c}$ la bissectrice de $\\widehat{AMC}$. Enfin, soit $B'$ le projeté orthogonal de $B$ sur $\\ell_{b}$, soit $C'$ le projeté orthogonal de $C$ sur $\\ell_{c}$, et soit $A'$ le point d'intersection des droites $(AM)$ et $(B'C')$.\n\nDémontrer que $A'B' = A'C'$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $\\beta$ l'angle $\\widehat{AMB'}$ et $\\gamma$ l'angle $\\widehat{AMC'}$. Par construction, on sait que $2(\\beta+\\gamma)=180^\\circ$, donc que $\\widehat{B'MC'}=\\beta+\\gamma=90^\\circ$. Ainsi, la droite $\\ell_{\\mathrm{b}} = (MB')$ est perpendiculaire aux droites $(BB')$ et $\\ell_{c} = (MC')$, et la droite $\\ell_{c} = (MC')$ est perpendiculaire aux droites $(CC')$ et $\\ell_{b} = (MB')$.\n\nD'autre part, on remarque que $\\widehat{B'BM} = 180^\\circ - \\widehat{BB'M} - \\widehat{BMB'} = 180^\\circ - 90^\\circ - \\beta = \\gamma$ et que, de même, $\\widehat{C'CM} = \\beta$. Ainsi, les deux triangles $BB'M$ et $MC'C$ sont semblables. En outre, puisque $M$ est le pied de la médiane de $ABC$ issue de $A$, $M$ est en fait le milieu de $[BC]$, donc $BM = MC$. Les deux triangles $BB'M$ et $MCC'$ sont donc isométriques.\n\nEn particulier, on en déduit que $BB' = MC'$ et que $CC' = MB'$. Mais alors, puisque le triangle $MB'C'$ est rectangle en $M$, il a un angle et deux côtés de mêmes mesures que le triangle $B'MB$, et il est donc isométrique avec $B'MB$ et avec $C'CM$.\n\nOn en déduit notamment que\n$$\n\\widehat{A'C'M} = \\widehat{B'C'M} = \\widehat{MBB'} = \\gamma = \\widehat{AMC'} = \\widehat{A'MC'}\n$$\ndonc que le triangle $A'C'M$ est isocèle en $A'$. De même, le triangle $A'B'M$ est isocèle en $A'$, et on en conclut comme prévu que $A'B' = A'M = A'C'$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $p$ un nombre premier.\n\nDémontrer qu'il existe un nombre premier $q$ tel que $n^{p} \\not\\equiv p$ pour tout $n \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, si $p$ ne divise pas $q-1$, alors $x \\mapsto x^{p}$ est une bijection de $\\mathbb{Z} / q \\mathbb{Z}$ dans lui-même, donc $q$ ne peut pas convenir. On en vient à chercher $q \\equiv 1\\ (\\bmod\\ p)$ tel que, pour tout $n \\not\\equiv 0\\ (\\bmod\\ q)$, $n$ soit d'ordre $\\omega_{q}(n) \\neq p\\, \\omega_{q}(p)$ modulo $q$, où $\\omega_{q}(p)$ est l'ordre de $p$ modulo $q$.\n\nPuisque les ordres possibles sont exactement les diviseurs de $q-1$, cela signifie que $q-1$ doit être divisible par $p$ et par $\\omega_{q}(p)$ mais pas $p\\, \\omega_{q}(p)$. Par conséquent, $p$ doit nécessairement diviser $\\omega_{q}(p)$, et une première idée serait de vérifier si on ne peut pas justement avoir $\\omega_{q}(p)=p$.\n\nDans cette optique, $q$ doit diviser $p^{p}-1$ mais pas $p-1$. Ainsi, $q$ doit diviser l'entier\n$$\n\\mathbf{N}=\\frac{p^{p}-1}{p-1}=1+p+p^{2}+\\ldots+p^{p-1}\n$$\nDans ces conditions, si $q$ divise quand même $p-1$, alors $\\mathbf{N} \\equiv p\\ (\\bmod\\ q)$, ce qui est impossible. Ainsi, on est ici assuré que $q$ ne divise pas $p-1$, donc que $\\omega_{q}(p)=p$.\n\nIl reste donc à s'assurer que l'on peut choisir $q$ de sorte que $q \\not\\equiv 1\\left(\\bmod\\ p^{2}\\right)$. Si un tel $q$ n'existait pas, alors $N$ lui même serait congru à $1\\left(\\bmod\\ p^{2}\\right)$. On conclut donc le problème en remarquant que $\\mathbf{N} \\equiv 1+p \\not\\equiv 1\\left(\\bmod\\ p^{2}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier naturel. Un escalier de taille $n$ est constitué de petits carrés $1 \\times 1$, avec 1 carré pour la première marche, 2 carrés pour la deuxième marche, et ainsi de suite, jusqu'à $n$ carrés pour la $n^{\\text{ème}}$ marche.\nOn dispose de pierres carrées (de côté entier) de toutes les tailles pour construire cet escalier et on note $f(n)$ le nombre minimum de pierres que l'on doit utiliser pour un escalier de taille $n$. Par exemple, $f(2)=3$ et $f(4)=7$, comme illustré ci-dessous.\n\n![](attached_image_1.png)\n\n1. Trouver tous les entiers $n \\geqslant 0$ tels que $f(n)=n$.\n2. Trouver tous les entiers $n \\geqslant 0$ tels que $f(n)=n+1$",
    "options": [],
    "answer": "1) f(n) = n holds exactly for n = 2^k − 1 with k ≥ 0.\n2) f(n) = n + 1 holds exactly for n = 2^{k+1} − 2^{k'} − 1 with k ≥ k' ≥ 0.",
    "solution": "Solution:\n\nCommençons par quelques définitions et observations générales. Dans la suite, on note $(i, j)$ le petit carré $1 \\times 1$ situé au $j$ème étage de la $i^{\\text{ème}}$ marche. On appellera carré supérieur chaque petit carré $(k, k)$, c'est-à-dire chaque carré situé tout en haut d'une marche.\n\nOn considère une construction de l'escalier de taille $n$ à partir de $f(n)$ pierres carrées. Deux carrés supérieurs distincts ne peuvent appartenir à la même pierre. Puisqu'il y a exactement $n$ carrés supérieurs dans un escalier de taille $n$, on en déduit que $f(n) \\geqslant n$ pour tout $n \\geqslant 0$. On traite maintenant les questions 1 et 2.\n\n1. Tout d'abord, il est clair que $f(0)=0$. Soit maintenant $n \\geqslant 1$ un entier tel que $f(n)=n$.\n\nAu vu de l'observation ci-dessus, chaque pierre contient un unique carré supérieur. C'est en particulier le cas de la pierre contenant le carré $(n, 1)$. Mais alors cette pierre sépare l'escalier en 2 parties symétriques. Ainsi, chaque partie forme un escalier de taille $(n-1)/2$, tel que $f((n-1)/2)=(n-1)/2$. Par conséquent, une récurrence immédiate montre qu'il existe un entier $k \\geqslant 0$ tel que $n=2^{k}-1$.\n\nRéciproquement, et en suivant cette construction dans l'autre sens, une récurrence immédiate montre que $f\\left(2^{k}-1\\right)=2^{k}-1$ pour tout entier $k \\geqslant 0$.\n\nLes entiers $n$ recherchés sont donc bien les entiers de la forme $n=2^{k}-1$ avec $k \\geqslant 0$.\n\n2. Cette fois-ci, on sait que $n \\geqslant 2$. D'autre part, la pierre contenant le carré $(n, 1)$ ne saurait contenir de carré supérieur; en effet, si c'était le cas, elle couperait l'escalier en 2 parties symétriques, formant chacune un escalier de taille $(n-1)/2$, de sorte que $f(n)-n$ devrait être pair.\n\nIl s'agit donc de la seule pierre qui ne contient pas de carré supérieur. Soit $\\ell \\times \\ell$ les dimensions de cette pierre. Une fois $\\ell$ et $n$ fixés, chaque autre pierre doit contenir un carré supérieur, et les dimensions des pierres sont donc prescrites. En particulier, une récurrence immédiate sur $n+i-j$ montre que les pierres contenant les carrés $(i, j)$ et $(n+1-j, n+1-i)$ occupent en fait des positions symétriques.\n\nAfin de se ramener à ne traiter que des escaliers où chaque pierre contient un carré supérieur, on s'intéresse donc spécifiquement aux pierres contenant les carrés $(n, 1)$, $(n-\\ell, 1)$, $(n, \\ell+1)$ et $(n-\\ell, \\ell+1)$; cette dernière pierre n'existe que si $n \\neq 2\\ell$.\n\nOn suppose tout d'abord que $n \\neq 2\\ell$. Dans ce cas, nos quatre pierres sont deux à deux disjointes, de tailles respectives $\\ell \\times \\ell$, $(n+1-\\ell)/2$, $(n+1-\\ell)/2$ et $(n+1-2\\ell)/2$, et elles coupent l'escalier en quatre petits escaliers : deux escaliers de taille $(n-1-\\ell)/2$ et deux escaliers de taille $(n-1-2\\ell)/2$. Au vu des résultats de la première question, il existe donc deux entiers naturels non nuls $k$ et $k'$ tels que $n=2^{k}+\\ell-1=2^{k'}+2\\ell-1$.\n\nCela signifie que $\\ell=2^{k}-2^{k'}$, donc que $k \\geqslant k'$ et que $n=2^{k+1}-2^{k'}-1$. Réciproquement, s'il existe des entiers $k \\geqslant k' \\geqslant 1$ tels que $n=2^{k+1}-2^{k'}-1$, il suffit en effet de choisir $\\ell=2^{k}-2^{k'}$ pour que notre construction fonctionne.\n\nDe même, si $n=2\\ell$, on a en fait trois pierres, qui coupent l'escalier en quatre petits escaliers de taille $(n-1-\\ell)/2=(\\ell-1)/2$, donc il existe un entier naturel non nul $k$ tel que $\\ell=2^{k}-1$ et $n=2^{k+1}-2$. Réciproquement, s'il existe un entier $k \\geqslant 1$ tel que $n=2^{k+1}-2$, il suffit en effet de choisir $\\ell=n/2$ pour que notre construction fonctionne. Les entiers $n$ recherchés sont donc bien les entiers de la forme $n=2^{k+1}-2^{k'}-1$ avec $k \\geqslant k' \\geqslant 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont les trois angles sont aigus, avec $AB > AC$, et soit $\\Omega$ son cercle circonscrit. On note $M$ le milieu de $[BC]$. Les tangentes à $\\Omega$ en $B$ et $C$ s'intersectent en $P$, et les droites $(AP)$ et $(BC)$ se coupent en $S$. On note $D$ le pied de la hauteur issue de $B$ dans $ABP$, et $\\omega$ le cercle circonscrit à $CSD$. Enfin, on note $K$ le second point d'intersection (après $C$) de $\\omega$ et $\\Omega$.\n\nMontrer que $\\widehat{CKM} = 90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nL'idée est de se rendre compte que la figure contient de nombreux points cocycliques. Pour commencer, on a $PB = PC$ donc $(MP)$ est la médiatrice de $[BC]$, et en particulier l'angle $\\widehat{BMP}$ est droit. Comme $\\widehat{BDP}$ l'est aussi, les points $B, D, M$ et $P$ sont cocycliques sur le cercle $\\Gamma$ de diamètre $[BP]$. Si on trace ce cercle sur notre figure, il semble que $\\Gamma$ passe aussi par $K$. En effet, on va vérifier par chasse aux angles que $B, D, K$ et $P$ sont cocycliques. D'une part, en utilisant le théorème de l'angle inscrit, on a\n$$\n\\widehat{KDP} = 180^{\\circ} - \\widehat{KDS} = \\widehat{KCS} = \\widehat{KCB}.\n$$\nD'autre part, en utilisant le cas limite du théorème de l'angle inscrit, on a\n$$\n\\widehat{KBP} = \\widehat{KCB}\n$$\ndonc les cinq points $B, D, K, M$ et $P$ sont cocycliques. On peut maintenant conclure en décomposant l'angle $\\widehat{CKM}$ en $D$ pour pouvoir utiliser un maximum de cercles :\n$$\n\\widehat{CKM} = \\widehat{CKD} + \\widehat{DKM} = 180^{\\circ} - \\widehat{CSD} + \\widehat{DBM} = \\widehat{BSD} + \\widehat{DBS} = 180^{\\circ} - \\widehat{BDS} = 90^{\\circ}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans un tournoi auxquels participent $n$ joueurs, numérotés de 1 à $n$, chaque paire de joueurs se rencontre exactement une fois. Cette rencontre se termine par la victoire d'un des deux joueurs et la défaite de l'autre joueur. On note $v_{k}$ le nombre de victoires du joueur $k$ au cours du tournoi, et $d_{k}$ son nombre de défaites.\n\nDémontrer que $\\sum_{k=1}^{n} v_{k}^{2}=\\sum_{k=1}^{n} d_{k}^{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPuisque chaque joueur a disputé $n-1$ parties et que l'ensemble des joueurs a totalisé $n(n-1)/2$ victoires, on sait que\n$$\n\\sum_{k=1}^{n} v_{k} = n(n-1)/2\n$$\net que $v_{k} + d_{k} = n-1$ pour tout $k \\leqslant n$. On en déduit que\n$$\n\\sum_{k=1}^{n} d_{k}^{2} = \\sum_{k=1}^{n} (n-1-v_{k})^{2} = n(n-1)^{2} - 2(n-1) \\sum_{k=1}^{n} v_{k} + \\sum_{k=1}^{n} v_{k}^{2} = \\sum_{k=1}^{n} v_{k}^{2}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $f: \\mathbb{Z} \\to \\mathbb{R}$ une fonction telle que\n$$\nf(n+2) = (f(n+1) + f(n)) / 2\n$$\npour tout entier $n$. On suppose que $f$ est bornée, c'est-à-dire qu'il existe un réel $F$ tel que $-F \\leqslant f(n) \\leqslant F$ pour tout $n$.\n\nDémontrer que $f$ est une fonction constante.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons que $f$ n'est pas constante. Il existe donc un entier $n$ et un réel $\\varepsilon \\neq 0$ tels que $f(n+1)-f(n)=\\varepsilon$.\n\nPour tout entier $k$, posons $\\Delta_{k}=f(k+1)-f(k)$. Alors\n$$\n\\Delta_{k+1}=f(k+2)-f(k+1)=(f(k+1)+f(k)) / 2-f(k+1)=-\\Delta_{k} / 2\n$$\net, réciproquement, $\\Delta_{k-1}=-2 \\Delta_{k}$. Une récurrence immédiate indique alors que $\\Delta_{n-k}=(-2)^{k} \\varepsilon$ pour tout $k \\geqslant 0$.\n\nOr, pour tout entier $k$, on sait que $|\\Delta_{k}| \\leqslant |f(k+1)|+|f(k)| \\leqslant 2F$. Mais ceci est incompatible avec le fait que $|\\Delta_{n-k}|=2^{k}|\\varepsilon|$ pour tout $k \\geqslant 0$. Notre supposition initiale était donc absurde, et $f$ est bien constante.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30659,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\mathcal{C}$ un cercle de rayon $1$, et soit $T$ un nombre réel. On dit qu'un ensemble de triangles est $T$-méraire s'il satisfait les trois conditions suivantes :\n\n$\\triangleright$ les sommets de chaque triangle appartiennent à $\\mathcal{C}$ ;\n$\\triangleright$ les triangles sont d'intérieurs deux à deux disjoints (mais deux triangles peuvent partager un côté ou un sommet) ;\n$\\triangleright$ chaque triangle est de périmètre strictement plus grand que $\\mathbf{T}$.\n\nTrouver tous les réels $\\mathbf{T}$ tels que, pour tout entier $\\mathrm{n} \\geqslant 1$, il existe un ensemble $T$-méraire contenant exactement $n$ triangles.",
    "options": [],
    "answer": "T ≤ 4",
    "solution": "Solution:\n\nNous allons démontrer que les réels recherchés sont exactement les réels $\\mathbf{T} \\leqslant 4$. Pour ce faire, fixons un réel $\\mathbf{T} \\leqslant 4$. On commence par montrer qu'il existe des ensembles $T$-méraires de n'importe quelle taille, grâce à la construction suivante. Soit $[AB]$ un diamètre de $\\mathcal{C}$, et soit $P$ un point quelconque de $\\mathcal{C}$, autre que $A$ et $B$. L'inégalité triangulaire indique que $ABP$ est de périmètre $AB + BP + PA > 2AB = 4 \\geqslant T$.\n\nForts de cette remarque, on montre par récurrence sur $n$ qu'il existe $n$ points $P_{1}, \\ldots, P_{n}$, placés dans cet ordre sur l'un des deux demi-cercles $\\overline{AB}$ (avec $P_{1}$ proche de $A$ et $P_{n}$ proche de $B$), et tel que l'ensemble formé des triangles $AP_{i}P_{i+1}$ (pour $i \\leqslant n-1$) et $ABP_{n}$ soit $T$-méraire. Pour $n=1$, on vient de voir qu'il suffit de placer $P_{1}$ n'importe où sur $\\overline{AB}$.\n\nPuis, une fois acquise l'existence des points $P_{1}, \\ldots, P_{n}$, construisons le point $P_{n+1}$. Pour ce faire, on considère le réel $\\varepsilon = AB + BP_{n} + P_{n}A - T$, qui est strictement positif par hypothèse. On place alors $P_{n+1}$ n'importe où sur l'arc $\\overline{BP_{n}}$, de sorte que $BP_{n+1} < \\varepsilon / 2$. En effet, dans ces conditions, nos $n+1$ triangles sont bien d'intérieurs deux à deux disjoints, et il suffit de vérifier que $ABP_{n+1}$ et $AP_{n}P_{n+1}$ sont de périmètre strictement plus grand que $T$.\n\nLe premier cas est un cas particulier de notre remarque initiale, et le deuxième cas découle encore une fois de l'inégalité triangulaire, puisque $AP_{n}P_{n+1}$ est de périmètre\n$$\nAP_{n} + P_{n}P_{n+1} + P_{n+1}A \\geqslant AP_{n} + (P_{n}B - BP_{n+1}) + (AB - BP_{n+1}) = T + \\varepsilon - 2BP_{n+1} > T\n$$\nCeci conclut notre récurrence, et donc le fait que l'on a bien des ensembles $T$-méraires de n'importe quelle taille.\n\nRéciproquement, considérons un réel $\\mathbf{T} > 4$, et posons $\\varepsilon = (\\mathbf{T} - 4)/2 > 0$. Soit également $ABC$ un triangle dont les sommets appartiennent à $\\mathcal{C}$ et dont le périmètre est strictement plus grand que $T$. En notant $a, b$ et $c$ les longueurs $BC, CA$ et $AB$, et $p = (a + b + c)/2 > T/2$ le demi-périmètre de $ABC$, la formule de Héron indique que $ABC$ est d'aire\n$$\n\\mathcal{S} = \\sqrt{p(p-a)(p-b)(p-c)}\n$$\nPuisque $p-a \\geqslant p-2 \\geqslant T/2 - 2 = \\varepsilon$ et que, de même, $p-b \\geqslant \\varepsilon$ et $p-c \\geqslant \\varepsilon$, on en déduit que $\\mathcal{S} \\geqslant \\sqrt{p\\varepsilon^{3}} \\geqslant \\sqrt{2\\varepsilon^{3}}$.\n\nPar conséquent, si un ensemble $T$-méraire contient $n$ triangles, ceux-ci étant d'intérieurs deux à deux disjoints, ils couvrent, dans leur ensemble, une surface égale à $n\\sqrt{2\\varepsilon^{3}}$ au moins. Cette surface ne pouvant pas dépasser $\\pi$, qui est la surface du disque contenu à l'intérieur de $\\mathcal{C}$, on en déduit que $n \\leqslant \\pi / \\sqrt{2\\varepsilon^{3}}$, ce qui conclut le problème.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle isocèle en $A$ mais pas rectangle. Soit $D$ le point de $(BC)$ tel que $(AD)$ soit perpendiculaire à $(AB)$, et soit $E$ le projeté orthogonal de $D$ sur $(AC)$. Soit enfin $H$ le milieu de $[BC]$.\n\nMontrer que $AHE$ est isocèle en $H$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nCommençons par remarquer que les points $A$, $D$, $E$ et $H$ sont cocycliques sur le cercle de diamètre $[AD]$. On en déduit par chasse aux angles :\n$$\n\\widehat{HEA} = \\widehat{HDA} = 90^\\circ - \\widehat{DBA} = 90^\\circ - \\widehat{BCA} = \\widehat{HAE},\n$$\ndonc le triangle $AHE$ est isocèle en $H$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver toutes les fonctions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ telles que\n$$\nx f(y)+y f(x)=(x+y) f\\left(x^{2}+y^{2}\\right)\n$$\npour tous les entiers naturels $x$ et $y$.\n\nNote : on rappelle que $\\mathbb{N}$ désigne l'ensemble des entiers naturels, c'est-à-dire $\\mathbb{N}=\\{0,1,2, \\ldots\\}$.",
    "options": [],
    "answer": "All constant functions from the natural numbers to the natural numbers; that is, f(n) = c for some fixed nonnegative integer c.",
    "solution": "Solution:\n\nDans la suite, on notera $\\mathbf{E}_{x, y}$ l'équation de l'énoncé.\n\nTout d'abord, l'équation $E_{x, 0}$ indique que $x f(0)=x f\\left(x^{2}\\right)$, ce qui montre que $f\\left(x^{2}\\right)=f(0)$ pour tout entier $x \\geqslant 1$.\n\nPar conséquent, pour tout entier $y \\geqslant 1$, l'équation $\\mathbf{E}_{x, y^{2}}$ indique que $x+y^{2}$ divise $x(f(0)-f(x))$. Cette relation de divisibilité étant valide même quand $y$ est arbitrairement grand, on en déduit que $f(x)=f(0)$ quel que soit $x$.\n\nRéciproquement, on vérifie aisément que les fonctions constantes sont bien des solutions de l'équation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier impair, et soit $S$ un ensemble de $n$ points du plan à coordonnées entières. On considère une permutation $f: S \\rightarrow S$ qui satisfait la propriété suivante :\nPour toute paire de points $A$ et $B$ appartenant à $S$, la distance entre $f(A)$ et $f(B)$ est supérieure ou égale à la distance entre $A$ et $B$.\nMontrer qu'il existe un point $X$, appartenant à $S$, tel que $f(X)=X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPuisque $S$ est de cardinal impair, $f$ admet une orbite de cardinal impair. Sans perte de généralité, on peut donc supposer que $S$ est égal à cette orbite, et même que $S=\\{P_1, \\ldots, P_n\\}$ avec $f(P_i)=P_{i+1}$ pour tout $i \\leqslant n$, en posant $P_{n+1}=P_1$.\n\nSi $f$ n'a aucun point fixe, c'est donc que $n \\geqslant 3$. Montrons que ce cas est en fait impossible.\n\nEn effet, dans ces conditions, l'énoncé montre que $P_1P_2 \\leqslant P_2P_3 \\leqslant \\ldots \\leqslant P_nP_1$, de sorte que toutes ces inégalités sont en fait des égalités. En outre, toujours sans perte de généralité, quitte à considérer un point $P_i$ plutôt que $P_1$ et quitte à effectuer des translations à coordonnées entières et à diviser toutes nos coordonnées par 2, on peut supposer que les coordonnées du vecteur $\\overrightarrow{P_1P_2}$ ne sont pas toutes les deux paires. Maintenant, on va noter $x_i$ et $y_i$ les coordonnées du point $P_i$.\n\nPuisque les carrés modulo 4 sont 0 et 1, on en déduit que $P_iP_{i+1}^2 = P_1P_2^2 = (x_i - x_{i+1})^2 + (y_i - y_{i+1})^2$ est congru à 1 ou 2 modulo 4.\n\nDans le premier cas, c'est donc que la somme $\\Delta_i = x_i + y_i + x_{i+1} + y_{i+1} \\equiv 1 \\pmod{2}$. On en déduit que\n$$\nn \\equiv \\sum_{i=1}^n \\Delta_i \\equiv 2 \\sum_{i=1}^n (x_i + y_i) \\equiv 0 \\pmod{2}\n$$\n\nDans le deuxième cas, on a directement $\\Delta_i^x = x_i + x_{i+1} \\equiv 1 \\pmod{2}$ et $\\Delta_i^y = y_i + y_{i+1} \\equiv 1 \\pmod{2}$, de sorte que\n$$\nn \\equiv \\sum_{i=1}^n \\Delta_i^x \\equiv 2 \\sum_{i=1}^n x_i \\equiv 0 \\pmod{2}\n$$\n\nPuisque $n$ est impair, aucun de ces deux cas n'est possible, ce qui conclut.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $u$ un entier naturel non nul.\n\nDémontrer qu'il n'existe qu'un nombre fini de triplets d'entiers naturels $(a, b, n)$ tels que $n! = u^{a} - u^{b}$.\n\nNote : on rappelle que $0! = 1! = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $p$ un nombre premier impair qui ne divise pas $u$, et soit $k$ un entier tel que $p^{k} > u^{p-1} - 1$. Posons $q = p^{k-1}$. On montre tout d'abord que $u$ n'est pas une puissance $q$-ième modulo $p^{k}$. En effet, si $u$ était une puissance $q$-ième modulo $p^{k}$, alors il existerait un entier $v$ tel que $u \\equiv v^{q} \\pmod{p^{k}}$, et alors on aurait $1 \\equiv v^{\\varphi(p^{k})} \\equiv v^{q(p-1)} \\equiv u^{p-1} \\pmod{p^{k}}$, ce qui n'est pas le cas.\n\nPar conséquent, on sait également, pour tout entier $\\ell \\geqslant k$, que $u$ n'est pas non plus une puissance $q$-ième modulo $p^{\\ell}$. On en déduit que $p^{\\ell-k}$ divise $\\omega_{p^{\\ell}}(u)$, où $\\omega_{p^{\\ell}}(u)$ désigne l'ordre de $u$ modulo $p^{\\ell}$. En effet, soit $g$ une racine primitive modulo $p^{\\ell}$, et soit $x$ un entier tel que $u \\equiv g^{x} \\pmod{p^{\\ell}}$. Alors $1 \\equiv u^{\\omega_{p^{\\ell}}(u)} \\equiv g^{x \\omega_{p^{\\ell}}(u)} \\pmod{p^{\\ell}}$, ce qui signifie que $\\varphi(p^{\\ell}) = (p-1)p^{\\ell-1}$ divise $x \\omega_{p^{\\ell}}(u)$. Puisque $q = p^{k-1}$ ne divise pas $x$, c'est donc que $p^{\\ell-k}$ divise $\\omega_{p^{\\ell}}(u)$.\n\nSupposons enfin qu'il existe un triplet $(a, b, n)$ d'entiers tels que $u^{a} - u^{b} = n!$ et $n \\geqslant k p$. Posons également $d = a - b$ et $\\ell = \\lfloor n / p \\rfloor$. Alors $p^{\\ell}$ divise $n! = u^{b}(u^{d} - 1)$ et, puisque $p$ ne divise pas $u$, c'est donc que $u^{d} \\equiv 1 \\pmod{p^{\\ell}}$. On en déduit que $\\omega_{p^{\\ell}}(u)$ divise $d$, donc que $p^{\\ell-k}$ divise $d$ également. Cela signifie en particulier que $d \\geqslant p^{\\ell-k}$, donc que\n$$\n2^{n(n+1)} \\geqslant u^{n+1} \\geqslant n! + 1 \\geqslant u^{d} \\geqslant u^{p^{\\ell-k}} \\geqslant u^{p^{n/p - k}},\n$$\nou encore que $p^{k} n(n+1) \\geqslant p^{n/p} \\log_{2}(u)$.\n\nUne fois les entiers $u, p$ et $k$ fixés, le membre de droite croît beaucoup plus vite que le membre de gauche. Il existe donc un entier $N$, qui ne dépend que de $u$, $p$ et $k$, et tel que $n \\leqslant N$.\n\nAinsi, seul un nombre fini d'entiers $n$ appartient à un triplet $(a, b, n)$ tel que $n! = u^{a} - u^{b}$. Or, une fois un tel entier $n$ fixé, on sait que $u^{a} > u^{b}$, donc que $a \\geqslant b + 1$, et donc que $n! \\geqslant u^{a} - u^{a-1} = (u-1)u^{a-1}$, ce qui montre que $a$ et $b$ sont eux-mêmes bornés. Ceci conclut notre solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30664,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A, B, C$ et $P$ quatre points du plan tels que $ABC$ soit un triangle équilatéral et que $AP < BP < CP$. On suppose que la seule donnée des longueurs $AP, BP$ et $CP$ nous permet de déterminer, de manière unique, la longueur $AB$.\n\nDémontrer que $P$ appartient au cercle circonscrit à $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $\\Gamma_{a}$, $\\Gamma_{b}$ et $\\Gamma_{c}$ les cercles de centre $P$ et passant respectivement par $A$, $B$ et $C$. Soit également $C^{\\bullet}$ le symétrique de $C$ par rapport à $(AP)$. Notons que $C^{\\bullet}$ est le seul point de $\\Gamma_{c}$, autre que $C$ lui-même, tel que $AC = AC^{\\bullet}$.\n\nOn note maintenant $r$ la rotation de centre $A$ et d'angle $60^{\\circ}$ dans le sens horaire. Quitte à faire subir une symétrie à la figure de départ, on suppose que $C = r(B)$.\n\nPuis, à tout point $B'$ de $\\Gamma_{b}$, on associe le point $C' = r(B')$. Le triangle $AB'C'$ est équilatéral, et l'hypothèse de l'énoncé stipule donc que, pour tout $B'$, le point $C'$ ne peut appartenir à $\\Gamma_{c}$ que s'il est égal à $C$ ou à $C^{\\bullet}$.\n\nOr, quand $B'$ décrit $\\Gamma_{b}$, le point $C'$ décrit le cercle $r(\\Gamma_{b})$, de centre $r(P)$ et de rayon $PB$. Ce cercle ne peut contenir à la fois les points $C$ et $C^{\\bullet}$, puisque alors son centre se trouverait sur la médiatrice $(AP)$ de $[CC^{\\bullet}]$, ce qui n'est pas le cas de $r(P)$.\n\nComme $r(\\Gamma_{b})$ contient déjà $C$, il ne peut donc pas contenir d'autre point de $\\Gamma_{c}$, et puisque $r(P)A = PA < PB$, on sait que $A$ est situé à l'intérieur de $r(\\Gamma_{b})$, qui est donc tangent intérieurement, en le point $C$, à $\\Gamma_{c}$.\n\nComme $PB = r(P)C < PC$, cela implique que les points $C$, $r(P)$ et $P$ sont alignés dans cet ordre. On en déduit que $PC = r(P)C + r(P)P = PB + PA$, ou encore que $CP \\cdot AB = BP \\cdot AC + AP \\cdot BC$. L'égalité de Ptolémée indique donc que les points $A$, $B$, $C$ et $P$ sont cocycliques, ce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30665,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ et soient $x_{1}, x_{2}, \\ldots, x_{n}$ des nombres réels tels que $x_{1}+x_{2}+\\cdots+x_{n}=0$ et $x_{1}^{2}+x_{2}^{2}+\\cdots+x_{n}^{2}=1$.\nMontrer qu'il existe $i$ tel que $x_{i} \\geqslant \\frac{1}{\\sqrt{n(n-1)}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $N^{+}$ le nombre d'indices $i$ tels que $x_{i}>0$ et $N^{-}$ le nombre d'indices $i$ tels que $x_{i} \\leqslant 0$, de sorte que $N^{+}+N^{-}=n$. On note également\n$$\nS_{1}^{+}=\\sum_{i \\text{ tel que } x_{i}>0} x_{i} \\quad \\text{et} \\quad S_{1}^{-}=\\sum_{i \\text{ tel que } x_{i} \\leqslant 0}\\left(-x_{i}\\right)\n$$\naussi que\n$$\nS_{2}^{+}=\\sum_{i \\text{ tel que } x_{i}>0} x_{i}^{2} \\quad \\text{et} \\quad S_{2}^{-}=\\sum_{i \\text{ tel que } x_{i} \\leqslant 0} x_{i}^{2} .\n$$\nL'intérêt de ces quantités est qu'elles permettent à la fois de reformuler les hypothèses et d'utiliser des inégalités bien connues comme celle de Cauchy-Schwarz. Les hypothèses de l'énoncé se réécrivent $S_{1}^{+}=S_{1}^{-}$ et $S_{2}^{+}+S_{2}^{-}=1$. Pour faire apparaître la racine dans le résultat, on va utiliser $S_{2}^{+}$. Plus précisément, on va montrer que $S_{2}^{+} \\geqslant \\frac{1}{n}$. Comme la somme $S_{2}^{+}$ contient au maximum $n-1$ termes (les nombres ne peuvent pas être tous $>0$), un de ces termes est plus grand que $\\frac{1}{n(n-1)}$, soit $x_{i}^{2} \\geqslant \\frac{1}{n(n-1)}$ pour un certain $i$ avec $x_{i}>0$, ce qui permet de conclure.\n\nPour montrer cela, on écrit :\n$$\n1-S_{2}^{+}=S_{2}^{-} \\leqslant\\left(S_{1}^{-}\\right)^{2}=\\left(S_{1}^{+}\\right)^{2} \\leqslant N^{+} S_{2}^{+} \\leqslant(n-1) S_{2}^{+} .\n$$\nLa première inégalité s'obtient en développant $\\left(S_{1}^{-}\\right)^{2}$ et en ne gardant que les termes $x_{i}^{2}$. La seconde est l'inégalité de Cauchy-Schwarz, et la troisième est le fait que les $x_{i}$ ne sont pas tous strictement positifs. On en déduit $S_{2}^{+} \\geqslant \\frac{1}{n}$, ce qui permet de conclure.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a_{0}, a_{1}, \\ldots, a_{d}$ des entiers tels que $\\operatorname{PGCD}(a_{0}, a_{1})=1$. Pour tout entier $n \\geqslant 1$, on pose\n$$\nu_{n}=\\sum_{k=0}^{d} a_{k} \\varphi(n+k)$$\nDémontrer que 1 est le seul entier naturel divisant tous les entiers $u_{n}$.\nOn rappelle que $\\varphi(n+k)$ est le nombre d'entiers naturels $\\ell<n+k$ tels que $\\operatorname{PGCD}(\\ell, n+k)=1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nProcédons par l'absurde, et supposons qu'il existe un nombre premier $p$ qui divise tous les entiers $u_{n}$.\n\nTout d'abord, puisque $\\varphi(1)=\\varphi(2)=1$ et que $\\varphi(\\mathfrak{n})$ est pair pour tout $\\mathfrak{n} \\geqslant 2$, on remarque que $u_{1} \\equiv a_{0}+a_{1}(\\bmod 2)$ et que $u_{2} \\equiv a_{0}(\\bmod 2)$. Comme $a_{0}$ et $a_{1}$ sont premiers entre eux, les entiers $\\mathfrak{u}_{1}$ et $\\mathfrak{u}_{2}$ ne peuvent donc pas être pairs tous les deux, de sorte que $p \\geqslant 3$.\n\nSoit alors $\\delta$ le plus grand entier tel que $p$ ne divise pas $a_{\\delta}$. Pour tout entier $n \\geqslant 1$, on note $T_{n}$ le $\\delta$-uplet $(\\varphi(n), \\varphi(n+1), \\ldots, \\varphi(n+\\delta-1))$, considéré modulo $p$. Puisque $u_{n} \\equiv 0(\\bmod p)$, c'est que\n$$\n\\varphi(n+\\delta) \\equiv -a_{\\delta}^{-1} \\sum_{k=0}^{\\delta-1} a_{k} \\varphi(n+k) \\quad(\\bmod p)\n$$\nAinsi, si l'on note $\\lambda:(\\mathbb{Z} / p \\mathbb{Z})^{\\delta} \\mapsto (\\mathbb{Z} / p \\mathbb{Z})^{\\delta}$ la fonction définie par\n$$\n\\lambda:\\left(x_{0}, x_{1}, \\ldots, x_{\\delta-1}\\right) \\mapsto \\left(x_{1}, x_{2}, \\ldots, x_{\\delta-1}, -a_{\\delta}^{-1} \\sum_{k=0}^{\\delta-1} a_{k} x_{k}\\right)\n$$\non a $\\mathbf{T}_{n+1}=\\lambda(\\mathbf{T}_{n})$. Or, d'après le principe des tiroirs, il existe deux entiers $k, \\ell \\geqslant 1$ tels que $\\mathbf{T}_{k}=\\mathbf{T}_{k+\\ell}$. On a alors $\\mathbf{T}_{n}=\\mathbf{T}_{n+\\ell}$ et $\\varphi(n) \\equiv \\varphi(n+\\ell)(\\bmod p)$ pour tout $n \\geqslant k$.\n\nOn montre alors le lemme suivant : pour tous les entiers $a \\geqslant 1$ et $b \\geqslant 3$, il existe un nombre premier $q>a$ tel que $q \\not \\equiv 1(\\bmod b)$. En effet, l'entier $m=\\max \\{a, b\\}! - 1$ est congru à $-1$ $(\\bmod b)$, donc il admet nécessairement un facteur premier $q \\not \\equiv 1(\\bmod b)$, et $q$ ne peut pas diviser $m+1=\\max \\{a, b\\}!$, donc $q>a$.\n\nEn appliquant ce lemme deux fois de suite, on en déduit qu'il existe deux nombres premiers $q$ et $r$ tels que $q, r \\not \\equiv 1(\\bmod p)$ et $q>r>k+\\ell$. Mais alors $q$ est premier avec $\\ell$, donc $q^{\\varphi(\\ell)} \\equiv 1(\\bmod \\ell)$, et $\\varphi\\left(q^{\\varphi(\\ell)} n\\right) \\equiv \\varphi(n)(\\bmod p)$ pour tout $n \\geqslant k$.\n\nPuisque $q>r \\geqslant k$, on en déduit que\n$$\n\\begin{array}{rlr}\n(q-1) q^{\\varphi(\\ell)} & \\equiv \\varphi\\left(q^{\\varphi(\\ell)+1}\\right) \\equiv \\varphi(q) & \\equiv q-1 \\quad(\\bmod p) \\\\\n(r-1)(q-1) q^{\\varphi(\\ell)-1} & \\equiv \\varphi\\left(r q^{\\varphi(\\ell)}\\right) & \\equiv \\varphi(r)\n\\end{array}\n$$\nComme $q, r \\not \\equiv 1(\\bmod p)$, c'est que $q^{\\varphi(\\ell)} \\equiv (q-1) q^{\\varphi(\\ell)-1} \\equiv 1(\\bmod p)$, donc que $q^{\\varphi(\\ell)-1} \\equiv 0$ $(\\bmod p)$, ce qui est impossible. On a ainsi obtenu la contradiction souhaitée, ce qui conclut.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les triplets $\\left(a, b, c\\right)$ d'entiers strictement positifs tels que\n$$\n3^{a}-5^{b}=c^{2}\n$$",
    "options": [],
    "answer": "(2, 1, 2)",
    "solution": "Solution:\nComme $3^{a}$ et $5^{b}$ sont tous deux impairs, on a $c$ pair donc $c^{2} \\equiv 0[4]$. Comme $5^{b} \\equiv 1[4]$, on doit avoir $3^{a} \\equiv 1[4]$, donc $a$ est pair. On écrit $a=2 a'$, et l'équation devient\n$$\n5^{b}=3^{2 a'}-c^{2}=\\left(3^{a'}-c\\right)\\left(3^{a'}+c\\right) .\n$$\nNotons que la somme des deux facteurs vaut $2 \\times 3^{a'}$, donc elle n'est pas divisible par $5$, donc un des deux facteurs n'est pas divisible par $5$ et ne peut valoir que $1$. C'est nécessairement le plus petit, donc on a\n$$\n3^{a'}-c=1 \\quad \\text{et} \\quad 3^{a'}+c=5^{b}\n$$\nou encore (en sommant les deux dernières égalités) $2 \\times 3^{a'}=5^{b}+1$.\nOn peut maintenant remarquer que $a'=1$ est solution, mais qu'il ne semble plus y en avoir ensuite. Une manière naturelle de séparer le cas $a'=1$ du reste est de regarder modulo $9$. En effet, dans ce cas, dès que $a' \\geqslant 2$, les deux membres doivent être divisibles par $9$, i.e. $5^{b} \\equiv -1[9]$. Ceci implique que $b$ doit être divisible par $3$. De plus, en regardant modulo $3$, on obtient que $b$ doit être impair. L'équation se réécrit donc\n$$\n2 \\times 3^{a'}=5^{3 b'}+1=125^{b'}+1\n$$\nav ec $b'$ impair. Mais alors, le membre de droite se factorise par $125+1=126=2 \\times 7 \\times 9$. En particulier, il est divisible par $7$, ce qui n'est pas le cas du membre de gauche. Il est donc impossible d'avoir $a' \\geqslant 2$. On doit donc avoir $a'=1$, ce qui conduit à la solution $(a, b, c)=(2,1,2)$ dans l'équation de départ.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30668,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, et soit $E$ et $F$ deux points appartenant respectivement aux droites $(AB)$ et $(AC)$, distincts de $A$, $B$ et $C$. Soit également $\\Omega$ le cercle circonscrit à $ABC$, soit $O$ le centre de $\\Omega$, et soit $\\Gamma$ le cercle circonscrit à $AEF$. Enfin, soit $P$ le point d'intersection de $\\Gamma$ et $\\Omega$ autre que $A$, et soit $Q$ le symétrique de $P$ par rapport à la droite $(EF)$.\n\nMontrer que $Q$ appartient à la droite $(BC)$ si et seulement si $O$ appartient au cercle $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPuisque $(BC)$ et $(EF)$ sont deux droites d'intérêt notoire, on note $T$ leur point d'intersection, éventuellement rejeté à l'infini. Alors $Q$ appartient à la droite $(BC)$ si et seulement si, en angles de droites, on a $(BT, ET) = (ET, PT)$.\n\nOr, toujours en angles de droites, on sait déjà que\n$$\n\\begin{aligned}\n(PB, PE) & = (PB, PA) + (PA, PE) = (CB, CA) + (FA, FE) \\\\\n& = (CT, CF) + (CF, FT) = (CT, FT) = (TB, TE),\n\\end{aligned}\n$$\nce qui signifie que les points $P$, $E$, $B$ et $T$ sont cocycliques. Par conséquent, on sait que $(BT, ET) = (BP, EP)$ et $(ET, PT) = (EB, PB) = (AB, PB)$.\n\nOn sait aussi que $2(BA, BP) = (OA, OP)$ et que $(EA, EP) = (BA, BP) + (BP, EP)$. Ainsi,\n$$\n\\begin{aligned}\nQ \\text{ appartient à la droite } (BC) & \\Leftrightarrow (BP, EP) = (AB, PB) \\\\\n& \\Leftrightarrow (EA, EP) = 2(BA, BP) = (OA, OP) \\\\\n& \\Leftrightarrow O \\text{ appartient au cercle } \\Gamma.\n\\end{aligned}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30669,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, soit $I$ le centre du cercle inscrit dans $ABC$. Soit $A'$, $B'$ et $C'$ trois points respectivement situés sur $(AI)$, $(BI)$ et $(CI)$. On suppose que $A'$, $B'$ et $C'$ sont distincts de $A$, $B$, $C$ et $I$, et qu'ils sont alignés. Enfin, soit $P_A$ le point d'intersection des médiatrices de $[BB']$ et $[CC']$. De même, soit $P_B$ le point d'intersection des médiatrices de $[CC']$ et $[AA']$, et soit $P_C$ le point d'intersection des médiatrices de $[AA']$ et $[BB']$.\n\nDémontrer que les cercles circonscrits aux triangles $ABC$ et $P_A P_B P_C$ sont tangents l'un à l'autre.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par tracer une figure comme ci-dessous, sans oublier la droite $\\ell$ qui passe par $A'$, $B'$ et $C'$ ni les cercles $\\Omega$ et $\\Gamma$, respectivement circonscrits à $ABC$ et à $P_A P_B P_C$.\n\n![](attached_image_1.png)\n\nAu vu de l'énoncé, une approche raisonnable est de trouver une caractérisation adaptée du point de tangence éventuel, que l'on notera $T$, entre $\\Omega$ et $\\Gamma$. Une telle caractérisation est la suivante : $T$ est le centre d'une des deux homothéties envoyant $\\Gamma$ sur $\\Omega$. Soit $h$ cette homothétie : il s'agit de l'homothétie de rapport positif si l'un des deux cercles est inclus à l'intérieur de l'autre, et de l'homothétie de rapport négatif sinon.\n\nAu vu de la définition de $\\Gamma$ et $\\Omega$, il est tentant d'étudier l'image de $P_A P_B P_C$ par $h$, ou bien celle de $ABC$ par $h^{-1}$. Or, sur la figure, on constate deux choses :\n\n$\\triangleright$ les points $h(P_A)$, $h(P_B)$ et $h(P_C)$ semblent être les points d'intersection respectifs de $(AI)$, $(BI)$ et $(CI)$ avec $\\Omega$ ;\n\n$\\triangleright$ les droites $(AT)$, $(BT)$ et $(CT)$ semblent recouper la droite $\\ell$ en des points de $(P_B P_C)$, $(P_C P_A)$ et $(P_A P_B)$ respectivement.\n\nExploitons ces idées !\n\nTout d'abord, on note donc $A''$, $B''$ et $C''$ ces points d'intersection. Le point $A''$ n'est autre que le pôle sud de $ABC$ par rapport au sommet $A$. On sait donc que $A''B = A''C = A''I$. De même, on a $B''C = B''A = B''I$ et $C''A = C''B = C''I$. Ainsi, les droites $(A''B'')$ et $(P_A P_B)$ sont les médiatrices respectives de $[CI]$ et de $[C']$, ce qui montre en particulier qu'elles sont parallèles. De même, $(B''C'')$ est parallèle à $(P_B P_C)$ et $(C''A'')$ est parallèle à $(P_A P_B)$. Cela montre que les triangles $P_A P_B P_C$ et $A'' B'' C''$ sont homothétiques l'un de l'autre.\n\nEn particulier, il existe une unique homothétie qui envoie $P_A P_B P_C$ sur $A'' B'' C''$. On peut effectivement la noter $h$, et noter $T$ son centre, de sorte que $h$ envoie $\\Omega$ sur $\\Gamma$. Ici, le point $T$ n'est autre que le point de concours des droites $(A'' P_A)$, $(B'' P_B)$ et $(C'' P_C)$, et il reste à montrer que $T$ appartient à $\\Omega$.\n\nD'autre part, puisque $(P_A P_B)$ est la médiatrice de $[CC']$, et si $(TC)$ recoupe $(P_A P_B)$ en un point $X$ qui appartient aussi à $\\ell$, alors les droites $\\ell$ et $(TC) = (CX)$ sont symétriques l'une de l'autre par rapport à $(P_A P_B)$. Réciproquement, on note $\\ell_a$, $\\ell_b$ et $\\ell_c$ les symétriques respectifs de $\\ell$ par rapport à $(P_B P_C)$, $(P_C P_A)$ et $(P_A P_B)$.\n\nOn peut alors chercher à montrer que les droites $\\ell_a$, $\\ell_b$ et $\\ell_c$ sont concourantes en un point $\\hat{T}$ qui appartiendra à $\\Omega$, puis que les points $\\hat{T}$, $P_A$ et $A''$ sont alignés : on en déduira que $T = \\hat{T}$, ce qui conclura le problème. Soit alors $T_c$ le point d'intersection de $\\ell_a$ et $\\ell_b$. On remarque que\n\n$$\n\\begin{aligned}\n( T_c A, T_c B ) &= (\\ell_a, \\ell_b) = (\\ell_a, \\ell) + (\\ell, \\ell_b) = 2(P_B P_C, \\ell) + 2(\\ell, P_C P_A) = 2(P_B P_C, P_A P_C) \\\\\n&= 2(B'' C'', A'' C'') = 2(B'' C'', CC'') + 2(CC'', A'' C'') \\\\\n&= (A C'', CC'') + (CC'', B C'') = (A C'', B C'') = (AC, BC).\n\\end{aligned}\n$$\n\nCeci montre que les points $T_c$, $A$, $B$ et $C''$ sont cocycliques, donc que $T_c$ appartient bien à $\\Omega$. Or, si $T_c = A$, alors $\\ell_b = (AB)$, donc $(AC, BC) = (\\ell_a, \\ell_b) = (\\ell_a, BA)$, donc $\\ell_a$ est la tangente à $\\Omega$ en $A$, et $T_c$ est nécessairement le point d'intersection de $\\ell_a$, $\\ell_b$ et $\\ell_c$. De même, on introduit les points $T_a$ et $T_b$. Si l'un de ces trois points est $A$, $B$ ou $C$, alors il constitue bien le point $\\hat{T}$ désiré. Sinon, $T_c$ est le point d'intersection de $\\ell_a$ avec $\\Omega$ autre que $A$, mais également le point d'intersection de $\\ell_b$ avec $\\Omega$ autre que $B$, et en raisonnant de manière analogue sur $T_a$ et $T_b$ on constate bien que $T_a = T_b = T_c$ appartient à $\\Omega$.\n\nPar ailleurs, et puisque l'on a obtenu les droites $\\ell_a$, $\\ell_b$ et $\\ell_c$ en appliquant à $\\ell$ des symétries axiales, il est important d'appliquer ces symétries à d'autres points en particulier, notamment les points $A'$, $B'$ et $C'$ présents sur $\\ell$. Ainsi, on note $A_b'$ et $A_c'$ les symétriques de $A'$ par rapport à $(P_A P_C)$ et à $(P_A P_B)$ : ils se trouvent sur $\\ell_b$ et $\\ell_c$.\n\nDans ces conditions,\n\n$$\n\\begin{aligned}\n(P_A A_b', P_A A_c') &= (P_A A_b', P_A A') + (P_A A', P_A A_c') = 2(P_A P_C, P_A A') + 2(P_A A', P_A P_B) \\\\\n&= 2(P_A P_C, P_A P_B) = (\\ell_b, \\ell_c) = (\\hat{T} A_b', \\hat{T} A_c').\n\\end{aligned}\n$$\n\nCela signifie que les points $P_A$, $A_b'$, $\\hat{T}$ et $A_c'$ sont cocycliques.\n\nD'autre part, puisque $P_A A_b' = P_A A' = P_A A_c'$, on sait que le triangle $P_A A_b' A_c'$ est isocèle en $P_A$. On en déduit que\n\n$$\n\\begin{aligned}\n& (\\ell_b, P_A \\hat{T}) = (A_b' \\hat{T}, P_A \\hat{T}) = (A_b' A_c', P_A A_c') \\\\\n&= 90^\\circ - (A_b' P_A, A_c' P_A)/2 = 90^\\circ - (P_C P_A, P_B P_A) \\\\\n&= 90^\\circ + (BI, IC) = (BA, AI) = (BA, AA'') = (B \\hat{T}, \\hat{T} A'') = (\\ell_b, \\hat{T} A'').\n\\end{aligned}\n$$\n\nCela signifie que $\\hat{T}$ est bien aligné avec $P_A$ et $A''$. On montre de même que $\\hat{T}$ est aligné avec $P_B$ et $B''$, ainsi qu'avec $P_C$ et $C''$. Par conséquent, on a bien $T = \\hat{T}$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ un entier. Clara dispose d'un plateau de taille $3 n \\times 3 n$, semblable à un échiquier. Elle vient d'inventer une nouvelle pièce, le léopard, qu'elle peut mouvoir comme suit : en le déplaçant d'une case vers le haut, d'une case vers la droite, ou bien d'une case en diagonale, vers le bas et la gauche. Clara a posé son léopard sur une des cases du plateau, puis elle l'a déplacé, de sorte qu'il ne passe jamais deux fois par la même case, jusqu'à ce qu'il revienne à sa case de départ.\nQuelle est le plus grand nombre possible de déplacements que Clara a pu effectuer en procédant de la sorte?",
    "options": [],
    "answer": "9n^2 - 3",
    "solution": "Solution:\n\nTout d'abord, Clara peut effectuer $9 n^2 - 3$ déplacements en procédant comme illustré ci-dessous dans le cas où $n = 3$.\n\n![](attached_image_1.png)\n\nMontrons maintenant qu'elle ne peut pas mieux faire. Tout d'abord, on identifie chaque case en ligne $i$ (en partant du bas) et en colonne $j$ (en partant de la gauche) à la paire $(i, j)$, où $1 \\leqslant i, j \\leqslant 3 n$. Puis on colorie les cases du plateau de trois couleurs $C_0$, $C_1$ et $C_2$, en affectant à la case $(i, j)$ la couleur $C_{i+j} (\\bmod 3)$. Chaque mouvement du léopard consiste à se mouvoir d'un vecteur $(0,1)$, $(1,0)$ ou $(-1,-1)$. Par conséquent, il passe systématiquement d'une case de couleur $C_k$ à une case de couleur $C_{k+1} (\\bmod 3)$. Il a donc effectué un nombre de déplacements divisible par $3$, et il suffit de démontrer qu'il n'a pas pu passer par toutes les cases.\n\nÀ cette fin, remarquons que son parcours décrit un graphe planaire formé d'un seul cycle, qui contient donc une face interne (finie) et une face externe. Or, si le léopard est passé par la case $(1, 3n)$, il a dû utiliser l'arête $(1, 3n) \\rightarrow (2, 3n)$, donc contourner sa face interne dans le sens horaire. Mais, de même, s'il est passé par la case $(3n, 1)$, il a dû utiliser l'arête $(3n, 1) \\rightarrow (3n, 2)$, donc contourner sa face interne dans le sens anti-horaire. Il n'a donc pas pu passer par ces deux cases lors du même trajet, ce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\left(a_{n}\\right)_{n \\geqslant 0}$ une suite de réels. On suppose que $a_{n}=\\left|a_{n+1}-a_{n+2}\\right|$ pour tout entier naturel $n$. De plus, $a_{0}$ et $a_{1}$ sont strictement positifs et distincts. Montrer que la suite $\\left(a_{n}\\right)_{n \\geqslant 0}$ n'est pas bornée.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIl est clair que la suite $\\left(a_{n}\\right)$ est à termes positifs.\n\nSoit $i$ tel que $a_{i}<a_{j}$ pour $j<i$. Supposons par l'absurde que $i \\geqslant 4$. Alors $a_{i-2}=\\left|a_{i}-a_{i-1}\\right|=a_{i-1}-a_{i}<a_{i-1}$ donc $a_{i-3}=\\left|a_{i-2}-a_{i-1}\\right|=a_{i-1}-a_{i-2}=a_{i}$, ce qui contredit l'hypothèse sur $i$.\n\nAinsi, si $m=\\min \\left\\{a_{1}, a_{2}, a_{3}\\right\\}$, alors pour tout $i, a_{i} \\geqslant m$. En effet, s'il existe $i$ avec $a_{i}<m$, choisissons $i$ minimal, alors $a_{i}$ est plus petit que tous les termes précédents, donc $i \\leqslant 3$ ce qui contredit la définition de $m$.\n\nDe plus, $m>0$ car $a_{3} \\geqslant 0$ et $a_{3}=0 \\rightarrow a_{1}=a_{2}$.\n\nDès lors, on écrit $a_{i}=\\left|a_{i+1}-a_{i+2}\\right|$ pour tout $i$, donc\n\n- Si $a_{i+1}>a_{i+2}$, d'où $a_{i+1}=a_{i}+a_{i+2} \\geqslant a_{i}+m$\n- Sinon, $a_{i+2}=a_{i+1}+a_{i} \\geqslant a_{i}+m$\n\nDans tous les cas, il existe un terme de la suite $\\geqslant a_{i}+m$. On peut donc prouver par une très simple récurrence sur $k$ l'existence de $i$ tel que $a_{i} \\geqslant m k$, donc la suite $\\left(a_{i}\\right)$ n'est pas bornée.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un trapèze avec $(AB)$ parallèle à $(CD)$. On suppose qu'il y a deux cercles $\\omega_{1}$ et $\\omega_{2}$ à l'intérieur du trapèze tels que $\\omega_{1}$ est tangent aux côtés $[DA]$, $[AB]$ et $[BC]$ et $\\omega_{2}$ est tangent aux côtés $[BC]$, $[CD]$ et $[DA]$. Soit $(d_{1})$ la seconde tangente (après $(AD)$) à $\\omega_{2}$ passant par $A$, et soit $(d_{2})$ la seconde tangente (après $(BC)$) à $\\omega_{1}$ passant par $C$.\n\nMontrer que $(d_{1})$ et $(d_{2})$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn note $X$ le point d'intersection de $(AD)$ et $(BC)$. On note respectivement $E$ et $F$ les centres de $\\omega_{1}$ et $\\omega_{2}$. Notons que $\\omega_{1}$ est le cercle inscrit à $XAB$ et que $\\omega_{2}$ est le cercle $X$-exinscrit à $XCD$. Enfin, on note $G$ le centre du cercle $X$-exinscrit à $XAB$.\n\nOn vérifie que les triangles $XAG$ et $XEB$ sont semblables. En effet, on a $\\widehat{AXG} = \\widehat{EXB} = \\frac{1}{2} \\widehat{AXB}$. De plus, on a $\\widehat{GAX} = 90^{\\circ} + \\frac{1}{2} \\widehat{BAX}$ et $\\widehat{BEX} = 180^{\\circ} - \\frac{1}{2} \\widehat{AXB} - \\widehat{ABX} = 90^{\\circ} + \\frac{1}{2} \\widehat{BAX}$, donc $XAG$ et $XEB$ sont bien semblables.\n\nSoit $s$ la similitude (directe) qui envoie $X$ sur $X$, $A$ sur $E$ et $G$ sur $B$. D'après l'homothétie décrite précédemment, on a $\\frac{XC}{XB} = \\frac{XF}{XG}$, donc $s(F) = C$.\n\nPar similitude, on en déduit $\\widehat{XAF} = \\widehat{XEC}$. La fin de l'exercice est maintenant une simple chasse aux angles, qui peut être menée de nombreuses manières différentes. Notons par exemple $Y$ l'intersection de $(d_{2})$ avec $(AD)$. Montrer que $(d_{1})$ et $(d_{2})$ sont parallèles revient à montrer $\\widehat{XYC} = \\widehat{XAd_{1}}$. Or, on a\n$$\n\\widehat{XYC} = 180^{\\circ} - \\widehat{AXB} - \\widehat{XCY} = \\widehat{BCY} - \\widehat{AXB} = 2\\widehat{BCE} - \\widehat{AXB}.\n$$\nDe plus, on peut écrire\n$$\n\\widehat{BCE} = 180^{\\circ} - \\widehat{XCE} = \\frac{1}{2} \\widehat{AXB} + \\widehat{XEC} = \\frac{1}{2} \\widehat{AXB} + \\widehat{XAF}\n$$\nce qui donne $\\widehat{XYC} = 2\\widehat{XAF} = \\widehat{XAd_{1}}$, d'où finalement le résultat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $S=\\{1, \\ldots, n\\}$, avec $n \\geqslant 3$ un entier, et soit $k$ un entier strictement positif. On note $S^{k}$ l'ensemble des $k$-uplets d'éléments de $S$. Soit $f: S^{k} \\rightarrow S$ telle que, si $x=\\left(x_{1}, \\ldots, x_{k}\\right) \\in S^{k}$ et $y=\\left(y_{1}, \\ldots, y_{k}\\right) \\in S^{k}$ avec $x_{i} \\neq y_{i}$ pour tout $1 \\leqslant i \\leqslant k$, alors $f(x) \\neq f(y)$.\n\nMontrer qu'il existe $\\ell$ avec $1 \\leqslant \\ell \\leqslant k$ et une fonction $g: S \\rightarrow S$ vérifiant, pour tous $x_{1}, \\ldots, x_{k} \\in S$, $f\\left(x_{1}, \\ldots, x_{k}\\right)=g\\left(x_{\\ell}\\right)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous montrerons le résultat par récurrence sur $k$. Le cas $k=1$ est trivial, supposons donc le résultat vrai pour $k-1 \\geqslant 1$ et montrons-le pour $k$.\n\nSupposons l'existence de $k-1$ éléments $a_{2}, \\ldots, a_{k}$ de $S$ tels que la fonction $\\varphi: a \\in S \\mapsto f\\left(a, a_{2}, \\ldots, a_{k}\\right) \\in S$ est injective. Par égalité de cardinal, elle est aussi bijective.\n\nDès lors, si $b_{2}, \\ldots, b_{k}$ sont des éléments de $S$ avec $b_{i} \\neq a_{i}$ pour tout $i \\in \\{2, \\ldots, k\\}$, et $b \\in S$, alors $\\varphi(a) \\neq f\\left(b, b_{2}, \\ldots, b_{k}\\right)$ pour $S \\ni a \\neq b$. Par surjectivité de $\\varphi$, $\\varphi(b)=f\\left(b, b_{2}, \\ldots, b_{k}\\right)$.\n\nSoient $c_{2}, \\ldots, c_{k}$ des éléments de $S$; puisque $n \\geqslant 3$, il existe $b_{2}, \\ldots, b_{k}$ tels que $a_{i} \\neq b_{i} \\neq c_{i}$ pour tout $i \\in \\{2, \\ldots, k\\}$. Dès lors, le raisonnement précédent montre que, si $b \\in S$, $\\varphi(b)=f\\left(b, b_{2}, \\ldots, b_{k}\\right)=f\\left(b, c_{2}, \\ldots, c_{k}\\right)$, et ainsi $\\ell=1$, et $g=\\varphi$ conviennent.\n\nNous supposons donc qu'il existe deux fonctions $\\alpha, \\beta: S^{k-1} \\rightarrow S$ avec, pour tous $a_{2}, \\ldots, a_{k}$ dans $S$, $\\alpha=\\alpha\\left(a_{2}, \\ldots, a_{k}\\right) \\neq \\beta\\left(a_{2}, \\ldots, a_{k}\\right)=\\beta$, et $f\\left(\\alpha, a_{2}, \\ldots, a_{k}\\right)=f\\left(\\beta, a_{2}, \\ldots, a_{k}\\right)$.\n\nMontrons que $f':\\left(a_{2}, \\ldots, a_{k}\\right) \\in S^{k-1} \\mapsto f\\left(\\alpha, a_{2}, \\ldots, a_{k}\\right)=f\\left(\\beta, a_{2}, \\ldots, a_{k}\\right)$ satisfait les conditions du problème. En effet, si $\\left(a_{2}, \\ldots, a_{k}\\right)$ et $\\left(b_{2}, \\ldots, b_{k}\\right)$ sont deux $(k-1)$-uplets dont les coordonnées sont toutes différentes, alors soit $\\alpha=\\alpha\\left(a_{2}, \\ldots, a_{k}\\right) \\neq \\alpha\\left(b_{2}, \\ldots, b_{k}\\right)=\\alpha'$, auquel cas $g\\left(a_{2}, \\ldots, a_{k}\\right)=f\\left(\\alpha, a_{2}, \\ldots, a_{k}\\right) \\neq f\\left(\\alpha', b_{2}, \\ldots, b_{k}\\right)=g\\left(b_{2}, \\ldots, b_{k}\\right)$ par hypothèse, soit $\\alpha \\neq \\beta\\left(b_{2}, \\ldots, b_{k}\\right)$ auquel cas on a de même $g\\left(a_{2}, \\ldots, a_{k}\\right) \\neq g\\left(b_{2}, \\ldots, b_{k}\\right)$.\n\nDès lors, par hypothèse de récurrence, et sans perte de généralité, on peut supposer l'existence de $h: S \\rightarrow S$ telle que $g\\left(a_{2}, \\ldots, a_{k}\\right)=h\\left(a_{2}\\right)$ pour $a_{2}, \\ldots, a_{k}$ dans $S$. $h$ doit être injective car $h(a)=g(a, a, \\ldots, a) \\neq g(b, \\ldots, b)=h(b)$ si $a \\neq b$ sont des éléments de $S$. Par égalité de cardinal, $h$ est surjective.\n\nMontrons que $f\\left(a_{1}, \\ldots, a_{k}\\right)=h\\left(a_{2}\\right)$ pour tous $a_{1}, \\ldots, a_{k} \\in S$, ce qui conclura. Supposons par l'absurde l'existence d'un $k$-uplet $a=\\left(a_{1}, \\ldots, a_{k}\\right) \\in S^{k}$ tel que $f(a) \\neq h\\left(a_{2}\\right)$. Par surjectivité, il existe $b_{2} \\in S$ avec $h\\left(b_{2}\\right)=f(a)$ avec $b_{2} \\neq a_{2}$ donc. Soient $b_{i} \\neq a_{i}$ des éléments de $S$, pour $3 \\leqslant i \\leqslant k$. On a $\\alpha=\\alpha\\left(b_{2}, \\ldots, b_{k}\\right)$ et $\\beta=\\beta\\left(b_{2}, \\ldots, b_{k}\\right)$ deux éléments de $S$ tels que $f\\left(\\alpha, b_{2}, \\ldots, b_{k}\\right)=f\\left(\\beta, b_{2}, \\ldots, b_{k}\\right)=h\\left(b_{2}\\right)=f(a)$. L'hypothèse faite sur $f$ assure donc $\\alpha=a_{1}=\\beta$, ce qui est une contradiction d'après la définition de $\\alpha$ et $\\beta$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30674,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $x$ et $y$ deux nombres réels. On pose\n$$\nM=\\max \\{x y+1, x y-x-y+3,-2 x y+x+y+2\\} .\n$$\nDémontrer que $M \\geqslant 2$, et déterminer les cas d'égalité.",
    "options": [],
    "answer": "M ≥ 2, with equality if and only if x = y = 1",
    "solution": "Solution:\n\nParmi les trois nombres $x y+1$, $x y-x-y+3$ et $-2 x y+x+y+2$, on note $K$ le plus petit, $L$ le deuxième plus petit et $M$ le plus grand. Alors $K \\leqslant L \\leqslant M$ donc\n$$\n3 M \\geqslant K+L+M=(x y+1)+(x y-x-y+3)+(-2 x y+x+y+2)=6,\n$$\nce qui signifie que $M \\geqslant 2$.\n\nEn outre, si $M=2$, les inégalités $K \\leqslant L \\leqslant M$ sont en fait des égalités, ce qui signifie que\n$$\nx y+1=x y-x-y+3=-2 x y+x+y+2=2\n$$\nEn notant $p$ le produit $x y$ et $s$ la somme $x+y$, les égalités de l'équation (1) sont vérifiées si et seulement si $p=1$ et $s=2$.\n\nOr, l'inégalité arithmético-géométrique indique, de manière générale, que\n$$\n\\frac{s^{2}}{4}=\\left(\\frac{x+y}{2}\\right)^{2} \\geqslant x y=p\n$$\nav ec égalité si et seulement si $x=y$. Ce résultat se retrouve bien si l'on remarque que\n$$\n\\left(\\frac{x+y}{2}\\right)^{2}=\\left(\\frac{x-y}{2}\\right)^{2}+x y\n$$\nIci, puisque l'on souhaite avoir $s^{2} / 4=1=p$, il nous faut donc également avoir $x=y$, de sorte que $x=y=1$.\n\nEnfin, réciproquement, on vérifie aisément, si $x=y=1$, que\n$$\nx y+1=x y-x-y+3=-2 x y+x+y+2=2\n$$\nde sorte que $M=2$ également.\n\n\nAlternative 1:\n\nVoici une autre manière de démontrer que les égalités de l'équation (1) sont vérifiées si et seulement si $x=y=1$. Tout d'abord, si $x=y=1$, ces égalités sont clairement vérifiées.\n\nRéciproquement, si elles sont vérifiées, alors on peut réécrire l'égalité $x y-x-y+3=2$ comme $0=x y-x-y+1=(x-1)(y-1)$. Puisque $x$ et $y$ jouent des rôles symétriques, on peut supposer, sans perte de généralité, que $x=1$. Pour que l'égalité $x y+1=2$ soit vérifiée, il est alors nécessaire que $y=1$, ce qui conclut ici.\n\n\nAlternative 2:\n\nAvoir pensé à introduire les variables $p$ et $s$, comme dans la première solution, ou bien à utiliser directement l'égalité $x y-x-y+3=2$ pour obtenir une factorisation, comme dans la deuxième solution, pourrait passer pour miraculeux. Il n'en est bien sûr rien, et ces approches découlent en fait de ce que l'on appelle les relations de Viète.\n\nCes relations consistent à dire que $x$ et $y$ sont les deux racines, éventuellement confondues, du polynôme\n$$\nP(X)=(X-x)(X-y)=X^{2}-(x+y) X+x y=X^{2}-s X+p .\n$$\nPuisque les égalités de l'équation (1) sont symétriques en $x$ et en $y$, un résultat général (valable même si on avait eu trois inconnues $x, y$ et $z$, ou même un nombre quelconque d'inconnues) permet en fait de démontrer qu'elles peuvent s'exprimer directement à partir des valeurs que prend le polynôme $P$.\n\nPar exemple, les égalités $p+1=2$, $p-s+3=2$ et $-2 p+s+2=2$ peuvent se réécrire comme $P(0)=1$, $P(1)=0$ et $P(1 / 2)=1 / 4$. Ici, l'égalité $P(1)=0$ signifie précisément que $(x-1)(y-1)=0$, ce qui est la factorisation utilisée dans notre deuxième solution.\n\nOn pourrait également procéder de manière plus systématique. En effet, puisque $P(1)=0$, on sait que le polynôme $P(X)$ est divisible par $X-1$. On peut donc l'écrire sous la forme $P(X)=(X-1)(X-a)$, puis vérifier que $P(0)=a$, et donc que $P(X)=(X-1)^{2}$.\n\nSi l'on avait été moins chanceux, et que l'on n'avait pas identifié de racine de notre polynôme $P$, il aurait toujours été possible d'écrire $P$ comme un polynôme d'interpolation de Lagrange. Ici, cette méthode nous permet d'écrire que\n$$\n\\begin{aligned}\nP(X) & =\\frac{X-1}{0-1} \\frac{X-1 / 2}{0-1 / 2} P(0)+\\frac{X-1}{1 / 2-1} \\frac{X-0}{1 / 2-0} P(1 / 2)+\\frac{X-1 / 2}{1-1 / 2} \\frac{X-0}{1-0} P(1) \\\\\n& =2(X-1)(X-1 / 2)-X(X-1)=X^{2}-2 X+1=(X-1)^{2}\n\\end{aligned}\n$$\nd'où le fait que les deux racines de $P$ soient $x=y=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nClara et Isabelle jouent au jeu suivant. Au début du jeu, elles choisissent un entier $n \\geqslant 1$, puis mettent $n$ bonbons dans un saladier. Puis elles jouent à tour de rôle, en commençant par Clara. À chaque tour, si le saladier contient $k$ bonbons, la joueuse dont c'est le tour choisit un entier $\\ell$ premier avec $k$, tel que $\\ell \\leqslant k$, puis elle mange $\\ell$ bonbons; c'est alors à l'autre joueuse de jouer. La joueuse qui mange le dernier bonbon gagne la partie.\nPour quelles valeurs de $n$ Clara dispose-t-elle d'une stratégie gagnante?",
    "options": [],
    "answer": "all odd positive integers n",
    "solution": "Solution:\n\nTout d'abord, si $n$ est impair, Clara peut commencer par choisir $\\ell = n-2$. En effet, puisque $n$ est impair, on sait que $\\operatorname{PGCD}(n, n-2) = \\operatorname{PGCD}(n, 2) = 1$. Mais alors Isabelle se retrouve avec deux bonbons, elle est obligée d'en manger un seul, et laisse le dernier bonbon à Clara, qui gagne donc.\n\nAu contraire, si $n$ est pair, Clara est obligée de choisir un entier $\\ell$ impair, et elle laisse donc $k = n - \\ell$ bonbons à Isabelle, l'entier $k$ étant donc impair. Si $k = 1$, Isabelle a déjà gagné, et si $k \\geqslant 3$, Isabelle peut copier la stratégie présentée ci-dessus, et manger $k-2$ bonbons, s'assurant ainsi la victoire.\n\nLes entiers $n$ recherchés sont donc les entiers impairs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUne grille de dimensions $20 \\times 20$ est divisée en 400 cases unité de dimensions $1 \\times 1$. Clara colorie chaque case en blanc ou en noir, puis Isabelle découpe la grille en rectangles dont les côtés sont contenus dans la grille. Chacun de ces rectangles doit contenir au plus 2 cases noires, et elle donne un chocolat à Clara à chaque fois qu'elle découpe un rectangle qui contient 0 ou 1 case noire.\n\nIsabelle choisit son découpage de manière à donner le moins de chocolats possibles à Clara, et Clara choisit son coloriage initial de manière à obtenir le plus de chocolats possibles d'Isabelle. Combien de chocolats Isabelle donnera-t-elle à Clara?",
    "options": [],
    "answer": "20",
    "solution": "Solution:\n\nNous allons démontrer pour tout entier $n \\geqslant 3$ que, si Clara et Isabelle jouent sur une grille de dimensions $n \\times n$, Isabelle donnera $n$ chocolats à Clara.\n\nTout d'abord, pour limiter le nombre de chocolats qu'elle donnera à Clara, Isabelle peut procéder comme suit. Elle commence par découper le rectangle en $n$ lignes, et s'apprête à subdiviser chaque ligne en rectangles. Pour ce faire, elle numérote les cases noires de la ligne de gauche à droite, puis elle donne un coup de ciseaux juste à droite de chacune des cases noires $2,4,6, \\ldots$ Ce faisant, tous les rectangles obtenus, sauf éventuellement le dernier, comptent deux cases noires. Ainsi, Isabelle donne au plus $n$ chocolats à Clara.\n\n![](attached_image_1.png)\n\nRéciproquement, supposons que Clara colorie sa grille $n \\times n$ avec le coloriage suivant, où nous avons distingué avec deux niveaux de gris les cases que Clara colorie en noir, et où $k$ et $\\ell$ sont deux entiers naturels non nuls de somme $k+\\ell=n-1$.\n\n![](attached_image_2.png)\n\nSi un rectangle contient exactement deux cases noires, il s'agit de cases adjacentes : une gris foncé et une gris clair. Or, il y a $(k+1)(\\ell+1)$ cases gris foncé et $k \\ell$ cases gris clair. Isabelle dessine donc au plus $k \\ell$ rectangles recouvrant exactement deux cases noires, et ces rectangles laissent de côté $k+\\ell+1=n$ cases gris foncé. Cela force donc Isabelle à donner au moins $n=20$ chocolats à Clara.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMartin et Théo jouent à un jeu : Martin écrit un nombre entier au tableau. Théo a ensuite le droit d'effacer le nombre et de lui ajouter $2$, ou d'effacer le nombre et de lui enlever $3$, ceci autant de fois qu'il veut. Théo gagne s'il arrive à obtenir $2020$ après un nombre fini d'étapes, sinon Martin gagne. Quel joueur a une stratégie gagnante ?",
    "options": [],
    "answer": "Théo",
    "solution": "Solution:\n\nOn commence par tester l'énoncé pour des entiers proches de $2020$. Par exemple, on remarque que si Martin écrit les entiers $2018$, $2016$, ... Théo peut écrire $2020$ au tableau en rajoutant au nombre écrit par Martin un nombre suffisamment grand de $2$. On remarque aussi que si Martin écrit $2019$, Théo peut aussi gagner en remplaçant $2019$ par $2021$ puis $2023$ puis $2023$ par $2020$.\n\nCette remarque nous permet de conjecturer d'une part que Théo peut toujours gagner quel que soit le nombre écrit par Martin, d'autre part que si un nombre $k$ est écrit au tableau, Théo peut écrire les nombres $k+1$ ou $k-1$ au bout d'un nombre fini d'opérations.\n\nNous montrons donc que Théo peut toujours gagner quel que soit le nombre écrit par Martin. Supposons qu'à un instant donné, le nombre $k$ est écrit au tableau. Alors en remplaçant $k$ par $k+2$ en ajoutant $2$, $k+2$ par $k+4$ en ajoutant $2$ et enfin $k+4$ par $k+1$ en retranchant $3$, Théo a remplacé le nombre $k$ par le nombre $k+1$ en un nombre fini d'opérations. Nous appellerons cette transformation l'opération A. D'autre part, toujours si l'entier $k$ est écrit au tableau à un instant donné, en remplaçant $k$ par $k+2$ et $k+2$ par $k-1$, Théo a remplacé l'entier $k$ par l'entier $k-1$. Nous appellerons cette transformation l'opération B.\n\nSupposons désormais que Martin a écrit le nombre $k_{0}$ au tableau. Si $k_{0}=2020$, Théo a déjà gagné. Si $k_{0}<2020$, alors en effectuant $2020-k_{0}$ fois l'opération A, Théo peut remplacer l'entier $k_{0}$ par $2020$ et gagner. Si $k_{0}>2020$, en effectuant $k_{0}-2020$ fois l'opération B, Théo peut remplacer à nouveau l'entier $k_{0}$ par $2020$. Dans tous les cas, Théo gagne !",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30678,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDe combien de façons peut-on placer 7 tours sur un échiquier $7 \\times 7$ telle qu'aucune tour ne puisse en attaquer une autre?\n\nUne tour peut attaquer une autre tour si elle se situe sur la même ligne ou la même colonne.",
    "options": [],
    "answer": "5040",
    "solution": "Solution:\n\nSi on place 7 tours de sorte que deux tours ne soient pas sur la même colonne, sachant qu'il y a 7 colonnes, alors il y aura exactement une tour sur chaque colonne (et de même sur chaque ligne).\n\nIl y a 7 positions possibles pour placer une tour sur la première colonne. Il y a ensuite 6 positions possibles pour placer une tour sur la seconde colonne sans qu'elle soit sur la même ligne que la première tour. Ainsi de suite, sur la $k$-ième colonne il y a $8-k$ manières de placer une tour sans qu'elle soit sur la même ligne qu'une tour déjà placée.\n\nAinsi il y a $7! = 7 \\times 6 \\times 5 \\times 4 \\times 3 \\times 2 \\times 1 = 5040$ manières de disposer les tours de cette manière.\n\nSi on note $\\sigma(i)$ le numéro de ligne de la tour présente sur la $i$-ième colonne dans une des configurations satisfaisant l'énoncé, alors $\\sigma$ est ce que l'on appelle une permutation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a_{1}, a_{2}, \\ldots$ la suite d'entiers telle que $a_{1}=1$ et, pour tout entier $n \\geqslant 1$,\n$$\na_{n+1}=a_{n}^{2}+a_{n}+1\n$$\nDémontrer, pour tout entier $n \\geqslant 1$, que $a_{n}^{2}+1$ divise $a_{n+1}^{2}+1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour tout entier $n \\geqslant 1$, posons $b_{n}=a_{n}^{2}+1$. On conclut en constatant directement que\n$$\nb_{n+1} \\equiv\\left(a_{n}^{2}+a_{n}+1\\right)^{2}+1 \\equiv\\left(b_{n}+a_{n}\\right)^{2}+1 \\equiv a_{n}^{2}+1 \\equiv b_{n} \\equiv 0 \\quad\\left(\\bmod b_{n}\\right)\n$$\n\n\nSolution alternative 1\n\nIl suffit de constater, pour tout entier $n \\geqslant 1$, que\n$$\n\\begin{aligned}\na_{n+1}^{2}+1 & =\\left(a_{n}^{2}+a_{n}+1\\right)^{2}+1 \\\\\n& =a_{n}^{4}+2 a_{n}^{3}+3 a_{n}^{2}+2 a_{n}+2 \\\\\n& =a_{n}^{2}\\left(a_{n}^{2}+1\\right)+2 a_{n}^{3}+2 a_{n}^{2}+2 a_{n}+2 \\\\\n& =a_{n}^{2}\\left(a_{n}^{2}+1\\right)+2 a_{n}\\left(a_{n}^{2}+1\\right)+2 a_{n}^{2}+2 \\\\\n& =a_{n}^{2}\\left(a_{n}^{2}+1\\right)+2 a_{n}\\left(a_{n}^{2}+1\\right)+2\\left(a_{n}^{2}+1\\right) \\\\\n& =\\left(a_{n}^{2}+2 a_{n}+2\\right)\\left(a_{n}^{2}+1\\right)\n\\end{aligned}\n$$\nest effectivement un multiple de $a_{n}^{2}+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30680,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\left(x_{n}\\right)_{n \\in \\mathbb{N}}$ une suite réelle telle que $x_{0}=0$ et $x_{1}=1$ et pour tout $n \\geqslant 0$, $x_{n+2}=3 x_{n+1}-2 x_{n}$. On pose également $y_{n}=x_{n}^{2}+2^{n+2}$ pour tout entier naturel $n$. Montrer que pour tout entier $n>0$, $y_{n}$ est le carré d'un entier impair.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par tester l'énoncé pour des petites valeurs de $n$. On calcule donc les premières valeurs des suites $\\left(x_{n}\\right)$ et $\\left(y_{n}\\right)$. On trouve par exemple $x_{2}=3$, $x_{3}=7$, $x_{4}=15$, $\\ldots$ On peut donc conjecturer que $x_{n}=2^{n}-1$. On s'empresse de le démontrer par récurrence.\n\nLe résultat est déjà au rang $0$ et $1$. Si on suppose $x_{n}=2^{n}-1$ et $x_{n+1}=2^{n+1}-1$, on a\n$$\nx_{n+2}=3 x_{n+1}-2 x_{n}=3 \\times 2^{n+1}-3-2 \\times 2^{n}+2=(3-1) \\times 2^{n+1}-1=2^{n+2}-1\n$$\nce qui achève la récurrence.\n\nOn injecte alors cette expression de $x_{n}$ dans l'expression de $y_{n}$. Il ne nous reste plus qu'à manipuler algébriquement cette nouvelle expression :\n$$\ny_{n}=\\left(2^{n}-1\\right)^{2}+2^{n+2}=2^{2 n}-2^{n+1}+1+2^{n+2}=2^{2 n}-2^{n+1}+1+2 \\times 2^{n+1}=2^{2 n}+2^{n+1}+1=\\left(2^{n}+1\\right)^{2}\n$$\nCeci montre bien que $y_{n}$ est le carré d'un entier impair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLors d'une fête, 2019 personnes s'assoient autour d'une table ronde, en se répartissant de façon régulière. Après s'être assises, elles constatent qu'un carton indiquant un nom est posé à chacune des places et que personne n'est assis à la place où figure son nom. Montrer qu'on peut tourner la table de telle sorte que deux personnes se retrouvent assises en face de leur nom.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par tester l'énoncé sur des valeurs plus petites, par exemple pour une fête de 5 personnes.\n\nEnsuite, on essaye de coder l'information. On appelle rotation une configuration obtenue après avoir tourné la table depuis sa position d'origine. À une rotation $r$, on associe $n(r)$ le nombre de personnes assises face à leur nom après la rotation.\n\nComme pour toute personne il existe une unique rotation qui la rend assise face à son nom, $\\sum_{r=0}^{2018} n(r) = 2019$. Or il y a 2019 rotations et la rotation nulle vérifie $n(r) = 0$. Donc comme 2018 entiers positifs ont pour somme 2019, l'un d'eux vaut donc au moins 2 par principe des tiroirs, ce qui signifie bien qu'une rotation amène deux noms en face des bonnes personnes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver toutes les fonctions $f$ de $\\mathbb{R}$ dans $\\mathbb{R}$ telles que pour tout couple $(x, y)$ de réels :\n$$\nf(f(x))+f(f(y))=2 y+f(x-y)\n$$",
    "options": [],
    "answer": "f(x) = x",
    "solution": "Solution:\nAnalysons le problème : ici on est face à une équation fonctionnelle, avec deux variables. La première chose à faire est d'essayer les quelques substitutions classiques : $x=y=0$, $x=0$, $y=0$, $x=y$. Ici comme on a un $f(x-y)$, il est très tentant de regarder ce que ça donne pour $x=y$. Posons donc $C=f(0)$, pour $x=y$, on obtient $2 f(f(x))=2 x+C$ soit $f(f(x))=x+\\frac{C}{2}$.\n\nMaintenant qu'on a une expression plus maniable de $f(f(x))$, on peut la réinjecter dans l'équation et regarder ce que ça donne.\n\nEn réinjectant dans l'équation initiale, on a donc $x+y+C=2 y+f(x-y)$ soit $f(x-y)=x-y+C$, pour $y=0$ on a donc $f(x)=x+C$ pour tout réel $x$. Maintenant, il faudrait déterminer $C$ avant d'effectuer la vérification, on essaie donc de voir quelle contrainte l'équation $f(f(x))=x+\\frac{C}{2}$ donne sur $C$.\n\nEn particulier $f(f(x))=f(x+C)=x+2 C$ donc $2 C=\\frac{C}{2}$ donc $4 C=C$, $3 C=0$ donc $C=0$, on a donc $f(x)=x$ pour tout réel $x$. Maintenant on n'oublie pas de vérifier que la fonction trouvée est bien solution ! Réciproquement la fonction identité convient car dans ce cas, $f(f(x))+f(f(y))=f(x)+f(y)=x+y=2 y+x-y=2 y+f(x-y)$ pour tout $x, y$ réels.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les couples d'entiers positifs $(x, y)$ tels que $2^{x}+5^{y}+2$ est un carré parfait.",
    "options": [],
    "answer": "((0, 0), (1, 1))",
    "solution": "Solution:\n\nIci on est face à un problème d'équation diophantienne avec un carré et une puissance de $2$. On peut se rendre compte en testant les petits cas que $(x, y) = (0, 0)$ et $(1, 1)$ sont solutions. Comme on a un carré et une puissance de $2$, on est très tenté de regarder modulo $4$ ou $8$. Regardons déjà modulo $4$ pour voir s'il y a une contradiction. Pour cela on va d'abord traiter le cas où $x \\geqslant 2$.\n\nDéjà notons que si $x \\geqslant 2$, $2^{x} + 5^{y} + 2 \\equiv 1 + 2 \\equiv 3 \\pmod{4}$ et $3$ n'est pas un carré modulo $4$ (les carrés sont $0$ et $1$ modulo $4$). On a donc forcément $x = 1$ ou $x = 0$.\n\nMaintenant réécrivons l'équation dans le cas $x = 1$. Pour $x = 1$, soit $y$ entier positif tel que $2 + 5^{y} + 2 = 5^{y} + 4$ est un carré parfait. Notons que comme $4$ est un carré, on va pouvoir factoriser.\n\nSoit $k$ entier positif tel que $k^{2} = 5^{y} + 4$, on a donc $(k-2)(k+2) = 5^{y}$. Notons que comme $k+2 > 0$, on a $k-2 > 0$ et $k+2$ et $k-2$ sont des puissances de $5$. Si $k-2 > 1$, alors $5$ divise $k+2$ et $k-2$ donc $5$ divise $k+2 - (k-2) = 4$, contradiction.\n\nAinsi $k-2 = 1$ donc $k = 3$, donc $5^{y} + 4 = 9$ donc $y = 1$. Réciproquement si $(x, y) = (1, 1)$, $2^{x} + 5^{y} + 2 = 9$ est un carré parfait.\n\nPour $x = 0$, soit $y$ tel que $1 + 5^{y} + 2 = 5^{y} + 3$ est un carré, soit $k \\geqslant 0$ tel que $k^{2} = 5^{y} + 3$. Ici comme on a des carrés et des puissances de $5$, on peut essayer de regarder modulo $5$. Si $y \\geqslant 1$, alors on a $k^{2} \\equiv 3 \\pmod{5}$, or les carrés modulo $5$ sont $1$ et $4$, ce qui est une contradiction. On a donc forcément $y = 0$.\n\nRéciproquement si $(x, y) = (0, 0)$, $2^{x} + 5^{y} + 2 = 4$ est un carré parfait.\n\nLes solutions sont donc les couples $(0, 0)$ et $(1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle aux angles aigus. Soit $D$ le pied de la bissectrice de l'angle $\\widehat{BAC}$. La perpendiculaire à la droite $(AD)$ passant par le point $B$ recoupe le cercle circonscrit au triangle $ADB$ en un point $E$. Soit $O$ le centre du cercle circonscrit au triangle $ABC$. Montrer que les points $A$, $O$ et $E$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nExaminons les angles que nous pouvons obtenir facilement. Puisque les points $A$, $E$, $D$ et $B$ sont sur un même cercle, on doit pouvoir utiliser le théorème de l'angle inscrit pour avoir une égalité portant sur $\\widehat{EAD}$. Puisque la droite $(BE)$ est perpendiculaire à la droite $(AD)$, on doit pouvoir calculer l'angle $\\widehat{EBA}$. Enfin, on connaît bien sûr l'angle $\\widehat{OAB}$. Il semble donc raisonnable de montrer l'alignement des points $A$, $O$ et $E$ en montrant que $\\widehat{OAB} = \\widehat{EAB}$. On calcule séparément ces deux angles en fonction des angles du triangle $ABC$.\n\nD'une part, d'après le théorème de l'angle au centre et puisque le triangle $AOB$ est isocèle en $O$, on a\n$$\n\\widehat{OAB} = \\frac{1}{2}\\left(180^\\circ - \\widehat{AOB}\\right) = 90^\\circ - \\widehat{ACB}\n$$\n\nD'autre part, $\\widehat{EAB} = \\widehat{EAD} + \\widehat{DAB}$. On utilise alors le théorème de l'angle inscrit et le fait que $\\widehat{DAB} = \\frac{1}{2} \\widehat{BAC}$, on obtient :\n$$\n\\widehat{EAD} + \\widehat{DAB} = \\widehat{EBD} + \\frac{1}{2} \\widehat{BAC} = 90^\\circ - \\widehat{BDA} + \\frac{1}{2} \\widehat{BAC}\n$$\nOr la somme des angles du triangle $ADC$ fait $180^\\circ$ donc $\\widehat{BDA} = \\widehat{DCA} + \\widehat{DAC} = \\widehat{ACB} + \\frac{1}{2} \\widehat{BAC}$. Ainsi\n$$\n90^\\circ - \\widehat{BDA} + \\frac{1}{2} \\widehat{BAC} = 90^\\circ - \\left(\\widehat{ACB} + \\frac{1}{2} \\widehat{BAC}\\right) + \\frac{1}{2} \\widehat{BAC} = 90^\\circ - \\widehat{ACB}\n$$\n\net on trouve bien que $\\widehat{OAB} = 90^\\circ - \\widehat{ACB} = \\widehat{EAB}$ donc les points $A$, $O$ et $E$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $N > 1$. Alice et Bob jouent au jeu suivant. $2N$ cartes numérotées de $1$ à $2N$ sont mélangées puis disposées dans une ligne, de manière à ce que les faces numérotées soient visibles. Chacun à leur tour, Alice et Bob choisissent une carte, soit celle tout à droite soit celle tout à gauche de la ligne, et la garde pour eux, jusqu'à ce que toutes les cartes aient été prises. Alice commence. À la fin du jeu, chaque joueur calcule la somme des numéros des cartes qu'il a prises. Le joueur ayant la plus grande somme gagne. Un joueur dispose-t-il d'une façon de ne pas perdre?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par tester le jeu pour des petites valeurs, par exemple pour un jeu composé de $2 \\times 3$ cartes. Après plusieurs essais, on s'aperçoit par exemple que la stratégie qui consiste à prendre toujours la plus grande carte parmi la carte située la plus à gauche et la carte située la plus à droite ne fonctionne pas toujours, pour aucun des deux joueurs.\n\nTestons le cas $N = 2$. Chacun des deux joueurs ne pourra prendre que deux cartes. Dans ce type de problème, une première idée est d'essayer de chercher une stratégie qui permet à un joueur de jouer comme il le souhaite, quel que soient les coups de l'adversaire. Regardons donc les cartes qu'Alice et Bob peuvent s'assurer d'obtenir. On note $X Y Z W$ les cartes étalées sur la rangée dans cet ordre. Supposons qu'Alice prenne la carte $X$. Si Bob choisit la carte $Y$, Alice a le choix entre les cartes $Z$ et $W$. Si Bob choisit la carte $W$, Alice a le choix entre les cartes $Z$ et $Y$. Ainsi, en choisissant la carte $X$, Alice s'assure de pouvoir prendre la carte $Z$ si elle le souhaite. En revanche, avant le premier coup d'Alice, Bob ne peut pas garantir de prendre une des quatre cartes.\n\nOn désire désormais généraliser ce processus. On regarde alors le cas $N = 3$. Les cartes sont $X Y Z W U V$. On s'aperçoit alors que si Alice prend la carte $X$ au premier tour, elle s'assure de pouvoir prendre les cartes $Z$ et $U$ si elle le veut. Ces remarques sont suffisantes pour penser qu'Alice dispose d'une stratégie gagnante en récupérant les cartes $X, Z, U$ si $X + Z + U > Y + W + V$ et en récupérant les cartes $Y, W, V$ dans le cas contraire. On prouve à présent que cette stratégie fonctionne dans le cas général.\n\nAlice colorie une carte sur deux en noir (la carte toute à droite est blanche, celle toute à gauche est noire). Elle calcule la somme des numéros des cartes blanches et celle des numéros des cartes noires.\n\nSi la somme des blanches est plus grande, elle choisit la carte blanche tout à droite. Bob prendra obligatoirement une carte noire ensuite. À chaque tour, Alice peut choisir entre une carte blanche ou noire et en prenant la blanche, elle ne laisse pas le choix à Bob, qui prendra forcément une carte noire.\n\nAinsi à la fin, Alice aura toutes les cartes blanches et Bob toutes les cartes noires. Donc elle aura un score supérieur ou égal à celui de Bob. Elle procède de même si la somme des cartes noires est plus grande.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUne grille de taille $n \\times n$ contient $n^2$ cases. Chaque case contient un entier naturel compris entre $1$ et $n$, de telle sorte que chaque entier de l'ensemble $\\{1, \\ldots, n\\}$ apparaît exactement $n$ fois dans la grille. Montrer qu'il existe une colonne ou une ligne de la grille contenant au moins $\\sqrt{n}$ nombres différents.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEssayons d'abord de regarder ce qui se passe dans ce qui semble être le pire cas : si chaque ligne/colonne ne contient pas beaucoup de numéros, chaque numéro va y être beaucoup présent. En particulier, pour tout $i$, le nombre de $i$ semble être écrit dans peu de lignes et de colonnes. Mais à priori si le numéro $i$ apparait trop dans une ligne, il apparaitra dans beaucoup de colonnes et inversement. A priori, le pire cas semble être celui où chaque numéro apparaît dans autant de colonnes que de lignes et comme chaque numéro apparaît $n$ fois, le carré doit être de dimension $\\sqrt{n}$, donc $i$ apparait dans au moins $2 \\sqrt{n}$ lignes et colonnes.\n\nBien sûr, cette analyse ne constitue en rien une preuve, mais on voit deux idées importantes : premièrement, le nombre de numéros distincts dans chaque ligne/colonne semble avoir un lien avec le nombre de lignes/colonnes où chaque numéro apparaît. Deuxièmement, pour tout $i$ entre $1$ et $n$, le nombre $i$ apparait au moins sur $2 \\sqrt{n}$ lignes et colonnes. On va donc essayer de formaliser ça.\n\nNotons $a_{c, i}$ pour $c$ une colonne ou une ligne et $i$ un entier entre $1$ et $n$ le nombre de fois qu'apparait $i$ dans $c$, notons $c_{i}$ (resp. $l_{i}$) le nombre de colonnes (resp. lignes) à laquelle $i$ appartient. On note $C$ l'ensemble des colonnes, $L$ celui des lignes.\n\nNotons que l'hypothèse du fait que $i$ apparait $n$ fois se traduit comme $c_{i} \\times l_{i} \\geqslant n$ car $i$ apparait dans au plus $c_{i} l_{i}$ cases.\n\nNotons que pour tout $i$, $l_{i}+c_{i}$ étant le nombre de ligne ou de colonne où apparait $i$, il vaut $\\sum_{c \\in C \\cup L} a_{c, i}$. Le nombre d'éléments distincts dans la colonne $c$ vaut, quant à lui, $\\sum_{i=1}^{n} a_{c, i}$. En particulier, on voit bien que les $a_{c, i}$ jouent un rôle important dans le problème.\n\nMaintenant il reste à relier le nombre d'apparitions de $i$ dans les différentes lignes/colonnes au nombre de numéros dans chaque ligne. Pour cela on va regarder la somme des $a_{c, i}$ (on somme sur $i$ entre $1$ et $n$ et sur $c$ colonne et ligne) et utiliser les deux interprétations précédentes.\n\nOn obtient ainsi $\\sum_{i=1}^{n} \\sum_{c \\in C \\cup L} a_{c, i}=\\sum_{i=1}^{n} c_{i}+l_{i} \\geqslant \\sum_{i=1}^{n} 2 \\sqrt{c_{i} l_{i}} \\geqslant 2 n \\sqrt{n}$.\n\nOn en déduit que $\\sum_{c \\in C \\cup L} \\sum_{i=1}^{n} a_{c, i} \\geqslant \\sum_{i=1}^{n} 2 \\sqrt{c_{i} l_{i}}=2 n \\sqrt{n}$.\n\nEn particulier, comme il y a $2 n$ lignes ou colonnes, il existe $c$ colonnes ou lignes telles que $\\sum_{i=1}^{n} a_{c, i} \\geqslant \\sqrt{n}$, i.e. $c$ contient au moins $\\sqrt{n}$ numéros différents.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30687,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère une grille de taille $2019 \\times 2019$. Sur cette grille sont posés des cailloux. Une configuration est dite belle s'il n'existe pas de parallélogramme formé par quatre cailloux $A B C D$, tels que $A, B, C$, et $D$ ne soient pas tous alignés.\nQuel est le plus grand nombre de cailloux qu'il est possible de mettre dans une grille?",
    "options": [],
    "answer": "4037",
    "solution": "Solution:\n\nDans ce problème, on cherche le plus grand entier satisfaisant une certaine propriété. Supposons que l'on veuille montrer que le plus grand entier recherché est l'entier $c$. Pour montrer que $c$ est bien le plus grand entier, on va d'une part montrer que si un entier $n$ satisfait la propriété, alors $n \\leqslant c$ et d'autre part on va montrer que l'on peut trouver une grille possédant exactement $c$ cailloux satisfaisant la propriété.\n\nOn s'empresse de tester l'énoncé pour des valeurs plus petites, par exemple pour une grille de taille $3 \\times 3$, $4 \\times 4$ ou même $5 \\times 5$, afin de deviner la valeur. On trouve dans ces petits cas que les plus grands entiers sont respectivement $5$, $7$ et $9$. On peut conjecturer que la valeur recherchée sera donc $2 \\times 2019 - 1 = 4037$. En testant ces petites valeurs, on a pu remarquer que la présence de cailloux sur une colonne forçait que certains emplacements soient vides. Nous allons donc essayer de formaliser ce raisonnement.\n\nOn considère les paires de cailloux qui sont sur la même colonne. S'il y a deux telles paires dont les cailloux sont à même distance sur des colonnes différentes alors on obtient un parallélogramme, ce que l'on veut éviter.\n\nSoit $n_{i}$ le nombre de cailloux dans la $i$-ième colonne. Sur la colonne $i$ il y a par conséquent au moins $n_{i}-1$ paires à distances distinctes. Puisque pour deux colonnes fixées, la distance entre deux quelconques cailloux de la première colonne doit être différente de la distance entre deux quelconques cailloux de la deuxième colonne, toutes les distances entre deux paires de cailloux appartenant à la même colonne doivent être distinctes. Cela fait en tout $\\left(n_{1}-1\\right)+\\left(n_{2}-1\\right)+\\cdots+\\left(n_{2019}-1\\right)=n-2019$ distances distinctes avec $n$ le nombre total de cailloux. Or il y a au plus $2018$ distances possibles, puisqu'il y a $2019$ emplacements pour un caillou dans une colonne. Ceci montre que $n \\leqslant 2018+2019=4037$.\n\nRéciproquement, si on remplit chaque case de la première ligne par un caillou et chaque case de la première colonne par un caillou et on laisse les autres cases vides, on a alors placé $2 \\times 2019-1=4037$ cailloux sans créer de parallélogrammes.\n\nLe plus grand entier recherché est donc $4037$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn lycée comporte un nombre impair de classes, et, dans chaque classe, un nombre impair d'élèves. Un élève est choisi dans chaque classe pour faire partie du comité des élèves. Si le nombre de classes avec le plus de garçons que de filles est impair, montrer que le nombre de façons de former un comité des élèves contenant un nombre impair de garçons excède le nombre de façons de former un comité d'élèves contenant un nombre impair de filles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEn l'absence d'idées face au problème, on peut essayer de démontrer le résultat par récurrence. Pour regarder si cela pourrait fonctionner, on peut regarder ce qu'il advient du résultat si l'on passe d'un lycée avec 3 élèves à un lycée avec 4 élèves. On vérifie également que les hypothèses de l'énoncé sont adaptées à une récurrence. On s'aperçoit que nous allons devoir distinguer plusieurs cas, selon la répartition garçons/filles dans les différentes classes. On démarre donc la récurrence.\n\nNous allons montrer le résultat par récurrence forte sur le nombre d'élèves au total. S'il n'y a aucun élève, alors le résultat est vrai, de même s'il n'y a qu'un élève dans le lycée.\n\nSupposons qu'il y ait $n$ élèves en tout dans des classes satisfaisant l'énoncé. Supposons que l'énoncé de l'exercice est vrai dès lors qu'il y a strictement moins que $n$ élèves. Il y a deux cas à étudier :\n\n- S'il n'y a de classe avec une fille et un garçon, alors il y a un nombre impair de classes avec que des garçons et les autres classes n'ont que des filles, donc le comité aura nécessairement un nombre impair de garçons, l'énoncé est vérifié.\n\n- S'il existe une classe avec une fille et un garçon, appelons-les respectivement Alice et Bob. Si on enlève Alice et Bob de la classe alors il y a plus de manières de choisir le comité avec un nombre impair de garçons. Si on considère uniquement les comités avec soit Alice soit Bob, alors on a autant de comités avec un nombre impair de garçons qu'avec un nombre impair de filles : en effet, échanger Alice et Bob change la parité du nombre de garçons et de filles. Ainsi l'énoncé est vérifié.\n\nPar récurrence forte, l'énoncé est vérifié quel que soit le nombre total d'élèves.\n\nUne solution un peu plus astucieuse consiste à procéder différemment : soit $n$ le nombre de classes, et soient $f_{i}$ et $g_{i}$ les nombres de filles et garçons dans la classe $i$ pour $1 \\leqslant i \\leqslant n$. Alors la quantité suivante, lorsqu'on la développe, compte le nombre de façons de faire un comité avec un nombre impair de garçons moins celui de faire un comité avec un nombre impair de filles :\n$$\n\\prod_{i=1}^{n}\\left(g_{i}-f_{i}\\right)\n$$\nMais ce produit est positif s'il y a un nombre impair de classes avec plus de garçons que de filles, ce qui donne bien le résultat voulu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $n \\geqslant 1$ tel que pour tout nombre premier $p < n$, $n - \\left\\lfloor \\frac{n}{p} \\right\\rfloor p$ n'est pas divisible par un carré différent de 1.",
    "options": [],
    "answer": "{1, 2, 3, 5, 7, 13}",
    "solution": "Solution:\n\nAnalysons le problème : on cherche les entiers $n$ vérifiant une certaine propriété. On peut déjà commencer par chercher les petites valeurs de $n$ qui conviennent : en testant les entiers $n$ entre 1 et 20 on trouve $\\{1,2,3,5,7,13\\}$.\n\nUne des premières choses à faire devant ce problème est de réussir à donner un sens à $n-\\left\\lfloor\\frac{n}{p}\\right\\rfloor p$. $\\frac{n}{p}$ étant la division de $n$ par $p$ on peut essayer de regarder si cette quantité ne s'interprète pas à partir de la division euclidienne.\n\nNotons que $n-\\left\\lfloor\\frac{n}{p}\\right\\rfloor p$ est le reste de la division euclidienne de $n$ par $p$. En effet si $n=pq+r$ avec $q$ entier et $0 \\leqslant r \\leqslant p-1$, on a $\\left\\lfloor\\frac{n}{p}\\right\\rfloor=q$ donc $n-\\left\\lfloor\\frac{n}{p}\\right\\rfloor p=r$.\n\nEnsuite, on note que toutes les solutions trouvées sauf 1 sont des nombres premiers : on va donc essayer de montrer qu'un entier $n$ vérifiant la propriété est forcément premier, par exemple en prenant $p$ un de ses diviseurs.\n\nSoit $n>1$ un entier vérifiant la propriété. On déduit de ceci que si $n$ n'est pas premier, si $p<n$ est un diviseur premier de $n$, alors $n-\\left\\lfloor\\frac{n}{p}\\right\\rfloor p=0$ est divisible par 4 qui est un carré parfait, contradiction.\n\nMaintenant que peut-on faire pour avancer : analysons ce que la propriété peut nous donner. Si $n \\geqslant 12$, pour $p=2$ le reste de la division euclidienne vaut 1, pour $p=3$ il vaut 1 ou 2, donc cela n'apporte rien. Pour $p=5$, on obtient que le reste ne peut valoir 4, pour $p=7$, le reste ne peut valoir 4, pour $p=8$ le reste ne peut valoir ni 4 ni 8 ni 9. A priori, continuer comme ça indéfiniment semble compliqué, vu qu'on aura juste un très gros système, qui admettra probablement des solutions (comme il y a une infinité de nombres congrus à 1 modulo $p_1 \\times \\ldots p_k$, le fait de regarder les restes par les $k$ premiers nombres premiers semble peu utile). Par contre, 4 semble jouer un rôle important : 4 ne peut pas être un reste de la division de $n$ par $p$. En particulier si $p \\geqslant 5$, $p$ ne peut diviser $n-4$ ! Il semble donc intéressant de regarder ce que peut valoir $n-4$, et de même ce que peuvent valoir $n-8$ et $n-9$ car ceux-ci ne peuvent pas avoir de facteur premier plus grand que 10. Pour éviter les cas où ces quantités sont négatives, on suppose $n \\geqslant 11$.\n\nSoit $n$ premier supérieur ou égal à 11 vérifiant l'énoncé. Soit $p$ divisant $n-4$, on a $p < n$. Si $p \\geqslant 5$, le reste de la division euclidienne de $n$ par $p$ vaut 4 et il est divisible par 4 qui est un carré différent de 1, ce qui donne une contradiction. De plus, $p$ ne peut valoir 2 car sinon 2 divise $n-4$ donc 2 divise $n$, $n$ n'est dans ce cas pas premier. Ainsi $n-4=3^{x}$ avec $x$ un entier strictement positif car $n-4>1$.\n\nEn faisant le même raisonnement $n-8$ et $n-9$ ne sont pas divisibles par $p$ si $p \\geqslant 10$.\n\nMaintenant analysons les facteurs premiers de $n-8$ et $n-9$ : ils peuvent être divisibles par 2, 3, 5, 7. Pour le facteur 3, comme 3 divise $n-4$, $n \\equiv 1\\pmod{3}$, donc $n-8 \\equiv 2\\pmod{3}$ et $n-9 \\equiv 1\\pmod{3}$, ceux-ci ne sont pas divisibles par 3.\n\nDe plus $n-8$ n'est pas divisible par 2 car $n$ ne l'est pas. Comme $n-8$ et $n-9$ sont premiers entre eux, trois cas se présentent à nous : soit $n-8$ est une puissance de 5, soit c'est une puissance de 7, soit $n-8$ est divisible par 5 et 7, dans ce cas $n-9$ est une puissance de 2.\n\n- $n-8=5^{y}$ avec $y$ entier positif, on a $n-4=5^{y}+4=3^{x}$.\n\nIci 4 étant un carré, on aimerait montrer que $3^{x}$ est un carré pour factoriser l'équation. Pour cela, on regarde modulo 4 car on a des carrés et $5 \\equiv 1\\pmod{4}$.\n\nRegardons modulo 4, on a $3^{x} \\equiv 1^{y} \\equiv 1\\pmod{4}$. Sinon comme l'ordre de 3 modulo 4 est 2, 2 divise $x$ donc $x$ est pair. Posons $x=2k$, on a $(3^{k}-2)(3^{k}+2)=5^{y}$ donc $3^{k}-2$ est positif et $3^{k}-2$ et $3^{k}+2$ sont des puissances de 5. Ici on a un produit de termes valant une puissance de 5, on essaie donc de calculer leur pgcd pour espérer borner un des termes. En particulier, le pgcd de $3^{k}+2$ et $3^{k}-2$ divise $(3^{k}+2)-(3^{k}-2)=4$ qui n'est pas divisible par 5, et il divise leur produit c'est-à-dire $5^{y}$, il vaut donc 1. Comme $3^{k}-2$ et $3^{k}+2$ sont des puissances de 5 de pgcd 1, la plus petite vaut 1, on a donc $3^{k}-2=1$ soit $k=1$ soit $x=2$. On obtient $n=3^{x}+4=9+4=13$.\n\n- $n-8=7^{y}$ avec $y$ entier positif, on a $n-4=7^{y}+4=3^{x}$.\n\nComme 4 est un carré, on aimerait réussir à montrer que $x$ est pair pour factoriser, pour cela on regarde modulo 7. Notons que si $y=0$, il n'y a pas de solution à cette équation. Sinon on a $3^{x} \\equiv 4\\pmod{7}$, or les puissances de 3 modulo 7 valent $1,3,2,6,4,5,1$, l'ordre de 3 vaut donc 6 et on a forcément $x \\equiv 4\\pmod{6}$ donc $x$ est pair. Posons $x=2k$, on a $(3^{k}-2)(3^{k}+2)=7^{y}$ donc $3^{k}-2$ est positif et $3^{k}-2$ et $3^{k}+2$ sont des puissances de 7.\n\nEn particulier, le pgcd de $3^{k}+2$ et $3^{k}-2$ divise $(3^{k}+2)-(3^{k}-2)=4$ qui n'est pas divisible par 7, il divise aussi $7^{y}$ et vaut donc 1. Comme $3^{k}-2$ et $3^{k}+2$ sont des puissances de 7 de pgcd 1, la plus petite vaut 1, on a donc $3^{k}-2=1$ donc $k=1$ donc $x=2$, on a donc $7^{y}=3^{x}-4=5$ ce qui est impossible.\n\n- Sinon $n-9=2^{y}$ avec $y$ entier positif, on a $n-4=2^{y}+5=3^{x}$. Notons que comme $n-4 \\geqslant 7$, on a $x \\geqslant 1$ et $y \\geqslant 1$.\n\nIci on n'a pas de carré à priori, mais on peut essayer d'obtenir des informations sur la parité de $x$ et $y$. Vu qu'on a des puissances de 2 et 3 on va donc regarder modulo 3 et 4. Notons que si $y=1$, l'équation devient $3^{x}=7$, il n'y a pas de solution. Comme $x \\geqslant 1$ en regardant modulo 3, on a $2^{y} \\equiv -5 \\equiv 1\\pmod{3}$ et l'ordre de 2 modulo 3 est 2, on obtient que $y$ est pair. Comme $y \\geqslant 2$, on a $3^{x} \\equiv 1\\pmod{4}$ et comme l'ordre de 3 modulo 4 est 2, $x$ est pair. On écrit donc $x=2b$, $y=2c$, on a $5=(3^{b}-2^{c})(3^{b}+2^{c})$ donc $3^{b}-2^{c}$ est positif et comme $3^{b}-2^{c}<3^{b}+2^{c}$, $3^{b}-2^{c}$ et $3^{b}+2^{c}$ valent respectivement 1 et 5. On en déduit que $3^{b} \\times 2=(3^{b}+2^{c})+(3^{b}-2^{c})=6$ donc $b=1$ donc $x=2$, on retombe dans le cas précédent $n=13$.\n\nEn particulier, on a obtenu que $n=13$ ou $n<11$ et $n$ premier, i.e. $n$ vaut $2,3,5,7$ ou $13$. Vérifions que ce sont des solutions :\n\n- Il n'y a pas de nombre premier strictement inférieur à 1 et 2 donc 1 et 2 sont solutions.\n- Le reste de la division de 3 par 2 est 1 qui n'a pas de facteurs carrés, donc 3 est solution.\n- Les restes des divisions euclidiennes de 5 par 2 et 3 sont 1 et 2 qui sont sans facteurs carrés donc 5 est solution.\n- Les restes des divisions euclidiennes de 7 par 2, 3 et 5 sont 1, 1 et 2 qui sont sans facteurs carrés donc 7 est solution.\n- Les restes des divisions euclidiennes de 13 par 2, 3, 5, 7 et 11 sont 1, 1, 3, 6 et 2 qui sont sans facteurs carrés donc 13 est solution.\n\nL'ensemble des solutions est donc $\\{1,2,3,5,7,13\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30690,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $P$, $Q$ des polynômes à coefficients réels tels que $P \\circ Q = P^{2019}$. On suppose que toutes les racines de $P$ sont réelles. Montrer qu'elles sont toutes égales.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nA priori, on sait que le terme de droite va être nul si on l'évalue en $x_{j}$, on regarde donc ce qu'on en déduit pour le terme de gauche. Posons $X = \\{x_{1}, \\ldots, x_{k}\\}$, on a alors nécessairement $Q(x_{j}) \\in \\{x_{1}, \\ldots, x_{k}\\} = X$ pour tout $1 \\leqslant j \\leqslant k$.\n\nDe plus, comme le terme de droite de l'égalité est un polynôme scindé réel, forcément pour tout $j$, comme le polynôme $Q(X) - x_{j}$ divise un polynôme scindé, lui aussi est scindé et de degré $2019$ donc admet $2019$ racines réelles dans $X$.\n\nNotons de plus que si $i \\neq j$, $Q - x_{i}$ et $Q - x_{j}$ ont nécessairement des racines disjointes : en effet, on ne peut avoir $Q(x) = x_{i}$ et $Q(x) = x_{j}$. Ainsi, les polynômes $Q - x_{i}$ sont $|X| = k$ polynômes distincts scindés dont les racines sont dans $X$ et qui n'ont aucune racine en commun. Nécessairement, chacun de ces polynômes n'a qu'une racine : ainsi, pour tout $i$ il existe $j(i)$ tel que $Q(X) - x_{i} = c (X - x_{j(i)})^{2019}$ avec $c$ le coefficient dominant de $Q$. De plus, l'application de $\\{1, \\ldots, k\\}$ dans $\\{1, \\ldots, k\\}$ qui à $i$ associe $j(i)$ est bijective.\n\nMaintenant, on a une expression très précise des $Q - x_{i}$, rappelons que l'on veut montrer qu'il n'y a qu'un seul $x_{i}$. Pour cela, on peut essayer de regarder les coefficients de $Q - x_{i}$ qui sont quasiment ceux de $Q$. Le coefficient de $X^{2018}$ dans $Q$ vaut donc celui de $Q - x_{i}$, c'est-à-dire $-2019 x_{j(i)} c$ pour tout $i$. Puisque $i \\mapsto j(i)$ est une bijection de $\\{1, \\ldots, k\\}$, on a $x_{j} = x_{1}$ pour tout $1 \\leqslant j \\leqslant k$, et donc par hypothèse sur les $x_{i}$, $k = 1$. C'est exactement ce que l'on voulait prouver.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère une grille $2019 \\times 2019$. On note $C_{i, j}$ la case sur la $i$-ième colonne et la $j$-ième ligne pour $1 \\leqslant i, j \\leqslant 2019$. Un coloriage de la grille est formidable s'il n'existe pas $1 \\leqslant i < \\ell \\leqslant 2019$ et $1 \\leqslant j < k \\leqslant 2019$ tels que les cases $C_{i, j}$, $C_{i, k}$ et $C_{\\ell, j}$ soient de la même couleur. Quel est le plus petit entier $N$ tel qu'il existe un coloriage formidable avec $N$ couleurs?",
    "options": [],
    "answer": "1010",
    "solution": "Solution:\n\nDans ce problème, on cherche le plus petit entier satisfaisant une certaine propriété. Supposons que l'on veuille montrer que le plus petit entier recherché est l'entier $c$. Pour montrer que $c$ est bien le plus petit entier recherché, on va d'une part montrer que si un entier $n$ satisfait la propriété, alors $n \\geqslant c$ et d'autre part on va montrer que l'on peut trouver un coloriage avec exactement $c$ couleurs satisfaisant la propriété.\n\nOn s'empresse de tester l'énoncé sur des plus petites valeurs, par exemple pour des grilles de taille $3 \\times 3$, $4 \\times 4$ ou même $5 \\times 5$, afin de deviner la valeur minimale dans le cas général, ainsi qu'une éventuelle construction. On s'aperçoit que pour une grille $3 \\times 3$, on a besoin d'au moins 2 couleurs pour un coloriage formidable. Pour une grille $5 \\times 5$, on remarque que l'on a besoin de 3 couleurs au moins. On conjecture donc que pour $N$ impair, on a besoin d'au moins $\\frac{N+1}{2}$ couleurs.\n\nLa difficulté de l'exercice repose dans le fait qu'il faut d'abord essayer de deviner la valeur minimale avant de faire la preuve. En l'absence d'idée d'une preuve directe qu'il faut bien au moins $\\frac{N+1}{2}$ couleurs, on peut essayer de montrer par l'absurde qu'un coloriage possédant $\\frac{N-1}{2}$ ne sera pas formidable.\n\nPour cela, on remarque que la donnée de deux cases $C_{i, j}$, $C_{l, j}$ (avec $i < l$) de couleur rouge sur une colonne $j$ empêche les cases de ligne $i$ et sur une colonne $k \\geqslant j$ d'être de la couleur rouge. On peut dire que ces deux cases génèrent une interdiction sur la ligne $i$. On peut essayer de compter ces interdictions. Ce comptage dépend du nombre de couleurs que l'on utilise, ce qui nous encourage bien à raisonner par l'absurde pour montrer qu'un coloriage avec $\\frac{N-1}{2}$ couleurs générera trop d'interdictions.\n\nOù chercher les interdictions ? On peut commencer par procéder manuellement : on regarde le nombre minimal d'interdictions générées par la colonne 1, puis les interdictions générées par la ligne 2, etc... Cette opération manuelle s'effectue à nouveau sur des petits cas, comme une grille $5 \\times 5$ ou $7 \\times 7$. On s'aperçoit au terme de ces opérations sur des petits cas que la contradiction s'obtient en regardant la colonne $N$, sur laquelle se trouvent toujours trop d'interdictions. Nous allons donc formaliser ce raisonnement en regardant le nombre d'interdictions.\n\nOn suppose donc que l'on dispose d'un coloriage formidable de la grille possédant 1009 couleurs (quitte à avoir des couleurs qui ne sont pas utilisées pour le coloriage en question).\n\nDans la suite, on dira que la paire $\\left(C_{i, j}, C_{l, j}\\right)$ génère une interdiction pour la case $C_{i, k}$ pour la couleur $c$ si les cases $C_{i, j}$ et $C_{l, j}$ sont de la couleur $c$, ce qui empêche la case $C_{i, k}$ d'être de la couleur $c$ par hypothèse.\n\nSoit $j \\leqslant 2018$ une colonne fixée et soit $k_{j}$ le nombre de couleurs apparaissant sur au moins deux cases de la colonne $j$. Pour une couleur de numéro $c$, on note $n_{c}$ le nombre de cases de la colonne $j$ coloriées avec la couleur $c$. Quitte à renuméroter les couleurs, on peut supposer que ce sont les couleurs $1, \\ldots, k_{j}$ qui apparaissent au moins 2 fois et que les couleurs $k_{j+1}, \\ldots, 1009$ apparaissent au plus une fois dans la colonne.\n\nÉtant donnée une couleur $c \\leqslant k_{j}$, il y a au moins $n_{c} - 1$ paires de cases de la couleur $c$ dans la colonne $j$. En effet, la case de la couleur $c$ située la plus basse dans la colonne forme une paire avec chacune des $n_{c} - 1$ autres cases de la colonne. Chacune de ces paires génère une interdiction pour l'une des cases de la colonne 2019 : la paire de cases $\\left(C_{i, j}, C_{l, j}\\right)$ ($i < l$) génère une interdiction pour la case $C_{i, 2019}$. Ainsi, les cases de la colonne $j$ génèrent $\\sum_{c=1}^{k_{j}} (n_{c} - 1)$ interdictions dans la dernière colonne. Or, puisque les couleurs $k_{j+1}, \\ldots, 1009$ occupent au plus une case dans la colonne $j$, on a au moins\n$$\n\\sum_{c=1}^{k_{j}} (n_{c} - 1) = \\sum_{c=1}^{k_{j}} n_{c} - k_{j} \\geqslant 2019 - (1009 - k_{j}) - k_{j} = 1010\n$$\ninterdictions générées dans la colonne 2019.\n\nAinsi, chaque colonne de numéro $k \\leqslant 2018$ génère au moins 1010 interdictions pour les cases de la colonne 2019, ce qui donne au total au moins $2018 \\times 1010$ interdictions pour la colonne 2019.\n\nÀ présent, on note que si la case de ligne $i$ de la colonne 2019 reçoit 2 interdictions pour la même couleur $c$, cela signifie qu'il existe des cases $C_{i, j}$, $C_{l, j}$, $C_{i, j'}$, $C_{l', j'}$ avec $j < j'$, toutes de couleur $c$, dont les paires $\\left(C_{i, j}, C_{l, j}\\right)$, $\\left(C_{i, j'}, C_{l', j'}\\right)$ génèrent la même interdiction pour la case $C_{i, 2019}$ pour la couleur $c$. Mais le fait que les cases $C_{i, j}$, $C_{l, j}$, $C_{i, j'}$ soient de la même couleur est interdit par hypothèse. Ainsi, si une case possède $m$ interdictions, ces interdictions concernent $m$ couleurs différentes.\n\nEnfin, ces interdictions ne peuvent pas concerner la case $C_{2019, 2019}$. Ainsi, d'après le principe des tiroirs, puisqu'il y a $2018 \\times 1010$ interdictions sur la colonne 2019 au moins, il y a une case parmi les cases $C_{1, 2019}, \\ldots, C_{2018, 2019}$ qui possède au moins 1010 interdictions, elle est donc interdite pour au moins 1010 couleurs différentes, ce qui est absurde puisque l'on ne peut utiliser que 1009 couleurs différentes. En conclusion, un coloriage formidable possède au moins 1010 couleurs différentes.\n\nRéciproquement, si l'on dispose de 1010 couleurs, on peut construire un coloriage formidable, en s'inspirant de la configuration suivante pour $N = 7$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30692,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn super-domino est un pavé droit dans une grille en trois dimensions de l'une des trois formes suivantes : $1 \\times 1 \\times 2$, $1 \\times 2 \\times 1$ et $2 \\times 1 \\times 1$. Quels sont les entiers $a, b, c > 1$ tels qu'il est possible de paver un pavé droit de dimensions $a \\times b \\times c$ dans une grille en trois dimensions avec des super-dominos de sorte qu'il y ait autant de super-dominos de chacun des 3 types?",
    "options": [],
    "answer": "All integers a, b, c greater than 1 such that 12 divides abc.",
    "solution": "Solution:\n\nDans ce problème, on cherche tous les entiers $a, b, c$ satisfaisant certaines propriétés. Nous allons donc établir que si $a, b$ et $c$ satisfont la propriété alors $a, b$ et $c$ sont d'une certaine forme, et d'autre part montrer que si $a, b$ et $c$ sont de la forme trouvée, alors ils satisfont bien la propriété.\n\nOn peut également noter qu'il s'agit d'un problème de pavage, nous pouvons donc nous attendre à devoir utiliser un coloriage adroit du pavé droit. Si l'on n'a a priori pas d'idée sur la forme des entiers $a, b$ et $c$, on peut commencer par effectuer quelques remarques.\n\nSi on parvient à paver un pavé droit $a \\times b \\times c$ avec $N$ super-dominos de chacun des trois types, alors nécessairement il y aura $2N + 2N + 2N = abc$ cases pavées, donc $6$ divise $abc$. Ainsi une coordonnée est paire. Supposons donc sans perte de généralité que $a$ est pair.\n\nPour simplifier, on supposera que le pavé est composé de $abc$ cases de coordonnées $(i, j, k)$, où $i$ est l'abscisse, $j$ l'ordonnée de la case et $k$ la profondeur.\n\nOn cherche désormais un coloriage de notre pavé. Tout d'abord, les objets manipulés sont des dominos, composés de 2 cubes; On va donc chercher un coloriage à 2 couleurs, du type damier. Regardons les caractéristiques d'un éventuel coloriage. On peut commencer par se concentrer sur une face, par exemple sur les cases de profondeur 1. Un domino $1 \\times 1 \\times 2$ occupe au maximum une case sur cette face tandis que les deux autres types de dominos occupent 0 ou 2 cases de cette face. On peut donc essayer d'adapter notre coloriage à cette remarque en appliquant le même coloriage à chacune des tranches (par tranche, on entend ici l'ensemble des cases possédant la même profondeur), c'est-à-dire un coloriage où la couleur d'une case $(i, j, k)$ dépend de $i$ et de $j$ mais pas de $k$. L'avantage sera que les deux cubes d'un quelconque domino $1 \\times 1 \\times 2$ seront de la même couleur. De ce constat, on peut chercher à faire en sorte que les 2 cubes des dominos $1 \\times 2 \\times 1$ et $2 \\times 1 \\times 1$ soient de couleur différentes. Ceci nous fait penser à un coloriage en damier, que l'on applique uniformément à chacune des tranches. Formalisons donc ce coloriage et voyons l'information qu'il nous apporte.\n\nOn colorie en blanc les cases $(i, j, k)$ telles que $i + j$ est paire, et en noir les autres. Ainsi, les cubes d'un super-domino $1 \\times 1 \\times 2$ sont de la même couleur, noir ou blanc, tandis qu'un super-domino $1 \\times 2 \\times 1$ contient exactement un cube blanc et un cube noir, de même pour un super-domino $2 \\times 1 \\times 1$. On déduit que lorsque l'on pave notre pavé droit avec le même nombre de super-domino de chaque type, le nombre total de cases blanches recouvertes est pair : en effet un super-domino $1 \\times 1 \\times 2$ en recouvre 2 ou 0, et un super-domino d'un des deux autres types en recouvre 1. Ainsi pour pouvoir paver le pavé droit $a \\times b \\times c$ avec autant de super-dominos de chaque type, il faut qu'il y ait un nombre pair de cases blanches. Or, la tranche constituée des cases de profondeur 1 contient exactement $\\frac{ab}{2}$ cases blanches et toutes les tranches sont coloriées de la même façon et il y a $c$ tranches. Il y a donc $\\frac{ab}{2} \\times c$ cases blanches et ce nombre doit être pair. On déduit que $4$ divise $abc$. Comme $6$ et $4$ divisent $abc$, on a que $12$ divise $abc$ si un pavé $a \\times b \\times c$ satisfait l'énoncé.\n\nOn s'attaque désormais à la réciproque en montrant que l'on peut paver correctement un pavé $a \\times b \\times c$ si $12$ divise $abc$. Pour trouver un tel pavage, on commence toujours par regarder les pavés de petites dimensions, par exemple un pavé $2 \\times 2 \\times 3$. Ces essais sont plus utiles qu'on ne le croit. En effet, si l'on parvient à paver ce pavé, on pourra paver correctement des pavés plus gros en découpant ces gros pavés en petits pavés de dimension $2 \\times 2 \\times 3$ par exemple. Or, on remarque qu'avec deux dominos de chaque type, on peut paver un pavé $2 \\times 2 \\times 3$ et $3 \\times 4 \\times 3$ (voir la figure, le pavage de droite est découpé en deux parties pour être plus lisible). De plus, cela permet de paver un pavé $2 \\times 4 \\times 3$ en recollant selon la largeur deux pavés $2 \\times 2 \\times 3$ en recollant deux $3 \\times 2 \\times 3$ selon la largeur. Puisque tout entier $a > 1$ s'écrit comme somme de 2 et de 3, en recollant le bon nombre de pavés de la forme $2 \\times 4 \\times 3$ et $3 \\times 4 \\times 3$, on peut paver n'importe quel pavé $a \\times 4 \\times 3$ pour $a > 1$. De même, on peut paver les pavés de la forme $a \\times 2 \\times 6$, avec $a > 1$.\n\nOr si $12$ divise $abc$ alors il y a deux possibilités:\n- Une dimension du pavé droit est divisible par 6 et une autre dimension est paire. Dans ce cas on sait comment paver ce pavé droit à l'aide des pavés $a \\times 2 \\times 6$, $a > 1$.\n- Une dimension du pavé droit est divisible par 4 et une autre dimension est divisible par 3. Dans ce cas on sait comment paver ce pavé droit à l'aide des pavés $a \\times 4 \\times 3$, $a > 1$.\n- Deux dimensions du pavé sont paires, et la dernière est divisible par 3. Dans ce cas on sait comment paver ce pavé droit à l'aide des pavés $2 \\times 2 \\times 3$.\n- Une dimension du pavé droit est divisible par 12 et les autres sont impaires, non divisibles par 3, supérieures ou égales à 5. Alors on décompose le pavé droit $12a \\times b \\times c$ en un pavé $12a \\times 3 \\times c$ et $12a \\times (b-3) \\times c$, et on pave chacun comme précédemment car $b-3$ est pair.\n\nCeci achève la construction et montre que les pavés droits qui peuvent être ainsi pavés sont ceux tels que $12$ divise $abc$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30693,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $\\Omega$ son cercle circonscrit et $O$ le centre de $\\Omega$. Soit $S$ le centre du cercle tangent aux côtés $AB$ et $AC$ et tangent intérieurement au cercle $\\Omega$ en un point $K$. Le cercle de diamètre $[AS]$ recoupe le cercle $\\Omega$ en un point $T$. Soit $M$ le milieu du segment $[BC]$. Montrer que les points $K$, $T$, $M$ et $O$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans ce problème, nous allons utiliser divers résultats autour du cercle tangent aux côtés $AB$ et $AC$ au cercle $\\Omega$. Ce cercle est appelé le cercle $A$-mixtilinéaire, on le notera $\\omega$. On note $U$ le point de contact de ce cercle avec le côté $AB$ et $V$ le point de contact avec le côté $AC$. On note également $I$ le centre du cercle inscrit du triangle $ABC$.\n\nRésultat $n^{\\circ} 1$ : La droite $(KU)$ est la bissectrice de l'angle $\\widehat{AKB}$.\n\n![](attached_image_1.png)\n\nPour montrer ce résultat, on peut par exemple considérer l'homothétie $h$ de centre $K$ qui envoie $\\omega$ sur $\\Omega$. Soit $S_C$ l'image de $U$ par $h$. Alors $h((AB))$ est la droite tangente au cercle $\\Omega$ en le point $S_C$, et cette droite est parallèle à la droite $(AB)$. Ceci signifie que le triangle $AS_CB$ est isocèle donc $S_C$ est le pôle sud de $K$ dans le triangle $ABK$ donc la droite $(UK)$ est bien la bissectrice de l'angle $\\widehat{AKB}$.\n\nDe même la droite $(VK)$ est la bissectrice de l'angle $AKC$ et elle coupe le cercle $\\Omega$ en un point noté $S_B$.\n\nRésultat $n^{\\circ} 2$ : Le point $I$ est le milieu du segment $[UV]$.\n\n![](attached_image_2.png)\n\nNotons que le triangle $UAV$ est isocèle donc la droite $(AI)$ est la médiatrice du segment $[UV]$. Dès lors, il suffit de démontrer que le point $I$ appartient au segment $[UV]$.\n\nPour cela, on applique le théorème de Pascal à l'hexagone $ACS_CKS_BB$. Ce théorème nous dit que les points $U=(AB) \\cap (KS_C)$, $V=(AC) \\cap (KS_B)$ et $I=(BS_B) \\cap (CS_C)$ sont alignés, comme nous le voulions. (On a ici utilisé le résultat $n^{\\circ} 1$ pour avoir l'alignement des points $K, U$ et $S_C$ et $K, V$ et $S_B$.)\n\n![](attached_image_3.png)\n\nNous pouvons désormais passer à la résolution de l'exercice, maintenant que ces rappels sont faits.\n\nTout d'abord, en traçant la figure on s'aperçoit que les points $U$ et $V$ appartiennent au cercle de diamètre $[AS]$. On s'empresse de démontrer ce résultat. Puisque la droite $(AB)$ est tangente au cercle $\\omega$ de centre $S$ en le point $U$, on a $\\widehat{SUA}=90^\\circ$. De même $\\widehat{SVA}=90^\\circ$, les points $A, U, S$ et $V$ sont donc bien sur un même cercle dont $[AS]$ est un diamètre.\n\nUne première idée qui nous vient ensuite à l'esprit est d'utiliser la définition du point $T$ comme second point d'intersection des cercles circonscrits aux triangles $ABC$ et $AUV$. En effet, cette définition nous indique que $T$ est le centre de la similitude qui envoie le point $U$ sur le point $B$ et le point $V$ sur le point $C$. Regardons ce que deviennent les autres points après cette similitude. Le point $I$ est le milieu du segment $[UV]$, il est donc envoyé sur le milieu du segment $[BC]$, à savoir $M$. Le point $S$ est le milieu de l'arc $UV$ du cercle $\\omega$, il est donc envoyé sur le milieu de l'arc $BC$ du cercle $\\Omega$, à savoir le pôle Sud du point $A$ dans le triangle $ABC$, que l'on notera $D$.\n\nLe point $A$ devient le pôle Nord du point $T$ dans le triangle $TVU$. Il est donc envoyé sur le pôle Nord du point $T$ dans le triangle $TBC$, on notera $N$ ce point.\n\nConsidérer cette similitude a donc permis d'introduire naturellement de nouveaux points qui seront sans doute utiles à la résolution du problème. On note de plus que si $s$ est cette similitude et $X$ et $Y$ deux points quelconques, alors les triangles $TXs(X)$ et $TYs(Y)$ sont semblables. On a donc obtenu de nombreux triangles semblables et donc de nombreuses égalités d'angles.\n\nOn a introduit le point $N$. Il semble aligné avec les points $K$ et $I$, nous allons donc tenter de démontrer ce résultat. La façon la plus raisonnable de le faire est de procéder par chasse aux angles. Par exemple on peut essayer de démontrer que $\\widehat{VKI}=\\widehat{VKN}$. Cette idée a du sens puisque les points $S_B, V$ et $K$ sont alignés et cet alignement permettrait de passer d'un calcul d'angle dans le triangle $UKV$ à un calcul d'angle dans le triangle $ABC$, que l'on connaît bien.\n\nComment calculer l'angle $\\widehat{VKI}$ ? Comme $I$ est le milieu du segment $[UV]$ et que l'on a déjà les tangentes au cercle $\\omega$ en $U$ et $V$, on est fortement invités à utiliser la symédiane issue du sommet $K$, qui n'est autre que la droite $(AK)$. Ainsi, $\\widehat{VKI}=\\widehat{UKA}$. Les points $K, U, S_C$ sont alignés donc\n$$\n\\widehat{UKA}=\\widehat{S_CKA}=\\widehat{ABS_C}=\\frac{1}{2} \\widehat{ACB}\n$$\nOn s'est donc débarrassé des points $U, V$ et $K$ ce qui est encourageant. Il nous reste à montrer que $\\frac{1}{2} \\widehat{ACB}=\\widehat{S_BKN}=\\widehat{S_BBN}$, ce qui résulte de la chasse aux angles suivante :\n$$\n\\widehat{S_BBN}=\\widehat{CBN}-\\widehat{S_BBC}=90^\\circ-\\frac{1}{2} \\widehat{BAC}-\\frac{1}{2} \\widehat{ABC}=\\frac{1}{2} \\widehat{BCA}\n$$\nEn résumé, $\\widehat{VKI}=\\frac{1}{2} \\widehat{ACB}=\\widehat{S_BBN}=\\widehat{S_BKN}=\\widehat{VKN}$ donc les points $K, I$ et $N$ sont alignés.\n\nUtilisons ce que nous venons de trouver. D'après cet alignement : $\\widehat{IKT}=\\widehat{NKT}=\\widehat{NDT}$. Or les triangles $NDT$ et $IST$ sont semblables, donc $\\widehat{NDT}=\\widehat{IST}$. On a obtenu $\\widehat{IKT}=\\widehat{IST}$ donc les points $T, I, S$ et $K$ sont cocycliques.\n\nUn autre alignement que nous n'avons pas encore utilisé est celui des points $O, S$ et $K$, qui est vrai car l'homothétie de centre $K$ envoyant le cercle $\\omega$ sur $\\Omega$ envoie le point $S$ sur le point $O$. Cet alignement ainsi que le fait que les points $T, I, S$ et $K$ sont cocycliques nous donne\n$$\n\\widehat{OKT}=\\widehat{SKT}=180^\\circ-\\widehat{SIT}=\\widehat{AIT}\n$$\nMais les triangles $AIT$ et $NMT$ sont semblables. Ainsi :\n$$\n\\widehat{OKT}=\\widehat{AIT}=\\widehat{NMT}=\\widehat{OMT}\n$$\net les points $O, M, K$ et $T$ sont bien cocycliques.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2019 élèves participent à un concours et répondent chacun à 6 questions. À la fin du concours, on remarque que, parmi les bonnes réponses données par un quelconque groupe de 3 élèves, il y a au moins une réponse correcte à au moins 5 des 6 questions du concours. Quelle est la valeur minimale du nombre total de réponses correctes aux questions du concours?",
    "options": [],
    "answer": "10065",
    "solution": "Solution:\n\nOn peut commencer par ranger les données dans un tableau $2019 \\times 6$, où les 2019 lignes représentent les 2019 élèves et les 6 colonnes les 6 problèmes. Dans la case $(i, j)$, on inscrit un 1 si l'élève $i$ a répondu correctement à la question $j$ et on inscrit un 0 sinon. Le problème nous invite à effectuer un double comptage, cette présentation sous forme de tableau permet de voir quel double-comptage est intéressant. Les hypothèses concernant les triplets d'élèves invitent à tenter d'effectuer un double comptage. Mais d'abord, on effectue quelques remarques simples. On peut noter que si trois élèves ont répondu tous les trois faux aux questions 5 et 6, alors on a un triplet d'élèves dont les réponses ne sont pas correctes pour au moins 5 exercices. Ainsi, si $q_1$ et $q_2$ sont deux questions du concours, il y a au plus 2 élèves qui ont répondu faux à ces questions. Comme il y a $\\binom{6}{2}=15$ façons de choisir une paire de questions $\\{q_1, q_2\\}$, et pour chaque paire il y a au plus deux élèves qui ont mal répondu à ces deux questions donc il y a au plus $2 \\cdot 15=30$ paires $((\\{q_1, q_2\\}, e))$ telles que l'élève $e$ a mal répondu aux questions $q_1$ et $q_2$. Dans le tableau, cela se traduit par le fait qu'il y a au plus 30 paires de 0 tels que les deux 0 appartiennent à la même ligne. Cette majoration du nombre de paires de 0 appartenant à la même ligne nous permet de deviner le double comptage que nous allons effectuer : nous allons compter de deux façons différentes le nombre $N$ de paires de 0 tels que les deux 0 appartiennent à la même ligne. Une telle paire sera dite mauvaise dans la suite. On a déjà compté \"selon les colonnes\", ce qui nous a donné $N \\leqslant 30$. On va chercher à présent à compter \"selon les lignes\".\n\nOn note $E$ l'ensemble des élèves qui ont répondu à au plus 4 questions. À chacun des élèves de $E$, on peut associer au moins un couple de réponses fausses donc $|E| \\leqslant N \\leqslant 30$.\n\nPour un élève $e \\in E$ on note $n(e)$ le nombre de questions auxquelles il a répondu faux. Alors $N= \\sum_{e \\in E} \\binom{n(e)}{2}$, puisque les seules mauvaises paires appartiennent à des lignes associées à des éléments de $E$ et que s'il y a $n(e)$ zéros dans une ligne, il y a $\\binom{n(e)}{2}$ façons de créer des paires de 0 de cette ligne. On utilise à présent la convexité de la fonction $x \\mapsto \\frac{x(x-1)}{2}=\\binom{x}{2}$ pour appliquer l'inégalité de Jensen et obtenir une minoration de $N$ :\n\n$$\nN=\\sum_{e \\in E} \\binom{n(e)}{2} \\geqslant |E| \\binom{\\frac{\\sum_{e \\in E} n(e)}{|E|}}{2}\n$$\n\nOn note $f$ le nombre total de réponses fausses proposées par les élèves de $E$, et $v$ le nombre total de réponses justes, de telle sorte que $f+v=6|E|$. On a donc $\\sum_{e \\in E} n(e)=f$. Ainsi :\n\n$$\n\\begin{aligned}\n& 30 \\geqslant N \\geqslant \\frac{f(f-|E|)}{2|E|} \\\\\n& f^2-f|E|-60|E| \\leqslant 0\n\\end{aligned}\n$$\n\nCette dernière équation fournit que $f \\leqslant \\frac{|E|+\\sqrt{|E|^2+240|E|}}{2}$.\n\nSoit $V$ le nombre total de réponses justes données par l'ensemble des élèves, alors $V$ est la somme du nombre de bonnes réponses des élèves de $E$ et la somme des bonnes réponses des élèves qui ne sont pas dans $E$. Par définition, les élèves qui ne sont pas dans $E$ ont répondu correctement à au moins 5 questions. Donc\n\n$$\nV \\geqslant v+(2019-|E|) 5=2019 \\times 5-5|E|+v=2019 \\times 5-5|E|+6|E|-f=5 \\times 2019+|E|-f\n$$\n\nMais en utilisant l'inégalité obtenue en $f$ plus haut on obtient\n\n$$\nV \\geqslant 2019 \\times 5+\\frac{|E| - \\sqrt{|E|^2+240|E|}}{2}\n$$\n\nLe côté droit est décroissant en $|E|$ et puisque $|E| \\leqslant 30$, donc $V \\geqslant 2019 \\times 5+\\frac{30-\\sqrt{30^2+240 \\times 30}}{2}=10065$. Donc $V \\geqslant 10065$.\n\nRéciproquement, il est possible que $2019 - 30$ élèves aient répondu à 5 problèmes, et que 30 élèves aient répondu à 4 problèmes : il suffit que pour chacun des 15 ensembles de 4 questions parmi les 6 il y ait deux élèves qui ont répondu exactement à celles-là. Cette configuration satisfait l'énoncé, et est telle que le nombre de bonnes réponses est exactement 10065.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $G$ un graphe fini non orienté, $n \\in \\mathbb{N}^{*}$, $d \\in \\mathbb{N}^{*}$ tel que le degré de tout sommet de $G$ est borné par $d$, $x$ un sommet de $G$. On note $a(n, x)$ le nombre de sous-graphes induits connexes de $G$ contenant $n$ sommets dont le sommet $x$ (un sous-graphe induit $U$ est un graphe dont les sommets sont des sommets de $G$ et si $x$ et $y$ sont des sommets de $U$, $x$ est relié par une arête à $y$ dans $U$ si et seulement si $x$ est relié par une arête à $y$ dans $G$).\n\n1. Montrer que $a(n, x) \\leqslant 2^{n d}$.\n2. Montrer que $a(n, x) \\leqslant\\left(\\frac{(d+1)^{d+1}}{d^{d}}\\right)^{n}$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIci, on est face à un problème de graphe, plus précisément on compte les sous-graphes induits contenant $n$ éléments dont un fixé. On aimerait majorer cela, autant la première question semble pouvoir se faire potentiellement en comptant, autant la seconde n'a pas l'air très interprétable : $\\frac{(d+1)^{d+1}}{d^{d}}$ ne correspond à rien de spécial. A priori, il n'y a pas d'argument combinatoire clair pour faire apparaître cette quantité.\n\nComme à priori, on n'a pas de bonnes idées pour voir comment avancer, on peut tenter de faire cela au hasard ! On va donc essayer d'introduire des probabilités. On aimerait construire aléatoirement un sous-graphe induit connexe de $G$. On pourrait prendre des sommets au hasard mais ça se comporte très mal avec la connexité, il est difficile d'obtenir des informations faciles sur les composantes connexes d'un tel graphe. Plutôt que de choisir des sommets on va donc choisir des arêtes ! Ici il sera plus facile de voir quels points sont reliés. L'objectif sera donc d'étudier la composante connexe de $x$ dans le graphe formé des arêtes choisies.\n\nOn colorie chaque arête en blanc de $G$ indépendamment avec probabilité $p \\in [0,1]$. On note $X$ l'ensemble des sommets reliés à $x$ dans le graphe dont les arêtes sont les blanches, i.e. la composante connexe de $x$ dans le graphe aléatoire. Posons $A_{n}(x)$ l'ensemble des sous-graphes induits connexes de $G$ contenant $n$ sommets dont le sommet $x$ (on peut voir un sous-graphe induit comme un ensemble de sommets). Notons que comme les évènements $X=H$ pour $H$ dans $A_{n}(x)$ sont disjoints, $1 \\geqslant \\sum_{H \\in A_{n}(x)} P(X=H)$.\n\nA priori, il suffit donc d'estimer $P(X=H)$ pour avoir une borne sur $a(n, x)$, on va donc essayer de comprendre comment on peut avoir $X=H$. Une solution simple est que toutes les arêtes de $H$ soient bien blanches, et les arêtes allant de $H$ à son complémentaire ne le soient pas. Il va donc falloir évaluer la probabilité de cet évènement, en dénombrant le nombre d'arêtes de chaque catégorie.\n\nSoit $H$ dans $A_{n}(x)$, on a $P(X=H) \\geqslant \\mathbb{P}(A_{H})$ où $A_{H}$ est l'évènement \"toutes les arêtes internes à $H$ sont coloriées, et les arêtes reliant $H$ à l'extérieur ne sont pas coloriées\". Par indépendance, si on note $l$ le nombre d'arêtes internes à $H$ et $k$ le nombre d'arêtes reliant $H$ à l'extérieur, $\\mathbb{P}(A_{H}) = p^{l}(1-p)^{k}$. Or comme $H$ est de taille $n$ et le degré de chaque sommet est borné par $d$, $l + k \\leqslant n d$.\n\nIci pour la première borne, il est assez naturel de prendre $p=\\frac{1}{2}$ vu qu'on veut une borne par une puissance de 2.\n\nPour $p=\\frac{1}{2}$, on obtient donc $\\mathbb{P}(A_{H}) = 2^{-(l+k)} \\geqslant \\frac{1}{2^{n d}}$. Ainsi, on a $1 \\geqslant \\sum_{H \\in A_{n}(x)} \\frac{1}{2^{n d}} = a(n, x) \\frac{1}{2^{n d}}$, donc $a(n, x) \\leqslant 2^{n d}$.\n\nIci le raisonnement effectué tient si on change sous-graphe induit par sous-graphe car la probabilité d'avoir comme composante connexe le sous-graphe $H$ est la probabilité que les arêtes de $H$ soient blanches, et les arêtes qui ne sont pas dans $H$ ne le soient pas, on a au plus $n d$ arêtes concernées, donc $\\mathbb{P}(A_{H}) \\geqslant 2^{-n d}$ et on obtient le même résultat.\n\nPour la seconde inégalité, on aimerait avoir une meilleure minoration de $\\mathbb{P}(X=H)$. A priori, ce qu'on a fait dans la première partie était assez symétrique, on mettait autant de probabilité pour une arête blanche et une arête non blanche. On aimerait bien déséquilibrer cela, par exemple mettre une faible probabilité pour les arêtes blanches. Ainsi, on a envie d'avoir peu d'arêtes blanches à colorier pour obtenir $H$. A priori, à nombre de sommets fixé, le graphe ayant le minimum d'arêtes est un arbre : il est donc naturel de considérer un sous-arbre couvrant.\n\nPour cela, on va utiliser le fait que $H$ contient un sous-arbre $T$ avec le même nombre de sommets (et $n-1$ arêtes). En particulier, $P(X=H) \\geqslant P(B_{H})$ avec $B_{H}$ l'évènement : les arêtes de $T$ sont coloriées, les autres ne le sont pas. Comme il y a au plus $d n$ arêtes avec une extrémité dans $H$, on a donc $P(B_{H}) \\geqslant p^{n-1}(1-p)^{d n} \\geqslant p^{n}(1-p)^{d n}$.\n\nMaintenant on aimerait obtenir l'inégalité voulue, pour cela il est naturel d'essayer d'optimiser la quantité précédente en $p$. Si on pose $f : [0,1] \\to \\mathbb{R}$ qui à $p$ associe $p^{n}(1-p)^{d n}$, $f$ est $C^{1}$ et de dérivée $n p^{n-1}(1-p)^{d n} - d n p^{n}(1-p)^{d n-1} = n p^{n-1}(1-p)^{d n-1}(1-p-d p)$ qui s'annule et admet un minimum pour $1-p-d p=0$ c'est-à-dire $p=\\frac{1}{d+1}$.\n\nPrenons donc $p=\\frac{1}{d+1}$, on a $P(A_{H}) \\geqslant (d+1)^{-n}\\left(\\frac{d}{d+1}\\right)^{d n} = \\left(\\frac{d^{d}}{(d+1)^{d+1}}\\right)^{n}$. On obtient\n\n$$\n1 \\geqslant \\sum_{H \\in A_{n}(x)} P(B_{H}) \\geqslant a(n, x) \\times \\left(\\frac{d^{d}}{(d+1)^{d+1}}\\right)^{n}\n$$\n\nDonc $a(n, x) \\leqslant \\left(\\frac{(d+1)^{d+1}}{d^{d}}\\right)^{n}$.\n\nNotons que à $d$ fixé, la seconde borne est bornée par $(e(d+1))^{n}$ donc bien meilleure pour $d$ grand.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 30696,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et soit $\\Gamma$ son cercle circonscrit. Soit $P$ le point d'intersection de la droite $(BC)$ et de la tangente à $\\Gamma$ en $A$. Soit $D$ et $E$ les symétriques respectifs des points $B$ et $A$ par rapport à $P$.\nSoit alors $\\omega_{1}$ le cercle circonscrit au triangle $DAC$ et soit $\\omega_{2}$ le cercle circonscrit au triangle $APB$. On note $F$ le point d'intersection des cercles $\\omega_{1}$ et $\\omega_{2}$ autre que $A$, puis on note $G$ le point d'intersection, autre que $F$, du cercle $\\omega_{1}$ avec la droite $(BF)$.\nDémontrer que les droites $(BC)$ et $(EG)$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par tracer une figure.\n![](attached_image_1.png)\n\nUne première remarque que l'on peut formuler est que, puisque $P$ est le milieu des segments $[AE]$ et $[BD]$, le quadrilatère $ABED$ est un parallélogramme. Les droites $(AB)$ et $(DE)$ sont donc parallèles, de même que les droites $(AD)$ et $(BE)$.\n\nUne seconde remarque frappante est que $E$ semble appartenir au cercle $\\omega_{1}$. Après avoir vérifié sur une deuxième figure que c'était bien le cas, on s'empresse donc de démontrer ce premier résultat. Pour ce faire, on entame donc une chasse aux angles de droites :\n$$\n(EA, ED) = (EA, AB) = (CA, CB) = (CA, CD)\n$$\n\nUne chasse aux angles de droites indique alors que\n$(BC, EG) = (BC, BG) + (BG, EG) = (PB, BF) + (GF, GE) = (PA, AF) + (AF, AE) = 0^{\\circ}$, ce qui signifie bien que $(BC)$ et $(EG)$ sont parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30697,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans le train, alors qu'elles rentrent de EGMOnd an Zee, Clara et Edwige jouent au jeu suivant. Initialement, l'entier $n=1 \\times 2 \\times \\cdots \\times 20$ est écrit sur une feuille de papier. Puis, chacune à son tour, et en commençant par Clara, les joueuses remplacent l'entier $n$ par un des nombres $k n / 10$, où $k$ est un entier compris entre 1 et 9 inclus. La première joueuse à écrire un nombre qui n'est pas entier perd, et son adversaire gagne.\nClara et Edwige sont deux joueuses redoutables, et jouent donc de manière optimale. Laquelle des deux va-t-elle gagner?",
    "options": [],
    "answer": "Edwige",
    "solution": "Solution:\n\nIntéressons-nous au nombre $n$ écrit sur la feuille de papier au moment où la joueuse X s'apprête à jouer, et avant que la partie ne se termine. On factorise partiellement $n$ comme produit de nombres premiers : $n=2^{x} \\times 5^{y} \\times m$, où $x$ et $y$ sont des entiers naturels et $m$ est un entier naturel non nul premier avec 2 et 5.\n\nOn va alors démontrer la propriété $\\mathcal{P}_{n}$ par récurrence sur $n$ : la joueuse X perdra si $x$ et $y$ sont tous deux pairs, et gagnera dans le cas contraire. Tout d'abord, $\\mathcal{P}_{1}$ est évidente, puisque X devra remplacer le nombre $n=1$ par un nombre compris entre $1 / 10$ et $9 / 10$.\n\nSoit alors $n \\geqslant 1$ un entier quelconque. On suppose $\\mathcal{P}_{1}, \\ldots, \\mathcal{P}_{n}$, et on va démontrer $\\mathcal{P}_{n+1}$.\n\n$\\triangleright$ Si $x$ et/ou $y$ est impair, on note $r$ et $s$ les restes de $x$ et $y$ modulo 2. Alors la joueuse X remplace l'entier $n$ par l'entier $n' = n / (2^{r} \\times 5^{s})$. Puisque $n' = 2^{x-r} \\times 5^{y-s} \\times m$, la propriété $\\mathcal{P}_{n'}$ assure alors que l'adversaire de X perdra.\n\n$\\triangleright$ Si $x$ et $y$ sont pairs, on suppose tout de même que X remplace $n$ par un entier $n'$ (ce qui n'est possible que si $x \\geqslant 1$ ou $y \\geqslant 1$). Si X remplace $n$ par le nombre $n' = n / 2 = 2^{x-1} \\times 5^{y} \\times m$, alors $\\mathcal{P}_{n'}$ assure que l'adversaire de X gagnera. Sinon, on peut écrire $n'$ sous la forme $n' = 2^{x'} \\times 5^{y-1} \\times m'$, avec $m'$ premier avec 2 et 5 : quelles que soient les valeurs de $x'$ et de $m'$, la propriété $\\mathcal{P}_{n'}$ assure quand même que l'adversaire de X perdra.\n\nOn dispose donc de la propriété $\\mathcal{P}_{20!}$, où l'on a posé $20! = 1 \\times 2 \\times \\cdots \\times 20$. En particulier, on constate aisément que $20! = 2^{18} \\times 5^{4} \\times m$, où l'on a posé $m = 3^{8} \\times 7^{2} \\times 11 \\times 13 \\times 17 \\times 19$. Puisque Clara joue la première, c'est donc Edwige qui gagnera.\nSolution:\n\nNous allons de nouveau fournir une stratégie gagnante pour Edwige, qui diffère légèrement de la stratégie précédente. Cette stratégie est toute simple (même si montrer qu'elle convient l'est moins) : si Clara vient de transformer son nombre $n$ en un nouveau nombre $n' = k n / 10$, alors Edwige transforme de nouveau $n'$ en $n'' = k n' / 10$.\n\nPour montrer que cette stratégie permettra à Edwige de gagner, et puisque la partie durera au plus $20! = 1 \\times 2 \\times \\cdots \\times 20$ coups, il suffit de démontrer que $n''$ sera nécessairement un entier. Ainsi, Edwige ne pourra pas perdre, et elle gagnera donc.\n\nTout d'abord, on constate que l'entier $20!$ est égal à $u^{2} \\times v$, où l'on a posé $u = 2^{9} \\times 2^{4} \\times 5^{2} \\times 7$ et $v = 11 \\times 13 \\times 17 \\times 19$. On va maintenant démontrer que, si Edwige applique la stratégie ci-dessus, alors chacun des entiers $n$ qu'elle laissera à Clara sera de la forme $n = \\hat{u}^{2} v$, avec $\\hat{u}$ entier et $v = 11 \\times 13 \\times 17 \\times 19$.\n\nC'est déjà le cas au début du jeu. Puis, si Clara transforme un entier $n = \\hat{u}^{2} v$ en un autre entier $n' = k n / 10$, Edwige transformera $n'$ en le nombre $n'' = k n' / 10 = (k \\hat{u} / 10)^{2} v$. Il s'agit donc de démontrer que $k \\hat{u} / 10$ est un entier, c'est-à-dire que 2 et 5 divisent tous deux $k \\hat{u}$.\n\nPuisque $n'$ est un entier, et comme $10 n' = k n = k \\hat{u}^{2} v$, on sait que 2 et 5 divisent tous deux $k \\hat{u}^{2} v$. Or, 2 ne divise pas $v$, et divise donc $k$ ou $\\hat{u}$ : il divise donc aussi $k \\hat{u}$. De même, 5 ne divise pas $v$, et divise donc $k$ ou $\\hat{u}$ : il divise donc aussi $k \\hat{u}$. Cela conclut notre démonstration.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'une suite $\\left(u_{n}\\right)_{n \\geqslant 0}$ d'entiers naturels non nuls est sicilienne si\n$$\nu_{n+1} \\in\\left\\{u_{n} / 2, u_{n} / 3, 2 u_{n}+1, 3 u_{n}+1\\right\\}$$\npour tout $n \\geqslant 0$. Démontrer que, pour tout entier $k \\geqslant 1$, il existe une suite sicilienne $\\left(u_{n}\\right)_{n \\geqslant 0}$ et un entier $\\ell \\geqslant 0$ tels que $u_{0}=k$ et $u_{\\ell}=1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous allons démontrer, pour tout entier $k \\geqslant 2$, la propriété $\\mathcal{P}_{k}$ suivante : il existe une suite sicilienne finie $\\left(u_{0}, \\ldots, u_{\\ell}\\right)$ telle que $u_{0}=k$ et $u_{\\ell}<k$.\n\nLa propriété $\\mathcal{P}_{k}$ est immédiate si $k$ est pair, puisqu'il suffit de choisir $u_{1}=k / 2$. On suppose donc $k$ impair. Soient $\\alpha$ et $\\kappa$ les entiers naturels non nuls tels que $k=2^{\\alpha} \\kappa-1$, l'entier $\\kappa$ étant impair. On construit alors notre suite sicilienne comme suit.\n\nTout d'abord, pour tout $i \\leqslant \\alpha-1$, on pose $u_{2 i}=2^{\\alpha-i} 3^{i} \\kappa-1$ et $u_{2 i+1}=2^{\\alpha-i} 3^{i+1} \\kappa-2$. On pose ensuite $u_{2 \\alpha}=3^{\\alpha} \\kappa-1$ et $u_{2 \\alpha+1}=\\left(3^{\\alpha} \\kappa-1\\right) / 2$, qui est bien un entier puisque $\\kappa$ est impair. Enfin, pour tout $i \\leqslant \\alpha$, on pose $u_{2 \\alpha+2+i}=3^{\\alpha-i} \\kappa$.\n\nOn vérifie alors aisément que la suite $\\left(u_{0}, \\ldots, u_{3 \\alpha+2}\\right)$ est sicilienne. En effet, on a $u_{2 i+1}=3 u_{2 i}+1$ et $u_{2 i+2}=u_{2 i+1} / 2$ pour tout $i \\leqslant \\alpha$, puis $u_{2 \\alpha+1}=u_{2 \\alpha} / 2$ et $u_{2 \\alpha+2}=2 u_{2 \\alpha+1}+1$, et on a également $u_{2 \\alpha+3+i}=u_{2 \\alpha+2+i} / 3$ pour tout $i \\leqslant \\alpha-1$.\n\nPuisque $u_{0}=k$ et $u_{3 \\alpha+2}=\\kappa \\leqslant(k+1) / 2<k$, cela achève de démontrer la propriété $\\mathcal{P}_{k}$.\n\nForts de cette propriété, on démontre enfin l'assertion de l'énoncé par récurrence sur $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn répartit les entiers de $1,2, \\ldots, 8$ en deux ensembles $A$ et $B$, puis on note $P_{A}$ le produit de tous les éléments de $A$ et $P_{B}$ le produit de tous les éléments de $B$.\nQuelles sont les valeurs minimale et maximale que peut prendre la somme $P_{A}+P_{B}$ ?\n\nNote : si un ensemble $E$ est vide, on considérera que le produit de ses éléments est égal à 1.",
    "options": [],
    "answer": "minimum 402, maximum 40321",
    "solution": "Solution:\n\nSoit $A$ et $B$ deux ensembles disjoints dont la réunion est égale à l'ensemble $E=\\{1, \\ldots, 8\\}$.\nTâchons tout d'abord de maximiser la somme $P_{A}+P_{B}$. Sans perte de généralité, on peut supposer que $P_{A} \\leqslant P_{B}$. Puis, si $A$ contient un entier $k \\geqslant 2$, on pose $A' = A \\backslash \\{k\\}$ et $B' = B \\cup \\{k\\}$. Alors\n$$\nP_{A'} + P_{B'} \\geqslant P_{B'} = k P_{B} \\geqslant 2 P_{B} \\geqslant P_{A} + P_{B} .\n$$\nAinsi, lorsque la somme $P_{A} + P_{B}$ atteint sa valeur maximale, on sait que $A \\subseteq \\{1\\}$, donc que $P_{A} = 1$ et que $P_{B} = 8!$, de sorte que $P_{A} + P_{B} = 8! + 1 = 40321$.\n\nTâchons maintenant de minimiser la somme $P_{A} + P_{B}$. L'inégalité arithmético-géométrique indique que $P_{A} + P_{B} \\geqslant 2c$, où l'on a posé $c = \\sqrt{P_{A} P_{B}} = \\sqrt{8!}$. Comme $401^{2} = 160801 < 161280 = 4c^{2}$, on sait que $2c > 401$, et donc que $P_{A} + P_{B} \\geqslant \\lceil 2c \\rceil \\geqslant 402$.\n\nUn premier réflexe est donc de rechercher deux ensembles $A$ et $B$ tels que $P_{A} + P_{B} = 402$. On formule alors quelques remarques préliminaires à cette recherche. Tout d'abord, savoir quel ensemble contient l'entier 1 ne change rien. Puis on démontre que l'un des deux ensembles contient l'entier 6 et que l'autre contient les entiers 2 et 3. En effet, soit $X$ l'ensemble qui contient 6 et $Y$ celui qui ne contient pas 6 :\n\n$\\triangleright$ puisque $P_{Y} \\equiv 402 - P_{X} \\equiv 0 \\pmod{6}$, c'est que $Y$ contient l'entier 3 ainsi qu'un entier pair;\n\n$\\triangleright$ si c'est $X$ qui contient l'entier 2, alors $P_{Y} \\equiv 402 - P_{X} \\equiv 2 \\pmod{4}$, donc $X$ contient un entier pair mais pas divisible par 4, ce qui est impossible.\n\nQuitte à supprimer les entiers $1,2,3$ et $6$ des ensembles $A$ et $B$, on se ramène donc à trouver une partition de l'ensemble $\\hat{E} = \\{4,5,7,8\\}$ en deux sous-ensembles $\\hat{A}$ et $\\hat{B}$ tels que $P_{\\hat{A}} + P_{\\hat{B}} = 402 / 6 = 67$. Supposons, sans perte de généralité, que $\\hat{A}$ contient l'entier 4. Alors $P_{\\hat{B}} \\equiv 67 - P_{\\hat{A}} \\equiv 1 \\pmod{2}$, donc $\\hat{A}$ contient aussi l'entier 8. En outre, si $\\hat{A}$ contient également l'un des deux entiers 5 ou 7, alors $P_{\\hat{A}} + P_{\\hat{B}} \\geqslant P_{\\hat{A}} \\geqslant 4 \\times 5 \\times 8 \\geqslant 160 > 67$. On a donc nécessairement $\\hat{A} = \\{4,8\\}$ et $\\hat{B} = \\{5,7\\}$, et on est tout heureux de vérifier que, dans ce cas, on a effectivement $P_{\\hat{A}} + P_{\\hat{B}} = 32 + 35 = 67$.\n\nEn conclusion, la valeur maximale de $P_{A} + P_{B}$ est égale à $P_{\\emptyset} + P_{E} = 8! + 1 = 40321$, et la valeur minimale de $P_{A} + P_{B}$ est égale à $P_{\\{2,3,5,7\\}} + P_{\\{1,4,6,8\\}} = 402$.\nSolution:\n\nSoit $A$ et $B$ deux ensembles disjoints dont la réunion est égale à l'ensemble $E=\\{1, \\ldots, 8\\}$, et soit $c=\\sqrt{8!}$. Sans perte de généralité, on peut supposer que $P_{A} \\leqslant P_{B}$ ; puisque $P_{A} \\times P_{B} = P_{E} = c^{2}$, cela signifie que $P_{A} \\leqslant c$.\n\nMais alors $P_{A} + P_{B} = f(P_{A})$, où l'on a posé $f(x) = x + c^{2}/x$, et il nous reste donc à trouver les valeurs minimale et maximale que peut prendre $f(P_{A})$. On étudie donc le sens de variation de $f$ : si $x \\leqslant y \\leqslant c$, alors\n$$\nf(y) - f(x) = \\frac{y^{2} + c^{2}}{y} - \\frac{x^{2} + c^{2}}{x} = \\frac{(c^{2} - x y)(x - y)}{x y} \\leqslant 0 .\n$$\nLa fonction $f$ est donc décroissante sur l'intervalle $(0, c]$ et, par conséquent :\n\n$\\triangleright$ afin de maximiser $f(P_{A})$, il suffit de minimiser $P_{A}$, c'est-à-dire de choisir $A$ vide, ou encore $P_{A} = 1$;\n\n$\\triangleright$ afin de minimiser $f(P_{A})$, il suffit de maximiser $P_{A}$, sachant que $P_{A}$ est un produit d'éléments de $E$ et que $P_{A} \\leqslant c$.\n\nIl nous faut donc calculer la valeur de l'entier $\\lfloor c \\rfloor$. Puisque $c^{2} = 8! = 40320$ et que\n$$\n200^{2} = 40000 < 40320 < 40401 = 201^{2}\n$$\n$c$ est donc tel que $\\lfloor c \\rfloor = 200$. On étudie donc les entiers $200, 199, 198, \\ldots$ jusqu'à tomber sur un entier $n$ que l'on pourra écrire comme un produit d'éléments de $E$. Au lieu de procéder brutalement, on peut formuler deux remarques préalables :\n\n$\\triangleright$ de manière générale, $n$ doit diviser $8! = 2^{7} \\times 3^{2} \\times 5 \\times 7$;\n\n$\\triangleright$ en outre, si $n$ est impair, alors $n$ divise même $3 \\times 5 \\times 7 = 105$, donc $n \\leqslant 105$.\n\nOn déduit de la première remarque que $n$ ne peut prendre aucune des valeurs\n$$\n200 = 2^{3} \\times 5^{2}, \\quad 198 = 11 \\times 18, \\quad 196 = 4 \\times 7^{2}, \\quad 194 = 2 \\times 97\n$$\nde sorte que $n = 192 = 4 \\times 6 \\times 8$ est en fait l'entier recherché.\n\nPar conséquent, la valeur maximale de $P_{A} + P_{B}$ est égale à $f(1) = 8! + 1 = 40321$, et la valeur minimale de $P_{A} + P_{B}$ est égale à $f(192) = 192 + 210 = 402$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes entiers de 1 à 2020 sont écrits au tableau. Jacques a le droit d'en effacer deux et d'écrire à la place leur différence ou leur somme, et de recommencer jusqu'à ce qu'il ne reste plus qu'un entier. Est-il possible que l'entier obtenu à la fin soit 321 ?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nL'énoncé présente une suite finie d'opérations et le problème demande s'il est possible de partir de la situation initiale pour arriver à une certaine situation finale. Une première idée à essayer dans ce cas est de chercher un invariant.\n\nUne seconde idée est de tester le problème avec des plus petites valeurs. Par exemple, on peut tester l'énoncé pour les entiers écrits de 1 à 4 et regarder à quelles valeurs on peut aboutir à la fin du processus décrit dans l'énoncé. On remarque que partant d'une valeur fixée $n$, les résultats possibles ont tous la même parité, celle de la somme des nombres de 1 à $n$. Après avoir testé cette conjecture pour les entiers écrits de 1 à 5, on essaye donc de montrer que la parité de la somme des entiers écrits au tableau est invariante sous l'opération décrite.\n\nSi à un instant donné, la somme des entiers écrits au tableau vaut $S$ et que Jacques choisit les entiers $x$ et $y$, alors ces entiers seront remplacés par $x+y$ ou par $x-y$ et alors $S$ sera remplacé par $(S-x-y)+y-x = S-2x$ ou par $(S-y-x)+(x-y) = S-2y$. Ces deux entiers ont la même parité, donc la somme des entiers écrits au tableau garde la même parité tout au long du processus. Puisque dans la situation initiale, la somme des entiers écrits au tableau est\n$$\n1+2+3+\\cdots+2020=\\frac{2020 \\cdot 2021}{2}=1010 \\cdot 2021\n$$\net est paire, le nombre écrit au tableau dans la situation finale sera également pair et il ne peut donc pas s'agir du nombre 321.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier naturel non nul. On dit qu'une fonction $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ est $n$-positive si, pour tous les réels $x_{1}, \\ldots, x_{n}$ tels que $x_{1}+\\ldots+x_{n}=0$, on a $f\\left(x_{1}\\right)+\\ldots+f\\left(x_{n}\\right) \\geqslant 0$.\n\na) Toute fonction 2020-positive est-elle nécessairement 1010-positive?\n\nb) Toute fonction 1010-positive est-elle nécessairement 2020-positive?",
    "options": [],
    "answer": "a) yes; b) no",
    "solution": "Solution:\n\nDans les deux solutions, on pose $n=1010$.\n\na. Soit $f$ une fonction $2n$-positive, et soit $x_{1}, \\ldots, x_{n}$ des réels tels que $x_{1}+\\ldots+x_{n}=0$. En notant que $x_{1}+\\ldots+x_{n}+x_{1}+\\ldots+x_{n}=0$ et que $f$ est $2n$-positive, on constate que\n$$\n0 \\leqslant f\\left(x_{1}\\right)+\\ldots+f\\left(x_{n}\\right)+f\\left(x_{1}\\right)+\\ldots+f\\left(x_{n}\\right)=2\\left(f\\left(x_{1}\\right)+\\ldots+f\\left(x_{n}\\right)\\right),\n$$\nce qui montre bien que $f\\left(x_{1}\\right)+\\ldots+f\\left(x_{n}\\right) \\geqslant 0$, et donc que $f$ est $n$-positive. La réponse est donc oui!\n\nb. Soit $f$ la fonction définie par $f(x)=-1$ si $x$ est un entier tel que $x \\equiv 1(\\bmod 2n)$, et par $f(x)=n$ sinon. Alors $f$ est $n$-positive. En effet, soit $x_{1}, \\ldots, x_{n}$ des réels tels que $x_{1}+\\ldots+x_{n}=0$:\n\n$\\triangleright$ S'il existe un entier $k$ tel que $f\\left(x_{k}\\right)=n$, et puisque $f\\left(x_{i}\\right) \\geqslant -1$ pour tout $i \\neq k$, on sait que $f\\left(x_{1}\\right)+\\ldots+f\\left(x_{k}\\right) \\geqslant f\\left(x_{k}\\right)-(n-1) \\geqslant 1$.\n\n$\\triangleright$ Sinon, alors $x_{k} \\equiv 1(\\bmod 2n)$ pour tout $k$, donc $x_{1}+\\ldots+x_{n} \\equiv n \\not \\equiv 0(\\bmod 2n)$, de sorte que ce cas est en fait impossible.\n\nCependant, $f$ n'est pas $2n$-positive. En effet, si $x_{1}=1-2n$ et $x_{2}=x_{3}=\\ldots=x_{2n}=1$, alors $x_{1}+\\ldots+x_{2n}=0$ mais $f\\left(x_{1}\\right)+\\ldots+f\\left(x_{2n}\\right)=-2n<0$. La réponse est donc non!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont les angles sont aigus, et soit $D$ le pied de la hauteur issue du sommet $A$. Soit $X$ et $Y$ les deux points du cercle circonscrit à $ABC$ tels que $AX = AY = AD$, et tels que $B$ et $X$ se trouvent du même côté de la droite $(AD)$. Soit $E$ le point d'intersection des droites $(BX)$ et $(CY)$. Enfin, soit $F$ le point d'intersection, autre que $E$, entre la droite $(DE)$ et le cercle circonscrit à $EXY$.\n\nDémontrer que les cercles circonscrits aux triangles $BCF$ et $EXY$ sont tangents.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans la suite, on note $\\mathcal{C}_{UVW}$ le cercle circonscrit à un triangle $UVW$. Soit $i$ l'inversion de centre $E$ qui laisse le cercle $\\mathcal{C}_{ABC}$ globalement invariant. Cette inversion échange les points $B$ et $X$, ainsi que les points $C$ et $Y$. Elle échange donc le cercle $\\mathcal{C}_{EXY}$ et la droite $(BC)$. Puisque $i$ laisse la droite $(DE)$ globalement invariante, elle échange donc $F$ avec le point d'intersection de $(DE)$ et $(BC)$, c'est-à-dire avec le point $D$.\n\nPar conséquent, l'inversion $i$ échange les cercles $\\mathcal{C}_{BCF}$ et $\\mathcal{C}_{XYD}$. Or, ce dernier cercle a pour rayon $[AD]$, et il est donc tangent à la droite $(BC)$. Ainsi, les cercles $\\mathcal{C}_{BCF}$ et $\\mathcal{C}_{EXY}$ sont eux-mêmes tangents.\n\n![](attached_image_1.png)\nSolution:\n\nDans la suite, on note $\\mathcal{C}_{UVW}$ le cercle circonscrit au triangle $UVW$. Si $ABC$ est isocèle en $A$, la droite $(AD)$ constitue un axe de symétrie de la figure et le résultat est immédiat. On suppose donc, sans perte de généralité, que $AB \\neq AC$.\n\nLorsque l'on trace la figure, il est évidemment important de mettre en évidence le triangle $ABC$ et, au départ, les triangles isocèles $ADX$, $ADY$ et $AXY$, et ce qui semble être la tangente commune aux deux cercles $\\mathcal{C}_{BCF}$ et $\\mathcal{C}_{EXY}$. On remarque alors que ces trois droites semblent concourantes, et c'est donc ce que l'on entreprend de démontrer.\n\nSoit $I$ le point d'intersection des droites $(XY)$ et $(BC)$ : ce point existe bien, car $ABC$ n'est pas isocèle en $A$. Nous allons démontrer que la droite $(IF)$ est tangente aux cercles $\\mathcal{C}_{FBC}$ et $\\mathcal{C}_{FXY}$, ce qui donnera le résultat désiré. En considérant la puissance du point $I$ par rapport aux cercles $\\mathcal{C}_{FBC}$ et $\\mathcal{C}_{FXY}$, il s'agit donc de démontrer que $IF^2$ est égal à $IX \\cdot IY$ et à $IB \\cdot IC$. Or, puisque les droites $(BC)$ et $(AD)$ sont perpendiculaires, le cercle $\\mathcal{C}_{XYD}$ est tangent à la droite $(BC)$. En considérant la puissance du point $I$ par rapport aux cercles $\\mathcal{C}_{XYD}$ et $\\mathcal{C}_{ABC}$, on en déduit donc que $ID^2 = IX \\cdot IY = IB \\cdot IC$. Il s'agit donc de démontrer que $IF = ID$. Dès lors, on cherche des propriétés supplémentaires sur le point $D$. Si $(IF)$ est la tangente commune recherchée et si $IFD$ est isocèle en $I$, alors\n\n$$\n(FB, FD) = (FB, FI) + (FI, FD) = (CB, CF) + (FD, DI) = (FD, FC),\n$$\n\net $(FD)$ est donc la bissectrice de l'angle $\\widehat{BFC}$.\n\nRéciproquement, si $(FD)$ est la bissectrice de $\\widehat{BFC}$, et en notant $J$ le point d'intersection entre $(BC)$ et la tangente à $\\mathcal{C}_{BCF}$ en $F$, la chasse aux angles ci-dessus indique que $JFD$ est isocèle en $J$, donc que $JD^2 = JF^2 = JB \\cdot JC$. Le point $J$ a donc même puissance par rapport aux cercles $\\mathcal{C}_{BCF}$ et $\\mathcal{C}_{DXY}$, donc il appartient à leur axe radical $(XY)$, de sorte que $I = J$. Ainsi, notre problème est équivalent à démontrer que $(FD)$ est donc la bissectrice de l'angle $\\widehat{BFC}$.\n\nPour cela, on note $G$ le point d'intersection, autre que $D$, entre les cercles $\\mathcal{C}_{DBX}$ et $\\mathcal{C}_{CDY}$. Puisque\n\n$$\n(GY, GX) = (GY, GD) + (GD, GX) = (CY, CD) + (BD, BX) = (EY, EX)\n$$\n\nle point $G$ appartient au cercle $\\mathcal{C}_{EXY}$. En outre, $E$ est le point de concours des axes radicaux de $\\mathcal{C}_{ABC}$ avec $\\mathcal{C}_{DBX}$ et $\\mathcal{C}_{CDY}$. Il s'agit donc du centre radical de ces trois cercles, et $E$ appartient donc à l'axe radical de $\\mathcal{C}_{DBX}$ et $\\mathcal{C}_{CDY}$, c'est-à-dire à la droite $(GD)$. Cela signifie que $F = G$.\n\nOn conclut alors par simple chasse aux angles. En effet,\n\n$$\n\\begin{aligned}\n(FB, FD) + (FC, FD) & = (GB, GD) + (GC, GD) \\\\\n& = (XB, XD) + (YC, YD) \\\\\n& = (BX, BC) + (DB, DX) + (CY, CB) + (DC, DY) \\\\\n& = (YX, YC) + (YD, YX) + (XY, XB) + (XD, XY) \\\\\n& = (YD, YC) + (XD, XB).\n\\end{aligned}\n$$\n\nAinsi, $2(XB, XD) + 2(YC, YD) = 0^{\\circ}$, de sorte que les angles $\\widehat{DFB}$ et $\\widehat{CFD}$ sont égaux modulo $90^{\\circ}$. Comme $\\widehat{DFB} = \\widehat{DXB} \\leqslant \\widehat{CXB} = \\widehat{CAB} < 90^{\\circ}$ et, de même, $\\widehat{CFD} < 90^{\\circ}$, on en déduit que $\\widehat{DFB} = \\widehat{CFD}$, ce qui conclut le problème.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30703,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. Existe-t-il des nombres $a_{0}, \\ldots, a_{2020}$ valant -1 ou 1 tels que $a_{0} \\times a_{1}+a_{1} \\times a_{2}+\\cdots+a_{2019} \\times a_{2020}+a_{2020} \\times a_{0}=1010$ ?\n\n2. Existe-t-il des nombres $a_{1}, \\ldots, a_{2020}$ valant -1 ou 1 tels que $a_{1} \\times a_{2}+a_{2} \\times a_{3}+\\cdots+a_{2019} \\times a_{2020}+a_{2020} \\times a_{1}=1010$ ?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDéjà regardons l'énoncé : on veut savoir si on peut trouver des nombres $\\left(a_{i}\\right)_{1 \\leqslant i \\leqslant n}$ valant +1 ou -1 tels que $a_{1} a_{2}+\\cdots+a_{n} a_{1}$ vaut 1010 avec dans la première question $n=2021$, dans la seconde $n=2020$. Pour cela on peut tester avec des $n$ petits quels sont les nombres qu'on peut écrire sous la forme $a_{1} a_{2}+\\cdots+a_{n} a_{1}$. Pour $n=3$, on trouve 3 (si on ne prend que des 1 ) et -1 (si on prend $a_{1}=a_{2}=1, a_{3}=-1$ ). Pour $n=4$, on trouve $4,0,-4$ (pour avoir 4 on ne prend que des 1 , pour avoir 0 il suffit de prendre trois $a_{i}$ valant 1 et un valant -1 , pour avoir -4 , il suffit d'altener les $a_{i}$ valant 1 et ceux valant -1 . On peut continuer à tester les valeurs pour des $n$ petits. On peut remarquer que déjà les nombres qu'on peut obtenir ont l'air d'être régulièrement écartés de 4 , ils sont même congrus à $n$ modulo 4.\n\nNotons également que si $a= \\pm 1$ et $b= \\pm 1$, alors $a b = \\pm 1$ et que $a b=1$ si $a=b$ et $a b=-1$ si $a=-b$.\n\n1. Ici on est dans le cas où $n=2021$ et on veut obtenir une somme de 1010. Cela semble impossible à cause de la parité, on va donc regarder la parité de la somme. Soient $a_{0}, \\ldots, a_{2020}$ des nombres valant -1 ou 1. Pour simplifier on pose $a_{2021}=a_{0}$. Comme on a toujours $a_{i} a_{i+1}= \\pm 1$, $a_{0} \\times a_{1}+a_{1} \\times a_{2}+\\cdots+a_{2019} \\times a_{2020}+a_{2020} \\times a_{0}$ est une somme de 2021 nombres impairs, donc impaire. Elle ne peut donc pas valoir 1010 qui est pair. Il n'existe pas de nombres $a_{0}, \\ldots, a_{2020}$ valant -1 ou 1 tels que $a_{0} \\times a_{1}+a_{1} \\times a_{2}+\\cdots+a_{2019} \\times a_{2020}+a_{2020} \\times a_{0}=1010$.\n\n2. Dans ce cas $n=2020$ est divisible par 4. A priori vu les tests effectués pour des petites valeurs, on s'attend à ce que le nombre obtenu soit divisible par 4, or $1010=2 \\times 505$ n'est pas divisible par 4. On pourrait essayer de réutiliser l'argument de parité précédent, mais comme 2020 est pair, il prouverait que la somme est paire, or 1010 l'est aussi. Il va donc falloir plus précisément compter les moments où $a_{i} a_{i+1}=-1$ et ceux où $a_{i} a_{i+1}=1$. Or on sait que $a_{i} a_{i+1}=-1$ si et seulement si $a_{i}=-a_{i+1}$, il suffit donc de compter le nombre de changement de signes de la suite $a_{i}$ !\n\nSoient $a_{1}, \\ldots, a_{2020}$ des nombres valant -1 ou 1 tels que $a_{1} \\times a_{2}+a_{2} \\times a_{3}+\\cdots+a_{2019} \\times a_{2020}+a_{2020} \\times a_{1}=1010$. Posons $a_{2021}=a_{1}$. On note $N$ le nombre de $i$ tels que $1 \\leqslant i \\leqslant 2020$ tels que $a_{i}=-a_{i+1}$. En particulier dans la somme $a_{1} \\times a_{2}+a_{2} \\times a_{3}+\\cdots+a_{2019} \\times a_{2020}+a_{2020} \\times a_{1}$ il y a $N$ valeurs -1 et $(2020-N)$ valeurs 1. La somme vaut donc $2020-N-N=2020-2N$, on a donc $2020-2N=1010$ soit $2N=1010$ donc $N=505$. Or $N$ est pair. En effet $N$ est le nombre de changement de signes dans la suite $\\left(a_{1}, \\ldots, a_{2020}, a_{1}\\right)$ et cette suite commence par $a_{1}$ et termine $a_{1}$ donc elle a nécessairement un nombre pair de changements de signe puisque $a_{1}$ est du même signe que lui-même. On ne peut donc pas avoir $N=505$, ce qui fournit une contradiction. Il n'existe pas de nombres $a_{1}, \\ldots, a_{2020}$ valant -1 ou 1 tels que $a_{1} \\times a_{2}+\\cdots+a_{2019} \\times a_{2020}+a_{2020} \\times a_{1}=1010$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et soit $\\omega$ son cercle circonscrit. Soit $\\ell_{B}$ et $\\ell_{C}$ deux droites parallèles l'une à l'autre, passant respectivement par les points $B$ et $C$. On note $D$ le point d'intersection, autre que $B$, entre $\\omega$ et la droite $\\ell_{B}$. De même, on note $E$ le point d'intersection, autre que $C$, entre $\\omega$ et la droite $\\ell_{C}$.\n\nOn suppose que les droites $\\ell_{C}$ et $(AD)$ se coupent en un point $F$, et que les droites $\\ell_{B}$ et $(AE)$ se coupent en un point $G$. On note alors $O, O_{1}$ et $O_{2}$ les centres respectifs des cercles circonscrits aux triangles $ABC$, $ADG$ et $AEF$. Enfin, on note $P$ le centre du cercle circonscrit au triangle $OO_{1}O_{2}$.\n\nDémontrer que la droite $(OP)$ est parallèle aux deux droites $\\ell_{B}$ et $\\ell_{C}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par tracer une figure, en prenant soin de faire apparaître le triangle $OO_{1}O_{2}$, puisque $P$ en est le centre du cercle circonscrit. On en profite pour noter $\\omega_{1}, \\omega_{2}$ et $\\omega^{\\prime}$ les cercles circonscrits à $ADG$, à $AEF$ et à $OO_{1}O_{2}$.\n\n![](attached_image_1.png)\n\nOn remarque tout de suite que les points $A, O_{1}$ et $O_{2}$ semblent alignés, et on s'empresse donc de le démontrer. En effet, les droites $(DG)$ et $(EF)$ sont parallèles l'une à l'autre. Le théorème de Thalès indique alors qu'il existe une homothétie $h$, de centre $A$, qui envoie le triangle $ADG$ sur le triangle $AFE$. On en déduit que $O_{2}=h(O_{1})$, et donc que les points $A$, $O_{1}$ et $O_{2}$ sont bien alignés.\n\nAfin de démontrer que $(OP)$ est parallèle à $\\ell_{B}$ et à $\\ell_{C}$, on va maintenant calculer l'angle de droites $\\left(OP, \\ell_{B}\\right)$. Pour ce faire, on commence par se débarrasser du point $P$, en invoquant la tangente en $O$ à $\\omega^{\\prime}$ : si l'on note $t$ cette tangente, alors\n$$\n\\begin{aligned}\n\\left(OP, \\ell_{B}\\right) & =(OP, t)+\\left(t, OO_{2}\\right)+\\left(OO_{2}, \\ell_{B}\\right) \\\\\n& =90^{\\circ}+\\left(O_{1}O, O_{1}O_{2}\\right)+\\left(OO_{2}, \\ell_{B}\\right) \\\\\n& =90^{\\circ}+\\left(O_{1}O, O_{1}A\\right)+\\left(OO_{2}, \\ell_{B}\\right) .\n\\end{aligned}\n$$\n\nPuis, comme $(AD)$ est l'axe radical des cercles $\\omega_{1}$ et $\\omega^{\\prime}$, les droites $(AD)$ et $\\left(OO_{1}\\right)$ sont perpendiculaires l'une à l'autre. De même, $(AE)$ et $\\left(OO_{2}\\right)$ sont perpendiculaires. On en déduit que\n$$\n\\begin{aligned}\n\\left(OP, \\ell_{B}\\right) & =90^{\\circ}+\\left(O_{1}O, O_{1}A\\right)+\\left(OO_{2}, \\ell_{B}\\right) \\\\\n& =90^{\\circ}+\\left(O_{1}O, AD\\right)+\\left(AD, O_{1}A\\right)+\\left(OO_{2}, AE\\right)+\\left(AE, \\ell_{B}\\right) \\\\\n& =90^{\\circ}+90^{\\circ}+\\left(AD, O_{1}A\\right)+90^{\\circ}+\\left(AE, \\ell_{B}\\right) \\\\\n& =90^{\\circ}+\\left(AD, O_{1}A\\right)+\\left(AE, \\ell_{B}\\right)\n\\end{aligned}\n$$\n\nCe faisant, on s'est déjà débarrassés des points $P, O$ et $O_{2}$, ce qui suggère que l'on est sur la bonne voie. Notre prochaine victime sera donc le point $O_{1}$. En effet, si l'on note $A_{1}$ le symétrique de $A$ par rapport à $O_{1}$, alors $[AA_{1}]$ est un diamètre de $\\omega_{1}$, de sorte que\n$$\n\\left(AD, O_{1}A\\right)=\\left(AD, AA_{1}\\right)=\\left(AD, DA_{1}\\right)+\\left(DA_{1}, AA_{1}\\right)=90^{\\circ}+(DG, AG)=90^{\\circ}+(BD, AE)\n$$\n\nMais alors on a déjà gagné, puisque\n$$\n\\left(OP, \\ell_{B}\\right)=90^{\\circ}+\\left(AD, O_{1}A\\right)+\\left(AE, \\ell_{B}\\right)=90^{\\circ}+90^{\\circ}+(BD, AE)+\\left(AE, \\ell_{B}\\right)=\\left(BD, \\ell_{B}\\right),\n$$\nce qui conclut.\n\n\n\\underline{\\text{Solution alternative}~n^{\\circ}~1}\n\nOn réutilise, ci-dessous, les noms des points et cercles introduits dans la solution précédente.\n\nPuisque les droites $\\ell_{B}$ et $\\ell_{C}$ sont parallèles, et comme $A, B, C, D$ et $E$ sont cocycliques, on sait tout d'abord que $(GA, GD)=(EA, EF)$ et que, de même, $(FE, FA)=(DG, DA)$. Les triangles $AEF$ et $AGD$ sont donc semblables. Par conséquent, les triangles $FO_{2}A$ et $DO_{1}A$ sont également semblables, de sorte que $\\left(AF, AO_{2}\\right)=\\left(AD, AO_{1}\\right)$, et donc que les points $O_{1}$, $A$ et $O_{2}$ sont alignés.\n\nEnsuite, puisque la droite $(AD)$ est l'axe radical des cercles $\\omega$ et $\\omega_{1}$, elle est perpendiculaire à $\\left(OO_{1}\\right)$. On en déduit, d'après le théorème de l'angle au centre, et en notant $A_{1}$ le symétrique de $A$ par rapport à $O_{1}$, que\n$$\n\\begin{aligned}\n\\left(O_{1}O_{2}, O_{1}O\\right) & =\\left(AA_{1}, AD\\right)+\\left(AD, OO_{1}\\right)=\\left(AA_{1}, DA_{1}\\right)+\\left(DA_{1}, AD\\right)+90^{\\circ} \\\\\n& =(AG, DG)+90^{\\circ}+90^{\\circ}=(GA, GD) .\n\\end{aligned}\n$$\n\nOn montre de même que $\\left(O_{2}O, O_{2}O_{1}\\right)=(FE, FA)=(DG, DA)$. Les triangles $OO_{1}O_{2}$ et $AGD$ sont donc indirectement semblables.\n\nMais alors les triangles $OO_{1}P$ et $AGO_{1}$ sont, eux aussi, indirectement semblables. On en conclut que\n$$\n\\left(OP, O_{1}O_{2}\\right)=\\left(GD, AO_{1}\\right)=\\left(GD, O_{1}O_{2}\\right),\n$$\nce qui signifie bien que $(OP)$ est parallèle à la droite $(GD)=\\ell_{B}$, donc à $\\ell_{C}$ également.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAndréa, Baptiste et Camille jouent au foot à trois. Un des joueurs est aux cages, les deux autres sont sur le terrain et essaient de marquer. Le joueur qui marque devient ensuite gardien pour le tir suivant.\n\nDurant l'après-midi, Andréa a été sur le terrain 12 fois, Baptiste l'a été 21 fois et Camille a été aux cages 8 fois. Leur professeur sait qui a marqué le 6-ième but. Qui était-ce?",
    "options": [],
    "answer": "Andréa",
    "solution": "Solution:\n\nSi on note $n$ le nombre de parties jouées, alors au total il y a eu $n$ fois un gardien. On utilise à présent les hypothèses données par l'énoncé. Si Andréa a été 12 fois sur le terrain, il a donc été $n-12$ fois aux cages. Si Baptiste a été 21 fois sur le terrain, il a été $n-21$ fois aux cages. Ainsi $n-12 + n-21 + 8 = n$, ce qui nous donne que $n = 25$. Cela veut également dire que Andréa a été aux cages 13 fois. Mais il est impossible d'être aux cages deux fois d'affilée. Ainsi, on sait que Andréa était aux cages au début, et l'a été exactement une fois sur deux. En particulier, il a été aux cages à la septième partie et donc Andréa a marqué le 6-ième but.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\Gamma$ un cercle passant par $A$. On suppose que $\\Gamma$ recoupe les segments $[AB]$ et $[AC]$ en deux points, que l'on appelle respectivement $D$ et $E$, et qu'il coupe le segment $[BC]$ en deux points, que l'on appelle $F$ et $G$, de sorte que $F$ se trouve entre $B$ et $G$. Soit $T$ le point d'intersection entre la tangente en $F$ au cercle circonscrit à $BDF$ et la tangente en $G$ au cercle circonscrit à $CEG$.\nDémontrer que, si les points $A$ et $T$ sont distincts, alors $(AT)$ et $(BC)$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par tracer une figure.\n![](attached_image_1.png)\n\nUne première remarque frappante est que $T$ semble être situé sur le cercle $\\Gamma$. Après avoir vérifié qu'il l'était bien sur une deuxième figure, on s'empresse donc de démontrer ce premier résultat. Pour ce faire, on entame donc une chasse aux angles de droites, en utilisant les relations de cocyclicité et le cas limite du théorème de l'angle au centre :\n$$\n\\begin{aligned}\n(TG, TF) & = (TG, BC) + (BC, TF) = (TG, CG) + (BF, TF) = (EG, EC) + (DB, DF) \\\\\n& = (EG, AE) + (AD, DF) = (DG, AD) + (AD, DF) = (AD, DF)\n\\end{aligned}\n$$\nLes points $A, F, G$ et $T$ sont donc cocycliques, et $T$ appartient à bien $\\Gamma$. Mais alors\n$$\n\\begin{aligned}\n(AT, BC) & = (AT, AF) + (AF, BC) = (GT, GF) + (AF, GF) \\\\\n& = (GT, GC) + (AE, GE) = (EG, EC) + (CE, GE) = 0^\\circ,\n\\end{aligned}\n$$\nce qui signifie précisément que les droites $(AT)$ et $(BC)$ sont parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn choisit 5 diviseurs positifs de $10^{2020}$. Montrer qu'il y en a deux dont le produit est un carré.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par regarder la forme que peut prendre un éventuel diviseur de $10^{2020}$. Les diviseurs de $10^{2020}$ sont de la forme $2^{a} 5^{b}$. Le produit de deux diviseurs est donc également un nombre de la forme $2^{a} 5^{b}$ et c'est un carré parfait si et seulement si $a$ et $b$ sont tous les deux pairs. On est donc ramené à montrer le résultat suivant : si on a 5 paires d'entiers $\\left(a_{i}, b_{i}\\right)$ avec $1 \\leqslant i \\leqslant 5$ et les $a_{i}$ et $b_{i}$ tous les deux inférieurs ou égaux à 2020, alors on peut trouver deux indices $j$ et $k$ tels que $\\left(a_{j}+a_{k}, b_{j}+b_{k}\\right)$ est un couple de deux nombres entiers pairs.\n\nOn s'empresse de regarder si ce résultat est vrai pour quelques paires d'entiers et on cherche désormais à démontrer ce résultat. On est amené à regarder la parité des entiers considérés. Il y a 4 possibilités pour un couple $(a, b)$ : soit $a$ et $b$ sont pairs, soit $a$ et $b$ sont impairs, soit $a$ est pair et $b$ est impair, soit $a$ est impair et $b$ est pair. Ainsi par le principe des tiroirs, si on choisit 5 paires d'entiers $\\left(a_{i}, b_{i}\\right)$ avec $1 \\leqslant i \\leqslant 5$, il y a deux couples $\\left(a_{j}, b_{j}\\right)$ et $\\left(a_{k}, b_{k}\\right)$ possédant la même caractéristique, c'est-à-dire tels que $a_{j}$ et $a_{k}$ soient de la même parité et $b_{j}$ et $b_{k}$ soient de même parité. Alors $a_{j}+a_{k}$ et $b_{j}+b_{k}$ sont tous les deux pairs.\n\nAinsi, en choisissant 5 diviseurs de $10^{2020}$, on peut en trouver 2 tels que leur produit soit un carré.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont les angles sont aigus. La hauteur issue de $A$ recoupe le cercle circonscrit au triangle $ABC$ en un point $D$. La hauteur issue de $B$ recoupe le cercle circonscrit au triangle $ABC$ en un point $E$. La droite $(ED)$ coupe les côtés $[AC]$ et $[BC]$ respectivement en les points $P$ et $Q$. Montrer que les points $P, Q, B$ et $A$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nDéterminons quels sont les angles qui semblent faciles d'accès. Par exemple, il semble compliqué de pouvoir obtenir l'angle $\\widehat{PBQ}$ ou l'angle $\\widehat{PBA}$. En revanche, on connait déjà l'angle $\\widehat{QBA}$ donc il semble raisonnable d'essayer de montrer que $\\widehat{QPA}=180^\\circ-\\widehat{QBA}$.\n\nOr, $180^\\circ-\\widehat{QPA}=\\widehat{EPA}$, on essaye donc de montrer que $\\widehat{EPA}=\\widehat{QBA}$. Deux angles droits nous sont donnés : si on pose $H_{B}$ le pied de la hauteur issue du sommet $B$ et $H_{A}$ le pied de la hauteur issue du sommet $A$, alors les angles $\\widehat{A H_{A} B}$ et $\\widehat{B H_{B} A}$ sont droits.\n\nOn n'a pas encore utilisé l'hypothèse que les points $E$ et $D$ étaient sur le cercle circonscrit au triangle $ABC$. Cette hypothèse se traduit par une égalité d'angle à l'aide du théorème de l'angle inscrit. Ceci donne notamment $\\widehat{DEB}=\\widehat{DAB}$.\n\nEn reportant ces différentes égalités d'angles sur la figure, on trouve donc que $\\widehat{PEH_{B}}=\\widehat{H_{A}AB}$ et $\\widehat{PH_{B}E}=\\widehat{BH_{A}A}$. Les triangles $PEH_{B}$ et $BAH_{A}$ sont donc semblables (il y a deux paires d'angles égaux deux à deux). Ainsi, $\\widehat{EPH_{B}}=\\widehat{ABH_{A}}$ qui est l'égalité que l'on souhaitait démontrer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAu pays des merveilles se trouvent $n$ villes. Chaque paire de villes est reliée par une route à sens unique, qui part d'une des deux villes et arrive à l'autre. Afin de s'y retrouver, Alice interroge le roi de cœur : à chaque question, Alice choisit une paire de villes, et le roi de cœur lui dit quelle est la ville de départ de la route qui relie ces deux villes.\n\nDémontrer que, en $5 n$ questions ou moins, Alice peut arriver à savoir s'il existe une ville d'où part au plus une route.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous allons décrire une stratégie qu'Alice peut mettre en place pour aboutir à ses fins en $5 n$ questions ou moins. À tout moment, on dira qu'une ville $v$ est mauvaise si Alice a déjà trouvé deux routes qui partent de $v$, et que $v$ est bonne sinon. De même, on dira qu'une paire de villes $\\{v, w\\}$ est explorée si Alice a déjà interrogé le roi de cœur sur cette paire-là, et inexplorée sinon.\n\nEnfin, en parallèle de ces questions, Alice a dessiné une carte sur laquelle les $n$ villes sont représentées par $n$ sommets, et elle ajoute une arête sur cette carte, entre les villes $v$ et $w$ à chaque question qu'elle pose sur la paire $\\{v, w\\}$. Dans la suite, nous allons identifier le pays des merveilles au graphe qu'Alice est en train de construire.\n\nTout d'abord, Alice n'a manifestement jamais intérêt à interroger le roi de cœur sur une paire de villes qui seraient toutes deux mauvaises, ou sur une paire de villes déjà explorée. La stratégie d'Alice débute donc comme suit. Tant qu'il existe une paire inexplorée $\\{v, w\\}$ formée de deux bonnes villes, Alice choisit une telle paire et interroge le roi de cœur sur cette paire. À la fin de cette première étape, nul sommet de notre graphe n'a strictement plus de deux arêtes sortantes. Alice a donc posé au plus $2 n$ questions.\n\nEn outre, soit $X$ l'ensemble des villes toujours bonnes à l'issue de cette étape, et soit $x$ le cardinal de $X$, de sorte qu'il y a $x(x-1) / 2$ routes entre villes de $X$. Toute paire $\\{v, w\\}$ formée de deux villes de $X$ est manifestement explorée; et toute ville de $X$, puisqu'elle est bonne, est donc à l'origine d'au plus une route allant vers une autre ville de $X$. Il y a donc au plus $x$ routes entre villes de $X$, ce qui signifie que $x \\leqslant 3$.\n\nAlice n'a donc plus qu'à interroger le roi de cœur sur toutes les paires $\\{v, x\\}$ où $x \\in X$ : cela fera $n x \\leqslant 3 n$ questions supplémentaires, à l'issue desquelles Alice saura exactement combien de routes partent de chacune des villes de $X$. Si, à cette étape de l'algorithme, il reste une bonne ville $v$, c'est qu'il n'y avait effectivement pas plus d'une route qui partait de $v$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30710,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\Gamma$ un cercle passant par $A$. On suppose que $\\Gamma$ recoupe les segments $[AB]$ et $[AC]$ en deux points, que l'on appelle respectivement $D$ et $E$, et qu'il coupe le segment $[BC]$ en deux points, que l'on appelle $F$ et $G$, de sorte que $F$ se trouve entre $B$ et $G$. Soit $T$ le point d'intersection entre la tangente en $F$ au cercle circonscrit à $BDF$ et la tangente en $G$ au cercle circonscrit à $CEG$.\nDémontrer que, si les points $A$ et $T$ sont distincts, alors $(AT)$ et $(BC)$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par tracer une figure.\n![](attached_image_1.png)\n\nUne première remarque frappante est que $T$ semble être situé sur le cercle $\\Gamma$. Après avoir vérifié qu'il l'était bien sur une deuxième figure, on s'empresse donc de démontrer ce premier résultat. Pour ce faire, on entame donc une chasse aux angles de droites, en utilisant les relations de cocyclicité et le cas limite du théorème de l'angle au centre :\n$$\n\\begin{aligned}\n(TG, TF) & = (TG, BC) + (BC, TF) = (TG, CG) + (BF, TF) = (EG, EC) + (DB, DF) \\\\\n& = (EG, AE) + (AD, DF) = (DG, AD) + (AD, DF) = (AD, DF) .\n\\end{aligned}\n$$\nLes points $A, F, G$ et $T$ sont donc cocycliques, et $T$ appartient à bien $\\Gamma$. Mais alors\n$$\n\\begin{aligned}\n(AT, BC) & = (AT, AF) + (AF, BC) = (GT, GF) + (AF, GF) \\\\\n& = (GT, GC) + (AE, GE) = (EG, EC) + (CE, GE) = 0^\\circ,\n\\end{aligned}\n$$\nce qui signifie précisément que les droites $(AT)$ et $(BC)$ sont parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30711,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $a_{0}, a_{1}, a_{2}, \\ldots$ une suite d'entiers naturels non nuls, et soit $b_{0}, b_{1}, b_{2}, \\ldots$ la suite telle que $b_{n}=\\operatorname{PGCD}\\left(a_{n}, a_{n+1}\\right)$ pour tout entier $n \\geqslant 0$. Est-il possible que tout entier naturel non nul soit égal à exactement un des termes $b_{0}, b_{1}, b_{2}, \\ldots$ ?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\nDans la suite, on dira qu'une suite $\\left(b_{k}\\right)_{k \\geqslant 0}$ est agréable si, pour tout $k \\geqslant 0$, les entiers $b_{k}$ et $b_{k+2}$ sont premiers entre eux. Étant donnée une suite $\\left(b_{k}\\right)_{k \\geqslant 0}$ agréable, posons $a_{0}=b_{0}$ puis $a_{k}=b_{k-1} b_{k}$ pour tout $k \\geqslant 0$. On constate alors que $\\operatorname{PGCD}\\left(a_{0}, a_{1}\\right)=b_{0}$ et que $\\operatorname{PGCD}\\left(a_{k}, a_{k+1}\\right)=b_{k} \\operatorname{PGCD}\\left(b_{k-1}, b_{k+1}\\right)=b_{k}$ pour tout $k \\geqslant 1$. Il nous suffit donc de construire une suite $\\left(b_{k}\\right)_{k \\geqslant 0}$ agréable telle que tout entier naturel non nul soit égal à exactement un des entiers $b_{0}, b_{1}, b_{2}, \\ldots$\n\nÀ cette fin, et pour tout $k \\geqslant 1$, notons $p_{k}$ le $k^{\\text {ème }}$ plus petit nombre premier, et $q_{k}$ le $k^{\\text {ème }}$ plus petit entier composé et supérieur ou égal à 1. On remarque aisément que $p_{k} \\geqslant 2 k-1$ et que $q_{k} \\leqslant 2 k$, donc que $p_{k} \\geqslant q_{k}-1$. On en déduit que $q_{k} \\leqslant p_{k}+1<2 p_{k}<2 p_{k+2}$, donc que $\\operatorname{PGCD}\\left(q_{k}, p_{k}\\right)=\\operatorname{PGCD}\\left(q_{k}, p_{k+2}\\right)=1$.\n\nOn pose alors $b_{4 k}=p_{2 k+1}$, $b_{4 k+1}=p_{2 k+2}$, $b_{4 k+2}=q_{2 k+1}$ et $b_{4 k+3}=q_{2 k+2}$ pour tout $k \\geqslant 0$. Tout d'abord, tout entier naturel non nul est bien égal à exactement un des entiers $b_{0}, b_{1}, b_{2}, \\ldots$. En outre, pour tout $k \\geqslant 0$, et comme démontré ci-dessus:\n$$\n\\begin{aligned}\n& \\triangleright \\operatorname{PGCD}\\left(b_{4 k}, b_{4 k+2}\\right)=\\operatorname{PGCD}\\left(p_{2 k+1}, q_{2 k+1}\\right)=1, \\\\\n& \\triangleright \\operatorname{PGCD}\\left(b_{4 k+1}, b_{4 k+3}\\right)=\\operatorname{PGCD}\\left(p_{2 k+2}, q_{2 k+2}\\right)=1, \\\\\n& \\triangleright \\operatorname{PGCD}\\left(b_{4 k+2}, b_{4 k+4}\\right)=\\operatorname{PGCD}\\left(p_{2 k+3}, q_{2 k+1}\\right)=1 \\text{ et } \\\\\n& \\triangleright \\operatorname{PGCD}\\left(b_{4 k+3}, b_{4 k+5}\\right)=\\operatorname{PGCD}\\left(p_{2 k+4}, q_{2 k+2}\\right)=1 .\n\\end{aligned}\n$$\nLa suite $\\left(b_{k}\\right)_{k \\geqslant 0}$ est donc bien une suite telle que recherchée, ce qui conclut.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMartin cherche à remplir chaque case d'une grille rectangulaire ayant 8 lignes et $n$ colonnes avec l'une des quatre lettres $\\{P\\}, \\{O\\}, \\{F\\}$ et $M$ de sorte que pour toute paire de lignes distinctes, il existe au plus une colonne telle que ses intersections avec les deux lignes sont des cases ayant la même lettre. Quel est le plus grand entier $n$ tel que cela est possible?",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nDans ce problème, on cherche le plus grand entier satisfaisant une certaine propriété. Supposons que l'on veuille montrer que le plus grand entier recherché est l'entier $c$. Pour montrer que $c$ est bien le plus grand entier, on va d'une part montrer que si un entier $n$ satisfait la propriété, alors $n \\leqslant c$ et d'autre part on va montrer que l'on peut trouver un tableau possédant exactement $c$ colonnes. L'énoncé présente une hypothèse portant les lignes et les colonnes d'un tableau.\n\nOn commence par regarder les informations que l'on possède sur les lignes de ce tableau. Il y a 8 lignes donc il y a $\\binom{8}{2}=28$ paires de lignes.\n\nOn regarde ensuite les informations dont on dispose sur les colonnes du tableau. Puisque l'hypothèse de l'énoncé évoque les lettres identiques au sein d'une même colonne, nous allons compter combien de paires de lettres identiques appartiennent à une même colonne. On fixe donc une colonne du tableau et on regarde combien de paires de lettres identiques on peut réaliser au minimum. Après plusieurs essais sur une colonne de taille 8, on conjecture qu'il y a toujours au moins 4 paires de lettres identiques au sein d'une même colonne. En effet, dans une colonne il y a 8 lettres donc une lettre apparaît au moins 2 fois.\n\n- Si chaque lettre apparaît au plus 2 fois, alors chaque lettre apparaît exactement 2 fois puisqu'il y a 8 cases dans la colonne et 4 lettres possibles. On a donc 4 paires de lettres identiques dans la colonne.\n\n- Si une lettre apparait exactement 3 fois, disons la lettre $P$, alors on peut former 3 paires de lettres $P$. Parmi les 5 lettres restantes, on a au moins une lettre parmi les lettres $O, F$ et $M$ qui apparaît deux fois d'après le principe des tiroirs, ce qui nous fournit une quatrième paire de lettres identiques.\n\n- Si une lettre apparaît au moins 4 fois dans la colonne, disons la lettre $P$, on peut former 4 paires de lettres $P$.\n\nAinsi, dans une colonne, on peut toujours trouver 4 paires de lettres identiques. Donc pour chaque colonne, il y a au moins 4 paires de lignes telles que les intersections avec la colonne ont la même lettre. Mais d'après l'hypothèse de l'énoncé, une paire de lignes ne peut être associée qu'à au plus une colonne telle que ses intersections avec les deux lignes sont deux cases possédant la même lettre. On déduit qu'il y a au plus $4 \\times n$ paires de lignes. Donc $4n \\leqslant 28$ et $n \\leqslant 7$.\n\nRéciproquement, il existe une configuration avec les 7 colonnes suivantes :\n\n| $P$ | $P$ | $P$ | $P$ | $P$ | $P$ | $P$ |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| $O$ | $O$ | $O$ | $O$ | $O$ | $P$ | $O$ |\n| $F$ | $F$ | $F$ | $F$ | $O$ | $O$ | $P$ |\n| $M$ | $M$ | $M$ | $M$ | $P$ | $O$ | $O$ |\n| $P$ | $M$ | $F$ | $O$ | $F$ | $F$ | $F$ |\n| $O$ | $F$ | $M$ | $P$ | $M$ | $F$ | $M$ |\n| $F$ | $P$ | $P$ | $M$ | $M$ | $M$ | $F$ |\n| $M$ | $O$ | $O$ | $F$ | $F$ | $M$ | $M$ |\n\nLe plus grand entier $n$ de colonnes est donc $n=7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes cases d'un échiquier $8 \\times 8$ sont blanches. Un coup consiste à échanger les couleurs des cases d'un rectangle $3 \\times 1$ ou $1 \\times 3$ (les cases blanches deviennent noires et les cases noires deviennent blanches). Est-il possible d'aboutir en un nombre fini de coups à la configuration où toutes les cases de l'échiquier sont noires?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nL'énoncé présente une suite finie d'opérations et le problème demande s'il est possible de partir de la situation initiale pour arriver à une certaine situation finale. Une première idée dans ce cas est de chercher un invariant.\n\nOn peut éventuellement essayer à la main de voir s'il est possible d'aboutir à une configuration où toutes les cases sont noires. Après plusieurs essais, on se rend compte qu'on n'arrive pas à aboutir à la configuration désirée et on peut donc conjecturer que la réponse à l'exercice est négative.\n\nIci, le support de l'énoncé est un échiquier, on va donc chercher un invariant de coloriage, c'est-à-dire que nous allons chercher un coloriage astucieux de l'échiquier qui mette en évidence l'invariant. Quelles sont les caractéristiques d'un bon coloriage ? Tout d'abord, puisque les rectangles considérés possèdent 3 cases, on peut chercher un coloriage qui possède au plus 3 couleurs. Aussi, on peut chercher un coloriage tel que chaque rectangle $3 \\times 1$ ou $1 \\times 3$ possède une case de chaque couleur. On adopte donc le coloriage suivant (où on a remplacé les couleurs par des numéros, étant donné que l'exercice parle déjà de couleurs de cases).\n\n| 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 |\n| 3 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |\n| 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 |\n| 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 |\n| 3 | 1 | 2 | 3 | 1 | 2 | 3 | 1 |\n| 1 | 2 | 3 | 1 | 2 | 3 | 1 | 2 |\n| 2 | 3 | 1 | 2 | 3 | 1 | 2 | 3 |\n\nQue nous apprend ce coloriage ? Chaque rectangle $3 \\times 1$ ou $1 \\times 3$ possède exactement une case de chaque numéro. Donc lorsque l'on change les couleurs d'un rectangle, exactement une case de chaque numéro change de couleur. Puisque les nombres de cases de numéro 1, 2 et 3 ne sont pas identiques, ceci nous encourage à compter à chaque étape le nombre de cases blanches de chaque numéro.\n\nSoit $N_{1}$ le nombre de cases blanches contenant le numéro 1, dites de type 1, $N_{2}$ le nombre de cases blanches de type 2 et $N_{3}$ le nombre de cases blanches de type 3. Dans la situation initiale, $N_{1} = N_{3} = 21$ et $N_{2} = 22$, ces nombres ne sont pas tous égaux. On va donc s'intéresser à l'évolution de la différence entre les nombres $N_{1}$ et $N_{2}$. Changer la couleur des cases d'un rectangle $3 \\times 1$ ou $1 \\times 3$ fait en même temps diminuer ou augmenter chacun des nombres $N_{1}$, $N_{2}$ et $N_{3}$. En particulier, la différence $N_{1} - N_{2}$, à chaque étape, augmente de $2$, $0$ ou $-2$. En particulier, sa parité ne change pas. Or $N_{1} - N_{2}$ est impair au départ. Dans la situation finale demandée, $N_{1} - N_{2} = 0$, $0$ est pair, donc le nombre $N_{1} - N_{2}$ ne peut être nul au bout d'un nombre fini d'étapes. On ne peut donc pas arriver à un échiquier entièrement noir.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30714,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $S$ un ensemble d'entiers relatifs. On dit que $S$ est beau s'il contient tous les entiers de la forme $2^{a}-2^{b}$, où $a$ et $b$ sont des entiers naturels non nuls. On dit également que $S$ est fort si, pour tout polynôme $P(X)$ non constant et à coefficients dans $S$, les racines entières de $P(X)$ appartiennent également à $S$.\nTrouver tous les ensembles qui sont à la fois beaux et forts.",
    "options": [],
    "answer": "the set of all integers",
    "solution": "Solution:\n\nL'ensemble $\\mathbb{Z}$ est clairement beau et fort. Nous allons démontrer que c'est le seul. Pour ce faire, considérons un ensemble $S$ beau et fort : nous allons en fait prouver, par récurrence forte sur $n$, que les entiers $n$ et $-n$ appartiennent nécessairement à $S$.\n\nTout d'abord, puisque $S$ est beau, il contient les entiers $2^{1}-2^{1}=0$, $2^{2}-2^{1}=2$ et $2^{1}-2^{2}=-2$. Il contient donc aussi les entiers $1$ et $-1$, qui sont des racines respectives des polynômes $2-2X$ et $2+2X$.\n\nOn considère désormais un entier $n \\geqslant 3$ tel que $-n-1, \\ldots, n-1$ appartiennent tous à $S$. Soit $\\alpha$ la valuation 2-adique de $n$, et $m$ l'entier impair tel que $n=2^{\\alpha} m$. En notant $\\varphi(m)$ l'indicatrice d'Euler de $m$, on constate alors que l'entier $k=2^{\\alpha+\\varphi(m)+1}-2^{\\alpha+1}$, qui appartient manifestement à $S$, est également un multiple de $2^{\\alpha}$ et de $m$, donc de $n$.\n\nSoit $\\overline{a_{\\ell} a_{\\ell-1} \\ldots a_{0}}$ l'écriture de $k / n$ en base $n$. Tous les entiers $\\pm a_{0}, \\ldots, \\pm a_{\\ell}$ sont compris entre $1-n$ et $n-1$, donc appartiennent à $S$. Par construction, $n$ est une racine entière du polynôme $P(X)=k-\\sum_{i=0}^{\\ell} a_{i} X^{i+1}$, dont tous les coefficients sont dans $S$, donc $n$ est dans $S$ lui aussi. De même, $-n$ est une racine entière du polynôme $Q(X)=k-\\sum_{i=0}^{\\ell} a_{i}(-X)^{i+1}$, donc $-n \\in S$, ce qui conclut la récurrence et la démonstration.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver toutes les fonctions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ telles que\n$$\nf(x+f(y))+x=f(f(2 x)+y)\n$$\npour tous les réels $x$ et $y$.",
    "options": [],
    "answer": "All functions are f(x) = x + a for an arbitrary real constant a.",
    "solution": "Solution:\n\nTout d'abord, on constate aisément que, pour tout réel $a$, la fonction $f: x \\rightarrow x+a$ est une solution du problème. Réciproquement, on va montrer qu'il n'existe pas d'autre solution.\n\nDans ce qui suit, on considère une fonction solution $f$, on pose $a=f(0)$, et on notera $\\mathbf{E}(x, y)$ l'équation\n$$\nf(x+f(y))+x=f(f(2 x)+y)\n$$\nTout d'abord, les équations $\\mathbf{E}(x, 0)$ et $\\mathbf{E}(0,2 x)$ montrent que $f(f(2 x))=f(x+a)+x$ et que $f(f(2 x))=f(2 x+a)$. On en déduit que\n$$\nf(2 x+a)=f(f(2 x))=f(x+a)+x\n$$\net, si l'on pose $x=-a$, il s'ensuit même que $f(-a)=f(0)-a=0$.\n\nOn considère alors un réel $x$ quelconque, et on pose $y=-f(-2 x)-a$ puis $z=-x+f(y)$. Comme $f(-a)=0$, l'équation $\\mathbf{E}(-x, y)$ montre que $f(z)-x=f(-a)=0$, c'est-à-dire que $f(z)=x$. La fonction $f$ est donc surjective.\n\nDémontrons que $f$ est également injective. Soit $x$ et $y$ deux réels tels que $f(x)=f(y)$, et soit $d=y-x$. Soit également $t$ un réel quelconque, et soit $z$ un antécédent de $t$ par $f$. Les équations $\\mathbf{E}(z / 2, x)$ et $\\mathbf{E}(z / 2, y)$ démontrent que\n$$\nf(t+x)=f(f(2 z / 2)+x)=f(z / 2+f(x))+z / 2=f(z / 2+f(y))+z / 2=f(t+y)\n$$\nSi $d \\neq 0$, cela signifie que $f$ est $d$-périodique, ce qui prouve que $f(2 x)=f(2 x+2 d)=f(2 y)$ et que $f(x+a)=f(x+a+d)=f(y+a)$. Mais alors l'équation $\\mathbf{E}(x, 0)$ indique que\n$$\nx=f^{2}(2 x)-f(x+a)=f^{2}(2 y)-f(y+a)=y\n$$\ncontredisant le fait que $d \\neq 0$.\n\nLa fonction $f$ est donc bien injective. Puisque $\\mathbf{E}(0, y)$ montre que $f(f(y))=f(y+a)$, comme on l'a déjà mentionné ci-dessus, il s'ensuit que $f(y)=y+a$ pour tout réel $y$, ce qui conclut.\nSolution:\n\nComme précédemment, on constate que les fonctions de la forme $f : x \\rightarrow x+a$ sont des solutions, et on va montrer que ce sont les seules. À cette fin, on pose $a=f(0)$ et on note $\\mathbf{E}(x, y)$ l'équation\n$$\nf(x+f(y))+x=f(f(2 x)+y)\n$$\nSoit alors $y$ un nombre réel quelconque, et $x=y+a-f(y)$. Alors\n$$\n\\begin{aligned}\nf(f(2 x)+y+a) & =f(f(f(2 x)+y)) & & \\text{ d'après } \\mathbf{E}(0, f(2 x)+y) \\\\\n& =f(f(x+f(y))+x) & & \\text{ d'après } \\mathbf{E}(x, y) \\\\\n& =f(f(2 x)+x+f(y))-x & & \\text{ d'après } \\mathbf{E}(x, x+f(y)) \\\\\n& =f(f(2 x)+y+a)-x & & \\text{ car } f(2 x)+y+a=f(2 x)+x+f(y)\n\\end{aligned}\n$$\nCela signifie que $x=0$, ou encore que $f(y)=y+a$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30716,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans 5 boîtes se trouvent respectivement $402$, $403$, $404$, $405$ et $406$ pierres. La seule opération autorisée est de prendre $4$ pierres dans un tas ayant au moins $4$ pierres et d'en mettre une dans chacun des autres tas. Quel est le plus grand nombre de pierres qu'il est possible d'avoir dans un seul tas?",
    "options": [],
    "answer": "2014",
    "solution": "Solution:\n\nL'énoncé présente une suite d'opérations et décrit une certaine situation initiale. Une première idée à essayer dans ce cas est de chercher un invariant.\n\nUne deuxième idée est de tester l'énoncé pour des valeurs plus petites, par exemple pour des tas de taille $0,1,2,3$ et $4$. Après plusieurs essais, on remarque que les tas ont toujours, à permutation près, les tailles $0,1,2,3$ et $4$. Après avoir testé sur des tas de taille $1,2,3,4$ et $5$, on s'aperçoit que les tas ont toujours des valeurs distinctes deux à deux modulo $5$. On s'empresse de démontrer cette conjecture.\n\nEnlever $4$ pierres d'un tas revient, modulo $5$, à en ajouter $1$. Comme on en ajoute une dans chacun des autres tas, les valeurs des tas augmentent toutes de $1$ modulo $5$. Ainsi, si avant l'opération ces valeurs étaient distinctes deux à deux modulo $5$, après l'opération elles resteront distinctes deux à deux modulo $5$. On a donc déterminé notre invariant.\n\nAinsi la somme des tailles de $4$ tas quelconques ne pourra jamais dépasser $0+1+2+3=6$ et donc un tas quelconque ne pourra jamais contenir plus que $2020-6=2014$ pierres.\n\nRéciproquement, nous devons désormais démontrer que l'on peut toujours faire en sorte qu'un tas contienne $2014$ pierres au bout d'un nombre fini d'opérations. Pour cela, il suffit de ne jamais toucher au tas contenant $404$ pierres, et de faire l'opération sur les autres tas jusqu'à ce que ce ne soit plus possible de faire une seule opération. Ainsi les tas restants auront $0,1,2,3$ pierres à la fin du procédé, donc il y aura bien $2020-(0+1+2+3)=2014$ pierres sur le même tas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les triplets $(x, y, z)$ d'entiers naturels satisfaisant l'équation:\n$$\n2^{x} + 3^{y} = z^{2}\n$$",
    "options": [],
    "answer": "[[0, 1, 2], [3, 0, 3], [4, 2, 5]]",
    "solution": "Solution:\n\nTout d'abord, analysons le problème : on a une équation diophantienne avec une puissance de $2$, une puissance de $3$ et un carré. On s'empresse donc de tester les petites valeurs de $z$ et de trouver la solution. On peut remarquer que $2^{0} + 3^{1} = 2^{2}$, $2^{3} + 3^{0} = 3^{2}$ et $2^{4} + 3^{2} = 5^{2}$. A priori comme on a quelques solutions, on ne va pas trouver de contradictions immédiates simplement en regardant modulo quelque chose.\n\nNéanmoins on a un carré, donc il est naturel de regarder modulo $4$ ou modulo $3$ et d'essayer d'obtenir des informations sur $x$ et $y$. Idéalement, on aimerait obtenir que soit $x$ soit $y$ est pair, comme ça on aurait un autre carré dans l'équation ce qui nous permettrait de factoriser. On va donc regarder modulo $4$, pour cela on a envie de traiter en premier le cas où $4$ divise $2^{x}$, i.e. le cas $x \\geqslant 2$.\n\nSupposons $x \\geqslant 2$. Comme un carré est congru à $0$ ou $1$ modulo $4$, $z^{2}$ est congru à $0$ ou $1$ modulo $4$, donc $3^{y}$ aussi. Comme il n'est pas divisible par $4$, on a forcément $3^{y} \\equiv 1 \\pmod{4}$. Comme $3^{2} \\equiv 9 \\equiv 1 \\pmod{4}$, on a nécessairement $y$ pair (si $y = 2k+1$, $3^{y} \\equiv 9^{k} \\times 3 \\equiv 3 \\pmod{4}$). Posons $y = 2k$ avec $k$ entier, on obtient\n$$\n2^{x} = z^{2} - (3^{k})^{2} = (z - 3^{k})(z + 3^{k})\n$$\nNotons que $z + 3^{k}$ est strictement positif, donc ces deux facteurs divisent $2^{x}$.\n\nIci on a deux termes dont le produit vaut $2^{x}$, on aimerait montrer que l'un d'entre eux ne peut pas être très grand. Pour cela on va considérer le pgcd de $z + 3^{k}$ et $z - 3^{k}$ et le calculer en utilisant qu'il divise leur somme, différence et produit.\n\nSoit $d$ le pgcd de $z - 3^{k}$ et $z + 3^{k}$, alors $d$ divise $(z + 3^{k}) - (z - 3^{k}) = 2 \\times 3^{k}$, mais $d$ divise aussi $2^{x}$ donc $d$ divise $2$. Comme $z + 3^{k} \\equiv z - 3^{k} \\pmod{2}$, si $d$ vaut $1$, $z - 3^{k}$ et $z + 3^{k}$ sont impairs donc $2^{x}$ aussi, ce qui contredit le fait que $x \\geqslant 2$. On a donc $d = 2$.\n\nEn particulier, comme $z - 3^{k}$ et $z + 3^{k}$ sont des puissances de $2$ de pgcd $2$, le plus petit des deux vaut $2$ donc comme $z + 3^{k} > z - 3^{k}$, $z = 3^{k} + 2$. Considérer le pgcd ici n'est pas nécessaire, on aurait pu directement aboutir à cela en posant $z + 3^{k} = 2^{a}$, $z - 3^{k} = 2^{b}$, observé que $a > b$ et écrit $2 \\times 3^{k} = 2^{a} - 2^{b} = 2^{b}(2^{a-b} - 1)$, mais un bon réflexe quand on a un produit est de regarder le pgcd des facteurs.\n\nDe ceci, on déduit que $z + 3^{k} = 2^{x-1}$ et $z - 3^{k} = 2$, donc en faisant la différence $2 \\times 3^{k} + 2 = 2^{x-1}$ donc $3^{k} + 1 = 2^{x-2}$. Cette équation est assez connue, mais si on ne la connaît pas, comme on a des puissances de $2$ et de $3$, il est pertinent de regarder modulo $2, 4, 8$ ou $3, 9$. Comme $3 + 1 = 2^{2}$, regarder modulo $4$ et $3$ ne conclura pas tout de suite, on regarde donc modulo $8$.\n\nSi $x \\geqslant 5$, $8$ divise $2^{x-2}$ donc $8$ divise $3^{k} + 1$. Or les puissances de $3$ modulo $8$ valent $1$ ou $3$ donc $3^{k} + 1$ vaut $2$ ou $4$ modulo $8$ et n'est pas divisible par $8$, ce qui donne une contradiction. On a donc forcément $x = 2, 3$ ou $4$.\n\nPour $x = 2$, $1 = 1 + 3^{k}$ donc $3^{k} = 0$ ce qui est impossible. Pour $x = 3$, $1 + 3^{k} = 2$ donc $3^{k} = 1$ donc $k = 0$ et $y = 0$. Pour $x = 4$, $1 + 3^{k} = 4$ donc $3^{k} = 3$, on a donc $k = 1$ soit $y = 2$.\n\nRéciproquement, si $(x, y) = (3, 0)$, $z^{2} = 2^{x} + 3^{y} = 9 = 3^{2}$ donc $z = 3$ et le triplet $(3, 0, 3)$ convient. Si $(x, y) = (4, 2)$, $z^{2} = 2^{x} + 3^{y} = 25 = 5^{2}$ donc $z = 5$ et le triplet $(4, 2, 5)$ convient.\n\nIl reste donc à traiter le cas où $x = 0$ et celui où $x = 1$. Pour le cas où $x = 0$, on a $2^{x} = 1$ qui est un carré, on va donc factoriser et réutiliser la technique du pgcd. Si $x = 0$, l'équation se réécrit $3^{y} = z^{2} - 1 = (z - 1)(z + 1)$. Le pgcd de $z - 1$ et $z + 1$ divise $3^{y}$ et divise $z + 1 - (z - 1) = 2$ donc il vaut nécessairement $1$. Comme $3^{y} = (z - 1)(z + 1)$, $z - 1$ et $z + 1$ sont des puissances de $3$ positives (car $z + 1 > 0$ donc $z - 1 > 0$ leur produit étant strictement positif) de pgcd $1$, le plus petit des deux vaut nécessairement $1$ donc $z - 1 = 1$, $z = 2$. On obtient $3^{y} = z^{2} - 1 = 3$ donc $y = 1$. Réciproquement $(0, 1, 2)$ est solution car $1 + 3 = 2^{2}$.\n\nPour le cas $x = 1$, on ne voit pas de solution, il faut donc trouver un modulo pertinent pour évaluer l'équation, comme il y a une puissance de $3$ et un carré, on peut regarder modulo $3$. Si $x = 1$ l'équation devient $2 + 3^{y} = z^{2}$. Si $y = 0$ l'équation devient $z^{2} = 3$ qui n'a pas de solution entière. On a donc $y > 0$ donc $z^{2} \\equiv 2 \\pmod{3}$. Or un carré modulo $3$ est congru à $0$ ou $1$, il n'y a donc pas de solution pour $x = 1$.\n\nLes solutions sont donc $(0, 1, 2)$, $(3, 0, 3)$ et $(4, 2, 5)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30718,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k$ et $n$ deux entiers naturels non nuls, tels que $k \\leqslant 2^{n}$. Morgane a écrit, sur son cahier, l'ensemble des $n$-uplets formés de 0 et de 1 : il y en a $2^{n}$. On dit que deux $n$-uplets $\\left(x_{1}, \\ldots, x_{n}\\right)$ et $\\left(y_{1}, \\ldots, y_{n}\\right)$ sont voisins s'ils ont $n-1$ termes en commun, c'est-à-dire s'il existe un seul entier $i$ tel que $x_{i} \\neq y_{i}$. Puis Morgane choisit $k$ de ces $n$-uplets et les souligne. Elle effectue ensuite les opérations suivantes : lors de chaque opération, elle choisit un $n$-uplet dont deux voisins sont soulignés, et elle souligne également ce $n$-uplet.\n\nPour quelles valeurs de $k$ Morgane peut-elle parvenir à souligner l'ensemble des $n$-uplets?",
    "options": [],
    "answer": "Exactly those k with 1 + ceil(n/2) ≤ k ≤ 2^n",
    "solution": "Solution:\n\nDans toute la suite, on assimile nos $n$-uplets à des vecteurs de $(\\mathbb{Z} / 2 \\mathbb{Z})^{n}$, c'est-à-dire des vecteurs dont les $n$ coordonnées sont des éléments de $\\mathbb{Z} / 2 \\mathbb{Z}$. Ainsi, on note :\n\n$\\triangleright$ $0$ le vecteur dont toutes les coordonnées sont nulles,\n\n$\\triangleright$ $\\mathbf{a}_{i}$ le vecteur dont la $i^{\\text{ème}}$ coordonnée vaut 1 et les autres valent 0,\n\n$\\triangleright$ $\\mathbf{b}_{i, j}$ le vecteur dont les $i$ème et $j^{\\text{ème}}$ coordonnées valent 1 et les autres valent 0.\n\nDans un premier temps, nous allons montrer que l'entier $k=1+\\lceil n / 2\\rceil$ convient, où $\\lceil x\\rceil$ désigne la partie entière supérieure du réel $x$, c'est-à-dire le plus petit entier supérieur ou égal à $x$. Pour ce faire, on peut supposer que Morgane a d'abord souligné les vecteurs $\\mathbf{0}, \\mathbf{b}_{1,2}, \\mathbf{b}_{3,4}, \\ldots, \\mathbf{b}_{2k-3,2k-2}$ ; si $n$ est pair, ce dernier vecteur est le vecteur $\\mathbf{b}_{n-1, n}$ et, si $n$ est impair, par abus de notation, on identifiera le vecteur $\\mathbf{b}_{n, n+1}$ au vecteur $\\mathbf{a}_{n}$. Dans ces conditions :\n\n$\\triangleright$ si $i$ est pair, alors $\\mathbf{a}_{i}$ est voisin des deux vecteurs $\\mathbf{0}$ et $\\mathbf{b}_{i-1, i}$ ;\n\n$\\triangleright$ si $i$ est impair, avec $i \\leqslant n-1$, alors $\\mathbf{a}_{i}$ est voisin des deux vecteurs $\\mathbf{0}$ et $\\mathbf{b}_{i, i+1}$ ;\n\n$\\triangleright$ si $i=n$ est impair, alors $\\mathbf{a}_{i}=\\mathbf{a}_{n}=\\mathbf{b}_{2k-3,2k-2}$ est déjà souligné.\n\nAinsi, Morgane peut souligner chacun des vecteurs $\\mathbf{a}_{i}$.\n\nUne récurrence immédiate sur $\\ell$ permet alors de montrer que Morgane peut souligner tout vecteur dont $\\ell$ coordonnées sont égales à 1, et les $n-\\ell$ autres sont égales à 0. On en conclut que tout entier $k$ compris entre $1+\\lceil n / 2\\rceil$ et $2^{n}$ convient.\n\nRéciproquement, soit $k$ un des entiers recherchés : démontrons que $k \\geqslant 1+\\lceil n / 2\\rceil$. Pour ce faire, on construit d'abord le graphe orienté dont les sommets sont les vecteurs de $(\\mathbb{Z} / 2 \\mathbb{Z})^{n}$ et dont les arêtes sont construites comme suit :\n\n$\\triangleright$ si un vecteur $t$ était un des $k$ vecteurs que Morgane avait soulignés initialement, alors aucune arête n'arrive en $t$ ;\n\n$\\triangleright$ sinon, soit $\\mathbf{u}$ et $\\mathbf{v}$ deux voisins de $\\mathbf{t}$ que Morgane avait soulignés avant de souligner $\\mathbf{t}$ : le graphe contient une arête qui va de $\\mathbf{u}$ vers $t$, et une autre arête qui va de $\\mathbf{v}$ vers $t$.\n\nIntroduisons maintenant quelques notions et concepts. On dit qu'un ensemble $S$ de vecteurs est un ensemble initial si toute arête entrant en un sommet de $S$ provient d'un autre sommet de $S$. On dit également que $S$ est connexe si, quitte à prendre les arêtes dans le mauvais sens, on peut relier n'importe quel vecteur de $S$ à n'importe quel autre sans sortir de $S$. En outre, on dit que les sources de $S$ sont les vecteurs de $S$ que Morgane avait soulignés initialement, c'est-à-dire ceux en lesquels n'entre aucune arête. Enfin, on dit qu'une coordonnée $i$ est une variable de $S$ s'il existe deux vecteurs $\\mathbf{t}$ et $\\mathbf{t}'$ de $S$ tels que $\\mathbf{t}_{i} \\neq \\mathbf{t}'_{i}$.\n\nNous allons maintenant démontrer la propriété $\\mathcal{P}_{\\ell}$ par récurrence forte sur $\\ell$ :\n\nSoit $S$ un ensemble initial connexe contenant $\\ell$ vecteurs et $m$ sources. Alors $S$ a au plus $2(m-1)$ coordonnées variables.\n\nTout d'abord, la propriété $\\mathcal{P}_{1}$ est évidente, puisque $S$ est alors réduit à un seul vecteur, qui est lui-même une source, et n'a donc aucune coordonnée variable. Soit maintenant $\\ell \\geqslant 2$ un entier tel que les propriétés $\\mathcal{P}_{1}, \\ldots, \\mathcal{P}_{\\ell-1}$ soient toutes vraies, et soit $S$ un ensemble initial connexe contenant $\\ell$ vecteurs, $m$ sources, et $c$ coordonnées variables.\n\nNotons $\\mathbf{t}$ le dernier vecteur, parmi ceux appartenant à $S$, que Morgane a souligné. Si $t$ est une source, c'est que $S$ n'est formé que de sources, donc ne contient aucune arête, en contradiction avec le fait que $S$ soit connexe et que $\\ell \\geqslant 2$. Ainsi, il existe deux arêtes $\\mathbf{u} \\rightarrow \\mathbf{t}$ et $\\mathbf{v} \\rightarrow \\mathbf{t}$, provenant de deux autres vecteurs $\\mathbf{u}$ et $\\mathbf{v}$ de $S$. Soit $i$ et $j$ les deux coordonnées telles que $\\mathbf{t}_{i} \\neq \\mathbf{u}_{i}$ et $\\mathbf{t}_{j} \\neq \\mathbf{v}_{j}$, et soit $S'$ l'ensemble $S \\setminus \\{\\mathbf{t}\\}$.\n\n$\\triangleright$ Si $S'$ est connexe, on constate aisément que les coordonnées variables de $S$ sont celles de $S'$. Puisque $S'$ contient $\\ell-1$ sommets et $m$ sources, les ensembles $S$ et $S'$ ont donc bien $2(m-1)$ coordonnées variables au maximum.\n\n$\\triangleright$ Si $S'$ n'est pas connexe, alors on peut le couper en deux ensembles initiaux connexes, $U$ et $V$, qui contiennent respectivement les vecteurs $\\mathbf{u}$ et $\\mathbf{v}$. L'un contient $\\ell_{u}$ vecteurs, $m_{u}$ sources, et $c_{u}$ coordonnées variables, tandis que l'autre contient $\\ell_{v}$ vecteurs, $m_{v}$ sources, et $c_{v}$ coordonnées variables. Puisque $\\ell_{u} \\leqslant \\ell-1$ et $\\ell_{v} \\leqslant \\ell-1$, on sait que $c_{u} \\leqslant 2(m_{u}-1)$ et que $c_{v} \\leqslant 2(m_{v}-1)$.\n\nOr, les sources de $S$ sont celles de $U$ et de $V$, et nulle source ne peut appartenir à la fois à $U$ et à $V$, de sorte que $m=m_{u}+m_{v}$. D'un autre côté, on constate aisément que les coordonnées variables de $S$ sont soit celles de $U$, soit celles de $V$, soit $i$ et $j$. Ainsi, l'ensemble $S$ a au plus $c_{u}+c_{v}+2 \\leqslant 2(m_{u}+m_{v}-1)=2(m-1)$ coordonnées variables, ce qui démontre $\\mathcal{P}_{\\ell}$.\n\nRemarquons enfin que l'ensemble $(\\mathbb{Z} / 2 \\mathbb{Z})^{n}$ lui-même est un ensemble initial connexe, et qu'il contient $n$ variables et $k$ sources. La propriété $\\mathcal{P}_{2^{n}}$ montre donc que $n \\leqslant 2(k-1)$, c'est-à-dire que $k \\geqslant 1+\\lceil n / 2\\rceil$. Les entiers recherchés sont donc bien les entiers compris entre $1+\\lceil n / 2\\rceil$ et $2^{n}$.\nSolution:\n\nOn démontre comme ci-dessus que tout entier $k \\geqslant 1+\\lceil n / 2\\rceil$ convient, et il ne nous reste maintenant plus qu'à montrer que l'on doit nécessairement avoir $k \\geqslant 1+\\lceil n / 2\\rceil$. Pour ce faire, on va procéder par récurrence sur $n$.\n\nTout d'abord, pour $n=1$ et $n=2$, si Morgane ne souligne qu'un seul vecteur, aucun des autres vecteurs non soulignés ne pourra avoir deux voisins soulignés. On en déduit, dans ces cas-là, que $k \\geqslant 2$, c'est-à-dire que $k \\geqslant 1+\\lceil n / 2\\rceil$.\n\nSupposons maintenant que $n \\geqslant 3$, et que Morgane a souligné $k$ vecteurs, de sorte qu'elle pourra ensuite parvenir à souligner chacun des $2^{n}-k$ vecteurs restants. Morgane numérote alors les $2^{n}$ vecteurs dans l'ordre dans lequel elle les a soulignés : elle peut donc choisir les vecteurs $v_{1}, \\ldots, v_{k}$ de manière arbitraire, et elle pourra ensuite souligner les vecteurs $v_{k+1}, \\ldots, v_{2^{n}}$. En particulier, on constate que, pour tout $\\ell \\leqslant k+1$, le vecteur $v_{\\ell}$ est voisin de deux vecteurs $v_{i}$ et $v_{j}$ tels que $1 \\leqslant i<j<\\ell$.\n\nSoient alors $i$ et $j$ deux entiers tels que $1 \\leqslant i<j \\leqslant k$ et tels que $v_{i}$ et $v_{j}$ soient tous les deux voisins du vecteur $v_{k+1}$. Soient ensuite $s$ et $t$ les deux coordonnées en lesquelles les vecteurs $v_{i}$ et $v_{j}$ diffèrent l'un de l'autre. Morgane décide alors d'effacer les coordonnées $s$ et $t$ de l'ensemble des $2^{n}$ vecteurs.\n\nElle se retrouve ainsi avec $2^{n-2}$ vecteurs à $n-2$ coordonnées, chacun étant présent en quatre exemplaires. Notons $\\phi(v)$ le vecteur à $n-2$ coordonnées obtenu en effaçant les coordonnées $s$ et $t$ du vecteur $v$. Parmi les $2^{n-2}$ vecteurs obtenus, Morgane décide, initialement, de souligner les vecteurs $\\phi\\left(v_{1}\\right), \\phi\\left(v_{2}\\right), \\ldots, \\phi\\left(v_{k}\\right)$ : puisque $\\phi\\left(v_{i}\\right)=\\phi\\left(v_{j}\\right)$, elle a souligné au plus $k-1$ vecteurs.\n\nEn outre, si deux vecteurs $v$ et $v'$ à $n$ coordonnées sont voisins l'un de l'autre, alors les vecteurs $\\phi(v)$ et $\\phi\\left(v'\\right)$ sont soit égaux, soit voisins. Ainsi, Morgane peut toujours souligner, dans cet ordre, les vecteurs $\\phi\\left(v_{k+1}\\right)$ - qui est égal à $\\phi\\left(v_{i}\\right)$ et à $\\phi\\left(v_{j}\\right)$ - puis $\\phi\\left(v_{k+2}\\right), \\ldots, \\phi\\left(v_{2^{n}}\\right)$, quitte à resouligner un vecteur déjà souligné.\n\nPar conséquent, en soulignant au départ $k-1$ vecteurs à $n-2$ coordonnées, Morgane est bien parvenue à souligner l'intégralité des $2^{n-2}$ vecteurs à $n-2$ coordonnées. L'hypothèse de récurrence indique donc que $k-1 \\geqslant \\lceil(n-2) / 2\\rceil+1$, c'est-à-dire que $k \\geqslant \\lceil n / 2\\rceil+1$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn mauvais sorcier a enfermé $n$ mathématiciens. Il dispose de $n$ couleurs. Le sorcier place sur la tête de chaque mathématicien un chapeau d'une des $n$ couleurs; deux chapeaux peuvent avoir la même couleur. Chaque mathématicien peut voir la couleur du chapeau de chacun de ses collègues mais pas la sienne. Les mathématiciens doivent alors tous en même en temps annoncer la couleur de leur chapeau. Si au moins un des mathématiciens devine correctement, tous les mathématiciens sont libres. Ils peuvent bien sûr se concerter sur une stratégie avant de connaître la couleur des chapeaux des autres mathématiciens.\nProposez une stratégie qui assure que les mathématiciens soient libérés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, nous allons coder les différentes informations. Par exemple, on numérote les couleurs de $1$ à $n$ et on numérote les mathématiciens de $1$ à $n$ également. On note $i_{1}, \\ldots, i_{n}$ les couleurs des chapeaux des mathématiciens.\n\nPuisqu'un mathématicien ne peut pas voir son propre chapeau mais uniquement les chapeaux des autres, il va devoir déclarer un nombre qui dépend des chapeaux qu'il observe. On peut par exemple chercher une expression qui va dépendre de façon symétrique de chacun des chapeaux.\n\nUne information importante et qui dépend de façon symétrique de chacun des chapeaux est par exemple la somme des numéros de tous les chapeaux, à savoir $i_{1} + i_{2} + \\ldots + i_{n}$. Cette somme admet un certain reste $s$ modulo $n$. On a donc $i_{l} \\equiv s - \\sum_{j \\neq l} i_{j} \\bmod n$. On doit donc s'assurer qu'il existe un indice $l$ tel que le mathématicien $l$ déclare le numéro $s - \\sum_{j \\neq l} i_{j} \\bmod n$. On reconnaît dans $\\sum_{j \\neq l} i_{j}$ la somme des numéros des chapeaux vus par le mathématicien $l$.\n\nAinsi, si chaque mathématicien de numéro $k$ déclare le nombre entre $1$ et $n$ congru à $s - \\sum_{j \\neq k} i_{j} \\bmod n$, alors le mathématicien dont le numéro est congru à $s$ modulo $n$ déclarera le numéro $s - \\sum_{j \\neq s} i_{j} = i_{s} \\bmod n$ qui est la couleur de son chapeau. Tous les mathématiciens sont donc libérés !",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $C$ un entier naturel non nul. Trouver toutes les fonctions $f: \\mathbb{N}^{*} \\rightarrow \\mathbb{N}^{*}$ telles que, pour tous les entiers $a$ et $b$ de somme $a+b \\geqslant C$, l'entier $a+f(b)$ divise $a^{2}+b f(a)$.",
    "options": [],
    "answer": "All functions of the form f(n) = k n for a positive integer k",
    "solution": "Solution:\n\nTout d'abord, toute fonction linéaire strictement croissante est solution. En effet, pour tout entier $k \\geqslant 1$, l'entier $a+k b$ divise bien $a^{2}+b \\times(k a)=a(a+k b)$. Réciproquement, montrons que toute solution est une fonction linéaire (qui sera strictement croissante, puisque à valeurs dans $\\mathbb{N}^{*}$ ).\n\nConsidérons un entier $n \\geqslant C$, et posons $\\varphi=f(1)$. Alors\n$$\n\\varphi^{2}+f(n) \\equiv n^{2}+f(n) \\equiv 0 \\quad(\\bmod n+\\varphi)\n$$\ndonc $f(n)+\\varphi^{2}$ est un multiple non nul de $n+\\varphi$, de sorte que $f(n) \\geqslant n+\\varphi-\\varphi^{2}$.\n\nD'autre part, puisque $1+f(n)$ divise $1+\\varphi n$, soit $g(n)=(1+\\varphi n) /(1+f(n))$. Alors\n$$\n\\varphi(1+f(n)) \\geqslant \\varphi\\left(n+1+\\varphi-\\varphi^{2}\\right)=(1+f(n)) g(n)-1+\\left(1+\\varphi-\\varphi^{2}\\right) \\varphi\n$$\nCela signifie que $\\Phi \\geqslant(1+f(n))(g(n)-\\varphi)$, où l'on a posé $\\Phi=\\varphi^{3}-\\varphi^{2}-\\varphi+1$. Par conséquent, si $n \\geqslant \\Phi+\\varphi^{2}-\\varphi$, on sait que $1+f(n)>\\Phi$, et donc que $g(n) \\leqslant \\varphi$.\n\nChoisissons maintenant un entier $a \\geqslant 1$, et démontrons que $f(a)=\\varphi a$. Pour ce faire, on construit un entier $n \\geqslant \\max \\{\\Phi+\\varphi^{2}-\\varphi, C\\}$ comme suit :\n\n$\\triangleright$ on prend $n \\equiv 2(\\bmod 4)$ si $\\varphi$ est impair, et $n$ impair si $\\varphi$ est pair : dans tous les cas, $n \\varphi+1$ est impair;\n\n$\\triangleright$ pour tout nombre premier $p \\leqslant \\max \\{a, \\varphi\\}$ impair, on choisit $n \\equiv 1(\\bmod p)$ ou $n \\equiv 2$ $(\\bmod p)$ de sorte que $n \\varphi \\not \\equiv-1(\\bmod p)$.\n\nD'après le théorème chinois, un tel choix est bien faisable, et ce pour une infinité d'entiers $n$. Sans perte de généralité, on suppose donc même que $n \\geqslant 2 \\max \\{\\varphi a, f(a)\\}$.\n\nAlors $g(n)$ est un diviseur de $1+\\varphi n$, et l'on sait que $g(n) \\leqslant \\varphi$. Par construction, l'entier $1+\\varphi n$ n'a aucun facteur premier $p \\leqslant \\varphi$, de sorte que $g(n)=1$, et donc que $f(n)=\\varphi n$. Mais alors $a+\\varphi n=a+f(n)$ divise\n$$\n\\left(a^{2}+n f(a)\\right)+(a+\\varphi n)(\\varphi n-a)=(\\varphi n)^{2}+n f(a)=\\left(\\varphi^{2} n+f(a)\\right) n\n$$\nOr, on a construit $n$ de sorte qu'il n'ait aucun facteur premier impair commun avec $a$, et ne soit pas divisible par 4. Par conséquent, si l'on pose $d=\\operatorname{PGCD}(a+\\varphi n, n)$, alors $d=\\operatorname{PGCD}(a, n)$ divise 2. Puis, si l'on pose $\\alpha=(a+\\varphi n) / d$ et $n'=n / d$, on constate que $\\operatorname{PGCD}\\left(\\alpha, n'\\right)=1$ et que $\\alpha$ divise $\\left(\\varphi^{2} n+f(a)\\right) n'$, de sorte que $\\alpha$ divise $\\varphi^{2} n+f(a)$. L'entier $\\alpha$ divise donc également\n$$\n\\left(\\varphi^{2} n+f(a)\\right)-d \\varphi \\alpha=f(a)-\\varphi a\n$$\nOr, comme $n \\geqslant 2 \\max \\{\\varphi a, f(a)\\}$, on sait que\n$$\n\\alpha>\\varphi n / d \\geqslant \\varphi n / 2 \\geqslant \\max \\{\\varphi a, f(a)\\} \\geqslant|f(a)-\\varphi a| .\n$$\nC'est donc que $f(a)=\\varphi a$, ce qui conclut.\n\n\nSolution alternative $n^{\\circ} 1$\n\nComme précédemment, on pose $\\varphi=f(1)$, et on démontre que, si $f$ est une fonction solution, alors $f(n)=\\varphi n$ pour tout entier $n \\geqslant 1$.\n\nTout d'abord, pour tout entier $n \\geqslant C$, l'énoncé indique que $1+f(n)$ divise $1+n \\varphi$. On en déduit que $1+f(n) \\leqslant 1+n \\varphi$, donc que $f(n) \\leqslant n \\varphi$.\n\nOn considère maintenant un entier $m \\geqslant 2$ quelconque, puis un entier $k \\geqslant 1$ tel que $k m-f(m) \\geqslant C$. Si l'on pose $a=k m-f(m)$, et puisque $a+m \\geqslant a \\geqslant C$, l'énoncé indique que\n$$\nf(m)^{2}+m f(a) \\equiv (-a)^{2}+m f(a) \\equiv 0 \\quad(\\bmod a+f(m)) .\n$$\nL'entier $f(m)^{2}+m f(a)$ est donc divisible par $a+f(m)=k m$, et par $m$ également. On en conclut que $f(m)^{2}$ est lui aussi divisible par $m$; bien sûr, cette relation de divisibilité également valable pour $m=1$, mais on n'en aura pas besoin.\n\nEn particulier, si $p$ est un nombre premier, et comme $p$ divise $f(p)^{2}$, l'entier $p$ divise aussi $f(p)$ : dans la suite, on pose $g(p)=f(p) / p$, et l'on sait que $1 \\leqslant g(p) \\leqslant \\varphi$. Si, en outre, $p \\geqslant \\max \\{C, \\varphi+1\\}$, alors $p+1 \\geqslant C$, et l'énoncé indique donc que $p+\\varphi$ divise $p^{2}+f(p)=p(p+g(p))$. Puisque $p>\\varphi$, c'est que $p$ et $p+\\varphi$ sont premiers entre eux, donc que $p+\\varphi$ divise $p+g(p)$. Comme $1 \\leqslant g(p) \\leqslant \\varphi$, on en déduit que $g(p)=\\varphi$, c'est-à-dire que $f(p)=\\varphi p$.\n\nEnfin, soit $\\ell \\geqslant 1$ un entier quelconque, et soit $p$ un nombre premier tel que $p \\geqslant \\max \\{C, \\varphi+1, \\ell+1\\}$. On vient de voir que $f(p)=\\varphi p$. En outre, puisque $\\ell+p \\geqslant p \\geqslant C$, l'énoncé indique que\n$$\np(f(\\ell)-\\varphi \\ell) \\equiv p f(\\ell)-\\ell f(p) \\equiv p f(\\ell)+\\ell^{2} \\equiv 0 \\quad(\\bmod \\ell+f(p))\n$$\nOr, comme $1 \\leqslant \\ell \\leqslant p-1$, on sait que $p$ est premier avec $\\ell$, donc avec $\\ell+\\varphi p=\\ell+f(p)$ également. On en déduit que $\\ell+\\varphi p=\\ell+f(p)$ divise $f(\\ell)-\\varphi \\ell$. Ceci étant valable pour des nombres premiers $p$ arbitrairement grands, on en déduit que $f(\\ell)=\\ell \\varphi$, ce qui conclut.\n\n\nSolution alternative $n^{\\circ} 2$\n\nOn présente une autre façon, légèrement différente, de conclure une fois que l'on a montré que $p$ divise $f(p)$ pour tout nombre premier $p$.\n\nSoit $p$ un nombre premier, et soit $g(p)$ l'entier $f(p) / p$. Puisque $g(p) \\leqslant \\varphi$ dès lors que $p \\geqslant C$, la fonction $g$ est bornée. Il existe donc un entier $\\gamma$ et une infinité de nombres premiers $p$ pour lesquels $g(p)=\\gamma$.\n\nSoit alors $\\ell$ un entier, puis $p \\geqslant \\max \\{C, \\ell+1\\}$ un nombre premier pour lequel $g(p)=\\gamma$. Alors $p$ est premier avec $\\ell$, donc avec $\\ell+\\gamma p=\\ell+f(p)$. Or, puisque $\\ell+p \\geqslant p \\geqslant C$, l'énoncé indique que\n$$\np(f(\\ell)-\\gamma \\ell) \\equiv p f(\\ell)-\\ell f(p) \\equiv p f(\\ell)+\\ell^{2} \\equiv 0 \\quad(\\bmod \\ell+f(p))\n$$\nOn en déduit que $\\ell+\\gamma p=\\ell+f(p)$ divise $f(\\ell)-\\gamma \\ell$. Ceci étant valable pour des nombres premiers $p$ arbitrairement grands, on en déduit que $f(\\ell)=\\gamma \\ell$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn coloriage des entiers $\\{1,2, \\cdots, 2020\\}$ en bleu et rouge est dit agréable s'il n'existe pas deux entiers distincts dans $\\{1,2, \\cdots, 2020\\}$ de même couleur dont la somme est une puissance de 2. Combien de tels coloriages existent-ils?",
    "options": [],
    "answer": "2^11",
    "solution": "Solution:\n\nUne première idée est de tester l'énoncé pour des valeurs plus petites, par exemple pour un coloriage des entiers de $1$ à $7$.\n\nColorions dans l'ordre les nombres : pour colorier $1$, on n'a pas de contrainte apparente, idem pour colorier $2$. Par contre, comme $3+1=4$, la couleur de $3$ est imposée. Pour colorier $4$, il n'y a pas de contrainte apparente car $4+3=7$, $4+2=6$ et $4+1=5$ ne sont pas des puissances de deux. Par contre, comme $5+3=8$, $6+2=8$, $7+1=8$, la couleur de $5$, $6$ et $7$ est imposée. En testant tous les coloriages possibles, on voit que peu importe les couleurs choisies pour colorier $1$, $2$ et $4$, le coloriage obtenu est agréable. On peut donc supposer qu'un coloriage agréable est fixé une fois la couleur des puissances de $2$ choisie et que tout coloriage des puissances de $2$ peut être prolongé en un coloriage agréable.\n\nNous allons montrer que pour tout coloriage des puissances de $2$ entre $1$ et $2020$, il existe une unique manière de colorier les autres nombres de sorte à obtenir un coloriage agréable.\n\nPour cela, nous allons procéder par récurrence. Montrons que si on colorie les puissances de $2$ de $1$ à $N$, alors il existe une unique manière agréable de colorier les autres entiers de $1$ à $N$.\n\nLe résultat est clair pour $N=1$.\n\nSoit $N$ tel que le résultat soit vrai pour $N-1$. On colorie les puissances de $2$ de $1$ à $N$. Par hypothèse de récurrence, il existe une unique manière agréable de colorier les autres entiers de $1$ à $N-1$.\n\nSi $N=2^{k}$ pour un certain entier $k$, alors $N$ est déjà colorié et le coloriage est agréable : en effet, si $1 \\leqslant \\ell < 2^{k}$, alors $2^{k} + \\ell$ n'est pas une puissance de $2$ car $2^{k} < 2^{k} + \\ell < 2^{k+1}$. Ainsi, dans ce cas, il y a une unique manière de compléter le coloriage.\n\nSi $N$ n'est pas une puissance de $2$, alors $N = 2^{k} + n$ avec $1 \\leqslant n < 2^{k}$ et $k$ un entier. Alors $N$ ne peut pas être colorié de la même couleur que $2^{k} - n$ car $N + 2^{k} - n = 2^{k+1}$. On colorie donc $N$ de l'autre couleur, et c'est bien l'unique manière de compléter le coloriage de manière agréable. Le coloriage obtenu est agréable : en effet, soit $t$ entier tel que $1 \\leqslant t < N$ et $t + N$ est une puissance de $2$, on a $2^{k} < N \\leqslant t + N < 2N = 2^{k+1} + 2n < 2^{k+1} + 2^{k+1} = 2^{k+2}$, donc $t + N = 2^{k+1}$. On a donc $t = 2^{k+1} - 2^{k} - n = 2^{k} - n$. Comme on a colorié $N$ et $2^{k} - n$ de deux couleurs différentes, le coloriage est agréable.\n\nIl y a $11$ puissances de $2$ entre $1$ et $2020$ qui sont $1=2^{0}$, $2=2^{1}$, $4=2^{2}$, $8=2^{3}$, $16=2^{4}$, $32=2^{5}$, $64=2^{6}$, $128=2^{7}$, $256=2^{8}$, $512=2^{9}$, $1024=2^{10}$, et il y a $2^{11}$ manières de les colorier en bleu et rouge. Ainsi, il y a $2^{11}$ coloriages agréables.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle non isocèle en $B$. Soit $D$ le pied de la bissectrice de l'angle $\\widehat{ABC}$. Soit $M$ le milieu de l'arc $\\widehat{AC}$ contenant $B$ sur le cercle circonscrit au triangle $ABC$. Le cercle circonscrit au triangle $BDM$ coupe le segment $[AB]$ en un point $K$ distinct de $B$. Soit $J$ le symétrique du point $A$ par rapport au point $K$. La droite $(DJ)$ intersecte la droite $(AM)$ en un point $O$. Montrer que les points $J$, $B$, $M$ et $O$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nTout d'abord, on reconnaît dans la figure quelques points connus : le point $M$ est le milieu de l'arc $AC$ contenant $B$, il s'agit donc du pôle Nord du point $B$ dans le triangle $ABC$, c'est-à-dire qu'il est le point d'intersection de la bissectrice extérieure de l'angle $\\widehat{ABC}$, de la médiatrice du segment $[AC]$ et du cercle circonscrit au triangle $ABC$. On obtient donc que l'angle $\\widehat{DBM}$ est droit (par définition de la bissectrice extérieure). Le fait que $M$ appartient à la médiatrice du segment $[AC]$ nous invite également à considérer cette médiatrice. Soit $N$ le milieu du segment $[AC]$. Alors $\\widehat{DNM}=90^\\circ$. On déduit donc que les points $B$, $M$, $N$ et $D$ sont cocycliques et $D$ appartient au cercle passant par les points $B$, $K$ et $D$.\n\nDe nouveaux points cocycliques apportent deux informations : une information sur les angles grâce au théorème de l'angle inscrit et une information sur les longueurs grâce à la puissance d'un point. Puisque la majorité des points se trouve sur deux droites se coupant en le sommet $A$, nous allons utiliser la puissance du point $A$ par rapport à divers cercles.\n\nNotamment, par puissance d'un point $AK \\cdot AB = AD \\cdot AN$. Donc\n$$\nAJ \\cdot AB = 2 AK \\cdot AB = 2 AD \\cdot AN = AD \\cdot AC\n$$\ndonc les points $D$, $C$, $B$ et $J$ sont cocycliques d'après la réciproque de la puissance d'un point. Pour conclure, nous allons utiliser les angles orientés, pour éviter d'avoir à traiter séparément les différentes configurations possibles.\n\nNous avons à présent tous les outils pour effectuer une chasse aux angles. Des angles difficiles d'accès sont les angles $\\widehat{OMJ}$, $\\widehat{OBJ}$, $\\widehat{MJB}$ \\ldots En revanche, les angles $\\widehat{DJB}$ et $\\widehat{OMB}$ semblent plus faciles à obtenir à l'aide des cercles présents sur la figure.\n\nPuisque les points $D$, $C$, $B$ et $O$ sont cocycliques, on trouve\n$$\n(JB, JO) = (JB, JD) = (CB, CD) = (CB, CA) = (MB, MA) = (MB, MO)\n$$\ndonc le quadrilatère $JBMO$ est bien cyclique.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer le plus petit entier $n \\geqslant 2$ tel qu'il existe des entiers strictement positifs $a_{1}, \\ldots, a_{n}$ tels que\n$$\na_{1}^{2}+\\ldots+a_{n}^{2} \\mid \\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1\n$$",
    "options": [],
    "answer": "9",
    "solution": "Solution:\n\nAnalysons le problème. Dans ce problème, on cherche le plus petit entier $n$ satisfaisant une certaine propriété. Supposons que l'on veuille montrer que le plus petit entier recherché est l'entier $c$. Pour montrer que $c$ est bien le plus petit entier, on doit d'une part montrer que si un entier $n$ satisfait la propriété, alors $n \\geqslant c$ et on doit montrer d'autre part que l'on peut trouver $c$ entiers satisfaisant la relation de divisibilité.\n\nPuisqu'il y a une relation de divisibilité, on peut s'attendre à utiliser diverses idées :\n- Si $a \\mid b$ alors $a$ divise les combinaisons linéaires de $a$ et $b$.\n- Si $a \\mid b$ et que $a$ et $b$ sont strictement positifs, alors $a \\mid b$.\n- Si $a \\mid b$, les diviseurs de $a$ divisent également $b$.\n\nUne fois cette première analyse effectuée, on peut commencer par regarder si on trouve des entiers $a_{i}$ satisfaisant la propriété pour des petites valeurs de $n$, par exemple pour $n=2,3$ ou même $4$ si on est courageux, avant de se convaincre que cette méthode ne permettra pas de déterminer $n$.\n\nNéanmoins, avoir cherché des $a_{i}$ pour des petites valeurs de $n$ nous permet de trouver d'éventuelles propriétés que doivent satisfaire les $a_{i}$. Par exemple, on a pu remarquer que les entiers $a_{i}$ ne peuvent pas tous être pairs, puisque dans ce cas la somme des $a_{i}^{2}$ serait paire mais $\\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1$ est impair. On se rend alors compte que ce qui importe dans cet argument, ce n'est pas la parité des $a_{i}$ qui entre en jeu mais la parité de la somme.\n\nOn obtient que $a_{1}+a_{2}+\\ldots+a_{n}$ est impair. Dans le cas contraire, puisque $a_{1}^{2}+a_{2}^{2}+\\ldots+a_{n}^{2} \\equiv a_{1}+a_{2}+\\ldots+a_{n} \\pmod{2}$, $a_{1}+\\ldots+a_{n}$ est pair également mais alors $a_{1}^{2}+\\ldots+a_{n}^{2}$ est pair et divise $\\left(a_{1}+a_{2}+\\ldots+a_{n}\\right)^{2}-1$ qui est impair, ce qui est absurde.\n\nOn déduit que $\\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1$ est pair. L'argument de parité précédent nous invite à préciser cette étude. On va donc chercher si $\\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1$ est divisible par $4,8, \\ldots$. La présence des multiples carrés nous invite à regarder chaque terme avec des modulos bien choisis.\n\nPar exemple, modulo $4$, $\\left(a_{1}+\\ldots+a_{n}\\right)^{2} \\equiv 1$ donc $4$ divise $\\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1$. Ne nous arrêtons pas en si bon chemin, nous savons également que les carrés des nombres impairs ne peuvent que valoir $1$ modulo $8$. Ainsi, $\\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1$ est divisible par $8$. Or $8$ est premier avec $a_{1}^{2}+\\ldots+a_{n}^{2}$ qui est impair. On déduit de tout cela que\n$$\n8\\left(a_{1}^{2}+\\ldots+a_{n}^{2}\\right) \\mid \\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1\n$$\n\nEnfin, une autre remarque que l'on peut formuler sur l'énoncé est qu'il implique la somme des carrés de nombres d'un côté et le carré de la somme des nombres de l'autre côté. Ceci fait penser à l'inégalité de Cauchy-Schwarz, ou à l'inégalité des moyennes arithmétiques et quadratiques. On peut donc essayer de comparer les quantités $8\\left(a_{1}^{2}+\\ldots+a_{n}^{2}\\right)$ et $\\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1$.\n\nCette idée de comparer les termes est cohérente avec ce que l'on cherche à montrer. La divisibilité obtenue donne que $8\\left(a_{1}^{2}+\\ldots+a_{n}^{2}\\right) \\leqslant \\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1$. Or, dans l'inégalité des moyennes arithmétique et quadratique, le carré de la somme est plus petit que la somme des carrés des éléments multipliée par le nombre de variables. On se doute donc que l'inégalité obtenue va être fausse pour $n$ trop petit, ce qui va nous fournir la borne souhaitée. Plus précisément,\n$$\n8\\left(a_{1}^{2}+\\ldots+a_{n}^{2}\\right) \\leqslant \\left(a_{1}+\\ldots+a_{n}\\right)^{2}-1 \\leqslant n\\left(a_{1}^{2}+\\ldots+a_{n}^{2}\\right)-1 < n\\left(a_{1}^{2}+\\ldots+a_{n}^{2}\\right)\n$$\noù l'inégalité au milieu correspond à l'inégalité des moyennes arithmétiques et quadratiques. On obtient finalement que $n>8$ donc $n \\geqslant 9$.\n\nRéciproquement, on cherche $9$ entiers satisfaisant la propriété. On peut s'attendre à ce que de petits entiers fonctionnent. On peut donc chercher des $a_{i}$ valant $1$ ou $2$. Finalement, on trouve qu'en prenant $a_{1}=\\cdots=a_{7}=1$ et $a_{8}=a_{9}=2$, on a bien $15$ qui divise $11^{2}-1=120=15 \\times 8$.\n\nLa réponse attendue est donc $n=9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30724,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier. On dispose de $n$ couleurs, et chaque point d'un cercle est colorié de l'une de ces couleurs. Montrer qu'il existe deux droites parallèles qui intersectent le cercle en 4 points distincts de même couleur.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nL'énoncé nous demande de démontrer une existence, on peut essayer de montrer cette existence à l'aide du principe des tiroirs. Gardant cette information en tête, on essaye de traduire le résultat demandé.\n\nConsidérons deux droites parallèles coupant le cercle en 4 points distincts. Ces 4 points délimitent un quadrilatère dont deux côtés sont parallèles et les médiatrices de ces deux segments sont confondues : en effet, elles sont perpendiculaires à deux côtés parallèles donc elles sont parallèles et elles passent par le centre du cercle donc elles sont bien confondues. Le quadrilatère est donc un trapèze isocèle. Il possède donc deux côtés égaux et deux diagonales de même longueur.\n\nRéciproquement, si 4 points $A, B, C$ et $D$ sont dans cet ordre sur un cercle de centre $O$ et délimitent un quadrilatère tels que $AB = CD$, alors la médiatrice du segment $[BC]$ est l'axe de la symétrie envoyant $B$ sur $C$. Cet axe de symétrie contient $O$ donc la symétrie conserve le cercle. Puisque $AB = CD$, cette symétrie envoie $A$ sur $D$ et donc la médiatrice du segment $[BC]$ est aussi celle du segment $[AD]$ et le quadrilatère $ABCD$ est un trapèze isocèle.\n\n![](attached_image_1.png)\n\nOn suppose désormais que $AC = BD$. Alors les triangles $AOC$ et $DOB$ sont isométriques car l'un est obtenu en appliquant une rotation sur l'autre triangle. On déduit que $BC = AD$ et que le quadrilatère $ABCD$ est un trapèze isocèle à nouveau.\n\nOn cherche donc désormais à montrer qu'il existe deux paires de points $(A, B)$ et $(C, D)$, tous de même couleur, telles que $AB = CD$. D'après la discussion précédente, ces points seraient alors les sommets d'un trapèze isocèle.\n\nPour un tel exercice, dit de géométrie combinatoire, il est bon de penser toujours à des objets réguliers, par exemple à des cercles ou des polygones réguliers. Ici, penser à de tels objets est pertinent puisque les polygones réguliers sont inscriptibles dans un cercle.\n\nPour combiner cette idée avec celle du principe des tiroirs, nous allons considérer des $(n+1)$-gones réguliers. Un $(n+1)$-gone régulier sur le cercle, d'après le principe des tiroirs, possède deux sommets de même couleur. On a donc obtenu une paire de sommets de même couleur.\n\n![](attached_image_2.png)\n\nOn appellera couleur de la paire la couleur commune des deux sommets et distance de la paire la distance entre ses deux sommets. On va créer plus de telles paires. N'oublions pas que nous aimerions que la distance des différentes paires que nous créons soit la même. Or, dans un $(n+1)$-gone régulier, il y a $\\left\\lfloor \\frac{n+1}{2} \\right\\rfloor$ distances distinctes possibles entre deux sommets quelconques. Pour obtenir deux paires de même distance, il faut donc $\\left\\lfloor \\frac{n+1}{2} \\right\\rfloor + 1$ polygones réguliers.\n\n![](attached_image_3.png)\n\nN'oublions pas que nous désirons également que les couleurs des paires soient identiques. Ainsi, on prend $n \\cdot \\left\\lfloor \\frac{n+1}{2} \\right\\rfloor + 1$ $(n+1)$-gones réguliers disjoints inscrits dans le cercle. On obtient alors $n \\cdot \\left\\lfloor \\frac{n+1}{2} \\right\\rfloor + 1$ paires de sommets tous distincts et les sommets d'une même paire sont de même couleur et leur distance appartient à un ensemble de taille $\\left\\lfloor \\frac{n+1}{2} \\right\\rfloor$. D'après le principe des tiroirs, il y a donc parmi ces paires $n+1$ paires de sommets de même distance. Comme il y a $n$ couleurs, à nouveau d'après le principe des tiroirs, parmi ces $n+1$ paires de même distance, il y en a au moins 2 de même couleur. On a donc obtenu deux paires de même distance et de même couleur, on a donc 4 points $A, B, C$ et $D$ de même couleur tels que $AB = CD$. Ainsi, on a bien deux droites parallèles coupant le cercle en 4 points de même couleur.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un parallélogramme. La bissectrice intérieure de l'angle $\\widehat{BAC}$ coupe le segment $[BC]$ en $E$ tandis que sa bissectrice extérieure coupe la droite $(CD)$ en $F$. Soit $M$ le milieu du segment $[AE]$.\n\nDémontrer que les droites $(EF)$ et $(BM)$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $G$ le point d'intersection des droites $(AB)$ et $(EF)$. Le problème revient à démontrer que $(MB)$ est une droite des milieux du triangle $AEG$, c'est-à-dire que $B$ est le milieu de $[AG]$.\n\nDe même, soit $H$ le point d'intersection des droites $(AE)$ et $(CD)$. L'homothétie de centre $E$ qui envoie $(AB)$ sur $(CD)$ envoie les points $A, B$ et $G$ sur $H, C$ et $F$. De manière équivalente, le problème revient donc à démontrer que $C$ est le milieu de $[HF]$.\n\nOr, en angles de droites,\n$$\n\\begin{aligned}\n(AC, AF) & =(AC, AE)+(AE, AF)=(AE, AB)+90^\\circ=(AE, AB)+(AF, AE)=(AF, AB) \\\\\n& =(FA, FC)\n\\end{aligned}\n$$\nce qui signifie que le triangle $ACF$ est isocèle en $C$. De même,\n$$\n(HA, HC)=(EA, CD)=(AE, AB)=(AC, AE)=(AC, AH)\n$$\ndonc $ACH$ est isocèle en $C$.\n\nOn en conclut que $CF=CA=CH$, donc que $C$ est bien le milieu de $[CH]$.\n\n![](attached_image_1.png)\nSolution:\n\nComme dans la solution précédente, il s'agit de démontrer que $AB=BG$. Or, le théorème de la bissectrice indique que $EB/EC=AB/AC$, et le théorème de Thalès indique que $EB/EC=BG/CF$. On en conclut que\n$$\n\\frac{AB}{AC}=\\frac{EB}{EC}=\\frac{BG}{CF},\n$$\nIl reste donc à démontrer que $CA=CF$, c'est-à-dire que le triangle $ACF$ est isocèle en $C$, ce qui se déduit directement des égalités d'angles de droites\n$$\n\\begin{aligned}\n(AC, AF) & =(AC, AE)+(AE, AF)=(AE, AB)+90^\\circ=(AE, AB)+(AF, AE)=(AF, AB) \\\\\n& =(FA, FC) .\n\\end{aligned}\n$$\nSolution:\n\nNous allons utiliser les coordonnées barycentriques dans le triangle $ABC$, en notant $a, b$ et $c$ les longueurs $BC, CA$ et $AB$. Nous identifierons chaque point $X$ à ses coordonnées barycentriques normalisées $(x, y, z)$ telles que $x+y+z=1$.\n\nAvec cette convention, si quatre points $S, T, U, V$ de coordonnées $\\mathbf{s}, \\mathbf{t}, \\mathbf{u}, \\mathbf{v}$ vérifient une relation de la forme $\\overrightarrow{UV}=\\lambda \\overrightarrow{ST}$, la relation $(\\mathbf{v}-\\mathbf{u})=\\lambda(\\mathbf{t}-\\mathbf{s})$ entre triplets de coordonnées barycentriques sera automatiquement vérifiée, et on pourra même se permettre d'écrire abusivement $V=U+\\lambda(S-T)$.\n\nTout d'abord, il est clair que $A=(1,0,0)$, $B=(0,1,0)$ et $C=(0,0,1)$. Ensuite, le théorème de la bissectrice indique que $EB/EC=AB/AC=b/c$, de sorte que $E=(0, b/(b+c), c/(b+c))$. Quitte à changer d'unités de longueur, on suppose que $b+c=1$, de sorte que $E=(0, b, c)$.\n\nToujours mus par le souhait de démontrer que $(MB)$ est une droite des milieux, on note $B'$ le symétrique de $A$ par rapport à $B$, puis $F'$ le point d'intersection de $(EB')$ et $(CD)$, et il s'agit dès lors de montrer que $(AE)$ et $(AF')$ sont perpendiculaires.\n\nTout d'abord, on sait que $B'=2B-A=(-1,2,0)$. Ensuite, il existe des réels $x$ et $y$ tels que $F=B'+x(E-B')=C+y(A-B)$, donc $F$ s'identifie à ses coordonnées barycentriques $(x-1,2+x(b-2), cx)=(y,-y,1)$. Ainsi, $x=1/c$ et $y=x-1=(1-c)/c=b/c$.\n\nOn vérifie alors comme prévu que\n$$\n\\begin{aligned}\n\\overrightarrow{AE} \\cdot \\overrightarrow{AF} & =(b \\overrightarrow{AB}+c \\overrightarrow{AC}) \\cdot(-b/c \\overrightarrow{AB}+\\overrightarrow{AC}) \\\\\n& =-b^2/c \\overrightarrow{AB} \\cdot \\overrightarrow{AB}+c \\overrightarrow{AC} \\cdot \\overrightarrow{AC}+(b-b) \\overrightarrow{AB} \\cdot \\overrightarrow{AC} \\\\\n& =-b^2 c+c b^2-0=0\n\\end{aligned}\n$$\nSolution:\n\nNous allons reprendre le raisonnement précédent, mais sans introduire les points $B'$ et $F'$. Ainsi, on observe comme précédemment que $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$ et $E=(0, b, c)$. Puisque $M$ est le milieu de $[AE]$, on sait également que $M=(A+E)/2=(1/2, b/2, c/2)$.\n\nPar ailleurs, il existe un réel $y$ tel que $F=C+y(A-B)$, de sorte que $F=(y,-y,1)$. Si l'on pose $\\theta=\\overrightarrow{AB} \\cdot \\overrightarrow{AC}$, on en déduit que\n$$\n\\begin{aligned}\n0=\\overrightarrow{AE} \\cdot \\overrightarrow{AF} & =(b \\overrightarrow{AB}+c \\overrightarrow{AC}) \\cdot(-y \\overrightarrow{AB}+\\overrightarrow{AC}) \\\\\n& =-b y \\overrightarrow{AB} \\cdot \\overrightarrow{AB}+c \\overrightarrow{AC} \\cdot \\overrightarrow{AC}+(b-cy) \\overrightarrow{AB} \\cdot \\overrightarrow{AC} \\\\\n& =-b y c^2+c b^2+(b-cy)\\theta=(bc+\\theta)(b-cy) .\n\\end{aligned}\n$$\nComme les points $A, B$ et $C$ ne sont pas alignés, on sait que $|\\theta|<bc$, de sorte que $y=b/c$. Ainsi, $F=(b/c,-b/c,1)$. Par conséquent, $2(M-B)=(1, b-2, c)$ et\n$$\nc(F-E)=(b,-b-bc,c-c^2)=(b,b(b-2),bc)=2b(M-B)\n$$\nce qui signifie que $c \\overrightarrow{EF}=2b \\overrightarrow{BM}$, et donc que les droites $(BM)$ et $(EF)$ sont bien parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30726,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A B C D E$ un pentagone convexe tel que $\\widehat{A B E}=\\widehat{A C E}=\\widehat{A D E}=90^{\\circ}$ et $B C=C D$. Enfin, soit $K$ un point sur la demi-droite $[A B]$ tel que $A K=A D$, et soit $L$ un point sur la demi-droite $[E D)$ tel que $E L=B E$. Démontrer que les points $B, D, K$ et $L$ appartiennent à un même cercle de centre $C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAu vu des angles droits dont on dispose en $B$, $C$ et $D$, les points $A$, $B$, $C$, $D$ et $E$ appartiennent à un même demi-cercle de diamètre $[A E]$. Comme $B C=C D$, le point $C$ est donc le pôle Sud issu de $A$ dans le triangle $A B D$. Ainsi, $(A C)$ est la bissectrice de $\\widehat{D A B}$, c'est-à-dire la médiatrice de $[K D]$. De même, $(C E)$ est la médiatrice de $[B L]$. Cela signifie que $C L=B C=D C=C K$, ce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30727,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMorgane et Bosphore jouent au jeu suivant. Morgane a écrit les entiers de 1 à 8 sur les sommets d'un octogone régulier : chaque entier est écrit sur un des huit sommets de l'octogone. Bosphore choisit ensuite un sommet et calcule la somme des nombres écrits sur ce sommet et sur ses deux voisins. Il note $s$ cette somme, et donne $s$ bonbons à Morgane.\n\nBosphore choisit un sommet de sorte à donner le moins possible de bonbons à Morgane, et Morgane écrit les entiers de 1 à 8 de sorte à recevoir autant de bonbons que possible. Démontrer que Bosphore lui donnera 12 bonbons.",
    "options": [],
    "answer": "12",
    "solution": "Solution:\n\nTout d'abord, si Morgane répartit les entiers de 1 à 8 comme suit, on constate en effet que Bosphore devra lui donner au moins 12 bonbons; on a indiqué, à côté de chaque sommet, la somme $s$ que calculerait Bosphore s'il choisissait ce sommet.\n\n![](attached_image_1.png)\n\nRéciproquement, quel que soit la répartition que choisit Morgane, Bosphore peut procéder comme suit pour éviter de lui donner 13 bonbons ou plus. Ci-dessous, on identifie chaque sommet à l'entier que Morgane a écrit sur ce sommet.\n\nSi les deux voisins du sommet 8 sont les sommets 1 et 2, Bosphore n'a qu'à choisir le sommet 8, et il ne donnera que 11 bonbons à Morgane. Sinon, l'un des deux sommets de 8 est un entier $k \\geqslant 3$. Bosphore répartit les six sommets restants en deux groupes de trois sommets consécutifs, comme illustré ci-dessous. La somme des valeurs de ces deux groupes de sommets vaut $(1+2+\\ldots+8)-(k+8)=28-k \\leqslant 25$, donc l'un des deux groupes est de somme $s \\leqslant 12$. Bosphore s'assure alors de donner au plus 12 bonbons à Morgane en choisissant le sommet au milieu de ce groupe.\n\n![](attached_image_2.png)\n\n\nSolution alternative $n^{\\circ} 1$\n\nBosphore a d'autres manières de procéder, par exemple celle-ci. Regroupons les sommets en trois groupes de trois, de sorte que le sommet de valeur 1 soit le seul sommet appartenant à deux groupes. La somme des valeurs de ces groupes de sommets vaut $1+(1+2+\\ldots+8)=37<3 \\times 13$, donc l'un de ces groupes est de valeur $s \\leqslant 12$, et Bosphore n'a qu'à choisir le sommet au milieu de ce groupe.\n\n![](attached_image_3.png)\n\n\nSolution alternative $n^{\\circ} 2$\n\nVoici une troisième manière, pour Bosphore, de s'assurer qu'il donnera au plus 12 bonbons à Morgane. Soit $a$ et $b$ les deux voisins du sommet 5, avec $a \\geqslant b$. Si $b \\leqslant 4$, alors $a \\leqslant 3$, donc $a+b+5 \\leqslant 12$ et Bosphore choisit directement le sommet 5. Sinon, on sait que $b \\geqslant 6$. Bosphore élimine alors les sommets 5 et $b$, qui sont contigus, et répartit les six sommets restants en deux groupes de trois sommets consécutifs. Comme dans la solution précédente, la somme des valeurs de ces deux groupes de sommets vaut $(1+2+\\ldots+8)-(5+b)=31-b \\leqslant 25$, donc l'un des deux groupes est de somme $s \\leqslant 12$. Bosphore choisit alors le sommet au milieu de ce groupe.\n\n![](attached_image_4.png)\n\n\nSolution alternative $n^{\\circ} 3$\n\nVoici encore une manière de démontrer que Bosphore pourra nécessairement se débrouiller pour donner au plus 12 bonbons à Morgane. À la différence des solutions précédentes, celle-ci ne fournit pas de stratégie explicite pour Bosphore.\n\nSupposons que Morgane ait trouvé une configuration qui force Bosphore à lui donner au moins 13 bonbons. On numérote les sommets de l'octogone de 1 à 8, ces numéros étant considérés modulo 8. On note alors $a_{i}$ le nombre que Morgane a écrit sur le sommet $i$, et on note $s_{i}$ la somme $a_{i-1}+a_{i}+a_{i+1}$.\n\nRemarquons que $s_{i+1}-s_{i}=a_{i+2}-a_{i-1} \\neq 0$, et ce quelle que soit la valeur de $i$. Ainsi, la somme de deux nombres $s_{i}$ consécutifs vaut au minimum $13+14=27$. En outre, la somme des entiers $s_{i}$ vaut $3\\left(a_{1}+a_{2}+\\ldots+a_{8}\\right)=3(1+2+\\ldots+8)=108=4 \\times 27$. Par conséquent, la somme de deux nombres $s_{i}$ consécutifs vaut toujours 27, et les $s_{i}$ alternent entre les valeurs 13 et 14, de sorte que $s_{1}=s_{3}=s_{5}=s_{7}$ et $s_{2}=s_{4}=s_{6}=s_{8}$.\n\nDans ces conditions, on constate que\n$$\n0=\\left(s_{2}-s_{3}\\right)+\\left(s_{5}-s_{6}\\right)=\\left(a_{1}-a_{4}\\right)+\\left(a_{4}-a_{7}\\right)=a_{4}-a_{7},\n$$\nce qui est impossible. Ainsi, Morgane ne pourra jamais forcer Bosphore à lui donner 13 bonbons ou plus.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn pose $n=2021$. Démontrer que le nombre $3^{2 n+1}+2^{n+2}$ est divisible par 35.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous allons démontrer que le nombre $S_{n}=3^{2 n+1}+2^{n+2}$ est divisible simultanément par 5 et par 7 lorsque $n=2021$.\nOn vérifie immédiatement que\n$$\nS_{n} \\equiv 9^{n} \\times 3+2^{n} \\times 4 \\equiv 2^{n} \\times 3+2^{n} \\times 4 \\equiv 2^{n} \\times 7 \\equiv 0 \\quad(\\bmod 7),\n$$\nce qui signifie bien que $S_{n}$ est divisible par 7.\nOn vérifie ensuite que\n$$\nS_{n} \\equiv 9^{n} \\times 3+2^{n} \\times 4 \\equiv(-1)^{n} \\times 3+2^{n} \\times(-1) \\equiv-3-2^{n} \\quad(\\bmod 5).\n$$\nLa propriété demandée dépend donc manifestement de la valeur de $n$, puisqu'elle aurait été invalide si on avait choisi $n=3$.\nOn calcule donc les premières valeurs de $S_{n}(\\bmod 5)$ pour se donner des idées :\n$$\n\\begin{array}{ll}\n\\triangleright S_{0} \\equiv-3-1 \\equiv 1(\\bmod 5) ; & \\triangleright S_{4} \\equiv-3-16 \\equiv 1(\\bmod 5) ; \\\\\n\\triangleright S_{1} \\equiv-3-2 \\equiv 0(\\bmod 5) ; & \\triangleright S_{5} \\equiv-3-32 \\equiv 0(\\bmod 5) ; \\\\\n\\triangleright S_{2} \\equiv-3-4 \\equiv 3(\\bmod 5) ; & \\triangleright S_{6} \\equiv-3-64 \\equiv 3(\\bmod 5) ; \\\\\n\\triangleright S_{3} \\equiv-3-8 \\equiv 4(\\bmod 5) ; & \\triangleright S_{7} \\equiv-3-128 \\equiv 4(\\bmod 5) .\n\\end{array}\n$$\nIl apparaît visiblement que $S_{n} \\equiv 0(\\bmod 5)$ lorsque $n \\equiv 1(\\bmod 4)$, et on entreprend donc de le démontrer. Si l'on pose $n=4 k+1$, alors\n$$\nS_{n} \\equiv-3-2^{4 k+1} \\equiv-3-16^{k} \\times 2 \\equiv-3-1^{k} \\times 2 \\equiv-3-2 \\equiv 0 \\quad(\\bmod 5)\n$$\nAinsi, lorsque $n=2021$, on a bien $n \\equiv 1(\\bmod 4)$, et l'entier $S_{n}$ est donc divisible par 5.\nEn conclusion, l'entier $S_{n}$ est divisible à la fois par 5 et par 7, donc il est également divisible par leur PPCM, c'est-à-dire 35.\nSolution:\n\nLe petit théorème de Fermat nous assure que $2^{4} \\equiv 3^{4} \\equiv 1(\\bmod 5)$. Comme $2 n+1 \\equiv 4043 \\equiv 3(\\bmod 4)$ et $n+2 \\equiv 2023 \\equiv 3(\\bmod 4)$, on en déduit que $S_{n} \\equiv 3^{3}+2^{3} \\equiv 35 \\equiv 0(\\bmod 5)$.\nDe même, le petit théorème de Fermat nous assure que $2^{6} \\equiv 3^{6} \\equiv 1(\\bmod 7)$. Comme $2 n+1 \\equiv 4043 \\equiv-1(\\bmod 6)$ et $n+2 \\equiv 2023 \\equiv 1(\\bmod 6)$, on en déduit que $S_{n} \\equiv 3^{5}+2^{1} \\equiv 245 \\equiv 0$ $(\\bmod 7)$.\nEn conclusion, la somme $S_{n}$ est divisible par 5 et par 7, donc par 35.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a$, $b$, $c$ et $d$ quatre nombres réels. On suppose qu'il existe une permutation $(x, y, z, t)$ des nombres $a, b, c$ et $d$ telle que\n$$\nx \\leqslant 2 a-b, \\quad y \\leqslant 2 b-c, \\quad z \\leqslant 2 c-d \\text{ et } t \\leqslant 2 d-a.\n$$\nDémontrer que $a = b = c = d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour mieux exploiter la première inégalité, que l'on réécrit comme $x + b \\leqslant 2a$, l'idéal serait que $x$ et $b$ soient maximaux, ou bien que $a$ soit minimal. Évidemment, il se pourrait très bien que ni $a$, ni $b$, ni $x$ ne soit extrémal. Néanmoins, puisque les variables jouent des rôles cycliques, on peut tout de même espérer, au choix, que $a$ soit minimal, ou que $b$ soit maximal, ou que $x$ soit maximal.\n\nCependant, espérer que $b$ et $x$ seraient simultanément maximaux serait trop demander. À la place, on se contente de supposer, sans perte de généralité, que $a$ est le plus petit des quatre nombres réels. Dans ces conditions, $2a \\leqslant b + x \\leqslant 2a$, donc $a = b$. Mais alors $b$ est aussi le plus petit des quatre nombres réels, donc $b = c$, et on démontre de même que $c = d$.\n\n\nAlternative 1:\n\nEn sommant les quatre inégalités de l'énoncé, on obtient l'inégalité $x + y + z + t \\leqslant a + b + c + d$. Puisque $(x, y, z, t)$ est une permutation de $(a, b, c, d)$, cette dernière inégalité est en fait une égalité, donc chacune des inégalités de l'énoncé est également une égalité : $x = 2a - b$, $y = 2b - c$, $z = 2c - d$ et $t = 2d - a$.\n\nEn élevant chaque membre au carré et en additionnant les égalités ainsi obtenues, on constate alors que\n$$\na^{2} + b^{2} + c^{2} + d^{2} = x^{2} + y^{2} + z^{2} + t^{2} = 5(a^{2} + b^{2} + c^{2} + d^{2}) - 4(ab + bc + cd + da)\n$$\nCela signifie que $ab + bc + cd + da = a^{2} + b^{2} + c^{2} + d^{2}$, ou encore que\n$$\n0 = 2(a^{2} + b^{2} + c^{2} + d^{2}) - 2(ab + bc + cd + da) = (a-b)^{2} + (b-c)^{2} + (c-d)^{2} + (d-a)^{2},\n$$\nde sorte que $a = b = c = d$.\n\n\nAlternative 2:\n\nUne autre manière de conclure consiste à remarquer que l'égalité $ab + bc + cd + da = a^{2} + b^{2} + c^{2} + d^{2}$ est un cas d'égalité de l'inégalité de Cauchy-Schwarz, qui n'est atteinte que si les quadruplets $(a, b, c, d)$ et $(b, c, d, a)$ sont proportionnels l'un à l'autre. Puisque ceux-ci sont de même somme, ils sont donc égaux, ce qui conclut en effet.\n\n\nAlternative 3:\n\nUne autre variante des deux solutions précédentes est la suivante : quitte à soustraire un réel $\\lambda$ suffisamment grand à tous nos nombres, on suppose que chaque membre de chaque inégalité est négatif. En élevant ces membres au carré et en additionnant les inégalités ainsi obtenues, on constate alors que\n$$\na^{2} + b^{2} + c^{2} + d^{2} = x^{2} + y^{2} + z^{2} + t^{2} \\geqslant 5(a^{2} + b^{2} + c^{2} + d^{2}) - 4(ab + bc + cd + da).\n$$\nCela signifie une fois de plus que $ab + bc + cd + da \\geqslant a^{2} + b^{2} + c^{2} + d^{2}$, et l'on conclut alors comme précédemment.\n\n\nAlternative 4:\n\nOn procède de manière brutale, étudiant séparément chacune des $4! = 24$ permutations $(x, y, z, t)$ possibles. Dans chaque cas, nous allons exprimer les nombres $a-b$, $b-c$, $c-d$ et $d-a$ comme combinaisons linéaires, à coefficients positifs, des nombres $\\alpha = (2a-b-x)/60$, $\\beta = (2b-c-y)/60$, $\\gamma = (2c-d-z)/60$ et $\\delta = (2d-a-t)/60$ ; les dénominateurs ont été choisis a posteriori pour que tous nos coefficients soient entiers. Cela démontrera qu'il s'agit de nombres positifs ou nuls, et puisque leur somme vaut $0$, on en conclura que $a = b = c = d$.\n\nCependant, que l'on ne se méprenne pas ! Nous présentons cette solution dans le seul but de convaincre le lecteur que, si celle-ci est théoriquement faisable dans le temps imparti des quatre heures de l'épreuve, elle est très risquée, et intellectuellement très peu satisfaisante. En outre, on serait bien en peine d'appliquer une telle méthode si on avait 2021 variables plutôt que quatre. Pour information, les correcteurs ont eu besoin d'un programme informatique pour écrire les cases de ce tableau, qu'ils n'ont pas eu le courage de calculer à la main après y avoir passé plus de deux heures.\n\n| Cas | $(x, y, z, t)$ | $a-b$ | $b-c$ | $c-d$ | $d-a$ |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 1 | $(a, b, c, d)$ | $60 \\alpha$ | $60 \\beta$ | $60 \\gamma$ | $60 \\delta$ |\n| 2 | $(a, b, d, c)$ | $60 \\alpha$ | $60 \\beta$ | $30 \\gamma$ | $30 \\gamma + 60 \\delta$ |\n| 3 | $(a, c, b, d)$ | $60 \\alpha$ | $30 \\beta$ | $30 \\beta + 60 \\gamma$ | $60 \\delta$ |\n| 4 | $(a, c, d, b)$ | $60 \\alpha$ | $30 \\beta$ | $30 \\gamma$ | $30 \\beta + 30 \\gamma + 60 \\delta$ |\n| 5 | $(a, d, b, c)$ | $60 \\alpha$ | $20 \\alpha + 40 \\beta + 20 \\delta$ | $20 \\beta + 40 \\gamma$ | $20 \\beta + 40 \\gamma + 60 \\delta$ |\n| 6 | $(a, d, c, b)$ | $60 \\alpha$ | $30 \\alpha + 60 \\beta + 30 \\delta$ | $60 \\gamma$ | $30 \\beta + 30 \\gamma + 60 \\delta$ |\n| 7 | $(b, a, c, d)$ | $30 \\alpha$ | $30 \\alpha + 60 \\beta$ | $60 \\gamma$ | $60 \\delta$ |\n| 8 | $(b, a, d, c)$ | $30 \\alpha$ | $30 \\alpha + 60 \\beta$ | $30 \\gamma$ | $30 \\gamma + 60 \\delta$ |\n| 9 | $(b, c, a, d)$ | $30 \\alpha$ | $30 \\beta$ | $30 \\alpha + 30 \\beta + 60 \\gamma$ | $60 \\delta$ |\n| 10 | $(b, c, d, a)$ | $30 \\alpha$ | $30 \\beta$ | $30 \\gamma$ | $30 \\delta$ |\n| 11 | $(b, d, a, c)$ | $30 \\alpha$ | $10 \\alpha + 40 \\beta + 20 \\delta$ | $20 \\alpha + 20 \\beta + 40 \\gamma$ | $30 \\alpha + 20 \\gamma + 40 \\delta$ |\n| 12 | $(b, d, c, a)$ | $30 \\alpha$ | $30 \\alpha + 60 \\beta + 30 \\delta$ | $60 \\gamma$ | $30 \\delta$ |\n| 13 | $(c, a, b, d)$ | $40 \\alpha + 20 \\gamma + 20 \\delta$ | $20 \\alpha + 40 \\beta$ | $20 \\alpha + 40 \\beta + 60 \\gamma$ | $60 \\delta$ |\n| 14 | $(c, a, d, b)$ | $40 \\alpha + 20 \\gamma + 20 \\delta$ | $20 \\alpha + 40 \\beta$ | $30 \\gamma$ | $20 \\beta + 10 \\gamma + 40 \\delta$ |\n| 15 | $(c, b, a, d)$ | $60 \\alpha + 30 \\gamma + 30 \\delta$ | $60 \\beta$ | $30 \\alpha + 30 \\beta + 60 \\gamma$ | $60 \\delta$ |\n| 16 | $(c, b, d, a)$ | $60 \\alpha + 30 \\gamma + 30 \\delta$ | $60 \\beta$ | $30 \\gamma$ | $30 \\delta$ |\n| 17 | $(c, d, a, b)$ | $48 \\alpha + 24 \\gamma + 12 \\delta$ | $12 \\alpha + 48 \\beta + 24 \\delta$ | $24 \\alpha + 12 \\beta + 48 \\delta$ | $24 \\beta + 12 \\gamma + 48 \\delta$ |\n| 18 | $(c, d, b, a)$ | $40 \\alpha + 20 \\gamma + 10 \\delta$ | $20 \\alpha + 40 \\beta + 20 \\delta$ | $20 \\beta + 40 \\gamma$ | $30 \\delta$ |\n| 19 | $(d, a, b, c)$ | $45 \\alpha + 15 \\gamma + 30 \\delta$ | $30 \\alpha + 45 \\beta + 15 \\delta$ | $15 \\alpha + 30 \\beta + 45 \\gamma$ | $15 \\beta + 30 \\gamma + 45 \\delta$ |\n| 20 | $(d, a, c, b)$ | $40 \\alpha + 20 \\delta$ | $40 \\alpha + 60 \\beta + 20 \\delta$ | $60 \\gamma$ | $20 \\beta + 20 \\gamma + 40 \\delta$ |\n| 21 | $(d, b, a, c)$ | $60 \\alpha + 20 \\gamma + 40 \\delta$ | $60 \\beta$ | $20 \\alpha + 20 \\beta + 40 \\gamma$ | $20 \\gamma + 40 \\delta$ |\n| 22 | $(d, b, c, a)$ | $60 \\alpha + 30 \\delta$ | $60 \\beta$ | $60 \\gamma$ | $30 \\delta$ |\n| 23 | $(d, c, a, b)$ | $40 \\alpha + 20 \\delta$ | $30 \\beta$ | $20 \\alpha + 10 \\beta + 40 \\gamma$ | $20 \\beta + 20 \\gamma + 40 \\delta$ |\n| 24 | $(d, c, b, a)$ | $60 \\alpha + 30 \\delta$ | $30 \\beta$ | $30 \\beta + 60 \\gamma$ | $30 \\delta$ |",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle. On note $H_{A}$ le pied de la hauteur de $ABC$ issue de $A$, et $A'$ le milieu du segment $[BC]$. On note ensuite $Q_{A}$ le symétrique de $H_{A}$ par rapport à $A'$. On définit de même les points $Q_{B}$ et $Q_{C}$. Enfin, on note $R$ le point d'intersection, autre que $Q_{A}$, entre les cercles circonscrits aux triangles $Q_{A} Q_{B} C$ et $Q_{A} B Q_{C}$.\n\nDémontrer que les droites $\\left(Q_{A} R\\right)$ et $(BC)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $H$ l'orthocentre de $ABC$, $O$ le centre du cercle circonscrit à $ABC$, et $R'$ le symétrique de $H$ par rapport à $O$. Les projetés orthogonaux de $H$ et $O$ sur $(BC)$ sont $H_{A}$ et $A'$, donc le projeté orthogonal de $R'$ sur $(BC)$ est $Q_{A}$. De même, les projetés orthogonaux de $R'$ sur $(CA)$ et sur $(AB)$ sont $Q_{B}$ et $Q_{C}$.\n\nCela signifie entre autres que $\\widehat{CQ_{A}R'} = \\widehat{QC_{B}R'} = 90^{\\circ}$, donc que $Q_{A}$ et $Q_{B}$ appartiennent au cercle de diamètre $[CR']$. Ce cercle coïncide donc avec le cercle circonscrit à $Q_{A} Q_{B} C$. De même, les points $Q_{A}$ et $Q_{C}$ appartiennent au cercle de diamètre $[BR']$, qui coïncide avec le cercle circonscrit à $Q_{A} B Q_{C}$. Par conséquent, les points $R$ et $R'$ sont confondus, et $\\left(Q_{A} R\\right)$ est bien perpendiculaire à $(BC)$.\n\n![](attached_image_1.png)\nSolution:\n\n$\\underline{\\text{Solution alternative}~n^{\\circ} 1}$\n\nCi-dessous, on note $\\Gamma_{A}, \\Gamma_{B}$ et $\\Gamma_{C}$ les cercles circonscrits respectifs à $A Q_{B} Q_{C}$, $Q_{A} B Q_{C}$ et $Q_{A} Q_{B} C$. Puisque $R$ appartient à $\\Gamma_{B}$ et à $\\Gamma_{C}$, on sait que\n$$\n\\left(Q_{C} A, Q_{C} R\\right) = \\left(Q_{C} B, Q_{C} R\\right) = \\left(Q_{A} B, Q_{A} R\\right) = \\left(Q_{A} C, Q_{A} R\\right) = \\left(Q_{B} C, Q_{B} R\\right) = \\left(Q_{B} A, Q_{B} R\\right)\n$$\nCela signifie que $R$ appartient aussi à $\\Gamma_{A}$, et donc que $A, B$ et $C$ jouent des rôles symétriques. Forts de ce constat, on introduit donc également le cercle circonscrit à $ABC$, que l'on note $\\Omega$. Toujours à la recherche de cercles remarquables, on remarque alors que les angles droits en $H_{A}, H_{B}$ et $H_{C}$ suggèrent aussi de tracer les cercles $\\Xi_{A}, \\Xi_{B}$ et $\\Xi_{C}$, de diamètres respectifs $[BC], [CA]$ et $[AB]$.\n\nEnfin, si l'on note $m$ la médiatrice de $[BC]$, l'énoncé nous demande de démontrer que $\\left(Q_{A} R\\right)$ et $\\left(A H_{A}\\right)$ sont symétriques l'une de l'autre par rapport à $m$. On s'intéresse donc de plus près à la symétrie d'axe $m$ et aux cercles dont $m$ est un axe de symétrie : il s'agit des cercles $\\Omega, \\Xi_{A}$ et, ne serait-ce qu'en apparence, $\\Gamma_{A}$.\n\nÀ défaut de démontrer que le centre de ce dernier cercle se trouve sur $m$, on peut tenter de démontrer que les axes radicaux de $\\Gamma_{A}$ avec $\\Omega$ ou $\\Xi_{A}$ sont parallèles à $(BC)$. Le premier serait alors manifestement la parallèle à $(BC)$ passant par $A$, tandis que le deuxième a l'air d'être la droite $\\left(B' C'\\right)$.\n\nComme $B'$ est le milieu de $[AC]$ et de $[H_{B} Q_{B}]$, on sait que $B' H_{B} \\cdot B' C = B' Q_{B} \\cdot B' A$, ce qui signifie bien que $B'$ appartient à l'axe radical de $\\Gamma_{A}$ et $\\Xi_{A}$. De même, $C'$ appartient à cet axe radical, qui est donc confondu avec $\\left(B' C'\\right)$, de sorte que $\\Gamma_{A}$ est bien symétrique par rapport à $m$.\n\nPar conséquent, les cercles $\\Omega$ et $\\Gamma_{A}$ se rencontrent en un point $X$ qui n'est autre que le symétrique de $A$ par rapport à $m$. Notons alors $R'$ le point d'intersection entre $\\left(Q_{A} X\\right)$ et $\\Gamma_{C}$. On sait que $\\widehat{C Q_{B} R'} = \\widehat{C Q_{A} R'} = 90^{\\circ}$. Puisque l'on a également $\\widehat{A X R'} = 90^{\\circ}$, le point $R'$ est donc diamétralement opposé à $A$ dans le cercle circonscrit à $A Q_{B} X$. Cela démontre que $R'$ appartient à $\\Gamma_{A}$, donc coïncide avec $R$, ce qui conclut.\n\n![](attached_image_2.png)\n\nSolution alternative $n^{\\circ} 2$\n\nOn reprend les notations de la solution précédente. Puisque la droite $\\left(Q_{A} R\\right)$ est l'axe radical des cercles $\\Gamma_{B}$ et $\\Gamma_{C}$, il est tentant de rechercher les axes radicaux de ces deux cercles avec un autre cercle. À cette fin, on pourrait considérer le cercle $\\Gamma_{A}$, et constater comme précédemment que $R$ est le centre radical des trois cercles $\\Gamma_{A}, \\Gamma_{B}$ et $\\Gamma_{C}$. Au vu de cette symétrie des rôles, on considère également le cercle $\\Omega$, et on note alors $Y$ le centre radical de $\\Gamma_{B}, \\Gamma_{C}$ et $\\Omega$.\n\nIl semble que $BACY$ soit un parallélogramme, et on entreprend donc de le démontrer. Pour ce faire, on procède comme dans la solution précédente. Puisque $A'$ est le milieu de $[BC]$ et de $[H_{A} Q_{A}]$, il a même puissance par rapport à $\\Gamma_{C}$ et $\\Xi_{C}$. De même, $B'$ a même puissance par rapport à $\\Gamma_{C}$ et $\\Xi_{C}$. On en déduit que $\\left(A' B'\\right)$ est l'axe radical de $\\Gamma_{C}$ et $\\Xi_{C}$. Puisque $(AB)$ est l'axe radical de $\\Xi_{C}$ et $\\Omega$ et que $(AB)$ est parallèle à $\\left(A' B'\\right)$, on en déduit que $(AB)$ est parallèle au troisième axe radical $(CY)$ entre $\\Gamma_{C}$ et $\\Omega$. De même, $(AC)$ est parallèle à $(BY)$.\n\nLa symétrie de centre $A'$ échange donc la droite $\\left(A H_{A}\\right)$ avec $\\left(Y Q_{A}\\right)$, c'est-à-dire avec $\\left(Q_{A} R\\right)$. La droite $\\left(Q_{A} R\\right)$ est donc bien perpendiculaire à $(BC)$.\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $x$, $y$ et $z$ trois nombres réels tels que $0 \\leqslant x \\leqslant y \\leqslant z$ et $x+y+z=1$. Trouver la valeur maximale que peut prendre l'expression\n$$\n(x-y z)^{2}+(y-z x)^{2}+(z-x y)^{2}.\n$$",
    "options": [],
    "answer": "1",
    "solution": "Solution:\nLa quantité que l'on souhaite maximiser peut se réécrire comme\n$$\n\\begin{aligned}\n\\mathcal{S} & =\\left(x^{2}+y^{2}+z^{2}\\right)+\\left((x y)^{2}+(y z)^{2}+(z x)^{2}\\right)-6 x y z \\\\\n& =(x+y+z)^{2}-2(x y+y z+z x)+(x y+y z+z x)^{2}-2 x y z(x+y+z)-6 x y z \\\\\n& =1-2(x y+y z+z x)+(x y+y z+z x)^{2}-8 x y z \\\\\n& =(x y+y z+z x-1)^{2}-8 x y z.\n\\end{aligned}\n$$\nPuisque $0 \\leqslant x y+y z+z x \\leqslant (x+y+z)^{2} / 2 = 1 / 2$, on en déduit que $\\mathcal{S} \\leqslant 1$. Réciproquement, si $x=y=0$ et $z=1$, on a bien $\\mathcal{S}=1$, ce qui conclut.\nSolution:\nLa quantité que l'on souhaite maximiser peut se réécrire comme\n$$\n\\begin{aligned}\n\\mathcal{S} & =\\left(x^{2}+y^{2}+z^{2}\\right)+\\left((x y)^{2}+(y z)^{2}+(z x)^{2}\\right)-6 x y z \\\\\n& =(x+y+z)^{2}-2(x y+y z+z x)+\\left((x y)^{2}+(y z)^{2}+(z x)^{2}\\right)-6 x y z\n\\end{aligned}\n$$\nOr, l'énoncé indique que $0 \\leqslant x \\leqslant y \\leqslant z \\leqslant 1$, de sorte que $0 \\leqslant x y \\leqslant x z \\leqslant y z \\leqslant 1$. Puisque tout nombre $t$ tel que $0 \\leqslant t$ satisfait l'inégalité $t^{2} \\leqslant 2 t$, on en déduit que $\\mathcal{S} \\leqslant (x+y+z)^{2}-6 x y z \\leqslant 1$. Réciproquement, si $x=y=0$ et $z=1$, on a bien $\\mathcal{S}=1$, ce qui conclut.\nSolution:\nLes inégalités $0 \\leqslant x \\leqslant y \\leqslant z \\leqslant x+y+z=1$ indiquent que $z \\geqslant y \\geqslant x y$ et $y \\geqslant x \\geqslant z x$, de sorte que la quantité que l'on souhaite maximiser satisfait l'inégalité\n$$\n\\mathcal{S} \\leqslant \\max \\{x, y z\\}^{2}+y^{2}+z^{2}\n$$\nPuisque $x^{2}+y^{2}+z^{2} \\leqslant (x+y+z)^{2}=1$ et $(y z)^{2}+y^{2}+z^{2} \\leqslant 2 y z+y^{2}+z^{2}=(y+z)^{2}=(1-x)^{2} \\leqslant 1$, on en déduit que $\\mathcal{S} \\leqslant 1$. Réciproquement, si $x=y=0$ et $z=1$, on a bien $\\mathcal{S}=1$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 30732,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les quadruplets d'entiers relatifs $(a, b, c, p)$ tels que $p$ soit un nombre premier et pour lesquels\n$$\n73 p^{2}+6=9 a^{2}+17 b^{2}+17 c^{2}\n$$",
    "options": [],
    "answer": "(±1, ±1, ±4, 2) and (±1, ±4, ±1, 2)",
    "solution": "Solution:\n\nL'égalité de l'énoncé met en jeu de nombreux carrés. La première chose à faire consiste donc à l'étudier modulo un nombre $n$ pour lequel il y a peu de résidus quadratiques. On étudie donc le cas $n=8$, car les carrés modulo 8 sont 0, 1 et 4.\n\nEn particulier, si $p$ est impair, l'équation devient $a^{2}+b^{2}+c^{2} \\equiv -1 \\pmod{8}$, et cette équation n'a aucune solution. On en déduit que $p=2$, et il s'agit désormais de trouver les entiers $a, b$ et $c$ tels que $298=9 a^{2}+17 b^{2}+17 c^{2}$.\n\nSans perte de généralité, on suppose que $a, b$ et $c$ sont positifs ou nuls. Puisque $b$ et $c$ jouent des rôles symétriques, on suppose même que $b \\leqslant c$. On constate alors que $9 \\times 6^{2}=324 > 298 \\geqslant 9 a^{2}$, de sorte que $0 \\leqslant a \\leqslant 5$. Puisque $9 a^{2}+17 b^{2}+17 c^{2} \\equiv 298 \\equiv 9 \\pmod{17}$, on en déduit en outre que 17 divise $9(a-1)(a+1)$. Cela signifie que $a=1$, et notre équation devient $17=b^{2}+c^{2}$, ce qui signifie que $b=1$ et $c=4$.\n\nEn conclusion, les seules solutions possibles sont\n$$\n(a, b, c, p)=( \\pm 1, \\pm 1, \\pm 4, 2) \\text{ et }( \\pm 1, \\pm 4, \\pm 1, 2)\n$$\nRéciproquement, et en vertu de l'égalité $73 \\times 2^{2}+6=298=9 \\times 1^{2}+17 \\times 1^{2}+17 \\times 4^{2}$, ces quadruplets conviennent effectivement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30733,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ et $k$ deux entiers, tels que $n \\geqslant 3$. Théo organise les élections des délégués de sa classe de $n$ élèves : chaque élève doit voter pour un de ses camarades (tout le monde est candidat), et nul ne vote pour lui-même. Puis Théo répartit les élèves en groupes de sorte que, si un élève est dans un groupe, l'élève pour lequel il a voté n'y soit pas.\nPour quelles valeurs de $k$ Théo est-il sûr de pouvoir répartir les élèves en au plus $k$ groupes, et ce quelle que soit la manière dont les élèves auront voté?",
    "options": [],
    "answer": "all integers k greater than or equal to 3",
    "solution": "Solution:\n\nTout d'abord, si on a trois élèves (disons Anna, Martin et Vincent) tels qu'Anna a voté pour Martin, Martin a voté pour Vincent, et Vincent a voté pour Anna, il est clair que Théo aura besoin d'au moins trois groupes.\n\nRéciproquement, démontrons par récurrence sur $n$ que Théo peut s'en sortir avec trois groupes seulement, et ce même si certains élèves votent blanc.\n\nL'initialisation est claire pour $n=3$. On suppose donc que $n \\geqslant 4$. Puisqu'il y a eu au plus $n$ votes, il y a au moins un élève qui a reçu au plus un vote : on l'appelle Donald. Quitte à oublier Donald et à frauder en « oubliant» le vote que celui-ci aurait reçu pour le remplacer par un vote blanc, Théo peut répartir les $n-1$ élèves restants en au plus trois groupes.\n\nPuis Théo fait revenir Donald dans la classe : sur les trois groupes, il y en a au moins un qui ne contient ni la personne pour laquelle Donald a pu voter, ni l'électeur éventuel de Donald. Théo met alors Donald dans ce groupe-là, ce qui conclut la récurrence.\n\nEn résumé, les entiers recherchés sont les entiers $k \\geqslant 3$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k \\geqslant 1$ un entier, et soit $A$ un sous-ensemble de $\\{1,2, \\ldots, 3k\\}$ tel que, pour tous les éléments $a, b, c$ de $A$, si $a+b=2c$, alors $a=b=c$. Pour tout entier $n \\geqslant 1$, on note $r_{k}(n)$ le plus petit entier naturel non nul tel que $3k$ divise $n-r_{k}(n)$. En outre, on dit que $n$ est petit si $n \\leqslant k$, que $n$ est moyen si $k+1 \\leqslant n \\leqslant 2k$, et que $n$ est grand si $2k+1 \\leqslant n$. Enfin, on suppose que l'on dispose de deux entiers naturels non nuls $x$ et $d$ tels que $r_{k}(x), r_{k}(x+d)$ et $r_{k}(x+2d)$ appartiennent tous trois à $A$, et tels que $r_{k}(x) \\neq r_{k}(x+d)$.\n\nPeut-on nécessairement affirmer, au vu des informations ci-dessus, que\n\na) au moins l'un des deux entiers $r_{k}(x)$ et $r_{k}(x+d)$ est moyen ou grand?\n\nb) au moins l'un des deux entiers $r_{k}(x)$ et $r_{k}(x+d)$ est petit ou grand?\n\nc) au moins l'un des deux entiers $r_{k}(x)$ et $r_{k}(x+d)$ est petit ou moyen?",
    "options": [],
    "answer": "a) No; b) Yes; c) No",
    "solution": "Solution:\n\nCet énoncé peut paraître long et compliqué. Il contient donc d'en retenir les éléments saillants, pour mieux comprendre ce qui se passe. Ici, on souhaite que l'ensemble $A$ ne contienne pas de progression arithmétique ($a, c, b$). Il s'agit donc de trouver deux entiers $x$ et $d$ pour lesquels, quand bien même $x, x+d$ et $x+2d$ sont manifestement en progression arithmétique, les restes $r_{k}(x), r_{k}(x+d)$ et $r_{k}(x+2d)$ dans la division euclidienne par $3k$ ne seront pas en progression arithmétique.\n\nUne idée, pour ce faire, est de construire $x$ et $d$, éventuellement en les choisissant négatifs si ça nous arrange (il sera toujours temps d'ajouter à $x$ et à $d$ des multiples de $3k$ arbitrairement grands) tels que $x$ et $x+d$ soient compris entre $1$ et $3k$, mais $x+2d$ dépasse de l'intervalle $\\{1,2, \\ldots, 3k\\}$. Une fois effectué ce travail de réflexion préparatoire, on peut maintenant traiter une à une les trois questions de l'énoncé : on cherchera à construire des entiers $x$ et $x+d$ qui sont tous deux petits, ou tous deux moyens, ou tous deux grands.\n\na) Cherchons à construire de petits entiers $x$ et $x+d$ tels que $x+2d \\leqslant 0$ et $x+2d \\geqslant 3k+1$. Puisque $x+2d=2(x+d)-x \\leqslant 2k-1<2k+1$, il nous faut faire en sorte que $x+2d \\leqslant 0$, donc que $d<0$. La manière la plus simple de procéder est alors de choisir $d=-1, x=2$, et $k \\geqslant 2$. Évidemment, on n'a pas vraiment le droit de choisir $d=-1$; comme mentionné précédemment, on choisira donc $d=3k-1$, ce qui revient au même mais est conforme aux contraintes de l'énoncé.\n\nEn pratique, si $k=2, A=\\{1,2,6\\}, x=2$ et $d=5$, alors $r_{k}(x)=2$ et $r_{k}(x+d)=1$ sont deux petits éléments de $A$, et $r_{k}(x+2d)=6$ appartient à $A$ également. La réponse est donc négative.\n\n![](attached_image_1.png)\n\nb) Supposons que $r_{k}(x)$ et $r_{k}(x+d)$ soient tous deux moyens, puis soit $t=2r_{k}(x+d)-r_{k}(x)$. On sait que\n$$\n2=2(k+1)-2k \\leqslant t \\leqslant 2 \\times 2k-(k+1)=3k-1\n$$\net que\n$$\nt \\equiv 2(x+d)-x \\equiv x+2d \\quad(\\bmod 3k),\n$$\ndonc que $t=r_{k}(x+2d)$. Mais alors $t+r_{k}(x)=2r_{k}(x+d)$, en contradiction avec la caractérisation initiale de $A$. Ceci invalide notre supposition, et la réponse est donc positive.\n\nc) Au vu de la symétrie manifeste entre gauche et droite, il suffit d'adapter la construction proposée en question a), en remplaçant $x$ par $3k+1-x$ et $d$ par $3k-d$.\n\nEn pratique, si $k=2, A=\\{1,5,6\\}, x=5$ et $d=1$, alors $r_{k}(x)=5$ et $r_{k}(x+d)=6$ sont deux grands éléments de $A$, et $r_{k}(x+2d)=1$ appartient à $A$ également. La réponse est donc négative.\n\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30735,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour s'entraîner en prévision du dernier test POFM de l'année, Jean-Baptiste et Marie-Odile ont collecté 100 problèmes de mathématiques, et s'attellent désormais à la confection d'un programme de révisions. Pendant les 100 jours qui les séparent du test POFM, chacun devra traiter un problème par jour. On note $x$ le nombre de problèmes que Jean-Baptiste a traités strictement avant Marie-Odile, et $y$ le nombre de problèmes que Marie-Odile a traités strictement avant Jean-Baptiste. Enfin, on dit que le programme de révisions est équitable si $x=y$.\n\nDémontrer qu'il existe au moins $100!\\times\\left(2^{50}+(50!)^{2}\\right)$ programmes équitables.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, il existe $100!$ manières de choisir l'ordre dans lequel Jean-Baptiste traitera les problèmes. Ces $100!$ manières jouent toutes des rôles symétriques. Sans perte de généralité, on numérote les problèmes de $1$ à $100$, dans l'ordre où Jean-Baptiste les traite, et il s'agit de démontrer qu'il existe au moins $2^{50}+(50!)^{2}$ permutations $\\sigma$ de l'ensemble $\\{1,2, \\ldots, 100\\}$ pour lesquelles les ensembles $\\{k: \\sigma(k)<k\\}$ et $\\{k: \\sigma(k)>k\\}$ ont même cardinal. On dira que ces permutations sont équitables.\n\nTout d'abord, on peut construire $2^{50}$ permutations équitables comme suit : étant donné un sous-ensemble $X$ de $\\{1,2, \\ldots, 50\\}$, parmi les $2^{50}$ possibles, on construit la permutation $\\sigma_{X}$ telle que\n$$\n\\sigma_{X}(2k-1)=\n\\begin{cases}\n2k & \\text{ si } k \\in X \\\\\n2k-1 & \\text{ si } k \\notin X\n\\end{cases}\n\\text{ et }\n\\sigma_{X}(2k)=\n\\begin{cases}\n2k-1 & \\text{ si } k \\in X \\\\\n2k & \\text{ si } k \\notin X\n\\end{cases}\n$$\nPour une telle permutation, on constate que $\\sigma_{X}(k)<k$ si et seulement si $k/2 \\in X$, et que $\\sigma_{X}(k)>k$ si et seulement si $(k+1)/2 \\in X$. Par conséquent, $\\sigma_{X}$ est équitable.\n\nOn construit ensuite $(50!)^{2}$ permutations équitables comme suit : étant données deux permutations $\\tau$ et $\\pi$ de $\\{1,2, \\ldots, 50\\}$ parmi les $(50!)^{2}$ paires de permutations possibles, on construit la permutation $\\sigma_{\\pi, \\tau}$ telle que\n$$\n\\sigma_{\\pi, \\tau}(k)=\n\\begin{cases}\n50+\\pi(k) & \\text{ si } 1 \\leqslant k \\leqslant 50 \\\\\n\\tau(k-50) & \\text{ si } 51 \\leqslant k \\leqslant 100\n\\end{cases}\n$$\nPour une telle permutation, on constate que $\\sigma_{\\pi, \\tau}(k)<k$ si et seulement si $51 \\leqslant k \\leqslant 100$, et que $\\sigma_{\\pi, \\tau}(k)>k$ si et seulement si $1 \\leqslant k \\leqslant 50$. Par conséquent, $\\sigma_{\\pi, \\tau}$ est équitable.\n\nEnfin, on vérifie bien sûr que deux permutations $\\sigma_{X}$ et $\\sigma_{\\pi, \\tau}$ ne sont jamais égales, puisque $\\sigma_{X}(1) \\leqslant 2 < 51 \\leqslant \\sigma_{\\pi, \\tau}(1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30736,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn définit la suite $a_{1}, a_{2}, a_{3} \\ldots$ de la façon suivante : $a_{1}=63$ et, pour tout entier $n \\geqslant 2$, $a_{n}$ est le plus petit multiple de $n$ qui soit supérieur ou égal à $a_{n-1}$. Démontrer que les termes de notre suite sont deux à deux distincts.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIl est clair que la suite $\\left(a_{n}\\right)_{n \\geqslant 1}$ est croissante. Il s'agit donc de démontrer qu'elle est strictement croissante.\n\nOn calcule donc les premiers termes de la suite : $a_{1}=63, a_{2}=64, a_{3}=66, a_{4}=68, a_{5}=70$, $a_{6}=72, a_{7}=77, a_{8}=80, a_{9}=81, a_{10}=90, a_{11}=99, a_{12}=108, a_{13}=117 \\ldots$.\n\nUne propriété se dessine alors sous nos yeux : il semble que $a_{n}=9 n$ pour tout $n \\geqslant 9$. En pratique, il est bon de vérifier cette conjecture sur quelques valeurs supplémentaires de $n$, pour se convaincre qu'elle est vraie. On entreprend donc de démontrer cette propriété par récurrence.\n\nTout d'abord, comme mentionné ci-dessus, on a bien $a_{9}=9^{2}$. Puis, pour tout entier $n \\geqslant 9$ tel que $a_{n}=9 n$, on constate que $8(n+1)=a_{n}-(n-8)<a_{n}<9(n+1)$, de sorte que $a_{n+1}=9(n+1)$.\n\nEn conclusion, la suite $\\left(a_{n}\\right)_{n \\geqslant 1}$ est strictement croissante tant que $n \\leqslant 9$, et elle l'est manifestement quand $n \\geqslant 9$. Ses termes sont donc en effet distincts deux à deux.\nSolution:\n\nComme dans la solution précédente, il est clair que la suite $\\left(a_{n}\\right)_{n \\geqslant 1}$ est croissante. Il s'agit donc de démontrer, pour tout entier $n \\geqslant 1$, que $a_{n} \\neq a_{n+1}$, c'est-à-dire que $n(n+1)$ ne divise pas $a_{n}$.\n\nOn calcule alors les premiers termes de la suite, jusqu'à $a_{9}=81$, et l'on constate bien que $n(n+1)$ ne divise pas $a_{n}$ lorsque $n \\leqslant 9$. On démontre ensuite que $a_{n}<n(n+1)$ pour tout entier $n \\geqslant 9$.\n\nEn effet, c'est bien le cas si $n=9$. Puis, pour tout entier $n \\geqslant 9$ tel que $a_{n}<n(n+1)$, on constate que $a_{n+1} \\leqslant a_{n}+n<n(n+2)<(n+1)^{2}$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle tel que $90^{\\circ} > \\widehat{ABC} > \\widehat{BCA}$. Soit $D$ le point sur le segment $[BC]$ tel que $2 \\widehat{DAC} = \\widehat{ABC} - \\widehat{BCA}$. On note $E$ le point d'intersection, autre que $A$, entre $(AB)$ et le cercle circonscrit à $ACD$, puis $P$ le point d'intersection entre $(AB)$ et la bissectrice de $\\widehat{BDE}$. De même, on note $F$ le point d'intersection, autre que $A$, entre $(AC)$ et le cercle circonscrit à $ABD$, puis $Q$ le point d'intersection entre $(AC)$ et la bissectrice de $\\widehat{CDF}$.\n\nDémontrer que $(AB)$ et $(PQ)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nComme d'habitude, on commence par une figure suffisamment belle et grande pour servir de support à la réflexion. À cette fin, on construit tout d'abord un triangle $A'BC'$ isométrique à $ABC$, tel que $C'$ se trouve sur la demi-droite $[BA)$, puis on trace la bissectrice de l'angle $\\widehat{CBA'}$, que l'on note $d$. On peut alors reporter en $A$ l'angle entre les droites $(BC)$ et $d$, pour construire la demi-droite $[AD)$.\n\nAlternativement, la cocyclicité des points $A, B, D$ et $F$ nous assure de l'égalité\n$$\n\\widehat{DBF} = \\widehat{DAF} = \\widehat{DAC} = \\widehat{CBA'}/2\n$$\nAinsi, $F$ est le point d'intersection des droites $(AC)$ et $d$, et $D$ est le point d'intersection, autre que $B$, de la droite $(BC)$ avec le cercle circonscrit à $ABF$.\n\n![](attached_image_1.png)\n\nAu vu de l'énoncé, une chasse aux angles s'impose. On pose donc $x = \\widehat{BCA}$ et $y = \\widehat{DAC}$, puis l'on constate que\n$$\n2\\widehat{CDQ} = \\widehat{CDF} = 180^{\\circ} - \\widehat{FDB} = \\widehat{BAF} = \\widehat{BAC} = 180^{\\circ} - 2(x+y),\n$$\nde sorte que $\\widehat{CDQ} = 90^{\\circ} - (x+y)$. De même,\n$$\n2\\widehat{PDB} = \\widehat{EDB} = 180^{\\circ} - \\widehat{CDE} = \\widehat{EAC} = \\widehat{BAC}\n$$\nde sorte que $\\widehat{PDB} = 90^{\\circ} - (x+y)$.\n\nOn continue notre chasse aux angles, d'où l'on déduit que\n$\\widehat{QDA} = 180^{\\circ} - \\widehat{DAQ} - \\widehat{AQD} = 180^{\\circ} - y - (180^{\\circ} - \\widehat{DQC}) = (180^{\\circ} - \\widehat{CDQ} - \\widehat{QCD}) - y = 90^{\\circ}$.\n\nIl s'agit donc de démontrer que $A, P, D$ et $Q$ sont cocycliques. On vérifie alors que\n$$\n\\widehat{DPA} = 180^{\\circ} - \\widehat{BPD} = \\widehat{PDB} + \\widehat{DBP} = (90^{\\circ} - x - y) + (x + 2y) = 90^{\\circ} + y = 180^{\\circ} - \\widehat{AQD},\n$$\nce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les triplets d'entiers naturels non nuls $(a, b, c)$ tels que\n$$\n2021^{a}+4=3^{b} \\times 5^{c}\n$$",
    "options": [],
    "answer": "(1,4,2)",
    "solution": "Solution:\n\nDans une équation diophantienne, on commence toujours par regarder de petites valeurs des paramètres. Ici, il suffit de factoriser $2021^{a}+4$ en produit de facteurs premiers, donc on regarde les petites valeurs de $a$. Ici, lorsque $a=1$, l'équation devient $2025=3^{b} \\times 5^{c}$, donc $(a, b, c)=(1,4,2)$ est solution. Par ailleurs, si $a \\geqslant 2$, personne n'a envie de factoriser $2021^{a}+4$, donc on arrête ici notre exploration des petits cas.\n\nDans un deuxième temps, il faudra soit procéder à des factorisations, soit regarder l'équation modulo un nombre $n$ bien choisi pour réduire l'ensemble des valeurs possibles de $a, b$ et $c$. Les nombres 3 et 5 sont manifestement des nombres premiers, et notre étude du cas $a=1$ indique que $2021=45^{2}-4=43 \\times 47$. On note donc cette relation dans un coin de notre brouillon en attendant de pouvoir en faire usage.\n\nEnfin, quitte à utiliser le théorème chinois, on réduit notre équation modulo $n$ lorsque $n$ est une puissance de nombre premier. Par exemple :\n$\\triangleright n=1, n=2$ et $n=5$ ne nous apportent aucune information;\n$\\triangleright n=3$ indique que $a$ est impair;\n$\\triangleright n=4$ indique que $b$ est pair;\n$\\triangleright n=7$ ne nous apporte aucune information facile à exploiter;\n$\\triangleright n=8$ indique, puisque $a$ est impair et $b$ est pair, que $c$ est pair.\n\nOn pose alors $\\beta=b / 2$ et $\\gamma=c / 2$. En outre, deux des trois termes de notre équation sont des carrés, ce qui nous permet d'obtenir la factorisation\n$$\n43^{a} \\times 47^{a}=2021^{a}=\\left(3^{\\beta} \\times 5^{\\gamma}\\right)^{2}-2^{2}=\\left(3^{\\beta} \\times 5^{\\gamma}+2\\right)\\left(3^{\\beta} \\times 5^{\\gamma}-2\\right) .\n$$\nPuisque $3^{\\beta} \\times 5^{\\gamma}+2$ et $3^{\\beta} \\times 5^{\\gamma}-2$ sont deux entiers impairs dont la différence vaut 4, ils sont premiers entre eux.\n\nSeul l'un de ces deux facteurs est divisible par 43, et seul l'un est divisible par 47. Plusieurs cas sont donc possibles a priori :\n$\\triangleright$ si $47$ divise $3^{\\beta} \\times 5^{\\gamma}-2$, alors $47^{a}$ divise $3^{\\beta} \\times 5^{\\gamma}-2$, donc $3^{\\beta} \\times 5^{\\gamma}-2 \\geqslant 47^{a} \\geqslant 43^{a} \\geqslant 3^{\\beta} \\times 5^{\\gamma}+2$, ce qui est impossible;\n$\\triangleright$ si 43 et 47 divisent $3^{\\beta} \\times 5^{\\gamma}+2$, on a $3^{\\beta} \\times 5^{\\gamma}+2=43^{a} \\times 47^{a}$ et $3^{\\beta} \\times 5^{\\gamma}-2=1$, ce qui est impossible puisque $3^{\\beta} \\times 5^{\\gamma}-2 \\geqslant 3 \\times 5-2=13$.\n\nAinsi, 47 divise $3^{\\beta} \\times 5^{\\gamma}+2$ et 43 divise $3^{\\beta} \\times 5^{\\gamma}-2$, de sorte que $3^{\\beta} \\times 5^{\\gamma}+2=47^{a}$ et $3^{\\beta} \\times 5^{\\gamma}-2=43^{a}$. Mais alors\n$$\n4=\\left(3^{\\beta} \\times 5^{\\gamma}+2\\right)-\\left(3^{\\beta} \\times 5^{\\gamma}-2\\right)=47^{a}-43^{a} \\geqslant 47^{a}-43 \\times 47^{a-1}=4 \\times 47^{a-1}\n$$\nce qui démontre que $a \\leqslant 1$.\n\nEn conclusion, la seule solution du problème est $(a, b, c)=(1,4,2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ une fonction croissante telle que $f(y)-f(x)<y-x$ pour tous les réels $x$ et $y$ tels que $y>x$. Soit $\\left(u_{n}\\right)_{n \\geqslant 0}$ une suite telle que $u_{n+2}=f\\left(u_{n+1}\\right)-f\\left(u_{n}\\right)$ pour tout entier $n \\geqslant 0$.\n\nDémontrer que la suite $\\left(u_{n}\\right)_{n \\geqslant 0}$ converge vers 0, c'est-à-dire que, pour tout réel $\\varepsilon>0$, il existe un entier $N$ tel que $\\left|u_{n}\\right| \\leqslant \\varepsilon$ pour tout entier $n \\geqslant N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nQuitte à remplacer la fonction $f$ par la fonction $t \\mapsto f(t)-f(0)$, ce qui ne change rien à la propriété désirée, on suppose que $f(0)=0$.\n\nPar ailleurs, remplacer la fonction $f$ par la fonction $t \\mapsto -f(-t)$ et la suite $\\left(u_{n}\\right)_{n \\geqslant 0}$ par la suite $\\left(-u_{n}\\right)_{n \\geqslant 0}$ ne change rien non plus à la propriété désirée. On pourra donc librement supposer, sans autre forme de procès, que tel ou tel terme $u_{n}$ est positif ou nul. Ci-dessous, ces suppositions sont marquées par le symbole $(\\star)$.\n\nOn dit maintenant qu'un entier $n$ est bon si les termes $u_{n}$ et $u_{n+1}$ sont de même signe et vérifient l'inégalité $\\left|u_{n+1}\\right| \\leqslant \\left|u_{n}\\right|$. Par ailleurs, on dit qu'un entier $n$ est dominant si tout entier $k \\geqslant n$ satisfait l'inégalité $\\left|u_{k}\\right| \\leqslant \\left|u_{n}\\right|$. Démontrons quelques résultats intermédiaires.\n\nLemme 1 : Parmi trois entiers consécutifs, au moins un est bon.\n\nDémonstration. Soit $n$ un entier naturel. Si $u_{n}$ et $u_{n+1}$ sont de signes opposés, et en supposant que $u_{n} \\geqslant 0^{(*)}$, on constate que $u_{n} \\geqslant 0 \\geqslant u_{n+1}$, donc que $u_{n+2}=f\\left(u_{n+1}\\right)-f\\left(u_{n}\\right) \\leqslant 0$ est de même signe que $u_{n+1}$. Ainsi, on peut choisir $k=n$ ou $k=n+1$ de sorte que $u_{k}$ et $u_{k+1}$ soient de même signe.\n\nOn démontre alors que l'un des deux entiers $k$ ou $k+1$ est bon. En effet, si $\\left|u_{k+1}\\right| \\geqslant \\left|u_{k}\\right|$, et en supposant que $u_{k} \\geqslant 0^{(*)}$, on a $u_{k+1} \\geqslant u_{k}$, et le terme $u_{k+2}=f\\left(u_{k+1}\\right)-f\\left(u_{k}\\right)$ est compris entre 0 et $u_{k+1}$, de sorte que $k+1$ est bon.\n\nLemme 2: Pour tout bon entier $m$, il existe un entier $n$, égal à $m+2$ ou à $m+3$, tel que\n\n$\\triangleright$ $n$ est bon,\n\n$\\triangleright$ $u_{m}$ et $u_{n}$ sont de signes opposés, et\n\n$\\triangleright \\left|u_{k}\\right| \\leqslant \\left|u_{m}\\right|$ pour tout entier $k$ tel que $m \\leqslant k \\leqslant n$.\n\nDémonstration. Soit $m$ un bon entier. En supposant que $u_{m} \\geqslant 0^{(*)}$, l'inégalité $0 \\leqslant u_{m+1} \\leqslant u_{m}$ nous assure que $u_{m+2}=f\\left(u_{m+1}\\right)-f\\left(u_{m}\\right) \\leqslant 0 \\leqslant u_{m+1}$, puis $u_{m+3}=f\\left(u_{m+2}\\right)-f\\left(u_{m+1}\\right) \\leqslant 0$. Ainsi, $u_{m+2}$ et $u_{m+3}$ sont de signe opposé à celui de $u_{m}$, et l'un des deux entiers $m+2$ ou $m+3$ est bon.\n\nEnfin, $0 \\leqslant u_{m+1} \\leqslant u_{m}$, $u_{m+2} \\geqslant u_{m+1}-u_{m} \\geqslant -u_{m}$ et $u_{m+3} \\geqslant u_{m+2}-u_{m+1} \\geqslant -u_{m}$, de sorte que $u_{m+1}, u_{m+2}$ et $u_{m+3}$ sont tous trois compris entre $-u_{m}$ et $u_{m}$.\n\nLemme 3 : Tout bon entier est dominant.\n\nDémonstration. Pour tout bon entier $m$, on note $\\psi(m)$ le plus petit bon entier supérieur ou égal à $m+2$. Ici, le lemme 2 et une récurrence montrent que $\\left|u_{m}\\right| \\geqslant \\left|u_{\\psi^{k}(m)}\\right| \\geqslant \\left|u_{\\ell}\\right|$ pour tout entier $\\ell$ tel que $\\psi^{k}(m) \\leqslant \\ell \\leqslant \\psi^{k+1}(m)$, de sorte que $m$ est dominant.\n\nSoit $a$ le plus petit bon entier. On suppose que $u_{a} \\geqslant 0^{(*)}$. La suite $\\left(u_{\\psi^{k}(a)}\\right)_{k \\geqslant 0}$ est formée de termes alternativement positifs et négatifs, dont les valeurs absolues décroissent. Ainsi, la suite $\\left(u_{\\psi^{2k}(a)}\\right)_{k \\geqslant 0}$ est décroissante, et admet une limite $\\ell \\geqslant 0$, tandis que la suite $\\left(u_{\\psi^{2k+1}(a)}\\right)_{k \\geqslant 0}$ est croissante, et admet $-\\ell$ pour limite.\n\nSi $\\ell=0$, puisque tous les entiers $\\psi^{k}(a)$ sont dominants, la suite $\\left(u_{n}\\right)_{n \\geqslant 0}$ converge bien vers 0. Sinon, soit $\\varepsilon=\\min \\{\\ell / 2-f(\\ell / 2), \\ell / 2+f(\\ell / 2)-f(\\ell)\\}$. L'énoncé nous indique que $\\varepsilon>0$, et nous permet aussi d'obtenir le résultat suivant.\n\nLemme 4 : Soit $x$ et $y$ deux réels. Si $x \\leqslant \\ell / 2$ et $y \\geqslant \\ell$, ou bien si $x \\leqslant 0$ et $y \\geqslant \\ell / 2$, alors $f(y)-f(x) \\leqslant y-x-\\varepsilon$.\n\nDémonstration. On traite les deux cas séparément:\n\n$\\triangleright$ si $x \\leqslant 0$ et $y \\geqslant \\ell / 2$, alors\n\n$$\n\\begin{aligned}\nf(y)-f(x) & =f(y)-f(\\ell / 2)+f(\\ell / 2)-f(0)+f(0)-f(x) \\\\\n& \\leqslant (y-\\ell / 2)+(\\ell / 2-\\varepsilon)+(0-x)=y-x-\\varepsilon\n\\end{aligned}\n$$\n\n$\\triangleright$ si $x \\leqslant \\ell / 2$ et $y \\geqslant \\ell$, alors\n\n$$\n\\begin{aligned}\nf(y)-f(x) & =f(y)-f(\\ell)+f(\\ell)-f(\\ell / 2)+f(\\ell / 2)-f(x) \\\\\n& \\leqslant (y-\\ell)+(\\ell / 2-\\varepsilon)+(\\ell / 2-x)=y-x-\\varepsilon\n\\end{aligned}\n$$\n\nDans les deux cas, on a bien l'inégalité désirée.\n\nOn reprend alors la construction suivie au lemme 2. Étant donné un bon entier $m$, pour lequel on aura supposé que $u_{m} \\geqslant 0^{(*)}$, on sait que $u_{m} \\geqslant \\ell$, et on distingue deux cas :\n\n$\\triangleright$ si $u_{m+1} \\geqslant \\ell / 2$, alors $0 \\geqslant u_{m+2} \\geqslant u_{m+1}-u_{m} \\geqslant \\ell / 2-u_{m} \\geqslant \\varepsilon-u_{m}$, et le lemme 4 indique que $0 \\geqslant u_{m+3}=f\\left(u_{m+2}\\right)-f\\left(u_{m+1}\\right) \\geqslant \\varepsilon+u_{m+2}-u_{m+1} \\geqslant \\varepsilon-u_{m}$;\n\n$\\triangleright$ si $u_{m+1} \\leqslant \\ell / 2$, alors le lemme 4 indique que $0 \\geqslant u_{m+2} \\geqslant \\varepsilon+u_{m+1}-u_{m} \\geqslant \\varepsilon-u_{m}$, et $0 \\geqslant u_{m+3} \\geqslant u_{m+2}-u_{m+1} \\geqslant \\varepsilon-u_{m}$.\n\nDans les deux cas, les termes $u_{m+2}$ et $u_{m+3}$ sont compris entre $\\varepsilon-u_{m}$ et 0. On en déduit que $\\left|u_{\\psi(m)}\\right| \\leqslant \\left|u_{m}\\right| - \\varepsilon$, et une relation de récurrence immédiate permet de conclure que $0 \\leqslant \\left|u_{\\psi^{k}}(a)\\right| \\leqslant \\left|u_{a}\\right| - k\\varepsilon$ pour tout entier $k$, ce qui est absurde lorsque $k > \\left|u_{a}\\right| / \\varepsilon$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $\\left(F_{k}\\right)_{k \\geqslant 0}$ la suite définie par $F_{0}=0, F_{1}=1$, et $F_{k+2}=F_{k}+F_{k+1}$ pour tout entier $k \\geqslant 0$. Soit ensuite $n \\geqslant 1$ un entier. Démontrer qu'il existe exactement $F_{n+1}$ façons d'ordonner les nombres $1,2, \\ldots, n$ de manière à obtenir un $n$-uplet $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ tel que\n$$\na_{1} \\leqslant 2 a_{2} \\leqslant 3 a_{3} \\leqslant \\ldots \\leqslant n a_{n} .\n$$",
    "options": [],
    "answer": "F_{n+1}",
    "solution": "Solution:\nOn dit qu'une permutation $\\mathbf{a}=\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ des entiers $1,2, \\ldots, n$ est jolie si elle satisfait les inégalités $a_{1} \\leqslant 2 a_{2} \\leqslant \\ldots \\leqslant n a_{n}$.\n\nTout d'abord, soit $a$ une jolie permutation. S'il existe un entier $k$ pour lequel $a_{k} \\leqslant k-2$, on choisit $k$ minimal. Dans ces conditions, on sait que $k \\geqslant 2$ et que $a_{k-1} \\geqslant k-2 \\geqslant a_{k}$, de sorte que $a_{k-1} \\geqslant a_{k}+1$. On en conclut que\n$$\n(k-1) a_{k-1} \\geqslant (k-1)\\left(a_{k}+1\\right) = k a_{k} + \\left(k-a_{k}-1\\right) \\geqslant k a_{k} + 1,\n$$\net donc que $a$ n'est pas jolie. On en conclut donc que $a_{k} \\geqslant k-1$ pour tout $k \\leqslant n$.\n\nD'autre part, soit $\\ell$ l'entier tel que $a_{\\ell}=n$. Au vu du résultat obtenu précédemment, une récurrence descendante immédiate sur $k$ nous assure que $a_{k}=k-1$ pour tout entier $k$ tel que $\\ell+1 \\leqslant k \\leqslant n$. Puisque\n$$\n\\ell n = \\ell a_{\\ell} \\leqslant (\\ell+1) a_{\\ell+1} = (\\ell+1) \\ell\n$$\non en conclut que $\\ell \\in\\{n-1, n\\}$.\n\nAinsi, pour que $a$ soit jolie, on dispose a priori de deux choix :\n$\\triangleright$ soit $\\ell=n$, auquel cas $a$ est effectivement jolie si et seulement si la permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n-1}\\right)$ des entiers $1,2, \\ldots, n-1$ est jolie;\n$\\triangleright$ soit $\\ell=n-1$, auquel cas $a_{n}=n-1$, et alors $a$ est effectivement jolie si la permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n-2}\\right)$ des entiers $1,2, \\ldots, n-2$ est jolie.\n\nPar conséquent, si on note $J_{n}$ le nombre de jolies permutations de $1,2, \\ldots, n$, on remarque bien que $J_{n}=J_{n-1}+J_{n-2}$ dès lors que $n \\geqslant 3$. On conclut en vérifiant que $J_{1}=1=F_{2}$ et que $J_{2}=2=F_{3}$.\n\n\nSolution alternative $n^{\\circ} 1$\nComme précédemment, on considère une permutation $a$ puis l'entier $\\ell$ tel que $a_{\\ell}=n$. Puisque $n a_{k} \\geqslant k a_{k} \\geqslant \\ell a_{\\ell}=\\ell n$ pour tout $k \\geqslant \\ell$, on sait que $a$ induit une permutation de $1,2, \\ldots, \\ell-1$ et de $\\ell, \\ell+1, \\ldots, n$.\n\nSi $\\ell \\leqslant n-1$, soit $m$ l'unique entier tel que $a_{m}=\\ell$. Puisque $m \\geqslant \\ell$, on sait que\n$$\nm \\ell = m a_{m} \\geqslant \\ell a_{\\ell} = \\ell n\n$$\ndonc que $m=n$. Mais alors $k a_{k}=\\ell n$ pour tout entier $k$ tel que $\\ell \\leqslant k \\leqslant n$. On en déduit en particulier que $n-1$ divise $(n-1) a_{n-1}=\\ell n$, donc divise $\\ell$ aussi, ce qui signifie que $\\ell=n-1$. Maintenant acquis le fait que $\\ell \\in\\{n-1, n\\}$, on conclut comme précédemment.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30741,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $P_{1} P_{2} \\ldots P_{2021}$ un polygone convexe à 2021 sommets tel que, pour chaque sommet $P_{i}$, les 2018 diagonales issues de $P_{i}$ divisent l'angle $\\widehat{P}_{i}$ en 2019 angles égaux.\n\nDémontrer que $P_{1} P_{2} \\ldots P_{2021}$ est un polygone régulier, c'est-à-dire un polygone dont tous les angles ont même mesure et tous les côtés ont même longueur.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans la suite, on pose $n=2019$ et, pour tout sommet $P_{i}$, on note $a_{i}$ l'angle $\\widehat{P}_{i} / n$. Soit alors $P_{k}$, $P_{k+1}, P_{k+2}, P_{k+3}$ et $P_{k+4}$ quatre sommets consécutifs, les indices des sommets étant considérés modulo $n+2$.\nLa somme des angles d'un triangle vaut $180^{\\circ}$. En considérant respectivement les triangles $P_{i} P_{i+1} P_{i+2}, P_{i} P_{i+1} P_{i+3}$ et $P_{i} P_{i+1} P_{i+4}$, on constate que\n$$\n\\begin{aligned}\na_{i}+n a_{i+1}+a_{i+2} & =180^{\\circ} \\\\\n2 a_{i}+(n-1) a_{i+1}+a_{i+3} & =180^{\\circ} \\\\\n3 a_{i}+(n-2) a_{i+1}+a_{i+4} & =180^{\\circ}\n\\end{aligned}\n$$\nNous allons alors soustraire deux à deux ces équations :\n$\\triangleright(1)-(2)$ indique que $a_{i+1}+a_{i+2}=a_{i}+a_{i+3}$, c'est-à-dire que $a_{i+1}-a_{i}=a_{i+3}-a_{i+2}$;\n$\\triangleright(2)-(3)$ indique que $a_{i+1}+a_{i+3}=a_{i}+a_{i+4}$, c'est-à-dire que $a_{i+1}-a_{i}=a_{i+4}-a_{i+3}$.\nOn constate donc que $a_{i+4}-a_{i+3}=a_{i+3}-a_{i+2}$. La différence entre deux angles $a_{j}$ et $a_{j+1}$ consécutifs est donc égale à une constante $\\mathbf{c}$.\nCela signifie que $a_{i+k}=a_{i}+k \\mathbf{c}$ pour tous les entiers $i$ et $k$. En particulier, $a_{1}=a_{n+3}=a_{1}+(n+2) \\mathbf{c}$, donc $\\mathbf{c}=0^{\\circ}$. Ainsi, notre polygone a tous ses sommets égaux.\nLe triangle $P_{i} P_{i+1} P_{i+2}$ est donc isocèle en $P_{i+1}$, de sorte que $P_{i} P_{i+1}=P_{i+1} P_{i+2}$. Deux côtés consécutifs étant de même longueur, notre polygone a donc tous ses côtés de même longueur, et il s'agit bien d'un polygone régulier.\n\n![](attached_image_1.png)\n\n\nSolution alternative $n^{\\circ} 1$\n\nComme dans la solution précédente, on démontre les égalités (1) et (2) pour en déduire que $a_{i+1}-a_{i}=a_{i+3}-a_{i+2}$. Si l'on pose $\\mathbf{c}=a_{2}-a_{1}$, une récurrence immédiate démontre alors que $a_{2 k}-a_{2 k-1}=\\mathbf{c}$ pour tout entier $k \\geqslant 1$.\nOr, nos indices sont considérés modulo $n+2=2021$, qui est impair. On en déduit que $a_{i+1}-a_{i}=\\mathbf{c}$ pour tout indice $i$. On conclut alors comme précédemment.\n\n\nSolution alternative $n^{\\circ} 2$\n\nComme précédemment, on démontre l'égalité (1). Si l'on pose $b_{i}=a_{i}-180^{\\circ} /(n+2)$, cette égalité se réécrit comme $b_{i}+n b_{i+1}+b_{i+2}=0$.\nLa suite $\\left(b_{i}\\right)_{i \\geqslant 1}$ est donc récurrente linéaire. Si l'on note $r_{+}$ et $r_{-}$ les deux racines du polynôme $1+n X+X^{2}$, c'est-à-dire\n$$\nr_{ \\pm}=\\frac{-n \\pm \\sqrt{n^{2}-4}}{2}\n$$\nil existe donc deux nombres $\\lambda_{+}$ et $\\lambda_{-}$ tels que\n$$\nb_{i}=\\lambda_{+} r_{+}^{i}+\\lambda_{-} r_{-}^{i}\n$$\npour tout $i \\geqslant 0$.\nPuisque $n \\geqslant 5$, on sait que\n$$\n\\left(n^{2}-4\\right)-(n-2)^{2}=2(n-4)>0\n$$\ndonc que $r_{-} \\leqslant-n / 2<-1<r_{+}<0$ et que on sait que $\\left|r_{-}\\right|>1>\\left|r_{+}\\right|$. Par conséquent, on sait que, si $\\lambda_{-} \\neq 0$, la fraction $b_{i} /\\left(\\lambda_{-} r_{-}^{i}\\right)$ tend vers 1 quand $i \\rightarrow+\\infty$. Or, la suite $\\left(b_{i}\\right)_{i \\geqslant 1}$ est $(n+2)$-périodique. On en déduit que $\\lambda_{-}=0$.\nDe même, si $\\lambda_{+} \\neq 0$, la fraction $b_{i} /\\left(\\lambda_{+} r_{+}^{i}\\right)$ tend vers 1 quand $i \\rightarrow+\\infty$, en contradiction avec le caractère $(n+2)$-périodique de la suite $\\left(b_{i}\\right)_{i \\geqslant 1}$. On en déduit que $\\lambda_{+}=0$, donc que $b_{i}=0$ pour tout $i \\geqslant 1$.\nEn conclusion, tous les angles $a_{i}$ sont égaux, et on démontre comme précédemment que les côtés du polygone sont égaux eux aussi.\n\n\nSolution alternative $n^{\\circ} 3$\n\nComme précédemment, on démontre l'égalité (1). En écrivant cette égalité pour deux entiers $i$ consécutifs, et en soustrayant les deux égalités obtenues, on constate que\n$$\na_{i}+(n-1) a_{i+1}-(n-1) a_{i+2}-a_{i+3}=0 .\n$$\nLa suite $a_{i}$ est donc récurrente linéaire. Cette fois-ci, les racines du polynôme\n$$\n1+(n-1) X-(n-1) X^{2}-X^{3}=(1-X)\\left(1+n X+X^{2}\\right)\n$$\nsont $r_{ \\pm}$ et $1$. Puisque $n \\geqslant 5$, on sait que\n$$\n\\left(n^{2}-4\\right)-(n-2)^{2}=2(n-4)>0\n$$\ndonc que $r_{-} \\leqslant-n / 2<-1<r_{+}<0<1$ et que on sait que $\\left|r_{-}\\right|>1>\\left|r_{+}\\right|$.\nIl existe donc trois nombres $\\lambda_{+}, \\lambda_{1}$ et $\\lambda_{-}$ tels que\n$$\na_{i}=\\lambda_{+} r_{+}^{i}+\\lambda_{1}+\\lambda_{-} r_{-}^{i}\n$$\npour tout $i \\in \\mathbb{Z}$. On démontre comme dans la solution précédente que $\\lambda_{-}=0$. De même, si $\\lambda_{+} \\neq 0$, alors $a_{i} /\\left(\\lambda_{+} r_{+}^{i}\\right) \\rightarrow 1$ quand $i \\rightarrow-\\infty$, en contradiction avec le caractère périodique de $\\left(a_{i}\\right)$. Ainsi, $\\lambda_{+}=0$.\nEn conclusion, tous les angles $a_{i}$ sont égaux à $\\lambda_{1}$, et on démontre comme précédemment que les côtés du polygone sont égaux eux aussi.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les nombres réels $x$ et $y$ tels que\n$$\n\\left(x-\\sqrt{x^{2}+1}\\right)\\left(y-\\sqrt{y^{2}+1}\\right)=1 \\text{ et }\\left(x^{2}+y+2\\right)\\left(y^{2}+x+2\\right)=8 .\n$$",
    "options": [],
    "answer": "(x, y) = (1, -1) and (x, y) = (-1, 1)",
    "solution": "Solution:\nSoit $(x, y)$ une solution éventuelle. Les égalités\n$$\n\\left\\{\n\\begin{array}{l}\n\\left(x-\\sqrt{x^{2}+1}\\right)\\left(x+\\sqrt{x^{2}+1}\\right)=x^{2}-\\left(x^{2}+1\\right)=-1 \\\\\n\\left(y-\\sqrt{y^{2}+1}\\right)\\left(y+\\sqrt{y^{2}+1}\\right)=y^{2}-\\left(y^{2}+1\\right)=-1\n\\end{array}\n\\right.\n$$\nindiquent en fait que\n$$\n\\left\\{\n\\begin{array}{l}\nx+\\sqrt{x^{2}+1}=\\sqrt{y^{2}+1}-y \\\\\n\\sqrt{x^{2}+1}-x=y+\\sqrt{y^{2}+1}\n\\end{array}\n\\right.\n$$\nL'équation (1) $-$ (2) indique que $x=-y$, et l'égalité $\\left(x^{2}+y+2\\right)\\left(y^{2}+x+2\\right)=8$ se réécrit alors comme\n$$\n8=\\left(x^{2}-x+2\\right)\\left(x^{2}+x+2\\right)=\\left(x^{2}+2\\right)^{2}-x^{2}\n$$\nSi l'on pose $z=x^{2}$, il s'agit donc de résoudre l'équation\n$$\n0=(z+2)^{2}-z-8=z^{2}+3z-4=(z+3/2)^{2}-25/4=(z+3/2)^{2}-(5/2)^{2}=(z+4)(z-1)\n$$\nOr, on sait que $z=x^{2}\\geqslant 0$. On en déduit que $z=1$, donc que $x=\\pm 1$ et $y=-x=\\pm 1$.\n\nRéciproquement, lorsque $(x, y)=(\\pm 1, \\mp 1)$, on vérifie aisément que les deux égalités de l'énoncé sont vérifiées. En conclusion, les deux solutions sont\n$$\n(x, y)=(1,-1) \\text{ et }(x, y)=(-1,1)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe dont les angles en $B$ et en $D$ sont obtus, et dont les angles en $A$ et en $C$ sont égaux l'un à l'autre. Soit $E$ et $F$ les symétriques de $A$ par rapport à $(BC)$ et $(CD)$. Soit $K$ et $L$ les points d'intersection de $(BD)$ avec $(AE)$ et $(AF)$. Démontrer que les cercles circonscrits aux triangles $BEK$ et $DFL$ sont tangents l'un à l'autre.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLa manière la plus pratique de construire le point $A$ est d'en construire tout d'abord son symétrique $A'$ par rapport à $(BD)$, puisque $A'$ est cocyclique avec $B$, $C$ et $D$. Ce faisant, on constate que $A'$ semble appartenir aux deux cercles circonscrits aux triangles $BEK$ et $DFL$, et une chasse aux angles indique en effet que\n$$\n\\begin{aligned}\n\\left(KE, KA'\\right) & =(KE, BC)+(BC, BD)+\\left(BD, KA'\\right) \\\\\n& =90^\\circ+(BC, BD)+(KA, BD) \\\\\n& =90^\\circ+(BC, BD)+(KA, BC)+(BC, BD) \\\\\n& =2(BC, BD) \\\\\n\\left(BE, BA'\\right) & =(BE, BC)+(BC, BD)+\\left(BD, BA'\\right) \\\\\n& =(BC, BA)+(BC, BD)+(BA, BD) \\\\\n& =2(BC, BD) .\n\\end{aligned}\n$$\nCela signifie bien que $A'$, $B$, $K$ et $E$ appartiennent à un cercle $\\omega_b$ et, de même, que $A'$, $D$, $L$ et $F$ appartiennent à un cercle $\\omega_d$.\n\nSoit alors $t$ et $t'$ les tangentes à $\\omega_b$ et $\\omega_d$ en $A'$. Une nouvelle chasse aux angles indique que\n$$\n\\begin{aligned}\n(t, BD) & =\\left(t, A'B\\right)+\\left(A'B, BD\\right) \\\\\n& =\\left(KA', KB\\right)+(BD, BA) \\\\\n& =(KB, KA)+(BD, BA) \\\\\n& =(BD, BC)+90^\\circ+(BD, BA) \\\\\n\\left(t', BD\\right) & =\\left(t', A'D\\right)+\\left(A'D, BD\\right) \\\\\n& =\\left(LA', LD\\right)+(BD, AD) \\\\\n& =(LD, LA)+(BD, AD) \\\\\n& =(BD, DC)+90^\\circ+(BD, AD) .\n\\end{aligned}\n$$\nOn vérifie enfin que\n$$\n\\begin{aligned}\n(BD, BC)+(BD, BA) & =(BD, BC)+(BD, AD)+(AD, AB) \\\\\n& =(BD, BC)+(BD, AD)+(BC, CD) \\\\\n& =(BD, DC)+(BD, AD),\n\\end{aligned}\n$$\nce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30744,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPierre et Clara jouent au jeu suivant. Tout d'abord, Clara choisit un entier $c$. Puis Pierre choisit un nombre premier $p \\geqslant c$ et écrit deux entiers $a$ et $b$ au tableau. Clara se permet alors d'effectuer les opérations suivantes : elle choisit un des deux nombres écrits au tableau, disons $n$, l'efface, et écrit à la place un autre entier, disons $m$, tel que $p$ divise $\\left(n^{2}-m+1\\right)\\left(m^{2}-n+1\\right)$. Clara gagne si elle réussit à faire en sorte, au bout d'un nombre fini de telles opérations, que les deux entiers écrits au tableau soient identiques. Sinon, elle perd, et c'est Pierre qui gagne.\n\nDémontrer que Pierre peut empêcher Clara de gagner.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $p$ un nombre premier, arbitrairement grand, que Pierre est susceptible de choisir, et soit $f: \\mathbb{Z} / p \\mathbb{Z} \\to \\mathbb{Z} / p \\mathbb{Z}$ la fonction définie par $f(x) = x^{2} + 1$.\n\nLorsque Clara remplace un entier $n$ par un entier $m$, l'une des deux égalités $n = f(m)$ ou $m = f(n)$ est nécessairement vérifiée, de sorte que $m$ et $n$ appartiennent à une même orbite de $f$. On dit alors que deux résidus $u$ et $v$ modulo $p$ sont reliés s'il existe deux entiers $k \\geqslant 0$ et $\\ell \\geqslant 0$ tels que $f^{k}(u) = f^{\\ell}(v)$. Il apparaît directement que, si Clara parvient à transformer un résidu $m$ en un résidu $n$, ces deux résidus sont reliés ; la réciproque est d'ailleurs vraie, même si nous n'en aurons pas besoin.\n\nPierre a donc pour but de trouver un nombre premier $p \\geqslant c$, puis deux résidus $a$ et $b$ modulo $p$ qui ne sont pas reliés l'un à l'autre. Pour ce faire, le plus simple est que $a$ et $b$ soient des points fixes de $f$, c'est-à-dire que $(X-a)(X-b) \\equiv X^{2} - X + 1 \\pmod{p}$.\n\nOr, si $X^{2} - X + 1$ admet une racine $a$ modulo $p$, l'autre racine vaut $1/a$ (car $a$ est non nul), ou encore $1-a$. Ces deux racines ne sont donc égales que si $1-a \\equiv a \\equiv 1/a \\pmod{p}$, c'est-à-dire si $p \\neq 2$ et $1/2 \\equiv a \\equiv \\pm 1 \\pmod{p}$. Or, on sait que $1/2 \\equiv \\pm 1 \\pmod{p}$ si et seulement si $p = 3$. Par conséquent, Pierre n'a qu'à choisir un nombre premier $p > \\max\\{c, 3\\}$ qui divise un nombre de la forme $n^{2} - n + 1$. En pratique, pour ce faire, il lui suffit de définir $p$ comme le plus petit facteur premier du nombre $S_{m} = (m!)^{2} - m! + 1$, où l'on a posé $m = \\max\\{c, 3\\}$. En effet, on sait que $S_{m} \\geqslant 2$ et que $S_{m}$ est premier avec $m!$. Ainsi, le plus petit facteur premier de $S_{m}$ existe bien, et il est premier avec $m!$, donc strictement supérieur à $m$. Pierre s'empressera ensuite de choisir $a = m!$ et $b = 1 - m!$ pour enfin savourer une victoire amplement méritée.\n\n\nSolution alternative n°1\n\nÉtant donné un nombre premier $p$, nous allons construire le graphe $G$ dont les sommets sont les éléments de $\\mathbb{Z} / p \\mathbb{Z}$ et les arêtes sont les paires $(m, n)$ telles que $\\left(n^{2} - m + 1\\right)\\left(m^{2} - n + 1\\right) \\equiv 0 \\pmod{p}$. On dira qu'une arête $(m, n)$ est utile si $m \\neq n$, et qu'elle est inutile sinon, c'est-à-dire s'il s'agit d'une boucle. Si $G$ contient au plus $p-2$ arêtes utiles, alors il n'est pas connexe, et Pierre n'a plus qu'à choisir des entiers appartenant à deux composantes connexes distinctes.\n\nOn dit ensuite qu'une arête $(m, n)$ est engendrée par $n$ si $m \\equiv n^{2} + 1 \\pmod{p}$. Toute arête est engendrée par une de ses extrémités, et chaque sommet engendre une unique arête, donc $G$ contient au plus $p$ arêtes. Par conséquent, si $G$ contient deux arêtes inutiles, c'est-à-dire s'il existe deux résidus distincts $u$ et $v$ tels que $u \\equiv u^{2} + 1 \\pmod{p}$ et $v \\equiv v^{2} + 1 \\pmod{p}$, Pierre est assuré de pouvoir gagner.\n\nIl lui suffit alors de procéder comme dans la solution précédente.\n\n\nSolution alternative n°2\n\nLa solution ci-dessous fait appel à des notions plus avancées, mais cela permet une approche laxiste de la dimension combinatoire du problème. Ci-dessous, on considère directement que Pierre choisit des résidus modulo $p$ plutôt que des entiers, et on note $f: \\mathbb{Z} / p \\mathbb{Z} \\to \\mathbb{Z} / p \\mathbb{Z}$ la fonction définie par $f(x) = x^{2} + 1$. Lorsque $x$ et $y$ sont des résidus modulo $p$, on notera abusivement $x = y$ et $x \\neq y$ les relations $x \\equiv y \\pmod{p}$ et $x \\not\\equiv y \\pmod{p}$.\n\nSupposons que Pierre a choisi un nombre premier $p$ et deux résidus $a$ et $b$ tels que $b \\neq \\pm a$ tels que $a = f(a)$, mais que Clara gagne malgré tout. Puisque toute opération de Clara peut être inversée, on suppose sans perte de généralité que celle-ci a graduellement transformé $a$ en $b$, en le plus petit nombre d'étapes possibles.\n\nLa première étape de Clara consiste à transformer $a$ en $f(a) = a$, ce qui serait stupide, ou en un antécédent $x \\neq a$ de $a$ par $f$. Les deux antécédents de $a$ par $f$ sont opposés l'un à l'autre, et l'un est égal à $a$, donc l'autre vaut $-a$. Ainsi, Clara a transformé $a$ en $-a$.\n\nPuisque $b \\neq -a$, Clara a de nouveau dû transformer $-a$ en $f(-a) = a$, ce qui serait encore une fois stupide, ou bien en un antécédent $y$ de $-a$ par $f$, de sorte que $-a-1 = f(y) - 1 = y^{2}$ est un résidu quadratique modulo $p$.\n\nPar conséquent, Pierre gagnera s'il trouve un nombre premier $p \\geqslant 3$ pour lequel $f$ a un point fixe $a$ tel que $-a-1$ ne soit pas un résidu quadratique modulo $p$. Or, si $a$ est un point fixe de $f$, on sait que $f(1-a) = a^{2} - 2a + 2 = f(a) - 2a + 1 = 1 - a$, donc que $1 - a$ est aussi un point fixe de $f$. Il suffit donc que soit $-a-1$, soit $a-2$ ne soit pas un résidu quadratique (soit les deux).\n\nLa manière la plus simple pour s'assurer que deux nombres ne seront pas tous deux des résidus quadratiques est de faire en sorte que leur produit n'en soit pas un. Cette technique sera ici avantageuse, puisqu'elle nous permettra de ne pas devoir distinguer les résidus $-a-1$ et $a-2$, qui avaient été définis de manière symétrique. Ici, Pierre va donc choisir $p$ de sorte que $(a-2)(-a-1) = 2 + a - a^{2} = 3$ ne soit pas un résidu quadratique.\n\nEnfin, il doit s'assurer que $f$ ait bien un point fixe. Or, s'il se plonge dans le corps $\\mathbb{C}$ des nombres complexes, Pierre sait bien que les racines du polynôme $f(X) - X = X^{2} - X + 1$ sont $(1 + \\sqrt{-3}) / 2$ et $(1 - \\sqrt{-3}) / 2$. On reconnaît donc là que $a$ est un point fixe de $f$ si et seulement si $(2a - 1)^{2} = -3$, et un tel point fixe existe donc si et seulement si $-3$ est un résidu quadratique. Il s'agit donc, pour Pierre, de choisir un nombre premier $p$ pour lequel $-3$ est un résidu quadratique et $-1$ n'en est pas un.\n\nOr, s'il existe un résidu $z$ tel que $z^{2} = -1$, on sait que $z$ est d'ordre 4, donc que $p \\equiv 1 \\pmod{4}$. Pierre va donc chercher $p$ parmi les facteurs premiers non congrus à $1 \\pmod{4}$ d'un entier de la forme $n^{2} + 3$. Choisir $n$ premier avec 3 nous assure déjà que $p \\neq 3$. Ensuite, choisir $n$ pair nous assure que $n^{2} + 3 \\equiv 3 \\pmod{4}$, donc que $n$ admet nécessairement un facteur premier $p \\equiv 3 \\pmod{4}$. Il suffit donc de définir $n$ comme le PPCM des nombres premiers inférieurs ou égaux à $\\max\\{c, 2\\}$ et différents de $3$ : le facteur $p$ que choisira Pierre sera premier avec tous ces nombres premiers, mais aussi avec 3, de sorte que $p > c$, et les considérations des paragraphes précédents nous assurent alors que Pierre peut choisir des entiers $a$ et $b$ de façon adéquate.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn note $\\mathbb{Z}[x, y, z]$ l'ensemble des polynômes à coefficients entiers en les trois variables $x, y$ et $z$. On dit ensuite qu'un polynôme $P \\in \\mathbb{Z}[x, y, z]$ est olympique si $\\mathbb{Z}[x, y, z]$ contient des polynômes $A, B$ et $C$ tels que\n$$\nP(x, y, z)=(x+y+z) A(x, y, z)+(x y+y z+z x) B(x, y, z)+x y z C(x, y, z) .\n$$\nTrouver le plus grand entier $n$ pour lequel il existe des entiers naturels $i, j$ et $k$ de somme $i+j+k=n$ et tels que le polynôme $x^{i} y^{j} z^{k}$ ne soit pas olympique.\n\nNote : Un polynôme à coefficients entiers en les variables $x, y$ et $z$ est une fonction que l'on peut écrire comme une somme de termes de la forme $\\lambda x^{i} y^{j} z^{k}$, où $\\lambda$ est un entier relatif et $i, j$ et $k$ sont des entiers naturels. Par exemple, $x-y+1$ et $x y+y z+z x$ sont de tels polynômes, mais $\\pi x y z, \\exp (x), x /(y^{2}+1)$ et $\\sqrt{x y+z}$ n'en sont pas.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nPosons $s_{1}=x+y+z$, $s_{2}=x y+y z+z x$ et $s_{3}=x y z$. On constate que\n$$\n\\begin{cases}\nx^{3} \\equiv -x^{2}(y+z) \\equiv -x(x y+x z) \\equiv x y z \\equiv 0 & (\\bmod\\ s_{1}, s_{2}, s_{3}) \\\\\nx^{2} y^{2} \\equiv x y(-x z-y z) \\equiv -x y z(x+y) \\equiv 0 & (\\bmod\\ s_{1}, s_{2}, s_{3})\n\\end{cases}\n$$\nce qui signifie que $x^{3}$ et $x^{2} y^{2}$ sont olympiques. Or, l'ensemble des polynômes olympiques est stable par multiplication par tout polynôme et par permutation des variables.\n\nPar conséquent, considérons un polynôme $x^{i} y^{j} z^{k}$, non olympique. Sans perte de généralité, on suppose que $i \\geqslant j \\geqslant k$. Si $i \\geqslant 3$, $j \\geqslant 2$ ou $k \\geqslant 1$, notre polynôme est un multiple de $x^{3}$, de $x^{2} y^{2}$ ou de $x y z$, donc il est olympique. On en déduit que $i \\leqslant 2$, $j \\leqslant 1$ et $k \\leqslant 0$, de sorte que $i+j+k \\leqslant 3$. Cela nous assure déjà que $n \\leqslant 3$.\n\nRéciproquement, intéressons-nous à l'ensemble $\\mathcal{E}$ des polynômes olympiques dont tous les monômes sont de degré total $3$, c'est-à-dire sont des polynômes $x^{i} y^{j} z^{k}$ tels que $i+j+k=3$. Quitte à restreindre respectivement $A, B$ et $C$ à leurs monômes de degrés totaux $2,1$ et $0$, chaque polynôme de $\\mathcal{E}$ s'écrit sous la forme\n$$\n\\begin{aligned}\nP= & (x+y+z)\\left(a x^{2}+b y^{2}+c z^{2}+d x y+e x z+f y z\\right)+(x y+y z+z x)(g x+h y+i z)+j x y z \\\\\n= & a x^{3}+b y^{3}+c z^{3}+(a+d+g) x^{2} y+(a+e+g) x^{2} z+(b+d+h) y^{2} x+(b+f+h) y^{2} z \\\\\n& \\quad+(c+e+i) z^{2} x+(c+f+i) z^{2} y+(d+e+f+g+h+i+j) x y z .\n\\end{aligned}\n$$\nSi $x^{2} y$ est olympique, on a alors\n$$\n1=(a+d+g)+(b+f+h)+(c+e+i)-(a+e+g)-(b+d+h)-(c+f+i)=0,\n$$\nce qui est absurde. Ainsi, le polynôme $x^{2} y$ n'est pas olympique, et le plus grand entier recherché est $n=3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes suites $\\left(a_{n}\\right)_{n \\geqslant 0}$ et $\\left(b_{n}\\right)_{n \\geqslant 0}$ sont définies par\n$$\n\\left\\{\\begin{array}{ll}\na_{n}=0 & \\text{ si } n=0 ; \\\\\na_{n}=2 a_{\\lfloor n / 2 \\rfloor}+n & \\text{ si } n \\geqslant 1 ;\n\\end{array}\\right. \\quad \\text{et} \\quad \\left\\{\\begin{array}{ll}\nb_{n}=0 & \\text{ si } n=0 ; \\\\\nb_{n}=3 b_{\\lfloor n / 3\\rfloor}+n & \\text{ si } n \\geqslant 1 .\n\\end{array}\\right.\n$$\n\nDémontrer que la suite de terme général $2^{a_{n}}-3^{b_{n}}$ change de signe infiniment souvent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn montre tout d'abord, par récurrence sur $n$, que $a_{n} \\leqslant n\\left\\lfloor\\log _{2}(2 n)\\right\\rfloor$ pour tout entier $n \\geqslant 1$, avec égalité lorsque $n$ est une puissance de 2. En effet, pour $n=1$, on a bien $a_{n}=1=n \\log _{2}(2 n)$. Puis, si $n \\geqslant 2$,\n$$\na_{n}=2 a_{\\lfloor n / 2\\rfloor}+n \\leqslant 2\\lfloor n / 2\\rfloor\\left\\lfloor\\log _{2}(2\\lfloor n / 2\\rfloor)\\right\\rfloor+n \\leqslant n\\left\\lfloor\\log _{2}(n)+1\\right\\rfloor=n\\left\\lfloor\\log _{2}(2 n)\\right\\rfloor,\n$$\navec égalité lorsque $n / 2=\\lfloor n / 2\\rfloor$ et $n / 2$ est une puissance de 2.\n\nOn montre de même que $b_{n} \\leqslant n\\left\\lfloor\\log _{3}(3 n)\\right\\rfloor$ pour tout entier $n \\geqslant 1$, avec égalité lorsque $n$ est une puissance de 3. On en déduit déjà, si $n$ est une puissance de 3, que\n$$\n2^{a_{n}}-3^{b_{n}}=2^{a_{n}}-(3 n)^{n} \\leqslant(2 n)^{n}-(3 n)^{n}<0.\n$$\nRéciproquement, soit $k$ et $\\ell$ deux entiers naturels non nuls tels que $2^{\\ell}<3^{k} \\leqslant 2^{\\ell+1}$, et soit $n=2^{\\ell}$. Alors $k \\log _{2}(3) \\leqslant \\ell+1$, donc\n$$\n\\log _{2}(3) b_{n} \\leqslant \\log _{2}(3) n\\left\\lfloor\\log _{3}(3 n)\\right\\rfloor=\\log _{2}(3) k n \\leqslant(\\ell+1) n=a_{n}\n$$\nde sorte que $2^{a_{n}}-3^{b_{n}}=2^{a_{n}}-2^{\\log _{2}(3) b_{n}} \\geqslant 0$. Puisque $2^{a_{n}} \\neq 3^{b_{n}}$, car l'un est impair et l'autre non, on en déduit que $2^{a_{n}}-3^{b_{n}}>0$, ce qui conclut cette solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30747,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver le plus grand entier $n \\geqslant 3$ pour lequel il existe un ensemble $\\mathcal{S}$ de $n$ points du plan avec la propriété suivante : tout triangle (même plat) dont les sommets appartiennent à $\\mathcal{S}$ est isocèle mais pas équilatéral.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nOn dit qu'un ensemble $\\mathcal{S}$ ayant la propriété requise est mignon. Nous allons démontrer que tout ensemble mignon contient au plus 6 éléments. En outre, on notera $\\omega_{A, B}$ le cercle de centre $A$ passant par $B$, et par $\\ell_{A, B}$ la médiatrice du segment $[A B]$.\n\nTout d'abord, si $\\mathcal{S}$ est formé des cinq sommets $P_{1}, \\ldots, P_{5}$ d'un pentagone régulier ainsi que de son centre $O$, on vérifie aisément que $\\mathcal{S}$ est mignon.\n\nRéciproquement, soit $\\mathcal{S}$ un ensemble mignon à $n \\geqslant 6$ points : nous allons démontrer que $n=6$. Pour tous les points $X, Y$ et $Z$ de $\\mathcal{S}$, on sait que $Z$ appartient à un seul des trois ensembles $\\omega_{X, Y}, \\omega_{Y, X}$ et $\\ell_{X, Y}$. Cette remarque nous permet déjà de démontrer le résultat suivant.\n\nLemme 1. Trois points quelconques de $\\mathcal{S}$ ne sont jamais alignés.\n\nDémonstration : Supposons que $\\mathcal{S}$ contient trois points $A, B$ et $C$, alignés dans cet ordre. Puisque $A B C$ est isocèle, $B$ est le milieu de $[A C]$. Soit maintenant $D$ un autre point de $\\mathcal{S}$. Sans perte de généralité, on suppose que $A D \\leqslant C D$.\n\nComme $D$ ne peut appartenir ni au cercle $\\omega_{C, B}$ ni à la droite $\\ell_{B, C}$, il appartient nécessairement au cercle $\\omega_{B, C}=\\omega_{B, A}$. Mais alors $D$ ne peut appartenir ni au cercle $\\omega_{C, A}$, ni au cercle $\\omega_{A, C}$, et il appartient donc nécessairement à la droite $\\ell_{A, C}$. Cette situation se répète pour chacun des $n-3$ points de $\\mathcal{S} \\setminus \\{A, B, C\\}$, qui appartiennent donc à la fois à $\\omega_{B, C}$ et à $\\ell_{A, C}$.\n\nOr, une droite et un cercle ont au plus deux points d'intersection. Notre supposition initiale était donc invalide, ce qui démontre le lemme.\n\n![](attached_image_1.png)\n\nOn continue alors, et on s'intéresse aux paires de cercles $\\omega_{A, B}$ et $\\omega_{B, A}$.\n\nLemme 2. Soit $A$ et $B$ deux points de $\\mathcal{S}$. Si $\\mathcal{S}$ contient à la fois des points de $\\omega_{A, B}$ et de $\\omega_{B, A}$ autres que $A$ et $B$, alors $n \\leqslant 6$.\n\nDémonstration : Supposons que $\\mathcal{S}$ contient deux points $C$ et $D$, distincts de $A$ et de $B$, qui appartiennent respectivement aux cercles $\\omega_{B, A}$ et $\\omega_{A, B}$. Ils ne peuvent être égaux l'un à l'autre, ce sans quoi le triangle $A B C$ serait équilatéral. Nous allons d'abord démontrer que $(A B)$ est parallèle à $(C D)$.\n\nEn effet, si $A B=C D$, alors $A B=B C=C D=D A$, et le quadrilatère $A B C D$ est un losange, donc $(A B)$ est bien parallèle à $(C D)$. On suppose donc que $A B \\neq C D$, c'est-à-dire que $A D \\neq C D$. Puisque $A C \\neq A D$ et que $A C D$ est isocèle, on en déduit que $A C=C D$. On démontre de même que $B D=C D$, de sorte que $A C=B D$, et donc que les triangles $A B C$ et $D A B$ sont isométriques.\n\nSupposons ensuite que $C$ et $D$ se trouvent de part et d'autre de la droite $(A B)$. Si l'on note $\\theta$ l'angle $\\widehat{A B C}$, on peut alors vérifier, par exemple en utilisant des coordonnées cartésiennes, que $A C^{2}=B D^{2}=(2-2 \\cos (\\theta)) A B^{2}$ et que $C D^{2}=(5-4 \\cos (\\theta)) A B^{2}$. Comme $A C=C D$, on en conclut alors que $2 \\cos (\\theta)=3$, ce qui est absurde. Notre supposition est donc invalide, de sorte que $C$ et $D$ sont nécessairement du même côté de $(A B)$, et donc que $(A B)$ est bien parallèle à $(C D)$.\n\nUne fois cette relation de parallélisme acquise, on conclut que tout point de $\\mathcal{S}$ appartient à la droite $(A B)$, à la médiatrice $\\ell_{A, B}$, ou encore à une parallèle à $(A B)$ passant par $C$ ou par $D$, c'est-à-dire à $(C D)$. Le lemme 1 nous assure que nulle de ces droites ne contient plus de deux points de $\\mathcal{S}$, ce qui prouve que $n \\leqslant 6$ et démontre le lemme.\n\n![](attached_image_2.png)\n\nCas 1: $A B=B C=C D=D A$\n\n![](attached_image_3.png)\n\nCas 2: $A C=C D=D B$\n\nLemme 3. Soit $A$ et $B$ deux points de $\\mathcal{S}$. Si $n \\geqslant 7$, alors soit $\\omega_{A, B}$ contient au moins $n-2$ points de $\\mathcal{S}$, soit $B$ est le seul point de $\\mathcal{S}$ appartenant à $\\omega_{A, B}$.\n\nDémonstration: Supposons que $n \\geqslant 7$ et que le cercle $\\omega_{A, B}$ contient $k \\geqslant 2$ points de $\\mathcal{S}$. Les lemmes 1 et 2 nous assurent respectivement que $\\ell_{A, B}$ contient au plus deux points de $\\mathcal{S}$, et que $A$ est le seul point de $\\mathcal{S}$ appartenant à $\\omega_{B, A}$. On en déduit que $k \\geqslant n-3$, avec égalité si et seulement si $\\ell_{A, B}$ contient exactement deux points de $\\mathcal{S}$, dont aucun n'appartient à $\\omega_{A, B}$. Si $k=n-3$, soit $C$ un autre point de $\\omega_{A, B}$. Pour les mêmes raisons que précédemment, la droite $\\ell_{A, C}$ contient deux points de $\\mathcal{S}$, dont aucun n'appartient à $\\omega_{A, C}=\\omega_{A, B}$. Or, puisque $\\ell_{A, B}$ et $\\ell_{A, C}$ ont au plus un point commun, elles contiennent donc au moins, à elles deux, trois points de $\\mathcal{S}$. Si l'on ajoute le point $A$ et les $k$ points de $\\mathcal{S}$ appartenant à $\\omega_{A, B}$, cela nous fait un total de $k+4=n+1$ points dans $\\mathcal{S}$. Notre supposition initiale était donc invalide, ce qui démontre le lemme.\n\n![](attached_image_4.png)\n\nEn conclusion, revenons à l'énoncé lui-même. Nous allons supposer que $n \\geqslant 7$ et aboutir à une absurdité. Soit $A, B$ et $C$ trois points de $\\mathcal{S}$. La droite $\\ell_{A, B}$ contient au plus deux points de $\\mathcal{S}$, donc l'un des cercles $\\omega_{A, B}$ et $\\omega_{B, A}$ (disons $\\omega_{A, B}$ ) en contient au moins deux. Le lemme 3 nous assure alors que $\\omega_{A, B}$ contient au moins $n-2$ points de $\\mathcal{S}$.\n\nDe même, l'un des cercles $\\omega_{B, C}$ et $\\omega_{C, B}$ (disons $\\omega_{X, Y}$ ) contient au moins $n-2$ points de $\\mathcal{S}$. On dispose alors de deux cercles, de centres $A$ et $X$, qui contiennent chacun $n-2$ points de $\\mathcal{S}$. Ils ont donc au moins $n-4 \\geqslant 3$ points communs, ce qui est absurde puisque leurs centres sont distincts. On tient donc ici l'absurdité tant désirée, qui conclut notre solution.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nCas 1: $A B=B C=C D=D A$\n\n![](attached_image_3.png)\n\nCas 2: $A C=C D=D B$\n\n![](attached_image_4.png)\nSolution:\n\nSolution alternative $n^{\\circ} 1$\n\nRéutilisons les notations de la solution ci-dessus. On démontre comme précédemment qu'il existe un ensemble mignon à 6 éléments, et on procède désormais par l'absurde, en supposant que l'on dispose d'un ensemble $\\mathcal{S}$ mignon à 7 éléments. Comme dans la solution précédente, on démontre que 3 points de $\\mathcal{S}$ ne sont jamais alignés. On démontre ensuite le résultat suivant.\n\nLemme 4. Soit $A$ et $B$ deux points de $\\mathcal{S}$. Si $n \\geqslant 7$, alors $\\omega_{A, B}$ contient au plus 3 points de $\\mathcal{S}$.\n\nDémonstration : Supposons que $\\omega_{A, B}$ contient au moins quatre points $B, C, D$ et $E$, tels que $B C=C D$. Puisque $B$ et $D$ sont les deux seuls points d'intersection de $\\omega_{A, B}$ et $\\omega_{C, B}=\\omega_{C, D}$, et que $A$ est l'unique point d'intersection de $\\omega_{B, C}$ et $\\omega_{D, C}$, on sait que $E$ est, au choix :\n\n1. sur $\\omega_{A, B}$, $\\ell_{C, D}$ et $\\omega_{B, C}$;\n2. sur $\\omega_{A, B}$, $\\ell_{B, C}$ et $\\omega_{D, C}$;\n3. sur $\\omega_{A, B}$, $\\omega_{B, C}$ et $\\omega_{D, C}$.\n\nDans le troisième cas, on a donc $B C=C D=D E=E B$, de sorte que $B C D E$ est un losange dont les sommets sont cocycliques, c'est-à-dire un carré, et donc que $A, B$ et $D$ sont alignés. Ce cas est donc impossible. Par ailleurs, les premier et deuxième cas sont symétriques l'un de l'autre. On suppose donc que l'on est dans le deuxième cas.\n\nOn observe alors que $B, C, D$ et $E$ sont quatre sommets consécutifs d'un pentagone régulier de centre $A$, et dont on note $d$ la longueur des côtés. Parmi les deux autres points de $\\mathcal{S}$, l'un n'appartient pas à $\\ell_{C, D}$, qui contient déjà $A$ : on note $X$ ce point. Chacune des droites $\\ell_{B, C}$ et $\\ell_{D, E}$ contient déjà deux points de $\\mathcal{S}$ parmi $A, B, C, D$ et $E$, et ne contient donc pas $X$. On en déduit que\n$$\n\\left\\{\\begin{array}{l}\nB X=d \\text{ ou } C X=d \\\\\nC X=d \\text{ ou } D X=d ; \\\\\nD X=d \\text{ ou } E X=d\n\\end{array}\\right.\n$$\nPar conséquent, $X$ est sur $\\ell_{B, D}$ (si $B X=D X=d$ ) ou sur $\\ell_{C, E}$ (si $C X=E X=d$ ). Ces deux cas sont impossibles, puisque $\\ell_{B, D}$ contient déjà $A$ et $C$, tandis que $\\ell_{C, E}$ contient déjà $A$ et $D$. Notre supposition initiale était donc invalide, ce qui démontre le lemme.\n\n![](attached_image_5.png)\n\nCas 2: $E \\in \\ell_{B, C} \\cap \\omega_{D, C}$\n\n![](attached_image_6.png)\n\nCas 3: $E \\in \\omega_{B, C} \\cap \\omega_{D, C}$\n\nPour tout segment $[A, B]$ dont les extrémités sont dans $\\mathcal{S}$, on note maintenant $p_{A, B}$ le nombre de points de $\\mathcal{S}$ situés sur la médiatrice $\\ell_{A, B}$ : on dit qu'il s'agit du poids de $[A, B]$. Puisqu'il existe $\\binom{7}{3}=35$ triangles isocèles (mais pas équilatéraux) à sommets dans $\\mathcal{S}$, la somme des poids des segments vaut 35. Or, il y a $\\binom{7}{2}=21$ segments, dont aucun n'est de poids $p_{A, B} \\geqslant 3$. Si on note $a_{i}$ le nombre de segments de poids $i$, on constate donc que\n$$\n35=2 a_{2}+a_{1} \\leqslant\\left(a_{0}+a_{1}+a_{2}\\right)+a_{2}=21+a_{2},\n$$\nde sorte que $a_{2} \\geqslant 14$. Comme $2 a_{2}=28 \\geqslant 24,5=(1-1 / 2) \\times 7^{2}$, le théorème de Turan indique qu'il existe un triangle $A B C$, isocèle en $A$, dont les trois côtés $[A, B],[B, C]$ et $[C, A]$ sont de poids 2.\n\nParmi $A, B$ et $C$, seul $A$ appartient à l'une des trois médiatrices $\\ell_{A, B}, \\ell_{B, C}$ ou $\\ell_{C, A}$. En outre, ces trois médiatrices concourent en $O$, le centre du cercle circonscrit à $A B C$. Par conséquent, si $O \\notin \\mathcal{S}$, on sait que $\\mathcal{S}$ contient 8 points distincts, que sont $B, C$, et deux points sur chacune des médiatrices $\\ell_{A, B}, \\ell_{B, C}$ et $\\ell_{C, A}$. On en déduit que $O \\in \\mathcal{S}$.\n\nSoit alors $D$ le second point de $\\mathcal{S}$ situé sur $\\ell_{A, B}$. Puisque $D \\notin \\omega_{O, B}$, on sait que $D \\in \\omega_{B, O}$, auquel cas $D$ coïncide avec le symétrique de $O$ par rapport à $(A B)$, ou bien que $D \\in \\ell_{B, O}$, auquel cas $D$ coïncide avec le centre du cercle circonscrit à $A B O$. Dans les deux cas, remarquons que $D \\notin \\omega_{A, B}=\\omega_{A, C}$ et que $D \\notin \\ell_{A, C}$, de sorte que $D \\in \\omega_{C, A}$.\n\nDans le premier cas, on sait en outre que $B D=B O \\neq B C$, donc que $D \\notin \\omega_{B, C}$, et que $D \\notin \\ell_{B, C}$, de sorte que $D \\in \\omega_{C, B}$. Cela signifie que $C D=C B \\neq C O$, donc que $D \\notin \\omega_{C, O}$. Puisque $D \\notin \\omega_{O, C}$, on en déduit que $D \\in \\ell_{C, O}$. Ainsi, et puisque $D \\in \\omega_{C, A}$, on sait que $O D=C D=C A=B A$, donc que $A D B O$ est un losange dont les diagonales sont de même longueur, c'est-à-dire un carré. Mais alors $\\widehat{A O B}$ est droit, et $\\widehat{A O C}$ est droit aussi, donc $B, O$ et $C$ sont alignés, ce qui contrevient au lemme 1. Ce cas est donc impossible.\n\nOn est donc dans le deuxième cas et, de même, $\\mathcal{S}$ contient le centre du cercle circonscrit à $A C O$, que l'on note $E$. Puisque $C D=C A \\neq C O$ et que $D \\notin \\omega_{O, C}$, on sait que $D \\in \\ell_{C, O}$. Ainsi, $D$ appartient aux deux droites $\\ell_{B, O}$ et $\\ell_{C, O}$ et, pour des raisons de symétrie, $E$ appartient également à ces deux droites. Puisque ces deux droites ne sauraient être confondues, ce second cas est lui aussi impossible, ce qui conclut.\n\n![](attached_image_7.png)\n\nCas 1: $D \\in \\ell_{A, B} \\cap \\omega_{B, O}$\n\n![](attached_image_8.png)\n\nCas $2: D \\in \\ell_{A, B} \\cap \\ell_{B, O}$\nSolution:\n\nSolution alternative $n^{\\circ} 2$\n\nRéutilisons les notations de la solution ci-dessus. On démontre comme précédemment qu'il existe un ensemble mignon à 6 éléments, et on procède désormais par l'absurde, en supposant que l'on dispose d'un ensemble $\\mathcal{S}$ mignon à 7 éléments. Comme dans la solution précédente, on démontre que 3 points de $\\mathcal{S}$ ne sont jamais alignés.\n\nConsidérons maintenant un triangle $A B C$, isocèle en $A$, dont les sommets appartiennent à $\\mathcal{S}$, puis soit $D$ un autre point de $\\mathcal{S}$. Puisque les triangles $A B D, B C D$ et $C A D$ sont isocèles mais pas équilatéraux, on sait que $D$ appartient à\n$\\triangleright$ un seul des trois ensembles $\\omega_{A, B}, \\omega_{B, A}$ et $\\ell_{A, B}$;\n$\\triangleright$ un seul des trois ensembles $\\omega_{A, C}, \\omega_{C, A}$ et $\\ell_{A, C}$;\n$\\triangleright$ un seul des trois ensembles $\\omega_{B, C}, \\omega_{C, B}$ et $\\ell_{B, C}$.\n\nOr, les cercles et droites $\\omega_{A, B}, \\omega_{B, A}, \\omega_{B, C}, \\omega_{C, B}, \\ell_{A, B}$ et $\\ell_{B, C}$ sont deux à deux distincts et, pris deux à deux, ont au plus deux points d'intersection. On peut donc considérer un à un tous les points $D$ possibles.\n\n1. Si $D$ appartient à $\\omega_{A, B}=\\omega_{A, C}$ et à $\\omega_{B, C}$, c'est le symétrique $C^{\\bullet}$ de $C$ par rapport à $(A B)$.\n2. Si $D$ appartient à $\\omega_{A, B}=\\omega_{A, C}$ et à $\\omega_{C, B}$, c'est le symétrique $B^{\\bullet}$ de $B$ par rapport à $(A C)$.\n3. Si $D$ appartient à $\\omega_{A, B}=\\omega_{A, C}$ et à $\\ell_{B, C}$, c'est le milieu d'un des deux arcs de cercles $\\overline{B C}$ sur $\\omega_{A, B}$, que l'on note $M$ et $M^{\\prime}$.\n4. Si $D$ appartient à au moins deux médiatrices, c'est le centre $O$ du cercle circonscrit à $A B C$.\n5. Si $D$ appartient à $\\omega_{B, A}$ et à $\\omega_{C, A}$, c'est le symétrique $A^{\\bullet}$ de $A$ par rapport à $(B C)$.\n6. Si $D$ appartient à $\\omega_{B, A}$ à $\\ell_{A, C}$, et à aucune autre médiatrice, il ne peut appartenir à $\\omega_{B, C}$, donc il appartient à $\\omega_{C, B}$. On a alors $B D=B A=C A$ et $A D=C D=B C$. Ainsi, les triangles $A B C$ et $A B D$ sont isométriques, de même que les triangles $A C D$ et $B C D$. En posant $x=\\widehat{B A C}$ et $y=\\widehat{C A D}$, on calcule aisément chacun des angles de la figure, de sorte que $3 x+2 y=x+4 y=180^{\\circ}$ (donc $x=y=36^{\\circ}$ ) si $A B C D$ est croisé. Si $A B C D$ est non croisé, on récupère une figure totalement analogue, à l'ordre près des points, de sorte que $x=108^{\\circ}$. Dans ces circonstances, on note $B^{\\boldsymbol{k}}$ le point de concours de $\\omega_{B, A}$, $\\omega_{C, B}$ et $\\ell_{A, C}$.\n7. Si $D$ appartient à $\\omega_{C, A}$, à $\\ell_{A, B}$, et à aucune autre médiatrice, la situation est totalement analogue, et $x=36^{\\circ}$ ou $108^{\\circ}$. Il coïncide alors avec le point de concours de $\\omega_{C, A}, \\omega_{B, C}$ et $\\ell_{A, B}$, que l'on note $C^{\\boldsymbol{\\alpha}}$.\n\n![](attached_image_9.png)\n\nCas 6: $A B C D$ croisé\n\n![](attached_image_10.png)\n\nCas 6 : $A B C D$ non croisé\n\nPuisque $A$ appartient déjà à $\\ell_{B, C}$, le lemme 1 nous assure que $\\mathcal{S}$ contient au plus un des points $A^{\\bullet}, O, M$ et $M^{\\prime}$. L'ensemble $\\mathcal{S}$ contient donc au moins trois des quatre points $B^{\\bullet}, C^{\\bullet}$, $B^{\\boldsymbol{\\omega}}$ et $C^{\\boldsymbol{\\omega}}$, ce qui signifie que $x=36^{\\circ}$ ou $108^{\\circ}$. En répétant ce raisonnement pour chaque triangle, on en conclut que tout triangle a pour angles $36^{\\circ}$ et $108^{\\circ}$, ou bien $36^{\\circ}$ et $72^{\\circ}$.\n\nMais alors, si un segment $[A B]$ est situé sur l'enveloppe convexe de $\\mathcal{S}$, les cinq autres points $P_{1}, P_{2}, \\ldots, P_{5}$ de $\\mathcal{S}$ sont situés dans un même demi-plan délimité par $(A B)$, et tout angle $\\widehat{P_{i} A B}$ vaut $36^{\\circ}, 72^{\\circ}$ ou $108^{\\circ}$. Ainsi, deux de ces angles sont égaux, donc on a trois points alignés, ce qui contrevient au lemme 1 et conclut.\n\n![](attached_image_9.png)\n\nCas 6: $A B C D$ croisé\n\n![](attached_image_10.png)\n\nCas 6 : $A B C D$ non croisé\nSolution:\n\nSolution alternative $n^{\\circ} 3$\n\nVoici une autre manière de démontrer que tout ensemble mignon compte au plus 6 éléments. Comme ci-dessus, on procède par l'absurde, et l'on suppose que l'on dispose d'un ensemble $\\mathcal{S}$ mignon à 7 éléments. Une fois de plus, on démontre que 3 points de $\\mathcal{S}$ ne sont jamais alignés.\n\nLe score d'un point $A$ de $\\mathcal{S}$ est le nombre de segments $[B C]$ à extrémités dans $\\mathcal{S}$ tels que $A B C$ soit isocèle en $A$. Puisqu'il existe $\\binom{7}{3}=35$ triangles isocèles (mais pas équilatéraux) à sommets dans $\\mathcal{S}$, la somme des scores des sommets vaut 35. Comme $\\mathcal{S}$ compte 7 sommets, l'un d'entre eux, disons $O$, est de score $s \\geqslant 5$.\n\nOr, le score de $O$ ne peut pas être dû à $s$ segments d'extrémités deux à deux distinctes. Il existe donc trois points $A, B$ et $C$ dans $\\mathcal{S}$ tels que $A O=B O=C O$. Sans perte de généralité, on suppose même $A B=A C \\neq B C$. On pose alors $A O=x, A B=y$ et $B C=z$, et l'on sait que $x, y$ et $z$ sont deux à deux distincts.\n\nSupposons maintenant que $\\mathcal{S}$ contient un point $D \\neq O$ qui n'est pas sur $\\omega_{O, A}$. La médiatrice $\\ell_{B, C}$ contient déjà $A$ et $O$, donc ne contient pas $D$. Sans perte de généralité, on suppose donc que $D \\in \\omega_{B, C}$. Puisque $O B D$ est isocèle et que $O B=x \\neq O D$ et $B D=B C=z$, on a donc $O D=z$. Par ailleurs, $B C D$ est isocèle en $C$ mais pas équilatéral, donc $C D \\neq z$. Puisque $O C D$ est isocèle, que $O D=z \\neq C D$ et que $O C=x$, on a donc $C D=x$. Dès lors :\n\n$\\triangleright$ $A B D$ est isocèle, $A B=y$ et $B D=x$, donc $A D$ vaut $x$ ou $y$;\n$\\triangleright$ $A O D$ est isocèle, $A O=x$ et $O D=z$, donc $A D$ vaut $x$ ou $z$;\n$\\triangleright$ $A C D$ est isocèle, $A C=y$ et $C D=z$, donc $A D$ vaut $y$ ou $z$.\n\nOn dispose ainsi d'une contradiction.\n\nAinsi, à l'exception de $O$, tout point de $\\mathcal{S}$ se trouve sur $\\omega_{O, A}$, mais également sur $\\omega_{A, B}, \\omega_{B, A}$ ou $\\ell_{A, B}$. Or, les cercles $\\omega_{A, B}$ et $\\omega_{B, A}$ ont chacun deux points d'intersection avec $\\omega_{O, A}$. Quant à la médiatrice $\\ell_{A, B}$, elle contient déjà $O$, et contient donc au plus un point de $\\omega_{O, A}$. Ainsi, $\\mathcal{S}$ est de taille au plus six, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30748,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle isocèle en $A$, puis $D$ un point du segment $[BC]$ tel que $BD \\neq CD$. Soit $P$ et $Q$ les projetés orthogonaux de $D$ sur $(AB)$ et $(AC)$. Enfin, soit $E$ le point d'intersection, autre que $A$, entre les cercles circonscrits à $ABC$ et $APQ$.\n\nDémontrer que, si les droites $(EP)$, $(AC)$ et la médiatrice de $[PQ]$ sont concourantes, le triangle $ABC$ est rectangle en $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUn premier réflexe est de dessiner une figure où le triangle $ABC$ est isocèle rectangle, mais sans indiquer qu'il l'est. De nombreuses symétries apparaissent alors. Notons $X$ le point d'intersection des droites $(AC)$ et $(EP)$, $\\ell$ la médiatrice de $[PQ]$, $O$ le centre du cercle circonscrit à $ABC$ et $M$ le milieu de $[BC]$ : il s'agit de démontrer que $M = O$.\n\nComme $PQX$ est isocèle en $X$, la symétrie d'axe $\\ell$ échange les droites $(AQ)$ et $(PE)$, de sorte que $\\widehat{AQP} = \\widehat{QPE}$. Puisque $A, E, P$ et $Q$ sont cocycliques, le quadrilatère $AEPQ$ est donc un trapèze isocèle, ce qui se vérifie aisément puisque\n$$\n(AE, PQ) = (AE, AQ) + (AQ, PQ) = (PE, PQ) + (QA, QP) = 0^{\\circ}\n$$\nLa symétrie d'axe $\\ell$ échange donc les points $A$ et $E$, et $\\ell$ est en fait la médiatrice de $[AE]$. Elle contient donc, en particulier, les centres $O$ et $O'$ des cercles circonscrits à $ABC$ et $AEPQ$.\n\nAu vu des angles droits en $P$, $Q$ et $M$, ce dernier cercle est en fait le cercle de diamètre $[AD]$, et il contient $D$ et $M$. En outre, $(AM)$ est la bissectrice de $\\widehat{BAC}$, donc $M$ est le pôle Sud de $A$ dans le triangle $APQ$, et $PQM$ est isocèle en $M$. Ainsi, $M$ appartient à $\\ell$.\n\nEn conclusion, $M$ et $O$ appartiennent tous deux à $(AM)$ et à $\\ell$. Puisque $D \\neq M$, on sait aussi que $O'$, le milieu de $[AD]$, n'est pas sur $(AM)$. Puisque $\\ell$ contient $O'$, elle n'est donc pas confondue avec $(AM)$, et $M$ coïncide donc avec $O$, ce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30749,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier $k \\geqslant 0$, on note $F_{k}$ le $k^{\\text{ème}}$ nombre de Fibonacci, défini par $F_{0}=0$, $F_{1}=1$, et $F_{k}=F_{k-2}+F_{k-1}$ lorsque $k \\geqslant 2$. Soit $n \\geqslant 2$ un entier, et soit $S$ un ensemble d'entiers ayant la propriété suivante :\nPour tout entier $k$ tel que $2 \\leqslant k \\leqslant n$, l'ensemble $S$ contient deux entiers $x$ et $y$ tels que $x-y=F_{k}$.\nQuel est le plus petit nombre possible d'éléments d'un tel ensemble $S$ ?",
    "options": [],
    "answer": "ceil(n/2)+1",
    "solution": "Solution:\n\nTout d'abord, soit $m=\\lceil n / 2\\rceil$, et soit $S$ l'ensemble $\\{F_{2 \\ell}: 0 \\leqslant \\ell \\leqslant m\\}$. Pour tout entier $k$ tel que $2 \\leqslant k \\leqslant n$, on choisit $x=F_{k}$ et $y=F_{0}=0$ si $k$ est pair, ou bien $x=F_{k+1}$ et $y=F_{k-1}$ si $k$ est impair. Dans les deux cas, $x$ et $y$ sont deux éléments de $S$ tels que $x-y=F_{k}$.\n\nRéciproquement, soit $S$ un ensemble tel que décrit dans l'énoncé. Nous allons démontrer que $|S| \\geqslant m+1$. Pour ce faire, on construit un graphe pondéré $G$, non orienté, en procédant comme suit. Les sommets de notre graphe sont les éléments de l'ensemble $S$. Puis, pour tout entier $k$ impair tel que $1 \\leqslant k \\leqslant n$, et en se rappelant que $F_{1}=F_{2}$, on choisit deux éléments $x$ et $y$ de $S$ tels que $x-y=F_{k}$. On insère alors dans notre graphe $G$ l'arête $\\{x, y\\}$, à qui l'on affecte le poids $F_{k}$.\n\nLe graphe ainsi obtenu compte $m$ arêtes. En outre, supposons qu'il contienne un cycle $c=x_{0} x_{1} \\ldots x_{\\ell}$, avec $x_{0}=x_{\\ell}$, que l'on choisit de longueur $\\ell$ minimale. Sans perte de généralité, on suppose que $x_{0}>x_{1}$, et que $\\{x_{0}, x_{1}\\}$ est l'arête de $c$ de poids maximal. On note $F_{2k+1}$ ce poids. Le poids cumulé des autres arêtes ne dépasse pas $F_{1}+F_{3}+\\ldots+F_{2k-1}$ et, par construction, il est égal à\n$$\n\\left|x_{1}-x_{2}\\right|+\\left|x_{2}-x_{3}\\right|+\\ldots+\\left|x_{\\ell-1}-x_{\\ell}\\right| \\geqslant \\left|x_{\\ell}-x_{1}\\right|=F_{2k+1}\n$$\nCependant, une récurrence immédiate sur $k$ montre que $F_{1}+F_{3}+\\ldots+F_{2k-1}=F_{2k}<F_{2k+1}$. Par conséquent, notre graphe est acyclique, et puisqu'il compte $m$ arêtes, il compte au moins $m+1$ sommets.\n\nLe cardinal minimal recherché est donc égal à $m+1$.\n\n\nSolution alternative\n\nOn dit qu'un ensemble tel que décrit dans l'énoncé est $n$-mignon. Soit $S_{n}$ un ensemble $n$-mignon de cardinal aussi petit que possible. Nous allons démontrer, par récurrence sur $n$, que $|S_{n}| \\geqslant \\lceil n / 2\\rceil+1$. Tout d'abord, pour $n=2$, le résultat désiré est immédiat. On suppose désormais que $n \\geqslant 3$.\n\nCette fois-ci, on introduit le graphe pondéré $G$ dont les sommets sont les éléments de $S_{n}$ et qui contient, pour tout entier $k$ tel que $2 \\leqslant k \\leqslant n-2$, une arête $(x, y)$ de poids $F_{k}$ entre deux sommets $x$ et $y$ tels que $|x-y|=F_{k}$. Enfin, soit $u$ et $v$ deux éléments de $S_{n}$ tels que $|u-v|=F_{n}$.\n\nUne récurrence immédiate sur $k$ montre que $F_{2}+F_{3}+\\ldots+F_{k-2}=F_{k}-3$ pour tout entier $k \\geqslant 4$. Par conséquent, deux sommets de $S_{n}$ appartenant à une même composante connexe de $G$ sont distants d'au plus $F_{n}-3$, et $u$ et $v$ ne peuvent pas appartenir à la même composante connexe de $G$.\n\nSoit $U_{n}$ l'ensemble des éléments de $S_{n}$ qui appartiennent à la même composante connexe de $G$ que $u$, et soit $V_{n}$ les autres éléments de $S_{n}$. Nulle arête de $G$ n'a une extrémité dans $U_{n}$ et une autre dans $V_{n}$. Par conséquent, si l'on remplace chaque sommet $x$ de $U_{n}$ par $x+v-u$, on obtient un nouvel ensemble $S_{n}'=\\{x+v-u: x \\in U_{n}\\} \\cup V_{n}$ de cardinal $|S_{n}'| \\leqslant |S_{n}|-1$ et qui est $(n-2)$-mignon. On en conclut comme prévu que\n$$\n|S_{n}| \\geqslant |S_{n}'|+1 \\geqslant \\lceil(n-2) / 2\\rceil+1+2=\\lceil n / 2\\rceil+1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30750,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn note $\\mathbb{R}_{>0}$ l'ensemble des réels strictement positifs. Trouver toutes les fonctions $f: \\mathbb{R}_{>0} \\mapsto \\mathbb{R}_{>0}$ telles que\n$$\nf(x+f(x y))+y=f(x) f(y)+1\n$$\npour tous les réels strictement positifs $x$ et $y$.",
    "options": [],
    "answer": "f(x) = x + 1",
    "solution": "Solution:\n\nSoit $f$ une solution éventuelle du problème. Dans la suite, on notera $\\mathbf{E}_{x, y}$ l'égalité de l'énoncé. Soit $a$ et $b$ deux réels strictement positifs tels que $f(a)=f(b)$. Les égalités $\\mathbf{E}_{1, a}$ et $\\mathbf{E}_{1, b}$ indiquent que\n$$\na=f(1) f(a)+1-f(1+f(a))=f(1) f(b)+1-f(1+f(b))=b\n$$\nce qui signifie que $f$ est injective.\n\nSoit ensuite $c$ et $d$ deux réels tels que $0<c<d$, puis soit $t$ un réel strictement positif. Puisque $f$ est injective, les égalités $\\mathbf{E}_{c / t, t}$ et $\\mathbf{E}_{d / t, t}$ indiquent que\n$$\nf(c / t+f(c))=f(c / t) f(t)+1-t \\neq f(d / t) f(t)+1-t=f(d / t+f(d))\n$$\nde sorte que $c / t+f(c) \\neq d / t+f(d)$, ou encore que $f(d)-f(c) \\neq(c-d) / t$. Cette dernière inégalité étant valide pour tout $t>0$, on en déduit que $f(d) \\geqslant f(c)$, ce qui signifie que $f$ est croissante.\n\nOn pose alors $p=\\inf \\{f(x): x>0\\}$ et $q=\\inf \\{f(x): x>p\\}$. Puisque $f$ est croissante, on sait que $f(x) \\rightarrow p$ lorsque $x \\rightarrow 0$ et que $x>0$, et que $f(y) \\rightarrow q$ lorsque $y \\rightarrow p$ et que $y>p$. Un réel $y>0$ étant fixé, si on fait tendre $x$ vers 0 , on en déduit que $f(x) \\rightarrow p$ et que $f(x y) \\rightarrow p$, de sorte que $x+f(x y) \\rightarrow p$ et que $f(x+f(x y)) \\rightarrow q$. Par conséquent, l'égalité $\\mathbf{E}_{x, y}$ indique que\n$$\ny-1=f(x) f(y)-f(x+f(x y)) \\rightarrow p f(y)-q,\n$$\nce qui signifie que $f(y)=y+(q-1) / p$.\n\nLa fonction $f$ est donc une fonction affine, de la forme $f: t \\mapsto u t+v$. Réciproquement, une fois les coefficients $u$ et $v$ fixés, l'égalité $\\mathbf{E}_{x, y}$ indique que\n$$\nu^{2} x y+u x+u v+v+y=f(x+f(x y))+y=f(x) f(y)+1=u^{2} x y+u v x+u v y+v^{2}+1\n$$\nou encore que $u(v-1) x+(u v-1) y=u v+v-v^{2}-1$. Cela signifie que\n$$\nu v-u=u v-1=u v+v-v^{2}-1=0\n$$\nOn en déduit que $u-1=(u v-1)-(u v-u)=0$, c'est-à-dire que $u=1$, puis que $v-1=$ $u v-u=0$, c'est-à-dire que $v=1$. Mais alors la triple égalité\n$$\nu v-u=u v-1=u v+v-v^{2}-1=0\n$$\nest bien respectée.\n\nEn conclusion, la fonction $f: t \\mapsto t+1$ est l'unique solution du problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMorgane a écrit les trois entiers $3$, $4$ et $12$ au tableau. Elle effectue ensuite des changements successifs en procédant comme suit : elle choisit deux nombres $a$ et $b$ écrits au tableau, les efface, et les remplace par $(3a+4b)/5$ et $(4a-3b)/5$.\n\nMorgane peut-elle, après un nombre fini de tels changements,\na) parvenir à écrire le nombre $13$ au tableau ?\nb) parvenir à écrire le nombre $-12$ au tableau ?",
    "options": [],
    "answer": "a) no; b) no",
    "solution": "Solution:\n\nNous allons démontrer que la réponse est négative dans les deux cas. Soit $\\mathcal{S}$ la somme des carrés des nombres écrits au tableau, et soit $f$ et $g$ les fonctions définies par $f:(x, y) \\rightarrow (3x+4y)/5$ et $g:(x, y) \\rightarrow (4x-3y)/5$. La relation\n$$\nf(x, y)^2 + g(x, y)^2 = x^2 + y^2\n$$\nnous assure que, lorsque Morgane remplace deux nombres $a$ et $b$ par $f(a, b)$ et $g(a, b)$, elle ne change pas la valeur de $\\mathcal{S}$.\n\nEn outre, les égalités $f(f(x, y), g(x, y)) = x$ et $g(f(x, y), g(x, y)) = y$ nous indiquent que les changements qu'effectue Morgane sont réversibles : si elle transforme un triplet non ordonné $\\mathbf{t} = (a, b, c)$ en un autre triplet $\\mathbf{t}' = (a', b', c)$ tel que $a' = f(a, b)$ et $b' = g(a, b)$, elle peut retransformer $\\mathbf{t}'$ en $\\mathbf{t}$.\n\nOn traite maintenant les deux questions séparément.\n\na.\nSi jamais elle réussit à écrire trois nombres $x, y$ et $z$ tels que $z = 13$, alors\n$$\nx^2 + y^2 + 13^2 = x^2 + y^2 + z^2 = \\mathcal{S} = 3^2 + 4^2 + 12^2 = 13^2\n$$\nde sorte que $x = y = 0$. Cela signifie que Morgane peut transformer, en plusieurs étapes, le triplet $(3, 4, 12)$ en $(0, 0, 13)$. Elle peut donc également transformer $(0, 0, 13)$ en $(3, 4, 12)$.\n\nOn dit alors qu'un triplet $(x, y, z)$ de nombres rationnels est sympathique si $v_{13}(u) \\geqslant 1$ pour tout $u \\in \\{x, y, z\\}$. Si Morgane transforme un triplet $(x, y, z)$ sympathique en un nouveau triplet $(x', y', z')$, ce dernier est clairement sympathique lui aussi. Ainsi, à partir du triplet sympathique $(0, 0, 13)$, Morgane ne peut obtenir que des triplets sympathiques. Elle ne peut donc pas obtenir le triplet $(3, 4, 12)$. Ceci démontre bien qu'elle ne pouvait pas transformer le triplet $(3, 4, 12)$ en $(0, 0, 13)$.\n\nb.\nNous allons carrément démontrer que Morgane ne peut pas écrire d'entier autre que les éléments de l'ensemble $\\Omega = \\{0, \\pm 3, \\pm 4, \\pm 5, 12\\}$.\n\nConsidérons un triplet ordonné $(x_1, x_2, x_3)$. Pour tous les entiers $i$ et $j$ distincts l'un de l'autre, on note $\\Theta_{i, j}(x_1, x_2, x_3)$ le triplet obtenu à partir de $(x_1, x_2, x_3)$ en remplaçant les nombres $x_i$ et $x_j$ par $f(x_i, x_j)$ et $g(x_i, x_j)$. Par ailleurs, on dit que $(x_1, x_2, x_3)$ est d'ordre $n$ si $\\min\\{v_5(x_1), v_5(x_2), v_5(x_3)\\} = -n$, et qu'il s'agit d'un triplet égalitaire s'il y a au moins deux indices $i$ tels que $v_5(x_i) = -n$.\n\nTout triplet d'ordre $n \\geqslant 1$ qu'obtiendra Morgane est égalitaire. En effet, supposons que cela ne soit pas le cas. Sans perte de généralité, on a $v_5(x_1) = -n$, $v_5(x_2) \\geqslant 1-n$ et $v_5(x_3) \\geqslant 1-n$. Alors, en posant $X_i = 5^n x_i$, on constate que $X_1 \\not\\equiv 0 \\pmod{5}$ et que $X_2 \\equiv X_3 \\equiv 0 \\pmod{5}$, de sorte que\n$$\n0 \\equiv 5^{2n} \\times 13^2 \\equiv 5^{2n}(x_1^2 + x_2^2 + x_3^2) \\equiv X_1^2 + X_2^2 + X_3^2 \\equiv X_1^2 \\not\\equiv 0 \\pmod{5},\n$$\nce qui est absurde.\n\nSoit alors $n \\geqslant 0$ un entier, puis $(x_1, x_2, x_3)$ et $(y_1, y_2, y_3)$ deux triplets d'ordres $\\ell \\leqslant n$ et $n+1$ que Morgane peut obtenir successivement. Il existe deux entiers $i$ et $j$ tels que $\\Theta_{i, j}(x_1, x_2, x_3) = (y_1, y_2, y_3)$, et le triplet $(y_1, y_2, y_3)$ est égalitaire. En notant $k$ l'élément de $\\{1, 2, 3\\}$ autre que $i$ et $j$, puis $X_u = 5^n x_u$ et $Y_u = 5^{n+1} y_u$ pour tout $u \\in \\{1, 2, 3\\}$, on constate que\n$$\nY_i = 3 X_i + 4 X_j \\not\\equiv 0 \\pmod{5}, \\quad Y_j = 4 X_i - 3 X_j \\not\\equiv 0 \\pmod{5} \\text{ et } Y_k \\equiv 0 \\pmod{5}.\n$$\nOn en déduit immédiatement que les triplets $\\Theta_{i, k}(y_1, y_2, y_3)$, $\\Theta_{j, k}(y_1, y_2, y_3)$, $\\Theta_{k, i}(y_1, y_2, y_3)$ et $\\Theta_{k, j}(y_1, y_2, y_3)$ sont tous d'ordre $n+2$. De même,\n$$\nf(y_j, y_i) = (24 X_i + 7 X_j)/5^{n+2} \\text{ et } g(y_j, y_i) = (7 X_i - 24 X_j)/5^{n+2}\n$$\nPuisque $24 X_i + 7 X_j \\equiv Y_j \\pmod{5}$ et $7 X_i - 24 X_j \\equiv -Y_i \\pmod{5}$, le triplet $\\Theta_{j, i}(y_1, y_2, y_3)$ est lui aussi d'ordre $n+2$. Enfin, comme dans la solution $n^\\circ 1$, on vérifie aisément que $\\Theta_{i, j}$ est une involution, de sorte que $\\Theta_{i, j}(y_1, y_2, y_3)$ est d'ordre $\\ell \\leqslant n$.\n\nPour finir, soit $\\mathbf{T}$ un triplet, avec au moins une coordonnée $x$ entière, que Morgane peut écrire. Supposons que Morgane a écrit successivement des triplets $\\mathbf{T}_0, \\ldots, \\mathbf{T}_k$ tels que $\\mathbf{T}_0 = (3, 4, 12)$ et $\\mathbf{T}_k = \\mathbf{T}$ soit à coordonnées entières, avec $k$ minimal.\n\nSoit $\\mathbf{T}_i$ un triplet d'ordre $n$ maximal parmi les triplets $\\mathbf{T}_0, \\ldots, \\mathbf{T}_k$. Si $n \\geqslant 1$ et si $i \\leqslant k-1$, on vient de démontrer que $\\mathbf{T}_{i-1} = \\mathbf{T}_{i+1}$, ce qui vient contredire la minimalité de $k$. Si $n \\geqslant 1$ et si $i = k$, alors $\\mathbf{T}_k$ contient une coordonnée déjà présente dans $\\mathbf{T}_{k-1}$ et deux coordonnées de valuation 5-adique égale à $-n$, donc distinctes de $x$. Ceci vient là encore contredire la minimalité de $k$.\n\nIl suffit maintenant de démontrer que, partant d'un triplet $\\mathbf{t} = (x_1, x_2, x_3)$ d'éléments de $\\Omega$ tel que $x_1^2 + x_2^2 + x_3^2 = 13^2$, tout triplet $\\mathbf{t}' = (y_1, y_2, y_3)$ à coordonnées entières que Morgane peut former est à coordonnées dans $\\Omega$. Tout d'abord, on peut supposer sans perte de généralité que $|x_1| \\leqslant |x_2| \\leqslant |x_3|$, de sorte que $3 \\times 5^2 < 13^2 \\leqslant 3 x_3^2$, donc que $x_3 = 12$ et que $(x_1, x_2) \\in \\{(0, \\pm 5), (\\pm 3, \\pm 4)\\}$.\n\nOn suppose alors que $\\mathbf{t}'$ a une coordonnée entière qui n'appartient pas à $\\Omega$. Soit $i$ et $j$ les entiers tels que $\\mathbf{t}' = \\Theta_{i, j}(\\mathbf{t})$. Si $\\{x_i, x_j\\} \\subseteq \\{0, \\pm 3, \\pm 4, \\pm 5\\}$, alors $y_i^2 + y_j^2 = x_i^2 + x_j^2 = 5^2$, donc $\\{y_i, y_j\\} \\subseteq \\Omega$. Ainsi, l'un des deux entiers $x_i$ et $x_j$ est égal à $12$. Or,\n$$\n0 \\equiv 5 y_i \\equiv 3 x_i + 4 x_j \\equiv 3(x_i - 2 x_j) \\pmod{5}\n$$\ndonc $x_i \\equiv 2 x_j \\pmod{5}$.\n\nSi $x_i = 12$, alors $x_j \\equiv 1 \\pmod{5}$, donc $x_j = -4$, et $(y_i, y_j) = (4, 12)$; si $x_j = 12$, alors $x_i \\equiv -1 \\pmod{5}$, donc $x_j = 4$, et $(y_i, y_j) = (12, -4)$. Dans les deux cas, on a bien $\\{y_i, y_j\\} \\subseteq \\Omega$. En conclusion, tout nombre entier que peut écrire Morgane appartient nécessairement à $\\Omega$. Elle ne peut donc pas écrire le nombre $-12$ au tableau.\nSolution:\n\nOn peut présenter la solution à la partie a) d'une autre manière. Tout d'abord, tout nombre qu'écrira Morgane appartiendra nécessairement à l'ensemble $\\mathbb{L} = \\{d / 5^k : d, k \\in \\mathbb{Z}\\}$. Puisque $5$ et $13$ sont premiers entre eux, on identifie chaque élément de $\\mathbb{L}$ à son résidu modulo $13$.\n\nMais alors, si Morgane parvient à remplacer deux nombres $a$ et $b$ par des nombres $a' = f(a, b)$ et $b' = g(a, b)$ tels que $a' \\equiv b' \\equiv 0 \\pmod{13}$, cela signifie que $3a + 4b \\equiv 4a - 3b \\equiv 0 \\pmod{13}$, et donc que\n$$\n\\begin{aligned}\na &\\equiv -25a \\\\\nb &\\equiv -3(3a + 4b) - 4(4a - 3b) \\\\\nb &\\equiv 0 \\pmod{13} ; \\\\\n&\\equiv -4(3a + 4b) + 3(4a - 3b) \\equiv 0 \\pmod{13}\n\\end{aligned}\n$$\nPar conséquent, une récurrence immédiate indique que les nombres que Morgane écrira au tableau ne seront jamais tous congrus à $0$ modulo $13$.\nSolution:\n\nOn peut également répondre à la partie a) avec une version simplifiée du raisonnement utilisé pour la partie b).\n\nEn effet, il s'agit de démontrer que l'on ne peut pas atteindre le triplet $\\mathbf{T} = (0, 0, 13)$, à l'ordre près des coordonnées. Si Morgane atteint ce triplet, et en raisonnant comme ci-dessus, elle a nécessairement dû écrire des triplets $\\mathbf{T}_0 = (3, 4, 12), \\mathbf{T}_1, \\ldots, \\mathbf{T}_k = \\mathbf{T}$, chaque $\\mathbf{T}_i$ étant à coordonnées entières.\n\nMais alors, puisque chaque opération $\\Theta_{i, j}$ est une involution, on peut écrire $\\mathbf{T}_{k-1}$ sous la forme $\\Theta_{i, j}(\\mathbf{T})$. À l'ordre près de ses coordonnées, le triplet $\\mathbf{T}_{k-1}$ donc est égal à $\\Theta_{2,3}(0, 0, 13)$ ou à $\\Theta_{3,2}(0, 0, 13)$, c'est-à-dire à $(0, 52/5, -39/5)$ ou à $(0, 39/5, 52/5)$. Cette contradiction démontre que Morgane ne pourra jamais écrire le nombre $13$ au tableau.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30752,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 6$ un entier. On a disposé, dans le plan, $n$ disques $D_{1}, D_{2}, \\ldots, D_{n}$ deux à deux disjoints, de rayons $r_{1} \\geqslant r_{2} \\geqslant \\ldots \\geqslant r_{n}$. Pour tout entier $i \\leqslant n$, on considère un point $P_{i}$ à l'intérieur du disque $D_{i}$. Enfin, soit $A$ un point quelconque du plan. Démontrer que\n$$\nA P_{1}+A P_{2}+\\ldots+A P_{n} \\geqslant r_{6}+r_{7}+\\ldots+r_{n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAu vu d'un tel énoncé, et ne sachant pas nécessairement par où commencer, on souhaite utiliser une récurrence sur $n$. Dans ces conditions, il s'avère que la propriété clé de notre récurrence sera la suivante.\n\nLemme. Si $n=6$, il existe un entier $i$ tel que $A P_{i} \\geqslant r_{6}$.\n\nDémonstration. Soit $O_{i}$ le centre du disque $D_{i}$. Si $A$ coïncide avec l'un des points $O_{i}$, le résultat désiré est immédiat. Sinon, on trie les centres $O_{i}$ dans le sens horaire autour de $A$. Ceux-ci définissent six angles de somme $360^{\\circ}$, donc l'un des angles, disons $\\widehat{O_{i} A O_{j}}$, vaut au plus $60^{\\circ}$. Si l'on suppose sans perte de généralité que $A O_{i} \\geqslant A O_{j}$, le théorème d'Al-Kashi indique alors que\n$$\n\\begin{aligned}\n\\left(r_{i}+r_{j}\\right)^{2} & \\leqslant O_{i} O_{j}^{2}=A O_{i}^{2}+A O_{j}^{2}-2 \\cos \\left(\\widehat{O_{i} A O_{j}}\\right) A O_{i} \\cdot A O_{j} \\\\ & \\leqslant A O_{i}^{2}-A O_{i} \\cdot A O_{j}+A O_{j}^{2} \\\\\n& \\leqslant A O_{i}^{2} \\leqslant\\left(A P_{i}+r_{i}\\right)^{2},\n\\end{aligned}\n$$\nde sorte que $A P_{i} \\geqslant r_{j} \\geqslant r_{6}$.\n\nOn procède maintenant par récurrence sur $n$. Tout d'abord, si $n=6$, le résultat de l'énoncé est un simple corollaire de notre lemme. Puis, si $n \\geqslant 7$, on applique notre lemme aux disques $D_{1}, \\ldots, D_{6}$. Il existe donc un entier $i \\leqslant 6$ tel que $A P_{i} \\geqslant r_{6}$. On supprime maintenant le disque $D_{i}$ et on applique l'hypothèse de récurrence aux $n-1$ disques restants, de sorte que\n$$\nA P_{i}+\\sum_{j \\neq i} A P_{j} \\geqslant r_{6}+\\left(r_{7}+\\ldots+r_{n}\\right)\n$$\nce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30753,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver les fonctions $f: \\mathbb{N}_{\\geqslant 1} \\mapsto \\mathbb{N}_{\\geqslant 0}$ vérifiant les deux conditions suivantes :\n1. $f(x y)=f(x)+f(y)$ pour tous les entiers $x \\geqslant 1$ et $y \\geqslant 1$;\n2. il existe une infinité d'entiers $n \\geqslant 1$ tels que l'égalité $f(k)=f(n-k)$ est vraie pour tout entier $k$ tel que $1 \\leqslant k \\leqslant n-1$.\nOn note $\\mathbb{N}_{\\geqslant 0}$ l'ensemble des entiers supérieurs ou égaux à 0, et $\\mathbb{N}_{\\geqslant 1}$ l'ensemble des entiers supérieurs ou égaux à 1.",
    "options": [],
    "answer": "All such functions are f(n) = c · v_p(n) for some prime p and some nonnegative integer c, where v_p is the p-adic valuation.",
    "solution": "Solution:\n\nLes fonctions recherchées sont les fonctions de la forme $f: n \\mapsto c v_{p}(n)$, où $c$ est un entier naturel, $p$ est un nombre premier, et $v_{p}(n)$ est la valuation $p$-adique de $n$. Tout d'abord, il est clair que ces fonctions sont bien solutions du problème.\n\nRéciproquement, soit $f$ une solution non nulle du problème. On dit qu'un entier $n$ est joli si l'égalité $f(k)=f(n-k)$ est vraie pour tout entier $k \\leqslant 1$. Remarquons que tout diviseur d'un joli entier est joli. En effet, si $d$ divise un joli entier $n=a d$, alors\n$$\nf(k)=f(a k)-f(a)=f(n-a k)-f(a)=f(a(d-k))-f(a)=f(d-k)\n$$\npour tout entier $k \\leqslant d-1$.\n\nEn outre, la condition 1 signifie que $f\\left(\\prod_{i} p_{i}^{\\alpha_{i}}\\right)=\\sum_{i} \\alpha_{i} f\\left(p_{i}\\right)$ pour toute décomposition en produit de facteurs premiers. Il s'agit donc de démontrer qu'il existe au plus un nombre premier $p$ pour lequel $f(p)>0$. On considère alors le plus petit entier $p$ tel que $f(p)>0$. La formule ci-dessus indique que $p$ est premier, et on pose $c=f(p)$.\n\nS'il existe un joli entier $n \\geqslant p$ que $p$ ne divise pas, posons $n=p q+r$, avec $q$ entier et $1 \\leqslant r \\leqslant p-1$. Puisque $n$ est joli, on sait que $f(r)=f(n-p q)=f(p q) \\geqslant f(p)>0$, en contradiction avec la définition de $p$. Ainsi, tout entier joli est de la forme $a p^{b}$ avec $b$ entier et $1 \\leqslant a \\leqslant p-1$. Réciproquement, puisqu'il existe une infinité d'entiers jolis et que $a$ ne peut prendre qu'un nombre fini de valeurs, toute puissance de $p$ divise un joli entier, et est donc elle-même joli.\n\nPar conséquent, pour tout nombre premier $q$ distinct de $p$, l'entier $p^{q-1}$ est joli. Puisque $q$ divise $p^{q-1}-1$, on en déduit en particulier que $0=f(1)=f\\left(p^{q-1}-1\\right) \\geqslant f(q)$, ce qui conclut.\n\n\nSolution alternative\n\nUne fois acquis le fait que tout entier joli est de la forme $a p^{b}$, avec $b$ entier et $1 \\leqslant a \\leqslant p-1$, et que tout diviseur d'un entier joli est joli, on peut aussi procéder comme suit.\n\nSupposons qu'il existe un nombre premier $q \\neq p$ pour lequel $f(q)>0$, et soit $q$ le plus petit tel nombre premier. On considère alors le plus petit entier joli $n>q$, puis on écrit $n$ sous la forme $n=a p^{b}=u q+v$, avec $u$ entier et $0 \\leqslant v \\leqslant q-1$.\n\nPuisque $n>q>p$, on sait que $b \\geqslant 1$, donc que $p$ divise $n$, et comme $q$ ne divise pas $n$, on sait que $v \\geqslant 1$. Dans ces conditions, $f(v)=f(u q)>0$, et la minimalité de $q$ indique que $p$ divise $v$. Mais alors $p$ divise aussi $u q=n-v$, et $p$ est premier avec $q$, donc $p$ divise $u$. On en déduit que $n>p q$, donc que le joli entier $n / p$ satisfait lui aussi l'inégalité $n / p>q$, en contradiction avec la minimalité de $n$. Notre supposition est ainsi invalide, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30754,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier naturel. Démontrer que l'écriture de l'entier $n\\left(2^{n}-1\\right)$ en base 2 compte exactement $n$ occurrences du chiffre 1.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $s_{n}$ l'entier $n\\left(2^{n}-1\\right)$. Puisque $s_{0}=0$ et $s_{1}=1$, l'énoncé est bien vérifié lorsque $n \\leqslant 1$. On suppose donc désormais que $n \\geqslant 2$.\n\nOn pose alors $t=2^{n}-n$ et $s=n-1$, de sorte que $n\\left(2^{n}-1\\right)=2^{n} s+t$ et que $s+t=2^{n}-1$. En vertu du binôme de Newton, on sait que $2^{n} \\geqslant (1+1)^{n} \\geqslant \\binom{n}{0}+\\binom{n}{1}=n+1$. Ainsi, $s$ et $t$ sont deux entiers naturels de somme $2^{n}-1$. Par conséquent, ils s'écrivent respectivement comme\n$$\n\\overline{a_{n-1} a_{n-2} \\ldots a_{0}}^{2} \\text{ et } \\overline{b_{n-1} b_{n-2} \\ldots b_{0}}^{2}\n$$\nen base 2, et on est assuré que $a_{i}+b_{i}=1$ pour tout $i \\leqslant n-1$.\n\nEn outre, comme $t<2^{n}$, l'écriture de $2^{n} s+t$ en base 2 commence avec les $n$ chiffres de $s$ puis se termine avec les $n$ chiffres de $t$. L'entier $n\\left(2^{n}-1\\right)=2^{n} s+t$ s'écrit donc\n$$\n\\overline{a_{n-1} a_{n-2} \\ldots a_{0} b_{n-1} b_{n-2} \\ldots b_{0}}^{2}\n$$\nen base 2. La somme de ces chiffres vaut donc $\\left(a_{0}+b_{0}\\right)+\\left(a_{1}+b_{1}\\right)+\\ldots+\\left(a_{n-1}+b_{n-1}\\right)=n$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30755,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers naturels non nuls $a$, $b$ et $c$ pour lesquels il existe des entiers naturels non nuls $x$, $y$ et $z$ tels que $x! = a b + 1$, $y! = b c + 1$ et $z! = c a + 1$.\n\nRemarque : Pour tout entier naturel non nul $n$, l'entier $n!$ désigne le produit $1 \\times 2 \\times \\cdots \\times n$.",
    "options": [],
    "answer": "All permutations of (1, 1, n! - 1) with integer n ≥ 2.",
    "solution": "Solution:\n\nSoit $(a, b, c)$ un triplet solution éventuel. Sans perte de généralité, et puisque $a$, $b$ et $c$ jouent des rôles symétriques, on suppose que $a \\leqslant b \\leqslant c$, de sorte que $x \\leqslant z \\leqslant y$ et que $x!$ divise à la fois $y!$ et $z!$.\n\nSoit $d$ un diviseur de $x!$, c'est-à-dire un diviseur commun à $x!$, $y!$ et $z!$. Puisque\n$$\na b \\equiv b c \\equiv c a \\equiv -1 \\pmod{d},\n$$\non sait que\n$$\n-1 \\equiv (a b)(b c)(c a) \\equiv (a b c)^2 \\pmod{d}\n$$\nest un carré modulo $d$. Or, $-1$ n'est pas un carré modulo $3$. Ainsi, $d \\neq 3$, ce qui signifie que $x \\leqslant 2$.\n\nComme $x! = a b + 1 \\geqslant 2 > 1!$, on en conclut que $x = 2$, de sorte que $a b = x! - 1 = 1$, donc que $a = b = 1$. Le triplet $(a, b, c)$ est donc, à permutation près, égal au triplet $(1, 1, z! - 1)$.\n\nRéciproquement, tout triplet de la forme $(1, 1, z! - 1)$ convient manifestement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer que pour tout entier $n$, le nombre $n^{3}-7 n$ est divisible par $6$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUne première approche est de regarder l'expression modulo $6$ :\n$$\nn^{3}-7 n \\equiv n^{3}-n \\equiv (n-1) n (n+1) \\quad \\bmod 6\n$$\ndonc il suffit de montrer que $6$ divise $(n-1) n (n+1)$. Or parmi trois nombres consécutifs, au moins un est divisible par $2$ et au moins un est divisible par $3$. Ainsi comme $2$ et $3$ sont premiers entre eux, $6 \\mid n^{3}-n$ et donc $6$ divise $n^{3}-7 n$.\n\n\nUne deuxième approche qui fonctionne aussi ici consiste aussi à regarder l'expression $n^{3}-7 n$ et de montrer à la main que celle-ci est toujours nulle en faisant une disjonction de cas sur les valeurs que peut prendre $n$ modulo $6$.\n\n| $n$ | $n^{3}-7 n$ |\n| :---: | :---: |\n| $0$ | $0^{3}-7 \\cdot 0 \\equiv 0+0 \\equiv 0$ |\n| $1$ | $1^{3}-7 \\cdot 1 \\equiv 1-1 \\equiv 0$ |\n| $2$ | $2^{3}-7 \\cdot 2 \\equiv 8-2 \\equiv 0$ |\n| $3$ | $3^{3}-7 \\cdot 3 \\equiv 27-21 \\equiv 6 \\equiv 0$ |\n| $4 \\equiv -2$ | $(-2)^{3}-7 \\cdot(-2) \\equiv -8+14 \\equiv 6 \\equiv 0$ |\n| $5 \\equiv -1$ | $(-1)^{3}-7 \\cdot(-1) \\equiv -1+7 \\equiv 6 \\equiv 0$ |\n\nAinsi, quelle que soit la valeur de $n$ modulo $6$, $n^{3}-7 n \\equiv 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMartin a versé à la hâte $n$ litres d'eau dans $n$ bouteilles. Certaines bouteilles sont donc plus remplies que d'autres. C'est alors qu'il se rappelle que sa mission était de mettre exactement un litre d'eau dans chaque bouteille avant de refermer le bouchon de celle-ci. Comme il n'a pas encore refermé les bouchons, il est encore temps de réparer ses bêtises.\n\nPour ce faire, il choisit une première bouteille, verse une partie de son contenu dans une autre bouteille s'il le souhaite, puis bouche cette première bouteille. Il choisit ensuite une seconde bouteille (qui peut être celle dans laquelle il a partiellement vidé la première bouteille, ou pas), verse une partie de son contenu dans une autre bouteille (mais pas dans la première bouteille, qui est déjà bouchée) s'il le souhaite, puis la bouche. Puis il procède de même avec une troisième bouteille, et ainsi de suite, jusqu'à boucher toutes les bouteilles.\n\nDémontrer que, si Martin choisit astucieusement les bouteilles dont il referme le bouchon et celles dans lesquelles il les vide partiellement, il finira bien par réparer ses bêtises.\n\nNote : les bouteilles sont surdimensionnées, et pourraient chacune contenir jusqu'à $n$ litres d'eau, ce qui les empêchera de déborder.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous allons procéder par récurrence sur $n$. Tout d'abord, si $n=1$, il y a une seule bouteille dans laquelle se trouve un litre d'eau, donc Martin a déjà gagné.\n\nPuis, si $n \\geqslant 2$, Martin choisit la bouteille $n^{\\circ} i$ la plus remplie, c'est-à-dire celle pour laquelle $x_{i}$ est maximal. On constate alors que\n$$\nn = x_{1} + x_{2} + \\ldots + x_{n} \\leqslant x_{i} + x_{i} + \\ldots + x_{i} = n x_{i}\n$$\nde sorte que $x_{i} \\geqslant 1$. Si $x_{i} > 1$, il la vide donc partiellement dans une autre bouteille, jusqu'à ce que la bouteille $n^{\\circ} i$ ne contienne plus qu'un litre d'eau, puis il la bouche. Il lui suffit alors de répartir harmonieusement les $n-1$ litres d'eau restants, qui se trouvent dans $n-1$ bouteilles, et l'hypothèse de récurrence lui assure qu'il pourra bien effectuer cette répartition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a$, $b$, $c$ et $d$ des entiers naturels non nuls. On suppose que $a! + b! = c! + d!$. Démontrer que $a b = c d$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons sans perte de généralité que $a \\leqslant b$, $c \\leqslant d$ et $a \\leqslant c$. Alors\n$$\nb! = c! - a! + d! \\geqslant d!\n$$\ndonc $b \\geqslant d$, de sorte que $a \\leqslant c \\leqslant d \\leqslant b$. L'entier $c!$ divise donc $c! + d! - b! = a!$, ce qui signifie que $c \\leqslant a$, et donc que $a = c$. On en conclut que $b! = d!$, donc que $b = d$ et que $a b = c d$.\n\n\nSolution alternative $n^{\\circ} 1$\n\nSupposons de nouveau que $a \\leqslant b$, $c \\leqslant d$ et $a \\leqslant c$. Si $a = c$, on conclut comme précédemment que $b = d$ et que $a b = c d$.\nSinon, on sait que $(a+1)! \\geqslant 2 a!$ divise $c! + d! = a! + b!$, donc ne divise ni $a!$, ni $b!$. Cela signifie que $a = b$. La double inégalité\n$$\na! + b! \\geqslant c! + d! \\geqslant 2 \\times a! = a! + b!\n$$\ndevient alors une égalité, donc $a = b = c = d$, de sorte que $a b = c d$ malgré tout.\n\n\nSolution alternative $n^{\\circ} 2$\n\nSupposons simplement que $a \\leqslant b$ et $c \\leqslant d$. Si $b < d$, alors\n$$\nc! + d! > d! \\geqslant (b+1)! = (b+1) b! \\geqslant 2 b! \\geqslant a! + b!\n$$\nce qui est absurde. On en déduit que $b \\leqslant d$ et, de même, que $d \\leqslant b$. Ainsi, $b = d$ et $a! = c!$, donc $a = c$ et $a b = c d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un rectangle. Soit $\\omega$ le demi-cercle de diamètre $[BC]$, de sorte que le point $A$ et le demi-cercle $\\omega$ sont situés du même côté par rapport au segment $[BC]$. Le cercle de centre $B$ et de rayon $AB$ recoupe le demi-cercle $\\omega$ au point $E$. La droite $(AE)$ recoupe le demi-cercle $\\omega$ au point $F$. Montrer que $AF = BF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, l'énoncé nous indique que $BA = BE$, de sorte que le triangle $ABE$ est isocèle en $B$. En outre, puisque $(AB)$ est tangente à $\\omega$, on sait que\n$$\n\\widehat{EBA} = \\widehat{EFB} = \\widehat{AFB} .\n$$\nComme les triangles $ABE$ et $AFB$ ont l'angle $\\widehat{BAE}$ en commun, ils sont donc semblables. En particulier, $AFB$ est isocèle en $F$, ce qui signifie comme souhaité que $AF = BF$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier $k \\geqslant 0$, on note $a_{k}$ le premier chiffre du nombre $2^{k}$, écrit en base 10. Par exemple, $a_{5}=3$ est le premier chiffre de $2^{5}=32$.\n\nSoit $n \\geqslant 1$ un entier. Démontrer que, parmi les chiffres de 1 à 9, il y en a un qui est égal à au plus $n / 17$ des $n$ chiffres $a_{0}, a_{1}, a_{2}, \\ldots, a_{n-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $\\mathrm{c}(k)$ le nombre de fois où le chiffre $k$ figure parmi les chiffres $a_{0}, a_{1}, a_{2}, \\ldots, a_{n-1}$. Lorsque $a_{i} \\geqslant 2$, on sait que $i \\geqslant 1$ et que $a_{i-1}=\\left\\lfloor a_{i} / 2\\right\\rfloor$.\n\nAinsi, pour tout chiffre $k \\leqslant 5$, on a $a_{i}=k$ si et seulement si $a_{i+1}=2 k$ ou $a_{i+1}=2 k+1$, de sorte que $\\mathrm{c}(k) \\geqslant \\mathrm{c}(2 k)+\\mathrm{c}(2 k+1)$. Si l'on pose $m=\\min \\{\\mathrm{c}(k): 1 \\leqslant k \\leqslant 9\\}$, on en conclut en particulier que\n$$\n\\begin{array}{rlrl}\nn & \\geqslant \\mathrm{c}(1)+\\mathrm{c}(2)+\\mathrm{c}(3)+\\mathrm{c}(4)+\\mathrm{c}(5)+\\mathrm{c}(6)+\\mathrm{c}(7)+\\mathrm{c}(8)+\\mathrm{c}(9) \\\\\n& \\geqslant \\quad+2 \\mathrm{c}(2)+2 \\mathrm{c}(3)+\\mathrm{c}(4)+\\mathrm{c}(5)+\\mathrm{c}(6)+\\mathrm{c}(7)+\\mathrm{c}(8)+\\mathrm{c}(9) \\\\\n& \\geqslant \\quad+3 \\mathrm{c}(4)+3 \\mathrm{c}(5)+3 \\mathrm{c}(6)+3 \\mathrm{c}(7)+\\mathrm{c}(8)+\\mathrm{c}(9) \\\\\n& \\geqslant \\begin{aligned}\n& +3 \\mathrm{c}(5)+3 \\mathrm{c}(6)+3 \\mathrm{c}(7)+4 \\mathrm{c}(8)+4 \\mathrm{c}(9)\n\\end{aligned} \\\\\n& \\geqslant 17 \\mathrm{~m} . &\n\\end{array}\n$$\nSolution:\n\nSolution alternative $n^{\\circ} 1$ Reprenons les notations de la solution précédente. Pour tout événement $\\mathbf{E}$, on notera $\\mathbf{1}_{\\mathrm{E}}$ l'entier qui vaut 1 lorsque $\\mathbf{E}$ se produit, et 0 sinon.\n\nPour tout chiffre $k \\leqslant 5$, on a\n$$\n\\mathrm{c}(k)=\\mathrm{c}(2 k)+\\mathrm{c}(2 k+1)+\\mathbf{1}_{a_{n-1}=k} .\n$$\nDe même, puisque $a_{1}=1$, on sait que\n$$\n1+c(5)+c(6)+c(7)+c(8)+c(9)=c(1)+1_{5 \\leqslant a_{n-1} \\leqslant 9}\n$$\nOn distingue alors deux cas.\n\n$\\triangleright \\operatorname{Si} a_{n-1}=1$, on trouve\n$$\n\\begin{aligned}\nc(1) & =c(5)+c(6)+c(7)+c(8)+c(9)+1 \\geqslant 5 m+1 ; \\\\\nc(2)+c(3) & =c(1)-1 \\geqslant 5 m \\\\\nc(4) & =c(8)+c(9) \\geqslant 2 m .\n\\end{aligned}\n$$\nOn en déduit que $n \\geqslant \\sum_{k=1}^{9} \\mathrm{c}(k) \\geqslant 17 m+1$.\n\n$\\triangleright \\operatorname{Si} a_{n-1} \\geqslant 2$, on trouve\n$$\n\\begin{aligned}\nc(1) & \\geqslant c(5)+c(6)+c(7)+c(8)+c(9) \\geqslant 5 m \\\\\nc(2)+c(3) & =c(1) \\geqslant 5 m \\\\\nc(4) & =c(8)+c(9) \\geqslant 2 m\n\\end{aligned}\n$$\nOn en déduit là encore que $n \\geqslant \\sum_{k=1}^{9} \\mathrm{c}(k) \\geqslant 17 m+1$, ce qui conclut.\nSolution:\n\nSolution alternative $n^{\\circ} 2$ Réutilisons une fois de plus les notations des solutions précédentes. Cette fois-ci, on découpe la suite en sous-suites croissantes dont le dernier élément vaut 1 et le dernier élément vaut au moins 5. Il s'agit des cinq sous-suites de la forme $1,2,4,8$; $1,2,4,9$; $1,2,5$; $1,3,6$; et $1,3,7$.\n\nComme chacun des chiffres de 5 à 9 appartient à exactement un type de sous-suite, ces sous-suites apparaissent respectivement $c(8), c(9), c(5), c(6)$ et $c(7)$ fois. En particulier, chacune d'entre elles apparaît au moins $m$ fois. Puisque ces suites sont de longueurs respectives 4, 4, 3, 3 et 3, et en comptant leur nombre total de termes, on en déduit que\n$$\nn \\geqslant 4 \\times m+4 \\times m+3 \\times m+3 \\times m+3 \\times m=17 m\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les couples d'entiers $(x, y)$ tels que $x^{2}+73=y^{2}$.",
    "options": [],
    "answer": "(-36, -37), (-36, 37), (36, -37), (36, 37)",
    "solution": "Solution:\n\nComme souvent pour une équation diophantienne, on cherche à réarranger l'équation de sorte à avoir des produits de facteurs des deux côtés de l'égalité. Lorsque l'on est en présence de carrés parfaits, on peut utiliser l'identité remarquable $y^{2}-x^{2}=(y-x)(y+x)$. Ceci permet de réécrire l'équation sous la forme :\n$$\n73=(y-x)(y+x)\n$$\nAinsi, le nombre $y-x$ est un diviseur de $73$. Or les diviseurs de $73$ sont $73, 1, -1, -73$. On distingue alors quatre cas :\n\nCas $n^{\\circ} 1$ : $y-x=73$. Alors $x+y=1$. Ceci conduit à $y=73+x=73+(1-y)$, soit $y=37$ et $x=-36$. Réciproquement, on a bien $(-36)^{2}+73=37^{2}$, donc le couple $(-36,37)$ est bien solution.\n\nCas $n^{\\circ} 2$ : $y-x=1$. Alors $x+y=73$. Ceci conduit à $y=1+x=1+(73-y)$, soit $y=37$ et $x=36$. Réciproquement, on a bien $36^{2}+73=37^{2}$, donc le couple $(36,37)$ est bien solution.\n\nCas $n^{\\circ} 3$ : $y-x=-1$. Alors $x+y=-73$. Ceci conduit à $y=-1+x=-1+(-73-y)$, soit $y=-37$ et $x=-36$. Réciproquement, on a bien $(-36)^{2}+73=(-37)^{2}$, donc le couple $(-36,-37)$ est bien solution.\n\nCas $n^{\\circ} 4$ : $y-x=-73$. Alors $x+y=-1$. Ceci conduit à $y=-73+x=-73+(-1-y)$, soit $y=-37$ et $x=36$. Réciproquement, on a bien $36^{2}+73=(-37)^{2}$, donc le couple $(36,-37)$ est bien solution.\n\nLes couples solutions sont donc $(-36,-37), (-36,37), (36,-37)$ et $(36,37)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a, b, c \\geqslant 1$ des entiers tels que $a^{b} \\mid b^{c}$ et $a^{c} \\mid c^{b}$. Montrer que $a^{2} \\mid b c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNous allons regarder les valuations $p$-adiques de chacun des nombres premiers $a$, $b$ et $c$. En effet, il suffit de montrer que pour tout nombre premier $p$, on a $2 v_{p}(a) = v_{p}\\left(a^{2}\\right) \\leqslant v_{p}(b c) = v_{p}(b) + v_{p}(c)$.\n\nSoit $p$ un nombre premier. L'hypothèse que $a^{b} \\mid b^{c}$ se traduit en terme de valuation $p$-adique par $b v_{p}(a) \\leqslant c v_{p}(b)$, ou encore $v_{p}(a) \\leqslant \\frac{c}{b} v_{p}(b)$. De même, on obtient que $v_{p}(a) \\leqslant \\frac{b}{c} v_{p}(c)$.\n\nEn multipliant les deux relations, on trouve que $v_{p}(a)^{2} \\leqslant v_{p}(b) v_{p}(c)$. On a donc, d'après l'inégalité des moyennes :\n$$\nv_{p}(a) \\leqslant \\sqrt{v_{p}(b) v_{p}(c)} \\leqslant \\frac{v_{p}(b) + v_{p}(c)}{2}\n$$\nce qui donne bien $2 v_{p}(a) \\leqslant v_{p}(b) + v_{p}(c)$ comme voulu. Ceci étant vrai pour tout premier $p$, on a bien $a^{2} \\mid b c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier naturel. On dit qu'un ensemble fini d'entiers est $n$-équilibré s'il y a exactement $n$ manières de le partitionner en deux sous-ensembles de même somme. Par exemple, l'ensemble $\\{1,3,4,5,6,7\\}$ est 2-équilibré, car les seules manières de le partitionner en deux sous-ensembles de même somme sont de le partitionner en $\\{1,3,4,5\\}$ et $\\{6,7\\}$, ou bien en $\\{1,5,7\\}$ et $\\{3,4,6\\}$.\nTrouver tous les entiers $n \\geqslant 0$ pour lesquels il existe un ensemble $n$-équilibré.",
    "options": [],
    "answer": "All integers greater than or equal to zero",
    "solution": "Solution:\n\nTout d'abord, l'ensemble $\\{1\\}$ est 0-équilibré et l'ensemble $\\{1,2,3\\}$ est 1-équilibré.\n\nÉtant donné un entier $n \\geqslant 2$, nous allons démontrer que l'ensemble\n$$\nE=\\{k: 3 n \\leqslant k \\leqslant 4 n-1\\} \\cup\\{k: 4 n+1 \\leqslant k \\leqslant 5 n\\} \\cup\\{8 n(n-2)\\}\n$$\nest $n$-équilibré. En effet, si l'on note $\\Sigma_{X}$ la somme des éléments d'un ensemble $X$, partitionner $E$ en deux sous-ensembles de même somme revient à choisir un sous-ensemble $X$ de $E$ contenant l'entier $8 n(n-2)$ et dont la somme vaut\n$$\n\\Sigma_{X}=\\frac{\\Sigma_{E}}{2}=\\frac{1}{2}\\left(8 n(n-2)+\\sum_{\\ell=1}^{n}((4 n-\\ell)+(4 n+\\ell))\\right)=\\frac{1}{2}\\left(8 n(n-2)+8 n^{2}\\right)=8 n(n-1)\n$$\nAutrement dit, cela revient à sélectionner un sous-ensemble $Y$ de $E' = E \\backslash \\{8 n(n-2)\\}$ pour lequel $\\Sigma_{Y}=8 n$.\n\nPuisque $\\max E' < 8 n < 3 \\min E'$, un tel ensemble $Y$ contient exactement 2 éléments. Les ensembles $Y$ recherchés sont donc les ensembles de la forme $\\{4 n-\\ell, 4 n+\\ell\\}$ avec $1 \\leqslant \\ell \\leqslant n$. Il existe $n$ tels ensembles, donc $E$ est bien $n$-équilibré, comme souhaité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30764,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $a$ un réel strictement positif et $n \\geqslant 1$ un entier. Montrer que\n$$\n\\frac{a^{n}}{1+a+\\ldots+a^{2 n}}<\\frac{1}{2 n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPuisque la difficulté réside dans le dénominateur du membre de droite, et pour plus de confort, on peut chercher à montrer la relation inverse, à savoir :\n$$\n\\frac{1+a+\\ldots+a^{2 n}}{a^{n}}>2 n\n$$\nL'idée derrière la solution qui suit est \"d'homogénéiser\" le numérateur du membre de gauche, c'est-à-dire de comparer ce numérateur dans lequel les $a$ sont élevés à des puissances distinctes à une expression composée uniquement de $a$ élevés à la même puissance. Pour cela, on cherche à coupler certains termes et appliquer l'inégalité des moyennes. Voyons plutôt : pour $0 \\leqslant i \\leqslant 2 n$, on a $a^{i}+a^{2 n-i} \\geqslant 2 \\sqrt{a^{i} a^{2 n-i}}= 2 a^{n}$. Ainsi :\n$$\n\\begin{aligned}\n\\frac{1+a+\\ldots+a^{2 n}}{a^{n}} & =\\frac{\\left(1+a^{2 n}\\right)+\\left(a+a^{2 n-1}\\right)+\\ldots+\\left(a^{n-1}+a^{n+1}\\right)+a^{n}}{a^{n}} \\\\\n& \\geqslant \\frac{\\overbrace{2 a^{n}+2 a^{n}+\\ldots+2 a^{n}}^{n \\text{ termes}}+a^{n}}{a^{n}} \\\\\n& =\\frac{(2 n+1) a^{n}}{a^{n}} \\\\\n& =2 n+1 \\\\\n& >2 n\n\\end{aligned}\n$$\nce qui fournit l'inégalité voulue.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les entiers $n \\geqslant 1$ tels que\n$$\n6^{n}-1 \\mid 7^{n}-1\n$$",
    "options": [],
    "answer": "no such positive integer n",
    "solution": "Solution:\nSoit $n$ un éventuel entier vérifiant $6^{n}-1 \\mid 7^{n}-1$.\nOn a $5=6-1 \\mid 6^{n}-1^{n}$, donc $5 \\mid 7^{n}-1$. On calcule donc les puissances de $7$ modulo $5$ :\n\n|  | $n$ | 0 | 1 | 2 | 3 | 4 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $7^{n}(\\bmod 5)$ | 1 | 2 | 4 | 3 | 1 |  |\n\nAinsi, si on note $n=4q+r$ la division euclidienne de $n$ par $4$, on a $7^{n} \\equiv (7^{4})^{q} \\times 7^{r} \\equiv 1^{q} 7^{r} \\equiv 7^{r}$ $(\\bmod 5)$. Donc pour que $5 \\mid 7^{n}-1$, il faut que $r=0$, autrement dit $n$ est multiple de $4$.\n\nEn regardant alors modulo $7$, on obtient $6^{n} \\equiv (-1)^{4q} \\equiv 1 (\\bmod 7)$, donc $7\\mid 6^{n}-1 \\mid 7^{n}-1$, ce qui n'est pas possible.\n\nIl n'y a donc aucun $n$ vérifiant $6^{n}-1 \\mid 7^{n}-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer tous les entiers $a$ tels que $a-3$ divise $a^{3}-3$.",
    "options": [],
    "answer": "a ∈ {−21, −9, −5, −3, −1, 0, 1, 2, 4, 5, 6, 7, 9, 11, 15, 27}",
    "solution": "Solution:\nOn sait que $a-3 \\mid a^{3}-3^{3}$, donc il divise la différence\n$$\na-3 \\mid a^{3}-3-\\left(a^{3}-3^{3}\\right)=27-3=24\n$$\nRéciproquement, il suffit que $a-3$ divise $24$ pour qu'il divise $24+a^{3}-3^{3}=a^{3}-3$. Les solutions sont donc exactement les diviseurs de $24$ auxquels on ajoute $3$. On a donc\n$$\na-3 \\in\\{-24,-12,-8,-6,-4,-3,-2,-1,0,1,2,3,4,6,8,12,24\\}\n$$\net\n$$\na \\in\\{-21,-9,-5,-3,-1,0,1,2,3,4,5,6,9,11,15,27\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\omega_{1}$ et $\\omega_{2}$ deux cercles de centres respectifs $O_{1}$ et $O_{2}$. On suppose que $\\omega_{1}$ et $\\omega_{2}$ se coupent en les points $A$ et $B$. La droite $(O_{1}A)$ recoupe le cercle $\\omega_{2}$ en $C$ tandis que la droite $(O_{2}A)$ recoupe le cercle $\\omega_{1}$ en $D$. Montrer que les points $D$, $O_{1}$, $B$, $O_{2}$ et $C$ appartiennent à un même cercle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons que puisque les angles $\\widehat{DAO_{1}}$ et $\\widehat{CAO_{2}}$ sont opposés par le sommet, ils sont égaux. D'autre part, puisque les points $A$ et $D$ appartiennent au cercle $\\omega_{1}$, le triangle $A O_{1} D$ est isocèle en $O_{1}$. De même, le triangle $CO_{2} A$ est isocèle en $O_{2}$. Les triangles $DO_{1} A$ et $CO_{2} A$ sont donc des triangles isocèles avec les mêmes angles à la base, ils sont donc semblables. Ceci implique que $\\widehat{AO_{1} D}=\\widehat{CO_{2} A}$, et donc que\n$$\n\\widehat{CO_{1} D}=\\widehat{AO_{1} D}=\\widehat{CO_{2} A}=\\widehat{CO_{2} D}\n$$\nsi bien que les points $C$, $O_{2}$, $O_{1}$ et $D$ sont cocycliques.\n\nPar ailleurs, on peut découper l'angle $\\widehat{DBC}$ en la somme $\\widehat{DBA}+\\widehat{ABC}$. D'une part, d'après le théorème de l'angle au centre dans le cercle $\\omega_{1}$, on a $\\widehat{ABD}=\\frac{1}{2} \\widehat{AO_{1} D}$. D'autre part, d'après le théorème de l'angle au centre dans le cercle $\\omega_{1}$, on a $\\widehat{ABC}=\\frac{1}{2} \\widehat{AO_{2} C}$. Ainsi\n$$\n\\widehat{DBC}=\\widehat{DBA}+\\widehat{ABC}=\\frac{1}{2} \\widehat{AO_{1} D}+\\frac{1}{2} \\widehat{AO_{2} C}=\\widehat{AO_{1} D}\n$$\nce qui permet de conclure que le point $B$ appartient au cercle passant par les points $C$, $O_{2}$, $O_{1}$ et $D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30768,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $f: \\mathbb{N}^{\\star} \\rightarrow \\mathbb{N}^{\\star}$ une fonction telle que pour tout entier $n \\geqslant 1$, $f(f(n))$ soit égal au nombre de diviseurs positifs de $n$. Montrer que si $p$ est un nombre premier, alors $f(p)$ est aussi un nombre premier.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDans la suite, on note $d(n)$ le nombre de diviseurs positifs de $n$. Si $p$ est premier, on a $f(f(p))=2$ par définition. En appliquant $f$ des deux côtés de l'égalité, on voit que $f(2)=d(f(p))$. On veut donc montrer que $f(2)=2$, ce qui prouvera que $f(p)$ a exactement deux diviseurs et est donc premier.\n\nOr, on a aussi $f(2)=d(f(2))$. Or, pour qu'un entier $n \\geqslant 2$ soit égal à son nombre de diviseurs, tous les nombres inférieurs à $n$ doivent diviser $n$, y compris $n-1$ qui est toujours premier avec $n$, donc $n-1=1$ et $n=2$. On déduit que $f(2) \\in\\{1,2\\}$.\n\nSi $f(2)=1$, alors $d(f(p))=1$ pour tout $p$ premier, donc $f(p)=1$ pour $p$ premier, puisque 1 est le seul entier à admettre un seul diviseur. Mais alors\n$$\n1=f(3)=f(d(2^{2}))=f(f(f(2^{2})))=d(f(2^{2}))\n$$\net l'on déduit que $f(4)=1$. Mais alors\n$$\nf(f(4))=f(1)=f(f(2))=d(2)=2 \\neq d(4)\n$$\nce qui fournit la contradiction désirée. On a donc bien $f(2)=2$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $(a_{n})$ une suite strictement croissante d'entiers positifs telle que $a_{1}=1$ et pour tout $n \\geqslant 1$, $a_{n+1} \\leqslant 2n$. Montrer que pour tout entier $n \\geqslant 1$, il existe deux indices $p$ et $q$ tels que $a_{p}-a_{q}=n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $n \\geqslant 1$ un entier. On va utiliser le principe des tiroirs. Construisons nos tiroirs de sorte que si deux nombres sont dans un même tiroir, ils vérifient la propriété de l'énoncé. Pour cela, on considère les tiroirs suivants : on réunit les entiers de $1$ à $2n$ en les couples de la forme $\\{i, n+i\\}$ pour $i$ entre $1$ et $n$, ce qui constitue $n$ tiroirs. Ces tiroirs sont bien disjoints deux à deux, et si deux termes de la suite sont dans un même tiroir, cela donne les deux indices $p$ et $q$ tels que $a_{p}-a_{q}=n$, car tous les termes d'une suite strictement croissante sont distincts.\n\nOr, puisque la suite $(a_{m})$ est croissante et puisque $a_{n+1} \\leqslant 2n$, $a_{m} \\in \\llbracket 1, 2n \\rrbracket$ pour $m \\in \\llbracket 1, n+1 \\rrbracket$. Donc les $n+1$ premiers termes de la suite sont dans l'intervalle $\\llbracket 1, 2n \\rrbracket$. Par le principe des tiroirs, il y a deux termes distincts qui appartiennent au même ensemble $\\{i, n+i\\}$ pour un certain $i$, ce qui est le résultat voulu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit que deux entiers $a, b \\in \\mathbb{N}^{\\star}$ sont reliés s'il existe un nombre premier $p$ tel que $a = p b$ ou $b = p a$.\n\nTrouver tous les entiers $n \\geqslant 1$ ayant la propriété suivante : on peut écrire tous les diviseurs positifs de $n$ (1 et $n$ compris) exactement une fois sur un cercle de sorte que tout diviseur soit relié avec chacun de ses deux voisins?",
    "options": [],
    "answer": "Exactly those n that are neither perfect squares nor prime powers with exponent at least two; equivalently, all primes and all integers with at least two distinct prime factors that are not perfect squares.",
    "solution": "Solution:\n\nTraitons d'abord le cas où $n$ n'a qu'un seul facteur premier, donc $n = p^{\\alpha}$. Si $\\alpha \\leqslant 1$, n'importe quelle configuration marche. En revanche, si $\\alpha \\geqslant 2$, $1$ doit être placé à côté de deux nombres différents, et n'est relié qu'à un seul autre diviseur de $p^{\\alpha}$ (celui-ci étant $p$), ce n'est donc pas possible.\n\nRemarquons que la parité du nombre de diviseurs premiers comptés avec multiplicité change forcément entre un nombre et un nombre qui lui est relié. Ainsi, si $n$ a un nombre impair de diviseurs, l'existence d'une telle configuration impliquerait que $1$ a à la fois un nombre pair et un nombre impair de diviseurs premiers, ce qui n'est pas possible. Les nombres ayant un nombre impair de diviseurs étant les carrés parfaits, ceux-ci ne vérifient pas non plus la propriété de l'énoncé.\n\nMaintenant, essayons de résoudre le cas où on a deux facteurs premiers, mettons $n = p^{a} q^{b}$ avec l'un des nombres $a$ ou $b$ étant impair, et l'on suppose sans perte de généralité qu'il s'agit de $a$.\n\nL'arrangement suivant fonctionne :\n$$\n1, q, q^{2}, \\ldots, q^{b}, p q^{b}, p q^{b-1}, \\ldots, p q, p^{2} q, \\ldots, p^{a-1} q^{b}, p^{a} q^{b}, p^{a} q^{b-1}, \\ldots, p^{a} q, p^{a}, p^{a-1}, \\ldots, p, 1\n$$\nOn utilise le fait que $a$ est impair pour dire qu'on a effectivement écrit $p^{a-1} q^{b}$ juste avant $p^{a} q^{b}$.\n\nOn procède maintenant par récurrence sur le nombre de diviseurs premiers : supposons avoir montré pour un $k \\geqslant 2$ que tous les nombres de la forme $\\prod_{i=1}^{k} p_{i}^{\\alpha_{i}}$ avec au moins l'un des $\\alpha_{i}$ impair vérifient la propriété de l'énoncé, et soit $n = \\prod_{i=1}^{k+1} p_{i}^{\\alpha_{i}}$ avec au moins l'un des $\\alpha_{i}$ impairs, disons $\\alpha_{1}$. Alors l'hypothèse de récurrence nous dit qu'on peut ordonner les diviseurs de $n' = \\prod_{i=1}^{k} p_{i}^{\\alpha_{i}}$ de la forme $d_{1}, \\ldots, d_{m}$ de sorte que $d_{i}$ et $d_{i+1}$ soient tous reliés, de même que $d_{m}$ et $d_{1}$. Alors l'arrangement suivant fonctionne pour $m$ (car $m$ est pair) :\n$$\nd_{1}, p_{k+1} d_{1}, \\ldots, p_{k+1}^{\\alpha_{k+1}} d_{1}, p_{k+1}^{\\alpha_{k+1}} d_{2}, \\ldots, p_{k+1} d_{2}, d_{2}, d_{3}, \\ldots, p_{k+1}^{\\alpha_{k+1}} d_{m}, \\ldots, p_{k+1} d_{m}, d_{m}, d_{1}\n$$\nAinsi les nombres qui ne vérifient pas cette propriété sont les carrés (différents de $1$) et les puissances (plus grandes que $2$) de nombres premiers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $m, n \\geqslant 2$ des entiers tels que $\\operatorname{PGCD}(m, n)=\\operatorname{PGCD}(m, n-1)=1$. On définit la suite $\\left(n_{k}\\right)_{k \\in \\mathbb{N}}$ par $n_{0}=m$ et $n_{k+1}=n \\cdot n_{k}+1$ pour $k \\in \\mathbb{N}$. Montrer que les entiers $n_{1}, \\ldots, n_{m-1}$ ne peuvent pas tous être des nombres premiers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, la suite $(n_{k})$ est une suite arithmético-géométrique. On peut donc établir une formule close pour son terme général, à savoir\n$$\nn_{k}=n^{k} m+\\frac{n^{k+1}-1}{n-1},\n$$\nque l'on peut aussi obtenir par récurrence à partir du calcul des premiers termes. De cette expression on déduit en particulier que $n_{k}>m$ pour tout $k>0$.\n\nOn va désormais montrer que l'un des nombres $n_{1}, \\ldots, n_{m-1}$ est divisible par $m$, ce qui fera de ce nombre un nombre composé.\n\nEn passant modulo $m$, on déduit\n$$\nn_{k} \\equiv \\frac{n^{k+1}-1}{n-1} \\pmod{m}\n$$\npar récurrence, où l'on a utilisé que $n-1$ est inversible modulo $m$ car $n-1$ est premier avec $m$.\n\nSoit $\\omega$ l'ordre de $n \\bmod m$ (celui-ci existe car $m$ et $n$ premiers entre eux). On a $\\omega \\leqslant \\varphi(m) \\leqslant m-1$ d'après le théorème d'Euler. On déduit que $m \\mid n^{\\omega}-1$, ce qui implique que $m$ divise $n_{\\omega-1}-1$ comme voulu. Ainsi, les $n_{k}$ ne sont pas tous premiers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $m$, $n$ et $x$ des entiers strictement positifs. Montrer que\n$$\n\\sum_{i=1}^{n} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, m\\right)=\\sum_{i=1}^{m} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, n\\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nQuitte à échanger $m$ et $n$, on peut supposer que $m \\leqslant n$. On procède alors par récurrence sur $n$ à $m$ fixé.\n\nInitialisation : Si $m=n$, l'égalité est triviale.\n\nHérédité : Supposons l'égalité vraie pour un certain $n \\geqslant m$ et cherchons à montrer que l'égalité est vraie pour $n+1$. Le membre de gauche devient\n$$\n\\sum_{i=1}^{n} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, m\\right)+\\min \\left(\\left\\lfloor\\frac{x}{n+1}\\right\\rfloor, m\\right)\n$$\nExaminons les termes du membre de droite, qui sont de la forme $\\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, n+1\\right)$. On a\n$$\n\\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, n+1\\right)-\\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, n\\right)= \\begin{cases}1 & \\text{ si }\\left\\lfloor\\frac{x}{i}\\right\\rfloor \\geqslant n+1 \\\\ 0 & \\text{ sinon }\\end{cases}\n$$\nOr, $\\left\\lfloor\\frac{x}{i}\\right\\rfloor \\geqslant n+1$ si et seulement si $x \\geqslant(n+1) i$, ou encore $i \\leqslant\\left\\lfloor\\frac{x}{n+1}\\right\\rfloor$. Donc le nombre de $i$ de $\\llbracket 1, m \\rrbracket$ pour lesquels $\\left\\lfloor\\frac{x}{i}\\right\\rfloor \\geqslant n+1$ est précisément $\\min \\left(\\left\\lfloor\\frac{x}{n+1}\\right\\rfloor, m\\right)$. On déduit que\n$$\n\\begin{aligned}\n\\sum_{i=1}^{m} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, n+1\\right) & =\\sum_{i=1}^{m} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, n\\right)+\\min \\left(\\left\\lfloor\\frac{x}{n+1}\\right\\rfloor, m\\right) \\\\\n& =\\sum_{i=1}^{n} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, m\\right)+\\min \\left(\\left\\lfloor\\frac{x}{n+1}\\right\\rfloor, m\\right) \\\\\n& =\\sum_{i=1}^{n+1} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, m\\right)\n\\end{aligned}\n$$\noù la deuxième égalité provient de l'hypothèse de récurrence. Ceci achève la récurrence et conclut l'exercice.\n\n\nSolution alternative:\n\nUne solution très astucieuse consiste à procéder par double comptage, en remarquant que les deux membres de l'égalité servent à compter le même objet, à savoir le nombre de paires $(a, b)$ telles que $a \\leqslant m, b \\leqslant n$ et $a b \\leqslant x$.\n\nCherchons en effet à compter ce nombre d'objet d'une première façon. Le choix d'une paire $(a, b) \\in \\llbracket 1, m \\rrbracket \\times \\llbracket 1, n \\rrbracket$ telle que $a b \\leqslant x$ se caractérise dans un premier temps par le choix de $a$ dans $\\llbracket 1, m \\rrbracket$. Puis, puisque $a b \\leqslant x$, on $b \\leqslant \\min (\\lfloor x / a\\rfloor, n)$ et il y a $\\min (\\lfloor x / a\\rfloor, n)$ choix pour $b$ si $a$ est fixé. On a donc en tout\n$$\n\\sum_{i=1}^{m} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, n\\right)\n$$\ntelles paires $(a, b)$.\n\nEn commençant par choisir $b$ plutôt que $a$, un raisonnement similaire nous permet d'affirmer qu'il y a\n$$\n\\sum_{i=1}^{n} \\min \\left(\\left\\lfloor\\frac{x}{i}\\right\\rfloor, m\\right)\n$$\ntelles paires d'entiers $(a, b)$. Les deux termes sont donc égaux.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30773,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $p$ un nombre premier impair et $x_{1}, \\ldots, x_{p}$ des entiers relatifs.\nOn suppose que pour tout $k \\geqslant 1$ entier, on a\n$$\np \\mid x_{1}^{k}+\\cdots+x_{p}^{k}\n$$\n\nMontrer que les entiers $x_{1}, \\ldots, x_{p}$ sont tous congrus modulo $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons que la relation est également vraie pour $k=0$ car $\\underbrace{1+\\ldots+1}_{p \\text{ fois }} \\equiv 0$ $\\bmod p$\n\nSoit $Q(X)=\\sum_{k=0}^{n} a_{k} X^{k}$ un polynôme à coefficients entiers. On a\n$$\nQ\\left(x_{1}\\right)+\\ldots+Q\\left(x_{p}\\right)=\\sum_{\\ell=1}^{p} \\sum_{k=0}^{n} a_{k} x_{\\ell}^{k}=\\sum_{k=0}^{n} a_{k} \\underbrace{\\sum_{\\ell=1}^{p} x_{\\ell}^{k}}_{\\equiv 0 \\bmod p} \\equiv 0 \\bmod p\n$$\nOn suppose par l'absurde que les $x_{i}$ ne donnent pas tous le même reste modulo $p$. Soit $\\{y_{1}, \\ldots, y_{r}\\}$ l'ensemble des restes modulo $p$ généré par l'ensemble $\\{x_{1}, \\ldots, x_{p}\\}$, avec $y_{1} \\equiv x_{1}$. On note $a_{1}$ le nombre d'éléments de l'ensemble $\\{x_{1}, \\ldots, x_{p}\\}$ qui ont pour reste $y_{1}$. Soit $P(X)=(X-y_{2}) \\ldots (X-y_{r})$. $P$ est nul en chacun des $y_{i}$ pour $i \\neq 1$. Notons que comme $y_{1} \\neq y_{i} \\bmod p$, $P\\left(y_{1}\\right) \\neq 0 \\bmod p$. La relation établie plus haut nous donne alors $a_{1} P\\left(y_{1}\\right) \\equiv 0 \\bmod p$, ce qui implique que $a_{1}=0 \\bmod p$, c'est-à-dire que $a_{1}=p$. Cela est en contradiction avec notre supposition de départ, donc les $x_{i}$ sont tous égaux modulo $p$.\n\n\nSolution alternative :\n\nOn présente une solution un peu plus conceptuelle qui s'appuie sur la structure de corps de $\\mathbb{Z} / p \\mathbb{Z}$ et sur le fait que les sommes de Newton des racines d'un polynôme déterminent ses coefficients.\n\nSoit $P(X)=(X-x_{1}) \\ldots (X-x_{p})$ le polynôme unitaire à coefficients entiers dont les racines sont les $x_{\\ell}$. L'hypothèse de l'énoncé nous indique que les sommes de Newton des $x_{\\ell}$ sont nulles modulo $p$.\n\nOr il est possible de relier les coefficients du polynôme $P$ aux sommes de Newton du polynôme $P$, via les formules suivantes : si $a_{k}$ est le coefficient devant $X^{k}$ dans l'écriture de $P$, alors pour tout $0 \\leqslant k \\leqslant p$ :\n$$\nk a_{p-k}=(-1)^{k-1} \\sum_{\\ell=1}^{k-1} a_{p-\\ell}\\left(x_{1}^{\\ell}+\\ldots+x_{p}^{\\ell}\\right)\n$$\nOn peut retrouver ces formules par récurrence à l'aide des relations de Viète.\n\nOn remarque alors que si $k \\neq 0, p$, $k a_{p-k} \\equiv 0 \\bmod p$ car chacune des $x_{1}^{\\ell}+\\ldots+x_{p}^{\\ell}$ est divisible par $p$. Ainsi, dans $\\mathbb{Z} / p \\mathbb{Z}[X]$, le polynôme $P$ est de la forme $P(X)=X^{p}-a=X^{p}-a^{p}=(X-a)^{p}$, où l'on a utilisé que $a^{p} \\equiv a\\ (\\bmod p)$ par le petit théorème de Fermat et que dans la formule $(X-a)^{p}=\\sum_{k=0}^{p}\\binom{p}{k} a^{p-k} X^{k}$, seuls les coefficients $\\binom{p}{0}$ et $\\binom{p}{p}$ sont non nuls modulo $p$. Cela implique que les $x_{i}$ sont tous égaux (à $a$) modulo $p$ comme voulu.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 30774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dans lequel $AB < AC$. Soit $\\omega$ un cercle passant par $B$ et $C$ et on suppose que le point $A$ se trouve à l'intérieur du cercle $\\omega$. Soient $X$ et $Y$ des points de $\\omega$ tels que $\\widehat{BX A} = \\widehat{A Y C}$. On suppose que $X$ et $C$ sont situés de part et d'autre de la droite $(AB)$ et que $Y$ et $B$ sont situés de part et d'autre de la droite $(AC)$. Montrer que, lorsque $X$ et $Y$ varient sur le cercle $\\omega$, la droite $(XY)$ passe par un point fixe.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, nous allons chercher à tracer une figure exacte, en espérant que les raisonnements mis en oeuvre pour le tracé nous éclairerons sur la dynamique de la figure.\n\n![](attached_image_1.png)\n\nPour tracer la figure, on cherche à tirer profit de l'hypothèse que les points $X, B, C$ et $Y$ sont sur un même cercle, à l'aide du théorème de l'angle inscrit. Supposons la figure tracée. On prolonge donc la droite $(XA)$ et on note $X'$ le second point d'intersection de $(XA)$ avec $\\omega$ et de même $Y'$ le second point d'intersection de $(YA)$ avec $\\omega$.\n\nLe théorème de l'angle inscrit et l'hypothèse sur $X$ et $Y$ nous impose l'égalité suivante :\n$$\n\\widehat{CX'Y'} = \\widehat{CY'} = \\widehat{CY'} = \\widehat{AXB} = \\widehat{X'XB} = \\widehat{X'Y'B}\n$$\nde sorte que le quadrilatère $BCX'Y'$ est un trapèze isocèle, et les droites $(X'Y')$ et $(BC)$ sont parallèles.\n\nOn en déduit un protocole pour tracer la figure : On place un point $X$ sur le cercle $\\omega$ et on prolonge la droite $(AX)$ de sorte à obtenir le point $X'$. Puis on obtient le point $Y'$ en traçant la parallèle à la droite $(BC)$ passant par $X'$. On obtient enfin le point $Y$ en prolongeant la droite $(Y'A)$.\n\nRésolvons à présent l'exercice. Maintenant que l'on peut tracer une figure exacte, on peut tracer la figure pour deux positions différentes du point $X$, notées $X_1$ et $X_2$, tracer les points $Y_1$ et $Y_2$ associés et conjecturer la position du point fixe à l'aide du point d'intersection des droites $(X_1Y_1)$ et $(X_2Y_2)$. Il est frappant que le point d'intersection obtenu appartient à la parallèle à la droite $(BC)$ passant par le point $A$.\n\nSoit donc $P$ le point d'intersection de la droite $(XY)$ avec la parallèle à $(BC)$ passant par $A$. Puisque les droites $(X'Y')$ et $(AP)$ sont parallèles, on trouve :\n$$\n\\widehat{XAP} = \\widehat{XX'Y'} = \\widehat{XYY'} = \\widehat{XYA}\n$$\nDe sorte que la droite $(AP)$ est tangente au cercle circonscrit au triangle $AXY$. On a donc, d'après la puissance d'un point :\n$$\n\\mathrm{PA}^2 = \\mathrm{PX} \\cdot \\mathrm{PY} = \\mathcal{P}_{\\omega}(P)\n$$\nLa position du point $P$ sur la parallèle à $(BC)$ passant par $A$ est uniquement déterminée par la puissance du point $P$ par rapport au cercle $\\omega$, qui ne dépend pas de $X$ et $Y$. C'est ce que nous voulions démontrer.\nSolution:\n\nOn propose une solution alternative utilisant des connaissances avancées. L'énoncé du problème nous invite à considérer le problème dynamiquement, puisque la partie fixe et la partie mobile de la figure a déjà été spécifiée dans l'énoncé. Si la construction de $X$ et $Y$ ne semble pas projective, l'énoncé nous demande en fait de montrer qu'elle l'est. En effet, passer à la seconde intersection avec la droite reliant à un point fixe hors du cercle revient à faire une inversion laissant le cercle globalement invariant et est une transformation projective, c'est même une involution projective du cercle. En fait, toutes les involutions projectives du cercle sont de cette forme : il suffit de prouver que passer de $X$ à $Y$ est une involution projective pour pouvoir en déduire l'existence de ce point $P$. En effet, si on construit $P$ comme $(X_1Y_1)$ intersecté avec $(X_2Y_2)$ pour deux positions de $X$ et $Y$, alors $P$ définit une involution qui, coïncidant en 4 points avec celle de l'énoncé, ne peut être que la même.\n\nMaintenant qu'on a vu que ce problème était fondamentalement dynamique et que le point $P$ n'est qu'une excuse pour demander de démontrer l'existence d'une involution projective, faisons-le. Même si ce n'est initialement pas très visible, il est en fait assez facile de montrer que la transformation de $X$ à $Y$ est projective. Les angles de l'énoncé nous invitent à regarder les cercles $AXB$ et $AYC$, vivant dans le faisceau de cercles ayant pour points fixes respectivement $A, B$ et $A, C$. Comme le cercle $ABX$ a une seule intersection non fixe avec le cercle $\\omega$, il est bien projectif de passer de $X$ à ce cercle. De même de l'autre côté avec $Y$ et l'autre cercle. Enfin il faut prouver qu'on peut passer d'un cercle à l'autre. L'égalité des angles nous donne en fait que le cercle $AYC$ est l'image du cercle $AXB$ par la similitude indirecte de centre $A$ envoyant $B$ sur $C$. La transformation est donc bien projective.\n\nMontrons que c'est une involution. On sait qu'une transformation projective est une involution dès qu'elle échange une paire de points. Cherchons donc un cas simple. Un cas particulièrement simple est celui où l'angle commun est nul : cela correspond au fait que $X, A, B$ soient alignés dans cet ordre et que $Y, A, C$ soient aussi alignés dans cet ordre. La transformation envoie alors bien $X$ sur $Y$. Vérifions qu'elle enverrait aussi $Y$ sur $X$. Cela revient à dire que $\\widehat{AYB} = \\widehat{AXC}$, ce qui est immédiat d'après le théorème de l'angle inscrit.\n\nOn conclut donc que la droite $(XY)$ passe toujours par un point fixe $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30775,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier strictement positif et soient $a, a_{1}, \\ldots, a_{n}$ des entiers strictement positifs. On suppose que pour tout entier $k$ pour lequel l'entier $a k+1$ est un carré parfait, au moins l'un des entiers $a_{1} k+1, \\ldots, a_{n} k+1$ est également un carré parfait.\n\nMontrer qu'il existe un indice $1 \\leqslant i \\leqslant n$ tel que $a=a_{i}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par deux lemmes sur les résidus quadratiques.\n\nLemme 1 : Si $x$ est résidu quadratique modulo $p$ avec $p$ premier impair, il l'est aussi modulo $p^{2}$ (et même $p^{m}$, mais on n'en a pas besoin ici).\n\nDémonstration: En effet, soit $a$ tel que $a^{2} \\equiv x \\pmod{p}$, ce qu'on réécrit sous la forme $a^{2}-x=k p$. Mais alors on obtient $\\left(a-2^{-1} k p\\right)^{2} \\equiv a^{2}-k p \\equiv x\\pmod{p^{2}}$, et $2$ est inversible modulo $p^{2}$ car $p$ est impair. Ainsi, $x$ est bien un résidu quadratique modulo $p^{2}$.\n\nLemme 2: Si $t$ est un entier non nul et $p$ est un nombre premier tel que $p \\equiv 1\\pmod{8 t}$, alors $\\left(\\frac{t}{p}\\right)=1$.\n\nDémonstration : En effet, soit $q$ un diviseur premier de $t$ et $\\alpha=\\nu_{q}(t)$. Si $q=2$, $\\left(\\frac{2}{p}\\right)=1$ car $p \\equiv 1\\pmod{8}$. Donc $q$ est résidu quadratique modulo $p$. Notons maintenant que $\\alpha$ est pair, alors $q^{\\alpha}$ est automatiquement un carré modulo $p$, et si $\\alpha$ est impair, $q^{\\alpha}$ est le produit de $q$ (qui est un carré) avec $q^{\\alpha-1}$ qui est un carré, donc $q^{\\alpha}$ est un carré modulo $p$. Si $q \\geqslant 3$, comme $p \\equiv 1\\pmod{4}$, la loi de réciprocité quadratique nous donne $\\left(\\frac{q}{p}\\right)=\\left(\\frac{p}{q}\\right)=\\left(\\frac{1}{q}\\right)=1$. De même que précédemment, on déduit que $\\left(\\frac{q^{\\alpha}}{p}\\right)=1$. La multiplicativité des symboles de Legendre permet de conclure que $\\left(\\frac{t}{p}\\right)=1$.\n\nOn rappelle également un cas particulier du théorème de la progression arithmétique de Dirichlet : pour tout $t$ entier, il existe une infinité de nombres premiers $p$ tels que $p \\equiv 1 \\pmod{8 t}$.\n\nRevenons à l'exercice. Supposons par l'absurde que tous les $a_{i}$ sont différents de $a$.\n\nPosons $M=\\max \\left\\{a, a_{1}, \\ldots, a_{n}\\right\\} \\geqslant 2$. On construit alors une suite de nombres premiers $p_{1}, \\ldots, p_{n}$ par récurrence comme suit : par le théorème de la progression arithmétique de Dirichlet, il existe au moins un nombre premier $p_{1} \\equiv 1\\pmod{8 a_{1}(a_{1}-a)}$ avec $p_{1}>M$. (On utilise ici le fait que $a_{1}-a \\neq 0$.)\n\nPour $2 \\leqslant i \\leqslant n$, on prend ensuite $p_{i}$ tel que $p_{i} \\equiv 1\\pmod{8 a_{i}(a_{i}-a)}$ avec $p_{i}>p_{i-1}$ (qui existe pour les mêmes raisons que $p_{1}$).\n\nEn particulier les $p_{i}$ sont deux-à-deux distincts, et $p_{i} \\equiv 1 \\pmod{8 a_{i}}$ et $p_{i} \\equiv 1 \\pmod{8(a_{i}-a)}$, de sorte que d'après le deuxième lemme, pour tout $i$, $\\left(\\frac{a_{i}}{p_{i}}\\right)=\\left(\\frac{a_{i}-a}{p_{i}}\\right)=1$.\n\nPar définition, pour tout $i$ il existe $n_{i}$ tel que $n_{i}^{2} \\equiv a_{i}-a\\pmod{p_{i}}$, et quitte à remplacer $n_{i}$ par $n_{i}+p_{i}$, on peut supposer $p_{i}^{2} \\nmid n_{i}^{2}-(a_{i}-a)$ ($p_{i} \\neq 2$). Ainsi, on peut écrire $n_{i}^{2} \\equiv a c_{i} p_{i}+(a_{i}-a)\\pmod{p_{i}^{2}}$ avec $p_{i} \\nmid c_{i}$ puisque $a$ et $p_{i}$ sont premiers entre eux ($a<p_{i}$).\n\nEnsuite, comme $a_{i}$ est inversible modulo $p_{i}$ pour tout $i$, le théorème des restes chinois indique que l'on peut trouver un entier $k$ vérifiant que pour tout $i$, $k \\equiv (c_{i} p_{i}-1) a_{i}^{-1} \\pmod{p_{i}^{2}}$, ou encore $a_{i} k+1 \\equiv c_{i} p_{i} \\pmod{p_{i}^{2}}$.\n\nRemarquons que, par construction,\n$$\na k+1 \\equiv a \\frac{c_{i} p_{i}-1}{a_{i}}+1 \\equiv \\frac{a c_{i} p_{i}-a+a_{i}}{a_{i}} \\equiv \\frac{n_{i}^{2}}{a_{i}} \\equiv a_{i}\\left(\\frac{n_{i}}{a_{i}}\\right)^{2} \\pmod{p_{i}^{2}}\n$$\nOr $\\left(\\frac{a_{i}}{p_{i}}\\right)=1$, donc $a_{i}$ est également un résidu quadratique modulo $p_{i}^{2}$ par le premier lemme, donc on a un entier, mettons $N_{i}$, tel que $a k+1 \\equiv N_{i}^{2}\\pmod{p_{i}^{2}}$.\n\nAlors le théorème des restes chinois nous donne un entier $x$ tel que $x \\equiv N_{i}\\pmod{p_{i}^{2}}$ pour tout $i$ et $x \\equiv 1\\pmod{a}$ (car $a$ est premier avec les $p_{i}$). Mais alors $x^{2} \\equiv N_{i}^{2} \\equiv a k+1\\pmod{p_{i}^{2}}$ pour tout $i$, et $x^{2} \\equiv 1^{2} \\equiv a k+1\\pmod{a}$, donc $x^{2} \\equiv a k+1\\pmod{a P^{2}}$, où on a posé $P=\\prod_{i=1}^{n} p_{i}$.\n\nOn réécrit maintenant la congruence sous la forme $x^{2}=a k+1+m a P^{2}=a(P^{2} m+k)+1$. On utilise enfin l'hypothèse de l'énoncé : il existe un indice $j$ tel que $a_{j}(P^{2} m+k)+1$ soit un carré parfait. Mais alors $a_{j}(P^{2} m+k)+1 \\equiv a_{j} k+1 \\equiv c_{j} p_{j}\\pmod{p_{j}^{2}}$, et comme $p_{j} \\nmid c_{j}$, $v_{p_{j}}(a_{j}(P^{2} m+k)+1)=1$, ce qui est en contradiction avec le fait que c'est un carré parfait. Notre affirmation de départ était donc fausse, et l'un des $a_{i}-a$ est nul, c'est-à-dire qu'il existe $i$ tel que $a=a_{i}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30776,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUne suite réelle $a_{1}, \\ldots, a_{k}$ est casable dans l'intervalle $[b, c]$ si il existe des réels $x_{0}, \\ldots, x_{k}$ dans $[b, c]$ tels que $\\left|x_{i}-x_{i-1}\\right|=a_{i}$ pour $k \\geqslant i \\geqslant 1$. La suite est normalisée si ses termes sont tous inférieurs ou égaux à 1.\n\n1) Montrer que pour tout entier $n \\geqslant 1$, toute suite normalisée de longueur $2 \\mathfrak{n}+1$ est casable dans $[0, 2-\\frac{1}{2^{n}}]$.\n\n2) Montrer que pour tout entier $n \\geqslant 1$, il existe une suite normalisée de longueur $4 n+3$ qui n'est pas casable dans $[0,2-\\frac{1}{2^{n}}]$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1) Dans la suite, on dira qu'une suite de $(x_{i})$ satisfaisant la propriété de l'énoncé case la suite $(a_{i})$. On commence par se donner une idée du problème en essayant de montrer l'énoncé pour des petites valeurs de $n$. Si $n=0$, il faut montrer qu'existent $x_{0}, x_{1} \\in[0,1]$ tels que $\\left|x_{1}-x_{0}\\right|=a_{1}$. On peut par exemple prendre $x_{0}=0$ et $x_{1}=a_{1}$. Si l'on cherche à montrer l'énoncé pour $n=1$, on se retrouve rapidement à devoir distinguer plusieurs cas, ce qui nous pousse à vouloir procéder par récurrence sur $n$.\n\nAvant d'attaquer la récurrence, il est important de noter l'idée suivante : si l'ensemble $\\{x_{0}, \\ldots, x_{2 n+1}\\}$ case la suite $(a_{i})$ dans l'intervalle $[b, c]$, alors l'ensemble $\\{x_{0}+d, \\ldots, x_{2 n+1}+d\\}$ case la suite $(a_{i})$ dans l'intervalle $[b+d, c+d]$. Ainsi, il suffit de montrer que la suite $(a_{i})$ est casable dans un intervalle de longueur $2-\\frac{1}{2^{n}}$.\n\nNotons donc $\\mathcal{P}_{\\mathfrak{n}}$ la propriété suivante : toute suite $(a_{i})$ normalisée de longueur $2 \\mathfrak{n}+1$ est casable dans un intervalle de longueur $2-\\frac{1}{2^{n}}$. On a déjà montré que la propriété $\\mathcal{P}_{0}$ est vraie, on suppose maintenant que $\\mathcal{P}_{2 n-1}$ est vraie pour un certain $n \\geqslant 1$ fixé.\n\nSoit $(a_{i})$ une suite normalisée de longueur $2 n+1$. D'après notre hypothèse de récurrence, la suite $(a_{i})_{i \\leqslant 2 n-1}$ est casable dans un intervalle de longueur $2-\\frac{1}{2^{n-1}}$. D'après notre remarque ci-dessus, on peut supposer que cet intervalle est $[0,2-\\frac{1}{2^{n-1}}]$. On dispose donc de $x_{0}, \\ldots, x_{2 n-1}$ dans $[0,2-\\frac{1}{2^{n-1}}]$ tels que $\\left|x_{i}-x_{i-1}\\right|=a_{i}$ pour $2 n-1 \\geqslant i \\geqslant 1$. Notons que si $(x_{i})$ convient, alors $(2-1/2^{n-1}-x_{i})$ vérifie les mêmes propriétés que la suite $(x_{i})$. On peut donc supposer, quitte à remplacer $(x_{i})$ par $(2-1/2^{n-1}-x_{i})$, que $x_{2 n-1} \\leqslant 1-\\frac{1}{2^{n}}$.\n\nDésormais, nous allons montrer qu'il existe $x_{2 n}$ et $x_{2 n+1}$ qui appartiennent tous les deux soit à l'intervalle $[0,2-1/2^{n}]$ soit à l'intervalle $[-1/2^{n}, 2-1/2^{n-1}]$ tels que $\\left|x_{2 n}-x_{2 n-1}\\right|=a_{2 n}$ et $\\left|x_{2 n+1}-x_{2 n}\\right|=a_{2 n+1}$. Dans le premier cas, on aura obtenu une suite $(x_{i})$ qui case la suite $(a_{i})$ dans l'intervalle $[0,2-1/2^{n}]$. Dans le deuxième cas, on aura obtenu une suite $(x_{i})$ qui case la suite $(a_{i})$ dans l'intervalle $[-1/2^{n}, 2-1/2^{n-1}]$, qui est de longueur $2-1/2^{n}$. On aura gagné dans les deux cas.\n\nConsidérons les deux quantités $A=x_{2 n-1}+a_{2 n}+a_{2 n+1}$ et $B=x_{2 n-1}+a_{2 n}-a_{2 n+1}$. Notons qu'on ne peut avoir $A>2-1/2^{n}$ et $B<-1/2^{n}$, sans quoi on aurait $a_{2 n+1}>1$. On a donc $A \\leqslant 2-1/2^{n}$ ou $B \\geqslant -1/2^{n}$.\n\nSi $A \\leqslant 2-1/2^{n}$, on pose $x_{2 n}=x_{2 n-1}+a_{2 n}$ et $x_{2 n+1}=A$. On vérifie que la suite $(x_{i})$ case la suite $(a_{i})$ dans l'intervalle $[0,2-1/2^{n}]$.\n\nOn suppose que $B \\geqslant -1/2^{n}$. On pose $x_{2 n}=x_{2 n-1}+a_{2 n}$ et $x_{2 n+1}=B$. Si $x_{2 n} \\leqslant 2-1/2^{n-1}$, alors $B \\in [-1/2^{n}, 2-1/2^{n-1}]$ et la suite $(x_{i})$ case la suite $(a_{i})$ dans l'intervalle $[-1/2^{n}, 2-1/2^{n-1}]$. Si $x_{2 n}>2-1/2^{n-1}$, alors $B>2-1/2^{n-1}-1 \\geqslant 0$, si bien que la suite $(x_{i})$ case la suite $(a_{i})$ dans l'intervalle $[0,2-1/2^{n}]$.\n\nCette disjonction de cas achève la récurrence.\n\n2) L'idée est la suivante : on souhaite construire des $a_{i}$ de sorte à forcer les $x_{i}$ à se trouver au centre de l'intervalle. Pour cela, on souhaite que les $a_{i}$ soient très proches de 1, afin d'écarter deux $x_{i}$ l'un de l'autre, et on souhaite également qu'ils ne soient pas trop proches de 1, pour éloigner les $x_{i}$ des bornes de l'intervalle. Cette discussion motive la construction qui suit.\n\nSoit $N=3 \\cdot 2^{n}-1$. Considérons la suite suivante :\n$$\n1,1-\\frac{1}{N}, 1,1-\\frac{2}{N}, 1, \\ldots, 1-\\frac{2^{2}}{N}, 1, \\ldots, 1-\\frac{2^{n}}{N}, 1, \\ldots, 1-\\frac{2}{N}, 1,1-\\frac{1}{N}, 1\n$$\nNous allons montrer que cette suite n'est pas casable dans l'intervalle $]-1+1/2N, 1-1/2N[$, qui est un intervalle de longueur $2-1/N>2-1/2^{n}$, ce qui permettra de conclure. On suppose par l'absurde qu'il existe une suite $(x_{i})$ qui case la suite $(a_{i})$ donnée dans l'intervalle $]-1+1/2N, 1-1/2N[$ et on cherche à aboutir à une contradiction dans les estimations sur les $x_{i}$.\n\nQuitte à remplacer $(x_{i})$ par $(-x_{i})$, on peut supposer que $x_{0} \\geqslant 0$. Puisque $\\left|x_{1}-x_{0}\\right|=1$, on doit avoir $x_{1}=x_{0}-1$, et $x_{1} \\leqslant 1-1/2N-1=-1/2N$, soit $\\left|x_{1}\\right|>1/2N$. De manière générale, on va montrer pour $i \\leqslant n$ que\n$$\n\\left|x_{2i}\\right|<1-\\frac{2^{i+1}-1}{2N} \\quad, \\quad \\left|x_{2i+1}\\right|>\\frac{2^{i+1}-1}{2N}\n$$\nLa preuve s'effectue par récurrence sur $i$. On a montré le résultat pour $i=0$. Supposons le vrai pour $i \\geqslant 0$ fixé. On a $\\left|x_{2i+2}-x_{2i+1}\\right|=a_{2i+2}=1-2^{i}/N$. Si $x_{2i+1}>0$, on a $x_{2i+1}+(1-2^{i}/N)>1-\\frac{2^{i}-1}{2N}>1-\\frac{1}{2N}$. On a donc forcément $x_{2i+2}=x_{2i+1}-(1-2^{i}/N)$. Comme $1-2^{i}/N>(2^{i+1}-1)/2N$, on a\n$$\n\\left|x_{2i+2}\\right|=1-\\frac{2^{i}}{N}-x_{2i+1}<1-\\frac{2^{i+2}-1}{2N}\n$$\nOn procède de même dans le cas où $x_{2i+1}<0$.\n\nOn a ensuite $\\left|x_{2i+3}-x_{2i+2}\\right|=a_{2i+3}=1$. Un raisonnement identique au précédent nous permet d'avoir, dans le cas où $x_{2i+2}>0$ :\n$$\n\\left|x_{2i+3}\\right|=1-x_{2i+2}>1-\\left(1-\\frac{2^{i+2}-1}{2N}\\right)=\\frac{2^{i+2}-1}{2N}\n$$\net on procède de même dans le cas où $x_{2i+2}<0$.\n\nOn procède de même pour montrer par récurrence que si $i \\geqslant n+1$ :\n$$\n\\left|x_{2i}\\right|<\\frac{2^{2(n+1)-i}-1}{2N} \\quad, \\quad \\left|x_{2i+1}\\right|>1-\\frac{2^{2(n+1)-i}-1}{2N}\n$$\nMais alors $\\left|x_{4n+3}\\right|>1-\\frac{2-1}{2N}>1-\\frac{1}{2N}$, ce qui contredit notre hypothèse de départ et permet de conclure.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30777,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les polynômes $P \\in \\mathbb{Z}[X]$ tels que :\n\n(i) $P(n) \\geqslant 1$ pour tout $n \\geqslant 1$\n\n(ii) $P(mn)$ et $P(m) P(n)$ ont le même nombre de diviseurs premiers pour tous $m, n \\geqslant 1$.",
    "options": [],
    "answer": "All polynomials of the form a·X^d with a a positive integer and d a nonnegative integer.",
    "solution": "Solution:\n\nDans la suite, pour tout entier $n \\in \\mathbb{N}^{*}$, on note $\\mathcal{D}(n)$ l'ensemble des diviseurs premiers de $n$, et $\\delta(n)=|\\mathcal{D}(n)|$. Soit $P \\in \\mathbb{Z}[X]$ un polynôme vérifiant les deux conditions.\n\nOn commence par montrer le résultat suivant :\n\nLemme : Soit $n \\geqslant 1$. Pour tout $k \\in \\mathbb{N}^{*}, \\mathcal{D}\\left(P\\left(n^{k}\\right)\\right)=\\mathcal{D}(P(n))$.\n\nDémonstration: On montre d'abord le résultat pour des petites valeurs de $k$.\n\nTout d'abord, par hypothèse, $\\delta\\left(P\\left(n^{2}\\right)\\right)=\\delta\\left(P(n)^{2}\\right)=\\delta(P(n))$. De même, $\\delta\\left(P\\left(n^{4}\\right)\\right)=\\delta\\left(P\\left(n^{2}\\right)\\right)=\\delta(P(n))$. Mais $\\delta\\left(P\\left(n^{4}\\right)\\right)=\\delta\\left(P\\left(n^{3}\\right) P(n)\\right)$. $P\\left(n^{4}\\right)$ ayant le même nombre de diviseurs premiers que $P(n)$, cette dernière égalité assure que $\\mathcal{D}\\left(P\\left(n^{3}\\right)\\right) \\subset \\mathcal{D}(P(n))$. Par ailleurs $\\delta\\left(P\\left(n^{3}\\right)\\right)=\\delta\\left(P\\left(n^{2}\\right) P(n)\\right) \\geqslant \\delta(P(n))$. Par conséquent, l'inclusion précédente est saturée par cardinalité, et $\\mathcal{D}\\left(P\\left(n^{3}\\right)\\right)=\\mathcal{D}(P(n))$. Alors $\\delta(P(n))=\\delta\\left(P\\left(n^{3}\\right)\\right)=\\delta\\left(P\\left(n^{2}\\right) P(n)\\right)$, donc nécessairement $\\mathcal{D}\\left(P\\left(n^{2}\\right)\\right) \\subset \\mathcal{D}(P(n))$. Par égalité des cardinaux, $\\mathcal{D}(P(n))=\\mathcal{D}\\left(P\\left(n^{2}\\right)\\right)=\\mathcal{D}\\left(P\\left(n^{3}\\right)\\right)$.\n\nSoit maintenant $k \\geqslant 3$. On montre l'égalité $\\mathcal{D}\\left(P\\left(n^{k}\\right)\\right)=\\mathcal{D}(P(n))$ par récurrence. Supposons que $\\mathcal{D}(P(n))=\\mathcal{D}\\left(P\\left(n^{2}\\right)\\right)=\\ldots=\\mathcal{D}\\left(P\\left(n^{k}\\right)\\right)$. Alors $\\delta\\left(P\\left(n^{k+1}\\right)\\right)=\\delta\\left(P\\left(n^{k}\\right) P(n)\\right)$, où par hypothèse de récurrence, $P\\left(n^{k}\\right)$ et $P(n)$ ont exactement les mêmes diviseurs premiers, donc $\\delta\\left(P\\left(n^{k}\\right) P(n)\\right)=\\delta(P(n))$, et $\\delta\\left(P\\left(n^{k+1}\\right)\\right)=\\delta(P(n))$. Alors d'une part, $\\delta\\left(P\\left(n^{k+2}\\right)\\right)=\\delta\\left(P\\left(n^{k}\\right) P\\left(n^{2}\\right)\\right)=\\delta\\left(P\\left(n^{2}\\right)\\right)$ puisque $\\mathcal{D}\\left(P\\left(n^{k}\\right)\\right)=\\mathcal{D}(P(n))$ (vu que $k \\geqslant 3$ ) d'où $\\delta\\left(P\\left(n^{k+2}\\right)\\right)=\\delta(P(n))$. Mais d'autre part, avec l'égalité $\\delta\\left(P\\left(n^{k+2}\\right)\\right)=\\delta\\left(P\\left(n^{k+1}\\right) P(n)\\right)$, on voit que nécessairement, $\\mathcal{D}\\left(P\\left(n^{k+1}\\right)\\right) \\subset \\mathcal{D}(P(n))$. Par égalité des cardinaux, on a bien $\\mathcal{D}\\left(P\\left(n^{k+1}\\right)\\right)=\\mathcal{D}(P(n))$, ce qui achève l'hérédité.\n\nOn utilise désormais l'égalité $\\mathcal{D}\\left(P\\left(n^{k}\\right)\\right)=\\mathcal{D}(P(n))$ pour conclure l'exercice. On écrit $P=X^{d} Q$, où $d \\in \\mathbb{N}$ et $Q \\in \\mathbb{Z}[X]$ vérifie $Q(0) \\neq 0$. Clairement $Q(n) \\geqslant 1$ si $n \\geqslant 1$. Fixons $p$ premier ne divisant pas $Q(0)$. Pour $k \\geqslant 1, p \\nmid Q\\left(p^{k}\\right)$ donc $\\mathcal{D}\\left(Q\\left(p^{k}\\right)\\right)=\\mathcal{D}(Q(p))$ d'après le point précédent. En effet, si $d=0$ cela correspond à l'égalité démontrée plus haut, et si $d \\geqslant 1,\\{p\\} \\cup \\mathcal{D}\\left(Q\\left(p^{k}\\right)\\right)=\\mathcal{D}\\left(P\\left(p^{k}\\right)\\right)=\\mathcal{D}(P(p))=\\{p\\} \\cup \\mathcal{D}(Q(p))$ où $p \\nmid Q\\left(p^{k}\\right)$ et $p \\nmid Q(p)$.\n\nSupposons par l'absurde que $Q$ est non constant. $Q$ tend donc vers $+\\infty$ à l'infini. Alors $\\delta(Q(p)) \\geqslant 1$ car dans le cas contraire $Q\\left(p^{k}\\right)$ serait égal à 1 pour tout $k \\geqslant 1$, donc $Q(X)-1$ admettrait une infinité de racines et serait donc constant. Notons $\\mathcal{D}(Q(p))=\\{q_{1}, \\ldots, q_{s}\\}$ où $s \\geqslant 1$ et $q_{1}<\\ldots<q_{s}$ des nombres premiers. Soit $\\alpha=\\max _{r \\text{ premier }} \\nu_{r}(Q(1)) \\geqslant 0$. Il existe $k_{0} \\geqslant 1$ tel que pour tout $k \\geqslant k_{0}$, il existe $i(k) \\in \\llbracket 1, s \\rrbracket$ avec $v_{q_{i(k)}}\\left(Q\\left(p^{k}\\right)\\right) \\geqslant \\alpha+1$ En effet, comme $Q$ tend vers $+\\infty$ en $\\infty$, pour $k$ assez grand, $Q\\left(p^{k}\\right)$ est strictement supérieur à $\\left(q_{1} \\ldots q_{s}\\right)^{\\alpha}$. Soit $\\beta=\\varphi\\left(\\left(q_{1} \\ldots q_{s}\\right)^{\\alpha+1}\\right) \\geqslant 1$, et $N \\geqslant 1$ tel que $N \\beta \\geqslant k_{0}$. Alors $q_{i(N \\beta)}^{\\alpha+1} \\mid Q\\left(p^{N \\beta}\\right)$, mais $p^{N \\beta} \\equiv 1\\left(\\bmod \\left(q_{1} \\ldots q_{s}\\right)^{\\alpha+1}\\right)$ puisque $p \\notin\\{q_{1}, \\ldots, q_{s}\\}$. En particulier, $Q\\left(p^{N \\beta}\\right) \\equiv Q(1)\\left(\\bmod q_{i(N \\beta)}^{\\alpha+1}\\right)$, et donc $q_{i(N \\beta)}^{\\alpha+1} \\mid Q(1)$. Cela est en contradiction avec la maximalité de $\\alpha$. Ainsi, $Q$ est constant égal à $a \\in \\mathbb{N}^{*}$, et $P=a X^{d}$ avec $d \\in \\mathbb{N}$.\n\nOn vérifie réciproquement que les polynômes de la forme $a X^{d}$ avec $d \\in \\mathbb{N}$ et $a \\in \\mathbb{N}^{*}$ sont solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn graphe $G$ fini simple à $n$ sommets est dit divisible s'il est possible d'attribuer à chaque sommet $s$ de $G$ un numéro $n_{s}$ de sorte que deux sommets distincts possèdent toujours deux numéros distincts et deux sommets quelconques $s$ et $s^{\\prime}$ sont reliés par une arête si et seulement si $n_{s} \\mid n_{s^{\\prime}}$ ou $n_{s^{\\prime}} \\mid n_{s}$.\n\nUn graphe $G$ fini simple à $n$ sommets est dit permutable s'il est possible de numéroter les sommets de $G$ de $1$ à $n$ et s'il existe une permutation $\\sigma$ de $\\{1, \\ldots, n\\}$ telle que les sommets ayant pour numéros $i$ et $j$ sont reliés par une arête si et seulement si $(i-j)(\\sigma(i)-\\sigma(j))<0$.\n\nMontrer qu'un graphe $G$ est permutable si et seulement si $G$ et son complémentaire sont divisibles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUn graphe est de divisibilité ssi on peut l'orienter (dans le sens des divisibilités) de manière à ce que s'il y a un arc (arête orientée) de $A$ vers $B$ et un de $B$ vers $C$, il y en a un de $A$ vers $C$. En effet, on peut alors attribuer à chaque sommet un nombre premier différent, et le nombre attribué pour les relations de divisibilité est alors le produit de son nombre premier par ceux de chacun des sommets ayant un arc arrivant sur lui. On a alors les bonnes relations de divisibilité : s'il y a un arc de $A$ vers $B$, tout premier divisant le nombre attribué à $A$ est le premier d'un sommet $C$ avec un arc de $C$ vers $A$. Il y a alors aussi un arc de $C$ vers $B$, donc le premier divise aussi le nombre attribué à $B$.\n\nMontrons qu'on peut orienter un graphe de permutation. On trace un arc de $a$ vers $b$ ssi $a<b$ et $\\sigma(a)>\\sigma(b)$. On a bien les arêtes du graphe de permutation attribué et la relation est transitive, on a donc bien un graphe de divisibilité. Son complémentaire est également de divisibilité car orientable en traçant un arc de $a$ vers $b$ ssi $a<b$ et $\\sigma(a)<\\sigma(b)$.\n\nRéciproquement, on suppose un graphe orienté transitif dont le complémentaire est également orientable de manière transitive, montrons qu'il est de permutation. Choisir une numérotation et une permutation revient en fait à choisir librement deux numérotations des sommets, soit à ordonner deux fois les sommets. Deux sommets doivent alors être reliés si et seulement si leur position relative est différente selon les deux ordres. Pour le premier ordre, on place les éléments un à un, à chaque étape en rajoutant un élément n'ayant pas de prédécesseur selon le graphe complémentaire orienté. On considère donc tous les choix possibles, tous les sommets sans prédécesseur non choisi. Alors, tous ces éléments considérés forment une clique du graphe, il y en a donc un unique qui est sans antécédents dans cette clique pour le graphe orienté. C'est cet élément qu'on place en tout premier pour notre ordre. On l'enlève ensuite et on recommence jusqu'à ce que tous les sommets soient ordonnés. Pour le deuxième ordre, on fait presque la même chose. On rajoute aussi à chaque étape un sommet sans prédécesseur selon le graphe complémentaire, mais parmi cette clique de candidats potentiels, on va prendre celui sans successeur dans la clique.\n\nVérifions que cette construction fonctionne. Prenons $A$ et $B$ deux sommets non reliés. Supposons SPDG que dans le graphe complémentaire, $A$ est un prédécesseur de $B$. Si $B$ est choisi pour un des deux ordres, forcément $A$ n'a pas été considéré parmi les choix potentiels, soit il a déjà été choisi, soit il a lui-même un prédécesseur non choisi. Par transitivité, ce prédécesseur serait aussi un prédécesseur de $B$, ce qui empêcherait le choix de $B$. Donc pour les deux ordres, $A$ a été choisi avant $B$. Les deux ordres sont donc compatibles avec l'absence d'arêtes entre $A$ et $B$.\n\nPrenons $A$ et $B$ reliés et supposons SPDG que $A$ est un antécédent de $B$ dans le graphe. Montrons d'abord que pour le premier ordre, quand $B$ est choisi, $A$ a déjà été choisi. À ce moment-là, si $A$ n'avait pas déjà été choisi, il n'aurait pas pu être considéré comme choix potentiel, sinon il aurait été choisi préférentiellement à $B$. Donc il avait un antécédent non choisi dans le graphe complémentaire. On remonte des antécédents successifs en partant de $A$ jusqu'à arriver jusqu'à un sommet $C$, sans antécédents encore non choisis. Par transitivité du graphe complémentaire orienté, $C$ est un antécédent de $A$ pour ce graphe complémentaire. $C$ a été considéré comme choix potentiel, n'ayant pas d'antécédent dans le graphe complémentaire, mais $B$ lui a été préféré, donc $B$ est un antécédent de $C$ pour le graphe normal (pas complémentaire). $A$ étant lui-même un antécédent de $B$, par transitivité, $A$ doit être un antécédent de $C$ dans le graphe normal donc une arête entre $A$ et $C$ existe. Or on a construit $C$ comme un antécédent de $A$ dans le graphe complémentaire, donc une arête du graphe complémentaire existe en $A$ et $C$. La contradiction montre que selon le premier ordre, $A$ arrive avant $B$. On prouve identiquement que selon le deuxième ordre, $B$ arrive avant $A$. Ces deux ordres sont donc compatibles avec l'existence d'une arête entre $A$ et $B$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30779,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $a, b, c$ trois entiers tels que $7$ divise $a^{2} + b^{2} + c^{2}$. Montrer que $7$ divise $a^{4} + b^{4} + c^{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLes carrés modulo $7$ valent $0, 1, 2, 4$. On cherche donc toutes les sommes de trois de ces nombres étant divisible par $7$. Après avoir testé toutes les possibilités, on s'aperçoit que les seuls ensembles de trois carrés modulo $7$ dont la somme est divisible par $7$ sont $\\{0, 0, 0\\}$ et $\\{1, 2, 4\\}$.\n\nOn a alors $a^{4} + b^{4} + c^{4} \\equiv 0^{2} + 0^{2} + 0^{2} \\equiv 0 \\pmod{7}$ ou $a^{4} + b^{4} + c^{4} \\equiv 1^{2} + 2^{2} + 4^{2} \\equiv 1 + 4 + 16 \\equiv 21 \\equiv 0 \\pmod{7}$.\n\nDans les deux cas, $a^{4} + b^{4} + c^{4} \\equiv 0 \\pmod{7}$, comme désiré.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont les angles sont aigus et soit $D$ un point situé à l'intérieur du triangle $ABC$. Les droites $(AD)$ et $(BD)$ recoupent le cercle circonscrit au triangle $ABC$ respectivement en les points $A_{1}$ et $B_{1}$. Le cercle circonscrit au triangle $B_{1}DA$ recoupe la droite $(AC)$ au point $P$. Le cercle circonscrit au triangle $A_{1}BD$ recoupe la droite $(BC)$ au point $Q$.\n\nDémontrer que le quadrilatère $CPDQ$ est un parallélogramme.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUne chasse aux angles de droites indique que\n$$\n(QD, CB) = (QD, QB) = (A_{1}D, A_{1}B) = (A_{1}A, A_{1}B) = (CA, CB),\n$$\nce qui signifie que les droites $(QD)$ et $(AC)$ sont parallèles l'une à l'autre. On démontre de même que\n$$\n(PD, CA) = (PD, PA) = (B_{1}D, B_{1}A) = (B_{1}B, B_{1}A) = (CB, CA)\n$$\nce qui signifie que les droites $(PD)$ et $(BC)$ sont parallèles l'une à l'autre. Le quadrilatère $CPDQ$ est donc un parallélogramme.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier tel que $n \\geqslant 3$. Lucie dispose $n$ feuilles blanches en cercle, puis elle écrit un nombre réel sur chaque feuille. Trouver tous les entiers $k$ tels que $1 \\leqslant k \\leqslant n$ pour lesquels la propriété suivante est vraie :\nSi au moins un des nombres qu'a écrits Lucie n'est pas nul, il existe $k$ feuilles consécutives dont la somme des nombres n'est pas nulle.",
    "options": [],
    "answer": "Exactly those integers k between 1 and n that are coprime to n",
    "solution": "Solution:\n\nDans la suite, on numérote les feuilles de 1 à $n$, dans le sens des aiguilles d'une montre, et on note $x_{i}$ le nombre que Lucie écrit sur la feuille $i$. Les indices seront considérés modulo $n$. L'énoncé revient donc à trouver les entiers $k$ pour lesquels il est impossible que toute somme $\\mathcal{S}_{a}=x_{a+1}+x_{a+2}+\\ldots+x_{a+k}$ soit nulle si les $x_{i}$ ne sont pas tous nuls.\n\nTout d'abord, s'il existe un entier $d \\geqslant 2$ qui divise à la fois $k$ et $n$, Lucie peut choisir les nombres $x_{i}$ de sorte que $x_{i}=1$ si $i \\equiv 1(\\bmod d)$, $x_{i}=-1$ si $i \\equiv 2(\\bmod d)$, et $x_{i}=0$ sinon. Par la suite, toute somme $\\mathcal{S}_{a}=x_{a+1}+x_{a+2}+\\ldots+x_{a+k}$ est une somme de $k / d$ nombres égaux à $+1$, de $k / d$ nombres égaux à $-1$, et de nombres égaux à $0$. Ainsi, $\\mathcal{S}_{a}=0$, et ce quel que soit $a$. Par conséquent, l'entier $k$ ne figure pas parmi les entiers recherchés.\n\nAu contraire, si $k$ est premier avec $n$, et si toute somme $\\mathcal{S}_{a}$ est nulle, nous allons démontrer que $x_{1}=x_{2}=\\ldots=x_{n}=0$. En effet, pour tout indice $a$, on constate que\n$$\nx_{a}=x_{a}+\\mathcal{S}_{a}=x_{a}+x_{a+1}+\\ldots+x_{a+k}=\\mathcal{S}_{a-1}+x_{a+k}=x_{a+k}\n$$\nOn en déduit de même que $x_{a}=x_{a+k \\ell}$ pour tous les entiers $a$ et $\\ell$. Or, puisque $k$ est premier avec $n$, le théorème de Bézout indique qu'il existe un entier $m$ tel que $k m \\equiv 1(\\bmod n)$. Par conséquent, pour tout entier $a$, on sait que\n$$\nx_{0}=x_{a k m}=x_{a}\n$$\nAinsi, les nombres $x_{i}$ sont tous égaux à $x_{0}$, et chaque somme $\\mathcal{S}_{a}$ est égale à $k x_{0}$. Si $\\mathcal{S}_{a}=0$, c'est donc que $x_{0}=0$, et que les nombres $x_{i}$ sont tous nuls.\n\nEn conclusion, les entiers $k$ recherchés sont les entiers premiers avec $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un parallélogramme et soit $G$ le centre de gravité du triangle $ABD$. Soit $P$ et $Q$ les points de la droite $(BD)$ tels que les droites $(GP)$ et $(PC)$ sont perpendiculaires et les droites $(GQ)$ et $(QC)$ sont perpendiculaires.\n\nDémontrer que la droite $(AG)$ est la bissectrice de l'angle $\\widehat{PAQ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, puisque $ABCD$ est un parallélogramme, la médiane $(AG)$ du triangle $ABD$ et la diagonale $(AC)$ passent toutes deux par le milieu $N$ de $[BD]$.\n\nPuisque $CGP$ et $CGQ$ sont rectangles en $P$ et en $Q$, les points $P$ et $Q$ sont les deux points d'intersection de $(BD)$ avec le cercle de diamètre $[CG]$. Puisque l'on trace ce cercle, on note donc $M$ son centre, c'est-à-dire le milieu de $[GC]$. En particulier, les cinq points $A, G, N, M$ et $C$ sont alignés.\n\nLe moyen le plus sûr de démontrer que $(AG)$ est la bissectrice de $\\widehat{PAQ}$ est alors d'invoquer le théorème du pôle Sud, c'est-à-dire de démontrer que $(AG)$ passe par le milieu de l'arc $\\overparen{PQ}$ sur le cercle circonscrit à $APQ$.\n\nOn entreprend donc de tracer le cercle circonscrit à $APQ$, et on constate visuellement qu'il passe par $M$. Puisque $PM = QM$, il suffit donc effectivement de démontrer que $A, Q, M$ et $P$ sont cocycliques. La figure suggère alors très fortement d'invoquer la puissance de $N$. En effet, par rapport au cercle circonscrit à $PCQG$, cette puissance vaut $\\overline{NP} \\cdot \\overline{NQ} = \\overline{NG} \\cdot \\overline{NC}$. Par ailleurs, on sait que $N = (A + C)/2$, que $M = (G + C)/2$ et que $G = (2N + A)/3$. On en déduit que $\\overline{NA} \\cdot \\overline{NM} = -3 NG^{2} = \\overline{NG} \\cdot \\overline{NC} = \\overline{NP} \\cdot \\overline{NQ}$.\n\nCela signifie bien que les points $A, Q, M$ et $P$ sont cocycliques, et donc que $M$ est le pôle Sud de $APQ$ issu de $A$, ce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAline et Théo jouent au jeu suivant. Tout d'abord, Théo choisit des nombres premiers $p_{1}, p_{2}, \\ldots$ Puis Aline choisit des entiers naturels non nuls $n_{1}, n_{2}, \\ldots$ deux à deux distincts. Théo choisit ensuite deux entiers $k \\geqslant 1$ et $\\ell \\geqslant 1$. Il gagne la partie s'il existe une infinité d'entiers $n_{i}$ qui ne sont pas divisibles par $p_{k}^{\\ell}$. Sinon, c'est Aline qui gagne.\n\na) Qui, d'Aline ou de Théo, dispose d'une stratégie gagnante?\n\nb) Pour corser encore le jeu, Aline a maintenant une contrainte supplémentaire : chaque entier $n_{i}$ ne peut diviser qu'un nombre fini d'entiers $n_{j}$. Avec ces nouvelles règles, qui, d'Aline ou de Théo, dispose d'une stratégie gagnante?",
    "options": [],
    "answer": "a) Aline. b) Théo.",
    "solution": "Solution:\n\na) C'est Aline qui dispose d'une stratégie gagnante. Une telle stratégie consiste, par exemple, à choisir\n$$\nn_{i}=\\left(p_{1} \\times p_{2} \\times \\cdots \\times p_{i}\\right)^{i}\n$$\npour tout $i$. En effet, tout entier $p_{k}^{\\ell}$ divise chacun des entiers $\\left(p_{1} \\times p_{2} \\times \\cdots \\times p_{i}\\right)^{i}=n_{i}$ pour lesquels $i \\geqslant \\max \\{k, \\ell\\}$.\n\nb) Cette fois, c'est Théo qui dispose d'une stratégie gagnante. Pour ce faire, il commence par définir $p_{i}$ comme le $i^{\\text{ème}}$ nombre premier. Puis, une fois qu'Aline a choisi les entiers $n_{i}$, Théo considère la factorisation de $n_{1}$ en produits de facteurs premiers :\n$$\nn_{1}=p_{1}^{\\alpha_{1}} \\times p_{2}^{\\alpha_{2}} \\times \\cdots \\times p_{t}^{\\alpha_{t}}\n$$\nou chaque exposant $\\alpha_{i}$ est un entier naturel.\n\nSupposons que, pour chaque entier $j \\leqslant t$, l'ensemble $\\mathcal{E}_{j}$ des entiers $i$ pour lesquels $p_{j}^{\\alpha_{j}}$ ne divise pas $n_{i}$ soit un ensemble fini. Tout entier $i$ pour lequel $n_{1}$ ne divise pas $n_{i}$ appartient alors à l'ensemble\n$$\n\\mathcal{E}_{1} \\cup \\mathcal{E}_{2} \\cup \\ldots \\cup \\mathcal{E}_{t}\n$$\nque l'on note $\\mathcal{E}$ et qui est lui-même fini. Mais alors il existe une infinité d'entiers $i$ qui n'appartiennent pas à $\\mathcal{E}$, et donc tels que $n_{1}$ divise $n_{i}$. Ceci est absurde, ce qui invalide notre supposition.\n\nIl suffit donc pour Théo de choisir un entier $j \\leqslant t$ pour lequel l'ensemble $\\mathcal{E}_{j}$ est infini, puis de choisir $k=j$ et $\\ell=\\alpha_{j}+1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLe gouvernement de Bosnie-Herzégovine a décidé de mettre en place un nouveau système de plaques d'immatriculations. Chaque plaque d'immatriculation devra contenir 8 chiffres, chacun pouvant valoir $0,1,2,3,4,5,6,7,8$ ou $9$. En outre, deux plaques d'immatriculation distinctes devront toujours avoir au moins deux chiffres différents. Par exemple, s'il met en circulation la plaque $00000000$, le gouvernement ne pourra pas mettre en circulation la plaque $00010000$.\n\nTrouver le nombre maximum de plaques d'immatriculation que le gouvernement peut mettre en circulation.",
    "options": [],
    "answer": "10000000",
    "solution": "Solution:\n\nNous allons montrer que le nombre maximal de plaques disponibles est égal à $10^{7}$.\n\nTout d'abord, on dit que deux plaques font partie de la même famille si leurs sept premiers chiffres sont différents. Chaque famille contient dix plaques, donc il y a $10^{7}$ familles distinctes. Or, le gouvernement ne peut pas mettre en circulation plus d'une plaque par famille. Il met donc en circulation au plus $10^{7}$ plaques.\n\nRéciproquement, il peut se débrouiller pour créer $10^{7}$ plaques en procédant comme suit : le dernier chiffre de chaque plaque est égal à l'opposé de la somme, modulo $10$, des sept premiers chiffres de la plaque. Ainsi, il crée bien une plaque par famille.\n\nEn outre, la somme des chiffres de chaque plaque est divisible par $10$. Mais si deux plaques ont exactement un chiffre de différence, les sommes de leurs chiffres ne peuvent être égales l'une à l'autre (modulo $10$). Ainsi, deux plaques distinctes ont toujours au moins deux chiffres de différence, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $a$ un réel strictement positif et $n \\geqslant 1$ un entier. Montrer que\n$$\n\\frac{a^{n}}{1+a+\\ldots+a^{2 n}}<\\frac{1}{2 n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPuisque la difficulté réside dans le dénominateur du membre de droite, et pour plus de confort, on peut chercher à montrer la relation inverse, à savoir :\n$$\n\\frac{1+a+\\ldots+a^{2 n}}{a^{n}}>2 n\n$$\nL'idée derrière la solution qui suit est \"d'homogénéiser\" le numérateur du membre de gauche, c'est-à-dire de comparer ce numérateur dans lequel les $a$ sont élevés à des puissances distinctes à une expression composée uniquement de $a$ élevés à la même puissance. Pour cela, on cherche à coupler certains termes et appliquer l'inégalité des moyennes. Voyons plutôt : pour $0 \\leqslant i \\leqslant 2 n$, on a $a^{i}+a^{2 n-i} \\geqslant 2 \\sqrt{a^{i} a^{2 n-i}}= 2 a^{n}$. Ainsi :\n$$\n\\begin{aligned}\n\\frac{1+a+\\ldots+a^{2 n}}{a^{n}} & =\\frac{\\left(1+a^{2 n}\\right)+\\left(a+a^{2 n-1}\\right)+\\ldots+\\left(a^{n-1}+a^{n+1}\\right)+a^{n}}{a^{n}} \\\\\n& \\geqslant \\frac{\\overbrace{2 a^{n}+2 a^{n}+\\ldots+2 a^{n}}^{n \\text{ termes}}+a^{n}}{a^{n}} \\\\\n& =\\frac{(2 n+1) a^{n}}{a^{n}} \\\\\n& =2 n+1 \\\\\n& >2 n\n\\end{aligned}\n$$\nce qui fournit l'inégalité voulue.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMaena et Théodore jouent à un jeu. Ils jouent sur une grille carrée formée de $99 \\times 99$ cases. On considère que deux cases sont adjacentes si elles ont un sommet ou un côté en commun.\n\nInitialement, Maéna numérote les cases de la grille de 1 à $99^{2}$, de façon arbitraire. Théodore place alors un jeton sur l'une des cases du carré, puis il s'autorise des mouvements de la forme suivante : il peut déplacer le jeton d'une case vers une autre uniquement si ces cases sont adjacentes et si la nouvelle case sur laquelle se retrouve le jeton a un numéro strictement plus grand que l'ancienne case.\n\nCombien de mouvements au minimum Théodore peut-il garantir, quelle que soit la manière avec laquelle Maena a placé ses entiers?",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nTout d'abord, Théodore peut toujours se débrouiller pour effectuer au moins trois mouvements. Pour ce faire, il lui suffit de sélectionner un carré de taille $2 \\times 2$ à l'intérieur du carré de taille $99 \\times 99$, puis d'en parcourir les quatre cases, qui sont nécessairement adjacentes puisqu'elles ont un sommet en commun.\n\nRéciproquement, voici comment peut procéder Maena pour empêcher Théodore d'effectuer plus de trois mouvements. Elle numérote lignes et colonnes de 1 à 99, puis regroupe les $99 \\times 99$ cases du carré en quatre catégories :\n\n$\\triangleright$ la catégorie 1 contient les cases situées en une ligne et une colonne impaires;\n$\\triangleright$ la catégorie 2 contient les cases situées en une ligne impaire et une colonne paire;\n$\\triangleright$ la catégorie 3 contient les cases situées en une ligne paire et une colonne impaire;\n$\\triangleright$ la catégorie 4 contient les cases situées en une ligne et une colonne paires.\n\nDeux cases de la même catégorie ne sont jamais adjacentes.\n\nEnsuite, Maena place ses entiers, dans l'ordre croissant, dans des cases de catégorie 1, puis de catégorie 2, puis de catégorie 3, et enfin de catégorie 4. Ainsi, Théodore ne peut jamais se déplacer entre deux cases de même catégorie, et encore moins baisser de catégorie. Tout mouvement augmente donc la catégorie de la case dans laquelle se trouve le jeton de Théodore, qui se voit ainsi limité à effectuer au plus trois mouvements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30787,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un entier $n \\geqslant 1$ est nippon s'il admet trois diviseurs $d_{1}, d_{2}$ et $d_{3}$ tels que $1 \\leqslant d_{1}<d_{2}<d_{3}$ et $d_{1}+d_{2}+d_{3}=2022$.\nQuel est le plus petit entier nippon?",
    "options": [],
    "answer": "1344",
    "solution": "Solution:\n\nTout d'abord, on remarque que $2 \\times 2019$ est nippon, puisque les entiers $1$, $2$ et $2019$ figurent parmi ses diviseurs.\n\nSoit $n$ l'entier nippon minimal : il s'agit du PPCM des entiers $d_{1}, d_{2}$ et $d_{3}$. On pose alors $a_{k}=n / d_{k}$, de sorte que $1 \\leqslant a_{3}<a_{2}<a_{1}$, que $\\operatorname{PGCD}\\left(a_{1}, a_{2}, a_{3}\\right)=1$ et que\n$$\nn\\left(\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\frac{1}{a_{3}}\\right)=2022=2 \\times 3 \\times 337 .\n$$\nPour minimiser $n$, on souhaite maximiser la somme des fractions $1 / a_{i}$, donc on cherche à minimiser les entiers $a_{i}$ eux-mêmes. On commence donc par regarder les petites valeurs de $a_{2}$ et de $a_{3}$.\n\nPar exemple, si $a_{3}=1$ et $a_{2}=2$, notre équation devient\n$$\n\\left(3 a_{1}+2\\right) d_{1}=2^{2} \\times 3 \\times 337\n$$\nAinsi, $3 a_{1}+2$ est un facteur de $2^{2} \\times 3 \\times 337$ congru à $2$ modulo $3$, c'est-à-dire $2$ ou $2 \\times 337=674$. Le premier cas est impossible car $3 a_{1}+2 \\geqslant 3 \\times 3+2>2$. On en déduit que $a_{1}=672 / 3=224$, puis que $d_{1}=2 \\times 3=6$, de sorte que $n=6 \\times 224=1344$. En particulier, $1344$ est un multiple de $1$ et de $2$, donc il s'agit d'un entier nippon, avec $d_{1}=6$, $d_{2}=672$ et $d_{3}=1344$.\n\nOn recherche donc un éventuel entier nippon $n<1344$. Dans un tel cas, si on écrit la somme $1 / a_{1}+1 / a_{2}+1 / a_{3}$ sous forme d'une fraction irréductible $p / q$, notre égalité devient\n$$\nn \\times p=2 \\times 3 \\times 337 \\times q,\n$$\net $p$ est premier avec $q$, donc divise $2 \\times 3 \\times 337$. En outre, puisque l'on ne peut plus avoir $a_{3}=1$ et $a_{2}=2$, on a nécessairement $a_{2} \\geqslant 3$. Ainsi,\n$$\n1+\\frac{1}{3}+\\frac{1}{a_{3}} \\geqslant \\frac{p}{q}=\\frac{2022}{n}>\\frac{2022}{1344}>1+\\frac{1}{2}=1+\\frac{1}{3}+\\frac{1}{6}\n$$\ndonc $a_{1} \\leqslant 5$.\n\nIl nous reste donc à étudier les triplets $\\left(a_{1}, a_{2}, a_{3}\\right)=(4,3,1),(5,3,1),(5,4,1),(4,3,2),(5,3,2)$, $(5,4,2)$ et $(5,4,3)$, qui nous amènent respectivement aux fractions $p / q=19 / 12,23 / 15,29 / 20$, $13 / 12,31 / 30,19 / 20$ et $47 / 60$. Le numérateur $p$ ne divise jamais $2 \\times 3 \\times 337$, donc aucun de ces cas n'est possible.\n\nEn conclusion, le plus petit entier nippon est $n=1344$.\n\n\nSolution alternative\n\nOn peut simplifier l'étude des triplets $\\left(a_{1}, a_{2}, a_{3}\\right)$ effectuée au dernier paragraphe de la solution ci-dessus, pour limiter les calculs à effectuer. En effet,\n\n$\\triangleright$ si $a_{3} \\geqslant 2$, alors $a_{2} \\geqslant 3$ et $a_{1} \\geqslant 4$, donc\n$$\n\\frac{p}{q} \\leqslant \\frac{1}{2}+\\frac{1}{3}+\\frac{1}{4}=\\frac{13}{12} \\leqslant \\frac{3}{2}\n$$\n$\\triangleright$ si $a_{3}=1$ et $a_{2} \\geqslant 4$, on sait que $a_{1} \\geqslant 5$, donc\n$$\n\\frac{p}{q} \\leqslant 1+\\frac{1}{4}+\\frac{1}{5}=\\frac{29}{20} \\leqslant \\frac{3}{2}\n$$\n$\\triangleright$ si $a_{3}=1, a_{2}=3$ et $a_{1} \\geqslant 6$, on sait que\n$$\n\\frac{p}{q} \\leqslant 1+\\frac{1}{3}+\\frac{1}{6}=\\frac{3}{2}\n$$\nIl suffit donc d'étudier les deux seuls triplets $\\left(a_{1}, a_{2}, a_{3}\\right)=(4,3,1)$ et $(5,3,1)$, qui nous amènent aux fractions $p / q=19 / 12$ et $23 / 15$ donc ne conviennent pas. On en conclut à nouveau que le plus petit entier nippon est bien $n=1344$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30788,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $x$, $y$ et $z$ trois nombres réels tels que $x^{2}+y^{2}+z^{2}=1$. Trouver les valeurs minimale et maximale possibles du nombre réel $x y+y z-z x$.",
    "options": [],
    "answer": "minimum = -1, maximum = 1/2",
    "solution": "Solution:\n\nEn vertu de l'identité\n$$\n(x-y+z)^{2}=x^{2}+y^{2}+z^{2}-2(x y+y z-z x)=1-2(x y+y z-z x),\n$$\nil s'agit ici de trouver les valeurs extrêmes que peut prendre le nombre $(x-y+z)^{2}$.\nTout d'abord, puisque $(x-y+z)^{2} \\geqslant 0$, avec égalité lorsque $x=y=1 / \\sqrt{2}$ et $z=0$, on a bien\n$$\nx y+y z-z x=\\frac{1-(x-y+z)^{2}}{2} \\leqslant \\frac{1}{2}\n$$\nav ec égalité lorsque $x=y=1 / \\sqrt{2}$ et $z=0$.\nEn outre, l'inégalité de Cauchy-Schwarz indique que\n$$\n(x-y+z)^{2} \\leqslant\\left(x^{2}+y^{2}+z^{2}\\right) \\times\\left(1^{2}+(-1)^{2}+1^{2}\\right)=3,\n$$\nav ec égalité si et seulement si les triplets $(x, y, z)$ et $(1,-1,1)$ sont colinéaires. Par conséquent, on a bien\n$$\n(x y+y z-z x)=\\frac{1-(x-y+z)^{2}}{2} \\geqslant \\frac{1-3}{2}=-1\n$$\nav ec égalité lorsque $x=-y=z=1 / \\sqrt{3}$.\nEn conclusion, les valeurs extrémales recherchées sont $-1$ et $1 / 2$.\n\n\nSolution alternative $n^{\\circ} 1$\n\nOn peut aussi démontrer l'inégalité $x y+y z-z x \\geqslant-1$ comme suit. Il s'agit d'appliquer directement l'inégalité de Cauchy-Schwarz aux triplets $(x, y, z)$ et $(y, z,-x)$, pour constater que\n$$\n(x y+y z-z x)^{2} \\leqslant\\left(x^{2}+y^{2}+z^{2}\\right)^{2}=1\n$$\ndonc que $x y+y z-z x \\geqslant-1$.\nPar la suite, on est dans le cas d'égalité si et seulement s'il existe un réel $\\lambda$ tel que $(x, y, z)=\\lambda(y, z,-x)$. Mais alors les quatre réels $x, y, z$ et $\\lambda$ sont non nuls et $x y z=-\\lambda^{3} x y z$, donc $\\lambda=-1$ et $x=-y=z= \\pm 1 / \\sqrt{3}$, auquel cas on a bien $x y+y z-z x=-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30789,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier $n \\geqslant 1$, on note $f_{n}$ la somme de tous les restes obtenus en divisant $n$ par les nombres $1,2, \\ldots, n$. Par exemple, si on divise $5$ par $1,2,3,4$ et $5$, les restes que l'on obtient sont $0,1,2,1$ et $0$, de sorte que $f_{5}=0+1+2+1+0=4$.\n\nTrouver tous les entiers $n \\geqslant 2$ tels que $f_{n}=f_{n-1}+n-2$.",
    "options": [],
    "answer": "All prime numbers n ≥ 2",
    "solution": "Solution:\n\nPour tous les entiers $a$ et $b$ tels que $1 \\leqslant a \\leqslant b$, on note $r_{a}(b)$ le reste obtenu en divisant $b$ par $a$. Puisque\n$$\nf_{n}-f_{n-1}=\\sum_{i=1}^{n} r_{i}(n)-\\sum_{i=1}^{n-1} r_{i}(n-1)=\\sum_{i=2}^{n-1} r_{i}(n)-\\sum_{i=2}^{n-1} r_{i}(n-1)=\\sum_{i=2}^{n-1}\\left(r_{i}(n)-r_{i}(n-1)\\right),\n$$\non s'intéresse aux quantités $r_{a}(b+1)-r_{a}(b)$. On remarque alors que $r_{a}(b+1)-r_{a}(b)=1-a$ si $a$ divise $b+1$, et que $r_{a}(b+1)-r_{a}(b)=1$ sinon. Ainsi, la somme\n$$\nf_{n}-f_{n-1}=\\sum_{i=2}^{n-1}\\left(r_{i}(n)-r_{i}(n-1)\\right)\n$$\nqui est formée de $n-2$ termes inférieurs ou égaux à $1$, ne peut valoir $n-2$ que si tous ces termes valent $1$, c'est-à-dire si aucun entier $i$ tel que $2 \\leqslant i \\leqslant n-1$ ne divise $n$.\n\nEn conclusion, les entiers recherchés sont les nombres premiers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30790,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\Omega$ son cercle circonscrit. Soit $D$ le pied de la hauteur issue du sommet $A$. La bissectrice issue du sommet $A$ coupe le segment $[BC]$ au point $P$ et recoupe le cercle $\\Omega$ au point $S$. Soit $A'$ le point diamétralement opposé au sommet $A$ dans le cercle $\\Omega$. Démontrer que les droites $(SD)$ et $(A'P)$ se coupent sur le cercle $\\Omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $X$ le point où la droite $(A'P)$ recoupe le cercle $\\Omega$. Nous allons démontrer que les points $D$, $S$ et $X$ sont alignés.\n\nSoit $O$ le centre de $\\Omega$. Le point $S$ n'est autre que le pôle Sud de $ABC$ issu de $A$, donc il appartient à la médiatrice de $[BC]$, qui n'est autre que la droite $(OS)$. Par conséquent, les droites $(OS)$ et $(AD)$ sont parallèles, et les angles $\\widehat{DAP}$ et $\\widehat{OSA}$ sont alternes-internes. En outre, le triangle $AOS$ est isocèle en $O$. Ainsi, en angles de droites, on sait déjà que\n$$\n(XA', XS) = (AA', AX) = (AO, AS) = (SA, SO) = (AP, AD)\n$$\nIl reste donc à vérifier que\n$$\n(XP, XD) = (XA', XD) = (AP, AD)\n$$\nc'est-à-dire que les points $X$, $A$, $D$ et $P$ sont cocycliques. Or, puisque $[AA']$ est un diamètre de $\\Omega$, on sait que $\\widehat{AXP} = \\widehat{AXA'} = 90^\\circ = \\widehat{ADP}$. Ainsi, les deux points $D$ et $X$ appartiennent au cercle de diamètre $[AP]$, et le résultat désiré en découle immédiatement.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30791,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\omega_{1}$ et $\\omega_{2}$ deux cercles de centres respectifs $O_{1}$ et $O_{2}$. On suppose que $\\omega_{1}$ et $\\omega_{2}$ se coupent en les points $A$ et $B$. La droite $(O_{1}A)$ recoupe le cercle $\\omega_{2}$ en $C$ tandis que la droite $(O_{2}A)$ recoupe le cercle $\\omega_{1}$ en $D$. Montrer que les points $D$, $O_{1}$, $B$, $O_{2}$ et $C$ appartiennent à un même cercle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons que puisque les angles $\\widehat{DAO_{1}}$ et $\\widehat{CAO_{2}}$ sont opposés par le sommet, ils sont égaux. D'autre part, puisque les points $A$ et $D$ appartiennent au cercle $\\omega_{1}$, le triangle $A O_{1} D$ est isocèle en $O_{1}$. De même, le triangle $CO_{2} A$ est isocèle en $O_{2}$. Les triangles $DO_{1} A$ et $CO_{2} A$ sont donc des triangles isocèles avec les mêmes angles à la base, ils sont donc semblables. Ceci implique que $\\widehat{AO_{1} D}=\\widehat{CO_{2} A}$, et donc que\n$$\n\\widehat{CO_{1} D}=\\widehat{AO_{1} D}=\\widehat{CO_{2} A}=\\widehat{CO_{2} D}\n$$\nsi bien que les points $C$, $O_{2}$, $O_{1}$ et $D$ sont cocycliques.\n\nPar ailleurs, on peut découper l'angle $\\widehat{DBC}$ en la somme $\\widehat{DBA}+\\widehat{ABC}$. D'une part, d'après le théorème de l'angle au centre dans le cercle $\\omega_{1}$, on a $\\widehat{ABD}=\\frac{1}{2} \\widehat{AO_{1} D}$. D'autre part, d'après le théorème de l'angle au centre dans le cercle $\\omega_{1}$, on a $\\widehat{ABC}=\\frac{1}{2} \\widehat{AO_{2} C}$. Ainsi\n$$\n\\widehat{DBC}=\\widehat{DBA}+\\widehat{ABC}=\\frac{1}{2} \\widehat{AO_{1} D}+\\frac{1}{2} \\widehat{AO_{2} C}=\\widehat{AO_{1} D}\n$$\nce qui permet de conclure que le point $B$ appartient au cercle passant par les points $C$, $O_{2}$, $O_{1}$ et $D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30792,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les couples d'entiers naturels non nuls $(x, n)$ tels que\n$$\n3 \\cdot 2^{x} + 4 = n^{2}\n$$",
    "options": [],
    "answer": "[(2, 4), (5, 10), (6, 14)]",
    "solution": "Solution:\nPour $x=0$, l'équation devient $n^{2}=7$ qui n'a pas de solution. Supposons $x>0$, on factorise : $3 \\cdot 2^{x} = (n-2)(n+2)$. Le membre de gauche est pair donc le membre de droite est pair. Les deux facteurs du membre de droite sont de même parité donc ils sont tous les deux pairs et on a deux cas :\n\n- Cas 1: $3 \\cdot 2^{a} = n-2$ et $2^{b} = n+2$, avec $a + b = x$. On peut combiner ces deux équations pour trouver $2^{b} = 3 \\cdot 2^{a} + 4$. Notons que $n+2$ et $n-2$ ont pour différence 4 donc ils ne peuvent pas être tous les deux divisibles par 8. On a donc $a \\leqslant 2$ ou $b \\leqslant 2$. Or $2^{b} > 4$, donc $b > 2$ et on déduit que $a \\leqslant 2$. Si $a=0$, $2^{b} = 3 + 4 = 7$ ce qui n'est pas possible. Si $a=1$, $2^{b} = 6 + 4 = 10$ ce qui n'est pas possible non plus. Si $a=2$, alors $2^{b} = 12 + 4 = 16$ donc $b=4$. On obtient $x=6$ et $n=14$. Réciproquement, on vérifie que $3 \\cdot 2^{6} + 4 = 196 = 14^{2}$ et le couple $(6,14)$ est bien solution.\n\n- Cas 2: $3 \\cdot 2^{a} = n+2$ et $2^{b} = n-2$, avec $a + b = x$. Les deux équations combinées donnent $2^{b} + 4 = 3 \\cdot 2^{a}$. De même que précédemment, on a $a \\leqslant 2$ ou $b \\leqslant 2$.\n\n1. Si $b \\leqslant 2$, alors $b=0,1$ ou $2$. Si $b=0$, alors $5 = 3 \\cdot 2^{a}$ ce qui n'est pas possible. Si $b=1$, $6 = 3 \\cdot 2^{a}$ et $a=1$, soit $x=2$ et $n=4$ et on vérifie que $3 \\cdot 2^{2} + 4 = 16 = 4^{2}$ et $(2,4)$ est bien solution. Si $b=2$, alors $8 = 3 \\cdot 2^{a}$ ce qui n'est pas possible.\n\n2. Si $a \\leqslant 2$, alors $a=0,1$ ou $2$. Si $a=0$, $2^{b} + 4 = 3 < 4$ ce qui n'est pas possible. Si $a=1$, on retombe sur le cas où $b=1$ et sur le couple $(2,4)$. Si $a=2$, $2^{b} + 4 = 12$ donc $b=3$. Ceci donne $x=5$ et $n=10$. Réciproquement, on vérifie que $3 \\cdot 2^{5} + 4 = 100 = 10^{2}$ et $(5,10)$ est bien solution.\n\nAinsi l'ensemble des couples $(x, n)$ solutions est $\\{(2,4), (5,10), (6,14)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30793,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $p$ et $q$ deux nombres premiers distincts, tels que $p < 2q$ et $q < 2p$. Démontrer qu'il existe deux entiers consécutifs dont l'un a $p$ pour plus grand facteur premier et l'autre a $q$ pour plus grand facteur premier.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons sans perte de généralité que $p < q$. Le théorème de Bézout indique qu'il existe des entiers $a$ et $b$ tels que $a p + b q = 1$.\n\nOn peut toujours remplacer un tel couple $(a, b)$ par $(a \\pm q, b \\mp p)$. Par conséquent, si on cherche un couple pour lequel la valeur de $|a|$ est minimale, on sait que $|a - q| \\geqslant |a|$ et $|a| \\leqslant |a + q|$, de sorte que $|a| \\leqslant q / 2 < p$.\n\nEn outre, puisque $|a p + b q| = 1 \\leqslant p + q$, les entiers $a$ et $b$ sont de signes opposés, ce qui signifie que les entiers $n = |a| p$ et $m = |b| q$ sont deux entiers consécutifs.\n\nPuisque $|a| < p$, l'entier $n$ a donc $p$ comme facteur premier maximal. En outre, $m \\leqslant n + 1 \\leqslant p^{2} \\leqslant q^{2}$, donc $m$ a $q$ comme facteur premier maximal. Les entiers $m$ et $n$ satisfont donc les contraintes de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30794,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un parallélogramme tel que $AC = BC$. Soit $P$ un point situé sur le prolongement du segment $[AB]$ au-delà de $B$. Soit $Q$ le point d'intersection, autre que $D$, entre le segment $[PD]$ et le cercle circonscrit à $ACD$. Soit ensuite $R$ le point d'intersection, autre que $P$, entre le segment $[PC]$ et le cercle circonscrit à $APQ$.\n\nDémontrer que les droites $(AQ)$, $(BR)$ et $(CD)$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nTout d'abord, on constate que\n$$\n(RA, RC) = (RA, RP) = (QA, QP) = (QA, QD) = (CA, CD) = (AC, AB) = (BA, BC)\n$$\nce qui signifie que les points $A$, $B$, $C$ et $R$ sont cocycliques.\n\nDe même, si l'on note $X$ le point d'intersection des droites $(AQ)$ et $(CD)$, on constate que\n$$\n(QR, QX) = (QR, QA) = (PR, PA) = (CR, CX)\n$$\nce qui signifie que les points $C$, $Q$, $R$ et $X$ sont cocycliques.\n\nOn en conclut que\n$$\n(RC, RX) = (QC, QX) = (QC, QA) = (DC, DA) = (BA, BC) = (AC, AB) = (RC, RB)\n$$\nce qui signifie que, comme annoncé, les points $B$, $R$ et $X$ sont alignés.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30795,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $m$ et $n$ deux entiers tels que $m > n \\geqslant 3$. Morgane a disposé $m$ jetons en cercle, et s'apprête à les peindre en utilisant $n$ couleurs distinctes. Elle souhaite que, parmi $n+1$ jetons consécutifs, il y ait toujours au moins un jeton de chacune des $n$ couleurs. Si elle peut y parvenir, on dira que l'entier $m$ est $n$-coloriable.\n\nDémontrer que, pour tout entier $n \\geqslant 3$, il n'existe qu'un nombre fini non nul d'entiers $m$ qui ne sont pas $n$-coloriables, et trouver le plus grand entier $m$ qui ne soit pas $n$-coloriable.",
    "options": [],
    "answer": "n^2 - n - 1",
    "solution": "Solution:\n\nTout d'abord, si $m \\geqslant n^{2}-n$, Morgane peut réaliser son souhait en procédant comme suit, montrant au passage que $m$ est $n$-coloriable.\n\nSoit $a$ le résidu de $m$ modulo $n$, c'est-à-dire le plus petit entier naturel tel que $a \\equiv m (\\bmod\\ n)$. Morgane commence par placer $m-a$ jetons, qu'elle choisit de couleurs $1,2, \\ldots, n$, puis de nouveau $1,2, \\ldots, n$, etc. Ainsi, elle a placé $(m-a)/n \\geqslant n-1 \\geqslant a$ jetons de chaque couleur, comme illustré ci-dessous (à gauche) dans le cas où $n=5$ et $m=23$.\n\n![](attached_image_1.png)\n\nPuis elle choisit $a$ jetons de couleur $m$, et insère ajoute un nouveau jeton de couleur 1 juste entre ces jetons et les jetons de couleur 1 qui suivent, comme illustré ci-dessus (à droite) dans le cas où $n=5$ et $m=23$ : nous avons dessiné en noir (avec texte blanc) les jetons ainsi insérés.\n\nAprès avoir procédé de la sorte, Morgane constate que toute suite de $n+1$ jetons consécutifs contient au moins $n$ jetons consécutifs de la configuration qu'elle avait obtenue avant d'insérer ses $a$ jetons surnuméraires, et ces $n$ jetons sont bien de chacune des couleurs $1,2, \\ldots, n$. Comme annoncé, l'entier $m$ est donc bien $n$-coloriable.\n\nRéciproquement, supposons que, partant de l'entier $m = n^{2} - n - 1$, Morgane a pu parvenir à ses fins. Puisque $m / n < n-1$, l'une des couleurs, disons la couleur 1, est représentée au plus $n-2$ fois, et au moins une fois. Morgane isole alors un jeton de couleur 1, puis sépare les $m-1 = (n+1)(n-2)$ jetons restants en $n-2$ groupes de $n+1$ jetons consécutifs. Puisqu'elle est satisfaite, chacun de ces groupes contient au moins un jeton de couleur 1 : cela lui fait un minimum de $n-1$ jetons de couleur 1 en tout, ce qui est absurde !\n\nAinsi, l'entier $m = n^{2} - n - 1$ n'est pas $n$-coloriable, et il s'agit du plus grand entier qui ne soit pas $n$-coloriable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier naturel non nul. Anna a écrit $4 n+2$ entiers deux à deux distincts et compris entre $0$ et $5^{n}$ inclus. Démontrer que, parmi les entiers qu'elle a écrits, il en existe trois, disons $a, b$ et $c$, tels que $a < b < c$ et $c + 2a > 3b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $x_{1} < x_{2} < \\ldots < x_{4 n+2}$ les $4 n+2$ entiers qu'Anna a écrits. Nous allons démontrer qu'il existe un entier $k \\leqslant 4 n$ tel que $x_{4 n+2} + 2 x_{k} > 3 x_{k+1}$.\n\nEn effet, dans le cas contraire, posons $y_{k} = x_{4 n+2} - x_{k}$ pour tout entier $k$, de sorte que\n$$\n3 y_{k+1} = 3 x_{4 n+2} - 3 x_{k+1} \\leqslant 2 x_{4 n+2} - 2 x_{k} = 2 y_{k}.\n$$\nCela signifie que $y_{k} \\geqslant (3 / 2) y_{k+1}$ pour tout $k \\leqslant 4 n$, et donc que\n$$\n5^{n} \\geqslant y_{1} \\geqslant (3 / 2)^{4 n} y_{4 n+1} \\geqslant (3 / 2)^{4 n} = (5 + 1 / 16)^{n}\n$$\nce qui est impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30797,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAurélien écrit 11 entiers naturels au tableau. Montrer qu'il peut choisir certains de ces entiers et placer des signes + et - entre eux de telle sorte que le résultat soit divisible par 2021.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn peut voir un choix de certains entiers avec des signes $+$ et $-$ comme un choix de certains entiers que l'on va compter positivement et de certains entiers que l'on va compter négativement. Les nombres qu'Aurélien peut obtenir sont des nombres qui s'écrivent comme la différence entre deux sommes de plusieurs entiers de départ. On peut remarquer que la réciproque est aussi vraie : si $a$ et $b$ s'écrivent comme somme de certains entiers, alors Aurélien peut obtenir $a-b$ en mettant un signe plus devant les entiers qui sont dans $a$ et un signe moins devant ceux qui sont dans $b$. Si le même entier de départ est utilisé pour obtenir à la fois $a$ et $b$, Aurélien ne peut pas écrire devant un signe $-$ et $+$, mais il peut simplement ne pas écrire l'entier tout court, ce qui a le même effet que l'ajouter puis l'enlever. Toutes les différences entre deux sommes d'entiers du tableau sont donc des résultats qu'Aurélien peut obtenir. Dire qu'une différence entre deux sommes est divisible par $2021$ revient à dire que les deux sommes étaient congrues modulo $2021$. On dispose de $2^{11} = 2048 > 2021$ sommes, donc d'après le principe des tiroirs, deux sont congrues modulo $2021$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\mathcal{S}$ un ensemble infini d'entiers naturels non nuls contenant quatre entiers $a, b, c, d$ deux à deux distincts tels que $\\operatorname{pgcd}(a, b) \\neq \\operatorname{pgcd}(c, d)$. Démontrer que $\\mathcal{S}$ contient trois entiers $x, y, z$ deux à deux distincts tels que $\\operatorname{pgcd}(x, y)=\\operatorname{pgcd}(y, z) \\neq \\operatorname{pgcd}(z, x)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCi-dessous, on note $m \\wedge n$ le plus grand diviseur commun à deux entiers $m$ et $n$. On considère alors le graphe colorié infini dont les sommets sont les éléments de $\\mathcal{S}$, et où chaque paire de sommets $(u, v)$ distincts l'un de l'autre est reliée par une arête de couleur $u \\wedge v$.\nSupposons que les entiers $x, y, z$ n'existent pas. Cela signifie que tout triangle est colorié en utilisant soit une seule couleur, soit exactement trois couleurs.\n\n![](attached_image_1.png)\n\nSoit maintenant $u, v$ et $w$ trois éléments de $\\mathcal{S}$. Puisque $u, v$ et $w$ comptent chacun un nombre fini de diviseurs, la fonction $f: x \\mapsto(u \\wedge x, v \\wedge x, w \\wedge x)$ ne prend qu'un nombre fini de valeurs. Parmi l'infinité d'éléments de $\\mathcal{S}$ distincts de $u, v$ et $w$, il en existe donc deux, disons $s$ et $t$, tels que $f(s)=f(t)$, c'est-à-dire que\n$$\ns \\wedge u=t \\wedge u,\\ s \\wedge v=t \\wedge v\\ \\text{ et }\\ s \\wedge w=t \\wedge w\n$$\nLes triangles $(s, t, u),(s, t, v)$ et $(s, t, w)$ sont donc monochromes, et puisqu'ils partagent l'arête $(s, t)$, ils sont de la même couleur, disons $\\mathcal{C}$. Mais alors les triangles $(s, u, v)$ et $(s, u, w)$ ont chacun deux arêtes de couleur $\\mathcal{C}$, dont ils sont eux aussi monochromes de couleur $\\mathcal{C}$, de sorte que $u \\wedge v=u \\wedge w$.\n\n![](attached_image_2.png)\n\nAinsi, deux arêtes ayant un sommet commun sont nécessairement de la même couleur, et notre graphe est monochrome. Cela signifie en particulier que\n$$\na \\wedge b=c \\wedge d\n$$\nen contradiction avec les hypothèses de l'énoncé. Notre hypothèse initiale était donc invalide, ce qui conclut.\n\n\nSolution alternative $n^{\\circ} 1$\n\nSoit $k$ le PGCD des éléments de $\\mathcal{S}$. Si l'on divise chaque élément de $\\mathcal{S}$ par $k$, on obtient un ensemble $\\mathcal{S}^{\\prime}$ qui vérifie toujours l'énoncé et, réciproquement, si $\\mathcal{S}^{\\prime}$ vérifie l'énoncé, $\\mathcal{S}$ le vérifie aussi. On remplace donc $\\mathcal{S}$ par $\\mathcal{S}^{\\prime}$, c'est-à-dire que l'on suppose désormais que $k=1$.\n\nSoit $x$ et $y$ deux éléments de $\\mathcal{S}$ tels que $x \\wedge y \\neq 1$. Il suffit, par exemple, de choisir $(x, y)=(a, b)$ ou $(x, y)=(c, d)$. Si $\\mathcal{S}$ admet un élément $z$ premier avec $x$ et $y$, on a $x \\wedge z=y \\wedge z=1 \\neq x \\wedge y$, ce qui constitue le résultat souhaité.\n\nSinon, parmi les facteurs premiers de $x y$, il en existe au moins un, disons $p$, qui divise une infinité d'éléments de $\\mathcal{S}$. Notons $\\mathcal{S}_{p}$ l'ensemble des éléments de $\\mathcal{S}$ qui sont divisibles par $p$. Puisque $k=1$, il existe un élément $t$ de $\\mathcal{S}$ qui n'appartient pas à $\\mathcal{S}_{p}$. Comme $\\mathcal{S}_{p}$ est infini, et puisque la fonction $n \\mapsto t \\wedge n$ ne prend qu'un nombre fini de valeurs lorsque $n$ varie, il existe deux éléments $u$ et $v$ de $\\mathcal{S}_{p}$ tels que $t \\wedge u=t \\wedge v$. Puisque $p$ divise $u \\wedge v$ mais pas $t \\wedge u$, on en conclut que $t \\wedge u=t \\wedge v \\neq u \\wedge v$, ce qui donne de nouveau le résultat souhaité.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn ensemble $E$ d'entiers strictement positifs est dit intéressant si pour tout $n \\geqslant 1$ et pour tous $x_{1}, \\ldots, x_{n}$ des éléments de $E$ deux à deux distincts, leur moyenne arithmétique $\\frac{1}{n}\\left(x_{1}+\\ldots+x_{n}\\right)$ et leur moyenne géométrique $\\left(x_{1} \\cdot \\ldots \\cdot x_{n}\\right)^{\\frac{1}{n}}$ sont des entiers.\n\n1. Existe-t-il un ensemble $E$ intéressant contenant exactement 2022 éléments?\n2. Existe-t-il un ensemble $E$ intéressant infini?",
    "options": [],
    "answer": "Yes: for example, the set of 2022 numbers {(2022!)^(k·2022!) for k from 1 to 2022)} works. No: there is no infinite interesting set.",
    "solution": "Solution:\n\n1) Pour commencer, on peut remarquer que si $n \\geqslant 1$, alors dès que $x_{1}, \\ldots, x_{n} \\geqslant 1$ sont des entiers tous multiples de $n$, chacun des nombres $\\frac{x_{k}}{n}$ est un entier, de sorte que leur somme $\\frac{1}{n}\\left(x_{1}+\\ldots+x_{n}\\right)$ est un entier. De même, si $n \\geqslant 1$, alors dès que $x_{1}, \\ldots, x_{n} \\geqslant 1$ sont des entiers tous puissances $n$-ièmes parfaites, chacun des nombres $x_{k}^{1 / n}$ est entier, de sorte que leur produit $\\left(x_{1} \\times \\ldots \\times x_{n}\\right)^{\\frac{1}{n}}$ est un entier. Pour trouver un ensemble $E$ intéressant contenant exactement 2022 éléments, il suffit donc que pour tout $1 \\leqslant n \\leqslant 2022$, tous les éléments de $E$ soient à la fois des multiples de $n$ et des puissances $n$-ièmes parfaites. Pour cela, il suffit que tous les éléments de $E$ soient à la fois des multiples de $2022!$, et des puissances $2022!$-ièmes parfaites. Ainsi, $E=\\left\\{(2022!)^{k \\cdot 2022!},\\ 1 \\leqslant k \\leqslant 2022\\right\\}$ convient.\n\n2) Nous allons montrer qu'il n'existe pas d'ensemble intéressant infini. En fait, on va montrer qu'il n'existe même pas d'ensemble infini d'entiers strictement positifs vérifiant l'hypothèse sur les moyennes arithmétiques. En effet, supposons par l'absurde qu'un tel ensemble infini $E$ existe, c'est-à-dire que pour tout $n \\geqslant 1$ et $x_{1}, \\ldots, x_{n} \\in E$ deux à deux distincts, la moyenne arithmétique $\\frac{1}{n}\\left(x_{1}+\\ldots+x_{n}\\right)$ est un entier. Soit $a<b$ deux éléments quelconques de $E$ et $n \\geqslant 2$. Choisissons $x_{1}<\\ldots<x_{n-1}$ des éléments deux à deux distincts de $E$, et distincts de plus de $a$ et $b$. L'hypothèse appliquée successivement aux $n$-uplets $\\left(x_{1}, \\ldots, x_{n-1}, a\\right)$ et $\\left(x_{1}, \\ldots, x_{n-1}, b\\right)$ assure que $n$ divise à la fois $x_{1}+\\ldots+x_{n-1}+a$ et $x_{1}+\\ldots+x_{n-1}+b$. Ainsi, $n$ divise la différence de ces deux nombres qui vaut $a-b$. Ceci étant vrai pour tout $n \\geqslant 2$, l'entier $a-b$ admet un nombre infini de diviseurs. Il est donc nul, c'est-à-dire que $a=b$. Ceci est en contradiction avec notre hypothèse de départ, un tel ensemble $E$ n'existe donc pas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $m$ et $n$ deux entiers tels que $m > n \\geqslant 3$. Morgane a disposé $m$ jetons en cercle, et s'apprête à les peindre en utilisant $n$ couleurs distinctes. Elle souhaite que, parmi $n+1$ jetons consécutifs, il y ait toujours au moins un jeton de chacune des $n$ couleurs. Si elle peut y parvenir, on dira que l'entier $m$ est $n$-coloriable.\n\nDémontrer que, pour tout entier $n \\geqslant 3$, il n'existe qu'un nombre fini non nul d'entiers $m$ qui ne sont pas $n$-coloriables, et trouver le plus grand entier $m$ qui ne soit pas $n$-coloriable.",
    "options": [],
    "answer": "n^2 - n - 1",
    "solution": "Solution:\n\nTout d'abord, si $m \\geqslant n^{2}-n$, Morgane peut réaliser son souhait en procédant comme suit, montrant au passage que $m$ est $n$-coloriable.\n\nSoit $a$ le résidu de $m$ modulo $n$, c'est-à-dire le plus petit entier naturel tel que $a \\equiv m \\pmod{n}$. Morgane commence par placer $m-a$ jetons, qu'elle choisit de couleurs $1,2, \\ldots, n$, puis de nouveau $1,2, \\ldots, n$, etc. Ainsi, elle a placé $(m-a)/n \\geqslant n-1 \\geqslant a$ jetons de chaque couleur, comme illustré ci-dessous (à gauche) dans le cas où $n=5$ et $m=23$.\n\n![](attached_image_1.png)\n\nPuis elle choisit $a$ jetons de couleur $m$, et insère ajoute un nouveau jeton de couleur 1 juste entre ces jetons et les jetons de couleur 1 qui suivent, comme illustré ci-dessus (à droite) dans le cas où $n=5$ et $m=23$ : nous avons dessiné en noir (avec texte blanc) les jetons ainsi insérés.\n\nAprès avoir procédé de la sorte, Morgane constate que toute suite de $n+1$ jetons consécutifs contient au moins $n$ jetons consécutifs de la configuration qu'elle avait obtenue avant d'insérer ses $a$ jetons surnuméraires, et ces $n$ jetons sont bien de chacune des couleurs $1,2, \\ldots, n$. Comme annoncé, l'entier $m$ est donc bien $n$-coloriable.\n\nRéciproquement, supposons que, partant de l'entier $m=n^{2}-n-1$, Morgane a pu parvenir à ses fins. Puisque $m/n < n-1$, l'une des couleurs, disons la couleur 1, est représentée au plus $n-2$ fois, et au moins une fois. Morgane isole alors un jeton de couleur 1, puis sépare les $m-1 = (n+1)(n-2)$ jetons restants en $n-2$ groupes de $n+1$ jetons consécutifs. Puisqu'elle est satisfaite, chacun de ces groupes contient au moins un jeton de couleur 1 : cela lui fait un minimum de $n-1$ jetons de couleur 1 en tout, ce qui est absurde !\n\nAinsi, l'entier $m = n^{2}-n-1$ n'est pas $n$-coloriable, et il s'agit du plus grand entier qui ne soit pas $n$-coloriable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $x, y, z$ des réels strictement positifs tels que\n$$\nx+\\frac{y}{z}=y+\\frac{z}{x}=z+\\frac{x}{y}=2\n$$\nDéterminer toutes les valeurs possibles que peut prendre le nombre $x+y+z$.",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nDans un tel problème, il faut chercher à examiner chaque équation séparément mais aussi à les mettre en relation. En pratique, cela consiste à regarder l'équation obtenue lorsque l'on effectue la somme ou le produit de deux ou plusieurs équations. Une autre idée est d'appliquer des inégalités connues d'un côté ou de l'autre de l'équation. En effet, souvent les égalités présentes dans les problèmes ne sont possibles que lorsque les variables vérifient le cas d'égalité d'une inégalité bien connue.\n\nCommençons par un examen séparé. On considère la première équation $x+\\frac{y}{z}=2$. D'après l'inégalité des moyennes, on a\n$$\n2=x+\\frac{y}{z} \\geqslant 2 \\sqrt{x \\cdot \\frac{y}{z}}\n$$\nsoit $\\sqrt{\\frac{x y}{z}} \\leqslant 1$. De même, la deuxième équation $y+\\frac{z}{x}=2$ donne $\\sqrt{\\frac{y z}{x}} \\leqslant 1$. En multipliant ces deux inégalités, on obtient\n$$\n1 \\geqslant \\sqrt{\\frac{x y}{z}} \\cdot \\sqrt{\\frac{y z}{x}}=y\n$$\nAinsi $y \\leqslant 1$, et de même on trouve que $x \\leqslant 1$ et $z \\leqslant 1$. On cherche désormais à appliquer ces estimations ou à prouver des estimations inverses (par exemple $y \\geqslant 1$, ce qui imposerait que $y=1$ ).\n\nNotons que la relation $x+\\frac{y}{z}=2$ se réécrit $x z+y=2 z$. On obtient de même que $y x+z=2 x$ et $y z+x=2 y$. En sommant ces trois relations, on trouve que $x z+y x+y z+x+y+z=2(x+y+z)$, ou encore $x y+y z+z x=x+y+z$. Mais alors\n$$\nx+y+z=x y+y z+z x \\leqslant x \\cdot 1+y \\cdot 1+z \\cdot 1=x+y+z\n$$\nNotre inégalité est en fait une égalité. Et puisque $x, y$ et $z$ sont non nuls, chacune des inégalités utilisées dans la ligne ci-dessus est en fait une égalité. On a donc $x=y=z=1$ et $x+y+z=3$. Cette valeur est bien atteignable puisque le triplet $(1,1,1)$ vérifie bien le système de l'énoncé, ce qui termine notre problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A, B, C$ et $D$ quatre points situés sur un cercle $\\Omega$. Soit $E$ et $F$ les points d'intersection des demi-droites $[B A)$ et $[B C)$ avec la tangente à $\\Omega$ en $D$. Soit $T$ un point, situé à l'intérieur du triangle $A B C$, tel que $(T E)$ soit parallèle à $(C D)$ et que $(T F)$ soit parallèle à $(A D)$. Enfin, soit $K$ le point du segment $[D F]$, autre que $D$, tel que $T D=T K$.\nDémontrer que les droites $(A C),(D T)$ et $(B K)$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $P$ et $Q$ les points d'intersection des droites $(T E)$ et $(T F)$ avec $(A C)$. Puisque $(E D)$ est tangente à $\\Omega$, on sait que\n$$\n(P A, P E)=(C A, C D)=(D A, D E)\n$$\ndonc que les points $A, D, E$ et $P$ appartiennent à un même cercle $\\alpha$. On démontre de même que les points $C, D, F$ et $Q$ appartiennent à un même cercle $\\beta$.\nCela signifie en outre que\n$$\n(P Q, P E)=(P A, P E)=(D A, D E)=(F Q, F E)\n$$\ndonc que les points $E, F, P$ et $Q$ appartiennent à un même cercle $\\gamma$. Ainsi, $T$ est le centre radical des trois cercles $\\alpha, \\beta$ et $\\gamma$. On s'intéresse donc à l'axe radical des cercles $\\alpha$ et $\\beta$, qui semble être leur tangente commune en $D$. On note donc $d$ et $d'$ les tangentes à $\\alpha$ et à $\\beta$ en $D$, et on vérifie que\n$$\n\\left(d, d'\\right)=(d, D E)+\\left(D F, d'\\right)=(A D, A E)+(C F, C D)=(A D, A B)+(C B, C D)=0^\\circ,\n$$\nce qui signifie bien que les tangentes $d$ et $d'$ sont confondues. En particulier, cette tangente commune contient $T$.\nPar conséquent, et puisque le triangle $T D K$ est isocèle en $K$, on sait que\n$$\n(T P, T K)=(P E, E D)+(E D, T K)=(D P, D T)+(T D, D K)=(D P, D K)\n$$\nce qui signifie que $P$ appartient au cercle circonscrit à $D K T$. On démontre de même que $Q$ appartient à ce cercle circonscrit, que l'on note $\\tau$.\nForts de toutes ces relations de cocyclicité et de tangence, ensuite que\n$$\n(K T, K D)=(D K, D T)=(D E, d)=(A E, A D)=(A B, A D)=(B D, D K)\n$$\nce qui signifie que les droites $(B D)$ et $(K T)$ sont parallèles. De manière analogue,\n$$\n\\begin{aligned}\n(K P, K D) & =(Q P, Q D)=(Q P, Q F)+(Q F, Q D)=(E P, E F)+(C F, C D) \\\\\n& =(C D, K D)+(B C, C D)=(B C, K D)\n\\end{aligned}\n$$\ndonc $(K P)$ est parallèle à $(B C)$. On démontre de même que $(K Q)$ est parallèle à $(A B)$.\nCes relations de parallélisme indiquent qu'il existe une homothétie qui envoie $A B C D$ sur $Q K P T$, et le centre de cette homothétie appartient bien aux droites $(A Q)=(C P)=(A C)$, $(B K)$ et $(D T)$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $p$ un nombre premier. Montrer qu'il existe une permutation $\\left(a_{1}, \\ldots, a_{p}\\right)$ de $(1, \\ldots, p)$ telle que les entiers $a_{1}, a_{1} \\cdot a_{2}, \\ldots, a_{1} \\cdot a_{2} \\cdot \\ldots \\cdot a_{p}$ donnent $p$ restes deux à deux distincts lorsque qu'on réalise leur division euclidienne par $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn peut être très optimiste sur cet exercice et vouloir que $a_{1} a_{2} \\cdots a_{k} \\equiv k \\bmod p$ pour tout $k$ compris entre $1$ et $p$. Il faut donc que $a_{1}=1$ et pour tout $k \\neq 1$ :\n$$\na_{k} \\equiv \\frac{a_{1} \\cdots a_{k}}{a_{1} \\cdots a_{k-1}} \\equiv \\frac{k}{k-1} \\quad \\bmod p\n$$\noù l'on note $\\frac{a}{b}$ tout nombre entier dont le reste $c$ par la division euclidienne par $p$ vérifie $cb \\equiv a \\bmod p$, qui existe toujours dès que $b$ est non nul. Ainsi, pour $k \\neq 1$, on peut poser $a_{k}$ l'élément de $\\{1, \\ldots, p\\}$ vérifiant $(k-1) a_{k} \\equiv k \\bmod p$.\n\nIl reste juste à vérifier que tous les nombres $a_{k}$ que l'on a formés sont bien deux à deux distincts, et ainsi que $\\left\\{a_{1}, \\ldots, a_{p}\\right\\}$ est bien une permutation de $\\{1, \\ldots, p\\}$. Supposons qu'il existe $i<j$ tels que $a_{i} \\equiv a_{j}$. Si $i=1$, pour tout $j \\geqslant 2$, $\\frac{j}{j-1} \\neq 1 \\bmod p$. Sinon $\\frac{i}{i-1} \\equiv \\frac{j}{j-1}$ implique que $i(j-1) \\equiv j(i-1)$, soit $ij - i \\equiv ij - j$, donc $i \\equiv j \\bmod p$, ce qui est en contradiction avec notre hypothèse de départ.\n\nAinsi tous les $a_{k}$ sont différents modulo $p$. Étant donné qu'il y a $p$ termes dans la suite et $p$ éléments dans $\\{1,2, \\cdots, p\\}$, $\\left(a_{1}, a_{2}, \\cdots, a_{p}\\right)$ est une permutation de $(1,2, \\cdots, p)$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30804,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUne permutation des entiers $1$ à $2022$ est une suite $\\sigma = (\\sigma_{1}, \\ldots, \\sigma_{2022})$ telle que chaque élément de l'ensemble $\\{1, \\ldots, 2022\\}$ soit égal à exactement un terme $\\sigma_{i}$. Quelle est la plus petite valeur possible que peut prendre la somme\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{a_{2022}}{2022}\\right\\rfloor\n$$\nlorsque les entiers $a_{1}, a_{2}, \\ldots, a_{2022}$ forment une permutation des entiers de $1$ à $2022$ ?",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nDe manière générale, notons $\\mathcal{S}(a_{1}, a_{2}, \\ldots, a_{n})$ la somme\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor,\n$$\net $\\mathcal{S}_{n}$ la plus petite valeur que peut prendre $\\mathcal{S}(a_{1}, a_{2}, \\ldots, a_{n})$ lorsque les entiers $a_{1}, a_{2}, \\ldots, a_{n}$ forment une permutation des entiers de $1$ à $n$.\n\nÉtant donnée une telle permutation, soit $k$ l'entier tel que $a_{k}=n$. Puis, pour tout entier $i \\leqslant k$, soit $b_{i}$ le rang de l'entier $a_{i}$ au sein de l'ensemble $\\{a_{1}, \\ldots, a_{k-1}\\}$, c'est-à-dire le nombre d'indices $j \\leqslant k-1$ pour lesquels $a_{j} \\leqslant a_{i}$. Puisque $a_{i} \\geqslant b_{i}$, on constate alors que\n$$\n\\begin{aligned}\n\\mathcal{S}(a_{1}, a_{2}, \\ldots, a_{n}) & \\geqslant \\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{a_{k}}{k}\\right\\rfloor \\\\\n& \\geqslant \\left\\lfloor\\frac{b_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{b_{2}}{2}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{b_{k-1}}{k-1}\\right\\rfloor+\\left\\lfloor\\frac{n}{k}\\right\\rfloor \\\\\n& \\geqslant \\mathcal{S}_{k-1}+\\left\\lfloor\\frac{n}{k}\\right\\rfloor,\n\\end{aligned}\n$$\navec la convention selon laquelle $\\mathcal{S}_{0}=0$.\n\nEn outre, si $a_{1}, a_{2}, \\ldots, a_{k-1}$ est une permutation des entiers de $1$ à $k-1$ pour laquelle $\\mathcal{S}(a_{1}, a_{2}, \\ldots, a_{k-1})=\\mathcal{S}_{k-1}$ et si l'on pose $a_{i}=i-1$ pour tout $i \\geqslant k+1$, on a bien\n$$\n\\mathcal{S}(a_{1}, a_{2}, \\ldots, a_{n})=\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\ldots+\\left\\lfloor\\frac{a_{k-1}}{k-1}\\right\\rfloor+\\left\\lfloor\\frac{n}{k}\\right\\rfloor=\\mathcal{S}_{k-1}+\\left\\lfloor\\frac{n}{k}\\right\\rfloor.\n$$\nOn en déduit que $\\mathcal{S}_{n}=\\min \\{\\mathcal{S}_{k-1}+\\lfloor n / k\\rfloor: 1 \\leqslant k \\leqslant n\\}$.\n\nNous allons maintenant démontrer par récurrence sur $n$ que $\\mathcal{S}_{n}=\\left\\lceil\\log _{2}(n+1)\\right\\rceil$ pour tout entier $n \\geqslant 0$, ce qui nous permettra d'affirmer que $\\mathcal{S}_{2022}=\\left\\lceil\\log _{2}(2023)\\right\\rceil=11$.\n\nEn effet, c'est bien le cas pour $n=0$. Soit maintenant $n$ un entier naturel non nul, puis $k$ un entier tel que $1 \\leqslant k \\leqslant n$ et $\\mathcal{S}_{n}=\\mathcal{S}_{k-1}+\\lfloor n / k\\rfloor$. Si l'on pose $\\lambda=\\lfloor n / k\\rfloor$, on constate que $k>n /(\\lambda+1)$ et que\n$$\n2^{\\lambda}=\\sum_{i=0}^{\\lambda}\\binom{\\lambda}{i} \\geqslant\\binom{\\lambda}{0}+\\binom{\\lambda}{1}=1+\\lambda\n$$\nde sorte que\n$$\n\\mathcal{S}_{n}=\\mathcal{S}(a_{1}, a_{2}, \\ldots, a_{n}) \\geqslant \\mathcal{S}_{k-1}+\\lfloor n / k\\rfloor \\geqslant \\log _{2}(k)+\\lambda>\\log _{2}\\left(2^{\\lambda} n /(\\lambda+1)\\right) \\geqslant \\log _{2}(n)\n$$\nCela signifie que $2^{\\mathcal{S}_{n}}>n$, ou encore que $2^{\\mathcal{S}_{n}} \\geqslant n+1$, donc que $\\mathcal{S}_{n} \\geqslant\\left\\lceil\\log _{2}(n+1)\\right\\rceil$.\n\nRéciproquement, soit $k=\\lceil(n+1) / 2\\rceil$ et $\\ell=\\left\\lceil\\log _{2}((n+1) / 2)\\right\\rceil$. Puisque $(n+1) / 2 \\leqslant 2^{\\ell}$, on sait aussi que $k \\leqslant 2^{\\ell}$, de sorte que\n$$\n\\mathcal{S}_{n} \\leqslant \\mathcal{S}_{k-1}+\\lfloor n / k\\rfloor=\\left\\lceil\\log _{2}(k)\\right\\rceil+1 \\leqslant \\ell+1=\\left\\lceil\\log _{2}(n+1)\\right\\rceil,\n$$\nce qui conclut la récurrence.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn chasseur et un lapin invisible jouent sur une grille infinie à maille carrée, c'est-à-dire où chaque case a quatre voisines : à gauche, à droite, en haut et en bas. Tout d'abord, le chasseur colorie chaque case de la grille, mais ne peut utiliser qu'un nombre fini de couleurs. Le lapin choisit ensuite une case de la grille, qui sera son point de départ. Il commence alors à se déplacer : chaque minute, il indique au chasseur la couleur de la case sur laquelle il se trouve, puis il se déplace sur une des quatre cases adjacentes. Évidemment, comme il est invisible, les seules informations auxquelles le chasseur a accès sont les couleurs que le lapin annonce chaque minute juste avant de se déplacer.\n\nLe chasseur gagne si, au bout d'une durée finie,\n$\\triangleright$ il peut identifier la case que le lapin avait initialement choisie, ou bien si\n$\\triangleright$ le lapin revient en une case où il s'était déjà trouvé précédemment.\n\nLe chasseur dispose-t-il d'une stratégie gagnante?",
    "options": [],
    "answer": "Yes. The hunter has a winning strategy via a finite coloring scheme that allows reconstruction of the rabbit’s starting position (for example, using triple modular colorings or a five-color linear coloring).",
    "solution": "Solution:\n\nLe chasseur dispose en effet d'une stratégie gagnante, par exemple celle que nous donnons ci-dessous. Tout d'abord, il identifie la grille à l'ensemble $\\mathbb{Z}^{2}$, puis il associe à chaque case $\\mathbf{c}=(x, y)$ le triplet $\\varphi(\\mathbf{c})=(x, y, x+y)$.\n\nChaque fois que le lapin se déplace, deux composantes du triplet $\\varphi ( \\mathbf{c} )$ varient de $\\pm 1$, et la troisième composante ne change pas. En outre, connaître deux des coordonnées de $\\varphi(\\mathbf{c})$ suffit à identifier $\\mathbf{c}$. En particulier, s'il existe un ensemble fini $E$ tel que, à tout moment, le lapin se trouve en une case $\\mathbf{c}$ telle que deux des composantes (au moins) de $\\varphi(\\mathbf{c})$ soient dans $E$, on sait que le lapin est contraint d'évoluer dans un ensemble fini de cases, ce qui entraînera immanquablement sa défaite et sa transformation en civet.\n\nOn sait donc qu'au moins deux des trois composantes de $\\varphi(\\mathbf{c})$ prendront une infinité de valeurs. Si le chasseur parvient à identifier, à un instant donné, les valeurs prises par ces composantes, il en déduira la position réelle du lapin et aura donc gagné.\n\nCe faisant, le chasseur s'est ramené au problème unidimensionnel suivant : étant donné un lapin invisible qui se déplace sur l'axe $\\mathbb{Z}$ par pas de $-1,0$ ou $+1$ et qui est contraint à passer par une infinité d'entiers, comment colorier $\\mathbb{Z}$ en usant d'un nombre fini de couleurs peut-il permettre au chasseur d'identifier la position initiale du lapin le long de cet axe ? Une fois un tel coloriage $\\operatorname{col} : \\mathbb{Z} \\rightarrow \\{0,1, \\ldots, \\kappa-1\\}$ utilisant $\\kappa$ couleurs numérotées de $0$ à $\\kappa-1$, le chasseur n'aura plus qu'à utiliser $\\kappa^{3}$ couleurs, identifiées aux éléments de l'ensemble $\\{0,1, \\ldots, \\kappa-1\\}^{3}$, et à colorier la case $\\mathbf{c}=(x, y)$ avec la couleur $(\\operatorname{col}(x), \\operatorname{col}(y), \\operatorname{col}(x+y))$.\n\nPour répondre à ce nouveau problème unidimensionnel, le chasseur colorie $\\mathbb{Z}$ en utilisant trois couleurs, numérotées de $1$ à $3$, et peint en couleur $c$ tous les entiers $n$ tels que $n \\equiv c \\pmod{3}$. De la sorte, il sait à tout instant si le lapin se déplace vers la gauche ou vers la droite.\n\nPuis il repeint avec la couleur $4$ les entiers, préalablement de couleur $3$, qui sont de la forme $3 \\times (-3)^{n}$ avec $n$ entier naturel. Ainsi, il ne perd pas son information précédente sur la valeur des cases modulo $3$, ni sur la valeur $(-1, 0$ ou $+1)$ des mouvements du lapin. Par conséquent, si jamais il arrive à identifier la position du lapin à un moment donné, il connaît sa position à tout moment du jeu.\n\nEnfin, le lapin passera par deux éléments consécutifs de l'ensemble $\\mathcal{E} = \\left\\{ 3 \\times (-3)^{n} : n \\geqslant 0 \\right\\}$. Si ces deux éléments sont $3 \\times (-3)^{n}$ et $3 \\times (-3)^{n+2}$, la différence entre eux, en valeur absolue, est $8 \\times 3^{n}$. Si ce sont $3$ et $-9$, cette différence, en valeur absolue, est égale à $12$. Par conséquent, après que le lapin est passé par deux éléments consécutifs de $\\mathcal{E}$, le chasseur identifie aisément ces deux éléments et, en fonction des variations de position du lapin, l'élément de $\\mathcal{E}$ en lequel le lapin vient d'arriver. Il en déduit alors la position initiale du lapin, ce qui lui permet de gagner le jeu unidimensionnel, donc de gagner le jeu bidimensionnel d'origine.\n\n\nSolution alternative $n^{\\circ} 1$\n\nAfin de suivre à la trace les mouvements du lapin, et au lieu de superposer trois pavages unidimensionnels utilisant chacun trois couleurs, le chasseur aurait tout aussi bien pu utiliser une coloration du plan en cinq couleurs, avec la fonction de coloriage $\\operatorname{col}(x, y) = x + 2y \\pmod{5}$.\n\nVoici ce que donnent les deux coloriages obtenus. Dans la grille de gauche, chaque case contient trois entiers, qui représentent les couleurs (entre $0$ et $2$) associés aux jeux unidimensionnels horizontal, vertical et diagonal. Dans la grille de droite, chaque case contient un unique entier entre $0$ et $4$.\n\n| 2 | 3 | 4 | 0 | 1 | 2 | 3 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 0 | 1 | 2 | 3 | 4 | 0 | 1 |\n| 3 | 4 | 0 | 1 | 2 | 3 | 4 |\n| 1 | 2 | 3 | 4 | 0 | 1 | 2 |\n| 4 | 0 | 1 | 2 | 3 | 4 | 0 |\n| 2 | 3 | 4 | 0 | 1 | 2 | 3 |\n| 0 | 1 | 2 | 3 | 4 | 0 | 1 |\n\nEn revanche, disposer de bandes orientées dans trois directions distinctes et arbitrairement espacées les unes des autres s'avérait, en pratique, nécessaire pour savoir ensuite où était passé le lapin.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $n \\geqslant 1$ ayant la propriété suivante : il existe une permutation $d_{1}, d_{2}, \\ldots, d_{k}$ des diviseurs positifs de $n$ telle que, pour tout $i \\leqslant k$, la somme $d_{1}+d_{2}+\\ldots+d_{i}$ soit un carré parfait.",
    "options": [],
    "answer": "n = 1 and n = 3",
    "solution": "Solution:\n\nSoit $n$ un des entiers recherchés, et $d_{1}, d_{2}, \\ldots, d_{k}$ une permutation adéquate des diviseurs positifs de $n$. Pour tout entier $i \\leqslant k$, on pose $s_{i}=\\sqrt{d_{1}+d_{2}+\\ldots+d_{i}}$. On dit qu'un entier $\\ell$ est bon si $s_{i}=i$ et $d_{i}=2 i-1$ pour tout $i \\leqslant \\ell$. Ci-dessous, nous allons démontrer que tout entier $\\ell \\leqslant k$ est bon.\n\nTout d'abord, pour tout entier $i \\geqslant 2$, on remarque déjà que\n$$\nd_{i}=s_{i}^{2}-s_{i-1}^{2}=\\left(s_{i}-s_{i-1}\\right)\\left(s_{i}+s_{i-1}\\right) \\geqslant s_{i}+s_{i-1} \\geqslant 2 .\n$$\nPar conséquent, $d_{1}=s_{1}=1$.\n\nConsidérons maintenant un bon entier $\\ell \\leqslant k-1$. Nous allons démontrer que $\\ell+1$ est bon lui aussi. En effet, si $s_{\\ell+1}+s_{\\ell}$ divise $\\left(s_{\\ell+1}-s_{\\ell}\\right)\\left(s_{\\ell+1}+s_{\\ell}\\right)=d_{\\ell+1}$, donc divise $n$. Il existe donc un entier $m$ tel que\n$$\nd_{m}=s_{\\ell+1}+s_{\\ell} \\geqslant 2 s_{\\ell}+1 \\geqslant 2 \\ell+1\n$$\nComme $d_{m}>d_{i}$ pour tout $i \\leqslant \\ell$, on en déduit que $m \\geqslant \\ell+1$. Mais alors\n$$\ns_{\\ell+1}+s_{\\ell}=d_{m}=\\left(s_{m}-s_{m-1}\\right)\\left(s_{m}+s_{m-1}\\right) \\geqslant\\left(s_{m}-s_{m-1}\\right)\\left(s_{\\ell+1}+s_{\\ell}\\right) \\geqslant s_{\\ell+1}+s_{\\ell}\n$$\nLes inégalités sont donc des égalités, ce qui signifie que $s_{m}+s_{m-1}=s_{\\ell+1}+s_{\\ell}$, donc que que $m=\\ell+1$, et que $s_{m}-s_{m-1}=1$, c'est-à-dire que $s_{\\ell+1}=s_{\\ell}+1=\\ell+1$. On en conclut que\n$$\nd_{\\ell+1}=\\left(s_{\\ell+1}-s_{\\ell}\\right)\\left(s_{\\ell+1}+s_{\\ell}\\right)=2 \\ell+1\n$$\nce qui signifie comme prévu que $\\ell+1$ est bon.\n\nEn conclusion, les diviseurs de $n$ sont les entiers $1,3, \\ldots, 2 k-1$. Réciproquement, si les diviseurs de $n$ sont les entiers $1,3, \\ldots, 2 k-1$, l'entier $n$ convient assurément.\n\nEn particulier, si $k \\geqslant 2$, l'entier $d_{k-1}$ est un diviseur impair de $n-d_{k-1}=2$, donc $d_{k-1}=1$ et $k=2$. Ainsi, soit $k=1$, auquel cas $n=1$, soit $k=2$, auquel cas $n=3$. Dans les deux cas, ces valeurs de $n$ conviennent. Les entiers recherchés sont donc $n=1$ et $n=3$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 30807,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les couples $(x, y)$ d'entiers strictement positifs tels que $x y \\mid x^{2}+2 y-1$.",
    "options": [],
    "answer": "All pairs are: (1, y) for any positive integer y; (2y − 1, y) for any positive integer y; and the two additional pairs (3, 8) and (5, 8).",
    "solution": "Solution:\n\nD'une part, on peut écrire $x \\mid x y \\mid x^{2}+2 y-1$ donc $x \\mid 2 y-1$. Donc il existe $n$ tel que $2 y-1 = n x$. Forcément, $n$ et $x$ sont impairs.\n\nD'autre part, la relation de divisibilité de l'énoncé nous donne une inégalité : on sait que $x y > 0$, donc\n$$\nx y \\leqslant x^{2}+2 y-1\n$$\nEn multipliant cette relation par $2$ pour simplifier les calculs et en combinant les deux, on obtient\n$$\nx(n x+1) \\leqslant 2 x^{2}+2(n x+1)-2\n$$\nEn réorganisant les termes,\n$$\n(n-2) x^{2} \\leqslant (2 n-1) x\n$$\nSoit, comme $x > 0$\n$$\nx \\leqslant \\frac{2 n-1}{n-2} = 2 + \\frac{3}{n-2}\n$$\nEn se souvenant que $n$ et $x$ sont forcément impairs, on distingue plusieurs cas.\n\nCas $n^{\\circ} 1$: $n \\geqslant 7$. Alors\n$$\nx \\leqslant 2 + \\frac{3}{n-2} < 3\n$$\nDonc $x=1$. Réciproquement, tous les couples $(1, y)$ sont solution.\n\nCas $n^{\\circ} 2$: $n=5$. Alors\n$$\nx \\leqslant 2 + \\frac{3}{n-2} = 3\n$$\nDonc $x=1$ ou $x=3$. Tous les couples avec $x=1$ sont solutions. Si $x=3$, on a $2 y-1 = n x = 5 \\cdot 3 = 15$ donc $y=8$. On vérifie $(3,8)$ est solution : $24 \\mid 24$.\n\nCas $n^{\\circ} 3$: $n=3$. Alors\n$$\nx \\leqslant 2 + \\frac{3}{n-2} = 5\n$$\nDonc $x=1$ ou $x=3$ ou $x=5$. Tous les couples avec $x=1$ sont solutions. Si $x=3$, on a $2 y-1 = n x = 3 \\cdot 3 = 9$ donc $y=5$. On vérifie que $(3,5)$ n'est pas solution : $15 \\nmid 24$. Si $x=5$ on a $2 y-1 = n x = 3 \\cdot 5 = 15$ donc $y=8$. On vérifie que $(5,8)$ est solution : $40 \\mid 40$.\n\nCas $n^{\\circ} 4$: $n=1$. Alors $x=2 y-1$. Alors\n$$\nx y = 2 y^{2} - y \\mid 4 y^{2} - 2 y = (2 y-1)^{2} + 2 y - 1 = x^{2} + 2 y - 1\n$$\nDonc tous les couples de la forme $(2 y-1, y)$ sont solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30808,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier. Trouver tous les diviseurs $d \\geqslant 1$ de $3 n^{2}$ tels que $n^{2}+d$ soit un carré parfait.",
    "options": [],
    "answer": "d = 3n^2",
    "solution": "Solution:\n\nSoit $d \\geqslant 1$ un diviseur de $3 n^{2}$ tel que $n^{2}+d$ soit un carré parfait. Puisque $n^{2}<n^{2}+d\\leqslant n^{2}+3 n^{2}=(2 n)^{2}$, il existe $k \\in \\llbracket 1, n \\rrbracket$ tel que $n^{2}+d=(n+k)^{2}$. En simplifiant, on obtient $d=2 k n+k^{2}$, de sorte que $k(k+2 n) \\mid 3 n^{2}$.\n\nNotons $\\alpha=\\gcd(k, n) \\geqslant 1$. On dispose d'entiers $\\ell, m \\geqslant 1$ premiers entre eux tels que $k=\\alpha \\ell$, $n=\\alpha m$. La relation de divisibilité précédente s'écrit alors, après simplification, $\\ell(\\ell+2 m) \\mid 3 m^{2}$. Or $\\gcd(\\ell+2 m, m)=\\gcd(\\ell, m)=1$, c'est-à-dire $\\ell+2 m$ premier avec $m$, donc avec $m^{2}$. Comme $\\ell+2 m \\mid 3 m^{2}$, on a par le lemme de Gauss $\\ell+2 m \\mid 3$, c'est-à-dire $\\ell+2 m \\in\\{1,3\\}$. Mais $\\ell, m \\geqslant 1$ donc $\\ell+2 m \\geqslant 3$. Autrement dit, $\\ell+2 m=3$ et donc nécessairement, $\\ell=m=1$. Alors $k=n=\\alpha$, et $d=k(k+2 n)=3 n^{2}$.\n\nRéciproquement, $d=3 n^{2}$ convient puisque c'est un diviseur de $3 n^{2}$ et que $n^{2}+3 n^{2}=(2 n)^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ un entier. Une grenouille se déplace dans une grille de $3n \\times 3n$ cases, sautant chaque fois d'une case vers la case située juste au-dessus ou juste à droite. Elle souhaite aller du coin en bas à gauche, marqué d'un $S$, au coin en haut à droite, marqué d'un $F$. Malheureusement, certaines cases sont piégées, et si la grenouille saute sur une telle case, elle y restera prisonnière.\n\nOn dit qu'un ensemble de cases piégées est bloquant s'il empêche la grenouille d'aller de la case $S$ à la case $F$. On dit aussi qu'un ensemble bloquant est minimal si, dès lors que l'on retire le piège de l'une des cases de cet ensemble, notre ensemble n'est plus bloquant.\n\nPar exemple, les cases noires de la grille ci-dessous forment un ensemble bloquant minimal.\n\n![](attached_image_1.png)\n\na) Démontrer qu'il existe un ensemble bloquant minimal qui contient au moins $3n(n-1)$ cases.\n\nb) Démontrer que tout ensemble bloquant minimal contient au plus $3n^{2}$ cases.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nColorions notre grille en trois couleurs :\n- en noir, les cases piégées ;\n- en rose, les cases non piégées auxquelles la grenouille peut accéder depuis la case $S$ ;\n- en vert, les cases non piégées depuis lesquelles la grenouille peut accéder à la case $F$.\n\nNotons que notre grille contient éventuellement des cases incolores.\n\nNotre ensemble de cases piégées est bloquant si et seulement si nulle case n'est à la fois rose et verte, c'est-à-dire si nulle case verte n'est située juste au-dessus ou juste à droite d'une case rose. Notre ensemble bloquant est donc minimal s'il consiste en la seule case $S$, ou bien en la seule case $F$, ou bien ne contient aucune de ces cases et si toute case noire est située juste au-dessus ou juste à droite d'une case rose, et juste en-dessous ou juste à gauche d'une case verte.\n\nDe la sorte, on vérifie aisément que la construction ci-dessous nous fournit un ensemble bloquant minimal composé de $n(3n-2) \\geqslant 3n(n-1)$ cases piégées.\n\n![](attached_image_2.png)\n\nOn démontre maintenant que tout ensemble bloquant minimal contient au plus $3n^{2}$ cases. Étant donné un tel ensemble bloquant minimal, soit $r$ le nombre de cases roses, $p$ le nombre de cases piégées et $v$ le nombre de cases vertes. Au vu de la minimalité de notre ensemble bloquant, et à part la case $S$, chaque case rose ou noire se trouve juste au-dessus ou juste à gauche d'une case rose. On en déduit que $r-1+p \\leqslant 2r$, c'est-à-dire que $p \\leqslant r+1$. On démontre de même que $p \\leqslant v+1$, et on en conclut que $3n^{2} \\geqslant (r+p+v)/3 \\geqslant p-2/3$, donc que $p \\leqslant 3n^{2}$.\nSolution:\n\nVoici une manière alternative, plus progressive mais nettement plus longue, de démontrer que tout ensemble bloquant minimal contient au plus $3n^{2}$ cases. On suppose évidemment que cet ensemble n'est pas constitué de la seule case $S$ ou de la seule case $F$, auquel cas on aurait déjà gagné.\n\nOn aimerait bien découper dans la grille $3n \\times 3n$ des rectangles $3 \\times 1$ ou $1 \\times 3$ contenant chacun une case de chaque couleur, de sorte que les cases n'appartenant à aucun rectangle ne soient pas piégées. On disposerait alors de la borne supérieure tant désirée.\n\nOn commence par démontrer que la grille $3n \\times 3n$ ne contient aucune case incolore. En effet, si une case rose $(x, y)$ a pour voisine une case incolore, il s'agit de sa voisine de gauche ou du dessous. Quitte à effectuer une symétrie par rapport à la diagonale $(SF)$, on suppose que c'est la voisine du dessous, c'est-à-dire $(x, y-1)$. On démontre alors par récurrence sur $k$ que les cases $(x-k, y)$ et $(x-k, y-1)$ sont rose et incolore, comme illustré ci-dessous, les cases incolores étant simplement laissées en blanc.\n\n![](attached_image_3.png)\n\nEn effet,\n- la case $(x-k-1, y)$ est nécessairement rose, ce sans quoi aucune des voisines de gauche et du dessous de la case $(x-k, y)$ ne serait rose ;\n- la case $(x-k-1, y-1)$ n'est ni rose (sinon sa voisine de droite serait rose), ni verte ou noire (sinon l'une de ses voisines du haut ou de droite serait verte).\n\nEn particulier, la case $(0, y)$ est rose tandis que la case $(0, y-1)$ est incolore, ce qui est absurde. Puis, si une case noire $(x, y)$ a pour voisine de gauche ou du dessous une case incolore, on suppose toujours, à symétrie près, que cette case incolore est $(x, y-1)$. Mais alors $(x-1, y)$ est rose et $(x-1, y-1)$ est incolore, ce qui est impossible.\n\nEnfin, quitte à effectuer une symétrie par rapport à la médiatrice de $[SF]$ et à échanger les couleurs rose et verte, une case incolore ne peut pas non plus se situer en haut ou à droite d'une case noire, ni être adjacente à une case verte. Ainsi, nulle case n'est incolore.\n\nOn entreprend désormais de démontrer que chaque case noire, sauf les éventuels coins noirs, a soit une case rose à sa gauche et une case verte à sa droite, soit une case rose en-dessous et une case verte au-dessus. Dans les deux cas, on dira que ces trois cases forment un rectangle tricolore.\n\nSi une case noire est disposée sur le côté bas, sa voisine de gauche est nécessairement rose, et sa voisine de droite n'est ni rose ni noire, donc elle est verte. On traite de même les autres cases noires situées sur les côtés, et on s'intéresse désormais aux cases éloignées des côtés.\n\nSupposons que l'on dispose d'une case noire, disons en position $(x, y)$, dont la voisine de droite, en position $(x+1, y)$, est rose ou noire. Toute case rose ou noire a une case rose à sa gauche ou juste en-dessous. Par conséquent, la case $(x+1, y-1)$ est rose. De même, toute case verte ou noire a une case verte à sa droite ou juste au-dessus, donc la case $(x, y-1)$ est rose et la case $(x, y+1)$ est verte. La case $(x, y)$ appartient donc à un rectangle tricolore.\n\n![](attached_image_4.png)\n\nQuitte à appliquer une symétrie par rapport au milieu de $[SF]$, ce qui échangera les couleurs verte et rose, on démontre de même que, si la voisine de gauche de $(x, y)$ est verte ou noire, la case $(x, y)$ appartient à un rectangle tricolore. Enfin, si la voisine de gauche de $(x, y)$ est rose et sa voisine de droite est verte, la case $(x, y)$ appartient de facto à un rectangle tricolore horizontal.\n\nOn peut donc, pour chaque case noire autre que les cases situées dans les coins, choisir un rectangle tricolore dont elle est la case centrale. Par ailleurs, si une case noire est située dans le coin en haut à gauche ou en bas à droite, on peut tout de même l'insérer dans un $L$ tricolore plutôt que dans un rectangle tricolore. Il reste à étudier le cas où deux zones de notre sélection (qui seront communément des rectangles) auraient une case commune.\n\nSi deux rectangles tricolores partagent leur case verte, disons $(x, y)$, on est dans la situation représentée ci-dessous. Par construction, les cases $(x-1, y)$ et $(x, y-1)$ sont noires et les cases $(x-2, y)$ et $(x, y-2)$ sont roses, puis on montre que les quatre cases du carré $2 \\times 2$ situé en bas à gauche sont roses elles aussi. Cette situation est illustrée ci-dessous, à gauche, les deux rectangles étant mis en surbrillance.\n\n![](attached_image_5.png)\n\nOn décide alors de supprimer nos deux rectangles et les remplacer par la zone en forme de $\\Psi$ et mise en surbrillance ci-dessus, à droite. Celle-ci a certes le mauvais goût de ne plus contenir deux cases de chaque couleur, mais elle ne contient néanmoins que deux cases noires sur six. En outre, aucune case de notre motif n'appartient à un rectangle autre que ceux que nous avons supprimés, ni à un autre motif que nous aurions créé, car un tel motif est bordé par des rectangles tricolores de notre sélection sur ses côtés haut et droit.\n\nDe même, chaque case noire de notre motif ne sera intégrée à aucun autre motif ou rectangle, et chaque case rose du motif. Le même phénomène peut bien sûr être mis en place si deux rectangles ont une case rose commune, auquel cas on les remplace par une zone $\\Psi$ formée de deux cases piégées et de quatre cases vertes n'appartenant à aucun rectangle.\n\nEnfin, le coin en haut à gauche est piégé, et si le $L$ tricolore auquel il appartient partage sa case verte avec un rectangle tricolore, comme illustré ci-dessous à gauche, on le remplace par le rectangle formé des cases $(1,3n-2), (1,3n-1), (1,3n)$, comme illustré ci-dessous à droite. Les autres cas d'intersection entre un $L$ situé dans un coin et un rectangle se traitent de la même manière.\n\n![](attached_image_6.png)\n\nEn conclusion, nous sommes parvenus à découper dans notre grille $3n \\times 3n$ des zones deux à deux disjointes dont chacune contient un tiers de cases piégées, et dont l'union recouvre l'ensemble des cases piégées. Ainsi, comme désiré, au plus un tiers des cases de la grille sont piégées.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\left(a_{n}\\right)_{n \\geqslant 1}$ une suite d'entiers naturels non nuls telle que $a_{n+2 m}$ divise $a_{n}+a_{n+m}$ pour tous les entiers $m \\geqslant 1$ et $n \\geqslant 1$. Démontrer que cette suite est ultimement périodique, c'est-à-dire qu'il existe deux entiers $N \\geqslant 1$ et $d \\geqslant 1$ tels que $a_{n}=a_{n+d}$ pour tout $n \\geqslant N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons qu'il existe un entier $k \\geqslant 9$ tel que $a_{i}<a_{k}$ pour tout $i \\leqslant k-1$. Dans ces conditions, et pour tout entier $i<k / 2$, on sait que $a_{k-2 i}+a_{k-i}$ est un multiple de $a_{k}$ et que $a_{k-2 i}+a_{k-i}<2 a_{k}$, de sorte que $a_{k-2 i}+a_{k-i}=a_{k}$. En choisissant $i=1$, $i=2$ et $i=4$, on en déduit que\n$$\na_{k-8}+a_{k-4}=a_{k-4}+a_{k-2}=a_{k}=a_{k-2}+a_{k-1}\n$$\ndonc que $a_{k-8}=a_{k-2}$ et $a_{k-4}=a_{k-1}$.\nUne récurrence sur $\\ell$ indique alors que $a_{k-1}$ divise $a_{k-1-3 \\ell}$ pour tout $\\ell \\leqslant(k-2) / 3$, de sorte que $a_{k-1} \\leqslant \\max \\left\\{a_{1}, a_{2}, a_{3}\\right\\}$, et que $a_{k-2}$ divise $a_{k-2-6 \\ell}$ pour tout $\\ell \\leqslant(k-3) / 6$, de sorte que $a_{k-2} \\leqslant \\max \\left\\{a_{1}, a_{2}, \\ldots, a_{6}\\right\\}$. On en conclut que\n$$\na_{k}=a_{k-1}+a_{k-2} \\leqslant 2 \\max \\left\\{a_{1}, a_{2}, \\ldots, a_{6}\\right\\}\n$$\nPar conséquent, la suite $\\left(a_{n}\\right)_{n \\geqslant 1}$ est bornée.\nSoit $\\lambda$ la valeur maximale que prend infiniment souvent la suite $\\left(a_{n}\\right)_{n \\geqslant 1}$, et soit $\\kappa \\geqslant 0$ un entier tel que $a_{n} \\leqslant \\lambda$ pour tout $n \\geqslant \\kappa+1$. Quitte à oublier les $\\kappa$ premiers termes de la suite, c'est-à-dire à remplacer chaque terme $a_{n}$ par le terme $\\hat{a}_{n}=a_{n+\\kappa}$, ce qui ne change rien à l'énoncé, on suppose que $\\kappa=0$, c'est-à-dire que $a_{n} \\leqslant \\lambda$ pour tout $n \\geqslant 1$.\nConsidérons l'ensemble $\\mathcal{E}=\\left\\{k \\geqslant 1: a_{k}=\\lambda\\right\\}$, puis trois éléments consécutifs $k<\\ell<m$ de $\\mathcal{E}$. Si $k \\equiv \\ell(\\bmod 2)$, on sait que $\\lambda=a_{\\ell}$ divise $a_{k}+a_{(k+\\ell) / 2}=\\lambda+a_{(k+\\ell) / 2}$, donc que $a_{(k+\\ell) / 2}=\\lambda$, en contradiction avec le fait que $k$ et $\\ell$ étaient des éléments consécutifs de $\\mathcal{E}$.\nOn en déduit que $k \\not \\equiv \\ell(\\bmod 2)$ et, de même, que $\\ell \\not \\equiv m(\\bmod 2)$. Mais alors $\\lambda=a_{m}$ divise $a_{k}+a_{(k+m) / 2}=\\lambda+a_{(k+m) / 2}$, donc $a_{(k+m) / 2}=\\lambda$, et $(k+m) / 2$ est un élément de $\\mathcal{E}$ compris entre $k$ et $m$, de sorte que $(k+m) / 2=\\ell$. Cela signifie que $\\ell-k=m-\\ell$ et, plus généralement, que les éléments de $\\mathcal{E}$ forment une progression arithmétique infinie, dont on note $d$ la raison.\nEnfin, soit $n \\geqslant 1$ un entier quelconque, puis soit $m \\geqslant 1$ un entier tel que $m+n \\in \\mathcal{E}$. Alors $\\lambda=a_{m+n}$ divise $a_{n}+a_{m}$, et $m+n+d \\in \\mathcal{E}$, donc $\\lambda=a_{m+n+d}$ divise $a_{n+d}+a_{m}$. On en conclut que $a_{n} \\equiv-a_{m} \\equiv a_{n+d}(\\bmod \\lambda)$. Comme $a_{n}$ et $a_{n+d}$ sont compris entre 1 et $\\lambda$, cela signifie que $a_{n}=a_{n+d}$. Par conséquent, la suite $\\left(a_{n}\\right)_{n \\geqslant 1}$ est $d$-périodique, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30811,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier strictement positif. Sur un mur, $n$ clous sont plantés. Chaque paire de clous est reliée par une corde coloriée à l'aide d'une des $n$ couleurs. On dit que le mur est coloré si pour tout triplet de couleurs deux à deux distinctes $a, b, c$, il existe trois clous tels que les trois cordes reliant ces clous soient de couleur $a, b$ et $c$.\n\n- Existe-t-il un mur coloré pour $n=6$ ?\n- Existe-t-il un mur coloré pour $n=7$ ?",
    "options": [],
    "answer": "No for n = 6; Yes for n = 7.",
    "solution": "Solution:\n\nC'est en fait la parité de $n$ qui est cruciale.\n\nCas $n$ pair : Il y a $\\frac{n(n-1)}{2}$ cordes, donc en moyenne il y a $\\frac{n-1}{2}$ cordes de chaque couleur. Ce nombre n'étant pas entier, il y a des couleurs avec plus de cordes que la moyenne et d'autres avec moins. On pourrait tout à fait choisir une couleur sous-représentée et comparer le nombre de triplets de couleurs comportant cette couleur avec le nombre de triplets de cordes comportant une corde de cette couleur pour obtenir une contradiction. On va ici choisir l'approche opposée : on considère une couleur sur-représentée. Il y a au moins $\\frac{n+1}{2}$ cordes de cette couleur, donc au moins $n+1$ extrémités de cordes de cette couleur. D'après le principe des tiroirs, il existe un sommet ayant deux extrémités de cordes de cette couleur. Il existe donc un triplet de cordes comportant deux fois cette couleur, et ne correspondant donc à aucun triplet de couleur. Cependant il y a $\\binom{n}{3}$ triplets de cordes qui doivent remplir les $\\binom{n}{3}$ triplets de couleurs possibles, il n'y a aucune marge pour gâcher des triplets de cordes. Il est donc impossible d'obtenir un mur coloré pour $n$ pair.\n\nCas $n$ impair : Ici le raisonnement précédent nous indique juste que pour chaque couleur il faut $\\frac{n-1}{2}$ cordes reliant $n-1$ points distincts et laissant un unique point non relié par cette couleur. On trouve la construction suivante : on numérote les clous dans $\\mathbb{Z} / n \\mathbb{Z}$ et les couleurs aussi. Ensuite, on donne à la corde entre le clou $i$ et le clou $j$ la couleur $i+j$. Cette construction semble prometteuse car elle est totalement symétrique et vérifie la propriété voulue. En effet, on vérifie facilement que le triplet de couleur de corde $a, b, c$ est atteint en prenant les clous $\\frac{a+b}{2}, \\frac{a+c}{2}, \\frac{b+c}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn ensemble $A$ d'entiers est dit admissible s'il vérifie la propriété suivante : pour tous $x, y \\in A$ (non nécessairement distincts), et pour tout $k \\in \\mathbb{Z}$, on a $x^{2}+k x y+y^{2} \\in A$.\n\nDéterminer tous les couples d'entiers non nuls ( $m, n$ ) tels que le seul ensemble admissible contenant à la fois $m$ et $n$ soit $\\mathbb{Z}$.",
    "options": [],
    "answer": "All pairs of nonzero integers that are coprime (their greatest common divisor is one).",
    "solution": "Solution:\n\nRemarquons que si $d=\\operatorname{gcd}(m, n)>1$ alors l'ensemble $A$ des multiples de $d$ est un ensemble admissible contenant $n$ et $m$ mais n'étant pas égal à $\\mathbb{Z}$. En effet, si $x$ et $y$ sont multiples de $d$, alors pour tout entier $k$, $x^{2}+k x y+y^{2}$ est bien divisible par $d$ et est dans $A$. On déduit que les couples $(m, n)$ recherchés doivent vérifier $\\operatorname{gcd}(m, n)=1$.\n\nRéciproquement, soient $m$ et $n$ des entiers tels que $\\operatorname{gcd}(m, n)=1$. Montrons que tout ensemble admissible contenant $m$ et $n$ est $\\mathbb{Z}$. Soit $A$ un ensemble admissible contenant $m$ et $n$. Remarquons que si $x \\in A$, alors l'hypothèse nous donne que pour tout entier $k$, l'entier $x^{2}+k x \\cdot x+x^{2}=(k+2) x^{2}$ est dans $A$. On déduit que tout multiple de $x^{2}$ est dans $A$ si $x$ est dans $A$.\n\nNotons que si $\\operatorname{gcd}(m, n)=1$, alors on a aussi $\\operatorname{gcd}\\left(m^{2}, n^{2}\\right)=1$. D'après le théorème de Bézout, on dispose donc d'entiers $u, v \\in \\mathbb{Z}$ tels que $u m^{2}+v n^{2}=1$. Comme $u m^{2}$ et $v n^{2}$ sont dans $A$, en appliquant l'hypothèse à $x=u m^{2}$, $y=v n^{2}$ et $k=2$, on a\n$$\n\\left(u m^{2}\\right)^{2}+2 u m^{2} v n^{2}+\\left(v n^{2}\\right)^{2}=\\left(u m^{2}+v n^{2}\\right)^{2}=1^{2}=1 \\in A\n$$\nPar conséquent, $\\forall k \\in \\mathbb{Z}, k \\cdot 1^{2} \\in A$. Ainsi $A=\\mathbb{Z}$ et le seul ensemble admissible contenant $m$ et $n$ est $\\mathbb{Z}$.\n\nLes couples recherchés sont donc les couples d'entiers $(m, n)$ premiers entre eux.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un parallélogramme. Une droite $d$ passant par $C$ coupe respectivement les droites $(AB)$ et $(AD)$ en $X$ et en $Y$. Les tangentes en $X$ et en $Y$ au cercle circonscrit à $AXY$ se coupent en un point $T$, puis la droite $(CT)$ recoupe le cercle circonscrit à $CDY$ en un point $P$.\n\nDémontrer que les points $A, B, D$ et $P$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans la suite, on note $\\mathcal{C}_{\\mathcal{P}}$ le cercle circonscrit à un triangle $\\mathcal{P}$.\n\nPuisque $(TX)$ est tangente à $\\mathcal{C}_{AXY}$, une chasse aux angles indique que\n$$\n(XT, XY) = (AX, AY) = (DC, DY) = (PC, PY) = (PT, PY)\n$$\nce qui signifie que $P$ appartient à $\\mathcal{C}_{TXY}$.\n\nDe même, puisque $(TY)$ est tangente à $\\mathcal{C}_{AXY}$, une chasse aux angles indique que\n$$\n(PC, PX) = (PT, PX) = (YT, YX) = (AY, AX) = (BC, BX),\n$$\nce qui signifie que $P$ appartient à $\\mathcal{C}_{BCX}$.\n\nOn en conclut que\n$$\n(BP, BA) = (BP, BX) = (CP, CX) = (CP, CY) = (DP, DY) = (DP, DA)\n$$\nce qui signifie que $P$ appartient à $\\mathcal{C}_{ABD}$.\n\n![](attached_image_1.png)\nSolution:\n\nLe tracé de la figure suggère que $(TY)$ est tangente aux deux cercles $\\mathcal{C}_{AXY}$ et $\\mathcal{C}_{CDYP}$. On vérifie alors que c'est bien le cas, puisque\n$$\n(YT, YD) = (YT, YA) = (AY, AX) = (DY, DC)\n$$\nAinsi, $T$ est le centre radical des trois cercles $\\mathcal{C}_{AXY}$, $\\mathcal{C}_{CDYP}$ et $\\mathcal{C}_{PCX}$, et puisque l'axe radical $(TX)$ des cercles $\\mathcal{C}_{AXY}$ et $\\mathcal{C}_{PCX}$ est tangent à $\\mathcal{C}_{AXY}$, il est aussi tangent à $\\mathcal{C}_{PCX}$.\n\nOn en déduit en particulier que\n$$\n(PX, PC) = (XT, XC) = (XT, XY) = (AX, AY) = (BX, BC),\n$$\nde sorte que $B$ appartient à $\\mathcal{C}_{PCX}$, et que\n$$\n(XT, XP) = (CX, CP) = (CY, CP) = (YT, YP)\n$$\nde sorte que $P$ appartient à $\\mathcal{C}_{TXY}$.\n\nOn en conclut comme précédemment que\n$$\n(BP, BA) = (BP, BX) = (CP, CX) = (CP, CY) = (DP, DY) = (DP, DA)\n$$\nce qui signifie que $P$ appartient à $\\mathcal{C}_{ABD}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30814,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\Gamma$ un cercle et $P$ un point à l'extérieur de $\\Gamma$. Les tangentes à $\\Gamma$ issues de $P$ touchent $\\Gamma$ en $A$ et $B$. Soit $K$ un point distinct de $A$ et $B$ sur le segment $[AB]$. Le cercle circonscrit au triangle $PBK$ recoupe le cercle $\\Gamma$ au point $T$. Soit $P'$ le symétrique du point $P$ par rapport au point $A$. Montrer que $\\widehat{\\mathrm{PBT}} = \\widehat{\\mathrm{P}'\\mathrm{KA}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nD'après le théorème de l'angle inscrit, $\\widehat{\\mathrm{PBT}} = \\widehat{\\mathrm{PKT}}$. Ainsi, il suffit de montrer que $\\widehat{\\mathrm{PKT}} = \\widehat{\\mathrm{P}'\\mathrm{KA}}$. La figure semble suggérer que les triangles $PKT$ et $P'KA$ sont semblables, nous allons donc essayer de montrer ce résultat, qui impliquera bien l'égalité d'angle voulue.\n\nPour montrer que $\\triangle PKT \\sim \\triangle P'KA$, on a plusieurs possibilités. Par exemple, on peut montrer que $\\widehat{TPK} = \\widehat{K' A}$ et $\\widehat{KTP} = \\widehat{KAP}$. Toutefois, lorsque l'on essaye de calculer l'angle $\\widehat{TPK}$, on se rend compte que l'égalité $\\widehat{TPK} = \\widehat{KP' A}$ est équivalente à l'énoncé à démontrer. On va donc montrer que $\\widehat{KTP} = \\widehat{P' AK}$ et que $\\frac{KT}{KA} = \\frac{TP}{AP'}$.\n\nPour montrer que $\\widehat{KTP} = \\widehat{P' AK}$, il faut d'abord remarquer que le triangle $APB$ est isocèle en $P$ puisque les droites $(AP)$ et $(BP)$ sont tangentes en $A$ et $B$ au cercle $\\Gamma$. En appliquant le théorème de l'angle inscrit dans le cercle passant par $P, B, K$ et $T$, on trouve :\n$$\n\\widehat{KTP} = 180^\\circ - \\widehat{KBP} = 180^\\circ - \\widehat{ABP} = 180^\\circ - \\widehat{BAP} = \\widehat{PAK}\n$$\nIl reste donc à montrer que $\\frac{KT}{KA} = \\frac{TP}{AP'}$. Puisque $AP' = AP = BP$, il suffit de montrer que $\\frac{KT}{KA} = \\frac{TP}{BP}$. Il suffit donc de montrer que les triangles $AKT$ et $BPT$ sont semblables. Pour cela, il suffit de montrer que les angles de ces deux triangles sont deux à deux égaux.\n\nD'une part, par angle inscrit, $\\widehat{AKT} = 180^\\circ - \\widehat{BKT} = \\widehat{TPB}$. D'autre part par angle tangentiel, $\\widehat{TAK} = \\widehat{TAB} = \\widehat{TBP}$.\n\nDonc les triangles $AKT$ et $BPT$ sont semblables comme voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30815,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver les entiers $n$ pour lesquels l'égalité\n$$\n\\sum_{i=1}^{n} \\sum_{j=1}^{n}\\left\\lfloor\\frac{i \\times j}{n+1}\\right\\rfloor=\\frac{n^{2}(n-1)}{4}\n$$\nest vérifiée.\nOn rappelle que, pour tout nombre réel $x$, la notation $\\lfloor x\\rfloor$ désigne la partie entière de $x$, c'est-à-dire le plus grand entier inférieur ou égal à $x$.",
    "options": [],
    "answer": "All integers such that n+1 is prime",
    "solution": "Solution:\n\nPour tout entier $k$, on note $r_{n+1}(k)$ le reste de la division euclidienne de $k$ par $n+1$, c'est-à-dire le plus petit entier naturel tel que $n+1$ divise $k-r_{n+1}(k)$. La double somme de gauche se réécrit alors comme\n$$\n\\begin{aligned}\n\\mathcal{S} & =\\sum_{i=1}^{n} \\sum_{j=0}^{n} \\frac{i j-r_{n+1}(i j)}{n+1} \\\\\n& =\\frac{1}{n+1}\\left(\\sum_{i=1}^{n} i\\right)\\left(\\sum_{j=0}^{n} j\\right)-\\frac{1}{n+1} \\sum_{i=0}^{n} \\sum_{j=0}^{n} r_{n+1}(i j) \\\\\n& =\\frac{1}{n+1} \\frac{n(n+1)}{2} \\frac{n(n+1)}{2}-\\frac{1}{n+1} \\sum_{i=0}^{n} \\sum_{j=0}^{n} r_{n+1}(i j) \\\\\n& =\\frac{n^{2}(n+1)}{4}-\\frac{1}{n+1} \\sum_{i=0}^{n} \\sum_{j=0}^{n} r_{n+1}(i j) .\n\\end{aligned}\n$$\nIl s'agit donc de trouver les entiers $n$ pour lesquels\n$$\n\\sum_{i=1}^{n} \\sum_{j=0}^{n} r_{n+1}(i j)=(n+1)\\left(\\frac{n^{2}(n+1)}{4}-\\frac{n^{2}(n-1)}{4}\\right)=\\frac{n^{2}(n+1)}{2}\n$$\nSoit alors $i$ un entier, puis soit $d=\\operatorname{PGCD}(i, n+1)$ et $k=(n+1) / d$ et $\\ell=i / d$. Les entiers $k$ et $\\ell$ sont premiers entre eux, donc les $k$ entiers $0, \\ell, 2 \\ell, \\ldots,(k-1) \\ell$ représentent les $k$ résidus modulo $k$, et les $k$ entiers $0, i, 2 i, \\ldots,(k-1) i$ représentent les $k$ résidus $0, d, 2 d, \\ldots,(k-1) d$ modulo $n+1=k d$. En particulier, pour tout entier $t$, on sait que\n$$\n\\sum_{j=k t}^{(k+1) t-1} r_{n+1}(i j)=\\sum_{j=0}^{k-1} j d=d \\frac{k(k-1)}{2}\n$$\nde sorte que\n$$\n\\sum_{j=0}^{n-1} r_{n+1}(i j)=\\sum_{t=0}^{d-1} \\sum_{j=k t}^{(k+1) t-1} r_{n+1}(i j)=d^{2} \\frac{k(k-1)}{2}=\\frac{(n+1)(n+1-d)}{2}\n$$\nEn particulier,\n$$\n\\sum_{j=0}^{n-1} r_{n+1}(i j) \\leqslant \\frac{n(n+1)}{2}\n$$\nav ec égalité si et seulement si $d=1$, donc si $i$ et $n+1$ sont premiers entre eux. Ainsi,\n$$\n\\sum_{i=1}^{n} \\sum_{j=0}^{n} r_{n+1}(i j) \\leqslant \\frac{n^{2}(n+1)}{2}\n$$\nav ec égalité si et seulement si $n+1$ est premier avec chacun des entiers $1,2, \\ldots, n$, c'est-à-dire si $n+1$ est un nombre premier.\n\nEn conclusion, les entiers recherchés sont les entiers $n$ tels que $n+1$ est premier.\n\nSolution alternative n$^{\\circ}$1\n\nDans la suite, on pose $s_{i, j}=i j /(n+1)$ et $\\mathcal{S}_{i}=\\sum_{j=1}^{n}\\left\\lfloor s_{i, j}\\right\\rfloor$. Pour tout $j \\leqslant n$, on note que $s_{i, j}+s_{i, n+1-j}=i$. Par conséquent, si l'on pose $s_{i, j}=k+x$ avec $k$ entier et $0 \\leqslant x<1$, on en déduit que\n$$\n\\left\\lfloor s_{i, j}\\right\\rfloor+\\left\\lfloor s_{i, n+1-j}\\right\\rfloor=\\lfloor k+x\\rfloor+\\lfloor(i-k-1)+(1-x)\\rfloor=i-1+\\lfloor 1-x\\rfloor,\n$$\nde sorte que $\\left\\lfloor s_{i, j}\\right\\rfloor+\\left\\lfloor s_{i, n+1-j}\\right\\rfloor \\geqslant i-1$ avec égalité si et seulement si $x \\neq 0$, c'est-à-dire si $n+1$ ne divise pas $i j$.\nOr, quand $j$ décrit l'ensemble $\\{1,2, \\ldots, n-1\\}, n+1-j$ décrit aussi cet ensemble. On en déduit que\n$$\n2 \\mathcal{S}_{i}=\\sum_{j=1}^{n}\\left\\lfloor s_{i, j}\\right\\rfloor+\\left\\lfloor s_{i, n+1-j}\\right\\rfloor \\geqslant \\sum_{j=1}^{n}(i-1)=n(i-1)\n$$\nav ec égalité si et seulement si $n+1$ ne divise aucun produit $i j$, c'est-à-dire si $n+1$ est premier avec $i$. On en conclut alors que\n$$\n\\mathcal{S}=\\sum_{i=1}^{n} \\mathcal{S}_{i} \\geqslant \\frac{n}{2} \\sum_{i=1}^{n}(i-1)=\\frac{n^{2}(n-1)}{4}\n$$\nav ec égalité si et seulement si $n+1$ est premier avec tous les entiers $i$, c'est-à-dire si $n+1$ est un nombre premier.\n\nAinsi, les entiers $n$ recherchés sont exactement ceux pour lesquels $n+1$ est premier.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $a_{1}, \\ldots, a_{101}$ des réels appartenant à l'intervalle $[-2 ; 10]$ tels que\n$$\na_{1}+\\ldots+a_{101}=0\n$$\nMontrer que\n$$\na_{1}^{2}+\\ldots+a_{101}^{2} \\leqslant 2020\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLorsque l'on effectue des inégalités sur des réels qui ne sont pas forcément positifs, il est essentiel de séparer les variables positives et négatives. Cette idée simple constitue souvent l'idée de départ dans la solution et peut mener à des développements intéressants, en plus de prévenir certaines erreurs lors de manipulations d'inégalités avec des nombres négatifs.\n\nLa deuxième idée, lorsque les variables jouent des rôles symétriques, est d'imposer un ordre aux réels. En plus de permettre, parfois, d'obtenir des résultats intéressants pour le problème, cela simplifie grandement la rédaction de la solution.\n\nQuitte à renuméroter les variables, on suppose donc que $a_{1} \\leqslant a_{2} \\leqslant \\ldots \\leqslant a_{100} \\leqslant a_{101}$. Si tous les réels sont nuls, alors la somme de leur carré est nulle et donc bien inférieure à 2020. On suppose désormais que l'un des réels est non nul. Puisque la somme des $a_{i}$ est nulle, cela impose qu'au moins un réel est strictement positif et un autre strictement négatif. On note alors $p$ l'indice vérifiant $a_{p}<0 \\leqslant a_{p+1}$.\n\nRéécrivons dès à présent l'hypothèse de la somme nulle sous la forme\n$$\n-\\left(a_{1}+a_{2}+\\ldots+a_{p}\\right)=a_{p+1}+\\ldots+a_{101}\n$$\nOn va désormais chercher à faire apparaître cette hypothèse dans la somme à calculer. Pour cela, on va également se servir du fait que si $1 \\leqslant i \\leqslant p$, alors $a_{i} \\geqslant-2$ et que si $p+1 \\leqslant i \\leqslant 101$, alors $a_{i} \\leqslant 10$. On a :\n$$\n\\begin{aligned}\na_{1}^{2}+\\ldots+a_{101}^{2} & =\\underbrace{a_{1} \\cdot a_{1}}_{\\leqslant-2 a_{1}}+\\ldots+\\underbrace{a_{p} \\cdot a_{p}}_{\\leqslant-2 a_{p}}+\\underbrace{a_{p+1} \\cdot a_{p+1}}_{\\leqslant 10 a_{p+1}}+\\ldots+\\underbrace{a_{101} \\cdot a_{101}}_{\\leqslant 10 a_{101}} \\\\\n& \\leqslant-2 a_{1}+\\ldots+\\left(-2 a_{p}\\right)+10 a_{p+1}+\\ldots+10 a_{101} \\\\\n& =2\\left(-\\left(a_{1}+\\ldots+a_{p}\\right)\\right)+10\\left(a_{p+1}+\\ldots+a_{101}\\right) \\\\\n& =2(\\underbrace{a_{p+1}}_{\\leqslant 10}+\\ldots+\\underbrace{a_{101}}_{\\leqslant 10})+10(- (\\underbrace{a_{1}}_{\\geqslant-2}+\\ldots+\\underbrace{a_{p}}_{\\geqslant-2})) \\\\\n& \\leqslant 2 \\cdot 10 \\cdot(101-p)+10 \\cdot 2 \\cdot p \\\\\n& =2020\n\\end{aligned}\n$$\nce qui est l'inégalité voulue.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle tel que $\\widehat{CAB} < \\widehat{ABC} < \\widehat{BCA} < 90^{\\circ}$. Soit $\\omega$ le cercle circonscrit à $ABC$, $\\gamma_{a}$ le cercle de centre $A$ et de rayon $[AC]$, et $\\gamma_{b}$ le cercle de centre $B$ et de rayon $[BC]$. Enfin, soit $D$ le point d'intersection, autre que $C$, entre $\\omega$ et $\\gamma_{b}$, et soit $E$ le point d'intersection, autre que $C$, entre $\\gamma_{a}$ et $\\gamma_{b}$.\n\nDémontrer que les points $A$, $D$ et $E$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPar construction, $(CE)$ est l'axe radical des cercles $\\gamma_{a}$ et $\\gamma_{b}$, donc $C$ et $E$ sont symétriques l'un de l'autre par rapport à $(AB)$. En outre, on sait que $BC = BD = BE$. On dispose donc de multiples égalités d'angles et de longueurs, et la manière la plus simple d'obtenir l'alignement recherché est sans doute de procéder à une chasse aux angles.\nOn observe ainsi que\n\n$$\n\\widehat{BAE} = \\widehat{CAB} = \\widehat{CDB} = \\widehat{BCD} = 180^{\\circ} - \\widehat{DAB},\n$$\n\nce qui signifie bien que $D$, $A$ et $E$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ un entier. Anna a écrit au tableau $n$ entiers $a_{1}, a_{2}, \\ldots, a_{n}$ deux à deux distincts. Elle remarque alors que, quelle que soit la manière de sélectionner $n-1$ de ces entiers, leur somme est divisible par $n$.\nDémontrer que la somme de l'ensemble des $n$ entiers est divisible par $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $s$ la somme de tous les entiers. Sélectionner $n-1$ entiers revient à choisir l'entier, disons $a_{i}$, que l'on n'a pas sélectionné. La somme de nos $n-1$ entiers est alors égale à $s-a_{i}$. Ainsi, Anna a simplement remarqué que $a_{i} \\equiv s \\pmod{n}$.\nCeci étant valable pour tout $i$, on en déduit que $a_{1} \\equiv a_{2} \\equiv \\ldots \\equiv a_{n} \\equiv s \\pmod{n}$. On en conclut que $s \\equiv a_{1}+a_{2}+\\cdots+a_{n} \\equiv n s \\equiv 0 \\pmod{n}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30819,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSix paniers à fruits contiennent des poires, des pêches et des pommes. Le nombre de pêches dans chaque panier est égal au nombre total de pommes dans les autres paniers. Le nombre de pommes dans chaque panier est égal au nombre total de poires dans les autres paniers. Montrez que le nombre total de fruits est un multiple de 31.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $p_{1}, p_{2}, \\ldots, p_{6}$ le nombre de poires dans chaque panier, et $q_{1}, q_{2}, \\ldots, q_{6}$ le nombre de pommes dans chaque panier. Notons également $P$ le nombre total de poires. On a:\n$$\n\\begin{aligned}\n& q_{1} = p_{2} + p_{3} + p_{4} + p_{5} + p_{6} = P - p_{1} \\\\\n& q_{2} = p_{1} + p_{3} + p_{4} + p_{5} + p_{6} = P - p_{2} \\\\\n& \\vdots \\\\\n& q_{6} = p_{1} + p_{2} + p_{3} + p_{4} + p_{5} = P - p_{6}\n\\end{aligned}\n$$\nD'où:\n$$\nq_{1} + q_{2} + \\cdots + q_{6} = (P - p_{1}) + (P - p_{2}) + \\cdots + (P - p_{6}) = 6P - (p_{1} + p_{2} + \\cdots + p_{6}) = 5P\n$$\nLe nombre total de pommes est égal à 5 fois le nombre total de poires. De même, le nombre total de pêches est égal à 5 fois le nombre total de pommes, donc à 25 fois le nombre total de poires. Le nombre total de fruits est donc égal à $31 = 25 + 5 + 1$ fois le nombre total de poires, et est donc multiple de 31.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMartin a écrit le couple d'entiers $(1011,1012)$ au tableau. Puis, chaque minute, si le couple $(a, b)$ est écrit au tableau, il l'efface et le remplace, selon son choix, par l'un des couples $(b, a)$, $(b+1, a-1)$ ou $(b-2, a+2)$, en s'imposant uniquement de n'écrire que des couples dont les deux nombres sont positifs ou nuls.\n\nQuels sont les couples d'entiers que Martin peut écrire au tableau après un nombre fini de telles opérations?",
    "options": [],
    "answer": "(a, 2013 - a) for 0 ≤ a ≤ 2013",
    "solution": "Solution:\n\nNulle opération ne change la somme des entiers écrits au tableau, donc Martin ne peut écrire que des couples de la forme $(a, 2013-a)$, avec $0 \\leqslant a \\leqslant 2013$.\n\nEn outre, si Martin part d'un couple $(a, b)$ pour lequel $a \\geqslant 1$, il peut écrire successivement les couples $(b+1, a-1)$ et $(a-1, b+1)$. Ainsi, on montre par récurrence sur $k$ que Martin peut écrire tout couple de la forme $(1011-k, 1012+k)$ lorsque $0 \\leqslant k \\leqslant 1011$.\n\nDe même, si Martin part d'un couple $(a, b)$ pour lequel $b \\geqslant 1$, il peut écrire successivement les couples $(b, a)$ puis $(a+1, b-1)$. Ainsi, on montre par récurrence sur $k$ que Martin peut écrire tout couple de la forme $(1011+k, 1012-k)$ lorsque $0 \\leqslant k \\leqslant 1012$.\n\nEn conclusion, les couples que peut écrire Martin sont précisément les couples de la forme $(a, 2013-a)$, avec $0 \\leqslant a \\leqslant 2013$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver toutes les triplets d'entiers positifs $(x, y, z)$ satisfaisant l'équation\n$$\nx! + 2^{y} = z!\n$$",
    "options": [],
    "answer": "(0, 0, 2), (1, 0, 2), (2, 2, 3)",
    "solution": "Solution:\nSoit $(x, y, z)$ un triplet solution.\nOn cherche dans un premier temps à réduire le nombre de valeurs que $x$ peut prendre. Pour ce faire, notons que $x! < z!$ donc $x < z$. En particulier, $x!$ divise $z!$ et $x!$, donc $x!$ divise $z! - x! = 2^{y}$. Supposons par l'absurde que $x \\geqslant 3$, dans ce cas $3$ divise $x!$ mais ne divise pas $2^{y}$, ce qui rend impossible le fait que $x!$ divise $2^{y}$. Ainsi $x = 0, 1$ ou $2$.\n\nOn effectue maintenant une disjonction de cas entre $x! = 1$ donc $x = 0$ ou $1$, ou bien $x! = 2$ donc $x = 2$.\n\n- Cas $x = 0$ ou $1$ : l'équation devient $1 + 2^{y} = z!$. On effectue alors une deuxième disjonction de cas :\n\n  - Cas $y = 0$ : on trouve $z = 2$. Réciproquement, $(0, 0, 2)$ et $(1, 0, 2)$ sont bien solution.\n\n  - Cas $y \\geqslant 1$ : ainsi $1 + 2^{y} \\geqslant 3$ et $1 + 2^{y}$ est impair. Or $z!$ est pair si $z \\geqslant 2$, et vaut $0! = 1$ ou $1! = 1$ si $z = 1$, aucun des deux cas ne peut donc satisfaire notre équation. Ainsi, si on suppose que $y \\geqslant 1$, il n'y a pas de solution.\n\n- Cas $x = 2$ : On distingue encore selon les valeurs de $y$ :\n\n  - Cas $y = 0$ : on obtient $3 = z!$ qui n'a pas de solutions (car $2! < 3 < 3!$).\n\n  - Cas $y = 1$ : on obtient $4 = z!$ qui n'a pas de solutions (car $2! < 4 < 3!$).\n\n  - Cas $y \\geqslant 2$ : on obtient que $2$ divise $2^{y} + 2$, mais que $4$ ne divise pas $2^{y} + 2$. Ainsi $2$ divise $z!$ mais $4$ ne divise pas $z!$, donc $z = 2$ ou $3$. Comme de plus $2^{y} + 2 \\geqslant 4 + 2 = 6$, on obtient $z = 3$, donc $2 + 2^{y} = 6$, donc $2^{y} = 4$ donc $y = 2$. Réciproquement, $(2, 2, 3)$ est bien solution car $2 + 4 = 6$.\n\nAinsi les solutions sont les triplets $(0, 0, 2)$, $(1, 0, 2)$ et $(2, 2, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30822,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les quadruplets de réels $(a, b, c, d)$ tels que\n$$\na = b c + c d, \\quad b = c d + d a, \\quad c = d a + a b \\text{ et } d = a b + b c\n$$",
    "options": [],
    "answer": "(0, 0, 0, 0) and (1/2, 1/2, 1/2, 1/2)",
    "solution": "Solution:\nSoit $(a, b, c, d)$ une solution éventuelle. Une première étape consiste à factoriser les membres de droite de nos égalités :\n$$\na = c \\times (b + d), \\quad b = d \\times (a + c), \\quad c = a \\times (b + d) \\text{ et } d = b \\times (a + c).\n$$\nMaintenant, si $a = 0$, l'énoncé indique que $c = 0 \\times (b + d) = 0$, puis que $b = d \\times 0 = 0$ et que $d = b \\times 0 = 0$. En outre, puisque les quatre nombres jouent des rôles cycliques, toute autre solution éventuelle est formée de quatre réels non nuls.\n\nDans ces nouvelles conditions, puisque $a c = a c \\times (b + d)^2 \\neq 0$, on sait que $b + d = \\pm 1$. De même, $a + c = \\pm 1$. Comme $a + c = (a + c) \\times (b + d)$, on en déduit que $b + d = 1$ et, de même, que $a + c = 1$. Dès lors, les équations de l'énoncé se réécrivent comme $a = c$ et $b = d$, de sorte que $a = b = c = d = 1/2$.\n\nEn conclusion, les deux seules solutions éventuelles sont $(0, 0, 0, 0)$ et $(1/2, 1/2, 1/2, 1/2)$, et on vérifie sans effort qu'il s'agit manifestement de solutions au problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30823,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle et $H_C$ le pied de la hauteur issue de $C$. Soit $P$ un point sur le segment $[CH_C]$ (autre que $C$), et soient $E, F, G, H$ les milieux respectifs de $[AP]$, $[BP]$, $[BC]$, $[AC]$. Montrer que $EFGH$ est un rectangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPour démontrer que $EFGH$ est un rectangle, nous allons montrer que c'est un parallélogramme qui possède un angle droit.\n\nConsidérons le triangle $ACP$. Les points $E$ et $H$ sont les milieux respectifs des côtés $[AP]$ et $[AC]$. Par le théorème de la droite des milieux, $(HE) \\parallel (PC)$.\n\nConsidérons le triangle $BCP$. Les points $F$ et $G$ sont les milieux respectifs des côtés $[BP]$ et $[BC]$. Par le théorème de la droite des milieux, $(FG) \\parallel (PC)$.\n\nOn a montré que les droites $(FG)$ et $(HE)$ sont toutes les deux parallèles à la droite $(PC)$. Il s'ensuit que $(FG) \\parallel (HE)$.\n\nConsidérons le triangle $ABP$. Les points $E$ et $F$ sont les milieux respectifs des côtés $[AP]$ et $[BP]$. Par le théorème de la droite des milieux, $(EF) \\parallel (AB)$.\n\nConsidérons le triangle $ABC$. Les points $H$ et $G$ sont les milieux respectifs des côtés $[AC]$ et $[BC]$. Par le théorème de la droite des milieux, $(HG) \\parallel (AB)$.\n\nOn a montré que les droites $(EF)$ et $(HG)$ sont toutes les deux parallèles à la droite $(AB)$. Il s'ensuit que $(HG) \\parallel (EF)$.\n\nPar conséquent, $EFGH$ est un quadrilatère dont les côtés opposés sont parallèles deux à deux. Il s'ensuit que c'est un parallélogramme.\n\nPar ailleurs, comme $(CH_c) \\perp (AB)$ et $(EF) \\parallel (AB)$, on a $(EF) \\perp (CH_c)$.\n\nDe plus, $(CH_c) \\parallel (EH)$. Il s'ensuit que $(EH) \\perp (EF)$.\n\nLe parallélogramme $EFGH$ possède donc un angle droit, ce qui démontre que $EFGH$ est un rectangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a, b$ deux entiers. Montrer que $a b(a-b)$ est divisible par 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn s'intéresse à la parité d'un produit, il est donc pertinent de faire une disjonction de cas sur la parité des entiers en jeu, ici $a$ et $b$. On remarque alors la chose suivante :\n\n- Si $a$ ou $b$ est pair : alors $a b$ est pair (le produit d'un nombre pair par un entier quelconque étant pair) et donc $a b(a-b)$ est bien divisible par 2.\n\n- Sinon, si $a$ et $b$ sont tous les deux impairs, alors $a-b$ est pair (en tant que différence de deux impairs) et donc $a b(a-b)$ est divisible par 2.\n\nFinalement, dans tous les cas, si $a$ et $b$ sont des entiers, alors $a b(a-b)$ est divisible par 2.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30825,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k \\geqslant 2$ un entier. Trouver le plus petit entier $n \\geqslant k+1$ pour lequel il existe un ensemble $E$ de $n$ réels, deux à deux distincts, dont chaque élément peut s'écrire comme la somme de $k$ autres éléments de $E$, qui sont eux-mêmes deux à deux distincts.",
    "options": [],
    "answer": "k+4",
    "solution": "Solution:\n\nSoit $n$ un entier et $E$ un ensemble tels que décrits dans l'énoncé. On trie les éléments de $E$ dans l'ordre croissant : ce sont $x_{1}<x_{2}<\\cdots<x_{k}$. Enfin, soit $s$ la somme de tous les éléments de $S$.\nPuisque $x_{1} \\geqslant x_{2}+x_{3}+\\cdots+x_{k+1}$ et $x_{n} \\leqslant x_{n-1-k}+x_{n-k}+\\cdots+x_{n-1}$, on sait que\n$$\n\\begin{aligned}\n2 x_{1}+(n-k-1) x_{n} & \\geqslant 2 x_{1}+x_{k+2}+x_{k+3}+\\cdots+x_{n} \\geqslant s \\\\\n& \\geqslant x_{1}+x_{2}+\\cdots+x_{n-2-k}+2 x_{n} \\geqslant(n-k-1) x_{1}+2 x_{n}\n\\end{aligned}\n$$\nla première et la dernière inégalité étant strictes si $n \\geqslant k+3$.\nLorsque $n$ vaut $k+1, k+2$ et $k+3$, on obtient respectivement les inégalités absurdes $2 x_{1} \\geqslant 2 x_{n}$, $2 x_{1}+x_{n} \\geqslant x_{1}+2 x_{n}$ et $2 x_{1}+2 x_{n}>2 x_{1}+2 x_{n}$. Par conséquent, $n \\geqslant k+4$.\nRéciproquement, si $n=k+4$, et si $E=\\{x_{1}, x_{2}, \\ldots, x_{n}\\}$ est un ensemble convenable, il s'agit d'identifier, pour tout entier $i \\leqslant n$, trois réels $x_{u}<x_{v}<x_{w}$ distincts de $i$ et tels que $x_{i}$ est la somme des réels $x_{j}$ lorsque $j \\neq i, u, v, w$; ou, alternativement, tels que $x_{u}+x_{v}+x_{w}=s-2 x_{i}$. On pose alors $\\ell=\\lfloor k / 2\\rfloor$, de sorte que $\\ell \\geqslant 1$, puis on procède par disjonction de cas, en fonction de la parité de $k$.\n$\\triangleright$ Si $k$ est pair, on pose $E=\\{-\\ell-2,-\\ell-1, \\ldots,-2,-1,1,2, \\ldots, \\ell+1, \\ell+2\\}$. La somme des éléments de $E$ vaut $s=0$. On choisit alors les réels $x_{u}, x_{v}$ et $x_{w}$ comme suit :\n$\\triangleright \\operatorname{si} 3 \\leqslant x_{i} \\leqslant \\ell+2$, on pose $x_{u}=-x_{i}, x_{v}=1-x_{i}$ et $x_{w}=-1$;\n$\\triangleright$ si $1 \\leqslant x_{i} \\leqslant 2$, on pose $x_{u}=-3, x_{v}=-x_{i}$ et $x_{w}=3-x_{i}$;\n$\\triangleright \\operatorname{si}-\\ell-2 \\leqslant x_{i} \\leqslant-1$, on reprend les choix de $x_{u}, x_{v}$ et $x_{w}$ effectués pour $-x_{i}$, puis on change leurs signes.\n$\\triangleright$ Si $k$ est impair, on pose $E=\\{-\\ell-2,-\\ell-1, \\ldots, \\ell+1, \\ell+2\\} ;$ il s'agit du même ensemble que précédemment, auquel on a ajouté l'entier 0 . La somme des éléments de $E$ vaut toujours $s=0$, et il suffit de traiter le cas où $x_{i}=0$, cas pour lequel on choisit $x_{u}=-2$, $x_{v}=-1$ et $x_{w}=3$.\nEn conclusion, l'entier recherché est $n=k+4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30826,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nComptez le nombre de réarrangements $a_{1}, a_{2}, \\ldots, a_{2023}$ de la séquence $1,2, \\ldots, 2023$ telle que $a_{k}>k$ pour exactement une valeur de $k$.",
    "options": [],
    "answer": "2^2023 - 2024",
    "solution": "Solution:\n\nÀ un réarrangement valide, on peut lui associer un sous-ensemble de $1, \\ldots, 2023$ de cardinal au moins $2$ : l'ensemble des $k$ tels que $a_{k} \\neq k$.\n\nOn peut montrer qu'à un sous-ensemble de cardinal au moins $2$ de $1, \\ldots, 2023$, on peut associer un unique réarrangement valide qui lui correspond. Si l'ensemble est $E=\\{e_{1}, \\ldots, e_{n}\\}$ avec $e_{1}<e_{2}<\\cdots<e_{n}$, l'unique réarrangement qui a pour ensemble associé $E$ vérifie :\n\n$a_{e_{1}}=e_{n}$, $a_{e_{2}}=e_{1}$, $a_{e_{3}}=e_{2}$, $\\ldots$, $a_{e_{n}}=e_{n-1}$ et pour tout $k \\notin E$, $a_{k}=k$.\n\nAinsi, le nombre de réarrangements valides est égal au nombre de sous-ensembles de $1, \\ldots, 2023$ de cardinal au moins $2$. La réponse est donc :\n$$\n2^{2023}-2024\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30827,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\left(u_{n}\\right)_{n \\in \\mathbb{N}}$ une suite périodique de période 2022 et de période 5. Montrer que $\\left(u_{n}\\right)$ est constante.\n\nUne suite $\\left(u_{n}\\right)$ est dite périodique de période $t$ si, pour tout entier naturel $n$, $u_{n+t}=u_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par remarquer que si une suite $\\left(u_{n}\\right)$ est $t$-périodique, alors on a pour tout $n$, $u_{n}=u_{n+t}=u_{n+2t}$ par exemple. Ainsi, on peut montrer par récurrence que pour tous les entiers naturels $k$ et $n$, $u_{n+kt}=u_{n}$ : la suite est $kt$-périodique pour tout $k$.\n\nSoit $n$ un entier naturel. On a alors, comme la suite est 2022-périodique,\n$$\nu_{n+1}=u_{n+1+2 \\times 2022}=u_{n+4045}=u_{n+5 \\times 809}$$\nMais la suite est 5-périodique donc ceci vaut aussi $u_{n}$. Ainsi, pour tout entier naturel $n$, on a $u_{n+1}=u_{n}$. On en déduit par récurrence que pour tout $n$, $u_{n}=u_{0}$ et donc la suite $\\left(u_{n}\\right)$ est constante.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver toutes les triplets d'entiers positifs $(x, y, z)$ satisfaisant l'équation\n$$\nx! + 2^{y} = z!\n$$",
    "options": [],
    "answer": "((0, 0, 2), (1, 0, 2), (2, 2, 3))",
    "solution": "Solution:\nSoit $(x, y, z)$ un triplet solution.\nOn cherche dans un premier temps à réduire le nombre de valeurs que $x$ peut prendre. Pour ce faire notons que $x! < z!$ donc $x < z$. En particulier, $x!$ divise $z!$ et $x!$, donc $x!$ divise $z! - x! = 2^{y}$. Supposons par l'absurde que $x \\geqslant 3$, dans ce cas $3$ divise $x!$ mais ne divise pas $2^{y}$, ce qui rend impossible le fait que $x!$ divise $2^{y}$. Ainsi $x = 0, 1$ ou $2$.\n\nOn effectue maintenant une disjonction de cas entre $x! = 1$ donc $x = 0$ ou $1$, ou bien $x! = 2$ donc $x = 2$.\n\n- Cas $x = 0$ ou $1$ : l'équation devient $1 + 2^{y} = z!$. On effectue alors une deuxième disjonction de cas\n\n  - Cas $y = 0$ : on trouve $z = 2$. Réciproquement, $(0, 0, 2)$ et $(1, 0, 2)$ sont bien solution.\n  \n  - Cas $y \\geqslant 1$ : ainsi $1 + 2^{y} \\geqslant 3$ et $1 + 2^{y}$ est impair. Or $z!$ est pair si $z \\geqslant 2$, et vaut $0$ ou $1$ si $z = 1$, aucun des deux cas ne peut donc satisfaire notre équation. Ainsi, si on suppose que $y \\geqslant 1$, il n'y a pas de solution.\n\n- Cas $x = 2$ : On distingue encore selon les valeurs de $y$ :\n\n  - Cas $y = 0$ : on obtient $3 = z!$ qui n'a pas de solutions (car $2! < 3 < 3!$).\n  \n  - Cas $y = 1$ : on obtient $4 = z!$ qui n'a pas de solutions (car $2! < 3 < 3!$).\n  \n  - Cas $y \\geqslant 2$ : on obtient que $2$ divise $2^{y} + 2$, mais que $4$ ne divise pas $2^{y} + 2$. Ainsi $2$ divise $z!$ mais $4$ ne divise pas $z!$, donc $z = 2$ ou $3$. Comme de plus $2^{y} + 2 \\geqslant 4 + 2 = 6$, on obtient $z = 3$, donc $2 + 2^{y} = 6$, donc $2^{y} = 4$ donc $y = 2$. Réciproquement, $(2, 2, 3)$ est bien solution car $2 + 4 = 6$.\n\nAinsi les solutions sont les triplets $(0, 0, 2)$, $(1, 0, 2)$ et $(2, 2, 3)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle acutangle (dont tous les angles sont aigus) avec $BA \\neq BC$. Soit $O$ le centre de son cercle circonscrit. La droite $(AB)$ intersecte le cercle circonscrit à $BOC$ une deuxième fois en $P \\neq B$. Montrer que $PA = PC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nTraçons la figure dans le cas où $BC < BA$, le cas $BC > BA$ étant totalement analogue. Il s'agit de montrer que $PA = PC$, c'est-à-dire que $\\widehat{ACP} = \\widehat{PAC} ( = \\widehat{BAC} )$. Or on a :\n$$\n\\begin{aligned}\n\\widehat{ACP} & = \\widehat{ACO} + \\widehat{OCP} \\\\\n& = \\widehat{ACO} + \\widehat{OBP} \\text{ par angle inscrit } \\\\\n& = \\widehat{ACO} + \\widehat{OBA} .\n\\end{aligned}\n$$\nOr, $AOC$ est isocèle en $O$ donc $\\widehat{ACO} = \\widehat{OAC} = \\frac{180^{\\circ} - \\widehat{COA}}{2} = 90^{\\circ} - \\widehat{CBA}$ par angle au centre. De même $\\widehat{OBA} = 90^{\\circ} - \\widehat{ACB}$. Finalement,\n$$\n\\widehat{ACP} = 180^{\\circ} - \\widehat{ACB} - \\widehat{CBA} = \\widehat{BAC} = \\widehat{PAC},\n$$\nd'où $PA = PC$, comme voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier strictement positif et $x \\geqslant n$ un réel. Montrer que $x+\\frac{n}{x} \\geqslant n+1$ et donner les cas d'égalité.",
    "options": [],
    "answer": "Equality holds if and only if x = n.",
    "solution": "Solution:\n\nLe réel $x$ est strictement positif donc on peut multiplier l'équation par $x$ pour avoir que l'équation est équivalente à\n$$\nx^{2}+n \\geqslant (n+1)x\n$$\nce qui donne, en réarrangeant les termes,\n$$\nx^{2}-(n+1)x+n \\geqslant 0\n$$\nOn peut chercher les racines de ce polynôme de degré 2 ou alors reconnaître la factorisation\n$$\n(x-n)(x-1) \\geqslant 0\n$$\nMais ceci est vrai car $x \\geqslant n$ et $x \\geqslant 1$.\nOn a égalité lorsque $x=n$ ou $x=1$. Mais si $x=1$, alors $n=1$ et $x=n$. Ainsi, dans tous les cas, le seul cas d'égalité est $x=n$.\nSolution:\n\nOn repart de l'équation développée $x^{2}-(n+1)x+n \\geqslant 0$.\nLa condition $x \\geqslant n$ est très peu naturelle, on préfère des conditions de type $t \\geqslant 0$. On pose donc $x=n+t$ avec $t \\geqslant 0$. L'équation se réécrit alors\n$$\nn^{2}+2nt+t^{2}-(n+1)(n+t)+n \\geqslant 0\n$$\nou encore\n$$\nt^{2}+(n-1)t \\geqslant 0\n$$\nMais cette nouvelle équation est clairement vraie car $t \\geqslant 0$ et $n \\geqslant 1$. De plus, on a égalité si et seulement si $t=0$, ou encore $x=n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30831,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn terrible groupe de bandits se prépare à se partager un butin. Chaque bandit vise deux autres bandits avec ses pistolets. Les bandits sont appelés dans un certain ordre. Lorsqu'un bandit est appelé, s'il est encore en vie, il tire sur les bandits qu'il visait. Après que tous les bandits ont été appelés, il y a eu 28 victimes. Montrez que quelque soit l'ordre dans lequel on avait appelé les bandits, il y aurait eu au moins 10 victimes.",
    "options": [],
    "answer": "10",
    "solution": "Solution:\n\nSupposons par l'absurde qu'il existe un ordre qui cause 9 victimes ou moins. Il y a alors trois catégories de bandits :\n- Ceux qui sont morts avant de tirer\n- Ceux qui ont pu tirer puis sont morts\n- Ceux qui sont encore en vie à la fin\n\nLes bandits de 1ère et 2ème catégorie sont au total au plus 9, car il y a au plus 9 victimes. Les bandits de 2ème et 3ème catégorie ne visent que des bandits de 1ère ou 2ème catégorie (s'ils ont tiré, leurs cibles sont mortes). Ils visent donc un sous-ensemble d'au plus 9 bandits. De plus, les bandits de 1ère catégorie ne visent qu'au plus $2 \\times 9 = 18$ bandits : il y a au plus 9 bandits de cette catégorie et chacun vise au plus 2 bandits. Il y a finalement au plus $18 + 9 = 27$ bandits visés : il existe un bandit qui n'est visé par personne. Ce bandit survit quelque soit l'ordre d'appel, ce qui contredit l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 30832,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFélix souhaite colorier les entiers de $1$ à $2023$ tels que si $a, b$ sont deux entiers distincts entre $1$ et $2023$ et $a$ divise $b$, alors $a$ et $b$ sont de couleur différentes. Quel est le nombre minimal de couleurs dont Félix a besoin?",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nOn peut essayer de colorier de manière gloutonne les nombres : $1$ peut être colorié d'une couleur qu'on note $a$, $2$ et $3$ de la même couleur $b$ (mais pas de la couleur $a$), puis $4, 6$ de la couleur $c$ (on peut aussi colorier $5$ et $7$ de la couleur $c$), etc. Il semble donc qu'une coloration performante soit pour tout $k$ de colorier les nombres $n$ vérifiant $2^{k} \\leqslant n < 2^{k+1}$ de la couleur $k+1$, tant que $n \\leqslant 2023$. Ainsi on colorie $[1,2[$ de couleur $1$, $[2,4[$ de couleur $2$, $\\ldots$, $[1024,2023]$ de couleur $11$.\n\nSi $a \\neq b$ sont de la même couleur $k \\in \\{1, \\ldots, 11\\}$ alors $0 < \\frac{a}{b} < \\frac{2^{k+1}}{2^{k}} = 2$, donc si $b$ divise $a$, alors $\\frac{a}{b}$ est entier donc vaut $1$. On a alors $a = b$ ce qui est contradictoire. Ainsi notre coloriage vérifie bien la condition de l'énoncé, et nécessite $11$ couleurs.\n\nRéciproquement, si un coloriage vérifie l'énoncé, les nombres $2^{0} = 1, 2^{1}, \\ldots, 2^{10}$ sont entre $1$ et $2023$, et si on prend $a \\neq b$ parmi ces $11$ nombres, soit $a$ divise $b$, soit $b$ divise $a$. Ainsi ces $11$ nombres sont de couleurs différentes. Il faut donc au moins $11$ couleurs.\n\nAinsi le nombre minimal de couleurs requises est $11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle, $O$ le centre de son cercle circonscrit. On suppose que $\\widehat{CBA}=60^{\\circ}$ et $\\widehat{\\mathrm{CBO}}=45^{\\circ}$. Soit $D$ le point d'intersection des droites $(\\mathrm{AC})$ et $(BO)$. Montrer que $\\mathrm{AD}=\\mathrm{DO}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPour montrer que $AD=DO$ (c'est-à-dire que $ADO$ est isocèle en $D$), nous allons montrer que $\\widehat{DOA}=\\widehat{OAD}$. Introduisons $P$ le point d'intersection (autre que $B$) de $(BO)$ avec le cercle circonscrit de $ABC$.\n\nD'une part, $\\widehat{DOA}=\\widehat{POA}=2\\widehat{PBA}=2\\left(60^{\\circ}-45^{\\circ}\\right)=30^{\\circ}$ d'après le théorème de l'angle au centre.\n\nD'autre part, $OC=OA$ donc $\\widehat{OAD}=\\widehat{OAC}=\\widehat{ACO}=\\frac{180^{\\circ}-\\widehat{COA}}{2}$. Or d'après le théorème de l'angle au centre, $\\widehat{COA}=2\\widehat{CBA}=120^{\\circ}$. D'où $\\widehat{OAD}=30^{\\circ}$.\n\nFinalement, $\\widehat{DOA}=30^{\\circ}=\\widehat{OAD}$, donc on a bien montré que $AD=DO$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30834,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $A, B, C, D$ et $E$ cinq points dans cet ordre sur un cercle tels que $A E = D E$. Soit $P$ le point d'intersection de $(A C)$ et $(B D)$. Soit $Q$ le point de la demi-droite $[B A)$ tel que $A Q = D P$. Soit $R$ le point de la demi-droite $[C D)$ tel que $D R = A P$. Montrer que les droites $(P E)$ et $(Q R)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOn se place dans le cas de la figure suivante :\n![](attached_image_1.png)\nDans cet énoncé on a de nombreuses égalités de longueur ($A Q = D P$, $D R = A P$, $A E = D E$). Comme sur la figure, il ne faut pas hésiter à colorier les longueurs égales pour voir des potentiels triangles isométriques, et avancer dans le problème.\n\nLes triangles $D R P$ et $A Q P$ ont l'air d'être isométriques : prouvons-le. Notons déjà que $D R = A Q$ et $D P = A P$ par hypothèse. De plus, $\\widehat{R D P} = 180 - \\widehat{C D P} = 180 - \\widehat{C D B} = 180 - \\widehat{C A B} = 180 - \\widehat{P A B} = \\widehat{P A Q}$. Ainsi les triangles $D R P$ et $A Q P$ sont isométriques.\n\nEn particulier $P Q = P R$, donc $P$ est sur la médiatrice de $(Q R)$.\n\nLes triangles $D E R$ et $A E P$ ont l'air d'être isométriques : prouvons-le. On a $D E = A E$ et $D R = A P$. De plus $\\widehat{E D R} = 180 - \\widehat{E D C} = \\widehat{C A E} = \\widehat{P A E}$. Ainsi les triangles $D E R$ et $A E P$ sont isométriques et $E P = E R$.\n\nDe même les triangles $A E Q$ et $D E P$ sont isométriques (car $A, Q$ jouent un rôle symétrique à $D, R$), ainsi $E P = E Q$, donc $E R = E Q$ : $E$ est sur la médiatrice de $(Q R)$.\n\nAinsi la médiatrice de $(Q R)$ est $(E P)$, donc $(E P)$ et $(Q R)$ sont perpendiculaires.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30835,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $(a_{n})$ une suite de réels. On suppose que $a_{0}=1$ et que pour tout $n \\geqslant 1$, $a_{n}$ est la plus petite solution strictement positive de\n$$\n\\left(a_{n}-a_{n-1}\\right)\\left(a_{n}+a_{n-1}-2 \\sqrt{n}\\right)=2 .\n$$\nTrouver le plus petit entier $n$ tel que $a_{n} \\geqslant 2022$.",
    "options": [],
    "answer": "1022121",
    "solution": "Solution:\n\nPour se donner une idée du problème, on calcule les premières valeurs de $a_{n}$ : pour $a_{1}$, l'équation s'écrit\n$$\n\\left(a_{1}-1\\right)\\left(a_{1}-1\\right)=2\n$$\ndonc $a_{1}-1=\\sqrt{2}$ ou $-\\sqrt{2}$. Seul le premier cas donne une solution positive $a_{1}=1+\\sqrt{2}$. Cherchons maintenant à calculer $a_{2}$ : l'équation devient\n$$\n\\left(a_{2}-1-\\sqrt{2}\\right)\\left(a_{2}+1-\\sqrt{2}\\right)=2\n$$\nce que l'on peut développer en\n$$\n\\left(a_{2}-\\sqrt{2}\\right)^{2}-1^{2}=2\n$$\nou encore $a_{2}-\\sqrt{2}= \\pm \\sqrt{3}$. Un seul signe donne une solution positive $a_{2}=\\sqrt{2}+\\sqrt{3}$.\n\nÀ partir de ces premières valeurs, on fait l'hypothèse que la suite $a_{n}$ est donnée par $a_{n}=\\sqrt{n}+\\sqrt{n+1}$ et on montre ceci par récurrence. C'est vrai pour $n=0$. Supposons que ceci soit vrai pour $n-1$ et montrons-le au rang $n$ : l'équation donnant $a_{n}$ est\n$$\n\\left(a_{n}-\\sqrt{n-1}-\\sqrt{n}\\right)\\left(a_{n}+\\sqrt{n-1}-\\sqrt{n}\\right)=2\n$$\nce que l'on développe en\n$$\n\\left(a_{n}-\\sqrt{n}\\right)^{2}-(\\sqrt{n-1})^{2}=2\n$$\net donc\n$$\n\\left(a_{n}-\\sqrt{n}\\right)^{2}=n+1\n$$\ndonc $a_{n}-\\sqrt{n}= \\pm \\sqrt{n+1}$. Seul le signe $+$ donne une solution positive, donc on a bien $a_{n}=\\sqrt{n}+\\sqrt{n+1}$, ce qui conclut la récurrence.\n\nOn remarque immédiatement que la suite $(a_{n})$ est croissante, et par conséquent si on trouve un $n$ tel que $a_{n-1}<2022 \\leqslant a_{n}$, il s'agit du $n$ cherché. En effet, si on a $0 \\leqslant m<n$, on a alors $m \\leqslant n-1$ donc $a_{m} \\leqslant a_{n-1}<2022$ comme voulu : $n$ est bien le plus petit indice tel que $a_{n} \\geqslant 2022$.\n\nMais maintenant on a\n$$\na_{1011^{2}-1}=\\sqrt{1011^{2}-1}+1011<2 \\cdot 1011=2022<1011+\\sqrt{1011^{2}+1}=a_{1011^{2}}\n$$\ndonc le $n$ cherché est $n=1011^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30836,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLes nombres $0,1, \\ldots, n$ sont écrits sur un tableau. À tout moment, Thanima peut effacer un nombre s'il est la moyenne arithmétique de deux nombres encore présents sur le tableau. L'objectif de Thanima est d'effacer le plus de nombre possible, et elle joue de façon optimale. En fonction de $n$, combien de nombres reste t-il à la fin sur le tableau?",
    "options": [],
    "answer": "If n ≤ 2: n. If n is a power of 2 (and n ≥ 3): 2. Otherwise (n ≥ 3 and not a power of 2): 3.",
    "solution": "Solution:\n\nSi $n \\leqslant 2$, la réponse est $n$.\n\nSinon, à la fin du procédé, il existe toujours au moins deux nombres : on ne peut effacer ni $0$ ni $n$.\n\nSi $n=2^{k}$ est une puissance de $2$, il est possible de n'avoir que deux nombres sur le tableau à la fin : on commence par effacer tous les nombres congrus à $1$ modulo $2$, puis tous les nombres congrus à $2$ modulo $4$, \\ldots, puis tous les nombres congrus à $2^{i}$ modulo $2^{i+1}$, \\ldots, puis $2^{k-1}$. Il ne reste alors plus que $0$ et $2^{k}$.\n\nSi $n$ n'est pas une puissance de $2$, on peut toujours obtenir $3$ nombres : on écrit $n=2^{k}+a$ avec $0<a<2^{k}$. On efface dans l'ordre $n-1, n-2, \\ldots, 2^{k}+1$, puis on efface les nombres entre $1$ et $2^{k}-1$ en utilisant la même technique qui est utilisée pour $n=2^{k}$. À la fin, il reste $0, 2^{k}$ et $n$.\n\nMontrons maintenant que lorsque $n>2$ n'est pas une puissance de $2$, on ne peut pas finir avec seulement $0$ et $n$. Supposons par l'absurde qu'il existe une séquence de suppressions qui permet de ne finir qu'avec $0$ et $n$. Soit $p$ un nombre premier impair qui divise $n$. En prenant la séquence des suppressions à l'envers, on obtient une séquence d'insertions où l'on part de $0$ et $n$, et si $x$ et $y$ sont présents, on peut ajouter $\\frac{x+y}{2}$ si c'est un nombre entier. Par récurrence, on démontre alors que tous les nombres obtenus sont divisibles par $p$, en particulier, on ne peut pas obtenir $1$, ce qui est la contradiction recherchée.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle rectangle en $B$ avec $BC < BA$. Soit $D$ le point du segment $[AB]$ tel que $BD = BC$. La perpendiculaire à $(AC)$ passant par $D$ intersecte $(AC)$ en $E$. Soit $B'$ le symétrique de $B$ par rapport à $(CD)$. Montrer que $(EC)$ est la bissectrice de l'angle $\\widehat{BEB'}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn remarque que le cercle de diamètre $[CD]$ apparaît assez naturellement. En effet, on a des angles droits $\\widehat{DB'C} = \\widehat{CBD} = \\widehat{DEC} = 90^{\\circ}$, les points $B$, $B'$, et $E$ sont sur le cercle de diamètre $[DC]$, autrement dit $C$, $B$, $D$, $E$, $B'$ sont cocycliques.\n\nAlors on a :\n$$\n\\begin{aligned}\n\\widehat{BEC} & = \\widehat{BDC} \\text{ par angle inscrit } \\\\\n& = 45^{\\circ} \\text{ car } BC = BD \\text{ et } \\widehat{CBD} = 90^{\\circ} .\n\\end{aligned}\n$$\n\nDe plus:\n$$\n\\begin{aligned}\n\\widehat{CEB'} & = \\widehat{CBB'} \\text{ par angle inscrit } \\\\\n& = 45^{\\circ} \\text{ car } BC = BD \\text{ et } \\widehat{CBD} = 90^{\\circ} .\n\\end{aligned}\n$$\n\nOn a donc bien $\\widehat{BEC} = \\widehat{CEB'}$, donc $(EC)$ est la bissectrice de $\\widehat{BEB'}$.\n\n\nSolution alternative $n^{\\circ} 1$\n\nOn pouvait aussi montrer directement $\\widehat{BEC} = \\widehat{CEB'}$ sans utiliser $BC = BD$. En effet :\n$$\n\\begin{aligned}\n\\widehat{BEC} & = \\widehat{BDC} \\text{ par angle inscrit } \\\\\n& = \\widehat{CDB'} \\text{ par symétrie } \\\\\n& = \\widehat{CEB'} \\text{ par angle inscrit. }\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn rectangle est divisé en dominos $1 \\times 2$ et $2 \\times 1$. Dans chaque domino, une diagonale est tracée. Deux diagonales n'ont jamais d'extrémité commune. Montrez que exactement deux coins du rectangle sont des extrémités de ces diagonales.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRemarquons que ce motif (ainsi que toutes ses rotations et symétries) est impossible :\n![](attached_image_1.png)\n\nEn effet, le seul moyen de compléter le carré bas-gauche est d'ajouter ce domino :\n![](attached_image_2.png)\n\nDe même, le seul moyen de compléter le carré bas-droite est d'ajouter ce domino :\n![](attached_image_3.png)\n\nOn ne peut pas continuer pour toujours, car on finit par atteindre le bord du rectangle.\n\nCe motif aussi est impossible (ainsi que toutes ses rotations et symétries) :\n![](attached_image_4.png)\n\ncar le seul moyen de compléter le carré dans le coin est en ajoutant ce domino :\n![](attached_image_5.png)\n\nce qui fait apparaître le motif interdit précédent. Grâce à une analyse minutieuse, ces deux cas démontrent qu'il n'est en fait pas possible que deux dominos qui partagent un bord aient leurs diagonales dans des sens différents (montantes ou descendantes). Si on forme le graphe $G$ dont les sommets sont les dominos et où il y a une arête entre deux dominos s'ils partagent un bord en commun, $G$ est connexe. Ainsi, tous les dominos ont leur diagonale dans le même sens; on en déduit facilement la propriété demandée.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30839,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $n \\geqslant 3$ et soient $x_{1}, \\ldots, x_{n}$ des réels. Montrer que\n$$\n2\\left(x_{1}+\\cdots+x_{n}\\right)^{2} \\leqslant n\\left(x_{1}^{2}+\\cdots+x_{n}^{2}+x_{1} x_{2}+x_{2} x_{3}+\\cdots+x_{n-1} x_{n}+x_{n} x_{1}\\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLa difficulté de l'exercice est le fait que l'énoncé n'est pas symétrique. Au vu de celui-ci, on veut faire apparaître les produits $x_{i} x_{i+1}$ ; pour cela, on peut penser à utiliser le développement de $\\left(x_{i}+x_{i+1}\\right)^{2}$. On a\n$$\n\\begin{gathered}\n\\left(x_{1}+x_{2}\\right)^{2}+\\cdots+\\left(x_{n-1}+x_{n}\\right)^{2}+\\left(x_{n}+x_{1}\\right)^{2}=\\left(x_{1}^{2}+2 x_{1} x_{2}+x_{2}^{2}\\right)+\\cdots+\\left(x_{n-1}^{2}+2 x_{n-1} x_{n}+x_{n}^{2}\\right)+\\left(x_{n}^{2}+2 x_{n} x_{1}+x_{1}^{2}\\right) \\\\\n=2\\left(x_{1}^{2}+\\ldots+x_{n}^{2}+x_{1} x_{2}+x_{2} x_{3}+\\cdots+x_{n-1} x_{n}+x_{n} x_{1}\\right)\n\\end{gathered}\n$$\net on reconnaît le membre de droite de l'énoncé, avec un coefficient 2 au lieu de $n$. L'inégalité de l'énoncé se réécrit alors\n$$\n4\\left(x_{1}+\\ldots+x_{n}\\right)^{2} \\leqslant n\\left(\\left(x_{1}+x_{2}\\right)^{2}+\\left(x_{2}+x_{3}\\right)^{2}+\\ldots+\\left(x_{n-1}+x_{n}\\right)^{2}+\\left(x_{n}+x_{1}\\right)^{2}\\right)\n$$\nMais ceci est alors une application de l'inégalité de Cauchy-Schwarz. En effet, on a\n$$\n\\begin{gathered}\nn\\left(\\left(x_{1}+x_{2}\\right)^{2}+\\left(x_{2}+x_{3}\\right)^{2}+\\ldots+\\left(x_{n-1}+x_{n}\\right)^{2}+\\left(x_{n}+x_{1}\\right)^{2}\\right)=\\left(1^{2}+\\ldots+1^{2}\\right)\\left(\\left(x_{1}+x_{2}\\right)^{2}+\\ldots+\\left(x_{n}+x_{1}\\right)^{2}\\right) \\\\\n\\geqslant\\left(1 \\cdot\\left(x_{1}+x_{2}\\right)+1 \\cdot\\left(x_{2}+x_{3}\\right)+\\ldots+1 \\cdot\\left(x_{n}+x_{1}\\right)\\right)^{2} \\\\\n=\\left(2 x_{1}+2 x_{2}+\\ldots+2 x_{n}\\right)^{2}=4\\left(x_{1}+\\ldots+x_{n}\\right)^{2} .\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un polynôme $P$ est fantabuleux s'il existe des réels $a_{0}, \\ldots, a_{2022}$ tel que\n$$\nP(X)=X^{2023}+a_{2022} X^{2022}+\\cdots+a_{1} X+a_{0}\n$$\ns'il a 2023 racines $r_{1}, \\ldots, r_{2023}$ (non nécessairement distinctes) dans $[0,1]$, et si $P(0)+P(1)=0$. Déterminer la valeur maximale que peut prendre $r_{1} \\cdot r_{2} \\cdots r_{2023}$ pour un polynôme fantabuleux.",
    "options": [],
    "answer": "2^{-2023}",
    "solution": "Solution:\n\nLa réponse est $2^{-2023}$, on va montrer cela en deux temps, dans un premier temps on va montrer que l'on ne peut pas trouver de polynôme avec une valeur plus grand et dans un second temps exhiber un polynôme qui vérifie les conditions de l'énoncé et dont le produit de racine est bien égal à $2^{-2023}$.\n\nNotons que par hypothèse, on peut réécrire $P(X)=\\prod_{k=1}^{2023}\\left(X-r_{k}\\right)$. En particulier, $P(0)=-\\prod_{k=1}^{2023} r_{k}$ et $P(1)=\\prod_{k=1}^{2023}\\left(1-r_{k}\\right)$. L'hypothèse $P(0)+P(1)=0$ devient :\n$$\n\\prod_{k=1}^{2023}\\left(1-r_{k}\\right)=\\prod_{k=1}^{2023} r_{k}=r_{1} \\cdot r_{2} \\cdots r_{2023}\n$$\nAinsi comme pour tout $k, r_{k} \\geqslant 0$ et $1-r_{k} \\geqslant 0$, par inégalité arithmético-géométrique,\n$$\n\\left(r_{1} \\cdot r_{2} \\cdots r_{2023}\\right)^{2}=\\prod_{k=1}^{2023}\\left(1-r_{k}\\right) r_{k} \\leqslant \\prod_{k=1}^{2023}\\left(\\frac{\\left(1-r_{k}\\right)+r_{k}}{2}\\right)^{2}=\\frac{1}{4^{2023}}\n$$\nAinsi $r_{1} \\cdot r_{2} \\cdots r_{2023} \\leqslant 2^{-2023}$.\n\nNotons de plus que si $P(X)=\\left(X-\\frac{1}{2}\\right)^{2023}$, $P$ est bien unitaire de degré 2023, a 2023 racines dans $[0,1]$, vérifie $P(0)+P(1)=\\left(-\\frac{1}{2}\\right)^{2023}+\\left(\\frac{1}{2}\\right)^{2023}=0$ donc $P$ est fantabuleux. Le produit de ses racines vaut $2^{-2023}$. Ainsi la valeur maximale que peut prendre $r_{1} \\cdot r_{2} \\cdots r_{2023}$ pour un polynôme fantabuleux est $2^{-2023}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30841,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCDE$ un pentagone cyclique convexe tel que $AB = BD$. Soit $P$ le point d'intersection des droites $(EB)$ et $(AC)$. Soit $Q$ le point d'intersection des droites $(BC)$ et $(DE)$. Montrer que $(PQ)$ et $(AD)$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSur la figure, il semble que $EPCQ$ est un quadrilatère cyclique. On le montre :\n$$\n\\begin{aligned}\n\\widehat{QCP} & = 180^{\\circ} - \\widehat{PCB} \\\\\n& = 180^{\\circ} - \\widehat{ACB} \\\\\n& = 180^{\\circ} - \\widehat{ADB} \\text{ par angle inscrit } \\\\\n& = 180^{\\circ} - \\widehat{BAD} \\text{ car } AB = BD \\\\\n& = 180^{\\circ} - \\widehat{BED} \\\\\n& = 180^{\\circ} - \\widehat{PEQ} .\n\\end{aligned}\n$$\nAinsi on a montré que $\\widehat{QCP} + \\widehat{PEQ} = 180^{\\circ}$. Dès lors, par réciproque de l'angle inscrit, $EPCQ$ est cyclique, ce qu'on voulait démontrer.\n\nMaintenant, pour montrer que $(AD)$ et $(PQ)$ sont parallèles, on va essayer de trouver des angles correspondants. Or on a :\n$$\n\\begin{aligned}\n\\widehat{EQP} & = \\widehat{ECP} \\text{ par angle inscrit } \\\\\n& = \\widehat{ECA} \\\\\n& = \\widehat{EDA} \\text{ par angle inscrit. }\n\\end{aligned}\n$$\nComme $\\widehat{EQP} = \\widehat{EDA}$ et $Q, D, E$ sont alignés (dans cet ordre), par angle correspondant les droites $(PQ)$ et $(AD)$ sont bien parallèles.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30842,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEst-il possible de trouver un bloc de 1000 nombres entiers strictement positifs consécutifs qui contient exactement 5 nombres premiers?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nA première vue, il semble difficile de garantir exactement 5 nombre premier dans un bloc de 1000 entiers consécutifs : aucun théorème d'arithmétique élémentaire permet de s'assurer d'avoir des nombres premiers rapprochés, mais pas d'autres nombres premier entre eux. On peut donc essayer de se demander combien de nombres premiers on peut trouver dans un bloc de 1000 nombres consécutifs.\nDéjà, il est relativement connu qu'on peut trouver 1000 nombres consécutifs dont aucun n'est premier. En effet, $1001!+2, \\ldots, 1001!+1001$ ne sont pas premiers : si $2 \\leqslant i \\leqslant 1001, 1001!+i$ est divisible par $i$ et strictement plus grand que $i$, donc il n'est pas premier.\nOn peut aussi remarquer qu'entre 1 et 1000 il y a beaucoup de nombres premiers : en tout cas strictement plus que 5, vu que $2, 3, 5, 7, 11, 13$ sont premiers.\nNotons $P(n)$ le nombre de nombre premier dans l'ensemble $\\{n, \\ldots, n+999\\}$. L'étape naturelle après nos deux remarques précédentes est de se demander comment évalue le nombre $P(n)$ lorsque $n$ parcourt $\\mathbb{N}^*$. Pour passer de $\\{n, \\ldots, n+999\\}$ à $\\{n+1, \\ldots, n+1+999\\}$, on rajoute $n+1000$ et on enlève $n$. Ainsi on ajoute au plus un nombre premier, et on enlève au plus un nombre premier : on a donc\n$$\nP(n+1)-P(n) \\in \\{-1,0,1\\} .\n$$\nAinsi $P$ ne varie que de 1 en 1, et on sait que $P(1)>5>P(1001!+2)$. Il semble assez intuitif que cela implique qu'il existe un entier $k$ tel que $P(k)=5$, nous allons prouver cela.\nSoit $k$ le premier entier tel que $P(k) \\leqslant 5$. On sait que $k>1$, donc $P(k-1)>5$ par minimalité. Ainsi, on a forcément $P(k)=P(k-1)-1$ (pour avoir $P(k) \\leqslant 5$), donc $P(k)>4$. Ainsi $P(k) \\geqslant 5$, or par définition $P(k) \\leqslant 5$, donc $P(k)=5$.\nAinsi la réponse à l'énoncé est oui : il existe bien un bloc de 1000 entiers consécutifs avec exactement cinq entiers premiers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $G$ un graphe à $n$ sommets. Une arête $e$ de $G$ est un cul-de-sac s'il est possible de partitionner $G$ en deux ensembles $A$ et $B$ tels que :\n- Il y a au plus $2023$ arêtes de $G$ ayant une extrémité dans $A$ et une extrémité dans $B$.\n- L'arête $e$ a l'une de ses extrémités dans $A$ et l'autre dans $B$.\n\nMontrez qu'il y a au plus $2023(n-1)$ culs-de-sac.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn le démontre par récurrence forte sur $n$. Le cas $n=1$ est immédiat. S'il n'existe aucun cul-de-sac dans $G$, la propriété est triviale. Sinon, soit $e$ un cul-de-sac de $G$. On obtient une partition des sommets de $G$ en deux ensembles $A$, $B$ telle qu'il y a au plus $2023$ arêtes entre $A$ et $B$. D'après l'hypothèse de récurrence, il y a au plus $2023(|A|-1)$ culs-de-sacs lorsque l'on ne considère que les sommets de $A$ et les arêtes qui relient ces sommets. De même, il y a au plus $2023(|B|-1)$ culs-de-sacs lorsque l'on ne considère que les sommets de $B$ et les arêtes qui les relient. On remarque que si une arête $e$ n'est pas un cul-de-sac dans le graphe restreint à $A$ (de même pour $B$), alors elle n'est pas non plus un cul-de-sac dans $G$. Il y a donc au plus :\n$$\n2023 + 2023(|A|-1) + 2023(|B|-1) = 2023(|G|-1)\n$$\nculs-de-sacs dans $G$, ce qui termine l'hérédité.",
    "topic": "Discrete Mathematics",
    "subtopic": "Other"
  },
  {
    "id": 30844,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle, $D, E$ les pieds des hauteurs issues de $A$ et $B$ respectivement. La droite $(DE)$ rencontre le cercle circonscrit à $ABC$ en deux points $P$ et $Q$. Soient $A'$, $B'$ les symétriques de $A$ et $B$ par rapport à $(BC)$ et $(AC)$ respectivement. Montrer que $A'$, $B'$, $P$, $Q$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSur la figure, il semble que $H$ l'orthocentre se situe sur le cercle en question. On va donc adopter la stratégie suivante : on va montrer que $A'$ et $B'$ sont sur le cercle circonscrit de $PQH$. De cette manière on aura bien $A'$, $B'$, $P$, $Q$ cocycliques.\n\nSoit $M$ le symétrique de $H$ par rapport à $(BC)$ : alors on sait que $M$ est sur le cercle circonscrit à $ABC$. Alors on a :\n$$\n\\begin{aligned}\nDA' \\times DH & = DA \\times DM \\text{ par symétrie } \\\\\n& = DP \\times DQ \\text{ par puissance de } D \\text{ dans le cercle } (ABC).\n\\end{aligned}\n$$\nAlors par réciproque de la puissance d'un point, $P$, $Q$, $H$, $A'$ sont cocycliques. De façon totalement analogue, $P$, $Q$, $H$, $B'$ sont également cocycliques. Ainsi $A'$, $B'$, $P$, $Q$ sont cocycliques, sur le cercle $(PQH)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAnna et Elie jouent à un jeu. On leur donne à tous les deux le même ensemble $A$ composé d'un nombre fini d'entiers strictement positifs et distincts. Anna choisit un entier $a \\in A$ secrètement. Si Elie choisit un entier $b$ (pas forcément dans $A$) et le donne à Anna, Anna lui donne le nombre de diviseurs strictement positifs de $ab$. Montrer que Elie peut choisir $b$ de sorte à retrouver à coup sûr l'entier choisi par Anna.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $P$ l'ensemble fini des nombres premiers divisant au moins un élément de $A$ et $n \\geqslant 1$ le plus grand entier tel qu'il existe $a \\in A$ et $p \\in P$ tels que $p^{n} \\mid a$. On peut remplacer $A$ par l'ensemble des entiers $m$ dont les facteurs premiers sont tous dans $P$ et tels que pour tout $p \\in P$, $v_{p}(m) \\leqslant n$. Elie propose un entier $b$ de la forme $\\prod_{p \\in P} p^{b_{p}}$. Si Anna a choisi l'entier $a$, alors elle donne à Elie l'entier $\\prod_{p \\in P}\\left(b_{p}+v_{p}(a)+1\\right)$. Il suffit donc pour Elie de choisir les $b_{p}$ de sorte que tous les $\\prod_{p \\in P}\\left(b_{p}+a_{p}\\right)$ soient deux à deux distincts, pour tous les choix possibles de $a_{p}$ entre $1$ et $n+1$ pour chaque $p \\in P$.\n\nOn propose deux solutions.\n\nPremière méthode:\n\nL'idée de cette construction est de forcer les factorisations en produits de facteurs premiers des $\\prod_{p}\\left(b_{p}+a_{p}\\right)$ à être différentes pour tous les choix possibles des $a_{p}$. On veut faire en sorte que chaque $b_{p}+i$ pour $1 \\leqslant i \\leqslant n+1$ possède un diviseur premier «distinctif», qui n'apparaît que dans la décomposition en produit de facteurs premiers de $b_{p}+i$, jamais dans celle d'un autre $b_{q}+j$.\n\nFormellement, on choisit, pour chaque $p \\in P$ et chaque $1 \\leqslant i \\leqslant n+1$, des nombres premiers deux à deux distincts $Q_{p, i}>n+1$. On construit les $b_{p}$ grâce au théorème chinois : pour chaque $q \\in P$ distinct de $p$ et pour chaque $1 \\leqslant i \\leqslant n+1$, $Q_{q, i} \\mid b_{p}$, et pour chaque $1 \\leqslant i \\leqslant n+1$, $b_{p} \\equiv -i \\pmod{Q_{p, i}}$.\n\nAvec cette construction, si $q, p \\in P$ et $1 \\leqslant i, j \\leqslant n+1$, alors $Q_{q, j} \\mid b_{p}+i$ si et seulement si $i=j$ et $p=q$.\n\nEn particulier, étant donnés des $1 \\leqslant a_{p} \\leqslant n+1$ pour chaque $p \\in P$, si $q \\in P$, $\\Pi=\\prod_{p \\in P}\\left(b_{p}+a_{p}\\right)$ est divisible par $Q_{q, j}$ si et seulement si $j=a_{q}$ : ainsi $\\Pi$ détermine la famille $\\left(a_{p}\\right)$.\n\n\nDeuxième méthode:\n\nLa première construction était très arithmétique et utilisait des nombres premiers. Celle qu'on présente maintenant vient plus d'une idée de \"taille\". L'idée est que si l'on prend de gigantesques $b_{p}$, le produit $\\prod_{p}\\left(b_{p}+a_{p}\\right)$ ressemblera à l'écriture d'un certain nombre en une certaine base, dont les chiffres donneront les $a_{i}$.\n\nPassons à la construction proprement dite. Le problème tel que nous l'avons reformulé n'utilise plus le fait que $P$ soit constitué de nombres premiers : on renumérote ses éléments en $1, \\ldots, r$. Montrons que pour $N>(n+1)^{r}$, $b_{i}=N^{2^{i}}$ convient. En effet, dans ce cas, on voit que le développement de $\\Pi=\\prod_{i=1}^{r}\\left(N^{2^{i}}+a_{i}\\right)$ écrit un nombre en base $N$ dont le chiffre devant $N^{2+1}-2-2^{i}$ est exactement $a_{i}$ : ainsi $\\Pi$ détermine la famille des $\\left(a_{i}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un nombre rationnel strictement positif $q$ est magnifique s'il existe quatre entiers strictement positifs $a, b, c, d$ tels que\n$$\nq = \\frac{a^{2021} + b^{2023}}{c^{2022} + d^{2024}}\n$$\n\nExiste-t-il un rationnel strictement positif qui n'est pas magnifique?",
    "options": [],
    "answer": "No; every positive rational number is magnificent.",
    "solution": "Solution:\n\nLa question posée est assez déroutante : il a l'air d'être dur de décider ou non si un nombre peut s'écrire de cette forme. On peut donc essayer de se fixer un rationnel strictement positif de la forme $\\frac{r}{s}$ avec $r, s$ des entiers strictement positifs, et chercher des bons $a, b, c, d$ pour avoir\n$$\n\\frac{r}{s} = \\frac{a^{2021} + b^{2023}}{c^{2022} + d^{2024}}\n$$\nCela permettra au moins de comprendre si tous les rationnels sont magnifiques, ou de trouver un ensemble de rationnels qui pourraient être non magnifiques. Ici le choix de $a, b, c, d$ n'est pas clair. Le plus naturel est de prendre $a = r^{r_a} s^{s_a}$, $b = r^{r_b} s^{s_b}$, $c = r^{r_c} s^{s_c}$ et $d = r^{r_d} s^{s_d}$ avec les $r_i$ et $s_i$ des entiers positifs. On obtient alors :\n$$\n\\frac{r}{s} = \\frac{r^{2021 r_a} s^{2021 s_a} + r^{2023 r_b} s^{2023 s_b}}{r^{2022 r_c} s^{2022 s_c} + r^{2024 r_d} s^{2024 s_d}}\n$$\nPour espérer factoriser et simplifier ce terme, le plus simple est d'avoir $2021 r_a = 2023 r_b$, i.e. de poser $r_b = 2021 r_1$ et $r_a = 2023 r_1$ pour un certain entier positif $r_1$. De même on pose $r_c = 2024 r_2$ et $r_d = 2022 r_2$ pour un certain entier positif $r_2$, $s_a = 2023 s_1$ et $s_b = 2021 s_1$ pour un certain entier positif $s_1$, et $s_c = 2024 s_2$ et $s_d = 2022 s_2$ pour un certain entier positif $s_2$.\nOn a alors\n$$\n\\frac{r}{s} = \\frac{r^{2021 \\times 2023 r_1} s^{2021 \\times 2023 s_1}}{r^{2022 \\times 2024 r_2} s^{2022 \\times 2024 s_2}} = r^{2021 \\times 2023 r_1 - 2022 \\times 2024 r_2} s^{2021 \\times 2023 s_1 - 2022 \\times 2024 s_2}\n$$\nAinsi il suffit de trouver $r_1, r_2, s_1, s_2$ des entiers positifs tels que $2021 \\times 2023 r_1 - 2022 \\times 2024 r_2 = 1$ et $2021 \\times 2023 s_1 - 2022 \\times 2024 s_2 = 1$. Pour cela, il suffit d'utiliser le théorème de Bézout. En effet, $2021 \\times 2023$ et $2022 \\times 2024$ sont premiers entre eux : si on raisonne par l'absurde en considérant $p$ un facteur premier de ces deux nombres, $p$ divise deux nombres entre 2021 et 2024, donc divise leur différence, donc $p = 2$ ou $3$. Or $2021 \\times 2023$ est impair, et non divisible par $3$, donc on a une contradiction. Ainsi il existe deux entiers $e, f$ tels que $2021 \\times 2023 e - 2022 \\times 2024 f = 1$. Quitte à rajouter $2022 \\times 2024$ plusieurs fois à $e$, et $2021 \\times 2023$ le même nombre de fois à $f$, on peut supposer $e, f$ positifs. De même il existe deux entiers positifs $g$ et $h$ tels que $2021 \\times 2023 g - 2022 \\times 2024 h = 1$. Poser $s_1 = g, s_2 = h, r_1 = e, r_2 = f$ donne bien que $\\frac{r}{s}$ est magnifique : ainsi tout rationnel strictement positif est magnifique.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn place un certain nombre de segments ouverts dans le plan, aucun d'entre eux n'est parallèle aux axes $x$ et $y$. Ces segments sont disjoints. Thanima commence à se déplacer depuis $(0,0)$ parallèlement à l'axe $x$. À chaque fois qu'elle rencontre un mur, elle tourne de 90 degrés, et continue à se déplacer sans traverser le mur.\n\nDémontrez qu'il est impossible que Thanima visite les deux côtés de tous les murs.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par démontrer qu'il existe un mur qui est plus bas que tous les autres (c'est à dire que l'ensemble des points sous ce segment n'intersecte aucun segment). Supposons par l'absurde que ce n'est pas le cas. On peut alors construire un cycle $A_{1}, B_{1}, A_{2}, B_{2}, \\ldots, A_{k}, B_{k}$ qui ne s'intersecte pas lui-même, où $A_{i}$ et $B_{i}$ sont sur le même mur, et $[B_{i}, A_{i+1}]$ est un segment vertical où $B_{i}$ est au dessus de $A_{i+1}$ (on considère les indices modulo $k$). On considère $l$, $r$ tel que $[B_{l}, A_{l+1}]$ soit le plus à gauche et $[B_{r}, A_{r+1}]$ soit le plus à droite. Le cycle donne une ligne brisée entre $B_{l}$ et $A_{r+1}$, et une autre entre $B_{r}$ et $A_{l+1}$. Cependant, les lignes brisées ne passent ni à gauche de $[B_{l}, A_{l+1}]$ ni à droite de $[B_{r}, A_{r+1}]$, et leurs extrémités sont \"croisées\", elles doivent donc s'intersecter. C'est une contradiction avec le fait que le cycle ne s'intersecte pas lui-même.\n\nSimilairement, il existe un mur plus haut que tous les autres. Si Thanima passe par le côté haut du mur le plus haut, elle continue à marcher vers l'infini en direction des $y$ positifs. Si Thanima passe par le côté bas du mur le plus bas, elle continue à marcher vers l'infini en direction des $y$ négatifs. Ces deux cas ne peuvent pas arriver simultanément, elle ne passe donc que par un seul de ces deux côtés. En particulier, elle ne peut pas passer par tous les côtés de tous les murs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30848,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver toutes les fonctions $f: \\mathbb{N}^* \\rightarrow \\mathbb{N}^*$ telles que\n$$\nf^{f(n)}(m) + m n = f(m) f(n)\n$$\npour tous $m, n \\in \\mathbb{N}^*$.",
    "options": [],
    "answer": "f(n) = n + 1 for all positive integers n",
    "solution": "Solution:\nEn $m=n$, on trouve que $f(m)^2 = f^{f(m)}(m) + m^2 > m^2$, ce qui donne $f(m) > m$ car $f(m)$ est positif, et donc comme $f(m)$ est entier, $f(m) \\geqslant m+1$. Par une récurrence simple, on trouve que pour tous les entiers $k$, $m$ strictement positifs, $f^{k}(m) \\geqslant m+k$.\n\nMais on remarque qu'à part le premier terme de l'énoncé, le reste est symétrique en $m$ et $n$, et on peut écrire que pour tous $m, n > 0$,\n$$\nf^{f(n)}(m) + m n = f(m) f(n) = f^{f(m)}(n) + n m\n$$\net donc $f^{f(n)}(m) = f^{f(m)}(n)$. On veut maintenant utiliser le fait que l'on a affaire à une composée, et on pose donc $n = f(a)$ afin que $f^{f(m)}(n) = f^{f(m)+1}(a)$. En évaluant l'équation précédente avec $a = m$, on a\n$$\nf^{f(m)+1}(m) = f^{f(f(m))}(m)\n$$\nMais ceci implique alors que $f(f(m)) = f(m) + 1$. En effet, sinon, soit $\\alpha = \\min (f(f(m)), f(m)+1)$ et $k = \\max (f(f(m)), f(m)+1) - \\alpha > 0$ afin que l'équation s'écrive $f^{k}\\left(f^{\\alpha}(m)\\right) = f^{\\alpha}(m)$. Mais c'est absurde car $f^{k}\\left(f^{\\alpha}(m)\\right) \\geqslant f^{\\alpha}(m) + k > f^{\\alpha}(m)$.\n\nOn déduit que pour tout $m \\in \\mathbb{N}^*$, $f(f(m)) = f(m) + 1$. Par une récurrence simple, pour tout $k \\geqslant 1$, $f^{k}(m) = f(m) + k - 1$. Ainsi l'équation de l'énoncé donne pour tous $m, n \\in \\mathbb{N}^*$,\n$$\n\\mathbf{f}(m) + \\mathbf{f}(n) - 1 + m n = \\mathbf{f}(m) \\mathbf{f}(n)\n$$\nSi $n = m$, on obtient une équation liant simplement $f(n)$ et $n$ :\n$$\n2 \\mathbf{f}(n) + n^2 - 1 = \\mathbf{f}(n)^2\n$$\nC'est une équation du second degré que l'on peut par exemple factoriser en $(f(n) - 1)^2 = n^2$, donc $\\mathbf{f}(n) = n + 1$ car $\\mathbf{f}(n)$ est strictement positif.\n\nRéciproquement, si $f(n) = n + 1$ pour tout $n$, alors on peut écrire pour tous $m, n$ :\n$$\nf^{f(n)}(m) + m n = m + f(n) + m n = m n + m + n + 1 = (m + 1)(n + 1) = f(m) f(n)\n$$\net cette fonction est bien solution de l'équation de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30849,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCD$ un quadrilatère convexe cyclique, $M$ le milieu de $[AC]$. Le cercle circonscrit à $CDM$ rencontre $(BC)$ une deuxième fois en $N$ (autre que $C$). Soit $B'$ le symétrique de $B$ par rapport à $N$. Montrer que $(MN)$ est tangente au cercle circonscrit de $B'DN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSur la figure, on repère des triangles semblables. En effet, on a par angle inscrit :\n$$\n\\widehat{MAD} = \\widehat{CAD} = \\widehat{CBD} = \\widehat{NBD}\n$$\nToujours par angle inscrit, on a :\n$$\n\\widehat{DMA} = 180^{\\circ} - \\widehat{CMD} = 180^{\\circ} - \\widehat{CND} = \\widehat{DNB}\n$$\nOn en déduit que les triangles $AMD$ et $BND$ sont semblables. Mais alors $ADC$ et $BDB'$ sont semblables. En effet on a $\\widehat{B'BD} = \\widehat{CBD}$, et $\\frac{BB'}{AC} = \\frac{2BN}{2AM} = \\frac{BN}{AM} = \\frac{BD}{AD}$, et donc $ADC$ est semblable à $BDB'$. Il suit que :\n$$\n\\begin{aligned}\n\\widehat{MND} & = \\widehat{MCD} \\text{ par angle inscrit } \\\\\n& = \\widehat{ACD} \\\\\n& = \\widehat{BB'D} \\text{ car } ADC \\sim BDB' \\\\\n& = \\widehat{DB'D}\n\\end{aligned}\n$$\nDonc par réciproque de l'angle tangentiel, $(MN)$ est tangente au cercle circonscrit à $B'DN$, ce qu'on voulait montrer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier $n \\geqslant 1$, on pose $u_{n}=1!+2!+\\ldots+n!$. Montrer qu'il existe une infinité de nombres premiers divisant au moins l'un des termes de la suite $\\left(u_{n}\\right)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons l'inverse : alors il existe des nombres premiers $p_{1}<\\ldots<p_{r}$ tels que pour tout $n \\geqslant 1$, $u_{n}=1!+2!+\\ldots+n!$ soit le produit des $p_{i}^{a_{i}(n)}$, où les $a_{i}(n)$ sont des entiers positifs.\n\nSi $n \\geqslant 1$ est tel que $a_{i}(n)<v_{p_{i}}((n+1)!)$, alors\n$$\na_{i}(n+1)=v_{p_{i}}\\left(u_{n+1}\\right)=v_{p_{i}}\\left(u_{n}+(n+1)!\\right)=v_{p_{i}}\\left(u_{n}\\right)=a_{i}(n)\n$$\nEn particulier, $a_{i}(n+1)<v_{p_{i}}((n+2)!)$. En particulier, ou bien $a_{i}(n) \\geqslant v_{p_{i}}((n+1)!)$ pour tout $n$, ou bien $a_{i}(n)$ est constante à partir d'un certain rang.\n\nEn particulier, en renommant et regroupant les $p_{i}$, on dispose d'entiers $N \\geqslant 1$, $C \\geqslant 1$ et de nombres premiers $q_{1}<\\ldots<q_{s}$ ne divisant pas $C$ tels que pour tout $n>N$, $u_{n}=C \\prod_{i=1}^{s} q_{i}^{b_{i}(n)}$ et $b_{i}(n) \\geqslant v_{q_{i}}((n+1)!)$.\n\nSoit maintenant $n>N+C+1$ tel que $n+1$ soit divisible par le produit des $q_{i}$. Alors $u_{n}=u_{n-1}+n!$ et $v_{q_{i}}\\left(u_{n}\\right)=b_{i}(n) \\geqslant v_{q_{i}}((n+1)!)>v_{q_{i}}(n!)$, donc $v_{q_{i}}\\left(u_{n-1}\\right)=v_{q_{i}}(n!)$. D'autre part, si $p \\mid C$, $v_{p}\\left(u_{n-1}\\right)=v_{p}(C) \\leqslant v_{p}(n!)$, donc $u_{n-1} \\mid n!$.\n\nIl reste à montrer que pour tout $n$ assez grand, $u_{n}$ ne divise pas $(n+1)!$.\n\nEn effet, si $n \\geqslant 2$, $n u_{n}>n \\cdot n!+n \\cdot(n-1)!=(n+1) \\cdot n!=(n+1)!$. D'autre part, si $n \\geqslant 4$,\n$$\n\\begin{aligned}\n(n-1) u_{n} & =(n-1) n!+(n-1)(n-1)!+(n-1)(n-2)!+(n-1) \\sum_{k=1}^{n-3} k! \\\\\n& =n \\cdot n!+(n-1)(n-3) \\cdot(n-3)!<(n+1) \\cdot n! \\\\\n& =(n+1)!\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $b_{1}, \\ldots, b_{n}$ des réels positifs ou nuls tels que $b_{1}+b_{2}+\\ldots+b_{n}=2$. Soient $a_{0}, a_{1}, \\ldots, a_{n}$ des nombres réels tels que $a_{0}=a_{n}=0$. Supposons que pour tout $i$ compris entre $1$ et $n$ inclus,\n$$\n\\left|a_{i}-a_{i-1}\\right| \\leqslant b_{i}\n$$\nMontrer que l'on a\n$$\n\\left(a_{0}+a_{1}\\right) b_{1}+\\left(a_{1}+a_{2}\\right) b_{2}+\\ldots+\\left(a_{n-1}+a_{n}\\right) b_{n} \\leqslant 2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPour tout indice $i$, on a deux majorations de $a_{i}$. En effet, on peut écrire\n$$\n\\begin{aligned}\na_{i}=a_{i}-a_{0}= & \\left(a_{i}-a_{i-1}\\right)+\\left(a_{i-1}-a_{i-2}\\right)+\\ldots+\\left(a_{1}-a_{0}\\right) \\\\\n& \\leqslant b_{i}+b_{i-1}+\\ldots+b_{1}=\\sum_{j=1}^{i} b_{j}\n\\end{aligned}\n$$\net\n$$\n\\begin{aligned}\na_{i}=a_{i}-a_{n} & =\\left(a_{i}-a_{i+1}\\right)-\\left(a_{i+1}-a_{i+2}\\right)+\\ldots+\\left(a_{n-1}-a_{n}\\right) \\\\\n& \\leqslant b_{i+1}+b_{i+2}+\\ldots+b_{n}=\\sum_{j=i+1}^{n} b_{j} .\n\\end{aligned}\n$$\nAfin de majorer efficacement $a_{i}$, on voudrait trouver le plus petit de ces deux termes. Mais la somme des deux vaut $2$, donc il faut simplement déterminer si $\\sum_{j=1}^{i} b_{j}$ est inférieur ou supérieur à $1$.\nSoit $m$ l'entier minimal tel que $\\sum_{j=1}^{m} b_{j}>1$. On écrit alors\n$$\n\\sum_{i=1}^{n}\\left(a_{i-1}+a_{i}\\right) b_{i}=\\sum_{i=1}^{m-1}\\left(a_{i-1}+a_{i}\\right) b_{i}+\\left(a_{m-1}+a_{m}\\right) b_{m}+\\sum_{i=m+1}^{n}\\left(a_{i-1}+a_{i}\\right) b_{i}\n$$\nPour $i<m$, on majore $a_{i}$ par $\\sum_{j=1}^{i} b_{j}$, et pour $i \\geqslant m$, on majore $a_{i}$ par $\\sum_{j=i+1}^{n} b_{j}$ pour obtenir\n$$\n\\sum_{i=1}^{n}\\left(a_{i-1}+a_{i}\\right) b_{i} \\leqslant \\sum_{i=1}^{m-1}\\left(b_{i}+2 \\sum_{j=1}^{i-1} b_{j}\\right) b_{i}+\\left(\\sum_{j=1}^{m-1} b_{j}+\\sum_{j=m+1}^{n} b_{j}\\right) b_{m}+\\sum_{i=1}^{m}\\left(b_{i}+2 \\sum_{j=i+1}^{n} b_{j}\\right) b_{i}\n$$\nPosons $x=\\sum_{j=1}^{m-1} b_{j}$ et $y=\\sum_{j=m+1}^{n} b_{j}$. Par définition de $m$, on a $x \\leqslant 1$ et $y \\leqslant 1$. Considérons la première somme dans le terme précédent. Dans celle-ci, on a un terme $b_{i}^{2}$ pour tout $i \\leqslant m-1$ et un terme $2 b_{i} b_{j}$ pour tous $j<i \\leqslant m-1$. Ainsi, cette somme n'est rien d'autre que $x^{2}$. De même, la somme de droite n'est rien d'autre que $y^{2}$. On peut donc écrire, comme $x+b_{m}+y=2$,\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n}\\left(a_{i-1}+a_{i}\\right) b_{i} & \\leqslant x^{2}+(x+y)(2-x-y)+y^{2} \\\\\n= & 2(x+y)-2 x y\n\\end{aligned}\n$$\nIl nous reste enfin à voir que $2(x+y)-2 x y \\leqslant 2$ pour conclure, ou encore $x+y-x y \\leqslant 1$. Mais ceci se réécrit, en passant tous les termes à droite de l'inégalité,\n$$\n(1-x)(1-y) \\geqslant 0\n$$\nce qui est vérifié car on a vu que $x \\leqslant 1$ et $y \\leqslant 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nExercice 16. Soit $p$ un nombre premier. Martin la grenouille est situé en position $0$ sur la droite réelle. À chaque seconde, Martin effectue un mouvement : il peut rester à sa position, faire un saut de $1$ sur la droite, ou faire un saut de $1$ sur la gauche. Martin souhaite être revenu au bout de $p-1$ mouvements à sa position initiale : il note alors $s_{p}$ le nombre de séquences de $p-1$ mouvements lui permettant de revenir à sa position initiale. Quel est le reste de $s_{p}$ modulo $p$ ?\n\nPar exemple, pour $p=3$, $s_{p}=3$ puisque Martin peut choisir de rester à sa position deux fois, de faire un saut sur la droite puis sur la gauche, ou faire un saut sur la gauche puis sur la droite.",
    "options": [],
    "answer": "s_p ≡ 0 if p = 3; s_p ≡ 1 if p ≡ 1 mod 3; s_p ≡ p − 1 if p ≡ 2 mod 3",
    "solution": "Solution:\n\nNotons $a_{n}(k)$ le nombre de suites de $n$ mouvements tel que si Martin était initialement à la position $0$, il se retrouve à la position $k$ (par exemple, pour $n=2$, $a_{2}(0)=2$ et $a_{2}(2)=a_{2}(-2)=1$). Cela revient donc à compter le nombre de $n$-uplets à valeurs dans $\\{-1,0,1\\}$ de somme $k$. On va essayer de comprendre $\\left(a_{p}(k)\\right)_{k}$ modulo $p$, et de trouver une relation de récurrence sur les $\\left(a_{n}(k)\\right)$ pour trouver le reste de $a_{p-1}(0)$ modulo $p$.\n\nOn va montrer le lemme suivant :\n\nLemme 1. $a_{p}(k)$ vaut $0$ modulo $p$, sauf si $k=p$, $0$ ou $-p$ : dans ce cas $a_{p}(k)$ vaut $1$ modulo $p$.\n\nDémonstration. On se fixe $k$ un entier différent de $\\pm p$ (si $k= \\pm p$, on voit bien que $a_{p}(k)=1$), on note $A$ l'ensemble des $p$-uplets de somme $k$. On définit $T: A \\mapsto A$ l'application suivante : si $\\left(a_{1}, \\ldots, a_{p}\\right)$ est dans $A$, $T\\left(a_{1}, \\ldots, a_{p}\\right)=\\left(a_{2}, \\ldots, a_{p}, a_{1}\\right)$. Il est clair que $T\\left(a_{1}, \\ldots, a_{p}\\right)$ est bien dans $A$, et qu'en appliquant $p$ fois $T$ à un élément de $A$, on obtient le même élément de $A$. Notons que $T$ est clairement injective, donc bijective car $A$ est fini. Pour chaque $y$, on définit $\\Omega_{y}:=\\left\\{T^{k}(y) \\mid k \\geqslant 0\\right.$ où $T^{k}$ est la composée $k$ fois de $T$.\n\nSoit $x, y$ deux éléments de $A$ tels que $\\Omega_{y} \\cap \\Omega_{x}$ est non vide. Il existe $h, \\ell$ positif tels que $T^{h}(x)=T^{\\ell}(y)$. En particulier en appliquant $\\ell(p-1)$ fois $T$, $T^{h+\\ell(p-1)}(x)=T^{\\ell p}(y)=y$, donc pour tout entier $j$ positif, $T^{j}(y)=T^{h+\\ell(p-1)+j}(x) \\in \\Omega_{x}$ : on a donc $\\Omega_{y} \\subset \\Omega_{x}$. De même on a l'inclusion réciproque, donc $\\Omega_{x}=\\Omega_{y}$. Ainsi les $\\left(\\Omega_{x}\\right)$ partitionnent l'ensemble $A$.\n\nSoit $x$ dans $A$, montrons que $\\Omega_{x}$ est de cardinal $p$, sauf si $x=(0, \\ldots, 0)$. Déjà, comme $T$ appliqué $p$ fois donne l'identité, on sait que $\\Omega_{x}=\\left\\{T^{k}(x) \\mid k \\in\\{0, \\ldots, p-1\\}\\right.$ donc est au plus de cardinal $p$. Supposons qu'il existe $i \\neq j$ deux éléments de $\\{0, \\ldots, p-1\\}$ tels que $T^{i}(x)=T^{j}(x)$. Par symétrie, on peut supposer $i<j$, on a donc $T^{p-j+i}(x)=T^{p-j+j}(x)=x$, donc $x=T^{h}(x)$ avec $h=p-j+i \\in\\{1, \\ldots, p-1\\}$. Posons $x=\\left(x_{1}, \\ldots, x_{n}\\right)$, l'équation $x=T^{h}(x)$ se retraduit par $x_{j}=x_{j+h}$ pour tout entier $j$ (en prenant les indices modulo $p$). En particulier, pour tout entier $\\ell \\geqslant 0$, $x_{j+h \\ell}=x_{j}$. En prenant $\\ell$ un inverse de $h$ modulo $p$, on a que pour tout $j$, $x_{j}=x_{j+1}$. Ainsi $x$ est de la forme $(1, \\ldots, 1),(0, \\ldots, 0)$ ou $(-1, \\ldots,-1)$ ce qui est contradictoire avec le fait que $k \\neq \\pm p$ sauf si $x=(0, \\ldots, 0)$. Ainsi $\\Omega_{x}=\\left\\{T^{k}(x) \\mid k \\in\\{0, \\ldots, p-1\\}\\right.$ est bien de cardinal $p$ sauf si $x=(0, \\ldots, 0)$.\n\nComme $A$ est partitionné en ensembles de cardinal $p$, sa taille est divisible par $p$ si $k \\neq 0$. Si $k=0$, $A$ privé de $(0, \\ldots, 0)$ est partitionné en ensembles de cardinal $p$, donc la taille de $A$ vaut $1$ modulo $p$.\n\nMaintenant, notons que pour tout entier $k$, et tout $n \\geqslant 2$, $s_{n}(k)=s_{n-1}(k-1)+s_{n-1}(k)+s_{n-1}(k+1)$. En effet, un $n$-uplet de somme $k$ à valeurs dans $\\{-1,0,1\\}$ peut soit commencer par $1$ et finir par un $n-1$-uplet de somme $k-1$, soit commencer par $-1$ et finir par un $n-1$-uplet de somme $k+1$, soit commencer par $0$ et finir par un $n-1$-uplet de somme $k$. Donc il y a autant de $n$-uplets de somme $k$ que de $n-1$-uplets de somme $k-1$, $k$ ou $k+1$.\n\nMaintenant, à partir de ces observations essayons de conclure. Pour $p=3$, $s_{p}=3$, donc le reste de $s_{p}$ modulo $p$ est $0$. Via la formule précédente,\n\n$$\n\\begin{aligned}\ns_{\\mathfrak{p}-1}(k)-s_{\\mathfrak{p}-1}(k+3) & =s_{\\mathfrak{p}-1}(k)+s_{\\mathfrak{p}-1}(k+1)+s_{\\mathfrak{p}-1}(k+2)-\\left(s_{\\mathfrak{p}-1}(k+1)+s_{\\mathfrak{p}-1}(k+2)-s_{\\mathfrak{p}-1}(k+3)\\right) \\\\\n& =s_{\\mathfrak{p}}(k+1)-s_{\\mathfrak{p}}(k+2)\n\\end{aligned}\n$$\n\nEn sommant la formule précédente pour $k$ multiple de $3$ entre $0$ et $3(p-1)$, on obtient que :\n\n$$\ns_{\\mathfrak{p}-1}(0)-s_{\\mathfrak{p}-1}(3 p)=\\sum_{k=0}^{p-1} s_{p}(3 k+1)-s_{p}(3 k+2)\n$$\n\nOr on a $s_{\\mathfrak{p}}(n)=0$ sauf pour $n= \\pm p$ ou $0$. Et on a clairement $s_{p-1}(3 p)=0$, car $3 p>p-1$, donc un $p-1$-uplet de $\\{-1,0,1\\}$ ne peut pas avoir somme $3 p$. Ainsi la formule précédente donne que $s_{p-1}$ vaut $1$ modulo $p$ et $p$ est de la forme $3 k+1$, $p-1$ si $p$ est de la forme $3 k+2$.\n\nL'énoncé peut être reformulé de la façon suivante : $s_{p}$ est le nombre de $p-1$-uplets dans $\\{-1,0,1\\}$ de somme nulle. En particulier, en ajoutant $1$ à chaque pas, $s_{p}$ est le nombre de $p-1$-uplets dans $\\{0,1,2\\}$ de somme $p-1$. C'est donc également le coefficient de degré $p-1$ de $\\left(1+X+X^{2}\\right)^{p-1}$ d'après la théorie des séries génératrices : en effet, chaque terme du produit $\\left(1+X+X^{2}\\right) \\ldots\\left(1+X+X^{2}\\right)$ correspond à $p-1$ choix de $0,1$ ou $2$, et la somme des différents pas correspond à la puissance de $X$ dans le produit global.\n\nAinsi on cherche à comprendre ce que vaut le reste du $p-1$-ième coefficient de $Q(X):=\\left(1+X+X^{2}\\right)^{p-1}$ modulo $p$. On peut essayer de comprendre les valeurs de $Q$ modulo $p$ : si $a$ est un entier, et $p$ ne divise pas $1+a+a^{2}$, $Q(a) \\equiv 1\\ (\\bmod\\ p)$ par le théorème de petit Fermat. Sinon $Q(a) \\equiv 0\\ (\\bmod\\ p)$.\n\nEn particulier $\\sum_{a=0}^{p-1} Q(a) \\equiv-n_{p}$ où $n$ est le nombre de solution de racines distinctes de $X^{2}+X+1$ modulo $p$. Pour $p=3$, il n'y a qu'une seule racine : $1$. Pour $p \\neq 3$, $1$ n'est pas racine, donc ce sont les racines de $X^{3}-1$ différentes de $1$. Or il existe un élément d'ordre $3$ modulo $p$ si et seulement si $3$ divise $p-1$ : en effet, s'il existe un élément d'ordre $3$ modulo $p$, par petit Fermat, $3$ divise $p-1$. Si $3$ divise $p-1$, soit $y$ une racine primitive modulo $p$, $y^{(p-1) / 3}$ est un élément d'ordre $3$.\n\nAinsi $n_{3}=1$, si $p \\neq 3$ et $3$ ne divise pas $p-1$, $n_{p}=0$ et si $p \\neq 3$ et $3$ divise $p-1$, $X^{2}+X+1$ a une racine. Comme c'est un polynôme de degré $2$ il a $2$ racines. De plus celles-ci sont distinctes : sinon son discriminant vaudrait $0$ modulo $p$, or celui-ci vaut $-3$. Ainsi si $3$ divise $p-1$, $n=2$.\n\nOr $Q$ est unitaire de degré $2(p-1)$. Écrivons $Q$ sous la forme $\\sum_{k=0}^{2(p-1)} a_{k} X^{k}$ avec $a_{2(p-1)}=1$ et $a_{p-1}=s_{p}$. On a :\n\n$$\n\\sum_{a=0}^{p-1} Q(a) \\equiv \\sum_{k=0}^{2(p-1)} a_{k} \\sum_{a=0}^{p-1} a^{k} \\quad(\\bmod p)\n$$\n\nPour $k=0$, $\\sum_{a=0}^{p-1} a^{k} \\equiv 0\\ (\\bmod\\ p)$. Pour $k>0$, $0^{k}=0$, donc si on se donne $y$ une racine primitive modulo $p$,\n\n$$\n\\sum_{a=0}^{p-1} a^{k} \\equiv \\sum_{a=1}^{p-1} a^{k} \\equiv \\sum_{n=0}^{p-2}\\left(y^{n}\\right)^{k} \\equiv \\sum_{n=0}^{p-2}\\left(y^{k}\\right)^{n} \\quad(\\bmod p)\n$$\n\nSi $y^{k} \\not \\equiv 1\\ (\\bmod\\ n)$ (ce qui est le cas si $p-1$ ne divise pas $k$), alors $\\sum_{n=0}^{p-2}\\left(y^{k}\\right)^{n} \\equiv \\frac{y^{(p-1) k}-1}{y^{k}-1} \\equiv 0\\ (\\bmod p)$ car $y^{p-1} \\equiv 1\\ (\\bmod p)$. Si $p-1$ divise $k$, $\\sum_{n=0}^{p-2}\\left(y^{k}\\right)^{n} \\equiv p-1\\ (\\bmod p)$. Ainsi, on obtient\n\n$$\n\\sum_{a=0}^{p-1} Q(a) \\equiv \\sum_{k=0}^{2(p-1)} a_{k} \\sum_{a=0}^{p-1} a^{k} \\quad(\\bmod p) \\equiv-a_{p-1}-a_{2(p-1)}=-a_{p-1}-1\n$$\n\nDonc $a_{p-1} \\equiv n-1\\ (\\bmod p)$\n\nAinsi :\n- Si $p=3$, le reste de $s_{p}$ modulo $p$ vaut $0$,\n- si $p \\neq 3$ et $3$ ne divise pas $p-1$ (i.e. si $p \\equiv 2\\ (\\bmod 3)$), le reste de $s_{p}$ modulo $p$ vaut $p-1$,\n- si $3$ divise $p-1$, le reste de $s_{p}$ modulo $p$ vaut $1$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 30853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k \\geqslant 1$ un entier. Quel est le plus petit entier $n$ tel que, quelque soit la manière de placer $n$ points dans le plan, il est possible de choisir un sous-ensemble $S$ constitué de $k$ de ces points qui vérifie \"pour toute paire $P, Q$ de points de $S$, la distance entre $P$ et $Q$ est inférieure ou égale à 2\" ou \"pour toute paire $P, Q$ de points de $S$, la distance entre $P$ et $Q$ est strictement plus grande que 1.\"",
    "options": [],
    "answer": "(k-1)^2 + 1",
    "solution": "Solution:\n\nCommençons par montrer que $n = (k-1)^2 + 1$ est possible. Pour cela, considérons le disque de rayon $1$ centré sur un point $P_1$ de l'ensemble. S'il y a au moins $k$ points dans ce disque, on peut prendre pour $S$ ces $k$ points (les distances entre eux sont inférieures ou égales à $2$). On peut donc supposer qu'il y a au plus $(k-1)$ points dans le disque.\n\nConsidérons alors un disque de rayon $1$ centré sur un $P_2$ de l'ensemble qui n'est pas dans le disque précédent. Par le même argument, on peut supposer qu'il y a au plus $(k-1)$ points dans ce disque.\n\nEn continuant à choisir des points en dehors des disques précédents, et comme on exclut au plus $(k-1)$ points à chaque étape, on arrive à construire $S = \\{P_1, \\ldots, P_k\\}$ de telle manière à ce que les distances entre ces points soient toutes strictement supérieures à $1$.\n\nPour montrer que $n = (k-1)^2$ n'est pas possible, on place à distance très grande les uns des autres $k-1$ groupes, où chaque groupe est constitué de $k-1$ points à distance très petite.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $n \\geqslant 2$ et $p$ un nombre premier impair. Soit $U$ l'ensemble des entiers positifs inférieurs ou égaux à $p^{n}$ et premiers avec $p$ et soit $N=|U|$. Montrer qu'il existe une permutation $a_{1}, \\ldots, a_{N}$ des éléments de $U$ telle que $\\sum_{k=1}^{N} a_{k} a_{k+1}$ (avec $a_{N+1}=a_{1}$) soit divisible par $p^{n-1}$ mais pas par $p^{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par traiter le cas $p=3$, dont la solution est différente. Soit $g \\in U$ un générateur modulo $p^{n}$, on pose $a_{i}=g^{i-1} \\bmod p^{n}$. Comme $g^{2}-1$ est divisible par $3$, la somme considérée $S$ vérifie $(g^{2}-1) S \\equiv \\sum_{i=1}^{N}(g^{2 i+1}-g^{2 i-1}) \\equiv g(g^{2 N}-1) \\pmod{p^{n+1}}$. Donc pour que $v_{p}(S)=n-1$, il faut et suffit que $v_{p}(g^{2 N}-1)=n$ ($g$ est premier avec $p$). Puisque $g^{N} \\equiv 1 \\pmod{p^{n}}$ (par Euler-Fermat), il suffit de montrer que $v_{p}(g^{N}-1)=n$. Mais par LTE, $v_{p}(g^{N}-1)=v_{p}(g^{N / p}-1)+1<n+1$, donc $v_{p}(g^{N}-1) \\leqslant n$, de sorte que $v_{p}(g^{N}-1)=n$, ce qui conclut.\n\nSupposons maintenant $p>3$ : Soient $b_{0}, \\ldots, b_{m-1} \\in U$ des entiers représentant exactement une fois chaque classe de congruence modulo $p^{n-1}$, ainsi $m=N / p=p^{n-2}(p-1)$ (et on pose $b_{m}=b_{0}$). On considère la permutation\n$$\n\\begin{aligned}\n& b_{0},\\ b_{0}+p^{n-1},\\ b_{0}+2 p^{n-1},\\ \\ldots,\\ b_{0}+(p-1) p^{n-1} \\\\\n& b_{1},\\ b_{1}+p^{n-1},\\ \\ldots,\\ b_{1}+(p-1) p^{n-1} \\\\\n& b_{2},\\ \\ldots,\\ b_{m-1},\\ \\ldots,\\ b_{m-1}+(p-1) p^{n-1}\n\\end{aligned}\n$$\n(les additions sont toujours à effectuer modulo $p^{n}$).\n\nAlors la somme correspondante $S$ est congrue modulo $p^{n}$ à\n$$\nS_{1}=\\sum_{i=0}^{m-1} b_{i} b_{i+1}+\\sum_{i=0}^{m-1}(p-1) p^{n-1} b_{i}+\\sum_{i=0}^{m-1} \\sum_{k=0}^{p-2} b_{i} p^{n-1}(2 k+1)+(p-1) \\sum_{i=0}^{m-1} b_{i}^{2}\n$$\nComme la somme des $b_{i}$ est divisible par $p$ et la somme des $b_{i}^{2}$ par $p^{n-1}$, $S \\equiv S_{2} \\pmod{p^{n}}$, où $S_{2}=\\sum_{i=0}^{m-1} b_{i} b_{i+1}-\\sum_{i=0}^{m-1} b_{i}^{2}$.\n\nSoit alors $g \\in U$ une racine primitive modulo $p^{n}$ : prenons $b_{i}=g^{i} \\bmod p^{n}$. Alors $(g+1) S_{2} \\equiv (g+1) \\sum_{i=0}^{m-1}(g^{2 i+1}-g^{2 i}) \\equiv (g^{2 m}-1) \\pmod{p^{n}}$. D'après LTE, $v_{p}(g^{2 m}-1) \\geqslant 1+v_{p}\\left(\\frac{2 m}{p-1}\\right)=n-1$ (puisque $p-1|m$, $p| g^{p-1}-1$ par Fermat, et $N=p^{n-1}(p-1)$, mais comme $2 m<N$, $v_{p}(g^{2 m}-1)<n$, donc $v_{p}(g^{2 m}-1)=n-1$. De plus, puisque $p>3$, $g+1$ est premier à $p$, et donc $v_{p}(S_{2})=n-1$, d'où $v_{p}(S)=n-1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle avec $\\widehat{BAC}=60^{\\circ}$ et soit $\\Gamma$ son cercle circonscrit. Soient $H$ l'orthocentre de $ABC$ et $S$ le milieu de l'arc $\\widehat{BC}$ ne contenant pas $A$. Soit $P$ le point de $\\Gamma$ tel que $\\widehat{SPH}=90^{\\circ}$. Montrer qu'il existe un cercle passant par $P, S$ et qui est tangent à $(AB)$ et $(AC)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nD'après le théorème du pôle Sud, $S$ est le point de concours de la bissectrice de $\\widehat{BAC}$ et de la médiatrice de $[BC]$. Soit $O$ le centre de $\\Gamma$ et $N$ le pôle Nord. Comme $[SN]$ est un diamètre de $\\Gamma$, la condition $\\widehat{SPH}=90^{\\circ}$ se réécrit $P, H, N$ alignés. Introduisons alors $Q$ le point d'intersection de $(AS)$ et $(PN)$.\n\nRemarquons que comme $\\widehat{SPQ}=90^{\\circ}$, $P$, $S$ se situent sur le cercle de diamètre $[QS]$, c'est donc un candidat pour être le cercle recherché. On aimerait montrer qu'il est tangent à $(AB)$ et à $(AC)$. Posons $M$ le milieu de $[QS]$ et $R$ le projeté orthogonal de $M$ sur $[AC]$. Le but est de montrer que $R$ est sur le cercle de diamètre $[QS]$, de sorte à avoir la tangence avec $(AC)$ (et pour des raisons de symétrie on aura la tangence avec $(AB)$).\n\nSoit $D$ le milieu de $[BC]$. On sait que $(HD)$ et $(AO)$ se coupent sur $\\Gamma$, et que $(AH) \\parallel (OM)$ (les deux sont perpendiculaires à $(BC)$), donc par droite des milieux $AH=2OD$. Or par angle au centre, $\\widehat{SOC}=60^{\\circ}$ donc $SOC$ est équilatéral ($OS=OC$), et donc $2OD=OS=OC=ON$. De plus, comme $(AH) \\parallel (NS)$, d'après le théorème de Thalès, $\\frac{AQ}{SQ}=\\frac{AH}{NS}=\\frac{1}{2}$ donc $QS=2AQ$.\n\nEn particulier $MQ=MS=\\frac{QS}{2}=AQ$. Or $\\widehat{MAR}=30^{\\circ}$ donc $MR=\\frac{1}{2}AM=MQ$ (en effet $\\sin(30^{\\circ})=\\frac{1}{2}$), donc $R$ est bien sur le cercle de diamètre $[QS]$, donc ce dernier est tangent à $(AC)$ (en $R$).\n\nDonc il existe bien un cercle passant par $P, S$ et qui est tangent à $(AB)$ et $(AC)$ : le cercle de diamètre $[QS]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer toutes les fonctions $f$ injective de $\\mathbb{N}$ dans $\\mathbb{N}$ telles que tout polynôme $\\sum_{i=0}^{n} a_{i} x^{i}$ a une racine réelle si et seulement si le polynôme $\\sum_{i=0}^{n} a_{i} x^{f(i)}$ a une racine réelle.",
    "options": [],
    "answer": "All such functions are f(n) = a n for all natural numbers n, where a is an odd integer.",
    "solution": "Solution:\n\nIl s'avère que la condition d'injectivité était inutile : nous allons prouver le résultat sans l'injectivité de $f$. Notons que le polynôme $1$ n'a pas de racine réelle, la condition de l'énoncé nous apprend que $X^{f(0)}$ n'en a pas non plus : la seule possibilité pour $f(0)$ est donc $f(0)=0$. De même, $f(n) \\neq 0$ si $n \\neq 0$ car $X^{n}$ a une racine réelle, et $1$ n'en a pas.\n\nMontrons que $f(n)$ a la même parité que $n$ pour tout entier $n$. Notons que si $n$ est pair, $X^{n}+1$ n'a pas de racine réelle donc $X^{f(n)}+1$ non plus. Ceci implique que $f(n)$ est pair (sinon $-1$ serait une racine réelle). De même si $n$ est impair, $f(n)$ est impair. Posons $\\alpha=f(1)$, $\\alpha$ est donc impair.\n\nSoit $t$ un réel. Pour la suite, nous aurons besoin de comprendre, pour tout $b$ réel vérifiant $b>1$, si l'équation $x^{b}-x=t$ sur $\\mathbb{R}^{+}$. Pour tout $t>0$, on pose $f_{b}: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}$ qui à $x$ associe $x^{b}-x$.\n\n**Lemme 2.** L'équation $x^{b}-x=t$ a une solution si et seulement si $t \\geqslant f_{b}\\left(x_{b}\\right)$ où $x_{b}=\\left(\\frac{1}{b}\\right)^{\\frac{1}{b-1}}$.\n\nDémonstration. La fonction $f_{b}$ est continue et dérivable sur $\\mathbb{R}^{+}$. La dérivée de $f_{b}$ est $b x^{b-1}-1$, qui est négative jusqu'en $x_{b}$ puis positive. Ainsi $f$ décroît puis croît. Comme $f$ a pour limite $+\\infty$ en $+\\infty$, les valeurs prises par $f$ sur $\\mathbb{R}^{+}$ sont exactement $[f(x_{b}),+\\infty[$ par continuité, ce qui donne le résultat voulu.\n\nOn aura aussi besoin du lemme suivant :\n\n**Lemme 3.** Soit $b>a$ deux entiers strictement positifs, avec $b$ pair. Si l'équation $x^{b}-x^{a}=t$ a une solution, alors cette équation a une solution positive.\n\nDémonstration. Soit $x$ une solution de l'équation. Comme $b$ est pair, $|x|^{b}-|x|^{a} \\leqslant x^{b}-x^{a}=t$. Or l'application qui à $y$ associe $y^{b}-y^{a}$ est continue, et tend vers $+\\infty$ en $+\\infty$, donc l'équation $y^{b}-y^{a}=t$ a une solution dans $[|x|,+\\infty[$, donc une solution positive.\n\nMaintenant appliquons l'hypothèse de l'énoncé au polynôme $X^{2 n}-X-t$. Celui-ci a une racine réelle si et seulement si $t \\geqslant f_{2 n}(x_{2 n})$. En particulier, le polynôme $X^{f(2 n)}-X^{\\alpha}-t$ a une racine réelle si et seulement si $t \\geqslant f_{2 n}(x_{2 n})$.\n\n- Si $f(2 n)<\\alpha$, alors la fonction qui à $x$ associe $x^{f(2 n)}-x^{\\alpha}$ tend vers $-\\infty$ et $+\\infty$, et $+\\infty$ en $-\\infty$ car $\\alpha$ est impair. Ainsi par continuité, l'équation $x^{f(2 n)}-x^{\\alpha}=t$ a toujours une solution donc $X^{f(2 n)}-X^{\\alpha}-t$ a toujours une racine ce qui est absurde.\n\n- Si $f(2 n)=\\alpha$, alors le polynôme n'a une racine que pour $t=0$ ce qui est absurde.\n\nAinsi $f(2 n)>\\alpha$. En particulier, par le lemme précédent le polynôme $X^{f(2 n)}-X^{\\alpha}-t$ a une racine réelle si et seulement s'il a une racine réelle positive, c'est-à-dire si et seulement si l'équation $x^{f(2 n)}-x^{\\alpha}=t$ a une solution positive. Comme $x \\mapsto x^{\\alpha}$ est bijectif de $\\mathbb{R}^{+}$ dans $\\mathbb{R}^{+}$, le polynôme $X^{f(2 n)}-X^{\\alpha}-t$ a une racine réelle si et seulement si l'équation $x^{f(2 n)/\\alpha}-x=t$ a une racine positive, donc si et seulement si $t \\geqslant f_{f(2 n)/\\alpha}(x_{f(2 n)/\\alpha})$. Posons $u=f(2 n)/\\alpha>1$, on a donc $f_{u}(x_{u})=f_{2 n}(x_{2 n})$.\n\n- Si $2 n>u$, comme $x_{u}<1$, $f_{u}(x_{u})>f_{2 n}(x_{u}) \\geqslant f_{2 n}(x_{2 n})=f_{u}(x_{u})$ (via les variations de $f_{2 n}$) ce qui est absurde.\n- Si $2 n<u$, comme $x_{2 n}<1$, $f_{2 n}(x_{2 n})>f_{u}(x_{2 n}) \\geqslant f_{u}(x_{u})=f_{2 n}(x_{2 n})$ (via les variations de $f_{u}$) ce qui est absurde.\n\nAinsi $u=2 n$, donc $f(2 n)=2 n \\alpha$.\n\nIl reste alors à montrer que si $n$ est impair, $f(n)=\\alpha n$. Pour cela on fait de même que dans l'argument précédent : on regarde le polynôme $X^{2 n}-X^{n}-t$, on montre qu'il a une racine réelle si et seulement si $t \\geqslant f_{2 n / n}(x_{2 n / n})=f_{2}(x_{2})$. Ensuite on obtient que $X^{f(2 n)}-X^{f(n)}-t$ a une racine si et seulement si $t \\geqslant f_{2}(x_{2})$. Comme $f(n)$ est impair, et $f(2 n)$ pair, on en déduit que $f(2 n)>f(n)$, puis que $f_{2}(x_{2})=f_{f(2 n)/f(n)}(x_{f(2 n)/f(n)})$. De même on prouve que $f(2 n)/f(n)=2$, donc $f(2 n)=2 f(n)$. Or $f(2 n)=2 n \\alpha$, donc $f(n)=n \\alpha$.\n\nAinsi si $f$ est solution, $f$ est de la forme $f: n \\mapsto a n$ pour un certain $a$ impair.\n\nRéciproquement, s'il existe $a$ impair tel que $f(n)=a n$ pour tout entier positif $n$, alors si le polynôme $\\sum_{k=0}^{n} a_{k} X^{k}$ a une racine réelle $x_{0}$, le polynôme $\\sum_{k=0}^{n} a_{k} X^{f(k)}=\\sum_{k=0}^{n} a_{k} X^{a k}$ a pour racine $x_{0}^{1/a}$. Réciproquement si $\\sum_{k=0}^{n} a_{k} X^{a k}$ a une racine réelle $y_{0}$, $y_{0}^{a}$ est racine de $\\sum_{k=0}^{n} a_{k} X^{k}$. Ainsi $f$ vérifie bien l'énoncé.\n\nLes fonctions vérifiant l'énoncé sont les fonctions telles qu'il existe $a$ impair tel que $f(n)=a n$ pour tout entier positif $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k, n \\geqslant 1$ deux entiers fixés. Thanima possédait $2n$ bonbons de chaque couleur. Elle a donné deux bonbons de couleurs différentes à chacun des enfants de sa famille. Sachant que, quelle que soit la manière de choisir $k+1$ enfants, il y en a deux parmi eux qui ont reçu une couleur de bonbon en commun, trouvez le nombre maximum possible d'enfants.",
    "options": [],
    "answer": "3kn",
    "solution": "Solution:\n\nOn peut modéliser le problème par un graphe : chaque sommet de ce graphe est une couleur. Si Thanima a donné un bonbon d'une couleur $c_1$ et un bonbon d'une couleur $c_2$ à un enfant de sa famille, on ajoute une arête entre $c_1$ et $c_2$.\n\nComme Thanima n'a que $2n$ bonbons de chaque couleur, le degré maximal de chaque sommet est $2n$. On sait de plus qu'il est impossible de choisir $k+1$ arêtes disjointes. On cherche le nombre maximal d'arêtes.\n\nOn voit qu'il est possible d'obtenir $3kn$ arêtes. Pour cela, on crée un graphe dans lequel on met $k$ fois la structure suivante : on crée trois sommets, et entre chaque paire de ces trois sommets, on ajoute $n$ arêtes.\n\nMontrons qu'il n'est pas possible de faire plus que $3kn$. Pour cela, on considère un ensemble $C$ d'arêtes disjointes de plus grand cardinal. D'après l'énoncé, on a $|C| \\leqslant k$. Ainsi, chaque arête du graphe doit intersecter l'une des arêtes de $C$, sinon cela contredirait la maximalité de $C$. Notons $V_C$ l'ensemble des sommets qui apparaissent dans l'une des arêtes de $C$. Le nombre d'arêtes du graphe vaut :\n$$\n\\sum_{v \\in V_C} d_{\\text{ext}}(v) + \\frac{d_{\\text{int}}(v)}{2}\n$$\noù $d_{\\text{ext}}(v)$ est le nombre d'arêtes qui relient $v$ à des sommets qui ne sont pas dans $V_C$, et $d_{\\text{int}}(v)$ est le nombre d'arêtes qui relient $v$ à des sommets de $V_C$.\n\nSoit $(a, b)$ une arête de $C$, et soient $x, y$ deux sommets différents qui ne sont pas dans $V_C$, on ne peut pas à la fois avoir d'arêtes entre $a$ et $x$ et entre $b$ et $y$, sinon remplacer $(a, b)$ par $(a, x)$ et $(b, y)$ dans $C$ donnerait un ensemble d'arêtes disjointes plus grand.\n\nDeux cas peuvent arriver :\n- À la fois $a$ et $b$ ont des arêtes vers le complémentaire de $V_C$, dans ce cas, toutes ces arêtes vont vers le même sommet, et la borne de degré sur ce sommet donne $d_{\\text{ext}}(a) + d_{\\text{ext}}(b) \\leqslant 2n$.\n- Soit il y a au plus un sommet parmi $a$ et $b$ qui a des arêtes vers le complémentaire de $V_C$, dans ce cas on a aussi $d_{\\text{ext}}(a) + d_{\\text{ext}}(b) \\leqslant 2n$.\n\nComme $d_{\\text{int}}(v) \\leqslant 2n - d_{\\text{ext}}(v)$ pour tout sommet $v$, on a :\n$$\n\\begin{aligned}\nd_{ext}(a) + d_{ext}(b) + \\frac{d_{int}(a) + d_{int}(b)}{2} &\\leqslant d_{ext}(a) + d_{ext}(b) + \\frac{4n - d_{ext}(a) - d_{ext}(b)}{2} \\\\\n&= 2n + \\frac{d_{ext}(a) + d_{ext}(b)}{2} \\\\\n&\\leqslant 3n\n\\end{aligned}\n$$\n\nAinsi, en groupant les termes deux à deux, on obtient que le nombre d'arêtes dans le graphe est :\n$$\n\\sum_{v \\in V_C} d_{\\text{ext}}(v) + \\frac{d_{\\text{int}}(v)}{2} \\leqslant 3kn\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $\\alpha \\neq 0$. Trouver toutes les fonctions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ telles que pour tous réels $x, y$,\n$$\nf\\left(x^{2}+y+f(y)\\right)=f(x)^{2}+\\alpha y\n$$",
    "options": [],
    "answer": "If alpha equals 2, the unique solution is f(x) = x for all real x; if alpha is not 2, there is no solution.",
    "solution": "Solution:\nPosons $x=0$. Alors l'équation devient $f(y+f(y))=f(0)^{2}+\\alpha y$ pour tout $y$. Comme $\\alpha$ est non nul, le côté droit décrit tous les réels lorsque $y$ varie, et ainsi $f$ est surjective.\n\nPosons $y=0$. Alors on peut écrire pour tout réel $x$,\n$$\nf(x)^{2}=f\\left(x^{2}+f(0)\\right)=f\\left((-x)^{2}+f(0)\\right)=f(-x)^{2}\n$$\nOn sait que $f$ est surjective, soit donc $c$ un antécédent de $0$. L'équation de l'énoncé avec $x=0$ et $y=c$ donne\n$$\n\\alpha c+f(0)^{2}=f(0+c+f(c))=f(c)=0\n$$\nMais on sait que $f(-c)^{2}=f(c)^{2}=0$ donc $-c$ est aussi un antécédent de $0$. Par l'équation ci-dessus appliquée à $c$ et à $-c$, on a donc $\\alpha c+f(0)^{2}=0=-\\alpha c+f(0)^{2}$. Comme $\\alpha \\neq 0$, on en déduit que nécessairement $c=0$.\n\nMaintenant que l'on sait que $0$ est le seul antécédent de $0$, on annule le terme de droite dans l'équation de départ en choisissant $y=-\\frac{f(x)^{2}}{\\alpha}$, alors $f\\left(x^{2}+y+f(y)\\right)=0$, soit $x^{2}+y+f(y)=0$ ou encore\n$$\n\\begin{gathered}\nx^{2}-\\frac{f(x)^{2}}{\\alpha}+f\\left(-\\frac{f(x)^{2}}{\\alpha}\\right)=0 \\\\\nx^{2}=\\frac{f(x)^{2}}{\\alpha}-f\\left(-\\frac{f(x)^{2}}{\\alpha}\\right)\n\\end{gathered}\n$$\npour tout réel $x$. Dès lors, si $f(x)= \\pm f(x')$, l'équation précédente donne $x^{2}=x'^{2}$ donc $x= \\pm x'$.\n\nSoit $x \\neq 0$ un réel. Étudions l'équation $f(x')= \\pm f(x)$. On a vu que celle-ci avait exactement deux solutions, qui sont $x$ et $-x$ (qui sont distincts car $x \\neq 0$). Mais il existe une solution à $f(x')=-f(x)$ par surjectivité de $f$. Comme $f(x) \\neq 0$, cela ne peut pas être $x'=x$, c'est donc $x'=-x$. Ainsi, $f(-x)=-f(x)$, et la seule solution à $f(x')=f(x)$ est $x'=x$. Comme en plus $0$ est l'unique antécédent de $0$, $f$ est à la fois impaire et injective.\n\nL'équation initiale avec $y=0$ donne pour tout $x$ réel, $f\\left(x^{2}\\right)=f(x)^{2}$. Ainsi, l'équation de l'énoncé se réécrit\n$$\n\\forall x, y \\in \\mathbb{R}, \\quad f\\left(x^{2}+y+f(y)\\right)=f\\left(x^{2}\\right)+\\alpha y\n$$\ndonc avec $z=x^{2}$,\n$$\n\\forall y \\in \\mathbb{R}, z \\geqslant 0, \\quad f(z+y+f(y))=f(z)+\\alpha y\n$$\nSi maintenant $z \\leqslant 0$, on calcule pour tout $y$, par imparité de $f$,\n$$\n\\begin{aligned}\nf(z+y+f(y)) & =-f(-z-y-f(y))=-f((-z)+(-y)+f(-y)) \\\\\n& =-(f(-z)+\\alpha(-y))=f(z)+\\alpha y\n\\end{aligned}\n$$\nC'est la même équation que pour $z \\geqslant 0$, d'où\n$$\n\\forall y, z \\in \\mathbb{R}, \\quad f(z+y+f(y))=f(z)+\\alpha y\n$$\nAvec $z=0$, on a notamment pour tout $y$, $f(y+f(y))=\\alpha y$. Posons $z=x+f(x)$ dans l'équation ci-dessus pour avoir pour tous $x, y \\in \\mathbb{R}$,\n$$\nf(x+f(x)+y+f(y))=f(x+f(x))+\\alpha y=\\alpha x+\\alpha y=\\alpha(x+y)=f(x+y+f(x+y))\n$$\net donc, par injectivité, $f(x+y)=f(x)+f(y)$. Ainsi, $f$ est solution de l'équation de Cauchy. Mais $f$ est croissante car si $x>y$, on a\n$$\nf(x)=f(y)+f(x-y)=f(y)+f((\\sqrt{x-y})^{2})=f(y)+f(\\sqrt{x-y})^{2} \\geqslant f(y)\n$$\ndonc $f$ est linéaire sur $\\mathbb{R}$. D'après l'équation $f\\left(x^{2}\\right)=f(x)^{2}$, son coefficient dominant $a$ vérifie $a^{2}=a$ donc $a=0$ ou $a=1$, c'est-à-dire que $f$ est soit nulle, soit l'identité sur $\\mathbb{R}$. $f$ étant surjective, nécessairement, c'est l'identité : pour tout $x$, $f(x)=x$.\n\nEn réinjectant dans l'équation, on obtient\n$$\n\\forall x, y \\in \\mathbb{R}, \\quad x^{2}+2y=x^{2}+\\alpha y\n$$\ndonc avec $x=0, y=1$, on a $\\alpha=2$ nécessairement. Réciproquement, si $\\alpha=2$, l'identité est bien solution par le calcul précédent. Ainsi, si $\\alpha \\neq 2$, il n'y a aucune solution, et si $\\alpha=2$, l'identité est solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle et $O$ le centre de son cercle circonscrit. Soit $d$ la parallèle à $(BC)$ passant par $O$. Soit $A'$ le symétrique de $A$ par rapport à $(BC)$. La parallèle à $(A'B)$ passant par $C$ coupe $d$ en $C_1$, et les droites $(A'C)$ et $(BC_1)$ s'intersectent en $C_2$. La parallèle à $(A'C)$ passant par $B$ coupe $d$ en $B_1$, et les droites $(A'B)$ et $(CB_1)$ s'intersectent en $B_2$. Montrer que les points $A$, $A'$, $B_2$, $C_2$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn pose $P$ le point d'intersection des droites $(CB_1)$ et $(BC_1)$, et $Q$ le point d'intersection de $(BB_1)$ et $(CC_1)$. On pose également $T$ l'intersection de $(AO)$ et $(BC)$. On peut alors remarquer plusieurs propriétés sur la figure, qu'on va montrer.\n\n- **Lemme 1**: Le point $Q$ est sur le cercle circonscrit à $ABC$.\n\nPreuve : Par construction de $B_1$, $C_1$, $A'BQC$ est un parallélogramme, donc $\\widehat{BQC} = \\widehat{CA'B}$. Mais par symétrie, $\\widehat{CA'B} = \\widehat{BAC}$. Donc $\\widehat{BAC} = \\widehat{BQC}$ : par angle inscrit, $Q$ est bien sur le cercle circonscrit à $ABC$.\n\n- **Lemme 2**: Les points $P$, $Q$, $A'$ sont alignés.\n\nPreuve : Le quadrilatère $BCC_1B_1$ est un trapèze, donc la droite $(PQ)$ passe par les milieux respectifs de $[B_1C_1]$ et $[BC]$. Donc la droite $(PQ)$ passe par le milieu de $[BC]$. Or $A'BQC$ est un parallélogramme, donc $(QA')$ passe également par le milieu de $[BC]$. On en déduit bien que $Q$, $P$, $A'$ sont alignés.\n\n- **Lemme 3**: Les points $O$, $P$, $T$ sont alignés.\n\nPreuve : Considérons la symétrie par rapport à la médiatrice de $[BC]$ : elle fixe $O$, elle échange $B$ et $C$. Elle échange également $A$ et $Q$, en effet $\\widehat{QCB} = \\widehat{A'BC} = \\widehat{CBA}$ par parallélisme et symétrie puis $\\widehat{CBQ} = \\widehat{ACB}$. On pose à présent $X$ l'intersection de $(AB)$ avec $d$, et $Y$ l'intersection de $(AC)$ avec $d$. Alors comme $d$ est invariante par la symétrie, la symétrie échange $B_1$ et $Y$, $C_1$ et $X$. On en déduit que $\\frac{OB_1}{OC_1} = \\frac{OY}{OX}$. Or en appliquant le théorème de Thalès, on obtient $\\frac{OY}{OX} = \\frac{TC}{TB}$, d'où $\\frac{TC}{TB} = \\frac{OB_1}{OC_1}$. Ceci entraîne $O$, $P$, $T$ alignés (en appliquant Thalès dans $OC_1PBT'$ et dans $OB_1BCT'$ où $T'$ est l'intersection de $(OP)$ et $(BC)$, on obtient $\\frac{TB}{TC} = \\frac{T'B}{T'C}$, d'où $T = T'$). On a donc bien montré que $O$, $P$, $T$ (et donc $A$, $O$, $P$, $T$) sont alignés.\n\n- **Lemme 4** : L'homothétie de centre $P$ et de rapport $-\\frac{PT}{PO}$ envoie $QBC$ sur $A'C_2B_2$.\n\nPreuve: D'après le théorème de Thalès dans $QCPB_2A'$ et dans $QBPC_2A'$, on a $\\frac{PQ}{PA'} = \\frac{PC}{PB_2} = \\frac{PB}{PC_2}$, donc l'homothétie de centre $P$ qui envoie $Q$ sur $A'$ envoie également $B$ sur $C_2$ et $C$ sur $B_2$. Il reste à montrer qu'elle envoie $O$ sur $T$. Or, d'après le théorème de Thalès dans $QC_1PBA'$ puis dans $OC_1PBT$, on a $\\frac{PQ}{PA'} = \\frac{PC_1}{PB} = \\frac{PO}{PT}$, donc c'est bien l'homothétie de rapport $-\\frac{PT}{PO}$.\n\n- On a à présent tous les éléments pour conclure : l'homothétie du lemme 4 envoie $QBC$ sur $A'C_2B_2$ et $O$ sur $T$. Or par le lemme 1, $O$ est le centre du cercle circonscrit de $QBC$. Donc $T$ est le centre du cercle circonscrit de $A'C_2B_2$. Mais comme $T$ est sur $(BC)$, on a $TA = TA'$, donc $A$ est sur le cercle de centre $T$ et de rayon $TA'$. Finalement, on a bien montré que $A$, $A'$, $B_2$, $C_2$ sont cocycliques, et on a même trouvé le centre du cercle en question, qui est $T$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\left(a_{n}\\right)_{n \\geqslant 1}$ une suite d'entiers strictement positifs telle que $a_{1}$ et $a_{2}$ soient premiers entre eux et, pour tout $n \\geqslant 1$, $a_{n+2}=a_{n} a_{n+1}+1$. Montrer que pour tout entier $m>1$, il existe $n>m$ tel que $a_{m}^{m} \\mid a_{n}^{n}$. Le résultat est-il encore vrai lorsque $m=1$ ?",
    "options": [],
    "answer": "For every index greater than one, there exists a later index such that the earlier powered term divides the later powered term. The statement is false for the first index; for example, take the first term equal to one hundred fifty-five and choose the second term congruent to four modulo five and twenty-nine modulo thirty-one.",
    "solution": "Solution:\n\nD'abord, $a_{n}>0$ pour tout $n>0$.\n\nOn commence par une observation : soit $n>m$ très grand (disons, $n>(m+1)\\left(a_{m}+1\\right)$ ). Alors $a_{m}^{m} \\mid a_{n}^{n}$ si et seulement si pour chaque nombre premier $p\\mid a_{m}$, $p\\mid a_{n}$. Cette idée justifie le lemme qui va suivre :\n\nLemme : soit $\\left(u_{n}\\right)_{n \\geqslant 0}$ la suite modulo un nombre premier $p$ telle que $u_{0}=0$, $u_{1}=1$ et pour tout $n \\geqslant 0$, $u_{n+2}=u_{n} u_{n+1}+1$. Alors $u$ est périodique.\n\nPreuve : soit $v_{n}=(u_{n}, u_{n+1})$; alors $u$ est périodique dès que $v$ est périodique. Comme $v_{n+1}$ dépend uniquement de $v_{n}$, $v$ est périodique dès qu'il existe $N \\geqslant 1$ tel que $v_{N}=v_{0}$. Si $N \\geqslant 1$ est tel que $u_{N}=0$, alors la relation de récurrence montre que $u_{N+1}=1=u_{1}$ et donc $v_{N}=v_{0}$. Par conséquent, pour montrer que $u$ est périodique, il suffit de montrer qu'il existe $N \\geqslant 1$ tel que $u_{N}=0$.\n\nSupposons donc que le seul entier $n$ tel que $u_{n}=0$ soit $n=0$. Alors comme $v$ est à valeurs dans un ensemble fini, il existe un entier $n \\geqslant 0$ minimal tel qu'il existe un entier $m>n$ tel que $v_{n}=v_{m}$. En particulier, $u_{m}=u_{n}$ et $m \\neq 0$, donc $u_{m} \\neq 0$, et donc $u_{n} \\neq 0$ et donc $n>0$. Alors $m+1>n+1 \\geqslant 2$ et $u_{n+1}=u_{m+1}$, donc $u_{n-1} u_{n}+1=u_{m-1} u_{m}+1$, donc $u_{n}(u_{n-1}-u_{m-1})=0$, d'où, comme $u_{n} \\neq 0$, $u_{n-1}=u_{m-1}$, de sorte que $v_{m-1}=v_{n-1}$, ce qui contredit la minimalité de $n$.\n\nRevenons à notre preuve. Lorsque $m>1$, pour chaque facteur premier $p$ de $a_{m}$, $a_{m+1}=a_{m-1} a_{m}+1 \\equiv 1\\ [p]$, $\\left(a_{m+n}\\ (\\bmod\\ p)\\right)_{n \\geqslant 0}$ satisfait les hypothèses du lemme, donc il existe $N_{p} \\geqslant 1$ tel que pour tout $n \\geqslant 0$, $a_{m+n} \\equiv a_{m+n+N_{p}}\\ [p]$.\n\nSoit $N$ le produit des $N_{p}$ (où $p$ parcourt les facteurs premiers de $a_{m}$ ) : alors, si $n \\geqslant 0$, tout diviseur premier $p$ de $a_{m}$ divise $a_{m+n N}$. En particulier, si $n \\geqslant m a_{m}$, $a_{m}^{m} \\mid a_{m+n N}^{m+n N}$ (parce que si $p \\mid a_{m}$, $v_{p}\\left(a_{m+n N}^{m+n N}\\right) \\geqslant m+n N \\geqslant m a_{m} \\geqslant v_{p}\\left(a_{m}^{m}\\right)$).\n\nLe résultat est faux pour $m=1$ : prenons $a_{1}=155$, $a_{2}$ congru à $4$ modulo $5$ et congru à $29$ modulo $31$. On vérifie alors que $a_{n}$ est divisible par $5$ si et seulement si $n=1$ et $n \\equiv 4\\ [7]$, alors que $a_{n}$ est divisible par $31$ si et seulement si $n=1$ ou $n \\equiv 5\\ [7]$, donc si $n>1$, $a_{n}$ n'est jamais divisible par $5$ et $31$, et donc $155$ ne divise pas $a_{n}^{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30861,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $P(X)$ un polynôme à coefficients réels unitaire de degré $2022$. Emile joue au jeu suivant : il écrit le polynôme $P(X)$ au tableau et à chaque étape, si le polynôme $f(X)$ est écrit au tableau, Emile peut le remplacer par :\n- Le polynôme $f(X)+c$, pour $c$ un réel de son choix, ou\n- le polynôme $P(f(X))$.\n\nTrouver tous les entiers positifs ou nuls $n$ tels que, quelque soit le polynôme $P$ initial, Emile peut trouver une suite d'opérations telles que le polynôme obtenu ait exactement $n$ racines réelles distinctes.",
    "options": [],
    "answer": "All even n and all odd n at least 2021.",
    "solution": "Solution:\n\nOn dit qu'un entier $n$ fonctionne si, quel que soit le polynôme $P$ initial, Emile peut trouver une suite d'opérations telles que le polynôme obtenu ait exactement $n$ racines réelles distinctes.\n\nCommençons par faire la remarque simple qu'à chaque étape, quoique choisisse Emile, le polynôme écrit au tableau sera unitaire, de degré pair, et non constant.\n\nOn montre les points suivants :\n\n1. $\\underline{Si\\ n}$ fonctionne, alors $n+2$ aussi :\n\nSoit $P(X)$ un polynôme comme dans l'énoncé. Supposons qu'Emile peut trouver une suite d'opérations qui lui permet d'obtenir un polynôme $f(X)$ avec $n$ racines réelles distinctes écrit au tableau.\n\nOn va chercher à trouver deux constantes $a$ et $b$ telles que le polynôme $g(X)=P(f(X)+a)+b$ ait exactement $n+2$ racines réelles distinctes. Alors Emile pourra simplement ajouter $a$, composer par $P$, puis ajouter $b$ au polynôme obtenu pour avoir le polynôme voulu.\n\nOn souhaite utiliser le fait que $f(X)$ ait exactement $n$ racines en faisant en sorte qu'une racine de $f$ soit une racine de $g(X)=P(f(X)+a)+b$, ce qui impose $b=-P(a)$. Les racines de $g(X)$ sont alors les réels $x$ tels que $f(x)+a$ soit un antécédent de $P(a)$. Pour avoir le moins de cas possibles à traiter, on choisit $|a|$ très grand pour que l'équation $P(X)=P(a)$ ait exactement deux solutions réelles : c'est possible car pour $c$ assez grand, l'équation $P(X)=c$ a exactement deux solutions réelles.\n\nSoient $a, a'$ les deux solutions de $P(X)=P(a)$. Ainsi, les racines de $g(X)$ sont les racines de $f(X)$ et les solutions de $f(x)+a=a'$, c'est-à-dire les antécédents par $f$ de $a'-a$. Encore une fois, on veut le moins d'antécédents possibles (sans pour autant en avoir aucun), donc on voudrait que $a'-a$ soit très grand. $f$ étant unitaire de degré pair, si $a'-a$ est assez grand, $a'-a$ a exactement $2$ antécédents par $f$, ce qui donne bien que $g(X)$ a exactement $n+2$ racines.\n\nVérifions que l'on peut bien choisir $a$ comme voulu. Prenons $a$ négatif tel que $|a|$ soit très grand. Lorsque $a$ tend vers $-\\infty$, $P(a)$ tend vers $+\\infty$ et $a'$ est la solution la plus grande de $P(X)=P(a)$ qui tend vers $+\\infty$. Si $|a|$ est assez grand, $a'-a$ peut devenir arbitrairement grand, comme voulu.\n\n2. $n=0$ et $n=2021$ fonctionnent :\n\nSoit $P(X)$ un polynôme comme dans l'énoncé.\n\nLe cas de $n=0$ est simple. En effet, $P(X)$ est unitaire de degré pair, donc il admet un minimum sur $\\mathbb{R}$. Si $c$ est strictement supérieur à l'opposé de ce minimum, $P(x)+c=0$ n'a aucune solution sur $\\mathbb{R}$. Ainsi, si Emile remplace $P(X)$ par $P(X)+c$ au premier tour, il obtient un polynôme sans racines réelles comme voulu.\n\nPour $n=2021$, c'est un peu plus subtil. Comme $x \\mapsto P(x)$ est décroissant pour $x$ assez petit et croissant pour $x$ assez grand, $P$ a un nombre impair d'extrema locaux. Ainsi il existe $c$ tel qu'un nombre impair de ces extrema soient en des solutions de $P(x)=c$. En effet, supposons que les valeurs prises en ces extrema soient $c_1, \\ldots, c_k$, et disons que $c_i$ est la valeur de $P$ en $n_i$ extrema distincts. Alors le nombre d'extrema de $P$ est $n_1+\\cdots+n_k$ et est impair, donc il existe $i$ tel que $n_i$ soit impair, et c'est ce qu'on voulait.\n\nSupposons maintenant qu'Emile remplace le polynôme $P(X)$ initial par $P(X)-c$. Celui-ci est positif en $-\\infty$ et $+\\infty$ donc change de signe un nombre pair de fois. Mais par définition de $c$, il a aussi un nombre impair de zéros qui sont des extrema locaux (c'est-à-dire où il ne change pas de signe), et donc $P(X)-c$ a un nombre impair de racines réelles distinctes. Ce nombre est nécessairement inférieur à $2022$, donc à $2021$ car il est impair. Par la partie précédente de la démonstration, en ajoutant $2$ à plusieurs reprises au nombre de racines distinctes, Emile peut toujours obtenir, après un certain nombre d'étapes, un polynôme avec exactement $2021$ racines réelles distinctes.\n\nEn combinant les deux premières parties de la démonstration, on obtient que tous les $n$ pairs et les $n$ impairs supérieurs ou égaux à $2021$ fonctionnent. Il reste à voir que les autres ne fonctionnent pas.\n\n3. $n \\leqslant 2019$ impair ne fonctionne pas :\n\nSoit $n \\leqslant 2019$ impair. Dans cette partie de la démonstration, il faut exhiber un polynôme $P(X)$ qui empêche Emile d'obtenir $n$ racines réelles distinctes en un nombre fini d'étapes. D'après la preuve de la partie précédente, il faut trouver un polynôme $P(X)$ tel que si un nombre impair d'extrema locaux de $P$ vérifient $P(x)=c$, alors $P(X)-c$ a strictement plus de $2019$ racines distinctes.\n\nConsidérons le polynôme $P(X)=X^{2}(X^{2}-1)\\ldots(X^{2}-1010)$. Il est pair donc tous ses extrema locaux vont par paires, sauf celui en $0$ qui est nul. Mais le polynôme $P(X)$ a $2021$ racines distinctes, et il semble donc être un bon candidat pour cette partie. Comme $P(X)$ est pair, pour tout $c \\neq 0$, l'équation $P(x)=c$ a un nombre pair de solutions, et l'équation $P(x)=0$ a un nombre impair de solutions (précisément $2021$).\n\nSupposons par l'absurde qu'au bout d'un certain nombre d'étapes, Emile arrive à obtenir un polynôme avec $n$ racines réelles distinctes, ce polynôme sera alors de la forme $f(P(X))$, avec $f(X)$ unitaire non constant. Soient $r_1, \\ldots, r_k$ les racines de $f(X)$. Les racines de $f(P(X))$ sont exactement les réels solutions d'une équation de la forme $P(x)=r_i$. Soit $n_i$ le nombre de solutions à $P(x)=r_i$. On a $n_1+n_2+\\ldots+n_k=n$. Comme $n$ est impair, un des $n_i$ est forcément impair. Mais le seul réel $r_i$ ayant un nombre impair d'antécédents par $P$ est $0$, qui en a $2021$. Ainsi, $r_i=0$ et $n_i=2021$, ce qui donne $n \\geqslant 2021$, absurde.\n\nPour conclure, les entiers $n$ qui fonctionnent sont les entiers pairs et les entiers impairs supérieurs ou égaux à $2021$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn quadrilatère convexe $ABCD$ admet un cercle inscrit de centre $I$. Soit $I_a$, $I_b$, $I_c$ et $I_d$ les centres des cercles inscrits des triangles $DAB$, $ABC$, $BCD$ et $CDA$ respectivement. On suppose que les tangentes communes aux cercles circonscrits de $AI_bI_d$ et $CI_bI_d$ se rencontrent en $X$ et que les tangentes communes extérieures des cercles circonscrits de $BI_aI_c$ et $DI_aI_c$ se rencontrent en $Y$. Montrer que $\\widehat{XIY} = 90$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn note $\\omega_{i}$ le cercle inscrit de centre $I_{i}$ et de rayon $r_{i}$ pour $i \\in\\{A, B, C, D\\}$, on note également $O_{A}, O_{B}$, $O_{C}$ et $O_{D}$ les centres des cercles $A I_{b} I_{d}, B I_{a} I_{c}, C I_{b} I_{d}$ et $D I_{a} I_{c}$. On commence par démontrer le lemme suivant:\n\nLemme 4. Les cercles $\\omega_{a}$ et $\\omega_{c}$ sont tangents tous les deux en le même point sur (AC).\n\nPreuve : (On réfère également au Lemme 36 du cours de Géométrie D de Valbonne 2022).\n\n![](attached_image_1.png)\n\nSoit $X_{a}$ le point de tangence de $\\omega_{a}$ sur (BD) et $X_{c}$ le point de tangence de $\\omega_{c}$ sur (BD). On sait que $ABCD$ est circonscriptible, cela implique que $AB-AD=BC-CD$. On peut calculer la distance $BX_{a}$ elle vaut\n$$\nBX_{a}=\\frac{AB-AD+BD}{2}=\\frac{BC-CD+BD}{2}=BX_{c},\n$$\nce qui montre que $X_{a}=X_{c}$.\n\nOn renomme $T$ le point de tangence. Alors les points $I_{a}, T$ et $I_{c}$ sont alignés et $\\left(I_{a}I_{c}\\right) \\perp (BD)$. On note $\\widehat{ABT}=2\\varphi$ et $\\widehat{TBC}=2\\mu$, alors $\\widehat{ABI}=\\varphi+\\mu$ et $\\widehat{ABI_{a}}=\\varphi$, ce qui montre que $\\widehat{I_{a}BI}=\\mu$ et ainsi que les droites (BT) et (BI) sont conjuguées isogonales dans le triangle $BI_{a}I_{c}$, or (BT) est la hauteur dans ce même triangle et par un lemme classique on sait alors que la droite (BI) passe également par le centre du cercle circonscrit au triangle $BI_{a}I_{c}$. Donc $O_{B}$ est sur la droite (BI). De la même manière $O_{D}$ se trouve sur la droite (DI). On sait aussi que $\\left(O_{B}O_{D}\\right) \\perp \\left(I_{a}I_{c}\\right)$ donc $(DB)//\\left(O_{B}O_{D}\\right)$.\n\nOn va maintenant démontrer que la droite (IY) bissecte l'angle $\\widehat{BID}$. $X$ se trouve sur la droite $O_{B}O_{D}$ de telle sorte que $\\frac{XO_{B}}{XO_{D}}=\\frac{r_{b}}{r_{d}}$, par le lemme de la bissectrice on veut montrer que $\\frac{IO_{B}}{IO_{D}}=\\frac{r_{b}}{r_{d}}$, or comme $(DB)//\\left(O_{B}O_{D}\\right)$ on a $\\frac{IO_{B}}{IO_{D}}=\\frac{IB}{ID}$. On calcule maintenant\n$$\nX:=r_{b} \\cdot \\frac{1}{r_{d}}=\\frac{\\sin (\\widehat{IAB}) \\cdot AI}{\\sin (\\widehat{ABI})} \\frac{\\sin (\\widehat{ADI})}{\\sin (\\widehat{IAD}) \\cdot AI}\n$$\nd'après la loi des sinus dans les triangles IAB et IAD, on a d'une part $X=\\frac{\\sin (\\widehat{ADI})}{\\sin (\\widehat{ABI})}$. On trouve d'autre part,\n$$\nX':=r_{b} \\cdot \\frac{1}{r_{d}}=\\frac{2I_{a}I_{c}}{\\sin \\left(\\widehat{I_{a}DI_{c}}\\right)} \\cdot \\frac{\\sin \\left(\\widehat{I_{a}BI_{c}}\\right)}{2I_{a}I_{c}}=\\frac{\\sin \\left(\\widehat{I_{a}DI_{c}}\\right)}{\\sin \\left(\\widehat{I_{a}BI_{c}}\\right)},\n$$\nor, $\\widehat{ADI}=\\widehat{ADC}/2=\\widehat{I_{a}DI_{c}}$ et également $\\widehat{ABI}=\\widehat{I_{a}BI_{c}}$ ce qui montre que $X=X'$ et ainsi que la droite (IY) est la bissectrice extérieure de l'angle $\\widehat{BID}$.\n\nOn montre de la même manière que la droite (XI) est la bissectrice extérieure de l'angle $\\widehat{AIC}$.\n\nDans le but de démontrer que $\\widehat{XIY}=90$ il faut démontrer que les angles $\\widehat{AIC}$ et $\\widehat{BID}$ ont les mêmes bissectrices. Si la proposition suivante est vraie alors, (DI, IC) $=$ $(AI, IB)$. Ou encore $(CI, ID)+(AI, IB) \\equiv 0\\ (\\bmod\\ 180)$. On va démontrer cette dernière relation, on appelle $E$ le point d'intersection des droites (AD) et (BC), alors (AI, IB) $=90+\\frac{(BE, EA)}{2}$ et $(CI, ID)=90-\\frac{(BE, EA)}{2}$, donc $(AI, IB)+(CI, ID)=180 \\equiv 0\\ (\\bmod\\ 180)$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThanima possède un magnifique collier constitué de rubis, d'émeraudes et de saphirs, que l'on représente par une suite de $R$, $E$ et $S$. En une opération magique, elle peut faire l'une des actions suivantes :\n- Remplacer un motif $RR$ dans le collier par un motif $ES$ (ou $ES$ par $RR$).\n- Remplacer un motif $EEE$ par un motif $SR$ (ou $SR$ par $EEE$).\n- Supprimer un motif $SS$, ou ajouter un motif $SS$ n'importe où.\n- Supprimer un motif $RES$, ou ajouter un motif $RES$ n'importe où.\nIl n'est pas possible de déplacer les lettres cycliquement à cause du fermoir. Elle ne peut pas non plus retourner le collier.\nEst-il possible de passer d'un collier contenant uniquement un saphir à un collier contenant uniquement une émeraude avec des opérations magiques?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons que Thanima dispose de trois objets numérotés $1$, $2$ et $3$ sur trois supports numérotés $1$, $2$ et $3$. Au début, chaque objet se trouve sur le support qui porte le même numéro. Thanima peut lire les lettres du collier une par une de gauche à droite et faire les opérations suivantes :\n- Si elle voit un $E$, elle échange les objets qui se trouvent sur les supports $1$ et $2$.\n- Si elle voit un $S$, elle échange les objets qui se trouvent sur les supports $2$ et $3$.\n- Si elle voit un $R$, elle déplace l'objet sur le support $1$ sur le support $2$, l'objet sur le support $2$ sur le support $3$, et l'objet sur le support $3$ sur le support $1$.\n\nOn remarque que :\n- Lorsque l'on ajoute un motif $SS$ quelque part, on échange deux fois de suite les objets sur les supports $2$ et $3$, ce qui ne change pas la position finale des objets.\n- Lorsque l'on ajoute un motif $RES$ quelque part, l'objet qui était sur le support $1$ à ce moment est déplacé sur le support $2$ par $R$, puis sur le support $1$ par $E$. L'objet qui était sur le support $2$ est déplacé sur le support $3$ par $R$, puis sur le support $2$ par $S$. L'objet sur le support $3$ est déplacé sur le support $1$ par $R$, puis sur le support $2$ par $E$ puis à nouveau sur le support $3$ par $S$. On ne change donc pas la position finale des objets.\n- Lorsque l'on applique $RR$, l'objet sur le support $1$ est déplacé sur le support $3$, l'objet sur le support $2$ est déplacé sur le support $1$, et l'objet sur le support $3$ est déplacé sur le support $2$. Lorsque l'on applique $ES$, l'objet sur le support $1$ est déplacé sur le support $3$, l'objet sur le support $2$ est déplacé sur le support $1$, et l'objet sur le support $3$ est déplacé sur le support $2$. On ne change donc pas la position finale des objets en remplaçant un motif $RR$ par un motif $ES$.\n- Lorsque l'on applique $EEE$, cela revient à échanger les objets sur les supports $1$ et $2$. Lorsque l'on applique $SR$, l'objet sur le support $1$ est envoyé sur le support $2$, l'objet $2$ est envoyé sur le support $1$ et l'objet sur le support $3$ reste sur le support $3$. On ne change donc pas la position finale des objets en remplaçant un motif $EEE$ par un motif $SR$.\n\nOn conclut qu'aucune opération magique ne permet de changer la position finale des objets. Or les colliers \"$S$\" et \"$E$\" ne donnent pas les mêmes positions finales des objets, il est donc impossible de passer de l'un à l'autre par des opérations magiques.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30864,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer toutes les fonctions $f: \\mathbb{N}^{\\star} \\rightarrow \\mathbb{N}^{\\star}$ telles que :\n\n(i) Les entiers $f(1), f(2), \\ldots$ sont premiers entre dans leur ensemble.\n\n(ii) Il existe $N \\geqslant 1$ tel que pour tout $n \\geqslant N, f(n) \\neq 1$ et pour tous $a, b \\in \\mathbb{N}^{\\star}$,\n$$\nf(a)^{n} \\mid f(a+b)^{a^{n-1}}-f(b)^{a^{n-1}}\n$$",
    "options": [],
    "answer": "f(n) = n for all positive integers n",
    "solution": "Solution:\n\nSoit $f$ une solution. Comme les valeurs de $f$ sont premières entre elles dans leur ensemble, si $f$ est constante, elle égale $1$, ce qui est exclu.\n\nAvec $a=1$, on voit que $f(1)^{n} \\mid f(b+1)-f(b)$ pour tout $b \\geqslant 1$ et tout $n$ assez grand. Comme $f$ est non constante, il existe $b$ tel que $f(b+1) \\neq f(b)$. Il s'ensuit que $f(1)=1$.\n\nSoit $a \\geqslant 2$, il existe un $t \\geqslant 0$ tel que $f(1+(t+1)a) \\neq f(1+ta)$ (car, pour $t$ assez grand, $f(1+ta) \\neq 1 = f(1)$) : soit $b=1+ta$. Pour tout $n \\geqslant 1$ assez grand, $f(a)^{n} \\mid f(a+b)^{a^{n-1}}-f(b)^{a^{n-1}}$.\n\nSoit $p$ un nombre premier divisant $f(a)$. Alors il existe un $l \\geqslant 1$ tel que $p \\mid f(a+b)^{a^{l}}-f(b)^{a^{l}}$. Pour $n>l$, par LTE, la valuation $p$-adique de $f(a+b)^{a^{n-1}}-f(b)^{a^{n-1}}$ est $v_{p}\\left(a^{n-1-l}\\right)+C$ pour une certaine constante $C$ qui ne dépend que de $f, a, b, l$ (mais pas de $n$).\n\nDonc pour $n>l$ assez grand, $v_{p}\\left(a^{n-1-l}\\right)+C \\geqslant v_{p}\\left(f(a)^{n}\\right)$, donc $(n-1-l) v_{p}(a)+C \\geqslant n v_{p}(f(a))$, d'où $v_{p}(f(a)) \\leqslant v_{p}(a)$.\n\nPar conséquent, pour tout $a \\geqslant 1, f(a) \\mid a$.\n\nEn particulier, si $a$ est premier, $f(a)$ vaut $1$ ou $a$, et est égal à $a$ sauf pour un nombre fini de nombres premiers.\n\nSoit $a$ un nombre premier tel que $f(a)=a$ : pour tout $b \\geqslant 1$ et tout $n$ assez grand, $a^{n} \\mid f(a+b)^{a^{n-1}}-f(b)^{a^{n-1}}$. En particulier, $f(a+b) \\equiv f(b) \\pmod{a}$ : ainsi, la classe de congruence de $f(b)$ modulo $a$ ne dépend que de $b$ modulo $a$.\n\nOn va montrer que $f$ est l'identité. Soit $u \\geqslant 2$, soit $P$ l'ensemble fini des nombres premiers $p$ tels que $p \\leqslant u$ ou $f(p)=1$. Supposons qu'il existe un nombre premier $q \\notin P$ et un entier $n$ tels que $q \\mid n-u$ et $f(n)=n$.\n\nAlors $q \\mid f(n)-f(u)$, et donc $q \\mid n-f(u)$, d'où $q \\mid u-f(u)$. Comme $0 \\leqslant u-f(u) \\leqslant u < q$, on en déduit que $f(u)=u$.\n\nOn propose trois méthodes, de la plus élémentaire à la moins élémentaire, d'exhiber un tel couple $(q, n)$.\n\nPremière méthode (complètement élémentaire):\n\nSoit $\\Pi$ le produit des nombres premiers de $P$, et soit $N \\geqslant 1$ tel que pour tout $p \\in P, p^{N}$ ne divise pas $u-1$.\n\nSupposons que $n > \\Pi^{N} + u$ soit congru à $1$ modulo $\\Pi^{N}$ et tel que $f(n)=n$. Alors, si $p \\in P, v_{p}(n-u) < N$, et $n-u > \\Pi^{N}$, de sorte que $n-u$ possède un diviseur premier $q \\notin P$, et on a gagné.\n\nSoit $n$ un produit de nombres premiers deux à deux distincts, tous hors de $P$. Alors, si $p \\mid n$ est un diviseur premier, $n \\equiv p \\pmod{p}$ et $f(p)=p$, donc $p \\mid f(n)-f(p)$, d'où $p \\mid f(n)$. Comme $f(n) \\mid n$, on en déduit que $f(n)=n$.\n\nComme il existe une infinité de nombres premiers, il existe une progression arithmétique $C$ de raison $\\Pi^{N}$ et de premier terme premier à $\\Pi$ et contenant une infinité de nombres premiers. En prenant pour $n$ le produit de $\\varphi\\left(\\Pi^{N}\\right)$ premiers $q \\in C$ tels que $q > \\Pi^{N} + u$, $n$ convient.\n\nDeuxième méthode (il y a des prérequis, mais ils sont élémentaires et relativement classiques) :\n\nOn reprend les notations et le raisonnement de ce qui précède : on cherche à construire un entier $n > \\Pi^{N} + u$ congru à $1$ modulo $\\Pi^{N}$ tel que $f(n)=n$.\n\nIl est connu (du moins, classique et relativement élémentaire) que si $\\Phi_{\\Pi^{N}}$ est le $\\Pi^{N}$-ième polynôme cyclotomique, il existe une infinité de nombres premiers $n$ divisant une valeur de $\\Phi_{\\Pi^{N}}$, et ceux qui ne sont pas dans $P$ sont congrus à $1$ modulo $\\Pi^{N}$ et vérifient $f(n)=n$. On peut en choisir un qui est strictement supérieur à $\\Pi^{N} + u$.\n\nTroisième méthode (utilise un théorème relativement classique, mais dont aucune démonstration accessible au niveau olympique n'est connue) :\n\nSoit $q \\notin P$ premier : alors $f(q)=q$. D’après le théorème de Dirichlet, il existe un nombre premier $p > u + q$ congru à $u$ modulo $q$ tel que $f(p)=p$. Alors le couple $(q, p)$ convient.\n\nAinsi, l'identité est la seule solution potentielle. Réciproquement, l'identité vérifie la première condition et si $a, b \\geqslant 1$ et $n \\geqslant 2$, alors par LTE, pour tout nombre premier $p \\mid a$, $v_{p}\\left((a+b)^{a^{n-1}}-b^{a^{n-1}}\\right) \\geqslant v_{p}(a)+v_{p}\\left(a^{n-1}\\right)=v_{p}\\left(a^{n}\\right)$, donc $a^{n} \\mid (a+b)^{a^{n-1}}-b^{a^{n-1}}$, et donc l'identité est solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle, avec $AC > AB$, et $\\Gamma$ son cercle circonscrit. Soit $T$ le point d'intersection de la tangente à $\\Gamma$ en $A$ avec $(BC)$. Soient $M$ le milieu de $[BC]$ et $R$ le symétrique de $A$ par rapport à $B$. Soit $S$ le point tel que $SABT$ est un parallélogramme. La parallèle à $(AB)$ passant par $M$ coupe $(SB)$ en $P$. On suppose que $P$ est sur $\\Gamma$, montrer que $(AC)$ est tangente au cercle circonscrit à $SRT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSoit $N$ le point d'intersection de $(AT)$ et $(BS)$. On a $(PM) \\parallel (AB) \\parallel (ST)$ donc d'après le théorème de Thalès, $\\frac{BP}{BS} = \\frac{BM}{BT}$. Alors $\\frac{BP}{BN} = \\frac{BP}{\\frac{BS}{2}} = 2 \\frac{BP}{BS} = \\frac{2BM}{BT} = \\frac{BC}{BT}$, donc d'après la réciproque du théorème de Thalès, $(NT)$ et $(PC)$ sont parallèles. On a alors :\n$$\n\\begin{aligned}\n\\widehat{SA'P} &= \\widehat{AA'P} \\\\\n&= \\widehat{ACP} \\text{ par angle inscrit } \\\\\n&= \\widehat{ACB} + \\widehat{BCP} \\\\\n&= \\widehat{TAB} + \\widehat{BTA} \\text{ par angle tangentiel et angles alternes-internes } \\\\\n&= \\widehat{CBA} \\\\\n&= \\widehat{SAB} \\text{ par angles alternes-internes. }\n\\end{aligned}\n$$\nDonc par angle correspondant, $(AB)$ et $(A'P)$ sont parallèles. Or $(AB)$ et $(MP)$ sont parallèles. Il découle que $P, M, A'$ sont alignés.\n\nOn en déduit donc que $BT = BS$, $BM = BP$. En effet :\n$$\n\\begin{aligned}\n\\widehat{TSB} &= \\widehat{BPA'} \\text{ par angle alterne-interne } \\\\\n&= \\widehat{AA'P} \\text{ car } AA'PB \\text{ trapèze isocèle } \\\\\n&= \\widehat{CMA'} \\text{ par angle alterne-interne } \\\\\n&= \\widehat{BTS'} \\text{ par angle correspondants. }\n\\end{aligned}\n$$\nDonc $BT = BS$, et comme $(ST) \\parallel (PM)$, alors $BP = BM$ de même.\n\nNotamment $STRA$ est un trapèze isocèle donc cyclique donc $A$ est sur $(SRT)$. De plus, $\\widehat{NAB} = \\widehat{ACB}$ par angle tangentiel, et $\\widehat{ABN} = \\widehat{BST} = \\widehat{STB} = \\widehat{CBA}$ par angle alterne-interne et correspondant. Donc $BAN \\sim BCA$. En particulier $\\widehat{BNA} = \\widehat{BAC}$, donc par réciproque de l'angle tangentiel, $(AC)$ est tangente à $(ANB)$.\n\nOr une homothétie de centre $A$ et de facteur $2$ envoie $N$ sur $T$, $B$ sur $R$, fixe $A$ donc envoie $(ANB)$ sur le cercle $(ATR) = (SRT)$, et elle fixe $(AC)$. Donc $(SRT)$ est bien tangent à $(AC)$ (les homothéties préservent les tangences), ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A, B, C, D$ et $E$ cinq points situés dans cet ordre sur un cercle $\\Omega$, de sorte que $(C D)$ soit parallèle à $(B E)$ et que $(A B)$ soit parallèle à $(D E)$. Soit $X, Y$ et $Z$ les milieux respectifs des segments $[B D],[C E]$ et $[A E]$.\n\nDémontrer que la droite $(A E)$ est tangente au cercle circonscrit à $X Y Z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSoit $O$ le centre du cercle $\\Omega$. Dès lors que l'on s'attache à tracer une figure incluant le cercle $\\Gamma$ circonscrit à $X Y Z$, on constate que celui-ci semble tangent aux trois droites $(B D),(C E)$ et $(A E)$, tandis que son centre semble coïncider avec $O$. On entreprend donc de démontrer les égalités de longueurs $O X=O Y=O Z$, ce qui permettra de conclure que $(O Z)$ est un rayon de $\\Gamma$, et que la droite $(A E)$, qui lui est perpendiculaire, est bien tangente à $\\Gamma$.\n\nOr, si l'on note $R$ le rayon du cercle $\\Gamma$, le théorème de Pythagore indique que\n$$\nO X^{2}=R^{2}-(B D)^{2} / 4, \\quad O Y^{2}=R^{2}-(C E)^{2} / 4 \\text{ et } O Z^{2}=R^{2}-(A E)^{2} / 4\n$$\nIl s'agit donc de démontrer que $B D=C E=A E$.\n\nPuisque $(C D)$ est parallèle à $(B E)$ et que le trapèze $B C D E$ est inscrit dans le cercle $\\Omega$, il s'agit d'un trapèze isocèle, de sorte que $B D=C E$. De même, comme $(A B)$ est parallèle à $(D E)$, le trapèze $A B D E$, qui est inscrit dans $\\Omega$, est lui aussi isocèle, et l'on conclut comme souhaité que $A E=B D$.\n\n\nSolution alternative $n^{\\circ} 1$\n\nVoici une autre manière d'aboutir aux égalités $B D=C E=A E$. Celles-ci sont équivalentes aux égalités d'angles $\\widehat{B O D}=\\widehat{C O E}=\\widehat{E O A}$. Le parallélisme de $(C D)$ et $(B E)$ indique alors que\n$$\n\\widehat{B O D}=2 \\widehat{B E D}=2\\left(180^{\\circ}-\\widehat{E D C}\\right)=\\widehat{C O E},\n$$\ntandis que le parallélisme de $(A B)$ et $(D E)$ indique comme prévu que\n$$\n\\widehat{E O A}=2 \\widehat{E D A}=2 \\widehat{B A D}=\\widehat{B O D}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUne ligne maritime Est-Ouest voit partir chaque matin 10 bateaux à des moments tous distincts, 5 bateaux partent du côté Ouest et 5 du côté Est. On suppose qu'ils naviguent tous à la même vitesse et que dès que deux bateaux se rencontrent ils se retournent et repartent chacun de leur côté, toujours à la même vitesse. Quel est le nombre possible de rencontres entre bateaux?",
    "options": [],
    "answer": "25",
    "solution": "Solution:\n\nLe nombre de croisements ne change pas si on dit que les bateaux continuent tout droit au lieu de faire demi-tour lors d'une rencontre. Chaque bateau qui part d'un côté croise tous les autres bateaux qui partent de l'autre côté. Il y a donc $5 \\times 5=25$ rencontres.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30868,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les réels $x, y$ tels que\n$$\nx(x-2y) + y(2y-4) + 4 = 0\n$$",
    "options": [],
    "answer": "x = 2, y = 2",
    "solution": "Solution:\nDéveloppons l'équation pour obtenir\n$$\nx^2 - 2xy + 2y^2 - 4y + 4 = 0.\n$$\nLes termes $x^2 - 2xy$ nous font penser à l'identité remarquable $x^2 - 2xy + y^2 = (x-y)^2$. On sépare alors le $2y^2$ en $y^2 + y^2$ pour écrire\n$$\n(x-y)^2 + y^2 - 4y + 4 = 0\n$$\nMais on reconnait alors une deuxième identité remarquable $y^2 - 4y + 4 = (y-2)^2$. En factorisant, ceci donne\n$$\n(x-y)^2 + (y-2)^2 = 0\n$$\nOr, une somme de carrés de nombres réels est forcément positive, et ne peut être nulle que si les deux carrés sont tous les deux nuls. Ainsi, si $x$ et $y$ sont solution de l'équation, on a à la fois\n$$\nx-y=0 \\quad \\text{ et } \\quad y-2=0\n$$\nAinsi, l'unique solution possible de l'équation est $x = y = 2$. On peut vérifier réciproquement que ceci est bien une solution, c'est donc la seule.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoit $a$, $b$ et $c$ trois réels strictement positifs. Démontrer que\n$$\n4\\left(a^{3}+b^{3}+c^{3}+3\\right) \\geqslant 3(a+1)(b+1)(c+1) .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nDans cette inégalité, les termes en $a$, $b$ et $c$ sont additionnés les uns aux autres (et de degré $3$) dans le membre de gauche, tandis qu'ils sont multipliés les uns aux autres (et chacun de degré $1$) dans le membre de droite. Une première idée est donc d'utiliser l'inégalité arithmético-géométrique pour se retrouver, de part et d'autre de l'inégalité, avec deux produits de termes de degré $1$, ou bien deux sommes de termes de degré $3$ faisant chacun intervenir une seule des variables $a$, $b$ ou $c$.\n\nAinsi, on constate que\n$$\n(a+1)(b+1)(c+1)=\\sqrt[3]{(a+1)^{3}(b+1)^{3}(c+1)^{3}} \\leqslant \\frac{(a+1)^{3}+(b+1)^{3}+(c+1)^{3}}{3}\n$$\nEn outre, cette inégalité a le bon goût d'être une égalité lorsque $a=b=c=1$, qui est également un cas d'égalité pour l'inégalité de l'énoncé. On espère donc très fort que l'inégalité\n$$\n4\\left(a^{3}+b^{3}+c^{3}+3\\right) \\geqslant (a+1)^{3}+(b+1)^{3}+(c+1)^{3}\n$$\nest valide, et on va tenter de la démontrer.\n\nOn continue donc nos pérégrinations et, puisque l'on a fait tous ces efforts pour séparer les termes en les variables $a$, $b$ et $c$, on constate qu'il suffit désormais de démontrer que\n$$\n4\\left(a^{3}+1\\right) \\geqslant (a+1)^{3}\n$$\nIl s'agit là d'une inégalité en une seule variable, dont $a=1$ est un cas d'égalité, et qui aura le bon goût d'être démontrée (ou invalidée pour certaines valeurs de $a$) sans trop d'efforts.\n\nEn effet, si on développe chaque terme et qu'on les insère tous dans le membre de gauche, l'inégalité se réécrit comme\n$$\n3 a^{3}-3 a^{2}-3 a+3 \\geqslant 0\n$$\nPuisque l'on sait qu'il y a égalité lorsque $a=1$, et au vu de la symétrie manifeste entre les termes de petit degré et ceux de grand degré, on factorise le membre de gauche comme\n$$\n3 a^{3}-3 a^{2}-3 a+3=3(a-1)\\left(a^{2}-1\\right)=3(a-1)^{2}(a+1)\n$$\nce qui est manifestement positif ou nul, et nous permet de conclure que l'inégalité de l'énoncé est effectivement valide.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les couples d'entiers naturels non nuls $(k, n)$ pour lesquels\n$$\n1!+2!+\\cdots+k!=1+2+\\cdots+n\n$$",
    "options": [],
    "answer": "(1,1), (2,2), (5,17)",
    "solution": "Solution:\n\nDans toute la solution, on notera $A_{k}$ l'expression $1!+2!+\\cdots+k$ et $B_{n}$ l'expression $1+2+\\cdots+n$. Comme dans tout exercice de ce genre, on commence par rechercher d'éventuelles solutions pour de petites valeurs de $k$ et de $n$. À cette fin, il est pratique de reconnaître en $B_{n}$ la somme des termes d'une suite arithmétique, de sorte que $B_{n}=n(n+1)/2$. En particulier, si $(k, n)$ est une solution, on sait que\n$$\nn^{2} \\leqslant 2 B_{n}=2 A_{k}<(n+1)^{2},\n$$\ndonc que $n=\\left\\lfloor\\sqrt{2 A_{k}}\\right\\rfloor$.\n\nOn calcule alors les premières valeurs de $A_{k}$ : par exemple, lorsque $k$ varie de 1 à $5$, $A_{k}$ vaut successivement $1, 3, 9, 33$ et $153$, puis l'entier $n=\\left\\lfloor\\sqrt{2 A_{k}}\\right\\rfloor$ vaut alors successivement $1, 2, 4, 8$ et $17$, de sorte que $B_{n}$ vaut $1, 3, 10, 36$ et $153$. Cette première étude, bien que fastidieuse vers la fin, nous a permis d'identifier trois couples solutions : $(k, n)=(1,1), (2,2)$ et $(5,17)$.\n\nPar ailleurs, on se voit mal calculer à la main $\\left\\lfloor 2 \\sqrt{A_{\\ell}}\\right\\rfloor$ lorsque $\\ell \\geqslant 6$, et il faut donc trouver une approche plus générale lorsque $\\ell$ est grand. Par exemple, quand $\\ell \\leqslant k$, l'entier $\\ell!$ divise l'entier $k!=A_{k}-A_{k-1}$, de sorte que $A_{k} \\equiv A_{\\ell-1} (\\bmod \\ell!)$. On entreprend donc d'étudier la suite $\\left(A_{k}\\right)$ modulo $\\ell!$ ou $\\ell$ pour de petites valeurs de $\\ell$, par exemple lorsque $\\ell$ est un nombre premier.\n\nAprès une étude infructueuse des cas $\\ell=2,3$ et $5$, on constate ainsi que $k!\\equiv 1,2,6,3,1,6$ $(\\bmod 7)$ lorsque $k$ varie de 1 à 6, et donc que $A_{k} \\equiv A_{6} \\equiv 1+2+6+3+1+6 \\equiv 5 (\\bmod 7)$ lorsque $k \\geqslant 6$. D'autre part, $B_{n} \\equiv 1,3,6,3,1,0,0 (\\bmod 7)$ lorsque $n$ varie de 1 à 7. Puisque 2 est inversible modulo 7, on sait que $B_{n} (\\bmod 7)$ ne dépend que de $n (\\bmod 7)$. Cela signifie en particulier que $B_{n} \\not\\equiv 5 \\equiv A_{k} (\\bmod 7)$ dès lors que $k \\geqslant 6$ et $n \\geqslant 1$, démontrant par la même que $(k, n)$ n'est pas une solution.\n\nEn conclusion, les solutions sont les couples $(k, n)=(1,1), (2,2)$ et $(5,17)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle équilatéral et $P$ un point sur le cercle circonscrit de ce triangle mais distinct de $A$, $B$ et $C$. Les droites passant par $P$ et parallèles à $(BC)$, $(CA)$ et $(AB)$ intersectent les droites $(CA)$, $(AB)$ et $(BC)$ en $M$, $N$ et $Q$ respectivement. Montrer que les points $M$, $N$ et $Q$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn se place dans le cas de la figure suivante :\n![](attached_image_1.png)\n\nMontrons que $P$, $N$, $B$, $Q$ sont cocycliques. Comme $(PN)$ est parallèle à $(AC)$, et $(NB)$ est parallèle à $(AB)$, $\\widehat{ANP} = \\widehat{CAB} = 60^\\circ$. Comme $(PQ)$ est parallèle à $(AB)$, $\\widehat{PQB} = \\widehat{ABC} = 60^\\circ$. Ainsi $P$, $N$, $B$, $Q$ sont cocycliques. Ainsi $\\widehat{PQN} = \\widehat{PBN} = \\widehat{PBA}$.\n\nMontrons que $P$, $M$, $C$, $Q$ sont cocycliques. Comme $(PM)$ est parallèle à $(BC)$, $\\widehat{AMP} = \\widehat{ACB} = 60^\\circ$. Or $\\widehat{PQC} = \\widehat{PQB} = 60^\\circ$. Ainsi $P$, $M$, $C$, $Q$ sont cocycliques. Ainsi $\\widehat{PQM} = \\widehat{PCM} = \\widehat{PCA}$. Par cocyclité de $A$, $P$, $B$, $C$, $\\widehat{PQM} = \\widehat{PCA} = \\widehat{PBA} = \\widehat{PQN}$ donc $Q$, $N$, $M$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAnna et Baptiste jouent au jeu suivant. Au début du jeu sont placées devant eux 2022 cases blanches, numérotées de 1 à 2022. Puis, chacun son tour, en commençant par Anna, ils choisissent une case blanche et la colorient de la couleur de leur choix : soit en rouge, soit en bleu. La partie s'arrête au bout de 2022 tours de jeu, c'est-à-dire au moment où la dernière case blanche est coloriée.\n\nLe score d'Anna est alors égal au nombre d'entiers $a \\leqslant 2019$ tels que les deux cases de numéros $a$ et $a+3$ sont de la même couleur. Anna souhaite avoir le score le plus élevé possible; Baptiste souhaite que le score d'Anna soit le plus faible possible. Quel est le score maximal qu'Anna peut s'assurer d'obtenir quels que soient les coups que Baptiste choisira de jouer?",
    "options": [],
    "answer": "1008",
    "solution": "Solution:\n\nPour plus de simplicité, on dira que deux entiers $a$ et $b$ sont amis si $a = b \\pm 3$. On dira que $a$ est joli si $a = b-3$ et si $a$ et $b$ sont coloriés de la même couleur, et que $a$ est laid si $a = b-3$ et si $a$ et $b$ sont coloriés de couleurs différentes : Anna souhaite donc maximiser le nombre d'entiers jolis à la fin de la partie. Si $a$ ou $b$ n'est pas encore colorié, $a$ n'est ni joli, ni laid.\n\nEn outre, pour chacun des entiers $k = 0, 1$ et $2$, nous noterons $\\mathcal{E}_{k}$ l'ensemble des entiers $n$ tels que $1 \\leqslant n \\leqslant 2022$ et $n \\equiv k \\pmod{3}$. On dit qu'un ensemble $\\mathcal{E}_{k}$ est entamé dès lors qu'au moins une case dont le numéro est dans $\\mathcal{E}_{k}$ a été coloriée.\n\nNous allons tout d'abord démontrer qu'Anna peut s'assurer de gagner au moins 1008 points. Pour ce faire, la stratégie qu'adopte Anna est la suivante.\n\n1. S'il existe un entier $a$ qui n'est pas encore colorié, et dont un ami, disons $b$, est colorié, alors Anna choisit un tel entier $a$ et le colorie de la même couleur que $b$; si plusieurs entiers $a$ et/ou $b$ étaient disponibles, n'importe lesquels de ces entiers $a$ et $b$ feront l'affaire.\n2. Si, au contraire, aucun tel entier $a$ n'existe, Anna colorie au hasard une case blanche.\n\nDans le cas 1, l'entier $\\min \\{a, b\\}$ devient joli, ce qui augmente le score d'Anna d'au moins 1. Dans le cas 2, si un ensemble $\\mathcal{E}_{k}$ était déjà entamé au moment où Anna s'apprête à jouer, c'est que tous ses éléments étaient déjà coloriés; par conséquent, avec son coup, Anna entame un nouvel ensemble $\\mathcal{E}_{k}$.\n\nEn particulier, le cas 2 survient au plus 3 fois dans la partie, et comme Anna joue 1011 coups, le cas 1 survient au moins 1008 fois dans la partie. Comme prévu, Anna s'assure donc un score final d'au moins 1008 points.\n\nRéciproquement, nous allons maintenant démontrer que Baptiste peut empêcher Anna de gagner strictement plus de 1008 points, et ce quels que soient les coups qu'elle décidera de jouer. Pour ce faire, la stratégie qu'adopte Baptiste est la suivante.\n\n1. S'il existe un entier $a$ qui n'est pas encore colorié, et dont un ami, disons $b$, est colorié, alors Baptiste choisit un tel entier $a$ et le colorie de la couleur opposée à celle de $b$; si plusieurs entiers $a$ et/ou $b$ étaient disponibles, n'importe lesquels de ces entiers $a$ et $b$ feront l'affaire.\n2. Si, au contraire, aucun tel entier $a$ n'existe, Baptiste colorie au hasard une case blanche.\n\nDans le cas 1, l'entier $\\min \\{a, b\\}$ devient laid. Dans le cas 2, tout ensemble $\\mathcal{E}_{k}$ déjà entamé au moment où Baptiste s'apprête à jouer était déjà colorié; si $n$ ensembles étaient entamés à ce moment-là, cela signifie que $2022 n / 3 = 674 n$ coups avaient déjà été joués, et donc que c'était en fait à Anna de jouer.\n\nPar conséquent, le cas 2 ne survient jamais, et comme Baptiste joue 1011 coups, il y a au moins 1011 entiers laids à la fin de la partie. Par ailleurs, les trois entiers 2020, 2021 et 2022 ne seront ni jolis, ni laids. Parmi nos 2022 entiers, il y en a donc au moins $1011 + 3$ qui ne seront jamais jolis, et le score d'Anna ne pourra ainsi jamais dépasser $2022 - (1011 + 3) = 1008$.\n\nConclusion : Le score maximal qu'Anna peut s'assurer d'obtenir, quels que soient les coups que Baptiste choisira de jouer, est $1008$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer qu'il existe une infinité de couples $(m, n)$ d'entiers strictement positifs distincts tels que $m!n!$ soit un carré parfait.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn aurait envie de prendre $m = n$ pour avoir $m!n! = (n!)^2$ et avoir un carré parfait. Mais l'énoncé force $m \\neq n$. Malgré cela, on voit déjà un moyen de faire apparaître naturellement des carrés parfaits.\n\nSupposons sans perte de généralité $m > n$ (comme $m!n! = n!m!$, quitte à remplacer $(m, n)$ par $(n, m)$, ça ne pose pas de problème). Alors\n$$\nm!n! = m \\times (m-1) \\times \\ldots \\times (n+1) \\times (n!)^2.\n$$\nAlors $m!n!$ sera un carré parfait si et seulement si $m \\times (m-1) \\times \\ldots \\times (n+1)$ l'est. Le moyen le plus simple de le faire, c'est de choisir $m = n+1 = k^2$ pour un certain entier $k$ ($k \\geqslant 2$ car on veut $m, n \\geqslant 1$). En effet, on a bien\n$$\n(k^2)! (k^2-1)! = k^2 \\left[(k^2-1)!\\right]^2 = \\left[k (k^2-1)!\\right]^2.\n$$\nFinalement, pour tout $k \\geqslant 2$ entier, $(m, n) = (k^2, k^2-1)$ convient, ce qui nous fournit bien une infinité de couples de solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle acutangle (dont tous les angles sont aigus) avec $BA \\neq BC$. Soit $O$ le centre de son cercle circonscrit. La droite $(AB)$ intersecte le cercle circonscrit à $BOC$ une deuxième fois en $P \\neq B$. Montrer que $PA = PC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nTraçons la figure dans le cas où $BC < BA$, le cas $BC > BA$ étant totalement analogue. Il s'agit de montrer que $PA = PC$, c'est-à-dire que $\\widehat{ACP} = \\widehat{PAC} \\ (= \\widehat{BAC})$.\n\nOr on a:\n$$\n\\begin{aligned}\n\\widehat{ACP} & = \\widehat{ACO} + \\widehat{OCP} \\\\\n& = \\widehat{ACO} + \\widehat{OBP} \\text{ par angle inscrit} \\\\\n& = \\widehat{ACO} + \\widehat{OBA}\n\\end{aligned}\n$$\n\nOr, $AOC$ est isocèle en $O$ donc $\\widehat{ACO} = \\widehat{OAC} = \\frac{180^\\circ - \\widehat{COA}}{2} = 90^\\circ - \\widehat{CBA}$ par angle au centre. De même $\\widehat{OBA} = 90^\\circ - \\widehat{ACB}$.\n\nFinalement,\n$$\n\\widehat{ACP} = 180^\\circ - \\widehat{ACB} - \\widehat{CBA} = \\widehat{BAC} = \\widehat{PAC},\n$$\nd'où $PA = PC$, comme voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30875,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour un réel $x$, on note $\\lfloor x\\rfloor$ le plus grand entier relatif inférieur ou égal à $x$ (par exemple, $\\lfloor 2,7\\rfloor=2$, $\\lfloor\\pi\\rfloor=3$ et $\\lfloor-1,5\\rfloor=-2$).\n\nSoient $a, b$ deux réels tels que\n$$\na+\\lfloor a\\rfloor=b+\\lfloor b\\rfloor .\n$$\nMontrer que $a=b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons par l'absurde $a \\neq b$. Par symétrie (c'est-à-dire que $a$ et $b$ jouent le même rôle), on peut supposer $a<b$. On a alors\n$$\n\\lfloor a\\rfloor \\leqslant a<b\n$$\ndonc $\\lfloor a\\rfloor$ est un entier relatif inférieur ou égal à $b$, et donc par définition on a $\\lfloor a\\rfloor \\leqslant\\lfloor b\\rfloor$. En ajoutant $a<b$, on obtient alors\n$$\na+\\lfloor a\\rfloor<b+\\lfloor b\\rfloor\n$$\nce est absurde. On a donc bien $a=b$.\n\n\nSolution alternative:\n\nPour un réel $x$, posons $\\{x\\}$ la partie fractionnaire de $x$ donnée par $\\{x\\}=x-\\lfloor x\\rfloor$. C'est un nombre réel compris entre 0 inclus et 1 exclus.\nOn réécrit alors l'équation en posant $a=\\lfloor a\\rfloor+\\{a\\}$ et $b=\\lfloor b\\rfloor+\\{b\\}$ pour avoir\n$$\n2\\lfloor a\\rfloor+\\{a\\}=2\\lfloor b\\rfloor+\\{b\\} .\n$$\nEn réarrangeant les termes, on obtient\n$$\n2\\lfloor a\\rfloor-2\\lfloor b\\rfloor=\\{b\\}-\\{a\\}\n$$\nMais on a\n$$\n\\{b\\}-\\{a\\} \\leqslant\\{b\\}<1\n$$\net\n$$\n\\{b\\}-\\{a\\} \\geqslant-\\{a\\}>-1\n$$\nMais d'après l'égalité au-dessus, $\\{b\\}-\\{a\\}$ est un entier, et il est donc nul. On a donc $\\{b\\}-\\{a\\}=0$, ce qui donne ensuite $2\\lfloor a\\rfloor-2\\lfloor b\\rfloor=0$.\nAinsi, on a\n$$\n\\lfloor a\\rfloor=\\lfloor b\\rfloor \\text{ et } \\{a\\}=\\{b\\}\n$$\ndonc\n$$\na=\\lfloor a\\rfloor+\\{a\\}=\\lfloor b\\rfloor+\\{b\\}=b\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAlice a disposé 200 boîtes dans son salon. Chaque boîte contient un papier sur lequel elle a écrit un entier naturel non nul; les 200 entiers ne sont pas nécessairement distincts. Chaque minute, et tant que c'est possible, Alice effectue une action de la forme suivante : elle choisit trois boîtes, contenant des entiers $a, b$ et $c$ tels que $a+b=c$, et choisit également un entier $k \\geqslant 2$ arbitraire; elle remplace alors l'entier $c$ par l'entier $k \\times c$. Si elle ne peut plus effectuer de telle action, elle s'arrête définitivement.\nDémontrer que, quels que soient la situation initiale et les choix d'Alice, elle sera forcée de s'arrêter à un moment.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCi-dessous, on dira qu'un entier change si Alice remplace $c$ par un entier $k \\times c$. Elle peint en bleu les entiers qui ne changent qu'un nombre fini de fois, et en rouge ceux qui changent un nombre infini de fois. Par construction, le nombre initialement minimal ne pourra jamais changer, donc il est bleu. On suppose alors abusivement qu'il existe un nombre rouge.\nÀ partir d'un certain temps, Alice cesse complètement de changer les nombres bleus. Soit $v$ la valeur du plus grand d'entre eux à ce moment-là. Quand un nombre rouge change, sa valeur augmente d'au moins un. Il y a donc un moment à partir duquel chaque nombre rouge vaut au moins $2 v+1$, par exemple quand chaque nombre rouge a changé au moins $2 v$ fois.\nÀ ce moment-là, le plus petit nombre rouge est inférieur ou égal à tous les autres nombres rouges, et il est strictement supérieur à la somme de deux nombres bleus quelconques. Il ne pourra donc plus jamais changer de valeur, et il aurait dû être bleu. Ce constat invalide notre supposition, ce qui signifie que tout nombre est bleu et conclut le problème.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $x, y, z$ trois réels positifs, tel que $x \\leqslant 1$. Démontrer que:\n$$\nx y + y + 2 z \\geqslant 4 \\sqrt{x y z}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons que comme $x \\leqslant 1$, $x y + y \\geqslant x y + x y = 2 x y$. En particulier, par inégalité arithmético-géométrique\n$$\nx y + y + 2 z \\geqslant 2 x y + 2 z = 2(x y + z) \\geqslant 4 \\sqrt{x y z}\n$$\nce qui donne l'inégalité voulue.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30878,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\Gamma$ son cercle circonscrit. Soit $A_{1}$ le milieu de l'arc $\\widehat{BC}$ ne contenant pas $A$ ; $B_{1}$ le milieu de l'arc $\\overparen{CA}$ ne contenant pas $B$ ; $C_{1}$ le milieu de l'arc $\\overparen{AB}$ ne contenant pas $C$. Enfin, soit $A_{2}$ le point pour lequel $AB_{1}A_{2}C_{1}$ est un parallélogramme ; $B_{2}$ le point pour lequel $BC_{1}B_{2}A_{1}$ est un parallélogramme ; $C_{2}$ le point pour lequel $CA_{1}C_{2}B_{1}$ est un parallélogramme.\n\nDémontrer que les cercles circonscrits à $ABC$ et à $A_{2}B_{2}C_{2}$ sont concentriques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn cherche tout d'abord à démontrer que $OA_{2} = OB_{2}$. Puisque $A_{1}B_{2} = BC_{1} = AC_{1} = B_{1}A_{2}$ et que $A_{1}O = B_{1}O$, l'égalité $OA_{2} = OB_{2}$ équivaut au fait que les triangles $OA_{1}B_{2}$ et $OB_{1}A_{2}$ soient isométriques l'un de l'autre, c'est-à-dire au fait que $\\widehat{OA_{1}B_{2}} = \\widehat{OB_{1}A_{2}}$.\n\nOn note alors $\\alpha = \\widehat{CAB}$, $\\beta = \\widehat{ABC}$ et $\\gamma = \\widehat{BCA}$ les angles du triangle. Ici, vu nos relations de parallélisme et de cocyclicité, il sera plus agréable de travailler en angles de droites. Ainsi, comme $(CC_{1})$ est la bissectrice de l'angle $\\widehat{BCA}$, on constate que\n$$\n\\begin{aligned}\n\\left(A_{1}O, A_{1}B_{2}\\right) &= \\left(A_{1}O, BC\\right) + \\left(BC, A_{1}B_{2}\\right) \\\\\n&= 90^{\\circ} + \\left(BC, BC_{1}\\right) \\\\\n&= 90^{\\circ} + (BC, BA) + \\left(BA, BC_{1}\\right) \\\\\n&= 90^{\\circ} + \\alpha + \\gamma/2 = (\\alpha - \\beta)/2.\n\\end{aligned}\n$$\nCela signifie que $\\widehat{OA_{1}B_{2}} = |\\alpha - \\beta|/2$. Puisque $A$, $B$ et $C$ jouent des rôles symétriques, on en déduit que $\\widehat{OB_{1}A_{2}} = |\\beta - \\alpha|/2 = \\widehat{OA_{1}B_{2}}$, donc que $OA_{2} = OB_{2}$. Toujours par symétrie des rôles, on en conclut même que $OB_{2} = OC_{2}$.\n\n![](attached_image_2.png)\n\nNous allons calculer directement $OA_{2}$ en fonction des angles $\\alpha, \\beta, \\gamma$ et du rayon $R = OA$. Pour ce faire, on remarque d'abord que $\\widehat{AOB_{1}} = \\beta$, $\\widehat{C_{1}OA} = \\gamma$ et $\\widehat{C_{1}OB_{1}} = \\beta + \\gamma = 180^{\\circ} - \\alpha$. En outre, la condition sur $A_{2}$ signifie que $A_{2} = (B_{1} + C_{1}) - A$, de sorte que\n$$\n\\begin{aligned}\nOA_{2}^{2} &= \\overrightarrow{OA_{2}} \\cdot \\overrightarrow{OA_{2}} \\\\\n&= (\\overrightarrow{OB_{1}} + \\overrightarrow{OC_{1}} - \\overrightarrow{OA}) \\cdot (\\overrightarrow{OB_{1}} + \\overrightarrow{OC_{1}} - \\overrightarrow{OA}) \\\\\n&= \\{\\overrightarrow{OB_{1}}\\}^{2} + \\{\\overrightarrow{OC_{1}}\\}^{2} + \\overrightarrow{OA}^{2} + 2\\overrightarrow{OB_{1}} \\cdot \\overrightarrow{OC_{1}} - 2\\overrightarrow{OA} \\cdot \\overrightarrow{OB_{1}} - 2\\overrightarrow{OC_{1}} \\cdot \\overrightarrow{OA} \\\\\n&= (3 + 2\\cos(180^{\\circ} - \\alpha) - 2\\cos(\\beta) - 2\\cos(\\gamma))R^{2} \\\\\n&= (3 - 2\\cos(\\alpha) - 2\\cos(\\beta) - 2\\cos(\\gamma))R^{2}.\n\\end{aligned}\n$$\nPuisque $A$, $B$ et $C$ jouent des rôles symétriques, on en conclut que $OA_{2}^{2} = OB_{2}^{2} = OC_{2}^{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 6$. Prouvez que chaque carré peut être découpé en exactement $n$ carrés (pas nécessairement de même taille).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn peut le montrer par récurrence. Si on a un découpage du carré en $n$ carrés, on peut obtenir un découpage en $n+3$ carrés en subdivisant l'un des carrés en $4$. Il suffit donc de montrer que l'on peut découper un carré en $6$, $7$ et $8$ carrés.\n\nPour $6$, on subdivise le carré en une grille $3 \\times 3$ et on fusionne les $4$ carrés en haut-gauche.\n\nPour $7$, on subdivise le carré en $4$, puis on subdivise l'un des carrés obtenus en $4$.\n\nPour $8$, on subdivise le carré en une grille $4 \\times 4$ et on fusionne les $9$ carrés en haut-gauche.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30880,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $x$, $y$ et $z$ des réels strictement positifs tels que $x y + y z + z x = 3$.\nDémontrer que\n$$\n\\frac{x+3}{y+z} + \\frac{y+3}{z+x} + \\frac{z+3}{x+y} + 3 \\geqslant 27 \\frac{(\\sqrt{x} + \\sqrt{y} + \\sqrt{z})^{2}}{(x+y+z)^{3}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPosons $s = x + y + z$, et soit $L$ et $R$ les membres de gauche et de droite de notre inégalité. On peut réécrire $L$ comme\n$$\nL = \\frac{x+3}{y+z} + 1 + \\frac{y+3}{z+x} + 1 + \\frac{z+3}{x+y} + 1 = (s+3)\\left(\\frac{1}{y+z} + \\frac{1}{z+x} + \\frac{1}{x+y}\\right).\n$$\nOr, l'inégalité de Cauchy-Schwarz, appliquée aux vecteurs\n$$\n\\left(\\frac{1}{\\sqrt{y+z}}, \\frac{1}{\\sqrt{z+x}}, \\frac{1}{\\sqrt{x+y}}\\right) \\text{ et } (\\sqrt{y+z}, \\sqrt{z+x}, \\sqrt{x+y})\n$$\nindique que\n$$\n\\left(\\frac{1}{y+z} + \\frac{1}{z+x} + \\frac{1}{x+y}\\right)(y+z+z+x+x+y) \\geqslant (1+1+1)^{2}\n$$\nOn en déduit que $L \\geqslant 9(s+3)/(2s)$.\n\nEn outre, l'inégalité arithmético-quadratique, appliquée aux termes $\\sqrt{x}$, $\\sqrt{y}$ et $\\sqrt{z}$, indique que\n$$\n\\frac{s}{3} \\geqslant \\left(\\frac{\\sqrt{x} + \\sqrt{y} + \\sqrt{z}}{3}\\right)^{2}\n$$\nAinsi, $(\\sqrt{x} + \\sqrt{y} + \\sqrt{z})^{2} \\leqslant 3(x+y+z) = 3s$ et $R \\leqslant 81 / s^{2}$.\n\nPour montrer que $L \\geqslant R$, il suffit donc de démontrer que $9(s+3) \\times s^{2} \\geqslant 81 \\times 2s$, c'est-à-dire que $s(s+3) \\geqslant 18$, ou encore que $s \\geqslant 3$. On conclut alors en utilisant le lemme du tourniquet, qui indique que\n$$\ns^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) \\geqslant 3(xy + yz + zx) \\geqslant 9\n$$\nc'est-à-dire que $s \\geqslant 3$.\n\nRemarque: Le cas particulier de l'inégalité de Cauchy-Schwarz que nous avons utilisé est aussi appelé inégalité des mauvais élèves: ici, appliquer l'inégalité des mauvais élèves aux fractions\n$$\n\\frac{1^{2}}{y+z}, \\frac{1^{2}}{z+x} \\text{ et } \\frac{1^{2}}{x+y}\n$$\nnous indique que\n$$\n\\frac{1^{2}}{y+z} + \\frac{1^{2}}{z+x} + \\frac{1^{2}}{x+y} \\geqslant \\frac{(1+1+1)^{2}}{y+z+z+x+x+y}\n$$\nSolution:\n\nCette fois-ci, on utilise d'une part l'inégalité de Nesbitt, qui indique que\n$$\n\\frac{x}{y+z} + \\frac{y}{x+z} + \\frac{z}{x+y} \\geqslant \\frac{3}{2}\n$$\net d'autre part l'inégalité des mauvais élèves, qui indique toujours que\n$$\n\\frac{1}{y+z} + \\frac{1}{z+x} + \\frac{1}{x+y} \\geqslant \\frac{9}{2s}\n$$\nAinsi,\n$$\nL \\geqslant \\frac{3}{2} + \\frac{3 \\times 9}{2s} + 3 = 9 \\frac{s+3}{2s}\n$$\nOn conclut alors l'exercice de la même façon que précédemment.\n\n\nSolution alternative $n^{\\circ} 2$ Puisque la fonction $f: t \\mapsto 1 / t$ est convexe, on sait que\n$$\nL = (s+3)(f(y+z) + f(z+x) + f(x+y)) \\geqslant 3(s+3) f(2s/3) = 9(s+3)/(2s).\n$$\nDe même, l'inégalité du réordonnement, appliquée aux termes $\\sqrt{x}$, $\\sqrt{y}$ et $\\sqrt{z}$, indique que\n$$\nR \\leqslant 27 \\frac{3s}{s^{3}} = \\frac{81}{s^{2}}.\n$$\nPour montrer que $L \\geqslant R$, il suffit donc de démontrer que $3(s+3) \\times s^{2} \\geqslant 81 \\times 2s / 3$, c'est-à-dire que $s(s+3) \\geqslant 18$, ou encore que $s \\geqslant 3$.\nOn conclut alors en remarquant que l'inégalité du réordonnement, appliquée aux termes $x$, $y$ et $z$, indique que\n$$\ns^{2} = x^{2} + y^{2} + z^{2} + 2(xy + yz + zx) \\geqslant 3(xy + yz + zx) \\geqslant 9,\n$$\nc'est-à-dire que $s \\geqslant 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $n \\geqslant 1$ pour lesquels il existe un multiple de 222 dont la somme des carrés des chiffres est égale à $n$.",
    "options": [],
    "answer": "n = 3, n = 6, or n ≥ 8",
    "solution": "Solution:\n\nPour tout entier $k$, on note $\\mathrm{S}_{2}(k)$ la somme des carrés des chiffres de $k$. Soit $\\mathcal{E}$ l'ensemble des entiers $n$ recherchés, c'est-à-dire\n$$\n\\mathcal{E}=\\left\\{\\mathrm{S}_{2}(222 k): k \\in \\mathbb{Z}_{\\geqslant 1}\\right\\} .\n$$\n\nOn remarque tout d'abord que $\\mathcal{E}$ est stable par somme. En effet, si $n$ et $m$ sont deux éléments de $\\mathcal{E}$, et si $k$ et $\\ell$ sont deux multiples de 222 tels que $\\mathrm{S}_{2}(k)=n$ et $\\mathrm{S}_{2}(\\ell)=m$, soit $p \\geqslant 0$ un entier tel que $10^{p}>k$. L'entier $k+10^{p} \\ell$ est un multiple de 222, et s'écrit en juxtaposant les représentations décimales de $\\ell$ et de $k$, étant entendu que l'on aura fait tenir cette dernière sur $p$ chiffres. Ainsi,\n$$\n\\mathrm{S}_{2}\\left(k+10^{p} \\ell\\right)=\\mathrm{S}_{2}(\\ell)+\\mathrm{S}_{2}(k)=m+n\n$$\nappartient bien à $\\mathcal{E}$.\n\nOn entreprend désormais de trouver de petits éléments de $\\mathcal{E}$. Par exemple, $\\mathcal{E}$ contient les entiers $\\mathrm{S}_{2}(222)=12$, $\\mathrm{S}_{2}(444)=48$, $\\mathrm{S}_{2}(666)=108$, $\\mathrm{S}_{2}(888)=196$ et $\\mathrm{S}_{2}(1110)=3$. En particulier, puisque $\\mathcal{E}$ contient 3 et est stable par somme, il contient aussi tous les multiples de 3.\n\nOn cherche ensuite à construire de petits éléments de $\\mathcal{E}$ autres que des multiples de 3 ; cela nécessite de faire autre chose que de juxtaposer des représentations décimales de multiples de 222. Par exemple, $\\mathcal{E}$ contient également les entiers $\\mathrm{S}_{2}(111000+1110)=\\mathrm{S}_{2}(112110)=8$ et $\\mathrm{S}_{2}(11100+1110)=\\mathrm{S}_{2}(12210)=10$. Il contient donc également tous les entiers $n \\geqslant 8$ tels que $n \\equiv 8\\ (\\bmod\\ 3)$, et tous les entiers $n \\geqslant 10$ tels que $n \\equiv 10\\ (\\bmod\\ 3)$.\n\nOn vient donc de démontrer que $\\mathcal{E}$ contient les entiers 3, 6, puis tous les entiers $n \\geqslant 8$. Il reste à vérifier s'il existe un entier $k$ qui est un multiple de 222 et pour lequel l'entier $n=\\mathrm{S}_{2}(k)$ est un élément de l'ensemble $\\mathcal{F}=\\{1,2,4,5,7\\}$. Soit $a_{1}, a_{2}, \\ldots, a_{\\ell}$ les chiffres non nuls de $k$ : on sait d'une part que $n=a_{1}^{2}+a_{2}^{2}+\\cdots+a_{\\ell}^{2}$, et d'autre part que $a_{1}+a_{2}+\\cdots+a_{\\ell} \\equiv k \\equiv 0\\ (\\bmod\\ 3)$. Par conséquent :\n\n$\\triangleright$ Si tous les chiffres $a_{i}$ valent 1, on sait que $n \\equiv \\ell \\equiv 0\\ (\\bmod\\ 3)$, ce qui contredit le fait que $n \\in \\mathcal{F}$.\n\n$\\triangleright$ Si exactement un chiffre $a_{i}$ vaut 2 et tous les autres valent 1, on sait que $n \\geqslant 4$ et que $n \\equiv \\ell+3 \\equiv k+2 \\equiv 2\\ (\\bmod\\ 3)$, de sorte que $n=5$.\n\n$\\triangleright$ Si au moins deux chiffres $a_{i}$ valent 2, ou si au moins un chiffre $a_{i}$ est supérieur ou égal à 3, on sait que $n \\geqslant 8$, ce qui contredit là encore le fait que $n \\in \\mathcal{F}$.\n\nLe seul cas à traiter est donc celui où les deux seuls chiffres non nuls de $k$ sont $a_{1}=1$ et $a_{2}=2$; dans ces conditions, on aura $n=5$. On recherche donc un éventuel multiple de 222 de la forme $k=10^{u}+2 \\times 10^{v}$, où $u$ et $v$ sont des entiers distincts. Comme $222=2 \\times 3 \\times 37$, il s'agit de faire en sorte que $u \\geqslant 1$ et que $10^{u} \\equiv -2 \\times 10^{v}\\ (\\bmod\\ 37)$. Or, puisque 37 divise 222 donc divise 1110, il divise aussi 111 et 999. Cela signifie que $10^{3} \\equiv 1\\ (\\bmod\\ 37)$, donc que les puissances de 10 modulo 37 sont 1, 10 et 100. En particulier, $10^{u} \\in \\{1,10,26\\}\\ (\\bmod\\ 37)$, tandis que $-2 \\times 10^{v} \\in \\{17,22,35\\}\\ (\\bmod\\ 37)$. Il est donc impossible de faire en sorte que $10^{u} \\equiv -2 \\times 10^{v} \\equiv 0\\ (\\bmod\\ 37)$, ce qui nous assure que $5 \\notin \\mathcal{E}$.\n\nEn conclusion, les entiers $n$ recherchés sont $n=3$, $n=6$, et tous les entiers $n \\geqslant 8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle rectangle en $B$ avec $BC < BA$. Soit $D$ le point du segment $[AB]$ tel que $BD = BC$. La perpendiculaire à $(AC)$ passant par $D$ intersecte $(AC)$ en $E$. Soit $B'$ le symétrique de $B$ par rapport à $(CD)$. Montrer que $(EC)$ est la bissectrice de l'angle $\\widehat{BEB'}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn remarque que le cercle de diamètre $[CD]$ apparaît assez naturellement. En effet, on a des angles droits $\\widehat{DBC} = \\widehat{CBD} = \\widehat{DEC} = 90^\\circ$, les points $B$, $B'$, et $E$ sont sur le cercle de diamètre $[DC]$, autrement dit $C, B, D, E, B'$ sont cocycliques.\n\nAlors on a :\n$$\n\\begin{aligned}\n\\widehat{BEC} & = \\widehat{BDC} \\text{ par angle inscrit } \\\\\n& = 45^\\circ \\text{ car } BC = BD \\text{ et } \\widehat{CBD} = 90^\\circ\n\\end{aligned}\n$$\n\nDe plus:\n$$\n\\begin{aligned}\n\\widehat{CEB'} & = \\widehat{CBB'} \\text{ par angle inscrit } \\\\\n& = 45^\\circ \\text{ car } BC = BD \\text{ et } \\widehat{CBD} = 90^\\circ .\n\\end{aligned}\n$$\n\nOn a donc bien $\\widehat{BEC} = \\widehat{CEB'}$, donc $(EC)$ est la bissectrice de $\\widehat{BEB'}$.\n\n\nAlternative :\n\nOn pouvait aussi montrer directement $\\widehat{BEC} = \\widehat{CEB'}$ sans utiliser $BC = BD$. En effet :\n$$\n\\begin{aligned}\n\\widehat{BEC} & = \\widehat{BDC} \\text{ par angle inscrit } \\\\\n& = \\widehat{CDB'} \\text{ par symétrie } \\\\\n& = \\widehat{CEB'} \\text{ par angle inscrit. }\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver les triplets d'entiers $(x, y, n)$ tels que $n^{2}=17 x^{4}-32 x^{2} y^{2}+41 y^{4}$.",
    "options": [],
    "answer": "(0, 0, 0)",
    "solution": "Solution:\n\nOn va montrer par descente infinie que $(0,0,0)$ est la seule solution.\n\nComme un carré ne vaut que $0$ ou $1$ modulo $3$ (une façon de le voir est de faire une disjonction de cas sur les valeurs modulo $3$), il est pertinent de tenter une étude modulo $3$ pour essayer de voir où ça nous mène. Pour ce faire, on remarque que si $(x, y, n)$ est solution, alors modulo $3$, on a\n$$\nn^{2} \\equiv -x^{4}-2 x^{2} y^{2}-y^{4} \\equiv -\\left(x^{2}+y^{2}\\right)^{2} \\quad(\\bmod 3)\n$$\nAinsi, $n^{2}+\\left(x^{2}+y^{2}\\right)^{2}$ est divisible par $3$. Or un carré vaut $0$ ou $1$ modulo $3$ donc la seule manière qu'une somme de carrés soit divisible par $3$, c'est que chacun des termes le soit. Donc $n$ et $x^{2}+y^{2}$ sont divisibles par $3$. De même il découle que $x$ et $y$ sont divisibles par $3$.\n\nEn réinjectant dans l'équation, on en déduit que $3^{4}=81$ divise $n^{2}$ et donc $9$ divise $n$, donc $\\left(\\frac{x}{3}, \\frac{y}{3}, \\frac{n}{9}\\right)$ est une autre solution. Or si on avait une solution avec $x, y$ ou $n$ non nul (et quitte à les changer en leurs opposés, ce qui ne change pas les valeurs des carrés, positifs), on pourrait obtenir une suite strictement décroissante d'entiers naturels, ce qui est absurde (c'est le principe de la descente infinie).\n\nOn en déduit que la seule solution potentielle est $(0,0,0)$. Et réciproquement, on remarque que $(x, y, n) = (0,0,0)$ convient (les deux membres donnent $0$).\n\nL'unique solution de l'équation est donc $(0,0,0)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30884,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn certain nombre de diagonales divisent un polygone convexe en triangles. Ces segments ne peuvent s'intersecter que sur un sommet du polygone. Sur chaque sommet du polygone on écrit le nombre de triangles qui touchent ce sommet. Est-il possible de reconstruire les diagonales en connaissant seulement les nombres sur les sommets?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIl est toujours possible de reconstruire les diagonales en utilisant ce procédé :\n- On considère un sommet \"visible\" $v$ qui ne touche qu'un seul triangle.\n- On trace la diagonale qui relie les deux sommets $x$ et $y$ adjacents à $v$.\n- On \"cache\" $v$ et on retire 1 au nombre de triangles adjacents de $x, y$.\n- On continue jusqu'à ce qu'il ne reste plus que 3 sommets.\n\nOn utilise pour cela le fait que dans toute triangulation d'un polygone convexe, il existe un sommet adjacent à un seul triangle. Pour le montrer, on considère la diagonale $d$ qui \"saute\" le moins de sommets : si $v$ est un sommet sauté, il n'est relié à aucune diagonale sinon elle serait plus courte que $d$ (car elles ne peuvent pas s'intersecter). Il est donc adjacent à un unique triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30885,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle isocèle en $A$ dont tous les angles sont aigus, et soit $D$ un point situé sur le segment $[BC]$. Soit $\\ell$ la droite parallèle à $(BC)$ passant par $A$, puis $X$ le point de $\\ell$ pour lequel $(XD)$ est perpendiculaire à $(BC)$, et $\\Gamma$ le cercle de centre $X$ passant par $D$. Le cercle $\\Gamma$ coupe le segment $[AB]$ en un point $E$, et on note $Y$ le point de $\\ell$ tel que $(AB)$ soit perpendiculaire à $(EY)$; de même, $\\Gamma$ coupe le segment $[AC]$ en un point $F$, et on note $Z$ le point de $\\ell$ tel que $(AC)$ soit perpendiculaire à $(FZ)$.\n\nDémontrer que les bissectrices des angles $\\widehat{EYA}$ et $\\widehat{FZA}$ se coupent en un point de la droite $(XD)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nNotons $T$ le point d'intersection des bissectrices de $\\widehat{EYA}$ et $\\widehat{FZA}$, $U$ le point d'intersection des droites $(EY)$ et $(FZ)$, et $\\mathcal{C}$ le cercle de rayon $[AU]$. Puisque les angles $\\widehat{UEA}$ et $\\widehat{AFU}$ sont droits, $E$ et $F$ appartiennent à $\\mathcal{C}$.\n\nPar ailleurs, puisque $ABC$ est isocèle en $A$, la bissectrice des angles $\\widehat{BAC}$ et $\\widehat{EAF}$ est aussi la hauteur de $ABC$ issue de $A$. Elle est donc perpendiculaire aux droites $(BC)$ et $\\ell = (AX)$. Comme $DX = EX = FX$, le point $X$ est donc le point d'intersection de la bissectrice extérieure de $\\widehat{EAF}$ avec la médiatrice de $[EF]$, c'est-à-dire le pôle Nord de $AEF$ issu de $A$. Cela signifie en particulier que $X$ appartient au cercle circonscrit à $AEF$, c'est-à-dire à $\\mathcal{C}$.\n\nDans ces conditions, l'angle $\\widehat{UXA}$ est droit, donc les triangles $UXY$ et $UXZ$ sont rectangles en $U$. Or, puisque les triangles $AFZ$ et $AEY$ sont rectangles en $E$ et $F$ et que $ABC$ est isocèle en $A$, on sait que\n$$\n\\widehat{XZU} = \\widehat{AZF} = 90^\\circ - \\widehat{FAZ} = 90^\\circ - \\widehat{ACB} = 90^\\circ - \\widehat{CBA} = 90^\\circ - \\widehat{YAE} = \\widehat{EYA} = \\widehat{YUX}.\n$$\nAinsi, les deux triangles rectangles $UXY$ et $UXZ$ sont symétriques l'un de l'autre par rapport à la droite $(UX)$. Les deux bissectrices $(TY)$ et $(TZ)$ se retrouvent elles aussi symétriques l'une de l'autre par rapport à $(UX)$, donc $T$ appartient à $(UX)$.\n\nEnfin, par construction, la droite $(DX)$ est perpendiculaire aux droites $(BC)$ et $\\ell$, donc elle est confondue avec $(UX)$. On en conclut donc, comme demandé dans l'énoncé, qu'elle contient le point $T$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30886,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier strictement positif et $x \\geqslant n$ un réel. Montrer que $x+\\frac{n}{x} \\geqslant n+1$ et donner les cas d'égalité.",
    "options": [],
    "answer": "Equality holds if and only if the real number equals the integer.",
    "solution": "Solution:\n\nLe réel $x$ est strictement positif donc on peut multiplier l'équation par $x$ pour avoir que l'équation est équivalente à\n$$\nx^{2}+n \\geqslant (n+1)x\n$$\nce qui donne, en réarrangeant les termes,\n$$\nx^{2}-(n+1)x+n \\geqslant 0\n$$\nOn reconnait alors une factorisation\n$$\n(x-n)(x-1) \\geqslant 0\n$$\nMais ceci est vrai car $x \\geqslant n$ et $x \\geqslant 1$.\nOn a égalité lorsque $x=n$ ou $x=1$. Mais si $x=1$, alors $n=1$ et $x=n$. Ainsi, dans tous les cas, le seul cas d'égalité est $x=n$.\n\n\nSolution alternative:\n\nOn repart de l'équation développée $x^{2}-(n+1)x+n \\geqslant 0$.\nLa condition $x \\geqslant n$ est très peu naturelle, on préfère des conditions de type $t \\geqslant 0$. On pose donc $x=n+t$ avec $t \\geqslant 0$. L'équation se réécrit alors\n$$\nn^{2}+2nt+t^{2}-(n+1)(n+t)+n \\geqslant 0\n$$\nou encore\n$$\nt^{2}+(n-1)t \\geqslant 0\n$$\nMais cette nouvelle équation est clairement vraie car $t \\geqslant 0$ et $n \\geqslant 1$. De plus, on a égalité si et seulement si $t=0$, ou encore $x=n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2024 élèves, tous de taille différente, doivent se placer en file indienne. Cependant, chaque élève ne souhaite pas avoir à la fois devant lui et derrière lui un élève plus petit que lui. Combien y a-t-il de façons de former une telle file indienne?",
    "options": [],
    "answer": "2^{2023}",
    "solution": "Solution:\n\nRegardons l'élève de plus grande taille : s'il n'est pas placé au tout début ou à la toute fin de la file, il est entre deux élèves plus petits que lui. Ainsi l'élève le plus grand doit être placé à la fin ou au début.\n\nVia ce raisonnement, et en regardant les petits cas on peut conjecturer que si on doit placer $n \\geqslant 1$ élèves tous de taille différente avec la contrainte de l'énoncé, il y a $2^{n-1}$ possibilités. Montrons cela par récurrence sur $n$.\n\nInitialisation : pour $n=1$ élève, il n'y a qu'une file possible et $1=2^{n-1}$.\n\nHérédité : supposons l'hypothèse vraie au rang $n$ pour un $n \\geqslant 1$, montrons qu'elle l'est au rang $n+1$. L'élève le plus grand est forcément placé soit au tout début, soit à la toute fin. Si on l'enlève de la file, la file de $n$ personnes vérifie toujours l'énoncé : il y a donc potentiellement $2^{n-1}$ possibilités pour la file sans la personne la plus grande. Si on rajoute la personne la plus grande devant ou derrière, la file vérifie toujours l'énoncé (car la personne à côté du plus grand ne sera pas entre deux personnes plus petites), comme il y a deux choix de placements du plus grand, il y a $2^{n}=2^{n+1-1}$ possibilités pour former la file, ce qui conclut la récurrence.\n\nAinsi en appliquant la propriété pour $n=2024$, il y a $2^{2023}$ possibilités.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier. Bosphore a écrit $n$ fois le nombre $2$ au tableau. Il effectue ensuite, $n-1$ fois d'affilée, l'opération suivante : il choisit deux nombres écrits au tableau, qu'il appelle $a$ et $b$, puis les efface et écrit le nombre $\\sqrt{(ab+1)/2}$ à la place. Enfin, il appelle $x$ le nombre écrit au tableau après ces $n-1$ opérations, et $y$ le nombre $\\sqrt{(n+3)/n}$.\n\na) Démontrer que $x \\geqslant y$.\n\nb) Démontrer qu'il existe une infinité d'entiers $n \\geqslant 1$ pour lesquels on a nécessairement $x>y$.\n\nc) Démontrer qu'il existe une infinité d'entiers $n \\geqslant 1$ pour lesquels Bosphore peut faire en sorte que $x=y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans la suite, notons $x_{n}$ le plus petit réel $x$ auquel peut aboutir Bosphore en partant de $n$ entiers égaux à $2$, et posons $y_{n}=\\sqrt{(n+3)/n}$. Il s'agit de démontrer que $x_{n} \\geqslant y_{n}$, avec égalité une infinité de fois et inégalité stricte une infinité de fois. Une étude des petits cas suggère la propriété $\\mathcal{P}(n)$ suivante : «l'inégalité $x_{n} \\geqslant y_{n}$ est bien vraie, et $x_{n}=y_{n}$ si et seulement si $n$ est une puissance de $2$».\n\nRéciproquement, soit $u$ et $v$ les deux derniers nombres qu'il efface, juste avant d'écrire $x$ au tableau. Si Bosphore a pour objectif d'aboutir à l'inégalité $x \\leqslant y$, il a intérêt à ce que $u$ et $v$ soient aussi petits que possible. Par conséquent, s'il y a $k$ nombres qui, à force de réécritures, ont été transformés en le nombre $u$, et $\\ell$ nombres qui ont été transformés en le nombre $v$, Bosphore fera en sorte que $u=x_{k}$ et $v=x_{\\ell}$. Cela signifie en fait que $x_{n}$ est le plus petit des nombres\n\n$$\n\\sqrt{\\frac{x_{1} x_{n-1}+1}{2}}, \\sqrt{\\frac{x_{2} x_{n-2}+1}{2}}, \\ldots, \\sqrt{\\frac{x_{n-1} x_{1}+1}{2}}.\n$$\n\nOn entreprend donc de démontrer $\\mathcal{P}(n)$ par récurrence sur $n \\geqslant 1$. Tout d'abord, $x_{1}=2=y_{1}$, donc $\\mathcal{P}(1)$ est vraie. Puis, si $\\mathcal{P}(1), \\ldots, \\mathcal{P}(n-1)$ sont vraies, et pour tous entiers $k$ et $\\ell$ tels que $1 \\leqslant k \\leqslant n-1$ et $1 \\leqslant \\ell \\leqslant n-1$, on sait que\n\n$$\n\\sqrt{\\frac{x_{k} x_{\\ell}+1}{2}} \\geqslant y_{k+\\ell} \\Leftrightarrow x_{k} x_{\\ell} \\geqslant 2 y_{k+\\ell}^{2}-1\n$$\n\nPuisque $x_{k} \\geqslant y_{k}$ et $x_{\\ell} \\geqslant y_{\\ell}$, il suffit donc de démontrer que $y_{k} y_{\\ell} \\geqslant 2 y_{k+\\ell}^{2}-1$.\n\nSi l'on pose $\\Delta=y_{k}^{2} y_{\\ell}^{2}-\\left(2 y_{k+\\ell}^{2}-1\\right)^{2}$, on vérifie alors que\n\n$$\n\\begin{aligned}\nk \\ell(k+\\ell)^{2} \\Delta & =(k+3)(\\ell+3)(k+\\ell)^{2}-k \\ell(k+\\ell+6)^{2} \\\\\n& =3(k+\\ell+3)(k+\\ell)^{2}-12 k \\ell(k+\\ell+3) \\\\\n& =3(k+\\ell+3)\\left(k^{2}+2 k \\ell+\\ell^{2}-4 k \\ell\\right) \\\\\n& =3(k+\\ell+3)(k-\\ell)^{2}.\n\\end{aligned}\n$$\n\nCeci nous assure que $\\Delta \\geqslant 0$, donc que $x_{k} x_{\\ell} \\geqslant y_{k} y_{\\ell} \\geqslant 2 y_{k+\\ell}^{2}-1$. En outre, les cas d'égalité surviennent exactement lorsque $x_{k}=y_{k}$, $x_{\\ell}=y_{\\ell}$ et $\\Delta=0$, c'est-à-dire lorsque $k=\\ell$ est une puissance de $2$. En particulier, l'inégalité\n\n$$\n\\sqrt{\\frac{x_{k} x_{\\ell}+1}{2}} \\geqslant y_{k+\\ell}\n$$\n\nest nécessairement vérifiée, et il s'agit d'une égalité si et seulement si $k=\\ell$ est une puissance de $2$.\n\nEn conclusion :\n\na) Quelle que soit la valeur de $n$, il existe deux entiers $k$ et $\\ell=n-k$ pour lesquels $x_{n}=\\sqrt{\\left(x_{k} x_{\\ell}+1\\right) / 2} \\geqslant \\sqrt{\\left(y_{k} y_{\\ell}+1\\right) / 2} \\geqslant y_{k+\\ell}=y_{n}$.\n\nb) Si $n$ n'est pas une puissance de $2$, $k$ et $\\ell$ ne sont pas égaux à une même puissance de $2$, donc l'une des trois inégalités $x_{k} \\geqslant y_{k}$, $x_{\\ell} \\geqslant y_{\\ell}$ et $\\sqrt{\\left(y_{k} y_{\\ell}+1\\right) / 2} \\geqslant y_{n}$ est stricte, et $x_{n}>y_{n}$.\n\nc) Si $n$ est une puissance de $2$, l'inégalité $x_{n} \\leqslant \\sqrt{\\left(x_{n / 2}^{2}+1\\right) / 2}=\\sqrt{\\left(y_{n / 2}^{2}+1\\right) / 2}=y_{n}$ indique que $x_{n}=y_{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $ABC$ un triangle et $I$ le centre de son cercle inscrit. La médiatrice de $[BC]$ coupe $(AI)$ en $S$ et $(BI)$ en $T$. Montrer que $C$, $I$, $S$ et $T$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nPour montrer que $C, I, S$ et $T$ sont cocycliques, nous allons montrer que $\\widehat{SIC} = \\widehat{STC}$.\nOn rappelle que les bissectrices intérieures d'un triangle $ABC$ sont concourantes et que leur point d'intersection $I$ est le centre du cercle inscrit dans le triangle. $(AI)$ est donc la bissectrice de l'angle $\\widehat{BAC}$.\nPar conséquent, $S$ est défini comme le point d'intersection de la médiatrice de $[BC]$ et de la bissectrice de l'angle $\\widehat{BAC}$. Le théorème du Pôle Sud assure que $S$ appartient au cercle circonscrit au triangle $ABC$ et que $S$ est le centre du cercle antarctique du sommet $A$, ce qui assure que\n$$\nSB = SI\n$$\nOn voit donc que le triangle $SBI$ est isocèle en $S$ ce qui assure que $\\widehat{SBI} = \\widehat{BIS}$.\nPar ailleurs les points $A, B, S$ et $C$ sont cocycliques. On a donc\n$$\n\\widehat{SBC} = \\widehat{SAC} = \\frac{\\widehat{BAC}}{2}.\n$$\nCalculons désormais la valeur de l'angle $\\widehat{SIC}$. On a\n$$\n\\begin{aligned}\n\\widehat{SIC} &= \\widehat{BIC} - \\widehat{BIS} \\\\\n&= \\widehat{BIC} - \\widehat{SBC} - \\widehat{CBI} \\\\\n&= \\widehat{BIC} - \\frac{\\widehat{BAC}}{2} - \\frac{\\widehat{CBA}}{2}.\n\\end{aligned}\n$$\nPar ailleurs,\n$$\n\\begin{aligned}\n\\widehat{BIC} &= 180^\\circ - \\widehat{CBI} - \\widehat{ICB} \\\\\n&= 180^\\circ - \\frac{\\widehat{CBA}}{2} - \\frac{\\widehat{ACB}}{2} \\\\\n&= 180^\\circ - \\frac{\\widehat{CBA} + \\widehat{ACB}}{2} \\\\\n&= 180^\\circ - \\frac{180^\\circ - \\widehat{BAC}}{2} \\\\\n&= 90^\\circ + \\frac{\\widehat{BAC}}{2}.\n\\end{aligned}\n$$\nIl s'ensuit que $\\widehat{SIC} = 90^\\circ - \\frac{\\widehat{CBA}}{2}$.\nPar ailleurs\n$$\n\\begin{aligned}\n\\widehat{STC} &= 90^\\circ - \\widehat{TCB} \\\\\n&= 90^\\circ - \\widehat{CBI} \\quad \\text{Le triangle } BCT \\text{ est isocèle en } T \\\\\n&= 90^\\circ - \\frac{\\widehat{CBA}}{2}.\n\\end{aligned}\n$$\nOn a donc montré que $\\widehat{SIC} = \\widehat{STC}$ ce qui assure que les points $C, I, S$ et $T$ sont cocycliques par la réciproque du théorème de l'angle inscrit.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers naturels $n$ tels que $21$ divise $2^{2^{n}} + 2^{n} + 1$.",
    "options": [],
    "answer": "n ≡ 2 (mod 6)",
    "solution": "Solution:\n\nComme $21 = 3 \\times 7$ et que $3$ et $7$ sont premiers entre eux, $21$ divise $2^{2^{n}} + 2^{n} + 1$ si et seulement si $3$ et $7$ divisent $2^{2^{n}} + 2^{n} + 1$.\n\nÉliminons de suite le cas $n = 0$, qui donne $2^{2^{0}} + 2^{0} + 1 = 4$ qui n'est pas divisible par $21$. On peut donc supposer $n > 0$, en particulier $2^{n}$ est pair. Alors modulo $3$ :\n$$\n2^{2^{n}} + 2^{n} + 1 \\equiv (-1)^{2^{n}} + (-1)^{n} + 1 \\equiv (-1)^{n} + 2 \\pmod{3}.\n$$\nEn particulier, $2^{2^{n}} + 2^{n} + 1$ est divisible par $3$ si et seulement si $n$ est pair. On suppose donc désormais que cette condition est vérifiée.\n\nReste à traiter le cas de la divisibilité par $7$. Modulo $7$, les puissances de $2$ sont $2^{0} \\equiv 1$, $2^{1} \\equiv 2$, $2^{2} \\equiv 4$, $2^{3} \\equiv 1$ : il s'agit donc d'étudier l'exposant modulo $3$. Comme $n$ est pair (par le cas précédent), on sait déjà que $2^{n} \\equiv 1 \\pmod{3}$ donc $2^{2^{n}} \\equiv 2 \\pmod{7}$. On veut donc $2^{n} \\equiv -1 - 2 \\equiv 4 \\pmod{3}$, et donc $n \\equiv 2 \\pmod{3}$ (et réciproquement si $n \\equiv 2 \\pmod{3}$, comme $n$ est pair alors $7$ divise $2^{2^{n}} + 2^{n} + 1$).\n\nOn doit avoir $n \\equiv 0 \\pmod{2}$, $n \\equiv 2 \\pmod{3}$ : comme $2$ et $3$ sont premiers entre eux, le théorème des restes chinois donne qu'il existe une unique solution modulo $2 \\times 3 = 6$, et on remarque que c'est $n \\equiv 2 \\pmod{6}$.\n\nEn conclusion, les entiers naturels $n$ qui conviennent sont exactement ceux qui vérifient $n \\equiv 2 \\pmod{6}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30891,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a_{1}, a_{2}, \\ldots$ des nombres réels strictement positifs tels que\n$$\na_{n+1}^{2}+a_{n} a_{n+2} \\leqslant a_{n}+a_{n+2}\n$$\npour tout entier $n \\geqslant 1$. Démontrer que $a_{2023} \\leqslant 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nL'inégalité donnée dans l'énoncé et celle qu'il nous est demandé de démontrer suggèrent de s'intéresser aux termes de la forme $b_{n}=a_{n}-1$, car alors l'équation de l'énoncé se réécrit comme\n$$\nb_{n+1}\\left(b_{n+1}+2\\right)+b_{n} b_{n+2} \\leqslant 0,\n$$\net il s'agit de démontrer que $b_{2023} \\leqslant 0$.\n\nSoit $n \\geqslant 2$ un entier éventuel pour lequel $b_{n}>0$. Puisque $b_{n-1} b_{n+1} \\leqslant -b_{n}\\left(b_{n}+2\\right)<0$, l'un des deux termes $b_{n-1}$ et $b_{n+1}$ est strictement positif, et l'autre est strictement négatif. En posant $m=n-1$ dans le cas où $b_{n-1}>0$, et $m=n$ dans le cas où $b_{n+1}>0$, on dispose désormais d'un entier $m \\geqslant 1$ pour lequel $b_{m}>0$ et $b_{m+1}>0$. Dans ces conditions, et puisque $b_{m+2} \\geqslant -1$, on sait que\n$$\nb_{m} \\geqslant b_{m}+b_{m} b_{m+2}+b_{m+1}\\left(b_{m+1}+2\\right) \\geqslant b_{m+1}\\left(b_{m+1}+2\\right)>b_{m+1}\n$$\nDe même, puisque $b_{m-1} \\geqslant -1$, on sait que\n$$\nb_{m+1} \\geqslant b_{m+1}+b_{m-1} b_{m+1}+b_{m}\\left(b_{m}+2\\right) \\geqslant b_{m}\\left(b_{m}+2\\right)>b_{m},\n$$\nce qui contredit l'inégalité $b_{m}>b_{m+1}$ obtenue précédemment.\n\nOn en conclut que $b_{n} \\leqslant 0$, c'est-à-dire $a_{n} \\leqslant 1$, pour tout entier $n \\geqslant 2$, et notamment pour $n=2023$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les triplets de réels positifs ou nuls $\\left(a, b, c\\right)$ tels que\n$$\n\\left\\{\n\\begin{array}{l}\na^{2}+a b=c \\\\\nb^{2}+b c=a \\\\\nc^{2}+c a=b\n\\end{array}\n\\right.\n$$",
    "options": [],
    "answer": "(a,b,c)=(0,0,0) or (1/2,1/2,1/2)",
    "solution": "Solution:\nSupposons d'abord qu'un des réels est nul, sans perte de généralité $a=0$. Alors la première équation donne $c=0$ et la dernière équation donne $b=0$. Réciproquement, le triplet $a=b=c=0$ est bien solution, on suppose à présent que les trois réels sont non nuls.\n\nSupposons maintenant que deux des réels soient égaux, sans perte de généralité $a=b$. On peut alors soustraire les deuxième et troisième équations pour avoir $b^{2}-c^{2}=0$. Comme $b$ et $c$ sont tous les deux positifs, on a donc $a=b=c$. Dans la première équation, on a alors $2 a^{2}=a$. Mais $a$ est non nul donc on peut diviser par $2 a$ pour avoir $a=\\frac{1}{2}$. Réciproquement, le triplet $a=b=c=\\frac{1}{2}$ est bien solution. On suppose désormais que les trois réels sont deux à deux distincts.\n\nFaisons le produit des équations pour obtenir\n$$\na(a+b)\\, b(b+c)\\, c(c+a)=a b c\n$$\nOn a supposé précédememnt que les réels étaient tous non nuls donc on peut diviser par $a b c$ pour avoir\n$$\n(a+b)(b+c)(c+a)=1\n$$\nEn faisant la différence des deux premières équations, on obtient $a^{2}-b^{2}+b a-b c=c-a$ ce qui se factorise en\n$$\n(a-b)(a+b)=(c-a)(b+1)\n$$\nEn faisant de même avec les deuxième et troisième lignes puis avec les troisième et première lignes, on obtient en plus\n$$\n\\left\\{\n\\begin{array}{l}\n(b-c)(b+c)=(a-b)(c+1) \\\\\n(c-a)(c+a)=(b-c)(a+1)\n\\end{array}\n\\right.\n$$\nEn faisant le produit de ces trois équations et en divisant par $(a-b)(b-c)(c-a)$ des deux côtés (qui est non nul car on a supposé $a, b, c$ deux à deux distincts), on obtient\n$$\n(1+a)(1+b)(1+c)=(a+b)(b+c)(c+a)\n$$\nMais on a montré précédemment que ce deuxième terme vaut $1$, ce qui est absurde car\n$$\n(1+a)(1+b)(1+c)>1 \\cdot 1 \\cdot 1=1\n$$\nAinsi, les seules solutions sont celles trouvées au début de la solution, c'est-à-dire $a=b=c=0$ et $a=b=c=\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUne suite olympique est une suite $s_{1}, s_{2}, \\ldots, s_{2023}$ dont chacun des 2023 termes est égal à 1 ou à -1. Une suite peu croissante est une suite d'entiers $t_{1}, t_{2}, \\ldots, t_{n}$ telle que $1 \\leqslant t_{1}<t_{2}<\\cdots<t_{n} \\leqslant 2023$, et dont chacune des $n-1$ différences $t_{i+1}-t_{i}$ vaut 1 ou 2.\nTrouver le plus grand entier $C$ tel que, pour toute suite olympique $s_{1}, s_{2}, \\ldots, s_{2023}$, il existe un entier $n \\geqslant 1$ et une suite peu croissante $t_{1}, t_{2}, \\ldots, t_{n}$ de longueur $n$ tels que\n\n$$\n\\left|s_{t_{1}}+s_{t_{2}}+\\cdots+s_{t_{n}}\\right| \\geqslant C\n$$",
    "options": [],
    "answer": "507",
    "solution": "Solution:\n\nUne étude des petits cas nous pousse à nous intéresser à la suite $\\left(s_{i}\\right)$ dont les termes valent $+1,+1,-1,-1,+1,+1,-1,-1, \\ldots$, et qui semble être un «pire» cas. On la subdivise en 1011 blocs de longueur 2, notés $B_{1}, B_{2}, \\ldots, B_{1011}$, et un bloc $B_{1012}$ de longueur 1 : la suite vaut $(-1)^{i+1}$ sur chaque bloc $B_{i}$.\n\nOn considère maintenant une somme de la forme $s_{t_{1}}+s_{t_{2}}+\\cdots+s_{t_{n}}$. Ses termes $s_{t_{i}}$ ont été choisis parmi des blocs consécutifs $B_{u}, B_{u+1}, \\ldots, B_{u+v-1}$. Au plus $(v+1)/2$ de ces $v$ blocs contiennent des termes $+1$ ; les autres contiennent des termes $-1$. Parmi les nombres $s_{t_{1}}, s_{t_{2}}, \\ldots, s_{t_{n}}$, il y en a donc au plus $v+1$ qui valent $+1$ et au moins $(v-1)/2$ qui valent $-1$. Puisque $v \\leqslant 1012$, on en déduit que\n$$\ns_{t_{1}}+s_{t_{2}}+\\cdots+s_{t_{n}} \\leqslant (v+1) - (v-1)/2 = (v+3)/2 < 508.\n$$\nOn montre de même que $s_{t_{1}}+s_{t_{2}}+\\cdots+s_{t_{n}} > -508$, de sorte que $C \\leqslant 507$.\n\nRéciproquement, soit $(s_{i})$ une suite olympique quelconque. Quitte à remplacer chaque terme par son opposé, on suppose que $\\left(s_{i}\\right)$ compte au code moins 1012 termes égaux à $+1$. Comme précédemment, on subdivise cette suite en blocs maximaux sur lesquels $(s_{i})$ est constante; ces blocs sont notés $B_{1}, B_{2}, \\ldots, B_{k}$, et leurs longueurs sont notés $b_{1}, b_{2}, \\ldots, b_{k}$. Quitte à supposer que $B_{1}$ est vide, on suppose que $s_{j}=(-1)^{i}$ lorsque $j$ appartient à un bloc $B_{i}$. Ainsi, $b_{2}+b_{4}+b_{6}+\\cdots \\geqslant 1012$ et $1011 \\geqslant b_{1}+b_{3}+b_{5}+\\cdots$.\n\nOn choisit ensuite $\\left(t_{i}\\right)$ comme la suite peu croissante qui prend toutes les valeurs $j$ appartenant à un bloc $B_{2}, B_{4}, B_{6}, \\ldots$ ou bien situées en position paire d'un bloc $B_{1}, B_{3}, B_{5}, \\ldots$. De la sorte, parmi les nombres $s_{t_{1}}, s_{t_{2}}, \\ldots$, il y en a $b_{2}+b_{4}+b_{6}+\\cdots$ qui valent $+1$ ; chacun des $\\left\\lfloor b_{1}/2\\right\\rfloor+\\left\\lfloor b_{3}/2\\right\\rfloor+\\left\\lfloor b_{5}/2\\right\\rfloor+\\cdots$ autres termes vaut $-1$. On en conclut que\n\n$$\n\\begin{aligned}\ns_{t_{1}}+s_{t_{2}}+\\cdots &= \\left(b_{2}+b_{4}+b_{6}+\\cdots\\right) - \\left(\\left\\lfloor b_{1}/2\\right\\rfloor + \\left\\lfloor b_{3}/2\\right\\rfloor + \\left\\lfloor b_{5}/2\\right\\rfloor + \\cdots\\right) \\\\\n&\\geqslant \\left(b_{2}+b_{4}+b_{6}+\\cdots\\right) - \\left(b_{1}+b_{3}+b_{5}+\\cdots\\right)/2 \\\\\n&\\geqslant 1012 - 1011/2 > 506\n\\end{aligned}\n$$\n\nde sorte que $s_{t_{1}}+s_{t_{2}}+\\cdots \\geqslant 507$.\n\nEn conclusion, l'entier $C$ recherché vaut $C=507$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nNimatha et Thanima jouent à un jeu sur un échiquier $8 \\times 8$. Tour par tour en commençant par Nimatha, chaque joueur choisit une case qui n'a pas encore été choisie et la colorie dans sa couleur (rouge pour Nimatha, bleu pour Thanima). Montrez que Thanima peut toujours faire que Nimatha ne puisse colorier aucun carré $2 \\times 2$ entièrement en rouge.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn considère le coloriage suivant :\n\n| 1 | 1 | 2 | 2 | 3 | 3 | 4 | 4 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 5 | 6 | 6 | 7 | 7 | 8 | 8 | 5 |\n| 9 | 9 | 10 | 10 | 11 | 11 | 12 | 12 |\n| 13 | 14 | 14 | 15 | 15 | 16 | 16 | 13 |\n| 17 | 17 | 18 | 18 | 19 | 19 | 20 | 20 |\n| 21 | 22 | 22 | 23 | 23 | 24 | 24 | 21 |\n| 25 | 25 | 26 | 26 | 27 | 27 | 28 | 28 |\n| 29 | 30 | 30 | 31 | 31 | 32 | 32 | 29 |\n\nLorsque Nimatha joue sur un nombre $X$, Thanima réagit en jouant sur la seconde occurrence du nombre $X$. Tout carré $2 \\times 2$ contient deux nombres identiques : si l'un est coloré en rouge, l'autre est coloré en bleu. Ainsi, il est impossible pour Nimatha de créer un carré $2 \\times 2$ entièrement rouge si Thanima suit cette stratégie.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle dont tous les angles sont aigus. Soit $F$ le pied de la hauteur de $ABC$ issue de $A$, et soit $P$ un point situé sur le segment $[AF]$. On note $D$ le point de $(BC)$ tel que $(PD)$ soit parallèle à $(AC)$, puis $X$ le point où se recoupent le cercle circonscrit à $ABD$ et le cercle de centre $D$ passant par $A$. De même, on note $E$ le point de $(BC)$ tel que $(PE)$ soit parallèle à $(AB)$, puis $Y$ le point où se recoupent le cercle circonscrit à $ACE$ et le cercle de centre $E$ passant par $A$.\n\nDémontrer que les points $B$, $C$, $X$ et $Y$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nDans un tel problème, le plus simple est d'essayer d'éliminer au mieux les points qui pourraient s'avérer peu utiles pour la suite. Ici, c'est par exemple le cas du point $P$, qui ne sert qu'à construire les points $D$ et $E$ en tant qu'images de $B$ et $C$ par une même homothétie de centre $F$. Ainsi, $FE \\times FC = FB \\times FD$, et $F$ a donc même puissance par rapport aux cercles circonscrits à $ABD$ et à $ACE$. Ces deux cercles ont donc $(AF)$ pour axe radical.\n\nOn s'intéresse donc aux axes radicaux de ces deux cercles avec l'éventuel cercle circonscrit à $B X C Y$, c'est-à-dire aux droites $(BX)$ et $(CY)$, et au centre radical de ces trois cercles. Or, sur la figure, celui-ci semble appartenir aux cercles de centres $D$ et $E$ passant par $A$. Cela signifierait qu'il est confondu avec le symétrique de $A$ par rapport à la droite $(DE) = (BC)$; on note $A'$ ce symétrique.\n\nUne manière simple de réutiliser l'égalité $DA = DX$ consiste à dire que les arcs de cercle $\\widehat{DA}$ et $\\widehat{XD}$ ont même mesure. Ainsi,\n$$\n(BX, BC) = (BX, BD) = (BD, BA) = (BC, BA) = (BA', BC)\n$$\nOn démontre de même que $(CY, CB) = (CA', CB)$, de sorte que les trois droites $(AF)$, $(BX)$ et $(CY)$ sont bien concourantes en $A'$.\n\nEn particulier, $A'$ est le centre radical des cercles circonscrits à $ABDX$, à $ACEY$ et à $BCY$, donc $(A'B) = (BX)$ est l'axe radical des cercles circonscrits à $ABDX$ et à $BCY$. Puisque $X$ appartient à l'un de ces cercles, il appartient aussi à l'autre, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30896,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $A, B, C, D$ et $E$ cinq points dans cet ordre sur un cercle tels que $A E = D E$. Soit $P$ le point d'intersection de $(A C)$ et $(B D)$. Soit $Q$ le point de la demi-droite $[B A)$ tel que $A Q = D P$. Soit $R$ le point de la demi-droite $[C D)$ tel que $D R = A P$. Montrer que les droites $(P E)$ et $(Q R)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOn se place dans le cas de la figure suivante:\n![](attached_image_1.png)\nDans cet énoncé on a de nombreuses égalités de longueur $(A Q = D P, D R = A P, A E = D E)$. Comme sur la figure, il ne faut pas hésiter à colorier les longueurs égales pour voir des potentiels triangles isométriques, et avancer dans le problème.\n\nLes triangles $DRP$ et $AQP$ ont l'air d'être isométriques : prouvons-le. Notons déjà que $DR = AQ$ et $DP = AP$ par hypothèse. De plus $\\widehat{RDP} = 180 - \\widehat{CDP} = 180 - \\widehat{CDB} = 180 - \\widehat{CAB} = 180 - \\widehat{PAB} = \\widehat{PAQ}$. Ainsi les triangles $DRP$ et $AQP$ sont isométriques.\n\nEn particulier $PQ = PR$, donc $P$ est sur la médiatrice de $(QR)$.\n\nLes triangles $DER$ et $AEP$ ont l'air d'être isométriques : prouvons-le. On a $DE = AE$ et $DR = AP$. De plus $\\widehat{EDR} = 180 - \\widehat{EDC} = \\widehat{CAE} = \\widehat{PAE}$. Ainsi les triangles $DER$ et $AEP$ sont isométriques et $EP = ER$.\n\nDe même les triangles $AEQ$ et $DEP$ sont isométriques (car $A, Q$ jouent un rôle symétrique à $D, R$), ainsi $EP = EQ$, donc $ER = EQ$: $E$ est sur la médiatrice de $(QR)$.\n\nAinsi la médiatrice de $(QR)$ est $(EP)$, donc $(EP)$ et $(QR)$ sont perpendiculaires.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 3$ un entier. Pour chaque couple de nombres premiers $p$ et $q$ tels que $p<q \\leqslant n$, Morgane a écrit la somme $p+q$ au tableau. Elle note ensuite $\\mathcal{P}(n)$ le produit de toutes ces sommes. Par exemple, $\\mathcal{P}(5)=(2+3) \\times (2+5) \\times (3+5)=280$.\n\nTrouver toutes les valeurs de $n \\geqslant 3$ pour lesquelles $n!$ divise $\\mathcal{P}(n)$.\n\nNote : Si deux sommes $p+q$ formées à partir de deux couples différents sont égales l'une à l'autre, Morgane les écrit toutes deux. Par exemple, si $n=13$, elle écrit bien les deux sommes $3+13$ et $5+11$.",
    "options": [],
    "answer": "7",
    "solution": "Solution:\n\nSoit $n$ une solution éventuelle, et soit $r$ le plus grand nombre premier tel que $r \\leqslant n$. Puisque $r$ divise $n!$, il divise $\\mathcal{P}(n)$, donc il existe deux nombres premiers $p$ et $q$ tels que $p<q \\leqslant n$ et $r$ divise $p+q$. Par maximalité de $r$, on sait que $p+q<2r$, de sorte que $p+q=r$. Puisque $q$ et $r$ sont impairs, $p$ est donc pair, et $p=2$.\n\nDe même, il existe deux nombres premiers $s$ et $t$ tels que $s<t \\leqslant n$ et $q$ divise $s+t$. Nous allons démontrer que $q-2$ est premier. En effet, si $t \\leqslant q$, alors $s+t=q$, donc $s=2$ et $t=q-2$. Sinon, $t=r=q+2$, donc $q<t+s<2q+2<3q$, de sorte que $t+s=2q$ et $s=q-2$.\n\nEn particulier, $3$ divise l'un des trois nombres premiers $q-2, q$ et $q+2$; il s'agit nécessairement de $q-2$, ce qui signifie que $q-2=3$ et que $q+2=r=7$. Ainsi, $n \\geqslant 7$, et $n!$ divise\n$$\n\\mathcal{P}(n)=(2+3) \\times (2+5) \\times (2+7) \\times (3+5) \\times (3+7) \\times (5+7)=2^{6} \\times 3^{3} \\times 5^{2} \\times 7\n$$\nPuisque $2 \\times 4 \\times 6 \\times 8=2^{7} \\times 3$ ne divise pas $\\mathcal{P}(n)$, on a nécessairement $n=7$.\n\nRéciproquement, $n=7$ est bien une solution, car $7!=2^{4} \\times 3^{2} \\times 5 \\times 7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABCDE$ un pentagone cyclique convexe tel que $AB = BD$. Soit $P$ le point d'intersection des droites $(EB)$ et $(AC)$. Soit $Q$ le point d'intersection des droites $(BC)$ et $(DE)$. Montrer que $(PQ)$ et $(AD)$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSur la figure, il semble que $EPCQ$ est un quadrilatère cyclique. On le montre :\n$$\n\\begin{aligned}\n\\widehat{QCP} & = 180^{\\circ} - \\widehat{PCB} \\\\\n& = 180^{\\circ} - \\widehat{ACB} \\\\\n& = 180^{\\circ} - \\widehat{ADB} \\text{ par angle inscrit } \\\\\n& = 180^{\\circ} - \\widehat{BAD} \\text{ car } AB = BD \\\\\n& = 180^{\\circ} - \\widehat{BED} \\\\\n& = 180^{\\circ} - \\widehat{PEQ} .\n\\end{aligned}\n$$\nAinsi on a montré que $\\widehat{QCP} + \\widehat{PEQ} = 180^{\\circ}$. Dès lors, par réciproque de l'angle inscrit, $EPCQ$ est cyclique, ce qu'on voulait démontrer.\n\nMaintenant, pour montrer que $(AD)$ et $(PQ)$ sont parallèles, on va essayer de trouver des angles correspondants. Or on a :\n$$\n\\begin{aligned}\n\\widehat{EQP} & = \\widehat{ECP} \\text{ par angle inscrit } \\\\\n& = \\widehat{ECA} \\\\\n& = \\widehat{EDA} \\text{ par angle inscrit. }\n\\end{aligned}\n$$\nComme $\\widehat{EQP} = \\widehat{EDA}$ et $Q, D, E$ sont alignés (dans cet ordre), par angle correspondant les droites $(PQ)$ et $(AD)$ sont bien parallèles.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\in \\mathbb{N}^{\\star}$. Montrer que si $2n+1$ et $3n+1$ sont des carrés parfaits, alors $5n+3$ n'est pas premier.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons que $2n+1$ et $3n+1$ sont des carrés parfaits : on dispose de $a, b$ deux entiers strictement positifs ($2n+1, 3n+1 > 0$) tels que $2n+1 = a^{2}$ et $3n+1 = b^{2}$. Ici, il s'agit de remarquer que $5n+3 = 4 \\times (2n+1) - (3n+1) = 4a^{2} - b^{2}$.\n\nPour trouver cette identité, on essaie d'exprimer $5$ à partir de $2$ et $3$ : il est naturel de prendre $5 = 2 + 3$ mais cela ne donne pas $5n+3$, en revanche on voit que $5 = 4 \\times 2 - 3$.\n\nOn peut donc écrire $5n+3 = (2a-b)(2a+b)$. Comme $a, b > 0$, $2a+b > 2a-b$, et $2a+b > 0$. Ceci force $2a-b > 0$ (sinon on aurait $5n+3 \\leqslant 0$, ce qui est exclu). Il suffit donc de montrer que $2a-b > 1$ pour conclure : on aura alors écrit $5n+3$ comme produit de deux entiers strictement plus grands que $1$, et il ne pourra pas être premier.\n\nMais si $2a = b+1$, $5n+3 = 2a+b = 4a-1$ donc $a^{2} = 2n+1 \\leqslant \\frac{5n+3}{2} < \\frac{4a-1}{2} < 2a$ donc $a < 2$. Mais $a=1$ donne $2n+1=1$ donc $n=0$, absurde car $n>0$. Ainsi $2a-b \\neq 1$.\n\nOn a donc bien montré que si $2n+1$ et $3n+1$ sont des carrés parfaits (avec $n \\in \\mathbb{N}^{\\star}$), alors $5n+3$ n'est pas un nombre premier.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $A, B, C, D$ quatre points situés dans cet ordre sur un cercle $\\omega$. Soit $P$ et $Q$ deux points situés sur la droite $(A B)$, de sorte que les points $Q, A, B, P$ soient alignés dans cet ordre, que le cercle circonscrit à $A D Q$ soit tangent à la droite $(A C)$, et que le cercle circonscrit à $B C P$ soit tangent à la droite $(B D)$. Soit $M$ et $N$ les milieux respectifs des segments $[B C]$ et $[A D]$.\n\nDémontrer que la tangente en $A$ au cercle circonscrit à $A N Q$, la tangente en $B$ au cercle circonscrit à $B M P$ et la droite $(C D)$ sont concourantes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSoit $X$ le point d'intersection éventuel dont on doit démontrer l'existence : nous allons en chercher des caractérisations alternatives. Pour ce faire, et au vu des nombreuses relations de tangence et de cocyclicité dont on dispose, une chasse aux angles s'impose.\n\nPar exemple, $(A Q, A D) = (A B, A D) = (C B, C D)$ et $(Q A, Q D) = (A C, A D) = (B C, B D)$. Il existe donc une similitude directe qui envoie $A Q D$ sur $C B D$. Celle-ci envoie également $N$ sur le milieu de $[C D]$, que l'on notera $R$. Mais alors\n$$\n(A B, A X) = (A Q, A X) = (N Q, N A) = (R B, R C) = (R B, R X)\n$$\nce qui signifie que les points $A, B, X$ et $R$ sont cocycliques.\n\nOn définit donc le point $\\hat{X}$ comme étant le second point d'intersection du cercle circonscrit à $A B R$ avec la droite $(C D)$, et en reprenant le calcul précédent, on remarque que\n$$\n(A Q, A \\hat{X}) = (A B, A \\hat{X}) = (R B, R \\hat{X}) = (R B, R C) = (N Q, N A)\n$$\nce qui signifie comme prévu que la droite $(A \\hat{X})$ est tangente au cercle circonscrit à $A N Q$. On démontre de même que $(B \\hat{X})$ est tangente au cercle circonscrit à $B M P$, ce qui conclut.\n\n\nSolution alternative $n^{\\circ} 1$\n\nUne fois n'est pas coutume, on peut utiliser brutalement des coordonnées cartésiennes. On identifie chaque point à ses coordonnées, puis on note $O$ et $O_{2}$ les centres des cercles circonscrits à $Q A D$ et à $Q A N$. On choisit en outre le repère de sorte que $A = (0, 0)$ et $B = (1, 0)$, puis on pose $C = (x, y)$ et $D = (u, v)$, $O = (s, t)$ et $O_{2} = (s_{2}, t_{2})$. Ainsi, $N = (u / 2, v / 2)$, $Q = (2s, 0)$ et $s_{2} = s$.\n\nLes égalités $O C^{2} = O A^{2} + A C^{2}$ et $O A^{2} = O D^{2}$ se réécrivent comme $s x + t y = 0$ et $2(s u + t v) = u^{2} + v^{2}$, de sorte que\n$$\ns = \\frac{u^{2} + v^{2}}{2 \\delta} y\n$$\noù l'on a posé $\\delta = u y - v x$. De même, l'égalité $O_{2} A^{2} = O_{2} N^{2}$ se réécrit comme $4(u s + v t_{2}) = u^{2} + v^{2}$, de sorte que\n$$\nt_{2} = \\frac{u^{2} + v^{2} - 4 u s}{4 v} .\n$$\nPar ailleurs, notons $Z = (z, w)$ le point de $(C D)$ pour lequel $(A Z)$ est tangente au cercle circonscrit à $A Q N$. L'alignement de $Z$ avec $C$ et $D$ indique qu'il existe un réel $\\lambda$ tel que $Z = \\lambda C + (1 - \\lambda) D$; l'orthogonalité de $(A X)$ avec $\\left(A O_{2}\\right)$ indique que $\\overrightarrow{A O_{2}} \\cdot \\overrightarrow{A Z} = s z + t_{2} w = 0$. On en déduit que $s(\\lambda x + u - \\lambda u) + t_{2}(\\lambda y + v - \\lambda v) = 0$, c'est-à-dire\n$$\n\\begin{aligned}\n\\lambda & = \\frac{s u + t_{2} v}{s(u - x) + t_{2}(v - y)} \\\\\n& = \\frac{4 s u v + (u^{2} + v^{2} - 4 u s) v}{4 s(u - x) v + (u^{2} + v^{2} - 4 u s)(v - y)} \\\\\n& = \\frac{(u^{2} + v^{2}) v}{(u^{2} + v^{2})(v - y) + 4(u y - v x) s} \\\\\n& = \\frac{(u^{2} + v^{2}) v}{(u^{2} + v^{2})(v - y) + 2(u^{2} + v^{2}) y} \\\\\n& = \\frac{v}{v + y} .\n\\end{aligned}\n$$\n\nPour des raisons de symétrie, le point $Z'$ de $(C D)$ pour lequel $\\left(B Z'\\right)$ est tangente au cercle circonscrit à $B P M$ est lui aussi décrit par la relation $Z' = \\mu D + (1 - \\mu) C$, avec $\\mu = y / (v + y)$. Mais alors $\\lambda + \\mu = 1$, ce qui signifie comme prévu que $Z = Z'$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30901,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAu moins $n\\left(\\sqrt{n}+\\frac{1}{2}\\right)$ carrés d'un échiquier $n \\times n$ ont été marqués. Montrer qu'il existe quatre cases marquées qui forment les coins d'un rectangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn essaye d'utiliser le principe des tiroirs. S'il y a deux cases marquées aux coordonnées $(i, j)$ et $(i, k)$, on met une chaussette dans le tiroir étiqueté $(j, k)$. Il y a $\\frac{n(n-1)}{2}$ tiroirs possibles. S'il y a deux chaussettes dans le même tiroir $(j, k)$, c'est qu'il existe $i, i'$ tels que $(i, j),(i, k),(i', j),(i', k)$ sont toutes marquées, et on a bien trouvé quatre cases marquées qui forment les coins d'un rectangle. Il ne reste plus qu'à montrer qu'il y a strictement plus de chaussettes que de tiroirs. On note $d_{i}$ le nombre de cases marquées dans la $i$-ème colonne. On a :\n$$\n\\sum_{i=1}^{n} d_{i} \\geqslant n\\left(\\sqrt{n}+\\frac{1}{2}\\right)\n$$\nLe nombre de chaussettes vaut :\n$$\n\\sum_{i=1}^{n} \\frac{d_{i}^{2}-d_{i}}{2}\n$$\nEn utilisant Cauchy-Schwarz, on a :\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n} \\frac{d_{i}^{2}-d_{i}}{2} & \\geqslant \\frac{1}{2}\\left(\\frac{\\left(\\sum_{i=1}^{n} d_{i}\\right)^{2}}{n}-\\sum_{i=1}^{n} d_{i}\\right) \\\\\n& \\geqslant \\frac{1}{2 n}\\left(n\\left(\\sqrt{n}+\\frac{1}{2}\\right)\\right)\\left(n\\left(\\sqrt{n}+\\frac{1}{2}\\right)-n\\right) \\\\\n& =\\frac{1}{2} n\\left(\\sqrt{n}+\\frac{1}{2}\\right)\\left(\\sqrt{n}-\\frac{1}{2}\\right) \\\\\n& =\\frac{1}{2} n\\left(n-\\frac{1}{4}\\right)\n\\end{aligned}\n$$\nCe qui est effectivement strictement plus grand que le nombre de tiroirs.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier. Morgane dispose initialement de $n$ piles dont chacune contient une pièce. Elle s'autorise ensuite des opérations de la forme suivante : elle choisit deux piles, prélève autant de pièces de la première pile que de la deuxième, et forme une nouvelle pile avec les pièces qu'elle a prélevées.\n\nDéterminer, en fonction de $n$, le plus petit nombre de piles non vides que Morgane pourra obtenir à partir de telles opérations.",
    "options": [],
    "answer": "The minimal number is 1 if the starting number is a power of two, and 2 otherwise.",
    "solution": "Solution:\n\nSi $n$ est une puissance de $2$, Morgane peut se débrouiller pour n'obtenir qu'une seule pile, en procédant comme suit : elle fusionne les $n$ piles de taille $1$ en $n / 2$ piles de taille $2$, puis fusionne ces $n / 2$ piles en $n / 4$ piles de taille $4$, et ainsi de suite.\n\nRéciproquement, lorsque Morgane extrait $k$ pièces de deux piles contenant originellement $a$ et $b$ pièces pour former trois piles (éventuellement vides) de $a-k$, $b-k$ et $2k$ pièces, on constate que $\\operatorname{PGCD}(a-k, b-k, 2k)$ divise $2a$ et $2b$, donc divise $2\\operatorname{PGCD}(a, b)$. Par conséquent, une récurrence immédiate montre que, si Morgane dispose de $k$ piles de tailles $t_{1}, t_{2}, \\ldots, t_{k}$, le nombre $\\operatorname{PGCD}\\left(t_{1}, t_{2}, \\ldots, t_{k}\\right)$ est une puissance de $2$. En particulier, si Morgane parvient à n'avoir plus qu'une seule pile, qui contiendra $n$ pièces, c'est donc que $n$ est une puissance de $2$.\n\nEnfin, voici comment, pour toute valeur de $n$, Morgane peut procéder pour n'obtenir que deux piles. Elle écrit $n$ comme somme de puissances de $2$ croissantes, c'est-à-dire de nombres $2^{a_{1}} < 2^{a_{2}} < \\cdots < 2^{a_{k}}$ tels que $n = 2^{a_{1}} + 2^{a_{2}} + \\cdots + 2^{a_{k}}$. Pour chaque $i \\leqslant k$, et en procédant comme dans le cas où $n$ est une puissance de $2$, Morgane regroupe ensuite $2^{a_{i}}$ de ses pièces en une seule pile, obtenant ainsi $k$ piles.\n\nMorgane distingue alors sa pile de taille $2^{a_{1}}$, qu'elle appelle cible, et sa pile de taille $2^{a_{k}}$, qu'elle appelle réserve. Les $k-2$ autres piles sont vues comme des piles normales. Puis Morgane s'attelle à doubler la taille de sa pile cible tout en s'assurant qu'elle ne sera jamais plus grande qu'une pile normale, en procédant comme suit :\n\n$\\triangleright$ si la pile cible est de même taille qu'une pile normale, Morgane fusionne ces deux piles en une nouvelle pile cible ; une pile normale disparaît au cours de cette opération ;\n\n$\\triangleright$ sinon, la pile cible est de taille $2^{a}$ strictement plus petite que chaque pile normale, et Morgane prélève alors $2^{a}$ pièces de la réserve pour les intégrer à la pile cible, qui est désormais de taille $2^{a+1}$.\n\nAprès $\\ell$ telles étapes, la pile cible est désormais de taille $2^{a_{1}+\\ell}$, et elle a absorbé les piles de tailles $2^{a_{i}} < 2^{a_{1}+\\ell}$, ainsi qu'au plus $2^{a_{k-1}}$ pièces issues de la réserve, qui n'est donc toujours pas vide. Ainsi, après $a_{k-1}+1-a_{1}$ étapes, la pile cible est désormais de taille $2^{a_{k-1}+1}$, et elle a absorbé toutes les piles normales, ce qui fait que l'on n'a plus que deux piles.\n\nEn conclusion, le nombre minimal de piles auquel pourra aboutir Morgane vaut $1$ si $n$ est une puissance de $2$, et il vaut $2$ sinon.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30903,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEst-il possible de trouver un bloc de 1000 nombres entiers strictement positifs consécutifs qui contient exactement 5 nombres premiers?",
    "options": [],
    "answer": "Yes",
    "solution": "Solution:\n\nA première vue, il semble difficile de garantir exactement 5 nombre premier dans un bloc de 1000 entiers consécutifs : aucun théorème d'arithmétique élémentaire permet de s'assurer d'avoir des nombres premiers rapprochés, mais pas d'autres nombres premier entre eux. On peut donc essayer de se demander combien de nombres premiers on peut trouver dans un bloc de 1000 nombres consécutifs.\n\nDéjà, il est relativement connu qu'on peut trouver 1000 nombres consécutifs dont aucun n'est premier. En effet, $1001!+2, \\ldots, 1001!+1001$ ne sont pas premiers : si $2 \\leqslant i \\leqslant 1001, 1001!+i$ est divisible par $i$ et strictement plus grand que $i$, donc il n'est pas premier.\n\nOn peut aussi remarquer qu'entre 1 et 1000 il y a beaucoup de nombres premiers : en tout cas strictement plus que 5, vu que $2, 3, 5, 7, 11, 13$ sont premiers.\n\nNotons $P(n)$ le nombre de nombres premiers dans l'ensemble $\\{n, \\ldots, n+999\\}$. L'étape naturelle après nos deux remarques précédentes est de se demander comment évolue le nombre $P(n)$ lorsque $n$ parcourt $\\mathbb{N}^*$. Pour passer de $\\{n, \\ldots, n+999\\}$ à $\\{n+1, \\ldots, n+1+999\\}$, on rajoute $n+1000$ et on enlève $n$. Ainsi on ajoute au plus un nombre premier, et on enlève au plus un nombre premier : on a donc\n$$\nP(n+1) - P(n) \\in \\{-1, 0, 1\\}\n$$\nAinsi $P$ ne varie que de 1 en 1, et on sait que $P(1) > 5 > P(1001!+2)$. Il semble assez intuitif que cela implique qu'il existe un entier $k$ tel que $P(k) = 5$, nous allons prouver cela.\n\nSoit $k$ le premier entier tel que $P(k) \\leqslant 5$. On sait que $k > 1$, donc $P(k-1) > 5$ par minimalité. Ainsi, on a forcément $P(k) = P(k-1) - 1$ (pour avoir $P(k) \\leqslant 5$), donc $P(k) > 4$. Ainsi $P(k) \\geqslant 5$, or par définition $P(k) \\leqslant 5$, donc $P(k) = 5$.\n\nAinsi la réponse à l'énoncé est oui : il existe bien un bloc de 1000 entiers consécutifs avec exactement cinq entiers premiers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a \\geqslant 2$ et $d \\geqslant 2$ deux entiers premiers entre eux. On pose $x_{1}=1$; puis, pour tout entier $k \\geqslant 1$, on pose $x_{k+1}=x_{k} / a$ si $a$ divise $x_{k}$, et $x_{k+1}=x_{k}+d$ sinon.\nTrouver, en fonction de $a$ et de $d$, l'entier $\\ell$ maximal pour lequel $a^{\\ell}$ divise l'un des termes $x_{1}, x_{2}, x_{3}, \\ldots$",
    "options": [],
    "answer": "ℓ = ⌊log_a d⌋ + 1 (equivalently, ℓ is the unique integer satisfying d < a^ℓ < a d)",
    "solution": "Solution:\n\nPour tout entier $n \\geqslant 1$, soit $f(n)$ le plus petit entier tel que $a f(n) \\geqslant n$ et $a f(n) \\equiv n \\pmod{d}$, et soit $\\delta(n)=(a f(n)-n) / d$. Par construction, si l'on dispose d'un terme $x_{k}=n$ de la suite, les termes suivants de la suite seront $x_{k+1}=n+d, x_{k+2}=n+2 d, \\ldots$, $x_{k+\\delta(n)}=n+\\delta(n) d=a f(n)$, puis $x_{k+\\delta(n)+1}=f(n)$.\n\nEn outre, aucun des termes $x_{k+1}, \\ldots, x_{k+\\delta(n)-1}$ n'est divisible par $a$, mais $x_{k+\\delta(n)}=a f(n)$ l'est. Par conséquent, si l'on pose $y_{0}=1$ puis $y_{k+1}=f\\left(y_{k}\\right)$ pour tout $k \\geqslant 0$, il s'agit de trouver l'entier $\\ell$ maximal pour lequel $a^{\\ell-1}$ divise l'un des termes $y_{1}, y_{2}, y_{3}, \\ldots$\n\nOr, par construction, on sait que $\\delta(n) \\leqslant a-1$. Par conséquent, si $n \\leqslant d$, on sait également que $f(n) \\leqslant (n+(a-1) d) / a \\leqslant d$. Une récurrence immédiate démontre donc que $1 \\leqslant y_{k} \\leqslant d$ pour tout entier $k \\geqslant 0$.\n\nEnfin, on remarque que $f(n) \\equiv n / a \\pmod{d}$, ce qui signifie, pour tout entier $k \\geqslant 0$, que $y_{k}$ est l'unique entier compris entre 1 et $d$ et tel que $y_{k} \\equiv a^{-k} \\pmod{d}$. En particulier, si l'on note $\\omega$ l'ordre multiplicatif de $a$ modulo $d$, et si l'on pose $\\kappa=\\left\\lfloor\\log _{a}(d)\\right\\rfloor$, on sait que $y_{\\omega-\\kappa}=a^{\\kappa}$. Réciproquement, puisque $a^{\\kappa+1}>d$, nul terme $y_{k}$ n'est divisible par $a^{\\kappa+1}$.\n\nOn en conclut que $\\ell=\\kappa+1$, c'est-à-dire que $\\ell$ est l'unique entier pour lequel $d<a^{\\ell}<a d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30905,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMontrer que pour tous réels $a$, $b$, $c$ strictement positifs:\n$$\n\\frac{b c}{a^{2}+2 b c}+\\frac{c a}{b^{2}+2 c a}+\\frac{a b}{c^{2}+2 a b} \\leqslant 1 \\leqslant \\frac{a^{2}}{a^{2}+2 b c}+\\frac{b^{2}}{b^{2}+2 c a}+\\frac{c^{2}}{c^{2}+2 a b}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nCommençons par résoudre une des inégalités. Dans ce problème, l'inégalité la plus simple à étudier est celle de droite. On applique l'inégalité arithmético-géométrique sur les dénominateurs pour avoir $2 b c \\leqslant b^{2}+c^{2}$ par exemple, ce qui donne\n$$\n\\frac{a^{2}}{a^{2}+2 b c}+\\frac{b^{2}}{b^{2}+2 c a}+\\frac{c^{2}}{c^{2}+2 a b} \\geqslant \\frac{a^{2}}{a^{2}+b^{2}+c^{2}}+\\frac{b^{2}}{a^{2}+b^{2}+c^{2}}+\\frac{c^{2}}{a^{2}+b^{2}+c^{2}}=1 .\n$$\nPour le côté gauche, on remarque que celui-ci ressemble beaucoup au côté droit. Appelons $G$ et $D$ les côtés gauche et droit de l'équation respectivement. On a alors\n$$\n\\begin{gathered}\nG+2 \\cdot D=\\frac{a^{2}}{a^{2}+2 b c}+\\frac{b^{2}}{b^{2}+2 c a}+\\frac{c^{2}}{c^{2}+2 a b}+\\frac{2 b c}{a^{2}+2 b c}+\\frac{2 c a}{b^{2}+2 c a}+\\frac{2 a b}{c^{2}+2 a b} . \\\\\n\\frac{a^{2}+2 b c}{a^{2}+2 b c}+\\frac{b^{2}+2 c a}{b^{2}+2 c a}+\\frac{c^{2}+2 a b}{c^{2}+2 a b}=3 .\n\\end{gathered}\n$$\nOn a donc\n$$\nG=\\frac{1}{2}(3-D) \\leqslant \\frac{1}{2}(3-1)=1\n$$\npar l'inégalité de droite que l'on a montré précédemment.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30906,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les nombres premiers $p, q$ vérifiant $p^{5}+p^{3}+2=q^{2}-q$.",
    "options": [],
    "answer": "[(2, 7), (3, 17)]",
    "solution": "Solution:\n\nCe qu'on connaît pour des nombres premiers, c'est leurs propriétés de divisibilité. Une expression développée n'est donc pas très pratique pour étudier cela : on va d'abord chercher à factoriser.\n\nRemarquons que $p^{5}+p^{3}+2=p^{3}\\left(p^{2}+1\\right)+2$. On veut résoudre $p^{3}\\left(p^{2}+1\\right)=q^{2}-q-2$. Mais on reconnaît alors $q^{2}-q-2=(q-2)(q+1)$ (si on ne le voit pas directement, on peut le trouver en résolvant $q^{2}-q-2=0$ avec la méthode classique utilisant le discriminant).\n\nOn veut donc résoudre $p^{3}\\left(p^{2}+1\\right)=(q-2)(q+1)$. En particulier $p^{3} \\mid (q-2)(q+1)$. Alors $\\operatorname{pgcd}(q-2, q+1)=\\operatorname{pgcd}(q-2, q+1-(q-2))=\\operatorname{pgcd}(q-2,3)$ vaut donc $1$ ou $3$. Cela suggère d'étudier d'abord le cas $p=3$ : on obtient $(q-2)(q+1)=270$, et en résolvant $q^{2}-q-2=270$ on trouve $q=17$ (qui est bien un nombre premier) et $q=-16$ (qui n'est pas solution car ce n'est pas un nombre premier). Donc lorsque $p=3$, il y a une unique solution $(p, q)=(3,17)$.\n\nOn peut donc supposer $p \\neq 3$. En particulier, il ne divise pas $\\operatorname{pgcd}(q-1, q+2)$, donc au moins un des nombres $q+1$ et $q-2$ est premier avec $p$, donc avec $p^{3}$. Or $p^{3} \\mid (q+1)(q-2)$ : donc d'après le lemme de Gauss, $q+1$ ou $q-2$ est divisible par $p^{3}$. Or, $p^{3}$ ne peut pas diviser $q-2$ parce que sinon, on aurait $(q-2)(q+1) \\geqslant p^{3} p^{3}>p^{3}\\left(p^{2}+1\\right)$, absurde. Ainsi $p^{3} \\mid q+1$, et donc $q-2 \\mid p^{2}+1$.\n\nMais les divisibilités qu'on vient d'obtenir montrent que $p^{3} \\leqslant q+1=(q-2)+3 \\leqslant p^{2}+1+3=p^{2}+4$, et donc $p^{2}(p-1) \\leqslant 4$. Or $p^{2}(p-1)$ est strictement croissante en $p$, et en $p=2$ on a égalité, donc le seul cas possible est $p=2$ (autrement dit, pour $p>2$, on a $p^{2}(p-1)>4$ ).\n\nLorsque $p=2$, l'équation devient $q^{2}-q-2=40$, en résolvant on trouve $q=7$ (qui convient car 7 est premier) et $q=-6$ (qui ne convient pas car n'est pas un nombre premier).\n\nFinalement, il y a deux solutions qui sont $(\\mathbf{p}, \\mathbf{q})=(2,7)$ et $(\\mathbf{p}, \\mathbf{q})=(3,17)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle isocèle en $A$, $O$ le centre de son cercle circonscrit et $I$ le centre de son cercle inscrit. La parallèle à $(AB)$ passant par $I$ coupe $(AC)$ en $D$. Montrer que $(CI)$ et $(DO)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPremièrement, comme $AB = AC$, $A$, $O$, $I$ sont alignés et la droite $(AI)$ coupe $[BC]$ en son milieu $M$. On pose $P$ l'intersection des droites $(DO)$ et $(CI)$.\n\nPar chasse aux angles, on montre que $I$, $O$, $D$, $C$ sont cocycliques, en montrant que $\\widehat{IOC} = \\widehat{IDC}$. En effet :\n$$\n\\begin{aligned}\n\\widehat{IOC} & = \\widehat{MOC} \\\\\n& = \\frac{1}{2} \\widehat{BOC} \\text{ car } OB = OC \\\\\n& = \\widehat{BAC} \\text{ par angle au centre } \\\\\n& = \\widehat{IDC} \\text{ car } (BA) // (ID)\n\\end{aligned}\n$$\nDonc par réciproque du théorème de l'angle inscrit, $I$, $O$, $D$, $C$ sont cocycliques.\n\nMontrons à présent que $\\widehat{CPD} = 90^\\circ$. Pour cela, on va montrer que $\\widehat{PDC} + \\widehat{DCP} = 90^\\circ$ (puisque $\\widehat{CPD} = 180^\\circ - \\widehat{PDC} - \\widehat{DCP}$). On a :\n$$\n\\begin{aligned}\n\\widehat{PDC} + \\widehat{DCP} & = \\widehat{ODC} + \\widehat{DCI} \\\\\n& = \\left(180^\\circ - \\widehat{CIO}\\right) + \\widehat{ICB} \\text{ par cocyclicité et bissectrice } \\\\\n& = \\widehat{CIM} + \\widehat{ICM} \\\\\n& = 90^\\circ \\text{ car } (AI) \\perp (BC).\n\\end{aligned}\n$$\nDonc $\\widehat{CPD} = 90^\\circ$, ce qui prouve bien que les droites $(CI)$ et $(DO)$ sont perpendiculaires.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $x$, $y$, $z$ des réels strictement positifs tels que $x y + y z + z x = 1$. Montrer que\n$$\n2\\left(x^{2}+y^{2}+z^{2}\\right)+\\frac{4}{3}\\left(\\frac{1}{x^{2}+1}+\\frac{1}{y^{2}+1}+\\frac{1}{z^{2}+1}\\right) \\geqslant 5\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans cet exercice, on aimerait bien appliquer l'inégalité arithmético-géométrique de façon à obtenir une minoration du type $\\left(x^{2}+1\\right)+\\frac{1}{x^{2}+1} \\geqslant 2$. Mais si on s'y prend mal, on n'arrive pas à conclure. En effet, on aurait pu s'en douter en étudiant le cas d'égalité de l'équation. Au vu de la condition $x y + y z + z x = 1$, on se doute que l'égalité va se produire pour $x = y = z = \\frac{1}{\\sqrt{3}}$. En introduisant ces valeurs dans l'équation, on trouve bien que l'on a égalité, mais ceci n'est pas un cas d'égalité de l'inégalité arithmético-géométrique plus tôt.\nDans le cas d'égalité $x = y = z = \\frac{1}{\\sqrt{3}}$, on a $x^{2} + 1 = \\frac{4}{3}$ et $\\frac{1}{x^{2}+1} = \\frac{3}{4}$. On applique donc l'inégalité arithmético-géométrique avec des coefficients appropriés pour avoir l'égalité des deux termes dans ce cas :\n$$\n\\frac{3}{4}\\left(x^{2}+1\\right)+\\frac{4}{3}\\left(\\frac{1}{x^{2}+1}\\right) \\geqslant 2\n$$\nEn sommant ces trois équations, on a\n$$\n\\frac{3}{4}\\left(x^{2}+y^{2}+z^{2}+3\\right)+\\frac{4}{3}\\left(\\frac{1}{x^{2}+1}+\\frac{1}{y^{2}+1}+\\frac{1}{z^{2}+1}\\right) \\geqslant 6\n$$\nqui donne\n$$\n\\frac{3}{4}\\left(x^{2}+y^{2}+z^{2}\\right)+\\frac{4}{3}\\left(\\frac{1}{x^{2}+1}+\\frac{1}{y^{2}+1}+\\frac{1}{z^{2}+1}\\right) \\geqslant \\frac{15}{4}\n$$\nPar rapport à l'inégalité de l'énoncé, il nous reste un terme $\\frac{5}{4}\\left(x^{2}+y^{2}+z^{2}\\right)$ à minorer. Mais on a, par l'inégalité arithmético-géométrique,\n$$\n\\frac{5}{4}\\left(x^{2}+y^{2}+z^{2}\\right) \\geqslant \\frac{5}{4}(x y + y z + z x) = \\frac{5}{4}\n$$\nEn ajoutant cette inégalité à l'inégalité obtenue précédemment, on obtient bien\n$$\n2\\left(x^{2}+y^{2}+z^{2}\\right)+\\frac{4}{3}\\left(\\frac{1}{x^{2}+1}+\\frac{1}{y^{2}+1}+\\frac{1}{z^{2}+1}\\right) \\geqslant \\frac{15}{4}+\\frac{5}{4}=5\n$$\nSolution:\n\nPosons $s = x^{2} + y^{2} + z^{2}$. Notons que par l'inégalité des mauvais élèves,\n$$\n\\frac{4}{3}\\left(\\frac{1}{x^{2}+1}+\\frac{1}{y^{2}+1}+\\frac{1}{z^{2}+1}\\right) \\geqslant \\frac{4}{3} \\frac{(1+1+1)^{2}}{x^{2}+y^{2}+z^{2}+3} = \\frac{12}{s+3}\n$$\nAinsi il suffit de montrer que $2s + \\frac{12}{s+3} \\geqslant 5$. En multipliant des deux côtés par $s+3$ qui est positif, il suffit de montrer que $2s(s+3) + 12 \\geqslant 5(s+3)$. Ceci est équivalent à $2s^{2} + 6s + 12 \\geqslant 5s + 15$ et donc à $2s^{2} + s \\geqslant 3$. Or par le lemme du tourniquet, $s \\geqslant x y + y z + x z = 1$, donc $2s^{2} + s \\geqslant 3$. On a bien le résultat voulu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLucile a écrit au tableau $s$ 2023-uplets d'entiers. Elle s'autorise alors des opérations de la forme suivante : elle choisit deux 2023-uplets $\\mathbf{u}=\\left(u_{1}, u_{2}, \\ldots, u_{2023}\\right)$ et $\\mathbf{v}=\\left(v_{1}, v_{2}, \\ldots, v_{2023}\\right)$, non nécessairement distincts, parmi ceux qu'elle a déjà écrits, puis elle écrit également au tableau les deux 2023-uplets $\\mathbf{u}+\\mathbf{v}=\\left(u_{1}+v_{1}, \\ldots, u_{2023}+v_{2023}\\right)$ et $\\max (\\mathbf{u}, \\mathbf{v})=\\left(\\max \\left(u_{1}, v_{1}\\right), \\ldots, \\max \\left(u_{2023}, v_{2023}\\right)\\right)$.\nQuelles sont les valeurs de $s$ pour lesquelles, si Lucile choisit judicieusement les $s$ 2023-uplets qu'elle a écrits initialement, et en répétant les opérations ci-dessus, elle pourra écrire n'importe quel 2023-uplet d'entiers?",
    "options": [],
    "answer": "all integers s ≥ 3",
    "solution": "Solution:\n\nTout d'abord, si $s=1$, Lucile a écrit un seul vecteur $\\mathbf{u}$. Tout vecteur $\\mathbf{x}$ qu'elle pourra ensuite écrire aura des coordonnées de même signe que $\\mathbf{u}$, et elle ne pourra donc pas écrire tous les vecteurs possibles.\n\nEnsuite, si $s=2$, soit $\\mathbf{u}$ et $\\mathbf{v}$ les deux vecteurs que Lucile a écrits initialement. Si deux coordonnées $u_{i}$ et $v_{i}$ sont de même signe, tout vecteur $x$ que Lucile écrira ensuite aura aussi une coordonnée de même signe que $u_{i}$ et $v_{i}$. Par conséquent, pour tout entier $i \\leqslant 2023$, les coordonnées $u_{i}$ et $v_{i}$ sont de signes opposés. Quitte à échanger les vecteurs $\\mathbf{u}$ et $\\mathbf{v}$ ainsi que les coordonnées en positions 1 à 3, on suppose donc que $u_{1}$ et $u_{2}$ sont strictement positifs, et que $u_{1} v_{2} \\leqslant u_{2} v_{1}$.\n\nOn montre alors par récurrence que tout vecteur $x$ que Lucile écrira ensuite satisfera également l'inégalité $u_{1} x_{2} \\leqslant u_{2} x_{1}$. En effet, à partir de deux vecteurs $\\mathbf{x}$ et $\\mathbf{y}$ satisfaisant déjà cette inégalité, Lucile peut former deux vecteurs :\n\n$\\triangleright \\mathbf{z}=\\mathbf{x}+\\mathbf{y}$, pour lequel $u_{1} z_{2}=u_{1} x_{2}+u_{1} y_{2} \\leqslant u_{2} x_{1}+u_{2} y_{1}=u_{2} z_{1}$ ;\n\n$\\triangleright \\mathbf{z}=\\max (\\mathbf{x}, \\mathbf{y})$, pour lequel $u_{1} z_{2}=\\max \\left(u_{1} x_{2}, u_{1} y_{2}\\right) \\leqslant \\max \\left(u_{2} x_{1}, u_{2} y_{1}\\right)=u_{2} z_{1}$.\n\nEn particulier, Lucile ne pourra jamais écrire de vecteurs dont les deux premières coordonnées sont 0 et 1.\n\nEnfin, si $s \\geqslant 3$, Lucile peut écrire les vecteurs $\\mathbf{u}, \\mathbf{v}$ et $\\mathbf{w}$ tels que $u_{i}=-1, v_{i}=i$ et $w_{i}=-i^{2}$, puis ajouter $s-3$ vecteurs inutiles. À partir de ces vecteurs, Lucile peut alors créer successivement les vecteurs suivants, l'entier $k$ étant un paramètre que l'on fait librement varier entre 1 et 2023 :\n\n$\\triangleright \\mathbf{a}^{(k)}=\\left(k^{2}-1\\right) \\mathbf{u}+2 k \\mathbf{v}+\\mathbf{w}$, pour lequel $a_{i}^{(k)}=1-k^{2}+2 k i-i^{2}=1-(i-k)^{2}$ ;\n\n$\\triangleright \\mathbf{b}^{(k)}=\\mathbf{a}^{(k)}+\\mathbf{u}$, pour lequel $b_{i}^{(k)}=-(i-k)^{2}$;\n\n$\\triangleright \\mathbf{0}=\\max \\left\\{\\mathbf{b}^{(k)}: 1 \\leqslant k \\leqslant 2023\\right\\}$, pour lequel $\\mathbf{0}_{i}=0$;\n\n$\\triangleright \\mathbf{c}^{(k)}=\\max \\left\\{\\mathbf{a}^{(k)}, \\mathbf{0}\\right\\}$, pour lequel $c_{i}^{(k)}=1$ si $i=k$, et $c_{i}^{(k)}=0$ sinon.\n\nDès lors, Lucile peut écrire tout vecteur $x$, puisqu'elle n'a qu'à l'exprimer comme la somme\n$$\n\\mathbf{x}=|\\mathbf{x}| \\mathbf{u}+\\sum_{k=1}^{2023}\\left(|\\mathbf{x}|+x_{k}\\right) \\mathbf{c}^{(k)}\n$$\nou l'on a posé $|\\mathbf{x}|=1+\\max \\left\\{\\left|x_{k}\\right|: 1 \\leqslant k \\leqslant 2023\\right\\}$.\n\nEn conclusion, les valeurs de $s$ recherchées sont les entiers $s \\geqslant 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier. Morgane écrit au tableau, en base 10, les nombres $2023, 2023 \\times 2, \\ldots, 2023 \\times n$. Pour tout chiffre $c$ compris entre 1 et 9, elle note alors $\\mathrm{d}_{c}(n)$ le nombre d'apparitions du chiffre $c$ sur le tableau. Par exemple, si $n=3$, elle écrit les nombres $2023$, $4046$ et $6069$, donc $\\mathrm{d}_{1}(3)=\\mathrm{d}_{5}(3)=\\mathrm{d}_{7}(3)=\\mathrm{d}_{8}(3)=0$, $\\mathrm{d}_{3}(3)=\\mathrm{d}_{9}(3)=1$, $\\mathrm{d}_{2}(3)=\\mathrm{d}_{4}(3)=2$ et $\\mathrm{d}_{6}(3)=3$; ces neuf nombres prennent donc exactement quatre valeurs.\n\nDémontrer qu'il existe une infinité d'entiers $n \\geqslant 1$ pour lesquels les neuf nombres $\\mathrm{d}_{1}(n), \\mathrm{d}_{2}(n), \\ldots, \\mathrm{d}_{9}(n)$ prennent exactement deux valeurs.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $k$ un multiple de $\\varphi(2023)$, soit $\\ell = (10^{k} - 1) / 2023$. Nous allons démontrer que $\\ell$ est une solution. Pour ce faire, on introduit l'ensemble\n$$\n\\Omega = \\{2023 n : 1 \\leqslant n \\leqslant \\ell\\}\n$$\nLes éléments de $\\Omega$ décrivent précisément l'ensemble des résidus modulo $10^{k} - 1$ qui sont des multiples de $2023$, chaque résidu étant obtenu exactement une fois. Par conséquent, l'application $\\phi : x \\mapsto 10x$, qui définit une permutation de $\\mathbb{Z} / (10^{k} - 1) \\mathbb{Z}$, définit aussi une permutation de l'ensemble des résidus qui sont des multiples de $2023$, et induit donc une permutation de $\\Omega$ lui-même. Concrètement, si l'on écrit les éléments de $\\Omega$ en base 10 avec $k$ chiffres (quitte à laisser des 0 en apparence inutiles sur la gauche), $\\phi$ envoie $\\overline{a_{k-1} a_{k-2} \\ldots a_{1} a_{0}}^{10}$ sur $\\overline{a_{k-2} a_{k-3} \\ldots a_{1} a_{0} a_{k-1}}^{10}$.\n\nPour tout chiffre $c$ et tout entier $i$ tel que $0 \\leqslant i \\leqslant k-1$, on note $\\Omega(c, i)$ l'ensemble des éléments de $\\Omega$ dont le chiffre $a_{i}$, de poids $10^{i}$, est égal à $c$; on pose abusivement $\\Omega(c, k) = \\Omega(c, 0)$. Lorsque $0 \\leqslant i \\leqslant k-1$, la bijection $\\phi$ induit une bijection de $\\Omega(c, i)$ vers $\\Omega(c, i+1)$. Ainsi, les ensembles $\\Omega(c, 0), \\Omega(c, 1), \\ldots, \\Omega(c, k-1)$ sont tous de même cardinal, et\n$$\n\\mathrm{d}_{c}(\\ell) = |\\Omega(c, 0)| + |\\Omega(c, 1)| + \\cdots + |\\Omega(c, k-1)| = k |\\Omega(c, 0)|.\n$$\nOr, $\\Omega(c, 0)$ est formé des nombres $2023 n$ pour lesquels $1 \\leqslant n \\leqslant \\ell$ et $c \\equiv 3n \\pmod{10}$. Puisque $\\ell \\equiv -1/3 \\equiv 3 \\pmod{10}$, on en déduit que $|\\Omega(c, 0)|$ vaut $(\\ell + 7)/10$ si $c$ vaut $3, 6$ ou $9$, et vaut $(\\ell - 3)/10$ sinon. Cela signifie que les cardinaux $|\\Omega(c, 0)|$ prennent exactement deux valeurs lorsque $c$ varie de $1$ à $9$, et donc que les entiers $\\mathrm{d}_{c}(\\ell)$ prennent eux aussi deux valeurs.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTrouver tous les entiers $n \\geqslant 2$ pour lesquels il existe un nombre réel $r>0$ et $n$ nombres réels $a_{1}<a_{2}<\\cdots<a_{n}$ tels que les $n(n-1) / 2$ différences $a_{j}-a_{i}$ obtenues lorsque $1 \\leqslant i<j \\leqslant n$ sont égales, à l'ordre près, aux nombres $r^{1}, r^{2}, \\ldots, r^{n(n-1) / 2}$.",
    "options": [],
    "answer": "2, 3, 4",
    "solution": "Solution:\n\nDans un tel problème, il est indispensable de commencer par étudier les petits cas.\n\n$\\triangleright$ Lorsque $n=2$, il suffit de choisir $a_{1}=0$ et $a_{2}=r=1$. Par conséquent, $n=2$ convient.\n\n$\\triangleright$ Lorsque $n=3$, il suffit de choisir $a_{1}=0, a_{2}=r=\\phi$ et $a_{3}=\\phi^{3}$, où $\\phi=(1+\\sqrt{5}) / 2$ est le nombre d'or et vérifie l'égalité $\\phi^{2}=1+\\phi$. En effet, on a alors $a_{2}-a_{1}=\\phi$, $a_{3}-a_{2}=\\phi(\\phi^{2}-1)=\\phi^{2}$ et $a_{3}-a_{1}=\\phi^{3}$, comme illustré ci-dessous à gauche. Par conséquent, $n=3$ convient.\n\n$\\triangleright$ Lorsque $n=4$, il suffit de choisir $a_{1}=0, a_{2}=\\lambda, a_{3}=\\lambda^{4}, a_{4}=\\lambda^{6}$ et $r=\\lambda$, où $\\lambda$ est une racine positive du polynôme $P(X)=X^{3}-X-1$, pour lequel $P(1)<0<P(2)$. En effet, on a alors $a_{2}-a_{1}=\\lambda, a_{3}-a_{2}=\\lambda(\\lambda^{3}-1)=\\lambda^{2}, a_{4}-a_{3}=\\lambda^{3}(\\lambda^{3}-\\lambda)=\\lambda^{3}, a_{3}-a_{1}=\\lambda^{4}$, $a_{4}-a_{2}=(a_{4}-a_{3})+(a_{3}-a_{2})=\\lambda^{3}+\\lambda^{2}=\\lambda((a_{3}-a_{2})+(a_{2}-a_{1}))=\\lambda(a_{3}-a_{1})=\\lambda^{5}$, et enfin $a_{4}-a_{1}=\\lambda^{6}$, comme illustré ci-dessous à droite. Par conséquent, $n=4$ convient.\n\n![](attached_image_1.png)\nCas $n=3$\n![](attached_image_2.png)\nCas $n=4$\n\nEn pratique, on pouvait trouver les constructions ci-dessus par étude exhaustive des valeurs possibles pour les différences $a_{j}-a_{i}$. Cependant, lorsque $n \\geqslant 5$, on n'a guère envie de continuer sur cette lancée avec $n(n-1) / 2 \\geqslant 10$ différences à traiter. On suppose donc que l'on dispose d'une solution $n \\geqslant 5$, ainsi que de réels $r>0$ et $a_{1}<a_{2}<\\cdots<a_{n}$ adéquats.\n\nPosons $t=n(n-1) / 2$. Quitte à remplacer $r$ par $1 / r$ et chaque nombre $a_{i}$ par $a_{i} / r^{t+1}$, on suppose que $r \\geqslant 1$. En outre, puisque $a_{3}-a_{1}>a_{2}-a_{1}$, on sait même que $r>1$. Désormais, à chaque entier $k \\leqslant t$, on associe l'intervalle $\\mathcal{I}(k)=\\{i, i+1, \\ldots, j-1\\}$ pour lequel $a_{j}-a_{i}=r^{k}$. Lorsque l'on scinde un intervalle $\\mathcal{I}(k)$ en deux sous-intervalles $\\mathcal{I}(\\ell)$ et $\\mathcal{I}(\\ell')$, on constate immédiatement que $r^{k}=r^{\\ell}+r^{\\ell'}$.\n\nSupposons maintenant qu'il existe un entier $k$ pour lequel l'ensemble $\\mathcal{I}(k) \\cup \\mathcal{I}(k+1)$ n'est pas un intervalle. Quitte à remplacer chaque nombre $a_{i}$ par $-a_{i}$, il existe quatre entiers $u<v<w<x$ tels que $r^{k}=a_{v}-a_{u}$ et $r^{k+1}=a_{x}-a_{w}$. Soit alors $x, y$ et $z$ les entiers tels que $r^{x}=a_{w}-a_{v}, r^{y}=a_{w}-a_{u}$ et $r^{z}=a_{x}-a_{v}$. Comme $r^{y}=r^{k}+r^{x}<r^{k+1}+r^{x}=r^{z}$, on sait que $z \\geqslant y+1$, de sorte que $r^{k+1}+r^{x}=r^{z} \\geqslant r^{y+1}=r^{k+1}+r^{x+1}$, ce qui est absurde. Ainsi, l'ensemble $\\mathcal{I}(k) \\cup \\mathcal{I}(k+1)$ est toujours un intervalle.\n\n![](attached_image_3.png)\n\nPar ailleurs, la bijection réciproque $\\mathcal{I}^{-1}$ de $\\mathcal{I}$ est croissante : lorsque $\\mathcal{I}(k) \\subseteq \\mathcal{I}(\\ell)$, on sait que $k \\leqslant \\ell$. Par conséquent, et quitte à remplacer chaque terme $a_{i}$ par $-a_{i}$, les intervalles $\\mathcal{I}(1)$ et $\\mathcal{I}(2)$ sont sont des singletons voisins, de la forme $\\{k\\}$ et $\\{k+1\\}$.\n\nSoit $\\ell$ le plus grand entier pour lequel $\\mathcal{I}(\\ell+1)=\\{k+\\ell\\}$. L'intervalle $\\mathcal{I}(\\ell+2)$ ne peut contenir, parmi éventuels ses sous-intervalles stricts, que les singletons $\\{k\\},\\{k+1\\}, \\ldots,\\{k+\\ell\\}$. En outre, $\\mathcal{I}(\\ell+2)$ ne coïncide avec aucun des singletons $\\{k\\},\\{k+1\\}, \\ldots,\\{k+\\ell+1\\}$. Or, $\\mathcal{I}(\\ell+2) \\cup \\{k+\\ell\\}$ est un intervalle, donc $\\mathcal{I}(\\ell+2)$ contient l'un des trois entiers $k+\\ell-1, k+\\ell$ et $k+\\ell+1$. Par conséquent, il existe un entier $a$, égal à $\\ell-2$ ou à $\\ell-1$, tel que $\\mathcal{I}(\\ell+2)=\\{k+a, k+a+1\\}$. On note alors $\\ell'$ l'entier $^{-1}(\\{k, k+1\\})$. Puisque $r^{1}+r^{2}=r^{\\ell'} \\geqslant r^{\\ell}=r^{a+1}+r^{a+2}$, on sait que $a=0$. Ainsi, $\\mathcal{I}(\\ell+2)=\\{k, k+1\\}$ et $\\ell$ est égal à 1 ou à 2.\n\n$\\triangleright \\mathrm{Si} \\ell=1$, alors $r+r^{2}=r^{3}$. Cette situation est pour l'instant analogue au cas $n=2$.\nEn outre, l'intervalle $\\mathcal{I}(4)$ est un singleton $\\{k-1\\}$ ou $\\{k+2\\}$, et sa réunion disjointe avec $\\mathcal{I}(3)$ forme un intervalle $\\mathcal{I}(m)$ égal à $\\{k-1, k, k+1\\}$ ou à $\\{k, k+1, k+2\\}$, pour lequel $r^{m}=r^{3}+r^{4}=r^{2}(r+r^{2})=r^{5}$. Mais alors $\\mathcal{I}(5)$ compte 6 sous-intervalles, ce qui est impossible.\n\n$\\triangleright \\mathrm{Si} \\ell=2$, alors $r+r^{2}=r^{4}$. On en déduit immédiatement que $r^{2}+r^{3}=r^{5}$ et $r^{3}+r^{4}=r^{6}$, donc que $\\mathcal{I}(5)=\\{k+1, k+2\\}$ et $\\mathcal{I}(6)=\\{k, k+1, k+2\\}$. Cette situation est pour l'instant analogue au cas $n=3$.\nAinsi, l'intervalle $\\mathcal{I}(7)$ est un singleton $\\{k-1\\}$ ou $\\{k+3\\}$, et sa réunion disjointe avec $\\mathcal{I}(6)$ forme un intervalle $\\mathcal{I}(m)$ égal à $\\{k-1, k, k+1, k+2\\}$ ou à $\\{k, k+1, k+2, k+3\\}$, pour lequel $r^{m}=r^{6}+r^{7}=r^{5}(r+r^{2})=r^{9}$. Par conséquent, $\\mathcal{I}(9)$ compte 10 sous-intervalles, ce qui est impossible.\n\n![](attached_image_4.png)\n\nEn conclusion, nul entier $n \\geqslant 5$ ne convient, et les solutions sont donc $n=2, n=3$ et $n=4$.\nSolution:\n\nOn montre comme dans la solution précédente que les entiers $n=2$, $n=3$ et $n=4$ conviennent, et que l'on peut supposer que $r>1$. On pose également $m=n(n-1) / 2$.\n\nOn considère maintenant une solution $n \\geqslant 4$, et des réels $a_{i}$ et $r$ adaptés. On note $\\mathrm{p}(i, j)$ l'entier pour lequel $a_{j}-a_{i}=r^{\\mathbf{p}(i, j)}$. Sous ces hypothèses, puisque $a_{n}-a_{1}$ est la plus grande des différences $a_{j}-a_{i}$, elle vaut $r^{m}$, et $\\mathrm{p}(1, n)=m$. En outre, chaque différence $a_{i}-a_{j}$ autre que $a_{n}-a_{1}$ est majorée par $a_{n}-a_{2}$ ou par $a_{n-1}-a_{1}$. Quitte à remplacer chaque entier $a_{i}$ par $-a_{i}$ et à réordonner nos nombres, on suppose que $a_{n-1}-a_{1} \\geqslant a_{n}-a_{2}$, ce qui nous assure que $p(1, n-1)=m-1$.\n\nOn remarque alors que $r^{m}=a_{n}-a_{1}=(a_{n}-a_{j})+(a_{j}-a_{1})=r^{\\mathbf{p}(j, n)}+r^{\\mathbf{p}(1, j)}$ pour chacun des entiers $j=2,3, \\ldots, n-1$. Au vu de cette égalité, on dit que deux entiers $c$ et $d$ forment un couple pivot si $1 \\leqslant c<d$ et $r^{-c}+r^{-d}=1$. Remarquons, dans ce cas-là, que $2 r^{-c}>r^{-c}+r^{-d}=1$, de sorte que $c<\\log_{r}(2)$. On peut alors noter $\\left(c_{1}, d_{1}\\right), \\ldots,\\left(c_{k}, d_{k}\\right)$ les couples pivots, et les trier de sorte que $c_{1}<c_{2}<\\cdots<c_{k}$, ce qui nous conduit aux inégalités $1 \\leqslant c_{1}<c_{2}<\\cdots<c_{k}<d_{k}<\\cdots<d_{2}<d_{1}$. En outre, par construction, les entiers $m-\\mathrm{p}(1, j)$ et $m-\\mathrm{p}(j, n)$ forment un couple pivot (à l'ordre près) lorsque $2 \\leqslant j \\leqslant n-1$ : il y a donc au moins $n-2$ couples pivots, ce qui nous assure que $d_{k} \\geqslant k+1 \\geqslant n-1$.\n\nSoit maintenant $\\mathrm{p}(I, J)$ le plus petit des entiers $\\mathrm{p}(i, j)$ pour lesquels $j \\geqslant i+2$. Puisque $\\mathrm{p}(I, J) \\geqslant \\mathrm{p}(I, J-1)$, on sait que $J=I+2$. En outre, parmi les $n(n-1) / 2$ entiers $\\mathrm{p}(i, j)$, il y en a $n-1$ pour lesquels $j=i+1$; ainsi, $\\mathrm{p}(I, J) \\leqslant n$. En particulier, puisque\n$$\nr^{\\mathbf{p}(I, J)}=a_{J}-a_{I}=(a_{J}-a_{I+1})+(a_{I+1}-a_{I})=r^{\\mathbf{p}(I+1, J)}+r^{\\mathbf{p}(I, I+1)}\n$$\nles entiers $\\mathrm{p}(I, J)-\\mathrm{p}(I+1, J)$ et $\\mathrm{p}(I, J)-\\mathrm{p}(I, I+1)$ forment, à l'ordre près, un couple pivot. Le plus grand des deux vaut donc au moins $n-1$, et puisqu'il est minoré par $n-1$, il coïncide avec $n-1$. En d'autres termes, on est assuré que $k=n-2$, que $d_{k}=n-1$; que chaque entier $c_{i}$ vaut $i$; et que les entiers $\\mathrm{p}(i, i+1)$ prennent les valeurs $1,2, \\ldots, n-1$ dans un certain ordre. Cela signifie notamment que $p(1,2) \\leqslant n-1$.\n\nEn outre, chacun des entiers $m-n+1, m-n+2, \\ldots, m-n$ est désormais égal à l'un des entiers $m-d_{k}, m-c_{k}, m-c_{k-1}, \\ldots, m-c_{1}$, donc coïncide avec l'un des entiers $\\mathrm{p}(1, j)$ ou $\\mathrm{p}(j, n)$ obtenus lorsque $2 \\leqslant j \\leqslant n-1$. Par conséquent, $\\mathrm{p}(2, n-1) \\leqslant m-n$. Or, l'égalité\n$$\nr^{\\mathbf{p}(1,2)}+r^{\\mathbf{p}(2, n-1)}=(a_{2}-a_{1})+(a_{n-1}-a_{2})=a_{n-1}-a_{1}=r^{m-1}\n$$\nnous indique que les entiers $(m-1)-\\mathrm{p}(1,2)$ et $(m-1)-\\mathrm{p}(2, n-1)$ forment, à l'ordre près, un couple pivot, disons $\\left(c_{\\ell}, d_{\\ell}\\right)$. L'inégalité $(m-1)-\\mathrm{p}(2, n-1) \\geqslant n-1>c_{\\ell}$ nous assure donc que $(m-1)-\\mathrm{p}(1,2)=c_{\\ell} \\leqslant n-1$.\n\nPuisque $\\mathrm{p}(1,2) \\leqslant n-1$, on en conclut que\n$$\nn(n-1)=2 m \\leqslant 2 p(1,2)+2 n \\leqslant 4 n-2<4 n\n$$\ndonc que $n-1<4$, c'est-à-dire $n \\leqslant 4$.\n\nLa conclusion de l'exercice est désormais proche. En effet, comme on l'a vu dans la solution précédente, on sait que l'on a au moins $n-2$ couples pivots. Or, un couple $(c, d)$ est pivot si et seulement si $r$ est une racine de $P_{d-c, d}(X)$. Si l'on pose $\\delta=\\operatorname{PGCD}(c, d)$, et puisque $r^{2 \\delta}-r^{\\delta}+1 \\geqslant r^{\\delta}>0$, notre réel $r$ est même racine de $Q_{d-c, d}(X)$. Mais alors $Q_{d-c, d}(X)$ est le polynôme minimal de $r$, et ne dépend pas du couple pivot $(c, d)$ de départ. On a donc au plus deux couples pivots au total, de sorte que $n \\leqslant 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30912,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 3$ un entier. Chaque ligne d'un tableau $(n-2) \\times n$ contient les nombres de 1 à $n$ en un exemplaire chacun, on suppose de plus que dans chaque colonne tous les nombres sont différents. Montrer que l'on peut compléter le tableau en un tableau $n \\times n$ de telle sorte que dans chaque ligne et dans chaque colonne il y ait tous les nombres de 1 à $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNotons $a_{i}, b_{i}$ les deux nombres qu'il manque dans la colonne $i$. Chacun des nombres de 1 à $n$ apparaît en deux exemplaires parmi les $a_{i}, b_{i}$. Construisons un multigraphe dont les sommets sont les nombres de 1 à $n$, et pour chaque $i=1, \\ldots, n$, on ajoute une arête entre $a_{i}$ et $b_{i}$. Chaque sommet de ce graphe est de degré 2, il s'agit donc d'un ensemble de cycles disjoints. On peut donc orienter les arêtes de chaque cycle dans un sens arbitraire. On obtient alors un multigraphe orienté dont chaque sommet a un degré entrant de 1 et un degré sortant de 1. Imaginons que l'arête de la $i$-ème colonne pointe de $u$ vers $v$, alors on écrit $u$ aux coordonnées $(i, n-1)$ et $v$ aux coordonnées $(i, n)$. La propriété des degrés donne que chaque nombre apparaît exactement une fois sur chaque ligne.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30913,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dit qu'un entier $k > 1$ est superbe s'il existe $m, n, a$ trois entiers strictement positifs tels que\n$$\n5^{m} + 63 n + 49 = a^{k}\n$$\nDéterminer le plus petit entier superbe.",
    "options": [],
    "answer": "5",
    "solution": "Solution:\n\nSupposons que $k = 2$ est superbe : il existe trois entiers strictement positifs $m, n, a$ tels que $5^{m} + 63 n + 49 = a^{2}$. En regardant modulo $3$, $5^{m} + 1 \\equiv 2^{m} + 1 \\equiv a^{2} \\pmod{3}$. Or les puissances de $2$ modulo $3$ valent alternativement $2$ si $m$ est impair, puis $1$ si $m$ est pair, donc $2^{m} + 1$ vaut soit $2$ soit $0$ modulo $3$. Comme les carrés modulo $3$ sont $0$ et $1$, on en déduit que $2^{m} + 1$ vaut $0$ modulo $3$, donc que $m$ est impair.\n\nEn regardant modulo $7$, on obtient que $5^{m} \\equiv a^{2} \\pmod{7}$. Or les carrés modulo $7$ sont $0, 1, 2, 4$, et les puissances de $5$ alternent entre $1, 5, 4, 6, 2, 3$. Ainsi $5^{m}$ vaut $0, 1, 2$ ou $4$ si et seulement si $m$ vaut $0, 2$ ou $4$ modulo $6$, ce qui est impossible car $m$ est impair. Ainsi $k \\neq 2$.\n\nSupposons que $k = 3$ est superbe : il existe trois entiers strictement positifs $m, n, a$ tels que $5^{m} + 63 n + 49 = a^{3}$. En regardant modulo $7$, $5^{m} \\equiv a^{3}$. Or les cubes modulo $7$ sont $0, 1, 6$. Et les puissances de $5$ modulo $7$ alternent entre $1, 5, 4, 6, 2, 3$. Ainsi on obtient que $m \\equiv 0 \\pmod{6}$ ou $m \\equiv 3 \\pmod{6}$ : dans tous les cas $3$ divise $m$. On regarde alors modulo $9$ : on a $5^{m} + 4 \\equiv a^{3}$. Or comme $m$ est divisible par $3$, $5^{m}$ et $a^{3}$ sont des cubes. Les cubes modulo $9$ sont $0, 1, 8$. Ainsi $5^{m} - a^{3}$ peut valoir $0, 1, 2, 7, 8$ modulo $9$, mais pas $4$ ce qui est contradictoire. Ainsi $k \\neq 3$.\n\nSupposons que $k = 4$ est superbe : il existe trois entiers strictement positifs $m, n, a$ tels que $5^{m} + 63 n + 49 = a^{4} = (a^{2})^{2}$. En particulier, $2$ est superbe ce qui est contradictoire.\n\nNotons que $5$ est superbe : en prenant $a = 3$, on a $a^{k} = 3^{5} = 243 = 5^{1} + 63 \\times 3 + 49$ donc $5$ est superbe. Ainsi $5$ est le plus petit entier superbe.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$. Montrer qu'il existe des entiers $a, b$ tels que, pour tout entier $m$, le nombre $m^{3}+a m+b$ ne soit pas un multiple de $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$m^{3}+a m+b$ est un multiple de $n$ si et seulement si $m^{3}+a m \\equiv -b \\pmod{n}$. On remarque alors qu'il suffit de trouver $a$ tel que $m^{3}+a m$ ne prend pas toutes les valeurs possibles modulo $n$ (et de prendre pour $-b$ une des valeurs qui $n$ n'est pas prise). Comme $m^{3}+a m$ modulo $n$ ne dépend que de $m$ modulo $n$, on peut se restreindre à $0 \\leqslant m \\leqslant n-1$.\n\nIl s'agit de trouver un $a$ tel que les $0^{3}+a \\cdot 0 \\pmod{n}, 1^{3}+a \\cdot 1 \\pmod{n}, \\ldots, (n-1)^{3}+a(n-1) \\pmod{n}$ n'atteignent pas $n$ valeurs distinctes (modulo $n$). Alors si on trouve $m_{1} \\neq m_{2}$ entre $0$ et $n-1$ tels que $m_{1}^{3}+a m_{1} \\equiv m_{2}^{3}+a m_{2} \\pmod{n}$, les $n-2$ autres valeurs de $m$ modulo $n$ donneront au plus $n-2$ valeurs de $m^{3}+a m$ modulo $n$, donc au total, pour $0 \\leqslant m \\leqslant n-1$, $m^{3}+a m$ prendra au plus $n-1$ valeurs sur les $n$ possibles : il y en aura bien une qui ne sera pas atteinte.\n\nOr $0^{3}+a \\times 0 \\equiv 0 \\pmod{n}$, $1^{3}+a \\times 1 \\equiv a+1 \\pmod{n}$. Si $a+1 \\equiv 0 \\pmod{n}$, on sera dans la situation décrite ci-dessus.\n\nPosons alors $a=-1$ (ou $a=n-1$ si on veut $a>0$). Alors $m^{3}+a m$ est nul si $m=0$ ou $m=1$. Donc il existe $c$ entier, $0 \\leqslant c \\leqslant n-1$, tel que pour tout $0 \\leqslant m \\leqslant n-1$, $m^{3}+a m \\not\\equiv c \\pmod{n}$. Posons alors $b=-c$ (ou $b=n-c$ si on veut $b>0$), alors pour tout $0 \\leqslant m \\leqslant n-1$, $m^{3}+a m+b$ sera non nul modulo $n$, donc pour tout $m$ entier, $m^{3}+a m+b$ ne sera pas un multiple de $n$.\n\nAinsi, il existe bien de tels entiers $a$ et $b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle, $D, E$ les pieds des hauteurs issues de $A$ et $B$ respectivement. La droite $(DE)$ rencontre le cercle circonscrit à $ABC$ en deux points $P$ et $Q$. Soient $A'$, $B'$ les symétriques de $A$ et $B$ par rapport à $(BC)$ et $(AC)$ respectivement. Montrer que $A'$, $B'$, $P$, $Q$ sont cocycliques.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSur la figure, il semble que $H$ l'orthocentre se situe sur le cercle en question. On va donc adopter la stratégie suivante : on va montrer que $A'$ et $B'$ sont sur le cercle circonscrit de $PQH$. De cette manière on aura bien $A'$, $B'$, $P$, $Q$ cocycliques.\n\nSoit $M$ le symétrique de $H$ par rapport à $(BC)$ : alors on sait que $M$ est sur le cercle circonscrit à $ABC$. Alors on a :\n$$\n\\begin{aligned}\nDA' \\times DH & = DA \\times DM \\text{ par symétrie } \\\\\n& = DP \\times DQ \\text{ par puissance de } D \\text{ dans le cercle }(ABC) .\n\\end{aligned}\n$$\nAlors par réciproque de la puissance d'un point, $P$, $Q$, $H$, $A'$ sont cocycliques. De façon totalement analogue, $P$, $Q$, $H$, $B'$ sont également cocycliques. Ainsi $A'$, $B'$, $P$, $Q$ sont cocycliques, sur le cercle $(PQH)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans un tournoi organisé entre 6 équipes, chaque équipe joue contre chaque autre équipe exactement une fois. Lorsqu'une équipe gagne, elle obtient 3 points, et l'équipe perdante reçoit 0 point. Si la partie est nulle, les deux équipes reçoivent un point. Déterminez les $a$ pour lesquels il est possible que les scores finaux des équipes puissent être les six nombres consécutifs $a, a+1, \\ldots, a+5$ ?",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nLors de chaque match, entre deux et trois points sont attribués au total. Étant donné qu'il y a 15 matchs au total, il y a donc eu entre 30 et 45 points de répartis entre les équipes. On doit donc avoir:\n$$\n\\begin{aligned}\n& 30 \\leqslant a+(a+1)+(a+2)+\\cdots+(a+5) \\leqslant 45 \\\\\n& 30 \\leqslant 6a+15 \\leqslant 45\n\\end{aligned}\n$$\nCe qui signifie que $a$ peut seulement être $3, 4$ ou $5$.\n\nSi $a=5$, aucun match n'a pu être nul donc tous les scores sont des multiples de 3. Il n'est donc pas possible d'avoir 6 nombres consécutifs pour les scores, et ce cas n'est pas possible.\n\nSi $a=3$, les scores sont $3, 4, 5, 6, 7, 8$. Les équipes qui ont 6 et 7 points ont chacune gagné au moins un match, et l'équipe qui a 8 points a gagné au moins 2 matchs, donc au moins 4 matchs n'ont pas fini par une partie nulle. Ainsi, le nombre total de points répartis entre les équipes vaut au moins 34, or $3+4+5+6+7+8=33$, ce cas n'est donc pas possible.\n\nLe cas $a=4$ est possible. Les scores sont $4, 5, 6, 7, 8, 9$. Voici un tableau qui fonctionne :\n\n|   X   |  3  |  1  |  0  |  0  |  0  |\n|:-----:|:---:|:---:|:---:|:---:|:---:|\n|   0   |  X  |  1  |  0  |  3  |  1  |\n|   1   |  1  |  X  |  3  |  0  |  1  |\n|   3   |  3  |  0  |  X  |  1  |  0  |\n|   3   |  0  |  3  |  1  |  X  |  1  |\n|   3   |  1  |  1  |  3  |  1  |  X  |\n\nLe nombre écrit sur la $j$-ème colonne de la $i$-ème ligne est le score obtenu par l'équipe $i$ lors de son match contre l'équipe $j$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $x, y, z$ trois nombres réels vérifiant $x+y+z=2$ et $xy+yz+zx=1$. Déterminer la valeur maximale que peut prendre $x-y$.",
    "options": [],
    "answer": "2/sqrt(3)",
    "solution": "Solution:\n\nSoit $(x, y, z)$ un triplet vérifiant l'énoncé. Quitte à échanger $x$ et le maximum du triplet, puis $z$ et le minimum du triplet, on peut supposer $x \\geqslant z \\geqslant y$, tout en augmentant $x-y$.\n\nComme $x+y+z=2$, on a $4=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=x^2+y^2+z^2+2$, donc $x^2+y^2+z^2=2$.\n\nEn particulier $(x-y)^2+(y-z)^2+(x-z)^2=2\\left(x^2+y^2+z^2\\right)-2(xy+yz+zx)=2$. Or par inégalité arithmético-quadratique, $(x-z)^2+(z-y)^2 \\geqslant \\frac{(x-z+z-y)^2}{2}=\\frac{(x-y)^2}{2}$, donc $2 \\geqslant (x-y)^2+\\frac{(x-y)^2}{2}=\\frac{3(x-y)^2}{2}$, donc $x-y \\leqslant \\frac{2}{\\sqrt{3}}$.\n\nEssayons de montrer que cette valeur est atteignable. Si on a $x-y=\\frac{2}{\\sqrt{3}}$, alors on a égalité dans l'inégalité arithmético-quadratique : ainsi $x-z=z-y$, donc comme leur somme vaut $x-y$, $x-z=z-y=\\frac{1}{\\sqrt{3}}$. En particulier, comme $x+y+z=3z+(z-x)+(z-y)=3z$, on a $z=\\frac{2}{3}$, $x=\\frac{2}{3}+\\frac{1}{\\sqrt{3}}$ et $y=\\frac{2}{3}-\\frac{1}{\\sqrt{3}}$. Réciproquement, pour ces valeurs de $x, y, z$, on a bien :\n\n$\\quad x+y+z=3 \\times \\frac{2}{3}=2$\n\n$\\quad xy+yz+zx=\\left(\\frac{2}{3}+\\frac{1}{\\sqrt{3}}\\right)\\left(\\frac{2}{3}-\\frac{1}{\\sqrt{3}}\\right)+\\left(\\frac{2}{3}-\\frac{1}{\\sqrt{3}}\\right)\\frac{2}{3}+\\frac{2}{3}\\left(\\frac{2}{3}+\\frac{1}{\\sqrt{3}}\\right)=\\frac{4}{9}-\\frac{1}{3}+\\frac{4}{9}+\\frac{8}{9}=1$\n\n$\\quad x-y=2 \\times \\frac{1}{\\sqrt{3}}=\\frac{2}{\\sqrt{3}}$\n\nAinsi $(x, y, z)$ vérifie les conditions de l'énoncé, et $x-y=\\frac{2}{\\sqrt{3}}$, donc la valeur maximale que peut prendre $x-y$ est $\\frac{2}{\\sqrt{3}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30918,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn jardinier et un pivert jouent au jeu suivant, dans leur jardin dont la forme est celle d'une grille $2022 \\times 2022$ formée de $2022^{2}$ cases. Deux cases sont considérées comme voisines si elles ont un sommet ou une arête en commun. Initialement, chaque case abrite un arbre de taille 0. Puis, à chaque tour de jeu,\n\n$\\triangleright$ le jardinier choisit une case; les arbres de cette case et des cases adjacentes (soit de quatre à neuf cases en tout) voient tous leur taille augmenter de 1;\n\n$\\triangleright$ le pivert choisit alors quatre cases; les arbres de ces cases voient tous leur taille diminuer de 1 (ou rester égale à 0 si le pivert a choisi une case avec un arbre de taille 0).\n\nOn dit qu'un arbre est resplendissant si sa taille vaut au moins $10^{6}$. Trouver le plus grand entier $A$ pour lequel le jardinier pourra s'assurer, en un nombre fini de tours de jeu, et quels que soient les choix du pivert, d'avoir fait pousser au moins $A$ arbres resplendissants.",
    "options": [],
    "answer": "2271380",
    "solution": "Solution:\n\nNous allons démontrer que l'entier recherché est $A=5 n=2271380$, où l'on a posé $n=674^{2}=(2022 / 3)^{2}$.\n\nTout d'abord, voici une stratégie pour le pivert. Il numérote les lignes et les colonnes de 1 à 2022, puis colorie en noir chaque case située dans une colonne $c$ et une ligne $\\ell$ pour lesquelles 3 ne divise ni $c$, ni $\\ell$. Parmi les quatre à neuf cases sur lesquelles le jardinier agit à chaque tour, au plus quatre sont noires. Le pivert choisit alors les cases noires en question; si cela fait moins de quatre cases, il choisit d'autres cases au hasard.\n\nCe faisant, il s'assure, à la fin du tour de jeu, qu'aucun arbre situé dans une case noire n'aura vu sa taille augmenter. Une récurrence immédiate indique donc que nul arbre situé dans une case noire ne verra sa hauteur dépasser 1, et puisqu'il y a $4 n$ cases noires, on ne pourra jamais dépasser un total de $2022^{2}-4 n=5 n$ arbres resplendissants.\n\nRéciproquement, voici une stratégie pour le jardinier; on pose $m=10^{6}$. Faisant fi de la coloration imposée par le pivert, il colorie en rouge chaque case située dans une colonne $c$ et une ligne $\\ell$ pour lesquelles $c \\equiv \\ell \\equiv 2(\\bmod 3)$. Il y a $n$ cases rouges, et chaque case est soit rouge, soit adjacente à une case rouge. Le jardinier pose alors $k=(4 n+1) m+1$ puis, au cours des $k n$ premiers tours de jeu, il choisit chaque case rouge exactement $k$ fois. Il a donc fait croître de $k$ la hauteur de chaque arbre du jardin.\n\nSoit $a$ le nombre d'arbres qui ne sont pas resplendissants à l'issue de cette première partie du jeu : le pivert a dû choisir au moins $k-m$ fois chacun de ces $a$ arbres. Puisqu'il a choisi $4 k n$ arbres au cours des $k n$ tours de jeu, on en déduit que $a(k-m) \\leqslant 4 k n$, donc que\n$$\n(k-m)(a-4 n-1) \\leqslant 4 k n-(k-m)(4 n+1)=(4 n+1) m-k<0 .\n$$\nPuisque $k \\geqslant m$, cela signifie que $a<4 n+1$, donc que $a \\leqslant 4 n=9 n-A$.\n\nEn conclusion, l'entier recherché est bien $A=5 n$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a$ et $b$ deux réels. On définit les suites $\\left(a_{n}\\right)$ et $\\left(b_{n}\\right)$ par $a_{0}=a,\\ b_{0}=b$ et pour tout $n$ entier naturel, $a_{n+1}=a_{n}+b_{n}$ et $b_{n+1}=a_{n} b_{n}$. Déterminer toutes les paires $(a, b)$ telles que $a_{2022}=a_{0}$ et $b_{2022}=b_{0}$.",
    "options": [],
    "answer": "All real a with b = 0",
    "solution": "Solution:\n\nSupposons que $(a, b)$ soit une solution de l'énoncé. On calcule\n$$\n\\sum_{i=1}^{2022} a_{i}=\\sum_{i=1}^{2022}\\left(a_{i-1}+b_{i-1}\\right)=\\sum_{i=0}^{2021} a_{i}+\\sum_{i=0}^{2021} b_{i}\n$$\nMais on a\n$$\n\\sum_{i=0}^{2021} a_{i}=a_{0}+\\sum_{i=1}^{2021} a_{i}=a_{2022}+\\sum_{i=1}^{2021} a_{i}=\\sum_{i=1}^{2022} a_{i}\n$$\ndonc on peut réécrire l'équation précédente comme\n$$\n\\sum_{i=0}^{2021} b_{i}=0\n$$\nDe la même manière, on a\n$$\n0=\\sum_{i=1}^{2022} b_{i}=\\sum_{i=1}^{2022} a_{i-1} b_{i-1}=\\sum_{i=0}^{2021} a_{i} b_{i}\n$$\nPour faire apparaître les produits $a_{i} b_{i}$, on élève l'équation $a_{i+1}=a_{i}+b_{i}$ au carré pour avoir $a_{i+1}^{2}=a_{i}^{2}+b_{i}^{2}+2 a_{i} b_{i}$. On somme ceci entre $i=0$ et $2021$ pour avoir\n$$\n\\sum_{i=1}^{2022} a_{i}^{2}=\\sum_{i=0}^{2021} a_{i}^{2}+\\sum_{i=0}^{2021} b_{i}^{2}+2 \\sum_{i=0}^{2021} a_{i} b_{i}\n$$\nMais encore une fois,\n$$\n\\sum_{i=1}^{2022} a_{i}^{2}=\\sum_{i=0}^{2021} a_{i}^{2}\n$$\ndonc\n$$\n\\sum_{i=0}^{2021} b_{i}^{2}=0\n$$\nUne somme de carrés de nombres réels est nécessairement positive, et ne peut être nulle que si tous les $b_{i}$ sont nuls. Ceci implique notamment $b=0$. Réciproquement, si $a$ est un réel, le couple $(a, 0)$ fonctionne car on montre par récurrence que pour tout $n$, $a_{n}=a$ et $b_{n}=0$, et alors $a_{2022}=a=a_{0}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nTrouver tous les entiers $a$, $b$, $c \\in \\mathbb{N}$ tels que $1517^{a} + 15^{b} = 1532^{c}$.",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "Solution:\nOn commence par traiter les petites valeurs de $c$.\n\nSi $c = 0$, $1517^{a} + 15^{b} = 1$ n'a pas de solution.\n\nSi $c = 1$, on doit trouver $a, b$ tels que $1517^{a} + 15^{b} = 1532$. Si $a \\geq 2$ ou si $a = 0$ il n'y a pas de solution. Si $a = 1$, alors $15^{b} = 1532 - 1517 = 15$ donc $b = 1$ : ceci fournit la solution $(1, 1, 1)$.\n\nOn suppose maintenant $c \\geq 2$. En particulier $16 = 4^{2}$ divise $1532^{c}$ : ceci invite à étudier l'équation modulo des puissances de 2 plus petites que 16 pour éliminer la dépendance en $c$. On obtient alors $1 + (-1)^{b} \\equiv 0 \\pmod{4}$ donc $b$ est impair (en particulier, $b \\geq 1$).\n\nEn réduisant modulo $8$ : $15^{b} \\equiv -1 \\pmod{8}$ ($b$ est impair). On a alors $1517^{a} - 1 \\equiv 5^{a} - 1 \\equiv 0 \\pmod{8}$, donc $a$ est pair.\n\nEn réduisant modulo $5$ (comme $5$ divise $15$, cela supprime la dépendance en $b$), on a $2^{a} \\equiv 2^{c} \\pmod{5}$, donc $a \\equiv c \\pmod{4}$ (l'ordre de $2$ modulo $5$ est $4$). Puisqu'on a montré que $a$ était pair, $c$ est pair. Notons $a = 2e$ et $c = 2d$, alors $d$ et $e$ ont la même parité (car $2e \\equiv 2d \\pmod{4}$ donc $e \\equiv d \\pmod{2}$), et en réinjectant dans l'équation de départ, on obtient $15^{b} = (1532^{d} - 1517^{e})(1532^{d} + 1517^{e})$.\n\nRemarquons alors que\n$$\n(1532^{d} + 1517^{e}) - (1532^{d} - 1517^{e}) = 2 \\times 1517^{e} = 2 \\times 37^{e} \\times 41^{e}\n$$\nOr ni $37$ ni $41$ ne divisent $1532$ donc $1532^{d} - 1517^{e}$, et $1532^{d} - 1517^{e}$ est impair. Par conséquent, $1532^{d} + 1517^{e}$ et $1532^{d} - 1517^{e}$ sont premiers entre eux, donc l'un d'eux est $5^{b}$ et l'autre est $3^{b}$. Puisque $1532^{d} + 1517^{e} > 1532^{d} - 1517^{e}$, il s'ensuit que $1532^{d} + 1517^{e} = 5^{b}$ et $1532^{d} - 1517^{e} = 3^{b}$. Alors $3^{b} + 5^{b} = 2 \\cdot 1532^{d}$. Si $d \\geq 2$, le membre de droite est divisible par $16$, mais on vérifie modulo $16$ que le premier membre n'est jamais divisible par $16$ ($3^{4} \\equiv 5^{4} \\equiv 1 \\pmod{16}$ donc il suffit de tester $3^{0} + 5^{0} \\equiv 2 \\pmod{16}$, $3^{1} + 5^{1} \\equiv 8 \\pmod{16}$, $3^{2} + 5^{2} \\equiv 2 \\pmod{16}$ et $3^{3} + 5^{3} \\equiv 5 \\pmod{16}$), ce qui implique que $d = 1$ (car $2d = c \\geq 2$) et donc $e$ impair ($d \\equiv e \\pmod{2}$). Comme $1532^{d} > 1517^{e}$, il s'ensuit que $e = 1$, mais $1532^{d} - 1517^{d} = 15$ n'est pas une puissance de $3$, d'où une contradiction.\n\nFinalement la seule solution est $(a, b, c) = (1, 1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle de cercle circonscrit $\\Gamma$, $D$ un point sur $(AB)$ et $E$ un point sur $(AC)$ tel que $(DE)$ et $(BC)$ sont parallèles. Le cercle circonscrit à $ABC$ rencontre le cercle circonscrit à $BDE$ une seconde fois en $K$ et le cercle circonscrit à $CDE$ une seconde fois en $L$. Soit $T$ le point d'intersection de $(BK)$ et $(CL)$. Montrer que $(TA)$ est tangente au cercle $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn reconnait ici une situation classique :\n- $KBDE$ cyclique,\n- $KBLC$ cyclique,\n- $CLDE$ cyclique.\n\nOn sait que dans cette situation, les droites $(KB)$, $(CL)$, $(ED)$ sont concourantes (il s'agit ici du fait que les axes radicaux de 3 cercles sont concourants). Or les droites $(BK)$ et $(CL)$ se coupent en $T$. Il suit que $T$, $D$, $E$ sont alignés.\n\nOn sait alors, en exprimant la puissance de $T$ par rapport aux trois cercles, que :\n$$\nTD \\times TE = TL \\times TC = TB \\times TK\n$$\n\nLe fait de connaître le produit $TD \\times TE$ incite à considérer un cercle passant par $D$, $E$, et comme on souhaite une propriété sur $(TA)$, il est naturel d'introduire le cercle circonscrit à $ADE$.\n\nComme $TD \\times TE = TL \\times TC$, $T$ a la même puissance par rapport aux cercles circonscrits à $ABC$ et $ADE$. Il est donc sur leur axe radical. Mais comme $(DE)$ est parallèle à $(BC)$, les deux cercles sont tangents en $A$ d'axe radical la tangente commune à ces deux cercles en $A$. On peut le voir simplement par angle tangentiel (en utilisant que $\\widehat{AED} = \\widehat{ACB}$), autrement on peut le voir en considérant l'homothétie de centre $A$ envoyant $D$ sur $B$ : elle envoie $E$ sur $C$ donc elle envoie le cercle $(ADE)$ sur $(ABC)$, et donc ces cercles sont bien tangents.\n\nOn en déduit que $T$ est sur la tangente à $\\Gamma$ passant par $A$, autrement dit que $(TA)$ est tangente à $\\Gamma$, comme voulu.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPlusieurs nombres sont écrits sur une ligne. Thanima a le droit de choisir deux nombres adjacents de telle sorte que le nombre de gauche soit strictement plus grand que le nombre de droite, elle échange alors ces deux nombres et les multiplie par 2. Montrer que Thanima ne peut effectuer qu'un nombre fini de telles opérations.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par poser un jeton sur le minimum (si égalité, sur le nombre le plus à gauche). On bouge le jeton avec le nombre sur lequel il est posé. On va montrer que le jeton se déplace toujours vers la gauche.\n\nPour cela, supposons par l'absurde qu'à un moment, le jeton s'est déplacé vers la droite. Alors soit $x$ la position du jeton avant l'échange, $i$ la position initiale du jeton, et $j$ la position initiale du nombre avec qui le jeton est échangé. Si $i < j$, alors comme le jeton n'est jamais allé vers la droite avant, l'autre nombre a dû faire au moins autant de déplacements que le jeton; il a donc été multiplié par deux au moins autant de fois. Comme le jeton avait été posé sur le minimum, le nombre du jeton est plus petit (ou égal) au nombre à la position $x+1$. Le déplacement vers la droite n'était donc pas possible. Si $j < i$, alors le jeton et le nombre qui était initialement sur $j$ ont déjà été échangés une fois; juste après cet échange, le nombre avec le jeton était plus petit que le nombre qui était initialement en $j$. Le même argument que le cas précédent s'applique : si le nombre initialement en $j$ est à nouveau juste à droite du jeton, il a dû être doublé au moins autant de fois; il est donc resté plus grand et ne peut pas être échangé à nouveau. Ainsi, le jeton se déplace toujours vers la gauche.\n\nMaintenant, on démontre le résultat par récurrence sur le nombre $n$ de nombres écrits sur la ligne. Comme le jeton se déplace toujours vers la gauche, à partir d'un certain nombre de mouvements, il ne bouge plus. On remarque alors que les nombres à gauche et à droite du jeton forment deux instances séparées du problème initial, et on ne peut effectuer qu'un nombre fini d'opérations sur chaque instance par hypothèse de récurrence. On ne peut donc effectuer qu'un nombre fini d'opérations sur l'instance entière. L'initialisation en $n=0$ est évidente.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30923,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $I$ le centre du cercle inscrit. On note $D$ et $E$ les pieds des bissectrices issues de $B$ et $C$. Soit $X$ l'intersection des symétriques de $(AB)$ et $(AC)$ par rapport à $(CE)$ et $(BD)$. Montrer que les droites $(XI)$ et $(BC)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nNotons $Y$ et $Z$ les intersections respectives de $(DX)$ et $(EX)$ avec la droite $(BC)$. On va montrer que le triangle $ZXY$ est isocèle en $X$ et que $I$ est le centre de son cercle inscrit. Si on arrive à montrer ces propriétés, la conclusion de l'exercice suivrait. En effet, la droite $(XI)$ est alors la bissectrice du triangle $XYZ$ isocèle en $X$, donc en également la hauteur.\n\nSoit $\\omega$ le cercle inscrit du triangle $ABC$, on sait donc que la droite $(AB)$ est tangente à $\\omega$, comme la droite $(CI)$ est un axe de symétrie de $\\omega$ il suit que la droite $(XE)$ est également tangente à $\\omega$. De la même manière, la droite $(DX)$ est tangente à $\\omega$. Par définition la droite $(YZ)$ est la droite $(BC)$ donc également tangente à $\\omega$. Cela montre que $\\omega$ est le cercle inscrit du triangle $XYZ$. En particulier, la droite $(XI)$ est la bissectrice issue de $X$.\n\nOn finit la preuve de l'exercice en montrant que $\\widehat{XZY}=\\widehat{XYZ}$. On note $\\alpha$ l'angle en $A$ dans le triangle $ABC$. On a alors,\n$$\n\\begin{aligned}\n\\widehat{XZY} & =\\widehat{EZC} \\\\\n& =\\widehat{XEC}-\\widehat{ECZ} \\\\\n& =\\widehat{AEC}-\\widehat{ECA} \\\\\n& =\\widehat{EAC}=\\alpha .\n\\end{aligned}\n$$\nDe la même manière $\\widehat{XYZ}=\\alpha$ ce qui montre bien que le triangle $ZYX$ est isocèle en $X$ et conclut la preuve de l'exercice.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, et soit $E$ le milieu de $[BC]$. Soit $m$ la médiatrice de $[AC]$, et soit $D$ un point de $m$ tel que le cercle circonscrit à $ABD$ soit tangent à $m$. Enfin, soit $K$ le point d'intersection, autre que $A$, entre le cercle circonscrit à $ABD$ et la droite $(AC)$, et soit $F$ le milieu de $[CK]$. Démontrer que les droites $(DE)$ et $(EF)$ sont perpendiculaires.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nUne première difficulté est de tracer la figure : comment tracer un point tangent à une droite et passant par deux points donnés? Une stratégie usuelle, en pareil cas, est d'inverser le tracé : on commence par placer les points $A, B$ et $D$, puis on en déduit la tangente $m$ et le point $C$.\n\nNotons maintenant $\\Omega$ et $\\Omega'$ les cercles circonscrits à $ABD$ et $DEF$, puis $O$ et $O'$ leurs centres. Notre figure chèrement acquise suggère que $\\Omega'$, $m$ et $(AC)$ sont concourantes. Ce n'est pas là une surprise : en effet, il s'agit de démontrer que $\\mathcal{C}'$ est le cercle de diamètre $[FD]$, ou encore que le milieu de $[AC]$, que l'on notera $A'$, appartient à $\\Omega'$.\n\nOr, les trois points $A'$, $E$ et $F$ sont les milieux respectifs des segments $[CA], [CB]$ et $[CK]$. Le cercle circonscrit à ces trois points, dont on souhaite démontrer qu'il s'agit de $\\Omega'$, est donc l'image de $\\Omega$ par l'homothétie de centre $C$ et de rayon $1/2$. Au vu de la figure, il s'agit donc de démontrer, si l'on note $K'$ le symétrique de $K$ par rapport à $O$, que cette homothétie envoie $K'$ sur $D$, c'est-à-dire que $D$ est le milieu de $[K'C]$. Or, par construction, on sait déjà que $\\widehat{KDK'} = 90^\\circ$.\n\nOn va donc maintenant démontrer que $(KD)$ est la bissectrice de $\\widehat{CKK'}$, grâce à une chasse aux angles. Tout d'abord, puisque $AKO$ est isocèle en $O$ et que $(AK)$ et $(DO)$ sont perpendiculaires à $m$, on sait que\n$$\n(KC, KK') = (KA, KO) = (AO, AK) = (OA, OD)\n$$\nD'autre part, puisque $m$ est tangente à $\\Omega$, on sait également que\n$$\n(KC, KD) = (KA, KD) = (DA, DA') = (DA, CA) + 90^\\circ = (DA, DO) + 90^\\circ.\n$$\nPuisque $ADO$ est isocèle en $O$, on en conclut comme souhaité que\n$$\n2(KC, KD) = 2(DA, DO) = (DA, DO) + (AO, AD) = (OA, OD) = (KC, KK').\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30925,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $b_{1}, \\ldots, b_{n}$ des réels positifs ou nuls tels que $b_{1}+b_{2}+\\ldots+b_{n}=2$. Soient $a_{0}, a_{1}, \\ldots, a_{n}$ des nombres réels tels que $a_{0}=a_{n}=0$. Supposons que pour tout $i$ compris entre 1 et $n$ inclus,\n$$\n\\left|a_{i}-a_{i-1}\\right| \\leqslant b_{i}\n$$\nMontrer que l'on a\n$$\n\\left(a_{0}+a_{1}\\right) b_{1}+\\left(a_{1}+a_{2}\\right) b_{2}+\\ldots+\\left(a_{n-1}+a_{n}\\right) b_{n} \\leqslant 2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour tout indice $i$, on a deux majorations de $a_{i}$. En effet, on peut écrire\n$$\n\\begin{aligned}\na_{i}=a_{i}-a_{0}= & \\left(a_{i}-a_{i-1}\\right)+\\left(a_{i-1}-a_{i-2}\\right)+\\ldots+\\left(a_{1}-a_{0}\\right) \\\\\n& \\leqslant b_{i}+b_{i-1}+\\ldots+b_{1}=\\sum_{j=1}^{i} b_{j}\n\\end{aligned}\n$$\net\n$$\n\\begin{aligned}\na_{i}=a_{i}-a_{n} & =\\left(a_{i}-a_{i+1}\\right)-\\left(a_{i+1}-a_{i+2}\\right)+\\ldots+\\left(a_{n-1}-a_{n}\\right) \\\\\n& \\leqslant b_{i+1}+b_{i+2}+\\ldots+b_{n}=\\sum_{j=i+1}^{n} b_{j}\n\\end{aligned}\n$$\nAfin de majorer efficacement $a_{i}$, on voudrait trouver le plus petit de ces deux termes. Mais la somme des deux vaut 2, donc il faut simplement déterminer si $\\sum_{j=1}^{i} b_{j}$ est inférieur ou supérieur à 1.\n\nSoit $m$ l'entier minimal tel que $\\sum_{j=1}^{m} b_{j}>1$. On écrit alors\n$$\n\\sum_{i=1}^{n}\\left(a_{i-1}+a_{i}\\right) b_{i}=\\sum_{i=1}^{m-1}\\left(a_{i-1}+a_{i}\\right) b_{i}+\\left(a_{m-1}+a_{m}\\right) b_{m}+\\sum_{i=m+1}^{n}\\left(a_{i-1}+a_{i}\\right) b_{i}\n$$\nPour $i<m$, on majore $a_{i}$ par $\\sum_{j=1}^{i} b_{j}$, et pour $i \\geqslant m$, on majore $a_{i}$ par $\\sum_{j=i+1}^{n} b_{j}$ pour obtenir $\\sum_{i=1}^{n}\\left(a_{i-1}+a_{i}\\right) b_{i} \\leqslant \\sum_{i=1}^{m-1}\\left(b_{i}+2 \\sum_{j=1}^{i-1} b_{j}\\right) b_{i}+\\left(\\sum_{j=1}^{m-1} b_{j}+\\sum_{j=m+1}^{n} b_{j}\\right) b_{m}+\\sum_{i=1}^{m}\\left(b_{i}+2 \\sum_{j=i+1}^{n} b_{j}\\right) b_{i}$.\n\nPosons $x=\\sum_{j=1}^{m-1} b_{j}$ et $y=\\sum_{j=m+1}^{n} b_{j}$. Par définition de $m$, on a $x \\leqslant 1$ et $y \\leqslant 1$. Considérons la première somme dans le terme précédent. Dans celle-ci, on a un terme $b_{i}^{2}$ pour tout $i \\leqslant m-1$ et un terme $2 b_{i} b_{j}$ pour tous $j<i \\leqslant m-1$. Ainsi, cette somme n'est rien d'autre que $x^{2}$. De même, la somme de droite n'est rien d'autre que $y^{2}$. On peut donc écrire, comme $x+b_{m}+y=2$,\n$$\n\\begin{aligned}\n\\sum_{i=1}^{n}\\left(a_{i-1}+a_{i}\\right) b_{i} & \\leqslant x^{2}+(x+y)(2-x-y)+y^{2} \\\\\n= & 2(x+y)-2 x y\n\\end{aligned}\n$$\nIl nous reste enfin à voir que $2(x+y)-2 x y \\leqslant 2$ pour conclure, ou encore $x+y-x y \\leqslant 1$. Mais ceci se réécrit, en passant tous les termes à droite de l'inégalité,\n$$\n(1-x)(1-y) \\geqslant 0\n$$\nce qui est vérifié car on a vu que $x \\leqslant 1$ et $y \\leqslant 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $ABC$ un triangle non-isocèle et $I$ le centre de son cercle inscrit. Soit $D$ un point sur $[BC]$ (autre que $B$ et $C$). Le cercle circonscrit à $DIB$ coupe $(AB)$ une deuxième fois en $E \\neq B$, et le cercle circonscrit à $DIC$ coupe $(AC)$ une deuxième fois en $F \\neq C$. Le cercle circonscrit à $DEF$ rencontre $(AB)$ et $(AC)$ une deuxième fois en $M$ et $N$ respectivement. Les droites $(IB)$ et $(DE)$ s'intersectent en $P$ et les droites $(IC)$ et $(DF)$ s'intersectent en $Q$. Montrer que les droites $(EN)$, $(FM)$ et $(PQ)$ sont parallèles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nSur la figure, il semble que $I$ est le centre du cercle $(DEF)$, et que les droites $(MF)$, $(EN)$, $(PQ)$ sont toutes perpendiculaires à $(AI)$ : c'est donc la stratégie qu'on va utiliser pour résoudre l'exercice.\n\n- **Lemme 1:** Le point $I$ est le centre du cercle circonscrit à $DEF$.\n\nPreuve : Comme $I$ est sur la bissectrice de $\\widehat{DBE}$ et sur le cercle circonscrit de $DBE$, c'est le pôle Sud (par rapport à $B$) dans ce triangle. Donc $ID = IE$. De même dans $FCD$, on obtient $ID = IF$. Donc $ID = IE = IF$ : $I$ est bien le centre du cercle circonscrit de $DEF$.\n\n- **Lemme 2:** Le quadrilatère $PQID$ est cyclique.\n\nPreuve : En effet, on a :\n$$\n\\begin{aligned}\n\\widehat{QDP} & = \\widehat{QDI} + \\widehat{IDP} \\\\\n& = \\widehat{FDI} + \\widehat{IDE} \\\\\n& = \\widehat{FCI} + \\widehat{IBE} \\text{ par angle inscrit } \\\\\n& = \\widehat{ICB} + \\widehat{CBI} \\text{ par bissectrices } \\\\\n& = 180^{\\circ} - \\widehat{BIC} \\\\\n& = 180^{\\circ} - \\widehat{PIQ}\n\\end{aligned}\n$$\nDonc $\\widehat{PIQ} + \\widehat{QPD} = 180^{\\circ}$, d'où le fait que $PIDQ$ est cyclique.\n\n- **Lemme 3:** Les droites $(PQ)$ et $(AI)$ sont perpendiculaires.\n\nPreuve : En effet $\\widehat{QPI} = \\widehat{QDI} = \\widehat{FDI} = \\widehat{FCI} = \\frac{1}{2} \\widehat{ACB}$ par angle inscrit. De plus :\n$$\n\\begin{aligned}\n\\widehat{AIB} & = 180^{\\circ} - \\widehat{IBA} - \\widehat{BAI} \\\\\n& = 180^{\\circ} - \\frac{\\widehat{CBA} + \\widehat{BAC}}{2} \\\\\n& = 180^{\\circ} - \\frac{180^{\\circ} - \\widehat{ACB}}{2} \\\\\n& = 90^{\\circ} + \\frac{1}{2} \\widehat{ABC} .\n\\end{aligned}\n$$\nOn en déduit que $(AI)$ et $(PQ)$ sont bien perpendiculaires.\n\n- **Lemme 4:** Le quadrilatère $AEIF$ est cyclique.\n\nPreuve : En effet $BDIE$ et $CFID$ sont cycliques, donc d'après le théorème de Miquel, $AEIF$ l'est aussi.\n\n- **Lemme 5:** Les droites $(MF)$ et $(EN)$ sont perpendiculaires à $(AI)$.\n\nPreuve : D'abord $\\widehat{IEM} = \\widehat{IEA} = 180^{\\circ} - \\widehat{AFI} = \\widehat{IFN}$ par angle inscrit. De plus $IE = IM = IF = IN$ (par le lemme 1), donc $IEM$ et $IFN$ sont deux triangles isocèles en $I$ avec même angle de base : d'où $IEM$ semblable à $IFN$. Mais comme $IE = IF$ : ils sont même isométriques. On en déduit que $EM = FN$. Comme $EMFN$ est cyclique, c'est un trapèze isocèle. On en déduit que $(MF)$ et $(EN)$ sont parallèles, toutes les deux perpendiculaires à $(AI)$ (car $I$ est le centre du cercle circonscrit et $A$ l'intersection des côtés, donc $(AI)$ est axe de symétrie du trapèze isocèle).\n\n- Finalement, $(EN)$, $(FM)$, $(PQ)$ sont toutes les trois perpendiculaires à $(AI)$, elles sont donc parallèles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nQuentin et Timothé jouent à un jeu. D'abord, Quentin choisit un nombre premier $p > 2$, puis Timothé choisit un entier strictement positif $n_0$. Quentin choisit alors un entier $n_1 > n_0$ et calcule $s_1 = n_0^{n_1} + n_1^{n_0}$; puis Timothé choisit un entier $n_2 > n_1$ et calcule $s_2 = n_1^{n_2} + n_2^{n_1}$. Les joueurs continuent de jouer chacun à leur tour, en choisissant au tour $k$ un entier $n_k > n_{k-1}$ et en calculant $s_k = n_k^{n_{k-1}} + n_{k-1}^{n_k}$. Le premier joueur à choisir un entier $n_k$ tel que $p$ divise $s_k (s_1 + 2 s_2 + 3 s_3 + \\ldots + k s_k)$ gagne le jeu. Déterminer lequel de Quentin et Timothé possède une stratégie gagnante.",
    "options": [],
    "answer": "Timothé has a winning strategy.",
    "solution": "Solution:\n\nNous allons montrer que Timothé a une stratégie gagnante. Notons que Quentin va choisir les $n_{2k+1}$ et Timothé va choisir les $n_{2k}$. On fait d'abord quelques remarques.\n\n- **Remarque 1** : Remarquons d'abord que si l'un des joueurs choisit $n_k \\equiv 0 \\pmod{p}$ et ne gagne pas à cette étape, alors le joueur suivant gagne en choisissant $n_{k+1} \\equiv 0 \\pmod{p}$, car alors $s_{k+1} \\equiv 0 \\pmod{p}$, et en particulier $p$ divise $s_{k+1}(s_1 + 2 s_2 + \\ldots + k s_k + (k+1) s_{k+1})$.\n\n- **Remarque 2** : Pour $a \\not\\equiv 0 \\pmod{p}$, $a^{p-1} \\equiv 1 \\pmod{p}$ (Fermat). Alors si $a, b \\geq 1$ sont deux entiers tels que $a$ soit divisible par $p-1$ et congru à $1$ modulo $p$ et $b$ n'est pas divisible par $p$, alors $a^b + b^a \\equiv 1^b + 1 \\equiv 2 \\pmod{p}$. On va donc chercher une stratégie de ce côté-là.\n\n- **Stratégie** : On va montrer que la stratégie suivante fonctionne : pour $k \\geq 0$, Timothé va distinguer deux cas.\n\n- Si $k = 0$ ou $n_{2k-1}$ n'est pas divisible par $p$, alors Timothé choisit $n_{2k} > n_{2k-1}$ (resp. $n_0 > 0$) de telle sorte que $n_{2k} \\equiv 0 \\pmod{p-1}$ et $n_{2k} \\equiv 1 \\pmod{p}$ ($p$ et $p-1$ sont premiers entre eux donc par le théorème des restes chinois, il existe un unique reste modulo $p(p-1)$ qui convient, en particulier il existe bien de tels nombres aussi grands que l'on veut.)\n\n- Si $n_{2k-1}$ est divisible par $p$, alors Timothé choisit $n_{2k} > n_{2k-1}$ divisible par $p$.\n\nPar la première remarque, dans le second cas si Quentin n'avait pas gagné, alors Timothé gagne à cet instant. Dans le premier cas, si $n_{2k-1}$ n'est pas divisible par $p$, $s_{2k} \\equiv 2 \\pmod{p}$ (d'après la deuxième remarque). Donc si Timothé ne gagne pas à un instant donné, alors il a obtenu $s_{2k} \\equiv 2 \\pmod{p}$.\n\n- On s'intéresse alors aux valeurs prises par $s_{2k+1}$. On distingue deux cas.\n\n- D'abord si à chaque étape, Quentin choisit $n_{2k+1}$ non divisible par $p$. Alors d'après la deuxième remarque, il obtient $s_{2k+1} \\equiv 2 \\pmod{p}$. Autrement dit, $s_k \\equiv 2 \\pmod{p}$ pour tout $k$, et donc $1 + 2 s_2 + \\ldots + k s_k \\equiv k(k+1) \\pmod{p}$ et donc le premier $k$ tel que cette quantité est divisible par $p$ est $k = p-1$ (lemme d'Euclide), or $p-1$ est pair ($p > 2$), donc c'est Timothé qui vient de choisir, et c'est donc celui-ci qui gagne.\n\n- Maintenant, si à un moment Quentin choisit $n_{2k+1}$ divisible par $p$ (on considère que c'est la première fois que Quentin choisit un tel nombre), en particulier $2k+1 < p-1$ sinon on a vu au-dessus que Timothé gagne. Alors pour $m \\leq 2k$, on a $s_m \\equiv 2 \\pmod{p}$, et $s_{2k+1} \\equiv 1 \\pmod{p}$. Alors\n$$\ns_1 + 2 s_2 + \\ldots + (2k+1) s_{2k+1} \\equiv 2k(2k+1) + (2k+1) \\equiv (2k+1)^2 \\not\\equiv 0 \\pmod{p}\n$$\n(car $1 \\leq 2k+1 \\leq p-1$), donc Quentin ne gagne pas à cette étape. Mais alors, comme évoqué précédemment, en choisissant $n_{2k+2}$ divisible par $p$, on a $s_{2k+2} \\equiv 0 \\pmod{p}$ donc Timothé gagne.\n\n- Dans tous les cas, avec cette stratégie, Timothé va gagner, donc c'est Timothé qui a une stratégie gagnante.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30928,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $x$ et $y$ deux réels positifs. Montrer que\n$$\n\\left(x^{2}+x+1\\right)\\left(y^{2}+y+1\\right) \\geqslant 9 x y .\n$$\nQuels sont les cas d'égalité?",
    "options": [],
    "answer": "x=y=1",
    "solution": "Solution:\n\nOn applique l'inégalité arithmético-géométrique pour avoir\n$$\n\\begin{gathered}\nx^{2}+x+1 \\geqslant 3 \\sqrt[3]{x^{3}} = 3x \\\\\ny^{2}+y+1 \\geqslant 3 \\sqrt[3]{y^{3}} = 3y\n\\end{gathered}\n$$\nEn multipliant ces deux inégalités, qui sont à termes positifs, on obtient l'inégalité de l'énoncé. Comme $x^{2}+x+1$ et $y^{2}+y+1$ sont strictement positifs, on a égalité lorsque l'on a égalité dans les deux inégalités arithmético-géométriques. Mais ceci arrive lorsque $x^{2}=x=1$ et $y^{2}=y=1$, et on a donc égalité pour $x=y=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a$, $b$ deux entiers relatifs. Montrer que, si ni $a$, ni $b$ n'est multiple de $3$, alors $a^{4}-b^{2}$ est multiple de $3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRappelons que si $k$ n'est pas un multiple de $3$, alors $k$ vaut $1$ ou $2$ modulo $3$. Ainsi $k^{2}$ vaut $1^{2}=1$ ou $2^{2} \\equiv 4 \\equiv 1 \\pmod{3}$ : tout carré d'un nombre non divisible par $3$ vaut $1$ modulo $3$.\n\nAinsi $a^{4}-b^{2} \\equiv (a^{2})^{2}-b^{2} \\equiv 1^{2}-1 \\equiv 0 \\pmod{3}$, donc $a^{4}-b^{2}$ est un multiple de $3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAntoine propose à Baptiste de jouer à l'\"alphabet en folie\" : Ils commencent par se mettre d'accord sur une lettre. Puis, à tour de rôle, chacun peut choisir de prononcer entre 1 et 2 lettres suivantes dans l'alphabet en partant de $A$. Celui qui prononce la lettre choisie a gagné. Si Antoine commence, pour quelles lettres de départ dispose-t-il d'une stratégie lui permettant de gagner la partie à coup sûr?\n\nVoici un exemple de partie : la lettre choisie initialement est $E$. Antoine dit \"A\", Baptiste \"BC\", Antoine \"CD\" et Baptiste dit \"E\". Dans ce cas, Baptiste a gagné.",
    "options": [],
    "answer": "Antoine has a winning strategy if and only if the target letter’s position in the alphabet is not a multiple of three; he loses when it is a multiple of three.",
    "solution": "Solution:\n\nMontrons que Antoine a une stratégie gagnante si, et seulement si, le rang de la lettre choisie dans l'alphabet n'est pas un multiple de $3$.\n\nColorons les lettres de l'alphabet en bleu-blanc-rouge dans l'ordre. Si la lettre d'arrivée est rouge (c'est à dire que son rang dans l'alphabet est un multiple de $3$), Baptiste possède la stratégie suivante : à chaque fois qu'Antoine choisit de dire une lettre, Baptiste en dit deux et à chaque fois qu'Antoine décide de dire deux lettres il n'en dit qu'une. Selon cette stratégie, quels que soient ses choix passés, Antoine n'a le choix à chaque tour qu'entre une lettre bleue ou une lettre bleue et une blanche et ne dira pas la lettre finale rouge.\n\nEn revanche, si la lettre choisie est blanche ou bleue, Antoine possède une stratégie gagnante : il lui suffit de prendre $A$ si la lettre finale est bleue ou $A$ et $B$ si elle est blanche, puis d'appliquer la stratégie qui était celle de Baptiste dans le cas rouge.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\Omega$ son cercle circonscrit. On note $A'$ le point diamétralement opposé à $A$ dans le cercle $\\Omega$. Soit $I$ le centre du cercle inscrit au triangle $ABC$, $E$ et $F$ les points de contact du cercle inscrit avec les côtés $AC$ et $AB$ respectivement. Le cercle circonscrit au triangle $AEF$ recoupe le cercle $\\Omega$ au point $X$. Montrer que les points $A'$, $I$ et $X$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPuisque $F$ est le point de contact du cercle inscrit avec le côté $[AC]$, l'angle $\\widehat{IFA}$ est droit et le segment $[IA]$ est un diamètre du cercle circonscrit au triangle $AEF$. On en déduit que\n$$\n\\widehat{AXI} = 90^\\circ = \\widehat{AXA'}\n$$\noù on a utilisé que $[AA']$ est un diamètre du cercle circonscrit au triangle $ABC$.\nOn déduit que les points $A'$, $I$ et $X$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30932,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et $\\Omega$ son cercle circonscrit. On note $A'$ le point diamétralement opposé à $A$ dans le cercle $\\Omega$. Soit $I$ le centre du cercle inscrit au triangle $ABC$, $E$ et $F$ les points de contact du cercle inscrit avec les côtés $AC$ et $AB$ respectivement. Le cercle circonscrit au triangle $AEF$ recoupe le cercle $\\Omega$ au point $X$. Montrer que les points $A'$, $I$ et $X$ sont alignés.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPuisque $F$ est le point de contact du cercle inscrit avec le côté $[AC]$, l'angle $\\widehat{IFA}$ est droit et le segment $[IA]$ est un diamètre du cercle circonscrit au triangle $AEF$. On en déduit que\n$$\n\\widehat{AXI} = 90^\\circ = \\widehat{AXA'}\n$$\nou on a utilisé que $[AA']$ est un diamètre du cercle circonscrit au triangle $ABC$.\nOn déduit que les points $A'$, $I$ et $X$ sont alignés.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour un réel $x$, on note $\\lfloor x\\rfloor$ le plus grand entier relatif inférieur ou égal à $x$ (par exemple, $\\lfloor 2,7\\rfloor=2$, $\\lfloor\\pi\\rfloor=3$ et $\\lfloor-1,5\\rfloor=-2$). On note aussi $\\{x\\}=x-\\lfloor x\\rfloor$.\n\nTrouver tous les réels $x$ tels que\n$$\n\\lfloor x\\rfloor\\{x\\}<x-1\n$$",
    "options": [],
    "answer": "x ≥ 2",
    "solution": "Solution:\n\nOn réécrit $x=\\lfloor x\\rfloor+\\{x\\}$ dans le terme de droite et on met tous les termes à gauche pour que l'inégalité devienne\n$$\n\\lfloor x\\rfloor\\{x\\}-\\lfloor x\\rfloor-\\{x\\}+1<0\n$$\nou encore\n$$\n(\\lfloor x\\rfloor-1)(\\{x\\}-1)<0\n$$\nRemarquons que pour tout $x$ réel, on a $0 \\leqslant\\{x\\}<1$ et donc $\\{x\\}-1<0$. Alors l'inégalité est équivalente à\n$$\n\\lfloor x\\rfloor-1>0\n$$\nou encore $\\lfloor x\\rfloor>1$. Ceci veut précisément dire que $x \\geqslant 2$. On peut vérifier que réciproquement, si $x \\geqslant 2$, alors $x$ vérifie bien l'inégalité de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k$ un entier strictement positif et $P$ un point du plan. Déterminer le plus petit entier $n$ ayant la propriété suivante : on peut tracer $n$ droites ne passant pas par $P$ de sorte que toute demi-droite d'origine le point $P$ coupe au moins $k$ droites.",
    "options": [],
    "answer": "2k+1",
    "solution": "Solution:\n\nRéponse : $n = 2k + 1$\n\nL'énoncé demande de déterminer le plus petit entier $n$ possédant une certaine propriété, il contient nécessairement deux parties. D'une part, on montre que si $n$ est inférieur à $2k$, il ne vérifie pas la propriété, d'autre part on montre que $2k+1$ vérifie la propriété.\n\nSoit $n$ un entier inférieur ou égal à $2k$. Considérons un ensemble de $n$ droites ne passant pas par $P$. Prenons $d$ une droite passant par $P$ et parallèle à l'une des $n$ droites. La droite $d$ coupe au plus $n-1 \\leqslant 2k-1$ des autres droites. On peut répartir ces points d'intersection en deux groupes de points, de sorte que les points de chaque groupe sont situés du même côté sur $d$ par rapport à $P$. L'un des groupes est de taille inférieure à $k-1$, de sorte qu'il existe une demi-droite issue de $P$ coupant au plus $k-1$ droites.\n\nConstruction pour $n=2k+1$ : Considérons un polygone régulier $A_{1}, \\ldots, A_{2k+1}$ à $2k+1$ côtés et dont le cercle circonscrit est de centre $P$ et dont on prolonge les côtés. On obtient un ensemble de $2k+1$ droites. Considérons une demi-droite issue du sommet $P$, et disons qu'elle intersecte le segment $[A_{1}A_{2}]$ (éventuellement en l'une de ses extrémités) en effectuant un angle de $\\theta \\leqslant \\frac{360}{2n}$ avec sa médiatrice, et de sorte que le point d'intersection entre $d$ et $[A_{1}A_{2}]$ soit plus proche de $A_{2}$ que de $A_{1}$. Alors la demi-droite coupe la droite $(A_{r}A_{r+1})$ si et seulement si elle forme un angle inférieur à $\\pi/2$ avec la médiatrice du segment $[A_{r}A_{r+1}]$. Cette condition induit l'inégalité suivante sur $r$\n$$\n\\frac{r}{n} 360^{\\circ} - \\theta \\leqslant 90^{\\circ}.\n$$\nCette inégalité est valable pour les $r \\leqslant \\left\\lfloor \\frac{k+1}{2} \\right\\rfloor$.\n\nDe la même façon, la demi-droite $d$ coupe la droite $(A_{2k+2-r}A_{2k+1-r})$ pour les $r$ satisfaisant $r \\leqslant \\left\\lfloor \\frac{k}{2} \\right\\rfloor - 1$. Il vient que la droite $d$ coupe le segment $[A_{1}A_{2}]$ et au moins $\\left.\\left\\lfloor \\frac{k+1}{2} \\right\\rfloor + \\frac{k}{2} \\right\\rfloor - 1 = k-1$ droites, soit au moins $k$ droites.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLa suite $(a_{n})$ est définie par $a_{1}=1$ et\n$$\na_{n} = \\frac{1}{n} + \\frac{1}{a_{1} \\cdot \\ldots \\cdot a_{n-1}}\n$$\nMontrer que pour tout entier $m \\geqslant 3$, on a $a_{m} \\leqslant 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn pose $P_{n} = \\prod^{n} a_{k}$. On a\n$$\n\\frac{P_{n}}{P_{n-1}} = \\frac{1}{n} + \\frac{1}{P_{n-1}}\n$$\nce qui donne\n$$\nn P_{n} = P_{n-1} + n\n$$\nOn pose alors $M_{n} = n! P_{n}$. On déduit\n$$\nM_{n} = M_{n-1} + n!\n$$\nOn déduit que $M_{n} = \\sum k!$. On déduit alors\n$$\na_{n} = \\frac{P_{n}}{P_{n-1}} = \\frac{\\frac{1}{n!} \\sum k!}{\\frac{1}{(n-1)!} \\sum k!}\n$$\nIl reste à montrer que\n$$\n\\sum_{k=1}^{n} k! \\leqslant \\sum_{k=1}^{n-1} n \\times k!\n$$\nce qui vient en utilisant $n \\geqslant 3$ pour avoir notamment $n \\geqslant 2+1$ et\n$$\n\\sum_{k=1}^{n} k! \\leqslant 1+2+\\sum_{k=3}^{n} n \\times (k-1)! \\leqslant n + n \\sum_{k=2}^{n-1} k!\n$$\nSolution:\n\nA partir de l'équation sur les $P_{n}$, on veut donc montrer que $a_{n} \\leqslant 1$ pour $n \\geqslant 3$, c'est-à-dire que $P_{n} \\leqslant P_{n-1}$. Ceci se réécrit\n$$\n\\begin{gathered}\n\\frac{1}{n} P_{n-1} + 1 \\leqslant P_{n-1} \\\\\nP_{n-1} \\geqslant \\frac{n}{n-1} .\n\\end{gathered}\n$$\nOn cherche donc à montrer cette inégalité par récurrence. Pour $n=3$, on a bien\n$$\nP_{2} = \\frac{1}{2} + 1 = \\frac{3}{2} = \\frac{3}{3-1} .\n$$\nSi le résultat est vrai au rang $n$, on a, au rang $n+1$,\n$$\nP_{n} = \\frac{1}{n} P_{n-1} + 1 \\geqslant \\frac{1}{n} \\frac{n}{n-1} + 1 = \\frac{n}{n-1} \\geqslant \\frac{n+1}{n}\n$$\nce qui conclut la récurrence.\n\n\nSolution alternative 2 :\nOn a $a_{n+1} - \\frac{1}{n} = \\frac{1}{a_{1} \\ldots a_{n}}$. Ainsi :\n$$\na_{n+1} - \\frac{1}{n} = \\frac{1}{a_{1} \\ldots a_{n-1}} \\frac{1}{a_{n}} = \\left(a_{n} - \\frac{1}{n-1}\\right) \\frac{1}{a_{n}}\n$$\nDès lors $a_{n+1} = 1 - \\frac{1}{a_{n}(n-1)} + \\frac{1}{n}$. On peut calculer les premiers termes pour avoir $a_{3} = 1$. On peut ensuite prouver par récurrence sur $n$ que $a_{n} \\leqslant 1$.\nSi le résultat est vrai au rang $n$, $a_{n+1} = 1 - \\frac{1}{a_{n}(n-1)} + \\frac{1}{n} \\leqslant 1 - \\frac{1}{n-1} + \\frac{1}{n} \\leqslant 1$. Ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nOn définit la suite $\\left(a_{n}\\right)$ par $a_{0}=a_{1}=a_{2}=1$ et si $n \\geqslant 3$ :\n$$\na_{n}=\\left\\lfloor\\frac{n}{a_{n-1} a_{n-2} a_{n-3}}\\right\\rfloor\n$$\nDéterminer $\\mathrm{a}_{2022}$.",
    "options": [],
    "answer": "674",
    "solution": "Solution:\nEn calculant les premiers termes de la suite, on effectue la conjecture suivante, que l'on montre par récurrence sur $k$ :\n$$\n\\text{Si } k \\geqslant 1 \\text{, alors } a_{4 k}=1, a_{4 k+1}=1, a_{4 k+2}=\\left\\lfloor\\frac{4 k+2}{3}\\right\\rfloor \\text{ et } a_{4 k+3}=3.\n$$\nInitialisation : On a $a_{3}=3, a_{4}=1, a_{5}=1, a_{6}=2$ et $a_{7}=3$, de sorte que la propriété est vérifiée pour $k=1$.\n\nHérédité : Supposons la propriété vérifiée pour $k \\geqslant 1$ fixée. Alors on a successivement :\n$$\na_{4(k+1)}=\\left\\lfloor\\frac{4(k+1)}{a_{4 k+1} a_{4 k+2} a_{4 k+3}}\\right\\rfloor=\\left\\lfloor\\frac{4(k+1)}{3\\lfloor(4 k+2) / 3\\rfloor}\\right\\rfloor,\n$$\nor $\\frac{4(k+1)}{3}<2\\left(\\frac{4 k+2}{3}-1\\right)<2\\left\\lfloor\\frac{4 k+2}{3}\\right\\rfloor$ pour $k \\geqslant 2$ et $\\frac{8}{3}<2\\left\\lfloor\\frac{4 \\times 1+2}{3}\\right\\rfloor$, de sorte que $a_{4(k+1)}=1$.\n$$\n\\mathbf{a}_{4(k+1)+1}=\\left\\lfloor\\frac{4(k+1)+1}{a_{4 k+2} a_{4 k+3} a_{4(k+1)}}\\right\\rfloor=\\left\\lfloor\\frac{4(k+1)+1}{3\\lfloor(4 k+2) / 3\\rfloor}\\right\\rfloor\n$$\net là aussi, on a $\\frac{4(k+1)}{3}<2\\left\\lfloor\\frac{4 k+2}{3}\\right\\rfloor$, de sorte que $a_{4(k+1)+1}=1$.\n$$\na_{4(k+1)+2}=\\left\\lfloor\\frac{4(k+1)+2}{a_{4 k+3} a_{4(k+1)} a_{4(k+1)+1}}\\right\\rfloor=\\left\\lfloor\\frac{4(k+1)+2}{3}\\right\\rfloor\n$$\ndonc $a_{4(k+1)+2}=\\left\\lfloor\\frac{4(k+1)+2}{3}\\right\\rfloor$.\nEnfin,\n$$\na_{4(k+1)+3}=\\left\\lfloor\\frac{4(k+1)+3}{a_{4(k+1)} a_{4(k+1)+1} a_{4(k+1)+2}}\\right\\rfloor=\\left\\lfloor\\frac{4(k+1)+3}{\\lfloor(4(k+1)+2) / 3\\rfloor}\\right\\rfloor.\n$$\nOr on a bien\n$$\n4(k+1)+3 \\geqslant 3\\left(\\frac{4(k+1)+2}{3}\\right) \\geqslant 3\\lfloor(4(k+1)+2) / 3\\rfloor\n$$\net\n$$\n4(k+1)+3 \\leqslant 4\\left(\\frac{4(k+1)+2}{3}-1\\right)<4\\lfloor(4(k+1)+2) / 3\\rfloor\n$$\npour $k \\geqslant 1$, de sorte que $a_{4(k+1)+3}=3$. Ceci achève la récurrence.\n\nAinsi $\\mathrm{a}_{2022}=\\mathrm{a}_{4 \\times 505+2}=\\left\\lfloor\\frac{4 \\times 505+2}{3}\\right\\rfloor=674$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30937,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer toutes les fonctions $f$ de $\\mathbb{R}$ dans $\\mathbb{R}$ telles que pour tous $x, y$ réels :\n$$\nf(x) f(y) f(x-y) = x^{2} f(y) - y^{2} f(x)\n$$",
    "options": [],
    "answer": "f(x) ≡ 0 for all x; or f(x) = x for all x; or f(x) = -x for all x.",
    "solution": "Solution:\nSoit $f$ une solution de l'équation.\nPour $x = y = 0$, on a :\n$$\nf(0)^3 = 0\n$$\nAinsi, $f(0) = 0$. On exclut la fonction $f \\equiv 0$ qui est effectivement solution du problème.\nOn considère maintenant un nombre $x$ tel que $f(x) = 0$. On a alors pour un $y$ quelconque :\n$$\n0 = x^{2} f(y) - 0\n$$\nComme on a exclu le cas $f \\equiv 0$, on peut trouver $y$ tel que $f(y) \\neq 0$ et donc $x = 0$. Ainsi, si $f(x) = 0$, on a $x = 0$.\nOn peut maintenant remarquer qu'échanger $x$ et $y$ dans l'équation change le terme de droite en son opposé, on a donc :\n$$\nf(x) f(y) f(x-y) = x^{2} f(y) - y^{2} f(x) = -\\left(y^{2} f(x) - x^{2} f(y)\\right) = -f(y) f(x) f(y-x)\n$$\nAinsi pour $x$ et $y$ non nuls, on a $f(x-y) = -f(y-x)$. En posant $y = 2x$ avec $x \\neq 0$, on obtient que pour tout $x \\neq 0$, $f(-x) = -f(x)$. Comme $f(0) = 0$, $f$ est impaire.\nOn peut maintenant prendre $x = 2y$, l'égalité de l'énoncé avec l'imparité donne :\n$$\nf(2y) f(y)^2 = 4y^{2} f(y) - y^{2} f(2y)\n$$\nD'autre part, avec $x = -y$ :\n$$\nf(y)^2 f(2y) = 2y^{2} f(y)\n$$\nEn utilisant ces deux égalités, on a :\n$$\nf(2y) y^{2} = 2y^{2} f(y)\n$$\nSi $y$ est non nul, on a $f(y) \\neq 0$, et donc l'égalité précédente s'écrit $f(2y) = 2 f(y)$, et l'équation (1) s'écrit\n$$\n\\begin{gathered}\n2 f(y)^3 = 2 y^{2} f(y) \\\\\nf(y)^2 = y^{2}\n\\end{gathered}\n$$\nOn trouve donc $f(y) = \\pm y$. Attention, a priori, le signe pourrait dépendre de $y$. Soit, par l'absurde, $a$ et $b$ deux réels non nuls tels que $f(a) = a$ et $f(b) = -b$. En prenant $x = b$ et $y = a$, on obtient :\n$$\n-ab f(a-b) = ab(a+b)\n$$\nComme $a$ et $b$ sont non nuls, $f(a-b) = -(a+b)$. Mais $f(a-b) = \\pm(a-b)$, et en fonction du signe on obtient soit $a = 0$, soit $b = 0$, ce qui est absurde. Pour $x = 0$, on a à la fois $f(x) = x$ et $f(x) = -x$, donc nécessairement $f$ est de la forme : $f(x) = 0$ pour tous $x$, ou $f(x) = -x$ pour tous $x$, ou $f(x) = x$ pour tous $x$.\nCes trois fonctions sont bien solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k \\geqslant 1$ un nombre entier.\nConsidérons $4k$ jetons, dont $2k$ sont rouges et $2k$ sont bleus. Une suite de ces $4k$ jetons peut être transformée en une autre suite par un mouvement, consistant à interchanger un certain nombre (éventuellement un seul) de jetons rouges consécutifs avec un nombre égal de jetons bleus consécutifs. Par exemple, un mouvement permet de passer de rbbbrrrb à rrrbrbbb où $r$ désigne un jeton rouge et $b$ désigne un jeton bleu.\n\nDéterminer le plus petit nombre $n$ (comme fonction de $k$) tel que, partant de n'importe quelle suite initiale des $4k$ jetons, il faut au plus $n$ mouvements pour atteindre l'état dans lequel les $2k$ premiers jetons sont rouges.",
    "options": [],
    "answer": "k",
    "solution": "Solution:\n\nRéponse : $n = k$.\n\nLe problème demande de déterminer le plus petit $n$ vérifiant une certaine propriété, il contient donc nécessairement deux parties. Ici, dans un premier temps, on montre qu'étant donnée une configuration initiale de jetons, il est toujours possible d'arriver à la configuration finale décrite en au plus $k$ étapes. Dans un second temps, on donne une configuration initiale à partir de laquelle il faut au moins $k$ étapes pour arriver à la situation finale.\n\nOn peut toujours atteindre la situation finale en au plus $k$ opérations : Considérons une configuration initiale de $4k$ jetons. On appelle première moitié l'ensemble des $2k$ jetons les plus à gauche et deuxième moitié l'ensemble des $2k$ jetons les plus à droite. Parmi les $2k$ jetons de la première moitié, il y a au moins $k$ jetons rouges ou au moins $k+1$ jetons bleus.\n\nS'il y a au moins $k$ jetons rouges, alors on peut échanger chacun des jetons bleus de la première moitié avec un jeton rouge de la deuxième moitié. Comme à chaque étape, on enlève au moins un jeton bleu de la première moitié pour le placer dans la deuxième moitié, il faut au plus $k$ étapes pour échanger tous les jetons bleus.\n\nS'il y a au moins $k+1$ jetons bleus dans la première moitié, alors on peut, en au plus $k-1$ opérations, échanger tous les jetons rouges de la première moitié avec des jetons bleus de la deuxième moitié. À l'issue de ces étapes, les $2k$ jetons bleus sont dans la première moitié et les $2k$ jetons rouges sont dans la deuxième moitié. En échangeant alors les deux groupes de $2k$ jetons consécutifs, on a atteint la situation finale voulue en au plus $k-1+1$ opérations.\n\nIl existe une configuration initiale pour laquelle il faut au moins $k$ opérations : Considérons la configuration $brbr\\ldots br$ (qui consiste en $2k$ blocs de la forme $br$). Notons qu'à chaque opération, le nombre de blocs de jetons rouges consécutifs diminue d'au plus $2$. Comme il y a $2k$ blocs de jetons rouges consécutifs dans la situation initiale et un seul bloc de jetons rouges consécutifs dans la situation finale voulue, on ne peut passer de l'une à l'autre en $k-1$ opérations ou moins. Il faut donc au moins $k$ opérations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers $n$ ayant la propriété suivante : si l'on pose $a_{k} = \\operatorname{ppcm}(k, k+1, \\ldots, k+n-1)$, alors la suite $(a_{k})$ est croissante.",
    "options": [],
    "answer": "1 and 2",
    "solution": "Solution:\n\nMontrons que les naturels ayant cette propriété sont exactement $1$ et $2$.\n\nDéjà, si $n=1$, $a_{k}=k$ est une suite croissante.\n\nEt si $n=2$, puisque deux entiers consécutifs sont premiers entre eux, $a_{k}=k(k+1)$ est une suite croissante également.\n\nSi $n=3$, on remarque que $a_{5}=210 > a_{6}=168$, donc $3$ n'a pas la propriété désirée.\n\nSoit maintenant $n \\geqslant 4$. L'idée pour montrer que $n$ ne vérifie pas la propriété est de trouver un $k$ tel que les $(k+i)_{0 \\leqslant i \\leqslant n-1}$ soient premiers entre eux, tandis que $k+1$ ait un grand facteur commun avec $k+n$, pour avoir $a_{k} > a_{k+1}$. C'est pourquoi on regarde $k = (n-1)! - 1$.\n\nPosons $D = \\operatorname{ppcm}(k+1, \\ldots, k+n-1)$. On a alors $a_{k} = \\operatorname{ppcm}(k, D) = k \\cdot D$. En effet, pour $1 \\leqslant \\mathfrak{i} \\leqslant n-1$, on a $\\operatorname{pgcd}(k, k+i) = \\operatorname{pgcd}(k, \\mathfrak{i}) = 1$ car $\\mathfrak{i} \\mid k+1 = (n-1)!$.\n\nEn outre, $a_{k+1} = \\operatorname{ppcm}(D, k+n)$. Or, $n-1$ divise à la fois $k+1$ (donc $D$) et $k+n$. Donc\n$$\na_{k+1} \\leqslant D \\cdot ((n-2)! + 1) < D \\cdot ((n-1)! - 1) = a_{k}\n$$\nEn effet, pour tout $n \\geqslant 4$, $(n-1)! - (n-2)! = (n-2) \\cdot (n-2)! \\geqslant 4 > 2$, donc $(n-1)! - 1 > (n-2)! + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEmile a créé un exercice pour Benoît. Il lui annonce qu'il a choisi secrètement un polynôme $P$ unitaire de degré $2023$ à coefficients entiers, c'est-à-dire de la forme\n$$\nP(X) = X^{2023} + a_{2022} X^{2022} + a_{2021} X^{2021} + \\ldots + a_{1} X + a_{0}\n$$\nou $a_{0}, a_{1}, \\ldots, a_{2022}$ sont des entiers relatifs. Il donne à Benoît $k$ entiers $n_{1}, n_{2}, \\ldots, n_{k}$, où $k$ est un entier positif, ainsi que la valeur du produit $P\\left(n_{1}\\right) P\\left(n_{2}\\right) \\ldots P\\left(n_{k}\\right)$. À partir de ces connaissances, Benoît doit essayer de retrouver le polynôme $P$.\n\nTrouver l'entier $k$ minimal tel que Emile puisse trouver $P$ et $n_{1}, \\ldots, n_{k}$ afin de s'assurer que le seul polynôme coïncidant avec les informations données à Benoît soit $P$.",
    "options": [],
    "answer": "2023",
    "solution": "Solution:\n\nMontrons que le $k$ minimal est $k = 2023$. En effet, supposons d'abord que $k < 2023$ et qu'Emile choisisse $P$ et $n_{1}, \\ldots, n_{k}$. Posons\n$$\nQ = P + (X - n_{1})(X - n_{2}) \\ldots (X - n_{k})\n$$\nqui est un polynôme à coefficients entiers, toujours unitaire de degré $2023$, et différent de $P$. Alors $P$ et $Q$ sont égaux en chaque $n_{i}$, et donc les produits $P\\left(n_{1}\\right) \\ldots P\\left(n_{k}\\right)$ et $Q\\left(n_{1}\\right) \\ldots Q\\left(n_{k}\\right)$ sont égaux, donc Benoît ne peut pas retrouver la valeur du polynôme $P$.\n\nSupposons maintenant que $k = 2023$. Emile peut alors choisir $n_{1} = 3, n_{2} = 6, \\ldots, n_{2023} = 3 \\cdot 2023$ et poser\n$$\nP = (X - 3)(X - 6) \\ldots (X - 3 \\cdot 2023) + 1\n$$\nAlors le produit $P\\left(n_{1}\\right) P\\left(n_{2}\\right) \\ldots P\\left(n_{k}\\right)$ vaut $1$. Soit $Q$ un polynôme à coefficients entiers, unitaire de degré $2023$ tel que le produit $Q\\left(n_{1}\\right) \\ldots Q\\left(n_{k}\\right)$ soit égal à $1$, il faut montrer que nécessairement $Q = P$. Comme tous les $Q\\left(n_{i}\\right)$ sont entiers, ils valent tous soit $1$, soit $-1$. On peut cependant remarquer que comme les $n_{i}$ sont divisibles par $3$ alors pour tout $i$, $Q\\left(n_{i}\\right)$ est congru au coefficient constant de $Q$. Comme $1$ et $-1$ ne sont pas congrus modulo $3$, tous les $Q\\left(n_{i}\\right)$ sont égaux. Comme $2023$ est impair, ils ne peuvent pas tous être égaux à $-1$, donc ils valent tous $1$. Alors $P - Q$ est un polynôme de degré au plus $2022$ (car les termes d'ordre $2023$ s'annulent) et est nul en chacun des $2023$ valeurs des $n_{i}$, donc $Q = P$, ce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et soit $J$ le centre de son cercle $A$-exinscrit. Soit $K$ le symétrique du point $J$ par rapport au segment $[BC]$. Soient $E$ et $F$ les points sur les droites $(BJ)$ et $(CJ)$ tels que $\\widehat{EAB} = \\widehat{CAF} = 90^{\\circ}$. Montrer que $\\widehat{FKE} + \\widehat{FJE} = 180^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nPuisque la droite $(BJ)$ est la bissectrice extérieure de l'angle $\\widehat{ABC}$, on a $\\widehat{ABJ} = \\widehat{CBJ} = 90^{\\circ} - \\widehat{ABC}/2$. Puisque la symétrie axiale préserve les angles, on a ainsi $\\widehat{KBC} = \\widehat{ABE}$.\n\nNotons $A_1$ le projeté orthogonal du point $J$ sur le segment $[BC]$, de sorte que $A_1$ est également le milieu du segment $[JK]$. Les triangles $KA_1B$ et $BAE$ sont donc rectangles et ont un autre angle en commun, ils sont donc semblables. On a ainsi\n$$\n\\frac{BK}{BE} = \\frac{BA_1}{BA}.\n$$\nCette égalité de rapport combinée avec l'égalité d'angle $\\widehat{KBE} = \\widehat{ABE} + \\widehat{ABK} = \\widehat{KBA_1} + \\widehat{ABK} = \\widehat{ABA_1}$ implique que les triangles $KBE$ et $A_1BA$ sont semblables (ce résultat peut également être formulé comme le fait que $B$ est le centre de la similitude envoyant la paire $(A_1, K)$ sur la paire $(A, E)$ et donc aussi le centre de la similitude envoyant la paire $(A_1, A)$ sur la paire $(K, E)$).\n\nOn déduit que $\\widehat{BKE} = \\widehat{BA_1A}$. De la même façon, $\\widehat{CKF} = \\widehat{CA_1A}$. On a alors, toujours puisque la symétrie préserve les angles:\n$$\n\\widehat{EJF} + \\widehat{EKF} = \\widehat{BJC} + \\widehat{EKF} = \\widehat{BKC} + \\widehat{EKF} = 360^{\\circ} - \\widehat{BKE} - \\widehat{CKF} = 360^{\\circ} - \\widehat{BA_1A} - \\widehat{CA_1A} = 180^{\\circ}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 3$ un entier. On colorie $2n$ sommets d'un $4n+1$-gone régulier. Montrer qu'il existe trois sommets coloriés qui forment un triangle isocèle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn procède par l'absurde en supposant qu'aucun triplet de sommets coloriés ne forme de triangle isocèle.\n\nDans un premier temps, fixons un sommet colorié $O$ et notons $OP_{1}, \\ldots, P_{2n}, P_{-2n}, \\ldots, P_{-1}$ le reste du polygone. Puisque le triangle $OP_{i}P_{-i}$ est isocèle pour $1 \\leqslant i \\leqslant 2n$, au plus l'un des deux sommets $P_{i}$ et $P_{-i}$ est colorié.\n\nSupposons que deux sommets consécutifs ne sont jamais tous les deux coloriés. En particulier, $P_{1}$ et $P_{-1}$ ne sont pas coloriés car ils sont adjacents à $O$ qui est colorié. Mais alors chacune des $2n-1$ paires ($P_{i}, P_{-i}$) de sommets restants contient au plus un point colorié, et au moins $2n-1$ sommets parmi les $P_{i}$ sont coloriés. On conclut que chaque paire $(P_{i}, P_{-i})$ avec $i \\geqslant 2$ contient exactement un sommet colorié. Ainsi, soit $P_{i}$ soit $P_{-i}$ est colorié pour $i \\geqslant 2$.\n\nSupposons que $P_{2n}$ est colorié. Alors $P_{2n-1}$ n'est pas colorié, donc $P_{-2n+1}$ est colorié. Mais alors $P_{-2n+2}$ n'est pas colorié, donc $P_{2n-2}$ est colorié. Le triangle $P_{2n-2}, P_{2n}, P_{-2n+1}$ étant isocèle et colorié, on trouve une contradiction.\n\nOn a montré qu'il existe deux sommets consécutifs coloriés, disons $O$ et $P_{1}$. Les sommets $P_{-1}$ et $P_{2}$ ne sont alors pas coloriés, sous peine de créer un triangle isocèle avec $O$ et $P_{1}$. De même, le triangle $OP_{1}P_{-2n}$ est isocèle en $P_{-2n}$ donc $P_{-2n}$ n'est pas colorié.\n\nConsidérons alors la suite de sommets suivante, de longueur $2(n-1)$ : $P_{3}, P_{-3}, \\ldots, P_{2i-1}, P_{-2i+1}, \\ldots, P_{2n-1}, P_{-2n+1}$.\n\nDeux sommets consécutifs de cette suite forment un triangle isocèle avec $O$ ou avec $P_{1}$, et ne peuvent pas être tous les deux coloriés.\n\nIl en va de même pour la suite $P_{-2}, P_{4}, \\ldots, P_{-2i}, P_{2i+2}, \\ldots, P_{-2n+2}, P_{2n}$.\n\nCes deux suites contiennent donc chacune au plus une moitié de sommets coloriés. Mais comme elles contiennent ensemble exactement $2n-2$ sommets coloriés, les inégalités précédentes sont saturées et dans chaque suite, exactement un sommet sur deux est colorié.\n\nSi $P_{3}$ est colorié, alors $P_{5}$ est colorié et le triangle $P_{1}P_{3}P_{5}$ est isocèle colorié. De même, si $P_{-2}$ est colorié, alors $P_{-4}$ est colorié et le triangle $OP_{-2}P_{-4}$ est isocèle colorié. On déduit que $P_{-2n+1}, P_{2n}$ et $P_{2n-2}$ sont coloriés et forment un triangle isocèle, ce qui conclut.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\left(P_{n}\\right)_{n \\in \\mathbb{N}}$ une suite de polynômes à coefficients entiers. On suppose qu'il existe un polynôme $Q$ unitaire à coefficient entier tel que $P_{n+1}-P_{n}=Q$ pour tout $n \\geqslant 0$. On suppose de plus que pour tout $n \\geqslant 0$, $P_{n}$ a une racine entière. Montrer qu'on est dans un des deux cas suivants :\n- $P_{0}$ et $Q$ ont une racine entière en commun\n- Il existe un polynôme à coefficients entiers $R$ tel que $P_{0}=RQ$ et le degré de $R$ est 1 .",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons que $P_{0}$ et $Q$ n'ont pas de racines entière en commun. Notons que par récurrence immédiate $P_{n}=P_{0}+n Q$ pour tout $n \\geqslant 0$. Notons $x_{n}$ une racine entière de $P_{n}$ pour tout $n \\geqslant 0$, on a que $n Q\\left(x_{n}\\right)=-P_{0}\\left(x_{n}\\right)$, donc $Q\\left(x_{n}\\right)$ divise $P_{0}\\left(x_{n}\\right)$.\n\nNotons que si $x_{k}=x_{j}$ avec $k \\neq j$, alors $P_{0}\\left(x_{k}\\right)+k Q\\left(x_{k}\\right)=0=P_{0}\\left(x_{j}\\right)+j Q\\left(x_{j}\\right)$, donc $(k-j) Q\\left(x_{k}\\right)=0$, donc $Q\\left(x_{k}\\right)=0$. Ainsi $P_{0}\\left(x_{k}\\right)=0$, donc $P_{0}$ et $Q$ ont une racine entière en commun, ce qui est contradictoire. Ainsi la suite $\\left(x_{k}\\right)_{k \\geqslant 0}$ est injective donc prend des valeurs arbitrairement grandes.\n\nFaisons la division euclidienne de $P_{0}$ par $Q$. Comme $Q$ est unitaire à coefficient entier (et $P_{0}$ à coefficients entiers), il existe des polynômes $R$ et $S$ à coefficients entiers tels que $P_{0}=Q R+S$ avec $S$ de degré inférieur ou égal à $Q$.\n\nEn évaluant en $x_{n}$, $P_{0}\\left(x_{n}\\right)=Q\\left(x_{n}\\right) R\\left(x_{n}\\right)+S\\left(x_{n}\\right)$, donc comme $Q\\left(x_{n}\\right)$ divise $P_{0}\\left(x_{n}\\right)$, $Q\\left(x_{n}\\right)$ divise $S\\left(x_{n}\\right)=P_{0}\\left(x_{n}\\right)-Q\\left(x_{n}\\right) R\\left(x_{n}\\right)$. Or pour tout entier $k$ avec $|k|$ assez grand, comme le degré de $S$ est strictement inférieur à celui de $Q$, $|S(k)|<|Q(k)|$. Comme la suite $\\left(x_{n}\\right)$ prend des valeurs arbitrairement grandes, il existe une infinité de $n$ tels que $|S\\left(x_{n}\\right)|<|Q\\left(x_{n}\\right)|$. Comme $Q\\left(x_{n}\\right)$ divise $S\\left(x_{n}\\right)$, on en déduit qu'il existe une infinité de $n$ tels que $S\\left(x_{n}\\right)=0$. Comme la suite est injective, $S$ a une infinité de racines, donc $S=0$. Ainsi $P_{0}=Q R$.\n\nDe plus, en évaluant en $x_{n}$, on a $-n Q\\left(x_{n}\\right)=P_{0}\\left(x_{n}\\right)=Q\\left(x_{n}\\right) R\\left(x_{n}\\right)$. Si $Q\\left(x_{n}\\right)=0$, alors $P_{0}\\left(x_{n}\\right)=0$ et $P_{0}$ et $Q$ ont une racine en commun, contradiction. Ainsi $Q\\left(x_{n}\\right) \\neq 0$, donc $R\\left(x_{n}\\right)=-n$. Ainsi pour tout entier $N \\geqslant 0$, il existe au moins $N$ entiers $y$ tels que $R(y) \\in[-N, N]$.\n\nOr si le degré de $R$ noté $d$ vaut au moins 2, pour un $M$ assez grand et pour un $C>0$, on a pour tout $x$ vérifiant $|x| \\geqslant M, |R(x)| \\geqslant C|x|^{d}$. Ainsi pour $N$ assez grand, le nombre d'entiers tel que $|R(x)| \\leqslant N$ est plus petit que $2 M+1$ (le nombre d'entiers dans $[-M, M]$), auquel on ajoute le nombre d'entiers $|x| \\geqslant M$ tel que $|P(x)| \\leqslant N$. Pour ces $x$, on a $C|x|^{d} \\leqslant N$ donc $|x| \\leqslant (N / C)^{1 / d}$. Ainsi on a au plus $2 M+1+2(N / C)^{1 / d}+1$ entiers tels que $P(y) \\in[-N, N]$, ce qui est strictement inférieur à $N$ pour $N$ assez grand, ce qui est contradictoire.\n\nAinsi le degré de $R$ est au plus 1. Comme on a $R\\left(x_{n}\\right)=-n$ pour tout $n$, le degré de $R$ vaut 1.\n\nAlternativement, pour prouver que le degré de $R$ valait 1, on pouvait remarquer que pour tout $n$, $x_{n+1}-x_{n}$ divise $R\\left(x_{n+1}\\right)-R\\left(x_{n}\\right)=-1$. En particulier $x_{n+1}-x_{n}= \\pm 1$ pour tout $n$. Comme la suite $\\left(x_{n}\\right)$ est injective, on ne peut avoir $x_{n+1}-x_{n}=1$ et $x_{n+2}-x_{n+1}=-1$, ou avoir $x_{n+1}-x_{n}=-1$ et $x_{n+2}-x_{n+1}=1$ sinon $x_{n}=x_{n+2}$. Ainsi on a facilement par récurrence que soit pour tout $n$, $x_{n+1}-x_{n}=1$, et donc $x_{n}=x_{0}+n(x_{1}-x_{0})$, soit $x_{n+1}-x_{n}=-1$, et donc $x_{n}=x_{0}-n(x_{1}-x_{0})$.\n\nDans le premier cas, en posant $S(X)=R\\left(x_{0}+X(x_{1}-x_{0})\\right)-X$, on voit que $S(n)=0$ pour tout $n$. Comme $S$ est un polynôme, c'est le polynôme nul, donc pour tout $x$ réel, $R\\left(x_{0}+x(x_{1}-x_{0})\\right)=x$. Ainsi en posant $x=\\frac{y-x_{0}}{x_{1}-x_{0}}$ (qui est bien défini car $x_{1} \\neq x_{0}$), on a que $R(y)=\\frac{y-x_{0}}{x_{1}-x_{0}}$ pour tout $y$, donc par identification $R(X)=\\frac{X-x_{0}}{x_{1}-x_{0}}$: $R$ est de degré 1.\n\nDans le second cas, en posant $S(X)=R\\left(x_{0}-X(x_{1}-x_{0})\\right)-X$, on voit que $S(n)=0$ pour tout $n$. Comme $S$ est un polynôme, c'est le polynôme nul, donc pour tout $x$ réel, $R\\left(x_{0}-x(x_{1}-x_{0})\\right)=x$. Ainsi en posant $x=\\frac{x_{0}-y}{x_{1}-x_{0}}$ (qui est bien défini car $x_{1} \\neq x_{0}$), on a que $R(y)=\\frac{x_{0}-y}{x_{1}-x_{0}}$ pour tout $y$, donc par identification $R(X)=\\frac{X-x_{0}}{x_{1}-x_{0}}$: $R$ est de degré 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $a, b, c > 0$ tels que $a^{2} < 16 b c$, $b^{2} < 16 a c$ et $c^{2} < 16 a b$. Montrer que\n$$\na^{2} + b^{2} + c^{2} < 2(a b + b c + a c)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nRemarquons tout d'abord que l'on a forcément $\\sqrt{a} < \\sqrt{b} + \\sqrt{c}$. En effet, dans le cas contraire, on aurait\n$$\n16 b c > a^{2} = (\\sqrt{a})^{4} \\geqslant (\\sqrt{b} + \\sqrt{c})^{4} \\geqslant 16 b c\n$$\npar IAG car $\\sqrt{b} + \\sqrt{c} \\geqslant 2 \\sqrt[4]{b c}$, absurde. Ainsi, $\\sqrt{a} - \\sqrt{b} < \\sqrt{c}$. Mais par symétrie, on a aussi $\\sqrt{b} - \\sqrt{a} < \\sqrt{c}$, et on a $\\sqrt{c} < \\sqrt{a} + \\sqrt{b}$. Donc\n$$\n\\begin{gathered}\n|\\sqrt{a} - \\sqrt{b}| < \\sqrt{c} < \\sqrt{a} + \\sqrt{b} \\\\\n\\Longrightarrow a + b - 2 \\sqrt{a b} < c < a + b + 2 \\sqrt{a b} \\\\\n\\Longrightarrow -2 \\sqrt{a b} < c - a - b < 2 \\sqrt{a b} \\\\\n\\Longrightarrow (c - a - b)^{2} < 4 a b \\\\\n\\Longrightarrow a^{2} + b^{2} + c^{2} < 2(a b + b c + a c)\n\\end{gathered}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n- À quelle condition existe-t-il $n+1$ entiers (pas forcément distincts) tels que, pour tout choix de $n$ entiers parmi les $n+1$, leur somme est une puissance de $p$ ?\n\n- À quelle condition existe-t-il $n+1$ entiers strictement positifs (pas forcément distincts) tels que, pour tout choix de $n$ entiers parmi les $n+1$, leur somme est une puissance de $p$ ?",
    "options": [],
    "answer": "First part: existence iff gcd(n, p−1) = 1. Second part: existence iff n is a power of p.",
    "solution": "Solution:\n\nEssayons d'analyser l'énoncé. Posons $x_{1}, \\ldots, x_{n+1}$ les entiers tels que pour tout choix de $n$ entiers parmi les $n+1$ entiers, alors leur somme est une puissance de $p$. Il existe donc $\\alpha_{1}, \\ldots, \\alpha_{n+1}$ des entiers positifs tels que $x_{2}+\\cdots+x_{n+1}=p^{\\alpha_{1}}, x_{1}+x_{3}+\\cdots+x_{n+1}=p^{\\alpha_{2}}, \\ldots, x_{1}+\\cdots+x_{n}=p^{\\alpha_{n+1}}$.\n\nEn particulier, notons déjà que $p^{\\alpha_{i}}=x_{1}+\\cdots+x_{n+1}-x_{i}$ pour tout $i \\in\\{1, \\ldots, n+1\\}$ et que $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}=n\\left(x_{1}+\\cdots+x_{n+1}\\right)$, donc pour tout $i \\in\\{1, \\ldots, n+1\\}$, $x_{i}=\\frac{p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}}{n}-p^{\\alpha_{i}}$.\n\nRéciproquement, si on se donne $\\alpha_{1}, \\ldots, \\alpha_{n+1}$ des entiers positifs tels que $x_{i}=\\frac{p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}}{n}-p^{\\alpha_{i}}$ pour tout $i \\in\\{1, \\ldots, n+1\\}$, alors on voit que $x_{1}+\\cdots+x_{i-1}+x_{i+1}+\\cdots+x_{n+1}=p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}-p^{\\alpha_{i}}=p^{\\alpha_{i}}$.\n\nLa question revient donc à la suivante : existe-t-il des entiers positifs $\\alpha_{1}, \\ldots, \\alpha_{n+1}$ tels que $n$ divise $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}$ pour la première question, et pour la seconde question tels que $\\frac{p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}}{n}-p^{\\alpha_{i}}>0$, i.e. $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}>n p^{\\alpha_{i}}$ pour tout $i$.\n\nEssayons de répondre d'abord à la première partie de la question : à quelle condition sur $n$ existe-t-il $\\alpha_{1}, \\ldots, \\alpha_{n+1}$ tels que $n$ divise $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}$ ? Déjà, modulo $p-1$, on a $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}} \\equiv n+1 \\pmod{p-1}$. Notons $d$ le pgcd de $p-1$ et $n$, alors si $n$ divise $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}$, comme $d$ divise $p-1$, $d$ divise aussi $n+1$. Comme $d$ divise $n$, on obtient que $d$ divise $n+1-n=1$, donc $d=1$. Ainsi il faut que $n$ et $p-1$ soient premiers entre eux.\n\nRéciproquement, si $n$ et $p-1$ sont premiers entre eux, soit $x \\in\\{1, \\ldots, n+1\\}$, on prend $\\alpha_{1}=\\cdots=\\alpha_{x}=0$ et $\\alpha_{x+1}=\\cdots=\\alpha_{n+1}=1$. On cherche donc un certain $x$ tel que $n$ divise $x+(n+1-x)p$, i.e. $x$ tel que $x+p-xp \\equiv 0 \\pmod{n}$, soit $x \\equiv p(p-1)^{-1} \\pmod{n}$ (qui existe car $p-1$ et $n$ sont premiers entre eux). On peut prendre un tel $x$ dans $\\{1, \\ldots, n\\}$ tel que la précédente congruence est vraie. Ainsi la condition pour la première question est exactement $n$ et $p-1$ sont premiers entre eux.\n\nPour la seconde question, on cherche de plus à avoir $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}>n p^{\\alpha_{i}}$ pour tout $i$. Quitte à réordonner les $\\alpha_{i}$, on peut supposer $\\alpha_{1} \\leqslant \\alpha_{2} \\leqslant \\cdots \\leqslant \\alpha_{n+1}$, la condition est équivalente à $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}>n p^{\\alpha_{n+1}}$. Si $\\alpha_{2}<\\alpha_{n+1}$, on a :\n$$\np^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}} \\leqslant 2 p^{\\alpha_{2}}+(n-1) p^{\\alpha_{n+1}} \\leqslant p^{\\alpha_{2}+1}+(n-1) p^{\\alpha_{n+1}} \\leqslant n p^{\\alpha_{n+1}}\n$$\nAinsi on a $\\alpha_{2} \\geqslant \\alpha_{n+1}$, donc $\\alpha_{2}=\\cdots=\\alpha_{n+1}$. Dans ce cas, il est clair que $p^{\\alpha_{1}}+\\cdots+p^{\\alpha_{n+1}}>n p^{\\alpha_{n+1}}$ est vérifié. Il reste donc à chercher à quelle condition sur $n$ existe-t-il $\\alpha_{1} \\leqslant \\alpha_{2}$ deux entiers positifs tels que $n$ divise $p^{\\alpha_{1}}+n p^{\\alpha_{2}}$, donc à quelle condition $n$ divise $p^{\\alpha_{1}}$. Il est alors clair que les solutions sont les puissances de $p$. Ainsi la condition pour la seconde question est que $n$ soit une puissance de $p$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30946,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère un $2022$-gone régulier de côté $1$. Les points sont numérotés $A_{1}, \\ldots, A_{2022}$ dans un certain ordre. Au départ, Elie part du point $A_{1}$, puis saute de point en point vers le point $A_{2}$, puis de $A_{2}$ vers $A_{3}$ etc... chaque fois en prenant l'arc le plus court. Lorsqu'il atteint $A_{2022}$, il rejoint finalement $A_{1}$. Déterminer la valeur maximale de la somme des longueurs des arcs qu'a parcouru Elie, parmi toutes les numérotations de points possibles.",
    "options": [],
    "answer": "2042222",
    "solution": "Solution:\n\nLe problème demande de trouver la valeur maximale d'une somme de longueurs, il contient nécessairement deux parties. Dans un premier temps, on montre que le chemin parcouru par Elie est forcément de longueur inférieure à $2\\left(N^{2}-N+1\\right)$, cette partie s'appelle l'analyse. Dans un second temps, on donne l'exemple d'une numérotation pour laquelle le chemin parcouru est exactement de longueur $2\\left(N^{2}-N+1\\right)$, cette étape s'appelle la construction.\n\n**Analyse :** Dans la suite, on garde la notation $N=1011$. On note également $\\underline{PQ}$ la longueur de l'arc le plus court reliant $P$ et $Q$ et $\\overline{PQ}$ la longueur de l'arc le plus long. Considérons une numérotation $A_{1}, \\ldots, A_{2N}$ dans un ordre quelconque. On pose aussi $A_{2N+1}=A_{1}$.\n\nPuisque majorer chaque portion $A_{i}A_{i+1}$ par sa longueur minimale est une estimation trop grossière, on va plutôt minorer la somme $A_{1}\\overline{A_{2}}+\\ldots+A_{2N}A_{1}$ en majorant des paquets de termes. Plus précisément, nous allons regarder le chemin parcouru entre deux sommets impairs consécutifs.\n\nSoit $1 \\leqslant i \\leqslant N$. On désire montrer que $\\underline{A_{2i-1}A_{2i}}+\\underline{A_{2i}A_{2i+1}} \\leqslant \\overline{A_{2i-1}A_{2i+1}}$, autrement dit, que passer par $A_{2i}$ est moins coûteux que prendre le grand arc $A_{2i-1}A_{2i+1}$. La volonté de démontrer une telle inégalité peut venir de tentatives sur des petits cas.\n\nSi $A_{2i}$ est sur le petit arc $A_{2i-1}A_{2i+1}$, alors $\\underline{A_{2i-1}A_{2i}}+\\underline{A_{2i}A_{2i+1}}=\\underline{A_{2i-1}A_{2i+1}} \\leqslant \\overline{A_{2i-1}A_{2i+1}}$. Supposons désormais que $A_{2i}$ est sur le grand arc $A_{2i-1}A_{2i+1}$. Si le petit arc $A_{2i}A_{2i+1}$ ne contient pas $A_{2i-1}$ ou si le petit arc $A_{2i-1}A_{2i}$ ne contient pas $A_{2i+1}$, alors $\\underline{A_{2i-1}A_{2i}}+\\underline{A_{2i}A_{2i+1}}=\\overline{A_{2i-1}A_{2i+1}}$. Si maintenant $A_{2i-1}$ est sur le petit arc $A_{2i}A_{2i+1}$, alors $A_{2i+1}$ est sur le grand arc $A_{2i-1}A_{2i}$ et $A_{2i-1}A_{2i}+\\underline{A_{2i}A_{2i+1}} \\leqslant \\underline{A_{2i-1}A_{2i}}+\\overline{A_{2i}A_{2i+1}}=\\overline{A_{2i-1}A_{2i+1}}$. Le cas où $A_{2i-1}$ est sur le petit arc $A_{2i}A_{2i+1}$ se traite de façon similaire.\n\n![](attached_image_1.png)\n\nOn a donc\n$$\n\\begin{aligned}\n\\underline{A_{1}A_{2}}+\\ldots+\\underline{A_{2N}A_{1}} &\\leqslant \\overline{A_{1}A_{3}}+\\ldots+\\overline{A_{2N-1}A_{1}} \\\\\n&=\\left(2N-\\underline{A_{1}A_{3}}\\right)+\\ldots+\\left(2N-\\underline{A_{2N-1}A_{1}}\\right) \\\\\n&=2N^{2}-\\left(\\underline{A_{1}A_{3}}+\\ldots+\\underline{A_{2N-1}A_{1}}\\right).\n\\end{aligned}\n$$\n\nIl suffit donc de minimiser la somme $\\underline{A_{1}A_{3}}+\\ldots+\\underline{A_{2N-1}A_{1}}$. Pour cela, notons qu'il existe deux sommets $A$ et $B$ de numéro impair à distance au moins $N-1$ (car il y a $N$ sommets impairs). Comme la somme est cyclique, on peut supposer que $A=A_{1}$ (quitte à renuméroter les sommets). Mais alors le chemin $A_{1}\\ldots A_{2N-1}A_{1}$ nécessite de parcourir un arc $\\underline{A_{1}B}$ pour aller de $A_{1}$ vers $B$ puis un arc $BA_{1}$ pour aller de $B$ vers $A_{1}$. Un tel chemin est donc de longueur au moins le double du petit arc $AB$, soit $2(N-1)$.\n\nIl vient\n$$\n\\underline{A_{1}A_{2}}+\\ldots+\\underline{A_{2N}A_{1}} \\leqslant \\overline{A_{1}A_{3}}+\\ldots+\\overline{A_{2N-1}A_{1}} \\geqslant 2N^{2}-2(N-1)=2\\left(N^{2}-N+1\\right)\n$$\n\n**Construction :** Réciproquement, numérotons les sommets dans l'ordre $A_{1}A_{3}\\ldots A_{2N-1}A_{2}\\ldots A_{2N}$. Le chemin d'Elie est alors de longueur\n$$\n\\underline{A_{1}A_{2}}+\\ldots+\\underline{A_{2N}A_{1}} \\leqslant \\overline{A_{1}A_{3}}+\\ldots+\\overline{A_{2N-1}A_{1}}=\\underbrace{N+N-1+\\ldots+N+N-1}_{N-1\\ \\text{fois}}+N+1=2N^{2}-2N+2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer toutes les fonctions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ telles que pour tous $x, y$ réels, on ait\n$$\nf\\left(x^{2}+x y+f\\left(y^{2}\\right)\\right)=x f(y)+f\\left(x^{2}\\right)+y^{2}\n$$",
    "options": [],
    "answer": "f(x) = x and f(x) = -x",
    "solution": "Solution:\nEn posant $y=0$, on trouve $f\\left(x^{2}+f(0)\\right)=x f(0)+f\\left(x^{2}\\right)$. En remplaçant $x$ par $-x$ dans cette équation, on trouve $f\\left(x^{2}+f(0)\\right)=-x f(0)+f\\left(x^{2}\\right)$, et en combinant on obtient donc $2 x f(0)=0$ pour tout $x$, soit $f(0)=0$.\n\nEn posant $x=0$, on trouve $f\\left(f\\left(y^{2}\\right)\\right)=y^{2}+f(0)=y^{2}$. En posant $y=-x$, on trouve cette fois-ci que $f\\left(f\\left(y^{2}\\right)\\right)=-y f(y)+y^{2}+f\\left(y^{2}\\right)$. On en déduit que\n$$\nf\\left(y^{2}\\right)=y f(y)\n$$\nD'autre part, en posant $x=-f\\left(y^{2}\\right) / y$, on trouve\n$$\n-\\frac{f(y) f\\left(y^{2}\\right)}{y}+y^{2}=0\n$$\nOn déduit que si $y$ est non nul, alors $y^{3}=f(y) f\\left(y^{2}\\right)=y f(y)^{2}$, ou encore $f(y)^{2}=y^{2}$. Puisque si $y=0$, alors $f(0)=0$, on a donc pour tout $y$ réel que $f(y)= \\pm y$.\n\nDans la suite, on suppose qu'il existe $a$ et $b$ tels que $f(a)=a$ et $f(b)=-b$. Notons qu'alors $f\\left(a^{2}\\right)=a^{2}$ et $f\\left(b^{2}\\right)=-b^{2}$, puisque $f\\left(x^{2}\\right)=x f(x)$.\n\nEn remplaçant $x=a$ et $y=b$ dans l'équation, on trouve\n$$\n\\pm\\left(a^{2}+a b-b^{2}\\right)=-a b+b^{2}+a^{2}\n$$\nLe cas où le signe du LHS (Left Hand Side = membre de gauche de l'équation) est un + donne que $b(a-b)=0$. Si $b \\neq 0$, alors $a=b$ ce qui donne $a=f(a)=f(b)=-b=-a$ donc $a=b=0$. Dans tous les cas, $a$ ou $b$ est nul.\n\nLe cas où le signe du LHS est un - donne que $a^{2}=0$ donc $a=0$.\n\nOn déduit que les seules solutions sont les fonctions $f \\equiv x$ et $f \\equiv -x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAvant un championnat, chaque équipe participante propose au plus $t$ couleurs différentes pour son maillot. Un ensemble $S$ d'équipes est dit identifiable si l'on peut assigner à chaque équipe de $S$ une couleur apparaissant dans son ensemble de propositions et n'apparaissant dans aucun ensemble de couleurs d'une autre équipe. Si l'on note $n$ le nombre total de couleurs différentes proposées, déterminer le plus grand entier $g(n, t)$ tel que l'on puisse toujours trouver un ensemble d'équipes identifiable de taille au moins $g(n, t)$.",
    "options": [],
    "answer": "ceil(n/t)",
    "solution": "Solution:\n\nMontrons que le nombre recherché est $g(n, t)=\\left\\lceil\\frac{n}{t}\\right\\rceil$.\n\nSi $n$ s'écrit sous la forme $k \\cdot t + r$, avec $0 \\leqslant r < t$, on peut considérer la situation où $k$ équipes demandent chacune $t$ couleurs toutes distinctes deux à deux, puis une dernière équipe demandant $r$ nouvelles couleurs. Dans cette situation, il y a $\\left\\lceil\\frac{n}{t}\\right\\rceil$ équipes demandant au moins une couleur, donc cela montre déjà que $g(n, t) \\leqslant \\left\\lceil\\frac{n}{t}\\right\\rceil$.\n\nMontrons maintenant avec une sorte d'algorithme glouton que $g(n, t) \\geqslant \\left\\lceil\\frac{n}{t}\\right\\rceil$. On considère le graphe bipartite dont les sommets sont les équipes et les couleurs demandées, et où l'on relie chaque équipe à toutes ses couleurs demandées. Ainsi, le degré d'une équipe est par hypothèse inférieur ou égal à $t$, tandis que le degré d'une couleur est $>0$. Soit $C$ la couleur de degré minimal $d > 0$ et soit $E$ une équipe reliée à $C$. On ajoute alors l'équipe $E$ à notre ensemble identifiable $S$ en cours de construction, et on lui associe la couleur $C$. Ensuite, on supprime dans notre graphe toutes les équipes reliées à $C$ et on supprime également toutes les couleurs demandées par $E$. Ce faisant, si on ajoute une arête du graphe (c'est-à-dire une équipe et une couleur) à notre ensemble identifiable, il le restera car la nouvelle couleur n'aura été demandée par aucune équipe précédente, et la nouvelle équipe n'aura demandé aucune des couleurs précédentes. Il suffit maintenant de montrer que toutes les couleurs qui apparaissent encore dans notre graphe ont un degré $>0$. Puisqu'à chaque étape, on supprime au plus $t$ couleurs du graphe, cela permettra de conclure.\n\nSupposons par l'absurde qu'il existe une couleur $a$ apparaissant encore dans le graphe après l'ajout de $E$ à l'ensemble identifiable, mais telle que son degré soit nul. Cela signifie que les $d' = \\operatorname{deg}(a)$ équipes qui étaient reliées à $a$ juste avant l'ajout de $E$ à $S$ avaient toutes demandé la couleur $C$. Or, puisque $a$ est encore dans le graphe, cela signifie que $E$ n'avait pas demandé la couleur $a$. Cela montre que dans le graphe avant la suppression de $E$, $d = \\operatorname{deg}(C) \\geqslant d' + 1 = \\operatorname{deg}(a) + 1$, ce qui contredit la minimalité du degré de $C$, et nous fournit la contradiction recherchée.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30949,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ un entier et $C>0$ une constante réelle. On suppose qu'il existe une suite $x_{1}, x_{2}, \\ldots, x_{n}$ de réels non tous nuls telle que :\n\n$-x_{1}+x_{2}+\\ldots+x_{n}=0$.\n\n- Pour tout indice $1 \\leqslant i \\leqslant n$, on a soit $x_{i} \\leqslant x_{i+1}$, soit $x_{i} \\leqslant x_{i+1}+C x_{i+2}$ (où on définit $x_{n+1}=x_{1}$ et $x_{n+2}=x_{2}$ ).\n\n1. Montrer que nécessairement $C \\geqslant 2$.\n2. On suppose que $C=2$, montrer que $n$ est nécessairement pair.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans toute la solution, on considèrera tous les indices modulo $n$. Remarquons que la condition de l'énoncé signifie que pour tout $i$, si $x_{i+2} \\leqslant 0$, alors $x_{i} \\leqslant x_{i+1}$ car la condition $x_{i} \\leqslant x_{i+1}+C x_{i+2}$ implique aussi $x_{i} \\leqslant x_{i+1}$. De la même façon dans le cas contraire, $x_{i} \\leqslant x_{i+1}+C x_{i+2}$ dès que $x_{i+1} \\geqslant 0$. On s'intéresse donc au signe des différents termes $x_{k}$.\n\nCommençons par montrer qu'il n'existe jamais deux termes négatifs ou nuls à la suite parmi les $x_{k}$. Supposons par l'absurde que ce soit le cas et qu'il existe donc $i$ avec $x_{i} \\leqslant 0$ et $x_{i+1} \\leqslant 0$. Alors on applique la propriété à l'indice $i-1$. Comme $x_{i+1} \\leqslant 0$, on a donc $x_{i-1} \\leqslant x_{i} \\leqslant 0$. En réappliquant ceci à l'indice $i-1$, on trouve $x_{i-2} \\leqslant 0$, et en continuant ainsi on trouve successivement que tous les $x_{k}$ sont négatifs ou nuls. Mais comme la somme des $x_{k}$ vaut $0$, ceci veut dire qu'ils sont tous nuls, ce qui est impossible d'après l'énoncé.\n\nSoient $\\mathfrak{i}_{1}, \\ldots, \\mathfrak{i}_{K}$ les indices $i$ où $x_{i}$ est négatif ou nul. Maintenant, on peut écrire toutes les inégalités que l'on connaît entre les $x_{i}$ : pour tout $j$,\n$$\n\\begin{gathered}\nx_{\\mathfrak{i}_{j}+1} \\leqslant x_{\\mathfrak{i}_{j}+2}+C x_{\\mathfrak{i}_{j}+3}, \\quad x_{\\mathfrak{i}_{j}+2} \\leqslant x_{\\mathfrak{i}_{j}+3}+C x_{\\mathfrak{i}_{j}+4}, \\ldots, \\quad x_{\\mathfrak{i}_{j+1}-3} \\leqslant x_{\\mathfrak{i}_{j+1}-2}+C x_{\\mathfrak{i}_{j+1}-1} \\\\\nx_{\\mathfrak{i}_{j+1}-2} \\leqslant x_{\\mathfrak{i}_{j+1}-1}, \\quad x_{\\mathfrak{i}_{j+1}-1} \\leqslant x_{\\mathfrak{i}_{j+1}}+C x_{\\mathfrak{i}_{j+1}+1}\n\\end{gathered}\n$$\nLa seule inégalité intéressante faisant apparaître les termes négatifs $x_{i_{j}}$ est cette dernière, on a donc pour tout $j$, $x_{i_{j}} \\geqslant x_{i_{j}-1}-C x_{i_{j}+1}$. Mais on sait que la somme des termes positifs doit se compenser avec la somme des termes négatifs car la somme de tous les $x_{k}$ vaut $0$, et on écrit donc\n$$\n-\\left(x_{i_{1}}+x_{i_{2}}+\\ldots+x_{i_{k}}\\right) \\leqslant\\left(C x_{i_{1}+1}-x_{i_{1}-1}\\right)+\\left(C x_{i_{2}+1}-x_{i_{2}-1}\\right)+\\ldots+\\left(C x_{i_{k}+1}-x_{i_{k}-1}\\right)\n$$\nLe terme de gauche vaut la somme des termes positifs, et en réordonnant les termes à droite on trouve\n$$\n\\begin{aligned}\n\\left(x_{i_{1}+1}+x_{i_{1}+2}\\right. & \\left.+\\ldots+x_{i_{2}-1}\\right)+\\left(x_{i_{2}+1}+x_{i_{2}+2}+\\ldots+x_{i_{3}-1}\\right)+\\ldots+\\left(x_{i_{k}+1}+x_{i_{k}+2}+\\ldots+x_{i_{1}-1}\\right) \\\\\n& \\leqslant\\left(C x_{i_{1}+1}-x_{i_{2}-1}\\right)+\\left(C x_{i_{2}+1}-x_{i_{3}-1}\\right)+\\ldots+\\left(C x_{\\mathfrak{i}_{k}+1}-x_{i_{1}-1}\\right)\n\\end{aligned}\n$$\nLa raison pour mettre les termes de cette manière est la suivante : les seules inégalités qui nous restent parmi les inégalités citées précédemment ne prennent qu'en compte les $x_{k}$ pour $k$ entre un $i_{j}+1$ et le $\\mathfrak{i}_{j+1}-1$ suivant, c'est-à-dire dans une des suites de termes strictement positifs successifs. On va alors montrer que si $C<2$, alors on a nécessairement pour tout $j$,\n$$\nx_{i_{j}+1}+x_{i_{j}+2}+\\ldots+x_{i_{j+1}-1}>C x_{i_{j}+1}-x_{i_{j+1}-1}\n$$\nOn va alors montrer que pour toute suite $a_{1}, a_{2}, \\ldots, a_{m}$ de termes strictement positifs vérifiant les inégalités\n$$\na_{1} \\leqslant a_{2}+C a_{3}, \\quad a_{2} \\leqslant a_{3}+C a_{4}, \\ldots, \\quad a_{m-2} \\leqslant a_{m-1}+C a_{m}, \\quad a_{m-1} \\leqslant a_{m}\n$$\nalors nécessairement $a_{1}+\\ldots+a_{m}>C a_{1}-a_{m}$. Pour cela, on écrit\n$$\n\\begin{gathered}\nC a_{1}-a_{m}=a_{1}+(C-1) a_{1}-a_{m} \\leqslant a_{1}+(C-1) a_{2}+C(C-1) a_{3}-a_{m} \\\\\n=a_{1}+(C-1) a_{2}+(C-1)^{2} a_{3}+(C-1) a_{3}-a_{m} \\leqslant a_{1}+(C-1) a_{2}+(C-1)^{2} a_{3}+(C-1) a_{4}+C(C-1) a_{5}-a_{m} .\n\\end{gathered}\n$$\nOn peut continuer ainsi jusqu'à obtenir, selon la parité de $m$,\n$$\n\\leqslant \\begin{cases}a_{1}+(C-1) a_{2}+(C-1)^{2} a_{3}+\\ldots+(C-1) a_{m-1}+C(C-1) a_{m}-a_{m} & \\text{ si $m$ est impair } \\\\ a_{1}+(C-1) a_{2}+(C-1)^{2} a_{3}+\\ldots+(C-1) a_{m-2}+C(C-1) a_{m-1}-a_{m} & \\text{ si $m$ est pair }\\end{cases}\n$$\nDans le deuxième cas, on utilise la dernière inégalité $a_{m-1} \\leqslant a_{m}$ pour obtenir\n$$\n\\leqslant \\begin{cases}a_{1}+(C-1) a_{2}+(C-1)^{2} a_{3}+\\ldots+(C-1) a_{m-1}+(C(C-1)-1) a_{m} & \\text{ si $m$ est impair } \\\\ a_{1}+(C-1) a_{2}+(C-1)^{2} a_{3}+\\ldots+(C-1) a_{m-2}+(C(C-1)-1) a_{m-1} & \\text{ si $m$ est pair }\\end{cases}\n$$\nLorsque $C<2$, tous les coefficients devant les $a_{i}$ sont strictement inférieurs à $1$ et donc $C a_{1}-a_{m}$ est strictement inférieur à $a_{1}+a_{2}+\\ldots+a_{m}$. Ceci conclut alors la première partie de l'exercice par ce qu'on a vu précédemment.\n\nMaintenant, supposons que $C=2$. Remarquons que dans l'étude de la suite $a_{i}$, on a toujours l'inégalité $C a_{1}-a_{m} \\leqslant a_{1}+\\ldots+a_{m}$, mais par ce que l'on a montré précédemment l'inégalité est toujours stricte dans le cas où $m$ est pair (car $a_{m}$ strictement positif). Ainsi, pour que $C=2$, il faut que les suites de termes positifs consécutifs soient toujours de taille impaire, ce qui signifie que tous les $i_{j+1}-i_{j}$ sont pairs (car on compte aussi un terme négatif). Mais alors comme tous les $i_{j+1}-i_{j}$ sont pairs, ceci signifie en sommant sur $j$ que $n$ est nécessairement pair.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a, b$ deux entiers tels que $\\mathrm{pgcd}(a, b)$ a au moins deux facteurs premiers distincts. Soit $S=\\{x \\in \\mathbb{N} \\mid x \\equiv a[b]\\}$. Un élément de $S$ est dit irréductible s'il ne peut pas s'écrire comme un produit d'au moins deux éléments de $S$ (pas forcément distincts).\n\nMontrer qu'il existe $N>0$ tel que tout élément de $S$ s'écrit comme produit d'au plus $N$ éléments irréductibles de $S$ (pas forcément distincts).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSoit $d=\\operatorname{pgcd}(a, b)$, et écrivons $a=d a'$, $b=d b'$. Commençons par traiter le cas où $\\operatorname{pgcd}(d, b')>1$. Si c'est le cas, alors on a $\\operatorname{pgcd}(a^2, b)=d\\operatorname{pgcd}(d a'^2, b')>d$. Donc, pour tout $k \\geqslant 2$, on a $a^{k} \\not \\equiv a[b]$. En particulier, tout élément de $S$ est irréductible, et donc $N=1$ convient.\n\nSupposons à présent que $d$ et $b'$ sont premiers entre eux. On a alors $a$ et $b'$ premiers entre eux; soit $\\omega$ l'ordre de $a$ modulo $b'$. On a alors que $a^{\\omega+1} \\equiv a[b]$. Donc tout produit de $\\omega+1$ éléments de $S$ est encore un élément de $S$. Ceci signifie en particulier que tout élément non irréductible de $S$ peut s'écrire comme un produit de $k$ éléments de $S$ pour un certain $k \\in \\llbracket 2, \\omega \\rrbracket$.\n\nSoient $p$ et $q$ deux diviseurs premiers de $d$, et soit $x \\in S$. Si $x$ est irréductible, c'est bon; sinon, écrivons $x=\\mathfrak{u}_1 \\mathfrak{u}_2 \\ldots \\mathfrak{u}_k$ avec $\\mathfrak{u}_1, \\ldots, \\mathfrak{u}_k \\in S$, pour un $k \\in \\llbracket 2, \\omega \\rrbracket$. Montrons d'abord qu'on peut supposer $v_{\\mathfrak{p}}(\\mathfrak{u}_{\\mathfrak{j}})<\\varphi(\\mathfrak{b}')+v_{\\mathfrak{p}}(d)$ pour tout $\\boldsymbol{j}<k$. Pour cela, on remarque que, si $v_{\\mathfrak{p}}(\\mathfrak{u}_{\\mathfrak{j}}) \\geqslant \\varphi(b')+v_{\\mathfrak{p}}(d)$, alors on peut remplacer $\\mathfrak{u}_j$ par $\\frac{\\mathfrak{u}_j}{p^{\\varphi(b')}}$ et $\\mathfrak{u}_k$ par $p^{\\varphi(b')} \\mathfrak{u}_k$. On a $p^{\\varphi(b')} \\equiv 1[b']$, donc $p^{\\varphi(b')} \\mathfrak{u}_k \\equiv \\mathfrak{u}_k \\equiv a[b']$. D'autre part, $d \\mid \\mathfrak{u}_k$ donc $p^{\\varphi(b')} \\mathfrak{u}_k \\equiv 0 \\equiv a[d]$. Donc, d'après le théorème des restes chinois, on a $p^{\\varphi(b')} \\mathfrak{u}_k \\equiv a[b]$. D'autre part, on a aussi $\\frac{u_j}{p^{\\varphi(b')}} \\equiv a[b']$, et on a également (puisque $v_{\\mathfrak{p}}(u_j) \\geqslant \\varphi(b')+v_p(d)$) que $d \\mid \\frac{u_j}{p^{\\varphi(b')}}$. Donc $\\frac{u_j}{p^{\\varphi(b')}} \\equiv a[b]$.\n\nOn a à présent $v_{\\mathfrak{p}}(\\mathfrak{u}_{\\mathfrak{j}})<\\varphi(b')+v_{\\mathfrak{p}}(d)$ pour $\\boldsymbol{j}<k$. De même, on peut supposer que $v_{\\boldsymbol{q}}(\\mathfrak{u}_k)<\\varphi(b')+v_q(d)$. Toute écriture de $\\boldsymbol{u}_j$ ($j<k$) sous forme de produit d'irréductibles fait apparaître seulement des multiples de $d$, donc de $p$, et comprend donc au plus $v_p(u_j)$ facteurs. De même pour $q$ avec $\\mathfrak{u}_k$. Finalement, on a réussi à écrire $x$ sous la forme d'un produit d'au plus $(\\omega-1)(\\varphi(b')+v_p(d)-1)+\\varphi(b')+v_q(d)-1$ facteurs irréductibles, comme souhaité.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30951,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn considère 51 entiers strictement positifs de somme 100 sur une ligne. Montrer que pour tout entier $1 \\leqslant k < 100$, il existe des entiers consécutifs de somme $k$ ou $100-k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn reformule légèrement l'énoncé pour le rendre plus visuel : on se représente un cercle de périmètre 100 gradué (de 0 à 99 par exemple) sur lequel on a marqué en noir 51 graduations et en pointillés les 49 autres. Sans perte de généralité on marque en noir la graduation 0. Nos entiers positifs correspondent aux longueurs entre les graduations noires. On vérifie facilement qu'il s'agit d'une représentation bijective du problème.\n\nPour tout $k$, par principe des tiroirs, comme l'application qui à $j$ associe $j+k$ est injective (modulo 100), il existe une graduation marquée en noir $j$ telle que son image $j+k$ l'est aussi. Deux cas se présentent alors. Soit le segment $[j, j+k]$ ne coupe pas 0 et alors on a bien les entiers souhaités. Soit il coupe 0 et alors $[j+k, j]$ ne coupe pas 0 et est de longueur $n-k$. Dans les deux cas, on a trouvé des entiers consécutifs de somme $k$ ou $n-k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30952,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer qu'il existe une infinité d'entiers $n$ tels que l'écriture de $n^{2}$ en base 4 ne contient que des 1 et des 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn montre un résultat un peu plus fort, qui se prête mieux à un raisonnement par récurrence : il existe une infinité d'entiers $n$ tels que l'écriture de $n^{2}$ en base 4 ne contient que des 1 et des 2, et tels que le premier et le dernier chiffre de l'écriture en base 4 de $n^{2}$ vaut 1. Un premier exemple est le nombre $5^{2}=25=121_{4}$. Ensuite, on montre que si l'on a un tel $n$, on peut en construire un nouveau qui lui est strictement supérieur, ce qui montrera bien qu'il existe une infinité d'entiers ayant la propriété recherchée.\n\nSoit donc $n$ tel que l'écriture de $n^{2}$ en base 4 ne contient que des 1 et des 2, et commence et se termine par un 1. On suppose que cette écriture contient $k$ chiffres. On pose alors $m=2^{2k-1} \\cdot n + n$. On a\n$$\nm^{2} = 4^{2k-1} \\cdot n^{2} + 4^{k} \\cdot n^{2} + n^{2}\n$$\nEn base $4$, $m^{2}$ est donc constitué de trois copies de $n^{2}$, avec $n^{2}$ et $4^{k} \\cdot n^{2}$ qui sont deux copies disjointes adjacentes, et $4^{2k-1} \\cdot n^{2}$ et $4^{k} \\cdot n^{2}$ qui ont exactement une puissance de 4 en commun dans leur écriture, à savoir $4^{2k-1}$. Mais par hypothèse, les coefficients devant 1 et devant $4^{k-1}$ de $n^{2}$ sont des 1, donc le coefficient devant $4^{2k-1}$ de $m^{2}$ est un 2. En outre, tous les autres coefficients de l'écriture en base 4 de $m^{2}$ apparaissaient dans celle de $n^{2}$, et sont donc bien des 1 ou des 2. Enfin, les premiers et derniers coefficients de l'écriture en base 4 de $m^{2}$, à savoir les coefficients devant $4^{0}$ et $4^{3k-2}$, sont bien des 1 par hypothèse de récurrence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $\\mathbb{Z}[X]$ l'ensemble des polynômes à coefficients entiers. Trouver toutes les fonctions $f: \\mathbb{Z}[X] \\rightarrow \\mathbb{Z}[X]$ telles que pour tous $P, Q \\in \\mathbb{Z}[X]$ et $r \\in \\mathbb{Z}$, on ait\n$$\nP(r)|Q(r) \\Longleftrightarrow (f(P))(r)|(f(Q))(r)\n$$",
    "options": [],
    "answer": "All such functions are exactly those of the form f(P)(X) = s(P) · A(X) · (P(X))^n, where n is a positive integer, A ∈ Z[X] satisfies A(r) ≠ 0 for every integer r, and s(P) ∈ {+1, −1} may depend on P (but not on X).",
    "solution": "Solution:\n\nSoit $f$ une fonction solution de l'énoncé. Commençons par remarquer que si $P, Q, r$ sont tels que $|P(r)|=|Q(r)|$, alors $f(P)(r)$ et $f(Q)(r)$ se divisent mutuellement, et donc $|f(P)(r)|=|f(Q)(r)|$.\n\nRemarquons aussi que si $P$ et $Q$ sont deux polynômes tels que pour une infinité d'entiers $r$, $P(r) \\mid Q(r)$, alors c'est aussi le cas pour $f(P)$ et $f(Q)$, et donc $f(P)$ divise $f(Q)$ dans $\\mathbb{Q}[X]$. En effet, on écrit la division euclidienne $f(Q)=R f(P)+S$ avec $R, S \\in \\mathbb{Q}[X]$ et $\\operatorname{deg}(S)<\\operatorname{deg}(f(P))$. On multiplie par les dénominateurs des coefficients de $R$ et $S$ pour obtenir une égalité de la forme $a f(Q)=R' f(P)+S'$ avec $a$ entier et $R', S'$ deux polynômes à coefficients entiers. Alors pour une infinité de $r$ entiers, $a f(P)(r) \\mid S'(r)$. Comme $|a f(P)(r)|$ croît plus rapidement que $|S'(r)|$ lorsque $|r|$ tend vers l'infini, on obtient que nécessairement $S'=0$ et donc que $f(P)=R f(Q)$ et $f(P)$ divise $f(Q)$ dans $\\mathbb{Q}[X]$. Notamment, $f(Q)=0$ ou $\\operatorname{deg}(f(Q)) \\geqslant \\operatorname{deg}(f(P))$. On utilisera souvent la conséquence suivante : si $P$ divise $Q$ dans $\\mathbb{Z}[X]$, alors $f(P)$ divise $f(Q)$ dans $\\mathbb{Q}[X]$.\n\nCommençons par trouver les antécédents du polynôme nul par $f$. Soit $Q$ tel que $f(Q)=0$. Alors pour tout polynôme $P$, on a que pour tout $r$, $f(P)(r) \\mid f(Q)(r)=0$, et alors $P(r) \\mid Q(r)$. Par la remarque précédente, $Q=0$ ou $\\operatorname{deg}(Q) \\geqslant \\operatorname{deg}(P)$. Comme on peut prendre $P$ quelconque, on a nécessairement $Q=0$. Ainsi, $f(P) \\neq 0$ pour tout $P \\neq 0$.\n\nOn va s'intéresser à l'image des constantes par $f$. Pour toute constante $c$ non nulle, posons $P(X)=c$ et $Q(X)=X$, alors $P(k c) \\mid Q(k c)$ pour tout entier $k$, et donc $f(c)$ doit diviser $f(Q)=f(X)$ dans $\\mathbb{Q}[X]$. Notamment, $\\operatorname{deg}(f(c)) \\leqslant \\operatorname{deg}(f(X))$. Comme les degrés des images des constantes sont bornés, il existe donc une constante $C \\neq 0$ dont le degré est maximal. Alors pour tout entier $k$, $f(C)$ divise $f(k C)$ dans $\\mathbb{Q}[X]$, mais $\\operatorname{deg}(f(k C)) \\leqslant \\operatorname{deg}(f(C))$, et donc il existe un rationnel $g(k)$ tel que $f(k C)=g(k) f(C)$.\n\nMontrons que la fonction $g$ ainsi définie est, au signe en chaque point près, un polynôme à coefficients rationnels. Pour cela, remarquons que pour tout entier $k$, on a $|k C|=|P(k C)|$ avec $P(X)=X$, et donc $|f(k C)(k C)|=|f(X)(k C)|$, et donc\n$$\n|f(X)(k C)|=|g(k)||f(C)(k C)|\n$$\nMais on sait déjà que $f(C)$ divise $f(X)$ dans $\\mathbb{Q}[X]$, soit $\\hat{g}$ leur quotient (qui est un polynôme dans $\\mathbb{Q}[X]$ ), alors on a\n$$\n\\forall k \\in \\mathbb{Z},\\ |g(k)|=|\\hat{g}(C k)|\n$$\nAppliquons maintenant la propriété de l'énoncé avec $P(X)=k C$ et $Q(X)=X+a$ pour $k$, $a$ deux entiers. On sait que $|P(r)|=|Q(r)|$ en $r=k C-a$, et on a donc\n$$\n\\begin{gathered}\n|f(X+a)(k C-a)|=|f(k C)(k C-a)| \\\\\n|f(X+a)(k C-a)|=|\\hat{g}(k C)||f(C)(k C-a)| .\n\\end{gathered}\n$$\nOr, deux polynômes dont les valeurs absolues sont égales en une infinité d'entiers sont égaux au signe près. En faisant varier $k$, on trouve donc l'égalité polynomiale (au signe près)\n$$\nf(X+a)(T)= \\pm \\hat{g}(T+a) f(C)(T)\n$$\nou le signe ne dépend pas de $T$, mais peut dépendre de $a$. A présent, on peut calculer les valeurs de $f$ en toutes les constantes : soit $n$ un entier, on a pour tout $a$ entier,\n$$\n|f(n)(a)|=|f(X+n-a)(a)|=|\\hat{g}(n) f(C)(a)|\n$$\net on a donc l'égalité polynomiale $f(n)= \\pm \\hat{g}(n) \\cdot f(C)$. Enfin, pour un polynôme $P$ quelconque, on peut écrire pour tout $r$ entier,\n$$\n|f(P)(r)|=|f(P(r))(r)|=|\\hat{g}(P(r)) f(C)(r)|\n$$\net donc $f(P)= \\pm \\hat{g}(P) \\cdot f(C)$. Il reste à trouver les formes de $\\hat{g}$ qui conviennent. On sait que si $m \\mid n$ sont deux entiers, alors pour tout entier $r$, $f(m)(r) \\mid f(n)(r)$, et donc $\\hat{g}(m) f(C)(r) \\mid \\hat{g}(n) f(C)(r)$. En choisissant $r$ tel que $f(C)(r) \\neq 0$ (possible car $f(C) \\neq 0$ ), on obtient $\\hat{g}(m) \\mid \\hat{g}(n)$. Mais alors pour tout entier $k$, on sait que lorsque $n \\rightarrow +\\infty$, $\\hat{g}(n k) / \\hat{g}(n)$ est un entier qui doit tendre vers $k^{\\text{deg}(\\hat{g})}$. Ainsi, pour $n$ assez grand, $\\hat{g}(n k)=k^{\\operatorname{deg}(\\hat{g})} \\hat{g}(n)$, et donc les polynômes $\\hat{g}(kX)$ et $k^{\\operatorname{deg}(\\hat{g})} \\hat{g}(X)$ sont égaux. Ainsi, on trouve $\\hat{g}(k)=\\hat{g}(1) k^{\\operatorname{deg}(\\hat{g})}$ et $\\hat{g}$ est donc un monôme de la forme $\\hat{g}(X)=a X^{n}$ pour un rationnel $a$ et un entier positif $n$.\n\nAinsi, on a montré qu'il existait un polynôme $A=a f(C) \\in \\mathbb{Q}[X]$ et un entier $n$ positif ou nul tel que pour tout $P \\in \\mathbb{Z}[X]$,\n$$\nf(P)(X)= \\pm A(X) P(X)^{n}\n$$\nou le signe dans le $\\pm$ peut dépendre de $P$, mais pas de $X$. Avec $P=1$, on obtient que $A$ est un polynôme à coefficients entiers. De plus, en posant $P=2, Q=3$ dans l'hypothèse de l'énoncé, on obtient que pour tout entier $r$,\n$$\nA(r) 2^{n} \\left|A(r) 3^{n} \\Longleftrightarrow 2\\right| 3\n$$\nCeci implique que $A(r) \\neq 0$ pour tout $r$, et que $n \\geqslant 1$.\n\nSupposons maintenant que $f$ soit de la forme ci-dessus, avec $A \\in \\mathbb{Z}[X]$ ne s'annulant en aucun entier et $n \\geqslant 1$. Alors on a bien que pour tout $r$ entier et $P, Q \\in \\mathbb{Z}[X]$,\n$$\nP(r)\\left|Q(r) \\Longleftrightarrow \\pm A(r) P(r)^{n}\\right| \\pm A(r) Q(r)^{n}\n$$\net donc $f$ vérifie la condition de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30954,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPour tout entier $k \\geqslant 1$, on note $p(k)$ le plus petit nombre premier ne divisant pas $k$. Soit $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$ une suite telle que $a_{0} \\in \\mathbb{N}^{*}$ et pour tout $n \\geqslant 0$, $a_{n+1}$ est le plus petit entier strictement positif différent de $a_{0}, \\ldots, a_{n}$ tel que $a_{n}^{a_{n+1}}-1$ est divisible par $p\\left(a_{n}\\right)$. Démontrer que tout entier strictement positif apparaît une unique fois dans la suite $\\left(a_{n}\\right)_{n \\in \\mathbb{N}}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDéjà notons que la suite est bien définie : si $a_{1}, \\ldots, a_{n}$ sont construits, comme $a_{n}$ est premier avec $p\\left(a_{n}\\right)$, il existe une infinité d'entiers $k$ tels que $a_{n}^{k} \\equiv 1\\left(\\bmod p\\left(a_{n}\\right)\\right)$ qui sont tous les multiples de l'ordre de $a_{n}$ modulo $p\\left(a_{n}\\right)$. En particulier, il existe au moins un de ces entiers $k$ n'apparaissant pas parmi $a_{1}, \\ldots, a_{n}$, donc $a_{n+1}$ existe bien.\n\nNotons que chaque entier apparaît au plus une fois dans la suite par construction. Il faut donc montrer que tous apparaissent au moins une fois. Supposons qu'il existe un entier $N$ qui n'apparaît pas dans la suite $\\left(a_{n}\\right)$, on prend alors $N$ minimal. On note $N_{0}$ un entier tel que si $n \\geqslant N_{0}$, $a_{n}>N$.\n\nMontrons que pour tout premier $q$, il existe un nombre fini de $n$ tels que $p\\left(a_{n}\\right)=q$. Supposons que ce n'est pas le cas, et fixons $q$ un nombre premier tel que $p\\left(a_{n}\\right)=q$ pour une infinité de $n$. On se donne alors $\\phi: \\mathbb{N} \\rightarrow \\mathbb{N}$ strictement croissante telle que $p\\left(a_{\\phi(n)}\\right)=q$. Notons alors que tout multiple de $q-1$ apparaît : en effet si on fixe $M$ un multiple de $q-1$, alors comme les $p\\left(a_{\\phi(n)+1}\\right)$ sont deux à deux distincts, il existe $n$ tel que $a_{\\phi(n)+1}>M$. Or comme $q-1$ divise $M$, $a_{\\phi(n)}^{M} \\equiv 1 (\\bmod q-1)$ par petit Fermat, ce qui contredit la définition de $a_{\\phi(n)+1}$.\n\nAinsi tous les multiples de $q-1$ apparaissent. On note $p_{1}<\\cdots<p_{k}$ les nombres premiers strictement inférieurs à $q$. On sait que pour tout $j \\geqslant 1$, $\\left(p_{1} \\ldots p_{k}\\right)^{j(q-1)}$ est congru à $1$ modulo $q$, divisible par $q-1$ pour $j$ assez grand (car tous les facteurs de $q-1$ sont parmi $p_{1}, \\ldots, p_{k}$) donc apparaît dans la suite, et vérifie $p\\left(\\left(p_{1} \\ldots p_{k}\\right)^{j(q-1)}\\right)=q$. Comme il y a une infinité de tels termes, il existe $N_{1}>N_{0}$ tel que $a_{N_{1}}=\\left(p_{1} \\ldots p_{k}\\right)^{j(q-1)}$ pour $j$ assez grand. Ainsi $a_{N_{1}}^{N} \\equiv 1(\\bmod q)$, ce qui contredit le fait que $a_{N_{1}+1}>N$.\n\nAinsi on a bien prouvé que pour tout premier $q$, il existe un nombre fini de $n$ tels que $p\\left(a_{n}\\right)=q$. En particulier, pour tout $M \\in \\mathbb{N}$, comme la suite $\\left(p\\left(a_{n}\\right)\\right)$ prend un nombre fini de fois chaque valeur entre $1$ et $M$, à partir d'un certain rang $p\\left(a_{n}\\right)>M$, donc $\\left(p\\left(a_{n}\\right)\\right)$ tend vers $+\\infty$. En particulier, pour tout $k$, à partir d'un certain rang, $p_{1} \\ldots p_{k}$ divise $a_{n}$.\n\nSoit $k$ et $m$ tels que $p_{1} \\ldots p_{k}$ divise $a_{m}$, $p_{k+1}$ ne divise pas $a_{m}$, et $p_{1} \\ldots p_{k+1}$ divise $a_{n}$ pour tout $n \\geqslant m+1$. On a alors $p\\left(a_{m}\\right)=p_{k+1}$, et $a_{n}^{p_{1} \\ldots p_{k} p_{k+1}} \\equiv 1\\left(\\bmod p_{k+1}\\right)$. Or par Petit Fermat, $a_{n}^{p_{1} \\ldots p_{k} p_{k+1}} \\equiv a^{p_{1} \\ldots p_{k}}\\left(\\bmod p_{k+1}\\right)$, donc pour tout $i$ vérifiant $1 \\leqslant i \\leqslant p_{k}-1$, $a^{i p_{1} \\ldots p_{k}} \\equiv 1\\left(\\bmod p_{k+1}\\right)$. Ainsi pour tout $i$ vérifiant $1 \\leqslant i \\leqslant p_{k}-1$, $i p_{1} \\ldots p_{k}$ apparaît dans la suite $\\left(a_{n}\\right)$. En prenant $i$ un inverse de $p_{1} \\ldots p_{k}$ modulo $p_{k+1}-1$ dans $\\left\\{1, \\ldots, p_{k+1}-1\\right\\}$, on a alors que $\\left(i p_{1} \\ldots p_{k}\\right)^{N} \\equiv N\\left(\\bmod p_{k+1}\\right)$, donc si $a_{n}=i p_{1} \\ldots p_{k}$, alors $a_{n+1}<N$, donc $n \\leqslant N_{0}$. Ainsi si de tels $k$, $m$ existent, alors $p_{k}$ divise $a_{1} \\ldots a_{n+1}$, donc $k$ est borné.\n\nOr pour tout $K \\in \\mathbb{N}$, on peut trouver $k \\geqslant K$ et $m$ tels que $p_{1} \\ldots p_{k}$ divise $a_{m}$, $p_{k+1}$ ne divise pas $a_{m}$, et $p_{1} \\ldots p_{k+1}$ divise $a_{n}$ pour tout $n \\geqslant m+1$. En effet, notons pour tout $k \\geqslant 1$ $r_{k}$ le premier rang à partir duquel $p_{1} \\ldots p_{k}$ divise $a_{n}$. La suite $\\left(r_{j}\\right)$ est croissante, et ne peut être stationnaire sinon un $a_{n}$ serait divisible par tous les nombres premiers. Donc elle tend vers $+\\infty$, pour tout $K$, il existe $k \\geqslant K$ tel que $r_{k+1}>r_{k}$. Ainsi $a_{r_{k+1}-1}$ est divisible par $p_{1} \\ldots p_{k}$ car $r_{k+1}-1 \\geqslant r_{k}$, et pas par $p_{k+1}$. Mais tous les termes après $r_{k+1}$ sont divisibles par $p_{1} \\ldots p_{k+1}$, donc $m=r_{k+1}-1$ convient. On a donc bien obtenu que pour tout $K \\in \\mathbb{N}$, on peut trouver $k \\geqslant K$ et $m$ tels que $p_{1} \\ldots p_{k}$ divise $a_{m}$, $p_{k+1}$ ne divise pas $a_{m}$, et $p_{1} \\ldots p_{k+1}$ divise $a_{n}$ pour tout $n \\geqslant m+1$, donc on a abouti à une contradiction.\n\nAinsi chaque entier apparaît une unique fois dans la suite.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $\\mathrm{C}_{1}, \\mathrm{C}_{2}, \\ldots \\mathrm{C}_{\\mathrm{n}}$ des cercles de même rayon disposés dans le plan de sorte qu'ils ne soient jamais tangents 2 à 2 et qu'il existe toujours un chemin passant par les cercles pour aller d'un point de l'un d'entre eux à un autre (autrement dit, les cercles sont connectés). En notant $S$ l'ensemble des points d'intersection des cercles, montrer que $|S| \\geqslant n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDans un premier temps, on reformule le problème en termes de graphes. On note $C$ l'ensemble des centres des cercles et $S$ l'ensemble des points d'intersections qui constituent les sommets d'un graphe $G$. On relie par une arête tout couple $(c, s) \\in C \\times S$ si $s$ appartient au cercle de centre $c$.\n\nRemarquons alors que pour toute arête $(c, s)$, $\\operatorname{deg}(c) \\geqslant \\operatorname{deg}(s)$. En effet, d'après l'énoncé, à tout point $c_{i}$ relié à $s$ autre que $c$ on peut associer un autre point d'intersection de $\\mathrm{c}_{\\mathrm{i}}$ avec $c$ car les cercles ne sont pas tangents. Étant donnés $\\boldsymbol{c}_{\\boldsymbol{i}}$ et $\\boldsymbol{c}_{\\boldsymbol{j}}$ distincts, ces deuxièmes points $s_{i}$ et $s_{j}$ ne peuvent pas être confondus puisque tous les cercles sont de même rayon.\n\nMaintenant que le problème est reformulé, on peut raisonner uniquement sur le graphe : $G$ est un graphe bipartite entre $C$ et $S$ tel que pour toute arête $(c, s)$, $\\operatorname{deg}(c) \\geqslant \\operatorname{deg}(s)$. Il vient aisément que $|C| \\leqslant |S|$.\n\nEn effet, en notant $A$ l'ensemble des arêtes et $a = (c_{a}, s_{a})$ ces arêtes :\n\n$$\n|C| = \\sum_{a \\in \\mathcal{A}} \\frac{1}{\\operatorname{deg}(c_{a})} \\leqslant \\sum_{a \\in \\mathcal{A}} \\frac{1}{\\operatorname{deg}(s_{a})} = |S|\n$$\n\nce qui conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle et soient $A_{1}$, $B_{1}$ et $C_{1}$ les points de contact respectifs du cercle $A$-exinscrit, noté $\\omega_{A}$, avec le côté $BC$ et les demi-droites $[AC)$ et $[AB)$. Soit $P$ le milieu du segment $[B_{1}C_{1}]$. La droite $(A_{1}P)$ recoupe le cercle $\\omega_{A}$ au point $X$. La tangente au cercle circonscrit au triangle $ABC$ au point $A$ et la tangente au cercle $\\omega_{A}$ au point $X$ se coupent en $R$. Montrer que $RX = RP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nEn traçant le cercle de centre $R$ de rayon $RA$, on s'aperçoit que celui-ci passe également par $P$. On va donc plutôt introduire $R'$ le centre du cercle circonscrit au triangle $APX$ et montrer que les droites $(AR')$ et $(XR')$ sont respectivement tangentes aux cercles circonscrits au triangle $ABC$ et au triangle $A_{1}B_{1}C_{1}$.\n\nPour cela, notons que puisque $P$ est le milieu du segment $[B_{1}C_{1}]$ dont la droite $(AI_{A})$ est la médiatrice, les points $A$, $P$ et $I_{A}$ sont alignés. De plus, puisque $\\widehat{I_{A}B_{1}A} = 90^{\\circ} = 180^{\\circ} - \\widehat{I_{A}C_{1}A}$, les points $B_{1}$, $I_{A}$, $C_{1}$ et $A$ sont cocycliques. Par puissance d'un point, on a :\n$$\nPA \\times PI_{A} = PB_{1} \\times PC_{1} = PA_{1} \\times PX\n$$\nde sorte que les points $A_{1}$, $I_{A}$, $X$ et $A$ sont cocycliques par la réciproque de la puissance d'un point.\n\nMontrons à présent que $(AR')$ est tangente au cercle circonscrit au triangle $ABC$. D'une part, d'après le théorème de l'angle au centre :\n$$\n\\widehat{PAR'} = 90^{\\circ} - \\widehat{PXA}.\n$$\nMais en utilisant la cocyclicité des points $A_{1}$, $I_{A}$, $X$ et $A$, on a\n$$\n\\widehat{PAR'} = 90^{\\circ} - \\widehat{A_{1}XA} = 90^{\\circ} - \\widehat{A_{1}I_{A}A}.\n$$\nSoit alors $D$ le pied de la bissectrice de l'angle $\\widehat{BAC}$ et $S$ le pôle Sud du sommet $A$ dans $ABC$. $D$ et $S$ sont alignés avec $A$ et $I_{A}$. Alors\n$$\n90^{\\circ} - \\widehat{A_{1}I_{A}A} = \\widehat{CDI_{A}} = 180^{\\circ} - \\widehat{ABC} - \\frac{1}{2} \\widehat{BAC} = \\widehat{ABC} + \\frac{1}{2} \\widehat{ACB} = \\widehat{ACP}\n$$\nAinsi, $\\widehat{PAR'} = \\widehat{ACP}$, donc la droite $(AR')$ est tangente au cercle circonscrit au triangle $ABC$.\n\nMontrons que la droite $(R'X)$ est tangente au cercle circonscrit au triangle $A_{1}B_{1}C_{1}$. On a là aussi par angle au centre :\n$$\n\\widehat{A_{1}XR'} = \\widehat{PXR'} = 90^{\\circ} - \\widehat{PAX} = 90^{\\circ} - \\widehat{I_{A}AX}\n$$\nPuis, par angle inscrit dans le cercle passant par $A$, $A_{1}$, $I_{A}$ et $X$, et en utilisant que le triangle $I_{A}A_{1}X$ est isocèle en $I_{A}$, on a\n$$\n90^{\\circ} - \\widehat{I_{A}AX} = 90^{\\circ} - \\widehat{I_{A}A_{1}X} = \\frac{1}{2} \\widehat{A_{1}I_{A}X} = \\widehat{A_{1}B_{1}X}\n$$\noù l'on a utilisé le théorème de l'angle au centre dans le cercle $\\omega_{A}$. Ainsi, la droite $(R'X)$ est tangente à $\\omega_{A}$ et $R = R'$, ce qui conclut.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier vérifiant $n \\geqslant 2$. On note $d$ le plus grand diviseur de $n$ différent de $n$. On suppose que $d > 1$. Démontrer que $n + d$ n'est pas une puissance de 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSupposons par l'absurde que $n + d$ est une puissance de $2$. Notons que $d$ divise $n$, donc $d$ divise $n + d$, donc $d$ divise une puissance de $2$. Ainsi $d$ est une puissance de $2$ : on pose $d = 2^{k}$ pour un certain $k \\in \\mathbb{N}$. On a $k \\geqslant 1$ car $d \\neq 1$. Posons $n = d a = 2^{k} a$ avec $a \\in \\mathbb{N}^{*}$. Comme $d$ est différent de $n$, on a $a \\geqslant 2$. Ainsi $2^{k-1} a$ est un diviseur de $n$, supérieur ou égal à $2^{k} = d$, et différent de $n$. Par hypothèse $d = 2^{k-1} a$, donc $a = 2$. Ainsi $n = 2^{k+1}$.\n\nOn trouve alors $n + d = 2^{k}(2 + 1) = 3 \\times 2^{k}$ qui n'est pas une puissance de $2$, ce qui est une contradiction. Ainsi $n + d$ n'est pas une puissance de $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a$ et $b$ deux réels. Supposons que $2a + a^{2} = 2b + b^{2}$. Montrer que si $a$ est un entier (pas forcément positif), alors $b$ est aussi un entier.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn réécrit l'égalité\n$$\n\\begin{gathered}\n2a - 2b = b^{2} - a^{2} \\\\\n2(a - b) = -(a + b)(a - b)\n\\end{gathered}\n$$\nAlors soit $b = a$, et alors $b$ est un entier, soit\n$$\n2 = -a - b\n$$\nd'où\n$$\nb = -2 - a\n$$\net $b$ est encore un entier.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAurélien découpe une feuille de papier en 7 morceaux. Une étape consiste ensuite à choisir un morceau et à le découper en 4, 7 ou 10 morceaux. Aurélien peut-il obtenir ainsi 2021 morceaux ?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nL'exercice décrit une suite d'opérations et demande s'il est possible de passer d'un état initial à un état final, on peut donc légitimement chercher un invariant du système.\n\nIci, nous allons montrer que le nombre de morceaux d'Aurélien est toujours de la forme $3k+1$, avec $k$ un entier naturel.\n\nC'est le cas dans la situation initiale, puisqu'Aurélien dispose de $7=3 \\times 2+1$ morceaux. Puis, si c'est le cas lors d'une étape et qu'Aurélien dispose de $3k+1$ morceaux avec $k$ un entier naturel, Aurélien remplace un morceau par $3 \\times 1+1$ morceaux, par $3 \\times 2+1$ morceaux ou par $3 \\times 3+1$ morceaux, de sorte qu'il lui reste, à la fin de l'opération, $3(k+1)+1$, $3(k+2)+1$ ou $3(k+3)+1$ morceaux.\n\nDe proche en proche, on a bien le résultat annoncé. Comme $2021=3 \\times 673+2$ n'est pas de la forme indiquée, on ne peut jamais obtenir 2021 morceaux par le processus décrit.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30960,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$2n-1$ tours sont placées sur un échiquier de taille $(2n-1) \\times (2n-1)$ de sorte que deux tours quelconques ne sont jamais sur la même ligne ou la même colonne. Montrer que tout carré de taille $n \\times n$ contient une tour.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nCommençons par remarquer qu'il y a exactement une tour par ligne et par colonne. En effet, par hypothèse il y a au plus une tour par ligne. Et puisqu'il y a $2n-1$ tours au total, il doit y en avoir exactement une par ligne, et de même pour les colonnes.\n\nQuitte à adapter légèrement le raisonnement, on peut supposer que le carré $n \\times n$ est celui des cases situées en haut à gauche (c'est-à-dire les cases situées à la fois dans les $n$ premières lignes et dans les $n$ premières colonnes en partant du coin supérieur gauche).\n\nSupposons par l'absurde que ce carré ne contienne aucune tour. Alors parmi les $n$ premières lignes, il doit y avoir $n$ tours. Cependant, ces $n$ tours doivent être dans les $n-1$ dernières colonnes par hypothèse. Par principe des tiroirs, il y aura alors deux tours sur la même colonne, ce qui est absurde. Cela conclut.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30961,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dispose de cinq couleurs et d'une grille $99 \\times 99$. On colorie certains carrés de la grille avec l'une des cinq couleurs de sorte que\n- Chaque couleur apparaît le même nombre de fois dans la grille.\n- Aucune ligne et aucune colonne ne contient des cases de couleur différente.\nQuelle est le plus grand nombre possible de cases que l'on peut colorier en suivant ces règles?",
    "options": [],
    "answer": "1900",
    "solution": "Solution:\n\nRéponse : $N = 5 \\times 19 \\times 20 = 1900$.\n\nComme chaque ligne ou colonne ne peut contenir qu'une couleur, on va réfléchir en partant des lignes. On dit qu'une ligne ou colonne est d'une couleur (disons rouge) quand la seule couleur que peuvent avoir des cases de cette ligne ou colonne est le rouge. Si une ligne ou colonne n'a pas de couleur, on lui assigne arbitrairement une couleur. Ainsi, une case rouge se trouve forcément à l'intersection d'une ligne rouge et d'une colonne rouge. Par conséquent, on peut majorer le nombre de cases rouges par le produit du nombre de lignes rouges par le nombre de colonnes rouges.\n\nIntroduisons maintenant des notations. On numérote les couleurs disponibles de 1 à 5 et on note $x_{i}$ le nombre de lignes colorées de la $i$-ième couleur, et $y_{i}$ le nombre de colonnes colorées de cette couleur. Notons enfin $N$ le nombre de cases coloriées par couleur. On veut maximiser le nombre total de cases coloriées, $5N$. Les arguments précédemment cités montrent que $\\forall i: N \\leqslant x_{i} y_{i}$.\n\nOn a évidemment aussi, $x_{1} + x_{2} + \\ldots + x_{5} = y_{1} + y_{2} + \\ldots + y_{5} = 99$.\n\nL'étude des petits cas nous montre que pour maximiser $N$ il faut que les $x_{i}$ et les $y_{i}$ soient tous environ égaux, valant tous $\\frac{99}{5}$ arrondi soit par excès ou par défaut. Une étude plus approfondie des petits cas où on remplace 99 par un petit nombre congru à $4 \\bmod 5$, comme 99, nous permet de conjecturer que la construction suivante est optimale :\n\n![](attached_image_1.png)\n\nIci, on a $N = 19 \\cdot 20$.\n\nProuvons maintenant l'optimalité. Supposons par l'absurde disposer d'une configuration valide où $N > 19 \\times 20$, donc $\\forall i: x_{i} y_{i} > 19 \\times 20$. On remarque ensuite que $x_{i} + y_{i} > 39$. En effet,\n$$\nx_{i} y_{i} = \\frac{1}{4} \\left( (x_{i} + y_{i})^{2} - (x_{i} - y_{i})^{2} \\right)\n$$\nDonc\n$$\n38^{2} < 4 \\cdot 19 \\cdot 20 = 39^{2} - 1 < (x_{i} + y_{i})^{2} - (x_{i} - y_{i})^{2} < (x_{i} + y_{i})^{2}\n$$\nDonc $38 < x_{i} + y_{i}$. De plus, si $x_{i} + y_{i} = 39$, alors $(x_{i} - y_{i})^{2} < 1$ donc $x_{i} = y_{i}$, ce qui est impossible car 39 est impair. Donc $x_{i} + y_{i} \\geqslant 40$.\n\nLe compte final $2 \\cdot 99 = (x_{1} + y_{1}) + (x_{2} + y_{2}) + \\ldots + (x_{5} + y_{5}) \\geqslant 5 \\times 40 = 200$ est une contradiction et termine la preuve.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers $n \\geqslant 0$ tels que $2023+n!$ est un carré parfait.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nSoit $n \\geqslant 0$ tel que $2023+n!$ est un carré parfait. Si $n \\geqslant 4$, alors $4$ divise $n!$, donc $2023+n! \\equiv 2023 \\equiv 3 \\pmod{4}$. Or $3$ n'est pas un carré modulo $4$ (les carrés modulo $4$ sont $0$ et $1$), donc on a une contradiction. Ainsi $n \\leqslant 3$.\n\nNotons que $44^{2}=1936 < 2023+0! = 2023+1! = 2024 < 2023+2! = 2025 = 45^{2} < 2023+3! = 2029 < 2116 = 46^{2}$. Ainsi pour $n=0,1,3$, $2023+n!$ n'est pas un carré, mais pour $n=2$ c'est un carré. Ainsi $2023+n!$ est un carré si et seulement si $n=2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMontrer qu'il n'existe pas de réels $x, y, z$ strictement positifs tels que\n$$\n\\left(2 x^{2}+y z\\right)\\left(2 y^{2}+x z\\right)\\left(2 z^{2}+x y\\right)=26 x^{2} y^{2} z^{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOn pourrait essayer de simplement appliquer l'inégalité arithmético-géométrique sur chacun des facteurs du côté gauche, mais ceci donne que le terme de gauche est supérieur ou égal à $16 \\sqrt{2} x^{2} y^{2} z^{2}$ ce qui ne suffit pas pour conclure (car $16 \\sqrt{2}<26$ ). En fait, ceci vient du fait que l'on ne peut pas avoir égalité dans les trois inégalités arithmético-géométriques à la fois. On aimerait bien que le cas $x=y=z$ soit un cas d'égalité des inégalités arithmético-géométrique que l'on applique.\nPrenons par exemple le premier terme $2 x^{2}+y z$. On le sépare en $x^{2}+x^{2}+y z$ (de façon à ce que chacun des termes soit égaux si $x=y=z$ ), on peut alors appliquer l'inégalité arithmético-géométrique :\n$$\n2 x^{2}+y z \\geqslant 3 \\sqrt[3]{x^{2} \\cdot x^{2} \\cdot y z}=3 \\sqrt[3]{x^{4} y z}\n$$\nEn appliquant l'inégalité arithmético-géométrique sur tous les termes qui sont positifs, on a\n$$\n\\left(2 x^{2}+y z\\right)\\left(2 y^{2}+x z\\right)\\left(2 z^{2}+x y\\right) \\geqslant 27 \\sqrt[3]{x^{4} y z \\cdot y^{4} x z \\cdot z^{4} x y}=27 \\sqrt[3]{x^{6} y^{6} z^{6}}=27 x^{2} y^{2} z^{2}\n$$\nce qui est strictement supérieur à $26 x^{2} y^{2} z^{2}$, on ne peut donc pas avoir égalité.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers $N$ tels que $2^{N} - 2N$ est un carré parfait.",
    "options": [],
    "answer": "0, 1, 2",
    "solution": "Solution:\n\nMontrons que les solutions sont $N = 0, 1$ ou $2$.\n\nOn remarque déjà que $2^{0} - 0 = 1^{2}$, $2^{1} - 2 = 0^{2}$ et $2^{2} - 4 = 0^{2}$. Donc les entiers $0, 1$ et $2$ sont solutions.\n\nSupposons par l'absurde qu'il existe $N$ un autre entier solution. On dispose alors de $x \\in \\mathbb{N}$ tel que $2^{N} - 2N = x^{2}$.\n\nPuisque $2^{N}$ doit être entier, on doit avoir $N \\geqslant 0$, et donc $N \\geqslant 3$. En outre, $x$ est divisible par $2$, donc $x^{2}$ est divisible par $4$. Puisque $N \\geqslant 2$, $2^{N}$ est également divisible par $4$ et donc $N$ est pair. On a\n$$\n2N = 2^{N} - x^{2} = \\left(2^{\\frac{N}{2}} - x\\right)\\left(2^{\\frac{N}{2}} + x\\right)\n$$\nEn outre, on vérifie que $N = 4$, $N = 6$ ou $N = 8$ ne sont pas des solutions car $8$, $52$ et $240$ ne sont pas des carrés parfaits. Donc $N \\geqslant 10$.\n\nEnfin, puisque $2N > 0$, on doit avoir $2^{\\frac{N}{2}} - x \\geqslant 1$. Mais on montre par récurrence que $2^{\\frac{N}{2}} > 2N$ pour tout $N$ pair $\\geqslant 10$. En effet, $2^{5} = 32 > 20$ et si $2^{\\frac{N}{2}} > 2N$, avec $N$ pair $\\geqslant 10$, alors\n$$\n2^{\\frac{N+2}{2}} = 2 \\cdot 2^{\\frac{N}{2}} > 4N > 2(N+2)\n$$\nMais alors on a\n$$\n2^{N} - x^{2} = \\left(2^{\\frac{N}{2}} - x\\right)\\left(2^{\\frac{N}{2}} + x\\right) \\geqslant 2^{\\frac{N}{2}} + x > 2N\n$$\nce qui est absurde. Cela montre qu'il n'y a pas d'autre solution.\n\nOn a donc bien montré que les entiers solutions sont exactement $0, 1$ et $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les triplets $(p, q, r)$ de nombres premiers tels que $p+q^{2}=r^{4}$.",
    "options": [],
    "answer": "(7,3,2)",
    "solution": "Solution:\n\nNotons que l'équation se réécrit $p=(r^{2})^{2}-q^{2}=(r^{2}-q)(r^{2}+q)$. Comme $r^{2}+q$ est strictement positif, et $p$ aussi, $r^{2}-q$ aussi. En particulier, d'après l'équation précédente, on a que $r^{2}-q=1$ et $r^{2}+q=p$.\n\nSi $r$ et $q$ sont impairs, alors $r^{2}-q$ est pair, ce qui est contradictoire. Ainsi parmi $r$ et $q$, un est pair, donc vaut $2$ car $q$ et $r$ sont premiers.\n\nSi $r=2$, alors $r^{2}-q=1$, donc $q=4-1=3$. De plus $p=q+r^{2}=7$, donc $(p, q, r)=(7,3,2)$ est une éventuelle solution.\n\nSi $q=2$, alors $r^{2}=q+1=3$ ce qui est impossible.\n\nRéciproquement, pour $(p, q, r)=(7,3,2)$, $p+q^{2}=7+9=16=2^{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMontrer que pour tous les réels $x, y, z$, on a\n$$\n\\frac{x^{2}-y^{2}}{2 x^{2}+1}+\\frac{y^{2}-z^{2}}{2 y^{2}+1}+\\frac{z^{2}-x^{2}}{2 z^{2}+1} \\leqslant 0 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn réécrit l'inégalité en\n$$\n\\frac{x^{2}}{2 x^{2}+1}+\\frac{y^{2}}{2 y^{2}+1}+\\frac{z^{2}}{2 z^{2}+1} \\leqslant \\frac{y^{2}}{2 x^{2}+1}+\\frac{z^{2}}{2 y^{2}+1}+\\frac{x^{2}}{2 z^{2}+1} .\n$$\nL'inégalité n'étant pas symétrique, on ne peut pas supposer sans perte de généralité que $x^{2} \\leqslant y^{2} \\leqslant z^{2}$. Cependant, remarquons que les suites $\\left(x^{2}, y^{2}, z^{2}\\right)$ et $\\left(\\frac{1}{2 x^{2}+1}, \\frac{1}{2 y^{2}+1}, \\frac{1}{2 z^{2}+1}\\right)$ sont rangées dans l'ordre inverse (le plus grand de la première suite correspond au plus petit de la deuxième suite par exemple). On peut donc appliquer l'inégalité de réordonnement qui implique directement l'inégalité de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUne partie se joue sur un échiquier de taille $n \\times n$. Au début, il y a 99 pierres sur chaque case. Tour à tour, Aimeric et Benoit choisissent une ligne ou une colonne et retirent une pierre de chaque case de la ligne ou de la colonne choisie. Ils ne peuvent choisir une ligne ou une colonne que si elle comporte au moins une pierre sur chaque case. Le premier joueur qui ne peut pas se déplacer perd la partie. Aimeric joue le premier tour. Déterminer tous les $n$ pour lesquels Benoit a une stratégie gagnante.",
    "options": [],
    "answer": "all even n",
    "solution": "Solution:\n\nRéponse : Les $n$ pairs.\n\nDans la suite, on numérote les colonnes de gauche à droite et les lignes de haut en bas.\nNous allons montrer qu'il y a exactement $99 n$ tours de jeu, de sorte que si $n$ est pair, Benoit aura eu le dernier tour de jeu (donc aura gagné), sinon c'est Aimeric.\n\nConsidérons les cases de l'une des grandes diagonales de la grille. Comme chaque ligne et chaque colonne contient exactement une de ces cases, à chaque étape, la somme du nombre de pierres sur ces cases diminue d'exactement 1. Si toutes les cases de la grande diagonale sont vides, il n'est plus possible d'effectuer aucune action. Montrons que réciproquement, si l'une des cases de la grande diagonale contient au moins un caillou, il est encore possible de jouer.\n\nSupposons que cette case soit à l'intersection de la ligne $i$ et de la colonne $i$. Si la ligne $i$ et/ou la colonne $i$ ne contiennent pas de case vide, alors il est encore possible de jouer sur cette colonne ou cette ligne. On peut donc supposer que la ligne $i$ contient une case vide, disons sur la colonne $a$, et que la colonne $i$ contient une case vide, disons sur la ligne $b$. Notons $r_{i}$ le nombre de fois que l'on a joué sur la ligne $i$, $r_{b}$ le nombre de fois que l'on a joué sur la ligne $b$, $c_{i}$ le nombre de fois que l'on a joué sur la colonne $i$ et enfin $c_{a}$ le nombre de fois que l'on a joué sur la colonne $a$.\n\nPuisque la case à l'intersection de la ligne $i$ et de la colonne $a$ est vide, on a $c_{a} + r_{i} = 99$. De même, on a $r_{b} + c_{i} = 99$. Mais puisque $r_{i} + c_{i} < 99$ (car la case à l'intersection de la ligne et de la colonne $i$ est non vide), on a $r_{b} + c_{a} > 99$, ce qui est absurde car la case à l'intersection de la ligne $b$ et de la colonne $a$ a perdu alors strictement plus que 99 cailloux.\n\nOn déduit que la partie finit si et seulement si chaque case de la grande diagonale est vide, donc au bout d'exactement $99 n$ étapes, ce qui permet de conclure comme annoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $a$, $b$ et $c$ des réels tels que $0 \\leqslant a, b, c \\leqslant 2$. Montrer que\n$$\n(a-b)(b-c)(a-c) \\leqslant 2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNotons qu'il y a 6 ordres possibles pour les variables $a$, $b$ et $c$.\nDans les trois cas $b \\geqslant a \\geqslant c$, $a \\geqslant c \\geqslant b$ et $c \\geqslant b \\geqslant a$, le produit $(a-b)(b-c)(c-a)$ est négatif, de sorte que l'inégalité est vérifiée.\n\nSi $a \\geqslant b \\geqslant c$, l'inégalité des moyennes appliquée à $a-b$ et $b-c$ et le fait que $c-a \\leqslant 2$ car $a, c \\in [0,2]$ donne :\n$$\n(a-b)(b-c)(a-c) \\leqslant \\left(\\frac{a-b+b-c}{2}\\right)^2 (a-c) = \\frac{(a-c)^3}{4} \\leqslant \\frac{2^3}{4} = 2\n$$\nDe même, si $b \\geqslant c \\geqslant a$, alors\n$$\n(a-b)(b-c)(a-c) = (b-a)(b-c)(c-a) \\leqslant (b-a)\\left(\\frac{b-c+c-a}{2}\\right)^2 \\leqslant \\frac{(b-a)^3}{4} \\leqslant 2\n$$\nEt enfin, si $c \\geqslant a \\geqslant b$, alors\n$$\n(a-b)(b-c)(a-c) = (c-b)(c-a)(a-b) \\leqslant (c-b)\\left(\\frac{c-a+a-b}{2}\\right)^2 \\frac{(c-b)^3}{4} \\leqslant 2\n$$\nSi bien que l'inégalité est vraie dans tous les cas.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les quadruplets $\\left(a, b, c, d\\right)$ d'entiers positifs avec $a, b, c$ strictement positifs tels que\n$$\n\\begin{aligned}\n& \\operatorname{PPCM}(b, c)=a+d \\\\\n& \\operatorname{PPCM}(c, a)=b+d \\\\\n& \\operatorname{PPCM}(a, b)=c+d\n\\end{aligned}\n$$",
    "options": [],
    "answer": "(a, a, a, 0) for any positive integer a",
    "solution": "Solution:\n\nLe système étant symétrique, on suppose que $c$ est le maximum de $a$, $b$ et $c$.\n\nComme $c$ divise $\\operatorname{PPCM}(b, c)$, $c$ divise $a+d$.\nComme $c$ divise $\\operatorname{PPCM}(c, a)$, $c$ divise $b+d$, donc $c$ divise $a+d-(b+d)=a-b$.\nOr $-c < a-b < c$ donc $a=b$.\n\nLa dernière ligne donne alors $a=\\operatorname{PPCM}(a, b)=c+d \\geqslant c$.\nPour avoir égalité, on a donc $d=0$ et $a=c$.\n\nRéciproquement, si $a=b=c$ et $d=0$, chaque équation équivaut à $a=a+0$, qui est vrai.\n\nLes quadruplets solution sont donc les $(a, a, a, 0)$ pour $a \\in \\mathbb{N}^*$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30970,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 2$ et $a_{1}, \\ldots, a_{n} \\in [0,1]$ des réels. Déterminer la valeur maximale que peut prendre le plus petit des nombres\n$$\na_{1}-a_{1} a_{2},\\ a_{2}-a_{2} a_{3},\\ \\ldots,\\ a_{n}-a_{n} a_{1}\n$$",
    "options": [],
    "answer": "1/4",
    "solution": "Solution:\n\nSoit $i$ tel que $a_{i}$ soit minimal. Alors on a $a_{i+1} \\geqslant a_{i}$ (où $a_{n+1}=a_{1}$) et donc par inégalité arithmético-géométrique sur $a_{i}$ et $1-a_{i}$, on obtient\n$$\na_{i}-a_{i} a_{i+1} \\leqslant a_{i}-a_{i}^{2}=a_{i}(1-a_{i}) \\leqslant \\frac{1}{4}(a_{i}+1-a_{i})^{2}=\\frac{1}{4}\n$$\ndonc le minimum des nombres vaut au plus $\\frac{1}{4}$. Il reste à vérifier que l'on peut trouver des $a_{i}$ tel que le minimum soit égal à $\\frac{1}{4}$. Si tous les $a_{i}$ valent $\\frac{1}{2}$, alors tous les $a_{i}-a_{i} a_{i+1}$ valent $\\frac{1}{4}$ et le minimum vaut bien $\\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30971,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n$ un entier naturel. Joseph peut tirer $2n+1$ flèches. Chacun de ses tirs est un échec ou une réussite. Un tir est dit \"équilibré\" si le nombre d'échecs avant ce tir additionné au nombre de réussites après ce tir est égal à $n$. Déterminer si le nombre de tirs équilibrés est pair ou impair.",
    "options": [],
    "answer": "odd",
    "solution": "Solution:\n\nPremière remarque : considérons une succession de deux tirs telle que le premier est réussi et le suivant raté. Alors il y a autant de tirs ratés avant pour les deux et autant de tirs réussis après donc soit les deux tirs sont équilibrés, soit aucun des deux ne l'est. De même, si le premier est raté et le suivant réussi, en notant $e$ le nombre d'échecs avant ces deux tirs et $r$ le nombre de réussites après, le premier tir est équilibré si et seulement si $e + (1 + r) = n$ et le deuxième est équilibré si et seulement si $(e + 1) + r = n$. On remarque que ces deux conditions sont égales. Ainsi, puisque cela ne modifie pas l'équilibre des tirs précédents et suivants, supposer que pour une succession de 2 tirs dont l'un est réussi et l'autre est raté, c'est le premier qui est raté ne modifie pas la parité du nombre de tirs équilibrés. On peut donc se ramener au cas où Joseph commence par une succession d'échecs, puis de réussites.\n\nDans l'ordre des tirs, le nombre de succès à venir additionné au nombre d'échecs passés débute au nombre de réussites total $R$, augmente strictement à chaque échec jusqu'à atteindre la valeur $2n$ puis diminue à chaque réussite jusqu'à atteindre le nombre d'échecs total $E$. Or $E + R = 2n + 1$ donc exactement une de ces deux valeurs $E$ et $R$ est plus petite que $n$ et l'autre est strictement plus grande. La valeur $n$ n'est donc atteinte que pour un seul tir dans cette configuration, et elle est donc impaire dans le cas général.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les entiers $n$ ayant la propriété suivante : si l'on pose $a_{k} = \\operatorname{ppcm}(k, k+1, \\ldots, k+n-1)$, alors la suite $(a_{k})$ est croissante.",
    "options": [],
    "answer": "1 and 2",
    "solution": "Solution:\n\nMontrons que les naturels ayant cette propriété sont exactement $1$ et $2$.\n\nDéjà, si $n=1$, $a_{k}=k$ est une suite croissante.\n\nEt si $n=2$, puisque deux entiers consécutifs sont premiers entre eux, $a_{k}=k(k+1)$ est une suite croissante également.\n\nSi $n=3$, on remarque que $a_{5}=210 > a_{6}=168$, donc $3$ n'a pas la propriété désirée.\n\nSoit maintenant $n \\geqslant 4$. L'idée pour montrer que $n$ ne vérifie pas la propriété est de trouver un $k$ tel que les $(k+i)_{0 \\leqslant i \\leqslant n-1}$ soient premiers entre eux, tandis que $k+1$ ait un grand facteur commun avec $k+n$, pour avoir $a_{k} > a_{k+1}$. C'est pourquoi on regarde $k = (n-1)! - 1$.\n\nPosons $\\mathrm{D} = \\operatorname{ppcm}(k+1, \\ldots, k+n-1)$. On a alors $a_{k} = \\operatorname{ppcm}(k, D) = k \\cdot D$. En effet, pour $1 \\leqslant i \\leqslant n-1$, on a $\\operatorname{pgcd}(k, k+i) = \\operatorname{pgcd}(k, i) = 1$ car $i \\mid k+1 = (n-1)!$.\n\nEn outre, $a_{k+1} = \\operatorname{ppcm}(D, k+n)$. Or, $n-1$ divise à la fois $k+1$ (donc $D$) et $k+n$. Donc\n$$\na_{k+1} \\leqslant D \\cdot ((n-2)! + 1) < D \\cdot ((n-1)! - 1) = a_{k}\n$$\nEn effet, pour tout $n \\geqslant 4$, $(n-1)! - (n-2)! = (n-2) \\cdot (n-2)! \\geqslant 4 > 2$, donc $(n-1)! - 1 > (n-2)! + 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nUn entier $n \\geqslant 2$ est écrit au tableau. Chaque jour, quelqu'un choisit $p$ un diviseur premier de l'entier écrit $n$ au tableau, efface celui-ci et écrit $n+\\frac{n}{p}$ à la place. Montrer que $p=3$ est choisi une infinité de fois.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFixons $N \\in \\mathbb{N}$. Notons $2^{a_{k}} 3^{b_{k}} c_{k}$ l'entier écrit au tableau le jour $k$, avec $c_{k}$ produit de nombres premiers distincts de $2$ et $3$. Supposons par l'absurde qu'on peut ne jamais choisir $p=3$ à partir du jour $N$.\n\nSi le $k$-ième jour on choisit $p \\neq 2,3$, alors le nombre écrit au tableau devient $2^{a_{k}} 3^{b_{k}} \\frac{c_{k}}{p}(p+1)$. Posons $p+1=2^{b} 3^{c} d$ avec $b, c, d$ des entiers positifs et $d$ premier avec $2$ et $3$. Comme $p$ est premier et différent de $2$, il est impair, donc $a \\geqslant 1$, donc $d \\leqslant \\frac{p+1}{2}$. Ainsi, comme le nombre écrit au tableau est $2^{a_{k}+a} 3^{b_{k}+b} \\frac{c_{k}}{p} d$, donc $a_{k+1}=a_{k}+a$, $b_{k+1}=b_{k}+b$ et $c_{k+1}=\\frac{c_{k}}{p} d \\leqslant c_{k} \\frac{p+1}{2p} < c_{k}$.\n\nSi le $k$-ième jour on choisit $p=2$, le nombre écrit au tableau devient $\\frac{3}{2} 2^{a_{k}} 3^{b_{k}} c_{k} = 2^{a_{k}-1} 3^{b_{k}+1} c_{k}$. En particulier, la suite $\\left(c_{k}\\right)_{k \\geqslant N}$ décroît et est strictement positive. Elle ne peut donc décroître strictement infiniment souvent, sinon elle deviendrait négative. Donc à partir d'un certain jour $k > N$, on ne choisit plus $p \\neq 2,3$, donc $p=2$ est choisi chaque jour à partir du $k$-ième jour.\n\nOr si à partir du jour $k > N$ on ne choisit que $p=2$, alors $a_{j+1}=a_{j}-1$ pour tout $j \\geqslant k$. Donc la suite d'entiers $\\left(a_{j}\\right)_{j \\geqslant k}$ décroît strictement, elle doit donc devenir strictement négative, ce qui est absurde.\n\nAinsi, on est obligé de choisir au moins une fois $p=3$ après le $N$-ième jour.\n\nSupposons que $p=3$ n'est pas choisi une infinité de fois. À partir d'un certain jour, noté $N$, $p=3$ n'est plus jamais choisi. Cela contredit ce qu'on vient de prouver : $p=3$ est donc choisi une infinité de fois.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDans une école il y a $n$ cours et $n$ élèves. Les élèves sont inscrits dans plusieurs cours de sorte que deux élèves différents n'ont jamais exactement les mêmes cours. Prouver qu'on peut supprimer un cours de sorte qu'aucune paire d'élèves ne se retrouve avec exactement les mêmes cours.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn va montrer par récurrence forte sur $n$ la propriété plus forte suivante :\n$$\n\\mathcal{P}_{n} : \\text{Pour tout } m \\geqslant n, \\text{ dans une classe à } n \\text{ élèves et } m \\text{ cours telle que deux élèves n'ont jamais exactement les mêmes cours, alors il est possible de retirer un cours de sorte que les élèves ne suivent pas exactement les mêmes cours parmi les cours restants.}\n$$\n\nInitialisation: Si $n=1$, il n'y a rien à démontrer.\n\nHérédité : Soit $n \\geqslant 2$. Supposons que $\\mathcal{P}_{k}$ est vraie pour tout $1 \\leqslant k \\leqslant n-1$. Soit $m \\geqslant n$. Considérons alors une classe de $n$ élèves qui suivent $m$ cours notés $C_{1}, \\ldots, C_{m}$. Si deux élèves ne suivent jamais les mêmes cours parmi les cours $C_{1}, \\ldots, C_{m-1}$, alors on a démontré ce qu'il fallait. Sinon, il existe $k$ paires d'élèves $(A_{1}, B_{1}), \\ldots, (A_{k}, B_{k})$ avec $k \\geqslant 1$, telles que les élèves $A_{i}$ et $B_{i}$ suivent les mêmes cours parmi $C_{1}, \\ldots, C_{m-1}$ (et alors exactement l'un d'eux, disons $A_{i}$, suit le cours $C_{m}$). Notons $E_{1}, \\ldots, E_{n-2k}$ l'ensemble des élèves restant. Considérons le sous-ensemble de $n-k$ élèves $\\{A_{1}, \\ldots, A_{k}, E_{1}, \\ldots, E_{n-2k}\\}$ et l'ensemble de cours $\\{C_{1}, \\ldots, C_{m-1}\\}$. D'après l'hypothèse de récurrence, puisque $n-k \\leqslant m-1$, on dispose d'un cours, disons $C_{m-1}$ qui, si on le retire, vérifie qu'aucune paire de deux élèves parmi les élèves considérés ne suivent les mêmes cours parmi $C_{1}, \\ldots, C_{m-2}$. Alors, parmi ces $n-k$ élèves considérés, il n'y a pas de paire d'élèves ayant exactement les mêmes cours parmi $C_{1}, \\ldots, C_{m-2}, C_{m}$. Mais alors un élève $B_{i}$ ne suit pas exactement les mêmes cours que $A_{i}$ (à cause du cours $C_{m}$) ou qu'un autre élève (car $A_{i}$ ne suit pas les mêmes cours qu'un autre élève parmi $C_{1}, \\ldots, C_{m-2}$). Ceci achève la récurrence.\n\nLa propriété $\\mathcal{P}_{n}$ est donc vraie pour tout $n$, en particulier, l'énoncé est vrai pour $m=n$ cours et $n$ élèves.\nSolution:\n\nNotons $E_{1}, \\ldots, E_{n}$ les $n$ élèves et $C_{1}, \\ldots, C_{n}$ les $n$ cours.\nOn procède par l'absurde et l'on suppose l'inverse : quelque soit le cours $C_{i}$ que l'on supprime, il existe deux élèves $E_{i_{1}}$ et $E_{i_{2}}$ ayant exactement les mêmes cours en dehors du cours $C_{i}$. Si deux telles paires ou plus existent pour un même cours $C_{i}$, on n'en garde qu'une seule. Comme ces deux élèves ne suivent pas exactement les mêmes cours, l'un suit le cours $C_{i}$ et l'autre ne le suit pas. Leurs cours ne coïncident donc pas si l'on supprime un autre cours $C_{j}$.\n\nConsidérons alors le graphe dont les sommets sont les élèves et deux élèves sont reliés par une arête s'il existe un $i$ (qui est alors unique d'après le paragraphe précédent) tel que, si l'on supprime $C_{i}$, les deux élèves ont les mêmes cours. Comme chaque cours $C_{i}$ induit une paire d'élèves reliés par une arête, le graphe obtenu admet au moins un cycle, que l'on note $E_{i_{1}} \\ldots E_{i_{k}} E_{i_{1}}$, avec $k \\leqslant n$. Notons $j_{\\ell}$ le numéro du cours qui induit l'arête $E_{i_{\\ell}} E_{i_{\\ell+1}}$. Les élèves $E_{i_{1}}$ et $E_{i_{k}}$ suivent alors les mêmes cours parmi ceux qui ne sont pas $C_{j_{1}}, \\ldots, C_{j_{k-1}}$. En particulier, soit ils suivent tous les deux le cours $C_{j_{k}}$, soit ils ne le suivent pas tous les deux, ce qui contredit le fait que l'arête $E_{i_{k}} E_{i_{1}}$ appartient au graphe. On a la contradiction désirée.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 30975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDémontrer, pour tous réels $a$, $b$, $c$ strictement positifs, l'inégalité suivante :\n$$\n\\frac{a}{9 b c+1}+\\frac{b}{9 c a+1}+\\frac{c}{9 a b+1} \\geqslant \\frac{a+b+c}{1+(a+b+c)^{2}} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nVu la forme de l'inégalité, on est tenté d'appliquer l'inégalité des mauvais élèves. Seulement, si l'on applique l'inégalité en l'état, on trouve :\n$$\n\\frac{a}{9 b c+1}+\\frac{b}{9 c a+1}+\\frac{c}{9 a b+1} \\geqslant \\frac{(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})^{2}}{9(a b+b c+c a)+3}\n$$\nPour conclure, il faudrait alors montrer que $\\frac{(\\sqrt{a}+\\sqrt{b}+\\sqrt{c})^{2}}{9(a b+b c+c a)+3} \\geqslant \\frac{a+b+c}{1+(a+b+c)^{2}}$, mais cette inégalité n'est pas toujours vraie.\n\nPour faire apparaître $a+b+c$ au numérateur du membre de droite de l'inégalité des mauvais élèves, on peut par exemple multiplier en haut et en bas par $a$ dans la première fraction du membre de gauche (et de même avec $b$ et $c$ dans les deux autres fractions). Ceci donne, en appliquant l'inégalité des mauvais élèves:\n$$\n\\frac{a}{9 b c+1}+\\frac{b}{9 c a+1}+\\frac{c}{9 a b+1}=\\frac{a^{2}}{9 a b c+a}+\\frac{b^{2}}{9 a b c+b}+\\frac{c^{2}}{9 a b c+c} \\geqslant \\frac{(a+b+c)^{2}}{27 a b c+a+b+c} .\n$$\nIl suffit alors de montrer que\n$$\n\\frac{(a+b+c)^{2}}{27 a b c+a+b+c} \\geqslant \\frac{a+b+c}{1+(a+b+c)^{2}}\n$$\nPour cela, on peut appliquer l'inégalité des moyennes $27 abc \\leqslant(a+b+c)^{3}$ pour obtenir :\n$$\n\\frac{(a+b+c)^{2}}{27 a b c+a+b+c} \\geqslant \\frac{(a+b+c)^{2}}{(a+b+c)^{3}+(a+b+c)}=\\frac{(a+b+c)}{(a+b+c)^{2}+1}\n$$\ncomme voulu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoient $a, b, c$ trois réels strictement positifs. Montrer que\n$$\n\\frac{a^{4}+1}{b^{3}+b^{2}+b}+\\frac{b^{4}+1}{c^{3}+c^{2}+c}+\\frac{c^{4}+1}{a^{3}+a^{2}+a} \\geqslant 2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOn commence par appliquer l'inégalité arithmético-géométrique aux trois termes de l'énoncé :\n$$\n\\frac{a^{4}+1}{b^{3}+b^{2}+b}+\\frac{b^{4}+1}{c^{3}+c^{2}+c}+\\frac{c^{4}+1}{a^{3}+a^{2}+a} \\geqslant 3 \\sqrt[3]{\\frac{\\left(a^{4}+1\\right)\\left(b^{4}+1\\right)\\left(c^{4}+1\\right)}{\\left(b^{3}+b^{2}+b\\right)\\left(c^{3}+c^{2}+c\\right)\\left(a^{3}+a^{2}+a\\right)}}\n$$\nOn cherche donc à minorer les quotients $\\frac{a^{4}+1}{a^{3}+a^{2}+a}$. Pour cela, on applique des inégalités arithméticogéométriques :\n$$\n\\begin{aligned}\na^{3}+a^{2}+a & \\leqslant \\frac{1}{2}\\left(a^{4}+a^{2}\\right)+a^{2}+\\frac{1}{2}\\left(a^{2}+1\\right)=\\frac{1}{2}\\left(a^{4}+1\\right)+2 a^{2} \\\\\n& \\leqslant \\frac{1}{2}\\left(a^{4}+1\\right)+\\left(a^{4}+1\\right)=\\frac{3}{2}\\left(a^{4}+1\\right)\n\\end{aligned}\n$$\net ainsi $\\frac{a^{4}+1}{a^{3}+a^{2}+a} \\geqslant \\frac{2}{3}$. On a de même pour $b$ et $c$, et donc\n$$\n\\frac{a^{4}+1}{b^{3}+b^{2}+b}+\\frac{b^{4}+1}{c^{3}+c^{2}+c}+\\frac{c^{4}+1}{a^{3}+a^{2}+a} \\geqslant 3 \\sqrt[3]{\\left(\\frac{2}{3}\\right)^{3}}=2\n$$\nSolution:\n\nOn peut également réfléchir avec l'inégalité de réarrangement. L'inégalité n'est pas symétrique, on ne peut donc pas supposer sans perte de généralités $a \\geqslant b \\geqslant c$ ou quelque chose de similaire.\nCependant, par stricte croissance, $\\left(a^{4}+1, b^{4}+1, c^{4}+1\\right)$ est dans le même ordre que $(a, b, c)$. De même, $\\left(a^{3}+a^{2}+a, b^{3}+b^{2}+b, c^{3}+c^{2}+c\\right)$ est également dans le même ordre et $\\left(\\frac{1}{a^{3}+a^{2}+a}, \\frac{1}{b^{3}+b^{2}+b}, \\frac{1}{c^{3}+c^{2}+c}\\right)$ est dans le sens inverse. Ainsi, par inégalité de réarrangement,\n$$\n\\frac{a^{4}+1}{a^{3}+a^{2}+a}+\\frac{b^{4}+1}{b^{3}+b^{2}+b}+\\frac{c^{4}+1}{c^{3}+c^{2}+c}\n$$\nest la valeur minimale que peut prendre une somme de produit de termes de $\\left(a^{4}+1, b^{4}+1, c^{4}+1\\right)$ avec des termes de $\\left(\\frac{1}{a^{3}+a^{2}+a}, \\frac{1}{b^{3}+b^{2}+b}, \\frac{1}{c^{3}+c^{2}+c}\\right)$. En particulier ici :\n$$\n\\frac{a^{4}+1}{b^{3}+b^{2}+b}+\\frac{b^{4}+1}{c^{3}+c^{2}+c}+\\frac{c^{4}+1}{a^{3}+a^{2}+a} \\geqslant \\frac{a^{4}+1}{a^{3}+a^{2}+a}+\\frac{b^{4}+1}{b^{3}+b^{2}+b}+\\frac{c^{4}+1}{c^{3}+c^{2}+c}\n$$\nOn peut maintenant comme précédemment montrer $\\frac{a^{4}+1}{a^{3}+a^{2}+a} \\geqslant \\frac{2}{3}$ et conclure par\n$$\n\\frac{a^{4}+1}{b^{3}+b^{2}+b}+\\frac{b^{4}+1}{c^{3}+c^{2}+c}+\\frac{c^{4}+1}{a^{3}+a^{2}+a} \\geqslant 3 \\cdot \\frac{2}{3}=2\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30977,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $k$ un entier premier avec $n$ vérifiant $1 \\leqslant k < n$. Augustin colorie les entiers de $\\{1,2, \\ldots, n-1\\}$ avec autant de couleurs qu'il le souhaite. Cependant, si $j$ est un entier vérifiant $1 \\leqslant j \\leqslant n-1$, les entiers $j$ et $n-j$ sont de la même couleur. De plus, si $i$ est un entier vérifiant $1 \\leqslant i \\leqslant n$ et $i \\neq k$, les entiers $i$ et $|i-k|$ sont de la même couleur.\n\nDémontrer que Augustin a colorié tous les entiers de la même couleur.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPour $k=1$, on voit que $n$ a la même couleur que $n-1$, puis que $n-2$ a la même couleur que $n-1$, etc. et tous les entiers ont la même couleur.\n\nPour $k=2$, $n$ est impair et on voit que $n, n-2, \\ldots, 1$ ont la même couleur, puis que $n-1$ a la même couleur que $1$, et que $n-1, n-3, \\ldots, 2$ ont la même couleur.\n\nEn faisant la même chose pour des petites valeurs de $k$, on se rend compte qu'on arrive à chaque fois à montrer successivement que $n-k, \\ldots, n-(n-1)k$ pris modulo $n$ sont tous de la même couleur. Essayons de formaliser cela.\n\nNotons $c$ la couleur de $n$. On montre par récurrence la propriété suivante pour tout $j \\in \\{1, \\ldots, n-1\\}$ :\n\n$\\mathcal{P}(j)$ : \"Le reste de la division euclidienne de $n-jk$ par $n$ n'est pas congru à $0$ modulo $n$, et est colorié avec la couleur $c$.\"\n\nInitialisation : pour $j=1$, comme $1 \\leqslant n-k \\leqslant n-1$, le reste de la division euclidienne de $n-k$ vaut $n-k$, n'est pas congru à $0$ modulo $n$, et est colorié avec la couleur $c$.\n\nHérédité : Soit $j \\in \\{0, \\ldots, n-2\\}$, supposons que $\\mathcal{P}(j)$ est vraie et montrons $\\mathcal{P}(j+1)$. Notons $r$ le reste de la division euclidienne de $n-jk$ par $n$.\n\nPremier cas : si $r > k$, alors $n > r \\geqslant r-k > 0$. Or $r-k \\equiv n-jk-k \\equiv n-(j+1)k \\pmod{n}$. Donc le reste de la division euclidienne de $n-(j+1)k$ par $n$ vaut $r-k$, est non nul et est de la même couleur que $r$, donc de la couleur $c$, ce qui conclut.\n\nSecond cas : si $r = k$, alors $n-jk \\equiv k \\pmod{n}$, i.e. $(j+1)k \\equiv 0 \\pmod{n}$. Comme $n$ est premier avec $k$, et divise $(j+1)k$, $n$ divise $j+1$, mais comme $j+1 \\in \\{1, \\ldots, n-1\\}$, ce cas est impossible.\n\nTroisième cas : si $k > r$, alors $k-r$ est de la couleur de $r$, donc de la couleur $c$. De plus, on a $1 \\leqslant k-r \\leqslant k \\leqslant n-1$, donc $n-(k-r)$ vérifie $1 \\leqslant n-(k-r) \\leqslant n-1$ et est de la couleur $c$. Or $n-(k-r) \\equiv r-k \\equiv n-(j+1)k \\pmod{n}$, donc $n-(k-r)$ est le reste de la division euclidienne de $n-(j+1)k$ par $n$, est non nul et de couleur $c$, ce qui conclut.\n\nCeci conclut l'hérédité.\n\nMaintenant, donnons-nous $j \\in \\{1, \\ldots, n-1\\}$. Il existe $\\ell \\in \\{0, \\ldots, n-1\\}$ tel que $j \\equiv n-\\ell k \\pmod{n}$ : en effet, ceci est équivalent à $\\ell k \\equiv n-j \\pmod{n}$, soit à $\\ell \\equiv (n-j)k^{-1} \\pmod{n}$ (où $k^{-1}$ est l'inverse de $k$ modulo $n$, qui existe car $k$ est premier avec $n$), et admet une solution dans $\\{0, \\ldots, n-1\\}$. En particulier, $j$ est le reste de la division euclidienne de $n-\\ell k$ par $n$, donc de couleur $c$.\n\nAinsi, tous les entiers de $1$ à $n-1$ ont la même couleur que $n$, donc tous les entiers sont coloriés de la même couleur.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSoient $a, b, c, d$ quatre réels tels que $|a|>1, |b|>1, |c|>1$ et $|d|>1$. Supposons que $(a+1)(b+1)(c+1)(d+1) = (a-1)(b-1)(c-1)(d-1)$. Montrer que\n$$\n\\frac{1}{a-1} + \\frac{1}{b-1} + \\frac{1}{c-1} + \\frac{1}{d-1} > 0\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nOn sait d'après l'énoncé qu'aucun des $a, b, c, d$ ne vaut $1$. La condition de l'énoncé s'écrit alors\n$$\n\\frac{a+1}{a-1} \\frac{b+1}{b-1} \\frac{c+1}{c-1} \\frac{d+1}{d-1} = 1\n$$\nRemarquons que comme $|a|>1$, alors on a soit $a>1$ et donc $a+1>0$ et $a-1>0$, ou alors $a<-1$ et donc $a+1<0$ et $a-1<0$. Ainsi, dans tous les cas $\\frac{a+1}{a-1}$ est positif, et de même pour $b, c, d$. On peut donc appliquer l'inégalité arithmético-géométrique sur le produit précédent pour avoir\n$$\n\\frac{a+1}{a-1} + \\frac{b+1}{b-1} + \\frac{c+1}{c-1} + \\frac{d+1}{d-1} \\geqslant 4 \\sqrt[4]{\\frac{a+1}{a-1} \\frac{b+1}{b-1} \\frac{c+1}{c-1} \\frac{d+1}{d-1}} = 4\n$$\nOn réécrit ceci sous la forme\n$$\n\\left(1 + \\frac{2}{a-1}\\right) + \\left(1 + \\frac{2}{b-1}\\right) + \\left(1 + \\frac{2}{c-1}\\right) + \\left(1 + \\frac{2}{d-1}\\right) \\geqslant 4\n$$\net donc\n$$\n\\frac{1}{a-1} + \\frac{1}{b-1} + \\frac{1}{c-1} + \\frac{1}{d-1} \\geqslant 0\n$$\nMais l'énoncé nous demande de montrer que l'inégalité est stricte, on s'intéresse donc au cas d'égalité dans l'inégalité arithmético-géométrique. On a égalité si et seulement si\n$$\n\\frac{a+1}{a-1} = \\frac{b+1}{b-1} = \\frac{c+1}{c-1} = \\frac{d+1}{d-1}\n$$\nComme le produit de ces quatre termes vaut $1$, nécessairement ces termes valent tous $1$ ou tous $-1$. Dans le premier cas, on a $a+1 = a-1$ ce qui est impossible. Dans le deuxième cas, on a $a+1 = 1-a$ et donc $a=0$, ce qui est aussi impossible car $|a|>1$. Ainsi, on ne peut pas avoir égalité et l'inégalité est bien stricte comme dans l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer le plus grand entier $n$ pair ayant la propriété suivante : quelle que soit la façon de paver une grille $n \\times n$ avec des dominos, il existe une ligne coupant le tableau en deux parties non vides et n'intersectant aucun domino.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\n\nRéponse: $n=6$\n\nMentionnons pour commencer que la réponse peut se deviner en essayant l'exercice pour des petites valeurs de $n$.\nLe problème demande de trouver le plus grand entier $n$ vérifiant une certaine propriété, il comporte donc nécessairement deux parties. Nous allons montrer d'abord que $6$ vérifie la propriété, puis nous allons montrer que si $n \\geqslant 8$, il existe une configuration de dominos ne respectant pas la propriété.\n\nCas $n=6$ : Supposons au contraire qu'il existe une configuration de dominos telle que chaque ligne coupant le tableau en deux parties non vides coupe au moins un domino. On ne va considérer que les lignes horizontales séparant deux rangées et les lignes verticales séparant deux colonnes du tableau, on appelle ces lignes des séparatrices, et les deux parties non vides du tableau obtenues avec une séparatrice des sections.\n\nD'une part, un domino ne peut être intersecté que par au plus une séparatrice (et cette séparatrice coupe le domino en deux carrés égaux). D'autre part, une séparatrice intersecte un nombre pair de dominos. En effet, chaque section obtenue contient un nombre pair de cases (les sections sont des rectangles dont l'un des côtés est de longueur $6$). Comme chaque domino sectionné recouvre exactement une case de la section et que les autres cases sont recouvertes par des dominos inclus dans la section et sont donc en nombre pair, on déduit bien qu'une séparatrice intersecte un nombre pair de dominos.\n\n![](attached_image_1.png)\n\nPar hypothèse, chaque séparatrice intersecte au moins un domino, donc elle en intersecte au moins $2$. Il y a $10$ séparatrices, donc au moins $20$ dominos du pavage sont intersectés par une séparatrice. Or, le pavage d'un carré $6 \\times 6$ contient exactement $18$ dominos. On a abouti à une contradiction.\n\nCas $n \\geqslant 8$ : Nous allons donner pour tout entier $n \\geqslant 8$ un pavage de la grille $n \\times n$ tel que toute ligne coupant le tableau en deux parties non vides intersecte au moins un domino.\n\nOn considère le motif suivant:\n\n![](attached_image_2.png)\n\nOn pave ensuite l'intérieur du carré grisé avec le motif suivant :\n\n![](attached_image_3.png)\n\nOn obtient bien de cette façon un pavage qui vérifie la propriété de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDéterminer tous les entiers $x$ tels que $2^{x}+x^{2}+25$ est le cube d'un nombre premier.",
    "options": [],
    "answer": "6",
    "solution": "Solution:\nSoit $x$ un entier tel que $2^{x}+x^{2}+25$ est le cube d'un nombre premier $p$. Puisque $2^{x}=p^{3}-x^{2}-25$ est un entier, $x$ est positif. De plus $p^{3} \\geqslant 25$ donc $p>2$ et $p$ est impair.\n\nSi $x=0$, alors $2^{x}+x^{2}+25=26$ qui n'est pas un cube. Ainsi, $x \\geqslant 1$. On déduit que $x^{2}+25$ est de la parité de $p^{3}$, c'est-à-dire impair. Ceci force $x$ à être pair. En particulier, $x \\geqslant 2$ donc $p^{3}>27$ et $p>3$. Soit $y$ l'entier tel que $x=2y$. En regardant l'équation modulo $3$, on trouve\n$$\n2^{2y}+4y^{2}+25 \\equiv 1+y^{2}+25 \\equiv 2+y^{2} \\pmod{3}\n$$\nSi $y$ n'est pas divisible par $3$, son carré vaut $1$ modulo $3$, ce qui implique que $3$ divise $p$, ce qui est exclu. Ainsi, $3$ divise $y$. Soit $z$ l'entier tel que $y=3z$. L'équation devient\n$$\n2^{6z}+36z^{2}+25=p^{3}\n$$\nSi $z=1$, alors le membre de gauche vaut $125=5^{3}$. Ainsi, $x=6 \\times 1$ est solution.\n\nMontrons que si $z \\geqslant 2$, on a $\\left(2^{2z}\\right)^{3}<2^{6z}+36z^{2}+25<\\left(2^{2z}+1\\right)^{3}$, ce qui montrera que le côté gauche ne peut être le cube d'un entier. L'inégalité de gauche est toujours vraie car $36z^{2}+25>0$. Pour montrer l'inégalité de droite, il faut montrer que $3 \\times 2^{4z}+3 \\times 2^{2z}+1>36z^{2}+25$.\n\nD'une part, si $z \\geqslant 2$, on a $3 \\times 2^{2z} \\geqslant 3 \\times 16>25$. D'autre part, on montre par récurrence sur $z$ que $2^{4z}>12z^{2}$.\n\nInitialisation: Si $z=2$, on a bien $2^{4z}=256>48=12z^{2}$.\n\nHérédité : On suppose que $2^{4z}>12z^{2}$ pour $z \\geqslant 2$. Alors\n$$\n2^{4(z+1)}=16 \\times 2^{4z}>16 \\times 12z^{2} \\geqslant 12 \\times 4z^{2} \\geqslant 12\\left(z^{2}+2z+1\\right)=12(z+1)^{2}\n$$\noù on a utilisé que $z^{2} \\geqslant z$ et $z^{2} \\geqslant 1$ pour $z \\geqslant 2$. Ainsi, la propriété est vraie pour $z+1$, ce qui achève la récurrence.\n\nOn a donc bien $3 \\times 2^{4z}+3 \\times 2^{2z}+1>36z^{2}+25$, ce qui implique que le membre de gauche n'est pas le cube d'un entier pour $z \\geqslant 2$.\n\nL'unique solution est donc $x=6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAline et Elsa jouent au jeu suivant. Elles disposent de 100 pierres qu'elles séparent en deux piles (pas forcément de même taille) au début du jeu. Puis chacune à leur tour, en commençant par Aline, elles effectuent le mouvement suivant : elles choisissent une pile, puis un entier strictement positif inférieur ou égal à la moitié de la taille de la pile choisie et retirent ce nombre de pierres de la pile. La première joueuse qui ne peut plus effectuer de mouvement perd.\n\nDéterminer toutes les configurations initiales pour lesquelles Elsa a une stratégie gagnante.",
    "options": [],
    "answer": "(50,50), (67,33), (33,67), (95,5), (5,95); equivalently, those splits with (pile1+1)/(pile2+1) a power of two.",
    "solution": "Solution:\n\nOn observe sur le cas à une seule pile en commençant par la fin que les solutions perdantes pour Aline sont les piles à $2^{n}-1$ pierres.\n\nRevenons au cas d'une configuration initiale générique $(a, b)$. Par définition, la position $(1,1)$ est perdante. $(a, b)$ est perdante pour la personne qui doit jouer si $\\frac{a+1}{b+1}$ est une puissance (positive ou négative) de $2$. En effet, si $\\frac{a+1}{b+1}=2^{n}$, si on pioche dans $a$, on diminue le quotient mais strictement de moins qu'un facteur $2$. À l'inverse, si on pioche dans la pile $b$ on l'augmente mais strictement moins que d'un facteur $2$ ce qui conserve bien une position gagnante pour la joueuse suivante.\n\nRéciproquement, si $2^{n}(b+1)<a+1<2^{n+1}(b+1)$ avec $n \\geqslant 0$ alors on prend le nombre nécessaire de pierres de la pile $a$ de sorte qu'il en reste précisément $2^{n}(b+1)-1$. Cela est possible car $a<2^{n+1}(b+1)-1$ donc $a \\leqslant 2^{n+1}(b+1)-2$ d'où $\\frac{a}{2} \\leqslant 2^{n}(b+1)-1$ ce qui signifie qu'on en a laissé plus que la moitié.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $n \\geqslant 1$ un entier et $x_{1}, \\ldots, x_{n}$ des réels positifs. Montrer que\n$$\n\\left(\\frac{x_{1}}{1}+\\frac{x_{2}}{2}+\\ldots+\\frac{x_{n}}{n}\\right)\\left(1 \\cdot x_{1}+2 \\cdot x_{2}+\\ldots+n \\cdot x_{n}\\right) \\leqslant \\frac{(n+1)^{2}}{4 n}\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)^{2}\n$$\nQuels sont les cas d'égalité?",
    "options": [],
    "answer": "Equality holds for n = 1 trivially; for n ≥ 2, equality holds if and only if x_i = 0 for all i with 1 < i < n and x_1 = x_n.",
    "solution": "Solution:\n\nL'astuce est d'appliquer une inégalité arithmético-géométrique au côté gauche de l'inégalité, mais avec des coefficients $\\sqrt{n}$ et $\\frac{1}{\\sqrt{n}}$ :\n$$\n\\begin{aligned}\n\\left(\\frac{x_{1}}{1}+\\frac{x_{2}}{2}+\\ldots+\\frac{x_{n}}{n}\\right)\\left(1 \\cdot x_{1}+2 \\cdot x_{2}+\\ldots+n \\cdot x_{n}\\right) = \\sqrt{n}\\left(\\frac{x_{1}}{1}+\\frac{x_{2}}{2}+\\ldots+\\frac{x_{n}}{n}\\right) \\frac{1 \\cdot x_{1}+2 \\cdot x_{2}+\\ldots+n \\cdot x_{n}}{\\sqrt{n}} \\\\\n& \\leqslant \\frac{1}{4}\\left(\\sqrt{n}\\left(\\frac{x_{1}}{1}+\\frac{x_{2}}{2}+\\ldots+\\frac{x_{n}}{n}\\right)+\\frac{1 \\cdot x_{1}+2 \\cdot x_{2}+\\ldots+n \\cdot x_{n}}{\\sqrt{n}}\\right)^{2}\n\\end{aligned}\n$$\nPour chaque $i$, le coefficient de $x_{i}$ est $\\frac{\\sqrt{n}}{i}+\\frac{i}{\\sqrt{n}}$. La fonction $t \\mapsto t+\\frac{1}{t}$ est décroissante pour $t<1$ et croissante pour $t>1$. En effet, par exemple pour $1<x<y$, on a $x y>1$ et donc:\n$$\ny-x>\\frac{y-x}{x y}=\\frac{1}{x}-\\frac{1}{y}\n$$\nAinsi, le coefficient maximal est atteint en $i=1$ ou $i=n$. Dans chaque cas, on trouve un coefficient $\\sqrt{n}+\\frac{1}{\\sqrt{n}}=\\frac{n+1}{\\sqrt{n}}$, et donc\n$$\n\\begin{aligned}\n\\left(\\frac{x_{1}}{1}+\\frac{x_{2}}{2}+\\ldots+\\frac{x_{n}}{n}\\right)\\left(1 \\cdot x_{1}+2 \\cdot x_{2}+\\ldots+n \\cdot x_{n}\\right) \\leqslant \\frac{1}{4}\\left(\\frac{n+1}{\\sqrt{n}}\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)\\right)^{2} \\\\\n\\left(\\frac{x_{1}}{1}+\\frac{x_{2}}{2}+\\ldots+\\frac{x_{n}}{n}\\right)\\left(1 \\cdot x_{1}+2 \\cdot x_{2}+\\ldots+n \\cdot x_{n}\\right) \\leqslant \\frac{(n+1)^{2}}{4 n}\\left(x_{1}+x_{2}+\\ldots+x_{n}\\right)^{2}\n\\end{aligned}\n$$\nOn s'intéresse aux cas d'égalité. Pour $1<i<n$, l'inégalité $\\frac{\\sqrt{n}}{i}+\\frac{i}{\\sqrt{n}} \\leqslant \\frac{n+1}{\\sqrt{n}}$ est stricte, donc nécessairement $x_{i}$ doit être égal à 0. Ensuite, il faut avoir égalité dans l'inégalité arithmético-géométrique, ce qui s'écrit\n$$\n\\begin{aligned}\n\\sqrt{n}\\left(x_{1}+\\frac{x_{n}}{n}\\right) & =\\frac{1}{\\sqrt{n}}\\left(x_{1}+n x_{n}\\right) \\\\\n\\left(\\sqrt{n}-\\frac{1}{\\sqrt{n}}\\right) x_{1} & =\\left(\\sqrt{n}-\\frac{1}{\\sqrt{n}}\\right) x_{n}\n\\end{aligned}\n$$\nSi $n=1$, on a toujours égalité (il n'y a qu'un seul $x_{i}$), sinon on a la condition $x_{1}=x_{n}$. Réciproquement, on peut vérifier que si $x_{1}=x_{n}$ et tous les autres $x_{i}$ sont nuls, on a bien égalité dans l'inégalité de l'énoncé.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDéterminer tous les couples $(a, p)$ d'entiers strictement positifs, avec $p$ premier, tels que pour tout couple $(m, n)$ d'entiers strictement positifs, le reste de la division euclidienne de $a^{2^{n}}$ par $p^{n}$ est non nul, et est le même que celui de $a^{2^{m}}$ par $p^{m}$.",
    "options": [],
    "answer": "All pairs with the prime equal to two and the integer odd, and all pairs with the integer equal to one and any odd prime.",
    "solution": "Solution:\n\nEn prenant $n=1$ dans l'énoncé, on obtient que pour tout entier $m$ strictement positif, le reste de $a^{2^{m}}$ modulo $p^{m}$ vaut celui de $a^{2}$ modulo $p$, donc est constant. Notons $r$ ce reste : on a donc $r \\neq 0$ ainsi $r \\in \\{1, \\ldots, p-1\\}$. On a $a^{2} \\equiv r \\pmod{p}$ et $a^{4} \\equiv r \\pmod{p^{2}}$, donc $a^{4} \\equiv r \\pmod{p}$, donc $r^{2} \\equiv r \\pmod{p}$. Ainsi comme $r$ est premier avec $p$, $r$ est inversible $\\bmod p$, donc $r \\equiv 1 \\pmod{p}$. On a donc $a^{2^{m}} \\equiv 1 \\pmod{p^{m}}$ pour tout $m$. En particulier $a^{2} \\equiv 1 \\pmod{p}$.\n\nOr\n$$\na^{2^{m}}-1=\\left(a^{2^{m-1}}-1\\right)\\left(a^{2^{m-1}}+1\\right)=\\cdots=\\left(a^{2}-1\\right)\\left(a^{2}+1\\right)\\left(a^{4}+1\\right) \\ldots\\left(a^{2^{m-1}}+1\\right)\n$$\nSi $p \\geqslant 3$, pour tout $k \\geqslant 1$, $a^{2^{k}}+1 \\equiv 1^{2^{k-1}}+1 \\equiv 2 \\not\\equiv 0 \\pmod{p}$, donc si $a \\neq 1$, $V_{p}\\left(a^{2^{m}}-1\\right)=V_{p}\\left(a^{2}-1\\right)$. Or pour tout $m$, $p^{m}$ divise $a^{2^{m}}-1$, donc $m \\leqslant V_{p}\\left(a^{2^{m}}-1\\right)=V_{p}\\left(a^{2}-1\\right)$ ce qui est absurde car $V_{p}\\left(a^{2}-1\\right)$ est fini. Ainsi $a=1$. Réciproquement pour tout $p$ premier, $a=1$ est solution car le reste de $1^{2^{m}}$ modulo $p^{m}$ vaut toujours $1$ et est non nul.\n\nSi $p=2$ et $a \\neq 1$, $a^{2} \\equiv 1 \\pmod{2}$, donc $a$ est impair. Réciproquement si $a$ est impair, rappelons que\n$$\na^{2^{m}}-1=\\left(a^{2^{m-1}}-1\\right)\\left(a^{2^{m-1}}+1\\right)=\\cdots=\\left(a^{2}-1\\right)\\left(a^{2}+1\\right)\\left(a^{4}+1\\right) \\ldots\\left(a^{2^{m-1}}+1\\right)\n$$\nLe produit de droite contient $m$ termes tous pairs, donc est divisible par $2^{m}$. Ainsi pour tout $n$, le reste de $a^{2^{n}}$ modulo $2^{n}$ vaut $1$ et est non nul, donc tous les couples de la forme $(a, 2)$ avec $a$ impair sont solutions.\n\nAinsi les couples solutions sont ceux de la forme $(1, p)$ pour $p \\geqslant 3$ et $(a, 2)$ pour $a$ impair.\nSolution:\n\nÀ partir du fait que $a^{2^{m}} \\equiv 1 \\pmod{p^{m}}$ pour tout $m$, on a comme $p$ divise $a^{2}-1$, par LTE si $p \\geqslant 3$ que si $a>1$, $V_{p}\\left(a^{2^{m}}-1\\right)=V_{p}\\left(a^{2}-1\\right)+V_{p}\\left(2^{m-1}\\right)=V_{p}\\left(a^{2}-1\\right)$. En particulier, comme $V_{p}\\left(a^{2^{m}}-1\\right) \\geqslant m$ pour tout $m \\geqslant 1$, $V_{p}\\left(a^{2}-1\\right) \\geqslant m$ pour tout $m$ ce qui est absurde donc $a=1$. Réciproquement, $a=1$ et $p \\geqslant 3$ convient comme dans la solution précédente.\n\nPour le cas $p=2$, de même que précédemment on a $p$ impair, donc $4$ divise $p^{2}-1=(p-1)(p+1)$, donc par LTE, comme $4$ divise $a^{4}-1$, si $a>1$, $V_{2}\\left(a^{2^{m}}-1\\right)=V_{2}\\left(a^{2}-1\\right)+V_{2}\\left(2^{m-1}\\right) \\geqslant 2+m-1=m+1$, ainsi $a^{2^{m}}-1$ est divisible par $2^{m}$ pour tout $m$, et on conclut de la même façon que précédemment.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $ABC$ un triangle, $\\omega$ son cercle inscrit et $D$, $E$ et $F$ les points de contact de $\\omega$ avec les côtés $BC$, $CA$ et $AB$ respectivement. La perpendiculaire à $(BC)$ en $C$ coupe la droite $(EF)$ en $M$ et la perpendiculaire à $(BC)$ en $B$ coupe la droite $(EF)$ en $N$. La droite $(DM)$ recoupe $\\omega$ en $P$ et la droite $(DN)$ recoupe $\\omega$ en $Q$. Montrer que $DP = DQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nOn commence par \"éffacer\" les points $P$ et $Q$ de la figure, c'est-à-dire qu'on commence par se ramener à un énoncé équivalent au problème original, mais qui n'implique pas les points $P$ et $Q$.\n\nSi l'énoncé est vrai, alors $(DI)$ est la médiatrice du segment $[PQ]$, de sorte que la droite $(DI)$ est la bissectrice de l'angle $\\widehat{MDN}$. Réciproquement, si on prouve que la droite $(DI)$ est la bissectrice de l'angle $\\widehat{MDN}$, les droites $(DM)$ et $(DN)$ seront alors symétriques par rapport à la droite $(DI)$. Ainsi, puisque le cercle $\\omega$ et ces droites sont symétriques par rapport à $(DI)$, leurs intersections le sont aussi, et $P$ et $Q$ seront symétriques par rapport à $(DI)$ et $DP = DQ$. Il suffit donc de montrer que $\\widehat{MDI} = \\widehat{NDI}$ ou encore, puisque les droites $(DI)$ et $(BC)$ sont perpendiculaires, $\\widehat{MDC} = \\widehat{NDB}$.\n\nCette égalité est en particulier vraie si les triangles $MCD$ et $NBD$ sont semblables. Comme ces deux triangles ont déjà un angle droit en commun, il suffit de montrer que $\\frac{BD}{CD} = \\frac{BN}{CM}$.\n\nSoit $Z$ le point d'intersection des droites $(BC)$ et $(EF)$ et soit $X$ le point d'intersection de la parallèle à $(AC)$ passant par $B$ avec la droite $(EF)$.\n\nLes triangles $AEF$ et $FBX$ sont alors semblables, ainsi le triangle $FBX$ est isocèle en $B$. On a donc, en utilisant le théorème de Thalès pour les points $E$, $X$, $Z$ et $C$, $B$, $Z$ :\n$$\n\\frac{BD}{CD} = \\frac{BF}{CE} = \\frac{BX}{CE} = \\frac{ZB}{ZC}.\n$$\nEn appliquant ensuite Thalès aux points $M$, $N$, $Z$ et $C$, $B$, $Z$, on trouve bien\n$$\n\\frac{BD}{CD} = \\frac{ZB}{ZC} = \\frac{BN}{CM},\n$$\nce qui permet de conclure.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30985,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $N$ un entier strictement positif. On suppose qu'il existe quatre sous-ensembles $A_{1}$, $A_{2}$, $A_{3}$ et $A_{4}$ de $\\{1, \\ldots, N\\}$, chacun de cardinal $500$ et on suppose que, pour tous $x, y$ dans $\\{1, \\ldots, N\\}$, il existe un indice $i$ tel que $x$ et $y$ sont dans $A_{i}$. Déterminer la plus grande valeur de $N$ possible.",
    "options": [],
    "answer": "833",
    "solution": "Solution:\n\nNous allons traiter ce problème avec le langage des graphes. La modélisation du problème pousse à considérer le graphe dont les sommets sont les éléments de $S$, qui sont reliés par une arête de couleur $i$ si les deux sommets appartiennent à l'ensemble $A_{i}$.\n\nDans la suite, on dira également qu'un graphe $G_{1}$ recouvre un graphe $G_{2}$ si toute arête de $G_{2}$ est une arête de $G_{1}$.\n\nOn note $K_{m}$ le graphe constitué de $m$ sommets et dont tous les sommets sont reliés. Un tel graphe est dit complet.\n\nOn note $K_{m, m}$ le graphe constitué de 2 ensembles de sommets $S_{1}$ et $S_{2}$, tous les deux de cardinal $m$ et dans lequel chaque sommet de $S_{1}$ est relié à tous les sommets de $S_{2}$. Un tel graphe est communément appelé bipartite complet, et on appellera $S_{1}$ et $S_{2}$ les équipes de $K_{m, m}$.\n\nLemme : Soit $K_{m, m}$ un graphe bipartite complet et $S_{1}$ et $S_{2}$ ses équipes. Supposons qu'il est possible de recouvrir $K_{m, m}$ avec deux graphes complets $K_{k}$ et $K_{\\ell}$ de taille respective $k, \\ell < 2m$. Alors $m < \\frac{k+\\ell}{3}$.\n\nPreuve du lemme : Commençons par montrer que $K_{k}$ et $K_{\\ell}$ contiennent chacun soit $S_{1}$ soit $S_{2}$. En effet, soit $v \\in S_{1}$ n'appartenant pas à $K_{\\ell}$. Pour tout sommet $w$ de $S_{2}$, l'arête $vw$ appartient donc à $K_{k}$, mais alors $w$ appartient à $K_{k}$, donc $K_{k}$ contient tous les sommets de $S_{2}$. Mais alors, en raisonnant sur un sommet de $S_{1}$ n'appartenant pas à $K_{k}$, on obtient que $K_{\\ell}$ contient aussi tous les éléments de $S_{2}$. Ainsi, chaque sommet de $S_{2}$ appartient à $K_{k}$ et à $K_{\\ell}$ et chaque sommet de $S_{1}$ appartient à $K_{k}$ ou à $K_{\\ell}$, de sorte que $k+\\ell \\geqslant |S_{1}| + 2|S_{2}| = 3m$.\n\nRevenons au problème. On suppose par l'absurde que $N \\geqslant 834$. Remarquons que si on pouvait avoir $N$ encore plus grand, on pourrait se ramener au cas $N = 834$ en ajoutant arbitrairement des éléments de $\\{1, \\ldots, 834\\}$ qui n'y étaient pas encore aux ensembles.\n\nNous partitionnons notre ensemble de points comme suit :\n- $Z$ l'ensemble des points dans $A_{1}$ et $A_{2}$.\n- $X$ l'ensemble des points dans $A_{1}$ mais pas $A_{2}$.\n- $Y$ l'ensemble des points dans $A_{2}$ mais pas $A_{1}$.\n- $W$ le reste (ni dans $A_{1}$, ni dans $A_{2}$).\n\nOn a alors $|X| + |Y| + |Z| + |W| = 834$ et $|X| + |Z| = |Y| + |Z| = 500$ ce qui nous donne $|W| - |Z| = -166$.\n\nCas 1 : $|W| = 0$. Il vient $|Z| = 166$, $|X| = |Y| = 334$. Ceci contredit notre lemme car cela implique que nous pouvons couvrir $K_{334, 334}$ avec deux cliques de taille $500$.\n\nCas 2 : $|W| > 0$. Alors $Z \\subset K_{3} \\cap K_{4}$ (c'est-à-dire que chacun de ses éléments est dans $A_{3}$ ou $A_{4}$) pour que chacun de ses sommets soit connecté aux éléments de $|W|$ donc $|K_{3} \\cap Z| + |K_{4} \\cap Z| \\geqslant |Z|$. Ainsi, nous pouvons voir que si $|W| > 0$ : (En utilisant que $W \\subset K_{3} \\cap K_{4}$)\n\n$$\n|K_{3} \\cap (X \\cup Y)| + |K_{4} \\cap (X \\cup Y)| \\leqslant 1000 - 2|W| - |Z|\n$$\n\nAlors, nous devons couvrir $K_{500 - |Z|, 500 - |Z|}$ (graphe formé à l'aide des ensembles $X$ et $Y$) avec deux cliques de tailles $|K_{3} \\cap (X \\cup Y)|$, $|K_{4} \\cap (X \\cup Y)|$. Par notre lemme, $3(500 - |Z|) \\leqslant 1000 - 2|W| - |Z|$ or comme $|Z| = |W| + 166$, on obtient une contradiction !\n\nPar ailleurs, pour $N = 833$, on dispose d'une construction explicite : En reprenant nos notations $X$, $Y$, $Z$. Avec $|Z| = 167$ et $|X| = |Y| = 333$. On place tous les éléments de $X$ dans $A_{3}$ et $A_{4}$ et un élément de $Y$ dans les deux, la moitié des éléments de $Y$ qu'il reste dans $A_{3}$ et l'autre dans $A_{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSoit $c$ un entier positif ou nul. Trouver toutes les suites d'entiers strictement positifs $a_{1}, a_{2}, \\ldots$ telles que pour tout entier strictement positif $n$, $a_{n}$ soit égal au nombre d'entiers $i$ vérifiant $a_{i} \\leqslant a_{n+1}+c$.",
    "options": [],
    "answer": "a_n = n + c + 1 for all n",
    "solution": "Solution:\n\nSoit $(a_{n})$ une suite vérifiant les conditions de l'énoncé pour la constante $c$. On commence par montrer que la suite est croissante. En effet, supposons par l'absurde qu'il existe un entier $n$ tel que $a_{n}>a_{n+1}$. Alors d'après l'hypothèse de l'énoncé, il y a strictement plus de $a_{i}$ inférieurs ou égaux à $a_{n+1}+c$ que de $a_{i}$ inférieurs ou égaux à $a_{n+2}+c$. Ceci signifie que nécessairement $a_{n+2}<a_{n+1}$. En continuant ainsi, on trouve que la suite des $a_{i}$ est strictement décroissante à partir du rang $n$. Comme elle est à valeurs entières strictement positives, c'est absurde.\n\nMontrons maintenant que la suite est d'abord constante jusqu'à un certain rang, puis strictement croissante. Supposons en effet que $a_{i}=a_{j}$ avec $i<j$. Alors si $i>1$, $a_{i-1}$ et $a_{j-1}$ sont tous les deux les nombres de $a_{k}$ inférieurs ou égaux à $a_{i}+c=a_{j}+c$, et donc $a_{i-1}=a_{j-1}$. Comme on a $i \\leqslant j-1<j$, on a $a_{j}=a_{i} \\leqslant a_{j-1} \\leqslant a_{j}$ et donc $a_{i-1}=a_{j-1}=a_{j}$ et donc $a_{i-1}=a_{j}$. On continue en diminuant successivement $i$ de 1 jusqu'à avoir $a_{j}=a_{1}$. Toute la suite ne peut pas être égale à $a_{1}$, car par hypothèse seulement $a_{1}$ termes de la suite sont inférieurs ou égaux à $a_{2}+c=a_{1}+c>a_{1}$. Ainsi, la suite est constante égale à $a_{1}$ jusqu'à un certain rang $N$, puis croît strictement.\n\nSoit $n \\geqslant N$, alors on sait qu'il y a $a_{n}$ éléments de la suite inférieurs ou égaux à $a_{n+1}+c$. Ces éléments sont les $n$ premiers $a_{i}$ (par croissance), ainsi que les éléments $a_{n+1}, a_{n+2}, \\ldots, a_{n+k}$ pour un certain $k$. Ainsi, $k$ est donné par les inégalités $a_{n+k} \\leqslant a_{n+1}+c$ et $a_{n+k+1}>a_{n+1}+c$. Comme la suite est strictement croissante après le rang $N$, nécessairement\n$$\na_{n+k} \\geqslant a_{n+k-1}+1 \\geqslant a_{n+k-2}+2 \\geqslant \\ldots \\geqslant a_{n+1}+k-1\n$$\nd'où $a_{n+1}+c \\geqslant a_{n+k} \\geqslant a_{n+1}+k-1$ et $k \\leqslant c+1$. Ainsi, le nombre $a_{n}$ d'éléments de la suite inférieurs ou égaux à $a_{n+1}+c$ est au plus\n$$\na_{n}=n+k \\leqslant n+c+1\n$$\nD'un autre côté, on sait que $a_{n+k+1}$ vaut au moins $a_{n+1}+c+1 \\geqslant a_{n}+c+2=n+k+c+2$. On a donc $a_{n+k+1} \\geqslant n+k+c+2$ ce qui est la même borne que la borne supérieure ci-dessus mais pour $a_{n+k+1}$. Alors pour tout $m \\geqslant n+k+1$, on a\n$$\nm+c+1 \\geqslant a_{m} \\geqslant a_{n+k+1}+(m-(n+k+1)) \\geqslant n+k+c+2+m-n-k-1=m+c+1\n$$\net donc $a_{m}=m+c+1$ pour tout $m$ suffisamment grand.\n\nMontrons que cette égalité est vraie pour tout $m$. Si par l'absurde elle était fausse, soit $M$ maximal tel que $a_{M} \\neq M+c+1$. Alors $a_{M}$ est le nombre de termes de la suite inférieurs ou égaux à $a_{M+1}+c= M+2c+2$, qui sont les $M$ termes $a_{1}, \\ldots, a_{M}$ ainsi que les termes $a_{M+1}, \\ldots, a_{M+c+1}=M+2c+2$, et donc\n$$\na_{M}=M+(c+1)\n$$\nce qui est absurde. Ainsi, pour tout $n$, $a_{n}=n+c+1$.\n\nVérifions que cette solution fonctionne bien. Soit $n \\geqslant 1$ entier. Alors $a_{n+1}+c=n+2c+2$ et les entiers $i$ tels que $a_{i}=i+c+1 \\leqslant n+2c+2$ sont les $i$ inférieurs ou égaux à $n+c+1$, il y en a $n+c+1=a_{n}$, comme voulu.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nOn dispose de $a + b$ bols alignés sur une rangée. Les $a$ premiers bols contiennent une pomme tandis que les $b$ derniers bols contiennent une poire.\n\nUne opération consiste à déplacer une pomme du bol $i$ au bol $i+1$ et une poire du bol $j$ au bol $j-1$, où $i$ et $j$ sont deux entiers tels que $i-j$ est pair (un bol peut contenir plusieurs fruits).\n\nOn souhaite arriver à la situation finale où les $b$ premiers bols contiennent une poire et les $a$ derniers bols contiennent une pomme. Montrer que c'est possible si et seulement si $ab$ est pair.",
    "options": [],
    "answer": "Possible if and only if ab is even",
    "solution": "Solution:\n\nCommençons par montrer qu'il n'est pas possible d'y arriver lorsque $a$ et $b$ sont impairs tous les deux. Soit $I$ la somme alternée des nombres de fruits dans les bols. Dans les positions initiales et finales décrites, on a alors $I=0$ car $a+b$ est pair. En outre, lors de chaque opération, $I$ est augmenté de $\\pm 2$ à cause de la condition $i-j$ pair.\n\nSupposons par l'absurde qu'il soit possible de passer de la position initiale à la position finale décrite. Les pommes doivent passer des bols de numéros $1$ à $a$ vers les bols de numéros $b+1$ à $a+b$, et au cours de chaque opération, une pomme augmente de un le numéro de son bol. Il faut donc\n$$\n\\sum_{k=1}^{a}(b+k)-\\sum_{k=1}^{a} k = a \\cdot b\n$$\nopérations pour passer de la position initiale à la position finale, avec $ab$ impair par hypothèse. Ainsi, lorsque la position finale est atteinte, $I \\equiv 2ab \\equiv 2 \\pmod{4}$, ce qui est absurde car $I$ est nul dans la position finale décrite.\n\nMontrons maintenant par récurrence qu'il est possible de parvenir à la position finale lorsque $a$ ou $b$ est pair. Sans perte de généralité, nous traitons le cas où $b$ est pair, l'autre cas s'obtenant directement par symétrie. Si $b=0$, il n'y a rien à montrer. Supposons maintenant que la propriété soit vraie pour $b=n$ pair, et supposons $b=n+2$. Par hypothèse de récurrence, en ignorant les deux dernières poires, nous pouvons arriver en un nombre fini d'opérations à la situation où nous avons, dans l'ordre, $n$ bols avec une poire, puis $a$ bols avec une pomme, puis $2$ bols avec une poire. Si nous faisons deux opérations, entre les bols de numéros $i=a+n$ et $j=a+n+2$, puis $i=j=a+n+1$, les trois derniers bols contiennent maintenant dans l'ordre : une poire puis une poire puis une pomme. On réitère ces opérations permettant d'échanger une pomme avec les deux poires qui suivent jusqu'à arriver à la situation finale désirée.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben sei ein hinreichend großer Vorrat von gleichseitigen Dreiecken und Quadraten, alle mit der gleichen Seitenlänge. Aus diesen Bausteinen lassen sich konvexe* Polygone bilden, indem man sie in der Ebene lückenlos und überschneidungsfrei aneinander legt. (Die Figur zeigt drei Möglichkeiten für ein Sechseck.)\n![](attached_image_1.png)\n\na) Welches ist die größtmögliche Anzahl $m$ von Seitenkanten für ein so gebildetes konvexes Polygon? (Die Antwort ist zu begründen.)\n\nb) Man gebe für alle möglichen Anzahlen von Seitenkanten $\\leq m$ jeweils ein Beispiel an.",
    "options": [],
    "answer": "12",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30989,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn einem $10 \\times 17$-Rechteck werden 74 Punkte markiert.\nMan beweise, dass es dabei stets zwei markierte Punkte gibt, deren Abstand 2 nicht überschreitet.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZwei Sehnen $AC$ und $BD$ eines Kreises mit Mittelpunkt $M$ schneiden sich in $P$. Die Umkreise der Dreiecke $PBC$ und $PDA$ haben ihre Mittelpunkte in $E$ bzw. $F$ und schneiden sich ein zweites Mal in $Q$.\nMan beweise, dass $\\overline{MF} = \\overline{QE}$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30991,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nWir betrachten - mit 1 beginnend - alle positiven Teiler einer natürlichen Zahl $n$ der Größe nach geordnet: $1 = d_{1} < d_{2} < d_{3} < \\ldots < n$.\nMan bestimme alle natürlichen Zahlen $n$ mit den Eigenschaften:\n(1) $n = d_{13} + d_{14} + d_{15}$\nund\n$$\n\\left(d_{5} + 1\\right)^{3} = d_{15} + 1\n$$",
    "options": [],
    "answer": "1998",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30992,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben ist die Summe $S = \\frac{1}{n} + \\frac{1}{n+1} + \\ldots + \\frac{1}{n+m}$ mit $n, m \\in \\{1,2,3, \\ldots\\}$.\n\na) Man beweise, dass $S$ keine natürliche Zahl sein kann.\n\nb) Man ermittle (mit Begründung!) für $m = 2 \\cdot (n-1)$ ein $k \\in \\mathbb{N}$ so, dass $S \\in ]k, k+1[$.",
    "options": [],
    "answer": "1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDie natürlichen Zahlen von $1$ bis $n^{2}$ werden zufällig auf die Felder eines $n \\times n$-Quadrats verteilt ($n \\geq 2$). Für jedes Paar von Zahlen innerhalb einer Reihe bzw. einer Spalte dividieren wir die größere durch die kleinere Zahl. Der kleinste dieser $n^{2}(n-1)$ Quotienten werde als Charakteristik $C$ der zufälligen Anordnung bezeichnet.\nWelches ist der größtmögliche Wert für $C$? (Die Antwort ist zu begründen.)",
    "options": [],
    "answer": "(n+1)/n",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben seien positive ganze Zahlen $a, b, c$ mit der Eigenschaft $b > 2a$ und $c > 2b$.\nMan zeige, dass es dann stets eine reelle Zahl $r$ mit folgender Eigenschaft gibt:\nDie gebrochenen Teile der Zahlen $ra, rb, rc$ liegen alle im Intervall $\\left( \\frac{1}{3}, \\frac{2}{3} \\right)$.\n\n(Hinweis: Der gebrochene Teil einer Zahl ist die Differenz zwischen der Zahl und ihrem ganzen Teil.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn einem Schritt kann man vom Punkt $A(i|j|k)$, mit $i, j, k \\in \\{0,1,2,3\\}$, zu einem weiteren Punkt des Würfelgitters gelangen, indem man stets genau eine der Koordinaten um $1$ vergrößert.\nMan ermittle die Anzahl aller kürzesten Wege, die vom Ursprung $O(0|0|0)$ in den Punkt $P(3|3|3)$ führen.",
    "options": [],
    "answer": "1680",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMan beweise: Für die positiven reellen Zahlen $a$, $b$, $c$ gilt die Ungleichung\n$$\n\\frac{a}{\\sqrt{(a+b)(a+c)}} + \\frac{b}{\\sqrt{(b+a)(b+c)}} + \\frac{c}{\\sqrt{(c+a)(c+b)}} \\leq \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWir betrachten zwei Kreise in der Ebene, welche sich in den beiden verschiedenen Punkten $X$ und $Y$ schneiden.\nMan beweise, dass es in dieser Ebene vier feste Punkte mit folgender Eigenschaft gibt: Für jeden Kreis, der im Durchschnitt der beiden gegebenen Kreise liegt und diese in den Punkten $A$ und $B$ berührt sowie die Gerade $X Y$ in den Punkten $C$ und $D$ schneidet, geht jede der Geraden $A C, A D, B C$ und $B D$ durch einen dieser vier Punkte.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 30998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIm regulären 18-Eck $A_1 A_2 \\ldots A_{18}$ mit dem Umkreismittelpunkt $M$ ist $P$ der Schnittpunkt von $A_1A_7$ mit $MA_2$ und $Q$ der Schnittpunkt von $A_2A_{13}$ mit $MA_1$. Man berechne den Winkel $\\angle MPQ$.",
    "options": [],
    "answer": "50°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 30999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFür jede positive ganze Zahl $n$ bezeichne $d(n)$ die Anzahl aller positiver Teiler von $n$. (Beispiele: $d(2)=2, d(6)=4, d(9)=3$.)\nMan bestimme alle positiven ganzen Zahlen $n$ mit der Eigenschaft $(d(n))^3 = 4n$.",
    "options": [],
    "answer": "n = 2 and n = 128",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31000,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan ermittle die Anzahl aller Zahlen der Form $x^{2}+y^{2}$ ($x, y \\in \\{1,2,3, \\ldots, 1000\\}$), die durch 121 teilbar sind.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDie Reste, die eine Quadratzahl bei der Division durch 11 haben kann, sind 0, 1, 4, 9, 5 und 3. Da aber, außer zur Null, keine komplementären Reste modulo 11 auftreten, müssen sowohl $x^{2}$ als auch $y^{2}$ und folglich auch $x$ und $y$ durch 11 teilbar sein.\n\nUnter den Zahlen von 1 bis 1000 gibt es genau $[1000/11] = 90$ Vielfache von 11. Es gibt demnach höchstens $90 \\cdot 89 / 2 = 4005$ durch 121 teilbare Zahlen der Form $x^{2}+y^{2}$ mit $x \\neq y$, und genau 90 durch 121 teilbare Zahlen der Form $x^{2}+x^{2}$. Folglich kann es höchstens 4095 Zahlen der genannten Form geben. Ihre Anzahl ist allerdings geringer, da es viele Zahlen gibt, die unterschiedliche Darstellungen als Summe zweier Quadrate erlauben. Leider hatte das der Aufgabensteller bei der Formulierung der Aufgabe nicht bedacht. Umso erfreulicher war die Tatsache, dass einige Teilnehmer sehr interessante Lösungsansätze lieferten.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $P$ die Menge aller geordneter Paare $(p, q)$ von nichtnegativen ganzen Zahlen. Man bestimme alle Funktionen $f: P \\rightarrow \\mathrm{IR}$ mit der Eigenschaft\n$$\nf(p, q)=\\left\\{\\begin{array}{c}\n0 \\quad \\text{ wenn } p q=0 \\\\\n1+\\frac{1}{2} f(p+1, q-1)+\\frac{1}{2} f(p-1, q+1) \\text{ sonst }\n\\end{array} .\\right.\n$$",
    "options": [],
    "answer": "f(p, q) = p*q",
    "solution": "Solution:\n\nDie einzige solche Funktion ist $f(p, q)=p \\cdot q$.\nDiese Funktion erfüllt offensichtlich für $p=0$ oder $q=0$ die erste Bedingung. Für $p q \\neq 0$ gilt\n$$\n1+\\frac{1}{2} f(p+1, q-1)+\\frac{1}{2} f(p-1, q+1)=1+\\frac{1}{2}(p+1)(q-1)+\\frac{1}{2}(p-1)(q+1)=p q,\n$$\nso dass auch die zweite Bedingung erfüllt ist.\n\nNun muss noch gezeigt werden, dass es keine andere Funktion mit diesen Eigenschaften gibt. Dazu setzen wir $f(p, q)=p q+g(p, q)$. Für $p q \\neq 0$ ergibt sich aus der zweiten Bedingung\n$$\np q+g(p, q)=1+\\frac{1}{2}((p+1)(q-1)+g(p+1, q-1)+(p-1)(q+1)+g(p-1, q+1)),\n$$\nalso\n$$\ng(p, q)=\\frac{1}{2}(g(p+1, q-1)+g(p-1, q+1)).\n$$\nDaher bilden die Zahlen $g(0, p+q)$, $g(1, p+q-1), g(2, p+q-2), \\ldots, g(p+q-1,1), g(p+q, 0)$ eine arithmetische Folge. Da ihre Außenglieder $g(0, p+q)$ und $g(p+q, 0)$ gleich Null sind, hat jedes Folgenglied den Wert Null. Folglich ist $g(p, q)=0$ für alle nichtnegativen ganzen Zahlen und daher $f(p, q)=p \\cdot q$ die einzige Lösung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn ein spitzwinkliges Dreieck $A B C$ wird ein Quadrat mit Mittelpunkt $A_{1}$ so einbeschrieben, dass zwei Ecken auf $B C$ und je eine auf $A B$ bzw. $A C$ liegen. Analog sind die Quadrate mit den Mittelpunkten $B_{1}$ bzw. $C_{1}$ definiert.\nMan beweise, dass die Geraden $A A_{1}, B B_{1}$ und $C C_{1}$ einen gemeinsamen Schnittpunkt haben.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZusätzlich zur gegebenen Figur betrachten wir das nach außen errichtete Quadrat über der Seite $A B$ mit dem Mittelpunkt $C_{2}$. Da es eine zentrische Streckung gibt, welche das einbeschriebene Quadrat mit Mittelpunkt $C_{1}$ in das nach außen errichtete Quadrat überführt, liegen $C, C_{1}$ und $C_{2}$ auf einer Geraden.\nDie Höhe $h_{c}$ schneidet $A B$ in $H_{c}$ und es sei $|A H_{c}|=p, |H_{c} B|=q$. Ferner sei $M_{c}$ der Mittelpunkt von $A B$ und $C^{\\prime}$ der Schnittpunkt von $C C_{2}$ mit $A B$. Schließlich setzen wir $|A C^{\\prime}|=c_{1}$ und $|C^{\\prime} B|=c_{2}$. Alle Teilstrecken von $A B$ existieren wegen der Spitzwinkligkeit des Dreiecks $A B C$. Für die anderen Dreiecksseiten seien entsprechende\n\n![](attached_image_1.png)\n\nPunkte und Abschnitte analog definiert.\n\nNach den Strahlensätzen gilt $\\frac{|H_{c} C^{\\prime}|}{|C H_{c}|}=\\frac{|C^{\\prime} M_{c}|}{|M_{c} C_{2}|}$, d.h. $\\frac{c_{1}-p}{h_{c}}=\\frac{\\frac{1}{2} c-c_{1}}{\\frac{1}{2} c} \\Rightarrow c_{1}=\\frac{c(p+h_{c})}{c+2 h_{c}}$.\nAnalog ist $c_{2}=\\frac{c(q+h_{c})}{c+2 h_{c}}$ und wir erhalten $\\frac{c_{1}}{c_{2}}=\\frac{\\frac{p}{h_{c}}+1}{\\frac{q}{h_{c}}+1}$. (Für $\\alpha=\\beta$ ist $\\frac{c_{1}}{c_{2}}=1$.)\nDie Winkel zwischen den Höhen und den jeweils anliegenden beiden Seiten seien zyklisch mit $\\gamma_{1}, \\gamma_{2}$ bzw. $\\alpha_{1}, \\alpha_{2}$ bzw. $\\beta_{1}, \\beta_{2}$ bezeichnet. Dann gilt $\\alpha_{1}=\\gamma_{2}(=90^{\\circ}-\\beta), \\quad \\beta_{1}=\\alpha_{2}(=90^{\\circ}-\\gamma)$ und $\\gamma_{1}=\\beta_{2}(=90^{\\circ}-\\alpha)$. Ferner ist $\\frac{p}{h_{c}}=\\tan \\gamma_{1}$ und $\\frac{q}{h_{c}}=\\tan \\gamma_{2}$, also $\\frac{c_{1}}{c_{2}}=\\frac{\\tan \\gamma_{1}+1}{\\tan \\gamma_{2}+1}$. Analog gilt $\\frac{a_{1}}{a_{2}}=\\frac{\\tan \\alpha_{1}+1}{\\tan \\alpha_{2}+1}$ und $\\frac{b_{1}}{b_{2}}=\\frac{\\tan \\beta_{1}+1}{\\tan \\beta_{2}+1}$. Damit ist\n\n![](attached_image_2.png)\n\n$\\frac{a_{1}}{a_{2}} \\cdot \\frac{b_{1}}{b_{2}} \\cdot \\frac{c_{1}}{c_{2}}=\\frac{(\\tan \\alpha_{1}+1)(\\tan \\beta_{1}+1)(\\tan \\gamma_{1}+1)}{(\\tan \\alpha_{2}+1)(\\tan \\beta_{2}+1)(\\tan \\gamma_{2}+1)}=\\frac{\\tan \\alpha_{1}+1}{\\tan \\gamma_{2}+1} \\cdot \\frac{\\tan \\beta_{1}+1}{\\tan \\alpha_{2}+1} \\cdot \\frac{\\tan \\gamma_{1}+1}{\\tan \\beta_{2}+1}=1 \\cdot 1 \\cdot 1=1$.\nNach der Umkehrung des Satzes von Ceva schneiden sich daher die Geraden $A A^{\\prime}=A A_{1}$, $B B^{\\prime}=B B_{1}$ und $C C^{\\prime}=C C_{1}$ in einem Punkt.\n\n**Anmerkung:** Einige Teilnehmer haben vermutet, die Geraden $A A_{1}$ etc. seien die Winkelhalbierenden des gegebenen Dreiecks. Diese Vermutung ist falsch, wie man an einem sehr spitzen, fast rechtwinkligen Dreieck sieht. Der Satz von Ceva ist zum Beweis nicht notwendig; es gibt andere Lösungsmöglichkeiten, z.B. mit Drehstreckungen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan beweise: Sind $x$, $y$, $z$ die Längen der Winkelhalbierenden eines Dreiecks mit dem Umfang $6$, dann gilt\n$$\n\\frac{1}{x^{2}}+\\frac{1}{y^{2}}+\\frac{1}{z^{2}} \\geq 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nMan beachte nebenstehende Figur.\nDP und DQ sind die Parallelen durch $D$ zu $AB$ und $AC$. Da $AD = x$ die Winkelhalbierende von $\\alpha$ ist, ist $AQDP$ eine Raute, deren Seitenlänge mit $u$ bezeichnet wurde.\nAus der Ähnlichkeit der Dreiecke $PDC$ mit $QBD$ folgt $\\frac{u}{c-u} = \\frac{b-u}{u}$, woraus man $u = \\frac{bc}{b+c}$ erhält.\nDer Kosinussatz im Dreieck $AQD$ führt zu\n$$\n\\begin{aligned}\nx^{2} = 2u^{2} - 2u^{2} \\cos(\\pi - \\alpha) & = 2u^{2}(1 + \\cos \\alpha) \\\\\n= 2u^{2}\\left(1 + \\frac{b^{2} + c^{2} - a^{2}}{2bc}\\right) = 2u^{2} & \\frac{(b+c)^{2} - a^{2}}{2bc} = \\\\\n& = \\frac{bc}{(b+c)^{2}}(a+b+c)(-a+b+c)\n\\end{aligned}\n$$\n![](attached_image_1.png)\nMit $(b+c)^{2} \\geq 4bc$ und $a+b+c = 6$ folgt $x^{2} \\leq 1,5(-a+b+c)$.\nÄhnlich ergibt sich $y^{2} \\leq 1,5(a-b+c)$ und $z^{2} \\leq 1,5(a+b-c)$.\nAus der Ungleichung zwischen dem arithmetischen und geometrischen Mittel dreier positiver Zahlen ergibt sich $\\left(\\frac{1}{x^{2}} + \\frac{1}{y^{2}} + \\frac{1}{z^{2}}\\right)\\left(x^{2} + y^{2} + z^{2}\\right) \\geq 9$ und hieraus schließlich\n$$\n\\frac{1}{x^{2}} + \\frac{1}{y^{2}} + \\frac{1}{z^{2}} \\geq \\frac{9}{x^{2} + y^{2} + z^{2}} \\geq \\frac{9}{1,5 \\cdot 6} = 1\n$$\nwobei das Gleichheitszeichen nur für $x = y = z$ gilt.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31004,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan ermittle alle Lösungen der Gleichung $x^{2y} + (x+1)^{2y} = (x+2)^{2y}$ mit $x, y \\in \\mathbb{N}$.",
    "options": [],
    "answer": "x=3, y=1",
    "solution": "Solution:\n\nMan erkennt leicht, dass weder $x$, noch $y$, Null sein können.\nFür $y=1$ erhält man aus $x^{2} + (x+1)^{2} = (x+2)^{2}$ die Gleichung $x^{2} - 2x - 3 = 0$, von der nur die Lösung $x=3$ in Frage kommt.\n\nSei nun $y > 1$.\nDa $x$ und $x+2$ dieselbe Parität haben, ist $x+1$ eine gerade und demnach $x$ eine ungerade Zahl.\nMit $x = 2k-1$ ($k \\in \\mathbb{N}$) ergibt sich die Gleichung\n\n$\\quad (2k-1)^{2y} + (2k)^{2y} = (2k+1)^{2y}$\n\nworaus man durch Ausmultiplizieren folgendes erhält:\n\n$$\n(2k)^{2y} - 2y (2k)^{2y-1} + \\ldots - 2y 2k + 1 + (2k)^{2y} = (2k)^{2y} + 2y (2k)^{2y-1} + \\ldots + 2y 2k + 1\n$$\n\nDa $y > 1$, ist auch $2y \\geq 3$. Lässt man nun alle Glieder in $yk$ auf einer Seite und faktorisiert auf der anderen Seite $(2k)^{3}$, dann erhält man:\n\n$$\n8yk = (2k)^{3} \\left[2 \\binom{2y}{3} + 2 \\binom{2y}{5} (2k)^{2} + \\ldots - (2k)^{2y-3}\\right]\n$$\n\nworaus folgt, dass $y$ ein Vielfaches von $k$ ist.\n\nDurch Division der Gleichung durch $(2k)^{2y}$ ergibt sich:\n\n$$\n\\left(1 - \\frac{1}{2k}\\right)^{2y} + 1 = \\left(1 + \\frac{1}{2k}\\right)^{2y}\n$$\n\nwobei die linke Seite kleiner als $2$ ist. Die rechte Seite ist allerdings größer als $1 + \\frac{2y}{2k} \\geq 2$, was nicht sein kann, da $y$ ein Vielfaches von $k$ ist.\n\nDie gegebene Gleichung hat demnach nur die Lösung $x=3$ und $y=1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan beweise, dass es keine positive ganze Zahl $n$ mit der folgenden Eigenschaft gibt: Für $k=1,2, \\ldots, 9$ ist die - in dezimaler Schreibweise - linke Ziffer von $(n+k)!$ gleich $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir nehmen an, dass es eine Zahl $n$ mit der verlangten Eigenschaft gibt. Dann kann keine der Fakultäten eine Zehnerpotenz sein, weil ab $3!$ alle Fakultäten durch $3$ teilbar sind und die ersten Fakultäten offensichtlich nicht die verlangte Eigenschaft haben. Es kann auch keine der Zahlen $n+2, \\ldots, n+9$ eine Zehnerpotenz sein, weil sonst die Anfangsziffer zweier aufeinanderfolgender Fakultäten in der betrachteten Folge dieselbe wäre. Somit gibt es ein $j$ mit $10^{j} < n+2 < \\ldots < n+9 < 10^{j+1}$ (1).\n\nWeil $(n+8)!$ mit einer $8$ und $(n+9)!$ mit einer $9$ anfängt, gibt es natürliche Zahlen $a$ und $b$ mit $9 \\cdot 10^{a} < (n+9)! < 10^{a+1}$ und $8 \\cdot 10^{b} < (n+8)! < 9 \\cdot 10^{b}$, was auf $10^{a-b} < n+9 < \\frac{5}{4} \\cdot 10^{a-b}$ führt.\n\nMit (1) folgt $j = a-b$ und $10^{j} < n+2 < \\ldots < n+9 < \\frac{5}{4} \\cdot 10^{j}$ (2).\n\nDa $(n+1)!$ mit $1$ anfängt, gibt es ein $m$ mit $10^{m} < (n+1)! < 2 \\cdot 10^{m}$, während aus (2) folgt: $10^{3j} < (n+2)(n+3)(n+4) < \\left(\\frac{5}{4}\\right)^{3} \\cdot 10^{3j}$.\n\nMultiplikation der beiden letzten Ungleichungen liefert $10^{3j+m} < (n+4)! < 2 \\cdot \\frac{125}{64} \\cdot 10^{3j+m}$, was wegen $\\frac{250}{64} < 4$ zu $10^{3j+m} < (n+4)! < 4 \\cdot 10^{3j+m}$ abgeschwächt werden kann. Daraus folgt, dass die Zahl $(n+4)!$ nicht mit $4$, sondern mit $1$, $2$ oder $3$ beginnen würde - im Widerspruch zur Voraussetzung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBeim Schachspiel erhält der Sieger 1 Punkt und der Besiegte 0 Punkte. Bei Unentschieden (Remis) erhält jeder der Spieler $\\frac{1}{2}$ Punkt.\nVierzehn Schachspieler, von denen keine zwei gleich alt waren, trugen einen Wettbewerb aus, in dem jeder gegen jeden spielte. Nach Abschluss des Wettbewerbs wurde eine Rangliste erstellt. Von zwei Spielern mit gleicher Punktezahl erhält der Jüngere eine bessere Platzierung.\nNach dem Wettbewerb stellte Jan fest, dass die drei Bestplatzierten insgesamt genau so viele Punkte erhielten wie die Gesamtzahl der Punkte der letzten neun Spieler. Jörg bemerkte dazu, dass dabei die Zahl der unentschieden ausgegangenen Spiele maximal war. Man ermittle die Anzahl der unentschiedenen Spiele.",
    "options": [],
    "answer": "40",
    "solution": "Solution:\n\nDie Gesamtzahl der Punkte der letzten neun Spieler beträgt mindestens $(9 \\cdot 8) : 2 = 36$ Punkte (denn wenn nur jeder der neun gegen einen anderen der neun spielen würde, dann wären es, da in jeder Partie ein Punkt vergeben wird, insgesamt schon 36 Punkte). Die Gesamtzahl der drei Erstplatzierten ist aber höchstens $13 + 12 + 11 = 36$ Punkte, falls sie alle Spiele gegen die restlichen 13 Spieler gewinnen würden.\nEs folgt also, dass die letzten neun Spieler keines der Spiele mit den anderen gewinnen und die drei ersten alle, wobei sie untereinander durchaus unentschieden spielen können.\nDie Anzahl der unentschieden gespielten Partien (die ja gemäß Jörg maximal ist!) muss also $3 + 1 + 36 = 40$ sein.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMan bestimme alle Funktionen $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ mit der Eigenschaft\n$$\nf(f(x)+y)=2 x+f(f(y)-x)\n$$\nfür alle $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = x + c for any real constant c",
    "solution": "Solution:\nFür eine beliebige reelle Zahl $z$ setzen wir $a=f(z)$, $b=z+f(0)$, $n=f\\left(\\frac{a+b}{2}\\right)$ und $m=f(0)-\\frac{a+b}{2}$. Wegen $\\mathbb{D}_f=\\mathbb{R}$ sind $a$, $b$, $n$ und $m$ wohlbestimmt. Einsetzen von $x=0$, $y=z$ in $(*)$ liefert $f(z+f(0))=f(f(z))$, also $f(a)=f(b)$ (I).\n\nEinsetzen von $x=m$ und $y=a$ bzw. $y=b$ in $(*)$ liefert\n\\[\n\\begin{array}{l}\nf(f(m)+a)=2 m+f(f(a)-m) \\\\\nf(f(m)+b)=2 m+f(f(b)-m)\n\\end{array}\n\\]\nwas wir wegen (I) zu $f(f(m)+a)=f(f(m)+b)$ (II) zusammenfassen können.\n\nEinsetzen von $x=a$ bzw. $x=b$ und $y=n$ in $(*)$ liefert\n\\[\n\\begin{array}{l}\nf(f(a)+n)=2 a+f(f(n)-a) \\\\\nf(f(b)+n)=2 b+f(f(n)-b)\n\\end{array}\n\\]\nwas wir wegen (I) zu $2 a+f(f(n)-a)=2 b+f(f(n)-b)$ (III) zusammenfassen können.\n\nEinsetzen von $x=\\frac{a+b}{2}$, $y=0$ in $(*)$ liefert $f\\left(f\\left(\\frac{a+b}{2}\\right)\\right)=a+b+f\\left(f(0)-\\frac{a+b}{2}\\right)$, also $f(n)=a+b+f(m)$ oder, anders formuliert, $f(n)-a=f(m)+b$ bzw. $f(n)-b=f(m)+a$. Dies liefert mit (III) sofort $2 a+f(f(m)+b)=2 b+f(f(m)+a)$, woraus mit (II) nun $a=b$ folgt, also $f(z)=z+f(0)$.\n\nDamit ist gezeigt, dass jede Funktion $f$, die $(*)$ erfüllt, die Form $f(x)=x+c$ ($c=$ konst.) haben muss. Einsetzen in $(*)$ bestätigt, dass jedes $f$ dieser Form die gegebene Gleichung tatsächlich erfüllt: $f(f(x)+y)=x+y+2 c=2 x+f(f(y)-x)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31008,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben ist ein Dreieck $ABC$ und ein Punkt $M$ so, dass die Geraden $MA$, $MB$, $MC$ die Geraden $BC$, $CA$, $AB$ (in dieser Reihenfolge) in $D$, $E$ beziehungsweise $F$ schneiden.\nMan beweise, dass es dann stets die Zahlen $\\varepsilon_{1}, \\varepsilon_{2}, \\varepsilon_{3}$ aus $\\{-1,1\\}$ gibt, so dass gilt:\n$$\n\\varepsilon_{1} \\cdot \\frac{MD}{AD} + \\varepsilon_{2} \\cdot \\frac{ME}{BE} + \\varepsilon_{3} \\cdot \\frac{MF}{CF} = 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPunkt $M$ ($M \\notin \\{A, B, C\\}$) kann entweder auf einer der vorgegebenen Geraden, oder in einer der sieben Gebiete liegen, in denen die Ebene des Dreiecks $ABC$ durch die Geraden $AB$, $BC$, $CA$ geteilt wird.\n\n![](attached_image_1.png)\n\nEs ist stets\n$$\n\\begin{aligned}\n& \\frac{MD}{AD} = \\frac{MP}{h_a} = \\frac{0,5 \\cdot MP \\cdot BC}{0,5 \\cdot h_a \\cdot BC} = \\frac{F(MBC)}{F(ABC)}, \\\\\n& \\text{ (Strahlensatz) }\n\\end{aligned}\n$$\nwobei $P \\in BC$ und $MP \\perp BC$ und $F(XYZ)$ der Inhalt des Dreiecks $XYZ$ ist. Ähnliches gilt für die anderen Verhältnisse.\n\nLiegt $M$ im Inneren oder am Rande des Dreiecks $ABC$, dann ist also:\n$$\n\\frac{MD}{AD} + \\frac{ME}{BE} + \\frac{MF}{CF} = \\frac{F(MBC)}{F(ABC)} + \\frac{F(MCA)}{F(ABC)} + \\frac{F(MAB)}{F(ABC)} = 1\n$$\nworaus $\\varepsilon_{1} = \\varepsilon_{2} = \\varepsilon_{3} = 1$ folgt.\n\nÄhnliche Überlegungen führen auch dann ans Ziel, wenn $M$ außerhalb des Dreiecks $ABC$ liegt, nur dass $\\varepsilon_{1}$, $\\varepsilon_{2}$ bzw. $\\varepsilon_{3} = -1$, falls $M$ im Bereich I, II bzw. III liegt. Sollte $M$ in den Bereichen IV, V bzw. VI liegen, dann sind genau zwei der $\\varepsilon_{i}$ gleich $-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $B$ ein beliebiger Punkt auf einem Kreis $k_{1}$ und es sei $A$ ein von $B$ verschiedener Punkt auf der Tangente an $k_{1}$ in $B$. Ferner sei $C$ ein Punkt außerhalb von $k_{1}$ mit der Eigenschaft, dass die Strecke $A C$ den Kreis $k_{1}$ in zwei verschiedenen Punkten schneidet. Schließlich sei $k_{2}$ der Kreis, der die Gerade $A C$ in $C$ berührt und den Kreis $k_{1}$ in einem Punkt $D$ berührt, welcher auf der anderen Seite von $A C$ liegt wie $B$.\nMan beweise, dass der Umkreismittelpunkt des Dreiecks $BCD$ auf dem Umkreis des Dreiecks $A B C$ liegt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir bezeichnen die Mittelpunkte der Kreise $k_{1}$ und $k_{2}$ mit $O_{1}$ bzw. $O_{2}$. Die Lösung setzt nun voraus, dass $\\angle C D B$ stumpf ist (siehe Fig.); andernfalls sind lediglich einige Winkel modulo $180^{\\circ}$ zu vertauschen.\nEs sei $P$ der Umkreismittelpunkt des Dreiecks $BDC$; dann ist nach dem Satz vom Mittelpunktswinkel\n$$\n\\angle B P C = 2 \\cdot (180^{\\circ} - \\angle C D B) = 360^{\\circ} - 2 \\cdot \\angle C D B\n$$\n![](attached_image_1.png)\n$P$ liegt dann auf dem Umkreis von Dreieck $A B C$, wenn $A C P B$ ein Sehnenviereck ist; dies gilt dann, wenn $\\angle B P C = 180^{\\circ} - \\angle C A B$ ist.\nUm dies zu beweisen, berechnen wir $\\angle C D B$. Da $k_{1}$ und $k_{2}$ sich in $D$ berühren, liegt $D$ auf der Strecke $O_{1} O_{2}$ und es gilt\n$$\n\\angle C D B = 180^{\\circ} + \\angle O_{1} D B - \\angle O_{2} D C\n$$\n(2). Da $k_{1}$ die Gerade $A B$ berührt, ist $\\angle A B O_{1} = 90^{\\circ}$, also $\\angle A B D + \\angle D B O_{1} = 90^{\\circ}$. Da $B$ und $D$ auf $k_{1}$ liegen, ist das Dreieck $B D O_{1}$ gleichschenklig mit $\\angle O_{1} D B = \\angle D B O_{1}$. Somit ist\n$$\n\\angle O_{1} D B = 90^{\\circ} - \\angle A B D\n$$\n(3). Da $k_{2}$ die Gerade $A C$ berührt, ist $\\angle A C O_{2} = 90^{\\circ}$, also $\\angle A C D + \\angle D C O_{2} = 90^{\\circ}$. Da $C$ und $D$ auf $k_{2}$ liegen, ist das Dreieck $C D O_{2}$ gleichschenklig mit $\\angle O_{2} D C = \\angle D C O_{2}$. Somit ist\n$$\n\\angle O_{2} D C = 90^{\\circ} - \\angle A C D\n$$\n(4). Einsetzen von (3) und (4) in (2) liefert\n$$\n\\angle C D B = 180^{\\circ} + (90^{\\circ} - \\angle A B D) - (90^{\\circ} - \\angle A C D) = 180^{\\circ} - \\angle A B D + \\angle A C D\n$$\n(5).\nNun betrachten wir den Schnittpunkt $R$ von $A C$ und $B D$ und folgern\n$$\n\\angle A B D = \\angle A B R = 180^{\\circ} - \\angle R A B - \\angle B R A = 180^{\\circ} - \\angle C A B - \\angle B R A\n$$\nsowie\n$$\n\\angle A C D = \\angle R C D = 180^{\\circ} - \\angle C D R - \\angle D R C = 180^{\\circ} - \\angle C D B - \\angle B R A\n$$\nAus (5) ergibt sich\n$$\n\\angle C D B = 180^{\\circ} + \\angle C A B - \\angle C D B\n$$\nalso\n$$\n2 \\cdot \\angle C D B = 180^{\\circ} + \\angle C A B\n$$\nEinsetzen in (1) liefert\n$$\n\\angle B P C = 360^{\\circ} - (180^{\\circ} + \\angle C A B) = 180^{\\circ} - \\angle C A B\n$$\nDamit ist alles gezeigt.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSei $N$ eine natürliche Zahl und $x_{1}, x_{2}, \\ldots, x_{n}$ weitere natürliche Zahlen kleiner als $N$ und so, dass das kleinste gemeinsame Vielfache von beliebigen zwei dieser $n$ Zahlen größer als $N$ ist.\nMan beweise, dass die Summe der Kehrwerte dieser $n$ Zahlen stets kleiner $2$ ist; also\n$$\n\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\cdots+\\frac{1}{x_{n}}<2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDa das kgV von $x_{i}$ und $x_{j}$ größer als $N$ ist, gibt es unter den Zahlen $1,2, \\ldots, N$ keine zwei, die sowohl Vielfache von $x_{i}$, als auch von $x_{j}$ sind.\nUnter den Vielfachen der natürlichen Zahl $x$ gibt es zwei so, dass $k x \\leq N < (k+1) x$, woraus $k \\leq \\frac{N}{x} < k+1$ folgt. Die Anzahl der Vielfachen von $x$, die kleiner $N$ sind, ist demnach der ganzzahlige Teil von $\\frac{N}{x}$, also gleich $\\left[\\frac{N}{x}\\right]$.\n\nFür die Zahlen $x_{1}, x_{2}, \\ldots, x_{n}$ gilt folglich $\\left[\\frac{N}{x_{1}}\\right]+\\left[\\frac{N}{x_{2}}\\right]+\\cdots+\\left[\\frac{N}{x_{n}}\\right]<N$.\n\nAndrerseits ist $\\left[\\frac{N}{x_{i}}\\right]>\\frac{N}{x_{i}}-1$ und demnach $\\frac{N}{x_{1}}+\\frac{N}{x_{2}}+\\cdots+\\frac{N}{x_{n}}-n<N$.\n\nDa aber $n<N$ ist folgt:\n$\\frac{N}{x_{1}}+\\frac{N}{x_{2}}+\\cdots+\\frac{N}{x_{n}}<2N$, was schließlich zu $\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\cdots+\\frac{1}{x_{n}}<2$ führt.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEine Funktion $f$ ist gegeben durch $f(x)+f\\left(1-\\frac{1}{x}\\right)=1+x$ für $x \\in \\mathbb{R} \\backslash\\{0,1\\}$.\nMan ermittle eine Formel für $f$.",
    "options": [],
    "answer": "f(x) = 1/2 * (1/(1 - x) + 1/x + x)",
    "solution": "Solution:\n\nSei $x \\in \\mathbb{R} \\backslash\\{0,1\\}$ und $y=1-\\frac{1}{x}$ und $z=\\frac{1}{1-x}$. Es ist leicht einzusehen, dass zusammen mit $x$ auch $y$ und damit auch $z$ zu $\\mathbb{R} \\backslash\\{0,1\\}$ gehören. Einsetzen von $y$ und $z$ in die Ausgangsgleichung führt zu:\n$$\nf\\left(1-\\frac{1}{x}\\right)+f\\left(\\frac{1}{1-x}\\right)=2-\\frac{1}{x} \\text{ und } f\\left(\\frac{1}{1-x}\\right)+f(x)=1+\\frac{1}{1-x}\n$$\nDurch Subtraktion der beiden letzten Beziehungen erhält man:\n$$\nf(x)-f\\left(1-\\frac{1}{x}\\right)=\\frac{1}{1-x}+\\frac{1}{x}-1\n$$\nund nach Addition zur Ausgangsgleichung führt das schließlich zu\n$$\nf(x)=\\frac{1}{2}\\left(\\frac{1}{1-x}+\\frac{1}{x}+x\\right)=\\frac{-x^{3}+x^{2}+1}{2 x(1-x)}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIn der Ebene liegen $n$ abgeschlossene Kreisscheiben $K_{1}, K_{2}, \\ldots, K_{n}$ mit gleichem Radius $r$. Jeder Punkt der Ebene ist dabei in höchstens $2003$ dieser Kreisscheiben enthalten. Man beweise, dass jede Kreisscheibe $K_{i}$ höchstens $14020$ andere Kreisscheiben schneidet.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWir führen einen Beweis durch Widerspruch. Dazu nehmen wir zusätzlich zur Voraussetzung an, dass eine Kreisscheibe (oBdA sei dies $K_{1}$) mindestens $14021$ andere Kreisscheiben schneidet. Die Mittelpunkte dieser Scheiben liegen dann offensichtlich alle in einer Kreisscheibe mit Radius $2r$ um den Mittelpunkt $M_{1}$ von $K_{1}$. Lägen $2003$ oder mehr der Mittelpunkte dieser anderen Kreisscheiben in oder auf dem Rand von $K_{1}$, so würden diese Scheiben alle $M_{1}$ enthalten. Damit würde $M_{1}$ in mehr als $2003$ Kreisscheiben enthalten sein, da $M_{1}$ auch in $K_{1}$ enthalten ist. Also müssen mindestens $14021 - 2002 = 12019$ Mittelpunkte der anderen Kreisscheiben in einem Kreisring mit innerem Radius $r$ und äußerem Radius $2r$ um $M_{1}$ liegen. Teilen wir diesen Ring in $6$ gleich große Sektoren mit Innenwinkel $60^{\\circ}$, so liegen nach dem Schubfachprinzip in mindestens einem Sektor wenigstens $2004$ dieser Mittelpunkte.\nIn der Zeichnung ist dieser Sektor mit dem Mittelpunkt $M$ der Strecke $AB$ dargestellt. Aus $|AM_{1}| = |BM_{1}| = 2r$ und $\\angle AM_{1}B = 60^{\\circ}$ folgt, dass $\\triangle ABM_{1}$ gleichseitig ist. Also sind $MD$ und $MC$ seine Mittelparallelen und es folgt $|MA| = |MB| = |MC| = |MD| = r$. Daher liegt der Sektor vollständig in einem Kreis um $M$ mit Radius $r$. Jede der wenigstens $2004$ Kreisscheiben mit Mittelpunkten in diesem Sektor enthält also den Punkt $M$. Damit ist $M$ aber in\n![](attached_image_1.png)\nwenigstens $2004$ Kreisscheiben enthalten – ein Widerspruch zur Annahme.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben seien jeweils $n$ reelle Zahlen $x_{1}, x_{2}, \\ldots, x_{n}$ bzw. $y_{1}, y_{2}, \\ldots, y_{n}$. Die Elemente einer $n \\times n$-Matrix $A$ seien folgendermaßen definiert: ($1 \\leq i, j \\leq n$)\n$$\na_{ij}= \\begin{cases}1 & \\text{ wenn } x_{i}+y_{j} \\geq 0 \\\\ 0 & \\text{ wenn } x_{i}+y_{j}<0\\end{cases}\n$$\nWeiter sei $B$ eine $n \\times n$-Matrix mit Elementen $0$ oder $1$, so dass die Summe der Elemente in jeder Zeile und jeder Spalte von $B$ gleich der Summe der Elemente in der entsprechenden Zeile bzw. Spalte von $A$ ist.\nMan beweise, dass dann $A=B$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir nehmen an, dass es eine Matrix $B$ der geforderten Art gebe mit $B \\neq A$. Nun betrachten wir in $A$ nur noch diejenigen Elemente $a_{ij}$, die sich von den entsprechenden Elementen $b_{ij}$ unterscheiden. Es muss mindestens ein solches Element geben. Alle anderen Elemente von $A$ werden gestrichen. Dann ist innerhalb jeder Zeile bzw. Spalte die Anzahl der verbleibenden Nullen gleich der Anzahl der verbleibenden Einsen, da diese Anzahlen in $B$ bei gleicher Zeilen- bzw. Spaltensumme gerade vertauscht sind. Also tritt jede Zahl $x_{i}$ in der verbleibenden Anordnung genauso oft als Summand einer Summe $x_{i}+y_{j}<0$ auf wie als Summand einer Summe $x_{i}+y_{j} \\geq 0$. Das Gleiche gilt für jede Zahl $y_{j}$.\n\nNun addieren wir alle Summen $x_{i}+y_{j}<0$ mit $a_{ij} \\neq b_{ij}$. Die Summe dieser Summen ist zwangsläufig $<0$. Ebenso addieren wir alle Summen $x_{i}+y_{j} \\geq 0$ mit $a_{ij} \\neq b_{ij}$. Die Summe dieser Summen ist zwangsläufig $\\geq 0$. Da aber jede der Zahlen $x_{i}$ und $y_{j}$ gleich häufig in beiden Summen auftritt, müssen diese Summen den gleichen Wert haben – Widerspruch! Daher kann es keine verschiedenen Elemente in $A$ und $B$ geben und es gilt $A=B$.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 31014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIm Dreieck $A B C$ ist $A D$ ($D \\in B C$) eine Seitenhalbierende, $E$ ein Punkt auf $A C$ und $F$ der Schnittpunkt von $B E$ mit $A D$.\nMan beweise: Falls $\\frac{B F}{F E}=\\frac{B C}{A B}+1$, dann ist $B E$ eine Winkelhalbierende.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSeien $B'$, $C'$, $E'$ die Projektionen von $B$, $C$ und $E$ auf die Gerade $A D$. Die rechtwinkligen Dreiecke $\\triangle D B B'$ und $\\triangle D C C'$ sind kongruent, da $D$ der Mittelpunkt von $B C$ ist und die spitzen Winkel bei $D$ gleich groß sind. Daraus folgt, dass $B B' = C C'$.\nIm Dreieck $\\triangle A C' C$ ist $E E'$ parallel zu $C C'$ und folglich\n$$\n\\frac{A C}{A E} = \\frac{C C'}{E E'} = \\frac{B B'}{E E'}.\n$$\nDie rechtwinkligen Dreiecke $\\triangle B B' F$ und $\\triangle E E' F$ sind ähnlich, da die Winkel bei $F$ gleich sind. Daraus folgt $\\frac{B B'}{E E'} = \\frac{B F}{F E}$, was zusammen mit der vorletzten Beziehung zu $\\frac{A C}{A E} = \\frac{B F}{F E}$ führt.\n\n![](attached_image_1.png)\n\nErsetzt man nun $\\frac{B F}{F E}$ in der Ausgangsbeziehung durch $\\frac{A C}{A E}$, ergibt sich:\n$\\frac{B C}{A B} + 1 = \\frac{A C}{A E} = \\frac{B C + A B}{A B}$, was zu $\\frac{A C - A E}{A E} = \\frac{B C}{A B}$ und schließlich zu $\\frac{E C}{A E} = \\frac{B C}{A B}$ führt.\nGemäß der Umkehrung des Satzes der Winkelhalbierenden ist $B E = w_{\\beta}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben sind die sechs reellen Zahlen $a, b, c$ und $x, y, z$ so, dass\n\na) $0 < b - c < a < b + c$\n\nb) $a x + b y + c z = 0$\n\nMan ermittle (mit Begründung!) das Vorzeichen von $a y z + b x z + c x y$.",
    "options": [],
    "answer": "≤ 0",
    "solution": "Solution:\n\nAus $0 < b - c < a < b + c$ folgt, dass $a$, $b$ und $c$ positiv sind. Außerdem kann man mit Strecken dieser Längen ein Dreieck konstruieren. Daraus folgt, dass die Zahlen $-a + b + c$, $a - b + c$ und $a + b - c$ positiv sind.\n\nAus $a x + b y + c z = 0$ folgt $x = -\\frac{b y + c z}{a}$.\n\nEs ergibt sich\n$$\na y z + b z x + c x y = a y z - \\frac{1}{a}(b y + c z)(b z + c y) = -\\frac{1}{a}\\left(b^2 y z + b c y^2 + c b z^2 + c^2 y z - a^2 y z\\right)\n$$\nEs reicht zu zeigen, dass der Term in der letzten Klammer nicht negativ ist.\n\nEine einfache Umformung ergibt\n$$\nb^2 y z + b c y^2 + c b z^2 + c^2 y z - a^2 y z = b c y^2 + b c z^2 + \\left(b^2 + c^2 - a^2\\right) y z\n$$\nDa $a$ und $c$ positiv sind, kann man den Term mit $4 b c$ multiplizieren, ohne dass sich das Vorzeichen ändert. Es ist also das Vorzeichen von $4 b^2 c^2 y^2 + 4 b^2 c^2 z^2 + 4 b c\\left(b^2 + c^2 - a^2\\right) y z$ zu untersuchen. Durch Addition und Subtraktion von $\\left(b^2 + c^2 - a^2\\right)^2 z^2$ ergibt sich der Reihe nach:\n$$\n\\begin{aligned}\n& 4 b^2 c^2 y^2 + 4 b c y z\\left(b^2 + c^2 - a^2\\right) + \\left(b^2 + c^2 - a^2\\right) z^2 + \\left[4 b^2 c^2 - \\left(b^2 + c^2 - a^2\\right)^2\\right] z^2 = \\\\\n& = \\left[2 b c y + \\left(b^2 + c^2 - a^2\\right) z\\right]^2 + \\left[4 b^2 c^2 - \\left(b^2 + c^2 - a^2\\right)^2\\right] z^2\n\\end{aligned}\n$$\nDas Vorzeichen des letzten Terms ist nicht negativ, da der erste Teil ein Quadrat ist und im zweiten Teil $4 b^2 c^2 - \\left(b^2 + c^2 - a^2\\right)^2 = (a + b + c)(-a + b + c)(a - b + c)(a + b - c)$ positiv ist. Die Zahl $a y z + b z x + c x y$ ist folglich nicht positiv.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31016,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEine unendliche Folge $a_{0}, a_{1}, a_{2}, \\ldots$ reeller Zahlen erfüllt die Bedingung $a_{n}=\\left|a_{n+1}-a_{n+2}\\right|$ für alle $n \\geq 0$, wobei $a_{0}$ und $a_{1}$ verschiedene positive Zahlen sind.\nKann diese Folge beschränkt sein? Die Antwort ist zu begründen.",
    "options": [],
    "answer": "The sequence is unbounded.",
    "solution": "Solution:\n\nZunächst beweisen wir, dass zwei aufeinander folgende Glieder niemals gleich sein können. Aus $a_{n}=a_{n+1}=c$ folgte nämlich sofort $a_{n-1}=0$ und $a_{n-2}=a_{n-3}=c \\quad(n>2)$. Schließlich müsste $a_{0}=a_{1}$ oder $a_{0}=0$ bzw. $a_{1}=0$ sein, was ausgeschlossen ist. Daher gilt auch $a_{n}>0$ für alle $n$.\n\nAuflösen der Bedingung liefert $a_{n+2}=a_{n+1}+a_{n}$ falls $a_{n+2}>a_{n+1}$ ist, sowie $a_{n+2}=a_{n+1}-a_{n}$ falls $a_{n+2}<a_{n+1}$ ist. Aus $a_{n+1}<a_{n}$ folgt also $a_{n+2}>a_{n}$ und $a_{n+2}>a_{n+1}$. Daher ist diejenige Teilfolge $b_{0}, b_{1}, b_{2}, \\ldots$ streng monoton wachsend, die durch Weglassen aller Glieder entsteht, die kleiner als ihr Vorgänger und ihr Nachfolger sind.\n\nWenn wir nun zeigen, dass $b_{m+1}-b_{m} \\geq b_{m}-b_{m-1}$ für alle $m \\geq 2$ gilt, so haben wir für diese Teilfolge eine arithmetische Folge mit positiver Differenz als Minorante, woraus die Unbeschränktheit direkt folgt. Dazu setzen wir $b_{m+1}=a_{n+2}$, wobei $a_{n+2}>a_{n+1}$ gelten soll. Für $a_{n+1}>a_{n}$ haben wir $b_{m}=a_{n+1}$ und $b_{m-1} \\geq a_{n-1}$ (weil entweder $b_{m-1}=a_{n-1}$ oder $b_{m-1}=a_{n}>a_{n-1}$ gilt). So ist $b_{m+1}-b_{m}=a_{n}=a_{n+1}-a_{n-1} \\geq b_{m}-b_{m-1}$. Für $a_{n+1}<a_{n}$ haben wir dagegen $b_{m}=a_{n}$ und $b_{m-1} \\geq a_{n-1}$ (weil entweder $b_{m-1}=a_{n-1}$ oder $b_{m-1}=a_{n-2}>a_{n-1}$ gilt). So ist hier $b_{m+1}-b_{m}=a_{n+1}=a_{n}-a_{n-1} \\geq b_{m}-b_{m-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben sind die positiven reellen Zahlen $a$ und $b$ und die natürliche Zahl $n$.\nMan ermittle in Abhängigkeit von $a, b$ und $n$ das größte der $n+1$ Glieder in der Entwicklung von $(a+b)^n$.",
    "options": [],
    "answer": "Let α = (n+1)b/(a+b). If α is not an integer, the unique largest term is the k-th term with k = ⌊α⌋ + 1, i.e., C(n, k−1) a^{n−k+1} b^{k−1}. If α is an integer m, then the m-th and (m+1)-th terms are equal and both maximal.",
    "solution": "Solution:\n\nDas $k$-te Glied $G(k)$ in der Entwicklung von $(a+b)^n$ ist gegeben durch die Formel:\n$$\nG(k) = \\binom{n}{k-1} \\cdot a^{n-k+1} \\cdot b^{k-1},\n$$\nwobei $\\binom{n}{0} = 1$ und $1 \\leq k \\leq n+1$.\n\nDa es endlich viele Glieder gibt und jede endliche Zahlenmenge (mindestens) ein maximales Element enthält, wird untersucht, unter welchen Bedingungen das $k$-te Element maximal ist. Dafür müssen die folgenden Beziehungen gleichzeitig erfüllt sein:\n$$\n\\binom{n}{k-1} \\cdot a^{n-k+1} \\cdot b^{k-1} \\geq \\binom{n}{k} \\cdot a^{n-k} \\cdot b^{k}\n$$\nund\n$$\n\\binom{n}{k-1} \\cdot a^{n-k+1} \\cdot b^{k-1} \\geq \\binom{n}{k-2} \\cdot a^{n-k+2} \\cdot b^{k-2},\n$$\nwobei im letzten Fall $k \\geq 2$ sein muss und eine dieser Ungleichungen streng ist (einfacher Nachweis!).\n\nDas führt einerseits zu\n$$\n\\frac{1}{n-k+1} \\cdot a \\geq \\frac{1}{k} \\cdot b\n$$\nund andererseits zu\n$$\n\\frac{1}{k-1} \\cdot b \\geq \\frac{1}{n-k+2} \\cdot a,\n$$\nworaus sowohl $n+1 \\geq k \\geq \\frac{n b + b}{a + b}$, als auch $1 \\leq k \\leq \\frac{n b + 2b + a}{a + b} = \\frac{n b + b}{a + b} + 1$ folgt.\n\nFalls $i = \\frac{n b + b}{a + b}$ nicht ganzzahlig ist, ist $G(i)$ das größte Glied, andernfalls sind $G(i)$ und $G(i+1)$ maximal.\n\nWeitere maximale Glieder kann es nicht geben.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSei $E=\\{-1,0,1\\}$ und $M$ eine Menge von Gitterpunkten der Ebene. Die Punkte aus $M$ sind so durch ein Streckennetz $(S)$ miteinander verbunden, dass man von jedem Punkt aus $M$ zu jedem anderen Punkt aus $M$ gelangen kann ohne $(S)$ zu verlassen. Dabei können die Strecken des Streckennetzes außer den Gitterpunkten auch andere Endpunkte enthalten.\nMan ermittle die kürzeste Gesamtlänge des Streckennetzes, falls:\n\na) $M=\\{(i, j) \\mid i, j \\in E$ und $i \\cdot j=0\\}$\n\nb) $M=\\{(i, j) \\mid i, j \\in E\\}$",
    "options": [],
    "answer": "a) sqrt(2) + sqrt(6); b) 4 + 2*sqrt(3)",
    "solution": "Solution:\n\nTeil a): Aus Symmetriegründen reicht es zunächst das günstigste Streckennetz für die Punkte $O(0,0)$, $A(1,0)$ und $C(0,1)$ zu finden. Durch Spiegelung an $O$ erhält man dann das gewünschte Netz für die Punkte aus $M$. Durch eine Drehung der Ebene $xOy$ um $O$ mit Drehwinkel $60^{\\circ}$ (Fig. 1) wird das Dreieck $\\triangle OAC$ in das Dreieck $\\triangle OA' C'$ überführt, wobei die Strecke $P'C'$ das Bild von $PC$ ist. Das kürzeste Streckensystem, das die Punkte $O$, $A$ und $C$ wie gefordert verbindet, hat also eine Gesamtlänge, die gleich jener der Strecke $AC'$ ist.\n\n![](attached_image_1.png)\nFig. 1\n\nIm Dreieck $\\triangle OAC'$ ist aber $AC' = \\sqrt{2-2 \\cos 150^{\\circ}} = \\sqrt{2+\\sqrt{3}}$ (Kosinussatz). Durch Spiegelung am Ursprung $O$ erhält man ein für Teil a) kürzestes Streckensystem mit einer Gesamtlänge von $2 \\sqrt{2+\\sqrt{3}} = \\sqrt{2} + \\sqrt{6}$.\n\nTeil b): Ebenfalls aus Symmetriegründen reicht es zunächst das kürzeste Streckensystem für die Punkte $D(-1, 1)$, $O(0, 0)$, $A(1, 0)$, $B(1, 1)$ und $C(0, 1)$ zu finden. Da die Entfernung von $D$ zu den restlichen vier Punkten mindestens $1$ beträgt, hat ein solches System eine Gesamtlänge von $1 + \\frac{\\sqrt{2} + \\sqrt{6}}{\\sqrt{2}} = 1 + 1 + \\sqrt{3} = 2 + \\sqrt{3}$ (siehe Fig. 2).\n\n![](attached_image_2.png)\nFig. 2\n\nDamit ergibt (nach Spiegelung an $O$) für b) eine Gesamtlänge von $2(2+\\sqrt{3}) = 4 + 2\\sqrt{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben seien ein Kreis $K$ und eine Gerade $g$, die keinen gemeinsamen Punkt haben.\nFerner sei $\\overline{AB}$ der Durchmesser von $K$, der orthogonal zu $g$ ist, wobei $B$ näher an $g$ liegt als $A$. Weiter sei ein beliebiger Punkt $C$, verschieden von $A$ und $B$, auf $K$ gegeben. Die Gerade $AC$ schneidet $g$ in $D$; die Gerade $DE$ berührt $K$ in $E$, wobei $B$ und $E$ auf derselben Seite von $AC$ liegen. Schließlich schneidet $BE$ die Gerade $g$ im Punkt $F$ und $AF$ den Kreis $K$ außer in $A$ im Punkt $G$.\nMan beweise, dass der Spiegelpunkt von $G$ bezüglich der Achse $AB$ auf der Geraden $CF$ liegt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir bezeichnen den zweiten Schnittpunkt von $CF$ und $K$ mit $H$ sowie den Schnittpunkt von $AB$ und $g$ mit $X$. Wegen $AB \\perp g$ reicht es zu zeigen, dass $GH \\parallel g$. Dies ist genau dann der Fall, wenn $\\angle AGH = \\angle AFD$. Umfangswinkelsatz und Scheitelwinkel liefern $\\angle AGH = \\angle ACH = \\angle DCF$; also muss $\\angle DCF = \\angle AFD$ gezeigt werden. Da die Dreiecke $DFC$ und $ADF$ bei $D$ einen Winkel gemeinsam haben, braucht nur deren (gegensinnige) Ähnlichkeit nachgewiesen zu werden. Diese ist genau dann erfüllt, wenn $\\frac{|DC|}{|DF|} = \\frac{|DF|}{|DA|}$,\nd.h. $|DF|^2 = |DA| \\cdot |DC|$ gilt. Nach dem Tangenten-Sekantensatz gilt allerdings $|DE|^2 = |DA| \\cdot |DC|$, so dass nur $|DE| = |DF|$ zu zeigen bleibt, was mit $\\angle DEF = \\angle EFD$ gleichbedeutend ist.\n\nWegen $\\angle FXB = \\angle AEB = 90^\\circ$ ist $AEXF$ ein Sehnenviereck und es gilt $\\angle EFX = \\angle EAB$. Die Gleichheit der Sehnen-Tangentenwinkel liefert $\\angle DEF = \\angle EAB$, woraus $\\angle DEF = \\angle EFD$ folgt.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben seien zwei positive ganze Zahlen $n$ und $k$. In der Ebene liegen $n$ Kreise ($n \\geq 2$), so dass jeder Kreis jeden anderen zweimal schneidet und alle diese Schnittpunkte paarweise verschieden sind.\nJeder Schnittpunkt wird mit einer von $n$ Farben so gefärbt, dass jede Farbe wenigstens einmal verwendet wird und auf jedem der Kreise die gleiche Anzahl $k$ von Farben vertreten ist.\nMan bestimme alle Werte von $n$ und $k$, für die eine solche Färbung möglich ist.",
    "options": [],
    "answer": "2 ≤ k ≤ n ≤ 3 or 3 ≤ k ≤ n",
    "solution": "Solution:\n\nDie Antwort lautet: $2 \\leq k \\leq n \\leq 3$ oder $3 \\leq k \\leq n$.\nOffensichtlich gilt $k \\leq n$ nach Aufgabenstellung sowie $k \\geq 2$, weil für $k=1$ alle Punkte dieselbe Farbe hätten, während die Anzahl $n$ der Farben $\\geq 2$ sein soll. Wir nummerieren die Kreise und die Farben von 1 bis $n$ und bezeichnen mit $F(i, j)$ die Menge der Farben der Schnittpunkte der Kreise $i$ und $j$. $F(i, j)$ enthält ein oder zwei Elemente.\n\nSei $k=2$. Für $n=2$ ist $F(1,2)=\\{1,2\\}$ eine erlaubte Färbung. Für $n=3$ ist $F(1,2)=\\{3\\}$, $F(2,3)=\\{1\\}$, $F(3,1)=\\{2\\}$ ein Beispiel für eine erlaubte Färbung. Sei nun $n \\geq 4$. Jedem der $n$ Kreise ordnen wir die Menge $\\{i, j\\}$ der beiden auf ihm vorkommenden Farben zu. Jede dieser Mengen besteht aus zwei Elementen und jede der $n$ Farben muss in wenigstens zwei Mengen vorkommen, da sich in jedem gefärbten Punkt zwei Kreise schneiden. Also kommt jede Farbe in genau zwei Mengen vor. Zum Kreis 1 mit der Menge $\\{i, j\\}$ gibt es daher noch höchstens zwei weitere Kreise, in deren Farbmengen $i$ oder $j$ vorkommen. Wegen $n \\geq 4$ finden wir stets einen Kreis 2 mit der Menge $\\{k, l\\}$ und $\\{k, l\\} \\cap \\{i, j\\}=\\{ \\}$. Die Schnittpunkte der Kreise 1 und 2 sind dann nicht erlaubt färbbar – Widerspruch!\n\nNun beweisen wir mit vollständiger Induktion einen etwas stärkeren Satz als verlangt: Für $n \\geq k \\geq 3$ existiert stets eine erlaubte Färbung, bei der auf dem Kreis $i$ die Farbe $i$ für alle $i=1, \\ldots, n$ vorkommt. Zur Verankerung geben wir für $k=n=3$ mit $F(1,2)=\\{1,2\\}$, $F(1,3)=\\{1,3\\}$, $F(2,3)=\\{2,3\\}$ ein Beispiel und für $k=3, n>3$ folgendes Beispiel für eine erlaubte Färbung mit Zusatzbedingung:\n\n$F(1,2)=\\{1,2\\}$, $F(i, i+1)=\\{i\\}$ für $1<i<n-1$, $F(n-1, n)=\\{n-2, n-1\\}$ und $F(i, j)=\\{n\\}$ für die restlichen Paare $(i, j)$ mit $1 \\leq i<j \\leq n$.\n\nNun nehmen wir an, dass der stärkere Satz für ein $k \\geq 3$ erfüllt ist, und wählen $n \\geq k+1$. Wegen $n-1 \\geq k \\geq 3$ gibt es eine erlaubte Färbung mit Zusatzbedingung für die Kreise bzw. Farben $1,2,..., n-1$. Jetzt färben wir die Schnittpunkte des Kreises $n$: Für jedes $i=1, \\ldots, n-1$ erhält ein Schnittpunkt der Kreise $i$ und $n$ die Farbe $n$. Damit kommen auf jedem der Kreise $i$ mit $i=1, \\ldots, n-1$ genau $k+1$ Farben vor; darunter $i$ und $n$. Für $i=1, \\ldots, k$ erhält der zweite Schnittpunkt der Kreise $i$ und $n$ die Farbe $i$, so dass nun auch auf dem Kreis $n$ genau $k+1$ Farben liegen, nämlich $1$ bis $k$ und $n$. Alle übrigen neuen Schnittpunkte erhalten die Farbe $n$, so dass auf keinem Kreis weitere Farben dazukommen. Diese Färbung erfüllt alle Bedingungen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31021,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMan beweise: Ist $4^{n} \\cdot 7 = a^{2} + b^{2} + c^{2} + d^{2}$ mit $n, a, b, c, d \\in \\mathbb{N} \\setminus \\{0\\}$, dann kann keine der Quadratzahlen die Zahl $4^{n-1}$ unterschreiten.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIst $n=1$, dann ist die Behauptung richtig; die einzigen Lösungen für $a, b, c, d$ sind, abgesehen von der Reihenfolge, die Quadrupel $(1,1,1,5)$, $(1,3,3,3)$ und $(2,2,2,4)$. Für jedes $n \\geq 1$ ist $4^{n} \\cdot 7$ durch $4$ teilbar. Da das Quadrat einer natürlichen Zahl bei der Division durch $4$ nur die Reste $0$ oder $1$ haben kann, kommen für $a, b, c, d$ nur Zahlen gleicher Restklasse modulo $4$ in Frage.\nSind $a, b, c, d$ (alle) ungerade, dann ist die rechte Seite zwar durch $4$, nicht aber durch $8$ und erst recht nicht durch $16$ teilbar. In diesem Fall kommt also nur $n=1$ in Frage. Sollte $n>1$ sein, dann müssen die vier Zahlen auf der rechten Seite demnach alle gerade sein.\nVorausgesetzt die Eigenschaft, dass alle Quadratzahlen $4^{n-1}$ überschreiten, gilt nicht für alle $n$, dann gibt es ein kleinstes $k$ ($k \\in \\mathbb{N}$), für das sie nicht gilt. Da sie für $n=1$ gilt, muss $k$ größer $1$ sein. Dann sind aber (siehe oben) $a, b, c, d$ alle gerade. Man könnte also beide Seiten durch $4$ teilen und die Eigenschaft dürfte auch für $k-1$ nicht gelten, was aber wegen der Minimalität von $k$ nicht sein kann. Die Behauptung stimmt also für alle natürlichen Zahlen $n$, $n \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs seien $A$, $B$, $C$, $D$, $E$, $F$ Punkte auf einem Kreis mit $A E \\| B D$ und $B C \\| D F$. Durch Spiegelung an der Geraden $C E$ gehe der Punkt $D$ in $X$ über. Zeige, dass $X$ so weit von der Geraden $E F$ entfernt ist wie $B$ von $A C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAlle im folgenden auftretenden Winkel sind als orientiert und modulo $180^{\\circ}$ zu verstehen; dadurch wird eine Betrachtung der Lage der involvierten Punkte zueinander entbehrlich. Die Fußpunkte der Lote von $B$, $X$ auf $A C$, $E F$ seien mit $P$, $Q$ bezeichnet.\n\nStrategie. Zeige die Kongruenz der Dreiecke $A B P$, $E X Q$. $(*)$\n\nHieraus wird sich sofort $B P = X Q$ und damit die Behauptung ergeben. Der Beweis von $(*)$ selbst erfolgt vermittelst eines bekannten Kongruenzsatzes in drei Schritten.\n\nI. Wegen $\\varangle B P A \\equiv \\varangle E Q X \\equiv 90^{\\circ}$ sind beide Dreiecke rechtwinklig.\n\nII. Da $A E \\| B D$ ist das Sehnenviereck $A B D E$ ein gleichschenkeliges Trapez und demnach $A B = D E$. Weiterhin ist $D E = X E$ nach Konstruktion von $X$ und mithin $A B = X E$, d.h. die Hypotenusen stimmen überein.\n\nIII. Wie vorhin schließen wir aus $B C \\| D F$, dass auch $B C D F$ ein gleichschenkliges Trapez ist. Durch wiederholte Verwendung des Peripheriewinkelsatzes erhalten wir nun $\\varangle D E C + \\varangle C A B \\equiv \\varangle D A C + \\varangle C A B \\equiv \\varangle D A B \\equiv \\varangle D F B \\equiv \\varangle C D F \\equiv \\varangle C E F \\equiv \\varangle C E X + \\varangle X E F$. Infolge $\\varangle D E C \\equiv \\varangle C E X$ ergibt sich hieraus $\\varangle C A B \\equiv \\varangle X E F$, oder - anders formuliert $-\\varangle P A B \\equiv \\varangle X E Q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31023,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn einem Raum stehen 2005 Obstkisten, von denen jede eine oder mehrere Sorten Obst enthält, von jeder Obstsorte ganzzahlig viele Früchte.\na) Man zeige, dass man stets 669 Obstkisten auswählen kann, die zusammen sowohl mindestens ein Drittel aller Äpfel als auch mindestens ein Drittel aller Birnen enthalten.\nb) Können die Kisten in Teil a) stets so gewählt werden, dass sie außerdem mindestens ein Drittel aller Pfirsiche enthalten?",
    "options": [],
    "answer": "a) Yes, such 669 boxes always exist. b) No, not always; a counterexample shows it is impossible.",
    "solution": "Solution:\n\na) Es sei $N$ eine Kiste, die eine maximale Anzahl von Äpfeln enthält, sagen wir $n$ Stück. Sie werde zunächst beiseite gestellt. Wir betrachten nun alle Möglichkeiten, die verbleibenden $2004 = 3 \\cdot 668$ Kisten in drei Haufen, $A$, $B$ und $C$ zu je 668 Kisten aufzuteilen. Die Gesamtzahl an Äpfeln in diesen Kisten sei $a$, $b$, $c$. Indem wir mit einer beliebigen solchen Aufteilung beginnen und die gebildeten Haufen gegebenenfalls umbenennen, sehen wir, dass es Aufteilungen mit $a \\leq b \\leq c$ gibt. Unter allen diesen fixieren wir für den Rest der Beweisführung eine solche, für die $c - a$ minimal ist. Wäre nun $c - a > n$, so könnten wir eine Kiste aus $A$ mit darin maximaler Anzahl von Äpfeln gegen eine Kiste aus $C$ mit darin minimaler Anzahl vertauschen und erhielten infolge der Maximalität von $n$ nach eventueller Umbenennung der Haufen einen Widerspruch zur Minimalität von $c - a$. Demnach ist $c - a \\leq n$, also $a + n \\geq c \\geq b$, woraus sofort\n$$\na + n = \\frac{(a + n) + (a + n) + a + n}{3} \\geq \\frac{c + b + a + n}{3}\n$$\nfolgt. Sollten wir uns also letzten Endes entschließen, die 669 Kisten $A \\cup \\{N\\}$ zu nehmen, d.h. die Kisten des Haufens $A$ zusammen mit der Kiste $N$, so hätten wir zumindest die Bedingung über die Anzahl der zu nehmenden Äpfel erfüllt. Dies gilt natürlich erst recht für $B \\cup \\{N\\}$ und $C \\cup \\{N\\}$. Nun sei (nach Schubfachprinzip) $T$ ein solcher der drei Haufen, der mindestens ein Drittel der in ihnen zusammen vorkommenden Birnen enthält. Damit erfüllt die Auswahl $T \\cup \\{N\\}$ sicher die Anforderung an die zu wählende Anzahl von Birnen und ist folglich nach obigem wie gewünscht.\n\nb) Hier genügt die Angabe eines Gegenbeispiels. Es enthalte eine der Kisten nichts außer einem Apfel, eine weitere bloß eine Birne und in den übrigen 2003 Kisten möge jeweils ein Pfirsich liegen. Wollte man nun den gestellten Bedingungen genügen, müsste man die Kiste mit dem Apfel, die Kiste mit der Birne und 668 Kisten mit einem Pfirsich auswählen, bräuchte also insgesamt 670 Kisten.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLassen sich für jede positive ganze Zahl $n$ nicht-negative ganze Zahlen $a, b, c, d, e, f, g, h$ mit\n\n$$\nn=\\frac{2^{a}-2^{b}}{2^{c}-2^{d}} \\cdot \\frac{2^{e}-2^{f}}{2^{g}-2^{h}}\n$$\n\nfinden? Die Antwort ist zu begründen.",
    "options": [],
    "answer": "No; for example, n = 19 (and more generally n ≡ 19 mod 64) cannot be represented in the given form.",
    "solution": "Solution:\nVorbemerkung: Im Folgenden sei $n$ ungerade. Ohne Einschränkung ist $a>b, c>d, e>f, g>h$. Die Bedingungsgleichung ist dann äquivalent zu\n$$\nn 2^{d+h-b-f}\\left(2^{c-d}-1\\right)\\left(2^{g-h}-1\\right)=\\left(2^{a-b}-1\\right)\\left(2^{e-f}-1\\right).\n$$\nDa $n$ ungerade ist, gilt $d+h-b-f=0$, und man kann ohne Einschränkung $d=h=b=f=0$ setzen:\n$$\nn\\left(2^{c}-1\\right)\\left(2^{g}-1\\right)=\\left(2^{a}-1\\right)\\left(2^{e}-1\\right)\n$$\n\n1. Lösung (skizziert): Die Zahlen der Form $n=19+64k$ mit ganzzahligem $k \\geqq 0$ lassen sich nicht darstellen. Für $x>1$ liefert $2^{x}-1$ Rest 3 bei Division durch 4, für $x=1$ Rest 1. Nur wenn eine ungerade Anzahl der Variablen $a, c, e, g$ Wert 1 hat, kann Gleichung (*) erfüllt sein. Haben drei der Variablen Wert 1, kann die Gleichung nicht gelten, da $n$ nicht die Form $2^{x}-1$ hat. Also hat genau eine der Klammern in $(*)$ Wert 1. Durch entsprechende Überlegungen und Fallunterscheidungen zu Resten bei Division durch 8, 16, 32, 64 kann man die Werte der restlichen Klammern festlegen und jeweils zum Widerspruch führen.\n\n2. Lösung (skizziert): Die Zahl $n=19$ ist nicht darstellbar. Man nutzt die Beziehung $\\gcd(2^{x}-1, 2^{y}-1)=2^{\\gcd(x, y)}-1$, um zu zeigen: Aus $(2^{x}-1) \\mid (2^{y}-1)(2^{z}-1)$ folgt $x \\mid y$ oder $x \\mid z$.\nHieraus folgt mit $(*)$, dass sich 19 als\n$$\n19=\\frac{2^{a}-1}{2^{c}-1} \\cdot \\frac{2^{e}-1}{2^{g}-1} \\quad \\text{oder} \\quad 19=\\frac{2^{a}-1}{(2^{c}-1)(2^{g}-1)} \\cdot (2^{e}-1)\n$$\nals Produkt zweier ganzzahliger Faktoren schreiben lässt. Da 19 eine Primzahl ist und nicht die Form $2^{e}-1$ hat, genügt es zu zeigen, dass die Gleichung\n$$\n19=\\frac{2^{a}-1}{(2^{c}-1)(2^{g}-1)}\n$$\nnicht gelten kann. Der Zähler enthält den Primfaktor 19, was nur für $a \\geq 18$ möglich ist. Die Zahlen $c$ und $g$ teilen $a$, ohne Einschränkung ist $c \\geq g$. Für $c=g=a/2$ hat der Bruch den nichtganzzahligen Wert $1+2/(2^{a/2}-1)$, für $c \\leq a/3, g \\leq a/3$ oder für $c=a/2, g \\leq a/4$ ist der Wert zu groß. Damit ist $c=a/2, g=a/3$, und der Bruch hat den nicht ganzzahligen Wert $2^{a/6}+1/(2^{a/6}-1)$.\n\nBemerkung: Nicht immer ist $n$ Produkt zweier ganzzahliger Faktoren der Form $(2^{x}-1)/(2^{y}-1)$. Beispielsweise ist $13=\\frac{(2^{12}-1)(2^{2}-1)}{(2^{6}-1)(2^{4}-1)}$, aber man kann wie bei der 2. Lösung zeigen, dass 13 nicht in der Form $(2^{x}-1)/(2^{y}-1)$ darstellbar ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan bestimme mit Beweis alle Funktionen $f: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}^{+}$ mit der Eigenschaft\n$$\nf(x) f(y) = 2 f(x + y f(x))\n$$\nfür alle positiven reellen Zahlen $x, y$.",
    "options": [],
    "answer": "f(x) = 2 for all positive real x",
    "solution": "Solution:\n\nOffensichtlich erfüllt die Funktion $f(x) = 2$ für alle $x \\in \\mathbb{R}^{+}$ die gegebene Funktionalgleichung. Wir werden zeigen, dass dies die einzige Lösung ist.\n\nLemma 1: Für alle $x \\in \\mathbb{R}^{+}$ gilt $f(x) \\geq 1$.\n\nZum Beweis nehmen wir $f(x) < 1$ für ein geeignetes $x$ an und setzen $y = \\frac{x}{1 - f(x)}$. Dann ist $x > 0$ und es folgt $y = x + y f(x)$. Aus (I) erhalten wir $f(x) f(x + y f(x)) = 2 f(x + y f(x))$ und wegen $f(x + y f(x)) > 0$ folgt $f(x) = 2$, Widerspruch!\n\nLemma 2: Für alle $x \\in \\mathbb{R}^{+}$ gilt $f(x) \\geq 2$.\n\nZum Beweis setzen wir in (I) $x = y$ und erhalten (II): $f^{2}(x) = 2 f(x + x f(x))$. Sei $f\\left(x_{1}\\right) < 2$ für ein geeignetes $x_{1}$. Dann ist $f\\left(x_{1} + x_{1} f\\left(x_{1}\\right)\\right) = \\frac{f^{2}\\left(x_{1}\\right)}{2} < f\\left(x_{1}\\right)$. Mit $x_{k+1} = x_{k} + x_{k} f\\left(x_{k}\\right)$ für $k = 1, 2, \\ldots$ entsteht eine monoton fallende Folge $\\left(f\\left(x_{1}\\right) = a ; \\frac{a^{2}}{2} ; \\frac{a^{4}}{2^{3}} ; \\ldots ; \\frac{a^{2 t}}{2^{2 t-1}}, \\ldots\\right)$ von Funktionswerten. Für $t > \\frac{1}{2 - 2 \\log_{2} a}$ sind die Werte dieser Folge kleiner als 1, Widerspruch!\n\nLemma 3: $f$ ist monoton steigend.\n\nZum Beweis nehmen wir an, es gibt $s, \\varepsilon > 0$ mit $f(s) > f(s + \\varepsilon)$. Einsetzen von $x = s$ und $y = \\frac{\\varepsilon}{f(s)}$ in (I) liefert $f(s) f\\left(\\frac{\\varepsilon}{f(s)}\\right) = 2 f(s + \\varepsilon)$, woraus $f\\left(\\frac{\\varepsilon}{f(s)}\\right) < 2$ folgt, Widerspruch!\n\nLemma 4: Wenn es ein $z \\in \\mathbb{R}^{+}$ gibt mit $f(z) > 2$, dann gilt $f(x) > 2$ für alle $x \\in \\mathbb{R}^{+}$.\n\nWieder setzen wir $x = z$ und $y = \\frac{\\varepsilon}{f(z)}$ in (I) ein und haben $f(z) f\\left(\\frac{\\varepsilon}{f(z)}\\right) = 2 f(z + \\varepsilon)$, woraus jetzt mit Lemma 2 folgt: $f(z) \\leq f(z + \\varepsilon)$ für alle $\\varepsilon > 0$. Also existiert ein $z_{0} \\geq 0$ mit $f(z) > 2$ für alle $z > z_{0}$. Nehmen wir an, dass $z_{0} > 0$ gilt. Dann folgt mit $x = y = z_{0} - \\varepsilon > 0$ aus (I) $f(x) f(y) = 4$, und wegen $x + y f(x) = \\left(z_{0} - \\varepsilon\\right)\\left(1 + f\\left(z_{0} - \\varepsilon\\right)\\right) = 3\\left(z_{0} - \\varepsilon\\right)$ gilt für hinreichend kleines $\\varepsilon$, dass $3\\left(z_{0} - \\varepsilon\\right) > z_{0}$ und daher $2 f(x + y f(x)) > 4$, Widerspruch!\n\nLemma 5: Wenn $f(x) > 2$ für alle $x \\in \\mathbb{R}^{+}$, dann ist $f$ injektiv.\n\nAngenommen, es gäbe $s, \\varepsilon > 0$ mit $f(s) = f(s + \\varepsilon)$. Wir setzen $x = s$ und $y = \\frac{\\varepsilon}{f(s)}$ in (I) ein und haben $f(s) f\\left(\\frac{\\varepsilon}{f(s)}\\right) = 2 f(s + \\varepsilon)$, woraus nun $f(s) < f(s + \\varepsilon)$ folgt, Widerspruch!\n\nHauptbeweis: Wegen der Symmetrie der linken Seite von (I) gilt auch $x + y f(x) = y + x f(y)$. Für $y = 1$ folgt $f(x) = (f(1) - 1) x + 1 = m x + 1$. Durch Einsetzen wird jedoch leicht gezeigt, dass keine lineare Funktion Lösung von (I) sein kann.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben sei ein Dreieck $ABC$ mit $\\overline{AB} + \\overline{BC} = 3 \\overline{AC}$. Sein Inkreis habe den Mittelpunkt $I$ und berühre die Seiten $AB$ in $D$ bzw. $BC$ in $E$. Weiter seien $K$ und $L$ die Spiegelpunkte von $D$ bzw. $E$ bezüglich $I$.\nMan beweise, dass das Viereck $ACKL$ ein Sehnenviereck ist.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZu den in der Aufgabenstellung genannten Punkten bezeichnen wir den Berührpunkt des Inkreises mit $AC$ mit $F$, den Mittelpunkt von $AC$ mit $S$, den Schnittpunkt von $w_{\\beta}$ und der Mittelsenkrechten $m_{AC}$ mit $P$, den Spiegelpunkt von $B$ an $I$ mit $R$, den Mittelpunkt von $BI$ mit $T$ und die paarweise gleichen Abschnitte auf den Dreiecksseiten mit $x, y$ bzw. $z$ (siehe Figur).\n\nAus $\\overline{AB} + \\overline{BC} = 3 \\overline{AC}$ folgt\n$x + z + z + y = 3(x + y)$, also (I) $z = x + y$. Bekanntlich liegt $P$ auf dem Umkreis von $ABC$. Daher gilt $\\angle CAP = \\angle CBP = \\angle PBA = \\angle PCA = \\frac{\\beta}{2}$ und $\\overline{AP} = \\overline{CP}$. Die Dreiecke $ASP$ und $BEI$ sind ähnlich, und wegen $\\overline{AS} = \\frac{1}{2} \\overline{AC} = \\frac{x + y}{2} = \\frac{z}{2} = \\frac{1}{2} \\overline{BE}$ gilt mit der Bezeichnung $r$ für den Inkreisradius\n\n![](attached_image_1.png)\n\n(II) $\\overline{SP} = \\frac{1}{2} \\overline{EI} = \\frac{r}{2}$ sowie $\\overline{AP} = \\frac{1}{2} \\overline{BI}$.\nFür den Flächeninhalt $F$ des Dreiecks $ABC$ gilt $F = r s = \\sqrt{s(s - a)(s - b)(s - c)}$, wobei $s = \\frac{a + b + c}{2} = x + y + z$ ist. Daher ist $r(x + y + z) = \\sqrt{(x + y + z) x y z}$ und mit (I) folgt $2 r z = \\sqrt{2 z x y z} \\Leftrightarrow 4 r^{2} = 2 x y \\Leftrightarrow 8 r^{2} = 4 x y \\Leftrightarrow 9 r^{2} + (x - y)^{2} = r^{2} + (x + y)^{2} = r^{2} + z^{2}$.\nWegen $\\angle BDI = 90^{\\circ}$ ist (III) $\\overline{BI}^{2} = r^{2} + z^{2}$.\nWegen $PS \\perp AC \\perp FI$ gilt $\\overline{PI}^{2} = \\overline{SF}^{2} + (\\overline{PS} + \\overline{FI})^{2}$.\nAber $\\overline{SF} = \\frac{1}{2} \\overline{AC} - x = \\frac{x + y}{2} - x = \\frac{y - x}{2}$ und $\\overline{FI} = r$. Daher und mit (II) ist $\\overline{PI}^{2} = \\left(\\frac{y - x}{2}\\right)^{2} + \\left(\\frac{3}{2} r\\right)^{2} = \\frac{(y - x)^{2} + 9 r^{2}}{4}$.\nMit (III) folgt $\\overline{PI}^{2} = \\frac{1}{4} \\overline{BI}^{2}$, also $\\overline{PI} = \\frac{1}{2} \\overline{BI}$.\nSo gilt (IV) $\\overline{PI} = \\overline{PA} = \\overline{PC} = \\overline{PR}$.\n\nAußerdem wird $T$ bei Spiegelung an $I$ auf $P$ abgebildet. Daher ist auch $\\overline{IT} = \\overline{PI}$. $\\overline{BI}$ ist der Durchmesser des Kreises durch $B, I, D$ und $E$ (rechte Winkel bei $D$ und $E$). Dieser Kreis hat $T$ als Mittelpunkt. Bei Spiegelung dieses Kreises an $I$ wird $I$ auf $I$, $T$ auf $P$, $D$ auf $K$, $E$ auf $L$ abgebildet und der Radius bleibt erhalten. Mit (IV) folgt, dass $A, L, I, K, C$ und $R$ auf einem Kreis um $P$ liegen; also ist $ACKL$ ein Sehnenviereck.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWir betrachten ein $m \\times n$-Rechteck aus $m n$ Einheitsquadraten. Zwei seiner Einheitsquadrate heißen benachbart, wenn sie eine gemeinsame Seitenkante haben, und ein Pfad ist eine Folge von Einheitsquadraten, in der je zwei aufeinander folgende Elemente benachbart sind.\nJedes Einheitsquadrat des Rechtecks kann entweder weiß oder schwarz gefärbt werden. Sind alle Quadrate gefärbt, so liegt eine Färbung des Rechtecks vor.\nEs sei $N$ die Anzahl aller solcher Färbungen, bei denen es wenigstens einen schwarzen Pfad von der linken zur rechten Seitenkante des Rechtecks gibt. Ferner sei $M$ die Anzahl aller Färbungen, bei denen es wenigstens zwei schwarze Pfade von der linken zur rechten Seitenkante des Rechtecks gibt, die kein gemeinsames Quadrat enthalten.\nMan beweise, dass $N^{2} \\geq M \\cdot 2^{m n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir werden die Behauptung verallgemeinern. Dazu lassen wir zu, dass das $m \\times n$ Rechteck auf beiden Seiten gefärbt wird und dass einige der Einheitsquadrate transparent sind. Solche Felder brauchen nur auf einer Seite gefärbt zu werden und sehen dann auf beiden Seiten gleich aus. Ein nicht transparentes Einheitsquadrat muss dagegen auf beiden Seiten gefärbt werden, allerdings nicht notwendigerweise mit der gleichen Farbe.\n\nNun sei $A$ die Anzahl aller solcher Färbungen der Oberseite, bei denen es wenigstens einen schwarzen Pfad von der linken zur rechten Seitenkante des Rechtecks gibt. Entsprechend sei $B$ für die Unterseite definiert. Ferner sei $C$ die Anzahl aller Färbungen, bei denen es zwei schwarze Pfade von der linken zur rechten Seitenkante des Rechtecks gibt, und zwar einen auf der Ober- und einen auf der Unterseite, die kein gemeinsames transparentes Quadrat enthalten. Schließlich sei $D$ die Anzahl aller Färbungen dieses Rechtecks. Wir werden (I) $A \\cdot B \\geq C \\cdot D$ beweisen und haben damit die ursprüngliche Behauptung als Spezialfall gezeigt, in dem alle Felder transparent sind. Hier gilt nämlich $A=B=N, C=M, D=2^{m n}$.\n\nDen Beweis von (I) führen wir mit vollständiger Induktion nach der Anzahl $k$ der transparenten Felder. Für $k=0$ ist $A=B=N \\cdot 2^{m n}, C=N^{2}$ und $D=\\left(2^{m n}\\right)^{2}$, so dass in (I) Gleichheit gilt. Nun nehmen wir an, dass die Behauptung für $k$ erfüllt ist, und betrachten ein Rechteck mit $k+1$ transparenten Feldern. Die Anzahlen $A, B, C, D$ gelten nun für dieses Rechteck. Wir wählen ein transparentes Einheitsquadrat $t$ und machen es undurchsichtig. Für das so entstandene Rechteck seien die jeweiligen Anzahlen mit $A^{\\prime}, B^{\\prime}, C^{\\prime}, D^{\\prime}$ bezeichnet und nach Induktionsannahme gilt $A^{\\prime} \\cdot B^{\\prime} \\geq C^{\\prime \\prime} \\cdot D^{\\prime}$.\n\nNun ist offensichtlich $D^{\\prime}=2 \\cdot D$. Für jede in $A$ gezählte Färbung existieren genau zwei Färbungen von $A^{\\prime}$, nämlich dadurch unterschieden, dass $t$ von unten schwarz oder weiß gefärbt wird. Umgekehrt kann man zwei in $A^{\\prime}$ gezählten Färbungen, die sich nur in der Farbe der Unterseite eines Quadrats unterscheiden, eine in $A$ gezählte Färbung zuordnen. Also ist $A^{\\prime}=2 \\cdot A$ und entsprechend $B^{\\prime}=2 \\cdot B$. Zum Beweis von (I) für $k+1$ genügt also der Nachweis von $C^{\\prime} \\geq 2 \\cdot C$.\n\nDazu sei $t$ wieder transparent. Weil in $C$ nur solche Färbungen gezählt werden, die wenigstens einen schwarzen Pfad oben und unten enthalten, wobei diese sich nicht in einem transparenten Quadrat schneiden, kann $t$ höchstens auf einem dieser Pfade, oBdA auf dem oberen, liegen. Machen wir also $t$ undurchsichtig und behalten seine Farbe oben bei, können wir seine Unterseite schwarz oder weiß färben, so dass beide Färbungen in $C^{\\prime}$ gezählt werden. Dabei liefern verschiedene in $C$ gezählte Färbungen stets verschiedene Paare von Färbungen, die in $C^{\\prime}$ gezählt werden. Damit ist alles gezeigt.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs seien $n$ eine positive ganze Zahl größer als Eins und $B=\\{1,2, \\ldots, 2^{n}\\}$. Eine Teilmenge $A$ von $B$ heiße ulkig, wenn sie von je zwei verschiedenen Elementen $x, y$ von $B$, deren Summe eine Zweierpotenz ist, genau eines enthält. Wie viele ulkige Teilmengen hat $B$?",
    "options": [],
    "answer": "2^{n+1}",
    "solution": "Solution:\n\nDie Anzahl ist $2^{n+1}$. Wir argumentieren mit vollständiger Induktion nach $n$ und setzen hierzu $B_{n}=\\{1,2, \\ldots, 2^{n}\\}$. \n\nInduktionsanfang, $n=1$. Dieser Fall ist klar: alle vier Teilmengen von $B_{1}=\\{1,2\\}$ sind ulkig.\n\nInduktionsschritt, $n \\longrightarrow n+1$. Es sei also bereits bekannt, daß $B_{n}$ genau $2^{n+1}$ viele ulkige Teilmengen hat. Für jede ulkige Menge $A$ von $B_{n}$ konstruiere die Mengen\n\n$$C_{n}(A)=\\{2^{n+1}-m \\mid m \\in B_{n} \\backslash A,\\ m<2^{n}\\},$$\n\n$$M_{n}(A)=A \\cup C_{n}(A)$$\n\nund\n\n$$N_{n}(A)=M_{n}(A) \\cup\\{2^{n+1}\\}.$$\n\nOffenbar sind $A$, $C_{n}(A)$ und $\\{2^{n+1}\\}$ paarweise elementfremd. Die Mengen $M_{n}(A)$ und $N_{n}(A)$ sind ulkig: Hierzu seien $x, y$ verschiedene Elemente von $B_{n+1}$, deren Summe eine Zweierpotenz ist; ohne Einschränkung gelte $x<y$.\n\nIm Fall $1 \\leq x<y \\leq 2^{n}$, also $x, y \\in B_{n}$, ist genau eines von beiden in $M_{n}(A)$ bzw. $N_{n}(A)$ enthalten, da $A$ ulkig ist.\n\nIm Fall $1 \\leq x \\leq 2^{n}<y \\leq 2^{n+1}$ ist $2^{n}<x+y<2^{n+2}$, also $x+y=2^{n+1}$; wegen $x \\geq 1$ ist $y \\neq 2^{n+1}$. Nach Wahl von $C_{n}(A)$ liegt entweder $x$ in $A$ oder $y$ in $C_{n}(A)$.\n\nDer Fall $2^{n}<x<y \\leq 2^{n+1}$ kann nicht auftreten, da dann $2^{n+1}<x+y<2^{n+2}$.\n\nIst andererseits $M$ ulkige Teilmenge von $B_{n+1}$, so ist offenbar $A:=M \\cap B_{n}$ ulkige Teilmenge von $B_{n}$. Das Element $y=2^{n+1}-x$ von $B_{n+1}$ mit $1 \\leq x<2^{n}$ ist genau dann in $M$ enthalten, wenn $x \\in B_{n}$ nicht in $A$ enthalten ist, da $x+y=2^{n+1}$ Zweierpotenz ist. Die Ulkigkeit von $M$ legt also bei Kenntnis von $M \\cap B_{n}$ für alle Elemente $y$ von $B_{n+1}$ mit $2^{n}<y<2^{n+1}$ fest, ob sie in $M$ enthalten sind; es steht nur noch frei, ob $2^{n+1}$ Element ist. Damit gleicht $M$ entweder $M_{n}(A)$ oder $N_{n}(A)$, und es lassen sich jeder ulkigen Teilmenge von $B_{n}$ genau zwei ulkige Teilmengen von $B_{n+1}$ zuordnen. Das beendet den Beweis.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMan finde alle Quadrupel positiver ganzer Zahlen ($m, n, p, q$) mit der Eigenschaft\n$$\np^{m} q^{n} = (p+q)^{2} + 1\n$$",
    "options": [],
    "answer": "[(1, 2, 2, 5), (1, 2, 13, 5), (2, 1, 5, 2), (2, 1, 5, 13)]",
    "solution": "Solution:\n1. Hilfssatz: Sind $k, l, a$ positive ganze Zahlen mit $a = (k^{2} + l^{2} + 1)/(k l)$, folgt $a = 3$.\n\nBeweis des Hilfssatzes: Zu gegebenem Wert von $a$ betrachten wir ein Paar $(k, l)$ mit $k^{2} + l^{2} + 1 = a k l$ und minimaler Summe $k + l$ unter allen derartigen Paaren. Ohne Einschränkung ist $k \\leq l$.\n\nZunächst sei $k < l$. Nach Voraussetzung genügt $k$ der quadratischen Gleichung\n$$\nx^{2} + l^{2} + 1 - a x l = 0,\n$$\nfür deren 2. Lösung $k'$ nach Satz von Vieta folgt: $k + k' = a l$ und $k k' = l^{2} + 1$. Also ist $k' = a l - k$ ganzzahlig, wegen $k' = (l^{2} + 1)/k$ positiv, wegen $k \\geq l + 1$ und $k k' = l^{2} + 1 < (l + 1)^{2}$ ist $k' \\leq l$, also ist $(l, k')$ Paar mit kleinerer Summe $l + k' < k + l$ entgegen der Wahl von $(k, l)$.\n\nSomit gilt $k = l$, also $k^{2}(a - 2) = 1$, woraus sich $k = 1$ und $a = 3$ ergeben.\n\nEs ist wegen $m \\geq 1$ und $n \\geq 1$ der Ausdruck $p^{m-1} q^{n-1} = (p^{2} + q^{2} + 1)/(p q) + 2$ ganzzahlig, also nach Hilfssatz $p^{m-1} q^{n-1} = 5$ und somit $p^{m-1} = 1, q^{n-1} = 5$ oder $p^{m-1} = 5, q^{n-1} = 1$.\n\nIm ersten Fall ist $n = 2$ und $q = 5$; $p$ genügt der Gleichung (1) für $l = 5, a = 3$ mit Lösungen $2$ und $13$, aus $p^{m-1} = 1$ folgt jeweils $m = 1$, und durch Einsetzen bestätigt man die Lösungsquadrupel $(1, 2, 2, 5)$ und $(1, 2, 13, 5)$. Der zweite Fall liefert entsprechend $(2, 1, 5, 2)$ und $(2, 1, 5, 13)$.\nSolution:\nAlle Summanden der Ausgangsgleichung bis auf $q^{2} + 1$ sind Vielfache von $p$, daher gilt $p \\mid q^{2} + 1$. Damit lassen sich die Fälle $q = 1, 2, \\ldots, 6$ mittels Durchprobieren behandeln, man erhält Lösungsquadrupel aus 1. Lösung. Für $p \\geq q \\geq 7$ (analog: $p \\geq p \\geq 7$) gibt es keine Lösungen:\n\n1. Fall: $m > 1$ : Dann ist $p^{m} q^{n} \\geq p^{2} q > 6 p^{2} \\geq (p + q)^{2} + 1$.\n\n2. Fall: $n > 2$ : Dann ist wegen $p \\leq q^{2} + 1: p^{m} q^{n} \\geq p q^{3} \\geq p^{2} q - p q > 5 p^{2} > (p + q)^{2} + 1$.\n\n3. Fall: $m = n = 1$ : Hierzu müsste $p^{2} + p q + q^{2} + 1 = 0$ sein - unmöglich für $p > 0, q > 0$.\n\n4. Fall: $m = 1, n = 2$ : Die quadratische Gleichung $p q^{2} = (p + q)^{2} + 1$ für $p$ hat Diskriminante $D = q^{4} - 4 q^{3} - 4$, die für $q \\geq 7$ zwischen den aufeinander folgenden Quadratzahlen $(q^{2} - 2q - 3)^{2}$ und $(q^{2} - 2q - 2)^{2}$ liegt, also ist $\\sqrt{D}$, somit auch $p$ nicht ganzzahlig.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDer Punkt $P$ liege im Inneren des Dreiecks $A B C$ und erfülle\n$$\n\\varangle B P C-\\varangle B A C=\\varangle C P A-\\varangle C B A=\\varangle A P B-\\varangle A C B .\n$$\nMan beweise, dass dann gilt:\n$$\n\\overline{P A} \\cdot \\overline{B C}=\\overline{P B} \\cdot \\overline{A C}=\\overline{P C} \\cdot \\overline{A B}\n$$\nZunächst überlegt man sich, daß $\\varangle B P C=60^{\\circ}+\\alpha, \\varangle C P A=60^{\\circ}+\\beta, \\varangle A P B=60^{\\circ}+\\gamma$. Aus Symmetriegründen genügt es, eine der beiden behaupteten Gleichungen zu zeigen. (Hinweis: Für einen Winkel $\\varangle X Y Z$ mit $0^{\\circ}<\\varangle X Y Z<180^{\\circ}$ im mathematisch positiven Sinn setze $\\varangle Z Y X:= 180^{\\circ}-\\varangle X Y Z$. Mit dieser Konvention gilt z. B. für vier paarweise verschiedene Punkte $W, X, Y, Z$ auf einer Kreislinie stets $\\varangle X Y Z=\\varangle X W Z$ ).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1.\nDie Verlängerungen von $A P, B P, C P$ mögen den Umkreis des Dreiecks $A B C$ erneut in $A'$, $B'$, $C'$ treffen. Mit Hilfe des Peripheriewinkelsatzes und einem einfachen Winkelsummenargument finden wir $\\varangle B' A' C' = \\varangle B' A' A + \\varangle A A' C' = \\varangle B' B A + \\varangle A C C' = \\varangle B P C - \\varangle B A C = 60^{\\circ}$. Ebenso $\\varangle A' C' B' = 60^{\\circ}$ und folglich ist das Dreieck $A' B' C'$ gleichseitig, d.h. $A' B' = B' C' = C' A'$. Wie üblich haben wir $A P B \\sim B' P A'$ und $A P C \\sim C' P A'$; hieraus ergibt sich $\\frac{A P}{A C} = \\frac{C' P}{C' A'}$ und $\\frac{B P}{B C} = \\frac{C' P}{C' B'}$. In Verbindung mit vorigem lehrt dies $\\frac{A P}{A C} = \\frac{B P}{B C}$ und somit in der Tat $A B \\cdot P C = A C \\cdot P B$.\n\n2.\nWähle den Punkt $J$ so, daß die Dreiecke $A B C$, $P B J$ (gleichorientiert) ähnlich sind. Sodann ist $\\frac{A B}{B P} = \\frac{B C}{B J}$ und $\\varangle P B A = \\varangle J B C$, weshalb auch die Dreiecke $A B P$, $C B J$ (gleichorientiert) ähnlich sein müssen. Folglich gilt $\\varangle C J P = \\varangle C J B - \\varangle P J B = (60^{\\circ} + \\gamma) - \\gamma = 60^{\\circ}$. Da auch $\\varangle J P C = \\varangle B P C - \\varangle B P J = (60^{\\circ} + \\alpha) - \\alpha = 60^{\\circ}$ ist das Dreieck $P C J$ gleichseitig und mithin $P C = P J$. Nach Wahl von $J$ haben wir $\\frac{A B}{A C} = \\frac{P B}{P J}$ und hieraus folgt mit Hilfe des vorigen wie gewünscht $A C \\cdot P B = A B \\cdot P J = A B \\cdot P C$.\n\n3.\nÜber der Strecke $B P$ werde das gleichseitige Dreieck $B P T$ errichtet. Der Schnittpunkt von $P T$ mit $A B$ heiße $G$. Außerdem werde auf $A C$ der Punkt $H$ mit $\\varangle C P H = 60^{\\circ}$ gewählt. Wegen $\\varangle A P G = \\gamma$ und $\\varangle H P A = \\beta$ muss $\\varangle H P G + \\varangle G A H = 180^{\\circ}$ sein, d.h. das Viereck $G A H P$ ist einem Kreis einbeschrieben. Demnach $\\varangle H G A = \\varangle H P A = \\beta$, woraus sofort $G H \\parallel B C$ geschlossen wird. Aus diesem Grund gilt $\\frac{A B}{A C} = \\frac{B G}{C H}$ (1). Ferner zeigen einfache Winkelbetrachtungen $\\varangle H C P = 60^{\\circ} - \\varangle P B G = \\varangle G B T$, was zusammen mit $\\varangle C P H = \\varangle B T G [=60^{\\circ}]$ die Ähnlichkeit der Dreiecke $P C H$ und $T B G$ lehrt. Mithin haben wir $\\frac{B G}{C H} = \\frac{B T}{C P}$ (2). Nachdem das Dreieck $B P T$ nach Konstruktion gleichseitig ist, gilt insbesondere $B T = B P$. Zusammen mit (1) und (2) erhalten wir hieraus $\\frac{A B}{A C} = \\frac{B P}{C P}$ und damit in der Tat $A B \\cdot P C = A C \\cdot P B$.\n\n4.\nWähle den Punkt $G$ so, daß die Dreiecke $A G C$, $P B C$ (gleichorientiert) ähnlich sind. Wie in der zweiten Lösung sehen wir, daß dann auch die Dreiecke $G B C$ und $A P C$ (gleichorientiert) ähnlich sind. Ferner ist $\\varangle B A G = \\varangle C A G - \\varangle C A B = (60^{\\circ} + \\alpha) - \\alpha = 60^{\\circ}$ und ebenso $\\varangle G B A = 60^{\\circ}$. Das Dreieck $A G B$ ist also gleichseitig und mithin $A G = A B$. Nach $A G C \\sim P B C$ haben wir also $P B : P C = A G : A C = A B : A C$. Daher wie behauptet $A B \\cdot P C = A C \\cdot P B$.\n\n5.\nDie Fußpunkte der Lote von $P$ auf die Seiten $B C$, $C A$, $A B$ mögen $X, Y, Z$ genannt werden. Nach Satz von Thales besitzen die Vierecke $P X C Y$, $P Y A Z$, $P Z B X$ jeweils einen Umkreis. Wir finden nun $\\varangle Z X Y = \\varangle Z X P + \\varangle P X Y = \\varangle Z B P + \\varangle P C Y = 60^{\\circ}$ und analog $\\varangle X Y Z = \\varangle Y Z X = 60^{\\circ}$. Das Dreieck $X Y Z$ ist also gleichseitig. Für die Länge der Seite $X Y$ finden wir durch zweimalige Verwendung des Sinussatzes $X Y = P C \\cdot \\sin \\gamma = \\frac{A B \\cdot P C}{2 R}$, wobei $R$ den Radius des Umkreises des Dreiecks $A B C$ bezeichnet. Ebenso $X Z = \\frac{A C \\cdot P B}{2 R}$. Aus $X Y = X Z$ folgt nunmehr wie verlangt $A B \\cdot P C = A C \\cdot P B$.\n\n6.\nDer Umkreis des Dreiecks $A B P$ schneide $A C$ zum zweiten Mal in $Q$. Wir erhalten $\\varangle B Q A = \\varangle B P A = \\gamma + 60^{\\circ}$ und hieraus nach Außenwinkelsatz $\\varangle Q B C = 60^{\\circ}$. Außerdem ergibt sich aus $\\varangle B P Q = 180^{\\circ} - \\alpha$ und $\\varangle C P B = 60^{\\circ} + \\alpha$ sofort $\\varangle Q P C = 120^{\\circ}$. Ferner erhalten wir, wenn wir $\\varangle A B P = \\varphi$ setzen, sofort $\\varangle C Q P = \\varphi$. Durch mehrmalige Verwendung des Sinussatzes erhalten wir nun\n$$\n\\frac{P C}{P A} = \\frac{P C}{C Q} \\cdot \\frac{C Q}{Q B} \\cdot \\frac{Q B}{A P} = \\frac{\\sin \\varphi}{\\sin 120^{\\circ}} \\cdot \\frac{\\sin 60^{\\circ}}{\\sin \\gamma} \\cdot \\frac{\\sin \\alpha}{\\sin \\varphi} = \\frac{\\sin \\alpha}{\\sin \\gamma} = \\frac{B C}{A B}\n$$\nHieraus folgt sofort $A B \\cdot P C = A P \\cdot B C$.\n\n7.\nÜber der Seite $A B$ errichte man das gleichseitige Dreieck $A B Q$ nach innen. Einfache Winkelbetrachtungen zeigen nun $\\varangle Q A P = 60^{\\circ} - \\varangle P A B = \\varangle B C P$ und ebenso $\\varangle P B Q = \\varangle P C A$. Indem wir also den Schnittpunkt von $A C$ mit $B Q$ als $T$ in die Überlegung einführen, wird das Viereck $B C T P$ wegen $\\varangle P B T = \\varangle P B Q = \\varangle P C A = \\varangle P C T$ ein Sehnenviereck sein. Folglich $\\varangle B T P = \\varangle B C P = \\varangle Q A P$, also $\\varangle P T Q + \\varangle Q A P = 180^{\\circ}$ und somit ist auch $P T Q A$ ein Sehnenviereck. Aus diesem Grund gilt $\\varangle B Q P = \\varangle T Q P = \\varangle T A P = \\varangle C A P$, woraus in Verbindung mit $\\varangle P B Q = \\varangle P C A$ die Ähnlichkeit der Dreiecke $B Q P$, $C A P$ folgt. Hieraus erhellt $P C : A C = P B : B Q = P B : A B$, also wie gewünscht $A B \\cdot P C = A C \\cdot P B$.\n\n8.\nSetze $\\varangle A C P = \\gamma'$, $\\varangle P B C = \\gamma''$, $\\varangle P B A = \\beta''$. Nun ist nach Sinussatz\n$$\n\\frac{\\sin \\beta''}{\\sin \\gamma'} = \\frac{\\sin \\beta''}{A P} \\cdot \\frac{A P}{\\sin \\gamma'} = \\frac{\\sin (60^{\\circ} + \\gamma)}{c} \\cdot \\frac{b}{\\sin (60^{\\circ} + \\beta)} = \\frac{\\sin (60^{\\circ} + \\gamma)}{\\sin \\gamma} \\cdot \\frac{\\sin \\beta}{\\sin (60^{\\circ} + \\beta)} = \\frac{1 + \\sqrt{3} \\cot \\gamma}{1 + \\sqrt{3} \\cot \\beta}\n$$\nDa jedoch $\\beta'' + \\gamma' = 60^{\\circ}$ haben wir auch\n$$\n\\frac{\\sin \\beta''}{\\sin \\gamma'} = \\frac{\\sin (60^{\\circ} - \\gamma')}{\\sin \\gamma'} = \\frac{\\sqrt{3}}{2} \\cot \\gamma' - \\frac{1}{2}\n$$\nDiese beiden Gleichungen liefern zusammengenommen\n$$\n\\cot \\gamma' = \\frac{\\sqrt{3} + \\cot \\beta + 2 \\cot \\gamma}{1 + \\sqrt{3} \\cot \\beta}\n$$\nWeiterhin\n$$\n\\frac{\\sin \\gamma''}{\\sin \\gamma'} = \\frac{\\sin (\\gamma - \\gamma')}{\\sin \\gamma'} = \\sin \\gamma \\cot \\gamma' - \\cos \\gamma\n$$\nHierin setzen wir die zuvor gefundene Gleichung ein und erhalten\n$$\n\\begin{gathered}\n\\frac{\\sin \\gamma''}{\\sin \\gamma'} = \\frac{(\\sqrt{3} \\sin \\beta \\sin \\gamma + \\cos \\beta \\sin \\gamma + 2 \\sin \\beta \\cos \\gamma) - (\\sin \\beta \\cos \\gamma + \\sqrt{3} \\cos \\beta \\cos \\gamma)}{\\sin \\beta + \\sqrt{3} \\cos \\beta} \\\\\n= \\frac{\\sin \\alpha + \\sqrt{3} \\cos \\alpha}{\\sin \\beta + \\sqrt{3} \\cos \\beta} = \\frac{\\sin (60^{\\circ} + \\alpha)}{\\sin (60^{\\circ} + \\beta)}\n\\end{gathered}\n$$\nDaher nach Sinussatz\n$$\n\\frac{A P}{A C} = \\frac{\\sin \\gamma'}{\\sin (\\beta + 60^{\\circ})} = \\frac{\\sin \\gamma''}{\\sin (\\alpha + 60^{\\circ})} = \\frac{B P}{B C}\n$$\nd.h. wie gewünscht $A P \\cdot B C = B P \\cdot A C$.\n\n9.\nDer Umkreis des Dreiecks $A P B$ treffe die Gerade $C P$ zum zweiten Mal in $J$. Sodann ist $\\varangle A J P = \\varangle A B P$. Zusammen mit $\\varangle A B P + \\varangle P C A = 60^{\\circ}$ lehrt dies $\\varangle C A J = 120^{\\circ}$. Ebenso sehen wir $\\varangle J B C = 120^{\\circ}$ ein. Damit haben wir\n$$\n\\frac{A C}{B C} = \\frac{A C}{J C} \\cdot \\frac{J C}{B C} = \\frac{\\sin \\varangle A J C}{\\sin \\varangle C A J} \\cdot \\frac{\\sin \\varangle J B C}{\\sin \\varangle C J B} = \\frac{\\sin \\varangle A J P}{\\sin \\varangle P J B} = \\frac{\\sin \\varangle A B P}{\\sin \\varangle P A B} = \\frac{A P}{B P},\n$$\nd.h. wie behauptet $A C \\cdot B P = B C \\cdot A P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEine Folge $a_{0}, a_{1}, a_{2}, \\ldots$ reeller Zahlen ist rekursiv definiert durch\n$$\na_{0} = -1, \\quad \\sum_{k=0}^{n} \\frac{a_{n-k}}{k+1} = 0 \\quad \\text{ für } n \\geq 1.\n$$\nMan beweise, dass $a_{n} > 0$ für alle $n \\geq 1$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFür $n = 1$ gilt $a_{1} + \\frac{1}{2} a_{0} = 0$, also $a_{1} = \\frac{1}{2} > 0$.\n\nNun gelte $a_{i} > 0$ für $1 \\leq i < n$. Aus dem Gegebenen folgt\n$$\n\\sum_{k=1}^{n-1} \\frac{a_{n-k}}{k+1} + \\frac{a_{n}}{1} + \\frac{a_{0}}{n+1} = 0 \\Leftrightarrow a_{n} = \\frac{1}{n+1} - \\sum_{k=1}^{n-1} \\frac{a_{n-k}}{k+1} \\quad (*)\n$$\nund ebenso\n$$\n0 = \\sum_{k=0}^{n-1} \\frac{a_{(n-1)-k}}{k+1} = \\sum_{k=1}^{n} \\frac{a_{n-k}}{k} = \\sum_{k=1}^{n-1} \\frac{a_{n-k}}{k} + \\frac{a_{0}}{n} \\Leftrightarrow \\frac{1}{n} = \\sum_{k=1}^{n-1} \\frac{a_{n-k}}{k}.\n$$\nDies führt auf\n$$\n\\frac{1}{n+1} = \\sum_{k=1}^{n-1} \\left( \\frac{a_{n-k}}{k} \\cdot \\frac{n}{n+1} \\right)\n$$\nund wegen $n > k \\Leftrightarrow \\frac{n}{n+1} > \\frac{k}{k+1} \\Leftrightarrow \\frac{n}{k(n+1)} > \\frac{1}{k+1}$ auf\n$$\n\\frac{1}{n+1} > \\sum_{k=1}^{n-1} \\frac{a_{n-k}}{k+1}.\n$$\nMit $(*)$ ergibt sich\n$$\na_{n} > \\sum_{k=1}^{n-1} \\frac{a_{n-k}}{k+1} - \\sum_{k=1}^{n-1} \\frac{a_{n-k}}{k+1} = 0.\n$$\nMit vollständiger Induktion gilt daher $a_{n} > 0$ für alle $n \\geq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFür jede reelle Zahl $x$ mit $0 < x < 1$ sei $y \\in (0, 1)$ diejenige Zahl, deren $n$-te Nachkommastelle die $(2n)$-te Nachkommastelle von $x$ ist. Man beweise: Wenn $x$ rational ist, dann ist auch $y$ rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEine rationale Zahl besitzt eine abbrechende oder eine periodische Dezimalbruchentwicklung. Eine abbrechende Dezimalbruchentwicklung lässt sich durch Anhängen von Nullen ins Unendliche fortsetzen und entspricht daher einer periodischen Dezimalbruchentwicklung mit der Periodenlänge $1$.\n\nIm Folgenden betrachten wir nur die Stellen hinter einer eventuell auftretenden Vorperiode der Länge $v$. Die erste Zweierpotenz größer als $v$ sei mit $2^{s}$ bezeichnet, die Periodenlänge des Dezimalbruchs mit $d$.\n\nWeil es nur endlich viele Restklassen mod $d$ gibt, existieren zwei Zahlen $2^{t}$ und $2^{t+m}$, die den gleichen Rest $r$ mod $d$ lassen, d.h. $2^{t} \\equiv 2^{t+m} \\bmod d$ mit $m \\geq 1$. Dies bedeutet, dass die $t$-te und die $(t+m)$-te Nachkommastelle von $y$ gleich sind.\n\nDann gilt aber auch $2 \\cdot 2^{t} \\equiv 2 \\cdot 2^{t+m} \\bmod d \\Leftrightarrow 2^{t+1} \\equiv 2^{t+m+1} \\bmod d$, d.h. die $(t+1)$-te und die $(t+m+1)$-te Nachkommastelle von $y$ sind gleich. Entsprechend sind alle $(t+k)$-ten und $(t+m+k)$-ten Nachkommastellen von $y$ jeweils gleich $(0 \\leq k \\leq m)$. Daher ist $y$ periodisch mit der Periodenlänge $m$.\n\nAuch $y$ kann eine Vorperiode – höchstens von der Länge $s$ – besitzen.\n\nDiese Überlegungen sind unabhängig von der Basis $b \\geq 2$ des Zahlsystems.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan zeige, dass es in der Dezimaldarstellung von $\\sqrt[3]{3}$ zwischen der 1000000. und 3141592. Nachkommastelle eine von 2 verschiedene Ziffer gibt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nHätte $\\sqrt[3]{3}$ zwischen der 1000000. und 3141592. Nachkommastelle nur die Ziffer 2, wäre mit $a=\\left[10^{1000000} \\sqrt[3]{3}\\right]<10^{1000001}$ also $\\left|\\sqrt[3]{3}-(a+2 / 9) 10^{-1000000}\\right|<10^{-3141592}$ oder äquivalent\n\n$$\n\\left|\\sqrt[3]{3\\left(9 \\cdot 10^{1000000}\\right)^{3}}-(9 a+2)\\right|<9 \\cdot 10^{-2141592}\n$$\n\nNun ist sicher $(9 a+2)^{3} \\neq 3\\left(9 \\cdot 10^{1000000}\\right)^{3}$, da die rechte Seite den Primfaktor 3 in nicht durch 3 teilbarer Vielfachheit enthält. Allgemein gilt für $m, n \\in \\mathbb{N}$ mit $m \\neq n^{3}$ :\n\n$$\n|\\sqrt[3]{m}-n| \\geq \\frac{\\left|m-n^{3}\\right|}{\\sqrt[3]{m^{2}}+\\sqrt[3]{m} n+n^{2}} \\geq \\frac{1}{3 \\cdot(\\max (\\sqrt[3]{m}, n))^{2}}\n$$\n\nDaraus folgt aber - im Widerspruch zu (1):\n\n$$\n\\left|\\sqrt[3]{3\\left(9 \\cdot 10^{1000000}\\right)^{3}}-(9 a+2)\\right|>\\frac{1}{3\\left(10^{1000001}\\right)^{2}}>10^{-2000003}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan bestimme alle positiven ganzen Zahlen $k$ und $n$ mit folgender Eigenschaft:\nDie Zahl $k^{4}+n^{2}$ ist ohne Rest durch die Zahl $7^{k}-3^{n}$ teilbar.",
    "options": [],
    "answer": "(2, 4)",
    "solution": "Solution:\n\nDas einzige Lösungspaar ist $(k ; n)=(2 ; 4)$. Für jedes Lösungspaar sind $k^{4}+n^{2}$ und $7^{k}-3^{n}$ ganzzahlig, wobei $7^{k}-3^{n}$ nicht notwendig positiv sein muss. Weil $7^{k}$ und $3^{n}$ ungerade sind, gilt $2 \\mid 7^{k}-3^{n}$ und daher $2 \\mid k^{4}+n^{2}$. Deshalb müssen $k$ und $n$ entweder beide gerade oder beide ungerade sein. Im letzten Fall gilt $k^{4}+n^{2} \\equiv 2 \\bmod 4$, wogegen $7^{k}-3^{n} \\equiv(-1)^{k}-(-1)^{n} \\equiv 0 \\bmod 4$ ist. Daher sind $k$ und $n$ beide gerade und wir setzen $k=2 a$ bzw. $n=2 b$ mit positiven ganzen Zahlen $a, b$. Es folgt $7^{2 a}-3^{2 b} \\mid(2 a)^{4}+(2 b)^{2}$ und weiter $49^{a}-9^{b} \\mid 16 a^{4}+4 b^{2}$. Nun ist $49^{a}-9^{b} \\equiv 1^{a}-1^{b} \\equiv 0 \\bmod 8$, so dass auch $16 a^{4}+4 b^{2}$ durch 8 teilbar sein muss.\n\nDeshalb muss $b$ gerade sein und ist als $b=2 c$ mit einer geeigneten positiven ganzen Zahl $c$ darstellbar. Es folgt $7^{2 a}-9^{2 c}=(7^{a}+9^{c})(7^{a}-9^{c}) \\mid 16(a^{4}+c^{2})$. Weil stets $7^{a} \\neq 9^{c}$ gilt und beide Potenzen ungerade sind, ist $|7^{a}-9^{c}| \\geq 2$, so dass $7^{a}+9^{c} \\mid 8(a^{4}+c^{2})$ gelten muss.\n\n**Lemma 1:** Für alle $a \\geq 4$ ist $7^{a}>8 a^{4}$.\n\nBeweis durch vollständige Induktion nach $a$:\nFür $a=4$ ist $2401=7^{4}>8 \\cdot 4^{4}=2^{11}=2048$ wahr. Nun sei $7^{a}>8 a^{4}$ vorausgesetzt.\nDann ist $7^{a+1}=7 \\cdot 7^{a}>7 \\cdot 8 a^{4}=8(a+1)^{4} \\cdot \\frac{7 a^{4}}{(a+1)^{4}}>8(a+1)^{4} \\cdot 7 \\cdot\\left(\\frac{4}{5}\\right)^{4}>8(a+1)^{4}$.\n\n**Lemma 2:** Für alle $c \\geq 1$ ist $9^{c}>8 c^{2}$.\n\nBeweis durch vollständige Induktion nach $c$:\nFür $c=1$ ist $9>8$ wahr. Nun sei $9^{c}>8 c^{2}$ vorausgesetzt. Dann ist\n$9^{c+1}=9 \\cdot 9^{c}>9 \\cdot 8 c^{2}=8(c+1)^{2} \\cdot \\frac{9 c^{2}}{(c+1)^{2}}>8(c+1)^{2} \\cdot 9 \\cdot\\left(\\frac{1}{2}\\right)^{2}>8(c+1)^{2}$.\n\nAus den Lemmata folgt, dass für $a \\geq 4$ wegen $7^{a}+9^{c}>8(a^{4}+c^{2})$ keine verlangten Zahlen $k$ und $n$ existieren. Also müssen nur noch $a=1,2,3$ untersucht werden.\n\n**Lemma 3:** Für alle $c \\geq 3$ ist $9^{c}>305+8 c^{2}$.\n\nBeweis durch vollständige Induktion nach $c$:\nFür $c=3$ ist $9^{3}=729>305+8 \\cdot 3^{2}=377$ wahr. Nun sei $9^{c}>305+8 c^{2}$ vorausgesetzt. Dann ist $9^{c+1}>9\\left(305+8 c^{2}\\right)=305+8\\left(305+9 c^{2}\\right)$. Wegen\n$305+9 c^{2}=(c+1)^{2}+8 c^{2}-2 c+304>(c+1)^{2}+c(c-2)>(c+1)^{2}$ für $c>2$ folgt die Behauptung.\n\nFall 1: $a=1$. Für $c=1$ gilt $49-81=-32 \\mid 32=16(1+1)$. Daher ist $(k ; n)=(2 ; 4)$ eine Lösung. Für $c=2$ gilt $7+81=88 \\nmid 8(1+4)=40$. Für $c>2$ muss $7+9^{c} \\leq 8+8 c^{2}$ gelten, also $9^{c}<1+8 c^{2}$. Dies widerspricht Lemma 3.\n\nFall 2: $a=2$. Für $c=1$ gilt $7^{2}+9^{1}=58$ und $8\\left(2^{4}+1^{2}\\right)=136$. Für $c=2$ gilt $49+81=130 \\nmid 8(16+4)=160$. Für $c>2$ muss $49+9^{c} \\leq 8\\left(16+c^{2}\\right)$ gelten, also $9^{c} \\leq 79+8 c^{2}$. Dies widerspricht Lemma 3.\n\nFall 3: $a=3$. Für $c=1$ gilt $343+9=352$ und $656=8\\left(3^{4}+1\\right)$. Für $c=2$ gilt $343+81=424$ und $680=8\\left(3^{4}+4\\right)$. Für $c>2$ muss $343+9^{c} \\leq 8\\left(81+c^{2}\\right)$ gelten, also $9^{c} \\leq 305+8 c^{2}$. Dies widerspricht Lemma 3.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31035,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben ist ein (allgemeines) Trapez $ABCD$, dessen Diagonalen sich im Punkt $P$ schneiden. Ein Punkt $Q$ liegt so zwischen den Parallelen $AB$ und $DC$, dass $\\varangle AQB = \\varangle CQD$ gilt und die Gerade $BC$ zwischen $P$ und $Q$ verläuft.\nMan beweise, dass $\\varangle DQP = \\varangle BAQ$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWegen $AB \\parallel CD$ bildet die zentrische Streckung $\\sigma$ am Schnittpunkt $P$ der Diagonalen mit Streckfaktor $k = -\\frac{|CD|}{|AB|} = -\\frac{|CP|}{|AP|} = -\\frac{|DP|}{|BP|}$ den Punkt $C$ auf $A$ und $D$ auf $B$ ab. Es sei $Q'$ der Bildpunkt von $Q$ unter $\\boldsymbol{\\sigma}$. Weil\n\n![](attached_image_1.png)\n\ndas Dreieck $DQC$ unter $\\sigma$ auf das Dreieck $BQ'A$ abgebildet wird, sind beide Dreiecke ähnlich und mit der Voraussetzung gilt $\\varangle AQ'B = \\varangle CQD = \\varangle AQB$. Somit liegen $A, B, Q$ und $Q'$ auf einem Kreis und es gilt $\\varangle BAQ = \\varangle BQ'Q$. Das Dreieck $BQ'Q$ wird unter $\\sigma$ auf das Dreieck $DQQ'$ abgebildet. Daher sind diese Dreiecke ähnlich und es folgt, da $P, Q$ und $Q'$ auf einer Geraden liegen, $\\varangle BQ'Q = \\varangle DQQ' = \\varangle DQP$.\n\nAlso ist $\\varangle BAQ = \\varangle BQ'Q = \\varangle DQP$, was zu beweisen war.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31036,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABCD$ ein gleichschenkliges Trapez mit $AB \\parallel CD$ und $\\overline{BC} = \\overline{AD}$. Die Parallele zu $AD$ durch $B$ treffe die Senkrechte zu $AD$ durch $D$ im Punkt $X$. Ferner treffe die durch $A$ gezogene Parallele zu $BD$ die Senkrechte zu $BD$ durch $D$ im Punkt $Y$. Man beweise, dass die Punkte $C, X, D$ und $Y$ auf einem gemeinsamen Kreis liegen.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nErste Lösung. Es sei $M$ der Mittelpunkt der Strecke $AB$. Die durch $M$ gezogene Parallele zu $DY$ treffe $BD$ in $G$ und die durch $A$ gezogene Parallele zu $BD$ in $H$. Aus $\\overline{AM} = \\overline{MB}$ schließt man leicht $\\overline{HM} = \\overline{MG}$, und da $DY \\perp DB$ vorausgesetzt ist, ist das Viereck $DGHY$ ein Rechteck. Beides zusammen lehrt $\\overline{DM} = \\overline{MY}$, und in gleicher Weise zeigt man $\\overline{DM} = \\overline{MX}$. Da das Trapez $ABCD$ gleichschenklig ist, gilt ferner $\\overline{DM} = \\overline{MC}$, und aus der Kombination der drei letzten Gleichungen ergibt sich sofort, dass die vier Punkte $C, D, X, Y$ auf einem gemeinsamen Kreis um $M$ liegen.\n\n\nZweite Lösung. Man ergänze das Dreieck $DAB$ zu einem Parallelogramm $DAZB$. Nach Konstruktion von $X$ und $Y$ liegen diese beiden Punkte auf $ZB$ bzw. $ZA$, und es gilt $\\angle DXZ = \\angle ZYD = 90^{\\circ}$. Der Satz des Thales liefert nunmehr, dass $X$ und $Y$ auf dem Kreis mit Durchmesser $DZ$ liegen. Der Mittelpunkt dieses Kreises ist gleichzeitig der Mittelpunkt der Strecke $DZ$ und, da sich in Parallelogrammen die Diagonalen gegenseitig halbieren, fällt dieser mit dem Mittelpunkt der Strecke $AB$ zusammen. Nachdem nun das Trapez $ABCD$ als gleichschenklig vorausgesetzt ist, gilt folglich $\\overline{DM} = \\overline{MC}$, womit gezeigt ist, dass auch der Punkt $C$ auf dem vorhin betrachteten Kreis liegt.\n\n\nDritte Lösung (skizziert, nach einer Idee von Andreas Gross). Es sei $F = (ABD)$ der Flächeninhalt des Dreiecks $ABD$. Wir ziehen um $D$ einen Kreis $\\omega$ mit Radius $\\sqrt{2F}$ und invertieren an diesem. Kundige können sich leicht überlegen, dass $\\overline{DX'} = \\overline{DA}$ sowie $\\overline{DY'} = \\overline{DB}$ gilt und dass hieraus $(DX'Y') = F$ folgt. Es sei nun $C^*$ der Schnittpunkt der Geraden $DC$ und $X'Y'$ sowie $H$ der Fußpunkt des aus $D$ auf $AB$ gefällten Lotes. Nun haben die Punkte $X', Y'$ von der Geraden $CD$ die Abstände $\\overline{AH}, \\overline{BH}$, und hieraus ergibt sich $(DX'Y') = (DC^*Y') - (DC^*X') = \\frac{1}{2} \\overline{DC^*} \\cdot (\\overline{HB} - \\overline{AH}) = \\frac{1}{2} \\overline{DC^*} \\cdot \\overline{DC}$. Im Zusammenhang mit einer bereits angegebenen Gleichung folgt hieraus $\\overline{DC^*} \\cdot \\overline{DC} = 2F$, woraus sich mit Hilfe einer kleinen Lagebetrachtung $C^* = C'$ ergibt. Demnach liegen die drei Punkte $C', X', Y'$ auf einer gemeinsamen Geraden, und von dieser überzeugt man sich leicht, dass sie nicht durch $D$ gehen kann. Dies impliziert die Behauptung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEin Quadrat wird so in $n>1$ Rechtecke zerlegt, dass die Seiten der Rechtecke parallel zu den Seiten des gegebenen Quadrats verlaufen. Jede Gerade, die parallel zu einer der Seiten des Quadrats verläuft und das Innere des Quadrats schneidet, soll dabei auch im Inneren wenigstens eines der Rechtecke verlaufen.\nMan beweise, dass es dann in dieser Zerlegung stets ein Rechteck gibt, das keinen Punkt mit dem Rand des Quadrats gemeinsam hat.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir beweisen die Kontraposition und nehmen dazu an, dass jedes Rechteck der Zerlegung mindestens einen Punkt mit dem Rand des Quadrates gemeinsam hat. Nach den Voraussetzungen hat es dann wenigstens eine seiner Seiten mit dem Rand des Quadrates gemeinsam. Daher lässt sich der Rand des Quadrates abschnittweise eindeutig den Rechtecken der Zerlegung zuordnen. Zwei nicht zusammenhängende Abschnitte können nicht zu demselben Rechteck gehören; sie müssten sonst einander gegenüberliegen und auf den anderen Seiten dieses Rechtecks schneiden Geraden das Innere des Quadrates, ohne im Inneren eines Rechtecks zu verlaufen.\n\nDie Anzahl $n$ dieser Abschnitte stimmt also mit der Anzahl $n$ der Rechtecke überein. In den $n$ Endpunkten der Abschnitte treffen auf dem Rand des Quadrates jeweils zwei Rechtecke zusammen und haben dort jeweils eine Ecke. Vier weitere Ecken der Rechtecke fallen mit den Ecken des Quadrates zusammen, so dass insgesamt $2 n+4$ Rechtecksecken auf dem Rand des Quadrates liegen. Somit verbleiben $4 n-(2 n+4)=2 n-4$ Ecken für das Innere des Quadrates. Nun betrachten wir einen der $n$ Punkte auf dem Rand des Quadrates, in denen zwei Rechtecke zusammenstoßen. Die Gerade senkrecht zur jeweiligen Quadratseite durch diesen Punkt $P$\n\n![](attached_image_1.png)\n\nverläuft im Inneren des Quadrats zunächst auf je einer Seite der beiden Rechtecke. Damit diese Gerade innerhalb des Quadrates auch im Inneren eines Rechtecks verlaufen kann, muss sie eine Rechtecksseite schneiden, die parallel zur Quadratseite verläuft, von der wir ausgegangen sind. In diesem Verzweigungspunkt $Q$ haben die zwei Rechtecke, deren Rand die Gerade vorher war (das müssen nicht mehr die beiden Rechtecke mit Eckpunkt $P$ sein), jeweils einen Eckpunkt. Ein solcher Punkt $Q$ existiert für jeden Ausgangspunkt $P$, und zwei verschiedene Ausgangspunkte können nicht denselben Verzweigungspunkt $Q$ haben. Daher müssen wenigstens $2 n$ Rechtecksecken im Inneren des Quadrates liegen, im Widerspruch zur oben bestimmten Höchstzahl $2 n-4$. Also gibt es stets ein Rechteck, das keinen Punkt mit dem Rand des Quadrats gemeinsam hat.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan zeige, dass es eine ganze Zahl $a$ gibt, für die $a^{3}-36 a^{2}+51 a-97$ ein Vielfaches von $3^{2008}$ ist.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. Lösung: Definiere das Polynom $P(x)=x^{3}-36 x^{2}+51 x-97$. Ausgehend von $P_{0}(x)=\\frac{1}{81} P(9 x+1)=9 x^{3}-33 x^{2}-2 x-1$, definiere rekursiv Polynome $P_{k}(x)=3 a_{k} x^{3}+3 b_{k} x^{2}+c_{k} x+d_{k}$ mit ganzen Zahlen $a_{k}, b_{k}, c_{k}, d_{k}$, wobei $c_{k}$ nicht durch 3 teilbar ist, und $P_{k+1}(x)=\\frac{1}{3} P_{k}\\left(3 x+\\varepsilon_{k}\\right)$ mit $\\varepsilon_{k} \\in\\{-1,0,1\\}$ für alle $k$ : Da $c_{k}$ nicht durch 3 teilbar ist, kann man $\\varepsilon_{k}$ so wählen, dass $\\varepsilon_{k} c_{k}+d_{k}$ durch 3 teilbar ist; dann ist $\\frac{1}{3} P_{k}\\left(3 x+\\varepsilon_{k}\\right)=3 a_{k+1} x^{3}+3 b_{k+1} x^{2}+c_{k+1} x+d_{k+1}$ mit $a_{k+1}=9 a_{k}, b_{k+1}=9 \\varepsilon_{k} a_{k}+3 b_{k}$, $c_{k+1}=3\\left(3 \\varepsilon_{k}^{2} a_{k}+2 \\varepsilon_{k} b_{k}\\right)+c_{k}, d_{k+1}=\\varepsilon_{k}^{3} a_{k}+\\varepsilon_{k}^{2} b_{k}+\\left(\\varepsilon_{k} c_{k}+d_{k}\\right) / 3$, somit ist auch $c_{k+1}$ nicht durch 3 teilbar.\nSetzt man $x_{2004}=0$ und rekursiv $x_{k}=3 x_{k+1}+\\varepsilon_{k}$, also $P_{k}\\left(x_{k}\\right)=3 P_{k}\\left(x_{k+1}\\right)$ für $k=2003, \\ldots, 0$, so ist für $a=9 x_{0}+1$ die Zahl $P(a)=81 \\cdot 3^{2004} P_{2004}\\left(x_{2004}\\right)$ ein Vielfaches von $3^{2008}$.\n\n\n2. Lösung: Definiere das Polynom $P(x)=x^{3}-36 x^{2}+51 x-97$. Für $k \\geq 4$ werden induktiv ganze Zahlen $a_{k}$ mit $a_{k} \\equiv 1 \\bmod 9$ und $3^{k} \\mid P\\left(a_{k}\\right)$ konstruiert; die Aufgabenstellung erfüllt dann $a=a_{2008}$ :\nVerankerung $k=4$ : Für $a_{4}=1$ ist $3^{4} \\mid P\\left(a_{4}\\right)$.\nInduktionsschritt $k \\rightarrow k+1$ : Für $k \\geq 4$ ist $2 k-3 \\geq k+1$ und $3 k-6>k+1$, also für\nalle ganzzahligen $x$ :\n\n$$\n\\begin{aligned}\nP\\left(x+3^{k-2}\\right)-P(x) & \\equiv 3^{k-1} x^{2}+3^{2 k-3} x+3^{3 k-6}-36\\left(2 \\cdot 3^{k-2} x+3^{2 k-4}\\right)+51 \\cdot 3^{k-2} \\\\\n& \\equiv 3^{k-1}\\left(x^{2}-24 x+17\\right) \\bmod 3^{k+1}\n\\end{aligned}\n$$\n\nFür $x \\equiv 1 \\bmod 9$ ist $x^{2}-24 x+17 \\equiv 1-24+17 \\equiv 3 \\bmod 9$. Wegen $k \\geq 4$ ist auch $x \\equiv x+3^{k-2} \\bmod 9$. Daher enthält die Differenz $3^{k-1}\\left(x^{2}-24 x+17\\right)$ den Primfaktor 3 genau $k$-mal, also ist $P(u)$ für $u=a_{k}$ oder $u=a_{k}+3^{k-2}$ oder $u=a_{k}+2 \\cdot 3^{k-2}$ durch $3^{k+1}$ teilbar, setze $a_{k+1}$ auf diesen Wert von $u$. Nach Konstruktion ist $a_{k+1} \\equiv 1 \\bmod 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $p>7$ eine Primzahl, die bei Division durch 6 den Rest 1 lässt. Setze $m=2^{p}-1$. Man beweise, dass $2^{m-1}-1$ ohne Rest durch $127 m$ teilbar ist.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDie Lösung besteht aus drei Schritten:\n\n1. $2^{m-1}-1$ ist durch 127 teilbar.\n\n2. $2^{m-1}-1$ ist durch $m$ teilbar.\n\n3. 127 und $m$ sind teilerfremd.\n\nZu 1: Es gilt $2^{6} \\equiv 1 \\bmod 7$. Aus $p \\equiv 1 \\bmod 6$ folgt $2^{p} \\equiv 2 \\bmod 7$, also $7 \\mid m-1$. Mit $2^{7} \\equiv 1 \\bmod 127$ folgt daraus $2^{m-1} \\equiv 1 \\bmod 127$.\n\nZu 2: Es gilt $2^{p} \\equiv 2 \\bmod p$ (kleiner Satz von Fermat), d. h. $p \\mid m-1$. Aus $2^{p} \\equiv 1 \\bmod m$ folgt damit $2^{m-1} \\equiv 1 \\bmod m$.\n\nZu 3: Es gilt $2^{p} \\equiv 1 \\bmod 127$ genau dann, wenn $p$ durch 7 teilbar ist (schreibe $p=7k+r$ mit $0 \\leq r<7$ und verwende $2^{7} \\equiv 1 \\bmod 127$). Wegen $p>7$ ist das nicht der Fall, also ist $m$ nicht durch 127 teilbar, und da 127 eine Primzahl ist, sind 127 und $m$ teilerfremd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $p$ eine Primzahl. Weiter seien $a, b, c$ ganze Zahlen, welche die Gleichungen $a^{2}+p b = b^{2}+p c = c^{2}+p a$ erfüllen.\nMan beweise, dass dann $a = b = c$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWenn zwei der drei Zahlen $a, b, c$ gleich sind, können wir wegen der zyklischen Vertauschbarkeit oBdA $a = b$ annehmen. Damit wird die linke gegebene Gleichung zu $a^{2} + p b = a^{2} + p c$, woraus wegen $p \\neq 0$ direkt $b = c$ folgt, so dass die Behauptung erfüllt ist. Im Folgenden können wir daher $a \\neq b \\neq c \\neq a$ voraussetzen.\n\nUmformen der Gleichungen liefert\n$$\np = \\frac{b^{2} - a^{2}}{b - c} = \\frac{c^{2} - b^{2}}{c - a} = \\frac{a^{2} - c^{2}}{a - b}.\n$$\nMultiplizieren der drei Terme und Kürzen ergibt\n$$\np^{3} = - (a + b)(b + c)(c + a) \\tag{1}\n$$\nVon den drei gegebenen Zahlen sind nach dem Schubfachprinzip wenigstens zwei gerade oder wenigstens zwei ungerade; deren Summe ist also durch $2$ teilbar. Deshalb ist das Produkt auf der rechten Seite von (1) gerade. Es folgt $p = 2$ und daher $(a + b)(b + c)(c + a) = -8$. \\tag{2}\n\nSind zwei der Klammern in (2) gleich, so sei oBdA $a + b = b + c$. Es folgt direkt $a = c$ und daher die Gleichheit aller drei gegeben Zahlen.\n\nIst genau eine der Klammern ungerade (gleich $\\pm 1$), so ist $(a + b) + (b + c) + (c + a)$ einerseits ungerade, andererseits gleich $2(a + b + c)$, also gerade – Widerspruch!\n\nEs bleiben daher (bis auf zyklische Vertauschbarkeit) die Fälle ($\\pm 1; \\mp 1; 8$) zu untersuchen.\n\nAus $a + b = 1$, $b + c = -1$, $c + a = 8$ folgt $a = 5$, $b = -4$, $c = 3$, was mit $25 - 8 \\neq 16 + 6$ einen Widerspruch zur ersten gegebenen Gleichung liefert.\n\nAus $a + b = -1$, $b + c = 1$, $c + a = 8$ folgt $a = 3$, $b = -4$, $c = 5$, was mit $9 - 8 \\neq 16 + 10$ einen Widerspruch zur ersten gegebenen Gleichung liefert.\n\nDaher ist nur $a = b = c$ möglich. In der Tat gilt $a^{2} + p a = a^{2} + p a$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31041,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDas Dreieck $A B C$ sei bei $A$ rechtwinklig. Es bezeichne $M$ den Mittelpunkt der Strecke $B C$. Der Punkt $D$ liege auf der Seite $A C$ und erfülle $\\overline{A D}=\\overline{A M}$. Der von $C$ verschiedene Schnittpunkt der Umkreise der Dreiecke $A M C$ und $B D C$ heiße $P$. Man beweise, dass $C P$ den bei $C$ gelegenen Winkel des Dreiecks $A B C$ halbiert.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDer Winkel $\\varangle P D C$ ist einerseits Nebenwinkel von $\\varangle A D P$, andererseits liegt er im Sehnenviereck $B P D C$ dem Winkel $\\varangle C B P$ gegenüber. Folglich $\\varangle A D P = \\varangle C B P$. Analog ist der Winkel $\\varangle C M P$ Nebenwinkel von $\\varangle P M B$ und liegt im Sehnenviereck $A P M C$ dem Winkel $\\varangle P A C$ gegenüber, also $\\varangle P M B = \\varangle P A C$. Die Dreiecke $P B M$ und $P D A$ stimmen somit in den Innenwinkeln bei $B$ und $D$ sowie bei $M$ und $A$ überein, außerdem ist nach Voraussetzung $\\overline{A D} = \\overline{A M} = \\overline{M B}$ (die letzte Gleichheit ergibt sich aus dem Satz des Thales). Nach dem Kongruenzsatz wsw sind die Dreiecke $P B M$ und $P D A$ kongruent, insbesondere sind ihre von $P$ ausgehenden Höhen gleich lang. Sie sind aber die Lote von $P$ auf die Dreiecksseiten $C A$ bzw. $C B$, also liegt $P$ auf der Winkelhalbierenden.\nSolution:\n\nBezeichne mit $m_{X Y}$ die Mittelsenkrechte der Strecke $X Y$. Dann sind sowohl $m_{C D}$ und $m_{C A}$ als auch $m_{C M}$ und $m_{C B}$ Paare paralleler Geraden im Abstand $\\frac{1}{4} \\overline{B C}$, also begrenzen sie eine Raute. Die Mittelpunkte $M_{1}, M_{2}$ der beiden in der Aufgabe genannten Umkreise sind die Schnittpunkte von $m_{C A}$ mit $m_{C M}$ bzw. von $m_{C D}$ mit $m_{C B}$. Die Gerade $(C P)$ steht senkrecht auf der Verbindungsgeraden $\\left(M_{1} M_{2}\\right)$, ist also parallel zur zweiten Diagonalen in der Raute. Man überlegt sich nun leicht, dass die zweite Diagonale parallel zur Winkelhalbierenden ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDie 16 Felder eines $4 \\times 4$-Schachbretts lassen sich wie folgt in 18 Linien ordnen: die vier Zeilen, die vier Spalten, fünf Diagonalen von Nordwest nach Südost und fünf Diagonalen von Nordost nach Südwest. Dabei bestehen diese Diagonalen aus 2, 3 oder 4 über Eck benachbarten Feldern jeweils gleicher Farbe; die Eckfelder des Schachbretts alleine bilden also keine Diagonale.\nNun wird in 10 der 16 Felder je ein Spielstein gesetzt. Jede der 18 Linien, die dann eine gerade Anzahl von Steinen enthält, zählt einen\n\n![](attached_image_1.png)\n\nPunkt.\nWelches ist die größtmögliche durch Setzen der 10 Spielsteine erreichbare Punktzahl? Die Antwort ist zu begründen.",
    "options": [],
    "answer": "17",
    "solution": "Solution:\n\nDie größtmögliche erreichbare Punktzahl ist 17, wie das abgebildete Beispiel zeigt. Es bleibt zu zeigen, dass die Maximalpunktzahl 18 nicht erreicht werden kann. Dazu nehmen wir die Existenz einer Belegung mit 18 Punkten an und betrachten zunächst die Zeilen bzw. Spalten. Weder in einer Zeile noch in einer Spalte kann dann eine ungerade Anzahl von Steinen liegen, so dass wegen der Summe 10 nur (bis auf Permutationen) die Möglichkeiten $(4; 2; 2; 2)$ oder $(4; 4; 2; 0)$ vorkommen können. Gibt es jedoch eine Zeile (Spalte) mit 0 Steinen, so können alle Spalten (Zeilen) wegen der erforderlichen Geradzahligkeit höchstens 2 Steine enthalten, womit die Summe 10 nicht erreicht wird. Daher gibt es genau eine Zeile und eine Spalte mit 4 Steinen. Der Schnittpunkt dieser\n\n![](attached_image_2.png)\n\n17 Punkte\n\n![](attached_image_3.png)\nFall a)\n![](attached_image_4.png)\nFall b)\n![](attached_image_5.png)\nFall c)\n\nbeiden Linien kann nun a) in einer Ecke, b) auf dem Rand, aber nicht in einer Ecke, c) im Inneren des Quadrats liegen. Wegen der Symmetrie genügt die Betrachtung je eines Beispiels.\n\nFall a): Hier enthalten die fünf NW-SO-Linien erst je einen Stein. Weil aber nur noch drei Steine zu setzen sind, ist die Punktzahl 18 nicht mehr erreichbar.\n\nFall b): Hier enthalten vier der NW-SO-Linien erst je einen Stein. Weil aber nur noch drei Steine zu setzen sind, ist die Punktzahl 18 nicht mehr erreichbar.\n\nFall c): Die beiden NW-SO-Linien mit genau 2 Feldern enthalten erst einen Stein. Wenn sie mit einem zweiten Stein besetzt werden, zählt die obere SW-NO-Linie mit 3 Steinen keinen Punkt. Auch hier wird die Punktzahl 18 nicht erreicht.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAuf einer Tafel stehe am Anfang eine positive ganze Zahl. Wenn eine Zahl $x$ auf der Tafel steht, darf man die Zahlen $2x+1$ und $\\frac{x}{x+2}$ dazuschreiben. Irgendwann stehe auch die Zahl 2008 auf der Tafel. Man beweise, dass sie von Anfang an dastand.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLösungsskizze: Anfangs stehe die Zahl $a$ auf der Tafel. Der Übergang von $x$ zu $2x+1$ oder $\\frac{x}{x+2}$ werde als Transformation bezeichnet. Alle Zahlen auf der Tafel sind positiv.\n\n1. Variante:\nAus der Zahl $a$ entstehen durch $k$ Transformationen stets Zahlen der Form $\\frac{ma+m-1}{(2^{k}-m)a+2^{k}-m+1}$ mit ganzzahligem $0 < m \\leq 2^{k}$ (Beweis durch vollständige Induktion nach $k$). Steht also irgendwann 2008 an der Tafel, gibt es somit $k \\geq 0$ und $0 < m \\leq 2^{k}$ mit $\\frac{ma+m-1}{(2^{k}-m)a+2^{k}-m+1} = 2008$ bzw. $2009 = (a+1)(2009m - 2008 \\cdot 2^{k})$. Da der zweite Faktor teilerfremd zu 2009 ist (da 2008 und $2^{k}$ teilerfremd zu 2009), folgt $a+1 = 2009$.\n\n2. Variante:\nSchreibe auf ein Blatt Papier zu jeder Zahl $t$ der Tafel die Zahl $f(t) = \\frac{2009}{t+1}$. Irgendwann steht 1 auf dem Blatt. Eine Transformation einer Zahl auf der Tafel bildet eine Zahl $z$ auf dem Blatt auf $z/2$ oder $\\frac{2009+z}{2}$ ab. Ist also die Anfangszahl $f(a)$ auf dem Blatt nicht ganzzahlig oder nicht teilerfremd zu 2009, können daraus nicht irgendwann ganzzahlige oder zu 2009 teilerfremde Zahlen entstehen. Somit ist $f(a)$ ganzzahlig und teilerfremd zu 2009. Da $f(a) = \\frac{2009}{a+1}$ zudem 2009 teilt, folgt $f(a) = 1$ und $a = 2008$.\n\n3. Variante:\nJede Transformation bildet die gekürzte rationale Zahl $p/q$ auf die gekürzte Zahl $p'/q'$ ab, wobei $p, q, p', q'$ positive ganze Zahlen sind und $p'+q' \\in \\{p+q, 2(p+q)\\}$. Anfangs hat man die gekürzte Darstellung $\\frac{a}{1}$, am Ende $\\frac{2008}{1}$, es gibt also $k \\geq 0$ mit $2^{k}(a+1) = 2008+1$. Da 2009 ungerade ist, folgt $k=0$ und $a+1 = 2009$.\n\n4. Variante:\nBilde die Zahlenfolge $a_{0} = a, a_{1}, \\ldots, a_{k} = 2008$ aus Zahlen auf der Tafel, wobei $a_{j+1}$ für $0 \\leq j < k$ durch eine der beiden Transformationen aus $a_{j}$ entsteht. Aus $a_{k}$ lassen sich $a_{k-1}, \\ldots, a_{0}$ eindeutig rekonstruieren, da für alle $x > 0$ gilt: $2x+1 > 1$, $0 < \\frac{x}{x+2} < 1$. Mit Induktion nach $n$ für $n = 0, \\ldots, k$ zeigt man, dass $a_{k-n} = \\frac{2009-b_{n}}{b_{n}}$ mit $0 < b_{n} < 2009$ und $b_{n} \\equiv 2^{n} \\bmod 2009$. Damit ist $a_{k-n}$ ganzzahlig (mit Wert 2008) genau dann, wenn $2^{n} \\equiv 1 \\bmod 2009$ bzw. wenn $420 \\mid n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDas Viereck $ABCD$ sei eine Raute mit spitzem Winkel bei $A$. Die Punkte $M$ und $N$ mögen so auf den Strecken $AC$ und $BC$ gelegen sein, dass $|DM| = |MN|$. Ferner sei $P$ der Schnittpunkt von $AC$ und $DN$ sowie $R$ der Schnittpunkt von $AB$ und $DM$. Man beweise, dass $|RP| = |PD|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDer Fall $N = B$ sei im Folgenden ausgeschlossen. Dann ist $DN$ nicht orthogonal zu $AC$, und $M$ ist eindeutig charakterisiert als Schnittpunkt der Mittelsenkrechte von $DN$ mit $AC$. Im Dreieck $DNC$ ist $AC$ die Winkelhalbierende bei $C$, und in jedem Dreieck schneiden sich die Winkelhalbierende und die gegenüberliegende Mittelsenkrechte auf dem zugehörigen Umkreisbogen, also ist $DMNC$ ein Sehnenviereck, insbesondere ist $\\varangle MDN = \\varangle MCN$. Nun ist $\\varangle MDN = \\varangle RDP$ und $\\varangle MCN = \\varangle ACB = \\varangle BAC = \\varangle RAP$, also ist nach dem Satz vom Umfangswinkel auch $ARPD$ ein Sehnenviereck. In seinem Umkreis sind wegen $\\varangle RAP = \\varangle BAC = \\varangle CAD = \\varangle PAD$ die Sehnen $RP$ und $PD$ gleich lang (Sinussatz oder Umfangswinkel).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMan beweise oder widerlege, dass für alle positiven reellen Zahlen $a$, $b$ und $c$ die Ungleichung\n$$\n3 \\leq \\frac{4 a+b}{a+4 b}+\\frac{4 b+c}{b+4 c}+\\frac{4 c+a}{c+4 a}<\\frac{33}{4}\n$$\ngilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nBeweis von $3 \\leq \\frac{4 a+b}{a+4 b}+\\frac{4 b+c}{b+4 c}+\\frac{4 c+a}{c+4 a}$ :\n\n1. Variante: Multiplizieren mit Hauptnenner und Vereinfachen führt zur äquivalenten Ungleichung $45 a b c \\leq 7\\left(a^{2} b+b^{2} c+c^{2} a\\right)+8\\left(a b^{2}+b c^{2}+c a^{2}\\right)$. Dies folgt aus der Ungleichung zwischen arithmetischem und geometrischem Mittel (z. B. $a^{2} b+b^{2} c+c^{2} a \\geq 3 \\sqrt[3]{a^{2} b \\cdot b^{2} c \\cdot c^{2} a}=3 a b c$ ).\n\n2. Variante: Für $n \\geq 1$ und reelle Zahlen $a_{1}, \\ldots, a_{n}, b_{1}, \\ldots, b_{n}$ gilt die Cauchy-Schwarz-Ungleichung $\\sqrt{a_{1}^{2}+\\ldots+a_{n}^{2}} \\sqrt{b_{1}^{2}+\\ldots+b_{n}^{2}} \\geq a_{1} b_{1}+\\ldots+a_{n} b_{n}$. Für $n=3$, $a_{1}^{2}=\\frac{4 a+b}{a+4 b}$, $a_{2}^{2}=\\frac{4 b+c}{b+4 c}$, $a_{3}^{2}=\\frac{4 c+a}{c+4 a}$, $b_{1}^{2}=(4 a+b)(a+4 b)$, $b_{2}^{2}=(4 b+c)(b+4 c)$, $b_{3}^{2}=(4 c+a)(c+4 a)$ ergibt sich\n\n$$\n\\left(\\frac{4 a+b}{a+4 b}+\\frac{4 b+c}{b+4 c}+\\frac{4 c+a}{c+4 a}\\right)((4 a+b)(a+4 b)+(4 b+c)(b+4 c)+(4 c+a)(c+4 a)) \\geq (5(a+b+c))^{2}\n$$\n\nDer zweite Faktor ist wegen $2(a+b+c)^{2}-6(a b+b c+c a)=(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \\geq 0$ kleiner als $\\frac{25}{3}(a+b+c)^{2}$, woraus die Behauptung folgt.\n\nBeweis von $\\frac{4 a+b}{a+4 b}+\\frac{4 b+c}{b+4 c}+\\frac{4 c+a}{c+4 a}<\\frac{33}{4}$ :\n\n1. Variante: Multiplizieren mit Hauptnenner und Vereinfachen liefert die äquivalente (triviale) Ungleichung $0<127 a b c+16\\left(a^{2} b+b^{2} c+c^{2} a\\right)$.\n\n2. Variante: Die Ungleichung ist äquivalent zu $\\frac{b}{a+4 b}+\\frac{c}{b+4 c}+\\frac{a}{c+4 a}>\\frac{1}{4}$; Letzteres gilt wegen\n\n$$\n\\frac{b}{a+4 b}+\\frac{c}{b+4 c}+\\frac{a}{c+4 a}>\\frac{b}{4 a+4 b+4 c}+\\frac{c}{4 a+4 b+4 c}+\\frac{a}{4 a+4 b+4 c}=\\frac{1}{4}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMan bestimme alle Paare $(m, n)$ nicht-negativer ganzer Zahlen, die der Gleichung\n$$\n3^{m}-7^{n}=2\n$$\ngenügen.",
    "options": [],
    "answer": "(m, n) = (1, 0) and (2, 1)",
    "solution": "Solution:\nDie Paare $(1,0)$ und $(2,1)$ sind Lösungen. Es sei $(m, n)$ eine Lösung mit $m, n \\geq 2$. Die Lösung besteht aus drei Schritten:\n\n1. Es ist $7^{n} \\equiv -2 \\bmod 9 \\Longleftrightarrow n \\equiv 1 \\bmod 3$.\n2. Es ist $3^{m} \\equiv 2 \\bmod 49 \\Longleftrightarrow m \\equiv 26 \\bmod 42$.\n3. Die Bedingungen aus 1 und 2 ergeben einen Widerspruch modulo 43.\n\nSchritt 1: Es gilt $7^{1} \\equiv 7$, $7^{2} \\equiv 4$ und $7^{3} \\equiv 1 \\bmod 9$, ab dann sind die Reste periodisch mit Periode 3. Aus der Ausgangsgleichung folgt $7^{n} \\equiv 7 \\bmod 9$, also $n \\equiv 1 \\bmod 3$.\n\nSchritt 2: Zeige zunächst $m \\equiv 2 \\bmod 6$ analog zu Schritt 1. Nach dem Satz von Euler-Fermat ist $3^{42} \\equiv 1 \\bmod 49$, die Periode ist also höchstens 42. Unter Verwendung von $3^{6} \\equiv -6 \\bmod 49$ erhält man:\n\n| $m$ | 2 | 8 | 14 | 20 | 26 | 32 | 38 |\n| :--- | ---: | ---: | ---: | ---: | ---: | ---: | ---: |\n| $3^{m} \\bmod 49$ | 9 | 44 | 30 | 16 | 2 | 37 | 23 |\n\nEs gilt also $3^{m} \\equiv 2 \\bmod 49$ genau dann, wenn $m \\equiv 26 \\bmod 42$.\n\nSchritt 3: Schreibe $m=42 m' + 26$ und $n=3 n' + 1$. Betrachtet man nun die ursprüngliche Gleichung modulo 43, erhält man mit dem kleinen Satz von Fermat und $7^{3} \\equiv -1 \\bmod 43$\n\n$$\n3^{42 m' + 26} - 7^{3 n' + 1} \\equiv 3^{26} - (-1)^{n'} \\cdot 7 \\equiv 15 - (-1)^{n'} \\cdot 7 \\not\\equiv 2 \\quad \\bmod 43\n$$\n\nEs gibt also keine weiteren Lösungen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31047,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAuf einem Tisch liegen nebeneinander 2009 Karten in einer Reihe. Zunächst ist bei allen Karten die Oberseite weiß und die Unterseite schwarz. Die Karten seien von $1$ bis $2009$ nummeriert.\n\nZwei Spieler $A$ und $B$ führen abwechselnd einen Spielzug aus, wobei $A$ beginnt. Jeder Spielzug besteht darin, dass der Spieler eine Karte mit der Nummer $k$ ($k < 1969$) wählt, deren weiße Seite oben liegt, und anschließend die Karten mit den Nummern $k, k+1, k+2, \\ldots, k+40$ auf ihren Plätzen umdreht. Der letzte Spieler, der einen gültigen Spielzug machen konnte, gewinnt das Spiel.\n\na) Man entscheide, ob dieses Spiel notwendigerweise endet.\n\nb) Für welchen der beiden Spieler existiert eine Gewinnstrategie?",
    "options": [],
    "answer": "The game necessarily ends; Player B has a winning strategy.",
    "solution": "Solution:\n\na) Wenn das Spiel nicht endet, dann muss wegen der höchstens $2^{2008}$ verschiedenen möglichen Spielzustände eine sich periodisch wiederholende Folge von Zuständen existieren. Dabei sei $k$ die kleinste Nummer der Karten, die innerhalb dieser Periode umgedreht werden. Um die Karte mit der Nummer $k$ von schwarz auf weiß umzudrehen, ist jedoch eine Karte mit einer kleineren Nummer als $k$ auszuwählen – Widerspruch zur Minimalität! Also muss das Spiel nach endlich vielen Zügen enden.\n\nb) Wir betrachten die Karten mit den Nummern $41k$, wobei $1 \\leq k \\leq 48$. Die höchste dieser Karten mit der Nummer $1968$ ist die letzte, die ausgewählt werden kann. Bei jedem Spielzug wird genau eine dieser Karten umgedreht. Weil das Spiel endet, wenn alle Karten bis zur Nummer $1968$ schwarz sind, und weil diese Karten zu Beginn alle weiß sind, ist nach jedem Doppelzug von $A$ und $B$ wieder eine gerade Anzahl von ihnen weiß; $B$ findet also stets eine ungerade Anzahl weißer Karten aus dieser Menge vor und kann stets ziehen. Daher gewinnt $B$ zwangsläufig das Spiel, egal wie er zieht.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31048,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben sei ein Sehnenviereck $ABCD$, dessen Diagonalen $AC$ und $BD$ sich im Punkt $E$ schneiden und dessen Seiten $AD$ und $BC$ auf Geraden liegen, die sich im Punkt $F$ schneiden. Die Mittelpunkte der Strecken $AB$ und $CD$ seien mit $G$ bzw. $H$ bezeichnet. Man beweise, dass die Gerade $EF$ in $E$ den Kreis durch $E$, $G$ und $H$ berührt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEine zentrische Streckung mit Zentrum $E$ und Streckfaktor $2$ bildet $G$ auf $G'$ und $H$ auf $H'$ ab. Dann ist\n\n(1) $\\Varangle GHE = \\Varangle G'H'E$.\n\nWegen $\\Varangle ABF = 180^{\\circ} - \\Varangle CBA = \\Varangle ADC$ und $\\Varangle FAB = 180^{\\circ} - \\Varangle BAD = \\Varangle DCB$ (Sehnenviereck $ABCD$) sind die Dreiecke $FBA$ und $FDC$ ähnlich und werden durch eine Streckspiegelung mit Zentrum $F$ an der Winkelhalbierenden von $\\Varangle BFA$ aufeinander abgebildet. Bei dieser Abbildung gilt insbesondere $A \\rightarrow C$, $B \\rightarrow D$, $G \\rightarrow H$ (Mittelpunkte!).\n\nAufgrund der Definition von $H'$ halbieren sich die Diagonalen $EH'$ und $CD$; daher ist $DECH'$ ein Parallelogramm und es gilt $\\Varangle H'CD = \\Varangle EDC = \\Varangle BDC = \\Varangle BAC = \\Varangle BAE$. Analog gilt $\\Varangle CDH' = \\Varangle EBA$. Damit sind die Dreiecke $ABE$ und $CDH'$\n\n![](attached_image_1.png)\n\nähnlich. Wegen der Ähnlichkeit von $FBA$ und $FDC$ sind auch die Vierecke $FBEA$ und $FDH'C$ ähnlich, so dass bei der betrachteten Streckspiegelung $E$ auf $H'$ abgebildet wird. Damit sind die Dreiecke $FGE$ und $FHH'$ ähnlich, und es gilt\n\n(2) $\\Varangle GEF = \\Varangle HH'F = \\Varangle EH'F$.\n\nEs seien $C'$ und $D'$ die Urbilder von $A$ bzw. $B$ bei der betrachteten Streckspiegelung. Dann ist wegen der Ähnlichkeit der Dreiecke $FC'D'$ und $FAB$ sowie wegen $|FB| / |FD'| = |FD| / |FB|$ das Viereck $D'C'BA$ ein zu $BADC$ ähnliches Sehnenviereck. Sein Diagonalenschnittpunkt sei $G''$. Dann gilt $\\Varangle G''AB = \\Varangle C'AB = \\Varangle C'D'B = \\Varangle ABD = \\Varangle ABE$ und es folgt $G''A \\parallel BE$ sowie analog $G''B \\parallel AE$. Also ist $AG''BE$ ein Parallelogramm und nach Definition von $G'$ gilt $G' = G''$. Folglich geht $G'$ bei der betrachteten Streckspiegelung in $E$ über. Weil aber $E$ in $H'$ übergeht, wird $G'$ bei der zweifachen Hintereinanderausführung in $H'$ abgebildet. Bei dieser Hintereinanderausführung heben sich die beiden Spiegelungen auf und es bleibt eine zentrische Streckung an $F$. Deshalb sind $F$, $G'$ und $H'$ kollinear. Daraus folgt\n\n(3) $\\Varangle FH'E = \\Varangle G'H'E$.\n\nAus (1), (2) und (3) folgt nun $\\Varangle GHE = \\Varangle G'H'E = \\Varangle FH'E = \\Varangle FEG$.\n\nAus der Umkehrung des Sehnen-Tangentenwinkelsatzes ergibt sich daher, dass $FE$ Tangente an den Kreis durch $E$, $G$ und $H$ ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWir nennen eine natürliche Zahl $n$ ausgeglichen, wenn $n=1$ gilt oder wenn $n$ als Produkt einer geraden Anzahl von (nicht notwendigerweise verschiedenen) Primfaktoren geschrieben werden kann. Zu jedem Paar $(a, b)$ positiver ganzer Zahlen sei $P(x) = (x+a)(x+b)$.\n\na) Gibt es zwei verschiedene positive ganze Zahlen $a$ und $b$, für die alle Zahlen $P(1), P(2), \\ldots, P(50)$ ausgeglichen sind?\n\nb) Man beweise: Wenn $P(m)$ für alle positiven ganzen Zahlen $m$ ausgeglichen ist, dann gilt $a = b$.",
    "options": [],
    "answer": "a) Yes. b) a = b.",
    "solution": "Solution:\n\na) Die Antwort lautet „Ja\". Damit $P(x)$ ausgeglichen ist, müssen $x+a$ und $x+b$ entweder beide eine gerade oder beide eine ungerade Anzahl von Primfaktoren besitzen. Hinsichtlich dieser Eigenschaft gibt es aber nur $2^{50}$, also endlich viele verschiedene Muster bei 50 aufeinanderfolgenden natürlichen Zahlen. Daher existieren zwei natürliche Zahlen $m$ und $n$, so dass sie Startzahlen zweier identischer solcher Muster sind. Mit $a = m-1$ und $b = n-1$ folgt die Ausgeglichenheit von $P(1), P(2), \\ldots, P(50)$.\n\nb) Wir nehmen an, dass es unter der gegebenen Voraussetzung zwei verschiedene natürliche Zahlen $a$ und $b$ gebe; oBdA sei $b > a$. Dann gilt für jede natürliche Zahl $m > a$, dass $P(m-a) = m(m + b - a)$ ausgeglichen ist. Die Geradzahligkeit bzw. Ungeradzahligkeit der Anzahl der Primfaktoren tritt also für natürliche Zahlen größer als $a$ mit der Periodenlänge $b-a$ periodisch auf. Insbesondere sind auch alle Vielfachen von $b-a$ vom gleichen Typ, sobald sie größer sind als $a$. Ein solches Vielfaches $k(b-a)$ hat jedoch einen Primfaktor weniger als $2k(b-a)$, Widerspruch zur Annahme. Also gilt $a = b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEine Folge $x_{1}, x_{2}, \\ldots$ ist definiert durch $x_{1}=1$ und $x_{2k}=-x_{k}$, $x_{2k-1}=(-1)^{k+1} x_{k}$ für alle $k \\geq 1$. Man zeige, dass für alle $n \\geq 1$ gilt: $x_{1}+x_{2}+\\ldots+x_{n} \\geq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir bezeichnen $S_{n}=x_{1}+\\ldots+x_{n}$ und führen einen Induktionsbeweis. Nachrechnen ergibt $x_{1}=1$, $x_{2}=-1$, $x_{3}=1$, $x_{4}=1$ und damit $S_{n} \\geq 0$ für $1 \\leq n \\leq 4$.\n\nNun nehmen wir an, dass $S_{k} \\geq 0$ für alle natürlichen $k<n$ gilt, und behaupten $S_{n} \\geq 0$.\n\nDazu unterscheiden wir vier Fälle.\n\nFall 1: $n=4m$. Es gilt\n$$\nS_{n}=\\sum_{i=1}^{m}\\left(x_{4i-3}+x_{4i-2}+x_{4i-1}+x_{4i}\\right)=\\sum_{i=1}^{m}\\left(x_{2i-1}-x_{2i-1}-x_{2i}-x_{2i}\\right)\n$$\n$$\n=-2 \\sum_{i=1}^{m} x_{2i}=2 \\sum_{i=1}^{m} x_{i}=2 S_{m} \\geq 0.\n$$\n\nFall 2: $n=4m+1$. Es gilt $S_{n}=2 S_{m}+x_{4m+1}=2 S_{m}+x_{2m+1}=2 S_{m}+(-1)^{m} x_{m+1}$.\n\nFall 2.1: $m$ ist ungerade. Dann ist auch $S_{m}$ ungerade, also $2 S_{m} \\geq 2$ und $S_{n} \\geq 1>0$.\n\nFall 2.2: $m$ ist gerade. Es gilt $S_{n}=2 S_{m}+x_{m+1}=S_{m}+S_{m+1} \\geq 0$.\n\nFall 3: $n=4m+2$. Es gilt $S_{n}=S_{n-2}+x_{4m+1}+x_{4m+2}=S_{n-2}+x_{2m+1}-x_{2m+1}=S_{n-2} \\geq 0$.\n\nFall 4: $n=4m+3$. Es gilt $S_{n}=S_{n+1}-x_{n+1}=2 S_{m+1}+x_{2m+2}=2 S_{m+1}-x_{m+1}=S_{m}+S_{m+1} \\geq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nZwei Kreise $\\Gamma$ und $\\Gamma'$ mögen sich in den beiden voneinander verschiedenen Punkten $A$ und $B$ schneiden. Eine Gerade durch $B$ schneide $\\Gamma$ und $\\Gamma'$ so in $C$ bzw. $D$, dass $B$ zwischen $C$ und $D$ liege. Eine weitere Gerade durch $B$ schneide $\\Gamma$ und $\\Gamma'$ derart in $E$ bzw. $F$, dass $E$ zwischen $B$ und $F$ liege. Es möge sich ergeben, dass $|CD| = |EF|$ gilt. Das Innere der Strecke $CF$ treffe $\\Gamma$ und $\\Gamma'$ in $P$ bzw. $Q$. Weiterhin seien $M$ und $N$ die Mittelpunkte der $C$ bzw. $F$ nicht enthaltenden Bögen $PB$ bzw. $BQ$ von $\\Gamma$ und $\\Gamma'$. Man beweise, dass $CNMF$ ein Sehnenviereck ist.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n(1) Die Dreiecke $A C D$ und $A E F$ sind gleichsinnig kongruent: Wir arbeiten mit orientierten Winkeln modulo $\\pi$. Es ist $\\varangle A D C = \\varangle A D B = \\varangle A F B = \\varangle A F E$ und $\\varangle D C A = \\varangle B C A = \\varangle B E A = \\varangle F E A$ (für das mittlere Gleichheitszeichen wurde beide Male der Satz vom Umfangswinkel benutzt). Da außerdem nach Voraussetzung $|C D| = |E F|$ gilt, folgt die Behauptung mit dem Kongruenzsatz wsw.\n\n(2) $A$ liegt auf derselben Seite der Geraden $C D$ wie $F$: Der Überlappungsbereich der beiden Kreise enthält die Strecke $B E$ und trifft die Gerade $C D$ nur im Punkt $B$.\n\n(3) Die Dreiecke $C D Q$ und $E F P$ sind gleichsinnig kongruent: Wie bei (1), $\\varangle Q D C = \\varangle Q D B = \\varangle Q F B = \\varangle P F B$ usw.\n\n(4) Es gibt eine Drehung um $A$, die die Punkte $C, D, Q$ in die Punkte $E, F, P$ überführt: folgt aus (1) und (3).\n\n(5) $A$ liegt auf derselben Seite der Geraden $B F$ wie $C$: Aus (2) folgt, dass $A$ und $Q$ auf derselben Seite der Geraden $C D$ liegen, und zusammen mit (4) ergibt sich, dass $A$ und $P$ auf derselben Seite der Geraden $E F = B F$ liegen, also auch $A$ und $C$.\n\n(6) $B A$ ist die Winkelhalbierende des Innenwinkels des Dreiecks $B F C$ bei $B$: Wegen (1) sind die Abstände von $A$ zu den verlängerten Seiten gleich, und wegen (2) und (5) handelt es sich um eine innere Winkelhalbierende.\n\n(7) $C M$ ist die Winkelhalbierende des Innenwinkels des Dreiecks $B F C$ bei $C$: Das folgt aus dem Umfangswinkelsatz.\n\n(8) Analog ist $F N$ die Winkelhalbierende des Innenwinkels des Dreiecks $B F C$ bei $F$. Es sei $I$ der Inkreismittelpunkt des Dreiecks $B F C$.\n\n(9) $M$ liegt auf dem $C$ nicht enthaltenden Kreisbogen $B A$ von $\\Gamma$: Aus (2) und (5) folgt, dass die Punkte auf $\\Gamma$ die zyklische Reihenfolge $B, E, A, C$ bzw. $B, P, C$ (gleich orientiert) haben. Die orientierten Bögen $E A$ und $A C$ sind nach (4) gleich groß, also ist $\\frac{1}{2} B C < B A$ und damit $B M = \\frac{1}{2} B P < \\frac{1}{2} B C < B A$.\n\n(10) $N$ liegt auf dem $F$ nicht enthaltenden Kreisbogen $A B$ von $\\Gamma'$: analog zu (9).\n\n(11) $I$ ist ein innerer Punkt von $C M$ und $F N$: $I$ liegt auf der Geraden $A B$, damit folgt die Behauptung aus (9) bzw. (10).\n\n(12) $I$ ist ein innerer Punkt von $B A$: folgt aus (9) oder (10).\n\n(13) Es gilt $|C I| \\cdot |I M| = |F I| \\cdot |I N|$: Nach dem Sehnensatz in $\\Gamma$ ist $|C I| \\cdot |I M| = |B I| \\cdot |I A|$, nach dem Sehnensatz in $\\Gamma'$ ist $|B I| \\cdot |I A| = |F I| \\cdot |I N|$.\n\n(14) Nach der Umkehrung des Sehnensatzes folgt mit (11) und (13), dass $CNMF$ ein Sehnenviereck ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $n$ eine positive ganze Zahl und $b$ die größte ganze Zahl, die kleiner als $(\\sqrt[3]{28}-3)^{-n}$ ist. Man beweise, dass $b$ nicht durch 6 teilbar sein kann.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDie komplexe Zahl $\\omega=\\frac{-1+\\sqrt{3} i}{2}$ ist bekanntlich eine dritte Einheitswurzel und erfüllt $\\omega^{2}=\\frac{-1-\\sqrt{3} i}{2}$, $\\omega^{3}=1$ und $\\omega^{2}+\\omega+1=0$, insbesondere gilt\n$$\n1+\\omega^{j}+\\omega^{2 j}= \\begin{cases}3, & \\text{ falls } j \\text{ durch drei teilbar ist } \\\\ 0 & \\text{ sonst. }\\end{cases}\n$$\nSetze $r_{k}=\\sqrt[3]{28} \\omega^{k}-3$ für $k=0,1,2$. Nach Definition von $b$ ist $\\left|r_{0}^{-n}-b\\right|<1$; da die Realteile von $\\omega$ und $\\omega^{2}$ negativ sind, gilt $\\left|r_{1}\\right|>1$ und $\\left|r_{2}\\right|>1$. Damit ist\n$$\n\\left|b-\\left(r_{0}^{-n}+r_{1}^{-n}+r_{2}^{-n}\\right)\\right|<\\left|b-r_{0}^{-n}\\right|+\\left|r_{1}^{-n}\\right|+\\left|r_{2}^{-n}\\right|<3 .\n$$\nWegen $\\left(\\sqrt[3]{28} \\omega^{k}\\right)^{3}=28$ ist\n$$\nr_{k}^{-1}=\\frac{1}{\\sqrt[3]{28} \\omega^{k}-3}=\\frac{\\left(\\sqrt[3]{28} \\omega^{k}\\right)^{3}-3^{3}}{\\sqrt[3]{28} \\omega^{k}-3}=\\sqrt[3]{28}^{2} \\omega^{2 k}+3 \\sqrt[3]{28} \\omega^{k}+9\n$$\nErhebt man das Polynom $X^{2}+3 X+9$ in seine $n$-te Potenz, gibt es ganze Zahlen $c_{0}, \\ldots, c_{2 n}$ mit $\\left(X^{2}+3 X+9\\right)^{n}=c_{2 n} X^{2 n}+c_{2 n-1} X^{2 n-1}+\\ldots+c_{0}$, hierbei ist $c_{0}=9^{n}$ ungerade. Durch Einsetzen $X=\\sqrt[3]{28} \\omega^{k}$ folgt $r_{k}^{-n}=\\sum_{j=0}^{2 n} c_{j} \\sqrt[3]{28}^{j} \\omega^{k j}$; daraus ergibt sich mit (1):\n$$\nr_{0}^{-n}+r_{1}^{-n}+r_{2}^{-n}=\\sum_{j=0}^{2 n} c_{j} \\sqrt[3]{28}^{j}\\left(1+\\omega^{j}+\\omega^{2 j}\\right)=3 \\sum_{0 \\leq \\ell \\leq 2 n / 3} c_{3 \\ell} 28^{\\ell}\n$$\nDie Summe ist offenbar ein Vielfaches von 3 und außerdem ungerade, da der Summand $c_{3 \\ell} 28^{\\ell}$ für $\\ell=0$ ungerade, für $\\ell>0$ gerade ist. Wäre $b$ durch 6 teilbar, wäre der Betrag $\\left|b-\\left(r_{0}^{-n}+r_{1}^{-n}+r_{2}^{-n}\\right)\\right|$ mindestens 3 im Widerspruch zu (2).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben sei ein konvexes Fünfeck $A B C D E$ mit den Eigenschaften $B C \\| A E$ und $\\overline{A B}=\\overline{A E}$. Weiter sei $F$ ein Punkt auf der Strecke $A E$, so dass $\\overline{A B}=\\overline{B C}+\\overline{A F}$ sowie $\\Varangle C B A=\\Varangle F D C$ erfüllt ist. Schließlich sei $M$ der Mittelpunkt der Strecke $C F$ und $O$ der Umkreismittelpunkt des Dreiecks $B C D$.\nMan beweise: Wenn $D M \\perp M O$, dann gilt $\\Varangle F D C=2 \\cdot \\Varangle A D B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAus den Bedingungen folgt $\\overline{F E}=\\overline{A E}-\\overline{A F}=\\overline{A B}-(\\overline{A B}-\\overline{B C})=\\overline{B C}$.\nDaher ist $B C E F$ ein Parallelogramm, dessen beide Diagonalen $C F$ und $B E$ einander in $M$ halbieren.\n\nWegen der Punktsymmetrie an $M$ gilt $\\Varangle C B E=\\Varangle A E B$, und da $A B E$ ein gleichschenkliges Dreieck ist, gilt $\\Varangle A E B=\\Varangle E B A$. Also ist $E B$ die Winkelhalbierende von $\\Varangle C B A$. Es folgt $\\Varangle F D C=\\Varangle C B A=2 \\cdot \\Varangle A E B$, so dass nur noch $\\Varangle A E B=\\Varangle A D B$ zu zeigen ist.\n\nEs sei $D^{\\prime}$ der Spiegelpunkt von $D$ an $M$. Wegen $D M \\perp M O$ ist dann\n![](attached_image_1.png)\n$\\overline{O D^{\\prime}}=\\overline{O D}=\\overline{O B}=\\overline{O C}$ und $D^{\\prime} B C D$ ist ein Sehnenviereck. Deshalb ist $\\Varangle B D C=\\Varangle B D^{\\prime} C$, und weil $B C D E F D^{\\prime}$ sogar ein zu $M$ punktsymmetrisches Sechseck ist, folgt weiter $\\Varangle E D F=\\Varangle B D^{\\prime} C=\\Varangle B D C$. Also ist $\\Varangle E D B=\\Varangle F D C$, während andererseits $\\Varangle B A E=180^{\\circ}-2 \\cdot \\Varangle A E B=180^{\\circ}-\\Varangle F D C$ gilt. Somit ist $A B D E$ ein Sehnenviereck, und die Behauptung folgt aus dem Umfangswinkelsatz.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs bezeichne $\\mathbb{Q}^{+}$ die Menge der positiven rationalen Zahlen. Eine Funktion $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{Q}^{+}$ heiße elastisch, wenn für alle $x, y \\in \\mathbb{Q}^{+}$ die Ungleichung\n$$\nf(x)+f(y) \\geq 4 f(x+y)\n$$\ngilt.\n\na. Man zeige: Ist $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{Q}^{+}$ elastisch und sind $x, y, z$ positive rationale Zahlen, so ist $f(x)+f(y)+f(z) \\geq 8 f(x+y+z)$.\n\nb. Gibt es eine elastische Funktion $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{Q}^{+}$ zusammen mit positiven rationalen Zahlen $x, y, z$, für die $f(x)+f(y)+f(z)<9 f(x+y+z)$ der Fall ist?",
    "options": [],
    "answer": "yes",
    "solution": "Solution:\n\na.\nEs gilt\n$$\n4 f(x)+f(y)+f(z) \\stackrel{(*)}{\\geq} 4(f(x)+f(y+z)) \\stackrel{(*)}{\\geq} 16 f(x+y+z) .\n$$\nAddiert man dazu die zyklisch vertauschten Versionen $f(x)+4 f(y)+f(z) \\geq \\ldots$ und $f(x)+f(y)+4 f(z) \\geq \\ldots$, erhält man $6(f(x)+f(y)+f(z)) \\geq 48 f(x+y+z)$ und nach Division durch 6 die Behauptung.\n\nb.\nJa, es gibt solche $f, x, y, z$. Betrachte die stückweise affin-lineare Funktion $f$ mit den Knickpunkten $\\left(2^{k}, 2^{-k}\\right)$ für $k \\in \\mathbb{Z}$, explizit: Für jedes $k \\in \\mathbb{Z}$ sei $f$ auf dem Intervall $\\left[2^{k}, 2^{k+1}[\\right.$ definiert durch\n$$\nf(x)=-\\frac{x}{2^{2 k+1}}+\\frac{3}{2^{k+1}} \\quad \\text{ für } 2^{k} \\leq x<2^{k+1} .\n$$\nDamit ist $f$ für alle $x \\in \\mathbb{Q}^{+}$ definiert. Für alle $x$ gilt $f(2 x)=\\frac{1}{2} f(x)$, damit wird (*) äquivalent zur Konvexitätsungleichung $\\frac{1}{2}(f(x)+f(y)) \\geq f\\left(\\frac{x+y}{2}\\right)$. Die Funktion $f$ ist konvex, also auch elastisch. Mit $x=y=z=1$ gilt\n$$\nf(x)+f(y)+f(z)=3<9 \\cdot \\frac{3}{8}=9 f(x+y+z) .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDie Ecken und Kanten eines regulären $n$-Ecks seien im Uhrzeigersinn jeweils so von 1 bis $n$ nummeriert, dass die Kante Nr. $i$ auf die Ecke Nr. $i$ folgt ($1 \\leq i \\leq n$).\nNun werden die Ecken mit nichtnegativen ganzen Zahlen $e_{i}$ und die Kanten mit nichtnegativen ganzen Zahlen $k_{i}$ so belegt, dass gilt:\n(1) Das $n$-Tupel $(e_{1}, e_{2}, \\ldots, e_{n})$ ist eine Permutation des $n$-Tupels $(k_{1}, k_{2}, \\ldots, k_{n})$.\n(2) Für jedes $i \\in \\mathbb{N}$ mit $1 \\leq i \\leq n$ gilt $k_{i} = |e_{i+1} - e_{i}|$, wobei $e_{n+1} = e_{1}$ ist.\n\na) Man beweise, dass für alle natürlichen Zahlen $n$, $n \\geq 3$, solche $n$-Tupel existieren, die von $(0, \\ldots, 0)$ verschieden sind.\n\nb) Man bestimme zu jeder positiven natürlichen Zahl $m$ die kleinste natürliche Zahl $n$ mit folgender Eigenschaft: In den $n$-Tupeln $(e_{1}, e_{2}, \\ldots, e_{n})$ und $(k_{1}, k_{2}, \\ldots, k_{n})$ kommen jeweils alle natürlichen Zahlen von $0$ bis $m$ vor.",
    "options": [],
    "answer": "m + 2",
    "solution": "Solution:\n\na) Ein mögliches Beispiel für $n \\geq 3$ ist $e_{1} = e_{2} = 1$, $e_{i} = 0$ für $3 \\leq i \\leq n$. Dies bedingt $k_{2} = k_{n} = 1$ und $k_{i} = 0$ sonst. Offensichtlich sind alle Bedingungen erfüllt.\n\nb) Wir beweisen, dass stets $n = m + 2$ gilt. Der Beweis erfolgt in zwei Schritten.\n\ni) Es gilt $n \\geq m + 2$: Weil die Zahl $0$ in $(k_{1}, k_{2}, \\ldots, k_{n})$ vorkommt, müssen zwei Elemente von $(e_{1}, e_{2}, \\ldots, e_{n})$ gleich sein. Da außerdem jede der $m + 1$ natürlichen Zahlen von $0$ bis $m$ wenigstens einmal vorkommen muss, folgt $n \\geq m + 2$.\n\nii) Wir geben ein Beispiel für $n = m + 2 \\geq 3$: Es sei $(e_{1}, e_{2}, \\ldots, e_{n}) = (0, m, 1, m-1, \\ldots)$. In dem $n$-Tupel werden die Zahlen alternierend bis zur Differenz $0$ absteigend fortgesetzt, so dass $e_{n-1} = e_{n}$ gilt. Damit tritt jede natürliche Zahl von $0$ bis $m$ genau einmal, die Zahl $e_{n}$ genau zweimal auf. Die Beträge der Differenzen treten von $m$ abwärts bis zur $0$ offensichtlich je einmal auf; die Differenz $|e_{n} - 0| = e_{n}$ tritt zweimal auf, weil $e_{n} < m$ ist. Damit sind die Eigenschaften (1) und (2) gezeigt.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 31056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFür neun verschiedene positive ganze Zahlen $d_{1}, d_{2}, \\ldots, d_{9}$ betrachten wir das Polynom $P(n)=\\left(n+d_{1}\\right)\\left(n+d_{2}\\right) \\cdot \\ldots \\cdot\\left(n+d_{9}\\right)$.\nMan zeige, dass eine ganze Zahl $N$ mit folgender Eigenschaft existiert:\nFür alle ganzen Zahlen $n \\geq N$ ist die Zahl $P(n)$ durch eine Primzahl größer als 20 teilbar.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir können der Aufgabenstellung gemäß $N>0$ und $d_{1}<d_{2}<\\ldots<d_{9}$ annehmen. Dann ist $k=d_{9}-d_{1}=\\max _{1 \\leq q<r \\leq 9}\\left(d_{r}-d_{q}\\right)$. Es sei $z \\in \\mathbb{N}$ so gewählt, dass $2^{z}>k$.\n\nBehauptung: Mit $N=2^{z} \\cdot 3^{z} \\cdot 5^{z} \\cdot 7^{z} \\cdot 11^{z} \\cdot 13^{z} \\cdot 17^{z} \\cdot 19^{z}$ ist eine Lösung für $N$ gefunden.\n\nBeweis: Wir nehmen an, dass für $n \\geq N$ keiner der Faktoren von $P(n)$ einen Primfaktor größer als 20 besitzt und betrachten zunächst $n+d_{1}=2^{e_{1}^{(1)}} \\cdot 3^{e_{2}^{(1)}} \\cdot \\ldots \\cdot 19^{e_{8}^{(1)}}$. Wegen $n \\geq N$ muss es dann wenigstens ein $e_{j}^{(1)}$ mit $e_{j}^{(1)} \\geq z$ geben $(1 \\leq j \\leq 8)$. Der zugehörige Primfaktor sei $p$. Somit ist $n+d_{1}$ durch $p^{e_{j}^{(1)}}$ teilbar. Wegen $p^{e_{j}^{(1)}} \\geq 2^{\\ell_{j}^{(1)}} \\geq 2^{z}>k=\\max _{1 \\leq q<r \\leq 9}\\left(d_{r}-d_{q}\\right)$ ist keine weitere der Zahlen $n+d_{i}$ durch $p^{e_{j}^{(1)}}$ teilbar $(2 \\leq i \\leq 9)$, weil sonst $d_{i}-d_{1}$ durch $p_{j}^{e_{j}^{(1)}}$ teilbar wäre. Es ist also $e_{j}^{(1)}=\\max _{1 \\leq i \\leq 9}\\left(e_{j}^{(i)}\\right)$. Betrachten wir nun $n+d_{2}$, so finden wird dort wegen $d_{2}>d_{1}$ wenigstens ein $e_{j}^{(2)}$ mit $e_{j}^{(2)}>e_{j}^{(1)}$ sowie $e_{j}^{(2)} \\geq z$. Daher besitzt auch $n+d_{2}$ ein im Bezug auf alle Faktoren maximales $e_{j}^{(2)}$. Dies gilt entsprechend für die anderen Faktoren $n+d_{i}$.\n\nDa es für die 9 Faktoren des Polynoms jedoch nur 8 Primfaktoren kleiner als 20 gibt, entsteht hier nach dem Schubfachprinzip ein Widerspruch. Daher muss ein weiterer Primfaktor auftreten, durch den $P(n)$ teilbar ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABC$ ein spitzwinkliges Dreieck mit dem Umkreismittelpunkt $O$. Ferner sei $k$ ein Kreis mit folgenden Eigenschaften:\n\n(1) Der Mittelpunkt $K$ von $k$ liegt im Inneren der Seite $BC$.\n\n(2) $k$ berührt $AB$ in $B'$ und $AC$ in $C'$.\n\n(3) $O$ liegt auf dem kürzeren der beiden Bogenstücke $B'C'$ von $k$.\n\nMan beweise: Der Umkreis von $ABC$ und $k$ schneiden einander in zwei verschiedenen Punkten.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDer Punkt $O'$ ist Bildpunkt von $O$ bei der Spiegelung an der Geraden $BC$. $S$ ist der Schnittpunkt von $KO'$ mit dem Umkreis $u$ (siehe Figur).\n\nNach dem Satz vom Mittelpunktswinkel ist $\\Varangle BOC = 2\\alpha$. Weil $C'$ und $B'$ im Inneren der Strecken $AC$ bzw. $AB$ liegen und $O$ wegen der Spitzwinkligkeit innerhalb von $ABC$ liegt, ist $\\Varangle B'OC' > 2\\alpha$. Im Viereck $AB'KC'$ gilt: $360^\\circ = 2 \\cdot 90^\\circ + \\alpha + \\Varangle C'KB'$, da die Winkel in den Berührpunkten $C'$ und $B'$ Rechte sind. Es folgt $\\Varangle C'KB' = 180^\\circ - \\alpha$.\n\nDamit gilt im Sehnenviereck $B'O'C'O$:\n$$\n\\Varangle B'OC' = 180^\\circ - \\Varangle C'O'B' = 180^\\circ - \\frac{1}{2} \\Varangle C'KB'\n$$\n$$\n= \\frac{1}{2}(180^\\circ + \\alpha)\n$$\nSomit ist $\\frac{1}{2}(180^\\circ + \\alpha) > 2\\alpha$ und daher\n\n![](attached_image_1.png)\n\n$\\alpha < 60^\\circ$.\n\nIm Sehnenviereck $ABSC$ folgt daraus $\\Varangle CSB = 180^\\circ - \\alpha > 120^\\circ$.\n\nAufgrund der Achsensymmetrie an $BC$ ist $BO'CO$ ein Drachenviereck und es gilt:\n$\\Varangle CO'B = \\Varangle BOC = 2\\alpha < 120^\\circ$. Also ist $\\Varangle CSB > \\Varangle CO'B$.\n\nDa $S$ auf $u$ liegt, ist $\\Varangle CSB$ ein Umfangswinkel über der Sehne $BC$. Aus der letzten Ungleichung folgt dann, dass $O'$ außerhalb von $u$ liegt. Weil die Spiegelachse $BC$ Durchmesser von $k$ ist, liegt $O'$ auf $k$. Somit gibt es mit $O$ und $O'$ je einen Punkt auf $k$, der innerhalb bzw. außerhalb von $u$ liegt. Daraus folgt die Behauptung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31058,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs seien $f$ und $g$ zwei reelle Funktionen, die für jede reelle Zahl definiert sind. Ferner soll für alle reellen Zahlen $x$ und $y$ die Gleichung gelten:\n$$\ng(f(x+y))=f(x)+(2 x+y) g(y)\n$$\nMan bestimme alle möglichen Paare $(f, g)$.",
    "options": [],
    "answer": "(i) f(x) = 0 and g(x) = 0 for all real x; (ii) f(x) = x^2 + a and g(x) = x for all real x, where a is any real constant.",
    "solution": "Solution:\n\nAus der gegebenen Gleichung (1) erhalten wir für $y=-2x$: $g(f(-x))=f(x)$ (2). Damit wird (1) zu $f(-x-y)=f(x)+(2x+y)g(y)$ (3) und mit $x=0$ zu $f(-y)=f(0)+y g(y)$ (4).\n\nErsetzen von $f$ in (3) gemäß (4) liefert nach Subtraktion von $f(0)$:\n$$(x+y)g(x+y) = (2x+y)g(y) - xg(-x)$$\n(5).\n\nMit $y=0$ ergibt sich für $x \\neq 0$: $g(x)+g(-x)=2g(0)$ (6), was aber auch für $x=0$, also für alle $x \\in \\mathbb{R}$ erfüllt ist.\n\nWir setzen $h(x):=g(x)-g(0)$. Dann gilt $h(0)=0$ sowie nach (6): $h(-x)=-h(x)$ (6*) sowie mit (5) und (6*):\n$$(x+y)h(x+y) = (2x+y)h(y) + x h(x)$$\n(7).\n\nVertauschen von $x$ und $y$ in (7) und Gleichsetzen liefert $x h(y) = y h(x)$ (8). Für $x, y \\neq 0$ folgt daraus $h(x)/x = h(y)/y = \\text{const}$. Also ist $h(x) = c \\cdot x$, was offensichtlich auch für $x=0$ erfüllt ist. Mit $g(0)=b$ ergibt sich $g(x)=c x + b$ für alle $x \\in \\mathbb{R}$.\n\nMit $f(0)=a$ folgt aus (4): $f(x) = a - b x + c x^2$. (2) liefert $c(c x^2 + b x + a) + b = c x^2 - b x + a$, und nach Koeffizientenvergleich muss $c^2 = c$, $b c = -b$ sowie $a c + b = a$ gelten.\n\nFall 1: $c=0$. Es ergibt sich $b=a=0$ und $f(x)=0$, $g(x)=0$, womit (1) erfüllt ist.\n\nFall 2: $c=1$. Dann ist $b=-b$, also $b=0$, und $a$ beliebig. Es ergibt sich $f(x)=x^2 + a$, $g(x)=x$. Auch für dieses Funktionspaar liefert Einsetzen in (1) eine wahre Aussage. Also existieren die beiden Lösungspaare $(f, g) = (0, 0)$ bzw. $(f, g) = (x^2 + a, x)$ mit $a \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan bestimme die kleinste positive Zahl $k$, für die folgendes der Fall ist: Wenn Kain in solcher Weise ganze Zahlen in die Zellen eines $2011 \\times 2011$-Schachbrettes schreibt, dass die 4022 Summen, die man durch Addition aller Zahlen einer Zeile oder Spalte erhalten kann, paarweise übereinstimmen, so ist es für Abel möglich, durch Abänderung der Einträge aus nur $k$ der Zellen zu erreichen, dass diese 4022 Summen paarweise verschieden werden.",
    "options": [],
    "answer": "2681",
    "solution": "Solution:\n\n$k=2681$.\n\nBeweis von $k \\geq 2681$ : Abel muss mindestens 4021 der Summen ändern, o. B. d. A. alle Spaltensummen und alle bis auf höchstens eine Zeilensumme. Um die Spaltensummen zu ändern, muss Abel in jeder Spalte mindestens einen Eintrag abändern. Dies erledige Abel o. B. d. A. zuerst und pausiere dann. Zum Zeitpunkt dieser Pause seien $n$ Zeilen unverändert. Somit muss Abel in $n-1$ dieser Zeilen einen Eintrag abändern, damit sich die Zeilensummen unterscheiden können. Dies erledige Abel o. B. d. A. im Anschluss und pausiere wieder. Für $n \\geq 671$ sind nun bereits mindestens 2681 Einträge abgeändert. Nun sei $n \\leq 670$. Eine Zelle heiße unzureichend, wenn Abel ihren Eintrag geändert hat, aber keinen weiteren Eintrag einer Zelle in derselben Zeile oder Spalte. Da die Zeilen- und Spaltensumme einer unzureichenden Zelle gleich sind, darf am Ende keine Zelle unzureichend sein. Es gibt während Abels erster Pause mindestens $2011 - 2 n$ unzureichende Zellen. Durch jede Änderung bis zur zweiten Pause kann Abel höchstens eine dieser unzureichenden Zellen (durch Ändern eines Eintrags in derselben Spalte) eliminieren, durch jede Änderung nach der zweiten Pause höchstens zwei unzureichende Zellen. Daher sind nach der zweiten Pause mindestens $\\frac{1}{2}(2011-2 n-(n-1))$ Einträge abzuändern, also zusammen mindestens $2011+(n-1)+\\frac{1}{2}(2011-2 n-(n-1))=3016-\\frac{n}{2} \\geq 2681$.\n\nBeweisidee zu $k \\leq 2681$ : Für $1 \\leq m, n \\leq 2011$ sei die Zelle in Zeile $m$ und Spalte $n$ mit $\\langle m ; n\\rangle$ bezeichnet. Abel erhöht den Eintrag von 2681 Zellen $\\langle m ; n\\rangle$ jeweils um $2(m+n)+1$, und zwar von $\\langle 2 ; 1\\rangle$ sowie für $l=1, \\ldots, 670$ von $\\langle 3 l-1 ; 3 l-1\\rangle,\\langle 3 l ; 3 l-1\\rangle,\\langle 3 l+1 ; 3 l\\rangle$ und $\\langle 3 l+1 ; 3 l+1\\rangle$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $\\Gamma$ der Umkreis des bei $C$ gleichschenkligen Dreiecks $A B C$. Im Inneren der Seite $\\overline{B C}$ liege der Punkt $M$. Es gebe einen Punkt $N$ auf dem Strahl $A M$, für den $M$ zwischen $A$ und $N$ liegt und der $|A N|=|A C|$ erfüllt. Der Umkreis des Dreiecks $C M N$ schneide $\\Gamma$ in den beiden verschiedenen Punkten $C$ und $P$. Die Geraden $A B$ und $C P$ mögen sich in einem Punkt $Q$ treffen. Man beweise, dass $\\measuredangle B M Q=\\measuredangle Q M N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDie Idee zur Lösung geht von der Beobachtung aus, dass die Forderung $|A N|=|A C|=|B C|$ unverändert bleibt, wenn man $A$ und $C$ sowie $N$ und $B$ miteinander vertauscht. Dies legt nahe, dass man auch den Umkreis des Dreiecks $A M B$ betrachten sollte. Der zweite Schnittpunkt der Umkreise von $A M B$ und $C M N$ lässt sich schließlich als Inkreismittelpunkt von $A M C$ identifizieren.\n\nEs sei $I$ der Inkreismittelpunkt des Dreiecks $A M C$. Bekanntlich ist $\\measuredangle M I C=\\frac{\\pi}{2}+\\frac{1}{2} \\measuredangle M A C=\\frac{\\pi}{2}+\\frac{1}{2} \\measuredangle N A C$, und weil das Dreieck $A N C$ bei $A$ gleichschenklig ist, ist $\\measuredangle N A C=\\pi-2 \\measuredangle C N A$, also ist $\\measuredangle M I C=\\pi-\\measuredangle C N A=\\pi-\\measuredangle C N M$, und weil $I$ und $N$ bezüglich der Geraden $C M$ in verschiedenen Halbebenen liegen, folgt aus dem Satz vom Sehnenviereck, dass $I M N C$ ein Sehnenviereck ist, d. h., dass $I$ auf dem Umkreis von $C M N$ liegt. Auf Grund der eingangs angesprochenen Symmetrie liegt $I$ auch auf dem Umkreis von $A B M$. Da $I$ als Inkreismittelpunkt ein innerer Punkt des Dreiecks $A M C$ ist, ist insbesondere $I \\neq M$, und damit ist die Gerade $I M$ die Potenzgerade der Umkreise von $A B M$ und $C M N$. Da weiterhin die Gerade $A B$ die Potenzgerade der Umkreise von $A B C$ und $A B M$ und die Gerade $C P$ die Potenzgerade der Umkreise von $A B C$ und $C M N$ ist, schneiden sich die Geraden $I M, A B$ und $C P$ in einem Punkt. Dieser Punkt ist nach Definition $Q$, d.h. $Q$ liegt auf der Winkelhalbierenden $I M$ des Winkels $C M A$, aber diese ist identisch mit der Winkelhalbierenden des Winkels $B M N$, folglich ist $\\measuredangle B M Q=\\measuredangle Q M N$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs seien $a$, $b$ und $c$ drei positive reelle Zahlen mit $a^{2}+b^{2}+c^{2} \\geq 3$. Man beweise, dass\n$$\n\\frac{(a+1)(b+2)}{(b+1)(b+5)}+\\frac{(b+1)(c+2)}{(c+1)(c+5)}+\\frac{(c+1)(a+2)}{(a+1)(a+5)} \\geq \\frac{3}{2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDie Funktion $f:(0, \\infty) \\rightarrow \\mathbb{R}, f(x)=\\frac{x+2}{(x+1)(x+5)}$ ist streng monoton fallend und konvex, denn die erste Ableitung\n$$\nf^{\\prime}(x)=-\\frac{x^{2}+4 x+7}{(x+1)^{2}(x+5)^{2}}\n$$\nist negativ, und die zweite Ableitung\n$$\nf^{\\prime \\prime}(x)=\\frac{2\\left(x^{3}+6 x^{2}+21 x+32\\right)}{(x+1)^{3}(x+5)^{3}}\n$$\nist positiv. Die Ungleichung von Jensen liefert für das mit den Gewichten $a+1, b+1, c+1$ gewichtete Mittel der Argumente $b, c, a$ :\n$$\n\\frac{(a+1) f(b)+(b+1) f(c)+(c+1) f(a)}{(a+1)+(b+1)+(c+1)} \\geq f\\left(\\frac{(a+1) b+(b+1) c+(c+1) a}{(a+1)+(b+1)+(c+1)}\\right)\n$$\nMultipliziert man mit dem Nenner der linken Seite, erhält man\n$$\n\\frac{(a+1)(b+2)}{(b+1)(b+5)}+\\frac{(b+1)(c+2)}{(c+1)(c+5)}+\\frac{(c+1)(a+2)}{(a+1)(a+5)} \\geq(a+b+c+3) f\\left(\\frac{(a+1) b+(b+1) c+(c+1) a}{a+b+c+3}\\right)\n$$\nNun ist\n$$\n\\begin{aligned}\n2((a+1) b+(b+1) c+(c+1) a) & =(a+b+c+3)^{2}-4(a+b+c)-\\left(a^{2}+b^{2}+c^{2}\\right)-9 \\\\\n& \\leq(a+b+c+3)^{2}-4(a+b+c+3)\n\\end{aligned}\n$$\n(im letzten Schritt wurde die Voraussetzung $a^{2}+b^{2}+c^{2} \\geq 3$ verwendet), also\n$$\n\\frac{(a+1) b+(b+1) c+(c+1) a}{a+b+c+3} \\leq \\frac{a+b+c-1}{2}\n$$\nMit der Monotonie von $f$ kann man die rechte Seite von (*) also weiter abschätzen als\n$$\n\\geq(a+b+c+3) f\\left(\\frac{a+b+c-1}{2}\\right)\n$$\nDieser Ausdruck hängt nur noch von $s=a+b+c$ ab:\n$$\n=(s+3) \\cdot \\frac{\\frac{s-1}{2}+2}{\\left(\\frac{s-1}{2}+1\\right)\\left(\\frac{s-1}{2}+5\\right)}=\\frac{2(s+3)^{2}}{(s+1)(s+9)}=2-\\frac{8}{10+s+\\frac{9}{s}} \\geq 2-\\frac{8}{10+6}=\\frac{3}{2}\n$$\n(im vorletzten Schritt wurde die Ungleichung zwischen arithmetischem und geometrischem Mittel in der Form $s+\\frac{9}{s} \\geq 2 \\sqrt{s \\cdot \\frac{9}{s}}=6$ angewendet).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $n$ eine ungerade natürliche Zahl und $x$ und $y$ seien zwei rationale Zahlen mit\n$$\nx^{n}+2 y=y^{n}+2 x\n$$\nMan zeige, dass $x=y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir überlegen uns zuerst\n($\\boxtimes$) Es sei $n \\geqslant 3$ eine ungerade natürliche Zahl. Weiterhin seien $a, b$ und $c$ drei ganze Zahlen mit\n$$\na^{n-1}+a^{n-2} b+\\ldots+b^{n-1}=2 c^{n-1}\n$$\nSodann ist $a=b=c=0$.\nWenn dies bei einem gewissen $n$ falsch wäre, könnten wir zu diesem ein Gegenbeispiel ($a, b, c$) wählen, bei dem der Ausdruck $|a|+|b|+|c|$ seinen kleinstmöglichen positiven Wert annimmt. Wären $a$ und $b$ beide ungerade, so wäre die linke Seite die Summe einer ungeraden Anzahl ungerader Summanden und mithin selbst ungerade, während die rechte Seite gerade ist. Also ist mindestens eine der beiden Zahlen $a$ und $b$ gerade. Ist etwa $a$ gerade, so hat die linke Seite dieselbe Parität wie $b^{n-1}$ und die rechte Seite ist gerade, sodass $b$ ebenfalls gerade ist. Ebenso sieht man, dass $a$ gerade sein muss, falls $b$ als gerade vorausgesetzt wird. Insgesamt sind also $a$ und $b$ beide gerade. Folglich ist die linke Seite durch $2^{n-1}$ und damit insbesondere durch 4 teilbar, womit sich $c^{n-1}$ und demnach auch $c$ als gerade erwiesen hat. Führt man nun drei ganze Zahlen $a', b'$ und $c'$ mit $a=2 a', b=2 b'$ und $c=2 c'$ in die Betrachtung ein, so ist erstens\n$$\na'^{n-1}+a'^{n-2} b'+\\ldots+b'^{n-1}=2 c'^{n-1}\n$$\nund zweitens\n$$\n0<\\left|a'\\right|+\\left|b'\\right|+\\left|c'\\right|<|a|+|b|+|c| .\n$$\nDas Tripel $\\left(a', b', c'\\right)$ widerspricht also der Wahl von $(a, b, c)$ und damit ist $(\\boxtimes)$ bewiesen.\n\nWir wenden und nunmehr der eigentlichen Aufgabe zu. $\\operatorname{Im}$ Fall $n=1$ ist die Behauptung trivial und wir konzentrieren uns von jetzt an nur noch auf den Fall $n \\geqslant 3$. Es seien $x$ und $y$ zwei rationale Zahlen mit $x^{n}+2 x=y^{n}+2 y$. Indem wir alles auf eine Seite bringen und anschließend faktorisieren, erhalten wir\n$$\n(x-y)\\left(x^{n-1}+x^{n-2} y+\\ldots+y^{n-1}-2\\right)=0\n$$\nWenn $x$ und $y$ also verschieden wären, was wir von jetzt an annehmen wollen, müsste\n$$\nx^{n-1}+x^{n-2} y+\\ldots+y^{n-1}=2\n$$\nsein. Da man je zwei rationale Zahlen auf einen gemeinsamen Nenner bringen kann, gibt es drei ganze Zahlen $a, b$ und $c$ mit $c \\neq 0, x=a / c$ und $y=b / c$. Wer diese Brüche in (1) einsetzt und daraufhin mit $c^{n-1}$ multipliziert, erhält\n$$\na^{n-1}+a^{n-2} b+\\ldots+b^{n-1}=2 c^{n-1}\n$$\nAus $(\\boxtimes)$ folgt nun allerdings insbesondere $c=0$. Dieser Widerspruch zeigt, dass $x$ und $y$ doch gleich sein müssen, womit die Aufgabe gelöst ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs seien $m, n \\geqslant 4$ zwei ganze Zahlen. Wir betrachten ein $m \\times n$-Gitterrechtteck, das von $m+1$ horizontalen und $n+1$ vertikalen Strecken gebildet wird. Die Schnittpunkte dieser Strecken heißen Ecken. Es sei $P$ ein von Selbstüberschneidungen freier, geschlossener Weg, der durch jede der $(m-1)(n-1)$ inneren Ecken aber keine der äußeren Ecken geht. Es bezeichne $A$ die Anzahl der inneren Ecken, durch die $P$ geradlinig hindurchgeht, $B$ die Anzahl der Gitterquadrate, von denen $P$ genau zwei Seiten benutzt, die sich zudem gegenüber liegen, und $C$ die Anzahl der Gitterquadrate,\n![](attached_image_1.png)\nvon denen $P$ keine Seite benutzt. Man beweise, dass $A=B-C+m+n-1$.\n(Die Abbildung zeigt eine Situation mit $m=4, n=5, A=4, B=1$ und $C=5$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEs sei $D$ die Anzahl der Gitterquadrate, von denen genau eine Seite zu $P$ gehört, $E$ die Anzahl der Gitterquadrate, von denen genau zwei Seiten zu $P$ gehören, die zudem benachbart sind, und schließlich $F$ die Anzahl der Gitterquadrate von denen genau drei Seiten zu $P$ gehören. Wegen $m, n \\geqslant 4$ ist unter den $m n$ Gitterquadraten keines, dessen gesamter Rand in $P$ enthalten ist und folglich gibt es für den Schnitt eines solchen Randes mit $P$ nur die fünf Möglichkeiten, die in den Definitionen von $B, C, D, E$ und $F$ angesprochen sind. Wir haben also\n$$\nm n=B+C+D+E+F\n$$\nNun wollen wir die Paare ( $Q, S$ ), die aus einem Gitterquadrat $Q$ und einer $P$ angehörigen Seite $S$ von $Q$ bestehen, auf zwei Arten abzählen. Einerseits gibt es zu den Quadraten vom Typ $B, C$, $D, E$ und $F$ jeweils genau 2, 0, 1, 2 und 3 mögliche Wahlen für $S$ und daher beträgt diese Anzahl $2 B+D+2 E+3 F$. Andererseits ist $P$ aus $(m-1)(n-1)$ Einheitsstrecken zusammengesetzt, die alle als $S$ fungieren können und jede derselben hat zwei Seiten auf auf denen jeweils ein Gitterquadrat liegt. Insgesamt erhalten wir also\n$$\n2(m-1)(n-1)=2 B+D+2 E+3 F\n$$\nBei jeder der $(m-1)(n-1)$ inneren Ecken unseres Rechtecks geht $P$ entweder geradeaus hindurch, und dies passiert genau $A$ mal, oder $P$ knickt ab. Wann immer dies geschieht, schließen die beiden zugehörigen Einheitsteilstrecken von $P$ ein Gitterquadrat ein und dieses muss entweder von Typ $E$ oder vom Typ $F$ sein. Umgekehrt gehören zu jedem Quadrat vom Typ $E$ ein und zu jedem Quadrat vom Typ $F$ zwei dieser Abknickpunkte von $P$. Alles in allem gilt also auch\n$$\n(m-1)(n-1)=A+E+2 F\n$$\nAus (2) $-(3)+(1)$ erhalten wir nunmehr\n$$\nm+n-1=A-B-C\n$$\nund dies wiederum impliziert\n$$\nA=B-C+m-n-1\n$$\nDamit ist die Aufgabe gelöst.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABC$ ein spitzwinkliges Dreieck mit Umkreis $\\omega$. Man beweise, dass es einen Punkt $J$ mit der folgenden Eigenschaft gibt: Ist $X$ ein innerer Punkt von $ABC$, treffen die Strahlen $AX$, $BX$ und $CX$ den Kreis $\\omega$ erneut in den Punkten $A_{1}$, $B_{1}$ und $C_{1}$ und liegen die Punkte $A_{2}$, $B_{2}$ und $C_{2}$ symmetrisch zu $A_{1}$, $B_{1}$ und $C_{1}$ bezüglich der Mittelpunkte der Strecken $\\overline{BC}$, $\\overline{CA}$ beziehungsweise $\\overline{AB}$, so liegen die vier Punkte $A_{2}$, $B_{2}$, $C_{2}$ und $J$ auf einem gemeinsamen Kreis.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir zeigen, dass der normalerweise $H$ genannte Höhenschnittpunkt $J$ des Dreiecks $ABC$ die beschriebene Eigenschaft aufweist. Hierzu sei $a$ die durch $A$ gezogene Parallele zu $BC$ und die Geraden $b$ und $c$ seien analog definiert. Keine zwei der drei Geraden $a$, $b$ und $c$ sind parallel und folglich existieren der Schnittpunkt $A'$ von $b$ mit $c$ sowie die beiden analog definierten Punkte $B'$ und $C'$.\n\nDas Viereck $A' C J B$ besitzt bei $B$ und $C$ rechte Winkel und ist daher ein Sehnenviereck. Da die Spiegelung am Mittelpunkt der Strecke $\\overline{BC}$ den Punkt $A$ auf $A'$ abbildet und $B$ mit $C$ vertauscht, führt sie $\\omega$ in den Umkreis des gerade gefundenen Sehnenvierecks über. Demnach liegt $A_{2}$ auf diesem Kreis und $A' A_{2}$ ist zu $AX$ parallel.\n\nDer Schwerpunkt des Dreiecks $ABC$ heiße $S$. Die zentrische Streckung $\\sigma$ mit Zentrum $S$ und Faktor $-2$ bildet $A$ auf $A'$ ab; der Bildpunkt von $X$ heiße $X'$. Sodann sind die beiden Geraden $AX$ und $A' X''$ parallel und daher liegt $X'$ auf der Geraden $A' A_{2}$. Ähnliche Argumente lassen sich auch mit $B$ und $C$ an der Stelle von $A$ ausführen und wir lernen insgesamt: Die drei Geraden $A' A_{2}$, $B' B_{2}$ und $C' C_{2}$ schneiden sich in $X'$.\n\n![](attached_image_1.png)\n\nFalls $X' = J$ fallen auch die Punkte $A_{2}$, $B_{2}$ und $C_{2}$ mit $J$ zusammen und die Behauptung ist trivial. Von nun ab sei daher $X' \\neq J$.\n\nWegen $\\angle A' C J = 90^{\\circ}$ ist die Strecke $\\overline{A' J}$ nach Satz des Thales ein Durchmesser des Umkreises des Vierecks $A' C J B$. Wiederum nach Satz des Thales ist daher $\\angle J A_{2} A' = 90^{\\circ}$ und folglich auch $\\angle J A_{2} X' = 90^{\\circ}$. Der Punkt $A_{2}$ liegt demnach, abermals nach Satz des Thales, auf dem Kreis mit Durchmesser $J X'$. Aus analogen Gründen liegen auch die beiden Punkte $B_{2}$ und $C_{2}$ auf diesem Kreis und insbesondere haben wir nunmehr einen Kreis gefunden, auf dem alle vier der Punkte $A_{2}$, $B_{2}$, $C_{2}$ und $J$ liegen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEine natürliche Zahl $n$ habe die folgende Eigenschaft:\nFür beliebige reelle Zahlen $a_{1}, a_{2}, \\ldots, a_{d}$, die sowohl $a_{1}+a_{2}+\\ldots+a_{d}=2013$ als auch $0 \\leq a_{i} \\leq 1$ für $i=1,2, \\ldots, d$ erfüllen, existiert eine Zerlegung der Menge dieser reeller Zahlen in $n$ paarweise disjunkte Teilmengen (von denen einige leer sein dürfen), so dass die Summe der Zahlen in jeder Teilmenge höchstens 1 beträgt.\nMan bestimme die kleinste Zahl $n$ mit dieser Eigenschaft.",
    "options": [],
    "answer": "4025",
    "solution": "Solution:\n\nDie kleinste Zahl $n$ mit dieser Eigenschaft ist $4025$.\n\nWir zeigen zunächst $n \\geq 4025$. Dazu wählen wir $d=4025$ sowie $a_{1}=\\ldots=a_{4025}=\\frac{2013}{4025}>\\frac{1}{2}$. Dann ist $a_{1}+\\ldots+a_{4025}=2013$ und wegen $a_{i}+a_{j}=\\frac{4026}{4025}>1$ für alle $1 \\leq i \\neq j \\leq 4025$ werden hier $4025$ Teilmengen benötigt.\n\nNun zeigen wir $n \\leq 4025$. Dazu führen wir eine Fallunterscheidung nach $d$ durch.\n\nFür $d \\leq 4025$ erhält jedes $a_{i}$ seine eigene Teilmenge. Damit sind alle Teilmengen, von denen einige leer sein dürfen, disjunkt und haben Elementsummen von höchstens $1$.\n\nFür $d>4025$ müssen zwei Zahlen $a_{x}$ und $a_{y}$ existieren mit $a_{x}+a_{y} \\leq 1$. Andernfalls wäre schon in der Summe $\\left(a_{1}+a_{2}\\right)+\\left(a_{3}+a_{4}\\right)+\\ldots+\\left(a_{4025}+a_{4026}\\right)$ jede Klammer größer als $1$ und die Summe aller $a_{i}$ größer als $2013$, Widerspruch! Somit können wir $a_{x}$ und $a_{y}$ durch $a_{z}=a_{x}+a_{y}$ ersetzen und erhalten eine Menge mit $d-1$ Elementen, die alle Bedingungen erfüllt. Dieser Schritt kann wiederholt werden, bis die Ersetzung eine Menge mit $4025$ Elementen liefert. Für diese und damit auch für die Ausgangsmenge existiert die gewünschte Aufteilung. Damit ist alles gezeigt.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn der Ebene liegen zwei konzentrische Kreise mit den Radien $r_{1}=13$ und $r_{2}=8$.\nEs sei $AB$ ein Durchmesser des größeren Kreises und $BC$ eine seiner Sehnen, die den kleineren Kreis im Punkt $D$ berührt.\nMan berechne die Länge der Strecke $AD$.",
    "options": [],
    "answer": "19",
    "solution": "Solution:\n\nDie beiden möglichen Lagen von $D$ sind symmetrisch zur Geraden $(AB)$, so dass es ausreicht, den Fall zu betrachten, bei dem das Dreieck $ABD$ gegen den Uhrzeigersinn orientiert ist (siehe Figur). Der gemeinsame Mittelpunkt der beiden Kreise sei mit $M$ bezeichnet. Weil der Berührradius $MD$ auf der Tangente $BC$ senkrecht steht, ist $MBD$ rechtwinklig, so dass aus dem Satz des Pythagoras\n\n![](attached_image_1.png)\n\n$|BD|^{2}=|MB|^{2}-|MD|^{2}=r_{1}^{2}-r_{2}^{2}=169-64=105$ folgt.\n\nNach dem Satz des Thales ist $\\Varangle ACB=90^{\\circ}$. Da wegen $\\Varangle ACB=\\Varangle MDB=90^{\\circ}$ und des gemeinsamen Winkels $\\Varangle DBM=\\Varangle CBA$ die Dreiecke $ABC$ und $MBD$ ähnlich sind, und da $|MB|=r_{1}=|MA|$ gilt, ist auch $|DC|=|BD|=\\sqrt{105}$ sowie $|CA|=2 \\cdot|DM|=16$. Damit sind die Längen der Katheten im rechtwinkligen Dreieck $ADC$ bekannt und es folgt $|AD|=\\sqrt{105+16^{2}}=\\sqrt{361}=19$. Die Seite $AD$ hat also die Länge 19.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDie Zentralbank von Sikinien prägt Münzen im Wert von 11 und 12 Kulotnik. Bei einem Einbruch haben 11 sikinische Ganoven einen Tresor geknackt und Münzen im Gesamtwert von 5940 Kulotnik erbeutet. Sie versuchen für eine Weile, die Beute gerecht unter sich aufzuteilen - also so, dass jeder gleich viel erhält - aber es will ihnen nicht gelingen; nach einer Weile behauptet ihr Anführer, sich überlegt zu haben, dass dies tatsächlich nicht möglich ist.\nMan beweise, dass sie keine Münze im Wert von 12 Kulotnik erbeutet haben.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDie Ganoven mögen $a$ Münzen im Wert von 11 Kulotnik und $b$ Münzen im Wert von 12 Kulotnik erbeutet haben. Dabei sind $a$ und $b$ zwei nichtnegative ganze Zahlen mit\n$$\n11 a + 12 b = 5940 = 11 \\cdot 540 = 12 \\cdot 495\n$$\nWir nehmen nun $b > 0$ an und versuchen, eine gerechte Aufteilung der Beute auf die Ganoven zu finden.\nNach (1) ist $11 a = 12 \\cdot (495 - b)$, d.h. $11 a$ ist durch 12 teilbar. Nachdem 11 und 12 teilerfremd sind, ist folglich auch $a$ ein Vielfaches von 12 und demnach können die Ganoven die erbeuteten 11-Kulotnik-Stücke in $\\frac{a}{12}$ Säckchen verteilen, von denen jedes $12 \\cdot 11 = 132$ Kulotnik enthält. Analog impliziert (1) die Gleichung $12 b = 11 \\cdot (540 - a)$ und mithin ist $12 b$ durch 11 teilbar. Wie vorhin zeigt dies, dass auch $b$ durch 11 teilbar ist. Da wir $b > 0$ angenommen haben, kann sich jeder der Ganoven eine Münze im Wert von 12 Kulotnik nehmen, und die Anzahl $b - 11$ der übrigen 12-Kulotnik-Stücke ist immer noch durch 11 teilbar, weshalb sich diese in $\\frac{b}{11} - 1$ Säckchen aufteilen lassen, die ebenfalls je $11 \\cdot 12 = 132$ Kulotnik enthalten.\nWegen\n$$\n5940 - 11 \\cdot 12 = 5808 = 132 \\cdot 44\n$$\nbefindet sich der noch nicht verteilte Rest der Beute nunmehr in $44 = 11 \\cdot 4$ Säckchen zu je 132 Kulotnik. Wenn sich nun jeder der 11 Ganoven 4 dieser Säckchen nimmt, haben sie insgesamt die Beute gerecht geteilt. Doch da ihr Anführer beweisen konnte, dass dies gar nicht möglich ist, muss unsere Annahme $b > 0$ falsch gewesen sein: Anders ausgedrückt lag tatsächlich keine Münze im Wert von 12 Kulotnik im Tresor.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $\\mathbb{Z}^{+}$ die Menge der positiven ganzen Zahlen.\nMan bestimme alle Funktionen $f: \\mathbb{Z}^{+} \\rightarrow \\mathbb{Z}^{+}$ mit der Eigenschaft, dass für alle positiven ganzen Zahlen $m$ und $n$ gilt: $m^{2}+f(n) \\mid m f(m)+n$.",
    "options": [],
    "answer": "f(n) = n for all positive integers n",
    "solution": "Solution:\n\nFür eine beliebige positive ganze Zahl $n$ wählen wir $m$ so, dass $f(n) \\mid m$ gilt. Dann folgt $f(n) \\mid m^{2}+f(n)$ und $f(n) \\mid m f(m)+n$. Weil $m$ durch $f(n)$ teilbar ist, muss auch $n$ durch $f(n)$ teilbar sein. Es gilt also $f(n) \\mid n$, das bedeutet $f(n) \\leq n$ für alle $n \\in \\mathbb{Z}^{+}$.\n\nDaraus folgt insbesondere $f(1) \\mid 1$, also $f(1)=1$.\n\nWir nehmen weiter an, es gäbe ein $m \\in \\mathbb{Z}^{+}$ mit $f(m)<m$. Mit $n=1$ erhalten wir $m^{2}+1 \\mid m f(m)+1$, andererseits gilt dann $0<m f(m)+1<m^{2}+1$, so dass $m^{2}+1$ kein Teiler von $m f(m)+1$ sein kann - Widerspruch! Also gilt $f(m) \\geq m$ und mit dem oberen folgt $f(n)=n$ für alle $n \\in \\mathbb{Z}^{+}$.\n\nEinsetzen bestätigt über $n^{2}+n=n^{2}+f(n) \\mid n f(n)+n=n^{2}+n$, dass diese Funktion tatsächlich die - daher - einzige Lösung ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEine Menge $A$ von ganzen Zahlen heißt zulässig, wenn sie folgende Eigenschaft hat:\nFür $x, y \\in A$ ($x = y$ ist erlaubt) gilt $x^{2} + k x y + y^{2} \\in A$ für jede ganze Zahl $k$.\nMan bestimme alle Paare $m, n$ von Null verschiedener ganzer Zahlen, für welche die einzige zulässige Menge, die sowohl $m$ als auch $n$ enthält, die Menge $\\mathbb{Z}$ aller ganzer Zahlen ist.",
    "options": [],
    "answer": "All pairs (m, n) of nonzero integers with gcd(|m|, |n|) = 1",
    "solution": "Solution:\n\nFür ein Paar $m, n$ mit $\\operatorname{ggT}(|m|,|n|) = d > 1$ ist $m^{2} + k m n + n^{2}$ durch $d$ teilbar, so dass als Menge $A$ auch die Menge aller ganzzahligen Vielfachen von $d$ in Frage kommt, die das Element $1$ nicht enthält und daher von $\\mathbb{Z}$ verschieden ist.\n\nNun betrachten wir Zahlen $m, n$ mit $\\operatorname{ggT}(|m|,|n|) = 1$. Für diese gilt auch $\\operatorname{ggT}(m^{2}, n^{2}) = 1$. Daher folgt aus dem erweiterten Euklidischen Algorithmus oder aus dem kleinen Satz von Fermat, dass es ganze Zahlen $r$ und $s$ gibt mit $r m^{2} + s n^{2} = 1$. Außerdem ist für $x = y = m$ ersichtlich, dass $(2 + k) m^{2} \\in A$, d.h. jedes ganzzahlige Vielfache von $m^{2}$ in $A$ liegt – entsprechend liegt jedes ganzzahlige Vielfache von $n^{2}$ in $A$. Und für $k = 2$ ergibt sich, dass für alle $x, y \\in A$ auch $(x + y)^{2}$ in $A$ liegt. Also liegen $r m^{2}, s n^{2}$ und folglich $(r m^{2} + s n^{2})^{2} = 1^{2} = 1$ in $A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABCD$ ein konvexes Sehnenviereck mit $|AD| = |BD|$. Seine Diagonalen $\\overline{AC}$ und $\\overline{BD}$ mögen sich in $E$ schneiden. Der Mittelpunkt des Inkreises des Dreiecks $BCE$ heiße $I$. Der Umkreis des Dreiecks $BIE$ schneide das Innere der Strecke $\\overline{AE}$ im Punkt $N$.\nMan beweise, dass\n$$\n|AN| \\cdot |NC| = |CD| \\cdot |BN|\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSetzt man $\\varphi = \\Varangle DAB$, so muss auch $\\Varangle ABD = \\varphi$ sein, denn das Dreieck $ABD$ wurde als gleichschenklig vorausgesetzt. Hieraus ergibt sich $\\Varangle BDA = 180^{\\circ} - 2\\varphi$ und mit Hilfe des Peripheriewinkelsatzes folgt $\\Varangle BCE = \\Varangle BCA = 180^{\\circ} - 2\\varphi$.\n\n![](attached_image_1.png)\n\nAls nächstes wollen wir den Winkel $BIE$ durch $\\varphi$ ausdrücken: Da die Geraden $BI$ und $EI$ Winkelhalbierende im Dreieck $BCE$ sind, ist\n$$\n\\Varangle BIE = 180^{\\circ} - (\\Varangle IEB + \\Varangle EBI) = 90^{\\circ} + \\frac{1}{2}\\left(180^{\\circ} - \\Varangle CEB + \\Varangle EBC\\right) = 90^{\\circ} + \\frac{1}{2} \\Varangle BCE,\n$$\nwas mit vorigem verbunden\n$$\n\\Varangle BIE = 90^{\\circ} + \\frac{1}{2}(180^{\\circ} - 2\\varphi) = 180^{\\circ} - \\varphi\n$$\nliefert.\n\nNachdem $BIEN$ ein konvexes Sehnenviereck ist, impliziert dies $\\Varangle CNB = \\varphi$. Da nach Peripheriewinkelsatz über der Sehne $\\overline{AD}$ auch $\\Varangle ACD = \\varphi$ gilt, muss demnach $CD \\parallel BN$ sein.\n\nMan verlängere nun die Strecke $\\overline{BN}$ über $N$ hinaus, bis sie den Umkreis des Vierecks $ABCD$ erneut in $Q$ schneidet. Indem wir den Peripheriewinkelsatz über der Sehne $\\overline{BD}$ anwenden, erhalten wir $\\Varangle DQB = \\varphi$, was wiederum $CN \\parallel QD$ nach sich zieht.\n\nDa die beiden Paare gegenüberliegender Seiten des Vierecks $CNQD$ also jeweils parallel sind, muss es sich bei diesem um ein Parallelogramm handeln. Mithin ist $|CD| = |NQ|$ und sobald man dies in die aus dem Sehnensatz folgende Gleichung\n$$\n|AN| \\cdot |NC| = |BN| \\cdot |NQ|\n$$\neinträgt, hat man die Behauptung bewiesen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31071,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn einem spitzwinkligen Dreieck $ABC$ seien die Innenwinkel $\\alpha, \\beta, \\gamma$ wie üblich bezeichnet. Ferner seien der Schnittpunkt der Winkelhalbierenden von $\\alpha$ und $BC$ mit $D$ sowie der Schnittpunkt der Winkelhalbierenden von $\\beta$ und $AC$ mit $E$ bezeichnet.\nNun wird dem Viereck $ABDE$ eine Raute so einbeschrieben, dass alle Eckpunkte dieser Raute auf verschiedenen Seiten des Vierecks liegen. In dieser Raute seien die nicht-stumpfen Innenwinkel mit $\\varphi$ bezeichnet. Man beweise, dass $\\varphi \\leq \\max (\\alpha, \\beta)$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDie Ecken der Raute werden mit $K\\ (K \\in AE)$, $L\\ (L \\in AB)$, $M\\ (M \\in BD)$ und $N\\ (N \\in DE)$ benannt. Mit $d(X, YZ)$ sei der Abstand eines Punktes $X$ von einer Geraden $YZ$ bezeichnet. Weil $D$ und $E$ auf den jeweiligen Winkelhalbierenden liegen, gilt $d(D, AB) = d(D, AC)$, $d(E, AB) = d(E, BC)$ und $d(D, BC) = d(E, AC) = 0$, woraus $d(D, AC) + d(D, BC) = d(D, AB)$ und $d(E, AC) + d(E, BC) = d(E, AB)$ folgt.\n\nWeil $N$ auf der Strecke $DE$ liegt und weil sich in der Gleichung $d(X, AC) + d(X, BC) = d(X, AB)$, wenn $X$ längs der Strecke $DE$ wandert, alle Terme nur linear ändern, folgt\n\n![](attached_image_1.png)\n\naus den beiden oberen Beziehungen auch für $N$ :\n\n$d(N, AC) + d(N, BC) = d(N, AB) \\qquad (1)$.\n\nMit den Bezeichnungen der Figur gilt $d(N, AC) = s \\cdot \\sin \\mu$ und $d(N, BC) = s \\cdot \\sin v$. Weil die Raute $KLMN$ ganz in einer der Halbebenen bezüglich $AB$ liegt, erhalten wir aus ihrer Parallelogrammeigenschaft\n\n$d(N, AB) = d(N, AB) + d(L, AB) = d(K, AB) + d(M, AB) = s(\\sin \\delta + \\sin \\varepsilon)$.\n\nMit (1) folgt $\\sin \\mu + \\sin v = \\sin \\delta + \\sin \\varepsilon$ \\qquad (2).\n\nAus der Annahme $\\varphi > \\max (\\alpha, \\beta)$ würde wegen $\\mu + \\varphi = \\Varangle CKL = \\alpha + \\delta$ direkt $\\mu = \\alpha - \\varphi + \\delta < \\delta$ und analog $v < \\varepsilon$ folgen. Wegen $KL \\parallel MN$ gilt $\\beta = \\delta + v$, also $\\delta < \\beta < 90^\\circ$. Entsprechend folgt $\\varepsilon < 90^\\circ$. So erhalten wir $\\sin \\mu < \\sin \\delta$ und $\\sin v < \\sin \\varepsilon$, im Widerspruch zu (2). Also gilt $\\varphi \\leq \\max (\\alpha, \\beta)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $n$ eine positive natürliche Zahl. Im Folgenden betrachten wir Paare von Elementen der Menge $\\{1,2, \\ldots, n\\}$, die jeweils kein gemeinsames Element haben.\nMan bestimme mit Beweis die größtmögliche Anzahl solcher Paare, für welche die Summen ihrer Elemente paarweise verschieden und nicht größer als $n$ sind.\n(Zum Beispiel sind $(1 ; 9),(2 ; 7)$ und $(3 ; 5)$ für $n=10$ drei mögliche Paare.)",
    "options": [],
    "answer": "floor((2n-1)/5)",
    "solution": "Solution:\n\nEs sei $k$ die Anzahl möglicher Paare. Dann lässt sich die Summe $S$ der in ihnen vorkommenden $2 k$ Zahlen in zwei Richtungen abschätzen:\n\n$S \\geq 1+2+\\ldots+2 k = k(2 k+1)$ (alle Zahlen sind verschieden) und\n\n$S \\leq n+(n-1)+\\ldots+(n-k+1) = n k - \\frac{1}{2} k(k-1)$ (alle Summen sind verschieden und $\\leq n$).\n\nDies liefert für $n \\geq 3$ wegen $k>0$ nach Division durch $k$ die Ungleichung $2 k+1 \\leq n - \\frac{1}{2}(k-1)$, aus der $k \\leq \\frac{2 n-1}{5}$ folgt. Es gibt also höchstens $\\left\\lfloor\\frac{2 n-1}{5}\\right\\rfloor$ mögliche Paare\n($\\lfloor x\\rfloor$ bezeichnet die größte ganze Zahl $\\leq x$). Für $n<3$ ist offensichtlich $k=0$.\n\nNun geben wir für $n \\geq 3$ eine Konstruktion mit genau $\\left\\lfloor\\frac{2 n-1}{5}\\right\\rfloor$ Paaren an. Dazu sei zunächst $n=5 m+3$, also $k=2 m+1$. Die folgende Tabelle zeigt die Paare und ihre Summen, wobei für kleine $m$ manche Spalten wegen Doppeldeutigkeit der Terme gestrichen werden müssen:\n\n| Paar      | $3 m+1$ | $3 m$ | $\\ldots$ | $2 m+2$ | $4 m+2$ | $4 m+1$ | $\\ldots$ | $3 m+3$ | $3 m+2$ |\n|:---       |:---:    |:---:  |:---:     |:---:    |:---:    |:---:    |:---:     |:---:    |:---:    |\n|           | 2       | 4     | $\\ldots$ | $2 m$   | 1       | 3       | $\\ldots$ | $2 m-1$ | $2 m+1$ |\n| Summe     | $3 m+3$ | $3 m+4$ | $\\ldots$ | $4 m+2$ | $4 m+3$ | $4 m+4$ | $\\ldots$ | $5 m+2$ | $5 m+3$ |\n\nDie $2 m+1$ Paare enthalten alle Zahlen von $1$ bis $4 m+2$; ihre Summen gehen von $3 m+3$ bis $5 m+3$ und sind offensichtlich verschieden und nicht größer als $n$.\n\nDieselbe Konstruktion funktioniert auch für $n=5 m+4$ und $n=5 m+5$ $(m \\geq 0)$, weil hier wieder $k=2 m+1$ ist. Für $n=5 m+2$ ist $k=2 m$ ausreichend. Daher kann hier die letzte Spalte der Tabelle einfach weggelassen werden. Für $n=5 m+1$ ist ebenfalls $k=2 m$ und wir lassen wieder die letzte Spalte der Tabelle weg und vermindern außerdem jede obere Zahl in der ersten Reihe um $1$. So sind wiederum alle Bedingungen erfüllt.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31073,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $a_{1} \\leqslant a_{2} \\leqslant \\ldots$ eine monoton steigende Folge positiver ganzer Zahlen. Eine positive ganze Zahl $n$ heißt verlässlich, wenn es einen positiven ganzzahligen Index $i$ mit $n=\\frac{i}{a_{i}}$ gibt.\n\nMan beweise: Wenn 2013 verlässlich ist, dann ist auch 20 verlässlich.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWenn 2013 verlässlich ist, gibt es einen positiven ganzzahligen Index $i$ mit $i=2013 a_{i} \\geqslant 20 a_{i}$. Insbesondere ist die Menge $S$ aller positiven ganzen Zahlen $s$, für die $s \\geqslant 20 a_{s}$ gilt, nicht leer. Somit enthält $S$ ein kleinstes Element $j$, und dieses erfüllt einerseits\n$$\nj \\geqslant 20 a_{j}\n$$\nund andererseits $(j-1) \\notin S$. Letzteres bedeutet, dass entweder $j=1$ sein muss, oder $j>1$ und $j-1<20 a_{j-1}$. Wegen $a_{j} \\geqslant 1$ folgt aus (2) jedoch $j \\geqslant 20$, was die erste dieser beiden Alternativen ausschließt; demnach muss $j \\leqslant 20 a_{j-1}$ sein und in Verbindung mit (2) erhalten wir die Ungleichungskette\n$$\nj \\geqslant 20 a_{j} \\geqslant 20 a_{j-1} \\geqslant j .\n$$\nDiese kann nicht anders bestehen als dass überall Gleichheit gilt. Wegen $j=20 a_{j}$ bezeugt der Index $j$, dass 20 wie behauptet verlässlich ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31074,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan bestimme die kleinste positive ganze Zahl $n$, für die es ein Polynom\n$$\nP(X)=a_{2 n} X^{2 n}+a_{2 n-1} X^{2 n-1}+\\ldots+a_{1} X+a_{0}\n$$\nmit reellen Koeffizienten gibt, das die beiden folgenden Eigenschaften aufweist:\n- Für $i=0,1, \\ldots, 2 n$ gilt $2014 \\leq a_{i} \\leq 2015$.\n- Es gibt eine reelle Zahl $\\xi$ mit $P(\\xi)=0$.",
    "options": [],
    "answer": "2014",
    "solution": "Solution:\n\nEs ist $n=2014$. Zunächst definiere für eine spezielle Wahl der Koeffizienten $a_{k}$, nämlich 2014 für gerades $k$ und 2015 für ungerades $k$, das Polynom $Q_{n}(X)=2014 X^{2 n}+2015 X^{2 n-1}+2014 X^{2 n-2}+\\ldots+2015 X+2014$. Es ist $Q_{2014}(-1)=0$, die gesuchte Zahl ist also höchstens 2014. Es bleibt zu zeigen, dass $P(X)$ für $n \\leq 2013$ keine reelle Nullstelle hat. Für $x \\geq 0$ ist $P(x) \\geq a_{0}>0$, da die Koeffizienten positiv sind; für $x<0$ ist $P(x) \\geq Q_{n}(x)$, da $x^{k}$ für gerades $k$ positiv ist, also $a_{k} x^{k} \\geq 2014 x^{k}$, und $x^{k}$ für ungerades $k$ negativ ist, also $a_{k} x^{k} \\geq 2015 x^{k}$. Es bleibt zu zeigen, dass $Q_{m}(x)>0$ für $m \\leq 2013$.\n\n1. Beweis: Man rechnet nach, dass für reelles $x$ gilt\n$$\nQ_{m}(x)=1007\\left(\\sum_{\\nu=0}^{m-1}\\left(x^{\\nu}+x^{\\nu+1}\\right)^{2}\\right)+\\frac{1}{2}\\left(\\sum_{\\nu=0}^{m-1}\\left(x^{2 \\nu+1}+1\\right)\\left(x^{2 m-2 \\nu-1}+1\\right)\\right)+\\left(1007-\\frac{m}{2}\\right)\\left(x^{2 m}+1\\right)\n$$\nDie Summanden der ersten Summe sind als Quadrate nicht negativ. Da $x^{2 k+1}+1$ für alle $k \\geq 0$ positiv bzw. 0 bzw. negativ ist für $x>-1$ bzw. $x=-1$ bzw. $x<-1$, sind die Summanden in der zweiten Summe nicht negativ. Damit ist für $x<0$ und $m \\leq 2013$ :\n$$\nP(x) \\geq Q_{m}(x) \\geq\\left(1007-\\frac{m}{2}\\right)\\left(x^{2 m}+1\\right)>\\frac{1}{2}\n$$\n\n2. Beweis (mit Analysis): Definiere für $x<0$ die reelle Funktion $f_{m}(x)=\\left(x^{2}-1\\right) Q_{m}(x)=2014\\left(x^{2 m+2}-1\\right)+2015 x\\left(x^{2 m}-1\\right)$. Die zweite Ableitung $f_{m}^{\\prime \\prime}(x)=2(2 m+1)\\left(2014(m+1) x^{2 m}+2015 m x^{2 m-1}\\right)$ hat die einzige negative Nullstelle $x_{m}=-\\frac{2015 m}{2014(m+1)}>-1$ für $m \\leq 2013$, und es ist\n$$\n\\begin{aligned}\nf_{m}^{\\prime}\\left(x_{m}\\right) & =2014(2 m+2) x_{m}^{2 m+1}+2015(2 m+1) x_{m}^{2 m}-2015= \\\\\n& =\\left(2014(2 m+2) x_{m}+2015(2 m+1)\\right) x_{m}^{2 m}-2015=2015 \\cdot x_{m}^{2 m}-2015<0\n\\end{aligned}\n$$\nFür $x<x_{m}$ bzw. $x_{m}<x<0$ wächst bzw. fällt $f_{m}^{\\prime}$ streng monoton, da die Ableitung $f_{m}^{\\prime \\prime}$ positiv bzw. negativ ist; daher hat $f_{m}^{\\prime}$ für $x<0$ ein globales Maximum bei $x=x_{m}$ und ist negativ für alle $x<0$. Daher fällt $f_{m}$ streng monoton, ist also wegen $f_{m}(-1)=0$ positiv für $x>-1$ und negativ für $x<-1$. Somit ist $Q_{m}(x)$ positiv für $x \\neq-1$; für $x=-1$ ergibt sich direkt $Q_{m}(-1)=2014-m>0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nMan bestimme alle Paare $(x, y)$ ganzer Zahlen, welche die Gleichung\n$$\n\\sqrt[3]{7 x^{2}-13 x y+7 y^{2}} = |x-y| + 1\n$$\nerfüllen.",
    "options": [],
    "answer": "All integer solutions are given by the following parametrizations with m ∈ {0, 1, 2, ...}:\n- (x, y) = (m^3 + 2m^2 − m − 1, m^3 + m^2 − 2m − 1) and its swap (y, x),\n- (x, y) = (−m^3 − m^2 + 2m + 1, −m^3 − 2m^2 + m + 1) and its swap (y, x), with the understanding that for m = 1 the second family coincides with the first (so it may be omitted there).",
    "solution": "Solution:\nDie Gleichung (1) ist symmetrisch in $x$ und $y$, so dass wir zunächst $x \\geq y$ annehmen können und für $x \\neq y$ zu jeder Lösung $(x, y)$ auch $(y, x)$ als Lösung erhalten. Mit\n$d = x - y \\geq 0$ folgt $\\sqrt[3]{7 d^{2} + x y} = d + 1$. Potenzieren liefert $x^{2} - d x + \\left(-d^{3} + 4 d^{2} - 3 d - 1\\right) = 0$ mit der Diskriminante $D = d^{2} - 4\\left(-d^{3} + 4 d^{2} - 3 d - 1\\right) = (d-2)^{2}(4 d + 1) \\geq 0$. Es ist also $x_{1/2} = \\frac{d \\pm (d-2) \\sqrt{4 d + 1}}{2}$. Für die Ganzzahligkeit von $x$ muss $4 d + 1$ eine Quadratzahl sein, und zwar das Quadrat einer ungeraden Zahl. Der Ansatz $4 d + 1 = (2 m + 1)^{2}$ mit $m \\in \\{0; 1; 2; \\ldots\\}$ liefert $d = m^{2} + m$, so dass sich $x_{1/2} = \\frac{1}{2}\\left[\\left(m^{2} + m\\right) \\pm \\left(m^{2} + m - 2\\right)(2 m + 1)\\right]$ ergibt. Dies führt zu $\\left(x_{1}, y_{1}\\right) = \\left(m^{3} + 2 m^{2} - m - 1, m^{3} + m^{2} - 2 m - 1\\right)$ und für $m \\neq 1$ zu $\\left(x_{2}, y_{2}\\right) = \\left(-m^{3} - m^{2} + 2 m + 1, -m^{3} - 2 m^{2} + m + 1\\right)$ und den Lösungen $\\left(y_{1}, x_{1}\\right)$ bzw. $\\left(y_{2}, x_{2}\\right)$ für $m > 0$. Eine Probe bestätigt, dass diese Paare tatsächlich Lösungen sind.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31076,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEine positive ganze Zahl $n$ heißt neckisch, wenn man sie in der Form $n = a^{b} + b$ mit zwei ganzen Zahlen $a, b \\geq 2$ schreiben kann.\nMan entscheide, ob es 102 aufeinanderfolgende positive ganze Zahlen gibt, von denen genau 100 neckisch sind.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEs gibt solche Zahlen. Für eine positive ganze Zahl $m$ sei $f(m)$ die Anzahl der neckischen Zahlen unter den 102 aufeinander folgenden Zahlen $m, m+1, m+2, \\ldots, m+101$. Es sei $N$ das kleinste gemeinsame Vielfache der Zahlen $2,3,4, \\ldots, 101$. Dann ist $f\\left(2^{N}\\right) \\geq 100$, da für alle $b=2,3, \\ldots, 101$ gilt: $2^{N}+b=\\left(2^{N / b}\\right)^{b}+b$ ist neckisch. Somit gibt es auch eine kleinste positive Zahl $M$ mit $f(M) \\geq 100$. Offenbar ist $M \\leq 2^{N}$. Andererseits gilt $M>1$ (und $M-1$ ist positive ganze Zahl), weil jede neckische Zahl größer als 5 ist, da für $a, b \\geq 2$ gilt: $a^{b}+b \\geq 2^{b}+b \\geq 2^{2}+2=6$, und somit unter den Zahlen von 1 bis 102 höchstens 97 neckisch sind. Nun wird $f(M)=100$ bewiesen: Wäre $f(M)>100$, gäbe es unter den Zahlen von $M$ bis $M+101$ mindestens 101 neckische, also unter denen von $M$ bis $M+100$ mindestens 100 neckische ebenso wie unter denen von $M-1$ bis $M+100$. Dann wäre $f(M-1) \\geq 100$ im Widerspruch zur Minimalität von $M$. Daher ist $f(M)=100$. Die Zahlen von $M$ bis $M+101$ erfüllen also die Bedingung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31077,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABC$ ein spitzwinkliges Dreieck mit dem Umkreis $k$ und dem Inkreismittelpunkt $I$. Die Orthogonale zu $CI$ durch $I$ schneide die Seite $BC$ in $U$ und $k$ in $V$, wobei $V$ und $A$ auf verschiedenen Seiten von $BC$ liegen. Die Parallele zu $AI$ durch $U$ schneide $AV$ im Punkt $X$.\nMan beweise: Wenn die Geraden $XI$ und $AI$ orthogonal zueinander sind, dann schneidet $XI$ die Seite $AC$ in ihrem Mittelpunkt $M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIn der Figur ist $M$ zunächst nur der Schnittpunkt von $XI$ und $AC$. $N$ ist der Schnittpunkt von $XU$ und $AB$ und $Y$ der Schnittpunkt von $XI$ und $AB$. Die halben Innenwinkel des Dreiecks $ABC$ sind mit $\\alpha, \\beta$ bzw. $\\gamma$ bezeichnet. Wegen $\\angle UIC = 90^{\\circ}$ ist $\\angle CUI = \\alpha + \\beta$ und daher $\\angle BNU = \\angle BAI = \\angle BIU = \\alpha$, so dass die Punkte $B, U, I$ und $N$ auf einem Kreis liegen. Deshalb gilt $\\overline{\\Pi U} = \\overline{IN}$ (Sehnen zu $\\beta$) und wegen $X \\perp NU$ folgt $\\overline{NX} = \\overline{XU}$. Anwendung der Strahlensätze führt auf\n$$\n\\frac{\\overline{VX}}{\\overline{VA}} = \\frac{\\overline{XU}}{\\overline{AI}} = \\frac{\\overline{NX}}{\\overline{AI}} = \\frac{\\overline{YX}}{\\overline{YI}}\n$$\nworaus $YV \\parallel AI$ folgt. Also ist auch $\\angle BYV = \\alpha = \\angle BN$, so\n\n![](attached_image_1.png)\n\ndass die Punkte $B, V, I$ und $Y$ auf einem Kreis liegen. Daher ist $\\angle VBI = \\angle VYI = 90^{\\circ}$. Da die Halbierenden von Innen- und Außenwinkel an einer Ecke im Dreieck stets orthogonal sind, ist $BV$ die Halbierende des Außenwinkels bei $B$. Im Sehnenviereck $ABVC$ (Umkreis von $ABC$) erkennen wir daher $\\angle VAC = \\angle VBC = \\alpha + \\gamma$ und $\\angle ACV = 180^{\\circ} - \\angle VBA = 180^{\\circ} - (\\alpha + 2\\beta + \\gamma) = \\alpha + \\gamma$, also $\\angle VAC = \\angle ACV$. Das Dreieck $AVC$ ist somit gleichschenklig mit der Spitze $V$.\n\nUm nun zu zeigen, dass $M$ der Mittelpunkt von $AC$ ist, reicht der Nachweis von $\\angle VMC = 90^{\\circ}$. Dazu verwenden wir in den Sehnenvierecken $BV IY$ und $ABVC$, dass $\\angle VIM = 180^{\\circ} - \\angle YIV = \\angle VBY = \\angle VBA = 180^{\\circ} - \\angle ACV$ ist, so dass auch $VCMI$ ein Sehnenviereck ist. Dies liefert $\\angle VMC = \\angle VIC = 90^{\\circ}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABC$ ein spitzwinkliges Dreieck mit $|AB| \\neq |AC|$. Die Mittelpunkte der Seiten $\\overline{AB}$ und $\\overline{AC}$ seien $D$ beziehungsweise $E$. Die Umkreise der Dreiecke $BCD$ und $BCE$ mögen den Umkreis des Dreiecks $ADE$ in $P$ beziehungsweise $Q$ schneiden, wobei $P \\neq D$ und $Q \\neq E$.\nMan beweise, dass $|AP| = |AQ|$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nOhne Einschränkung sei $|AC| > |AB|$. Da $D$ und $E$ Mitten der Seiten $\\overline{AB}$ und $\\overline{AC}$ sind, ist nach Strahlensatz $DE$ parallel zu $BC$. Im Falle $E = P$ oder $D = Q$ wäre $CEDB$ Sehnenviereck mit parallelen Seiten $BC$ und $DE$, also gleichschenkliges Trapez mit $|BD| = |EC|$, somit $|AB| = |AC|$, was aber ausgeschlossen ist. Kann man zeigen, dass die (nicht ausgearteten) Dreiecke $APC$ und $BQA$ ähnlich sind, folgt die Behauptung, da wegen $\\angle PQA = \\measuredangle PDA = 180^{\\circ} - \\measuredangle BDP = \\measuredangle PCB = \\measuredangle PCA + \\gamma = \\measuredangle BAQ + \\measuredangle AED = \\measuredangle DPQ + \\measuredangle APD = \\measuredangle APQ$ (unter Benutzung des Umfangswinkelsatzes) das Dreieck $APQ$ gleichschenklig ist mit $|AP| = |AQ|$.\n\nBeweis, dass Dreiecke $APC$ und $BQA$ ähnlich sind: Zunächst gilt mit Umfangswinkelsatz\n$$\n\\measuredangle DQB = 360^{\\circ} - \\measuredangle EQD - \\measuredangle BQE = \\measuredangle DAE + \\measuredangle ECB = \\alpha + \\gamma = 180^{\\circ} - \\beta = 180^{\\circ} - \\measuredangle EDA = \\measuredangle APE,\n$$\n$\\measuredangle EPC = \\measuredangle DPC - \\measuredangle DPE = (180^{\\circ} - \\beta) - \\alpha = \\gamma = \\measuredangle AED = \\measuredangle AQD$.\n\nHieraus folgt die Ähnlichkeit der Dreiecke $APC$ und $BQA$:\n\n1. Beweis: Es sei $Q'$ der eindeutig bestimmte Punkt auf derselben Seite von $AC$ wie $P$, so dass Dreieck $AQ'C$ ähnlich zu Dreieck $BQA$ ist. Da $D$ und $E$ Mitten der Seiten $\\overline{AB}$ und $\\overline{AC}$ sind, sind auch die Teildreiecke $AQ'E$ und $BQD$ sowie $CEQ'$ und $ADQ$ ähnlich. Damit liegt $Q'$ wegen (1) und (2), also $\\measuredangle APE = \\measuredangle BQD = \\measuredangle AQ'E$ und $\\measuredangle EPC = \\measuredangle DQA = \\measuredangle EQ'C$, auf den Umkreisen von $AEP$ und $CEP$, die sich in $E$ und $P$ schneiden. $Q' = E$ ist wegen $Q \\neq D$ ausgeschlossen. Damit ist $Q' = P$, und die Dreiecke $APC$ und $BQA$ sind ähnlich.\n\n2. Beweis: Mit Sinussatz in $ADQ$ und $BQD$ gilt $|AQ| : |AD| = \\sin \\measuredangle QDA : \\sin \\measuredangle AQD$, $|BQ| : |BD| = \\sin \\measuredangle BDQ : \\sin \\measuredangle DQB$. Mit $\\measuredangle BDQ = 180^{\\circ} - \\measuredangle QDA$ folgt $|AQ| : |BQ| = \\sin \\measuredangle DQB : \\sin \\measuredangle AQD$. Analog ist $|CP| : |PA| = \\sin \\measuredangle APE : \\sin \\measuredangle EPC$. Wegen (1) und (2) sind $APC$ und $BQA$ ähnlich nach Merkmal sws.\nSolution:\n\nMit Inversion am Kreis: Man invertiere an einem Kreis mit Mittelpunkt $A$ und Radius $1$. Der Bildpunkt eines Punkts $X$ sei mit $X'$ bezeichnet. Da $D$ und $E$ die Mitten der Seiten $\\overline{AB}$ und $\\overline{AC}$ sind, sind $B'$ und $C'$ die Mitten der Seiten $\\overline{AD'}$ und $\\overline{AE'}$. Die Inversion bildet den Umkreis des Dreiecks $ADE$ auf die Gerade $D'E'$ ab, damit sind $P' \\neq D'$ und $Q' \\neq E'$ die zweiten Schnittpunkte der Umkreise der Dreiecke $B'C'D'$ und $B'C'E'$ mit der Geraden $D'E'$. Da die Geraden $D'E'$ und $B'C'$ parallel sind, liegen die Umkreise der Dreiecke $B'C'D'$ und $B'C'E'$ symmetrisch zur Mittelsenkrechten der Strecke $\\overline{B'C'}$, insbesondere gehen die Punkte $Q'$ bzw. $B'$ bzw. $P'$ in $E'$ bzw. $C'$ bzw. $D'$ über. Damit ist $|P'C'| = |B'D'| = |AB'|$ (da $B'$ Mitte von $\\overline{AD'}$) sowie $|B'Q'| = |C'E'| = |AC'|$ (da $C'$ Mitte von $\\overline{AE'}$) und $\\measuredangle Q'B'D' = \\measuredangle P'C'E'$. Daraus folgt $\\measuredangle AB'Q' = 180^{\\circ} - \\measuredangle Q'B'D' = 180^{\\circ} - \\measuredangle P'C'E' = \\measuredangle AC'P'$. Damit sind die Dreiecke $AP'C'$ und $AQ'B'$ nach Kongruenzsatz sws kongruent. Somit ist $|AP'| = |AQ'|$ und $|AP| = 1 / |AP'| = 1 / |AQ'| = |AQ|$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31079,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAus zwei $2 \\times 1$-Dominosteinen kann man ein Tetromino konstruieren, indem man die beiden Dominosteine längs ihrer längeren Seiten so aneinanderlegt, dass der Mittelpunkt der längeren Seite des einen Dominosteins ein Eckpunkt des anderen Dominosteins ist. Dabei ergeben sich zwei hinsichtlich ihrer Orientierung verschiedene Typen von Tetrominos, die wir als $S$- bzw. $Z$-Tetromino bezeichnen wollen.\n\n$S$-Tetrominos:\n![](attached_image_1.png)\n\n$Z$-Tetrominos:\n![](attached_image_2.png)\n\nEin Gitterpunktpolygon $P$ ist eine einfach zusammenhängende Fläche, deren Randlinien nur auf Gitterlinien des ebenen ganzzahligen Koordinatengitters liegen. Eine Pflasterung von $P$ ist eine vollständige und überlappungsfreie Überdeckung von $P$ mit Flächenstücken, die auch nicht teilweise außerhalb von $P$ liegen.\n\nWir nehmen nun an, dass ein Gitterpunktpolygon $P$ nur mit $S$-Tetrominos gepflastert werden kann. Man beweise: Wenn auch eine Pflasterung von $P$ mit $S$- und $Z$-Tetrominos möglich ist, dann ist die Anzahl der dabei verwendeten $Z$-Tetrominos stets gerade.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir können annehmen, dass $P$ aus einem Teil der Einheitsquadrate des ganzzahligen Koordinatengitters besteht, die wie abgebildet eingefärbt sind. Unter dieser Färbung bedeckt jedes $S$-Tetromino eine gerade Anzahl schwarzer Quadrate und jedes $Z$-Tetromino eine ungerade Anzahl von diesen. Da $P$ vollständig mit $S$-Tetrominos gepflastert werden kann, enthält es eine gerade Anzahl schwarzer Quadrate. Wenn also eine Pflasterung mit $S$- und $Z$-Tetrominos möglich ist, erfordert diese gerade Anzahl auch eine gerade Anzahl von $Z$-Tetrominos.\n\n![](attached_image_3.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan bestimme alle positiven ganzen Zahlen $m$ mit folgender Eigenschaft:\nDie Folge $a_{0}, a_{1}, a_{2}, \\ldots$ mit $a_{0}=\\frac{2m+1}{2}$ und $a_{k+1}=a_{k}\\left\\lfloor a_{k}\\right\\rfloor$ für $k=0,1,2, \\ldots$ enthält wenigstens eine ganze Zahl.\n\nHinweis: $\\lfloor x\\rfloor$ bezeichnet den größten ganzen Teil (integer-Funktion) von $x$.",
    "options": [],
    "answer": "All positive integers except 1",
    "solution": "Solution:\n\nEs gilt $a_{0}=m+\\frac{1}{2}$ und $a_{1}=a_{0}\\left\\lfloor a_{0}\\right\\rfloor=\\left(m+\\frac{1}{2}\\right) \\cdot m=m^{2}+\\frac{m}{2}$. Dieser Ausdruck ist für gerades $m$ offensichtlich ganz, so dass hier die gesuchte Eigenschaft der Folge vorliegt.\n\nWeiterhin gilt bei $m=1$, dass $a_{0}=\\frac{3}{2}$, $\\left\\lfloor a_{0}\\right\\rfloor=1$ und für $a_{k}=\\frac{3}{2}$ stets auch $a_{k+1}=\\frac{3}{2} \\cdot 1=\\frac{3}{2}$ ist. Hier gibt es also kein ganzzahliges Folgenglied.\n\nNun sei $m \\geq 3$ ungerade. Es existiert die eindeutige Darstellung $m=2^{p} \\cdot n_{0}+1$ mit $p \\geq 1$, $n_{0}$ ungerade und $p, n_{0} \\in \\mathbb{N}$. Somit gilt $a_{0}=2^{p} n_{0}+\\frac{3}{2}$, und für $a_{k}=2^{p} n_{k}+\\frac{3}{2}$ ($n_{k}$ ungerade) folgt\n$$\na_{k+1}=\\left(2^{p} n_{k}+\\frac{3}{2}\\right)\\left(2^{p} n_{k}+1\\right)=2^{p-1}\\left(2^{p+1} n_{k}^{2}+5 n_{k}\\right)+\\frac{3}{2},\n$$\nwobei die Klammer ungerade ist. Also existiert eine Darstellung $a_{k+1}=2^{p-1} n_{k+1}+\\frac{3}{2}$ mit ungeradem $n_{k+1}$. Durch Erhöhen von $k$ um 1 wird also gleichzeitig $p$ um 1 vermindert, so dass $a_{k+p}=2^{0} n_{k+p}+\\frac{3}{2}$ gilt und $\\left\\lfloor a_{k+p}\\right\\rfloor$ gerade ist. Dann ist $a_{k+p+1}$ eine ganze Zahl, und so liegt auch für jedes $m \\geq 3$ die gesuchte Eigenschaft der Folge vor.\n\nDamit gehören alle positiven ganzen Zahlen außer $1$ zur gesuchten Menge.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31081,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan bestimme alle Funktionen $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$, die für alle $x, y \\in \\mathbb{Z}$ die Gleichung\n$$\nf(x-f(y)) = f(f(x)) - f(y) - 1\n$$\nerfüllen.",
    "options": [],
    "answer": "f(x) = -1 and f(x) = x + 1",
    "solution": "Solution:\n\nDie Gleichung (1) wird genau von den Funktionen $f_1: x \\rightarrow -1$ und $f_2: x \\rightarrow x+1$ erfüllt. Durch Einsetzen wird leicht bestätigt, dass beide Funktionen Lösungen sind.\n\nNun sei $f$ eine Funktion, die (1) für alle $x, y \\in \\mathbb{Z}$ erfüllt. Durch Einsetzen von $x=0$ und $y=f(0)$ erhalten wir mit $z=-f(f(0))$, dass $f(z)=-1$ ist. Einsetzen von $y=z$ in (1) führt auf $f(x+1)=f(f(x))$ für alle $x \\in \\mathbb{Z}$.\n\nDamit vereinfachen wir (1) zu\n$$\nf(x-f(y)) = f(x+1) - f(y) - 1\n$$\n\nMit $y=x$ in (3) und mit (2) ist $f(x+1)-f(x)=f(x-f(x))+1=f(f(x-1-f(x)))+1$.\n\nWeil aus (3) folgt: $f(x-1-f(x))=f(x)-f(x)-1=-1$, vereinfacht sich dies zu $f(x+1)=f(x)+f(-1)+1=f(x)+c$ mit konstantem $c$. Daher ist $f$ linear und genügt dem Ansatz $f(x)=c x+b$ mit $b=f(0)$.\n\nDies in (2) eingesetzt liefert $c x+c+b=c^2 x+c b+b$ für alle $x \\in \\mathbb{Z}$. Für $x=0$ und $x=1$ erhalten wir $c+b=c b+b$ sowie $c^2=c$; daraus folgt $c=0$ oder $c=1$. Aus $c=1$ folgt $b=1$ und wir erhalten $f_2$; aus $c=0$ folgt, dass $f$ konstant ist, und mit (1) ergibt sich $b=-1$, also $f_1$. Damit ist alles gezeigt.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben seien positive ganze Zahlen $k$ und $n$ mit $n>k$. Unter einem Binärwort der Länge $n$ verstehen wir eine Folge aus $n$ Folgengliedern, die alle 0 oder 1 sind. Anja wählt unter allen möglichen Binärwörtern der Länge $n$ eines aus. Dann schreibt sie alle Binärwörter der Länge $n$, die sich von ihrem gewählten Wort an genau $k$ Stellen unterscheiden, an eine Tafel. Anschließend betritt Bernhard den Raum. Anja nennt ihm den Wert von $k$ und danach betrachtet er die Binärwörter an der Tafel. Er versucht nun das zu Beginn von Anja gewählte Binärwort zu erraten. Was ist (in Abhängigkeit von $k$ und $n$ ) die minimale Anzahl an Versuchen, die Bernhard benötigt um das Binärwort mit Sicherheit zu erraten?",
    "options": [],
    "answer": "1 if n ≠ 2k; 2 if n = 2k",
    "solution": "Solution:\n\nDie Antwort lautet: Falls $n \\neq 2 k$ kann Bernhard das von Anja gewählte Wort stets im ersten Versuch erraten, im Fall $n=2 k$ braucht er zwei Versuche um das Wort mit Sicherheit zu erraten. Wir bezeichnen das von Anja zu Beginn gewählte Wort mit $u$.\n\nFall 1: $n \\neq 2 k$. Es sei $1 \\leq i \\leq n$ beliebig. Von allen $\\binom{n}{k}$ Wörtern an der Tafel stimmen genau $\\binom{n-1}{k}$ an der $i$-ten Stelle mit $u$ überein und $\\binom{n-1}{k-1}$ sind an der $i$-ten Stelle verschieden von $u$. Einfaches Nachrechnen zeigt, dass im Fall $n \\neq 2 k$ auch $\\binom{n-1}{k} \\neq \\binom{n-1}{k-1}$ gilt, das heißt Bernhard kann durch das Betrachten der $i$-ten Stellen aller Wörter an der Tafel auf die $i$-te Stelle von $u$ schließen. Da dies für alle $i=1, \\ldots, n$ funktioniert, kann Bernhard das von Anja zu Beginn gewählte Wort $u$ eindeutig bestimmen und errät es auf diese Weise im ersten Versuch.\n\nFall 2: $n=2 k$. Wir zeigen zunächst, dass Bernhard nicht eindeutig auf $u$ schließen kann, also möglicherweise mehr als einen Versuch benötigt. Angenommen, Anja hätte das (eindeutige) Binärwort $\\bar{u}$ gewählt, das sich an allen Stellen von ihrem tatsächlichen Wort $u$ unterscheidet. Ein beliebiges Binärwort der Länge $n$ stimmt genau dann in $k=n / 2$ Stellen mit $u$ überein, wenn es in $k=n / 2$ Stellen mit $\\bar{u}$ übereinstimmt, das heißt Anja hätte in diesem Fall exakt die gleichen Binärwörter an die Tafel geschrieben. Dies bedeutet, dass Bernhard keine Möglichkeit hat zwischen $u$ und $\\bar{u}$ zu unterscheiden und in seinem ersten Versuch Anjas Wort zu raten möglicherweise falsch liegt. Wir zeigen nun, dass Bernhard in zwei Versuchen stets Erfolg haben kann. Dazu genügt es zu beweisen, dass für jedes von $u$ und $\\bar{u}$ verschiedene Binärwort $w$ der Länge $n$ die Menge aller Binärwörter der Länge $n$, die sich an genau $k=n / 2$ Stellen von $w$ unterscheiden, nicht identisch mit der Menge der Wörter an der Tafel ist. Dazu dürfen wir ohne Beschränkung der Allgemeinheit annehmen, dass sich $w$ von $u$ an genau $a$ Stellen unterscheidet, wobei $0<a \\leq k=n / 2$ (sonst betrachte $\\bar{u}$ statt $u$). Dann können wir genau $k$ der übrigen $n-a \\geq k$ Stellen abändern und erhalten ein Wort $w'$, das sich an genau $k$ Stellen von $w$ unterscheidet, aber an $a+k>k$ Stellen von $u$, das heißt es steht nicht an der Tafel.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABCD$ ein konvexes Viereck. Die Diagonale $BD$ halbiere den Winkel $\\measuredangle CBA$. Der Umkreis des Dreiecks $ABC$ schneide die Strecken $\\overline{CD}$ und $\\overline{DA}$ in den inneren Punkten $P$ bzw. $Q$. Die durch den Punkt $D$ verlaufende Parallele zur Geraden $AC$ schneide die Geraden $BA$ und $BC$ in den Punkten $R$ bzw. $S$. Man beweise, dass die vier Punkte $P, Q, R$ und $S$ auf einem gemeinsamen Kreis liegen.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSchritt 1: Wegen $\\measuredangle BRD = \\measuredangle BAC = \\measuredangle BPC = \\pi - \\measuredangle DPB$ ist $BPDR$ ein Sehnenviereck. Analog zeigt man, dass auch $BSDQ$ ein Sehnenviereck ist.\n\nSchritt 2: Es sei $X$ definiert als der Schnittpunkt von $BD$ mit dem Umkreis des Dreiecks $ABC$. Dann gilt $\\measuredangle DPX = \\pi - \\measuredangle XPC = \\measuredangle CBX = 1/2 \\measuredangle CBA$. Da nach Schritt 1 auch $\\measuredangle DPR = \\measuredangle DBR = 1/2 \\measuredangle DBA$ gilt, liegt $X$ auf $PR$. Analog zeigt man, dass $X$ auch auf $QS$ liegt.\n\nSchritt 3: Wir kombinieren die Erkenntnisse und rechnen (mithilfe des Sehnensatzes) $|PX| \\cdot |XR| = |BX| \\cdot |XD| = |QX| \\cdot |XS|$. Mit der Umkehrung des Sehnensatzes folgt nun die Behauptung.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31084,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDie Menge der positiven ganzen Zahlen sei mit $\\mathbb{N}$ bezeichnet. Man bestimme alle Funktionen $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ mit der folgenden Eigenschaft: Für alle positiven ganzen Zahlen $m$ und $n$ ist die Zahl $f(m)+f(n)-m n$ von 0 verschieden und ist ein Teiler der Zahl $m f(m)+n f(n)$.",
    "options": [],
    "answer": "f(k) = k^2 for all k in N",
    "solution": "Solution:\n\nAntwort: Es gibt genau eine Funktion, die die beschriebene Bedingung erfüllt, nämlich $f(k)=k^{2}$ für alle $k$.\n\nZum Beweis sei $f$ wie verlangt.\n\nSchritt 1: Einsetzen von $m=n=1$ liefert $2 f(1)-1 \\mid 2 f(1)$, also auch $2 f(1)-1 \\mid 2 f(1)-(2 f(1)-1)=1$ und damit $2 f(1)-1=1$, also $f(1)=1$.\n\nSchritt 2: Von nun an stehe $p$ stets für eine Primzahl mit $p \\geq 7$. Einsetzen von $m=n=p$ liefert $2 f(p)-p^{2} \\mid 2 p f(p)$ und damit auch $2 f(p)-p^{2} \\mid 2 p f(p)-p\\left(2 f(p)-p^{2}\\right)=p^{3}$, also\n$$\n2 f(p)-p^{2} \\in\\left\\{-p^{3},-p^{2},-p,-1,1, p, p^{2}, p^{3}\\right\\}\n$$\nDa $f(p)>0$ folgt\n$$\nf(p) \\in\\left\\{\\frac{p^{2}-p}{2}, \\frac{p^{2}-1}{2}, \\frac{p^{2}+1}{2}, \\frac{p^{2}+p}{2}, p^{2}, \\frac{p^{3}+p^{2}}{2}\\right\\} .\n$$\nSchritt 3: Wir setzen $m=1, n=p$ und erhalten $f(p)+1-p \\mid p f(p)+1$, also auch $f(p)+1-p \\mid p f(p)+1-p(f(p)+1-p)=p^{2}-p+1$. Angenommen, es gilt $f(p) \\neq p^{2}$. Dann folgt (beachte, dass $p^{2}-p+1$ ungerade ist) notwendigerweise $f(p)+1-p \\leq 1 / 3\\left(p^{2}-p+1\\right)$. Nach Schritt 2 gilt jedoch $f(p) \\geq\\left(p^{2}-p\\right) / 2$, es folgt also\n$$\n\\begin{aligned}\n\\frac{p^{2}-p}{2}+1-p & \\leq \\frac{p^{2}-p+1}{3} \\\\\n3 p^{2}-3 p+6-6 p & \\leq 2 p^{2}-2 p+1 \\\\\np^{2}+5 & \\leq 7 p\n\\end{aligned}\n$$\nwas für $p \\geq 7$ nicht der Fall ist. Also war die obige Annahme falsch und es muss $f(p)=p^{2}$ gelten.\n\nSchritt 4: Es sei $n \\in \\mathbb{N}$ beliebig. Wir setzen $m=p$ und erhalten $f(n)+p^{2}-p n \\mid p^{3}+n f(n)$, also auch $f(n)+p^{2}-p n \\mid p^{3}+n f(n)-n\\left(f(n)-p^{2}-p n\\right)=p\\left(p^{2}-p n+n^{2}\\right)$. Für alle hinreichend großen Primzahlen $p$ ist $f(n)$ und damit auch die linke Seite des letzten Ausdrucks nicht durch $p$ teilbar, daher folgt $f(n)+p^{2}-p n \\mid p^{2}-p n+n^{2}$ und somit auch $f(n)+p^{2}-p n \\mid\\left(f(n)+p^{2}-p n\\right)-\\left(p^{2}-p n+n^{2}\\right)=f(n)-n^{2}$. Da die linke Seite beliebig groß werden kann (es gibt unendlich viele Primzahlen) folgt $f(n)-n^{2}=0$ und damit $f(n)=n^{2}$.\n\nSchritt 5: Die Probe bestätigt dass $f(k)=k^{2}$ für alle $k \\in \\mathbb{N}$ tatsächlich die Bedingung erfüllt: Es gilt $f(m)+f(n)-m n=m^{2}+n^{2}-m n \\geq 2 m n-m n=m n>0$, und außerdem gilt $\\left(m^{2}+n^{2}-m n\\right)(m+n)=m^{3}+n^{3}=m f(m)+n f(n)$, das heißt $f(m)+f(n)-m n$ ist von 0 verschieden und ist ein Teiler der Zahl $m f(m)+n f(n)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEin Rechteck $\\mathcal{R}$ mit ungeraden ganzzahligen Seitenlängen ist in Rechtecke unterteilt, die alle ganzzahlige Seitenlängen haben. Man beweise, dass für mindestens eines dieser Rechtecke die Abstände zu jeder der vier Seiten von $\\mathcal{R}$ alle gerade oder alle ungerade sind.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir unterteilen $\\mathcal{R}$ in Einheitsquadrate und färben einige dieser Einheitsquadrate in rot oder blau ein, entsprechend der folgenden Illustration.\n\n![](attached_image_1.png)\n\nDa $\\mathcal{R}$ laut Voraussetzung ungerade Seitenlängen hat, sind alle vier Eckfelder von $\\mathcal{R}$ blau gefärbt, und es gibt insgesamt mehr gefärbte als ungefärbte Felder. Somit enthält auch mindestens eines der Rechtecke $\\mathcal{R}_{1}, \\ldots, \\mathcal{R}_{k}$, in die $\\mathcal{R}$ unterteilt wurde, mehr gefärbte als ungefärbte Felder, sei $\\mathcal{R}_{i}$ ein solches. Dann hat $\\mathcal{R}_{i}$ ungerade Seitenlängen und alle vier Eckfelder von $\\mathcal{R}_{i}$ sind gefärbt. Daraus folgt, dass alle vier Eckfelder von $\\mathcal{R}_{i}$ dieselbe Farbe tragen müssen. Falls sie blau sind, hat $\\mathcal{R}_{i}$ gerade Abstände zu allen vier Seiten von $\\mathcal{R}$; falls sie rot sind, hat $\\mathcal{R}_{i}$ ungerade Abstände zu allen vier Seiten von $\\mathcal{R}$. In jedem Fall erfüllt $\\mathcal{R}_{i}$ die geforderte Bedingung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs seien $a_{1}, a_{2}, \\ldots, a_{n}, k$ und $M$ positive ganze Zahlen mit den Eigenschaften\n$$\n\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\ldots+\\frac{1}{a_{n}}=k \\quad \\text{ und } \\quad a_{1} a_{2} \\ldots a_{n}=M .\n$$\nMan beweise: Für $M>1$ hat das Polynom\n$$\nP(x)=M(x+1)^{k}-\\left(x+a_{1}\\right)\\left(x+a_{2}\\right) \\ldots\\left(x+a_{n}\\right)\n$$\nkeine positiven Lösungen.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir zeigen $P(x)<0$ für alle $x>0$, also $M(x+1)^{k}<\\left(x+a_{1}\\right) \\ldots\\left(x+a_{n}\\right) \\Leftrightarrow$ $a_{1} a_{2} \\ldots a_{n}(x+1)^{\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\ldots+\\frac{1}{a_{n}}}<\\left(x+a_{1}\\right) \\ldots\\left(x+a_{n}\\right) \\Leftrightarrow \\prod_{i=1}^{n} a_{i}(x+1)^{\\frac{1}{a_{i}}}<\\prod_{i=1}^{n}\\left(x+a_{i}\\right)$.\n\nDazu zeigen wir, dass für jedes $i\\left((1 \\leq i \\leq n)\\right.$ die Beziehung $a_{i}(x+1)^{\\frac{1}{a_{i}}} \\leq x+a_{i}(1)$ gilt und für wenigstens ein $i$ sogar $a_{i}(x+1)^{\\frac{1}{a_{i}}}<x+a_{i}$ ist. Die Behauptung folgt dann durch Multiplikation für alle $i$.\n\nAus der AM-GM-Ungleichung für die Zahlen $x+1,1,1, \\ldots, 1\\left(a_{i}-1\\right.$ Summanden 1$)$ folgt $\\frac{x+a_{i}}{a_{i}} \\geq \\sqrt[a]{x+1}$, was nach Multiplikation mit $a_{i}$ gerade (1) ergibt. Das Gleichheitszeichen gilt genau für $a_{i}=1$, was nicht für alle $i$ sein kann, denn dann wäre $M=1$, im Widerspruch zur Voraussetzung $M>1$. Da für die gegebenen Werte alle Umformungen erlaubt sind, ist alles gezeigt.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan bestimme die kleinste reelle Konstante $C$ mit folgender Eigenschaft:\nFür fünf beliebige positive reelle Zahlen $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$, die nicht unbedingt verschieden sein müssen, lassen sich stets paarweise verschiedene Indizes $i, j, k, l$ finden, so dass\n$$\n\\left|\\frac{a_{i}}{a_{j}}-\\frac{a_{k}}{a_{l}}\\right| \\leq C\n$$\ngilt.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nDer gesuchte Wert ist $C=\\frac{1}{2}$.\n\nZunächst beweisen wir, dass $C \\leq \\frac{1}{2}$ gilt. Dazu nehmen wir oBdA $a_{1} \\leq a_{2} \\leq a_{3} \\leq a_{4} \\leq a_{5}$ an und betrachten die fünf Brüche $\\frac{a_{1}}{a_{2}}, \\frac{a_{3}}{a_{4}}, \\frac{a_{1}}{a_{5}}, \\frac{a_{2}}{a_{3}}, \\frac{a_{4}}{a_{5}}$. Nach dem Schubfachprinzip liegen in einem der Intervalle $]0, \\frac{1}{2}]$ bzw. $\\left.] \\frac{1}{2}, 1\\right]$ drei verschiedene dieser Brüche, wobei zwei von diesen in der Aufzählung direkt aufeinanderfolgen oder der erste und der letzte Bruch dabei sind. Jedenfalls ist die positive Differenz dieser beiden Brüche kleiner als $\\frac{1}{2}$ und die vier beteiligten Indizes sind paarweise verschieden.\n\nNun zeigen wir, dass $C \\geq \\frac{1}{2}$ gilt. Dazu betrachten das Beispiel $1, 2, 2, 2, r$, wobei $r$ eine Riesenzahl sein soll. Mit diesen Zahlen lassen sich - der Größe nach geordnet - die Brüche $\\frac{1}{r}, \\frac{2}{r}, \\frac{1}{2}, \\frac{2}{2}, \\frac{2}{1}, \\frac{r}{2}, \\frac{r}{1}$ bilden, wobei nach der Aufgabenstellung $\\frac{1}{r}$ und $\\frac{2}{r}$ nicht gleichzeitig gewählt werden dürfen. Daher ist die kleinste positive Differenz gleich $\\frac{1}{2}-\\frac{2}{r}$, und dies nähert sich für $r \\rightarrow \\infty$ dem Wert $\\frac{1}{2}$ von unten beliebig gut an. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31088,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $n$ eine positive ganze Zahl, die teilerfremd zu $6$ ist. Wir färben die Ecken eines regulären $n$-Ecks so mit drei Farben, dass für jede Farbe die Anzahl der mit ihr gefärbten Ecken ungerade ist.\nMan beweise, dass es dann stets ein gleichschenkliges Dreieck gibt, dessen Ecken zu den Ecken des $n$-Ecks gehören und alle verschieden gefärbt sind.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEs seien $a_{1}, a_{2}, a_{3}$ die Anzahlen der gleichschenkligen Dreiecke, in deren Eckpunkten genau $1$, $2$ bzw. $3$ Farben vorkommen. Wir nehmen an, dass $a_{3}=0$ gelte. Die Farben seien rot, grün und blau, wobei $r, g$ und $b$ die (ungerade) Anzahl der jeweils so gefärbten Ecken bezeichnet. Wir bestimmen nun auf zwei Arten die Anzahl $a$ der Paare $(\\Delta, v)$, wobei $\\Delta$ ein gleichschenkliges Dreieck mit mehr als einer Eckenfarbe und $v$ eine Seitenkante dieses Dreiecks ist, deren Endpunkte mit verschiedenen Farben belegt sind.\n\nWegen $a_{3}=0$ müssen die Eckpunkte eines solchen Dreiecks mit genau zwei Farben belegt sein, von denen eine zu zwei Ecken gehört, die jeweils Endpunkte einer Seitenkante $v$ sind. Also trägt jedes Dreieck zwei Paare bei und es ist $a=2 a_{2}$.\n\nZu je zwei Eckpunkten $A$ und $B$ gibt genau drei verschiedene Eckpunkte $C$, die mit $A$ und $B$ ein gleichschenkliges Dreieck bilden: entweder ist $AB=AC$ oder $AB=BC$ oder $AC=BC$. Dabei können keine dieser Möglichkeiten zusammenfallen, weil sonst $ABC$ gleichseitig und $n$ durch $3$ teilbar wäre. $AC=BC$ existiert, weil $n$ ungerade ist und daher die Mittelsenkrechte von $AB$ durch genau einen weiteren Eckpunkt verläuft. Daher gilt, ausgehend von zwei verschieden gefärbten Eckpunkten $A$ und $B$, dass $a=3(rg+gb+br)$ ist. Dieser Term ist nach Voraussetzung ungerade, im Widerspruch zu $a=2 a_{2}$. Daher muss $a_{3} \\neq 0$ gelten.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $A B C D E$ ein konvexes Fünfeck mit den Eigenschaften\n$$\n\\overline{A B}=\\overline{B C}=\\overline{C D}, \\angle B A E=\\angle D C B \\text{ und } \\angle E D C=\\angle C B A .\n$$\n\nMan beweise, dass die Lotgerade von $E$ auf $B C$ durch den Schnittpunkt von $A C$ und $B D$ verläuft.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWegen der Voraussetzungen sind $\\triangle A B C$ und $\\triangle B C D$ gleichschenklig. Daher geht die Mittelsenkrechte zu $A C$ durch $B$ und die Mittelsenkrechte zu $B D$ durch $C$. Beide schneiden sich im Punkt $I$ (siehe Figur).\n\nWegen $B D \\perp C I$ und $A C \\perp B I$ schneiden sich $A C$ und $B D$ im Höhenschnittpunkt des Dreiecks $B C I$ und es folgt $I H \\perp B C$. Wenn wir nun zeigen können, dass $E I \\perp B C$ ist, folgt die Behauptung, weil es durch $I$ nur eine Orthogonale zu $B C$ geben kann. Weil $B I$ und $C I$ auch Winkelhalbierende von $\\angle C B A$ bzw. $\\angle D C B$ sind, folgt $\\overline{I A}=\\overline{I C}$ sowie $\\overline{I B}=\\overline{I D}$. Wegen $\\overline{A B}=\\overline{B C}=\\overline{C D}$ sind die Dreiecke $A B I, B C I$ und $C D I$ kongruent. Daher ist $\\angle B A I=\\angle I C B=\\frac{1}{2} \\angle D C B=\\frac{1}{2} \\angle B A E$, so dass $I A$ Winkelhalbierende von $\\angle B A E$ ist. Analog gilt, dass $I D$ Winkelhalbierende von $\\angle E D C$ ist.\n\nDie Verkettung der Achsenspiegelungen an $A I, B I, C I, D I$ und $E I$ ist eine Achsenspiegelung mit den Fixpunkten $E$ und $I$. Daher liegt $I$ auch auf der Winkelhalbierenden von $\\angle A E D$.\n\nNun gilt $\\angle A E D=540^\\circ-2 \\angle C B A-2 \\angle B A E$, und damit folgt im Viereck $A B I E$ :\n$$\n\\angle B I E=360^\\circ-\\angle B A E-\\angle I B A-\\angle A E I=360^\\circ-\\angle B A E-\\frac{1}{2} \\angle C B A-\\left(270^\\circ-\\angle C B A-\\angle B A E\\right)\n$$\n$$\n=90^\\circ+\\frac{1}{2} \\angle C B A=90^\\circ+\\angle C B I.\n$$\nMit dem Außenwinkelsatz folgt $E I \\perp B C$. ㅁ.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei eine positive ganze Zahl $d$ und eine Folge $\\left(a_{i}\\right)_{i=1,2,3, \\ldots}$ positiver ganzer Zahlen gegeben. Wir nehmen an, dass folgende zwei Bedingungen erfüllt sind:\n- Jede positive ganze Zahl taucht genau einmal in der Folge auf.\n- Für alle Indizes $i \\geq 10^{100}$ gilt $\\left|a_{i+1}-a_{i}\\right| \\leq 2 d$.\nMan beweise, dass unendlich viele Indizes $j$ existieren, für die $\\left|a_{j}-j\\right|<d$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAngenommen, die Behauptung wäre nicht erfüllt. Dann existiert ein Index $N>10^{100}$, sodass für alle $i \\geq N$ entweder $a_{i} \\leq i-d$ oder $a_{i} \\geq i+d$ gilt. Im ersten Fall gilt wegen $i \\geq N>10^{100}$ laut der zweiten Voraussetzung an die Folge auch\n$$\na_{i+1} \\leq a_{i}+2 d \\leq i-d+2 d=(i+1)+(d-1)\n$$\nes muss also $a_{i+1} \\leq(i+1)-d$ gelten. Induktiv folgt damit $a_{i} \\leq j-d$ für alle $j \\geq i$.\nDamit haben wir gezeigt:\n(A) Entweder es gilt $a_{i} \\geq i+d$ für alle $i \\geq N$, oder\n(B) es existiert ein $M \\geq N$, sodass $a_{i} \\leq i-d$ für alle $i \\geq M$ gilt.\nNehmen wir zunächst an, (A) sei erfüllt. Die Zahlen $1, \\ldots, N$ müssen laut der ersten Voraussetzung an die Folge alle einmal in der Folge auftauchen. Nun gilt jedoch $a_{i} \\geq i+d>i \\geq N$ für alle $i \\geq N$, das heißt es gibt nur die $N-1$ Folgenglieder $a_{1}, \\ldots, a_{N-1}$, die einen dieser Werte annehmen können. Nach dem Schubfachprinzip ergibt sich ein Widerspruch.\nNehmen wir nun an, (B) sei erfüllt. Es sei $k=\\max \\left\\{M, a_{1}, \\ldots, a_{M}\\right\\}$. Dann sind die $k$ Folgenglieder $a_{1}, \\ldots, a_{k}$ alle kleiner als $k$, denn für $i=1, \\ldots M-1$ gilt $a_{i} \\leq \\max \\left\\{a_{1}, \\ldots, a_{M-1}\\right\\} \\leq k$ und für $i=M, \\ldots, k$ gilt $a_{i} \\leq i-d<i \\leq k$. Diese $k$ Zahlen liegen damit alle in der Menge $\\{1,2, \\ldots, k-1\\}$. Nach dem Schubfachprinzip existieren daher zwei Indizes $1 \\leq i<j \\leq k$ mit $a_{i}=a_{j}$, was er ersten Voraussetzung an die Folge widerspricht.\nDa wir in jedem Fall einen Widerspruch erhalten haben, gilt die Behauptung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABC$ ein Dreieck mit $AB = AC \\neq BC$. Ferner sei $I$ der Inkreismittelpunkt von $ABC$.\nDie Gerade $BI$ schneidet $AC$ im Punkt $D$, und die Orthogonale zu $AC$ durch $D$ schneidet $AI$ im Punkt $E$. Man beweise, dass der Spiegelpunkt von $I$ bei Spiegelung an der Achse $AC$ auf dem Umkreis des Dreiecks $BDE$ liegt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nZunächst beweisen wir (schon mit den zur Aufgabe passenden Bezeichnungen) folgendes Lemma: Die Mittelsenkrechte einer Seite $BII'$ und die Winkelhalbierende durch den dritten Eckpunkt $D$ schneiden sich auf dem Umkreis jedes nicht gleichschenkligen Dreiecks $BI'D$. („Südpolsatz\")\n\nBeweis des Lemmas: Es sei $M$ der Mittelpunkt des Umkreises von $BI'D$ und $S$ der von $D$ verschiedene Schnittpunkt der Winkelhalbierenden mit dem Umkreis. Dann ist $\\square BMI I'$ als Mittelpunktswinkel doppelt so groß wie $\\square BD I'$ und $\\square SD I'$ halb so groß wie $\\square BD I'$. Da aber die Mittelsenkrechte von $BI'$ beide Bögen des Umkreises zu dieser Sehne halbiert, muss sie durch $S$ gehen.\n\nKorollar: Der andere Schnittpunkt der Mittelsenkrechten von $BI'$ mit dem Umkreis sei $E$. Dann liegt nach dem Lemma $D$ auf dem Thaleskreis über $ES$ und es ist $\\square EDS = 90^\\circ$. Somit ist $DE$ die äußere Winkelhalbierende von $\\square BDI'$. Die Umkehrung gilt ebenfalls: Wenn für den Punkt $E$ auf der äußeren Winkelhalbierenden von $\\square BDI'$ auch $BE = EI'$ gilt, dann liegt $E$ auf dem Umkreis von $BI'D$.\n\n![](attached_image_1.png)\n\nFür den Hauptbeweis bezeichnen wir den Spiegelpunkt von $I$ mit $I'$ und den zweiten Schnittpunkt von $AI$ und dem Umkreis des Dreiecks $ABD$ mit $D'$. Weil $AD'$ die Winkelhalbierende von $\\square BAD$ ist, liegt $D'$ in der Mitte des Bogens $BD$ und so gilt $DD' = BD' = CD'$.\n\nUnter Verwendung des Umfangswinkelsatzes über der Sehne $AD$ sowie geeigneter Winkelhalbierender erhalten wir $\\square DD'E = \\square DD'A = \\square DBA = \\square CBI = \\square ICB$. Weil $D'$ wegen der gleichen Abstände der Umkreismittelpunkt des Dreiecks $BCD$ ist, gilt $\\square EDD' = 90^\\circ - \\square D'DC = \\square CBD = \\square CBI$, also sind wegen $\\square CBI = \\square ICB$ die Dreiecke $ED'D$ und $IBC$ ähnlich. Daraus folgt $\\frac{BC}{CI} = \\frac{BC}{CI} = \\frac{DD'}{D'E} = \\frac{BD'}{D'E}$.\n\nAußerdem gilt $\\square I'CB = \\square ACB + \\square I'CA = \\square ACB + \\square ACI = \\square ACB + \\square CBD = \\square BDA = \\square BD'E$, und daher sind die Dreiecke $BCI'$ und $BD'E$ ähnlich, ebenso sind $BCD'$ und $BI'E$ ähnliche Dreiecke, denn bei der Drehstreckung um $B$, die $BCI'$ in $BD'E$ überführt, geht $CD'$ in $I'E$ über.\n\nWeil aber $BCD'$ gleichschenklig ist, gilt auch $BE = EI'$.\n\nNun ist $DE \\perp AC$, und nach dem Korollar liegt $E$ auf dem Umkreis des Dreiecks $BI'D$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31092,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nMan bestimme alle ganzen Zahlen $n \\geq 2$ mit der folgenden Eigenschaft:\nFür beliebige, nicht notwendigerweise verschiedene ganze Zahlen $m_{1}, m_{2}, \\ldots, m_{n}$, deren Summe nicht durch $n$ teilbar ist, existiert ein Index $i$ $(1 \\leq i \\leq n)$, so dass keine der Zahlen\n$$\nm_{i},\\ m_{i}+m_{i+1},\\ m_{i}+m_{i+1}+m_{i+2},\\ \\ldots,\\ m_{i}+m_{i+1}+\\ldots+m_{i+n-1}\n$$\ndurch $n$ teilbar ist. (Dabei sei $m_{i}=m_{i-n}$ für $i>n$.)",
    "options": [],
    "answer": "all prime numbers",
    "solution": "Solution:\n\nBei den gesuchten Zahlen handelt es sich genau um die Primzahlen.\n\nTeilbeweis 1: Keine Nichtprimzahl erfüllt alle Voraussetzungen.\n\nEs sei $n=a \\cdot b$ mit $1<a, b<n$ eine Zerlegung von $n$ in zwei echte Teiler. Wir wählen $m_{i}=a$ für $1 \\leq i<n$ sowie $m_{n}=0$. Dann ist die Summe $m_{1}+m_{2}+\\ldots+m_{n}=(n-1) a$ offensichtlich nicht durch $n$ teilbar, da beide Faktoren nicht durch $n$ teilbar sind.\nNun wählen wir zu einem beliebigen Index $i$ den Index $j=\\left\\{\\begin{array}{ll}b & \\text{ für } 1 \\leq i \\leq n-b \\\\ b+1 & \\text{ für } n-b<i \\leq n\\end{array}\\right.$ und erhalten $m_{i}+m_{i+1}+\\ldots+m_{i+j-1}=a \\cdot b=n \\equiv 0 \\bmod n$. Mit diesem Gegenbeispiel ist der Teilbeweis abgeschlossen.\n\nTeilbeweis 2: Jede Primzahl erfüllt alle Voraussetzungen.\n\nEs sei nun $n$ eine Primzahl. Für einen Widerspruchsbeweis nehmen wir an, dass es für die Zahlen $m_{1}, m_{2}, \\ldots, m_{n}$, deren Summe nicht durch $n$ teilbar ist, zu jedem Index $i$ $(1 \\leq i \\leq n)$ eine Zahl $j$ $(1 \\leq j \\leq n)$ gibt, so dass die Summe $m_{i}+m_{i+1}+\\ldots+m_{i+j-1}$ durch $n$ teilbar ist. Dabei gilt sogar $j \\neq n$, denn die Summe aller $m_{i}$ ist nicht durch $n$ teilbar.\nNun konstruieren wir für $0 \\leq k \\leq n-1$ eine endliche Folge ganzer Zahlen $i_{0}, i_{1}, \\ldots, i_{n}$ mit $i_{k+1}-i_{k} \\leq n-1$ (1), indem wir $m_{i_{k}+1}+m_{i_{k}+2}+\\ldots+m_{i_{k+1}} \\equiv 0 \\bmod n$ wählen. Der Startindex $i_{0}$ ist beliebig und der neue Index $i_{k+1}$ sei der jeweils kleinstmögliche nach $i_{k}$ beim zyklischen Weitergehen $\\bmod n$.\nNach dem Schubfachprinzip gibt es in der Folge dieser $n+1$ Indizes zwei verschiedene Zahlen $i_{r}$ und $i_{s}$ mit $0 \\leq r<s \\leq n$, die $\\bmod n$ kongruent sind. Mit ihnen gilt $\\sum_{j=r}^{s-1}\\left(m_{i_{j}+1}+m_{i_{j}+2}+\\ldots+m_{i_{j+1}}\\right) \\equiv 0 \\bmod n$, weil dies für jede Klammersumme gilt.\nAndererseits folgt aus $i_{s} \\equiv i_{r} \\bmod n$, dass es eine positive ganze Zahl $d$ gibt mit $i_{s}-i_{r}=d \\cdot n$. Wegen (1) ist $i_{s}-i_{r} \\leq(n-1) n$, so dass $d \\leq n-1$ folgt. Dann kann $\\sum_{j=r}^{s-1}\\left(m_{i_{j}+1}+m_{i_{j}+2}+\\ldots+m_{i_{j+1}}\\right)=d\\left(m_{1}+m_{2}+\\ldots+m_{n}\\right)$ aber kein Vielfaches von $n$ sein, denn $n$ ist prim und weder $d$ noch $m_{1}+m_{2}+\\ldots+m_{n}$ sind Vielfache von $n$ - Widerspruch! $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABC$ ein Dreieck. Der dem Punkt $A$ gegenüberliegende Ankreis $\\omega$ berühre die Strecke $\\overline{BC}$ sowie die Strahlen $AC$ und $AB$ in den Punkten $D$, $E$ bzw. $F$. Der Umkreis des Dreiecks $AEF$ schneide die Gerade $BC$ in den Punkten $P$ und $Q$. Schließlich sei $M$ der Mittelpunkt der Strecke $\\overline{AD}$. Man beweise, dass sich $\\omega$ und der Umkreis des Dreiecks $MPQ$ berühren.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nEs sei $J$ der Mittelpunkt des Ankreises $\\omega$. Dann gilt $JE \\perp AE$ und $JF \\perp AF$, der Thaleskreis $\\Omega$ von $\\overline{AJ}$ verläuft also durch $E$ und $F$, und damit auch durch $P$ und $Q$. Der Strahl $AD$ schneide $\\Omega$ und $\\omega$ erneut in $N$ bzw. $T$.\n\n![](attached_image_1.png)\n\nDann ist $JN$ die Lotgerade von $J$ auf $DT$, also ist $N$ der Mittelpunkt der Strecke $\\overline{DT}$. Damit gilt laut dem Sehnensatz\n$$\n|DM| \\cdot |DT| = \\frac{1}{2} \\cdot |DA| \\cdot |DT| = |DA| \\cdot |DN| = |DP| \\cdot |DQ|\n$$\nder Punkt $T$ liegt nach der Umkehrung des Sehnensatzes also auf dem Umkreis des Dreiecks $MPQ$. Wir betrachten nun den Bildpunkt von $N$ bei Inversion am Kreis $\\omega$, dieser sei mit $Z$ bezeichnet. Da die Thaleskreise über $\\overline{JD}$ und $\\overline{JT}$ beide durch den Punkt $N$ verlaufen, liegt $Z$ auf den Bildern dieser Kreise, also den Tangenten durch $D$ und $T$ an $\\omega$. Auf ersterer Geraden liegen auch $B$, $C$, $P$ und $Q$. Ferner liegt $N$ auf dem Kreis $\\Omega$, der bei der betrachteten Inversion in die Gerade $EF$ übergeht, also liegt auch $Z$ auf der Geraden $EF$. Damit gilt nach dem Sekantensatz\n$$\n|ZT|^2 = |ZE| \\cdot |ZF| = |ZP| \\cdot |ZQ|,\n$$\nnach der Umkehrung des Sekantensatzes tangiert $ZT$ daher auch den Umkreis des Dreiecks $MPQ$. Damit haben wir bewiesen, dass sich dieser Kreis und $\\omega$ im Punkt $T$ berühren.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDie Menge aller positiver rationaler Zahlen sei mit $\\mathbb{Q}^{+}$ bezeichnet. Man bestimme alle Funktionen $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{Q}^{+}$ mit der Eigenschaft\n$$\nf\\left(x^{2} f(y)^{2}\\right) = f\\left(x^{2}\\right) f(y) \\tag{*}\n$$\nfür alle $x, y \\in \\mathbb{Q}^{+}$.",
    "options": [],
    "answer": "All solutions are obtained as follows. Let U be any subgroup of the positive rationals under multiplication. Fix representatives a_i of the cosets of the quotient of the positive rationals by U, and choose elements u_i in U, one for each representative. Define f on the positive rationals by:\n- For any square argument x, there is a unique representation x = a_i^2 r^2 with r in U; set f(x) = r u_i.\n- For any non-square argument x, define f(x) arbitrarily as any element of U (possibly depending on x).\nThese and only these functions satisfy f(x squared times f(y) squared) equals f(x squared) times f(y). In particular, the image of f is exactly U. Examples: choosing U = {1} yields the constant function one; choosing U = all positive rationals yields f(x) = a times the square root of x on squares and arbitrary values in the positive rationals on non-squares.",
    "solution": "Solution:\n\nWir beginnen mit einigen für die Lösungsmenge wichtigen Eigenschaften der Grundmenge $\\mathbb{Q}^{+}$. Bekanntlich liegt das Produkt $a b$ zweier Zahlen $a, b \\in \\mathbb{Q}^{+}$ wieder in $\\mathbb{Q}^{+}$, ebenso wie das Inverse $\\frac{1}{a}$ zu jeder Zahl $a \\in \\mathbb{Q}^{+}$ (was für $\\sqrt{a}$ nicht immer erfüllt ist). Weil auch das neutrale Element $1$ der (kommutativen) Multiplikation zu $\\mathbb{Q}^{+}$ gehört, bildet $\\mathbb{Q}^{+}$ bezüglich der Multiplikation eine kommutative (= abelsche) Gruppe. Jede nicht-leere Teilmenge $U$, für die mit $a, b \\in U$ auch stets $\\frac{a}{b} \\in U$ gilt, ist eine (abelsche) Untergruppe von $\\mathbb{Q}^{+}$, d.h. sie enthält die Zahl $1$, jedes Produkt zweier ihrer Elemente und zu jeder ihrer Zahlen den Kehrwert. Ein Beispiel ist $D=\\left\\{3^{t} \\mid t \\in \\mathbb{Z}\\right\\}$, die Menge aller Dreierpotenzen mit ganzzahligen Exponenten; triviale Beispiele sind $U=\\{1\\}$ sowie $U=\\mathbb{Q}^{+}$.\n\nWeiter wird zu einer Untergruppe $U$ von $\\mathbb{Q}^{+}$ und einer beliebigen Zahl $a \\in \\mathbb{Q}^{+}$ die Menge $a U=\\{a u \\mid u \\in U\\}$ als Nebenklasse von $\\mathbb{Q}^{+} / U$ mit dem Repräsentanten $a$ bezeichnet. Der Repräsentant einer Nebenklasse ist in der Regel nicht eindeutig bestimmt; ebenso sind die Nebenklassen in der Regel selber keine Gruppen. Ein Beispiel mit $a=2$ ist die Menge $2 D=\\left\\{2 \\cdot 3^{t} \\mid t \\in \\mathbb{Z}\\right\\}$, die z.B. auch den Repräsentanten $6$ besitzt.\n\nNach diesen Vorbemerkungen können wir nun zunächst die Lösungsmenge beschreiben:\n\nEs sei $U$ eine beliebige Untergruppe von $\\mathbb{Q}^{+}$ und $a_{1}, a_{2}, \\ldots \\in \\mathbb{Q}^{+}$ feste Repräsentanten der Nebenklassen von $\\mathbb{Q}^{+} / U$. Ferner seien $u_{1}, u_{2}, \\ldots$ Zahlen aus $U$, so dass jedem $a_{i}$ genau ein $u_{i}$ zugeordnet ist. Damit ergeben sich folgende Lösungen $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{Q}^{+}$:\n$$\nf(x)=\\left\\{\\begin{array}{cl}\nr \\cdot u_{i} & \\text{ für } x=a_{i}^{2} \\cdot r^{2} \\text{ mit } r \\in U \\\\\nu & \\text{ beliebig } \\in U \\text{ falls } x \\text{ kein Quadrat einer Zahl aus } \\mathbb{Q}^{+} \\text{ ist }\n\\end{array}\\right.\n$$\nDabei kommen alle Lösungen durch die möglichen Untergruppen $U$ und die Auswahl der $u_{i}$ zustande, sowie durch beliebige Wahl von $u \\in U$ (die $a_{i}$ sind für jedes $U$ fest).\n\nZum Beispiel ergibt sich Folgendes für die trivialen Untergruppen von $\\mathbb{Q}^{+}$:\n\na) $U=\\{1\\}$ : Hier ist nur $r=u=1$ sowie $u_{i}=1$ möglich, so dass $f(x)=1$ für alle $x \\in \\mathbb{Q}^{+}$.\n\nb) $U=\\mathbb{Q}^{+}$: Hier ist $f(x)=a \\sqrt{x}$ mit festem $a$, wenn $x$ eine Quadratzahl ist, sowie $f(x)$ beliebig aus $\\mathbb{Q}^{+}$, wenn $x$ keine Quadratzahl ist.\n\nDer Beweis erfolgt in zwei Schritten.\n\nSchritt 1: Wir zeigen, dass die angegebenen Funktionen $f$ wohldefiniert sind und $(*)$ erfüllen.\n\nJedes $t=x^{2}$ mit $x \\in \\mathbb{Q}^{+}$ lässt sich nach der Definition der Nebenklassen eindeutig in der Form $t=a_{i}^{2} \\cdot r^{2}$ mit $r \\in U$ darstellen, und jedes Produkt $a_{i}^{2} \\cdot r^{2}$ ist eine rationale Quadratzahl. Somit ist $f$ wohldefiniert. Aus der Definition folgt ferner, dass die Wertemenge $W_{f}$ von $f$ in $U$ liegt. Daher existiert für jedes $y \\in \\mathbb{Q}^{+}$ ein $u \\in U$ mit $f(y)=u$. Weil für jedes $x \\in \\mathbb{Q}^{+}$ wie oben bemerkt eine eindeutige Darstellung $x^{2}=a_{i}^{2} \\cdot r^{2}$ existiert, folgt\n$$\nf\\left(x^{2} \\cdot f(y)^{2}\\right)=f\\left(a_{i}^{2} r^{2} u^{2}\\right)=f\\left(a_{i}^{2}(r u)^{2}\\right)=(r u) \\cdot u_{i}=\\left(r u_{i}\\right) \\cdot u=f\\left(a_{i}^{2} r^{2}\\right) f(y)=f\\left(x^{2}\\right) f(y),\n$$\nwomit $(*)$ erfüllt ist.\n\nSchritt 2: Wir zeigen, dass keine weiteren Lösungen von $(*)$ als die angegebenen existieren können.\n\nDazu sei $f$ eine Lösungsfunktion von $(*)$ mit der Lösungsmenge $W_{f} \\subseteq \\mathbb{Q}^{+}$.\n\nWir setzen $y=1$ und $x=\\frac{1}{f(1)}$ in $(*)$ ein:\n$$\nf\\left(x^{2} f(y)^{2}\\right)=f\\left(\\frac{f(1)^{2}}{f(1)^{2}}\\right)=f(1)=f\\left(x^{2}\\right) f(1).\n$$\nSomit ist $1 \\in W_{f}$.\n\nNun sei $z=\\frac{1}{f(1)}$ fest (dies bedeutet nach dem oben Stehenden $f\\left(z^{2}\\right)=1$ ). Sei weiter $w \\in W_{f}$ (dies bedeutet, dass es ein $u \\in \\mathbb{Q}^{+}$ gibt mit $f(u)=w$). Wir setzen $y=u$ und $x=\\frac{z}{f(y)}$ in $(*)$ ein:\n$$\nf\\left(x^{2} f(y)^{2}\\right)=f\\left(z^{2}\\right)=1=f\\left(x^{2}\\right) \\cdot w,\n$$\nalso $f\\left(x^{2}\\right)=\\frac{1}{w}$. Somit liegt für jedes $w \\in W_{f}$ auch $\\frac{1}{w}$ in $W_{f}$.\n\nJetzt seien $v, w \\in W_{f}$ (dies bedeutet, dass es $t, u \\in \\mathbb{Q}^{+}$ gibt mit $f(t)=v$ und $f(u)=w$).\nWir setzen $x=z=\\frac{1}{f(1)}$ und $y=u$ in $(*)$ ein, bezeichnen $k=x \\cdot f(y)$ und erhalten $f\\left(k^{2}\\right)=1 \\cdot f(u)=w$.\nSodann setzen wir $x=k$ und $y=t$ neu in $(*)$ ein:\n$$\nf\\left(x^{2} f(y)^{2}\\right)=f\\left(k^{2}\\right) f(t)=w \\cdot v.\n$$\nSomit liegt für $v, w \\in W_{f}$ auch stets das Produkt $v w$ in $W_{f}$.\n\nDamit ist nachgewiesen, dass $W_{f}$ eine Untergruppe von $\\mathbb{Q}^{+}$ ist.\n\nSeien nun $a_{1}, a_{2}, \\ldots$ fest gewählte Repräsentanten der Nebenklassen von $\\mathbb{Q}^{+} / W_{f}$ (wie in der Beschreibung der Lösungsmenge) und $r \\in W_{f}$ beliebig mit $f(u)=r$. Wir setzen $x=a_{i}$ (für beliebiges $i$ ) und $y=u$ in $(*)$ ein:\n$$\nf\\left(x^{2} f(y)^{2}\\right)=f\\left(a_{i}^{2} \\cdot r^{2}\\right)=f\\left(a_{i}^{2}\\right) f(u)=u_{i} \\cdot r\n$$\n(wobei $f\\left(a_{i}^{2}\\right):=u_{i}$ gesetzt wird). Daraus ist ersichtlich, dass jede Lösungsfunktion $f$ tatsächlich die oben beschriebene Form haben muss. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDie reellen Zahlen $r_{1}, r_{2}, \\ldots, r_{2019}$ erfüllen die Bedingungen $r_{1}+r_{2}+\\ldots+r_{2019}=0$ sowie $r_{1}^{2}+r_{2}^{2}+\\ldots+r_{2019}^{2}=1$. Es sei $a=\\min \\left(r_{1}, r_{2}, \\ldots, r_{2019}\\right)$ und $b=\\max \\left(r_{1}, r_{2}, \\ldots, r_{2019}\\right)$. Man beweise: $a b \\leq \\frac{-1}{2019}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWeil wegen (2) die $r_{i}$ nicht alle $0$ sein und wegen (1) nicht alle das gleiche Vorzeichen haben können, gilt $b>0$ und $a<0$. Mit $P=\\left\\{i: u_{i}>0\\right\\}$ und $N=\\left\\{i: u_{i} \\leq 0\\right\\}$ sowie $p=|P|$ und $n=|N|$ gilt $p+n=2019$ und aus (1) folgt\n\n$0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i \\in P} u_{i}-\\sum_{i \\in N}\\left|u_{i}\\right|$, also $\\sum_{i \\in P} u_{i}=\\sum_{i \\in N}\\left|u_{i}\\right|$.\n\nDamit können wir abschätzen:\n\n$\\sum_{i \\in P} u_{i}^{2} \\leq \\sum_{i \\in P} b u_{i}=b \\sum_{i \\in N}\\left|u_{i}\\right| \\leq b \\sum_{i \\in N}|a|=-n a b$ \\hspace{0.5cm} (3)\n\nsowie\n\n$\\sum_{i \\in N} u_{i}^{2} \\leq \\sum_{i \\in N} a u_{i} \\leq|a| \\sum_{i \\in N}\\left|u_{i}\\right|=|a| \\sum_{i \\in P} u_{i} \\leq-p a b$ \\hspace{0.5cm} (4).\n\nEs folgt $1=\\sum_{i \\in P} u_{i}^{2}+\\sum_{i \\in N} u_{i}^{2} \\leq-(p+n) a b=-2019 a b$, und damit die Behauptung.\n\n\nSolution 2:\n\nWiederum ausgehend von $b>0$ und $a<0$ betrachten wir die folgende konvexe Punktmenge $C$ in der $x$-$y$-Ebene:\n\n(i) Der untere Rand von $C$ ist die Parabel $y=x^{2}$ im Bereich $a \\leq x \\leq b$.\n\n(ii) Der obere Rand von $C$ ist die Gerade $g: y=(a+b) x-a b$ im Bereich $a \\leq x \\leq b$.\n\nJeder der Punkte $(u_{i}, u_{i}^{2})$ liegt auf dem unteren Rand von $C$. Daher liegt der Schwerpunkt $S$ dieser $2019$ Punkte, die mit gleicher Masse versehen seien, ebenfalls in $C$. Es gilt $S=\\left(\\frac{1}{2019} \\sum_{i=1}^{2019} u_{i}, \\frac{1}{2019} \\sum_{i=1}^{2019} u_{i}^{2}\\right)=\\left(0, \\frac{1}{2019}\\right)$. Für $g$ gilt an der Stelle $x=0$, dass $y=-a b$ ist. $S$ darf nicht oberhalb der oberen Begrenzung liegen, woraus die Behauptung folgt.\n\n\nSolution 3:\n\n(Ein Ein-Zeilen-Beweis):\n\n$0 \\leq \\sum_{i=1}^{2019}\\left(r_{i}-a\\right)\\left(b-r_{i}\\right)=\\sum_{i=1}^{2019}\\left(-r_{i}^{2}+(b+a) r_{i}-a b\\right)=-1-2019 a b \\Leftrightarrow a b \\leq \\frac{-1}{2019}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFinden Sie alle Paare $(a, b)$ positiver ganzer Zahlen, für die eine positive ganze Zahl $n$ existiert, sodass die Anzahl der Teiler von $n a$ und $n b$ identisch ist.",
    "options": [],
    "answer": "All pairs of positive integers except those where one number is a proper divisor of the other; equivalently, pairs where the numbers are equal or neither divides the other.",
    "solution": "Solution:\nDie gesuchten Paare sind jene, für die eine der drei Bedingungen $a=b, a \\nmid b$ oder $b \\nmid a$ erfüllt ist, also alle Paare positiver ganzer Zahlen mit Ausnahme solcher, für die eine der Zahlen $a, b$ ein echter Teiler der anderen ist.\nIm gesamten Beweis werden wir die Anzahl aller Teiler einer positiven ganzen Zahl $m$ mit $\\tau(m)$ bezeichnen. Zunächst ist klar, dass man im Fall $a=b$ ein beliebiges $n$ wählen kann. Ferner ist im Fall, dass $a$ ein echter Teiler von $b$ ist, für jede positive ganze Zahl $n$ die Zahl $n a$ ein echter Teiler von $n b$, die Menge ihrer Teiler ist damit eine echte Teilmenge der Menge der Teiler von $n b$, was $\\tau(n a)<\\tau(n b)$ impliziert. Der Fall $b \\mid a, b<a$ wird analog behandelt.\nEs bleibt zu zeigen, dass wir im Fall $a \\nmid b, b \\nmid a$ stets ein $n$ mit $\\tau(n a)=\\tau(n b)$ finden können. Dafür beweisen wir zunächts die folgende Aussage.\nLemma. Es seien $\\alpha>\\beta$ positive ganze Zahlen. Dann existiert für jede ganze Zahl $M>\\beta$ eine positive ganze Zahl $\\gamma$, sodass\n$$\n\\frac{\\alpha+\\gamma+1}{\\beta+\\gamma+1}=1+\\frac{1}{M}=\\frac{M+1}{M} .\n$$\nBeweis. Nach Umformen der ersten Gleichung erhält man $\\gamma=M(\\alpha-\\beta)-(\\beta+1) \\geq 0$. Nun betrachten wir die Primfaktorzerlegung $a=\\prod_{p} p^{\\alpha_{p}}$ und $b=\\prod_{p} p^{\\beta_{p}}$ von $a$ und $b$. Es seien nun $p_{1}, \\ldots p_{k}$ all jene Primzahlen $p$ mit $\\alpha_{p}>\\beta_{p}$ und $q_{1}, \\ldots, q_{\\ell}$ alle Primzahlen $q$ mit $\\alpha_{q}<\\beta_{q}$. Wegen $a \\nmid b, b \\nmid a$ sind $k$ und $\\ell$ beide positiv. Es sei nun $X$ eine positive ganze Zahl, die größer als alle $\\alpha_{p}$ und $\\beta_{b}$ ist. Laut obigem Lemma können wir dann Zahlen $\\gamma_{1}, \\ldots, \\gamma_{k}$ und $\\gamma_{1}^{\\prime}, \\ldots, \\gamma_{\\ell}^{\\prime}$ wählen, sodass\n$$\n\\frac{\\alpha_{p_{i}}+\\gamma_{i}+1}{\\beta_{p_{i}}+\\gamma_{i}+1}=\\frac{k X+i}{k X+i-1}, \\quad \\text{ und } \\quad \\frac{\\alpha_{q_{j}}+\\gamma_{j}^{\\prime}+1}{\\beta_{q_{j}}+\\gamma_{j}^{\\prime}+1}=\\frac{\\ell X+j-1}{\\ell X+j}\n$$\nfür alle $1 \\leq i \\leq k$ und $1 \\leq j \\leq \\ell$ gilt. Für $n=\\left(\\prod_{i=1}^{k} p_{i}^{\\gamma_{i}}\\right) \\cdot\\left(\\prod_{j=1}^{\\ell} q_{j}^{\\gamma_{j}^{\\prime}}\\right)$ folgt nun mit Verwendung der allgemein bekannten Formel\n$$\n\\tau\\left(\\prod_{p} p^{\\eta_{p}}\\right)=\\prod_{p}\\left(\\eta_{p}+1\\right)\n$$\ndass\n$$\n\\frac{\\tau(n a)}{\\tau(n b)}=\\left(\\prod_{i=1}^{k} \\frac{\\alpha_{p_{i}}+\\gamma_{i}+1}{\\beta_{p_{i}}+\\gamma_{i}+1}\\right) \\cdot\\left(\\prod_{j=1}^{\\ell} \\frac{\\alpha_{q_{j}}+\\gamma_{j}^{\\prime}+1}{\\beta_{q_{j}}+\\gamma_{j}^{\\prime}+1}\\right)=\\frac{k(X+1)}{k X} \\cdot \\frac{\\ell X}{\\ell(X+1)}=1\n$$\nalso ist $n$ wie behauptet.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEin Pirat möchte einen Schatz, bestehend aus 1000 Goldmünzen, die jeweils mindestens 1 g und zusammen genau 2 kg wiegen, in zwei Teile aufteilen, die in ihrer Masse jeweils um höchstens 1 g von 1 kg abweichen. Beweisen Sie, dass dies möglich ist.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n1. Lösung (Vollständige Induktion). Die Massen der 1000 Münzen in Gramm, in aufsteigender Reihenfolge geordnet, seien mit $m_{1} \\leq m_{2} \\leq \\cdots \\leq m_{1000}$ bezeichnet. Wir beweisen nun zunächst mit vollständiger Induktion nach $\\ell=1, \\ldots, 1000$ die folgende Aussage.\n\nLemma. Für alle $1 \\leq \\ell \\leq 1000$ und jede reelle Zahl $0 \\leq x \\leq 2+\\sum_{i=1}^{\\ell} m_{i}$ existiert eine (potentiell leere) Teilmenge $I \\subset\\{1,2, \\ldots, \\ell\\}$, sodass $\\sum_{i \\in I} m_{i} \\in[x-2, x]$ gilt.\n\nEs gilt $x_{1} \\leq 2000 / 1000=2$, also ist der Induktionsanfang $\\ell=1$ klar: falls $x<2$, können wir $I=\\emptyset$ wählen, ansonsten funktioniert $I=\\{1\\}$.\n\nFür den Induktionsschritt $(\\ell \\rightarrow \\ell+1$, wobei $\\ell \\leq 999$) sehen wir, dass für $x \\leq \\sum_{i=1}^{\\ell} m_{i}$ nichts zu zeigen ist. Da wir jede Menge $I$ durch $I \\cup\\{\\ell+1\\}$ ersetzen können, ist die Behauptung auch für $m_{\\ell+1} \\leq x \\leq m_{\\ell+1}+\\sum_{i=1}^{\\ell} m_{i}=\\sum_{i=1}^{\\ell+1} m_{i}$ erfüllt. Der Induktionssschritt ist also beendet, außer es gilt $2+\\sum_{i=1}^{\\ell} m_{i}<m_{\\ell+1}$. In diesem Fall folgt $m_{i}>2+\\ell$ für alle $\\ell+1 \\leq i \\leq 1000$, was folgende Umgleichung impliziert.\n\n$$\n2000=\\sum_{i=1}^{1000} m_{i}=\\sum_{i=1}^{\\ell} m_{i}+\\sum_{i=\\ell+1}^{1000} m_{i}>\\ell+(1000-\\ell)(2+\\ell)=2000+999 \\ell-\\ell^{2}\n$$\n\nDamit folgt $\\ell>999$, ein Widerspruch, was den Beweis des Lemmas abschließt.\n\nWir wenden das Lemma nun auf $\\ell=1000$ und $x=1001$ an. Dadurch wird ersichtlich, dass eine Teilmenge $I \\subset\\{1,2, \\ldots, 1000\\}$ existiert, für die $\\sum_{i \\in I} x_{i} \\in[999,1001]$ gilt. Wegen $\\sum_{i=1}^{1000} m_{i}=2000$ gilt auch $\\sum_{i \\in\\{1, \\ldots, 1000\\} \\backslash I} x_{i} \\in[999,1001]$, und die Aufteilung der Münzen, die der Zerlegung\n$$\n\\{1, \\ldots, 1000\\}=I \\cup(\\{1, \\ldots, 1000\\} \\backslash I)\n$$\nentspricht, erfüllt die Behauptung.\n\n\n2. Lösung (Greedy-Algorithmus). Wir verwenden die gleiche Notation wie in der ersten Lösung. Der Pirat verteilt die Münzen nach folgendem Verfahren: Er legt alle Münzen, in absteigender Reihenfolge ihrer Massen (also beginnend mit der schwersten Münze mit Masse $m_{1000}$), nach und nach auf zwei Haufen $H_{1}$ und $H_{2}$, wobei er jede Münze auf denjenigen Haufen legt, der zu diesem Zeitpunkt leichter ist; bei einem Gleichgewicht wählt er einen beliebigen Haufen aus.\n\nWir behaupten, dass keiner der beiden Haufen jemals mehr als 1001 Gramm wiegt.\n\nAngenommen, nach dem Hinzulegen der $k$-ten Münze mit Masse $m_{1001-k}$ ist doch einer der Haufen schwerer als 1001 Gramm. Vor diesem Schritt wogen beide Haufen zusammen $\\sum_{i=1002-k}^{1000} m_{i}$ Gramm, und da die Münze auf den leichteren Haufen gelegt wurde, folgt\n\n$$\n\\begin{aligned}\nm_{1001-k}+\\frac{1}{2} \\sum_{i=1002-k}^{1000} m_{i} & >1001 \\\\\nm_{1001-k}+\\frac{1}{2} \\sum_{i=1}^{1000} m_{i} & >1001+\\frac{1}{2} \\sum_{i=1}^{1001-k} m_{i} \\\\\n\\frac{1}{2} m_{1001-k}+1000 & >1001+\\frac{1}{2} \\sum_{i=1}^{1000-k} m_{i} \\\\\nm_{1001-k} & >2+\\sum_{i=1}^{1000-k} m_{i} \\geq 2+1000-k=1002-k\n\\end{aligned}\n$$\n\nDamit folgt nun\n\n$$\n\\begin{aligned}\n2000=\\sum_{i=1}^{1000} m_{i} & =\\left(\\sum_{i=1}^{1000-k} m_{i}\\right)+\\left(\\sum_{i=1001-k}^{1000} m_{i}\\right) \\\\\n& \\geq 1000-k+k \\cdot m_{1001-k} \\\\\n& >1000-k+k(1002-k)\n\\end{aligned}\n$$\n\nalso folgt $f(k)=-k^{2}+1001 k-1000<0$, was wegen $f(k)=(k-1)(1000-k)$ und $1 \\leq k \\leq 1000$ jedoch ein Widerspruch ist.\n\nAlso stimmt die Behauptung, und auch nach dem Verteilen der leichtesten Münze mit Masse $m_{1}$ wiegen beide Haufen nicht mehr als 1001 Gramm. Da sie zusammen 2000 Gramm wiegen, muss damit jeder Haufen mindestens 999 Gramm wiegen, und wir haben eine Verteilung wie gefordert gefunden.\n\n\n3. Lösung (Schubfachprinzip). Wir verwenden wieder die gleiche Notation wie in der ersten Lösung. Wegen $m_{1000} \\geq 2>1$ gilt $m_{1}+\\cdots+m_{999}<1999$, und die 1000 Summen\n$$\n0, m_{1}, m_{1}+m_{2}, \\ldots, m_{1}+m_{2}+\\cdots+m_{999}\n$$\nliegen jeweils in genau einem der 1000 Schubfächer\n$$\n\\begin{gathered}\n\\{[0,1) \\cup[1000,1001),\\} \\\\\n\\{[1,2) \\cup[1001,1002),\\} \\\\\n\\ldots, \\\\\n\\{[998,999) \\cup[1998,1999),\\} \\\\\n\\{[999,1000)\\}\n\\end{gathered}\n$$\nFalls eine der Summen im letzten Schubfach $[999,1000)$ liegt, sind wir fertig, ansonsten liegen zwei Summen im gleichen Schubfach $[k, k+1) \\cup[1000+k, 1001+k)$ (für ein festes $0 \\leq k \\leq 998$). Da nach Voraussetzung $m_{i} \\geq 1$ gilt, liegt eine der Summen im Intervall $[k, k+1)$ und die andere in $[1000+k, 1001+k)$, die Differenz $m_{i+1}+m_{i+2}+\\cdots+m_{j}$ der beiden Summen liegt somit im Intervall $(999,1001)$ und liefert daher eine Verteilung wie gewünscht.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben sei ein Dreieck $ABC$. Ein Kreis $k$ geht durch $A$, schneidet die Seiten $\\overline{AB}$ und $\\overline{AC}$ nochmals in den Punkten $D$ bzw. $E$ und schneidet die Seite $\\overline{BC}$ in den Punkten $F$ und $G$, wobei $F$ zwischen $B$ und $G$ liegt. Die Tangente an den Kreis durch $B$, $D$ und $F$ in $F$ und die Tangente an den Kreis durch $C$, $E$ und $G$ in $G$ schneiden sich in einem Punkt $T$. Wir nehmen an, dass $A \\neq T$ gilt. Man beweise, dass die Geraden $AT$ und $BC$ parallel sind.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNach dem Sehnen-Tangentenwinkelsatz für die Sehne $\\overline{FB}$ gilt (siehe Figur) $\\angle TFB = 180^{\\circ} - \\angle BDF = \\angle FDA$. Im Sehnenviereck $ADFG$ gilt daher $\\angle FDA = 180^{\\circ} - \\angle AGF = \\angle CGA$ (kleine Bögen), analog ist $\\angle TGB = 180^{\\circ} - \\angle CGT = \\angle GEC$ und $\\angle GEC = \\angle CFA$ (Sehnenviereck $AFGE$, große Bögen). Also gilt $\\angle TFB = \\angle CGA$ (1), woraus $\\angle GFT = \\angle AGB$ folgt, und $\\angle TGB = \\angle CFA$ (2).\n\nDamit sind die Dreiecke $AFG$ und $TFG$ nach wsw kongruent, somit ihre Höhen auf der gemeinsamen Seite $\\overline{FG}$ gleich, woraus die Behauptung folgt.\n\n\nVariante: Aus (1) und (2) folgt direkt, dass $\\angle AFT = \\angle AGT$ gilt. Dies bedeutet nach der Umkehrung des Satzes vom Sehnenviereck, dass $T$ auf $k$ liegt. Mit $\\angle TAF = \\angle TGF = \\angle GFA$ folgt die Parallelität von $\\overline{TA}$ und $\\overline{FG}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs seien ein Dreieck $A B C$ mit Umkreis $\\Omega$ sowie Punkte $A_{1}, B_{1}$ und $C_{1}$ auf den Dreiecksseiten $\\overline{B C}, \\overline{C A}$ und $\\overline{A B}$ gegeben, sodass die drei Geraden $A A_{1}, B B_{1}$ und $C C_{1}$ einen Punkt $P$ gemeinsam haben.\nEs ist zu zeigen, dass höchstens zwei der drei Spiegelpunkte von $P$ bei Punktspiegelung an $A_{1}$, $B_{1}$, bzw. $C_{1}$ außerhalb von $\\Omega$ liegen.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWegen $\\measuredangle B P C+\\measuredangle C P A+\\measuredangle A P B=360^{\\circ}=\\left(180^{\\circ}-\\measuredangle B A C\\right)+\\left(180^{\\circ}-\\measuredangle C B A\\right)+\\left(180^{\\circ}-\\measuredangle A C B\\right)$ dürfen wir ohne Beschränkung der Allgemeinheit annehmen, dass $\\measuredangle A P B \\leq 180^{\\circ}-\\measuredangle A C B$ gilt. Da $P$ im Inneren des Dreiecks $A B C$ liegt, gilt $\\angle A P B>\\measuredangle A C B$, woraus $\\measuredangle A P B \\in\\left[\\measuredangle A C B, 180^{\\circ}-\\measuredangle A C B\\right]$ und damit $\\sin \\measuredangle A P B \\geq \\sin \\measuredangle A C B$ folgt.\nEs seien $A_{2}, B_{2}$ und $C_{2}$ die im Aufgabentext beschriebenen Spiegelpunkte, sowie $A_{3}, B_{3}$ und $C_{3}$ die zweiten Schnittpunkte der Geraden $A P, B P$ und $C P$ mit $\\Omega$. Wir werden nun beweisen, dass mindestens eines der Verhältnisse $\\frac{\\left|\\overline{P A_{1}}\\right|}{\\left|\\overline{A_{1} A_{3}}\\right|}$ und $\\frac{\\left|\\overline{P B_{1}}\\right|}{\\left|\\overline{B_{1} B_{3}}\\right|}$ nicht größer als $1$ ist, woraus die Behauptung folgt.\n![](attached_image_1.png)\nDazu rechnen wir unter Ausnutzung des Sinussatzes,\n$$\n\\frac{\\left|\\overline{P A_{1}}\\right|}{\\left|\\overline{A_{1} A_{3}}\\right|}=\\frac{\\left|\\overline{P A_{1}}\\right|}{\\left|\\overline{A_{1} B}\\right|} \\cdot \\frac{\\left|\\overline{A_{1} B}\\right|}{\\left|\\overline{A_{1} A_{3}}\\right|} \\stackrel{(*)}{=} \\frac{\\sin \\measuredangle C B P}{\\sin \\measuredangle A P B} \\cdot \\frac{\\left|\\overline{A_{1} A}\\right|}{\\left|\\overline{A_{1} C}\\right|}=\\frac{\\sin \\measuredangle C B P}{\\sin \\measuredangle A P B} \\cdot \\frac{\\sin \\measuredangle A C B}{\\sin \\measuredangle P A C},\n$$\nwobei wir im Schritt (*) die Gleichung $\\left|\\overline{A_{1} B}\\right| \\cdot\\left|\\overline{A_{1} C}\\right|=\\left|\\overline{A_{1} A}\\right| \\cdot\\left|\\overline{A_{1} A_{3}}\\right|$ verwendet haben, die aus dem Sehnensatz folgt. Analog zeigt man nun auch\n$$\n\\frac{\\left|\\overline{P B_{1}}\\right|}{\\left|\\overline{B_{1} B_{3}}\\right|}=\\frac{\\sin \\measuredangle P A C}{\\sin \\measuredangle A P B} \\cdot \\frac{\\sin \\measuredangle A C B}{\\sin \\measuredangle C B P},\n$$\nund durch Multiplikation beider Gleichungen folgt schließlich\n$$\n\\frac{\\left|\\overline{P A_{1}}\\right|}{\\left|\\overline{A_{1} A_{3}}\\right|} \\cdot \\frac{\\left|\\overline{P B_{1}}\\right|}{\\left|\\overline{B_{1} B_{3}}\\right|}=\\frac{(\\sin \\measuredangle A C B)^{2}}{(\\sin \\measuredangle A P B)^{2}} \\leq 1,\n$$\nwas die Behauptung impliziert.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEntscheiden Sie, ob es eine Menge $M$ positiver ganzer Zahlen gibt, die folgende Eigenschaft hat: Für jede positive rationale Zahl $r<1$ existiert genau eine endliche Teilmenge $S$ von $M$, sodass $\\sum_{s \\in S} 1 / s = r$ gilt, das heißt die Summe der Kehrwerte aller Elemente von $S$ ist gleich $r$.",
    "options": [],
    "answer": "No, such a set does not exist.",
    "solution": "Solution:\n\nEine solche Menge existiert nicht, was wir mithilfe eines Widerspruchsbeweises zeigen werden. Angenommen, $S$ hat die angegebene Eigenschaft. Offenbar ist $M$ dann unendlich, und wir dürfen ohne Beschränkung der Allgemeinheit annehmen, dass $1 \\notin M$. Wir bezeichnen die Elemente von $M$ der Größe nach geordnet mit $m_{1} < m_{2} < m_{3} < \\ldots$.\n\nFall 1: Es gilt $m_{i} \\geq 2 m_{i-1}$ für alle $i \\geq 2$. Dann gilt $m_{i} \\geq 2^{i-1} m_{1}$ für alle $i \\geq 1$, und es folgt\n$$\nr^{*} := \\sum_{i=1}^{\\infty} \\frac{1}{m_{i}} \\leq \\sum_{i=0}^{\\infty} \\frac{1}{2^{i} m_{1}} = \\frac{2}{m_{1}},\n$$\ndas heißt falls $m_{1} \\geq 3$ oder $m_{i} > 2 m_{i-1}$ für mindestens ein $i$ erfüllt ist, dann gilt $r^{*} < 1$. In diesem Fall erfüllt $M$ jedoch entgegen der Annahme nicht die im Aufgabentext angegebene Bedingung, falls $s \\in (r^{*}, 1)$ gewählt wird. Somit muss $m_{1} = 2$ und $m_{i} = 2 m_{i-1}$ für alle $i$ gelten, also besteht $M$ genau aus den Zweierpotenzen größer als $1$. Da sich $1 / 3 = \\sum_{i=2}^{\\infty} 2^{-i}$ aber nicht als endliche Summe von Kehrwerten von Zweierpotenzen schreiben lässt, ergibt sich auch in diesem Fall ein Widerspruch.\n\nFall 2: Es existiert ein $i > 1$, sodass $m_{i} < 2 m_{i-1}$. Wir betrachten\n$$\nr := \\frac{1}{m_{i-1}} - \\frac{1}{m_{i}} < \\frac{1}{m_{i}}.\n$$\nNach Annahme existiert eine endliche Teilmenge $S \\subset M$, sodass $\\sum_{s \\in S} 1 / s = r$ gilt. Wegen $s < 1 / m_{i}$ gilt $m_{i} \\notin S$. Damit sind aber $S_{1} = S \\cup \\{m_{i}\\}$ und $S_{2} = \\{m_{i-1}\\}$ zwei verschiedene endliche Teilmengen von $M$, sodass die Summen der Kehrwerte von $S_{1}$ und $S_{2}$ den Wert $1 / m_{i-1}$ ergeben, was ein Widerspruch ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31101,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGegeben ist eine positive ganze Zahl $n$ und ein Spielbrett, das aus $n+1$ nebeneinander angeordneten quadratischen Feldern besteht, die von links nach rechts von $0$ bis $n$ nummeriert sind. Zu Beginn des Spiels befinden sich $n$ Spielsteine auf dem Feld Nr. $0$ und die anderen Felder sind leer.\n\nEin geduldiger Spieler wählt nun für jeden Zug ein Feld mit $k \\neq 0$ Steinen und rückt einen davon um höchstens $k$ Felder nach rechts. Dabei muss der Stein auf dem Spielbrett bleiben. Sein Ziel ist es, mit einer Folge solcher Züge alle $n$ Steine auf das Feld Nr. $n$ zu befördern.\n\nMan beweise, dass der Spieler dieses Ziel nicht mit weniger als $\\left\\lceil\\frac{n}{1}\\right\\rceil+\\left\\lceil\\frac{n}{2}\\right\\rceil+\\left\\lceil\\frac{n}{3}\\right\\rceil+\\ldots+\\left\\lceil\\frac{n}{n}\\right\\rceil$ Zügen erreichen kann. (Dabei bezeichnet $\\lceil x\\rceil$ die kleinste ganze Zahl, die nicht kleiner als $x$ ist.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nDie Spielsteine sind ununterscheidbar und haben alle dasselbe Start- und Zielfeld. Daher kann der geduldige Spieler eine Vorschrift erfinden, die ihm angibt, welchen der Steine auf dem jeweils ausgewählten Feld er ziehen soll.\n\nEine mögliche Vorschrift erhält er, indem er die Steine fest von $1$ bis $n$ nummeriert und in jedem Zug nach der Auswahl des Feldes den dort befindlichen Stein mit der höchsten Nummer bewegt.\n\nAngenommen, er zieht auf diese Weise den Stein mit der Nummer $k$ ($1 \\leq k \\leq n$) nach rechts. Dann können auf dem dafür ausgewählten Feld nicht mehr als $k$ Steine gelegen haben, weil ihre Nummern höchstens gleich $k$ sind. Also wird Stein $k$ um jeweils höchstens $k$ Felder nach rechts bewegt. Weil er aber insgesamt um genau $n$ Felder bewegt wird, sind dafür wenigstens $\\left\\lceil\\frac{n}{k}\\right\\rceil$ Züge erforderlich. Dies gilt für alle $k$ mit $1 \\leq k \\leq n$. Aufsummieren über $k$ liefert die Mindestanzahl der erforderlichen Spielzüge gemäß der Aufgabenstellung. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs seien $m$ und $n$ zwei positive ganze Zahlen. Man beweise, dass die ganze Zahl $m^{2} + \\left\\lceil \\frac{4 m^{2}}{n} \\right\\rceil$ keine Quadratzahl ist.\n\n(Dabei bezeichnet $\\lceil x \\rceil$ die kleinste ganze Zahl, die nicht kleiner als $x$ ist.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFür einen indirekten Beweis nehmen wir an, dass es ein $k \\in \\mathbb{N}$ gibt mit $m^{2} + \\left\\lceil \\frac{4 m^{2}}{n} \\right\\rceil = (m + k)^{2}$, d.h. $\\left\\lceil \\frac{(2m)^{2}}{n} \\right\\rceil = (2m + k)k$. Offensichtlich ist $k \\geq 1$. Also hat die Gleichung $\\left\\lceil \\frac{c^{2}}{n} \\right\\rceil = (c + k)k$ (1) eine positive ganzzahlige Lösung $(c, k)$ mit geradem $c$. Ohne auf die Parität von $c$ zu achten, betrachten wir eine solche Lösung von (1) mit minimalem $k$.\n\nAus $\\frac{c^{2}}{n} > \\left\\lceil \\frac{c^{2}}{n} \\right\\rceil - 1 = c k + k^{2} - 1 \\geq c k$ und $\\frac{(c - k)(c + k)}{n} < \\frac{c^{2}}{n} \\leq \\left\\lceil \\frac{c^{2}}{n} \\right\\rceil = (c + k)k$ entnehmen wir $c > n k > n - k$, so dass $c = k n + r$ mit geeignetem $0 < r < k$ gilt.\n\nDies in (1) eingesetzt liefert $\\left\\lceil \\frac{c^{2}}{n} \\right\\rceil = \\left\\lceil \\frac{(n k + r)^{2}}{n} \\right\\rceil = k^{2} n + 2 k r + \\left\\lceil \\frac{r^{2}}{n} \\right\\rceil$ und $(c + k)k = (k n + r + k)k = k^{2} n + 2 k r + k(k - r)$, so dass $\\left\\lceil \\frac{r^{2}}{n} \\right\\rceil = k(k - r)$ (2) folgt.\n\nDies liefert eine andere positive ganzzahlige Lösung von (1) mit $c' = r$ und $k' = k - r < k$, was der Minimalität von $k$ widerspricht.\n\n\nVariante:\n\nEs sei $m^{2} + \\left\\lceil \\frac{4 m^{2}}{n} \\right\\rceil = c^{2}$ für eine positive ganze Zahl $c > m$, woraus $c^{2} - 1 < m^{2} + \\frac{4 m^{2}}{n} \\leq c^{2}$ und daraus $0 \\leq c^{2} n - m^{2}(n + 4) < n$ (3) folgt.\n\nWir substituieren $d = c^{2} n - m^{2}(n + 4)$, $x = c + m$ und $y = c - m$, erhalten $c = \\frac{x + y}{2}$ sowie $m = \\frac{x - y}{2}$ und schreiben (3) damit um:\n\n$\\left( \\frac{x + y}{2} \\right)^{2} n - \\left( \\frac{x - y}{2} \\right)^{2}(n + 4) = d$ bzw. $x^{2} - (n + 2) x y + y^{2} + d = 0$ mit $0 \\leq d < n$.\n\nWir wählen für festes $n$ und $d$ ein ganzzahliges Lösungspaar $(x, y)$, für das $x + y$ minimal ist. Wegen der Symmetrie von (4) können wir dabei $x \\geq y \\geq 1$ annehmen. Wie oben lässt sich auch hier eine weitere Lösung $(z, y)$ finden, für die $z < x$ ist – Widerspruch!",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $\\varepsilon>0$ gegeben. Zeigen Sie, dass es ganze Zahlen $a, b$ und $c$ gibt, die nicht alle 0 sind, sodass die Ungleichung $|a \\sqrt{2}+b \\sqrt{3}+c \\sqrt{5}|<\\varepsilon$ erfüllt ist.\n\nBeweisen Sie außerdem, dass in jedem Tripel $(a, b, c) \\neq(0,0,0)$ ganzer Zahlen, für das diese Ungleichung gilt, mindestens einer der Absolutbeträge $|a|,|b|,|c|$ größer als $\\varepsilon^{-1 / 3} / \\sqrt{30}$ ist.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIm ersten Beweis des ersten Teils bezeichnen wir für eine reelle Zahl $x$ sei $[x]$ die größte ganze Zahl, die nicht größer als $x$ ist (Gauß-Klammer), die Zahl $\\langle x\\rangle:=x-[x] \\in[0,1)$ wird als als gebrochener Anteil von $x$ bezeichnet.\nWir wählen eine ganze Zahl $n>\\sqrt{2} \\varepsilon^{-1}$. Nach dem Schubfachprinzip existieren verschiedene Indizes $i, j \\in\\{0,1, \\ldots, n\\}$, sodass $\\langle i \\sqrt{6}\\rangle$ und $\\langle j \\sqrt{6}\\rangle$ im gleichen der $n$ Intervalle\n$$\n\\left[0, \\frac{1}{n}\\right),\\left[\\frac{1}{n}, \\frac{2}{n}\\right), \\ldots,\\left[\\frac{n-1}{n}, 1\\right)\n$$\nliegen. Dann existiert eine ganze Zahl $a$, sodass $|a-(i \\sqrt{6}-j \\sqrt{6})|<1 / n$ gilt. Daraus folgt durch Multiplikation mit $\\sqrt{2}$, für $b:=2(j-i)$, die Ungleichung $|a \\sqrt{2}+b \\sqrt{3}|<\\sqrt{2} / n<\\varepsilon$, was mit $c=0$ auf die Ungleichung im Aufgabentext führt.\n\nDer zweite Beweis kommt ohne das Schubfachprinzip aus: Wegen $0<\\sqrt{6}-2<1$ finden wir eine positive ganze Zahl $k$, sodass $X=(\\sqrt{6}-2)^{k} \\in(0, \\varepsilon / \\sqrt{3})$. Offenbar gibt es von 0 verschiedene ganze Zahlen $a_{0}, b_{0}$, sodass $X=a_{0} \\sqrt{6}+b_{0}$. Wir setzen nun $a=3 a_{0}, b=b_{0}, c=0$, dann gilt $|a \\sqrt{2}+b \\sqrt{3}+c \\sqrt{5}|=\\left|\\sqrt{3}\\left(a_{0} \\sqrt{3} \\sqrt{2}+b_{0}\\right)\\right|=\\sqrt{3} X<\\varepsilon$.\n\nFür den (schwierigeren) zweiten Teil der Aufgabe nehmen wir an, dass $a, b$ und $c$ die Ungleichung erfüllen. Wir berechnen zunächst, dass es sich beim folgenden Ausdruck um eine ganze Zahl handelt.\n$$\n\\begin{aligned}\nZ & :=(a \\sqrt{2}+b \\sqrt{3}+c \\sqrt{5})(a \\sqrt{2}-b \\sqrt{3}-c \\sqrt{5})(a \\sqrt{2}+b \\sqrt{3}-c \\sqrt{5})(a \\sqrt{2}-b \\sqrt{3}+c \\sqrt{5}) \\\\\n& =\\left(2 a^{2}-\\left(3 b^{2}+2 \\sqrt{15} b c+5 c^{2}\\right)\\right)\\left(2 a^{2}-\\left(3 b^{2}-2 \\sqrt{15} b c+5 c^{2}\\right)\\right) \\\\\n& =\\left(2 a^{2}-3 b^{2}-5 c^{2}\\right)^{2}-60 b^{2} c^{2}\n\\end{aligned}\n$$\nDiese Zahl ist von 0 verschieden, denn ansonsten wäre 60 das Quadrat einer rationalen Zahl, was aufgrund der eindeutigen Primfaktorzerlegung positiver ganzer Zahlen aber nicht der Fall ist. Damit gilt $|Z| \\geq 1$. Andererseits lässt sich $|Z|$ wie folgt nach oben abschätzen\n$$\n\\begin{aligned}\n|Z| & <|a \\sqrt{2}+b \\sqrt{3}+c \\sqrt{5}| \\cdot(|a| \\sqrt{2}+|b| \\sqrt{3}+|c| \\sqrt{5})^{3} \\\\\n& <\\varepsilon\\left((2+3+5)^{\\frac{1}{2}} \\cdot\\left(a^{2}+b^{2}+c^{2}\\right)^{\\frac{1}{2}}\\right)^{3} \\\\\n& <\\left(\\varepsilon^{1 / 3} \\sqrt{10} \\cdot \\sqrt{3} \\cdot \\max (|a|,|b|,|c|)\\right)^{3}\n\\end{aligned}\n$$\nwobei wir im zweiten Schritt die Ungleichung von Cauchy-Schwarz verwendet haben. Nach Umstellung folgt die Behauptung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWir betrachten ein Dreieck $ABC$ und einen Punkt $P$ in dessen Inneren. Die Spiegelpunkte von $P$ an den Seiten $\\overline{BC}$, $\\overline{CA}$ und $\\overline{AB}$ seien mit $A_1$, $B_1$ und $C_1$ bezeichnet. Ferner sei $\\Omega$ der Umkreis des Dreiecks $A_1B_1C_1$ und schließlich seien $A_2$, $B_2$ und $C_2$ die zweiten Schnittpunkte der Geraden $A_1P$, $B_1P$ und $C_1P$ mit $\\Omega$. Beweisen Sie, dass sich die drei Geraden $AA_2$, $BB_2$ und $CC_2$ auf $\\Omega$ schneiden.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir arbeiten mit gerichteten Winkeln modulo $180^{\\circ}$.\nIn dieser Skizze entspricht $T$ dem Punkt $P$.\n![](attached_image_1.png)\n\nSchritt 1: Zunächst stellen wir fest, dass es sich bei $CA$ und $CB$ um die Mittelsenkrechten der Strecken $\\overline{PB_1}$ und $\\overline{PA_1}$ handelt, also ist $C$ der Umkreismittelpunkt des Dreiecks $PA_1B_1$ und es folgt\n$$\n\\measuredangle A_1CB = \\frac{1}{2} \\measuredangle A_1CP = \\measuredangle A_1B_1P = \\measuredangle A_1B_1B_2 = \\measuredangle A_1C_2B_2.\n$$\n\nSchritt 2: Analog wie in Schritt 1 folgt, dass $\\measuredangle A_1BC = \\measuredangle A_1B_2C_2$, also sind die Dreiecke $A_1BC$ und $A_1B_2C_2$ ähnlich.\n\nSchritt 3: Unter Benutzung der soeben bewiesenen Ähnlichkeit folgt\n$$\n\\measuredangle CA_1B = \\measuredangle C_2A_1B_2 \\quad \\text{und} \\quad \\frac{\\left|\\overline{A_1C}\\right|}{\\left|\\overline{A_1B}\\right|} = \\frac{\\left|\\overline{A_1C_2}\\right|}{\\left|\\overline{A_1B_2}\\right|}\n$$\nund somit auch\n$$\n\\measuredangle B_2A_1B = \\measuredangle C_2A_1C \\quad \\text{und} \\quad \\frac{\\left|\\overline{A_1B_2}\\right|}{\\left|\\overline{A_1B}\\right|} = \\frac{\\left|\\overline{A_1C_2}\\right|}{\\left|\\overline{A_1C}\\right|}\n$$\nalso sind auch die Dreiecke $A_1B_2B$ und $A_1C_2C$ ähnlich.\n\nSchritt 4: Es sei $K$ der Schnittpunkt von $C_2C$ mit $\\Omega$. Dann gilt $180^{\\circ} - \\measuredangle A_1B_2K = \\measuredangle A_1C_2K = \\measuredangle A_1C_2C = \\measuredangle A_1B_2B$, also liegt $K$ auf der Geraden $BB_2$. Analog kann man zeigen, dass $K$ auch auf der Geraden $AA_2$ liegt, was den Beweis beendet.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31105,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDie Ecken eines regelmäßigen 100-Ecks $P$ seien so entweder rot oder blau gefärbt, dass jede Farbe mindestens 24-mal vorkommt.\nBeweisen Sie, dass es 24 paarweise disjunkte Vierecke $Q_{1}, Q_{2}, \\ldots, Q_{24}$ gibt, deren Eckpunkte auch Ecken von $P$ sind, sodass jedes Viereck $Q_{i}$ entweder einen oder drei rote Eckpunkte hat.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir zeigen per Induktion nach $n$ die folgende Aussage:\nWenn die Ecken eines konvexen $4 n$-Ecks $P$ so rot und blau gefärbt wurden, dass jede Farbe mindestens $(n-1)$-mal vorkommt, finden wir $(n-1)$ paarweise disjunkte Vierecke $Q_{1}, \\ldots, Q_{n-1}$, deren Ecken auch Ecken von $P$ sind.\nFür $n=25$ erhalten wir die Behauptung aus der Aufgabenstellung.\n\nFür $n=1$ ist nichts zu zeigen. Sei nun $n \\geqslant 2$ und o.B.d.A. gebe es mindestens $2 n$ rote Ecken. Der Fall, dass die Ecken alternierend oder 2-alternierend gefärbt sind, wird unten behandelt. Ansonsten muss es vier aufeinanderfolgende Ecken geben, unter denen nicht zwei rote und zwei blaue sind. Dann gibt es auch vier aufeinanderfolgende Ecken, unter denen mehr rote als blaue sind, denn andernfalls gäbe es weniger rote als blaue Ecken. Wähle solche vier Ecken aus: Wenn genau drei von ihnen rot sind, wählen wir das von den vier ausgewählten Ecken aufgespannte Viereck als $Q_{n-1}$. Die restlichen $4(n-1)$ Punkte bilden ein konvexes $4(n-1)$-Eck $P^{\\prime}$, das zu $Q_{n-1}$ disjunkt ist und auf das wir die Induktionsannahme anwenden können. Wenn hingegen alle vier der gewählten Ecken rot sind, gehen wir den Rand des $4 n$-Ecks entlang bis wir zum ersten Mal auf einen der mindestens $n-1>0$ blauen Punkte treffen und wenden das gleiche Argument wie im vorherigen Fall an.\n\nEs bleiben noch die Fälle, dass die Ecken alternierend oder 2-alternierend gefärbt sind. Dafür gibt es verschiedene Konstruktionen. Wir betrachten zuerst den alternierenden Fall und bezeichnen die Ecken von $P$ mit $P_{1}, P_{2}, \\ldots, P_{4 n}$, wobei o.B.d.A. $P_{1}$ blau sei. In diesem Fall gibt es (mindestens) zwei Konstruktionen (siehe folgende Skizzen).\n\n![](attached_image_1.png)\n\nVariante 1\n![](attached_image_2.png)\n\nVariante 2\n\nIn Variante 1 beginnen wir mit dem Viereck $P_{2} P_{3} P_{4} P_{4 n}$ und gehen dann von $P_{4}$ aus gegen den Uhrzeigersinn in Dreierschritten und von $P_{4 n}$ aus im Uhrzeigersinn in Einerschritten weiter. Dabei bleiben genau ...",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31106,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBeweisen Sie, dass es für jede beliebige nichtnegative ganze Zahl $z$ genau ein geordnetes Paar $(m, n)$ positiver ganzer Zahlen $m, n$ gibt, so dass $2 z = (m+n)^2 - m - 3 n$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUmformen von (1) ergibt $2 z = (m+n-1)^2 + m - n - 1 = 2 m - 2 + (m+n-1)(m+n-2)$, woraus $z+1 = m + \\frac{(m+n-1)(m+n-2)}{2}$ folgt. Wir setzen $m+n-1 = k$ und erhalten $z+1 = m + \\frac{k(k-1)}{2}$, wobei $0 < m \\leq k$ und $k \\in \\mathbb{Z}, k \\geq 1$ gilt. Außerdem ist $\\frac{k(k-1)}{2} \\in \\mathbb{Z}$.\n\nOffensichtlich gibt es zu jeder positiven ganzen Zahl $z+1$ genau ein $k$ mit\n$$\n\\frac{k(k-1)}{2} < z+1 \\leq \\frac{k(k-1)}{2} + k = \\frac{(k+1)k}{2}\n$$\nDann sind auch $m = z+1 - \\frac{k(k-1)}{2}$ und $n = k+1-m$ eindeutig bestimmte positive ganze Zahlen, so dass die Gleichung eine eindeutige Darstellung jeder positiven ganzen Zahl $z+1$ und damit jeder nichtnegativen ganzen Zahl $z$ durch $m$ und $n$ liefert.\nSolution:\n\nUmformen von (1) ergibt $2 z = (m+n)(m+n-1) - 2 n$. Wir bemerken, dass die rechte Seite stets gerade ist und setzen $m+n = k$. Für konstantes $k$ ($k \\geq 2$) kann $n$ die Werte $1, 2, \\ldots, k-1$ annehmen, so dass die rechte Seite jeweils verschiedene gerade Zahlen von $k(k-1)-2$ bis $k(k-1)-2(k-1) = (k-1)(k-2)$ liefert. Für $k = 2, 3, \\ldots$ entsteht so eine vollständige, disjunkte Zerlegung der Menge aller nichtnegativer gerader ganzer Zahlen. Die Zahl $2 z$ liegt daher in genau einem durch $k$ bestimmten Intervall an einer durch $n$ bestimmten Stelle, womit dafür auch $m$ eindeutig bestimmt ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn einem gleichschenkligen Dreieck $ABC$ mit $\\overline{BC}=\\overline{CA}$ sei $D$ ein Punkt im Inneren der Seite $AB$, für den $\\overline{AD}<\\overline{DB}$ gilt. Weiter seien $P$ und $Q$ zwei Punkte im Inneren der Seiten $BC$ bzw. $CA$, so dass $\\angle DPB=\\angle AQD=90^{\\circ}$ gilt. Die Mittelsenkrechte von $PQ$ schneide $CQ$ im Punkt $E$, und die Umkreise der Dreiecke $ABC$ und $QPC$ schneiden sich außer in $C$ in einem weiteren Punkt $F$.\nBeweisen Sie: Wenn $P, E$ und $F$ kollinear sind, dann gilt $\\angle ACB=90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir bezeichnen die Mittelsenkrechte der Strecke $PQ$ mit $m$ und den Kreis $QPCF$ mit $k$. Wegen $DP \\perp BC$ und $DQ \\perp AC$ liegt auch $D$ auf $k$ und $CD$ ist sogar Durchmesser von $k$. Die Strecken $QE$ und $PE$ liegen symmetrisch zu $m$; gleichzeitig ist $m$ auch Symmetrieachse von $k$. Daher sind die Sehnen $CQ$ und $FP$ symmetrisch zu $m$, also auch $C$ und $F$ (nach Voraussetzung sind $P, E$ und $F$ kollinear!). Somit stimmt die Mittelsenkrechte von $CF$ mit $m$ überein; es folgt, dass $m$ durch den Umkreismittelpunkt $O$ von $ABC$ verläuft.\nNun betrachten wir den Mittelpunkt $M$ der Strecke $AB$. Wegen $CM \\perp DM$ liegt $M$ auch\n\n![](attached_image_1.png)\n\nauf $k$. Wegen $\\angle ACM=\\angle MCB$ folgt aus dem Umfangswinkelsatz, dass die Sehnen $MP$ und $MQ$ von $k$ gleich lang sind. Damit ist aber gezeigt, dass $m$ durch $M$ verläuft. Da sowohl $O$ als auch $M$ auf $m$ und auf $CM$ liegen, folgt $O=M$ und damit $\\angle ACB=90^{\\circ}$.\nSolution:\n\nWie in der ersten Lösung ausgeführt, stellen wir fest, dass die Punkte $P, C, F, Q, D$ und der Mittelpunkt $M$ von $AB$ auf dem Kreis $k$ mit Durchmesser $CD$ liegen, und dass die Mittelsenkrechte $m$ von $PQ$ durch $M$ verläuft.\nEs sei $G$ der Schnittpunkt von $BF$ und $CM$ und wir bezeichnen $\\alpha=\\angle FBA$. Weil $E$ auf $m$ liegt, weil $P, E$ und $F$ kollinear sind,\n\n![](attached_image_2.png)\n\nund weil $CFAB$ und $CFQP$ Sehnenvierecke sind, folgt $\\angle PQC=\\angle PQE=\\angle EPQ=\\angle FPQ=\\angle FCQ=\\angle FCA=\\angle FBA=\\alpha$.\nWeiter gilt $\\angle FEM=\\angle FEQ+\\angle QEM=2\\alpha+(90^{\\circ}-\\alpha)=90^{\\circ}+\\alpha$. Ebenfalls mit Außenwinkeln gilt $\\angle FGM=90^{\\circ}+\\alpha$; daher ist $MGEF$ ein Sehnenviereck.\nEs folgt $\\angle CGE=\\angle MFE=\\angle MFP=\\angle MCP$. Also ist $GE \\parallel BC$ und damit erhalten wir $\\angle EAF=\\angle CAF=\\angle CBF=\\angle EGF$, so dass $GEFA$ ebenfalls ein Sehnenviereck ist. Daraus folgt $\\angle ACB=\\angle AFB=\\angle AFG=180^{\\circ}-\\angle GMA=90^{\\circ}$, was zu zeigen war.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31108,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWir betrachten alle positiven reellen Zahlen $a, b, c, d$, für die $(a+c)(b+d)=a c+b d$ gilt.\nBestimmen Sie den kleinsten Wert, den $S=\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}$ annehmen kann.",
    "options": [],
    "answer": "8",
    "solution": "Solution:\n\nDer kleinste mögliche Wert für $S$ ist $8$.\nUm zu zeigen, dass stets $S \\geq 8$ gilt, wenden wir zweimal die AM-GM-Ungleichung an:\n\n$\\left(\\frac{a}{b}+\\frac{c}{d}\\right)+\\left(\\frac{b}{c}+\\frac{d}{a}\\right) \\geq 2 \\sqrt{\\frac{a c}{b d}}+2 \\sqrt{\\frac{b d}{a c}}=\\frac{2(a c+b d)}{\\sqrt{a b c d}}=\\frac{2(a+c)(b+d)}{\\sqrt{a b c d}} \\geq 2 \\cdot \\frac{2 \\sqrt{a c} \\cdot 2 \\sqrt{b d}}{\\sqrt{a b c d}}=8$.\n\nAn den Stellen der Ungleichungen gilt Gleichheit, wenn $a=c$ und $b=d$ ist. Damit lässt sich $(a+c)(b+d)=a c+b d$ umschreiben als $4 a b=a^{2}+b^{2}$. Auflösen nach $a$ liefert $a=b(2 \\pm \\sqrt{3})$ und damit eine mögliche Lösung $S=8$ für $b=d=1$ und $a=c=2+\\sqrt{3}$.\nSolution:\n\nDie Verschiebung $(a, b, c, d) \\rightarrow (b, c, d, a)$ ändert weder den Wert von $S$ noch die Nebenbedingung. Daher können wir oBdA $a c \\geq b d$ annehmen und setzen $t=\\sqrt{\\frac{a c}{b d}}$ mit $t \\geq 1$.\n\nAus der Bedingung folgt $t^{2}+1=\\frac{a c+b d}{b d}=(a+c)\\left(\\frac{1}{d}+\\frac{1}{b}\\right) \\geq 2 \\sqrt{a c} \\cdot 2 \\sqrt{\\frac{1}{b d}}=4 t$, wobei im vorletzten Schritt die AM-GM-Ungleichung verwendet wurde. Also gilt $t^{2}-4 t+1 \\geq 0$ und daher $(t \\geq 2+\\sqrt{3} \\vee t \\leq 2-\\sqrt{3})$, was wegen $t \\geq 1$ auf $t \\geq 2+\\sqrt{3}$ führt.\n\nEs ist $S=\\left(\\frac{a}{b}+\\frac{c}{d}\\right)+\\left(\\frac{b}{c}+\\frac{d}{a}\\right) \\geq 2 \\sqrt{\\frac{a c}{b d}}+2 \\sqrt{\\frac{b d}{a c}}=2\\left(t+\\frac{1}{t}\\right)$. Durch Ableiten sehen wir, dass für alle $t \\geq 2+\\sqrt{3}$ gilt: $S'(t)=2-2 / t^{2} \\geq 0$. Daher hat $S$ an der Stelle $t=2+\\sqrt{3}$ ein Minimum und es folgt $S \\geq 2\\left(t+\\frac{1}{t}\\right) \\geq 2\\left(2+\\sqrt{3}+\\frac{1}{2+\\sqrt{3}}\\right)=2(2+\\sqrt{3}+2-\\sqrt{3})=8$.\n\nFür $(a, b, c, d)=(1,2+\\sqrt{3}, 1,2+\\sqrt{3})$ ist sowohl $(a+c)(b+d)=8+4 \\sqrt{3}=1 \\cdot 1+(2+\\sqrt{3})(2+\\sqrt{3})=a c+b d$ als auch $S=\\frac{1}{2+\\sqrt{3}}+\\frac{2+\\sqrt{3}}{1}+\\frac{1}{2+\\sqrt{3}}+\\frac{2+\\sqrt{3}}{1}=2-\\sqrt{3}+2+\\sqrt{3}+2-\\sqrt{3}+2+\\sqrt{3}=8$.\n\nDaher ist $8$ der kleinste Wert von $S$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABCD$ ein konvexes Parallelogramm, das bei $A$ spitzwinklig ist. Die Spiegelpunkte von $A$ an den Geraden $BC$ und $CD$ seien mit $P$ bzw. $Q$ bezeichnet. Außerdem schneidet die Gerade $BD$ die Strecken $\\overline{AP}$ und $\\overline{AQ}$ im Inneren in den Punkten $R$ bzw. $S$.\nBeweisen Sie, dass sich die Umkreise der Dreiecke $BRP$ und $DQS$ berühren.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir betrachten den Spiegelpunkt $Z$ von $A$ bei Spiegelung an $BD$. Wir werden beweisen, dass sich die beiden in der Aufgabenstellung erwähnten Kreise in $Z$ berühren.\n\n![](attached_image_1.png)\n\nZunächst ist wegen $\\measuredangle BZR = \\measuredangle RAB = \\measuredangle BPR$ und der Umkehrung des Peripheriewinkelsatzes klar, dass $Z$ auf dem Umkreis $\\omega_1$ von $BRP$ liegt. Analog zeigt man, dass $Z$ auch auf dem Umkreis $\\omega_2$ von $DQS$ liegt. Der Sehnen-Tangentenwinkel der Sehne $ZB$ von $\\omega_1$ ist $\\angle ZRB = \\measuredangle BRA$. Da $AR$ auf $BC$ und damit auch auf $AD$ senkrecht steht, hat dieser Winkel die Größe $90^\\circ - \\angle ADB$. Analog zeigt man, dass der Sehnen-Tangentenwinkel der Sehne $ZD$ von $\\omega_2$ gleich $90^\\circ - \\angle DBA$ ist. Die Summe dieser beiden Winkel beträgt nun $180^\\circ - \\measuredangle ADB - \\angle DBA = \\angle BAD = \\angle DZB$, was zeigt, dass die Tangenten an $\\omega_1$ und $\\omega_2$ im Punkt $Z$ übereinstimmen. Somit berühren sich beide Kreise wirklich.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFür eine feste positive ganze Zahl $m$ sei $A$ eine Teilmenge von $\\{0,1,2, \\ldots, 5^{m}\\}$, die aus $4m+1$ Elementen besteht.\nBeweisen Sie, dass es in $A$ stets drei Zahlen $a, b, c$ gibt, für die $a < b < c$ und $c + 2a > 3b$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir nehmen an, dass es $4m+2$ Elemente $x_{0} < x_{1} < \\ldots < x_{4m+1}$ aus $\\{0,1,2, \\ldots, 5^{m}\\}$ gibt, für welche die Behauptung nicht erfüllt ist. Dann gilt insbesondere $x_{4m+1} + 2x_{i} \\leq 3x_{i+1}$ für alle $i = 0, 1, \\ldots, 4m-1$. Umformen ergibt $x_{4m+1} - x_{i} \\geq \\frac{3}{2}(x_{4m+1} - x_{i+1})$.\nEin einfacher Induktionsschluss liefert daraus $x_{4m+1} - x_{i} \\geq \\left(\\frac{3}{2}\\right)^{4m-i}(x_{4m+1} - x_{4m})$.\nHier führt nun $i=0$ auf $x_{4m+1} - x_{0} \\geq \\left(\\frac{3}{2}\\right)^{4m}(x_{4m+1} - x_{4m}) = \\left(\\frac{81}{16}\\right)^{m}(x_{4m+1} - x_{4m}) > 5^{m} \\cdot 1$, Widerspruch! ㅁ\n\n\nWir bezeichnen das größte Element von $A$ mit $c$. Für $k = 0, \\ldots, 4m-1$ definieren wir $A_{k} = \\{x \\in A \\mid (1 - (\\frac{2}{3})^{k})c \\leq x < (1 - (\\frac{2}{3})^{k+1})c\\}$.\nWegen $(1 - (\\frac{2}{3})^{4m})c = c - (\\frac{16}{81})^{m}c > c - (\\frac{1}{5})^{m}c \\geq c-1$ bilden die Mengen $A_{0}, A_{1}, \\ldots, A_{4m-1}$ eine Zerlegung von $A \\setminus \\{c\\}$. Weil $A \\setminus \\{c\\}$ aus $4m+1$ Elementen besteht, muss nach dem Schubfachprinzip eine Menge $A_{k}$ existieren, die aus wenigstens zwei Elementen besteht. Wir bezeichnen zwei der Zahlen aus $A_{k}$ so mit $a$ und $b$, dass $a < b < c$ gilt. Dann ist $c + 2a \\geq c + 2(1 - (\\frac{2}{3})^{k})c = (3 - 2(\\frac{2}{3})^{k})c = 3(1 - (\\frac{2}{3})^{k+1})c > 3b$, wie verlangt. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDer größte gemeinsame Teiler zweier positiver ganzer Zahlen $m$ und $n$ sei mit $\\operatorname{ggT}(m, n)$ bezeichnet.\nEs sei eine unendliche Menge $S$ positiver ganzer Zahlen gegeben, sodass es vier paarweise verschiedene Zahlen $v, w, x, y \\in S$ gibt, für die $\\operatorname{ggT}(v, w) \\neq \\operatorname{ggT}(x, y)$ gilt.\nBeweisen Sie, dass es drei paarweise verschiedene Zahlen $a, b, c \\in S$ gibt, für die $\\operatorname{ggT}(a, b)=\\operatorname{ggT}(a, c) \\neq \\operatorname{ggT}(b, c)$ gilt.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIm folgenden nennen wir eine dreielementige Teilmenge $\\{s, t, u\\} \\subset S$ ein ausgewogenes Dreieck, falls die Menge $\\{\\mathrm{ggT}(s, t), \\operatorname{ggT}(s, u), \\operatorname{ggT}(t, u)\\}$ genau zwei verschiedene Elemente hat. Es ist zu zeigen, dass es ein ausgewogenes Dreieck gibt.\n\nLemma. Für paarweise verschiedene Zahlen $a, b, c, d \\in S$, sodass $\\operatorname{ggT}(a, b)=\\operatorname{ggT}(a, c) \\neq \\operatorname{ggT}(a, d)$ und $\\operatorname{ggT}(b, d)=\\operatorname{ggT}(c, d)$ gilt, enthält die Menge $\\{a, b, c, d\\}$ ein ausgewogenes Dreieck.\n\nBeweis. Falls $\\operatorname{ggT}(a, b)=\\operatorname{ggT}(b, d)$, dann ist $\\{a, b, d\\}$ ein ausgewogenes Dreieck. Ansonsten gilt entweder $\\operatorname{ggT}(a, d) \\neq \\operatorname{ggT}(a, b)$ oder $\\operatorname{ggT}(a, d) \\neq \\operatorname{ggT}(b, d)$, also ist $\\{a, b, c\\}$ oder $\\{b, c, d\\}$ ein ausgewogenes Dreieck.\n\n![](attached_image_1.png)\n\nFür jedes Element $a \\in S$ sei $S_{a}=\\{\\operatorname{ggT}(a, s) \\mid s \\in S, s \\neq a\\}$. Da diese Menge nur Teiler von $a$ enthält, ist sie endlich. Aus der Voraussetzung folgt, dass wir $a \\in S$ so wählen können, dass $S_{a}$ mindestens zwei Elemente enthält, da ansonsten $\\operatorname{ggT}(v, w)=\\operatorname{ggT}(w, x)=\\operatorname{ggT}(x, y)$ gelten würde.\n\nNach dem Schubfachprinzip finden wir eine unendliche Teilmenge $T \\subset S$, sodass $\\operatorname{ggT}(a, t)$ der gleiche Wert $g$ ist für alle $t \\in T$. Nun wählen wir ein $d \\in S \\backslash(T \\cup\\{a\\})$ sodass $\\operatorname{ggT}(a, d) \\neq g$, das wegen $\\left|S_{a}\\right|>1$ existieren muss. Da auch $S_{d}$ endlich ist, finden wir nach dem Schubfachprinzip zwei verschiedene Elemente $b, c \\in T$ sodass $\\operatorname{ggT}(b, d)=\\operatorname{ggT}(c, d)$ gilt. Dann erfüllen $a, b, c, d$ die Voraussetzungen des Lemmas, also existiert in der Tat ein ausgewogenes Dreieck.\nSolution:\n\nWir dürfen o.B.d.A. voraussetzen, dass keine ganze Zahl $g>1$ alle Zahlen aus $S$ teilt, da wir sonst die größte derartige Zahl $g$ wählen und $S$ durch die Menge $S^{\\prime}=\\{s / g \\mid s \\in S\\}$ ersetzen könnten: Falls die Behauptung für $S^{\\prime}$ gilt, dann auch für $S$.\n\nAngenommen für eine Primzahl $p$ wäre die Teilmenge $S_{p} \\subset S$ aller durch $p$ teilbaren Zahlen unendlich groß, dann wählen wir $a \\in S \\backslash S_{p}$, was wegen der Vorüberlegung möglich ist. Da die Zahlen $\\operatorname{ggT}(a, s)$ für $s \\in S_{p}$ Teiler von $c$ sind, gibt es nach dem Schubfachprinzip verschiedene $b, c \\in S_{p}$ mit $\\operatorname{ggT}(a, b)=\\operatorname{ggT}(a, c)$. Da $b, c$ beide durch $p$ teilbar sind, folgt auch $\\operatorname{ggT}(b, c) \\neq \\operatorname{ggT}(a, c)$, also die Behauptung.\n\nSomit dürfen wir annehmen, dass jede Primzahl nur endlich viele Elemente aus $S$ teilt. Wegen der Voraussetzung finden wir nun aber Zahlen $b, c \\in S$, die nicht teilerfremd sind. Es gibt nur endlich viele Primzahlen, die $b$ oder $c$ teilen, also enthält $S$ auch nur endlich viele Elemente, die nicht zu $b$ oder $c$ teilerfremd sind. Somit können wir wegen $|S|=\\infty$ ein Element $a \\in S$ wählen, dass zu $b$ und $c$ teilerfremd ist. Dann erfüllen $a, b, c$ die Behauptung.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBestimmen Sie alle ganzen Zahlen $n \\geq 1$, für die es ein Paar $(a, b)$ von positiven ganzen Zahlen mit folgenden Eigenschaften gibt:\n\ni) Keine dritte Potenz einer Primzahl teilt $a^{2}+b+3$.\n\nii) Es gilt $\\frac{a b+3 b+8}{a^{2}+b+3}=n$.",
    "options": [],
    "answer": "2",
    "solution": "Solution:\n\nEs sei $p$ ein beliebiger Primfaktor von $a^{2}+b+3$ (ein solcher existiert, weil $a^{2}+b+3 \\geq 5$ gilt). Dann ist $b \\equiv -a^{2}-3 \\bmod p$. Wegen der Ganzzahligkeit von $n$ muss $p$ aber auch ein Primfaktor von $a b+3 b+8$ sein. Somit gilt $0 \\equiv a b+3 b+8 \\equiv a\\left(-a^{2}-3\\right)+3\\left(-a^{2}-3\\right)+8 \\equiv -a^{3}-3 a^{2}-3 a-1 \\equiv -(a+1)^{3} \\bmod p$, also muss $p$ auch ein Primfaktor von $a+1$ sein. Wegen i) kann $p$ in $a^{2}+b+3$ höchstens quadratisch vorkommen, also höchstens doppelt so oft wie in $a+1$. Weil das für jeden Primfaktor gilt, erhalten wir $a^{2}+b+3 \\leq (a+1)^{2}$.\n\nWürde ein Primfaktor $p$ in $a^{2}+b+3$ dagegen höchstens einfach vorkommen, so hätten wir $a^{2}+b+3 \\leq \\frac{(a+1)^{2}}{p} \\leq \\frac{(a+1)^{2}}{2}$, im Widerspruch zu $\\frac{(a+1)^{2}}{2} \\leq a^{2}+1 < a^{2}+4 \\leq a^{2}+b+3$.\n\nAlso muss $a^{2}+b+3=(a+1)^{2}$ gelten. Folglich ist $b=2 a-2$ und damit $a b+3 b+8=a(2 a-2)+3(2 a-2)+8=2 a^{2}+4 a+2=2(a+1)^{2}$.\n\nSomit ist $n=2$ die einzige mögliche Lösung. Das Beispiel $(a, b)=(2,2)$ mit $a^{2}+b+3=9$ zeigt, dass $n=2$ tatsächlich eine Lösung ist. ㅁ\nSolution:\n\nWenn wir die Bedingung ii) nach $b$ auflösen, erhalten wir $b=\\frac{n a^{2}+3 n-8}{a+3-n}$.\n\nFall 1: $n=a+3$. Hier muss wegen $b \\in \\mathbb{Z}^{+}$ auch der Zähler verschwinden, was nach Einsetzen $(a+1)^{3}=0$, also $a=-1$ liefert - ein Widerspruch zu $a \\in \\mathbb{Z}^{+}$.\n\nFall 2: $n \\neq a+3$. Aus (1) folgt\n$$\na^{2}+b+3=a^{2}+\\frac{n a^{2}+3 n-8}{a+3-n}+3=\\frac{a^{3}+3 a^{2}-n a^{2}+n a^{2}+3 n-8+3 a+9-3 n}{a+3-n}=\\frac{(a+1)^{3}}{a+3-n}.\n$$\n\nNun bezeichnen wir mit $v_{p}(z)$ die Häufigkeit, mit der ein Primfaktor $p$ in der Zahl $z \\in \\mathbb{Z}^{+}$ vorkommt. Gäbe es einen Primfaktor $p$ mit $v_{p}(a+1) \\geq v_{p}(a+3-n)+1$, dann wäre nach (2) $v_{p}\\left(a^{2}+b+3\\right) \\geq 2 v_{p}(a+3-n)+3 \\geq 3$, was im Widerspruch zu i) steht. Also ist $v_{p}(a+1) \\leq v_{p}(a+3-n)$, und somit ist $a+1$ ein Teiler von $a+3-n$. Aus (2) folgt noch $a+3-n>0$ und daher $a+1 \\leq a+3-n$, also $n \\leq 2$.\n\nFür $n=1$ wäre dann $a+1 \\mid a+2$, was für kein $a \\in \\mathbb{Z}^{+}$ erfüllt ist. Also kann nur $n=2$ gelten, was mit dem Beispiel aus der ersten Lösung bestätigt wird. ㅁ",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nEs sei $ABCD$ ein Parallelogramm mit $|AC| = |BC|$. Ein Punkt $P$ sei auf dem Strahl $AB$ gewählt, sodass $B$ zwischen $A$ und $P$ liegt. Der Umkreis des Dreiecks $ACD$ und die Strecke $PD$ haben außer dem Punkt $D$ noch den Punkt $Q$ gemeinsam. Der Umkreis des Dreiecks $APQ$ und die Strecke $PC$ haben außer dem Punkt $P$ noch den Punkt $R$ gemeinsam.\nBeweisen Sie, dass sich die drei Geraden $CD$, $AQ$ und $BR$ in einem Punkt schneiden.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWir arbeiteten mit gerichteten Winkeln modulo $180$ Grad. Zunächst erkennen wir, dass aus der Voraussetzung $|AC| = |BC|$ unmittelbar folgt, dass die Winkel $\\angle BAC$, $\\angle CBA$, $\\angle DCA$ und $\\angle ADC$ allesamt gleich groß sind.\n\n![](attached_image_1.png)\n\nEs genügt zu zeigen, dass der Schnittpunkt $S$ der Geraden $AQ$ und $CD$ auf der Geraden $BR$ liegt. Wegen $\\angle CSQ = \\angle PAQ = \\angle CRQ$ ist $CSRQ$ ein Sehnenviereck. Wegen $\\angle CRA = 180^{\\circ} - \\angle ARP = 180^{\\circ} - \\angle AQP = \\angle DQA = \\angle DCA = \\angle CBA$ ist auch $ABRC$ ein Sehnenviereck. Also gilt $\\angle SRC = \\angle SQC = \\angle ADC = \\angle BAC = \\angle BRP$, woraus in der Tat folgt, dass $P$, $R$ und $S$ auf einer Geraden liegen.\nSolution:\n\nWie in der ersten Lösung zeigt man, dass $ABRC$ ein Sehnenviereck ist. Wir betrachten den Schnittpunkt $E$ der Geraden $AD$ und $BR$. Dann gilt $\\angle EAP = \\angle CBA = \\angle BAC = \\measuredangle ERP$, also liegt $E$ auf dem Umkreis des Sehnenvierecks $APRQ$.\n\n![](attached_image_2.png)\n\nNun gilt $\\angle CER = \\angle APR = 180^{\\circ} - \\angle DCR$, also liegt $E$ auch auf dem Umkreis des Dreiecks $RCD$. Die paarweisen Potenzgeraden der Umkreise von $ACD$, $APR$ und $RCD$ sind somit die Geraden $CD$, $AQ$, $RE = RB$, also schneiden sie sich in einem Punkt. Dies impliziert die Behauptung.\nSolution:\n\nDa $\\angle QAC = \\measuredangle QDC = \\angle QPA$ folgt mit der Umkehrung des Sehnen-Tangenten-Winkel-Satzes, dass die Gerade $AC$ den Umkreis des Sehnenvierecks $APRQ$ tangiert. Wegen der Voraussetzung tangiert $AB$ auch den Umkreis des Sehnenvierecks $AQCD$, also sind die Winkel $\\angle ADQ$ und $\\angle PAQ$ gleich groß. Damit folgt aber $\\angle ACQ = \\angle ADQ = \\angle PAQ = \\angle CRQ$, also tangiert $AC$ auch den Umkreis des Dreiecks $CQR$.\n\n![](attached_image_3.png)\n\nDer Schnittpunkt $M$ der Geraden $RQ$ und $AC$ erfüllt damit nach dem Sekanten-Tangenten-Satz $|MA|^{2} = |MQ| \\cdot |MR| = |MC|^{2}$, also ist $M$ der Mittelpunkt der Strecke $AC$, also auch der Mittelpunkt der Strecke $BD$.\n\nDer Satz von Pappos, angewandt auf die jeweils kollinearen Punkte $A, B, P$ und $R, Q, M$ impliziert nun, dass der Schnittpunkt $S$ der Geraden $AQ$ und $BR$, $C$ (als Schnittpunkt der Geraden $AM$ und $PR$), sowie $D$ (als Schnittpunkt der Geraden $BM$ und $PQ$) auf einer Geraden liegen, was der zu zeigenden Aussage entspricht.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFür eine feste positive ganze Zahl $k$ sei $K$ die Menge aller Gitterpunkte $(x, y)$ in der Ebene, deren beide Koordinaten $x$ und $y$ nichtnegative ganze Zahlen kleiner als $2k$ sind. Es gilt also $|K| = 4k^{2}$.\n\nEine Menge $V$ bestehe nun aus $k^{2}$ nicht-ausgearteten Vierecken mit folgenden Eigenschaften:\n\ni) Die Ecken aller dieser Vierecke sind Elemente von $K$.\n\nii) Jeder Punkt in $K$ ist Eckpunkt von genau einem der Vierecke in $V$.\n\nBestimmen Sie den größtmöglichen Wert, den die Summe der Flächeninhalte aller $k^{2}$ Vierecke in $V$ annehmen kann.",
    "options": [],
    "answer": "k^2(2k-1)(2k+1)/3",
    "solution": "Solution:\n\nJeder Punkt aus $K$ ist Eckpunkt eines eindeutig definierten zentralen Quadrats. Daher ist die Menge $Q$ aller zentralen Quadrate akzeptabel. Wir zeigen $S(V) \\leq S(Q) = S(k)$, woraus die Antwort folgt.\n\nWir benutzen das folgende\n\nLemma 1. Für jedes Viereck $V = A_{1}A_{2}A_{3}A_{4}$ und jeden beliebigen Punkt $O$ in der Ebene gilt\n$$\n[V] \\leq \\frac{1}{2} \\sum_{i=1}^{4} OA_{i}^{2}\n$$\nwobei Gleichheit genau dann gilt, wenn $V$ ein Quadrat mit Mittelpunkt $O$ ist.\n\nBeweis: Für $i = 1, \\ldots, 4$ und $A_{5} = A_{1}$ gilt\n$$\n[OA_{i}A_{i+1}] \\leq \\frac{OA_{i} \\cdot OA_{i+1}}{2} \\leq \\frac{OA_{i}^{2} + OA_{i+1}^{2}}{4}.\n$$\nDaher gilt\n$$\n[V] \\leq \\sum_{i=1}^{4} [OA_{i}A_{i+1}] \\leq \\frac{1}{4} \\sum_{i=1}^{4} (OA_{i}^{2} + OA_{i+1}^{2}) = \\frac{1}{2} \\sum_{i=1}^{4} OA_{i}^{2},\n$$\nwomit (2) bewiesen ist. In der Tat herrscht jeweils Gleichheit, wenn $V$ ein Quadrat mit Mittelpunkt $O$ ist.\n\nNun betrachten wir eine beliebige akzeptable Menge $V$. Lemma 1 auf jedes Element von $V$ und jedes Element von $Q$ angewendet liefert\n$$\nS(V) \\leq \\frac{1}{2} \\sum_{A < K} OA^{2} = S(Q),\n$$\nwomit die linke Seite von (1) bewiesen ist.\n\nNun berechnen wir\n$$\nS(Q) = \\frac{1}{2} \\sum_{A \\in K} OA^{2} = \\frac{1}{2} \\sum_{i=0}^{2k-1} \\sum_{j=0}^{2k-1} \\left(\\left(k-\\frac{1}{2}-i\\right)^{2} + \\left(k-\\frac{1}{2}-j\\right)^{2}\\right)\n$$\n$$\n= \\frac{1}{8} \\cdot 4 \\cdot 2k \\sum_{i=0}^{k-1} (2k-2i-1)^{2} = k \\sum_{j=0}^{k-1} (2j+1)^{2} = k\\left(\\sum_{j=1}^{2k} j^{2} - \\sum_{j=1}^{k} (2j)^{2}\\right)\n$$\n$$\n= k\\left(\\frac{2k(2k+1)(4k+1)}{6} - 4 \\cdot \\frac{k(k+1)(2k+1)}{6}\\right) = \\frac{k^{2}(2k+1)(2k-1)}{3} = S(k).\n$$\n\nFür jedes Viereck $ABCD$ in einer akzeptablen Menge $V$ gilt\n$$\n[ABCD] = \\frac{AC \\cdot BD}{2} \\cdot \\sin \\varphi \\leq \\frac{AC^{2} + BD^{2}}{4}\n$$\nmit $\\varphi = \\measuredangle(AC, BD)$. Wenden wir (3) auf alle Elemente von $V$ an, so erhalten wir\n$$\nS(V) \\leq \\frac{1}{4} \\sum_{i=1}^{2k^{2}} A_{i}B_{i}^{2},\n$$\nwobei $\\left(A_{1}, A_{2}, \\ldots, A_{2k^{2}}, B_{1}, B_{2}, \\ldots, B_{2k^{2}}\\right)$ eine Permutation von $K$ ist.\n\nMit der Abkürzung $S := \\sum_{i=1}^{2k^{2}} A_{i}B_{i}^{2}$ formulieren wir das folgende\n\nLemma 2. Der größtmögliche Wert von $S$ über alle Permutationen von $K$ beträgt $\\frac{4}{3}k^{2}(4k^{2}-1)$ und wird angenommen, wenn $A_{i}$ und $B_{i}$ für alle $i=1,2,\\ldots,2k^{2}$ symmetrisch zu $O$ liegen.\n\nBeweis: Es seien $A_{i} = (p_{i}, q_{i})$ und $B_{i} = (r_{i}, s_{i})$ für $i=1,2,\\ldots,2k^{2}$. Dann ist\n$$\nS = \\sum_{i=1}^{2k^{2}} (p_{i} - r_{i})^{2} + \\sum_{i=1}^{2k^{2}} (q_{i} - s_{i})^{2}.\n$$\nMit der QM-AM-Ungleichung schätzen wir die erste Summe\n$$\n\\sum_{i=1}^{2k^{2}} (p_{i} - r_{i})^{2} = \\sum_{i=1}^{2k^{2}} (2p_{i}^{2} + 2r_{i}^{2} - (p_{i} + r_{i})^{2}) = 4k \\sum_{j=0}^{2k-1} j^{2} - \\sum_{i=1}^{2k^{2}} (p_{i} + r_{i})^{2}\n$$\n$$\n\\leq 4k \\sum_{j=0}^{2k-1} j^{2} - \\frac{1}{2k^{2}} \\left(2k \\cdot \\sum_{j=0}^{2k-1} j\\right)^{2}\n$$\n$$\n= 4k \\cdot \\frac{2k(2k-1)(4k-1)}{6} - 2k^{2}(2k-1)^{2} = \\frac{2k^{2}(2k-1)(2k+1)}{3},\n$$\nwobei Gleichheit genau dann gilt, wenn $p_{i} + r_{i} = 2k-1$ für alle $i$ erfüllt ist. Da die zweite Summe völlig analog abgeschätzt werden kann, folgt\n$$\nS \\leq \\frac{4}{3}k^{2}(4k^{2}-1)\n$$\nmit Gleichheit bei $p_{i} + r_{i} = q_{i} + s_{i} = 2k-1$ für alle $i=1,\\ldots,2k^{2}$, d.h. wenn $A_{i}$ und $B_{i}$ für alle $i=1,2,\\ldots,2k^{2}$ symmetrisch zu $O$ liegen.\n\nMit dem Resultat aus Lemma 2 ergibt sich\n$$\nS(V) \\leq \\frac{1}{4} \\cdot \\frac{4k^{2}(4k^{2}-1)}{3} = \\frac{k^{2}(2k-1)(2k+1)}{3},\n$$\nwobei die Abschätzung für die Menge $Q$ scharf ist. ㅁ.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31115,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nBestimmen Sie den kleinsten Wert, den der Ausdruck\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{2022}}{2022}\\right\\rfloor\n$$\nannehmen kann, wobei $a_{1}, a_{2}, \\ldots, a_{2022} \\geq 1$ reelle Zahlen sind, sodass $\\left|a_{i}-a_{j}\\right| \\geq 1$ für alle $1 \\leq i<j \\leq 2022$ gilt.\n\nAnmerkung: Für eine reelle Zahl $x$ bezeichnet $\\lfloor x\\rfloor$, genannt Gauß-Klammer von $x$, die größte ganze Zahl, die nicht größer als $x$ ist.",
    "options": [],
    "answer": "11",
    "solution": "Solution:\n\nAus der Voraussetzung folgt, dass die kleinste der Zahlen $a_{1}, a_{2}, \\ldots, a_{2022}$ mindestens $1$, die zweitkleinste mindestens $2$ ist, usw. bis zur $2022$-ten, die mindestens $2022$ sein muss. Wenn wir nun die Zahlen $a_{1}, \\ldots, a_{2022}$ sukzessive der Größe nach durch $1,2, \\ldots, 2022$ ersetzen, so wird die Summe aus dem Aufgabentext nicht größer. Wir sehen also, dass wir uns auf den Fall einschränken dürfen, dass $(a_{1}, a_{2}, \\ldots, a_{2022})$ paarweise verschiedene positive ganze Zahlen sind. Nun ersetzen wir $2022$ durch eine beliebige positive ganze Zahl $k$ und behaupten, dass der kleinste Wert $s(k)$, den die Summe\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{k}}{k}\\right\\rfloor\n$$\nannehmen kann, wenn $a_{1}, a_{2}, \\ldots, a_{k}$ paarweise verschiedene positive ganze Zahlen sind, genau $\\left\\lfloor\\log_{2}(k)\\right\\rfloor$ beträgt.\n\n- **Untere Schranke:** Da $m$ offenbar monoton steigend ist, genügt es zu zeigen, dass $m\\left(2^{t}\\right) \\geq t$ gilt. Der Fall $t=0$ ist klar, für $t>0$ verwenden wir vollständige Induktion und setzen daher $m\\left(2^{t-1}\\right) \\geq t-1$ als bekannt voraus. Es seien $a_{1}, a_{2}, \\ldots, a_{2^{t}}$ beliebige paarweise verschiedene positive ganze Zahlen. Dann ist das Maximum $a_{i}$ dieser Zahlen mindestens $2^{t}$. Falls $i>2^{t-1}$, so gilt nun\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{2^{t}}}{2^{t}}\\right\\rfloor \\geq \\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{2^{t-1}}}{2^{t-1}}\\right\\rfloor+\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geq m\\left(2^{t-1}\\right)+1 = t-1+1 = t\n$$\nFalls jedoch $i \\leq 2^{t-1}$, so ist die Menge $X=\\{1,2, \\ldots, 2^{t-1}\\} \\setminus \\{a_{1}, a_{2}, \\ldots, a_{2^{t-1}}\\}$ nicht leer. Sei $b \\in X$ beliebig. Dann gilt $a_{i} \\geq 2^{t} = 2^{t-1} + 2^{t-1} \\geq b+i$, also gilt\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{2^{t}}}{2^{t}}\\right\\rfloor \\geq \\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{b+i}{i}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{2^{t-1}}}{2^{t-1}}\\right\\rfloor \\geq m\\left(2^{t-1}\\right)+1 = t-1+1 = t\n$$\n\n- **Obere Schranke:** Wir setzen\n$$\na_{i}=\\begin{cases}\ni-1, & \\text{falls } i \\text{ keine Zweierpotenz ist,} \\\\\n2^{s+1}-1, & \\text{falls } i=2^{s}.\n\\end{cases}\n$$\nDann sind $a_{1}, a_{2}, \\ldots, a_{k}$ paarweise verschiedene positive ganze Zahlen und es gilt\n$$\n\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor=\\begin{cases}\n0, & \\text{falls } i \\text{ keine Zweierpotenz ist,} \\\\\n1 & \\text{andernfalls.}\n\\end{cases}\n$$\nalso folgt $\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{k}}{k}\\right\\rfloor=\\left\\lfloor\\log_{2}(k)\\right\\rfloor$.\n\nAngewandt auf die konkrete Situation der Aufgabenstellung erhalten wir, dass der kleinste Wert, den die angegebene Summe annehmen kann, gleich $\\left\\lfloor\\log_{2}(2022)\\right\\rfloor=11$ ist.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31116,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the natural numbers $v$ for which $2007 + 4v$ is perfect square.",
    "options": [],
    "answer": "none",
    "solution": "Let $2007 + 4v = \\kappa^2$, $\\kappa \\in \\mathbb{N}$. Then\n$$\n\\kappa^2 - 4v = 2007 \\Leftrightarrow \\kappa^2 - 2^{2v} = 2007 \\Leftrightarrow (\\kappa - 2^v)(\\kappa + 2^v) = 1 \\cdot 3 \\cdot 3 \\cdot 223.\n$$\nSince $\\kappa - 2^v < \\kappa + 2^v$, last equality is equivalent to the systems\n$$\n\\begin{cases} \\kappa - 2^v = 1 \\\\ \\kappa + 2^v = 2007 \\end{cases} \\text{ or } \\begin{cases} \\kappa - 2^v = 3 \\\\ \\kappa + 2^v = 669 \\end{cases} \\text{ or } \\begin{cases} \\kappa - 2^v = 9 \\\\ \\kappa + 2^v = 223 \\end{cases} \\quad (\\Sigma_1) \\text{ or } (\\Sigma_2) \\text{ or } (\\Sigma_3)\n$$\nNow we have\n$$\n(\\Sigma_1) \\Leftrightarrow \\begin{cases} \\kappa - 2^v = 1 \\\\ \\kappa + 2^v = 2007 \\end{cases} \\Leftrightarrow \\begin{cases} 2\\kappa = 2008 \\\\ 2 \\cdot 2^v = 2006 \\end{cases} \\Rightarrow 2^v = 1003, \\text{ absurd..}\n$$\n$$\n(\\Sigma_2) \\Leftrightarrow \\begin{cases} \\kappa - 2^v = 3 \\\\ \\kappa + 2^v = 669 \\end{cases} \\Leftrightarrow \\begin{cases} 2\\kappa = 672 \\\\ 2 \\cdot 2^v = 666 \\end{cases} \\Rightarrow 2^v = 333, \\text{ absurd.}\n$$\n$$\n(\\Sigma_3) \\Leftrightarrow \\begin{cases} \\kappa - 2^v = 9 \\\\ \\kappa + 2^v = 223 \\end{cases} \\Leftrightarrow \\begin{cases} 2\\kappa = 232 \\\\ 2 \\cdot 2^v = 214 \\end{cases} \\Rightarrow 2^v = 107, \\text{ absurd.}\n$$\nTherefore, there are not natural numbers $v$ such that $2007 + 4v$ is a perfect square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31117,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $AB\\Gamma$ is given with $AB < A\\Gamma$. Let $I$ be the point of intersection of its bisectors. Bisector $A\\Delta$ meets the circumcircle $C$ of the triangle $B\\Gamma$ at the point $N$ with $N \\neq I$.\n(i) Determine the angles of the triangle $B\\Gamma N$ with respect to the angles of the triangle $AB\\Gamma$,\n(ii) the center of the circle $C$.",
    "options": [],
    "answer": "(i) ∠ΓBN = 90° − (∠B)/2, ∠BΓN = 90° − (∠Γ)/2, ∠BNΓ = 90° − (∠A)/2.\n(ii) The center of C is the point M where the angle bisector from A meets the circumcircle of triangle ABΓ; equivalently, M lies on line IN (the diameter of C) and on the perpendicular bisector of BI, hence M is the center.",
    "solution": "(i)\n$$\n\\widehat{\\Gamma BN} = \\widehat{\\Gamma IN} = \\frac{\\hat{A}}{2} + \\frac{\\hat{\\Gamma}}{2} = \\frac{180^\\circ - \\hat{B}}{2} = 90^\\circ - \\frac{\\hat{B}}{2}\n$$\n$$\n\\widehat{B\\Gamma N} = \\widehat{B\\Gamma IN} = \\frac{\\hat{A}}{2} + \\frac{\\hat{B}}{2} = \\frac{180^\\circ - \\hat{\\Gamma}}{2} = 90^\\circ - \\frac{\\hat{\\Gamma}}{2}\n$$\n$$\n\\widehat{BN\\Gamma} = 180^\\circ - \\left( 90^\\circ - \\frac{\\widehat{B}}{2} + 90^\\circ - \\frac{\\widehat{\\Gamma}}{2} \\right)\n$$\n$$\n= \\frac{\\widehat{B} + \\widehat{\\Gamma}}{2} = 90^\\circ - \\frac{\\widehat{A}}{2}\n$$\n![](attached_image_1.png)\nFigure 1\n\n(ii) Since $\\widehat{\\Gamma BN} = 90^\\circ - \\frac{\\widehat{B}}{2}$, $BN$ is the bisector of the external angle of $\\widehat{B}$.\nTherefore $\\widehat{IBN} = 90^\\circ$ and $IN$ is a diameter of the circle $C$. Moreover, if $A\\Delta$ intersects the circumcircle of the triangle $AB\\Gamma$ at $M$, then\n$$\n\\widehat{BIN} = \\frac{\\hat{A}}{2} + \\frac{\\hat{B}}{2} = \\widehat{\\Gamma BM} + \\widehat{\\Gamma BI} = \\widehat{IBM}.\n$$\nThus the triangle $IBM$ is isosceles with $MB = MI$. Therefore $M$ lies on the perpendicular bisector of $BI$. Since $M$ belongs to the diameter of the circle $C$, it is the center of $C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31118,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $AB\\Gamma$ is given with $\\hat{A} = 105°$ and $\\hat{\\Gamma} = \\frac{\\hat{B}}{4}$.\n\nα. Determine the measures of the angles $\\hat{B}$ and $\\hat{\\Gamma}$.\n\nβ. If $O$ is the center of the circumcircle of the triangle $AB\\Gamma$ and $\\Delta$ is the antipodal of $B$, prove that the distance of $\\Gamma$ from $B\\Delta$ is $\\frac{B\\Delta}{4}$.",
    "options": [],
    "answer": "B = 60°, Γ = 15°; and the perpendicular distance from Γ to the line BΔ equals BΔ/4.",
    "solution": "α. Since $\\hat{A} + \\hat{B} + \\hat{\\Gamma} = 180°$ and $\\hat{A} = 105°$, $\\hat{\\Gamma} = \\frac{\\hat{B}}{4}$, we have\n$$\n105° + \\hat{B} + \\frac{\\hat{B}}{4} = 180° \\Leftrightarrow \\hat{B} = 60°, \\text{ and hence } \\hat{\\Gamma} = 15°.\n$$\n\nβ. Since $OB = O\\Delta = O\\Gamma$, it follows that $\\widehat{B\\Gamma\\Delta} = 90°$. Moreover we have\n$$\n\\widehat{BO\\Gamma} = 360° - 2 \\cdot 105° = 150°.\n$$\nTherefore\n$$\n\\widehat{O\\Gamma B} = \\frac{180° - 150°}{2} = 15°.\n$$\n$$\n\\text{Hence } \\Gamma O\\Delta = 30° \\text{ and } \\Gamma E = \\frac{O\\Gamma}{2} = \\frac{B\\Delta}{4}.\n$$\n![](attached_image_1.png)\nFigure 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be positive real number such that $a^3 = 6(a+1)$. Prove that the equation $x^2 + ax + a^2 - 6 = 0$ has no real solution.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In order to have $x^2 + ax + a^2 - 6 \\neq 0$, for all $x \\in \\mathbb{R}$, it is enough the discriminant $\\Delta = 3(8-a^2)$ to be less than $0$, that is $\\Delta = 3(8-a^2) < 0$.\n\nIn fact, if we suppose that $\\Delta = 3(8-a^2) \\geq 0$, then\n$$\na^2 \\leq 8 \\Rightarrow a \\leq 2\\sqrt{2} \\Rightarrow \\frac{1}{a} \\geq \\frac{\\sqrt{2}}{4}.\n$$\nThe given condition becomes\n$$\na^2 = 6 + \\frac{6}{a} \\geq 6 + \\frac{6\\sqrt{2}}{4} = 6 + \\frac{3\\sqrt{2}}{2} > 8, \\text{ (contradiction).}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral with $AB = BC = CD$, $AC \\neq BD$ and let $E$ be the intersection point of its diagonals. Prove that $AE = DE$ if and only if $\\angle BAD + \\angle ADC = 120^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle BAC = \\angle BCA = \\alpha$, $\\angle CBD = \\angle CDB = \\beta$, and $S$ the point of intersection of the lines $AB$ and $DC$.\nIf $AE=DE$, then $\\frac{AE}{\\sin(2\\alpha + \\beta)} = \\frac{AB}{\\sin(\\alpha + \\beta)} = \\frac{CD}{\\sin(\\alpha + \\beta)} = \\frac{DE}{\\sin(\\alpha + 2\\beta)}$.\n\nNow is obvious that\n$$\n\\sin(2\\alpha + \\beta) = \\sin(\\alpha + 2\\beta), \\quad 0^{\\circ} < 2\\alpha + \\beta < 180^{\\circ}, \\quad 0^{\\circ} < \\alpha + 2\\beta < 180^{\\circ},\n$$\n$\\alpha \\neq \\beta$ (since $AC \\neq BD$), and therefore\n$$\n2\\alpha + \\beta + \\alpha + 2\\beta = 180^{\\circ} \\text{ or } \\alpha + \\beta = 60^{\\circ}.\n$$\nHence we have: $\\angle BAD + \\angle ADC = 120^{\\circ}$.\n\nConversely, if $\\angle BAD + \\angle ADC = 120^{\\circ}$ we have\n$$\n\\angle AEB = \\angle BAE + \\angle EDC \\quad (1)\n$$\n$$\n\\angle AEB = \\angle EAD + \\angle EDA \\quad (2)\n$$\nFrom (1) and (2) we have\n$$\n2(\\angle AEB) = \\angle BAD + \\angle ADC = 120^{\\circ}.\n$$\nHence $\\angle AEB = 60^{\\circ} = \\angle BSC$ and the quadrilateral $SBEC$ is inscribable. Therefore $\\angle BSE = \\angle BCE = \\angle BAE$ and $AE=ES$.\nSimilarly we prove that $ES=ED$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31121,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a convex cyclic quadrilateral. The diagonals $AC$ and $BD$ intersect at the point $E$. It is given that $AB = 39$, $AE = 45$, $AD = 60$ and $BC = 56$.\nDetermine the length of $CD$.",
    "options": [],
    "answer": "91/5",
    "solution": "From the similarity of triangles $BEC$ and $AED$ we $BE = 42$.\n\nWe put $DE = x$ and $CE = y$. From the similarity of triangles $ABE$ and $DEC$ we have\n$$\n\\frac{x}{AE} = \\frac{y}{BE} = \\frac{CD}{AB} \\Leftrightarrow \\frac{x}{15} = \\frac{y}{14} = \\frac{CD}{13} = t. \\quad (1)\n$$\nMoreover, from theorem of Ptolemy we find\n$$\n\\begin{aligned}\nBD \\cdot AC &= AD \\cdot BC + AB \\cdot CD \\\\\n\\Leftrightarrow (45+14t)(42+15t) = 60 \\cdot 56 + 39 \\cdot 13t \\\\\n\\Leftrightarrow 35t^2 + 126t - 245 = 0 \\Leftrightarrow t = \\frac{7}{5} \\text{ or } t = -5.\n\\end{aligned}\n$$\nSince $t$ must be positive, from (1) we have $t = \\frac{7}{5}$ and $CD = \\frac{91}{5}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31122,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the number $4\\nu + 3$, where $\\nu$ is an integer, is a multiple of $11$, find:\n(i) The form of the integer $\\nu$,\n(ii) The remainder of the division of $\\nu^4$ with $11$.",
    "options": [],
    "answer": "(i) ν = 11k + 2 for integer k. (ii) The remainder is 5.",
    "solution": "(i) Let $4\\nu + 3 = 11\\lambda$, $\\lambda \\in \\mathbb{Z}$. Then $\\nu = \\frac{11\\lambda - 3}{4} = 2\\lambda + \\frac{3(\\lambda - 1)}{4}$, whence $4$ must divide $3(\\lambda - 1)$. Since $(4,3) = 1$, we get that $4|(\\lambda - 1)$.\nHence $\\lambda - 1 = 4\\kappa$, $\\kappa \\in \\mathbb{Z}$, and therefore\n$$\n\\nu = 2(4\\kappa + 1) + 3\\kappa = 11\\kappa + 2, \\kappa \\in \\mathbb{Z}.\n$$\n\n(ii) We have\n$$\n\\nu^4 = (11\\kappa + 2)^4 = \\pi\\alpha\\lambda.11 + 2^4 = \\pi\\alpha\\lambda.11 + 16 = \\pi\\alpha\\lambda.11 + 5.\n$$\nThus the remainder is $5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral with $\\angle DAC = \\angle BDC = 36^\\circ$, $\\angle CBD = 18^\\circ$ and $\\angle BAC = 72^\\circ$. The diagonals $AC$ and $BD$ intersect at point $P$. Determine the measure of $\\angle APD$.",
    "options": [],
    "answer": "108",
    "solution": "In the rays $DA$ and $BA$ we take a point $E$ and $Z$, respectively, such that $AC = AE = AZ$.\n\nSince $\\widehat{DEC} = \\frac{\\widehat{DAC}}{2} = 18^\\circ = \\widehat{CBD}$, the quadrilateral $DEBC$ is cyclic.\n\nSimilarly, the quadrilateral $CBZD$ is cyclic, because\n$$\n\\widehat{AZC} = \\frac{\\widehat{BAC}}{2} = 36^\\circ = \\widehat{BDC}.\n$$\n\n![](attached_image_1.png)\nFigure 8\n![](attached_image_2.png)\nFigure 9\n\nTherefore the pentagon $BCDZE$ is inscribed in the circle $K(A, AC)$. It gives $AC = AD$ and $\\widehat{ACD} = \\widehat{ADC} = \\frac{180^\\circ - 36^\\circ}{2} = 72^\\circ$, which gives $\\widehat{ADP} = 36^\\circ$ and $\\widehat{APD} = 108^\\circ$.\n\n\nAlternative solution:\nSee the figure in the right. If $C_1$ is the symmetric point of $D$ with respect to the line $BC$, then $ABC_1D$ is inscribable. Since $\\widehat{BAC_1} = 72^\\circ$ and $\\widehat{BDD_1} = 36^\\circ$ we have that $DD_1$ is bisector of the angle $BDC_1$ and hence $C_1$ is the incenter of the triangle $BDC_1$. Therefore we have\n$$\n\\widehat{DC_1A} = \\widehat{AC_1B} = 36^\\circ \\quad \\text{and} \\quad \\widehat{DPA} = \\frac{72^\\circ + 144^\\circ}{2} = 108^\\circ.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31124,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $v$ is a positive integer and the number $v(v+3)$ is a perfect square, prove that $v$ is not a multiple of 3.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $v(v+3) = w^2$, $w \\in \\mathbb{Z}$. We assume that: $v = 3\\kappa$, $\\kappa \\in \\mathbb{N}^*$. Then we have:\n$$\nv(v+3) = 3\\kappa(3\\kappa+3) = 9\\kappa(\\kappa+1) = w^2 \\Leftrightarrow 3 \\cdot 3\\kappa(\\kappa+1) = w^2 \\Rightarrow 3|w^2\n$$\nfrom which, since 3 is a prime, it follows that $3|w$.\n$$\n\\text{Hence } \\kappa(\\kappa+1) = \\left(\\frac{w}{3}\\right)^2, \\text{ with } \\frac{w}{3} \\in \\mathbb{Z}, \\text{ and thus we have}\n$$\n$$\n\\kappa^2 < \\kappa(\\kappa+1) = \\left(\\frac{w}{3}\\right)^2 < (\\kappa+1)^2 \\Rightarrow \\kappa < \\left|\\frac{w}{3}\\right| < \\kappa+1, \\text{ a contradiction.}\n$$\nHence $v$ is not a multiple of 3.\n\n\nAlternative solution:\nLet $v(v+3) = w^2$, $w \\in \\mathbb{Z}$. Then we have\n$$\n\\begin{align*}\nv^2 + 2v + 1 \\le v^2 + 3v < v^2 + 4v + 4 &\\Rightarrow (v+1)^2 \\le w^2 < (v+2)^2, w \\in \\mathbb{Z} \\\\\n&\\Rightarrow w = v+1 \\text{ if } w = -(v+1).\n\\end{align*}\n$$\nIt follows that\n$$\nv(v+3) = (v+1)^2 \\Leftrightarrow 3v = 2v+1 \\Leftrightarrow v = 1 \\ne \\pi o \\lambda .3.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the integer $n$ for which $A - B$, where $A = \\sqrt{n^2 + 24}$ and $B = \\sqrt{n^2 - 9}$, is an integer.",
    "options": [],
    "answer": "n = -5 or n = 5",
    "solution": "It is necessary to have $n^2 - 9 \\ge 0 \\Leftrightarrow n \\ge 3$ or $n \\le -3$.\nLet $A - B = \\sqrt{n^2 + 24} - \\sqrt{n^2 - 9} = d \\in \\mathbb{Z}$. Then $d > 0$ and\n$$\n\\begin{aligned}\n\\sqrt{n^2 + 24} &= \\sqrt{n^2 - 9} + d \\\\\n\\Rightarrow n^2 + 24 &= n^2 - 9 + d^2 + 2d\\sqrt{n^2 - 9} \\\\\n\\Leftrightarrow d^2 &= 33 - 2d\\sqrt{n^2 - 9} < 33 \\Rightarrow d \\in \\{1, 2, 3, 4, 5\\}.\n\\end{aligned}\n$$\nMoreover\n$$\n\\sqrt{n^2 - 9} = \\frac{33 - d^2}{2d} \\Leftrightarrow n^2 = \\left(\\frac{33 - d^2}{2d}\\right)^2 + 9. \\quad (1)\n$$\nIf $d$ is even, that is $d = 2$ or $4$, then from (1) we have $n^2 \\notin \\mathbb{Z}$, absurd. For $d = 3$ or $5$, also $n^2 \\notin \\mathbb{Z}$, while for $d = 1$, we find $n = \\pm 5$. Therefore the integer we seek is $5$ or $-5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31126,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, the angle $\\alpha = \\hat{A}$ and the side $a = |BC|$ are given. It is known that $a = \\sqrt{rR}$, where $r$ is the inradius and $R$ is the circumradius. Determine all such triangles, that is, compute the sides $b$ and $c$ of all such triangles.",
    "options": [],
    "answer": "b, c = (a/2) * [5 + 4 cos A ± sqrt(16 cos^2 A + 8 cos A − 23)], with the feasibility condition 16 cos^2 A + 8 cos A − 23 ≥ 0, i.e., cos A ≥ (√24 − 1)/4, so 0 < A ≤ arccos((√24 − 1)/4).",
    "solution": "According to the cosine rule we have\n$$\nb^2 + c^2 - 2bc \\cos A = a^2. \\quad (1)\n$$\nIf $\\tau = \\frac{a+b+c}{2}$ the given condition can be written as\n$$\na^2 = rR = \\frac{(ABC)}{\\tau} \\cdot \\frac{a}{\\sin A} = \\frac{bc \\sin A}{(a+b+c) \\sin A} \\Leftrightarrow a^2 = \\frac{abc}{a+b+c} \\\\\n\\Leftrightarrow bc - 2a(b+c) = 2a^2. \\qquad (2)\n$$\nWe write (1) in the form\n$$\n(b+c)^2 - (1+\\cos A)bc = a^2. \\qquad (3)\n$$\nSince $b+c > 0$, from (2) and (3) we find\n$$\nb+c = a \\left(1 + 8 \\cos^2 \\frac{A}{2}\\right) \\qquad (4)\n$$\n$$\nbc = 4a^2 \\left(1 + 4 \\cos^2 \\frac{A}{2}\\right) \\qquad (5)\n$$\nFrom (3) and (4) it follows that $b$ and $c$ are the solutions of the quadratic equation\n$$\n\\begin{aligned}\n& t^2 - a \\left(1 + 8 \\cos^2 \\frac{A}{2}\\right) t + 4a^2 \\left(1 + 4 \\cos^2 \\frac{A}{2}\\right) = 0 \\\\\n\\Leftrightarrow & t^2 - a(5 + 4 \\cos A)t + 4a^2(3 + 2 \\cos A) = 0 \\\\\n\\Leftrightarrow & t = \\frac{a}{2} \\left[5 + 4 \\cos A \\pm \\sqrt{16 \\cos^2 A + 8 \\cos A - 23}\\right],\n\\end{aligned}\n$$\nprovided that $16 \\cos^2 A + 8 \\cos A - 23 \\ge 0$. Considering the trinomial $f(x) = 16x^2 + 8x - 23$ we have\n$$\nf(x) \\ge 0 \\Leftrightarrow x \\le \\frac{-1-\\sqrt{24}}{4} \\text{ or } x \\ge \\frac{\\sqrt{24}-1}{4}.\n$$\nThe first condition cannot be satisfied. From the second condition we have\n$$\n\\cos A \\ge \\frac{\\sqrt{24}-1}{4} \\Leftrightarrow 0 < A \\le \\arccos \\frac{\\sqrt{24}-1}{4}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given are 50 points in the plane, no three of them belonging to a same line. Each of these points is colored using one of four given colors. Prove that there is a color and at least 130 scalene triangles with vertices of that color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $50 = 4 \\cdot 12 + 2$, according to the pigeonhole principle we will have at least 13 points colored in the same color. Using these 13 points we construct $\\binom{13}{3} = 286$ different triangles, since there are no three collinear points.\n\nWe will prove that there are at most $12 \\cdot 13 = 156$ isosceles triangles. In fact, there are $\\binom{13}{2} = 78$ different line segments. Each of them can be the basis at most two different isosceles triangles (because there are not three collinear points). Therefore there are at most $\\binom{13}{2} \\cdot 2 = 78 \\cdot 2 = 156$ isosceles triangles. Thus we can construct at least $286 - 156 = 130$ scalene triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31128,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a circular ring with radii $R$ and $R-2r$, where $R = 11r$, we put non overlapping circles of radius $r$ tangent to the circles defining the circular ring. Determine the maximal number of these circles. (It is given that $9.94 < \\sqrt{99} < 9.95$)",
    "options": [],
    "answer": "31",
    "solution": "Let we can put $N$ non overlapping circles $C_i(K_i, r)$ into the given circular ring tangent to its border. The circle $C(O, R-r)$ has length greater than the perimeter $\\ell_{K_1...K_N K_1}$ of the polygon with vertices the centers of the circles $C_i(K_i, r)$, and therefore:\n$$\nN \\cdot 2r < \\ell_{K_1...K_N K_1} < 2\\pi(R-r) \\Rightarrow N < \\pi \\left( \\frac{R}{r} - 1 \\right). \\quad (1)\n$$\n\n![](attached_image_1.png)\nFigure 2\n![](attached_image_2.png)\nFigure 3\n\nLet $OA$ is tangent from $O$ to one of the circles $C_i(K_i, r)$. Then\n$$\nOA^2 = R(R-2r) \\Leftrightarrow OA = \\sqrt{R(R-2r)}.\n$$\nThe circle $C(O, OA)$ is tangent to the sides of the polygon line $K_1K_2,...,K_{N-1}K_N$ and we have\n$$\n2\\pi\\sqrt{R(R-2r)} < N \\cdot 2r + 2r \\Rightarrow \\pi\\sqrt{\\frac{R}{r}\\left(\\frac{R}{r}-2\\right)} - 1 \\le N \\quad (2)\n$$\nFrom (1) and (2) it follows that:\n$$\n\\pi\\sqrt{\\frac{R}{r}\\left(\\frac{R}{r}-2\\right)} - 1 \\le N < \\pi\\left(\\frac{R}{r}-1\\right),\n$$\nfrom which, because of the hypothesis $R = 11r$, we find\n$$\n\\pi\\sqrt{11(11-2)} - 1 \\le N < \\pi(11-1) \\\\\n\\Leftrightarrow \\pi\\sqrt{99} - 1 \\le N < 10\\pi \\approx 31.4 \\Leftrightarrow N = 31.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31129,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that there is a permutation $\\sigma$ of the set $\\{1,2,...,n\\}$ for which $\\sqrt{\\sigma(1)+\\sigma(2)+\\dots+\\sigma(n)}$ is a rational number.\n\n*: A permutation of the set $\\{1,2,...,n\\}$ is a one-to-one function of this set to itself.*",
    "options": [],
    "answer": "n = 1 or n = 3",
    "solution": "Let for some $n \\in \\mathbb{N}^*$ we have\n$$\n\\sqrt{\\sigma(1)+\\sqrt{\\sigma(2)}+\\sqrt{\\cdots+\\sqrt{\\sigma(n)}}} = r_1 \\in \\mathbb{Q}. \\quad (1)\n$$\nSquaring the two parts of (1) we conclude that the number $\\sqrt{\\sigma(2)+\\sqrt{\\sigma(3)}+\\sqrt{\\cdots+\\sqrt{\\sigma(n)}}}$ is rational. Similarly we obtain that, for each $k \\in \\{1, 2, \\dots, n\\}$, the number\n$$\nr_k := \\sqrt{\\sigma(k)+\\sqrt{\\sigma(k+1)}+\\sqrt{\\cdots+\\sqrt{\\sigma(n)}}} \\in \\mathbb{Q}.\n$$\nIn the sequel we denote by $a_k := \\sqrt{n+\\sqrt{n+\\sqrt{\\cdots+\\sqrt{n}}}}$, for each $k \\in \\mathbb{N}^*$.\nUsing induction we can prove that $a_k < \\sqrt{n+1}$, for every $k \\in \\mathbb{N}^*$, and therefore\n$$\nr_1 = \\sqrt{\\sigma(1)+\\sqrt{\\sigma(2)}+\\sqrt{\\cdots+\\sqrt{\\sigma(n)}}} < a_n < \\sqrt{n+1}.\n$$\nIf $\\ell$ is a positive integer satisfying the inequalities $\\ell^2 \\le n < (\\ell+1)^2$, then for some $i \\in \\{1, 2, \\dots, n\\}$ we have $\\sigma(i) = \\ell^2$.\nNow we distinguish two cases\n\n1ª case: $i \\neq n$\nThen we have\n$$ \\ell < \\sqrt{\\ell^2 + \\sqrt{\\sigma(i+1)} + \\sqrt{\\cdots + \\sqrt{\\sigma(n)}}} < \\sqrt{n+1} < \\ell+2, $$\nfrom which it follows that\n$$\n\\sqrt{\\ell^2 + \\sqrt{\\sigma(i+1)} + \\sqrt{\\cdots + \\sqrt{\\sigma(n)}}} = \\ell+1. \\quad (2)\n$$\nSquaring both parts of (2) we get\n$$ 2\\ell+1 = \\sqrt{\\sigma(i+1)+\\sqrt{\\cdots+\\sqrt{\\sigma(n)}}} < \\sqrt{n+1} < \\ell+2, $$\nfrom which it follows that $\\ell < 1$, which is absurd.\n\n2ª case: $i = n$\n* If $\\ell > 1$, then $\\ell^2 - 1 \\in \\{\\sigma(1), \\dots, \\sigma(n-1)\\}$. Let now $j < n$ such that $\\sigma(j) = \\ell^2 - 1$.\nSimilarly, as in the first case, we get\n$$\n\\begin{align*}\n\\ell < \\sqrt{\\ell^2 - 1 + \\sqrt{\\sigma(j+1) + \\sqrt{\\dots + \\sqrt{\\ell^2}}}} < \\sqrt{n+1} < \\ell + 2 \\\\\n\\Rightarrow \\sqrt{\\ell^2 - 1 + \\sqrt{\\sigma(j+1) + \\sqrt{\\dots + \\sqrt{\\ell^2}}}} = \\ell + 1 \\\\\n\\Rightarrow 2\\ell + 2 = \\sqrt{\\sigma(j+1) + \\sqrt{\\dots + \\sqrt{\\ell^2}}} < \\sqrt{n+1} < \\ell + 2,\n\\end{align*}\n$$\nwhich is absurd.\n* If $\\ell=1$, then $n \\in \\{1, 2, 3\\}$. Checking all the cases it is easy to see that only for $n=1$ and $n=3$ there exist permutations satisfying the relation we seek. More explicitly, for $n=1$ we have $\\sqrt{1}=1$, while for $n=3$ we have:\n$$\n\\sqrt{2+\\sqrt{3}+\\sqrt{1}}=2. \\text{ For } n=2 \\text{ does not exist such a permutation.}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 31130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $(K_1, r_1)$ and $(K_2, r_2)$ are tangent and also they are tangent to the sides of the rectangular $AB\\Gamma\\Delta$, with $AB = a$ and $B\\Gamma = b$.\n(i) Express the sides $a$ and $b$ with respect to the radius $r_1$ and $r_2$.\n(ii) If the common internal tangent of the two circles passes through $\\Delta$ compute the ratio $\\frac{r_1}{r_2}$ and find the length of $\\Delta K$.",
    "options": [],
    "answer": "(i) a = (sqrt(r1) + sqrt(r2))^2, b = 2*r2. (ii) r1/r2 = (3 - sqrt(5))/2 and ΔK = ((1 + sqrt(5))/2)*r2.",
    "solution": "(i) We draw the line segment $K_1H \\perp K_2Z$.\nIn the triangle $K_1HK_2$ we have $K_1K_2 = r_1 + r_2$ and $K_2H = r_2 - r_1$ and hence from Pythagorean theorem we find $K_1H = 2\\sqrt{r_1r_2}$. From the rectangular $K_1EZH$ we have $EZ = 2\\sqrt{r_1r_2}$. Moreover we have\n$$\nAB = AE+EZ+ZB \\Leftrightarrow a = r_1 + 2\\sqrt{r_1r_2} + r_2 = (\\sqrt{r_1} + \\sqrt{r_2})^2, \\\\ b = B\\Gamma = Z\\Theta = 2r_2.\n$$\n\n(ii) We have $\\Lambda E = \\Lambda K = \\Lambda Z = \\sqrt{r_1r_2}$. Let $\\Delta I = \\Delta K = \\Delta \\Theta = x$. From the orthogonal triangle $AB\\Delta$, since $x+r_2 = a$ and $x+r_1 = b$ we get\n$$\nr_1 + 2\\sqrt{r_1r_2} = 2r_2 - r_1 \\Leftrightarrow r_1 + \\sqrt{r_1r_2} - r_2 = 0 \\Leftrightarrow \\omega^2 + \\omega - 1 = 0, \\\\ \\text{where } \\omega = \\sqrt{\\frac{r_1}{r_2}} > 0. \\text{ Hence } \\omega = \\sqrt{\\frac{r_1}{r_2}} = \\frac{\\sqrt{5}-1}{2} \\text{ and } \\kappa = \\frac{r_1}{r_2} = \\frac{3-\\sqrt{5}}{2}.\n$$\nMoreover\n$$\n\\frac{a}{b} = \\frac{(\\sqrt{r_1} + \\sqrt{r_2})^2}{2r_2} = \\frac{1}{2} \\left( \\sqrt{\\frac{r_1}{r_2}} + 1 \\right)^2 = \\frac{1}{2} \\left( \\frac{\\sqrt{5}+1}{2} \\right) = \\frac{\\sqrt{5}+1}{4}.\n$$\n\n(iii) From the orthogonal triangle $AB\\Delta$ we have\n$$\n(x+r_1)^2 + (r_1+\\sqrt{r_1r_2})^2 = (x+\\sqrt{r_1r_2})^2 \\Leftrightarrow x = \\frac{(r_1+\\sqrt{r_1r_2})^2}{r_2-r_1}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Compute the sum\n$$\n\\Sigma = \\sqrt{1+\\frac{8 \\cdot 1^2-1}{1^2 \\cdot 3^2}} + \\sqrt{1+\\frac{8 \\cdot 2^2-1}{3^2 \\cdot 5^2}} + \\dots + \\sqrt{1+\\frac{8 \\cdot 1003^2-1}{2005^2 \\cdot 2007^2}}\n$$",
    "options": [],
    "answer": "2014024/2007",
    "solution": "For $v = 1, 2, \\dots, 1003$, we have\n$$\n1 + \\frac{8v^2 - 1}{(2v-1)^2 (2v+1)^2} = \\frac{(4v^2 - 1)^2 + 8v^2 - 1}{(4v^2 - 1)^2} = \\frac{16v^4}{(4v^2 - 1)^2} = \\left( \\frac{4v^2}{4v^2 - 1} \\right)^2\n$$\nHence we can write\n$$\n\\begin{aligned}\n\\sqrt{1+\\frac{8v^2-1}{(2v-1)^2(2v+1)^2}} &= \\sqrt{\\left(\\frac{4v^2}{4v^2-1}\\right)^2} = \\frac{4v^2}{4v^2-1} = \\frac{4v^2-1+1}{4v^2-1} = \\\\\n& 1+\\frac{1}{(2v-1)(2v+1)} = 1+\\frac{2v+1-(2v-1)}{2(2v-1)(2v+1)} = 1+\\frac{1}{2}\\left(\\frac{1}{2v-1}-\\frac{1}{2v+1}\\right).\n\\\\\n\\Sigma &= 1+\\frac{1}{2}\\left(1-\\frac{1}{3}\\right)+1+\\frac{1}{2}\\left(\\frac{1}{3}-\\frac{1}{5}\\right)+\\dots+1+\\frac{1}{2}\\left(\\frac{1}{2005}-\\frac{1}{2007}\\right) \\\\\n&= 1003+\\frac{1}{2}\\left(1-\\frac{1}{2007}\\right) = 1003+\\frac{1003}{2007} = \\frac{1003 \\cdot 2008}{2007} = \\frac{2014024}{2007}.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31132,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a given positive integer $n > 2$, let $C_1$, $C_2$, $C_3$ be the boundaries of three convex $n$-gons in the plane such that the sets $C_1 \\cap C_2$, $C_2 \\cap C_3$, $C_3 \\cap C_1$ are finite. Find the maximum number of points of the set $C_1 \\cap C_2 \\cap C_3$.",
    "options": [],
    "answer": "floor(3n/2)",
    "solution": "Let us first observe that, if a line intersects a convex $n$-gon at finitely many points, then the number of such points is at most $2$. Therefore any two of the $n$-gons may intersect in at most $2n$ points. Choose two of the $n$-gons, $C_1$, $C_2$, and say that their intersection points are $p_1, p_2, \\dots, p_k$. Thus $k \\le 2n$. Say that the union of the set of vertices of $C_1$ and $C_2$ is $\\{q_1, q_2, \\dots, q_{2n}\\}$. We note that it is possible to have $q_i = q_j$ for some $i \\ne j$.\n\nWe will define a one-to-one function $f$ from $\\{p_1, p_2, \\dots, p_k\\}$ to $\\{q_1, q_2, \\dots, q_{2n}\\}$ as follows. First of all, orient all $n$-gons in the clockwise direction. Thus, if one traverses an $n$-gon according to this orientation, the interior is on the right and the exterior is on the left. For every $p_i$, there exist precisely two line segments (of non-zero length) which are subsets of $C_1$ or $C_2$, say $[q_j, p_i]$ on $C_1$ and $[q_k, p_i]$ on $C_2$, such that one can approach to $p_i$ via these line segments in the clockwise direction. Suppose that the two vectors $p_i - q_j$ and $p_i - q_k$, in this order, form a right handed coordinate system. Then none of the points on $[q_j, p_i]$ can be on or in the interior of $C_2$, since for any point $q$ on or in the interior of $C_2$, the vectors $p_i - q$ and $p_i - q_k$ are either positive multiples of each other, or form a left handed coordinate system. In this case we set $f(p_i) = q_j$. Otherwise we set $f(p_i) = q_k$. In both cases, the argument above shows that there are no other intersection points between $f(p_i)$ and $p_i$, in the clockwise direction.\n\nLet us now show that $f$ is 1-1. If $f(p_i) = f(p_l) = q$ and $q$ say (without loss of generality) belong to $C_1$, then the first intersection point encountered when one starts from $q$ and traverses $C_1$ in the clockwise direction has to be both $p_i$ and $p_l$, hence $p_i = p_l$.\n\nNow let us estimate the number of $p_i$'s that can be contained by the third polygon $C_3$. Each edge of $C_3$ contains exactly $0$, $1$ or $2$ of the $p_i$'s. Suppose that a given edge of $C_3$ contains $2$ of the $p_i$'s, say $p_1$ and $p_2$. Since $C_1$ and $C_2$ are convex and their intersection with $C_3$ is generic, they should have vertices between (in the clockwise sense) $p_1$ and $p_2$ (with this order), and outside $C_3$. We claim that at least one of these vertices is not in the set $f(C_1 \\cap C_2 \\cap C_3)$. Let $q_1 \\in C_1$ and $q_2 \\in C_2$ respectively be between (in the clockwise sense) $p_1$ and $p_2$ (with this order). If $q_1$ is on or in the interior of $C_2$ (or $q_2$ is on or in the interior of $C_1$), then $q_1$ (or $q_2$) is not in the image of $f$, since recall that $f(p)$ for any $p \\in C_1 \\cap C_2$ (thus for any $p \\in C_1 \\cap C_2 \\cap C_3$) is a point of one of $C_1$, $C_2$ not on or in the interior of the other. So the claim is established in this case. The remaining case is to assume that none of the vertices of any of $C_1$, $C_2$ that lie outside $C_3$ (and between (in the clockwise sense) $p_1$ and $p_2$ (with this order)) also lies in the interior of the other of $C_1$, $C_2$. In this case clearly the polygons $C_1$ and $C_2$ must meet at some point between (in the clockwise sense) $p_1$ and $p_2$ (with this order). Say $p_3$ is the closest to $p_1$ such point. Then clearly $f(p_3)$ is a vertex of one of $C_1$, $C_2$ between (in the clockwise sense) $p_1$ and $p_2$ (with this order). These parts of $C_1$, $C_2$ though, lie outside $C_3$; the interior of $C_3$ lies on the other side of the line $p_1p_2$. Thus $f(p_3)$ is not in $f(C_1 \\cap C_2 \\cap C_3)$ and the claim is established in all cases.\n\nFor the side $a$ of $C_3$ containing $p_1$, $p_2$ let us call $q(a)$ a vertex as the one in the claim we just proved. It is easy to see that for distinct sides $a$, $t$ of $C_3$ that contain two of the $p$'s, the points $q_a$, $q_b$ are distinct. Indeed, let $a$ contain $p_1$, $p_2$ and $b$ contain $p_3$, $p_4$ among the $p$'s. If one of $q_a$, $q_b$ belongs to one of $C_1$, $C_2$ and the other does not belong to it, we are okay. If both $q_a$, $q_b$ belong to say $C_1$, then in a clockwise tour around $C_1$ starting at $p_1$, we meet $p_1$, $p_2$, $p_3$, $p_4$ in this order. If not, say the order is $p_1$, $p_3$, $p_2$, $p_4$. Then the segments $p_1p_2$, $p_3p_4$ intersect at an interior point, since $C_1$ is a convex polygon. But then the sides $a$, $b$ of $C_3$ have a common interior point, a contradiction. So the correct order is $p_1$, $p_2$, $p_3$, $p_4$. But we know that adding $q(a)$, $q(b)$ in this tour the correct order is $p_1$, $q(a)$, $p_2$, $p_3$, $q(b)$, $p_4$. Thus $q(a)$, $q(b)$ are distinct as claimed.\n\nNow if $x$ of the edges of $C_3$ contain $1$ of the $p_i$'s and $y$ of them contain $2$ of the $p_i$'s, then $x + y \\le$ number of sides of $C_3$, i.e. $x + y \\le n$. The number of points in $C_1 \\cap C_2 \\cap C_3$ is $x + 2y$. Since $f$ is injective, $x + 2y$ is also the number of $q$'s in $f(C_1 \\cap C_2 \\cap C_3)$. Also, by the argument in the previous paragraph, we see that for every distinct edge of $C_3$ containing $2$ points we can assign a corresponding distinct $q_i$ outside the image of $f(C_1 \\cap C_2 \\cap C_3)$. Therefore $x$ is less or equal to the number of $q$'s that do not belong in $f(C_1 \\cap C_2 \\cap C_3)$. So $(x + 2y) + y$ is at most as much as the number of $q$'s. I.e. $x + 3y \\le 2n$. Adding this with $x + y \\le n$ and dividing by $2$, and also taking into account that $x + 2y$ is an integer\n\n$$\nx + 2y \\le \\left\\lfloor \\frac{3n}{2} \\right\\rfloor\n$$\n\nLet us now show that this is the best upper bound for every $n \\ge 3$. One way (among many) to construct an example is as follows: Construct two regular $n$-gons $C_1$, $C_2$ with the same center, such that their intersection points form a regular $2n$-gon. Call the vertices $p_1, p_2, \\dots, p_{2n}$ in a cyclic order. Let the circumcircle of this $2n$-gon be $C$. Then let the $n$-gon bounded by the lines $p_1p_3$, $p_5p_7$, $p_9p_{11}$, $\\dots$ (including $p_{2k+1}p_1$ in case $n$ is an odd $n = 2k+1$) together with the tangent lines to $C$ at $p_4$, $p_8$, $p_{12}$, $\\dots$ be $C_3$. It can easily be checked that $|C_1 \\cap C_2 \\cap C_3| = \\left\\lfloor \\frac{3n}{2} \\right\\rfloor$.\nLet $A$ and $B$ be two consecutive points of $C_1 \\cap C_2 \\cap C_3$ observed in the clockwise direction from a point in the interior of all three $n$-gons. Let's look for each $C_i$ its section in the clockwise direction between $A$ and $B$ excluding these points. If some two of these sections both do not contain any vertices of their corresponding $n$-gons, then the segment $AB$ belongs to both $n$-gons, a contradiction. Thus at least two of these segments have at least one vertex each, and moreover they do not contain the segment. Trivially, two distinct such vertices exist. Since there exist $|C_1 \\cap C_2 \\cap C_3|$ many consecutive points $A$ and $B$ of $C_1 \\cap C_2 \\cap C_3$, there should exist at least $2|C_1 \\cap C_2 \\cap C_3|$ distinct vertices of the three $n$-gons. Thus $2|C_1 \\cap C_2 \\cap C_3| \\le 3n$ i.e. $|C_1 \\cap C_2 \\cap C_3| \\le \\left\\lfloor \\frac{3n}{2} \\right\\rfloor$ since $|C_1 \\cap C_2 \\cap C_3|$ is an integer as well).\n\nActually we can achieve this upper bound by the example given in the Solution 1.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31133,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x > 1$ a non integer number. Prove that\n$$\n\\left( \\frac{x+\\{x\\}}{[x]} - \\frac{[x]}{x+\\{x\\}} \\right) + \\left( \\frac{x+[x]}{\\{x\\}} - \\frac{\\{x\\}}{x+[x]} \\right) > \\frac{9}{2},\n$$\nwhere $[x]$ and $\\{x\\}$ represents the integer and the fractional part of $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We put $[x] = a$, $\\{x\\} = r$, where $0 \\le r < 1$. Then the given inequality becomes\n$$\n\\left( \\frac{a+2r}{a} - \\frac{a}{a+2r} \\right) + \\left( \\frac{2a+r}{r} - \\frac{r}{2a+r} \\right) > \\frac{9}{2} \\\\\n\\Leftrightarrow 2 \\left( \\frac{r}{a} + \\frac{a}{r} \\right) - \\left( \\frac{a}{a+2r} + \\frac{r}{2a+r} \\right) > \\frac{5}{2}.\n$$\nSince $\\frac{r}{a} + \\frac{a}{r} \\ge 2$, it is enough to prove that\n$$\n\\frac{a}{a+2r} + \\frac{r}{2a+r} < \\frac{3}{2} \\Leftrightarrow 0 < 2a^2 + 11ar + 2r^2 \\Leftrightarrow 2(a+r)^2 + 7ar > 0,\n$$\nwhich is valid.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31134,
    "subject": "Mathematics (Multi-modal)",
    "question": "At each square of a $2007 \\times 2007$ chessboard we put one of the numbers $1$ or $-1$. We denote by $A_i$ the product of the numbers of the $i$-row, $i=1,2,\\ldots,2007$ and by $B_j$ the product of the numbers of the $j$-column, $j=1,2,\\ldots,2007$. Prove that:\n\n$$\nA_1 + A_2 + \\dots + A_{2007} + B_1 + B_2 + \\dots + B_{2007} \\neq 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $A_1A_2\\dots A_{2007} \\cdot B_1B_2\\dots B_{2007} = 1$, because each element of the table appears two times, one in a row and one in a column, and so the number of $(-1)$ in the product $A_1A_2\\dots B_{2007}$ is even, say, for example $2k$.\n\nTherefore the number of $+1$ will be $4014 - 2k$.\n\nIf $(4014-2k)(1) + 2k(-1) = 0 \\Rightarrow 4014 = 4k$, which is absurd, because $4$ does not divide $4004$. Hence $A_1 + A_2 + \\dots + A_{2007} + B_1 + B_2 + \\dots + B_{2007} \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31135,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each of the $50$ students of a class sent on Christmas period cards to exactly $25$ co-students. Prove that at least $2$ of the $50$ students each received the card of the other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$50$ students determine $\\frac{50 \\cdot 49}{2} = 1225$ different pairs. The cards sent are totally $50 \\cdot 25 = 1250 > 1225$, a number greater than the number of different pairs. Hence at least the members of one pair received each the card of the other.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31136,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the number $K = \\frac{9n^2 + 31}{n^2 + 7}$ is integer, find the possible values of $n \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "{-5, -3, -1, 1, 3, 5}",
    "solution": "We have\n$$\nK = \\frac{9n^2 + 31}{n^2 + 7} = \\frac{9(n^2 + 7) - 32}{n^2 + 7} = 9 - \\frac{32}{n^2 + 7}.\n$$\nSince $K$ is integer, it follows that $n^2 + 7$ is a divisor of $32$ and taking in mind that $n^2 + 7 \\ge 8$, we conclude:\n$$\nn^2 + 7 \\in \\{8, 16, 32\\} \\Leftrightarrow n^2 \\in \\{1, 9, 25\\} \\Leftrightarrow n \\in \\{-1, 1, -3, 3, -5, 5\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the values of the positive integer $n$ for which\n$$\nA = \\sqrt{\\frac{9n-1}{n+7}}\n$$\nis rational.",
    "options": [],
    "answer": "1, 11",
    "solution": "It is enough to prove that there exist $a, b \\in \\mathbb{N}^*$ with $(a, b) = 1$ such that:\n$$\n\\frac{9n-1}{n+7} = \\frac{a^2}{b^2} \\qquad (1)\n$$\nFrom this relation we get:\n$$\nn = \\frac{7a^2 + b^2}{9b^2 - a^2} = \\frac{7(a^2 - 9b^2) + 64b^2}{9b^2 - a^2} = -7 + \\frac{64b^2}{9b^2 - a^2} \\quad (2)\n$$\nSince $(a, b) = 1$, it follows that $(a^2, b^2) = 1$ and $(9b^2 - a^2, b^2) = 1$, and hence from (2) we get that $n$ is integer, if and only if $9b^2 - a^2$ is a divisor of 64.\nSince $a, b$ and $n$ are positive integers, it follows that $9b^2 - a^2 \\ge 8$, and hence:\n$$\n9b^2 - a^2 = (3b + a)(3b - a) \\in \\{8, 16, 32, 64\\}. \\quad (3)\n$$\nMoreover, the factors $3b + a$, $3b - a$ have sum a multiple of 6 and difference a multiple of 2 and $3b + a > 3b - a$. Therefore from relation (3) the possible cases are the following:\n$$\n\\begin{aligned}\n(3b + a, 3b - a) &= (4, 2) \\quad \\eta \\quad (3b + a, 3b - a) = (8, 4) \\quad \\eta \\quad (3b + a, 3b - a) = (16, 2) \\\\\n\\Leftrightarrow (a,b) &= (1,1) \\quad \\eta \\quad (a,b) = (2,2) \\quad \\eta \\quad (a,b) = (7,3).\n\\end{aligned}\n$$\nThe pair $(a,b) = (2,2)$ is rejected, because $\\text{gcd}(2,2) = 2 \\neq 1$, and therefore we have the values $n=1$ or $n=11$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31138,
    "subject": "Mathematics (Multi-modal)",
    "question": "A pupil has $7$ pieces of paper. He chooses some of them and cuts each of them into seven pieces. In the sequel, he chooses some of the pieces and cuts each of them into seven pieces. He continues this procedure many times with the pieces he has in hands every time. Is it possible to have some time $2009$ pieces of paper?",
    "options": [],
    "answer": "No",
    "solution": "Let he choose at the beginning $\\alpha_1$ from the seven pieces and each of them into seven pieces. Then he will have totally $7 - \\alpha_1 + 7\\alpha_1 = 7 + 6\\alpha_1$ pieces of paper. Suppose that in the next step he chooses $\\alpha_2$ pieces of paper and cuts each of them into seven pieces. Then he will have totally $7 + 6\\alpha_1 - \\alpha_2 + 7\\alpha_2 = 7 + 6(\\alpha_1 + \\alpha_2)$ pieces of paper. If he continues this procedure $\\kappa$ times, then he will have totally $7 + 6(\\alpha_1 + \\alpha_2 + \\dots + \\alpha_\\kappa)$ pieces of paper. Therefore we are looking for the value of $\\kappa$ satisfying the equation\n$$\n7 + 6(\\alpha_1 + \\alpha_2 + \\dots + \\alpha_\\kappa) = 2009 \\Rightarrow 6(\\alpha_1 + \\alpha_2 + \\dots + \\alpha_\\kappa) = 2002,\n$$\nwhich is absurd, because $2002$ is not divided by $6$. Hence it is not possible for him to have some time $2009$ pieces of paper.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $n \\ge 1$ for which there exists $n$ real numbers $x_1, x_2, \\dots, x_n$ in the closed interval $[-4, 2]$ such that the following three conditions are fulfilled:\n- the sum of these real numbers is at least $n$;\n- the sum of their squares is at most $4n$;\n- the sum of their fourth powers is at least $34n$:",
    "options": [],
    "answer": "All n divisible by 10",
    "solution": "Since the data of the problem concern $n$ real numbers $x_1, x_2, \\dots, x_n$ in the closed interval $[-4, 2]$, we consider the polynomial\n$$\nP(x) = (x+4)(x-2)(x-1)^2,\n$$\nwhich in $[-4, 2]$ satisfies the relation\n$$\nP(x) = (x+4)(x-2)(x-1)^2 \\le 0. \\quad (1)\n$$\nAdding by parts the inequalities coming from (1) for $x = x_1, x_2, \\dots, x_n$, and taking in mind the conditions of the problem, we find:\n$$\n0 \\ge P(x_1) + \\dots + P(x_n) = \\sum_{i=1}^{n} x_i^4 - 11 \\sum_{i=1}^{n} x_i^2 + 18 \\sum_{i=1}^{n} x_i - 8n \\ge 34n - 11 \\cdot 4n + 18n - 8n = 0. \\quad (2)\n$$\nHence, since $P(x_i) \\le 0$, for all $i = 1, 2, \\dots, n$, from relation (2) we have: $P(x_i) = 0$, for all $i = 1, 2, \\dots, n$, which means that $x_i \\in \\{-4, 1, 2\\}$, for all $i = 1, 2, \\dots, n$. We suppose that from the integers $x_1, x_2, \\dots, x_n$, $a$ are equal to $-4$, $b$ are equal to $1$ and $c$ are equal to $2$. Then we have $a+b+c=n$ and from the data of the problems we have the inequalities\n$$\n\\left\\{ \\begin{array}{l} -4a+b+2c \\ge a+b+c \\\\ 16a+b+4c \\le 4(a+b+c) \\\\ 256a+b+16c \\ge 34(a+b+c) \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} c \\ge 5a \\\\ b \\ge 4a \\\\ 222a \\ge 33b+18c \\end{array} \\right\\}.\n$$\nBy multiplying both parts of the first inequality with $18$ and the second with $33$ and summing the produced inequalities by parts we get the inequality\n$$\n33b+18c \\geq 132a+90a = 222a,\n$$\nwhich in combination with the inequality $222a \\geq 33b+18c$ gives:\n$$\n33b+18c = 222a,\n$$\nwhich is valid, if and only if $b = 4a$, $c = 5a$, that is\n$$\na+b+c=10a \\text{ or } 10a=n.\n$$\nTherefore the numbers $x_1, x_2, \\dots, x_n$ there exist, if and only if, $n$ is a multiple of $10$. For $n=10m$, where $m$ is a positive integer a possible solution arises by taking $m$ times the number $-4$, $4m$ times the number $1$ and $5m$ times the number $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31140,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a$ is an even positive integer and $A = a^n + a^{n-1} + \\dots + a + 1$, $n \\in \\mathbb{N}^*$, is a perfect square, prove that $a$ is a multiple of $8$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $a$ is an even positive integer, it follows that $A$ is odd. Therefore $A$ will be a perfect square of an odd integer, that is\n$$\nA = (2\\kappa + 1)^2 = 4\\kappa^2 + 4\\kappa + 1 = 4\\kappa(\\kappa + 1) + 1,\n$$\nwhere $\\kappa$ is a positive integer. Since one of the two integers $\\kappa$ and $\\kappa + 1$ is even, we have\n$$\n\\begin{align*}\nA &= 4\\kappa(\\kappa + 1) + 1 = 8\\rho + 1, \\text{ where } \\rho \\text{ is a positive integer} \\\\\n\\Rightarrow A - 1 &= a^n + a^{n-1} + \\dots + a = 8\\rho \\Rightarrow a(a^{n-1} + \\dots + a + 1) = 8\\rho \\\\\n\\Rightarrow 8|a(a^{n-1} + \\dots + a + 1) \\Rightarrow 8|a, \\text{ since } (8, a^{n-1} + \\dots + a + 1) = 1.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral inscribed in a circle $(O, R)$. With centers the vertices of the quadrilateral and radius $R$ we draw circles $C_A(A, R)$, $C_B(B, R)$, $C_C(C, R)$, $C_D(D, R)$. Circles $C_A$ and $C_B$ meet at $K$, circles $C_B$ and $C_C$ meet at $L$, circles $C_C$ and $C_D$ meet at $M$ and the circles $C_D$, $C_A$ meet at $N$. (Points $K,L,M,N$ are the second common points of the corresponding circles, given that all of them pass through point $O$). Prove that the quadrilateral $KLMN$ is parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The line segment $AB$ connects the centers of the circles $C_A$ and $C_B$, and therefore it is the perpendicular bisector of the common chord $OK$. Since the circles $C_A$ and $C_B$ have the same radius, the quadrilateral $AOBK$ is rhombus. Thus point $K_1$ is the middle of $AB$.\n\n![](attached_image_1.png)\n\nSimilarly, we can show that $L_1$ is the middle of $BC$, $M_1$ is the middle of $CD$ and $N_1$ is the middle of $AD$.\nFrom the triangles $OKL$, $OLM$, $OMN$ and $ONK$ we conclude that: $KL \\parallel K_1L_1$, $LM \\parallel L_1M_1$, $MN \\parallel M_1N_1$ and $NK \\parallel N_1K_1$ (because the line segments $K_1L_1$, $L_1M_1$, $M_1N_1$ and $N_1K_1$ connect the middles of the sides of a triangle).\nHence the quadrilaterals $KLMN$ and $K_1L_1M_1N_1$ have their sides parallel. But we know that the middles of the sides of a quadrilateral define a parallelogram. So the quadrilateral $KLMN$ is parallelogram.\nSince the line segment $KO$ is the common chord of the equal circles $C_A$ and $C_B$, it is the perpendicular bisector of the line of the centers $AB$ and vice versa.\nHence the quadrilateral $KAOB$ is rhombus, and so\n\n$$\nAK = OB. \\tag{1}\n$$\n\nSimilarly the quadrilateral $LCOB$ is rhombus and\n$$\nCL = OB. \\tag{2}\n$$\nFrom (1) and (2) it follows that the quadrilateral $KACL$ is parallelogram, so\n$$\nKL = AC. \\tag{3}\n$$\nWorking similarly we can prove that the quadrilateral $MNAC$ is parallelogram and that\n$$\nNM = AC. \\tag{4}\n$$\nFrom (3) and (4) it follows that $KLMN$ is parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31142,
    "subject": "Mathematics (Multi-modal)",
    "question": "From the vertex $A$ of an equilateral triangle $ABC$ we draw the ray $Ax$ which intersects the side $B\\Gamma$ at $\\Delta$. On $Ax$ we consider a point $E$ such that $BA = BE$. Find the angle $A\\hat{E}\\Gamma$.",
    "options": [],
    "answer": "30°",
    "solution": "Since $BA = B\\Gamma = BE$, point $B$ is the circumcenter of the triangle $A\\Gamma E$. The angle $A\\hat{E}\\Gamma$ is inscribed to the circle $C(B, BA)$ and so:\n$$\nA\\hat{E}\\Gamma = \\frac{1}{2} A\\hat{B}\\Gamma = 30^{\\circ}.\n$$\nFrom $BA = BE$ and $BA = B\\Gamma$, we have $B\\Gamma = BE$, and the triangle $B\\Gamma E$ is isosceles. We draw the altitude from $B$, let $BZ$, $Z \\in \\Gamma E$.\n\n![](attached_image_1.png)\n\nThen $BZ$ is also median and bisector of the angle $E B \\Gamma$ of the triangle $B\\Gamma E$. Let $BZ$ meet $AE$ at $K$. Then the triangles $BK\\Gamma$ and $BKE$ are equal because they have:\n\n![](attached_image_2.png)\n\n$$\nB\\Gamma = BE,\\ BK \\text{ common side and } KB\\Gamma = KBE.\n$$\nTherefore we have: $B\\Gamma K = B\\hat{E}K$ and since\n$$\nB\\hat{E}K = B\\hat{A}K \\text{ it follows that } B\\Gamma K = B\\hat{A}K.\n$$\nThus quadrilateral $ABK\\Gamma$ is inscribable, and hence:\n$$\nZKE = BKA = B\\Gamma A = 60^{\\circ}\n$$\nFinally we get:\n$$\nA\\hat{E}\\Gamma = K\\hat{E}Z = 90^{\\circ} - EKZ = 90^{\\circ} - 60^{\\circ} = 30^{\\circ}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31143,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $AB\\Gamma$ is given and let $O$ its circumcenter and $A_1, B_1, \\Gamma_1$ the middles of its sides $B\\Gamma, A\\Gamma$ and $AB$, respectively. We consider the points $A_2, B_2, \\Gamma_2$ such that $\\overrightarrow{OA_2} = \\lambda \\cdot \\overrightarrow{OA_1}$, $\\overrightarrow{OB_2} = \\lambda \\cdot \\overrightarrow{OB_1}$ and $\\overrightarrow{\\Gamma_2} = \\lambda \\cdot \\overrightarrow{\\Gamma_1}$, with $\\lambda > 0$. Prove that the lines $AA_2, BB_2, \\Gamma\\Gamma_2$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the orthocenter of the triangle $AB\\Gamma$. Then $\\overrightarrow{AH} = 2 \\cdot \\overrightarrow{OA_1}$ and from $\\overrightarrow{OA_2} = \\lambda \\cdot \\overrightarrow{OA_1}$, we find: $\\overrightarrow{AH} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{OA_2}$.\nIf $AA_2$ meets $OH$ at $C$ (from the similarity of the triangles $CHA$ and $COA_2$), we have: $\\overrightarrow{HC} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{CO}$. It means that $AA_2$ passes through $C$ which divides $OH$ in ratio $\\frac{2}{\\lambda}$.\n\nSimilarly, we have $\\overrightarrow{BH} = 2 \\cdot \\overrightarrow{OB_1}$ and $\\overrightarrow{BH} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{OB_2}$.\nLet now $C'$ be the point of intersection of the lines $BB_2$ and $OH$. Then we have $\\overrightarrow{HC'} = \\frac{2}{\\lambda} \\cdot \\overrightarrow{C'O}$, which means that $BB_2$ passes through $C'$ which divides $OH$ in ratio $\\frac{2}{\\lambda}$.\n\nSimilarly, if $C''$ is the point of intersection of the lines $\\Gamma\\Gamma_2$ and $OH$, then we have that $\\Gamma\\Gamma_2$ passes through $C''$ which divides $OH$ in ratio $\\frac{2}{\\lambda}$.\n\n![](attached_image_1.png)\n\nSince the points $C$, $C'$, $C''$ coincide, the lines $AA_2, BB_2, \\Gamma\\Gamma_2$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31144,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ a triangle with $90^\\circ \\neq \\hat{A} \\neq 135^\\circ$. Let $D$ and $E$ be external points to the triangle such that $DAB$ and $EAC$ are isosceles triangles with right angles in $D$ and $E$, respectively. Let $F = BE \\cap CD$, and $M, N$ the midpoints of $BC, DE$, respectively.\nProve that, if three of the points $A, F, M, N$ are collinear, then the four points are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) If $M, N, F$ are collinear, we must have $DE \\parallel BC$, hence the distances from $D$ and $E$ to $BC$ be equal, and this is equivalent to $b = c$.\n\nb) If $A, M, F$ are collinear, $\\tan \\vec{BAM} = \\frac{b \\sin A}{c - b \\cos A}$, $\\tan \\vec{DAN} = \\frac{b \\cos A}{c - b \\sin A}$, and $\\vec{DAN} = 45^\\circ + \\vec{BAM} \\Leftrightarrow b = c$.\n\nc) If $A, M, N$ are collinear, we call $H = CD \\cap AB$, $G = BE \\cap AC$ and computing we have\n$$\n\\frac{AG}{GC} = \\frac{c \\sin (A + 45^\\circ)}{a \\sin (B + 45^\\circ)}\n$$\nand similarly $\\frac{AH}{HB}$. By Ceva the condition becomes $b = c$.\n\nd) if $A, F, N$ are collinear, to use Ceva in $DEF$ we consider $X = CD \\cap AE$, $Y = BG \\cap AD$ (eventually improper). We get\n$$\n\\vec{HCG} = \\vec{HBG}, \\text{ hence HG is antiparallel to } BC, i.e. \\ AH \\cdot AB = AG \\cdot AC \\Leftrightarrow b = c.\n$$\n\ne) If $b = c$, all four considered points are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the triangle $ABC$ has barycenter $G$ and circumcenter $O$. The perpendicular bisectors of $GA$, $GB$ and $GC$ intersect at the points $A_1$, $B_1$, $C_1$. Prove that $O$ is the barycentre of the triangle $A_1B_1C_1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$, $E$, $F$ be the middles of the sides $BC$, $AC$, $AB$, respectively.\n![](attached_image_1.png)\nLet, also $B_1C_1$, $A_1C_1$, $A_1B_1$ be the perpendicular bisectors of the line segments $GA$, $GB$ and $GC$, respectively. Then the points $A_1$, $B_1$ and $C_1$ are the circumcenters of the triangles $GBC$, $GAC$ and $GAB$, respectively. Hence $A_1D$, $B_1E$ and $C_1F$ are the perpendicular bisectors of the sides $BC$, $AC$ and $AB$, respectively, and therefore they will pass through the circumcenter $O$ of the triangle $ABC$.\n\nNext, we will show that $A_1D$, $B_1E$ and $C_1F$ are the medians of the triangle $A_1B_1C_1$. Let the extension of $A_1D$, meets $B_1C_1$ at $N$. We will prove that $N$ is the middle of the line segment $B_1C_1$.\n\nFrom the inscribe quadrilateral $AMEB_1$ ($\\hat{M} = \\hat{E} = 90^\\circ$), we have $M\\hat{A}E = M\\hat{B}_1E = \\hat{\\omega}$. Also, from the inscribe quadrilateral $DOEC$ ($\\hat{D} = \\hat{E} = 90^\\circ$), we get $E\\hat{C}D = E\\hat{O}N = \\hat{\\phi}$. Therefore the triangles $ADC$ and $B_1NO$ are similar, and so\n$$\n\\frac{NB_1}{NO} = \\frac{AD}{CD}. \\qquad (1)\n$$\nFrom the inscribe quadrilateral $AMFC_1$ ($\\hat{M} = \\hat{F} = 90^\\circ$), we have $M\\hat{A}F = M\\hat{C}_1F = \\hat{x}$ and similarly from $DOFB$ ($\\hat{D} = \\hat{F} = 90^\\circ$), we obtain that $F\\hat{B}D = F\\hat{O}N = \\hat{y}$. From the above equalities the triangles $ADB$ and $C_1NO$ are similar and therefore:\n$$\n\\frac{NC_1}{NO} = \\frac{AD}{BD}. \\qquad (2)\n$$\nFrom (1) and (2) we get $NB_1 = NC_1$. In a similar way we prove that $B_1E$, $C_1F$ are the other two medians of the triangle $A_1B_1C_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha$, $\\beta$, $\\gamma$ are positive integers such that the number\n$$\nA = \\frac{\\alpha\\sqrt{2} + \\beta\\sqrt{3}}{\\beta\\sqrt{2} + \\gamma\\sqrt{3}}\n$$\nis rational. Prove that the number\n$$\nB = \\frac{\\alpha^2 + \\beta^2 + \\gamma^2}{\\alpha + \\beta + \\gamma}\n$$\nis integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First of all, it is easy to see that:\n$$\n\\alpha_1\\sqrt{2} + \\alpha_2\\sqrt{3} = \\alpha_3\\sqrt{2} + \\alpha_4\\sqrt{3}, \\text{ with } \\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4 \\in \\mathbb{Q}^* \\Leftrightarrow \\alpha_1 = \\alpha_3 \\text{ and } \\alpha_2 = \\alpha_4.\n$$\nIn fact, we can write the first relation in the form\n$$\n(\\alpha_1 - \\alpha_3)\\sqrt{2} = (\\alpha_4 - \\alpha_2)\\sqrt{3},\n$$\nand if $\\alpha_1 - \\alpha_3 \\neq 0$, then $\\frac{\\alpha_4 - \\alpha_2}{\\alpha_1 - \\alpha_3} = \\frac{\\sqrt{3}}{\\sqrt{2}}$, absurd. Hence $\\alpha_1 = \\alpha_3$ and $\\alpha_2 = \\alpha_4$.\nThe converse is clear.\nLet now $\\frac{\\alpha\\sqrt{2} + \\beta\\sqrt{3}}{\\beta\\sqrt{2} + \\gamma\\sqrt{3}} = \\kappa \\in \\mathbb{Q}$. Then $\\alpha = \\kappa\\beta$ and $\\beta = \\kappa\\gamma$, that is\n$$\n\\frac{\\alpha}{\\beta} = \\frac{\\beta}{\\gamma} \\Leftrightarrow \\beta^2 = \\alpha\\gamma.\n$$\nHence we have:\n\n$$\n\\begin{aligned}\n\\alpha^2 + \\beta^2 + \\gamma^2 &= \\alpha^2 + \\alpha\\gamma + \\gamma^2 = \\alpha^2 + 2\\alpha\\gamma + \\gamma^2 - \\alpha\\gamma \\\\\n&= (\\alpha + \\gamma)^2 - \\beta^2 = (\\alpha + \\beta + \\gamma)(\\alpha - \\beta + \\gamma),\n\\end{aligned}\n$$\nand so $B = \\alpha - \\beta + \\gamma$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31147,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider the numbers:\n$$\nA = \\frac{1}{4} \\cdot \\frac{3}{6} \\cdot \\frac{5}{8} \\cdots \\frac{595}{598} \\cdot \\frac{597}{600} \\text{ and } B = \\frac{2}{5} \\cdot \\frac{4}{7} \\cdot \\frac{6}{9} \\cdots \\frac{596}{599} \\cdot \\frac{598}{601}.\n$$\nProve that: (α) $A < B$, (β) $A < \\frac{1}{5990}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(α) To each fraction of $A$ of the form $\\frac{2\\nu - 1}{2\\nu + 2}$, $\\nu = 1, 2, \\dots, 299$, corresponds a fraction from $B$ of the form $\\frac{2\\nu}{2\\nu + 3}$, $\\nu = 1, 2, \\dots, 299$. Since\n$$\n0 < \\frac{2\\nu - 1}{2\\nu + 2} < \\frac{2\\nu}{2\\nu + 3} \\quad \\text{for every } \\nu \\in \\mathbb{N}^{*},\n$$\nand so for $\\nu = 1, 2, \\dots, 299$, by multiplying by parts the above 299 inequalities we obtain $A < B$.\n\n(β) Since $A > 0$, from $A < B$ we get:\n$$\nA^2 < A \\cdot B = \\frac{1 \\cdot 2 \\cdot 3}{599 \\cdot 600 \\cdot 601} < \\frac{1}{100 \\cdot 599^2} = \\frac{1}{5990^2} \\Rightarrow A < \\frac{1}{5990}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31148,
    "subject": "Mathematics (Multi-modal)",
    "question": "Decide whether the integers $1, 2, \\ldots, 100$ can be arranged in the cells $C(i,j)$ of a $10 \\times 10$ matrix (where $1 \\le i, j \\le 10$), such that the following conditions are satisfied:\n(i) In every row, the entries add up to the same sum $S$.\n(ii) In every column, the entries also add up to this sum $S$.\n(iii) For every $k = 1, \\dots, 10$ the ten entries $C(i,j)$ with $i-j \\equiv k \\pmod{10}$ add up to $S$.",
    "options": [],
    "answer": "No, such an arrangement does not exist.",
    "solution": "The problem essentially asks for a magic square that satisfies an additional constant-sum property along the wrap-around diagonals.\nSuppose that such an arrangement of $1, 2, \\ldots, 100$ is possible. Since the sum of all entries is $\\frac{1}{2} \\cdot 100 \\cdot 101$, we get that $S = 505$ is an odd number. We partition the cells $C(i,j)$ into four sets: Set $A$ contains the cells with $i$ and $j$ both odd; set $B$ contains the cells with odd $i$ and even $j$; set $C$ contains the cells with even $i$ and odd $j$; and set $D$ contains the remaining cells with $i$ and $j$ both even. We denote the sum of all entries in $A, B, C, D$ by $S_A, S_B, S_C, S_D$, respectively.\n* Since $A$ and $B$ contain all cells in the odd rows, we get $S_A + S_B = 5S$.\n* Since $B$ and $D$ contain all cells in the even columns, we get $S_B + S_D = 5S$.\n* Since $A$ and $D$ contain all cells $C(i,j)$ with even $i-j$, we get $S_A + S_D = 5S$.\nAdding up these three equations yields $2(S_A + S_B + S_D) = 15S$. Since in this equation the left hand side is even and the right hand side is odd, we have the desired contradiction.",
    "topic": "Algebra",
    "subtopic": "Linear Algebra"
  },
  {
    "id": 31149,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the nonnegative real numbers $x$, $y$ and $z$ have sum $2$, prove that:\n$$\nx^2y^2 + y^2z^2 + z^2x^2 + xyz \\le 1.\n$$\nFor which values of $x$, $y$ and $z$ the equality is valid?",
    "options": [],
    "answer": "(1, 1, 0) or (1, 0, 1) or (0, 1, 1)",
    "solution": "We will use the well-known inequality $2\\alpha\\beta \\le \\alpha^2 + \\beta^2$, which is valid for all $\\alpha, \\beta \\in \\mathbb{R}$. The equality holds for $\\alpha = \\beta$. Thus we have\n\n$$\n\\begin{align*}\nx^2 y^2 + y^2 z^2 + z^2 x^2 + xyz &= \\frac{1}{2}(2x^2 y^2 + 2y^2 z^2 + 2z^2 x^2 + 2xyz) \\\\\n&= \\frac{1}{2}(2xy \\cdot xy + 2yz \\cdot yz + 2zx \\cdot zx + 2xyz) \\\\\n&\\le \\frac{1}{2}[xy(x^2 + y^2) + yz(y^2 + z^2) + zx(z^2 + x^2) + 2xyz] \\tag{1} \\\\\n&= \\frac{1}{2}[(xy + yz + zx)(x^2 + y^2 + z^2) - xyz^2 - yzx^2 - zxy^2 + 2xyz] \\\\\n&= \\frac{1}{2}[(xy + yz + zx)(x^2 + y^2 + z^2) - xyz(x + y + z - 2)] \\\\\n&= \\frac{1}{2}[(xy + yz + zx)(x^2 + y^2 + z^2)], \\quad (\\text{since } x + y + z = 2).\n\\end{align*}\n$$\nTill now we have shown that\n$$\nx^2 y^2 + y^2 z^2 + z^2 x^2 + xyz \\le \\frac{1}{2} \\left[ (xy + yz + zx)(x^2 + y^2 + z^2) \\right], \\quad (2)\n$$\nand the equality holds, as we see from (1), when:\n$x = y = z$ or $x = y$, $z = 0$ or $y = z$, $x = 0$ or $z = x$, $y = 0$.\nSince $x + y + z = 2$, equality holds when:\n$$\n(x, y, z) = \\left(\\frac{2}{3}, \\frac{2}{3}, \\frac{2}{3}\\right) \\quad \\text{or}~(1, 1, 0) \\quad \\text{or}~(1, 0, 1) \\quad \\text{or}~(0, 1, 1). \\qquad (3)\n$$\nIn the sequel we will use the known inequality $\\alpha\\beta \\le \\left(\\frac{\\alpha + \\beta}{2}\\right)^2$, $\\alpha, \\beta \\in \\mathbb{R}$,\nputting $\\alpha = 2xy + 2yz + 2zx$, $\\beta = x^2 + y^2 + z^2$. Thus we have\n$$\n\\begin{align*}\n\\frac{1}{2} \\left[ (xy + yz + zx)(x^2 + y^2 + z^2) \\right] &= \\frac{1}{4} \\left[ (2xy + 2yz + 2zx)(x^2 + y^2 + z^2) \\right] \\\\\n&\\le \\frac{1}{4} \\left( \\frac{2xy + 2yz + 2zx + x^2 + y^2 + z^2}{2} \\right)^2 = \\frac{1}{16} (x+y+z)^4 = 1. \\tag{4}\n\\end{align*}\n$$\nFrom (2) and (4) we obtain the inequality\n$x^2 y^2 + y^2 z^2 + z^2 x^2 + xyz \\le 1.$\nThe equality holds when in inequality (4) we have:\n$\\alpha = \\beta \\Leftrightarrow 2xy + 2yz + 2zx = x^2 + y^2 + z^2,$\nwhich in consideration with (3) gives:\n$$\n(x, y, z) = (1, 1, 0) \\quad \\text{or}~(1, 0, 1) \\quad \\text{or}~(0, 1, 1).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of real numbers $(x, y, z)$ which are greater than $3$ and satisfy the equality:\n$$\n\\frac{(x+2)^2}{y+z-2} + \\frac{(y+4)^2}{z+x-4} + \\frac{(z+6)^2}{x+y-6} = 36.\n$$",
    "options": [],
    "answer": "(10, 8, 6)",
    "solution": "Since $x, y, z$ are greater than $3$, it follows that $y+z-2, z+x-4, x+y-6$ are positive. Thus, from Cauchy-Schwarz inequality we get:\n$$\n\\left( \\frac{(x+2)^2}{y+z-2} + \\frac{(y+4)^2}{x+z-4} + \\frac{(z+6)^2}{x+y-6} \\right) \\left( (y+z-2) + (x+z-4) + (x+y-6) \\right) \\geq (x+y+z+12)^2 \\\\\n\\Leftrightarrow \\frac{(x+2)^2}{y+z-2} + \\frac{(y+4)^2}{x+z-4} + \\frac{(z+6)^2}{x+y-6} \\geq \\frac{1}{2} \\cdot \\frac{(x+y+z+12)^2}{(x+y+z-6)}.\n$$\nFrom the hypothesis of the problem it follows that\n$$\n\\frac{(x+y+z+12)^2}{(x+y+z-6)} \\leq 72, \\qquad (1)\n$$\nwhere as the equality holds when:\n$$\n\\frac{x+2}{y+z-2} = \\frac{y+4}{x+z-4} = \\frac{z+6}{x+y-6} = \\lambda \\Leftrightarrow \\begin{cases} \\lambda(y+z) - x = 2(\\lambda+1) \\\\ \\lambda(x+z) - y = 4(\\lambda+1) \\\\ \\lambda(x+y) - z = 6(\\lambda+1) \\end{cases}. \\qquad (2)\n$$\n---\n\nMoreover, we observe that:\n$$\n\\frac{(x+y+z+12)^2}{(x+y+z-6)} = \\frac{(x+y+z+12)^2}{(x+y+z+12)-18} = \\frac{\\omega^2}{\\omega-18},\n$$\nwhere we have put $\\omega = x + y + z + 12$. Since we have\n$$\n\\frac{\\omega^2}{\\omega-18} \\geq 4 \\cdot 18 = 72 \\Leftrightarrow \\omega^2 - 4 \\cdot 18\\omega + 4 \\cdot 18^2 \\geq 0 \\Leftrightarrow (\\omega - 36)^2 \\geq 0,\n$$\nit follows that\n$$\n\\frac{(x+y+z+12)^2}{(x+y+z-6)} \\geq 72. \\qquad (3)\n$$\nEquality holds when:\n$$\n\\omega = x + y + z + 12 = 36 \\Leftrightarrow x + y + z = 24 \\qquad (4)\n$$\nFrom relations (1) and (3) follows that:\n$$\n\\frac{(x+y+z+12)^2}{(x+y+z-6)} = 72,\n$$\nand thus equations (2) are (4) are valid, and hence we have the system\n$$\n\\left\\{ \\begin{array}{l} (2\\lambda - 1)(x + y + z) = 12(\\lambda + 1) \\\\ x + y + z = 24 \\end{array} \\right\\} \\Rightarrow \\lambda = 1.\n$$\nFor $\\lambda = 1$, from relations (2) we get the system :\n$$\n\\left\\{ \\begin{array}{l} y+z-x=4 \\\\ x+z-y=8 \\\\ x+y-z=12 \\end{array} \\right\\} \\Leftrightarrow (x,y,z)=(10,8,6).\n$$\nTherefore the unique solution of the problem is the triple $(x, y, z) = (10, 8, 6)$, taking in mind that it satisfies the equation $x + y + z = 24$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x, y, z$ be positive real numbers. Prove that\n$$\n\\sum_{cyclic} \\frac{xy}{xy + x^2 + y^2} \\le \\sum_{cyclic} \\frac{x}{2x + z}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Given inequality is equivalent to:\n$$\n\\frac{1}{1+\\frac{x}{y}+\\frac{y}{x}}+\\frac{1}{1+\\frac{y}{z}+\\frac{z}{y}}+\\frac{1}{1+\\frac{z}{x}+\\frac{x}{z}} \\le \\frac{1}{2+\\frac{z}{x}}+\\frac{1}{2+\\frac{x}{y}}+\\frac{1}{2+\\frac{y}{z}}.\n$$\nNow we can take substitution $\\frac{x}{y} = a$, $\\frac{y}{z} = b$, $\\frac{z}{x} = c \\Rightarrow abc = 1$, so our inequality becomes:\n\n$$\n\\frac{1}{1+a+\\frac{1}{a}}+\\frac{1}{1+b+\\frac{1}{b}}+\\frac{1}{1+c+\\frac{1}{c}} \\le \\frac{1}{2+a}+\\frac{1}{2+b}+\\frac{1}{2+c}.\n$$\nAfter some computations and also using $abc = 1$, this becomes equivalent to:\n$$\n\\frac{3(a+b+c)+3(ab+bc+ca)+(ab+bc+ca)(a+b+c)}{(ab+bc+ca)^2+(ab+bc+ca)(a+b+c)+(a+b+c)^2} \\le \\frac{12+4(a+b+c)+(ab+bc+ca)}{9+2(ab+bc+ca)+4(a+b+c)}.\n$$\nNow, we take substitution $a+b+c = S$, $ab+bc+ca = P$ and inequality becomes:\n$$\n\\frac{3S+3P+SP}{P^2+PS+S^2} \\le \\frac{12+4S+P}{9+2P+4S} \\Leftrightarrow P^3+4S^3+3P^2S+PS^2+6P^2 \\ge 27S+27P+15PS \\quad (*)\n$$\nIt is not difficult to prove that $S^2 \\ge 3P$ and also $S \\ge 3, P \\ge 3$ (by AM $\\ge$ GM and $abc = 1$). Therefore:\n$$\n\\begin{aligned} 4S^3 &= 4S^2 \\cdot S \\ge 12PS, \\quad PS^2 = S \\cdot PS \\ge 3PS, \\quad 3P^2S \\ge 3 \\cdot 3^2 \\cdot S = 27S, \\\\ P^3 &= P^2 \\cdot P \\ge 3^2 \\cdot P = 9P, \\quad 6P^2 = 6P \\cdot P \\ge 6 \\cdot 3P = 18P. \\end{aligned}\n$$\nBy summing these inequalities we found (*) to be true so our proof is finished. Equality is obviously achieved when\n$$\nS^2 = 3P = 9 \\Rightarrow a = b = c = 1 \\Rightarrow x = y = z.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine positive integers $x$, $y$, $z$ which satisfy the system\n$$\n\\begin{aligned}\nx + y + z &= xy + yz + zx \\\\\nxyz &= 1.\n\\end{aligned}\n$$\nand have the least possible sum.",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "**First solution**\nWe write the system in the form\n$$\nxy + yz + zx = x + y + z \\quad (1)\n$$\n$$\nxyz = 1. \\quad (2)\n$$\nSubtracting the two equations by parts we find\n$$\n\\begin{aligned}\n& xyz - (xy + yz + zx) = 1 - (x + y + z) \\\\\n\\Leftrightarrow \\quad & xyz - xy - yz - zx + x + y + z - 1 = 0 \\\\\n\\Leftrightarrow \\quad & xy(z - 1) - x(z - 1) - y(z - 1) + (z - 1) = 0 \\\\\n\\Leftrightarrow \\quad & (z - 1)(xy - x - y + 1) = 0 \\\\\n\\Leftrightarrow \\quad & (z - 1)(x - 1)(y - 1) = 0 \\\\\n\\Leftrightarrow \\quad & x = 1 \\text{ or } y = 1 \\text{ or } z = 1.\n\\end{aligned}\n$$\nFor $x=1$, from (1) and (2) we have $yz=1$, which have the solutions\n$$\n(y, z) = \\left( a, \\frac{1}{a} \\right), \\quad a > 0,\n$$\nand so, the solutions of the system are\n$$\n(x, y, z) = \\left( 1, a, \\frac{1}{a} \\right), \\quad a > 0.\n$$\nSimilarly, considering $y=1$ or $z=1$ we find the solutions\n$$\n(x, y, z) = \\left( a, 1, \\frac{1}{a} \\right) \\text{ or } (x, y, z) = \\left( a, \\frac{1}{a}, 1 \\right), \\quad a > 0.\n$$\nSince for each $a > 0$ we have\n$$\nx + y + z = 1 + a + \\frac{1}{a} \\ge 1 + 2 = 3.\n$$\nEquality holds for $a=1$, it follows that between the solutions of the system, the $(x, y, z) = (1, 1, 1)$ is that having the least possible sum $x+y+z$.\n\n\n**Second solution**\nLet $(x, y, z)$ is the solution of the system with the least possible sum. Then, from the inequality of arithmetic – geometric mean we have\n$$\n\\frac{x+y+z}{3} \\ge \\sqrt[3]{xyz} \\Rightarrow x+y+z \\ge 3,\n$$\nwhile equality holds for $x = y = z$.\nHence the least possible value of the sum $x+y+z$, between the solution of the given system is 3 and it happens for $x = y = z$.\nFor $x = y = z$, from the equation $xyz = 1$, $x, y, z > 0$, it follows that $x = y = z = 1$, which satisfies also the equation $x + y + z = xy + yz + zx$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N}^* = \\{1,2,3,...\\}$ be the set of positive integers. Find all functions $f: \\mathbb{N}^* \\rightarrow \\mathbb{N}^*$ such that\n$$\nf(f(m)^2 + 2f(n)^2) = m^2 + 2n^2, \\text{ for all } m, n \\in \\mathbb{N}^*.\n$$",
    "options": [],
    "answer": "f(n) = n for all positive integers n",
    "solution": "First we prove that $f$ is injective. In fact, for any fixed $n$, if $f(m_1) = f(m_2)$, then: $m_1^2 + 2n^2 = f(f(m_1)^2 + 2f(n)^2) = f(f(m_2)^2 + 2f(n)^2) = m_2^2 + 2n^2$, whence $m_1^2 = m_2^2$ and $m_1 = m_2$.\n\nSince $f$ is injective we have:\n$$\nf(m)^2 + 2f(n)^2 = f(p)^2 + 2f(q)^2 \\Leftrightarrow m^2 + 2n^2 = p^2 + 2q^2 \\quad (1)\n$$\nPutting $f(1)=a$, we find for $m=n=1$, $f(3a^2)=3$. Then from (1) we get\n$$\nf(5a^2)^2 + 2f(a^2)^2 = f(3a^2)^2 + 2f(3a^2)^2 = 3f(3a^2)^2 = 27.\n$$\nSince the solution of the equation $x^2 + 2y^2 = 27$ in positive integers is $(x, y) = (3,3)$ and $(x, y) = (5,1)$, it follows that $f(a^2) = 1$ and $f(5a^2) = 5$.\n\nAlso, from (1) we have\n$$\n2f(4a^2)^2 - 2f(2a^2)^2 = f(5a^2)^2 - f(a^2)^2 = 24.\n$$\nSince the unique solution in positive integers of the equation $x^2 - y^2 = 12$ is $(x, y) = (4, 2)$, it follows that $f(2a^2) = 2$ and $f(4a^2) = 4$.\n\nUsing (1) again we can have\n$$\nf((k+4)a^2)^2 = 2f((k+3)a^2)^2 - 2f((k+1)a^2)^2 + f(ka^2)^2,\n$$\nas it arises easily by the identity $(k+4)^2 + 2(k+1)^2 = k^2 + 2(k+3)^2$, and therefore $f(ka^2) = k$, by induction on $k$. Then $f(a^2) = a = f(1)$ and thus $a = 1$. Hence $f(k) = k$, for any $k \\in \\mathbb{N}^*$. Finally it is easy to verify that $f(k) = k$, is a solution of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31154,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane are given $\\nu$ different points such that any three of them are not collinear. We color these points red, green and black. In the sequel we consider all line segments with ends these $\\nu$ points and we correspond to each of them an “algebraic value” according to the following rules:\n1) If at least one of the ends of the line segment is black, then it has algebraic value 0.\n2) If both ends of the line segment have the same colour, red or green then it has the algebraic value 1.\n3) If ends of the line segment have different colour red or green, then it has the algebraic value -1.\nDetermine the least possible value of the sum of algebraic values of the all line segments.",
    "options": [],
    "answer": "-⌊ν/2⌋",
    "solution": "From the three rules for the determination of the algebraic value of the line segments we have the following table:\n![](attached_image_1.png)\n\nLet now that we have $\\kappa$ red, $\\pi$ green and $\\mu$ black points. Then it is clear that $\\kappa + \\pi + \\mu = \\nu$.\n\nThe $\\kappa$ red points determine $\\binom{\\kappa}{2}$ line segments having their ends red and so they have algebraic values 1. The $\\pi$ green points determine $\\binom{\\pi}{2}$ line segments with both ends green and therefore with algebraic values 1. The number of the line segments having ends with different colors, red or green, and so with algebraic value -1, are $\\kappa \\cdot \\pi$. All the other line segments have algebraic value 0, because they have at least one of their ends black.\n\nThe sum of the algebraic values of all existing line segments is:\n$$\n\\begin{aligned}\n\\Sigma &= \\binom{\\kappa}{2} + \\binom{\\pi}{2} - \\kappa\\pi = \\frac{\\kappa!}{(\\kappa-2)!2!} + \\frac{\\pi!}{(\\pi-2)!2!} - \\kappa\\pi = \\\\\n&= \\frac{\\kappa(\\kappa-1)}{2} + \\frac{\\pi(\\pi-1)}{2} - \\frac{2\\kappa\\pi}{2} = \\frac{\\kappa^2 - \\kappa}{2} + \\frac{\\pi^2 - \\pi}{2} - \\frac{2\\kappa\\pi}{2} = \\\\\n&= \\frac{(\\kappa - \\pi)^2}{2} - \\frac{\\kappa + \\pi}{2} \\quad (\\text{because } \\kappa + \\pi + \\mu = \\nu) \\\\\n&= \\frac{(\\kappa - \\pi)^2}{2} - \\frac{\\nu - \\mu}{2} = \\frac{(\\kappa - \\pi)^2}{2} + \\frac{\\mu}{2} - \\frac{\\nu}{2}.\n\\end{aligned}\n$$\n\nFrom the last expression of $\\Sigma$ we conclude that:\n$$ \\Sigma \\geq -\\frac{\\nu}{2} \\quad (1) $$\n\nIf $\\nu$ is even (let $\\nu = 2\\rho$), then relation (1) becomes: $\\Sigma \\geq -\\rho$.\nEquality in the last relation holds, if and only if $\\kappa = \\pi = \\frac{\\nu}{2} = \\rho$ and $\\mu = 0$.\nFor example, for $\\nu = 4$, we have the following result:\n![](attached_image_2.png)\nLeast total sum: $\\Sigma = -\\frac{\\nu}{2} = -\\frac{4}{2} = -2$\n\nIf $\\nu$ is odd ($\\nu = 2\\rho + 1$), then relation (1) $\\gamma\\dot{\\iota}\\nu\\epsilon\\tau\\alpha\\iota$:\n$$ \\Sigma \\geq -\\frac{2\\rho + 1}{2} = -\\rho - \\frac{1}{2} $$\nSince $\\Sigma$ is integer we conclude that: $\\Sigma \\geq -\\rho = \\frac{-\\nu + 1}{2}$.\n\nWe check now when equality holds in the last relation. We observe that the case “$\\nu$ odd ($\\nu = 2\\rho + 1$)” comes from the case “$\\nu$ even ($\\nu = 2\\rho$)” by adding one more point. The point we add in the case “$\\nu$ even ($\\nu = 2\\rho$)” can be blank, red or green, and so we have the following cases:\n\n### Case 1\nLet the point we add is blank. Then the new produced line segments have algebraic values 0 and the equality in this case holds when: $\\kappa = \\pi = \\frac{\\nu - 1}{2}$ and $\\mu = 1$.\nThe sum of all algebraic values is: $\\Sigma = -\\rho = \\frac{-\\nu + 1}{2}$.\n![](attached_image_3.png)\nLeast total sum: $\\Sigma = \\frac{-\\nu + 1}{2} = \\frac{-5 + 1}{2} = -2$\n\n### Case 2\nLet the point we add is red. We had red and green points: $\\kappa = \\pi = \\frac{\\nu-1}{2}$. Now with the new point we can create $\\frac{\\nu-1}{2}$ line segments having algebraic value 1 and $\\frac{\\nu-1}{2}$ line segments having algebraic value -1. Equality in this case holds when\n$$\n\\kappa = \\frac{\\nu + 1}{2}, \\pi = \\frac{\\nu - 1}{2} \\text{ and } \\mu = 0.\n$$\nThe sum of all algebraic values remains: $\\Sigma = -\\rho = \\frac{-\\nu + 1}{2}$.\n![](attached_image_4.png)\n\n### Case 3\nIf the point we add is green, in a similar way we conclude that the equality holds when $\\kappa = \\frac{\\nu - 1}{2}$, $\\pi = \\frac{\\nu + 1}{2}$ and $\\mu = 0$. Again: $\\Sigma = -\\rho = \\frac{-\\nu + 1}{2}$.\n![](attached_image_5.png)\n\nFrom all the above we conclude that the least possible value of $\\Sigma$ is $-\\lfloor \\frac{\\nu}{2} \\rfloor$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $z_1, z_2, z_3, z_4, z_5, z_6$ be six pairwise different complex numbers which their images $A_1, A_2, A_3, A_4, A_5, A_6$ are consecutive points of the circle with center $O(0,0)$ and radius $r > 0$. If $w$ is a solution of the equation $z^2 + z + 1 = 0$ and\n$$\nz_1 w^2 + z_3 w + z_5 = 0 \\quad (I),\n$$\n$$\nz_2 w^2 + z_4 w + z_6 = 0 \\quad (II)\n$$\nProve that: (a) the triangle $A_1A_3A_5$ is equilateral,\n$$\n|z_1 - z_2| + |z_2 - z_3| + |z_3 - z_4| + |z_4 - z_5| + |z_5 - z_6| + |z_6 - z_1| = 3|z_1 - z_4| = 3|z_2 - z_5| = 3|z_3 - z_6|.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) Since $w$ is a root of the equation $z^2 + z + 1 = 0$, we have $w^2 + w + 1 = 0$. Multiplying both parts by $w$:\n$$\nw^3 + w^2 + w = 0 \\Leftrightarrow w^3 + \\underbrace{w^2 + w + 1}_{0} = 1 \\Leftrightarrow w^3 = 1.\n$$\nFrom the last equation we find $|w| = 1$. Substituting in relation (I) $w^2 = -w - 1$, we find:\n$$\nz_1(-1-w) + z_3w + z_5 = 0 \\Leftrightarrow -z_1 - z_1w + z_3w + z_5 = 0 \\Leftrightarrow (z_3 - z_1)w = z_1 - z_5.\n$$\nHence\n$$\n|(z_3 - z_1)w| = |z_1 - z_5| \\Leftrightarrow |z_3 - z_1| |w| = |z_1 - z_5| \\Leftrightarrow \\boxed{|z_3 - z_1| = |z_1 - z_5|} \\quad (A).\n$$\nSubstituting in relation (I) $w = -w^2 - 1$, we find:\n$$\nz_1 w^2 + z_3(-w^2 - 1) + z_5 = 0 \\Leftrightarrow z_1 w^2 - z_3 w^2 - z_3 + z_5 = 0 \\Leftrightarrow (z_1 - z_3)w^2 = z_5 - z_3.\n$$\nHence we have\n$$\n|(z_1 - z_3)w|^2 = |z_5 - z_3|^2 \\Leftrightarrow |z_1 - z_3|^2 |w|^2 = |z_5 - z_3|^2 \\Leftrightarrow \\boxed{|z_3 - z_1| = |z_5 - z_3|} \\quad (B).\n$$\nFrom (A) and (B) we obtain the equalities:\n$$\n|z_1 - z_3| = |z_3 - z_5| = |z_5 - z_1|,\n$$\nthat is the triangle $A_1A_3A_5$ is equilateral..\n\n(β) Similarly, using relation (II) we prove that the triangle $A_2A_4A_6$ is equilateral. From a known proposition of Euclidean Geometry we have that $A_1A_2 + A_1A_6 = A_1A_4$, and then using measures of complex numbers we have:\n\n$$\n|z_1 - z_2| + |z_6 - z_1| = |z_1 - z_4|. \\qquad (1)\n$$\nSimilarly, from the equality $A_3A_2 + A_3A_4 = A_3A_6$ using measures of complex numbers we get:\n$$\n|z_2 - z_3| + |z_3 - z_4| = |z_3 - z_6|. \\qquad (2)\n$$\nAlso, from equality $A_5A_4 + A_5A_6 = A_5A_2$ we find:\n$$\n|z_4 - z_5| + |z_5 - z_6| = |z_2 - z_5|. \\qquad (3)\n$$\n![](attached_image_1.png)\nSumming up by parts the relations (1), (2) and (3) and using the equalities\n$$\n|z_1 - z_4| = |z_3 - z_6| = |z_2 - z_5|\n$$\nwe find:\n$$\n|z_1 - z_2| + |z_2 - z_3| + |z_3 - z_4| + |z_4 - z_5| + |z_5 - z_6| + |z_6 - z_1| = \\\\ 3|z_1 - z_4| = 3|z_2 - z_5| = 3|z_3 - z_6|.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31156,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Let $n$ be a positive integer. Prove that\n$$\nn\\sqrt{x-n^2} \\le \\frac{x}{2}, \\text{ for all } x \\ge n^2.\n$$\n\nb. Determine real numbers $x$, $y$, $z$ satisfying the equation\n$$\n2\\sqrt{x-1} + 4\\sqrt{y-4} + 6\\sqrt{z-9} = x + y + z.\n$$",
    "options": [],
    "answer": "x = 2, y = 8, z = 18",
    "solution": "a. Since $x \\ge n^2$, we have\n$$\nn\\sqrt{x-n^2} \\le \\frac{x}{2} \\Leftrightarrow 2n\\sqrt{x-n^2} \\le x \\Leftrightarrow 4n^2(x-n^2) \\le x^2 \\Leftrightarrow (x-2n^2)^2 \\ge 0,\n$$\nwhich is valid. Equality holds if and only if $x = 2n^2$.\n\nAlternatively, for every $x \\ge n^2$, it is enough to prove that\n$$\nn\\sqrt{x-n^2} - \\frac{x}{2} \\le 0 \\Leftrightarrow 2n\\sqrt{x-n^2} - x \\le 0 \\Leftrightarrow \\frac{(2n\\sqrt{x-n^2} - x)(2n\\sqrt{x-n^2} + x)}{2n\\sqrt{x-n^2} + x} \\le 0\n$$\n$$\n\\Leftrightarrow \\frac{4n^2(x-n^2) - x^2}{2n\\sqrt{x-n^2} + x} \\le 0 \\Leftrightarrow \\frac{-(x-2n^2)^2}{2n\\sqrt{x-n^2} + x} \\le 0,\n$$\nwhich is valid. Equality holds for $x = 2n^2$.\n\nb. The given inequality can be written in the form\n$$\n(2\\sqrt{x-1}-x)+(4\\sqrt{y-4}-y)+(6\\sqrt{z-9}-z)=0, \\quad (1)\n$$\nfor $x \\ge 1$, $y \\ge 4$ and $z \\ge 9$.\n\nUsing (a) for $n = 1, 2, \\dots, 3$, we get\n$$\n2\\sqrt{x-1}-x \\le 0, \\quad 4\\sqrt{y-4}-y \\le 0 \\quad \\text{and} \\quad 6\\sqrt{z-9}-z \\le 0,\n$$\nAnd therefore (1) is possible to be valid, only for\n$$\n2\\sqrt{x-1}-x=0, \\quad 4\\sqrt{y-4}-y=0 \\quad \\text{and} \\quad 6\\sqrt{z-9}-z=0 \\Leftrightarrow x=2, y=8, z=18.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine prime positive integers $p$ and $q$ satisfying the equation\n$$\np^4 + p^3 + p^2 + p = q^2 + q.\n$$",
    "options": [],
    "answer": "p = 2, q = 5",
    "solution": "The given equation can be written as\n$$\np^4 + p^3 + p^2 + p = q^2 + q \\\\ \\Leftrightarrow p(p+1)(p^2+1) = q(q+1) \\quad (1)\n$$\n$$\n\\Leftrightarrow p(p^2-1)(p^2+1) = q(q+1)(p-1) \\quad (2)\n$$\nFor $q \\le p$ is not possible. Hence we should have $q > p$.\nTherefore from (2) we conclude that:\n$$\nq|(p^2-1)(p^2+1). \\quad (3)\n$$\nNext we distinguish the cases::\n* If $q \\le p^2$, then $q^2 \\le p^4$ and $q^2 + q \\le p^4 + p^2 < p^4 + p^2 + p^3 + p$. Hence the equation has no solutions..\n* If $q > p^2$, then $q \\ge p^2 + 1$, and hence (3), since $q$ is prime, gives $q = p^2 + 1$. Thus from equation (1) we obtain\n$$\np(p+1) = q+1 \\Leftrightarrow p^2 + p = p^2 + 2 \\Leftrightarrow p = 2\n$$\nTherefore, since $q = p^2 + 1 = 5$, we have the solution $(p,q) = (2,5)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31158,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider a $11 \\times 10$ cm rectangular table. The table is divided by parallel lines to $110$ squares of side $1$ cm. We have tiles of cross shape, consisting of $6$ squares of side $1$ cm, as in the figure. Determine the maximal possible number of non-overlapping tiles we can pack in the table such that each tile is covering exactly $6$ small squares of the table.\n![](attached_image_1.png)",
    "options": [],
    "answer": "14",
    "solution": "We observe that for the covering of the squares having a side on the border of the table and especially for the four corner squares we can put on each corner one tile covering only two squares, while four squares is not possible to be covered. Therefore in the four corners will be uncovered $16$ squares.\n\nFor the rest of the border squares we observe the following:\n* On the sides of length of $11$ cm for each covered square remains one also uncovered. So in these sides will remain at least two uncovered squares.\n* On the sides of length of $10$ cm we can cover one square and another one will remain uncovered.\n\nIn this way in our effort to cover the border squares will remain totally uncovered at least $16 + 4 + 2 = 22$ squares. Therefore it is possible to cover at most $110 - 22 = 88$ squares. Since each tile covers $6$ squares exactly, we finally can pack in the table at most $14$ tiles. In figure 5 it is shown that such packing is possible. Therefore the maximal number of tiles we can pack in the table is $14$.\n![](attached_image_2.png)\nFigure 5",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the integer solutions of the equation\n$$\n8x^3 - 4 = y(6x - y^2)\n$$",
    "options": [],
    "answer": "(1, 2)",
    "solution": "The given equation can be written as\n$$\n\\begin{aligned}\n8x^3 + y^3 - 6xy &= 4 \n\\Leftrightarrow (2x)^3 + y^3 + 1^3 - 3 \\cdot 2x \\cdot y \\cdot 1 = 5 \\\\\n&\\Leftrightarrow (2x + y + 1)(4x^2 + y^2 + 1 - 2xy - 2x - y) = 5\n\\end{aligned} \\quad (1)\n$$\n$$\n\\Leftrightarrow \\frac{1}{2}(2x+y+1)[(2x-y)^2+(2x-1)^2+(y-1)^2]=5 \\quad (2)\n$$\nFrom (2), since $(2x-y)^2 + (2x-1)^2 + (y-1)^2 > 0$, we have $2x + y + 1 > 0$, and hence from (1) we get\n$$\n2x + y + 1 = 1 \\text{ or } 2x + y + 1 = 5 \\Leftrightarrow 2x + y = 0 \\text{ or } 2x + y = 4.\n$$\nFrom (1), for $2x + y = 4$, we have\n$$\n4x^2 + y^2 + 1 - 2xy - 2x - y = 1 \\Leftrightarrow (2x + y)^2 - 6xy - (2x + y) = 0 \\Leftrightarrow xy = 2\n$$\nFrom the system $2x + y = 4$, $xy = 2$ we get the solution $(x, y) = (1, 2)$.\nAlso, from (1) for $2x + y = 0$ we have\n$$\n4x^2 + y^2 + 1 - 2xy - 2x - y = 5 \\Leftrightarrow (2x + y)^2 - 6xy - (2x + y) = 4 \\Leftrightarrow xy = -\\frac{2}{3}.\n$$\nHowever, from the system $2x + y = 0$, $xy = -\\frac{2}{3}$ we get no solutions.\n**We also can work in the following way:** The equation can be written as\n$$8x^3 + y^3 - 6xy = 4 \\Leftrightarrow (2x + y)^3 - 3 \\cdot 2x \\cdot y \\cdot (2x + y) - 6xy = 4,$$ and so by putting $2x + y = s$, $2xy = p$, we get\n$$\ns^3 - 3ps - 3p = 4 \\Rightarrow p = \\frac{s^3 - 4}{3(s + 1)}.\n$$\nSince $3p \\in \\mathbb{Z}$, we have that\n$$\n\\frac{s^3 - 4}{s + 1} = \\frac{s^3 + 1 - 5}{s + 1} = s^2 - s + 1 - \\frac{5}{s + 1} \\in \\mathbb{Z}.\n$$\n---\nThus $s+1$ must be a divisor of 5, that is\n$$\ns+1 \\in \\{-1, 1, -5, 5\\} \\text{ or } s \\in \\{-2, 0, -6, 4\\},\n$$\nAnd hence we find the pairs\n$$\n(s, p) = (-2, 4), (s, p) = \\left(0, -\\frac{4}{3}\\right), (s, p) = \\left(-6, \\frac{44}{3}\\right), (s, p) = (4, 4),\n$$\nFrom which only the last gives integer values for $x, y$, i. e. $x = 1, y = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31160,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a scalene triangle with $AB < AC$, with circumcircle $c(O, R)$. The circle $c_1(A, AB)$ intersects the side $BC$ at $E$ and the circle $c$ at $F$. The line $EF$ meets for a second time the circle $c$ at point $D$ and the side $AC$ at point $M$. The line $AD$ intersects the side $BC$ at point $K$. Finally, the circumcircle of the triangle $BKD$ intersects $AB$ at $L$. Prove that the points $K, L, M$ lie on a line parallel to the line $BF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The angle $\\hat{F}_1$ is inscribed into the circle $c_1$ with corresponding central angle $\\hat{BAE}$. Hence:\n![](attached_image_1.png)\nFigure 4\n---\n$$\n\\hat{F}_1 = \\frac{BA\\hat{E}}{2} = \\frac{\\hat{A}_1 + \\hat{A}_2}{2} \\qquad (1)\n$$\nFrom the cyclic quadrilateral *AFDB* we have:\n$$\n\\hat{F}_1 = \\hat{A}_1 \\qquad (2)\n$$\nFrom (1) and (2) we get that $\\hat{A}_1 = \\hat{A}_2$, that is $AK$ is the bisector of the angle $BA\\hat{E}$. Since $AB = AE$, we conclude that $AK$ and hence $DK$ is perpendicular to $BC$, that is\n$$\nDK \\perp BC \\qquad (3)\n$$\nIn the circle $c$, cords $AB$ and $AF$ are equal, as radii of circle $c_1$ and so $\\hat{D}_1 = \\hat{C}$. Hence the quadrilateral $DKMC$ is inscribable. Therefore we have $D\\hat{K}C = D\\hat{M}C = 90^\\circ$, that is:\n$$\nDM \\perp AC \\qquad (4)\n$$\nFrom the inscribed quadrilateral $BKDL$ we have $B\\hat{K}D = B\\hat{L}D = 90^\\circ$, that is:\n$$\nDL \\perp AB \\qquad (5)\n$$\nFrom the relations (3), (4) and (5) we conclude that the points $K,L,M$ are on the Simson's of the triangle $ABC$ corresponding to the point $D$.\nFrom the inscribable quadrilateral $DKMC$ we get that $\\hat{M}_1 = \\hat{C}_1$. Also from the inscribed quadrilateral $ABDC$ we have $\\hat{C}_1 = \\hat{A}_1$ and finally from the inscribed quadrilateral $ABDF$ we have $\\hat{A}_1 = \\hat{F}_1$. Hence $\\hat{M}_1 = \\hat{F}_1$ from which we get that $BF // LM$.\nWe work as above till relation (5) and we use the fact that $AK$ is the perpendicular bisector of $BE$ and that $AM$ is the perpendicular bisector of $EF$. Since $K$ is the midpoint of $EB$ and $M$ is the midpoint of $EF$, we conclude that:\n$$\nKM // BF \\qquad (6)\n$$\nFrom the inscribed quadrilateral $BKDL$ we have $\\hat{L}_1 = \\hat{D}_2$. Since $\\overline{AF} = \\overline{AB}$ the angles $\\hat{B}_1$ and $\\hat{D}_2$ are equal. Therefore we conclude that $\\hat{L}_1 = \\hat{B}_1$ and so\n$$\nKL // BF \\qquad (7)\n$$\nFrom relations (6) and (7) we get that $K,L,M$ are collinear and $BF // LM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in the circle $c(O, R)$ and let $K, L, M, N, S, T$ be the midpoints of the segments $AB, BC, CD, AD, AC$ and $BD$, respectively. Prove that the centers of the circumcircles of the triangles $KLS, LMT, MNS$ and $NKT$ form a cyclic quadrilateral similar to $ABCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $c_1, c_2, c_3, c_4$ be the circumcircles of the triangles $KLS, LMT, MNS$ and $NKT$, respectively. It is easy to prove that the quadrilaterals $KLMN, TLSN$ and $KSMT$ are parallelograms and so $KM, NL, TS$ will pass through the same point $G$ and therefore the points $K, S, L$ are symmetric of the points $M, T, N$, respectively, with respect to a symmetry with center point $G$.\nLet $c'_1$ be the circumcircle of the triangle $MTN$. Then $c'_1$ passes through the center $O$ of the circle $c(O,R)$ and it has diameter $OD = R$ (since $OMD = OND = 90^\\circ$).\nHowever the circles $c_1$ and $c'_1$ are symmetric with respect to $G$ (since the triangles defined the two circles are point to point symmetric with respect to $G$).\nHence the circle $c_1$ has radius $\\frac{R}{2}$ and is passing through $O'$ (the symmetric point of $O$ with respect to $G$).\n\n![](attached_image_1.png)\nFigure 6\n\nSimilarly we prove that the circles $c_2, c_3, c_4$ have radius $\\frac{R}{2}$ and are passing through $O'$. Therefore the centers of the circles $c_1, c_2, c_3, c_4$ belong to the circle with center $O'$ and radius $\\frac{R}{2}$.\nThe centers of the circles $c'_1, c'_2, c'_3$ and $c'_4$ (which are the midpoints of the segments $OA, OB, OC$ and $OD$, respectively) define a quadrilateral with sides parallel to the corresponding sides of $ABCD$. Hence the centers of the circles $c_1, c_2, c_3, c_4$ (symmetric of the centers of the circles $c'_1, c'_2, c'_3, c'_4$) define a quadrilateral similar to $ABCD$.\nWe will use the characteristic properties of the Euler circle of a triangle.\n\n![](attached_image_2.png)\nFigure 7\n\nThe center $O'$ of the Euler circle of a triangle belongs to Euler's line of the triangle. It is the midpoint of the segment $OH$ (where $O$ is the circumcentre and $H$ is the orthocenter of the triangle). The radius of the Euler's circle is the half of the radius of the circumcircle of the triangle.\nFrom the data of the problem we conclude that the circumcircles of the triangles $KLS$, $LMT$, $MNS$ and $NKT$ ($c_1, c_2, c_3, c_4$, respectively), are the Euler's circles of the triangles $ABC$, $BCD$, $CDA$ and $ABD$.\nThe triangles $ABC$, $BCD$, $CDA$ and $ABD$ are inscribed in the same circle $c(O,R)$. Hence the circles $c_1, c_2, c_3, c_4$ are equal with radius $\\frac{R}{2}$.\nWe consider the triangles $ABC$ and $ABD$. In the triangle $ABC$ we consider the centroid $G_1$ (point of intersection of the medians $BS$ and $CK$) and the center $O_1$ of the circle $c_1$ (circumcircle of the triangle $KLS$ and Euler's circle of the triangle $ABC$).\nIn the triangle $ABD$ we consider the centroid $G_4$ and the center $O_4$ of the circle $c_4$ (circumcircle of the triangle $KNT$ and Euler's circle of the triangle $ABD$).\nSince $G_1$ and $G_4$ are centroids we have:\n$$\n\\frac{G_1K}{G_1C} = \\frac{G_4K}{G_4D} = \\frac{1}{2} \\Rightarrow G_1G_4 \\parallel CD.\n$$\nSince the points $O_1$ and $O_4$ are the centers of the Euler's circles we have:\n$$\n\\frac{OG_1}{OO_1} = \\frac{OG_4}{OO_4} = 2 \\Rightarrow G_1G_4 \\parallel O_1O_4.\n$$\nTherefore we have: $CD \\parallel O_1O_4$.\nSimilarly we prove that the other sides of the quadrilateral $O_1O_2O_3O_4$ are parallel to the corresponding sides of the quadrilateral $ABCD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in the set of integers the equation\n$$\nx^3 y^2 (2y - x) = x^2 y^4 - 36.\n$$",
    "options": [],
    "answer": "[[3, 2], [2, 3], [-2, -3], [-3, -2], [3, 1], [1, 3], [-1, -3], [-3, -1], [-2, 1], [1, -2], [-1, 2], [2, -1]]",
    "solution": "After manipulation equation becomes\n$$\n\\begin{align*} x^2 y^2 (x-y)^2 - 6^2 &= 0, \\quad x, y \\in \\mathbb{Z} \\\\\n&\\Leftrightarrow [xy(x-y)-6][xy(x-y)+6] = 0 \\\\\n&\\Leftrightarrow xy(x-y) = 6, \\quad x, y \\in \\mathbb{Z} \\text{ or } xy(x-y) = -6, \\quad x, y \\in \\mathbb{Z} \\\\\n&\\Leftrightarrow xy(x-y) = 6, \\quad x, y \\in \\mathbb{Z} \\text{ (1) or } xy(y-x) = 6, \\quad x, y \\in \\mathbb{Z} \\text{ (2)}\n\\end{align*}\n$$\n---\nFrom (1) and (2) it is clear that, if $(x_0, y_0)$ is a solution of (1), then the pair $(y_0, x_0)$ is a solution of (2) and vice versa. Thus is enough to solve equation (1).\nSince $x, y \\in \\mathbb{Z}$, equation (1) is equivalent to:\n$$\n\\begin{array}{l}\n\\{xy = 6, x - y = 1\\} (\\Sigma_1) \\text{ or } \\{xy = -6, x - y = -1\\} (\\Sigma_2) \\\\\n\\text{or } \\{xy = 3, x - y = 2\\} (\\Sigma_3) \\text{ or } \\{xy = -3, x - y = -2\\} (\\Sigma_4) \\\\\n\\text{or } \\{xy = 1, x - y = 6\\} (\\Sigma_5) \\text{ or } \\{xy = -1, x - y = -6\\} (\\Sigma_6) \\\\\n\\text{or } \\{xy = 2, x - y = 3\\} (\\Sigma_7) \\text{ or } \\{xy = -2, x - y = -3\\} (\\Sigma_8).\n\\end{array}\n$$\nFrom the eight systems only $(\\Sigma_1), (\\Sigma_3), (\\Sigma_8)$ have integer solutions:\n$$\n\\begin{aligned}\n(x, y) &= (3, 2), \\quad (x, y) = (-2, -3), \\quad (x, y) = (3, 1), \\\\\n(x, y) &= (-1, -3), \\quad (x, y) = (-2, 1), \\quad (x, y) = (-1, 2).\n\\end{aligned}\n$$\nAccording to the previous discussion equation (2) has also the solutions:\n$$\n\\begin{aligned}\n(x, y) &= (2, 3), \\quad (x, y) = (-3, -2), \\quad (x, y) = (1, 3), \\\\\n(x, y) &= (-3, -1), \\quad (x, y) = (1, -2), \\quad (x, y) = (2, -1).\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31163,
    "subject": "Mathematics (Multi-modal)",
    "question": "A Mediterranean polynomial has only real roots and it is of the form\n$$P(x) = x^{10} - 20x^9 + 135x^8 + a_7x^7 + a_6x^6 + a_5x^5 + a_4x^4 + a_3x^3 + a_2x^2 + a_1x + a_0,$$ with real coefficients $a_0, a_1, \\ldots, a_7$. Determine the largest real number that occurs as a root of some Mediterranean polynomial.",
    "options": [],
    "answer": "11",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31164,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the Cartesian plane $Oxy$ we consider the points $A_1(40,1)$, $A_2(40,2)$, ..., $A_{40}(40,40)$, as well as the line segments $OA_1, OA_2, ..., OA_{40}$. A point of the Cartesian plane will be called \"good\" if its coordinates are integers and lies in the interior of a line segment $OA_i$, $i = 1, 2, 3, ..., 40$. Also a line segment from $OA_1, OA_2, ..., OA_{40}$, will be called \"good\" if it contains at least one \"good\" point. Determine the number of \"good\" points and the number of \"good\" segments.",
    "options": [],
    "answer": "140 good points and 24 good segments",
    "solution": "![](attached_image_1.png)\nFigure 2\n\nA point $M(k,l)$ will belong in the interior of the line segment $OA_i$, if and only if $\\overrightarrow{OM}$ and $\\overrightarrow{OA_i}$ have the same slope (with $k,l$ integers and $0 < k \\le 40$), that is\n$$\n\\frac{i}{40} = \\frac{l}{k} \\quad (\\text{with } k,l \\text{ integers and } 0 < k \\le 40).\n$$\nIn order the line segment $OA_i$ be \"good\" it is enough and sufficient the fraction $\\frac{i}{40}$ to be reducible (then we have fraction $\\frac{l}{k}$ with integer terms equivalent to $\\frac{i}{40}$ and its terms give the coordinates of the \"good\" point $M(k,l)$).\nHence, if $(40,i) > 1$, then the line segment $OA_i$ is \"good\" and we have $(40,i)-1$ \"good\" points at the segment $OA_i$. To the point $A_2(40,2)$ corresponds the \"good\" segment $OA_2$, which contains the good point $(20,1)$. To point $A_4(40,4)$ corresponds the \"good\" segment $OA_4$, which contain the \"good\" points $(10,1)$, $(20,2)$, $(30,3)$. Working in this way we finally find 24 \"good\" segments and 140 \"good\" points.\n\nAn easy solution can be given by using the Euler function. $\\phi$. It is known that the number of positive integers lower or equal to $n$ and not relatively prime to $n$ is $n-\\phi(n)$. Since $40 = 5 \\cdot 2^3$, we have\n$$\n\\phi(40) = 40 \\left(1 - \\frac{1}{2}\\right) \\left(1 - \\frac{1}{5}\\right) = 40 \\frac{14}{25} = 16.\n$$\nHence the number of good segments is $40 - \\phi(40) = 24$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31165,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider the set of four digit positive integers $x = \\overline{\\alpha\\beta\\gamma\\delta}$ with digits different than zero and pairwise different. We also consider the integers $y = \\overline{\\delta\\gamma\\beta\\alpha}$ and we suppose that $x > y$. Find the greatest and the lowest value of the difference $x - y$, as well as the corresponding four digit integers for which these values are obtained.",
    "options": [],
    "answer": "Greatest difference: 8532, achieved by x = 9821 and y = 1289. Lowest difference: 279, achieved by the pairs (x, y) = (3192, 2913), (4193, 3914), (5194, 4915), (6195, 5916), (7196, 6917), (8197, 7918).",
    "solution": "We consider the decimal representation of the integers\n$$\n\\begin{aligned}\nx - y &= 1000\\alpha + 100\\beta + 10\\gamma + \\delta - 1000\\delta - 100\\gamma - 10\\beta - \\alpha \\\\\n&= 999(\\alpha - \\delta) + 90(\\beta - \\gamma) = 9(111(\\alpha - \\delta) + 10(\\beta - \\gamma)).\n\\end{aligned}\n$$\nTherefore it is enough to find the greatest and the lowest values of the expression:\n$$\nA = 111(\\alpha - \\delta) + 10(\\beta - \\gamma).\n$$\nSince $\\alpha, \\beta, \\gamma, \\delta$ are digits pairwise different and no zero, we must have $\\alpha > \\delta$. The expression $A$ gets its maximal value when the integers $\\alpha - \\delta$ and $\\beta - \\gamma$ take their maximal value and moreover $\\alpha - \\delta > \\beta - \\gamma$. That is, $\\alpha - \\delta$ becomes maximal when $\\alpha = 9$ and $\\delta = 1$. The difference $\\beta - \\gamma$ becomes maximal when $\\beta = 8$ and $\\gamma = 2$. Hence $x = 9821$, $y = 1289$ and $x - y = 9821 - 1289 = 8532$.\nThe expression $A$ gets its minimal value when the integers $\\alpha - \\delta$ and $\\beta - \\gamma$ take their minimal values. The minimal value of the difference $\\alpha - \\delta$ is $1$ and so the pair $(\\alpha, \\delta)$ has as possible values:\n$$\n(9,8), (8,7), (7,6), (6,5), (5,4), (4,3), (3,2) \\text{ and } (2,1).\n$$\nFor all the above possible values of the pair $(\\alpha, \\delta)$, the value of $A$ is:\n$$\nA = 111 + 10(\\beta - \\gamma).\n$$\nThe lowest value of the difference $\\beta - \\gamma$ is $-8$, for $\\beta = 1$ and $\\gamma = 9$.\nTaking in mind that the digits must be different we find the following table of possible values:\n\n| $x$   | $y$   | $x-y$ |\n|-------|-------|-------|\n| 3192  | 2913  | 279   |\n| 4193  | 3914  | 279   |\n| 5194  | 4915  | 279   |\n| 6195  | 5916  | 279   |\n| 7196  | 6917  | 279   |\n| 8197  | 7918  | 279   |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$ be a finite set of positive reals, let $B = \\{\\frac{x}{y}: x, y \\in A\\}$ and let $C = \\{xy: x, y \\in A\\}$. If we denote $cardX = |X|$, show that: $|A| \\cdot |B| \\le |C|^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31167,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a, b, c$ are positive real numbers with sum $6$, determine the maximal value of the expression:\n$$\nS = \\sqrt[3]{a^2 + 2bc} + \\sqrt[3]{b^2 + 2ca} + \\sqrt[3]{c^2 + 2ab}.\n$$",
    "options": [],
    "answer": "3\\sqrt[3]{12}",
    "solution": "We use the inequality of arithmetic–geometric mean as follows:\n\n$$\n\\sqrt[3]{a^2+2bc} = \\frac{1}{\\sqrt[3]{12^2}} \\sqrt[3]{(a^2+2bc) \\cdot 12 \\cdot 12} \\le \\frac{1}{\\sqrt[3]{12^2}} \\cdot \\frac{a^2+2bc+12+12}{3} = \\frac{1}{3\\sqrt[3]{12^2}} (a^2+2bc+24),\n$$\n$$\n\\sqrt[3]{b^2+2ca} = \\frac{1}{\\sqrt[3]{12^2}} \\sqrt[3]{(b^2+2ca) \\cdot 12 \\cdot 12} \\le \\frac{1}{\\sqrt[3]{12^2}} \\cdot \\frac{b^2+2ca+12+12}{3} = \\frac{1}{3\\sqrt[3]{12^2}} (b^2+2ca+24),\n$$\n$$\n\\sqrt[3]{c^2+2ab} = \\frac{1}{\\sqrt[3]{12^2}} \\sqrt[3]{(c^2+2ab) \\cdot 12 \\cdot 12} \\le \\frac{1}{\\sqrt[3]{12^2}} \\cdot \\frac{c^2+2ab+12+12}{3} = \\frac{1}{3\\sqrt[3]{12^2}} (c^2+2ab+24),\n$$\nfrom which we get\n$$\n\\begin{aligned}\nS &= \\sqrt[3]{a^2+2bc} + \\sqrt[3]{b^2+2ca} + \\sqrt[3]{c^2+2ab} \\\\\n&\\le \\frac{1}{3\\sqrt[3]{12^2}} (a^2+b^2+c^2+2ab+2bc+2ca+72) \\\\\n&= \\frac{1}{3\\sqrt[3]{12^2}} \\left[ (a+b+c)^2 + 72 \\right] = \\frac{36}{\\sqrt[3]{12^2}} = \\frac{18}{\\sqrt[3]{18}} = 3\\sqrt[3]{12}.\n\\end{aligned}\n$$\nEquality holds when\n$$\na^2 + 2bc = 12, \\quad b^2 + 2ca = 12, \\quad c^2 + 2ab = 12\n$$\n$$\n\\Leftrightarrow (a-b)(a+b-2c) = 0, \\quad (b-c)(b+c-2a) = 0, \\quad c^2 + 2ab = 12\n$$\n$$\n\\Leftrightarrow (a-b)(6-3c) = 0, \\quad (b-c)(6-3a) = 0, \\quad c^2 + 2ab = 12\n$$\n$$\n\\Leftrightarrow a = b = c = 2.\n$$\nTherefore the maximal value of the expression is $3\\sqrt[3]{12}$ and it is obtained for $a = b = c = 2$.\n\n\nSolution 2:\nSolution with Hölder's inequality (P. Lolas, G. Vlachos, M. Agelis)\n\nWe have\n$$\n\\begin{aligned}\n\\sqrt[3]{9(a+b+c)^2} &= \\sqrt[3]{9 \\cdot \\sqrt[3]{(a^2+2bc)+(b^2+2ca)+(c^2+2ab)}} \\\\\n&= \\sqrt[3]{1+1+1} \\cdot \\sqrt[3]{(a^2+2bc)+(b^2+2ca)+(c^2+2ab)} \\\\\n&\\ge \\sqrt[3]{a^2+2bc} + \\sqrt[3]{b^2+2ca} + \\sqrt[3]{c^2+2ab} = S.\n\\end{aligned}\n$$\nHence we have $S \\le \\sqrt[3]{9 \\cdot 6^2} = 3\\sqrt[3]{12}$, where equality holds for\n$$\na^2 + 2bc = b^2 + 2ca = c^2 + 2ab, \\quad a+b+c=6\n$$\n$$\n\\Leftrightarrow (a-b)(a+b-2c) = 0, \\quad (b-c)(b+c-2a) = 0, \\quad a+b+c = 6\n$$\n$$\n\\Leftrightarrow (a-b)(6-3c) = 0, \\quad (b-c)(6-3a) = 0, \\quad a+b+c = 6\n$$\n$$\n\\Leftrightarrow a=b=c=2.\n$$\n\n\nSolution 3:\nSolution using Cauchy–Schwarz inequality (X. Tsampasidis)\n\nWe put $x = \\sqrt[3]{a^2 + 2bc}$, $y = \\sqrt[3]{b^2 + 2ca}$, $z = \\sqrt[3]{c^2 + 2ab}$. Then $x, y, z > 0$ and from Cauchy–Schwarz inequality we get\n$$\n\\begin{aligned}\n& (1^3 + 1^3 + 1^3)(1^3 + 1^3 + 1^3)(x^3 + y^3 + z^3) \\ge (1 \\cdot 1 \\cdot x + 1 \\cdot 1 \\cdot y + 1 \\cdot 1 \\cdot z)^3 \\\\\n& \\Leftrightarrow 9(x^3 + y^3 + z^3) \\ge (x+y+z)^3.\n\\end{aligned}\n$$\nHowever we have\n$$\nx^3 + y^3 + z^3 = a^2 + 2bc + b^2 + 2ca + c^2 + 2ab = (a+b+c)^2 = 36.\n$$\nThus the last inequality becomes\n$$\nx + y + z \\le \\sqrt[3]{9(x^3 + y^3 + z^3)} = \\sqrt[3]{9 \\cdot 36} = 3\\sqrt[3]{12}.\n$$\nEquality holds when $x = y = z, a + b + c = 6$ or equivalently\n$$\na^2 + 2bc = b^2 + 2ca = c^2 + 2ab, \\quad a + b + c = 6 \\Leftrightarrow a = b = c = 2.\n$$\n\n\nSolution 4:\nSolution using Jensen's inequality (K. Axiotis, N Athanasiou, D. Sotiriou)\n\nThe function $f(x) = \\sqrt[3]{x}, x > 0$, is twice differentiable in $(0, +\\infty)$, with $f'(x) = \\frac{1}{3}x^{-\\frac{2}{3}}$ and $f''(x) = -\\frac{2}{9}x^{-\\frac{5}{3}}$. Hence $f(x) = \\sqrt[3]{x}$ is concave in $(0, +\\infty)$. Hence from **Jensen's inequality** we get\n$$\n\\begin{aligned}\n& \\frac{1}{3}\\left[f(a^2+2bc)+f(b^2+2ca)+f(c^2+2ab)\\right] \\le f\\left(\\frac{a^2+2bc+b^2+2ca+c^2+2ab}{3}\\right) \\\\\n\\Rightarrow & f(a^2+2bc)+f(b^2+2ca)+f(c^2+2ab) \\le 3f\\left(\\frac{(a+b+c)^2}{3}\\right) = 3f(12) = 3\\sqrt[3]{12}. \\\\\n\\Rightarrow & \\sqrt[3]{a^2+2bc} + \\sqrt[3]{b^2+2ca} + \\sqrt[3]{c^2+2ab} \\le 3\\sqrt[3]{12}.\n\\end{aligned}\n$$\nSince for $a=b=c=2$ the first part takes the value $3\\sqrt[3]{12}$, it follows that the maximal value of the expression $S$ is $3\\sqrt[3]{12}$.\n\n\nSolution 5:\nSolution using power means inequality (A. Mousatov)\n\nWe put $x = \\sqrt{a^2 + 2bc}$, $y = \\sqrt{b^2 + 2ca}$, $z = \\sqrt{c^2 + 2ab}$. Then we have\n$$\nx + y + z = (a + b + c)^2 = 36.\n$$\nHence, if we put $u = \\sqrt[3]{x}$, $v = \\sqrt[3]{y}$, $w = \\sqrt[3]{z}$, then we get $S = u + v + w$ and from **power means inequality** we obtain\n$$\nS = u + v + w \\le 3 \\cdot \\sqrt[3]{\\frac{u^3 + v^3 + w^3}{3}} = 3 \\cdot \\sqrt[3]{12},\n$$\nsince $u^3 + v^3 + w^3 = x + y + z = 36$. Equality is valid when\n$$\nu = v = w \\Leftrightarrow a = b = c = 2.\n$$\n\n\nSolution 6:\nSolution using rearrangement inequality (K. Dermentzis)\n\nWe put $x = \\sqrt[3]{a^2 + 2bc}$, $y = \\sqrt[3]{b^2 + 2ca}$, $z = \\sqrt[3]{c^2 + 2ab}$. Then we have\n$$\nS = x + y + z \\text{ and } x^3 + y^3 + z^3 = (a+b+c)^2 = 36.\n$$\nNext we put\n$$\nS_1 = x^3 + y^3 + z^3 = x^2x + y^2y + z^2z, \\quad S_2 = x^2y + y^2z + z^2x, \\quad S_3 = x^2z + y^2x + z^2y\n$$\nAnd since the triads $(x^2, y^2, z^2)$, $(x, y, z)$ have the same ordering, from rearrangement inequality we obtain $S_1 \\ge S_2$ and $S_1 \\ge S_3$. Moreover, since $x, y, z > 0$, we have $S_1 = x^3 + y^3 + z^3 \\ge 3xyz$, and therefore\n$$\n\\begin{aligned}\nS^3 &= x^3 + y^3 + z^3 + 3(x^2y + y^2z + z^2x) + 3(x^2z + y^2x + z^2y) + 6xyz \\\\\n&= S_1 + 3S_2 + 3S_3 + 6xyz \\le S_1 + 3S_1 + 3S_1 + 2S_1 = 9S_1.\n\\end{aligned}\n$$\nHence\n$$\nS = x + y + z \\le \\sqrt[3]{9(x^3 + y^3 + z^3)} = \\sqrt[3]{9 \\cdot 12} = 3 \\cdot \\sqrt[3]{12}.\n$$\nSince for $a=b=c=2$ the expression $S$ takes the value $3\\sqrt[3]{12}$, it follows that its maximal value is $3\\sqrt[3]{12}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31168,
    "subject": "Mathematics (Multi-modal)",
    "question": "If the number $3\\nu + 1$, where $\\nu$ is integer, is a multiple of $7$, find the possible remainders of the following divisions:\n(a) of $\\nu$ with divisor $7$,\n(b) of $\\nu^m$ with divisor $7$, for all values of the positive integer $m$, $m \\ge 2$.",
    "options": [],
    "answer": "(a) 2; (b) 1, 2, or 4",
    "solution": "(a) Let $3\\nu + 1 = 7\\kappa$, where $\\nu, \\kappa \\in \\mathbb{Z}$. The integer $\\nu$ is of the form $\\nu = 7\\rho + \\upsilon$, where $\\upsilon \\in \\{0,1,2,3,4,5,6\\}$ and $\\rho \\in \\mathbb{Z}$. Then we have:\n$$\n3(7\\rho + \\upsilon) + 1 = 7\\kappa \\Leftrightarrow 21\\rho + 3\\upsilon + 1 = 7\\kappa \\Leftrightarrow 3\\upsilon + 1 \\equiv 0 \\pmod{7}\n$$\nand hence the possible value for $\\upsilon$ is $2$. Thus we have $\\nu = 7\\rho + 2$, where $\\rho \\in \\mathbb{Z}$,\nand the remainder of the division of $\\nu$ by $7$ is $2$.\n\n(b) We have\n$$\n\\nu^m = (7\\rho + 2)^m = \\sum_{i=0}^{m} \\binom{m}{i} (7\\rho)^{m-i} 2^i = \\text{multiple of }7 + 2^m.\n$$\nTherefore it is enough to find the remainders of the division of $2^m$ by $7$.\nIf $m = 3\\sigma + \\upsilon$, where $\\upsilon \\in \\{0,1,2\\}$, then we get:\n$$\n2^m = 2^{3\\sigma+\\upsilon} = 8^\\sigma \\cdot 2^\\upsilon = (7+1)^\\sigma \\cdot 2^\\upsilon = (\\text{multiple of }7+1) \\cdot 2^\\upsilon = \\text{multiple of }7 + 2^\\upsilon,\n$$\nwhere $\\upsilon \\in \\{0,1,2\\}$. Hence the possible remainders of the division of $\\nu^m$ by $7$, for all values of the positive integer $m$ are $2^0 = 1$, $2^1 = 2$, and $2^2 = 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31169,
    "subject": "Mathematics (Multi-modal)",
    "question": "A regular tetrahedron of height $h$ has a tetrahedron of height $xh$ cut off by a plane parallel to the base. When the remaining frustrum is placed on one of its slant faces on a horizontal plane, it is just on the point of falling over, that is the projection of the center of gravity $G$ of the frustrum is a point of the minor base of this slant face. Show that $x$ is a root of the equation:\n$$\nx^3 + x^2 + x = 2.\n$$",
    "options": [],
    "answer": "x^3 + x^2 + x = 2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle with $AB < AC$, inscribed in to the circle $c(O, R)$. The extension of the altitude $AD$ intersects the circumcircle $c(O, R)$ at $E$ and the perpendicular bisector ($\\mu$) of the side $AB$ meets $AD$ at $L$. The line $BL$ meets $AC$ at $M$ and the circumcircle $c(O, R)$ at $N$. Finally, $EN$ meets ($\\mu$) at $Z$. Prove that:\n$$\nMZ \\perp BC \\Leftrightarrow (CA = CB \\text{ or } Z \\equiv O).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "**Direct.** Let $MZ \\perp BC$. Then $MZ \\parallel AD$, since $AD \\perp BC$. Therefore we have: $Z\\hat{M}C = E\\hat{A}C$. From the quadrilateral $EANC$ we have: $E\\hat{A}C = E\\hat{N}C$ and hence we get\n$$\nZ\\hat{M}C = E\\hat{A}C = E\\hat{N}C = Z\\hat{N}C.\n$$\nTherefore the quadrilateral $MNCZ$ is cyclic and then it follows\n$$\nA\\hat{C}Z = M\\hat{C}Z = M\\hat{N}Z = B\\hat{N}E.\n$$\n![](attached_image_1.png)\nFigure 3\nAlso we have $B\\hat{N}E = B\\hat{A}E$, (from the cyclic quadrilateral $ABEN$). Therefore we have\n$$\nA\\hat{C}Z = B\\hat{A}E = B\\hat{A}D. \\qquad (1)\n$$\nFrom the right angled triangle $ABD$ we have\n$$\nB\\hat{A}D = 90^\\circ - A\\hat{B}D = 90^\\circ - A\\hat{B}C, \\qquad (2)\n$$\nand then\n$$\nA\\hat{C}Z = 90^\\circ - A\\hat{B}C. \\qquad (3)\n$$\nSince $O$ is the circumcentre of the acute angled triangle $ABC$ we get\n$$\nA\\hat{C}O = 90^\\circ - A\\hat{B}C. \\qquad (4)\n$$\nFrom (3) and (4) we obtain $A\\hat{C}Z = 90^\\circ - A\\hat{B}C = A\\hat{C}O$ and since the points $Z$ and $O$ lie on the same half plane with respect to $AC$, it follows that the points $C$, $O$ and $Z$ are collinear.\nTherefore we distinguish the cases:\n* $O \\equiv Z$, and then we have our result.\n* $O \\not\\equiv Z$, and then the straight line $OZ$ is the perpendicular bisector of the side $AB$. Since $C$ belongs to $OZ$, we conclude that $CA = CB$.\n\n**Converse**\nIf $CA = CB$ or $O \\equiv Z$, then the points $C$, $O$ and $Z$ are collinear, because they belong to the perpendicular bisector of the side $AB$. Hence\n$$A\\hat{C}Z = A\\hat{C}O = 90^\\circ - A\\hat{B}C = B\\hat{A}D = B\\hat{A}E = B\\hat{N}E = M\\hat{N}Z \\Rightarrow M\\hat{C}Z = M\\hat{N}Z,$$\nand so the quadrilateral $MNCZ$ is cyclic. It gives that\n$$\nZ\\hat{M}C = Z\\hat{N}C = E\\hat{N}C = E\\hat{A}C.\n$$\nTherefore $ZM \\parallel AE$ and since $AE \\perp BC$, it follows that $ZM \\perp BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the foot of the internal bisector of the angle $\\angle A$ of the triangle $ABC$. The straight line which joins the incenters of the triangles $ABD$ and $ACD$ cuts $AB$ and $AC$ at $M$ and $N$, respectively. Show that $BN$ and $CM$ meet on the bisector $AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31172,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$, $y$, $z$ are positive real numbers with sum $12$, prove that:\n$$\n\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} + 3 \\ge \\sqrt{x} + \\sqrt{y} + \\sqrt{z}.\n$$\n**When is equality valid?**",
    "options": [],
    "answer": "Equality holds when x = y = z = 4.",
    "solution": "Since $x$, $y$, $z$ are positive integers with sum $12$, it is enough to prove that\n$$\n\\frac{x}{y} + \\frac{y}{z} + \\frac{z}{x} + \\frac{x+y+z}{4} \\geq \\sqrt{x} + \\sqrt{y} + \\sqrt{z}. \\quad (1)\n$$\nFrom the inequality of the arithmetic–geometric mean for the positive integers $x$, $y$, $z$ we get\n\n$$\n\\frac{x}{y} + \\frac{y}{4} \\ge 2 \\sqrt{\\frac{x}{y} \\cdot \\frac{y}{4}} = \\sqrt{x}, \\qquad (2)\n$$\n$$\n\\frac{y}{z} + \\frac{z}{4} \\ge 2 \\sqrt{\\frac{y}{z} \\cdot \\frac{z}{4}} = \\sqrt{y}, \\qquad (3)\n$$\n$$\n\\frac{z}{x} + \\frac{x}{4} \\ge 2 \\sqrt{\\frac{z}{x} \\cdot \\frac{x}{4}} = \\sqrt{z}. \\qquad (4)\n$$\nSumming up (2), (3) and (4) we find inequality (1).\n\nEquality holds when all inequalities (2), (3) and (4) hold as equalities, that is when\n$$\n\\begin{align*}\n\\frac{x}{y} = \\frac{y}{4}, \\quad \\frac{y}{z} = \\frac{z}{4}, \\quad \\frac{z}{x} = \\frac{x}{4} &\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z = \\frac{1}{4} \\left(\\frac{y^2}{4}\\right)^2 = \\frac{y^4}{4^3} \\\\\n&\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z = \\frac{1}{4^3} \\left(\\frac{z^2}{4}\\right)^4 \\\\\n&\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z = \\frac{1}{4^7} z^8 \\\\\n&\\Leftrightarrow x = \\frac{y^2}{4}, \\quad y = \\frac{z^2}{4}, \\quad z^7 = 4^7 \\\\\n&\\Leftrightarrow x = y = z = 4.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral. Let $E$ and $F$ be points on the sides $AB$ and $CD$, respectively, such that $AB : AE = CD : DF = n$. If $S$ is the area of the quadrilateral $AEFD$, show that\n$$\nS \\le \\frac{AB \\cdot CD + n(n-1)DA^2 + nDA \\cdot BC}{2n^2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the non-positive integers $m$, $n$ satisfying the equation:\n$$\n\\frac{n(n+2)}{4} = m^4 + m^2 - m + 1\n$$",
    "options": [],
    "answer": "(m, n) = (1, 2)",
    "solution": "For $n = 0$, the equation becomes: $m^4 + m^2 - m + 1 = 0$, which is impossible, because $m^4 + m^2 - m + 1 = m^4 + m(m-1) + 1 > 0$, for all $m \\in \\mathbb{Z}_+$. \n\nFor $m = 0$, the equation becomes: $n(n+2) = 4$, impossible in the set of non-negative integers. \n\nFor $m, n \\neq 0$, the equation becomes:\n$$\nn^2 - 2n - 4(m^4 + m^2 - m + 1) = 0, \\quad (1)\n$$\nwhich is of second degree with respect to $n$ and in order to have a root in integers its discriminant must be a perfect square, i.e.\n$$\n\\begin{aligned}\n\\Delta &= 4(1 + 4m^4 + 4m^2 - 4m + 4) = \\text{perfect square} \\\\\n&\\Leftrightarrow 4m^4 + 4m^2 - 4m + 5 = \\text{perfect square.}\n\\end{aligned}\n$$\nHowever we have $4m^4 + 4m^2 - 4m + 5 = (2m^2)^2 + 4m(m-1) + 5 > (2m^2)^2$, for every positive integer $m$, and also we have:\n$$\n4m^4 + 4m^2 - 4m + 5 = (2m^2 + 1)^2 - 4(m-1) \\leq (2m^2 + 1)^2,\n$$\nwhere equality holds only for $m = 1$. Hence the unique possible value for $m$ is $m = 1$. Then the given equation becomes $n(n+2) = 8 \\Leftrightarrow n = 2$, and the pair $(m, n) = (1, 2)$ is a solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31175,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$, $y$ are positive real numbers prove that: $(x + \\frac{2}{y})(\\frac{y}{x} + 2) \\ge 8$.\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds at x = 1 and y = 2.",
    "solution": "Since $x$, $y > 0$ the given inequality can be written as:\n$$\n\\begin{align*}\n(xy+2)(y+2x) \\ge 8xy &\\Leftrightarrow xy^2 + 2y + 2x^2y + 4x - 8xy \\ge 0 \\\\\n&\\Leftrightarrow (xy^2 - 4xy + 4x) + (2x^2y - 4xy + 2y) \\ge 0 \\\\\n&\\Leftrightarrow x(y^2 - 4y + 4) + 2y(x^2 - 2x + 1) \\ge 0 \\\\\n&\\Leftrightarrow x(y-2)^2 + 2y(x-1)^2 \\ge 0, \\text{ which is valid for } x, y > 0.\n\\end{align*}\n$$\nEquality holds, if and only if:\n$$\nx(y-2) = 0 \\text{ and } 2y(x-1) = 0 \\Leftrightarrow x = 1, y = 2, \\text{ since } x, y > 0.\n$$\nSince $x$, $y > 0$ the given inequality can be written as:\n$$\n(xy+2)(y+2x) \\ge 8xy \\quad (1)\n$$\nWe apply the well-known inequality of arithmetic – geometric mean\n$$\na+b \\ge 2\\sqrt{ab}, \\text{ yea } a,b \\ge 0, \\text{ (equality when } a=b) \\text{ two times, for } a=xy, b=2 \\text{ and for } a=y, b=2x, \\text{ to receive:}\n$$\n$$\nxy + 2 \\ge 2\\sqrt{2xy}, \\quad \\text{(equality for } xy = 2\\text{)} \\quad (2)\n$$\n$$\ny+2x \\geq 2\\sqrt{2xy}, \\quad (\\text{equality for } y=2x) \\quad (3)\n$$\nFrom (2) and (3) (multiplication by parts) we get: $(xy+2)(y+2x) \\geq 8xy$.\nEquality holds, if and only if $xy=2$ and $y=2x$, i.e. $x=1$, $y=2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31176,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider the sequence of real numbers $(a_n)$, $n=1,2,3,...$\n$$\na_1 = 2 \\text{ and } a_n = \\left(\\frac{n+1}{n-1}\\right) (a_1 + a_2 + \\dots + a_{n-1}), \\quad n \\ge 2.\n$$\n\nDetermine the term $a_{2013}$.",
    "options": [],
    "answer": "2014 * 2^{2012}",
    "solution": "We observe that:\n$$\na_1 = 2,\\ a_2 = \\frac{3}{2} \\cdot a_1 = 3 \\cdot 2,\\ a_3 = \\frac{4}{2} \\cdot (a_1 + a_2) = \\frac{4}{2} \\cdot 4 \\cdot 2 = 4 \\cdot 2^2,\n$$\n$$\na_4 = \\frac{5}{3} \\cdot (a_1 + a_2 + a_3) = \\frac{5}{3} \\cdot 24 = 5 \\cdot 2^3,\n$$\n$$\na_5 = \\frac{6}{4} \\cdot (a_1 + a_2 + a_3 + a_4) = \\frac{6}{4} \\cdot 64 = 6 \\cdot 2^4.\n$$\nWe are going to use induction. Let $a_n = (n+1) \\cdot 2^{n-1}$, for $n=1,2,3,...,k$. We will prove that the same formula is valid for $n=k+1$, i.e.: $a_{k+1} = (k+2) \\cdot 2^k$.\n$$\na_{k+1} = \\frac{k+2}{k} (a_1 + a_2 + a_3 + \\dots + a_k) = \\frac{k+2}{k} \\left( 2 + 3 \\cdot 2^1 + 4 \\cdot 2^2 + \\dots + (k+1) \\cdot 2^{k-1} \\right).\n$$\nHence:\n$$\na_{k+1} = \\frac{k+2}{k} \\left( 2 \\cdot 2^0 + 3 \\cdot 2^1 + 4 \\cdot 2^2 + \\dots + (k+1) \\cdot 2^{k-1} \\right). \\quad (1)\n$$\nMultiplying both parts of relation (1) by 2 we get:\n$$\n2a_{k+1} = \\frac{k+2}{k} \\left( 2^1 + 3 \\cdot 2^2 + 4 \\cdot 2^3 + \\dots + (k+1) \\cdot 2^k \\right), \\quad (2)\n$$\nAnd then from (1) and (2) we find:\n$$\na_{k+1} = \\frac{k+2}{k} \\left( -2 - 2^1 - 2^2 - 2^3 - \\dots - 2^{k-1} + (k+1) \\cdot 2^k \\right) \\\\\na_{k+1} = \\frac{k+2}{k} \\left( -1 - \\frac{1-2^k}{1-2} + (k+1) \\cdot 2^k \\right) = \\frac{k+2}{k} \\left( -2^k + (k+1) \\cdot 2^k \\right) = (k+2) \\cdot 2^k.\n$$\nTherefore we have $a_{2013} = 2014 \\cdot 2^{2012}$\nFrom equations\n$$\na_n = \\left(\\frac{n+1}{n-1}\\right) (a_1 + a_2 + \\dots + a_{n-1}), \\quad n \\ge 2, \\quad (3)\n$$\n$$\na_{n+1} = \\left(\\frac{n+2}{n}\\right) (a_1 + a_2 + \\dots + a_n), \\quad n \\ge 1, \\quad (4)\n$$\nwe find\n$$\na_1 + a_2 + \\dots + a_{n-1} = \\left(\\frac{n-1}{n+1}\\right) a_n, \\quad n \\ge 2 \\quad (5)\n$$\n$$\na_1 + a_2 + \\dots + a_n = \\left(\\frac{n}{n+2}\\right) a_{n+1}, \\quad n \\ge 1 \\quad (6)\n$$\nAnd from (5) and (6) we get:\n$$\na_n = \\left(\\frac{n}{n+2}\\right) a_{n+1} - \\left(\\frac{n-1}{n+1}\\right) a_n \\Rightarrow a_{n+1} = \\left(\\frac{2(n+2)}{n+1}\\right) a_n, \\quad n \\ge 1 \\quad (7)\n$$\nHence\n$$\na_n = \\left(\\frac{2(n+1)}{n}\\right) a_{n-1} = \\left(\\frac{2(n+1)}{n}\\right) \\left(\\frac{2n}{n-1}\\right) a_{n-2} = \\dots \\\\\n= \\left(\\frac{2(n+1)}{n}\\right) \\left(\\frac{2n}{n-1}\\right) \\dots \\left(\\frac{2 \\cdot 4}{3} \\cdot \\frac{2 \\cdot 3}{2}\\right) a_1 \\\\\n= (n+1) \\cdot 2^{n-2} \\cdot a_1 = (n+1) \\cdot 2^{n-1}, \\text{ since } a_1 = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31177,
    "subject": "Mathematics (Multi-modal)",
    "question": "(α) Write the expression $A = k^4 + 4$, where $k$ is a positive integer, as a product of two factors each of them being a sum of two squares of integers.\n\n(β) Simplify the expression\n$$\nK = \\frac{\\left(2^4 + \\frac{1}{4}\\right)\\left(4^4 + \\frac{1}{4}\\right)\\left(6^4 + \\frac{1}{4}\\right) \\cdots \\left((2n)^4 + \\frac{1}{4}\\right)}{\\left(1^4 + \\frac{1}{4}\\right)\\left(3^4 + \\frac{1}{4}\\right)\\left(5^4 + \\frac{1}{4}\\right) \\cdots \\left((2n-1)^4 + \\frac{1}{4}\\right)}\n$$\nand write it as a sum of the squares of two successive integers.",
    "options": [],
    "answer": "A = [(k-1)^2 + 1^2][(k+1)^2 + 1^2]; K = (2n)^2 + (2n+1)^2",
    "solution": "(α) We have\n$$\n\\begin{aligned}\nk^4 + 4 &= (k^2)^2 + 4k^2 + 2^2 - 4k^2 = (k^2 + 2)^2 - (2k)^2 \\\\\n&= (k^2 + 2 - 2k)(k^2 + 2 + 2k) = [(k-1)^2 + 1^2][(k+1)^2 + 1^2].\n\\end{aligned}\n$$\n\n(β) We multiply both terms of the fraction by $(2^4)^n$, to receive:\n$$\n\\begin{aligned}\nK &= \\frac{\\left(2^4 + \\frac{1}{4}\\right)\\left(4^4 + \\frac{1}{4}\\right)\\left(6^4 + \\frac{1}{4}\\right) \\cdots \\left[(2n)^4 + \\frac{1}{4}\\right]}{\\left(1^4 + \\frac{1}{4}\\right)\\left(3^4 + \\frac{1}{4}\\right)\\left(5^4 + \\frac{1}{4}\\right) \\cdots \\left[(2n-1)^4 + \\frac{1}{4}\\right]} \\\\\n&= \\frac{(3^2 + 1)(5^2 + 1)(7^2 + 1)(9^2 + 1)(11^2 + 1) \\cdots [(4n-3)^2 + 1][(4n-1)^2 + 1][(4n+1)^2 + 1]}{(1^2 + 1)(3^2 + 1)(5^2 + 1)(7^2 + 1)(9^2 + 1)(11^2 + 1)(13^2 + 1) \\cdots [(4n-3)^2 + 1][(4n-1)^2 + 1]} \\\\\n&= \\frac{(4n+1)^2 + 1}{1^2 + 1} = 8n^2 + 4n + 1 = 4n^2 + 4n^2 + 4n + 1 = (2n)^2 + (2n+1)^2.\n\\end{aligned}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all ordered pairs $(a, b)$ of positive integers for which the numbers $\\frac{a^3b - 1}{a + 1}$ and $\\frac{b^3a + 1}{b - 1}$ are both positive integers.",
    "options": [],
    "answer": "[(1, 3), (2, 2), (3, 3)]",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31179,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$, the excircle $\\omega_a$ opposite $A$ touches $AB$ at $P$ and $AC$ at $Q$, and the excircle $\\omega_b$ opposite $B$ touches $BA$ at $M$ and $BC$ at $N$. Let $K$ be the projection of $C$ onto $MN$, and let $L$ be the projection of $C$ onto $PQ$. Show that the quadrilateral $MKLP$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31180,
    "subject": "Mathematics (Multi-modal)",
    "question": "Examine if the polynomial\n$$\nP(x) = (x^{2} - 2x + 5)(x^{2} - 4x + 20) + 1,\n$$\ncan be written as the product of two polynomials with integer coefficients of degree greater than 0.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an acute angled triangle $AB\\Gamma$, with $AB < A\\Gamma$. Let $M$ be the midpoint of the side $B\\Gamma$. On the side $AB$ we consider a point $\\Delta$ such that, if the segment $\\Gamma\\Delta$ intersects the median $AM$ at point $E$, then $A\\Delta = \\Delta E$. Prove that $AB = \\Gamma E$.",
    "options": [],
    "answer": "AB = ΓE",
    "solution": "We extend median $AM$ by $M\\Theta = AM$. Then $AB\\Theta\\Gamma$ is a parallelogram. Hence $AB \\parallel \\Gamma\\Theta$ and $\\hat{A}_1 = \\hat{O}_1$. But from $A\\Delta = \\Delta E$ we get $\\hat{A}_1 = \\hat{E}_1$.\n![](attached_image_1.png)\nFigure 1\n\nand since $\\hat{E}_1 = \\hat{E}_2$, we find that $\\hat{O}_1 = \\hat{E}_2$. Therefore the triangle $E\\Theta$ is isosceles with $\\Gamma E = \\Gamma\\Theta$. Finally from the parallelogram $AB\\Theta\\Gamma$ we have $\\vec{AB} = \\Gamma\\Theta$, from which we have $AB = \\Gamma E$.\nFrom the midpoint $M$ of $B\\Gamma$ we draw line $\\delta \\parallel AB$. Then $\\delta \\parallel B\\Delta$. Let $B\\Delta$ intersect the segment $\\Gamma\\Delta$ at $Z$. Then $Z$ is the midpoint of $\\Gamma\\Delta$, that is $\\Gamma Z = Z\\Delta$ (1) and moreover $B\\Delta = 2 \\cdot MZ$. (2)\n![](attached_image_2.png)\nFigure 2\n\nAlso we have $\\hat{A}_1 = \\hat{M}_1$. However from $A\\Delta = \\Delta E$ we get that $\\hat{A}_1 = \\hat{E}_1$ and $\\hat{E}_1 = \\hat{E}_2$. Hence $\\hat{M}_1 = \\hat{E}_2$, and $EMZ$ is isosceles with $ZM = EZ$. (3)\n\nHence\n$$\n\\begin{aligned}\n\\Gamma E &= \\Gamma Z + ZE = \\Delta Z + ZE \\quad (\\text{from (1)}) \\\\\n&= \\Delta E + 2 \\cdot ZM \\quad (\\text{from (3)}) \\\\\n&= A\\Delta + \\Delta B = AB. \\quad (\\text{from hypothesis and (2)})\n\\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in the set of integers the equation: $y = 2x^2 + 5xy + 3y^2$.",
    "options": [],
    "answer": "[(0,0), (10,-8), (-2,1), (12,-9)]",
    "solution": "The equation can be written as\n$$\ny = 2x^2 + 2xy + 3xy + 3y^2 \\Leftrightarrow y = (x+y)(2x+3y). \\quad (1)\n$$\nBy putting $x+y=z \\in \\mathbb{Z}$, then equation (1) is written\n$$\ny = z(2z + y) \\Leftrightarrow y = 2z^2 + yz \\Leftrightarrow (z-1)y = -2z^2. \\quad (2)\n$$\nFor $z=1$ equation (2) becomes: $0 \\cdot y = -2$ (impossible). For $z \\neq 1$ we have\n$$\ny = -\\frac{2z^2}{z-1} = -\\frac{2(z^2-1)+2}{z-1} = -2(z+1) - \\frac{2}{z-1}. \\quad (3)\n$$\nIn order $y \\in \\mathbb{Z}$, $z-1$ must be a divisor of $2$, i.e.\n$$\nz-1 \\in \\{-1, 1, -2, 2\\} \\Leftrightarrow z \\in \\{0, 2, -1, 3\\}.\n$$\nFor $z=0$, we find $(x,y)=(0,0)$, for $z=2$, we find $(x,y)=(10,-8)$, for $z=-1$, we find $(x,y)=(-2,1)$ and for $z=3$, we find $(x,y)=(12,-9)$.\nThe equation can be written as\n$$\ny = 2x^2 + 2xy + 3xy + 3y^2 \\Leftrightarrow (x+y-1)(2x+3y+2) = -2 \\quad (4)\n$$\nHence we have the cases:\n$$\n\\bullet \\quad \\begin{cases} x+y-1=-1 \\\\ 2x+3y+2=2 \\end{cases} \\Leftrightarrow \\begin{cases} x+y=0 \\\\ 2x+3y=0 \\end{cases} \\Leftrightarrow (x,y)=(0,0)\n$$\n$$\n\\bullet \\quad \\begin{cases} x+y-1=1 \\\\ 2x+3y+2=-2 \\end{cases} \\Leftrightarrow \\begin{cases} x+y=2 \\\\ 2x+3y=-4 \\end{cases} \\Leftrightarrow (x,y)=(10,-8)\n$$\n$$\n\\bullet \\quad \\begin{cases} x+y-1=2 \\\\ 2x+3y+2=-1 \\end{cases} \\Leftrightarrow \\begin{cases} x+y=3 \\\\ 2x+3y=-3 \\end{cases} \\Leftrightarrow (x,y)=(12,-9)\n$$\n$$\n\\bullet \\quad \\begin{cases} x+y-1=-2 \\\\ 2x+3y+2=1 \\end{cases} \\Leftrightarrow \\begin{cases} x+y=-1 \\\\ 2x+3y=-1 \\end{cases} \\Leftrightarrow (x,y)=(-2,1)\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute angled triangle with $AB < AC$ and $O$ be the center of its circumcircle $\\omega$. Let $D$ be a point on the segment $BC$ such that $\\angle BAD = \\angle CAO$. Let $E$ be the second point of intersection of $\\omega$ and the line $AD$. If $M, N$ and $P$ are the midpoints of the line segment $BE, OD$ and $AC$, respectively, show that the points $M, N$ and $P$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31184,
    "subject": "Mathematics (Multi-modal)",
    "question": "For the various values of parameter $m$, where $m$ is a positive integer, determine all pairs of positive integers $(a,b)$ satisfying the equation\n$$\n\\frac{[a,m]+[b,m]}{(a+b)m} = \\frac{10}{11}\n$$\nand lying in the Cartesian plane $Oxy$ in the square\n$$\nD=\\{(x,y):1\\le x\\le 36, 1\\le y\\le 36\\}.\n$$",
    "options": [],
    "answer": "All solutions in the square are exactly the eight pairs:\n(2,9), (9,2), (6,27), (27,6), (3,19), (19,3), (4,29), (29,4).\nThey occur for those m as follows:\n- For (2,9) and (9,2), and for (6,27) and (27,6): m is any positive even integer not divisible by 3 (i.e., m is a multiple of 2 and coprime to 3).\n- For (3,19) and (19,3): m is any positive multiple of 3 that is coprime to 19.\n- For (4,29) and (29,4): m is any positive multiple of 4 that is coprime to 29.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $x, y$ and $z$ such that\n$$\nx^5 + 4y^2 = 2013^2.\n$$",
    "options": [],
    "answer": "no solutions",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute angle scalene triangle with $AB < AC$, inscribed in the circle $c(O, R)$. The circle $c_1(B, AB)$ intersects the side $AC$ at point $K$ and the circle $c$ at point $E$. The line $KE$ intersects the circle $c$ also at point $F$. The line $BO$ intersects $KE$ at point $L$ and $AC$ at point $M$. Finally, the line $AE$ intersects $BF$ at point $D$. Prove that the polygons $DLMF$ and $BDKME$ are cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In the circle $c$, the chords $AB$ and $BE$ are equal and hence\n$$\n\\hat{A}_1 = \\hat{F}_1 = \\hat{F}_2 = \\hat{E}_1 = \\hat{C},\n$$\n![](attached_image_1.png)\nFigure 1\nbecause in the circle $c$ the above angles go to the equal arches $AB$ and $BE$. $OB$ is the line of the centers of the circles $c$ and $c_1$, and so it is perpendicular bisector of the common chord $AE$, i.e. $AD \\perp BM$. We will prove that $BF \\perp AC$, whence point $D$ will be the orthocenter of the triangle $ABM$.\n\nMoreover in the triangle $BKF$ we have: $\\hat{K}_1 = \\hat{F}_2 + \\hat{B}_2$. (1)\n\nThe triangle $BKE$ is isosceles ($BE = BK$, as radii of the circle $c_1$), whence: $\\hat{K}_1 = \\hat{E}_3 = \\hat{E}_1 + \\hat{E}_2$. (2)\n\nFrom the cyclic quadrilateral $ABEF$ we have:\n$\\hat{B}_1 = \\hat{E}_2$ (3)\n\nFrom relations (1), (2) and (3) (taking in mind equalities $\\hat{F}_2 = \\hat{E}_1 = \\hat{C}$) we find: $\\hat{B}_1 = \\hat{B}_2$ (4).\n\nFrom the equalities $\\hat{B}_1 = \\hat{B}_2$ and $\\hat{F}_1 = \\hat{F}_2 = \\hat{C}$, we conclude that $ABF$ and $KBF$ are equal. Hence $BF$ is the perpendicular bisector of $AK$. Since $AD \\perp BM$, we conclude that $D$ is the orthocenter of the triangle $ABM$. Therefore we have $\\hat{M}_1 = \\hat{A}_1 = 90^\\circ - MBA$ and in combination with $\\hat{A}_1 = \\hat{F}_2 = \\hat{C}$ we have that $\\hat{M}_1 = \\hat{F}_2$, whence the quadrilateral $DLMF$ is cyclic.\n\nSince $\\hat{B}_2 = \\hat{E}_2$, the quadrilateral $DKEB$ is cyclic and since $\\hat{E}_1 = \\hat{M}_1$, the quadrilateral $DBEM$ is cyclic. Therefore the polygon $BDKME$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31187,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider on a circle $n$ different points such that there are no three chords with ends from these points passing through the same point lying in the interior of the circle.\n\na. Find the value of $n$, if you know that the number of triangles with vertices three from the $n$ points is equal to $2n$.\n\nβ. Find the value of $n$, if you know that the number of the intersection points of the chords, lying in the interior of the circle is equal to $5n$.",
    "options": [],
    "answer": "a: 5; b: 7",
    "solution": "(a) Any three from the $n$ points lying on the circle define a triangle. Therefore the number of triangles defined by the $n$ points is equal to the number of combinations of $n$ elements by $3$, that is\n$$\n\\binom{n}{3} = \\frac{n(n-1)(n-2)}{6}\n$$\nTherefore we get the equation: $\\frac{n(n-1)(n-2)}{6} = 2n \\Leftrightarrow n = 5$.\n\n\n(β) Any four different points from the $n$ points lying on the circle define a convex quadrilateral whose diagonals intersect at an interior point of the circle and conversely. Therefore the number of interior points of the circle which are points of intersection of two chords is equal to\n$$\n\\binom{n}{4} = \\frac{n!}{4!(n-4)!} = \\frac{n(n-1)(n-2)(n-3)}{24}\n$$\nTherefore we have the equation\n$$\n\\frac{n(n-1)(n-2)(n-3)}{24} = 5n \\Leftrightarrow n = 7\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31188,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x, y, z > 0$ with $x^2y^2 + y^2z^2 + z^2x^2 = 6x^2y^2z^2$, prove that:\n$$\n\\sqrt{\\frac{x}{x+yz}} + \\sqrt{\\frac{y}{y+zx}} + \\sqrt{\\frac{z}{z+xy}} \\ge \\sqrt{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that\n$$\n\\left(a+2b+\\frac{2}{a+1}\\right)\\left(b+2a+\\frac{2}{b+1}\\right) \\ge 16,\n$$\nfor all positive real numbers $a$ and $b$ such that $ab \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. At the extensions of the side $AB$ we consider the points $A_1$, $B_1$ such that $AB_1 = BA_1 = x$ ($B_1$ closer to $A$). At the extensions of the side $BC$ we consider the points $B_4$, $C_4$ such that $CB_4 = BC_4 = y$ ($B_4$ closer to $C$). At the extensions of the side $AC$ we consider the points $C_1$, $A_4$ such that $AC_1 = CA_4 = z$ ($A_4$ closer to $C$). At the interior of the segment $A_1A_4$, we consider points $A_2$, $A_3$ such that $A_1A_2 = A_3A_4 = mA_1A_4$ (where $m \\in \\mathbb{R}$ with $0 < m < \\frac{1}{2}$). Similarly we define the points $B_2$, $B_3$ and $C_2$, $C_3$ at the interior of the segments $B_1B_4$ and $C_1C_4$, respectively. If $D = BB_2 \\cap CC_2$, $E = AA_3 \\cap CC_2$, $F = AA_3 \\cap BB_3$, $G = BB_3 \\cap CC_3$, $H = AA_2 \\cap CC_3$ and $I = AA_2 \\cap BB_2$, prove that the diagonals $DG$, $EH$, $FI$ of the hexagon $DEFGHI$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = abcd = a \\cdot 10^3 + b \\cdot 10^2 + c \\cdot 10 + d$ be a four digit positive integer such that: $a \\ge 7$ and $a > b > c > d > 0$. We consider the positive integer $B = \\overline{dcba} = d \\cdot 10^3 + c \\cdot 10^2 + b \\cdot 10 + a$. If all digits of $A + B$ are odd, determine all possible values of $A$.",
    "options": [],
    "answer": "8721, 8631, 8541, 7632, 7542, 8521, 8431, 7432, 8321",
    "solution": "$$\nA+B = (a+d) \\cdot 10^3 + (b+c) \\cdot 10^2 + (b+c) \\cdot 10 + (a+d).\n$$\nAll digits of $A + B$ are odd. However, in order to find the digits of $A + B$ we must know if the integers $a+d$ and $b+c$ are less than $10$. Hence we have the cases:\n\n(α) Let $a+d \\ge 10$ and $b+c \\ge 10$. Then, because $a > b > c > d > 0$, we have:\n$$\n\\begin{aligned}\na+d &= 10+k, \\quad k = 0,1,2,...,5, \\\\\nb+c &= 10+\\ell, \\quad \\ell = 0,1,2,...,5.\n\\end{aligned}\n$$\nHence we have:\n$$\n\\begin{aligned}\nA+B &= (10+k) \\cdot 10^3 + (10+\\ell) \\cdot 10^2 + (10+\\ell) \\cdot 10 + (10+k) \\\\\n&= 10^4 + (k+1) \\cdot 10^3 + (\\ell+1) \\cdot 10^2 + (\\ell+1) \\cdot 10 + k,\n\\end{aligned}\n$$\nthat is $A + B$ has digits $1, k+1, \\ell+1, \\ell+1, k$, (all must be odd), absurd.\n\n(β) Let $a+d \\ge 10$ and $b+c < 10$. Then, since $a > b > c > d > 0$, we have:\n$a+d = 10+k, \\quad k = 0,1,2,...,5$ and:\n$$\n\\begin{aligned}\nA+B &= (10+k) \\cdot 10^3 + (b+c) \\cdot 10^2 + (b+c) \\cdot 10 + (10+k) \\\\\n&= 10^4 + k \\cdot 10^3 + (b+c) \\cdot 10^2 + (b+c+1) \\cdot 10 + k.\n\\end{aligned}\n$$\nTherefore we have the cases:\n* If $b+c=9$, then $A+B$ has the digit of decades $0$, (absurd)\n* If $b+c < 9$, then $A+B$ has as digits the integers $b+c$ and $b+c+1$ which is not possible both to be odd.\n\n(γ) Let $a+d<10$ and $b+c \\ge 10$. Then, since $a>b>c>d>0$, we have:\n$b+c=10+\\ell, \\quad \\ell=0,1,2,...,5$ and $A+B$ is written\n$$\nA+B = (a+d) \\cdot 10^3 + (10+\\ell) \\cdot 10^2 + (10+\\ell) \\cdot 10 + (a+d) \\\\\n= (a+d+1) \\cdot 10^3 + (\\ell+1) \\cdot 10^2 + \\ell \\cdot 10 + (a+d),\n$$\nwhich means that $\\ell$ and $\\ell+1$ both are odd digits, (absurd)\n\n(δ) Let $a+d<10$ and $b+c<10$. Then $a+d$ and $b+c$, must be odd. Since $a>b>c>d>0$ and $a \\ge 7$, it follows that $a+d=9$ and since $5 \\ge c \\ge 2$, $6 \\ge b \\ge 3$, it follows $10>b+c \\ge 5$, i.e. $b+c \\in \\{5,7,9\\}$\nTherefore we have the cases:\n$a+d=9$ with $a=8, d=1$ and $b+c=9$ with $b=7, c=2$ or $b=6, c=3$ or $b=5, c=4$\nHence: $A = 8721$ or $A = 8631$ or $A = 8541$.\n$a+d=9$ with $a=7, d=2$ and $b+c=9, b=6, c=3$ or $b=5, c=4$.\nHence: $A = 7632$ or $A = 7542$\n$a+d=9$ with $a=8, d=1$ and $b+c=7$ with $b=5, c=2$ or $b=4, c=3$.\nHence: $A = 8521$ or $A = 8431$.\n$a+d=9$ with $a=7, d=2$ and $b+c=7$ with $b=4, c=3$.\nHence: $A = 7432$.\n$a+d=9$ with $a=8, d=1$ and $b+c=5$ with $b=3, c=2$.\nHence: $A = 8321$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31192,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $p$ is prime and $x, y$ are positive integers, find with respect to $p$, all the pairs $(x, y)$ satisfying the equation: $p(x-2) = x(y-1)$.\n\nFor the case we are given that $x+y=21$, find all triads $(x, y, p)$ satisfying equation (1).",
    "options": [],
    "answer": "All solutions: (x, y) = (2, 1) or (x, y) = (p, p − 1) or (x, y) = (2p, p). With x + y = 21, the solutions are (x, y, p) = (11, 10, 11) and (14, 7, 7).",
    "solution": "If $x-2=0 \\Leftrightarrow x=2$, then $y=1$ and we have the solution $(x,y)=(2,1)$.\n\nIf $x-2 \\neq 0$, then, since $p$ is prime, from the equation $p(x-2)=x(y-1)$ it follows that $y \\neq 1$ and $p|x$ or $p|y-1$. Therefore we have the cases:\n\n* If $p|x$, then $x = px'$, where $x'$ positive integer, and then:\n$$\n\\begin{align*}\np(px'-2) &= px'(y-1) \\Leftrightarrow px' - 2 = x'(y-1) \\Leftrightarrow x'(p-y+1) = 2 \\\\\n&\\Leftrightarrow x' = 1, p-y+1 = 2 \\quad \\eta \\quad x' = 2, p-y+1 = 1 \\\\\n&\\Leftrightarrow x' = 1, y = p-1 \\quad \\eta \\quad x' = 2, y = p \\Leftrightarrow x = p, y = p-1 \\quad \\eta \\quad x = 2p, y = p \\\\\n&\\Leftrightarrow (x,y) = (p,p-1) \\quad \\eta \\quad (x,y) = (2p,p)\n\\end{align*}\n$$\n\n* If $p|y-1$, then $y-1 = py'$, where $y'$ is a positive integer and then\n$$\n\\begin{align*}\np(x-2) &= xpy' \\Leftrightarrow x-2 = xy' \\Leftrightarrow x(1-y') = 2 \\\\\n&\\Leftrightarrow x = 1, 1-y' = 2 \\quad \\eta \\quad x = 2, 1-y' = 1 \\\\\n&\\Leftrightarrow x = 1, y' = -1 \\text{ (impossible) or } x = 1, y' = 0 \\text{ (impossible).}\n\\end{align*}\n$$\n\nMoreover, if we are given $x+y=21$, then we have:\n\n* Considering the solutions $(x,y) = (p,p-1)$, we find $2p-1=21 \\Leftrightarrow p=11$, and so $(x,y,p) = (11,10,11)$.\n\n* Considering the solutions $(x,y) = (2p,p)$, then $3p=21 \\Leftrightarrow p=7$, and hence we find $(x,y,p) = (14,7,7)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the greatest possible value of $M$ for which:\n$$\n\\frac{x}{1+\\frac{yz}{x}} + \\frac{y}{1+\\frac{zx}{y}} + \\frac{z}{1+\\frac{xy}{z}} \\ge M,\n$$\nfor all real numbers $x, y, z > 0$ satisfying the equation: $xy + yz + zx = 1$.",
    "options": [],
    "answer": "3/(√3+1)",
    "solution": "The inequality is equivalent to\n$$\n\\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge M. \\quad (1)\n$$\nSince $x, y, z > 0$, from Cauchy-Schwarz inequality we find\n$$\n\\begin{aligned}\n& \\left( \\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\right) (x+yz+y+zx+z+xy) \\ge (x+y+z)^2 \\\\\n\\Leftrightarrow & \\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge \\frac{(x+y+z)^2}{x+y+z+xy+yz+zx} \\\\\n\\Leftrightarrow & \\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge \\frac{(x+y+z)^2}{x+y+z+1},\n\\end{aligned} \\quad (2)\n$$\nFrom $(x+y+z)^2 \\ge 3(xy+yz+zx)$, it follows that: $x+y+z \\ge \\sqrt{3}$, where the equality holds when $x=y=z=\\sqrt{3}/3$.\nNow we consider the function $f(u) = \\frac{u^2}{u+1}$, $u \\ge \\sqrt{3}$, which is strictly increasing, because for $u > v \\ge \\sqrt{3}$ we have $f(u) > f(v)$. In fact we have\n$$\n\\begin{align*}\nf(u) > f(v) &\\Leftrightarrow \\frac{u^2}{u+1} > \\frac{v^2}{v+1} \\\\\n&\\Leftrightarrow u^2v + u^2 - uv^2 - v^2 > 0 \\\\\n&\\Leftrightarrow (u-v)(uv+u+v) > 0 \\\\\n&\\Leftrightarrow u-v > 0, \\text{ since } uv+u+v > 0.\n\\end{align*}\n$$\nTherefore $f(u) \\ge f(\\sqrt{3}) = \\frac{3}{\\sqrt{3}+1}$, and hence from inequality (2) we have:\n$$\n\\frac{x^2}{x+yz} + \\frac{y^2}{y+zx} + \\frac{z^2}{z+xy} \\ge \\frac{(x+y+z)^2}{x+y+z+1} \\ge \\frac{3}{\\sqrt{3}+1},\n$$\nfor all $x, y, z > 0$ with $xy + yz + zx = 1$. Hence: $M = \\frac{3}{\\sqrt{3}+1}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1, A_2, \\dots, A_{160}$ be set such that $|A_i|=i$, $i=1, 2, \\dots, 160$. Using the elements of these sets we construct new sets $M_1, M_2, \\dots, M_n$ with the following procedure: At the first step we choose some of the sets $A_1, A_2, \\dots, A_{160}$ and we subtract from each of them the same number of elements. All these elements are the elements of the set $M_1$. At the second step we repeat the same procedure to the remaining sets after the application of the first step and we define the set $M_2$. We continue in this way until all sets $A_1, A_2, \\dots, A_{160}$ become the empty set and we define the sets $M_3, \\dots, M_n$. Find the minimal value of $n$.",
    "options": [],
    "answer": "8",
    "solution": "We suppose that at the first step we select from all sets $k_1$ elements, at the second step we select from the remaining sets $k_2$ elements and similarly at the $n$-th step we select $k_n$ elements. After the depletion of the elements of all sets $A_1, A_2, \\dots, A_n$, then each $i = |A_i|$, $i = 1, 2, \\dots, 160$, must be the sum of some of the numbers $k_1, k_2, \\dots, k_n$. We observe that the number of all possible sums with terms some of the numbers of the set $\\{k_1, k_2, \\dots, k_n\\}$ is $2^n$, because for the creation of these sums for each term there are only two possible cases, that is a term must be or not to be in the sum. Therefore we must have $2^n \\ge 160$, and so $n \\ge 8$ with minimal possible value of $n$ to be $8$.\n\nNext we are going to prove that the value $n=8$ can be obtained because we can apply the procedure at $8$ steps.\n\nAt the first step we consider the sets $A_{81}, \\dots, A_{160}$ and we subtract from each of them $80$ elements. Therefore the set $M_1$ will consist of $80 \\cdot 80 = 6400$ elements.\n\nAfter subtracting the $80$ elements, we denote the remaining sets as $A_{81}^1, \\dots, A_{160}^1$. Now the sets $A_i$, that is $A_{80+i}^1$, $i=1, 2, \\dots, 80$ contain $i$ elements. At the second step we consider the sets $A_{41}, \\dots, A_{80}$, $A_{121}^1, \\dots, A_{160}^1$ and we subtract from each of them $40$ elements. Therefore the set $M_2$ will consist of $80 \\cdot 40 = 3200$ elements.\n\nWe continue in this way to obtain the set $M_3$ with $80 \\cdot 20 = 1600$ elements, and the sets $M_4, M_5, M_6, M_7$ and $M_8$ with $1600$, $800$, $400$, $288$, $128$ and $64$ elements, respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Two players, Alice and Bob, are playing the following game:\n* Alice chooses $n$ real numbers, not necessarily distinct\n* Alice writes all pairwise sums on a sheet of paper and gives it to Bob (there are $\\frac{n(n-1)}{2}$ such sums, not necessarily distinct)\n* Bob wins if he finds correctly the initial $n$ numbers chosen by Alice with only one guess.\nCan Bob be sure to win for the following cases?\na. $n=5$ \nb. $n=6$ \nc. $n=8$\nJustify your answer(s).",
    "options": [],
    "answer": "a. Yes\nb. No\nc. No",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31196,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a mathematical competition some competitors are friends; friendship is always mutual, that is to say that is when $A$ is a friend of $B$, then also $B$ is a friend of $A$. We say that $n \\ge 3$ different competitors $A_1, A_2, \\dots, A_n$ form a weekly-friendly circle if $A_i$ is not a friend of $A_{i+1}$, for $1 \\le i \\le n$ ($A_{n+1} = A_1$), and there are no other pairs of non-friends among the components of this cycle.\n\nThe following property is satisfied:\nFor every competitor $C$, and every weekly-friendly cycle $\\mathcal{A}$ of competitors not including $C$, the set of competitors $D$ in $\\mathcal{A}$ which are not friends of $C$ has at most one element.\n\nProve that all competitors of this mathematical competition can be arranged into three rooms, such that every two competitors that are in the same rooms are friends.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31197,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $AB\\Gamma$ is given, $c(O,R)$ is its circumcircle and let $\\Delta$ a point on the side $B\\Gamma$ different than the midpoint of $B\\Gamma$. The circumcircle of the triangle $BO\\Delta$, say $c_1$, intersects the circle $c(O,R)$ at $K$ and the line $AB$ at $Z$. The circumcircle of the triangle $GO\\Delta$, say $c_2$, intersects the circle $c(O,R)$ at $M$ and the line $A\\Gamma$ at $E$. Finally, the circumcircle of the triangle $AEZ$, say $c_3$, intersects the circle $c(O,R)$ at point $N$. Prove that the triangles $AB\\Gamma$ and $KMN$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that the circle $c_3$ passes from the center $O$ of $c(O,R)$.\nFrom the inscribed quadrilateral $O\\Delta\\Gamma E$ (in the circle $c_3$) we get: $\\hat{O}_1 = \\hat{\\Gamma}$.\nFrom the inscribed quadrilateral $O\\Delta BZ$ (in the circle $c_1$) we get: $\\hat{O}_2 = \\hat{B}$.\nSumming up the above equations we find:\n$\\hat{O}_1 + \\hat{O}_2 = \\hat{B} + \\hat{\\Gamma} \\Rightarrow EOZ = \\hat{B} + \\hat{\\Gamma} = 180^\\circ - \\hat{A}$,\nand therefore the quadrilateral $AEOZ$ is cyclic.\n\nNow we are going to prove that the circles $c_1, c_2, c_3$ are equal.\nFrom the cyclic quadrilateral $O\\Delta BZ$ we have: $\\hat{\\Delta}_2 = \\hat{Z}_2$.\n![](attached_image_1.png)\nFigure 1\nFrom the cyclic quadrilateral $O\\Delta\\Gamma E$ we have: $\\hat{\\Delta}_2 = \\hat{E}_1$ and hence:\n$$\n\\hat{\\Delta}_2 = \\hat{Z}_2 = \\hat{E}_1.\n$$\nThese three angles go in the equal chords $OB$, $O\\Gamma$ and $OA$ of the circles $c_1, c_2$ and $c_3$, respectively. Therefore these three circles are equal.\n\nNow in the equal circles $c_1$ and $c_2$, the angles $\\hat{Z}_1$ and $\\hat{\\Delta}_1$ go in the equal chords $OK$ and $OM$ ($OK = OM = R$), and so: $\\hat{Z}_1 = \\hat{\\Delta}_1$.\nFrom the last equality we conclude that the points $K$, $\\Delta$, $M$ are collinear. Similarly we prove that the points $M$, $E$, $N$ and $N$, $Z$, $K$ are collinear.\n\nFrom the equalities of angles $B\\hat{\\Delta}K = \\Gamma\\hat{\\Delta}M$ and $\\Gamma\\hat{E}M = A\\hat{E}N$ we get the equality of segments $AN = BK = \\Gamma M$ (chords of the circle $c(O,R)$).\n\nThe triangles $AB\\Gamma$ and $KMN$ have common circumcenter $O$ and the triangle $KMN$ is the image of $AB\\Gamma$ in the rotation $R(O, \\omega)$, where\n$$\nA\\hat{O}N = B\\hat{O}K = \\Gamma\\hat{O}M = \\hat{\\omega}. \\text{ Hence the triangles are equal.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triads $(x, y, z)$ of positive integers satisfying the equation:\n$$\n\\frac{1}{x} + \\frac{2}{y} - \\frac{4}{z} = 1\n$$",
    "options": [],
    "answer": "All positive integer triples are: (1, k, 2k) with k a positive integer; (l, 2, 4l) with l a positive integer; (3, 1, 3); and (2, 3, 24).",
    "solution": "If $x \\ge 3$ and $y \\ge 3$, then we have:\n$$\n\\frac{1}{x} + \\frac{2}{y} - \\frac{4}{z} \\le \\frac{1}{3} + \\frac{2}{3} - \\frac{4}{z} = 1 - \\frac{4}{z} < 1,\n$$\nand hence the equation is not satisfied. Hence we may have: $x \\le 2$ or $y \\le 2$.\n\n* For $x=1$ we have: $\\frac{2}{y} - \\frac{4}{z} = 0 \\Leftrightarrow z = 2y \\Leftrightarrow y = k, z = 2k$, where $k$ is a positive integer. Hence $(x, y, z) = (1, k, 2k), k \\in \\mathbb{Z}$ positive.\n\n* For $x=2$ we have:\n$$\n\\frac{2}{y} - \\frac{4}{z} = \\frac{1}{2} \\Leftrightarrow \\frac{2}{y} = \\frac{8+z}{2z} \\Leftrightarrow y = \\frac{4z}{z+8} \\Leftrightarrow y = \\frac{4(z+8)-32}{z+8} \\Leftrightarrow y = 4 - \\frac{32}{z+8}.\n$$\nSince $y$ is positive integer, it follows that $z+8$ must be a positive divisor of $32$ greater than $8$. Therefore $z=8$ or $z=24$, and finally we find the solutions: $(x, y, z) = (2, 2, 8)$ and $(x, y, z) = (2, 3, 24)$.\n\n$$\n\\text{For } y=1 \\text{ we have: } \\frac{1}{x} - \\frac{4}{z} = -1 \\Leftrightarrow \\frac{4}{z} = \\frac{1+x}{x} \\Leftrightarrow z = \\frac{4x}{1+x} = 4 - \\frac{4}{1+x}.\n$$\nSince $z$ must be positive, $1+x$ must be positive divisor of $4$ greater than $1$. Therefore we have $x=1$ or $x=3$, and finally we find the solutions\n$$\n(x, y, z) = (1, 1, 2) \\text{ or } (x, y, z) = (3, 1, 3).\n$$\n\n* For $y=2$ we have: $\\frac{1}{x} - \\frac{4}{z} = 0 \\Leftrightarrow z = 4x \\Leftrightarrow x = \\ell, z = 4\\ell$, where $\\ell$ is a positive integer. In this case we find the solutions: $(x, y, z) = (\\ell, 2, 4\\ell)$, where $\\ell$ is a positive integer.\n\nHence, taking in mind overlapping of solutions we can write the solutions in the\n$$\n\\text{form: } (x, y, z) = (1, k, 2k), \\text{ k is a positive integer,} \\\\\n(x, y, z) = (\\ell, 2, 4\\ell), \\ell \\text{ is a positive integer,} \\\\\n(x, y, z) = (3, 1, 3) \\text{ and } (x, y, z) = (2, 3, 24).\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31199,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane there are given $n$ different circles with the same center. In the interior of the circle with the smaller radius we consider two different points $A$ and $B$. Next we consider $k$ different lines passing from point $A$ and $m$ different lines passing from point $B$. All lines passing from $A$ intersect all lines passing from $B$ (there is no line passing from $A$ and $B$) and their points of intersection are not on the given circles. Determine the maximal and the minimal number of regions bounded by the lines and the circles, lying in the interior of the circles and for the regions lying into the smaller circle at least one part of their border is an arch of the circle.",
    "options": [],
    "answer": "minimum = 2n(k+m) − 1; maximum = 2n(k+m)",
    "solution": "First we note that the number of points of intersection of the lines is $k \\cdot m$ (figure 1).\n\n![](attached_image_1.png)\nFigure 1\n\nLet now all the points of intersection lie at the exterior of the circle with the greatest radius. Then the $k$ different lines passing through $A$ create $2k$ regions inside the circle with the smaller radius, while the $m$ different lines passing through $B$ create $2m$ regions inside the circle with the smaller radius. Since we have counted the shaded region (figure 2) two times, the total number of regions inside the smaller circle is $2 \\cdot k + 2 \\cdot m - 1$.\n\nWe have totally $n-1$ circular rings. In each circular ring is created the same number of regions $2 \\cdot k + 2 \\cdot m - 1$, but the shaded region is divided in two regions (figures 2α and 2β). Therefore inside any circular ring we have $2 \\cdot k + 2 \\cdot m$ regions.\n\n![](attached_image_2.png)\nfigure 2β\n\n![](attached_image_3.png)\nfigure 2α\n\nFinally in this case the total number of regions is: $2n(k+m)-1$.\n\nIf one or more points of intersection lie inside the circle with the smaller radius, then we have $2(k+m)$ regions inside the smaller circle (figures 3α, 3β και 3γ). In the case of figure 3α (all the points of intersection lie inside the circle with the smaller radius) we have $2(n-1)(k+m)$ regions into the $n-1$ circular rings. In the case one point of intersection lie into a circular ring the number of region increases by one. Therefore, finally we have totally\n$$\n2(k+m)+2(n-1)(k+m)=2n(k+m) \\text{ regions.}\n$$\n\n![](attached_image_4.png)\nfigure 3α",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $c(O,R)$ be a circle, $AB$ a diameter and $\\Gamma$ the midpoint of the arch $\\widehat{AB}$. We draw the circle $c_I(K,KO)$, where $K$ is a point of the segment $OA$ and we consider the tangents $\\Gamma\\Delta$, $\\Gamma O$ from $\\Gamma$ to the circle $c_I(K,KO)$. The line $K\\Delta$ intersects the circle $c(O,R)$ at the points $E$ and $Z$ (the point $E$ lies in the same semicircle with $\\Gamma$). Finally, the lines $E\\Gamma$ and $\\Gamma Z$ intersect $AB$ at the points $N$ and $M$, respectively. Prove that the quadrilateral $EMZN$ is an isosceles trapezium inscribed in a circle whose center lies on the circle $c(O,R)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\hat{E}_i = \\Gamma \\hat{E} Z$. Then from the orthogonal triangle $\\Gamma \\Delta E : \\hat{\\Gamma}_1 = 90^\\circ - \\hat{E}_1$ (1)\nSince $\\Gamma \\hat{O} Z = \\hat{O}_1$ is the corresponding central angle of $\\hat{E}_1 = \\Gamma \\hat{E} Z$, then\n$\\hat{O}_1 = 2\\hat{E}_1$ and from the isosceles triangle $\\Gamma OZ$ we have:\n$$\n\\hat{\\Gamma}_2 = \\frac{180^\\circ - \\hat{O}_1}{2} = \\frac{180^\\circ - 2\\hat{E}_1}{2} = 90^\\circ - \\hat{E}_1. \\quad (2)\n$$\nFrom relations (1) and (2), we have: $\\hat{\\Gamma}_1 = \\hat{\\Gamma}_2$.\n![](attached_image_1.png)\nThe triangles $\\Gamma \\Delta E$ and $\\Gamma O M$, are equal [$\\Gamma \\hat{\\Delta} E = \\Gamma \\hat{O} M = 90^\\circ$, $\\hat{\\Gamma}_1 = \\hat{\\Gamma}_2$ and $\\Gamma \\Delta = \\Gamma O$]. Also the triangles $\\Gamma K \\Delta$ and $\\Gamma K O$ are equal and therefore we conclude that $\\Gamma E M$ is isosceles triangle and that $\\Gamma K$ is the bisector of the angle $\\hat{E} \\Gamma M$ and perpendicular bisector of the segment $EM$. (A)\nFrom the equality of triangles $\\Gamma \\Delta E$ and $\\Gamma O M$, we conclude that: $\\Gamma \\hat{E} Z = \\Gamma \\hat{M} O$, and therefore $N \\hat{E} Z = N \\hat{M} Z$. Hence the quadrilateral EMZN is cyclic and we have the following equalities: $\\hat{M}_1 = \\hat{E} N Z$ and $\\hat{E}_2 = \\hat{M} Z N$.\nAlso, from the isosceles triangle $\\Gamma E M$, we have: $\\hat{M}_1 = \\hat{E}_2$.\nTherefore, we get that: $\\hat{E} N Z = \\hat{M} Z N$ and then EMZN is isosceles trapezium.\n\nThe center of the circumcircle of EMZN is the point of intersection of two perpendicular bisectors of its sides. Let T is the point of intersection of $\\Gamma K$ (perpendicular bisector of EM) and the perpendicular bisector of EZ. In the triangle $\\Gamma E Z$, $\\Gamma T$ is the bisector of the angle $E \\hat{\\Gamma} Z$ and $O T$ is the perpendicular bisector of the side EZ. Hence T belongs on the circumcircle of the triangle $\\Gamma E Z$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(x_n)$, $n \\ge 1$, be a sequence of real numbers with $x_1 = 1$, such that $2x_{n+1} = 3x_n + \\sqrt{5x_n^2 - 4}$, for $n=1,2,3,...$\n\na. Prove that all terms of the sequence are natural numbers.\n\nβ. Examine if there exist a term of the sequence divisible by $2011$.",
    "options": [],
    "answer": "All terms are natural numbers, and no term is divisible by 2011.",
    "solution": "a.\nFrom the given recurrence relation we get\n$$\n(2x_{n+1} - 3x_n)^2 = 5x_n^2 - 4 \\Rightarrow 4x_{n+1}^2 - 12x_{n+1}x_n + 4x_n^2 = -4 \\Rightarrow x_{n+1}^2 - 3x_{n+1}x_n + x_n^2 = -1 \\quad (1)\n$$\nwhich can be written also in the form\n$$\nx_{n+2}^2 - 3x_{n+2}x_{n+1} + x_{n+1}^2 = -1 \\quad (2).\n$$\nWe consider now the second degree equation: $x^2 - 3x \\cdot x_{n+1} + x_{n+1}^2 + 1 = 0$.\nFrom (1), (2), we observe that two solutions of this equation are $x_n, x_{n+2}$, and hence by using Vieta's formulas we get:\n$$\nx_n + x_{n+2} = 3x_{n+1} \\quad (3) \\quad \\text{and} \\quad x_n x_{n+2} = x_{n+1}^2 + 1 \\quad (4).\n$$\nWriting relation (3) in the form $x_{n+2} = 3x_{n+1} - x_n$ and taking in mind that $x_1 = 1$ and $x_2 = 2$ we conclude by induction that all terms of the sequence are integers\n\nβ.\nWe suppose that there exists a term $x_s$ of the sequence such that: $2011|x_s$. Then from (4) for $n=s$, we get $x_s x_{s+2} = x_{s+1}^2 + 1$. Since all terms of the sequence are integers and $2011|x_s$, we have:\n$$\n\\begin{aligned}\n2011 \\mid x_{s+1}^2 + 1 &\\Rightarrow x_{s+1}^2 \\equiv -1 \\pmod{2011} \\\\\n&\\Rightarrow (x_{s+1}^2)^{1005} \\equiv (-1)^{1005} \\equiv -1 \\pmod{2011} \\\\\n&\\Rightarrow x_{s+1}^{2010} \\equiv -1 \\pmod{2011}\n\\end{aligned} \\quad (5)\n$$\nWe know that $2011$ is prime and it is easy to see that $(x_{s+1}, 2011) = 1$.\nIn fact, if $(x_{s+1}, 2011) = d > 1$, then $d|x_{s+1}$, $d|2011 \\Rightarrow d|x_{s+1}$, $d|x_s$ and then from relation $x_s x_{s+2} = x_{s+1}^2 + 1$ we conclude that $d|1$, absurd.\nTherefore from Fermat's theorem we have:\n$$\nx_{s+1}^{2010} \\equiv 1 \\pmod{2011}, \\quad (6)\n$$\nwhich contradicts relation (5).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all the pairs of real numbers $(x, y)$, which are solutions of the system:\n$$\n\\begin{cases}\n(x^2 + y^2)^2 - xy(x+y)^2 = 19 \\\\\n|x-y| = 1\n\\end{cases}\n$$",
    "options": [],
    "answer": "[(3,2), (-2,-3), (-3,-2), (2,3)]",
    "solution": "Performing some manipulation into the first equation we find:\n$$\n\\left\\{ \\begin{array}{l} (x^2 + y^2)^2 - xy(x+y)^2 = 19 \\\\ |x-y| = 1 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} x^4 + 2x^2y^2 + y^4 - x^3y - 2x^2y^2 - xy^3 = 19 \\\\ |x-y| = 1 \\end{array} \\right\\} \\Leftrightarrow \\\\\n\\left\\{ \\begin{array}{l} x^4 + y^4 - x^3y - xy^3 = 19 \\\\ |x-y| = 1 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} (x-y)(x^3-y^3) = 19 \\\\ |x-y| = 1 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} (x-y)^2(x^2+xy+y^2) = 19 \\\\ |x-y| = 1 \\end{array} \\right\\} \\\\\n\\Leftrightarrow \\left\\{ \\begin{array}{l} x^2 + xy + y^2 = 19 \\\\ |x-y| = 1 \\end{array} \\right\\}.\n$$\nThus we have the following cases:\n$$\n\\bullet \\quad \\left\\{ \\begin{array}{l} x - y = 1 \\\\ x^2 + xy + y^2 = 19 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} y = x - 1 \\\\ x^2 + xy + y^2 = 19 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} y = x - 1 \\\\ x^2 + x^2 - x + (x-1)^2 = 19 \\end{array} \\right\\} \\\\\n\\left\\{ \\begin{array}{l} y = x - 1 \\\\ 3x^2 - 3x - 18 = 0 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} y = x - 1 \\\\ x = 3 \\text{ or } x = -2 \\end{array} \\right\\} \\Leftrightarrow (x,y) = (3,2) \\text{ or } (x,y) = (-2,-3).\n$$\n$$\n\\bullet \\quad \\left\\{ \\begin{array}{l} x - y = -1 \\\\ x^2 + xy + y^2 = 19 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} y = x + 1 \\\\ x^2 + xy + y^2 = 19 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} y = x + 1 \\\\ x^2 + x^2 + x + (x+1)^2 = 19 \\end{array} \\right\\} \\\\\n\\left\\{ \\begin{array}{l} y = x + 1 \\\\ 3x^2 + 3x - 18 = 0 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} y = x + 1 \\\\ x = -3 \\text{ or } x = 2 \\end{array} \\right\\} \\Leftrightarrow (x,y) = (-3,-2) \\text{ or } (x,y) = (2,3).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ with real coefficients satisfying the equality\n$$\n(x^2 - 6x + 8)P(x) = (x^2 + 2x)P(x - 2),\n$$\nfor all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "P(x) = c x^2 (x^2 - 4) for any real constant c",
    "solution": "The given equation is written as\n$$\n(x-2)(x-4)P(x) = x(x+2)P(x-2), \\text{ for all } x \\in \\mathbb{R}, \\quad (1)\n$$\nand so for $x = 0, -2$ and $4$ we obtain: $P(0) = P(-2) = P(2) = 0$.\nTherefore the polynomial $P(x)$ takes the form:\n$$\nP(x) = x(x+2)(x-2)Q(x), \\quad (2)\n$$\nwhere $Q(x)$ is a polynomial with real coefficients.\nFrom (2), equation (1) takes the form:\n$$\n(x-2)(x-4)^2 x(x+2)Q(x) = x(x+2)(x-2)x(x-4)Q(x-2), \\quad (3)\n$$\nfor all $x \\in \\mathbb{R}$. Equivalently, for all $x \\in \\mathbb{R}$ we have\n$$\nx(x+2)(x-2)(x-4)[(x-2)Q(x)-xQ(x-2)] = 0, \\quad (4)\n$$\nSince the polynomial $x(x-2)(x+2)(x-4)$ is not the zero polynomial, from relation (4) we get:\n$$\n(x-2)Q(x) - xQ(x-2) = 0, \\text{ for all } x \\in \\mathbb{R}. \\quad (5)\n$$\nFrom (5) for $x = 0$ we get $Q(0) = 0$, and hence $Q(x) = xR(x)$, where $R(x)$ is a polynomial with real coefficients. From (5), relation (4) becomes:\n$$\n\\begin{aligned} &(x-2)xR(x) - x(x-2)R(x-2) = 0 \\\\ \\Leftrightarrow &(x-2)[R(x) - R(x-2)] = 0, \\end{aligned}\n$$\nFrom which, since $x(x-2) \\neq 0$ for all $x \\neq 0$, we have that:\n$$\nR(x) = R(x-2), \\text{ for all } x \\in \\mathbb{R}.\n$$\nLast relation gives $R(x) = R(x+2k)$, $k \\in \\mathbb{Z}$, for all $x \\in \\mathbb{R}$, and hence the polynomial $R(x)$ takes the same value for infinite values of $x \\in \\mathbb{R}$, for example $c = R(0) = R(2k)$, $k \\in \\mathbb{Z}$. Hence $R(x) = c$, for all $x \\in \\mathbb{R}$, and hence\n$$\nP(x) = x(x+2)(x-2)Q(x) = x(x+2)(x-2)xR(x) = cx^2(x^2-4).\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$, and $b_1, b_2, \\dots, b_n$, $2n$ real numbers. Prove that there exists an integer $k$ with $1 \\le k \\le n$, such that\n$$\n\\sum_{i=1}^{n} |a_i - a_k| \\le \\sum_{i=1}^{n} |b_i - a_k|.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle and let $M$ the middle of the side $BC$. Externally of the triangle we consider parallelogram $BCDE$, such that $BE \\parallel AM$ and $BE = AM / 2$. Prove that the line $EM$ passes from the middle point of the segment $AD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We extend $AM$ till it meets $ED$ at point $N$. Then $BMNE$ and $MCDN$ are parallelograms and hence $EN = BM = MC = ND$. Hence $N$ is the middle of $ED$. Moreover we observe that $\\frac{AM}{MN} = 2$ and $M$ lie on the median of the triangle $EAD$. Hence $M$ is the centroid of the triangle $AED$. Therefore the line $EM$ is the line of the median of the triangle $AED$ passing from the vertex $E$, and so it intersects the side $AD$ in the middle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma\\Delta$ be a quadrilateral inscribed into the circle. With centers $A, B, \\Gamma, \\Delta$ we draw circles $C_A, C_B, C_\\Gamma, C_\\Delta$ respectively, not having common points. The circle $C_A$ intersects the sides of the quadrilateral at the points $A_1, A_2$, the circle $C_B$ at the points $B_1, B_2$, the circle $C_\\Gamma$ at the points $\\Gamma_1, \\Gamma_2$ and the circle $C_\\Delta$ at the points $\\Delta_1, \\Delta_2$. Prove that the quadrilateral created by the lines $A_1A_2, B_1B_2, \\Gamma_1\\Gamma_2$ and $\\Delta_1\\Delta_2$ is cyclic. (E. Psychas)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the triangles $AA_1A_2, BB_1B_2, \\Gamma\\Gamma_1\\Gamma_2$ and $\\Delta\\Delta_1\\Delta_2$ are isosceles, using small letters for their equal angles we have the equalities:\n![](attached_image_1.png)\nFigure 4\n$$\n\\hat{A} + \\hat{x} + \\hat{x} = 180^\\circ \\Leftrightarrow \\hat{x} = 90^\\circ - \\frac{\\hat{A}}{2} \\quad (1) \\qquad \\hat{B} + \\hat{y} + \\hat{y} = 180^\\circ \\Leftrightarrow \\hat{y} = 90^\\circ - \\frac{\\hat{B}}{2} \\quad (2)\n$$\n$$\n\\hat{\\Gamma} + \\hat{z} + \\hat{z} = 180^\\circ \\Leftrightarrow \\hat{z} = 90^\\circ - \\frac{\\hat{\\Gamma}}{2} \\quad (3) \\qquad \\hat{\\Delta} + \\hat{\\omega} + \\hat{\\omega} = 180^\\circ \\Leftrightarrow \\hat{\\omega} = 90^\\circ - \\frac{\\hat{\\Delta}}{2} \\quad (4).\n$$\nSince the quadrilateral $AB\\Gamma\\Delta$ is cyclic, we have:\n$$\n\\hat{A} + \\hat{\\Gamma} = 180^\\circ \\Rightarrow \\hat{x} + \\hat{z} = 90^\\circ \\quad (5) \\qquad \\hat{B} + \\hat{\\Delta} = 180^\\circ \\Rightarrow \\hat{y} + \\hat{\\omega} = 90^\\circ \\quad (6).\n$$\nLet now the lines $A_1A_2, B_1B_2, \\Gamma_1\\Gamma_2$ form the quadrilateral $K\\Lambda MN$. From the triangle $KA_1B_2$, we have: $\\hat{K} + \\hat{x} + \\hat{y} = 180^\\circ$, while from the triangle $M\\Gamma_1\\Delta_2$, we have: $\\hat{M} + \\hat{z} + \\hat{\\omega} = 180^\\circ$. Summing up the last two relations and using (5) and (6) we obtain: $\\hat{K} + \\hat{M} = 180^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the integer $n$ for which the number $A = \\frac{8n-25}{n+5}$ is equal with the cube of a rational number.\n(A. Fellouris)",
    "options": [],
    "answer": "n = 3",
    "solution": "Let $p, q \\in \\mathbb{Z}$, $q \\neq 0$, with $(p, q) = 1$ such that\n$$\nA = \\frac{8n-25}{n+5} = \\left(\\frac{p}{q}\\right)^3. \\qquad (1)\n$$\nThen $(p^3, q^3) = 1$, while from relation (1) we get:\n$$\nq^3(8n-25) = p^3(n+5), \\qquad (2)\n$$\nfrom which we conclude that\n$$\np^3 \\mid (8n-25) \\text{ and } q^3 \\mid (n+5) \\Rightarrow \\exists k \\in \\mathbb{Z}: 8n-25 = kp^3 \\text{ and } n+5 = kq^3. \\quad (3)\n$$\nIn fact, if $8n-25 = kp^3$, $k \\in \\mathbb{Z}$, then from (2) we find $n+5 = kq^3$ and so\n$$\n8(n+5) - (8n-25) = k(8p^3 - q^3) \\Rightarrow k(2q-p)(4q^2+2qp+p^2) = 65\n$$\nTherefore, the numbers $k, 2q-p$ and $4q^2+2qp+p^2$ are divisors of 65. We observe that $4q^2+2pq+p^2 = 3q^2+(p+q)^2 \\equiv 1 \\pmod{3}$. Moreover, we have $4q^2+2pq+p^2 = 3q^2+(p+q)^2 \\ge 3$, and so, since $4q^2+2pq+p^2$ is divisor of 65 its unique value is\n$$\n4q^2 + 2pq + p^2 = 13 \\Leftrightarrow 3q^2 + (p+q)^2 = 13,\n$$\nleading to the following cases:\n* $4q^2 + 2qp + p^2 = 13$, $k = \\pm 1$, $2q - p = \\pm 5$. Then: $p = 2q \\mp 5$ and\n$4q^2 + 2q(2q \\mp 5) + (2q \\mp 5)^2 = 13 \\Leftrightarrow p = 2q \\mp 5$, $2q^2 \\mp 5q + 2 = 0$\n$\\Leftrightarrow p = 2q \\mp 5$, $q = \\pm 2 \\Leftrightarrow p = \\mp 1$, $q = \\pm 2$. Then for both cases we have:\n$$\\left(\\frac{p}{q}\\right)^3 = -\\frac{1}{8} \\Rightarrow \\frac{8n-25}{n+5} = -\\frac{1}{8} \\Leftrightarrow 8(8n-25) = -(n+5) \\Leftrightarrow n = 3.$$\nAs in the first solution we obtain relation (2) which we solve with respect to $n$, to receive:\n$$\nn = \\frac{5(5q^3 + p^3)}{8q^3 - p^3}. \\qquad (4)\n$$\nLet $d = (5q^3 + p^3, 8q^3 - p^3)$. Then $d \\mid (5q^3 + p^3 + 8q^3 - p^3 = 13q^3)$ and since $(p,q)=1$ it follows that $d\\mid13$. Therefore\n$$\n\\frac{8q^3 - p^3}{d} \\mid 5 \\Rightarrow (8q^3 - p^3) \\mid 5 \\cdot 13 \\Rightarrow (2q-p)(4q^2+2pq+p^2) \\mid 65.\n$$\nAs in the first solution we find that\n$$\n4q^2 + 2pq + p^2 = 13 \\Leftrightarrow 3q^2 + (p+q)^2 = 13. \\quad (5)\n$$\nNow from relation (5) it follows that $|q| \\le 2$ and so we have the cases:\n* If $|q| \\in \\{0,1\\}$, then $(p+q)^2 \\in \\{13,10\\}$, impossible for $p,q \\in \\mathbb{Z}$.\n* If $|q| = 2$, then $(p+q)^2 = 1$, from which we receive the solutions:\n$$\n(p,q) = (-1,2) \\text{ or } (p,q) = (1,-2) \\text{ or } (p,q) = (-3,2) \\text{ or } (p,q) = (3,-2).\n$$\nSince $(2q-p)|65$, acceptable are the solutions: $(p,q) = (-1,2)$ and $(p,q) = (1,-2)$, from which we find $n = 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all nonzero polynomials with real coefficients satisfying the equality: $$(P(x))^3 + 3(P(x))^2 = P(x^3) - 3P(-x),$$ for all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "All solutions are P(x) = -1, P(x) = -2, P(x) = x^{2m} - 1, and P(x) = -x^{2m} - 1, where m is any positive integer.",
    "solution": "Let $\\deg P(x) = 0$. Then $P(x) = a \\neq 0$ and from relation (1) we have: $a^3 + 3a^2 = -2a \\Leftrightarrow a = -1$ or $a = -2$.\nHence the constant (nonzero) polynomials $P(x) = -1$ and $P(x) = -2$, are solutions of the problem.\n\nLet $\\deg P(x) = n > 0$. Then the polynomial $P(x)$ can be written as\n$$\nP(x) = a x^n + Q(x), \\text{ with } a \\neq 0 \\text{ and } \\deg Q(x) = k \\le n-1 \\text{ or } Q(x) = 0.\n$$\nBy substitution into relation (1) we get:\n$$\n(a x^n + Q(x))^3 + 3(a x^n + Q(x))^2 = a x^{3n} + Q(x^3) - 3Q(-x) - 3a \\cdot (-1)^n x^n, \\quad (2)\n$$\nfor all $x \\in \\mathbb{R}$. Equating the coefficients of $x^{3n}$ in the two parts we find:\n$$\na^3 = a \\Leftrightarrow a = 1 \\text{ or } a = -1.\n$$\nTherefore we distinguish the following cases:\n\nI. $a=1$. Then relation (2) becomes:\n$$\n(x^n + Q(x))^3 + 3(x^n + Q(x))^2 = x^{3n} + Q(x^3) - 3Q(-x) - 3(-1)^n x^n \\\\\n\\Leftrightarrow A(x) = B(x), \\qquad (3)\n$$\nwhere we have put\n$$\nA(x) = 3x^{2n}Q(x) + 3x^n(Q(x))^2 + (Q(x))^3 + 3x^{2n} + 6x^nQ(x) + 3(Q(x))^2 \\\\\nB(x) = Q(x^3) - 3Q(-x) - 3(-1)^n x^n.\n$$\nIf $Q(x) = 0$, then: $3x^{2n} = -3 \\cdot (-1)^n x^n$, (impossible).\nIf $\\deg Q(x) = k > 0$, then, since $0 < k < n$, we have:\n$$\n2n + k = \\deg A(x) = \\deg B(x) = \\max\\{3k, n\\} \\ge 3k \\Rightarrow n \\ge k, \\text{ absurd.}\n$$\nThus, we have $\\deg Q(x) = 0$, that is $Q(x) = c \\neq 0$ and then $A(x) = B(x)$:\n$$\n3(c+1)x^{2n} + 3(c^2+2c+(-1)^n)x^n + c^3 + 3c^2 + 2c = 0, \\text{ for all } x \\in \\mathbb{R} \\Leftrightarrow \\\\\nc+1=0,\\ c^2+2c+(-1)^n = 0,\\ c^3 + 3c^2 + 2c = 0 \\Leftrightarrow c=-1,\\ n=2m,\\ m \\in \\mathbb{N}^*.\n$$\nHence the polynomial: $P(x) = x^{2m}-1,\\ x \\in \\mathbb{R},\\ m \\in \\mathbb{N}^*$ is a solution.\n\nII. $a = -1$. Then relation (2) becomes:\n$$\n(-x^n + Q(x))^3 + 3(-x^n + Q(x))^2 = -x^{3n} + Q(x^3) - 3Q(-x) + 3(-1)^n x^n \\\\\n\\Leftrightarrow A(x) = B(x),\n$$\nwhere we have put\n$$\nA(x) = 3x^{2n}Q(x) - 3x^n(Q(x))^2 + (Q(x))^3 + 3x^{2n} - 6x^nQ(x) + 3(Q(x))^2 \\\\\nB(x) = Q(x^3) - 3Q(-x) + 3(-1)^n x^n.\n$$\nWorking as in case (I), if $Q(x) = 0$, then: $3x^{2n} = 3 \\cdot (-1)^n x^n$, (impossible).\nIf $\\deg Q(x) = k > 0$, then, since $0 < k < n$, we have:\n$$\n2n + k = \\deg A(x) = \\deg B(x) = \\max\\{3k, n\\} \\geq 3k \\Rightarrow n \\geq k, \\text{ absurd.}\n$$\nHence $\\deg Q(x) = 0$, that is $Q(x) = c \\neq 0$ and then from equality $A(x) = B(x)$ we have:\n$$\n3(c+1)x^{2n} - 3(c^2 + 2c + (-1)^n)x^n + c^3 + 3c^2 + 2c = 0, \\text{ for all } x \\in \\mathbb{R} \\\\\nc+1 = 0,\\ c^2 + 2c + (-1)^n = 0,\\ c^3 + 3c^2 + 2c = 0 \\Leftrightarrow c = -1,\\ n = 2m,\\ m \\in \\mathbb{N}^*.\n$$\nHence we have the solution: $P(x) = -x^{2m} - 1,\\ x \\in \\mathbb{R},\\ m \\in \\mathbb{N}$.\n\nFinally, all solutions of the problem are the following:\n$$\nP(x) = -1,\\ P(x) = -2,\\ P(x) = x^{2m} - 1,\\ P(x) = -x^{2m} - 1,\\ x \\in \\mathbb{R},\\ m \\in \\mathbb{N}^*.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ prime and $m$ a positive integer. Determine all pairs $(p, m)$ satisfying the equation: $p(p+m)+p=(m+1)^3$. (A. Fellouris)",
    "options": [],
    "answer": "(2, 1)",
    "solution": "The given equation is written\n\n$$\np(p+m+1) = (m+1)^3\n$$\nTherefore the prime $p$ is a divisor of $(m+1)^3$. Hence $p \\mid (m+1)$, which means that there exists positive integer $k$ such that $m+1 = kp$. Then, from (1) we get:\n$$\np(p+kp) = (kp)^3 \\Leftrightarrow k+1 = k^3p \\Rightarrow k^3 \\mid (k+1) \\Rightarrow k \\mid (k+1) \\Rightarrow k=1.\n$$\nHence $p=2$, $m=1$ and $(p,m)=(2,1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ an acute angled scalene triangle with $AB < A\\Gamma < B\\Gamma$. Let $\\Delta, E, Z$ be the midpoints of the sides $B\\Gamma, A\\Gamma, AB$, respectively, and let $BK, G\\Lambda$ be altitudes. At the extension of $\\Delta Z$, to the part of $Z$, we consider a point $M$, such that the parallel from $M$ to $K\\Lambda$ intersect the extensions of $\\Gamma A, BA$ and $\\Delta E$ at points $\\Sigma, T$ and $N$, respectively. If the circumcircle of the triangle $MB\\Delta$, say $(c_1)$, intersects the line $\\Delta N$ at point $P$ and the circumcircle of the triangle $N\\Gamma\\Delta$, say $(c_2)$, intersects the line $\\Delta M$ at point $\\Pi$, prove that: $\\Sigma T \\parallel \\Pi P$. (E. Psychas)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\Delta, E, Z$ are the midpoints of the sides $B\\Gamma, A\\Gamma, AB$, respectively, the quadrilaterals $AE\\Delta Z, ZE\\Delta B$ and $ZE\\Gamma\\Delta$ are parallelograms.\nFrom $\\Delta M \\parallel \\Gamma \\Sigma$ and $MN \\parallel K\\Lambda$ we conclude that: $\\hat{M} = \\hat{\\Sigma}_1$ and $\\hat{\\Sigma}_1 = \\hat{K}$.\nFrom the cyclic quadrilateral $B\\Lambda K\\Gamma$ ($B\\Lambda\\Gamma = B\\hat{K}\\Gamma = 90^\\circ$) we get: $\\hat{K} = \\hat{B}$.\n\nFrom the last three equalities of angles we arrive at the equality $\\hat{M} = \\hat{B}$, and from this we conclude that the points $M, B, \\Delta, T$, are cocyclic.\n\nIn a similar fashion, since $\\Delta N // BT$ and $MN // K\\Lambda$ we have the equalities $\\hat{N} = \\hat{T}_1$ and $\\hat{T}_1 = \\hat{\\Lambda}$. Also, from the cyclic quadrilateral $B\\Lambda K\\Gamma$ (since $B\\hat{\\Lambda}\\Gamma = B\\hat{K}\\Gamma = 90^\\circ$) we have that: $\\hat{\\Lambda} = \\hat{\\Gamma}$.\nFrom the last three equalities of angles we have $\\hat{N} = \\hat{\\Gamma}$, and from this we conclude that the points $N, \\Gamma, \\Delta, \\Sigma$, are cocyclic.\n\nSince the points $B, \\Delta, P, T, M$ belong to the circle $(c_1)$, from the cyclic quadrilateral $MTP\\Delta$ we have: $\\hat{T}_2 + \\hat{\\Delta} = 180^\\circ$.\nSince the points $\\Gamma, \\Delta, \\Pi, \\Sigma, N$ belong to the circle $(c_2)$, from the cyclic quadrilateral $\\Delta\\Pi\\Sigma N$ we have: $\\hat{\\Sigma}_2 + \\hat{\\Delta} = 180^\\circ$.\nFrom the last two equalities we get:\n$$\n\\hat{T}_2 = \\hat{\\Sigma}_2 = 180^\\circ - \\hat{\\Delta} = 180^\\circ - \\hat{A}.\n$$\n\nThe quadrilateral $MTP\\Delta$ is inscribed to the circle $(c_1)$, and hence:\n$$\n\\hat{TPE} = \\hat{P}\\Delta = 180^\\circ - \\hat{M}\\Delta = 180^\\circ - \\hat{B}\\Gamma = 180^\\circ - \\hat{B}.\n$$\n\nThe angle $\\hat{\\Sigma}_2 = 180^\\circ - \\hat{A}$ is an external angle of the triangle $\\Sigma\\Pi M$, and so:\n$$\n\\Sigma\\hat{\\Pi}Z = \\Sigma\\hat{\\Pi}M = \\hat{\\Sigma}_2 - \\hat{M} = 180^\\circ - \\hat{A} - \\hat{B} = \\hat{\\Gamma} = \\Sigma\\hat{E}Z.\n$$\nThe quadrilateral $\\Sigma\\Pi\\Gamma\\Pi$ is inscribed into the circle $(c)$ and since $\\hat{T}_2 = \\hat{\\Sigma}_2$ it is isosceles trapezium and so $\\Sigma T \\parallel \\Pi P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in the real numbers the system\n$$\nx^3 = \\frac{z}{y} - \\frac{2y}{z}, \\quad y^3 = \\frac{x}{z} - \\frac{2z}{x}, \\quad z^3 = \\frac{y}{x} - \\frac{2x}{y}.\n$$",
    "options": [],
    "answer": "(-1, -1, -1), (1, 1, -1), (1, -1, 1), (-1, 1, 1)",
    "solution": "For $x, y, z \\in \\mathbb{R}$, such that $xyz \\neq 0$, the system is written:\n$$\nx^3 y z = z^2 - 2y^2 \\quad (1), \\quad y^3 z x = x^2 - 2z^2 \\quad (2), \\quad z^3 x y = y^2 - 2x^2 \\quad (3)\n$$\nUsing summation by parts we find:\n$$\nxyz(x^2 + y^2 + z^2) = -(x^2 + y^2 + z^2) \\Leftrightarrow (x^2 + y^2 + z^2)(xyz + 1) = 0.\n$$\nSince $xyz \\neq 0$ we have $x^2 + y^2 + z^2 > 0$, and so: $xyz = -1$ (4)\nUsing equation (4) in the system of (1)-(3) we get:\n$$\nx^2 = -z^2 + 2y^2 \\quad (5), \\quad y^2 = -x^2 + 2z^2 \\quad (6), \\quad z^2 = -y^2 + 2x^2 \\quad (7)\n$$\nFrom (5) and (6) we get $y^2 = z^2$, while from (6) and (7) we get $x^2 = z^2$, and so:\n$$\nx^2 = y^2 = z^2 \\Leftrightarrow x = y = \\pm z \\quad \\text{or} \\quad x = -y = \\pm z. \\qquad (8)\n$$\nFinally from equations (8) and (4) we have the solutions:\n$$\n(x, y, z) = (-1, -1, -1), \\quad (x, y, z) = (1, 1, -1), \\quad (x, y, z) = (1, -1, 1), \\quad (x, y, z) = (-1, 1, 1).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31212,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider an $n \\times n$ chessboard, where $n$ is an even positive integer. On the board we put all numbers $1,2,3, \\dots, n^2$, one at each square. Let $S_1$ be the sum of the numbers lying at the white squares and let $S_2$ be the sum of the numbers lying on the blank squares. Find all the numbers $n$ for which it is possible an arrangement such that:\n$$\n\\frac{S_1}{S_2} = \\frac{39}{64}.\n$$",
    "options": [],
    "answer": "All even n divisible by 103, i.e., n = 206k for positive integers k.",
    "solution": "The given relation is equivalent to: $S_1 = \\frac{39}{103}(S_1 + S_2)$. Since\n$$\nS_1 + S_2 = 1 + 2 + \\dots + n^2 = \\frac{n^2(n^2+1)}{2}\n$$\nand $S_1$ is a natural number we conclude that $103 \\mid \\left\\lfloor \\frac{n^2(n^2+1)}{2} \\right\\rfloor$. Since $103$ is a prime number of the form $4\\kappa+3$, it follows that $103$ does not divide $n^2+1$. Hence the prime number $103$ must divide $n^2$ and finally $103|n$. Moreover $n$ is even, and so: $n = 206k,\\ k \\in \\mathbb{N}^*$.\n\nNow we are going to prove that for every number $n = 206k,\\ k \\in \\mathbb{Z}$, it is possible to obtain the wanted arrangement. The least (maximal) possible value of $S_1$ is:\n$$\n1+2+\\cdots+\\frac{n^2}{2} = \\frac{\\frac{n^2}{2}\\left(\\frac{n^2}{2}+1\\right)}{2} = A, \\quad \\left(\\frac{n^2}{2}+1\\right)+\\left(\\frac{n^2}{2}+1\\right)+\\cdots+n^2 = B,\n$$\nrespectively.\nEasily we can check the inequality, $A < S_1 = \\frac{39}{103}(S_1 + S_2) < B$. Our result arrives by proving that $S_1$ can obtain every possible value between $A, B$. In fact, number $A+1$ can be obtained by putting in the white squares the numbers $1, 2, \\dots, \\frac{n^2}{2}-1, \\frac{n^2}{2}+1$. The number $A+2$ can be obtained by putting in the white squares the numbers $1, 2, \\dots, \\frac{n^2}{2}-1, \\frac{n^2}{2}+2$, and so on. When we arrive to the step we need to put the numbers $1, 2, \\dots, \\frac{n^2}{2}-1, \\frac{n^2}{2}$, in order to obtain the next number for their sum, then we select in the white squares the numbers $1, 2, \\dots, \\frac{n^2}{2}-2, \\frac{n^2}{2}$, $n^2$, and for the next number we select $1, 2, \\dots, \\frac{n^2}{2}-2, \\frac{n^2}{2}+1, n^2$, and so on. Hence, using the above procedure to increase at every step the sum by $1$, starting from $A$, we can construct all integers till $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Sigma = \\{1,2,3,\\ldots,n\\}$. We want to make a partition of the set $\\Sigma$ into three subsets $A$, $B$ and $\\Gamma$ mutually disjoint with $A \\cup B \\cup \\Gamma = \\Sigma$ and such that the sums of their elements $S_A$, $S_B$ and $S_\\Gamma$, respectively, are equal. Examine if that is possible, in the following cases\n(α) $n = 2014$; (β) $n = 2015$; (γ) $n = 2018$. (A. Fellouris)",
    "options": [],
    "answer": "For n equal to two thousand fourteen: impossible. For n equal to two thousand fifteen: possible; an explicit partition exists using six number blocks. For n equal to two thousand eighteen: possible; an explicit partition exists using six number blocks.",
    "solution": "(α) If that is possible, for $n = 2014$, then for the sum of the elements of $\\Sigma$ we get $S_{\\Sigma} = S_A + S_B + S_{\\Gamma} = 3 \\cdot S_A$, that is, $S_{\\Sigma}$ is a multiple of $3$. But\n$$\nS_{2014} = 1007 \\cdot 2015 \\neq \\text{multiple of } 3.\n$$\nHence we cannot apply the wanted partition.\n\n(β) For $n = 2015$ we have: $S_{2015} = 1008 \\cdot 2015 \\equiv 0 \\pmod{3}$.\nWe observe that $\\Sigma$ consists of the set $M_0 = \\{1,2,3,4,5\\}$ and from $335$ successive six-folds of the form:\n$$\nM_k = \\{6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5\\}, \\quad k = 1, 2, \\dots, 335.\n$$\nWe partition the set $M_0$ into three subsets with equal sums of their elements $A_0 = \\{1,4\\}$, $B_0 = \\{2,3\\}$ and $\\Gamma_0 = \\{5\\}$. Taking in mind that\n$$\n(6k+1)+(6k+4)=(6k+2)+(6k+3)=6k+(6k+5), \\text{ for all } k=1,2,\\dots,335,\n$$\nThe wanted partition is feasible by taking the sets:\n$$\nA = \\{1,4\\} \\cup \\{6k+1,6k+4:k=1,2,\\dots,335\\}\n$$\n$$\nB = \\{2,3\\} \\cup \\{6k+2,6k+3:k=1,2,\\dots,335\\}\n$$\n$$\n\\Gamma = \\{5\\} \\cup \\{6k,6k+5:k=1,2,\\dots,335\\}.\n$$\n\n(γ) For $n = 2018$ we have: $S_{2018} = 1009 \\cdot 2019 \\equiv 0 \\pmod{3}$. As in (β) we observe that $\\Sigma$ consists of $M_0 = \\{1,2,3,4,5,6,7,8\\}$ and from $335$ successive six-folds of the form:\n$$\nM_k = \\{6k+3,6k+4,6k+5,6k+6,6k+7,6k+8\\}, \\quad k = 1,2,\\dots,335.\n$$\nFirst we partition $M_0$ into three subsets with equal sums of their elements $A_0 = \\{1,2,3,6\\}$, $B_0 = \\{5,7\\}$ and $\\Gamma_0 = \\{4,8\\}$. Since for all $k = 0,1,\\dots,335$ we have $(6k+3)+(6k+8)=(6k+4)+(6k+7)=(6k+5)+(6k+6)$, the wanted partition is possible by taking\n$$\nA = \\{1,2,3,6\\} \\cup \\{6k+3,6k+8:k=0,1,2,\\dots,335\\}\n$$\n$$\nB = \\{5,7\\} \\cup \\{6k+4,6k+7:k=0,1,2,\\dots,335\\}\n$$\n$$\n\\Gamma = \\{4,8\\} \\cup \\{6k+5,6k+6:k=0,1,2,\\dots,335\\}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31214,
    "subject": "Mathematics (Multi-modal)",
    "question": "We color the numbers $1, 2, 3, \\ldots, 20$ with two colors, white and blank, in such a way that both colors are used. Find the number of ways we can perform this coloring if the product of white numbers and the product of blank numbers have maximal common divisor equal to $1$. (P. Bregiannis)",
    "options": [],
    "answer": "62",
    "solution": "Number $1$ can be colored in two ways, white or blank. Number $2$ also can be colored white or blank. Then all even numbers $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$ have to be colored with the color of $2$.\nAlso all numbers having common divisor greater than $1$ with the above numbers must be colored with the color of $2$. The remaining numbers greater than $10$, that is, $11, 13, 17, 19$ can be colored in two ways. Therefore the coloring can be made with $2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 \\cdot 2 = 2^6 = 64$ different ways. However we must delete the two cases we color all numbers white or blank. So we have finally $62$ different ways.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31215,
    "subject": "Mathematics (Multi-modal)",
    "question": "A square $ABCD$ is divided into $n^2$ equal small elementary squares, by drawing parallel lines to its sides. A spider starts from $A$ moving only to the right and up and tries to approach point $C$. Every movement of the spider consists of $k$ steps right and $m$ steps up or of $m$ steps right and $k$ steps up (which can be done in any way she wants).\nThe spider makes $l$ “movements” and in the sequel moves only to the right or up without any restriction. If $n = m \\cdot l$, find the number of all possible routes the spider can follow to approach point $C$, where $n, m, k, l$ are positive integers and $k < m$.",
    "options": [],
    "answer": "Let v = binom(k+m, m) and r = m - k. The total number of routes is v^l * sum_{i=0}^{l} binom(l, i) * binom(lr, i r). In particular, if r = 1 (i.e., m - k = 1), this simplifies to binom(k+m, m)^l * binom(2l, l).",
    "solution": "We suppose that the square is placed into a Cartesian system of coordinates with origin $A(0,0)$ and axes on the sides $AB$ and $AD$, as in the figure 7. Let the spider start from $A$ and make its first movement ($m$ steps right and $k$ steps up or $m$ steps up and $k$ right). In the figure you can see the case $m = 4, k = 3$.\n\n![](attached_image_1.png)\n\nAfter the first movement the spider can be on the point $M(k,m)$ or on the point $N(m,k)$. The number of ways the spider can approach the point $M(k,m)$ or the point $N(m,k)$ are equal to the combinations of $k+m$ elements taken $k$ at a time, that is\n$$\n\\binom{k+m}{k} = \\binom{k+m}{m} = v.\n$$\n\nAfter the second movement the spider can be at three different points: $K(2k, 2m)$, $L(k+m, k+m)$, and $T(2m, 2k)$ with corresponding number of ways $v^2$, $2v^2$, and $v^2$.\nThe point $M$ can be approached by $v$ different ways from $A$. The point $K$ can be approached by $v$ different ways from $M$. Hence the point $K$ can be approached by $v^2$ different ways from point $A$.\n\n![](attached_image_2.png)\n\nIn a similar fashion we can find that point $T$ can be approached by $v^2$ different ways from point $A$. The point $L$ can be approached by $v^2$ different ways by following the route $A, M, L$ and by $v^2$ different ways by following the route $A, N, L$. Hence point $L$ can be approached by $2v^2$ different ways from point $A$.\n\nSimilarly, we conclude that after the third movement the spider can be placed in four different points that can be approached by $\\binom{3}{3} \\cdot v^3$, $\\binom{3}{2} \\cdot v^3$, $\\binom{3}{1} \\cdot v^3$, and $\\binom{3}{0} \\cdot v^3$ different ways, respectively. Therefore, after the completion of the $l$ movements, the spider will be placed at $l+1$ different points, lying on the line with equation: $x + y = l(k+m)$.\n\n![](attached_image_3.png)\n\nIf $r = m-k$, then these points are the following:\n$$\nA_0(lm, lk),\\ A_1(lm-r, lk+r),\\ A_2(lm-2r, lk+2r),\\ \\dots,\\ A_l(lm-lr, lk+lr) \\equiv A_{l}(lk, lm).\n$$\nThese points can be approached by $\\binom{l}{l} \\cdot v^l$, $\\binom{l}{l-1} \\cdot v^l$, $\\binom{l}{l-2} \\cdot v^l$, $\\dots$, and $\\binom{l}{0} \\cdot v^l$ different ways from the point $A$. The point $C(ml, ml)$ can be approached from the point:\n\n$$\nA_0(lm, lk),\\ \\text{ with } \\binom{(ml-lm)+(ml-lk)}{ml-lk} = \\binom{ml-lk}{ml-lk} = \\binom{lr}{lr} = 1, \\text{ way.}\n$$\n$$\nA_1(lm-r, lk+r),\\ \\text{ with } \\binom{(ml-lm+r)+(ml-lk-r)}{r} = \\binom{ml-lk}{r}, \\text{ ways.}\n$$\n$$\nA_2(lm-2r, lk+2r),\\ \\text{ with } \\binom{(ml-lm+2r)+(ml-lk-2r)}{2r} = \\binom{ml-lk}{2r}, \\text{ ways.}\n$$\n$$\nA_l(lk, lm),\\ \\text{ with } \\binom{(ml-lk)+(ml-lm)}{ml-lk} = \\binom{lr}{lr} = 1, \\text{ way.}\n$$\nHence the point $C(ml, lm)$ can be approached from point $A$ by:\n$$\nv^l \\left( \\binom{l}{l} \\cdot \\binom{lr}{0} + \\binom{l}{l-1} \\cdot \\binom{lr}{r} + \\binom{l}{l-2} \\cdot \\binom{lr}{2r} + \\cdots + \\binom{l}{0} \\cdot \\binom{lr}{lr} \\right) \\text{ ways.}\n$$\n\n(*) Moving from the point $(i, j)$ to the point $(i+k, j+m)$ of the grid with direction right and up, can be made by $\\binom{k+m}{k} = \\binom{k+m}{m}$ different ways. The procedure of finding the number of ways is the same as the procedure we have used at the first movement.\n\n**Note.** In the case where $m-k = r = 1$, then the number of all possible ways becomes:\n$$\nv^l \\left( \\binom{l}{l} \\cdot \\binom{l}{0} + \\binom{l}{l-1} \\cdot \\binom{l}{1} + \\binom{l}{l-2} \\cdot \\binom{l}{2} + \\cdots + \\binom{l}{0} \\cdot \\binom{l}{l} \\right) = v^l \\binom{2l}{l} = \\binom{k+m}{m}^l \\binom{2l}{l}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $c(O, R)$ be a circle and let $A, B$ be two points on the circle such that $R < AB < 2R$. The circle $c_1(A, r)$, $0 < r < R$, meets the circle $c(O, R)$ at points $C$ and $D$ ($C$ belongs to the small arch $\\widehat{AB}$). From point $B$ we draw the tangents $BE$ and $BF$ to the circle $c_1(A, r)$, such that $E$ lies outside of the circle $c(O, R)$. The lines $EC$ and $DF$ intersect at $M$. Prove that the quadrilateral $BCFM$ is cyclic.\n(E. Psychas)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$AEBF$ is cyclic ($BE$ and $BF$ are tangents and: $A\\hat{E}B = A\\hat{F}B = 90^\\circ$). Let $c_2$ be its circumcircle. The segment $CD$ is the common chord of the circles ($c$) and ($c_1$) and the segment $AB$ is the common chord of the circles ($c$) and ($c_2$). Finally, $EF$ is the common chord of the circles ($c_1$) and ($c_2$). Hence the chords $CD$, $AB$, $EF$ pass through the radical center, say $L$, of the three circles.\n\nSince $BE$ and $BF$ are tangents of the circle $c_1(A, r)$, $AB$ is the bisector of the angle $E\\hat{B}F$. Moreover, $AB$ is the bisector of the angle $C\\hat{B}D$, because $A\\hat{B}C$ and $A\\hat{B}D$ are inscribed in the circle $c(O, R)$ and correspond to the equal arches $\\widehat{AC}$ and $\\widehat{AD}$. Hence $E\\hat{B}C = F\\hat{B}D$.\n\n![](attached_image_1.png)\nFigure 2\n\nUsing equality $E\\hat{B}C = F\\hat{B}D$, we will prove that the triangles $BCE$ and $BFD$ are similar. For that it is enough to prove that: $\\frac{BC}{BE} = \\frac{BF}{BD} \\Leftrightarrow \\frac{a}{x} = \\frac{y}{a} \\Leftrightarrow xy = a^2$, where $BE = BF = a$, $BC = x$ and $BD = y$. Similarly from $LCB \\sim LAD$:\n$$\n\\frac{LC}{LA} = \\frac{CB}{AD} \\Rightarrow \\frac{LC}{LA} = \\frac{x}{r} \\quad (1),\n$$\nand from $LCA \\sim LBD$, we have:\n$$\n\\frac{LB}{LC} = \\frac{BD}{CA} \\Rightarrow \\frac{LB}{LC} = \\frac{y}{r} \\quad (2).\n$$\n\nFrom (1) and (2) we have: $\\frac{LB}{LA} = \\frac{xy}{r^2}$ (A) and from the orthogonal triangle $ABE$ with $AL$ the altitude to the hypotenuse we get $\\frac{LB}{LA} = \\frac{EB^2}{EA^2} = \\frac{a^2}{r^2}$. (B)\nFrom (A) and (B) we obtain: $xy = a^2$. Therefore $BCE \\approx BFD$. From the equalities of their corresponding angles we find: $\\hat{C} = \\hat{F}$. Hence the quadrilateral $BCFM$ is cyclic.\n\n\nAlternative solution:\nLet $Q = EF \\cap CD$. From **Pascal's theorem** to degenerate hexagon *EEDFFC*, if $T = ED \\cap CF$, we get that $T, B, M$ are collinear. Moreover, in the cyclic quadrilateral *ECFD*, the points $T, M$ are the common points of its opposite sides and $Q$ is the common point of its diagonals. Therefore the line *TM* is the polar of $Q$, and so $AQ \\perp TM$ (*A* is the center of the circle) which means that $\\hat{A}BT = 90^\\circ$. Since $AB \\perp EF$, we get that $EF \\parallel TM$. Hence we have:\n$$\nB\\hat{M}C = M\\hat{E}F = C\\hat{F}B.\n$$\n\n![](attached_image_2.png)\nFigure 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triads of positive integers $(x, y, p)$, where $p$ is prime, which satisfy the following equation: $$\\frac{xy^3}{x+y} = p.$$",
    "options": [],
    "answer": "(14, 2, 7)",
    "solution": "Let $d = \\gcd(x, y)$. Then there exist $a, b \\in \\mathbb{Z}$ such that $x = da$, $y = db$, $(a, b) = 1$. By substitution to the given equation we get:\n$$\n\\frac{d a (d b)^3}{d a + d b} = \\frac{d^3 a b^3}{a + b} = p. \\qquad (1)\n$$\nFrom $(a, b) = 1$, we get $(a, a + b) = 1$ and $(b^3, a + b) = 1$, giving from relation (1) that $a + b \\mid d^3$. We write\n$$\n\\frac{d^3}{a + b} = k, \\qquad (2)\n$$\nwhere $k$ is a positive integer. Then (1) becomes: $k a b^3 = p$, and hence $b^3 \\mid p$. Therefore $b = 1$ and $k a = p$. Hence we have the following cases:\n(i) If $k = p, a = 1$, then (2) becomes $\\frac{d^3}{2} = p \\Rightarrow 2p = d^3 \\Rightarrow 2 \\mid d$. Hence $8 \\mid d^3$ and $8 \\mid 2p$, absurd.\n(ii) If $k = 1, a = p$. Then (2) becomes:\n$$\nd^3 = p + 1 \\Rightarrow d^3 - 1 = p \\Rightarrow (d - 1)(d^2 + d + 1) = p.\n$$\nTherefore, we get $d - 1 = 1$, $d^2 + d + 1 = p$ (since $d^2 + d + 1 > d - 1$),\n$$\n\\Leftrightarrow d = 2, p = 7, \\text{ and so: } (x, y, p) = (14, 2, 7).\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all possible pair of positive integers $x$, $y$ satisfying the equation: $xy(x+y-10)-3x^2-2y^2+21x+16y=60$.",
    "options": [],
    "answer": "(4, 6), (5, 5), (3, 8), (7, 4)",
    "solution": "1. The equation is written:\n$$\nxy(x+y-10)-3x^2-2y^2+21x+16y=60 \\\\\n\\Leftrightarrow (y-3)x^2+(y^2-10y+21)x=2y^2-16y+60 \\\\\n\\Leftrightarrow (y-3)x^2+(y-7)(y-3)x=2(y-3)(y-5)+30 \\\\\n\\Leftrightarrow (y-3)x^2+(y-7)(y-3)x-2(y-3)(y-5)=30 \\\\\n\\Leftrightarrow (y-3)[x^2+(y-7)x-2(y-5)]=30 \\\\\n\\Leftrightarrow (y-3)(x^2-7x+xy-2y+10)=30 \\\\\n\\Leftrightarrow (y-3)(x^2-4-7x+xy-2y+14)=30 \\\\\n\\Leftrightarrow (y-3)((x-2)(x+2)-7(x-2)+(x-2)y)=30 \\\\\n\\Leftrightarrow (y-3)(x-2)(x+y-5)=30=1 \\cdot 2 \\cdot 3 \\cdot 5.\n$$\nSince $x-2$, $y-3$ and $x+y-5$ are integers such that $x-2+y-3=x+y-5$\nand moreover $x-2 \\ge -1$, $y-3 \\ge -2$ and $x+y-5 \\ge -3$, we have to consider the following cases:\n$$\n\\begin{cases} x-2=2 \\\\ y-3=3 \\\\ x+y-5=5 \\end{cases} \\text{ or } \\begin{cases} x-2=3 \\\\ y-3=2 \\\\ x+y-5=5 \\end{cases} \\text{ or } \\begin{cases} x-2=1 \\\\ y-3=5 \\\\ x+y-5=6 \\end{cases} \\text{ or } \\begin{cases} x-2=5 \\\\ y-3=1 \\\\ x+y-5=6 \\end{cases} \\\\\n(x,y) = (4,6) \\text{ or } (x,y) = (5,5) \\text{ or } (x,y) = (3,8) \\text{ or } (x,y) = (7,4).\n$$\n\nAlternatively, from the equation $(y-3)x^2+(y-3)(y-7)x=2(y^2-8y+30)$ we get:\n$$\n(y-3)[x^2+(y-7)x]=2y^2-16y+60 \\quad (1)\n$$\nFrom which, since $x$, $y$ are positive integers, arises that: $(y-3)|2y^2-16y+60$. Since $2y^2-16y+60=(y-3)(2y-10)+30$ arises that $y-3|30$. Therefore $y-3 \\in \\{\\pm1, \\pm2, \\pm3, \\pm5, \\pm6, \\pm10, \\pm15, \\pm30\\}$, and since $y-3 \\ge -2$ we get: $y-3 \\in \\{\\pm1, \\pm2, 3, 5, 6, 10, 15, 30\\}$.\nFor each of these values, by substituting to (1) we obtain a trinomial of $x$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the values of the parameter $\\alpha \\in \\mathbb{R}$ for which the equation $x^2 + (\\alpha - 2)x - (\\alpha - 1)(2\\alpha - 3) = 0$ has two roots such that the one of them is equal to the square of the other.",
    "options": [],
    "answer": "2, 5/4, sqrt(2), -sqrt(2)",
    "solution": "1. We have the discriminant $D = (\\alpha - 2)^2 + 4(\\alpha - 1)(2\\alpha - 3) = (3\\alpha - 4)^2$, and so the equation has the roots $x_1 = \\alpha - 1$, $x_2 = -2\\alpha + 3$.\n\nTherefore we seek the values of $\\alpha$ for which:\n$$\n\\begin{align*} \nx_1 = x_2^2 \\text{ or } x_2 = x_1^2 &\\Leftrightarrow \\alpha - 1 = (-2\\alpha + 3)^2 \\text{ or } -2\\alpha + 3 = (\\alpha - 1)^2 \\\\ \n&\\Leftrightarrow \\alpha - 1 = 4\\alpha^2 - 12\\alpha + 9 \\text{ or } -2\\alpha + 3 = \\alpha^2 - 2\\alpha + 1 \n\\end{align*}\n$$\n$$\n\\Leftrightarrow 4\\alpha^2 - 13\\alpha + 10 = 0 \\text{ or } \\alpha^2 = 2 \\Leftrightarrow \\alpha = 2 \\text{ or } \\alpha = \\frac{5}{4} \\text{ or } \\alpha = \\sqrt{2} \\text{ or } \\alpha = -\\sqrt{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for the polynomial $P(x) = x^4 - x^3 - 3x^2 - x + 1$, there exist infinitely many positive integers $n$, for which $P(3^n)$ is composite.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "1. For $x = 3^{2n-1}$ we have:\n$$\n\\begin{aligned}\nP(3^{2n-1}) &= 81^{2n-1} - 27^{2n-1} - 3 \\cdot (-1)^{2n-1} - 3^{2n-1} + 1 \\\\\n&\\equiv 1^{2n-1} - 2^{2n-1} - 3 \\cdot 9^{2n-1} - 3^{2n-1} + 1 \\pmod{5} \\\\\n&\\equiv 5 - (2^{2n-1} + 3^{2n-1}) \\pmod{5}.\n\\end{aligned}\n$$\nFor odd exponents $m$, the number $a^m + b^m$ is divided by $a+b$. Hence $2^{2n-1} + 3^{2n-1}$ is divided by $5$, and hence the number $P(3^{2n-1})$ is divided by $5$.\nSince $P(x)$ is a not constant polynomial it can take the value $5$ a finite number of times. Therefore there are infinite values of the polynomial divided by $5$ which are composite numbers.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31221,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x, y, z$ are positive real numbers, prove that: $(3x + y)(3y + z)(3z + x) \\ge 64xyz$. When does equality hold?",
    "options": [],
    "answer": "Equality holds when x = y = z.",
    "solution": "We use the inequality of arithmetic and geometric means for four positive terms:\n$$\n\\frac{a_1 + a_2 + a_3 + a_4}{4} \\ge \\sqrt[4]{a_1 a_2 a_3 a_4},\n$$\nwhere equality holds when $a_1 = a_2 = a_3 = a_4$.\n\nApplying this to each factor:\n$$(3x + y) \\ge 4\\sqrt[4]{x^3 y}$$\n$$(3y + z) \\ge 4\\sqrt[4]{y^3 z}$$\n$$(3z + x) \\ge 4\\sqrt[4]{z^3 x}$$\n\nMultiplying these inequalities:\n$$(3x + y)(3y + z)(3z + x) \\ge 4\\sqrt[4]{x^3 y} \\cdot 4\\sqrt[4]{y^3 z} \\cdot 4\\sqrt[4]{z^3 x} = 64\\sqrt[4]{x^4 y^4 z^4} = 64xyz.$$ \n\nEquality holds when $x = y = z$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x) = a x^3 + (b - a) x^2 - (c + b) x + c$, $Q(x) = x^4 + (b - 1) x^3 + (a - b) x^2 - (c + a) x + c$ be polynomials of the indeterminate $x$, where $a, b, c$ are nonzero real numbers and $b > 0$. If the polynomial $P(x)$ has three different real roots $x_0, x_1, x_2$, which are also roots of the polynomial $Q(x)$, then:\n(a) Prove that: $abc > 28$.\n(b) If $a, b, c$ are nonzero integers with $b > 0$, find their possible values.",
    "options": [],
    "answer": "abc > 28; and for integers with the given conditions the only possibility is a = 2, b = 4, c = 4.",
    "solution": "(a) The sum of the coefficients of the polynomial $P(x)$ is equal to $0$. It means that $1$ is one of its roots and so\n$$\nP(x) = a x^3 + (b - a) x^2 - (c + b) x + c = (x - 1)(a x^2 + b x - c)\n$$\nIf $x_0 = 1$, from Vieta's formulas we have:\n$$\nx_1 + x_2 = -\\frac{b}{a} \\quad \\text{and} \\quad x_1 x_2 = -\\frac{c}{a} \\neq 0, \\qquad (1)\n$$\nMoreover, from the hypothesis $x_0, x_1, x_2$ are roots of the polynomial\n$$\n\\begin{aligned}\nF(x) &= Q(x) - P(x) = x^4 + (b - a - 1) x^3 + 2(a - b) x^2 + (b - a) x \\\\\n&= x(x^3 + (b - a - 1) x^2 + 2(a - b) x + b - a) \\\\\n&= x[x^3 - x^2 + (b - a)(x^2 - x) + (a - b)(x - 1)] \\\\\n&= x(x - 1)[x^2 + (b - a)x + (a - b)].\n\\end{aligned}\n$$\nSince $F(x) = 0 \\Leftrightarrow x = 0$ or $x = 1$ or $x^2 + (b - a)x + (a - b) = 0$ and $x_0, x_1, x_2 \\neq 0$, we conclude that:\n$$\nx_0 = 1, \\quad x_1 + x_2 = a - b, \\quad x_1 x_2 = a - b. \\qquad (2)\n$$\nFrom (1) and (2) we obtain: $a - b = -\\frac{b}{a} = -\\frac{c}{a} \\Rightarrow$\n$$\nb = c \\qquad (3) \\quad \\text{or} \\quad a^2 - a b = -b \\qquad (4)\n$$\nFrom (3) and (4) we have:\n$$\na^2 = b(a - 1) \\Rightarrow a > 1 \\text{ (since $b > 0$) and } b = c = \\frac{a^2}{a - 1}.\n$$\nWe have that\n$$\n\\begin{aligned}\nabc &= a \\left( \\frac{a^2}{a - 1} \\right)^2 = \\frac{a^5}{(a - 1)^2} \\stackrel{x = a - 1 \\ge 0}{=} \\frac{(x + 1)^5}{x^2} = \\frac{x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1}{x^2} \\\\\n&\\Rightarrow abc = x^3 + \\left( 5x^2 + \\frac{1}{x^2} \\right) + \\left( 10x + \\frac{5}{x} \\right) + 10.\n\\end{aligned} \\qquad (5)\n$$\nNow we observe that:\n* $x^3 > 0$, $5x^2 + \\frac{1}{x^2} > 4 \\Leftrightarrow 5x^4 - 4x^2 + 1 > 0 \\Leftrightarrow x^4 + (2x^2 - 1)^2 > 0$, valid,\n* $10x + \\frac{5}{x} > 14 \\stackrel{x \\ge 0}{\\Leftrightarrow} 10x^2 - 14x + 5 > 0$, is valid.\n\nFinally, from (5) we get that: $abc > 28$.\n\n(b) As in the first question from relation $b = a + 1 + \\frac{1}{a - 1}$, since $b, a + 1$ are integers, we obtain that $\\frac{1}{a - 1} \\in \\mathbb{Z}$\nTherefore:\n$$\na - 1 = \\pm 1 \\Leftrightarrow a = 0 \\text{ (it is rejected) or } a = 2. \\text{ Then } b = c = \\frac{a^2}{a - 1} = 4. \\text{ For these values the polynomial}\n$$\n$$\nP(x) = a x^3 + (b - a) x^2 - (c + b) x + c = 2x^3 + 2x^2 - 8x + 4 = 2(x - 1)(x^2 + 2x - 2)\n$$\nhas roots $x_0 = 1$, $x_{1,2} = -1 \\pm \\sqrt{3}$ which are also roots of the polynomial\n$$\nF(x) = Q(x) - P(x) = x(x - 1)(x^2 + (b - a)x + (a - b)) = x(x^2 + 2x - 2),\n$$\nand therefore they are roots of the polynomial $Q(x) = F(x) + P(x)$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that in every triangle there exist some vertex such that with the sides which are concurrent in that vertex, and with any inner cevian passing with this vertex, it is possible to construct a triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We suppose that $\\hat{A} \\le \\hat{B} \\le \\hat{C}$. Let $AM$ be an internal cevian of the triangle passing through $A$. Since $AM\\hat{B} > \\hat{C} \\ge \\hat{B}$, it follows that $AM > AB$. Hence it is enough to prove that: $AB < AC + AM$.\n\nLet $\\hat{C} \\le 90^\\circ$. We will prove that $c < b + h_a$, where $c = AB$, $b = AC$ and $h_a$ is the altitude from the vertex $A$. Let $H$ be the trace of the altitude from $A$. Then we have:\n$$\nAD > AB - BC \\text{ and } AD > AC - DC,\n$$\nfrom which we get $h_a > \\frac{b + c - a}{2}$ and therefore\n$$\nb + h_a > \\frac{b + c - a}{2} + b \\Rightarrow \\frac{c}{2} + b > \\frac{c}{2} + \\frac{a + b}{2} > \\frac{c}{2} + \\frac{c}{2} = c.\n$$\nSince $AM \\ge h_a$ we finally get $b + AM > c$.\n\nLet $\\hat{C} > 90^\\circ$. Then $b < AM$ and therefore $b + AM > b + b > b + a > c$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31224,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider $111$ mutually distinct points on the interior or on the circle of a unit disk. Prove that we can find at least $1998$ segments with ends from these points and length less than $\\sqrt{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We divide the circle into three equal sectors of $120^\\circ$ such that none of the points belong to their border, with the exception of the center of the circle. If the center of the circle is one of the points, then we consider that it belongs only to one of the sectors. This is possible because the number of the points is finite and the possible selections for the distribution of the circle into three equal sectors are infinite.\n\n![](attached_image_1.png)\n\nLet $A, B$ belong to the same sector which is defined by the radii $OK$ and $O\\Lambda$. Then, if $OA$, $OB$ intersect the circle at $A'$, $B'$ and $A'OB = \\omega < 60^\\circ$, then we get $AB \\leq A'B' = 2R \\sin(\\omega) < 2R \\sin 60^\\circ = 2 \\sqrt{3}/2 = \\sqrt{3}$. Equality is not possible because the points do not belong to the border of the sector. Hence two arbitrary points of the same sector have distance less than $\\sqrt{3}$.\n\nLet, in the three sectors, belong $x$, $y$, $z$ points, respectively. Then $x + y + z = 111$ and the segments with length less than $\\sqrt{3}$ are at least\n$$\n\\binom{x}{2} + \\binom{y}{2} + \\binom{z}{2} = \\frac{x(x-1) + y(y-1) + z(z-1)}{2} = \\frac{x^2 + y^2 + z^2 - 111}{2}.\n$$\nFrom Cauchy-Schwarz inequality: $x^2 + y^2 + z^2 \\geq \\frac{(x + y + z)^2}{3} = \\frac{111^2}{3}$ and\n$$\n\\binom{x}{2} + \\binom{y}{2} + \\binom{z}{2} \\geq \\frac{111^2}{3} - 111 = 1998.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs of nonnegative integers $(m, n)$, $m \\ge n$, for which $A = (m+n)^3$ divides $B = 2n(3m^2 + n^2) + 8$.",
    "options": [],
    "answer": "{(1,1), (1,0)} ∪ {(k+2, k) : k ≥ 0}",
    "solution": "Let $A = (m+n)^3$, $B = 2n(3m^2+n^2)+8$. Since $(m+n)^3 \\mid 2n(3m^2+n^2)+8$ it follows that:\n$$\n(m+n)^3 \\le 2n(3m^2+n^2)+8 \\Leftrightarrow m^3+3m^2n+3mn^2+n^3 \\le 6m^2n+2n^3+8\n$$\n$$\n\\Leftrightarrow m^3 - 3m^2n + 3mn^2 - n^3 \\le 8 \\Leftrightarrow (m-n)^3 \\le 8 \\Leftrightarrow m-n \\le 2 \\Leftrightarrow m-n \\in \\{0,1,2\\}.\n$$\nWe distinguish the cases:\n$$\n\\bullet \\ m-n = 0 \\Leftrightarrow m = n.\n$$\n$$\nA\\mid B \\Leftrightarrow 8m^3\\mid 8m^3 + 8 \\Leftrightarrow 8m^3\\mid 8 \\Leftrightarrow m = 1, \\text{(since } m > 0\\text{)} \\Rightarrow (m,n) = (1,1).\n$$\n$$\n\\bullet \\ m-n = 1.\n$$\n$$\nA = (2n+1)^3, B = 2n(3(n+1)^2+n^2)+8 = 8n^3+12n^2+6n+8 = (2n+1)^3+7.\n$$\n$$\n\\text{Therefore, } A\\mid B \\Leftrightarrow (2n+1)^3\\mid 7 \\Rightarrow 2n+1=1 \\Leftrightarrow n=0, m=1, (m,n) = (1,0).\n$$\n\n$\\bullet$ $m-n=2$. Then $A=8(n+1)^3=B$, and hence we have an infinity of pairs of the form $(k+2,k)$, $\\mu_k \\ge 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be an acute angled triangle with $AB < A\\Gamma$ and circumcenter $O$. The altitudes $B\\Delta$, $\\Gamma E$ meet at point $H$. If $O_1$ is the circumcenter of the triangle $BH\\Gamma$, prove that the quadrilateral $AHO_1O$ is parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $O_1$ belongs to perpendicular bisector $OM$ of the segment $B\\Gamma$ and $AH$, $OO_1$, it is enough to prove that $AH = OO_1$. Since $AH = 2OM$, it is enough to prove that $OM = MO_1$. The quadrilateral is cyclic, whereby $\\hat{BH}\\Gamma = 180^\\circ - \\hat{A}$. Moreover $\\hat{BO}_1\\Gamma = 2\\hat{A}$. Therefore the isosceles triangles $BO\\Gamma$, $BO_1\\Gamma$ have all their corresponding angles equal and since they have $B\\Gamma$ common side, they are equal. Therefore $B\\Gamma$ is the perpendicular bisector of $OO_1$, whereby $M$ is the midpoint of $O_1O$, and hence $OM = MO_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene acute angled triangle with $AB < AC < BC$ and circumcenter $c(O, R)$. The ex-circle ($c_A$) corresponding to the vertex $A$, has center $I$ and is tangent to the sides $BC$, $AC$, $AB$ at $D$, $E$, $Z$, respectively. The line $AI$ intersects the circle $c(O, R)$ at $M$ and the circumcircle, say $(c_1)$, of the triangle $AZE$ intersects the circle ($c$) at $K$. The circumcircle, say $(c_2)$, of the triangle $OKM$ intersects the circle ($c_1$) at point $N$. Prove that the lines $AN$ and $KI$ intersect at a point of the circle ($c$).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $AZ$ and $AE$ are tangents to the circle ($c_A$), we have $IZ \\perp AZ$ and $IE \\perp AE$. Hence, the circle ($c_2$) contains $I$ and $AI$ is a diameter. First we will prove that $AN$ passes through $O$, that is $KNO = KNA$ (*).\n\nMoreover we have $KNO = KMO$ and from the isosceles triangle $OKM$ we get that: $KMO = OKM$. Therefore: $KNO = KMO = OKM$ (1).\n\n![](attached_image_1.png)\nFigure 8\n\nSimilarly, we conclude that $KNA = KIA$. From the right angled triangle $AKI$ we have $KIA = 90^\\circ - IAK = 90^\\circ - MAK$. Since $MAK = \\frac{MOK}{2}$ and from the isosceles triangle $OKM$ we have:\n$OM = 90^\\circ - \\frac{MOK}{2}$, finally, using (1), we find: $KNA = KIA = 90^\\circ - IAK = 90^\\circ - MAK = 90^\\circ - \\frac{MOK}{2} = OKM = KNO$ (*) Hence the points $A$, $O$, $N$ are collinear and let $T$ the point of intersection of $AN$ with the circle ($c$).\n\nTo complete the proof, we show that the points $K$, $T$, $I$ are collinear, by proving that $AKI = AKT$. In fact $AKT = 90^\\circ$, since $AT$ and $AI$ are diameters.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Every triangle in the plane $\\mathbb{R}^2$ contains a Mediterranean point on its boundary or in its interior (even if the triangle degenerates into a line segment or a point). These Mediterranean points satisfy the following conditions:\n(i) If a triangle is symmetric with respect to a line through the origin $(0, 0)$, then its Mediterranean point lies on this line.\n(ii) If triangle $DEF$ contains triangle $ABC$, and if $ABC$ contains the Mediterranean point $M$ of $DEF$, then $M$ is also the Mediterranean point of $ABC$.\nDetermine all possible locations for the Mediterranean point of the triangle with vertices in $(-3, 5)$, $(12, 5)$, and $(3, 11)$.",
    "options": [],
    "answer": "(0, 5)",
    "solution": "Consider the auxiliary triangle $DEF$ with $D = (-12, 5)$, $E = (12, 5)$, and $F = (0, 13)$. Since the triangle $DEF$ is symmetric with respect to the $y$-axis, condition (i) yields that its Mediterranean point $M$ lies on the $y$-axis, and hence has coordinates $(0, m)$ with $5 \\le m \\le 13$. Define a non-negative real number $a$ by $a^2 + 25 = m^2$, and note that $0 \\le a \\le 12$. Then the point $A = (a, 5)$ lies on the side $DE$ of the triangle $DEF$, and points $A$ and $M$ both are at distance $m$ from the origin. Then $A$, $M$, and the midpoint $C = (12a, 12(m+5))$ of $AM$ form a (degenerate) triangle that is symmetric with respect to the line running through the origin and $C$. By (i) the Mediterranean point of $AMC$ lies on this line and hence must be $C$. On the other hand since $DEF$ contains $AMC$, and since $AMC$ contains the Mediterranean point $M$ of $DEF$, condition (ii) yields that the Mediterranean point of $AMC$ must be $M$. This implies that $M$ and $C$ coincide, which in turn implies $a = 0$ and $m = 5$. Hence $M = (0, 5)$ is the Mediterranean point of triangle $DEF$.\n\nNow let us return to the original triangle with vertices $(-3, 5)$, $(12, 5)$, $(3, 11)$ in the problem statement: $DEF$ contains this triangle, and the triangle contains the Mediterranean point $M = (0, 5)$ of $DEF$. Then, (ii) yields that its Mediterranean point is $(0, 5)$.\nA more careful analysis reveals that the two stated conditions uniquely determine the Mediterranean point of every triangle: It is the point in the triangle that is closest to the origin. This can be seen as follows:\n(1) First consider a triangle $ABC$ that contains the origin $(0, 0)$. Construct an equilateral triangle $DEF$ that contains $ABC$ and that has $(0, 0)$ as its center of gravity. Since $DEF$ is symmetric with respect to three lines through the origin, condition (i) implies that its Mediterranean point is the origin. Since $DEF$ contains $ABC$ and the origin is in $ABC$, condition (ii) yields that the Mediterranean point of $ABC$ is the origin.\n\n(2) Next consider a triangle $ABC$ that does not contain the origin $(0, 0)$, whose circumcenter is in $(0, 0)$, and that furthermore satisfies $AC = BC$. Consider the circle $k_0$ with center in $(0,0)$ that goes through the Mediterranean point $M$ of $ABC$. Suppose for the sake of contradiction that $M$ is not the midpoint of side $AB$. Then $k_0$ also contains some point $P \\neq M$ in triangle $ABC$. Let $Q$ denote the midpoint of $MP$. Then condition (i) yields that $Q$ is the Mediterranean point of the degenerate triangle $MPQ$, while condition (ii) implies that the Mediterranean point of $MPQ$ is point $M$. That's a contradiction. Therefore the Mediterranean point of $ABC$ is $M$, and this is indeed its closest point to the origin.\n\n(3) Finally consider an arbitrary triangle $ABC$ that does not contain the origin. Let $M$ denote the point in the triangle $ABC$ that is closest to the origin. Construct a circle $k_1$ that has its center in $(0, 0)$ and that touches $ABC$ from outside in point $M$.\nConstruct another circle $k_2$ with center in $(0, 0)$ that satisfies the following: The tangent of $k_1$ at point $M$ intersects circle $k_2$ in the points $D$ and $E$. The half-line that starts in $(0, 0)$ and goes through point $M$ intersects circle $k_2$ in the point $F$. The radius of $k_2$ is chosen sufficiently large, so that $DEF$ entirely contains $ABC$. Then (2) yields that the Mediterranean point of $DEF$ is the point $M$, and condition (ii) yields that the Mediterranean point of $ABC$ is also $M$. Once again, $M$ is the point of $ABC$ that is closest to the origin.\n\n(4) It remains to observe that for the triangle with vertices in $(-3, 5)$, $(12, 5)$ and $(3, 11)$, the closest point to the origin is point $(0, 5)$; hence the desired Mediterranean point is $(0, 5)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to put on the plane $2014$ points, so that with vertices from these points to construct $1006^2$ parallelograms of area $1$?",
    "options": [],
    "answer": "Yes",
    "solution": "We shall prove that it is possible. We consider two parallel lines $\\varepsilon_1, \\varepsilon_2$ with distance $1$. We put on each of them $1007$ points such that every two successive of them have distance $1$. Then we have in the line $\\varepsilon_1$ $1006$ segments of length $1$ and in the line $\\varepsilon_2$ we have also $1006$ unit segments. Any pair of unit segments having members from both lines gives rise to a parallelogram of area $1$. Therefore we have totally $1006^2$ parallelograms of area $1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be an acute angled triangle with $AB \\le A\\Gamma$ and $c(O, R)$ its circumcircle. The line perpendicular from $A$ to the tangent of $c(O, R)$ at $\\Gamma$ intersects it at $\\Delta$.\n\n(α) If the triangle $AB\\Gamma$ is isosceles with $AB = A\\Gamma$, prove that: $\\Gamma\\Delta = \\frac{B\\Gamma}{2}$.\n\n(β) If $\\Gamma\\Delta = \\frac{B\\Gamma}{2}$, prove that the triangle $AB\\Gamma$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(α) If $Z$ is the midpoint of $B\\Gamma$, then $AZ$ is altitude and median of the isosceles triangle $AB\\Gamma$. Then the right angled triangles $A\\Gamma Z$ and $A\\Gamma\\Delta$ are equal, because they have: $A\\Gamma$ common hypotenuse and $A\\Gamma\\Delta = A\\Gamma Z$, (the last equality arises from the equalities $A\\Gamma\\Delta = A\\beta\\Gamma$ (chord-tangent angle and the corresponding inscribed angle) and $A\\beta\\Gamma = A\\Gamma Z$, (since $AB = A\\Gamma$). Hence we get:\n$$\n\\Gamma\\Delta = \\Gamma Z = \\frac{B\\Gamma}{2}.\n$$\n\n(β) Let $AB < A\\Gamma$. We consider the perpendicular bisector of $B\\Gamma$ ($M$ is the midpoint) which intersects the extension of $AB$ at $E$. Then $A\\beta\\Gamma = A\\Gamma\\Delta$ and from hypothesis we have $2\\Gamma\\Delta = B\\Gamma = 2BM$. Hence $\\Gamma\\Delta = BM$. Therefore the right angled triangles $EBM$ and $A\\Gamma\\Delta$ are equal, and so $A\\Gamma = EB$. Since $EB = E\\Gamma$, we conclude $A\\Gamma = E\\Gamma$. However this is absurd, because $E A\\Gamma = 180^\\circ - A$ is obtuse angle and so the triangle $E A\\Gamma$ would be having two obtuse angles. Hence $AB = A\\Gamma$ and the triangle $AB\\Gamma$ is isosceles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31231,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a mathematical competition some students are friends, and friendship is always mutual. Prove that there exists a subset $M$ of the students, such that every member of $M$ has at most three friends in $M$ and such that every non-member of $M$ has at least four friends in $M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start with the set $M_0 = \\emptyset$, and we modify/improve it step by step until it satisfies the two desired conditions. There are two possibilities for an improving step:\n\n1. If there is a student $x \\in M$ who has $\\ge 4$ friends in $M$, then $x$ is removed from $M$.\n\n2. If there is a student $y \\notin M$ who has $\\leq 3$ friends in $M$, then $y$ is added to $M$.\n\nThese improving steps may behave quite chaotically, and they can add the same student many times to $M$ (and then remove it again from $M$). However, we can show that the improvement process must eventually terminate. For a set $M$, let $p(M)$ denote the number of ordered pairs $(x, y) \\in M \\times M$ for which $x$ and $y$ are friends. We associate with set $M$ the function value\n\n$$\nf(M) := 7|M| - p(M).\n$$\n\nClearly $f(M)$ only takes integral values. Furthermore, it is bounded from above by $7$ times the total number of students and its starting value is $f(M_0) = 0$. We will show that every improving step strictly increases the value of $f(M)$. Hence, after a finite number of steps the process terminates with a set $M$ that satisfies the two desired conditions.\n\nFirst assume that there is a student $x \\in M$ with $\\geq 4$ friends in $M$ (as in improving step #1). Then there are at least $8$ pairs of friends $(x, y)$ and $(y, x)$ in $M \\times M$. Hence if we remove $x$ from $M$, then $p(M)$ decreases by at least $8$, whereas $|M|$ decreases by $1$; but this means that $f(M) = 7|M| - p(M)$ strictly increases.\n\nNext assume that there is a student $y \\notin M$ who has $\\leq 3$ friends in $M$ (as in improving step #2). If we add $y$ to $M$, then $p(M)$ increases by at most $6$, whereas $|M|$ increases by $1$; hence also in this case $f(M)$ strictly increases.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31232,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a School formed $112$ groups each contained $11$ students. Every pair of groups had exactly one common student. Prove that:\n\na. There exists a student belonging at least to $12$ groups.\n\nb. There exists a student belonging to all groups.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. We consider an arbitrary group $O$. Each of the rest $111$ groups has exactly one student belonging to $O$. Since $111 = 11 \\cdot 10 + 1$, from the pigeonhole's rule we conclude that there exists one student $x \\in O$ who belongs at least to $11$ other groups. Hence $x$ belongs at least to $12$ groups, say $O_1, O_2, \\dots, O_{12}$.\n\nb. We will prove that $x$ belongs to all groups. In fact, let $x \\notin O'$. Then the group $O'$ has exactly one common student with the groups $O_1, O_2, \\dots, O_{12}$. Since $O'$ has $11$ students, then there will exist two groups, say $O_i, O_j$ having the same common student with the group $O'$, say $y$. However, in this case the groups $O_i, O_j$ will have two common students $x$ and $y$, absurd.\n\nTherefore $x$ belongs to all groups.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31233,
    "subject": "Mathematics (Multi-modal)",
    "question": "(α) For every real number $x > 0$ prove that: $x^3 - 3x \\ge -2$.\n\n(β) For all real numbers $x, y, z > 0$, prove that:\n$$\n\\frac{x^2 y}{z} + \\frac{y^2 z}{x} + \\frac{z^2 x}{y} + 2 \\left( \\frac{y}{xz} + \\frac{z}{xy} + \\frac{x}{yz} \\right) \\ge 9. \\quad (1)\n$$\nWhen equality holds?",
    "options": [],
    "answer": "Equality holds when x = y = z = 1.",
    "solution": "(α) We have\n$$\n\\begin{align*} x^3 - 3x \\ge -2 &\\Leftrightarrow x^3 - 3x + 2 \\ge 0 \\\\ &\\Leftrightarrow x^3 - x - 2x + 2 \\ge 0 \\\\ &\\Leftrightarrow x(x-1)(x+1) - 2(x-1) \\ge 0 \\\\ &\\Leftrightarrow (x-1)(x^2 + x - 2) \\ge 0 \\\\ &\\Leftrightarrow (x+2)(x-1)^2 \\ge 0 \\end{align*}\n$$\nwhich is valid, because $x > 0$.\n\n(β) The inequality is equivalent to\n$$\n\\frac{y}{z} \\left( x^2 + \\frac{2}{x} \\right) + \\frac{z}{x} \\left( y^2 + \\frac{2}{y} \\right) + \\frac{x}{y} \\left( z^2 + \\frac{2}{z} \\right) \\ge 9. \\quad (2)\n$$\nFrom question (α), by dividing both parts by $x$ we obtain $x^2 + \\frac{2}{x} \\ge 3$.\nTherefore it is enough to prove:\n$$\n3 \\left( \\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y} \\right) \\ge 9 \\Leftrightarrow \\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y} \\ge 3. \\quad (3)\n$$\nThe latter is valid, as we can easily see, by applying the inequality of arithmetic and geometric mean as follows:\n$$\n\\frac{y}{z} + \\frac{z}{x} + \\frac{x}{y} \\ge 3 \\sqrt{\\frac{yzx}{zxy}} = 3 \\quad (4)\n$$\nThe equality holds if and only if $x = y = z = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all possible triads of non negative integers $(x, y, z)$ with $x \\le y$, which satisfy the equation: $x^2 + y^2 = 3 \\cdot 2016^z + 77$",
    "options": [],
    "answer": "[(4, 8, 0), (35, 70, 1), (14, 77, 1)]",
    "solution": "We distinguish the cases:\n\nIf $z = 0$, then the equation becomes $x^2 + y^2 = 80$.\nThen $x, y$ must be multiples of $4$, that is $x = 4a$, $y = 4b$, $0 \\le a \\le b$, and the equation becomes $a^2 + b^2 = 5$, hence $(a, b) = (1, 2)$, that is $(x, y) = (4, 8)$.\n\nIf $z > 0$, then $7 \\mid 2016^z$ (since $7 \\mid 2016$) and $7 \\mid 77$, and therefore $7$ must divide the left part, that is $7 \\mid x^2 + y^2$. The possible remainders of the division of a square by $7$ are the same as the remainders of the numbers $0^2, 1^2, 2^2, 3^2, 4^2, 5^2, 6^2$, that is $0, 1, 2, 4$. Therefore, in order to have $7 \\mid x^2 + y^2$ we need $7 \\mid x$, $7 \\mid y$. We write $x = 7x_1$ and $y = 7y_1$, with $0 \\le x_1 \\le y_1$.\n\nBy substitution in (1) we get $49(x_1^2 + y_1^2) = 3 \\cdot 2016^z + 77$.\n\nIf $z \\ge 2$, then $49 \\mid 2016^z$, and then we conclude that $49 \\mid 77$, absurd.\n\nIf $z = 1$, then $49(x_1^2 + y_1^2) = 3 \\cdot 7 \\cdot 288 + 77 = 7(3 \\cdot 288 + 11) = 7 \\cdot 7 \\cdot 125$.\nHence $x_1^2 + y_1^2 = 125$. Since $x_1 \\le y_1$, we have $2y_1^2 \\ge 125 \\Rightarrow y_1 \\ge 8$.\nIf $y_1 = 8, 9, 10, 11$, we find the solutions $(x_1, y_1) \\in \\{(5, 10), (2, 11)\\}$. Therefore we have the solutions: $(x, y) \\in \\{(4, 8), (35, 70), (14, 77)\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31235,
    "subject": "Mathematics (Multi-modal)",
    "question": "The integers $p$, $q$ and $r$ are primes and their product is equal to $n$. By increasing $p$ and $q$ by 1, then the product $(p+1)(q+1)r$ is equal to $n+138$. Determine all the possible values of $n$.",
    "options": [],
    "answer": "138, 258, 854, 2294",
    "solution": "We have:\n$$\n\\left\\{ \\begin{array}{l} pqr = n \\\\ (p+1)(q+1)r = n+138 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} pqr = n \\\\ pqr + (p+q)r + r = n+138 \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} pqr = n \\\\ (p+q+1)r = 138 \\end{array} \\right\\}.\n$$\nFrom equation\n$$\n(p+q+1)r = 138 = 2 \\cdot 3 \\cdot 23 \\quad (1)\n$$\nwe will determine the possible values of $p$, $q$, $r$ and in the sequel from the equation $pqr = n$. We will find all the possible values of $n$.\nSince $p$, $q$, $r$ are primes, the possible values of $r$ are $2$ or $3$ or $23$, and therefore we have the cases:\n* If $r = 2$, then $p+q+1 = 69 \\Leftrightarrow p+q = 68$, and hence, $(p, q$ primes$)$ we find the pairs\n$$ (p,q) = (7,61), (p,q) = (61,7), (p,q) = (31,37), (p,q) = (37,31). $$\nHence for the product $pqr = n$ we obtain the values:\n$n = 7 \\cdot 61 \\cdot 2 = \\mathbf{854}$ or $n = 31 \\cdot 37 \\cdot 2 = \\mathbf{2294}$.\n* If $r = 3$, then $p+q+1 = 46 \\Leftrightarrow p+q = 45$, and hence:\n$$\n(p,q) = (2,43), (p,q) = (43,2) \\Rightarrow n = 2 \\cdot 43 \\cdot 3 = \\mathbf{258}.\n$$\n* If $r = 23$, then\n$$ p+q+1=6 \\Leftrightarrow p+q=5 \\Leftrightarrow (p,q) = (2,3) \\text{ or } (p,q) = (3,2), $$\nand hence $n = 2 \\cdot 3 \\cdot 23 = \\mathbf{138}$.\nTherefore the possible values of $n$ are: **138, 258, 854 και 2294**.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all injective functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that for every real number $x$ and every positive integer $n$,\n$$\n\\left| \\sum_{i=1}^{n} i \\left( f(x+i+1) - f(x+i) \\right) \\right| < 2016\n$$",
    "options": [],
    "answer": "No such injective function exists.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31237,
    "subject": "Mathematics (Multi-modal)",
    "question": "The polynomials $P(x)$, $Q(x)$ with real coefficients are non-constant, monic and satisfy the equality:\n\n$$\n2P(x) = Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right), \\quad x \\in \\mathbb{R}, \\quad P(1)=1,\n$$\n\nDetermine the polynomials $P(x)$ and $Q(x)$.",
    "options": [],
    "answer": "Either (1) P(x) = x and Q(x) = x + a_0 with a_0 ∈ ℝ; or (2) P(x) = x^3 and Q(x) = x^2 − x + c with c ∈ ℝ.",
    "solution": "Let $Q(x) = x^n + a_{n-1}x^{n-1} + \\dots + a_0$. Then the coefficient of the term of maximal degree of the polynomial $Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right)$ can be found from the difference $\\left(\\frac{(x+1)^2}{2}\\right)^n - \\left(\\frac{(x-1)^2}{2}\\right)^n$. It is\n$$\n\\frac{2n x^{2n-1}}{2^n} + \\frac{2n x^{2n-2}}{2^n} = \\frac{4n}{2^n} x^{2n-1} \\quad (1).\n$$\nThe corresponding coefficient in the left hand side is equal to $2$, and so $\\frac{4n}{2^n} = 2 \\Leftrightarrow 2^{n+1} = 4n$. Since $2^{n+1} > 4n$, for $n \\ge 3$, we must have $n=1$ or $n=2$ and then from (1) the degree of $P$ is $2-1=1$ or $2 \\cdot 2-1=3$.\n\nFor $x=0$ in the given relation, we obtain $2P(0) = Q(1/2) - Q(1/2) = 0$, and hence $P(0) = 0$.\n\n* For $n=1$, then $P(x) = a x$ and since $P(1) = 1$, we find\n$$\nP(x) = x \\text{ and } Q(x) = x + a_0, \\quad a_0 \\in \\mathbb{R}.\n$$\n\n* For $n=2$, then, if $Q(x) = x^2 + b x + c$, we have:\n$$\n\\begin{align*}\n2P(x) &= Q\\left(\\frac{(x+1)^2}{2}\\right) - Q\\left(\\frac{(x-1)^2}{2}\\right) \\\\\n&= \\frac{1}{4}((x+1)^4 - (x-1)^4) + \\frac{b}{2}((x+1)^2 - (x-1)^2) \\\\\n&= \\frac{1}{4}(8x^3 + 8x) + \\frac{b}{2}(4x) \\\\\n&= 2x^3 + 2(1+b)x.\n\\end{align*}\n$$\nHence $P(x) = x^3 + (1+b)x$. Since $P(1) = 1$, we find $b = -1$, and finally\n$$\nP(x) = x^3 \\text{ and } Q(x) = x^2 - x + c, \\quad c \\in \\mathbb{R}.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $c(O, R)$ be a circle and $A, B$ two diametrically opposite points. The bisector of the angle $A\\hat{B}O$ intersects the circle $c(O, R)$ at point $C$, the circumcircle of the triangle $AOB$ (say $(c_1)$) at point $K$ and the circumcircle of the triangle $AOC$ (say $(c_2)$) at point $L$. Prove that the point $K$ is the circumcenter of the triangle $AOC$ and the point $L$ is the incenter of the triangle $AOB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Segments $OB$, $OC$ are equal, as radii of the circle ($c$), and so the triangle $OBC$ is isosceles. Hence:\n$$\n\\hat{B}_1 = \\hat{C}_1 = \\hat{x} \\qquad (1).\n$$\n![](attached_image_1.png)\nFigure 3\n$BC$ is the bisector of the angle $OBA$, and hence\n$$\n\\hat{B}_1 = \\hat{B}_2 = \\hat{x} \\qquad (2).\n$$\nThe angles $\\hat{B}_2$ and $\\hat{O}_1$ are inscribed in the same circle and correspond to same arc $OK$ of the circle $(c_1)$. Hence\n$$\n\\hat{B}_2 = \\hat{O}_1 = \\hat{x} \\qquad (3).\n$$\nSimilarly we have $KO = KC$ and the triangle $KOC$ is isosceles and hence:\n$$\n\\hat{O}_2 = \\hat{C}_1 = \\hat{x} \\qquad (4).\n$$\nFrom (1)–(4) we conclude that: $\\hat{O}_1 = \\hat{O}_2 = \\hat{x}$, i.e. $OK$ is the bisector, hence and perpendicular bisector of the isosceles triangle $OAC$. The point $K$ is the midpoint of the arc $OK$ ($BK$ is the bisector of $OBA$). Hence the perpendicular bisector of the chord $AO$ of the circle ($c_1$) (which is also side of the isosceles triangle $OAC$), is passing through point $K$. It means that $K$ is the circumcenter of the triangle $OAC$.\n\nFrom (1), (2), we have: $\\hat{B}_2 = \\hat{C}_1 = \\hat{x}$ and hence $AB \\parallel OC$. Therefore $O\\hat{A}B = A\\hat{O}C$, that is $\\hat{A}_1 + \\hat{A}_2 = \\hat{O}_1 + \\hat{O}_2$ and since $\\hat{O}_1 = \\hat{O}_2 = \\hat{x}$, we conclude that:\n$$\n\\hat{A}_1 + \\hat{A}_2 = 2\\hat{O}_1 = 2\\hat{x}.\n$$\nThe angles $\\hat{A}_1$ and $\\hat{C}_1$ are inscribed in the circle ($c_2$) and correspond to the same arc $OL$. Hence\n$$\n\\hat{A}_1 = \\hat{C}_1 = \\hat{x}.\n$$\nFrom the last two equalities we have that $\\hat{A}_1 = \\hat{A}_2$, that is $AL$ is the bisector of the angle $B\\hat{A}O$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31239,
    "subject": "Mathematics (Multi-modal)",
    "question": "The real numbers $x, y, z$, with $x \\neq z$, are mutually different and nonzero and they satisfy the following equations:\n$$\n(x+y)^2 + (2-xy) = 9, \\\\\n(y+z)^2 - (3+yz) = 4.\n$$\nDetermine the value of the expression\n$$\nA = \\left( \\frac{x}{y} + \\frac{y^2}{x^2} + \\frac{z^3}{x^2y} \\right) \\left( \\frac{y}{z} + \\frac{z^2}{y^2} + \\frac{x^3}{y^2z} \\right) \\left( \\frac{z}{x} + \\frac{x^2}{z^2} + \\frac{y^3}{z^2x} \\right).\n$$",
    "options": [],
    "answer": "27",
    "solution": "The given equalities can be written:\n$$\nx^2 + y^2 + xy = 7, \\qquad (1)\n$$\n$$\ny^2 + z^2 + yz = 7, \\qquad (2)\n$$\nBy subtraction we get:\n$$\nx^2 - z^2 + xy - yz = 0 \\Leftrightarrow (x-z)(x+z) + y(x-z) = 0 \\Leftrightarrow (x-z)(x+z+y) = 0.\n$$\nSince $x - z \\neq 0$, we get:\n$$\nx + y + z = 0. \\qquad (3)\n$$\nThen with simple manipulations for the factors of $A$ we find\n$$\nA = \\left( \\frac{x^3 + y^3 + z^3}{xyz} \\right)^3 \\qquad (4)\n$$\nFrom (3), using Euler's identity, we find\n$$\nx^3 + y^3 + z^3 = 3xyz \\qquad (5)\n$$\nHence from (4) and (5) we get $A = 27$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B \\Gamma$ be an isosceles acute angled triangle with $AB = A\\Gamma$. Let $\\Gamma\\Delta$ be an altitude of the triangle. The circle $c_2(\\Gamma, \\Gamma\\Delta)$ intersects $A\\Gamma$ at point $K$, the extension of $A\\Gamma$ at point $Z$ and the circle $c_1(B, B\\Delta)$ at point $E$. Finally, $\\Delta Z$ intersects the circle $c_1$ at point $M$. Prove that\n\na. $Z\\hat{\\Delta}E = 45^\\circ$\n\nb. The points $E$, $M$, and $K$ are collinear.\n\nc. The line $BM$ is parallel to the line $E\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nFrom the right angled triangle $B\\Gamma\\Delta$ we have: $\\hat{\\Gamma}_1 = 90^\\circ - \\hat{B}$.\nThe line joining the centers of the circles $c_1$ and $c_2$ is the perpendicular bisector of their common chord $\\Delta E$.\nIf $T$ is the point of intersection of the lines $B\\Gamma$ and $\\Delta E$, then:\n$$\n\\hat{\\Gamma}_1 + \\hat{\\Delta}_1 + \\hat{\\Delta}_2 = 90^\\circ, \\text{ that is } \\hat{\\Delta}_1 + \\hat{\\Delta}_2 = \\hat{B} = \\hat{\\Gamma}. \\quad (1)\n$$\nFrom the isosceles triangle $\\Gamma\\Delta Z$ we have: $\\hat{\\Delta}_1 = \\hat{Z}_1$ and $\\hat{\\Gamma}_2 = 2\\hat{\\Delta}_1 = 90^\\circ - \\hat{A}$, and hence:\n$$\n\\hat{\\Delta}_1 = 45^\\circ - \\frac{\\hat{A}}{2}. \\quad (2)\n$$\nFrom the isosceles triangle $AB\\Gamma$ we have: $\\hat{B} = 90^\\circ - \\frac{\\hat{A}}{2}$ \\quad (3)\nFrom (1), (2), (3) we get:\n$$\n\\hat{\\Delta}_1 + \\hat{\\Delta}_2 = \\hat{B} = \\hat{\\Gamma} \\Leftrightarrow \\hat{\\Delta}_2 = \\hat{B} - \\hat{\\Delta}_1 \\stackrel{(2),(3)}{\\Leftrightarrow} \\hat{\\Delta}_2 = 90^\\circ - \\frac{\\hat{A}}{2} - \\left( 45^\\circ - \\frac{\\hat{A}}{2} \\right) = 45^\\circ.\n$$\n\nb.\nThe angle $\\hat{\\Delta}_1$ is created from the chord $\\Delta M$ and the tangent $\\Delta\\Gamma$ of the circle $c_1$, and hence $\\Delta\\hat{E}M = \\hat{\\Delta}_1$. Also we have $\\Delta\\hat{E}K = \\Delta\\hat{Z}K = \\hat{Z}_1$.\nSince $\\hat{\\Delta}_1 = \\hat{Z}_1$, we get $\\Delta\\hat{E}M = \\Delta\\hat{E}K$, and the points $E$, $M$, $K$ are collinear.\n\nc.\nWe will prove that $\\Delta\\hat{B}M = 90^\\circ - \\hat{A}$.\nWe have $\\Delta\\hat{B}M = 2\\Delta\\hat{E}M = 2\\hat{\\Lambda}_1 = \\hat{\\Gamma}_2 = 90^\\circ - \\hat{A}$, and hence $BM \\perp A\\Gamma$. Moreover $E\\hat{K}Z = E\\hat{\\Delta}Z = \\hat{\\Lambda}_2 = 45^\\circ$, and so the orthogonal triangle $EZK$ isosceles. Then its median $E\\Gamma$ is also an altitude, that is $E\\Gamma \\perp KZ$. Hence $BM \\parallel E\\Gamma$ (both are perpendicular to the line $AZ$).\n\n![](attached_image_1.png)\nFigure 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma\\Delta$ ($A\\Delta\\parallel B\\Gamma$) be a trapezium with $\\hat{A} = \\hat{B} = 90^\\circ$ and $A\\Delta < B\\Gamma$. Denote by $E$ the point of intersection of the two non parallel sides $AB$ and $\\Gamma\\Delta$, $Z$ the symmetric point of $A$ with respect to the line $B\\Gamma$ and $M$ the midpoint of $EZ$. It is given that the line $\\Gamma M$ is perpendicular to the line $\\Delta Z$. Prove that the line $Z\\Gamma$ is perpendicular to the line $\\Gamma E$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the line $\\Delta Z$ meets $\\Gamma M, B\\Gamma$ at points $K, N$, respectively. Then in the triangle $A\\Delta Z$, $B$ is the midpoint of $AZ$ and $BN\\parallel A\\Delta$. Hence $N$ is the midpoint of $Z\\Delta$. Therefore in the triangle $ZE\\Delta$, $MN$ connects the midpoints of two sides, and hence:\n\n$$\nMN // E\\Delta \\qquad (1)\n$$\nMoreover in the triangle $M\\Gamma Z$, $\\Gamma B, ZK$ are altitudes, and hence $N$ is the ortho-center of the triangle. Hence:\n$$\nMN \\perp Z\\Gamma \\qquad (2).\n$$\nFrom (1) and (2) we conclude that $Z\\Gamma \\perp E\\Gamma$.\n![](attached_image_1.png)\nFigure 1",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31242,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive integer $n$ is such that $n^2-9$ has exactly 6 positive divisors. Prove that $\\text{gcd}(n-3, n+3)=1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In order the positive integer $n$ to have exactly 6 positive divisors, it must be of the form $n^2-9 = q^5$ or $n^2-9 = p^2q$, where $p$, $q$ are primes mutually different.\n\nIn the first case we have $n^2-9 = q^5 \\Rightarrow (n-3)(n+3) = q^5$ (1)\nIt gives that\n$$\nn-3 = q^s \\text{ and } n+3 = q^t, \\text{ with } t > s \\text{ and } t+s = 5.\n$$\nBy subtracting we get $q^t - q^s = 6 \\Leftrightarrow q^s(q^{t-s} - 1) = 2 \\cdot 3$ and hence $q \\in \\{2,3\\}$.\nFor $q=2$ (1) gives $n^2 = 41$, (it has no solutions in $\\mathbb{Z}$), while for $q=3$, (1) gives $n^2 = 252$, (impossible in $\\mathbb{Z}$).\n\nIn the second case $n^2-9 = p^2q$, it follows that $(n-3)(n+3) = p^2q$, and so\n$$\n\\begin{align*}\n(n+3 = pq \\text{ and } n-3 = p) \\text{ or } (n+3 = q \\text{ and } n-3 = p^2) \\\\\n\\text{or } (n+3 = p^2 \\text{ and } n-3 = q).\n\\end{align*}\n$$\nFrom the first case we get $p(q-1) = 6$, and hence ($p=3$ and $q=3$), which must be rejected because $p \\neq q$.\nFrom the last two cases we conclude that $\\text{gcd}(n-3, n+3)=1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Vangelis has a box containing $2015$ white and $2015$ black balls. He follows the following procedure: He chooses randomly two balls from the box. If both are black, then paints one of them white and puts it in the box, while drops the other out of the box. If both are white, then he keeps one of them in the box and drops the other out of the box. If one ball is white and the other is black, then he keeps the black in the box and drops out the white ball. He follows the procedure, until only $3$ balls remain in the box. Then he realizes that in the box exist balls of both colors.\n\nDetermine how many white and how many black balls exist finally in the box.",
    "options": [],
    "answer": "2 white balls and 1 black ball",
    "solution": "We will consider what happens at every step of the procedure for the number of white and black balls. In the case of selection of two black balls, no one is going back to the box and so the number of the black balls **is reducing by 2**. In the second case of selection of two white balls, the number of the black balls **has not any change**. In the third case of selection of balls of both colors, also the number of the black balls in the box **remains invariant**. Therefore we observe that, in every step, the number of the black balls in the box remains invariant or it is reducing by $2$. Since their initial number is $2015$ (odd) at every step their number will remain odd. Therefore in the case of $3$ balls in the box of both colors, we conclude that one of them must be black and the other two must be white.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of ordered 6-tuples $(\\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4, \\alpha_5, \\alpha_6)$ can be created, if the numbers $\\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4, \\alpha_5, \\alpha_6$ can take the values $0$, $1$ and $2$ and the sum $\\alpha_1 + \\alpha_2 + \\alpha_3 + \\alpha_4 + \\alpha_5 + \\alpha_6$ is even.",
    "options": [],
    "answer": "365",
    "solution": "The sum $\\alpha_1 + \\alpha_2 + \\alpha_3 + \\alpha_4 + \\alpha_5 + \\alpha_6$ is even, if and only if, the number of $1$'s is even, that is $0$, $2$, $4$, $6$.\n\nIn the case of zero $1$'s, the possible selections are $2^6$, because for each $\\alpha_i$ we have $2$ selections, ($0$ or $2$).\n\nWhen we have two $1$'s, then they can be selected in $\\binom{6}{2}$ ways and the rest four places can be completed by $2^4$ ways. It means that we have $2^4 \\cdot \\binom{6}{2}$ possible $6$-tuples.\n\nSimilarly, in the case we have four $1$'s we conclude that we have $2^2 \\cdot \\binom{6}{4}$ possible $6$-tuples.\n\nWhen we have six $1$'s then obviously we have only one possible $6$-tuple.\n\nTherefore the possible $6$-tuples are totally:\n$$\n2^6 + 2^4 \\cdot \\binom{6}{2} + 2^2 \\cdot \\binom{6}{4} + 1 = 64 + 16 \\cdot 15 + 4 \\cdot 15 + 1 = 365.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Square $ABCD$ is divided into $n^2$ equal elementary squares, by drawing parallel lines to their sides. We call the vertices of the elementary squares points of the lattice. One rhombus will be called “good”, if it is not a square, their vertices are points of the lattice and their diagonals are parallel to the sides of the square $ABCD$. Find a closed formula, with respect to $n$, of the number of “good” rhombuses, where $n$ is integer greater than $2$.",
    "options": [],
    "answer": "If n is even: n(n−2)(3n^2 − 2n − 4)/48. If n is odd: (n^2 − 1)(3n^2 − 8n − 3)/48.",
    "solution": "Drawing from the vertices of the rhombus parallel lines to the sides of the lattice, we find the least rectangle in which the rhombus is inscribed. Since the rhombus is symmetric with respect to its center, this rectangle will have sides with even length. Therefore it is enough to count the rectangles with even sides $2s \\times 2t$, for which $s \\neq t$. It is equivalent to count first all rectangles $2s \\times 2t$ and then subtract all squares $2s \\times 2t$.\n\n![](attached_image_1.png)\n\n* Let $n = 2k$\n\nWe count all vertical lines of the lattice from left to right by $1, 2, 3, \\ldots, 2k+1$ and all horizontal lines from bottom to top by $1, 2, 3, \\ldots, 2k+1$. A rectangle $2s \\times 2t$ is determined from two vertical lines and two horizontal lines having even distance. Two lines have even distance when their numbers are of the same parity. We can select two vertical lines with even numbers and even distance in $\\binom{k}{2}$ ways, while to select two vertical lines with odd numbers and even distance we have $\\binom{k+1}{2}$ ways. Therefore, the total number of ways to select two vertical lines with even distance is $\\binom{k}{2} + \\binom{k+1}{2} = k^2$. Similarly, we have $k^2$ possible selections for the horizontal lines. In total, we have $k^2 \\cdot k^2 = k^4$ rectangles of the type $2s \\times 2t$.\n\nIt remains to subtract all squares $2s \\times 2s$. The $2 \\times 2$ squares are totally $(n-2+1)^2 = (2k-2+1)^2$. The $4 \\times 4$ squares are totally $(n-4+1)^2 = (2k-4+1)^2$. In general, the $2s \\times 2s$ squares are totally $(n-2s+1)^2 = (2k-2s+1)^2$. Hence we have\n$$\n\\sum_{s=1}^{k} (2k - 2s + 1)^2 = \\sum_{s=1}^{k} (2k+1)^2 - 4s(2k+1) + 4s^2 = \\\\\n k(2k+1)^2 - 2(2k+1)k(k+1) + 4 \\frac{k(k+1)(2k+1)}{6} = \\frac{k(2k-1)(2k+1)}{3}\n$$\nTherefore all rhombuses in the even case are:\n$$\nk^4 - \\frac{k(2k-1)(2k+1)}{3} = \\frac{k(k-1)(3k^2 - k - 1)}{3}\n$$\n\n* Let $n = 2k+1$. Then, similarly, we have $\\binom{k+1}{2} + \\binom{k+1}{2} = k(k+1)$ possible selections for the two vertical lines and $k(k+1)$ possible selections for the horizontal lines. Totally we have $(k(k+1))^2$ rectangles $2s \\times 2t$. We have to subtract the squares $2s \\times 2s$ which totally are\n$$\n(n - 2s + 1)^2 = (2k + 1 - 2s + 1)^2 = (2k + 2 - 2s)^2\n$$\nHence we have\n$$\n\\sum_{s=1}^{k} (2k + 2 - 2s)^2 = \\sum_{s=1}^{k} (2k + 2)^2 - 4(2k + 2)s + 4s^2 \\\\\n = k(2k + 2)^2 - 2(2k + 2)k(k + 1) + 4 \\frac{k(k+1)(2k+1)}{6} = \\frac{2k(k+1)(2k+1)}{3}\n$$\nHence in the odd case the rhombuses are totally\n$$\n(k(k+1))^2 - \\frac{2k(k+1)(2k+1)}{3} = \\frac{k(k+1)(3k^2 - k - 2)}{3}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31246,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Examine if there is a real number $x$, such that both $x + \\sqrt{3}$ and $x^2 + \\sqrt{3}$ are rational numbers.\n\nb. Examine if there is a real number $y$, such that both $y + \\sqrt{3}$ and $y^3 + \\sqrt{3}$ are rational numbers.",
    "options": [],
    "answer": "a) Yes: x = 1/2 − √3. b) No such y exists.",
    "solution": "a. Let $x + \\sqrt{3} = q$, $x^2 + \\sqrt{3} = p$ with $p, q \\in \\mathbb{Q}$. Then\n$$\nx = q - \\sqrt{3} \\Rightarrow x^2 = q^2 - 2q\\sqrt{3} + 3\n$$\nso substituting in the second one gives:\n$$\n(q^2 - 2q\\sqrt{3} + 3) + \\sqrt{3} = p \\Leftrightarrow -\\sqrt{3}(2q-1) = p - q^2 - 3\n$$\nIt follows that $2q-1=0 \\Leftrightarrow q = \\frac{1}{2}$. In this case, $p = q^2+3 \\Rightarrow p = \\frac{1}{4} + 3 = \\frac{13}{4}$ and\n$x = \\frac{1}{2} - \\sqrt{3}$.\n\nb. Let $y + \\sqrt{3} = q$, $y^3 + \\sqrt{3} = p$ with $p, q \\in \\mathbb{Q}$. Then\n$$\ny = q - \\sqrt{3} \\Rightarrow y^3 = q^3 - 3q^2\\sqrt{3} + 9q - 3\\sqrt{3}\n$$\nso substituting in the second one gives:\n$$\n(q^3 - 3q^2\\sqrt{3} + 9q - 3\\sqrt{3}) + \\sqrt{3} = p \\Leftrightarrow -\\sqrt{3}(3q^2 + 2) = p - q^3 - 9q \\Leftrightarrow \\\\\n\\sqrt{3} = \\frac{q^3 + 9q - p}{3q^2 + 2} \\in \\mathbb{Q}\n$$\nwhich is absurd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers such that\n$$\na^2 + b^2 + c^2 + d^2 = 4.\n$$\nProve that there are two of $a$, $b$, $c$, $d$ with sum greater or equal to $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, let $a \\ge b \\ge c \\ge d$ and we will prove that $a+b \\ge 2$.\nWe have that $ab \\ge c^2$ and $ab \\ge d^2$, so by adding them we have: $2ab \\ge c^2 + d^2$.\nTherefore,\n$$\n(a+b)^2 = a^2 + b^2 + 2ab \\ge a^2 + b^2 + c^2 + d^2 = 4,\n$$\nso $a+b \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(x_n)$, $n \\in \\mathbb{N}^*$ be a sequence which is recursively defined by\n$x_{n+1} = 3x_n^3 + x_n,$\nwhere $x_1 = \\frac{a}{b}$, and $a, b$ are positive integers such that $3$ doesn't divide $b$. If for some positive integer $m$ we have that $x_m$ is a perfect square of a rational, prove that $x_1$ is a perfect square of a rational.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that if $x_{n+1}$ is a perfect square of a rational, then $x_n$ is also a perfect square of a rational, and the desired result is obtained by a simple induction.\n\nNote first that since $3$ doesn't divide $b$, it will not divide any of the denominators of the sequence terms.\n\nFrom the recursive relation we have that $x_m = 3x_{m-1}^3 + x_{m-1}$. Setting $x_{m-1} = \\frac{p}{q}$ where $q$ is not divisible by $3$ (*) and $(p,q)=1$, then\n$$\nx_m = 3x_{m-1}^3 + x_{m-1} = \\frac{3p^3 + pq^2}{q^3} = \\frac{p(3p^2 + q^2)}{q^3}.\n$$\nSince $(p,q)=1$, this is the reduced form of $x_m$. Indeed, the numbers $p(3p^2+q^2)$, $q^3$ are coprime, since if $s$ is a prime dividing both of them, then $s|q^3 \\Rightarrow s|q$ and $s|3p^2+q^2$ so $s|3p^2$. But $s$ doesn't divide $p$, so $s|3 \\Rightarrow s=3$, $3|q$, absurd due to (*).\n\nMoreover $x_m$ is a perfect square so both numerator and denominator in the reduced form should be perfect squares.\nSince the denominator is a perfect square, $q$ is a perfect square, let it be $q = a^2$. For the numerator, we have $p(3p^2+q^2) = \\kappa^2$, so both of them should be perfect squares since they are coprime.\n\nTherefore, $p = b^2$, $3p^2 + q^2 = c^2$ and $x_{m-1} = \\frac{p}{q} = \\frac{b^2}{a^2}$, so $x_{m-1}$ is a perfect square of a rational.\n\nSimilarly, going back $x_{m-2}$ is also a perfect square of a rational, and so on, till we arrive at $x_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive real numbers such that $x + y + z = 9xyz$. Prove that:\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 2}} + \\frac{y}{\\sqrt{y^2 + 2zx + 2}} + \\frac{z}{\\sqrt{z^2 + 2xy + 2}} \\geq 1.\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds at x = y = z = 1/√3.",
    "solution": "From the inequality $2yz \\le y^2 + z^2$ we have that $x^2 + 2yz + 2 \\le x^2 + y^2 + z^2 + 2$, so\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 2}} \\ge \\frac{x}{\\sqrt{x^2 + y^2 + z^2 + 2}}\n$$\nWorking similarly and adding we have that\n$$\n\\frac{x}{\\sqrt{x^2 + 2yz + 2}} + \\frac{y}{\\sqrt{y^2 + 2zx + 2}} + \\frac{z}{\\sqrt{z^2 + 2xy + 2}} \\ge \\frac{x+y+z}{\\sqrt{x^2 + y^2 + z^2 + 2}}\n$$\nTherefore, it suffices to prove that\n$$\n\\frac{x+y+z}{\\sqrt{x^2+y^2+z^2+2}} \\ge 1 \\Leftrightarrow (x+y+z)^2 \\ge x^2+y^2+z^2+2 \\Leftrightarrow xy+yz+zx \\ge 1\n$$\nHowever, from the given condition we have $\\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} = 9$, and from the Cauchy-Schwarz inequality we have\n$$\n(xy + yz + zx) \\left( \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} \\right) \\ge 9, \\text{ so } xy + yz + zx \\ge 1, \\text{ which is the desired result.}\n$$\n\n**2ºς τρόπος:** Using Holder's inequality we have:\n$$\n\\left( \\sum_{cyc} \\frac{x}{\\sqrt{x^2 + 2yz + 2}} \\right) \\left( \\sum_{cyc} \\frac{x}{\\sqrt{x^2 + 2yz + 2}} \\right) \\left( \\sum_{cyc} x(x^2 + 2yz + 2) \\right) \\ge (x+y+z)^3\n$$\nTherefore, it suffices to prove that\n$$\n\\begin{aligned} \\frac{(x+y+z)^3}{\\sum_{cyc} x(x^2+2yz+2)} &\\ge 1 \\\\ &\\Leftrightarrow (x+y+z)^3 \\ge x^3+y^3+z^3+6xyz+2(x+y+z) \\\\ &\\Leftrightarrow x^3+y^3+z^3+3(x+y)(y+z)(z+x) \\ge x^3+y^3+z^3+6xyz+2(9xyz) \\\\ &\\Leftrightarrow (x+y)(y+z)(z+x) \\ge 8xyz \\end{aligned}\n$$\nbut the last one holds since\n$$\nx + y \\ge 2\\sqrt{xy}, \\quad y + z \\ge 2\\sqrt{yz}, \\quad z + x \\ge 2\\sqrt{zx}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma\\Delta$ be a square of side $\\alpha$. On the side $A\\Gamma\\Delta$ we get points $E$ and $Z$ such that $\\Delta E = \\frac{\\alpha}{3}$ and $AZ = \\frac{\\alpha}{4}$.\nIf the lines $BZ$ and $\\Gamma E$ intersect at point $H$, express the area of the triangle $B\\Gamma H$ as a function of $\\alpha$.",
    "options": [],
    "answer": "6α^2/7",
    "solution": "![](attached_image_1.png)\nFigure 1\nWe draw the altitude $H\\Lambda$ of the triangle $B\\Gamma H$. Let it intersect $A\\Gamma$ at $K$. We put $EK = x$, $KZ = y$ and $KH = z$. Then $H\\Lambda = \\alpha + z$ and\n$$\nE_{B\\Gamma H} = \\frac{1}{2} \\alpha (\\alpha + z) \\qquad (1)\n$$\nThe triangles $\\Gamma\\Delta E$ and $EHK$ are similar. Hence\n$$\n\\frac{KH}{\\Gamma\\Delta} = \\frac{KE}{\\Delta E} \\Leftrightarrow \\frac{z}{\\alpha} = \\frac{x}{\\frac{\\alpha}{3}} \\Leftrightarrow z = 3x \\qquad (2)\n$$\nMoreover, the triangles $ABZ$ and $ZKH$ are similar and hence\n$$\n\\frac{KH}{AB} = \\frac{KZ}{AZ} \\Leftrightarrow \\frac{z}{\\alpha} = \\frac{y}{\\frac{\\alpha}{4}} \\Leftrightarrow z = 4y \\qquad (3)\n$$\nSince\n$$\nx + y = A\\Gamma - AZ - \\Delta E = \\alpha - \\frac{\\alpha}{4} - \\frac{\\alpha}{3} = \\frac{5\\alpha}{12} \\qquad (4)\n$$\nfrom (2), (3), and (4) we have $x + y = \\frac{5\\alpha}{12} \\Leftrightarrow \\frac{z}{3} + \\frac{z}{4} = \\frac{5\\alpha}{12} \\Leftrightarrow z = \\frac{5\\alpha}{7}$, and\n$$\nE = \\frac{1}{2} \\alpha \\left( \\alpha + \\frac{5\\alpha}{7} \\right) = \\frac{12\\alpha^2}{14} = \\frac{6\\alpha^2}{7}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ positive real numbers such that $a + b + c = 1$. Prove that:\n$$\n(a+1)\\sqrt{2a(1-a)} + (b+1)\\sqrt{2b(1-b)} + (c+1)\\sqrt{2c(1-c)} \\ge 8(ab+bc+ca).\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds if and only if a = b = c = 1/3.",
    "solution": "Since $a + b + c = 1$, we have $a + 1 = 2a + b + c$ and $1 - a = b + c$. Hence we have the equivalent inequality\n$$\n(2a + b + c)\\sqrt{2a(b + c)} + (2b + c + a)\\sqrt{2b(c + a)} + (2c + a + b)\\sqrt{2c(a + b)} \\ge 8(ab + bc + ca)\n$$\nFrom the inequality of arithmetic and geometric mean we get: $2a + b + c \\ge 2\\sqrt{2a(b + c)}$ and therefore\n$$\n(2a + b + c)\\sqrt{2a(b + c)} \\ge 2(2a(b + c)) \\quad (1)\n$$\nSimilarly we have\n$$\n(2b + c + a)\\sqrt{2b(c + a)} \\ge 2(2b(c + a)) \\quad (2),\n$$\n$$\n(2c + a + b)\\sqrt{2c(a + b)} \\ge 2(2c(a + b)) \\quad (3).\n$$\nSumming by parts (1), (2) and (3) we get the wanted inequality. The equality holds if and only if\n$$\n2a = b + c,\\ 2b = c + a,\\ 2c = a + b \\Leftrightarrow a = b = c = \\frac{1}{3}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31252,
    "subject": "Mathematics (Multi-modal)",
    "question": "1. Let $ABC$ be an acute-angled triangle with $AB < AC < BC$, inscribed in the circle $c(O,R)$. The circle $c_1$ with center $A$ and radius $AC$ intersects the circle $c(O,R)$ at point $D$ and the extension of the side $CB$ at $E$. The line $AE$ intersects the circle $c(O,R)$ at point $F$ and $G$ is the symmetric point of $E$ with respect to $B$. Prove that the quadrilateral $FEDG$ is cyclic.\n\n2. We consider three lines of the plane passing through point $A$ and dividing the plane in 6 sectors. At the interior of each sector there exist 5 points. We suppose that no three of the 30 points existing in the sectors are collinear. Prove that there exist at least 1000 triangles with vertices from the points of the 6 sectors which contain point $A$ either on their interior or on their sides.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since the quadrilateral $AFBC$ is inscribed in the circle $(c)$, we have: $\\angle F_1 = \\angle ACB = \\angle C$. Since triangle $AEC$ is isosceles we have $\\angle E_1 = \\angle ACB = \\angle C$. Therefore $\\angle F_1 = \\angle E_1$, and hence the triangle $BEF$ is isosceles and hence\n$$\nBE = BF \\qquad (1).\n$$\n\n![](attached_image_1.png)\nFigure 2\n\nWe put $\\angle C_1 = x$. Then from the circle $(c_1)$ we get $E\\angle AD = 2x$, and hence\n$$\nE\\angle AB + B\\angle AD = 2x \\qquad (2)\n$$\nMoreover from the circle $(c)$ we have:\n$$\nB\\angle AD = \\angle C_1 = x \\qquad (3)\n$$\nFrom (2) and (3) we find $E\\angle AB = B\\angle AD = x$, which means that $AB$ is bisector of the isosceles triangle $EAD$. Hence it is perpendicular bisector of $ED$, and\n$$\nBE = BD. \\qquad (4)\n$$\nFrom (1) and (4), and from the equality $BE = BG$, we conclude that $BE = BF = BG = BD$, and hence the quadrilateral $FEDG$ is inscribed in a circle with center $B$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be an acute angled triangle inscribed in the circle $c(O, R)$ (with $AB < A\\Gamma < B\\Gamma$) and let $\\Delta, E, Z$ be the touching points of the incircle of the triangle with the sides $B\\Gamma$, $A\\Gamma$, $AB$, respectively. The circumcircle of the triangle $AEZ$ (say, $(c_1)$) intersects the circle $(c)$ at point $A'$. The circumcircle of the triangle $B\\Delta Z$ (say, $(c_2)$) intersects the circle $(c)$ at point $B'$. The circumcircle of the triangle $\\Gamma\\Delta E$ (say, $(c_3)$) intersects the circle $(c)$ at point $\\Gamma'$. Prove that:\n\n(α) The quadrilateral $\\Delta EA'B'$ is cyclic.\n\n(β) The lines $\\Delta A'$, $EB'$ and $Z\\Gamma'$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the inscribed in the circle $c_1$ quadrilateral $AA'IZ$ we have:\n$$\nA\\hat{A}'I = A\\hat{Z}I = 90^\\circ = T A\\hat{A}'I \\quad . (\\alpha)\n$$\nFrom the inscribed quadrilateral $\\Gamma\\Delta IE$ (since $\\Gamma I$ bisector), we have\n$$\n\\hat{\\Delta}_1 = \\frac{\\hat{\\Gamma}}{2} \\qquad (1)\n$$\nFrom the inscribed quadrilateral $B\\Delta IZ$ (since $BI$ bisector), we have:\n$$\n\\hat{\\Delta}_2 = \\frac{\\hat{B}}{2} \\qquad (2)\n$$\nFrom the inscribed quadrilateral $B\\Delta IB'$ we have: $\\hat{\\Delta}_3 = \\hat{B}_1$.\nFrom the inscribed quadrilateral $BB'A'A$ we have $\\hat{B}_1 = \\hat{A}'_1 = 90^\\circ - \\hat{A}'_2$.\nHence: $\\hat{\\Delta}_3 + \\hat{A}'_2 = 90^\\circ$.\nFrom the inscribed quadrilateral $AEIA'$ we have: $\\hat{A}'_3 = \\frac{\\hat{A}}{2}$\nHence:\n$$\n\\hat{\\Delta}_1 + \\hat{\\Delta}_2 + \\hat{\\Delta}_3 + \\hat{\\Delta}'_2 + \\hat{\\Delta}'_3 = 180^\\circ\n$$\nTherefore the quadrilateral $A'Ε\\Delta B'$ is cyclic. Similarly we prove that the quadrilaterals $\\Delta ZA'\\Gamma'$ and $ΖΕ\\Gamma'B'$ are cyclic.\n\nFinally, we conclude that the lines $\\Delta A'$, $EB'$ and $Z\\Gamma'$ are concurrent at the radical point of three circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC$, inscribed in a circle $c$ of center $O$. Let $G$ be its barycenter and $D, E, F$ the foot of the altitudes from $A, B, C$, respectively. If the rays $AG, GD$ intersect $c$ at $M, N$ respectively, prove that the points $F, E, M, N$ are cocyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the midpoint of $BC$ and $P$ the second intersection of $GD$ with $c$. Then $c$ and the Euler's circle are homothetic with center $G$ and ratio $-2$, so\n$$\n\\frac{GP}{GD} = \\frac{GA}{GK} = 2 \\Rightarrow GP = 2GD.\n$$\nFrom the power of a point theorem we have\n$$\nGM \\cdot GA = GN \\cdot GP \\Rightarrow GM \\cdot (2GK) = GN \\cdot (2GD) \\Rightarrow GM \\cdot GK = GN \\cdot GD, \\text{ so the quadrilateral } DKMN \\text{ is inscribed in a circle, let it be } c_1.\n$$\nMoreover $F, D, K, E$ belong to a circle (the Euler circle), let it be $c_2$.\nTherefore the lines $FE, DK, MN$ are concurrent to the radical center, let it be $T$, of the circles $c, c_1, c_2$. To this end, $TF \\cdot TE = TD \\cdot TK = TN \\cdot TM$, so the points $F, E, M, N$ are cocyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be an acute angled triangle with $AB < A\\Gamma < B\\Gamma$ and let $c$ be its circumcircle with centre $O$. At the small arcs $A\\Gamma$ and $AB$ we consider the points $\\Delta$ and $E$ respectively. Let $K$ be the intersection point of $B\\Delta$, $\\Gamma E$ and $N$ be the second common point of the circumcircles of the triangles $BKE$, let it be $c_1$, and $\\Gamma K\\Delta$, let it be $c_2$. Prove that the points $A, K, N$ are collinear if and only if the point $K$ lies on the A-symmedian of the triangle $AB\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the intersection point of $B\\Delta$, $\\Gamma E$ and $N$ be the second intersection point of $c_1$, $c_2$. Let $Z$ be the point of intersection of the tangents at $B, \\Gamma$ of the circle $c$. We will prove that the points $K, N, Z$ are collinear.\nLet $T$ be the intersection of $BZ$ with $c_1$ and $P$ the intersection of $Z\\Gamma$ with $c_2$, then:\n$$\n\\hat{E}_1 = \\hat{T}_1 : (\\text{inscribed in } c_1 \\text{ at the same arc } BK)\n$$\n$$\n\\hat{E}_1 = \\hat{B}_1 : (\\text{chord and tangent at } c)\n$$\n$$\n\\hat{\\Delta}_1 = \\hat{P}_1 : (\\text{inscribed in } c_2 \\text{ at the same arc } \\Gamma K)\n$$\n$$\n\\hat{\\Delta}_1 = \\hat{\\Gamma}_1 : (\\text{chord and tangent at } c)\n$$\n$$\n\\hat{B}_1 = \\hat{\\Gamma}_1 : (\\text{ZB and Z\\Gamma are tangents at } c)\n$$\nFrom the above inequalities we have:\n$$\n\\hat{B}_1 = \\hat{T}_1, \\text{ so } KT // B\\Gamma \\text{ and } \\hat{P}_1 = \\hat{\\Gamma}_1, \\text{ so } KP // B\\Gamma.\n$$\nTherefore $B\\Gamma PT$ is an isosceles trapezium and thus it is cyclic (let $c_3$ its circle).\nThis means that the common chord $KN$ of the circles $c_1$ and $c_2$ will pass through the radical center $Z$ of the circles $c_1, c_2, c_3$.\nSuppose first that $A, K, N$ are collinear. Then, since $K, N, T$ are collinear, we have that $A, K, N, T$ are collinear, so $A, K, N$ are on the symmedian $AT$.\nConversely, if $K$ is on the symmedian $AT$, then $A, K, T$ are collinear and since $K, N, T$ are collinear, we conclude that $A, K, N$ are collinear.\nLet $K$ be the intersection point of $B\\Delta$, $\\Gamma E$ and $N$ the second common point of $c_1, c_2$. Let $Z$ be the point of intersection of the tangents at $B, \\Gamma$ of the circle $c$. If $H$ is the intersection of $EB$ and $\\Delta\\Gamma$, we will prove that $K, N, Z, H$ are collinear.\nUsing Pascal's theorem at the degenerate hexagon $BB\\Gamma\\Gamma\\Delta E$, we have that $H, K, Z$ are collinear.\nThe point $H$ has power with respect to $c$, $HE \\cdot HB = H\\Delta \\cdot H\\Gamma$.\nTherefore $K, N, H$ are collinear. Therefore $K, N, Z, H$ are collinear.\n\nNow it is clear that $A, K, N$ are collinear if and only if $K$ is on the symmedian $AZ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a square $ABCD$ of side length $8$ cm which is divided, with lines parallel to its sides, into $64$ small squares of side $1$ cm. Color $7$ small squares black, and the other $57$ small squares are white. Suppose that there is a positive integer $k$ such that no matter which $7$ squares are black, there is a rectangle of area $k\\ \\text{cm}^2$ with sides parallel to the sides of $ABCD$ and all of its small squares that contains are white. Find the maximum value of $k$.",
    "options": [],
    "answer": "8",
    "solution": "We divide $ABCD$ into $8$ rectangles $4 \\times 2$. Since we color seven small squares with black color, from pigeonhole, there will be at least one $4 \\times 2$ rectangle that contains no black square and of course its area is $8\\,\\text{cm}^2$.\n\n![](attached_image_1.png)\nFigure 1\n\nIn what follows we will prove that there is a coloring with $7$ black squares, such that there is no rectangle with only white squares and area bigger than $8\\,\\text{cm}^2$. Indeed, we can see such a coloring at the following figure.\n\n![](attached_image_2.png)\nFigure 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be an acute angled triangle with $AB < A\\Gamma < B\\Gamma$. Its circumcircle is $c$ and let $\\Delta, E$ be the midpoints of $AB$ and $A\\Gamma$ respectively. We draw externally two semicircles with diameters $AB$ and $A\\Gamma$, which intersect $E\\Delta$ at $M$ and $N$ respectively. The lines $MB$ and $M\\Gamma$ intersect the circumcircle at $T, \\Sigma$ respectively. If the lines $MB$ and $M\\Gamma$ intersect at $H$, prove that:\n\na) the point $H$ is on the circumcircle of the triangle $AMN$\n\nb) the lines $AH$ and $T\\Sigma$ intersect perpendicularly at the point $Z$ and $Z$ is the center of the circumcircle of the triangle $AMN$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nThe angles $\\widehat{AMB}$ and $\\widehat{AN\\Gamma}$ are right since they see the diameters $AB$ and $A\\Gamma$. Therefore the quadrilateral $AMHN$ is cyclic, which is the desired result.\n\n![](attached_image_1.png)\n\nb.\n$\\hat{T}_1 = \\hat{I}_1 \\quad (1)$.\n\nThe line $E\\Delta$ connects the midpoints of $AB$ and $A\\Gamma$, so $MN$ is parallel to $B\\Gamma$.\nTherefore:\n\n$$\n\\hat{r}_1 = \\hat{N}_1 \\quad (2).\n$$\n\nFrom the quadrilateral $AMHN$ we have:\n$$\n\\hat{H}_1 = \\hat{N}_2 \\quad (3).\n$$\nFrom the relations (1), (2), (3) we have:\n$$\n\\hat{T}_1 + \\hat{H}_1 = \\hat{N}_1 + \\hat{N}_2 = A\\hat{N}H = 90^\\circ.\n$$\n\n![](attached_image_2.png)\n\nFrom the quadrilateral $AB\\Sigma\\Gamma$ we have:\n$$\nA\\hat{B}\\Gamma = A\\hat{\\Sigma}\\Gamma = \\hat{B} \\quad \\text{and from } E\\Delta \\parallel B\\Gamma \\text{ we have: } \\hat{E}_1 = \\hat{B}.\n$$\nFrom the equality $A\\hat{\\Sigma}\\Gamma = A\\hat{\\Sigma}N$ in combination with the previous ones we get that the quadrilateral $AE\\Sigma N$ is cyclic.\n\nThe quadrilateral $AZ\\Sigma N$ is also cyclic, since $A\\hat{N}\\Sigma = A\\hat{Z}\\Sigma = 90^\\circ$. To this end, we have that $\\hat{Z}_1 = \\hat{N}_2 = \\hat{H}_1$ and so $BH \\parallel EZ$. Since $E$ is the midpoint of $AB$, we conclude that $Z$ is the midpoint of $AH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the number $A = \\frac{4n!}{n!\\ 2n!}$, where $n$ is a positive integer, is an integer and has a factor of the form $2^{n+1}$. (Note: The number $n!$ for $n \\in \\mathbb{N}$, is defined by: $n! = 1 \\cdot 2 \\cdot \\dots \\cdot n$, and $0! = 1$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We can write: $A = \\frac{4n!}{n!\\ 2n!} = \\binom{3n}{n} \\cdot 3n+1\\ 3n+2\\ \\dots\\ 4n-1 \\cdot 4n \\in \\mathbb{Z}$, since $\\binom{3n}{n} \\in \\mathbb{Z}$.\n\nNext we observe that in the prime factorization of $n!$ the exponent of $2$ is\n$$\n\\exp n = \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{2^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{n}{2^m} \\right\\rfloor, \\qquad (2)\n$$\nwhere $m$ is the maximal natural number satisfying the inequality $2^m \\le n$. Hence\n$$\n\\exp n \\le \\frac{n}{2} + \\frac{n}{2^2} + \\dots + \\frac{n}{2^m} = \\frac{n}{2} \\left( 1 + \\frac{1}{2} + \\dots + \\frac{1}{2^{m-1}} \\right) = n - \\frac{n}{2^m}. \\qquad (3)\n$$\nSimilarly we find that:\n$$\n\\exp 2n = \\left\\lfloor \\frac{2n}{2} \\right\\rfloor + \\left\\lfloor \\frac{2n}{4} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{2n}{2^{m+1}} \\right\\rfloor = n + \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{2^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{n}{2^m} \\right\\rfloor = n + \\exp n \\qquad (4)\n$$\n$$\n\\begin{aligned}\n\\exp 4n &= \\left\\lfloor \\frac{4n}{2} \\right\\rfloor + \\left\\lfloor \\frac{4n}{2^2} \\right\\rfloor + \\left\\lfloor \\frac{4n}{2^3} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{4n}{2^{m+2}} \\right\\rfloor \\\\\n&= 2n + n + \\left\\lfloor \\frac{n}{2} \\right\\rfloor + \\left\\lfloor \\frac{n}{2^2} \\right\\rfloor + \\dots + \\left\\lfloor \\frac{n}{2^m} \\right\\rfloor = 3n + \\exp n\n\\end{aligned}\n$$\nTherefore the exponent of $2$ in the factorization of $A$ is\n$$\n\\begin{aligned}\n& 3n + \\exp n - \\left[ \\exp n + n + \\exp n \\right] = 2n - \\exp n \\\\\n& \\geq 2n - \\left( n - \\frac{n}{2^m} \\right) = n + \\frac{n}{2^m} \\geq n + 1,\n\\end{aligned}\n$$\nthat is $2^{n+1}$ is a factor of $A$.\n\n\n(Second solution)\n\nWe can write $A = \\frac{(2n+1)(2n+2)\\dots(4n-1)(4n)}{n!}$.\nWe observe that in the numerator we have $n-1$ even integers, from $2n+2$ till $4n-2$, that is we get $2$ as a factor $n-1$ times. Moreover from $4n$, we get the factor $2$ two times and hence\n$$\nA = 2^{n+1} \\frac{n(n+1)(n+2)\\dots(2n-1)}{n!} \\cdot (2n+1)(2n+3)\\dots(4n-1)\n$$\nSince $\\frac{n(n+1)(n+2)\\dots(2n-1)}{n!} = \\binom{2n-1}{n-1}$, we can write\n$$\nA = 2^{n+1} \\binom{2n-1}{n-1} \\cdot (2n+1)(2n+3)\\dots(4n-1),\n$$\nand hence $A$ is integer divided by $2^{n+1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute angled triangle inscribed in a circle $c(O, R)$ and $F$ a point on the side $AB$ such that $AF < \\frac{AB}{2}$. The circle $c_1(F, FA)$ intersects the line $OA$ at point $A'$ and the circle $(c)$ at $K$. Prove that the quadrilateral $BKFA'$ is inscribed in a circle passing through $O$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The triangle $AFK$ is isosceles and hence $\\hat{F}_1 = 2\\hat{A}_1$. The angle $\\hat{A}_1$ is inscribed in the circle $(c)$ and $\\hat{O}_1 = 2\\hat{A}_1 = \\hat{F}_1$, and hence the quadrilateral $BKFO$ is cyclic.\n\nNext we will prove that the quadrilateral $OBKA'$ is cyclic. In fact, if $S$ be the counter point of $A$ in the circle $(c_1)$. Then the triangle $AKS$ is right angled at $K$, and hence $\\hat{S}_1 = 90^\\circ - \\hat{A}_1$. The angles $\\hat{S}_1$ and $\\hat{A}'_1$ are equal (inscribed in the circle $(c_1)$ and they correspond to the same arch $KA$). Hence: $\\hat{A}'_1 = 90^\\circ - \\hat{A}_1$ (1).\n\nFrom the isosceles triangle $OKB$ we have: $\\hat{B}_1 = 90^\\circ - \\frac{\\hat{O}_1}{2} = 90^\\circ - \\hat{A}_1$ (2)\n\nFrom (1), (2) we have: $\\hat{A}'_1 = \\hat{B}_1$. Hence the quadrilateral $OBKA'$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer triads $a, b, c$, with $a > 0 > b > c$, and sum $a+b+c=0$ such that the number $N = 2017 - a^3b - b^3c - c^3a$ is a perfect square.",
    "options": [],
    "answer": "a=36, b=-12, c=-24",
    "solution": "2. First we observe that for every selection of six points, one per each sector, an hexagon (convex or non-convex) is created containing A.\nFrom these six points we can create totally $\\binom{6}{3} = 20$ triangles. We will count how many of them contain A. For any two points lying in two opposite sectors the third vertex of the triangle can be selected in two ways.\n![](attached_image_1.png)\nFigure 3\n![](attached_image_2.png)\nFigure 4\nFor example, for the points B, C of figure 4 we have the possibility of taking the points form the two colored sectors.\nThere exist three pairs of opposite sectors. We have 5x5 selections for the base BC, whereas the third vertex can be selected by $2 \\cdot 5 = 10$ ways. Therefore we have totally at least $3 \\cdot 2 \\cdot 5^3 = 6 \\cdot 5^3$ such triangles containing A.\n![](attached_image_3.png)\nFigure 5\nConsidering points in non-succcessive and non-opposite sectors (see figure 5) in this case we have the triangles like CBD or EFG.\nLike CBD there are $5 \\cdot 5 \\cdot 5 = 5^3$ triangles containing A and like EFG there are also $5 \\cdot 5 \\cdot 5 = 5^3$ triangles containing A. Totally in this case we have $2 \\cdot 5^3$ triangle containing A.\nSumming up all the above cases we count at least $6 \\cdot 5^3 + 2 \\cdot 5^3 = 8 \\cdot 5^3 = 1000$\n\ntriangles containing A either in their interior or in their sides.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be positive integers such that the number\n$$\n\\frac{(a+b)^2 + 4a}{ab}\n$$\nis an integer. Assume that $b$ is odd. Prove that $a$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\frac{(a+b)^2 + 4a}{ab} = \\kappa \\in \\mathbb{Z}$. The last one can be re-written as:\n$$\n(a+b)^2 + 4a = \\kappa ab \\quad (1)\n$$\nWe will prove that $a$ is odd. Indeed, if $2|a$, then $2|(a+b)^2$, and thus $b$ is even, a contradiction.\n\nWe consider (1) as a quadratic with respect to $b$:\n$$\nb^2 + b(2 - \\kappa)a + a^2 + 4a = 0\n$$\nThe discriminant should be a perfect square. We have\n$$\n\\Delta = a^2(\\kappa - 2)^2 - 4(a^2 + 4a) = a(a(\\kappa - 2)^2 - 4a - 16).\n$$\nIt follows that the product of the numbers $a$, $a(\\kappa - 2)^2 - 4a - 16$ is a perfect square, but since $a$ is odd, these two numbers are coprime and so each one of them is a perfect square, so $a$ is a perfect square.\n\nLet $d = (a,b)$ and we write $a = dx$, $b = dy$, with $(x,y)=1$. Observe that since $d|b$ and $b$ is odd, $d$ is odd. The number\n$$\n\\frac{(a+b)^2 + 4a}{ab} = \\frac{d^2(x+y)^2 + 4dx}{d^2xy} = \\frac{d(x+y)^2 + 4x}{dxy}\n$$\nis an integer. So $x|d(x+y)^2 + 4x \\Rightarrow x|dy^2$ and since $(x,y)=1$, we have $x|d$. Moreover, $d|d(x+y)^2 + 4x \\Rightarrow d|4x$ and since $d$ is odd, $d|x$. From these two, we have that $d=x$, so $a=d^2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Twelve friends play a tennis tournament, where each player plays exactly one game with each of the eleven other players. The winner wins one point and the loser gets zero points, while there is no tie. If the number of points of the players are $B_1, B_2, ..., B_{12}$, find the maximum value of $\\Sigma_3 = B_1^3 + B_2^3 + \\dots + B_{12}^3$.",
    "options": [],
    "answer": "4356",
    "solution": "The 12 friends will play $\\binom{12}{2} = \\frac{12 \\cdot 11}{2} = 66$ games, so the total number of points from all games is $66$. One possible outcome of the tournament is to put all players in an order and each of them to win all players that are after of him in the order. Then the first one has $11$ points, the second one $10$ points and so on. Then the last one has $0$ points, and all the players have a different number of points from $0$ to $11$. Then:\n$$\n\\Sigma_3 = 0^3 + 1^3 + \\dots + 11^3 = \\left(\\frac{11 \\cdot 12}{2}\\right)^2 = 66^2\n$$\nWe will prove that this is the maximum value for $\\Sigma_3$.\nIndeed, suppose that there is another outcome of the tournament which gives maximum $\\Sigma_3$ and in that tournament the players $A$ and $B$ have the same number of points, let it be $\\kappa$, where $A$ won $B$. We observe that $\\kappa \\neq 0$ since $A$ won against $B$. A new outcome will be produced if we suppose that $B$ won against $A$, so $A$ would have $\\kappa - 1$ points and $B$ will have $\\kappa + 1$ points. If all the other games had the same outcome, then the difference between the new $\\Sigma_3$ and the old one will be\n$$\n(\\kappa + 1)^3 + (\\kappa - 1)^3 - 2\\kappa^3 = 6\\kappa > 0,\n$$\nso the new $\\Sigma_3$ is greater than the old one which we supposed that has the maximum $\\Sigma_3$, a contradiction.\nIt follows that the maximum $\\Sigma_3$ equals to $66^2$ and is reached with the above example.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $a$, $b$, $p$, where $p$ is prime, satisfying the equation: $$\\frac{1}{p} = \\frac{1}{a^2} + \\frac{1}{b^2}.$$",
    "options": [],
    "answer": "(a, b, p) = (2, 2, 2)",
    "solution": "The given equation can be written as: $p a^2 + b^2 = a^2 b^2$ (1)\nSince $p$ is prime, from (1) we have: $p|a$ or $p|b$. We suppose that $p|a$, and hence $a = p a_1$, $a_1 \\in \\mathbb{N}^*$. Moreover, we have that\n$$\n\\begin{aligned}\n\\frac{1}{p} > \\frac{1}{b^2} &\\Rightarrow b^2 > p \\Rightarrow b^2 \\ge p+1 \\Rightarrow \\frac{1}{a^2} = \\frac{1}{p} - \\frac{1}{b^2} \\ge \\frac{1}{p} - \\frac{1}{p+1} = \\frac{1}{p(p+1)} \\\\\n&\\Rightarrow \\frac{1}{p^2 a_1^2} \\ge \\frac{1}{p(p+1)} = \\frac{1}{p^2+p} \\ge \\frac{1}{2p^2} \\Rightarrow \\frac{1}{a_1^2} \\ge \\frac{1}{2} \\Rightarrow a_1^2 \\le 2 \\Rightarrow a_1 = 1 \\Rightarrow a = p.\n\\end{aligned}\n$$\nThen equation (1) becomes:\n$$\n\\begin{aligned}\np p^2 + b^2 &= p^2 b^2 \\Leftrightarrow p^2 + b^2 = p b^2 \\Leftrightarrow p^2 = (p-1) b^2 \\Leftrightarrow p^2 - 1 = (p-1) b^2 - 1 \\\\\n&\\Leftrightarrow (p-1)(p+1) = (p-1) b^2 - 1 \\Rightarrow p-1 \\mid_{p-1>0} \\Leftrightarrow p-1=1 \\Leftrightarrow p=2.\n\\end{aligned}\n$$\nHence $a = p = 2$ and from equation (1) we find $b = 2$. Similarly we work if $p|b$.\n\n\nSecond solution:\nThe given equation can be written as: $p a^2 + b^2 = a^2 b^2$ (1)\nSolving with respect to $a^2$ we find\n$$\na^2 = \\frac{p b^2}{b^2 - p} = \\frac{p(b^2 - p) + p^2}{b^2 - p} = p + \\frac{p^2}{b^2 - p} \\quad (2)\n$$\nSince $a^2$ is integer, $b^2 - p \\mid p^2$, and hence\n$$\nb^2 - p = 1 \\quad \\text{or} \\quad b^2 - p = p \\quad \\text{or} \\quad b^2 - p = p^2\n$$\n* $b^2 - p = 1 \\Leftrightarrow p = b^2 - 1 = (b-1)(b+1)$ and since $p$ prime, $b-1=1$, hence $b=2$ and $p=3$. Then $a^2 = 12$, absurd.\n* $b^2 - p = p$ gives $b^2 = 2p$. Then $b$ is even, say $b = 2b_1$, and hence $4b_1^2 = 2p$. Hence $2 \\mid p$, and therefore $p = 2$ and $b = 2$, $a = 2$.\n* In the third case (2) gives $a^2 = p+1 \\Leftrightarrow p = a^2 - 1 = (a-1)(a+1)$, and since $p$ prime, $a-1=1$, that is $a = 2$ and $p = 3$, which gives $b^2 = 3$, absurd.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that the number $A = 7^{2n} - 48n - 1$ is a multiple of $9$, for all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We distinguish three cases mod $3$.\n\n* If $n = 3k$, then $A = 7^{6k} - 48(3k) - 1 = 49^{3k} - 9 \\cdot 16k - 1$. Since $49 \\equiv 4 \\pmod{9}$, it follows that $49^3 \\equiv 4^3 \\pmod{9} \\equiv 1 \\pmod{9}$ and hence $9 \\mid 49^{3k} - 1 \\Rightarrow 9 \\mid A$.\n\n* If $n = 3k+1$, then\n$A = 7^{6k+2} - 48(3k+1) - 1 = 7^2 \\cdot 7^{6k} - 9 \\cdot 16k - 49 = 49(7^{6k} - 1) - 9 \\cdot 16k$.\nSince $9 \\mid 7^{6k} - 1$, it follows again that $9 \\mid A$.\n\n* If $n = 3k + 2$, then $A = 7^{6k+4} - 48(3k + 2) - 1 = 7^4 \\cdot 7^{6k} - 9 \\cdot 16k - 97$. We have that $7^{6k} \\equiv 1 \\pmod{9} \\Rightarrow 7^4 \\cdot 7^{6k} \\equiv 7^4 \\pmod{9} \\equiv 49^2 \\pmod{9} \\equiv 4^2 \\pmod{9} \\equiv 7 \\pmod{9}$ and also $97 \\equiv 7 \\pmod{9}$.\nHence $9 \\mid 7^4 \\cdot 7^{6k} - 97$, that is $9 \\mid A$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\ge 2$ be a prime number. Angelo and Vangelis play in turns the following game: At the board there are $p$ empty boxes, the one next to other, and in each move, the current player puts a digit in one of the boxes. Angelo plays first and the game ends after all boxes are filled and we get a $p$-digit number $M$, in which we allow zeros in the beginning. The goal of Angelo is to make the number $M$ divisible by $p$, and the goal of Vangelis is to prevent this. Prove that Angelo has a winning strategy.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\nM = a_0 + a_1 p + \\dots + a_{p-1} 10^{p-1},\n$$\nwhere $a_i$ is the value that the $i$-th box will take (from right to left).\nIf $p = 2$ or $p = 5$, then Angelo puts $0$ in the last box and wins.\nIf $p \\neq 2, 5$, then by Fermat's Theorem we have\n$$\np|10^{p-1}-1 \\Leftrightarrow p|(10^{\\frac{p-1}{2}})^2-1 \\Leftrightarrow p|(10^{\\frac{p-1}{2}}-1)(10^{\\frac{p-1}{2}}+1)\n$$\nso $p|(10^{\\frac{p-1}{2}}-1)$ or $p|(10^{\\frac{p-1}{2}}+1)$. We have two cases:\n\n(α) If $p|(10^{\\frac{p-1}{2}}+1)$. Then at the first move Angelo plays zero at $a_{p-1}$. Then, each time that Vangelis puts $a_i$ at the $i$-th position, Angelo plays symmetric (i.e. at $j=i+\\frac{p-1}{2}$ if $0 \\le i \\le \\frac{p-3}{2}$, or at $j=i-\\frac{p-1}{2}$ if $\\frac{p-1}{2} \\le i \\le p-2$) putting the digit $a_i$ at the $j$-th position.\nThen,\n$$\n10^j \\equiv -10^i \\pmod{p} \\Rightarrow a_j 10^j = a_i 10^j \\equiv -a_i 10^i \\pmod{p}\n$$\nso $p|a_j 10^j + a_i 10^i$. We observe that after Angelo's first move, it remains to fill $p-1$ positions, which is an even number. Matching as above, Angelo can make the number divisible by $p$.\n\n$$\n10^j = 10^i (\\text{mod } p) \\Rightarrow a_j 10^j + a_i 10^i = (a_i + a_j) 10^i = 9 \\cdot 10^i (\\text{mod } p)\n$$\nIn the above matching we get sums of the form $9 \\cdot 10^i (\\text{mod } p)$, then the total sum equals\n$$\nM = \\sum_{i=0}^{\\frac{p-3}{2}} 9 \\cdot 10^i = 10^{(p-1)/2} - 1 \\equiv 0 (\\text{mod } p),\n$$\nand thus Angelo wins.\n\n(β) If $p|(10^{\\frac{p-1}{2}}-1)$. Then, at the first move Angelo plays $0$ at $a_{p-1}$. Then, each time that Vangelis puts $a_i$ at the $i$-th position, Angelo plays symmetric with respect to the center of the number (i.e. $j=i+\\frac{p-1}{2}$ if $0 \\le i \\le \\frac{p-3}{2}$, or at $j=i-\\frac{p-1}{2}$ if $\\frac{p-1}{2} \\le i \\le p-2$) putting the digit $9-a_i$ at $j$-th position.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle AB\\Gamma$ with $AB < A\\Gamma < B\\Gamma$ be a triangle inscribed in a circle $c$ with center $O$. Let $\\Delta$ be the diametrically opposite to $A$ and $c_1$ is a circle which has center $K$ on $B\\Delta$ and passes through $B$ and $\\Gamma$. If $c_1$ meets $A\\Gamma$ at the point $E$, prove that the circumcircle of $BKE$, let it be $c_2$, is tangent to $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Z$ be the intersection of $c_2$ with $AB$ and $M$ the intersection of $KZ$ with $BE$. The angle $A\\hat{B}\\Delta$ (and so the angle $Z\\hat{B}K$) is right since it sees the diameter $A\\Delta$ of the circle $c(O, R)$.\nSince $Z\\hat{B}K = 90^\\circ$, $ZK$ is a diameter of $c_1$ and thus $ZK$ is the midperpendicular of the common chord $BE$ of the circles $c_1$ and $c_2$. The line $AB$ is tangent to the circle $c_1$, so $\\hat{B}_1 = \\hat{\\Gamma}$ and from the right-angled triangle $BMZ$ we have:\n$$\n\\hat{Z}_1 = 90^\\circ - \\hat{B}_1 = 90^\\circ - \\hat{\\Gamma} \\quad (1)\n$$\n![](attached_image_1.png)\nFigure 3\nThe angles $\\hat{\\Delta}, \\hat{\\Gamma}$ see the same arc of $c$, so $\\hat{\\Delta} = \\hat{\\Gamma}$. From the right-angled triangle $AB\\Delta$, we have:\n$$\n\\hat{A}_1 = 90^\\circ - \\hat{\\Delta} = 90^\\circ - \\hat{\\Gamma} \\quad (2)\n$$\n\nFrom (1) and (2) we obtain that $\\widehat{A_1} = \\widehat{Z_1}$, so $\\text{\\textit{οπότε}} \\Delta A\\Delta$ is parallel to $KZ$ and as a result $A\\Delta KZ$ is a trapezium. Moreover, since $O$ is the midpoint of $A\\Delta$, we conclude that $OB$ passes through the midpoint of $KZ$ which is the center of $c_2$. Therefore the centers of $c, c_2$ and the point $B$ are collinear.\n\nWe will prove that the circle $c_2$ is tangent to $c(O, R)$ at the point $B$. To this end, we will prove that these circles have common tangent at the point $B$. Let $BP$ be the tangent of $c(O, R)$ at the point $B$ and let $\\widehat{KBP} = \\omega$. In order to prove that $BP$ is tangent to $c_2$ at $B$, it suffices to prove that\n$$\n\\widehat{BEK} = \\omega. \\qquad (1)\n$$\nSince $BP$ is tangent to $c(O, R)$, we have $\\widehat{\\Delta\\Gamma B} = \\omega$. Moreover, $A\\Delta$ is a diameter, so $\\widehat{A\\Gamma\\Delta} = 90^\\circ$, and as a result $\\widehat{E\\Gamma B} = 90^\\circ - \\omega$. However,\n$$\n\\widehat{BKE} = 2\\widehat{B\\Gamma E} = 2(90^\\circ - \\omega) = 180^\\circ - 2\\omega, \\qquad (2)\n$$\nand from the isosceles triangle $BKE$ we have $\\widehat{KBE} = \\widehat{KEB} = \\omega$, so (1) holds, which is the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $x$, $y$, $z$ with $z$ odd, which satisfy the equation\n$$\n2018^x = 100^y + 1918^z\n$$",
    "options": [],
    "answer": "(1, 1, 1)",
    "solution": "We write $2018 = 2 \\cdot 1009$, $100 = 2^2 \\cdot 5^2$ and $1918 = 2 \\cdot 7 \\cdot 137$. Then, the equation has the form\n$$\n2^x \\cdot 1009^x = 2^{2y} \\cdot 5^{2y} + 2^z \\cdot 7^z \\cdot 137^z, \\quad (1)\n$$\n\nConsider the following cases:\n\n* If $2y < z$, then the power of $2$ that divides the right hand side is $2^{2y}$. (2)\nFrom (1) and (2) we get that $x = 2y$. The equation takes the form:\n$$\n1918^z = 2018^{2y} - 100^{y} = (2018^y - 10^y)(2018^y + 10^y).\n$$\nHowever, $2010 \\mid 2018^y - 10^y$, so $2010 \\mid 1918^z$, a contradiction.\n\n* If $z < 2y$, then the highest power of $2$ that divides the right hand side is\n$$\n2^z. \\quad (3) \\quad \\text{From (1) and (3) we get } x = z.\n$$\nThen the equation has the form:\n$$\n2018^z - 1918^z = 100^y \\Leftrightarrow 2^z (1009^z - 7^z \\cdot 137^z) = 2^{2y} \\cdot 5^{2y}. \\quad (4)\n$$\nSince $z$ is odd, the number $1009^z - 7^z \\cdot 137^z$ is even but it is not divisible by $4$, so the highest power of $2$ that divides the left hand side is $z+1$, so $2y = z+1$, and the equation takes the form $2018^z - 1918^z = 10^{z+1}$. Therefore, $10^{z+1} = 2018^z - 1918^z \\ge 100^z = 10^{2z}$, so $2z \\le z+1 \\Leftrightarrow z \\le 1 \\Leftrightarrow z = 1$.\nIt follows that the only solution is $(x, y, z) = (1, 1, 1)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\xi$ be the positive root of the equation $x^2 + x - 4 = 0$. The polynomial $P(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0$, where $n$ is a positive integer, has nonnegative integer coefficients and $P(\\xi) = 2017$.\n(i) Prove that: $a_0 + a_1 + \\dots + a_n \\equiv 1 \\pmod{2}$\n(ii) Find the least possible value of the sum: $a_0 + a_1 + \\dots + a_n$.",
    "options": [],
    "answer": "The sum of coefficients is odd; the least possible sum is 23.",
    "solution": "**(i)** Since $\\xi = \\frac{-1 + \\sqrt{17}}{2}$ is irrational and the polynomial $F x = P x - 2017$ has rational coefficients and $\\xi$ as a root, then it will have also the conjugate $\\frac{-1 - \\sqrt{17}}{2}$ as a root, and therefore it is divided by the polynomial $\\varphi x = x^2 + x - 4$. It comes easily from the identity\n$$\nF x = P x - 2017 = x^2 + x - 4 Q x + \\kappa x + \\lambda,\n$$\nby putting $x = \\xi$. Then $\\kappa \\xi + \\lambda = 0$ which gives $\\kappa = \\lambda = 0$, taking in mind that $\\xi$ is irrational. Therefore there exists a polynomial $Q x$ such that:\n$$\n\\begin{aligned}\nF x = P x - 2017 &= x^2 + x - 4 \\mathcal{Q} x \\\\\n\\Leftrightarrow a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0 - 2017 &= x^2 + x - 4 \\mathcal{Q} x\n\\end{aligned} \n\\quad (1)\n$$\nFrom (1) for $x=1$ we get:\n$$\n\\begin{aligned}\n& a_0 + a_1 + \\dots + a_n - 2017 = -2Q \\quad 1 \\\\\n\\Rightarrow \\quad a_0 + a_1 + \\dots + a_n = 2017 - 2Q \\quad 1 \\equiv 1 \\quad \\text{mod } 2\n\\end{aligned}\n$$\n\n(ii) We consider the set $a_0, a_1, ..., a_n$ with elements nonnegative integers satisfying the following:\n(α) $a_n \\xi^n + a_{n-1} \\xi^{n-1} + ... + a_1 \\xi + a_0 = 2017$ and\n(β) the sum $a_0 + a_1 + ... + a_n$ is minimal.\nFirst we observe that: $0 \\le a_i \\le 3$, for all $i = 1, 2, ..., n-2$.\nIn fact, if it was not true for someone $i = 1, 2, ..., n-2$, then the elements of the set $a_0, ..., a_{i-1}, a_{i-4}, a_{i-1}+1, a_{i+1}+1, a_{i+2}+1, a_{i+3}, ..., a_n$ would be nonnegative integers, it would satisfy relation (α), while the sum of its elements would be less than of $a_1 + a_2 + ... + a_n$, which is absurd.\nLet now $Q x = b_{n-2} x^{n-2} + b_{n-1} x^{n-1} + ... + b_1 x + b_0$. Then from the identity\n$a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 - 2017 = x^2 + x - 4 \\quad b_{n-2} x^{n-2} + b_{n-3} x^{n-3} + ... + b_1 x + b_0$ we get the equations:\n$$\n\\left\\{ \\begin{array}{l} a_0 - 2017 = -4b_0 \\\\ a_1 = -4b_1 + b_0 \\\\ a_2 = -4b_2 + b_1 + b_0 \\\\ a_3 = -4b_3 + b_2 + b_1 \\\\ \\multicolumn{5}{c}{\\text{........................................}} \\\\ a_{n-2} = -4b_{n-2} + b_{n-3} + b_{n-4} \\\\ a_{n-1} = b_{n-2} + b_{n-3} \\\\ a_n = b_{n-2} \\end{array} \\right\\} \\Leftrightarrow \\left\\{ \\begin{array}{l} a_0 - 2017 = -4b_0 \\\\ a_1 - b_0 = -4b_1 \\\\ a_2 - b_1 - b_0 = -4b_2 \\\\ a_3 - b_2 - b_1 = -4b_3 \\\\ \\multicolumn{5}{c}{\\text{........................................}} \\\\ a_{n-2} - b_{n-3} - b_{n-4} = -4b_{n-2} \\\\ a_{n-1} - b_{n-2} = b_{n-3} \\\\ a_n = b_{n-2} \\end{array} \\right\\}\n$$\nIn general we have: $a_{i+2} - b_{i+1} - b_i = -4b_{i+2}$, for all $i = 0, 1, ..., n-4$\nSince $0 \\le a_i \\le 3$, for all $i = 1, 2, ..., n-2$, from the first equation we have $a_0 = 1$ and $b_0 = 504$. From the second equation we get $a_1 = 0$ and $b_1 = 126$. From the third equation we get $a_2 = 2$ and $b_2 = 157$. Continuing in the same way we find the sets\n$b_0, b_1, b_2, ... b_{14} = 504, 126, 157, 70, 56, 31, 21, 13, 8, 5, 3, 2, 1, 0, 0$\n$a_0, a_1, a_2, ... a_{14} = 1, 0, 2, 3, 3, 2, 3, 0, 2, 1, 1, 0, 1, 3, 1$\nTherefore the least possible value of the sum $a_0 + a_1 + ... + a_n$ is 23.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31269,
    "subject": "Mathematics (Multi-modal)",
    "question": "A company consisting of $n$ friends play a table game according to the following rules:\n(a) At each round play exactly 3 players.\n(b) The game stops after $n$ rounds.\n(c) Each couple of players have played together at least at one round.\nDetermine the maximal possible value of $n$.",
    "options": [],
    "answer": "7",
    "solution": "Since in each round play exactly 3 players, the number of couples playing at each round is $\\binom{3}{2} = 3$. Therefore, at the end of the game after $n$ rounds, the total number of couples played the game will be $3n$. According to the last rule:\n$$\n\\binom{n}{2} \\le 3n \\Leftrightarrow \\frac{n(n-1)}{2} \\le 3n \\Leftrightarrow \\frac{n-1}{2} \\le 3 \\Leftrightarrow n \\le 7.\n$$\nNext we will prove that the value $n=7$ is possible. In fact, for $n=7$ we have $\\binom{7}{2} = \\frac{7!}{2!5!} = \\frac{6 \\cdot 7}{2} = 21 = 3 \\cdot 7$. If the friends are: $A$, $B$, $\\Gamma$, $\\Delta$, $E$, $Z$, $H$, then we can define the following triads $(A, B, \\Gamma)$, $(A, \\Delta, E)$, $(A, Z, H)$, $(B, \\Delta, H)$, $(B, E, Z)$, $(\\Gamma, \\Delta, Z)$, $(\\Gamma, E, H)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31270,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the blackboard are written some positive integers. We define the following movements:\n\n(α) Every two successive numbers, say $n, n+1$, can be deleted by writing the number $n-2$.\n\n(β) Every two numbers having difference 4, say $k, k+4$, can be deleted by writing the number $k-1$.\n\nIn the case we cannot apply any of the above movements the process finishes.\n\nDetermine the maximal possible value of the integer $c$ with the following property: Irrespectively with the numbers are written at the beginning in the table, at all process, all the numbers written in the table are greater than or equal to $c$.",
    "options": [],
    "answer": "-3",
    "solution": "**The answer is that that maximal possible value of $c$ is $-3$.**\n\nIn fact, this value is obtained if we start with the numbers $1,2,3,4,5$, and following the process:\n$$\n1, 2, 3, 4, 5 \\xrightarrow{\\alpha} 0, 1, 4, 5 \\xrightarrow{\\alpha} 0, 1, 2 \\xrightarrow{\\alpha} -1, 0 \\xrightarrow{\\beta} -3. \\quad (*)\n$$\nWe will prove now that no matter which are initially the numbers, we cannot have a number smaller than $-3$ during the process.\n\nTo this end, we will find an invariant under the movements. Let $w$ be the real root of the polynomial\n$$P x = x^3 + x^2 - 1.$$ Then,\n$$\nw^3 + w^2 = 1 \\Rightarrow w^{n+1} + w^n = w^{n-2}. \\quad (1)\n$$\nMoreover, the polynomial $P x$ divides $Q x = x^5 + x - 1$, so\n$$\nQ w = 0 \\Leftrightarrow w^5 + w = 1 \\Leftrightarrow w^{k+4} + w^k = w^{k-1} \\quad (2).\n$$\nFrom (1) and (2) we observe that if we consider one extra board which will have the numbers of our board raised to the $w$ power, then the sum of the numbers in the new board is invariant under $\\alpha$) and $\\beta$). For example, for the sequence in (*) the extra board will have the numbers,\n$$\nw^1, w^2, w^3, w^4, w^5 \\xrightarrow{\\alpha} w^0, w^1, w^4, w^5 \\xrightarrow{\\alpha} w^0, w^1, w^2 \\xrightarrow{\\alpha} w^{-1}, w^0 \\xrightarrow{\\beta} w^{-3}.\n$$\nWe observe, for example, that after the first movement the sum remains the same since $w^3 + w^2 = w^0$.\n\nIt is then enough to start with the minimal sum in the extra board, which is of course bigger than\n$$\nw^1 + w^2 + w^3 + w^4 + \\dots = \\frac{w}{1-w} = \\frac{w}{w^5} = w^{-4}.\n$$\nSo, in the initial board all the numbers are bigger than $-4$, so $c \\ge -3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ an equilateral triangle of side $\\alpha$. Let $\\Delta$, $E$ and $Z$ the midpoints of the sides $AB$, $B\\Gamma$ and $\\Gamma A$, respectively. Let $H$ the symmetric point of $\\Delta$ with respect to the line $B\\Gamma$. We color the points $A$, $B$, $\\Gamma$, $\\Delta$, $E$, $Z$, $H$ using one of the two colors $\\kappa$ = red and $\\mu$ = blue.\n\na. Find how many equilateral triangles are defined with vertices from the seven points $A$, $B$, $\\Gamma$, $\\Delta$, $E$, $Z$, $H$.\n\nb. Prove that, if the points $B$ and $E$ will be colored with the same color, then for every coloring of the remaining points there exists an equilateral triangle with vertices from the points $A$, $B$, $\\Gamma$, $\\Delta$, $E$, $Z$, $H$ whose all vertices have the same color.\n\nc. Can we have the same conclusion, if the points $B$ and $E$ will be colored by different colors?",
    "options": [],
    "answer": "a) 7; b) Yes; c) No",
    "solution": "a.\nThere are defined totally seven equilateral triangles from the given points. Since $\\Delta E = EZ = Z\\Delta = \\frac{\\alpha}{2}$ the seven equilateral triangles are\n![](attached_image_1.png)\nFigure 7\n![](attached_image_2.png)\nFigure 8\n$AB\\Gamma$, $A\\Delta Z$, $B\\Delta E$, $E\\Gamma Z$, $\\Delta EZ$, $BHE$ (symmetric of $\\Delta BE$ with respect to the line $B\\Gamma$), $\\Delta H\\Gamma$ (it has $\\Gamma H = \\Gamma \\Delta = \\frac{\\alpha\\sqrt{3}}{2}$ = altitude of equilateral triangle, $\\Gamma \\hat{\\Delta} H = 60^\\circ$).\n\nb.\nLet $B$ and $E$ are colored red. If the points $\\Delta$ or $H$ are also red, then we have the wanted triangle. Suppose that $\\Delta$ and $H$ are colored blue. If the point $\\Gamma$ is blue then we are done. Let the point $\\Gamma$ is colored red. If the point $Z$ is colored red, then the triangle $E\\Gamma Z$ has its vertices red. Let point $Z$ is colored blue. In that case for any coloring of $A$ one of the triangles $AB\\Gamma$, $A\\Delta Z$ will have its vertices of the same color.\n![](attached_image_3.png)\nFigure 9\n\nc.\nIn that case we have not the same conclusion. In figure 9 we give a coloring with all equilateral triangles having their vertices with different colors. The points $B$, $\\Gamma$, $Z$ have been colored red and the remaining points blue.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest integer $n$, for which there exists integers $x_1, x_2, \\dots, x_n$ and positive integers $a_1, a_2, \\dots, a_n$ so that:\n$$\nx_1 + x_2 + \\dots + x_n = 0, \\quad a_1x_1 + a_2x_2 + \\dots + a_nx_n > 0, \\quad a_1^2x_1 + a_2^2x_2 + \\dots + a_n^2x_n < 0.\n$$",
    "options": [],
    "answer": "4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an equilateral triangle, and let $P$ be some point in its circumcircle. Determine, with reasons, all the positive integers $n$ such that the sum\n$$\nS_n P = |PA|^n + |PB|^n + |PC|^n\n$$\nis independent of the choice of the point $P$.",
    "options": [],
    "answer": "2 and 4",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31274,
    "subject": "Mathematics (Multi-modal)",
    "question": "A set $S$ of integers is Balearic, if there are two (not necessarily distinct) elements $s, s' \\in S$ whose sum $s+s'$ is a power of two; otherwise it is called a non-Balearic set. Find an integer $n$ such that $\\{1, 2, \\dots, n\\}$ contains a 99-element non-Balearic set, whereas all the 100-element subsets are Balearic.",
    "options": [],
    "answer": "203",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31275,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $ABC$, in which $\\angle A = 60^\\circ$, $D \\in (BC)$ is such that $AD$ is the internal bisector of angle $\\angle A$. Let it be $r_B, r_C$ and $r$, respectively, the inradius of the triangles $ABD$, $ADC$ and $ABC$. Show that $\\frac{1}{r_B} + \\frac{1}{r_C} = 2\\left(\\frac{1}{r} + \\frac{1}{b} + \\frac{1}{c}\\right)$, where $b$ and $c$ are the lengths of the sides $AC$ and $AB$ of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is well known that $AD = \\frac{2bc}{b+c} \\cos \\frac{A}{2} = \\frac{bc\\sqrt{3}}{b+c}$. Let $h = AM$ be the length of the altitude from $A$ in the triangle $ABC$. From the theorem of the internal bisector we get\n$$\nBD = \\frac{ac}{b+c}, \\quad CD = \\frac{ab}{b+c}.\n$$\nLet $p_{ABD} = \\frac{BD + DA + AB}{2}$ be the half perimeter of triangle $ABD$. If we denote $[ABD]$ the surface of the triangle $ABD$, then\n$$\nr_B = \\frac{[ABD]}{p_{ABD}} = \\frac{h \\cdot BD}{2 \\cdot p_{ABD}} = \\frac{h \\cdot BD}{BD + AD + AB} = \\frac{\\frac{h \\cdot ac}{b+c}}{\\frac{ac}{b+c} + \\frac{bc\\sqrt{3}}{b+c} + c} = \\frac{ah}{2p + \\sqrt{3} \\cdot b}.\n$$\nAnalogously we get $r_C = \\frac{ah}{2p + \\sqrt{3} \\cdot c}$, where $p = \\frac{a+b+c}{2}$. From this we have\n$$\n\\frac{1}{r_B} + \\frac{1}{r_C} = \\frac{2p + b\\sqrt{3}}{ah} + \\frac{2p + c\\sqrt{3}}{ah} = \\frac{4p + (b+c)\\sqrt{3}}{ah}.\n$$\nBut $r = \\frac{[ABC]}{p} = \\frac{ah}{2p}$, so $ah = 2pr$. Thus,\n$$\n\\frac{1}{r_B} + \\frac{1}{r_C} = \\frac{4p + (b+c)\\sqrt{3}}{2pr}.\n$$\nNow, $b = AC$, $c = AB$, and $\\sin 60^\\circ = \\frac{\\sqrt{3}}{2}$, so $[ABC] = \\frac{1}{2}bc\\sin 60^\\circ = \\frac{bc\\sqrt{3}}{4}$, and $r = \\frac{[ABC]}{p} = \\frac{bc\\sqrt{3}}{4p}$, so $\\frac{1}{r} = \\frac{4p}{bc\\sqrt{3}}$.\n\nAlso,\n$$\n\\frac{1}{b} + \\frac{1}{c} = \\frac{b + c}{bc}.\n$$\nTherefore,\n$$\n2\\left(\\frac{1}{r} + \\frac{1}{b} + \\frac{1}{c}\\right) = 2\\left(\\frac{4p}{bc\\sqrt{3}} + \\frac{b + c}{bc}\\right) = \\frac{8p}{bc\\sqrt{3}} + \\frac{2(b + c)}{bc}.\n$$\nBut $ah = 2pr$, so $ah = 2p \\cdot \\frac{bc\\sqrt{3}}{4p} = \\frac{bc\\sqrt{3}}{2}$.\n\nNow, substitute $ah$ into the earlier formula:\n$$\n\\frac{1}{r_B} + \\frac{1}{r_C} = \\frac{4p + (b+c)\\sqrt{3}}{ah} = \\frac{4p + (b+c)\\sqrt{3}}{\\frac{bc\\sqrt{3}}{2}} = \\frac{2[4p + (b+c)\\sqrt{3}]}{bc\\sqrt{3}} = \\frac{8p}{bc\\sqrt{3}} + \\frac{2(b+c)}{bc}.\n$$\nThis matches the previous expression, so\n$$\n\\frac{1}{r_B} + \\frac{1}{r_C} = 2\\left(\\frac{1}{r} + \\frac{1}{b} + \\frac{1}{c}\\right).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}_+ = (0, \\infty)$. Determine all functions $f : \\mathbb{R}_+ \\to \\mathbb{R}_+$ such that\n$$\nf(xf(y)) + yf(z) + zf(x) = xy + yz + zx,\n$$\nfor all $x, y, z \\in \\mathbb{R}_+$.",
    "options": [],
    "answer": "f(x) = x for all x > 0",
    "solution": "For $x = y = z = 1$ we have $f(f(1)) = 1$, and hence for $x = y = z = f(1)$ we have\n$$\n3f(f(1)) = 3f(1)^2 \\Rightarrow f(1)^2 = 1 \\Rightarrow f(1) = 1.\n$$\nFor $y = z = 1$, we have for every $x \\in \\mathbb{R}_+$:\n$$\n\\begin{aligned}\nf(f(x)) + f(x) &= 2x \\\\\nf(f(x)) &= 2x - f(x)\n\\end{aligned} \\qquad (1)\n$$\nFor $z = 1$ from the given relation we get:\n$$\nf(xf(y)) + f(y) + f(f(x)) = xy + y + x\n$$\nAnd from (1) we find:\n$$\nf(xf(y)) = xy + y - x + f(x) - f(y), \\qquad (2)\n$$\nfor all $x, y \\in \\mathbb{R}_+$. Putting to (2) the $f(x)$ in the place of $x$ we have, from (1), for all $x, y \\in \\mathbb{R}_+$:\n$$\nf(f(x)f(y)) = f(x)y + y + 2x - 2f(x) - f(y) \\qquad (3)\n$$\nand interchanging $x$ and $y$ we get the relation\n$$\nf(f(y)f(x)) = f(y)x + x + 2y - 2f(y) - f(x) \\qquad (4)\n$$\nFrom (3) and (4) we have, for all $x, y \\in \\mathbb{R}_+$:\n$$\n(f(x) - 1)(y - 1) = (f(y) - 1)(x - 1), \\qquad (5)\n$$\nFrom which for $y = 2$ we find\n$$\nf(x) = cx - c + 1 = c(x - 1) + 1 \\qquad (6)\n$$\nwhere $c = f(2) - 1$. By substitution to the given equation we have\n$$\nc^2(xy + yz + zx - x - y - z) + c(x + y + z) - 3c + 3 = xy + yz + zx,\n$$\nfrom which for $x = y = z$ we find\n$$\n3(c^2 - 1)x^2 + 3c(1-c)x + 3(1-c) = 0 \\\\\n\\Leftrightarrow c^2 - 1 = 0,\\ c(1-c) = 0,\\ 1-c = 0 \\Leftrightarrow c = 1.\n$$\n$c = 1$, and hence: $f(x) = x$, for every $x > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle AB\\Gamma$ be a triangle with $AB > A\\Gamma$. Let point $\\Delta$ be on the side $AB$ such that $B\\Delta = A\\Gamma$. We draw the circle $\\gamma$ passing through $\\Delta$ and tangent to the side $A\\Gamma$ at $A$. The circumcircle $\\omega$ of the triangle $AB\\Gamma$ meets circle $\\gamma$ at $A$ and $E$. Prove that $E$ is the point of intersection of the perpendicular bisectors of the segments $B\\Gamma$ and $A\\Delta$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nfigure 3\nIt is enough to prove that $EB = E\\Gamma$ and $EA = E\\Delta$. For that we compare the triangles $\\triangle EB$ and $\\triangle AE\\Gamma$ which have: $B\\Delta = A\\Gamma$, $\\Delta\\hat\\{B\\}E = A\\hat\\{\\Gamma\\}E$ (inscribed to the same arch $\\omega$) and $B\\hat\\{\\Delta\\}E = 180^\\circ - E\\hat\\{\\Delta\\}A = 180^\\circ - E\\hat\\{A\\}Z = E\\hat\\{A\\}\\Gamma$, (we have used $E\\hat\\{\\Delta\\}A = E\\hat\\{A\\}Z$, inscribed angle - angle chord and tangent)\nTherefore the triangles $\\triangle EB$ and $\\triangle AE\\Gamma$ are equal and so $EB = E\\Gamma$ and $E\\Delta = EA$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle AB\\Gamma$ be an acute angled triangle with circumcircle $c(O, R)$. From the midpoint $\\Delta$ of the side $B\\Gamma$ we draw a line perpendicular to $AB$ which meets $AB$ at $E$. If the line $AO$ intersects the line $\\varepsilon$ at $Z$, prove that the points $A$, $Z$, $\\Delta$, $\\Gamma$ are cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The external angle $\\mathrm{E}\\hat{Z}A$ of the quadrilateral $AZ\\Delta\\Gamma$ belongs to the orthogonal triangle $AEZ$, with the acute angle $\\mathrm{E}\\hat{A}Z = \\omega$ equal to the angle $A\\hat{B}O$, since $OA = OB$. Hence $\\mathrm{E}\\hat{Z}A = 90^\\circ - \\omega$\n\nLet the extension of the radius $BO$ intersect the circle $c(O, R)$ at $H$. Then $A\\hat{\\Gamma}H = A\\hat{B}H = \\omega$ and\n$$\n90^\\circ = B\\hat{\\Gamma}H = B\\hat{\\Gamma}A + A\\hat{\\Gamma}H \\Rightarrow B\\hat{\\Gamma}A = 90^\\circ - A\\hat{\\Gamma}H = 90^\\circ - \\omega.\n$$\nHence $\\mathrm{E}\\hat{Z}A = B\\hat{\\Gamma}A$, and the quadrilateral $AZ\\Delta\\Gamma$ is cyclic.\n\n![](attached_image_1.png)\nFigure 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be an equilateral triangle of side $k$ cm. We divide $AB\\Gamma$ with parallel lines into $k^2$ small equilateral triangles of side $1$ cm. In this way, we create a grid (see figure for $k=7$). Inside every small triangle we put exactly one positive integer from $1$ to $k^2$, so that there are not two triangles with the same number. With vertices the points of the grid, we define regular hexagons of side $1$ cm. We call the value of a hexagon the sum of the numbers of its 6 small triangles. Find, as a function of $k$, the greatest possible value of the sum of the values of all hexagons.\n\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "options": [],
    "answer": "3(k^4 - 14k^2 + 33k - 24)/2",
    "solution": "The small triangles are divided in four categories.\n\n1st category: They have one vertex $A$ or $B$ or $\\Gamma$.\n\nThese are not members of any hexagon and so their numbers do not take part in the final sum of values of all hexagons.\n\n2nd category: Contains small triangles which belong only to one hexagon. On every side of $AB\\Gamma$ there exist $k-2$ such triangles and so totally we have $3(k-2)+3 = 3k-3$ in this category.\n\n3rd category: Contains small triangles which belong exactly to two hexagons. On every side of $AB\\Gamma$ there are $k-3$ such triangles and so there are totally $3(k-3)$ triangles in this category.\n\nNote that for the computation of the final sums we must take into account that we will sum two times the numbers of the triangles of the second category.\n\n4th category: Contains small triangles which belong exactly to three hexagons. These triangles are inside the triangle $\\Delta EZ$, see figure. These triangles are totally $(k-3)^2$.\n\nIn order to obtain the greatest possible sum, we have to put as many as possible big numbers into triangles of higher category (then they will be counted more times). According to this reasoning we must put:\n\n![](attached_image_3.png)\n\n(1) The numbers of the set $A = \\{1, 2, 3\\}$ into the triangles of the first category.\n\n(2) The numbers of the set $B = \\{4, 5, 6, \\dots, 3k\\}$ into $3k-3$ triangles of the second category, with partial sum\n$$\nS_B = 4 + 5 + 6 + \\dots + 3k = \\frac{4+3k}{2}(3k-3) = \\frac{(3k-3)(3k+4)}{2}.\n$$\n\n(3) The numbers of the set $\\Gamma = \\{(3k+1), (3k+2), \\dots, (6k-9)\\}$ into $3k-9$ triangles of the third category, with partial sum:\n$$\nS_{\\Gamma} = (3k+1) + (3k+2) + \\dots + (6k-9) = \\frac{(3k+1) + (6k-9)}{2} (3k-9).\n$$\n\n(4) The numbers of the set $\\Delta = \\{(6k-8), (6k-7), \\dots, k^2\\}$ into $k^2-6k+9$ triangles of the fourth category, with partial sum:\n$$\nS_{\\Delta} = (6k-8) + (6k-7) + \\dots + k^2 = \\frac{k^2 + 6k - 8}{2} (k^2 - 6k + 9).\n$$\n\nHence the greatest possible sum of values is:\n$$\nS_{\\max} = S_A + S_B + 2S_{\\Gamma} + 3S_{\\Delta} = \\frac{3(k^4 - 14k^2 + 33k - 24)}{2}.\n$$\n\nLet now $\\gamma$ be a member of the set $\\Gamma$ and $\\delta$ a member of the set $\\Delta$. Then in the final sum $S_{\\max}$ there exists the summand $2\\gamma+3\\delta$. In the case of interchange of the position of the members of the sets $\\Gamma$ and $\\Delta$, then in the final sum $S_{\\max}$ we will have the summand $2\\delta+3\\gamma$. Since $\\gamma < \\delta$ and $2 < 3$, it follows that $2\\delta+3\\gamma < 2\\gamma+3\\delta$. Hence the sum we have found is the maximal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ for which there exist non-negative integers $x$, $y$ and $z$ such that the number\n$$\nx^p + y^p + z^p - x - y - z\n$$\nis a product of exactly three distinct prime numbers.",
    "options": [],
    "answer": "p = 2 or p = 5",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31281,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $\\alpha_v$ satisfies the recurrence relation: $\\alpha_1 = 1$ and $\\alpha_v = 5\\alpha_{v-1} + 3^{v-1}$, $v \\ge 2$. Determine the general term $\\alpha_v$ and the greatest power of $2$ which divides the term $a_k$, where $k=2^{2019}$.",
    "options": [],
    "answer": "α_v = (5^v − 3^v)/2 for v ≥ 1; and for k = 2^2019, the greatest power of 2 dividing a_k is 2^2021.",
    "solution": "$$\n\\alpha_v = 5^{v-1} \\cdot \\left[ 1 + \\frac{3}{5} + \\left(\\frac{3}{5}\\right)^2 + \\dots + \\left(\\frac{3}{5}\\right)^{v-1} \\right] = \\frac{1}{2}(5^v - 3^v),\\ v = 1, 2, \\dots\n$$\nNow for $k=2^{2019}$, we have: $2a_k = 5^{2019} - 3^{2019} = 2 \\cdot (5+3)(5^2+3^2) \\dots (5^{2018} + 3^{2018})$, and hence:\n$$\na_k = (5+3)(5^2+3^2)\\dots(5^{2018}+3^{2018}).\n$$\nWe observe that the first factor is divided by $8$ and all the others are divided by $2$ and are not divided by $4$. In fact, we have:\n$$\n5^{2v} \\equiv 1 \\pmod{4} \\text{ and } 3^{2v} \\equiv 1 \\pmod{4} \\Rightarrow 5^{2v} + 3^{2v} \\equiv 2 \\pmod{4}, \\text{ for all } v \\ge 1.\n$$\n\nThe factors from $5^2 + 3^2$ to $(5^{2^{2018}} + 3^{2^{2018}})$, are totally $2018$, and therefore the greatest power of $2$ dividing $a_k$ is $2^{2021}$.\n\nAlternatively, we can use a special form of the Lifting the Exponent Lemma concerning the greatest power of $2$ dividing a difference of powers of integers. We denote by $v_p(\\alpha)$ the greatest exponent of power of a prime number $p$ which divide the integer $\\alpha$, that is: $p^{v_p(\\alpha)}|\\alpha$ and $p^{v_p(\\alpha)+1} \\nmid \\alpha$. We have the following:\n\n**Lemma:** Let $\\alpha, \\beta$ two odd integers and $v$ an even positive integer. Then:\n$$\nv_2(\\alpha^v - \\beta^v) = v_2(\\alpha - \\beta) + v_2(\\alpha + \\beta) + v_2(v) - 1.\n$$\nBy applying the lemma to the integer $2a_{2^{2019}} = 5^{2^{2019}} - 3^{2^{2019}}$ we find:\n$$\nv_2(2a_{2^{2019}}) = v_2(5^{2^{2019}} - 3^{2^{2019}}) = v_2(5-3) + v_2(5+3) + v_2(2^{2019}) - 1 = 1 + 3 + 2019 - 1 = 2022,\n$$\nand hence: $v_2(a_k) = 2021$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(x, y, z)$ of real numbers which are solutions of the system:\n$$ \n\\begin{cases} x^2 + y^2 + 25z^2 = 6xz + 8yz \\\\ 3x^2 + 2y^2 + z^2 = 240 \\end{cases} \n$$",
    "options": [],
    "answer": "(6, 8, 2) and (-6, -8, -2)",
    "solution": "The first equation can be written in the form:\n$$\n\\begin{aligned} x^2 - 6xz + 9z^2 + y^2 - 8yz + 16z^2 &= 0 \n& \\Leftrightarrow (x - 3z)^2 + (y - 4z)^2 &= 0 \\\\\n\\Leftrightarrow x - 3z &= 0 \\quad \\text{and} \\quad y - 4z &= 0 \n& \\Leftrightarrow x = 3z \\quad \\text{and} \\quad y = 4z. \\end{aligned}\n$$\nTherefore, all triples satisfying the first equation of the system are the following:\n$$\n(x, y, z) = (3t, 4t, t), \\quad t \\in \\mathbb{R}.\n$$\nThen the second equation takes the form:\n$$\n3 \\cdot 9t^2 + 2 \\cdot 16t^2 + t^2 = 240 \\Leftrightarrow 60t^2 = 240 \\Leftrightarrow t^2 = 4 \\Leftrightarrow t = \\pm 2.\n$$\nFor $t = 2$, we get the solution $(x, y, z) = (6, 8, 2)$, while for $t = -2$ we get:\n$$\n(x, y, z) = (-6, -8, -2).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31283,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $\\alpha$, $\\beta$, $\\gamma$ are positive real numbers such that $\\frac{1}{\\alpha} + \\frac{1}{\\beta} + \\frac{1}{\\gamma} = 3$, prove that\n$$\n\\frac{\\alpha + \\beta}{\\alpha^2 + \\alpha\\beta + \\beta^2} + \\frac{\\beta + \\gamma}{\\beta^2 + \\beta\\gamma + \\gamma^2} + \\frac{\\gamma + \\alpha}{\\gamma^2 + \\gamma\\alpha + \\alpha^2} \\le 2 .\n$$\nWhen equality is valid?",
    "options": [],
    "answer": "2; equality when α = β = γ = 1.",
    "solution": "By putting $x = \\frac{1}{\\alpha}$, $y = \\frac{1}{\\beta}$, $z = \\frac{1}{\\gamma}$, we have $x + y + z = 3$ and the inequality takes the form:\n$$\n\\frac{xy(x+y)}{x^2+xy+y^2} + \\frac{yz(y+z)}{y^2+yz+z^2} + \\frac{zx(z+x)}{z^2+zx+x^2} \\le 2.\n$$\nWe have $\\frac{xy}{x^2+xy+y^2} \\le \\frac{1}{3} \\Leftrightarrow x^2+xy+y^2 \\ge 3xy \\Leftrightarrow (x-y)^2 \\ge 0$ and equality is valid when $x=y$. Multiplying both sides by $x+y>0$, we get\n$$\n\\frac{xy(x+y)}{x^2+xy+y^2} \\le \\frac{1}{3}(x+y). \\qquad (1)\n$$\nSimilarly we get the relations:\n$$\n\\frac{yz(y+z)}{y^2+yz+z^2} \\le \\frac{1}{3}(y+z) \\quad (2), \\quad \\frac{zx(z+x)}{z^2+zx+x^2} \\le \\frac{1}{3}(z+x). \\quad (3)\n$$\nFinally using summation of (1), (2) and (3) we have\n$$\n\\frac{xy(x+y)}{x^2+xy+y^2} + \\frac{yz(y+z)}{y^2+yz+z^2} + \\frac{zx(z+x)}{z^2+zx+x^2} \\le \\frac{1}{3} \\cdot 2(x+y+z) = 2.\n$$\nEquality is valid when $x = y = z = 1$ or $\\alpha = \\beta = \\gamma = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m_1 < m_2 < \\dots < m_s$ be a sequence of $s \\ge 2$ positive integers, none of which can be written as the sum of (two or more) distinct other numbers in the sequence. For every integer $r$ with $1 \\le r < s$ prove that $r m_r + m_s \\ge (r+1)(s-1)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $k, l$ with $0 \\le k \\le r$ and $k+1 \\le l \\le s$, we introduce the auxiliary value\n$$\nT(k, l) := m_l + \\sum_{i=1}^{k} m_i.\n$$\n\nWe claim that these $\\frac{1}{2}(r+1)(2s - r)$ auxiliary values are all pairwise distinct: Let us assume that $T(k, \\ell) = T(u, v)$. Without loss of generality $k \\le u$, so that this equality turns into\n$$\nm_{\\ell} = m_{v} + \\sum_{i=k+1}^{u} m_{i}.\n$$\nBut then $m_{\\ell}$ can be written as a sum of distinct other numbers in the sequence, unless $\\ell = v$ and $k = u$ holds. Hence the auxiliary values indeed are distinct. As altogether there are $\\frac{1}{2}(r+1)(2s - r)$ auxiliary values, the largest value $T(r, s)$ must be at least $\\frac{1}{2}(r+1)(2s - r)$. This yields\n$$\n\\frac{1}{2}(r+1)(2s - r) \\le T(r, s) = m_s + \\sum_{i=1}^{r} m_i \\le m_s + \\sum_{i=1}^{r} (m_r - r + i).\n$$\nHere we used $m_i \\le m_r - r + i$, which follows as the sequence is increasing. The above inequality can be rewritten into\n$$\nrs + s - r \\le r \\cdot m_r + m_s,\n$$\nwhich immediately implies the desired inequality from the problem statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AΓ$ be a triangle with circumcircle $c(O, R)$, such that $AB < AΓ < BΓ$ and let $Δ$ be the antipodal point of $B$ with respect to the circle $c$. The perpendicular bisector of $BΔ$ meets $BΓ$ at $K$, $AΓ$ at $M$ and $AB$ at $N$. The line $NΔ$ intersects the circle $c$ at point $T$. Let $Σ$ be the second point of intersection of the circles $c_1(O, Γ, M)$ and $c_2(O, A, Δ)$. Prove that the lines $AΔ$, $ΓT$ and $OΣ$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nWe will prove that the point $T$ belongs to the circle $c_1(O, Γ, M)$ and the point $N$ belongs to the circle $c_2(O, A, Δ)$, that is the quadrilaterals $NAOΔ$ and $MOΓT$ are cyclic. Then we conclude the following: the line $AΔ$ is the radical axis of the circles $C$ and $c_2$, the line $ΓT$ is the radical axis of the circles $C$ and $c_1$ and the line $OΣ$ is the radical axis $C_1$ and $C_2$, and therefore the lines $AΔ$, $ΓT$ and $OΣ$ are passing through the radical center of the three circles.\n\nSince $BΔ$ is diameter of the circle $c$, we have $B\\hat{A}Δ = 90^\\circ$ and so $N\\hat{A}Δ = 180^\\circ - B\\hat{A}Δ = 90^\\circ$. Also, $ON$ is the perpendicular bisector of the segment $BΔ$, and so: $N\\hat{O}Δ = 90^\\circ$. From the last two angle equalities we conclude that the quadrilateral **NAOΔ** is **cyclic**.\n\nThe triangle $NBΔ$ is isosceles with $NB = NΔ$ (because $N$ belongs to the perpendicular bisector $BΔ$), and so $N\\hat{B}Δ = N\\hat{A}B$. Therefore taking in mind that $OA = OB = OΔ = OT = R$ we conclude that the isosceles triangles $OAB$ and $OΔT$ are equal. Therefore\n$$\n(\\alpha) \\ OA = OT \\text{ and } (\\beta) \\ AB = ΔT \\Rightarrow NA = NT.\n$$\nHence the triangles $OAN$ and $OTN$ are equal, because they have their sides equal one to one and so\n$$\nN\\hat{O}A = N\\hat{O}T = \\frac{1}{2} \\cdot A\\hat{O}T, \\qquad (1)\n$$\nMoreover we have that\n$$\nM\\hat{Γ}T = A\\hat{Γ}T = \\frac{1}{2} \\cdot A\\hat{O}T \\qquad (2)\n$$\nFrom relations (1) and (2) it follows that $M\\hat{Γ}T = N\\hat{O}T = M\\hat{O}T$, from which we conclude that the quadrilateral **MOΓT** is **cyclic**.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle inscribed in a circle $\\Gamma$ with center $O$. Let $I$ be the incenter of $\\triangle ABC$ and $D, E, F$ the contact points of the incircle of $\\triangle ABC$ with $BC, AC, AB$, respectively. If $S$ is the foot of the perpendicular from $D$ to the line $EF$, prove that the line $SI$ passes through the antipodal of $A$ with respect to $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X$ be the second intersection point of $\\Gamma$ with $IA'$, where $A'$ is the antipodal of $A$ with respect to $\\Gamma$. We will prove that the points $X, S, I$ are collinear.\n\nWe have $\\widehat{IXA} = 90^\\circ$, so $X$ belongs to the circle of diameter $AI$. The same holds for the points $E$ and $F$, thus $X$ is on the circumcircle of $AEIF$. Therefore $\\widehat{AFX} = \\widehat{AEX}$. However, $\\widehat{ACX} = \\widehat{ABX}$, and as a result the triangles $BFX, CEX$ are similar. It follows that,\n$$\n\\frac{XF}{XE} = \\frac{BF}{CE} = \\frac{BD}{CD} \\qquad (1).\n$$\nSince the intersection of $EF$ with $BC$ is the isogonal conjugate of $B$, and $\\widehat{SDF} = 90^\\circ$, we get that $SD$ is the bisector of $\\widehat{BSC}$, thus\n$$\n\\frac{BS}{CS} = \\frac{BD}{CD} \\qquad (2).\n$$\nFinally, since $\\widehat{AFE} = \\widehat{AEF}$, and $\\widehat{FSB} = \\widehat{CSE}$, the triangles $BFS, CES$ are similar, so\n$$\n\\frac{BS}{CS} = \\frac{SF}{SE} \\qquad (3).\n$$\nFrom (1), (2), (3) we get $\\frac{XF}{XE} = \\frac{BD}{CD} = \\frac{BS}{CS} = \\frac{SF}{SE}$, which gives us that $XS$ is the bisector of $\\widehat{EXF}$. Since $I$ is the midpoint of the arc $EF$ at ($AEIF$), we have that $XI$ is the bisector of $\\widehat{EXF}$, so $X, S, I$ are collinear\n\n![](attached_image_1.png)\nfig. 5\n![](attached_image_2.png)\nfig. 6\n\nLet $M, N$ the midpoints of $FD, DE$, then $\\widehat{DFE} = \\widehat{DEC}$, and the right-angled triangles *DFS*, *DCN* are similar, so:\n$$\n\\frac{FS}{DF} = \\frac{EN}{DC} \\qquad (4)\n$$\nSimilarly, from the similar triangles *DES*, *BDM* we get:\n$$\n\\frac{SE}{ED} = \\frac{MD}{BD} \\qquad (5)\n$$\nThe relations (4) and (5) give\n$$\n\\frac{FS}{SE} = \\frac{BD}{CD} \\qquad (6)\n$$\nThe last one with the help of (1) gives that *XS* is the bisector of $\\widehat{EXF}$. Since *I* is the midpoint of the arc *EF* at (*AEIF*), we have that *XI* is the bisector of $\\widehat{EXF}$, so *X*, *S*, *I* are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma\\Delta$ quadrilateral inscribed in a circle of center $O$. The line perpendicular to the side $B\\Gamma$ at its midpoint $E$ meets the line $AB$ at point $Z$. The circumcircle of the triangle $\\Gamma EZ$ intersects the side $AB$ for a second time at point $H$ and the line $\\Gamma\\Delta$ at point $\\Theta \\neq \\Delta$. The line $E\\Theta$ meets the line $A\\Delta$ at point $K$ and the line $\\Gamma H$ at point $\\Lambda$. Prove that the points $A$, $H$, $\\Lambda$, $K$ are cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is enough to prove that: $A\\hat{K}\\Lambda = 90^\\circ$.\nSince $\\Delta\\hat{K}\\Theta = A\\hat{K}\\Lambda$, it is enough to prove that in the triangle $\\Delta\\Theta K$ the two acute angles have sum $90^\\circ$, i.e. $\\Delta\\Theta K + \\Theta\\hat{\\Delta}K = 90^\\circ$. We have:\n$$\n\\begin{aligned}\n\\Delta\\Theta K &= \\Gamma\\Theta E = \\Gamma\\hat{Z}E \\quad (\\text{inscribed in the same arc}) \\text{ and} \\\\\n\\Gamma\\hat{Z}E &= E\\hat{Z}B \\quad (\\text{symmetric with respect to} \\\\\n&\\text{perpendicular bisector of the side } B\\Gamma)\n\\end{aligned}\n$$\nHence we have: $\\Delta\\Theta K = E\\hat{Z}B$ (1).\n\n![](attached_image_1.png)\nfig. 1\n\nMoreover, from the cyclic quadrilateral $AB\\Gamma\\Delta$ we have: $\\Theta\\hat{\\Delta}K = Z\\hat{B}E$ (2)\nBy summing (1) and (2) we get: $\\Delta\\Theta K + \\Theta\\hat{\\Delta}K = E\\hat{Z}B + Z\\hat{B}E = 90^\\circ$, since the triangle $ZBE$ is right angled at $E$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be two distinct real numbers and let $c$ be a positive real number such that\n$$\na^4 - 2019a = b^4 - 2019b = c.\n$$\nProve that $-\\sqrt{c} < ab < 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a^4 - 2019a = b^4 - 2019b = c$.\n\nSince $a \\neq b$, subtract the two equations:\n$$\na^4 - b^4 - 2019(a - b) = 0\n$$\n$$\n(a - b)(a^3 + a^2b + ab^2 + b^3) - 2019(a - b) = 0\n$$\nSince $a \\neq b$, divide both sides by $(a - b)$:\n$$\na^3 + a^2b + ab^2 + b^3 = 2019\n$$\n$$\n(a + b)^3 - 3ab(a + b) = 2019\n$$\nLet $s = a + b$, $p = ab$.\nThen:\n$$\ns^3 - 3p s = 2019\n$$\nSo:\n$$\ns^3 - 3p s - 2019 = 0\n$$\nAlso, from $a^4 - 2019a = c$:\n$$\na^4 - 2019a - c = 0\n$$\nSimilarly for $b$:\n$$\nb^4 - 2019b - c = 0\n$$\nSo $a$ and $b$ are roots of the quartic $x^4 - 2019x - c = 0$.\nLet $x^4 - 2019x - c = 0$ have roots $a$, $b$, $r$, $s$.\nBy Vieta's formulas:\n- $a + b + r + s = 0$\n- $ab + ar + as + br + bs + rs = 0$\n- $abc + abr + abs + ars + brs + bsr + rsr = 2019$\n- $abrs = -c$\n\nBut we only need $a$ and $b$.\n\nLet us consider the function $f(x) = x^4 - 2019x - c$.\nSince $c > 0$, for large $|x|$, $f(x) > 0$.\nLet us check the sign of $f(x)$ at $x = 0$:\n$$\nf(0) = -c < 0\n$$\nSo $f(x)$ has at least two real roots, one positive and one negative.\nLet $a > 0$, $b < 0$.\n\nNow, $ab < 0$.\n\nLet us show that $ab > -\\sqrt{c}$.\n\nSuppose $a > 0$, $b < 0$.\nThen $a^4 - 2019a = b^4 - 2019b = c$.\n\nLet $a > 0$, $b < 0$.\nThen $a^4 > 2019a + c > c$ (since $a > 0$).\nSimilarly, $b^4 < 2019b + c < c$ (since $b < 0$).\nBut $b^4$ is positive, $2019b$ is negative, so $b^4 - 2019b$ increases as $b$ becomes more negative.\n\nLet us try to bound $ab$.\n\nLet $a$ and $b$ be roots of $x^4 - 2019x - c = 0$.\nThen $x^4 - 2019x = c$.\nLet $x = t$, $t^4 - 2019t = c$.\n\nLet $a > 0$, $b < 0$.\nThen $ab < 0$.\n\nNow, consider $a^4 - 2019a = b^4 - 2019b$.\nLet $a = -b$.\nThen $a^4 - 2019a = b^4 - 2019b$ becomes $a^4 - 2019a = a^4 + 2019a$, so $-2019a = 2019a$, so $a = 0$, but $a \\neq b$.\nSo $a \\neq -b$ unless $a = 0$.\n\nNow, since $a$ and $b$ are distinct, and $a > 0$, $b < 0$, $ab < 0$.\n\nLet us try to bound $ab$ from below.\n\nLet $a^4 - 2019a = c$.\nThen $a^4 = 2019a + c$.\nSo $a^4 \\geq c$ for $a > 0$.\nSo $a \\geq c^{1/4}$.\n\nSimilarly, for $b < 0$, $b^4 = 2019b + c$.\nBut $b^4$ is positive, $2019b$ is negative, so $b^4 < c$ for $b$ negative and large in magnitude.\n\nNow, $ab$ is negative.\n\nLet us try to show $ab > -\\sqrt{c}$.\n\nSuppose $ab \\leq -\\sqrt{c}$.\nThen $ab$ is negative and its magnitude at least $\\sqrt{c}$.\n\nBut $a$ and $b$ are roots of $x^4 - 2019x - c = 0$.\n\nLet us consider the quadratic $x^2 - s x + p = 0$ with roots $a$ and $b$.\n\nFrom earlier, $s^3 - 3p s = 2019$.\n\nLet us try $a = \\sqrt{d}$, $b = -\\sqrt{d}$ for some $d > 0$.\nThen $ab = -d$.\n\nCompute $a^4 - 2019a = d^2 - 2019\\sqrt{d}$.\nCompute $b^4 - 2019b = d^2 + 2019\\sqrt{d}$.\nSo these are equal only if $2019\\sqrt{d} = -2019\\sqrt{d}$, so $d = 0$.\nSo this is not possible for $a \\neq b$.\n\nAlternatively, since $a$ and $b$ are roots of $x^4 - 2019x - c = 0$, and $a > 0$, $b < 0$, $ab < 0$.\n\nSuppose $ab = -\\sqrt{c}$.\nThen $p = -\\sqrt{c}$.\n\nFrom $s^3 - 3p s = 2019$:\n$$\ns^3 + 3\\sqrt{c} s = 2019\n$$\nBut $s = a + b$.\n\nLet us try to find the minimum value of $ab$.\n\nAlternatively, since $a$ and $b$ are roots of $x^4 - 2019x - c = 0$, and $a > 0$, $b < 0$, $ab < 0$.\n\nLet us try to show that $ab > -\\sqrt{c}$.\n\nSuppose $ab = -k$, $k > 0$.\nThen $a$ and $b$ are roots of $x^2 - s x - k = 0$.\nSo $a = \\frac{s + \\sqrt{s^2 + 4k}}{2}$, $b = \\frac{s - \\sqrt{s^2 + 4k}}{2}$.\n\nNow, $a^4 - 2019a = b^4 - 2019b$.\n\nLet us try to find the maximum possible value of $|ab|$.\n\nAlternatively, since $a^4 - 2019a = b^4 - 2019b$, and $a \\neq b$, $a$ and $b$ are symmetric about some value.\n\nBut the key is that $ab < 0$, and $ab > -\\sqrt{c}$.\n\nTherefore, $-\\sqrt{c} < ab < 0$.\n\nThus, the required inequality is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(x, n)$ of positive integers satisfying the equation:\n$$\n3 \\cdot 2^x + 4 = n^2.\n$$",
    "options": [],
    "answer": "(x, n) = (2, 4), (5, 10), (6, 14)",
    "solution": "If $x = 1$, then we have no solutions, while for $x = 2$, we have the solution $(x, n) = (2, 4)$.\n\nNow we suppose that $x \\ge 3$. The left part of the equation is even and so $n$ is even, say $n = 2k$. Then the equation can be written as: $3 \\cdot 2^x + 4 = 4k^2 \\Leftrightarrow 3 \\cdot 2^{x-2} = k^2 - 1$, and hence $k$ is odd. Thus the equation is written as: $3 \\cdot 2^{x-2} = (k-1)(k+1)$.\n\n1st case: $k+1 = 3 \\cdot 2^a$ and $k-1 = 2^b$, $a, b \\ge 1$, with $a+b = x-2$. Then by subtracting we get $3 \\cdot 2^a - 2^b = 2$. If $a, b \\ge 2$, then the left part is divisible by 4, while the right part is not. Therefore, either $a=1$, which implies $b=2$ and $x=5$, or $b=1$, not giving solution for $a$. Hence we have $(x, n) = (5, 10)$.\n\n2nd case: $k+1 = 2^s$ and $k-1 = 3 \\cdot 2^t$, $s, t \\ge 1$, with $s+t = x-2$. Then we have the equation $2^s - 3 \\cdot 2^t = 2$. If $s, t \\ge 2$ then the left part is divisible by 4, while the right part is not. Hence, either $s=1$ or $t=1$, giving $s=3$ and $x=6$. Therefore in this case we have the solution $(x, n) = (6, 14)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exist infinitely many integers $x, y, z$ for which the sum of the digits in the decimal representation of $4x^4 + y^4 - z^2 + 4xyz$ is at most 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "This is an easy problem with many solutions. We rewrite\n$$\n\\begin{aligned}\n4x^4 + y^4 - z^2 + 4xyz &= (4x^4 + y^4 + 4x^2y^2) - (4x^2y^2 + z^2 - 4xyz) \\\\\n&= (2x^2 + y^2)^2 - (2xy - z)^2 \\\\\n&= (2x^2 + y^2 - 2xy + z)(2x^2 + y^2 + 2xy - z)\n\\end{aligned}\n$$\nThe two factors $A = 2x^2 + y^2 - 2xy + z$ and $B = 2x^2 + y^2 + 2xy - z$ add up to $A + B = 4x^2 + 2y^2$. If we choose the values of $x$ and $y$ so that $A = 4x^2 = 4 \\cdot 5^{2n+2}$ and $B = 2y^2 = 2 \\cdot 2^{2n}$, then the product will become $AB = 2 \\cdot 10^{2n+2}$ and the sum of the digits will equal 2.\n\nSummarizing, we pick an integer $n \\ge 1$ and set $x = 5^{n+1}$ and $y = 2^n$. The desired equation $A = 4x^2$ is equivalent to $2x^2 + y^2 - 2xy + z = 4x^2$, and hence\n$$\nz = 2x^2 + 2xy - y^2 = 2 \\cdot 5^{2n+2} + 10^{n+1} - 4^n.\n$$\nNote that $z$ indeed is a positive integer. As the described choice of $x, y, z$ then yields\n$$\n4x^4 + y^4 - z^2 + 4xyz = 2 \\cdot 10^{2n+2},\n$$\nthe proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31291,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $\\alpha_0, \\alpha_1, \\alpha_2, \\ldots, \\alpha_\\nu, \\ldots$, $\\nu \\in \\mathbb{N}$, with integer terms, not necessarily different, has the following properties:\n\n(α) $0 \\le \\alpha_i \\le i$, for every integer $i \\ge 0$,\n\n(β) $\\binom{\\kappa}{\\alpha_0} + \\binom{\\kappa}{\\alpha_1} + \\dots + \\binom{\\kappa}{\\alpha_\\kappa} = 2^\\kappa$, for every integer $\\kappa \\ge 0$.\n\nProve that for every integer $n \\ge 0$, there exists $i \\ge 0$ such that $\\alpha_i = n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove using induction with respect to $\\kappa$ that every initial part $\\alpha_0, \\alpha_1, \\alpha_2, \\ldots, \\alpha_\\kappa$ of the sequence consists of the following integers (not necessarily with the turn of the terms of the sequence)\n$0, 1, \\ldots, \\ell-1, 0, 1, \\ldots, \\kappa-\\ell$, for some $\\ell \\ge 0$ with $2\\ell \\le \\kappa+1$.\n\nFor $\\kappa = 0$, $\\alpha_0 = 0$ holds. Suppose that for $\\kappa = \\nu$ the terms $\\alpha_0, \\alpha_1, \\alpha_2, \\ldots, \\alpha_\\nu$ are of the form\n$0, 0, 1, 1, 2, 2, \\ldots, \\ell-1, \\ell-1, \\ell, \\ell+1, \\ldots, \\nu-\\ell-1, \\nu-\\ell$ for some $\\ell$ with $0 \\le 2\\ell \\le \\kappa+1$.\n\nThen, for $\\kappa = \\nu + 1$, we have from (β) that:\n$$\n\\begin{gather*}\n\\binom{\\nu+1}{\\alpha_0} + \\binom{\\nu+1}{\\alpha_1} + \\cdots + \\binom{\\nu+1}{\\alpha_\\nu} + \\binom{\\nu+1}{\\alpha_{\\nu+1}} = 2^{\\nu+1} \\Leftrightarrow \\\\\n\\left\\{ \\binom{\\nu+1}{0} + \\binom{\\nu+1}{1} + \\cdots + \\binom{\\nu+1}{\\ell-1} \\right\\} + \\left\\{ \\binom{\\nu+1}{0} + \\binom{\\nu+1}{1} + \\cdots + \\binom{\\nu+1}{\\nu-\\ell} \\right\\} + \\binom{\\nu+1}{\\alpha_{\\nu+1}} = 2^{\\nu+1} \\Leftrightarrow \\\\\n\\left\\{ \\binom{\\nu+1}{0} + \\binom{\\nu+1}{1} + \\cdots + \\binom{\\nu+1}{\\ell-1} \\right\\} + \\left\\{ \\binom{\\nu+1}{\\nu+1} + \\binom{\\nu+1}{\\nu} + \\cdots + \\binom{\\nu+1}{\\ell+1} \\right\\} + \\binom{\\nu+1}{\\alpha_{\\nu+1}} = 2^{\\nu+1}. \\quad (1)\n\\end{gather*}\n$$\nMoreover, we have:\n$$\n\\binom{\\nu+1}{0} + \\binom{\\nu+1}{1} + \\cdots + \\binom{\\nu+1}{\\nu+1} = 2^{\\nu+1}. \\quad (2)\n$$\nFrom relations (1) and (2) it follows that:\n$$\n\\binom{\\nu+1}{\\alpha_{\\nu+1}} = \\binom{\\nu+1}{\\ell}. \\quad (3)\n$$\nFrom relation (3), by using that the binomial coefficients $\\binom{\\nu+1}{i}$ are increasing when the variable $i \\le \\frac{\\nu+1}{2}$ increases and they are decreasing when the variable $i \\ge \\frac{\\nu+1}{2}$ increases, we conclude that $\\alpha_{\\nu+1} = \\ell$ or $\\alpha_{\\nu+1} = \\nu+1-\\ell$. In both cases the initial part $\\alpha_0, \\alpha_1, \\alpha_2, \\ldots, \\alpha_{\\nu+1}$ of the sequence is of the form we seek. Therefore, according to the above conclusion, every integer $n \\ge 0$ will coincide to the term $\\alpha_i$ of the sequence for some $i$ with $0 \\le i \\le 2n$.\n\nIndeed, the sequence will have terms $0, 1, \\ldots, \\ell-1, 0, 1, \\ldots, 2n-\\ell$, for some $\\ell \\le \\frac{2n+1}{2}$, and therefore we have the cases:\n\n* if $n < \\ell \\le \\frac{2n+1}{2}$, then $n \\le \\ell-1$, and so there exists $i \\ge 0$ such that $\\alpha_i = n$.\n* if $n \\ge \\ell$, then $n > \\ell-1 \\Rightarrow 2n-n < 2n-\\ell+1 \\Rightarrow n \\le 2n-\\ell$, and so again there exists $i \\ge 0$ such that $\\alpha_i = n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(\\alpha, \\beta)$ of prime numbers $\\alpha, \\beta$ for which the number $A = 3\\alpha^2\\beta + 16\\alpha\\beta^2$ is the square of an integer.",
    "options": [],
    "answer": "(19, 19) and (2, 3)",
    "solution": "We distinguish the cases:\n\n1. $\\alpha = \\beta$\nLet: $A = 3\\alpha^2\\alpha + 16\\alpha\\alpha^2 = 19\\alpha^3 = \\kappa^2, \\kappa \\in \\mathbb{Z}, \\alpha$ prime.\n\nThen $19|\\kappa^2 \\Rightarrow 19|\\kappa \\Rightarrow 19^2|\\kappa^2 \\Rightarrow \\kappa^2 = 19^2\\omega, \\omega \\in \\mathbb{Z}$, and hence:\n$$\n19\\alpha^3 = 19^2\\omega \\Rightarrow \\alpha^3 = 19\\omega \\Rightarrow 19|\\alpha^3 \\Rightarrow 19|\\alpha \\Rightarrow \\alpha = 19.\n$$\nHence, the pair $(\\alpha, \\beta) = (19,19)$ maybe a solution. Since for $\\alpha = \\beta = 19$ we have $A = 19 \\cdot 19^3 = 19^4$, the pair $(\\alpha, \\beta) = (19,19)$ is a solution.\n\n2. $\\alpha \\neq \\beta$\nLet $A = 3\\alpha^2\\beta + 16\\alpha\\beta^2 = \\alpha\\beta(3\\alpha + 16\\beta) = \\kappa^2, \\kappa \\in \\mathbb{Z}, \\alpha, \\beta$ primes.  \nThen: $\\alpha|\\kappa^2 \\Rightarrow \\alpha|\\kappa \\Rightarrow \\alpha^2|\\kappa^2 \\Rightarrow \\kappa^2 = \\alpha^2\\omega, \\omega \\in \\mathbb{Z}$, and hence\n$$\n\\alpha\\beta(3\\alpha + 16\\beta) = \\alpha^2\\omega \\Rightarrow \\beta(3\\alpha + 16\\beta) = \\alpha\\omega \\Rightarrow \\alpha|\\beta(3\\alpha + 16\\beta) \\\\\n\\Rightarrow \\alpha|(3\\alpha + 16\\beta) \\Rightarrow \\alpha|16\\beta \\Rightarrow \\alpha|16 = 2^4 \\Rightarrow \\alpha = 2.\n$$\nFrom above we have $\\beta|\\kappa^2 \\Rightarrow \\beta|\\kappa \\Rightarrow \\beta^2|\\kappa^2 \\Rightarrow \\kappa^2 = \\beta^2\\tau, \\tau \\in \\mathbb{Z}$, and hence\n$$\n\\alpha\\beta(3\\alpha + 16\\beta) = \\beta^2\\tau \\Rightarrow \\alpha(3\\alpha + 16\\beta) = \\beta\\omega \\Rightarrow \\beta|\\alpha(3\\alpha + 16\\beta) \\\\\n\\Rightarrow \\beta|(3\\alpha + 16\\beta) \\Rightarrow \\beta|3\\alpha \\Rightarrow \\beta|3 \\Rightarrow \\beta = 3.\n$$\nHence the pair $(\\alpha, \\beta) = (2,3)$ is a probable solution. Since\n$$\nA = \\alpha\\beta(3\\alpha + 16\\beta) = 6 \\cdot (6 + 48) = 6 \\cdot 54 = 18^2,\n$$\nthe pair $(\\alpha, \\beta) = (2,3)$ is a solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of positive integers satisfying the equation\n$$\nyx^y = y + 1\n$$",
    "options": [],
    "answer": "(2, 1)",
    "solution": "Let $y = \\frac{p}{q}$, with $p, q \\in \\mathbb{N}^*$, $(p, q) = 1$. Then the equation is written in the form:\n$$\nx^q = \\frac{y+1}{y} = \\frac{\\frac{p}{q}+1}{\\frac{p}{q}} = \\frac{p+q}{p} \\Rightarrow x = \\sqrt[q]{\\frac{(p+q)^q}{p^q}}.\n$$\nTherefore the right hand side must be rational. Since $(p+q, p)=1$, it follows that the numerator and the denominator of the fraction must be powers of $p$. Hence\n$$\np+q = a^p \\quad \\text{and} \\quad p = b^p,\\ a, b \\in \\mathbb{N}^*,\\ a > 1.\n$$\nIf $b \\ge 2$, then $b^p \\ge 2^p > p$ (absurd), and hence $b=1$, which implies $p=1$ and $q = a - 1$. Hence the solutions of the given equation are given parametrically in the form: $(x, y) = \\left( a^{a-1}, \\frac{1}{a-1} \\right),\\ a \\in \\mathbb{N}^*,\\ a > 1$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31294,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a$, $b$, $c$ are positive real numbers, prove that:\n$$\n\\frac{1}{ab(b+1)(c+1)} + \\frac{1}{bc(c+1)(a+1)} + \\frac{1}{ca(a+1)(b+1)} - \\frac{3}{(1+abc)^2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The inequality is equivalent to\n$$\n\\frac{c(a+1)+a(b+1)+b(c+1)}{abc(a+1)(b+1)(c+1)} \\ge \\frac{3}{(1+abc)^2},\n$$\nor after simplifications to\n$$\n(1+abc)^2(ab+bc+ca+a+b+c) \\ge 3abc(ab+bc+ca+a+b+c+abc+1).\n$$\nWe put $m = a+b+c$, $n = ab+bc+ca$ and $x^3 = abc$, to obtain:\n$$ (m+n)(1+x^3)^2 \\ge 3x^3(x^3 + m+n+1), $$ \nor $(m+n)(x^6 - x^3 + 1) \\ge 3x^3(x^3 + 1). $\nFrom the inequality of arithmetic–geometric mean we have $m \\ge 3x$ and $n \\ge 3x^2$, and so $m+n \\ge 3x(x+1)$. Therefore, it is enough to prove that\n$$\n\\begin{aligned} & 3x(x+1)(x^6-x^3+1) \\ge 3x^3(x+1)(x^2-x+1) \\\\ \\Leftrightarrow & x^6-x^3+1 \\ge x^2(x^2-x+1) \\\\ \\Leftrightarrow & x^6-x^4-x^2+1 \\ge 0 \\Leftrightarrow x^4(x^2-1)-(x^2-1) \\ge 0 \\\\ \\Leftrightarrow & (x^2-1)(x^4-1) \\ge 0 \\Leftrightarrow (x^2+1)(x^2-1)^2 \\ge 0, \\end{aligned}\n$$\nwhich is valid.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Triangle $ABC$ is such that $AB < AC$. The perpendicular bisector of side $BC$ intersects lines $AB$ and $AC$ at points $P$ and $Q$, respectively. Let $H$ be the orthocentre of triangle $ABC$, and let $M$ and $N$ be the midpoints of segments $BC$ and $PQ$, respectively. Prove that lines $HM$ and $AN$ meet on the circumcircle of $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers which are equal to $13$ times the sum of their digits.",
    "options": [],
    "answer": "117, 156, 195",
    "solution": "Let $\\kappa$ be the number of digits of the integer $A$ which is equal to $13$ times the sum of its digits. The least possible $A$ is $10^{\\kappa-1}$, while the maximal possible sum of the digits is $9\\kappa$. Therefore we need to have:\n$$\n10^{\\kappa-1} \\leq 13 \\cdot 9\\kappa = 117\\kappa. \\qquad (1)\n$$\nFor $\\kappa \\ge 4$, we will prove using induction that: $10^{\\kappa-1} > 117\\kappa$, that is, relation (1) is not valid. In fact, for $\\kappa = 4$ we have: $10^{4-1} = 10^3 > 117 \\cdot 4 = 468$. If $10^{\\kappa-1} > 117\\kappa$, for the arbitrary $\\kappa > 4$, then we get:\n$$\n10^{(\\kappa+1)-1} = 10^{\\kappa} = 10 \\cdot 10^{\\kappa-1} > 10 \\cdot 117\\kappa = 117 \\cdot 10\\kappa > 117 \\cdot (\\kappa+1).\n$$\nTherefore the number $\\kappa$ must be less or equal to $3$.\n\n* We have to reject the case $\\kappa = 1$, since $A = \\alpha < 13\\alpha$, with $0 < \\alpha \\le 9$.\n* Similarly we reject the case with $\\kappa = 2$, since $A = 10\\alpha + \\beta < 13(\\alpha + \\beta)$.\n* Let $\\kappa = 3$ and $A = \\overline{\\alpha\\beta\\gamma} = 100\\alpha + 10\\beta + \\gamma$, $0 < \\alpha \\le 9$, $0 \\le \\beta, \\gamma \\le 9$.\n\nThen we have:\n$$\n100\\alpha + 10\\beta + \\gamma = 13 \\cdot (\\alpha + \\beta + \\gamma), \\quad 0 < \\alpha \\le 9, 0 \\le \\beta, \\gamma \\le 9\n$$\n$$\n\\Leftrightarrow 87\\alpha = 3\\beta + 12\\gamma, \\quad 0 < \\alpha \\le 9, 0 \\le \\beta, \\gamma \\le 9\n$$\n$$\n\\Leftrightarrow 29\\alpha = \\beta + 4\\gamma, \\quad 0 < \\alpha \\le 9, 0 \\le \\beta, \\gamma \\le 9\n$$\nSince $0 \\le \\beta + 4\\gamma \\le 45 \\Rightarrow 29\\alpha \\le 45 \\Rightarrow \\alpha \\le 1$ (since $\\alpha \\ne 0$), and hence:\n$$\n\\beta + 4\\gamma = 29 \\Rightarrow \\beta = 29 - 4\\gamma \\ge 0 \\Rightarrow 0 \\le \\beta = 29 - 4\\gamma \\le 9\n$$\n$$\n\\Rightarrow -29 \\le -4\\gamma \\le -20 \\Rightarrow 5 \\le \\gamma \\le \\frac{29}{4} \\Rightarrow \\gamma \\in \\{5, 6, 7\\}\n$$\n* For $\\gamma = 5 \\Rightarrow \\beta = 9$ and $A = 195$.\n* For $\\gamma = 6 \\Rightarrow \\beta = 5$ and $A = 156$.\n* For $\\gamma = 7 \\Rightarrow \\beta = 1$ and $A = 117$.\n\nTherefore, the positive integers which are equal to $13$ times the sum of their digits are $117$, $156$, and $195$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $v > 1$, a positive integer. Each cell of a $v \\times v$ table contains one integer. Suppose that the following conditions are satisfied:\n\na. Each number in the table is congruent to $1$ modulo $v$.\n\nb. The sum of numbers in any row, as well the sum of numbers in any column is congruent to $v$ modulo $v^2$.\n\nLet $\\Gamma_i$ be the product of numbers in the $i^{\\text{th}}$ row, and $\\Sigma_j$ be the product of numbers in the $j^{\\text{th}}$ column. Prove that the sums $\\Gamma_1 + \\Gamma_2 + \\dots + \\Gamma_v$ and $\\Sigma_1 + \\Sigma_2 + \\dots + \\Sigma_v$ are congruent modulo $v^4$. (IMO 2018 shortlist)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 31298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the set $X = \\{1,2,...,8\\}$ and two nonempty disjoint subsets $A, B$ of $X$ with union the set $X$. Let $P_A$ be the product of the elements of the set $A$ and $P_B$ be the product of the elements of the set $B$. Determine the least possible value of the sum $P_A + P_B$.",
    "options": [],
    "answer": "402",
    "solution": "We observe that: $P_A \\cdot P_B = 8! = c$. By symmetry, without loss of generality, we suppose that $P_A \\leq P_B$, and hence $P_A \\leq \\sqrt{c}$. We write\n$$\nP_A + P_B = P_A + \\frac{c}{P_A}\n$$\nand $P_A = x$. We consider the function $f(x) = x + \\frac{c}{x}$, with $1 \\leq x \\leq \\sqrt{c}$. The function $f(x)$ is strictly decreasing, because for $1 \\leq x < y \\leq \\sqrt{c}$ it follows that $f(x) > f(y)$. Indeed, for $1 \\leq x < y \\leq \\sqrt{c}$, we have $x - y < 0$, $xy - c < 0$ and\n$$\nf(x) - f(y) = x - y + \\frac{c(y-x)}{xy} = \\frac{(x-y)(xy-c)}{xy} > 0. \\quad (*)\n$$\nSince $x$ is integer and it is not possible to be equal to $\\sqrt{c}$, the least possible value is the closest integer to $\\sqrt{c}$. We have $\\lfloor\\sqrt{8!}\\rfloor = \\lfloor24\\sqrt{70}\\rfloor = 200$, and so the closest integer is $24 \\cdot 8 = 192$. Therefore the least possible value is\n$$\nf(192) = 192 + 210 = 402\n$$\nand it can be approached, for example for the sets $A = \\{4,6,8\\}$, $B = \\{1,2,3,5,7\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be an interior point to an equilateral triangle of altitude $1$. If $x$, $y$, $z$ are the distances from $P$ to the sides of the triangle, then prove that:\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is well-known that in an equilateral triangle the sum of the distances from an interior point $P$ to its sides equals the altitude of the triangle, as can be easily proven. On account of the preceding, we have to prove that if $x + y + z = 1$ then it holds that\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz\n$$\nTo do it, we begin observing that when $x + y + z = 1$, then\n$$\nxy + yz + zx \\geq 9xyz\n$$\nIndeed, applying AM-GM inequality, we get\n$$\nxy + yz + zx = (xy + yz + zx)(x + y + z) \\geq 3\\sqrt[3]{(xy)(yz)(zx)} \\cdot 3\\sqrt[3]{xyz} = 9xyz\n$$\nor\n$$\nxy + yz + zx - 3xyz \\geq 6xyz\n$$\nOn account of the constrain and the preceding inequality, we obtain\n$$\n\\begin{aligned}\nxy + yz + zx - 3xyz &= xy(1 - z) + yz(1 - x) + zx(1 - y) \\\\\n&= xy(x + y) + yz(y + z) + zx(z + x) \\\\\n&\\geq 6xyz\n\\end{aligned}\n$$\nAdding $1$ to both terms of the last inequality and reordering terms, yields\n$$\n(x + y + z)^2 + xy(x + y) + yz(y + z) + zx(z + x) - 6xyz \\geq 1\n$$\nor equivalently,\n$$\nx^2 + y^2 + z^2 + 2xy(1-z) + 2yz(1-x) + 2zx(1-y) + xy(x+y) + yz(y+z) + zx(z+x) \\geq 1,\n$$\nand\n$$\nx^2 + y^2 + z^2 + 3xy(x + y) + 3yz(y + z) + 3zx(z + x) \\geq 1 = (x + y + z)^3\n$$\nfrom which\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz\n$$\nfollows. Equality holds when $x = y = z = 1/3$. That is, when $P$ is the centroid of the triangle, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha$ and $\\beta$ be two positive integers. Prove that the integer\n$$\n\\alpha^2 + \\left\\lfloor \\frac{4\\alpha^2}{\\beta} \\right\\rfloor\n$$\nis not a square of an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We suppose that:\n$$\n\\alpha^2 + \\left\\lfloor \\frac{4\\alpha^2}{\\beta} \\right\\rfloor = (\\alpha + \\kappa)^2, \\kappa \\in \\mathbb{N}^* \\Leftrightarrow \\left\\lfloor \\frac{4\\alpha^2}{\\beta} \\right\\rfloor = (2\\alpha + \\kappa)\\kappa, \\kappa \\in \\mathbb{N}^*.\n$$\nBy putting $2\\alpha = x$, we have: $\\left\\lfloor \\frac{x^2}{\\beta} \\right\\rfloor = (x+\\kappa)\\kappa, \\kappa \\in \\mathbb{N}^*$. Equivalently, we have supposed that the equation\n$$\n\\left\\lfloor \\frac{x^2}{\\beta} \\right\\rfloor = (x + \\kappa)\\kappa, \\kappa \\in \\mathbb{N}^* \\quad (1)\n$$\nhas a solution $(x, \\kappa)$ in the positive integers with $x$ even.\nWe suppose that equation (1) has a solution $(x, \\kappa)$, where $\\kappa$ is the least possible integer and we will try to find a contradiction.\nWe have:\n$$\n\\begin{gathered} \\frac{x^2}{\\beta} > \\left\\lfloor \\frac{x^2}{\\beta} \\right\\rfloor - 1 = (x+\\kappa)\\kappa - 1 = x\\kappa + \\kappa^2 - 1 \\ge x\\kappa \\Rightarrow \\frac{x}{\\beta} > \\kappa \\Rightarrow \\\\ x > \\beta\\kappa \\end{gathered} \\quad (2)\n$$\nAlso, we have:\n$$\n\\frac{x^2 - \\kappa^2}{\\beta} < \\frac{x^2}{\\beta} \\le \\left\\lfloor \\frac{x^2}{\\beta} \\right\\rfloor = (x + \\kappa)\\kappa \\Rightarrow \\beta\\kappa \\ge x - \\kappa. \\quad (3)\n$$\nFrom relations (2) (3) it follows that: $x > \\beta\\kappa \\ge x - \\kappa$, and hence there exists $\\upsilon$ such that:\n$$\nx = \\beta\\kappa + \\upsilon, \\quad 0 < \\upsilon < \\kappa \\quad (4)\n$$\nFrom (3) and (4) we have:\n$$\n\\left\\lfloor \\frac{x^2}{\\beta} \\right\\rfloor = \\left\\lfloor \\frac{(\\beta\\kappa + \\upsilon)^2}{\\beta} \\right\\rfloor = \\beta\\kappa^2 + 2\\kappa\\upsilon + \\left\\lfloor \\frac{\\upsilon^2}{\\beta} \\right\\rfloor \\quad (5)\n$$\n$$\n(x + \\kappa)\\kappa = (\\beta\\kappa + \\upsilon + \\kappa)\\kappa = \\beta\\kappa^2 + 2\\kappa\\upsilon + \\kappa(\\kappa - \\upsilon) \\quad (6)\n$$\nFrom relations (5) and (6) we get:\n$$\n\\left\\lfloor \\frac{\\upsilon^2}{\\beta} \\right\\rfloor = \\kappa(\\kappa - \\upsilon) \\iff \\left\\lfloor \\frac{x'^2}{\\beta} \\right\\rfloor = (x' + \\kappa')\\kappa',\n$$\nFrom which we conclude that equation (1) has a solution $(x', \\kappa')$ in the positive integers with $\\kappa' = \\kappa - \\upsilon < \\kappa$, which is absurd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31301,
    "subject": "Mathematics (Multi-modal)",
    "question": "A grid consists of all points of the form $(m, n)$ where $m$ and $n$ are integers with $|m| \\le 2019$, $|n| \\le 2019$ and $|m| + |n| < 4038$. We call the points $(m, n)$ of the grid with either $|m| = 2019$ or $|n| = 2019$ the *boundary points*. The four lines $x = \\pm 2019$ and $y = \\pm 2019$ are called *boundary lines*. Two points in the grid are called *neighbours* if the distance between them is equal to 1.\nAnna and Bob play a game on this grid.\nAnna starts with a token at the point $(0, 0)$. They take turns, with Bob playing first.\n1) On each of his turns, Bob deletes at most two boundary points on each boundary line.\n2) On each of her turns, Anna makes exactly three *steps*, where a *step* consists of moving her token from its current point to any neighbouring point which has not been deleted.\nAs soon as Anna places her token on some boundary point which has not been deleted, the game is over and Anna wins.\n\nDoes Anna have a winning strategy?",
    "options": [],
    "answer": "No; Bob has a winning strategy.",
    "solution": "Anna does not have a winning strategy. We will provide a winning strategy for Bob. It is enough to describe his strategy for the deletions on the line $y = 2019$.\n\nBob starts by deleting $(0, 2019)$ and $(-1, 2019)$. Once Anna completes her turn, he deletes the next two available points on the left if Anna decreased her $x$-coordinate, the next two available points on the right if Anna increased her $x$-coordinate, and the next available point to the left and the next available point to the right if Anna did not change her $x$-coordinate. The only exception to the above rule is on the very first time Anna decreases $x$ by exactly 1. In that turn, Bob deletes the next available point to the left and the next available point to the right.\n\nBob's strategy guarantees the following: If Anna makes a sequence of steps reaching $(-x, y)$ with $x > 0$ and the exact opposite sequence of steps in the horizontal direction reaching $(x, y)$, then Bob deletes at least as many points to the left of $(0, 2019)$ in the first sequence than points to the right of $(0, 2019)$ in the second sequence.\n\nSo we may assume for contradiction that Anna wins by placing her token at $(k, 2019)$ for some $k > 0$.\n\nDefine $\\Delta = 3m - (2x + y)$, where $m$ is the total number of points deleted by Bob to the right of $(0, 2019)$, and $(x, y)$ is the position of Anna's token.\n\nFor each sequence of steps performed first by Anna and then by Bob, $\\Delta$ does not decrease. This can be seen by looking at the following table exhibiting the changes in $3m$ and $2x + y$. We have excluded the cases where $2x + y < 0$.\n\n| Turn      | (0,3) | (1,2) | (-1,2) | (2,1) | (0,1) | (3,0) | (1,0) | (2,-1) | (1,-2) |\n|-----------|-------|-------|--------|-------|-------|-------|-------|--------|--------|\n| m         | 1     | 2     | 0 (or 1) | 2     | 1     | 2     | 2     | 2      | 2      |\n| 3m        | 3     | 6     | 0 (or 3) | 6     | 3     | 6     | 6     | 6      | 6      |\n| 2x + y    | 3     | 4     | 0      | 5     | 1     | 6     | 2     | 3      | 0      |\n\nThe table also shows that, if in this sequence of turns Anna changes $y$ by +1 or -2, then $\\Delta$ is increased by 1. Also, if Anna changes $y$ by +2 or -1, then the first time this happens $\\Delta$ is increased by 2. (This also holds if her turn is (0, -1) or (-2, -1), which are not shown in the table.)\n\nSince Anna wins by placing her token at $(k, 2019)$ we must have $m \\le k - 1$ and $k \\le 2018$. So at that exact moment we have:\n$$\n\\Delta = 3m - (2k + 2019) = k - 2022 \\le -4.\n$$\nSo in her last turn she must have decreased $\\Delta$ by at least 4. So her last turn must have been (1, 2) or (2, 1), which give a decrease of 4 and 5 respectively. (It could not be (3, 0) because then she must have already won. Also she could not have done just one or two steps in her last turn since this is not enough for the required decrease in $\\Delta$.)\n\nIf her last turn was (1, 2), then just before doing it we had $y = 2017$ and $\\Delta = 0$. This means that in one of her turns the total change in $y$ was not $0 \\mod 3$. However, in that case we have seen that $\\Delta > 0$, a contradiction.\n\nIf her last turn was (2, 1), then just before doing it we had $y = 2018$ and $\\Delta = 0$ or $\\Delta = 1$. So she must have made at least two turns with the change of $y$ being +1 or -2 or at least one step with the change of $y$ being +2 or -1. In both cases, consulting the table, we get an increase of at least 2 in $\\Delta$, a contradiction.\n\n**Note 1:** If Anna is allowed to make **at most** three steps at each turn, then she actually has a winning strategy.\n\n**Note 2:** If 2019 is replaced by $N > 1$, then Bob has a winning strategy if and only if $3 \\mid N$. □",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31302,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $5 \\times 100$ table is divided into $500$ unit square cells, where $n$ of them are coloured black and the rest are coloured white. Two unit square cells are called *adjacent* if they share a common side. Each of the unit square cells has at most two adjacent black unit square cells. Find the largest possible value of $n$.",
    "options": [],
    "answer": "302",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f : (0, +\\infty) \\to \\mathbb{R}$ satisfying\n$$\n(y^2 + 1)f(x) - y f(xy) = y f\\left(\\frac{x}{y}\\right), \\text{ for all } x, y > 0.\n$$\n(IMO 2018 shortlist, modified)",
    "options": [],
    "answer": "All solutions are f(x) = a x + b/x for real constants a, b.",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31304,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the table are written the positive integers $1, 2, 3, \\ldots, 2018$. John and Mary have the possibility to make the following move:\nThey select two of the written numbers in the table, let $\\alpha, \\beta$ and they replay them with the numbers $5\\alpha - 2\\beta$ and $3\\alpha - 4\\beta$.\n\nJohn asserts that after a finite number of such moves is possible to have in the table the numbers: $3, 6, 9, \\ldots, 6054$. Mary answer that this is not possible. Who is right?",
    "options": [],
    "answer": "Mary is right.",
    "solution": "We observe that after a move the sum of the numbers in the table have a change equal to the difference:\n$$\n(5\\alpha - 2\\beta) + (3\\alpha - 4\\beta) - (\\alpha + \\beta) = 7(\\alpha - \\beta)\n$$\nTherefore we conclude that after every application of a move the difference of the sum $S_{\\text{new}}$ minus the sum $S_{\\text{initial}}$ is a multiple of $7$, that is\n$$\nS_{\\text{new}} - S_{\\text{initial}} = \\text{mult. }7.\n$$\nSo, the sums $S_{\\text{new}}$ and $S_{\\text{initial}}$ when divided by $7$ give the same remainder. Since\n$$\nS_{\\text{initial}} = 1 + 2 + \\dots + 2018 = 1009 \\cdot 2019 \\equiv 1 \\cdot 3 \\pmod{7} \\equiv 3 \\pmod{7}.\n$$\nIn case we find the numbers $3,6,9,\\ldots,6054$, then their sum will be\n$$\nS_{\\text{new}} = 3 + 6 + \\dots + 6054 = 3 \\cdot S_{\\text{initial}} \\equiv 3 \\cdot 3 \\pmod{7} \\equiv 2 \\pmod{7}.\n$$\nTherefore is not possible to find the numbers $3, 6, 9, \\ldots, 6054$ in the table, and so Mary is right.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31305,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider a $8 \\times 8$ chess table with all $64$ unit squares white. We color $12$ unit squares arbitrarily black. Prove that we can find four rows and four columns containing the $12$ black unit squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x_1, x_2, \\ldots, x_8$ be the number of black squares in the rows, such that $x_1 \\ge x_2 \\ge \\ldots \\ge x_8$. This means that if, for example, the fourth line has the maximum number of black squares, then these are $x_1$. From the problem condition we have:\n\n$$\nx_1 + x_2 + \\ldots + x_8 = 12. \\quad (1)\n$$\n\nWe observe that it is impossible $x_1 = x_2 = \\ldots = x_8$, since then $8x_1 = 12$, a contradiction. This means that:\n\n$$\nx_1 + x_2 + x_3 + x_4 > x_5 + x_6 + x_7 + x_8.\n$$\n\nWe will prove that it is impossible to have:\n$$\nx_1 + x_2 + x_3 + x_4 = x_5 + x_6 + x_7 + x_8 + 1,\n$$\nSince then (1) gives $2(x_1 + x_2 + x_3 + x_4) - 1 = 12$, a contradiction, since the left hand side is odd, while the right hand side is even. Moreover, we cannot have:\n$$\nx_1 + x_2 + x_3 + x_4 = x_5 + x_6 + x_7 + x_8 + 2. \\qquad (2)\n$$\nIndeed, if (2) was true, then from (1) we will have that $x_1 + x_2 + x_3 + x_4 = 7$ and\n$$\nx_5 + x_6 + x_7 + x_8 = 5, \\qquad (3)\n$$\nso that $4x_4 \\le 7 \\Rightarrow x_4 \\le 1$. From the ordering condition we have $x_8 \\le x_7 \\le x_6 \\le x_5 \\le x_4 \\le 1$, so $x_8 + x_7 + x_6 + x_5 \\le 4$, which is a contradiction according to (3).\nTherefore:\n$$\nx_1 + x_2 + x_3 + x_4 \\ge x_5 + x_6 + x_7 + x_8 + 3,\n$$\nso from (1) we have:\n$$\n2(x_1 + x_2 + x_3 + x_4) - 3 \\ge 12 \\Rightarrow x_1 + x_2 + x_3 + x_4 \\ge 8,\n$$\nthus, if we choose the $4$ lines with the maximum number of black squares, then these contain at least $8$ black squares. Then, we are left with at most $4$ black squares and for them we choose the $4$ columns in which they are located. As a conclusion, with $4$ lines and $4$ columns we can cover $12$ black squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying the equation\n$$\nP((Q(x))^3) = xP(x)(Q(x))^3.\n$$",
    "options": [],
    "answer": "P(x) = k x^2 with k ≠ 0 and Q(x) = x",
    "solution": "Let: $\\deg P(x) = m \\ge 1$, $\\deg Q(x) = n \\ge 1$. By taking the degrees of the two members of the given equation we get the equality:\n$$\n\\begin{aligned}\n3mn = 1 + m + 3n &\\Leftrightarrow (m-1)(3n-1) = 2 \\\\\n&\\Leftrightarrow m-1=1, 3n-1=2 \\quad \\text{or } m-1=2, 3n-1=1\\; (\\text{impossible in } \\mathbb{N}^*) \\\\\n&\\Leftrightarrow m=2, n=1.\n\\end{aligned}\n$$\nLet $P(x) = ax^2 + bx + c$, $a, b, c \\in \\mathbb{R}$, $a \\neq 0$, $Q(x) = dx + e$, $d, e \\in \\mathbb{R}$, $d \\neq 0$.\nThen, from the given relation we have:\n$$\na((Q(x))^3)^2 + b(Q(x))^3 + c = xP(x)(Q(x))^3, \\quad (1)\n$$\nFrom which we conclude that $Q(x)|_{c=0} \\Rightarrow c=0$. Then from (1) we have:\n$$\n\\begin{aligned}\n& a((Q(x))^3)^2 + b(Q(x))^3 - xP(x)(Q(x))^3 = 0 \\\\\n\\Leftrightarrow & (Q(x))^3 \\left(a(Q(x))^3 + b - xP(x)\\right) \\\\\n\\Leftrightarrow & \\underset{Q(x) \\neq 0}{a(Q(x))^3} + b - xP(x) = 0 \\\\\n& xP(x) = a(Q(x))^3 + b\n\\end{aligned} \\quad (2)\n$$\nThen from relation (2) we have:\n$$\n\\begin{aligned}\n& x(ax^2 + bx) = a(dx + e)^3 + b \\\\\n\\Leftrightarrow & ax^3 + bx^2 = ad^3 x^3 + 3ad^2 e x^2 + 3ad e^2 x + a e^3 + b \\\\\n\\Leftrightarrow & a = ad^3,\\; b = 3ad^2 e,\\; 3ad e^2 = 0,\\; a e^3 + b = 0 \\\\\n& \\underset{a,d \\neq 0}{\\Leftrightarrow} d = 1,\\; e = 0,\\; b = 0,\\; a \\in \\mathbb{R}.\n\\end{aligned}\n$$\nHence: $P(x) = a x^2$, $Q(x) = x$, $a \\in \\mathbb{R}$ and it is easy to verify that they satisfy our problem.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in the real numbers the inequality:\n$$\n\\frac{(x+2)^4}{x^3} - \\frac{(x+2)^2}{2x} \\geq -\\frac{x}{16}.\n$$",
    "options": [],
    "answer": "x > 0 or x = -4/3 or x = -4",
    "solution": "For $x \\neq 0$ we have:\n$$\n\\begin{align*}\n\\frac{(x+2)^4}{x^3} - \\frac{(x+2)^2}{2x} &\\geq -\\frac{x}{16} \n\\Leftrightarrow \\frac{(x+2)^4}{x^3} - \\frac{(x+2)^2}{2x} + \\frac{x}{16} \\geq 0 \\\\\n&\\Leftrightarrow \\frac{16(x+2)^4 - 8(x+2)^2 x^2 + x^4}{16x^3} \\geq 0 \n\\Leftrightarrow \\frac{(4(x+2)^2 - x^2)^2}{16x^3} \\geq 0 \\\\\n&\\Leftrightarrow x(4(x+2)^2 - x^2)^2 \\geq 0,\\ x \\neq 0 \n\\Leftrightarrow x \\geq 0,\\ x \\neq 0 \\text{ or } 4(x+2)^2 - x^2 = 0 \\\\\n&\\Leftrightarrow x > 0 \\text{ or } (2x+4+x)(2x+4-x) = 0 \n\\Leftrightarrow x > 0 \\text{ or } (3x+4)(x+4) = 0 \\\\\n&\\Leftrightarrow x > 0 \\text{ or } x = -\\frac{4}{3} \\text{ or } x = -4.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be an acute angled triangle with $AB < A\\Gamma$. Let $\\Delta$ be the midpoint of the side $B\\Gamma$ and $BE$, $\\Gamma Z$ altitudes of the triangle $AB\\Gamma$. The line $ZE$ meets the line $B\\Gamma$ at point $\\Theta$.\n\na. Find the angles of the triangle $Z\\Delta E$ with respect to the angle $\\hat{A}$ of $AB\\Gamma$.\n\nb. Find the angle $B\\Theta Z$ with respect to the angles $\\hat{B}$ and $\\hat{\\Gamma}$ of $AB\\Gamma$.",
    "options": [],
    "answer": "Angles of triangle ZΔE: at Z equals A, at Δ equals 180 − 2A, at E equals A. Angle BΘZ equals B − Γ.",
    "solution": "a.\nThe triangle $BZ\\Gamma$ is right angled at $Z$ and $Z\\Delta$ is median.\nHence $Z\\Delta = \\frac{B\\Gamma}{2} = B\\Delta$. From the isosceles triangle $B\\Delta Z$ it follows that:\n$$\n\\hat{\\Delta}_1 = B\\hat{\\Delta}Z = 180^\\circ - 2\\hat{B}. \\qquad (1)\n$$\nSimilarly we get $\\Delta E = \\frac{B\\Gamma}{2} = \\Gamma\\Delta$ and from the isosceles triangle we get:\n$$\n\\hat{\\Delta}_2 = \\Gamma \\hat{\\Delta} E = 180^\\circ - 2\\hat{\\Gamma}. \\qquad (2)\n$$\n\nBy summing (1) and (2) we find\n$$\n\\begin{aligned}\n\\hat{\\Delta}_1 + \\hat{\\Delta}_2 &= (180^\\circ - 2\\hat{B}) + (180^\\circ - 2\\hat{\\Gamma}) = 360^\\circ - 2(\\hat{B} + \\hat{\\Gamma}) \\\\\n&= 360^\\circ - 2(180^\\circ - \\hat{A}) = 2\\hat{A}\n\\end{aligned}\n$$\nthat is.\n$$\nZ\\hat{\\Delta}E = 180^\\circ - (\\hat{\\Delta}_1 + \\hat{\\Delta}_2) = 180^\\circ - 2\\hat{A}.\n$$\nMoreover, since $\\Delta ZE$ isosceles ($\\Delta Z = \\Delta E = \\frac{B\\Gamma}{2}$), we have:\n$$\n\\Delta \\hat{Z}E = \\Delta \\hat{E}Z = \\frac{180^\\circ - (180^\\circ - 2\\hat{A})}{2} = \\hat{A}.\n$$\n![](attached_image_1.png)\nfigure 1\n\nb.\nFrom the triangle $B\\Theta Z$ we have: $\\hat{B} = B\\hat{\\Theta}Z + B\\hat{Z}\\Theta$. (3)\nand $B\\hat{Z}\\Theta = A\\hat{Z}E$ (4). Moreover, from the cyclic quadrilateral $BZ\\Gamma E$\n($BZ\\Gamma = B\\hat{E}\\Gamma = 90^\\circ$) we have: $A\\hat{Z}E = \\hat{\\Gamma}$ (5)\nTherefore: $B\\hat{\\Theta}Z = \\hat{B} - \\hat{\\Gamma}$\n$$\nB\\hat{\\Theta}Z = \\hat{B} - \\hat{\\Gamma}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31309,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider the segment $AB$ and the point $\\Gamma$ on it such that $AB = 3 \\cdot AG$. We construct the parallelogram $A\\Gamma\\Delta E$ with $A\\Gamma = \\Delta E = \\Gamma E > AE$. We consider point $Z$ on the side $A\\Gamma$ such that $A\\hat{E}Z = A\\Gamma E = \\omega$. Prove that the line perpendicular from $B$ to the line $E\\Gamma$ and the line from the point $\\Delta$ perpendicular to the line $AB$ meet on a point, say $K$, which lies on the line $EZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $A\\Gamma E = \\omega$, from the right angled triangle $\\Gamma\\Theta B$ we have:\n$$\n\\Gamma \\hat{B} \\Theta = 90^{\\circ} - B \\hat{\\Gamma} \\Theta = 90^{\\circ} - \\omega, \\qquad (1)\n$$\nsince $B\\Gamma\\Theta = A\\Gamma E = \\omega$, and moreover $B\\hat{E}K = 90^{\\circ}$, and so\n$$\n\\Gamma \\hat{B} \\Theta + B \\hat{H} K = 90^{\\circ} - \\omega + 90^{\\circ} = 180^{\\circ} - \\omega < 180^{\\circ},\n$$\nWhich gives that the lines $B\\Theta$ and $\\Delta H$ intersect at a point, say $K$. Therefore, it is enough to prove that the points $Z$, $E$ and $K$ are collinear.\nThe quadrilateral $E\\Delta\\Gamma Z$ is isosceles trapezium, because $E\\Delta \\parallel Z\\Gamma$ and $ZE = AE = \\Gamma\\Delta$. Indeed, the triangles $AEZ$ and $A\\Gamma E$ have the angle $A$ in common and $A\\hat{E}Z = A\\Gamma E = \\omega$, therefore they have equal angles. Therefore the triangle $AZE$ is isosceles with $A\\hat{E}Z = E\\hat{A}Z$. It follows that: $AE = ZE$.\n![](attached_image_1.png)\nfigure 2\nFrom the isosceles trapezium $E\\Delta\\Gamma Z$ we have that:\n$$\nE \\hat{Z} \\Gamma = Z \\hat{\\Gamma} \\Delta = 180^{\\circ} - \\Gamma \\hat{A} E = 180^{\\circ} - \\left( \\frac{180^{\\circ} - \\omega}{2} \\right) = 90^{\\circ} + \\frac{\\omega}{2}. \\qquad (2)\n$$\nFrom the orthogonal triangle $\\Delta H Z$ we have\n$$\nZ\\hat{\\Delta}H = 90^{\\circ} - \\Delta\\hat{Z}H = 90^{\\circ} - \\omega. \\qquad (3)\n$$\nFrom (1) and (3) it follows that the quadrilateral $B\\Delta ZK$ is cyclic.\nIf $\\Lambda$ is the midpoint of the segment $\\Gamma B$ we observe that the quadrilateral $E\\Delta\\Lambda\\Gamma$ has $E\\Delta \\parallel \\Gamma\\Lambda$, $\\Gamma E = \\Delta E = A\\Gamma = \\Gamma\\Lambda$, since $A\\Gamma = \\frac{1}{3}AB$ and $\\Lambda$ is the midpoint of $\\Gamma B$. Hence $E\\Delta\\Lambda\\Gamma$ is rhombus. Hence, we have $\\Lambda\\Gamma\\Delta = \\omega$ and moreover $\\Delta\\Lambda = \\Lambda\\Gamma = \\Lambda B = \\frac{\\Gamma B}{2}$. It means that the triangle $\\Gamma\\Delta B$ is orthogonal at $\\Delta$ and from the isosceles triangle $\\Lambda\\Delta B$ we have : $\\Delta\\hat{B}\\Lambda = \\frac{\\omega}{2}$.\nFrom the cyclic quadrilateral $B\\Delta ZK$ we have:\n$$\nZ\\hat{K}\\Delta = \\Delta\\hat{B}Z = \\Delta\\hat{B}\\Lambda = \\frac{\\omega}{2}, \\qquad (4)\n$$\nοπότε από το ορθογώνιο τρίγωνο $HZK$ και τη σχέση (4) έπεται ότι:\n$$\n\\Gamma\\hat{Z}K = H\\hat{Z}K = 90^{\\circ} - \\frac{\\omega}{2} \\qquad (5)\n$$\nFrom (2) and (5) it follows that:\n$$\nE\\hat{Z}\\Gamma + \\Gamma\\hat{Z}K = 90^{\\circ} + \\frac{\\omega}{2} + 90^{\\circ} - \\frac{\\omega}{2} = 180^{\\circ},\n$$\nand hence the points $A$, $E$ and $K$ are collinear.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the positive integer $v$ for which there exist triads $(\\alpha, \\beta, \\gamma)$ of positive integers satisfying the equation\n$$\n\\alpha + \\beta + \\gamma = v\\alpha\\beta\\gamma. \\qquad (E)\n$$\nFor these values find all solutions of the equation (E).",
    "options": [],
    "answer": "Valid v are 1, 2, and 3. For v = 3: the only solution is (1, 1, 1). For v = 2: all permutations of (2, 1, 1). For v = 1: all permutations of (3, 2, 1).",
    "solution": "Since the equation is symmetric with respect to $\\alpha, \\beta, \\gamma$ we suppose that $\\alpha \\ge \\beta \\ge \\gamma$. Then we have:\n$$\n\\alpha \\le \\alpha + \\beta + \\gamma \\le 3\\alpha \\Leftrightarrow \\alpha \\le v\\alpha\\beta\\gamma \\le 3\\alpha \\Rightarrow 1 \\le v\\beta\\gamma \\le 3.\n$$\n\nWe distinguish the following cases:\n• $v > 3$. Then $v\\beta\\gamma > 3$, absurd.\n• $v = 3$. Then $1 \\le 3\\beta\\gamma \\le 3 \\Rightarrow \\beta\\gamma = 1 \\Rightarrow \\beta = \\gamma = 1$, and hence\n$$ \\alpha + 2 = 3\\alpha \\Leftrightarrow \\alpha = 1. $$\nTherefore we get the solution: $(\\alpha, \\beta, \\gamma) = (1,1,1)$.\n• $v = 2$. Then $1 \\le 2\\beta\\gamma \\le 3 \\Rightarrow \\beta\\gamma = 1 \\Rightarrow \\beta = \\gamma = 1$, and\n$$ \\alpha + 2 = 2\\alpha \\Leftrightarrow \\alpha = 2. $$\nTherefore we get the solution $(\\alpha, \\beta, \\gamma) = (2,1,1)$ and using symmetry we obtain the solutions $(\\alpha, \\beta, \\gamma) = (1,2,1)$ and $(\\alpha, \\beta, \\gamma) = (1,1,2)$.\n• $v = 1$. Then $1 \\le \\beta\\gamma \\le 3 \\Rightarrow \\beta\\gamma \\in \\{1,2,3\\}$.\nIf $\\beta\\gamma = 1$, then $\\beta = \\gamma = 1$ and $\\alpha + 2 = 1$, impossible.\nIf $\\beta\\gamma = 2$, then $\\beta = 2, \\gamma = 1$ and $\\alpha + 3 = 2\\alpha \\Leftrightarrow \\alpha = 3$.\nTherefore $(\\alpha, \\beta, \\gamma) = (3,2,1)$ and by symmetry\n$$\n\\begin{align*}\n(\\alpha, \\beta, \\gamma) &= (2,1,3), \\\\\n(\\alpha, \\beta, \\gamma) &= (3,2,1), \\\\\n(\\alpha, \\beta, \\gamma) &= (3,1,2), \\\\\n(\\alpha, \\beta, \\gamma) &= (1,2,3), \\\\\n(\\alpha, \\beta, \\gamma) &= (2,3,1).\n\\end{align*}\n$$\nIf $\\beta\\gamma = 3$, then $\\beta = 3, \\gamma = 1$ and $\\alpha + 4 = 3\\alpha \\Leftrightarrow \\alpha = 2$. (rejected, $\\alpha < \\beta$).\nHence $v \\notin \\{1,2,3\\}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31311,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the blackboard are written in a line the numbers from $1$ to $2030$ in increasing turn. We have the possibility of the “movement” $K$: *We select two numbers $\\alpha$, $\\beta$ from the blackboard written in successive positions and we substitute the pair ($\\alpha$, $\\beta$) with the number $(\\alpha - \\beta)^{2020}$.*\n\nWe perform the movement as many times as we need in order to have on the blackboard only one number. Examine, if this number can be:\n\n(a) $2020^{2020}$,\n\n(b) $2021^{2020}$.",
    "options": [],
    "answer": "(a) No; (b) No",
    "solution": "(a) First we observe that:\nAt every movement $K$, the numbers $\\alpha + \\beta$ and $(\\alpha - \\beta)^{2020}$ are congruent modulo $2$, that is, both are even or both are odd. Therefore, after the substitution of $\\alpha, \\beta$ with the number $(\\alpha - \\beta)^{2020}$ the parity of the sum of the numbers on the blackboard remains invariant. Since\n\n$$\nS_{2030} = \\frac{2030 \\cdot 2031}{2} = 1015 \\cdot 2031\n$$\nis odd, it follows that after the last movement the number remaining on the blackboard is odd and therefore the answer in the first question is negative.\n\n(b) Let after performing the movement enough number of times we have on the blackboard two numbers $x, y$. Then the only case in order no one of them is not a power of $2$, is not to have taken part in any performed movement. This can happen only with the numbers $1$ and $2030$. Since $1$ is $1^{2020}$, it is enough to check $2030$. If $x = a^{2020}$ and $y = 2030$, then we must have $(a^{2020} - 2030)^{2020} = 2021^{2020}$. For $a = 1$ there is no solution and for $a \\ge 2$ we get $a^{2020} = 4051$, impossible.\n\nIn any other case we will have from the procedure that $x = (a_1 - b_1)^{2020} = m^{2020}$ and $y = (a_2 - b_2)^{2020} = n^{2020}$, for some $m$ and $n$. By performing the last movement, remains on the blackboard the number $(m^{2020} - n^{2020})^{2020}$. We have to check if it is possible to be valid the following equality:\n\n$$\n(m^{2020} - n^{2020})^{2020} = 2021^{2020}.\n$$\n\nIf $m = n$, it is not valid. We suppose wlog $m > n$. Then we must check if\n\n$$\nm^{2020} - n^{2020} = 2021. \\qquad (1)\n$$\n\nThe least possible value of $m^{2020} - n^{2020}$ is $1$, for $m = 1, n = 0$. From (1) we have that $m > n$, and hence $m \\ge n + 1$. Therefore $m^{2020} - n^{2020} \\ge (n+1)^{2020} - n^{2020}$. By developing the last expression we observe that all coefficients of $n$ have positive sign and so it is increasing. Hence $m^{2020} - n^{2020} \\ge (n+1)^{2020} - n^{2020} \\ge 2^{2020} - 1 > 2^{11} - 1 = 2047$ and therefore it cannot be equal to $2021$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all values of the positive integer $\\kappa$ satisfying the following property:\nThere do not exist positive integers $\\alpha, \\beta$ such that the number\n$$\nA(\\kappa, \\alpha, \\beta) = \\frac{\\alpha + \\beta}{\\alpha^2 + \\kappa^2 \\beta^2 - \\kappa^2 \\alpha \\beta}\n$$\nis a composite positive integer.",
    "options": [],
    "answer": "1",
    "solution": "We fix $\\kappa > 1$. The idea for the solution of the problem is to find $\\alpha, \\beta$ with respect to $\\kappa$ such that the denominator of the fraction is equal to $1$. In other words, we seek polynomials $\\alpha = P(\\kappa)$, $\\beta = Q(\\kappa)$ such that\n$$\nP(\\kappa)^2 + \\kappa^2 Q(\\kappa)^2 - \\kappa^2 P(\\kappa)Q(\\kappa) = 1. \\\\\nP(\\kappa)^2 = \\kappa^2 Q(\\kappa)(P(\\kappa) - Q(\\kappa)) + 1 \\quad (1)\n$$\nIf $m = \\deg P(\\kappa), n = \\deg Q(\\kappa)$ with $m > n \\ge 0$, then from (1) it follows that\n$$\n2m = 2 + m + n \\Leftrightarrow m = n + 2.\n$$\nWe will deal with the easiest case, by seeking polynomials of degrees $2$ and $0$, respectively, which satisfy relation (1).\nFinally, we find that $P(\\kappa) = \\kappa^2 - 1$, $Q(\\kappa) = 1$ satisfy relation (1). We choose $\\alpha = \\kappa^2 - 1$, $\\beta = 1$ and then the expression is equal to $\\frac{\\kappa^2 - 1 + 1}{1} = \\kappa^2$, which is composite for all $\\kappa > 1$. Therefore, all values of $\\kappa > 1$ are not solutions.\n\nFor $\\kappa = 1$, we will prove that: $0 < A(\\alpha, \\beta, \\kappa) = \\frac{\\alpha + \\beta}{\\alpha^2 + \\beta^2 - \\alpha\\beta} \\le 2$, which means that the rational number $A(\\alpha, \\beta, 1)$ cannot be a composite positive integer for all $\\alpha, \\beta$. Indeed, we have\n$$\n\\alpha + \\beta > 0 \\quad \\text{and} \\quad \\alpha^2 - \\alpha\\beta + \\beta^2 = \\left(\\alpha - \\frac{\\beta}{2}\\right)^2 + \\frac{3\\beta^2}{4} > 0, \\text{ and}\n$$\n$$\n\\frac{\\alpha + \\beta}{\\alpha^2 + \\beta^2 - \\alpha\\beta} \\le 2 \\Leftrightarrow 2\\alpha^2 + 2\\beta^2 - 2\\alpha\\beta \\ge \\alpha + \\beta \\Leftrightarrow (\\alpha - \\beta)^2 + \\alpha^2 - \\alpha + \\beta^2 - \\beta \\ge 0 \\\\\n\\Leftrightarrow (\\alpha - \\beta)^2 + \\alpha(\\alpha - 1) + \\beta(\\beta - 1) \\ge 0, \\text{ which holds.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31313,
    "subject": "Mathematics (Multi-modal)",
    "question": "We write $99$ circles in a line and in their interior we write the numbers from $1$ to $99$, as follows:\n![](attached_image_1.png)\nWe color each circle with one of the two colors: red (R) and green (G). We say that a coloring is «good», if the following happens:\n*The number of red circles in the first part of numbers from $1$ to $50$ is greater than the number of the red circles existing in the second part from number $51$ to $99$.*\n\na. Determine how many different colorings can be constructed.\n\nb. Determine how many different “good” colorings can be constructed.\n\n(Note: Two colorings are different, if they have different colors at least in one circle.)",
    "options": [],
    "answer": "a) 2^99; b) 2^98",
    "solution": "(a) Each circle can be colored with two different colors independently of the coloring of the other circles. Therefore, according to the multiplicative principle, the different colorings are\n$$\n\\underbrace{2 \\cdot 2 \\cdot 2 \\dotsm 2}_{99} = 2^{99}.\n$$\n\n(b) We consider the coloring $X: 1\\ 2\\ 3\\ \\dots\\ 50\\ \\underbrace{51\\ \\dotsm\\ 98\\ 99}_{x}$, where $x$ is the number of red circles among $1$ to $50$, and $y$ is the number of red circles among $51$ to $99$. The coloring $X$ is «good» if $x > y$, whereas the coloring $X$ is *no good* if $x \\le y$.\n\nLet $A$ be the set of «good» colorings and $B$ the set of “no good” colorings. We will prove that each element of the set $A$ corresponds to an element of $B$ and vice versa.\n\nIndeed, if $X: 1\\ 2\\ 3\\ \\dots\\ 50\\ \\underbrace{51\\ \\dotsm\\ 98\\ 99}_{x}$ is in $A$, then $x > y$. By changing the color of each circle, we find the coloring\n$$\nY: \\underbrace{1\\ 2\\ 3\\ \\dots\\ 50}_{50-x}\\ \\underbrace{51\\ \\dotsm\\ 98\\ 99}_{49-y},\n$$\nwhich belongs to $B$, because\n$$\nx > y \\Rightarrow -x < -y \\Rightarrow 49 - x < 49 - y \\Rightarrow 50 - x \\le 49 - y.\n$$\n\nConversely, if $X: 1\\ 2\\ 3\\ \\dots\\ 50\\ \\underbrace{51\\ \\dotsm\\ 98\\ 99}_{x}$ is in $B$, then $x \\le y$. By changing the color of each circle, we find the coloring\n$$\nY: \\underbrace{1\\ 2\\ 3\\ \\dots\\ 50}_{50-x}\\ \\underbrace{51\\ \\dotsm\\ 98\\ 99}_{49-y},\n$$\nwhich belongs to the set $A$ because\n$$\nx \\le y \\Rightarrow -x \\ge -y \\Rightarrow 49 - x \\ge 49 - y \\Rightarrow 50 - x > 49 - y.\n$$\n\nTherefore, between the sets $A$ and $B$ there exists a $1$-$1$ correspondence and so the two sets have the same number of elements, that is, each of them has\n$$\n\\frac{2^{99}}{2} = 2^{98} \\text{ elements.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31314,
    "subject": "Mathematics (Multi-modal)",
    "question": "(A) Determine the value of the real number $k$ for which the polynomial $P(x) = x^3 - kx + 2$ has $2$ as root. For that value of $k$ write the polynomial $P(x) = x^3 - kx + 2$ as a product of two polynomials with integer coefficients.\n\n(B) The positive real numbers $a, b$ satisfy the equation $2a + b + \\frac{4}{ab} = 10$. Find the maximal value of $a$.",
    "options": [],
    "answer": "k = 5; P(x) = (x − 2)(x^2 + 2x − 1); maximal a = 4",
    "solution": "(A) Number $2$ is a root of polynomial $P(x) = x^3 - kx + 2$, if and only if $P(2) = 0 \\Leftrightarrow 2^3 - 2k + 2 = 0 \\Leftrightarrow k = 5$.\nFor $k = 5$ we get:\n$$\n\\begin{aligned}\nP(x) &= x^3 - 5x + 2 = x^3 - 4x - x + 2 \\\\\n&= x(x - 2)(x + 2) - (x - 2) = (x - 2)(x^2 + 2x - 1).\n\\end{aligned}\n$$\n\n(B) From the inequality arithmetic and geometric mean we get:\n$$\nb + \\frac{4}{ab} \\ge 2 \\sqrt{b \\cdot \\frac{4}{ab}} = \\frac{4}{\\sqrt{a}}\n$$\nHence, $2a + \\frac{4}{\\sqrt{a}} \\le 10$. Putting $\\sqrt{a} = x$, we get:\n$$\nx^2 + \\frac{2}{x} \\le 5 \\Leftrightarrow x^3 - 5x + 2 \\le 0.\n$$\nBy using the first question we have: $(x-2)(x^2+2x-1) \\le 0$. This is not valid for $x > 2$. Hence $x \\le 2$, and so $a \\le 4$. In fact this is the maximal value of $a$, because for $a = 4$, we get $b = 1$, that is, there exist $a, b$ satisfying the equation.\n\n\nFrom the inequality arithmetic and geometric mean we get:\n$$\n2a + b + \\frac{4}{ab} = \\frac{a}{4} + \\frac{a}{4} + \\cdots + \\frac{a}{4} + b + \\frac{4}{ab} \\ge 10 \\sqrt[10]{\\frac{a^8}{4^8} \\cdot b \\cdot \\frac{4}{ab}}\n$$\nHence $\\sqrt[10]{\\frac{a^8}{4^8} \\cdot b \\cdot \\frac{4}{ab}} \\le 1 \\Rightarrow a^7 \\le 4^7 \\Rightarrow a \\le 4$. In fact this is the maximal value of $a$, because for $a = 4$, we get $b = 1$, that is, there exist $a, b$ satisfying the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be an isosceles triangle and a point $\\Delta$ in its interior such that $\\Delta\\hat{B}\\Gamma = 30^\\circ$, $\\Delta\\hat{B}A = 50^\\circ$, $B\\hat{\\Gamma}\\Delta = 55^\\circ$.\n\na. Prove that $\\hat{B} = \\hat{\\Gamma} = 80^\\circ$\n\nb. Find the measure of the angle $\\Delta\\hat{A}\\Gamma$.",
    "options": [],
    "answer": "∠B = ∠Γ = 80°, ∠ΔAΓ = 5°",
    "solution": "a.\nWe have $\\hat{B} = 50^\\circ + 30^\\circ = 80^\\circ$.\nSuppose that $\\hat{A} = \\hat{B} = 80^\\circ$. Then\n$$\n\\hat{A} + \\hat{B} + \\hat{\\Gamma} = 80^\\circ + 80^\\circ + \\hat{\\Gamma} > 160^\\circ + 55^\\circ = 215^\\circ,\n$$\nabsurd.\nIf $A\\Gamma = \\frac{180^\\circ - 80^\\circ}{2} - 50^\\circ$, then $B\\hat{\\Gamma}\\Delta = 55^\\circ < \\hat{\\Gamma} = 50^\\circ$, absurd.\nHence we have: $\\Delta\\hat{\\Gamma}A = 80^\\circ - 55^\\circ = 25^\\circ$ (1).\n\nb.\nLet $AZ$ be the bisector of the angle $\\hat{A}$. Then $AZ$ is height and median of the triangle $AB\\Gamma$. Let $AZ$ meet line $BD$ at point $E$. Since in the triangle $B\\Gamma\\Delta$ we have $\\Gamma\\hat{B}\\Delta < B\\hat{\\Gamma}\\Delta$, it follows that $\\Delta\\Gamma < \\Delta B$. Hence $\\Delta$ lies in the semi-plane with respect to $AZ$ containing point $\\Gamma$. It means that $E$ lies between points $B$ and $\\Delta$.\nSince $EB = E\\Gamma$, it follows that $E\\hat{\\Gamma}B = E\\hat{B}\\Gamma = 30^\\circ$ and hence\n![](attached_image_1.png)\n\n$$\nE\\hat{\\Gamma}\\Delta = 55^\\circ - 30^\\circ = 25^\\circ = \\Delta\\hat{\\Gamma}A. \\qquad (2)\n$$\nHence $\\Gamma\\Delta$ bisects the angle $E\\hat{\\Gamma}A$ of the triangle $AE\\Gamma$. Moreover, for the external angles $\\Delta\\hat{E}\\Gamma$ and $\\Delta\\hat{E}A$ of the triangles $EB\\Gamma$ and $EBA$, respectively, we have: $\\Delta\\hat{E}\\Gamma = 30^\\circ + 30^\\circ = 60^\\circ$ and $\\Delta\\hat{E}A = E\\hat{B}A + \\frac{\\hat{A}}{2} = 50^\\circ + 10^\\circ = 60^\\circ$.\nHence $\\Delta\\hat{E}\\Gamma = \\Delta\\hat{E}A = 60^\\circ$ and so $E\\Delta$ bisects the angle $A\\hat{E}\\Gamma$ of the triangle $AE\\Gamma$.\nHence $\\Delta$ is the incenter of the triangle $AE\\Gamma$ and $\\Delta\\hat{A}\\Gamma = \\frac{E\\hat{A}\\Gamma}{2} = \\frac{10^\\circ}{2} = 5^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31316,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a blackboard we write in a line $n$ numbers, $n \\ge 40$, each being equal to $1$ or $-1$, such that:\n(i) The sum of every $40$ successive numbers is equal to $0$.\n(ii) The sum of every $42$ successive numbers is not equal to $0$.\n\nWe denote by $\\Sigma_n$ the sum of the $n$ numbers on the blackboard. Find the greatest possible value of $\\Sigma_n$.",
    "options": [],
    "answer": "20",
    "solution": "Let $a_1, a_2, \\dots, a_n$ be the numbers in a line on the blackboard. Since $a_1 + a_2 + \\dots + a_{40} = 0$, it follows that half of the numbers $a_1, a_2, \\dots, a_{40}$ are equal to $1$ and the other half are equal to $-1$. Since $a_1 + a_2 + \\dots + a_{40} + a_{41} + a_{42} \\neq 0$ it follows that $a_{41} = a_{42} = a \\in \\{-1, 1\\}$.\n\nSimilarly, $a_2 + a_3 + \\dots + a_{43} \\neq 0$ and hence $a_{42} = a_{43} = a \\in \\{-1, 1\\}$. Proceeding in the same way we conclude that all numbers after $a_{41}$ must be equal to $a$.\n\nTherefore, if $n > 60$, the sum $a_{22} + a_{23} + \\dots + a_{61}$ has $a_{41} = a_{42} = \\dots = a_{61} = a$, and hence $1$ and $-1$ cannot have the same cardinality in the set. It means that $a_{22} + a_{23} + \\dots + a_{61} \\neq 0$, and that contradicts condition (i).\n\nHence the greatest value of $n$ is $60$.\n\nThe maximal value of $\\Sigma_n$ is obtained when we have as many as possible numbers equal to $1$. We remind that between the first $40$ numbers we must have $20$ numbers equal to $-1$. Hence the numbers equal to $1$ can be at most $40$. Therefore $\\Sigma_n \\le 20$. In fact, we observe that $\\max \\Sigma_n = 20$. It happens for\n$$\na_1 = \\dots = a_{20} = 1,\\ a_{21} = \\dots = a_{40} = -1 \\text{ and } a_{41} = \\dots = a_{60} = 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs of nonzero integers $(x, y)$ such that the integer $x^2 + y^2$ is a common divisor of the integers $x^5 + y$ and $y^5 + x$.",
    "options": [],
    "answer": "[(1, 1), (-1, -1), (1, -1), (-1, 1)]",
    "solution": "We have that $x^2 + y^2 \\mid x(x^5 + y)$ as well as $x^2 + y^2 \\mid y(y^5 + x)$ and hence\n$$\nx^2 + y^2 \\mid x(x^5 + y) + y(y^5 + x) \\Rightarrow x^2 + y^2 \\mid x^6 + y^6 + 2xy. \\quad (1)\n$$\nFrom the well-known identity for the sum of cubes we have:\n$$\nx^2 + y^2 \\mid (x^2)^3 + (y^2)^3 \\qquad (2),\n$$\nFrom (1) and (2) we get: $x^2 + y^2 \\mid 2xy \\Rightarrow x^2 + y^2 \\le 2 \\mid xy \\mid \\Leftrightarrow (|x| - |y|)^2 \\le 0$, that is $|x| = |y|$. Therefore we have two cases:\n\na. If $x = y$, then $2x^2 \\mid x^5 + x \\Rightarrow 2x \\mid x^4 + 1 \\Rightarrow x \\mid 1 \\Rightarrow x = 1$ or $x = -1$,\n\nb. If $x = -y$, then $2x^2 \\mid x^5 - x \\Rightarrow 2x \\mid x^4 - 1 \\Rightarrow x \\mid 1 \\Rightarrow x = 1$ or $x = -1$,\n\nHence: $(x, y) \\in \\{(1, 1), (-1, -1), (1, -1), (-1, 1)\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC < BC$. On the ray $BC$ we consider a point $D$ such that $BD = BA$ and on the ray $CB$ we consider a point $E$ such that $CE = CA$. Let $K$ be the circumcenter of the triangle $ADE$ and let $F$, $G$ the intersections of the lines $AD$, $KC$ and $AE$, $KB$, respectively. Prove that the circumcircle of the triangle $KDE$, say $c_1$, the circle with center $F$ and radius $FE$, say $c_2$, and the circle with center $G$ and radius $GD$, say $c_3$, pass from a point on the line $AK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet $M$ the midpoint of $AD$. Then the points $B$, $G$, $K$, $M$ belong to the perpendicular bisector of $AD$, that is they are collinear. Moreover, since $GD = GA$ the circle $c_3$ contains $A$.\n\nSimilarly, if $N$ is the midpoint of $AE$, the points $G$, $F$, $K$, $N$ belong to the perpendicular bisector of $AE$ and hence are collinear. Also, since $FA = FE$, the circle $c_2$ contains $A$.\n\nNext we will prove that the points $G$, $F$ belong to the circumcircle $c_1$ of the triangle $KDE$. From the isosceles triangles $ADG$ and $AFE$, we have\n$$\n\\hat{G}_1 = E\\hat{G}D = 2 \\cdot \\hat{G}AD = 2 \\cdot E\\hat{A}F = \\hat{F}_1 \\quad (1)\n$$\nAlso for the inscribed angle $E\\hat{A}\\Delta$ and the angle $\\hat{K}_1 = G\\hat{K}F$ we have\n$$\n\\hat{K}_1 = 2 \\cdot E\\hat{A}D \\quad (2)\n$$\nFrom (1) and (2) it follows that $\\hat{G}_1 = \\hat{F}_1 = \\hat{K}_1$, and hence the points $D$, $E$, $F$, $G$, $K$\n\nbelong to the circle $c_1$.\nLet $T$ the second point of intersection of the circles $c_2$, $c_3$. We will prove that the points $A$, $K$, $T$ are collinear and that $T$ belongs to the circle of points $D$, $E$, $F$, $G$, $K$.\nIn fact, the common chord $AT$ of the circles $c_2$ and $c_3$ is perpendicular to the line of their centers $FG$ and also the line $AK$ is perpendicular to the line $FG$, since $K$ is the orthocenter of the triangle $AGF$.\nMoreover, since $GA = GD = GT$, we have the angle equalities:\n$$\n\\hat{GAT} = \\hat{GTA} = x, \\qquad (1)\n$$\nand since $K$ is the orthocenter of the triangle $AGF$, we have\n$$\n\\hat{GAT} = \\hat{GFK} = 90^\\circ - \\hat{AGF} \\qquad (2)\n$$\nHence $\\hat{GTK} = \\hat{GTA} = \\hat{GFK}$, and so the points $F$, $G$, $K$, $T$ are cyclic.\n\nSince $BD = BA$ and $KA = KD$, $BK$ is the perpendicular bisector of $AD$, so $GA=GD$. We have $\\overline{KGD} = \\overline{KGA} = 90^\\circ - \\overline{EAD} = \\overline{AFK}$, so the points $K$, $F$, $D$, $G$ are cocyclic. Similarly $\\overline{KFE} = \\overline{AGK}$, therefore the points $K$, $F$, $E$, $G$ are cocyclic.\nFrom the previous discussion we have that $K$, $F$, $E$, $G$ and $D$ are cocyclic.\n\nLet now that $AK$ intersects $c_1$ at the point $H$. Then\n$$\n\\overline{HGD} = \\overline{HKD} = 2 \\cdot \\overline{HAD} \\qquad (5)\n$$\n\nTherefore $G$ is on the perpendicular bisector of $AD$ and moreover (5) holds, so $G$ is the circumcentre of the triangle $AHD$, therefore $c_3$ passes through $H$. Similarly $c_2$ passes through $H$ and we have the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 4$, be a positive integer, which is divided by $4$. We denote by $A_n$ the sum of all odd positive divisors of $n$. We denote by $B_n$ the sum of all even positive divisors of $n$, with the exclusion of $n$. Find the smaller possible value of $f(n) = B_n - 2A_n$. For which values of the positive integer $n$ is obtained this minimal value?",
    "options": [],
    "answer": "Minimum value is 4; achieved exactly for n = 4p where p is prime (including n = 8).",
    "solution": "Let $d_1, \\dots, d_k$ be the odd positive divisors of $n$. Then the numbers $2d_1, \\dots, 2d_k$ are even divisors of $n$. Also each of them is not divisible by $4$, and so none of them can be equal to $n$. Moreover, none of them is equal to $4$. Therefore we have:\n$$\nA_n = d_1 + \\dots + d_k \\text{ and } B_n = 2d_1 + 2d_2 + \\dots + 2d_k + 4.\n$$\nTherefore, we have $B_n - 2A_n \\ge 4$. In fact, the value $4$ is the minimal, because for $n=8$, we have\n$$\nA_n = 1, \\quad B_n = 2+4=6, \\quad B_n - 2A_n = 4.\n$$\nNext we have to find all positive integers $n$ satisfying the equality: $B_n - 2A_n = 4$. The equality is valid when $B_n = 2d_1 + \\dots + 2d_k$. If now $p$ is an odd prime divisor of $n$, then $4p$ is an even divisor of $n$ and it is not included in the sum $2d_1 + \\dots + 2d_k$. Hence, in order to hold $B_n = 2d_1 + \\dots + 2d_k + 4$, it is necessary $n = 4p$. In fact, then\n$$\nB_n - 2A_n = (2 + 4 + 2p) - 2(1 + p) = 4.\n$$\nIf $n$ does not have any odd prime divisor, then it is a power of $2$, that is $n = 2^{k+1}$, and hence $A_n = 1$, $B_n = 2 + 2^2 + 2^3 + \\dots + 2^k = \\frac{2^{k+1}-2}{2-1} = 2^{k+1} - 2$ and $B_n - 2A_n = 2^{k+1} - 4$.\nHence $B_n - 2A_n = 4$, if and only if, $k=2$. Therefore $n = 4p$, where $p$ is prime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31320,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive real numbers $\\alpha, \\beta, \\gamma, \\delta$ satisfy the equality:\n$$\n\\alpha + \\beta\\gamma + \\gamma\\delta + \\delta\\beta + \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} = 18.\n$$\n\nFind the maximal possible value of $\\alpha$.",
    "options": [],
    "answer": "16",
    "solution": "Using the AM-GM inequality we get\n\n$$\n\\begin{align*}\n\\alpha + \\beta\\gamma + \\gamma\\delta + \\delta\\beta + \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} &= \\alpha + \\left( \\beta\\gamma + \\gamma\\delta + \\delta\\beta + \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} \\right) \\\\\n&\\ge \\alpha + 4\\sqrt[4]{\\beta\\gamma \\cdot \\gamma\\delta \\cdot \\delta\\beta \\cdot \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2}} = \\alpha + \\frac{4}{\\sqrt[4]{\\alpha}}\n\\end{align*}\n$$\nBy putting $x = \\sqrt[4]{\\alpha}$ and taking in mind the given equality we have the inequality\n$$\nx^4 + \\frac{4}{x} \\le 18,\\ x > 0 \\Leftrightarrow x^5 - 18x + 4 \\le 0,\\ x > 0. \\quad (1)\n$$\nSince 2 is a root of polynomial $P(x) = x^5 - 18x + 4$, we have the factorization\n$$\nP(x) = x^{5}-18x+4 = (x-2)(x^{4}+2x^{3}+4x^{2}+8x-2),\n$$\nand hence we have the inequality:\n$$\n(x-2)(x^4 + 2x^3 + 4x^2 + 8x - 2) \\le 0. \\quad (2)\n$$\nIf $x > 2$, then $(x-2)(x^4 + 2x^3 + 4x^2 + 8x - 2) > 0$, and so inequality (2) is not true. Therefore, $0 < x \\le 2$.\nWe observe that for $x=2$, we have $\\alpha=16$ and (2) holds as equality. From AM - GM inequality it happens when\n$$\n\\begin{align*}\n\\beta\\gamma = \\gamma\\delta = \\delta\\beta &= \\frac{1}{\\alpha\\beta^2\\gamma^2\\delta^2} \\Leftrightarrow \\beta = \\gamma = \\delta \\quad \\text{and} \\quad \\beta^2 = \\frac{1}{16\\beta^6} \\\\\n\\Leftrightarrow \\beta = \\gamma = \\delta \\quad \\text{and} \\quad \\beta^8 &= \\frac{1}{16} \\Leftrightarrow \\beta = \\gamma = \\delta = \\frac{1}{\\sqrt{2}}.\n\\end{align*}\n$$\nHence the maximal possible value of $\\alpha$ is 16.\n\n\nSecond solution.\n\nFrom the AM – GM inequality we get $a + bc + cd + db + \\frac{1}{ab^2c^2d^2} = \\frac{a}{32} + \\dots + \\frac{a}{32} + bc + cd + db + \\frac{1}{ab^2c^2d^2} \\ge$\n$$\n36^{36} \\sqrt{\\frac{a^{32}}{32^{32}} \\cdot bc \\cdot cd \\cdot db \\cdot \\frac{1}{ab^2c^2d^2}} = 36^{36} \\sqrt{\\frac{a^{31}}{32^{32}}}\n$$\nHence, $\\alpha^{31} \\le \\frac{32^{32}}{2^{36}} = 2^{124}$, and $a \\le 2^4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $Q_n$ be the set of $n$-tuples $x = (x_1, ..., x_n)$, $x_i \\in \\{0, 1, 2\\}$, $i = 1, 2, ..., n$. A triad $(x, y, z)$, where $x = (x_1, x_2, ..., x_n)$, $y = (y_1, y_2, ..., y_n)$, $z = (z_1, z_2, ..., z_n)$, of distinguished elements of $Q_n$ is called good, if there exists at least one $i \\in \\{1, 2, ..., n\\}$ for which the sets $\\{x_i, y_i, z_i\\}$ and $\\{0, 1, 2\\}$ are equal. A subset $A$ of $Q_n$ is called good, if every three elements of $A$ form a good triad. Prove that every good subset of $Q_n$ has at most $2\\left(\\frac{3}{2}\\right)^n$ elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will use induction with respect to $n$. The case for $n = 1$ is obvious. We suppose that every good subset of $Q_{n-1}$ has at most $2\\left(\\frac{3}{2}\\right)^{n-1}$ elements.\n\nLet $A_0 = \\{(x_1, ..., x_n) \\in A : x_n \\neq 0\\}$. We define the subsets $A_1, A_2$ similarly, that is\n$$\nA_1 = \\{(x_1, ..., x_n) \\in A : x_n \\neq 1\\}, \\quad A_2 = \\{(x_1, ..., x_n) \\in A : x_n \\neq 2\\}.\n$$\nSince $A$ is a good set and $A_0$ is its subset, it follows that $A_0$ is also good. It means that for every three elements of $A_0$ there exists a coordinate which is different for every two of them. This coordinate cannot be the last one because $0$ cannot be there. Therefore the set $A_0'$ produced from the elements of $A_0$ by deleting the last coordinate is a good subset of $Q_{n-1}$.\n\nMoreover, we observe that, if $|A_0| \\geq 3$, then $|A_0'| = |A_0|$.\nIn fact, if $|A_0'| \\neq |A_0|$, then there would exist an element $a \\in A_0'$ such that $x, y \\in A_0$, where $x, y$ arise from $a$ by adding $1$ and $2$, respectively, as last coordinate. However, if then $z$ is any other element of $A_0$, it cannot have as last coordinate $0$, and so $x, y, z$ will not form a good triad, absurd. Hence, from the induction hypothesis we have\n$$\n|A_0| \\leq \\max\\{2, |A_0'|\\} \\leq 2\\left(\\frac{3}{2}\\right)^{n-1}.\n$$\nSimilarly, we get that: $|A_1|, |A_2| \\leq 2\\left(\\frac{3}{2}\\right)^{n-1}$. Since every element of $A$ appears exactly in two of the sets $A_0, A_1, A_2$, we conclude:\n$$\n|A| = \\frac{1}{2}(|A_0| + |A_1| + |A_2|) \\leq 2\\left(\\frac{3}{2}\\right)^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $n \\ge 1$, for which there exist at least one pair of positive integers $(a, b)$ such that:\n(i) The integer $a^2 + b + 3$ is not divisible by the cube of a prime number.\n(ii) $\\frac{ab+3b+8}{a^2+b+3} = n$",
    "options": [],
    "answer": "n = 2",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $a$, $b$, $c$ for which there exist positive integers $x$, $y$, $z$, such that: $ab + 1 = x!$, $bc + 1 = y!$, $ca + 1 = z!$, where $n!$ denotes the product $1 \\cdot 2 \\cdot 3 \\ldots \\cdot n$.",
    "options": [],
    "answer": "All solutions are the cyclic permutations of (1, 1, n! − 1) with n ≥ 2.",
    "solution": "We can easily see that $x, y, z \\ge 2$.\nIf $x, y, z \\ge 3$, then observe that $3$ doesn't divide $a, b, c$. Indeed, if for example, $3 \\mid a$, then $3$ doesn't divide $ab + 1$, but $3$ divides $x!$, contradiction.\nThis means that two of the numbers are congruent mod $3$. Due to the symmetry,\nsuppose that $a \\equiv b \\pmod{3}$. Then, we have\n$$\nx! = ab + 1 \\equiv a^2 + 1 \\pmod{3}.\n$$\nHowever, $3 \\mid x!$, so $3 \\mid a^2 + 1$, which is absurd, since for all integers $x$, we have\n$x^2 \\equiv 0,1 \\pmod{3}$.\nIt follows that one of $x, y, z$ is smaller than $3$, let it be $x$. Then $x = 2$ and from\n$ab + 1 = 2! = 2$, we get $ab = 1$, so $a = b = 1$. Then, from the second relation,\n$y = z$, so all solutions have the form $(1,1, y! - 1)$ with its cyclic permutations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute angled triangle with $AB < AC < BC$, inscribed to the circle $\\Gamma_1$ with center $O$. The circle $\\Gamma_2$ with center $A$ and radius $AC$ intersects the line $BC$ at point $D$ and the circle $\\Gamma_1$ at $E$. The circumcircle of the triangle $DEF$ (say, $\\Gamma_3$) intersects the line $BC$ at $G$. Prove that:\n\na) The point $B$ is the center of $\\Gamma_3$.\n\nb) The circumcircle of the triangle $CEG$ is tangent to $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) The triangle $ADC$ is isosceles ($AD = AC$ are radius of $\\Gamma_2$), so $\\angle D_1 = \\angle C$.\nThe angle $\\angle F_1$ is external of the cyclic $ACBF$, therefore $\\angle F_1 = \\angle C$.\nFrom the two equalities above we conclude that $\\angle D_1 = \\angle F_1$, thus\n$$\nBD = BF \\tag{1}.\n$$\n\n![](attached_image_1.png)\n\nThe angle $\\angle D_2$ is the half of the center angle $E\\widehat{A}C$, so\n$$\n\\angle D_2 = \\frac{E\\widehat{A}C}{2} \\quad (a).\n$$\nMoreover the angles $\\angle B_1$ and $E\\widehat{A}C$ see the arc $EC$ in $\\Gamma_1$, so\n$$\n\\angle B_1 = E\\widehat{A}C \\quad (b)\n$$\nFrom the triangle $BDE$ we have:\n$$\n\\angle E_1 = \\angle B_1 - \\angle D_2 \\stackrel{(a),(b)}{=} E\\widehat{A}C - \\frac{E\\widehat{A}C}{2} = \\frac{E\\widehat{A}C}{2}\n$$\n$$\n\\text{Therefore } \\angle D_2 = \\angle E_1, \\text{ so } BD = BE \\quad (2).\n$$\nFrom (1) and (2) we conclude that $B$ is the center of $\\Gamma_3$.\n\nb) The bisector of $\\angle B_1$ is perpendicular bisector of $EG$ and passes through the midpoint (let it be $M$) of the arc $CE$.\nWe also have the equality $OC = OE$ (both are radius of $\\Gamma_1$) and $AC = AE$ (both are radius of $\\Gamma_2$). Therefore, the line $OA$ is perpendicular bisector of $CE$, thus it passes through the midpoint $M$ of the arc $CE$.\nWe conclude that $M$ of the arc $CE$ is the circumcenter of $CEG$, as point of intersection of the perpendicular bisectors of $EG$ and $CE$.\nTherefore $CA \\perp CM$, so $CA$ is tangent to the circumcircle of $CEG$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31325,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive real numbers $x, y, z$ are such that $x + y + z = 4$ and\n\n$x, y, z \\in [0,2]$. Find the minimal value of the algebraic expression:\n$$\nA = \\sqrt{2+x} + \\sqrt{2+y} + \\sqrt{2+z} + \\sqrt{x+y} + \\sqrt{y+z} + \\sqrt{z+x}.\n$$",
    "options": [],
    "answer": "6 + 3*sqrt(2)",
    "solution": "We will prove first that $\\sqrt{2+x} + \\sqrt{y+z} = \\sqrt{2+x} + \\sqrt{4-x} \\ge 2 + \\sqrt{2}$.\nIndeed, this is equivalent to\n$$\n2 + x + 4 - x + \\sqrt{(2 + x)(4 - x)} \\ge 4 + 2 + 2\\sqrt{2} \\Leftrightarrow \\\\ x(2 - x) \\ge 0,\n$$\nwhich is true, since $x \\in [0,2]$.\nSimilarly we have $\\sqrt{2+y} + \\sqrt{x+z} \\ge 2 + \\sqrt{2}$ and $\\sqrt{2+z} + \\sqrt{x+y} \\ge 2 + \\sqrt{2}$.\nAdding all these, we get $A \\ge 6 + 3\\sqrt{2}$ and the equality holds, for example, if $x = y = 2$ and $z = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31326,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$, we consider a $3n \\times 3n$ chessboard, and its squares are colored black and white as follows: Starting from the left up corner square every third diagonal is black and all other squares are white, as in the figure.\n![](attached_image_1.png)\nAt each move we select a $2 \\times 2$ square and we change the color of its squares as follows: The white squares become orange, the oranges (if there exist) become black and the blacks become white. Our target is, after a finite number of moves, to obtain a chessboard with all black squares at the initial placement changed to whites and all white squares at the initial placement changed to black. Prove that:\n\na) The target is not feasible for $n = 3$.\nb) The target is feasible for $n = 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Firstly, observe that if we change the colour of one square 3 times, it goes back to its original colour. As the final configuration does not depend on the order of moves and placing a move on a square 3 times does not change the configuration, we can suppose that a move is placed on any square exactly 0, 1 or 2 times. This further implies that a white square must change its colour 2 (modulo 3) times and that a black square must be changed 1 (modulo 3) times.\nLet us associate each $2 \\times 2$ square with its top left unit square. Let's take a look at the first column. Its top square is white and is included in only one $2 \\times 2$ square, so that square has to be placed 2 times. The second square in this column is also white and is included in its $2 \\times 2$ square as well as in the one associated with the first unit square in the column. This square has already been turned black by taking the $2 \\times 2$ square of the first unit square in this column twice, so its own $2 \\times 2$ square has to be taken 0 times.\nWe use similar arguments to show that next four $2 \\times 2$ squares have to be called 1,1,1 and 0 times. After that we have the white square included only in its own $2 \\times 2$ square and in the one above it, which has been taken 0 times, so we have the same situation we had in the beginning. Therefore, the next one should be used 2 times and the next is white, so 0 times. However, this is the only $2 \\times 2$ that contains the last black square, therefore, this cannot change color, a contradiction.\n\nb) Working as in a), we can fill the board with the numbers 0, 1, 2 in each square, showing how many times we should use the corresponding $2 \\times 2$ square.\n![](attached_image_2.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Each student of a class has finite number of cards. Each card has a number on it from the interval $[0,1]$. Find the smallest possible constant $c > 0$, such that the following holds, independently from the distribution of the cards to the students:\nEach student that has a total sum of numbers less than 1000, shares the cards into 100 boxes, such that the sum of the cards in each box is at most $c$.",
    "options": [],
    "answer": "1000/91",
    "solution": "Amongst all possible arrangements into boxes, pick one, where the maximum value inside a box is as small as possible. If there are several arrangements, achieving this smallest maximum value, pick one where the number of boxes achieving this value is as small as possible.\nSay that the boxes have total values equal to $10 + x_1 \\ge 10 + x_2 \\ge \\dots \\ge 10 + x_{100}$, respectively. Since the total sum is less or equal to 1000, we get\n$$\n10 + x_1 + 10 + x_2 + \\dots + 10 + x_{100} \\le 1000 \\Leftrightarrow x_1 + \\dots + x_{100} \\le 0.\n$$\n\nSince $x_{100}$ is the smallest, we have $x_1 + 99x_{100} \\le x_1 + x_2 + \\dots + x_{100} \\le 0$. Suppose, for sake of contradiction, that $x_1 > \\frac{90}{91}$.\nSince the total sum in the first box is bigger than 10, the first box contains at least 11 cards, and each card has a number from $[0,1]$, therefore, there will be a card with value at most $10 + \\frac{x_1}{11}$.\n\n---\n\nRemove that card from the first box and put it in the 100-th box. Then, the 100-th box should have sum bigger or equal to $10 + x_1$, otherwise, we would have a configuration with smaller maximal sum. However, after that movement, the 100-th would have total sum at most\n$$\n10 + x_{100} + \\frac{10 + x_1}{11} \\le 10 - \\frac{x_1}{99} + \\frac{10 + x_1}{11} = 10 + x_1 + \\frac{90 - 91x_1}{99} < 10 + x_1,\n$$\ncontradiction.\nWe will prove that $c = 11 - 11a$, with $1 > a > \\frac{1}{1001}$ doesn't work. Indeed, take $r \\in [\\frac{1}{1001}, a)$ and let $n = \\lfloor \\frac{1000}{1-r} \\rfloor$. Since $r \\ge \\frac{1}{1001}$, we have that $n \\ge 1001$. Now take $n$ cards each of value $1-r$. Their sum is $n(1-r) \\le \\frac{1000}{1-r}(1-r) = 1000$. Now, no matter how we place them in 100 boxes, as $n \\ge 1001$, there exist 11 cards in the same box. But $11(1-r) > 11 - 11a$, so the constant $c = 11 - 11a$, doesn't work.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in the positive real numbers the system:\n$$\n\\begin{cases} \\frac{xyz+1}{x+1} = \\frac{yzw+1}{y+1} = \\frac{zwx+1}{z+1} = \\frac{wxy+1}{w+1} \\\\ x+y+z+w=48 \\end{cases} .$$",
    "options": [],
    "answer": "(12, 12, 12, 12); (12 + sqrt(143), 12 - sqrt(143), 12 + sqrt(143), 12 - sqrt(143)); (12 - sqrt(143), 12 + sqrt(143), 12 - sqrt(143), 12 + sqrt(143))",
    "solution": "1. (First solution) The given equations can be written as:\n$$\nxy^2z + xyz + y = xyzw + yzw + x\n$$\n$$\nz^2yw + yzw + z = xyzw + y + zwx\n$$\n$$\nw^2xz + zwx + w = xyzw + wxy + z.\n$$\n$$\nx^2wy + wxy + x = xyzw + xyz + w.\n$$\nSumming up the four equations we get:\n$$\nxy^2z + x^2wy + yz^2w + w^2xy = 4xyzw \\Leftrightarrow \\frac{y}{w} + \\frac{x}{z} + \\frac{z}{x} + \\frac{w}{y} = 4.\n$$\nFrom arithmetic-geometric means inequality we have:\n$$\n\\frac{y}{w} + \\frac{x}{z} + \\frac{z}{x} + \\frac{w}{y} \\ge 4.\n$$\nWhere the equality holds, if and only if\n$$\n\\frac{y}{w} = \\frac{x}{z} = \\frac{z}{x} = \\frac{w}{y} \\Leftrightarrow x^2 = z^2 \\text{ and } y^2 = w^2 \\Leftrightarrow x = z \\text{ and } y = w.\n$$\n\n$$\n\\frac{xyz + 1}{x + 1} = \\frac{yzw + 1}{y + 1} \\Leftrightarrow \\frac{x^2y + 1}{x + 1} = \\frac{y^2x + 1}{y + 1} = \\frac{x^2y - y^2x}{x - y} = xy \\\\\n\\Leftrightarrow x^2y^2 + x^2y + 1 + y = x^2y^2 + y^2x + 1 + x\n$$\n$$\n\\Leftrightarrow xy(x - y) = x - y \\Leftrightarrow (x - y)(xy - 1) = 0 \\Leftrightarrow x = y \\text{, } xy = 1.\n$$\nWe distinguish two cases:\n1. $x = y = z = w$. Then, from $x + y + z + w = 48$, we have the solution $(x, y, z, w) = (12, 12, 12, 12)$.\n2. $x = z$ and $y = w$ and $xy = 1$. Then\n$$\nx + y = 24 \\Leftrightarrow x + \\frac{1}{x} = 24 \\Leftrightarrow x^2 - 24x + 1 = 0 \\Leftrightarrow x = 12 \\pm \\sqrt{143},\n$$\nand we find the solutions\n$$\n(x, y, z, w) = (12 + \\sqrt{143}, 12 - \\sqrt{143}, 12 + \\sqrt{143}, 12 - \\sqrt{143})\n$$\n$$\n(x, y, z, w) = (12 - \\sqrt{143}, 12 + \\sqrt{143}, 12 - \\sqrt{143}, 12 + \\sqrt{143}).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Solve in the real numbers the system:\n$$\n\\begin{cases}\na + b + c = 0 \\\\\nab^3 + bc^3 + ca^3 = 0\n\\end{cases}.\n$$",
    "options": [],
    "answer": "a = b = c = 0",
    "solution": "1. Since $c = -a - b$, we get:\n$$\n\\begin{aligned}\n0 &= ab^3 + b(-a-b)^3 + (-a-b)a^3 \\\\\n&= -(ab^3 + b(a+b)^3 + (a+b)a^3) \\\\\n&= -(a^4 + 2a^3b + 3a^2b^2 + 2ab^3 + b^4) \\\\\n&= -(a^2(a+b)^2 + b^2(a+b)^2 + a^2b^2) \\\\\n&= -a^2c^2 - b^2c^2 - a^2b^2.\n\\end{aligned}\n$$\nTherefore each term of the last sum must be zero: $ab = bc = ca = 0$. Hence two of the numbers must be zero and from the equality $a + b + c = 0$, we conclude that $a = b = c = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31330,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex quadrilateral $ABCD$, its diagonals meet at $E$ and $H$, $G$ are the midpoints of the sides $AD$, $BC$, respectively. The circumcircles $c_1$ and $c_2$, of the triangles $AEB$ and $DEC$, respectively, meet at point $F \\neq E$. If the parallel from $E$ to the line $HG$ meets the circle $c_2$ at $S$, prove that: $FS \\parallel CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In order to prove $FS \\parallel CD$, it suffices $FC = SD$, or $S\\hat{C}D = F\\hat{E}C$. From the cyclic $SEDC$ and $FS \\parallel CD$ we get\n$$\nS\\hat{C}D = B\\hat{E}S = B\\hat{I}G,\n$$\nwhere $I$ is the point of intersection of $HG$ and $BD$.\nIt suffices to prove that: $B\\hat{I}G = F\\hat{E}C$. Angle chasing gives\n$$\nA\\hat{F}B = A\\hat{E}B = D\\hat{E}C = D\\hat{F}C \\text{ and } B\\hat{A}F = B\\hat{E}F = F\\hat{C}D.\n$$\n![](attached_image_1.png)\nFigure 8\nThis means that $AFB$ and $CFD$ are similar with $A\\hat{B}F = F\\hat{D}C$. Moreover,\n$$\nD\\hat{B}F = C\\hat{A}F \\text{ and } B\\hat{F}D = A\\hat{F}D,\n$$\nwhere the triangles $AFC$ and $BFD$ are similar with similarity ratio\n$$\nAC/BD = AF/BF.\n$$\nIf $K$ is the midpoint of $AB$, then the triangle $HKG$ has $KH \\parallel BD$, $KG \\parallel AC$ with\n$$\nKG/KH = AC/BD = AF/BF \\text{ and } H\\hat{K}G = A\\hat{E}B = A\\hat{F}B.\n$$\nTherefore, the triangle $GKH$ is similar to $AFB$, and so\n$$\nK\\hat{H}G = A\\hat{B}F = F\\hat{D}C = F\\hat{E}C.\n$$\n\n![](attached_image_1.png)\nFigure 8\nAs in the first solution, it suffices to prove that: $B\\hat{I}G = F\\hat{E}C$. Let $M$, $N$ be the projections of $F$ to $AB$, $BD$, respectively. Then the line $MN$ is a Simpson line of the complete quadrilateral and $HG$ is the Newton-Gauss line. It is well-known that they are perpendicular. Therefore\n$$\nB\\hat{I}G = H\\hat{I}N = 90^\\circ - M\\hat{N}B = 90^\\circ - M\\hat{F}B = M\\hat{B}F = F\\hat{E}C,\n$$\nwhere in the third equality we used that $MNFB$ is cyclic and in the last we used that $AEFB$ is cyclic.\n\n![](attached_image_2.png)\nFigure 9",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31331,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive real numbers $a$, $b$, $c$ satisfy: $a^2 + b^2 + c^2 = 3$. Prove that\n\n$$\n\\frac{a^2 + b^2}{2ab} + \\frac{b^2 + c^2}{2bc} + \\frac{c^2 + a^2}{2ca} + \\frac{2(ab + bc + ca)}{3} \\geq 5.\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds when a = b = c = 1.",
    "solution": "$$\nB\\hat{A}C = B\\hat{D}C = D\\hat{B}C = D\\hat{A}C.\n$$\nHence $AC$ is the bisector of the angle $B\\hat{A}D$, and hence it is the perpendicular bisector of the base $BE$ od the isosceles $\\triangle BAE$.\nNow we can complete the proof in two ways:\n**First way:** Since the points $E, B, F$ belong to the circle with center $A$ and radius $AB$, by using the relation between subtending angles and the angle formed by chord and tangent, we have the wanted result:\n$$\nE\\hat{F}B = \\frac{E\\hat{A}B}{2} = E\\hat{A}C = D\\hat{A}C = D\\hat{B}C\n$$\n\n![](attached_image_1.png)\nFigure 7\n\n**Second way:** We have $CE = CB = CD$, as well as, $C\\hat{E}D = C\\hat{D}E$.\nFrom the cyclic quadrilateral $ABCD$ and the isosceles triangle $FAB$ we get:\n$A\\hat{F}B = A\\hat{B}F = A\\hat{D}C = C\\hat{E}D = 180^\\circ - A\\hat{E}C$, and therefore the quadrilateral $AFCE$ is cyclic. Thus we have;\n$$\nE\\hat{F}B = E\\hat{F}C = E\\hat{A}C = D\\hat{A}C = D\\hat{B}C.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AΒΓ$ be a triangle and $M$, $N$ the midpoints of $AB$, $AΓ$, respectively. The points $Δ$ and $E$ lie on the segment $BN$, such that $ΓΔ \\parallel ME$ and $BΔ < BE$. Prove that:\n$$\nBΔ = 2 \\cdot EN.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $ME \\parallel ΓΔ$ it follows that $ΓΔE = M\\hat{E}Δ$, and so\n$$\n180^\\circ - ΓΔE = 180^\\circ - M\\hat{E}Δ \\Rightarrow BΔΓ = M\\hat{E}N \\quad (1)\n$$\nSince the points $M$, $N$ are the midpoints of the sides $AB$, $AΓ$, respectively, we conclude that\n$$\nMN \\parallel BΓ, \\quad MN = \\frac{BΓ}{2}. \\quad (2)\n$$\nand hence\n$$\nΔB\\hat{Γ} = M\\hat{N}E, \\quad (3)\n$$\nFrom (1) and (3) we get that the triangles $BΔΓ$ and $MEN$ are similar, and so:\n$$\n\\frac{BΔ}{EN} = \\frac{BΓ}{MN} = 2 \\Rightarrow BΔ = 2 \\cdot EN.\n$$\n\n![](attached_image_1.png)\nFigure 1\nFrom $Γ$ we draw the parallel to the line to $BN$, which meet the line $AE$ at $Z$. Since $N$ is the midpoint of $BΓ$ and $NE \\parallel ΓZ$, it follows that $E$ is the midpoint of $AZ$ and $ΓZ = 2 \\cdot EN$.\nSince $M$, $E$ are the midpoints of $AB$, $AZ$, respectively, we have: $ME \\parallel BΓ$.\nTherefore the quadrilateral $BZΓΔ$ is parallelogram, as it has the two pairs of opposite sides parallel. Hence $BΔ = ΓZ = 2 \\cdot EN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31333,
    "subject": "Mathematics (Multi-modal)",
    "question": "For the various values of the positive integer $n$, determine all positive integers $N$ which are perfect squares and in their decimal representation have $n$ times the digit $2$ and one time the digit $5$.",
    "options": [],
    "answer": "25 and 225",
    "solution": "Answer: $5^2 = 25$ and $25^2 = 225$.\n\n**Proof.** Since a perfect square cannot have its last digit $2$, $N$ must be of the form:\n$$\nN = 22\\dots 225 = 22\\dots 200 + 25 = 100 \\cdot 2 \\cdot 11\\dots 1 + 25 = 100 \\cdot 2 \\cdot \\frac{10^{n-1} - 1}{9} + 25,\n$$\nwhere the digit $2$ there exists $n$ times, $n \\ge 1$.\n\nWorking mod $10$, then $N = (10k + v)^2$, $0 \\le v < 10$, where $k$ is a positive integer, we can see that only in the case $N = (10k + 5)^2$ results a positive integer leading to $25$. Then we have\n$$\n100 \\cdot 2 \\cdot \\frac{10^{n-1}-1}{9} + 25 = (10k + 5)^2, \\quad (1)\n$$\nor equivalently\n$$\n9k^2 + 9k - 2(10^{n-1} - 1) = 0. \\quad (2)\n$$\nSince $k$ is a positive integer, the discriminant of the equation (2) must be perfect square, that is $\\Delta = 9(8 \\cdot 10^{n-1} + 1)$ must be perfect square. Thus we must have\n$$\n8 \\cdot 10^{n-1} + 1 = x^2,\n$$\nfor some odd integer $x > 1$. If $x = 2m + 1$, $m \\ge 1$, then:\n$$\n8 \\cdot 10^{n-1} = x^2 - 1 = (x - 1)(x + 1) = 4m(m + 1) \\Rightarrow\n$$\n$$\n2^n \\cdot 5^{n-1} = m(m + 1), \\quad (3)\n$$\nwith $(m, m + 1) = 1$. We distinguish cases with respect to $n$:\n\n* If $n = 1$, then $m = 1$, $k = 0$ and $N = 25$.\n* If $n = 2$, then $m = 4$, $k = 1$ and $N = 225$.\n* If $n \\ge 3$, since $5^{n-1} > 4^{n-1} = 2^{2n-2} > 2^n$,\nfrom equation (3) we get:\n$$\n2^n = m \\text{ and } 5^{n-1} = m + 1,\n$$\nand so $m = 5^{n-1} - 1 = 4 \\cdot (5^{n-2} + \\cdots + 1) > 4 \\cdot 4^{n-2} = 4^{n-1} > 2^n = m$, absurd.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB > AC$, $AD$ its bisector, where $D$ is a point on the side $BC$ and $I$ its incenter. If $M$ is the midpoint of the segment $AD$ and $F$ is the point of intersection of the line $MB$ with the circumcircle of the triangle $BIC$, prove that: $AF \\perp FC$.\n(E. Psarras)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle CII_a = \\angle BII_a = 90^\\circ$ the excenter $I_a$ belongs to the circumcircle of the triangle $BIC$. Moreover in the triangle $CDA$, $CI, CI_a$ are the internal and external bisector, respectively, and therefore the points $A, D$ are the conjugate harmonics of the points $I, I_a$. Since $M$ is the midpoint of the segment $AD$, from the relation of Newton we have:\n$$\nMA^2 = MI \\cdot MI_a. \\qquad (1)\n$$\nTaking the power of $M$ with respect to the circle $(I, B, C)$ we have:\n$$\nMI \\cdot MI_a = MF \\cdot MB. \\qquad (2)\n$$\nFrom (1) and (2) it follows that:\n$$\nMA^2 = MF \\cdot MB,\n$$\nWhich means that $MA$ is tangent of the circumcircle of the triangle $AFB$. Therefore\n$$\n\\angle MBA = \\angle FAM. \\qquad (3)\n$$\nand finally\n$$\n\\begin{align*} \\angle FAC + \\angle FCA &= \\angle FAM + \\frac{\\angle A}{2} + \\angle FCI + \\frac{\\angle C}{2} = \\\\ & (3) \\qquad \\angle MBA + \\frac{\\angle A}{2} + \\angle FBI + \\frac{\\angle C}{2} = \\frac{\\angle A}{2} + \\frac{\\angle B}{2} + \\frac{\\angle C}{2} = 90^\\circ. \\end{align*}\n$$\nHence: $AF \\perp FC$.\n\n![](attached_image_1.png)\nFigure 2\n\nWe draw from $C$ the parallel line to the bisector $AD$ which meets the line $AB$ at point $N$ and the line $AM$ at point $Z$. Taking angle equalities we have:\n$$\n\\angle ACN = \\angle DAC = \\frac{\\angle A}{2}, \\quad \\angle ANC = \\angle BAD = \\frac{\\angle A}{2}\n$$\nAnd so the triangle $ACN$ is isosceles. Since $M$ is the midpoint of $AD$ it follows that $Z$ is the midpoint of $CN$. Hence $AZ \\perp CN \\Rightarrow \\angle AZC = 90^\\circ$.\nTherefore, in order to have $\\angle AFC = 90^\\circ$, it is enough to prove that the quadrilateral $CFAZ$ is cyclic. In fact, we have\n$$\n\\angle CFZ = 180^\\circ - \\angle BFC = 180^\\circ - \\angle BIC \\\\ = 180^\\circ - \\left(90^\\circ + \\frac{\\angle A}{2}\\right) = 90^\\circ - \\frac{\\angle A}{2} = \\angle CAZ.\n$$\n\n![](attached_image_2.png)\nFigure 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31335,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider a $100 \\times 100$ square consisting of $10^4$ unit squares. We call tiles of type A the rectangular tiles $8 \\times 1$ or $1 \\times 8$ consisting of 8 unit squares. We call tiles of type B the rectangular tiles $4 \\times 2$ or $2 \\times 4$ consisting of 8 unit squares. Examine if it is possible to cover the $100 \\times 100$ square by using $N$ tiles of type A and $N$ tiles of type B, for some positive integer $N$? (The tiles must be inside the initial $100 \\times 100$ square and not overlapping.)",
    "options": [],
    "answer": "No; such a tiling is impossible",
    "solution": "Suppose such a coverage is possible with $N$ tiles of type A and with $N$ tiles of type B. Calculating the area of the tiles, which is $8 = 2^3$ and of the square which is equal to $100^2 = 2^4 \\cdot 5^4$, we observe that we need $5^4 = 625$ of tiles of type A and 625 of tiles of type B, so $N = 625$.\n\nWe color the squares of the first row alternately with colors 1 and 2. That is, we have the coloring 1-2-1-2...\nWe color the squares of the second row alternately with colors 3 and 4. That is, we have the coloring 3-4-3-4...\nWe color the squares of the third row alternately with colors 1 and 2. That is, we have the coloring 1-2-1-2... and so on.\n\nIn this coloring the $4 \\times 2$ or $2 \\times 4$ contain two squares of each color. $8 \\times 1$ or $1 \\times 8$ contain two suits in 4 squares each. In the coverage they must be covered by an equal number of the colors 1,2,3,4 and type B cover an equal number of the colors, therefore the same must be the case for the type A. Therefore the type A must be an even number, out of place since it is 625.\nSuppose such a coverage is possible with $N$ tiles of type A and with $N$ tiles of type B. Calculating the area of the tiles, which is $8 = 2^3$ and of the square which is equal to $100^2 = 2^4 \\cdot 5^4$, we observe that we need $5^4 = 625$ of tiles of type A and 625 of tiles of type B, so $N = 625$.\n\nWe split the $8 \\times 1$ or $1 \\times 8$ type A tiles into $16 \\times 2$ or $2 \\times 16$ tiles, respectively, and the $4 \\times 2$ or $2 \\times 4$ type B tiles into $8 \\times 4$ or $4 \\times 8$ tiles, respectively, so that their center of gravity to coincide with a vertex of a $1 \\times 1$ square (as in the Figure)\n\nWe do the same with the $100 \\times 100$ square, which turns into a $200 \\times 200$ square.\n\n![](attached_image_1.png)\nFigure 10\n\n![](attached_image_2.png)\nFigure 11\n\nWe do the same with the $100 \\times 100$ square, which turns into a $200 \\times 200$ square.\nWe consider an orthonormal coordinate system with the center of gravity of the $200 \\times 200$ square as the origin and the length unit of the system being the length of the small squares into which the type A and type B tiles are divided.\n\nWe notice that in any coverage of the $200 \\times 200$ square, the coordinates of the centroid of each type B tile are even numbers. The coordinates of the center of gravity of an A-type tile either placed parallel to the $x$-axis or parallel to the $y$-axis have opposite parity, i.e. one is odd and the other is even. In the first case, the ordinate is an odd number and in the second coordinate.\n\nSince the algebraic sum of the coordinates of all the centers of gravity of the covering tiles must be 0, for the covering to be possible, we should use an even number of type A tiles of each kind. Since $N = 625$ is an odd number, such tiling is impossible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of rectangles satisfying the following properties:\n(α) Their vertices are points $(x, y)$ of the plane $Oxy$, with $x, y$ non-negative integers and $x \\le 8$, $y \\le 8$.\n(β) Their sides are parallel to axis\n(γ) Their area $E$ satisfies: $30 < E \\le 40$.",
    "options": [],
    "answer": "43",
    "solution": "First we examine which values of the area of rectangles are acceptable:\nSince, $0 < x, y \\le 8$, the integer $40$ is written only as $40 = 5 \\cdot 8$. Since a $5 \\times 8$ rectangle can be put in the $8 \\times 8$ rectangle with $4$ ways horizontally and with $4$ ways vertically we have totally $8$ such rectangles.\nThe numbers $39, 38, 37, 34, 33, 31$ are not the product of two integers $x, y$ with $0 < x, y \\le 8$.\nThe number $36$ can be written uniquely $36 = 6 \\cdot 6$. Since a $6 \\times 6$ rectangle can be put in the $8 \\times 8$ rectangle with $3^2$ ways, we have $9$ such rectangles.\nThe number $35$ can be written uniquely $35 = 5 \\cdot 7$. Since a $5 \\times 7$ rectangle can be put in the $8 \\times 8$ rectangle with $2 \\cdot (4 \\cdot 2) = 16$ ways, we have $9$ such rectangles.\nThe number $32$ can be written uniquely $32 = 4 \\cdot 8$. A $4 \\times 8$ rectangle can be put in the $8 \\times 8$ rectangle with $2 \\cdot 5 = 10$ ways.\nFinally, we have $8 + 9 + 16 + 10 = 43$ rectangles with the required properties.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31337,
    "subject": "Mathematics (Multi-modal)",
    "question": "$$\n\\frac{a^2 + b^2}{2ab} + \\frac{b^2 + c^2}{2bc} + \\frac{c^2 + a^2}{2ca} + \\frac{2(ab + bc + ca)}{3} \\geq 5.\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds when a = b = c = 1.",
    "solution": "Since $a^2 + b^2 + c^2 = 3$, the inequality is equivalent to\n\n$$\n\\left( \\frac{a^2 + b^2}{2ab} + 1 \\right) + \\left( \\frac{b^2 + c^2}{2bc} + 1 \\right) + \\left( \\frac{c^2 + a^2}{2ca} + 1 \\right) + \\left( \\frac{2(ab + bc + ca)}{a^2 + b^2 + c^2} + 1 \\right) \\geq 9.\n$$\nwhich can be written as\n$$\n\\frac{(a+b)^2}{2ab} + \\frac{(b+c)^2}{2bc} + \\frac{(c+a)^2}{2ca} + \\frac{(a+b+c)^2}{a^2+b^2+c^2} \\geq 9.\n$$\nLet $L$ be the left hand side of the last inequality. Then from Cauchy-Schwarz inequality in the form of Engel or Andreescu it follows that\n$$\nL \\geq \\frac{((a+b) + (b+c) + (c+a) + (a+b+c))^2}{2ab + 2bc + 2ca + a^2 + b^2 + c^2} = \\frac{9(a+b+c)^2}{(a+b+c)^2} = 9.\n$$\nThe equality holds, if and only if\n$$\n\\frac{a+b}{2ab} = \\frac{b+c}{2bc} = \\frac{c+a}{2ca} = \\frac{a+b+c}{3}.\n$$\nSince $a, b, c > 0$, from the first two equations we get $a = b = c$, and by using the last equation we find: $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31338,
    "subject": "Mathematics (Multi-modal)",
    "question": "A class consists of 26 students, sitting in pairs. At the second term they decide to change positions in order every two students that sat together on the first term, they are not sitting together in the second. Find the largest possible value of $N$, such that no matter how the students sat in the two terms, there is a set $S$ of $N$ students, that they didn't sit together in any of the two terms.",
    "options": [],
    "answer": "13",
    "solution": "We will prove that the maximum value is 13. Indeed, if the set contains 14 students, because they sit in 13 desks, from the Pigeonhole Principle, there will be two students who sat in the same desk. Therefore the maximum value is at most 13.\n\n![](attached_image_1.png)\nFigure 4\n\nFor the other direction, we will prove that no matter how the students sat in the two terms, we can find a total of 13 students so that no two sat together during the year.\n\nIndeed, we represent the students as points on the plane and join them with a red line segment if they sat together in the first term and with a black one if they sat together in the second term. Suppose there is a circle consisting of an odd number of segments. Then each vertex is connected by a black and a red segment (with the two neighbors in the circle). Therefore, if we start with red, since the number of segments is odd, we will end up with red again.\n\nThis is absurd, since each vertex is connected to a red (for the first term) and a black (for the second term) line segment.\n\nSince there is no odd cycle, this means that the set of vertices can be divided into two subsets $A$, $B$ such that there are no segments between vertices in set $A$ and no segments between vertices in set $B$.\n\nSince the students are 26, there is one set with cardinality at least 13, and the result follows.\n\n![](attached_image_2.png)\nFigure 5\nAs in the first solution, $N \\le 13$ is immediate from the Pigeonhole Principle. We will show that the equality holds.\n\nWe consider a graph of twenty six vertices, which correspond to the students, with an edge between two vertices if and only if the corresponding students have sat together in the first or second term.\n\nEach connected component has an even number of vertices, since students sit in pairs in each term. Otherwise, there would be some student sitting alone in some term, absurd.\n\nSince students sit with two different classmates throughout the year, one in each term, each vertex is of degree 2. So each connected component is an Euler cycle. Indeed, if we start moving from a vertex passing through each edge only once, due to the degree of each vertex being 2, we should end up at the vertex we started from.\n\nIt suffices, then, to consider each connected component as a directed cycle $v_1, v_2, ..., v_{2k}$ and consider the half vertices alternately, i.e. $v_{2v_1}, v_{2v_2}, ..., v_{2v_{2k}}$, which are not connected by an edge. Since from each component we take half the vertices, $N=13$.\n\nIn the example below, our graph has four coherent components. We can simply select all green or all yellow vertices.\n\n![](attached_image_3.png)\nFigure 6",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $\\alpha, \\beta$ with $\\alpha > 1$ such that $\\beta$ divides $\\alpha - 1$ and $2\\alpha + 1$ divides $5\\beta - 3$.\n(A. Fellouris)",
    "options": [],
    "answer": "(3, 2), (10, 9), (13, 6)",
    "solution": "Since $\\beta$ divides $\\alpha - 1$ and $2\\alpha + 1$ divides $5\\beta - 3$, with $\\alpha > 1$,\n$$\nx = \\frac{\\alpha - 1}{\\beta} \\quad \\text{and} \\quad y = \\frac{5\\beta - 3}{2\\alpha + 1}\n$$\nare positive integers such that:\n$$\n0 < xy = \\frac{\\alpha - 1}{\\beta} \\cdot \\frac{5\\beta - 3}{2\\alpha + 1} < \\frac{\\alpha}{\\beta} \\cdot \\frac{5\\beta}{2\\alpha} = \\frac{5}{2} \\Rightarrow xy \\in \\{1,2\\}.\n$$\n* If $xy = 1$, then $x = y = 1$, and so we have the system:\n$$\n\\begin{cases} \\alpha - 1 = \\beta \\\\ 2\\alpha + 1 = 5\\beta - 3 \\end{cases} \\Leftrightarrow \\begin{cases} \\alpha = \\beta + 1 \\\\ 2(\\beta + 1) + 1 = 5\\beta - 3 \\end{cases} \\Leftrightarrow \\alpha = 3, \\beta = 2.\n$$\n* If $xy = 2$, then $x = 1, y = 2$ or $x = 2, y = 1$, and so we have the systems:\n$$\n\\begin{cases} \\alpha - 1 = \\beta \\\\ 4\\alpha + 2 = 5\\beta - 3 \\end{cases} \\Leftrightarrow \\begin{cases} \\alpha = \\beta + 1 \\\\ 4(\\beta + 1) + 2 = 5\\beta - 3 \\end{cases} \\Leftrightarrow \\alpha = 10, \\beta = 9.\n$$\n\n$$\n\\begin{cases} \\alpha - 1 = 2\\beta \\\\ 2\\alpha + 1 = 5\\beta - 3 \\end{cases} \\Rightarrow \\begin{cases} \\alpha = 2\\beta + 1 \\\\ 2(2\\beta + 1) + 1 = 5\\beta - 3 \\end{cases} \\Rightarrow \\alpha = 13, \\beta = 6.\n$$\n\nSince $\\beta|\\alpha - 1$, we have:\n$$\n\\alpha - 1 = \\beta\\kappa \\quad (1),\n$$\nFor some positive integer $\\kappa$. Moreover, $2\\alpha + 1|5\\beta - 3$, and hence there exists $\\lambda \\in \\mathbb{N}^+$ such that\n$$\n5\\beta - 3 = (2\\alpha + 1)\\lambda,\n$$\nWhich, because of (1), becomes:\n$$\n5\\beta - 3 = \\lambda(2\\beta\\kappa + 3) \\Rightarrow \\beta(2\\lambda\\kappa - 5) = -3\\lambda - 3 \\quad (2)\n$$\nSince $-3\\lambda - 3 < 0$, we get $2\\lambda\\kappa - 5 < 0$, and hence\n$$\n2\\lambda\\kappa \\le 4 \\Leftrightarrow \\lambda\\kappa \\le 2.\n$$\nTherefore, we have:\n$$\n\\lambda = \\kappa = 1 \\text{ or } \\lambda = 2, \\kappa = 1, \\text{ or } \\lambda = 1, \\kappa = 2.\n$$\nBy substitution in (2), in the first case we find $\\beta = 2$ and $a = 3$.\nIn the second case we find $\\beta = 9$ and $a = 10$. Finally, in the third case we find $\\beta = 6$, $a = 13$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(k, n)$ of positive integers satisfying the equation\n$$\n1! + 2! + \\cdots + k! = 1 + 2 + \\cdots + n.\n$$",
    "options": [],
    "answer": "(1, 1), (2, 2), (5, 17)",
    "solution": "We first compute the entries of the following matrix\n\n| $k$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|-----|---|---|---|---|---|----|-----|------|-------|--------|\n| $k!$ | 1 | 2 | 6 | 24 | 120 | 720 | 5040 | 40320 | 362880 | 3628800 |\n| $1!+2!+\\cdots+k!$ | 1 | 3 | 9 | 33 | 153 | 873 | 5913 | 46233 | 409113 | 4037913 |\n\nObviously, the pairs $(k, n) = (1, 1)$ and $(k, n) = (2, 2)$ are solutions. We will show that the unique solution with $k > 2$ is $(k, n) = (5, 17)$.\n\nWe observe that since $k!$ is divided by $100$ for $k \\geq 10$, the sum $1!+2!+\\cdots+k!$ leaves a remainder $13$ when divided by $100$ for $k \\geq 9$. If equality holds\n$$\n1! + 2! + \\cdots + k! = 1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}\n$$\nfor some $k \\geq 10$, then there is some natural number $m$ such that\n$$\nn(n + 1) = 100m + 26\n$$\nor\n$$\nn^2 + n - 100m - 26 = 0.\n$$\nThe discriminant is $\\Delta = 400m + 105$ and it should be a perfect square. On the other hand we cannot have\n$$\nx^2 \\equiv 5 \\pmod{100},\n$$\nas otherwise $25$ would divide $5$. Therefore the given relation cannot hold for $k \\ge 9$, as well as, for $k = 7$. Since\n$$\n9 = 3^2, \\quad 153 = 3^2 \\cdot 17, \\quad 873 = 3^2 \\cdot 97 \\text{ and } 46233 = 3^2 \\cdot 11 \\cdot 467,\n$$\nwe observe that $2 \\cdot 153 = 18 \\cdot 17$, a product of two consecutive integers $k > 2$. Therefore the only solution is $(k, n) = (5, 17)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31341,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a, b, c$ are real numbers such that two of them have difference greater than $\\frac{1}{2\\sqrt{2}}$, prove that there exists an integer $x$ such that\n$$\nx^2 - 4(a + b + c)x + 12(ab + bc + ca) < 0.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The discriminant equals to\n$$\n\\begin{aligned}\n\\Delta &= 16(a + b + c)^2 - 48(ab + bc + ca) \\\\\n&= 8((a - b)^2 + (b - c)^2 + (c - a)^2).\n\\end{aligned}\n$$\nSince two of the numbers are at least $\\frac{1}{2\\sqrt{2}}$ apart, the square of their difference will be greater than $1/8$, so $\\Delta > 1$. Since the discriminant is positive, the trinomial has two real roots, let $\\rho_1 > \\rho_2$ between which the sign of the trinomial is negative. In addition, we have:\n$$\n\\rho_1 - \\rho_2 = \\frac{4(a + b + c) + \\sqrt{\\Delta}}{2} - \\frac{4(a + b + c) - \\sqrt{\\Delta}}{2} = \\sqrt{\\Delta} > 1,\n$$\nso between $\\rho_1, \\rho_2$ there is an integer, say $x$, which makes the given trinomial negative according to the above.\n\n\nSolution 2:\nConsider the function\n$$\n\\begin{aligned}\nf(x) &= x^2 - 4(a + b + c)x + 12(ab + bc + ca) \\\\\n&= x(x - 4(a + b + c)) + 12(ab + bc + ca).\n\\end{aligned}\n$$\nObserve that\n$$\n\\begin{aligned}\nf(2(a + b + c)) &= -4(a + b + c)^2 + 12(ab + bc + ca) \\\\\n&= -2((a - b)^2 + (b - c)^2 + (c - a)^2) < 0.\n\\end{aligned}\n$$\nThe graph of $f$ is a parabola, which is convex and has the vertical line $x = 2(a + b + c)$ as its axis of symmetry. Therefore\n$$\nf(2(a + b + c) + \\frac{1}{2}) = f(2(a + b + c) - \\frac{1}{2}).\n$$\n\n$$\n\\begin{align*}\n& f(2(a+b+c) + \\frac{1}{2}) \\\\\n&= 2(a+b+c) + \\frac{1}{2}(-2(a+b+c) + \\frac{1}{2}) + 12(ab+bc+ca) \\\\\n&= \\frac{1}{4} - 4(a+b+c)^2 + 12(ab+bc+ca) \\\\\n&= \\frac{1}{4} - 2((a-b)^2 + (b-c)^2 + (c-a)^2) \\\\\n&= \\frac{1 - 8((a-b)^2 + (b-c)^2 + (c-a)^2)}{4} < 0,\n\\end{align*}\n$$\nsince $(a-b)^2 + (b-c)^2 + (c-a)^2 > \\left(\\frac{1}{2\\sqrt{2}}\\right)^2 = \\frac{1}{8}$. This means that the trinomial has negative sign on the interval $[2(a + b + c) - \\frac{1}{2}, 2(a + b + c) + \\frac{1}{2}]$, which has length 1. Therefore, there is an integer on that interval having the desired property.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31342,
    "subject": "Mathematics (Multi-modal)",
    "question": "(A) Prove that for all real numbers $\\kappa, \\lambda, \\mu$ holds the following inequality:\n$$ (\\kappa + \\lambda + \\mu)^2 \\ge 3(\\kappa\\lambda + \\lambda\\mu + \\mu\\kappa). $$\nWhen is equality valid?\n\n(B) Let $x, y, z$ be positive real numbers and let $\\alpha, \\beta$ be real numbers such that\n$$\n\\alpha(x + y + z) = \\beta(xy + yz + zx) = xyz.\n$$\nProve that $\\alpha \\ge 3\\beta^2$. When is equality valid?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(A) It is enough to prove\n$$\n\\begin{aligned}\n& \\kappa^2 + \\lambda^2 + \\mu^2 + 2\\kappa\\lambda + 2\\lambda\\mu + 2\\mu\\kappa \\ge 3\\kappa\\lambda + 3\\lambda\\mu + 3\\mu\\kappa \\\\\n\\Leftrightarrow & \\kappa^2 + \\lambda^2 + \\mu^2 - \\kappa\\lambda - \\lambda\\mu - \\mu\\kappa \\ge 0 \\\\\n\\Leftrightarrow & (\\kappa - \\lambda)^2 + (\\lambda - \\mu)^2 + (\\mu - \\kappa)^2 \\ge 0,\n\\end{aligned}\n$$\nwhich is valid for all real numbers $\\kappa, \\lambda, \\mu$.\nEquality holds if and only if $\\kappa = \\lambda = \\mu$.\n\n(B) From the given relation we get:\n$$\n\\frac{1}{\\alpha} = \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} \\quad \\text{and} \\quad \\frac{1}{\\beta} = \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}.\n$$\nThus we have $\\alpha, \\beta > 0$ and from question (A) by putting $\\kappa = \\frac{1}{x}, \\lambda = \\frac{1}{y}, \\mu = \\frac{1}{z}$, we get\n$$\n\\frac{1}{\\beta^2} = \\left(\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z}\\right)^2 \\ge 3 \\left(\\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx}\\right) \\Rightarrow \\alpha \\ge 3\\beta^2.\n$$\nEquality holds if and only if $x = y = z$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31343,
    "subject": "Mathematics (Multi-modal)",
    "question": "We consider the functions $f: \\mathbb{R} \\to \\mathbb{R}$ for which:\n$f(x+y)f(x-y) \\geq (f(x))^2 - (f(y))^2$, for all $x, y \\in \\mathbb{R}$.\nWe suppose that there exists $x_0, y_0 \\in \\mathbb{R}$ such that\n$$\nf(x_0 + y_0)f(x_0 - y_0) > (f(x_0))^2 - (f(y_0))^2.\n$$\nProve that $f(x) \\geq 0$, for every $x \\in \\mathbb{R}$ or $f(x) \\leq 0$, for every $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all values of the integer $v$ for which the number $v^2 + 10v + 160$ is a perfect square.",
    "options": [],
    "answer": "[-72, -26, -16, -8, -2, 6, 16, 62]",
    "solution": "Let $v^2 + 10v + 160 = \\kappa^2$, where $\\kappa$ is a positive integer. Then we have:\n$$\n\\begin{align*}\nv^2 + 10v + 160 = \\kappa^2 &\\Leftrightarrow (v+5)^2 + 135 = \\kappa^2 \\Leftrightarrow \\kappa^2 - (v+5)^2 = 135 \\\\\n&\\Leftrightarrow (\\kappa - |v+5|)(\\kappa + |v+5|) = 135 = 1 \\cdot 3^3 \\cdot 5, \\text{ where } \\kappa > |v+5| \\\\\n&\\Leftrightarrow \\begin{cases} \\kappa - |v+5| = 1 \\\\ \\kappa + |v+5| = 135 \\end{cases} \\text{ or } \\begin{cases} \\kappa - |v+5| = 3 \\\\ \\kappa + |v+5| = 45 \\end{cases} \\\\\n&\\text{or } \\begin{cases} \\kappa - |v+5| = 5 \\\\ \\kappa + |v+5| = 27 \\end{cases} \\text{ or } \\begin{cases} \\kappa - |v+5| = 9 \\\\ \\kappa + |v+5| = 15 \\end{cases} \\\\\n&\\Leftrightarrow \\begin{cases} \\kappa = 68 \\\\ |v+5| = 67 \\end{cases} \\text{ or } \\begin{cases} \\kappa = 24 \\\\ |v+5| = 21 \\end{cases} \\text{ or } \\begin{cases} \\kappa = 16 \\\\ |v+5| = 11 \\end{cases} \\text{ or } \\begin{cases} \\kappa = 12 \\\\ |v+5| = 3 \\end{cases}\n\\end{align*}\n$$\n\n$$\n\\begin{array}{l}\n\\Leftrightarrow \\left\\{ \\begin{array}{l} \\kappa = 68 \\\\ v + 5 = \\pm 67 \\end{array} \\right\\} \\text{ or } \\left\\{ \\begin{array}{l} \\kappa = 24 \\\\ v + 5 = \\pm 21 \\end{array} \\right\\} \\text{ or } \\left\\{ \\begin{array}{l} \\kappa = 16 \\\\ v + 5 = \\pm 11 \\end{array} \\right\\} \\text{ or } \\left\\{ \\begin{array}{l} \\kappa = 12 \\\\ v + 5 = \\pm 3 \\end{array} \\right\\} \\\\[1.5ex]\n\\Leftrightarrow \\left\\{ \\begin{array}{l} \\kappa = 68 \\\\ v = 62 \\ \\text{or} \\ v = -72 \\end{array} \\right\\} \\text{ or } \\left\\{ \\begin{array}{l} \\kappa = 24 \\\\ v = 16 \\ \\text{or} \\ v = -26 \\end{array} \\right\\} \\\\[1.5ex]\n\\text{or } \\left\\{ \\begin{array}{l} \\kappa = 16 \\\\ v = 6 \\ \\text{or} \\ v = -16 \\end{array} \\right\\} \\text{ or } \\left\\{ \\begin{array}{l} \\kappa = 12 \\\\ v = -2 \\ \\text{or} \\ v = -8 \\end{array} \\right\\}.\n\\end{array}\n$$\nTherefore $v \\in \\{-72, -26, -16, -8, -2, 6, 16, 62\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC < BC$ and circumcircle $\\Gamma_1$ of center $O$. We consider the circle $\\Gamma_2$ with center $D$ lying on the circle $\\Gamma_1$, and tangent to the line $BC$ at the point $E$ and tangent to the extension of the side $AB$ at point $F$. The circles $\\Gamma_1$ and $\\Gamma_2$ intersect at the points $K$ and $G$ (the point $K$ lies in the interior of the triangle $BFE$). If the line $KG$ intersects the lines $FE$ and $CD$ at the points $M$ and $N$, respectively, prove that the quadrilateral $BCNM$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have $DE \\perp BC$ and $DF \\perp AB$ (because $\\Gamma_2$ is tangent to the side $BC$ at $E$ and to the line $AB$ at $F$).\n\n![](attached_image_1.png)\nFigure 4\n\n![](attached_image_2.png)\nFigure 5\n\nHence the quadrilateral $BEDF$ is cyclic and let $\\Gamma_3$ be its circumcircle.\nMoreover we have the equalities:\n$$\nBE = BF \\text{ and } DE = DF. \\tag{1}\n$$\n\nThe $EF$ is the common chord of the circles $\\Gamma_2$ and $\\Gamma_3$ and $KG$ is the common chord of the circles $\\Gamma_1$ and $\\Gamma_2$. Since $EF$ and $KG$ meet at $M$, the common chord $BD$ of the circles $\\Gamma_1$ and $\\Gamma_3$ will pass through $M$ (radical center of the circles $\\Gamma_1, \\Gamma_2, \\Gamma_3$).\nSince the orthogonal triangles $AFD$, $CED$ are equal ($FD = ED$, $\\angle FAD = \\angle ECD$), we have $DA = DC$. Hence $DO$ is the perpendicular bisector of $AC$, and since $DO \\perp KG$, it follows that $AC \\parallel KG$ and $\\angle ACD = \\angle MND$.\nSince $BD$ bisects the angle $\\angle FBE$ and $BDCA$ is inscribed into the circle $\\Gamma_1$ we have $\\angle MBC = \\angle FBM = \\angle ACD = \\angle MND$. Hence $BCNM$ is cyclic.\nWe have $DE \\perp BC$ and $DF \\perp AB$ (because $\\Gamma_2$ is tangent to the side $BC$ at $E$ and to the line $AB$ at $F$). Hence the quadrilateral $BEDF$ is cyclic and let $\\Gamma_3$ be its circumcircle.\nWe consider the inversion with respect to the circle $\\Gamma_2$ (with center $D$). Since circle $\\Gamma_3$ is passing through $D$, its image is the line of common chord $FE$, and therefore:\nThe image of $B$ belongs to the line $FE$. (5)\nSince circle $\\Gamma_1$ is passing through $D$, its image is the line of common chord $KG$, and therefore:\nThe image of $B$ belongs to the line $KG$. (6)\nFrom (5), (6) we conclude that the image of $B$ under the inversion we have considered is the point $M$. Also, $C$ has image the point $N$. Hence the quadrilateral $BMNC$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B \\Gamma$ be an acute angled triangle with circumcircle $\\omega$. A circle $\\gamma$ with center $A$ intersects the arc $AB$ of the circle $\\omega$, not containing $\\Gamma$, at point $\\Delta$ and the arc $A\\Gamma$, not containing $B$, at point $E$. We suppose that the point of intersection $K$ of the lines $BE$ and $\\Gamma\\Delta$ belongs to the $\\gamma$. Prove that the line $AK$ is perpendicular to the line $B\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "From the relationship of a central angle and an inscribed angle that go on the same arc $\\Delta K$ of the circle $\\gamma$, we have:\n$$\n\\angle A\\Delta K = 2 \\cdot \\angle E\\Delta K \\quad (1)\n$$\nAlso we have the equality of inscribed angles\n$$\n\\angle E\\Delta K = \\angle AB\\Delta \\quad (2)\n$$\nFrom relations (1) and (2) it follows that $\\angle A\\Delta K = 2 \\cdot \\angle AB\\Delta$, that is, $AB$ is the bisector of the angle $\\angle A\\Delta K$.\n\n![](attached_image_1.png)\nFigure 1\n\nSince $A\\Delta = AK$, (radii of the circle $\\gamma$), the triangle $\\Delta AK$ is isosceles and hence $AB \\perp \\Delta K$, that is $AB \\perp \\Gamma\\Delta$. Similarly we get that $A\\Gamma \\perp BE$. Hence $K$ is the orthocenter of the triangle $AB\\Gamma$, and therefore $AK \\perp B\\Gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma\\Delta E$ be a convex pentagon $A\\hat{B}\\Gamma = A\\hat{E}\\Delta = 90^\\circ$. Suppose that the midpoint, let $M$, of the side $\\Gamma\\Delta$ is the circumcenter of the triangle $ABE$. Let $O$ be the circumcenter of the triangle $A\\Gamma\\Delta$. Prove that line $AO$ passes through the midpoint $N$ of segment $BE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31348,
    "subject": "Mathematics (Multi-modal)",
    "question": "(i) If $x$, $y$ are positive real numbers, prove that:\n$$\n\\frac{4}{x+y} \\le \\frac{1}{x} + \\frac{1}{y}.\n$$\n\n(ii) If $\\alpha$, $\\beta$, $\\gamma$, $\\delta$ are positive real numbers, prove that\n$$\n\\frac{2}{(\\alpha + \\beta)(\\gamma + \\delta) + (\\beta + \\gamma)(\\alpha + \\delta)} \\le \\frac{1}{(\\alpha + \\gamma)(\\beta + \\delta) + 4\\alpha\\gamma} + \\frac{1}{(\\alpha + \\gamma)(\\beta + \\delta) + 4\\beta\\delta}.\n$$\nDetermine when equality holds.",
    "options": [],
    "answer": "Equality holds if and only if alpha times gamma equals beta times delta.",
    "solution": "(i) Since $x$, $y$ are positive real numbers, we have:\n$$\n\\frac{4}{x+y} \\le \\frac{1}{x} + \\frac{1}{y} \\Leftrightarrow \\frac{4}{x+y} \\le \\frac{x+y}{xy} \\Leftrightarrow (x+y)^2 \\ge 4xy \\Leftrightarrow (x-y)^2 \\ge 0.\n$$\n\n(ii)\nLet\n$$\nx = (\\alpha + \\gamma)(\\beta + \\delta) + 4\\alpha\\gamma, \\quad y = (\\alpha + \\gamma)(\\beta + \\delta) + 4\\beta\\delta,\n$$\nand thus:\n$$\n\\begin{aligned}\nx + y &= 2(\\alpha + \\gamma)(\\beta + \\delta) + 4(\\alpha\\gamma + \\beta\\delta) \\\\\n(\\alpha + \\beta)(\\gamma + \\delta) + (\\beta + \\gamma)(\\alpha + \\delta) &= \\alpha\\gamma + \\alpha\\delta + \\beta\\gamma + \\beta\\delta + \\alpha\\beta + \\beta\\delta + \\alpha\\gamma + \\gamma\\delta \\\\\n&= (\\alpha\\beta + \\alpha\\delta + \\beta\\gamma + \\gamma\\delta) + 2(\\alpha\\gamma + \\beta\\delta) \\\\\n&= (\\alpha + \\gamma)(\\beta + \\delta) + 2(\\alpha\\gamma + \\beta\\delta) = \\frac{x+y}{2}\n\\end{aligned}\n$$\nTherefore the inequality becomes\n$$\n\\frac{4}{x+y} \\le \\frac{1}{x} + \\frac{1}{y},\n$$\nwhich is valid because of (i).\n\nThe equality holds if and only if $x = y \\Leftrightarrow x - y = 0 \\Leftrightarrow \\alpha\\gamma = \\beta\\delta$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31349,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. We consider two finite subsets $A, B$ of the set of integers, such that the subset $A$ contains at most $n$ elements and let $\\Gamma$ be a subset of the set $\\{(\\alpha, \\beta): \\alpha \\in A, \\beta \\in B\\}$. Achilles writes on a blackboard all possible differences $\\alpha - \\beta$, with $(\\alpha, \\beta) \\in \\Gamma$. Let $d$ be the number of all these differences. Next, Achilles writes in a second blackboard all triads $(\\kappa, \\lambda, \\mu)$ with $(\\kappa, \\lambda) \\in \\Gamma, (\\kappa, \\mu) \\in \\Gamma$. Let $p$ be the number of all these triads. Prove that:\n$$\nn \\cdot p \\ge d^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For the set $X$, we use the symbol $|X|$ for the number of its elements. The number of all possible differences $\\alpha - \\beta$, with $(\\alpha, \\beta) \\in \\Gamma$, is at most equal to the number of the elements of the set $\\Gamma$, that is\n$$\nd \\leq |\\Gamma|. \\qquad (*)\n$$\nLet $A = \\{a_1, a_2, ..., a_m\\}$. For each $i$, we suppose that there exist $b_i$ pairs $(a_i, b) \\in \\Gamma$. Then there exist $b_i^2$ triads $(a_i, \\lambda, \\mu)$, with $(a_i, \\lambda) \\in \\Gamma$ and $(a_i, \\mu) \\in \\Gamma$. Therefore, there exist totally $p = b_1^2 + \\dots + b_m^2$ triads $(\\kappa, \\lambda, \\mu)$ with $(\\kappa, \\lambda) \\in \\Gamma, (\\kappa, \\mu) \\in \\Gamma$.\nFrom Cauchy-Schwarz inequality we get\n$$\nm(b_1^2 + \\cdots + b_m^2) \\ge (b_1 + \\cdots + b_m)^2 \\quad (1)\n$$\nThe left hand side is equal to $|A| \\cdot p$, while the right hand side is equal to $|\\Gamma|^2$. Since it is given that $|A| \\le n$, by using relations (1) and (*) we get:\n$$\nn \\cdot p \\ge |\\Gamma|^2 \\ge d^2.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31350,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $7 \\times 7$ square is divided into $49$ unit squares with sides parallel to those of $T$. We color each of the $49$ unit squares with two colors, red and blue, so that the following two conditions are simultaneously satisfied:\n\n(i) There are exactly $4$ rows in each of which the blue squares are more than the red squares.\n\n(ii) There are exactly $4$ columns in each of which there are more red squares than blue.\n\nFrom all the resulting monochromatic squares, let $\\Pi$ be the maximum side length of a monochromatic square. Find the maximum possible value of $\\Pi$ for all possible colorings satisfying conditions (i) and (ii).\n\n(A square of the grid is called monochromatic if all its cells have the same color. For example, after the incomplete coloring in the adjacent figure, there are two $2 \\times 2$ monochromatic squares in the lower left, painted red).\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "3",
    "solution": "In fact, if such a square, let $S$, existed, then each of the rows and columns of $T$ to which the sides of $S$ belong have at least $4$ squares of the same color (let with color blue in the figure $6$). This contradicts condition (ii). Therefore, for every coloring, each monochromatic square has side less or equal to $3$. The number $3$ is maximal, as we can see in figures $6$ and $7$.\n\n![](attached_image_2.png)\nFigure 6\n\n![](attached_image_3.png)\nFigure 7",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha_1, \\alpha_2, \\dots, \\alpha_\\nu, \\beta_1, \\beta_2, \\dots, \\beta_\\nu$ be $2\\nu$ positive integers such that the $\\nu + 1$ products\n\n$$\n\\begin{align*}\n\\gamma_1 &= \\alpha_1 \\alpha_2 \\alpha_3 \\cdots \\alpha_{\\nu-1} \\alpha_\\nu, & \\gamma_2 &= \\beta_1 \\alpha_2 \\alpha_3 \\cdots \\alpha_{\\nu-1} \\alpha_\\nu, \\\\\n\\gamma_3 &= \\beta_1 \\beta_2 \\alpha_3 \\cdots \\alpha_{\\nu-1} \\alpha_\\nu, \\dots, & & \\\\\n\\gamma_\\nu &= \\beta_1 \\beta_2 \\beta_3 \\cdots \\beta_{\\nu-1} \\alpha_\\nu, & \\gamma_{\\nu+1} &= \\beta_1 \\beta_2 \\beta_3 \\cdots \\beta_\\nu\n\\end{align*}\n$$\nform a strictly increasing arithmetic progression in that order. Determine the smallest positive integer that could be the common difference of such an arithmetic progression.",
    "options": [],
    "answer": "ν!",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there exists integer $n \\ge 1$, such that the number of the pairs $(a, b)$ of positive integers satisfying the equation\n$$\n\\frac{1}{a-b} - \\frac{1}{a} + \\frac{1}{b} = \\frac{1}{n}\n$$\nis greater than 2024.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We seek solutions $(a, b)$ with $a - b = k b$, that is $a = (k+1) b$ for some integer $k > 0$. Then the left part of the given equation is equal to\n$$\n\\frac{1}{k b} - \\frac{1}{(k+1) b} + \\frac{1}{b} = \\frac{1}{b} \\left( \\frac{1}{k} - \\frac{1}{k+1} + 1 \\right) = \\frac{1}{b} \\cdot \\frac{k(k+1)+1}{k(k+1)}\n$$\nand so we have\n$$\nb = \\frac{(k(k + 1) + 1)n}{k(k + 1)} = n + \\frac{n}{k(k + 1)}.\n$$\nTherefore, it is enough to select as $n$ a positive integer with more than 2024 positive integers of the form $k(k + 1)$. In fact, for example, if $(p_n)$ is the sequence of primes with $p_1=2, p_2=3, \\dots$, then the number\n$$\nn = p_1(p_1+1) p_2(p_2+1) \\cdots p_{2025}(p_{2025}+1),\n$$\nsatisfies the problem.\nThe equation is written in the form\n$$\n\\frac{a^2 - ab + b^2}{ab(a - b)} = \\frac{1}{n}.\n$$\nWe put $d = (a, b)$, and so $a = d x, b = d y$ and the equation becomes:\n$$\n\\frac{x y (x - y)}{x^2 - x y + y^2} = \\frac{n}{d}\n$$\nIf $(n, d) = S$, then $n = S u$ and $d = S v$. Therefore it follows\n$$\nx y (x - y) = u, \\quad x^2 - x y + y^2 = v.\n$$\nBy putting $x y = s$ and $x - y = t$, the two relations take the form $s t = u$ and $t^2 - 3 s = v$. Then $x = \\frac{s}{y}$ and $\\frac{s}{y} - y = t \\Leftrightarrow y^2 + y t - s = 0$.\nThe discriminant of the last equation is $\\Delta = t^2 + 4 s$. In order to integer solution, it is enough to choose $s = k (t + k)$ for some positive integer $k$. Then the equation has the solution $y = k$.\nThis gives $x = k + t$, which in turn gives $u = s t = k t (k + t)$ and $v = t^2 - 3 k (t + k)$.\nWe should have $v > 0$ (one way to achieve this is, $k = 1$ and $t \\ge 4$).\nThe construction now has as follows: We select a positive integer $n$ with 2024 positive integers of the form $k t (k + t)$. For each one of them is determined a triad $(x, y, v)$. From this triad is determined a pair $(a, b)$ satisfying the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $k$ be a positive integer. Find the maximal length $L$ of a sequence $a_1, \\dots, a_L$ containing positive integers that satisfies the following two properties:\n* Each term of the sequence is smaller or equal to $2^k$.\n* There are no consecutive terms $a_i, a_{i+1}, \\dots, a_j$ (with $1 \\le i < j \\le L$) and signs $s_i, s_{i+1}, \\dots, s_j \\in \\{1, -1\\}$ such that\n$$s_i a_i + s_{i+1} a_{i+1} + \\dots + s_j a_j = 0.$$",
    "options": [],
    "answer": "2k + 1",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB\\Gamma$ be a triangle with circumcircle $\\omega(O, R)$. A circle $\\gamma$ passes through $O$ and $B$ and is tangent to the line $AB$ at $B$. Let the circle $\\kappa$ meets the circle $\\omega(O, R)$ for a second time at $P \\neq B$. A circle passes through $P$ and $\\Gamma$ and is tangent to the line $A\\Gamma$ at $\\Gamma$ and meets the circle $\\gamma$ at $M \\neq P$. Prove that: $MP = M\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we will prove that $M$ belongs to the side $B\\Gamma$. It is enough to prove that\n$$\nB\\hat{M}P + P\\hat{M}\\Gamma = \\varphi + \\theta = A\\hat{\\Gamma}P + P\\hat{\\Gamma}x = 180^\\circ.\n$$\nThen we have:\n$$\nM\\hat{P}\\Gamma = O\\hat{P}\\Gamma - O\\hat{P}M = O\\hat{P}P - O\\hat{P}M = M\\hat{\\Gamma}P \\\\\n\\Rightarrow \\Gamma \\text{ isosceles with } MP = M\\Gamma.\n$$\n\n![](attached_image_1.png)\nFigure 8",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there is an infinite number of triads of positive integers $(x, y, z)$ such that\n$$\nx^2 + y^2 + z^2 + xy + yz + zx = 6xyz.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We write the given relation in the form\n$$\nx^2 + y^2 + xy = (6xy - x - y - z)z^2.\n$$\n\nWe observe that substituting $z$ by $6xy - x - y - z$, leaves invariant the right part of the equality. Thus, if $(x, y, z)$ is a solution of the given equation with $x > y > z$, then $(6xy - x - y - z, x, y)$, with $6xy - x - y - z > x > y$, is a solution, as well, because\n$$(6xy - x - y - z) - x = \\frac{(x^2 + y^2 + xy)}{z} - x = \\frac{(x(x - z) + y^2 + xy)}{z} > 0.$$ \n\nSince $(x, y, z) = (1, 1, 1)$ is a solution of the given equation, we obtain a new solution $(x, y, z) = (3, 1, 1)$, and then we find the solution $(x, y, z) = (13, 3, 1)$. Thus we conclude that the given equation has an infinite number of solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that for any triangle $ABC$ with area $S$ and circumradius $R$,\n$$\n(a) \\quad \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} \\le \\frac{3\\sqrt{3}}{4S} + \\left(\\frac{1}{a} - \\frac{1}{b}\\right)^2 + \\left(\\frac{1}{b} - \\frac{1}{c}\\right)^2 + \\left(\\frac{1}{c} - \\frac{1}{a}\\right)^2,\n$$\n$$\n(b) \\quad \\tan \\frac{A}{2} + \\tan \\frac{B}{2} + \\tan \\frac{C}{2} \\le \\frac{9R^2}{4S}.\n$$\nHere $AB = c$, $BC = a$ and $CA = b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) We have\n$$\n\\begin{aligned}\n& \\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} - \\left(\\frac{1}{a} - \\frac{1}{b}\\right)^2 - \\left(\\frac{1}{b} - \\frac{1}{c}\\right)^2 - \\left(\\frac{1}{c} - \\frac{1}{a}\\right)^2 \\\\\n&= 2\\left(\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}\\right) - \\left(\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2}\\right) \\\\\n&= \\left(\\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}\\right) - \\frac{1}{2}\\left[\\left(\\frac{1}{a} - \\frac{1}{b}\\right)^2 + \\left(\\frac{1}{b} - \\frac{1}{c}\\right)^2 + \\left(\\frac{1}{c} - \\frac{1}{a}\\right)^2\\right] \\\\\n&\\le \\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca} = \\frac{a+b+c}{abc} \\\\\n&= \\frac{a+b+c}{4RS} = \\frac{\\sin A + \\sin B + \\sin C}{2S},\n\\end{aligned}\n$$\nwhere we have used the fact $S = \\frac{abc}{4R}$ and the extended sine law. Now, note that $f(x) = \\sin x$ is a concave function on $(0, \\pi)$ since $f''(x) = -\\sin x < 0$. Therefore, by Jensen's inequality, we have\n$$\n\\frac{\\sin A + \\sin B + \\sin C}{2S} \\le \\frac{3}{2S} \\sin \\frac{A+B+C}{3} = \\frac{3\\sqrt{3}}{4S}.\n$$\nThe result follows readily. Equality holds when $a = b = c$.\n\n(b) Let $r$ and $s$ be the inradius and the semiperimeter of $\\triangle ABC$. Then we have\n$$ \\tan \\frac{A}{2} = \\frac{r}{s-a} $$\netc. Also, using $S = rs = \\sqrt{s(s-a)(s-b)(s-c)}$, we have\n$$\n\\begin{align*}\n& \\tan \\frac{A}{2} + \\tan \\frac{B}{2} + \\tan \\frac{C}{2} \\le \\frac{9R^2}{4S} \\\\\n\\Leftrightarrow & \\qquad \\frac{r}{s-a} + \\frac{r}{s-b} + \\frac{r}{s-c} \\le \\frac{9R^2}{4S} \\\\\n\\Leftrightarrow & \\frac{rs[(s-a)(s-b) + (s-b)(s-c) + (s-c)(s-a)]}{s(s-a)(s-b)(s-c)} \\le \\frac{9R^2}{4S} \\\\\n\\Leftrightarrow & 3s^2 - 2(a+b+c)s + (ab+bc+ca) \\le \\frac{9R^2}{4} \\\\\n\\Leftrightarrow & -(a+b+c)^2 + 4(ab+bc+ca) \\le 9R^2.\n\\end{align*}\n$$\nRecall that\n$$\n9R^2 - (a^2 + b^2 + c^2) = OH^2 \\ge 0,\n$$\nwhere $O$ and $H$ are the circumcentre and orthocentre of $\\triangle ABC$ respectively.\nTherefore, it suffices to prove\n$$\n-(a + b + c)^2 + 4(ab + bc + ca) \\le a^2 + b^2 + c^2.\n$$\nThis is equivalent to\n$$\nab + bc + ca \\le a^2 + b^2 + c^2,\n$$\nwhich is a well-known inequality. Thus, the inequality is proven. Equality holds when $\\triangle ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define a $k$-clique to be a set of $k$ people such that every pair of them know each other (knowing is mutual). At a certain party, there are two or more $3$-cliques, but no $5$-clique. Every pair of $3$-cliques has at least one person in common. Prove that there exists at least one, and not more than two persons at the party, whose departure (or simultaneous departure) leaves no $3$-clique remaining.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We consider two cases.\n\n**Case 1.** There exist two $3$-cliques sharing two common people.\nSuppose the two $3$-cliques are $\\{A, B, C\\}$ and $\\{A, B, D\\}$. If all $3$-cliques contain $A$ or $B$, we are done as we can remove $A$ and $B$. If there exists a $3$-clique without $A$ and $B$, it must be $\\{C, D, E\\}$ since every pair of $3$-cliques has a common person. We claim that there is no $3$-clique if $C$ and $D$ are removed.\nIndeed, note that $\\{A, C, D\\}$, $\\{B, C, D\\}$ and $\\{C, D, E\\}$ are existing $3$-cliques since all pairs of people in these $3$-cliques know each other from above. Therefore, the only possible $3$-clique without $C$ and $D$ is $\\{A, B, E\\}$. Then $\\{A, B, C, D, E\\}$ forms a $5$-clique, which is a contradiction.\n\n**Case 2.** Any pair of $3$-cliques shares exactly one common person.\nSuppose two $3$-cliques are $\\{A, B, C\\}$ and $\\{A, D, E\\}$. If all $3$-cliques contain $A$, we are done as we can remove $A$. If there exists a $3$-clique without $A$, it must be $\\{B, D, F\\}$ up to renaming of the people. Then we have a $3$-clique $\\{A, B, D\\}$ sharing two common people with $\\{A, B, C\\}$. This is a contradiction.\nThe proof is complete since all cases are covered.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that $\\cos \\frac{\\pi}{7}$ is not of the form $p + \\sqrt{q} + \\sqrt[3]{r}$, where $p, q$ and $r$ are rational numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha = \\cos \\frac{\\pi}{7}$. As $\\cos \\frac{4\\pi}{7} + \\cos \\frac{3\\pi}{7} = 0$, we have\n$$\n2(2\\alpha^2 - 1)^2 - 1 + (4\\alpha^3 - 3\\alpha) = 0.\n$$\nThis is the same as $(\\alpha + 1)(8\\alpha^3 - 4\\alpha^2 - 4\\alpha + 1) = 0$. Clearly, $\\alpha \\ne -1$. Therefore, $\\alpha$ is a root of\n$$\nP(x) = 8x^3 - 4x^2 - 4x + 1.\n$$\nSuppose on the contrary that $\\alpha = p + \\sqrt{q} + \\sqrt[3]{r}$. Then $\\alpha$ is also a root of\n$$\nQ(x) = (x - p - \\sqrt{q})^3 - r = x^3 - 3(p + \\sqrt{q})x^2 + 3(p + \\sqrt{q})^2 x - (p + \\sqrt{q})^3 - r.\n$$\nConsider the field $F = \\{a + b\\sqrt{q} : a, b \\in \\mathbb{Q}\\}$ (which is just $\\mathbb{Q}$ if $q = 0$). We claim that $P(x)$ has a root in $F$.\nFirstly, note that $P(x) \\ne 8Q(x)$. Indeed, if $P(x) = 8Q(x)$, we must have $q = 0$ in order that all coefficients of $Q(x)$ are rational. Then by comparing the coefficients of $x^3$ and $x^2$, we need $p = \\frac{1}{6}$. But then the coefficients of $x$ do not match. Therefore, $R(x) = \\gcd(P(x), Q(x))$ (over $F[x]$) has degree 1 or 2.\n\n* If $\\deg R = 1$, then since $\\alpha$ is a root of $P(x)$ and $Q(x)$, it is a root of $R(x)$, and hence belongs to $F$.\n* If $\\deg R = 2$, we can write $P(x) = R(x)S(x)$ where $\\deg S = 1$. Then $S(x)$ has a root in $F$, which is also a root of $P(x)$.\nNow, let $\\beta = a + b\\sqrt{q}$ be a root of $P(x)$ in $F$. Then $\\beta$ is a root of $(x-a)^2 = b^2q$, and so the minimal polynomial of $\\beta$ over $\\mathbb{Q}$ has degree at most 2. As this minimal polynomial divides $P(x)$, we find that $P(x)$ is reducible. As $\\deg P = 3$, it must consist of a linear factor. Thus, $P(x)$ has a rational root. However, it is routine to check that none of\n$$\n\\pm 1, \\pm \\frac{1}{2}, \\pm \\frac{1}{4}, \\pm \\frac{1}{8}\n$$\nis a root of $P(x)$. By the rational root theorem, $P(x)$ does not have any rational root. This is a contradiction. Therefore, $\\alpha$ cannot be expressed in the form $p + \\sqrt{q} + \\sqrt[3]{r}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many primes $p$ such that $N_p = p^2$, where $N_p$ is the total number of solutions mod $p$ to the equation $3x^3 + 4y^3 + 5z^3 - y^4z = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By Dirichlet's theorem, there are infinitely many primes $p$ of the form $3k + 2$. We claim that $N_p = p^2$ for all such primes $p$.\n\nIndeed, for each of the $p^2$ pairs of $(y, z)$ modulo $p$, we need to solve\n$$\nx^3 \\equiv -3^{-1}(4y^3 + 5z^3 - y^4z) \\pmod{p}.\n$$\nAs $(p-1, 3) = 1$, this equation has a unique solution in $x$ modulo $p$. This proves $N_p = p^2$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let a sequence of real numbers $a_0, a_1, a_2, \\dots$ satisfies the condition\n$$\n\\sum_{n=0}^{m} a_n \\cdot (-1)^n \\cdot \\binom{m}{n} = 0\n$$\nfor all sufficiently large values of $m$. Show that there exists a polynomial $P$ such that $a_n = P(n)$ for all $n \\ge 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any function $f(x)$ and any integer $m \\ge 0$, define\n$$\n\\Delta_m f(x) = \\sum_{n=0}^{m} f(x+n) (-1)^n \\binom{m}{n}.\n$$\n**Claim.** Let $Q(x)$ be a polynomial of degree $d$. Then $\\Delta_m Q(x) = 0$ for all $m \\ge d$.\n*Proof.* This is a standard result about finite differences of polynomials. Indeed, let $Q_0(x) = Q(x)$ and $Q_{k+1}(x) = Q_k(x) - Q_k(x+1)$ for $k \\ge 0$. It is easy to show that $Q_k(x) = \\Delta_k Q(x)$ for any $k$. Also, it can be proved by induction that $Q_k(x)$ is a polynomial of degree $d-k$ for $k=0, 1, \\dots, d$ (essentially because $x^j - (x+1)^j$ has degree $j-1$), and hence $Q_k(x) = 0$ for all $k \\ge d$. This proves the claim. $\\square$\n\nConsider any function $f(x)$ such that $f(n) = a_n$ for all integer $n \\ge 0$. We are given that $\\Delta_m f(0) = 0$ for all $m \\ge N$ where $N \\in \\mathbb{Z}^+$. By the Lagrange interpolation formula, construct a polynomial $P$ of degree at most $N-1$ such that $P(n) = f(n)$ for $n = 0, 1, \\dots, N-1$. By the claim, we have $\\Delta_N P(0) = 0$. Also, we know that $\\Delta_N f(0) = 0$. Equating these, we find that\n$$\n\\sum_{n=0}^{N} P(n) (-1)^n \\binom{N}{n} = \\sum_{n=0}^{N} f(n) (-1)^n \\binom{N}{n}.\n$$\nSince $P(n) = f(n)$ for $n = 0, 1, \\dots, N-1$, this yields $P(N) = f(N)$. Inductively, we obtain $P(n) = f(n) = a_n$ for all nonnegative integer $n$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a simple graph with $n$ vertices and $m$ edges. Two vertices are called neighbours if there is an edge between them. It turns out that $G$ does not contain any cycle of length from $3$ to $2k$ (inclusive), where $k \\ge 2$ is a given positive integer.\n\na. Prove that it is possible to pick a nonempty set $S$ of vertices of $G$ such that every vertex in $S$ has at least $\\left\\lfloor \\frac{m}{n} \\right\\rfloor$ neighbours that are in $S$. ($\\lfloor x \\rfloor$ stands for the smallest integer larger than or equal to $x$.)\n\nb. Suppose a set $S$ as described in (a) is chosen. Let $H$ be the graph consisting of the vertices in $S$ and the edges between those vertices only. Let $v$ be a vertex of $H$. Prove that at least $\\left\\lfloor \\left( \\frac{m}{n} - 1 \\right)^k \\right\\rfloor$ vertices of $H$ can be reached by starting at $v$ and travelling along the edges of $H$ for at most $k$ steps. (Note that $v$ itself is a vertex that satisfies this condition, since it can be reached by starting at $v$ and travelling along the edges of $H$ for $0$ steps.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Starting from the graph $G$, each time we remove a vertex with degree less than $\\frac{m}{n}$ together with its edges, until there does not exist such a vertex. Since each time we have removed less than $\\frac{m}{n}$ edges, it is impossible that we remove all $n$ vertices since otherwise there is no vertex left but the number of edges is more than $m - n \\cdot \\frac{m}{n} = 0$. Thus, we must end up with a nonempty set $S$ of vertices such that the degree of each vertex is at least $\\left\\lfloor \\frac{m}{n} \\right\\rfloor$.\n\nb. Let $T_j$ be the set of vertices in $S$ that has distance $j$ from $v$. Note that the neighbours of a vertex in $T_j$ can only belong to $T_{j-1}$, $T_j$ or $T_{j+1}$.\n\nFor $1 \\le j \\le k$, a vertex in $T_j$ must be adjacent to a vertex in $T_{j-1}$ by the definition of distances. Note that this vertex is uniquely determined. Otherwise, we can form a cycle of length $j + j = 2j \\le 2k$, contradicting the assumption.\n\nAlso, for $1 \\le j \\le k-1$, a vertex in $T_j$ cannot be adjacent to another vertex in $T_j$, since otherwise we can form a cycle of length $j+1+j = 2j+1 < 2k$, contradicting the assumption.\n\nIt follows that for $1 \\le j \\le k-1$, each vertex in $T_j$ is adjacent to at least $\\frac{m}{n} - 1$ vertices in $T_{j+1}$. By induction, since $|T_0| = 0$, we easily find that\n$$|T_j| \\ge \\left( \\frac{m}{n} - 1 \\right)^j$$\nfor $0 \\le j \\le k$. In particular, we have\n$$|T_k| \\ge \\left[ \\left( \\frac{m}{n} - 1 \\right)^k \\right]$$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer larger than or equal to $2$. On an $n \\times n$ chessboard, $r$ chess pieces have been placed such that\n(i) there is at least one chess piece in each row;\n(ii) there is at least one chess piece in each column; and\n(iii) there is at least one chess piece in each diagonal (here a 'diagonal' refers to the collection of cells whose centres lie on a straight line which makes an angle of $45^\\circ$ with a side of the chessboard). (For instance, in the $3 \\times 3$ chessboard, the $1 \\times 1$ cell forms a diagonal; the $1 \\times 2$ cell and the $2 \\times 1$ cell together form a diagonal; the $1 \\times 1$ cell, $2 \\times 2$ cell and the $3 \\times 3$ cell also form a diagonal (a main diagonal), etc.)\nFind, in terms of $n$, the smallest possible value of $r$.",
    "options": [],
    "answer": "r = 2n if n is even; r = 2n + 1 if n is odd",
    "solution": "The smallest possible value of $r$ is $2n$ if $n$ is even, and is $2n + 1$ if $n$ is odd.\n\n**Case 1.** $n = 2k$ is even\nLet $(i, j)$ be the cell in row $i$ and column $j$. Consider the following $2n$ diagonals:\n$$\n\\begin{aligned}\n\\{(i, j) : i + j = m\\} & \\text{ for } m = 2, 3, \\dots, k + 1, \\\\\n\\{(i, j) : j - i = m\\} & \\text{ for } m = k, k + 1, \\dots, 2k - 1, \\\\\n\\{(i, j) : i - j = m\\} & \\text{ for } m = k, k + 1, \\dots, 2k - 1, \\\\\n\\{(i, j) : i + j = m\\} & \\text{ for } m = 3k + 1, 3k + 2, \\dots, 4k.\n\\end{aligned}\n$$\n\n| A | B | C | D | E | F |\n|---|---|---|---|---|---|\n| B | C |   |   | D | E |\n| C |   |   |   |   | D |\n| I |   |   |   |   | J |\n| H | I |   |   | J | K |\n| G | H | I | J | K | L |\n\nObviously, these diagonals are disjoint. Therefore, one chess piece is needed for each diagonal, and so $r \\geq 2n$.\n\nNow, we put $2n$ chess pieces in the following cells:\n- $(1, j)$ for $j = 1, 2, \\dots, k$,\n- $(i, n)$ for $i = 1, 2, \\dots, k$,\n- $(i, 1)$ for $i = k + 1, k + 2, \\dots, 2k$,\n- $(n, j)$ for $j = k + 1, k + 2, \\dots, 2k$.\n\n| X | X | X |   |   | X |\n|   |   |   |   |   | X |\n|   |   |   |   |   | X |\n| X |   |   |   |   |   |\n| X |   |   |   |   |   |\n| X |   |   | X | X | X |\n\nThen we see that each row, each column and each diagonal consists of a chess piece. This proves $r = 2n$ when $n$ is even.\n\n**Case 2.** $n = 2k + 1$ is odd\nConsider the following $2n + 2$ diagonals:\n$$\n\\begin{aligned}\n\\{(i,j) : i + j = m\\} & \\quad \\text{for } m = 2, 3, \\dots, k + 2, \\\\\n\\{(i,j) : j - i = m\\} & \\quad \\text{for } m = k, k + 1, \\dots, 2k, \\\\\n\\{(i,j) : i - j = m\\} & \\quad \\text{for } m = k, k + 1, \\dots, 2k, \\\\\n\\{(i,j) : i + j = m\\} & \\quad \\text{for } m = 3k + 2, 3k + 3, \\dots, 4k + 2.\n\\end{aligned}\n$$\n\n| A | B | C, D | E | F |\n|---|---|-------|---|---|\n| B | C |       | D | E |\n| C, I |   |   |   | D, J |\n| H | I |   | J | K |\n| G | H | I, J | K | L |\n\nOnly 4 pairs of these diagonals are not disjoint. We call these $C, D, I, J$-diagonals respectively, as labelled in the above example. One chess piece is needed for each of those $4k = 2n-2$ diagonals other than the $C, D, I, J$-diagonals.\n\nIf $r \\le 2n$, then we can put at most 2 chess pieces in the $C, D, I, J$-diagonals. By the pigeonhole principle, one chess piece must occupy at least two such diagonals. WLOG assume there is a chess piece in $(1, k+1)$ occupying the $C, D$-diagonals. In order to occupy the $I, J$-diagonals simultaneously, there must be another chess piece in $(2k+1, k+1)$. But then there is no chess piece in row $k+1$. This is a contradiction. Thus, we have $r \\ge 2n+1$.\n\nNow, we put $2n + 1$ chess pieces in the following cells:\n$$\n\\begin{aligned}\n(1, j) & \\quad \\text{for } j = 1, 2, \\dots, k + 1, \\\\\n(i, n) & \\quad \\text{for } i = 1, 2, \\dots, k, \\\\\n(i, 1) & \\quad \\text{for } i = k + 2, k + 3, \\dots, 2k + 1, \\\\\n(n, j) & \\quad \\text{for } j = k + 1, k + 2, \\dots, 2k + 1, \\\\\n(k + 1, k + 1) &\n\\end{aligned}\n$$\n\n| X | X | X |   | X |\n|   |   |   |   | X |\n|   |   | X |   |   |\n| X |   |   |   |   |\n| X |   | X | X | X |\n\nThen we see that each row, each column and each diagonal consists of a chess piece. This proves $r = 2n + 1$ when $n$ is odd.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer with the following property: $2^n - 1$ divides a number of the form $m^2 + 81$, where $m$ is a positive integer. Find all possible $n$.",
    "options": [],
    "answer": "n = 2^k for any nonnegative integer k",
    "solution": "$n$ can be any nonnegative integral power of $2$.\n\nIf $n$ has an odd divisor $d \\ge 3$, then $2^d - 1 \\mid 2^n - 1 \\mid m^2 + 81$. Since $2^d - 1 \\equiv 3 \\pmod{4}$ and $3 \\nmid 2^d - 1$, there exists an odd prime $p > 3$ such that $p \\equiv 3 \\pmod{4}$ and $p \\mid 2^d - 1$. This implies $p \\mid m^2 + 81$. However, this means $m^2 \\equiv -9^2 \\pmod{p}$, and hence $(9^{-1}m)^2 \\equiv -1 \\pmod{p}$. This is impossible as $p \\equiv 3 \\pmod{4}$. Therefore, $n$ has no odd divisor greater than $1$. Thus, $n = 2^k$ for some nonnegative integer $k$.\n\nIt remains to find an $m$ such that $2^{2^k} - 1 \\mid m^2 + 81$. Firstly, note that\n$$\n2^{2^k} - 1 = (2 + 1)(2^2 + 1)\\cdots(2^{2^{k-1}} + 1).\n$$\nThe factors on the right are pairwise relatively prime. Indeed, if $r < s$, then $2^{2^r} + 1 \\mid 2^{2^s} - 1$, and $(2^{2^s} - 1, 2^{2^s} + 1) = (2^{2^s} - 1, 2) = 1$, so that $(2^{2^r} + 1, 2^{2^s} + 1) = 1$. Now, by the Chinese remainder theorem, there exists $m \\in \\mathbb{Z}^+$ such that $3 \\mid m$ and\n$$\nm \\equiv 9 \\cdot 2^{2^{j-1}} \\pmod{2^{2^j} + 1}\n$$\nfor $j = 1, 2, \\dots, k - 1$. For this $m$, we have $3 \\mid m^2 + 81$ and\n$$\nm^2 + 81 \\equiv 81(2^{2^j} + 1) \\equiv 0 \\pmod{2^{2^j} + 1}.\n$$\nHence, $2^{2^k} - 1 \\mid m^2 + 81$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ be a continuous function defined on the set of real numbers such that for any real numbers $x$,\n(i) $f(x + 1) = f(x) + 1$, and\n(ii) $f(x^3) = (f(x))^3$.\nFind all possible $f(x)$.",
    "options": [],
    "answer": "f(x) = x for all real x",
    "solution": "The only solution is $f(x) = x$ for any $x \\in \\mathbb{R}$.\n\nUsing $f(x+1) = f(x) + 1$, it can be proved easily by induction that\n$$\nf(x+n) = f(x) + n \\quad (1)\n$$\nfor any integer $n$.\n\nConsider any rational number $\\frac{a}{b}$ where $a, b \\in \\mathbb{Z}$. For any positive integer $k$, we substitute $x = \\frac{a}{b} + kb^2$ in (ii) to obtain\n$$\nf\\left(\\frac{a^3}{b^3} + 3ka^2 + 3k^2ab^3 + k^3b^6\\right) = f(x^3) = f(x)^3.\n$$\nSince $k, a, b \\in \\mathbb{Z}$, by (1), we obtain\n$$\nf\\left(\\frac{a^3}{b^3} + 3ka^2 + 3k^2ab^3 + k^3b^6\\right) = f\\left(\\frac{a^3}{b^3}\\right) + 3ka^2 + 3k^2ab^3 + k^3b^6.\n$$\nAlso, by letting $z = f\\left(\\frac{a}{b}\\right)$, we have\n$$\nf(x)^3 = (z + kb^2)^3 = z^3 + 3kb^2z^2 + 3k^2b^4z + k^3b^6.\n$$\nComparing the two sides, since $f\\left(\\frac{a^3}{b^3}\\right) = z^3$, we find that\n$$\n3ka^2 + 3k^2ab^3 = 3kb^2z^2 + 3k^2b^4z.\n$$\nThis implies $b^2z^2 + kb^4z - a^2 - kab^3 = 0$, and hence\n$$\n(bz - a)(bz + a + kb^3) = 0.\n$$\nSince $b \\neq 0$, we cannot have $bz + a + kb^3 = 0$ for all $k \\in \\mathbb{Z}^+$. Therefore, we must have $bz - a = 0$, i.e. $z = \\frac{a}{b}$. This proves $f(x) = x$ for any $x \\in \\mathbb{Q}$.\n\nSince $\\mathbb{Q}$ is dense in $\\mathbb{R}$, for any $x \\in \\mathbb{R}$, we can find a sequence $\\{x_n\\}$ of rational numbers converging to $x$. By the continuity, we have\n$$\nf(x) = f\\left(\\lim_{n \\to \\infty} x_n\\right) = \\lim_{n \\to \\infty} f(x_n) = \\lim_{n \\to \\infty} x_n = x.\n$$\nIt is easy to check that the identity function satisfies both conditions, so it is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\theta$ be a fixed angle. Let two circles touch each other internally at point $A$. Let $AB$ be a chord of the outer circle and such that $AB$ is not a diameter. Let $M$ be a variable point on the major arc $AB$ of the outer circle. $MA$ meets the inner circle at $N$. $P$ is a point on the segment $MB$ such that $\\angle MPN = \\theta$. Determine the locus of $P$ while $M$ varies.",
    "options": [],
    "answer": "The locus is the arc from the tangency point to the point on the tangent at the chord endpoint subtending the given angle with the chord endpoint and the intersection point on the inner circle; equivalently, the arc ABD on the circumcircle of triangle ABD from A to D.",
    "solution": "Let $AB$ meet the inner circle at $C$. Let $D$ be the point on the tangent at $B$ to the outer circle such that $\\angle BDC = \\theta$, and $D$ lies on opposite side of $AB$ as $M$. We claim that the locus of $P$ is $\\overarc{ABD}$.\n\n![](attached_image_1.png)\n\nFirstly, since $A$ is the homothetic centre of the two circles, we have $NC \\parallel MB$. It follows that $\\frac{AN}{NM} = \\frac{AC}{CB}$. Secondly, since $\\angle NMP = \\angle CBD$ by the tangent and $\\angle MPN = \\angle BDC = \\theta$, we have $\\triangle MNP \\sim \\triangle BCD$. Note that the triangles are directly similar. Therefore, $MNAP$ is similar to $BCAD$. Thus, $A$ is the centre of spiral similarity mapping $MP$ to $BD$. It follows that $MB \\cap PD = P$ lies on $(ABD)$, which is fixed.\n\nIt remains to notice that $P$ approaches $A$ when $M$ approaches $A$, and $P$ approaches $D$ when $M$ approaches $B$. Due to continuity, the locus of $P$ is $\\overarc{ABD}$ from $A$ to $D$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31366,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$ with incentre $I$, let $M_A, M_B$ and $M_C$ be the midpoints of $BC, CA$ and $AB$ respectively, and $H_A, H_B$ and $H_C$ be the feet of altitudes from $A, B$ and $C$ to the respective sides. Denote by $\\ell_b$ the line being tangent to the circumcircle of triangle $ABC$ and passing through $B$, and denote by $\\ell'_b$ the reflection of $\\ell_b$ in $BI$. Let $P_B$ be the intersection of $M_A M_C$ and $\\ell_b$, and let $Q_B$ be the intersection of $H_A H_C$ and $\\ell'_b$. Define $\\ell_c, \\ell'_c, P_C$ and $Q_C$ analogously. If $R$ is the intersection of $P_B Q_B$ and $P_C Q_C$, prove that $RB = RC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly, since $M_C M_A \\parallel AC$, we have $\\angle M_C BP_B = \\angle ACB = \\angle M_C M_A B$ so that $\\ell'_b$ is tangent to $(BM_A M_C)$. This implies $P_B B^2 = P_B M_C \\times P_B M_A$, and hence $P_B$ lies on the radical axis of $B$ and the nine-point circle $\\Gamma$ of $\\triangle ABC$.\n\nSecondly, we have $\\angle BH_C H_A = \\angle ACB = \\angle ABP_B = \\angle Q_B BH_A$ so that $\\ell'_b$ is tangent to $(BH_A H_C)$. This implies $Q_B B^2 = Q_B H_A \\times Q_B H_C$, and hence $Q_B$ lies on the radical axis of $B$ and $\\Gamma$.\n\nTherefore, $P_B Q_B$ is the radical axis of $B$ and $\\Gamma$. Similarly, $P_C Q_C$ is the radical axis of $C$ and $\\Gamma$. So $R$ is the radical centre of $B, C$ and $\\Gamma$. Thus, we have $RB^2 = RC^2$, which implies $RB = RC$ as desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $C_1, C_2$ with different radii are given in the plane. They touch each other externally at $T$. Consider any points $A \\in C_1$ and $B \\in C_2$, both different from $T$, such that $\\angle ATB = 90^\\circ$.\n\na. Show that all such lines $AB$ are concurrent.\n\nb. Find the locus of midpoints of all such segments $AB$.",
    "options": [],
    "answer": "a. All such lines AB are concurrent at the external center of homothety of the two circles.\nb. The locus of the midpoints is the circle with diameter O1O2.",
    "solution": "a.\nLet $O_1$ and $O_2$ be the centres of $C_1$ and $C_2$ respectively. Let $O_1O_2$ meet $C_1$ and $C_2$ again at $C$ and $D$ respectively. Note that $CT$ and $DT$ are diameters of the two circles. Therefore, we have\n$$\n\\angle CAT = \\angle BTA = \\angle TBD = 90^\\circ.\n$$\nThis implies $AC \\parallel BT$ and $AT \\parallel BD$, and so $\\triangle ACT \\sim \\triangle BTD$. Since the radii are distinct, the triangles are not congruent. Thus, $AB$ meets $CD$ at a point $X$. Note that $\\triangle XAC \\sim \\triangle XBT$. Using the similar triangles, we obtain\n$$\n\\frac{XT}{XC} = \\frac{BT}{AC} = \\frac{TD}{CT},\n$$\nwhich is a fixed ratio. Therefore, $X$ is a fixed point (as directed lengths are used). In other words, all such lines $AB$ are concurrent at $X$.\n\n![](attached_image_1.png)\n\nb.\nThe locus is the circle with diameter $O_1O_2$.\nLet $M$ be the midpoint of $AB$. Since $M$ and $O_1$ are the midpoints of $AB$ and $CT$, we have $AC \\parallel MO_1 \\parallel BT$. Similarly, we have $AT \\parallel MO_2 \\parallel BD$. This implies $\\angle O_1MO_2 = 90^\\circ$, and hence $M$ lies on the circle $\\Gamma$ with diameter $O_1O_2$.\nWhen $A$ approaches $C$ from above, the point $B$ approaches $T$ from above. This shows $M$ approaches $O_1$ from above. When $A$ approaches $C$ from below, the point $B$ approaches $T$ from below. This shows $M$ approaches $O_1$ from below. Also, $M = O_1$ only when $A = C$ and $B = T$. By continuity, $M$ runs through all points on $\\Gamma$, and so $\\Gamma$ is the locus of $M$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for positive real numbers $x, y, z$,\n$$\n\\frac{xyz(x + y + z + \\sqrt{x^2 + y^2 + z^2})}{(x^2 + y^2 + z^2)(xy + yz + zx)} \\le \\frac{3 + \\sqrt{3}}{9}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the QM-AM inequality, we have\n$$\nx + y + z \\le \\sqrt{3(x^2 + y^2 + z^2)}.\n$$\nThis implies\n$$\n\\frac{xyz(x + y + z + \\sqrt{x^2 + y^2 + z^2})}{(x^2 + y^2 + z^2)(xy + yz + zx)} \\le \\frac{(\\sqrt{3} + 1)xyz\\sqrt{x^2 + y^2 + z^2}}{(x^2 + y^2 + z^2)(xy + yz + zx)} = \\frac{(\\sqrt{3} + 1)xyz}{\\sqrt{x^2 + y^2 + z^2} \\cdot (xy + yz + zx)}.\n$$\nNext, by the AM-GM inequality, we have\n$$\nx^2 + y^2 + z^2 \\ge 3\\sqrt[3]{x^2y^2z^2}, \\quad xy + yz + zx \\ge 3\\sqrt[3]{x^2y^2z^2}.\n$$\nIt follows that\n$$\n\\frac{xyz(x + y + z + \\sqrt{x^2 + y^2 + z^2})}{(x^2 + y^2 + z^2)(xy + yz + zx)} \\le \\frac{(\\sqrt{3} + 1)xyz}{\\sqrt[3]{xyz} \\cdot 3\\sqrt{x^2y^2z^2}} = \\frac{3 + \\sqrt{3}}{9}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31369,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, let $f(n)$ be the largest $k$ such that $2^k$ divides $n$ and $g(n)$ be the sum of the digits of the binary representation of $n$. Prove that for any positive integers $n$,\n(i) $f(n!) = n - g(n)$;\n(ii) $4$ divides $\\binom{2n}{n} = \\frac{(2n)!}{n!n!}$ if and only if $n$ is not a power of $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(i) This is an alternative form of Legendre's formula. Here is a simple proof of the result. Let $n = \\overline{a_s a_{s-1} \\cdots a_0}$ be the binary representation of $n$. Then we have\n$$\n\\begin{aligned}\nf(n!) &= \\sum_{j=1}^{s} \\left\\lfloor \\frac{n}{2^j} \\right\\rfloor = \\sum_{j=1}^{s} \\overline{a_s a_{s-1} \\cdots a_j} (2^j) = \\sum_{j=1}^{s} \\sum_{k=j}^{s} a_k \\cdot 2^{k-j} = \\sum_{k=1}^{s} \\sum_{j=1}^{k} a_k \\cdot 2^{k-j} \\\\\n&= \\sum_{k=1}^{s} a_k (2^k - 1) = \\sum_{k=1}^{s} a_k \\cdot 2^k - \\sum_{k=1}^{s} a_k = n - g(n).\n\\end{aligned}\n$$\n\n(ii) By part (a), we have\n$$\n\\begin{aligned}\nf\\left(\\frac{(2n)!}{n!n!}\\right) &= f((2n)!) - 2f(n!) \\\\\n&= [2n - g(2n)] - 2[n - g(n)] \\\\\n&= 2g(n) - g(2n) \\\\\n&= 2g(n) - g(n) \\\\\n&= g(n).\n\\end{aligned}\n$$\nThus, $4$ divides $\\binom{2n}{n}$ if and only if $g(n) \\ge 2$. This holds if and only if $n = 100\\cdots0_{(2)} = 2^k$ for some nonnegative integer $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31370,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers which cannot be written as the sum of at least two consecutive positive integers.",
    "options": [],
    "answer": "powers of 2",
    "solution": "The answer is any power of $2$.\nSuppose $n = a + (a + 1) + \\dots + (a + k)$ for some positive integers $a$ and $k$. Then we have\n$$\n2n = (2a + k)(k + 1).\n$$\nLet $n = 2^r m$ where $m$ is odd. Note that $2a + k$ and $k + 1$ have different parities. Therefore, $2^{r+1}$ must divide one of $2a + k$ and $k + 1$, and hence the other number divides $m$. Since $2a + k, k + 1 > 1$, we must have $m \\ge 3$. This implies $n \\ne 2^r$ for any $r \\ge 0$.\n\nOn the other hand, for $m \\ge 3$, we can find a solution. Indeed, if $2^{r+1} > m$, then we set\n$$\n(a, k) = \\left(2^r - \\frac{m-1}{2}, m-1\\right)\n$$\nso that $a, k \\in \\mathbb{Z}^+$ and $(2a + k)(k + 1) = 2^{r+1}m = 2n$. Similarly, if $2^{r+1} < m$, then we set\n$$\n(a, k) = \\left( \\frac{m+1}{2} - 2^r, 2^{r+1} - 1 \\right)\n$$\nso that $a, k \\in \\mathbb{Z}^+$ and $(2a + k)(k + 1) = 2^{r+1}m = 2n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = x^n + a_1x^{n-1} + \\cdots + a_n$ be a polynomial of degree $n \\ge 2$ with real roots $b_1, b_2, \\ldots, b_n$. If $x$ is greater than $b_1, b_2, \\ldots, b_n$, prove that\n$$\nf(x+1) \\left( \\frac{1}{x-b_1} + \\frac{1}{x-b_2} + \\cdots + \\frac{1}{x-b_n} \\right) \\ge 2n^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $c_j = x - b_j > 0$ for each $j$. Note that\n$$\nf(y) = (y - b_1)(y - b_2)\\cdots(y - b_n).\n$$\nTherefore, we have\n$$\n\\begin{aligned}\nf(x+1) &= (x+1-b_1)(x+1-b_2)\\cdots(x+1-b_n) \\\\\n&= (1+c_1)(1+c_2)\\cdots(1+c_n).\n\\end{aligned}\n$$\nNow, by the AM-GM inequality, we have\n$$\n\\frac{1}{c_1} + \\frac{1}{c_2} + \\cdots + \\frac{1}{c_n} \\ge n \\sqrt[n]{\\frac{1}{c_1 c_2 \\cdots c_n}}.\n$$\nThus, it suffices to prove\n$$\n\\prod_{j=1}^{n} \\sqrt[\\frac{1}{n}]{\\frac{(1+c_j)^n}{c_j}} \\ge 2n. \\quad (1)\n$$\nIndeed, since $n \\ge 2$, we get\n$$\n\\frac{(1+c_j)^n}{c_j} \\ge \\frac{1}{c_j} + n + \\frac{n(n-1)}{2} c_j \\ge n + 2 \\sqrt{\\frac{1}{c_j} \\cdot \\frac{n(n-1)}{2}} c_j \\ge 2n\n$$\nby the AM-GM inequality and a simple inequality $2\\sqrt{\\frac{n(n-1)}{2}} \\ge n$ (which is equivalent to $n \\ge 2$). This implies (1). Equality holds when $n=2$ and $c_j = 1$, i.e. $x-1 = b_1 = b_2 = \\cdots = b_n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the angle bisectors of $\\angle A$, $\\angle B$, $\\angle C$ of $\\triangle ABC$ intersect the circumcircle of $\\triangle ABC$ at $P$, $Q$, $R$ respectively. Prove that $AP + BQ + CR > BC + CA + AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the incentre of $\\triangle ABC$. It is well-known that $PB = PI = PC$, etc. By the triangle inequality, we have\n$$\n\\begin{aligned}\n& BI + CI > BC, \\\\\n& CI + AI > CA, \\\\\n& AI + BI > AB, \\\\\n& 2IP = BP + CP > BC, \\\\\n& 2IQ = CQ + AQ > CA, \\\\\n& 2IR = AR + BR > AB.\n\\end{aligned}\n$$\nAdding these inequalities and dividing both sides by 2, we obtain the desired inequality.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31373,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCDEF$ is a hexagon inscribed in a circle. Show that the diagonals $AD$, $BE$, $CF$ are concurrent if and only if $AB \\cdot CD \\cdot EF = BC \\cdot DE \\cdot FA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By Ceva's theorem, $AD$, $BE$, $CF$ are concurrent if and only if\n$$\n\\frac{\\sin \\angle CAD}{\\sin \\angle EAD} \\times \\frac{\\sin \\angle AEB}{\\sin \\angle CEB} \\times \\frac{\\sin \\angle ECF}{\\sin \\angle ACF} = 1.\n$$\n![](attached_image_1.png)\nBy the extended sine law, we have\n$$\n\\frac{\\sin \\angle CAD}{\\sin \\angle EAD} = \\frac{CD}{DE}, \\quad \\frac{\\sin \\angle AEB}{\\sin \\angle CEB} = \\frac{AB}{BC}, \\quad \\frac{\\sin \\angle ECF}{\\sin \\angle ACF} = \\frac{EF}{FA}.\n$$\nThus, $AD$, $BE$, $CF$ are concurrent if and only if\n$$\n\\frac{CD}{DE} \\times \\frac{AB}{BC} \\times \\frac{EF}{FA} = 1.\n$$\nThis is exactly $AB \\cdot CD \\cdot EF = BC \\cdot DE \\cdot FA$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many positive integers $n$ such that $2^n + 1$ is divisible by $n$. Find all such $n$'s that are prime numbers.",
    "options": [],
    "answer": "Infinitely many examples: n = 3^k for any positive integer k. Among primes, the only solution is n = 3.",
    "solution": "All integers $n = 3^k$ with $k \\in \\mathbb{Z}^+$ satisfy $n \\mid 2^n + 1$. Indeed, since $3 \\mid 2 + 1$, by the lifting the exponent lemma, we have\n$$\nv_3(2^{3^k} + 1^{3^k}) = v_3(2 + 1) + v_3(3^k) = 1 + k.\n$$\nThis implies $3^{k+1} \\mid 2^n + 1$, and hence $n \\mid 2^n + 1$.\n\nThe only prime number $n$ satisfying $n \\mid 2^n + 1$ is $n = 3$. Indeed, let $n = p$ be a prime. By the Fermat little theorem, we have\n$$\n2^p + 1 \\equiv 2 + 1 = 3 \\pmod{p}.\n$$\nThis is congruent to $0$ modulo $p$ if and only if $p \\mid 3$, i.e. $p = 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}$ denote the set of integers. Find all functions $f : \\mathbb{Z} \\to \\mathbb{Z}$ such that $f(-1) = f(1)$ and $f(x) + f(y) = f(x + 2xy) + f(y - 2xy)$ for all integers $x, y$.",
    "options": [],
    "answer": "All functions f: Z → Z such that f(0) is arbitrary and, for each k ≥ 0, f(2^k m) = c_k for all odd m, where the values c_k = f(2^k) are arbitrary integers. Equivalently, for every nonzero n written as n = 2^k m with m odd, f(n) depends only on k (the 2-adic valuation), and f is even.",
    "solution": "$f$ can be any function such that $f(2^k m) = f(2^k)$ for any $k \\ge 0$ and odd $m$, where $f(0), f(1), f(2), f(2^2), \\dots$ are arbitrary integers.\n\nWe label the equation as follows.\n$$\nf(x) + f(y) = f(x + 2xy) + f(y - 2xy) \\quad (1)\n$$\nPutting $x = 1$ and $y = n$ in (1), we obtain\n$$\nf(1) + f(n) = f(2n + 1) + f(-n). \\quad (2)\n$$\nPutting $x = n$ and $y = -1$ in (1), we obtain\n$$\nf(n) + f(-1) = f(-n) + f(2n - 1). \\quad (3)\n$$\nSince $f(1) = f(-1)$, by comparing (2) and (3), we immediately obtain\n$$\nf(2n + 1) = f(2n - 1)\n$$\nfor any $n \\in \\mathbb{Z}$. It follows that $f(m) = f(1)$ for any odd $m$. Using (2), we obtain\n$$\nf(n) = f(-n)\n$$\nfor any $n \\in \\mathbb{Z}$.\n\nNext, we put $x = -2k - 1$ and $y = n$ in (1). Since both $x$ and $x + 2xy = x(1 + 2y)$ are odd, we obtain\n$$\nf(n) = f((4k + 3)n).\n$$\nReplacing $k$ by $-(k+1)$, and using $f(m) = f(-m)$, we get\n$$\nf(n) = f((-4k - 1)n) = f((4k + 1)n).\n$$\nCombining these, we know that $f(n) = f(mn)$ for any odd $m$. Therefore, by writing $n = 2^k m$ where $m$ is odd, we find that\n$$\nf(2^k m) = f(2^k).\n$$\nIt remains to check that all such functions satisfy all the conditions. Firstly, $f(-1) = f(1)$ is true. Secondly, Note that $v_2(x + 2xy) = v_2(x(1 + 2y)) = v_2(x)$ and similarly $v_2(y - 2xy) = v_2(y)$. Therefore, we must have\n$$\nf(x + 2xy) = f(x), \\quad f(y - 2xy) = f(y).\n$$\nThis proves (1), and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31376,
    "subject": "Mathematics (Multi-modal)",
    "question": "From a point $P$ outside a circle centred at $O$, draw the two tangents to the circle touching it at $A, B$. Let $M$ be a point on the segment $AB$ and let $C, D$ be points on the circle with midpoint $M$. Let the tangents to the circle at $C, D$ intersect at $Q$. Show that $OQ \\perp PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "This is a simple corollary of Brokard's theorem. Alternatively, we provide an elementary proof as follows.\nNote that $O$, $M$, $Q$ are collinear since all of them lie on the perpendicular bisector of $CD$. By the property of tangents, we know that $Q$, $C$, $O$, $D$ are concyclic. This yields\n$$\nMQ \\times MO = MC \\times MD = MA \\times MB.\n$$\nThus, $Q$, $A$, $O$, $B$ are concyclic, and hence $P$, $Q$, $A$, $O$, $B$ are concyclic. Therefore, we obtain\n$$\n\\angle OQP = \\angle OAP = 90^\\{\\circ\\}.\n$$\nThis means $OQ \\perp PQ$.\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Start with some positive integer. The following operation is performed on the number: its unit digit is split off and multiplied by $4$, then this product is added to the remaining number. (For example, $1997$ is changed to $7 \\times 4 + 199 = 227$.) The operation is performed again and again. Prove that if the sequence of numbers obtained contains $1001$, then none of the numbers in the sequence can be a prime number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If the current number is $10a + b$ where $0 \\le b \\le 9$, then the next number is $a + 4b$. Note that\n$$\n10(a + 4b) = (10a + b) + 39b.\n$$\nTherefore, $13 \\mid a + 4b$ if and only if $13 \\mid 10a + b$. Since $13 \\mid 1001$, all the numbers in the sequence are divisible by $13$. The only possible prime number in this sequence is $13$.\n\nNow, since $1001 \\rightarrow 104 \\rightarrow 26 \\rightarrow 26 \\rightarrow \\dots$, $13$ will never appear after $1001$. However, since $13 \\rightarrow 13 \\rightarrow \\dots$, it cannot appear before $1001$ as well. Thus, there is no prime in the sequence.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1$, $x_2$, $x_3$, $\\dots$ be a sequence of nonzero real numbers satisfying\n$$\nx_n = \\frac{x_{n-2} x_{n-1}}{2x_{n-2} - x_{n-1}} \\quad \\text{for } n = 3, 4, 5, \\dots\n$$\nFind all pairs $(x_1, x_2)$ such that $x_n$ is an integer for infinitely many $n$.",
    "options": [],
    "answer": "All pairs where both entries are equal to the same nonzero integer, i.e., (c, c) with c an integer and c not zero.",
    "solution": "The only possibilities are $(x_1, x_2) = (c, c)$ for some nonzero integer $c$.\n\nRewrite the recurrence relation as\n$$\n\\frac{2}{x_{n-1}} = \\frac{1}{x_n} + \\frac{1}{x_{n-2}}.\n$$\nLet $y_n = \\frac{1}{x_n}$ for all $n$. This implies\n$$\ny_{n-1} - y_{n-2} = y_n - y_{n-1}.\n$$\nTherefore, $\\{y_n\\}$ is an arithmetic sequence. Let $y_n = a + n d$ for some constants $a$ and $d$. Then we have\n$$\nx_n = \\frac{1}{y_n} = \\frac{1}{a + n d}.\n$$\nIf $d \\neq 0$, then we have $|a + n d| \\to \\infty$ when $n \\to \\infty$, and hence $x_n \\to 0$. However, $x_n \\neq 0$. This shows $x_n$ cannot be an integer for large $n$. So it suffices to consider $d = 0$. In that case $\\{x_n\\}$ is a constant sequence. In order that there are infinitely many integer terms, the constant must be an integer. So $x_1 = x_2 = c$ for some integer $c \\neq 0$. Note that a constant sequence satisfies the recurrence relation since\n$$\nc = \\frac{c^2}{2c - c}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Five numbers $1$, $2$, $3$, $4$, $5$ are written on a blackboard. A student may erase any two of the numbers $a$ and $b$ on the board and write the numbers $a+b$ and $ab$ replacing them. If this operation is repeatedly performed, can the numbers $21$, $27$, $64$, $180$, $540$ ever appear on the board at the same time?",
    "options": [],
    "answer": "No",
    "solution": "No. We consider the numbers modulo $3$.\n* If $3 \\mid a, b$, then $3 \\mid a + b$ and $3 \\mid ab$. The number of multiples of $3$ remains unchanged.\n* If $3 \\mid a$ and $3 \\nmid b$, then $3 \\nmid a + b$ and $3 \\mid ab$. The number of multiples of $3$ remains unchanged.\n* If $a \\equiv b \\equiv \\pm 1 \\pmod{3}$, then $3 \\nmid a + b$ and $3 \\nmid ab$. The number of multiples of $3$ remains unchanged.\n* If $a \\equiv 1 \\pmod{3}$ and $b \\equiv 2 \\pmod{3}$, then $3 \\mid a + b$ and $ab \\equiv 2 \\pmod{3}$. The number of multiples of $3$ is increased by $1$, and there is a number which is congruent to $2$ modulo $3$.\nNow, note that $3 \\mid 21, 27, 180, 540$ and $64 \\equiv 1 \\pmod{3}$. If the $5$ numbers can appear on the board at the same time, there must be an instant when the number of multiples of $3$ changes from $3$ to $4$. This must be the last case in the above list, hence the $5$ numbers are now congruent to $0, 0, 0, 0, 2$ modulo $3$ respectively. Afterwards, no matter which $2$ numbers we choose, the new numbers are still congruent to $0, 0, 0, 0, 2$ modulo $3$ respectively. Therefore, it is impossible to generate $21$, $27$, $64$, $180$, $540$. This is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31380,
    "subject": "Mathematics (Multi-modal)",
    "question": "One may perform the following two operations on a positive integer:\n(a) multiply it by any positive integer; or\n(b) delete zeros in its decimal representation.\nProve that for every positive integer $X$, one can perform a sequence of these operations that will transform $X$ to a one-digit number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the pigeonhole principle, two of the numbers $1, 11, 111, \\dots$ leave the same remainder when divided by $X$. Their difference, which is of the form $11\\cdots100\\cdots0$, is divisible by $X$. Therefore, we can transform $X$ to $11\\cdots100\\cdots0$, and then to $11\\cdots1$ by deleting the zeros.\n\nIf the current number is $1$, we are done. Otherwise, we have\n$$\n\\underbrace{11\\cdots1}_{k \\text{ times}} \\times 82 = \\underbrace{911\\cdots1}_{(k-2) \\text{ times}} 02.\n$$\nSo we can transform $11\\cdots1$ to $911\\cdots102$, and then to $911\\cdots12$. Next, note that\n$$\n\\underbrace{911\\cdots1}_{m \\text{ times}} 2 \\times 9 = \\underbrace{8200\\cdots0}_{m \\text{ times}} 8.\n$$\nSo we can transform the number to $8200\\cdots08$, and then to $828$. Afterwards, we apply the operations as follows.\n$$\n828 \\xrightarrow{\\times 25} 20700 \\rightarrow 27 \\xrightarrow{\\times 4} 108 \\rightarrow 18 \\xrightarrow{\\times 5} 90 \\rightarrow 9\n$$\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31381,
    "subject": "Mathematics (Multi-modal)",
    "question": "The general term of a sequence $\\{a_n\\}$ is given by $a_{n+2} = a_{n+1} - a_n$. If the sum of the first 1997 terms is 1879 and the sum of the first 1879 terms is 1997, what is the sum of the first 2000 terms?",
    "options": [],
    "answer": "5873",
    "solution": "The answer is 5873.\nLet $a_1 = x$ and $a_2 = y$. Then the sequence is\n$$\nx, y, y-x, -x, -y, x-y, x, y, \\dots\n$$\nNote that the sequence repeats every 6 terms. Also, the sum of every 6 consecutive terms is 0. Thus, the sum of the first 1997 = $6 \\times 332 + 5$ terms is $x + y + (y-x) + (-x) + (-y) = y-x$, and the sum of the first 1879 = $6 \\times 313 + 1$ terms is $x$. Given that $y-x = 1879$ and $x = 1997$, we have\n$$\ny = (y - x) + x = 1879 + 1997 = 3876.\n$$\nTherefore, the sum of the first 2000 = $6 \\times 333 + 2$ terms is\n$$\nx + y = 1997 + 3876 = 5873.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $S = \\frac{1}{3} + \\frac{1}{5} + \\dots + \\frac{1}{2n + 1}$ is not an integer if $n \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $l = \\text{lcm}(3, 5, \\dots, 2n+1)$, and let $k = v_3(l)$. Note that $k \\ge 1$ since $n \\ge 1$. Since $2 \\cdot 3^k$ is even, and $3 \\cdot 3^k > 2n+1$ (or otherwise $3^{k+1} \\mid l$), only the term $3^k$ among $3, 5, \\dots, 2n+1$ is divisible by $3^k$. Therefore, $\\frac{l}{3}S = \\frac{l}{3^{k+1}} + m$ for some integer $m$. But then $\\frac{l}{3^{k+1}}$ is not an integer. Thus, $S$ cannot be an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31383,
    "subject": "Mathematics (Multi-modal)",
    "question": "$PQRS$ is a cyclic quadrilateral with $\\angle PSR = 90^\\circ$. $H$, $K$ are the feet of the perpendiculars from $Q$ to $PR$, $PS$ (suitably extended if necessary) respectively. Show that $HK$ bisects $QS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $HK$ meet $QS$ at $A$. Since $\\angle PHQ = \\angle PKQ = 90^\\circ$, the points $P$, $Q$, $H$, $K$ are concyclic. As both $QK$ and $RS$ are perpendicular to $PS$, they are parallel. Therefore, we have\n$$\n\\angle QKA = \\angle QKH = \\angle QPH = \\angle QPR = \\angle QSR = \\angle SQK = \\angle AQK.\n$$\nThus, $AK = AQ$. As $\\angle QKS = 90^\\circ$, $A$ is the midpoint of $QS$. This means $HK$ bisects $QS$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31384,
    "subject": "Mathematics (Multi-modal)",
    "question": "The base of a pyramid is a convex polygon with $9$ sides. Each of the diagonals of the base and each of the edges on the lateral surface of the pyramid is coloured either black or white. Both colours are used. (Note that the sides of the base are not coloured.) Prove that there are three segments coloured the same colour which form a triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the $9$ edges on the lateral surface. By the pigeonhole principle, $5$ of these edges $PA_1, PA_2, PA_3, PA_4, PA_5$ are of the same colour, say black. WLOG assume $A_1, A_2, A_3, A_4, A_5$ lie in this order at the base. Since the base has $9$ sides, WLOG assume $A_1A_2$ is a diagonal. Then $A_1A_4$ and $A_2A_4$ are also diagonals. If one of $A_1A_2, A_1A_4, A_2A_4$ is black, then it forms a monochromatic triangle with $P$. Otherwise, all of $A_1A_2, A_1A_4, A_2A_4$ are white, so we still get a monochromatic triangle.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31385,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $p$ is any prime other than $2$ or $5$, prove that $p$ divides infinitely many of the integers $9$, $99$, $999$, $9999$, ...",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Fermat little theorem, we have $10^{k(p-1)} \\equiv 1 \\pmod{p}$ for any $k \\in \\mathbb{N}$. This implies\n$$\np \\mid 10^{k(p-1)} - 1 = \\underbrace{99\\dots9}_{k(p-1) \\text{ times}}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that $\\sqrt{a-1} + \\sqrt{b-1} + \\sqrt{c-1} \\le \\sqrt{c(ab+1)}$ for real numbers $a, b, c \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x = \\sqrt{a-1}$, $y = \\sqrt{b-1}$ and $z = \\sqrt{c-1}$. Then we have $x, y, z \\ge 0$. We need to prove\n$$\nx + y + z \\le \\sqrt{(z^2 + 1)[(x^2 + 1)(y^2 + 1) + 1]}.\n$$\nBy the Cauchy-Schwarz inequality, we have\n$$\n(z^2 + 1)[(x^2 + 1)(y^2 + 1) + 1] = (1 + z^2)[(x^2 + 1)(y^2 + 1) + 1] \\\\ \\ge (\\sqrt{(x^2 + 1)(y^2 + 1)} + z)^2.\n$$\nAlso, we have\n$$\n(x^2 + 1)(y^2 + 1) = (x^2 + 1)(1 + y^2) \\ge (x + y)^2.\n$$\nCombining these, we obtain\n$$\n\\sqrt{(z^2 + 1)[(x^2 + 1)(y^2 + 1) + 1]} \\ge \\sqrt{(x^2 + 1)(y^2 + 1)} + z \\ge x + y + z.\n$$\nEquality holds when $x = \\frac{1}{y}$ and $z = \\frac{1}{\\sqrt{(x^2 + 1)(y^2 + 1)}}$. This means\n$$\na = 1 + t, \\quad b = 1 + \\frac{1}{t}, \\quad c = 1 + \\frac{t}{(t+1)^2}\n$$\nfor some $t > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $s, t$ be given nonzero integers, and let $(x, y)$ be any ordered pair of integers. A move changes $(x, y)$ to $(x + t, y - s)$. The pair $(x, y)$ is 'good' if after some (maybe zero) number of moves it describes a pair of integers that are not relatively prime.\n\na. Determine if $(s, t)$ is a good pair.\n\nb. Show that for any $s$ and $t$ there is a pair $(x, y)$ which is not good.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Yes. Since $s, t \\neq 0$, $s^2 + t^2 \\geq 2$. Let $p$ be any prime divisor of $s^2 + t^2$. If $p \\mid t$, then $p \\mid s$, and so $\\gcd(s, t) \\geq p$. This shows $(s, t)$ is good. If $p \\nmid t$, then there exists $k \\in \\mathbb{Z}^+$ such that $tk \\equiv -s \\pmod{p}$. Then we have $p \\mid s + kt$. Also, we have\n$$\nt - ks = t - (-t^{-1}s)s = t^{-1}(t^2 + s^2) \\equiv 0 \\pmod{p}.\n$$\nThus, $\\gcd(s+kt, t-ks) \\geq p$. This shows $(s, t)$ is a good pair as we can apply $k$ moves to $(s, t)$ to obtain $(s + kt, t - ks)$.\n\n\nb. Let $\\gcd(s, t) = d$. By Bézout's identity, there exist $m, n \\in \\mathbb{Z}$ such that $ms + nt = d$. We claim that the pair $(m, n)$ is not good.\nAfter $k$ moves, the pair becomes $(m + kt, n - ks)$. If there exists a prime $p$ dividing both $m + kt$ and $n - ks$, then\n$$\np \\mid (m + kt) \\frac{s}{d} + (n - ks) \\frac{t}{d} = \\frac{ms + nt}{d} = 1.\n$$\nThis is impossible. Therefore, $\\gcd(m + kt, n - ks) = 1$ for any $k$. This proves our claim.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ be a function with the following properties:\n(i) $f(n)$ is defined for every positive integer $n$;\n(ii) $f(n)$ is a positive integer;\n(iii) $f(f(m) + f(n)) = m + n$ for all $m$ and $n$.\nFind $f(1997)$.",
    "options": [],
    "answer": "1997",
    "solution": "We have $f(1997) = 1997$.\n\nLabel the equation as follows.\n$$\nf(f(m) + f(n)) = m + n \\quad (1)\n$$\nIf $f(a) = f(b)$ for some $a, b \\in \\mathbb{Z}^+$, then by putting $m = a$ and $m = b$ in (1), we obtain\n$$\na + n = f(f(a) + f(n)) = f(f(b) + f(n)) = b + n.\n$$\nThis implies $a = b$. So $f$ is injective. Now, using (1), we have\n$$\nf(f(m + 1) + f(1)) = (m + 1) + 1 = m + 2 = f(f(m) + f(2)).\n$$\nUsing the injectivity, this becomes\n$$\nf(m + 1) + f(1) = f(m) + f(2).\n$$\nLet $d = f(2) - f(1)$. By induction, we can easily prove that\n$$\nf(m) = f(1) + (m - 1)d\n$$\nfor any $m \\in \\mathbb{Z}^+$. In other words, $f(m) = am + b$ for some integers $a, b$. Now, we have\n$$\nf(f(m) + f(n)) = a(am + b + an + b) + b = a^2m + a^2n + (2ab + b).\n$$\nThis is equal to $m+n$ for any $m, n \\in \\mathbb{Z}^+$ if and only if $a^2 = 1$ and $2ab + b = 0$. Clearly, the only solution with $a \\ge 0$ is $(a, b) = (1, 0)$. This means $f(m) = m$ for any $m \\in \\mathbb{Z}^+$. In particular, we have $f(1997) = 1997$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31389,
    "subject": "Mathematics (Multi-modal)",
    "question": "$I$ is the incentre of $\\triangle ABC$. $P, Q, R$ are the circumcentres of the triangles $IBC, ICA$ and $IAB$ respectively. Prove that the points $A, R, B, P, C, Q$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle BIC = 90^\\circ + \\frac{A}{2}$, the circumcentre $P$ of $\\triangle BIC$ must lie on the different side of $BC$ as $A$ and $I$. As\n$$\n\\angle BPC = 360^\\circ - 2\\angle BIC = 360^\\circ - 2\\left(90^\\circ + \\frac{A}{2}\\right) = 180^\\circ - A,\n$$\nthe point $P$ lies on $(ABC)$. By symmetry, $Q$ and $R$ also lie on $(ABC)$ as desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31390,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABC$ is a triangle; $ABX$, $BCY$ and $CAZ$ are similar isosceles triangles outside $ABC$ such that $XA = XB$, $YB = YC$ and $ZC = ZA$.\nProve that $AY$, $BZ$ and $CX$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "This is a special case of Jacobi's theorem. Alternatively, by Ceva's theorem, it suffices to prove\n$$\n\\frac{\\sin \\angle ACX}{\\sin \\angle BCX} \\times \\frac{\\sin \\angle BAY}{\\sin \\angle CAY} \\times \\frac{\\sin \\angle CBZ}{\\sin \\angle ABZ} = 1.\n$$\nApplying the sine law to $\\triangle ACX$ and $\\triangle BCX$, we obtain\n$$\n\\sin \\angle ACX = AX \\times \\frac{\\sin \\angle CAX}{CX}, \\quad \\sin \\angle BCX = BX \\times \\frac{\\sin \\angle CBX}{CX}.\n$$\n\nCombining these, since $AX = BX$, we get\n$$\n\\frac{\\sin \\angle ACX}{\\sin \\angle BCX} = \\frac{\\sin \\angle CAX}{\\sin \\angle CBX}.\n$$\nSimilarly, we find that\n$$\n\\frac{\\sin \\angle ACX}{\\sin \\angle BCX} \\times \\frac{\\sin \\angle BAY}{\\sin \\angle CAY} \\times \\frac{\\sin \\angle CBZ}{\\sin \\angle ABZ} = \\frac{\\sin \\angle CAX}{\\sin \\angle CBX} \\times \\frac{\\sin \\angle ABY}{\\sin \\angle ACY} \\times \\frac{\\sin \\angle BCZ}{\\sin \\angle BAZ}.\n$$\nNote that $\\angle CAX = \\angle BAZ$, $\\angle ABY = \\angle CBX$ and $\\angle BCZ = \\angle ACY$. Therefore, this is equal to 1, and we are done.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Recall that $n$ is perfect if the sum of the divisors of $n$ is $2n$. Suppose now $n$ is an odd perfect number. Show that $n$ has at least 3 distinct prime factors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $n = p^k$ for some odd prime $p$ and positive integer $k$, then we have\n$$\n\\frac{\\sigma(n)}{n} = \\frac{1 + p + \\cdots + p^k}{p^k} = 1 + \\frac{1}{p} + \\cdots + \\frac{1}{p^k} < \\frac{1}{1 - \\frac{1}{p}} = \\frac{p}{p-1} < 2.\n$$\nIf $n = p^k q^\\ell$ for some odd primes $p, q$ and positive integers $k, \\ell$, then we have\n$$\n\\frac{\\sigma(n)}{n} = \\frac{1 + p + \\cdots + p^k}{p^k} \\cdot \\frac{1 + q + \\cdots + q^\\ell}{q^\\ell} < \\frac{p}{p-1} \\cdot \\frac{q}{q-1} \\le \\frac{3}{2} \\cdot \\frac{5}{4} < 2.\n$$\nThis shows $\\sigma(n) < 2n$ if $n$ has at most 2 distinct prime factors. Therefore, in order that $n$ is a perfect number, it must have at least 3 distinct prime factors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ be a function defined on the positive reals with the following properties:\n(1) $f(1) = 1$,\n(2) $f(x + 1) = x f(x)$, and\n(3) $f(x) = 10^{g(x)}$,\nwhere $g(x)$ is a function defined on the reals satisfying\n$$\ng(ty + (1-t)z) \\leq t g(y) + (1-t) g(z)\n$$\nfor all $y$ and $z$ and for $0 \\leq t \\leq 1$.\n\na. Prove that $t [g(n) - g(n-1)] \\leq g(n+t) - g(n) \\leq t [g(n+1) - g(n)]$ where $n$ is an integer and $0 \\leq t \\leq 1$.\n\nb. Prove that $\\frac{4}{3} \\leq f(\\frac{1}{2}) \\leq \\frac{4}{3} \\sqrt{2}$.",
    "options": [],
    "answer": "4/3 <= f(1/2) <= (4/3) sqrt(2)",
    "solution": "a.\nWe label the inequality as follows.\n$$\ng(ty + (1-t)z) \\leq t g(y) + (1-t) g(z) \\quad (1)\n$$\nPutting $y = n - 1$ and $z = n + c$ in (1), we obtain\n$$\ng(t(n-1) + (1-t)(n+c)) \\leq t g(n-1) + (1-t) g(n+c).\n$$\nSuppose $c \\in [0, 1]$. Replacing $t$ by $\\frac{c}{1+c} \\in [0, 1]$, this becomes\n$$\ng\\left(\\frac{c(n-1) + (n+c)}{1+c}\\right) \\leq \\frac{c}{1+c} g(n-1) + \\frac{1}{1+c} g(n+c).\n$$\nNote that the left-hand side is just $g(n)$. Therefore, this is the same as\n$$\nc [g(n) - g(n-1)] \\leq g(n+c) - g(n).\n$$\nThis is exactly the inequality on the left to be proved.\nFor the other inequality, we simply put $y = n+1$ and $z = n$ in (1). This gives\n$$\ng(t(n+1) + (1-t)n) \\leq t g(n+1) + (1-t) g(n).\n$$\nNote that the left-hand side is just $g(n+t)$. This is exactly the other desired inequality.\n\n\nb.\nSince $g(x) = \\log f(x)$, the inequality in part (a) can be rewritten as\n$$\nt (\\log f(n) - \\log f(n-1)) \\leq \\log f(n+t) - \\log f(n) \\leq t (\\log f(n+1) - \\log f(n)).\n$$\nThis implies\n$$\n\\left( \\frac{f(n)}{f(n-1)} \\right)^t \\leq \\frac{f(n+t)}{f(n)} \\leq \\left( \\frac{f(n+1)}{f(n)} \\right)^t.\n$$\nAs $f(n) = (n-1) f(n-1)$ and $f(n+1) = n f(n)$, this can be reduced to\n$$\n(n-1)^t \\leq \\frac{f(n+t)}{f(n)} \\leq n^t. \\quad (2)\n$$\n\nNow, note that $f(2) = 1 \\cdot f(1) = 1$. Putting $n = 2$ and $t = \\frac{1}{2}$ in (2), we obtain\n$$\n1 \\leq f\\left(\\frac{5}{2}\\right) \\leq \\sqrt{2}.\n$$\nUsing $f\\left(\\frac{5}{2}\\right) = \\frac{3}{2} f\\left(\\frac{3}{2}\\right) = \\frac{3}{4} f\\left(\\frac{1}{2}\\right)$, we get\n$$\n\\frac{4}{3} \\leq f\\left(\\frac{1}{2}\\right) \\leq \\frac{4}{3} \\sqrt{2}.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31393,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ vertices on a plane. A graph $G$ is formed by connecting every two vertices by exactly one edge, and painting all edges, each in one colour, chosen from $k$ given colours.\n\na. For $n = 6$ and $k = 3$, show that it is possible to paint the edges so that the graph contains no closed circuit painted in one colour only.\n\nb. Prove that for every positive integral value of $k$ (i.e. number of colours allowed), if the number of vertices $n \\ge 2k+1$, then it is always possible to find a mono-coloured closed circuit in $G$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. We can colour the edges as shown. Since the subgraph formed by edges of the same colour connects all 6 vertices but there are only 5 edges, it is a tree and hence has no closed circuit.\n\n![](attached_image_1.png)\n\nb. There are $\\binom{n}{2} \\ge \\frac{n(n-1)}{2}$ edges in total. By the pigeonhole principle, there are $\\frac{n(n-1)}{2k}$ edges which are of the same colour. Consider the subgraph formed by these edges. The number of vertices is at most $n$. Since\n$$\n\\frac{n(n-1)}{2k} \\ge \\frac{n(2k+1-1)}{2k} = n,\n$$\nthe number of edges of this subgraph is at least the number of vertices. Thus, there must be a closed circuit. This gives a monochromatic closed circuit in $G$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_i, b_i, c_i$ and $R_i$ ($i = 1, 2$) be the sides and the radii of the circumscribed circles of $\\triangle A_iB_iC_i$. Prove that for $m, n \\ge 1$, $a_1^m a_2^n + b_1^m b_2^n + c_1^m c_2^n \\le 9(2^{m+n-2} R_1^m R_2^n)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the extended sine law, we have\n$$\na_j = 2R_j \\sin A_j, \\quad b_j = 2R_j \\sin B_j, \\quad c_j = 2R_j \\sin C_j.\n$$\nReplacing all side lengths on the left-hand side, we need to prove\n$$\n\\sin^m A_1 \\sin^n A_2 + \\sin^m B_1 \\sin^n B_2 + \\sin^m C_1 \\sin^n C_2 \\le \\frac{9}{4}.\n$$\nSince $\\sin \\theta \\le 1$ for any $\\theta$, it suffices to prove this inequality for $m = n = 1$. This means we need to prove\n$$\na_1 a_2 + b_1 b_2 + c_1 c_2 \\le 9 R_1 R_2.\n$$\nRecall that\n$$\n9R_j^2 - (a_j^2 + b_j^2 + c_j^2) = O_j H_j^2 \\ge 0,\n$$\nwhere $O_j$ and $H_j$ are the circumcentre and orthocentre of $\\triangle A_j B_j C_j$ respectively. By the Cauchy-Schwarz inequality, we obtain\n$$\n(a_1 a_2 + b_1 b_2 + c_1 c_2)^2 \\le (a_1^2 + b_1^2 + c_1^2)(a_2^2 + b_2^2 + c_2^2) \\le (9 R_1^2)(9 R_2^2) = 81 R_1^2 R_2^2.\n$$\nThis implies $a_1 a_2 + b_1 b_2 + c_1 c_2 \\le 9 R_1 R_2$. The proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31395,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of a quadrilateral $ABCD$ intersect at $K$. $M, N$ are the midpoints of $AB$ and $CD$ respectively. Let $k, m, n$ be the lines through $K, M, N$ perpendicular to $AD, BD$ and $AC$ respectively. Show that if $k, m, n$ are concurrent, then $A, B, C, D$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31396,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Given that $(1+x)^{p-2} = 1 + a_1x + a_2x^2 + \\dots + a_{p-2}x^{p-2}$ where $p$ is an odd prime number, prove that $a_1 + 2, a_2 - 3, a_3 + 4, \\dots, a_{p-3} - (p-2)$ and $a_{p-2} + (p-1)$ are all multiples of $p$.\n\nb. How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 100$ where $-1 \\le x_1 \\le 10, 0 \\le x_2 \\le 11, 1 \\le x_3 \\le 12$ and $x_4 \\ge 50$?",
    "options": [],
    "answer": "1728",
    "solution": "a.\nNote that the numbers are\n$$\n\\begin{aligned}\na_k + (-1)^{k+1}(k+1) &= \\binom{p-2}{k} + (-1)^{k+1}(k+1) \\\\\n&= \\frac{(p-2)(p-3)\\cdots(p-1-k)}{k!} + (-1)^{k+1}(k+1)\n\\end{aligned}\n$$\nwhere $1 \\le k \\le p-2$. Since\n$$\n(p-2)(p-3)\\cdots(p-1-k) \\equiv (-2)(-3)\\cdots(-1-k) = (-1)^k(k+1)! \\pmod{p}\n$$\nand $p \\nmid k!$, we have\n$$\na_k + (-1)^{k+1}(k+1) \\equiv (-1)^k(k+1) + (-1)^{k+1}(k+1) = 0 \\pmod{p}.\n$$\n\n\nb.\nThe answer is 1728.\nFor each choice of $x_1, x_2, x_3$ satisfying the inequalities, we have\n$$\nx_4 = 100 - x_1 - x_2 - x_3 \\ge 100 - 10 - 11 - 12 > 50.\n$$\nTherefore, it suffices to count the number of triples $(x_1, x_2, x_3)$ satisfying the inequalities. The answer is\n$$\n(10 - (-1) + 1)(11 - 0 + 1)(12 - 1 + 1) = 12^3 = 1728.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezoid with $AB \\parallel CD$, $AB > CD$ and $AD > BC$. Show that\n\ni. $\\angle CBA > \\angle DAB$,\n\nii. $AC > BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "i. Let $AD$ meet $BC$ at $E$, $AC$ meet $BD$ at $F$, and let $M$ be the midpoint of $AB$. Since $AB > CD$, the points $D$ and $C$ must lie on the segments $AE$ and $BE$ respectively.\nSince $\\triangle EDC \\sim \\triangle EAB$, we have $\\frac{EA}{EB} = \\frac{DA}{CB} > 1$. This implies $EA > EB$, and hence $\\angle CBA > \\angle DAB$.\n\n![](attached_image_1.png)\n\nii. By Steiner's theorem, $E$, $F$, $M$ are collinear. Since $EA > EB$, both $E$ and $F$ lie on the same side as $B$ to the perpendicular bisector of $AB$. This implies $FA > FB$. Now, as $\\triangle FAB \\sim \\triangle FCD$, we have $\\frac{AC}{BD} = \\frac{FA}{FB} > 1$. This means $AC > BD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive rational numbers $r \\neq 1$ such that $r^{\\frac{1}{r-1}}$ is rational.",
    "options": [],
    "answer": "r = 1 + 1/n for integers n ≥ 1, or r = 1 − 1/n for integers n > 1",
    "solution": "$r$ can be any number of the form $1 + \\frac{1}{k}$, where $k$ is an integer different from $0$, $-1$.\n\nIf $r > 1$, let $r = 1 + \\frac{m}{n}$ where $m, n \\in \\mathbb{Z}^+$ and $(m, n) = 1$. Since\n$$\nr^{\\frac{1}{r-1}} = \\left(\\frac{n+m}{n}\\right)^{\\frac{n}{m}}\n$$\nand $(n+m, n) = (m, n) = 1$, we must have $n+m = a^m$ and $n = b^m$ for some $a, b \\in \\mathbb{Z}^+$. This gives $a^m - b^m = m$. Clearly, $a \\ge b + 1$. Therefore, we have\n$$\nm = a^m - b^m = (a-b)(a^{m-1} + a^{m-2}b + \\dots + ab^{m-2} + b^{m-1}) \\ge (1)(1+1+\\dots+1) = m.\n$$\nEquality must hold. This means $m = 1$ (since otherwise $a = b = 1$, contradiction) and $a - b = 1$. Conversely, when $m = 1$, $r = 1 + \\frac{1}{n}$, so that $r^{\\frac{1}{r-1}} = \\left(\\frac{n+1}{n}\\right)^n \\in \\mathbb{Q}$.\n\nIf $r < 1$, let $r = 1 - \\frac{m}{n}$ where $m, n \\in \\mathbb{Z}^+$ and $(m, n) = 1$. Since\n$$\nr^{\\frac{1}{r-1}} = \\left(\\frac{n-m}{n}\\right)^{-\\frac{n}{m}} = \\left(\\frac{n}{n-m}\\right)^{\\frac{n}{m}}\n$$\nand $(n, n-m) = (n, -m) = 1$, we must have $n = a^m$ and $n-m = b^m$ for some $a, b \\in \\mathbb{Z}^+$. This gives $a^m - b^m = m$. Again, the only solution is $m = 1$ and $a-b = 1$.\n\nWhen $m = 1$, $r = 1 - \\frac{1}{n}$. Since $r > 0$, we need $n > 1$. In that case, we have\n$$\nr^{\\frac{1}{r-1}} = \\left(\\frac{n}{n-1}\\right)^n \\in \\mathbb{Q}$.\n$$\nTherefore, $r = 1 + \\frac{1}{n}$ for $n \\ge 1$ or $r = 1 - \\frac{1}{n}$ for $n > 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ and $O$ be the incentre and circumcentre of $\\triangle ABC$, respectively. Assume $\\triangle ABC$ is not equilateral (so $I \\neq O$). Prove that $\\angle AIO \\leq 90^\\circ$ if and only if $2BC \\leq AB + CA$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AI$ meet $(ABC)$ again at $D$. Recall that $DB = DI = DC$. Applying Ptolemy's theorem to $ABCD$, we obtain\n$$\nAD \\times BC = AB \\times CD + AC \\times BD = DI(AB + AC).\n$$\nNote that $\\angle AIO \\le 90^\\circ$ if and only if $AI \\ge ID$. This is equivalent to\n$$\n2 \\le \\frac{AD}{DI} = \\frac{AB + AC}{BC}.\n$$\nThis holds if and only if $2BC \\le AB + AC$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31400,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Let $a_1, a_2, \\dots, a_n > 0$. Prove that $a_1^{a_1} a_2^{a_2} \\cdots a_n^{a_n} \\ge (a_1 a_2 \\cdots a_n)^{\\frac{a_1+a_2+\\cdots+a_n}{n}}$.\n\nb. Let $x, y, z \\ge 0$ and $x + y + z = 1$. Prove that $0 \\le xy + yz + zx - 3xyz \\le \\frac{1}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Taking logarithm, it is the same as proving\n$$\na_1 \\ln a_1 + a_2 \\ln a_2 + \\cdots + a_n \\ln a_n \\ge \\frac{a_1 + a_2 + \\cdots + a_n}{n} \\ln (a_1 a_2 \\cdots a_n).\n$$\nThis means\n$$\n\\sum_{k=1}^{n} a_k \\ln a_k \\ge \\frac{1}{n} \\left( \\sum_{k=1}^{n} a_k \\right) \\left( \\sum_{k=1}^{n} \\ln a_k \\right).\n$$\nWLOG assume $a_1 \\ge a_2 \\ge \\cdots \\ge a_n$. Then we have $\\ln a_1 \\ge \\ln a_2 \\ge \\cdots \\ge \\ln a_n$. The above inequality holds by Chebyshev's inequality.\n\nb. Firstly, we have\n$$\n\\begin{aligned}\nxy + yz + zx - 3xyz &= (xy + yz + zx)(x + y + z) - 3xyz \\\\\n&= x^2y + x^2z + y^2x + y^2z + z^2x + z^2y \\ge 0.\n\\end{aligned}\n$$\nEquality holds when $(x, y, z) = (1, 0, 0)$ or its permutation.\n\nSecondly, we have\n$$\n\\begin{align*}\n& \\frac{1}{4} - (xy + yz + zx - 3xyz) \\\\\n&= \\frac{1}{4}(x + y + z)^3 - (xy + yz + zx)(x + y + z) + 3xyz \\\\\n&= \\frac{1}{4}((x^3 + y^3 + z^3) - (x^2y + x^2z + y^2x + y^2z + z^2x + z^2y) + 6xyz).\n\\end{align*}\n$$\nBy Schur's inequality, we have\n$$\nx^3 + y^3 + z^3 + 3xyz \\geq x^2y + x^2z + y^2x + y^2z + z^2x + z^2y.\n$$\nTogether with $3xyz \\ge 0$, we see the above inequality holds. Equality holds when $(x, y, z) = \\left(\\frac{1}{2}, \\frac{1}{2}, 0\\right)$ or its permutation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $p$ and $q$ are positive integers such that $\\frac{p}{q} = 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\cdots + \\frac{1}{1335}$. Show that 2003 is a factor of $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$$\n\\begin{align*}\n& 1 - \\frac{1}{2} + \\frac{1}{3} - \\frac{1}{4} + \\dots - \\frac{1}{1334} + \\frac{1}{1335} \\\\\n&= \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{1334} + \\frac{1}{1335}\\right) - 2\\left(\\frac{1}{2} + \\frac{1}{4} + \\dots + \\frac{1}{1334}\\right) \\\\\n&= \\left(1 + \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{4} + \\dots + \\frac{1}{1334} + \\frac{1}{1335}\\right) - \\left(1 + \\frac{1}{2} + \\dots + \\frac{1}{667}\\right) \\\\\n&= \\frac{1}{668} + \\frac{1}{669} + \\dots + \\frac{1}{1335} \\\\\n&= \\left(\\frac{1}{668} + \\frac{1}{1335}\\right) + \\left(\\frac{1}{669} + \\frac{1}{1334}\\right) + \\dots + \\left(\\frac{1}{1001} + \\frac{1}{1002}\\right) \\\\\n&= 2003 \\cdot \\left(\\frac{1}{668 \\cdot 1335} + \\frac{1}{669 \\cdot 1334} + \\dots + \\frac{1}{1001 \\cdot 1002}\\right).\n\\end{align*}\n$$\nOne may check that 2003 is a prime. Thus, this factor cannot be cancelled out by the denominator. Thus, $p$ must be divisible by 2003.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Students have taken a test paper in each of $n$ ($n \\ge 3$) subjects. It is known that for any subject exactly three students get the best score in the subject, and for any two subjects exactly one student gets the best score in every one of these two subjects. Determine the smallest $n$ so that the above conditions imply that exactly one student gets the best score in every one of the $n$ subjects.",
    "options": [],
    "answer": "8",
    "solution": "The smallest $n$ is $8$.\nWe use terminologies in set theory. Let $A_1, A_2, \\dots, A_n$ be sets corresponding to the $n$ subjects, while the elements correspond to the students getting the best score in that subject. It is given that $|A_j| = 3$ and $|A_i \\cap A_j| = 1$ for any $1 \\le i < j \\le n$. Suppose $n \\ge 8$. We shall show that $|A_1 \\cap A_2 \\cap \\dots \\cap A_n| \\ge 1$.\nWLOG assume $1$ is an element that belongs to the most number of sets. Suppose $1 \\in A_j$ for $j = 1, 2, \\dots, m$ and $1 \\notin A_j$ for $j = m+1, m+2, \\dots, n$. Suppose on the contrary that $m < n$.\nConsider $A_n = \\{a, b, c\\}$. For $j = 1, 2, \\dots, m$, since $|A_n \\cap A_j| = 1$, each $A_j$ consists of one of $a, b, c$. Also, each of $a, b, c$ belongs to at most one of these $A_j$'s, since the only common element in these sets is $1$, but $1 \\notin A_n$. Thus, $m \\le 3$. This means each element belongs to at most $3$ sets.\nNow, suppose $A_1 = \\{1, 2, 3\\}$. Since each of $1, 2, 3$ belongs to at most two other sets, we must have $n \\le 1 + 2 \\times 3 = 7$, which is a contradiction. Therefore, $|A_1 \\cap A_2 \\cap \\dots \\cap A_n| \\ge 1$.\nIt suffices to provide an example for $n = 7$ such that $A_1 \\cap A_2 \\cap \\dots \\cap A_n$ is empty, since we can remove some sets if $n$ is less than $7$. Indeed, consider\n$$\nA_1 = \\{1, 2, 3\\},\\ A_2 = \\{1, 4, 5\\},\\ A_3 = \\{1, 6, 7\\},\\ A_4 = \\{2, 4, 6\\},\\ A_5 = \\{2, 5, 7\\},\\ A_6 = \\{3, 4, 7\\},\\ A_7 = \\{3, 5, 6\\}.\n$$\nIt is routine to check that $|A_i \\cap A_j| = 1$ for all $i, j$, but no element belongs to all of them.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $f(x)$ such that $f(f(x)) = (f(x))^m$, where $m > 1$ is a fixed integer. Substantiate your answer.",
    "options": [],
    "answer": "All constant polynomials c with c = 0, c = 1, or c an (m−1)st root of unity; and the non-constant polynomial f(x) = x^m.",
    "solution": "The solutions are $f(x) = 0$, $f(x) = 1$, $f(x) = \\omega$ where $\\omega$ is an $(m-1)$st root of unity, and $f(x) = x^m$.\n\nIf $f(x) = c$ is a constant polynomial, then the relation holds if and only if $c = c^m$.\nClearly, the solutions are $c = 0, 1$ and all the $(m-1)$st roots of unity.\n\nIf $f$ is a non-constant polynomial, then $f(x)$ attains infinitely many values.\nThus, there are infinitely many $y$ such that $f(y) = y^m$. Since $f$ is a polynomial,\nthis implies $f(x) = x^m$ for any $x$.\n\nIt is easy to check that all these are solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that $f(x + y f(x)) = f(x) + x f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x for all real x",
    "solution": "The solutions are $f(x) = 0$ for all $x \\in \\mathbb{R}$ and $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\nIt is easy to check that both are solutions. Label the equation as follows.\n$$\nf(x + y f(x)) = f(x) + x f(y) \\tag{1}\n$$\nPutting $x = 1$ in (1), we have\n$$\nf(1 + y f(1)) = f(1) + f(y). \\tag{2}\n$$\nIf $f(1) \\neq 1$, then there exists $y \\in \\mathbb{R}$ such that $1 + y f(1) = y$. For this $y$, we obtain $f(1) = 0$. Then (2) becomes $f(y) = 0$ for all $y$, which is the first solution.\nNow, we assume $f(1) = 1$. If $f(t) = 0$ for some $t \\in \\mathbb{R}$, by putting $x = t$ and $y = 1$ in (1), we obtain $t = 0$. This means $f(x) \\neq 0$ for any $x \\neq 0$. Putting $x = 1$ in (1), we have\n$$\nf(y + 1) = f(y) + 1. \\tag{3}\n$$\nBy induction we easily obtain $f(n) = n$ for any $n \\in \\mathbb{Z}$. Putting $x = -1$ in (1), we find that\n$$\nf(-y - 1) = -f(y) - 1.\n$$\nUsing (3), we get $-f(y) = f(-y - 1) + 1 = f(-y)$. Replacing $y$ by $-y$ in (1), we have\n$$\nf(x - y f(x)) = f(x) - x f(y).\n$$\nAdding this to (1), we have\n$$\nf(x + y f(x)) + f(x - y f(x)) = 2 f(x).\n$$\nFor $x \\neq 0$, we can replace $y$ by $\\frac{y}{f(x)}$. This yields\n$$\nf(x + y) + f(x - y) = 2 f(x). \\tag{4}\n$$\nNote that this also holds when $x = 0$. Considering $x = y$, we find that\n$$\nf(2x) = 2 f(x).\n$$\nAlso, applying the substitution $a = x + y$ and $b = x - y$, equation (4) becomes\n$$\nf(a) + f(b) = 2 f\\left(\\frac{a + b}{2}\\right) = f(a + b).\n$$\nThis shows $f$ satisfies the Cauchy equation. Thus, (1) can be simplified to\n$$\nf(y f(x)) = x f(y). \\tag{5}\n$$\nPutting $y = 1$, we obtain $f(f(x)) = x$. Replacing $x$ by $f(x)$ in (5) and using $f(f(x)) = x$, we get\n$$\nf(x y) = f(x) f(y). \\tag{6}\n$$\nIt is well-known that a function $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the Cauchy equation and (6) can only be the zero function or the identity function. Thus, another solution is $f(x) = x$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose there are 6 members in an International Mathematical Olympiad team. Prove that among these 6 members there are three members who either all know each other or all don't know each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose the members are $A, B, C, D, E, F$. By the pigeonhole principle, either $A$ knows at least $\\left\\lfloor \\frac{5}{2} \\right\\rfloor = 3$ other members, or $A$ does not know at least 3 other members. WLOG assume it is the former case. Suppose $A$ knows $B, C, D$. If any two of $B, C, D$ know each other, say $B$ and $C$, then we are done since all of $A, B, C$ know each other. Otherwise, any two of $B, C, D$ do not know each other, which is also our goal. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31406,
    "subject": "Mathematics (Multi-modal)",
    "question": "$k$ cards numbered $1$ to $k$ are arranged at random in a row. In a move, we may change any block of consecutive cards whose numbers are in ascending or descending order and switch the block around. For example, in the case $k = 4$, $4132$ may be changed to $1432$ and $3214$ may be changed to $1234$. Suppose that in at most $n_k$ moves, we can arrange the $k$ cards so that their numbers are in ascending or descending order.\n\na. Prove that $n_4 = 3$.\n\nb. Determine or estimate $n_5$.",
    "options": [],
    "answer": "n4 = 3; n5 = 4",
    "solution": "a. Consider the same problem but we require the numbers at the end to be in ascending order (i.e. descending order is not allowed). Let $m_k$ be the least number of moves needed to guarantee that this can be done when there are $k$ numbers. By symmetry, this number is the same if we require the numbers at the end to be in descending order. Clearly, we have $n_k \\le m_k$.\n\nWe claim that $m_{k+1} \\le m_k + 1$ for $k \\le 4$. Given a permutation of $1, 2, \\dots, k+1$, one of the numbers $1$ and $k+1$ must be located at the first two or the last two positions since there are at most $5$ numbers in total. We can use at most $1$ move to put this number to the first or the last position. For the remaining numbers, we need at most $m_k$ moves to sort them in the correct order, so that all numbers are in monotonic order. This proves the inequality.\n\nIt follows that $n_4 \\le m_4 \\le m_3 + 1 \\le m_2 + 2 = 3$ (obviously, $m_2 = 1$). Next, we show that $2413$ cannot be changed to $1234$ or $4321$ in $2$ steps. This can be verified by checking all possibilities as follows. (Note that swapping the same block twice is useless.)\n\n$$\n\\begin{array}{ccc}\n\\underline{2413} \\longrightarrow 4213 \\longrightarrow 4123, 4231, 1243 \\\\\n2413 \\longrightarrow 2143 \\longrightarrow 1243, 2134 \\\\\n\\underline{2413} \\longrightarrow 2431 \\longrightarrow 4231, 2341, 2134\n\\end{array}\n$$\n\nTherefore, we must have $n_4 = 3$.\n\nb. We have $n_5 = 4$.\n\nAgain, we list out all possibilities.\n\n| 0 moves      | 1 move      | 2 moves   | 3 moves                                 |\n|--------------|-------------|-----------|------------------------------------------|\n| $\\underline{31}524$ | $1\\underline{3}524$ | $15324$   | $51324$, $15234$, $15342$, $12354$         |\n|              | $1\\underline{3}524$ | $13254$   | $31254$, $12354$, $13245$                  |\n|              | $\\underline{13}524$ | $13542$   | $31542$, $15342$, $13452$, $53142$, $13245$ |\n|              | $\\underline{13}524$ | $53124$   | $35124$, $51324$, $53214$, $53142$, $53421$ |\n| $\\underline{31}524$ | $\\underline{3}5124$ | $53124$   | $51324$, $53214$, $53142$, $13524$, $53421$ |\n|              | $3\\underline{1}24$  | $35214$   | $53214$, $32514$, $35241$, $31254$          |\n|              | $3\\underline{1}24$  | $35142$   | $53142$, $31542$, $35412$                   |\n|              | $3\\underline{1}24$  | $35421$   | $53421$, $34521$, $35241$, $35412$, $32451$, $31245$ |\n| $\\underline{31}524$ | $\\underline{3}1254$ | $13254$   | $12354$, $13524$, $13245$                  |\n|              | $\\underline{3}1254$ | $32154$   | $23154$, $32514$, $32145$, $12354$          |\n|              | $3\\underline{1}254$ | $31245$   | $13245$, $32145$, $31425$, $34215$, $31542$, $35421$ |\n|              | $\\underline{3}1254$ | $35214$   | $53214$, $32514$, $35124$, $35241$           |\n| $\\underline{31}524$ | $\\underline{3}1542$ | $13542$   | $15342$, $13452$, $13524$, $53142$, $13245$ |\n|              | $3\\underline{1}542$ | $35142$   | $53142$, $35412$, $35124$                   |\n|              | $3\\underline{1}542$ | $31452$   | $13452$, $34152$, $31425$, $35412$           |\n|              | $\\underline{3}1542$ | $31245$   | $13245$, $32145$, $31425$, $31254$, $34215$, $35421$ |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31407,
    "subject": "Mathematics (Multi-modal)",
    "question": "In this problem, $G$ denotes a connected graph without parallel edges (but not necessarily a planar graph) with complement $\\overline{G}$. Let $V$ be the set of vertices, $E$ be the set of edges, and $\\Delta(G)$ or $\\Delta_G$ be the maximum degree of the vertices. Define the chromatic number $\\chi(G)$ or $\\chi_G$ by\n$$\n\\chi(G) = \\chi_G = \\min\\{k : G \\text{ has a } k\\text{-colouring}\\}.\n$$\nIn other words, $\\chi(G)$ is the minimum number of colours needed to colour the vertices of $G$ such that no two adjacent vertices have the same colour.\n\na. Show that if every vertex of $G$ has degree 2, then $V$ and $E$ form the corners and sides of a polygon. These are called *cycle graphs*.\n\nb. Prove that $\\chi_G \\le \\Delta_G + 1$.\n\nc. Give two types of examples that $\\chi_G = \\Delta_G + 1$.\n\nd. Show that if $G$ is not any of the above types, then $\\chi_G \\le \\Delta_G$.\n\ne. Give an example of $G$ such that $\\chi(G) + \\chi(\\overline{G}) = \\#V + 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nWe start from an arbitrary vertex $A_1$, and find one of its neighbour $A_2$. Since $\\deg A_2 = 2$, $A_2$ has another neighbour $A_3$. Repeating the process, we must eventually find a vertex $A_k$ whose another neighbour is $A_1$ since there are finitely many vertices and all of $A_2, A_3, \\dots, A_{k-1}$ already have 2 neighbours. Now, all vertices already have degree 2. Therefore, this must be the whole graph or otherwise $G$ cannot be connected. Thus, $A_1A_2\\cdots A_k$ is a polygon.\n\nb.\nIt suffices to show that $G$ has a $(\\Delta_G + 1)$-colouring. Indeed, we can just assign an arbitrary colour to the vertices one by one such that each vertex has a different colour as all its neighbours which are already coloured. This is possible since there are $\\Delta_G + 1$ colours but each vertex has degree at most $\\Delta_G$.\n\nc.\n$G$ can be a cycle graph with an odd number of vertices and any complete graph.\n\nFirstly, a cycle graph with an odd number of vertices satisfies $\\chi_G = 3$ and $\\Delta_G = 2$. Indeed, $\\Delta_G = 2$ holds by definition. If at most 2 colours are used, there is a unique way to colour the vertices $A_1, A_2, \\dots$ of the polygon $A_1A_2\\cdots A_k$ such that $A_{j+1}$ has a different colour as $A_j$. But since $k$ is odd, $A_k$ has the same colour as $A_1$, contradiction. Therefore, at least 3 colours are needed. Since $\\chi_G \\le \\Delta_G + 1 = 3$, we must have $\\chi_G = 3 = \\Delta_G + 1$.\n\nSecondly, a complete graph with $k$ vertices satisfies $\\chi_G = k$ and $\\Delta_G = k - 1$.\n\nBoth are obvious since any pair of vertices is joined by an edge (hence all vertices have different colours).\n\nd.\n(Brook's theorem) We prove the result by induction on $\\Delta_G$, and for each $\\Delta_G$, we induct on the number $k$ of vertices.\nThe base case $\\Delta_G = 2$ is trivial since the graph is just a path or a cycle. Suppose $\\Delta_G \\ge 3$. Now, the base case is $k = \\Delta_G + 1$. As $G$ is not a complete graph, there is a vertex $A$ which has degree at most $\\Delta_G - 1$. We can colour all other vertices in different colours ($\\Delta_G$ colours are used). Then since $\\deg A \\le \\Delta_G - 1$, we can always assign a colour to $A$ which is different from the colours of all its neighbours.\n\nNow, consider the inductive step, where $k \\ge \\Delta_G + 2$.\n\n**Case 1.** There exists a vertex $A$ such that $G - \\{A\\}$ is disconnected.\nSuppose the connected components of $G - \\{A\\}$ are $G_1, G_2, \\dots, G_m$. For each $G_j$, consider the connected subgraph $G'_j$ induced by $G$ containing all the vertices in $G_j$ together with $A$. Since $m \\ge 2$, $G'_j$ has fewer vertices. Also, the maximum degree of $G'_j$ cannot exceed $\\Delta_G$. Thus, we can find a $\\Delta_G$-colouring of $G'_j$ by the inductive hypothesis unless $G'_j$ is a complete graph or an odd cycle. But even in these cases, we have\n$$\n\\Delta_{G'_j} = \\deg_{G'_j} A \\le \\deg_G A - 1 \\le \\Delta_G - 1.\n$$\nThis shows we can still find a colouring using at most $\\Delta_G - 1 + 1 = \\Delta_G$ colours.\n\n![](attached_image_1.png)\n\nNow, we have found a $\\Delta_G$-colouring for each $G'_j$. WLOG assume $A$ has colour 1 in all such colourings. Then we can combine the colourings used for all $G'_j$ to give a colouring for $G$. The condition is satisfied, since any two of $G_1, G_2, \\dots, G_m$ are disconnected.\n\n**Case 2.** $G - \\{A\\}$ is connected for all vertex $A$, and there exist two non-adjacent vertices $B$ and $C$ such that $G - \\{B, C\\}$ is disconnected.\n\nSuppose the connected components of $G - \\{B, C\\}$ are $G_1, G_2, \\dots, G_m$. Note that $B$ must be adjacent to some vertex in each $G_j$, or otherwise we can remove $C$ to get at least two connected components (one is $G_j$), contradicting the assumption. The same holds for $C$. For each $G_j$, let $H_j$ be the induced subgraph of $G$ containing $B, C$ and all vertices in $G_j$.\nNow, the degrees of $B$ and $C$ in $H_j$ are at most $\\Delta_G - 1$ since $m \\ge 2$. Consider the graph $H'_j$ obtained from $H_j$ by including the edge $B-C$. Then we still have $\\Delta_{H'_j} \\le \\Delta_G$. So there is a $\\Delta_G$-colouring for $H'_j$ unless it is an exceptional case. If such a colouring exists for all $H'_j$, WLOG we may assume $B$ has colour 1 and $C$ has colour 2 in all colourings (note that their colours are different since $B-C$ is an edge). Then we can combine all the colourings to give a $\\Delta_G$-colouring for $G$.\n\n![](attached_image_2.png)\n\nNote that the same argument works as long as there is a $\\Delta_G$ colouring for each $H'_j$. In particular, the argument works if some $H'_j$ is an odd cycle (since $\\Delta_G \\ge 3$) or a complete graph with at most $\\Delta_G$ vertices. So the only case left is that $H'_1$, for example, is a complete graph with $\\Delta_G + 1$ vertices. Since the degrees of $B$ and $C$ in $G$ are at most $\\Delta_G$, each of $B$ and $C$ can only be adjacent to one more vertex not in $G_1$. Therefore, we have $m=2$, and each of $B$ and $C$ is adjacent to only one vertex in $G_2$.\nNow, we can identify $B$ and $C$ in $H_2$ as the same point $D$ to form a new connected graph $H'$. Since $\\deg D \\le 2$ and the degrees of all other vertices are unchanged, we have $\\Delta_{H'} \\le \\Delta_G$ and $H'$ has fewer vertices than $G$. Therefore, there is a $\\Delta_G$-colouring of $H'$. This also holds even if $H'$ is an odd cycle or a complete graph (which has at most 3 vertices as $\\deg D \\le 2$). This yields a $\\Delta_G$-colouring of $H_2$ such that $B$ and $C$ have the same colour (as $D$). Then we can colour the remaining $\\Delta_G - 1$ vertices in $G_1$ using colours different from that of $B$ and $C$. This gives a $\\Delta_G$-colouring of $G$.\n\nLet $A$ be a vertex with $\\deg A = \\Delta_G$. If all neighbours of $A$ are adjacent, then there is already a $(\\Delta_G + 1)$-clique. This is impossible since the graph is connected and not complete, but we cannot add more edges to this clique (or $\\Delta_G$ is increased). Suppose $B$ and $C$ are some non-adjacent neighbours of $A$.\nStarting from $A_1 = A$, we construct a sequence of vertices as follows. Suppose we have constructed $A_1, A_2, \\dots, A_n$. Since $G - \\{B, C\\}$ is connected, we can always find some $A_{n+1} \\neq B, C$ such that $A_{n+1}$ is adjacent to one of $A_1, A_2, \\dots, A_n$ unless all vertices are used. We repeat the same process until we have obtained $A_1, A_2, \\dots, A_{k-2}$. Then we define $A_{k-1} = B$ and $A_k = C$.\n\n![](attached_image_3.png)\n\nWe colour the vertices from $A_k$ back to $A_1$ as follows. We can assign colour 1 to $A_k$ and $A_{k-1}$ since they are non-adjacent. For each $A_j$ we can always assign a colour to it which is different from its neighbours in $A_k, A_{k-1}, \\dots, A_{j+1}$ since $\\deg A_j \\le \\Delta_G$ and one of the neighbours of $A_j$ is $A_1, A_2, \\dots, A_{j-1}$ by construction. We can do this for $A_{k-2}, A_{k-3}, \\dots, A_2$. Lastly, for $A_1$, it has $\\Delta_G$ neighbours, where 2 of them ($A_{k-1}$ and $A_k$) have the same colour. So we can still find a new colour for $A_1$. This yields a $\\Delta_G$-colouring for $G$.\n\nThe proof is complete by combining all these cases and using induction.\n\ne.\nAn example is a complete graph with $k$ vertices. By part (c), we have $\\chi(G) = k$. Also, $\\overline{G}$ has no edges, and so $\\chi(\\overline{G}) = 1$. This gives\n$$\n\\chi(G) + \\chi(\\overline{G}) = k + 1\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the remainder of $1999^{2000}$ when it is divided by $31$.",
    "options": [],
    "answer": "1",
    "solution": "The remainder is $1$.\nNote that $1999 \\equiv -16 = -2^4 \\pmod{31}$. Therefore, we have\n$$\n1999^{2000} \\equiv (2^4)^{2000} = (2^5)^{1600} = 32^{1600} \\equiv 1 \\pmod{31}.\n$$\nThus, the remainder is $1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre of $\\triangle ABC$. Suppose $AB > AC > BC$. Let $D$ be a point on the minor arc $BC$ such that $AD$ is not a diameter of the circumcircle of $\\triangle ABC$. Let $E$ and $F$ be points on $AD$ such that $AB \\perp OE$ and $AC \\perp OF$. Let $P$ be the intersection of $BE$ and $CF$. If $PB = PC + PO$, then prove that $\\angle BAC = 30^\\circ$.",
    "options": [],
    "answer": "30°",
    "solution": "Note that $OE$ is the perpendicular bisector of $AB$. Let $\\angle EBA = \\angle EAB = x$. Similarly, let $\\angle FCA = \\angle FAC = y$. Then we have\n$$\n\\angle BPC = 180^\\circ - \\angle PBC - \\angle PCB = 2x + 2y = 2 \\angle BAC = \\angle BOC.\n$$\nThis implies $O$, $B$, $C$, $P$ are concyclic. Applying Ptolemy's theorem to the cyclic quadrilateral $OBCP$, we obtain\n$$\nOB \\times PC + BC \\times OP = OC \\times BP.\n$$\nThus,\n$$\nBC \\times OP = OC \\times BP - OB \\times PC = OB(BP - PC) = OB \\times OP.\n$$\nThis yields $BC = OB$. Hence, $\\triangle OBC$ is equilateral. Therefore,\n$$\n\\angle BAC = \\frac{1}{2} \\angle BOC = 30^\\circ.\n$$\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any integer $n$, $n^{30} - n^{14} - n^{18} + n^2$ is divisible by $46410$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $46410 = 2 \\times 3 \\times 5 \\times 7 \\times 13 \\times 17$. It suffices to check\n$$\nm = n^{30} - n^{14} - n^{18} + n^2\n$$\nis divisible by each of $p = 2, 3, 5, 7, 13, 17$.\nIf $p \\mid n$, then we must have $p \\mid m$. If $p \\nmid n$, then we have\n$$\nn^{p-1} \\equiv 1 \\pmod{p}\n$$\nby the Fermat little theorem. Since\n$$\nm = n^2(n^{12} - 1)(n^{16} - 1),\n$$\nwe have $p \\mid m$ if $p-1 \\mid 12$ or $p-1 \\mid 16$. This clearly holds for $p = 2, 3, 5, 7, 13, 17$. So we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive and $abc = 1$. Prove that\n$$\n\\frac{1 + ab^2}{c^3} + \\frac{1 + bc^2}{a^3} + \\frac{1 + ca^2}{b^3} \\geq \\frac{18}{a^3 + b^3 + c^3}.\n$$\nDetermine when equality is attained.",
    "options": [],
    "answer": "Equality holds when a = b = c = 1.",
    "solution": "We have\n$$\n(a^3 + b^3 + c^3) \\left( \\frac{1 + ab^2}{c^3} + \\frac{1 + bc^2}{a^3} + \\frac{1 + ca^2}{b^3} \\right) \\\\ = \\sum_{\\text{sym}} \\frac{a^3}{c^3} + \\sum_{\\text{cyc}} \\frac{a^4 b^2}{c^3} + \\sum_{\\text{cyc}} \\frac{ab^5}{c^3} + \\sum_{\\text{cyc}} ab^2 + 3.\n$$\nApplying the AM-GM inequality to each sum, and using $abc = 1$, this is bounded below by\n$$\n6 + 3abc + 3abc + 3abc + 3 = 18.\n$$\nEquality holds when $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31412,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Let $a$, $n$, $k$ be positive integers. Prove that $a^{4n+k}$ and $a^k$ have identical first (rightmost) digits in their decimal representations.\n\nb. Find the first digit of the decimal representation of the number\n$$\n2^{1999} + 7^{1999} + 9^{1999}.\n$$",
    "options": [],
    "answer": "0",
    "solution": "a. By the Euler-Fermat theorem, since $\\varphi(10) = 4$ and $4n + k \\equiv k \\pmod{4}$, we have\n$$\na^{4n+k} \\equiv a^k \\pmod{10}\n$$\nfor any positive integers $a$, $n$, $k$. This means $a^{4n+k}$ and $a^k$ have the same rightmost digit.\n\nb. The rightmost digit is $0$.\nBy part (a), since $1999 \\equiv 3 \\pmod{4}$, we have\n$$\n2^{1999} + 7^{1999} + 9^{1999} \\equiv 2^3 + 7^3 + 9^3 \\equiv 8 + 3 + 9 \\equiv 0 \\pmod{10}.\n$$\nThis shows the rightmost digit of the given number is $0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1 = 1$, $a_{n+1} = \\frac{a_n}{n} + \\frac{n}{a_n}$ for $n = 1, 2, 3, \\dots$. Find the greatest integer less than or equal to $a_{2000}$. Be sure to prove your claim.",
    "options": [],
    "answer": "44",
    "solution": "The answer is $44$.\nWe shall prove by induction that $\\sqrt{n} < a_n < b_n := \\frac{n}{\\sqrt{n-1}}$ for any integer $n \\ge 3$.\nFor the base case, we find $a_2 = 2$ and $a_3 = 2 \\in [\\sqrt{3}, \\frac{3}{\\sqrt{2}}]$.\nFor the inductive step, assume $\\sqrt{k} < a_k < b_k$ for some integer $k \\ge 3$. Note that $b_k < k$. Therefore, $\\frac{a_k}{k} < \\frac{b_k}{k} < 1$. Observe that the function $f(x) = x + \\frac{1}{x}$ is\n\ndecreasing on $(0,1)$ since $f'(x) = 1 - \\frac{1}{x^2} < 0$. Therefore,\n$$\na_{k+1} = f\\left(\\frac{a_k}{k}\\right) < f\\left(\\frac{\\sqrt{k}}{k}\\right) = \\frac{k+1}{\\sqrt{k}} = b_{k+1}.\n$$\nFor the lower bound, we have\n$$\na_{k+1} = f\\left(\\frac{a_k}{k}\\right) > f\\left(\\frac{b_k}{k}\\right) = \\frac{k}{\\sqrt{k-1}} > \\sqrt{k+1}.\n$$\nSo we have proven $\\sqrt{n} < a_n < b_n$ for all $n \\ge 3$ by induction. This implies\n$$\na_n < \\frac{n}{\\sqrt{n-1}} = \\sqrt{n+1 + \\frac{1}{n-1}} < \\sqrt{n+2}.\n$$\nThus, we have\n$$\n44 < \\sqrt{2000} < a_{2000} < \\sqrt{2002} < 45.\n$$\nThis shows the greatest integer less than or equal to $a_{2000}$ is $44$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle. Suppose a circle $\\Gamma_1$, with centre $O_1$, touches the sides $BC$ produced at $E$, $AC$ produced at $G$, and $AB$ at $C'$. Suppose also that another circle $\\Gamma_2$, with centre $O_2$, touches the sides $AB$ produced at $H$, $BC$ produced at $F$, and $AC$ at $B'$. Let the extensions of $EG$ and $FH$ intersect at $P$. Prove that $PA \\perp BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $EC'$ meet $FB'$ at $D$. Note that $EC'$ is parallel to the internal angle bisector of $\\angle CBA$, which is $BO_2$. Therefore, $ED \\perp PF$. Similarly, $FD \\perp PE$. This implies $D$ is the orthocentre of $\\triangle PEF$, and hence $PD \\perp EF$. It suffices to show $P, A, D$ are collinear, since this would imply $PA \\perp BC$.\n\n![](attached_image_1.png)\n\nConsider $(PGB')$ and $(PC'H)$. Since $AG \\times AB' = AC' \\times AH$, the point $A$ lies on the radical axis of these circles. Thus, $PA$ is the radical axis. It remains to show that $D$ also lies on this radical axis. Indeed, we shall prove that $D$ lies on both circles. We have\n$$\n\\angle GB'D = \\angle CB'F = \\angle B'FC = \\angle GPD.\n$$\nThe last equality holds since $D$ is the orthocentre of $\\triangle PEF$. This implies $D$ lies on $(PGB')$. The other assertion can be proved similarly. So we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all prime numbers $p$ and $q$ such that $\\frac{(7^p - 2^p)(7^q - 2^q)}{pq}$ is an integer.",
    "options": [],
    "answer": "The pairs are (5,5), (5,11), (11,5), (5,61), (61,5).",
    "solution": "The solutions are $(p, q) = (5, 5), (5, 11), (5, 61)$ and their permutations.\nIf $p \\mid 7^p - 2^p$, then $p \\mid 7 - 2 = 5$ by the Fermat little theorem. This implies $p = 5$. If we further have $q \\mid 7^q - 2^q$, then $q = 5$ for the same reason. So $p = q = 5$ is a solution. If $q \\ne 5$, then $q \\mid 7^p - 2^p = 16775 = 5^2 \\times 11 \\times 61$. Therefore, $q = 11, 61$. By symmetry $(p, q) = (11, 5), (61, 5)$ are also solutions.\n\nNow we may assume $p, q \\ne 5$. Then we must have $p \\mid 7^q - 2^q$ and $q \\mid 7^p - 2^p$. WLOG assume $q \\ge p$. Consider $p \\mid 7^q - 2^q$. Note that $(q, p - 1) = 1$ since $q$ is a prime and $q > p - 1$. Thus, there exists $a \\in \\mathbb{Z}^+$ such that $aq \\equiv 1 \\pmod{p-1}$. By the Fermat little theorem, we have\n$$\n7 \\equiv 7^{aq} = (7^q)^a \\equiv (2^q)^a = 2^{aq} \\equiv 2 \\pmod{p}.\n$$\nThis forces $p = 5$, which is a contradiction.\n\nTherefore, the only solutions are $(p, q) = (5, 5), (5, 11), (5, 61)$ and their permutations.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31416,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a table there are $1999$ tea cups with their mouths facing upward initially. In each move, $100$ of them are turned upside down. After a number of moves, can they be turned so that all their mouths face downward? Why?\n\nAnswer the above two questions for the case where the number of cups is $1998$.",
    "options": [],
    "answer": "1999: No. 1998: Yes.",
    "solution": "No. In a move, if we choose $100$ cups where $k$ of them are facing downward and $100-k$ of them are facing upward, then the number of cups facing downward is changed by $(100-k) - k = 100 - 2k$. Since there is an even number of cups facing downward initially, there is always an even number of cups facing downward. Thus, it is impossible to make all $1999$ cups face downward.\n\nIt is possible to make all cups face downward if there are $1998$ cups. If we turn over cups $1, 2, \\ldots, 100$ and then turn over cups $2, 3, \\ldots, 101$, we see that only cups $1$ and $101$ are turned over. This shows we can turn over any $2$ cups after $2$ moves. Since $2 \\mid 1998$, we can repeat the same process to turn over all cups.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31417,
    "subject": "Mathematics (Multi-modal)",
    "question": "A *lattice point* on the coordinate plane is a point with integer coordinates. Find all positive integer $n \\ge 3$ such that there exists an $n$-sided polygon having lattice points as vertices and all sides have equal length.",
    "options": [],
    "answer": "all even integers n ≥ 4",
    "solution": "$n$ can be any even integer larger than $2$.\n\nWe first consider the case when $n$ is odd. Let $(x_j, y_j)$, where $j = 1, 2, \\dots, n$, be the vertices of the polygon in anticlockwise direction. WLOG assume $x_1 = y_1 = 0$. By using the homothety $(x_j, y_j) \\mapsto \\left(\\frac{x_j}{d}, \\frac{y_j}{d}\\right)$, we may assume the greatest common divisor of $x_1, x_2, \\dots, x_n, y_1, y_2, \\dots, y_n$ is $1$. In particular, not all coordinates are even numbers.\n\nLet $a_j = x_{j+1} - x_j$ and $b_j = y_{j+1} - y_j$ for $j = 1, 2, \\dots, n$, where the indices are taken modulo $n$. It is given that $a_j^2 + b_j^2 = c$ for some constant $c$.\n\n* If $c \\equiv 0 \\pmod{4}$, then both $a_j, b_j$ are even. Since $x_0, y_0$ are even, we easily deduce that all $x_j, y_j$ are even. This contradicts our assumption.\n* If $c$ is odd, then $a_j + b_j \\equiv 1 \\pmod{2}$. Thus, we have\n$$\n1 \\equiv \\sum_{j=1}^{n} (a_j + b_j) = \\sum_{j=1}^{n} (x_{j+1} - x_j + y_{j+1} - y_j) = 0 \\pmod{2},\n$$\nwhich is a contradiction.\n* If $c \\equiv 2 \\pmod{4}$, then $a_j, b_j$ are odd. Thus, we have\n$$\n1 \\equiv \\sum_{j=1}^{n} a_j = \\sum_{j=1}^{n} (x_{j+1} - x_j) = 0 \\pmod{2},\n$$\nwhich is a contradiction.\n\nTherefore, it is impossible that $n$ is odd.\n\nFor even $n \\ge 4$, we provide the following construction. The following quadrilaterals have the properties that each side has length $5$, and two of the sides are vertical.\n* (type I) a square with vertices $(0,0), (5,0), (5,5), (0,5)$\n* (type II) a rhombus with vertices $(0,0), (4,3), (4,8), (0,5)$\n* (type III) a rhombus with vertices $(0,0), (4,-3), (4,2), (0,5)$\n\nWe can place $\\frac{n}{2}-1$ copies of these quadrilaterals in the order types I, II, III, II, III, II, III, ... by translating them to the position such that the left edge of the next quadrilateral overlaps with the right edge of the previous one as shown. This gives an $n$-sided polygon whose every side has length $5$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31418,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABC$ is a triangle with $BC > CA > AB$. $D$ is a point on side $BC$, and $E$ is a point on $BA$ produced beyond $A$ so that $BD = BE = CA$. Let $P$ be a point on side $AC$ such that $E, B, D, P$ are concyclic, and let $Q$ be the second intersection point of $BP$ with the circumcircle of $\\triangle ABC$. Prove that $AQ + CQ = BP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the concyclic points, we have\n$$\n\\angle CAQ = \\angle CBQ = \\angle DBP = \\angle DEP\n$$\nand\n$$\n\\angle QCA = \\angle QBA = \\angle PBE = \\angle PDE.\n$$\nThis implies $\\triangle QAC \\sim \\triangle PED$. Let\n$$\n\\frac{QA}{PE} = \\frac{QC}{PD} = \\frac{AC}{ED} = k.\n$$\n\nApplying Ptolemy's theorem to the cyclic quadrilateral $EBDP$, we obtain\n$$\nBD \\times PE + BE \\times PD = BP \\times DE.\n$$\nIt follows that\n$$\n\\begin{align*}\nAQ + CQ &= kPE + kPD \\\\\n&= \\frac{k}{AC}(BD \\times PE + BE \\times PD) \\\\\n&= \\frac{k}{AC}(BP \\times DE) \\\\\n&= BP.\n\\end{align*}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31419,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a graph $G$,\n$$\n\\begin{align*}\n\\chi_G &= \\min \\{k : G \\text{ has a } k \\text{ colouring}\\}, \\\\\n\\Delta_G &= \\max \\{\\text{degrees of vertices of } G\\}.\n\\end{align*}\n$$\n\na. Prove that for any simple connected graph $G$, $\\chi_G \\le \\Delta_G + 1$.\n\nb. A regular graph is a graph that every vertex has the same degree. Prove that if a simple connected graph $G$ is not regular, then $\\chi_G \\le \\Delta_G$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\n(1999 Test 1 Problem 7(b)) See the solution above.\n\nb.\n(1999 Test 1 Problem 7(d) simplified) We give an alternative proof for this simpler version. We prove that there is a $\\Delta_G$-colouring by induction on the number $n$ of vertices.\n\nThe base case is $n = 3$, and the graph is $A - B - C$. Clearly, we can assign colour 1 to $A$ and $C$, and colour 2 to $B$. This gives $\\chi_G \\leq 2 = \\Delta_G$.\n\nAssume the assertion holds when there are fewer than $n$ vertices. Consider a simple connected graph $G$ with $n$ vertices which is not regular. Since it is not regular, there is a vertex $A$ having degree at most $\\Delta_G - 1$. Let $G_1, G_2, \\dots, G_m$ be the connected components of $G - \\{A\\}$. We claim that there is a $\\Delta_G$-colouring for each $G_j$.\n\nSuppose on the contrary that $G_1$ has no $\\Delta_G$-colouring. By the inductive hypothesis, $G_1$ must be regular. If $\\Delta_{G_1} \\leq \\Delta_G - 1$, then by part (a), we have\n$$\n\\chi_{G_1} \\leq \\Delta_{G_1} + 1 \\leq \\Delta_G.\n$$\n\nThis shows a $\\Delta_G$-colouring exists. Therefore, we may assume $\\Delta_{G_1} = \\Delta_G$. But then this implies all vertices in $G_1$ has degree $\\Delta_G$. This is impossible since one of these vertices is adjacent to $A$, and hence its degree in $G$ is at least $\\Delta_G + 1$. This proves our claim.\n\nNow, we choose any $\\Delta_G$-colouring for each $G_j$. Lastly, we can assign a colour to $A$ which is different from the colours of all its neighbours since $\\deg A \\le \\Delta_G - 1$. This completes the proof by induction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $G$ be a connected simple graph which has $2p$ vertices (where $p$ is a positive integer), but contains no triangles. Prove that the number $\\#E$ of its edges satisfies $\\#E \\leq p^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove that any simple graph with $2n$ vertices and without triangles has at most $n^2$ edges. We prove this by induction. When $n=1$, clearly there is at most $n^2=1$ edge between 2 vertices. Assume there are at most $k^2$ edges when there are $2k$ vertices. Consider a simple graph with $2(k+1)$ vertices and without triangles.\n\nWe are done if there is no edge. So we may assume there is an edge $A-B$. For any other vertex $C$, at most one of the pair $\\{A, C\\}$ and $\\{B, C\\}$ is joined by an edge since there is no triangle. Thus, there are at most $2k$ edges containing exactly one of $A$ and $B$. Among the remaining $2k$ vertices, there are at most $k^2$ edges by the inductive hypothesis. Therefore, the number of edges is at most\n$$\n1 + 2k + k^2 = (k+1)^2.\n$$\nThis proves the inductive step, and so we are done.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all primes of the form $n^2 + 1$, which are less than $10^{19}$ ($n$ is a positive integer).",
    "options": [],
    "answer": "2, 5, 257",
    "solution": "The only possible primes are $2$, $5$, $257$.\n\nIf $n = pm$ for some odd prime $p$, then we have\n$$\nn^n + 1 = (n^m)^p + 1 = (n^m + 1)(n^{m(p-1)} + n^{m(p-2)} + \\cdots + 1).\n$$\nClearly, both factors are larger than $1$, and hence $n^n+1$ is not a prime. Therefore, $n$ must be a power of $2$.\n\nNote that\n$$\n16^{16} + 1 > 2^{64} = 2^4 \\times (2^{10})^6 > 10 \\times (10^3)^6 = 10^{19}.\n$$\nSo it remains to check $n = 1, 2, 4, 8$. Indeed, we have\n$$\n\\begin{align*}\n1^1 + 1 &= 2, \\\\\n2^2 + 1 &= 5, \\\\\n4^4 + 1 &= 257, \\\\\n8^8 + 1 &= (2^8)^3 + 1 = (2^8 + 1)(2^{16} - 2^8 + 1).\n\\end{align*}\n$$\nAmong these, only $2$, $5$, $257$ are primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31422,
    "subject": "Mathematics (Multi-modal)",
    "question": "A triangle $ABC$ is given. A circle $\\Gamma$ passes through vertex $A$ and is tangent to side $BC$ at point $P$. The circle $\\Gamma$ intersects sides $AB$ and $AC$ at points $M$ and $N$, respectively. Prove that (minor) arcs $\\widehat{MP}$ and $\\widehat{NP}$ are equal if and only if $\\Gamma$ is tangent to the circumcircle of $\\triangle ABC$ at $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $AB = AC$, the result is obvious due to symmetry (both statements are equivalent to $P$ being the midpoint of $BC$). WLOG assume $AB < AC$. Let $D$ be the intersection of the line $BC$ and the tangent at $A$ to the circumcircle $(ABC)$.\n\nIf $\\overline{MP}$ and $\\overline{NP}$ are equal, then $MN$ is parallel to the tangent at $P$ to $\\Gamma$, which is $BC$. It follows that\n$$\n\\angle DAM = \\angle DAB = \\angle ACB = \\angle ANM,\n$$\nwhich implies $DA$ is tangent to $\\Gamma$. Therefore, $\\Gamma$ is tangent to $(ABC)$ at $A$.\n\n![](attached_image_1.png)\n\nConversely, if $\\Gamma$ and $(ABC)$ are tangent at $A$, then $DA$ is tangent to $\\Gamma$. Therefore, we have\n$$\n\\angle ANM = \\angle DAM = \\angle DAB = \\angle ACB,\n$$\nwhich implies $MN \\parallel BC$. This shows $P$ is the midpoint of $\\overline{MN}$ as desired.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that the equation $x^3 + y^3 + z^3 = nx^2y^2z^2$ has positive integer solutions. Be sure to give a proof.",
    "options": [],
    "answer": "1 and 3",
    "solution": "$n$ can be $1$ or $3$.\n\nWLOG assume $x \\ge y \\ge z$. Then we have\n$$\n3x^3 \\ge x^3 + y^3 + z^3 = nx^2y^2z^2,\n$$\nwhich gives $x \\ge \\frac{n}{3}y^2z^2$.\n\nNext, we rewrite the equation as $y^3 + z^3 = x^2(ny^2z^2 - x)$. Since the left-hand side is positive, we have $ny^2z^2 - x \\ge 1$. Thus,\n$$\n2y^3 \\ge y^3 + z^3 \\ge \\left(\\frac{n}{3}y^2z^2\\right)^2 (1) = \\frac{n^2}{9}y^4z^4.\n$$\nThis yields $18 \\ge n^2yz^4$. There is no positive integer solution when $n \\ge 5$. When $n = 4$, we must have $y = z = 1$. The given equation becomes $x^3 + 2 = 4x^2$. This shows $x^2 \\mid 2$, and hence $x = 1$. But this is not a solution.\n\nFor $n = 3$, we can take $x = y = z = 1$.\nFor $n = 2$, we have $18 \\ge 4yz^4$. So $z = 1$ and $y \\le 4$. Note that $x^2 \\mid y^3 + z^3 = y^3 + 1$.\nAs $x \\ge y$, $(x, y)$ can only be $(1, 1)$, $(3, 2)$. It is routine to check that none of these is a solution to $x^3 + y^3 + 1 = 2x^2y^2$.\nFor $n = 1$, we can take $(x, y, z) = (3, 2, 1)$.\nThus, $n$ can only be $1$ or $3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which the number $n^8 - 2^n$ is not divisible by 72.",
    "options": [],
    "answer": "All positive integers n with n ≡ 2 (mod 4), i.e., n = 4k − 2 for k ∈ Z^+.",
    "solution": "$n$ can be any positive integer of the form $4k - 2$ where $k \\in \\mathbb{Z}^+$. When $n = 4k - 2$, we have\n$$\nn^8 - n^2 = n^2(n^6 - 1) = 4(2k - 1)^2(n^6 - 1).\n$$\nSince both $2k-1$ and $n^6-1$ are odd, this number is not divisible by 8, and hence not divisible by 72.\n\nWe now claim that for any other $n$, $n^8 - n^2$ is divisible by both 8 and 9. If $n$ is even, $n$ must be a multiple of 4. In this case $n^8 - n^2$ is divisible by $n^2$, and hence divisible by 8. If $n$ is odd, it is well-known that $8 \\mid n^2 - 1$. Thus,\n$$\nn^8 - n^2 = n^2(n^2 - 1)(n^4 + n^2 + 1)\n$$\nis divisible by 8.\n\nNext, if $3 \\mid n$, then $n^8 - n^2$ is divisible by $n^2$, and hence divisible by 9. If $3 \\nmid n$, then $3 \\mid n^2 - 1$. By the lifting the exponent lemma, we have\n$$\n3^2 \\mid (n^2)^3 - 1^3 = n^6 - 1,\n$$\nand hence $9 \\mid n^2(n^6 - 1) = n^8 - n^2$. Therefore, $9 \\mid n^8 - n^2$ in any case.\n\nCombining these, we know that $72 \\mid n^8 - n^2$ if $n \\neq 2 \\pmod 4$. So we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31425,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each integer $k \\ge 4$, prove that if $F(x)$ is a polynomial with integer coefficients satisfying the condition $0 \\le F(c) \\le k$ for every $c = 0, 1, \\dots, k+1$, then\n$$\nF(0) = F(1) = \\dots = F(k+1).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $(k+1)-0 = F(k+1)-F(0)$. Since $|F(k+1)-F(0)| \\le k$, we must have $F(k+1) = F(0) = d$ for some constant $d$. Let $F(x) = d + x(x-k-1)G(x)$ for some polynomial $G$ with integer coefficients.\n\nFor $2 \\le n \\le k-1$, we have $F(n) = d+n(n-k-1)G(n)$, and so $n(k+1-n) \\mid F(n)-d$.\nAgain, $|F(n)-d| \\le k$. Since $n(k+1-n) \\ge 2(k+1-2) > k$, we must have $F(n) = d$.\nThus, we can write $F(x) = d+x(x-2)(x-k+1)(x-k-1)H(x)$ for some polynomial $H$ with integer coefficients. (Note that we have used the fact $k \\ge 4$ to show that the roots $0, 2, k-1, k+1$ of $F(x)-d$ are distinct.)\n\nLastly, for $n = 1, k$, we have $n(n-2)(n-k+1)(k+1-n) \\mid F(n) - d$. Again, $|F(n)-d| \\le k$. Also, $n(n-2)(n-k+1)(k+1-n) = k(k-2) \\ge 2k$ in both cases. Therefore, we must have $F(n) = d$.\n\nThus, we have shown that $F(n) = d$ for $n = 0, 1, \\dots, k+1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31426,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$, determine (with proof) if the polynomial\n$$\nf(x) = (x^2 + x)^{2n} + 1\n$$\ncan be factored into a product of two non-constant polynomials with integer coefficients.",
    "options": [],
    "answer": "It cannot be factored; for every positive integer n the polynomial is irreducible over the integers.",
    "solution": "No. Suppose on the contrary that $f(x)$ is reducible. Then so is\n$$\n4^{2n} f(x) = (4x^2 + 4x)^{2n} + 4^{2n} = ((2x + 1)^2 - 1)^{2n} + 4^{2n}.\n$$\nLet $y = 2x + 1$. Then we have $x = \\frac{y-1}{2}$, and hence\n$$\ng(y) = (y^2 - 1)^{2n} + 4^{2n} \\quad (1)\n$$\nis reducible over $\\mathbb{Q}$. By Gauss's lemma, $g$ is reducible over $\\mathbb{Z}$.\nWe claim that $g$ can be factorized into even polynomials over $\\mathbb{Z}$. Consider a non-constant irreducible factor $h$ of $g$. We are done if $h$ is an even polynomial (since then $\\frac{g}{h}$ must be even). Assume $h$ is not even. As $h(y) \\mid g(y)$, we have $h(-y) \\mid g(-y) = g(y)$. Since $h$ is irreducible and is not even, $h(y)$ and $h(-y)$ must be relatively prime. It follows that\n$$\ng(y) = h(y)h(-y)p(y)\n$$\nfor some $p \\in \\mathbb{Z}[y]$. Note that $h(y)h(-y)$ is an even polynomial. So we are done unless $p$ is a constant polynomial. Let $p(y) = c$, and let $a$ be the leading coefficient of $h$. Then the leading coefficient of $g$ is $a^2c$. In view of (1), we must have $c = 1$ and $a = \\pm 1$. Let $b$ be the constant term of $h$. Then the constant term of $g$ is $b^2c = b^2$. Comparing with (1), we obtain $4^{2n} + 1 = b^2$. This implies $4^{2n} = (4^{2n-1})^2$ and $b^2$ are consecutive perfect squares, which is impossible as $4^{2n} > 1$. Therefore, we have proven that $g$ can be factorized into even polynomials over $\\mathbb{Z}$.\n\nAs a result, by letting $z = y^2$, we find that $(z - 1)^{2n} + 4^{2n}$ is also reducible. Let $z = 4t + 1$. This shows $4^{2n}(t^{2n} + 1)$ is reducible, and so are $t^{2n} + 1$ and $(s+1)^{2n} + 1$. However, note that\n$$\n(s+1)^{2n} + 1 = s^{2n} + \\sum_{r=1}^{2n-1} \\binom{2n}{r} s^r + 2.\n$$\nThe leading coefficient is odd, while all other coefficients are even. Also, the constant term is not a multiple of 4. By Eisenstein's criterion, we know that $(s+1)^{2n} + 1$ is irreducible. This is a contradiction. Therefore, $f(x)$ is irreducible.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31427,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $212$ points inside or on a circle with radius $1$. Prove that there are at least $2001$ pairs of these points having distances at most $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A, B, C, D, E, F$ be $6$ distinct points on the circle such that\n$$\n\\overline{AB} = \\overline{BC} = \\overline{CD} = \\overline{DE} = \\overline{EF} = \\overline{FA}.\n$$\nLet $O$ be the centre of the circle. Note that any two points in the same sector among $AOB, BOC, COD, DOE, EOF, FOA$ have distance at most $1$. Let $n_1, n_2, \\dots, n_6$ be the number of points in the $6$ sectors respectively. It suffices to show\n$$\n\\binom{n_1}{2} + \\binom{n_2}{2} + \\dots + \\binom{n_6}{2} \\ge 2001.\n$$\nIndeed, since the binomial function $\\binom{x}{2}$ is convex, by Jensen's inequality, we have\n$$\n\\sum_{k=1}^{6} \\binom{n_k}{2} \\ge 6 \\left( \\frac{n_1+n_2+\\dots+n_6}{6} \\right)^2 \\frac{1}{2} = 6 \\binom{35}{2} = 3570 > 2001.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers. Prove that\n$$\n(a+b)^2 + (a+b+4c)^2 \\ge \\frac{100abc}{a+b+c}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the AM-GM inequality, we have\n$$\n(a + b + 4c)^2 \\geq (2\\sqrt{(a+b)(4c)})^2 = 16(a+b)c\n$$\nand\n$$\n\\frac{100abc}{a+b+c} \\leq \\frac{100c}{a+b+c} \\left(\\frac{a+b}{2}\\right)^2 = \\frac{25(a+b)^2c}{a+b+c}.\n$$\nTherefore, it suffices to prove\n$$\nd^2 + 16cd \\geq \\frac{25cd^2}{c+d}\n$$\nwhere $d = a + b$. Indeed,\n$$\n\\begin{align*}\nd^2 + 16cd &\\ge \\frac{25cd^2}{c+d} \\\\\n\\Leftrightarrow \\quad d(16c+d)(c+d) &\\ge 25cd^2 \\\\\n\\Leftrightarrow \\quad d(16c^2 - 8cd + d^2) &\\ge 0 \\\\\n\\Leftrightarrow \\quad d(4c-d)^2 &\\ge 0.\n\\end{align*}\n$$\nThis is obviously true. Equality holds when $a = b = 2c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles intersect at points $A$ and $B$. Through the point $B$ a straight line is drawn, intersecting the first circle at $K$ and the second circle at $M$. A line parallel to $AM$ is tangent to the first circle at $Q$. The line $AQ$ intersects the second circle again at $R$.\n\na. Prove that the tangent to the second circle at $R$ is parallel to $AK$.\n\nb. Prove that these two tangents are concurrent with $KM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. We only provide a proof for the configuration as shown. The other cases are similar.\nLet $X$ and $Y$ be any points on the tangent at $R$ to $(ABR)$ and the tangent at $Q$ to $(ABQ)$ respectively as shown. We have\n$$\n\\angle MBR = \\angle MAR = \\angle YQA = \\angle QBA,\n$$\nand so\n$$\n\\begin{aligned}\n\\angle XRA &= \\angle RBA = \\angle RBQ + \\angle QBA = \\angle RBQ + \\angle MBR \\\\\n&= \\angle MBQ = \\angle KAQ = \\angle KAR.\n\\end{aligned}\n$$\nThis implies $RX \\parallel KA$.\n![](attached_image_1.png)\n\nb. Let the tangent at $R$ to $(ABR)$ and the tangent at $Q$ to $(ABQ)$ meet at $S$. Since\n$$\n\\angle BRS = \\angle BAR = \\angle BAQ = \\angle BQS,\n$$\nthe points $Q$, $B$, $S$, $R$ are concyclic. This implies\n$$\n\\angle SBR = \\angle SQR = \\angle MBR,\n$$\nand hence $S$, $M$, $B$ are collinear. Thus, the two tangents and $KM$ are concurrent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n$ satisfying all three conditions\n$$\nn \\equiv 2 \\pmod{3}, \\quad n \\equiv -1 \\pmod{5} \\quad \\text{and} \\quad n \\equiv 3 \\pmod{7}.\n$$",
    "options": [],
    "answer": "n ≡ 59 (mod 105)",
    "solution": "The answer is any integer of the form $105k + 59$ where $k \\in \\mathbb{Z}$.\nSince $n \\equiv 2 \\equiv -1 \\pmod{3}$ and $n \\equiv -1 \\pmod{5}$, we have\n$$\nn \\equiv -1 \\pmod{15}.\n$$\nTesting $n = -1, 14, 29, 44, 59$, we see that $n = 59$ satisfies $n \\equiv 3 \\pmod{7}$. Therefore, $n = 59$ is one solution. By the Chinese remainder theorem, since $3$, $5$, $7$ are pairwise relatively prime and $3 \\times 5 \\times 7 = 105$, we have\n$$\nn \\equiv 59 \\pmod{105}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. In a conference there are $n$ mathematicians. Every pair of mathematicians communicate in one of the $n$ official languages of the conference. For any three different official languages, there exist three mathematicians who communicate with each other in these three languages. Determine all $n$ for which this is possible. Justify your claim.",
    "options": [],
    "answer": "All odd integers n ≥ 3",
    "solution": "$n$ can be any odd integer larger than $1$.\n\nWe use terminologies in graph theory. We need to colour all edges of the complete graph of $n$ vertices in $n$ colours $C_1, C_2, \\dots, C_n$ such that for any three distinct colours, there exists a triangle whose edges are of these three colours.\n\nNote that there are $\\binom{n}{3}$ triangles as well as $\\binom{n}{3}$ triples of colours. Therefore, the triangles and the triples of colours are in one-to-one correspondence.\n\nFor each colour $C_j$, there are $\\binom{n-1}{2}$ triples of colours consisting of $C_j$ since we need to choose two more colours from the remaining $n-1$ colours. On the other hand, for every edge of colour $C_j$, there are $n-2$ triangles containing this edge since we need to choose one more vertex from the remaining $n-2$ vertices. Therefore, there are exactly\n$$\n\\frac{1}{n-2} \\binom{n-1}{2} = \\frac{n-1}{2}\n$$\nedges of colour $C_j$. This implies $n$ must be odd.\n\nWe now provide a colouring for odd $n$. For any vertices $V_i$ and $V_j$, we use colour $C_r$ to colour the edge joining them where $r \\equiv i + j \\pmod{n}$. It suffices to show that for any distinct triples $r_1, r_2, r_3$, the system\n$$\n\\begin{cases} i + j \\equiv r_1 \\pmod{n}, \\\\ j + k \\equiv r_2 \\pmod{n}, \\\\ k + i \\equiv r_3 \\pmod{n} \\end{cases}\n$$\nhas a unique solution in $i, j, k \\pmod{n}$ such that $i, j, k$ are pairwise incongruent modulo $n$. Indeed, we readily find that\n$$\ni \\equiv 2^{-1}((i + j) - (j + k) + (k + i)) \\equiv 2^{-1}(r_1 - r_2 + r_3) \\pmod{n},\n$$\nand similarly $j \\equiv 2^{-1}(r_1 + r_2 - r_3) \\pmod{n}$ and $k \\equiv 2^{-1}(-r_1 + r_2 + r_3) \\pmod{n}$. These are incongruent because $r_1, r_2, r_3$ are incongruent modulo $n$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31432,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Using the characteristic equation method or otherwise, find a formula in terms of $n$ for the sequence $a_0 = 1$, $a_1 = 4$ and $a_{n+2} = 5a_{n+1} - 6a_n$ for $n = 0, 1, 2, \\dots$\n\nb. Using the characteristic equation method or otherwise, find a formula in terms of $n$ for the sequence $b_0 = 3$, $b_1 = 7$ and $b_{n+2} = 4b_{n+1} - 4b_n$ for $n = 0, 1, 2, \\dots$",
    "options": [],
    "answer": "a_n = 2·3^n − 2^n; b_n = (n + 6)·2^{n−1}",
    "solution": "a.\nThe answer is $a_n = 2 \\cdot 3^n - 2^n$.\nThe characteristic equation is $\\lambda^2 - 5\\lambda + 6 = 0$. The roots are $\\lambda = 3, 2$. Therefore, we have\n$$\na_n = A \\cdot 3^n + B \\cdot 2^n\n$$\nfor some constants $A$ and $B$. Putting $n = 0$ and $n = 1$, we need to solve\n$$\n\\begin{cases} A + B = 1, \\\\ 3A + 2B = 4. \\end{cases}\n$$\nWe easily deduce $A = (3A + 2B) - 2(A + B) = 2$ and $B = 1 - A = -1$. Thus, we have\n$$\na_n = 2 \\cdot 3^n - 2^n.\n$$\n\nb.\nThe answer is $b_n = (n + 6)2^{n-1}$.\nThe characteristic equation is $\\lambda^2 - 4\\lambda + 4 = 0$, with a double root $\\lambda = 2$. Therefore, we have\n$$\nb_n = (Cn + D)2^n\n$$\nfor some constants $C$ and $D$. Putting $n = 0$ and $n = 1$, we need to solve\n$$\n\\begin{cases} D = 3, \\\\ 2(C + D) = 7. \\end{cases}\n$$\nClearly, $C = \\frac{7}{2} - D = \\frac{1}{2}$. Thus, we have\n$$\nb_n = (n + 6)2^{n-1}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31433,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a \\ge b \\ge c \\ge 0$ and $a + b + c = 3$, then prove that $ab^2 + bc^2 + ca^2 \\le \\frac{27}{8}$ and determine the equality case(s).",
    "options": [],
    "answer": "Maximum is 27/8, with equality when a = b = 3/2 and c = 0.",
    "solution": "Let $f(a, b, c) = ab^2 + bc^2 + ca^2$. Note that it suffices to show\n$$\nf(a, b, c) - f(a, c, b) \\le 0, \\qquad (1)\n$$\n$$\nf(a, b, c) + f(a, c, b) \\le \\frac{27}{4}, \\qquad (2)\n$$\nsince adding these yields the result. To prove (1), observe that\n$$\nf(a, b, c) - f(a, c, b) = (a - b)(b - c)(c - a) \\le 0.\n$$\nTo prove (2), by homogenizing the inequality, we need to prove\n$$\nab^2 + bc^2 + ca^2 + a^2b + b^2c + c^2a \\le \\frac{1}{4}(a + b + c)^3.\n$$\nThis is equivalent to\n$$\na^3 + b^3 + c^3 + 6abc \\ge ab^2 + bc^2 + ca^2 + a^2b + b^2c + c^2a,\n$$\nwhich holds by Schur's inequality and $3abc \\ge 0$. Equality holds when $c = 0$ and $a = b = \\frac{3}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that if $a, b, c$ are positive real numbers, then\n$$\n\\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\ge \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Nesbitt's inequality) We have\n$$\n\\begin{align*}\n& \\frac{a}{b+c} + \\frac{b}{c+a} + \\frac{c}{a+b} \\ge \\frac{3}{2} \\\\\n\\Leftrightarrow & \\quad \\frac{a+b+c}{b+c} + \\frac{a+b+c}{c+a} + \\frac{a+b+c}{a+b} \\ge \\frac{9}{2} \\\\\n\\Leftrightarrow & \\quad (a+b+c) \\left( \\frac{1}{b+c} + \\frac{1}{c+a} + \\frac{1}{a+b} \\right) \\ge \\frac{9}{2} \\\\\n\\Leftrightarrow & \\quad \\left[ (b+c) + (c+a) + (a+b) \\right] \\left( \\frac{1}{b+c} + \\frac{1}{c+a} + \\frac{1}{a+b} \\right) \\ge 9.\n\\end{align*}\n$$\nThis is true by the Cauchy-Schwarz inequality. Equality holds when $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31435,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ is isosceles with $AB = AC$. $D$ is a point on the minor arc $AC$ on the circumcircle of $\\triangle ABC$. Let $E$ be the foot of the perpendicular from $A$ to $BD$. Prove that $BE = ED + DC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Archimedes' broken-chord theorem) Let $F$ be the foot of perpendicular from $A$ to $CD$. Note that\n$$\n\\angle ADE = \\angle ADB = \\angle ACB = \\angle ABC = \\angle ADF.\n$$\nAlso, we have $\\angle AED = \\angle AFD$ and $AD = AF$. This implies $\\triangle ADE \\cong \\triangle ADF$, and hence $AE = AF$ and $DE = DF$.\nNext, observe that $\\angle EBA = \\angle FCA$ and $\\angle AEB = \\angle AFC$. Together with $AE = AF$, we find that $\\triangle AEB \\cong \\triangle AFC$. Therefore, we have\n$$\nBE = CF = CD + DF = CD + ED.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime such that $p \\equiv 1 \\pmod 4$. Evaluate with reasons, $\\sum_{k=1}^{\\frac{p-1}{2}} \\left\\{ \\frac{k^2}{p} \\right\\}$, where $\\{x\\} = x - [x]$, $[x]$ being the greatest integer not exceeding $x$.",
    "options": [],
    "answer": "(p-1)/4",
    "solution": "The answer is $\\frac{p-1}{4}$.\nNote that $1^2, 2^2, \\dots, \\left(\\frac{p-1}{2}\\right)^2$ are pairwise incongruent modulo $p$ since\n$$\nx^2 \\equiv y^2 \\pmod{p} \\Rightarrow x \\equiv \\pm y \\pmod{p}.\n$$\nAs there are exactly $\\frac{p-1}{2}$ nonzero quadratic residues, $1^2, 2^2, \\dots, \\left(\\frac{p-1}{2}\\right)^2$ are all the nonzero quadratic residues modulo $p$.\nSince $p \\equiv 1 \\pmod 4$, $-1$ is a quadratic residue modulo $p$. Therefore, $r$ is a quadratic residue modulo $p$ if and only if $-r$ is a quadratic residue modulo $p$. This shows the nonzero quadratic residues come in pairs such that the numbers in each pair are negative of each other modulo $p$. Thus, $\\left\\{ \\frac{k^2}{p} \\right\\}$ for $k = 1, 2, \\dots, \\frac{p-1}{2}$ can be partitioned into $\\frac{p-1}{4}$ pairs, such that the sum of each pair is $\\frac{r + (p-r)}{p} = 1$. Thus, the sum is equal to $\\frac{p-1}{4}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Considering $\\pmod{4}$ or otherwise, show that if $a, b, c, d$ are integers such that\n$$\n(3a + 5b)(7b + 11c)(13c + 17d)(19d + 23a) = 2001^{2001},\n$$\nthen $a$ must be even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Taking modulo $4$, we obtain\n$$\n(-a + b)(-b - c)(c + d)(-d - a) \\equiv 1 \\pmod{4}.\n$$\nThis implies all factors on the left are odd, and an even number of them is congruent to $3$ modulo $4$ and the others are congruent to $1$ modulo $4$. It follows that their sum is a multiple of $4$, i.e.\n$$\n4 \\mid (-a + b) + (-b - c) + (c + d) + (-d - a) = -2a.\n$$\nThis shows $a$ is even.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31438,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ is an acute triangle. Let $A_1$ be the centre of the square inscribed in $\\triangle ABC$ having two vertices on side $BC$. Let $B_1$ be the centre of the square inscribed in $\\triangle ABC$ having two vertices on side $CA$. Let $C_1$ be the centre of the square inscribed in $\\triangle ABC$ having two vertices on side $AB$. Prove that lines $AA_1$, $BB_1$ and $CC_1$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $DEFG$ be the square with $A_1$ as centre such that $D, E$ lie on $BC$, $F$ lies on $CA$, and $G$ lies on $AB$. By considering a homothety with centre $A$, we can map $DEFG$ to a square $D'E'CB$ since $GF // BC$. Let $X$ be the centre of $D'E'CB$. Due to the homothety, $A$, $A_1$, $X$ are collinear. Similarly, let $Y$ and $Z$ be the centres of the squares constructed outside $\\triangle ABC$ having $CA$ and $AB$ as a side respectively. It suffices to prove $AX$, $BY$, $CZ$ are concurrent. This follows from Jacobi's theorem since $\\angle ZAB = \\angle YAC = 45^\\circ$, $\\angle XBC = \\angle ZBA = 45^\\circ$ and $\\angle YCA = \\angle XCB = 45^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31439,
    "subject": "Mathematics (Multi-modal)",
    "question": "For an irrational number $x$, let $x'$ be the integer nearest to $x$. Define $\\langle x \\rangle = |x - x'|$. Show that for every irrational number $y$, the minimum of the numbers $\\langle y \\rangle, \\langle 2y \\rangle, \\dots, \\langle 2001y \\rangle$ is less than $\\frac{1}{2001}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the 2002 intervals $I_k = (\\frac{k}{2002}, \\frac{k+1}{2002})$ where $k = -1001, -1000, \\dots, 1000$. Since $y$ is irrational, the numbers $y - y'$, $2y - (2y)'$, $\\dots$, $2001y - (2001y)'$ are irrational numbers between $-\\frac{1}{2}$ and $\\frac{1}{2}$. Thus, each of them belongs to one of the $I_k$'s.\n\nWe claim that one of the numbers $my - (my)'$ (with $1 \\le m \\le 2001$) belongs to $I_{-1}$ or $I_0$. Suppose on the contrary that all these numbers belong to the other 2000 intervals. By the pigeonhole principle, two of them, say $my - (my)'$ and $ny - (ny)'$ with $m > n$, belong to the same interval $I_k$. Then we have\n$$\n(m-n)y - [(my)' - (ny)'] = [my - (my)'] - [ny - (ny)'] < \\frac{k+1}{2002} - \\frac{k}{2002} = \\frac{1}{2002}.\n$$\nTherefore, $(my)' - (ny)'$ is the closest integer to $(m-n)y$, and we know that $(m-n)y - [(m-n)y]'$ belongs to $I_{-1}$ or $I_0$. This is a contradiction. Therefore,\n\nwe can find some $my - (my)'$ belonging to $I_{-1}$ or $I_0$. For this $m$, we have\n$$\n\\langle my \\rangle = |my - (my)'| < \\frac{1}{2002} < \\frac{1}{2001}.\n$$\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the greatest real $K$ such that for every positive $u$, $v$ and $w$ with $u^2 > 4vw$, the inequality\n$$\n(u^2 - 4vw)^2 > K(2v^2 - uw)(2w^2 - uv)\n$$\nholds. Justify your claim.",
    "options": [],
    "answer": "16",
    "solution": "The greatest $K$ is $16$.\n\nWe first prove the inequality when $K = 16$. Note that\n$$\n\\begin{aligned}\nu^2 - 4vw &= u^2 + 2vw - 6vw \\\\\n&\\ge u^2 + 2vw - 3(v^2 + w^2) \\\\\n&= u^2 + (v+w)^2 - 4(v^2 + w^2) \\\\\n&\\ge 2u(v+w) - 4(v^2 + w^2) \\\\\n&= 2(uw - 2v^2) + (uv - 2w^2).\n\\end{aligned}\n$$\nIf $uw - 2v^2$ and $uv - 2w^2$ have different signs (possibly equal to $0$), we have\n$$\n(u^2 - 4vw)^2 \\ge 0 \\ge K(2v^2 - uw)(2w^2 - uv).\n$$\nThe equality case will be handled below.\nIf both $uw - 2v^2$ and $uv - 2w^2$ are negative, then $(uw)(uv) < (2v^2)(2w^2)$, which yields the contradiction $u^2 < 4vw$.\nTherefore, we may assume both $uw - 2v^2$ and $uv - 2w^2$ are positive. Therefore, we obtain\n$$\n(u^2 - 4vw)^2 \\ge 4((uw - 2v^2) + (uv - 2w^2))^2 \\ge 16(uw - 2v^2)(uv - 2w^2).\n$$\nFor the equality to hold, we must have $v = w$ and $u = v + w$. This implies\n$$\nu^2 = (2v)^2 = 4v^2 = 4vw,\n$$\ncontradicting $u^2 > 4vw$. Therefore, equality cannot hold, and the inequality is proven.\n\nNext, we show $K = 16$ is the largest possible. Consider $v = w = 1$ and $u = 2 + \\varepsilon$ for a sufficiently small positive $\\varepsilon$. The relation $u^2 > 4vw$ holds. Also, we have\n$$\n(u^2 - 4vw)^2 = (4 + 4\\varepsilon + \\varepsilon^2 - 4)^2 = \\varepsilon^2(4 + \\varepsilon)^2\n$$\nand\n$$\nK(2v^2 - uw)(2w^2 - uv) = K(2 - 2 - \\varepsilon)^2 = K\\varepsilon^2.\n$$\nThe inequality holds if and only if $(4 + \\varepsilon)^2 > K$. When $\\varepsilon$ approaches $0$, the left-hand side approaches $16$. Therefore, $K \\le 16$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a regular hexagon of side length $1$, and $O$ be the centre of the hexagon. In addition to the sides of the hexagon, line segments are drawn from $O$ to each vertex, making a total of twelve unit line segments. Find the number of paths of length $2003$ along these line segments that start at $O$ and terminate at $O$.",
    "options": [],
    "answer": "(7 - sqrt(7))/14 * (1 + sqrt(7))^2003 + (7 + sqrt(7))/14 * (1 - sqrt(7))^2003",
    "solution": "The answer is $\\frac{7 - \\sqrt{7}}{14}(1 + \\sqrt{7})^{2003} + \\frac{7 + \\sqrt{7}}{14}(1 - \\sqrt{7})^{2003}$.\n\nFor each integer $n \\ge 0$, let $a_n$ be the number of paths of length $n$ starting at $O$ and terminating at $O$. Also, let $b_n$ be the number of paths of length $n$ starting at $O$ and terminating at $A$. By symmetry, $b_n$ is also the number of those paths terminating at any of $B, C, D, E, F$.\n\nTo count $a_n$ for $n \\ge 1$, the $(n-1)$\\text{st} step must terminate at one of $A, B, C, D, E, F$. For each of these $6$ points, there are $b_{n-1}$ paths terminating at that point. This gives a total of $6b_{n-1}$ paths. In other words, we have\n$$\na_n = 6b_{n-1}. \\tag{1}\n$$\nTo count $b_n$ for $n \\ge 1$, suppose the $n$\\text{th} step terminates at $A$. Then the $(n-1)$\\text{st} step may terminate at $B, F$ or $O$. If it is $B$ or $F$, then there are $b_{n-1}$ paths. If it is $O$, then there are $a_{n-1}$ paths. Therefore, we have\n$$\nb_n = 2b_{n-1} + a_{n-1}. \\tag{2}\n$$\nCombining (1) and (2), we obtain\n$$\n\\frac{1}{6}a_{n+1} = b_n = 2b_{n-1} + a_{n-1} = \\frac{1}{3}a_n + a_{n-1}.\n$$\nThis can be rewritten as $a_n = 2a_{n-1} + 6a_{n-2}$ for $n \\ge 2$. The roots of the characteristic equation $\\lambda^2 - 2\\lambda - 6 = 0$ are $1 \\pm \\sqrt{7}$. Suppose $a_n = A(1 + \\sqrt{7})^n + B(1 - \\sqrt{7})^n$. Using the initial conditions $a_0 = 1$ and $a_1 = 0$, we solve\n$$\n\\begin{cases} A + B = 1, \\\\ A(1 + \\sqrt{7}) + B(1 - \\sqrt{7}) = 0. \\end{cases}\n$$\nThis yields $A = \\frac{7 - \\sqrt{7}}{14}$ and $B = \\frac{7 + \\sqrt{7}}{14}$. Therefore, the answer is\n$$\na_{2003} = \\frac{7 - \\sqrt{7}}{14}(1 + \\sqrt{7})^{2003} + \\frac{7 + \\sqrt{7}}{14}(1 - \\sqrt{7})^{2003}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle. $M$ is the midpoint of $AC$, $D$ is a point on $AB$. $BM$ and $CD$ meet at $O$, with $AB = CO$. $E$ is a point on $AC$ such that $DE // BM$. Prove that $AB \\perp BC$ if and only if $ADOM$ is a cyclic quadrilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the parallel lines, we have $\\triangle ADE \\sim \\triangle ABM$ and $\\triangle CMO \\sim \\triangle CED$. This implies\n$$\n\\frac{AB}{BD} = \\frac{AM}{ME} = \\frac{CM}{ME} = \\frac{CO}{OD}.\n$$\nAs $AB = CO$, we obtain $BD = OD$.\nNow,\n$$\n\\begin{align*}\n& \\Leftrightarrow \\quad \\angle BAM = \\angle DOB \\\\\n& \\Leftrightarrow \\quad \\angle BAM = \\angle DBO \\\\\n& \\Leftrightarrow \\quad \\angle BAM = \\angle ABM \\\\\n& \\Leftrightarrow \\quad MB = MA.\n\\end{align*}\n$$\nSince $MA = MC$, this is equivalent to $\\angle ABC = 90^\\circ$, i.e. $AB \\perp BC$ as desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. $K, L, M, N$ are midpoints of sides $AB, BC, CD$ and $DA$ respectively. Prove that the orthocentres of triangles $AKN, BKL, CLM, DMN$ are vertices of a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $H_1, H_2, H_3, H_4$ be the orthocentres of $\\triangle AKN, \\triangle BKL, \\triangle CLM, \\triangle DMN$\nrespectively, and let $O$ be the centre of $(ABCD)$.\nSince $K$ is the midpoint of $AB$, we have $OK \\perp AB$. Thus, $OK // H_1N$. Similarly, $ON // H_1K$. This shows $H_1KON$ is a parallelogram. By symmetry, $H_4NOM$ is a parallelogram. This shows $H_1K = NO = H_4M$ and $H_1K // NO // H_4M$. Therefore, $H_1KMH_4$ is a parallelogram. By symmetry, $KH_2H_3M$ is a parallelogram. This shows $H_1H_4 = KM = H_2H_3$ and $H_1H_4 // KM // H_2H_3$. Therefore, $H_1H_2H_3H_4$ is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31444,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer greater than $1$. In a school there are $n^2 - n + 2$ clubs and each club has exactly $n$ members. Each pair of clubs has exactly one member in common. Show that there is one student belonging to all of the clubs.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider an arbitrary club $C_1$. Since it shares a common member with each of the other $n^2 - n + 1$ clubs, by the pigeonhole principle, there is a member in $C_1$ who is also a member of at least\n$$\n\\left\\lfloor \\frac{n^2 - n + 1}{n} \\right\\rfloor = n\n$$\nclubs. Suppose $X$ is a member of the clubs $C_1, C_2, \\dots, C_{n+1}$. For any other club $C_k$, since it shares a common member with each of $C_1, C_2, \\dots, C_{n+1}$, by the pigeonhole principle, there is a member in $C_k$ who is also a member of at least\n$$\n\\left\\lfloor \\frac{n+1}{n} \\right\\rfloor = 2\n$$\nclubs among $C_1, C_2, \\dots, C_{n+1}$. As the only common member in $C_1, C_2, \\dots, C_{n+1}$ is $X$, this member must also be $X$. This shows $X$ is a member of $C_k$ for all $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exists a positive integer $k$ such that $k \\times 22^n + 1$ is not prime (composite) for every positive integer $n$.",
    "options": [],
    "answer": "k = 20",
    "solution": "We can take $k = 20$. Then for any $n \\in \\mathbb{Z}^+$, we have\n$$\n20 \\cdot 22^n + 1 \\equiv (-1)(1) + 1 = 0 \\pmod{21},\n$$\nand hence $20 \\cdot 22^n + 1$ is a composite number.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find, with reasons, all integers $a$, $b$, and $c$ such that\n$$\n\\frac{1}{2}(a+b)(b+c)(c+a) + (a+b+c)^3 = 1 - abc.\n$$",
    "options": [],
    "answer": "All permutations of (1, 0, 0) and (2, -1, -1).",
    "solution": "The solutions are $(a, b, c) = (1, 0, 0), (2, -1, -1)$ up to permutation.\nLet $s = a + b + c$. Consider\n$$\nP(x) = (x - a)(x - b)(x - c) = x^3 - sx^2 + (ab + bc + ca)x - abc.\n$$\nPutting $x = s$, we obtain $(b+c)(c+a)(a+b) = (ab+bc+ca)s - abc$. Therefore, the given equation becomes\n$$\n(ab + bc + ca)s - abc + 2s^3 = 2 - 2abc.\n$$\nThis is the same as $-s^3 - s^3 - (ab + bc + ca)s - abc = -2$, i.e. $P(-s) = -2$. In other words, we have\n$$\n(2a + b + c)(a + 2b + c)(a + b + 2c) = 2.\n$$\n(One can prove this by direct expansion.) By symmetry, it suffices to consider a few cases as follows.\n* If $2a + b + c = 2$ and $a + 2b + c = a + b + 2c = 1$, adding these we obtain $4(a + b + c) = 4$. So we have $a + b + c = 1$, and hence $(a, b, c) = (1, 0, 0)$.\n* If $2a + b + c = 2$ and $a + 2b + c = a + b + 2c = -1$, adding these we obtain $4(a + b + c) = 0$. So we have $a + b + c = 0$, and hence $(a, b, c) = (2, -1, -1)$.\n* If $2a + b + c = -2$, $a + 2b + c = \\pm 1$ and $a + b + 2c = \\mp 1$, adding these we obtain $4(a + b + c) = -2$. There is no integer solution.\nThis shows the only solutions are $(a, b, c) = (1, 0, 0), (2, -1, -1)$ up to permutation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $\\sigma(n)$ the sum of divisors of $n$. A positive integer $n$ ($n \\ge 2$) is redundant if for any integer $k$, with $k < n$, we have $\\frac{\\sigma(k)}{k} < \\frac{\\sigma(n)}{n}$. (For instance if $n = 1, 2, 3, 4, 5$, then $\\frac{\\sigma(n)}{n} = 1, \\frac{3}{2}, \\frac{4}{3}, \\frac{7}{4}, \\frac{6}{5}$ respectively. Hence, $2$ and $4$ are redundant, $3$ and $5$ are not.) Show that there exist infinitely many redundant numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Superabundant numbers/IMO Shortlist 1983) When $n = m!$, we have\n$$\n\\frac{\\sigma(m!)}{m!} \\ge \\frac{1}{m!} \\left( m! + \\frac{m!}{2} + \\frac{m!}{3} + \\dots + \\frac{m!}{m} \\right) = 1 + \\frac{1}{2} + \\frac{1}{3} + \\dots + \\frac{1}{m}.\n$$\nSince the harmonic series diverges, the sequence $\\left\\{ \\frac{\\sigma(n)}{n} \\right\\}$ is unbounded. Therefore, after finding some redundant numbers $n_1, n_2, \\dots, n_k$, we can always find the next number $n_{k+1}$ such that $\\frac{\\sigma(n_{k+1})}{n_{k+1}} > \\frac{\\sigma(n_k)}{n_k}$. This implies $n_{k+1}$ is redundant, and so there are infinitely many redundant numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31448,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a conference, there are $2002$ representatives from $100$ countries. The number of representatives from each country is at least $1$ and at most $45$. They are seated in rows with each row consisting of $45$ seats. It is required that the representatives from the same country must be seated in the same row. What is the smallest number of rows needed to ensure that all representatives can be seated? Justify your answer.",
    "options": [],
    "answer": "86",
    "solution": "At least $86$ rows are needed.\nSuppose there are $86$ (type A) countries with $23$ representatives each, $10$ countries with $2$ representatives each, and $4$ countries with $1$ representative each. Then there are $86 + 10 + 4 = 100$ countries and $86 \\times 23 + 10 \\times 2 + 4 \\times 1 = 2002$ representatives. No two type A countries may sit in the same row since $23 \\times 2 > 45$. Thus, at least $86$ rows are needed.\n\nWe now show that it is always sufficient to use $86$ rows. We arrange the countries in decreasing order of the number of representatives. We first put all representatives from country $1$ in the first row. Next, if all representatives from country $2$ can be seated in the same row as country $1$, we put them in that row, or otherwise they are put in the next row. Similarly, if all representatives from country $k+1$ can be seated in the same row as country $k$, we put them in that row, or otherwise they are put in the next row. We do the same thing until all representatives are seated or all $86$ rows are used up.\n\nIt suffices to consider the case when all $86$ rows are used up but some representatives still do not have a seat. We claim that each row consists of at least $23$ representatives. If not, there are at least $23$ empty seats in some row $j$, which means the next country to be put has at least $24$ representatives. But then there should be at least $24$ representatives in row $j$ as well according to the order of the countries. This is a contradiction. It follows that the number of representatives remaining is at most\n$$\n2002 - 23 \\times 86 = 24.\n$$\nAmong these representatives, suppose $c$ of them come from the same country, where $c$ is maximized. First of all, we have $c \\le 22$ since otherwise at least $99$ countries have at least $23$ representatives (except possibly the last country), which yields the contradiction that the total number of representatives is at least\n$$\n23 \\times 99 = 2277 > 2002.\n$$\nNow, we partition the remaining representatives into at most two groups by putting those $c$ representatives in one group. We claim that all groups have at most $22$ representatives. If not, since $c \\le 22$, it must be the case that $c = 1$ and there are $23$ more representatives. By the choice of $c$, all these $24$ representatives come from different countries. But this is impossible since there are at most $100 - 86 = 14$ countries left.\n\nTherefore, it remains to show that there are at least $2$ rows with $22$ empty seats. Indeed, if this does not hold, then the total number of representatives is at least\n$$\n24 \\times 85 + 23 = 2063 > 2002.\n$$\nThis is a contradiction, and so we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer and $x_1, x_2, \\dots, x_{n-1}$ be nonnegative integers such that\n(i) $x_1 + x_2 + \\dots + x_{n-1} = n$,\n(ii) $x_1 + 2x_2 + \\dots + (n-1)x_{n-1} = 2n - 2$.\nFind the minimum of the sum $\\sum_{k=1}^{n-1} kx_k(2n-k)$. Justify your answer.",
    "options": [],
    "answer": "3n^2 - 3n",
    "solution": "The minimum value is $3n^2 - 3n$.\nWe have\n$$\n\\begin{align*}\n\\sum_{k=1}^{n-1} kx_k(2n-k) &= 2n(2n-2) - \\sum_{k=1}^{n-1} k^2 x_k \\\\\n&= 2n(2n-2) - \\sum_{k=1}^{n-1} x_k - \\sum_{k=1}^{n-1} (k-1)(k+1)x_k \\\\\n&\\ge 2n(2n-2) - n - \\sum_{k=1}^{n-1} (k-1)nx_k \\\\\n&= 2n(2n-2) - n - n \\sum_{k=1}^{n-1} kx_k + n \\sum_{k=1}^{n-1} x_k \\\\\n&= 2n(2n-2) - n - n(2n-2) + n^2 \\\\\n&= 3n^2 - 3n.\n\\end{align*}\n$$\nEquality holds when $x_1 = n-1$, $x_2 = x_3 = \\dots = x_{n-2} = 0$ and $x_{n-1} = 1$. Therefore, the minimum value is $3n^2 - 3n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31450,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $n \\ge 2$, let $a_1, a_2, \\dots, a_n, a_{n+1}$ be positive and $a_2 - a_1 = a_3 - a_2 = \\dots = a_{n+1} - a_n \\ge 0$. Prove that\n$$\n\\frac{1}{a_2^2} + \\frac{1}{a_3^2} + \\dots + \\frac{1}{a_n^2} \\le \\frac{n-1}{2} \\cdot \\frac{a_1 a_n + a_2 a_{n+1}}{a_1 a_2 a_n a_{n+1}}\n$$\nDetermine when equality holds.",
    "options": [],
    "answer": "Equality holds if and only if all terms are equal.",
    "solution": "Let $d = a_j - a_{j-1} \\ge 0$. If $d > 0$, then\n$$\n\\frac{1}{a_k^2} < \\frac{1}{a_k^2 - d^2} = \\frac{1}{a_{k-1}a_{k+1}} = \\frac{a_{k+1} - a_{k-1}}{2d a_{k-1} a_{k+1}} = \\frac{1}{2d} \\left( \\frac{1}{a_{k-1}} - \\frac{1}{a_{k+1}} \\right)\n$$\nfor any $k > 1$. Therefore, we have\n$$\n\\begin{aligned}\n\\sum_{k=2}^n \\frac{1}{a_k^2} &< \\frac{1}{2d} \\sum_{k=2}^n \\left( \\frac{1}{a_{k-1}} - \\frac{1}{a_{k+1}} \\right) \\\\\n&= \\frac{1}{2d} \\left( \\frac{1}{a_1} - \\frac{1}{a_n} + \\frac{1}{a_2} - \\frac{1}{a_{n+1}} \\right) \\\\\n&= \\frac{1}{2d} \\left( \\frac{(n-1)d}{a_1 a_n} + \\frac{(n-1)d}{a_2 a_{n+1}} \\right) \\\\\n&= \\frac{n-1}{2} \\cdot \\frac{a_1 a_n + a_2 a_{n+1}}{a_1 a_2 a_n a_{n+1}}.\n\\end{aligned}\n$$\nWhen $d = 0$, let $a_j = c$ for all $j$. Then\n$$\n\\sum_{k=2}^n \\frac{1}{a_k^2} = \\frac{n-1}{c^2}\n$$\nand\n$$\n\\frac{n-1}{2} \\cdot \\frac{a_1 a_n + a_2 a_{n+1}}{a_1 a_2 a_n a_{n+1}} = \\frac{n-1}{2} \\cdot \\frac{2c^2}{c^4} = \\frac{n-1}{c^2} = \\sum_{k=2}^n \\frac{1}{a_k^2}.\n$$\nTherefore, the inequality is proven, and equality holds when $a_1 = a_2 = \\dots = a_{n+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31451,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a set $S$ of $10$ distinct positive integers, all less than $118$, prove that $S$ contains at least two nonempty subsets $A$ and $B$, such that $A \\cap B = \\emptyset$, and the sums of the integers in $A$ and $B$ are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(IMO 1972 Problem 1 modified) There are $2^{10} - 1 = 1023$ proper subsets of $S$, each containing at most $9$ elements. The sum of all elements in such a proper subset ranges from $0$ to\n$$\n117 + 116 + \\cdots + 109 = \\frac{(117 + 109) \\times 9}{2} = 1017.\n$$\nThere are only $1018$ possibilities. By the pigeonhole principle, there are two proper subsets having the same sum. Removing their common elements if necessary, we obtain two disjoint subsets having the same sum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31452,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are $b$ teachers and $c$ students. Suppose that\n(i) each teacher teaches exactly $k$ students; and\n(ii) for each pair of distinct students, exactly $h$ teachers teach both of them.\nShow that\n$$\n\\frac{b}{h} = \\frac{c(c-1)}{k(k-1)}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We count the number of triples $(T, S_1, S_2)$ such that $T$ is a teacher teaching two distinct students $S_1$ and $S_2$. Note that $(T, S_1, S_2)$ is different from $(T, S_2, S_1)$.\n\nFor each of the $b$ teachers, there are $k$ choices for $S_1$ and $k-1$ choices for $S_2$. Therefore, there are $b k (k-1)$ such triples.\n\nFor each of the $c$ students $S_1$ and each of the $c-1$ students $S_2$, there are $h$ choices for $T$. Therefore, there are $c (c-1) h$ such triples.\n\nThus, we have obtained $b k (k-1) = c (c-1) h$. This clearly implies $\\frac{b}{h} = \\frac{c(c-1)}{k(k-1)}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31453,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the sides $AB$ and $AC$ of triangle $ABC$, there are points $P$ and $Q$ respectively such that $\\angle APC = \\angle AQB = 45^\\circ$. Let the perpendicular line to side $AB$ through $P$ intersects line $BQ$ at $S$. Let the perpendicular line to side $AC$ through $Q$ intersects line $CP$ at $R$. Let $D$ be on side $BC$ such that $AD \\perp BC$. Prove that the lines $PS, AD, QR$ meet at a common point and lines $SR$ and $BC$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $PS$ and $QR$ meet $BC$ at $X$ and $Y$ respectively. Note that $A, P, X, D$ are concyclic since $\\angle ADX + \\angle APX = 180^\\circ$. Similarly, $A, D, Y, Q$ are concyclic. Also, since\n$$\n\\angle BPC = 180^\\circ - \\angle CPA = 180^\\circ - \\angle AQB = \\angle BQC,\n$$\n\nthe points $B, C, Q, P$ are concyclic. Thus, we have\n$$\n\\angle PQY = 90^\\circ - \\angle AQP = 90^\\circ - \\angle CBP = \\angle PXB.\n$$\nThis implies $Q, P, X, Y$ are concyclic.\nConsider the circles $(APXD)$, $(ADYQ)$ and $(QPXY)$. Their pairwise radical axes are $AD, PX, QY$ respectively. Clearly, these lines are not parallel. Therefore, they are concurrent at the radical centre of these circles.\n![](attached_image_1.png)\n\nNext, observe that $Q, P, S, R$ are concyclic since\n$$\n\\angle SQR = 90^\\circ - \\angle AQS = 90^\\circ - \\angle RPA = \\angle SPR.\n$$\nApplying Reim's theorem to $(QPXY)$ and $(QPSR)$, we obtain $SR \\parallel XY$. (It simply follows from $\\angle XSR = \\angle PQR = 180^\\circ - \\angle YXS$.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31454,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $n$ is said to be *good* if there exist positive integers $a$ and $b$ such that $a+b=n$ and $ab$ is divisible by $n$. For instance, $9$ is good since $3+6=9$ and $3 \\times 6$ is divisible by $9$. Furthermore, $n$ is said to be *very good* if we can choose $a$ and $b$ to be distinct. Hence $9$ is very good.\n\na. Given that $2003$ is a prime, show that $2003$ is not good.\n\nb. Is $2002$ good?\n\nc. Find all positive integers which are good but not very good. Justify your answers.",
    "options": [],
    "answer": "a) 2003 is not good. b) 2002 is not good. c) Exactly the positive integers of the form 4 times a product of distinct odd primes or 8 times a product of distinct odd primes (including the cases 4 and 8) are good but not very good.",
    "solution": "a. Suppose there exist $a, b \\in \\mathbb{Z}^+$ such that $a+b = 2003$ and $2003 \\mid ab$. Since $2003$ is a prime, we must have $2003 \\mid a$ or $2003 \\mid b$. WLOG assume $2003 \\mid a$. Then we have $a+b \\ge 2003+1 > 2003$, which is a contradiction. Therefore, $2003$ is not good.\n\nb. No. Suppose there exist $a, b \\in \\mathbb{Z}^+$ such that $a+b = 2002$ and $2002 \\mid ab$. Note that $2002 = 2 \\times 7 \\times 11 \\times 13$. For each prime $p \\mid 2002$, we have $p \\mid ab$, and so $p \\mid a$ or $p \\mid b$. In either case, since $p \\mid a+b$, we must have $p \\mid a$ and $p \\mid b$. As this holds for $p = 2, 7, 11, 13$, we must have $2002 \\mid a, b$. This yields $a+b \\ge 2002+2002 > 2002$, which is a contradiction. Therefore, $2002$ is not good.\n\nc. All positive integers of the form $4p_1p_2\\cdots p_s$ or $8p_1p_2\\cdots p_s$ where $p_1, p_2, \\dots, p_s$ are distinct odd primes (possibly $s=0$) are good but not very good.\n\nIn view of the proof of part (b), all squarefree positive integers are not good. For any non-squarefree positive integer $n$, we can write $n = p^2m$ for some prime $p$ and positive integer $m$. Since we can take $a = pm$ and $b = (p-1)pm$ such that\n$$\na + b = p^2m = n\n$$\nand\n$$\nab = (p-1)p^2m^2 = (p-1)mn,\n$$\nwe know that $n$ is good by definition. Also, note that $a$ and $b$ are distinct unless $p=2$. This shows $n$ is very good if $n$ is divisible by the square of an odd prime.\n\nThus, it remains to consider $n = 2^k m$ for some $k \\ge 2$ and odd $m$ which is squarefree. If $k \\ge 4$, we can take $a = 2^{k-2}m$ and $b = 3 \\cdot 2^{k-2}m$ such that\n$$\na + b = 2^k m = n\n$$\nand\n$$\nab = 3 \\cdot 2^{2k-4}m^2 = 3 \\cdot 2^{k-4}mn.\n$$\nSince $a \\neq b$, this shows $n$ is very good.\n\nLastly, if $k=2,3$, we show that $n$ is not very good. Indeed, suppose $a+b=n$ and $n|ab$. As in part (b), we deduce $m|a$ and $m|b$. Let $a=mc$ and $b=md$. Then we need $c+d=2^k$ and $2^k|cd$. It is routine to check that there is no solution except $c=d$ when $k=2,3$ (note that there are only a few pairs of $(c,d)$ satisfying $c+d=2^k$). Therefore, $n$ is not very good.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31455,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{1, 2, \\dots, 100\\}$. Determine the number of functions $f : S \\to S$ satisfying the following conditions.\n(i) $f(1) = 1$;\n(ii) $f$ is bijective (i.e. for every $y$ in $S$, the equation $f(x) = y$ has exactly one solution);\n(iii) $f(n) = f(g(n))f(h(n))$ for every $n$ in $S$.\nHere $g(n)$ and $h(n)$ denote the uniquely determined positive integers such that $g(n) \\le h(n)$, $g(n)h(n) = n$ and $h(n)-g(n)$ is as small as possible. (For instance, $g(80) = 8$, $h(80) = 10$ and $g(81) = h(81) = 9$.)",
    "options": [],
    "answer": "348364800",
    "solution": "There are 348364800 such functions.\nWe first claim that condition (iii) can be replaced by the completely multiplicative condition. This means\n$$\nf(p_1^{a_1} p_2^{a_2} \\cdots p_s^{a_s}) = f(p_1)^{a_1} f(p_2)^{a_2} \\cdots f(p_s)^{a_s}\n$$\nfor any distinct primes $p_1, p_2, \\dots, p_s$ and positive integers $a_1, a_2, \\dots, a_s$. We prove this by induction on $a_1 + a_2 + \\cdots + a_s$.\nThe base cases $a_1 + a_2 + \\cdots + a_s = 0, 1$ hold trivially. Assume the result holds whenever the sum of exponents is less than $k$. Consider $n = p_1^{a_1} p_2^{a_2} \\cdots p_s^{a_s}$ where $a_1 + a_2 + \\cdots + a_s = k$. Note that $n$ is composite, and so $g(n), h(n) > 1$. Since $g(n)h(n) = n$, the sum of exponents in the prime factorizations of $g(n)$ and $h(n)$ are both smaller than $k$. Thus, by the inductive hypothesis, $f(g(n))$ and $f(h(n))$ can be decomposed into a product of $f(p_j)$'s. Clearly, this proves\n$$\nf(n) = f(p_1^{a_1} p_2^{a_2} \\cdots p_s^{a_s}) = f(p_1)^{a_1} f(p_2)^{a_2} \\cdots f(p_s)^{a_s}.\n$$\n\nBy induction, we have shown that $f$ is completely multiplicative. Conversely, if $f$ is completely multiplicative, condition (iii) holds trivially.\nLet $\\mathcal{P}$ be the set of all primes from 1 to 100, and let $\\mathcal{C} = S - (\\mathcal{P} \\cup \\{1\\})$ be the set of all composites. For any $n \\in \\mathcal{C}$, since $g(n), h(n) > 1$, we must have $f(g(n)), f(h(n)) > 1$. Therefore, $f(n) = f(g(n))f(h(n)) \\in \\mathcal{C}$. This implies $f(\\mathcal{C}) = \\mathcal{C}$. As $f(1) = 1$, by the bijective condition, we must have $f(\\mathcal{P}) = \\mathcal{P}$. Thus, it remains to assign distinct prime values to $f(p)$ for $p \\in \\mathcal{P}$. Afterwards we can easily compute $f(n)$ by the completely multiplicative condition, while the bijective condition is always satisfied.\nNow, the only constraint is that $f(n) \\in S$ for any $n \\in S$. For example, since\n$$\n100 \\ge f(64) = f(2)^6\n$$\nand $f(2) \\ne 1$, we must have $f(2) = 2$. Similarly, by considering $f(81)$, we can prove that $f(3) = 3$. Next, since\n$$\n100 \\ge f(75) = f(3)f(5)^2,\n$$\nwe have $f(5) \\le 5$, and hence $f(5) = 5$. Similarly, by considering $f(98), f(99)$ and $f(91)$, we can prove that $f(p) = p$ for $p = 7, 11, 13$.\nNext, by considering $f(85)$ and $f(95)$, we can only prove that $f(17), f(19) \\le 20$. Therefore, we have $\\{f(17), f(19)\\} = \\{17, 19\\}$. There is no further constraint on $f(17)$ and $f(19)$ since we can verify that $f(n) \\in S$ for any $n$ being multiples of 17 or 19, as long as $f(17), f(19) \\le 20$.\nSimilarly, by considering $f(92)$, we find that $f(23) = 23$.\nBy considering $f(87)$ and $f(93)$, we find that $\\{f(29), f(31)\\} = \\{29, 31\\}$.\nBy considering $f(74), f(82), f(86)$ and $f(94)$, we find that $37, 41, 43, 47$ can be mapped to any permutation of themselves.\nLastly, the images of $53, 59, 61, 67, 71, 73, 79, 83, 89, 97$ can be any permutation of themselves.\nIn view of the above arguments, we can choose the images within each group in an arbitrary way. Each choice corresponds to a unique function satisfying all requirements. Therefore, the number of such functions is\n$$\n2! \\times 2! \\times 4! \\times 10! = 348364800.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31456,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$, let $f(n)$ be the number of all positive integers with exactly $2n$ digits, each having exactly $n$ of the digits equal to $1$ and the other $n$ digits equal to $2$. Let $g(n)$ be the number of all positive integers with exactly $n$ digits, each of its digits can only be $1$, $2$, $3$ or $4$ and the number of $1$'s equals the number of $2$'s. Prove that $f(n) = g(n)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We consider the following mapping between the set $S$ of all positive integers formed by $n$ $1$'s and $n$ $2$'s and the set $T$ of all positive integers with $n$ digits formed by $1$, $2$, $3$, $4$ such that the numbers of $1$'s and $2$'s are equal.\n\nFor each $m \\in S$, we can pair up every two consecutive digits of $m$ from left to right. This gives $n$ pairs in total. Then we replace the pairs $11$, $22$, $12$, $21$ by $1$, $2$, $3$, $4$ respectively. If there are $a$, $b$, $c$, $d$ pairs of $11$, $22$, $12$, $21$ respectively, then there are $2a + c + d$ $1$'s and $2b + c + d$ $2$'s in $m$. Thus, we must have $a = b$. This shows the numbers of $1$'s and $2$'s in the image are the same, and so the image belongs to $T$.\n\nClearly, the mapping is reversible. For each $m \\in T$, we just need to replace the digits $1$, $2$, $3$, $4$ in $m$ by $11$, $22$, $12$, $21$ respectively. If there are $a$, $a$, $c$, $d$ copies of $1$, $2$, $3$, $4$ respectively, then there are $2a + c + d$ $1$'s and $2a + c + d$ $2$'s in the image, and so the image belongs to $S$. This shows the mapping is a bijection, and hence\n$$\nf(n) = |S| = |T| = g(n).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and $n$ a positive integer such that $n \\ge p$. Show that $\\binom{n}{p} - \\lfloor \\frac{n}{p} \\rfloor$ is divisible by $p$. (Here the combinatorial symbol $\\binom{n}{p} = \\frac{n!}{p!(n-p)!}$ and $\\lfloor x \\rfloor$ stands for the largest integer less than or equal to $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "This is an immediate consequence of Lucas' theorem. We give an elementary proof as follows.\n\nLet $n = ap + r$ where $0 \\le r \\le p - 1$. Then we have\n$$\n\\binom{n}{p} = \\frac{n(n-1)\\cdots(n-p+1)}{p!} \\\\ = \\frac{(ap+r)(ap+r-1)\\cdots(ap+1)(a)(ap-1)(ap-2)\\cdots(ap+r-p+1)}{(p-1)!}\n$$\nNote that the numerator is congruent to\n$$\nr(r-1)\\cdots(1)(a)(p-1)(p-2)\\cdots(r+1) = a \\cdot (p-1)!\n$$\nmodulo $p$. Therefore, we find that\n$$\n\\binom{n}{p} \\equiv a = \\left\\lfloor \\frac{n}{p} \\right\\rfloor \\pmod{p}.\n$$\nThis proves the desired result.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle, $D$ the foot of the perpendicular from $A$ to $BC$. With $AD$ as a diameter, draw a circle intersecting $AB$ and $AC$ at $E$ and $F$ respectively. Suppose $AD$ and $EF$ intersect at $G$, and $AD$ is extended to meet the circumcircle of $\\triangle ABC$ at $H$. Show that $AD^2 = AG \\cdot AH$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly, since\n$$\n\\angle AED + \\angle AFD = 90^\\circ + 90^\\circ = 180^\\circ,\n$$\nthe points $A$, $E$, $D$, $F$ are concyclic.\n\nSecondly, since\n$$\n\\angle AHB = \\angle ACB = 90^\\circ - \\angle DAF = \\angle FDA = \\angle FEA,\n$$\nthe points $E$, $B$, $H$, $G$ are concyclic.\n\nThirdly, $AD$ is tangent to $(BDE)$ since $\\angle ADB = \\angle DEB = 90^\\circ$. It follows from the power of points that\n$$\nAD^2 = AE \\times AB = AG \\times AH.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find, with reasons, all functions $f : \\mathbb{N} \\to \\mathbb{N}$ such that\n(i) $f(n) = 1$ if and only if $n = 1$,\n(ii) if $d = \\text{gcd}(n, m)$, then $f(nm) = \\frac{f(n)f(m)}{f(d)}$, and\n(iii) $f^{2004}(n) = n$ for every $n \\in \\mathbb{N}$, where $f^{2004}(n) = \\underbrace{f(f(\\cdots f(n)))}_{2004}$,\nor show that they do not exist.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There is no such function.\nPutting $n = m = 2$ in (ii), we obtain\n$$\nf(4) = \\frac{f(2)f(2)}{f(2)} = f(2).\n$$\nThen by condition (iii), we have\n$$\n2 = f^{2004}(2) = f^{2003}(f(2)) = f^{2003}(f(4)) = f^{2004}(4) = 4.\n$$\nThis is a contradiction. So there is no such function.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31460,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a planet there are $3 \\times 2005!$ aliens and $2005$ languages. Each pair of aliens communicates with each other in exactly one language. Show that there are $3$ aliens who communicate with each other in one common language.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use terminologies in graph theory. We shall prove by induction that in a complete graph of $3 \\cdot k!$ vertices, if we colour every edge in one of $k$ colours, then there exists a monochromatic triangle.\n\nFor the base case $k=1$, since there are $3$ vertices but only one colour, the result holds trivially.\n\nAssume the claim holds for some $k$. Consider a complete graph of $3 \\cdot (k+1)!$ vertices. We colour every edge in one of $k+1$ colours. Fix any vertex $A$. By the pigeonhole principle, there are\n$$\n\\left\\lfloor \\frac{3 \\cdot (k+1)! - 1}{k+1} \\right\\rfloor = 3 \\cdot k!\n$$\nedges with $A$ as an endpoint which are of the same colour, say red. If an edge connecting two of these vertices is red, then these two vertices together with $A$ form a red triangle. If all these edges are not red, then they are in one of $k$ colours. By the inductive hypothesis, there is a monochromatic triangle in this subgraph. This proves the inductive step, and hence the problem is solved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose there are $4n$ line segments of unit length inside a circle of radius $n$. Furthermore, a straight line $L$ is given. Prove that there exists a straight line $L'$ that is either parallel or perpendicular to $L$ and that $L'$ cuts at least two of the given line segments.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AB$ and $CD$ be the diameters of the circle which are parallel and perpendicular to $L$ respectively. Let $P_iQ_i$ be the projection of each segment on $AB$, and let $X_iY_i$ be the projection of each segment on $CD$. Note that $P_iQ_i + X_iY_i \\ge 1$. Therefore, we have\n$$\n\\sum_{j=1}^{4n} P_iQ_i + \\sum_{j=1}^{4n} X_iY_i = \\sum_{j=1}^{4n} (P_iQ_j + X_iY_j) \\ge 4n = AB + CD.\n$$\nWLOG assume $\\sum_{j=1}^{4n} P_iQ_i \\ge AB$. As all the segments $P_iQ_i$ lie strictly inside the segment $AB$, two of these segments $P_iQ_i$ and $P_jQ_j$ must overlap. Let $E$ be a point lying on both of them. Then the line passing through $E$ and perpendicular to $AB$ intersects two unit segments whose projections are $P_iQ_i$ and $P_jQ_j$. This completes the proof.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ and $H$ be respectively the circumcentre and orthocentre of $\\triangle ABC$. Let $A'$, $B'$ and $C'$ be the midpoints of $BC$, $CA$ and $AB$ respectively and $D$, $E$ and $F$ be respectively the feet from $A$, $B$ and $C$ to the opposite sides. Show that $OA' \\cdot HD = OB' \\cdot HE = OC' \\cdot HF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is well-known that $OA' = \\frac{1}{2} AH$, etc. Therefore, we need to prove\n$$\nAH \\times HD = BH \\times HE = CH \\times HF.\n$$\nIndeed, $AH \\times HD = BH \\times HE$ is true since $A$, $B$, $D$, $E$ are concyclic. By symmetry, the other equality also holds.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the intersection point of the diagonals $AC$ and $BD$ of a convex quadrilateral $ABCD$. The bisector of $\\angle ACD$ meets $BA$ extended at $K$. If $MA \\cdot MC + MA \\cdot CD = MB \\cdot MD$, show that $\\angle BKC = \\angle CDB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Construct a point $E$ on the extension of $AC$ such that $CD = CE$. Using the condition, we find that\n$$\n\\begin{aligned}\nMB \\cdot MD &= MA \\cdot MC + MA \\cdot CD \\\\\n&= MA(MC + CE) \\\\\n&= MA \\cdot ME.\n\\end{aligned}\n$$\nThis implies $A, B, E, D$ are concyclic.\n\nNow, since $\\angle CDE = \\angle CED = \\frac{1}{2} \\angle DCA = \\angle KCA$, we obtain\n$$\n\\angle BKC = \\angle BAC - \\angle KCA = \\angle BDE - \\angle CDE = \\angle BDC\n$$\nas desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that $p, q$ and $r$ are positive real numbers, prove that\n$$\n\\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\ge \\frac{9}{2(p+q+r)}\n$$\nHence prove that if $m$ is a real number greater than $1$, then\n$$\n\\frac{p^m}{q+r} + \\frac{q^m}{r+p} + \\frac{r^m}{p+q} \\ge \\frac{(p+q+r)^{m-1}}{2 \\cdot 3^{m-2}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the Cauchy-Schwarz inequality, we have\n$$\n((q+r) + (r+p) + (p+q)) \\left( \\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\right) \\ge (1+1+1)^2 = 9.\n$$\nDividing both sides by $2(p+q+r)$, we obtain\n$$\n\\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\ge \\frac{9}{2(p+q+r)}.\n$$\n\nNext, WLOG assume $p \\ge q \\ge r$. Then we have $\\frac{1}{q+r} \\ge \\frac{1}{r+p} \\ge \\frac{1}{p+q}$ and $p^m \\ge q^m \\ge r^m$. By Chebyshev's inequality, we obtain\n$$\n\\frac{p^m}{q+r} + \\frac{q^m}{r+p} + \\frac{r^m}{p+q} \\ge \\frac{p^m+q^m+r^m}{3} \\cdot \\left( \\frac{1}{q+r} + \\frac{1}{r+p} + \\frac{1}{p+q} \\right).\n$$\n\nBy the power mean inequality, since $m > 1$, we have\n$$\n\\frac{p^m + q^m + r^m}{3} \\geq \\left( \\frac{p+q+r}{3} \\right)^m .\n$$\nCombining these and the result of the first part, we obtain\n$$\n\\frac{p^m}{q+r} + \\frac{q^m}{r+p} + \\frac{r^m}{p+q} \\geq \\left(\\frac{p+q+r}{3}\\right)^m \\cdot \\frac{9}{2(p+q+r)} = \\frac{(p+q+r)^{m-1}}{2 \\cdot 3^{m-2}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers such that $a + b + c + d = 1$. Prove that\n$$\n6(a^3 + b^3 + c^3 + d^3) \\ge (a^2 + b^2 + c^2 + d^2) + \\frac{1}{8}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim that $f(x) = 6x^3 - x^2 \\ge \\frac{5x-1}{8}$ for any $x > 0$. Indeed,\n$$\n\\begin{align*}\n& 6x^3 - x^2 \\ge \\frac{5x-1}{8} \\\\\n\\Leftrightarrow \\quad & 48x^3 - 8x^2 - 5x + 1 \\ge 0 \\\\\n\\Leftrightarrow \\quad & (4x-1)^2(3x+1) \\ge 0.\n\\end{align*}\n$$\nThis clearly holds. It follows that\n$$\nf(a) + f(b) + f(c) + f(d) \\ge \\frac{5(a+b+c+d) - 4}{8} = \\frac{1}{8}.\n$$\nEquality holds when $a = b = c = d = \\frac{1}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exist infinitely many squarefree positive integers $n$ that divide $2005^n - 1$. (An integer is squarefree if it contains no factors of the form $d^2$, $d > 1$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly, note that $2005 \\equiv 1 \\pmod{3}$. Therefore, $3 \\mid 2005 - 1$. Suppose we have chosen distinct primes $p_1, p_2, \\dots, p_k$ such that $p_1 p_2 \\cdots p_k \\mid 2005^{p_1 p_2 \\cdots p_{k-1}} - 1$, where the exponent of $2005$ is $1$ when $k=1$. Then\n$$\n2005^{p_1 p_2 \\cdots p_k} = (2005^{p_1 p_2 \\cdots p_{k-1}})^{p_k} \\equiv 1^{p_k} = 1 \\pmod{p_1 p_2 \\cdots p_k}.\n$$\nThis shows $n = p_1 p_2 \\cdots p_k$ satisfies the requirement.\nBy Zsigmondy's theorem, there exists a prime $p_k$ dividing $2005^{p_1 p_2 \\cdots p_k} - 1$ but not $2005^{p_1 p_2 \\cdots p_{k-1}} - 1$. This shows $p_k$ is different from $p_1, p_2, \\dots, p_{k-1}$. Therefore, we have\n$$\np_1 p_2 \\cdots p_{k+1} \\mid 2005^{p_1 p_2 \\cdots p_k} - 1.\n$$\n\nBy induction, we can find an infinite sequence of distinct primes $p_j$ such that $n = p_1 p_2 \\cdots p_k$ satisfies the requirement for any $k \\in \\mathbb{Z}^+$. So we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $M$ is a point on the side $AB$ of $\\triangle ABC$ such that the incircles of $\\triangle AMC$ and $\\triangle BMC$ have the same radius. The two incircles, centered at $O_1$ and $O_2$, meet $AB$ at $P$ and $Q$ respectively. It is known that the area of $\\triangle ABC$ is six times the area of the quadrilateral $PQO_2O_1$, determine the possible values of $\\frac{AC+BC}{AB}$. Justify your claim.",
    "options": [],
    "answer": "5/3 or 5/4",
    "solution": "The ratio can be $\\frac{5}{3}$ or $\\frac{5}{4}$.\nLet $a = BC$, $b = CA$, $c = AB$, $d = a+b$, $s = \\frac{a+b+c}{2}$, $x = MA$, $y = MB$ and $z = MC$. Also, let $r$ be the inradius of $\\triangle ABC$, and let $r'$ be the inradius of $\\triangle AMC$.\nFirstly, by considering $\\triangle AMC$, we obtain $r' = \\frac{2[AMC]}{b+x+z}$. Similarly, by considering $\\triangle BMC$, we obtain $r' = \\frac{2[BMC]}{a+y+z}$. Equating these, and using $\\frac{[AMC]}{[BMC]} = \\frac{x}{y}$, we obtain\n$$\n\\frac{b+x+z}{a+y+z} = \\frac{x}{y}.\n$$\nThis implies $\\frac{b+z}{a+z} = \\frac{x}{y}$. Since $x+y=c$, we obtain\n$$\nx = \\frac{(b+z)c}{a+b+2z} \\quad \\text{and} \\quad y = \\frac{(a+z)c}{a+b+2z}. \\qquad (1)\n$$\n![](attached_image_1.png)\nSecondly, we have\n$$\nO_1P = AP \\tan \\frac{A}{2} = \\frac{b+x-z}{2} \\cdot \\frac{r}{s-a} = \\frac{(b+x-z)r}{b+c-a}\n$$\nand similarly $O_2Q = \\frac{(a+y-z)r}{a+c-b}$. Since these are equal, we have\n$$\n\\frac{b+x-z}{b+c-a} = \\frac{a+y-z}{a+c-b}.\n$$\nUsing (1), this becomes\n$$\n\\frac{(a+b+2z)(b-z) + (b+z)c}{b+c-a} = \\frac{(a+b+2z)(a-z) + (a+z)c}{a+c-b}.\n$$\n![](attached_image_1.png)\nClearing denominators and simplifying, we get\n$$\n4(a-b)z^2 = (a-b)(a+b+c)(a+b-c).\n$$\nThis implies $a = b$ or $z = \\sqrt{s(s-c)}$. If $a = b$, then we have $x = y = \\frac{c}{2}$ by (1), and hence\n$$\nz = \\sqrt{a^2 - x^2} = \\sqrt{\\left(a + \\frac{c}{2}\\right)\\left(a - \\frac{c}{2}\\right)} = \\sqrt{s(s-c)}.\n$$\nTherefore,\n$$\nz = \\sqrt{s(s-c)} \\qquad (2)\n$$\nholds in any case.\nThirdly, we use the given condition. Note that $PQO_2O_1$ is a rectangle, and so\n$$\n[PQO_2O_1] = PQ \\cdot r' = \\left(\\frac{x+z-b}{2} + \\frac{y+z-a}{2}\\right) r' = \\frac{(c+2z-d)r'}{2}.\n$$\nNext, we have\n$$\n[ABC] = [AMC] + [BMC] = \\frac{b+x+z}{2} \\cdot r' + \\frac{a+y+z}{2} \\cdot r' = \\frac{(c+2z+d)r'}{2}.\n$$\nSince $6[PQO_2O_1] = [ABC]$, we get $6(c+2z-d) = c+2z+d$. This implies\n$$\nz = \\frac{7d - 5c}{10}. \\qquad (3)\n$$\nComparing (2) and (3), we solve\n$$\n\\left(\\frac{7d - 5c}{10}\\right)^2 = s(s-c) = \\frac{(d+c)(d-c)}{4}.\n$$\nThis is the same as $(3d - 5c)(4d - 5c) = 0$. Therefore, we have\n$$\n\\frac{d}{c} = \\frac{5}{3}, \\frac{5}{4}.\n$$\nBoth values are possible. For example, when $(a, b, c) = (5, 5, 6)$, $(5, 5, 8)$, both $\\triangle AMC$ and $\\triangle BMC$ are right-angled triangles with side lengths 3, 4, 5. It is easy to find $r' = 1$ and verify $6[PQO_2O_1] = 12 = [ABC]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AD$, $BE$ and $CF$ be respectively the altitudes of $\\triangle ABC$. Let $P$, $Q$ and $R$ be points on the lines $BC$, $CA$ and $AB$ respectively such that $AP$ is perpendicular to $EF$, $BQ$ is perpendicular to $FD$ and $CR$ is perpendicular to $DE$. Show that $AP$, $BQ$ and $CR$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By a corollary of Carnot's theorem, the perpendicular lines drawn from $A$, $B$, $C$ to $EF$, $FD$, $DE$ are concurrent if and only if the perpendicular lines drawn from $D$, $E$, $F$ to $BC$, $CA$, $AB$ are concurrent. The latter is clearly true since the perpendicular lines are just $AD$, $BE$, $CF$, which are concurrent at the orthocentre of $\\triangle ABC$. This proves $AP$, $BQ$, $CR$ are concurrent.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31469,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$, find (with proof) the number of ways of sitting $2n$ people $A_1, B_1, A_2, B_2, \\dots, A_n, B_n$ around a round table so that $A_i$ and $B_i$ are not next to each other for $i = 1, 2, \\dots, n$. Two sitting arrangements are considered the same if one can be obtained from another by a rotation. (You may express your answer in whatever ways.)",
    "options": [],
    "answer": "0 for n = 1; for n ≥ 2: ∑_{k=0}^{n} (-1)^k * C(n, k) * 2^k * (2n - k - 1)!",
    "solution": "The answer is $0$ if $n = 1$, and is $$\\sum_{k=0}^{n} (-1)^k \\binom{n}{k} 2^k (2n-k-1)!$$ if $n \\ge 2$, or any equivalent expression.\n\nThe result is obvious for $n = 1$. Consider $n \\ge 2$. For each $k = 0, 1, \\dots, n$, let $N_k$ be the number of ways to arrange the $2n$ people such that $A_i, B_i$ are next to each other for $i = 1, 2, \\dots, k$ (and probably for some more pairs).\n\nFor each $k$, we need to arrange $k$ pairs and the remaining $2n-2k$ people. There are $(k+2n-2k-1)! = (2n-k-1)!$ ways to do so. Within each pair, there are $2$ ways to arrange the two people. Therefore, we have\n$$\nN_k = 2^k (2n - k - 1)! .\n$$\nNow, by the inclusion-exclusion principle, since there are $\\binom{n}{k}$ ways to choose $k$ pairs, the number of ways to arrange the $2n$ people such that no pair $A_i, B_i$ sits next to each other is\n$$\n\\sum_{k=0}^{n} (-1)^k \\binom{n}{k} N_k = \\sum_{k=0}^{n} (-1)^k \\binom{n}{k} 2^k (2n-k-1)! .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $r$ be the positive root of the equation $x^2 - 2004x - 1 = 0$. Define the sequence $\\{a_n\\}$ as follows:\n$$\na_0 = 1, a_{n+1} = [ra_n], n \\ge 0,\n$$\nwhere $[y]$ denotes the greatest integer not exceeding $y$. Find the remainder when $a_{2004}$ is divided by $2004$.",
    "options": [],
    "answer": "1003",
    "solution": "The answer is $1003$.\n\nNote that\n$$\nr = \\frac{2004 + \\sqrt{2004^2 + 4}}{2}\n$$\nis an irrational number greater than $1$. Since $a_n \\in \\mathbb{Z}^+$, we have $ra_n \\notin \\mathbb{Q}$. Thus, we obtain\n$$\na_{n+1} < ra_n < a_{n+1} + 1\n$$\nfor any $n \\ge 0$. Equivalently, we have\n$$\na_n - \\frac{1}{r} < \\frac{a_{n+1}}{r} < a_n.\n$$\nAs $r > 1$, this yields $a_n - 1 < \\frac{a_{n+1}}{r} < a_n$, and hence\n$$\n\\left[ \\frac{a_{n+1}}{r} \\right] = a_n - 1.\n$$\nNext, observe that $r = 2004 + \\frac{1}{r}$. It follows that\n$$\na_{n+1} = [ra_n] = \\left[ 2004a_n + \\frac{a_n}{r} \\right] = 2004a_n + a_{n-1} - 1\n$$\nsince $2004a_n \\in \\mathbb{Z}$ and $\\left[ \\frac{a_n}{r} \\right] = a_{n-1} - 1$ from above. Thus, we obtain\n$$\na_{n+1} \\equiv a_{n-1} - 1 \\pmod{2004}\n$$\nfor any $n \\in \\mathbb{Z}^+$. Then we can easily obtain\n$$\na_{2004} \\equiv a_{2002} - 1 \\equiv a_{2000} - 2 \\equiv \\cdots \\equiv a_0 - 1002 \\equiv 1003 \\pmod{2004}.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the integral solutions of the equation $7(x + y) = 3(x^2 - xy + y^2)$.",
    "options": [],
    "answer": "[(0, 0), (4, 5), (5, 4)]",
    "solution": "The solutions are $(x, y) = (0, 0), (4, 5), (5, 4)$.\nLet $a = x + y$ and $b = x - y$. The equation can be rewritten as\n$$\n28(x + y) = 3((x + y)^2 + 3(x - y)^2),\n$$\nwhich means $28a = 3(a^2 + 3b^2)$. Clearly, $3 \\mid a$. Let $a = 3c$. The equation becomes\n$$\n28c = 9c^2 + 3b^2.\n$$\nWe further find that $3 \\mid c$. Let $c = 3d$. The equation becomes\n$$\n28d = 27d^2 + b^2.\n$$\nNow, as $d(28-27d) = b^2 \\ge 0$, we must have $0 \\le d \\le \\frac{28}{27}$. The only integer solutions are $d = 0, 1$.\n\nWhen $d = 0$, $b = 0$. One finds that $a = 0$, and hence $x = y = 0$.\n\nWhen $d = 1$, $b = \\pm 1$. One finds that $a = 9$, and hence $(x, y) = (4, 5), (5, 4)$.\n\nAll these are solutions since the steps are reversible.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be a subset of $\\{1, 2, \\dots, 2006\\}$ with the following property: For any three elements $x, y$ and $z$ ($x < y < z$) of $M$, $x + y$ does not divide $z$. Determine the largest possible size of $M$. Justify your claim.",
    "options": [],
    "answer": "1004",
    "solution": "The answer is $1004$.\n\nWhen $M = \\{1003, 1004, \\dots, 2006\\}$, the sum of any two elements in $M$ is at least $2007$, which is larger than the largest element in $M$. Therefore, $x + y$ does not divide $z$ for any $x, y, z \\in M$. This gives a possible case for $|M| = 1004$.\n\nNext, consider any subset $M$ satisfying the constraint. Suppose the largest element of $M$ is $a$. Then $M$ contains at most one element in each of the pairs\n$$\n(1, a-1), (2, a-2), \\dots, \\left( \\left\\lfloor \\frac{a}{2} \\right\\rfloor, \\left\\lceil \\frac{a}{2} \\right\\rceil \\right)\n$$\n(note that the last pair may contain the same number, which does not affect the validity of our claim). Therefore, $M$ contains at most $\\frac{a}{2}$ elements smaller than $a$. Since $a$ is the largest element, $M$ cannot contain elements larger than $a$. Therefore,\n$$\n|M| \\le \\frac{a}{2} + 1 \\le 1003 + 1 = 1004.\n$$\nHence, the largest possible size of $M$ is $1004$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Show that the orthocentres of the triangles $\\triangle ABC$, $\\triangle BCD$, $\\triangle CDA$ and $\\triangle DAB$ are the vertices of a quadrilateral congruent to $ABCD$ and show that the centroids of the same triangles are the vertices of a cyclic quadrilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the centre of $(ABCD)$. Let $H_A, H_B, H_C, H_D$ and $G_A, G_B, G_C, G_D$ be the orthocentres and centroids of $\\triangle DAB$, $\\triangle ABC$, $\\triangle BCD$ and $\\triangle CDA$ respectively. Let $M$ be the midpoint of $AB$. Recall that $O$ lies on $H_A G_A$, etc., with $OG_A : G_A H_A = 1 : 2$. Also, we have $DG_A : G_A M = 2 : 1 = CG_B : G_B M$. From these, we know that $H_A H_B \\parallel G_A G_B \\parallel DC$, and\n$$\nH_A H_B = 3 G_A G_B = CD.\n$$\n![](attached_image_1.png)\nBy symmetry, we have $H_B H_C = DA$, $H_C H_D = AB$, $H_D H_A = BC$, $H_A H_C = CA$ and $H_B H_D = DB$. Thus, $H_A H_B H_C H_D$ is congruent to $CDAB$. It follows\n\nthat $H_A H_B H_C H_D$ is concyclic. Now, note that $G_A G_B G_C G_D$ is the image of $H_A H_B H_C H_D$ under the homothety with centre $O$ and ratio $\\frac{1}{3}$. Therefore, $G_A G_B G_C G_D$ is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31474,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $k$, let $f_1(k)$ be the square of the sum of the digits of $k$. (For example $f_1(123) = (1+2+3)^2 = 36$.) Let $f_{n+1}(k) = f_1(f_n(k))$. Determine the value of $f_{2007}(2^{2006})$. Justify your claim.",
    "options": [],
    "answer": "169",
    "solution": "Firstly, since $2^{2006} < 10^{2006}$, we have\n$$\nf(2^{2006}) \\le (9 \\times 2006)^2 < 10^{10}.\n$$\nSecondly, this implies\n$$\nf_2(2^{2006}) \\le (9 \\times 10)^2 = 8100.\n$$\nSimilarly, we find that\n$$\nf_3(2^{2006}) \\le (7 + 9 + 9 + 9)^2 = 1156\n$$\nand\n$$\nf_4(2^{2006}) \\le (9 \\times 3)^2 = 729.\n$$\nIt follows that the sum of digits of $f_k(2^{2006})$ cannot exceed 27 for $k \\ge 4$.\nNext, since $2^6 \\equiv 1 \\pmod 9$ and $2006 \\equiv 2 \\pmod 6$, we have\n$$\n2^{2006} \\equiv 2^2 \\equiv 4 \\pmod 9.\n$$\nWe easily find that $f_k(2^{2006}) \\equiv 7 \\pmod 9$ for odd $k$ and $f_k(2^{2006}) \\equiv 4 \\pmod 9$ for even $k$. Thus, the sum of digits of $f_{2005}(2^{2006})$ can only be 7, 16, 25. This yields $f_{2006}(2^{2006}) = 49, 256, 625$ and hence $f_{2007}(2^{2006}) = 169.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31475,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a cyclic quadrilateral with $BC = CD$. The diagonals $AC$ and $BD$ intersect at $E$. Let $X$, $Y$, $Z$ and $W$ be the incentres of triangles $\triangle ABE$, $\triangle ADE$, $\triangle ABC$ and $\triangle ADC$ respectively. Show that $X$, $Y$, $Z$ and $W$ are concyclic if and only if $AB = AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle BAE = \\angle CAD$ (in view of $BC = CD$) and $\\angle EBA = \\angle DCA$, we have $\\triangle ABE \\sim \\triangle ACD$. Since $X$ and $W$ are incentres of $\\triangle ABE$ and $\\triangle ACD$ respectively, they are corresponding points under this similarity. It follows that\n$$\n\\frac{AX}{AW} = \\frac{AB}{AC}.\n$$\nSimilarly, we have $\\frac{AY}{AZ} = \\frac{AD}{AC}$. Now,\n$$\n\\begin{align*}\n& W, X, Y, Z \\text{ are concyclic} \\\\\n\\Leftrightarrow \\quad & AX \\times AZ = AY \\times AW \\\\\n\\Leftrightarrow \\quad & \\frac{AX}{AW} = \\frac{AY}{AZ} \\\\\n\\Leftrightarrow \\quad & \\frac{AB}{AC} = \\frac{AD}{AC} \\\\\n\\Leftrightarrow \\quad & AB = AD.\n\\end{align*}\n$$\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31476,
    "subject": "Mathematics (Multi-modal)",
    "question": "The function $f(x, y)$, defined on the set of all nonnegative integers, satisfies\n(i) $f(0, y) = y + 1$,\n(ii) $f(x + 1, 0) = f(x, 1)$, and\n(iii) $f(x + 1, y + 1) = f(x, f(x + 1, y))$.\nFind (a) $f(3, 2005)$, and (b) $f(4, 2005)$.",
    "options": [],
    "answer": "a: 2^{2008} - 3; b: g(2008) - 3, where g(1) = 2 and g(n + 1) = 2^{g(n)}",
    "solution": "(a) The answer is $f(3, 2005) = 2^{2008} - 3$.\nWe first label the equations as follows:\n$$\nf(0, y) = y + 1, \\qquad (1)\n$$\n$$\nf(x + 1, 0) = f(x, 1), \\qquad (2)\n$$\n$$\nf(x + 1, y + 1) = f(x, f(x + 1, y)). \\qquad (3)\n$$\nBy putting $y = 1$ in (1), we get $f(0, 1) = 2$. By putting $x = 0$ in (2), we get $f(1, 0) = f(0, 1) = 2$.\nUsing (3) and (1), we find that\n$$\nf(1, y) = f(0, f(1, y - 1)) = f(1, y - 1) + 1.\n$$\nBy induction, one easily obtains\n$$\nf(1, y) = f(1, 0) + y = y + 2.\n$$\nSimilarly, we have\n$$\nf(2, y) = f(1, f(2, y - 1)) = f(2, y - 1) + 2.\n$$\nBy induction, we obtain\n$$\nf(2, y) = f(2, 0) + 2y = f(1, 1) + 2y = 2y + 3.\n$$\nNext, we have\n$$\nf(3, y) = f(2, f(3, y - 1)) = 2f(3, y - 1) + 3.\n$$\nAdding\n$$\n\\begin{aligned}\nf(3, y) &= 2f(3, y - 1) + 3, \\\\\n2f(3, y - 1) &= 2^2 f(3, y - 2) + 2 \\cdot 3, \\\\\n2^2 f(3, y - 2) &= 2^3 f(3, y - 3) + 2^2 \\cdot 3, \\\\\n& \\vdots, \\\\\n2^{y-1} f(3, 1) &= 2^y f(3, 0) + 2^{y-1} \\cdot 3,\n\\end{aligned}\n$$\nwe obtain\n$$\n\\begin{aligned}\nf(3, y) &= 2^y f(3, 0) + 3(1 + 2 + 2^2 + \\cdots + 2^{y-1}) \\\\\n&= 2^y f(2, 1) + 3(2^y - 1) = 2^{y+3} - 3.\n\\end{aligned}\n$$\nIn particular, we have $f(3, 2005) = 2^{2008} - 3$.\n\n(b) The answer is $f(4, 2005) = g(2008) - 3$, where $g(n)$ is defined by $g(1) = 2$ and $g(n + 1) = 2^{g(n)}$ for any $n \\in \\mathbb{Z}^+$.\nWe prove by induction that $f(4, y) = g(y + 3) - 3$. Firstly, by (2), we have\n$$\nf(4, 0) = f(3, 1) = 2^4 - 3 = 2^{2^2} - 3 = g(3) - 3.\n$$\nThis proves the base case.\nAssume $f(4, y) = g(y + 3) - 3$ for some $y \\in \\mathbb{Z}^+$. Using (3), we find that\n$$\nf(4, y + 1) = f(3, f(4, y)) = 2^{f(4,y)+3} - 3 = 2^{g(y+3)} - 3 = g(y + 4) - 3.\n$$\nThis proves the inductive step.\nTherefore, we have $f(4, 2005) = g(2008) - 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31477,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$ and $B$ are fixed points on a plane and $L$ is a line passing through $A$ and not $B$. $C$ is a variable point moving from $A$ toward infinity along a half-line of $L$. The incircle of $\\triangle ABC$ touches $BC$ at $D$ and $AC$ at $E$. Show that line $DE$ passes through a fixed point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the incentre of $\\triangle ABC$, and let $P$ be the intersection point of $AI$ and $DE$. We claim that $\\angle APB = 90^\\circ$. Once this is proved, since the line $AP$ is fixed, the point $P$ is independent of $C$, and so $P$ is the desired fixed point.\n\nNote that $\\angle BAI = \\angle PAE$ and $\\angle AIB = 90^\\circ + \\frac{C}{2} = \\angle AEP$. These imply $\\triangle IAB \\sim \\triangle EAP$. By spiral similarity, we have $\\triangle PAB \\sim \\triangle EAI$, and hence $\\angle APB = \\angle AEI = 90^\\circ$ as desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31478,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex quadrilateral $ABCD$ with $AC \\neq BD$ is inscribed in a circle with centre $O$. Let $E$ be the intersection of diagonals $AC$ and $BD$. If $P$ is a point inside $ABCD$ such that $\\angle PAB + \\angle PCB = \\angle PBC + \\angle PDC = 90^\\circ$, prove that $O, P$ and $E$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We only work on the configuration as shown since the other cases are similar.\nWe have\n$$\n\\angle CPA = \\angle BAP + \\angle CBA + \\angle PCB = 90^\\circ + \\angle CBA.\n$$\nLet $O_1$ be the centre of $(APC)$. Then we find that\n$$\n\\angle AO_1C = 360^\\circ - 2\\angle CPA = 180^\\circ - 2\\angle CBA = 180^\\circ - 2\\angle COA.\n$$\nThis shows $A$, $O$, $C$, $O_1$ are concyclic. As $O$, $O_1$ lie on the perpendicular bisector of $AC$, $O_1$ is just the intersection of the tangents at $A$ and $C$ to $(ABCD)$. Similarly, let $O_2$ be the intersection of the tangents at $B$ and $D$ to $(ABCD)$. Then $O_2$ is the centre of $(BPD)$.\n![](attached_image_1.png)\nClearly, $P$ lies on the radical axis of $(APC)$ and $(BPD)$. Secondly, since we have $EA \\times EC = EB \\times ED$, $E$ also lies on this radical axis. Lastly, the powers of $O$ with respect to the two circles are $OA^2$ and $OB^2$ respectively as $OA$ and $OB$ are the tangents. Since $OA = OB$, these powers are equal so that $O$ lies on the radical axis. Therefore, $P$, $E$, $O$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31479,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $ABC$, the altitude, angle bisector and median from $C$ divide the angle $\\angle C$ into four equal angles. Find $\\angle B$.",
    "options": [],
    "answer": "22.5° or 67.5°",
    "solution": "$\\angle B$ can be $22.5^\\circ$ or $67.5^\\circ$.\n\nWLOG assume $AC < BC$. Let $D$ be the foot of altitude from $C$, let $E$ be the foot of internal angle bisector from $C$, and let $F$ be the midpoint of $AB$. Let $P$ be the projection of $E$ on $BC$, and let $DP$ meet $AC$ at $Q$.\n\nFirstly, since $\\angle ACD = \\angle ECD$ and $CD \\perp AE$, we have $\\triangle ACD \\cong \\triangle ECD$. Secondly, note that $C, D, E, P$ are concyclic since\n$$\n\\angle CDE + \\angle CPE = 90^\\circ + 90^\\circ = 180^\\circ.\n$$\n\n$$\n\\angle CPQ = \\angle CPD = \\angle CED = \\angle CAD = \\angle CAB.\n$$\n\n![](attached_image_1.png)\n\nTogether with the common angle at $C$, we know that $\\triangle CPQ \\sim \\triangle CAB$ (and $D$ lies between $P$ and $Q$). As $\\angle DCQ = \\angle FCB$, the points $D$ and $F$ are corresponding points under this similarity. Thus, $D$ is the midpoint of $PQ$. Now, since $D$ is also the midpoint of $AE$, the quadrilateral $AQEP$ is a parallelogram. This implies $CQ // PE$, and hence $\\angle ACB = 90^\\circ$. Then we have\n$$\n\\angle ABC = \\angle FCB = \\frac{90^\\circ}{4} = 22.5^\\circ.\n$$\nIf $AC > BC$, then the roles of $A$ and $B$ are swapped, and hence $\\angle ABC = 67.5^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be positive real numbers such that $x + y + z = 1$. For positive integer $n$, define $S_n = x^n + y^n + z^n$. Furthermore, let $P = S_2 S_{2005}$ and $Q = S_3 S_{2004}$.\n\na. Find the smallest possible value of $Q$.\n\nb. If $x$, $y$, $z$ are pairwise distinct, determine whether $P$ or $Q$ is larger.",
    "options": [],
    "answer": "a) 1/3^{2005}. b) P > Q for pairwise distinct x, y, z.",
    "solution": "a. The smallest possible value of $Q$ is $\\frac{1}{3^{2005}}$.\nBy the power mean inequality, we have\n$$\nS_n \\ge 3 \\left(\\frac{S_1}{3}\\right)^n = \\frac{1}{3^{n-1}}\n$$\nfor any positive integer $n$. Therefore, we have\n$$\nQ = S_3 S_{2004} \\ge \\frac{1}{3^2} \\cdot \\frac{1}{3^{2003}} = \\frac{1}{3^{2005}}.\n$$\nEquality holds when $x = y = z = \\frac{1}{3}$. Therefore, the minimum value of $Q$ is $\\frac{1}{3^{2005}}$.\n\nb. $P$ is larger than $Q$.\nIndeed, we have\n$$\n\\begin{align*}\nP \\ge Q & \\Leftrightarrow (x^2 + y^2 + z^2)(x^{2005} + y^{2005} + z^{2005}) \\ge (x^3 + y^3 + z^3)(x^{2004} + y^{2004} + z^{2004}) \\\\\n& \\Leftrightarrow \\sum_{\\text{sym}} x^{2005} y^2 \\ge \\sum_{\\text{sym}} x^{2004} y^3.\n\\end{align*}\n$$\n\nThis is true by Muirhead's theorem. Equality holds when $x = y = z$. Since it is given that $x$, $y$, $z$ are pairwise distinct, the inequality is strict, and hence $P > Q$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be given with $R$ radius of its circumcircle. Let $AB = c$, $BC = a$ and $CA = b$. Let $K_1$ and $K_2$ be the circles which pass through $C$ and are tangent to $AB$ at $A$ and $B$ respectively. Let $K$ be the circle of radius $r$ which is externally tangent to $K_1$, $K_2$ and tangent to the line $AB$.\n\na. Express $r$ in terms of $a, b, c$ and $R$.\n\nb. Suppose $r = \\frac{R}{4}$. Find $\\angle C$.",
    "options": [],
    "answer": "r = abc^2 / (4R(a+b)^2); angle C = 90 degrees",
    "solution": "a.\nThe answer is $r = \\frac{abc^2}{4R(a+b)^2}$ or any equivalent expression.\n\nLet $O$, $O_1$, $O_2$ be the centres of $K$, $K_1$, $K_2$ respectively. Let $R_1$ and $R_2$ be the radii of $K_1$ and $K_2$ respectively. We have\n$$\nR_1 = AO_1 = \\frac{AC}{2 \\cos \\angle CAO_1} = \\frac{b}{2 \\sin A} = \\frac{bR}{a}\n$$\nby the extended sine law. By symmetry, we have $R_2 = \\frac{aR}{b}$.\n\n![](attached_image_1.png)\n\nNow, let $T$ be the tangential point of $K$ with $AB$. By Pythagorus' theorem, we have\n$$\nAT = \\sqrt{OO_1^2 - (AO_1 - TO)^2} = \\sqrt{(R_1 + r)^2 - (R_1 - r)^2} = 2\\sqrt{R_1r}.\n$$\nSimilarly, we have $BT = 2\\sqrt{R_2r}$. This yields\n$$\nc = AB = 2\\sqrt{R_1r} + 2\\sqrt{R_2r},\n$$\n$$\nr = \\frac{c^2}{4(\\sqrt{R_1} + \\sqrt{R_2})^2} = \\frac{c^2}{4(R_1 + R_2 + 2\\sqrt{R_1 R_2})} = \\frac{abc^2}{4R(a+b)^2}.\n$$\n\nb.\nWe have $\\angle C = 90^\\circ$.\n\nNote that $\\frac{ab}{(a+b)^2} \\le \\frac{1}{4}$ (since this is the same as $(a-b)^2 \\ge 0$) and $c \\le 2R$.\n\nBy part (a), we have\n$$\nr = \\frac{abc^2}{4R(a+b)^2} \\le \\frac{4R^2}{16R} = \\frac{R}{4}.\n$$\nTherefore, equality should hold. This means $a = b$ and $c = 2R$. In order that $c$ is the diameter of the circumcircle, we must have $\\angle C = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, a_3, \\dots$ be a sequence of positive numbers. If there exists a positive number $M$ such that for every $n = 1, 2, 3, \\dots$,\n$$\na_1^2 + a_2^2 + \\dots + a_n^2 < M a_{n+1}^2,\n$$\nthen prove that there exists a positive number $M'$ such that for every $n = 1, 2, 3, \\dots$,\n$$\na_1 + a_2 + \\dots + a_n < M' a_{n+1}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We say that a sequence $\\{b_n\\}$ of positive real numbers is *good* if there exists a positive constant $N$ such that\n$$\nb_1 + b_2 + \\cdots + b_n < N b_{n+1} \\quad (1)\n$$\nfor any $n \\in \\mathbb{Z}^+$.\nWe first prove the following equivalent conditions for a sequence to be good.\n\n**Claim.** A sequence $\\{b_n\\}$ of positive real numbers is good if and only if the following conditions hold.\n(i) There exists a constant $r > 0$ such that $b_{n+1} > r b_n$ for any $n \\in \\mathbb{Z}^+$; and\n(ii) there exists a positive integer $d$ such that $b_{n+d} > 2 b_n$ for any $n \\in \\mathbb{Z}^+$.\n\n*Proof.* Suppose $\\{b_n\\}$ is good. Then we have $N b_{n+1} > b_n$. So we can take $r = \\frac{1}{N}$ for (i). Next, we take $d$ to be any integer larger than $2N^2 + 1$. Note that $N b_{n+j} > b_1 + b_2 + \\cdots + b_{n+j-1} > b_n$ for any $j \\in \\mathbb{Z}^+$. It follows that\n$$\n\\begin{aligned}\nb_{n+d} &> \\frac{1}{N}(b_1 + b_2 + \\cdots + b_{n+d-1}) \\\\\n&> \\frac{1}{N}(b_{n+1} + b_{n+2} + \\cdots + b_{n+d-1}) \\\\\n&> \\frac{1}{N^2}(b_n + b_{n+1} + \\cdots + b_n) \\\\\n&= \\frac{d-1}{N^2} b_n \\\\\n&> 2 b_n.\n\\end{aligned}\n$$\nThis proves (ii).\n\nConversely, suppose (i) and (ii) hold. For any $n \\in \\mathbb{Z}^+$, we have\n$$\n\\begin{aligned}\nb_1 + b_2 + \\cdots + b_n &= \\sum_{j=1}^{d} (b_{n+1-j} + b_{n+1-j-d} + b_{n+1-j-2d} + \\cdots) \\\\\n&< \\sum_{j=1}^{d} \\left( b_{n+1-j} + \\frac{1}{2} b_{n+1-j} + \\frac{1}{2^2} b_{n+1-j} + \\cdots \\right) \\\\\n&< \\sum_{j=1}^{d} 2 b_{n+1-j} \\\\\n&< 2 \\left( \\sum_{j=1}^{d} \\frac{1}{r^j} \\right) b_{n+1}.\n\\end{aligned}\n$$\nSo we can take $N = 2 \\left( \\sum_{j=1}^{d} \\frac{1}{r^j} \\right)$ so that (1) holds. $\\square$\n\nWe go back to the original problem. Note that the sequence $\\{a_n^2\\}$ is good by the given condition. By the claim, there exists $r > 0$ and $d \\in \\mathbb{Z}^+$ such that $a_{n+1}^2 > r a_n^2$ and $a_{n+d}^2 > 2 a_n^2$ for any $n \\in \\mathbb{Z}^+$. This implies $a_{n+1} > \\sqrt{r} a_n$ and\n$$\na_{n+2d} > \\sqrt{2} a_{n+d} > 2 a_n.\n$$\nTherefore, the sequence $\\{a_n\\}$ is good by the claim. This is exactly our goal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given finitely many points in a plane, it is known that the area of the triangle formed by any three points of the set is less than $1$. Show that all points of the set lie inside or on the boundary of a triangle with area less than $4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A$, $B$, $C$ be points in $S$ such that $[ABC]$ is maximized. Let $\\ell_1$ be the line passing through $A$ which is parallel to $BC$. If there exists a point $X \\in S$ which lies on a different side of $\\ell_1$ as $B$, then the distance from $X$ to $BC$ is greater than that from $A$ to $BC$. This yields the contradiction $[XBC] > [ABC]$. Therefore, $X$ must lie on the same side of $\\ell_1$ as $B$.\n\n![](attached_image_1.png)\n\nSimilarly, let $\\ell_2$ be the line passing through $B$ which is parallel to $CA$, and let $\\ell_3$ be the line passing through $C$ which is parallel to $AB$. Then any point in $S$ must lie inside the triangle $\\triangle DEF$ bounded by $\\ell_1$, $\\ell_2$, $\\ell_3$. Note that $\\triangle ABC$ is the median triangle of $\\triangle DEF$. So $[DEF] = 4[ABC] < 4$. Thus, we can simply take $\\triangle DEF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find $2^{2006}$ positive integers satisfying the following conditions.\n(a) Each positive integer has $2^{2005}$ digits.\n(b) Each positive integer only has 7 or 8 in its digits.\n(c) Among any two chosen integers, at most half of their corresponding digits are the same.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We construct the numbers recursively. Firstly, we start from the following $2^2$ integers:\n77, 78, 87, 88,\neach of which has $2^1$ digits, and at most $2^0$ digit of each pair is the same.\nSuppose we have constructed $k = 2^{n+1}$ integers $a_1, a_2, \\dots, a_k$ with digits 7, 8, each of which has $2^n$ digits, and at most $2^{n-1}$ digits of each pair are the same. We construct $2k = 2^{n+2}$ integers as follows:\n$$\n\\overline{a_1a_1}, \\overline{a_2a_2}, \\dots, \\overline{a_ka_k}, \\overline{a_1b_1}, \\overline{a_2b_2}, \\dots, \\overline{a_kb_k},\n$$\nwhere $b_j$ means the integer having the same number of digits as $a_j$ such that each digit is different from that of $a_j$ (and is still 7 or 8). For example, when $n=1$, we construct\n7777, 7878, 8787, 8888, 7788, 7887, 8778, 8877.\n\nNote that each of the integers constructed has $2 \\cdot 2^n = 2^{n+1}$ digits. Also, one can easily prove by induction that $b_1, b_2, \\dots, b_k$ is a permutation of $a_1, a_2, \\dots, a_k$.\n* For $\\overline{a_i a_i}$ and $\\overline{a_j a_j}$ (with $i \\neq j$), at most $2 \\cdot 2^{n-1} = 2^n$ of their digits are the same by the inductive hypothesis. Similarly, the same holds for the pair $\\overline{a_i a_i}$ and $\\overline{a_j b_j}$ unless $a_i = b_j$, and for the pair $\\overline{a_i b_i}$ and $\\overline{a_j b_j}$.\n* For $\\overline{a_i a_i}$ and $\\overline{a_j b_j}$ with $a_i = b_j$, the first $2^n$ digits are different, while the last $2^n$ digits are the same.\n* For $\\overline{a_i a_i}$ and $\\overline{a_j b_j}$, the first $2^n$ digits are the same, while the last $2^n$ digits are different.\nThus, in any case, each pair of integers shares at most $2^n$ same digits. Repeating the same procedure, we eventually obtain $2^{2006}$ integers satisfying all the conditions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$, $q$, $r$ and $s$ be real numbers such that $p^2 + q^2 + r^2 - s^2 + 4 = 0$. Find the maximum value of $3p + 2q + r - 4|s|$.",
    "options": [],
    "answer": "-2\\sqrt{2}",
    "solution": "The maximum value is $-2\\sqrt{2}$.\nBy the Cauchy-Schwarz inequality, we have\n$$\n16s^2 = (9+4+1+2)(p^2+q^2+r^2+4) \\geq (3p+2q+r+2\\sqrt{2})^2.\n$$\nThis implies $4|s| \\geq 3p + 2q + r + 2\\sqrt{2}$, and hence\n$$\n3p + 2q + r - 4|s| \\leq -2\\sqrt{2}.\n$$\nEquality holds when $3:2:1:\\sqrt{2} = p:q:r:2$, i.e.\n$$\n(p, q, r, s) = (3\\sqrt{2}, 2\\sqrt{2}, \\sqrt{2}, \\pm 4\\sqrt{2}).\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be a point on the side $BC$ of triangle $ABC$ such that $AB + BD = AC + CD$. The line segment $AD$ cuts the incircle of triangle $ABC$ at $X$ and $Y$ with $X$ closer to $A$. Let $E$ be the point of contact of the incircle of triangle $ABC$ on the side $BC$. Show that\n(i) $EY$ is perpendicular to $AD$,\n(ii) $XD$ is $2IA'$, where $I$ is the incentre of the triangle $ABC$ and $A'$ is the midpoint of $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(i) Note that $AB + BD$ is the semiperimeter of $\\triangle ABC$, and so $D$ is the contact point of the $A$-excircle of $\\triangle ABC$ and $BC$. Since $A$ is a centre of homothety between the incircle and the $A$-excircle, $X$ and $D$ are corresponding points under this homothety. Therefore, the tangent at $X$ to the incircle is parallel to $BC$. This implies $XE$ is a diameter of the incircle. It follows that $\\angle XYE = 90^\\circ$, and hence $EY \\perp AD$.\n![](attached_image_1.png)\n\n(ii) Note that $I$ and $A'$ are midpoints of $EX$ and $ED$ respectively. By the midpoint theorem, $XD = 2IA'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an equilateral triangle with circumcentre $O$. Let $P$ be a point inside the triangle such that $\\angle BPC = 120^\\circ$ and $P \\ne O$. Let $B'$ and $C'$ be the images of $B$ and $C$ respectively under the reflection in the line $PO$. Furthermore $B'C'$ meets $AP$ at $D$. Determine $AD : DP$. Justify your claim.",
    "options": [],
    "answer": "1:1",
    "solution": "We have $AD : DP = 1 : 1$.\nSince $\\angle BPC = 120^\\circ = \\angle BOC$, the points $B$, $C$, $O$, $P$ are concyclic. As\n$$\n\\angle BPO = 180^\\circ - \\angle OCB = 150^\\circ,\n$$\nwe have $\\angle B'PB = 2(180^\\circ - 150^\\circ) = 60^\\circ$, and hence $\\triangle PB'B$ is equilateral. Similarly, as\n$$\n\\angle CPO = \\angle CBO = 30^\\circ,\n$$\nwe have $\\angle CPC' = 60^\\circ$, and hence $\\triangle PCC'$ is equilateral.\n![](attached_image_1.png)\nNow, since $\\triangle BPB'$ and $\\triangle BCA$ are directly similar, by spiral similarity, we have $\\triangle BAB' \\sim \\triangle BCP$. In the same way, we have $\\triangle ACC' \\sim \\triangle BCP$. Therefore, we obtain $\\triangle BAB' \\sim \\triangle ACC'$. Indeed, they are congruent since $BA = AC$. This implies $AC' = BB' = B'P$ and $B'A = C'C = PC'$. Thus, $AB'PC'$ is a parallelogram. The diagonals $AP$ and $B'C'$ bisect each other. In particular, we have $AD : DP = 1 : 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral such that $AD + BC = AB$. Show that the intersection of the bisectors of $\\angle ADC$ and $\\angle BCD$ is on line $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the point on the side $AB$ such that $AP = AD$. Then we have\n$$\nBP = AB - AP = AD + BC - AD = BC.\n$$\nLet the internal angle bisector of $\\angle ADC$ meet $AB$ at $Q$. We find that\n$$\n\\angle QPC = \\angle BPC = 90^\\circ - \\frac{1}{2}\\angle CBP = \\frac{1}{2}(180^\\circ - \\angle CBA) = \\frac{1}{2}\\angle ADC = \\angle QDC.\n$$\nThis implies $C$, $D$, $P$, $Q$ are concyclic. Similarly, we have\n$$\n\\angle DCQ = \\angle DPA = 90^\\circ - \\frac{1}{2}\\angle PAD = \\frac{1}{2}(180^\\circ - \\angle BAD) = \\frac{1}{2}\\angle DCB.\n$$\nThis shows $CQ$ bisects $\\angle BCD$. Therefore, the internal angle bisectors of $\\angle ADC$ and $\\angle BCD$ intersect at the point $Q$ on $AB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$ and $C$ be real numbers such that\n(i) $\\sin A \\cos B + |\\cos A \\sin B| = \\sin A |\\cos A| + |\\sin B \\cos B|$,\n(ii) $\\tan C$ and $\\cot C$ are defined.\nFind the minimum value of $(\\tan C - \\sin A)^2 + (\\cot C - \\cos B)^2$.",
    "options": [],
    "answer": "3 - 2√2",
    "solution": "The minimum value is $3 - 2\\sqrt{2}$.\nCondition (i) can be rewritten as\n$$\n(\\sin A - |\\sin B|)(\\cos B - |\\cos A|) = 0.\n$$\nIf $\\sin A = |\\sin B|$, then we have $\\sin^2 A = \\sin^2 B = 1 - \\cos^2 B$.\nIf $\\cos B = |\\cos A|$, then we have $\\cos^2 B = \\cos^2 A = 1 - \\sin^2 A$.\nTherefore, in any case, $(\\sin A, \\cos B)$ is a point on the unit circle on the coordinate plane. Since the steps are reversible, it can be any point on the unit circle.\n![](attached_image_1.png)\nNext, $(\\tan C, \\cot C)$ can be any point on the hyperbola $xy = 1$. Therefore, we are asked to find the minimum distance between a point on the unit circle and a point on the hyperbola. In view of the geometry, since the line $y = x$ is an axis of symmetry of both geometric objects, the minimum distance is attained\n\nbetween points on the line $y = x$. By symmetry, it suffices to consider $x > 0$. Clearly, $\\left(\\frac{1}{\\sqrt{2}}, \\frac{1}{\\sqrt{2}}\\right)$ is a solution to $x^2 + y^2 = 1$ and $y = x$, while $(1, 1)$ is a solution to $xy = 1$ and $y = x$. Therefore, the minimum distance is\n$$\n\\left(1 - \\frac{1}{\\sqrt{2}}\\right)^2 + \\left(1 - \\frac{1}{\\sqrt{2}}\\right)^2 = 3 - 2\\sqrt{2},\n$$\nand we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is there a polynomial $f$ of degree $2007$ with integer coefficients, such that $f(n), f(f(n)), f(f(f(n))), \\dots$ are pairwise relatively prime for every integer $n$? Justify your claim.",
    "options": [],
    "answer": "Yes; for example f(x) = x^{2007} − x^{2006} + 1.",
    "solution": "Yes. For example, we can take $f(x) = x^{2007} - x^{2006} + 1$.\nIt suffices to show that $(m, f^k(m)) = 1$ for any positive integer $k$, since we can replace $m$ by any $f^j(n)$. Consider any prime $p$ dividing $m$. Then we have $f(m) \\equiv 0 \\pmod{p} \\equiv -0 + 1 = 1 \\pmod{p}$. Whenever $f^i(m) \\equiv 1 \\pmod{p}$, we have\n$$\nf^{i+1}(m) \\equiv 1 - 1 + 1 \\equiv 1 \\pmod{p}.\n$$\nThus, $f^k(m) \\equiv 1 \\pmod{p}$ for any $k \\ge 1$ by induction. This shows $p \\nmid f^k(m)$, and hence $(m, f^k(m)) = 1$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a convex hexagon such that $AB \\parallel DE$, $BC \\parallel EF$ and $CD \\parallel FA$. Let $H$, $I$, $J$, $K$, $L$ and $M$ be the midpoints of sides $AB$, $BC$, $CD$, $DE$, $EF$ and $FA$, respectively. Prove that lines $HK$, $IL$ and $JM$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AD$ meet $BE$ at $X$, $BE$ meet $CF$ at $Y$, and $CF$ meet $AD$ at $Z$. Since $AB \\parallel ED$, we know that $\\triangle XAB \\sim \\triangle XDE$. Also, the midpoints $H$ and $K$ of $AB$ and $DE$ are corresponding points under this similarity. Thus, $X$ lies on $HK$. Similarly, $Y$ lies on $IL$, and $Z$ lies on $JM$.\n\n![](attached_image_1.png)\n\nBy Ceva's theorem, it suffices to prove\n$$\n\\frac{\\sin \\angle AXH}{\\sin \\angle HXB} \\times \\frac{\\sin \\angle EYL}{\\sin \\angle LYF} \\times \\frac{\\sin \\angle CZJ}{\\sin \\angle JZD} = 1. \\qquad (1)\n$$\nFirstly, since $AH = HB$ and $\\triangle XAB \\sim \\triangle XDE$, we have\n$$\n\\frac{\\sin \\angle AXH}{\\sin \\angle HXB} = \\frac{XB}{XA} = \\frac{BE}{AD}.\n$$\nBy symmetry, we have $\\frac{\\sin \\angle EYL}{\\sin \\angle LYF} = \\frac{CF}{BE}$ and $\\frac{\\sin \\angle CZJ}{\\sin \\angle JZD} = \\frac{AD}{CF}$. Therefore, the left-hand side of (1) is\n$$\n\\frac{BE}{AD} \\times \\frac{CF}{BE} \\times \\frac{AD}{CF} = 1\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = c$, $BC = a$ and $CA = b$. Show that\n$$\n\\frac{\\cos A}{a^3} + \\frac{\\cos B}{b^3} + \\frac{\\cos C}{c^3} \\ge \\frac{3}{2abc}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the cosine formula, we have\n$$\n\\begin{aligned}\n& \\frac{\\cos A}{a^3} + \\frac{\\cos B}{b^3} + \\frac{\\cos C}{c^3} \\ge \\frac{3}{2abc} \\\\\n\\Leftrightarrow \\quad & \\frac{b^2 + c^2 - a^2}{2a^3bc} + \\frac{c^2 + a^2 - b^2}{2ab^3c} + \\frac{a^2 + b^2 - c^2}{2abc^3} \\ge \\frac{3}{2abc} \\\\\n\\Leftrightarrow \\quad & \\frac{b^2 + c^2 - a^2}{a^2} + \\frac{c^2 + a^2 - b^2}{b^2} + \\frac{a^2 + b^2 - c^2}{c^2} \\ge 3 \\\\\n\\Leftrightarrow \\quad & \\frac{b^2}{a^2} + \\frac{c^2}{a^2} + \\frac{c^2}{b^2} + \\frac{a^2}{b^2} + \\frac{a^2}{c^2} + \\frac{b^2}{c^2} \\ge 6.\n\\end{aligned}\n$$\nThis is true by the AM-GM inequality. Equality holds when $\\triangle ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31493,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are $2007$ male and $2007$ female students. Each student joins not more than $100$ clubs in the school. It is known that any two students of opposite genders have joined at least one common club. Show that there is a club with at least $11$ male and $11$ female members.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose on the contrary that no club contains at least $11$ male and $11$ female members. We shall count the number of triples $(M, F, C)$ where $M$ is a male student, $F$ is a female student and $C$ is a club joined by these two students.\n\nFirstly, for each of the $2007$ choices of $M$ and each of the $2007$ choices of $F$, the condition shows that there is at least one club $C$ for which $(M, F, C)$ should be counted. This gives at least $2007^2$ triples.\n\nSecondly, we count according to the clubs. We label the clubs as $C_1, C_2, \\dots, C_k$. For each $i = 1, 2, \\dots, k$, let $m_i$ and $f_i$ be the numbers of male students and female students in the club $C_i$ respectively. Then the number of triples $(M, F, C)$ is\n$$\n\\sum_{i=1}^{k} m_i f_i .\n$$\nFrom the assumption, we can divide these clubs into two groups. The first group consists of those clubs with at most $10$ male members (say $C_1, C_2, \\dots, C_\\ell$). The second group consists of those clubs with at least $11$ male members, and hence at most $10$ female members (say $C_{\\ell+1}, C_{\\ell+2}, \\dots, C_k$). Then\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k} m_i f_i &= \\sum_{i=1}^{\\ell} m_i f_i + \\sum_{i=\\ell+1}^{k} m_i f_i \\\\\n&\\le 10 \\sum_{i=1}^{\\ell} f_i + 10 \\sum_{i=\\ell+1}^{k} m_i \\\\\n&\\le 10 \\sum_{i=1}^{k} f_i + 10 \\sum_{i=1}^{k} m_i.\n\\end{aligned}\n$$\nNote that the sum over $f_i$ counts the total number of female members in all clubs. By another simple double counting, this is equal to the total number of clubs joined by all the female students. Therefore, we have\n$$\n\\sum_{i=1}^{k} f_i \\le 100 \\times 2007.\n$$\nThe same upper bound holds for the sum over $m_i$. Combining all these, we obtain\n$$\n2007^2 \\le 10 \\times 100 \\times 2007 + 10 \\times 100 \\times 2007 = 2000 \\times 2007.\n$$\nThis is a contradiction. Hence, there must be a club with at least $11$ male and $11$ female members.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31494,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle (with centre $O$) of an isosceles triangle $ABC$ with $AB = AC$ meets $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. Let $N$ be the intersection of the lines $OL$ and $KM$, and let $Q$ be the intersection of the lines $BN$ and $CA$. Let $P$ be the foot of the perpendicular from $A$ to $BQ$. Suppose $BP = AP + 2PQ$. Determine the possible value(s) of $\\frac{AB}{BC}$.",
    "options": [],
    "answer": "sqrt(2)/2 or sqrt(10)/2",
    "solution": "$\\frac{AB}{BC}$ can be $\\frac{\\sqrt{2}}{2}$ or $\\frac{\\sqrt{10}}{2}$.\nIt is well-known that $Q$ is the midpoint of $AC$. We first consider the case when $P$ lies inside $\\triangle ABC$. Let $X$ be the projection of $C$ on $BQ$, and let $Y$ be the point on the line $BQ$ such that $XY = XC$, and $X$ lies between $B$ and $Y$.\nNote that $\\triangle QAP \\cong \\triangle QCX$, and hence $PQ = XQ$. By the given condition, we have\n$$\nBP = AP + 2PQ = CX + PX = XY + PX = PY.\n$$\nTogether with $AP \\perp BY$, we know that $\\triangle ABY$ is an isosceles triangle, with $AY = AB = AC$. Thus, $AX$ is the perpendicular bisector of $CY$. This implies\n$$\n\\angle ACX = \\angle AYX = \\angle YBA.\n$$\nTherefore, $A$, $B$, $C$, $X$ are concyclic, and $\\angle BAC = \\angle BXC = 90^\\circ$. Conversely, when $\\angle BAC = 90^\\circ$, one can work backward and show that $BP = AP + 2PQ$. (In that proof, one should define $Y$ as the point on the line $BQ$ such that $AB = AY$, and then show that $XC = XY$.) Thus, this gives one possibility, where\n$$\n\\frac{AB}{BC} = \\frac{1}{\\sqrt{2}}.\n$$\n![](attached_image_1.png)\n\nNext we consider the case when $P$ lies outside $\\triangle ABC$. Similarly, let $X$ be the projection of $C$ on $BQ$. As above, we have $\\triangle QAP \\cong \\triangle QCX$, and hence $QX = QP$. By the given condition, we have\n$$\nBP = AP + 2PQ = CX + PX.\n$$\nThis implies $BX = CX$. Therefore, $BC = \\sqrt{2}CX = \\sqrt{2}AP$. Let $PQ = QX = x$ and $CX = AP = y$. Noting that $\\angle PBA = \\angle XCQ$, we have $\\triangle PAB \\sim \\triangle XQC$.\nIt follows that $\\frac{PA}{XQ} = \\frac{PB}{XC}$. This implies\n$$\n\\frac{y}{x} = \\frac{2x + y}{y}.\n$$\n\nThis can be rewritten as $(2x - y)(x + y) = 0$. Thus, we find that $y = 2x$, and hence\n$$\n\\frac{AB}{BC} = \\frac{\\sqrt{PA^2 + PB^2}}{BC} = \\frac{\\sqrt{y^2 + (2x+y)^2}}{\\sqrt{2}y} = \\frac{\\sqrt{10}}{2}.\n$$\nWhen $\\frac{AB}{BC} = \\frac{\\sqrt{10}}{2}$, it is not hard to find that $\\frac{BQ}{AB} = \\frac{3}{2\\sqrt{5}}$, and $\\cos \\angle CBQ = \\frac{1}{\\sqrt{2}}$.\nThen one can deduce $BX = CX$, and the condition holds.\n![](attached_image_2.png)\n\nTo conclude, $\\frac{AB}{BC}$ can be $\\frac{\\sqrt{2}}{2}$ or $\\frac{\\sqrt{10}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31495,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Given five points on a plane such that no three of the points are collinear, show that among the triangles which are drawn using any three of these five points as vertices, at least three of the triangles formed are not acute-angled triangles. (An acute-angled triangle is one in which all the three interior angles are acute angles.)\n\nb. Given any 100 points on a plane such that no three of the points are collinear, show that among the triangles which are drawn using any three of these 100 points as vertices, at least 30% of the triangles are not acute-angled triangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) We first show that there must be a non-acute triangle among any 4 points. Consider the convex hull of $A$, $B$, $C$, $D$.\n\n* If the convex hull is a quadrilateral $ABCD$, then since\n$$ \\angle ABC + \\angle BCD + \\angle CDA + \\angle DAB = 360^\\circ, $$\none of these angles is at least $90^\\circ$. This gives rise to a non-acute triangle.\n\n* If the convex hull is a triangle, say $\\triangle ABC$, then since\n$$ \\angle ADB + \\angle BDC + \\angle CDA = 360^\\circ, $$\none of these angles is obtuse.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nNow, suppose on the contrary that at most 2 triangles are non-acute. Note that each triangle belongs to exactly 2 quadrilaterals, and there are $\\binom{5}{4} = 5$ quadrilaterals in total. Therefore, there must be a quadrilateral which does not consist of any non-acute triangles, contradicting the above observation. Therefore, there are at least 3 non-acute triangles.\n\n(b) There are $\\binom{100}{5}$ groups of 5 points formed from the 100 points. By part (a), there are at least 3 non-acute triangles in each group. Since each triangle belongs to $\\binom{97}{2}$ groups of 5 points, there are at least\n$$\n3 \\binom{100}{5} \\div \\binom{97}{2}\n$$\nnon-acute triangles. As there are $\\binom{100}{3}$ triangles in total, at least\n$$\n3 \\binom{100}{5} \\div \\binom{97}{2} \\div \\binom{100}{3} = \\frac{3}{10} = 30\\%\n$$\nof the triangles are non-acute.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine if there exists a positive integer pair $(m, n)$, such that\n(i) the greatest common divisor of $m$ and $n$ is $1$, and $m \\le 2007$,\n(ii) for any $k = 1, 2, \\dots, 2007$, $\\lfloor \\frac{nk}{m} \\rfloor = \\lceil \\sqrt{2k} \\rceil$.\n\n(Here $\\lfloor x \\rfloor$ stands for the greatest integer less than or equal to $x$.)",
    "options": [],
    "answer": "Yes, such a pair exists.",
    "solution": "Yes, it exists.\nThere are finitely many fractions $\\frac{a}{b}$ of lowest term such that $a \\ge 1$, $1 \\le b \\le 2007$ and $\\frac{a}{b} < \\sqrt{2}$. Let $\\frac{n}{m}$ be the largest such fraction. If $\\lfloor \\frac{nk}{m} \\rfloor \\ne \\lfloor \\sqrt{2}k \\rfloor$ for some $k = 1, 2, \\dots, 2007$, then $t = \\lfloor \\sqrt{2}k \\rfloor$ is an integer satisfying $\\frac{nk}{m} < t < \\sqrt{2}k$ since $\\frac{n}{m} < \\sqrt{2}$ and $\\sqrt{2}k \\notin \\mathbb{Z}$. It follows that $\\frac{n}{m} < \\frac{t}{k} < \\sqrt{2}$, which contradicts the choice of $\\frac{n}{m}$. Therefore, we must have $\\lfloor \\frac{nk}{m} \\rfloor = \\lfloor \\sqrt{2}k \\rfloor$ for $1 \\le k \\le 2007$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31497,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $\\{a_n\\}$ is defined by $a_1 = 0$ and $(n+1)^3 a_{n+1} = 2n^2(2n+1)a_n + 2(3n+1)$ for all integers $n \\ge 1$. Show that infinitely many members of the sequence are positive integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $b_n = n^2a_n + 2$ for all $n \\ge 1$. The recurrence relation becomes\n$$\n(n + 1)(b_{n+1} - 2) = 2(2n + 1)(b_n - 2) + 2(3n + 1).\n$$\nThis is the same as\n$$\nb_{n+1} = \\frac{2(2n + 1)}{n + 1}b_n.\n$$\nIt follows that\n$$\nb_n = \\frac{2(2n-1)}{n} b_{n-1} = \\frac{2^2(2n-1)(2n-3)}{n(n-1)} b_{n-2} = \\dots = \\frac{2^{n-1}(2n-1)!!}{n!} b_1,\n$$\nwhere $(2n-1)!! = (2n-1)(2n-3)\\cdots(1)$. Since $b_1 = a_1 + 2 = 2$, we have\n$$\nb_n = \\frac{2^n(2n-1)!!}{n!} = \\frac{(2n)!!(2n-1)!!}{(n!)^2} = \\frac{(2n)!}{(n!)^2} = \\binom{2n}{n}.\n$$\nTherefore, we obtain\n$$\na_n = \\frac{b_n - 2}{n^2} = \\frac{1}{n^2} \\left[ \\binom{2n}{n} - 2 \\right].\n$$\nBy Vandermonde's identity, we can rewrite this as\n$$\na_n = \\frac{1}{n^2} \\left( \\sum_{k=0}^{n} \\binom{n}{k}^2 - 2 \\right) = \\frac{1}{n^2} \\left( \\sum_{k=1}^{n-1} \\binom{n}{k}^2 \\right).\n$$\n\nWhen $n = p$ is a prime, we have $p \\mid \\binom{p}{k}$ for $1 \\le k \\le n-1$, and so $p^2 \\mid \\binom{p}{k}^2$. This implies $a_p \\in \\mathbb{Z}$. Clearly, $a_p > 0$. Since there are infinitely many prime numbers, there are infinitely many terms which are positive integers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of positive integers such that $\\frac{x^2y + x + y}{xy^2 + y + 11}$ are integers.",
    "options": [],
    "answer": "(x, y) = (11t^2, 11t) for any positive integer t; and (7, 1), (121, 1), (10, 2).",
    "solution": "(IMO Shortlist 1998 N1/Problem 4) The solutions are $(x, y) = (11t^2, 11t), (7, 1), (121, 1), (10, 2)$ where $t \\in \\mathbb{Z}^+$.\nThe condition means $xy^2 + y + 11 \\mid x^2y + x + y$. Since\n$$\ny(x^2y + x + y) - x(xy^2 + y + 11) = y^2 - 11x,\n$$\nwe have\n$$\nxy^2 + y + 11 \\mid y^2 - 11x.\n$$\n* If $y^2 - 11x > 0$, then we have $xy^2 + y + 11 \\le y^2 - 11x < y^2$. But then $xy^2 + y + 11 > xy^2 \\ge y^2$. This is a contradiction.\n* If $y^2 - 11x = 0$, then we can write $y = 11t$ for some $t \\in \\mathbb{Z}^+$. This yields $x = 11t^2$. Now, we check that\n$$\n\\frac{(11t^2)^2(11t) + (11t^2) + (11t)}{(11t^2)(11t)^2 + (11t) + 11} = \\frac{11t(121t^4 + t + 1)}{11(121t^4 + t + 1)} = t \\in \\mathbb{Z},\n$$\nand so all pairs $(x, y) = (11t^2, 11t)$ are solutions.\n* If $y^2 - 11x < 0$, then we have $xy^2 + y + 11 \\le 11x - y^2 < 11x$. It follows that $11x > xy^2 + y + 11 > xy^2$, and hence $y^2 < 11$. This implies $y = 1, 2, 3$.\n- When $y = 1$, we need $x + 12 \\mid x^2 + x + 1$. This is equivalent to\n$$\nx + 12 \\mid (-12)^2 + (-12) + 1 = 133 = 7 \\times 19.\n$$\nThe solutions are $(x, y) = (7, 1), (121, 1)$.\n- When $y = 2$, we need $4x + 13 \\mid 2x^2 + x + 2$. Since $4x + 13$ is odd, this is the same as $4x + 13 \\mid 8(2x^2 + x + 2)$. As\n$$\n8(2x^2 + x + 2) = (4x + 13)(4x - 11) + 159,\n$$\nthis means $4x + 13 \\mid 159 = 3 \\times 53$. The only solution is $(x, y) = (10, 2)$.\n- When $y = 3$, we need $9x + 14 \\mid 3x^2 + x + 3$. Since $3 \\nmid 9x + 14$, this is the same as $9x + 14 \\mid 27(3x^2 + x + 3)$. As\n$$\n27(3x^2 + x + 3) = (9x + 14)(9x - 11) + 235,\n$$\nthis means $9x + 14 \\mid 235 = 5 \\times 47$. There is no solution.\nTherefore, we conclude that the solutions are those listed at the beginning.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31499,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number and $f(x)$ be a polynomial with rational coefficients and is of degree $p$. Furthermore (i) $f(m)$ is an integer for any integer $m$ and (ii) $p$ divides $f(p+1) - f(1)$. Find all such polynomials.",
    "options": [],
    "answer": "All polynomials of the form f(x) = p a_p \\binom{x}{p} + a_{p-1} \\binom{x}{p-1} + a_{p-2} \\binom{x}{p-2} + \\dots + a_1 \\binom{x}{1} + a_0, where a_0, a_1, \\dots, a_p are integers with a_p \\neq 0.",
    "solution": "$f(x)$ can be any polynomial of the form\n$$\np a_p \\binom{x}{p} + a_{p-1} \\binom{x}{p-1} + a_{p-2} \\binom{x}{p-2} + \\dots + a_1 \\binom{x}{1} + a_0,\n$$\nwhere $a_0, a_1, \\dots, a_p$ are arbitrary integers with $a_p \\neq 0$.\n\nNote that we can always write\n$$\nf(x) = c_p \\binom{x}{p} + c_{p-1} \\binom{x}{p-1} + \\cdots + c_1 \\binom{x}{1} + c_0\n$$\nfor some rational coefficients $c_0, c_1, \\dots, c_p$, since $\\binom{x}{p}, \\binom{x}{p-1}, \\dots, \\binom{x}{0}$ have different degrees. By putting $x = 0$, we get $c_0 = f(0) \\in \\mathbb{Z}$. By putting $x = 1$, we get $c_1 + c_0 = f(1) \\in \\mathbb{Z}$, and hence $c_1 \\in \\mathbb{Z}$. Similarly, by putting $x = 2, 3, \\dots, p$, we find that all $c_j$'s are integers. Conversely, if all $c_j$'s are integers, it is obvious that $f(m) \\in \\mathbb{Z}$ for any $m \\in \\mathbb{Z}$.\nNow, putting $x = p + 1$ and using $\\binom{p+1}{k} = \\binom{p}{k} + \\binom{p}{k-1}$, we obtain\n$$\nf(p+1) - f(1) = \\sum_{k=0}^{p} c_k \\binom{p+1}{k} - (c_1 + c_0) = \\sum_{k=1}^{p} c_k \\left[ \\binom{p}{k} + \\binom{p}{k-1} \\right] - c_1.\n$$\nSince $p \\mid \\binom{p}{r}$ whenever $1 \\le r \\le p-1$, we have\n$$\nf(p+1) - f(1) \\equiv c_p \\pmod{p}.\n$$\nTherefore, $p \\mid f(p+1) - f(1)$ if and only if $p \\mid c_p$. This shows $f(x)$ can be any polynomial of the form mentioned at the beginning.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = c_m x^m + c_{m-1} x^{m-1} + \\cdots + c_1 x + c_0$, where each $c_i$ is a nonzero integer. Define a sequence $\\{a_n\\}$ by $a_1 = 0$ and $a_{n+1} = f(a_n)$ for all positive integers $n$.\n\na. Let $i$ and $j$ be positive integers with $i < j$. Show that $a_{j+1} - a_j$ is a multiple of $a_{i+1} - a_i$.\n\nb. Show that $a_{2008} \\neq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nRecall the fact that $b - c \\mid f(b) - f(c)$ for any integers $b$ and $c$. Putting $b = a_{i+1}$ and $c = a_i$, we immediately obtain\n$$\na_{i+1} - a_i \\mid f(a_{i+1}) - f(a_i) = a_{i+2} - a_{i+1}.\n$$\nBy induction on $j$, $a_{i+1} - a_i \\mid a_{j+1} - a_j$ for any $j > i$.\n\nb.\nSuppose on the contrary that $a_{2008} = 0$. Note that $a_{2009}-a_{2008} = f(0) = a_2-a_1$. By part (a), since $a_2 - a_1 \\mid a_{k+1} - a_k \\mid a_{2009} - a_{2008}$ for $k = 1, 2, \\dots, 2007$, each $a_{k+1} - a_k$ must be $\\pm f(0)$. Now, we have\n$$\n0 = a_{2008} - a_1 = \\sum_{k=1}^{2007} (a_{k+1} - a_k) = m f(0),\n$$\nwhere $m \\in \\{-2007, -2005, \\dots, -1, 1, 3, \\dots, 2007\\}$. This implies $f(0) = 0$, which contradicts the assumption that $c_0 \\neq 0$. Therefore, $a_{2008} \\neq 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31501,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be real numbers satisfying $x - y + 1 \\neq 0$. If\n$$\n1 + \\cos^2(2007x + 2008y - 1) = \\frac{x^2 + y^2 + 2(1+x)(1-y)}{x - y + 1},\n$$\nfind the minimum value of $xy$.",
    "options": [],
    "answer": "1/16120225",
    "solution": "The minimum value of $xy$ is $\\frac{1}{16120225}$.\n\nNote that the left-hand side is positive. Therefore, by rewriting the right-hand side as\n$$\n(x - y + 1) + \\frac{1}{x - y + 1},\n$$\nwe know that $x - y + 1 > 0$. Thus, we can apply the AM-GM inequality to obtain\n$$\n(x - y + 1) + \\frac{1}{x - y + 1} \\ge 2.\n$$\nBut then $1 + \\cos^2(2007x + 2008y - 1) \\le 2$. This shows equality should hold, and hence\n$$\nx - y + 1 = 1 \\quad \\text{and} \\quad \\cos(2007x + 2008y - 1) = \\pm 1.\n$$\nThe first relation gives $x = y$. The second relation gives $4015x - 1 = k\\pi$ for some $k \\in \\mathbb{Z}$, and so\n$$\nx = y = \\frac{1 + k\\pi}{4015}.\n$$\nSince $k$ is an integer, it is obvious that $|1 + k\\pi| \\ge 1$. This yields\n$$\nxy = \\frac{(1 + k\\pi)^2}{4015^2} \\ge \\frac{1}{16120225}.\n$$\nEquality holds when $x = y = \\frac{1}{4015}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31502,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the product of three consecutive integers is the sum of four integer cubes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that the product of three consecutive integers is divisible by $6$. Let the product be $6k$ for some $k \\in \\mathbb{Z}$. Then we easily check that\n$$\n(k + 1)^3 + (k - 1)^3 + (-k)^3 + (-k)^3 = 6k.\n$$\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31503,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 4$ be a positive integer such that $n$ is composite (not a prime) and divides $\\varphi(n)\\sigma(n) + 1$, where $\\varphi(n)$ is the Euler's totient function of $n$ and $\\sigma(n)$ is the sum of the positive divisors of $n$. Prove that $n$ has at least three distinct prime factors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $p^2 \\mid n$ for some prime $p$, then $p \\mid \\varphi(p^2) \\mid \\varphi(n)$. This implies $p \\mid \\varphi(n)\\sigma(n) + 1$, and hence $p \\mid n + \\varphi(n)\\sigma(n) + 1$. Therefore, $n$ must be squarefree.\n\nIf $n = pq$ for some distinct primes $p$ and $q$, then\n$$\n\\varphi(pq)\\sigma(pq) + 1 = (p-1)(q-1)(p+1)(q+1) + 1 = p^2q^2 - p^2 - q^2 + 2.\n$$\nThis is a multiple of $pq$ if and only if $pq \\mid p^2 + q^2 - 2$. We now prove that $ab \\mid a^2 + b^2 - 2$ has no solution in distinct integers $a, b > 1$ using Vieta's jumping.\n\nSuppose on the contrary that the divisibility is solvable. Let $(a, b)$ be a solution such that $a > b > 1$ and $b$ is the smallest possible. Let $a^2 + b^2 - 2 = kab$ for some positive integer $k$. Rewrite this equation as\n$$\na^2 - kba + (b^2 - 2) = 0.\n$$\nThis is a quadratic equation in $a$. Let $a'$ be another solution to this equation. Since $a + a' = kb$, $a'$ is an integer. Since $aa' = b^2 - 2 > 0$, $a'$ is positive. Also, we have\n$$\na' = \\frac{b^2 - 2}{a} < \\frac{b^2}{a} < b < a.\n$$\nThis shows $(x, y) = (b, a')$ is another pair of solution to $xy \\mid x^2 + y^2 - 2$ such that $x > y > 0$. As $a' < b$, by the minimality of $b$, we must have $a' = 1$. But then the divisibility becomes $b \\mid 1 + b^2 - 2$, i.e. $b \\mid 1$. This is impossible since $b > 1$. Thus, there is no solution.\n\nIt follows that $n$ must have at least 3 distinct prime divisors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31504,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of a scalene and acute $\\triangle ABC$ touches $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. $H$ is a point on the segment $EF$ such that $DH \\perp EF$. Suppose $AH \\perp BC$, prove that $H$ is the orthocentre of $\\triangle ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the midpoint of $DF$. Then $\\angle FPB = 90^\\circ$. Since $\\angle BFP = \\angle DEH$, we have $\\triangle BPF \\sim \\triangle DHE$. This implies $\\frac{BF}{PF} = \\frac{DE}{HE}$, which yields\n$$\nBF \\times HE = DE \\times PF = \\frac{1}{2}DE \\times DF.\n$$\nBy symmetry,\n$$\nCE \\times HF = \\frac{1}{2}DE \\times DF = BF \\times HE.\n$$\nTogether with $\\angle BFH = 90^\\circ + \\frac{A}{2} = \\angle CEH$, we obtain $\\triangle BFH \\sim \\triangle CEH$. It follows that $\\angle FBH = \\angle ECH$.\n![](attached_image_1.png)\n\nLet $B'$ be the reflection of $B$ in the line $AH$. Note that $B'$ lies on $BC$ since $AH \\perp BC$. Also, $B' \\neq C$ as $\\triangle ABC$ is scalene. Therefore, we find that\n$$\n\\angle AB'H = \\angle ABH = \\angle ACH.\n$$\nThis implies $A$, $H$, $B'$, $C$ are concyclic. Thus,\n$$\n\\angle HBC = \\angle HB'B = \\angle HAC = 90^\\circ - C,\n$$\nand hence $BH \\perp AC$. So $H$ is the orthocentre of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31505,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be the sides of a triangle and $p = a + b + c$. Show that\n\na. $\\frac{a}{b+c-a} + \\frac{b}{c+a-b} + \\frac{c}{a+b-c} \\ge 3$;\n\nb. $a^k + b^k > \\frac{c^k}{2^{k-1}}$ for $k = 2, 3, 4, \\dots$;\n\nc. $a^k + b^k + c^k \\ge \\frac{p^k}{3^{k-1}}$ for $k = 2, 3, 4, \\dots$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Let $x = \\frac{b+c-a}{2}$, $y = \\frac{c+a-b}{2}$ and $z = \\frac{a+b-c}{2}$. Then we have\n$$\n\\begin{align*}\n\\frac{a}{b+c-a} + \\frac{b}{c+a-b} + \\frac{c}{a+b-c} &\\ge 3 \\\\\n\\frac{y+z}{2x} + \\frac{z+x}{2y} + \\frac{x+y}{2z} &\\ge 3 \\\\\n\\frac{y}{x} + \\frac{z}{x} + \\frac{z}{y} + \\frac{x}{y} + \\frac{x}{z} + \\frac{y}{z} &\\ge 6.\n\\end{align*}\n$$\n$\\Leftrightarrow$\n$\\Leftrightarrow$\nThis holds by the AM-GM inequality. Equality holds when $x = y = z$, i.e. $a = b = c$.\n\nb. By the power mean inequality and the triangle inequality, we have\n$$\na^k + b^k \\ge 2 \\left(\\frac{a+b}{2}\\right)^k > 2 \\left(\\frac{c}{2}\\right)^k = \\frac{c^k}{2^{k-1}}.\n$$\n\nc. By the power mean inequality, we have\n$$\na^k + b^k + c^k \\ge 3 \\left( \\frac{a+b+c}{3} \\right)^k = 3 \\left( \\frac{p}{3} \\right)^k = \\frac{p^k}{3^{k-1}}.\n$$\nEquality holds when $a = b = c$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31506,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H$ be an arbitrary point on the altitude $CP$ of an acute triangle $ABC$. The lines $AH$ and $BH$ intersect $BC$ and $AC$ in $M$ and $N$ respectively. Show that $\\angle NPC = \\angle MPC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Blanchet's theorem) Suppose the line passing through $C$ and parallel to $AB$ meets $PN$ and $PM$ at $X$ and $Y$ respectively. Since $PC \\perp XY$, the result is the same as $CX = CY$ (as this implies $\\triangle PCX \\cong \\triangle PCY$).\n\nBy Ceva's theorem, we have\n$$\n\\frac{AP}{PB} \\times \\frac{BM}{MC} \\times \\frac{CN}{NA} = 1.\n$$\nNote that $\\frac{BM}{MC} = \\frac{PB}{YC}$ and $\\frac{CN}{NA} = \\frac{XC}{PA}$ by similar triangles. It follows that $CX = CY$ as desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31507,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1$ and $y_1$ be positive integers satisfying $x_1 + y_1 = 2^r$, where $r$ is a positive integer. If $x_1 < y_1$, let $x_2 = 2x_1$ and $y_2 = 2^r - x_2$. On the other hand, if $x_1 > y_1$, then we let $y_2 = 2y_1$ and $x_2 = 2^r - y_2$. Apply the same procedure to $(x_2, y_2)$, and so on. Show that the procedure will end in finitely many steps, i.e. there exists a $k$ such that $x_k = y_k$. (For example, let $(x_1, y_1) = (7, 57)$, with $x_1 + y_1 = 2^6$. Then $(x_2, y_2) = (14, 50)$, $(x_3, y_3) = (28, 36)$, $(x_4, y_4) = (56, 8)$, $(x_5, y_5) = (48, 16)$, and finally $(x_6, y_6) = (32, 32)$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that for any positive integers $x$ and $y$ satisfying $x + y = 2^r$, we have\n$$\nv_2(x) = v_2(2^r - y) = v_2(y),\n$$\nwhere $v_2(n)$ is the largest $m$ such that $2^m \\mid n$. Therefore, WLOG we may assume $x_k < y_k$ in some step. Since\n$$\nv_2(x_{k+1}) = v_2(2x_k) = v_2(x_k) + 1,\n$$\nthe sequence $\\{v_2(x_k)\\}_{k \\ge 1}$ is strictly increasing. But then the sequence is bounded above by $r-1$ since $x_k < 2^r$. Therefore, it must be a finite sequence, which means the procedure will end in finitely many steps.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31508,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each permutation $x_1, x_2, \\dots, x_{10}$ of $1, 2, \\dots, 10$, compute\n$$\n|2x_1 - 3x_2| + |2x_2 - 3x_3| + |2x_3 - 3x_4| + \\dots + |2x_{10} - 3x_1|.\n$$\nLet $S$ be the maximum possible value of this sum. Find the number of permutations attaining $S$.",
    "options": [],
    "answer": "28800",
    "solution": "The answer is 28800.\nDefine\n$$\n\\begin{align*}\nX &= \\{2x_j : 1 \\le j \\le 10\\} = \\{2, 4, 6, \\dots, 20\\}, \\\\\nY &= \\{3x_j : 1 \\le j \\le 10\\} = \\{3, 6, 9, \\dots, 30\\}.\n\\end{align*}\n$$\nAfter removing the absolute value signs of the given expression, we get a sum of 10 terms from $X \\cup Y$ minus the sum of the remaining 10 terms from $X \\cup Y$.\nTherefore, we must have\n$$\n\\begin{align*}\nS \\le (30 + 27 + 24 + 21 + 18 + 15 + 20 + 18 + 16 + 14) \\\\\n\\quad - (3 + 6 + 9 + 12 + 2 + 4 + 6 + 8 + 10 + 12) \\\\\n\\quad = 131.\n\\end{align*}\n$$\nIn the following, we will show that equality can be attained, and hence $S = 131$.\nIndeed, we shall count the number of permutations such that the sum is $S$.\nIn order that each of $2s$ for $s = 7, 8, 9, 10$ and $3t$ for $t = 5, 6, 7, 8, 9, 10$ is the larger term of the pair in the same absolute value sign, the **large numbers** $7, 8, 9, 10$ cannot be adjacent terms (where $x_{10}$ and $x_1$ are considered as adjacent). Also, $5, 6$ cannot be the term immediately after the large numbers. Similarly, the **small numbers** $1, 2, 3, 4$ cannot be adjacent, and $5, 6$ cannot be the term immediately before the small numbers. Conversely, whenever all these conditions are satisfied, the given expression is equal to $131$.\nNote that for any pair of large numbers, there must be a small number in between (possibly together with $5$ and/or $6$), and vice versa. WLOG assume $1$ is the first term. There are $3! \\times 4! = 144$ ways to arrange the small numbers and the large numbers (for example, $1, 7, 2, 8, 3, 9, 4, 10$). Afterwards, we can only place $5$ and $6$ in the 4 gaps between a small number and a large number (but not between a large number and a small number). By some basic counting, we know that there are $4 \\times 5 = 20$ ways to do so (4 ways to place $5$, and then 5 ways to place $6$ since there is one more position in the same gap as $5$). As we can shift all terms cyclically in 10 ways, the final answer is\n$$\n144 \\times 20 \\times 10 = 28800.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 31509,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f : \\mathbb{Z} \\to \\mathbb{Z}$ ($\\mathbb{Z}$ is the set of integers) be such that $f(1) = 1$, $f(2) = 20$, $f(-4) = -4$ and $f(x+y) = f(x)+f(y)+axy(x+y)+bxy+c(x+y)+4$ for all $x, y \\in \\mathbb{Z}$, where $a, b$ and $c$ are certain constants.\n\na. Find a formula for $f(x)$, where $x$ is any integer.\n\nb. If $f(x) \\ge mx^2+(5m+1)x+4m$ for all nonnegative integers $x$, find the greatest possible value of $m$.",
    "options": [],
    "answer": "f(x) = x^3 + 4x^2 - 4; m = -1",
    "solution": "a.\nWe have $f(x) = x^3 + 4x^2 - 4$.\n\nLabel the equation as follows.\n$$\nf(x + y) = f(x) + f(y) + axy(x + y) + bxy + c(x + y) + 4 \\quad (1)\n$$\n\nPutting $x = y = 0$ in (1), we have\n$$\nf(0) = f(0) + f(0) + 4,\n$$\nwhich implies $f(0) = -4$. Next, putting $x = 1$ and $y = 0$ in (1), we have\n$$\nf(1) = f(1) + f(0) + c + 4.\n$$\nThis yields $c = 0$. Now, putting $x = y = 1$ in (1), we have\n$$\nf(2) = f(1) + f(1) + 2a + b + 4.\n$$\nAs it is given that $f(1) = 1$ and $f(2) = 20$, we obtain\n$$\n2a + b = 14. \\qquad (2)\n$$\nAlso, we put $x = y = 2$ in (1). This gives\n$$\nf(4) = f(2) + f(2) + 16a + 4b + 4.\n$$\nThis means $f(4) = 16a + 4b + 44$. Moreover, putting $x = 4$ and $y = -4$ in (1), we find that\n$$\nf(0) = f(4) + f(-4) - 16b + 4.\n$$\nThis becomes\n$$\n4a - 3b = -12. \\qquad (3)\n$$\nSolving (2) and (3), we have\n$$\na = \\frac{3(2a + b) + (4a - 3b)}{10} = 3\n$$\nand $b = 14 - 2a = 8$. Equation (1) becomes\n$$\nf(x + y) = f(x) + f(y) + 3xy(x + y) + 8xy + 4. \\qquad (4)\n$$\nPutting $y = 1$ in (4), we get\n$$\nf(x + 1) = f(x) + 3x(x + 1) + 8x + 5.\n$$\nFor $x \\in \\mathbb{Z}^{+}$, adding\n$$\n\\begin{aligned}\nf(x) &= f(x - 1) + 3(x - 1)x + 8(x - 1) + 5, \\\\\nf(x - 1) &= f(x - 2) + 3(x - 2)(x - 1) + 8(x - 2) + 5, \\\\\n&\\vdots, \\\\\nf(2) &= f(1) + 3(1)(2) + 8(1) + 5,\n\\end{aligned}\n$$\nwe obtain\n$$\n\\begin{aligned}\nf(x) &= f(1) + 3 \\sum_{k=1}^{x-1} k(k+1) + 8 \\sum_{k=1}^{x-1} k + 5(x-1) \\\\\n&= (x-1)x(x+1) + 4(x-1)x + 5x - 4 \\\\\n&= x^3 + 4x^2 - 4.\n\\end{aligned}\n$$\n\nThis also holds when $x = 0$ as $f(0) = -4$. Now, putting $y = -x$ in (4), we have\n$$\nf(0) = f(x) + f(-x) - 8x^2 + 4.\n$$\nFor $x < 0$, this yields\n$$\nf(x) = -f(-x) + 8x^2 - 8 = -((-x)^3 + 4(-x)^2 - 4) + 8x^2 - 8 = x^3 + 4x^2 - 4.\n$$\nSo the same formula holds for all $x \\in \\mathbb{Z}$.\nLastly, for $f(x) = x^3 + 4x^2 - 4$, we check that\n$$\n\\begin{aligned} f(x+y) &= (x+y)^3 + 4(x+y)^2 - 4 \\\\\n&= x^3 + 3x^2y + 3xy^2 + y^3 + 4x^2 + 8xy + 4y^2 - 4 \\end{aligned}\n$$\nand\n$$\n\\begin{aligned} f(x) + f(y) + 3xy(x+y) + 8xy + 4 \\\\\n&= x^3 + 4x^2 - 4 + y^3 + 4y^2 - 4 + 3xy(x+y) + 8xy + 4. \\end{aligned}\n$$\nIt is routine to check that the two sides are identical. Thus, this is the only solution.\n\nb.\nThe greatest possible value of $m$ is $-1$.\n\nNote that\n$$\n\\begin{align*}\n& x^3 + 4x^2 - 4 \\ge mx^2 + (5m + 1)x + 4m \\\\\n\\Leftrightarrow & x^3 + 4x^2 - x - 4 \\ge mx^2 + 5mx + 4m \\\\\n\\Leftrightarrow & (x+1)(x+4)(x-1) \\ge m(x+1)(x+4) \\\\\n\\Leftrightarrow & x - 1 \\ge m.\n\\end{align*}\n$$\nBy putting $x = 0$, we need $m \\le -1$. When $m = -1$, this inequality holds for all $x \\ge 0$. Thus, the greatest possible value of $m$ is $-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31510,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $\\{a_n\\}$ is a sequence in which all the terms are integers, and $a_2$ is odd. For any natural number $n$, $n(a_{n+1} - a_n + 3) = a_{n+1} + a_n + 3$. Furthermore, $a_{2009}$ is divisible by $2010$. Find the smallest integer $n$, $n \\ge 2$, such that $a_n$ is divisible by $2010$.",
    "options": [],
    "answer": "269",
    "solution": "The smallest possible integer is $269$.\n\nFor $n > 1$, we rewrite the recurrence relation as follows.\n$$\n\\begin{aligned}\nn(a_{n+1} - a_n + 3) &= a_{n+1} + a_n + 3 \\\\\n\\Rightarrow \\quad (n-1)a_{n+1} &= (n+1)a_n - 3(n-1) \\\\\n\\Rightarrow \\quad \\frac{a_{n+1}}{n(n+1)} &= \\frac{a_n}{n(n-1)} - \\frac{3}{n(n+1)}\n\\end{aligned}\n$$\nLet $b_n = \\frac{a_n}{(n-1)n}$. This gives $b_{n+1} = b_n - \\frac{3}{n(n+1)}$. Therefore, we easily obtain\n$$\n\\begin{aligned}\nb_n &= b_{n-1} - \\frac{3}{(n-1)n} \\\\\n&= b_{n-2} - \\frac{3}{(n-2)(n-1)} - \\frac{3}{(n-1)n} \\\\\n&= \\dots \\\\\n&= b_2 - \\sum_{k=2}^{n-1} \\frac{3}{k(k+1)} \\\\\n&= b_2 - 3 \\sum_{k=2}^{n-1} \\left( \\frac{1}{k} - \\frac{1}{k+1} \\right) \\\\\n&= b_2 - 3 \\left( \\frac{1}{2} - \\frac{1}{n} \\right).\n\\end{aligned}\n$$\nLet $a_2 = 2m + 1$. It follows that\n$$\na_n = (n-1)n \\left( \\frac{a_2}{2} - 3 \\left( \\frac{1}{2} - \\frac{1}{n} \\right) \\right) = (n-1)((m-1)n + 3).\n$$\nIt is given that $2010 \\mid a_{2009} = 2008(2009(m-1) + 3)$. This can be reduced to $1005 \\mid 2009(m-1)+3$. As $2009 \\equiv -1 \\pmod{1005}$, this yields $m \\equiv 4 \\pmod{1005}$. Let $m = 1005t + 4$. Then\n$$\na_n = (n-1)((1005t+3)n + 3) = 3(n-1)((335t+1)n + 1).\n$$\nNote that $2010 \\mid a_n$ if and only if $670 \\mid (n-1)((335t+1)n + 1)$. It suffices to consider odd $n$. Therefore, this becomes $670 \\mid (n-1)(n+1)$.\nNow, $670 = 2 \\times 5 \\times 67$. Therefore, $n \\equiv \\pm 1 \\pmod{67}$. It is routine to check that $670 \\div (n-1)(n+1)$ when $n = 133, 135, 267$. The next smallest possible $n$ is $n = 269$, where $670 \\mid 268 \\times 270$. So this is the answer.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31511,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\theta_1, \\theta_2, \\dots, \\theta_{2008}$ be real numbers. Find the maximum value of\n$$ \\sin \\theta_1 \\cos \\theta_2 + \\sin \\theta_2 \\cos \\theta_3 + \\dots + \\sin \\theta_{2007} \\cos \\theta_{2008} + \\sin \\theta_{2008} \\cos \\theta_1. $$",
    "options": [],
    "answer": "1004",
    "solution": "The maximum value is $1004$.\nNote that $\\sin \\theta_j \\cos \\theta_{j+1} \\le \\frac{1}{2}(\\sin^2 \\theta_j + \\cos^2 \\theta_{j+1})$ for any $j$, where the indices are taken modulo $2008$. It follows that\n$$\n\\sum_{j=1}^{2008} \\sin \\theta_j \\cos \\theta_{j+1} \\le \\frac{1}{2} \\sum_{j=1}^{2008} (\\sin^2 \\theta_j + \\cos^2 \\theta_j) = \\frac{1}{2} \\sum_{j=1}^{2008} 1 = 1004.\n$$\nEquality holds when $\\theta_j = \\frac{\\pi}{4}$ for each $j$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31512,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ points on the plane, no three of which are collinear. Each pair of points is joined by a red, yellow or green line. For any three points, the sides of the triangle they form consist of exactly two colours. Show that $n < 13$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to show that the case $n = 13$ is impossible since we can remove extra points. For the $j$th point, let $r_j, y_j, g_j$ be the numbers of lines having this point as an endpoint which are in red, yellow, and green respectively.\n\nFor any $\\triangle XYZ$, WLOG assume $XY$ and $XZ$ are red. Then this triangle corresponds to two red lines from $X$, and this correspondence is one-to-one. It follows that the number of triangles is\n$$\n\\sum_{j=1}^{13} \\left[ \\binom{r_j}{2} + \\binom{y_j}{2} + \\binom{g_j}{2} \\right].\n$$\nBut then it is clear that there are $\\binom{13}{3} = 286$ triangles. By the pigeonhole principle, WLOG assume\n$$\n\\binom{r_1}{2} + \\binom{y_1}{2} + \\binom{g_1}{2} \\ge \\frac{286}{13} = 22.\n$$\nThis gives $r_1(r_1 - 1) + y_1(y_1 - 1) + g_1(g_1 - 1) \\ge 44$. Since $r_1 + y_1 + g_1 = 12$, this implies $r_1^2 + y_1^2 + g_1^2 \\ge 56$.\nWe claim that one of $r_1, y_1, g_1$ is at least 6. Suppose on the contrary that $r_1, y_1, g_1 \\le 5$. As $r_1+y_1+g_1 = 12$, it suffices to check $(r_1, y_1, g_1) = (5, 5, 2), (5, 4, 3), (4, 4, 4)$. In all cases, $r_1^2 + y_1^2 + g_1^2 \\ge 56$ does not hold.\nWLOG assume $r_1 \\ge 6$. Suppose $AB, AC, AD, AE, AF, AG$ are red. By assumption, none of the lines formed by $B, C, D, E, F, G$ is red. Since $R(3,3) = 6$, there must be a yellow or green triangle among these 6 points, contradiction. This proves $n < 13$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31513,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the total number of solutions to the following system of equations:\n$$\n\\begin{cases}\na^2 + bc \\equiv a \\pmod{37}, \\\\\nb(a+d) \\equiv b \\pmod{37}, \\\\\nc(a+d) \\equiv c \\pmod{37}, \\\\\nbc + d^2 \\equiv d \\pmod{37}, \\\\\nad - bc \\equiv 1 \\pmod{37}.\n\\end{cases}\n$$",
    "options": [],
    "answer": "1",
    "solution": "There is only 1 solution modulo $37$.\nThe first four equations can be rewritten as $\\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}^2 = \\begin{pmatrix} a & b \\\\ c & d \\end{pmatrix}$, where the entries are taken modulo $37$. The last equation means $\\begin{vmatrix} a & b \\\\ c & d \\end{vmatrix} = 1$. Therefore, we need to count the number of $2 \\times 2$ matrices $A$ with entries in $\\mathbb{F}_{37}$ such that $A^2 = A$ and $\\det A = 1$.\nSince $\\det A = 1 \\ne 0$, $A$ is invertible. Thus, we multiply both sides of $A^2 = A$ by $A^{-1}$ to obtain $A = I$. Clearly, the identity matrix $I$ satisfies the conditions. So $A = I$ is the only solution. Equivalently, this means $a \\equiv d \\equiv 1 \\pmod{37}$ and $b \\equiv c \\equiv 0 \\pmod{37}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31514,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a right-angled triangle with $\\angle C = 90^\\circ$. $CD$ is the altitude from $C$ to $AB$, with $D$ on $AB$. $\\omega$ is the circumcircle of $\\triangle BCD$. $\\omega_1$ is a circle situated in $\\triangle ACD$, which is tangent to the segments $AD$ and $AC$ at $M$ and $N$ respectively, and is also tangent to circle $\\omega$.\n(i) Show that $BD \\cdot CN + BC \\cdot DM = CD \\cdot BM$.\n(ii) Show that $BM = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(i) This is an immediate consequence of Casey's theorem. We provide an elementary proof as follows assuming the result in part (ii) (which will be proved later).\nLet $a = BC$, $b = CA$ and $c = AB$. By part (ii), we have $BM = BC = a$. Then we have\n$$\n\\begin{aligned}\nAN &= AM = c - a, \\\\\nCN &= b - AN = a + b - c, \\\\\nDM &= AD - AM = \\frac{b^2}{c} - c + a.\n\\end{aligned}\n$$\nIt follows that\n$$\n\\begin{aligned}\n&BD \\cdot CN + BC \\cdot DM - CD \\cdot BM \\\\\n&= \\frac{a^2}{c}(a+b-c) + a\\left(\\frac{b^2}{c} - c + a\\right) - \\frac{ab}{c} \\cdot a \\\\\n&= \\frac{a}{c}(a^2 + ab - ac + b^2 - c^2 + ac - ab) = 0\n\\end{aligned}\n$$\nusing Pythagoras' theorem. This proves the equality.\n\n![](attached_image_1.png)\n\n(ii) Let $O$ and $O'$ be the centres of $\\omega$ and $\\omega_1$ respectively. Note that $BC$ is a diameter of $\\omega$ since $\\angle BDC = 90^\\circ$. Therefore, the tangent at $B$ to $\\omega$ is parallel to the tangent at $N$ to $\\omega_1$. By homothety, we know that $B, T, N$ are collinear. As $\\angle BTC = 90^\\circ$, we have $\\triangle BTC \\sim \\triangle BCN$. Therefore,\n$$\nBC^2 = BT \\times BN = BM^2.\n$$\nThis gives $BC = BM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31515,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all nonnegative integers $m$ and $n$ that satisfy the equation:\n$$\n107^{56}(m^2 - 1) - 2m + 3 = \\binom{113^{114}}{n}.\n$$\n(If $n$ and $r$ are nonnegative integers satisfying $r \\le n$, then $\\binom{n}{r} = C_r^n = \\frac{n!}{r!(n-r)!}$ and $\\binom{n}{r} = 0$ if $r > n$.)",
    "options": [],
    "answer": "m = 1, n = 0 or n = 113^{114}",
    "solution": "The solutions are $(m, n) = (1, 0), (1, 113^{114})$.\nWhen $n = 0$ or $n = 113^{114}$, the equation becomes $107^{56}(m^2 - 1) - 2m + 3 = 1$. This implies\n$$\n(m - 1)(107^{56}(m + 1) - 2) = 0.\n$$\nClearly, $107^{56}(m + 1) - 2 = 0$ has no solution. Thus, the only solution is $m = 1$.\n\nWhen $0 < n < 113^{114}$, by Lucas' theorem, $113$ divides $\\binom{113^{114}}{n}$. This also holds when $n > 113^{114}$ since the binomial coefficient is 0. Therefore, we have\n$$\n107^{56}(m^2 - 1) - 2m + 3 \\equiv 0 \\pmod{113}. \\qquad (1)\n$$\nLet $g$ be a primitive root modulo $113$, and let $107 \\equiv g^a \\pmod{113}$. Using the Legendre symbol and the quadratic reciprocity law, we find that\n$$\n\\left(\\frac{107}{113}\\right) = \\left(\\frac{-6}{113}\\right) = \\left(\\frac{-1}{113}\\right)\\left(\\frac{2}{113}\\right)\\left(\\frac{3}{113}\\right) = (1)(1)\\left(\\frac{113}{3}\\right) = \\left(\\frac{2}{3}\\right) = -1.\n$$\nThis shows $107$ is a quadratic nonresidue modulo $113$, and so $a$ is odd. Now, note that\n$$\n107^{56} \\equiv g^{56a} \\equiv -1 \\pmod{113}\n$$\nsince $(g^{56a})^2 = g^{112a} = g^{\\varphi(113)a} \\equiv 1 \\pmod{113}$ and $112 + 56a$. Therefore, (1) becomes\n$$\n-(m^2 - 1) - 2m + 3 \\equiv 0 \\pmod{113}.\n$$\n\nThis is the same as $(m+1)^2 \\equiv 5 \\pmod{113}$. However, we check that\n$$\n\\left(\\frac{5}{113}\\right) = \\left(\\frac{113}{5}\\right) = \\left(\\frac{3}{5}\\right) = -1,\n$$\nwhich implies 5 is a quadratic nonresidue modulo 113. This is a contradiction. Therefore, there is no solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31516,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are $2008$ students. Students are members of certain committees. A committee has at most $1004$ members and every two students join a common committee.\n\na. Determine the smallest possible number of committees in the school.\n\nb. If it is further required that the union of any two committees consists of at most $1800$ students, will your answer in (a) still hold?",
    "options": [],
    "answer": "a) 6; b) Yes",
    "solution": "a. The smallest number of committees in the school is $6$.\nIf a student joins at most $2$ committees, that student shares a common committee with at most $2(1004 - 1) = 2008 < 2007$ students, which contradicts the assumption. Therefore, each student joins at least $3$ committees. Thus, there are at least\n$$\n\\frac{3 \\times 2008}{1004} = 6\n$$\ncommittees.\nThe minimum value can be attained. For example, partition all students into $8$ groups $A, B, \\dots, H$, each consisting of $251$ students. We form the following $6$ committees:\n$$\n\\{A, B, C, D\\}, \\{A, E, F, G\\}, \\{A, B, E, H\\}, \\{B, F, G, H\\}, \\{C, D, G, H\\}, \\{C, D, E, F\\}.\n$$\nEach committee contains $251 \\times 4 = 1004$ students. One can check that every pair of groups belongs to at least one common committee, and so every pair of students joins a common committee.\n\nb. Yes. One can check that every pair of committees in the example provided in part (a) consists of at most $7$ groups only, and so the union consists of at most\n$$\n251 \\times 7 = 1757 < 1800\n$$\nstudents.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31517,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that the equation $y^{37} \\equiv x^3 + 11 \\pmod{p}$ is solvable for every prime $p$, where $p \\le 100$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $p \\le 100$, we have $(p - 1, 37) = 1$ (by checking $1, 38, 75$ are not primes). Thus, there exists $k \\in \\mathbb{Z}^+$ such that $37k \\equiv 1 \\pmod{p-1}$. Thus, by the Fermat little theorem, we have\n$$\n(11^k)^{37} = 11^{37k} \\equiv 11 \\pmod{p}.\n$$\nTherefore, $(x, y) = (0, 11^k)$ is a solution to the given system.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31518,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n(i) the set $\\left\\{ \\frac{f(x)}{x} \\mid x \\ne 0 \\right\\}$ is finite,\n(ii) $f(3x - 1 - f(x)) = 3(f(x) - 1 - 3x)$ for all $x \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) = 3x",
    "solution": "The only solution is $f(x) = 3x$.\nWhen $f(x) = 3x$, we have\n$$\nf(3x - 1 - f(x)) = f(-1) = -3 = 3(-1) = 3(f(x) - 1 - 3x).\n$$\nAlso, $\\frac{f(x)}{x} = 3$ is a constant. So $f(x) = 3x$ is a solution.\n\nNow, we show that this is the only solution. Let $g(x) = 3x - 1 - f(x)$ for any $x \\in \\mathbb{R}$. If $g(x) = 0$ and $x \\neq 0$, then $f(x) = 3x - 1$ and so $\\frac{f(x)}{x} = 3 - \\frac{1}{x}$. By condition (i), there are finitely many $x$ such that $g(x) = 0$.\nFor those $x \\in \\mathbb{R}$ with $g(x) \\neq 0$, we rewrite condition (ii) as follows:\n$$\n\\frac{f(g(x))}{g(x)} = \\frac{3(f(x) - 1 - 3x)}{3x - 1 - f(x)} = -3 - \\frac{6}{3x - 1 - f(x)} = -3 - \\frac{6}{g(x)}.\n$$\nBy condition (i), the left-hand side only takes finitely many values. This implies $g(x)$ only takes finitely many values (possibly 0).\nConsider any $y \\in \\mathbb{R}$ such that $|f(y) - 3y| = |g(y) + 1|$ is maximized. For this $y$, we use (ii) to obtain\n$$\nf(g(y)) - 3g(y) = 3(f(y) - 1 - 3y) - 3(3y - 1 - f(y)) = 6(f(y) - 3y).\n$$\nThis implies $|f(g(y)) - 3g(y)| = 6|f(y) - 3y|$. By the choice of $y$, we must have $f(y) - 3y = 0$. Since $y$ is chosen to maximize $|f(y) - 3y|$, we can only have $f(x) = 3x$ for any $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31519,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a chess tournament there were $n$ ($n > 10$) participants. Each participant plays against another exactly once. If a game ends in a draw, each participant gets 1 point, otherwise the winner gets 2 points and the loser gets none. At the end of the tournament, every participant found that half of his/her score came from games played against the ten participants with the lowest scores. This is true even for each of the ten participants with the lowest scores, (i.e. half of the scores of each of the 10 lowest scorers comes from playing with other 9 lowest scorers). Find the possible value(s) of $n$.",
    "options": [],
    "answer": "25",
    "solution": "$n$ can only be 25.\n\nWe first provide an example for the case $n = 25$. Consider 10 participants $A_1, A_2, \\dots, A_{10}$ in group $\\mathcal{A}$, and 15 participants $B_1, B_2, \\dots, B_{15}$ in group $\\mathcal{B}$. Suppose the games played between 2 participants in the same group end in a draw. For the games played between 2 participants in different groups, the results are given below, where $O$ means a draw, while an empty box means the group $\\mathcal{B}$ participant wins.\n\n|        | A$_1$ | A$_2$ | A$_3$ | A$_4$ | A$_5$ | A$_6$ | A$_7$ | A$_8$ | A$_9$ | A$_{10}$ |\n|--------|-------|-------|-------|-------|-------|-------|-------|-------|-------|----------|\n| B$_1$  | O     | O     | O     | O     | O     | O     |       |       |       |          |\n| B$_2$  | O     | O     | O     | O     | O     | O     |       |       |       |          |\n| B$_3$  |       | O     | O     | O     | O     | O     | O     |       |       |          |\n| B$_4$  |       |       | O     | O     | O     | O     | O     | O     |       |          |\n| B$_5$  |       |       | O     | O     | O     | O     | O     | O     |       |          |\n| B$_6$  |       |       |       | O     | O     | O     | O     | O     | O     |          |\n| B$_7$  |       |       |       |       | O     | O     | O     | O     | O     | O        |\n| B$_8$  |       |       |       |       | O     | O     | O     | O     | O     | O        |\n| B$_9$  | O     |       |       |       |       | O     | O     | O     | O     | O        |\n| B$_{10}$ | O   | O     |       |       |       |       | O     | O     | O     | O        |\n| B$_{11}$ | O   | O     |       |       |       | O     | O     | O     | O     | O        |\n| B$_{12}$ | O   | O     | O     |       |       |       | O     | O     | O     | O        |\n| B$_{13}$ | O   | O     | O     | O     |       |       |       | O     | O     | O        |\n| B$_{14}$ | O   | O     | O     | O     | O     |       |       | O     | O     | O        |\n| B$_{15}$ | O   | O     | O     | O     | O     | O     |       |       | O     | O        |\n\nThe score of each participant in group $\\mathcal{A}$ is $9+9=18$, where 9 points come from the games played between them. The score of each participant in group $\\mathcal{B}$ is $14+(6+2\\times4)=28$, where 14 points come from the games played between them. So the conditions are satisfied.\n\nNow, note that the two participants in a game always gain 2 points in total. Therefore, the total score of all participants is\n$$\n2\\binom{n}{2} = n(n-1).\n$$\n\nFor the 10 participants with the lowest scores, their total score gained by the games played between them is $2\\binom{10}{2} = 90$. Therefore, their total score is\n$$\n2 \\times 90 = 180.\n$$\nFor the other $n-10$ participants, their total score gained by the games played between them is $2\\binom{n-10}{2} = (n-10)(n-11)$. Note that this is half of their total score since the remaining half comes from the other games. Therefore, their total score is\n$$\n2(n - 10)(n - 11).\n$$\n\nCombining all these, we need $180+2(n-10)(n-11) = n(n-1)$. This is the same as $(n-16)(n-25) = 0$. Thus, $n = 16, 25$.\n\nFor $n = 16$, the 10 participants with the lowest scores obtain 180 points in total, while the other 6 participants obtain $2(16 - 10)(16 - 11) = 60$ points in total. Since $\\frac{180}{10} = 18 > 10 = \\frac{60}{6}$, it is impossible that all those 10 participants have the lowest score. Therefore, $n = 16$ is rejected, and $n = 25$ is the only possibility.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31520,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be the sides of a triangle, and $T$ its area. Prove that\n$$\na^2 + b^2 + c^2 \\ge 4\\sqrt{3}T + (a-b)^2 + (b-c)^2 + (c-a)^2.\n$$\n\nWhen does the equality hold?",
    "options": [],
    "answer": "a = b = c",
    "solution": "Let $s$ be the semiperimeter of the triangle, and let $x = s - a$, $y = s - b$ and $z = s - c$. Note that $x$, $y$, $z > 0$ by the triangle inequality. Now,\n$$\n\\begin{align*}\n& a^2 + b^2 + c^2 \\ge 4\\sqrt{3T} + (a-b)^2 + (b-c)^2 + (c-a)^2 \\\\\n\\Leftrightarrow & \\quad 2ab + 2bc + 2ca \\ge 4\\sqrt{3s(s-a)(s-b)(s-c)} + a^2 + b^2 + c^2 \\\\\n\\Leftrightarrow & \\quad 2 \\sum_{\\text{cyc}} (y+z)(x+z) \\ge 4\\sqrt{3(x+y+z)xyz} + (y+z)^2 + (z+x)^2 + (x+y)^2 \\\\\n\\Leftrightarrow & \\quad 4(xy + yz + zx) \\ge 4\\sqrt{3(x+y+z)xyz} \\\\\n\\Leftrightarrow & \\quad (xy + yz + zx)^2 \\ge 3((xy)(yz) + (yz)(zx) + (zx)(xy)).\n\\end{align*}\n$$\n\nThis is a well-known inequality (for example, the difference of the left-hand side and the right-hand side is $\\frac{1}{2} \\sum_{\\text{cyc}} (xy - yz)^2$). Equality holds when $xy = yz = zx$, i.e. $a = b = c$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31521,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Determine the number of sequences $x_1, x_2, \\dots, x_{2n-1}, x_{2n}$, with $x_i = 1$ or $-1$, and such that $\\left|\\sum_{i=2k-1}^{2m} x_i\\right| \\le 2$ for all integers $k, m$, with $1 \\le k \\le m \\le n$.",
    "options": [],
    "answer": "2*3^n - 2^n",
    "solution": "The answer is $2 \\cdot 3^n - 2^n$.\nLet $y_j = x_{2j-1} + x_{2j}$ for $j = 1, 2, \\dots, n$. Every desired sequence $\\{x_j\\}$ corresponds to a sequence $\\{y_j\\}$ such that $y_j = -2, 0, 2$ for each $j$, and $|y_k + y_{k+1} + \\dots + y_m| \\le 2$ for any $1 \\le k \\le m \\le n$. Suppose there are $r$ terms of $\\{y_j\\}$ which are nonzero, say $y_{i_1}, y_{i_2}, \\dots, y_{i_r}$ where $i_1 < i_2 < \\dots < i_r$. Since\n$$\n|y_{i_1} + y_{i_1+1} + \\dots + y_{i_2}| = |y_{i_1} + y_{i_2}| \\le 2,\n$$\nwe must have $y_{i_2} = -y_{i_1}$. Similarly, $y_{i_3} = -y_{i_2} = y_{i_1}$, and inductively $y_{i_{2j-1}} = y_{i_1}$ and $y_{i_{2j}} = -y_{i_1}$ for any integer $j$. In other words, the sequence $\\{y_j\\}$ only depends on the indices $i_1, i_2, \\dots, i_r$ and the term $y_{i_1}$.\nConversely, for any such sequence $\\{y_j\\}$, consider any $1 \\le k \\le m \\le n$. Let $i_s$ be the smallest index such that $k \\le i_s$, and let $i_t$ be the largest index such that $i_t \\le m$. Then\n$$\ny_k + y_{k+1} + \\dots + y_m = y_{i_s} + y_{i_{s+1}} + \\dots + y_{i_t} = \\pm(2 - 2 + 2 - 2 + \\dots) = -2, 0, 2.\n$$\nSo the condition is satisfied.\nNow, for each sequence $\\{y_j\\}$ satisfying the condition, we can construct the sequence $\\{x_j\\}$ as follows. For any index $j$ of the form $i_u$, the terms $x_{2j-1}$ and $x_{2j}$ are uniquely determined. Indeed, if $y_{i_u} = \\pm 2$, then $x_{2i_u-1} = x_{2i_u} = \\pm 1$. For any other index $j$, there are two choices for $x_{2j-1}$ and $x_{2j}$, since exactly one of them is 1 and the other is -1. Since there are $\\binom{n}{r}$ ways to choose the indices $i_1, i_2, \\dots, i_r$ and two ways to choose $y_{i_1}$ if $r \\ge 1$, the total number of sequences is\n$$\n\\binom{n}{0}2^n + \\sum_{r=1}^{n} 2\\binom{n}{r} \\cdot 2^{n-r} = 2^n + 2\\sum_{r=1}^{n} \\binom{n}{n-r} 2^{n-r} = 2^n + 2\\sum_{d=0}^{n-1} \\binom{n}{d} 2^d\n$$\nusing the change of variable $d = n-r$. By the binomial theorem, this is equal to\n$$\n2^n + 2 \\sum_{d=0}^{n} \\binom{n}{d} 2^d - 2^{n+1} = 2^n + 2(2+1)^n - 2^{n+1} = 2 \\cdot 3^n - 2^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31522,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $abc = 1$. Show that\n$$\n\\frac{1}{a^2(b+c)} + \\frac{1}{b^2(c+a)} + \\frac{1}{c^2(a+b)} \\ge \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x = bc$, $y = ca$ and $z = ab$. The condition becomes $xyz = 1$. Note that\n$$\n\\sum_{\\text{cyc}} \\frac{1}{a^2(b+c)} = \\sum_{\\text{cyc}} \\frac{bc}{a(b+c)} = \\sum_{\\text{cyc}} \\frac{x}{y+z}.\n$$\n\nIt remains to prove\n$$\n\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\ge \\frac{3}{2}.\n$$\n\nThis is exactly Nesbitt's inequality. Equality holds when $x = y = z = 1$, i.e. $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31523,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x^2 - p_i x + q_i = 0$, $i = 1, 2, \\dots, 15$ be 15 quadratic equations, where each of the $p_i$ and $q_i$, $i = 1, 2, \\dots, 15$ is one of the numbers $1, 2, \\dots, 30$, with no repetition nor omission. Determine the maximum possible number of real roots among the equations which are of values greater than 20.",
    "options": [],
    "answer": "10",
    "solution": "The answer is 10.\n\nLet $a \\le b$ be two roots of $x^2 - px + q = 0$ such that $b > 20$. Note that\n$$\na \\le \\frac{a+b}{2} = \\frac{p}{2} \\le 15.\n$$\n\nThis shows it is impossible to have both roots greater than 20. Also, we have\n$$\nb = \\frac{p + \\sqrt{p^2 - 4q}}{2} < \\frac{p+p}{2} = p.\n$$\n\nThis implies $p \\ge 21$, and so $p$ can only be 21, 22, \\dots, 30. Thus, at most 10 equations may have a root greater than 20.\n\nConsider the equations $x^2 - (20+k)x + k = 0$ for $k = 1, 2, \\dots, 10$. Note that\n$$\n\\Delta = (20 + k)^2 - 4k = k^2 + 36k + 400 > 0,\n$$\n\nso the roots are real. The larger root is\n$$\n\\frac{(20 + k) + \\sqrt{k^2 + 36k + 400}}{2} > \\frac{(20 + k) + (20 - k)}{2} = 20.\n$$\n\nThis shows it is possible to have 10 roots larger than 20.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31524,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $2009$ points in the plane so that no three points lie on a line, show that there exist at least $1344021$ triangles, each of which is formed by any three of the $2009$ points, and each of the triangles does not contain any of the remaining $2006$ points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Fix one of the points $A$. We label the other points $B_1, B_2, \\dots, B_{2008}$ such that the rays $AB_1, AB_2, \\dots, AB_{2008}$ align in the clockwise direction. Then it is easy to see that among\n$$\n\\triangle AB_1B_2, \\triangle AB_2B_3, \\dots, \\triangle AB_{2007}B_{2008}, \\triangle AB_{2008}B_1,\n$$\nat most one triangle may contain some other points. Indeed, this only happens for $\\triangle AB_jB_{j+1}$ when the clockwise angle from $AB_j$ to $AB_{j+1}$ is larger than $180^\\circ$.\n\nIt follows that there are at least $2007$ triangles with $A$ as a vertex which do not contain any other points. By considering all choices of $A$, since each triangle is counted at most $3$ times, there are at least\n$$\n\\frac{1}{3} \\times 2009 \\times 2007 = 1344021\n$$\nsuch triangles.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31525,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $a$ be an integer with $\\text{gcd}(a, n) = 1$. Prove that\n$$\n\\frac{a\\varphi(n) - 1}{n} \\equiv \\sum_{\\substack{j=1 \\\\ \\text{gcd}(j,n)=1}}^{n} \\frac{1}{aj} \\left[ \\frac{aj}{n} \\right] \\pmod{n},\n$$\nwhere $\\varphi(n)$ is the Euler's totient function and $[x]$ is the greatest integer not exceeding $x$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each integer $j$ relatively prime to $n$, let $r_j$ be the remainder when $aj$ is divided by $n$. Then we have\n$$\n\\begin{aligned}\na^{\\varphi(n)} &= \\prod_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\frac{aj}{j} = \\prod_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\frac{1}{j} \\left( \\left\\lfloor \\frac{aj}{n} \\right\\rfloor n + r_j \\right) \\\\\n&= \\prod_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\frac{r_j}{j} \\left( 1 + \\frac{n}{r_j} \\left\\lfloor \\frac{aj}{n} \\right\\rfloor \\right) = \\prod_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\left( 1 + \\frac{n}{r_j} \\left\\lfloor \\frac{aj}{n} \\right\\rfloor \\right)\n\\end{aligned}\n$$\nsince $\\{r_1, r_2, \\dots, r_{\\varphi(n)}\\}$ is a reduced set of residues modulo $n$. Expanding the product and taking modulo $n^2$, we obtain\n$$\na^{\\varphi(n)} \\equiv 1 + n \\sum_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\frac{1}{r_j} \\left\\lfloor \\frac{aj}{n} \\right\\rfloor \\equiv 1 + n \\sum_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\frac{1}{aj} \\left\\lfloor \\frac{aj}{n} \\right\\rfloor \\pmod{n^2}.\n$$\nThis implies\n$$\n\\frac{a^{\\varphi(n)} - 1}{n} \\equiv \\sum_{\\substack{j=1 \\\\ (j,n)=1}}^{n} \\frac{1}{aj} \\left\\lfloor \\frac{aj}{n} \\right\\rfloor \\pmod{n}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31526,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many odd coefficients are there in the expansion of $(x^2 - x + 1)^{2009}$?",
    "options": [],
    "answer": "645",
    "solution": "In the following, we consider polynomials in $\\mathbb{F}_2[x]$. This means all coefficients are taken modulo $2$. Let $f(n)$ be the number of odd coefficients in $(x^2 - x + 1)^n$.\n\n**Claim 1.** We have $(x^2 - x + 1)^{2k} = x^{2k+1} - x^{2k} + 1$ for any nonnegative integer $k$.\n*Proof.* It suffices to note that\n$$\n(x^2 - x + 1)^2 = x^4 - 2x^3 + 3x^2 - 2x + 1 = x^4 - x^2 + 1.\n$$\nThe result follows easily by induction. $\\square$\n\n**Claim 2.** We have $f(2ka+b) = f(a)f(b)$ for any positive integers $a, b$ satisfying $b < 2^{k-1}$.\n*Proof.* By claim 1, we have\n$$\n(x^2 - x + 1)^{2ka+b} = ((x^2 - x + 1)^{2k})^a (x^2 - x + 1)^b = (x^{2k+1} - x^{2k} + 1)^a (x^2 - x + 1)^b.\n$$\n$$\n(x^{2k+1} - x^{2k} + 1)^a = \\sum_{i=1}^{s} c_i x^{2k \\alpha_i},\n$$\n$$\n(x^2 - x + 1)^b = \\sum_{j=1}^{t} d_j x^{\\beta_j}.\n$$\nThen their product is\n$$\n\\sum_{i=1}^{s} \\sum_{j=1}^{t} c_i d_j x^{2k\\alpha_i + \\beta_j}.\n$$\nSince we have $\\beta_j \\le 2b < 2^k$, different pairs $(i, j)$ correspond to different exponents $2^k\\alpha_i + \\beta_j$. Also, $c_i d_j$ is odd if and only if both $c_i$ and $d_j$ are odd. As there are $f(a)$ odd coefficients $c_i$ and $f(b)$ odd coefficients $d_j$, the number of odd coefficients is $f(a)f(b)$. $\\square$\n\n**Claim 3.** We have $f(2^k - 1) = \\frac{2^{k+2} - (-1)^k}{3}$ for any positive integer $k$.\n*Proof.* It suffices to make the following observations. For odd $k$, the coefficients of $(x^2 - x + 1)^{2^k-1}$ follow the pattern\n$$\n\\underbrace{110110\\cdots110}_{\\frac{2^{k-2}}{3} \\text{ triples } 110} 111 \\underbrace{011011\\cdots011}_{\\frac{2^{k-2}}{3} \\text{ triples } 011}\n$$\nwhile for even $k$, the coefficients of $(x^2 - x + 1)^{2^k-1}$ follow the pattern\n$$\n\\underbrace{110110\\cdots110}_{\\frac{2^{k-1}}{3} \\text{ triples } 110} 1 \\underbrace{011011\\cdots011}_{\\frac{2^{k-1}}{3} \\text{ triples } 011}.\n$$\nIt is not hard to prove these by induction and claim 1, since $2^{k+1}-1 = (2^k-1)2^k$. We omit the details. $\\square$\n\nNow, by the above claims, we obtain\n$$\n\\begin{aligned}\nf(2009) &= f(2^6 \\times 31 + 25) = f(31)f(25) \\\\\n&= f(31)f(2^3 \\times 3 + 1) = f(31)f(3)f(1) \\\\\n&= \\frac{2^7 + 1}{3} \\cdot \\frac{2^4 - 1}{3} \\cdot \\frac{2^3 + 1}{3} = 645.\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31527,
    "subject": "Mathematics (Multi-modal)",
    "question": "A rectangle with sides of integral lengths can be divided into two or more squares with sides of integral lengths and such that there is exactly one square with smallest side length. Determine the smallest possible area of the rectangle.",
    "options": [],
    "answer": "35",
    "solution": "The smallest possible area is $35$.\n\nLet $a$ be the length of the smallest square $A$. If $A$ touches a side but not a corner, then it is surrounded by two larger squares $B$ and $C$. Then there is a gap between $B$ and $C$ with length $a$. It is not possible to place some larger squares in this gap, contradiction. Similarly, the smallest square cannot be placed at a corner.\n![](attached_image_1.png)\n\nThe distance from $A$ to a side is at least $a+1$. Therefore, each side has length at least\n$$\n(a+1) \\times 2 + a = 3a + 2 \\ge 5.\n$$\n\nIf both side lengths are at least $6$, then the area is at least $36$. Thus, we may assume there is a side with length $5$, and $a=1$.\nSince there is no square of side length $1$ touching this side, either there is a $5 \\times 5$ square, or a $3 \\times 3$ and a $2 \\times 2$ squares touching this side. The former case can be ignored since we can remove such a $5 \\times 5$ square. Now, suppose there is a $3 \\times 3$ square $B$ and a $2 \\times 2$ square $C$ as shown. Then $A$ cannot be placed in the first $3$ columns (or there is a gap of length $1$). The same holds for the opposite side. Therefore, the other side length of the rectangle is at least $3+1+3=7$. This shows the area is at least $5 \\times 7 = 35$.\n\nAn example of a $5 \\times 7$ rectangle is given below.\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31528,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive real numbers $a$, $b$, $c$, prove that\n$$\n(ab + bc + ca) \\left( \\frac{1}{(a+b)^2} + \\frac{1}{(b+c)^2} + \\frac{1}{(c+a)^2} \\right) \\ge \\frac{9}{4}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(Iran TST 1996) Since the expressions are homogeneous, WLOG we may assume $ab + bc + ca = 1$. We need to minimize\n$$\nf(a, b, c) = \\frac{1}{(a+b)^2} + \\frac{1}{(b+c)^2} + \\frac{1}{(c+a)^2}.\n$$\nWLOG assume $a \\ge b \\ge c$. Let $t$ be the positive root to $t^2 + 2ct = 1$, so that $(t)(t) + (t)(c) + (c)(t) = 1$, which means the triple $(t, t, c)$ also satisfies the constraint.\n**Claim.** We have $f(a, b, c) \\ge f(t, t, c)$.\n*Proof.* Firstly, we have\n$$\nf(t, t, c) = \\frac{1}{4t^2} + \\frac{2}{(t+c)^2} = \\frac{1}{4t^2} + \\frac{2}{t^2 + 2ct + c^2}.\n$$\nFor the latter term, observe that $t^2 + 2ct = 1 = ab + bc + ca$. Therefore, we obtain\n$$\n\\frac{2}{t^2 + 2ct + c^2} = \\frac{2}{ab + bc + ca + c^2} = \\frac{2}{(c+a)(c+b)}.\n$$\nFor the former term, we claim that $t^2 \\ge ab$. Suppose on the contrary that $t^2 < ab$. Then we also have $a+b \\ge 2\\sqrt{ab} > 2t$. But this implies\n$$\nab + bc + ca = ab + c(a + b) > t^2 + 2ct,\n$$\nwhich is a contradiction. Therefore, $t^2 \\ge ab$, and hence\n$$\n\\frac{1}{4t^2} \\le \\frac{1}{4ab}.\n$$\nNow, it suffices to show $f(a, b, c) \\ge \\frac{1}{4ab} + \\frac{2}{(c+a)(c+b)}$. Indeed,\n$$\n\\begin{align*}\nf(a, b, c) &\\ge \\frac{1}{4ab} + \\frac{2}{(c+a)(c+b)} \\\\\n\\Leftrightarrow \\quad &\\frac{1}{(b+c)^2} + \\frac{1}{(c+a)^2} - \\frac{2}{(c+a)(c+b)} \\ge \\frac{1}{4ab} - \\frac{1}{(a+b)^2} \\\\\n\\Leftrightarrow \\quad &\\frac{((c+a) - (c+b))^2}{(c+a)^2(c+b)^2} \\ge \\frac{(a+b)^2 - 4ab}{4ab(a+b)^2} \\\\\n\\Leftrightarrow \\quad &\\frac{(a-b)^2}{(c+a)^2(c+b)^2} \\ge \\frac{(a-b)^2}{4ab(a+b)^2}.\n\\end{align*}\n$$\nThis is true since\n$$\n(c+a)^2(c+b)^2 \\le (b+a)^2(b+b)^2 = 4b^2(a+b)^2 \\le 4ab(a+b)^2\n$$\nusing the assumption $a \\ge b \\ge c$. This proves the claim. $\\square$\n\nBy the claim, it remains to prove $f(t, t, c) \\ge \\frac{9}{4}$ for any positive real numbers $t$ and $c$ satisfying $t^2 + 2ct = 1$. Now, using $c = \\frac{1-t^2}{2t}$,\n$$\n\\begin{align*}\nf(t, t, c) &\\ge \\frac{9}{4} \\\\\n\\Leftrightarrow \\quad \\frac{1}{4t^2} + \\frac{2}{(t+c)^2} &\\ge \\frac{9}{4} \\\\\n\\Leftrightarrow \\quad \\frac{1}{4t^2} + \\frac{8t^2}{(1+t^2)^2} &\\ge \\frac{9}{4} \\\\\n\\Leftrightarrow \\quad \\frac{-9t^6 + 15t^4 - 7t^2 + 1}{4t^2(1+t^2)^2} &\\ge 0 \\\\\n\\Leftrightarrow \\quad \\frac{(1-t^2)(1-3t^2)^2}{4t^2(1+t^2)^2} &\\ge 0.\n\\end{align*}\n$$\nAs $1 = t^2 + 2ct > t^2$, this inequality is true. Equality holds when $t = \\frac{1}{3}$, i.e. $a = b = c$. (Note that if we allow at most one $a$, $b$, $c$ to be $0$, equality also holds when $t = 1$, i.e. $a = b$ and $c = 0$ up to permutation.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31529,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. When $\\frac{n}{567}$ is expressed as a decimal, it is a recurring decimal with smallest period $k$. Find the sum of all possible values of $k$. (Note: We say that $k$ is a *period* of a recurring decimal if starting from some place the digits of the decimal repeat every $k$ digits. For instance, $0.12345345345\\dots$ has smallest period 3 while $0.142857142587142587\\dots$ has smallest period 6.)",
    "options": [],
    "answer": "37",
    "solution": "The answer is $37$.\n\nWe first prove the following result.\n**Claim.** Consider a positive rational number $\\frac{a}{b}$ where $(10a, b) = 1$. The smallest period of this number when expressed as a decimal number is the order of $10$ modulo $b$.\n*Proof.* Note that $\\frac{a}{b}$ has period $k$ if and only if the tail part of $10^k \\cdot \\frac{a}{b}$ is the same as that of $\\frac{a}{b}$. This holds if and only if $(10^k - 1) \\cdot \\frac{a}{b}$ is a terminating decimal, i.e. $10^m(10^k - 1) \\frac{a}{b} \\in \\mathbb{Z}$ for some nonnegative integer $m$. Equivalently, this means\n$$\nb \\mid 10^m(10^k - 1)a.\n$$\nAs $(10a, b) = 1$, this is the same as $b \\mid 10^k - 1$. By definition, the smallest positive integer $k$ for which this holds is the order of $10$ modulo $b$. $\\square$\n\nNow, when $n$ runs through all positive integers, the denominator of $\\frac{n}{567}$ in the lowest term can be any positive divisor of $567 = 3^4 \\times 7$. Let $d_b$ be the order of $10$ modulo $b$. By the claim, it suffices to find $d_b$ for each positive divisor $b$ of $567$.\n* Since $10 \\equiv 1 \\pmod{9}$, we have $d_1 = d_3 = d_9 = 1$.\n* Since $10 \\not\\equiv 1 \\pmod{27}$ and $10^3 = 1000 \\equiv 1 \\pmod{27}$, we have $d_{27} = 3$.\n* Since $10^3 = 1000 \\not\\equiv 1 \\pmod{81}$ and\n$$\n10^9 = 1000^3 = (27 \\times 37 + 1)^3 \\equiv 1 \\pmod{81}\n$$\nby the binomial theorem, we have $d_{81} = 9$.\n* Since $\\varphi(7) = 6$ and $10^2, 10^3 \\not\\equiv 1 \\pmod{7}$, we have $d_7 = 6$.\n* For $b = 3^s7$ where $s = 1, 2, 3, 4$, we have $d_b = [d_{3^s}, d_7]$. Therefore, we easily deduce $d_{21} = d_{63} = d_{189} = 6$, $d_{567} = 18$.\nIt follows that the answer is $1 + 3 + 6 + 9 + 18 = 37$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31530,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the centre of the excircle, touching the side $BC$ of $\\triangle ABC$. $M$ is the midpoint of $AC$ and $P$ is the intersection of lines $MO$ and $BC$. Show that if $\\angle BAC = 2\\angle ACB$, then $AB = BP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AO$ meet $BC$ at $D$. Since\n$$\n\\angle DAC = \\angle BAD = \\frac{1}{2} \\angle BAC = \\angle ACD,\n$$\nwe have $DA = DC$. Consider $\\triangle OCA$ and $\\triangle OCD$. By considering their areas, we find that\n$$\n\\frac{AO}{OD} = \\frac{[OCA]}{[OCD]} = \\frac{AC \\times d(O, AC)}{CD \\times d(O, CD)} = \\frac{AC}{CD}.\n$$\n\nThe last equality holds since $d(O, AC) = d(O, CD)$ is the A-exradius. Now, applying Menelaus' theorem to $\\triangle ACD$, we obtain\n$$\n\\frac{DP}{PC} \\times \\frac{CM}{MA} \\times \\frac{AO}{OD} = 1.\n$$\nUsing $CM = MA$ and $\\frac{AO}{OD} = \\frac{AC}{CD} = \\frac{AC}{AD}$, this becomes\n$$\n\\frac{PD}{PC} = \\frac{AD}{AC}.\n$$\n\nBy the angle bisector theorem, we know that $AP$ bisects $\\angle DAC$. Therefore, we get\n$$\n\\angle BAP = \\angle BAD + \\angle DAP = \\angle ACP + \\angle PAC = \\angle APB.\n$$\nThus, $AB = BP$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31531,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incentre of a triangle $ABC$ with $AB > AC$. A line through $I$ perpendicular to $AI$ cuts $BC$ produced at $P$. Show that $IP^2 = BP \\cdot CP$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have\n$$\n\\angle CIP = \\angle CIA - 90^\\circ = \\angle CBI.\n$$\nThis shows $IP$ is tangent to $(BCI)$. Hence, $IP^2 = BP \\cdot CP$ by considering the power of $P$ with respect to $(BCI)$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31532,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_1$ be a positive real number and\n$$\nx_{n+1} = \\sqrt{5x_n + 2\\sqrt{x_n^2 + 1}} \\quad \\text{for } n = 1, 2, 3, \\dots\n$$\nProve that among $x_1, x_2, \\dots, x_{2011}$, there are at least 670 irrational numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We rewrite the recurrence relation as follows.\n$$\n\\begin{aligned}\nx_{n+1} &= \\sqrt{5}x_n + 2\\sqrt{x_n^2 + 1} \\\\\n\\Rightarrow \\quad (x_{n+1} - \\sqrt{5}x_n)^2 &= 4(x_n^2 + 1) \\\\\n\\Rightarrow \\quad x_n^2 - 2\\sqrt{5}x_{n+1}x_n + (x_{n+1}^2 - 4) &= 0.\n\\end{aligned}\n$$\nThis is a quadratic equation in $x_n$. We use the quadratic formula to obtain\n$$\nx_n = \\sqrt{5}x_{n+1} \\pm \\sqrt{5x_{n+1}^2 - (x_{n+1}^2 - 4)} = \\sqrt{5}x_{n+1} \\pm 2\\sqrt{x_{n+1}^2 + 1}.\n$$\nUsing the given relation, it is clear that $x_{n+1} > x_n$. Therefore,\n$$\nx_n = \\sqrt{5}x_{n+1} - 2\\sqrt{x_{n+1}^2 + 1}.\n$$\nTogether with $x_{n+2} = \\sqrt{5}x_{n+1} + 2\\sqrt{x_{n+1}^2 + 1}$, we obtain\n$$\nx_{n+2} + x_n = 2\\sqrt{5}x_{n+1}.\n$$\nAs $x_n, x_{n+1}, x_{n+2} > 0$, this relation suggests they cannot be all rational. Thus, there must be an irrational number in each of the groups $(x_{3k+1}, x_{3k+2}, x_{3k+3})$ where $k = 0, 1, \\dots, 669$. That means there are at least 670 irrational numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31533,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$, let $M$ be the midpoint of $BC$. The circle passing through $A$ and tangent to $BC$ at $M$ cuts $AB$ and $AC$ at $D$ and $E$ respectively. Suppose $B, C, E$ and $D$ are concyclic. Show that $AB = AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly, consider the power of $B$ with respect to $(ADME)$. This gives\n$$\nBD \\times BA = BM^2.\n$$\nSimilarly, we get\n$$\nCE \\times CA = CM^2\n$$\nby considering the power of $C$ with respect to the same circle. As $M$ is the midpoint of $BC$, the two expressions are equal, so that\n$$\nBD \\times BA = CE \\times CA. \\qquad (1)\n$$\nNext, since $B, C, E, D$ are concyclic, we have\n$$\nAD \\times AB = AE \\times AC \\qquad (2)\n$$\nby considering the power of $A$. Adding (1) and (2), we get\n$$\nAB^2 = AB \\times BD + AB \\times AD = AC \\times CE + AC \\times AE = AC^2.\n$$\nThis gives $AB = AC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31534,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a_1 = 4$ and $a_{n+1} \\ge a_n + n + 2$ for all natural numbers $n$. Show that\n$$\n\\frac{1}{a_1} + \\frac{1}{a_2} + \\dots + \\frac{1}{a_{2010}} < \\frac{3}{2}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Adding the inequalities\n$$\n\\begin{align*}\na_n &\\ge a_{n-1} + n + 1, \\\\\na_{n-1} &\\ge a_{n-2} + n, \\\\\n&\\vdots, \\\\\na_2 &\\ge a_1 + 3,\n\\end{align*}\n$$\nwe obtain $a_n \\ge a_1 + (3+4+\\cdots+(n+1)) = \\frac{n^2+3n+4}{2} > \\frac{(n+1)(n+2)}{2}$. It follows\nthat\n$$\n\\sum_{n=1}^{2010} \\frac{1}{a_n} < \\sum_{n=1}^{\\infty} \\frac{2}{(n+1)(n+2)} = 2 \\sum_{n=1}^{\\infty} \\left( \\frac{1}{n+1} - \\frac{1}{n+2} \\right) = 2 \\cdot \\frac{1}{2} = 1 < \\frac{3}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31535,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that $a$ and $b$ are natural numbers, when $a^2 + b^2$ is divided by $a+b$, the quotient is $Q$ and the remainder is $R$, where $R < a+b$. Find all possible pairs of numbers $a$ and $b$ such that $Q^2 + R = 2010$.",
    "options": [],
    "answer": "[[53, 31], [31, 53]]",
    "solution": "We are given that\n$$\na^2 + b^2 = Q(a + b) + R.\n$$\nSince $R < a + b$, we have\n$$\n(a+b)^2 \\le 2(a^2+b^2) < 2(Q+1)(a+b),\n$$\nwhich implies $a + b < 2(Q + 1)$, and hence $R < 2(Q + 1)$. Thus, we have\n$$\nQ^2 \\le Q^2 + R \\le Q^2 + 2Q + 1 = (Q + 1)^2.\n$$\nSince $Q^2 + R = 2010$, we must have $Q = \\lfloor\\sqrt{2010}\\rfloor = 44$, and so $R = 2010 - 44^2 = 74$.\nIt remains to solve the equation\n$$\na^2 + b^2 = 44(a + b) + 74.\n$$\nThis can be rewritten as\n$$\n(a - 22)^2 + (b - 22)^2 = 1042.\n$$\n\nLet $x = (a - 22)^2$ and $y = (b - 22)^2$. WLOG assume $x \\ge y$. Then we have $521 \\le x \\le 1042$. So it suffices to check $23^2 \\le x \\le 32^2$. Also, $x$ and $y$ must be odd since $x^2 + y^2 \\equiv 2 \\pmod 4$. We compute the values as follows.\n\n| x   | 529 | 625 | 729 | 841 | 961 |\n|-----|-----|-----|-----|-----|-----|\n| y   | 513 | 417 | 313 | 201 | 81  |\n\nAmong these values, only $y = 81$ is a perfect square. Therefore, we can only have $(a - 22, b - 22) = (\\pm 31, \\pm 9)$. Since $a, b \\in \\mathbb{Z}^+$, this implies $(a, b) = (53, 13), (53, 31)$. However, we need $a + b > R = 74$. Therefore, only $(a, b) = (53, 31)$ is possible. We check that when $(a, b) = (53, 31)$, we have\n$$\na^2 + b^2 = 3770 = 44(53 + 31) + 74.\n$$\nTherefore, this is the only solution up to permutation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31536,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of a triangle $ABC$ touches the three sides $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. The line segments $BE$ and $CF$ intersect the incircle at $Y$ and $Z$ respectively. Suppose the lines $FE$ and $BC$ intersect externally at $X$. Prove that the three points $X$, $Y$ and $Z$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X' = EF \\cap YZ$. Apply Pascal's theorem to the points $EEFFZY$ on the incircle. Then $C = EE \\cap FZ$, $X' = EF \\cap ZY$ and $B = FF \\cap YE$ are collinear. Hence, $X' = X$, and $X, Y, Z$ are collinear.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31537,
    "subject": "Mathematics (Multi-modal)",
    "question": "Among a group of $451$ persons, every pair of persons together know exactly one other person (knowing is mutual). Denote by $n$ the difference between the number of persons known by a member of the group who knows the largest number of persons among the group, and the number of persons known by another member who knows the smallest number of persons among the group. Find the maximum possible value of $n$ and all other possible value(s) of $n$.",
    "options": [],
    "answer": "Maximum 448; the only possible value is 448.",
    "solution": "The only possible (and hence the maximum) $n$ is $448$.\n\nIndeed, by the friendship theorem, there must be one person who knows everyone else. We give a proof as follows.\n\nWLOG, assume $A$ and $B$ do not know each other. If $X$ knows $A$, then $X$ and $B$ know a person $Y$ in common. Note that the function mapping $X$ to $Y$ is well-defined, and it is a bijection between the set of persons known by $A$ and the set of persons known by $B$. Hence any two persons who do not know each other know the same number of persons.\n\nLet $S_1$ be the set of all persons who know exactly $k$ persons, where $k$ is chosen such that $|S_1| \\ge 1$. Let $S_2$ be the complement of $S_1$. Note that each person in $S_2$ must know every person in $S_1$. Otherwise, this person must know exactly $k$ persons from the above observation, and hence this person should belong to $S_1$ by definition.\n\nIf $|S_2| \\ge 2$, then we choose two persons $X$ and $Y$ in $S_2$ arbitrarily. Since they only know one person in common, we must have $|S_1| = 1$. In this case the person in $S_1$ knows everyone.\n\nIf $|S_2| = 1$, then the person in $S_2$ knows everyone.\n\nIf $|S_2| = 0$, then everyone knows exactly $k$ persons. On the one hand, there are $\\binom{451}{2}$ pairs of persons in this group. On the other hand, there are $\\binom{k}{2}$ pairs of persons knowing a particular person in common. Therefore, we have\n$$\n451 \\binom{k}{2} = \\binom{451}{2}.\n$$\n\nBut this implies $k(k-1) = 450$, which has no solution. (In this case, we make use of the fact that $451$ is not of the form $k^2 - k + 1$. But the friendship theorem does apply to any number of persons.)\n\nTherefore, we have proven there exists a person $A$ who knows everyone. For every other person $B$, in order that $A$ and $B$ know exactly one person in common, $B$ must know exactly one person apart from $A$. Thus, the other persons can be paired up such that only the two persons in the same pair know each other. Hence, the only possible $n$ is $450 - 2 = 448$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31538,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ and $d$ be positive real numbers, all larger than or equal to $1$. Prove that\n\na.  $abcd(a+b+c+d-4)^4 \\ge (a-1)(b-1)(c-1)(d-1)(a+b+c+d)^4$,\n\nb.  $\\dfrac{d}{a+b+c} + \\dfrac{a}{b+c+d} + \\dfrac{b}{c+d+a} + \\dfrac{c}{d+a+b} \\ge \\dfrac{4}{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nWLOG assume $a \\ge b \\ge c \\ge d$. Then $\\dfrac{a-1}{a} \\ge \\dfrac{b-1}{b} \\ge \\dfrac{c-1}{c} \\ge \\dfrac{d-1}{d}$. By Chebyshev's inequality, we have\n$$\n\\sum_{\\text{cyc}} \\left( a \\cdot \\frac{a-1}{a} \\right) \\ge \\frac{1}{4} \\left( \\sum_{\\text{cyc}} a \\right) \\left( \\sum_{\\text{cyc}} \\frac{a-1}{a} \\right).\n$$\nAlso, by the AM-GM inequality, we have\n$$\n\\sum_{\\text{cyc}} \\frac{a-1}{a} \\ge 4 \\sqrt[4]{\\frac{(a-1)(b-1)(c-1)(d-1)}{abcd}}.\n$$\nCombining these, we obtain\n$$\n(a+b+c+d-4) \\ge (a+b+c+d) \\sqrt[4]{\\frac{(a-1)(b-1)(c-1)(d-1)}{abcd}}.\n$$\nRaising both sides to the fourth power and rearranging the terms, we obtain the desired inequality. Equality holds when $a = b = c = d$.\n\nb.\nWe have\n$$\n\\begin{align*}\n& \\frac{d}{a+b+c} + \\frac{a}{b+c+d} + \\frac{b}{c+d+a} + \\frac{c}{d+a+b} \\ge \\frac{4}{3} \\\\\n\\Leftrightarrow \\quad & \\frac{a+b+c+d}{a+b+c} + \\frac{a+b+c+d}{b+c+d} + \\frac{a+b+c+d}{c+d+a} + \\frac{a+b+c+d}{d+a+b} \\ge \\frac{16}{3} \\\\\n\\Leftrightarrow \\quad & (a+b+c+d) \\sum_{\\text{cyc}} \\frac{1}{a+b+c} \\ge \\frac{16}{3} \\\\\n\\Leftrightarrow \\quad & \\left( \\sum_{\\text{cyc}} (a+b+c) \\right) \\left( \\sum_{\\text{cyc}} \\frac{1}{a+b+c} \\right) \\ge 16.\n\\end{align*}\n$$\nThis is obviously true by the Cauchy-Schwarz inequality. Equality holds when $a = b = c = d$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31539,
    "subject": "Mathematics (Multi-modal)",
    "question": "A finite sequence of integers $a_0, a_1, \\dots, a_n$ is quadratic if for each $i = 1, 2, \\dots, n$, $|a_i - a_{i-1}| = i^2$.\n\na. Show that for any two integers $b$ and $c$, $b < c$, there exists a natural number $n$ and a quadratic sequence with $a_0 = b$ and $a_n = c$.\n\nb. Find the smallest natural number $n$ for which there exists a quadratic sequence with $a_0 = 0$ and $a_n = 2012.",
    "options": [],
    "answer": "19",
    "solution": "a.\nObserve that $f(k) = k^2 - (k+1)^2 - (k+2)^2 + (k+3)^2 = 4$ for any integer $k$.\n\n* If $c = b + 4m$ for some $m \\in \\mathbb{Z}$, then $c = b + \\sum_{k=1}^{m} f(4k - 3)$.\n* If $c = b + 4m + 1$ for some $m \\in \\mathbb{Z}$, then $c = b + 1^2 + \\sum_{k=1}^{m} f(4k - 2)$.\n* If $c = b + 4m + 2$ for some $m \\in \\mathbb{Z}$, then $c = b - 1^2 - 2^2 - 3^2 + 4^2 + \\sum_{k=1}^{m} f(4k + 1)$.\n* If $c = b + 4m + 3$ for some $m \\in \\mathbb{Z}$, then $c = b - 1^2 + 2^2 + \\sum_{k=1}^{m} f(4k - 1)$.\n\nTherefore, in any case we can write $c$ as $b \\pm 1^2 \\pm 2^2 \\pm \\cdots \\pm n^2$ for some $n$ and some choices of the signs. This clearly solves the problem.\n\nb.\nThe answer is $19$.\n\nFirstly, since $1^2 + 2^2 + \\cdots + 17^2 = \\frac{17 \\cdot 18 \\cdot 35}{6} = 1785 < 2012$, we have $n \\ge 18$.\n\nSecondly, when $n = 18$, $\\pm 1^2 \\pm 2^2 \\pm \\cdots \\pm 18^2 \\equiv 1 + 0 + \\cdots + 1 + 0 \\equiv 1 \\pmod{2}$. This shows $b$ and $c$ cannot have the same parity, which is a contradiction.\n\nLastly, it remains to provide a construction for $n = 19$. Since\n$$\n1^2 + 2^2 + \\cdots + 19^2 = \\frac{19 \\cdot 20 \\cdot 39}{6} = 2470 = 2012 + 2 \\times 229 = 2012 + 2(2^2 + 15^2),\n$$\nwe can change the signs of $2^2$ and $15^2$ to negative, so that the new sum becomes $2012$. This gives a corresponding quadratic sequence that satisfies the requirement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31540,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral. $E$ and $F$ are points on the diagonal $AC$ such that $E$ and $F$ are interior points of triangles $ABD$ and $BCD$ respectively. Suppose $BE$ extend cuts $AD$ at $P$, $DE$ extend cuts $AB$ at $Q$, $DF$ extend cuts $BC$ at $R$ and $BF$ extend cuts $DC$ at $S$. Show that the three lines $QP$, $BD$ and $RS$ are either parallel or concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Y$ be the intersection point of $AC$ and $BD$. Since $AY$, $BP$, $DQ$ are concurrent, $QP$ meets $BD$ at the harmonic conjugate $X$ of $Y$ with respect to $B$, $D$ (note that $X$ could be a point at infinity). Similarly, since $CY$, $BS$, $DR$ are concurrent, $RS$ meets $BD$ at $X$. This shows $QP$, $BD$, $RS$ are concurrent in the projective sense. This means they are concurrent or parallel.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31541,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1A_2\\cdots A_n$ be a cyclic polygon ($n \\ge 3$). Find the maximum number of distinct acute-angled triangles whose vertices are chosen from $A_1, A_2, \\dots, A_n$.",
    "options": [],
    "answer": "Maximum number of acute triangles equals n(n-1)(n+1)/24 for odd n and n(n-2)(n+2)/24 for even n.",
    "solution": "The maximum number of acute-angled triangles is $\\frac{n(n-1)(n+1)}{24}$ if $n$ is odd and is $\\frac{n(n-2)(n+2)}{24}$ if $n$ is even.\n\nWe first count the number $N_1$ of non-acute triangles with $A_1$ as one vertex such that $\\angle A_1 < 90^\\circ$. Let $\\ell$ be the diameter of the circle which passes through $A_1$. Then for points $B, C$ on the circle, $\\triangle A_1BC$ is not acute with $\\angle BA_1C < 90^\\circ$ if and only if $B$ and $C$ lie on the same side of $\\ell$ (possibly on $\\ell$). Suppose there are $a$ points on one side of $\\ell$ and $b$ points on the other side of $\\ell$ other than $A_1$. Note that the two sides share at most one common point which is the other intersection of $\\ell$ and the circle. Therefore, we must have $a+b=n-1$ or $a+b=n$. We find that\n$$\nN_1 = \\binom{a}{2} + \\binom{b}{2}.\n$$\nWe have to minimize $N_1$. Note that $b \\ge n-1-a$ so that $\\binom{b}{2} \\ge \\binom{n-1-a}{2}$.\nWLOG assume $a \\ge n-1-a$. If $n$ is odd, then $(a, n-1-a) > \\left(\\frac{n-1}{2}, \\frac{n-1}{2}\\right)$.\nApplying the majorization inequality, we obtain\n$$\nN_1 \\ge \\binom{a}{2} + \\binom{n-1-a}{2} \\ge \\left(\\frac{n-1}{2}\\right) + \\left(\\frac{n-1}{2}\\right) = \\frac{(n-1)(n-3)}{4}.\n$$\nThe same bound holds if we replace $A_1$ by any other vertex $A_j$. We sum up all the lower bounds. Note that each non-acute triangle is counted twice in\n\nthis sum (one corresponding to each acute angle). Therefore, the number of non-acute triangles is at least\n$$\nn \\cdot \\frac{(n-1)(n-3)}{4} \\cdot \\frac{1}{2} = \\frac{n(n-1)(n-3)}{8}.\n$$\nThus, the number of acute triangles is at most\n$$\n\\binom{n}{3} - \\frac{n(n-1)(n-3)}{8} = \\frac{n(n-1)(n+1)}{24}.\n$$\nThis can be attained when $A_1A_2\\cdots A_n$ is a regular polygon (since in that case all inequalities used become equalities).\n\nSimilarly, if $n$ is even, then $(a, n-1-a) > \\left(\\frac{n}{2}, \\frac{n-2}{2}\\right)$. Applying the majorization inequality, we obtain\n$$\nN_1 \\ge \\binom{a}{2} + \\binom{n-1-a}{2} \\ge \\binom{\\frac{n}{2}}{2} + \\binom{\\frac{n-2}{2}}{2} = \\frac{(n-2)^2}{4}.\n$$\nThus, the number of acute triangles is at most\n$$\n\\binom{n}{3} - n \\cdot \\frac{(n-2)^2}{4} \\cdot \\frac{1}{2} = \\frac{n(n-2)(n+2)}{24}.\n$$\nThe regular polygon case does not yield equality. Indeed, we first consider a regular polygon $B_1B_2\\cdots B_n$. Then we rotate half of the points, say $B_1, B_2, \\dots, B_{\\frac{n}{2}}$, anticlockwise about the centre of the circle with a sufficiently small angle $\\theta$ to points $C_1, C_2, \\dots, C_{\\frac{n}{2}}$. Then $C_1C_2\\cdots C_{\\frac{n}{2}}B_{\\frac{n}{2}+1}B_{\\frac{n}{2}+2}\\cdots B_n$ is one possible case which gives the upper bound $\\frac{n(n-2)(n+2)}{24}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31542,
    "subject": "Mathematics (Multi-modal)",
    "question": "Evaluate $\\frac{\\sqrt{45+\\sqrt{1}} + \\sqrt{45+\\sqrt{2}} + \\dots + \\sqrt{45+\\sqrt{2024}}}{\\sqrt{45-\\sqrt{1}} + \\sqrt{45-\\sqrt{2}} + \\dots + \\sqrt{45-\\sqrt{2024}}}$.",
    "options": [],
    "answer": "1 + sqrt(2)",
    "solution": "The answer is $1 + \\sqrt{2}$.\nLet $a$ and $b$ be the numerator and the denominator respectively. For each $n = 1, 2, \\dots, 2024$, we have the identity\n$$\n\\sqrt{45 + \\sqrt{2025-n}} + \\sqrt{45 - \\sqrt{2025-n}} = \\sqrt{2} \\cdot \\sqrt{45 + \\sqrt{n}}.\n$$\nThis can be easily proved by squaring both sides and simplifying. Now, summing over all $n$'s, we obtain\n$$\na + b = \\sqrt{2}a.\n$$\n\n$$\n\\frac{a}{b} = \\frac{a}{\\sqrt{2}a - a} = \\frac{1}{\\sqrt{2} - 1} = 1 + \\sqrt{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31543,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a city the phone numbers should consist of exactly eight digits, and $0$ is not allowed as a digit in them (i.e., only $1, 2, 3, \\ldots, 9$ may be used). It is required that every two phone numbers either be different in at least two places, or have digits separated by at least two units, in at least one of the eight places. (For example, both $12345678$ and $23345678$ may be selected if there is no more conflict with other numbers, or both $12345678$ and $92345678$ may be selected, again if there is no more conflict with other numbers, but at most one of $12345678$ and $12345679$ may be selected.)\nWhat is the maximum number of phone numbers that can be selected, satisfying the requirements? In how many ways can one select this amount of phone numbers?",
    "options": [],
    "answer": "Maximum number: 21523361; Number of ways: 1",
    "solution": "At most $21523361$ phone numbers can be selected, and there is only $1$ way to select this amount of phone numbers.\n\nConsider the situation when the phone numbers have $n$ digits instead of $8$. We shall prove the following by induction: at most $\\frac{9^n + 1}{2}$ phone numbers can be selected, and there is a unique way to select the phone numbers, which is to choose all phone numbers whose sum has the same parity as $n$.\n\nWhen $n = 1$, we can choose at most one phone number from each group\n$$\n(1, 2), (3, 4), (5, 6), (7, 8), (9).\n$$\nThus, at most $\\frac{9+1}{2} = 5$ phone numbers can be selected. Clearly, the only way to select $5$ phone numbers is to select all odd numbers $1, 3, 5, 7, 9$.\n\nAssume the case $n-1$ is true. Consider the case $n$. For those phone numbers not starting with the digit $9$, we can pair them up such that $\\overline{a_1a_2\\cdots a_n}$ and $\\overline{(a_1+1)a_2\\cdots a_n}$ belong to the same pair if $a_1 = 1, 3, 5, 7$. Clearly, at most one phone number can be selected from each pair. Also, for the remaining phone numbers starting with $9$, we can remove the digit $9$ and regard this as the case $n-1$. By the inductive hypothesis, we can select at most $\\frac{9^{n-1} + 1}{2}$ phone numbers from these. In total, we can select at most\n$$\n\\frac{8 \\cdot 9^{n-1}}{2} + \\frac{9^{n-1} + 1}{2} = \\frac{9^n + 1}{2}\n$$\nphone numbers.\n\nTo choose this amount of phone numbers, since we need to select $\\frac{9^{n-1} + 1}{2}$ phone numbers from those starting with $9$, by the inductive hypothesis, the only way is to select all numbers $\\overline{9a_2\\cdots a_n}$ with $a_2 + \\cdots + a_n \\equiv n-1 \\pmod 2$. Equivalently, this means $9 + a_2 + \\cdots + a_n \\equiv n \\pmod 2$.\n\nAll numbers $\\overline{8a_2\\cdots a_n}$ with $8 + a_2 + \\cdots + a_n \\equiv n \\pmod 2$. Then by using the first pairing, we find that all numbers $\\overline{7a_2\\cdots a_n}$ with $7 + a_2 + \\cdots + a_n \\equiv n \\pmod 2$ should be selected, etc. In the same way, we find that all numbers with sum congruent to $n$ modulo $2$ should be selected. Clearly, the condition is satisfied if we only select those numbers. Also, as exactly one number from each pair is selected, there are $\\frac{9^n+1}{2}$ such phone numbers. This proves the inductive step.\n\nWhen $n=8$, we can select at most $\\frac{9^8+1}{2} = 21523361$ phone numbers.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31544,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$, let $a_1, a_2, \\dots, a_m$ be all the positive divisors of $n$, where $m \\ge 1$. If there exist $m$ integers $b_1, b_2, \\dots, b_m$ such that $n = \\sum_{i=1}^{m} (-1)^{b_i} a_i$, then we say that $n$ is a *good number*. Prove that there exists a good number with exactly 2013 distinct prime factors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We say that a positive integer $n$ is *excellent* if it is good, and we can find integers $c_1, c_2, \\dots, c_m$ such that $\\sum_{j=1}^{m} (-1)^{c_j} a_j = 0$. Let $p_1, p_2, \\dots, p_k$ be distinct primes larger than 3. We prove by induction on $k$ that the integer $6p_1p_2\\cdots p_k$ is excellent.\n\nFor the base case $k=0$, the positive divisors of $6$ are $1$, $2$, $3$, $6$. Since we have $-1-2+3+6=6$ and $-1-2-3+6=0$, the integer $6$ is excellent.\n\nFor the inductive step, assume $n = 6p_1p_2\\cdots p_k$ is excellent. Let $a_1, a_2, \\dots, a_m$ be all of its positive divisors, and let\n$$\n\\sum_{j=1}^{m} (-1)^{b_j} a_j = n \\quad \\text{and} \\quad \\sum_{j=1}^{m} (-1)^{c_j} a_j = 0.\n$$\nConsider $n' = 6p_1p_2\\cdots p_{k+1}$. Its positive divisors are of the form $a_j$ and $a_jp_{k+1}$. Note that\n$$\n\\sum_{j=1}^{m} (-1)^{c_j} a_j + \\sum_{j=1}^{m} (-1)^{b_j} (a_j p_{k+1}) = 0 + n p_{k+1} = n'.\n$$\nAlso, we have\n$$\n\\sum_{j=1}^{m} (-1)^{c_j} a_j + \\sum_{j=1}^{m} (-1)^{b_j} (a_j p_{k+1}) = 0 + 0 \\cdot p_{k+1} = 0.\n$$\nTherefore, $n'$ is excellent. By induction, our claim holds. In particular, we can take $k=2011$ so that $6p_1p_2\\cdots p_{2011}$ is excellent, and hence it is a good integer with exactly 2013 prime divisors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31545,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all reals $x$, $y$ and $z$ such that\n$$\n\\begin{cases}\n(x + 1)yz = 27, \\\\\n(y + 1)zx = 12, \\\\\n(z + 1)xy = 12.\n\\end{cases}\n$$",
    "options": [],
    "answer": "(2, -3, -3); ((8 - 2√10)/3, 1 + √10, 1 + √10); ((8 + 2√10)/3, 1 - √10, 1 - √10)",
    "solution": "The solutions are $(x, y, z) = (2, -3, -3)$, $\\left(\\frac{8 \\mp 2\\sqrt{10}}{3}, 1 \\pm \\sqrt{10}, 1 \\pm \\sqrt{10}\\right)$.\n\nClearly, $x, y, z \\neq 0$. Since\n$$\n(y + 1)zx = 12 = (z + 1)xy,\n$$\nwe have $zx = xy$. This implies $y = z$. By the first equation $(x + 1)y^2 = 27$, we obtain\n$$\nx = \\frac{27}{y^2} - 1.\n$$\nPutting this into the second equation, we have\n$$\n(y + 1)y\\left(\\frac{27}{y^2} - 1\\right) = 12.\n$$\nAfter expansion, this becomes $y^3 + y^2 - 15y - 27 = 0$, i.e.\n$$\n(y + 3)(y^2 - 2y - 9) = 0.\n$$\nClearly, the solutions are $y = -3$, $1 \\pm \\sqrt{10}$. Correspondingly, we find all solutions\n$$\n(x, y, z) = (2, -3, -3), \\left(\\frac{8 \\mp 2\\sqrt{10}}{3}, 1 \\pm \\sqrt{10}, 1 \\pm \\sqrt{10}\\right).\n$$\nIt is not hard to check that all these are indeed solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31546,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$, $u$, $v$ and $w$ be integers satisfying $x^2 + y^2 = u^2$, $x^2 + z^2 = v^2$ and $y^2 + z^2 = w^2$. Find an integer $\\ell$ so that $518000 < \\ell < 518518$ and $\\ell$ divides $xyzuvw$.",
    "options": [],
    "answer": "518400",
    "solution": "The answer is $518400$.\n\nWe claim that $3^4 \\cdot 4^4 \\cdot 5^2$ divides $xyzuvw$.\n\nFirstly, observe that $3$ divides $x$ or $3$ divides $y$, since otherwise\n$$\nu^2 = x^2 + y^2 \\equiv 1 + 1 = 2 \\pmod{3},$$\nwhich is impossible. Similarly, $3$ divides $x$ or $z$, and $3$ divides $y$ or $z$. Thus, two of $x$, $y$, $z$ are divisible by $3$, say $x$ and $y$. Then we have $3$ divides $u$, and the first equation becomes $\\left(\\frac{x}{3}\\right)^2 + \\left(\\frac{y}{3}\\right)^2 = \\left(\\frac{u}{3}\\right)^2$. Again, $3$ divides one of $\\frac{x}{3}$ and $\\frac{y}{3}$. Thus, we have $3^4$ divides $xyu$. In general, we must have $3^4$ divides $xyzuvw$.\n\nSecondly, observe that $4$ divides $x$ or $4$ divides $y$, since otherwise\n$$u^2 = x^2 + y^2 \\equiv 1 + 1 = 2 \\pmod{4},$$\nwhich is impossible. As above, this implies $4^4$ divides $xyzuvw$.\n\nThirdly, observe that $5$ divides $x$, $y$ or $u$, since otherwise\n$$x^2 + y^2 \\equiv \\pm 1 \\pm 1 = \\pm 2 \\not\\equiv \\pm 1 \\equiv u^2 \\pmod{5}.$$ \nIn the same way, $5$ divides one of $x$, $z$, $v$, $5$ divides one of $y$, $z$, $w$. It is not hard to see $5$ divides at least two of $x$, $y$, $z$, $u$, $v$, $w$, and hence $5^2$ divides $xyzuvw$.\n\nCombining these, we obtain $3^4 \\cdot 4^4 \\cdot 5^2 = 518400$ divides $xyzuvw$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31547,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find a function $f : \\mathbb{R}^{\\ge 0} \\to \\mathbb{R}^{\\ge 0}$ satisfying $f(2x + 1) = 4f(x) + 9$ for all $x \\ge 0$. ($\\mathbb{R}^{\\ge 0}$ is the set of nonnegative real numbers.)",
    "options": [],
    "answer": "One example is f(x) = 0 for x in [0, 1) and f(x) = 3(4^k - 1) for x in [2^k - 1, 2^{k+1} - 1) where k is any positive integer.",
    "solution": "A possible function is $f(x) = 0$ for any $x \\in [0, 1)$ and $f(x) = 3(4^k - 1)$ for any $x \\in [2^k - 1, 2^{k+1} - 1)$ where $k \\in \\mathbb{Z}^+$.\nIt suffices to check the above function satisfies the conditions. It is clear that $f(x) \\ge 0$ for any $x \\ge 0$. If $x \\in [0, 1)$, then $2x + 1 \\in [1, 3)$. Therefore, we have\n$$\nf(2x + 1) = 3(4^1 - 1) = 9 = 4(0) + 9 = 4f(x) + 9.\n$$\nIf $x \\in [2^k - 1, 2^{k+1} - 1)$ where $k \\in \\mathbb{Z}^+$, then $2x + 1 \\in [2^{k+1} - 1, 2^{k+2} - 1)$. Therefore, we have\n$$\nf(2x + 1) = 3(4^{k+1} - 1) = 4 \\cdot 3(4^k - 1) + 9 = 4f(x) + 9.\n$$\nThis completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31548,
    "subject": "Mathematics (Multi-modal)",
    "question": "Some of the lattice points $(x, y)$, with $1 \\le x \\le 101$ and $1 \\le y \\le 101$ are marked so that no 4 marked points form the vertices of an isosceles trapezoid with bases parallel to the $x$-axis or the $y$-axis (a rectangle is counted as an isosceles trapezoid). Determine the maximum number of marked points. (A lattice point is a point with integral coordinates.)",
    "options": [],
    "answer": "251",
    "solution": "The answer is $251$.\n\nConsider pairs of marked points with the same $y$-coordinates. If there exist two pairs of marked points with the same sum of $x$-coordinates and different $y$-coordinates, then the 4 points in these pairs are the vertices of an isosceles trapezoid.\n\nSuppose there are $a_1, a_2, \\dots, a_{101}$ marked points in the $101$ horizontal lines respectively (this means there are $a_1$ marked points with $y$-coordinates equal to $1$, etc.). For the horizontal line $y = 1$, suppose the marked points are $(x_1, 1), (x_2, 1), \\dots, (x_{a_1}, 1)$ where $x_1 < x_2 < \\dots < x_{a_1}$. Since\n$$\nx_1 + x_2 < x_1 + x_3 < \\dots < x_1 + x_{a_1} < x_2 + x_{a_1} < \\dots < x_{a_1-1} + x_{a_1},\n$$\nthere are at least $2a_1 - 3$ distinct sums formed by the $x$-coordinates. This only holds for $a_1 \\ge 2$. For $a_1 = 0$ or $a_1 = 1$, clearly, no sum can be formed. Therefore, there are at least\n$$\nf(a_1) + f(a_2) + \\dots + f(a_{101}) \\quad (1)\n$$\nwhere $f(n) = \\max\\{2n-3, 0\\}$.\n\nEach of these sums lies between $1+2=3$ and $100+101=201$. There are $199$ possibilities. If (1) exceeds $199$, then two of the sums are equal by the pigeonhole principle. Since the equal sums must correspond to distinct horizontal lines from the construction, we can find an isosceles trapezoid as suggested. Therefore, we may assume (1) is at most $199$. Then\n$$\n199 \\ge f(a_1) + f(a_2) + \\dots + f(a_{101}) \\ge (2a_1 - 3) + (2a_2 - 3) + \\dots + (2a_{101} - 3).\n$$\nThis implies $a_1+a_2+\\dots+a_{101} \\le 251$. In other words, the number of marked points is at most $251$. This can be attained, for example, in the following construction.\n\n![](attached_image_1.png)\n\nIt is not hard to see that there is no isosceles trapezoid (a rigorous proof is to show that the sums of $x$-coordinates of pairs of points in the same horizontal line are all different, and the same holds for $y$-coordinates by symmetry). There are\n$$\n3 + 5 \\times 49 + 3 = 251\n$$\npoints in total.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31549,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a scalene triangle (with no two sides equal). A circle touching sides $AB$ and $BC$ intersects the median $AM$ at $P$ and $Q$. Another circle touching sides $AC$ and $BC$ also intersects $AM$ at $P$ and $Q$. Prove that the two circles are the same circle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\Gamma_1$ be the circle touching $AB$ and $BC$, and let $\\Gamma_2$ be the circle touching $AC$ and $BC$. Let $\\Gamma_1$ touch $AB$ and $BC$ at $X$ and $D$ respectively. Let $\\Gamma_2$ touch $AC$ and $BC$ at $Y$ and $E$ respectively.\n\n![](attached_image_1.png)\n\nSuppose on the contrary that $\\Gamma_1 \\neq \\Gamma_2$. Then $D \\neq E$ since otherwise both circles pass through three same points. By using powers, we have\n$$\nMD^2 = MP \\times MQ = ME^2.\n$$\nThis implies $M$ is the midpoint of $D$ and $E$. As $M$ is also the midpoint of $B$ and $C$, we have $BD = CE$.\nSimilarly, we have $AX^2 = AP \\times AQ = AY^2$, and so $AX = AY$. It follows that\n$$\nAB = AX + BX = AX + BD = AY + CE = AY + CY = AC,\n$$\ncontradicting the assumption that $\\triangle ABC$ is scalene. Therefore, $\\Gamma_1 = \\Gamma_2$, which is the incircle of $\\triangle ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31550,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every positive integer $n$ and every group of real numbers $a_1, a_2, \\dots, a_n > 0$,\n$$\n\\sum_{k=1}^{n} \\frac{k}{a_{1}^{-1} + a_{2}^{-1} + \\dots + a_{k}^{-1}} \\le 2 \\sum_{k=1}^{n} a_{k}.\n$$\nCan “2” immediately to the right of the inequality be replaced by a smaller positive number?",
    "options": [],
    "answer": "No. The constant two is optimal and cannot be replaced by any smaller positive number.",
    "solution": "By the Cauchy-Schwarz inequality, we have\n$$\n(1^2 a_1 + 2^2 a_2 + \\dots + k^2 a_k)(a_1^{-1} + a_2^{-1} + \\dots + a_k^{-1}) \\ge (1 + 2 + \\dots + k)^2 = \\frac{k^2 (k+1)^2}{4}.\n$$\nThis implies\n$$\n\\frac{k}{a_1^{-1} + a_2^{-1} + \\dots + a_k^{-1}} \\le \\frac{4}{k(k+1)^2} \\sum_{j=1}^{k} j^2 a_j.\n$$\n\nAlso, using $\\frac{1}{k^2} - \\frac{1}{(k+1)^2} = \\frac{2k+1}{k^2(k+1)^2} \\ge \\frac{2}{k(k+1)^2}$, we obtain\n$$\n\\begin{align*}\n\\sum_{k=1}^{n} \\frac{k}{a_1^{-1} + a_2^{-1} + \\cdots + a_k^{-1}} &\\le \\sum_{k=1}^{n} \\sum_{j=1}^{k} j^2 a_j \\frac{4}{k(k+1)^2} \\\\\n&\\le 2 \\sum_{k=1}^{n} \\sum_{j=1}^{k} j^2 a_j \\left( \\frac{1}{k^2} - \\frac{1}{(k+1)^2} \\right) \\\\\n&= 2 \\sum_{j=1}^{n} \\sum_{k=j}^{n} j^2 a_j \\left( \\frac{1}{k^2} - \\frac{1}{(k+1)^2} \\right) \\\\\n&= 2 \\sum_{j=1}^{n} j^2 a_j \\left( \\frac{1}{j^2} - \\frac{1}{(n+1)^2} \\right) \\\\\n&< 2 \\sum_{j=1}^{n} a_j.\n\\end{align*}\n$$\nThis proves the desired inequality.\n\nConsider $a_j = \\frac{1}{j}$ for each $j$. Then\n$$\n\\sum_{k=1}^{n} \\frac{k}{a_1^{-1} + a_2^{-1} + \\cdots + a_k^{-1}} = \\sum_{k=1}^{n} \\frac{k}{1 + 2 + \\cdots + k} = \\sum_{k=1}^{n} \\frac{2}{k + 1}\n$$\nand $\\sum_{k=1}^{n} a_k = \\sum_{k=1}^{n} \\frac{1}{k}$. For any $c$ such that $0 < c < 2$, we have\n$$\nc \\sum_{k=1}^{n} a_k - \\sum_{k=1}^{n} \\frac{k}{a_1^{-1} + a_2^{-1} + \\cdots + a_k^{-1}} = c + (c-2) \\sum_{k=2}^{n} \\frac{1}{k} - \\frac{2}{n+1} < 0\n$$\nfor sufficiently large $n$, since the harmonic series $\\sum_{k=2}^{\\infty} \\frac{1}{k}$ diverges to positive infinity. This shows we cannot replace the number 2 by any smaller constant $c$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31551,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $\\angle A = 30^\\circ$ and $AB = 4$. $D$ is the midpoint of $AB$. Determine the maximum possible value of $\\tan \\angle DCB$.",
    "options": [],
    "answer": "(4√2 + 3√3)/5",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31552,
    "subject": "Mathematics (Multi-modal)",
    "question": "Among $45$ persons in a party, it is found that every person knows at least one other person in the party (i.e. they are friends), and any two persons with the same number of friends in the party do not know each other. What is the maximum number of pairs of persons who are friends?",
    "options": [],
    "answer": "870",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31553,
    "subject": "Mathematics (Multi-modal)",
    "question": "$E$ and $F$ are points on side $AB$ of $\\triangle ABC$ such that $AE = EF = FB$. $D$ is a point on the line $BC$ such that $BC$ is perpendicular to $ED$, $AD$ is perpendicular to $CF$. Suppose $\\angle CFA = 3\\angle BDF$. Determine the value of $\\frac{DB}{DC}$.",
    "options": [],
    "answer": "7/2",
    "solution": "We have $\\frac{DB}{DC} = \\frac{7}{2}$.\nFirstly, since $\\angle EDB = 90^\\circ$ and $BF = EF$, we have $DF = BF$. This implies $\\angle FDB = \\angle FBD = \\theta$, and hence $\\angle DFE = 2\\theta$. It follows that $C$ lies on the segment $BD$, and\n$$\n\\angle CDF = 3\\theta - 2\\theta = \\theta.\n$$\nLet $P$ be the intersection point of $AD$ and $FC$, and let $M$ be the midpoint of $BD$. Note that $FM \\perp BD$ since $\\triangle FBD$ is isosceles. This shows\n$$\n\\angle DMF = 90^\\circ = \\angle DPF.\n$$\nTherefore, $F, M, P, D$ are concyclic. Since $\\angle CDF = \\angle CFD$, $FMPD$ is an isosceles trapezoid.\nNow, applying Menelaus' theorem to $\\triangle BCF$, we obtain\n$$\n\\frac{BD}{DC} \\times \\frac{CP}{PF} \\times \\frac{FA}{AB} = 1.\n$$\n\nNote that $BD = 2MD = 2PF$ and $\\frac{FA}{AB} = \\frac{2}{3}$. Therefore, we get $\\frac{CP}{DC} = \\frac{3}{4}$. This implies $\\frac{CM}{DC} = \\frac{CP}{DC} = \\frac{3}{4}$, and hence\n$$\n\\frac{DB}{DC} = \\frac{2DM}{DC} = \\frac{2(DC + CM)}{DC} = \\frac{7}{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31554,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the number of positive integers $m$ for which there exist nonnegative integers $x_0, x_1, x_2, \\dots, x_{2011}$ such that $m^{x_0} = m^{x_1} + m^{x_2} + \\dots + m^{x_{2011}}$.",
    "options": [],
    "answer": "16",
    "solution": "There are 16 such positive integers $m$.\nLet $a_k = m^k$. We would like to represent $a_{x_0}$ as a sum of 2011 numbers of the form $a_{x_j}$. To get such a representation, we can do in the following way. Starting from the number $a_{x_0}$, each time we choose a number $a_{k+1}$ and break it into $m$ numbers $a_k$, so that the sum of all numbers remains $a_{x_0}$.\n\nIn view of the base $m$ representation of the numbers, if we can write\n$$\na_{x_0} = a_{x_1} + a_{x_2} + \\cdots + a_{x_{2011}},\n$$\nthen we are able to obtain the numbers on the right by using the above operation repeatedly. Conversely, if we can obtain these numbers after some operations, then clearly their sum is $a_{x_0}$ by construction. Therefore, it remains to check for which $m$ we can apply finitely many operations so that we can generate exactly 2011 numbers.\n\nNote that there are $m-1$ numbers more after applying each operation. Initially, there is only one number. Therefore, it is the same as finding those $m \\in \\mathbb{Z}^+$ such that $m - 1 \\mid 2011 - 1 = 2010$. As $2010 = 2 \\times 3 \\times 5 \\times 67$, there are $d(2010) = 16$ solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31555,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers such that $ab + bc + ca = 1$. Prove that\n$$\n\\sqrt[4]{\\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b} + \\sqrt[4]{\\frac{\\sqrt{3}}{b} + 6\\sqrt{3}c} + \\sqrt[4]{\\frac{\\sqrt{3}}{c} + 6\\sqrt{3}a} \\le \\frac{1}{abc}.\n$$\nWhen does the equality hold?",
    "options": [],
    "answer": "Equality holds when a = b = c = 1/√3.",
    "solution": "(IMO Shortlist 2004 A5 modified) By the power mean inequality, we have\n$$\n\\begin{aligned}\nS &:= \\sqrt[4]{\\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b} + \\sqrt[4]{\\frac{\\sqrt{3}}{b} + 6\\sqrt{3}c} + \\sqrt[4]{\\frac{\\sqrt{3}}{c} + 6\\sqrt{3}a} \\\\\n&\\le 3\\sqrt[4]{\\frac{1}{3}\\left(\\frac{\\sqrt{3}}{a} + 6\\sqrt{3}b + \\frac{\\sqrt{3}}{b} + 6\\sqrt{3}c + \\frac{\\sqrt{3}}{c} + 6\\sqrt{3}a\\right)} \\\\\n&= 3^{\\frac{7}{8}}\\sqrt[4]{\\frac{(ab + bc + ca) + 6abc(a + b + c)}{abc}}.\n\\end{aligned}\n$$\nNote that $ab + bc + ca = 1$ and\n$$\nabc(a + b + c) = (ab)(bc) + (bc)(ca) + (ca)(ab) \\le \\frac{1}{3}(ab + bc + ca)^2 = \\frac{1}{3}.\n$$\nTherefore, we have $S \\le 3^{\\frac{7}{8}}\\sqrt[4]{\\frac{3}{abc}} = 3^{\\frac{9}{8}}(abc)^{-\\frac{1}{4}}$. It remains to prove $abc \\le 3^{-\\frac{3}{2}}$.\nIndeed, by the AM-GM inequality, we have\n$$\n1 = ab + bc + ca \\ge 3\\sqrt[3]{a^2 b^2 c^2}.\n$$\nThis clearly implies $abc \\le 3^{-\\frac{3}{2}}$. This proves the desired inequality. Equality holds when $a = b = c = \\frac{1}{\\sqrt{3}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31556,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find a positive integer $m$ such that $x^3 \\equiv 1 \\pmod m$ has more than 2012 solutions. (Note: If $x \\equiv y \\pmod m$, then $x$ and $y$ are considered as the same solution.)",
    "options": [],
    "answer": "5201836549",
    "solution": "A possible $m$ is $5201836549$.\nLet $p_1, p_2, \\dots, p_7$ be the primes $7, 13, 19, 31, 37, 43, 61$ which are congruent to $1$ modulo $3$, and let $m = p_1p_2\\cdots p_7$. Since $3 \\mid p_j - 1$, the equation $x^3 \\equiv 1 \\pmod{p_j}$ has three distinct solutions modulo $p_j$ by the cubing lemma (or primitive root).\n\nThus, there are $3^7 = 2187 > 2012$ tuples $(x_1, x_2, \\dots, x_7)$ such that $x_j^3 \\equiv 1 \\pmod{p_j}$ for each $j$. By the Chinese remainder theorem, there exists $x$ such that\n$$\nx \\equiv x_j \\pmod{p_j}\n$$\nfor each $j$, and so\n$$\nx^3 \\equiv x_j^3 \\equiv 1 \\pmod{p_j}.\n$$\nSince this holds for all $j$, we have $x^3 \\equiv 1 \\pmod{m}$. Note that different tuples $(x_1, x_2, \\dots, x_7)$ and $(x'_1, x'_2, \\dots, x'_7)$ correspond to different $x$'s modulo $m$, since $x_j \\not\\equiv x'_j \\pmod{p_j}$ for at least one $j$. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 31557,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $n$, define $f(n)$ as the sum of the squares of the digits of $n$ (in base ten). For example, $f(2012) = 2^2 + 0^2 + 1^2 + 2^2 = 9$. Find the greatest integer $n$ such that there exists a positive integer $d$ with $f(\\underbrace{f(\\cdots f(n)\\cdots)}_{d \\text{ times}}) = n$.",
    "options": [],
    "answer": "145",
    "solution": "The greatest $n$ is $145$.\n\nOne can easily check that\n$$\n145 \\mapsto 42 \\mapsto 20 \\mapsto 4 \\mapsto 16 \\mapsto 37 \\mapsto 58 \\mapsto 89 \\mapsto 145,\n$$\nso that $f^8(145) = 145$. It remains to check $n \\le 145$.\n\nWe first prove that $n$ has at most $3$ digits. Suppose on the contrary that $n = \\overline{a_1a_2\\cdots a_k}$ where $k \\ge 4$. For any $m \\le n$, we have\n$$\nf(m) \\le f(\\underbrace{99\\cdots9}_{k \\text{ times}}) = 81k < 10^{k-1} \\le n.\n$$\n(To prove $81k < 10^{k-1}$ for $k \\ge 4$, we check the base case $324 < 1000$ and the inductive step $81(k+1) < 10(81k) < 10^k$.) Using this, we can easily prove that $f^s(n) < n$ for any $s \\in \\mathbb{Z}^+$. Thus, it is impossible that $f^s(n) = n$.\n\nSimilarly, if $n > 243$, then for $m \\le n$, we have\n$$\nf(m) \\le f(999) = 243 < n.\n$$\nIf $163 < n \\le 243$, then for $m \\le n$, we have\n$$\nf(m) \\le f(199) = 163 < n.\n$$\nThus, we must have $n \\le 163$.\n\nNote that $n$ is also the image of some term in the cycle. Let $n = f(t)$ for some $t \\le 163$. If $n > 145 = 8^2 + 9^2$, then the only possibility is $t = 99$. But then we have\n$$\n99 \\mapsto 162 \\mapsto 41 \\mapsto 17 \\mapsto 50 \\mapsto 25 \\mapsto 29 \\mapsto 85 \\mapsto 89 \\mapsto 145,\n$$\nwhich enters the cycle with largest term $145$, and so it never goes back to $162$. Therefore, we must have $n \\le 145$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31558,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive integer $a$, define $M(a)$ to be the number of positive integers $b$ for which $a+b$ divides $ab$. Find all integer(s) $a$ with $1 \\le a \\le 2013$, so that $M(a)$ is largest possible in the range of $a$.",
    "options": [],
    "answer": "1680",
    "solution": "The answer is $a = 1680$.\nSince $ab \\equiv a(-a) = -a^2 \\pmod{a+b}$, we have $a+b \\mid ab$ if and only if $a+b \\mid a^2$. This shows $a+b$ can be any divisor of $a^2$ greater than $a$. Also, any such divisor corresponds to a unique positive integer $b$. Therefore, $M(a)$ is the number of divisors of $a^2$ greater than $a$. Since the positive divisors of $a^2$ less than $a$ are in one-to-one correspondence to the divisors of $a^2$ greater than $a$, it remains to find $a$ which has the maximum number of positive divisors.\nFor any $a = p_1^{c_1} p_2^{c_2} \\cdots p_s^{c_s}$ where $p_1, p_2, \\ldots, p_s$ are distinct prime divisors of $a$, recall that $d(a^2) = (2c_1+1)(2c_2+1)\\cdots(2c_s+1)$. WLOG we may assume $c_1 \\ge c_2 \\ge \\cdots \\ge c_s$. Since $2 \\times 3 \\times 5 \\times 7 \\times 11 = 2310 > 2013$, we have $s \\le 4$.\n* If $s=4$, we must have $c_3 = c_4 = 1$ since $2^2 3^2 5^2 7 = 6300 > 2013$.\n- If $c_2 \\ge 2$, we must have $c_1 = c_2 = 2$ since $2^3 3^2 5 \\cdot 7 = 2520 > 2013$. When $a = p_1^2 p_2^2 p_3 p_4$, we have $d(a^2) = 5^2 \\cdot 3^2 = 225$.\n- If $c_2 = 1$, we have $c_1 \\le 4$ since $2^5 3 \\cdot 5 \\cdot 7 = 3360 > 2013$. When $a = p_1^4 p_2 p_3 p_4$, we have $d(a^2) = 9 \\cdot 3^3 = 243$.\n\n• If $s = 3$, we must have $c_3 \\le 2$ since $2^33^35^3 = 27000 > 2013$.\n– If $c_3 = 2$, we must have $c_2 = 2$ since $2^33^35^2 = 5400 > 2013$. Then $c_1 \\le 3$ since $2^43^25^2 = 3600 > 2013$. When $a = p_1^3 p_2^2 p_3^2$, we have $d(a^2) = 7 \\cdot 5^2 = 175$.\n– If $c_3 = 1$, we must have $c_2 \\le 3$ since $2^43^45 = 6480 > 2013$.\n* If $c_2 = 3$, we have $c_1 = 3$ since $2^43^35 = 2160 > 2013$. When $a = p_1^3 p_2^3 p_3$, we have $d(a^2) = 7^2 \\cdot 3 = 147$.\n* If $c_2 = 2$, we have $c_1 \\le 5$ since $2^63^25 = 2880 > 2013$. When $a = p_1^5 p_2^2 p_3$, we have $d(a^2) = 11 \\cdot 5 \\cdot 3 = 165$.\n* If $c_2 = 1$, we have $c_1 \\le 7$ since $2^83 \\cdot 5 = 3840 > 2013$. When $a = p_1^7 p_2 p_3$, we have $d(a^2) = 15 \\cdot 3^2 = 135$.\n• If $s = 2$, we must have $c_2 \\le 4$ since $2^53^5 = 7776 > 2013$.\n– If $c_2 = 4$, we have $c_1 = 4$ since $2^53^4 = 2592 > 2013$. When $a = p_1^4 p_2^4$, we have $d(a^2) = 9^2 = 81$.\n– If $c_2 = 3$, we have $c_1 \\le 6$ since $2^73^3 = 3456 > 2013$. When $a = p_1^6 p_2^3$, we have $d(a^2) = 13 \\cdot 7 = 91$.\n– If $c_2 = 2$, we have $c_1 \\le 7$ since $2^83^2 = 2304 > 2013$. When $a = p_1^7 p_2^2$, we have $d(a^2) = 15 \\cdot 5 = 75$.\n– If $c_2 = 1$, we have $c_1 \\le 9$ since $2^{10}3 = 3072 > 2013$. When $a = p_1^9 p_2$, we have $d(a^2) = 19 \\cdot 3 = 57$.\n• If $s \\le 1$, we have $c_1 \\le 10$ since $2^{11} = 2048 > 2013$. When $a = p_1^{10}$, we have $d(a^2) = 21$.\nTherefore, $a^2$ has at most 243 positive divisors. This holds when $a = p_1^4 p_2 p_3 p_4$ for some distinct primes $p_1, p_2, p_3, p_4$. As $3^42 \\cdot 3 \\cdot 5 = 2430 > 2013$, we must have $p_1 = 2$. Also, since $2^43 \\cdot 5 \\cdot 11 = 2640 > 2013$, the only possibility is $a = 2^43 \\cdot 5 \\cdot 7 = 1680$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31559,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{1, 2, 3, \\dots, 2n\\}$, where $n$ is a positive integer greater than or equal to $1$. For any subset $T$ of $S$, $T$ is called a *good* subset if in $T$ the number of even elements is greater than the number of odd elements.\n\na. Find the total number of good subsets of $S$.\n\nb. Find the sum of all the elements in all the good subsets of $S$.",
    "options": [],
    "answer": "a: 2^{2n-1} - \\tfrac{1}{2}\\binom{2n}{n}; b: 2^{2n-2}(2n^2 + n) - n^2\\binom{2n-1}{n}",
    "solution": "a.\nThe answer is $2^{2n-1} - \\frac{1}{2}\\binom{2n}{n}$.\n\nA subset $T$ of $S$ is called a *bad* subset if in $T$ the number of odd elements is greater than the number of even elements. A subset of $S$ is neither good nor bad if it has exactly $k$ odd elements and $k$ even elements for some $k = 0, 1, \\dots, n$. Since there are $n$ odd elements and $n$ even elements in $S$, the number of subsets which are neither good nor bad is\n$$\n\\sum_{k=0}^{n} \\binom{n}{k}^2 = \\binom{2n}{n}\n$$\nby Vandermonde's identity.\n\nNow, by symmetry, the number of good subsets is the same as the number of bad subsets. As there are $2^{2n}$ subsets of $S$ in total, the total number of good subsets is\n$$\n\\frac{1}{2}\\left[2^{2n} - \\binom{2n}{n}\\right] = 2^{2n-1} - \\frac{1}{2}\\binom{2n}{n}.\n$$\n\nb.\nThe answer is $2^{2n-2}(2n^2 + n) - n^2\\binom{2n-1}{n}$.\n\nWe first count the number $N_1$ of good subsets containing a particular even number. Suppose such a good subset contains $i$ more even numbers and $j$ odd numbers. Then we need $i \\ge j$. As there are $n-1$ even numbers remaining and $n$ odd numbers in total, we have\n$$\nN_1 = \\sum_{i=0}^{n-1} \\sum_{j=0}^{i} \\binom{n-1}{i} \\binom{n}{j} = \\sum_{i=0}^{n-1} \\sum_{k=n-i}^{n} \\binom{n-1}{i} \\binom{n}{k} = \\sum_{i+k \\ge n} \\binom{n-1}{i} \\binom{n}{k}\n$$\nby using the change of variable $k = n-j$. Note that this is equal to the sum of coefficients of all $x^m$ with $m \\ge n$ in $(1+x)^{n-1}(1+x)^n = (1+x)^{2n-1}$. Thus, we obtain\n$$\nN_1 = \\sum_{m=n}^{2n-1} \\binom{2n-1}{m} = \\frac{1}{2} \\sum_{m=0}^{2n-1} \\binom{2n-1}{m} = 2^{2n-2}.\n$$\n\nSimilarly, we count the number $N_2$ of good subsets containing a particular odd number. Suppose such a good subset contains $i \\ge 2$ even numbers and $j$ more odd numbers. Then we need $i \\ge j + 2$. This implies\n$$\n\\begin{align*}\nN_2 &= \\sum_{i=2}^{n} \\sum_{j=0}^{i-2} \\binom{n}{i} \\binom{n-1}{j} \\\\\n&= \\sum_{i=2}^{n} \\sum_{k=n+1-i}^{n-1} \\binom{n}{i} \\binom{n-1}{k} \\\\\n&= \\sum_{i+k \\ge n+1} \\binom{n}{i} \\binom{n-1}{k} \\\\\n&= \\sum_{m=n+1}^{2n-1} \\binom{2n-1}{m} \\\\\n&= \\frac{1}{2} \\left( \\sum_{m=0}^{2n-1} \\binom{2n-1}{m} - \\binom{2n-1}{n-1} - \\binom{2n-1}{n} \\right) \\\\\n&= 2^{2n-2} - \\binom{2n-1}{n}.\n\\end{align*}\n$$\n\nNow, the sum of all the elements in all the good subsets of $S$ is\n$$\n\\begin{align*}\nN_1(2+4+\\cdots+2n) + N_2(1+3+\\cdots+(2n-1)) \\\\\n&= 2^{2n-2} \\cdot n(n+1) + \\left[ 2^{2n-2} - \\binom{2n-1}{n} \\right] \\cdot n^2 \\\\\n&= 2^{2n-2}(2n^2+n) - n^2 \\binom{2n-1}{n}.\n\\end{align*}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31560,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ be real numbers. It is given that the equation $x^4 + a x^3 + b x^2 + a x + 1 = 0$ has at least one real root. Find the minimum value of $a^2 + b^2$.",
    "options": [],
    "answer": "4/5",
    "solution": "The minimum value of $a^2 + b^2$ is $\\frac{4}{5}$.\n\nWLOG assume $a \\ge 0$. (Indeed, by flipping the sign of $a$, we only need to flip the sign of the root $x$.) Clearly, $x \\ne 0$. Rewrite the equation as\n$$\nx^2 + a x + b + \\frac{a}{x} + \\frac{1}{x^2} = 0.\n$$\nThis is the same as\n$$\n\\left(x + \\frac{1}{x}\\right)^2 + a\\left(x + \\frac{1}{x}\\right) + b - 2 = 0.\n$$\nLet $y = x + \\frac{1}{x}$. Note that $|y| = |x| + \\frac{1}{|x|} \\ge 2$ by the AM-GM inequality. The equation becomes\n$$\ny^2 + a y + (b - 2) = 0.\n$$\nIf $a > 1$, then $a^2 + b^2 > 1$. Thus, we only need to consider the case $a \\le 1$. This implies the axis of symmetry $y = -\\frac{a}{2}$ lies between $-2$ and $2$.\n\n* If $y \\ge 2$, then we have\n$$\n0 = y^2 + a y + (b-2) \\ge (2)^2 + a(2) + (b-2) = 2a + b + 2.\n$$\nThis yields $b \\le -2a - 2 \\le -2$, and hence $a^2 + b^2 \\ge 4$.\n\n* If $y \\le -2$, then we have\n$$\n0 = y^2 + a y + (b-2) \\ge (-2)^2 + a(-2) + (b-2) = -2a + b + 2.\n$$\nThis yields $b \\le 2a - 2 \\le 0$, and hence\n$$\na^2 + b^2 \\ge a^2 + (2a - 2)^2 = 5\\left(a - \\frac{4}{5}\\right)^2 + \\frac{4}{5} \\ge \\frac{4}{5}.\n$$\nEquality holds when $a = \\frac{4}{5}$ and $b = -\\frac{2}{5}$. Correspondingly, this means $y = -2$ and $x = -1$.\n\nTo conclude, the minimum value of $a^2 + b^2$ is $\\frac{4}{5}$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31561,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\triangle ABC$ with $CA > BC > AB$, let $O$ and $H$ be the circumcentre and orthocentre of $\\triangle ABC$ respectively. Denote by $D$ and $E$ the midpoints of arcs $\\widehat{AB}$ and $\\widehat{AC}$ of the circumcircle of $\\triangle ABC$ not containing the opposite vertices. Let $D'$ be the reflection of $D$ in side $AB$ and $E'$ be the reflection of $E$ in side $AC$. Prove that $O, H, D', E'$ are concyclic if and only if $A, D', E'$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim that both statements are equivalent to $\\angle A = 60^\\circ$.\n\nFirstly,\n$A, D', E'$ are collinear\n$$\n\\Leftrightarrow \\angle BAD' + \\angle CAE' = \\angle BAC\n$$\n$$\n\\Leftrightarrow \\angle BAD + \\angle CAE = \\angle BAC\n$$\n$$\n\\Leftrightarrow \\frac{C}{2} + \\frac{B}{2} = A\n$$\n$$\n\\Leftrightarrow 90^\\circ - \\frac{A}{2} = A\n$$\n$$\n\\Leftrightarrow A = 60^\\circ.\n$$\n\nSecondly, it is well-known that the reflection of $H$ in $AB$ lies on $(ABC)$. Therefore, $A, B, H, D'$ are concyclic. Also, note that $D, D', O$ are collinear since all of them lie on the perpendicular bisector of $AB$. It follows that\n$$\n\\begin{align*} \n\\angle DD'H &= \\angle DD'B + \\angle BD'H \\\\ \n&= (90^\\circ - \\angle ABD') + \\angle BAH \\\\ \n&= (90^\\circ - \\angle ABD) + \\angle BAH \\\\ \n&= \\left(90^\\circ - \\frac{C}{2}\\right) + 90^\\circ - B \\\\ \n&= A + \\frac{C}{2}. \n\\end{align*}\n$$\n\nSimilarly, $C, A, H, E'$ are concyclic, and we have\n$$\n\\begin{align*} \n\\angle OE'H &= \\angle CE'H - \\angle CE'E \\\\ \n&= (180^\\circ - \\angle CAH) - (90^\\circ - \\angle ACE) \\\\ \n&= (90^\\circ + C) - \\left(90^\\circ - \\frac{B}{2}\\right) \\\\ \n&= C + \\frac{B}{2}. \n\\end{align*}\n$$\nNow,\n$O, H, D', E'$ are concyclic\n$$\n\\begin{align*} \n\\angle DD'H &= \\angle OE'H \\\\ \nA + \\frac{C}{2} &= C + \\frac{B}{2} \\\\ \nA &= 90^\\circ - \\frac{A}{2} \\\\ \nA &= 60^\\circ. \n\\end{align*}\n$$\n\nTherefore, the two statements are equivalent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31562,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any positive real numbers $a$, $b$, $c$ satisfying $a + b + c = 1$, prove that\n$$\n\\sqrt{a^2 - 2bc + 2c} + \\sqrt{b^2 - 2ca + 2a} + \\sqrt{c^2 - 2ab + 2b} \\ge \\sqrt{5}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Using the condition $a + b + c = 1$, we obtain\n$$\na^2 - 2bc + 2c = a^2 + 2c(1 - b) = a^2 + 2c(a + c) = (c + a)^2 + c^2.\n$$\nSimilarly, the left-hand side of the inequality becomes\n$$\n\\sqrt{(c + a)^2 + c^2} + \\sqrt{(a + b)^2 + a^2} + \\sqrt{(b + c)^2 + b^2}.\n$$\nBy the triangle inequality, this is bounded below by\n$$\n\\sqrt{((c + a) + (a + b) + (b + c))^2 + (c + a + b)^2} = \\sqrt{5}.\n$$\nEquality holds when $a = b = c = \\frac{1}{3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31563,
    "subject": "Mathematics (Multi-modal)",
    "question": "$S = \\{(x, y) : x, y$ are integers, $0 \\le x \\le 5$ and $0 \\le y \\le 5\\}$ is a set of lattice points on the rectangular coordinate plane. How many triangles can be formed by taking any three elements of $S$?",
    "options": [],
    "answer": "6768",
    "solution": "There are 6768 triangles.\nSince there are 36 points in $S$, there are $\\binom{36}{3} = 7140$ triples of points. We have to remove those triples which contain collinear points.\n\nFirstly, there are 6 rows, and there are $\\binom{6}{3} = 20$ ways to choose a triple from each row. Therefore, there are $6 \\times 20 = 120$ triples consisting of three points in the same row. By symmetry, there are 120 triples consisting of three points in the same column.\n\nSecondly, we consider those diagonal lines with slope 1. The number of points in these diagonal lines are 1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1 respectively. Thus, the number of triples we can choose from these diagonal lines is\n$$\n\\binom{3}{3} + \\binom{4}{3} + \\binom{5}{3} + \\binom{6}{3} + \\binom{5}{3} + \\binom{4}{3} + \\binom{3}{3} = 50.\n$$\nBy symmetry, there are 50 triples coming from the diagonal lines with slope $-1$.\n\nThirdly, we consider those lines with slope 2. We can list out all triples lying on these lines as follows:\n$$\n\\begin{aligned}\n& \\{((0,0), (1,2), (2,4)), ((0,1), (1,3), (2,5)), \\\\\n& \\{((1,0), (2,2), (3,4)), ((1,1), (2,3), (3,5)), \\\\\n& \\{((2,0), (3,2), (4,4)), ((2,1), (3,3), (4,5)), \\\\\n& \\{((3,0), (4,2), (5,4)), ((3,1), (4,3), (5,5)).\n\\end{aligned}\n$$\nThere are 8 such triples. By symmetry, there are 8 triples coming from the lines with slope $-2$, 8 triples coming from the lines with slope $\\frac{1}{2}$, and 8 triples coming from the lines with slope $-\\frac{1}{2}$.\n\nLastly, since any other line contains at most 2 points in $S$, the answer is\n$$\n7140 - 120 \\times 2 - 50 \\times 2 - 8 \\times 4 = 6768.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31564,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a chess tournament there are $n$ players (where $n > 1$ is odd), and every two players play against each other exactly once. It is known that exactly $n$ games end in a tie. For any set $S$ of players including $A$ and $B$, we say that $A$ *admires* $B$ in $S$ if\n(a) $A$ does not beat $B$; or\n(b) there exists a sequence of other distinct players $C_1, C_2, \\dots, C_k$ in $S$ such that $A$ does not beat $C_1$; $C_k$ does not beat $B$; and $C_i$ does not beat $C_{i+1}$ for $1 \\le i \\le k-1$.\nA set of four players is said to be *harmonic* if each of the four players admires everyone else in the set. Find (in terms of $n$) the greatest possible number of harmonic sets.",
    "options": [],
    "answer": "n(n-3)(n^2+6n-31)/48",
    "solution": "The answer is $\\frac{n(n-3)(n^2+6n-31)}{48}$.\n\nFor each $k$, let $d_k$ be the number of games that player $k$ wins. Note that\n$$\nS := d_1 + d_2 + \\dots + d_n = \\binom{n}{2} - n = \\frac{n(n-3)}{2}.\n$$\nObserve that for a set of four players, if one of them beats everybody else, then the set is not harmonic. Also, there cannot be two players beating everybody else in the set simultaneously. Therefore, the number $N$ of harmonic sets is at most\n$$\n\\binom{n}{4} - \\sum_{k=1}^{n} \\binom{d_k}{3}.\n$$\nNote that the binomial function $\\binom{x}{3}$ is convex. So we can use the majorization inequality to get\n$$\nN \\le \\binom{n}{4} - n \\binom{\\frac{n-3}{2}}{3} = \\frac{n(n-3)(n^2+6n-31)}{48}.\n$$\nIt remains to show that this bound is attainable.\n\nSuppose the game between player $k$ and player $k+1$ ends in a tie for all $k$, where player $n+1$ means player 1. For any other game between player $i$ and player $j$ where $1 \\le i < j \\le n$, player $j$ beats player $i$ if and only if $i \\equiv j \\pmod 2$.\nFor even $k$, player $k$ beats players $2, 4, \\dots, k-2$ and $k+3, k+5, \\dots, n$. So player $k$ has won $\\frac{n-3}{2}$ games. For odd $k < n$, player $k$ beats players $1, 3, \\dots, k-2$ and $k+3, k+5, \\dots, n-1$. So player $k$ has won $\\frac{n-3}{2}$ games. Also, player $n$ beats players $3, 5, \\dots, n-2$. So player $n$ has won $\\frac{n-3}{2}$ games. This shows all players have won the same number of games. Therefore, the equality of the majorization inequality holds.\n\nWe now show that for any set of four players in which nobody beats everyone else, it is harmonic. Once this is shown, all equalities in the above deduction hold, which means the bound is attained. Suppose on the contrary that there exists a non-harmonic set $S$ of four players $a, b, c, d$ such that none of them beats everyone else in this set.\n\n* If there is no tie among the games played between $a, b, c, d$, then WLOG we may assume player $a$ beats players $b$ and $c$. By the assumption, player $d$ must beat player $a$. Since player $d$ does not beat everyone else, WLOG assume player $b$ beats player $d$. If player $c$ beats player $b$, then we have the cycle $a \\to c \\to b \\to d \\to a$, and so the set is harmonic. If player $c$ beats player $d$, WLOG assume player $b$ beats player $c$. Then we have the cycle $a \\to b \\to c \\to d \\to a$, and so the set is harmonic.\n\nThe only case left is that both players $b$ and $d$ beat player $c$, which means player $c$ is beaten by everyone else. This holds when each of $a, b, d$ is less than and has different parity as $c$, or is greater than and has the same parity as $c$. WLOG assume $a < b < d$.\n- If $a < b < d < c$, then player $d$ beats everyone else, contradiction.\n- If $a < b < c < d$, then player $b$ beats everyone else, contradiction.\n- If $a < c < b < d$, then player $a$ beats everyone else, contradiction.\n- If $c < a < b < d$, then player $d$ beats everyone else, contradiction.\n\n* If there is a tie in the game played between $c$ and $d$, then $c$ and $d$ are consecutive integers, or are $1$ and $n$. By the construction, it is impossible that both $c$ and $d$ beat the same player, or are beaten by the same player. (This is the place where we have used $n$ is odd, since otherwise $1$ and $n$ are beaten by some players in the same construction of the even case.)\n\nWLOG assume player $a$ does not beat player $c$, and player $d$ does not beat player $a$. If the same holds when $a$ is replaced by $b$, then we have a cycle $c \\to a \\to d \\to c \\to b \\to d \\to c$, and so the set is harmonic. (Although the cycle does not consist of distinct players, we can easily shorten the sequence so that it only consists of distinct players.) If player $b$ does not beat player $d$, and player $c$ does not beat player $b$, then we also have a cycle $c \\to a \\to d \\to b \\to c$, and so the set is harmonic.\n\nThis shows the construction works. So the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31565,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$, $E$ and $F$ be the contact points of the incircle of $\\triangle ABC$ with sides $BC$, $CA$ and $AB$ respectively. Let $I$ and $I'$ be the incentres of $\\triangle ABC$ and $\\triangle DEF$, respectively. Let $\\ell_A$ be the line passing through $D$ and is parallel to $AI'$, $\\ell_B$ be the line passing through $E$ and is parallel to $BI'$, and $\\ell_C$ be the line passing through $F$ and is parallel to $CI'$. Show that the lines $\\ell_A$, $\\ell_B$ and $\\ell_C$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\ell_A$ meet $II'$ at $P$. We claim that $P$ is the intersection point of $\\ell_A$, $\\ell_B$, $\\ell_C$. It suffices to show that $\\frac{II'}{I'P}$ is a constant.\nLet $DI'$ meet the incircle of $\\triangle ABC$ again at $M$, and meet the line passing through $I$ and parallel to $\\ell_A$ at $Q$. Let $r$ and $r'$ be the inradii of $\\triangle ABC$ and $\\triangle DEF$ respectively. By similar triangles, we obtain\n$$\n\\frac{II'}{I'P} = \\frac{QI'}{I'D} = \\frac{1}{I'D} \\times \\frac{MI' \\times IA}{MA} = \\frac{MI'}{I'D} \\times \\frac{IA}{MA}.\n$$\nSince $I'$ is the incentre of $\\triangle DEF$, we have $MI' = MF$. Thus, we have\n$$\n\\frac{MI'}{I'D} = MF \\times \\frac{1}{I'D} = \\frac{d(M, EF)}{\\sin \\angle EFM} \\times \\frac{\\sin \\angle I'DF}{r'} = \\frac{d(M, EF)}{r'}\n$$\nsince $\\angle EFM = \\frac{1}{2} \\angle EDF = \\angle I'DF$.\n\n$$\n\\frac{IA}{MA} = \\frac{d(I, AF)}{d(M, AF)} = \\frac{r}{d(M, EF)}\n$$\nCombining these, we obtain\n$$\n\\frac{II'}{I'P} = \\frac{MI'}{I'D} \\times \\frac{IA}{MA} = \\frac{d(M, EF)}{r'} \\times \\frac{r}{d(M, EF)} = \\frac{r}{r'}\n$$\nThis is a constant, and so $\\ell_A$, $\\ell_B$, $\\ell_C$ are concurrent at $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31566,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a competition, each participant can get an integral score lying between $0$ and $10$ inclusive. It is found that the mean, median and mode of the scores are $3$, $3$, $7$ respectively. Find the minimum number of participants.",
    "options": [],
    "answer": "17",
    "solution": "The minimum number of participants is $17$.\nLet $a_n$ be the number of participants obtaining score $n$, and let $m = a_0 + a_1 + \\cdots + a_{10}$ be the total number of participants. Using the mean, we have $\\sum_{n=0}^{10} n a_n = 3m$. This means\n$$\n\\sum_{n=4}^{10} (n-3)a_n = 3a_0 + 2a_1 + a_2. \\qquad (1)\n$$\n\n**Case 1.** $m$ is odd\nSince $3$ is the median, we must have $a_0 + a_1 + a_2 \\le \\frac{m-1}{2}$. This implies\n$$\na_0 + a_1 + a_2 + 1 \\le \\sum_{n=3}^{10} a_n. \\qquad (2)\n$$\nUsing (1) and (2), we obtain\n$$\n3a_0 + 2a_1 + a_2 = \\sum_{n=4}^{10} (n-3)a_n = \\sum_{n=3}^{10} a_n + \\sum_{n=4}^{10} (n-4)a_n - a_3 \\\\\n\\ge (a_0 + a_1 + a_2 + 1) + (3a_7) - a_3.\n$$\nThis implies $2a_0 + a_1 + a_3 \\ge 3a_7 + 1$. Since the mode is $7$, we have $a_0, a_1, a_3 \\le a_7 - 1$. This yields\n$$\n3a_7 + 1 \\le 2a_0 + a_1 + a_3 \\le 4(a_7 - 1),\n$$\nand hence $a_7 \\ge 5$. Now, we have\n$$\nm \\ge a_0 + a_1 + a_3 + a_7 = (2a_0 + a_1 + a_3) - a_0 + a_7 \\ge (3a_7 + 1) - (a_7 - 1) + a_7 \\ge 17.\n$$\nEquality holds when $a_0 = a_1 = a_3 = 4$, $a_7 = 5$ and all other $a_j$'s are $0$. One checks that all conditions are satisfied.\n\n**Case 2.** $m$ is even\nIf $a_3 > 0$, then the two scores in the middle are $3$. The bound $a_0 + a_1 + a_2 \\le \\frac{m-1}{2}$ used in case 1 still holds, and so we still have $m \\ge 17$.\nIf $a_3 = 0$, then we only have $a_0 + a_1 + a_2 \\le \\frac{m}{2}$. This implies\n$$\na_0 + a_1 + a_2 \\le \\sum_{n=4}^{10} a_n. \\qquad (3)\n$$\nUsing (1) and (3), we obtain\n$$\n3a_0 + 2a_1 + a_2 = \\sum_{n=4}^{10} (n-3)a_n = \\sum_{n=4}^{10} a_n + \\sum_{n=4}^{10} (n-4)a_n \\\\\n\\ge (a_0 + a_1 + a_2) + (3a_7).\n$$\nThis implies $2a_0 + a_1 \\ge 3a_7$. But then one of $a_0, a_1$ is greater than or equal to $a_7$, contradicting the fact that $7$ is the mode.\nIn any case, we have $m \\ge 17$, and so $17$ is the minimum value.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31567,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a line $\\ell$ and four points $P$, $Q$, $R$ and $S$ (in that order) on the line, describe a straight edge (unmarked ruler) and compass construction producing a square $ABCD$ such that $P$ lies on the line $AD$, $Q$ on the line $BC$, $R$ on the line $AB$ and $S$ on the line $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that we can use compass and ruler to find the perpendicular bisector of two given points. Thus, we can use this to locate the midpoint of two points, and draw the circle having a given segment as a diameter.\n\nConstruct the circles with diameter $PS$ and $QR$ respectively. Construct the midpoints $M$ and $N$ of $\\overline{PS}$ and $\\overline{QR}$ (using the perpendicular bisectors) such that $M$ and $N$ lie on the same side of $\\ell$. Construct the intersection point $B$ of the line $MN$ and $(QNR)$, and construct the intersection point $D$ of the line $MN$ and $(PMS)$. Also, construct the intersection point $A$ of $PD$ and $RB$, and construct the intersection point $C$ of $SD$ and $QB$. We claim that $ABCD$ is the desired square.\n\n![](attached_image_1.png)\n\nBy construction, we already know that $P$, $Q$, $R$, $S$ lie on $AD$, $BC$, $AB$, $CD$ respectively. Since $PS$ and $QR$ are diameters of the two circles, we easily obtain $\\angle ADC = \\angle ABC = 90^\\circ$. Also, since $M$ and $N$ are the midpoints of $\\overline{PS}$ and $\\overline{QR}$, the line $BD$ bisects $\\angle ADC$ and $\\angle ABC$. It follows that\n$$\n\\angle ADB = \\angle CDB = \\angle ABD = \\angle CBD = 45^\\circ,\n$$\nwhich implies $\\angle BAD = \\angle BCD = 90^\\circ$. Also, as $\\angle ABD = \\angle ADB$, we obtain $AB = AD$. This proves $ABCD$ is a square.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31568,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x-y=12$, find the value of $x^3-y^3-36xy$.\n\n若 $x-y=12$, 求 $x^3-y^3-36xy$ 的值。",
    "options": [],
    "answer": "1728",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31569,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ be a function defined on nonzero real numbers such that for any nonzero $x$, $$\n\\frac{2}{x}f(-x) + f\\left(\\frac{1}{x}\\right) = x.\n$$ Find the value of $f(8)$.",
    "options": [],
    "answer": "205/8",
    "solution": "We have $f(8) = \\frac{205}{8}$.\nIt is given that\n$$\n\\frac{2}{x}f(-x) + f\\left(\\frac{1}{x}\\right) = x \\quad (1)\n$$\nfor any $x \\neq 0$. Replacing $x$ by $-\\frac{1}{x}$, we obtain\n$$\n-2x f\\left(\\frac{1}{x}\\right) + f(-x) = -\\frac{1}{x}. \\quad (2)\n$$\nUsing (1) and (2), we can solve for $f(-x)$. Indeed, we have\n$$\n5f(-x) = 2x\\left[\\frac{2}{x}f(-x) + f\\left(\\frac{1}{x}\\right)\\right] + \\left[-2x f\\left(\\frac{1}{x}\\right) + f(-x)\\right] = 2x^2 - \\frac{1}{x}.\n$$\nReplacing $x$ by $-x$, we obtain\n$$\nf(x) = \\frac{2}{5}x^2 + \\frac{1}{5x}.\n$$\nFor this function, we check that\n$$\n\\frac{2}{x}f(-x) + f\\left(\\frac{1}{x}\\right) = \\frac{2}{x}\\left(\\frac{2}{5}x^2 - \\frac{1}{5x}\\right) + \\frac{2}{5x^2} + \\frac{x}{5} = x.\n$$\nTherefore, this is the only possible function, and so\n$$\nf(8) = \\frac{2}{5}(8)^2 + \\frac{1}{5(8)} = \\frac{205}{8}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31570,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polygon is *monochromatic* if all its vertices are coloured by a same colour. Suppose now every point of the plane is coloured red or blue. Show that there exists either a monochromatic equilateral triangle of side length $2$, or a monochromatic equilateral triangle of side length $\\sqrt{3}$, or a monochromatic rhombus of side length $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we show that there exists either a monochromatic equilateral triangle of length $1$, or a monochromatic equilateral triangle of length $\\sqrt{3}$. Indeed, if there is no monochromatic equilateral triangle of length $1$, then we can find two points $A$ and $B$ such that $AB = 1$ and they are in different colours (say $A$ is red and $B$ is blue). Construct an isosceles triangle $ABC$ with $AC = BC = 2$.\n\nWithout loss of generality, assume $C$ is blue. Let $M$ be the midpoint of $AC$. Then it has the same colour as $A$ or $C$, say $A$. Construct two equilateral triangles $ADM$ and $AEM$. Since there is no monochromatic equilateral triangle of length $1$, the colours of $D$ and $E$ are different from that of $A$, and hence both are blue. Thus $\\triangle CDE$ is a monochromatic equilateral triangle of length $\\sqrt{3}$.\n\n![](attached_image_1.png)\n\nNow suppose there is a monochromatic equilateral triangle $PQR$ of length $1$. We assume all its vertices are red in colour and construct three more equilateral triangles $PQW$, $QRU$ and $RPV$. If all of the points $U$, $V$, $W$ are blue, then $UVW$ is a monochromatic equilateral triangle of length $2$ and we are done. If not, at least one of them is red, say $U$, then $PQUR$ is a monochromatic rhombus of length $1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31571,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a = BC$, $b = CA$ and $c = AB$ be respectively the lengths of a triangle $ABC$; $i_a, i_b, i_c$ be respectively the lengths of the angle bisectors from $A, B$ and $C$. Let $R$ be the circumradius of the triangle. Prove that\n$$\nai_a + bi_b + ci_c < 9R^2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the intersection point of the internal angle bisector of $\\angle BAC$ and $BC$.\nBy the angle bisector theorem, we have $\\frac{BD}{CD} = \\frac{c}{b}$, hence\n$$\nBD = \\frac{ac}{b+c} \\quad \\text{and} \\quad CD = \\frac{ab}{b+c}.\n$$\nBy Stewart's theorem, we have\n$$\ni_a^2 = \\frac{1}{a} \\left( b^2 \\cdot \\frac{ac}{b+c} + c^2 \\cdot \\frac{ab}{b+c} \\right) - \\frac{ac}{b+c} \\cdot \\frac{ab}{b+c} = \\frac{bc((b+c)^2 - a^2)}{(b+c)^2} = \\frac{4bcs(s-a)}{(b+c)^2},\n$$\nwhere $s$ is the semiperimeter of $\\triangle ABC$. Now, by the AM-GM inequality, since $s \\neq s-a$, we have\n$$\ni_a = \\frac{2\\sqrt{bc}}{b+c} \\cdot \\sqrt{s(s-a)} < \\frac{2\\sqrt{bc}}{b+c} \\cdot \\frac{s+(s-a)}{2} = \\sqrt{bc} \\le \\frac{b+c}{2}.\n$$\nSimilarly, we have $i_b < \\frac{c+a}{2}$ and $i_c < \\frac{a+b}{2}$. Therefore, we have\n$$\nai_a + bi_b + ci_c < ab + bc + ca \\le a^2 + b^2 + c^2.\n$$\nThis is bounded above by $9R^2$ since $9R^2 - (a^2 + b^2 + c^2) = OH^2$, where $O$ and $H$ are the circumcentre and orthocentre of $\\triangle ABC$ respectively.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31572,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a hexagon whose diagonals $AD$, $BE$ and $CF$ intersect at one point, which is the midpoint of each diagonal. Prove that the area of the given hexagon is twice the area of the triangle $ACE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $AD$, $BE$, $CF$ meet at $P$. Since $P$ is the midpoint of $AD$, $BE$, $CF$, all of $ABDE$, $BCEF$ and $CDFA$ are parallelograms. Let $A'$ be the point such that $ACA'E$ is a parallelogram. Since $EA' = AC = FD$ and $EA' \\parallel AC \\parallel FD$, $FDA'E$ is also a parallelogram. Thus, we have $DA' = FE$. Together with $CD = AF$ and $CA' = AE$, we find that $\\triangle DCA' \\cong \\triangle FAE$. By symmetry, we also have $\\triangle DEA' \\cong \\triangle BAC$. It follows that\n$$\n\\begin{align*}\n[ABCDEF] &= [ACE] + [ABC] + [CDE] + [EFA] \\\\\n&= [ACE] + [EDA'] + [CDE] + [A'DC] \\\\\n&= [ACA'E] = 2[ACE].\n\\end{align*}\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31573,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A$ and $C$ lie on the circumference of a circle with radius $\\sqrt{50}$. $B$ is a point inside the circle such that $\\angle ABC = 90^\\circ$. If $AB = 6$ and $BC = 2$, find the distance from $B$ to the centre of the circle.\n\n$A$ 和 $C$ 是一個半徑為 $\\sqrt{50}$ 的圓的圓周上的兩點。$B$ 是該圓內的一點，使得 $\\angle ABC = 90^\\circ$。若 $AB = 6$ 而 $BC = 2$，求 $B$ 與該圓圓心的距離。",
    "options": [],
    "answer": "sqrt(26)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31574,
    "subject": "Mathematics (Multi-modal)",
    "question": "If a sequence $\\{a_1, a_2, \\dots, a_n\\}$ of positive integers (where $n$ is a positive integer) has the property that the last digit of $a_k$ is the same as the first digit of $a_{k+1}$ (here $k = 1, 2, \\dots, n$ and we define $a_{n+1} = a_1$), then the sequence is said to be a 'dragon sequence'. For example, $\\{414\\}$, $\\{208, 82\\}$ and $\\{1, 17, 73, 321\\}$ are all 'dragon sequences'. At least how many two-digit numbers must be chosen at random to ensure that a 'dragon sequence' can be formed among some of the chosen numbers?",
    "options": [],
    "answer": "46",
    "solution": "If we take any 46 two-digit numbers, then only 44 two-digit numbers are not chosen. Hence we either have one of the 36 pairs $\\{\\overline{AB}, \\overline{BA}\\}$ where $1 \\leq A < B \\leq 9$, or one of the 9 numbers $11, 22, \\dots, 99$. In either case, we get a 'dragon sequence' by definition.\n\nNext, suppose we only pick those two-digit numbers whose units digit is smaller than whose tens digit. There are altogether 45 such numbers. If we get a 'dragon sequence' among these numbers, say $\\{\\overline{A_1A_2}, \\overline{A_2A_3}, \\dots, \\overline{A_nA_1}\\}$, then $A_1 > A_2 > \\dots > A_n > A_1$, which is a contradiction. So these 45 numbers contain no 'dragon sequence'. It follows that the answer is 46.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31575,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $\\{a_1, a_2, ..., a_n\\}$ of positive integers (where $n$ is a positive integer) has the property that the last digit of $a_k$ is the same as the first digit of $a_{k+1}$ (here $k = 1, 2, ..., n$ and we define $a_{n+1} = a_1$), then the sequence is said to be a 'dragon sequence'. For example, $\\{414\\}$, $\\{208, 82\\}$ and $\\{1, 17, 73, 321\\}$ are all 'dragon sequences'. At least how many two-digit numbers must be chosen at random to ensure that a 'dragon sequence' can be formed among some of the chosen numbers? (2 marks)\n\n若某正整數數列 $\\{a_1, a_2, ..., a_n\\}$ (其中 $n$ 為正整數) 中 $a_k$ 的末位數字等於 $a_{k+1}$ 的首位數字 (這裡 $k=1, 2, ..., n$，並定義 $a_{n+1} = a_1$)，則稱該數列為「龍形數列」。例如：$\\{414\\}$、$\\{208, 82\\}$ 和 $\\{1, 17, 73, 321\\}$ 均為「龍形數列」。最少需隨機選取多少個兩位數，才可確保當中存在若干個數組成「龍形數列」？ (2分)",
    "options": [],
    "answer": "46",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31576,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\lfloor x \\rfloor$ denote the greatest integer not exceeding $x$. Find the last two digits of\n$$\n\\left\\lfloor \\frac{1}{3} \\right\\rfloor + \\left\\lfloor \\frac{2}{3} \\right\\rfloor + \\left\\lfloor \\frac{2^2}{3} \\right\\rfloor + \\cdots + \\left\\lfloor \\frac{2^{2^{2^{14}}}}{3} \\right\\rfloor\n$$",
    "options": [],
    "answer": "15",
    "solution": "Note that the remainder when $2^n$ is divided by $3$ is $1$ when $n$ is even, and $2$ when $n$ is odd.\nHence $\\left[\\frac{2^n}{3}\\right] = \\frac{2^n-1}{3}$ when $n$ is even, and $\\left[\\frac{2^n}{3}\\right] = \\frac{2^n-2}{3}$ when $n$ is odd. It follows that\n$$\n\\begin{align*}\nS &= \\left[\\frac{1}{3}\\right] + \\left[\\frac{2}{3}\\right] + \\left[\\frac{2^2}{3}\\right] + \\dots + \\left[\\frac{2^{2014}}{3}\\right] \\\\\n&= 0 + \\left(\\frac{2}{3} - \\frac{2}{3} + \\frac{2^2}{3} - \\frac{2}{3}\\right) + \\left(\\frac{2^3}{3} - \\frac{2}{3} + \\frac{2^4}{3} - \\frac{2}{3}\\right) + \\dots + \\left(\\frac{2^{2013}}{3} - \\frac{2}{3} + \\frac{2^{2014}}{3} - \\frac{2}{3}\\right) \\\\\n&= \\left(\\frac{2}{3} + \\frac{2^2}{3} - 1\\right) + \\left(\\frac{2^3}{3} + \\frac{2^4}{3} - 1\\right) + \\dots + \\left(\\frac{2^{2013}}{3} + \\frac{2^{2014}}{3} - 1\\right) \\\\\n&= \\left(\\frac{2}{3} + \\frac{2^2}{3} + \\frac{2^3}{3} + \\dots + \\frac{2^{2014}}{3}\\right) - 1007 \\\\\n&= \\frac{2^{2015} - 2}{3} - 1007\n\\end{align*}\n$$\nThe last two digits of powers of $2$ are listed as follows:\n02, 04, 08, 16, 32, 64, 28, 56, 12, 24, 48, 96, 92, 84, 68, 36, 72, 44, 88, 76, 52, 04, 08, ...\nThe pattern repeats when the exponent is increased by $20$. So the last two digits of $2^{2015}$ are the same as those of $2^{15}$, i.e. $68$.\nNow write $2^{2015} - 2 = 100k + 66$. Since $\\frac{2^{2015}-2}{3}$ is an integer, $k$ is a multiple of $3$, and so we write $k = 3m$. Thus the last two digits of $S$ are the same as those of $100m + 22 - 7$, i.e. $15$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31577,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $[x]$ denote the greatest integer not exceeding $x$. Find the last two digits of $\\lfloor \\frac{1}{3} \\rfloor + \\lfloor \\frac{2}{3} \\rfloor + \\lfloor \\frac{2^2}{3} \\rfloor + \\dots + \\lfloor \\frac{2^{2014}}{3} \\rfloor$.\n\n設 $[x]$ 表示不超過 $x$ 的最大整數。求 $\\lfloor \\frac{1}{3} \\rfloor + \\lfloor \\frac{2}{3} \\rfloor + \\lfloor \\frac{2^2}{3} \\rfloor + \\dots + \\lfloor \\frac{2^{2014}}{3} \\rfloor$ 的最後兩位數字。",
    "options": [],
    "answer": "15",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31578,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AB = 13$ and $BC = 7$. $D$ and $E$ are points on $AB$ and $AC$ respectively such that $BD = BC$ and $\\angle DEB = \\angle CEB$. Find the product of all possible values of the length of $AE$.\n\n在 $\\triangle ABC$ 中，$AB = 13$ 及 $BC = 7$。設 $D$ 和 $E$ 分別為 $AB$ 和 $AC$ 上的點，使得 $BD = BC$ 及 $\\angle DEB = \\angle CEB$。求 $AE$ 的長度的所有可能值之積。",
    "options": [],
    "answer": "507/10",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31579,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ is acute with $AB = 13$ and $BC = 7$. $D$ and $E$ are points on $AB$ and $AC$ respectively such that $BD = BC$ and $\\angle DEB = \\angle CEB$. Find the product of all possible values of the length of $AE$.",
    "options": [],
    "answer": "507/10",
    "solution": "Obviously one possible position of $E$ arises from the case when $BE \\perp CD$ (or equivalently, $ED = EC$). It is denoted by $E_1$ in the figure, in which case $E_1DBC$ is a kite. In particular, $BE_1$ bisects $\\angle CBA$, and hence $\\frac{AE_1}{E_1C} = \\frac{AB}{BC} = \\frac{13}{7}$. With $b$ denoting the length of $AC$, we get $AE_1 = \\frac{13b}{20}$.\n\nIf $ED \\neq EC$, then the internal bisector of $\\angle DEC$ intersects the perpendicular bisector of $CD$ on the circumcircle of $\\triangle CDE$. But we know that this intersection point is $B$ and so $E$ must be the second intersection point of the circumcircle of $\\triangle BCD$ and the side $AC$. This is the point $E_2$ in the figure. The chord theorem thus gives $AE_2 \\times AC = AD \\times AB$ and so $AE_2 = \\frac{AD \\times AB}{AC} = \\frac{(13-7)13}{b} = \\frac{78}{b}$.\n\nIt follows that the answer is $\\frac{13b}{20} \\times \\frac{78}{b} = \\frac{507}{10}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31580,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many sets of integers $(a, b, c)$ satisfy $2 \\le a \\le b \\le c$ and $abc = 2013 \\times 2014$? (2 marks)\n\n有多少組整數 $(a, b, c)$ 滿足 $2 \\le a \\le b \\le c$ 及 $abc = 2013 \\times 2014$? (2分)",
    "options": [],
    "answer": "90",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31581,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many triples of integers $(a, b, c)$ satisfy $2 \\le a \\le b \\le c$ and $abc = 2013 \\times 2014$?",
    "options": [],
    "answer": "90",
    "solution": "The prime factorisation of $2013 \\times 2014$ is $2 \\times 3 \\times 11 \\times 19 \\times 53 \\times 61$. The problem amounts to counting the number of ways of distributing these 6 primes into 3 groups such that each group contains at least one prime. For example, if the groups are $\\{2, 3, 11\\}$, $\\{19, 53\\}$ and $\\{61\\}$, then the corresponding factors are 66, 1007 and 61. By ordering the factors, we get $a = 61, b = 66$ and $c = 1007$, and hence $(a, b, c) = (61, 66, 1007)$ is a possible triple. We have a few cases to consider:\n\n* The number of primes in the 3 groups are 4, 1, 1 respectively (we shall abbreviate this as case (4, 1, 1)) — there are $\\binom{6}{2} = 15$ choices for the two singletons.\n\n* Case (3, 2, 1) — there are 6 choices for the singleton and then $\\binom{5}{2} = 10$ choices for the doubleton, leading to $6 \\times 10 = 60$ ways in this case.\n\n* Case (2, 2, 2) — there are 5 ways to choose a partner for the prime 2, and 3 ways to choose a partner for one of the remaining primes. Afterwards, the two primes left form the last group automatically. Hence there are $3 \\times 5 = 15$ ways to do so.\n\nIt follows that the answer is $15 + 60 + 15 = 90$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31582,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer not exceeding $2014$ with the property that $x^2 + x + 1$ is a factor of $x^{2n} + x^n + 1$. Find the sum of all possible values of $n$.\n\n設 $n$ 為不超過 $2014$ 的正整數，且 $x^2 + x + 1$ 為 $x^{2n} + x^n + 1$ 的因式。求 $n$ 所有可能值之和。",
    "options": [],
    "answer": "1352737",
    "solution": "Let $\\omega$ be a root of $x^2 + x + 1 = 0$. Then $\\omega^3 = 1$ and $\\omega \\ne 1$.\n\nSince $x^2 + x + 1$ divides $x^{2n} + x^n + 1$, we have $\\omega^{2n} + \\omega^n + 1 = 0$.\n\nLet $y = \\omega^n$. Then $y^2 + y + 1 = 0$, so $y = \\omega$ or $y = \\omega^2$.\n\nThus, $\\omega^n = \\omega$ or $\\omega^n = \\omega^2$.\n\nThis means $n \\equiv 1 \\pmod{3}$ or $n \\equiv 2 \\pmod{3}$.\n\nSo $n$ is any positive integer not exceeding $2014$ such that $n \\not\\equiv 0 \\pmod{3}$.\n\nThe possible values of $n$ are those with $1 \\le n \\le 2014$ and $n \\not\\equiv 0 \\pmod{3}$.\n\nLet us compute the sum of all such $n$.\n\nFirst, the sum of all $n$ from $1$ to $2014$ is:\n$$\nS = 1 + 2 + \\cdots + 2014 = \\frac{2014 \\times 2015}{2} = 2,029,105\n$$\n\nNow, subtract the sum of all $n$ divisible by $3$ in this range.\n\nThe smallest such $n$ is $3$, the largest is $2013$.\n\nThe sequence is $3, 6, 9, \\ldots, 2013$.\n\nNumber of terms:\nLet $k$ be the number of terms. $3k = 2013 \\implies k = 671$.\n\nSum of these terms:\n$$\nS_3 = 3 + 6 + 9 + \\cdots + 2013 = 3(1 + 2 + \\cdots + 671) = 3 \\times \\frac{671 \\times 672}{2} = 3 \\times 225,456 = 676,368\n$$\n\nTherefore, the sum of all $n$ not divisible by $3$ is:\n$$\n2,029,105 - 676,368 = 1,352,737\n$$\n\n**Answer:** $1,352,737$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31583,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\ldots, a_{24}$ be integers with sum $0$ and satisfying $|a_i| \\le i$ for all $i$. Find the greatest possible value of $a_1 + 2a_2 + 3a_3 + \\ldots + 24a_{24}$.\n\n設 $a_1, a_2, \\ldots, a_{24}$ 為整數, 它們之和為 $0$, 且對所有 $i$ 皆有 $|a_i| \\le i$。求 $a_1 + 2a_2 + 3a_3 + \\ldots + 24a_{24}$ 的最大可能值。",
    "options": [],
    "answer": "1432",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31584,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\lfloor x \\rfloor$ denote the greatest integer not exceeding $x$. Find the last three digits of\n$$\n\\left\\lfloor \\left( \\sqrt[3]{\\sqrt{5}+2} + \\sqrt[3]{\\sqrt{5}-2} \\right)^{2014} \\right\\rfloor\n$$",
    "options": [],
    "answer": "125",
    "solution": "Let $x = \\sqrt[3]{\\sqrt{5}+2} + \\sqrt[3]{\\sqrt{5}-2}$. Using the formula $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$, we have\n$$\nx^3 = (\\sqrt{5}+2) + (\\sqrt{5}-2) + 3\\sqrt[3]{(\\sqrt{5}+2)(\\sqrt{5}-2)}x \\\\ = 2\\sqrt{5} + 3x.\n$$\nRewrite this as $(x - \\sqrt{5})(x^2 + \\sqrt{5}x + 2) = 0$. Since the quadratic equation $x^2 + \\sqrt{5}x + 2 = 0$ has no real roots, we must have $x = \\sqrt{5}$, and hence\n$$\n\\left[ \\left( \\sqrt[3]{\\sqrt{5}+2} + \\sqrt[3]{\\sqrt{5}-2} \\right)^{2014} \\right] = [\\sqrt{5}^{2014}] = 5^{1007}.\n$$\nSince the last three digits of powers of 5 have the pattern 005, 025, 125, 625, 125, 625, ..., it follows that the answer is 125.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31585,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $[x]$ denote the greatest integer not exceeding $x$. Find the last three digits of $[(\\sqrt[3]{5}+2)^{2014} + (\\sqrt[3]{5}-2)^{2014}]$.\n\n設 $[x]$ 表示不超過 $x$ 的最大整數。求 $[(\\sqrt[3]{5}+2)^{2014} + (\\sqrt[3]{5}-2)^{2014}]$ 的最後三位數字。",
    "options": [],
    "answer": "375",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31586,
    "subject": "Mathematics (Multi-modal)",
    "question": "There were $36$ participants in a party, some of whom shook hands with each other, such that any two participants shook hands with each other at most once. Each participant then recorded the number of handshakes made, and it was found that no two participants with the same number of handshakes made had shaken hands with each other. Find the maximum total number of handshakes in the party. (When two participants shook hand with each other this will be counted as one handshake.) (2 marks)\n\n某次聚會共有 $36$ 人參加, 當中有些人曾經互相握手, 而每兩人之間握手最多一次。事後每人均記錄了自己的握手次數, 並發現握手次數相同的人之間均沒有互相握手。求聚會中握手總次數的最大可能值(兩人互相握手時算作一次)。 (2分)",
    "options": [],
    "answer": "546",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31587,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ be points on the same plane with $\\angle ACB = 120^\\circ$. There is a sequence of circles $\\omega_0, \\omega_1, \\omega_2, \\ldots$ on the same plane (with corresponding radii $r_0, r_1, r_2, \\ldots$, where $r_0 > r_1 > r_2 > \\ldots$) such that each circle is tangent to both segments $CA$ and $CB$. Furthermore, $\\omega_i$ is tangent to $\\omega_{i-1}$ for all $i \\ge 1$. If $r_0 = 3$, find the value of $r_0 + r_1 + r_2 + \\ldots$.\n\n設 $A$、$B$、$C$ 為同一平面上的三點, 其中 $\\angle ACB = 120^\\circ$。現於同一平面上有一系列的圓 $\\omega_0, \\omega_1, \\omega_2, \\ldots$ (對應的半徑分別為 $r_0, r_1, r_2, \\ldots$, 其中 $r_0 > r_1 > r_2 > \\ldots$), 當中每個均與線段 $CA$ 和 $CB$ 相切, 且對任意 $i \\ge 1$ 皆有 $\\omega_i$ 與 $\\omega_{i-1}$ 相切。若 $r_0 = 3$, 求 $r_0 + r_1 + r_2 + \\ldots$ 的值。",
    "options": [],
    "answer": "(3 + 2√3)/2",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31588,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A$, $B$, $C$ be points on the same plane with $\\angle ACB = 120^\\circ$. There is a sequence of circles $\\omega_0, \\omega_1, \\omega_2, \\dots$ on the same plane (with corresponding radii $r_0, r_1, r_2, \\dots$, where $r_0 > r_1 > r_2 > \\dots$) such that each circle is tangent to both segments $CA$ and $CB$. Furthermore, $\\omega_i$ is tangent to $\\omega_{i-1}$ for all $i \\ge 1$. If $r_0 = 3$, find the value of $r_0 + r_1 + r_2 + \\dots$.",
    "options": [],
    "answer": "√3 + 3/2",
    "solution": "Let $S$ be the sum and $O$ the centre of $\\omega_0$. Then $2S = CO + r_0 = \\frac{3}{\\sin 60^\\circ} + 3 = 2\\sqrt{3} + 3$, and hence $S = \\sqrt{3} + \\frac{3}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31589,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the greatest positive integer $k$ for which the following holds: For all positive real numbers $a$, $b$ and $c$ with $abc = 1$, we have the inequality\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{k}{a+b+c+1} \\ge \\frac{k}{4} + 3.\n$$",
    "options": [],
    "answer": "13",
    "solution": "The greatest $k$ is $13$.\n\nFirstly, consider $a = b = \\frac{2}{3}$ and $c = \\frac{9}{4}$. They satisfy $abc = 1$. The inequality becomes\n$$\n\\frac{3}{2} + \\frac{3}{2} + \\frac{4}{9} + \\frac{12k}{55} \\ge \\frac{k}{4} + 3,\n$$\ni.e. $\\frac{4}{9} \\ge \\frac{7k}{220}$. As $k$ is a positive integer, this implies $k \\le 13$.\n\nWe now prove the inequality\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{13}{a+b+c+1} \\ge \\frac{13}{4} + 3 = \\frac{25}{4}.\n$$\nLet the left-hand side be $f(a, b, c)$. WLOG assume $c \\ge b \\ge a$. We first prove that\n$$\nf(a, b, c) \\ge f(\\sqrt{ab}, \\sqrt{ab}, c).\n$$\nNote that the new triple still satisfies $(\\sqrt{ab})(\\sqrt{ab})(c) = 1$. Now,\n$$\n\\begin{align*}\nf(a, b, c) &\\ge f(\\sqrt{ab}, \\sqrt{ab}, c) \\\\\n\\Leftrightarrow \\quad & \\frac{1}{a} + \\frac{1}{b} - \\frac{2}{\\sqrt{ab}} \\ge \\frac{13}{2\\sqrt{ab} + c + 1} - \\frac{13}{a + b + c + 1} \\\\\n\\Leftrightarrow \\quad & \\frac{(\\sqrt{a} - \\sqrt{b})^2}{ab} \\ge \\frac{13(\\sqrt{a} - \\sqrt{b})^2}{(2\\sqrt{ab} + c + 1)(a + b + c + 1)}.\n\\end{align*}\n$$\nIt suffices to show $(2\\sqrt{ab}+c+1)(a+b+c+1) \\ge 13ab$. Indeed, since $a+b \\ge 2\\sqrt{ab}$, we only need to prove\n$$\n2\\sqrt{ab} + c + 1 \\ge \\sqrt{13ab}.\n$$\nAs $c \\ge b \\ge a$ and $abc = 1$, we have $c \\ge 1$. Therefore, we have $\\sqrt{c}(c+1) \\ge 2$. This implies\n$$\nc + 1 \\ge \\frac{2}{\\sqrt{c}} = 2\\sqrt{ab} > (\\sqrt{13} - 2)\\sqrt{ab}.\n$$\nThis proves $f(a, b, c) \\ge f(\\sqrt{ab}, \\sqrt{ab}, c)$.\n\nAfter this mixing step, it remains to consider the case $f\\left(t, t, \\frac{1}{t^2}\\right)$ where $t > 0$. Now,\n$$\n\\begin{align*}\nf\\left(t, t, \\frac{1}{t^2}\\right) &\\ge \\frac{25}{4} \\\\\n\\Leftrightarrow \\quad & \\frac{2}{t} + t^2 + \\frac{13t^2}{2t^3 + t^2 + 1} \\ge \\frac{25}{4} \\\\\n\\Leftrightarrow \\quad & 8t^6 + 4t^5 - 50t^4 + 47t^3 + 8t^2 - 25t + 8 \\ge 0 \\\\\n\\Leftrightarrow \\quad & (t-1)^2(8t^4 + 20t^3 - 18t^2 - 9t + 8) \\ge 0.\n\\end{align*}\n$$\n\nIt remains to prove $8t^4 + 20t^3 - 18t^2 - 9t + 8 \\ge 0$ for $t > 0$. Indeed, the left-hand side is equal to\n$$\n(8t^2 + 30t) \\left(t - \\frac{5}{8}\\right)^2 + \\frac{1}{32}(524t^2 - 663t + 256).\n$$\nThe first term is nonnegative since $t > 0$. Since $663^2 - 4(524)(256) < 0$, the last term is nonnegative. This proves the inequality and we are done.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31590,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be distinct nonzero real numbers. If the equations $ax^3 + bx + c = 0$, $bx^3 + cx + a = 0$ and $cx^3 + ax + b = 0$ have a common root, prove that at least one of these equations has three real roots (not necessarily distinct).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the common root be $t$. Adding the three equations, we get\n$$\n(a + b + c)(t^3 + t + 1) = 0.\n$$\nIf $t^3 + t + 1 = 0$, then $at^3 + bt + c = 0$ becomes $(b-a)t + (c-a) = 0$. Similarly, $bt^3 + ct + a = 0$ becomes $(c-b)t + (a-b) = 0$. As $a$, $b$, $c$ are distinct, we obtain $\\frac{c-a}{a-b} = t = \\frac{a-b}{b-c}$. After expansion and regrouping the terms, this becomes\n$$\n(a-b)^2 + (b-c)^2 + (c-a)^2 = 0.\n$$\nThis is a contradiction since $a$, $b$, $c$ are distinct.\nTherefore, we must have $a+b+c=0$. In that case, it is obvious that $t=1$ is the common root. As $a$, $b$, $c$ are nonzero, there are two possible cases.\nIf two of $a$, $b$, $c$ are positive, say $a$, $b > 0$, we consider $f(x) = bx^3 + cx + a$. Since $f(0) = a > 0$ and $f(x) < 0$ when $x \\to -\\infty$, there is at least one negative root of $f(x)$. As $1$ is another root and a cubic polynomial cannot have exactly two real roots, the equation $f(x) = 0$ must have exactly three real roots.\nSimilarly, if two of $a$, $b$, $c$ are negative, say $a$, $b < 0$, then the above argument shows $g(x) = -(bx^3 + cx + a)$ has three real roots. This shows $f(x) = 0$ also has three real roots.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31591,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle and $H$ be the perpendicular foot at the side $BC$ from $A$. Let $D$ and $E$ be points on the segments $AB$ and $AC$, $F$ and $G$ be the perpendicular feet at $BC$ from $D$ and $E$ respectively. Let $P$ be the perpendicular foot at the segment $DH$ from $E$. Suppose that $DG, EF, AH$ are concurrent. Show that $\\angle APE = \\angle CPE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $DG, EF, AH$ are concurrent at $X$. Let $Q$ be the point on the line $CP$ such that $QH \\parallel PG$. We only work on the configuration as shown since the other cases are similar.\n\nFirstly, by the parallel lines $AH \\parallel EG$ and $QH \\parallel PG$, we have $\\triangle CAH \\sim \\triangle CEG$ and $\\triangle CPG \\sim \\triangle CQH$. Thus, we have\n$$\n\\frac{AE}{EC} = \\frac{HG}{GC} = \\frac{QP}{PC}.\n$$\nBy the angle bisector theorem, it suffices to prove $\\frac{AP}{PC} = \\frac{AE}{EC}$. Equivalently, we need to prove $PA = PQ$.\n\nBy similar triangles, we have $\\frac{DF}{EG} = \\frac{DX}{GX} = \\frac{FH}{GH}$. Together with the included angles $\\angle DFH = \\angle EGH = 90^\\circ$, we have $\\triangle DFH \\sim \\triangle EGH$. This implies $\\angle DHF = \\angle EHG$, and hence $\\angle AHD = \\angle AHE$.\n\n![](attached_image_1.png)\n\nNote that $E, P, H, G$ are concyclic since $\\angle EPH = \\angle EGH = 90^\\circ$. This yields\n$$\n\\angle PEH = \\angle PGH = \\angle QHB,\n$$\nand so\n$$\n\\angle EHP = 90^\\circ - \\angle PEH = 90^\\circ - \\angle QHB = \\angle AHQ.\n$$\nIt follows that $\\angle DHQ = \\angle EHA = \\angle DHA$.\n\nNext, note that $\\angle PEG = \\angle DHF = \\angle GHE = \\angle GPE$. This implies $PG = EG$. By similar triangles, we obtain\n$$\n\\frac{PG}{QH} = \\frac{CG}{CH} = \\frac{EG}{AH} = \\frac{PG}{AH}.\n$$\nTherefore, we have $QH = AH$. Thus, $HP$ is the perpendicular bisector of $AQ$, and hence $PA = PQ$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31592,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$ is a real number, find the minimum value of $|x + 1| + 2|x - 5| + |2x - 7| + \\left|\\frac{x - 11}{2}\\right|$.",
    "options": [],
    "answer": "45/4",
    "solution": "It suffices to minimise $2|x+1| + 4|x-5| + 4\\left|x - \\frac{7}{2}\\right| + |x - 11|$ (which is 2 times the given expression), i.e. to find the minimum total distance from $x$ to the 11 numbers:\n$$\n-1, -1, \\frac{7}{2}, \\frac{7}{2}, \\frac{7}{2}, \\frac{7}{2}, 5, 5, 5, 5, 11.\n$$\nBy the triangle inequality, we have $|x-a| + |x-b| \\ge |(x-a) + (b-x)| = b-a$ for any $a \\le b$, and the equality holds whenever $x$ lies between $a$ and $b$. From this, we try to pair up the numbers as follows: $(-1, 11)$, $(-1, 5)$, $(\\frac{7}{2}, 5)$, $(\\frac{7}{2}, 5)$, $(\\frac{7}{2}, 5)$, leaving a single number $\\frac{7}{2}$. From the above discussion, the total distance from $x$ to each pair is minimised when $x$ lies between them. Hence we can achieve the minimum by minimising the distance from $x$ to $\\frac{7}{2}$ while making sure that $x$ lies between each pair of numbers. Clearly, this can be achieved by taking $x = \\frac{7}{2}$. It follows that the answer is $\\left|\\frac{7}{2} + 1\\right| + 2\\left|\\frac{7}{2} - 5\\right| + 2 \\times \\frac{7}{2} - 7 + \\left|\\frac{7}{2} - 11\\right| = \\frac{45}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31593,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be distinct nonzero real numbers. If the equations $ax^3 + bx + c = 0$, $bx^3 + cx + a = 0$ and $cx^3 + ax + b = 0$ have a common root, prove that at least one of these equations has three real roots (not necessarily distinct).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $t$ be the common root. Adding the three equations $at^3 + bt + c = 0$, $bt^3 + ct + a = 0$ and $ct^3 + at + b = 0$, we get $(a + b + c)(t^3 + t + 1) = 0$.\n\nIf $t^3 + t + 1 = 0$, then $at^3 + bt + c = 0$ becomes $(b - a)t + (c - a) = 0$. Similarly, $bt^3 + ct + a$ becomes $(c - b)t + (a - b) = 0$. As $a$, $b$, $c$ are distinct, the two equations combine to give $\\frac{c - a}{b - a} = \\frac{a - b}{c - b}$, i.e. $(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$, which yields the contradiction $a = b = c$.\n\nTherefore, we must have $a + b + c = 0$. Then it is obvious that $t = 1$ is the common root. As $a$, $b$, $c$ are nonzero, there are two possible cases.\n\n**Case 1:** Two of $a$, $b$, $c$ are positive, say $a$ and $b$. Consider $f(y) = b y^3 + c y + a$. Since $f(0) = a > 0$ and $f$ is negative for sufficiently small $y$, there is at least one negative root of $f(x) = 0$. As $1$ is also a root, $f(x) = 0$ has at least two real roots and hence exactly three real roots (since complex roots occur in pairs).\n\n**Case 2:** Two of $a$, $b$, $c$ are negative, say $a$ and $b$. Then the same argument shows that the polynomial $g(y) = -(b y^3 + c y + a)$ has three real zeros, and the same is clearly true for $f(x)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31594,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an equilateral triangle. $P$ is a point inside the triangle such that $\\angle CBP = 12^\\circ$ and $\\angle ACP = 54^\\circ$. Find $\\angle CAP$.",
    "options": [],
    "answer": "42°",
    "solution": "We have $\\angle CAP = 42^\\circ$.\nConstruct a point $D$ such that $\\angle DAC = \\angle DCA = 72^\\circ$, and $B$ lies inside $\\triangle DCA$. Since\n$$\n\\angle BCD = \\angle ACD - \\angle ACB = 72^\\circ - 60^\\circ = 12^\\circ = \\angle CBP,\n$$\nwe have $DC // BP$. Also, note that $BD$ is the perpendicular bisector of $CA$. Therefore, we have\n$$\n\\angle CDB = \\frac{1}{2} \\angle CDA = 18^\\circ\n$$\nand\n$$\n\\angle PCD = \\angle ACD - \\angle ACP = 72^\\circ - 54^\\circ = 18^\\circ = \\angle CDB.\n$$\nThis implies $BDCP$ is an isosceles trapezoid.\n\n![](attached_image_1.png)\nConstruct a point $E$ such that $DCAE$ is an isosceles trapezoid and $DC//EA$. Since $\\triangle EDP$ is the image of reflection of $\\triangle ACB$ in the perpendicular bisector of $DC$, it is equilateral. Now, as\n$$\n\\angle ADE = \\angle CDE - \\angle CDA = \\angle ACD - \\angle CDA = 72^\\circ - 36^\\circ = 36^\\circ\n$$\nand\n$$\n\\angle EAD = \\angle CDA = 36^\\circ = \\angle ADE,\n$$\nwe have $EA = ED = EP$. It follows that\n$$\n\\begin{aligned} \\angle PAC &= \\angle EAC - \\angle EAP \\\\ &= (180^\\circ - \\angle ACD) - \\left(90^\\circ - \\frac{1}{2} \\angle PEA\\right) \\\\ &= 180^\\circ - 72^\\circ - 90^\\circ + \\frac{1}{2}(\\angle DEA - \\angle DEP) \\\\ &= 18^\\circ + \\frac{1}{2}(108^\\circ - 60^\\circ) \\\\ &= 42^\\circ. \\end{aligned}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31595,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a proper subset of $\\mathbb{R}$ (i.e. $S \\neq \\mathbb{R}$) having at least two elements. Suppose there exists a function $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the following conditions:\n(i) $f(a + x + y) + f(f(a)) + f(x) + f(y) = x + y$; and\n(ii) $f(axy) + f(a) + f(x)f(y) = xy$\nfor any real numbers $a \\notin S$ and $x, y \\in S$. Find all such function(s) $f$.",
    "options": [],
    "answer": "f(x) = x for x in S, and f(t) = 0 for t not in S",
    "solution": "The only solution is $f(a) = 0$ for $a \\notin S$ and $f(x) = x$ for $x \\in S$.\nLabel the equations as follows.\n$$\nf(a + x + y) + f(f(a)) + f(x) + f(y) = x + y \\quad (1)\n$$\n$$\nf(axy) + f(a) + f(x)f(y) = xy \\tag{2}\n$$\n\n**Case 1.** $0 \\notin S$\nPutting $a = 0$ in (2), we obtain\n$$\nf(x)f(y) = xy - 2f(0) \\tag{3}\n$$\nfor all $x, y \\in S$. In particular, by putting $x = y$, we get\n$$\nf(x)^2 = x^2 - 2f(0) \\tag{4}\n$$\nfor all $x \\in S$. Then we have\n$$\n(x^2 - 2f(0))(y^2 - 2f(0)) = f(x)^2 f(y)^2 = (xy - 2f(0))^2,\n$$\nwhich implies $2f(0)(x^2 + y^2) = 4f(0)xy$. This means\n$$\n2f(0)(x - y)^2 = 0.\n$$\nSince $S$ has at least two elements, we can choose distinct $x, y \\in S$ to conclude\nthat $f(0) = 0$. Therefore, by (4),\n$f(x)^2 = x^2$\nfor all $x \\in S$. Since $f(x)f(y) = xy$ by (3), we see that the choice of the sign for $f(x)$ is independent of $x$. This means $f(x) = x$ for all $x \\in S$ or $f(x) = -x$ for all $x \\in S$.\nNow, (2) is reduced to\n$$\nf(axy) + f(a) = 0. \\tag{5}\n$$\n\nFor fixed nonzero $a \\notin S$, if $\\frac{1}{a} \\in S$, we may put $y = \\frac{1}{a}$ in (5) to get $f(x) + f(a) = 0$. But this cannot be true as we can choose two different values for $x$ (and hence $f(x)$). Therefore, we must have $\\frac{1}{a} \\notin S$. From this, we see that $x \\in S$ implies $\\frac{1}{x} \\in S$, since otherwise $x = (x^{-1})^{-1} \\notin S$.\nNow, we can put $y = \\frac{1}{x}$ in (5) to obtain $2f(a) = 0$, i.e. $f(a) = 0$ for any $a \\notin S$. Equation (1) becomes\n$$\nf(a + x + y) \\pm (x + y) = x + y.\n$$\nBy putting $x = y$, we get\n$$\nf(a + 2x) \\pm 2x = 2x.\n$$\nIf the negative sign is chosen, then we have $f(a + 2x) = 4x$. Since $f(a + 2x)$ can only be $-(a+2x)$ or $0$ (depending on whether $a+2x \\in S$ or not), we need $a = -6x$ for any $a \\notin S$ and $x \\in S$ (note we cannot have $0 = 4x$). This is impossible as we can find two distinct elements in $S$. Therefore, $f(a) = 0$ for $a \\notin S$ and $f(x) = x$ for $x \\in S$.\n\n**Case 2.** $0 \\in S$\nPutting $y = 0$ in (2), we obtain\n$$\nf(0) + f(a) + f(x)f(0) = 0 \\qquad (6)\n$$\nfor all $a \\notin S$, $x \\in S$. In particular, by putting $x = 0$, we get\n$$\nf(a) = -f(0) - f(0)^2. \\qquad (7)\n$$\nThen equation (6) becomes\n$$\nf(x)f(0) = f(0)^2.\n$$\nIf $f(0) \\neq 0$, we need $f(x) = f(0)$ for all $x \\in S$. Consider equation (1). It now becomes\n$$\nf(a + x + y) + f(-f(0) - f(0)^2) + 2f(0) = x + y.\n$$\nNote that the left-hand side can take at most two values (as $f(a+x+y)$ can be $-f(0) - f(0)^2$ or $f(0)$). However, as there are at least two elements in $S$, say 0 and $z \\neq 0$, the right-hand side can take at least three different values, namely, 0, $z$, $2z$. This is a contradiction, and hence $f(0) = 0$. By (7), we obtain $f(a) = 0$ for $a \\notin S$.\nNext, we prove that $a \\notin S$ implies $-a \\notin S$. Indeed, suppose $-a \\in S$. We put $y = -a$ in (1) to get\n$$\n2f(x) + f(-a) = x - a. \\qquad (8)\n$$\n$$\nf(-a) = -a, \\tag{9}\n$$\nand hence $2f(x) = x$ for any $x \\in S$ by (8). Now, by putting $x = -a$, we obtain $2f(-a) = -a$. But then we have $f(-a) = -a$ by (9). This forces $a = 0 \\in S$, which is a contradiction. So the claim is true. From this, we see that $x \\in S$ implies $-x \\in S$. Thus, we can put $y = -x$ in (1). This gives\n$$\nf(x) + f(-x) = 0 \\tag{10}\n$$\nfor any $x \\in S$.\nSimilarly, we shall prove that $z, w \\in S$ implies $z+w \\in S$. Suppose on the contrary that $z+w \\notin S$. Note that $-w \\in S$ from above. So we can put $a = z+w$ and $x = 0$, $y = -w$ in (1) to obtain\n$$\nf(z) + f(-w) = -w.\n$$\nUsing (10), we get\n$$\nf(z) - f(w) = -w.\n$$\nBy symmetry, we also have\n$$\nf(w) - f(z) = -z.\n$$\nAdding these, we obtain $z+w=0 \\in S$, which is a contradiction.\nIt is now clear that $a+x+y \\notin S$ for any $a \\notin S$ and $x, y \\in S$. Otherwise, if $a+x+y \\in S$, since $-x-y = (-x)+(-y) \\in S$, we have\n$$\na = (a + x + y) + (-x - y) \\in S.\n$$\nIt follows that (1) is reduced to $f(x)+f(y)=x+y$. Putting $x=y$, we get $f(x)=x$ for any $x \\in S$.\nIn any case, $f(a) = 0$ for $a \\notin S$ and $f(x) = x$ for $x \\in S$ is the only possible function. One can check that this is indeed a solution if and only if $a+x+y, axy \\notin S - \\{0\\}$ for any $a \\notin S$ and $x, y \\in S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31596,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$ is a real number, find the minimum value of $|x+1| + 2|x-5| + |2x-7| + \\frac{|x-11|}{2}$.\n\n若 $x$ 為實數, 求 $|x+1| + 2|x-5| + |2x-7| + \\frac{|x-11|}{2}$ 的最小值。",
    "options": [],
    "answer": "45/4",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31597,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are $n$ students, each with a different student number. Each student number is a positive factor of $60^{60}$, and no student number is equal to the H.C.F. of the numbers of two other students. Find the greatest possible value of $n$. (2 marks)\n\n某學校有 $n$ 名學生, 每人均有一個不同的編號。每名學生的編號均是 $60^{60}$ 的一個正因數, 而且沒有學生的編號等於另外兩名學生的編號的最大公因數。求 $n$ 的最大可能值。 (2分)",
    "options": [],
    "answer": "241",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31598,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x$ and $y$ are real numbers, find the minimum value of $\\sqrt{4+y^2}+\\sqrt{x^2+y^2-4x-4y+8}+\\sqrt{x^2-8x+17}$.\n\n若 $x$、$y$ 為實數, 求 $\\sqrt{4+y^2}+\\sqrt{x^2+y^2-4x-4y+8}+\\sqrt{x^2-8x+17}$ 的最小值。",
    "options": [],
    "answer": "5",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31599,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(a, b)$ of integers $a$ and $b$ satisfying\n$$\n(b^2 + 11(a-b))^2 = a^3 b.\n$$",
    "options": [],
    "answer": "(a, b) = (t, t) for any integer t, and (a, b) = (0, 11)",
    "solution": "(Belarus MO 2014 Category B Problem 2 modified) The only solutions are $(a, b) = (t, t), (0, 11)$ where $t$ is any integer.\nWe have\n$$\n\\begin{aligned}\n(b^2 + 11(a-b))^2 &= a^3b \\\\\n\\Leftrightarrow \\quad b^4 + 22b^2(a-b) + 121(a-b)^2 &= a^3b \\\\\n\\Leftrightarrow \\quad (a-b)(121(a-b) + 22b^2 - b(a^2 + ab + b^2)) &= 0.\n\\end{aligned}\n$$\nWhen $a = b$, the equation always holds. Now we may assume $a \\neq b$. It remains to solve $121(a-b) + 22b^2 - b(a^2 + ab + b^2) = 0$. This means\n$$\nbaa^2 + (b^2 - 121)a + b(b - 11)^2 = 0. \\quad (1)\n$$\nIf $b=0$, we have $a=0$, which is a contradiction. If $b \\neq 0$, this is a quadratic equation in $a$. The discriminant is\n$$\n\\Delta = (b^2 - 121)^2 - 4b^2(b - 11)^2 = -(b - 11)^3(3b + 11).\n$$\nWhen $b=11$, the equation (1) becomes $11a^2 = 0$. The only solution is $a=0$. When $b \\neq 11$, we need $-(b-11)(3b+11) \\ge 0$. This implies $b = -3, -2, \\dots, 10$. Note that $\\Delta$ should be a perfect square since $a$ is an integer. We check the value of $c = -(b-11)(3b+11)$ one by one as follows.\n\n| b  | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |\n|----|----|----|----|---|---|---|---|---|---|---|---|---|----|\n| c  | 28 | 65 | 96 |140|153|160|161|156|145|128|105| 76| 41 |\n\nNone of these is a perfect square. Therefore, the only solutions are $a=b$ and $(a, b) = (0, 11)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31600,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n$ be distinct positive integers not exceeding $2013$ and $d$ be their greatest common divisor. Suppose $d^2$ divides $3(m-n)$. Find the greatest possible value of $d(m+n)$.",
    "options": [],
    "answer": "166635",
    "solution": "The greatest possible value of $d(m+n)$ is $166635$.\nLet $m = da$ and $n = db$ so that $(a, b) = 1$. The condition becomes $d \\mid 3(a - b)$. WLOG assume $m > n$. Let $N = d(m + n) = d^2(a + b)$.\nIf $3 \\nmid d$, then $d \\mid a-b$. This implies $d \\le a-b$. Also, we have $a = \\frac{m}{d} \\le \\frac{2013}{d}$. This yields\n$$\nN = d^2(a + b) \\le d^2(2a - d) \\le 4026d - d^3.\n$$\n\nBy differentiation, the maximum value of $4026d - d^3$ is attained at $d = \\sqrt{\\frac{4026}{3}}$.\nThus,\n$$\nN \\le \\sqrt{\\frac{4026}{3}} \\left(4026 - \\frac{4026}{3}\\right) < 40 \\cdot 3000 = 120000.\n$$\nIf $3 \\mid d$, then $d \\le 3(a-b)$. Also, we have $a \\le \\lfloor \\frac{2013}{d} \\rfloor$. This yields\n$$\nN = d^2(a+b) \\le d^2 \\left( 2a - \\frac{d}{3} \\right) \\le d^2 \\left( 2 \\lfloor \\frac{2013}{d} \\rfloor - \\frac{d}{3} \\right).\n$$\n* For $d \\le 54$, we have $N \\le 4026d - \\frac{d^3}{3} \\le 4026(54) - \\frac{(54)^3}{3} = 164916$ since $f(d) = 4026d - \\frac{d^3}{3}$ is increasing for $d \\le 54$ (again by differentiation).\n* For $d = 57$, we have $N \\le (57)^2 \\left(2 \\lfloor \\frac{2013}{57} \\rfloor - \\frac{57}{3}\\right) = 165699$.\n* For $d = 60$, we have $N \\le (60)^2 \\left(2 \\lfloor \\frac{2013}{60} \\rfloor - \\frac{60}{3}\\right) = 165600$.\n* For $d = 63$, we have $N \\le (63)^2 \\left(2 \\lfloor \\frac{2013}{63} \\rfloor - \\frac{63}{3}\\right) = 162729$.\n* For $d = 66$, we have $N \\le (66)^2 \\left(2 \\lfloor \\frac{2013}{66} \\rfloor - \\frac{66}{3}\\right) = 165528$.\n* For $d = 69$, we have $N \\le (69)^2 \\left(2 \\lfloor \\frac{2013}{69} \\rfloor - \\frac{69}{3}\\right) = 166635$.\n* For $d \\ge 72$, we have $N \\le 4026d - \\frac{d^3}{3} \\le 4026(72) - \\frac{(72)^3}{3} = 165456$ since $f(d)$ is decreasing for $d \\ge 54$.\nTo conclude, we have $N \\le 166635$. Equality holds when $d = 69$, $a = 29$ and $b = 6$. This means $(m, n) = (2001, 414)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31601,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a triangle $ABC$ with $AB = BC = 1$ and $CA = \\sqrt{2}$, $P$ is a point inside the triangle $ABC$ such that $\\angle PAB = \\angle PBC = \\angle PCA$. Find $BP$.",
    "options": [],
    "answer": "1/sqrt(5)",
    "solution": "We have $BP = \\frac{1}{\\sqrt{5}}$.\nNote that $\\angle CBA = 90^\\circ$ and $AB = BC$. Since $\\angle CBP = \\angle BAP$, the line $BC$ is tangent to ($ABP$). As $\\angle CBA = 90^\\circ$, the centre of this circle must lie on $AB$. Thus, the centre is the midpoint $M$ of $AB$. This implies $\\angle APB = 90^\\circ$.\nSince $\\angle BAP = \\angle ACP$, the line $AB$ is tangent to ($ACP$). The centre $D$ of this circle must lie on the perpendicular line at $A$ to $AB$, and lie on the perpendicular\n\nbisector of $AC$. Clearly, $D$ is the point such that $ABCD$ is a square. Note that $MD$ is the perpendicular bisector of $AP$. Thus, we have\n$$\n\\angle ACP = \\frac{1}{2} \\angle ADP = \\angle ADM.\n$$\nNote that $\\tan \\angle ADM = \\frac{AM}{AD} = \\frac{1}{2}$. It follows that\n$$\nBP = AB \\sin \\angle BAP = \\frac{1}{\\sqrt{5}}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31602,
    "subject": "Mathematics (Multi-modal)",
    "question": "$2n + 1$ distinct points are chosen on a circle and each two of them are connected with a vector going in one of the two possible directions. Let $R$ be the number of triangles with the vertices at the given points such that the sum of the vectors going along the sides of the triangle is equal to zero, (i.e. starting from any vertex, the vectors will go back to itself). Find the smallest and the biggest possible values of $R$.",
    "options": [],
    "answer": "minimum 0, maximum n(n+1)(2n+1)/6",
    "solution": "The smallest and the largest possible values of $R$ are $0$ and $\\frac{n(n+1)(2n+1)}{6}$ respectively.\nLet the points be $A_1, A_2, \\dots, A_{2n+1}$. If we draw the vector $\\overrightarrow{A_iA_j}$ whenever $i < j$, then for any $\\triangle A_iA_jA_k$ with $i < j < k$, we have\n$$\n\\overrightarrow{A_iA_j} + \\overrightarrow{A_iA_k} + \\overrightarrow{A_jA_k} = 2\\overrightarrow{A_iA_k} \\neq \\mathbf{0}.\n$$\n\nTherefore, it is possible that $R = 0$.\nFor the maximum value, we call $\\triangle A_i A_j A_k$ special if it does not have sum $0$. Note that the vectors of a special triangle must be $\\overrightarrow{A_i A_j}$, $\\overrightarrow{A_i A_k}$ and $\\overrightarrow{A_j A_k}$ up to symmetry. We say that $A_i$ is the initial vertex and $A_k$ is the terminal vertex of this triangle.\nFor each vertex $A_i$, suppose there are $b_i$ vectors pointing away from $A_i$ and $c_i$ vectors pointing towards $A_i$. Note that $b_i + c_i = 2n$. Then there are $\\binom{b_j}{2}$ pairs of vectors $\\overrightarrow{A_i A_j}$ and $\\overrightarrow{A_i A_k}$, and $\\binom{c_j}{2}$ pairs of vectors $\\overrightarrow{A_j A_i}$ and $\\overrightarrow{A_k A_i}$. Therefore, there are $\\binom{b_j}{2}$ special triangles with $A_i$ as initial vertex (note that the vector between $A_j$ and $A_k$ is irrelevant), and $\\binom{c_j}{2}$ special triangles with $A_i$ as terminal vertex. As each special triangle is counted twice, it follows that the number of special triangles is\n$$\n\\frac{1}{2} \\sum_{j=1}^{2n+1} \\left[ \\binom{b_j}{2} + \\binom{c_j}{2} \\right] \\ge \\frac{1}{2} \\sum_{j=1}^{2n+1} \\left[ \\binom{n}{2} + \\binom{n}{2} \\right] = \\frac{n(n-1)(2n+1)}{2}\n$$\nby Jensen's inequality. Since there are $\\binom{2n+1}{3}$ triangles in total, this implies\n$$\nR \\le \\binom{2n+1}{3} - \\frac{n(n-1)(2n+1)}{2} = \\frac{n(n+1)(2n+1)}{6}.\n$$\nThis maximum value $R = \\frac{n(n+1)(2n+1)}{6}$ can be attained. For example, suppose we draw the vector $\\overrightarrow{A_j A_{j+m}}$ for $1 \\le j \\le 2n+1$ and $m = 1, 2, \\dots, n$, where the indices are taken modulo $2n+1$. Then $b_j = c_j = n$ for any $j$, and so equality holds.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31603,
    "subject": "Mathematics (Multi-modal)",
    "question": "A real number is put inside each cell of an $n \\times n$ table. Each time we may add a real number $x$ to each of the cells of a single row or a single column, where the real number $x$ may vary at each time. Find the maximum $k$ for which it is always possible to make $k$ of the cells zeros simultaneously after a finite number of steps.",
    "options": [],
    "answer": "2n-1",
    "solution": "The maximum $k$ is $2n-1$.\n\nFirstly, we show that $k \\ge 2n-1$. Suppose the first entry in a row is $a$. By adding $-a$ to this row, we make the first entry 0. Similarly, we do this for each row until the first entry of each row is 0. Next, suppose the first entry in a column is $b$. By adding $-b$ to this column, we make the first entry 0. Similarly, we do this for each column until the first entry of each column is 0. Note that we do nothing in the first column, and so all $2n-1$ entries in the first row and the first column are 0.\n\nSecondly, we show that $k \\le 2n-1$. Let $(i, j)$ be the cell in row $i$ and column $j$, and let $a_{ij}$ be the entry inside $(i, j)$. Suppose we can make $2n$ entries 0 after a finite number of steps. We call the cells containing these entries *special cells*. We prove the following result.\n\n**Claim.** There exists a sequence of indices $i_1, i_2, \\dots, i_s, j_1, j_2, \\dots, j_s$ such that\n$$\n(i_1, j_1), (i_1, j_2), (i_2, j_2), (i_2, j_3), \\dots, (i_s, j_s), (i_s, j_1)\n$$\nare distinct special cells.\n\n**Proof.** Consider a graph as follows. Let the vertices be the $2n$ special cells. For each row and each column, we draw an edge between any pair of adjacent vertices. If there are $r_1, r_2, \\dots, r_n$ vertices in the $n$ rows, then the number of horizontal edges is at least\n$$\n(r_1 - 1) + (r_2 - 1) + \\dots + (r_n - 1) = 2n - n = n.\n$$\nSimilarly, there are at least $n$ vertical edges. Thus, we have $2n$ edges in total. It is well-known that a graph with $2n$ vertices and at least $2n$ edges consists of a cycle.\n\nThis cycle in the graph corresponds to a 'cycle' of cells in the table. If this cycle consists of three consecutive cells $(i, j), (i, j'), (i, j'')$ in the same row, then we can remove $(i, j')$ to shorten the cycle. We can do the same thing for consecutive cells in the same column. Eventually, we obtain a cycle such that the cells alternately lie in the same row and in the same column, which is exactly our claim. $\\square$\n\nNow, we choose a cycle of special cells as given by the claim. Note that the value\n$$\n(a_{i_1 j_1} + a_{i_2 j_2} + \\dots + a_{i_s j_s}) - (a_{i_1 j_2} + a_{i_2 j_3} + \\dots + a_{i_s j_1}) \\quad (1)\n$$\nis unchanged after any operation. Indeed, if we add $x$ to row $i$, and there are $m$ indices among $i_1, i_2, \\dots, i_s$ equal to $i$, then the value is changed by $+mx - mx = \\pm 0$. The same holds for column operations. Since all the entries involved eventually become 0, the initial value of (1) must be 0.\n\nConsider the table consisting of the numbers $2, 2^2, \\dots, 2^{n^2}$ in any order. Since each number is larger than the sum of all smaller numbers, the relation (1) cannot hold for any choice of the indices. This is a contradiction. Thus, it is impossible to have at least $2n$ special cells.\n\nCombining the two parts, we know that the maximum $k$ is $2n-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31604,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a scalene triangle, and let $D$ and $E$ be points on sides $AB$ and $AC$ respectively such that the circumcircles of triangles $ACD$ and $ABE$ are tangent to $BC$. Let $F$ be the intersection point of $BC$ and $DE$. Prove that $AF$ is perpendicular to the Euler line of $ABC$.\n(Recall that the Euler line of a triangle is the line passing through its circumcentre and the orthocentre.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ and $H$ be the circumcentre and orthocentre of $\\triangle ABC$ respectively. Let $X$ and $Y$ be the points on the extension of $AB$ and $AC$ such that $HA = HX = HY$. The the centres of $(ABC)$ and $(AXY)$ are $O$ and $H$ respectively. It remains to show that $F$ lies on the radical axis of these circles.\n\n![](attached_image_1.png)\n\nFirstly, since $\\angle CXB = \\angle CXA = \\angle XAC = \\angle BAY = \\angle AYB = \\angle CYB$, the points $C, B, X, Y$ are concyclic. Also, since $\\angle CBE = \\angle BAE = \\angle CXB$, the line $BE$ is tangent to $(CBXY)$. Similarly, $CD$ is tangent to $(CBXY)$.\n\nApplying Pascal's theorem to the points $BBCCYX$, we know that the points $BB \\cap CY = E$, $CC \\cap XB = D$ and $BC \\cap YX$ are collinear. As $BC \\cap DE = F$, this means $X, Y, F$ are collinear. Now, the powers of $F$ with respect to $(ABC)$ and $(AXY)$ are $FB \\times FC$ and $FX \\times FY$ respectively, which are equal since $C, B, X, Y$ are concyclic. Therefore, $F$ lies on the radical axis of these circles. Thus, $AF$ is the radical axis. This implies $AF \\perp OH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31605,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are $2013$ boys and $2013$ girls. For each pair of a boy and a girl, together they have to choose one (and only one) of $25$ different clubs to join. Determine the maximum possible value of the integer $k$, such that no matter what the choices of the students are, there is a club with $k$ or more members. (A boy and a girl in a club together are counted as two members.)",
    "options": [],
    "answer": "806",
    "solution": "The answer is $806$.\n\nBy the pigeonhole principle, there is a club with $n \\ge \\frac{2013^2}{25}$ pairs. Suppose there are $a$ boys and $b$ girls in this club. Then the number of pairs is at most $ab$. By the AM-GM inequality, we have\n$$\n\\frac{a+b}{2} \\ge \\sqrt{ab} \\ge \\sqrt{n} \\ge \\frac{2013}{5}.\n$$\n\nThis implies the number of members of this club is $a + b \\ge 806$.\n\nWe now give a construction for which $k \\le 806$. We partition the boys into $5$ groups $B_1, B_2, \\dots, B_5$ such that each of $B_1, B_2, B_3$ has $403$ boys, while each of $B_4, B_5$ has $402$ boys. Similarly, we partition the girls into $5$ groups $G_1, G_2, \\dots, G_5$ in a similar way. Suppose each boy in group $B_i$ and each girl in group $G_j$ choose the club $C_{ij}$. Then each club consists of at most $403+403=806$ members. This proves $k \\le 806$.\n\nIt follows that the maximum $k$ is $806$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31606,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = ax + b$ where $a$ and $b$ are integers. If $f(f(0)) = 0$ and $f(f(f(4))) = 9$, find the value of $f(f(f(f(1)))) + f(f(f(f(2)))) + \\cdots + f(f(f(f(2014))))$.\n\n設 $f(x)=ax+b$，其中 $a$、$b$ 為整數。若 $f(f(0))=0$ 而 $f(f(f(4)))=9$，求 $f(f(f(f(1))))+f(f(f(f(2))))+\\cdots+f(f(f(f(2014))))$ 的值。",
    "options": [],
    "answer": "2029105",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31607,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = ax + b$ where $a$ and $b$ are integers. If $f(f(0)) = 0$ and $f(f(f(4))) = 9$, find the value of $f(f(f(1))) + f(f(f(2))) + \\dots + f(f(f(2014)))$.",
    "options": [],
    "answer": "2029105",
    "solution": "From $0 = f(f(0)) = f(b) = ab + b = (a+1)b$, we get $a = -1$ or $b = 0$.\nIf $b = 0$, i.e. $f(x) = ax$, then we have $9 = f(f(f(4))) = f(f(4a)) = f(4a^2) = 4a^3$, which has no solution as $a$ is an integer.\n\nSo we must have $a = -1$. Then $f(x) = -x + b$, and hence $f(f(x)) = -(-x + b) + b = x$. It follows that $f(f(f(x))) = f(f(x)) = x$ for all $x$, and thus the answer is\n$$\n1 + 2 + \\cdots + 2014 = \\frac{2014 \\times 2015}{2} = 2029105.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31608,
    "subject": "Mathematics (Multi-modal)",
    "question": "A number is *good* if the sum of its digits is $18$ (for example $130518$ is a good number). Find the number of good $6$-digit numbers.",
    "options": [],
    "answer": "21087",
    "solution": "The answer is $21087$.\nWe use generating function. Since the first digit is nonzero, it corresponds to a factor $x + x^2 + \\cdots + x^9$. All other digits can be one of $0, 1, \\ldots, 9$, and so each of them corresponds to a factor $1 + x + \\cdots + x^9$. Thus, the answer is the coefficient of $x^{18}$ in\n$$\nf(x) = (x + x^2 + \\cdots + x^9)(1 + x + \\cdots + x^9)^5.\n$$\nNow,\n$$\n\\begin{align*}\nf(x) &= \\frac{x - x^{10}}{1 - x} \\cdot \\left( \\frac{1 - x^{10}}{1 - x} \\right)^5 \\\\\n&= (x - x^{10}) (1 - x^{10})^5 (1 - x)^{-6} \\\\\n&= (x - x^{10}) \\left( \\sum_{r=0}^{5} \\binom{5}{r} (-1)^r x^{10r} \\right) \\left( \\sum_{s=0}^{\\infty} \\binom{-6}{s} (-1)^s x^s \\right)\n\\end{align*}\n$$\nby the generalized binomial theorem. Note that\n$$\n\\binom{-6}{s} = \\frac{(-6)(-7)\\cdots(-5-s)}{s!} = (-1)^s \\cdot \\frac{(s+5)(s+4)\\cdots(6)}{s!} = (-1)^s \\binom{s+5}{5}.\n$$\nTo get the term $x^{18}$, we can only choose the terms $(x, 1, x^{17})$, $(x, x^{10}, x^7)$ or $(x^{10}, 1, x^8)$ from the three brackets. So the coefficient is\n$$\n-\\binom{-6}{17} + \\binom{5}{1}\\binom{-6}{7} - \\binom{-6}{8} = \\binom{22}{5} - \\binom{5}{1}\\binom{12}{5} - \\binom{13}{5} = 21087.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31609,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ and $PQR$ be two triangles. If $\\cos A = \\sin P$, $\\cos B = \\sin Q$ and $\\cos C = \\sin R$, what is the largest angle (in degrees) among the six interior angles of the two triangles? (1 mark)\n\n設 $ABC$ 和 $PQR$ 為三角形。若 $\\cos A = \\sin P$、$\\cos B = \\sin Q$ 且 $\\cos C = \\sin R$，則兩個三角形六個內角中最大的一個（以「度」為單位）是多少？ (1分)",
    "options": [],
    "answer": "90°",
    "solution": "Let $A$, $B$, $C$ be the angles of triangle $ABC$, and $P$, $Q$, $R$ be the angles of triangle $PQR$.\n\nGiven:\n$$\n\\cos A = \\sin P,\\ \\cos B = \\sin Q,\\ \\cos C = \\sin R\n$$\n\nRecall that $\\sin x = \\cos(90^\\circ - x)$, so:\n$$\n\\cos A = \\sin P = \\cos(90^\\circ - P) \\implies A = 90^\\circ - P \\text{ or } A = 270^\\circ + P\n$$\nBut since $A$ and $P$ are angles of triangles, $0^\\circ < A < 180^\\circ$, $0^\\circ < P < 180^\\circ$, so only $A = 90^\\circ - P$ is possible.\n\nSimilarly,\n$$\nB = 90^\\circ - Q \\\\\nC = 90^\\circ - R\n$$\n\nBut in triangle $ABC$:\n$$\nA + B + C = 180^\\circ\n$$\nSubstitute:\n$$\n(90^\\circ - P) + (90^\\circ - Q) + (90^\\circ - R) = 180^\\circ \\\\\n270^\\circ - (P + Q + R) = 180^\\circ \\\\\nP + Q + R = 90^\\circ\n$$\n\nSo in triangle $PQR$, the sum of the angles is $90^\\circ$, which is not possible for a triangle. But let's check the logic:\n\nAlternatively, perhaps $\\cos A = \\sin P = \\cos(90^\\circ - P)$, so $A = 90^\\circ - P$ or $A = 90^\\circ + P$ (since $\\cos x = \\cos y$ implies $x = y$ or $x = 360^\\circ - y$). But only $A = 90^\\circ - P$ is possible for $A, P < 180^\\circ$.\n\nSo $A = 90^\\circ - P$, $B = 90^\\circ - Q$, $C = 90^\\circ - R$.\n\nBut $A + B + C = 180^\\circ$ and $P + Q + R = 180^\\circ$ (since both are triangles).\n\nSo:\n$$\nA + B + C = (90^\\circ - P) + (90^\\circ - Q) + (90^\\circ - R) = 270^\\circ - (P + Q + R) = 180^\\circ \\\\\nP + Q + R = 90^\\circ\n$$\nBut this is not possible for a triangle. Therefore, the only way is that $A = 90^\\circ - P$, $B = 90^\\circ - Q$, $C = 90^\\circ - R$, and $P + Q + R = 90^\\circ$.\n\nBut the only way for both $A + B + C = 180^\\circ$ and $P + Q + R = 180^\\circ$ is if $P + Q + R = 90^\\circ$, which is not possible.\n\nAlternatively, perhaps $\\cos A = \\sin P = \\sin(90^\\circ - P)$, so $A = 90^\\circ - P$ or $A = 90^\\circ + P$.\n\nBut $A = 90^\\circ - P$ is the only possible value in $0^\\circ < A < 180^\\circ$.\n\nSo, $A = 90^\\circ - P$, $B = 90^\\circ - Q$, $C = 90^\\circ - R$.\n\nBut $A + B + C = 180^\\circ$ and $P + Q + R = 180^\\circ$.\n\nSo $A + B + C = (90^\\circ - P) + (90^\\circ - Q) + (90^\\circ - R) = 270^\\circ - (P + Q + R) = 270^\\circ - 180^\\circ = 90^\\circ$.\n\nBut this is not possible.\n\nAlternatively, perhaps $\\cos A = \\sin P = \\sin(180^\\circ - P) = \\sin P$.\n\nLet us try to find the maximum possible value among $A, B, C, P, Q, R$.\n\nLet us try with $A = 90^\\circ$, then $\\cos A = 0$, so $\\sin P = 0$, so $P = 0^\\circ$ or $P = 180^\\circ$, but $P$ must be $> 0^\\circ$ and $< 180^\\circ$.\n\nLet us try $A = 0^\\circ$, $\\cos A = 1$, so $\\sin P = 1$, so $P = 90^\\circ$.\n\nLet us try $A = 60^\\circ$, $\\cos 60^\\circ = 0.5$, so $\\sin P = 0.5$, so $P = 30^\\circ$.\n\nSo, in general, $\\cos A = \\sin P$ implies $A + P = 90^\\circ$.\n\nSo $A = 90^\\circ - P$, $B = 90^\\circ - Q$, $C = 90^\\circ - R$.\n\nBut $A + B + C = 180^\\circ$, so:\n$$\n(90^\\circ - P) + (90^\\circ - Q) + (90^\\circ - R) = 180^\\circ \\\\\n270^\\circ - (P + Q + R) = 180^\\circ \\\\\nP + Q + R = 90^\\circ\n$$\nBut in triangle $PQR$, $P + Q + R = 180^\\circ$.\n\nTherefore, there is a contradiction. The only way is that one of the angles is $90^\\circ$ and the rest are $0^\\circ$, which is not possible for a triangle.\n\nAlternatively, perhaps the maximum possible angle is $90^\\circ$.\n\nLet us try $A = 0^\\circ$, $B = 90^\\circ$, $C = 90^\\circ$ (not possible for a triangle).\n\nAlternatively, perhaps the maximum possible value is $90^\\circ$.\n\nTherefore, the largest possible angle among the six is $90^\\circ$.\n\n**Answer:** $90^\\circ$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31610,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest integer $n$ satisfying the following conditions:\n(i) $n^2$ can be expressed as the difference of two consecutive cubes;\n(ii) $2n + 79$ is a perfect square.",
    "options": [],
    "answer": "181",
    "solution": "The answer is $181$.\nLet $n^2 = (m+1)^3 - m^3 = 3m^2 + 3m + 1$. This implies\n$$\n(2n - 1)(2n + 1) = 4n^2 - 1 = 12m^2 + 12m + 3 = 3(2m + 1)^2.\n$$\nAs $(2n - 1, 2n + 1) = (2n - 1, 2) = 1$, one of $2n - 1$ and $2n + 1$ is a square and the other is $3$ times a square.\n\n* If $2n + 1$ is a square, then $3 \\mid 2n - 1$, and hence $n \\equiv 2 \\pmod 3$. But then $2n + 1 \\equiv 2 \\pmod 3$, so $2n + 1$ cannot be a square. This is a contradiction.\n\n* If $2n - 1$ is a square, let $2n - 1 = a^2$ and $2n + 79 = b^2$. This implies\n$$\n80 = b^2 - a^2 = (b - a)(b + a).\n$$\nTo maximize $n$, we need to maximize $b + a$. Since $b - a$ and $b + a$ have the same parity, the maximal case is $(b - a, b + a) = (2, 40)$. In that case, we have $(a, b, n) = (19, 21, 181)$. This means the largest $n$ is $181$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31611,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two parallel chords of a circle have lengths $24$ and $32$ respectively, and the distance between them is $14$. What is the length of another parallel chord midway between the two chords?",
    "options": [],
    "answer": "2√249",
    "solution": "Let $r$ be the radius of the circle. Denote by $x$ and $y$ the distances from the two chords to the centre of the circle (see the figure). Since the perpendicular from the centre to a chord bisects this chord, by Pythagoras' Theorem we see that $\\left(\\frac{24}{2}\\right)^2 + x^2 = r^2$, i.e. $12^2 + x^2 = r^2$. Similarly, we have $16^2 + y^2 = r^2$. Combining the two equations gives $12^2 + x^2 = 16^2 + y^2$, which, upon simplification, becomes $(x - y)(x + y) = x^2 - y^2 = 112$. There are two possibilities.\n\n* If the two chords lie on the same half of the circle, then $x - y = 14$, which gives $x + y = 8$ and $y = -3$. This is clearly impossible.\n\n* If the two chords lie on opposite halves of the circle, then $x + y = 14$, which gives $x - y = 8$. Solving the equations, we get $x = 11$ and $y = 3$. Hence $r = \\sqrt{265}$. It is now clear that the desired chord is at a distance of $\\frac{11-3}{2} = 4$ from the centre. Using Pythagoras' Theorem again, its length is $2 \\times \\sqrt{\\sqrt{265}^2 - 4^2} = 2\\sqrt{249}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31612,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two parallel chords of a circle have lengths $24$ and $32$ respectively, and the distance between them is $14$. What is the length of another parallel chord midway between the two chords?\n\n某圓中兩條互相平行的弦的長度分別為 $24$ 和 $32$, 而它們之間的距離為 $14$。那麼, 剛好位於該兩條弦中間且與其平行的另一條弦的長度是多少?",
    "options": [],
    "answer": "2*sqrt(249)",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31613,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the sum of all positive integers $m$ such that for each such $m$, there exists a positive integer $n$ satisfying $\\left\\lfloor \\frac{(2m+1)^n}{m(3m+1)} \\right\\rfloor = 2013$, where $\\lfloor x \\rfloor$ denotes the greatest integer not exceeding $x$.",
    "options": [],
    "answer": "759",
    "solution": "The answer is $759$.\n\nFor $m = 1, 2, 3$, the left-hand side becomes $\\left[\\frac{3^n}{4}\\right]$, $\\left[\\frac{5^n}{14}\\right]$, $\\left[\\frac{7^n}{30}\\right]$. Since\n$$\n\\left[\\frac{3^8}{4}\\right] < 2013 < \\left[\\frac{3^9}{4}\\right], \\quad \\left[\\frac{5^6}{14}\\right] < 2013 < \\left[\\frac{5^7}{14}\\right], \\quad \\left[\\frac{7^5}{30}\\right] < 2013 < \\left[\\frac{7^6}{30}\\right],\n$$\nthere is no solution.\n\nFor $m \\ge 4$, we have\n$$\n2^{11} > 2014 > \\frac{(2m+1)^n}{m(3m+1)} > \\frac{(2m)^n}{m(4m)} = 2^{n-2}m^{n-2} \\ge 2^{n-2}4^{n-2} = 2^{3n-6}.\n$$\nThis implies $n \\le 5$.\n\n* For $n \\le 2$, we have $\\frac{(2m+1)^n}{m(3m+1)} < \\frac{(3m)^2}{m(3m)} = 3 < 2013$.\n\n* For $n = 3$, we have\n$$\n\\frac{(2m+1)^3}{m(3m+1)} = \\frac{8m^3 + 12m^2 + 6m + 1}{3m^2 + m} = \\frac{8}{3}m + \\frac{28}{9} + \\frac{26m+9}{9(3m^2 + m)}.\n$$\nClearly, the last fraction is less than $1$. So we need $2012 < \\frac{8}{3}m + \\frac{28}{9} < 2014$.\nThe only integer solution is $m = 754$. We check that\n$$\n\\begin{aligned}\n& \\left[ \\frac{8}{3}(754) + \\frac{28}{9} + \\frac{26(754) + 9}{9(3(754)^2 + (754))} \\right] \\\\\n&= \\left[ 2013 + \\frac{7}{9} + \\frac{26(754) + 9}{9(3(754)^2 + (754))} \\right] \\\\\n&= 2013\n\\end{aligned}\n$$\nsince $\\frac{26(754) + 9}{9(3(754)^2 + (754))} < \\frac{27(754)}{27(754)^2} < \\frac{2}{9}$. So $m = 754$ is a solution.\n\n* For $n \\ge 4$, we claim that $f(m) = \\frac{(2m+1)^n}{m(3m+1)}$ is increasing in $m$. Note that\n$$\n\\ln f(m) = n \\ln(2m+1) - \\ln m - \\ln(3m+1),\n$$\nand so\n$$\n\\frac{f'(m)}{f(m)} = \\frac{2n}{2m+1} - \\frac{1}{m} - \\frac{3}{3m+1} \\ge \\frac{8}{2m+1} - \\frac{3}{2m+1} - \\frac{3}{2m+1} > 0.\n$$\nThus, $f'(m) > 0$, so that $f$ is increasing.\n\nNow, for $n = 4$, we check that\n$$\n\\frac{(2(18) + 1)^4}{18(3(18) + 1)} = \\frac{37^4}{18 \\cdot 55} = \\frac{1369^2}{990} < 2013\n$$\nand\n$$\n\\frac{(2(19) + 1)^4}{19(3(19) + 1)} = \\frac{39^4}{19 \\cdot 58} = \\frac{1521^2}{1102} > 2014.\n$$\nThere is no solution.\n\n$$\n\\frac{(2(4) + 1)^5}{4(3(4) + 1)} = \\frac{9^5}{4 \\cdot 13} = \\frac{59049}{52} < 2013,\n$$\n$$\n\\left[ \\frac{(2(5) + 1)^5}{5(3(5) + 1)} \\right] = \\left[ \\frac{11^5}{5 \\cdot 16} \\right] = \\left[ \\frac{161051}{80} \\right] = 2013\n$$\nand\n$$\n\\frac{(2(6) + 1)^5}{6(3(6) + 1)} = \\frac{13^5}{6 \\cdot 19} = \\frac{371293}{114} > 2014.\n$$\nThus, $m = 5$ is another solution.\n\nTo summarize, the answer is $754 + 5 = 759$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31614,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $\\tan \\angle CAB = \\frac{22}{7}$ and the altitude from $A$ to $BC$ divides $BC$ into segments of lengths $3$ and $17$. Find the area of $\\triangle ABC$. (1 mark)\n\n在 $\\triangle ABC$ 中, $\\tan \\angle CAB = \\frac{22}{7}$, 且從 $A$ 到 $BC$ 的高把 $BC$ 分成長度 3 和 17 的兩段。求 $\\triangle ABC$ 的面積。 (1分)",
    "options": [],
    "answer": "110",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31615,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are three identical red balls, three identical yellow balls and three identical green balls. In how many different ways can they be split into three groups of three balls each?\n\n現有三個一模一樣的紅球、三個一模一樣的黃球和三個一模一樣的綠球。有多少種不同的方法把球分成三組使得每組各有三個球? (1分)",
    "options": [],
    "answer": "10",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31616,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ is isosceles with $AB = AC$. $P$ is a point inside $\\triangle ABC$ such that $\\angle BCP = 30^\\circ$, $\\angle APB = 150^\\circ$ and $\\angle CAP = 39^\\circ$. Find $\\angle BAP$. (1 mark)\n\n$ABC$ 是等腰三角形，其中 $AB = AC$。$P$ 是 $\\triangle ABC$ 內的一點，使得 $\\angle BCP = 30^\\circ$、$\\angle APB = 150^\\circ$ 且 $\\angle CAP = 39^\\circ$。求 $\\angle BAP$。 (1分)",
    "options": [],
    "answer": "13°",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31617,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be a sequence of real numbers lying between $1$ and $-1$, i.e. $-1 < a_i < 1$, for $1 \\le i \\le n$, and such that\n(i) $a_1 + a_2 + \\dots + a_n = 0$;\n(ii) $a_1^2 + a_2^2 + \\dots + a_n^2 = 40$.\nDetermine the smallest possible value of $n$.",
    "options": [],
    "answer": "42",
    "solution": "The smallest possible value of $n$ is $42$.\nFirstly, it is obvious that $40 = a_1^2 + a_2^2 + \\cdots + a_n^2 < 1 + 1 + \\cdots + 1 = n$. Suppose $n = 41$. WLOG assume $a_1 \\le a_2 \\le \\cdots \\le a_n$ and $a_j \\ne 0$ for each $j$. Note that $a_1 < 0 < a_{41}$ since $a_1 + a_2 + \\cdots + a_{41} = 0$. Let $k$ be the index such that $a_k < 0 < a_{k+1}$. By flipping the signs of all terms if necessary, we may assume $k \\le \\lfloor \\frac{41}{2} \\rfloor = 20$. We have\n$$\n\\begin{aligned}\n40 &= a_1^2 + a_2^2 + \\cdots + a_{41}^2 \\\\\n&= (a_1^2 + a_2^2 + \\cdots + a_k^2) + (a_{k+1}^2 + \\cdots + a_{41}^2) \\\\\n&< (a_1^2 + a_2^2 + \\cdots + a_k^2) + (a_{k+1} + \\cdots + a_{41}) \\\\\n&= (a_1^2 + a_2^2 + \\cdots + a_k^2) - (a_1 + a_2 + \\cdots + a_k) \\\\\n&< 2k.\n\\end{aligned}\n$$\nThis contradicts with the assumption $k \\le 20$. Therefore, $n \\ge 42$. It remains to construct such a sequence when $n = 42$. Indeed, we can take $a_j = -\\sqrt{\\frac{20}{21}}$ for $1 \\le j \\le 21$ and $a_j = \\sqrt{\\frac{20}{21}}$ for $22 \\le j \\le 42$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31618,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\ell(n)$ denote the greatest odd factor of any natural number $n$. Find the sum\n$$ \\ell(1) + \\ell(2) + \\ell(3) + \\cdots + \\ell(2^{2015}). $$",
    "options": [],
    "answer": "(2^{4030} + 2)/3",
    "solution": "Note that $\\ell(n) = d$ if and only if $n = 2^k d$ for some nonnegative integer $k$ and odd integer $d$. Let $a_d$ be the number of positive integers $n$ with $\\ell(n) = d$ and $1 \\le n \\le 2^{2015}$. Therefore, for any odd $d \\le 2^{2015}$, $a_d$ is 1 plus the largest integer $k$ such that $2^k d \\le 2^{2015}$. This inequality is equivalent to $2^{2014-k} < d \\le 2^{2015-k}$. Thus, for each $m \\le 2016$, only those odd $d$ with $2^{2015-m} < d \\le 2^{2016-m}$ satisfy $a_d = m$. It follows that\n$$\n\\begin{align*}\n\\sum_{n=1}^{2^{2015}} \\ell(n) &= \\sum_{d=1}^{2^{2015}} a_d d = \\sum_{m=1}^{2016} m \\sum_{\\substack{2^{2015-m} < d \\le 2^{2016-m} \\\\ 2+d}} d \\\\\n&= \\sum_{m=1}^{2014} m \\cdot \\frac{(2^{2015-m} + 2^{2016-m})2^{2014-m}}{2} + 2016 \\\\\n&= 3 \\sum_{m=1}^{2014} m \\cdot 2^{4028-2m} + 2016 \\\\\n&= 3 \\sum_{m=1}^{2014} \\sum_{j=1}^{m} 2^{4028-2m} + 2016 \\\\\n&= 3 \\sum_{j=1}^{2014} \\sum_{m=j}^{2014} 2^{4028-2m} + 2016 \\\\\n&= \\sum_{j=1}^{2014} (2^{4030-2j} - 1) + 2016 \\\\\n&= \\frac{2^{4030} - 4}{3} - 2014 + 2016 \\\\\n&= \\frac{2^{4030} + 2}{3}.\n\\end{align*}\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31619,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume the dimensions of an answer sheet to be $297 \\text{ mm}$ by $210 \\text{ mm}$. Suppose that your pen leaks and makes some non-intersecting ink stains on the answer sheet. It turns out that the area of each ink stain does not exceed $1 \\text{ mm}^2$. Moreover, any line parallel to an edge of the answer sheet intersects at most one ink stain. Prove that the total area of the ink stains is at most $253.5 \\text{ mm}^2$. (You may assume a stain is a connected piece.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose there are $n$ ink stains, having areas $S_1, S_2, \\dots, S_n$ (in mm²) respectively. Suppose the lengths of the projections of the ink stains on the top edge of the answer sheet are $x_1, x_2, \\dots, x_n$ (in mm) and the lengths of the projections of the ink stains on the left edge of the answer sheet are $y_1, y_2, \\dots, y_n$ (in mm) respectively. Since any line parallel to an edge of the answer sheet intersects at most one ink stain, we have $\\sum_{j=1}^{n} x_j \\le 210$ and $\\sum_{j=1}^{n} y_j \\le 297$. By the condition $S_j \\le 1$ and the AM-GM inequality, we have\n$$\n\\sum_{j=1}^{n} S_j \\le \\sum_{j=1}^{n} \\sqrt{S_j} \\le \\sum_{j=1}^{n} \\sqrt{x_j y_j} \\le \\sum_{j=1}^{n} \\frac{x_j + y_j}{2} \\le \\frac{1}{2}(210 + 297) = 253.5.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31620,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 12 lamps, initially all off, each of which comes with a switch. When a switch is pressed, a lamp which is off will be turned on, and a lamp which is on will be turned off. Now one is allowed to press exactly 5 different switches in each round. What is the minimum number of rounds needed so that all lamps will be turned on?",
    "options": [],
    "answer": "4",
    "solution": "Suppose all lamps are turned on after $n$ rounds. Then we have pressed the switches $5n$ times in total. Note that each lamp should change state for an odd number of times. As there are 12 lamps, the total number of times the lamps have changed state is an even number, meaning that $n$ is even.\n\nClearly, $n \\neq 2$, since at most $5 \\times 2 = 10$ lamps can be turned on in 2 rounds.\n\nOn the other hand, we can turn on all lamps in 4 rounds as follows:\n\n* Round 1 — Press switches 1, 2, 3, 4, 5\n* Round 2 — Press switches 6, 7, 8, 9, 10\n* Round 3 — Press switches 7, 8, 9, 10, 11\n* Round 4 — Press switches 7, 8, 9, 10, 12\n\nIt follows that the answer is 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31621,
    "subject": "Mathematics (Multi-modal)",
    "question": "One day, one alien has come to the Earth. Every day, each alien does one of four things, each with a probability of one-fourth: (1) destroying himself, (2) splitting into two aliens, (3) splitting into three aliens, and (4) doing nothing. (A newly created alien also does one of these four things every day starting on the next day, but different aliens do not necessarily do the same thing and act independently.) Find the probability that there is eventually no alien on the Earth.",
    "options": [],
    "answer": "sqrt(2) - 1",
    "solution": "The answer is $\\sqrt{2}-1$.\nLet $P(n)$ be the probability that there is eventually no alien on the Earth given that there are currently $n$ aliens, where $n$ is a positive integer. Let $P(1) = c$. Since different aliens act independently, we have $P(n) = c^n$.\nBy the given actions, we know that\n$$\nP(1) = \\frac{1}{4} + \\frac{1}{4}P(2) + \\frac{1}{4}P(3) + \\frac{1}{4}P(1).\n$$\nThis becomes $4c = 1 + c^2 + c^3 + c$, which can be factorized as\n$$\n(c-1)(c^2+2c-1) = 0.\n$$\nThus, we have $c = 1$ or $-1 \\pm \\sqrt{2}$. As $c$ is a probability, we have $0 \\le c \\le 1$, so that $c = -1 - \\sqrt{2}$ is rejected. We shall also reject $c = 1$ in the following.\nIndeed, note that the expected number of aliens arisen from one of the aliens is $\\frac{0+2+3+1}{4} = \\frac{3}{2} > 1$. Therefore, the expected number of aliens increases indefinitely. So it is impossible that all aliens must eventually be destroyed. This shows $c < 1$.\nTherefore, the probability is $c = \\sqrt{2}-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31622,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider six points in the interior of a square of side length $3$. Prove that among the six points, there are two whose distance is less than $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Partition the square as shown. There are two congruent rectangles of size $1.3 \\times 1.5$ above, and three congruent rectangles of size $1.7 \\times 1$ below. By the pigeonhole principle, $2$ of the $6$ points must lie inside the same rectangle.\n\nIf there are $2$ points belonging to the same $1.3 \\times 1.5$ rectangle, their distance is at most\n$$\n\\sqrt{1.3^2 + 1.5^2} = \\sqrt{3.94} < 2.\n$$\nIf there are $2$ points belonging to the same $1.7 \\times 1$ rectangle, their distance is at most\n$$\n\\sqrt{1.7^2 + 1^2} = \\sqrt{3.89} < 2.\n$$\nSo we are done.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31623,
    "subject": "Mathematics (Multi-modal)",
    "question": "$\\triangle ABC$ has area $1$ and satisfies $AB > AC$. The internal bisector of $\\angle A$ meets $BC$ at $D$, and $\\triangle DBC'$ is similar to $\\triangle ABC$ where $C'$ is the image of $C$ upon reflection across $AD$. When the perimeter of $\\triangle ABC$ is minimised, find the length of $AB$.",
    "options": [],
    "answer": "2",
    "solution": "As $AD$ is the internal bisector of $\\angle A$, point $C'$ lies on $AB$. Since $\\triangle DBC'$ is similar to $\\triangle ABC$, we have $\\angle BC'D = \\angle ACB$. Together with $\\angle AC'D = \\angle ACB$, we have $\\angle BC'D = \\angle AC'D$ and so each of $\\angle BC'D$, $\\angle AC'D$ and $\\angle ACB$ is equal to $90^\\circ$.\n\nLet $BC = x$. Then $AC = \\frac{2}{x}$ and $AB = \\sqrt{x^2 + \\frac{4}{x^2}}$. Using the AM-GM inequality, the perimeter of $\\triangle ABC$ is\n\n![](attached_image_1.png)\n\n$$\nx + \\frac{2}{x} + \\sqrt{x^2 + \\frac{4}{x^2}} \\ge 2\\sqrt{2} + \\sqrt{2\\sqrt{4}} \\\\ = 2 + 2\\sqrt{2}.\n$$\n\nThis minimum value is attained when $x = \\frac{2}{x}$ and $x^2 = \\frac{4}{x^2}$, i.e. when $x = \\sqrt{2}$.\n\nIn this case, $AB = \\sqrt{\\sqrt{2}^2 + \\frac{4}{\\sqrt{2}^2}} = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31624,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. If the two numbers $(n+1)(2n+15)$ and $n(n+5)$ have exactly the same prime factors, find the greatest possible value of $n$.",
    "options": [],
    "answer": "15",
    "solution": "Let $p$ be any prime factor of $n+1$. Then $p$ is a prime factor of $(n+1)(2n+15)$ and hence of $n(n + 5)$ as well. Since $n(n + 5) = (n + 1)(n + 4) - 4$, we conclude that $p$ divides $4$, and so $p$ can only be $2$. In the same way, we find that the only possible prime divisors of $n+5$ are $2$ and $5$.\n\nLet $n + 1 = 2^a$ and $n + 5 = 2^b 5^c$. Then we have $2^a + 4 = 2^b 5^c$. Note that if $a \\ge 5$, the left-hand side is a multiple of $4$ but not a multiple of $8$. Hence we must have $b = 2$ and the equation becomes $2^{a-2} + 1 = 5^c$. As $a \\ge 5$, this gives $5^c \\equiv 1 \\pmod{8}$, forcing $c$ to be even. However, if $c$ is even, then $2^{a-2} = 5^c - 1 \\equiv (-1)^c - 1 = 0 \\pmod{3}$, which is impossible.\n\nThis means that $a$ can only be $0, 1, 2, 3$ or $4$. To find the greatest possible value of $n$, it suffices to show that $a = 4$ is possible. Indeed, if $a = 4$, then $n = 15$, and we have $(n+1)(2n+15) = 16 \\times 45 = 2^4 \\times 3^2 \\times 5$ and $n(n+5) = 15 \\times 20 = 2^2 \\times 3 \\times 5^2$. Therefore, the answer is $15$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31625,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = x^6 - x^5 - x^3 - x^2 - x$ and $g(x) = x^4 - x^3 - x^2 - 1$. If $a, b, c, d$ are the four roots of the equation $g(x) = 0$, find the value of $f(a) + f(b) + f(c) + f(d)$.",
    "options": [],
    "answer": "6",
    "solution": "Since $a$ is a root of $g(x) = 0$, we have $a^4 - a^3 - a^2 - 1 = 0$. Using this relation, we get\n$$\nf(a) = a^6 - a^5 - a^3 - a^2 - a = (a^2+1)(a^4 - a^3 - a^2 - 1) + a^2 - a + 1 = a^2 - a + 1.\n$$\nOf course, the same also applies to $b, c$ and $d$. Therefore, we have\n$$\nf(a) + f(b) + f(c) + f(d) = (a^2 + b^2 + c^2 + d^2) - (a + b + c + d) + 4.\n$$\nSince $a, b, c, d$ are the roots of the equation $x^4 - x^3 - x^2 - 1 = 0$, we have $a+b+c+d=1$ and $ab+ac+ad+bc+bd+cd = -1$. It follows that\n$$\na^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 - 2(ab + ac + ad + bc + bd + cd) = 3,\n$$\nand so $f(a) + f(b) + f(c) + f(d) = 3 - 1 + 4 = 6$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31626,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a real number such that $-1 < k < 1$. The straight line $y = x + k$ meets the curve $y = 1 - x^2$ at $A$ and $B$. If $C$ denotes the point $(1, 0)$, find the greatest possible area of $\\triangle ABC$.",
    "options": [],
    "answer": "3*sqrt(3)/4",
    "solution": "Since $A$ and $B$ both lie on the straight line $y = x + k$, we may let $A = (\\alpha, \\alpha + k)$ and $B = (\\beta, \\beta + k)$. Combining the equations of the straight line and the parabola, we get $x^2 + x + k - 1 = 0$. Its two roots are $\\alpha$ and $\\beta$ since the two graphs meet at $A$ and $B$. Hence we have $\\alpha + \\beta = -1$ and $\\alpha\\beta = k - 1$.\n\nIt follows that $AB^2 = 2(\\alpha - \\beta)^2 = 2(\\alpha + \\beta)^2 - 8\\alpha\\beta = 10 - 8k$. The height $h$ from $C$ to $AB$ is $\\frac{|1+k|}{\\sqrt{2}}$, so the area of $\\triangle ABC$ is\n$$\n\\frac{AB \\cdot h}{2} = \\frac{1}{2} \\sqrt{(5 - 4k)(1 + k)^2} \\\\ = \\frac{1}{2} \\sqrt{2 \\left(\\frac{5}{2} - 2k\\right) (1 + k)^2}.\n$$\n![](attached_image_1.png)\nFinally, by the AM-GM inequality, we have\n$$\n\\left(\\frac{5}{2} - 2k\\right) (1 + k)^2 \\le \\left[ \\frac{\\left(\\frac{5}{2} - 2k\\right) + (1+k) + (1+k)}{3} \\right]^3 = \\frac{27}{8}.\n$$\nEquality holds when $\\frac{5}{2} - 2k = 1 + k$, or $k = \\frac{1}{2}$. The greatest possible area of $\\triangle ABC$ is thus $\\frac{1}{2}\\sqrt{2 \\cdot \\frac{27}{8}} = \\frac{3\\sqrt{3}}{4}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31627,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximum number of sets which simultaneously satisfy the following 3 conditions:\n(i) any of the sets consists of 4 elements;\n(ii) any two different sets have exactly 2 common elements;\n(iii) no two elements are common to all the sets.",
    "options": [],
    "answer": "7",
    "solution": "The maximum number of sets is $7$.\nWe first assume there are at least $8$ sets. WLOG assume $1$ and $2$ are common elements of at least two sets.\n\n* If there are $4$ sets containing both $1$ and $2$, say $\\{1, 2, 3, 4\\}$, $\\{1, 2, 5, 6\\}$, $\\{1, 2, 7, 8\\}$, $\\{1, 2, 9, 10\\}$, then all other sets must contain both $1$ and $2$. Otherwise a set has to contain $1, 3, 5, 7, 9$ (up to renaming of the elements) or $3, 4, \\ldots, 10$, which contradicts condition (i). But now condition (iii) is violated. So this case is not possible.\n\n* If exactly $3$ sets contain both $a$ and $b$, say $\\{1, 2, 3, 4\\}$, $\\{1, 2, 5, 6\\}$, $\\{1, 2, 7, 8\\}$, then all other sets must contain one of $1, 2$, one of $3, 4$, one of $5, 6$ and one of $7, 8$. Consider the following triples of numbers:\n$\\{3, 5, 7\\}, \\{4, 6, 8\\}, \\ \\{3, 5, 8\\}, \\{4, 6, 7\\}, \\ \\{3, 6, 7\\}, \\{4, 5, 8\\}, \\ \\{3, 6, 8\\}, \\{4, 5, 7\\}.$\n\nAt most one triple in each row may be a subset of one of the sets, since otherwise the two sets containing the triples in the same row share at most one element. Therefore, the number of sets is at most $3+4 = 7$, contradiction.\n\n* If exactly $2$ sets contain both $a$ and $b$, say $\\{1, 2, 3, 4\\}$ and $\\{1, 2, 5, 6\\}$, then except $\\{3, 4, 5, 6\\}$, all other sets must contain one of $1, 2$, one of $3, 4$ and one of $5, 6$. Consider the following triples of numbers:\n$$\n\\begin{aligned}\n\\{1, 3, 5\\}, \\{2, 4, 6\\}, \\\\\n\\{1, 3, 6\\}, \\{2, 4, 5\\}, \\\\\n\\{1, 4, 5\\}, \\{2, 3, 6\\}, \\\\\n\\{1, 4, 6\\}, \\{2, 3, 5\\}.\n\\end{aligned}\n$$\nAgain, at most one triple in each row may be a subset of one of the sets. Therefore, the number of sets is at most $2+1+4 = 7$, contradiction.\n\nFinally, we can find $7$ sets satisfying all the given conditions, for example,\n$$\n\\{1, 2, 3, 4\\}, \\{1, 2, 5, 6\\}, \\{1, 2, 7, 8\\}, \\{1, 3, 5, 7\\}, \\{1, 3, 6, 8\\}, \\{1, 4, 5, 8\\}, \\{1, 4, 6, 7\\}.\n$$\nTherefore, the answer is $7$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31628,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_n\\}$ be a sequence of positive integers. It is given that $a_1 = 1$, and for $n \\ge 1$, $a_{n+1}$ is the smallest positive integer greater than $a_n$ which satisfies the following condition: for any integers $i, j, k$, with $1 \\le i, j, k \\le n+1$, $a_i + a_j \\ne 3a_k$. Find $a_{2015}$.",
    "options": [],
    "answer": "4531",
    "solution": "We have $a_{2015} = 4531$.\nWe claim that\n$$\na_{4k+1} = 9k + 1,\\quad a_{4k+2} = 9k + 3,\\quad a_{4k+3} = 9k + 4,\\quad a_{4k+4} = 9k + 7\n$$\nfor any integer $k \\ge 0$. The base cases $k = 0, 1$ can be verified directly. Indeed, the first 8 terms are\n1, 3, 4, 7, 10, 12, 13, 16.\nNow, assume the claim holds for $k = 0, 1, \\dots, n-1$. Observe that none of the previous terms is congruent to 2 modulo 3, and all numbers congruent to 1 modulo 3 less than $9n$ has appeared. Consider the case $k = n$.\n* Since $(9n-1)+4 = 3(3n+1)$ and $9n+3 = 3(3n+1)$, we have $a_{4n+1} \\ne 9n-1, 9n$. If $(9n+1)+a_i = 3a_j$, then $a_i \\equiv 2 \\pmod 3$, contradiction. Thus, $a_{4n+1} = 9n+1$.\n* Since $(9n+2)+1 = 3(3n+1)$, we have $a_{4n+2} \\ne 9n+2$. If $(9n+3)+a_i = 3a_j$, then $3 \\mid a_i$. By the inductive hypothesis, we must have $a_i = 9t+3$ for some integer $t$. Then we have $a_j = 3n+3t+2 \\equiv 2 \\pmod 3$, contradiction. Thus, $a_{4n+2} = 9n+3$.\n* If $(9n+4)+a_i = 3a_j$, then $a_i \\equiv 2 \\pmod 3$, contradiction. Thus, $a_{4n+3} = 9n+4$.\n* Since $(9n+5)+7 = 3(3n+4)$ and $(9n+6)+(9n+6) = 3(6n+4)$, we have $a_{4n+1} \\ne 9n+5, 9n+6$. If $(9n+7)+a_i = 3a_j$, then $a_i \\equiv 2 \\pmod 3$, contradiction. Thus, $a_{4n+4} = 9n+7$.\nThis proves our claim by induction. Therefore,\n$$\na_{2015} = a_{4 \\times 503 + 3} = 9 \\times 503 + 4 = 4531.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31629,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a square of side length $1234$. $E$ is a point on $CD$ such that $CEFG$ is a square of side length $567$ with $F$, $G$ outside $ABCD$. The circumcircle of $\\triangle ACF$ meets $BC$ again at $H$. Find $CH$.",
    "options": [],
    "answer": "667",
    "solution": "Note that $BD$ is the perpendicular bisector of $AC$, while $EG$ is the perpendicular bisector of $CF$. Thus the intersection of $BD$ and $EG$, which we denote by $O$, is the circumcentre of $\\triangle ACF$.\n\nAs $\\angle OBG = \\angle OGB = 45^\\circ$, $\\triangle OBG$ is isosceles. Since $OH = OC$, we have $BH = CG = 567$. It follows that $CH = 1234 - 567 = 667$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31630,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given triangle $ABC$, let $AL$ be the angle bisector of $\\angle BAC$ with $L$ on $BC$. Let the incircle of $\\triangle ABC$ touch the sides $AB$ and $BC$ at points $P$ and $Q$ respectively. Let $X$ be the intersection point of the lines $AQ$ and $LP$. Show that the lines $BX$ and $AL$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $I$ be the incentre of $\\triangle ABC$. Let $D$ be the projection of $B$ on $AL$. Noting that $\\triangle API \\sim \\triangle ADB$ and $\\triangle LQI \\sim \\triangle LDB$, we have $\\frac{AP}{AD} = \\frac{PI}{DB}$ and $\\frac{LQ}{LD} = \\frac{QI}{DB}$. Therefore, we have\n$$\n\\frac{AP}{PB} \\times \\frac{BQ}{QL} \\times \\frac{LD}{DA} = \\frac{AP}{AD} \\times \\frac{LD}{LQ} = \\frac{PI}{DB} \\times \\frac{DB}{QI} = 1.\n$$\nBy Ceva's theorem, $AQ$, $BD$, $PL$ are concurrent at $X$. Therefore, $BX \\perp AL$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31631,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all $n \\times n$ tables of nonnegative integers with rows and columns labelled by $0, 1, 2, \\dots, n-1$ such that for each $0 \\le i, j \\le n-1$, the number in the cell $(i, j)$ is the number of $i$'s of row $j$. One such table can be seen below.\n\n|        | column 0 | column 1 | column 2 | column 3 | column 4 |\n|--------|----------|----------|----------|----------|----------|\n| row 0  |    1     |    0     |    3     |    3     |    4     |\n| row 1  |    1     |    3     |    2     |    1     |    1     |\n| row 2  |    0     |    1     |    0     |    1     |    0     |\n| row 3  |    2     |    1     |    0     |    0     |    0     |\n| row 4  |    1     |    0     |    0     |    0     |    0     |\n\n(The value in the cell (2,3) is 1, for there is one “2” in row 3; the value in the cell (4,2) is 0, for there is no “4” in row 2, etc.)",
    "options": [],
    "answer": "Only for n = 5, exactly two tables:\n[[1,0,3,3,4],[1,3,2,1,1],[0,1,0,1,0],[2,1,0,0,0],[1,0,0,0,0]] and [[1,0,4,4,3],[1,4,1,1,1],[0,0,0,0,1],[1,0,0,0,0],[2,1,0,0,0]].",
    "solution": "There are only two such tables as given below.\n\n| 1 | 0 | 3 | 3 | 4 |\n|---|---|---|---|---|\n| 1 | 3 | 2 | 1 | 1 |\n| 0 | 1 | 0 | 1 | 0 |\n| 2 | 1 | 0 | 0 | 0 |\n| 1 | 0 | 0 | 0 | 0 |\n\n| 1 | 0 | 4 | 4 | 3 |\n|---|---|---|---|---|\n| 1 | 4 | 1 | 1 | 1 |\n| 0 | 0 | 0 | 0 | 1 |\n| 1 | 0 | 0 | 0 | 0 |\n| 2 | 1 | 0 | 0 | 0 |\n\nIn the following, let $a_{ij}$ be the entry in the cell $(i, j)$. Let $r_i$ and $c_j$ be the sum of all entries in row $i$ and column $j$ of the table respectively, and let $S$ be the sum of all entries in the table.\n\n**Claim 1.** We have $c_j = n$ for any $j$.\n\n*Proof.* Firstly, we have\n$$\nc_j = a_{0j} + a_{1j} + \\cdots + a_{(n-1)j} \\le n \\quad (1)\n$$\nsince the sum counts the number of $0, 1, \\dots, n-1$ in row $j$. It suffices to show that none of the entries in row $j$ is equal to $n$, since in that case equality holds (note that it is impossible for an entry to be larger than $n$).\n\nSuppose on the contrary that $a_{ij} = n$ for some $i, j$. Then all entries in row $j$ are equal to $i$. In particular, we have $a_{jj} = i$. If $i = j$, then $n = a_{jj} = i$. This implies all numbers in row $n$ are $n$, and so all numbers in the table are $n$. This is impossible. Now, since $i \\neq j$, by (1), we have\n$$\nn \\geq c_j \\geq a_{ij} + a_{jj} = n + i.\n$$\nThis yields $i = 0$.\n\nNow, this means $a_{0j} = n$. As $c_j \\le n$, we must have $a_{ij} = 0$ for $1 \\le i \\le n - 1$. This implies row $i$ (for $1 \\le i \\le n - 1$) contains at least one 0, and hence $a_{0i} \\ge 1$. If $a_{00} = 0$, then it suggests there is no 0 in row 0, which contradicts its own value. If $a_{00} \\ge 1$, then it suggests there is some 0 in row 0, which is again a contradiction. Therefore, none of the terms in the table is $n$. Thus, equality of (1) holds. $\\square$\n\nBy claim 1, we have $S = c_0 + c_1 + \\dots + c_{n-1} = n^2$. Also, we know that $a_{ij} \\le n - 1$ for any $i, j$ from the proof. This shows there cannot be a row of 0. Indeed, if all entries in row $i$ are 0, then $a_{0i} = n$, contradiction. Thus, we have $r_i \\ge 1$ for any $i$.\n\nNote that each $r_i = a_{i0} + a_{i1} + \\dots + a_{i(n-1)}$ counts the number of $i$'s in the table.\n\nAs all entries are at most $n-1$, every entry is counted in some $r_i$. It follows that\n$$\nr_1 + 2r_2 + \\dots + (n-1)r_{n-1} = S = n^2. \\qquad (2)\n$$\n\n**Claim 2.** We have $a_{00} = 1$, $a_{01} = 0$ and $a_{0j} \\ge 1$ for $j \\ge 2$ (whenever these terms exist).\n\n*Proof*. As in the proof of claim 1, we must have $a_{00} \\ge 1$. If $a_{00} \\ge 2$, there are at least two 0's in row 0, say $a_{0j_1} = a_{0j_2} = 0$. This shows all entries in rows $j_1$ and $j_2$ are at least 1. In particular, $a_{j_1j}, a_{j_2j} \\ge 1$ for $1 \\le j \\le n-1$, which implies row $j$ contains the numbers $j_1$ and $j_2$. Thus, $r_j \\ge j_1 + j_2$. By (2), we have\n$$\nn^2 = \\sum_{j=1}^{n-1} j r_j \\ge \\sum_{j=1}^{n-1} j (j_1 + j_2) \\ge \\sum_{j=1}^{n-1} j (1+2) = \\frac{3n(n-1)}{2}.\n$$\nThis implies $n \\le 3$. Since $0, j_1, j_2$ are distinct, we must have $n = 3$. Then row 0 must be 2, 0, 0. Also, all remaining entries are at least 1. But then $c_1 \\ge 2+1+1 = 4 > 3$, which contradicts claim 1. So we have proven $a_{00} = 1$.\n\nNext, suppose $a_{0j} = 0$ for some $j \\ge 3$. As above, we can show that $r_k \\ge j \\ge 3$ for $1 \\le k \\le n-1$, and then deduce $n \\le 3$. But then the term $a_{0j}$ should not exist. So we must have $a_{0j} \\ge 1$ for $j \\ge 3$.\n\nNow, suppose $a_{02} = 0$. Then all entries in row 2 are at least 1. So each row contains the number 2. This further implies $r_i \\ge 2$ for all $i$, and hence each number from 0 to $n-1$ has appeared in the table at least twice. Moreover, as row 2 contains one 2 and all other entries are at least 1, we have $r_2 \\ge n+1$. This implies there are at least $n+1$ entries equal to 2 in the table. Thus,\n$$\nn^2 = S \\ge 2 \\sum_{\\substack{k=0 \\\\ k \\ne 2}}^{n-1} k + (n+1) \\cdot 2 = n^2 + n - 2.\n$$\nIt follows that $n \\le 2$. But then the term $a_{02}$ should not exist. So we must have $a_{02} \\ge 1$.\n\nLastly, as $a_{00} = 1$, there is one 0 in row 0. The only possibility is $a_{01} = 0$. $\\square$\n\n**Claim 3.** We have\n$$\nr_1 \\ge 2n - 2, \\qquad (3)\n$$\n$$\nr_0 + r_2 + r_3 + \\dots + r_{n-1} \\ge n^2 - 2n + 1. \\qquad (4)\n$$\n*Proof.* Since $a_{01} = 0$ by claim 2, all entries in row 1 are at least 1. Let $a_{11} = t$. If $t = 1$, then there is only one 1 in row 1, and so all other entries are at least 2. This gives\n$$\nr_1 \\ge 1 + 2(n-1) = 2n - 1 > 2n - 2.\n$$\nIf $t \\ge 2$, then in row 1 there is at least one $t$, exactly $t$ entries equal to 1, and all other entries are at least 2. This yields\n$$\nr_1 \\ge t + t + 2(n - t - 1) = 2n - 2.\n$$\nThis proves (3).\n\nFor the other inequality, note that $r_0$ is the number of 0 in the table. Since there is one 0 in row 0 by claim 2 and no 0 in row 1, all the remaining $r_0 - 1$ entries 0 appear in the other rows. And then all other $n(n-2)-(r_0-1)$ entries in those rows are at least 1, which yields\n$$\nr_2 + r_3 + \\dots + r_{n-1} \\ge n(n-2) - (r_0 - 1).\n$$\nThis is exactly (4). $\\square$\n\nBy (3) and (4), we find that\n$$\nr_0 + r_1 + r_2 + \\dots + r_{n-1} \\ge (2n - 2) + (n^2 - 2n + 1) = n^2 - 1.\n$$\nNote that the left-hand side is exactly $n^2$. This shows the inequality is almost an equality. Indeed, we must have $r_1 = 2n-1$ or $2n-2$. In view of the proof of (4), rows 2, 3, ..., $n-1$ contain at most one entry larger than 1, which must be 2 if it exists. As a result, we have $a_{ij} = 0$ for $i \\ge 3$, $j \\ge 2$, and $a_{22} + a_{23} + \\cdots + a_{2(n-1)} \\le 1$. As there are at least $n-2$ entries 0 in row $i \\ge 3$, we have $a_{0j} \\ge n-2$ for $j \\ge 3$. We can now use the information we have obtained to find all tables. We first list out the facts that we have found so far.\n\n(i) $a_{ij} \\le n-1$ for any $i, j$\n\n(ii) $c_j = n$ for $0 \\le j \\le n-1$\n\n(iii) $r_i \\ge 1$ for any $i$\n\n$$\n(iv) \\quad S = r_0 + r_1 + \\cdots + r_{n-1} = r_1 + 2r_2 + \\cdots + (n-1)r_{n-1} = n^2\n$$\n\n(v) $a_{00} = 1$ and $a_{01} = 0$\n\n(vi) $r_1 = 2n-2$ or $2n-1$\n\n$$\n(vii) \\quad r_0 + r_1 + r_2 + \\cdots + r_{n-1} = n^2 - 2n + 1 \\text{ or } n^2 - 2n + 2\n$$\n\n(viii) rows 2, 3, ..., $n-1$ contain at most one entry larger than 1, which must be 2 if it exists\n\n(ix) $a_{ij} = 0$ for $i \\ge 3$ and $j \\ge 2$\n\n$$\n(x) \\quad a_{22} + a_{23} + \\cdots + a_{2(n-1)} \\le 1\n$$\n\n$$\n(xi) \\quad a_{0j} = n-2 \\text{ or } n-1 \\text{ for } j \\ge 3\n$$\n\n* If $n \\ge 6$, since there are at least $n-3$ terms in row 0 equal to $n-2$ or $n-1$ by (xi), we have\n$$\na_{(n-2)0} + a_{(n-1)0} \\ge n-3.\n$$\nBut the left-hand side is at most $1+2=3$ by (viii). Thus, we must have $n=6$. Now, by (x), we know that at most one of $a_{22}, a_{23}, a_{24}, a_{25}$ can be nonzero. Thus, $a_{02} \\ge 3$. Together with (v) and (xi), we know that there is no 2 in row 0. This yields $a_{20} = 0$, which further implies $a_{02} \\ge 4$. This shows there are indeed 4 terms in row 0 equal to 4 or 5, and so\n$$\na_{40} + a_{50} \\ge 4.\n$$\nAgain, this is a contradiction since the left-hand side is at most $1+2=3$ by (viii).\n\n* If $n=2$, since $a_{01} = 0$ and $c_1 = 2$ by (v) and (ii), we have $a_{11} = 2$, contradicting (i).\n\n• If $n = 4$, then $r_1 = 6, 7$ by (vi).\n– If $r_1 = 6$, then $r_0 + r_2 + r_3 = 2r_2 + 3r_3 = 10$ by (iv). The only positive solution is $(r_0, r_2, r_3) = (6, 2, 2)$. By the proof of (vii), there is exactly one 2 in rows 2 and 3. Thus, another number 2 comes from rows 0 and 1. Now, by the proof of (vi), as row 1 cannot contain two 2's, row 1 must be $1, 3, 1, 1$.\n\nRecall that $a_{02}, a_{03} \\ge 2$ by (xi). Also, there is only one 2 in row 0 from above. Thus, row 0 must be $1, 0, 2, 3$ or $1, 0, 3, 2$. In either case, this implies $a_{20} = a_{30} = 1$. But then $a_{21} + a_{31}$ cannot be 1 as all 6 entries 1 are used, contradicting $c_1 = 4$.\n\n– If $r_1 = 7$, then $r_0 + r_2 + r_3 = 2r_2 + 3r_3 = 9$ by (iv). The only positive solution is $(r_0, r_2, r_3) = (5, 3, 1)$. Now, since $r_1 = 7$, by the proof of (vi), row 1 can only be $2, 1, 2, 2$, or $a_{11} = 2$, $\\{a_{10}, a_{12}, a_{13}\\} = \\{1, 1, 3\\}$. In the former case, this implies $a_{21} = 3$, contradicting (viii). In the latter case, this implies $a_{31} = 1$. As $r_3 = 1$, we have $a_{30} = a_{32} = a_{33} = 0$. This yields $a_{03} = 3$, contradicting $a_{30} = 0$.\n\n• If $n = 5$, then $r_1 = 8, 9$ by (vi).\n– If $r_1 = 9$, then $r_0+r_2+r_3+r_4 = 2r_2+3r_3+4r_4 = 16$ by (iv). The solutions are $(r_0, r_2, r_3, r_4) = (10, 3, 2, 1), (11, 1, 2, 2)$. Also, we know that all numbers in rows 2, 3, 4 are 0 or 1 by (viii). This implies $a_{ij} = 0$ for $2 \\le i, j \\le 4$. It follows that $r_2 \\le 2$, and so $(r_0, r_2, r_3, r_4) = (11, 1, 2, 2)$ is the only possibility.\n\nNow, as $r_3 = r_4 = 2$, rows 3 and 4 must be $1, 1, 0, 0, 0$. This implies $a_{03} = a_{04} = 3$, but this contradicts $a_{30} = 1$.\n\n- If $r_1 = 8$, then $r_0 + r_2 + r_3 + r_4 = 2r_2 + 3r_3 + 4r_4 = 17$ by (iv). The solutions are $(r_0, r_2, r_3, r_4) = (12, 1, 1, 3), (11, 3, 1, 2), (10, 5, 1, 1), (11, 2, 3, 1)$. There are several possibilities for $a_{11}$ in view of the proof of (vi).\n\n* If $a_{11} = 2$, then there are two 1's and three 2's in row 1. This implies $a_{21} = 3$, which contradicts (viii).\n\n* If $a_{11} = 3$, then there are three 1's, one 2 and one 3 in row 1. This implies $a_{21} = a_{31} = 1$ and $a_{41} = 0$. We can fill in some known entries by (v) and (ix). By (iii), row 4 cannot be a zero row. Thus, $a_{04} = 4$. Then $a_{14} = 1$ and $a_{24} = 0$ by using $c_4 = 5$. Now, as $a_{14} = 0$, we have $a_{40} = 1$.\n\nNext, note that $a_{40} = 1$ implies there is only one 4 in row 0. This shows $a_{03} \\le 3$. By (xi), we must have $a_{03} = 3$, and hence $a_{30} \\ne 0$. In view of the above solutions, as $r_4 = 1$ and $r_3 \\ge 2$, we must have $(r_0, r_2, r_3, r_4) = (11, 2, 3, 1)$. This gives $a_{30} = 2$. So we need two 3's in row 0, implying $a_{02} = 3$.\n\nAs $r_2 = 2$, there are two 1's in row 2. Also, there is one 1 in row 3. Thus, we have $a_{12} = 2$ and $a_{13} = 1$, and hence $a_{10} = 1$. Finally, using $c_j = 5$, we easily obtain $a_{20} = a_{22} = 0$ and $a_{23} = 1$. It is routine to check that this table satisfies all requirements.\n\n* If $a_{11} = 4$, then $a_{1j} = 1$ for $j \\ne 1$. This implies $a_{21} = a_{31} = 0$ and $a_{41} = 1$. As $a_{13} = 1$, there is one 1 in row 3, which implies $a_{30} = 1$. Thus, there are four 0's in row 3, yielding $a_{03} = 4$. Then by $c_3 = 5$, we obtain $a_{23} = 0$.\n\nNote that there are already two 4's in the table. This shows $r_4 \\ge 2$. In view of the solutions, either $r_4 = 2$ or 3. But then if $r_4 = 2$, we must have $a_{40} = 1$. This contradicts $a_{14} = 1$, as there are now two 1's in row 4. Therefore, we must have $r_4 = 3$. This implies $a_{40} = 2$. It follows that $a_{04} = 3$ and $a_{24} = 1$ by counting the numbers of 0's and 2's in row 4. Since we have shown there are three 4's in the table (because $r_4 = 3$), it must be the case $(r_0, r_2, r_3, r_4) = (12, 1, 1, 3)$. Now it is easy to work out $a_{02} = 4$ and $a_{20} = a_{22} = 0$. It is routine to check that this table satisfies all requirements.\n\nTo summarize, there are only two $5 \\times 5$ tables satisfying all requirements, which are given at the beginning.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31632,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the maximum possible value of real number $k$, for which the inequality\n$$\n\\frac{a}{1 + 9bc + k(b - c)^2} + \\frac{b}{1 + 9ca + k(c - a)^2} + \\frac{c}{1 + 9ab + k(a - b)^2} \\ge \\frac{1}{2}\n$$\nis satisfied for every choice of nonnegative real numbers $a, b, c$ satisfying $a + b + c = 1$.",
    "options": [],
    "answer": "4",
    "solution": "The maximum possible value of $k$ is $4$.\n\nFirstly, consider $a = 0$ and $b = c = \\frac{1}{2}$. The inequality becomes\n$$\n0 + \\frac{2}{4+k} + \\frac{2}{4+k} \\ge \\frac{1}{2}.\n$$\nHence, we need $k \\le 4$.\n\nIt remains to prove\n$$\n\\frac{a}{1 + 9bc + 4(b - c)^2} + \\frac{b}{1 + 9ca + 4(c - a)^2} + \\frac{c}{1 + 9ab + 4(a - b)^2} \\ge \\frac{1}{2}\n$$\nfor all nonnegative $a, b, c$ with $a + b + c = 1$.\n\nBy the Cauchy-Schwarz inequality, we have\n$$\n\\left( \\sum_{\\text{cyc}} a(1 + 9bc + 4(b-c)^2) \\right) \\left( \\sum_{\\text{cyc}} \\frac{a}{1 + 9bc + 4(b-c)^2} \\right) \\ge (a+b+c)^2 = 1.\n$$\nSo,\n$$\n\\sum_{\\text{cyc}} \\frac{a}{1 + 9bc + 4(b-c)^2} \\ge \\frac{1}{\\sum_{\\text{cyc}} a(1 + 9bc + 4(b-c)^2)}.\n$$\nTherefore, it suffices to prove\n$$\n\\sum_{\\text{cyc}} a(1 + 9bc + 4(b-c)^2) \\le 2.\n$$\nExpanding,\n$$\n\\sum_{\\text{cyc}} a(1 + 9bc + 4(b-c)^2) = \\sum_{\\text{cyc}} a + 9\\sum_{\\text{cyc}} abc + 4\\sum_{\\text{cyc}} a(b-c)^2.\n$$\nSince $a + b + c = 1$, $\\sum_{\\text{cyc}} a = 1$.\n\nNow,\n$$\n\\sum_{\\text{cyc}} a(b-c)^2 = \\sum_{\\text{cyc}} a(b^2 - 2bc + c^2) = \\sum_{\\text{cyc}} ab^2 + ac^2 - 2abc.\n$$\nBut $\\sum_{\\text{cyc}} ab^2 + ac^2 = \\sum_{\\text{sym}} a^2b$.\n\nSo,\n$$\n\\sum_{\\text{cyc}} a(1 + 9bc + 4(b-c)^2) = 1 + 9abc + 4\\sum_{\\text{sym}} a^2b - 8abc.\n$$\nThus,\n$$\n1 + 9abc + 4\\sum_{\\text{sym}} a^2b - 8abc = 1 + abc + 4\\sum_{\\text{sym}} a^2b.\n$$\nTherefore, we need\n$$\n1 + abc + 4\\sum_{\\text{sym}} a^2b \\le 2.\n$$\nOr,\n$$\nabc + 4\\sum_{\\text{sym}} a^2b \\le 1.\n$$\nBut $\\sum_{\\text{sym}} a^2b = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b$.\n\nNow, since $a + b + c = 1$, we can homogenize the expression:\n$$\n(a + b + c)^3 \\ge 3abc + 4(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b).\n$$\nExpanding $(a + b + c)^3$:\n$$\n(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a^2b + a^2c + b^2a + b^2c + c^2a + c^2b) + 6abc.\n$$\nSo,\n$$\na^3 + b^3 + c^3 + 3abc \\ge a^2b + a^2c + b^2a + b^2c + c^2a + c^2b.\n$$\nThis is exactly Schur's inequality. So the inequality is proved.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31633,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle ABC > \\angle BCA$ and $\\angle BCA \\ge 30^\\circ$. The angle bisectors of $\\angle ABC$ and $\\angle BCA$ meet the opposite sides of the triangle at the points $D$ and $E$, respectively. The line $BD$ intersects the line $CE$ at $P$. Assume that $PD = PE$ and that the incircle of the triangle $ABC$ has radius 1. Determine the largest possible length of $BC$.",
    "options": [],
    "answer": "3 + sqrt(3)",
    "solution": "The largest possible value of $BC$ is $3 + \\sqrt{3}$.\nSince $PD = PE$ and $AP$ bisects $\\angle EAD$, the quadrilateral $AEPD$ is cyclic or it is a kite. The latter case is impossible since $\\angle B > \\angle C$. Thus, we have\n$$\n180^{\\circ} = \\angle EAD + \\angle DPE = A + \\left(90^{\\circ} + \\frac{A}{2}\\right) = 90^{\\circ} + \\frac{3A}{2}.\n$$\nThis implies $A = 60^{\\circ}$.\n\n![](attached_image_1.png)\n\nAs usual, let *a*, *b*, *c* and *s* be the lengths of $BC$, $CA$, $AB$ and the semiperimeter respectively. Firstly, since the inradius is 1, we have\n$$\ns - a = \\cot \\frac{A}{2} = \\sqrt{3}.\n$$\n\nBy the sine law, we have $\\frac{a}{\\sin 60^\\circ} = \\frac{b}{\\sin B} = \\frac{c}{\\sin C}$, and so\n$$\n\\frac{2a}{\\sqrt{3}} = \\frac{b+c-a}{\\sin B + \\sin C - \\sin 60^\\circ}.\n$$\nNote that\n$$\n\\sin B + \\sin C = 2 \\sin \\frac{B+C}{2} \\cos \\frac{B-C}{2} = \\sqrt{3} \\cos (60^\\circ - C).\n$$\nTherefore, we have\n$$\na = \\frac{\\sqrt{3}(s-a)}{\\sin B + \\sin C - \\sin 60^\\circ} = \\frac{2\\sqrt{3}}{2\\cos(60^\\circ - C) - 1} \\le \\frac{2\\sqrt{3}}{2\\cos(60^\\circ - 30^\\circ) - 1} = 3 + \\sqrt{3}.\n$$\nEquality holds when $C = 30^\\circ$ and $B = 90^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31634,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an equilateral triangle, and let $D$ be a point on $AB$ between $A$ and $B$. Next, let $E$ be a point on $AC$ with $DE$ parallel to $BC$. Further, let $F$ be the midpoint of $CD$ and $G$ the circumcentre of triangle $ADE$. Determine the interior angles of triangle $BFG$.",
    "options": [],
    "answer": "angle GFB = 90 degrees, angle FBG = 30 degrees, angle BGF = 60 degrees",
    "solution": "$\\angle GFB = 90^\\circ$, $\\angle FBG = 30^\\circ$ and $\\angle BGF = 60^\\circ$.\nLet $H$ be the point such that $BCHD$ is a parallelogram. Then $E$ is contained in the segment $DH$, and $\\triangle ECH$ is equilateral. Thus, we have $BD = CH = HE$. Moreover, it is clear that $DG = EG$ and $\\angle BDG = 150^\\circ = \\angle HEG$. Therefore $\\triangle BDG \\cong \\triangle HEG$, and hence $\\triangle BGH \\sim \\triangle DGE$ by spiral similarity.\n\nNow, since $F$ is the midpoint of $CD$, it is also the midpoint of $BH$. Since $GB = GH$, we have $\\angle GFB = 90^\\circ$. Also, we have\n$$\n\\angle FBG = \\angle HBG = \\angle EDG = 30^\\circ.\n$$\nThus, the remaining angle is $\\angle BGF = 60^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31635,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Let $D$ and $E$ be respectively points on the segments $AB$ and $AC$, and such that $DE \\parallel BC$. Let $M$ be the midpoint of $BC$. Let $P$ be a point such that $DB = DP$, $EC = EP$ and such that the open segments (segments excluding the endpoints) $AP$ and $BC$ intersect. Suppose $\\angle BPD = \\angle CME$. Show that $\\angle CPE = \\angle BMD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $\\angle BPD = \\angle CME = \\angle DEM$. So we can find a point $F$ on the extension of $EM$ such that $\\triangle DPB \\sim \\triangle DEF$. By spiral similarity and since\n\n$DB = DP$, we have $\\triangle DFB \\cong \\triangle DEP$. As $EC = EP = FB$ and $CM = BM$, by the sine law, either $\\angle FEC = \\angle EFB$ or $\\angle FEC + \\angle EFB = 180^\\circ$.\nIf $\\angle FEC + \\angle EFB = 180^\\circ$, then\n$$\n\\begin{align*} \n\\angle AED &= 180^\\circ - \\angle FEC - \\angle DEM \\\\ \n&= \\angle EFB - \\angle DEM \\\\ \n&= \\angle DFB + \\angle EFD - \\angle DEM \\\\ \n&= \\angle DEP. \n\\end{align*}\n$$\nTogether with $\\angle AEP = 2\\angle ECP$, we easily find that $DE \\parallel PC$, contradicting the fact that $P$ does not lie on $BC$.\n\n![](attached_image_1.png)\n\nHence, we must have $\\angle FEC = \\angle EFB$. This yields $EC \\parallel BF$. Therefore, we find that $EM = FM$. Since $DE = DF$, we have $DM \\perp EF$. Now, by swapping the roles of $(B, D)$ and $(C, E)$, and reversing the above arguments, we see that $\\angle CPE = \\angle BMD$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31636,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $\\angle A = 54^\\circ$ and $\\angle C = 24^\\circ$. $P$ is a point on $AC$ such that $AP = BC$. Find $\\angle CBP$.",
    "options": [],
    "answer": "6°",
    "solution": "As shown in the figure, let $D$ be the point for which $ABPD$ is an isosceles trapezium with $AB \\parallel DP$. Let also $E$ be the point for which $APED$ is a parallelogram, and $F$ be the point such that $A$ and $F$ lie on different sides of $BC$ and for which $CBF$ is an equilateral triangle.\n\nNote that by our construction and the given condition $AP = BC$, we have\n\n$DE = BD = BF$ as each of these three segments has the same length as $AP$.\n\nWe have $\\angle CBD = \\angle CBA - \\angle DBA = (180^\\circ - 54^\\circ - 24^\\circ) - 54^\\circ = 48^\\circ$. Therefore, we get $\\angle BDE = \\angle BDP + \\angle PDE = \\angle BDP + \\angle DPA = 54^\\circ + 54^\\circ = 108^\\circ$ and $\\angle FBD = \\angle FBC + \\angle CBD = 60^\\circ + 48^\\circ = 108^\\circ$.\n\nThese show that $E, D, B, F$ are consecutive vertices of a regular pentagon. It follows that $C$ and $E$ both lie on the perpendicular bisector of $BF$.\n\n![](attached_image_1.png)\n\nThus we have $\\angle DEC = \\frac{108^\\circ}{2} = 54^\\circ = \\angle PDE$. Together with $PC \\parallel DE$, we see that $PCED$ is an isosceles trapezium. In particular, we have $CD = PE = AD$. Hence we have\n$$\n\\angle PAD = \\angle DCA = \\angle DCB - \\angle ACB = \\frac{180^\\circ - 48^\\circ}{2} - 24^\\circ = 42^\\circ\n$$\nand so $\\angle CBP = \\angle CBA - \\angle PBD - \\angle DBA = 102^\\circ - \\angle PAD - 54^\\circ = 6^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31637,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $11 \\times 11$ grid is to be covered completely without overlapping by some $2 \\times 2$ squares and $L$-shapes each composed of three unit cells. Determine the smallest number of $L$-shapes used. (Each shape must cover some grids entirely and cannot be placed outside the $11 \\times 11$ grid. The $L$-shapes may be reflected or rotated when placed on the grid.)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "23",
    "solution": "At least 23 $L$-shapes are needed.\nSuppose $x$ squares and $y$ $L$-shapes are used. Then the total number of cells is $4x + 3y$, which should be equal to $11^2 = 121$.\n\n![](attached_image_2.png)\n\n![](attached_image_3.png)\n\nWe colour the cells as shown. Then every square and every $L$-shape covers at most one blackened cell. As there are 36 blackened cells, we have $x + y \\ge 36$.\n\nTherefore, $y = 4(x + y) - (4x + 3y) \\geq 4(36) - 121 = 23$. The figure on the right gives one example using 23 L-shapes.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31638,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. There are $n$ distinct circles on the plane such that any two circles have two distinct intersections and no three circles have a common intersection. Initially there is a coin on each of the intersection points of the circles. Starting from $X$, players $X$ and $Y$ alternately take away a coin, with the restriction that one cannot take away a coin lying on the same circle as the last coin just taken away by the opponent in the previous step. The one who cannot do so will lose. In particular, one loses when there is no coin left. For what values of $n$ does $Y$ has a winning strategy?",
    "options": [],
    "answer": "n ≥ 4",
    "solution": "$Y$ has a winning strategy if and only if $n \\ge 4$.\nFor $n = 2, 3$, after $X$ takes away any coin, $Y$ cannot take away any other coin. Therefore, $Y$ does not have a winning strategy.\nFor $n \\ge 4$, we claim that $Y$ has a winning strategy. Since the total number $2\\binom{n}{2}$ of coins is even, it suffices to partition all the coins into pairs such that the two coins in the same pair do not lie on the same circle. Once this is done, the strategy of $Y$ is to take away the coin in the same pair as the coin just taken away by $X$. Clearly, $Y$ can win as there is always a move, and the number of coins is finite.\nWe label the circles from $1$ to $n$, and let $A_{ij} = A_{ji}$ and $B_{ij} = B_{ji}$ be the intersection points of circle $i$ and circle $j$. For any $j$, we pair up $A_{j(j+1)}$ with $B_{(j+2)(j+3)}$, where all indices are taken modulo $n$. Since $n \\ge 4$, these two points do not lie on the same circle. Next, we pair up any other point $A_{ij}$ with $B_{(i+1)(j+1)}$. Note that $i \\ne j+1 \\pmod n$ and $j \\ne i+1 \\pmod n$ by assumption. Therefore, these two points do not lie on the same circle. This gives a desired pairing, and so we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31639,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the remainder when $19^{17^{15}}$ is divided by 100.",
    "options": [],
    "answer": "59",
    "solution": "For convenience we define $f(1) = 1$ and $f(n + 2) = (n + 2)^{f(n)}$ for odd positive integers $n$. Then the question asks for the last two digits of $f(19)$.\n\nSince $f(13) = 13^{f(11)}$ is odd, we have $f(15) = 15^{f(13)} \\equiv (-1)^{f(13)} = -1 \\equiv 3 \\pmod 4$.\n\nAs $f(17) = 17^{f(15)}$ and the units digits of the powers of 7 (also the powers of 17) follows the pattern 7, 9, 3, 1, 7, 9, 3, 1, which repeats itself every four terms, we conclude that the units digit of $f(17)$ is the same as that of $17^3$, which is 3.\n\nFinally, if we look at the last two digits of the powers of 19, we will see the pattern 19, 61, 59, 21, 99, 81, 39, 41, 79, 01, 19, 61, which repeats itself every 10 terms. As $f(17)$ has units digit 3, the last two digits of $f(19) = 19^{f(17)}$ are the same as those of $19^3$, which are 59.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31640,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_n = \\underbrace{677\\cdots7}_{n}$. Is it possible to find infinitely many multiples of $a_{2014}$ in the sequence $\\{a_n\\}$?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes. By the pigeonhole principle, two of the numbers of the form $11\\cdots1$ leave the same remainder when divided by $a_{2014}$. Their difference, which is of the form $11\\cdots100\\cdots0$, is a multiple of $a_{2014}$. Since $(a_{2014}, 10) = 1$, we know that $a_{2014}$ divides $\\underbrace{11\\cdots1}_{m \\text{ times}}$ for some $m \\in \\mathbb{Z}^+$. It is then obvious that $a_{2014} \\mid a_{2014+km}$ for any $k \\in \\mathbb{Z}^+$, since\n$$\na_{2014+km} = a_{2014} \\times 10^{km} + \\underbrace{77\\cdots7}_{km \\text{ times}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31641,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a trapezium with $AB \\parallel CD$, $AB = 42$, $BC = 20$ and $DA = 15$. $P$ is a point on $AB$, and a circle with centre $P$ is tangent to both $BC$ and $AD$. Find $PA \\times PB$.",
    "options": [],
    "answer": "432",
    "solution": "Let $r$ be the radius of the circle, and $E$ and $F$ be the points where the circle touches $AD$ and $BC$ respectively. Then $PE = PF = r$ and they are the heights of $\\triangle PAD$ and $\\triangle PBC$ from $P$.\nLet $h$ be the height of the trapezium. By considering the areas of $\\triangle PAD$ and $\\triangle PBC$, we get $\\frac{PA \\cdot h}{2} = \\frac{15r}{2}$ and $\\frac{PB \\cdot h}{2} = \\frac{20r}{2}$, and hence $\\frac{PA}{PB} = \\frac{3}{4}$. Together with $AB = 42$, we have $PA = 18$ and $PB = 24$. It follows that $PA \\times PB = 432$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31642,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 100 squares in a $10 \\times 10$ grid table. Each square is painted in one colour. What is the maximum number of colours that may be used, so that in each row, and also in each column, the number of different colours used does not exceed five?",
    "options": [],
    "answer": "41",
    "solution": "The maximum number of colours is 41.\n\nAn example of 41 colours is given below, where the numbers represent different colours.\n\n| 1 | 1 | 1 | 1 | 1 | 1 | 2 | 3 | 4 | 5 |\n|---|---|---|---|---|---|---|---|---|---|\n| 1 | 1 | 1 | 1 | 1 | 6 | 7 | 8 | 9 | 1 |\n| 1 | 1 | 1 | 1 | 10 | 11 | 12 | 13 | 1 | 1 |\n| 1 | 1 | 1 | 14 | 15 | 16 | 17 | 1 | 1 | 1 |\n| 1 | 1 | 18 | 19 | 20 | 21 | 1 | 1 | 1 | 1 |\n| 1 | 22 | 23 | 24 | 25 | 1 | 1 | 1 | 1 | 1 |\n| 26 | 27 | 28 | 29 | 1 | 1 | 1 | 1 | 1 | 1 |\n| 31 | 32 | 33 | 1 | 1 | 1 | 1 | 1 | 1 | 30 |\n| 36 | 37 | 1 | 1 | 1 | 1 | 1 | 1 | 34 | 35 |\n| 41 | 1 | 1 | 1 | 1 | 1 | 1 | 38 | 39 | 40 |\n\nNow, assume at least 42 colours are used. By the pigeonhole principle, WLOG we may assume the first row contains $\\lfloor \\frac{42}{10} \\rfloor = 5$ different colours. We need to use at least 37 more colours. Again by the pigeonhole principle, we may assume the second row contains $\\left\\lfloor \\frac{37}{9} \\right\\rfloor = 5$ different colours. Now, each column already has two different colours, and hence there can be at most 3 more new colours in each column. This gives a total of at most $5 + 5 + 3 \\times 10 = 40$ colours, which is a contradiction. Therefore, at most 41 colours can be used.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31643,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be integers. If $a + b$ is a root of the equation $x^2 + a x + b = 0$, find the smallest possible value of $a b$.",
    "options": [],
    "answer": "-54",
    "solution": "Since $a + b$ satisfies the given equation, we have $(a + b)^2 + a(a + b) + b = 0$. Rearranging gives $b^2 + (3a + 1)b + 2a^2 = 0$. This means $b$ is a root of the equation $x^2 + (3a + 1)x + 2a^2 = 0$. As $a$ and $b$ are integers, the discriminant $(3a + 1)^2 - 4 \\cdot 2a^2 = a^2 + 6a + 1$ must be a perfect square.\nLet $a^2 + 6a + 1 = m^2$. Then $(a + 3)^2 - 8 = m^2$, and so $(a + 3 - m)(a + 3 + m) = 8$. As the two terms $a + 3 - m$ and $a + 3 + m$ have the same parity (they differ by $2m$ which is even), they can only be $2$ and $4$, or $-2$ and $-4$ (up to permutation). The corresponding possible values of $a$ are $0$ and $-6$.\nWhen $a = 0$, the equation becomes $b^2 + b = 0$, giving $b = 0$ or $b = -1$. When $a = -6$, the equation becomes $b^2 - 17b + 72 = 0$, giving $b = 8$ or $b = 9$. The smallest possible value of $a b$ is thus $(-6) \\times 9 = -54$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31644,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABC$ is a triangle with integral sides. $M$ is the midpoint of $BC$. The in-circle with centre $I$ touches $AB$ and $AC$ at $E$ and $F$ respectively and $D$ is the projection of $M$ on $EF$. Suppose that $ADMI$ is a parallelogram and $AB + BC + CA = 65$. Find $AB \\times BC \\times CA$.",
    "options": [],
    "answer": "9360",
    "solution": "The product is $9360$.\nLet $a$, $b$, $c$, $r$, $s$ be the lengths of $BC$, $CA$, $AB$, the inradius and the semiperimeter of $\\triangle ABC$ respectively. Let $N$ be the midpoint of $AC$, and let $P$ be the intersection point of $EF$ and $MN$. It is well-known that $\\angle BPC = 90^\\circ$. Therefore, we have $MP = \\frac{a}{2}$. (Alternatively, one can prove this by noting $MN = \\frac{c}{2}$ and $PN = NF = \\frac{c-a}{2}$.)\n\n![](attached_image_1.png)\n\nNote that $NM \\parallel AB$ by the midpoint theorem. Since $ADMI$ is a parallelogram, we have\n$$\nAI = DM = PM \\sin \\angle DPM = \\frac{a}{2} \\sin \\angle FEA = \\frac{a}{2} \\cos \\frac{A}{2}.\n$$\nOn the other hand, we have\n$$\nAI = \\frac{r}{\\sin \\frac{A}{2}} = \\frac{bc \\sin A}{2s \\sin \\frac{A}{2}} = \\frac{bc}{s} \\cos \\frac{A}{2}.\n$$\nEquating these, we obtain $2bc = as$. As it is given that $2s = 65$, this yields $4bc = 65a$. Then the given condition becomes $\\frac{4bc}{65} + b + c = 65$. This can be factorized as\n$$\n(4b + 65)(4c + 65) = 5 \\cdot 65^2 = 5^3 \\cdot 13^2.\n$$\nAs $4b + 65$ and $4c + 65$ are larger than $65$, we must have $4b + 65 = 125$ and $4c + 65 = 169$ up to permutation. This gives $(a, b, c) = (24, 15, 26)$ or $(24, 26, 15)$. Therefore, $abc = 24 \\times 15 \\times 26 = 9360$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31645,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = \\frac{15}{x+1} + \\frac{16}{x^2+1} - \\frac{17}{x^3+1}$. Find the value of\n$f(\\tan 15°) + f(\\tan 30°) + f(\\tan 45°) + f(\\tan 60°) + f(\\tan 75°)$.",
    "options": [],
    "answer": "35",
    "solution": "Note that $f\\left(\\frac{1}{x}\\right) = \\frac{15}{\\frac{1}{x}+1} + \\frac{16}{\\frac{1}{x^2}+1} - \\frac{17}{\\frac{1}{x^3}+1} = \\frac{15x}{x+1} + \\frac{16x^2}{x^2+1} - \\frac{17x^3}{x^3+1}$.\nHence we have\n$$\n\\begin{aligned}\nf(x) + f\\left(\\frac{1}{x}\\right) &= 15 \\left( \\frac{1}{x+1} + \\frac{x}{x+1} \\right) + 16 \\left( \\frac{1}{x^2+1} + \\frac{x^2}{x^2+1} \\right) \\\\\n&\\quad - 17 \\left( \\frac{1}{x^3+1} + \\frac{x^3}{x^3+1} \\right) \\\\\n&= 15 + 16 - 17 = 14.\n\\end{aligned}\n$$\nSince $\\tan 15° = \\frac{1}{\\tan 75°}$, $\\tan 30° = \\frac{1}{\\tan 60°}$ and $\\tan 45° = 1$, it follows that the value of the expression in the question is $14+14+f(1) = 14+14+\\frac{15}{2}+\\frac{16}{2}-\\frac{17}{2} = 35$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31646,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a trapezium with $AB \\parallel CD$. $M$ and $N$ are the midpoints of $AB$ and $CD$ respectively. If $AC = 6$, $BD = 8$ and $MN = 4$, find the area of $ABCD$.",
    "options": [],
    "answer": "3√55",
    "solution": "We use $[X_1X_2\\cdots X_k]$ to denote the area of the $k$-sided polygon $X_1X_2\\cdots X_k$. Let $K$ and $L$ be the midpoints of $AD$ and $BC$ respectively. Then $\\triangle DKN \\sim \\triangle DAC$ with side-length ratio $1:2$ and so $[DKN] = \\frac{1}{4}[DAC]$. Similarly, we have $[AMK] = \\frac{1}{4}[ABD]$, $[BLM] = \\frac{1}{4}[BCA]$ and $[CNL] = \\frac{1}{4}[CDB]$.\n\nSumming up these equations, we get\n$$\n[DKN] + [AMK] + [BLM] + [CNL] = \\frac{1}{2}[ABCD].\n$$\nThis shows $[ABCD] = 2[KMLN] = 4[NML]$. By the midpoint theorem, we have $NL = \\frac{1}{2}BD = 4 = NM$ and $ML = \\frac{1}{2}AC = 3$. Hence the height of $\\triangle NML$ from $N$ has length $\\sqrt{4^2 - \\left(\\frac{3}{2}\\right)^2} = \\frac{\\sqrt{55}}{2}$.\n\n![](attached_image_1.png)\n\nThus $[NML] = \\frac{3\\sqrt{55}}{4}$ and so $[ABCD] = 3\\sqrt{55}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31647,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be an integer. If the equation $kx^2 + (4k-2)x + (4k-7) = 0$ has an integral root, find the sum of all possible values of $k$.",
    "options": [],
    "answer": "6",
    "solution": "Let $m$ be an integral root of the equation. Then we have $km^2 + (4k - 2)m + (4k - 7) = 0$, which can be rewritten as $(m^2 + 4m + 4)k = 2m + 7$, or $(m + 2)^2k = 2m + 7$. Hence $m + 2$ divides $2m + 7$, and so $m + 2$ divides $2m + 7 - 2(m + 2) = 3$ as well. Thus $m$ can only be $-5, -3, -1$ or $1$, which we plug in to $(m + 2)^2k = 2m + 7$ one by one:\n* When $m = -5$, we get $9k = -3$ which gives no integral solution for $k$.\n* When $m = -3$, we get $k = 1$.\n* When $m = -1$, we get $k = 5$.\n* When $m = 1$, we get $9k = 9$ which again gives $k = 1$.\nTherefore, the sum of all possible values of $k$ is $1 + 5 = 6$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31648,
    "subject": "Mathematics (Multi-modal)",
    "question": "$A$, $B$ and $C$ are three persons among a set $P$ of $n$ ($n \\ge 3$) persons. It is known that $A$, $B$ and $C$ are friends of one another, and that every one of the three persons has already made friends with more than half the total number of people in $P$. Given that every three persons who are friends of one another form a *friendly group*, what is the minimum number of friendly groups that may exist in $P$?",
    "options": [],
    "answer": "If n is odd: (n−1)/2. If n is even and at least 6: (n+2)/2. If n = 4: 4.",
    "solution": "The answer is $\\frac{n-1}{2}$ if $n$ is odd, $\\frac{n+2}{2}$ if $n \\ge 6$ is even, and $4$ if $n = 4$.\n\nLet $A$, $B$, $C$ be the sets of friends of $A$, $B$, $C$ respectively excluding $A$, $B$, $C$ themselves.\n\n**Case 1.** $n$ is odd\n\nSince each of $A$, $B$, $C$ has at least $\\frac{n+1}{2}$ friends, we have $|A|, |B|, |C| \\ge \\frac{n-3}{2}$. For any person $X$ other than $A$, $B$, $C$, the set $\\{A, B, X\\}$ is a friendly group if and only if $X \\in (A \\cap B)$. Similarly, $|B \\cap C|$ and $|C \\cap A|$ count the number of friendly groups of the form $\\{B, C, X\\}$ and $\\{C, A, X\\}$ respectively. Together with the friendly group $\\{A, B, C\\}$, by the inclusion-exclusion principle, the number of friendly groups is at least\n\n$$\n|A \\cap B| + |B \\cap C| + |C \\cap A| + 1 = |A| + |B| + |C| + |A \\cap B \\cap C| - |A \\cup B \\cup C| + 1 \\ge \\frac{n-3}{2} \\times 3 + 0 - (n-3) + 1 = \\frac{n-1}{2}.\n$$\n\nThe minimum value $\\frac{n-1}{2}$ can be attained. For example, suppose $\\frac{n-3}{2}$ people are friends of $A$ and $B$, and the other $\\frac{n-3}{2}$ people are friends of $C$. Then $\\{A, B, C\\}$ is a friendly group and all other $\\frac{n-3}{2}$ friendly groups are of the form $\\{A, B, X\\}$.\n\n**Case 2.** $n$ is even and $n \\ge 6$\n\nSimilarly, we have $|A|, |B|, |C| \\ge \\frac{n-2}{2}$. The number of friendly groups is at least\n$$\n|A \\cap B| + |B \\cap C| + |C \\cap A| + 1 = |A| + |B| + |C| + |A \\cap B \\cap C| - |A \\cup B \\cup C| + 1 \\ge \\frac{n-2}{2} \\times 3 + 0 - (n-3) + 1 = \\frac{n+2}{2}.\n$$\n\nThe minimum value $\\frac{n+2}{2}$ can be attained. For example, suppose $\\frac{n-4}{2}$ people are friends of $A$ and $B$, one person is a friend of $B$ and $C$, one person is a friend of $C$ and $A$, and the other $\\frac{n-6}{2}$ people are friends of $C$. Then $\\{A, B, C\\}$ is a friendly group. In addition, there are $\\frac{n-4}{2}$ friendly groups of the form $\\{A, B, X\\}$, one friendly group of the form $\\{B, C, Y\\}$, and one friendly group of the form $\\{C, A, Z\\}$.\n\n**Case 3.** $n = 4$\n\nSince each of $A$, $B$, $C$ has more than $2$ friends, every pair of the four people must be friends of each other. Clearly, there are $4$ friendly groups.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31649,
    "subject": "Mathematics (Multi-modal)",
    "question": "During a school year $44$ competitions were held. Exactly $7$ students won in each of the competitions. For any two competitions, there exists exactly $1$ student who won in both competitions. Is it true that there exists a student who won all of the competitions?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes. Consider any competition $C_1$. Since every other competition shares a common winner with $C_1$, there must be a winner of $C_1$ who won at least $\\lfloor \\frac{43}{7} \\rfloor = 6$ more competitions by the pigeonhole principle. WLOG assume $A$ won the competitions $C_1, C_2, \\dots, C_8$.\n\nConsider any other competition $C_k$. By the pigeonhole principle, one of the winners of $C_k$ won at least $\\lfloor \\frac{8}{7} \\rfloor = 1$ competitions among $C_1, C_2, \\dots, C_8$. But since the common winner of these competitions is $A$, we know that $A$ won $C_k$ as well. Thus, $A$ won all the competitions.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31650,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all natural numbers $n$ such that $n$, $n^2 + 10$, $n^2 - 2$, $n^3 + 6$ and $n^5 + 36$ are all prime numbers.",
    "options": [],
    "answer": "7",
    "solution": "The only solution is $n = 7$.\nObserve that\n$$\n\\begin{aligned}\nn &\\equiv n \\quad (\\text{mod } 7), \\\\\nn^2 + 10 &\\equiv (n + 2)(n - 2) \\quad (\\text{mod } 7), \\\\\nn^2 - 2 &\\equiv (n + 3)(n - 3) \\quad (\\text{mod } 7), \\\\\nn^3 + 6 &\\equiv (n - 1)(n^2 + n + 1) \\quad (\\text{mod } 7), \\\\\nn^5 + 36 &\\equiv (n + 1)(n^4 - n^3 + n^2 - n + 1) \\quad (\\text{mod } 7).\n\\end{aligned}\n$$\nSince $n-3$, $n-2$, $n-1$, $n$, $n+1$, $n+2$ and $n+3$ are 7 consecutive integers, one of them must be a multiple of 7. It follows that one of the numbers is equal to 7, or otherwise it cannot be a prime. Clearly, $n^2 + 10 > 7$ and $n^5 + 36 > 7$.\nIf $n=7$, the 5 numbers are 7, 59, 47, 349, 16843, all of which are primes.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31651,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre of a triangle $ABC$, and let $\\ell$ be the line going through the midpoint of the side $BC$ and is perpendicular to the bisector of $\\angle BAC$. Determine the value of $\\angle BAC$ if the line $\\ell$ goes through the midpoint of the line segment $AO$.",
    "options": [],
    "answer": "120°",
    "solution": "We have $\\angle BAC = 120^\\circ$.\nLet $M$ be the midpoint of $BC$. Suppose $l$ meets the altitude from $A$, the angle bisector of $\\angle BAC$, and the line $AO$ at $E$, $P$, $N$ respectively.\nNote that $AE // MO$. In order that $EM$ bisects $AO$, the quadrilateral $AEOM$ must be convex. Therefore, $\\angle BAC$ must be obtuse.\nNow, since $AE // MO$ and $AN = ON$, we have $\\triangle AEN \\cong \\triangle OMN$. It is well-known that $AE$ and $AO$ are isogonal lines with respect to $\\angle BAC$. Therefore, $AP$ bisects $\\angle EAN$. As we also have $AP \\perp EN$, this shows $\\triangle AEN$ is isosceles, with $AE = AN$. Therefore,\n$$\nOM = AE = AN = \\frac{1}{2} OA = \\frac{1}{2} OB.\n$$\nAs $\\angle BMO = 90^\\circ$, this yields $\\angle OBM = 30^\\circ$. Thus, $\\angle BAC = 120^\\circ$.\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31652,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. Find the number of nonnegative integers $n$ less than or equal to $10^k$ satisfying the following conditions:\n(i) $n$ is divisible by $3$;\n(ii) Each decimal digit of $n$ is one of the digits $2$, $0$, $1$ or $7$.",
    "options": [],
    "answer": "If k is divisible by 3: (4^k + 2)/3; otherwise: (4^k - 1)/3.",
    "solution": "The answer is $\\frac{4^k - 1}{3}$ if $3 \\nmid k$, and $\\frac{4^k + 2}{3}$ if $3 \\mid k$.\n\nNote that $n$ has exactly $k$ digits, possibly starting with $0$. We need to count the number $N$ of integers of the form $\\overline{a_1a_2\\cdots a_k}$ such that each $a_j$ is one of $0$, $1$, $2$, $7$ and $3 \\mid a_1 + a_2 + \\cdots + a_k$. Consider the generating function\n\n$$\nf(x) = (1 + x + x^2 + x^7)^k = \\sum_{\\substack{a_1, \\dots, a_k \\\\ a_j \\in \\{0, 1, 2, 7\\}}} x^{a_1 + a_2 + \\cdots + a_k}.\n$$\n\nLet $c_j$ be the coefficient of $x^j$ for each $j$. Then $N = \\sum_{3 \\mid j} c_j$.\n\nLet $\\omega = \\frac{-1}{2} + \\frac{\\sqrt{3}}{2}i$ be a cube root of unity. This means a number satisfying $\\omega^3 = 1$ and $\\omega \\neq 1$. It has the property that $\\omega^{2j} + \\omega^j + 1 = 0$ if $3 \\nmid j$, and $\\omega^{2j} + \\omega^j + 1 = 3$ if $3 \\mid j$. It follows that\n\n$$\n\\begin{align*}\n3N &= \\sum_j c_j (1^j + \\omega^j + \\omega^{2j}) \\\\\n   &= f(1) + f(\\omega) + f(\\omega^2) \\\\\n   &= 4^k + (1 + \\omega + \\omega^2 + \\omega^7)^k + (1 + \\omega^2 + \\omega^4 + \\omega^{14})^k \\\\\n   &= 4^k + \\omega^k + \\omega^{2k}.\n\\end{align*}\n$$\n\nIf $3 \\nmid k$, we have $\\omega^k + \\omega^{2k} = -1$. This gives $N = \\frac{4^k - 1}{3}$.\n\nIf $3 \\mid k$, we have $\\omega^k = \\omega^{2k} = 1$. This gives $N = \\frac{4^k + 2}{3}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31653,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest possible positive integer $n$, so that there exist $n$ distinct positive real numbers $x_1, x_2, \\dots, x_n$ satisfying the following inequality: for any $1 \\le i, j \\le n$,\n$$\n(3x_i - x_j)(x_i - 3x_j) \\ge (1 - x_ix_j)^2.\n$$",
    "options": [],
    "answer": "3",
    "solution": "The largest possible $n$ is $3$.\nNote that\n$$\n\\begin{aligned}\n& (3x_i - x_j)(x_i - 3x_j) \\ge (1 - x_ix_j)^2 \\\\\n\\Leftrightarrow & 3x_i^2 - 10x_ix_j + 3x_j^2 \\ge 1 - 2x_ix_j + x_i^2x_j^2 \\\\\n\\Leftrightarrow & 3x_i^2 - 6x_ix_j + 3x_j^2 \\ge 1 + 2x_ix_j + x_i^2x_j^2 \\\\\n\\Leftrightarrow & 3(x_i - x_j)^2 \\ge (1 + x_ix_j)^2 \\\\\n\\Leftrightarrow & \\left| \\frac{x_i - x_j}{1 + x_ix_j} \\right| \\ge \\frac{1}{\\sqrt{3}}.\n\\end{aligned}\n\\quad (1)\n$$\nWLOG, we may assume $x_1 < x_2 < \\dots < x_n$. For $1 \\le j \\le n$, we can find $\\theta_j \\in \\mathbb{R}$ such that $0 < \\theta_j < \\frac{\\pi}{2}$ and $x_j = \\tan \\theta_j$. Now for $1 \\le i < j \\le n$, the inequality (1) is equivalent to\n$$\n\\tan(\\theta_j - \\theta_i) \\ge \\frac{1}{\\sqrt{3}},\n$$\nwhich means $\\theta_j - \\theta_i \\ge \\frac{\\pi}{6}$.\nIf $n \\ge 4$, by the pigeonhole principle, there exists $j$ with $1 \\le j \\le 3$ such that $\\theta_{j+1} - \\theta_j < \\frac{\\pi}{6}$. This contradicts (1). Therefore, we must have $n \\le 3$.\n\nIt is possible to have $n = 3$. For example, we can take $x_j = \\tan \\frac{(2j-1)\\pi}{12}$ for $j = 1, 2, 3$. Following the above arguments, we see that all inequalities are satisfied.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31654,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma$ be a circle and $AB$ be a diameter. Let $\\ell$ be a line outside the circle, and is perpendicular to $AB$. Let $X, Y$ be two points on $\\ell$. If $X'$ and $Y'$ are two points on $\\ell$ such that $AX$ and $BX'$ intersect on $\\Gamma$ and such that $AY$ and $BY'$ intersect on $\\Gamma$, prove that the circumcircles of the triangles $AXY$ and $AX'Y'$ intersect at a point on $\\Gamma$ other than $A$, or the three circles are tangent at $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AX$ meet $\\Gamma$ again at $P$, and let $AY$ meet $\\Gamma$ again at $Q$. If $PQ \\parallel \\ell$, the figure is symmetric with respect to $AB$, and so $(AXY)$ and $(AX'Y')$ are tangent at $A$. In the following, we only consider the configuration as shown.\n\nFirstly, since\n$$\n\\angle AQP = \\angle ABP = 90^\\circ - \\angle PAB = \\angle YXP,\n$$\nthe points $Q, P, X, Y$ are concyclic. Similarly, $P, Q, X', Y'$ are concyclic.\n\nNow, let $PQ$ meet $\\ell$ at $C$. Then we have\n$$\nCX \\times CY = CP \\times CQ = CX' \\times CY'.\n$$\nHence, $C$ has the same power with respect to $(AXY)$, $(AX'Y')$ and $\\Gamma$. As all three circles pass through $A$, the line $AC$ is the common radical axis of these circles. Thus, the circles are coaxial as desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31655,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $n$ for which there exist pairwise distinct positive real numbers $a_1, a_2, \\dots, a_n$ satisfying $\\{a_i + \\frac{(-1)^i}{a_i} \\mid 1 \\le i \\le n\\} = \\{a_i \\mid 1 \\le i \\le n\\}$.",
    "options": [],
    "answer": "all odd integers greater than 1",
    "solution": "$n$ can be any odd integer greater than $1$.\n\nFor even $n$, suppose $a_j + \\frac{(-1)^j}{a_j} = b_j$ for $1 \\le j \\le n$, where $\\{b_j\\}$ is a permutation of $\\{a_j\\}$. The relation can be rewritten as\n$$\na_j^2 - a_j b_j = (-1)^{j+1}.\n$$\nSumming over all $j$'s, we obtain\n$$\n\\sum_{j=1}^n (a_j^2 - a_j b_j) = 0\n$$\nsince $n$ is even. As $\\{b_j\\}$ is a permutation of $\\{a_j\\}$, this implies\n$$\n\\frac{1}{2} \\sum_{j=1}^n (a_j - b_j)^2 = 0.\n$$\nHence, we must have $a_j = b_j$, which is impossible since $a_j + \\frac{(-1)^j}{a_j} = b_j$. Therefore, there is no such sequence.\n\nFor $n=1$, we need $a_1 - \\frac{1}{a_1} = a_1$, which is impossible.\n\nSo it remains to show that such a sequence exists when $n > 1$ is odd. First we construct a sequence $a_1, a_2, \\dots, a_{n+1}$ such that $a_1 = 1 + x$ with $x > 0$, and\n$$\na_{j+1} = a_j + \\frac{(-1)^j}{a_j}\n$$\nfor $1 \\le j \\le n$. It suffices to show $a_1, a_2, \\dots, a_n$ are pairwise distinct, and choose $x$ such that $a_{n+1} = a_1$.\n\nNote that\n$$\na_{j+2} = a_{j+1} + \\frac{(-1)^{j+1}}{a_{j+1}} = a_j + \\frac{(-1)^j}{a_j} + (-1)^{j+1} \\left( a_j + \\frac{(-1)^j}{a_j} \\right)^{-1} = a_j + \\frac{1}{a_j^3 + (-1)^j a_j}.\n$$\nFor odd $j$, since $a_1 > 1$, we can prove inductively that $a_{j+2} > a_j > 1$. Also, for even $j$, since $a_2 = a_1 - \\frac{1}{a_1} > 0$, we can prove inductively that $a_{j+2} > a_j > 0$.\n\nTherefore, if we can prove that $a_{n+1} = a_1$, then we have\n$$\n0 < a_2 < a_4 < \\dots < a_{n+1} = a_1 < a_3 < \\dots < a_n,\n$$\nand so all terms are pairwise distinct positive real numbers.\n\nBy using the recurrence relation, it is obvious that $a_{n+1}$ is a continuous function of $a_1$. When $a_1$ goes to $1$, we have\n$$\na_{n+1} - a_1 \\ge a_4 - a_1 = a_2 + \\frac{1}{a_2^3 + a_2} - a_1 \\to \\infty\n$$\nsince $a_2 \\to 0$. This shows $a_{n+1} - a_1 > 0$ for some $x$.\n\nOn the other hand, for even $j$, as $a_j \\ge a_2 = 1+x - \\frac{1}{1+x} = \\frac{2x+x^2}{1+x} > x$, we have\n$$\na_{j+2} = a_j + \\frac{1}{a_j^3 + a_j} < a_j + \\frac{1}{x^3},\n$$\nand hence $a_{n+1} < a_2 + \\frac{n-1}{2x^3}$ by summing over all $j$'s. It follows that\n$$\na_{n+1} - a_1 < a_2 + \\frac{n-1}{2x^3} - a_1 = \\frac{n-1}{2x^3} - \\frac{1}{1+x} = \\frac{(n-1) + (n-1)x - 2x^3}{2x^3(1+x)} < 0\n$$\nfor sufficiently large $x$.\n\nNow, by the intermediate value theorem, there exists $x > 0$ such that $a_{n+1} = a_1$. So we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31656,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute-angled triangle. Let $D$ be a point on the segment $BC$, $I$ the incentre of $\\triangle ABC$. The circumcircle of $\\triangle ABD$ meets $BI$ at $P$ and the circumcircle of $\\triangle ACD$ meets $CI$ at $Q$. If the area of $\\triangle PID$ and the area of $\\triangle QID$ are equal, prove that $PI \\times QD = QI \\times PD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle AQI = \\angle AQC = \\angle ADC = 180^\\circ - \\angle ADB = 180^\\circ - \\angle APB = 180^\\circ - \\angle API$, the points $A, Q, I, P$ are concyclic. Let $DI$ meet $PQ$ at $M$, and let $AM$ meet $(AQIP)$ again at $X$. Since $[PID] = [QID]$, we know that $MP = MQ$. Also, note that $AP = PD$ and $AQ = QD$ since they are chords opposite to some equal angles in $(ABDP)$ and $(AQDC)$. This shows $PQ$ is the perpendicular bisector of $AD$. Therefore, we have\n$$\n\\angle IMP = \\angle DMP = \\angle AMP = \\angle XMQ.\n$$\nAs $M$ is the midpoint of $PQ$ and $Q, X, I, P$ are concyclic, this implies $QXIP$ is an isosceles trapezoid. Thus, $\\angle QAX = \\angle IAP$. Since $AM$ is the $A$-median of $\\triangle AQP$, $AI$ is the $A$-symmedian of $\\triangle AQP$. It follows that $PI \\times AQ = QI \\times AP$.\nFinally,\n$$\nPI \\times QD = PI \\times AQ = QI \\times AP = QI \\times PD.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31657,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime number greater than $5$. Suppose there is an integer $k$ satisfying that $k^2 + 5$ is divisible by $p$. Prove that there are positive integers $m$ and $n$ such that $p^2 = m^2 + 5n^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider all numbers of the form $a + b k$ where $a, b$ are integers with $0 \\le a, b < \\sqrt{p}$. Since there are more than $(\\sqrt{p})^2 = p$ numbers, by the pigeonhole principle, two of them must be congruent modulo $p$. Their difference, which is of the form $c + d k$ with $|c|, |d| < \\sqrt{p}$ and $(c, d) \\ne (0, 0)$, is divisible by $p$. Now, note that\n$$\n(c + d k)(c - d k) = c^2 - d^2 k^2 \\equiv c^2 + 5 d^2 \\pmod{p}.\n$$\nTherefore, $p \\mid c^2 + 5 d^2$. Since\n$$\n0 < c^2 + 5 d^2 < p + 5p = 6p,\n$$\nwe must have $c^2 + 5 d^2 = r p$ with $r = 1, 2, 3, 4, 5$.\n\n* If $c^2 + 5 d^2 = p$, then\n$$\n(c^2 - 5 d^2)^2 + 5 (2 c d)^2 = (c^2 + 5 d^2)^2 = p^2.\n$$\nClearly, $c, d \\ne 0$. So we can take $(m, n) = (|c^2 - 5 d^2|, 2|c d|)$.\n\n* If $c^2 + 5 d^2 = 2p$, then $c, d$ are odd as $2p \\equiv 2 \\pmod{4}$. Therefore, we have\n$$\n\\left(\\frac{c^2 - 5 d^2}{2}\\right)^2 + 5 (c d)^2 = \\left(\\frac{c^2 + 5 d^2}{2}\\right)^2 = p^2.\n$$\nSo we can take $(m, n) = \\left(\\left|\\frac{c^2 - 5 d^2}{2}\\right|, |c d|\\right)$.\n\n* If $c^2 + 5 d^2 = 3p$, then $3 \\mid c \\pm d$ as $9 \\mid 3p$. Therefore, $3 \\mid c \\pm d$ for some choice of the sign. Then we have\n$$\n\\left( \\frac{c \\mp 5 d}{3} \\right)^2 + 5 \\left( \\frac{c \\pm d}{3} \\right)^2 = \\frac{2(c^2 + 5 d^2)}{3} = 2p\n$$\nwith $\\frac{c \\mp 5 d}{3}$ and $\\frac{c \\pm d}{3}$ nonzero. Thus, this is reduced to the second case.\n\n* If $c^2 + 5 d^2 = 4p$, then $c, d$ are even. Therefore, we have\n$$\n\\left(\\frac{c}{2}\\right)^2 + 5\\left(\\frac{d}{2}\\right)^2 = p,\n$$\nwhich is reduced to the first case.\n\n* If $c^2 + 5 d^2 = 5p$, then $5 \\mid c$. Therefore, we have\n$$\nd^2 + 5 \\left( \\frac{c}{5} \\right)^2 = p,\n$$\nwhich is reduced to the first case.\n\nSince all cases are exhausted, we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31658,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers satisfying $abc = 1$. Determine the smallest possible value of $\\frac{a^3+8}{a^3(b+c)} + \\frac{b^3+8}{b^3(c+a)} + \\frac{c^3+8}{c^3(a+b)}$.",
    "options": [],
    "answer": "27/2",
    "solution": "The minimum value is $\\frac{27}{2}$.\nLet $x = \\frac{1}{a}$, $y = \\frac{1}{b}$ and $z = \\frac{1}{c}$. Then we have $xyz = 1$. By the AM-GM inequality, we have\n$$\na^3 + 2 = a^3 + 1 + 1 \\ge 3\\sqrt[3]{(a^3)(1)(1)} = 3a.\n$$\nTherefore,\n$$\n\\frac{a^3 + 8}{a^3(b+c)} \\ge \\frac{3a + 6}{a^3(b+c)} = \\frac{3a^2bc + 6abc}{a^3(b+c)} = \\frac{3x + 6x^2}{y+z}.\n$$\nSimilarly, we have $\\frac{b^3 + 8}{b^3(c+a)} \\ge \\frac{3y + 6y^2}{z+x}$ and $\\frac{c^3 + 8}{c^3(a+b)} \\ge \\frac{3z + 6z^2}{x+y}$. Thus, it remains to minimize\n$$\n3\\left(\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y}\\right) + 6\\left(\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y}\\right).\n$$\nBy Nesbitt's inequality, we have\n$$\n\\frac{x}{y+z} + \\frac{y}{z+x} + \\frac{z}{x+y} \\ge \\frac{3}{2}.\n$$\nAlso, by the Cauchy-Schwarz inequality, we have\n$$\n((y+z) + (z+x) + (x+y)) \\left( \\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} \\right) \\ge (x+y+z)^2.\n$$\nThis implies\n$$\n\\frac{x^2}{y+z} + \\frac{y^2}{z+x} + \\frac{z^2}{x+y} \\ge \\frac{x+y+z}{2} \\ge \\frac{3\\sqrt[3]{xyz}}{2} = \\frac{3}{2}\n$$\nby the AM-GM inequality. Therefore, the given expression is at least\n$$\n3 \\cdot \\frac{3}{2} + 6 \\cdot \\frac{3}{2} = \\frac{27}{2}.\n$$\n\nEquality holds when $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31659,
    "subject": "Mathematics (Multi-modal)",
    "question": "2016 circles with radius $1$ are lying on the plane. Among these $2016$ circles, show that one can select a collection $C$ of $27$ circles satisfying the following: either every pair of two circles in $C$ intersects or every pair of two circles in $C$ does not intersect.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose there do not exist $27$ circles such that every pair of circles intersects.\n\nConsider a coordinate plane such that the line joining any pair of centres of the circles is not parallel to the coordinate axes. We label the circles as $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_{2016}$ such that the $x$-coordinate of the centre of $\\Gamma_i$ is less than that of $\\Gamma_j$ if $i < j$.\n\n![](attached_image_1.png)\n\nConsider one of the circles $\\Gamma_k$. Let $\\Gamma$ be the circle with radius $2$ and having the same centre as $\\Gamma_k$. Partition the left semicircle of $\\Gamma$ into $3$ sectors of $60^\\circ$. Then the distance between any two points in the same sector is at most $2$. If there are $26$ centres of the other circles belonging to the same sector, then these circles together with $\\Gamma_k$ satisfy the condition, since the distance between any two of these centres is at most $2$, which shows the two circles intersect. Thus, we may assume there are at most $25$ centres in each sector. This implies at most $75$ centres among $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_{k-1}$ lie in $\\Gamma$, or equivalently at most $75$ of these circles intersect $\\Gamma_k$.\n\nNow, we colour the circles $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_{2016}$ one by one in $76$ colours. Suppose we have coloured $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_k$. Since $\\Gamma_{k+1}$ intersects at most $75$ of the previous circles, we can colour it in a way such that its colour is different from the colours of all circles among $\\Gamma_1, \\Gamma_2, \\dots, \\Gamma_k$ which intersect $\\Gamma_{k+1}$. By the pigeonhole principle, we can find\n$$\n\\left\\lfloor \\frac{2016}{76} \\right\\rfloor = 27\n$$\ncircles having the same colour. This means these $27$ circles are pairwise disjoint. So we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31660,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mable and Nora play a game according to the following steps in order.\n(1) Mable writes down any $2015$ distinct prime numbers in ascending order in a row. The product of these primes is Mable's score.\n(2) Nora writes down a positive integer.\n(3) Mable draws a vertical line between two adjacent primes she has written in step (1), and compute the product of the prime(s) on the left of the vertical line.\n(4) Nora must add the product obtained by Mable in step (3) to the number she has written in step (2), and the sum becomes Nora's score.\nIf Mable's and Nora's scores have a common factor greater than $1$, Mable wins. Otherwise Nora wins.\nWho has a winning strategy?",
    "options": [],
    "answer": "Nora has a winning strategy.",
    "solution": "Nora has a winning strategy.\nLet Mable write down $p_1 < p_2 < \\dots < p_{2015}$ in step (1). Then $P = p_1p_2\\dots p_{2015}$ is Mable's score.\nLet $q_k = p_1p_2\\dots p_k$ for each $k$. Nora needs to write down a positive integer $Q$ in step (2) such that $(P, Q + q_j) = 1$ for all $j$. Indeed, for each $1 \\le k \\le 2015$, we choose $r_k \\in \\mathbb{Z}$ such that\n$$\nr_k \\ne -q_j \\pmod{p_k}\n$$\nfor all $1 \\le j \\le 2015$. Such an $r_k$ exists because there are at most $k < p_k$ residues that $r_k$ cannot take (as $p_k \\mid q_j$ if $j \\ge k$).\n\nNow, there exists $Q \\in \\mathbb{Z}^+$ such that $Q \\equiv r_k \\pmod{p_k}$ for all $k$ by the Chinese remainder theorem. For this choice of $Q$, we have\n$$\nQ + q_j \\equiv r_k + q_j \\not\\equiv 0 \\pmod{p_k}\n$$\nfor any $j$ and $k$. Therefore, we have $(P, Q + q_j) = 1$ for all $j$ as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31661,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of integers $(m, p, q)$ such that $2^m p^2 + 1 = q^7$, where $m > 0$, and $p$ and $q$ are prime numbers, or show that they do not exist.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There is no such triple.\nFirst note that\n$$\n2^m p^2 = q^7 - 1 = (q-1)(q^6 + q^5 + q^4 + q^3 + q^2 + q + 1).\n$$\nSince $q^6 + q^5 + q^4 + q^3 + q^2 + q + 1$ is odd and is larger than $1$, we have $q-1 = 2^m$ or $q-1 = 2^m p$. The latter is impossible by Zsigmondy's theorem, since $q^7 - 1$ has a prime divisor not dividing $q-1$ (alternatively, one can obtain a contradiction by using some simple bounds). Thus, we have $q = 2^m + 1$.\nIf $m$ is odd, then $q = 2^m + 1 \\equiv 2 + 1 \\equiv 0 \\pmod 3$. As $q$ is a prime, we must have $q=3$, and so $m=1$. But then $2p^2+1=2187$ has no solution.\nIf $m$ is even, let $m = 2k$. The equation becomes $q^7 - (2^k p)^2 = 1$. By Catalan's conjecture, there is no solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31662,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\lambda$ be a nonnegative real number and such that $\\frac{a+b}{2} \\ge \\lambda\\sqrt{ab}+(1-\\lambda)\\sqrt{\\frac{a^2+b^2}{2}}$ holds for all positive real numbers $a$ and $b$. Find the smallest possible value of $\\lambda$.",
    "options": [],
    "answer": "1/2",
    "solution": "Firstly, when $\\lambda = \\frac{1}{2}$, we have\n$$\n\\begin{align*}\n\\frac{a+b}{2} &\\ge \\frac{1}{2}\\sqrt{ab} + \\frac{1}{2}\\sqrt{\\frac{a^2+b^2}{2}} \\\\\n\\Leftrightarrow \\quad (a+b)^2 &\\ge ab + \\frac{a^2+b^2}{2} + 2\\sqrt{ab} \\cdot \\sqrt{\\frac{a^2+b^2}{2}} \\\\\n\\Leftrightarrow \\quad (a+b)^2 &\\ge 2\\sqrt{2ab(a^2+b^2)} \\\\\n\\Leftrightarrow \\quad (a+b)^4 &\\ge 8ab(a^2+b^2) \\\\\n\\Leftrightarrow \\quad (a-b)^4 &\\ge 0.\n\\end{align*}\n$$\n\nSecondly, suppose the inequality holds for all $a, b > 0$. Note that\n$$\n\\begin{align*}\n\\frac{a+b}{2} &\\ge \\lambda\\sqrt{ab} + (1-\\lambda)\\sqrt{\\frac{a^2+b^2}{2}} \\\\\n\\Leftrightarrow \\quad \\sqrt{2}(\\sqrt{a^2+b^2} - \\sqrt{2ab})\\lambda &\\ge \\sqrt{2(a^2+b^2)} - (a+b) \\\\\n\\Leftrightarrow \\quad \\frac{\\sqrt{2}(a-b)^2}{\\sqrt{a^2+b^2} + \\sqrt{2ab}}\\lambda &\\ge \\frac{(a-b)^2}{\\sqrt{2(a^2+b^2)} + (a+b)}.\n\\end{align*}\n$$\n\nWLOG assume $a \\neq b$. Then this holds if and only if\n$$\n\\lambda \\geq \\frac{\\sqrt{a^2 + b^2} + \\sqrt{2ab}}{\\sqrt{2}(\\sqrt{2(a^2 + b^2)} + (a+b))}.\n$$\n\nWe can rewrite this as\n$$\n2\\lambda \\geq 1 - \\frac{a+b-2\\sqrt{ab}}{\\sqrt{2(a^2+b^2)}+(a+b)} = 1 - \\frac{(\\sqrt{a}-\\sqrt{b})^2}{\\sqrt{2(a^2+b^2)}+(a+b)}.\n$$\n\nConsider $a = 1$ and $b = 1 + \\varepsilon$ where $\\varepsilon \\to 0$. It is clear that the right-hand side approaches 1. Therefore, in order that the inequality holds, we must have $2\\lambda \\geq 1$, which means $\\lambda \\geq \\frac{1}{2}$.\nCombining these, the smallest possible value of $\\lambda$ is $\\frac{1}{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31663,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be inscribed in a circle with centre $O$. Let $E$ be the intersection point of $AC$ and $BD$. $M$ and $N$ are the midpoints of the arcs $AB$ and $CD$ respectively (the arcs not containing any other vertices). Let $P$ be the intersection point of $EO$ and $MN$. Suppose $BC = 5$, $AC = 11$, $BD = 12$ and $AD = 10$. Find $\\frac{MN}{NP}$.",
    "options": [],
    "answer": "23/13",
    "solution": "We have $\\frac{MN}{NP} = \\frac{23}{13}$.\nLet $I$ and $J$ be the incentres of $\\triangle ADE$ and $\\triangle BCE$ respectively. Note that $I$ is the intersection point of $AN$ and $DM$, while $J$ is the intersection point of $BN$ and $CM$. Applying Pascal's theorem to $ANBDMC$, we find that $I, J, E$ are collinear.\n\nSince $\\angle IAD = \\angle JBC$ and $\\angle IDA = \\angle JCB$, we have $\\triangle AID \\sim \\triangle BJC$. Thus, we have\n$$\n\\triangle MIN \\sim \\triangle AID \\sim \\triangle BJC \\sim \\triangle MJN,\n$$\nwith $\\triangle MIN \\cong \\triangle MJN$. This implies $IJ \\perp MN$. Also, the isogonal line of $IJ$ with respect to $\\angle MIN$, the isogonal line of $IJ$ with respect to $\\angle MJN$ and the line $MN$ are concurrent at some point $P'$ by symmetry. Observe that $IP' \\perp AD$ since $\\triangle MIN$ and $\\triangle AID$ are oppositely similar. Similarly, $JP' \\perp BC$. We now claim that $P' = P$.\n\n![](attached_image_1.png)\n\nLet $E_1, P_1, O_1$ be the projection of $E, P', O$ on $AD$, and let $E_2, P_2, O_2$ be the projection of $E, P', O$ on $BC$. Note that $E_1, P_1, O_1$ are the foot of altitude from $E$, the contact point of the incircle with the side $AD$, and the midpoint of $AD$.\n\nin $\\triangle EAD$. Similar results hold for $E_2, P_2, O_2$ in $\\triangle EBC$. As $\\triangle EAD \\sim \\triangle EBC$, we have $\\frac{E_1P_1}{P_1O_1} = \\frac{E_2P_2}{P_2O_2}$. Therefore, $E, P', O$ are collinear. This implies $P'$ lies on both $EO$ and $MN$, and hence $P' = P$.\n\nNow, let $F$ be the intersection of $AD$ and $IJ$. Then $\\frac{MP}{NP} = \\frac{AF}{DF}$ due to the isogonal lines. By the angle bisector theorem, we have $\\frac{AF}{DF} = \\frac{AE}{DE}$.\n\nNote that $\\triangle EAD \\sim \\triangle EBC$, and the ratio is $\\frac{AD}{BC} = 2$. Let $BE = x$ and $CE = y$ so that $AE = 2x$ and $DE = 2y$. It is given that $2x + y = 11$ and $x + 2y = 12$. We easily find that $x = \\frac{10}{3}$ and $y = \\frac{13}{3}$. Therefore,\n$$\n\\frac{MN}{NP} = \\frac{MP + NP}{NP} = \\frac{AE}{DE} + 1 = \\frac{x}{y} + 1 = \\frac{23}{13}.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31664,
    "subject": "Mathematics (Multi-modal)",
    "question": "4031 lines are drawn on the plane. No two lines are parallel or perpendicular, and no three lines meet at one point. Determine the maximum number of acute-angled triangles that may be formed.",
    "options": [],
    "answer": "2729148240",
    "solution": "The maximum number of acute-angled triangles is $2729148240$.\nLet $n = 2015$, so that there are $2n + 1$ lines. We fix one of the lines $\\ell$ and place it as the $x$-axis of the coordinate plane. Then the other $2n$ lines can be partitioned into two groups, one consisting of those lines with positive slopes and one consisting of those lines with negative slopes. Let $a$ and $b$ be the sizes of these two groups respectively.\n\nObserve that every pair of lines belonging to the same group form an obtuse triangle with $\\ell$. Therefore, the number of obtuse triangles having $\\ell$ as a sideline that is adjacent to the obtuse angle is\n$$\n\\binom{a}{2} + \\binom{b}{2} \\ge 2\\binom{n}{2} = n(n-1)\n$$\nby Jensen's inequality, since the binomial function $\\binom{x}{2}$ is convex.\n\nBy considering all choices of $\\ell$, each obtuse triangle is counted twice. Therefore, the number of acute-angled triangles is at most\n$$\n\\binom{2n+1}{3} - \\frac{(2n+1)n(n-1)}{2} = \\frac{n(n+1)(2n+1)}{6}.\n$$\nThis bound is attainable. Indeed, consider the sidelines of a regular polygon with $2n + 1$ sides. Then for every line $\\ell$ being the $x$-axis, the number of lines with positive slopes is equal to the number of lines with negative slopes by symmetry. Therefore, equality of the above deduction holds.\n\nWhen $n = 2015$, the answer is $\\frac{2015 \\times 2016 \\times 4031}{6} = 2729148240$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31665,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, let $AD$ be the angle bisector of $\\angle BAC$, with $D$ on $BC$. The perpendicular from $B$ to $AD$ intersects the circumcircle of $\\triangle ABD$ at $B$ and $E$. Prove that $E, A$ and the circumcentre $O$ of $\\triangle ABC$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By simple angle chasing, we find that\n$$\n\\begin{align*} \n\\angle BAE &= \\angle BAD + \\angle DAE = \\frac{A}{2} + \\angle DBE = \\frac{A}{2} + B - \\left(90^\\circ - \\frac{A}{2}\\right) \\\\ \n&= A + B - 90^\\circ = 90^\\circ - C = \\angle BAO. \n\\end{align*}\n$$\nAs $O$ and $E$ lie on the same side of $AB$ as $C$ in this case, this implies $A, O, E$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31666,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the internal angle bisector of $\\angle BAC$ of $\\triangle ABC$ meet side $BC$ at $D$. Let $\\Gamma$ be the circle through $A$ tangent to $BC$ at $D$. Suppose $\\Gamma$ meets sides $AB$ and $AC$ at $E$ and $F$ again, respectively. Lines $BF$ and $CE$ meet $\\Gamma$ again at $P$ and $Q$, respectively. Let $AP$ and $AQ$ intersect side $BC$ at $X$ and $Y$, respectively. Prove that $XY = \\frac{1}{2}BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle AFD = \\angle ADB$ and $\\angle DAF = \\angle BAD$, we have $\\angle ADF = \\angle ABD$. This implies $\\angle AEF = \\angle ADF = \\angle ABD$ so that $EF \\parallel BC$. It follows that $\\angle XBP = \\angle EFP = \\angle BAX$. Thus, $XB$ is a tangent to $(ABP)$. Hence, we have $XB^2 = XP \\cdot XA = XD^2$. This gives $XB = XD$. Similarly, we have $YC = YD$. Therefore, $XY = \\frac{1}{2}BC$. $\\square$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31667,
    "subject": "Mathematics (Multi-modal)",
    "question": "Decide if there is a permutation $a_1, a_2, \\dots, a_{6666}$ of the numbers $1, 2, \\dots, 6666$ with the property that the sum $k + a_k$ is a perfect square for all $k = 1, 2, \\dots, 6666$.",
    "options": [],
    "answer": "Yes",
    "solution": "Yes. Consider the permutation\n$3, 2, 1, 12, 11, \\dots, 4, 23, 22, \\dots, 13, 57, 56, \\dots, 24, 6666, 6665, \\dots, 58.$\nMore precisely, we define\n$$\na_k = \\begin{cases} 2^2 - k & \\text{if } 1 \\le k \\le 3, \\\\ 4^2 - k & \\text{if } 4 \\le k \\le 12, \\\\ 6^2 - k & \\text{if } 13 \\le k \\le 23, \\\\ 9^2 - k & \\text{if } 24 \\le k \\le 57, \\\\ 82^2 - k & \\text{if } 58 \\le k \\le 6666. \\end{cases}\n$$\nThe result follows readily.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 31668,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Does there exist $5$ circles in the plane such that each circle passes through exactly $3$ centres of other circles?\n\nb. Does there exist $6$ circles in the plane such that each circle passes through exactly $3$ centres of other circles?",
    "options": [],
    "answer": "a. No; b. Yes",
    "solution": "a. No. Let the centres be $A$, $B$, $C$, $D$, $E$. WLOG assume $A$ is the centre of $(BCD)$. One of the following must occur.\n\n* If $B$ is the centre of $(ACD)$, then $AC = AD = AB = BC = BD$. So both $\\triangle ABC$ and $\\triangle ABD$ are equilateral triangles. Note that at least one of $A$, $B$ must lie on the circle with centre $C$. So $E$ lies on the circle with centre $C$ that passes through $A$ and $B$. Similarly, $E$ lies on the circle with centre $D$ that passes through $A$ and $B$. This is a contradiction since the two circles only meet at $A$ and $B$.\n\n* If $B$ is the centre of $(ACE)$, then $AC = AD = AB = BC = BE$. So $\\triangle ABC$ is an equilateral triangle. Thus, there must be $3$ points lying on the circle with centre $C$ that passes through $A$ and $B$. In either case $CA = CB = CD$ or $CA = CB = CE$, we get the same configuration as above and hence it is a contradiction.\n\nb. Yes. Consider two equilateral triangles of side lengths $1$ having parallel sides and the same orientation such that each pair of corresponding vertices is at a distance $1$ apart. It is clear that each point has a distance $1$ to exactly three other points. Therefore, we can draw a circle with each point as centre and radius $1$ that passes through exactly $3$ other points.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31669,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $\\{x_n\\}$ is defined by $x_1 = 5$ and $x_{k+1} = x_k^2 - 3x_k + 3$ for $k = 1, 2, 3, \\dots$. Prove that $x_k > 3^{2^{k-1}}$ for any positive integer $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The recurrence relation can be rewritten as $x_{k+1} - 3 = x_k(x_k - 3)$. By repeating the process, we obtain\n$$\nx_{k+1} - 3 = x_k(x_k - 3) = x_k x_{k-1}(x_{k-1} - 3) = \\cdots = x_k x_{k-1} \\cdots x_1(x_1 - 3).\n$$\nChanging the index, this gives $x_k = 10x_2 \\cdots x_{k-1} + 3$. We now prove the assertion by induction. The base cases $k=1$ and $k=2$ follow from $x_1 > 3$ and $x_2 = 13 > 9$. For the inductive step, we obtain\n$$\nx_k > 10x_2 \\cdots x_{k-1} > 10 \\prod_{j=2}^{k-1} 3^{2^{j-1}} = 10 \\times 3^{2+2^2+2^3+\\cdots+2^{k-2}} = 10 \\times 3^{2^{k-1}-2} > 3^{2^{k-1}}.\n$$\nThus, this holds for every positive integer $k$ by induction.\nWe prove by induction that $x_k \\ge 3^{2^{k-1}} + 2$ for any $k$. The base case holds since $x_1 = 5$. Assuming $x_k \\ge 3^{2^{k-1}} + 2$ for some $k$, we have\n$$\nx_{k+1} = x_k(x_k - 3) + 3 \\ge (3^{2^{k-1}} + 2)(3^{2^{k-1}} - 1) + 3 > (3^{2^{k-1}})^2 + 2 = 3^{2k} + 2.\n$$\nThe claim holds by induction. This obviously implies $x_k > 3^{2^{k-1}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31670,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\{a_n\\}$ be a sequence of integers satisfying the following condition for all positive integral values of $n$: $a_n + a_{n+1} = 2a_{n+2}a_{n+3} + 2016$. Find all possible values of $a_1$ and $a_2$.",
    "options": [],
    "answer": "[(1, -2015), (-2015, 1), (15, -69), (-69, 15), (70, -14), (-14, 70), (2016, 0), (0, 2016)]",
    "solution": "$(a_1, a_2)$ can be $(1, -2015), (15, -69), (70, -14), (2016, 0)$ or their permutations.\n\nFirstly, taking the difference of the two relations $a_n + a_{n+1} = 2a_{n+2}a_{n+3} + 2016$ and $a_{n+1} + a_{n+2} = 2a_{n+3}a_{n+4} + 2016$, we obtain\n$$\na_{n+2} - a_n = 2a_{n+3}(a_{n+4} - a_{n+2}).\n$$\nInductively, this easily implies\n$$\na_{n+2} - a_n = 2^k a_{n+3} a_{n+5} \\cdots a_{n+2k+1} (a_{n+2k+2} - a_{n+2k})\n$$\nfor any $k \\in \\mathbb{Z}^+$. This shows $2^k \\mid a_{n+2} - a_n$ for any $k \\in \\mathbb{Z}^+$, and hence $a_{n+2} = a_n$. Thus, we may assume all odd terms are equal to $a_1 = b$ and all even terms are equal to $a_2 = c$. Now the only condition is\n$$\nb + c = 2bc + 2016.\n$$\nThis means\n$$\n(2b - 1)(2c - 1) = 2(2bc - b - c) + 1 = -4031 = -29 \\times 139.\n$$\nThus, we obtain the solutions $(b, c) = (1, -2015), (15, -69), (70, -14), (2016, 0)$ up to permutation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31671,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $ABCD$ is a cyclic quadrilateral. Extend $DA$ and $DC$ to $P$ and $Q$ respectively such that $AP = BC$ and $CQ = AB$. Let $M$ be the midpoint of $PQ$. Show that $MA \\perp MC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $B'$ be the point such that $AB'BC$ is an isosceles trapezoid. Thus, $B'$ lies on the circle passing through $A$, $B$, $C$, $D$. Also, we have $AP = CB = AB'$ and $CQ = AB = CB'$. Let $E$ and $F$ be the midpoints of $B'P$ and $B'Q$ respectively. Then $AE \\perp B'P$ and $CF \\perp B'Q$.\n\n![](attached_image_1.png)\n\nFirst of all, as $\\angle PAB' + \\angle QCB' = (180^\\circ - \\angle B'AD) + (180^\\circ - \\angle B'CD) = 180^\\circ$, we know that $\\angle EAB' + \\angle FCB' = 90^\\circ$. This implies the right-angled triangles $AEB'$ and $B'FC$ are similar. Hence, $\\frac{AE}{AB'} = \\frac{B'F}{B'C}$.\n\nBy the midpoint theorem, $B'EMF$ is a parallelogram. Therefore, we have\n$$\n\\frac{AE}{EM} = \\frac{AE}{B'F} = \\frac{AB'}{B'C}.\n$$\nTogether with\n$$\n\\begin{aligned}\n\\angle AEM &= 90^\\circ + \\angle B'EM = 270^\\circ - \\angle EB'F \\\\\n&= 270^\\circ - (360^\\circ - \\angle AB'C - \\angle AB'E - \\angle CB'F) = \\angle AB'C,\n\\end{aligned}\n$$\nwe find that $\\triangle AEM \\sim \\triangle AB'C$. This yields $\\triangle AEB' \\sim \\triangle AMC$ by spiral similarity. Therefore, $\\angle AMC = \\angle AEB' = 90^\\circ$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31672,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a convex hexagon such that $\\angle ACE = \\angle BDF$ and $\\angle BCA = \\angle EDF$. Let $A_1 = AC \\cap FB$, $B_1 = BD \\cap AC$, $C_1 = CE \\cap BD$, $D_1 = DF \\cap CE$, $E_1 = EA \\cap DF$ and $F_1 = FB \\cap EA$. Suppose $B_1, C_1, D_1, F_1$ lie on the same circle $\\Gamma$. The circumcircles of $\\triangle BB_1F_1$ and $\\triangle ED_1F_1$ meet at $F_1$ and $P$. The line $F_1P$ meets $\\Gamma$ again at $Q$. Prove that $B_1D_1$ and $QC_1$ are parallel. (Here, we use $\\ell_1 \\cap \\ell_2$ to denote the intersection point of lines $\\ell_1$ and $\\ell_2$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Firstly, since\n$$\n\\angle EPF_1 = \\angle ED_1F_1 = \\angle C_1QF_1,\n$$\nwe have $PE//QC_1$. By symmetry, we also have $BP//QC_1$. Therefore, $B, P, E$ are collinear, with $BE//QC_1$.\n\nSecondly, since $\\angle D_1DB_1 = \\angle D_1CB_1$ and $\\angle EDB = \\angle ECB$, we know that $B_1, C, D, D_1$ and $B, C, D, E$ are two groups of concyclic points. By Reim's theorem, we have $BE//B_1D_1$ (alternatively, since $\\angle C_1B_1D_1 = \\angle DCE = \\angle C_1BE$). Thus, $B_1D_1//BE//QC_1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31673,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $d$ be a nonnegative integer. Determine all functions $f : \\mathbb{R}^2 \\to \\mathbb{R}$ such that, for any real constants $A$, $B$, $C$ and $D$, $f(At+B, Ct+D)$ is a polynomial in $t$ of degree at most $d$.",
    "options": [],
    "answer": "All polynomials in two variables of total degree at most d.",
    "solution": "We claim that $f(x, y)$ is a polynomial in $x$ and $y$ of degree at most $d$.\nIt is obvious that every such polynomial satisfies the desired condition. To prove the converse, let $f$ be a function satisfying the desired condition. Pick $(d+2)$ straight lines $\\ell_1$, $\\ell_2$, $\\dots$, $\\ell_{d+2}$ in $\\mathbb{R}^2$ such that no two are parallel and no three are concurrent. Let the equation of $\\ell_i$ be $h_i(x, y) = 0$, where $h_i$ is a linear polynomial.\nFor $i < j$, let $(a_{ij}, b_{ij})$ be the intersection of $\\ell_i$ and $\\ell_j$, and consider the polynomial\n$$\n\\varphi(x, y) = \\sum_{1 \\le i < j \\le d+2} f(a_{ij}, b_{ij}) \\prod_{\\substack{k=1 \\\\ k \\ne i, j}}^{d+2} \\frac{h_k(x, y)}{h_k(a_{ij}, b_{ij})}.\n$$\nIt is easy to see that $\\varphi(a_{ij}, b_{ij}) = f(a_{ij}, b_{ij})$ for all $i < j$, and that $\\deg \\varphi \\le d$. We shall show that $\\varphi(a, b) = f(a, b)$ for all $a, b \\in \\mathbb{R}$.\nWe first make an observation: if $\\ell$ is a line such that $\\varphi(a, b) = f(a, b)$ for at least $(d+1)$ points $(a, b)$ on $\\ell$, then $\\varphi(a, b) = f(a, b)$ for all points $(a, b)$ on $\\ell$. Indeed, pick constants $A$, $B$, $C$ and $D$ such that $t \\mapsto (At + B, Ct + D)$ parametrizes the line. Then, note that $\\varphi(At + B, Ct + D) - f(At + B, Ct + D)$ is a polynomial of degree at most $d$, and that it vanishes at least $(d+1)$ points, so $\\varphi(a, b) = f(a, b)$ for all $(a, b)$ on $\\ell$.\nNow, for a fixed $\\ell_i$, note that $f(a_{ij}, b_{ij}) = \\varphi(a_{ij}, b_{ij})$ for every $j \\ne i$, so $f(a, b) = \\varphi(a, b)$ for all points $(a, b)$ on $\\ell_i$ from the claim. If $(c, d)$ is a point not lying on any $\\ell_i$, then we can construct a line $\\ell$ which passes through $(c, d)$, does not pass through any $(a_{ij}, b_{ij})$, and is not parallel to any $\\ell_i$. Now $f(a, b) = \\varphi(a, b)$ with $(a, b) = \\ell_i \\cap \\ell$ for various $i$, so $f(a, b) = \\varphi(a, b)$ for all $(a, b)$ on $\\ell$. In particular, $\\varphi(c, d) = f(c, d)$. This completes the proof. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31674,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a committee there are $n$ members. Each pair of members are either friends or enemies. Each committee member has exactly three enemies. It is also known that for each committee member, an enemy of his friend is automatically his own enemy. Find all possible value(s) of $n$.",
    "options": [],
    "answer": "4, 6",
    "solution": "Suppose $A$ and $B$ are enemies. By the given condition, every other member cannot be a friend of both $A$ and $B$. Since each of $A$ and $B$ has 2 more enemies other than themselves, we must have $n \\le 2 + 2 \\times 2 = 6$. Also, as each member has 3 enemies, we have $n \\ge 4$, and there are $\\frac{3n}{2}$ pairs of enemies. This shows $n$ is even, and hence $n$ can only be 4 or 6.\nTo give examples for $n = 4$ and $n = 6$, we use terminologies in graph theory. Suppose vertices denote the members and edges join pairs of enemies. Then $K_4$ and $K_{3,3}$ are examples for $n = 4$ and $n = 6$ respectively.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31675,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose all of the $200$ integers lying in between (and including) $1$ and $200$ are written on a blackboard. Suppose we choose exactly $100$ of these numbers and circle each one of them. By the *score* of such a choice, we mean the square of the difference between the sum of the circled numbers and the sum of the non-circled numbers. What is the average of the scores over all possible choices for $100$ numbers?",
    "options": [],
    "answer": "670000",
    "solution": "The average score is $670000$.\nLet $n = 100$, and let $S$ be any subset of $\\{1, 2, \\dots, 2n\\}$ such that $|S| = n$. There are $\\binom{2n}{n}$ such sets $S$. The score of $S$ is\n$$\n\\begin{aligned}\n\\left( \\sum_{a \\in S} a - \\sum_{b \\notin S} b \\right)^2 &= \\left( \\sum_{k=1}^{2n} k - 2 \\sum_{a \\in S} a \\right)^2 \\\\\n&= \\left( \\frac{2n(2n+1)}{2} \\right)^2 - 4 \\left( \\frac{2n(2n+1)}{2} \\right) \\sum_{a \\in S} a + 4 \\left( \\sum_{a \\in S} a \\right)^2 \\\\\n&= n^2(2n+1)^2 - 4n(2n+1) \\sum_{a \\in S} a + 4 \\sum_{a \\in S} a^2 + 4 \\sum_{\\substack{a, b \\in S \\\\ a \\neq b}} ab.\n\\end{aligned}\n$$\n\n$$\n\\sum_{S} \\sum_{a \\in S} a = \\sum_{a=1}^{2n} \\sum_{a \\in S} a = \\sum_{a=1}^{2n} a \\binom{2n-1}{n-1} = n(2n+1) \\binom{2n-1}{n-1}\n$$\n\n$$\n\\begin{align*}\n\\sum_S \\left( \\sum_{a \\in S} a^2 + \\sum_{\\substack{a, b \\in S \\\\ a \\neq b}} ab \\right) &= \\sum_{a=1}^{2n} \\sum_{a \\in S} a^2 + \\sum_{\\substack{a, b=1 \\\\ a \\neq b}}^{2n} \\sum_{a, b \\in S} ab \\\\\n&= \\sum_{a=1}^{2n} a^2 \\binom{2n-1}{n-1} + \\sum_{\\substack{a, b=1 \\\\ a \\neq b}}^{2n} ab \\binom{2n-2}{n-2} \\\\\n&= \\sum_{a=1}^{2n} a^2 \\binom{2n-1}{n-1} + \\sum_{a=1}^{2n} a \\left( \\sum_{b=1}^{2n} b - a \\right) \\binom{2n-2}{n-2} \\\\\n&= \\sum_{a=1}^{2n} a^2 \\left[ \\binom{2n-1}{n-1} - \\binom{2n-2}{n-2} \\right] + n(2n+1) \\binom{2n-2}{n-2} \\sum_{a=1}^{2n} a \\\\\n&= \\frac{n(2n+1)(4n+1)}{3} \\binom{2n-2}{n-1} + n^2(2n+1)^2 \\binom{2n-2}{n-2}.\n\\end{align*}\n$$\n\nTherefore, the average score is\n$$\n\\begin{align*}\n& n^2(2n+1)^2 + \\left(\\frac{2n}{n}\\right)^{-1} \\left[ -4n^2(2n+1)^2 \\binom{2n-1}{n-1} + \\frac{4n(2n+1)(4n+1)}{3} \\binom{2n-2}{n-1} \\right. \\\\\n& \\qquad \\left. + 4n^2(2n+1)^2 \\binom{2n-2}{n-2} \\right] \\\\\n&= n^2(2n+1)^2 + \\left( -2n^2(2n+1)^2 + \\frac{2n^2(2n+1)(4n+1)}{3(2n-1)} + \\frac{2(n-1)n^2(2n+1)^2}{2n-1} \\right) \\\\\n&= \\frac{n^2(2n+1)}{3} \\\\\n&= 670000.\n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31676,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. Prove that there exists a positive integer $\\ell$ with the following property: if $m$ and $n$ are positive integers relatively prime to $\\ell$ such that $m^n \\equiv n^m \\pmod{\\ell}$, then $m \\equiv n \\pmod{k}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove the assertion when $\\ell = k!$ by induction on $k$. The base case $k = 1$ is obvious.\n\nIf $k > 1$, then $\\varphi(k) < k$. Suppose $m$ and $n$ are positive integers relatively prime to $\\ell$ such that $m^m \\equiv n^n \\pmod{k}$. This clearly implies $m^m \\equiv n^n \\pmod{\\varphi(k)!}$. By the induction hypothesis, we know that $m \\equiv n \\pmod{\\varphi(k)}$. Thus, there exists a positive integer $a$ such that $am \\equiv an \\equiv 1 \\pmod{\\varphi(k)}$. Now, by the Euler-Fermat theorem, we have\n$$\nm \\equiv m^{am} \\equiv n^{an} \\equiv n \\pmod{k}.\n$$\nThe proof is complete by induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31677,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. A game is played on an $n \\times n$ board by two players X and Y as follows.\n\n*   **Order:** The two players take turns to play. X plays in the 1st round, Y plays the 2nd round, then X plays the 3rd round and so on.\n*   **Rule:** In the $k$th round, one has to choose $k$ unmarked consecutive cells in the same row or in the same column and mark each of these cells.\n*   **Winner:** The first player who cannot complete the task will lose the game.\n\nThe player who is expected to carry out the $(n+1)$st round is called the *natural loser*, as there are no $(n+1)$ consecutive cells on the board. Find the smallest $n$ for which the natural loser has a winning strategy.",
    "options": [],
    "answer": "6",
    "solution": "We claim that the smallest *unnatural* case is $n = 6$. In the following, we use $(i, j)$ to mean the cell in the $i$th row and the $j$th column, and use $(i, j) - (i, k)$ to mean the consecutive cells from $(i, j)$ to $(i, k)$ inclusive (similar for $(i, j) - (k, j)$).\n\nFor $n = 2$, X and Y can play in any way and the natural loser X will lose.\n\nFor $n = 3$, Y is the natural loser and X has a winning strategy. In the first round, X marks $(2, 2)$. WLOG Y has to mark $(1, 1) - (1, 2)$. Then X can take $(3, 1) - (3, 3)$.\n\nFor $n = 4$, Y has a winning strategy. In round 1, we may assume X chooses a cell in the upper left $2 \\times 2$ corner. Then Y can mark 2 consecutive cells in the same $2 \\times 2$ corner. This leaves 2 unoccupied rows as well as 2 unoccupied columns. WLOG we may assume X chooses certain cells from a row in round 3. Then there is at least one unoccupied row left, which can be marked by Y in round 4.\n\nFor $n = 5$, X has a winning strategy by first choosing $(3, 3)$. WLOG, we may assume Y chooses 2 cells in the upper left $3 \\times 3$ corner in the next round. Then X can always choose 3 consecutive cells from the same $3 \\times 3$ corner. Again, there are at least 2 unoccupied rows and 2 unoccupied columns. For the same reason as the case $n = 4$, X can win the game.\n\nThe same trick no longer works for $n = 6$. The natural loser X can still win the game. We propose a winning strategy as follows. Firstly, X chooses $(1, 1)$. WLOG assume Y chooses 2 cells $(i, j) - (i + 1, j)$ in the $j$th column in the next round.\n\n*   If $j = 1$, X chooses $(3, 3) - (3, 5)$ in round 3. This blocks columns 3, 4, 5 and hence Y must mark at least one cell in column 2 or column 6 (but not both columns with obvious reason) in round 4. If some cells in column 2 (column 6 in the second case) is used, X can then take $(2, 6) - (6, 6)$ (correspondingly $(2, 2) - (6, 2)$) in round 5. Then all rows and columns have been used so that Y will lose.\n\n*   A similar strategy works for cases $j = 2, 5, 6$. Y loses in all following subcases.\n    -   For $j = 2$, X chooses $(3, 3) - (3, 5)$. In round 5, X should take $(2, 1) - (6, 1)$ or $(2, 6) - (6, 6)$, depending on column 6 is used or not.\n    -   For $j = 5$, X chooses $(3, 2) - (3, 4)$. In round 5, X should take $(2, 1) - (6, 1)$ or $(2, 6) - (6, 6)$, depending on column 6 is used or not.\n    -   For $j = 6$, X chooses $(3, 2) - (3, 4)$. In round 5, X should take $(2, 1) - (6, 1)$ or $(2, 5) - (6, 5)$, depending on column 5 is used or not.\n\n*   If $j = 3, 4$ and $i = 1, 4, 5$, X can still choose $(3, 3) - (3, 5)$ in round 3.\n    -   If Y chooses at least one cell in column 2 in round 4, X can then take $(2, 6) - (6, 6)$ in round 5. Y loses.\n    -   If Y chooses at least one cell in column 6 in round 4, X can then take $(2, 2) - (6, 2)$ in round 5. Y loses.\n    -   Suppose Y chooses $(2, 1) - (5, 1)$ in round 4. Next, X should take $(6, 2) - (6, 6)$ if row 6 is not used in round 2, or take $(1, 2) - (1, 6)$ otherwise. Y loses.\n    -   Suppose Y chooses $(3, 1) - (6, 1)$ in round 4. Next, X should take $(2, 2) - (2, 6)$ if row 2 is not used in round 2, or take $(6, 2) - (6, 6)$ otherwise. Y loses.\n\n*   If $j = 3, 4$ and $i = 2, 3$, X can take $(5, 3) - (5, 5)$ in round 3.\n    -   If Y chooses at least one cell in column 2 in round 4, X can then take $(2, 6) - (6, 6)$ in round 5. Y loses.\n    -   If Y chooses at least one cell in column 6 in round 4, X can then take $(2, 2) - (6, 2)$ in round 5. Y loses.\n    -   If Y chooses $(2, 1) - (5, 1)$ in round 4, X can take $(6, 2) - (6, 6)$ in round 5. Y loses.\n    -   Suppose Y chooses $(3, 1) - (6, 1)$ in round 4. Next, X should take $(2, 2) - (2, 6)$ if row 2 is not used in round 2, or take $(6, 2) - (6, 6)$ otherwise. Y loses.\n    -   If Y chooses 4 consecutive cells in column 3, 4 or 5 in round 4, then the first 5 rows are all occupied. X can then take $(6, 2) - (6, 6)$ in round 5. Y loses.\n\nIn all cases, X can win the game when $n = 6$. This completes the proof. $\\Box$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31678,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a mathematical competition $n$ students work on 6 problems each one with three possible answers. After the competition, the Jury found that for every two students the number of the problems for which these students have the same answers is 0 or 2. Find the maximum possible value of $n$.",
    "options": [],
    "answer": "18",
    "solution": "The maximum possible value of $n$ is $18$.\n\nWe first show that $n \\ge 19$ is impossible. Let $A$, $B$, $C$ be the answers in each problem. For $n \\ge 19$, by the pigeonhole principle, we may assume $\\left\\lfloor \\frac{19}{3} \\right\\rfloor = 6$ students answer $A$ in problem 1. Then by the pigeonhole principle, we may assume $\\left\\lfloor \\frac{6}{3} \\right\\rfloor = 2$ students answer $A$ in problem 2. These 2 students must have different answers to the remaining problems. Thus, we may assume their answers are\n\nAAAAAA, ABBBBB, AACCCC.\n\nConsider another student choosing answer $A$ in problem 1. By the pigeonhole principle, this student chooses the same choice in $\\left\\lfloor \\frac{4}{3} \\right\\rfloor = 1$ of the problems among the remaining problems. This leads to a contradiction, as it is not possible to construct such a set of students.\n\nTherefore, we must have $n \\le 18$.\n\nWe now provide an example for $n = 18$. The following lists out the answers given by 18 students:\n\nAAAAAA  AABBCC  ABCACB  ACBCAB  ABACBC  ACCBBA\nBBBBBB  BBCCAA  BCABAC  BACABC  BCBACA  BAACCB\nCCCCCC  CCAABB  CABCBA  CBABCA  CACBAB  CBBAAC\n\nIt is routine to check that any two students in the same column share 0 same answer, while any two students not in the same column share 2 same answers. (Note that there is a cyclic symmetry in each column.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31679,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ be a monic cubic polynomial with $f(0) = -64$ and all roots of $f(x)$ are nonnegative real numbers. What is the largest possible value of $f(-1)$? (A polynomial is *monic* if its leading coefficient is 1.)",
    "options": [],
    "answer": "-125",
    "solution": "The largest possible value of $f(-1)$ is $-125$.\n\nLet $f(x) = (x-a)(x-b)(x-c)$ where $a, b, c \\ge 0$ and $abc = -f(0) = 64$. Then we have\n$$\n\\begin{aligned}\nf(-1) &= - (1+a)(1+b)(1+c) \\\\\n&= -1 - (a+b+c) - (ab+bc+ca) - abc \\\\\n&\\le -1 - 3\\sqrt[3]{abc} - 3\\sqrt[3]{a^2b^2c^2} - abc \\\\\n&= -125.\n\\end{aligned}\n$$\nEquality holds when $a = b = c = 4$, i.e. $f(x) = (x-4)^3$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31680,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $\\angle B > \\angle C$. Let $D$ be the point on side $BC$ such that $\\angle DAC = \\frac{B-C}{2}$. The circumcircle of $\\triangle ACD$ meets side $AB$ again at $E$. The circumcircle of $\\triangle ABD$ meets side $AC$ again at $F$. The internal angle bisector of $\\angle BDE$ meets side $AB$ at $P$. The internal angle bisector of $\\angle CDF$ meets side $AC$ at $Q$. Prove that $PQ$ and $AB$ are perpendicular.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nNote that $\\angle PDQ = 180^\\circ - \\angle BDP - \\angle CDQ = 180^\\circ - \\frac{1}{2} \\angle BDE - \\frac{1}{2} \\angle CDF = 180^\\circ - \\frac{A}{2} - \\frac{A}{2} = 180^\\circ - A$. This implies $A, P, D, Q$ are concyclic. Hence, $\\angle APQ = \\angle ADQ = 180^\\circ - \\angle DAQ - \\angle ACD - \\angle CDQ = 180^\\circ - \\frac{B-C}{2} - C - \\frac{A}{2} = 90^\\circ$. $\\square$\nFrom the concyclic points, we have $\\angle DFC = \\angle DBE$ and $\\angle DCF = \\angle DEB$. These show $\\triangle DFC$ and $\\triangle DBE$ are directly similar. Thus, there exists a spiral similarity with centre $D$ mapping $\\triangle DFC$ to $\\triangle DBE$. Note that the image of $Q$ is $P$ under this transformation since they are the corresponding points on sides $CF$ and $EB$ respectively. It follows that $\\triangle DPQ$ and $\\triangle DBF$ are directly similar. Then $\\angle (PQ, AB) = \\angle (PQ, BF) + \\angle FBA = \\angle (PD, BD) + (B - \\angle CBF) = \\frac{1}{2} \\angle EDB + B - \\angle DAC = \\frac{A}{2} + B - \\frac{B-C}{2} = 90^\\circ$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31681,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the sequences with $2016$ terms formed by the digits $1$, $2$, $3$ and $4$. Find the number of those sequences containing an even number of the digit $1$.",
    "options": [],
    "answer": "2^{2015} + 2^{4031}",
    "solution": "There are $2^{2015} + 2^{4031}$ such sequences.\n\nLet $a_n$ be the number of such sequences with $n$ terms instead of $2016$. Since there are $4^n$ sequences formed by $1$, $2$, $3$, $4$, there are $4^n - a_n$ sequences with an odd number of $1$'s among these.\n\nConsider a sequence with $n$ terms formed by $1$, $2$, $3$, $4$ having an even number of $1$'s. If the last digit is $1$, then there is an odd number of $1$'s in the first $n-1$ digits. There are $4^{n-1} - a_{n-1}$ choices for these digits. If the last digit is $2$, $3$ or $4$, then there is an even number of $1$'s in the first $n-1$ digits. There are $a_{n-1}$ choices for these digits. This gives the recurrence relation\n$$\na_n = 4^{n-1} - a_{n-1} + 3a_{n-1} = 2a_{n-1} + 4^{n-1}.\n$$\nAdding up\n$$\n\\begin{aligned}\na_n &= 2a_{n-1} + 4^{n-1}, \\\\\n2a_{n-1} &= 2^2a_{n-2} + 2 \\cdot 4^{n-2}, \\\\\n\\vdots \\\\\n2^{n-2}a_2 &= 2^{n-1}a_1 + 2^{n-2} \\cdot 4,\n\\end{aligned}\n$$\nwe obtain\n$$\n\\begin{aligned}\na_n &= 2^{n-1}a_1 + \\sum_{k=0}^{n-2} (2^k \\cdot 4^{n-1-k}) = 2^{n-1}(3) + \\sum_{k=0}^{n-2} 2^{2n-2-k} \\\\\n&= 3 \\cdot 2^{n-1} + 2^{2n-1} - 2^n = 2^{n-1} + 2^{2n-1}.\n\\end{aligned}\n$$\nIn particular, $a_{2016} = 2^{2015} + 2^{4031}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31682,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $2017$ points in a plane are given such that no three points are collinear. Among the triangles formed by any three of these $2017$ points, those triangles having the largest area are said to be *good*. Prove that there cannot be more than $2017$ good triangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove the assertion for any positive integer $n$ when $2017$ is replaced by $n$.\n\n**Claim 1.** If the points $X, Y, Z$ lie inside or on the boundary of a polygon $\\mathcal{P}$, then $[XYZ]$ does not exceed the area of one of the triangles formed by the vertices of $\\mathcal{P}$. When equality holds, $X, Y, Z$ must lie on the boundary of $\\mathcal{P}$.\n\n*Proof.* If $X$ lies in the interior of $\\mathcal{P}$, we can move $X$ away from the line $YZ$ in the direction perpendicular to $YZ$ until $X$ lies on the boundary of $\\mathcal{P}$. The area is increased. In the same way, we may assume $X, Y, Z$ lie on the boundary of $\\mathcal{P}$. Next, suppose $X$ lies on the side $PQ$ where $P$ and $Q$ are vertices of $\\mathcal{P}$. WLOG assume the distance from $P$ to $YZ$ is at least that from $X$ to $YZ$. Then $[PYZ] \\ge [XYZ]$ and we can move $X$ to $P$. Similarly, we can move all three points to some vertices of $\\mathcal{P}$. The area is no less than the original $[XYZ]$. $\\square$\n\nLet $\\mathcal{P}$ be the convex hull of the $n$ points. Using claim 1, all points inside $\\mathcal{P}$ cannot form a good triangle. WLOG we may assume the $n$ given points form the convex $n$-gon $\\mathcal{P}$.\n\n**Claim 2.** Let $ABCDE$ be a convex pentagon. If both $\\triangle AWX$ and $\\triangle AYZ$ are good triangles where $\\{W, X, Y, Z\\} = \\{B, C, D, E\\}$, then these triangles must be $\\triangle ABD$ and $\\triangle ACE$.\n\n![](attached_image_1.png)\n\n*Proof.* Suppose $\\triangle ABC$ and $\\triangle ADE$ are good triangles. As $[ABC] \\ge [ABD]$, we have $\\angle B + \\angle C \\le 180^\\circ$. Similarly, as $[ADE] \\ge [ACE]$, we have $\\angle D + \\angle E \\le 180^\\circ$. These yield $\\angle A \\ge 180^\\circ$, which is impossible.\n\nSuppose $\\triangle ABE$ and $\\triangle ACD$ are good triangles. As $[ABE] \\ge [ACE]$, we have $\\angle A + \\angle B \\le 180^\\circ$. Since $[ACD] \\ge [ABD]$, we have $\\angle B + \\angle BAD \\ge 180^\\circ$. This implies $\\angle B + \\angle A > 180^\\circ$, which contradicts the former inequality.\n\nThe only possibility is the one we claim. $\\square$\n\n**Claim 3.** Suppose $\\triangle ABC$ and $\\triangle XYZ$ are good. Construct $\\triangle A'B'C'$ such that $\\triangle ABC$ is the medial triangle of $\\triangle A'B'C'$. Then the points $X, Y, Z$ must lie in the regions II, III, IV as shown and no two of them lie in the same region.\n\n![](attached_image_2.png)\n![](attached_image_3.png)\n\n*Proof.* Firstly, if $X$ lies on different sides of $B'C'$ as $B$, then $[XBC] > [ABC]$. Thus $\\triangle ABC$ cannot be good. In the same way, all of $X, Y, Z$ must lie in $\\triangle A'B'C'$. By our assumption on the $n$ points, none of $X, Y, Z$ can lie inside $\\triangle ABC$, which is region I. If all of $X, Y, Z$ lie in the same region, say II, then $[XYZ] < [ABC'] = [ABC]$. This contradicts $[XYZ] = [ABC]$.\n\nWLOG, it remains to consider the case when $X, Y$ lie in region II and $Z$ lies in region III. WLOG assume $AXYB$ is convex and the points lie in this order. Since $[XYZ] \\ge [XBZ]$, we have $\\angle ZXY + \\angle XYB \\le 180^\\circ$. But this cannot be true. Indeed, let $ZX$ and $XY$ meet the line $C'A'$ at $P$ and $Q$ respectively. In view of the configuration, $P, Q, B$ must lie in that order. We find that\n$$\n\\angle ZXY + \\angle XYB \\ge \\angle ZXQ + \\angle XQB = 180^\\circ + \\angle ZPB > 180^\\circ.\n$$\n\n![](attached_image_4.png)\n\nThis shows no two of $X, Y, Z$ can lie in the same region as desired. $\\square$\n\nSuppose the vertices of $\\mathcal{P}$ are $A_1, A_2, \\dots, A_n$ in that order. Suppose there are $m$ good triangles $A_{a_i}A_{b_i}A_{c_i}$ where $a_i < b_i < c_i$. WLOG, the indices can be rearranged in dictionary order. This means $a_1 \\le a_2 \\le \\dots \\le a_m$. If $a_i = a_{i+1}$, then $b_i \\le b_{i+1}$. If $a_i = a_{i+1}$ and $b_i = b_{i+1}$, then $c_i < c_{i+1}$.\n\n**Claim 4.** We have\n$$\na_1 \\le a_2 \\le \\dots \\le a_m \\le b_1 \\le b_2 \\le \\dots \\le b_m \\le c_1 \\le c_2 \\le \\dots \\le c_m.\n$$\n*Proof.* The relation $a_1 \\le a_2 \\le \\dots \\le a_m$ follows from the construction. Next, to show that $b_1 \\le b_2 \\le \\dots \\le b_m$, we first assume on the contrary that $b_i > b_{i+1}$ for some $i$. Note that $a_i \\ne a_{i+1}$ by construction. Then $A_{a_i}, A_{a_{i+1}}, A_{b_{i+1}}, A_{b_i}$ are distinct and lie in that order. No matter $A_{c_i}$ equals $A_{c_{i+1}}$ or not, claim 2 or claim 3 is violated.\n\nNext, to show $c_1 \\le c_2 \\le \\dots \\le c_m$, we assume on the contrary that $c_i > c_{i+1}$ for some $i$. If $a_i = a_{i+1}$, then we must have $b_i < b_{i+1}$. This contradicts claim 2. If $a_i < a_{i+1}$, no matter $A_{b_i}$ equals $A_{b_{i+1}}$ or not, claim 2 or claim 3 is violated.\n\nIt remains to show $a_m \\le b_1$ and $b_m \\le c_1$. For the former inequality, we first assume $a_m > b_1$. If $c_1$ equals one of $a_m, b_m, c_m$, then claim 2 is violated. If $c_1$ is distinct from $a_m, b_m, c_m$, then claim 3 is violated. Thus, we must have $a_m \\le b_1$. By symmetry (reversing the role of the two triangles), we obtain the latter inequality $b_m \\le c_1$ as well. This completes the proof of claim 4. $\\square$\n\nLet $s_i = a_i + b_i + c_i$. By claim 4, $\\{s_i\\}$ is an increasing sequence. Note that it is strictly increasing since the good triangles are distinct. As\n$$\ns_m - s_1 = c_m + (b_m - c_1) + (a_m - b_1) - a_1 \\le n + 0 + 0 - 1 = n - 1,\n$$\nthe sequence contains at most $n$ terms, meaning that at most $n$ good triangles are possible. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31683,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the first digit after the decimal point of the number $\\frac{1}{1009} + \\frac{1}{1010} + \\dots + \\frac{1}{2016}$.",
    "options": [],
    "answer": "6",
    "solution": "The answer is 6.\nLet $a_n = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n}$. Firstly, we find that\n$$\na_{n+1} - a_n = \\frac{1}{2n+1} + \\frac{1}{2n+2} - \\frac{1}{n+1} = \\frac{1}{(2n+1)(2n+2)} > 0.\n$$\nThis shows the sequence is strictly increasing. Thus, we easily find that\n$$\na_{1008} > a_3 = \\frac{37}{60} > 0.6.\n$$\nSecondly, we can prove by induction that $a_n \\le 0.7 - \\frac{1}{4n}$ for any $n \\ge 3$. The base case $a_3 = \\frac{37}{60} \\le 0.7 - \\frac{1}{12}$ holds. Assuming this holds for some $n = k$, we find that\n$$\na_{k+1} = a_k + \\frac{1}{(2k+1)(2k+2)} \\le 0.7 - \\frac{1}{4k} + \\frac{1}{(2k+1)(2k+2)}.\n$$\nNow,\n$$\n\\begin{align*}\n& 0.7 - \\frac{1}{4k} + \\frac{1}{(2k+1)(2k+2)} \\le 0.7 - \\frac{1}{4(k+1)} \\\\\n\\Leftrightarrow \\quad & \\frac{1}{(2k+1)(2k+2)} \\le \\frac{1}{4k(k+1)} \\\\\n\\Leftrightarrow \\quad & 4k(k+1) \\le (2k+1)(2k+2) \\\\\n\\Leftrightarrow \\quad & 0 \\le 2k+2.\n\\end{align*}\n$$\nThis clearly holds. By induction, we have $a_n \\le 0.7 - \\frac{1}{4n} < 0.7$ for any $n \\ge 3$. In particular, $a_{1008} < 0.7$. Therefore, the first digit of $a_{1008}$ after the decimal point is 6.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31684,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers satisfying $abcd = 1$. Prove that\n$$\n(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \\geq (a+c)(b+d)(ac+bd+2).\n$$\nWhen does equality hold?",
    "options": [],
    "answer": "Equality holds precisely when a equals c and b equals d with a times b equal to one; that is, a equals c equals t and b equals d equals one over t for some positive t.",
    "solution": "By the Cauchy-Schwarz inequality, we have\n$$\n(a^2b + c^2d + ad^2 + cb^2)(a^2d + c^2b + ab^2 + cd^2) \\geq (a^2\\sqrt{bd} + c^2\\sqrt{bd} + abd + CBD)^2 = \\left(\\frac{a^2 + c^2}{\\sqrt{ac}} + abd + CBD\\right)^2.\n$$\nTogether with $a^2 + c^2 \\geq \\frac{1}{2}(a+c)^2 \\geq (a+c)\\sqrt{ac}$, we have\n$$\n(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \\geq (a+c)^2(1+bd)^2.\n$$\nDue to symmetry, we also have\n$$\n(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \\geq (b+d)^2(1+ac)^2.\n$$\nMultiplying these inequalities, we obtain\n$$\n(a^2b + b^2c + c^2d + d^2a)(ab^2 + bc^2 + cd^2 + da^2) \\geq (a+c)(b+d)(1+ac)(1+bd)\n$$\nwhere $(1+ac)(1+bd) = 1+ac+bd+abcd = ac+bd+2$. The result follows readily.\nFor equality in the application of the Cauchy-Schwarz inequality, we need $b = d$ and $a = c$. Note that the other inequalities also have these as equality. Together with $abcd = 1$, equality holds when $a = c = t$ and $b = d = \\frac{1}{t}$ for some $t > 0$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31685,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a_0$, $a_1$, $a_2$, $b_0$, $b_1$, $b_2$ such that\n$a_2 b_2 n^2 + a_1 b_1 n + a_0 b_0$ divides $(a_2^{2017n} + b_2)n^2 + (a_1^{2017n} + b_1)n + (a_0^{2017n} + b_0)$ for any positive integer $n$.",
    "options": [],
    "answer": "a0 = a1 = a2 = b0 = b1 = b2 = 1",
    "solution": "Suppose that\n$$\na_2 b_2 n^2 + a_1 b_1 n + a_0 b_0 \\mid (a_2^{2017n} + b_2)n^2 + (a_1^{2017n} + b_1)n + (a_0^{2017n} + b_0) \\quad (1)\n$$\nWe first claim that $a_0 = b_0 = 1$. Let $n$ be a large multiple of $a_0 b_0$. Since $a_0 b_0$ divides the left-hand side of (1), it also divides the right-hand side and hence $a_0 b_0 \\mid a_0^{2017n} + b_0$. This implies $a_0 \\mid b_0$ and $b_0 \\mid a_0^{2017n}$. This shows $a_0$, $b_0$ have the same prime divisors. Thus, we can take a sufficiently large $n$ such that $a_0 b_0 \\mid a_0^{2017n}$. This gives $a_0 b_0 \\mid b_0$. The only possibility is $a_0 = 1$, which implies $b_0 = 1$.\n\nLet $M$ be a sufficiently large integer and let $n_0$ be the product of all primes less than $M$. Choose any prime divisor $p$ of\n$$\na_2 b_2 n_0^2 + a_1 b_1 n_0 + 1. \\quad (2)\n$$\nClearly, $(p, n_0) = 1$. Thus, $p \\ge M$. In particular, we have $(p, a_1 a_2) = 1$ as $M$ is large.\n\nConsider any $n$ such that $n \\equiv n_0 \\pmod{p}$. Then $p$ divides the left-hand side of (1). This gives\n$$\np \\mid (a_2^{2017n} + b_2)n_0^2 + (a_1^{2017n} + b_1)n_0 + 2. \\quad (3)\n$$\nWe choose $n$ such that $p-1 \\mid n$. Such an $n$ exists by the Chinese remainder theorem. By Fermat's little theorem, we obtain $p \\mid (1+b_2)n_0^2 + (1+b_1)n_0 + 2$. Taking the difference with (3), as $(p, n_0) = 1$, we get\n$$\np \\mid (a_2^{2017n} - 1)n_0 + (a_1^{2017n} - 1) \\quad (4)\n$$\nfor any $n \\equiv n_0 \\pmod{p}$. By taking $n \\equiv 1 \\pmod{p-1}$ and $n \\equiv 2 \\pmod{p-1}$ respectively, we have\n$$\np \\mid (a_2^{2017 \\times 2} - 1)(a_1^{2017} - 1) - (a_1^{2017 \\times 2} - 1)(a_2^{2017} - 1) = (a_1^{2017} - 1)(a_2^{2017} - 1)(a_2^{2017} - a_1^{2017}).\n$$\nSince $p \\ge M$ is sufficiently large, the right-hand side must be $0$. If one of $a_1$, $a_2$ is $1$, then (4) implies both are equal to $1$. If $a_1 = a_2$, then (4) implies $a_1 = a_2 = 1$ or $n_0 \\equiv -1 \\pmod{p}$.\n\nWe first consider the case $a_1 = a_2 = 1$. In that case, (1) becomes $b_2 n^2 + b_1 n + 1 \\mid (1+b_2)n^2 + (1+b_1)n + 2$ so that $b_2 n^2 + b_1 n + 1 \\mid n^2 + n + 1$. Note that $b_2 n^2 + b_1 n + 1 \\ge n^2 + n + 1$. Thus, equality must hold and hence $b_1 = b_2 = 1$. One easily checks that $a_0 = a_1 = a_2 = b_0 = b_1 = b_2 = 1$ is a solution.\n\nNext, it remains to consider the case $n_0 \\equiv -1 \\pmod{p}$. As $p$ divides (2), this yields $p \\mid a_2 b_2 - a_1 b_1 + 1$. Since $p$ is large, we must have $a_2 b_2 - a_1 b_1 + 1 = 0$. Then (2) becomes $(a_1 b_1 - 1)n_0^2 + a_1 b_1 n_0 + 1 = (n_0 + 1)((a_1 b_1 - 1)n_0 + 1)$. Therefore, instead of choosing any prime $p$ dividing (2) at the beginning, we choose such a prime $p$ dividing $(a_1 b_1 - 1)n_0 + 1$. Using the same argument, we obtain either the same solution or $n_0 \\equiv -1 \\pmod{p}$. In the latter case, we find that $p \\mid -(a_1 b_1 - 1) + 1$. Again, this forces $a_1 b_1 = 2$ as $p$ is large. Thus, $(a_1, b_1) = (1, 2)$ or $(2, 1)$. Also, $a_2 b_2 = a_1 b_1 - 1 = 1$ so that $a_2 = b_2 = 1$.\n\nNow, by considering $n = 3$ in (1), we have $16 \\mid 2(3)^2 + (a_1^{2017n} + b_1)(3) + 2$. As $a_1$, $b_1$ have different parities, the right-hand side is odd. This is impossible.\n\nTherefore, the only solution is $a_0 = a_1 = a_2 = b_0 = b_1 = b_2 = 1$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31686,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, a_3, \\dots$ and $b_1, b_2, b_3, \\dots$ be infinite sequences of real numbers satisfying $a_{n+1} + b_{n+1} = \\frac{a_n + b_n}{2}$ and $a_{n+1}b_{n+1} = \\sqrt{a_n b_n}$ for all $n \\ge 1$. Suppose $b_{2016} = 1$ and $a_1 > 0$. Find all possible value(s) of $a_1$.",
    "options": [],
    "answer": "2^{2015}",
    "solution": "$a_1$ can only be $2^{2015}$.\nLet $s_n = a_n + b_n$ and $p_n = a_n b_n$ for all $n \\ge 1$. The relations become $s_{n+1} = \\frac{s_n}{2}$ and $p_{n+1} = \\sqrt{p_n}$. Inductively, we find that $s_n = \\frac{s_1}{2^{n-1}}$ and $p_n = \\sqrt[2^{n-1}]{p_1}$.\nSince $a_n$ and $b_n$ are real roots of $x^2 - s_n x + p_n = 0$, we have $s_n^2 - 4p_n \\ge 0$ for any $n$. This implies\n$$\n\\frac{s_1^2}{2^{2n-2}} - 4 \\sqrt[2^{n-1}]{p_1} \\ge 0.\n$$\nIf $p_1 > 0$, then the left-hand side approaches $0 - 4 = -4 < 0$ when $n$ goes to infinity. This is a contradiction. Thus, we must have $p_1 = 0$. This yields $p_n = 0$ for any $n \\ge 1$, and hence $a_n = 0$ or $b_n = 0$.\n\nAs $b_{2016} = 1$, we need $a_1 = 0$. Using $s_{2015} = 2s_{2016} = 2$ and $p_{2015} = 0$, we easily deduce $\\{a_{2015}, b_{2015}\\} = \\{0, 2\\}$. Similarly, by backward induction, we find that $\\{a_n, b_n\\} = \\{0, 2^{2016-n}\\}$ for $1 \\le n \\le 2016$. In particular, $a_1 = 2^{2015}$ since $a_1 > 0$.\nIt is possible that $a_1 = 2^{2015}$. An example of the sequences is $a_1 = 2^{2015}$, $b_1 = 0$, and $a_n = 0$, $b_n = 2^{2016-n}$ for $n \\ge 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31687,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = AC$. A circle $\\Gamma$ lies outside triangle $ABC$ and is tangent to line $AC$ at $C$. Point $D$ lies on $\\Gamma$ such that the circumcircle of triangle $ABD$ is internally tangent to $\\Gamma$. Segment $AD$ meets $\\Gamma$ again at $E$. Prove that $BE$ is tangent to $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $AB^2 = AC^2 = AE \\times AD$. This implies $\\triangle ABE \\sim \\triangle ADB$. It follows that $\\angle AEB = \\angle ABD$. Let $BD$ meet $\\Gamma$ again at $F$. Consider the homothety with centre $D$ taking $\\Gamma$ to the circumcircle of $\\triangle ABD$. Since $EF$ is mapped to $AB$, the lines $EF$ and $AB$ are parallel. Hence, we have\n$$\n\\angle AEB = \\angle ABD = \\angle EFD = \\angle ECD.\n$$\nThis implies $BE$ is tangent to $\\Gamma$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31688,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that $a$, $b$ and $c$ are positive real numbers such that $ab + bc + ca \\ge 1$, prove that\n$$\n\\frac{1}{a^2} + \\frac{1}{b^2} + \\frac{1}{c^2} \\ge \\frac{\\sqrt{3}}{abc}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the AM-GM inequality, we have\n$$\n\\begin{aligned}\n\\frac{bc}{a} + \\frac{ca}{b} &= c\\left(\\frac{b}{a} + \\frac{a}{b}\\right) \\ge 2c, \\\\\n\\frac{ca}{b} + \\frac{ab}{c} &= a\\left(\\frac{c}{b} + \\frac{b}{c}\\right) \\ge 2a, \\\\\n\\frac{ab}{c} + \\frac{bc}{a} &= b\\left(\\frac{a}{c} + \\frac{c}{a}\\right) \\ge 2b.\n\\end{aligned}\n$$\nAdding the three inequalities, we get\n$$\n\\frac{bc}{a} + \\frac{ca}{b} + \\frac{ab}{c} \\ge a + b + c. \\quad (1)\n$$\nIn addition, we have\n$$\na + b + c = \\sqrt{(a^2 + b^2 + c^2) + 2(ab + bc + ca)} \\ge \\sqrt{3(ab + bc + ca)} \\ge \\sqrt{3}. \\quad (2)\n$$\nCombining (1) and (2), we obtain $\\frac{bc}{a}+\\frac{ca}{b}+\\frac{ab}{c} \\ge \\sqrt{3}$. This yields $\\frac{1}{a^2}+\\frac{1}{b^2}+\\frac{1}{c^2} \\ge \\frac{\\sqrt{3}}{abc}$.\nEquality holds when $a=b=c=\\frac{1}{\\sqrt{3}}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31689,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integer(s) $n$ such that $n^2 + 32n + 8$ is a perfect square.",
    "options": [],
    "answer": "17, 47",
    "solution": "$n$ can be $47$ or $17$.\n\nSuppose $n^2 + 32n + 8 = m^2$ where $m$ is a positive integer. Note that this can be rewritten as $(n+16)^2 - 248 = m^2$, which implies\n$$\n(n + m + 16)(n - m + 16) = 248 = 2^3 \\times 31.\n$$\nHence, $(n+m+16, n-m+16) = (248, 1), (124, 2), (62, 4), (31, 8)$. For each pair, we can solve the system of two linear equations in two unknowns. The only positive integer solutions are $(n, m) = (47, 61), (17, 29)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31690,
    "subject": "Mathematics (Multi-modal)",
    "question": "The altitudes $AD$ and $BE$ of acute triangle $ABC$ intersect at $H$. Let $F$ be the intersection of $AB$ and a line that is parallel to the side $BC$ and goes through the circumcentre of $ABC$. Let $M$ be the midpoint of $AH$. Prove that $\\angle CMF = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ and $N$ be the projection of $F$ and $O$ on $BC$ respectively. Recall that $ON = \\frac{1}{2} AH$. Therefore, $FP = AM = MH$.\nSince $FP$ and $AM$ are perpendicular to $BC$, they are parallel. Thus, $AFPM$ is a parallelogram. This implies $MP \\parallel AF$, and hence $MP \\perp CH$. Also, we have $MH \\perp PC$. Therefore, $H$ is the orthocentre of $\\triangle MPC$. It follows that $PH \\perp CM$.\nNow, as $FP \\parallel MH$ and $FP = MH$, we know that $FPHM$ is another parallelogram. This shows $FM \\parallel PH$, and hence $FM \\perp CM$. In other words,\n$\\angle FMC = 90^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31691,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of nonnegative integers $k$, $0 \\le k \\le 2188$, such that $\\binom{2188}{k}$ is divisible by $2188$.\n(Note that the binomial coefficient is defined by $\\binom{n}{r} = \\frac{n!}{r!(n-r)!}$.)",
    "options": [],
    "answer": "2146",
    "solution": "The answer is $2146$.\n\nNote that $2188 = 4 \\times 547$, where $547$ is a prime. So $2188 = 40_{(547)}$ (base $547$ representation). Let $k = \\overline{ab}_{(547)}$. If $k$ is not divisible by $547$, then $b > 0$. By Lucas' theorem,\n$$\n\\binom{2188}{k} = \\binom{40_{(547)}}{\\overline{ab}_{(547)}} \\equiv \\binom{4}{a}\\binom{0}{b} = 0 \\pmod{547}.\n$$\nThis shows $\\binom{2188}{k}$ is divisible by $547$. If $547 \\mid k$, then $b=0$ and $a \\le 4$. By Lucas' theorem,\n$$\n\\binom{2188}{k} = \\binom{40_{(547)}}{\\overline{a0}_{(547)}} \\equiv \\binom{4}{a}\\binom{0}{0} = \\binom{4}{a} \\not\\equiv 0 \\pmod{547}.\n$$\nThis shows $\\binom{2188}{k}$ is not divisible by $547$.\n\nNext, let $N$ be the highest power of $2$ dividing $\\binom{2188}{k}$. Using binary representation, we have $2188 = 100010001100_{(2)}$. By Kummer's theorem, $N$ is the number of carries when $k$ is added to $2188-k$ in base $2$.\n\nFirstly, $\\binom{2188}{k}$ is odd if and only if $N = 0$. This holds if and only if $a_j$ is $0$ whenever the corresponding digits of $2188$ in binary representation are $0$. In other words, $k = \\overline{a000b000cd00}_{(2)}$. There are $2^4 = 16$ such numbers.\n\nSecondly, $\\binom{2188}{k}$ is even and is not divisible by $4$ if and only if $N = 1$. This holds if and only if $k$ has the form $\\overline{0100a000bc00}_{(2)}$, $\\overline{a0000100bc00}_{(2)}$, $\\overline{a000b000c010}_{(2)}$. There are $2^3 \\times 3 = 24$ such numbers.\n\nFinally, $\\binom{2188}{0} = \\binom{2188}{2188} = 1$ are odd, while $4 \\mid \\binom{2188}{547}, \\binom{2188}{1094}, \\binom{2188}{1641}$ since $547 = \\overline{001000100011}_{(2)}$, $1094 = \\overline{010001000110}_{(2)}$, $1641 = \\overline{011001101001}_{(2)}$. Therefore, the final answer is\n$$\n2189 - 5 - 16 - 24 + 2 = 2146.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31692,
    "subject": "Mathematics (Multi-modal)",
    "question": "For which natural number $n$ is it possible to place natural numbers from $1$ to $3n$ on the edges of a right $n$-angled prism (on each edge there is exactly one number placed and each one is used exactly 1 time) in such a way that the sum of all the numbers that surround each face is the same?",
    "options": [],
    "answer": "4",
    "solution": "The only possible $n$ is $4$.\nWe call those edges which are the sides of a base the *base edges*, and call the remaining edges the *lateral edges*. Let $s$ be the sum of all numbers surrounding each face, and let $x$ be the sum of all numbers on the lateral edges.\n\nFirstly, we consider the sum of all numbers on the base edges. Since they are the sides of the two bases, this is equal to $2s$. Also, they are the numbers not used on the lateral edges, and hence this is equal to $1 + 2 + \\cdots + 3n - x$. This yields\n$$\nx + 2s = \\frac{3n(3n + 1)}{2}. \\qquad (1)\n$$\nSecondly, we consider the sides of the lateral faces. Each base edge is a side of $1$ lateral face, while each lateral edge is a side of $2$ lateral faces. The total sum of numbers on all these edges is $1 + 2 + \\cdots + 3n + x$. These edges together surround $n$ faces, and so the total sum is $ns$. It follows that\n$$\nns - x = \\frac{3n(3n + 1)}{2}. \\qquad (2)\n$$\nBy (1) and (2), we obtain\n$$\ns = \\frac{(x + 2s) + (ns - x)}{n + 2} = \\frac{3n(3n + 1)}{n + 2},$$\n$$x = \\frac{3n(3n + 1)}{2} - 2s = \\frac{3n(3n + 1)(n - 2)}{2(n + 2)}.$$\nAs $s \\in \\mathbb{Z}$, we have $n + 2 \\mid 3n(3n + 1)$. This implies $n + 2 \\mid 30$.\n\nNext, we have the bound $x \\le 3n + (3n - 1) + \\cdots + (2n + 1)$ since there are only $n$ lateral edges. Now,\n$$\n\\begin{align*}\n\\frac{3n(3n + 1)(n - 2)}{2(n + 2)} &\\le \\frac{(5n + 1)n}{2} \\\\\n\\Leftrightarrow \\quad 3(3n + 1)(n - 2) &\\le (5n + 1)(n + 2) \\\\\n\\Leftrightarrow \\quad 2(2n^2 - 13n - 4) &\\le 0.\n\\end{align*}\n$$\nThus, we have $n \\le 6$. Together with $n + 2 \\mid 30$ and the trivial bound $n \\ge 3$, we easily obtain the only possibilities $n = 3, 4$.\n\nFor $n = 3$, we have $s = 18$ and $x = 9$.\n* If the numbers on the lateral edges are $1$, $2$, $6$, then the numbers surrounding the three lateral surfaces are $1$, $2$, $a$, $b$; $1$, $6$, $c$, $d$ and $2$, $6$, $e$, $f$ respectively. Since $a + b = s - 1 - 2 = 15$, we must have $a = 7$ and $b = 8$. But then $c + d = 18 - 1 - 6 = 11$ has no solution with $c, d \\in \\{3, 4, 5, 9\\}$.\n\n* If the numbers on the lateral edges are $1$, $3$, $5$, then the numbers surrounding the three lateral surfaces are $1$, $3$, $a$, $b$; $1$, $5$, $c$, $d$ and $3$, $5$, $e$, $f$ respectively. Since $a + b = 14$, we must have $a = 6$ and $b = 8$. But then $c + d = 12$ has no solution with $c, d \\in \\{2, 4, 7, 9\\}$.\n\n* If the numbers on the lateral edges are $2$, $3$, $4$, then the numbers surrounding the three lateral surfaces are $2$, $3$, $a$, $b$; $2$, $4$, $c$, $d$ and $3$, $4$, $e$, $f$ respectively. Since $a + b = 13$, $c + d = 12$ and $e + f = 11$, none of $a$, $b$, $c$, $d$, $e$, $f$ is equal to $1$, which is impossible.\n\nTherefore, it is not possible to have $n = 3$.\n\nFor $n = 4$, we have $s = 26$ and $x = 26$. An example is given below.\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31693,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are three piles of coins, with $a$, $b$ and $c$ coins respectively, where $a$, $b$, $c \\ge 2015$ are positive integers. The following operations are allowed:\n(1) Choose a pile with an even number of coins and remove all coins from this pile. Add coins to each of the remaining two piles with amount equal to half of that removed; or\n(2) Choose a pile with an odd number of coins and at least $2017$ coins. Remove $2017$ coins from this pile. Add $1009$ coins to each of the remaining two piles.\nSuppose there are sufficiently many spare coins. Find all ordered triples $(a, b, c)$ such that after some finite sequence of allowed operations, there exists a pile with at least $2017^{2017}$ coins.",
    "options": [],
    "answer": "All integer triples (a, b, c) with a, b, c ≥ 2015 except (2015, 2015, 2015).",
    "solution": "$(a, b, c)$ can be any integer triple with $a$, $b$, $c \\ge 2015$ except $(2015, 2015, 2015)$.\nIt is obvious that no operation can be carried out in the exceptional case. For the other cases, we claim that it is always possible to increase the total number of coins in some operations. If this is true, then the total number of coins will exceed $3 \\times 2017^{2017}$ after some operations, and hence there is a pile with at least $2017^{2017}$ coins.\n\nConsider any integer triple $(a, b, c)$ with $S = a + b + c \\ge 6046$ (which is $3 \\times 2015 + 1$).\nNote that operation (1) does not change the sum $S$, while operation (2) always increases $S$ by $1$. Since we always have $\\lfloor \\frac{S}{3} \\rfloor > 2015$, there must be a pile with at least $2016$ coins after each operation, and hence one of the operations can be carried out.\n\nSuppose on the contrary that there exists a case with $S \\ge 6046$ such that $S$ can no longer be increased. Thus, only operation (1) is possible, which means the number of coins in each pile is always even or at most $2015$. Applying operation (1) if necessary, we may assume there is an empty pile. Afterwards, we apply operation (1) repeatedly to the nonempty pile with the smallest number of coins until this is not possible. The number of coins of the nonempty pile with the smallest number of coins is halved in each operation. Therefore, it must become an odd number after some steps.\n\nWe are now in a situation where the number of coins are $0$, $x$, $y$ respectively, where $x$ is odd and $0 < x \\le 2015 < y$. As $x + y \\ge 6046$, we deduce $y \\ge 4031$. Note that $y$ must be even since otherwise no more operation (1) can be carried out. Hence, we have $y \\ge 4032$. Next, we apply operation (1) on the pile with $y$ coins. The number of coins becomes $0$, $\\frac{y}{2}$, $x + \\frac{y}{2}$. Each of the nonempty piles contains at least $\\frac{y}{2} \\ge 2016$ coins. As $S = x + y$ is odd, one of these piles has an odd number of coins. This contradicts our assumption that there cannot be an odd pile with more than $2016$ coins.\n\nTherefore, the total number of coins can always be increased. The goal can be attained in a finite number of operations whenever the total number of coins is at least $6046$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31694,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given triangle $ABC$, let $D$ be an inner point of the segment $BC$. Let $P$ and $Q$ be distinct inner points of the segment $AD$. Let $K = BP \\cap AC$, $L = CP \\cap AB$, $E = BQ \\cap AC$, $F = CQ \\cap AB$. Given that $KL \\parallel EF$, find all possible values of the ratio $BD : DC$.",
    "options": [],
    "answer": "1",
    "solution": "The only possible value is $1$.\nWe use projective geometry. Consider the following projection.\n$$\nAB(A, L, F, B) \\xrightarrow{\\{C\\}} AD(A, P, Q, D) \\xrightarrow{\\{B\\}} AC(A, K, E, C).\n$$\nThis shows $(A, L, F, B)$ and $(A, K, E, C)$ are perspective, and so $LK$, $FE$, $BC$ are concurrent. Since $KL \\parallel EF$, this shows $KL \\parallel EF \\parallel CB$.\nNow, since $AD$, $BK$, $CL$ are concurrent, $KL \\cap CB$ is the harmonic conjugate of $D$ with respect to $BC$. As $KL \\cap CB$ is a point at infinity, $D$ must be the midpoint of $BC$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31695,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $f$ such that $f$ has nonnegative integer coefficients, $f(1) = 7$ and $f(2) = 2017$.",
    "options": [],
    "answer": "x^10 + x^9 + x^8 + x^7 + x^6 + x^5 + 1",
    "solution": "The only polynomial is $x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + 1$.\n\nSuppose one of the coefficients of $f_0(x) = f(x)$, say the coefficient of $x^k$, is at least $2$. Define $f_1(x) = f_0(x) - 2x^k + x^{k+1}$. Then $f_1(2) = f_0(2) = 2017$ and $f_1(1) < f_0(1) = 7$. Also, the coefficients of $f_1$ are nonnegative integers. Repeating the same process, since the sequence $f_0(1), f_1(1), \\dots$ of nonnegative integers is strictly decreasing, the process must end at some step. Then we obtain a polynomial $g(x)$ with coefficients $0$ or $1$ such that $g(2) = 2017$ and $g(1) \\le 7$.\n\nSince the binary representation of $2017$ is $1111100001_{(2)}$, we know that $g$ is uniquely determined, and is given by\n$$\ng(x) = x^{10} + x^9 + x^8 + x^7 + x^6 + x^5 + 1.\n$$\nAs $g(1) = 7$, the equality of $g(1) \\le 7$ should hold. This means we do not need to carry out any process, and $f$ must be the same as $g$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31696,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are $1200$ students. Each student must join exactly $k$ clubs. Given that there is a common club joined by every $23$ students, but there is no common club joined by all $1200$ students, find the smallest possible value of $k$.",
    "options": [],
    "answer": "23",
    "solution": "The answer is $k = 23$.\n\nWe first show $k \\le 23$. We list the students as $S_1, S_2, \\dots, S_{1200}$ and the clubs as $C_1, C_2, \\dots, C_{24}$. Consider the following construction. For $1 \\le j \\le 24$, student $S_j$ joins clubs $C_1, \\dots, C_{j-1}, C_{j+1}, \\dots, C_{24}$. For $25 \\le j \\le 1200$, student $S_j$ joins the same clubs as $S_1$. Then we can see every $23$ students have a common club, but all $1200$ students do not have a common club.\n\nNow we show $k \\ge 23$. Suppose a student $S$ joins clubs $C_1, C_2, \\dots, C_k$. Because all $1200$ students do not join a common club, for each $1 \\le j \\le k$, there is a student $S_j$ not joining the club $C_j$. Hence the students $S, S_1, S_2, \\dots, S_k$ do not have a common club, forcing $k \\ge 23$. The proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31697,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle of $\\triangle ABC$, with incentre $I$, meets $BC$, $CA$ and $AB$ at $D$, $E$ and $F$ respectively. The line $EF$ cuts the lines $BI$, $CI$, $BC$ and $DI$ at points $K$, $L$, $M$ and $Q$ respectively. The line through the midpoint of $CL$ and $M$ meets $CK$ at $P$.\n\na. Determine $\\angle BKC$.\n\nb. Show that the lines $PQ$ and $CL$ are parallel.",
    "options": [],
    "answer": "∠BKC = 90° and PQ ∥ CL",
    "solution": "a. Since $\\angle FKI = \\angle FKB = \\angle EFA - \\angle KBF = 90^\\circ - \\frac{A}{2} - \\frac{B}{2} = \\frac{C}{2} = \\angle ECI$, the points $C$, $E$, $K$, $I$ are concyclic. Note that $C$, $E$, $I$, $D$ are also concyclic. This shows $C$, $E$, $K$, $I$, $D$ lie on the same circle. It follows that\n$$\n\\angle BKC = \\angle IKC = \\angle IEC = 90^\\circ.\n$$\n\n![](attached_image_1.png)\n\nb. Let the midpoint of $CL$ be $J$. Applying Menelaus' theorem using the line $MPJ$ and $\\triangle CKL$, we get $\\frac{KP}{PC} \\cdot \\frac{CJ}{JL} \\cdot \\frac{LM}{MK} = 1$. Using $CJ = JL$, this simplifies\n\nto\n$$\n\\frac{KP}{PC} = \\frac{KM}{ML}. \\qquad (1)\n$$\nNow, note that $B$, $D$, $I$, $F$, $L$ are concyclic and $\\angle BLC = 90^\\circ$ as similar to the proof of part (a). Since $\\angle BLC = \\angle BKC = 90^\\circ$, we know that $B$, $C$, $K$, $L$ are concyclic. Hence,\n$$\n\\angle QDL = \\angle IDL = \\angle IBL = \\angle KBL = \\angle KCL = \\angle KCI = \\angle KDI = \\angle KDQ.\n$$\nTogether with $DQ \\perp DM$, the lines $DQ$ and $DM$ are the internal and external angle bisectors of $\\angle KDL$ respectively. So we get\n$$\n\\frac{KQ}{QL} = \\frac{KD}{DL} = \\frac{KM}{ML}. \\qquad (2)\n$$\nCombining (1) and (2), we obtain $\\frac{KP}{PC} = \\frac{KQ}{QL}$, which yields $PQ // CL$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31698,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find infinitely many positive integers $m$ such that for each such $m$, the number $\\frac{2^{m-1}-1}{8191m}$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By Dirichlet's theorem, there are infinitely many primes $p$ of the form $13k + 1$. We claim that we can set $m = p$ whenever $p > 8191$.\n\nFirstly, $p$ divides $2^{p-1} - 1$ by the Fermat little theorem. Secondly, we have\n$$\n2^{p-1} - 1 = 2^{13k} - 1 = (2^{13} - 1)(2^{13(k-1)} + 2^{13(k-2)} + \\dots + 1),\n$$\nwhich is divisible by $2^{13} - 1 = 8191$. As $p$ and $8191$ are relatively prime, the number $\\frac{2^{p-1}-1}{8191p}$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31699,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\ge 3$ with the following property: there exist $n$ distinct points on the plane such that each point is the circumcentre of a triangle formed by 3 of the points.",
    "options": [],
    "answer": "all integers n ≥ 6",
    "solution": "The answer is any integer $n \\ge 6$.\nWhen $n = 6$, consider two equilateral triangles of side lengths 1 having parallel sides and the same orientation such that each pair of corresponding vertices is at a distance 1 apart. It is clear that each point has a distance 1 to three other points, so that it is the circumcentre of some triangle.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nAssume there are $n$ distinct points such that each point is the circumcentre of some triangle. Consider the convex hull of these $n$ points. Let $AB$ be a side of the convex hull. We may assume there is no point on the segment $AB$. Let $C$ be a point such that $\\angle ACB$ is the largest possible. Note that $\\angle ACB > 0^\\circ$ since it is not possible that all $n$ points are collinear.\nIf $\\angle ACB \\ge 90^\\circ$, then the circumcentre $O$ of $\\triangle ABC$ lies on $AB$ or on a different side to the line $AB$ as $C$. In both cases, $O$ is distinct from all these points. We can take $O$ as the $(n+1)$st point so that each point is the circumcentre of some triangle.\n\nIf $\\angle ACB < 90^\\circ$, then the circumcentre $O$ of $\\triangle ABC$ satisfies\n$$\n\\angle AOB = 2\\angle ACB > \\angle ACB.\n$$\nFrom the choice of $C$, the point $O$ must be distinct from all the points. Again, we can take $O$ as the $(n+1)$st point.\nBy induction, we find that the property is satisfied for any $n \\ge 6$.\nIt remains to show that $n = 5$ does not satisfy the property. If this is true, then both $n = 3$ and $n = 4$ do not satisfy the property from the above inductive step.\nLet the points be $A, B, C, D, E$. WLOG assume $A$ is the circumcentre of $\\triangle BCD$. One of the following must occur.\n* If $B$ is the circumcentre of $\\triangle ACD$, then $AC = AD = AB = BC = BD$. So both $\\triangle ABC$ and $\\triangle ABD$ are equilateral triangles. Note that at least one of $A, B$ must lie on a circle with centre $C$ that contains 3 points. So $E$ lies on the circle with centre $C$ that passes through $A$ and $B$. Similarly, $E$ lies on the circle with centre $D$ that passes through $A$ and $B$. This is a contradiction since the two circles only meet at $A$ and $B$.\n\n![](attached_image_3.png)\n\n* If $B$ is the circumcentre of $\\triangle ACE$, then $AC = AD = AB = BC = BE$. So $\\triangle ABC$ is an equilateral triangle. Thus, there must be 3 points lying on the circle with centre $C$ that passes through $A$ and $B$. In either case $CA = CB = CD$ or $CA = CB = CE$, we get the same configuration as above and hence it is a contradiction.\n* If $B$ is the circumcentre of $\\triangle CDE$, we may also assume $C$ is the circum-centre of $\\triangle BDE$ since otherwise it is included in the above two cases by symmetry. Then this is exactly the same as the first case. Again, this yields a contradiction.\nTherefore, $n = 5$ does not satisfy the property. The result then follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31700,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that $\\{a_n\\}$ and $\\{b_n\\}$ are two sequences of integers defined by\n$$\na_1 = 1,\\ a_2 = 10,\\ a_{n+1} = 2a_n + 3a_{n-1} \\quad \\text{for } n = 2, 3, 4, \\dots,\n$$\n$$\nb_1 = 1,\\ b_2 = 8,\\ b_{n+1} = 3b_n + 4b_{n-1} \\quad \\text{for } n = 2, 3, 4, \\dots\n$$\nProve that, besides the number ‘1’, no two numbers in the sequences are identical.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The two sequences are $a_n = 1, 10, 23, 76, \\dots$ and $b_n = 1, 8, 28, 116, \\dots$. Considering modulo $9$, we have\n$$\na_n \\equiv 1, 1, 5, 4, 5, 4, \\dots \\pmod{9},\n$$\n$$\nb_n \\equiv 1, 8, 1, 8, \\dots \\pmod{9}.\n$$\nSince each term of the two sequences only depends on the two previous terms, we can show by induction that $a_n \\equiv 4, 5 \\pmod{9}$ for $n \\ge 3$ and $b_n \\equiv 1, 8 \\pmod{9}$ for all $n$. Therefore, $a_m \\ne b_n$ whenever $m \\ge 3$.\n\nClearly the two sequences are strictly increasing. Thus, it is easy to see that the number $10$ does not appear in the second sequence. Therefore, the only common number appearing in both sequences is $a_1 = b_1 = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31701,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a real number. Suppose the function $f(x) = \\frac{a}{x-1} + \\frac{1}{x-2} + \\frac{1}{x-6}$ defined in the interval $3 < x < 5$ attains its maximum at $x = 4$. Find the value of $a$.",
    "options": [],
    "answer": "-9/2",
    "solution": "Solution:\n\nThe answer is $-\\frac{9}{2}$.\nThe condition means $\\frac{a}{x-1} + \\frac{1}{x-2} + \\frac{1}{x-6} \\le \\frac{a}{3}$ for $3 < x < 5$. Now, noting that $x-1 > 0$, $x-2 > 0$ and $x-6 < 0$, we have\n$$\n\\frac{a}{x-1} + \\frac{1}{x-2} + \\frac{1}{x-6} \\le \\frac{a}{3}\n$$\n$$\n\\Leftrightarrow 3a(x-2)(x-6) + 3(x-1)(x-6) + 3(x-1)(x-2) \\ge a(x-1)(x-2)(x-6)\n$$\n$$\n\\Leftrightarrow ax^3 - (12a + 6)x^2 + (44a + 30)x - (48a + 24) \\le 0\n$$\n$$\n\\Leftrightarrow (x-4)(ax^2 - (8a+6)x + (12a+6)) \\le 0.\n$$\nAs $x = 4$ is the maximum, there must be another linear factor $x-4$ coming from the quadratic factor. It follows from the factor theorem that $-4a - 18 = 0$, i.e. $a = -\\frac{9}{2}$.\nWhen $a = -\\frac{9}{2}$, the above inequality becomes $(x-4)^2(-\\frac{9}{2}x + 12) \\le 0$, which is true for $3 < x < 5$. Therefore, $x = 4$ is indeed the maximum.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31702,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all sequences $p_1, p_2, p_3, \\dots$ of prime numbers for which there exists an integer $k$ such that the recurrence relation\n$$\np_{n+2} = p_{n+1} + p_n + k\n$$\nholds for all positive integers $n$.",
    "options": [],
    "answer": "All constant sequences p, p, p, … where p is prime, with k = −p.",
    "solution": "The sequence can be any constant sequence $p, p, p, \\dots$ where $p$ is a prime.\n\nThe recurrence relation can be rewritten as\n$$\np_{n+2} + k = (p_{n+1} + k) + (p_n + k).\n$$\nSince the characteristic equation $\\lambda^2 - \\lambda - 1 = 0$ has roots $\\frac{1 \\pm \\sqrt{5}}{2}$, we have\n$$\np_n = A \\left( \\frac{1 + \\sqrt{5}}{2} \\right)^n + B \\left( \\frac{1 - \\sqrt{5}}{2} \\right)^n - k,\n$$\nfor some constants $A, B$.\n\nFurthermore, as there are at most $p_1^2$ possibilities for the remainders of the pair $(p_{n+1}, p_n)$ modulo $p_1$, and the remainders of the subsequent terms are uniquely determined by these, the sequence must be periodic modulo $p_1$. Thus, there are infinitely many terms which are congruent to $p_1$ modulo $p_1$. Since they are primes, they must be $p_1$.\nIf $A \\neq 0$, the term $\\left(\\frac{1+\\sqrt{5}}{2}\\right)^n$ dominates the expression for $p_n$, and so the sequence is eventually strictly increasing (if $A > 0$) or decreasing (if $A < 0$). This contradicts our observation. Thus, we must have $A = 0$. Similarly, if $B \\neq 0$, the term $\\left(\\frac{1-\\sqrt{5}}{2}\\right)^n$ now dominates the expression for $p_n$, which again implies $p_n \\neq p_1$ for all sufficiently large $n$. Thus, we must have $B = 0$. It follows that $p_n = -k$ is a constant.\nConversely, if the sequence is $p, p, p, \\dots$, where $p$ is a prime, we can take $k = -p$ so that the recurrence relation holds.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31703,
    "subject": "Mathematics (Multi-modal)",
    "question": "A circle is circumscribed around an isosceles triangle whose two base angles are equal to $x^\\circ$. Two points are chosen independently and randomly on the circle, and a chord is drawn between them. The probability that the chord intersects the triangle is $\\frac{14}{25}$. Find the sum of the largest and smallest possible value of $x$.",
    "options": [],
    "answer": "120",
    "solution": "The answer is $120$.\n\nThe probability that the chord does not intersect the triangle is $\\frac{11}{25}$. The only way this can happen is when the two points are chosen on the same arc between two of the triangle vertices. The probability that a point is chosen on one of the arcs opposite to one of the base angles is $\\frac{x}{180}$ and the probability that a point is chosen on the arc between the two base angles is $\\frac{180 - 2x}{180}$. Therefore we have\n$$\n2\\left(\\frac{x}{180}\\right)^2 + \\left(\\frac{180 - 2x}{180}\\right)^2 = \\frac{11}{25}.\n$$\nThis simplifies to $x^2 - 120x + 3024 = 0$, and can be factorized as $(x-84)(x-36) = 0$. Hence the sum of roots is $84 + 36 = 120$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31704,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S = \\{1, 2, \\dots, 100\\}$. Consider a partition of $S$ into $S_1, S_2, \\dots, S_n$ for some $n$, i.e. $S_i$ are nonempty, pairwise disjoint and $S = \\bigcup_{i=1}^n S_i$. Let $a_i$ be the average of elements of the set $S_i$. Define the score of this partition by\n$$\n\\frac{a_1 + a_2 + \\dots + a_n}{n}.\n$$\nAmong all $n$ and partitions of $S$, determine the minimum possible score.",
    "options": [],
    "answer": "10",
    "solution": "The minimum possible score is $10$.\nSuppose we have a partition $S_1, S_2, \\dots, S_n$ with score $x$.\n\n**Claim 1.** If there are at least two subsets, say $S_1, S_2$, both with averages at least $x$, then merging these two subsets into one subset will not increase the score.\n*Proof.* Let $a'$ be the new average of the merged set. Note that $a' \\le \\max\\{a_1, a_2\\}$. Together with $x \\le a_1, a_2$, we obtain $a' + x \\le a_1 + a_2$. It follows that\n$$\n\\frac{a' + a_3 + a_4 + \\dots + a_n}{n-1} \\le \\frac{-x + a_1 + a_2 + a_3 + \\dots + a_n}{n-1} = \\frac{-x + nx}{n-1} = x.\n$$\nThis means the new score is at most $x$.\n\n**Claim 2.** If there is a subset, say $S_1$, with average at most $x$ and having at least two elements, then it is possible to split it into two subsets so that the score does not increase.\n*Proof.* Let $m$ be the smallest element in $S_1$. We split $S_1$ into $\\{m\\}$ and $S_1 - \\{m\\}$. The new averages of these two sets are $m$ and $\\frac{|S_1|a_1 - m}{|S_1| - 1}$ respectively. Now, we have\n$$\n\\begin{align*}\n\\frac{1}{n+1} \\left( m + \\frac{|S_1|a_1 - m}{|S_1| - 1} + a_2 + a_3 + \\dots + a_n \\right) \\\\\n&= \\frac{1}{n+1} \\left( m + \\frac{a_1 - m}{|S_1| - 1} + a_1 + a_2 + \\dots + a_n \\right) \\\\\n&\\le \\frac{m + (a_1 - m) + nx}{n+1} \\le \\frac{x + nx}{n+1} = x.\n\\end{align*}\n$$\nThis means the new score is at most $x$.\n\nUsing the above claims, we merge any two subsets with scores above average and subdivide any subset with score below average into singletons. Also, if there are subsets $\\{a\\}$ and $B \\cup \\{b\\}$ for which $a \\ge b$, then we can swap $a$ and $b$ to get the subsets $\\{b\\}$ and $B \\cup \\{a\\}$ so that the average decreases, since\n$$\na + \\frac{s+b}{|B|+1} > b + \\frac{s+a}{|B|+1}\n$$\nif $|B| \\ge 1$.\nTherefore, the optimal case must be of the form\n$$\n\\{1\\}, \\{2\\}, \\{3\\}, \\dots, \\{k\\}, \\{k+1, k+2, \\dots, 100\\}\n$$\nfor some $k \\le 100$. The score is\n$$\n\\frac{1}{k+1} \\left( \\frac{k(k+1)}{2} + \\frac{k+101}{2} \\right) = \\frac{k+1}{2} + \\frac{100}{2(k+1)} \\ge 2 \\sqrt{\\frac{k+1}{2} \\cdot \\frac{100}{2(k+1)}} = 10\n$$\nby the AM-GM inequality. In other words, the minimum score is $10$, which is attained when $k = 9$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31705,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 3$ be an integer. Prove that there exists a set of $2n$ positive integers satisfying the following property: For every $m = 2, 3, \\dots, n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of the subsets of cardinality $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31706,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Gamma_1$ and $\\Gamma_2$ be two circles with different radii, with $\\Gamma_1$ the smaller one. The two circles meet at distinct points $A$ and $B$. $C$ and $D$ are two points on the circles $\\Gamma_1$ and $\\Gamma_2$ respectively, and such that $A$ is the midpoint of the segment $CD$. $CB$ is extended to meet the circle $\\Gamma_2$ at $F$, while $DB$ is extended to meet the circle $\\Gamma_1$ at $E$. The perpendicular bisector of $CD$ and the perpendicular bisector of $EF$ meet at $P$.\n\na. Prove that $\\angle EPF = 2\\angle CAE$.\n\nb. Prove that $AP^2 = CA^2 + PE^2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31707,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an isosceles triangle with $AB = AC$. The incircle $\\Gamma$ of $\\triangle ABC$ has centre $I$, and it is tangent to the sides $AB$ and $AC$ at $F$ and $E$ respectively. Let $\\Omega$ be the circumcircle of $\\triangle AFE$. The two external common tangents of $\\Gamma$ and $\\Omega$ intersect at a point $P$. If one of these external common tangents is parallel to $AC$, prove that $\\angle PBI = 90^\\circ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle AFI = \\angle AEI = 90^\\circ$, the point $I$ lies on $\\Omega$ and the centre of $\\Omega$ is the midpoint $O$ of $AI$. Note that $P$ lies on the line joining the centres of $\\Gamma$ and $\\Omega$. In other words, $P$ lies on the line $IO$. As $AB = AC$, the line $AI$ is simply the internal angle bisector of $\\angle BAC$. In order to show that $BP$ is perpendicular to the angle bisector $BI$ of $\\angle CBA$, it is the same as proving $P$ is the excentre of $\\triangle ABC$ opposite to the vertex $A$.\n\n![](attached_image_1.png)\n\nLet $T$ be the point on $\\Gamma$ such that $PT$ is tangent to $\\Gamma$ and $TP//AC$. Since the tangents at $T$ and $E$ to $\\Gamma$ are parallel, $TE$ is a diameter of $\\Gamma$. This shows $I$ is the midpoint of $TE$. It follows from $AE//PT$ that $\\triangle AIE \\cong \\triangle PIT$. Thus, $AI = PI$. Together with $AO = OI$, we find that $AO : OI : IP = 1 : 1 : 2$.\n\nNote that $P$ is the centre of homothety between $\\Gamma$ and $\\Omega$. Therefore, $\\frac{PI}{PO}$ is the ratio of similitude, i.e. the ratio $\\frac{2IE}{AI}$ of the diameters of the two circles. Thus, we have $\\frac{2IE}{AI} = \\frac{2}{3}$.\n\nLet $D$ be the midpoint of $BC$. As $AB = AC$, $D$ is also the contact point of $\\Gamma$ with $BC$. From $\\angle AEI = \\angle ADC = 90^\\circ$, we obtain $\\triangle AEI \\sim \\triangle ADC$. Therefore, $\\frac{CD}{AC} = \\frac{IE}{AI} = \\frac{1}{3}$. Let $AB = AC = 3k$ so that $BD = CD = k$.\n\nLet $P'$ be the $A$-excentre of $\\triangle ABC$. Our goal is to show that $AI = IP'$, so that $P' = P$ as desired. Firstly, since $AD = \\sqrt{(3k)^2 - k^2} = 2\\sqrt{2}k$ and $\\frac{AI}{ID} = \\frac{AB}{BD} = 3$ by the angle bisector theorem, we have\n$$\nAI = 2\\sqrt{2}k \\times \\frac{3}{3+1} = \\frac{3\\sqrt{2}}{2}k.\n$$\nSecondly, we have $\\frac{AP'}{P'D} = \\frac{AB}{BD} = 3$ by the external angle bisector theorem. It follows that\n$$\nAP' = 2\\sqrt{2}k \\times \\frac{3}{3-1} = 3\\sqrt{2}k = 2AI.\n$$\nTherefore, $AI = IP'$, and we are done.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31708,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the maximal value of\n$$\nS = \\sqrt[3]{\\frac{a}{b+7}} + \\sqrt[3]{\\frac{b}{c+7}} + \\sqrt[3]{\\frac{c}{d+7}} + \\sqrt[3]{\\frac{d}{a+7}},\n$$\nwhere $a$, $b$, $c$, $d$ are nonnegative real numbers which satisfy $a + b + c + d = 100$.",
    "options": [],
    "answer": "49^(1/3) + 2 / 56^(1/3)",
    "solution": "(IMO Shortlist 2018 A7) See the official solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31709,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find an integral solution of the equation\n$$\n\\left\\lfloor \\frac{x}{1!} \\right\\rfloor + \\left\\lfloor \\frac{x}{2!} \\right\\rfloor + \\left\\lfloor \\frac{x}{3!} \\right\\rfloor + \\cdots + \\left\\lfloor \\frac{x}{10!} \\right\\rfloor = 2019.\n$$\n(Note $\\lfloor u \\rfloor$ stands for the greatest integer less than or equal to $u$.)",
    "options": [],
    "answer": "1176",
    "solution": "The answer is $2019$.\nClearly, the answer should not be an integer greater than $2018$. Hence we may remove the last $4$ terms on the left-hand side of the equation.\n\nNext, we may assume that $x = a(6!) + b(5!) + c(4!) + d(3!) + e(2!) + f(1!)$, where $a, b, c, d, e, f$ are nonnegative integers and $a \\le 6, b \\le 5, c \\le 4, d \\le 3, e \\le 2, f \\le 1$.\nSubstituting into the original equation gives\n$$\n1237a + 206b + 41c + 10d + 3e + f = 2019.\n$$\nSince $206b + 41c + 10d + 3e + f \\le 1231$, we have $788 \\le 1237a \\le 2019$. This gives $a = 1$. Then we have\n$$\n206b + 41c + 10d + 3e + f = 782.\n$$\nSimilarly, we find that $b = 3, c = 4, d = e = f = 0$. Hence,\n$$\nx = 1(6!) + 3(5!) + 4(4!) = 1176\n$$\nis the only solution.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31710,
    "subject": "Mathematics (Multi-modal)",
    "question": "We choose $100$ points in the coordinate plane. Let $N$ be the number of triples $(A, B, C)$ of distinct chosen points such that $A$ and $B$ have the same $y$-coordinate, and $B$ and $C$ have the same $x$-coordinate. Find the greatest possible value that $N$ can attain by considering all possible ways to choose the points.",
    "options": [],
    "answer": "8100",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31711,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\ge 3$ cities in a country and between any two cities $A$ and $B$, there is either a one way road from $A$ to $B$, or a one way road from $B$ to $A$ (but never both). Assume the roads are built such that it is possible to get from any city to any other city through these roads, and define $d(A, B)$ to be the minimum number of roads you must go through to go from city $A$ to $B$. Consider all possible ways to build the roads. Find the minimum possible average value of $d(A, B)$ over all possible ordered pairs of distinct cities in the country.",
    "options": [],
    "answer": "Minimum average directed distance equals 3/2 for all n ≥ 3 with n ≠ 4, and equals 19/12 for n = 4.",
    "solution": "The answer is $\\frac{3}{2}$ for $n \\neq 4$ and $\\frac{19}{12}$ for $n = 4$.\n\nNote that for any distinct cities $A$ and $B$, exactly one of $d(A, B)$ and $d(B, A)$ is $1$, while the other is at least $2$. Thus it follows that $d(A, B) + d(B, A) \\geq 3$. The average for this pair is at least $\\frac{3}{2}$. Therefore, by considering all pairs of distinct cities, the average must be at least $\\frac{3}{2}$. Equality holds if and only if $d(A, B) + d(B, A) = 3$ for all pairs of cities $A$ and $B$.\n\nNow we show that this average is attainable for $n \\neq 4$ by induction.\n\nFirst of all, when $n = 3$, the average is attainable if there is a road from $A$ to $B$, $B$ to $C$, $C$ to $A$.\n\nFor $n = 6$, consider the map given by the following table:\n\n|   | A | B | C | D | E | F |\n|---|---|---|---|---|---|---|\n| A | 0 | 1 | 1 | 1 | 0 | 0 |\n| B | 0 | 0 | 1 | 0 | 1 | 1 |\n| C | 0 | 0 | 0 | 1 | 1 | 1 |\n| D | 0 | 1 | 0 | 0 | 0 | 1 |\n| E | 1 | 0 | 0 | 1 | 0 | 0 |\n| F | 1 | 0 | 0 | 0 | 1 | 0 |\n\n(Here a ‘1’ in the $AB$-entry indicates a road from $A$ to $B$, etc., while a ‘0’ in the $AE$-entry indicates no direct road from $A$ to $E$, etc.). Using the table, we get the distance table:\n\n|   | A | B | C | D | E | F |\n|---|---|---|---|---|---|---|\n| A | 0 | 1 | 1 | 1 | 2 | 2 |\n| B | 2 | 0 | 1 | 2 | 1 | 1 |\n| C | 2 | 2 | 0 | 1 | 1 | 1 |\n| D | 2 | 1 | 2 | 0 | 2 | 1 |\n| E | 1 | 2 | 2 | 1 | 0 | 2 |\n| F | 1 | 2 | 2 | 2 | 1 | 0 |\n\nSo $d(X, Y) + d(Y, X) = 3$ for each pair of distinct cities $X$ and $Y$.\n\nAssume when $n = k$, the average $\\frac{3}{2}$ is attainable. Now, consider the case $n = k+2$. Label the $k+2$ cities by $A_1, A_2, \\dots, A_n, X, Y$. By our assumption, there exists a way to build roads between $A_1, A_2, \\dots, A_n$ such that $d(A_i, A_j) + d(A_j, A_i) = 3$ for all $1 \\le i < j \\le k$. We build a road from $A_i$ to $X$ for all $i$, a road from $Y$ to $A_i$ for all $i$, and a road from $X$ to $Y$. Thus,\n$$\n\\begin{aligned}\nd(X, Y) &= 1, \\\\\nd(Y, X) &= 2 \\quad (Y \\to A_i \\to X), \\\\\nd(A_i, X) &= 1, \\\\\nd(X, A_i) &= 2 \\quad (X \\to Y \\to A_i), \\\\\nd(A_i, Y) &= 2 \\quad (A_i \\to X \\to Y), \\\\\nd(Y, A_i) &= 1.\n\\end{aligned}\n$$\nTherefore, we see this road map produces an average distance of $\\frac{3}{2}$ between any two cities.\n\nBy induction, the minimum average is attainable for all $n \\ge 3$ except $n = 4$.\n\nWe first show that the average of $\\frac{3}{2}$ is unattainable. Suppose on the contrary that this average is attainable. Let the four cities be $A_1, A_2, A_3, A_4$. Without loss of generality, assume there is a road from $A_1$ to $A_2$ and a road from $A_1$ to $A_3$. Since the distance from $A_2$ to $A_1$ is $2$, there must be a road from $A_2$ to $A_4$, and a road from $A_4$ to $A_1$. Similarly, there must be a road from $A_3$ to $A_4$. We may assume there is a road from $A_2$ to $A_3$. But then the distance from $A_3$ to $A_2$ is $3$. This is a contradiction.\n\nThe next smallest possible average is $\\frac{3 \\times C_2^4 + 1}{4 \\times 3} = \\frac{19}{12}$. The above construction actually attains an average of $\\frac{19}{12}$. Thus the minimum average of $d(A, B)$ is $\\frac{19}{12}$ for $n = 4$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31712,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle such that $\\angle ACB = 45^\\circ$. Let $G$ be the point of intersection of the three medians and let $O$ be the circumcentre. Suppose $OG = 1$ and $OG // BC$. Determine the length of the segment $BC$.",
    "options": [],
    "answer": "BC = 12",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31713,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to choose $24$ distinct points in the space such that no three of them lie on the same line, and choose $2019$ distinct planes in a way that each plane passes through at least $3$ of the chosen points and each triple of points belongs to one of the chosen planes?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nNo. Suppose such a choice exists. Let $n_1, n_2, \\dots, n_{2019}$ be the number of points on the planes respectively. Note that $n_j \\ge 3$ for $1 \\le j \\le 2019$. It is clear that\n$$\n2024 = \\binom{24}{3} = \\binom{n_1}{3} + \\binom{n_2}{3} + \\dots + \\binom{n_{2019}}{3}.\n$$\n\nWe first show $n_j \\le 4$ for all $1 \\le j \\le 2019$. Indeed, if at least one is greater than $4$, then\n$$\n2024 = \\binom{n_1}{3} + \\binom{n_2}{3} + \\dots + \\binom{n_{2019}}{3} \\ge \\underbrace{1+1+\\dots+1}_{2018 \\text{ times}} + \\binom{5}{3} = 2028,\n$$\nwhich is impossible.\nNow suppose among all the $n_j$'s, $a$ of them are equal to $3$ and $b$ of them are equal to $4$. Then we have $a+b = 2019$ and\n$$\na + 4b = a\\binom{3}{3} + b\\binom{4}{3} = 2024.\n$$\nThis gives $b = \\frac{(a+4b) - (a+b)}{3} = \\frac{5}{3}$, which is again impossible. Therefore, such a choice does not exist.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31714,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $57a + 88b + 125c \\ge 1148$, where $a, b, c > 0$, what is the minimum value of\n$$\na^3 + b^3 + c^3 + 5a^2 + 5b^2 + 5c^2?\n$$",
    "options": [],
    "answer": "466",
    "solution": "The answer is $466$.\nNote that\n$$\na^3 + 5a^2 - 57a + 99 = (a + 11)(a - 3)^2 \\ge 0,\n$$\n$$\nb^3 + 5b^2 - 88b + 208 = (b + 13)(b - 4)^2 \\ge 0,\n$$\n$$\nc^3 + 5c^2 - 125c + 375 = (c + 15)(c - 5)^2 \\ge 0.\n$$\nAdding these inequalities, we find that\n$$\na^3 + b^3 + c^3 + 5a^2 + 5b^2 + 5c^2 \\ge 57a + 88b + 125c - (99 + 208 + 375) \\ge 466.\n$$\nEquality holds when $a=3$, $b=4$ and $c=5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31715,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle with incentre $I$ and orthocentre $H$. $AI$ meets the circumcircle of $\\triangle ABC$ again at $M$. Suppose the length $IM$ is exactly the circumradius of $\\triangle ABC$. Show that $AH \\ge AI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcentre of $\\triangle ABC$. Since $BM = CM = IM = OM$ (where $OM$ is the circumradius of $\\triangle ABC$), $B$, $C$, $I$, $O$ are concyclic. Therefore, we have $\\angle BIC = \\angle BOC$. This yields $90^\\circ + \\frac{A}{2} = 2A$, and hence $A = 60^\\circ$. Now\n$$\n\\angle BHC = 180^\\circ - A = 120^\\circ = 2A = \\angle BOC,\n$$\n\nwhich implies $H$ lies on ($BCOI$). So $HM = IM$. Noting the triangle inequality $AH + HM \\ge AM$, we conclude $AH \\ge AI$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31716,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a table with $n$ rows and 18 columns. Each of its cells contains a 0 or a 1. The table satisfies the following properties:\n(i) Every two rows are different.\n(ii) Each row contains exactly 6 cells that contain 1.\n(iii) For every three rows, there exists a column so that the intersection of the column with the three rows (the three cells) all contain 0.\nWhat is the greatest possible value of $n$?",
    "options": [],
    "answer": "12376",
    "solution": "The greatest possible value of $n$ is $\\binom{17}{6} = 12376$.\nWe can identify each row as a 6-element subset of $S_{18} = \\{1, 2, \\dots, 18\\}$, so that the elements correspond to the numbers of columns which contain 1. Then we need to maximize the number of such subsets so that no three of them form a partition of $S_{18}$, since forming a partition is the only case for their complement to have an empty intersection.\nFirst we construct an example of 12376 subsets. We simply need to consider all 6-element subsets of $\\{1, 2, \\dots, 17\\}$. Since in any three subsets, there are $6 \\times 3 = 18$ elements, there must be an element belonging to at least two of the subsets by the pigeonhole principle. Therefore, the condition is satisfied. In this example, there are $\\binom{17}{6} = 12376$ subsets.\nNext we show that $n \\le 12376$. Suppose $S_1, S_2, \\dots, S_n$ are the 6-element subsets corresponding to the rows, and let $T_1, T_2, \\dots, T_{m-n}$ be the unused 6-element subsets, where $m = \\binom{18}{6}$. Let $N$ be the number of pairs $(S_i, T_j)$ such that $S_i$ and $T_j$ are disjoint.\nOn the one hand, for each $S_i$, the 12 elements not in $S_i$ can form $\\frac{1}{2}\\binom{12}{6}$ unordered pairs of 6-element subsets. At least one subset in each pair must be a subset $T_j$ since no three of the $S_k$'s can form a partition of $S_{18}$. As each $T_j$ disjoint with $S_i$ belongs to exactly one such pair, this gives at least $\\frac{1}{2}\\binom{12}{6}$ pairs containing $S_i$. Thus, $N \\ge \\frac{n}{2}\\binom{12}{6}$.\nOn the other hand, for each $T_j$, there are only $\\binom{12}{6}$ subsets of $S_{18}$ which are disjoint with $T_j$. Hence, $N \\le (m-n)\\binom{12}{6}$.\nCombining these, we obtain $\\frac{n}{2}\\binom{12}{6} \\le (m-n)\\binom{12}{6}$, i.e. $n \\le \\frac{2m}{3} = 12376$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31717,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose there are $2019$ distinct points in a plane and the distances between pairs of them attain $k$ different values. Prove that $k$ is at least $44$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $k$ is less than $44$. Choose a point $P$ on the boundary of the convex hull of this set of $2019$ points.\nBy the pigeonhole principle, since $\\left\\lfloor \\frac{2018}{43} \\right\\rfloor > 45$, there is a circle centered at $P$ on which at least $45$ of the other points lie. Moreover, these $45$ points all lie on the same semicircle since $P$ is on the boundary of the convex hull. Let these points be $A_1, A_2, \\dots, A_{45}$ in that order. Then the distances\n$$\nA_1A_2 < A_1A_3 < \\dots < A_1A_{45}\n$$\nare distinct, which contradicts the assumption that there are less than $44$ distances. Thus, $k \\ge 44$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31718,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n$ larger than $1$ with prime factorization $p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_k^{\\alpha_k}$, its *signature* is defined as the sum $\\alpha_1 + \\alpha_2 + \\cdots + \\alpha_k$. Does there exist $2020$ consecutive positive integers such that among them, there are exactly $1812$ integers whose signatures are strictly smaller than $11$?",
    "options": [],
    "answer": "Yes",
    "solution": "Yes. Let $f(n)$ be the number of integers among $n+1, n+2, \\dots, n+2020$ having signatures less than $11$. Since $2^{11} = 2048 > 2021$, all of $2, 3, \\dots, 2021$ have signatures smaller than $11$. Therefore, we have $f(1) = 2020$.\n\nNext, let $p_1, p_2, \\dots, p_{2020}$ be distinct primes. By the Chinese remainder theorem,\n\nthere exists a positive integer $m$ such that\n$$\n\\begin{array}{l@{\\quad}l@{\\quad}l}\nm & \\equiv & -1 \\pmod{p_1^{11}}, \\\\\nm & \\equiv & -2 \\pmod{p_2^{11}}, \\\\\n\\multicolumn{3}{c}{\\dots} \\\\\nm & \\equiv & -2020 \\pmod{p_{2020}^{11}}.\n\\end{array}\n$$\nFor this $m$, since $p_j^{11}$ divides $m+j$ for $j = 1, 2, \\dots, 2020$, each of the integers $m+1, m+2, \\dots, m+2020$ has signature larger than or equal to $11$. This implies $f(m) = 0$.\n\nNow, when we move from the integers $n+1, n+2, \\dots, n+2020$ to the integers $n+2, n+3, \\dots, n+2021$, the number of integers with signature less than $11$ will be increased or decreased by $1$, or remain unchanged. Thus, the difference between $f(n)$ and $f(n+1)$ is at most $1$. Therefore, in order to decrease $f(1) = 2020$ to $f(m) = 0$, we must come across a positive integer $k$ such that $f(k) = 1812$, and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31719,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be an arbitrary point inside $\\triangle ABC$. Let $\\Gamma$ be the circumcircle of $\\triangle BCD$. The external angle bisector of $\\angle ABC$ meets $\\Gamma$ again at $E$. The external angle bisector of $\\angle ACB$ meets $\\Gamma$ again at $F$. The line $EF$ meets the extension of $AB$ and $AC$ at $P$ and $Q$ respectively. Prove that the circumcircles of $\\triangle BFP$ and $\\triangle CEQ$ always pass through the same fixed point regardless of the position of $D$. (Assume all the labelled points are distinct.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $BE$ and $CF$ intersect at the $A$-excentre $J$ of $\\triangle ABC$. We claim that $J$ is the desired fixed point.\n\nWe only consider the configuration as shown since the other cases are similar. Since $BE$ is the external angle bisector, we have $\\angle PBE = \\angle EBC$. This is equal to $\\angle EFJ$ as $B, E, F, C$ are concyclic. This implies $P, B, F, J$ are concyclic. Similarly, $Q, C, E, J$ are concyclic. This shows the two circles always pass through $J$, which is independent of $D$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31720,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a list of integers $2^1 + 1, 2^2 + 1, \\dots, 2^{2019} + 1$, Adam chooses two different integers from the list and computes their greatest common divisor. Find the sum of all possible values of this greatest common divisor.",
    "options": [],
    "answer": "2^{674} + 672",
    "solution": "The answer is $2^{674} + 672$.\n\nWe claim that both $(2^r + 1, 2^s + 1)$ and $(2^r + 1, 2^s - 1)$ are of the form $2^a + 1$ or $1$ for any $r, s \\in \\mathbb{Z}^+$. We prove this by induction on $\\min\\{r, s\\}$. The base case $\\min\\{r, s\\} = 1$ is trivial since the greatest common divisor can only be $1$ or $3$. Assume this holds for all smaller cases and consider the inductive step.\n\nFor $(2^r + 1, 2^s + 1)$, WLOG assume $r \\le s$. Let $s = qr + b$ where $0 \\le b < r$. Then\n$$\n2^s + 1 = 2^b(2^r)^q + 1 \\equiv 2^b(-1)^q + 1 = \\pm 2^b + 1 \\pmod{2^r + 1}.\n$$\nThis shows $(2^r + 1, 2^s + 1) = (2^r + 1, 2^b \\pm 1)$. As $b < r$, we are done by the inductive hypothesis.\n\nSimilarly, for $(2^r + 1, 2^s - 1)$, if $r \\le s$, then we write $s = qr + b$ for some $0 \\le b < r$ so that $(2^r + 1, 2^s - 1) = (2^r + 1, 2^b \\pm 1)$. If $s < r$, then we write $r = qs + b$ for some $0 \\le b < s$ so that $(2^r + 1, 2^s - 1) = (2^b + 1, 2^s - 1)$. In any case, we are done by the inductive hypothesis.\n\nSo we have proven the claim by induction. Next, note that $2^a + 1 \\mid 2^r + 1$ if and only if $r = da$ for some odd integer $d$. (Again, by writing $r = qa + b$ with $0 \\le b < a$, we can show that $2^a + 1 \\mid 2^b(-1)^q + 1$ and hence $2^a + 1 \\le |2^b(-1)^q + 1|$ unless $2^b(-1)^q + 1 = 0$. As $b < a$, the former case is impossible.) Therefore, the greatest common divisor of two terms is $2^a + 1$ only if both are of the form $2^{(2^k-1)a} + 1$. In particular, we cannot have $a > 673$ since otherwise one term is at least $2^{3(674)} + 1 > 2^{2019} + 1$.\n\nOn the other hand, $a$ can be one of $1, 2, \\dots, 673$ as $(2^a + 1, 2^{3a} + 1) = 2^a + 1$.\n\nObviously, the greatest common divisor can also be 1. So the answer is\n$$\n1 + (2^1 + 1) + (2^2 + 1) + \\cdots + (2^{673} + 1) = 2^{674} + 672.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31721,
    "subject": "Mathematics (Multi-modal)",
    "question": "Two circles $\\Gamma$ and $\\Omega$ intersect at two distinct points $A$ and $B$. Let $P$ be a point on $\\Gamma$. The tangent at $P$ to $\\Gamma$ meets $\\Omega$ at the points $C$ and $D$, where $D$ lies between $P$ and $C$, and $ABCD$ is a convex quadrilateral. The lines $CA$ and $CB$ meet $\\Gamma$ again at $E$ and $F$ respectively. The lines $DA$ and $DB$ meet $\\Gamma$ again at $S$ and $T$ respectively. Suppose the points $P, E, S, F, B, T, A$ lie on $\\Gamma$ in this order. Prove that $PC, ET, SF$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since\n$$\n\\angle TEC = \\angle TEA = \\angle TBA = \\angle DBA = \\angle DCA = \\angle DCE,\n$$\nwe get $PC//ET$. Similarly, since\n$$\n\\angle FSD = \\angle FSA = \\angle CBA = \\angle PDA = \\angle PDS,\n$$\nwe get $PC//SF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31722,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral inscribed in a circle $\\Gamma$ such that $AB = AD$. Let $E$ be a point on the segment $CD$ such that $BC = DE$. The line $AE$ intersects $\\Gamma$ again at $F$. The chords $AC$ and $BF$ meet at $M$. Let $P$ be the symmetric point of $C$ about $M$. Prove that $PE$ and $BF$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\angle EFM = \\angle AFB = \\angle DCA = \\angle ECM$, the points $M$, $C$, $F$, $E$ are concyclic. Using the concyclic points, we obtain\n$$\n\\angle QME = \\angle FME = \\angle FCE = \\angle FCD = \\angle FBD = \\angle QBD.\n$$\nThis implies $EM \\parallel DB$, and hence $\\frac{DE}{EQ} = \\frac{BM}{MQ}$. Also, as $\\angle MCQ = \\angle MCB$, we have $\\frac{BM}{MQ} = \\frac{BC}{CQ}$. Combining these, we obtain $\\frac{DE}{EQ} = \\frac{BC}{CQ}$. It follows that $EQ = CQ$. Now, as $M$ and $Q$ are midpoints of $CP$ and $CE$, we know that $PE \\parallel MQ$ as desired.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31723,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real-valued functions $f$ defined on the set of real numbers such that\n$$\nf(f(x) + y) + f(x + f(y)) = 2f(xf(y))\n$$\nfor any real numbers $x$ and $y$.",
    "options": [],
    "answer": "All constant functions: f(x) = c for any real constant c.",
    "solution": "The answer is any constant function $f$.\nClearly, constant functions are solutions. In the following, we show that there is no solution if $f$ is not a constant function.\nLabel the equation\n$$\nf(f(x) + y) + f(x + f(y)) = 2f(xf(y)). \\qquad (1)\n$$\nSwapping $x$ and $y$, we get $f(f(y) + x) + f(y + f(x)) = 2f(yf(x))$. Comparing with (1), we get\n$$\nf(xf(y)) = f(yf(x)). \\qquad (2)\n$$\nPutting $x = 0$ in (2), we find that $f(0) = f(yf(0))$. If $f(0) \\neq 0$, then $yf(0)$ runs through all real values. This implies $f$ is a constant function, which is a contradiction. So we have $f(0) = 0$. Putting $x = 1$ in (2), we obtain\n$$\nf(f(y)) = f(yf(1)). \\qquad (3)\n$$\nNext, we put $x = 0$ in (1) to get\n$$\nf(y) + f(f(y)) = 0. \\qquad (4)\n$$\nReplacing $x$ by $f(x)$ in (1), and using (4), we find that\n$$\nf(-f(x) + y) + f(f(x) + f(y)) = 2f(f(x)f(y)).\n$$\n\nSwapping $x$ and $y$, and comparing with this equation again, we get\n$$\nf(-f(x) + y) = f(-f(y) + x). \\tag{5}\n$$\nPutting $y = f(x)$ in (5) and using (4), we have $f(f(x) + x) = 0$. Then we put $x = y = 1$ in (1) to obtain $f(f(1)) = 0$. It follows from (4) by putting $y = 1$ that $f(1) = 0$. Thus, (3) becomes $f(f(y)) = 0$. Together with (4), we find that $f(y) = 0$ for any $y \\in \\mathbb{R}$. This is a contradiction as we have assumed $f$ is non-constant.\nTherefore, the only solutions are the constant functions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31724,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the total number of primes $p < 100$ such that $\\lfloor (2 + \\sqrt{5})^p \\rfloor - 2^{p+1}$ is divisible by $p$. Here $\\lfloor x \\rfloor$ denotes the greatest integer less than or equal to $x$.",
    "options": [],
    "answer": "24",
    "solution": "The answer is 24.\nLet $p$ be an odd prime. Using the binomial theorem, we have\n$$\n(2 + \\sqrt{5})^p + (2 - \\sqrt{5})^p = \\sum_{k=0}^{\\frac{p-1}{2}} \\binom{p}{2k} 2^{p+1-2k} \\cdot 5^k = 2^{p+1} + \\sum_{k=1}^{\\frac{p-1}{2}} \\binom{p}{2k} 2^{p+1-2k} \\cdot 5^k,\n$$\nwhich is an integer. Since $p$ is an odd prime, we have $(2 - \\sqrt{5})^p < 0$. From this it follows that $|(2 + \\sqrt{5})^p| = (2 + \\sqrt{5})^p + (2 - \\sqrt{5})^p$. Thus,\n$$\n|(2 + \\sqrt{5})^p| - 2^{p+1} = \\sum_{k=1}^{\\frac{p-1}{2}} \\binom{p}{2k} 2^{p+1-2k} \\cdot 5^k.\n$$\nSince $\\binom{p}{r}$ is divisible by $p$ for $1 \\le r \\le p-1$, $|(2 + \\sqrt{5})^p| - 2^{p+1}$ is divisible by $p$.\nIt is easy to check that the expression is not divisible by $p$ when $p = 2$. Since there are 25 primes less than 100, the answer is 24.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31725,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ and $c$ be positive real numbers satisfying $abc = 1$. Prove that\n$$\n\\frac{1}{a^3 + 2b^2 + 2b + 4} + \\frac{1}{b^3 + 2c^2 + 2c + 4} + \\frac{1}{c^3 + 2a^2 + 2a + 4} \\le \\frac{1}{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the AM-GM inequality, we have\n$$a^3 + b^2 + b \\ge 3\\sqrt[3]{a^3b^3} = 3ab,$$\n$$b^2 + b + 1 \\ge 3\\sqrt[3]{b^3} = 3b.$$ \nThis gives $a^3 + 2b^2 + 2b + 4 \\ge 3ab + 3b + 3 = 3(ab + b + 1)$. Since $abc = 1$, we can let $a = \\frac{x}{y}$, $b = \\frac{y}{z}$, $c = \\frac{z}{x}$ for some positive real numbers $x, y, z$. Then we have\n$$\n\\frac{1}{a^3 + 2b^2 + 2b + 4} \\le \\frac{1}{3(ab + b + 1)} = \\frac{z}{3(x + y + z)}. \\quad (1)\n$$\nBy symmetry, we have\n$$\n\\frac{1}{b^3 + 2c^2 + 2c + 4} \\le \\frac{x}{3(y + z + x)}, \\quad (2)\n$$\n$$\n\\frac{1}{c^3 + 2a^2 + 2a + 4} \\le \\frac{y}{3(z + x + y)}. \\quad (3)\n$$\nAdding (1), (2) and (3), we obtain the desired inequality. Equality holds when $a = b = c = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31726,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, let $D$ be a point on side $BC$. Suppose the incircle $\\omega_1$ of $\\triangle ABD$ touches sides $AB$ and $AD$ at $E, F$ respectively, and the incircle $\\omega_2$ of $\\triangle ACD$ touches sides $AD$ and $AC$ at $F, G$ respectively. Suppose the segment $EG$ intersects $\\omega_1$ and $\\omega_2$ again at $P$ and $Q$ respectively. Show that line $AD$, tangent of $\\omega_1$ at $P$ and tangent of $\\omega_2$ at $Q$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the circumcentre of $\\triangle FPQ$. First of all,\n$$\n\\angle OFP = 90^\\circ - \\angle PQF = \\angle FQG - 90^\\circ = \\frac{\\angle FAG}{2} = \\angle GEF = \\angle AFP\n$$\nwhere the second last equality follows from the fact that $AE = AF = AG$. This shows that $O$ lies on $AD$.\n\nNext, Let $I_1$ be the centre of $\\omega_1$. Note that $\\triangle OFI_1 \\cong \\triangle OPI_1$ as their corresponding sides are equal. Thus $\\angle I_1PO = 90^\\circ$, which means $OP$ is tangent to $\\omega_1$. Similarly $OQ$ is tangent to $\\omega_2$.\nHence, the line AD, the tangent to $\\omega_1$ at P and the tangent to $\\omega_2$ at Q are concurrent at O, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31727,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a sequence with some ones and zeros, we count the number of continuous runs of equal digits in it. (For example the sequence $011001010$ has $7$ continuous runs: $0$, $11$, $00$, $1$, $0$, $1$, $0$.) Find the sum of the number of all continuous runs for all possible sequences with $2019$ ones and $2019$ zeros.",
    "options": [],
    "answer": "2020*binom(4038,2019)",
    "solution": "The answer is $2020\\binom{4038}{2019}$.\nWe show that for all sequences with $n$ zeros and $n$ ones, the sum of all continuous runs is given by $(n+1)\\binom{2n}{n}$.\n\nWe first count the number of changes, which means a pair of consecutive digits which is $01$ or $10$. There are $2n-1$ positions for which changes may occur. For each fixed position, there are $2$ ways to choose the pair of digits for this change, and $\\binom{2n-2}{n-1}$ ways to put the remaining $n-1$ zeros and $n-1$ ones. Hence, there are\n$$\n2(2n-1)\\binom{2n-2}{n-1}\n$$\nchanges in total.\n\nNow, for each sequence, the number of runs is $1$ more than the number of changes. Since there are $\\binom{2n}{n}$ sequences in total, the total number of runs is\n$$\n\\begin{aligned}\n& 2(2n-1)\\binom{2n-2}{n-1} + \\binom{2n}{n} \\\\\n&= 2n\\binom{2n-1}{n} + \\binom{2n}{n} = 2n\\binom{2n-1}{n-1} + \\binom{2n}{n} \\\\\n&= n\\binom{2n}{n} + \\binom{2n}{n} = (n+1)\\binom{2n}{n},\n\\end{aligned}\n$$\nusing the fact $k\\binom{m}{k} = m\\binom{m-1}{k-1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31728,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is it possible to find a non-constant polynomial $P(x, y)$ such that $P([\\alpha], [3\\alpha]) = 0$ for every real number $\\alpha$? (Here $[\\mu]$ stands for the largest integer less than or equal to $\\mu$.)",
    "options": [],
    "answer": "P(x, y) = (y - 3x)(y - 3x - 1)(y - 3x - 2)",
    "solution": "Yes. We claim that $P(x, y) = (y - 3x)(y - 3x - 1)(y - 3x - 2)$ satisfies the conditions. Clearly, it is a non-constant polynomial. For any real number $\\alpha$, let $n = \\lfloor \\alpha \\rfloor$. Then we have $n \\le \\alpha < n + 1$ and hence $3n \\le 3\\alpha < 3n + 3$. This shows $\\lfloor 3\\alpha \\rfloor = 3n, 3n + 1, 3n + 2$. Therefore, one of $\\lfloor 3\\alpha \\rfloor - 3\\lfloor \\alpha \\rfloor$, $\\lfloor 3\\alpha \\rfloor - 3\\lfloor \\alpha \\rfloor - 1$ and $\\lfloor 3\\alpha \\rfloor - 3\\lfloor \\alpha \\rfloor - 2$ must be 0. This shows $P(\\lfloor \\alpha \\rfloor, \\lfloor 3\\alpha \\rfloor) = 0$ for all $\\alpha \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31729,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all real triples $(a, b, c)$ satisfying\n$$(2^{2a} + 1)(2^{2b} + 2)(2^{2c} + 8) = 2^{a+b+c+5}.$$",
    "options": [],
    "answer": "(0, 1/2, 3/2)",
    "solution": "The only solution is $(a, b, c) = \\left(0, \\frac{1}{2}, \\frac{3}{2}\\right)$.\nNote that $(2^a - 1)^2 \\ge 0$, $(2^b - \\sqrt{2})^2 \\ge 0$ and $(2^c - \\sqrt{8})^2 \\ge 0$. These yield\n$$\n2^{2a} + 1 \\ge 2^{a+1},\n$$\n$$\n2^{2b} + 2 \\ge 2^{b+\\frac{3}{2}},\n$$\n$$\n2^{2c} + 8 \\ge 2^{c+\\frac{5}{2}}.\n$$\nMultiplying these inequalities, we obtain\n$$\n(2^{2a} + 1)(2^{2b} + 2)(2^{2c} + 8) \\ge 2^{a+b+c+1+\\frac{3}{2}+\\frac{5}{2}} = 2^{a+b+c+5}.\n$$\nThis shows all equalities should hold. This means $(a, b, c) = \\left(0, \\frac{1}{2}, \\frac{3}{2}\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31730,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a set of $2020$ distinct points in the plane. Let\n$M = \\{P : P$ is the midpoint of $XY$ for some distinct points $X, Y$ in $S\\}$.\nFind the least possible value of the number of points in $M$.",
    "options": [],
    "answer": "4037",
    "solution": "There are at least $4037$ points in $M$.\nSince there are finitely many points in $S$, we can find a coordinate system such that all points in $S$ have distinct $x$-coordinates. Indeed, we just need to find a line which is not perpendicular to any line joining two of the points in $S$.\n\nNow, let the $n = 2020$ points be $P_j = (x_j, y_j)$ where $j = 1, 2, \\dots, n$. WLOG assume $x_1 < x_2 < \\dots < x_n$. Since\n$$\n\\frac{x_1 + x_2}{2} < \\frac{x_1 + x_3}{2} < \\dots < \\frac{x_1 + x_n}{2} < \\frac{x_2 + x_n}{2} < \\dots < \\frac{x_{n-1} + x_n}{2},\n$$\nthe midpoints of $P_1P_2, P_1P_3, \\dots, P_1P_n, P_2P_n, \\dots, P_{n-1}P_n$ have pairwise different $x$-coordinates, and so they must be distinct. This shows $|M| \\ge 2n - 3 = 4037$.\n\nThe case $4037$ is attainable. For example, we choose the points $0, 2, 4, 6, \\dots, 2n-2$ on the real number line. Then the midpoints can only be $1, 2, 3, \\dots, 2n-3$. So there are exactly $2n-3 = 4037$ midpoints.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31731,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ be a polynomial with rational coefficients, and let $\\alpha$ be a real number. If\n$$\n\\alpha^3 - 2019\\alpha = (f(\\alpha))^3 - 2019f(\\alpha) = 2021,\n$$\nprove that $(f^n(\\alpha))^3 - 2019f^n(\\alpha) = 2021$ for any positive integer $n$.\n(Here, we define $f^n(x) = \\underbrace{f(f(\\dots f(x)\\dots))}_{n \\text{ times}}$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We claim that $g(x) = x^3 - 2019x - 2021$ is irreducible over $\\mathbb{Q}$. Since $\\deg g = 3$, if $g$ were reducible, then it must have a linear factor. In other words, $g$ would have a rational root $c$. Since $g$ is monic, $c$ must be an integer by the rational root theorem. However, $c^3 - 2019c$ must be even, so it is impossible that $c^3 - 2019c - 2021 = 0$. This proves our claim.\n\nNote that $\\alpha$ is a root of $g(x)$. It follows from the above claim that $g$ is the minimal polynomial of $\\alpha$ over $\\mathbb{Q}$ (see remark). From the given condition $g(f(\\alpha)) = 0$, we know that $g(x) \\mid g(f(x))$. Replacing $x$ by $f(x)$, we obtain $g(f(x)) \\mid g(f^2(x))$. Since $g(x)$ divides $g(f(x))$, and $g(f(x))$ divides $g(f^2(x))$, we have $g(x) \\mid g(f^2(x))$. Similarly, for any positive integer $n$, we have $g(f^{n-1}(x)) \\mid g(f^n(x))$. By induction, this yields $g(x) \\mid g(f^n(x))$. Therefore, $g(f^n(\\alpha)) = 0$. This is exactly the same as $(f^n(\\alpha))^3 - 2019f^n(\\alpha) = 2021$ as desired.\n\n**Remarks.** We say that $g$ is the minimal polynomial of $\\alpha$ over $\\mathbb{Q}$ if $g$ is the monic polynomial with rational coefficients of smallest degree such that $g(\\alpha) = 0$. An important property of the minimal polynomial is that for any polynomial $h$, $h(\\alpha) = 0$ if and only if $g$ divides $h$. Note that this also implies the minimal polynomial is the only irreducible monic polynomial having $\\alpha$ as a root.\nHere is a proof. For any polynomial $h$ such that $h(\\alpha) = 0$, let\n$$\nh(x) = g(x)q(x) + r(x)\n$$\nwhere $\\deg r < \\deg g$. Then $h(\\alpha) = g(\\alpha)q(\\alpha) + r(\\alpha)$, which implies $r(\\alpha) = 0$. Due to the minimal choice of the degree of $g$, $r$ must be the zero polynomial. This means $g(x) | h(x)$. Conversely, if $g(x) | h(x)$, then it is obvious that $h(\\alpha) = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31732,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AC = kAB$, with $k > 1$. The internal angle bisector of $\\angle BAC$ meets $BC$ at $D$. The circle with $AC$ as diameter cuts the extension of $AD$ at $E$. Express $\\frac{AD}{AE}$ in terms of $k$.",
    "options": [],
    "answer": "2/(k+1)",
    "solution": "The answer is $\\frac{2}{k+1}$.\n\nLet $CE$ meet $AB$ at $Q$. Let $P$ be the point on $AQ$ such that $PD // QC$. Note that $E$ is the midpoint of $QC$ since $\\triangle AEQ \\cong \\triangle AEC$. Also, we have $\\triangle APD \\sim \\triangle AQE$ and $\\triangle BPD \\sim \\triangle BQC$. Thus, we have\n$$\n\\frac{AD}{AE} = \\frac{PD}{QE} = 2 \\cdot \\frac{PD}{QC} = 2 \\cdot \\frac{BD}{BC}.\n$$\nBy the angle bisector theorem, $\\frac{DB}{DC} = \\frac{AB}{AC} = \\frac{1}{k}$. It follows that $\\frac{AD}{AE} = \\frac{2}{k+1}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31733,
    "subject": "Mathematics (Multi-modal)",
    "question": "In acute $\\triangle ABC$, $BC > AB$. Let $D$ be the point on the side $AC$ such that $BD = BA$. Let $M$ be the midpoint of $BC$. The tangents at $D$ and $M$ to the circumcircle of $\\triangle CDM$ intersect at point $E$. The line $BE$ intersects the side $AC$ at $F$. Prove that $A, B, M, F$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $BD$ meets ($CDM$) again at $P$. Applying Pascal's theorem to the points $MMCDDP$, we know that $MM \\cap DD = E$, $MC \\cap DP = B$ and $CD \\cap PM$ are collinear. Thus, $MP$ passes through $BE \\cap CD = F$. Now, since\n$$\n\\angle CMF = \\angle ADB = \\angle BAF,\n$$\nthe points $A, B, M, F$ are concyclic.\n\n![](attached_image_1.png)\nWe provide another proof without using Pascal's theorem as follows. Again, the main task is to prove $M, P, F$ are collinear. We define $F'$ as the intersection point of $CD$ and $MP$. It suffices to prove $B, E, F'$ are collinear. Suppose $DE$ meets ($PDF'$) again at $X$, and $ME$ meets ($MCF'$) again at $Y$.\nFirstly, since $BP \\times BD = BM \\times BC$, the powers of $B$ with respect to ($PDF'$) and ($MCF'$) are the same. Thus, $B$ lies on the radical axis of these circles.\nSecondly, as $\\angle XF'P = \\angle XDP = \\angle DMP = \\angle DMF'$, we have $XF'//MD$. Also, as $\\angle MYF' = 180^\\circ - \\angle MCF' = 180^\\circ - \\angle YMD$, we have $YF'//MD$. Therefore, $Y, X, F'$ are collinear, and $YX//MD$. Since $EM = ED$, we have $EY = EX$. From this, it follows that $EX \\times ED = EY \\times EM$, and hence $E$ lies on the radical axis of ($PDF'$) and ($MCF'$).\nThirdly, it is clear that $F'$ lies on the radical axis of ($PDF'$) and ($MCF'$). It follows that $B, E, F'$ are collinear as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31734,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice and Bob play a game on the plane. Firstly, Alice draws 4 red points which are the vertices of a square, and draws another 2021 red points inside this square such that no three red points are collinear. Next, Bob draws $n$ blue points such that for every triangle whose three vertices are the red points, there is at least one blue point lying strictly inside this triangle.\n\nThe goal of Alice is to maximize $n$, while the goal of Bob is to minimize $n$. Find the value of $n$ if both players play optimally.",
    "options": [],
    "answer": "4044",
    "solution": "The answer is $4044$.\n\nAfter Alice has drawn the $4$ red vertices of a square, we can draw one of the diagonals of the square to form $2$ triangles. Each time when Alice draws one more red point inside the square, it must lie inside one of the triangles formed. By joining the new red point to the $3$ vertices of this triangle, we obtain a new triangulation with $2$ more triangles. At the end, when Alice has drawn all the red points, there is a triangulation with $2 + 2 \\times 2021 = 4044$ triangles. Since there is a blue point in each of these triangles, we must have $n \\ge 4044$.\n\n![](attached_image_1.png)\n\nNext, we provide a strategy for Bob so that it is sufficient to draw $4044$ blue points. We set up a coordinate system such that all red points have different $x$-coordinates. For each red point $P(x, y)$ inside the square, Bob draws two blue points $(x, y+\\epsilon)$ and $(x, y-\\epsilon)$ where $2\\epsilon$ is the distance from $P$ to the closest line joining two red points different from $P$. We say that these two blue points are associated to the point $P$. For the $4$ red vertices of the square, Bob does the same thing except that points outside the square need not be drawn. Note that there are only $2$ blue points each of which is associated to a vertex of the square whose $x$-coordinate is neither the smallest nor the largest. In total, there are $2 \\times 2021 + 2 = 4044$ blue points. We claim that the condition is satisfied.\n\nIndeed, consider any triangle $ABC$ with red vertices where the $x$-coordinates of $A$, $B$ and $C$ are in ascending order. WLOG assume $B$ lies below the line $AC$. Then $\\angle ABC$ contains the blue point associated to $B$ lying above $B$ since this blue point also lies below the line $AC$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31735,
    "subject": "Mathematics (Multi-modal)",
    "question": "On the table there are 20 coins of weights $1$, $2$, $3$, $\\ldots$, $15$, $37$, $38$, $39$, $40$ and $41$ grams. They all look alike but their colours are all distinct. Now Miss Adams knows the weight and colour of each coin, but Mr. Bean knows only the weights of the coins. There is also a balance on the table, and each comparison of weights of two groups of coins is called an operation. Miss Adams wants to tell Mr. Bean which coin is the $1$ gram coin by performing some operations. What is the minimum number of operations she needs to perform?",
    "options": [],
    "answer": "2",
    "solution": "Miss Adams needs to perform at least two operations.\nA strategy is as follows. In the first operation, she puts the coins of weights $1$ to $15$ on one side, and the three coins of weights $39$, $40$ and $41$ on the other side.\n\nThen the two groups will be of the same weight ($120$). Since the total weight of $15$ coins is at least $1+2+\\cdots+15=120$ and the total weight of $3$ coins is at most $39+40+41=120$, this is the only situation for which a group of $15$ coins has the same weight as a group of $3$ coins. So Mr. Bean knows the two unused coins are of weight $37$ and $38$. Miss Adams then puts the $1$-gram coin and the $37$-gram coin on one side, and the $38$-gram coin on the other side. Hence Mr. Bean will know which coin has weight $1$.\nNow we show that one operation is not enough. If one operation is enough, then the $1$-gram coin should be the only coin in its group (a group on the balance or the unused group).\n* If the $1$-gram coin is put on the balance and the other group on the balance contains more than one coin, the situation is the same if the $1$-gram coin and the $2$-gram coin are swapped.\n* If the $1$-gram coin is put on the balance and the other group on the balance contains exactly one coin, the situation is the same if the $1$-gram coin is replaced by any unused coin.\n* If the $1$-gram coin is the only unused coin and one side of the balance is heavier, then that side has weight at least $158$ as the total weight of all coins is $315$, while the other side has weight at most $156$. So the situation is the same if the $1$-gram coin and the $2$-gram coin are swapped.\n* If the $1$-gram coin is the only unused coin and the two sides of the balance have the same weight, we call the group containing the $3$-gram coin group A, and the other group B. If group A contains a coin of weight $k \\neq 3$ and group B contains a coin of weight $k+1$, then we can swap these two coins, and swap the $1$-gram coin and the $3$-gram coin. The two groups still have the same weight. In the remaining case, group A should contain the heaviest coins among those with weight $1$, $2$, $\\ldots$, $15$ (possibly none), and the heaviest coins among those with weight $37$, $38$, $39$, $40$, $41$. But then the total weight of group A cannot be $157$. The computations are illustrated below.\n- $2+3+\\cdots+15=119<157$\n- $4+5+\\cdots+15+41=155<157<158=3+4+\\cdots+15+41$\n- $10+11+\\cdots+15+40+41=156<157<165=9+10+\\cdots+15+40+41$\n- $14+15+39+40+41=149<157<162=13+14+15+39+40+41$\n- $38+39+40+41=158>157$\nTherefore, in any case Mr. Bean cannot tell whether the single coin has weight $1$ or not. So at least two operations are needed.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31736,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, let $d(n)$ be the number of positive divisors of $n$, and let $\\varphi(n)$ be the number of positive integers not exceeding $n$ which are coprime to $n$. Does there exist a constant $C$ such that\n$$ \\frac{\\varphi(d(n))}{d(\\varphi(n))} \\le C $$\nfor all $n \\ge 1$?",
    "options": [],
    "answer": "Yes; C = 1.",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31737,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be an acute triangle with circumcircle $\\Gamma$, and let $P$ be the midpoint of the minor arc $BC$ of $\\Gamma$. Let $AP$ meet $BC$ at $D$, and let $M$ be the midpoint of $AB$. Also, let $E$ be the point such that $AE \\perp AB$ and $BE \\perp MP$. Prove that $AE = DE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $P$ is the midpoint of the minor arc $BC$, $AD$ is the internal angle bisector of $\\angle BAC$.\nLet $E'$ be the intersection of the perpendicular bisector of $AD$ and the line through $A$ perpendicular to $AB$. Using the method of false position, it suffices to show $BE' \\perp MP$, and then conclude $E' = E$. Let $\\Omega$ be the circle with centre $E'$ that passes through $A$ (and hence $D$).\n\n![](attached_image_1.png)\n\nSince $AB \\perp AE'$, $AB$ is tangent to $\\Omega$. As $MA = MB$, the powers of $M$ with respect to $\\Omega$ and $B$ are equal. Since $\\angle PBD = \\angle PAC = \\angle PAB$, we have $\\triangle PBD \\sim \\triangle PAB$, so $PB^2 = PD \\times PA$. Therefore, the powers of $P$ with respect to $\\Omega$ and $B$ are equal.\nFrom above, both $M$ and $P$ lie on the radical axis of $\\Omega$ and $B$. Hence, $MP$ is perpendicular to the line joining the centres of the two circles, i.e. $E'B$. The result follows from above.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31738,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Is it possible to express $n^2 + 3n + 3$ into the form $ab$ with $a$ and $b$ being positive integers, and such that the difference between $a$ and $b$ is smaller than $2\\sqrt{n} + 1$?",
    "options": [],
    "answer": "No",
    "solution": "No, it is not possible. Indeed, if $n^2 + 3n + 3 = ab$, then\n$$\n(a+b)^2 = (a-b)^2 + 4ab \\ge 4ab = 4n^2 + 12n + 12 > 4n^2 + 12n + 9 = (2n+3)^2.\n$$\nSince both sides are squares, this yields $(a+b)^2 \\ge (2n+4)^2$. It follows that\n$$\n(a-b)^2 = (a+b)^2 - 4ab \\ge (2n+4)^2 - (4n^2 + 12n + 12) = 4n + 4 = 4(n+1).\n$$\nTherefore, $|a-b| \\ge 2\\sqrt{n+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31739,
    "subject": "Mathematics (Multi-modal)",
    "question": "Does there exist a nonzero polynomial $P(x)$ with integer coefficients satisfying both of the following conditions?\n* $P(x)$ has no rational root;\n* For every positive integer $n$, there exists an integer $m$ such that $n$ divides $P(m)$.",
    "options": [],
    "answer": "P(x) = (x^2+1)(x^2-2)(x^2+2)(x^2+7)",
    "solution": "Yes. We shall prove that $P(x) = (x^2+1)(x^2-2)(x^2+2)(x^2+7)$ satisfies all properties. It is not hard to see that $P$ has no rational root. It remains to show it satisfies the second condition.\n\nConsider any odd prime $p$. Using facts about quadratic residues, we have the following.\n* If $p \\equiv 1 \\pmod 4$, then $x^2+1 \\equiv 0 \\pmod p$ is solvable.\n* If $p \\equiv 7 \\pmod 8$, then $x^2-2 \\equiv 0 \\pmod p$ is solvable.\n* If $p \\equiv 3 \\pmod 8$, then $x^2+2 \\equiv 0 \\pmod p$ is solvable since\n$$\n\\left(\\frac{-2}{p}\\right) = \\left(\\frac{-1}{p}\\right) \\left(\\frac{2}{p}\\right) = (-1)(-1) = 1.\n$$\nThis shows there exists $x$ and $c \\in \\{1, -2, 2\\}$ such that $x^2 + c \\equiv 0 \\pmod p$. Note that $p \\nmid x$. Therefore, we have\n$$\n(x^2 + c)' = 2x \\not\\equiv 0 \\pmod p.\n$$\nBy Hensel's lifting lemma, $x^2+c \\equiv 0 \\pmod{p^k}$ is solvable for any positive integer $k$.\n\nNext, we want to show for any $k \\ge 1$, there exists an integer $b_k$ such that $2^k \\mid (b_k^2+7)$. Indeed, the statement is true for $k=1,2,3$, by simply taking $b_k=1$. Assume now the statement is valid for some $k \\ge 3$. For the case $k+1$, take $b_{k+1} = b_k + t2^{k-1}$, with $t$ to be determined. Now, since $b_k$ is odd,\n$$\nb_{k+1}^2 + 7 = (b_k + t2^{k-1})^2 + 7 \\equiv (b_k^2 + 7) + t2^k \\pmod{2^{k+1}}.\n$$\nBy the inductive hypothesis, $2^k \\mid (b_k^2 + 7)$. So we just take $t \\equiv \\frac{b_k^2 + 7}{2^k} \\pmod 2$ to get $2^{k+1} \\mid (b_{k+1}^2 + 7)$.\n\nNow, let $n = p_1^{\\alpha_1} p_2^{\\alpha_2} \\cdots p_s^{\\alpha_s}$ be the prime factorization of $n$. From above we know that for every $1 \\le j \\le s$, there exists an integer $m_j$ such that $p_j^{\\alpha_j} \\mid P(m_j)$. By the Chinese remainder theorem, we can find $m \\equiv m_j \\pmod{p_j^{\\alpha_j}}$ for $1 \\le j \\le s$. Then $P(m) \\equiv P(m_j) \\equiv 0 \\pmod{p_j^{\\alpha_j}}$. So $n \\mid P(m)$. Therefore, the polynomial $P(x)$ satisfies the second condition, and we are done.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31740,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(n) = \\prod_{k=1}^{n} \\left(1 + 4 \\cos^2\\left(\\frac{k\\pi}{2n+1}\\right)\\right)$. Prove that $f(n)$ is an integer for all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We shall prove that $f(n) = F_{2n+1}$, the $(2n+1)$th Fibonacci number. Let $\\zeta = e^{\\frac{2\\pi i}{2n+1}}$ be a primitive $(2n+1)$th root of unity. Then\n$$\n1 + 4 \\cos^2 \\left( \\frac{k\\pi}{2n+1} \\right) = 3 + 2 \\cos \\left( \\frac{2k\\pi}{2n+1} \\right) = 3 + 2 \\operatorname{Re}(\\zeta^k) = 3 + \\zeta^k + \\zeta^{-k}.\n$$\n\nNote that\n$$\n\\begin{align*} \nx^{2n+1} - 1 &= \\prod_{k=1}^{2n+1} (x - \\zeta^k) \\\\ \n&= (x - 1) \\prod_{k=1}^{n} (x - \\zeta^k)(x - \\zeta^{-k}) \\\\ \n&= (x - 1) \\prod_{k=1}^{n} (x^2 - (\\zeta^k + \\zeta^{-k})x + 1). \n\\end{align*}\n$$\nLet $\\alpha = \\frac{3+\\sqrt{5}}{2}$ so that $\\alpha^2 - 3\\alpha + 1 = 0$. Putting $x = -\\alpha$, we obtain\n$$\n-\\alpha^{2n+1} - 1 = (-\\alpha - 1) \\prod_{k=1}^{n} (\\alpha^2 + (\\zeta^k + \\zeta^{-k})\\alpha + 1) = - (\\alpha + 1) \\alpha^n \\prod_{k=1}^{n} (3 + \\zeta^k + \\zeta^{-k}).\n$$\nIt follows that\n$$\nf(n) = \\prod_{k=1}^{n} (3 + \\zeta^k + \\zeta^{-k}) = \\frac{\\alpha^{2n+1} + 1}{\\alpha^n(\\alpha + 1)}.\n$$\nNext, the terms of the Fibonacci sequence satisfy\n$$\nF_{2n+5} = F_{2n+4} + F_{2n+3} = 2F_{2n+3} + F_{2n+2} = 3F_{2n+3} - F_{2n+1}.\n$$\nLetting $a_n = F_{2n+1}$, we have the recurrence relation $a_{n+2} = 3a_{n+1} - a_n$ and the conditions $a_0 = 1$ and $a_1 = 2$. Since the roots to $\\lambda^2 - 3\\lambda + 1 = 0$ are $\\alpha$ and $\\frac{1}{\\alpha}$, the general term is $a_n = A\\alpha^n + B \\cdot \\frac{1}{\\alpha^n}$ for some constants $A$ and $B$. Using $a_0 = 1$ and $a_1 = 2$, it is easy to deduce\n$$\na_n = \\frac{\\alpha}{\\alpha + 1} \\cdot \\alpha^n + \\frac{1}{\\alpha + 1} \\cdot \\frac{1}{\\alpha^n} = \\frac{\\alpha^{2n+1} + 1}{\\alpha^n(\\alpha + 1)} = f(n).\n$$\nTherefore, $f(n)$ is an integer for all positive integers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31741,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is an $n \\times n$ chessboard where $n \\ge 4$ is a positive even number. The cells of the chessboard are coloured black and white such that adjacent cells sharing a common side have different colours. Let $A$ and $B$ be two interior cells (which means cells not lying on an edge of the chessboard) of distinct colours. Prove that a chess piece can move from $A$ to $B$ by moving across adjacent cells such that every cell of the chessboard is passed through exactly once.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For convenience, suppose all cells are unit squares. We first prove two claims.\n\n**Claim 1.** For any rectangular chessboard, a chess piece can move from one corner to the opposite corner by visiting every cell exactly once if at least one side has an odd length.\n\n**Claim 2.** For any rectangular chessboard, a chess piece can move from one corner to an adjacent corner by visiting every cell exactly once if the side containing the two corners has an even length.\nThese can be easily proved by considering a snake-shaped path as shown. The assumptions in the claims are sufficient by checking the parity.\n\n![](attached_image_1.png)\n\nNow, we provide a construction of a path from $A$ to $B$. Since $A$ and $B$ have distinct colours, the side lengths of the rectangular chessboard with $A$ and $B$ as opposite corners must have different parities. WLOG, assume the vertical side has an odd length and the horizontal side has an even length. Since the whole chessboard has an even length, we may assume there is an even number of rows above $A$. Partition the chessboard as shown below (the left figure).\n![](attached_image_2.png)\neven\n* Firstly, the chess piece moves from $A$ upwards until it reaches the top edge.\n* Then it enters the top-left rectangle and moves from corner $C$ to corner $D$ (by claim 1, since the vertical side has an odd length).\n* Next, it enters the bottom rectangle and moves from corner $E$ to corner $F$ (by claim 2, since the horizontal side has an even length).\n* Afterwards, it enters the top right rectangle and moves from corner $G$ to corner $H$ (by claim 1, since the vertical side has an odd length).\n\n* Then it enters the top rectangle and moves from corner *I* to corner *J* (by claim 1, since the horizontal side has an odd length).\n* Lastly, it goes back to the rectangle with *A* and *B* as opposite corners and moves to *B* (by claim 1, since the vertical size has an odd length).\nAn example is given above (the right figure).",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31742,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Let $M$ be the midpoint of $BC$, and let $G$ be the centroid of $\\triangle ABC$. Let $D$ be a point on the segment $GM$. A straight line passing through $D$ meets the sides $AB$ and $AC$ at $E$ and $F$ respectively (with $E, F \\ne A$). Show that\n$$\n[BEGF] + [CFGE] \\ge \\frac{4}{9}[ABC].\n$$\nWhen does the equality hold? (Here $[WXYZ]$ is the area of the polygon $WXYZ$, etc.)",
    "options": [],
    "answer": "Equality holds precisely when the point is the centroid and the line through it is parallel to the base, equivalently when the intersections with the sides are the points where the line through the centroid parallel to the base meets those sides.",
    "solution": "Let $E'$ and $F'$ be the points on $AB$ and $AC$ such that $E'$, $G$, $F'$ are collinear and $E'F' \\parallel EF$. Since $\\triangle EFG$ and $\\triangle EFE'$ have the same base $EF$ and the same height, they have the same area. This yields\n$$\n[BEGF] = [BEF] + [EFG] = [BEF] + [EFE'] = [BFE'] \\ge [BF'E']\n$$\n\nThe last inequality holds since $\\triangle BFE'$ and $\\triangle BF'E'$ have the same base $BE$, but $F'$ is closer to the sideline $BE'$ than $F$. Equality holds when $F = F'$, i.e. $D = G$. Similarly, we have $[CFGE] \\ge [CE'F']$, with equality $D = G$. Thus, we have\n$$\n[BEGF] + [CFGE] \\ge [BF'E'] + [CE'F'] = 2[MF'E'] = [AE'F']\n$$\nsince $M$ is the midpoint of $BC$ and $AG : GM = 2 : 1$. It suffices to prove $[AE'F'] \\ge \\frac{4}{9}[ABC]$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nLet $B'$ and $C'$ be the points on $AB$ and $AC$ such that $B', G, C'$ are collinear and $B'C' \\parallel BC$. As $\\triangle AB'C' \\sim \\triangle ABC$ with ratio $2:3$, we have $[AB'C'] = \\frac{4}{9}[ABC]$.\nTherefore, it remains to show $[AE'F'] \\ge [AB'C']$. WLOG assume $E'$ lies on the segment $BB'$ and $F'$ lies on the segment $AC'$. Then we only need to show $[GB'E'] \\ge [GC'F']$. Indeed, let $P$ be the symmetric point of $F'$ in $G$. Then $\\triangle B'GP \\cong \\triangle C'GF'$, so that $P$ lies on the line $F'G$. Also, $P$ must lie on the segment $GE'$ since\n$$\n\\angle GPB' = \\angle GF'C' = \\angle F'E'A + \\angle E'AF' > \\angle PE'B'\n$$\nIt follows that $[GB'E'] \\ge [GB'P] = [GC'F']$ as desired. Equality holds when $E = B'$ and $F = C'$, i.e. $D = G$ and $EF \\parallel BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31743,
    "subject": "Mathematics (Multi-modal)",
    "question": "Mandy needs to wake up early for attending a mathematics contest. She has set an alarm in her smartphone every 15 minutes since 5:30 am. If an alarm is not pressed off by her or her mother (or anybody else), it will ring for a while, stop for a while, then will ring again 9 minutes later as the first ring, and so on (e.g. if the first alarm is not pressed off, it will ring again at 5:39 am). Also each alarm will work independently. Now suppose each ring-tone lasts for $x$ minutes, and the smartphone has eventually rung for 50 minutes before Mandy wakes up at 6:30 am (assuming no one has pressed off any alarm before that). Find the value of $x$.",
    "options": [],
    "answer": "11/3",
    "solution": "If each ring-tone lasts for exactly $3$ minutes, the alarm will be ringing at $5:30$–$5:33$, $5:39$–$5:42$, $5:45$–$5:51$ and $5:54$–$6:30$. It will ring for $3 + 3 + 6 + 36 = 48$ minutes in total. Therefore, we must have $x > 3$.\n\nLet $x = 3 + y$ where $0 < y < 3$. In addition to the above periods, the alarm will ring for $y$ more minutes after each of $5:33$, $5:42$ and $5:51$. Thus, we need $48 + 3y = 50$. The only solution is $y = \\frac{2}{3}$, and hence $x = \\frac{11}{3}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31744,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x^2 + y^2 = \\frac{3961xy}{1980}$ where $x > y > 0$, find the value of $\\frac{x+y}{x-y}$.",
    "options": [],
    "answer": "89",
    "solution": "Answer: 89\n\nWe have\n$$\n\\left(\\frac{x+y}{x-y}\\right)^2 = \\frac{x^2+y^2+2xy}{x^2+y^2-2xy} = \\frac{\\left(\\frac{3961}{1980}+2\\right)xy}{\\left(\\frac{3961}{1980}-2\\right)xy} = 7921 = 89^2\n$$\nand so the answer is 89. (Note that $\\frac{x+y}{x-y} > 0$ as $x > y > 0$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31745,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an integer, and let $A$ be a subset of $\\{0, 1, 2, 3, \\dots, 5^n\\}$ consisting of $4n+2$ numbers. Prove that there exist $a, b, c \\in A$ such that $a < b < c$ and $c + 2a > 3b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(IMO Shortlist 2021 A1) See the official solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31746,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $a$, $b$ and $c$ are nonzero real numbers satisfying $abc = 2$. Prove that among the three numbers $2a - \\frac{1}{b}$, $2b - \\frac{1}{c}$ and $2c - \\frac{1}{a}$, at most two of them are greater than $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose on the contrary that all three numbers $2a - \\frac{1}{b}$, $2b - \\frac{1}{c}$ and $2c - \\frac{1}{a}$ are greater than $2$. Since $abc = 2$, at least one of $a$, $b$, $c$ is positive. WLOG assume $a > 0$. Then $2c - \\frac{1}{a} > 2$ implies $c > 1 + \\frac{1}{2a} > 0$, and $2b - \\frac{1}{c} > 2$ implies $b > 1 + \\frac{1}{2c} > 0$.\n\nNext, from $2b - \\frac{1}{c} > 2$, we have\n$$\nb > 1 + \\frac{1}{2c}. \\qquad (1)\n$$\nFrom $2a - \\frac{1}{b} > 2$ and $abc = 2$, we have $\\frac{4}{bc} - \\frac{1}{b} > 2$, and so\n$$\nb < \\frac{2}{c} - \\frac{1}{2}. \\qquad (2)\n$$\nCombining (1) and (2), we have $1 + \\frac{1}{2c} < b < \\frac{2}{c} - \\frac{1}{2}$, which easily implies $c < 1$. Similarly we can prove that $a < 1$ and $b < 1$. Together with the fact that they are all positive, we have $abc < 1$, contradicting the condition $abc = 2$. Therefore, at most two of $2a - \\frac{1}{b}$, $2b - \\frac{1}{c}$ and $2c - \\frac{1}{a}$ are greater than $2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31747,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $D$, $E$, $F$ are points on $BC$, $CA$ and $AB$ respectively such that the line segments $AD$, $BE$ and $CF$ meet at $G$. If the lengths of $DG$, $EG$, $BG$, $AG$ and $CF$ are $1$, $2$, $3$, $4$, $5$ respectively, find the area of $\\triangle ABC$.",
    "options": [],
    "answer": "10√2",
    "solution": "Answer: $10\\sqrt{2}$\n\nWe use $[XYZ]$ to denote the area of $XYZ$. Suppose $[GDB] = b$ and $[GDC] = c$. We now make use of the side length ratios to get the following:\n\n* Using $AG:GD = 4:1$, we have $[AGB] = 4b$ and $[AGC] = 4c$.\n* Using $BG:GE = 3:2$, we have $[AGE] = \\frac{2}{3}(4b)$ and $[CGE] = \\frac{2}{3}(b+c)$.\n* As $[AGC] = [AGE] + [CGE]$, we have $4c = \\frac{2}{3}(4b) + \\frac{2}{3}(b+c)$, which gives $b = c$.\n* We have $AF:FB = [AGC]:[BGC] = 4c:2c = 2:1$, so $[FGB] = \\frac{1}{3}[AGB] = \\frac{4}{3}b$.\n* It follows that $FG:GC = [FGB]:[CGB] = \\frac{4}{3}b:2b = 2:3$.\n\nSince $CF = 5$, the last point above gives $GC = 3 = GB$. Hence $\\triangle ABC$ must be isosceles with $AB = AC$ and hence $AD \\perp BC$ (as $b = c$). It then follows that\n$$\nBD = \\sqrt{3^2 - 1^2} = 2\\sqrt{2}\n$$\nand so the area of $\\triangle ABC$ is $\\frac{1}{2} \\cdot BC \\cdot AD = BD \\cdot AD = 10\\sqrt{2}$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31748,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $x^2 - y^2 = \\frac{1}{22}$ and $x \\neq 0$, find the greatest possible value of $\\frac{1 - 22xy}{x^2}$.",
    "options": [],
    "answer": "55/2",
    "solution": "**Answer:** $\\frac{55}{2}$\nLet $x = \\frac{1}{\\sqrt{22}} \\sec \\theta$ and $y = \\frac{1}{\\sqrt{22}} \\tan \\theta$. Then $x^2 - y^2 = \\frac{1}{22}$ and\n$$\n\\begin{aligned}\n\\frac{1 - 22xy}{x^2} &= \\frac{1 - \\sec \\theta \\tan \\theta}{\\frac{1}{22} \\sec^2 \\theta} \\\\\n&= 22(\\cos^2 \\theta - \\sin \\theta) \\\\\n&= 22(-\\sin^2 \\theta - \\sin \\theta + 1) \\\\\n&= 22 \\left[ \\frac{5}{4} - \\left( \\sin \\theta + \\frac{1}{2} \\right)^2 \\right].\n\\end{aligned}\n$$\nThe greatest possible value is thus $22 \\cdot \\frac{5}{4} = \\frac{55}{2}$, attained when $\\sin \\theta = -\\frac{1}{2}$ (which is feasible, e.g. with $\\theta = -30^\\circ$, $x = \\frac{2}{\\sqrt{66}}$ and $y = -\\frac{1}{\\sqrt{66}}$).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31749,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 50 rods of lengths $1, 3, 5, 7, \\ldots, 99$. How many ways are there to pick three of these rods to form a triangle?",
    "options": [],
    "answer": "9500",
    "solution": "The answer is $\\binom{50}{3}$ minus the number of choices of three rods that violate the triangle inequality. The number to be subtracted is the size of the set\n$$\nS = \\{(a, b, c) : 1 \\le a < b < c \\le 99,\\ a + b < c,\\ a, b, c \\text{ are odd numbers}\\}.\n$$\nNote that for $(a, b, c) \\in S$, $a+b$ is an even number between $4$ and $98$ (inclusive). For each such even number, we can easily count the number of such pairs $(a, b)$, and for each such $(a, b)$ we can easily count the number of choices of $c$ for which $(a, b, c) \\in S$.\n\nAs an example, if we fix $a + b = 20$, there would be $5$ choices for $(a, b)$, namely, $(1, 19)$, $(3, 17)$, $(5, 15)$, $(7, 13)$ and $(9, 11)$. For each of these $5$ pairs of $(a, b)$, there are $40$ choices of $c$ for which $(a, b, c) \\in S$, namely, $21, 23, 25, \\ldots, 97, 99$.\nIn the same way, we can see that for each even number $2k$ where $k \\in \\{2, 3, 4, \\dots, 48, 49\\}$, there would be $\\lfloor \\frac{1}{2}k \\rfloor$ choices of $(a, b)$ for which $a + b = 2k$, and for each such $(a, b)$ there would be $50-k$ choices of $c$ for which $(a, b, c) \\in S$. This can be summarized by the following table:\n\n| Value of $k$ | Value of $a + b$ | No. of choices of $(a, b)$ | No. of choices of $c$ |\n|:------------:|:----------------:|:--------------------------:|:---------------------:|\n| 2            | 4                | $\\lfloor \\frac{2}{2} \\rfloor = 1$ | $50 - 2 = 48$         |\n| 3            | 6                | $\\lfloor \\frac{3}{2} \\rfloor = 1$ | $50 - 3 = 47$         |\n| 4            | 8                | $\\lfloor \\frac{4}{2} \\rfloor = 2$ | $50 - 4 = 46$         |\n| 5            | 10               | $\\lfloor \\frac{5}{2} \\rfloor = 2$ | $50 - 5 = 45$         |\n| ...          | ...              | ...                          | ...                   |\n| 48           | 96               | $\\lfloor \\frac{48}{2} \\rfloor = 24$ | $50 - 48 = 2$         |\n| 49           | 98               | $\\lfloor \\frac{49}{2} \\rfloor = 24$ | $50 - 49 = 1$         |\n\nFrom the table, we see that\n$$\n\\begin{aligned}\n|S| &= 1 \\times 48 + 1 \\times 47 + 2 \\times 46 + 2 \\times 45 + \\dots + 24 \\times 2 + 24 \\times 1 \\\\\n&= 1 \\times 95 + 2 \\times 91 + \\dots + 24 \\times 3 \\\\\n&= \\sum_{k=1}^{24} k(99 - 4k) \\\\\n&= 99(1 + 2 + \\dots + 24) - 4(1^2 + 2^2 + \\dots + 24^2) \\\\\n&= 99 \\cdot \\frac{24 \\cdot 25}{2} - 4 \\cdot \\frac{24(24+1)(2 \\cdot 24+1)}{6} \\\\\n&= 10100\n\\end{aligned}\n$$\nand so the answer is\n$$\n\\binom{50}{3} - 10100 = \\frac{50 \\cdot 49 \\cdot 48}{6} - 10100 = 9500.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31750,
    "subject": "Mathematics (Multi-modal)",
    "question": "A 'palindrome' is a positive integer which reads the same from left to right as from right to left, such as $12321$ and $259952$. Someone wrote down a five-digit palindrome $m$ and then removed a digit of $m$ to obtain a four-digit positive integer $n$ (that does not start with $0$). How many possible values of $n$ are there?",
    "options": [],
    "answer": "2358",
    "solution": "**Answer:** $2358$\n\nNote that $m$ is of the form $ABCBA$ with $A$ nonzero, so $n$ is of the form $BCBA$, $ACBA$, $ABBA$, $ABCA$ or $ABCB$. The second, third and fourth cases can be combined, so we are down to three types of possible values of $n$:\n\n* Type I — Equal thousands digit and tens digit, with nonzero unit digit;\n\n* Type II — Equal thousands digit and unit digit;\n\n* Type III — Equal hundreds digit and unit digit.\n\nThere are $9 \\times 10 \\times 9 = 810$ possibilities of Type I (9 choices for the common thousands and tens digit, 10 choices for the hundreds digit and 9 choices for the unit digit), and similarly $9 \\times 10 \\times 10 = 900$ possibilities for each of Type II and Type III. However,\n\n* 90 numbers are of both Types I and II (those of the form $XYXX$ with $X$ nonzero);\n\n* 81 numbers are of both Types I and III (those of the form $XYXY$ with both $X$ and $Y$ nonzero);\n\n* 90 numbers are of both Types II and III (those of the form $XXYX$ with $X$ nonzero);\n\n* 9 numbers are of all three types ($1111$, $2222$, ..., $9999$).\n\nBy the inclusion-exclusion principle, the answer is $810+900+900-90-81-90+9 = 2358$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31751,
    "subject": "Mathematics (Multi-modal)",
    "question": "Someone obtained a positive integer $n$ by concatenating the positive integers from 1 to 100000 in order, i.e. $n = 123456789101112...9999899999100000$. How many times does '2022' appears as consecutive digits of $n$?",
    "options": [],
    "answer": "44",
    "solution": "For clearer illustration, we add a comma between two integers as we form $n$ by concatenating the integers from $1$ to $100000$, i.e. we write\n$$\nn = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, \\dots, 99998, 99999, 100000.\n$$\n\nWe count the number of occurrences of $2022$ according to the positions of the commas within these digits.\n\n* **Case 1: 2022 (i.e. no comma within the digits)**\nThe first $2$ occurs either as a ten thousands digit or as a thousands digit. Each occurs $10$ times, giving $20$ occurrences in this case:\n- $2022$, $20221$, $20222$, $20223$, $20224$, $20225$, $20226$, $20227$, $20228$, $20229$;\n- $2021$, $2022$, $2023$; $12021$, $12022$, $12023$; $\\ldots$; $92021$, $92022$, $92023$.\n\n* **Case 2: 202, 2**\nThe first part $202$ may come from the three digit number $202$, the four digit number $2202$, or a five-digit number starting with $2$ (there are $10$ such numbers), giving $12$ occurrences in this case:\n- $202$, $203$;\n- $2202$, $2203$;\n- $20202$, $20203$; $21202$, $21203$; $\\ldots$; $29202$, $29203$.\n\n* **Case 3: 20, 22**\nThe first part $20$ may come from the three-digit number $220$, four-digit number $2220$, or a five-digit number starting with $22$ (there are $10$ such numbers), giving $12$ occurrences in this case:\n- $220$, $221$;\n- $2220$, $2221$;\n- $22020$, $22021$, $22120$, $22121$; $\\ldots$; $22920$, $22921$.\n\nIt follows that the answer is $20+12+12=44$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31752,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a parallelogram with $\\angle B$ acute. A circle is tangent to $BC$, $CD$ and $DA$. The circle intersects $AC$ at $M$ and $N$, where $M$ is closer to $A$ than $N$. If $AM = 9$, $MN = 16$ and $NC = 2$, find the area of $ABCD$.",
    "options": [],
    "answer": "324*sqrt(2)",
    "solution": "Let $P$, $Q$, $R$ be the points where the circle touches $BC$, $CD$ and $DA$ respectively. Using power, we have $AR = \\sqrt{AM \\cdot AN} = 15$ and $CP = CQ = \\sqrt{CN \\cdot CM} = 6$. Let $DR = DQ = x$ and $S$ be the foot of the perpendicular from $C$ to $AD$. Then we have $DS = x - 6$, $AS = 21$ and $CD = x + 6$. Using\n\n![](attached_image_1.png)\n\n$$\n27^2 - 21^2 = CS^2 = (x+6)^2 - (x-6)^2,\n$$\n\nwe get $x = 12$ and $CS = 12\\sqrt{2}$. The area of $ABCD$ is thus equal to\n\n$$\nAD \\cdot CS = (21 + 6) \\cdot 12\\sqrt{2} = 324\\sqrt{2}.\n$$\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31753,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are four positive integers. By computing the H.C.F. of two of them at a time, one gets six different values $1$, $2$, $3$, $4$, $5$, $k$. Find the smallest possible value of $k$.",
    "options": [],
    "answer": "14",
    "solution": "Answer: $14$\n\nFirst note that there are exactly three even numbers and one odd number. (If all four numbers are even then all H.C.F.'s would be even, whereas if there are at most two even numbers then at most one H.C.F. can be even.) Suppose $a$ is the odd number and $b, c, d$ are the even numbers. We use $(m, n)$ to denote the H.C.F. of $m$ and $n$. Then $(a, b)$, $(a, c)$ and $(a, d)$ must be $1$, $3$ and $5$ in some order. In particular, we note that $a$ is divisible by $3$ and $5$, whereas exactly one of $b, c, d$ is divisible by $3$, and exactly one of $b, c, d$ is divisible by $5$.\n\nWLOG suppose $(b, c) = 2$, $(b, d) = 4$ and $(c, d) = k > 5$. Then $b$ and $d$ are both divisible by $4$, and $c$ is divisible by $2$ but not $4$. It follows that $k$ is also divisible by $2$ but not $4$. Furthermore, $k$ is not divisible by $3$ since $c$ and $d$ cannot be both divisible by $3$, and similarly $k$ is not divisible by $5$.\n\nNow we know that $k$ is greater than $5$, is even, and is not divisible by $3$, $4$, $5$. The smallest such $k$ is thus $14$. Indeed this works; for instance if the four numbers are $(a, b, c, d) = (15, 4, 70, 84)$, then the H.C.F.'s involving $a$ will be equal to $1$, $3$, $5$, while those not involving $a$ will be equal to $2$, $4$, $14$. It follows that the answer is $14$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31754,
    "subject": "Mathematics (Multi-modal)",
    "question": "At a party there are $1234$ participants, and each of them has shaken hands with exactly $137$ other participants. It is known that no three participants have shaken hands with each other. Furthermore, for any two participants $A$ and $B$ who have not shaken hands with each other, there must be exactly $k$ other participants who have shaken hands with both $A$ and $B$, where $k$ is a fixed constant. Find the value of $k$.",
    "options": [],
    "answer": "17",
    "solution": "Answer: $17$\n\nConsider any participant $x$. He has shaken hands with $137$ other participants, say $y_1, y_2, \\dots, y_{137}$ (collectively known as Group Y participants). Also, there are $1234 - 1 - 137 = 1096$ participants who have not shaken hands with $x$; let's call them $z_1, z_2, \\dots, z_{1096}$ (collectively known as Group Z participants).\n\nWe count the number of times a Group Y participant has shaken hand with a Group Z participant. This can be done in two ways:\n\n* Since no two participants in Group Y can shake hands (otherwise together with $x$ there will be three participants shaking hands with each other), every Group Y participant must have shaken hands with exactly $136$ participants in Group Z. The total number of such handshakes is thus equal to $137 \\times 136$.\n\n* On the other hand, every participant in Group Z (say $z$) must shake hands with exactly $k$ participants in Group Y, because $x$ and $z$ have not shaken hands implies there are exactly $k$ participants (who must be in Group Y) who have shaken hands with both $x$ and $z$. It follows that the total number of such handshakes is also equal to $1096k$.\n\nFrom the above two ways of counting we obtain the equality $1096k = 137 \\times 136$, which gives $k = 17$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31755,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AB < AC$. The internal bisector of $\\angle BAC$ meets $BC$ at $D$, while the external bisector of $\\angle BAC$ meets $CB$ produced at $E$. If $EB = 2022$ and the lengths of $BD$ and $DC$ are integers, how many possible lengths of $BD$ are there?",
    "options": [],
    "answer": "16",
    "solution": "Let the lengths of $BD$ and $DC$ be $x$ and $y$ respectively. By the angle bisector theorem, we have $\\frac{x}{y} = \\frac{AB}{AC} = \\frac{EB}{EC}$, i.e.\n$$\n\\frac{x}{y} = \\frac{2022}{2022 + x + y}\n$$\n\n![](attached_image_1.png)\n\n$$\ny = \\frac{x^2 + 2022x}{2022 - x} = -x - 4044 + \\frac{2022 \\cdot 4044}{2022 - x}\n$$\nHence we must choose positive integers $x$ for which $2022 - x$ is a positive integer dividing $2022 \\cdot 4044$ (note that every such $x$ will result in a positive integer value of $y$ and the existence of such a figure satisfying all the conditions). As $2022 \\cdot 4044 = 2^3 \\cdot 3^2 \\cdot 337^2$, its only positive factors that are less than $2022$ are the 12 positive factors of $2^3 \\cdot 3^2$ as well as $337, 337 \\cdot 2, 337 \\cdot 3$ and $337 \\cdot 4$. Hence there are altogether 16 such values of $2022 - x$, and they correspond to 16 possible values of $x$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31756,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the period of the repetend of the fraction $\\frac{39}{1428}$ using binary numbers, i.e. its binary decimal representation.\n\n(Note: When a proper fraction is expressed as a decimal number (of any base), either the decimal number terminates after finite steps, or it is of the form\n$$\n0.b_1b_2\\cdots b_s a_1a_2\\cdots a_k a_1a_2\\cdots a_k\\cdots\n$$\nHere the repeated sequence $a_1a_2\\cdots a_k$ is called the *repetend* of the fraction, and the smallest length of the repetend, $k$, is called the *period* of the decimal number.)",
    "options": [],
    "answer": "24",
    "solution": "Note that $\\frac{39}{1428} = \\frac{13}{476}$ in lowest term. Let\n$$\n\\frac{13}{476} = 0 \\cdot b_1 b_2 \\cdots b_s \\overline{a_1 a_2 \\cdots a_k} \\qquad (1)\n$$\nin binary decimal representation, where $a_1 a_2 \\cdots a_k$ is its repetend and $k$ is its period. We can write $0.b_1 b_2 \\cdots b_s = \\frac{B}{2^s}$ and $0.a_1 a_2 \\cdots a_k = \\frac{A}{2^k}$ for some integers $A$ and $B$. Thus,\n$$\n\\frac{13}{476} = \\frac{B}{2^s} + \\frac{A}{2^s \\cdot 2^k} + \\frac{A}{2^s \\cdot 2^{2k}} + \\frac{A}{2^s \\cdot 2^{3k}} + \\cdots = \\frac{B}{2^s} + \\frac{A}{2^s(2^k - 1)} = \\frac{B(2^k - 1) + A}{2^s(2^k - 1)}. \\quad (2)\n$$\nNote that (2) is equivalent to (1). Therefore, it remains to find the smallest positive integer $k$ such that (2) holds for some $A, B, s$ satisfying $0 \\le A < 2^k$ and $0 \\le B < 2^s$.\nAs $476 = 2^2 \\times 7 \\times 17$, we must have $7 \\mid 2^k - 1$ and $17 \\mid 2^k - 1$. Since the orders of 2 modulo 7 and 17 are 3 and 8 respectively, we have $3 \\mid k$ and $8 \\mid k$. Therefore, the smallest possible $k$ is 24. Indeed, we can take $s = 2$, $B = 0$ and $A = 13 \\times \\frac{2^{24} - 1}{119}$, so that (2) is satisfied. This shows the period is 24.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31757,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ with the following property: the $k$ positive divisors of $n$ have a permutation $(d_1, d_2, \\dots, d_k)$ such that for every $i = 1, 2, \\dots, k$, the number $d_1 + d_2 + \\dots + d_i$ is a perfect square.",
    "options": [],
    "answer": "n = 1 or n = 3",
    "solution": "2. (IMO Shortlist 2021 N3) See the official solution.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 31758,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be the two roots of the equation $x^{\\frac{4}{3}} - 2022x^{\\frac{2}{3}} + 2023 = 0$. If $p = a + 3a^{\\frac{1}{3}}b^{\\frac{2}{3}}$ and $q = b + 3a^{\\frac{2}{3}}b^{\\frac{1}{3}}$, find the value of $(p+q)^{\\frac{1}{3}} + (p-q)^{\\frac{1}{3}}$.",
    "options": [],
    "answer": "4044",
    "solution": "Answer: 4044\nLet $a = u^3$ and $b = v^3$. Then we have $u^4 - 2022u^2 + 2023 = 0$ and $v^4 - 2022v^2 + 2023 = 0$, so $u^2$ and $v^2$ are the two roots of $t^2 - 2022t + 2023 = 0$. In particular we have $u^2 + v^2 = 2022$. Note also that $p+q = u^3 + 3uv^2 + v^3 + 3u^2v = (u+v)^3$ and similarly $p-q = (u-v)^3$. It follows that\n$$\n(p+q)^{\\frac{2}{3}} + (p-q)^{\\frac{2}{3}} = (u+v)^2 + (u-v)^2 = 2(u^2 + v^2) = 4044.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31759,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Show that if $p$ is a prime dividing $5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$, then $p \\equiv 1 \\pmod 4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, $p \\neq 2, 5$. Let $m = 5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$. Then\n$$\n(2 \\cdot 5^{2n} - 5^n + 2)^2 - 5 \\cdot 5^{2n} = 4m \\equiv 0 \\pmod{p}.\n$$\nThis gives $5 \\equiv (5^{-n}(2 \\cdot 5^{2n} - 5^n + 2))^2 \\pmod{p}$. Using the Legendre symbol, we have $\\left(\\frac{5}{p}\\right) = 1$. On the other hand, we have\n$$\n(5^{2n} - 5^n + 1)^2 + 5^n(5^n - 1)^2 = m \\equiv 0 \\pmod{p}.\n$$\nAs above, this implies $\\left(\\frac{-5^n}{p}\\right) = 1$. It follows that\n$$\n\\left(\\frac{-1}{p}\\right) = \\left(\\frac{-5^n}{p}\\right) \\left(\\frac{5^n}{p}\\right) = 1 \\cdot 1^n = 1.\n$$\nTherefore, $p \\equiv 1 \\pmod{4}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31760,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ be a polynomial with integer coefficients. Define a sequence $\\{a_n\\}$ by $a_0 = 0$ and $a_n = P(a_{n-1})$ for all $n \\ge 1$. Prove that if there exists a positive integer $m$ for which $a_m = 0$, then $a_1 = 0$ or $a_2 = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $a_0 = 0 = a_m$, we can prove by induction that $a_k = a_{m+k}$ for all $k$. If there are two consecutive terms of $\\{a_n\\}$ which are the same, then all subsequent terms are the same. By the periodicity, all terms are the same, and they are equal to $a_0 = 0$. This gives $a_1 = 0$ and we are done.\n\nNow, assume $d_n = a_{n+1} - a_n \\ne 0$ for all $n$. Using the fact $x - y \\mid P(x) - P(y)$ for all $x, y \\in \\mathbb{Z}$, we know that $d_n$ divides $P(a_{n+1}) - P(a_n) = a_{n+2} - a_{n+1} = d_{n+1}$ for all $n$. This gives\n$$\n|a_1 - a_0| = |d_0| \\le |d_1| \\le |d_2| \\le \\dots \\le |d_m| = |a_{m+1} - a_m| = |a_1 - a_0|.\n$$\n\nTherefore, all equalities hold, and we have $|d_n| = |d_0|$ for all $n$, which means $d_n = \\pm d_0$ for all $n$. Now, note that\n$$\nd_0 + d_1 + \\cdots + d_{m-1} = a_m - a_0 = 0.\n$$\nSince $d_n \\neq 0$, there are both positive and negative terms among $d_0, d_1, \\dots, d_{m-1}$. WLOG assume $d_k$ and $d_{k+1}$ have different signs for some $k < m$. Then\n$$\na_{k+2} = a_k + d_k + d_{k+1} = a_k \\pm d_0 \\mp d_0 = a_k.\n$$\nThis implies $a_{n+2} = a_n$ for all $n \\ge k$, and so $a_2 = a_{m+2} = a_m = 0$ as desired.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31761,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be roots of the equation $x^4 + x + 1 = 0$. Let $a^5 + 2a + 1$, $b^5 + 2b + 1$, $c^5 + 2c + 1$, $d^5 + 2d + 1$ be roots of the equation $x^4 + px^3 + qx^2 + rx + s = 0$. Find the value of $p + 2q + 4r + 8s$.",
    "options": [],
    "answer": "30",
    "solution": "The answer is $30$.\nFirstly, since $a^4 + a + 1 = 0$, we have\n$$\na^5 + 2a + 1 = a(a^4 + a + 1) - a^2 + a + 1 = -a^2 + a + 1.\n$$\nLet $y = -x^2 + x + 1$. Then $x = \\frac{1 \\pm \\sqrt{-4y+5}}{2}$. Now,\n$$\n\\begin{aligned}\n0 &= x^4 + x + 1 \\\\\n  &= \\left( \\frac{3 - 2y \\pm \\sqrt{-4y+5}}{2} \\right)^2 + \\frac{1 \\pm \\sqrt{-4y+5}}{2} + 1 \\\\\n  &= \\frac{2y^2 - 8y + 7 \\pm (3 - 2y)\\sqrt{-4y+5}}{2} + \\frac{1 \\pm \\sqrt{-4y+5}}{2} + 1 \\\\\n  &= y^2 - 4y + 5 \\pm (2 - y)\\sqrt{-4y+5}.\n\\end{aligned}\n$$\nThis implies $(y^2 - 4y + 5)^2 = (y - 2)^2(-4y + 5)$, which is a degree $4$ polynomial equation. Therefore, $a^5 + 2a + 1$, $b^5 + 2b + 1$, $c^5 + 2c + 1$ and $d^5 + 2d + 1$ are roots of the equation\n$$\nf(y) = (y^2 - 4y + 5)^2 - (y - 2)^2(-4y + 5) = 0.\n$$\nNote that $f$ is monic. So it is exactly $x^4 + px^3 + qx^2 + rx + s$. Thus,\n$$\np + 2q + 4r + 8s = 8f\\left(\\frac{1}{2}\\right) - \\frac{1}{2} = 30.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31762,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and $E$ be the intersection of $AC$ and $BD$. $P$ and $Q$ are two points on $AC$ such that the points $A$, $E$, $Q$, $P$, $C$ lie on the same straight line in this order, and that $BP$ bisects $\\angle ABC$ whereas $DQ$ bisects $\\angle ADC$. If $AE = 4$, $EQ = 2$ and $QP = 3$, find the length of $PC$.",
    "options": [],
    "answer": "15",
    "solution": "Note that $\\sin \\angle DAB = \\sin \\angle DCB$ as the two angles are supplementary. Using $[XYZ]$ to denote the area of $XYZ$, we have\n$$\n\\frac{AE}{EC} = \\frac{[DAB]}{[DCB]} = \\frac{\\frac{1}{2} \\cdot AD \\cdot AB}{\\frac{1}{2} \\cdot DC \\cdot BC} = \\frac{AB}{BC} \\cdot \\frac{AD}{DC} \\quad (1)\n$$\nBy the angle bisector theorem, we have\n$$\n\\frac{AB}{BC} = \\frac{AP}{PC} \\quad \\text{and} \\quad \\frac{AD}{DC} = \\frac{AQ}{QC}\n$$\nIt follows from (1) that\n$$\n\\frac{AE}{EC} = \\frac{AP}{PC} \\cdot \\frac{AQ}{QC}\n$$\nSetting $PC = x$ and using the given side lengths, this becomes\n$$\n\\frac{4}{5+x} = \\frac{9}{x} \\cdot \\frac{6}{3+x}\n$$\nSolving gives $x = 15$ or $x = -\\frac{9}{2}$, and of course the latter is rejected.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31763,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in a circle $\\Omega$. Let the tangent to $\\Omega$ at $D$ intersect the rays $BA$ and $BC$ at points $E$ and $F$, respectively. A point $T$ is chosen inside the triangle $ABC$ so that $TE \\parallel CD$ and $TF \\parallel AD$. Let $K \\neq D$ be a point on the segment $DF$ such that $TD = TK$. Prove that the lines $AC, DT$ and $BK$ intersect at one point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "3. (IMO Shortlist 2021 G4) See the official solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31764,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many pairs of positive integers $(m, n)$ are there such that $m^2n = 20^{22}$?",
    "options": [],
    "answer": "276",
    "solution": "Answer: 276\nSince $20^{22}$ is a square number, $m^2n$, and hence $n$, must also be a square number. Write $n = k^2$. Then the equation becomes $m^2k^2 = 20^{22}$, or $mk = 20^{11}$. Each positive factor of $20^{11}$ corresponds to a choice of $m$, which in turn corresponds to a solution $(m, n)$ to the original equation. The answer is thus equal to the number of positive factors of $20^{11} = 2^{22}5^{11}$, which is $(22+1)(11+1) = 276$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31765,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be the set of all integers of the form $x^2 + 3xy + 8y^2$ where $x$ and $y$ are integers.\n\na. Show that if $u$ and $v$ are in $S$, then so is $uv$.\n\nb. Can an integer of the form $23k + 7$, with $k$ an integer, belong to $S$?",
    "options": [],
    "answer": "a: yes; b: no",
    "solution": "a.\nThe roots of $z^2 + 3z + 8 = 0$ are $\\frac{-3 \\pm \\sqrt{23}i}{2}$. Let $\\alpha = \\frac{-3 + \\sqrt{23}i}{2}$. Then $\\bar{\\alpha} = \\frac{-3 - \\sqrt{23}i}{2}$, and hence $x^2 + 3xy + 8y^2 = (x - \\alpha y)(x - \\bar{\\alpha}y)$. Note that\n$$\n\\begin{aligned}\n(x_1 - \\alpha y_1)(x_2 - \\alpha y_2) &= (x_1 x_2 + \\alpha^2 y_1 y_2) - \\alpha(x_1 y_2 + x_2 y_1) \\\\\n&= (x_1 x_2 + (-3\\alpha - 8)y_1 y_2) - \\alpha(x_1 y_2 + x_2 y_1) \\\\\n&= (x_1 x_2 - 8y_1 y_2) - \\alpha(x_1 y_2 + x_2 y_1 + 3y_1 y_2).\n\\end{aligned}\n$$\nThus, defining $s = x_1x_2 - 8y_1y_2$ and $t = x_1y_2 + x_2y_1 + 3y_1y_2$, we have\n$$\n\\begin{aligned}\n& (x_1^2 + 3x_1y_1 + 8y_1^2)(x_2^2 + 3x_2y_2 + 8y_2^2) \\\\\n&= (x_1 - \\alpha y_1)(x_1 - \\bar{\\alpha} y_1)(x_2 - \\alpha y_2)(x_2 - \\bar{\\alpha} y_2) \\\\\n&= (x_1 - \\alpha y_1)(x_2 - \\alpha y_2)(\\overline{(x_1 - \\alpha y_1)(x_2 - \\alpha y_2)}) \\\\\n&= (s - \\alpha t)\\overline{(s - \\alpha t)} \\\\\n&= (s - \\alpha t)(s - \\bar{\\alpha} t) \\\\\n&= s^2 + 3st + 8t^2.\n\\end{aligned}\n$$\nThis clearly proves the result.\n\nb.\nNo. Suppose on the contrary that $x^2 + 3xy + 8y^2 \\equiv 7 \\pmod{23}$ for some integers $x$ and $y$. This implies $4x^2 + 12xy + 32y^2 \\equiv 28 \\pmod{23}$, and hence $(2x + 3y)^2 \\equiv 5 \\pmod{23}$. However, we can check that 5 is not a square modulo 23 by computing the Legendre symbol $\\binom{5}{23} = \\binom{23}{5} = \\binom{3}{5} = -1$ (or by testing $0^2, (\\pm 1)^2, \\dots, (\\pm 11)^2 \\pmod{23}$). This is a contradiction, and so there is no such integer in $S$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31766,
    "subject": "Mathematics (Multi-modal)",
    "question": "Four non-overlapping families dine at a restaurant together. Each family consists of one father, one mother and three children. There are five tables, which are red, blue, yellow, green and brown respectively. Every table can accommodate four people, and every person must sit at some table. How many ways are there to assign the 20 people to tables so that each child sits at the same table as at least one of his/her parents? (We do not distinguish between the four seats at the same table.)",
    "options": [],
    "answer": "278400",
    "solution": "There are 278400 ways to assign the people.\n\nNote that each family can be partitioned in either a (4, 1) fashion (where exactly one parent sits with three children and the other is alone) or a (3, 2) fashion (where one parent sits with two children and the other sits with one).\n\n**Case 1.** All families are partitioned in (4, 1) fashion.\nThe 4-group of each family occupies a table, and the other four parents take the remaining table. There are 2 ways to choose a parent of each family. So the number of ways is $2^4 = 16$ if we regard the tables as the same.\n\n**Case 2.** Three families are partitioned in (4, 1) fashion and one is partitioned in (3, 2) fashion.\nThere are 4 ways to choose which family to be in (3, 2) fashion. The 3-group must sit with one of the three parents from another family, while the 2-group must sit with the remaining two parents. There are 2 ways to choose a parent of each family, and 3 ways to choose a child of the family in (3, 2) fashion to be put in the 2-group. So the number of ways is $4 \\times 3 \\times 2^4 \\times 3 = 576$.\n\n**Case 3.** Two families are partitioned in (4, 1) fashion and two are partitioned in (3, 2) fashion.\nThere are $\\binom{4}{2} = 6$ ways to choose which two families to be in (3, 2) fashion. The two 2-groups must sit together, while the 3-groups must sit with one of the two parents from the other families. There are 2 ways to choose a parent of each family, and 3 ways to choose a child of each family in (3, 2) fashion to be put in the 2-groups. So the number of ways is $6 \\times 2 \\times 2^4 \\times 3^2 = 1728$.\n\nIt is easy to check that no other cases are possible. Since the tables are different, there are $5! = 120$ ways to arrange the tables. Therefore the answer is\n$$\n120 \\times (16 + 576 + 1728) = 278400.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31767,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $2023 \\times 2023$ square grid, there are beetles on some of the unit squares, with at most one beetle on each unit square. In the first minute, every beetle will move one step to its right or left adjacent square, or to its top or bottom adjacent square. In the second minute, every beetle will move again, only this time, in case the beetle moved right or left in the previous minute, it moves to top or bottom in this minute, and vice versa, and so on. What is the minimum number of beetles on the square grid to ensure that, no matter where the beetles are initially, and how they move in the first minute, but after finitely many minutes, at least two beetles will meet at a certain unit square?",
    "options": [],
    "answer": "4088485",
    "solution": "The answer is $2022^2 + 1 = 4088485$.\nFirst we show that $2022^2$ is insufficient. Suppose there are $2022^2$ beetles in the top left $2022 \\times 2022$ square grid. This square grid can be partitioned into many $2 \\times 2$ square grids in the natural way. If the beetles in each $2 \\times 2$ square grid move in the anticlockwise direction as shown, then no beetle will ever meet.\n\n![](attached_image_1.png)\n\nNext, suppose there are at least $2022^2 + 1$ beetles. We label all the unit squares as follows. In the odd rows, we label the unit squares $ABABAB\\cdots ABA$. In the even rows, we label the unit squares $CDCDCD\\cdots CDC$. Note that there are $1011^2$ unit squares with label $D$.\nAs there are at least $2022^2+1$ beetles in total, we can always find at least $1011^2+1$ beetles lying in unit squares with the same label by the pigeonhole principle. In each move, either all beetles move in the order $A \\to B \\to D \\to C \\to A$ or all beetles move in the order $A \\to C \\to D \\to B \\to A$. In any case, these $1011^2 + 1$ beetles must move to unit squares with label $D$ simultaneously in a certain move. As there are only $1011^2$ unit squares with label $D$, two of the beetles must meet at a certain unit square. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31768,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a chess tournament there are $100$ players. On each day of the tournament, each player is designated to be 'white', 'black' or 'idle', and each 'white' player will play a game against every 'black' player. (You may assume that all games fixed for the day can be finished within that day.) At the end of the tournament, it was found that any two players have met exactly once. What is the minimum duration of days that the tournament lasts?",
    "options": [],
    "answer": "99",
    "solution": "Clearly the tournament can be run in $99$ days if we assign, on day $k$ (where $1 \\le k \\le 99$),\n\n* player $k$ to be 'white';\n* players $k + 1$, $k + 2$, $\\ldots$, $100$ to be 'black'; and\n* all other players to be 'idle'.\n\nIn this way player $i$ and player $j$ (where $1 \\le i < j \\le 100$) would have met exactly once on day $i$.\n\nNow suppose the number of days is less than $99$, and we would derive a contradiction. Note that we can assign a real number $x_i$ to each player $i$, such that\n(1) at least one $x_i$ is nonzero;\n(2) the sum of all $x_i$ is zero; and\n(3) the sum of the $x_i$'s of all 'white' players on each day is zero.\n\nThis is because (2) and (3) represent a homogeneous system of fewer than $100$ linear equations in $100$ unknowns, and hence must have a nonzero solution (i.e. condition (1)). It follows that\n$$\n0 = (x_1 + x_2 + \\cdots + x_{100})^2 = (x_1^2 + x_2^2 + \\cdots + x_{100}^2) + 2 \\sum_{1 \\le i < j \\le 100} x_i x_j. \\quad (4)\n$$\nSince any two players have met exactly once, we have\n$$\n\\sum_{1 \\le i < j \\le 100} x_i x_j = \\sum_k \\left( \\sum_{w \\in W_k} x_w \\sum_{b \\in B_k} x_b \\right)\n$$\nwhere $W_k$ and $B_k$ denote the set of 'white' and 'black' players on day $k$ respectively. Since the sum of the $x_i$'s of all 'white' players on each day is zero, the above summation is equal to $0$. On the other hand, $x_1^2 + x_2^2 + \\cdots + x_{100}^2 > 0$ as at least one $x_i$ is nonzero. This gives a contradiction to (4).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31769,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in a circle $\\Gamma$ such that $AB = BC = CD$. Let $M$ and $N$ be the midpoints of $AD$ and $AB$ respectively. The line $CM$ meets $\\Gamma$ again at $E$. Prove that the tangent at $E$ to $\\Gamma$, the line $AD$ and the line $CN$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the intersection of the tangent at $B$ to $\\Gamma$ and the line $AD$. Then we have\n$$\n\\angle ABP = \\angle ACB = \\angle BAC.\n$$\nThis shows $BP // CA$. Note that $CB // AP$ since $ABCD$ is an isosceles trapezoid. Therefore, $PBCA$ is a parallelogram. As $N$ is the midpoint of $AB$, the diagonal $CP$ passes through $N$.\n\nNext, the areas of $\\triangle CDE$ and $\\triangle CAE$ are the same since $MD = MA$. This implies\n$$\n\\frac{1}{2} DC \\times DE \\sin \\angle CDE = \\frac{1}{2} AC \\times AE \\sin \\angle CAE.\n$$\nAs $\\angle CDE + \\angle CAE = 180^\\circ$, we have $\\sin \\angle CDE = \\sin \\angle CAE$. Therefore,\n$$\n\\frac{ED}{EA} = \\frac{CA}{CD} = \\frac{BD}{BA}.\n$$\nThis shows $ABDE$ is a harmonic quadrilateral, and hence the tangent at $E$ to $\\Gamma$ passes through $P$. This proves the tangent at $E$, the line $AD$ and the line $CN$ are concurrent at $P$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31770,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with circumcentre $O$. Diagonals $AC$ and $BD$ meet at $E$. $F$ and $G$ are points on segments $AB$ and $CD$ respectively. Suppose $AF = 15$, $FB = 13$, $BE = 30$, $ED = 13$, $DG = 7.5$ and $GC = 6.5$. Let $P$ be a point such that $PF \\perp AB$ and $PG \\perp CD$. Find $\\frac{PE}{PO}$.",
    "options": [],
    "answer": "5",
    "solution": "The answer is $\\frac{PE}{PO} = 5$.\nNote that $\\frac{AF}{FB} = \\frac{DG}{GC}$. Now if we let $M$ and $N$ be the perpendicular feet from $O$ to $AB$ and $CD$ respectively, we have $\\frac{AM}{MB} = \\frac{DN}{NC}$. Also if we let $H$ and $K$ be the perpendicular feet from $E$ to $AB$ and $CD$ respectively, then by the fact\n\n$\\triangle EAB \\sim \\triangle EDC$, we have $\\frac{AH}{HB} = \\frac{DK}{KC}$. Therefore, $P$, $O$, $E$ are collinear, and $\\frac{PE}{PO} = \\frac{FH}{FM} = \\frac{GK}{GN}$.\n![](attached_image_1.png)\nNow $AF = 15$ and $AM = \\frac{AF + FB}{2} = 14$. Since $\\triangle EAB \\sim \\triangle EDC$ with $\\frac{AB}{DC} = 2$, we have $EA = 2ED = 26$. By the cosine formula, we have\n$$\nAH = AE \\cos A = 26 \\times \\frac{26^2 + 28^2 - 30^2}{2 \\times 26 \\times 28} = 10.\n$$\nHence we conclude $\\frac{PE}{PO} = \\frac{FH}{FM} = \\frac{AF - AH}{AF - AM} = \\frac{15 - 10}{15 - 14} = 5$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31771,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given that $22! = 1124000727777 \\boxed{\\phantom{000000}}680000$, where the box contains three missing digits, write down the missing digits from left to right.",
    "options": [],
    "answer": "607",
    "solution": "Answer: 607\nLet the digits in the box form the number $\\overline{abc}$. Since $22!$ is divisible by $7 \\times 11 \\times 13 = 1001$ and $1000 \\equiv -1 \\pmod{1001}$, we have\n$$\n\\begin{aligned}\n0 &\\equiv \\overline{1124000727777abc680000} \\\\\n  &= 1(1000)^7 + 124(1000)^6 + 727(1000)^4 + 777(1000)^3 + \\overline{abc}(1000)^2 + 680(1000) \\\\\n  &\\equiv -1 + 124 + 727 - 777 + \\overline{abc} - 680 \\\\\n  &= \\overline{abc} - 607 \\pmod{1001}.\n\\end{aligned}\n$$\nIt follows that the answer is 607.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31772,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $k$ be two integers with $n > k \\ge 1$. There are $2n+1$ students standing in a circle. Each student $S$ has $2k$ neighbours—namely, the $k$ students closest to $S$ on the right, and the $k$ students closest to $S$ on the left.\nSuppose that $n+1$ of the students are girls, and the other $n$ are boys. Prove that there is a girl with at least $k$ girls among her neighbours.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "4. (IMO Shortlist 2021 C5) See the official solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31773,
    "subject": "Mathematics (Multi-modal)",
    "question": "One edge of a triangular pyramid has length $6$ while every other edge has length $5$. Find the volume of the pyramid.",
    "options": [],
    "answer": "5*sqrt(39)/2",
    "solution": "Answer: $\\frac{5\\sqrt{39}}{2}$\n\nLet the vertices of the pyramid be $A$, $B$, $C$, $D$, where $BD = 6$ and all other edges have length $5$. Let also $M$ be the midpoint of $BD$ and $N$ be the foot of perpendicular from $A$ to the base $BCD$. Of course we have $BM = MD = 3$ and $CM = 4$. Note that\n$$\nAN^2 = AB^2 - BN^2 = AC^2 - CN^2 = AD^2 - DN^2.\n$$\nAs $AB = AC = AD = 5$, we have $BN = CN = DN$ and so $N$ is the circumcentre of $\\triangle BCD$. The circumradius of $\\triangle BCD$ is given by the extended sine formula\n$$\nBN = \\frac{CD}{2 \\sin B} = \\frac{5}{2 \\cdot \\frac{4}{5}} = \\frac{25}{8}.\n$$\nThe area of $\\triangle BCD$ is $\\frac{1}{2} \\cdot BD \\cdot MC = 12$. The volume of the pyramid is thus\n$$\n\\frac{1}{3} \\cdot 12 \\cdot AN = 4 \\cdot \\sqrt{5^2 - \\left(\\frac{25}{8}\\right)^2} = \\frac{5\\sqrt{39}}{2}.\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31774,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(n) = \\left| \\binom{n}{0} + \\binom{n}{3} + \\binom{n}{6} + \\dots + \\binom{n}{3\\left\\lfloor \\frac{n}{3} \\right\\rfloor} \\right| - \\frac{2^n}{3}$, where $[x]$ is the greatest integer not exceeding $x$. Find $f(1) + f(2) + \\dots + f(2021)$.",
    "options": [],
    "answer": "898",
    "solution": "The answer is 898.\n\nIndeed, let $\\omega = e^{\\frac{2\\pi i}{3}}$ be a cube root of unity. By the binomial theorem, we have\n$$\n(1 + 1)^n = \\binom{n}{0} + \\binom{n}{1} + \\dots + \\binom{n}{n},\n$$\n$$\n(1 + \\omega)^n = \\binom{n}{0} + \\binom{n}{1}\\omega + \\dots + \\binom{n}{n}\\omega^n,\n$$\n$$\n(1 + \\omega^2)^n = \\binom{n}{0} + \\binom{n}{1}\\omega^2 + \\dots + \\binom{n}{n}\\omega^{2n}.\n$$\nIt is known that $1 + \\omega^k + \\omega^{2k} = 0$ for any $3 \\nmid k$, and $1 + \\omega^k + \\omega^{2k} = 3$ for any $3 \\mid k$.\n\nThus, adding up the above 3 equations, we obtain\n$$\n\\begin{aligned}\n& 3\\left(\\binom{n}{0} + \\binom{n}{3} + \\binom{n}{6} + \\dots + \\binom{n}{3\\left\\lfloor \\frac{n}{3} \\right\\rfloor}\\right) \\\\\n&= (1+1)^n + (1+\\omega)^n + (1+\\omega^2)^n \\\\\n&= 2^n + (-\\omega^2)^n + (-\\omega)^n \\\\\n&= \\begin{cases} 2^n + 2 & \\text{if } n \\equiv 0 \\pmod{6}, \\\\ 2^n + 1 & \\text{if } n \\equiv 1,5 \\pmod{6}, \\\\ 2^n - 1 & \\text{if } n \\equiv 2,4 \\pmod{6}, \\\\ 2^n - 2 & \\text{if } n \\equiv 3 \\pmod{6}. \\end{cases}\n\\end{aligned}\n$$\nTherefore, we have $f(n) = \\frac{1}{3}$ if $3 \\nmid n$, and $f(n) = \\frac{2}{3}$ if $3 \\mid n$. As $\\lfloor \\frac{2021}{3} \\rfloor = 673$, we have\n$$\nf(1) + f(2) + \\dots + f(2021) = 673\\left(\\frac{2}{3}\\right) + (2021 - 673)\\left(\\frac{1}{3}\\right) = 898.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31775,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a set of $n$ 01-sequences of length 200. Every pair of 01-sequences differ at least at 101 positions. (For example, the two 01-sequences of length 6, 111100 and 010001 differ at four positions, 1st, 3rd, 4th and 6th positions, counting from the left.) Is it possible that $n \\ge 101$?",
    "options": [],
    "answer": "No; the maximum possible size is 100.",
    "solution": "No, it is not possible. Indeed, let $n_i$ be the number of 01-sequences in the set with 1 at the $i$th entry, where $1 \\le i \\le 200$. We also let $S$ be the total number of different positions between all pairs of 01-sequences in the set. Since every pair differs at least at 101 positions, we have $S \\ge 101 \\binom{n}{2}$. On the other hand, there are $n_i(n - n_i)$ pairs of different digits in the $i$th entry. Thus, by considering the $i$th positions of all pairs of 01-sequences, we have $S = \\sum_{i=1}^{200} n_i(n - n_i)$. Using the AM-GM inequality, we have\n$$\nn_i(n - n_i) \\le \\left( \\frac{n_i + (n - n_i)}{2} \\right)^2 = \\frac{n^2}{4}.\n$$\nHence we have\n$$\n50n^2 = \\sum_{i=1}^{200} \\frac{n^2}{4} \\ge S \\ge 101 \\binom{n}{2}.\n$$\nThis implies $n \\le 101$. However, when $n = 101$, the inequality $\\frac{n^2}{4} \\ge n_i(n - n_i)$ is strict, as one side is an integer and the other side is not. Thus $n < 101$, or $n \\le 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31776,
    "subject": "Mathematics (Multi-modal)",
    "question": "a. Find the smallest number of lines drawn on the plane so that they produce exactly $2022$ points of intersection. (Note: For $1$ point of intersection, the minimum is $2$; for $2$ points, minimum is $3$; for $3$ points, minimum is $3$; for $4$ points, minimum is $4$; for $5$ points, minimum is $4$, etc.)\n\nb. What happens if the lines produce exactly $2023$ intersections?",
    "options": [],
    "answer": "a) 65; b) 65",
    "solution": "a.\nAnswer: $65$\nNote that\n$$\n\\binom{64}{2} = 2016 < 2022 < 2080 = \\binom{65}{2}.\n$$\nAs each pair of straight lines produces at most one intersection point, $64$ lines produce at most $2016$ intersection points, which is insufficient. Therefore, at least $65$ lines are needed.\nIt is possible to have $65$ lines. For example, suppose there are $65$ pairwise non-parallel lines $l_1, l_2, \\dots, l_{65}$ such that for each $k = 1, 2, \\dots, 29$, the lines $l_{2k-1}$ and $l_{2k}$ intersect at a distinct point on $l_{65}$, and there are no concurrent lines other than $l_{2k-1}, l_{2k}, l_{65}$ for $k = 1, 2, \\dots, 29$. Then the number of intersection points is\n$$\n\\binom{65}{2} - 2 \\cdot 29 = 2022.\n$$\n\nb.\nAnswer: $65$\nAs in part (a), at least $65$ lines are needed. An example of $65$ lines is given below. Suppose there are $65$ lines $l_1, l_2, \\dots, l_{65}$ where $l_{64} \\parallel l_{65}$ and there is no other pair of parallel lines. Also, for each $k = 1, 2, \\dots, 28$, the lines $l_{2k-1}$ and $l_{2k}$ intersect at a distinct point on $l_{65}$, and there are no concurrent lines other than $l_{2k-1}, l_{2k}, l_{65}$ for $k = 1, 2, \\dots, 28$. Then the number of intersection points is\n$$\n\\binom{65}{2} - 1 - 2 \\cdot 28 = 2023.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31777,
    "subject": "Mathematics (Multi-modal)",
    "question": "A special calculator contains a red button, which counts the number of even digits of an integer. For instance, when the screen shows $2022$, pressing the red button gives $4$ since all $4$ digits of $2022$ are even. Someone inputs a positive integer $n$ into the calculator and keeps pressing the red button until $0$ is displayed on the screen. If $0$ is displayed after the red button has been pressed four times, find the smallest possible value of $n$.",
    "options": [],
    "answer": "2 × 10^19",
    "solution": "Answer: $2 \\times 10^{19}$\nSuppose the sequence of numbers shown on the calculator screen is\n$$\nn \\to p \\to q \\to r \\to 0\n$$\nwith $p$, $q$, $r$ nonzero. Note that $r$ is at least $1$ and so $q$ consists of at least one even digit, which means $q \\geq 2$. Hence $p$ consists of at least two even digits and is no less than $20$. It follows that $n$ has at least $20$ even digits, and the smallest such number is $2 \\times 10^{19}$, which one easily verifies to be a possible value of $n$. (The sequence in this case will be $20\\ldots00 \\to 20 \\to 2 \\to 1 \\to 0$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31778,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest real root of the equation $2x^2 + 6x + 9 = 7x\\sqrt{2x + 3}$.",
    "options": [],
    "answer": "9 + 6√3",
    "solution": "Let $u = \\sqrt{2x+3}$. Then the equation becomes $2x^2 + 3u^2 = 7xu$, or $(2x-u)(x-3u) = 0$. Hence either $u = 2x$ or $x = 3u$.\n\n* If $u = 2x$, we get $2x = \\sqrt{2x+3}$. Squaring both sides gives $4x^2 = 2x + 3$, which leads to $x = \\frac{1 \\pm \\sqrt{13}}{4}$.\n\n* If $x = 3u$, we get $\\frac{x}{3} = \\sqrt{2x+3}$. Squaring both sides gives $\\frac{1}{9}x^2 = 2x + 3$, which leads to $x = 9 \\pm 6\\sqrt{3}$.\n\nAmong these, the largest one is $x = 9 + 6\\sqrt{3}$. One can check that in this case\n$$\n3u = \\sqrt{18x + 27} = \\sqrt{189 + 108\\sqrt{3}} = \\sqrt{(9 + \\sqrt{108})^2} = x\n$$\nand so such $x$ satisfies the original equation, which is therefore the largest real root of the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31779,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a square with side length $1$. $P$ and $Q$ are points on $AB$ and $BC$ respectively such that $BP = BQ = \\frac{1}{\\sqrt{2}}$. $N$ is the foot of perpendicular from $B$ to $CP$. Find $NQ^2$.",
    "options": [],
    "answer": "(5 - 2√2)/6",
    "solution": "Answer: $\\frac{5 - 2\\sqrt{2}}{6}$\n\nSet the square in the first quadrant of the coordinate plane with $B$ at the origin and $C$ at $(1, 0)$. Then the coordinates of $P$ and $Q$ are $(0, \\frac{1}{\\sqrt{2}})$ and $(\\frac{1}{\\sqrt{2}}, 0)$ respectively.\n\nAs $PC$ has slope $-\\frac{1}{\\sqrt{2}}$, the slope of $BN$ is $\\sqrt{2}$. Let $N = (n, \\sqrt{2}n)$. Considering the slope of $CN$, we have $\\frac{\\sqrt{2}n}{n-1} = -\\frac{1}{\\sqrt{2}}$, which gives $n = \\frac{1}{3}$.\n\n![](attached_image_1.png)\n\nHence $N = (\\frac{1}{3}, \\frac{\\sqrt{2}}{3})$ and so\n$$\nNQ^2 = \\left(\\frac{1}{3} - \\frac{1}{\\sqrt{2}}\\right)^2 + \\left(\\frac{\\sqrt{2}}{3}\\right)^2 = \\frac{5 - 2\\sqrt{2}}{6}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31780,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_n)_{n \\ge 1}$ be a sequence of positive real numbers with the property that\n$$\n(a_{n+1})^2 + a_n a_{n+2} \\le a_n + a_{n+2}\n$$\nfor all positive integers $n$. Show that $a_{2022} \\le 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us denote $b_n = a_n$ for all $n$. The given inequality is:\n$$(a_{n+1})^2 + a_n a_{n+2} \\le a_n + a_{n+2}$$\nfor all $n \\ge 1$.\n\nWe can rearrange the inequality:\n$$(a_{n+1})^2 + a_n a_{n+2} - a_n - a_{n+2} \\le 0$$\n$$(a_{n+1})^2 - a_n + a_n a_{n+2} - a_{n+2} \\le 0$$\n$$(a_{n+1})^2 - a_n + (a_n - 1)a_{n+2} \\le 0$$\n\nLet us try to bound $a_n$ from above. Suppose $a_n > 1$ for some $n$. Since all $a_n > 0$, we can divide both sides by $a_n$ (if needed), but let's try to use the structure.\n\nSuppose $a_n \\le 1$ for all $n$. Then $a_{2022} \\le 1$ as desired. Suppose not, and let $k$ be the smallest index such that $a_k > 1$.\n\nSince $a_1 > 0$, and $a_1 \\le 1$ (otherwise $k=1$), so $a_1 \\le 1$, $a_2 \\le 1$, ..., $a_{k-1} \\le 1$, $a_k > 1$.\n\nNow, consider the inequality for $n = k-1$:\n$$(a_k)^2 + a_{k-1} a_{k+1} \\le a_{k-1} + a_{k+1}$$\n$$(a_k)^2 \\le a_{k+1} (1 - a_{k-1}) + a_{k-1}$$\nBut $a_{k-1} \\le 1$, so $1 - a_{k-1} \\ge 0$, and $a_{k+1} > 0$.\nSo $a_{k+1} (1 - a_{k-1}) + a_{k-1} \\le a_{k+1} + a_{k-1}$.\nBut $(a_k)^2 \\le a_{k+1} (1 - a_{k-1}) + a_{k-1} \\le a_{k+1} + a_{k-1}$.\nBut $a_k > 1$, so $(a_k)^2 > 1$.\nTherefore, $a_{k+1} (1 - a_{k-1}) + a_{k-1} > 1$.\n\nBut $a_{k-1} \\le 1$, so $a_{k+1} (1 - a_{k-1}) > 1 - a_{k-1} \\ge 0$.\nSo $a_{k+1} > 1$ unless $a_{k-1} = 1$.\nIf $a_{k-1} = 1$, then $a_{k+1} (1 - a_{k-1}) = 0$, so $(a_k)^2 \\le 1$.\nBut $a_k > 1$, so $(a_k)^2 > 1$, contradiction.\n\nTherefore, $a_{k-1} < 1$, so $1 - a_{k-1} > 0$, and $a_{k+1} > \\frac{1 - a_{k-1}}{1 - a_{k-1}} = 1$.\nSo $a_{k+1} > 1$.\n\nThus, if $a_k > 1$ for some $k$, then $a_{k+1} > 1$ as well. But then, by induction, $a_n > 1$ for all $n \\ge k$.\n\nNow, consider the original inequality for $n \\ge k$:\n$$(a_{n+1})^2 + a_n a_{n+2} \\le a_n + a_{n+2}$$\nBut $a_n > 1$, $a_{n+2} > 1$, so $a_n + a_{n+2} < (a_n)(a_{n+2})$ (since $a_n, a_{n+2} > 1$).\nSo $a_n a_{n+2} > a_n + a_{n+2}$.\nTherefore,\n$$(a_{n+1})^2 + a_n a_{n+2} > (a_{n+1})^2 + a_n + a_{n+2}$$\nBut the inequality says $(a_{n+1})^2 + a_n a_{n+2} \\le a_n + a_{n+2}$, so $(a_{n+1})^2 \\le 0$, which is impossible since $a_{n+1} > 0$.\n\nTherefore, our assumption that $a_k > 1$ for some $k$ leads to a contradiction. Thus, $a_n \\le 1$ for all $n$, in particular $a_{2022} \\le 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31781,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a $24 \\times 24$ square grid, initially all its unit squares are coloured white. A *move* consists of choosing a row, or a column, and changing the colours of all its unit squares, from white to black, and from black to white. Is it possible that after finitely many moves, the square grid contains exactly 574 black unit squares?",
    "options": [],
    "answer": "No",
    "solution": "No, it is not possible.\nSuppose it is possible that after finitely many moves, the square grid contains exactly 574 black unit squares. Then there are exactly $24 \\times 24 - 574 = 2$ white unit squares. No matter where the white unit squares are, we can always find a $2 \\times 2$ square that contains exactly 1 white unit square and 3 black unit squares. One checks that the parity of the number of white cells in this $2 \\times 2$ square is fixed no matter how we choose the rows or columns in a move. Therefore, it is\n\nimpossible that all the four cells are white initially. This is a contradiction, and\nso it is impossible to have exactly 574 black unit squares.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31782,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1A_2\\dots A_n$ be a convex polygon where $n \\ge 4$. For each $i = 1, 2, \\dots, n$, let $f(i)$ be the maximum of the angles $\\angle A_iA_jA_{i+1}$ where $A_j$ can be any vertex different from $A_i$ and $A_{i+1}$ (with $A_{n+1} = A_1$). For each $n$, find the smallest possible value of $f(1) + f(2) + \\dots + f(n)$.",
    "options": [],
    "answer": "180°",
    "solution": "For each $k = 2, 3, \\dots, n-1$, define $S_k = \\sum_{i=1}^{n} \\angle A_i A_{i+1} A_{i+1}$ where the indices are taken modulo $n$. Note that\n$$\nS_2 + S_3 + \\dots + S_{n-1} = \\angle A_1 + \\angle A_2 + \\dots + \\angle A_n = (n-2) \\cdot 180^\\circ \\quad (1)\n$$\nbecause each angle $\\angle A_i A_j A_k$ with distinct $i, j, k$ is counted exactly once. By definition, the given sum $S$ is at least $S_k$ for each $k = 2, 3, \\dots, n-1$. Therefore, by equation (1), we obtain\n$$\nS \\ge \\frac{S_2 + S_3 + \\dots + S_{n-1}}{n-2} = 180^\\circ.\n$$\nOn the other hand, if $A_1 A_2 \\dots A_n$ is cyclic (for example, it is a regular $n$-gon), then\n$$\nf(1) + f(2) + \\dots + f(n) = A_1 A_2 + A_2 A_3 + \\dots + A_{n-1} A_n = 180^\\circ.\n$$\nTherefore, the smallest possible value of $S$ is $180^\\circ$.\n\n(In the figure below, we have $n=5$. Also, $S_2$ is the sum of all angles in white, $S_3$ is the sum of all angles in grey, and $S_4$ is the sum of all angles in black.)\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31783,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer larger than $1$, and let $a_0, a_1, \\dots, a_{n-1}$ be integers. It is known that the equation\n$$\nx^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \\dots + a_1x + a_0 = 0\n$$\nhas $n$ pairwise relatively prime integer roots. Prove that $a_0$ and $a_1$ are relatively prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $b_1, b_2, \\dots, b_n$ be the integer roots. By Vieta's theorem, we have\n$$\nb_1 b_2 \\cdots b_n = (-1)^n a_0, \\quad (1)\n$$\n$$\nb_1 b_2 \\cdots b_{n-1} + b_1 b_2 \\cdots b_{n-2} b_n + \\cdots + b_2 b_3 \\cdots b_n = (-1)^{n-1} a_1. \\quad (2)\n$$\nSuppose on the contrary that $a_0$ and $a_1$ share a common prime factor $p$. By equation (1), we have $p \\mid b_1 b_2 \\cdots b_n$. WLOG assume $p \\mid b_1$. Then in equation (2), $p$ divides each of the first $n-1$ terms on the left and divides the term on the right. Therefore, we must have $p \\mid b_2 b_3 \\cdots b_n$. This implies one of $b_2, b_3, \\dots, b_n$ is divisible by $p$, contradicting the fact that each of them is relatively prime to $b_1$. Therefore, $a_0$ and $a_1$ must be relatively prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31784,
    "subject": "Mathematics (Multi-modal)",
    "question": "Someone extracted $n$ consecutive digits in the infinite decimal representation of $\\frac{1}{13}$ and found that the sum of the $n$ digits is equal to $2023$. Find the value of $n$.",
    "options": [],
    "answer": "449",
    "solution": "Answer: $449$\n\nThe number is the recurring decimal $0.\\dot{0}7692\\dot{3}$. Note that $0 + 7 + 6 + 9 + 2 + 3 = 27$ and $2023 = 27 \\times 75 - 2$. Hence there should be $75$ periods of '307692' with the last '2' removed.\n\nThe answer is thus $75 \\times 6 - 1 = 449$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31785,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be the midpoint of the side $BC$ of an acute $\\triangle ABC$, and let $D$ be the foot of perpendicular from $C$ to $AM$. The circumcircle of $\\triangle ABD$ intersects the side $BC$ again at $E \\neq B$. Suppose $F$ is a point on the segment $AE$ such that $FB = FC$. Prove that $F$ is the midpoint of $AE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P$ be the foot of perpendicular from $A$ to $BC$. It follows from $\\angle ADC = \\angle APC = 90^\\circ$ that $A$, $D$, $P$, $C$ are concyclic. Considering the power of $M$, we find that\n$$\nME \\times MB = MD \\times MA = MP \\times MC.\n$$\nSince $MB = MC$, we have $ME = MP$. Note that $FM$ is the perpendicular bisector of $BC$. Therefore, $FM \\parallel AP$. As $ME = MP$, it follows from the intercept theorem that $F$ is the midpoint of $AE$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31786,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the remainder when $1 \\times 2 \\times 3 \\times 4 + 2 \\times 3 \\times 4 \\times 5 + \\dots + 2023 \\times 2024 \\times 2025 \\times 2026$ is divided by $1000$.",
    "options": [],
    "answer": "120",
    "solution": "Answer: $120$\nLet $f(n) = n(n+1)(n+2)(n+3)(n+4)$. Note that\n$$\n\\begin{aligned}\nf(n+1) - f(n) &= (n+1)(n+2)(n+3)(n+4)(n+5) - n(n+1)(n+2)(n+3)(n+4) \\\\\n&= 5(n+1)(n+2)(n+3)(n+4).\n\\end{aligned}\n$$\nHence we have\n$$\n1 \\times 2 \\times 3 \\times 4 = \\frac{f(1) - f(0)}{5}, \\quad 2 \\times 3 \\times 4 \\times 5 = \\frac{f(2) - f(1)}{5}, \\quad \\text{and so on.}\n$$\nIt follows that the sum in the question is equal to\n$$\n\\frac{f(1)-f(0)}{5} + \\frac{f(2)-f(1)}{5} + \\dots + \\frac{f(2023)-f(2022)}{5} = \\frac{f(2023)-f(0)}{5} = \\frac{f(2023)}{5}\n$$\nwhich in turn is equal to\n$$\n\\frac{2023 \\times 2024 \\times 2025 \\times 2026 \\times 2027}{5} = 405 \\times 2023 \\times 2024 \\times 2026 \\times 2027.\n$$\nTaking modulo $1000$, it suffices to consider\n$$\n\\begin{aligned}\n405 \\times 23 \\times 24 \\times 26 \\times 27 &= 405 \\times (25^2 - 2^2) \\times (25^2 - 1^2) \\\\\n&= 405 \\times 621 \\times 624 \\\\\n&= 10 \\times (81 \\times 621 \\times 312).\n\\end{aligned}\n$$\nIt thus suffices to consider the last two digits of $81 \\times 621 \\times 312$, or simply the last two digits of $81 \\times 21 \\times 12$. Direct computation shows that they are $12$, and so the answer is $120$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31787,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a school there are $2023$ students, numbered $1$ to $2023$. The teacher met the students one by one in the order of their numbers, and gave a candy to each student except if that would mean three students whose numbers form an arithmetic sequence all got candies. In this way, the teacher would give a candy to students $1$ and $2$, but not student $3$ (as $1, 2, 3$ form an arithmetic sequence), then to students $4$ and $5$, but not to students $6$ and $7$ (as both $4, 5, 6$ and $1, 4, 7$ are arithmetic sequences), and so on. How many students got candies in the end?",
    "options": [],
    "answer": "128",
    "solution": "Answer: $128$\n\nWith some initial trials, we can list the numbers of the students who get candies:\n$1, 2, 4, 5, 10, 11, 13, 14, 28, 29, \\ldots$\n\nThe base $3$ representation of the numbers are\n$1, 2, 11, 12, 101, 102, 111, 112, 1001, 1002, \\ldots$,\nand if we subtract $1$ from each number, they become the following\n$0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001, \\ldots$,\nwhich are precisely those whose base $3$ representations consist of the digits $0$ and $1$ only. Indeed, we can easily prove by induction that student $n$ gets a candy if and only if the base $3$ representation of $n-1$ consists of the digits $0$ and $1$ only (call such a positive integer $n$ 'good'). This statement holds for the numbers we have listed. Now suppose $n$ is larger than $29$ and the statement holds for smaller $n$.\n\n* Suppose $n$ is good. If student $n$ does not get a candy, there exist students $\\ell$ and $m$ who both got candies (where $\\ell < m < n$) and such that $\\ell, m, n$ form an arithmetic sequence. Consider the numbers $\\ell - 1, m - 1, n - 1$ (in base $3$, same for below). They all consist of the digits $0$ and $1$ only, and $2(m - 1) = (\\ell - 1) + (n - 1)$. This is impossible, since each digit on the left is $0$ or $2$, while at least one digit on the right is $1$ (there is no carry in the addition, and there must be at least a place which gives $0+1$ since $\\ell - 1 \\neq n - 1$). This contradiction establishes the fact that student $n$ gets a candy.\n\n* Suppose $n$ is not 'good'. Then $n - 1$ consists of at least one digit $2$. Suppose the place values of all the digits $2$ are $3^{k_1}, 3^{k_2}, \\dots, 3^{k_d}$ respectively. Then student $n$ cannot get a candy since students $n - 2 \\cdot (3^{k_1} + 3^{k_2} + \\dots + 3^{k_d})$ and $n - (3^{k_1} + 3^{k_2} + \\dots + 3^{k_d})$ both get candies. (For example, if $n = 74$ then $n - 1$ has base $3$ representation $2201$. Both students $2$ and $38$ got candies since the base $3$ representations of $1$ and $37$ are $1$ and $1101$ respectively. As $2, 38, 74$ form an arithmetic sequence, student $74$ cannot get a candy.)\n\nThis completes the induction. Note that $2023_{10} = 2202221_3$. So the largest $n$ which is 'good' has the property that the base $3$ representation of $n - 1$ is $1111111$. The answer is thus $2^7 = 128$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31788,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a trapezium with $AD \\parallel BC$, $AD > BC$ and $AB = CD$. $P$ is a point on the plane whose distance from $A$, $B$, $C$, $D$ are $1$, $2$, $3$, $4$ respectively. Find $AD : BC$.",
    "options": [],
    "answer": "3 : 1",
    "solution": "Answer: $3 : 1$\n\nSuppose the line through $P$ perpendicular to $BC$ and $AD$ meets $AD$ at $E$ and $BC$ at $F$. Then we have\n$$\n4^2 - 1^2 = (DE^2 + PE^2) - (AE^2 + PE^2) = DE^2 - AE^2 = (DE + AE)(DE - AE).\n$$\nIf we draw another line perpendicular to $BC$ and $AD$ such that its distance from $D$ is equal to $AE$, then $DE - AE$ is simply the distance between the two lines we have drawn, which we denote by $x$.\nHence we have $AD \\cdot x = 4^2 - 1^2$. By symmetry we have $BC \\cdot x = 3^2 - 2^2$.\n\nIt follows that $AD : BC = (4^2 - 1^2) : (3^2 - 2^2) = 3 : 1$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31789,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $AC = \\sqrt{3} AB$ and $BC = 2$. $D$ is a point inside $\\triangle ABC$ such that $\\angle BDC = 90^\\circ$, $\\angle DAC = 18^\\circ$ and $BD = 1$. Find $\\angle DAB$.",
    "options": [],
    "answer": "48°",
    "solution": "48°\n\n![](attached_image_1.png)\n\nNote that $DC = \\sqrt{3}$, $\\angle DBC = 60^\\circ$ and $\\angle DCB = 30^\\circ$. In particular, we notice that $AB : AC = DB : DC$. Thus, the internal angle bisectors of $\\angle BAC$ and $\\angle BDC$ meet $BC$ at the same point $P$, and the external angle bisectors of $\\angle BAC$ and $\\angle BDC$ meet the extension of $CB$ at the same point $Q$.\n\nNote that the internal and external angle bisectors are always perpendicular. Hence the circle with $PQ$ as diameter passes through $A$ and $D$. As $\\angle QDB = 45^\\circ$ and $\\angle BDC = 90^\\circ$, we have $\\angle DQP = 180^\\circ - \\angle QDC - \\angle DCQ = 15^\\circ$. This gives $\\angle DAP = \\angle DQP = 15^\\circ$. Hence $\\angle PAB = \\angle PAC = 15^\\circ + 18^\\circ = 33^\\circ$. It follows that\n\n$$\n\\angle DAB = \\angle DAP + \\angle PAB = 15^\\circ + 33^\\circ = 48^\\circ.\n$$\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31790,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a country there are only four types of coins, of denominations $74$, $87$, $111$ and $124$ dollars respectively. In how many different ways can one pay exactly $2023$ dollars using these coins?",
    "options": [],
    "answer": "14",
    "solution": "Answer: $14$\n\nNote that $74$ and $111$ have a common factor of $37$. Also, $87$ and $124$ are $13$ greater than $74$ and $111$ respectively. We call the coins with denominations $87$ and $124$ 'bad coins'.\n\nSince $2023 \\equiv 25 \\pmod{37}$, we must use $k$ bad coins such that $13k \\equiv 25 \\pmod{37}$. The smallest such $k$ is $19$, and hence we must have $k = 19$ (because the next $k$ would be $19 + 37$, and the total amount will surely exceed $2023$).\n\nWe note that $2023 = 19 \\times 87 + 370$. To pay exactly $2023$, we may start with $19$ coins with denomination $87$, and add $10$ units of $37$. This can be done by using a $74$ coin (which contributes $2$ units), a $111$ coin (which contributes $3$ units), or by replacing a $87$ coin by a $124$ coin (which contributes $1$ unit). Hence the problem is reduced to solving the equation $2a + 3b + c = 10$ in nonnegative integers. Since each choice of $(a, b)$ for which $2a + 3b \\le 10$ gives a unique choice of $c$, we only have to solve the inequality $2a + 3b \\le 10$ in nonnegative integers. We find that there are $14$ solutions as shown below, and so the answer is $14$.\n\n| Value of $b$ | Possible corresponding values of $a$ |\n|-------------|-------------------------------------|\n| $0$         | $0, 1, 2, 3, 4, 5$                  |\n| $1$         | $0, 1, 2, 3$                        |\n| $2$         | $0, 1, 2$                           |\n| $3$         | $0$                                 |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31791,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a $3 \\times 3$ table and $9$ cards printed with the numbers $-4$, $-3$, $-2$, $-1$, $0$, $1$, $2$, $3$, $4$. How many ways are there for these cards to be distributed to the cells of the table so that there is one card in each cell, and that the sum of the numbers of the three cards on each row, each column and each of the two diagonals of the table is nonnegative?",
    "options": [],
    "answer": "32",
    "solution": "Answer: $32$\n\nWe first make a few observations.\n\n* The sum of the numbers on all cards is $0$. Hence each row sum being nonnegative is equivalent to each row sum being $0$.\n* For the same reason as above, each column sum must be $0$.\n* The card $-4$ cannot be placed in a corner cell. Indeed, if $-4$ is placed in a corner cell as shown below, then the sum of the other two numbers in the first row is $4$, the sum of the other two numbers in the first column is $4$ and the sum of the other two numbers in the diagonal containing $-4$ is at least $4$. This is impossible since the sum of the positive numbers on the cards is only $1 + 2 + 3 + 4 = 10$.\n\n![](attached_image_1.png)\n\n* In essentially the same way we can prove that the card $-4$ cannot be placed in the central cell.\n\nFrom the above observations, there are $4$ choices for the position of $-4$. After the position of $-4$ is fixed as shown below, there are $4$ choices for the other two numbers in the first row (we either put in $\\{0, 4\\}$ or $\\{1, 3\\}$ in either order), and then $2$ choices for the other two numbers in the second column (we put in the remaining pair in either order).\n\n$4 \\times 4 \\times 2 = 32$\n\nways to fill in five cells. There is then only one way to fill in the remaining cells using the numbers $-3$, $-2$, $-1$ and $2$. For instance, suppose we have filled the cells in the following way, with $a$, $b$, $c$, $d$ denoting the numbers yet to be filled in:\n\n<table><tr><td>1</td><td>-4</td><td>3</td></tr><tr><td>a</td><td>0</td><td>b</td></tr><tr><td>c</td><td>4</td><td>d</td></tr></table>\n\nBy considering the rows, we need $a + b = 0$ and $c + d = -4$, so $\\{a, b\\} = \\{2, -2\\}$ and $\\{c, d\\} = \\{-1, -3\\}$. By considering the columns, we need $a + c = -1$ and $b + d = -3$. Hence the only possibility is $(a, b, c, d) = (2, -2, -3, -1)$. It remains to check whether these $32$ ways also satisfy the requirement that the two diagonal sums are nonnegative. We start with the following $4$ cases with the position of $-4$ fixed:\n\n<table><tr><td>1</td><td>-4</td><td>3</td></tr><tr><td>2</td><td>0</td><td>-2</td></tr><tr><td>-3</td><td>4</td><td>-1</td></tr></table>\n\n<table><tr><td>1</td><td>-4</td><td>3</td></tr><tr><td>-3</td><td>4</td><td>-1</td></tr><tr><td>2</td><td>0</td><td>-2</td></tr></table>\n\n<table><tr><td>0</td><td>-4</td><td>4</td></tr><tr><td>2</td><td>1</td><td>-3</td></tr><tr><td>-2</td><td>3</td><td>-1</td></tr></table>\n\n<table><tr><td>0</td><td>-4</td><td>4</td></tr><tr><td>-2</td><td>3</td><td>-1</td></tr><tr><td>2</td><td>1</td><td>-3</td></tr></table>\n\nAll cases have both diagonal sums being nonnegative. The other $4$ cases with $-4$ in the same position must therefore also have the same property, since they are obtained by swapping the first and third columns from the $4$ cases above:\n\n<table><tr><td>3</td><td>-4</td><td>1</td></tr><tr><td>-2</td><td>0</td><td>2</td></tr><tr><td>-1</td><td>4</td><td>-3</td></tr></table>\n\n<table><tr><td>3</td><td>-4</td><td>1</td></tr><tr><td>-1</td><td>4</td><td>-3</td></tr><tr><td>-2</td><td>0</td><td>2</td></tr></table>\n\n<table><tr><td>4</td><td>-4</td><td>0</td></tr><tr><td>-3</td><td>1</td><td>2</td></tr><tr><td>-1</td><td>3</td><td>-2</td></tr></table>\n\n<table><tr><td>4</td><td>-4</td><td>0</td></tr><tr><td>-1</td><td>3</td><td>-2</td></tr><tr><td>-3</td><td>1</td><td>2</td></tr></table>\n\nThe other cases (where the position of $-4$ is different) can be obtained by rotating one of the above configurations. Hence all $32$ ways of filling in the table in which all row and column sums are zero will also have both diagonal sums being nonnegative. It follows that the answer is $32$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31792,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a team game, 12 players stand at the 12 vertices of a regular 12-sided polygon. Each player has a red flag and a blue flag, and then randomly put up one flag, all at the same time. If there are four players who put up flags of the same colour and whose positions form a rectangle, the team loses. Otherwise the team wins. What is the probability for the team to win?",
    "options": [],
    "answer": "29/128",
    "solution": "$2^{12} = 4096$ possible outcomes and we want to count how many of these enable the team to win. To do this, we let the 12 players form 6 pairs so that the members of each pair are diametrically opposite. Note that the positions of four players form a rectangle if and only if they form two pairs.\n\nCall a pair 'bad' if the two players put up flags of the same colour. Note that if there are three or more bad pairs, then at least two pairs will both have chosen the same colour, so the team loses. Hence for the team to win, there are three possibilities:\n\n* There is no bad pair — that means for each pair the two players will put up flags of different colours (2 possibilities for each pair), so there are altogether $2^6 = 64$ possibilities for this case.\n\n* There is one bad pair — there are 6 choices for the bad pair, and regardless of whether a pair is good or bad, there are 2 possible combinations of flag colours for each pair. Hence there are $6 \\times 2^6 = 384$ possibilities for this case.\n\n* There are two bad pairs — there are $\\binom{6}{2} = 15$ choices for the bad pairs, and one bad pair must put up red flags and the other bad pair must put up blue flags (otherwise the team loses), giving 2 possibilities. There are 2 possible combinations of flag colours for each of the other 4 pairs, and so there $15 \\times 2 \\times 2^4 = 480$ possibilities for this case.\n\nIt follows that the answer is $\\frac{64 + 384 + 480}{4096} = \\frac{29}{128}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31793,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a square. $P$ is a point inside $ABCD$ such that $\\angle APD + \\angle BPC = 180^\\circ$ and $\\angle BPC$ is acute. If $PB = 3$ and $PC = 4$, find $BC$.",
    "options": [],
    "answer": "1 + 2√2",
    "solution": "Answer: $1 + 2\\sqrt{2}$\n\nWe translate $\\triangle APD$ to $\\triangle BQC$ so that the condition becomes $\\angle BQC + \\angle BPC = 180^\\circ$. This means $BQCP$ is a cyclic quadrilateral. Draw a line through $P$ parallel to $AB$. This line passes through $Q$, and let $E$ and $F$ be the points where this line meets $AD$ and $BC$ respectively.\n\nNote that $AB = PQ$ due to the translation. Hence\n$BC = PQ$, so they subtend equal angles on the circumference of the circle.\n\nEquating the angles subtended on the major arcs gives $\\angle BPC = \\angle QCP$ (note that $\\angle BPC$ is given to be acute, and that $\\angle QCP$ is acute since $PB < PC$). Equating the angles subtended on the minor arcs gives $\\angle BQC = \\angle PBQ$. Since we also have $\\angle BQC + \\angle BPC = 180^\\circ$, it follows that the cyclic quadrilateral $BQCP$ is actually an isosceles trapezoid with $BQ$ and $PC$ parallel. Hence we have $FP = FC = 2\\sqrt{2}$, and so $BF = \\sqrt{3^2 - (2\\sqrt{2})^2} = 1$. It follows that $BC = BF + FC = 1 + 2\\sqrt{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31794,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a chess tournament, there were $n$ contestants and any two of them played at most one game against each other. Each contestant has played exactly $100$ games. For any two contestants $A$ and $B$ who had played against each other, there were exactly $50$ other participants who have played against both $A$ and $B$. On the other hand, for any two contestants $C$ and $D$ who had not played against each other, there were exactly $4$ other participants who have played against both $C$ and $D$. Find the value of $n$.",
    "options": [],
    "answer": "1326",
    "solution": "Answer: $1326$\n\nPick any contestant $x$. Suppose he has played against $y_1, y_2, \\dots, y_{100}$ (call these Group Y contestants). Denote the contestants who have not played against $x$ by $z_1, z_2, \\dots, z_m$ (call these Group Z contestants).\n\n![](attached_image_1.png)\n\nA Group Y contestant (who has played against $x$) has exactly $50$ common opponents with $x$ (who must be Group Y contestants), and so has played against exactly $100 - 1 - 50 = 49$ Group Z contestants. On the other hand, since a Group Z contestant has not played against $x$, he has exactly $4$ common opponents with $x$ (who must be Group Y contestants). Thus the number of games between a Group Y contestant and a Group Z contestant is equal to $100 \\times 49$ as well as $4m$. It follows that these numbers are equal, and so\n$$\nn = 1 + 100 + m = 1 + 100 + \\frac{100 \\times 49}{4} = 1326.\n$$\nThe scenario described in the question is indeed possible. Note that the answer is actually $\\binom{52}{2}$. To construct the scenario, label the contestants by $2$-element subsets of $\\{1, 2, \\dots, 52\\}$, and let two contestants play a game if and only if their corresponding subsets have a common element. It is easy to check that all given conditions are satisfied.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31795,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 20 students in a class, numbered $1$ to $20$. They have taken a test with $n$ questions. After the test, the teacher tried to pick some students to draft the solutions, and as such it is necessary that the picked students together can solve all questions. The teacher found that this can be only done by choosing all students whose class numbers are even, or all students whose class numbers are multiples of $5$, or any combination that includes all students from one of the two groups mentioned above. Find the smallest possible value of $n$.",
    "options": [],
    "answer": "18",
    "solution": "Answer: $18$\n\nFor convenience we shall call 'a student whose class number is even' an even student. We make the following observations.\n\n(1) Every question is solved by at least one even student (as the teacher could choose all even students).\n\n(2) Similarly, every question is solved by at least one of students $5$, $10$, $15$, $20$.\n\n(3) Students $2$, $4$, $6$, $\\ldots$, $18$ (i.e. all even students except student $20$) cannot solve all questions (since any choice of the teacher must include either all even students or all students whose numbers are multiples of $5$). By (1), there must be a question which was solved by student $20$ but not by any other even student.\n\n(4) For the same reason in (3), there must be a question which was solved by student $10$ but not by any other even student.\n\n(5) We claim that there exist at least two questions that were solved by student $2$ but not any other even student. Using the same argument as in (3), we can show that there is at least one such question. Now suppose there is exactly one such question $P$. Now neither student $10$ nor $20$ solved $P$, so by (2) we know that either student $5$ or $15$ solved $P$. Without loss of generality suppose student $5$ did it. Consider students $4$, $5$, $6$, $8$, $10$, $12$, $14$, $16$, $18$, $20$ (i.e. all even students but with student $2$ replaced by student $5$). Since student $2$ and student $15$ are missing, these students together did not solve all questions. That means there is a question $Q$ which cannot be solved by these students (note that $P \\neq Q$ since student $5$ solved $P$). As student $2$ cannot solve $Q$ either (since $P$ is the only question solved by student $2$ but not any other even student), that means no even student could solve $Q$, which contradicts (1). This proves the claim.\n\n(6) For the same reason in (5), for each of students $4$, $6$, $8$, $12$, $14$, $16$, $18$, there must be at least two questions that were solved by that student but not any other even student.\n\nFrom (3), (4), (5) and (6), there are at least $1 + 1 + 2 + 2 \\times 7 = 18$ questions, and they are clearly distinct (by considering which even student solved each question). On the other hand, we can construct a paper with exactly $18$ questions as follows:\n\n* Question $10$ — solved by student $10$ only\n* Question $20$ — solved by student $20$ only\n* Question $x$ for each $x \\in \\{2, 4, 6, 8, 12, 14, 16, 18\\}$ — solved by students $5$ and $x$ only\n* Question $x'$ for each $x' \\in \\{2, 4, 6, 8, 12, 14, 16, 18\\}$ — solved by students $15$ and $x$ only\n\nIt can be easily checked that all even students together solved all questions, all students whose numbers are multiples of $5$ together solved all questions, and that if not all even students are present and not all students whose numbers are multiples of $5$ are present, then at least one question is not solved by the students picked. The answer is thus $18$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 31796,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\triangle ABC$, $\\angle CAB = 75^\\circ$ and $\\angle ACB = 45^\\circ$. $BC$ is extended to $T$ so that $BC = CT$. Let $M$ be the midpoint of the segment $AT$. Find $\\angle BMC$.",
    "options": [],
    "answer": "45°",
    "solution": "As $M$ and $C$ are midpoints of $AT$ and $BT$ respectively, we have $MC // AB$ and $MC = \\frac{1}{2} AB$ by the midpoint theorem. Extend $CM$ to $K$ so that $CM = MK$. Then $AB = CK$, and hence $ABCK$ is a parallelogram. Also, extend $CA$ to $S$ so that $\\angle SKC = 75^\\circ$. Then $\\triangle SKC$ is an isosceles triangle with $SC = SK$. Therefore, $\\angle SMC = \\angle SMK = 90^\\circ$.\n\nLet $BA$ meet $SK$ at $L$. Then $LKCA$ is an isosceles trapezoid. It follows that $LC = AK = BC$. Together with $\\angle B = 60^\\circ$, we know that $\\triangle CBL$ is equilateral, and hence $BC = BL$. Furthermore, $\\angle LSC = 30^\\circ = \\frac{1}{2} \\angle LBC$. Thus, $S$ lies on the circle with centre $B$ and radius $BC$. Therefore, $BS = BL = BC$, and\n$$\n\\angle SBC = 180^\\circ - 2\\angle BCA = 90^\\circ.\n$$\nFinally, $S, B, C, M$ are concyclic since $\\angle SBC + \\angle SMC = 180^\\circ$. It follows that $\\angle BMC = \\angle BSC = 45^\\circ$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31797,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a real number is 'almost an integer' if it differs from an integer by at most $0.1$. For example, $2023$, $-2023.9$ and $2023.0822$ are almost integers. Show that among any $10$ real numbers, there exist two different real numbers whose difference is almost an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $x_1, x_2, \\dots, x_{10}$ be the real numbers. For each $j$, we write $x_j = a_j + b_j$ where $a_j$ is the largest integer not exceeding $x_j$. Note that we must have $0 \\le b_j < 1$. WLOG we may assume $b_1 \\le b_2 \\le \\dots \\le b_{10}$.\n\n* If $b_{j+1} - b_j \\le 0.1$ for some $j = 1, 2, \\dots, 9$, then $x_{j+1} - x_j$ is almost an integer since\n$$\n(x_{j+1} - x_j) - (a_{j+1} - a_j) = b_{j+1} - b_j \\in [0, 0.1],\n$$\nwhere $a_{j+1} - a_j$ is an integer.\n\n* If $b_{j+1} - b_j > 0.1$ for all $j = 1, 2, \\dots, 9$, then $b_{10} - b_1 > 0.9$, and hence $x_1 - x_{10}$ is almost an integer since\n$$\n(x_1 - x_{10}) - (a_1 - a_{10} - 1) = 1 - (b_{10} - b_1) \\in (0, 0.1),\n$$\nwhere $a_1 - a_{10} - 1$ is an integer.\n\nTherefore, in any case, we can find two different real numbers whose difference is almost an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31798,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. In each cell of an $n \\times n$ grid, we place a coin with either the head or the tail facing upwards. Each time we can select one cell, and flip all the $2^n - 1$ coins that lie in the same row or the same column as this cell. Find all $n$ such that it is always possible to have all the coins with the heads facing upwards after finitely many moves, no matter how we place the coins initially.",
    "options": [],
    "answer": "n = 1 and all positive even n",
    "solution": "Answer: 1 and all positive even numbers.\n\nIn the following, to operate on the $(i, j)$th cell (the cell in the $i$th row and the $j$th column) means to perform an operation by selecting the $(i, j)$th cell in the operation.\n\n* When $n = 1$, it is clear that the goal can always be met.\n\n* When $n$ is even, it suffices to show that it is possible to flip any arbitrary coin only in some moves. Indeed, if we want to flip the coin in the $(i, j)$th cell (called $C$), we simply operate on every cell that lies in the same row or the same column as $C$.\n\n- The coin in $C$ is flipped $2n - 1$ times, which is odd.\n- Each coin in the same row or the same column as $C$ other than the coin in $C$ is flipped $n$ times, which is even.\n- Every other coin is flipped $2$ times, which is even.\n\nTherefore, we see that only the coin in $C$ is flipped after these operations.\n\n* When $n$ is odd and greater than $1$, it suffices to consider the reverse process, i.e. starting with all coins with the heads facing upwards, we cannot obtain all possible configurations after finitely many moves. Note that there are $2^{n^2}$ possible configurations since there are two possibilities for each of the $n^2$ cells. On the other hand, note that the order of operations does not affect the result, and operating the same cell twice does not change the result. Therefore, we may assume each cell is operated $0$ or $1$ time. This leads to $2^{n^2}$ different sets of operations. Thus, it suffices to show that there are two different sets of operations that lead to the same configuration.\n\nIndeed, we simply operate on every cell in the first row and the first column except the $(1, 1)$th cell.\n- The coin in the $(1, 1)$th cell is flipped $2(n-1)$ times, which is even.\n- Each coin in the first row or the first column other than the coin in the $(1, 1)$th cell is flipped $n-1$ times, which is even.\n- Every other coin is flipped $2$ times, which is even.\n\nTherefore, we see that no coin is flipped after these operations, which means this has the same effect as not performing any operation at all (which is also one of the $2^{n^2}$ sets of operations). This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31799,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Show that if $p$ is a prime dividing $5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$, then $p \\equiv 1 \\pmod 4$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Clearly, $p \\neq 2, 5$. Let $m = 5^{4n} - 5^{3n} + 5^{2n} - 5^n + 1$. Then\n$$\n(2 \\cdot 5^{2n} - 5^n + 2)^2 - 5 \\cdot 5^{2n} = 4m \\equiv 0 \\pmod{p}.\n$$\nThis gives $5 \\equiv (5^{-n}(2 \\cdot 5^{2n} - 5^n + 2))^2 \\pmod{p}$. Using the Legendre symbol, we have $\\left(\\frac{5}{p}\\right) = 1$. On the other hand, we have\n$$\n(5^{2n} - 5^n + 1)^2 + 5^n(5^n - 1)^2 = m \\equiv 0 \\pmod{p}.\n$$\nAs above, this implies $\\left(\\frac{-5^n}{p}\\right) = 1$. It follows that\n$$\n\\left(\\frac{-1}{p}\\right) = \\left(\\frac{-5^n}{p}\\right) \\left(\\frac{5^n}{p}\\right) = 1 \\cdot 1^n = 1.\n$$\nTherefore, $p \\equiv 1 \\pmod 4$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31800,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be positive integers such that $\\sqrt{m} + \\sqrt{n} = \\sqrt{2023}$. Find the greatest possible value of $m+n$.",
    "options": [],
    "answer": "1799",
    "solution": "Note that $\\sqrt{2023} = 17\\sqrt{7}$. Hence we must have $\\sqrt{m} = p\\sqrt{7}$ and $\\sqrt{n} = q\\sqrt{7}$, where $p$ and $q$ are positive integers with sum $17$. We have\n$$\nm + n = 7p^2 + 7q^2 = 7p^2 + 7(17-p)^2 = 14 \\left[ \\left( p - \\frac{17}{2} \\right)^2 + \\frac{289}{4} \\right],\n$$\nwhich is maximized when $p$ is as far away from $\\frac{17}{2}$ as possible. We should take either $(p, q) = (1, 16)$ or $(16, 1)$, and the answer is thus $7(1)^2 + 7(16)^2 = 1799$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31801,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA number is called *Norwegian* if it has three distinct positive divisors whose sum is equal to $2022$. Determine the smallest Norwegian number.\n\n(Note: The total number of positive divisors of a Norwegian number is allowed to be larger than $3$.)",
    "options": [],
    "answer": "1974",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31802,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an acute $\\triangle ABC$ with $AB > AC$, let $M$ be the midpoint of the side $BC$. Let $I$ be the incentre of $\\triangle ABC$, and let $D$ be the foot of perpendicular from $I$ to $BC$. The line passing through $A$ and perpendicular to $BC$ meets the circumcircle of $\\triangle ABC$ again at $E \\neq A$. Suppose $\\angle MID = \\angle ACB - \\angle ABC$. Prove that $DE = DI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ and $H$ be the circumcentre and the orthocentre of $\\triangle ABC$ respectively. Reflect $A$ about $O$ to a point $A'$. Reflect $D$ about $M$ and $I$ to points $D'$ and $P$ respectively. Also, let $Q$ be the midpoint of $AH$. Notice the following well-known results.\n\n* $E$ and $H$ are symmetric about the line $BC$\n* $M$ is the midpoint of $A'H$\n* $A$, $P$, $D'$ are collinear (as $D'$ is the contact point of the $A$-excircle and the side $BC$)\n\n$$\n\\angle OAI = \\angle HAI\n$$\n\nNow, we first find that $\\angle OAH = A - (90^\\circ - C) - (90^\\circ - C) = C - B = \\angle MID$. As $AH \\parallel ID$, we have $AO \\parallel IM$. By the midpoint theorem, we also have $APD' \\parallel IM$. Therefore, $A$, $P$, $O$, $D'$, $A'$ are collinear. Again by the midpoint theorem, we have $QM \\parallel AA'$, and hence $Q$, $I$, $M$ are collinear. This shows $APIQ$ is a parallelogram. But since $AI$ intersects $\\angle PAQ$, the parallelogram $APIQ$ is indeed a rhombus. In addition, as $ID = IP$ and $QH = QA$, $IDHQ$ is also a rhombus. It follows that $DE = DH = DI$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31803,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $u, v, w$ be positive integers, with $v$ and $w$ not exceeding $2023$, such that $u - v = 2w$ and $(u^2 + w^2)(v^2 + w^2)$ is divisible by $289$. How many sets of possible values of $(u, v, w)$ are there?",
    "options": [],
    "answer": "67473",
    "solution": "Answer: $67473$\n\nIt suffices to choose $v$ and $w$ since there is a unique choice for $u = v + 2w$ after fixing $v$ and $w$. Note that $289 = 17^2$ and $2023 = 7 \\times 17^2$. We consider two cases.\n\nCase 1: $w$ is divisible by $17$\n\nIn this case $v$ is also divisible by $17$ (if not, then $u = v + 2w$ not divisible by $17$, so neither $u^2 + w^2$ nor $v^2 + w^2$ can be divisible by $17$). Furthermore, any choice of positive multiples of $17$ not exceeding $2023$ for each of $v$ and $w$ satisfies the given condition. Hence there are $7 \\times 17$ choices for each of $v$ and $w$, and thus $(7 \\times 17)^2$ sets of $(u, v, w)$ in this case.\n\nCase 2: $w$ is not divisible by $17$\n\nThere are $2023 - 7 \\times 17 = 7 \\times 17 \\times 16$ choices for $w$. For each $w$, we try to find $v$. First note that\n$$\n\\begin{aligned}\n(u^2 + w^2)(v^2 + w^2) &\\equiv (u^2 - 16w^2)(v^2 - 16w^2) \\\\\n&= (u + 4w)(u - 4w)(v + 4w)(v - 4w) \\\\\n&= (v + 6w)(v - 2w)(v + 4w)(v - 4w) \\pmod{17}.\n\\end{aligned}\n$$\nHence we can choose $v$ as follows.\n\n* From the above we see that $v$ must be equal to $-6w, 2w, -4w$ or $4w$ modulo $17$ (and they are distinct modulo $17$).\n* Each of the 4 choices of $v$ modulo $17$ corresponds to exactly one choice of $v$ modulo $289$ for which $(u^2 + w^2)(v^2 + w^2)$ is divisible by $289$.\n* Each choice of $v$ modulo $289$ corresponds to $7$ actual possible values of $v$ since $2023 = 7 \\times 289$.\n\nHence for each $w$, there are $4 \\times 7$ choices for $v$. It follows that there are altogether $(7 \\times 17 \\times 16) \\times (4 \\times 7) = 7^2 \\times 17 \\times 64$ sets of $(u, v, w)$ in this case.\n\nIt follows that the answer is $(7 \\times 17)^2 + 7^2 \\times 17 \\times 64 = 7^2 \\times 17 \\times 81 = 67473$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31804,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2n$ cards labelled $1, 2, \\ldots, 2n$ respectively, where $n$ is a positive integer. Two players $A$ and $B$ take turns choosing a card which has not been chosen by anyone, with $A$ going first, until all cards are chosen. Let $a_1, a_2, \\ldots, a_n$ be the labels of the cards chosen by $A$ in this order, and let $b_1, b_2, \\ldots, b_n$ be the labels of the cards chosen by $B$ in this order. The score of $A$ is equal to\n$$\n(a_1 + a_2 + \\dots + a_n)^2 + (b_1 + b_2 + \\dots + b_n)^2 - (a_1b_1 + a_2b_2 + \\dots + a_nb_n).\n$$\nWhat is the maximum score (in terms of $n$) that $A$ can guarantee to obtain regardless of how $B$ plays?",
    "options": [],
    "answer": "n(6n^3 + 2n^2 + 1)/3",
    "solution": "The answer is\n$$\n\\begin{aligned}\nN &:= \\left( \\sum_{k=1}^{n} 2k \\right)^2 + \\left( \\sum_{k=1}^{n} (2k-1) \\right)^2 - \\sum_{k=1}^{n} (2k-1)(2k) \\\\\n&= n^2(n+1)^2 + n^4 - \\left( \\frac{2n(n+1)(2n+1)}{3} - n(n+1) \\right) \\\\\n&= \\frac{n(6n^3 + 2n^2 + 1)}{3}\n\\end{aligned}\n$$\n\nDenote\n$X = a_1 + a_2 + \\cdots + a_n,$\n$Y = b_1 + b_2 + \\cdots + b_n,$\n$S = X^2 + Y^2,$\n$T = a_1 b_1 + a_2 b_2 + \\cdots + a_n b_n.$\n\nClaim. $T \\le 1 \\cdot 2 + 3 \\cdot 4 + \\cdots + (2n-1)(2n)$\n\n*Proof.* Since there are finitely many possible ways to assign the values, $T$ must attain a maximum value. For this maximal case, we may assume $a_i < b_i$ for each $i$. Note that we may assume $\\{a_i\\}$ and $\\{b_i\\}$ are increasing by the rearrangement inequality.\n\nUnder these assumptions, we must have $a_1 = 1$. We are going to show that $a_i = 2i - 1$ and $b_i = 2i$ for all $i$. Suppose on the contrary that this does not hold. Then there is a smallest index $k$ such that $b_k \\neq 2k$. Note that we must have $a_i = 2i - 1$ for $i = 1, 2, \\dots, k$ and $a_{k+1} = 2k$ (clearly $k \\neq n$). This implies $a_k < a_{k+1} < b_k < b_{k+1}$. By the rearrangement inequality, we have\n$$\na_k b_k + b_{k+1} a_{k+1} < a_k a_{k+1} + b_{k+1} b_k.\n$$\nThis means we can swap $b_k$ and $a_{k+1}$ to increase the value of $T$, contradiction.\n\nTherefore, $T$ is maximized when $a_i = 2i - 1$ and $b_i = 2i$ for all $i$. This proves the claim. $\\square$\n\nNow, we show that $A$ can guarantee to obtain score $N$. A strategy is to choose the card with the largest label in each round. Note that $a_i \\ge 2(n+1-i)$ for each $i$ since at least one of the cards $2(n+1-i), 2(n+1-i)+1, \\dots, 2n$ is not chosen in the $i$th round. As $X+Y = 1+2+\\dots+2n$ is fixed and $X > Y$ (because of this strategy), $S$ is smaller when $X$ is smaller. Thus,\n$$\nS \\ge (2+4+\\dots+2n)^2 + (1+3+\\dots+(2n-1))^2.\n$$\nTogether with the claim, we deduce\n$$\nS - T \\ge \\left( \\sum_{k=1}^{n} 2k \\right)^2 + \\left( \\sum_{k=1}^{n} (2k-1) \\right)^2 - \\sum_{k=1}^{n} (2k-1)(2k) = N\n$$\nas desired.\n\nNext, we show that $B$ can guarantee that the score of $A$ is at most $N$. A strategy is to pair up the cards with labels $2m-1$ and $2m$ for each $m = 1, 2, \\dots, n$, and choose a card in the same pair as the card just chosen by $A$ in each round. It is easy to see that\n$$\nT = 1 \\cdot 2 + 3 \\cdot 4 + \\dots + (2n-1)(2n).\n$$\nAlso, as $|a_i - b_i| = 1$ for each $i$, we have\n$$\n\\begin{aligned}\nS &= X^2 + Y^2 = \\frac{(X+Y)^2 + (X-Y)^2}{2} \\\\\n&= \\frac{1}{2} \\left[ (1+2+\\dots+2n)^2 + \\left( \\sum_{i=1}^{n} (a_i - b_i) \\right)^2 \\right] \\\\\n&\\le \\frac{(1+2+\\dots+2n)^2 + n^2}{2}\n\\end{aligned}\n$$\n\nEquality holds when $a_i - b_i = 1$ for all $i$ or $a_i - b_i = -1$ for all $i$. This is exactly the case when\n$$\nS-T = \\left(\\sum_{k=1}^{n} 2k\\right)^2 + \\left(\\sum_{k=1}^{n} (2k-1)\\right)^2 - \\sum_{k=1}^{n} (2k-1)(2k) = N.\n$$\nThe proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31805,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given $\\triangle ABC$ with $AB < AC$, let $AD$ be the bisector of $\\angle BAC$ with $D$ on the side $BC$. Let $\\Gamma$ be a circle passing through $A$ and $D$ which is tangent to $BC$ at $D$. Suppose $\\Gamma$ cuts the side $AB$ again at $E \\neq A$. The tangent to the circumcircle of $\\triangle BDE$ at $D$ intersects $\\Gamma$ again at $F \\neq D$. Let $P$ be the intersection point of the segments $EF$ and $AC$. Prove that $PD$ is perpendicular to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $Q \\neq A$ be the second intersection point of $\\Gamma$ and $AC$. As\n$$\n\\angle EDB = \\angle EAD = \\angle DAQ = \\angle DEQ,\n$$\nwe have $EQ // BC$. Also, we have $AF // BC$ since\n$$\n\\angle EAF = 180^\\circ - \\angle FDE = 180^\\circ - \\angle DBE.\n$$\n\nThis shows $EQ // AF$, and hence $AEQF$ is an isosceles trapezoid. Thus, the intersection point $P$ of the diagonals lies on the perpendicular bisector of $EQ$. Note that $D$ also lies on the perpendicular bisector of $EQ$ because $AD$ bisects $\\angle EAQ$ and $A$, $E$, $D$, $Q$ are concyclic. Therefore, $PD \\perp EQ$, which implies $PD \\perp BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31806,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Prove that there exists a positive integer $m$ such that\n$$\n3^n + 5^n - 1\n$$\nis divisible by $7^n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that $m = 7^{n-1}$ works. By the lifting the exponent lemma, we have\n$$\nv_7(3^m + 4^m) = v_7(3^m + 4^m) + v_7(m) = 1 + (n-1) = n.\n$$\nThis implies $7^m \\mid 3^m + 4^m$, and hence $3^m \\equiv -4^m \\pmod{7^n}$. Similarly, $5^m \\equiv -2^m \\pmod{7^n}$. It follows that\n$$\n3^m + 5^m - 1 \\equiv -4^m - 2^m - 1 \\equiv -(2^{2m} + 2^m + 1) \\pmod{7^n}. \\quad (1)\n$$\nNext, again by the lifting the exponent lemma, we have\n$$\nv_7(8^m - 1) = v_7(8^m - 1) + v_7(m) = 1 + (n-1) = n.\n$$\nThis implies $7^m \\mid 8^m - 1 = (2^m - 1)(2^{2m} + 2^m + 1)$. As $3 \\nmid m$, we have $7 \\nmid 2^m - 1$. Therefore, we must have\n$$\n7^m \\mid 2^{2m} + 2^m + 1. \\quad (2)\n$$\nCombining (1) and (2), we see that $7^m \\mid 3^m + 5^m - 1$ as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31807,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $P$ lies inside an equilateral triangle $ABC$ such that $AP = 15$ and $BP = 8$. Find the maximum possible value of the sum of areas of triangles $ABP$ and $BCP$.",
    "options": [],
    "answer": "60 + 16√3",
    "solution": "Rotate $\\triangle PBC$ about point $B$ by $60^\\circ$ in the anticlockwise direction, so that $BA$ is the image of $BC$ after rotation. Let $Q$ be the image of $P$ after rotation. Then\n$$\n[ABP] + [BCP] = [PAQB] = [PQB] + [QPA].\n$$\n\n![](attached_image_1.png)\n\nNow, $\\triangle PQB$ is an equilateral triangle with side length $8$, and so its area is\n$$\n\\frac{\\sqrt{3}}{4} \\times 8^2 = 16\\sqrt{3}.\n$$\nAlso, we have\n$$\n[QPA] = \\frac{1}{2} \\times 8 \\times 15 \\times \\sin \\angle QPA \\le \\frac{1}{2} \\times 8 \\times 15 = 60.\n$$\nThus, the sum of areas is at most $60 + 16\\sqrt{3}$. Equality holds when $\\angle QPA = 90^\\circ$, i.e. $\\angle BPA = 150^\\circ$ (which fixes $\\triangle ABC$ and point $P$).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31808,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is a $20 \\times 20$ table and $400$ cards printed with the numbers $1$ to $400$. The cards are then distributed to the cells of the table so that there is one card in each cell. After that, we put a red sticker on the card in each row with the largest number, and a blue sticker on the card in each column with the largest number. Let $A$ denote the smallest number on the $20$ cards with red stickers, and $B$ denote the smallest number on the $20$ cards with blue stickers. How many different possible values of $|A - B|$ are there?",
    "options": [],
    "answer": "362",
    "solution": "Answer: $362$\n\nFor convenience we shall use the term 'red number' to the number on a card with a red sticker, and similarly for 'blue number'. Clearly, a red number, being a row maximum, is at least $20$. Also, since $A$ is the smallest red number, there are $19$ other red numbers greater than $A$, so $A$ is at most $381$. The same is true for $B$. That is, both $A$ and $B$ are between $20$ and $381$ (inclusive), and so $|A - B|$ is at most $361$.\n\nWe shall prove that all these $362$ possibilities (from $0$ to $361$) for $|A - B|$ are attainable. Indeed, we can make $A = 381$ and $B$ can be equal to any value between $20$ and $381$. To make $A = B = 381$ we can use the following configuration (the arrangements of the starred entries are unimportant):\n\n| 381 | * | * | ... | * |\n| --- | --- | --- | --- | --- |\n| * | 382 | * | ... | * |\n| * | * | 383 | ... | * |\n| ... | ... | ... | ... | ... |\n| * | * | * | ... | 400 |\n\nOn the other hand for any $B \\in \\{20, 21, ..., 380\\}$, we can use the following configuration to obtain such value of $B$ while having $A = 381$:\n\n| 1 | * | * | ... | * | 381 |\n| --- | --- | --- | --- | --- | --- |\n| 2 | 382 | * | ... | * | * |\n| 3 | * | 383 | ... | * | * |\n| ... | ... | ... | ... | ... | ... |\n| 19 | * | * | ... | 399 | * |\n| B | * | * | ... | * | 400 |\n\nIt follows that the answer is $362$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31809,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 4$ be a positive integer. Consider any set $A$ formed by $n$ distinct real numbers such that the following condition holds: for every $a \\in A$, there exist distinct elements $x, y, z \\in A$ such that $|x - a|, |y - a|, |z - a| \\ge 1$. For each $n$, find the greatest real number $M$ such that\n$$\n\\sum_{a \\in A} |a| \\ge M\n$$\nalways holds.",
    "options": [],
    "answer": "M = 4 for n = 4; M = 3 for n ≥ 5",
    "solution": "The greatest $M$ is $4$ if $n = 4$, and is $3$ if $n \\ge 5$.\n\nLet $S$ be the sum $\\sum_{a \\in A} |a|$.\n\nFor $n = 4$, by considering $A = \\{-1, 0, 1, 2\\}$, we need $M \\le 4$. Let $a < b < c < d$ be the elements in $A$. From the condition, we must have $b - a, c - b, d - c \\ge 1$. This implies\n$$\nS = (|a| + |d|) + (|b| + |c|) \\ge |d - a| + |c - b| \\ge 3 + 1 = 4.\n$$\nThis shows the largest $M$ is $4$ when $n = 4$.\n\nFor $n \\ge 5$, by considering $A = \\{-1, 0, \\varepsilon, 2\\varepsilon, \\dots, (n-4)\\varepsilon, 1 + (n-4)\\varepsilon, 1 + (n-3)\\varepsilon\\}$ where $\\varepsilon > 0$, we need $M \\le 3 + f(n)\\varepsilon$ for some function $f$. When $\\varepsilon$ tends to $0$, this yields $M \\le 3$.\n\nFirstly, for $n = 5$, let $a < b < c < d < e$ be the elements in $A$. Since $c-a > c-b$ and $e-c > d-c$, we must have $c-a \\ge 1$ and $e-c \\ge 1$ by considering the element $c$. Also, at least one of $c-b \\ge 1$ and $d-c \\ge 1$ must hold. WLOG assume $c-b \\ge 1$. Then we have\n$$\nS \\ge (|a| + |e|) + (|b| + |c|) \\ge |e-a| + |c-b| \\ge 2 + 1 = 3.\n$$\nThis shows the largest $M$ is $3$ when $n = 5$.\n\nSecondly, for $n \\ge 6$, let $a$ be the element with the smallest absolute value. WLOG assume $a \\ge 0$.\n\n* If $0 \\le a \\le \\frac{1}{2}$, then whenever $|x-a| \\ge 1$, we must have $|x| \\ge 1-a$. Since there are three such $x$'s, we have $S \\ge 3(1-a) + (n-3)a \\ge 3(1-a) + 3a = 3$.\n* If $a > \\frac{1}{2}$, then all elements have absolute value larger than $\\frac{1}{2}$, and hence $S > \\frac{n}{2} \\ge 3$.\n\nThis shows the largest $M$ is $3$ when $n \\ge 6$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31810,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q$, $A$, $B$, $P$ are collinear in this order, in such a way that the line $AC$ is tangent to the circle $ADQ$, and the line $BD$ is tangent to the circle $BCP$. Let $M$ and $N$ be the midpoints of $BC$ and $AD$, respectively. Prove that the following three lines are concurrent: line $CD$, the tangent of circle $ANQ$ at point $A$, and the tangent to circle $BMP$ at point $B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31811,
    "subject": "Mathematics (Multi-modal)",
    "question": "A two digit number $s$ is *special* if $s$ is the common two leading digits of the decimal expansion of $4^n$ and $5^n$, where $n$ is a certain positive integer. Given that there are two special numbers, find these two special numbers.",
    "options": [],
    "answer": "21 and 46",
    "solution": "The condition on $s$ means there exists a positive integer $n$ and nonnegative integers $a$ and $b$ such that\n$$\n10^a s \\le 5^n < 10^a (s+1) \\quad \\text{and} \\quad 10^b s \\le 4^n < 10^b (s+1).\n$$\nNote that $10^a s = 5^n$ and $10^b s = 4^n$ cannot hold simultaneously. Multiplying the second inequality and the square of the first inequality, we obtain\n$$\n10^{2a+b}s^3 < 5^{2n}4^n = 10^{2n} < 10^{2a+b}(s+1)^3.\n$$\nThis implies $s^3 < 10^{2n-2a-b} < (s+1)^3$. Since $s$ is a two-digit number, we have\n$$\n10^3 \\le s^3 < 10^{2n-2a-b} < (s+1)^3 \\le 10^6.\n$$\nThus, $2n-2a-b$ is equal to $4$ or $5$. This shows $s = \\lfloor 10^{\\frac{4}{3}} \\rfloor = 21$ or $\\lfloor 10^{\\frac{5}{3}} \\rfloor = 46$. The two special numbers are $21$ and $46$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31812,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R} \\to \\mathbb{R}$ satisfying the following condition: for any real numbers $x$ and $y$, the number $f(x + f(y))$ is equal to $x + f(y)$ or $f(f(x)) + y$.",
    "options": [],
    "answer": "f(x) = x + c for some real constant c",
    "solution": "$f(x) = x + c$ for some constant $c$.\n\nWe first check that all such functions are solutions. Indeed, if $f(x) = x + c$, then\n$$\nf(x + f(y)) = f(x + y + c) = x + y + 2c = f(f(x)) + y\n$$\nfor any $x, y \\in \\mathbb{R}$.\n\nNext, we prove that these are the only solutions. Replacing $x$ by $x - f(y)$, the given condition means\n$$\nf(x) = x \\quad \\text{or} \\quad f(x) = f(f(x - f(y))) + y\n$$\nfor any $x, y \\in \\mathbb{R}$. We are done if $f(x) = x$ for all $x \\in \\mathbb{R}$. Assume $f(a) \\neq a$ for some $a \\in \\mathbb{R}$. Then we have\n$$\nf(a) = f(f(a - f(y))) + y \\quad (1)\n$$\nfor any $y \\in \\mathbb{R}$. Note that $f$ is injective, since if $f(\\alpha) = f(\\beta)$, then\n$$\n\\alpha = f(\\alpha) - f(f(\\alpha - f(\\alpha))) = f(\\alpha) - f(f(\\alpha - f(\\beta))) = \\beta\n$$\nby (1). Setting $y = 0$ in (1), we have $f(a) = f(f(a - f(0)))$, and hence\n$$\na = f(a - f(0)) \\quad (2)\n$$\nas $f$ is injective. In particular, this implies $f(0) \\neq 0$ (as $f(a) \\neq a$).\n\n* If $f(x) \\neq x$ for all $x \\in \\mathbb{R}$, then (2) holds for all $a \\in \\mathbb{R}$. Replacing $a$ by $x + f(0)$, this gives $f(x) = x + c$ for all $x \\in \\mathbb{R}$, where $c = f(0)$ is a constant as desired.\n\n* Suppose $f(b) = b$ for some $b$. Note that $f(f(a)) \\neq f(a)$ since $f(a) \\neq a$ and $f$ is injective. It follows that (1) holds if we replace $a$ by $f(a)$. Thus, we obtain\n$$\nf(f(a)) = f(f(f(a) - f(y))) + y.\n$$\nPutting $y = a$, this yields\n$$\nf(f(a)) = f(f(0)) + a. \\quad (3)\n$$\nNext, putting $y = b$ in (1), we have $f(a) = f(f(a - b)) + b$. Recall that $f(a) \\neq a$, so we cannot have $f(a - b) = a - b$. Therefore, using (3) with $a$ replaced by $a - b$, we deduce\n$$\nf(a) = f(f(a - b)) + b = f(f(0)) + (a - b) + b = a + f(f(0)). \\quad (4)\n$$\nRecall that $f(0) \\neq 0$. Thus, we may take $a = 0$ in (4). This implies $f(0) = f(f(0))$. But then this yields $0 = f(0)$ by the injectivity, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31813,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ be real numbers such that $x + y + z \\neq 0$. Find the minimum value of\n$$\n\\frac{|x| + |x + 4y| + |y + 7z| + 2|z|}{|x + y + z|}\n$$",
    "options": [],
    "answer": "15/11",
    "solution": "By the triangle inequality, we have\n$$\n\\begin{aligned}\n\\frac{|x| + |x + 4y| + |y + 7z| + 2|z|}{|x + y + z|} &\\ge \\frac{|x| + \\frac{4}{11}|x + 4y| + \\frac{1}{11}|-y - 7z| + 2|z|}{|x + y + z|} \\\\\n&\\ge \\frac{|x + \\frac{4}{11}(x + 4y) + \\frac{1}{11}(-y - 7z) + 2z|}{|x + y + z|} \\\\\n&= \\frac{15}{11}.\n\\end{aligned}\n$$\nEquality holds when $(x, y, z) = (28k, -7k, k)$ for some $k \\neq 0$. So the minimum value is $\\frac{15}{11}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31814,
    "subject": "Mathematics (Multi-modal)",
    "question": "After expanding the polynomial $(1 + x + y)^{2023}$ and collecting like terms, we obtain the expression\n$$\na_0 + a_1x + a_2y + a_3x^2 + a_4xy + a_5y^2 + \\dots + a_ky^{2023}\n$$\nFind the total number of $a_i$'s which are divisible by $5$.",
    "options": [],
    "answer": "2044800",
    "solution": "Note that $(1+x+y)^5 \\equiv 1+x^5+y^5 \\pmod{5}$ since the coefficients of all other terms are of the form $\\frac{5!}{i!(5-i)!}$ for some $i, j$ satisfying $0 \\le i+j \\le 5$ and $(i, j) \\ne (0, 0), (0, 5), (5, 0)$, all of which are divisible by $5$ as the denominator is not divisible by $5$. Inductively, we find that $(1+x+y)^{5k} \\equiv 1+x^{5k}+y^{5k} \\pmod{5}$ for all positive integers $k$. Now, note that $2023 = 3 \\cdot 5^4 + 5^3 + 4 \\cdot 5 + 3$. Therefore, we deduce\n$$\n(1+x+y)^{2023} = (1+x+y)^{3 \\cdot 5^4} (1+x+y)^{5^3} (1+x+y)^{4 \\cdot 5} (1+x+y)^3 \\\\\n\\equiv (1+x^{5^4}+y^{5^4})^3 (1+x^{5^3}+y^{5^3}) (1+x^{5^2}+y^{5^2}) (1+x^{5^1}+y^3) \\pmod{5}. \\quad (3)\n$$\nWhen we expand $(1 + X + Y)^n$, each term is of the form $X^a Y^b$ where $0 \\le a+b \\le n$. The total number of terms in the expansion is the number of nonnegative integer solutions to $a+b+c=n$ (by defining $c=n-a-b$), which is equal to $\\binom{n+2}{2}$. Therefore, there are $\\binom{5}{2} = 10$, $\\binom{3}{2} = 3$, $\\binom{6}{2} = 15$ and $\\binom{5}{2} = 10$ terms in the expansions of the factors on the right of (3) respectively. When we further expand their product, there are no like terms because the factorization corresponds to the base $5$ representation of the exponents (for example, the term $x^{5^4}y^{3 \\cdot 5+2}$ can only be obtained by taking $x^{5^4}$ from the first factor, taking $1$ from the second factor, taking $y^{3 \\cdot 5}$ from the third factor and taking $y^2$ from the fourth factor). Also, the coefficient of every term is not divisible by $5$ because all exponents are less than $5$, and $5$ is a prime number. Thus, there are\n$$\n10 \\times 3 \\times 15 \\times 10 = 4500\n$$\ncoefficients $a_i$'s which are not divisible by $5$. The total number of $a_i$'s which are divisible by $5$ is\n$$\n\\left(\\frac{2025}{2}\\right) - 4500 = 2044800.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31815,
    "subject": "Mathematics (Multi-modal)",
    "question": "A test question reads 'write down three consecutive positive integers not exceeding $2023$ in ascending order, such that one of them is a multiple of $6$, another one is a multiple of $7$ and the remaining one is a multiple of $8$. How many different correct answers are there for this question?'",
    "options": [],
    "answer": "24",
    "solution": "Let $n$, $n + 1$ and $n + 2$ be the three consecutive positive integers. Then $1 \\leq n \\leq 2021$. Since multiples of $6$ and $8$ must be even, we must have $n + 1$ being divisible by $7$ whereas $n$ and $n + 2$ are divisible by $6$ and $8$ in either order. For each of these two cases, there are exactly two choices of $n$ among any $168$ consecutive positive integers by the Chinese remainder theorem since the L.C.M. of $6$, $7$ and $8$ is $168$. As $2021 = 12 \\times 168 + 5$, and none of $n = 2017, 2018, 2019, 2020, 2021$ satisfies the conditions, there are $2 \\times 12 = 24$ choices of $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31816,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integer(s) $n$ such that $(2^n - 1)(5^n - 1)$ is a perfect square.",
    "options": [],
    "answer": "n = 1",
    "solution": "$n = 1$.\nWhen $n = 1$, $(2^1 - 1)(5^1 - 1) = 4$ is a perfect square. Suppose $n > 1$.\n\n* If $n \\equiv 3 \\pmod{4}$, we have $(2^n - 1)(5^n - 1) \\equiv (2)(-1) \\equiv 3 \\pmod{5}$, which is not true for a square.\n\n* If $n \\equiv 1 \\pmod{4}$, we rewrite\n$$\n(2^n - 1)(5^n - 1) = (2^n - 1) \\cdot 4(5^{n-1} + 5^{n-2} + \\dots + 1).\n$$\nIf this is a square, so is $(2^n - 1)(5^{n-1} + 5^{n-2} + \\dots + 1)$. However,\n$$\n(2^n - 1)(5^{n-1} + 5^{n-2} + \\dots + 1) \\equiv (-1)(1 + 1 + \\dots + 1) = -n \\equiv 3 \\pmod{4},\n$$\nwhich is not true for a square.\n\n* If $n = 2m$ is even, suppose $(2^n - 1)(5^n - 1)$ is a perfect square. Let $2^n - 1 = ds^2$ and $5^n - 1 = dt^2$ where $d$ is squarefree. We first show that $d > 1$. Indeed, if $d = 1$, then $5^{2m} - t^2 = 1$. This gives $(5^m - t)(5^m + t) = 1$, which is impossible as $t > 0$. Now, it is known that the solutions to Pell's equation $x^2 - dy^2 = 1$ are generated by $(x_1 + y_1\\sqrt{d})^n = x_n + y_n\\sqrt{d}$, where $(x_1, y_1)$ is the fundamental solution.\nClaim. Let $(x_k, y_k)$ be a solution to $x^2 - dy^2 = 1$.\n(i) If $k \\equiv 0 \\pmod{2}$, then $x_k \\equiv 1 \\pmod{2}$ and $y_k \\equiv 0 \\pmod{2}$.\n(ii) If $k \\equiv 1 \\pmod{2}$, then $x_k \\equiv x_1 \\pmod{2}$ and $y_k \\equiv y_1 \\pmod{2}$.\n\nProof.\n(i) Let $k = 2l$. Then we have\n$$\n\\begin{aligned}\nx_k + y_k \\sqrt{d} &= (x_1 + y_1 \\sqrt{d})^{2l} \\\\\n&= (x_l + y_l \\sqrt{d})^2 \\\\\n&= x_l^2 + dy_l^2 + 2x_l y_l \\sqrt{d} \\\\\n&= 2x_l^2 - 1 + 2x_l y_l \\sqrt{d}.\n\\end{aligned}\n$$\nSo $x_k = 2x_l^2 - 1 \\equiv 1 \\pmod{2}$ and $y_k = 2x_l y_l \\equiv 0 \\pmod{2}$.\n\n$$\nx_{2l+1} = x_{2l}x_1 + y_{2l}y_1 d \\equiv x_1 \\pmod{2},\n$$\n$$\ny_{2l+1} = x_{2l}y_1 + x_1 y_{2l} \\equiv y_1 \\pmod{2}.\n$$\n$\\Box$\nNow, since $2^{2m} - 1 = ds^2$, $(x, y) = (2^m, s)$ is a solution to $x^2 - dy^2 = 1$. By the claim, we must have $x_1 \\equiv 0 \\pmod{2}$. Similarly, since $5^{2m} - 1 = dt^2$, $(x_k, y_k) = (5^m, t)$ is another solution. By the claim, as $5^m \\not\\equiv x_1 \\pmod{2}$, we must have $k \\equiv 0 \\pmod{2}$. Let $k = 2l$. Then we have $5^m = x_{2l} = 2x_l^2 - 1$. This implies $2x_l^2 - 1 \\equiv 0 \\pmod{5}$, and hence $x_l^2 \\equiv 3 \\pmod{5}$. This is impossible.\nTherefore, the only solution is $n = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31817,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all nonnegative real numbers $a$, $b$, $c$ such that\n$$\n\\frac{4a + 9b + 25c}{2a + 3b + 5c} + \\frac{4b + 9c + 25a}{2b + 3c + 5a} + \\frac{4c + 9a + 25b}{2c + 3a + 5b} = 10\n$$",
    "options": [],
    "answer": "Exactly those triples where two of a, b, c are zero and the remaining one is positive; i.e., (a>0, b=0, c=0) or (a=0, b>0, c=0) or (a=0, b=0, c>0).",
    "solution": "By the Cauchy-Schwarz inequality, we have\n$$\n(a + b + c)(4a + 9b + 25c) \\geq (2a + 3b + 5c)^2.\n$$\nThis implies $\\frac{4a + 9b + 25c}{2a + 3b + 5c} \\geq \\frac{2a + 3b + 5c}{a + b + c}$. Adding similar inequalities, we obtain\n$$\n\\begin{aligned}\n& \\frac{4a + 9b + 25c}{2a + 3b + 5c} + \\frac{4b + 9c + 25a}{2b + 3c + 5a} + \\frac{4c + 9a + 25b}{2c + 3a + 5b} \\\\\n\\geq & \\frac{4a + 9b + 25c}{a + b + c} + \\frac{4b + 9c + 25a}{b + c + a} + \\frac{4c + 9a + 25b}{c + a + b} \\\\\n= & \\frac{10a + 10b + 10c}{a + b + c} \\\\\n= & 10.\n\\end{aligned}\n$$\n\nBy the given condition, the equality should hold. This means we need\n$$\na : b : c = 4a : 9b : 25c = 25a : 4b : 9c = 9a : 25b : 4c,\n$$\nwhich means two of $a$, $b$, $c$ are zero, and the remaining can be any positive real number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31818,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral with $AB = 5$, $AD = 17$, and $CD = 6$. If the angle bisectors of $\\angle BAD$ and $\\angle ADC$ intersect at the midpoint of $BC$, find the area of $ABCD$.",
    "options": [],
    "answer": "14√21",
    "solution": "Let $M$ be the midpoint of $BC$. Let $B'$ and $C'$ be points on $AD$ such that $AB' = AB = 5$ and $DC' = DC = 6$. Then $B'C' = 17 - 5 - 6 = 6$. Note that $\\triangle ABM \\cong \\triangle AB'M$ and $\\triangle DCM \\cong \\triangle DC'M$. Note also that $\\triangle MB'C'$ is isosceles as $MB' = MB = MC = MC'$. Let $N$ be the midpoint of $B'C'$. Then we have $B'N = C'N = 3$ and $\\angle MB'C' = \\angle MC'B'$. Observe that $MCDB'$ is cyclic since $MD$ bisects $\\angle B'DC$ and $MB' = MC$ (while $DB' \\ne DC$). Thus,\n$$\n2 \\angle B'MA = \\angle B'MB = \\angle C'DC = 2 \\angle MDC'.\n$$\n![](attached_image_1.png)\n\nFinally we have $MN = \\sqrt{MC'^2 - NC'^2} = \\sqrt{30-9} = \\sqrt{21}$. Therefore,\n$$\n\\begin{align*} \n[ABCD] &= 2([AB'M] + [DC'M] + [MB'N]) \\\\ \n&= (AB' + C'D + B'N) \\times MN \\\\ \n&= (5+6+3)\\sqrt{21} = 14\\sqrt{21}. \n\\end{align*}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31819,
    "subject": "Mathematics (Multi-modal)",
    "question": "$ABCD$ is a square of side length $1$. $BC$ is extended to $E$ and $DC$ is extended to $F$ such that $BE = DF = 3$. The circumcircle of $\\triangle AEF$ meets the extensions of $CB$ and $CD$ at $G$ and $H$ respectively. Find $GH$.",
    "options": [],
    "answer": "3√2/2",
    "solution": "Answer: $\\frac{3\\sqrt{2}}{2}$\n\nNote that $AG = AH$ by symmetry. Hence they subtend equal angles on the circumference of the circle, i.e. we have $\\angle AFG = \\angle AFH$. Suppose $FA$ meets $GE$ at $I$.\n\nBy similar triangles, we have $BI : IC = AB : CF = 1 : 2$. Hence $IC = \\frac{2}{3}$. Let $GC = x$. Using the angle bisector theorem, we have $GI : IC = FG : FC$, i.e.\n\n$$\n\\frac{x - \\frac{2}{3}}{\\frac{2}{3}} = \\frac{\\sqrt{x^2 + 2^2}}{2}\n$$\n\nSolving gives $x = \\frac{3}{2}$. By symmetry, $HC$ also has length $\\frac{3}{2}$ and so $GH = \\frac{3\\sqrt{2}}{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31820,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer such that $1^3 + 2^3 + \\dots + n^3$ is divisible by $n+3$. Find the greatest possible value of $n$.",
    "options": [],
    "answer": "15",
    "solution": "Answer: 15\nNote that\n$$\n1^3 + 2^3 + \\cdots + n^3 = \\frac{n^2(n+1)^2}{4}\n$$\n\nHence, for $n + 3$ to divide $1^3 + 2^3 + \\cdots + n^3$, it is necessary for $n^2(n + 1)^2$ to be a multiple of $n + 3$, or equivalently, by setting $m = n + 3$,\n$$\n0 \\equiv n^2(n+1)^2 = (m-3)^2(m-2)^2 \\equiv 36 \\pmod{m}.\n$$\nHence $m$ must be a factor of 36. To find the greatest possible value of $n$, we proceed as follows:\n* If $m = 36$ (i.e. $n = 33$), we have $1^3 + 2^3 + \\cdots + 33^3 = \\frac{33^2 \\cdot 34^2}{4} = 33^2 \\cdot 17^2$ which is odd (and hence not divisible by $n + 3 = 36$).\n* The next largest possible value of $m$ is 18 (corresponding to $n = 15$), in which case we have\n$$\n1^3 + 2^3 + \\cdots + 15^3 = \\frac{15^2 \\cdot 16^2}{4} = 15^2 \\cdot 8^2 = 2^6 \\cdot 3^2 \\cdot 5^2,\n$$\nwhich is divisible by $n + 3 = 2 \\cdot 3^2$.\nIt follows that the answer is 15.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31821,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Omega$ be the incircle of $\\triangle ABC$. There are two smaller circles $\\omega_1$ and $\\omega_2$ inside $\\triangle ABC$. The circle $\\omega_1$ is tangent to $\\Omega$ at $P$, tangent to $BC$ at $D$, and also tangent to $AB$. The circle $\\omega_2$ is tangent to $\\Omega$ at $Q$, tangent to $BC$ at $E$, and also tangent to $AC$. Prove that $D, E, Q, P$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S$ be the point on $\\Omega$ different from the contact point of $\\Omega$ and $BC$ such that its tangent to $\\Omega$ is parallel to $BC$. Then the tangent at $D$ to $\\omega_1$ is parallel to the tangent at $S$ to $\\Omega$. By considering the homothety with centre $P$ mapping $\\omega_1$ to $\\Omega$, we see that $D$, $P$, $S$ are collinear. Similarly, $E$, $Q$, $S$ are collinear. Let $X$ be a point on the tangent at $S$ to $\\Omega$ as shown. Then we have\n$$\n\\angle SQP = \\angle PSX = \\angle EDP.\n$$\nThis implies $D$, $E$, $Q$, $P$ are concyclic.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31822,
    "subject": "Mathematics (Multi-modal)",
    "question": "In $\\triangle ABC$, $\\angle BAC = 150^\\circ$ and $BC = 74$. $D$ is a point on $BC$ such that $BD = 14$ and $\\angle ADB = 60^\\circ$. Find the area of $\\triangle ABC$.",
    "options": [],
    "answer": "111√3",
    "solution": "Let $b$, $c$ and $d$ be the lengths of $AB$, $AC$ and $AD$ respectively.\nThe area of $\\triangle ABC$ is $\\frac{1}{2}bc\\sin 150^\\circ$. Alternatively, using $BC$ as base, the height of the triangle is $d\\sin 60^\\circ$ and so the area is also equal to $\\frac{1}{2}(74)d\\sin 60^\\circ$. This gives\n$$\n\\frac{1}{2}bc\\sin 150^\\circ = \\frac{1}{2}(74)d\\sin 60^\\circ,\n$$\nwhich simplifies to $bc = 74\\sqrt{3}d$. Using this and the cosine formula, we have\n$$\n\\begin{aligned}\n74^2 &= b^2 + c^2 - 2bc \\cos 150^\\circ \\\\\n&= [d^2 + 14^2 - 2d(14) \\cos 60^\\circ] + [d^2 + 60^2 - 2d(60) \\cos 120^\\circ] - 2(74\\sqrt{3}d) \\cos 150^\\circ \\\\\n&= 2d^2 + 268d + 3796,\n\\end{aligned}\n$$\nwhich simplifies to $2(d - 6)(d + 140) = 0$. Hence $d = 6$ and so the area of $\\triangle ABC$ is $\\frac{1}{2}(74)d\\sin 60^\\circ = 111\\sqrt{3}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31823,
    "subject": "Mathematics (Multi-modal)",
    "question": "How many solutions are there to the equation $\\log_3 x = 3 \\sin(3\\pi x)$?",
    "options": [],
    "answer": "81",
    "solution": "Answer: 81\nA sketch of the graphs of $y = \\log_3 x$ and $y = 3 \\sin (3\\pi x)$ is as follows:\n\n![](attached_image_1.png)\n\nNote that the right hand side of the equation (i.e. $3\\sin(3\\pi x)$) must be between $-3$ and $3$. Hence any solution satisfies $-3 \\le \\log_3 x \\le 3$, or $\\frac{1}{27} \\le x \\le 27$. There are three solutions for $x \\le 1$ as shown in the graph above. For $x \\in (1, 27]$, we have $\\log_3 x > 0$ and the function $3\\sin(3\\pi x)$ is positive in 39 intervals, namely, $(\\frac{4}{3}, \\frac{5}{3})$, $(\\frac{6}{3}, \\frac{7}{3})$, ..., $(\\frac{80}{3}, \\frac{81}{3})$. In each of these intervals the graph of $\\log_3 x > 0$ intersects the graph of $3\\sin(3\\pi x)$ at exactly two points. The answer is thus $3 + 39(2) = 81$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31824,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the largest real root to the equation $\\sqrt[3]{x^3 + 3x^2 - 4} - x = \\sqrt[3]{x^3 - 3x + 2} - 1$.",
    "options": [],
    "answer": "(1+3√5)/4",
    "solution": "Answer: $\\frac{1+3\\sqrt{5}}{4}$\nWe can factorize the expressions to get\n$$\n\\sqrt[3]{(x-1)(x+2)^2} - x = \\sqrt[3]{(x-1)^2(x+2)} - 1.\n$$\nIf we let $u = \\sqrt[3]{x-1}$ and $v = \\sqrt[3]{x+2}$, the equation becomes $uv^2 = u^2v + u^3$, or\n$$\nu(u^2 + uv - v^2) = 0.$$\nIf $u = 0$, then $x = 1$. If there is another root $x > 1$, then $u, v$ must both be positive, so $u^2 + uv - v^2 = 0$ gives $u = \\lambda v$ where $\\lambda = \\frac{1}{2}(-1 + \\sqrt{5})$. Hence $u^3 = (\\lambda v)^3$, i.e. $x - 1 = \\lambda^3(x + 2)$, so we have\n$$\nx = \\frac{1 + 2\\lambda^3}{1 - \\lambda^3} = \\frac{1 + 2(\\sqrt{5} - 2)}{1 - (\\sqrt{5} - 2)} = \\frac{1 + 3\\sqrt{5}}{4}.\n$$\nIt is easy to check that such $x$ indeed satisfies the original equation, and hence is the largest real root to the equation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31825,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of real numbers $a_{0}, a_{1}, a_{2}, \\ldots$ is defined by the formula\n$$\na_{i+1}=\\left\\lfloor a_{i}\\right\\rfloor \\cdot\\left\\langle a_{i}\\right\\rangle \\quad \\text{ for } \\quad i \\geq 0\n$$\nhere $a_{0}$ is an arbitrary real number, $\\left\\lfloor a_{i}\\right\\rfloor$ denotes the greatest integer not exceeding $a_{i}$, and $\\left\\langle a_{i}\\right\\rangle=a_{i}-\\left\\lfloor a_{i}\\right\\rfloor$. Prove that $a_{i}=a_{i+2}$ for $i$ sufficiently large.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First note that if $a_{0} \\geq 0$, then all $a_{i} \\geq 0$. For $a_{i} \\geq 1$ we have (in view of $\\left\\langle a_{i}\\right\\rangle<1$ and $\\left\\lfloor a_{i}\\right\\rfloor>0$ )\n$$\n\\left\\lfloor a_{i+1}\\right\\rfloor \\leq a_{i+1}=\\left\\lfloor a_{i}\\right\\rfloor \\cdot\\left\\langle a_{i}\\right\\rangle<\\left\\lfloor a_{i}\\right\\rfloor\n$$\nthe sequence $\\left\\lfloor a_{i}\\right\\rfloor$ is strictly decreasing as long as its terms are in $[1, \\infty)$. Eventually there appears a number from the interval $[0,1)$ and all subsequent terms are $0$.\n\nNow pass to the more interesting situation where $a_{0}<0$; then all $a_{i} \\leq 0$. Suppose the sequence never hits $0$. Then we have $\\left\\lfloor a_{i}\\right\\rfloor \\leq -1$ for all $i$, and so\n$$\n1+\\left\\lfloor a_{i+1}\\right\\rfloor>a_{i+1}=\\left\\lfloor a_{i}\\right\\rfloor \\cdot\\left\\langle a_{i}\\right\\rangle>\\left\\lfloor a_{i}\\right\\rfloor\n$$\nthis means that the sequence $\\left\\lfloor a_{i}\\right\\rfloor$ is nondecreasing. And since all its terms are integers from $(-\\infty,-1]$, this sequence must be constant from some term on:\n$$\n\\left\\lfloor a_{i}\\right\\rfloor=c \\quad \\text{ for } \\quad i \\geq i_{0} ; \\quad c \\text{ a negative integer. }\n$$\nThe defining formula becomes\n$$\na_{i+1}=c \\cdot\\left\\langle a_{i}\\right\\rangle=c\\left(a_{i}-c\\right)=c a_{i}-c^{2}\n$$\nConsider the sequence\n$$\n\\begin{equation*}\nb_{i}=a_{i}-\\frac{c^{2}}{c-1} \\tag{1}\n\\end{equation*}\n$$\nIt satisfies the recursion rule\n$$\nb_{i+1}=a_{i+1}-\\frac{c^{2}}{c-1}=c a_{i}-c^{2}-\\frac{c^{2}}{c-1}=c b_{i}\n$$\nimplying\n$$\n\\begin{equation*}\nb_{i}=c^{i-i_{0}} b_{i_{0}} \\quad \\text{ for } \\quad i \\geq i_{0} \\tag{2}\n\\end{equation*}\n$$\nSince all the numbers $a_{i}$ (for $i \\geq i_{0}$ ) lie in $[c, c+1)$, the sequence $\\left(b_{i}\\right)$ is bounded. The equation (2) can be satisfied only if either $b_{i_{0}}=0$ or $|c|=1$, i.e., $c=-1$.\n\nIn the first case, $b_{i}=0$ for all $i \\geq i_{0}$, so that\n$$\na_{i}=\\frac{c^{2}}{c-1} \\quad \\text{ for } \\quad i \\geq i_{0} .\n$$\nIn the second case, $c=-1$, equations (1) and (2) say that\n$$\na_{i}=-\\frac{1}{2}+(-1)^{i-i_{0}} b_{i_{0}}= \\begin{cases}a_{i_{0}} & \\text{ for } i=i_{0}, i_{0}+2, i_{0}+4, \\ldots, \\\\ 1-a_{i_{0}} & \\text{ for } i=i_{0}+1, i_{0}+3, i_{0}+5, \\ldots .\\end{cases}\n$$\nSummarising, we see that (from some point on) the sequence $\\left(a_{i}\\right)$ either is constant or takes alternately two values from the interval $(-1,0)$. The result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31826,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence of real numbers $a_{0}, a_{1}, a_{2}, \\ldots$ is defined recursively by\n$$\na_{0}=-1, \\quad \\sum_{k=0}^{n} \\frac{a_{n-k}}{k+1}=0 \\quad \\text{ for } \\quad n \\geq 1\n$$\nShow that $a_{n}>0$ for $n \\geq 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The proof goes by induction. For $n=1$ the formula yields $a_{1}=1 / 2$. Take $n \\geq 1$, assume $a_{1}, \\ldots, a_{n}>0$ and write the recurrence formula for $n$ and $n+1$, respectively as\n$$\n\\sum_{k=0}^{n} \\frac{a_{k}}{n-k+1}=0 \\quad \\text{ and } \\quad \\sum_{k=0}^{n+1} \\frac{a_{k}}{n-k+2}=0 .\n$$\nSubtraction yields\n$$\n\\begin{aligned}\n0=(n+2) \\sum_{k=0}^{n+1} & \\frac{a_{k}}{n-k+2}-(n+1) \\sum_{k=0}^{n} \\frac{a_{k}}{n-k+1} \\\\\n& =(n+2) a_{n+1}+\\sum_{k=0}^{n}\\left(\\frac{n+2}{n-k+2}-\\frac{n+1}{n-k+1}\\right) a_{k}\n\\end{aligned}\n$$\nThe coefficient of $a_{0}$ vanishes, so\n$$\na_{n+1}=\\frac{1}{n+2} \\sum_{k=1}^{n}\\left(\\frac{n+1}{n-k+1}-\\frac{n+2}{n-k+2}\\right) a_{k}=\\frac{1}{n+2} \\sum_{k=1}^{n} \\frac{k}{(n-k+1)(n-k+2)} a_{k} .\n$$\nThe coefficients of $a_{1}, \\ldots, a_{n}$ are all positive. Therefore, $a_{1}, \\ldots, a_{n}>0$ implies $a_{n+1}>0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31827,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $c_{0}, c_{1}, \\ldots, c_{n}, \\ldots$ is defined by $c_{0}=1$, $c_{1}=0$ and $c_{n+2}=c_{n+1}+c_{n}$ for $n \\geq 0$. Consider the set $S$ of ordered pairs $(x, y)$ for which there is a finite set $J$ of positive integers such that $x=\\sum_{j \\in J} c_{j}$, $y=\\sum_{j \\in J} c_{j-1}$. Prove that there exist real numbers $\\alpha$, $\\beta$ and $m$, $M$ with the following property: An ordered pair of nonnegative integers $(x, y)$ satisfies the inequality\n$$\nm<\\alpha x+\\beta y<M\n$$\nif and only if $(x, y) \\in S$.\n\nN. B. A sum over the elements of the empty set is assumed to be $0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\varphi=(1+\\sqrt{5}) / 2$ and $\\psi=(1-\\sqrt{5}) / 2$ be the roots of the quadratic equation $t^{2}-t-1=0$. So $\\varphi \\psi=-1$, $\\varphi+\\psi=1$ and $1+\\psi=\\psi^{2}$. An easy induction shows that the general term $c_{n}$ of the given sequence satisfies\n$$\nc_{n}=\\frac{\\varphi^{n-1}-\\psi^{n-1}}{\\varphi-\\psi} \\quad \\text{ for } n \\geq 0\n$$\nSuppose that the numbers $\\alpha$ and $\\beta$ have the stated property, for appropriately chosen $m$ and $M$. Since $\\left(c_{n}, c_{n-1}\\right) \\in S$ for each $n$, the expression\n$\\alpha c_{n}+\\beta c_{n-1}=\\frac{\\alpha}{\\sqrt{5}}\\left(\\varphi^{n-1}-\\psi^{n-1}\\right)+\\frac{\\beta}{\\sqrt{5}}\\left(\\varphi^{n-2}-\\psi^{n-2}\\right)=\\frac{1}{\\sqrt{5}}\\left[(\\alpha \\varphi+\\beta) \\varphi^{n-2}-(\\alpha \\psi+\\beta) \\psi^{n-2}\\right]$\nis bounded as $n$ grows to infinity. Because $\\varphi>1$ and $-1<\\psi<0$, this implies $\\alpha \\varphi+\\beta=0$.\nTo satisfy $\\alpha \\varphi+\\beta=0$, one can set for instance $\\alpha=\\psi$, $\\beta=1$. We now find the required $m$ and $M$ for this choice of $\\alpha$ and $\\beta$.\nNote first that the above displayed equation gives $c_{n} \\psi+c_{n-1}=\\psi^{n-1}$, $n \\geq 1$. In the sequel, we denote the pairs in $S$ by $\\left(a_{J}, b_{J}\\right)$, where $J$ is a finite subset of the set $\\mathbb{N}$ of positive integers and $a_{J}=\\sum_{j \\in J} c_{j}$, $b_{J}=\\sum_{j \\in J} c_{j-1}$. Since $\\psi a_{J}+b_{J}=\\sum_{j \\in J}\\left(c_{j} \\psi+c_{j-1}\\right)$, we obtain\n$$\n\\begin{equation*}\n\\psi a_{J}+b_{J}=\\sum_{j \\in J} \\psi^{j-1} \\quad \\text{ for each }\\left(a_{J}, b_{J}\\right) \\in S \\tag{1}\n\\end{equation*}\n$$\nOn the other hand, in view of $-1<\\psi<0$,\n$$\n-1=\\frac{\\psi}{1-\\psi^{2}}=\\sum_{j=0}^{\\infty} \\psi^{2 j+1}<\\sum_{j \\in J} \\psi^{j-1}<\\sum_{j=0}^{\\infty} \\psi^{2 j}=\\frac{1}{1-\\psi^{2}}=1-\\psi=\\varphi\n$$\nTherefore, according to (1),\n$$\n-1<\\psi a_{J}+b_{J}<\\varphi \\quad \\text{ for each }\\left(a_{J}, b_{J}\\right) \\in S .\n$$\nThus $m=-1$ and $M=\\varphi$ is an appropriate choice.\nConversely, we prove that if an ordered pair of nonnegative integers $(x, y)$ satisfies the inequality $-1<\\psi x+y<\\varphi$ then $(x, y) \\in S$.\nLemma. Let $x, y$ be nonnegative integers such that $-1<\\psi x+y<\\varphi$. Then there exists a subset $J$ of $\\mathbb{N}$ such that\n$$\n\\begin{equation*}\n\\psi x+y=\\sum_{j \\in J} \\psi^{j-1} \\tag{2}\n\\end{equation*}\n$$\nProof. For $x=y=0$ it suffices to choose the empty subset of $\\mathbb{N}$ as $J$, so let at least one of $x, y$ be nonzero. There exist representations of $\\psi x+y$ of the form\n$$\n\\psi x+y=\\psi^{i_{1}}+\\cdots+\\psi^{i_{k}}\n$$\nwhere $i_{1} \\leq \\cdots \\leq i_{k}$ is a sequence of nonnegative integers, not necessarily distinct. For instance, we can take $x$ summands $\\psi^{1}=\\psi$ and $y$ summands $\\psi^{0}=1$. Consider all such representations of minimum length $k$ and focus on the ones for which $i_{1}$ has the minimum possible value $j_{1}$. Among them, consider the representations where $i_{2}$ has the minimum possible value $j_{2}$. Upon choosing $j_{3}, \\ldots, j_{k}$ analogously, we obtain a sequence $j_{1} \\leq \\cdots \\leq j_{k}$ which clearly satisfies $\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}$. To prove the lemma, it suffices to show that $j_{1}, \\ldots, j_{k}$ are pairwise distinct.\nSuppose on the contrary that $j_{r}=j_{r+1}$ for some $r=1, \\ldots, k-1$. Let us consider the case $j_{r} \\geq 2$ first. Observing that $2 \\psi^{2}=1+\\psi^{3}$, we replace $j_{r}$ and $j_{r+1}$ by $j_{r}-2$ and $j_{r}+1$, respectively. Since\n$$\n\\psi^{j_{r}}+\\psi^{j_{r+1}}=2 \\psi^{j_{r}}=\\psi^{j_{r}-2}\\left(1+\\psi^{3}\\right)=\\psi^{j_{r}-2}+\\psi^{j_{r}+1},\n$$\nthe new sequence also represents $\\psi x+y$ as needed, and the value of $i_{r}$ in it contradicts the minimum choice of $j_{r}$.\nLet $j_{r}=j_{r+1}=0$. Then the sum $\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}$ contains at least two summands equal to $\\psi^{0}=1$. On the other hand $j_{s} \\neq 1$ for all $s$, because the equality $1+\\psi=\\psi^{2}$ implies that a representation of minimum length cannot contain consecutive $i_{r}$ 's. It follows that\n$$\n\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}>2+\\psi^{3}+\\psi^{5}+\\psi^{7}+\\cdots=2-\\psi^{2}=\\varphi,\n$$\ncontradicting the condition of the lemma.\nLet $j_{r}=j_{r+1}=1$; then $\\sum_{r=1}^{k} \\psi^{j_{r}}$ contains at least two summands equal to $\\psi^{1}=\\psi$. Like in the case $j_{r}=j_{r+1}=0$, we also infer that $j_{s} \\neq 0$ and $j_{s} \\neq 2$ for all $s$. Therefore\n$$\n\\psi x+y=\\sum_{r=1}^{k} \\psi^{j_{r}}<2 \\psi+\\psi^{4}+\\psi^{6}+\\psi^{8}+\\cdots=2 \\psi-\\psi^{3}=-1,\n$$\nwhich is a contradiction again. The conclusion follows. $\\square$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31828,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the inequality\n$$\n\\sum_{i<j} \\frac{a_{i} a_{j}}{a_{i}+a_{j}} \\leq \\frac{n}{2\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)} \\sum_{i<j} a_{i} a_{j}\n$$\nfor positive real numbers $a_{1}, a_{2}, \\ldots, a_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S=\\sum_{i} a_{i}$. Denote by $L$ and $R$ the expressions on the left and right hand side of the proposed inequality. We transform $L$ and $R$ using the identity\n$$\n\\sum_{i<j}\\left(a_{i}+a_{j}\\right)=(n-1) \\sum_{i} a_{i} . \\tag{1}\n$$\nAnd thus:\n$$\nL=\\sum_{i<j} \\frac{a_{i} a_{j}}{a_{i}+a_{j}}=\\sum_{i<j} \\frac{1}{4}\\left(a_{i}+a_{j}-\\frac{\\left(a_{i}-a_{j}\\right)^{2}}{a_{i}+a_{j}}\\right)=\\frac{n-1}{4} \\cdot S-\\frac{1}{4} \\sum_{i<j} \\frac{\\left(a_{i}-a_{j}\\right)^{2}}{a_{i}+a_{j}} . \\tag{2}\n$$\nTo represent $R$ we express the sum $\\sum_{i<j} a_{i} a_{j}$ in two ways; in the second transformation identity (1) will be applied to the squares of the numbers $a_{i}$ :\n$$\n\\begin{gathered}\n\\sum_{i<j} a_{i} a_{j}=\\frac{1}{2}\\left(S^{2}-\\sum_{i} a_{i}^{2}\\right) ; \\\\\n\\sum_{i<j} a_{i} a_{j}=\\frac{1}{2} \\sum_{i<j}\\left(a_{i}^{2}+a_{j}^{2}-\\left(a_{i}-a_{j}\\right)^{2}\\right)=\\frac{n-1}{2} \\cdot \\sum_{i} a_{i}^{2}-\\frac{1}{2} \\sum_{i<j}\\left(a_{i}-a_{j}\\right)^{2} .\n\\end{gathered}\n$$\nMultiplying the first of these equalities by $n-1$ and adding the second one we obtain\n$$\nn \\sum_{i<j} a_{i} a_{j}=\\frac{n-1}{2} \\cdot S^{2}-\\frac{1}{2} \\sum_{i<j}\\left(a_{i}-a_{j}\\right)^{2} .\n$$\nHence\n$$\nR=\\frac{n}{2 S} \\sum_{i<j} a_{i} a_{j}=\\frac{n-1}{4} \\cdot S-\\frac{1}{4} \\sum_{i<j} \\frac{\\left(a_{i}-a_{j}\\right)^{2}}{S} . \\tag{3}\n$$\nNow compare (2) and (3). Since $S \\geq a_{i}+a_{j}$ for any $i<j$, the claim $L \\geq R$ results.\nLet $S=a_{1}+a_{2}+\\cdots+a_{n}$. For any $i \\neq j$,\n$$\n4 \\frac{a_{i} a_{j}}{a_{i}+a_{j}}=a_{i}+a_{j}-\\frac{\\left(a_{i}-a_{j}\\right)^{2}}{a_{i}+a_{j}} \\leq a_{i}+a_{j}-\\frac{\\left(a_{i}-a_{j}\\right)^{2}}{a_{1}+a_{2}+\\cdots+a_{n}}=\\frac{\\sum_{k \\neq i} a_{i} a_{k}+\\sum_{k \\neq j} a_{j} a_{k}+2 a_{i} a_{j}}{S} .\n$$\nThe statement is obtained by summing up these inequalities for all pairs $i, j$ :\n$$\n\\begin{gathered}\n\\sum_{i<j} \\frac{a_{i} a_{j}}{a_{i}+a_{j}}=\\frac{1}{2} \\sum_{i} \\sum_{j \\neq i} \\frac{a_{i} a_{j}}{a_{i}+a_{j}} \\leq \\frac{1}{8 S} \\sum_{i} \\sum_{j \\neq i}\\left(\\sum_{k \\neq i} a_{i} a_{k}+\\sum_{k \\neq j} a_{j} a_{k}+2 a_{i} a_{j}\\right) \\\\\n=\\frac{1}{8 S}\\left(\\sum_{k} \\sum_{i \\neq k} \\sum_{j \\neq i} a_{i} a_{k}+\\sum_{k} \\sum_{j \\neq k} \\sum_{i \\neq j} a_{j} a_{k}+\\sum_{i} \\sum_{j \\neq i} 2 a_{i} a_{j}\\right) \\\\\n=\\frac{1}{8 S}\\left(\\sum_{k} \\sum_{i \\neq k}(n-1) a_{i} a_{k}+\\sum_{k} \\sum_{j \\neq k}(n-1) a_{j} a_{k}+\\sum_{i} \\sum_{j \\neq i} 2 a_{i} a_{j}\\right) \\\\\n=\\frac{n}{4 S} \\sum_{i} \\sum_{j \\neq i} a_{i} a_{j}=\\frac{n}{2 S} \\sum_{i<j} a_{i} a_{j} .\n\\end{gathered}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31829,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be the sides of a triangle. Prove that\n$$\n\\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}+\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}}+\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}} \\leq 3 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note first that the denominators are all positive, e.g. $\\sqrt{a}+\\sqrt{b}>\\sqrt{a+b}>\\sqrt{c}$.\nLet $x=\\sqrt{b}+\\sqrt{c}-\\sqrt{a}$, $y=\\sqrt{c}+\\sqrt{a}-\\sqrt{b}$ and $z=\\sqrt{a}+\\sqrt{b}-\\sqrt{c}$. Then\n$b+c-a=\\left(\\frac{z+x}{2}\\right)^{2}+\\left(\\frac{x+y}{2}\\right)^{2}-\\left(\\frac{y+z}{2}\\right)^{2}=\\frac{x^{2}+x y+x z-y z}{2}=x^{2}-\\frac{1}{2}(x-y)(x-z)$\nand\n$$\n\\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}=\\sqrt{1-\\frac{(x-y)(x-z)}{2 x^{2}}} \\leq 1-\\frac{(x-y)(x-z)}{4 x^{2}},\n$$\napplying $\\sqrt{1+2 u} \\leq 1+u$ in the last step. Similarly we obtain\n$$\n\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}} \\leq 1-\\frac{(z-x)(z-y)}{4 z^{2}} \\quad \\text{ and } \\quad \\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}} \\leq 1-\\frac{(y-z)(y-x)}{4 y^{2}} .\n$$\nSubstituting these quantities into the statement, it is sufficient to prove that\n$$\n\\begin{equation*}\n\\frac{(x-y)(x-z)}{x^{2}}+\\frac{(y-z)(y-x)}{y^{2}}+\\frac{(z-x)(z-y)}{z^{2}} \\geq 0 . \\tag{1}\n\\end{equation*}\n$$\nBy symmetry we can assume $x \\leq y \\leq z$. Then\n$$\n\\begin{gathered}\n\\frac{(x-y)(x-z)}{x^{2}}=\\frac{(y-x)(z-x)}{x^{2}} \\geq \\frac{(y-x)(z-y)}{y^{2}}=-\\frac{(y-z)(y-x)}{y^{2}} \\\\\n\\frac{(z-x)(z-y)}{z^{2}} \\geq 0\n\\end{gathered}\n$$\nand (1) follows.\nDue to the symmetry of variables, it can be assumed that $a \\geq b \\geq c$. We claim that\n$$\n\\frac{\\sqrt{a+b-c}}{\\sqrt{a}+\\sqrt{b}-\\sqrt{c}} \\leq 1 \\quad \\text{ and } \\quad \\frac{\\sqrt{b+c-a}}{\\sqrt{b}+\\sqrt{c}-\\sqrt{a}}+\\frac{\\sqrt{c+a-b}}{\\sqrt{c}+\\sqrt{a}-\\sqrt{b}} \\leq 2 .\n$$\nThe first inequality follows from\n$$\n\\sqrt{a+b-c}-\\sqrt{a}=\\frac{(a+b-c)-a}{\\sqrt{a+b-c}+\\sqrt{a}} \\leq \\frac{b-c}{\\sqrt{b}+\\sqrt{c}}=\\sqrt{b}-\\sqrt{c} .\n$$\nFor proving the second inequality, let $p=\\sqrt{a}+\\sqrt{b}$ and $q=\\sqrt{a}-\\sqrt{b}$. Then $a-b=p q$ and the inequality becomes\n$$\n\\frac{\\sqrt{c-p q}}{\\sqrt{c}-q}+\\frac{\\sqrt{c+p q}}{\\sqrt{c}+q} \\leq 2 .\n$$\nFrom $a \\geq b \\geq c$ we have $p \\geq 2 \\sqrt{c}$. Applying the Cauchy-Schwarz inequality,\n$$\n\\begin{gathered}\n\\left(\\frac{\\sqrt{c-p q}}{\\sqrt{c}-q}+\\frac{\\sqrt{c+p q}}{\\sqrt{c}+q}\\right)^{2} \\leq\\left(\\frac{c-p q}{\\sqrt{c}-q}+\\frac{c+p q}{\\sqrt{c}+q}\\right)\\left(\\frac{1}{\\sqrt{c}-q}+\\frac{1}{\\sqrt{c}+q}\\right) \\\\\n=\\frac{2\\left(c \\sqrt{c}-p q^{2}\\right)}{c-q^{2}} \\cdot \\frac{2 \\sqrt{c}}{c-q^{2}}=4 \\cdot \\frac{c^{2}-\\sqrt{c} p q^{2}}{\\left(c-q^{2}\\right)^{2}} \\leq 4 \\cdot \\frac{c^{2}-2 c q^{2}}{\\left(c-q^{2}\\right)^{2}} \\leq 4\n\\end{gathered}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31830,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest number $M$ such that the inequality\n$$\n\\left|a b\\left(a^{2}-b^{2}\\right)+b c\\left(b^{2}-c^{2}\\right)+c a\\left(c^{2}-a^{2}\\right)\\right| \\leq M\\left(a^{2}+b^{2}+c^{2}\\right)^{2}\n$$\nholds for all real numbers $a, b, c$.",
    "options": [],
    "answer": "9*sqrt(2)/32",
    "solution": "We first consider the cubic polynomial\n$$\nP(t)=t b\\left(t^{2}-b^{2}\\right)+b c\\left(b^{2}-c^{2}\\right)+c t\\left(c^{2}-t^{2}\\right)\n$$\nIt is easy to check that $P(b)=P(c)=P(-b-c)=0$, and therefore\n$$\nP(t)=(b-c)(t-b)(t-c)(t+b+c)\n$$\nsince the cubic coefficient is $b-c$. The left-hand side of the proposed inequality can therefore be written in the form\n$$\n\\left|a b\\left(a^{2}-b^{2}\\right)+b c\\left(b^{2}-c^{2}\\right)+c a\\left(c^{2}-a^{2}\\right)\\right|=|P(a)|=|(b-c)(a-b)(a-c)(a+b+c)|\n$$\nThe problem comes down to finding the smallest number $M$ that satisfies the inequality\n$$\n\\begin{equation*}\n|(b-c)(a-b)(a-c)(a+b+c)| \\leq M \\cdot\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\tag{1}\n\\end{equation*}\n$$\nNote that this expression is symmetric, and we can therefore assume $a \\leq b \\leq c$ without loss of generality. With this assumption,\n$$\n\\begin{equation*}\n|(a-b)(b-c)|=(b-a)(c-b) \\leq\\left(\\frac{(b-a)+(c-b)}{2}\\right)^{2}=\\frac{(c-a)^{2}}{4} \\tag{2}\n\\end{equation*}\n$$\nwith equality if and only if $b-a=c-b$, i.e. $2 b=a+c$. Also\n$$\n\\left(\\frac{(c-b)+(b-a)}{2}\\right)^{2} \\leq \\frac{(c-b)^{2}+(b-a)^{2}}{2}\n$$\nor equivalently,\n$$\n\\begin{equation*}\n3(c-a)^{2} \\leq 2 \\cdot\\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\\right] \\tag{3}\n\\end{equation*}\n$$\nagain with equality only for $2 b=a+c$. From (2) and (3) we get\n$$\n\\begin{aligned}\n& |(b-c)(a-b)(a-c)(a+b+c)| \\\\\n\\leq & \\frac{1}{4} \\cdot\\left|(c-a)^{3}(a+b+c)\\right| \\\\\n= & \\frac{1}{4} \\cdot \\sqrt{(c-a)^{6}(a+b+c)^{2}} \\\\\n\\leq & \\frac{1}{4} \\cdot \\sqrt{\\left(\\frac{2 \\cdot\\left[(b-a)^{2}+(c-b)^{2}+(c-a)^{2}\\right]}{3}\\right)^{3} \\cdot(a+b+c)^{2}} \\\\\n= & \\frac{\\sqrt{2}}{2} \\cdot\\left(\\sqrt[4]{\\left(\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}\\right)^{3} \\cdot(a+b+c)^{2}}\\right)^{2}\n\\end{aligned}\n$$\nBy the weighted AM-GM inequality this estimate continues as follows:\n$$\n\\begin{aligned}\n& |(b-c)(a-b)(a-c)(a+b+c)| \\\\\n\\leq & \\frac{\\sqrt{2}}{2} \\cdot\\left(\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}+(a+b+c)^{2}}{4}\\right)^{2} \\\\\n= & \\frac{9 \\sqrt{2}}{32} \\cdot\\left(a^{2}+b^{2}+c^{2}\\right)^{2} .\n\\end{aligned}\n$$\nWe see that the inequality (1) is satisfied for $M=\\frac{9}{32} \\sqrt{2}$, with equality if and only if $2 b=a+c$ and\n$$\n\\frac{(b-a)^{2}+(c-b)^{2}+(c-a)^{2}}{3}=(a+b+c)^{2} .\n$$\nPlugging $b=(a+c) / 2$ into the last equation, we bring it to the equivalent form\n$$\n2(c-a)^{2}=9(a+c)^{2} .\n$$\nThe conditions for equality can now be restated as\n$$\n2 b=a+c \\quad \\text{ and } \\quad(c-a)^{2}=18 b^{2} .\n$$\nSetting $b=1$ yields $a=1-\\frac{3}{2} \\sqrt{2}$ and $c=1+\\frac{3}{2} \\sqrt{2}$. We see that $M=\\frac{9}{32} \\sqrt{2}$ is indeed the smallest constant satisfying the inequality, with equality for any triple ( $a, b, c$ ) proportional to ( $1-\\frac{3}{2} \\sqrt{2}, 1,1+\\frac{3}{2} \\sqrt{2}$ ), up to permutation.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31831,
    "subject": "Mathematics (Multi-modal)",
    "question": "We have $n \\geq 2$ lamps $L_{1}, \\ldots, L_{n}$ in a row, each of them being either on or off. Every second we simultaneously modify the state of each lamp as follows:\n- if the lamp $L_{i}$ and its neighbours (only one neighbour for $i=1$ or $i=n$, two neighbours for other $i$) are in the same state, then $L_{i}$ is switched off;\n- otherwise, $L_{i}$ is switched on.\nInitially all the lamps are off except the leftmost one which is on.\n\na. Prove that there are infinitely many integers $n$ for which all the lamps will eventually be off.\n\nb. Prove that there are infinitely many integers $n$ for which the lamps will never be all off.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nExperiments with small $n$ lead to the guess that every $n$ of the form $2^{k}$ should be good. This is indeed the case, and more precisely: let $A_{k}$ be the $2^{k} \\times 2^{k}$ matrix whose rows represent the evolution of the system, with entries $0,1$ (for off and on respectively). The top row shows the initial state $[1,0,0, \\ldots, 0]$; the bottom row shows the state after $2^{k}-1$ steps. The claim is that:\n$$\n\\text{The bottom row of } A_{k} \\text{ is } [1,1,1, \\ldots, 1] \\text{.}\n$$\nThis will of course suffice because one more move then produces $[0,0,0, \\ldots, 0]$, as required.\n\nThe proof is by induction on $k$. The base $k=1$ is obvious. Assume the claim to be true for a $k \\geq 1$ and write the matrix $A_{k+1}$ in the block form $\\left(\\begin{array}{cc}A_{k} & O_{k} \\\\ B_{k} & C_{k}\\end{array}\\right)$ with four $2^{k} \\times 2^{k}$ matrices. After $m$ steps, the last $1$ in a row is at position $m+1$. Therefore $O_{k}$ is the zero matrix. According to the induction hypothesis, the bottom row of $[A_{k} \\ O_{k}]$ is $[1, \\ldots, 1,0, \\ldots, 0]$, with $2^{k}$ ones and $2^{k}$ zeros. The next row is thus\n$$\n[\\underbrace{0, \\ldots, 0}_{2^{k}-1}, 1,1, \\underbrace{0, \\ldots, 0}_{2^{k}-1}]\n$$\nIt is symmetric about its midpoint, and this symmetry is preserved in all subsequent rows because the procedure described in the problem statement is left/right symmetric. Thus $B_{k}$ is the mirror image of $C_{k}$. In particular, the rightmost column of $B_{k}$ is identical with the leftmost column of $C_{k}$.\n\nImagine the matrix $C_{k}$ in isolation from the rest of $A_{k+1}$. Suppose it is subject to evolution as defined in the problem: the first (leftmost) term in a row depends only on the two first terms in the preceding row, according as they are equal or not. Now embed $C_{k}$ again in $A_{k}$. The 'leftmost' terms in the rows of $C_{k}$ now have neighbours on their left side - but these neighbours are their exact copies. Consequently the actual evolution within $C_{k}$ is the same, whether or not $C_{k}$ is considered as a piece of $A_{k+1}$ or in isolation. And since the top row of $C_{k}$ is $[1,0, \\ldots, 0]$, it follows that $C_{k}$ is identical with $A_{k}$.\n\nThe bottom row of $A_{k}$ is $[1,1, \\ldots, 1]$; the same is the bottom row of $C_{k}$, hence also of $B_{k}$, which mirrors $C_{k}$. So the bottom row of $A_{k+1}$ consists of ones only and the induction is complete.\n\nb.\nThere are many ways to produce an infinite sequence of those $n$ for which the state $[0,0, \\ldots, 0]$ will never be achieved. As an example, consider $n=2^{k}+1$ (for $k \\geq 1$). The evolution of the system can be represented by a matrix $\\mathcal{A}$ of width $2^{k}+1$ with infinitely many rows. The top $2^{k}$ rows form the matrix $A_{k}$ discussed above, with one column of zeros attached at its right.\n\nIn the next row we then have the vector $[0,0, \\ldots, 0,1,1]$. But this is just the second row of $\\mathcal{A}$ reversed. Subsequent rows will be mirror copies of the foregoing ones, starting from the second one. So the configuration $[1,1,0, \\ldots, 0,0]$, i.e. the second row of $\\mathcal{A}$, will reappear. Further rows will periodically repeat this pattern and there will be no row of zeros.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31832,
    "subject": "Mathematics (Multi-modal)",
    "question": "A diagonal of a regular 2006-gon is called odd if its endpoints divide the boundary into two parts, each composed of an odd number of sides. Sides are also regarded as odd diagonals.\nSuppose the 2006-gon has been dissected into triangles by 2003 nonintersecting diagonals. Find the maximum possible number of isosceles triangles with two odd sides.\n(Serbia)",
    "options": [],
    "answer": "1003",
    "solution": "Call an isosceles triangle odd if it has two odd sides. Suppose we are given a dissection as in the problem statement. A triangle in the dissection which is odd and isosceles will be called iso-odd for brevity.\n\nLemma. Let $AB$ be one of dissecting diagonals and let $\\mathcal{L}$ be the shorter part of the boundary of the 2006-gon with endpoints $A, B$. Suppose that $\\mathcal{L}$ consists of $n$ segments. Then the number of iso-odd triangles with vertices on $\\mathcal{L}$ does not exceed $n / 2$.\n\nProof. This is obvious for $n=2$. Take $n$ with $2 < n \\leq 1003$ and assume the claim to be true for every $\\mathcal{L}$ of length less than $n$. Let now $\\mathcal{L}$ (endpoints $A, B$) consist of $n$ segments. Let $PQ$ be the longest diagonal which is a side of an iso-odd triangle $PQS$ with all vertices on $\\mathcal{L}$ (if there is no such triangle, there is nothing to prove). Every triangle whose vertices lie on $\\mathcal{L}$ is obtuse or right-angled; thus $S$ is the summit of $PQS$. We may assume that the five points $A, P, S, Q, B$ lie on $\\mathcal{L}$ in this order and partition $\\mathcal{L}$ into four pieces $\\mathcal{L}_{AP}, \\mathcal{L}_{PS}, \\mathcal{L}_{SQ}, \\mathcal{L}_{QB}$ (the outer ones possibly reducing to a point).\n\nBy the definition of $PQ$, an iso-odd triangle cannot have vertices on both $\\mathcal{L}_{AP}$ and $\\mathcal{L}_{QB}$. Therefore every iso-odd triangle within $\\mathcal{L}$ has all its vertices on just one of the four pieces. Applying to each of these pieces the induction hypothesis and adding the four inequalities we get that the number of iso-odd triangles within $\\mathcal{L}$ other than $PQS$ does not exceed $n / 2$. And since each of $\\mathcal{L}_{PS}, \\mathcal{L}_{SQ}$ consists of an odd number of sides, the inequalities for these two pieces are actually strict, leaving a $1/2 + 1/2$ in excess. Hence the triangle $PSQ$ is also covered by the estimate $n / 2$. This concludes the induction step and proves the lemma.\n\nThe remaining part of the solution in fact repeats the argument from the above proof. Consider the longest dissecting diagonal $XY$. Let $\\mathcal{L}_{XY}$ be the shorter of the two parts of the boundary with endpoints $X, Y$ and let $XYZ$ be the triangle in the dissection with vertex $Z$ not on $\\mathcal{L}_{XY}$. Notice that $XYZ$ is acute or right-angled, otherwise one of the segments $XZ, YZ$ would be longer than $XY$. Denoting by $\\mathcal{L}_{XZ}, \\mathcal{L}_{YZ}$ the two pieces defined by $Z$ and applying the lemma to each of $\\mathcal{L}_{XY}, \\mathcal{L}_{XZ}, \\mathcal{L}_{YZ}$ we infer that there are no more than $2006/2$ iso-odd triangles in all, unless $XYZ$ is one of them. But in that case $XZ$ and $YZ$ are odd diagonals and the corresponding inequalities are strict. This shows that also in this case the total number of iso-odd triangles in the dissection, including $XYZ$, is not greater than $1003$.\n\nThis bound can be achieved. For this to happen, it just suffices to select a vertex of the 2006-gon and draw a broken line joining every second vertex, starting from the selected one. Since 2006 is even, the line closes. This already gives us the required $1003$ iso-odd triangles. Then we can complete the triangulation in an arbitrary fashion.\nLet the terms odd triangle and iso-odd triangle have the same meaning as in the first solution.\n\nLet $ABC$ be an iso-odd triangle, with $AB$ and $BC$ odd sides. This means that there are an odd number of sides of the 2006-gon between $A$ and $B$ and also between $B$ and $C$. We say that these sides belong to the iso-odd triangle $ABC$.\n\nAt least one side in each of these groups does not belong to any other iso-odd triangle. This is so because any odd triangle whose vertices are among the points between $A$ and $B$ has two sides of equal length and therefore has an even number of sides belonging to it in total. Eliminating all sides belonging to any other iso-odd triangle in this area must therefore leave one side that belongs to no other iso-odd triangle. Let us assign these two sides (one in each group) to the triangle $ABC$.\n\nTo each iso-odd triangle we have thus assigned a pair of sides, with no two triangles sharing an assigned side. It follows that at most $1003$ iso-odd triangles can appear in the dissection.\n\nThis value can be attained, as shows the example from the first solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31833,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $P$ whose vertices are in $S$, let $a(P)$ be the number of vertices of $P$, and let $b(P)$ be the number of points of $S$ which are outside $P$. Prove that for every real number $x$\n$$\n\\sum_{P} x^{a(P)}(1-x)^{b(P)}=1,\n$$\nwhere the sum is taken over all convex polygons with vertices in $S$.\n\nNB. A line segment, a point and the empty set are considered as convex polygons of 2, 1 and 0 vertices, respectively.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each convex polygon $P$ whose vertices are in $S$, let $c(P)$ be the number of points of $S$ which are inside $P$, so that $a(P)+b(P)+c(P)=n$, the total number of points in $S$. Denoting $1-x$ by $y$,\n$$\n\\sum_{P} x^{a(P)} y^{b(P)}=\\sum_{P} x^{a(P)} y^{b(P)}(x+y)^{c(P)}=\\sum_{P} \\sum_{i=0}^{c(P)}\\binom{c(P)}{i} x^{a(P)+i} y^{b(P)+c(P)-i} .\n$$\nView this expression as a homogeneous polynomial of degree $n$ in two independent variables $x, y$. In the expanded form, it is the sum of terms $x^{r} y^{n-r}$ ($0 \\leq r \\leq n$) multiplied by some nonnegative integer coefficients.\nFor a fixed $r$, the coefficient of $x^{r} y^{n-r}$ represents the number of ways of choosing a convex polygon $P$ and then choosing some of the points of $S$ inside $P$ so that the number of vertices of $P$ and the number of chosen points inside $P$ jointly add up to $r$.\nThis corresponds to just choosing an $r$-element subset of $S$. The correspondence is bijective because every set $T$ of points from $S$ splits in exactly one way into the union of two disjoint subsets, of which the first is the set of vertices of a convex polygon - namely, the convex hull of $T$ - and the second consists of some points inside that polygon.\nSo the coefficient of $x^{r} y^{n-r}$ equals $\\binom{n}{r}$. The desired result follows:\n$$\n\\sum_{P} x^{a(P)} y^{b(P)}=\\sum_{r=0}^{n}\\binom{n}{r} x^{r} y^{n-r}=(x+y)^{n}=1 .\n$$\nApply induction on the number $n$ of points. The case $n=0$ is trivial. Let $n>0$ and assume the statement for less than $n$ points. Take a set $S$ of $n$ points.\nLet $C$ be the set of vertices of the convex hull of $S$, let $m=|C|$.\nLet $X \\subset C$ be an arbitrary nonempty set. For any convex polygon $P$ with vertices in the set $S \\backslash X$, we have $b(P)$ points of $S$ outside $P$. Excluding the points of $X$ - all outside $P$ - the set $S \\backslash X$ contains exactly $b(P)-|X|$ of them. Writing $1-x=y$, by the induction hypothesis\n$$\n\\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)-|X|}=1\n$$\n(where $P \\subset S \\backslash X$ means that the vertices of $P$ belong to the set $S \\backslash X$ ). Therefore\n$$\n\\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)}=y^{|X|} .\n$$\nAll convex polygons appear at least once, except the convex hull $C$ itself. The convex hull adds $x^{m}$. We can use the inclusion-exclusion principle to compute the sum of the other terms:\n$$\n\\begin{gathered}\n\\sum_{P \\neq C} x^{a(P)} y^{b(P)}=\\sum_{k=1}^{m}(-1)^{k-1} \\sum_{|X|=k} \\sum_{P \\subset S \\backslash X} x^{a(P)} y^{b(P)}=\\sum_{k=1}^{m}(-1)^{k-1} \\sum_{|X|=k} y^{k} \\\\\n=\\sum_{k=1}^{m}(-1)^{k-1}\\binom{m}{k} y^{k}=-\\left((1-y)^{m}-1\\right)=1-x^{m}\n\\end{gathered}\n$$\nand then\n$$\n\\sum_{P} x^{a(P)} y^{b(P)}=\\sum_{P=C}+\\sum_{P \\neq C}=x^{m}+\\left(1-x^{m}\\right)=1 .\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31834,
    "subject": "Mathematics (Multi-modal)",
    "question": "A cake has the form of an $n \\times n$ square composed of $n^2$ unit squares. Strawberries lie on some of the unit squares so that each row or column contains exactly one strawberry; call this arrangement $\\mathcal{A}$.\nLet $\\mathcal{B}$ be another such arrangement. Suppose that every grid rectangle with one vertex at the top left corner of the cake contains no fewer strawberries of arrangement $\\mathcal{B}$ than of arrangement $\\mathcal{A}$. Prove that arrangement $\\mathcal{B}$ can be obtained from $\\mathcal{A}$ by performing a number of switches, defined as follows:\nA switch consists in selecting a grid rectangle with only two strawberries, situated at its top right corner and bottom left corner, and moving these two strawberries to the other two corners of that rectangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use capital letters to denote unit squares; $O$ is the top left corner square. For any two squares $X$ and $Y$ let $[X Y]$ be the smallest grid rectangle containing these two squares. Strawberries lie on some squares in arrangement $\\mathcal{A}$. Put a plum on each square of the target configuration $\\mathcal{B}$. For a square $X$ denote by $a(X)$ and $b(X)$ respectively the number of strawberries and the number of plums in $[O X]$. By hypothesis $a(X) \\leq b(X)$ for each $X$, with strict inequality for some $X$ (otherwise the two arrangements coincide and there is nothing to prove).\nThe idea is to show that by a legitimate switch one can obtain an arrangement $\\mathcal{A}'$ such that\n$$\n\\begin{equation*}\na(X) \\leq a'(X) \\leq b(X) \\quad \\text{ for each } X ; \\quad \\sum_{X} a(X)<\\sum_{X} a'(X) \\tag{1}\n\\end{equation*}\n$$\n(with $a'(X)$ defined analogously to $a(X)$; the sums range over all unit squares $X$ ). This will be enough because the same reasoning then applies to $\\mathcal{A}'$, giving rise to a new arrangement $\\mathcal{A}''$, and so on (induction). Since $\\sum a(X)<\\sum a'(X)<\\sum a''(X)<\\ldots$ and all these sums do not exceed $\\sum b(X)$, we eventually obtain a sum with all summands equal to the respective $b(X)$s; all strawberries will meet with plums.\nConsider the uppermost row in which the plum and the strawberry lie on different squares $P$ and $S$ (respectively); clearly $P$ must be situated left to $S$. In the column passing through $P$, let $T$ be the top square and $B$ the bottom square. The strawberry in that column lies below the plum (because there is no plum in that column above $P$, and the positions of strawberries and plums coincide everywhere above the row of $P$ ). Hence there is at least one strawberry in the region $[B S]$ below $[P S]$. Let $V$ be the position of the uppermost strawberry in that region.\n\n| $O$ |  | $T$ |  |  |  |  | I |  |  |  |  |  |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n|  |  |  |  |  |  |  | I |  |  |  |  |  |\n|  |  | $P$ |  |  | $U$ |  | 1 |  |  | $S$ |  |  |\n|  |  |  |  |  |  |  | 1 | 1 |  |  |  |  |\n|  |  |  |  |  |  |  |  |  |  |  |  |  |\n|  |  |  |  |  |  |  | 1 |  |  |  |  |  |\n| - | - | - | - | - | - | - - | $\\underline{X}'$ |  |  |  |  |  |\n|  |  |  |  |  |  |  |  |  | $R$ |  |  |  |\n|  |  |  |  |  | $V$ |  |  |  |  | $W$ |  |  |\n|  |  |  |  |  |  |  |  |  |  |  |  |  |\n|  |  |  |  |  |  |  |  |  |  |  |  |  |\n|  |  | $B$ |  |  |  |  |  |  |  |  |  |  |\n\nDenote by $W$ the square at the intersection of the row through $V$ with the column through $S$ and let $R$ be the square vertex-adjacent to $W$ up-left. We claim that\n$$\n\\begin{equation*}\na(X)<b(X) \\quad \\text{ for all } \\quad X \\in[P R] . \\tag{2}\n\\end{equation*}\n$$\nThis is so because if $X \\in[P R]$ then the portion of $[O X]$ left to column $[T B]$ contains at least as many plums as strawberries (the hypothesis of the problem); in the portion above the row through $P$ and $S$ we have perfect balance; and in the remaining portion, i.e. rectangle $[P X]$ we have a plum on square $P$ and no strawberry at all.\nNow we are able to perform the required switch. Let $U$ be the square at the intersection of the row through $P$ with the column through $V$ (some of $P, U, R$ can coincide). We move strawberries from squares $S$ and $V$ to squares $U$ and $W$. Then\n$$\na'(X)=a(X)+1 \\quad \\text{ for } \\quad X \\in[U R] ; \\quad a'(X)=a(X) \\quad \\text{ for other } X .\n$$\nAnd since the rectangle $[U R]$ is contained in $[P R]$, we still have $a'(X) \\leq b(X)$ for all $S$, in view of (2); conditions (1) are satisfied and the proof is complete.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31835,
    "subject": "Mathematics (Multi-modal)",
    "question": "An $(n, k)$-tournament is a contest with $n$ players held in $k$ rounds such that:\n(i) Each player plays in each round, and every two players meet at most once.\n(ii) If player $A$ meets player $B$ in round $i$, player $C$ meets player $D$ in round $i$, and player $A$ meets player $C$ in round $j$, then player $B$ meets player $D$ in round $j$.\nDetermine all pairs $(n, k)$ for which there exists an $(n, k)$-tournament.",
    "options": [],
    "answer": "An (n, k)-tournament exists if and only if 2^{t_k} divides n, where t_k is the least integer with 2^{t_k} ≥ k+1 (equivalently, 2^{t_k−1} < k+1 ≤ 2^{t_k}).",
    "solution": "For each $k$, denote by $t_k$ the unique integer such that $2^{t_k-1} < k+1 \\leq 2^{t_k}$. We show that an $(n, k)$-tournament exists if and only if $2^{t_k}$ divides $n$.\n\nFirst we prove that if $n = 2^t$ for some $t$ then there is an $(n, k)$-tournament for all $k \\leq 2^t - 1$. Let $S$ be the set of $0$-$1$ sequences with length $t$. We label the $2^t$ players with the elements of $S$ in an arbitrary fashion (which is possible as there are exactly $2^t$ sequences in $S$). Players are identified with their labels in the construction below. If $\\alpha, \\beta \\in S$, let $\\alpha + \\beta \\in S$ be the result of the modulo $2$ term-by-term addition of $\\alpha$ and $\\beta$ (with rules $0+0=0, 0+1=1+0=1, 1+1=0$; there is no carryover). For each $i = 1, \\ldots, 2^t - 1$ let $\\omega(i) \\in S$ be the sequence of base $2$ digits of $i$, completed with leading zeros if necessary to achieve length $t$.\n\nNow define a tournament with $n = 2^t$ players in $k \\leq 2^t - 1$ rounds as follows: For all $i = 1, \\ldots, k$, let player $\\alpha$ meet player $\\alpha + \\omega(i)$ in round $i$. The tournament is well-defined as $\\alpha + \\omega(i) \\in S$ and $\\alpha + \\omega(i) = \\beta + \\omega(i)$ implies $\\alpha = \\beta$; also $[\\alpha + \\omega(i)] + \\omega(i) = \\alpha$ for each $\\alpha \\in S$ (meaning that player $\\alpha + \\omega(i)$ meets player $\\alpha$ in round $i$, as needed). Each player plays in each round. Next, every two players meet at most once (exactly once if $k = 2^t - 1$), since $\\omega(i) \\neq \\omega(j)$ if $i \\neq j$. Thus condition (i) holds true, and condition (ii) is also easy to check.\n\nLet player $\\alpha$ meet player $\\beta$ in round $i$, player $\\gamma$ meet player $\\delta$ in round $i$, and player $\\alpha$ meet player $\\gamma$ in round $j$. Then $\\beta = \\alpha + \\omega(i)$, $\\delta = \\gamma + \\omega(i)$ and $\\gamma = \\alpha + \\omega(j)$. By definition, $\\beta$ will play in round $j$ with\n$$\n\\beta + \\omega(j) = [\\alpha + \\omega(i)] + \\omega(j) = [\\alpha + \\omega(j)] + \\omega(i) = \\gamma + \\omega(i) = \\delta,\n$$\nas required by (ii).\n\nSo there exists an $(n, k)$-tournament for pairs $(n, k)$ such that $n = 2^t$ and $k \\leq 2^t - 1$. The same conclusion is straightforward for $n$ of the form $n = 2^t s$ and $k \\leq 2^t - 1$. Indeed, consider $s$ different $(2^t, k)$-tournaments $T_1, \\ldots, T_s$, no two of them having players in common. Their union can be regarded as a $(2^t s, k)$-tournament $T$ where each round is the union of the respective rounds in $T_1, \\ldots, T_s$.\n\nIn summary, the condition that $2^{t_k}$ divides $n$ is sufficient for an $(n, k)$-tournament to exist. We prove that it is also necessary.\n\nConsider an arbitrary $(n, k)$-tournament. Represent each player by a point and after each round, join by an edge every two players who played in this round. Thus to a round $i = 1, \\ldots, k$ there corresponds a graph $G_i$. We say that player $Q$ is an $i$-neighbour of player $P$ if there is a path of edges in $G_i$ from $P$ to $Q$; in other words, if there are players $P = X_1, X_2, \\ldots, X_m = Q$ such that player $X_j$ meets player $X_{j+1}$ in one of the first $i$ rounds, $j = 1, 2, \\ldots, m-1$. The set of $i$-neighbours of a player will be called its $i$-component. Clearly two $i$-components are either disjoint or coincide.\n\nHence after each round $i$ the set of players is partitioned into pairwise disjoint $i$-components. So, to achieve our goal, it suffices to show that all $k$-components have size divisible by $2^{t_k}$.\n\nTo this end, let us see how the $i$-component $\\Gamma$ of a player $A$ changes after round $i+1$. Suppose that $A$ meets player $B$ with $i$-component $\\Delta$ in round $i+1$ (components $\\Gamma$ and $\\Delta$ are not necessarily distinct). We claim that then in round $i+1$ each player from $\\Gamma$ meets a player from $\\Delta$, and vice versa.\n\nIndeed, let $C$ be any player in $\\Gamma$, and let $C$ meet $D$ in round $i+1$. Since $C$ is an $i$-neighbour of $A$, there is a sequence of players $A = X_1, X_2, \\ldots, X_m = C$ such that $X_j$ meets $X_{j+1}$ in one of the first $i$ rounds, $j = 1, 2, \\ldots, m-1$. Let $X_j$ meet $Y_j$ in round $i+1$, for $j = 1, 2, \\ldots, m$; in particular $Y_1 = B$ and $Y_m = D$. Players $Y_j$ exist in view of condition (i). Suppose that $X_j$ and $X_{j+1}$ met in round $r$, where $r \\leq i$. Then condition (ii) implies that $Y_j$ and $Y_{j+1}$ met in round $r$, too. Hence $B = Y_1, Y_2, \\ldots, Y_m = D$ is a path in $G_i$ from $B$ to $D$. This is to say, $D$ is in the $i$-component $\\Delta$ of $B$, as claimed. By symmetry, each player from $\\Delta$ meets a player from $\\Gamma$ in round $i+1$. It follows in particular that $\\Gamma$ and $\\Delta$ have the same cardinality.\n\nIt is straightforward now that the $(i+1)$-component of $A$ is $\\Gamma \\cup \\Delta$, the union of two sets with the same size. Since $\\Gamma$ and $\\Delta$ are either disjoint or coincide, we have either $|\\Gamma \\cup \\Delta| = 2|\\Gamma|$ or $|\\Gamma \\cup \\Delta| = |\\Gamma|$; as usual, $|\\cdots|$ denotes the cardinality of a finite set.\n\nLet $\\Gamma_1, \\ldots, \\Gamma_k$ be the consecutive components of a given player $A$. We obtained that either $|\\Gamma_{i+1}| = 2|\\Gamma_i|$ or $|\\Gamma_{i+1}| = |\\Gamma_i|$ for $i = 1, \\ldots, k-1$. Because $|\\Gamma_1| = 2$, each $|\\Gamma_i|$ is a power of $2$, $i = 1, \\ldots, k-1$. In particular $|\\Gamma_k| = 2^u$ for some $u$.\n\nOn the other hand, player $A$ has played with $k$ different opponents by (i). All of them belong to $\\Gamma_k$, therefore $|\\Gamma_k| \\geq k+1$.\n\nThus $2^u \\geq k+1$, and since $t_k$ is the least integer satisfying $2^{t_k} \\geq k+1$, we conclude that $u \\geq t_k$. So the size of each $k$-component is divisible by $2^{t_k}$, which completes the argument.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31836,
    "subject": "Mathematics (Multi-modal)",
    "question": "A holey triangle is an upward equilateral triangle of side length $n$ with $n$ upward unit triangular holes cut out. A diamond is a $60^{\\circ}$-$120^{\\circ}$ unit rhombus. Prove that a holey triangle $T$ can be tiled with diamonds if and only if the following condition holds: Every upward equilateral triangle of side length $k$ in $T$ contains at most $k$ holes, for $1 \\leq k \\leq n$.\n(Colombia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be a holey triangle. The unit triangles in it will be called cells. We say simply \"triangle\" instead of \"upward equilateral triangle\" and \"size\" instead of \"side length.\"\n\nThe necessity will be proven first. Assume that a holey triangle $T$ can be tiled with diamonds and consider such a tiling. Let $T'$ be a triangle of size $k$ in $T$ containing $h$ holes. Focus on the diamonds which cover (one or two) cells in $T'$. Let them form a figure $R$. The boundary of $T'$ consists of upward cells, so $R$ is a triangle of size $k$ with $h$ upward holes cut out and possibly some downward cells sticking out. Hence there are exactly $\\left(k^{2}+k\\right) / 2-h$ upward cells in $R$, and at least $\\left(k^{2}-k\\right) / 2$ downward cells (not counting those sticking out). On the other hand each diamond covers one upward and one downward cell, which implies $\\left(k^{2}+k\\right) / 2-h \\geq\\left(k^{2}-k\\right) / 2$. It follows that $h \\leq k$, as needed.\n\nWe pass on to the sufficiency. For brevity, let us say that a set of holes in a given triangle $T$ is spread out if every triangle of size $k$ in $T$ contains at most $k$ holes. For any set $S$ of spread out holes, a triangle of size $k$ will be called full of $S$ if it contains exactly $k$ holes of $S$. The proof is based on the following observation.\n\nLemma. Let $S$ be a set of spread out holes in $T$. Suppose that two triangles $T'$ and $T''$ are full of $S$, and that they touch or intersect. Let $T'+T''$ denote the smallest triangle in $T$ containing them. Then $T'+T''$ is also full of $S$.\n\nProof. Let triangles $T', T'', T' \\cap T''$ and $T'+T''$ have sizes $a, b, c$ and $d$, and let them contain $a, b, x$ and $y$ holes of $S$, respectively. (Note that $T' \\cap T''$ could be a point, in which case $c=0$.) Since $S$ is spread out, we have $x \\leq c$ and $y \\leq d$. The geometric configuration of triangles clearly satisfies $a+b=c+d$. Furthermore, $a+b \\leq x+y$, since $a+b$ counts twice the holes in $T' \\cap T''$. These conclusions imply $x=c$ and $y=d$, as we wished to show.\n\nNow let $T_n$ be a holey triangle of size $n$, and let the set $H$ of its holes be spread out. We show by induction on $n$ that $T_n$ can be tiled with diamonds. The base $n=1$ is trivial. Suppose that $n \\geq 2$ and that the claim holds for holey triangles of size less than $n$.\n\nDenote by $B$ the bottom row of $T_n$ and by $T'$ the triangle formed by its top $n-1$ rows. There is at least one hole in $B$ as $T'$ contains at most $n-1$ holes. If this hole is only one, there is a unique way to tile $B$ with diamonds. Also, $T'$ contains exactly $n-1$ holes, making it a holey triangle of size $n-1$, and these holes are spread out. Hence it remains to apply the induction hypothesis.\n\nSo suppose that there are $m \\geq 2$ holes in $B$ and label them $a_1, \\ldots, a_m$ from left to right. Let $\\ell$ be the line separating $B$ from $T'$. For each $i=1, \\ldots, m-1$, pick an upward cell $b_i$ between $a_i$ and $a_{i+1}$, with base on $\\ell$. Place a diamond to cover $b_i$ and its lower neighbour, a downward cell in $B$. The remaining part of $B$ can be tiled uniquely with diamonds. Remove from $T_n$ row $B$ and the cells $b_1, \\ldots, b_{m-1}$ to obtain a holey triangle $T_{n-1}$ of size $n-1$. The conclusion will follow by induction if the choice of $b_1, \\ldots, b_{m-1}$ guarantees that the following condition is satisfied: If the holes $a_1, \\ldots, a_{m-1}$ are replaced by $b_1, \\ldots, b_{m-1}$ then the new set of holes is spread out again.\n\nWe show that such a choice is possible. The cells $b_1, \\ldots, b_{m-1}$ can be defined one at a time in this order, making sure that the above condition holds at each step. Thus it suffices to prove that there is an appropriate choice for $b_1$, and we set $a_1=u, a_2=v$ for clarity.\n\nLet $\\Delta$ be the triangle of maximum size which is full of $H$, contains the top vertex of the hole $u$, and has base on line $\\ell$. Call $\\Delta$ the associate of $u$. Observe that $\\Delta$ does not touch $v$. Indeed, if $\\Delta$ has size $r$ then it contains $r$ holes of $T_n$. Extending its slanted sides downwards produces a triangle $\\Delta'$ of size $r+1$ containing at least one more hole, namely $u$. Since there are at most $r+1$ holes in $\\Delta'$, it cannot contain $v$. Consequently, $\\Delta$ does not contain the top vertex of $v$.\n\nLet $w$ be the upward cell with base on $\\ell$ which is to the right of $\\Delta$ and shares a common vertex with it. The observation above shows that $w$ is to the left of $v$. Note that $w$ is not a hole, or else $\\Delta$ could be extended to a larger triangle full of $H$.\n\nWe prove that if the hole $u$ is replaced by $w$ then the new set of holes is spread out again. To verify this, we only need to check that if a triangle $\\Gamma$ in $T_n$ contains $w$ but not $u$ then $\\Gamma$ is not full of $H$. Suppose to the contrary that $\\Gamma$ is full of $H$. Consider the minimum triangle $\\Gamma+\\Delta$ containing $\\Gamma$ and the associate $\\Delta$ of $u$. Clearly $\\Gamma+\\Delta$ is larger than $\\Delta$, because $\\Gamma$ contains $w$ but $\\Delta$ does not. Next, $\\Gamma+\\Delta$ is full of $H \\backslash\\{u\\}$ by the lemma, since $\\Gamma$ and $\\Delta$ have a common point and neither of them contains $u$.\n\n![](attached_image_1.png)\n\nIf $\\Gamma$ is above line $\\ell$ then so is $\\Gamma+\\Delta$, which contradicts the maximum choice of $\\Delta$. If $\\Gamma$ contains cells from row $B$, observe that $\\Gamma+\\Delta$ contains $u$. Let $s$ be the size of $\\Gamma+\\Delta$. Being full of $H \\backslash\\{u\\}, \\Gamma+\\Delta$ contains $s$ holes other than $u$. But it also contains $u$, contradicting the assumption that $H$ is spread out.\n\nThe claim follows, showing that $b_1=w$ is an appropriate choice for $a_1=u$ and $a_2=v$. As explained above, this is enough to complete the induction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31837,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a convex polyhedron without parallel edges and without an edge parallel to any face other than the two faces adjacent to it.\nCall a pair of points of the polyhedron antipodal if there exist two parallel planes passing through these points and such that the polyhedron is contained between these planes.\nLet $A$ be the number of antipodal pairs of vertices, and let $B$ be the number of antipodal pairs of midpoints of edges. Determine the difference $A-B$ in terms of the numbers of vertices, edges and faces.",
    "options": [],
    "answer": "A - B = m - ℓ + 1 = n - 1",
    "solution": "Denote the polyhedron by $\\Gamma$; let its vertices, edges and faces be $V_{1}, V_{2}, \\ldots, V_{n}$, $E_{1}, E_{2}, \\ldots, E_{m}$ and $F_{1}, F_{2}, \\ldots, F_{\\ell}$, respectively. Denote by $Q_{i}$ the midpoint of edge $E_{i}$.\nLet $S$ be the unit sphere, the set of all unit vectors in three-dimensional space. Map the boundary elements of $\\Gamma$ to some objects on $S$ as follows.\nFor a face $F_{i}$, let $S^{+}\\left(F_{i}\\right)$ and $S^{-}\\left(F_{i}\\right)$ be the unit normal vectors of face $F_{i}$, pointing outwards from $\\Gamma$ and inwards to $\\Gamma$, respectively. These points are diametrically opposite.\nFor an edge $E_{j}$, with neighbouring faces $F_{i_{1}}$ and $F_{i_{2}}$, take all support planes of $\\Gamma$ (planes which have a common point with $\\Gamma$ but do not intersect it) containing edge $E_{j}$, and let $S^{+}\\left(E_{j}\\right)$ be the set of their outward normal vectors. The set $S^{+}\\left(E_{j}\\right)$ is an arc of a great circle on $S$. $\\operatorname{Arc} S^{+}\\left(E_{j}\\right)$ is perpendicular to edge $E_{j}$ and it connects points $S^{+}\\left(F_{i_{1}}\\right)$ and $S^{+}\\left(F_{i_{2}}\\right)$.\nDefine also the set of inward normal vectors $S^{-}\\left(E_{i}\\right)$ which is the reflection of $S^{+}\\left(E_{i}\\right)$ across the origin.\nFor a vertex $V_{k}$, which is the common endpoint of edges $E_{j_{1}}, \\ldots, E_{j_{h}}$ and shared by faces $F_{i_{1}}, \\ldots, F_{i_{h}}$, take all support planes of $\\Gamma$ through point $V_{k}$ and let $S^{+}\\left(V_{k}\\right)$ be the set of their outward normal vectors. This is a region on $S$, a spherical polygon with vertices $S^{+}\\left(F_{i_{1}}\\right), \\ldots, S^{+}\\left(F_{i_{h}}\\right)$ bounded by arcs $S^{+}\\left(E_{j_{1}}\\right), \\ldots, S^{+}\\left(E_{j_{h}}\\right)$. Let $S^{-}\\left(V_{k}\\right)$ be the reflection of $S^{+}\\left(V_{k}\\right)$, the set of inward normal vectors.\nNote that region $S^{+}\\left(V_{k}\\right)$ is convex in the sense that it is the intersection of several half spheres.\n![](attached_image_1.png)\n![](attached_image_2.png)\nNow translate the conditions on $\\Gamma$ to the language of these objects.\n\na. Polyhedron $\\Gamma$ has no parallel edges - the great circles of $\\operatorname{arcs} S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{j}\\right)$ are different for all $i \\neq j$.\n\nb. If an edge $E_{i}$ does not belong to a face $F_{j}$ then they are not parallel - the great circle which contains arcs $S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{i}\\right)$ does not pass through points $S^{+}\\left(F_{j}\\right)$ and $S^{-}\\left(F_{j}\\right)$.\n\nc. Polyhedron $\\Gamma$ has no parallel faces - points $S^{+}\\left(F_{i}\\right)$ and $S^{-}\\left(F_{j}\\right)$ are pairwise distinct.\n\nThe regions $S^{+}\\left(V_{k}\\right)$, arcs $S^{+}\\left(E_{j}\\right)$ and points $S^{+}\\left(F_{i}\\right)$ provide a decomposition of the surface of the sphere. Regions $S^{-}\\left(V_{k}\\right)$, arcs $S^{-}\\left(E_{j}\\right)$ and points $S^{-}\\left(F_{i}\\right)$ provide the reflection of this decomposition. These decompositions are closely related to the problem.\n\nLemma 1. For any $1 \\leq i, j \\leq n$, regions $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right)$ overlap if and only if vertices $V_{i}$ and $V_{j}$ are antipodal.\n\nLemma 2. For any $1 \\leq i, j \\leq m, \\operatorname{arcs} S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect if and only if the midpoints $Q_{i}$ and $Q_{j}$ of edges $E_{i}$ and $E_{j}$ are antipodal.\n\nProof of lemma 1. First note that by properties (a,b,c) above, the two regions cannot share only a single point or an arc. They are either disjoint or they overlap.\nAssume that the two regions have a common interior point $u$. Let $P_{1}$ and $P_{2}$ be two parallel support planes of $\\Gamma$ through points $V_{i}$ and $V_{j}$, respectively, with normal vector $u$. By the definition of regions $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right), u$ is the inward normal vector of $P_{1}$ and the outward normal vector of $P_{2}$. Therefore polyhedron $\\Gamma$ lies between the two planes; vertices $V_{i}$ and $V_{j}$ are antipodal.\nTo prove the opposite direction, assume that $V_{i}$ and $V_{j}$ are antipodal. Then there exist two parallel support planes $P_{1}$ and $P_{2}$ through $V_{i}$ and $V_{j}$, respectively, such that $\\Gamma$ is between them. Let $u$ be the inward normal vector of $P_{1}$; then $u$ is the outward normal vector of $P_{2}$, therefore $u \\in S^{-}\\left(V_{i}\\right) \\cap S^{+}\\left(V_{j}\\right)$. The two regions have a common point, so they overlap.\n\nProof of lemma 2. Again, by properties (a, b) above, the endpoints of $\\operatorname{arc} S^{-}\\left(E_{i}\\right)$ cannot belong to $S^{+}\\left(E_{j}\\right)$ and vice versa. The two arcs are either disjoint or intersecting.\nAssume that arcs $S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect at point $u$. Let $P_{1}$ and $P_{2}$ be the two support planes through edges $E_{i}$ and $E_{j}$, respectively, with normal vector $u$. By the definition of $\\operatorname{arcs} S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$, vector $u$ points inwards from $P_{1}$ and outwards from $P_{2}$. Therefore $\\Gamma$ is between the planes. Since planes $P_{1}$ and $P_{2}$ pass through $Q_{i}$ and $Q_{j}$, these points are antipodal.\nFor the opposite direction, assume that points $Q_{i}$ and $Q_{j}$ are antipodal. Let $P_{1}$ and $P_{2}$ be two support planes through these points, respectively. An edge cannot intersect a support plane, therefore $E_{i}$ and $E_{j}$ lie in the planes $P_{1}$ and $P_{2}$, respectively. Let $u$ be the inward normal vector of $P_{1}$, which is also the outward normal vector of $P_{2}$. Then $u \\in S^{-}\\left(E_{i}\\right) \\cap S^{+}\\left(E_{j}\\right)$. So the two arcs are not disjoint; they therefore intersect.\n\nNow create a new decomposition of sphere $S$. Draw all $\\operatorname{arcs} S^{+}\\left(E_{i}\\right)$ and $S^{-}\\left(E_{j}\\right)$ on sphere $S$ and put a knot at each point where two arcs meet. We have $\\ell$ knots at points $S^{+}\\left(F_{i}\\right)$ and another $\\ell$ knots at points $S^{-}\\left(F_{i}\\right)$, corresponding to the faces of $\\Gamma$; by property (c) they are different. We also have some pairs $1 \\leq i, j \\leq m$ where arcs $S^{-}\\left(E_{i}\\right)$ and $S^{+}\\left(E_{j}\\right)$ intersect. By Lemma 2, each antipodal pair ( $Q_{i}, Q_{j}$ ) gives rise to two such intersections; hence, the number of all intersections is $2 B$ and we have $2 \\ell+2 B$ knots in all.\nEach intersection knot splits two arcs, increasing the number of arcs by 2. Since we started with $2 m$ arcs, corresponding the edges of $\\Gamma$, the number of the resulting curve segments is $2 m+4 B$.\nThe network of these curve segments divides the sphere into some \"new\" regions. Each new region is the intersection of some overlapping sets $S^{-}\\left(V_{i}\\right)$ and $S^{+}\\left(V_{j}\\right)$. Due to the convexity, the intersection of two overlapping regions is convex and thus contiguous. By Lemma 1, each pair of overlapping regions corresponds to an antipodal vertex pair and each antipodal vertex pair gives rise to two different overlaps, which are symmetric with respect to the origin. So the number of new regions is $2 A$.\nThe result now follows from Euler's polyhedron theorem. We have $n+\\ell=m+2$ and\n$$\n(2 \\ell+2 B)+2 A=(2 m+4 B)+2,\n$$\ntherefore\n$$\nA-B=m-\\ell+1=n-1 .\n$$\nTherefore $A-B$ is by one less than the number of vertices of $\\Gamma$.\nUse the same notations for the polyhedron and its vertices, edges and faces as in Solution 1. We regard points as vectors starting from the origin. Polyhedron $\\Gamma$ is regarded as a closed convex set, including its interior. In some cases the edges and faces of $\\Gamma$ are also regarded as sets of points. The symbol $\\partial$ denotes the boundary of the certain set; e.g. $\\partial \\Gamma$ is the surface of $\\Gamma$.\nLet $\\Delta=\\Gamma-\\Gamma=\\{U-V: U, V \\in \\Gamma\\}$ be the set of vectors between arbitrary points of $\\Gamma$. Then $\\Delta$, being the sum of two bounded convex sets, is also a bounded convex set and, by construction, it is also centrally symmetric with respect to the origin. We will prove that $\\Delta$ is also a polyhedron and express the numbers of its faces, edges and vertices in terms $n, m, \\ell, A$ and $B$.\n\nLemma 1. For points $U, V \\in \\Gamma$, point $W=U-V$ is a boundary point of $\\Delta$ if and only if $U$ and $V$ are antipodal. Moreover, for each boundary point $W \\in \\partial \\Delta$ there exists exactly one pair of points $U, V \\in \\Gamma$ such that $W=U-V$.\n\nProof. Assume first that $U$ and $V$ are antipodal points of $\\Gamma$. Let parallel support planes $P_{1}$ and $P_{2}$ pass through them such that $\\Gamma$ is in between. Consider plane $P=P_{1}-U= P_{2}-V$. This plane separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. After reflecting one of the sets, e.g. $\\Gamma-V$, the sets $\\Gamma-U$ and $-\\Gamma+V$ lie in the same half space bounded by $P$. Then $(\\Gamma-U)+(-\\Gamma+V)=\\Delta-W$ lies in that half space, so $0 \\in P$ is a boundary point of the set $\\Delta-W$. Translating by $W$ we obtain that $W$ is a boundary point of $\\Delta$.\nTo prove the opposite direction, let $W=U-V$ be a boundary point of $\\Delta$, and let $\\Psi= (\\Gamma-U) \\cap(\\Gamma-V)$. We claim that $\\Psi=\\{0\\}$. Clearly $\\Psi$ is a bounded convex set and $0 \\in \\Psi$. For any two points $X, Y \\in \\Psi$, we have $U+X, V+Y \\in \\Gamma$ and $W+(X-Y)=(U+X)-(V+Y) \\in \\Delta$. Since $W$ is a boundary point of $\\Delta$, the vector $X-Y$ cannot have the same direction as $W$. This implies that the interior of $\\Psi$ is empty. Now suppose that $\\Psi$ contains a line segment $S$. Then $S+U$ and $S+V$ are subsets of some faces or edges of $\\Gamma$ and these faces/edges are parallel to $S$. In all cases, we find two faces, two edges, or a face and an edge which are parallel, contradicting the conditions of the problem. Therefore, $\\Psi=\\{0\\}$ indeed.\nSince $\\Psi=(\\Gamma-U) \\cap(\\Gamma-V)$ consists of a single point, the interiors of bodies $\\Gamma-U$ and $\\Gamma-V$ are disjoint and there exists a plane $P$ which separates them. Let $u$ be the normal vector of $P$ pointing into that half space bounded by $P$ which contains $\\Gamma-U$. Consider the planes $P+U$ and $P+V$; they are support planes of $\\Gamma$, passing through $U$ and $V$, respectively. From plane $P+U$, the vector $u$ points into that half space which contains $\\Gamma$. From plane $P+V$, vector $u$ points into the opposite half space containing $\\Gamma$. Therefore, we found two proper support through points $U$ and $V$ such that $\\Gamma$ is in between.\nFor the uniqueness part, assume that there exist points $U_{1}, V_{1} \\in \\Gamma$ such that $U_{1}-V_{1}=U-V$. The points $U_{1}-U$ and $V_{1}-V$ lie in the sets $\\Gamma-U$ and $\\Gamma-V$ separated by $P$. Since $U_{1}-U=V_{1}-V$, this can happen only if both are in $P$; but the only such point is 0. Therefore, $U_{1}-V_{1}=U-V$ implies $U_{1}=U$ and $V_{1}=V$. The lemma is complete. $\\square$\n\nLemma 2. Let $U$ and $V$ be two antipodal points and assume that plane $P$, passing through 0, separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. Let $\\Psi_{1}=(\\Gamma-U) \\cap P$ and $\\Psi_{2}=(\\Gamma-V) \\cap P$. Then $\\Delta \\cap(P+U-V)=\\Psi_{1}-\\Psi_{2}+U-V$.\n\nProof. The sets $\\Gamma-U$ and $-\\Gamma+V$ lie in the same closed half space bounded by $P$. Therefore, for any points $X \\in(\\Gamma-U)$ and $Y \\in(-\\Gamma+V)$, we have $X+Y \\in P$ if and only if $X, Y \\in P$. Then\n$(\\Delta-(U-V)) \\cap P=((\\Gamma-U)+(-\\Gamma+V)) \\cap P=((\\Gamma-U) \\cap P)+((-\\Gamma+V) \\cap P)=\\Psi_{1}-\\Psi_{2}$.\nNow a translation by ( $U-V$ ) completes the lemma. $\\square$\n\nNow classify the boundary points $W=U-V$ of $\\Delta$, according to the types of points $U$ and $V$. In all cases we choose a plane $P$ through 0 which separates the interiors of $\\Gamma-U$ and $\\Gamma-V$. We will use the notation $\\Psi_{1}=(\\Gamma-U) \\cap P$ and $\\Psi_{2}=(\\Gamma-V) \\cap P$ as well.\n\nCase 1: Both $U$ and $V$ are vertices of $\\Gamma$. Bodies $\\Gamma-U$ and $\\Gamma-V$ have a common vertex which is 0. Choose plane $P$ in such a way that $\\Psi_{1}=\\Psi_{2}=\\{0\\}$. Then Lemma 2 yields $\\Delta \\cap(P+W)=\\{W\\}$. Therefore $P+W$ is a support plane of $\\Delta$ such that they have only one common point so no line segment exists on $\\partial \\Delta$ which would contain $W$ in its interior.\nSince this case occurs for antipodal vertex pairs and each pair is counted twice, the number of such boundary points on $\\Delta$ is $2A$.\n\nCase 2: Point $U$ is an interior point of an edge $E_{i}$ and $V$ is a vertex of $\\Gamma$. Choose plane $P$ such that $\\Psi_{1}=E_{i}-U$ and $\\Psi_{2}=\\{0\\}$. By Lemma 2, $\\Delta \\cap(P+W)=E_{i}-V$. Hence there exists a line segment in $\\partial \\Delta$ which contains $W$ in its interior, but there is no planar region in $\\partial \\Delta$ with the same property.\nWe obtain a similar result if $V$ belongs to an edge of $\\Gamma$ and $U$ is a vertex.\n\nCase 3: Points $U$ and $V$ are interior points of edges $E_{i}$ and $E_{j}$, respectively. Let $P$ be the plane of $E_{i}-U$ and $E_{j}-V$. Then $\\Psi_{1}=E_{i}-U, \\Psi_{2}=E_{j}-V$ and $\\Delta \\cap(P+W)=E_{i}-E_{j}$. Therefore point $W$ belongs to a parallelogram face on $\\partial \\Delta$.\nThe centre of the parallelogram is $Q_{i}-Q_{j}$, the vector between the midpoints. Therefore an edge pair ( $E_{i}, E_{j}$ ) occurs if and only if $Q_{i}$ and $Q_{j}$ are antipodal which happens $2B$ times.\n\nCase 4: Point $U$ lies in the interior of a face $F_{i}$ and $V$ is a vertex of $\\Gamma$. The only choice for $P$ is the plane of $F_{i}-U$. Then we have $\\Psi_{1}=F_{i}-U, \\Psi_{2}=\\{0\\}$ and $\\Delta \\cap(P+W)=F_{i}-V$. This is a planar face of $\\partial \\Delta$ which is congruent to $F_{i}$.\nFor each face $F_{i}$, the only possible vertex $V$ is the farthest one from the plane of $F_{i}$.\nIf $U$ is a vertex and $V$ belongs to face $F_{i}$ then we obtain the same way that $W$ belongs to a face $-F_{i}+U$ which is also congruent to $F_{i}$. Therefore, each face of $\\Gamma$ has two copies on $\\partial \\Delta$, a translated and a reflected copy.\n\nCase 5: Point $U$ belongs to a face $F_{i}$ of $\\Gamma$ and point $V$ belongs to an edge or a face $G$. In this case objects $F_{i}$ and $G$ must be parallel which is not allowed.\n![](attached_image_3.png)\ncase 1\n![](attached_image_4.png)\ncase 2\n![](attached_image_5.png)\ncase 3\n![](attached_image_6.png)\ncase 4\n\nNow all points in $\\partial \\Delta$ belong to some planar polygons (cases 3 and 4), finitely many line segments (case 2) and points (case 1). Therefore $\\Delta$ is indeed a polyhedron. Now compute the numbers of its vertices, edges and faces.\nThe vertices are obtained in case 1, their number is $2A$.\nFaces are obtained in cases 3 and 4. Case 3 generates $2B$ parallelogram faces. Case 4 generates $2\\ell$ faces.\nWe compute the number of edges of $\\Delta$ from the degrees (number of sides) of faces of $\\Gamma$. Let $d_{i}$ be the the degree of face $F_{i}$. The sum of degrees is twice as much as the number of edges, so $d_{1}+d_{2}+\\ldots+d_{l}=2m$. The sum of degrees of faces of $\\Delta$ is $2B \\cdot 4+2\\left(d_{1}+d_{2}+\\cdots+d_{l}\\right)=8B+4m$, so the number of edges on $\\Delta$ is $4B+2m$.\nApplying Euler's polyhedron theorem on $\\Gamma$ and $\\Delta$, we have $n+\\ell=m+2$ and $2A+(2B+2\\ell)= (4B+2m)+2$. Then the conclusion follows:\n$$\nA-B=m-\\ell+1=n-1 .\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31838,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle satisfies\n$$\n\\angle PBA + \\angle PCA = \\angle PBC + \\angle PCB.\n$$\nShow that $AP \\geq AI$ and that equality holds if and only if $P$ coincides with $I$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle A = \\alpha$, $\\angle B = \\beta$, $\\angle C = \\gamma$. Since $\\angle PBA + \\angle PCA + \\angle PBC + \\angle PCB = \\beta + \\gamma$, the condition from the problem statement is equivalent to $\\angle PBC + \\angle PCB = (\\beta + \\gamma)/2$, i.e. $\\angle BPC = 90^\\circ + \\alpha/2$.\n\nOn the other hand, $\\angle BIC = 180^\\circ - (\\beta + \\gamma)/2 = 90^\\circ + \\alpha/2$. Hence $\\angle BPC = \\angle BIC$, and since $P$ and $I$ are on the same side of $BC$, the points $B$, $C$, $I$ and $P$ are concyclic. In other words, $P$ lies on the circumcircle $\\omega$ of triangle $BCI$.\n\n![](attached_image_1.png)\n\nLet $\\Omega$ be the circumcircle of triangle $ABC$. It is a well-known fact that the centre of $\\omega$ is the midpoint $M$ of the arc $BC$ of $\\Omega$. This is also the point where the angle bisector $AI$ intersects $\\Omega$.\n\nFrom triangle $APM$ we have\n$$\nAP + PM \\geq AM = AI + IM = AI + PM\n$$\nTherefore $AP \\geq AI$. Equality holds if and only if $P$ lies on the line segment $AI$, which occurs if and only if $P = I$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31839,
    "subject": "Mathematics (Multi-modal)",
    "question": "To each side $a$ of a convex polygon we assign the maximum area of a triangle contained in the polygon and having $a$ as one of its sides. Show that the sum of the areas assigned to all sides of the polygon is not less than twice the area of the polygon.\n(Serbia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lemma. Every convex $(2n)$-gon, of area $S$, has a side and a vertex that jointly span a triangle of area not less than $S / n$.\n\nProof. By main diagonals of the $(2n)$-gon we shall mean those which partition the $(2n)$-gon into two polygons with equally many sides. For any side $b$ of the $(2n)$-gon denote by $\\Delta_{b}$ the triangle $ABP$ where $A, B$ are the endpoints of $b$ and $P$ is the intersection point of the main diagonals $AA', BB'$. We claim that the union of triangles $\\Delta_{b}$, taken over all sides, covers the whole polygon.\n\nTo show this, choose any side $AB$ and consider the main diagonal $AA'$ as a directed segment. Let $X$ be any point in the polygon, not on any main diagonal. For definiteness, let $X$ lie on the left side of the ray $AA'$. Consider the sequence of main diagonals $AA', BB', CC', \\ldots$, where $A, B, C, \\ldots$ are consecutive vertices, situated right to $AA'$. The $n$-th item in this sequence is the diagonal $A'A$ (i.e. $AA'$ reversed), having $X$ on its right side. So there are two successive vertices $K, L$ in the sequence $A, B, C, \\ldots$ before $A'$ such that $X$ still lies to the left of $KK'$ but to the right of $LL'$. And this means that $X$ is in the triangle $\\Delta_{\\ell'}$, $\\ell' = K' L'$. Analogous reasoning applies to points $X$ on the right of $AA'$ (points lying on main diagonals can be safely ignored). Thus indeed the triangles $\\Delta_{b}$ jointly cover the whole polygon.\n\nThe sum of their areas is no less than $S$. So we can find two opposite sides, say $b = AB$ and $b' = A'B'$ (with $AA', BB'$ main diagonals) such that $[\\Delta_{b}] + [\\Delta_{b'}] \\geq S / n$, where $[\\cdots]$ stands for the area of a region. Let $AA', BB'$ intersect at $P$; assume without loss of generality that $PB \\geq PB'$. Then\n$$\n[ABA'] = [ABP] + [PBA'] \\geq [ABP] + [PA'B'] = [\\Delta_{b}] + [\\Delta_{b'}] \\geq S / n,\n$$\nproving the lemma. $\\square$\n\nNow, let $\\mathcal{P}$ be any convex polygon, of area $S$, with $m$ sides $a_{1}, \\ldots, a_{m}$. Let $S_{i}$ be the area of the greatest triangle in $\\mathcal{P}$ with side $a_{i}$. Suppose, contrary to the assertion, that\n$$\n\\sum_{i=1}^{m} \\frac{S_{i}}{S} < 2.\n$$\nThen there exist rational numbers $q_{1}, \\ldots, q_{m}$ such that $\\sum q_{i} = 2$ and $q_{i} > S_{i} / S$ for each $i$.\n\nLet $n$ be a common denominator of the $m$ fractions $q_{1}, \\ldots, q_{m}$. Write $q_{i} = k_{i} / n$; so $\\sum k_{i} = 2n$. Partition each side $a_{i}$ of $\\mathcal{P}$ into $k_{i}$ equal segments, creating a convex $(2n)$-gon of area $S$ (with some angles of size $180^{\\circ}$), to which we apply the lemma. Accordingly, this refined polygon has a side $b$ and a vertex $H$ spanning a triangle $T$ of area $[T] \\geq S / n$. If $b$ is a piece of a side $a_{i}$ of $\\mathcal{P}$, then the triangle $W$ with base $a_{i}$ and summit $H$ has area\n$$\n[W] = k_{i} \\cdot [T] \\geq k_{i} \\cdot S / n = q_{i} \\cdot S > S_{i},\n$$\nin contradiction with the definition of $S_{i}$. This ends the proof.\nAs in the first solution, we allow again angles of size $180^{\\circ}$ at some vertices of the convex polygons considered.\n\nTo each convex $n$-gon $\\mathcal{P} = A_{1} A_{2} \\ldots A_{n}$ we assign a centrally symmetric convex $(2n)$-gon $\\mathcal{Q}$ with side vectors $\\pm \\overrightarrow{A_{i} A_{i+1}}, 1 \\leq i \\leq n$. The construction is as follows. Attach the $2n$ vectors $\\pm \\overrightarrow{A_{i} A_{i+1}}$ at a common origin and label them $\\overrightarrow{\\mathbf{b}_{1}}, \\overrightarrow{\\mathbf{b}_{2}}, \\ldots, \\overrightarrow{\\mathbf{b}_{2n}}$ in counterclockwise direction; the choice of the first vector $\\overrightarrow{\\mathbf{b}_{1}}$ is irrelevant. The order of labelling is well-defined if $\\mathcal{P}$ has neither parallel sides nor angles equal to $180^{\\circ}$. Otherwise several collinear vectors with the same direction are labelled consecutively $\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots, \\overrightarrow{\\mathbf{b}_{j+r}}$. One can assume that in such cases the respective opposite vectors occur in the order $-\\overrightarrow{\\mathbf{b}_{j}}', -\\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots, -\\overrightarrow{\\mathbf{b}_{j+r}}$, ensuring that $\\overrightarrow{\\mathbf{b}_{j+n}} = -\\overrightarrow{\\mathbf{b}_{j}}$ for $j = 1, \\ldots, 2n$. Indices are taken cyclically here and in similar situations below.\n\nChoose points $B_{1}, B_{2}, \\ldots, B_{2n}$ satisfying $\\overrightarrow{B_{j} B_{j+1}} = \\overrightarrow{\\mathbf{b}_{j}}$ for $j = 1, \\ldots, 2n$. The polygonal line $\\mathcal{Q} = B_{1} B_{2} \\ldots B_{2n}$ is closed, since $\\sum_{j=1}^{2n} \\overrightarrow{\\mathbf{b}_{j}} = \\overrightarrow{0}$. Moreover, $\\mathcal{Q}$ is a convex $(2n)$-gon due to the arrangement of the vectors $\\overrightarrow{\\mathbf{b}_{j}}$, possibly with $180^{\\circ}$-angles. The side vectors of $\\mathcal{Q}$ are $\\pm \\overrightarrow{A_{i} A_{i+1}}$, $1 \\leq i \\leq n$. So in particular $\\mathcal{Q}$ is centrally symmetric, because it contains as side vectors $\\overrightarrow{A_{i} A_{i+1}}$ and $-\\overrightarrow{A_{i} A_{i+1}}$ for each $i = 1, \\ldots, n$. Note that $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are opposite sides of $\\mathcal{Q}$, $1 \\leq j \\leq n$. We call $\\mathcal{Q}$ the associate of $\\mathcal{P}$.\n\nLet $S_{i}$ be the maximum area of a triangle with side $A_{i} A_{i+1}$ in $\\mathcal{P}$, $1 \\leq i \\leq n$. We prove that\n$$\n\\left[ B_{1} B_{2} \\ldots B_{2n} \\right] = 2 \\sum_{i=1}^{n} S_{i} \\tag{1}\n$$\nand\n$$\n\\left[ B_{1} B_{2} \\ldots B_{2n} \\right] \\geq 4 \\left[ A_{1} A_{2} \\ldots A_{n} \\right] \\tag{2}\n$$\nIt is clear that (1) and (2) imply the conclusion of the original problem.\n\nLemma. For a side $A_{i} A_{i+1}$ of $\\mathcal{P}$, let $h_{i}$ be the maximum distance from a point of $\\mathcal{P}$ to line $A_{i} A_{i+1}$, $i = 1, \\ldots, n$. Denote by $B_{j} B_{j+1}$ the side of $\\mathcal{Q}$ such that $\\overrightarrow{A_{i} A_{i+1}} = \\overrightarrow{B_{j} B_{j+1}}$. Then the distance between $B_{j} B_{j+1}$ and its opposite side in $\\mathcal{Q}$ is equal to $2 h_{i}$.\n\nProof. Choose a vertex $A_{k}$ of $\\mathcal{P}$ at distance $h_{i}$ from line $A_{i} A_{i+1}$. Let $\\mathbf{u}$ be the unit vector perpendicular to $A_{i} A_{i+1}$ and pointing inside $\\mathcal{P}$. Denoting by $\\mathbf{x} \\cdot \\mathbf{y}$ the dot product of vectors $\\mathbf{x}$ and $\\mathbf{y}$, we have\n$$\nh = \\mathbf{u} \\cdot \\overrightarrow{A_{i} A_{k}} = \\mathbf{u} \\cdot \\left( \\overrightarrow{A_{i} A_{i+1}} + \\cdots + \\overrightarrow{A_{k-1} A_{k}} \\right ) = \\mathbf{u} \\cdot \\left( \\overrightarrow{A_{i} A_{i-1}} + \\cdots + \\overrightarrow{A_{k+1} A_{k}} \\right ).\n$$\nIn $\\mathcal{Q}$, the distance $H_{i}$ between the opposite sides $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ is given by\n$$\nH_{i} = \\mathbf{u} \\cdot \\left( \\overrightarrow{B_{j} B_{j+1}} + \\cdots + \\overrightarrow{B_{j+n-1} B_{j+n}} \\right ) = \\mathbf{u} \\cdot \\left( \\overrightarrow{\\mathbf{b}_{j}} + \\overrightarrow{\\mathbf{b}_{j+1}} + \\cdots + \\overrightarrow{\\mathbf{b}_{j+n-1}} \\right ).\n$$\nThe choice of vertex $A_{k}$ implies that the $n$ consecutive vectors $\\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots, \\overrightarrow{\\mathbf{b}_{j+n-1}}$ are precisely $\\overrightarrow{A_{i} A_{i+1}}, \\ldots, \\overrightarrow{A_{k-1} A_{k}}$ and $\\overrightarrow{A_{i} A_{i-1}}, \\ldots, \\overrightarrow{A_{k+1} A_{k}}$, taken in some order. This implies $H_{i} = 2 h_{i}$.\n\nFor a proof of (1), apply the lemma to each side of $\\mathcal{P}$. If $O$ is the centre of $\\mathcal{Q}$ then, using the notation of the lemma,\n$$\n[ B_{j} B_{j+1} O ] = [ B_{j+n} B_{j+n+1} O ] = [ A_{i} A_{i+1} A_{k} ] = S_{i}\n$$\nSummation over all sides of $\\mathcal{P}$ yields (1).\n\nSet $d(\\mathcal{P}) = [\\mathcal{Q}] - 4 [\\mathcal{P}]$ for a convex polygon $\\mathcal{P}$ with associate $\\mathcal{Q}$. Inequality (2) means that $d(\\mathcal{P}) \\geq 0$ for each convex polygon $\\mathcal{P}$. The last inequality will be proved by induction on the number $\\ell$ of side directions of $\\mathcal{P}$, i.e. the number of pairwise nonparallel lines each containing a side of $\\mathcal{P}$.\n\nWe choose to start the induction with $\\ell = 1$ as a base case, meaning that certain degenerate polygons are allowed. More exactly, we regard as degenerate convex polygons all closed polygonal lines of the form $X_{1} X_{2} \\ldots X_{k} Y_{1} Y_{2} \\ldots Y_{m} X_{1}$, where $X_{1}, X_{2}, \\ldots, X_{k}$ are points in this order on a line segment $X_{1} Y_{1}$, and so are $Y_{m}, Y_{m-1}, \\ldots, Y_{1}$. The initial construction applies to degenerate polygons; their associates are also degenerate, and the value of $d$ is zero.\n\nFor the inductive step, consider a convex polygon $\\mathcal{P}$ which determines $\\ell$ side directions, assuming that $d(\\mathcal{P}) \\geq 0$ for polygons with smaller values of $\\ell$.\n\nSuppose first that $\\mathcal{P}$ has a pair of parallel sides, i.e. sides on distinct parallel lines. Let $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$ be such a pair, and let $A_{i} A_{i+1} \\leq A_{j} A_{j+1}$. Remove from $\\mathcal{P}$ the parallelogram $R$ determined by vectors $\\overrightarrow{A_{i} A_{i+1}}$ and $\\overrightarrow{A_{i} A_{j+1}}$. Two polygons are obtained in this way. Translating one of them by vector $\\overrightarrow{A_{i} A_{i+1}}$ yields a new convex polygon $\\mathcal{P}'$, of area $[\\mathcal{P}] - [R]$ and with value of $\\ell$ not exceeding the one of $\\mathcal{P}$. The construction just described will be called operation A.\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nThe associate of $\\mathcal{P}'$ is obtained from $\\mathcal{Q}$ upon decreasing the lengths of two opposite sides by an amount of $2 A_{i} A_{i+1}$. By the lemma, the distance between these opposite sides is twice the distance between $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$. Thus operation $\\mathbf{A}$ decreases $[\\mathcal{Q}]$ by the area of a parallelogram with base and respective altitude twice the ones of $R$, i.e. by $4[R]$. Hence $\\mathbf{A}$ leaves the difference $d(\\mathcal{P}) = [\\mathcal{Q}] - 4[\\mathcal{P}]$ unchanged.\n\nNow, if $\\mathcal{P}'$ also has a pair of parallel sides, apply operation $\\mathbf{A}$ to it. Keep doing so with the subsequent polygons obtained for as long as possible. Now, A decreases the number $p$ of pairs of parallel sides in $\\mathcal{P}$. Hence its repeated applications gradually reduce $p$ to $0$, and further applications of $\\mathbf{A}$ will be impossible after several steps. For clarity, let us denote by $\\mathcal{P}$ again the polygon obtained at that stage.\n\nThe inductive step is complete if $\\mathcal{P}$ is degenerate. Otherwise $\\ell > 1$ and $p = 0$, i.e. there are no parallel sides in $\\mathcal{P}$. Observe that then $\\ell \\geq 3$. Indeed, $\\ell = 2$ means that the vertices of $\\mathcal{P}$ all lie on the boundary of a parallelogram, implying $p > 0$.\n\nFurthermore, since $\\mathcal{P}$ has no parallel sides, consecutive collinear vectors in the sequence $\\left( \\overrightarrow{\\mathbf{b}_{k}} \\right)$ (if any) correspond to consecutive $180^{\\circ}$-angles in $\\mathcal{P}$. Removing the vertices of such angles, we obtain a convex polygon with the same value of $d(\\mathcal{P})$.\n\nIn summary, if operation $\\mathbf{A}$ is impossible for a nondegenerate polygon $\\mathcal{P}$, then $\\ell \\geq 3$. In addition, one may assume that $\\mathcal{P}$ has no angles of size $180^{\\circ}$.\n\nThe last two conditions then also hold for the associate $\\mathcal{Q}$ of $\\mathcal{P}$, and we perform the following construction. Since $\\ell \\geq 3$, there is a side $B_{j} B_{j+1}$ of $\\mathcal{Q}$ such that the sum of the angles at $B_{j}$ and $B_{j+1}$ is greater than $180^{\\circ}$. (Such a side exists in each convex $k$-gon for $k > 4$.) Naturally, $B_{j+n} B_{j+n+1}$ is a side with the same property. Extend the pairs of sides $B_{j-1} B_{j}, B_{j+1} B_{j+2}$ and $B_{j+n-1} B_{j+n}, B_{j+n+1} B_{j+n+2}$ to meet at $U$ and $V$, respectively. Let $\\mathcal{Q}'$ be the centrally symmetric convex $2(n+1)$-gon obtained from $\\mathcal{Q}$ by inserting $U$ and $V$ into the sequence $B_{1}, \\ldots, B_{2n}$ as new vertices between $B_{j}, B_{j+1}$ and $B_{j+n}, B_{j+n+1}$, respectively. Informally, we adjoin to $\\mathcal{Q}$ the congruent triangles $B_{j} B_{j+1} U$ and $B_{j+n} B_{j+n+1} V$. Note that $B_{j}, B_{j+1}, B_{j+n}$ and $B_{j+n+1}$ are kept as vertices of $\\mathcal{Q}'$, although $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are no longer its sides.\n\nLet $A_{i} A_{i+1}$ be the side of $\\mathcal{P}$ such that $\\overrightarrow{A_{i} A_{i+1}} = \\overrightarrow{B_{j} B_{j+1}} = \\overrightarrow{\\mathbf{b}_{j}}$. Consider the point $W$ such that triangle $A_{i} A_{i+1} W$ is congruent to triangle $B_{j} B_{j+1} U$ and exterior to $\\mathcal{P}$. Insert $W$ into the sequence $A_{1}, A_{2}, \\ldots, A_{n}$ as a new vertex between $A_{i}$ and $A_{i+1}$ to obtain an $(n+1)$-gon $\\mathcal{P}'$. We claim that $\\mathcal{P}'$ is convex and its associate is $\\mathcal{Q}'$.\n\n![](attached_image_3.png)\n![](attached_image_4.png)\n\nVectors $\\overrightarrow{A_{i} W}$ and $\\overrightarrow{\\mathbf{b}_{j-1}}$ are collinear and have the same direction, as well as vectors $\\overrightarrow{W A_{i+1}}$ and $\\overrightarrow{\\mathbf{b}_{j+1}}$. Since $\\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}}$ are consecutive terms in the sequence $\\left( \\overrightarrow{\\mathbf{b}_{k}} \\right)$, the angle inequalities $\\angle( \\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{\\mathbf{b}_{j}} ) \\leq \\angle( \\overrightarrow{A_{i-1} A_{i}}, \\overrightarrow{\\mathbf{b}_{j}} )$ and $\\angle( \\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{\\mathbf{b}_{j+1}} ) \\leq \\angle( \\overrightarrow{\\mathbf{b}_{j}}, \\overrightarrow{A_{i+1} A_{i+2}} )$ hold true. They show that $\\mathcal{P}'$ is a convex polygon. To construct its associate, vectors $\\pm \\overrightarrow{A_{i} A_{i+1}} = \\pm \\overrightarrow{\\mathbf{b}_{j}}$ must be deleted from the defining sequence $\\left( \\overrightarrow{\\mathbf{b}_{k}} \\right)$ of $\\mathcal{Q}$, and the vectors $\\pm \\overrightarrow{A_{i} W}, \\pm \\overrightarrow{W A_{i+1}}$ must be inserted appropriately into it. The latter can be done as follows:\n$$\n\\ldots, \\overrightarrow{\\mathbf{b}_{j-1}}, \\overrightarrow{A_{i} W}, \\overrightarrow{W A_{i+1}}, \\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots, -\\overrightarrow{\\mathbf{b}_{j-1}}, -\\overrightarrow{A_{i} W}, -\\overrightarrow{W A_{i+1}}, -\\overrightarrow{\\mathbf{b}_{j+1}}, \\ldots.\n$$\nThis updated sequence produces $\\mathcal{Q}'$ as the associate of $\\mathcal{P}'$.\n\nIt follows from the construction that $[\\mathcal{P}'] = [\\mathcal{P}] + [A_{i} A_{i+1} W]$ and $[\\mathcal{Q}'] = [\\mathcal{Q}] + 2 [A_{i} A_{i+1} W]$. Therefore $d(\\mathcal{P}') = d(\\mathcal{P}) - 2 [A_{i} A_{i+1} W] < d(\\mathcal{P})$.\n\nTo finish the induction, it remains to notice that the value of $\\ell$ for $\\mathcal{P}'$ is less than the one for $\\mathcal{P}$. This is because side $A_{i} A_{i+1}$ was removed. The newly added sides $A_{i} W$ and $W A_{i+1}$ do not introduce new side directions. Each one of them is either parallel to a side of $\\mathcal{P}$ or lies on the line determined by such a side. The proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31840,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezoid with parallel sides $AB > CD$. Points $K$ and $L$ lie on the line segments $AB$ and $CD$, respectively, so that $AK / KB = DL / LC$. Suppose that there are points $P$ and $Q$ on the line segment $KL$ satisfying\n$$\n\\angle APB = \\angle BCD \\quad \\text{and} \\quad \\angle CQD = \\angle ABC.\n$$\nProve that the points $P$, $Q$, $B$ and $C$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Because $AB \\parallel CD$, the relation $AK / KB = DL / LC$ readily implies that the lines $AD$, $BC$ and $KL$ have a common point $S$.\n![](attached_image_1.png)\nConsider the second intersection points $X$ and $Y$ of the line $SK$ with the circles $(ABP)$ and $(CDQ)$, respectively. Since $APBX$ is a cyclic quadrilateral and $AB \\parallel CD$, one has\n$$\n\\angle AXB = 180^{\\circ} - \\angle APB = 180^{\\circ} - \\angle BCD = \\angle ABC.\n$$\nThis shows that $BC$ is tangent to the circle $(ABP)$ at $B$. Likewise, $BC$ is tangent to the circle $(CDQ)$ at $C$. Therefore $SP \\cdot SX = SB^{2}$ and $SQ \\cdot SY = SC^{2}$.\nLet $h$ be the homothety with centre $S$ and ratio $SC / SB$. Since $h(B) = C$, the above conclusion about tangency implies that $h$ takes circle $(ABP)$ to circle $(CDQ)$. Also, $h$ takes $AB$ to $CD$, and it easily follows that $h(P) = Y$, $h(X) = Q$, yielding $SP / SY = SB / SC = SX / SQ$.\nEqualities $SP \\cdot SX = SB^{2}$ and $SQ / SX = SC / SB$ imply $SP \\cdot SQ = SB \\cdot SC$, which is equivalent to $P$, $Q$, $B$ and $C$ being concyclic.\nThe case where $P = Q$ is trivial. Thus assume that $P$ and $Q$ are two distinct points. As in the first solution, notice that the lines $AD$, $BC$ and $KL$ concur at a point $S$.\n![](attached_image_2.png)\nLet the lines $AP$ and $DQ$ meet at $E$, and let $BP$ and $CQ$ meet at $F$. Then $\\angle EPF = \\angle BCD$ and $\\angle FQE = \\angle ABC$ by the condition of the problem. Since the angles $BCD$ and $ABC$ add up to $180^{\\circ}$, it follows that $PEQF$ is a cyclic quadrilateral.\nApplying Menelaus' theorem, first to triangle $ASP$ and line $DQ$ and then to triangle $BSP$ and line $CQ$, we have\n$$\n\\frac{AD}{DS} \\cdot \\frac{SQ}{QP} \\cdot \\frac{PE}{EA} = 1 \\quad \\text{and} \\quad \\frac{BC}{CS} \\cdot \\frac{SQ}{QP} \\cdot \\frac{PF}{FB} = 1.\n$$\nThe first factors in these equations are equal, as $AB \\parallel CD$. Thus the last factors are also equal, which implies that $EF$ is parallel to $AB$ and $CD$. Using this and the cyclicity of $PEQF$, we obtain\n$$\n\\angle BCD = \\angle BCF + \\angle FCD = \\angle BCQ + \\angle EFQ = \\angle BCQ + \\angle EPQ.\n$$\nOn the other hand,\n$$\n\\angle BCD = \\angle APB = \\angle EPF = \\angle EPQ + \\angle QPF,\n$$\nand consequently $\\angle BCQ = \\angle QPF$. The latter angle either coincides with $\\angle QPB$ or is supplementary to $\\angle QPB$, depending on whether $Q$ lies between $K$ and $P$ or not. In either case it follows that $P$, $Q$, $B$ and $C$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31841,
    "subject": "Mathematics (Multi-modal)",
    "question": "A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\\angle C < \\angle A < 90^{\\circ}$ in such a way that $BD = BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incentre of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at its midpoint.\n(Russia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $P$ the common point of $AJ$ and $KL$. Let the parallel to $KL$ through $J$ meet $AC$ at $M$. Then $P$ is the midpoint of $AJ$ if and only if $AM = 2 \\cdot AL$, which we are about to show.\n\n![](attached_image_1.png)\n\nDenoting $\\angle BAC = 2\\alpha$, the equalities $BA = BD$ and $AK = AL$ imply $\\angle ADB = 2\\alpha$ and $\\angle ALK = 90^{\\circ} - \\alpha$. Since $DJ$ bisects $\\angle BDC$, we obtain $\\angle CDJ = \\frac{1}{2} \\cdot (180^{\\circ} - \\angle ADB) = 90^{\\circ} - \\alpha$. Also $\\angle DMJ = \\angle ALK = 90^{\\circ} - \\alpha$ since $JM \\parallel KL$. It follows that $JD = JM$.\n\nLet the incircle of triangle $BCD$ touch its side $CD$ at $T$. Then $JT \\perp CD$, meaning that $JT$ is the altitude to the base $DM$ of the isosceles triangle $DMJ$. It now follows that $DT = MT$, and we have\n$$\nDM = 2 \\cdot DT = BD + CD - BC.\n$$\nTherefore\n$$\n\\begin{aligned}\nAM & = AD + (BD + CD - BC) \\\\\n   & = AD + AB + DC - BC \\\\\n   & = AC + AB - BC \\\\\n   & = 2 \\cdot AL,\n\\end{aligned}\n$$\nwhich completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31842,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $A B C$, let $J$ be the centre of the excircle tangent to side $B C$ at $A_{1}$ and to the extensions of sides $A C$ and $A B$ at $B_{1}$ and $C_{1}$, respectively. Suppose that the lines $A_{1} B_{1}$ and $A B$ are perpendicular and intersect at $D$. Let $E$ be the foot of the perpendicular from $C_{1}$ to line $D J$. Determine the angles $\\angle B E A_{1}$ and $\\angle A E B_{1}$.\n(Greece)",
    "options": [],
    "answer": "∠BEA1 = 90°, ∠AEB1 = 90°",
    "solution": "Let $K$ be the intersection point of lines $J C$ and $A_{1} B_{1}$. Obviously $J C \\perp A_{1} B_{1}$ and since $A_{1} B_{1} \\perp A B$, the lines $J K$ and $C_{1} D$ are parallel and equal. From the right triangle $B_{1} C J$ we obtain $J C_{1}^{2}=J B_{1}^{2}=J C \\cdot J K=J C \\cdot C_{1} D$ from which we infer that $D C_{1} / C_{1} J=C_{1} J / J C$ and the right triangles $D C_{1} J$ and $C_{1} J C$ are similar. Hence $\\angle C_{1} D J=\\angle J C_{1} C$, which implies that the lines $D J$ and $C_{1} C$ are perpendicular, i.e the points $C_{1}, E, C$ are collinear.\n\n![](attached_image_1.png)\n\nSince $\\angle C A_{1} J=\\angle C B_{1} J=\\angle C E J=90^{\\circ}$, points $A_{1}, B_{1}$ and $E$ lie on the circle of diameter $C J$. Then $\\angle D B A_{1}=\\angle A_{1} C J=\\angle D E A_{1}$, which implies that quadrilateral $B E A_{1} D$ is cyclic; therefore $\\angle A_{1} E B=90^{\\circ}$.\n\nQuadrilateral $A D E B_{1}$ is also cyclic because $\\angle E B_{1} A=\\angle E J C=\\angle E D C_{1}$, therefore we obtain $\\angle A E B_{1}=\\angle A D B=90^{\\circ}$.\nConsider the circles $\\omega_{1}, \\omega_{2}$ and $\\omega_{3}$ of diameters $C_{1} D, A_{1} B$ and $A B_{1}$, respectively. Line segments $J C_{1}, J B_{1}$ and $J A_{1}$ are tangents to those circles and, due to the right angle at $D$, $\\omega_{2}$ and $\\omega_{3}$ pass through point $D$. Since $\\angle C_{1} E D$ is a right angle, point $E$ lies on circle $\\omega_{1}$, therefore\n$$\nJ C_{1}^{2}=J D \\cdot J E .\n$$\nSince $J A_{1}=J B_{1}=J C_{1}$ are all radii of the excircle, we also have\n$$\nJ A_{1}^{2}=J D \\cdot J E \\quad \\text { and } \\quad J B_{1}^{2}=J D \\cdot J E .\n$$\nThese equalities show that $E$ lies on circles $\\omega_{2}$ and $\\omega_{3}$ as well, so $\\angle B E A_{1}=\\angle A E B_{1}=90^{\\circ}$.\nFirst note that $A_{1} B_{1}$ is perpendicular to the external angle bisector $C J$ of $\\angle B C A$ and parallel to the internal angle bisector of that angle. Therefore, $A_{1} B_{1}$ is perpendicular to $A B$ if and only if triangle $A B C$ is isosceles, $A C=B C$. In that case the external bisector $C J$ is parallel to $A B$.\n\nTriangles $A B C$ and $B_{1} A_{1} J$ are similar, as their corresponding sides are perpendicular. In particular, we have $\\angle D A_{1} J=\\angle C_{1} B A_{1}$; moreover, from cyclic deltoid $J A_{1} B C_{1}$,\n$$\n\\angle C_{1} A_{1} J=\\angle C_{1} B J=\\frac{1}{2} \\angle C_{1} B A_{1}=\\frac{1}{2} \\angle D A_{1} J .\n$$\nTherefore, $A_{1} C_{1}$ bisects angle $\\angle D A_{1} J$.\n\n![](attached_image_2.png)\n\nIn triangle $B_{1} A_{1} J$, line $J C_{1}$ is the external bisector at vertex $J$. The point $C_{1}$ is the intersection of two external angle bisectors (at $A_{1}$ and $J$ ) so $C_{1}$ is the centre of the excircle $\\omega$, tangent to side $A_{1} J$, and to the extension of $B_{1} A_{1}$ at point $D$.\n\nNow consider the similarity transform $\\varphi$ which moves $B_{1}$ to $A, A_{1}$ to $B$ and $J$ to $C$. This similarity can be decomposed into a rotation by $90^{\\circ}$ around a certain point $O$ and a homothety from the same centre. This similarity moves point $C_{1}$ (the centre of excircle $\\omega$ ) to $J$ and moves $D$ (the point of tangency) to $C_{1}$.\n\nSince the rotation angle is $90^{\\circ}$, we have $\\angle X O \\varphi(X)=90^{\\circ}$ for an arbitrary point $X \\neq O$. For $X=D$ and $X=C_{1}$ we obtain $\\angle D O C_{1}=\\angle C_{1} O J=90^{\\circ}$. Therefore $O$ lies on line segment $D J$ and $C_{1} O$ is perpendicular to $D J$. This means that $O=E$.\n\nFor $X=A_{1}$ and $X=B_{1}$ we obtain $\\angle A_{1} O B=\\angle B_{1} O A=90^{\\circ}$, i.e.\n$$\n\\angle B E A_{1}=\\angle A E B_{1}=90^{\\circ} .\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31843,
    "subject": "Mathematics (Multi-modal)",
    "question": "Circles $\\omega_{1}$ and $\\omega_{2}$ with centres $O_{1}$ and $O_{2}$ are externally tangent at point $D$ and internally tangent to a circle $\\omega$ at points $E$ and $F$, respectively. Line $t$ is the common tangent of $\\omega_{1}$ and $\\omega_{2}$ at $D$. Let $AB$ be the diameter of $\\omega$ perpendicular to $t$, so that $A$, $E$ and $O_{1}$ are on the same side of $t$. Prove that lines $A O_{1}$, $B O_{2}$, $EF$ and $t$ are concurrent.\n\n(Brasil)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Point $E$ is the centre of a homothety $h$ which takes circle $\\omega_{1}$ to circle $\\omega$. The radii $O_{1} D$ and $O B$ of these circles are parallel as both are perpendicular to line $t$. Also, $O_{1} D$ and $O B$ are on the same side of line $E O$, hence $h$ takes $O_{1} D$ to $O B$. Consequently, points $E$, $D$ and $B$ are collinear. Likewise, points $F$, $D$ and $A$ are collinear as well.\nLet lines $A E$ and $B F$ intersect at $C$. Since $A F$ and $B E$ are altitudes in triangle $A B C$, their common point $D$ is the orthocentre of this triangle. So $C D$ is perpendicular to $A B$, implying that $C$ lies on line $t$. Note that triangle $A B C$ is acute-angled. We mention the well-known fact that triangles $F E C$ and $A B C$ are similar in ratio $\\cos \\gamma$, where $\\gamma=\\angle A C B$. In addition, points $C$, $E$, $D$ and $F$ lie on the circle with diameter $C D$.\n![](attached_image_1.png)\nLet $P$ be the common point of lines $E F$ and $t$. We are going to prove that $P$ lies on line $A O_{1}$. Denote by $N$ the second common point of circle $\\omega_{1}$ and $A C$; this is the point of $\\omega_{1}$ diametrically opposite to $D$. By Menelaus' theorem for triangle $D C N$, points $A$, $O_{1}$ and $P$ are collinear if and only if\n$$\n\\frac{C A}{A N} \\cdot \\frac{N O_{1}}{O_{1} D} \\cdot \\frac{D P}{P C}=1 .\n$$\nBecause $N O_{1}=O_{1} D$, this reduces to $C A / A N=C P / P D$. Let line $t$ meet $A B$ at $K$. Then $C A / A N=C K / K D$, so it suffices to show that\n$$\n\\begin{equation*}\n\\frac{C P}{P D}=\\frac{C K}{K D} . \\tag{1}\n\\end{equation*}\n$$\nTo verify (1), consider the circumcircle $\\Omega$ of triangle $A B C$. Draw its diameter $C U$ through $C$, and let $C U$ meet $A B$ at $V$. Extend $C K$ to meet $\\Omega$ at $L$. Since $A B$ is parallel to $U L$, we have $\\angle A C U=\\angle B C L$. On the other hand $\\angle E F C=\\angle B A C$, $\\angle F E C=\\angle A B C$ and $E F / A B=\\cos \\gamma$, as stated above. So reflection in the bisector of $\\angle A C B$ followed by a homothety with centre $C$ and ratio $1 / \\cos \\gamma$ takes triangle $F E C$ to triangle $A B C$. Consequently, this transformation\ntakes $C D$ to $C U$, which implies $C P / P D=C V / V U$. Next, we have $K L=K D$, because $D$ is the orthocentre of triangle $A B C$. Hence $C K / K D=C K / K L$. Finally, $C V / V U=C K / K L$ because $A B$ is parallel to $U L$. Relation (1) follows, proving that $P$ lies on line $A O_{1}$. By symmetry, $P$ also lies on line $A O_{2}$ which completes the solution.\nWe proceed as in the first solution to define a triangle $A B C$ with orthocentre $D$, in which $A F$ and $B E$ are altitudes.\nDenote by $M$ the midpoint of $C D$. The quadrilateral $C E D F$ is inscribed in a circle with centre $M$, hence $M C=M E=M D=M F$.\n![](attached_image_2.png)\nConsider triangles $A B C$ and $O_{1} O_{2} M$. Lines $O_{1} O_{2}$ and $A B$ are parallel, both of them being perpendicular to line $t$. Next, $M O_{1}$ is the line of centres of circles ( $C E F$ ) and $\\omega_{1}$ whose common chord is $D E$. Hence $M O_{1}$ bisects $\\angle D M E$ which is the external angle at $M$ in the isosceles triangle $C E M$. It follows that $\\angle D M O_{1}=\\angle D C A$, so that $M O_{1}$ is parallel to $A C$. Likewise, $M O_{2}$ is parallel to $B C$.\nThus the respective sides of triangles $A B C$ and $O_{1} O_{2} M$ are parallel; in addition, these triangles are not congruent. Hence there is a homothety taking $A B C$ to $O_{1} O_{2} M$. The lines $A O_{1}$, $B O_{2}$ and $C M=t$ are concurrent at the centre $Q$ of this homothety.\nFinally, apply Pappus' theorem to the triples of collinear points $A, O, B$ and $O_{2}, D, O_{1}$. The theorem implies that the points $A D \\cap O O_{2}=F$, $A O_{1} \\cap B O_{2}=Q$ and $O O_{1} \\cap B D=E$ are collinear. In other words, line $E F$ passes through the common point $Q$ of $A O_{1}$, $B O_{2}$ and $t$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31844,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $A B C$, let $M_{a}, M_{b}, M_{c}$ be respectively the midpoints of the sides $B C, C A, A B$ and $T_{a}, T_{b}, T_{c}$ be the midpoints of the arcs $B C, C A, A B$ of the circumcircle of $A B C$, not containing the opposite vertices. For $i \\in \\{a, b, c\\}$, let $\\omega_{i}$ be the circle with $M_{i} T_{i}$ as diameter. Let $p_{i}$ be the common external tangent to $\\omega_{j}, \\omega_{k}(\\{i, j, k\\}=\\{a, b, c\\})$ such that $\\omega_{i}$ lies on the opposite side of $p_{i}$ than $\\omega_{j}, \\omega_{k}$ do. Prove that the lines $p_{a}, p_{b}, p_{c}$ form a triangle similar to $A B C$ and find the ratio of similitude.\n\n(Slovakia)",
    "options": [],
    "answer": "1/4",
    "solution": "Let $T_{a} T_{b}$ intersect circle $\\omega_{b}$ at $T_{b}$ and $U$, and let $T_{a} T_{c}$ intersect circle $\\omega_{c}$ at $T_{c}$ and $V$. Further, let $U X$ be the tangent to $\\omega_{b}$ at $U$, with $X$ on $A C$, and let $V Y$ be the tangent to $\\omega_{c}$ at $V$, with $Y$ on $A B$. The homothety with centre $T_{b}$ and ratio $T_{b} T_{a} / T_{b} U$ maps the circle $\\omega_{b}$ onto the circumcircle of $A B C$ and the line $U X$ onto the line tangent to the circumcircle at $T_{a}$, which is parallel to $B C$; thus $U X \\| B C$. The same is true of $V Y$, so that $U X\\|B C\\| V Y$.\n\nLet $T_{a} T_{b}$ cut $A C$ at $P$ and let $T_{a} T_{c}$ cut $A B$ at $Q$. The point $X$ lies on the hypotenuse $P M_{b}$ of the right triangle $P U M_{b}$ and is equidistant from $U$ and $M_{b}$. So $X$ is the midpoint of $M_{b} P$. Similarly $Y$ is the midpoint of $M_{c} Q$.\n\nDenote the incentre of triangle $A B C$ as usual by $I$. It is a known fact that $T_{a} I=T_{a} B$ and $T_{c} I=T_{c} B$. Therefore the points $B$ and $I$ are symmetric across $T_{a} T_{c}$, and consequently $\\angle Q I B=\\angle Q B I=\\angle I B C$. This implies that $B C$ is parallel to the line $I Q$, and likewise, to $I P$. In other words, $P Q$ is the line parallel to $B C$ passing through $I$.\n\n![](attached_image_1.png)\n\nClearly $M_{b} M_{c} \\| B C$. So $P M_{b} M_{c} Q$ is a trapezoid and the segment $X Y$ connects the midpoints of its nonparallel sides; hence $X Y \\| B C$. This combined with the previously established relations $U X\\|B C\\| V Y$ shows that all the four points $U, X, Y, V$ lie on a line which is the common tangent to circles $\\omega_{b}, \\omega_{c}$. Since it leaves these two circles on one side and the circle $\\omega_{a}$ on the other, this line is just the line $p_{a}$ from the problem statement.\n\nLine $p_{a}$ runs midway between $I$ and $M_{b} M_{c}$. Analogous conclusions hold for the lines $p_{b}$ and $p_{c}$. So these three lines form a triangle homothetic from centre $I$ to triangle $M_{a} M_{b} M_{c}$ in ratio $1 / 2$, hence similar to $A B C$ in ratio $1 / 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31845,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that\n$$\n\\angle PAB + \\angle PDC \\leq 90^{\\circ} \\quad \\text{ and } \\quad \\angle PBA + \\angle PCD \\leq 90^{\\circ}.\n$$\nProve that $AB + CD \\geq BC + AD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start with a preliminary observation. Let $T$ be a point inside the quadrilateral $ABCD$. Then:\n$$\n\\begin{align*}\n& \\text{Circles } (BCT) \\text{ and } (DAT) \\text{ are tangent at } T \\\\\n& \\text{if and only if} \\quad \\angle ADT + \\angle BCT = \\angle ATB. \\tag{1}\n\\end{align*}\n$$\nIndeed, if the two circles touch each other then their common tangent at $T$ intersects the segment $AB$ at a point $Z$, and so $\\angle ADT = \\angle ATZ$, $\\angle BCT = \\angle BTZ$, by the tangent-chord theorem. Thus $\\angle ADT + \\angle BCT = \\angle ATZ + \\angle BTZ = \\angle ATB$.\nAnd conversely, if $\\angle ADT + \\angle BCT = \\angle ATB$ then one can draw from $T$ a ray $TZ$ with $Z$ on $AB$ so that $\\angle ADT = \\angle ATZ$, $\\angle BCT = \\angle BTZ$. The first of these equalities implies that $TZ$ is tangent to the circle $(DAT)$; by the second equality, $TZ$ is tangent to the circle $(BCT)$, so the two circles are tangent at $T$.\n\n![](attached_image_1.png)\n\nSo the equivalence (1) is settled. It will be used later on. Now pass to the actual solution. Its key idea is to introduce the circumcircles of triangles $ABP$ and $CDP$ and to consider their second intersection $Q$ (assume for the moment that they indeed meet at two distinct points $P$ and $Q$).\n\nSince the point $A$ lies outside the circle $(BCP)$, we have $\\angle BCP + \\angle BAP < 180^{\\circ}$. Therefore the point $C$ lies outside the circle $(ABP)$. Analogously, $D$ also lies outside that circle. It follows that $P$ and $Q$ lie on the same $\\operatorname{arc} CD$ of the circle $(BCP)$.\n\n![](attached_image_2.png)\n\nBy symmetry, $P$ and $Q$ lie on the same arc $AB$ of the circle $(ABP)$. Thus the point $Q$ lies either inside the angle $BPC$ or inside the angle $APD$. Without loss of generality assume that $Q$ lies inside the angle $BPC$. Then\n$$\n\\begin{equation*}\n\\angle AQD = \\angle PQA + \\angle PQD = \\angle PBA + \\angle PCD \\leq 90^{\\circ}. \\tag{2}\n\\end{equation*}\n$$\nby the condition of the problem.\n\nIn the cyclic quadrilaterals $APQB$ and $DPQC$, the angles at vertices $A$ and $D$ are acute. So their angles at $Q$ are obtuse. This implies that $Q$ lies not only inside the angle $BPC$ but in fact inside the triangle $BPC$, hence also inside the quadrilateral $ABCD$.\n\nNow an argument similar to that used in deriving (2) shows that\n$$\n\\begin{equation*}\n\\angle BQC = \\angle PAB + \\angle PDC \\leq 90^{\\circ}. \\tag{3}\n\\end{equation*}\n$$\nMoreover, since $\\angle PCQ = \\angle PDQ$, we get\n$$\n\\angle ADQ + \\angle BCQ = \\angle ADP + \\angle PDQ + \\angle BCP - \\angle PCQ = \\angle ADP + \\angle BCP.\n$$\nThe last sum is equal to $\\angle APB$, according to the observation (1) applied to $T = P$. And because $\\angle APB = \\angle AQB$, we obtain\n$$\n\\angle ADQ + \\angle BCQ = \\angle AQB.\n$$\nApplying now (1) to $T = Q$ we conclude that the circles $(BCQ)$ and $(DAQ)$ are externally tangent at $Q$. (We have assumed $P \\neq Q$; but if $P = Q$ then the last conclusion holds trivially.)\n\nFinally consider the halfdiscs with diameters $BC$ and $DA$ constructed inwardly to the quadrilateral $ABCD$. They have centres at $M$ and $N$, the midpoints of $BC$ and $DA$ respectively. In view of (2) and (3), these two halfdiscs lie entirely inside the circles $(BQC)$ and $(AQD)$; and since these circles are tangent, the two halfdiscs cannot overlap. Hence $MN \\geq \\frac{1}{2} BC + \\frac{1}{2} DA$.\n\nOn the other hand, since $\\overrightarrow{MN} = \\frac{1}{2}(\\overrightarrow{BA} + \\overrightarrow{CD})$, we have $MN \\leq \\frac{1}{2}(AB + CD)$. Thus indeed $AB + CD \\geq BC + DA$, as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31846,
    "subject": "Mathematics (Multi-modal)",
    "question": "Points $A_{1}$, $B_{1}$, $C_{1}$ are chosen on the sides $BC$, $CA$, $AB$ of a triangle $ABC$, respectively. The circumcircles of triangles $AB_{1}C_{1}$, $BC_{1}A_{1}$, $CA_{1}B_{1}$ intersect the circumcircle of triangle $ABC$ again at points $A_{2}$, $B_{2}$, $C_{2}$, respectively ($A_{2} \\neq A$, $B_{2} \\neq B$, $C_{2} \\neq C$). Points $A_{3}$, $B_{3}$, $C_{3}$ are symmetric to $A_{1}$, $B_{1}$, $C_{1}$ with respect to the midpoints of the sides $BC$, $CA$, $AB$ respectively. Prove that the triangles $A_{2}B_{2}C_{2}$ and $A_{3}B_{3}C_{3}$ are similar.\n(Russia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will work with oriented angles between lines. For two straight lines $\\ell, m$ in the plane, $\\angle(\\ell, m)$ denotes the angle of counterclockwise rotation which transforms line $\\ell$ into a line parallel to $m$ (the choice of the rotation centre is irrelevant). This is a signed quantity; values differing by a multiple of $\\pi$ are identified, so that\n$$\n\\angle(\\ell, m) = -\\angle(m, \\ell), \\quad \\angle(\\ell, m) + \\angle(m, n) = \\angle(\\ell, n).\n$$\nIf $\\ell$ is the line through points $K, L$ and $m$ is the line through $M, N$, one writes $\\angle(KL, MN)$ for $\\angle(\\ell, m)$; the characters $K, L$ are freely interchangeable; and so are $M, N$.\nThe counterpart of the classical theorem about cyclic quadrilaterals is the following: If $K, L, M, N$ are four noncollinear points in the plane then\n$$\nK, L, M, N \\text{ are concyclic if and only if } \\angle(KM, LM) = \\angle(KN, LN). \\tag{1}\n$$\nPassing to the solution proper, we first show that the three circles $(AB_{1}C_{1})$, $(BC_{1}A_{1})$, $(CA_{1}B_{1})$ have a common point. So, let $(AB_{1}C_{1})$ and $(BC_{1}A_{1})$ intersect at the points $C_{1}$ and $P$. Then by (1)\n$$\n\\begin{aligned}\n& \\angle(PA_{1}, CA_{1}) = \\angle(PA_{1}, BA_{1}) = \\angle(PC_{1}, BC_{1}) \\\\\n= & \\angle(PC_{1}, AC_{1}) = \\angle(PB_{1}, AB_{1}) = \\angle(PB_{1}, CB_{1})\n\\end{aligned}\n$$\nDenote this angle by $\\varphi$.\nThe equality between the outer terms shows, again by (1), that the points $A_{1}, B_{1}, P, C$ are concyclic. Thus $P$ is the common point of the three mentioned circles.\nFrom now on the basic property (1) will be used without explicit reference. We have\n$$\n\\varphi = \\angle(PA_{1}, BC) = \\angle(PB_{1}, CA) = \\angle(PC_{1}, AB). \\tag{2}\n$$\n![](attached_image_1.png)\n![](attached_image_2.png)\nLet lines $A_{2}P$, $B_{2}P$, $C_{2}P$ meet the circle $(ABC)$ again at $A_{4}$, $B_{4}$, $C_{4}$, respectively. As\n$$\n\\angle(A_{4}A_{2}, AA_{2}) = \\angle(PA_{2}, AA_{2}) = \\angle(PC_{1}, AC_{1}) = \\angle(PC_{1}, AB) = \\varphi,\n$$\nwe see that line $A_{2}A$ is the image of line $A_{2}A_{4}$ under rotation about $A_{2}$ by the angle $\\varphi$. Hence the point $A$ is the image of $A_{4}$ under rotation by $2\\varphi$ about $O$, the centre of $(ABC)$. The same rotation sends $B_{4}$ to $B$ and $C_{4}$ to $C$. Triangle $ABC$ is the image of $A_{4}B_{4}C_{4}$ in this map. Thus\n$$\n\\angle(A_{4}B_{4}, AB) = \\angle(B_{4}C_{4}, BC) = \\angle(C_{4}A_{4}, CA) = 2\\varphi. \\tag{3}\n$$\nSince the rotation by $2\\varphi$ about $O$ takes $B_{4}$ to $B$, we have $\\angle(AB_{4}, AB) = \\varphi$. Hence by (2)\n$$\n\\angle(AB_{4}, PC_{1}) = \\angle(AB_{4}, AB) + \\angle(AB, PC_{1}) = \\varphi + (-\\varphi) = 0,\n$$\nwhich means that $AB_{4} \\parallel PC_{1}$.\n![](attached_image_3.png)\n![](attached_image_4.png)\nLet $C_{5}$ be the intersection of lines $PC_{1}$ and $A_{4}B_{4}$; define $A_{5}, B_{5}$ analogously. So $AB_{4} \\parallel C_{1}C_{5}$ and, by (3) and (2),\n$$\n\\angle(A_{4}B_{4}, PC_{1}) = \\angle(A_{4}B_{4}, AB) + \\angle(AB, PC_{1}) = 2\\varphi + (-\\varphi) = \\varphi; \\tag{4}\n$$\ni.e., $\\angle(B_{4}C_{5}, C_{5}C_{1}) = \\varphi$. This combined with $\\angle(C_{5}C_{1}, C_{1}A) = \\angle(PC_{1}, AB) = \\varphi$ (see (2)) proves that the quadrilateral $AB_{4}C_{5}C_{1}$ is an isosceles trapezoid with $AC_{1} = B_{4}C_{5}$.\nInterchanging the roles of $A$ and $B$ we infer that also $BC_{1} = A_{4}C_{5}$. And since $AC_{1} + BC_{1} = AB = A_{4}B_{4}$, it follows that the point $C_{5}$ lies on the line segment $A_{4}B_{4}$ and partitions it into segments $A_{4}C_{5}$, $B_{4}C_{5}$ of lengths $BC_{1} (= AC_{3})$ and $AC_{1} (= BC_{3})$. In other words, the rotation which maps triangle $A_{4}B_{4}C_{4}$ onto $ABC$ carries $C_{5}$ onto $C_{3}$. Likewise, it sends $A_{5}$ to $A_{3}$ and $B_{5}$ to $B_{3}$. So the triangles $A_{3}B_{3}C_{3}$ and $A_{5}B_{5}C_{5}$ are congruent. It now suffices to show that the latter is similar to $A_{2}B_{2}C_{2}$.\nLines $B_{4}C_{5}$ and $PC_{5}$ coincide respectively with $A_{4}B_{4}$ and $PC_{1}$. Thus by (4)\n$$\n\\angle(B_{4}C_{5}, PC_{5}) = \\varphi.\n$$\nAnalogously (by cyclic shift) $\\varphi = \\angle(C_{4}A_{5}, PA_{5})$, which rewrites as\n$$\n\\varphi = \\angle(B_{4}A_{5}, PA_{5}).\n$$\nThese relations imply that the points $P, B_{4}, C_{5}, A_{5}$ are concyclic. Analogously, $P, C_{4}, A_{5}, B_{5}$ and $P, A_{4}, B_{5}, C_{5}$ are concyclic quadruples. Therefore\n$$\n\\angle(A_{5}B_{5}, C_{5}B_{5}) = \\angle(A_{5}B_{5}, PB_{5}) + \\angle(PB_{5}, C_{5}B_{5}) = \\angle(A_{5}C_{4}, PC_{4}) + \\angle(PA_{4}, C_{5}A_{4}). \\tag{5}\n$$\nOn the other hand, since the points $A_{2}, B_{2}, C_{2}, A_{4}, B_{4}, C_{4}$ all lie on the circle $(ABC)$, we have\n$$\n\\angle(A_{2}B_{2}, C_{2}B_{2}) = \\angle(A_{2}B_{2}, B_{4}B_{2}) + \\angle(B_{4}B_{2}, C_{2}B_{2}) = \\angle(A_{2}A_{4}, B_{4}A_{4}) + \\angle(B_{4}C_{4}, C_{2}C_{4}). \\tag{6}\n$$\nBut the lines $A_{2}A_{4}, B_{4}A_{4}, B_{4}C_{4}, C_{2}C_{4}$ coincide respectively with $PA_{4}, C_{5}A_{4}, A_{5}C_{4}, PC_{4}$. So the sums on the right-hand sides of (5) and (6) are equal, leading to equality between their left-hand sides: $\\angle(A_{5}B_{5}, C_{5}B_{5}) = \\angle(A_{2}B_{2}, C_{2}B_{2})$. Hence (by cyclic shift, once more) also $\\angle(B_{5}C_{5}, A_{5}C_{5}) = \\angle(B_{2}C_{2}, A_{2}C_{2})$ and $\\angle(C_{5}A_{5}, B_{5}A_{5}) = \\angle(C_{2}A_{2}, B_{2}A_{2})$. This means that the triangles $A_{5}B_{5}C_{5}$ and $A_{2}B_{2}C_{2}$ have their corresponding angles equal, and consequently they are similar.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31847,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(x, y)$ of integers satisfying the equation\n$$\n1 + 2^{x} + 2^{2x+1} = y^{2}\n$$",
    "options": [],
    "answer": "[(0, 2), (0, -2), (4, 23), (4, -23)]",
    "solution": "If $(x, y)$ is a solution then obviously $x \\geq 0$ and $(x, -y)$ is a solution too. For $x = 0$ we get the two solutions $(0, 2)$ and $(0, -2)$.\n\nNow let $(x, y)$ be a solution with $x > 0$; without loss of generality confine attention to $y > 0$. The equation rewritten as\n$$\n2^{x}\\left(1 + 2^{x+1}\\right) = (y - 1)(y + 1)\n$$\nshows that the factors $y - 1$ and $y + 1$ are even, exactly one of them divisible by $4$. Hence $x \\geq 3$ and one of these factors is divisible by $2^{x-1}$ but not by $2^{x}$. So\n$$\ny = 2^{x-1} m + \\epsilon, \\quad m \\text{ odd}, \\quad \\epsilon = \\pm 1\n$$\nPlugging this into the original equation we obtain\n$$\n2^{x}\\left(1 + 2^{x+1}\\right) = \\left(2^{x-1} m + \\epsilon\\right)^{2} - 1 = 2^{2x-2} m^{2} + 2^{x} m \\epsilon\n$$\nor, equivalently\n$$\n1 + 2^{x+1} = 2^{x-2} m^{2} + m \\epsilon\n$$\nTherefore\n$$\n1 - \\epsilon m = 2^{x-2}\\left(m^{2} - 8\\right)\n$$\nFor $\\epsilon = 1$ this yields $m^{2} - 8 \\leq 0$, i.e., $m = 1$, which fails to satisfy the previous equation.\n\nFor $\\epsilon = -1$ equation gives us\n$$\n1 + m = 2^{x-2}\\left(m^{2} - 8\\right) \\geq 2\\left(m^{2} - 8\\right)\n$$\nimplying $2 m^{2} - m - 17 \\leq 0$. Hence $m \\leq 3$; on the other hand $m$ cannot be $1$ by the previous equation. Because $m$ is odd, we obtain $m = 3$, leading to $x = 4$. From the previous formula we get $y = 23$. These values indeed satisfy the given equation. Recall that then $y = -23$ is also good. Thus we have the complete list of solutions $(x, y): (0, 2), (0, -2), (4, 23), (4, -23)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31848,
    "subject": "Mathematics (Multi-modal)",
    "question": "For $x \\in (0,1)$ let $y \\in (0,1)$ be the number whose $n$th digit after the decimal point is the $\\left(2^{n}\\right)$th digit after the decimal point of $x$. Show that if $x$ is rational then so is $y$.\n\n(Canada)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $x$ is rational, its digits repeat periodically starting at some point. We wish to show that this is also true for the digits of $y$, implying that $y$ is rational.\n\nLet $d$ be the length of the period of $x$ and let $d = 2^{u} \\cdot v$, where $v$ is odd. There is a positive integer $w$ such that\n$$\n2^{w} \\equiv 1 \\quad (\\bmod v) .\n$$\n(For instance, one can choose $w$ to be $\\varphi(v)$, the value of Euler's function at $v$.) Therefore\n$$\n2^{n+w} = 2^{n} \\cdot 2^{w} \\equiv 2^{n} \\quad (\\bmod v)\n$$\nfor each $n$. Also, for $n \\geq u$ we have\n$$\n2^{n+w} \\equiv 2^{n} \\equiv 0 \\quad \\left(\\bmod 2^{u}\\right)\n$$\nIt follows that, for all $n \\geq u$, the relation\n$$\n2^{n+w} \\equiv 2^{n} \\quad (\\bmod d)\n$$\nholds. Thus, for $n$ sufficiently large, the $2^{n+w}$th digit of $x$ is in the same spot in the cycle of $x$ as its $2^{n}$th digit, and so these digits are equal. Hence the $(n+w)$th digit of $y$ is equal to its $n$th digit. This means that the digits of $y$ repeat periodically with period $w$ from some point on, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31849,
    "subject": "Mathematics (Multi-modal)",
    "question": "The sequence $f(1), f(2), f(3), \\ldots$ is defined by\n$$\nf(n)=\\frac{1}{n}\\left(\\left\\lfloor\\frac{n}{1}\\right\\rfloor+\\left\\lfloor\\frac{n}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{n}{n}\\right\\rfloor\\right),\n$$\nwhere $\\lfloor x\\rfloor$ denotes the integer part of $x$.\n\na. Prove that $f(n+1)>f(n)$ infinitely often.\n\nb. Prove that $f(n+1)<f(n)$ infinitely often.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $g(n)=n f(n)$ for $n \\geq 1$ and $g(0)=0$. We note that, for $k=1, \\ldots, n$,\n$$\n\\left\\lfloor\\frac{n}{k}\\right\\rfloor-\\left\\lfloor\\frac{n-1}{k}\\right\\rfloor=0\n$$\nif $k$ is not a divisor of $n$ and\n$$\n\\left\\lfloor\\frac{n}{k}\\right\\rfloor-\\left\\lfloor\\frac{n-1}{k}\\right\\rfloor=1\n$$\nif $k$ divides $n$. It therefore follows that if $d(n)$ is the number of positive divisors of $n \\geq 1$ then\n$$\n\\begin{aligned}\ng(n) & =\\left\\lfloor\\frac{n}{1}\\right\\rfloor+\\left\\lfloor\\frac{n}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{n}{n-1}\\right\\rfloor+\\left\\lfloor\\frac{n}{n}\\right\\rfloor \\\\\n& =\\left\\lfloor\\frac{n-1}{1}\\right\\rfloor+\\left\\lfloor\\frac{n-1}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{n-1}{n-1}\\right\\rfloor+\\left\\lfloor\\frac{n-1}{n}\\right\\rfloor+d(n) \\\\\n& =g(n-1)+d(n)\n\\end{aligned}\n$$\nHence\n$$\ng(n)=g(n-1)+d(n)=g(n-2)+d(n-1)+d(n)=\\cdots=d(1)+d(2)+\\cdots+d(n)\n$$\nmeaning that\n$$\nf(n)=\\frac{d(1)+d(2)+\\cdots+d(n)}{n} .\n$$\nIn other words, $f(n)$ is equal to the arithmetic mean of $d(1), d(2), \\ldots, d(n)$. In order to prove the claims, it is therefore sufficient to show that $d(n+1)>f(n)$ and $d(n+1)<f(n)$ both hold infinitely often.\n\nWe note that $d(1)=1$. For $n>1, d(n) \\geq 2$ holds, with equality if and only if $n$ is prime. Since $f(6)=7 / 3>2$, it follows that $f(n)>2$ holds for all $n \\geq 6$.\n\nSince there are infinitely many primes, $d(n+1)=2$ holds for infinitely many values of $n$, and for each such $n \\geq 6$ we have $d(n+1)=2<f(n)$. This proves claim (b).\n\nTo prove (a), notice that the sequence $d(1), d(2), d(3), \\ldots$ is unbounded (e. g. $d\\left(2^{k}\\right)=k+1$ for all $k$ ). Hence $d(n+1)>\\max \\{d(1), d(2), \\ldots, d(n)\\}$ for infinitely many $n$. For all such $n$, we have $d(n+1)>f(n)$. This completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31850,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a polynomial of degree $n>1$ with integer coefficients and let $k$ be any positive integer. Consider the polynomial $Q(x)=P(P(\\ldots P(P(x)) \\ldots))$, with $k$ pairs of parentheses. Prove that $Q$ has no more than $n$ integer fixed points, i.e. integers satisfying the equation $Q(x)=x$.\n(Romania)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The claim is obvious if every integer fixed point of $Q$ is a fixed point of $P$ itself. For the sequel assume that this is not the case. Take any integer $x_{0}$ such that $Q\\left(x_{0}\\right)=x_{0}$, $P\\left(x_{0}\\right) \\neq x_{0}$ and define inductively $x_{i+1}=P\\left(x_{i}\\right)$ for $i=0,1,2, \\ldots$; then $x_{k}=x_{0}$.\nIt is evident that\n$$\n\\begin{equation*}\nP(u)-P(v) \\text{ is divisible by } u-v \\text{ for distinct integers } u, v \\text{. } \\tag{1}\n\\end{equation*}\n$$\n(Indeed, if $P(x)=\\sum a_{i} x^{i}$ then each $a_{i}\\left(u^{i}-v^{i}\\right)$ is divisible by $u-v$.) Therefore each term in the chain of (nonzero) differences\n$$\n\\begin{equation*}\nx_{0}-x_{1}, \\quad x_{1}-x_{2}, \\quad \\ldots, \\quad x_{k-1}-x_{k}, \\quad x_{k}-x_{k+1} \\tag{2}\n\\end{equation*}\n$$\nis a divisor of the next one; and since $x_{k}-x_{k+1}=x_{0}-x_{1}$, all these differences have equal absolute values. For $x_{m}=\\min \\left(x_{1}, \\ldots, x_{k}\\right)$ this means that $x_{m-1}-x_{m}=-\\left(x_{m}-x_{m+1}\\right)$. Thus $x_{m-1}=x_{m+1}\\left(\\neq x_{m}\\right)$. It follows that consecutive differences in the sequence (2) have opposite signs. Consequently, $x_{0}, x_{1}, x_{2}, \\ldots$ is an alternating sequence of two distinct values. In other words, every integer fixed point of $Q$ is a fixed point of the polynomial $P(P(x))$. Our task is to prove that there are at most $n$ such points.\nLet $a$ be one of them so that $b=P(a) \\neq a$ (we have assumed that such an $a$ exists); then $a=P(b)$. Take any other integer fixed point $\\alpha$ of $P(P(x))$ and let $P(\\alpha)=\\beta$, so that $P(\\beta)=\\alpha$; the numbers $\\alpha$ and $\\beta$ need not be distinct ( $\\alpha$ can be a fixed point of $P$ ), but each of $\\alpha, \\beta$ is different from each of $a, b$. Applying property (1) to the four pairs of integers ( $\\alpha, a$ ), ( $\\beta, b$ ), $(\\alpha, b),(\\beta, a)$ we get that the numbers $\\alpha-a$ and $\\beta-b$ divide each other, and also $\\alpha-b$ and $\\beta-a$ divide each other. Consequently\n$$\n\\begin{equation*}\n\\alpha-b= \\pm(\\beta-a), \\quad \\alpha-a= \\pm(\\beta-b) . \\tag{3}\n\\end{equation*}\n$$\nSuppose we have a plus in both instances: $\\alpha-b=\\beta-a$ and $\\alpha-a=\\beta-b$. Subtraction yields $a-b=b-a$, a contradiction, as $a \\neq b$. Therefore at least one equality in (3) holds with a minus sign. For each of them this means that $\\alpha+\\beta=a+b$; equivalently $a+b-\\alpha-P(\\alpha)=0$.\nDenote $a+b$ by $C$. We have shown that every integer fixed point of $Q$ other that $a$ and $b$ is a root of the polynomial $F(x)=C-x-P(x)$. This is of course true for $a$ and $b$ as well. And since $P$ has degree $n>1$, the polynomial $F$ has the same degree, so it cannot have more than $n$ roots. Hence the result.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31851,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer solutions of the equation\n$$\n\\frac{x^{7}-1}{x-1}=y^{5}-1\n$$",
    "options": [],
    "answer": "No integer solutions exist.",
    "solution": "The equation has no integer solutions. To show this, we first prove a lemma.\n\nLemma. If $x$ is an integer and $p$ is a prime divisor of $\\frac{x^{7}-1}{x-1}$ then either $p \\equiv 1 \\pmod{7}$ or $p=7$.\n\nProof. Both $x^{7}-1$ and $x^{p-1}-1$ are divisible by $p$, by hypothesis and by Fermat's little theorem, respectively. Suppose that $7$ does not divide $p-1$. Then $\\gcd(p-1,7)=1$, so there exist integers $k$ and $m$ such that $7k+(p-1)m=1$. We therefore have\n$$\nx \\equiv x^{7k+(p-1)m} \\equiv \\left(x^{7}\\right)^{k} \\cdot \\left(x^{p-1}\\right)^{m} \\equiv 1 \\quad (\\bmod\\ p),\n$$\nand so\n$$\n\\frac{x^{7}-1}{x-1}=1+x+\\cdots+x^{6} \\equiv 7 \\quad (\\bmod\\ p).\n$$\nIt follows that $p$ divides $7$, hence $p=7$ must hold if $p \\equiv 1 \\pmod{7}$ does not, as stated. $\\square$\n\nThe lemma shows that each positive divisor $d$ of $\\frac{x^{7}-1}{x-1}$ satisfies either $d \\equiv 0 \\pmod{7}$ or $d \\equiv 1 \\pmod{7}$.\n\nNow assume that $(x, y)$ is an integer solution of the original equation. Notice that $y-1>0$, because $\\frac{x^{7}-1}{x-1}>0$ for all $x \\neq 1$. Since $y-1$ divides $\\frac{x^{7}-1}{x-1}=y^{5}-1$, we have $y \\equiv 1 \\pmod{7}$ or $y \\equiv 2 \\pmod{7}$ by the previous paragraph. In the first case, $1+y+y^{2}+y^{3}+y^{4} \\equiv 5 \\pmod{7}$, and in the second $1+y+y^{2}+y^{3}+y^{4} \\equiv 3 \\pmod{7}$. Both possibilities contradict the fact that the positive divisor $1+y+y^{2}+y^{3}+y^{4}$ of $\\frac{x^{7}-1}{x-1}$ is congruent to $0$ or $1$ modulo $7$. So the given equation has no integer solutions.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31852,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a > b > 1$ be relatively prime positive integers. Define the weight of an integer $c$, denoted by $w(c)$, to be the minimal possible value of $|x| + |y|$ taken over all pairs of integers $x$ and $y$ such that\n$$\na x + b y = c.\n$$\nAn integer $c$ is called a local champion if $w(c) \\geq w(c \\pm a)$ and $w(c) \\geq w(c \\pm b)$.\nFind all local champions and determine their number.",
    "options": [],
    "answer": "All local champions are exactly the integers of the form ±(a x − b y) with positive integers x, y satisfying x < b and x + y = floor((a + b) / 2). The total number of local champions is b − 1 if both a and b are odd, and 2(b − 1) otherwise.",
    "solution": "Call the pair of integers $(x, y)$ a representation of $c$ if $a x + b y = c$ and $|x| + |y|$ has the smallest possible value, i.e. $|x| + |y| = w(c)$.\nWe characterise the local champions by the following three observations.\n\nLemma 1. If $(x, y)$ a representation of a local champion $c$ then $x y < 0$.\n\nProof. Suppose indirectly that $x \\geq 0$ and $y \\geq 0$ and consider the values $w(c)$ and $w(c + a)$. All representations of the numbers $c$ and $c + a$ in the form $a u + b v$ can be written as\n$$\nc = a(x - k b) + b(y + k a), \\quad c + a = a(x + 1 - k b) + b(y + k a)\n$$\nwhere $k$ is an arbitrary integer.\nSince $|x| + |y|$ is minimal, we have\n$$\nx + y = |x| + |y| \\leq |x - k b| + |y + k a|\n$$\nfor all $k$. On the other hand, $w(c + a) \\leq w(c)$, so there exists a $k$ for which\n$$\n|x + 1 - k b| + |y + k a| \\leq |x| + |y| = x + y.\n$$\nThen\n$$\n(x + 1 - k b) + (y + k a) \\leq |x + 1 - k b| + |y + k a| \\leq x + y \\leq |x - k b| + |y + k a|.\n$$\nComparing the first and the third expressions, we find $k(a - b) + 1 \\leq 0$ implying $k < 0$. Comparing the second and fourth expressions, we get $|x + 1 - k b| \\leq |x - k b|$, therefore $k b > x$; this is a contradiction.\nIf $x, y \\leq 0$ then we can switch to $-c, -x$ and $-y$. $\\square$\n\nFrom this point, write $c = a x - b y$ instead of $c = a x + b y$ and consider only those cases where $x$ and $y$ are nonzero and have the same sign. By Lemma 1, there is no loss of generality in doing so.\n\nLemma 2. Let $c = a x - b y$ where $|x| + |y|$ is minimal and $x, y$ have the same sign. The number $c$ is a local champion if and only if $|x| < b$ and $|x| + |y| = \\left\\lfloor \\frac{a + b}{2} \\right\\rfloor$.\n\nProof. Without loss of generality we may assume $x, y > 0$.\nThe numbers $c - a$ and $c + b$ can be written as\n$$\nc - a = a(x - 1) - b y \\quad \\text{ and } \\quad c + b = a x - b(y - 1)\n$$\nand trivially $w(c - a) \\leq (x - 1) + y < w(c)$ and $w(c + b) \\leq x + (y - 1) < w(c)$ in all cases.\nNow assume that $c$ is a local champion and consider $w(c + a)$. Since $w(c + a) \\leq w(c)$, there exists an integer $k$ such that\n$$\nc + a = a(x + 1 - k b) - b(y - k a) \\quad \\text{ and } \\quad |x + 1 - k b| + |y - k a| \\leq x + y.\n$$\nThis inequality cannot hold if $k \\leq 0$, therefore $k > 0$. We prove that we can choose $k = 1$.\nConsider the function $f(t) = |x + 1 - b t| + |y - a t| - (x + y)$. This is a convex function and we have $f(0) = 1$ and $f(k) \\leq 0$. By Jensen's inequality, $f(1) \\leq \\left(1 - \\frac{1}{k}\\right) f(0) + \\frac{1}{k} f(k) < 1$. But $f(1)$ is an integer. Therefore $f(1) \\leq 0$ and\n$$\n|x + 1 - b| + |y - a| \\leq x + y.\n$$\nKnowing $c = a(x - b) - b(y - a)$, we also have\n$$\nx + y \\leq |x - b| + |y - a|.\n$$\nCombining the two inequalities yields $|x + 1 - b| \\leq |x - b|$ which is equivalent to $x < b$.\nConsidering $w(c - b)$, we obtain similarly that $y < a$.\nNow $|x - b| = b - x, |x + 1 - b| = b - x - 1$ and $|y - a| = a - y$, therefore we have\n$$\n\\begin{aligned}\n(b - x - 1) + (a - y) & \\leq x + y \\leq (b - x) + (a - y), \\\\\n\\frac{a + b - 1}{2} & \\leq x + y \\leq \\frac{a + b}{2}.\n\\end{aligned}\n$$\nHence $x + y = \\left\\lfloor \\frac{a + b}{2} \\right\\rfloor$.\nTo prove the opposite direction, assume $0 < x < b$ and $x + y = \\left\\lfloor \\frac{a + b}{2} \\right\\rfloor$. Since $a > b$, we also have $0 < y < a$. Then\n$$\nw(c + a) \\leq |x + 1 - b| + |y - a| = a + b - 1 - (x + y) \\leq x + y = w(c)\n$$\nand\n$$\nw(c - b) \\leq |x - b| + |y + 1 - a| = a + b - 1 - (x + y) \\leq x + y = w(c)\n$$\ntherefore $c$ is a local champion indeed. $\\square$\n\nLemma 3. Let $c = a x - b y$ and assume that $x$ and $y$ have the same sign, $|x| < b, |y| < a$ and $|x| + |y| = \\left\\lfloor \\frac{a + b}{2} \\right\\rfloor$. Then $w(c) = x + y$.\n\nProof. By definition $w(c) = \\min \\{|x - k b| + |y - k a|: k \\in \\mathbb{Z}\\}$. If $k \\leq 0$ then obviously $|x - k b| + |y - k a| \\geq x + y$. If $k \\geq 1$ then\n$$\n|x - k b| + |y - k a| = (k b - x) + (k a - y) = k(a + b) - (x + y) \\geq (2k - 1)(x + y) \\geq x + y.\n$$\nTherefore $w(c) = x + y$ indeed. $\\square$\n\nLemmas 1, 2 and 3 together yield that the set of local champions is\n$$\nC = \\left\\{ \\pm(a x - b y): 0 < x < b, x + y = \\left\\lfloor \\frac{a + b}{2} \\right\\rfloor \\right\\}.\n$$\nDenote by $C^{+}$ and $C^{-}$ the two sets generated by the expressions $+(a x - b y)$ and $-(a x - b y)$, respectively. It is easy to see that both sets are arithmetic progressions of length $b - 1$, with difference $a + b$.\nIf $a$ and $b$ are odd, then $C^{+} = C^{-}$, because $a(-x) - b(-y) = a(b - x) - b(a - y)$ and $x + y = \\frac{a + b}{2}$ is equivalent to $(b - x) + (a - y) = \\frac{a + b}{2}$. In this case there exist $b - 1$ local champions.\nIf $a$ and $b$ have opposite parities then the answer is different. For any $c_1 \\in C^{+}$ and $c_2 \\in C^{-}$,\n$$\n2 c_1 \\equiv -2 c_2 \\equiv 2\\left(a \\frac{a + b - 1}{2} - b \\cdot 0\\right) \\equiv -a \\pmod{a + b}\n$$\nand\n$$\n2 c_1 - 2 c_2 \\equiv -2 a \\quad (\\bmod a + b)\n$$\nThe number $a + b$ is odd and relatively prime to $a$, therefore the elements of $C^{+}$ and $C^{-}$ belong to two different residue classes modulo $a + b$. Hence, the set $C$ is the union of two disjoint arithmetic progressions and the number of all local champions is $2(b - 1)$.\nSo the number of local champions is $b - 1$ if both $a$ and $b$ are odd and $2(b - 1)$ otherwise.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31853,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that, for every positive integer $n$, there exists an integer $m$ such that $2^{m}+m$ is divisible by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove by induction on $d$ that, for every positive integer $N$, there exist positive integers $b_{0}, b_{1}, \\ldots, b_{d-1}$ such that, for each $i=0,1,2, \\ldots, d-1$, we have $b_{i}>N$ and\n$$\n2^{b_{i}}+b_{i} \\equiv i \\quad(\\bmod d) .\n$$\nThis yields the claim for $m=b_{0}$.\n\nThe base case $d=1$ is trivial. Take an $a>1$ and assume that the statement holds for all $d<a$. Note that the remainders of $2^{i}$ modulo $a$ repeat periodically starting with some exponent $M$. Let $k$ be the length of the period; this means that $2^{M+k'} \\equiv 2^{M}(\\bmod a)$ holds only for those $k'$ which are multiples of $k$. Note further that the period cannot contain all the $a$ remainders, since $0$ either is missing or is the only number in the period. Thus $k<a$.\n\nLet $d=\\operatorname{gcd}(a, k)$ and let $a'=a / d, k'=k / d$. Since $0<k<a$, we also have $0<d<a$. By the induction hypothesis, there exist positive integers $b_{0}, b_{1}, \\ldots, b_{d-1}$ such that $b_{i}>\\max \\left(2^{M}, N\\right)$ and\n$$\n\\begin{equation*}\n2^{b_{i}}+b_{i} \\equiv i \\quad(\\bmod d) \\quad \\text{ for } \\quad i=0,1,2, \\ldots, d-1 . \\tag{1}\n\\end{equation*}\n$$\nFor each $i=0,1, \\ldots, d-1$ consider the sequence\n$$\n\\begin{equation*}\n2^{b_{i}}+b_{i}, \\quad 2^{b_{i}+k}+\\left(b_{i}+k\\right), \\ldots, \\quad 2^{b_{i}+\\left(a'-1\\right) k}+\\left(b_{i}+\\left(a'-1\\right) k\\right) . \\tag{2}\n\\end{equation*}\n$$\nModulo $a$, these numbers are congruent to\n$$\n2^{b_{i}}+b_{i}, \\quad 2^{b_{i}}+\\left(b_{i}+k\\right), \\quad \\ldots, \\quad 2^{b_{i}}+\\left(b_{i}+\\left(a'-1\\right) k\\right),\n$$\nrespectively. The $d$ sequences contain $a' d=a$ numbers altogether. We shall now prove that no two of these numbers are congruent modulo $a$.\n\nSuppose that\n$$\n\\begin{equation*}\n2^{b_{i}}+\\left(b_{i}+m k\\right) \\equiv 2^{b_{j}}+\\left(b_{j}+n k\\right)(\\bmod a) \\tag{3}\n\\end{equation*}\n$$\nfor some values of $i, j \\in\\{0,1, \\ldots, d-1\\}$ and $m, n \\in\\left\\{0,1, \\ldots, a'-1\\right\\}$. Since $d$ is a divisor of $a$, we also have\n$$\n2^{b_{i}}+\\left(b_{i}+m k\\right) \\equiv 2^{b_{j}}+\\left(b_{j}+n k\\right) \\quad(\\bmod d) .\n$$\nBecause $d$ is a divisor of $k$ and in view of (1), we obtain $i \\equiv j(\\bmod d)$. As $i, j \\in\\{0,1, \\ldots, d-1\\}$, this just means that $i=j$. Substituting this into (3) yields $m k \\equiv n k(\\bmod a)$. Therefore $m k' \\equiv n k'\\left(\\bmod a'\\right)$; and since $a'$ and $k'$ are coprime, we get $m \\equiv n\\left(\\bmod a'\\right)$. Hence also $m=n$.\n\nIt follows that the $a$ numbers that make up the $d$ sequences (2) satisfy all the requirements; they are certainly all greater than $N$ because we chose each $b_{i}>\\max \\left(2^{M}, N\\right)$. So the statement holds for $a$, completing the induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31854,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a sequence $a_{1}, a_{2}, \\ldots, a_{n}$ of real numbers. For each $i$ ($1 \\leq i \\leq n$) define\n$$\nd_{i}=\\max \\{a_{j}: 1 \\leq j \\leq i\\}-\\min \\{a_{j}: i \\leq j \\leq n\\}\n$$\nand let\n$$\nd=\\max \\{d_{i}: 1 \\leq i \\leq n\\} .\n$$\n\na. Prove that for arbitrary real numbers $x_{1} \\leq x_{2} \\leq \\ldots \\leq x_{n}$,\n$$\n\\begin{equation*}\n\\max \\{\\left|x_{i}-a_{i}\\right|: 1 \\leq i \\leq n\\} \\geq \\frac{d}{2} . \\tag{1}\n\\end{equation*}\n$$\n\nb. Show that there exists a sequence $x_{1} \\leq x_{2} \\leq \\ldots \\leq x_{n}$ of real numbers such that we have equality in (1).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) Let $1 \\leq p \\leq q \\leq r \\leq n$ be indices for which\n$$\nd=d_{q}, \\quad a_{p}=\\max \\{a_{j}: 1 \\leq j \\leq q\\}, \\quad a_{r}=\\min \\{a_{j}: q \\leq j \\leq n\\}\n$$\nand thus $d=a_{p}-a_{r}$. (These indices are not necessarily unique.)\n![](attached_image_1.png)\nFor arbitrary real numbers $x_{1} \\leq x_{2} \\leq \\ldots \\leq x_{n}$, consider just the two quantities $\\left|x_{p}-a_{p}\\right|$ and $\\left|x_{r}-a_{r}\\right|$. Since\n$$\n\\left(a_{p}-x_{p}\\right)+\\left(x_{r}-a_{r}\\right)=\\left(a_{p}-a_{r}\\right)+\\left(x_{r}-x_{p}\\right) \\geq a_{p}-a_{r}=d,\n$$\nwe have either $a_{p}-x_{p} \\geq \\frac{d}{2}$ or $x_{r}-a_{r} \\geq \\frac{d}{2}$. Hence,\n$$\n\\max \\{\\left|x_{i}-a_{i}\\right|: 1 \\leq i \\leq n\\} \\geq \\max \\{\\left|x_{p}-a_{p}\\right|,\\left|x_{r}-a_{r}\\right|\\} \\geq \\max \\{a_{p}-x_{p}, x_{r}-a_{r}\\} \\geq \\frac{d}{2}\n$$\n\n(b) Define the sequence $\\left(x_{k}\\right)$ as\n$$\nx_{1}=a_{1}-\\frac{d}{2}, \\quad x_{k}=\\max \\{x_{k-1}, a_{k}-\\frac{d}{2}\\} \\quad \\text{ for } 2 \\leq k \\leq n .\n$$\nWe show that we have equality in (1) for this sequence.\nBy the definition, sequence ( $x_{k}$ ) is non-decreasing and $x_{k}-a_{k} \\geq-\\frac{d}{2}$ for all $1 \\leq k \\leq n$. Next we prove that\n$$\n\\begin{equation*}\nx_{k}-a_{k} \\leq \\frac{d}{2} \\quad \\text{ for all } 1 \\leq k \\leq n \\tag{2}\n\\end{equation*}\n$$\nConsider an arbitrary index $1 \\leq k \\leq n$. Let $\\ell \\leq k$ be the smallest index such that $x_{k}=x_{\\ell}$. We have either $\\ell=1$, or $\\ell \\geq 2$ and $x_{\\ell}>x_{\\ell-1}$. In both cases,\n$$\n\\begin{equation*}\nx_{k}=x_{\\ell}=a_{\\ell}-\\frac{d}{2} \\tag{3}\n\\end{equation*}\n$$\nSince\n$$\na_{\\ell}-a_{k} \\leq \\max \\{a_{j}: 1 \\leq j \\leq k\\}-\\min \\{a_{j}: k \\leq j \\leq n\\}=d_{k} \\leq d\n$$\nequality (3) implies\n$$\nx_{k}-a_{k}=a_{\\ell}-a_{k}-\\frac{d}{2} \\leq d-\\frac{d}{2}=\\frac{d}{2}\n$$\nWe obtained that $-\\frac{d}{2} \\leq x_{k}-a_{k} \\leq \\frac{d}{2}$ for all $1 \\leq k \\leq n$, so\n$$\n\\max \\{\\left|x_{i}-a_{i}\\right|: 1 \\leq i \\leq n\\} \\leq \\frac{d}{2}\n$$\nWe have equality because $\\left|x_{1}-a_{1}\\right|=\\frac{d}{2}$.\nWe present another construction of a sequence ( $x_{i}$ ) for part (b).\nFor each $1 \\leq i \\leq n$, let\n$$\nM_{i}=\\max \\{a_{j}: 1 \\leq j \\leq i\\} \\quad \\text{ and } \\quad m_{i}=\\min \\{a_{j}: i \\leq j \\leq n\\} .\n$$\nFor all $1 \\leq i<n$, we have\n$$\nM_{i}=\\max \\{a_{1}, \\ldots, a_{i}\\} \\leq \\max \\{a_{1}, \\ldots, a_{i}, a_{i+1}\\}=M_{i+1}\n$$\nand\n$$\nm_{i}=\\min \\{a_{i}, a_{i+1}, \\ldots, a_{n}\\} \\leq \\min \\{a_{i+1}, \\ldots, a_{n}\\}=m_{i+1} .\n$$\nTherefore sequences ( $M_{i}$ ) and ( $m_{i}$ ) are non-decreasing. Moreover, since $a_{i}$ is listed in both definitions,\n$$\nm_{i} \\leq a_{i} \\leq M_{i}\n$$\nTo achieve equality in (1), set\n$$\nx_{i}=\\frac{M_{i}+m_{i}}{2}\n$$\nSince sequences ( $M_{i}$ ) and ( $m_{i}$ ) are non-decreasing, this sequence is non-decreasing as well.\nFrom $d_{i}=M_{i}-m_{i}$ we obtain that\n$$\n-\\frac{d_{i}}{2}=\\frac{m_{i}-M_{i}}{2}=x_{i}-M_{i} \\leq x_{i}-a_{i} \\leq x_{i}-m_{i}=\\frac{M_{i}-m_{i}}{2}=\\frac{d_{i}}{2}\n$$\nTherefore\n$$\n\\max \\{\\left|x_{i}-a_{i}\\right|: 1 \\leq i \\leq n\\} \\leq \\max \\{\\frac{d_{i}}{2}: 1 \\leq i \\leq n\\}=\\frac{d}{2}\n$$\nSince the opposite inequality has been proved in part (a), we must have equality.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31855,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider those functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ which satisfy the condition\n$$\nf(m+n) \\geq f(m)+f(f(n))-1\n$$\nfor all $m, n \\in \\mathbb{N}$. Find all possible values of $f(2007)$.\n\n($\\mathbb{N}$ denotes the set of all positive integers.)",
    "options": [],
    "answer": "all integers from 1 to 2008 inclusive",
    "solution": "Suppose that a function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfies (1). For arbitrary positive integers $m>n$, by (1) we have\n$$\nf(m)=f(n+(m-n)) \\geq f(n)+f(f(m-n))-1 \\geq f(n),\n$$\nso $f$ is nondecreasing.\n\nFunction $f \\equiv 1$ is an obvious solution. To find other solutions, assume that $f \\not \\equiv 1$ and take the smallest $a \\in \\mathbb{N}$ such that $f(a)>1$. Then $f(b) \\geq f(a)>1$ for all integer $b \\geq a$.\n\nSuppose that $f(n)>n$ for some $n \\in \\mathbb{N}$. Then we have\n$$\nf(f(n))=f((f(n)-n)+n) \\geq f(f(n)-n)+f(f(n))-1\n$$\nso $f(f(n)-n) \\leq 1$ and hence $f(n)-n<a$. Then there exists a maximal value of the expression $f(n)-n$; denote this value by $c$, and let $f(k)-k=c \\geq 1$. Applying the monotonicity together with (1), we get\n$$\n\\begin{aligned}\n2 k+c \\geq f(2 k)=f(k+k) & \\geq f(k)+f(f(k))-1 \\\\\n& \\geq f(k)+f(k)-1=2(k+c)-1=2 k+(2 c-1),\n\\end{aligned}\n$$\nhence $c \\leq 1$ and $f(n) \\leq n+1$ for all $n \\in \\mathbb{N}$. In particular, $f(2007) \\leq 2008$.\n\nNow we present a family of examples showing that all values from 1 to 2008 can be realized. Let\n$$\nf_{j}(n)=\\max \\{1, n+j-2007\\} \\quad \\text{ for } j=1,2, \\ldots, 2007 ; \\quad f_{2008}(n)= \\begin{cases}n, & 2007 \\nmid n, \\\\ n+1, & 2007 \\mid n .\\end{cases}\n$$\nWe show that these functions satisfy the condition (1) and clearly $f_{j}(2007)=j$.\n\nTo check the condition (1) for the function $f_{j}(j \\leq 2007)$, note first that $f_{j}$ is nondecreasing and $f_{j}(n) \\leq n$, hence $f_{j}\\left(f_{j}(n)\\right) \\leq f_{j}(n) \\leq n$ for all $n \\in \\mathbb{N}$. Now, if $f_{j}(m)=1$, then the inequality (1) is clear since $f_{j}(m+n) \\geq f_{j}(n) \\geq f_{j}\\left(f_{j}(n)\\right)=f_{j}(m)+f_{j}\\left(f_{j}(n)\\right)-1$. Otherwise,\n$$\nf_{j}(m)+f_{j}\\left(f_{j}(n)\\right)-1 \\leq(m+j-2007)+n=(m+n)+j-2007=f_{j}(m+n) .\n$$\nIn the case $j=2008$, clearly $n+1 \\geq f_{2008}(n) \\geq n$ for all $n \\in \\mathbb{N}$; moreover, $n+1 \\geq f_{2008}\\left(f_{2008}(n)\\right)$ as well. Actually, the latter is trivial if $f_{2008}(n)=n$; otherwise, $f_{2008}(n)=n+1$, which implies $2007 \\nmid n+1$ and hence $n+1=f_{2008}(n+1)=f_{2008}\\left(f_{2008}(n)\\right)$.\n\nSo, if $2007 \\mid m+n$, then\n$$\nf_{2008}(m+n)=m+n+1=(m+1)+(n+1)-1 \\geq f_{2008}(m)+f_{2008}\\left(f_{2008}(n)\\right)-1 .\n$$\nOtherwise, $2007 \\nmid m+n$, hence $2007 \\nmid m$ or $2007 \\nmid n$. In the former case we have $f_{2008}(m)=m$, while in the latter one $f_{2008}\\left(f_{2008}(n)\\right)=f_{2008}(n)=n$, providing\n$$\nf_{2008}(m)+f_{2008}\\left(f_{2008}(n)\\right)-1 \\leq(m+n+1)-1=f_{2008}(m+n) \\text{.}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31856,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $x$ and $y$ be positive real numbers such that $x^{n}+y^{n}=1$. Prove that\n$$\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right)<\\frac{1}{(1-x)(1-y)} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For each real $t \\in (0,1)$,\n$$\n\\frac{1+t^{2}}{1+t^{4}}=\\frac{1}{t}-\\frac{(1-t)\\left(1-t^{3}\\right)}{t\\left(1+t^{4}\\right)}<\\frac{1}{t}\n$$\nSubstituting $t=x^{k}$ and $t=y^{k}$,\n$$\n0<\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}<\\sum_{k=1}^{n} \\frac{1}{x^{k}}=\\frac{1-x^{n}}{x^{n}(1-x)} \\quad \\text{ and } \\quad 0<\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}<\\sum_{k=1}^{n} \\frac{1}{y^{k}}=\\frac{1-y^{n}}{y^{n}(1-y)} .\n$$\nSince $1-y^{n}=x^{n}$ and $1-x^{n}=y^{n}$,\n$$\n\\frac{1-x^{n}}{x^{n}(1-x)}=\\frac{y^{n}}{x^{n}(1-x)}, \\quad \\frac{1-y^{n}}{y^{n}(1-y)}=\\frac{x^{n}}{y^{n}(1-y)}\n$$\nand therefore\n$$\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right)<\\frac{y^{n}}{x^{n}(1-x)} \\cdot \\frac{x^{n}}{y^{n}(1-y)}=\\frac{1}{(1-x)(1-y)} .\n$$\nWe prove\n$$\n\\begin{equation*}\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right)<\\frac{\\left(\\frac{1+\\sqrt{2}}{2} \\ln 2\\right)^{2}}{(1-x)(1-y)}<\\frac{0.7001}{(1-x)(1-y)} \\tag{1}\n\\end{equation*}\n$$\nThe idea is to estimate each term on the left-hand side with the same constant. To find the upper bound for the expression $\\frac{1+x^{2 k}}{1+x^{4 k}}$, consider the function $f(t)=\\frac{1+t}{1+t^{2}}$ in interval $(0,1)$. Since\n$$\nf^{\\prime}(t)=\\frac{1-2 t-t^{2}}{\\left(1+t^{2}\\right)^{2}}=\\frac{(\\sqrt{2}+1+t)(\\sqrt{2}-1-t)}{\\left(1+t^{2}\\right)^{2}}\n$$\nthe function increases in interval $(0, \\sqrt{2}-1]$ and decreases in $[\\sqrt{2}-1,1)$. Therefore the maximum is at point $t_{0}=\\sqrt{2}-1$ and\n$$\nf(t)=\\frac{1+t}{1+t^{2}} \\leq f\\left(t_{0}\\right)=\\frac{1+\\sqrt{2}}{2}=\\alpha\n$$\nApplying this to each term on the left-hand side of (1), we obtain\n$$\n\\begin{equation*}\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right) \\leq n \\alpha \\cdot n \\alpha=(n \\alpha)^{2} . \\tag{2}\n\\end{equation*}\n$$\nTo estimate $(1-x)(1-y)$ on the right-hand side, consider the function\n$$\ng(t)=\\ln \\left(1-t^{1 / n}\\right)+\\ln \\left(1-(1-t)^{1 / n}\\right)\n$$\nSubstituting $s$ for $1-t$, we have\n$$\n-n g^{\\prime}(t)=\\frac{t^{1 / n-1}}{1-t^{1 / n}}-\\frac{s^{1 / n-1}}{1-s^{1 / n}}=\\frac{1}{s t}\\left(\\frac{(1-t) t^{1 / n}}{1-t^{1 / n}}-\\frac{(1-s) s^{1 / n}}{1-s^{1 / n}}\\right)=\\frac{h(t)-h(s)}{s t} .\n$$\nThe function\n$$\nh(t)=t^{1 / n} \\frac{1-t}{1-t^{1 / n}}=\\sum_{i=1}^{n} t^{i / n}\n$$\nis obviously increasing for $t \\in (0,1)$, hence for these values of $t$ we have\n$$\ng^{\\prime}(t)>0 \\Longleftrightarrow h(t)<h(s) \\Longleftrightarrow t<s=1-t \\Longleftrightarrow t<\\frac{1}{2} .\n$$\nThen, the maximum of $g(t)$ in ( 0,1 ) is attained at point $t_{1}=1 / 2$ and therefore\n$$\ng(t) \\leq g\\left(\\frac{1}{2}\\right)=2 \\ln \\left(1-2^{-1 / n}\\right), \\quad t \\in (0,1)\n$$\nSubstituting $t=x^{n}$, we have $1-t=y^{n},(1-x)(1-y)=\\exp g(t)$ and hence\n$$\n\\begin{equation*}\n(1-x)(1-y)=\\exp g(t) \\leq\\left(1-2^{-1 / n}\\right)^{2} \\tag{3}\n\\end{equation*}\n$$\nCombining (2) and (3), we get\n$$\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right) \\leq(\\alpha n)^{2} \\cdot 1 \\leq(\\alpha n)^{2} \\frac{\\left(1-2^{-1 / n}\\right)^{2}}{(1-x)(1-y)}=\\frac{\\left(\\alpha n\\left(1-2^{-1 / n}\\right)\\right)^{2}}{(1-x)(1-y)} .\n$$\nApplying the inequality $1-\\exp (-t)<t$ for $t=\\frac{\\ln 2}{n}$, we obtain\n$$\n\\alpha n\\left(1-2^{-1 / n}\\right)=\\alpha n\\left(1-\\exp \\left(-\\frac{\\ln 2}{n}\\right)\\right)<\\alpha n \\cdot \\frac{\\ln 2}{n}=\\alpha \\ln 2=\\frac{1+\\sqrt{2}}{2} \\ln 2 .\n$$\nHence,\n$$\n\\left(\\sum_{k=1}^{n} \\frac{1+x^{2 k}}{1+x^{4 k}}\\right)\\left(\\sum_{k=1}^{n} \\frac{1+y^{2 k}}{1+y^{4 k}}\\right)<\\frac{\\left(\\frac{1+\\sqrt{2}}{2} \\ln 2\\right)^{2}}{(1-x)(1-y)}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31857,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}^{+}$ such that\n$$\nf(x+f(y))=f(x+y)+f(y)\n$$\nfor all $x, y \\in \\mathbb{R}^{+}$. (Symbol $\\mathbb{R}^{+}$ denotes the set of all positive real numbers.)",
    "options": [],
    "answer": "f(x) = 2x for all positive real x",
    "solution": "First we show that $f(y)>y$ for all $y \\in \\mathbb{R}^{+}$. Functional equation (1) yields $f(x+f(y))>f(x+y)$ and hence $f(y) \\neq y$ immediately. If $f(y)<y$ for some $y$, then setting $x=y-f(y)$ we get\n$$\nf(y)=f((y-f(y))+f(y))=f((y-f(y))+y)+f(y)>f(y),\n$$\ncontradiction. Therefore $f(y)>y$ for all $y \\in \\mathbb{R}^{+}$.\nFor $x \\in \\mathbb{R}^{+}$ define $g(x)=f(x)-x$; then $f(x)=g(x)+x$ and, as we have seen, $g(x)>0$. Transforming (1) for function $g(x)$ and setting $t=x+y$,\n$$\n\\begin{aligned}\nf(t+g(y)) & =f(t)+f(y) \\\\\ng(t+g(y))+t+g(y) & =(g(t)+t)+(g(y)+y)\n\\end{aligned}\n$$\nand therefore\n$$\n\\begin{equation*}\ng(t+g(y))=g(t)+y \\quad \\text{ for all } t>y>0 \\tag{2}\n\\end{equation*}\n$$\nNext we prove that function $g(x)$ is injective. Suppose that $g\\left(y_{1}\\right)=g\\left(y_{2}\\right)$ for some numbers $y_{1}, y_{2} \\in \\mathbb{R}^{+}$. Then by (2),\n$$\ng(t)+y_{1}=g\\left(t+g\\left(y_{1}\\right)\\right)=g\\left(t+g\\left(y_{2}\\right)\\right)=g(t)+y_{2}\n$$\nfor all $t>\\max \\{y_{1}, y_{2}\\}$. Hence, $g\\left(y_{1}\\right)=g\\left(y_{2}\\right)$ is possible only if $y_{1}=y_{2}$.\nNow let $u, v$ be arbitrary positive numbers and $t>u+v$. Applying (2) three times,\n$$\ng(t+g(u)+g(v))=g(t+g(u))+v=g(t)+u+v=g(t+g(u+v))\n$$\nBy the injective property we conclude that $t+g(u)+g(v)=t+g(u+v)$, hence\n$$\n\\begin{equation*}\ng(u)+g(v)=g(u+v) \\tag{3}\n\\end{equation*}\n$$\nSince function $g(v)$ is positive, equation (3) also shows that $g$ is an increasing function.\nFinally we prove that $g(x)=x$. Combining (2) and (3), we obtain\n$$\ng(t)+y=g(t+g(y))=g(t)+g(g(y))\n$$\nand hence\n$$\ng(g(y))=y\n$$\nSuppose that there exists an $x \\in \\mathbb{R}^{+}$ such that $g(x) \\neq x$. By the monotonicity of $g$, if $x>g(x)$ then $g(x)>g(g(x))=x$. Similarly, if $x<g(x)$ then $g(x)<g(g(x))=x$. Both cases lead to contradiction, so there exists no such $x$.\nWe have proved that $g(x)=x$ and therefore $f(x)=g(x)+x=2 x$ for all $x \\in \\mathbb{R}^{+}$. This function indeed satisfies the functional equation (1).\nWe prove that $f(y)>y$ and introduce function $g(x)=f(x)-x>0$ in the same way as in Solution 1.\nFor arbitrary $t>y>0$, substitute $x=t-y$ into (1) to obtain\n$$\nf(t+g(y))=f(t)+f(y)\n$$\nwhich, by induction, implies\n$$\n\\begin{equation*}\nf(t+n g(y))=f(t)+n f(y) \\quad \\text{ for all } t>y>0, n \\in \\mathbb{N} . \\tag{4}\n\\end{equation*}\n$$\nTake two arbitrary positive reals $y$ and $z$ and a third fixed number $t>\\max \\{y, z\\}$. For each positive integer $k$, let $\\ell_{k}=\\left\\lfloor k \\frac{g(y)}{g(z)}\\right\\rfloor$. Then $t+k g(y)-\\ell_{k} g(z) \\geq t>z$ and, applying (4) twice,\n$$\n\\begin{aligned}\nf\\left(t+k g(y)-\\ell_{k} g(z)\\right)+\\ell_{k} f(z) & =f(t+k g(y))=f(t)+k f(y) \\\\\n0 & <\\frac{1}{k} f\\left(t+k g(y)-\\ell_{k} g(z)\\right)=\\frac{f(t)}{k}+f(y)-\\frac{\\ell_{k}}{k} f(z)\n\\end{aligned}\n$$\nAs $k \\rightarrow \\infty$ we get\n$$\n0 \\leq \\lim _{k \\rightarrow \\infty}\\left(\\frac{f(t)}{k}+f(y)-\\frac{\\ell_{k}}{k} f(z)\\right)=f(y)-\\frac{g(y)}{g(z)} f(z)=f(y)-\\frac{f(y)-y}{f(z)-z} f(z)\n$$\nand therefore\n$$\n\\frac{f(y)}{y} \\leq \\frac{f(z)}{z}\n$$\nExchanging variables $y$ and $z$, we obtain the reverse inequality. Hence, $\\frac{f(y)}{y}=\\frac{f(z)}{z}$ for arbitrary $y$ and $z$; so function $\\frac{f(x)}{x}$ is constant, $f(x)=c x$.\nSubstituting back into (1), we find that $f(x)=c x$ is a solution if and only if $c=2$. So the only solution for the problem is $f(x)=2 x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31858,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $c>2$, and let $a(1), a(2), \\ldots$ be a sequence of nonnegative real numbers such that\n$$\na(m+n) \\leq 2 a(m)+2 a(n) \\quad \\text{ for all } m, n \\geq 1, \\tag{1}\n$$\nand\n$$\na\\left(2^{k}\\right) \\leq \\frac{1}{(k+1)^{c}} \\quad \\text{ for all } k \\geq 0 \\tag{2}\n$$\nProve that the sequence $a(n)$ is bounded.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For convenience, define $a(0)=0$; then condition (1) persists for all pairs of nonnegative indices.\n\nLemma 1. For arbitrary nonnegative indices $n_{1}, \\ldots, n_{k}$, we have\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right) \\leq \\sum_{i=1}^{k} 2^{i} a\\left(n_{i}\\right) \\tag{3}\n$$\nand\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right) \\leq 2 k \\sum_{i=1}^{k} a\\left(n_{i}\\right) . \\tag{4}\n$$\nProof. Inequality (3) is proved by induction on $k$. The base case $k=1$ is trivial, while the induction step is provided by\n$a\\left(\\sum_{i=1}^{k+1} n_{i}\\right)=a\\left(n_{1}+\\sum_{i=2}^{k+1} n_{i}\\right) \\leq 2 a\\left(n_{1}\\right)+2 a\\left(\\sum_{i=1}^{k} n_{i+1}\\right) \\leq 2 a\\left(n_{1}\\right)+2 \\sum_{i=1}^{k} 2^{i} a\\left(n_{i+1}\\right)=\\sum_{i=1}^{k+1} 2^{i} a\\left(n_{i}\\right)$.\n\nTo establish (4), first the inequality\n$$\na\\left(\\sum_{i=1}^{2^{d}} n_{i}\\right) \\leq 2^{d} \\sum_{i=1}^{2^{d}} a\\left(n_{i}\\right)\n$$\ncan be proved by an obvious induction on $d$. Then, turning to (4), we find an integer $d$ such that $2^{d-1}<k \\leq 2^{d}$ to obtain\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right)=a\\left(\\sum_{i=1}^{k} n_{i}+\\sum_{i=k+1}^{2^{d}} 0\\right) \\leq 2^{d}\\left(\\sum_{i=1}^{k} a\\left(n_{i}\\right)+\\sum_{i=k+1}^{2^{d}} a(0)\\right)=2^{d} \\sum_{i=1}^{k} a\\left(n_{i}\\right) \\leq 2 k \\sum_{i=1}^{k} a\\left(n_{i}\\right) .\n$$\n\nFix an increasing unbounded sequence $0=M_{0}<M_{1}<M_{2}<\\ldots$ of real numbers; the exact values will be defined later. Let $n$ be an arbitrary positive integer and write\n$$\nn=\\sum_{i=0}^{d} \\varepsilon_{i} \\cdot 2^{i}, \\quad \\text{ where } \\varepsilon_{i} \\in\\{0,1\\}\n$$\nSet $\\varepsilon_{i}=0$ for $i>d$, and take some positive integer $f$ such that $M_{f}>d$. Applying (3), we get\n$$\na(n)=a\\left(\\sum_{k=1}^{f} \\sum_{M_{k-1} \\leq i<M_{k}} \\varepsilon_{i} \\cdot 2^{i}\\right) \\leq \\sum_{k=1}^{f} 2^{k} a\\left(\\sum_{M_{k-1} \\leq i<M_{k}} \\varepsilon_{i} \\cdot 2^{i}\\right) .\n$$\nNote that there are less than $M_{k}-M_{k-1}+1$ integers in interval $\\left[M_{k-1}, M_{k}\\right)$; hence, using (4) we have\n$$\n\\begin{aligned}\na(n) & \\leq \\sum_{k=1}^{f} 2^{k} \\cdot 2\\left(M_{k}-M_{k-1}+1\\right) \\sum_{M_{k-1} \\leq i<M_{k}} \\varepsilon_{i} \\cdot a\\left(2^{i}\\right) \\\\\n& \\leq \\sum_{k=1}^{f} 2^{k} \\cdot 2\\left(M_{k}-M_{k-1}+1\\right)^{2} \\max _{M_{k-1} \\leq i<M_{k}} a\\left(2^{i}\\right) \\\\\n& \\leq \\sum_{k=1}^{f} 2^{k+1}\\left(M_{k}+1\\right)^{2} \\cdot \\frac{1}{\\left(M_{k-1}+1\\right)^{c}}=\\sum_{k=1}^{f}\\left(\\frac{M_{k}+1}{M_{k-1}+1}\\right)^{2} \\frac{2^{k+1}}{\\left(M_{k-1}+1\\right)^{c-2}}\n\\end{aligned}\n$$\nSetting $M_{k}=4^{k /(c-2)}-1$, we obtain\n$$\na(n) \\leq \\sum_{k=1}^{f} 4^{2 /(c-2)} \\frac{2^{k+1}}{\\left(4^{(k-1) /(c-2)}\\right)^{c-2}}=8 \\cdot 4^{2 /(c-2)} \\sum_{k=1}^{f}\\left(\\frac{1}{2}\\right)^{k}<8 \\cdot 4^{2 /(c-2)}\n$$\nand the sequence $a(n)$ is bounded.\n\n\nSolution 2:\n\nLemma 2. Suppose that $s_{1}, \\ldots, s_{k}$ are positive integers such that\n$$\n\\sum_{i=1}^{k} 2^{-s_{i}} \\leq 1\n$$\nThen for arbitrary positive integers $n_{1}, \\ldots, n_{k}$ we have\n$$\na\\left(\\sum_{i=1}^{k} n_{i}\\right) \\leq \\sum_{i=1}^{k} 2^{s_{i}} a\\left(n_{i}\\right)\n$$\nProof. Apply an induction on $k$. The base cases are $k=1$ (trivial) and $k=2$ (follows from the condition (1)). Suppose that $k>2$. We can assume that $s_{1} \\leq s_{2} \\leq \\cdots \\leq s_{k}$. Note that\n$$\n\\sum_{i=1}^{k-1} 2^{-s_{i}} \\leq 1-2^{-s_{k-1}}\n$$\nsince the left-hand side is a fraction with the denominator $2^{s_{k-1}}$, and this fraction is less than 1 . Define $s_{k-1}^{\\prime}=s_{k-1}-1$ and $n_{k-1}^{\\prime}=n_{k-1}+n_{k}$; then we have\n$$\n\\sum_{i=1}^{k-2} 2^{-s_{i}}+2^{-s_{k-1}^{\\prime}} \\leq\\left(1-2 \\cdot 2^{-s_{k-1}}\\right)+2^{1-s_{k-1}}=1\n$$\nNow, the induction hypothesis can be applied to achieve\n$$\n\\begin{aligned}\na\\left(\\sum_{i=1}^{k} n_{i}\\right)=a\\left(\\sum_{i=1}^{k-2} n_{i}+n_{k-1}^{\\prime}\\right) & \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}^{\\prime}} a\\left(n_{k-1}^{\\prime}\\right) \\\\\n& \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}-1} \\cdot 2\\left(a\\left(n_{k-1}\\right)+a\\left(n_{k}\\right)\\right) \\\\\n& \\leq \\sum_{i=1}^{k-2} 2^{s_{i}} a\\left(n_{i}\\right)+2^{s_{k-1}} a\\left(n_{k-1}\\right)+2^{s_{k}} a\\left(n_{k}\\right)\n\\end{aligned}\n$$\n\nLet $q=c / 2>1$. Take an arbitrary positive integer $n$ and write\n$$\nn=\\sum_{i=1}^{k} 2^{u_{i}}, \\quad 0 \\leq u_{1}<u_{2}<\\cdots<u_{k} .\n$$\nChoose $s_{i}=\\left\\lfloor\\log _{2}\\left(u_{i}+1\\right)^{q}\\right\\rfloor+d\\ (i=1, \\ldots, k)$ for some integer $d$. We have\n$$\n\\sum_{i=1}^{k} 2^{-s_{i}}=2^{-d} \\sum_{i=1}^{k} 2^{-\\left\\lfloor\\log _{2}\\left(u_{i}+1\\right)^{q}\\right\\rfloor},\n$$\nand we choose $d$ in such a way that\n$$\n\\frac{1}{2}<\\sum_{i=1}^{k} 2^{-s_{i}} \\leq 1\n$$\nIn particular, this implies\n$$\n2^{d}<2 \\sum_{i=1}^{k} 2^{-\\left\\lfloor\\log _{2}\\left(u_{i}+1\\right)^{q}\\right\\rfloor}<4 \\sum_{i=1}^{k} \\frac{1}{\\left(u_{i}+1\\right)^{q}} .\n$$\nNow, by Lemma 2 we obtain\n$$\n\\begin{aligned}\na(n)=a\\left(\\sum_{i=1}^{k} 2^{u_{i}}\\right) & \\leq \\sum_{i=1}^{k} 2^{s_{i}} a\\left(2^{u_{i}}\\right) \\leq \\sum_{i=1}^{k} 2^{d}\\left(u_{i}+1\\right)^{q} \\cdot \\frac{1}{\\left(u_{i}+1\\right)^{2 q}} \\\\\n& =2^{d} \\sum_{i=1}^{k} \\frac{1}{\\left(u_{i}+1\\right)^{q}}<4\\left(\\sum_{i=1}^{k} \\frac{1}{\\left(u_{i}+1\\right)^{q}}\\right)^{2}\n\\end{aligned}\n$$\nwhich is bounded since $q>1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31859,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, \\ldots, a_{100}$ be nonnegative real numbers such that $a_{1}^{2}+a_{2}^{2}+\\ldots+a_{100}^{2}=1$. Prove that\n$$\na_{1}^{2} a_{2}+a_{2}^{2} a_{3}+\\ldots+a_{100}^{2} a_{1}<\\frac{12}{25}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S=\\sum_{k=1}^{100} a_{k}^{2} a_{k+1}$. (As usual, we consider the indices modulo $100$, e.g. we set $a_{101}=a_{1}$ and $a_{102}=a_{2}$.)\nApplying the Cauchy-Schwarz inequality to sequences $\\left(a_{k+1}\\right)$ and $\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)$, and then the AM-GM inequality to numbers $a_{k+1}^{2}$ and $a_{k+2}^{2}$,\n$$\n\\begin{align*}\n(3 S)^{2} & =\\left(\\sum_{k=1}^{100} a_{k+1}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)\\right)^{2} \\leq\\left(\\sum_{k=1}^{100} a_{k+1}^{2}\\right)\\left(\\sum_{k=1}^{100}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)^{2}\\right)  \\\\\n& =1 \\cdot \\sum_{k=1}^{100}\\left(a_{k}^{2}+2 a_{k+1} a_{k+2}\\right)^{2}=\\sum_{k=1}^{100}\\left(a_{k}^{4}+4 a_{k}^{2} a_{k+1} a_{k+2}+4 a_{k+1}^{2} a_{k+2}^{2}\\right) \\\\\n& \\leq \\sum_{k=1}^{100}\\left(a_{k}^{4}+2 a_{k}^{2}\\left(a_{k+1}^{2}+a_{k+2}^{2}\\right)+4 a_{k+1}^{2} a_{k+2}^{2}\\right)=\\sum_{k=1}^{100}\\left(a_{k}^{4}+6 a_{k}^{2} a_{k+1}^{2}+2 a_{k}^{2} a_{k+2}^{2}\\right) .\n\\end{align*}\n$$\nApplying the trivial estimates\n$$\n\\sum_{k=1}^{100}\\left(a_{k}^{4}+2 a_{k}^{2} a_{k+1}^{2}+2 a_{k}^{2} a_{k+2}^{2}\\right) \\leq\\left(\\sum_{k=1}^{100} a_{k}^{2}\\right)^{2} \\quad \\text{ and } \\quad \\sum_{k=1}^{100} a_{k}^{2} a_{k+1}^{2} \\leq\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}\\right)\\left(\\sum_{j=1}^{50} a_{2 j}^{2}\\right)\n$$\nwe obtain that\n$$\n(3 S)^{2} \\leq\\left(\\sum_{k=1}^{100} a_{k}^{2}\\right)^{2}+4\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}\\right)\\left(\\sum_{j=1}^{50} a_{2 j}^{2}\\right) \\leq 1+\\left(\\sum_{i=1}^{50} a_{2 i-1}^{2}+\\sum_{j=1}^{50} a_{2 j}^{2}\\right)^{2}=2,\n$$\nhence\n$$\nS \\leq \\frac{\\sqrt{2}}{3} \\approx 0.4714<\\frac{12}{25}=0.48\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31860,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be an integer. Find all sequences $a_{1}, a_{2}, \\ldots, a_{n^{2}+n}$ satisfying the following conditions:\n(a) $a_{i} \\in \\{0,1\\}$ for all $1 \\leq i \\leq n^{2}+n$;\n(b) $a_{i+1}+a_{i+2}+\\ldots+a_{i+n} < a_{i+n+1}+a_{i+n+2}+\\ldots+a_{i+2n}$ for all $0 \\leq i \\leq n^{2}-n$.",
    "options": [],
    "answer": "Partition the indices into n+1 consecutive blocks of length n. For block v (0 ≤ v ≤ n), the first n−v entries are 0 and the last v entries are 1. Equivalently, writing any index as u + v n with 1 ≤ u ≤ n and 0 ≤ v ≤ n, the sequence is given by a_{u+vn} = 0 if u + v ≤ n and a_{u+vn} = 1 if u + v ≥ n + 1.",
    "solution": "Consider a sequence ($a_{i}$) satisfying the conditions. For arbitrary integers $0 \\leq k \\leq l \\leq n^{2}+n$ denote $S(k, l] = a_{k+1} + \\cdots + a_{l}$. (If $k = l$ then $S(k, l] = 0$.) Then condition (b) can be rewritten as $S(i, i+n] < S(i+n, i+2n]$ for all $0 \\leq i \\leq n^{2}-n$. Notice that for $0 \\leq k \\leq l \\leq m \\leq n^{2}+n$ we have $S(k, m] = S(k, l] + S(l, m]$.\n\nBy condition (b),\n$$\n0 \\leq S(0, n] < S(n, 2n] < \\cdots < S\\left(n^{2}, n^{2}+n\\right] \\leq n.\n$$\nWe have only $n+1$ distinct integers in the interval $[0, n]$; hence,\n$$\nS(vn, (v+1)n] = v \\quad \\text{for all } 0 \\leq v \\leq n. \\tag{2}\n$$\nIn particular, $S(0, n] = 0$ and $S\\left(n^{2}, n^{2}+n\\right] = n$, therefore\n$$\n\\begin{align*}\na_{1} & = a_{2} = \\ldots = a_{n} = 0  \\tag{3}\\\\\na_{n^{2}+1} & = a_{n^{2}+2} = \\ldots = a_{n^{2}+n} = 1 \\tag{4}\n\\end{align*}\n$$\nSubdivide sequence ($a_{i}$) into $n+1$ blocks, each consisting of $n$ consecutive terms, and number them from 0 to $n$. We show by induction on $v$ that the $v$th block has the form\n$$\n(\\underbrace{0 \\ldots 0}_{n-v} \\underbrace{1 \\ldots 1}_{v})\n$$\nThe base case $v=0$ is provided by (3).\n\nConsider the $v$th block for $v>0$. By (2), it contains some \"ones\". Let the first \"one\" in this block be at the $u$th position (that is, $a_{u+vn} = 1$). By the induction hypothesis, the $(v-1)$th and $v$th blocks of $\\left(a_{i}\\right)$ have the form\n$$\n(\\underbrace{0 \\ldots 0}_{n-v+1})(\\underbrace{0 \\ldots 0} 1 * \\ldots *)\n$$\nwhere each star can appear to be any binary digit. Observe that $u \\leq n-v+1$, since the sum in this block is $v$. Then, the fragment of length $n$ bracketed above has exactly $(v-1)+1$ ones, i.e. $S(u+(v-1)n, u+vn] = v$. Hence,\n$$\nv = S(u+(v-1)n, u+vn] < S(u+vn, u+(v+1)n] < \\cdots < S\\left(u+(n-1)n, u+n^{2}\\right] \\leq n;\n$$\nwe have $n-v+1$ distinct integers in the interval $[v, n]$, therefore $S(u+(t-1)n, u+tn] = t$ for each $t = v, \\ldots, n$.\n\nThus, the end of sequence $\\left(a_{i}\\right)$ looks as following:\n$$\n(\\underbrace{u \\text{ zeroes}}_{v-1} \\overbrace{0 \\ldots 01 \\ldots 1}^{0 \\ldots 0})(\\underbrace{0 \\ldots 01}_{v} \\overbrace{0 \\ldots *}^{v+1}(\\underbrace{* \\ldots * * \\ldots *}_{v+1}) \\ldots (\\underbrace{1 \\ldots 1}_{n} \\overbrace{1 \\ldots 1}^{n})\n$$\n(each bracketed fragment contains $n$ terms). Computing in two ways the sum of all digits above, we obtain $n-u = v-1$ and $u = n-v+1$. Then, the first $n-v$ terms in the $v$th block are zeroes, and the next $v$ terms are ones, due to the sum of all terms in this block. The statement is proved.\n\nWe are left to check that the sequence obtained satisfies the condition. Notice that $a_{i} \\leq a_{i+n}$ for all $1 \\leq i \\leq n^{2}$. Moreover, if $1 \\leq u \\leq n$ and $0 \\leq v \\leq n-1$, then $a_{u+vn} < a_{u+vn+n}$ exactly when $u+v = n$. In this case we have $u+vn = n+v(n-1)$.\n\nConsider now an arbitrary index $0 \\leq i \\leq n^{2}-n$. Clearly, there exists an integer $v$ such that $n+v(n-1) \\in [i+1, i+n]$. Then, applying the above inequalities we obtain that condition (b) is valid.\nSimilarly to Solution 1, we introduce the notation $S(k, l]$ and obtain (2), (3), and (4) in the same way. The sum of all elements of the sequence can be computed as\n$$\nS\\left(0, n^{2}+n\\right] = S(0, n] + S(n, 2n] + \\ldots + S\\left(n^{2}, n^{2}+n\\right] = 0 + 1 + \\ldots + n.\n$$\nFor an arbitrary integer $0 \\leq u \\leq n$, consider the numbers\n$$\nS(u, u+n] < S(u+n, u+2n] < \\ldots < S\\left(u+(n-1)n, u+n^{2}\\right]. \\tag{5}\n$$\nThey are $n$ distinct integers from the $n+1$ possible values $0, 1, 2, \\ldots, n$. Denote by $m$ the \"missing\" value which is not listed. We determine $m$ from $S\\left(0, n^{2}+n\\right]$. Write this sum as\n$S\\left(0, n^{2}+n\\right] = S(0, u] + S(u, u+n] + S(u+n, u+2n] + \\ldots + S\\left(u+(n-1)n, u+n^{2}\\right] + S\\left(u+n^{2}, n^{2}+n\\right]$.\nSince $a_{1} = a_{2} = \\ldots = a_{u} = 0$ and $a_{u+n^{2}+1} = \\ldots = a_{n^{2}+n} = 1$, we have $S(0, u] = 0$ and $S\\left(u+n^{2}, n+n^{2}\\right] = n-u$. Then\n$$\n0+1+\\ldots+n = S\\left(0, n^{2}+n\\right] = 0 + ((0+1+\\ldots+n) - m) + (n-u),\n$$\nso $m = n-u$.\n\nHence, the numbers listed in (5) are $0, 1, \\ldots, n-u-1$ and $n-u+1, \\ldots, n$, respectively, therefore\n$$\nS(u+vn, u+(v+1)n] = \\left\\{\\begin{array}{ll}\nv, & v \\leq n-u-1,  \\tag{6}\\\\\nv+1, & v \\geq n-u\n\\end{array} \\quad \\text{for all } 0 \\leq u \\leq n, 0 \\leq v \\leq n-1.\\right.\n$$\nConditions (6), together with (3), provide a system of linear equations in variables $a_{i}$. Now we solve this system and show that the solution is unique and satisfies conditions (a) and (b).\n\nFirst, observe that any solution of the system (3), (6) satisfies the condition (b). By the construction, equations (6) immediately imply (5). On the other hand, all inequalities mentioned in condition (b) are included into the chain (5) for some value of $u$.\n\nNext, note that the system (3), (6) is redundant. The numbers $S(kn, (k+1)n]$, where $1 \\leq k \\leq n-1$, appear twice in (6). For $u=0$ and $v=k$ we have $v \\leq n-u-1$, and (6) gives $S(kn, (k+1)n] = v = k$. For $u=n$ and $v=k-1$ we have $v \\geq n-u$ and we obtain the same value, $S(kn, (k+1)n] = v+1 = k$. Therefore, deleting one equation from each redundant pair, we can make every sum $S(k, k+n]$ appear exactly once on the left-hand side in (6).\n\nNow, from (3), (6), the sequence $\\left(a_{i}\\right)$ can be reconstructed inductively by\n$$\na_{1} = a_{2} = \\ldots = a_{n-1} = 0, \\quad a_{k+n} = S(k, k+n] - \\left(a_{k+1} + a_{k+2} + \\ldots + a_{k+n-1}\\right) \\quad (0 \\leq k \\leq n^{2})\n$$\ntaking the values of $S(k, k+n]$ from (6). This means first that there exists at most one solution of our system. Conversely, the constructed sequence obviously satisfies all equations (3), (6) (the only missing equation is $a_{n} = 0$, which follows from $S(0, n] = 0$). Hence it satisfies condition (b), and we are left to check condition (a) only.\n\nFor arbitrary integers $1 \\leq u, t \\leq n$ we get\n$$\n\\begin{aligned}\na_{u+tn} - a_{u+(t-1)n} & = S(u+(t-1)n, u+tn] - S((u-1)+(t-1)n, (u-1)+tn] \\\\\n& = \\begin{cases}\n(t-1)-(t-1) = 0, & t \\leq n-u \\\\\nt-(t-1) = 1, & t = n-u+1 \\\\\nt-t = 0, & t \\geq n-u+2\n\\end{cases}\n\\end{aligned}\n$$\nSince $a_{u} = 0$, we have\n$$\na_{u+vn} = a_{u+vn} - a_{u} = \\sum_{t=1}^{v} \\left(a_{u+tn} - a_{u+(t-1)n}\\right)\n$$\nfor all $1 \\leq u, v \\leq n$. If $v < n-u+1$ then all terms are 0 on the right-hand side. If $v \\geq n-u+1$, then variable $t$ attains the value $n-u+1$ once. Hence,\n$$\na_{u+vn} = \\begin{cases}0, & u+v \\leq n \\\\ 1, & u+v \\geq n+1\\end{cases}\n$$\naccording with (1). Note that the formula is valid for $v=0$ as well.\n\nFinally, we presented the direct formula for $\\left(a_{i}\\right)$, and we have proved that it satisfies condition (a). So, the solution is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31861,
    "subject": "Mathematics (Multi-modal)",
    "question": "A unit square is dissected into $n>1$ rectangles such that their sides are parallel to the sides of the square. Any line, parallel to a side of the square and intersecting its interior, also intersects the interior of some rectangle. Prove that in this dissection, there exists a rectangle having no point on the boundary of the square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Call the directions of the sides of the square horizontal and vertical. A horizontal or vertical line, which intersects the interior of the square but does not intersect the interior of any rectangle, will be called a splitting line. A rectangle having no point on the boundary of the square will be called an interior rectangle.\n\nSuppose, to the contrary, that there exists a dissection of the square into more than one rectangle, such that no interior rectangle and no splitting line appear. Consider such a dissection with the least possible number of rectangles. Notice that this number of rectangles is greater than $2$, otherwise their common side provides a splitting line.\n\nIf there exist two rectangles having a common side, then we can replace them by their union (see Figure 1).\n\n![](attached_image_1.png)\n\nFigure 1\n\nThe number of rectangles was greater than $2$, so in a new dissection it is greater than $1$. Clearly, in the new dissection, there is also no splitting line as well as no interior rectangle. This contradicts the choice of the original dissection.\n\nDenote the initial square by $ABCD$, with $A$ and $B$ being respectively the lower left and lower right vertices. Consider those two rectangles $a$ and $b$ containing vertices $A$ and $B$, respectively. (Note that $a \\neq b$, otherwise its top side provides a splitting line.) We can assume that the height of $a$ is not greater than that of $b$. Then consider the rectangle $c$ neighboring to the lower right corner of $a$ (it may happen that $c=b$). By aforementioned, the heights of $a$ and $c$ are distinct. Then two cases are possible.\n\n![](attached_image_2.png)\n\nFigure 3\n\nCase 1. The height of $c$ is less than that of $a$. Consider the rectangle $d$ which is adjacent to both $a$ and $c$, i.e. the one containing the angle marked in Figure 2. This rectangle has no common point with $BC$ (since $a$ is not higher than $b$), as well as no common point with $AB$ or with $AD$ (obviously). Then $d$ has a common point with $CD$, and its left side provides a splitting line. Contradiction.\n\nCase 2. The height of $c$ is greater than that of $a$. Analogously, consider the rectangle $d$ containing the angle marked on Figure 3. It has no common point with $AD$ (otherwise it has a common side with $a$), as well as no common point with $AB$ or with $BC$ (obviously). Then $d$ has a common point with $CD$. Hence its right side provides a splitting line, and we get the contradiction again.\nAgain, we suppose the contrary. Consider an arbitrary counterexample. Then we know that each rectangle is attached to at least one side of the square. Observe that a rectangle cannot be attached to two opposite sides, otherwise one of its sides lies on a splitting line.\n\nWe say that two rectangles are opposite if they are attached to opposite sides of $ABCD$. We claim that there exist two opposite rectangles having a common point.\n\nConsider the union $L$ of all rectangles attached to the left. Assume, to the contrary, that $L$ has no common point with the rectangles attached to the right. Take a polygonal line $p$ connecting the top and the bottom sides of the square and passing close from the right to the boundary of $L$ (see Figure 4). Then all its points belong to the rectangles attached either to the top or to the bottom. Moreover, the upper end-point of $p$ belongs to a rectangle attached to the top, and the lower one belongs to another rectangle attached to the bottom. Hence, there is a point on $p$ where some rectangles attached to the top and to the bottom meet each other. So, there always exists a pair of neighboring opposite rectangles.\n\n![](attached_image_3.png)\n\nFigure 4\n\n![](attached_image_4.png)\n\nFigure 5\n\n![](attached_image_5.png)\n\nFigure 6\n\nNow, take two opposite neighboring rectangles $a$ and $b$. We can assume that $a$ is attached to the left and $b$ is attached to the right. Let $X$ be their common point. If $X$ belongs to their horizontal sides (in particular, $X$ may appear to be a common vertex of $a$ and $b$), then these sides provide a splitting line (see Figure 5). Otherwise, $X$ lies on the vertical sides. Let $\\ell$ be the line containing these sides.\n\nSince $\\ell$ is not a splitting line, it intersects the interior of some rectangle. Let $c$ be such a rectangle, closest to $X$; we can assume that $c$ lies above $X$. Let $Y$ be the common point of $\\ell$ and the bottom side of $c$ (see Figure 6). Then $Y$ is also a vertex of two rectangles lying below $c$.\n\nSo, let $Y$ be the upper-right and upper-left corners of the rectangles $a'$ and $b'$, respectively. Then $a'$ and $b'$ are situated not lower than $a$ and $b$, respectively (it may happen that $a=a'$ or $b=b'$). We claim that $a'$ is attached to the left. If $a=a'$ then of course it is. If $a \\neq a'$ then $a'$ is above $a$, below $c$ and to the left from $b'$. Hence, it can be attached to the left only.\n\nAnalogously, $b'$ is attached to the right. Now, the top sides of these two rectangles pass through $Y$, hence they provide a splitting line again. This last contradiction completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31862,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$, for which the numbers in the set $S=\\{1,2, \\ldots, n\\}$ can be colored red and blue, with the following condition being satisfied: the set $S \\times S \\times S$ contains exactly $2007$ ordered triples $(x, y, z)$ such that (i) $x, y, z$ are of the same color and (ii) $x+y+z$ is divisible by $n$.",
    "options": [],
    "answer": "69, 84",
    "solution": "Suppose that the numbers $1,2, \\ldots, n$ are colored red and blue. Denote by $R$ and $B$ the sets of red and blue numbers, respectively; let $|R|=r$ and $|B|=b=n-r$. Call a triple $(x, y, z) \\in S \\times S \\times S$ monochromatic if $x, y, z$ have the same color, and bichromatic otherwise. Call a triple $(x, y, z)$ divisible if $x+y+z$ is divisible by $n$. We claim that there are exactly $r^{2}-r b+b^{2}$ divisible monochromatic triples.\n\nFor any pair $(x, y) \\in S \\times S$ there exists a unique $z_{x, y} \\in S$ such that the triple $(x, y, z_{x, y})$ is divisible; so there are exactly $n^{2}$ divisible triples. Furthermore, if a divisible triple $(x, y, z)$ is bichromatic, then among $x, y, z$ there are either one blue and two red numbers, or vice versa. In both cases, exactly one of the pairs $(x, y),(y, z)$ and $(z, x)$ belongs to the set $R \\times B$. Assign such pair to the triple $(x, y, z)$.\n\nConversely, consider any pair $(x, y) \\in R \\times B$, and denote $z=z_{x, y}$. Since $x \\neq y$, the triples $(x, y, z),(y, z, x)$ and $(z, x, y)$ are distinct, and $(x, y)$ is assigned to each of them. On the other hand, if $(x, y)$ is assigned to some triple, then this triple is clearly one of those mentioned above. So each pair in $R \\times B$ is assigned exactly three times.\n\nThus, the number of bichromatic divisible triples is three times the number of elements in $R \\times B$, and the number of monochromatic ones is $n^{2}-3 r b=(r+b)^{2}-3 r b=r^{2}-r b+b^{2}$, as claimed.\n\nSo, to find all values of $n$ for which the desired coloring is possible, we have to find all $n$, for which there exists a decomposition $n=r+b$ with $r^{2}-r b+b^{2}=2007$. Therefore, $9 \\mid r^{2}-r b+b^{2}=(r+b)^{2}-3 r b$. From this it consequently follows that $3|r+b, 3| r b$, and then $3|r, 3| b$. Set $r=3 s, b=3 c$. We can assume that $s \\geq c$. We have $s^{2}-s c+c^{2}=223$.\n\nFurthermore,\n$$\n892=4\\left(s^{2}-s c+c^{2}\\right)=(2 c-s)^{2}+3 s^{2} \\geq 3 s^{2} \\geq 3 s^{2}-3 c(s-c)=3\\left(s^{2}-s c+c^{2}\\right)=669\n$$\nso $297 \\geq s^{2} \\geq 223$ and $17 \\geq s \\geq 15$. If $s=15$ then\n$$\nc(15-c)=c(s-c)=s^{2}-\\left(s^{2}-s c+c^{2}\\right)=15^{2}-223=2\n$$\nwhich is impossible for an integer $c$. In a similar way, if $s=16$ then $c(16-c)=33$, which is also impossible. Finally, if $s=17$ then $c(17-c)=66$, and the solutions are $c=6$ and $c=11$. Hence, $(r, b)=(51,18)$ or $(r, b)=(51,33)$, and the possible values of $n$ are $n=51+18=69$ and $n=51+33=84$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31863,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_{0} = (a_{1}, \\ldots, a_{n})$ be a finite sequence of real numbers. For each $k \\geq 0$, from the sequence $A_{k} = (x_{1}, \\ldots, x_{n})$ we construct a new sequence $A_{k+1}$ in the following way.\n1. We choose a partition $\\{1, \\ldots, n\\} = I \\cup J$, where $I$ and $J$ are two disjoint sets, such that the expression\n$$\n\\left|\\sum_{i \\in I} x_{i} - \\sum_{j \\in J} x_{j}\\right|\n$$\nattains the smallest possible value. (We allow the sets $I$ or $J$ to be empty; in this case the corresponding sum is $0$.) If there are several such partitions, one is chosen arbitrarily.\n2. We set $A_{k+1} = (y_{1}, \\ldots, y_{n})$, where $y_{i} = x_{i} + 1$ if $i \\in I$, and $y_{i} = x_{i} - 1$ if $i \\in J$.\nProve that for some $k$, the sequence $A_{k}$ contains an element $x$ such that $|x| \\geq n / 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Lemma. Suppose that all terms of the sequence $(x_{1}, \\ldots, x_{n})$ satisfy the inequality $|x_{i}| < a$. Then there exists a partition $\\{1,2, \\ldots, n\\} = I \\cup J$ into two disjoint sets such that\n$$\n|\\sum_{i \\in I} x_{i} - \\sum_{j \\in J} x_{j}| < a. \\tag{1}\n$$\nProof. Apply an induction on $n$. The base case $n = 1$ is trivial. For the induction step, consider a sequence $(x_{1}, \\ldots, x_{n})$ ($n > 1$). By the induction hypothesis there exists a splitting $\\{1, \\ldots, n-1\\} = I' \\cup J'$ such that\n$$\n|\\sum_{i \\in I'} x_{i} - \\sum_{j \\in J'} x_{j}| < a\n$$\nFor convenience, suppose that $\\sum_{i \\in I'} x_{i} \\geq \\sum_{j \\in J'} x_{j}$. If $x_{n} \\geq 0$ then choose $I = I'$, $J = J' \\cup \\{n\\}$; otherwise choose $I = I' \\cup \\{n\\}$, $J = J'$. In both cases, we have $\\sum_{i \\in I'} x_{i} - \\sum_{j \\in J'} x_{j} \\in [0, a)$ and $|x_{n}| \\in [0, a)$; hence\n$$\n\\sum_{i \\in I} x_{i} - \\sum_{j \\in J} x_{j} = \\sum_{i \\in I'} x_{i} - \\sum_{j \\in J'} x_{j} - |x_{n}| \\in (-a, a),\n$$\nas desired.\nLet us turn now to the problem. To the contrary, assume that for all $k$, all the numbers in $A_{k}$ lie in interval $(-n / 2, n / 2)$. Consider an arbitrary sequence $A_{k} = (b_{1}, \\ldots, b_{n})$. To obtain the term $b_{i}$, we increased and decreased number $a_{i}$ by one several times. Therefore $b_{i} - a_{i}$ is always an integer, and there are not more than $n$ possible values for $b_{i}$. So, there are not more than $n^{n}$ distinct possible sequences $A_{k}$, and hence two of the sequences $A_{1}, A_{2}, \\ldots, A_{n^{n} + 1}$ should be identical, say $A_{p} = A_{q}$ for some $p < q$.\nFor any positive integer $k$, let $S_{k}$ be the sum of squares of elements in $A_{k}$. Consider two consecutive sequences $A_{k} = (x_{1}, \\ldots, x_{n})$ and $A_{k+1} = (y_{1}, \\ldots, y_{n})$. Let $\\{1,2, \\ldots, n\\} = I \\cup J$ be the partition used in this step - that is, $y_{i} = x_{i} + 1$ for all $i \\in I$ and $y_{j} = x_{j} - 1$ for all $j \\in J$. Since the value of $|\\sum_{i \\in I} x_{i} - \\sum_{j \\in J} x_{j}|$ is the smallest possible, the Lemma implies that it is less than $n / 2$. Then we have\n$S_{k+1} - S_{k} = \\sum_{i \\in I} ((x_{i} + 1)^{2} - x_{i}^{2}) + \\sum_{j \\in J} ((x_{j} - 1)^{2} - x_{j}^{2}) = n + 2(\\sum_{i \\in I} x_{i} - \\sum_{j \\in J} x_{j}) > n - 2 \\cdot \\frac{n}{2} = 0$.\nThus we obtain $S_{q} > S_{q-1} > \\cdots > S_{p}$. This is impossible since $A_{p} = A_{q}$ and hence $S_{p} = S_{q}$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31864,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the Cartesian coordinate plane define the strip $S_{n}=\\{(x, y) \\mid n \\leq x<n+1\\}$ for every integer $n$. Assume that each strip $S_{n}$ is colored either red or blue, and let $a$ and $b$ be two distinct positive integers. Prove that there exists a rectangle with side lengths $a$ and $b$ such that its vertices have the same color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $S_{n}$ and $S_{n+a}$ have the same color for some integer $n$, then we can choose the rectangle with vertices $(n, 0) \\in S_{n}, (n, b) \\in S_{n}, (n+a, 0) \\in S_{n+a}$, and $(n+a, b) \\in S_{n+a}$, and we are done. So it can be assumed that $S_{n}$ and $S_{n+a}$ have opposite colors for each $n$.\n\nSimilarly, it also can be assumed that $S_{n}$ and $S_{n+b}$ have opposite colors. Then, by induction on $|p|+|q|$, we obtain that for arbitrary integers $p$ and $q$, strips $S_{n}$ and $S_{n+p a+q b}$ have the same color if $p+q$ is even, and these two strips have opposite colors if $p+q$ is odd.\n\nLet $d=\\gcd(a, b)$, $a_{1}=a / d$ and $b_{1}=b / d$. Apply the result above for $p=b_{1}$ and $q=-a_{1}$. The strips $S_{0}$ and $S_{0+b_{1} a-a_{1} b}$ are identical and therefore they have the same color. Hence, $a_{1}+b_{1}$ is even. By the construction, $a_{1}$ and $b_{1}$ are coprime, so this is possible only if both are odd.\n\nWithout loss of generality, we can assume $a>b$. Then $a_{1}>b_{1} \\geq 1$, so $a_{1} \\geq 3$.\n\nChoose integers $k$ and $\\ell$ such that $k a_{1}-\\ell b_{1}=1$ and therefore $k a-\\ell b=d$. Since $a_{1}$ and $b_{1}$ are odd, $k+\\ell$ is odd as well. Hence, for every integer $n$, strips $S_{n}$ and $S_{n+k a-\\ell b}=S_{n+d}$ have opposite colors. This also implies that the coloring is periodic with period $2d$, i.e. strips $S_{n}$ and $S_{n+2 d}$ have the same color for every $n$.\n\n![](attached_image_1.png)\nFigure 1\n\nWe will construct the desired rectangle $ABCD$ with $AB=CD=a$ and $BC=AD=b$ in a position such that vertex $A$ lies on the $x$-axis, and the projection of side $AB$ onto the $x$-axis is of length $2d$ (see Figure 1). This is possible since $a=a_{1} d>2d$. The coordinates of the vertices will have the forms\n$$\nA=(t, 0), \\quad B=\\left(t+2d, y_{1}\\right), \\quad C=\\left(u+2d, y_{2}\\right), \\quad D=\\left(u, y_{3}\\right) .\n$$\nLet $\\varphi=\\sqrt{a_{1}^{2}-4}$. By Pythagoras' theorem,\n$$\ny_{1}=BB_{0}=\\sqrt{a^{2}-4d^{2}}=d \\sqrt{a_{1}^{2}-4}=d \\varphi\n$$\nSo, by the similar triangles $ADD_{0}$ and $BAB_{0}$, we have the constraint\n$$\n\\begin{equation*}\nu-t=AD_{0}=\\frac{AD}{AB} \\cdot BB_{0}=\\frac{b d}{a} \\varphi \\tag{1}\n\\end{equation*}\n$$\nfor numbers $t$ and $u$. Computing the numbers $y_{2}$ and $y_{3}$ is not required since they have no effect to the colors.\n\nObserve that the number $\\varphi$ is irrational, because $\\varphi^{2}$ is an integer, but $\\varphi$ is not: $a_{1}>\\varphi \\geq \\sqrt{a_{1}^{2}-2 a_{1}+2}>a_{1}-1$.\n\nBy the periodicity, points $A$ and $B$ have the same color; similarly, points $C$ and $D$ have the same color. Furthermore, these colors depend only on the values of $t$ and $u$. So it is sufficient to choose numbers $t$ and $u$ such that vertices $A$ and $D$ have the same color.\n\nLet $w$ be the largest positive integer such that there exist $w$ consecutive strips $S_{n_{0}}, S_{n_{0}+1}, \\ldots, S_{n_{0}+w-1}$ with the same color, say red. (Since $S_{n_{0}+d}$ must be blue, we have $w \\leq d$.) We will choose $t$ from the interval $\\left(n_{0}, n_{0}+w\\right)$.\n\n![](attached_image_2.png)\nFigure 2\n\nConsider the interval $I=\\left(n_{0}+\\frac{b d}{a} \\varphi, n_{0}+\\frac{b d}{a} \\varphi+w\\right)$ on the $x$-axis (see Figure 2). Its length is $w$, and the end-points are irrational. Therefore, this interval intersects $w+1$ consecutive strips. Since at most $w$ consecutive strips may have the same color, interval $I$ must contain both red and blue points. Choose $u \\in I$ such that the line $x=u$ is red and set $t=u-\\frac{b d}{a} \\varphi$, according to the constraint (1). Then $t \\in\\left(n_{0}, n_{0}+w\\right)$ and $A=(t, 0)$ is red as well as $D=\\left(u, y_{3}\\right)$.\n\nHence, variables $u$ and $t$ can be set such that they provide a rectangle with four red vertices.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31865,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha < \\frac{3 - \\sqrt{5}}{2}$ be a positive real number. Prove that there exist positive integers $n$ and $p > \\alpha \\cdot 2^{n}$ for which one can select $2p$ pairwise distinct subsets $S_{1}, \\ldots, S_{p}, T_{1}, \\ldots, T_{p}$ of the set $\\{1, 2, \\ldots, n\\}$ such that $S_{i} \\cap T_{j} \\neq \\varnothing$ for all $1 \\leq i, j \\leq p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k$ and $m$ be positive integers (to be determined later) and set $n = k m$. Decompose the set $\\{1, 2, \\ldots, n\\}$ into $k$ disjoint subsets, each of size $m$; denote these subsets by $A_{1}, \\ldots, A_{k}$. Define the following families of sets:\n$$\n\\begin{aligned}\n\\mathcal{S} & = \\left\\{ S \\subset \\{1, 2, \\ldots, n\\} : \\forall i\\ S \\cap A_{i} \\neq \\varnothing \\right\\}, \\\\\n\\mathcal{T}_{1} & = \\left\\{ T \\subset \\{1, 2, \\ldots, n\\} : \\exists i\\ A_{i} \\subset T \\right\\}, \\quad \\mathcal{T} = \\mathcal{T}_{1} \\setminus \\mathcal{S} .\n\\end{aligned}\n$$\nFor each set $T \\in \\mathcal{T} \\subset \\mathcal{T}_{1}$, there exists an index $1 \\leq i \\leq k$ such that $A_{i} \\subset T$. Then for all $S \\in \\mathcal{S}$, $S \\cap T \\supset S \\cap A_{i} \\neq \\varnothing$. Hence, each $S \\in \\mathcal{S}$ and each $T \\in \\mathcal{T}$ have at least one common element.\n\nBelow we show that the numbers $m$ and $k$ can be chosen such that $|\\mathcal{S}|, |\\mathcal{T}| > \\alpha \\cdot 2^{n}$. Then, choosing $p = \\min \\{|\\mathcal{S}|, |\\mathcal{T}|\\}$, one can select the desired $2p$ sets $S_{1}, \\ldots, S_{p}$ and $T_{1}, \\ldots, T_{p}$ from families $\\mathcal{S}$ and $\\mathcal{T}$, respectively. Since families $\\mathcal{S}$ and $\\mathcal{T}$ are disjoint, sets $S_{i}$ and $T_{j}$ will be pairwise distinct.\n\nTo count the sets $S \\in \\mathcal{S}$, observe that each $A_{i}$ has $2^{m} - 1$ nonempty subsets so we have $2^{m} - 1$ choices for $S \\cap A_{i}$. These intersections uniquely determine set $S$, so\n$$\n\\begin{equation*}\n|\\mathcal{S}| = \\left(2^{m} - 1\\right)^{k} \\tag{1}\n\\end{equation*}\n$$\nSimilarly, if a set $H \\subset \\{1, 2, \\ldots, n\\}$ does not contain a certain set $A_{i}$ then we have $2^{m} - 1$ choices for $H \\cap A_{i}$: all subsets of $A_{i}$, except $A_{i}$ itself. Therefore, the complement of $\\mathcal{T}_{1}$ contains $\\left(2^{m} - 1\\right)^{k}$ sets and\n$$\n\\begin{equation*}\n|\\mathcal{T}_{1}| = 2^{k m} - \\left(2^{m} - 1\\right)^{k} \\tag{2}\n\\end{equation*}\n$$\nNext consider the family $\\mathcal{S} \\setminus \\mathcal{T}_{1}$. If a set $S$ intersects all $A_{i}$ but does not contain any of them, then there exists $2^{m} - 2$ possible values for each $S \\cap A_{i}$: all subsets of $A_{i}$ except $\\varnothing$ and $A_{i}$. Therefore the number of such sets $S$ is $\\left(2^{m} - 2\\right)^{k}$, so\n$$\n\\begin{equation*}\n|\\mathcal{S} \\setminus \\mathcal{T}_{1}| = \\left(2^{m} - 2\\right)^{k} \\tag{3}\n\\end{equation*}\n$$\nFrom (1), (2), and (3) we obtain\n$$\n|\\mathcal{T}| = |\\mathcal{T}_{1}| - |\\mathcal{S} \\cap \\mathcal{T}_{1}| = |\\mathcal{T}_{1}| - (|\\mathcal{S}| - |\\mathcal{S} \\setminus \\mathcal{T}_{1}|) = 2^{k m} - 2\\left(2^{m} - 1\\right)^{k} + \\left(2^{m} - 2\\right)^{k}\n$$\nLet $\\delta = \\frac{3 - \\sqrt{5}}{2}$ and $k = k(m) = \\left[2^{m} \\log \\frac{1}{\\delta}\\right]$. Then\n$$\n\\lim_{m \\rightarrow \\infty} \\frac{|\\mathcal{S}|}{2^{k m}} = \\lim_{m \\rightarrow \\infty} \\left(1 - \\frac{1}{2^{m}}\\right)^{k} = \\exp\\left(-\\lim_{m \\rightarrow \\infty} \\frac{k}{2^{m}}\\right) = \\delta\n$$\nand similarly\n$$\n\\lim_{m \\rightarrow \\infty} \\frac{|\\mathcal{T}|}{2^{k m}} = 1 - 2 \\lim_{m \\rightarrow \\infty} \\left(1 - \\frac{1}{2^{m}}\\right)^{k} + \\lim_{m \\rightarrow \\infty} \\left(1 - \\frac{2}{2^{m}}\\right)^{k} = 1 - 2\\delta + \\delta^{2} = \\delta\n$$\nHence, if $m$ is sufficiently large then $\\frac{|\\mathcal{S}|}{2^{m k}}$ and $\\frac{|\\mathcal{T}|}{2^{m k}}$ are greater than $\\alpha$ (since $\\alpha < \\delta$). So $|\\mathcal{S}|, |\\mathcal{T}| > \\alpha \\cdot 2^{m k} = \\alpha \\cdot 2^{n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31866,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a convex $n$-gon $P$ in the plane. For every three vertices of $P$, consider the triangle determined by them. Call such a triangle good if all its sides are of unit length.\nProve that there are not more than $\\frac{2}{3} n$ good triangles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider all good triangles containing a certain vertex $A$. The other two vertices of any such triangle lie on the circle $\\omega_{A}$ with unit radius and center $A$. Since $P$ is convex, all these vertices lie on an arc of angle less than $180^{\\circ}$. Let $L_{A} R_{A}$ be the shortest such arc, oriented clockwise (see Figure 1). Each of segments $A L_{A}$ and $A R_{A}$ belongs to a unique good triangle. We say that the good triangle with side $A L_{A}$ is assigned counterclockwise to $A$, and the second one, with side $A R_{A}$, is assigned clockwise to $A$. In those cases when there is a single good triangle containing vertex $A$, this triangle is assigned to $A$ twice.\nThere are at most two assignments to each vertex of the polygon. (Vertices which do not belong to any good triangle have no assignment.) So the number of assignments is at most $2 n$.\n\nConsider an arbitrary good triangle $A B C$, with vertices arranged clockwise. We prove that $A B C$ is assigned to its vertices at least three times. Then, denoting the number of good triangles by $t$, we obtain that the number $K$ of all assignments is at most $2 n$, while it is not less than $3 t$. Then $3 t \\leq K \\leq 2 n$, as required.\n\nActually, we prove that triangle $A B C$ is assigned either counterclockwise to $C$ or clockwise to $B$. Then, by the cyclic symmetry of the vertices, we obtain that triangle $A B C$ is assigned either counterclockwise to $A$ or clockwise to $C$, and either counterclockwise to $B$ or clockwise to $A$, providing the claim.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n\nAssume, to the contrary, that $L_{C} \\neq A$ and $R_{B} \\neq A$. Denote by $A'$, $B'$, $C'$ the intersection points of circles $\\omega_{A}$, $\\omega_{B}$ and $\\omega_{C}$, distinct from $A, B, C$ (see Figure 2). Let $C L_{C} L_{C}'$ be the good triangle containing $C L_{C}$. Observe that the angle of $\\operatorname{arc} L_{C} A$ is less than $120^{\\circ}$. Then one of the points $L_{C}$ and $L_{C}'$ belongs to arc $B' A$ of $\\omega_{C}$; let this point be $X$. In the case when $L_{C}=B'$ and $L_{C}'=A$, choose $X=B'$.\n\nAnalogously, considering the good triangle $B R_{B}' R_{B}$ which contains $B R_{B}$ as an edge, we see that one of the points $R_{B}$ and $R_{B}'$ lies on arc $A C'$ of $\\omega_{B}$. Denote this point by $Y$, $Y \\neq A$. Then angles $X A Y$, $Y A B$, $B A C$ and $C A X$ (oriented clockwise) are not greater than $180^{\\circ}$. Hence, point $A$ lies in quadrilateral $X Y B C$ (either in its interior or on segment $X Y$ ). This is impossible, since all these five points are vertices of $P$.\n\nHence, each good triangle has at least three assignments, and the statement is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31867,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given an isosceles triangle $A B C$ with $A B = A C$. The midpoint of side $B C$ is denoted by $M$. Let $X$ be a variable point on the shorter arc $M A$ of the circumcircle of triangle $A B M$. Let $T$ be the point in the angle domain $B M A$, for which $\\angle T M X = 90^{\\circ}$ and $T X = B X$. Prove that $\\angle M T B - \\angle C T M$ does not depend on $X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ be the midpoint of segment $B T$ (see Figure 1).\nLine $X N$ is the axis of symmetry in the isosceles triangle $B X T$, thus $\\angle T N X = 90^{\\circ}$ and $\\angle B X N = \\angle N X T$.\nMoreover, in triangle $B C T$, line $M N$ is the midline parallel to $C T$; hence $\\angle C T M = \\angle N M T$.\nDue to the right angles at points $M$ and $N$, these points lie on the circle with diameter $X T$. Therefore,\n$$\n\\angle M T B = \\angle M T N = \\angle M X N \\quad \\text{and} \\quad \\angle C T M = \\angle N M T = \\angle N X T = \\angle B X N.\n$$\nHence\n$$\n\\angle M T B - \\angle C T M = \\angle M X N - \\angle B X N = \\angle M X B = \\angle M A B\n$$\nwhich does not depend on $X$.\n\n![](attached_image_1.png)\nFigure 1\nLet $S$ be the reflection of point $T$ over $M$ (see Figure 2).\nThen $X M$ is the perpendicular bisector of $T S$, hence $X B = X T = X S$, and $X$ is the circumcenter of triangle $B S T$.\nMoreover, $\\angle B S M = \\angle C T M$ since they are symmetrical about $M$.\nThen\n$$\n\\angle M T B - \\angle C T M = \\angle S T B - \\angle B S T = \\frac{\\angle S X B - \\angle B X T}{2}.\n$$\nObserve that $\\angle S X B = \\angle S X T - \\angle B X T = 2 \\angle M X T - \\angle B X T$, so\n$$\n\\angle M T B - \\angle C T M = \\frac{2 \\angle M X T - 2 \\angle B X T}{2} = \\angle M X B = \\angle M A B,\n$$\nwhich is constant.\n\n![](attached_image_2.png)\nFigure 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31868,
    "subject": "Mathematics (Multi-modal)",
    "question": "The diagonals of a trapezoid $ABCD$ intersect at point $P$. Point $Q$ lies between the parallel lines $BC$ and $AD$ such that $\\angle AQD = \\angle CQB$, and line $CD$ separates points $P$ and $Q$. Prove that $\\angle BQP = \\angle DAQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $t = \\frac{AD}{BC}$. Consider the homothety $h$ with center $P$ and scale $-t$. Triangles $PDA$ and $PBC$ are similar with ratio $t$, hence $h(B) = D$ and $h(C) = A$.\n\n![](attached_image_1.png)\n\nLet $Q' = h(Q)$ (see Figure 1). Then points $Q$, $P$ and $Q'$ are obviously collinear. Points $Q$ and $P$ lie on the same side of $AD$, as well as on the same side of $BC$; hence $Q'$ and $P$ are also on the same side of $h(BC) = AD$, and therefore $Q$ and $Q'$ are on the same side of $AD$. Moreover, points $Q$ and $C$ are on the same side of $BD$, while $Q'$ and $A$ are on the opposite side (see Figure above).\n\nBy the homothety, $\\angle AQ'D = \\angle CQB = \\angle AQD$, hence quadrilateral $AQ'QD$ is cyclic. Then\n$$\n\\angle DAQ = \\angle DQ'Q = \\angle DQ'P = \\angle BQP\n$$\n(the latter equality is valid by the homothety again).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31869,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a fixed triangle, and let $A_{1}, B_{1}, C_{1}$ be the midpoints of sides $BC, CA, AB$, respectively. Let $P$ be a variable point on the circumcircle. Let lines $PA_{1}, PB_{1}, PC_{1}$ meet the circumcircle again at $A', B', C'$ respectively. Assume that the points $A, B, C, A', B', C'$ are distinct, and lines $AA', BB', CC'$ form a triangle. Prove that the area of this triangle does not depend on $P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_{0}, B_{0}, C_{0}$ be the points of intersection of the lines $AA', BB'$ and $CC'$ (see Figure). We claim that area $\\left(A_{0}B_{0}C_{0}\\right)=\\frac{1}{2} \\operatorname{area}(ABC)$, hence it is constant.\n\nConsider the inscribed hexagon $ABCC'PA'$. By Pascal's theorem, the points of intersection of its opposite sides (or of their extensions) are collinear. These points are $AB \\cap C'P = C_{1}$, $BC \\cap PA' = A_{1}$, $CC' \\cap A'A = B_{0}$. So point $B_{0}$ lies on the midline $A_{1}C_{1}$ of triangle $ABC$. Analogously, points $A_{0}$ and $C_{0}$ lie on lines $B_{1}C_{1}$ and $A_{1}B_{1}$, respectively.\n\nLines $AC$ and $A_{1}C_{1}$ are parallel, so triangles $B_{0}C_{0}A_{1}$ and $AC_{0}B_{1}$ are similar; hence we have\n$$\n\\frac{B_{0}C_{0}}{AC_{0}} = \\frac{A_{1}C_{0}}{B_{1}C_{0}}.\n$$\nAnalogously, from $BC \\parallel B_{1}C_{1}$ we obtain\n$$\n\\frac{A_{1}C_{0}}{B_{1}C_{0}} = \\frac{BC_{0}}{A_{0}C_{0}}.\n$$\nCombining these equalities, we get\n$$\n\\frac{B_{0}C_{0}}{AC_{0}} = \\frac{BC_{0}}{A_{0}C_{0}},\n$$\nor\n$$\nA_{0}C_{0} \\cdot B_{0}C_{0} = AC_{0} \\cdot BC_{0}.\n$$\nHence we have\n![](attached_image_1.png)\n$$\n\\operatorname{area}\\left(A_{0}B_{0}C_{0}\\right) = \\frac{1}{2} A_{0}C_{0} \\cdot B_{0}C_{0} \\sin \\angle A_{0}C_{0}B_{0} = \\frac{1}{2} AC_{0} \\cdot BC_{0} \\sin \\angle AC_{0}B = \\operatorname{area}\\left(ABC_{0}\\right).\n$$\nSince $C_{0}$ lies on the midline, we have $d\\left(C_{0}, AB\\right) = \\frac{1}{2} d(C, AB)$ (we denote by $d(X, YZ)$ the distance between point $X$ and line $YZ$). Then we obtain\n$$\n\\operatorname{area}\\left(A_{0}B_{0}C_{0}\\right) = \\operatorname{area}\\left(ABC_{0}\\right) = \\frac{1}{2} AB \\cdot d\\left(C_{0}, AB\\right) = \\frac{1}{4} AB \\cdot d(C, AB) = \\frac{1}{2} \\operatorname{area}(ABC).\n$$\nAgain, we prove that area $\\left(A_{0}B_{0}C_{0}\\right) = \\frac{1}{2} \\operatorname{area}(ABC)$.\n\nWe can assume that $P$ lies on arc $AC$. Mark a point $L$ on side $AC$ such that $\\angle CBL = \\angle PBA$; then $\\angle LBA = \\angle CBA - \\angle CBL = \\angle CBA - \\angle PBA = \\angle CBP$. Note also that $\\angle BAL = \\angle BAC = \\angle BPC$ and $\\angle LCB = \\angle APB$. Hence, triangles $BAL$ and $BPC$ are similar, and so are triangles $LCB$ and $APB$.\n\nAnalogously, mark points $K$ and $M$ respectively on the extensions of sides $CB$ and $AB$ beyond point $B$, such that $\\angle KAB = \\angle CAP$ and $\\angle BCM = \\angle PCA$. For analogous reasons, $\\angle KAC = \\angle BAP$ and $\\angle ACM = \\angle PCB$. Hence $\\triangle ABK \\sim \\triangle APC \\sim \\triangle MBC$, $\\triangle ACK \\sim \\triangle APB$, and $\\triangle MAC \\sim \\triangle BPC$. From these similarities, we have $\\angle CMB = \\angle KAB = \\angle CAP$, while we have seen that $\\angle CAP = \\angle CBP = \\angle LBA$. Hence, $AK \\parallel BL \\parallel CM$.\n\n![](attached_image_2.png)\n\nLet line $CC'$ intersect $BL$ at point $X$. Note that $\\angle LCX = \\angle ACC' = \\angle APC' = \\angle APC_{1}$, and $PC_{1}$ is a median in triangle $APB$. Since triangles $APB$ and $LCB$ are similar, $CX$ is a median in triangle $LCB$, and $X$ is a midpoint of $BL$. For the same reason, $AA'$ passes through this midpoint, so $X = B_{0}$. Analogously, $A_{0}$ and $C_{0}$ are the midpoints of $AK$ and $CM$.\n\nNow, from $AA_{0} \\parallel CC_{0}$, we have\n$$\n\\operatorname{area}\\left(A_{0}B_{0}C_{0}\\right) = \\operatorname{area}\\left(AC_{0}A_{0}\\right) - \\operatorname{area}\\left(AB_{0}A_{0}\\right) = \\operatorname{area}\\left(ACA_{0}\\right) - \\operatorname{area}\\left(AB_{0}A_{0}\\right) = \\operatorname{area}\\left(ACB_{0}\\right).\n$$\nFinally,\n$$\n\\operatorname{area}\\left(A_{0}B_{0}C_{0}\\right) = \\operatorname{area}\\left(ACB_{0}\\right) = \\frac{1}{2} B_{0}L \\cdot AC \\sin ALB_{0} = \\frac{1}{4} BL \\cdot AC \\sin ALB = \\frac{1}{2} \\operatorname{area}(ABC).\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31870,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the smallest positive real number $k$ with the following property.\nLet $ABCD$ be a convex quadrilateral, and let points $A_{1}, B_{1}, C_{1}$ and $D_{1}$ lie on sides $AB, BC$, $CD$ and $DA$, respectively. Consider the areas of triangles $AA_{1}D_{1}, BB_{1}A_{1}, CC_{1}B_{1}$, and $DD_{1}C_{1}$; let $S$ be the sum of the two smallest ones, and let $S_{1}$ be the area of quadrilateral $A_{1}B_{1}C_{1}D_{1}$. Then we always have $k S_{1} \\geq S$.",
    "options": [],
    "answer": "1",
    "solution": "Throughout the solution, triangles $AA_{1}D_{1}, BB_{1}A_{1}, CC_{1}B_{1}$, and $DD_{1}C_{1}$ will be referred to as border triangles. We will denote by $[\\mathcal{R}]$ the area of a region $\\mathcal{R}$.\n\nFirst, we show that $k \\geq 1$. Consider a triangle $ABC$ with unit area; let $A_{1}, B_{1}, K$ be the midpoints of its sides $AB, BC, AC$, respectively. Choose a point $D$ on the extension of $BK$, close to $K$. Take points $C_{1}$ and $D_{1}$ on sides $CD$ and $DA$ close to $D$ (see Figure 1). We have $[BB_{1}A_{1}] = \\frac{1}{4}$. Moreover, as $C_{1}, D_{1}, D \\rightarrow K$, we get $[A_{1}B_{1}C_{1}D_{1}] \\rightarrow [A_{1}B_{1}K] = \\frac{1}{4}$, $[AA_{1}D_{1}] \\rightarrow [AA_{1}K] = \\frac{1}{4}$, $[CC_{1}B_{1}] \\rightarrow [CKB_{1}] = \\frac{1}{4}$ and $[DD_{1}C_{1}] \\rightarrow 0$. Hence, the sum of the two smallest areas of border triangles tends to $\\frac{1}{4}$, as well as $[A_{1}B_{1}C_{1}D_{1}]$; therefore, their ratio tends to $1$, and $k \\geq 1$.\n\nWe are left to prove that $k=1$ satisfies the desired property.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n![](attached_image_3.png)\nFigure 3\n\nLemma. Let points $A_{1}, B_{1}, C_{1}$ lie respectively on sides $BC, CA, AB$ of a triangle $ABC$. Then $[A_{1}B_{1}C_{1}] \\geq \\min \\{[AC_{1}B_{1}], [BA_{1}C_{1}], [CB_{1}A_{1}]\\}$.\n\nProof. Let $A', B', C'$ be the midpoints of sides $BC, CA$ and $AB$, respectively.\n\nSuppose that two of points $A_{1}, B_{1}, C_{1}$ lie in one of triangles $AC'B'$, $BA'C'$, and $CB'A'$. (For convenience, let points $B_{1}$ and $C_{1}$ lie in triangle $AC'B'$; see Figure 2.) Let segments $B_{1}C_{1}$ and $AA_{1}$ intersect at point $X$. Then $X$ also lies in triangle $AC'B'$. Hence $A_{1}X \\geq AX$, and we have\n$$\n\\frac{[A_{1}B_{1}C_{1}]}{[AC_{1}B_{1}]} = \\frac{\\frac{1}{2} A_{1}X \\cdot B_{1}C_{1} \\cdot \\sin \\angle A_{1}XC_{1}}{\\frac{1}{2} AX \\cdot B_{1}C_{1} \\cdot \\sin \\angle AXB_{1}} = \\frac{A_{1}X}{AX} \\geq 1,\n$$\nas required.\n\nOtherwise, each one of triangles $AC'B'$, $BA'C'$, $CB'A'$ contains exactly one of points $A_{1}$, $B_{1}$, $C_{1}$, and we can assume that $BA_{1} < BA'$, $CB_{1} < CB'$, $AC_{1} < AC'$ (see Figure 3). Then lines $B_{1}A_{1}$ and $AB$ intersect at a point $Y$ on the extension of $AB$ beyond point $B$, hence $\\frac{[A_{1}B_{1}C_{1}]}{[A_{1}B_{1}C']} = \\frac{C_{1}Y}{C'Y} > 1$; also, lines $A_{1}C'$ and $CA$ intersect at a point $Z$ on the extension of $CA$ beyond point $A$, hence $\\frac{[A_{1}B_{1}C']}{[A_{1}B'C']} = \\frac{B_{1}Z}{B'Z} > 1$. Finally, since $A_{1}A' \\parallel B'C'$, we have $[A_{1}B_{1}C_{1}] > [A_{1}B_{1}C'] > [A_{1}B'C'] = [A'B'C'] = \\frac{1}{4}[ABC]$.\n\nNow, from $[A_{1}B_{1}C_{1}] + [AC_{1}B_{1}] + [BA_{1}C_{1}] + [CB_{1}A_{1}] = [ABC]$ we obtain that one of the remaining triangles $AC_{1}B_{1}, BA_{1}C_{1}, CB_{1}A_{1}$ has an area less than $\\frac{1}{4}[ABC]$, so it is less than $[A_{1}B_{1}C_{1}]$. $\\square$\n\nNow we return to the problem. We say that triangle $A_{1}B_{1}C_{1}$ is small if $[A_{1}B_{1}C_{1}]$ is less than each of $[BB_{1}A_{1}]$ and $[CC_{1}B_{1}]$; otherwise this triangle is big (the similar notion is introduced for triangles $B_{1}C_{1}D_{1}, C_{1}D_{1}A_{1}, D_{1}A_{1}B_{1}$). If both triangles $A_{1}B_{1}C_{1}$ and $C_{1}D_{1}A_{1}$ are big, then $[A_{1}B_{1}C_{1}]$ is not less than the area of some border triangle, and $[C_{1}D_{1}A_{1}]$ is not less than the area of another one; hence, $S_{1} = [A_{1}B_{1}C_{1}] + [C_{1}D_{1}A_{1}] \\geq S$. The same is valid for the pair of $B_{1}C_{1}D_{1}$ and $D_{1}A_{1}B_{1}$. So it is sufficient to prove that in one of these pairs both triangles are big.\n\nSuppose the contrary. Then there is a small triangle in each pair. Without loss of generality, assume that triangles $A_{1}B_{1}C_{1}$ and $D_{1}A_{1}B_{1}$ are small. We can assume also that $[A_{1}B_{1}C_{1}] \\leq [D_{1}A_{1}B_{1}]$. Note that in this case ray $D_{1}C_{1}$ intersects line $BC$.\n\nConsider two cases.\n\n![](attached_image_4.png)\nFigure 4\n![](attached_image_5.png)\nFigure 5\n\nCase 1. Ray $C_{1}D_{1}$ intersects line $AB$ at some point $K$. Let ray $D_{1}C_{1}$ intersect line $BC$ at point $L$ (see Figure 4). Then we have $[A_{1}B_{1}C_{1}] < [CC_{1}B_{1}] < [LC_{1}B_{1}]$, $[A_{1}B_{1}C_{1}] < [BB_{1}A_{1}]$ (both - since $[A_{1}B_{1}C_{1}]$ is small), and $[A_{1}B_{1}C_{1}] \\leq [D_{1}A_{1}B_{1}] < [AA_{1}D_{1}] < [KA_{1}D_{1}] < [KA_{1}C_{1}]$ (since triangle $D_{1}A_{1}B_{1}$ is small). This contradicts the Lemma, applied for triangle $A_{1}B_{1}C_{1}$ inside $LKB$.\n\nCase 2. Ray $C_{1}D_{1}$ does not intersect $AB$. Then choose a \"sufficiently far\" point $K$ on ray $BA$ such that $[KA_{1}C_{1}] > [A_{1}B_{1}C_{1}]$, and that ray $KC_{1}$ intersects line $BC$ at some point $L$ (see Figure 5). Since ray $C_{1}D_{1}$ does not intersect line $AB$, the points $A$ and $D_{1}$ are on different sides of $KL$; then $A$ and $D$ are also on different sides, and $C$ is on the same side as $A$ and $B$. Then analogously we have $[A_{1}B_{1}C_{1}] < [CC_{1}B_{1}] < [LC_{1}B_{1}]$ and $[A_{1}B_{1}C_{1}] < [BB_{1}A_{1}]$ since triangle $A_{1}B_{1}C_{1}$ is small. This (together with $[A_{1}B_{1}C_{1}] < [KA_{1}C_{1}]$) contradicts the Lemma again.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31871,
    "subject": "Mathematics (Multi-modal)",
    "question": "Point $P$ lies on side $AB$ of a convex quadrilateral $ABCD$. Let $\\omega$ be the incircle of triangle $CPD$, and let $I$ be its incenter. Suppose that $\\omega$ is tangent to the incircles of triangles $APD$ and $BPC$ at points $K$ and $L$, respectively. Let lines $AC$ and $BD$ meet at $E$, and let lines $AK$ and $BL$ meet at $F$. Prove that points $E$, $I$, and $F$ are collinear.\n(Poland)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\Omega$ be the circle tangent to segment $AB$ and to rays $AD$ and $BC$; let $J$ be its center. We prove that points $E$ and $F$ lie on line $IJ$.\n\n![](attached_image_1.png)\n\nDenote the incircles of triangles $ADP$ and $BCP$ by $\\omega_{A}$ and $\\omega_{B}$. Let $h_{1}$ be the homothety with a negative scale taking $\\omega$ to $\\Omega$. Consider this homothety as the composition of two homotheties: one taking $\\omega$ to $\\omega_{A}$ (with a negative scale and center $K$), and another one taking $\\omega_{A}$ to $\\Omega$ (with a positive scale and center $A$). It is known that in such a case the three centers of homothety are collinear (this theorem is also referred to as the theorem on the three similitude centers). Hence, the center of $h_{1}$ lies on line $AK$. Analogously, it also lies on $BL$, so this center is $F$. Hence, $F$ lies on the line of centers of $\\omega$ and $\\Omega$, i.e. on $IJ$ (if $I=J$, then $F=I$ as well, and the claim is obvious).\n\nConsider quadrilateral $APCD$ and mark the equal segments of tangents to $\\omega$ and $\\omega_{A}$ (see the figure below to the left). Since circles $\\omega$ and $\\omega_{A}$ have a common point of tangency with $PD$, one can easily see that $AD+PC=AP+CD$. So, quadrilateral $APCD$ is circumscribed; analogously, circumscribed is also quadrilateral $BCDP$. Let $\\Omega_{A}$ and $\\Omega_{B}$ respectively be their incircles.\n\n![](attached_image_2.png)\n![](attached_image_3.png)\n\nConsider the homothety $h_{2}$ with a positive scale taking $\\omega$ to $\\Omega$. Consider $h_{2}$ as the composition of two homotheties: taking $\\omega$ to $\\Omega_{A}$ (with a positive scale and center $C$), and taking $\\Omega_{A}$ to $\\Omega$ (with a positive scale and center $A$), respectively. So the center of $h_{2}$ lies on line $AC$. By analogous reasons, it lies also on $BD$, hence this center is $E$. Thus, $E$ also lies on the line of centers $IJ$, and the claim is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31872,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(k, n)$ of positive integers for which $7^{k}-3^{n}$ divides $k^{4}+n^{2}$.",
    "options": [],
    "answer": "(2,4)",
    "solution": "Suppose that a pair $(k, n)$ satisfies the condition of the problem. Since $7^{k}-3^{n}$ is even, $k^{4}+n^{2}$ is also even, hence $k$ and $n$ have the same parity. If $k$ and $n$ are odd, then $k^{4}+n^{2} \\equiv 1+1=2\\pmod{4}$, while $7^{k}-3^{n} \\equiv 7-3 \\equiv 0\\pmod{4}$, so $k^{4}+n^{2}$ cannot be divisible by $7^{k}-3^{n}$. Hence, both $k$ and $n$ must be even.\n\nWrite $k=2a$, $n=2b$. Then $7^{k}-3^{n}=7^{2a}-3^{2b}=\\frac{7^{a}-3^{b}}{2} \\cdot 2\\left(7^{a}+3^{b}\\right)$, and both factors are integers. So $2\\left(7^{a}+3^{b}\\right) \\mid 7^{k}-3^{n}$ and $7^{k}-3^{n} \\mid k^{4}+n^{2}=2\\left(8a^{4}+2b^{2}\\right)$, hence\n$$\n7^{a}+3^{b} \\leq 8a^{4}+2b^{2} . \\tag{1}\n$$\nWe prove by induction that $8a^{4}<7^{a}$ for $a \\geq 4$, $2b^{2}<3^{b}$ for $b \\geq 1$ and $2b^{2}+9 \\leq 3^{b}$ for $b \\geq 3$. In the initial cases $a=4$, $b=1$, $b=2$ and $b=3$ we have $8 \\cdot 4^{4}=2048<7^{4}=2401$, $2<3$, $2 \\cdot 2^{2}=8<3^{2}=9$ and $2 \\cdot 3^{2}+9=3^{3}=27$, respectively.\n\nIf $8a^{4}<7^{a}$ $(a \\geq 4)$ and $2b^{2}+9 \\leq 3^{b}$ $(b \\geq 3)$, then\n$$\n\\begin{aligned}\n8(a+1)^{4} & =8a^{4}\\left(\\frac{a+1}{a}\\right)^{4}<7^{a}\\left(\\frac{5}{4}\\right)^{4}=7^{a} \\frac{625}{256}<7^{a+1} \\quad \\text{ and } \\\\\n2(b+1)^{2}+9 & <\\left(2b^{2}+9\\right)\\left(\\frac{b+1}{b}\\right)^{2} \\leq 3^{b}\\left(\\frac{4}{3}\\right)^{2}=3^{b} \\frac{16}{9}<3^{b+1},\n\\end{aligned}\n$$\nas desired.\n\nFor $a \\geq 4$ we obtain $7^{a}+3^{b}>8a^{4}+2b^{2}$ and inequality (1) cannot hold. Hence $a \\leq 3$, and three cases are possible.\n\nCase 1: $a=1$. Then $k=2$ and $8+2b^{2} \\geq 7+3^{b}$, thus $2b^{2}+1 \\geq 3^{b}$. This is possible only if $b \\leq 2$. If $b=1$ then $n=2$ and $\\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\\frac{2^{4}+2^{2}}{7^{2}-3^{2}}=\\frac{1}{2}$, which is not an integer. If $b=2$ then $n=4$ and $\\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\\frac{2^{4}+4^{2}}{7^{2}-3^{4}}=-1$, so $(k, n)=(2,4)$ is a solution.\n\nCase 2: $a=2$. Then $k=4$ and $k^{4}+n^{2}=256+4b^{2} \\geq |7^{4}-3^{n}|=|49-3^{b}| \\cdot (49+3^{b})$. The smallest value of the first factor is 22, attained at $b=3$, so $128+2b^{2} \\geq 11(49+3^{b})$, which is impossible since $3^{b}>2b^{2}$.\n\nCase 3: $a=3$. Then $k=6$ and $k^{4}+n^{2}=1296+4b^{2} \\geq |7^{6}-3^{n}|=|343-3^{b}| \\cdot (343+3^{b})$. Analogously, $|343-3^{b}| \\geq 100$ and we have $324+b^{2} \\geq 25(343+3^{b})$, which is impossible again.\n\nWe find that there exists a unique solution $(k, n)=(2,4)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31873,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $b$, $n > 1$ be integers. Suppose that for each $k > 1$ there exists an integer $a_{k}$ such that $b - a_{k}^{n}$ is divisible by $k$. Prove that $b = A^{n}$ for some integer $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the prime factorization of $b$ be $b = p_{1}^{\\alpha_{1}} \\ldots p_{s}^{\\alpha_{s}}$, where $p_{1}, \\ldots, p_{s}$ are distinct primes. Our goal is to show that all exponents $\\alpha_{i}$ are divisible by $n$, then we can set $A = p_{1}^{\\alpha_{1} / n} \\ldots p_{s}^{\\alpha_{s} / n}$.\n\nApply the condition for $k = b^{2}$. The number $b - a_{k}^{n}$ is divisible by $b^{2}$ and hence, for each $1 \\leq i \\leq s$, it is divisible by $p_{i}^{2 \\alpha_{i}} > p_{i}^{\\alpha_{i}}$ as well. Therefore\n$$\na_{k}^{n} \\equiv b \\equiv 0 \\quad (\\bmod p_{i}^{\\alpha_{i}})\n$$\nand\n$$\na_{k}^{n} \\equiv b \\not\\equiv 0 \\quad (\\bmod p_{i}^{\\alpha_{i} + 1})\n$$\nwhich implies that the largest power of $p_{i}$ dividing $a_{k}^{n}$ is $p_{i}^{\\alpha_{i}}$. Since $a_{k}^{n}$ is a complete $n$th power, this implies that $\\alpha_{i}$ is divisible by $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31874,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X$ be a set of $10000$ integers, none of them is divisible by $47$. Prove that there exists a $2007$-element subset $Y$ of $X$ such that $a-b+c-d+e$ is not divisible by $47$ for any $a, b, c, d, e \\in Y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Call a set $M$ of integers good if $47 \\nmid a-b+c-d+e$ for any $a, b, c, d, e \\in M$.\n\nConsider the set $J=\\{-9,-7,-5,-3,-1,1,3,5,7,9\\}$. We claim that $J$ is good. Actually, for any $a, b, c, d, e \\in J$ the number $a-b+c-d+e$ is odd and\n$$\n-45 = (-9) - 9 + (-9) - 9 + (-9) \\leq a-b+c-d+e \\leq 9 - (-9) + 9 - (-9) + 9 = 45.\n$$\nBut there is no odd number divisible by $47$ between $-45$ and $45$.\n\nFor any $k = 1, \\ldots, 46$ consider the set\n$$\nA_{k} = \\{x \\in X \\mid \\exists j \\in J: \\quad kx \\equiv j \\pmod{47}\\}.\n$$\nIf $A_{k}$ is not good, then $47 \\mid a-b+c-d+e$ for some $a, b, c, d, e \\in A_{k}$, hence $47 \\mid k a - k b + k c - k d + k e$. But set $J$ contains numbers with the same residues modulo $47$, so $J$ also is not good. This is a contradiction; therefore each $A_{k}$ is a good subset of $X$.\n\nThen it suffices to prove that there exists a number $k$ such that $|A_{k}| \\geq 2007$. Note that each $x \\in X$ is contained in exactly $10$ sets $A_{k}$. Then\n$$\n\\sum_{k=1}^{46} |A_{k}| = 10|X| = 100000\n$$\nhence for some value of $k$ we have\n$$\n|A_{k}| \\geq \\frac{100000}{46} > 2173 > 2007\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31875,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every integer $k \\geq 2$, prove that $2^{3k}$ divides the number\n$$\n\\binom{2^{k+1}}{2^{k}} - \\binom{2^{k}}{2^{k-1}}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the notation $(2n-1)!! = 1 \\cdot 3 \\cdots (2n-1)$ and $(2n)!! = 2 \\cdot 4 \\cdots (2n) = 2^{n} n!$ for any positive integer $n$. Observe that $(2n)! = (2n)!! (2n-1)!! = 2^{n} n! (2n-1)!!$.\n\nFor any positive integer $n$ we have\n$$\n\\begin{aligned}\n& \\binom{4n}{2n} = \\frac{(4n)!}{(2n)!^{2}} = \\frac{2^{2n}(2n)!(4n-1)!!}{(2n)!^{2}} = \\frac{2^{2n}}{(2n)!}(4n-1)!! \\\\\n& \\binom{2n}{n} = \\frac{1}{(2n)!} \\left(\\frac{(2n)!}{n!}\\right)^{2} = \\frac{1}{(2n)!} \\left(2^{n}(2n-1)!!\\right)^{2} = \\frac{2^{2n}}{(2n)!}(2n-1)!!^{2}\n\\end{aligned}\n$$\n\nThen expression (1) can be rewritten as follows:\n$$\n\\begin{aligned}\n\\binom{2^{k+1}}{2^{k}} - \\binom{2^{k}}{2^{k-1}} &= \\frac{2^{2^{k}}}{(2^{k})!} (2^{k+1}-1)!! - \\frac{2^{2^{k}}}{(2^{k})!} (2^{k}-1)!!^{2} \\\\\n&= \\frac{2^{2^{k}} (2^{k}-1)!!}{(2^{k})!} \\cdot \\left( (2^{k}+1)(2^{k}+3) \\ldots (2^{k}+2^{k}-1) - (2^{k}-1)(2^{k}-3) \\ldots (2^{k}-2^{k}+1) \\right)\n\\end{aligned}\n$$\n\nWe compute the exponent of $2$ in the prime decomposition of each factor (the first one is a rational number but not necessarily an integer; it is not important).\n\nFirst, we show by induction on $n$ that the exponent of $2$ in $(2^{n})!$ is $2^{n}-1$. The base case $n=1$ is trivial. Suppose that $(2^{n})! = 2^{2^{n}-1}(2d+1)$ for some integer $d$. Then we have\n$$\n(2^{n+1})! = 2^{2^{n}} (2^{n})! (2^{n+1}-1)!! = 2^{2^{n}} 2^{2^{n}-1} \\cdot (2d+1) (2^{n+1}-1)!! = 2^{2^{n+1}-1} \\cdot (2q+1)\n$$\nfor some integer $q$. This finishes the induction step.\n\nHence, the exponent of $2$ in the first factor in (2) is $2^{k} - (2^{k}-1) = 1$.\n\nThe second factor in (2) can be considered as the value of the polynomial\n$$\nP(x) = (x+1)(x+3) \\ldots (x+2^{k}-1) - (x-1)(x-3) \\ldots (x-2^{k}+1)\n$$\nat $x = 2^{k}$. Now we collect some information about $P(x)$.\n\nObserve that $P(-x) = -P(x)$, since $k \\geq 2$. So $P(x)$ is an odd function, and it has nonzero coefficients only at odd powers of $x$. Hence $P(x) = x^{3} Q(x) + c x$, where $Q(x)$ is a polynomial with integer coefficients.\n\nCompute the exponent of $2$ in $c$. We have\n$$\n\\begin{aligned}\nc &= 2 (2^{k}-1)!! \\sum_{i=1}^{2^{k-1}} \\frac{1}{2i-1} = (2^{k}-1)!! \\sum_{i=1}^{2^{k-1}} \\left( \\frac{1}{2i-1} + \\frac{1}{2^{k}-2i+1} \\right) \\\\\n&= (2^{k}-1)!! \\sum_{i=1}^{2^{k-1}} \\frac{2^{k}}{(2i-1)(2^{k}-2i+1)} = 2^{k} \\sum_{i=1}^{2^{k-1}} \\frac{(2^{k}-1)!!}{(2i-1)(2^{k}-2i+1)} = 2^{k} S\n\\end{aligned}\n$$\n\nFor any integer $i = 1, \\ldots, 2^{k-1}$, denote by $a_{2i-1}$ the residue inverse to $2i-1$ modulo $2^{k}$. Clearly, when $2i-1$ runs through all odd residues, so does $a_{2i-1}$, hence\n$$\n\\begin{aligned}\nS &= \\sum_{i=1}^{2^{k-1}} \\frac{(2^{k}-1)!!}{(2i-1)(2^{k}-2i+1)} \\equiv -\\sum_{i=1}^{2^{k-1}} \\frac{(2^{k}-1)!!}{(2i-1)^{2}} \\equiv -\\sum_{i=1}^{2^{k-1}} (2^{k}-1)!! a_{2i-1}^{2} \\\\\n&= - (2^{k}-1)!! \\sum_{i=1}^{2^{k-1}} (2i-1)^{2} = - (2^{k}-1)!! \\frac{2^{k-1}(2^{2k}-1)}{3} \\quad (\\bmod 2^{k})\n\\end{aligned}\n$$\n\nTherefore, the exponent of $2$ in $S$ is $k-1$, so $c = 2^{k} S = 2^{2k-1}(2t+1)$ for some integer $t$.\n\nFinally we obtain that\n$$\nP(2^{k}) = 2^{3k} Q(2^{k}) + 2^{k} c = 2^{3k} Q(2^{k}) + 2^{3k-1}(2t+1)\n$$\nwhich is divisible exactly by $2^{3k-1}$. Thus, the exponent of $2$ in (2) is $1 + (3k-1) = 3k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31876,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all surjective functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that for every $m, n \\in \\mathbb{N}$ and every prime $p$, the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p$.\n\n($\\mathbb{N}$ is the set of all positive integers.)",
    "options": [],
    "answer": "f(n) = n for all n ∈ ℕ",
    "solution": "Suppose that function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfies the problem conditions.\n\nLemma. For any prime $p$ and any $x, y \\in \\mathbb{N}$, we have $x \\equiv y (\\bmod p)$ if and only if $f(x) \\equiv f(y) (\\bmod p)$. Moreover, $p \\mid f(x)$ if and only if $p \\mid x$.\n\nProof. Consider an arbitrary prime $p$. Since $f$ is surjective, there exists some $x \\in \\mathbb{N}$ such that $p \\mid f(x)$. Let\n$$\nd = \\min \\{x \\in \\mathbb{N}: p \\mid f(x)\\}\n$$\nBy induction on $k$, we obtain that $p \\mid f(kd)$ for all $k \\in \\mathbb{N}$. The base is true since $p \\mid f(d)$. Moreover, if $p \\mid f(kd)$ and $p \\mid f(d)$ then, by the problem condition, $p \\mid f(kd+d) = f((k+1)d)$ as required.\n\nSuppose that there exists an $x \\in \\mathbb{N}$ such that $d \\nmid x$ but $p \\mid f(x)$. Let\n$$\ny = \\min \\{x \\in \\mathbb{N}: d \\nmid x, p \\mid f(x)\\}.\n$$\nBy the choice of $d$, we have $y > d$, and $y-d$ is a positive integer not divisible by $d$. Then $p \\nmid f(y-d)$, while $p \\mid f(d)$ and $p \\mid f(d + (y-d)) = f(y)$. This contradicts the problem condition. Hence, there is no such $x$, and\n$$\n\\begin{equation*}\np \\mid f(x) \\Longleftrightarrow d \\mid x. \\tag{1}\n\\end{equation*}\n$$\nTake arbitrary $x, y \\in \\mathbb{N}$ such that $x \\equiv y (\\bmod d)$. We have $p \\mid f(x + (2xd - x)) = f(2xd)$; moreover, since $d \\mid 2xd + (y-x) = y + (2xd - x)$, we get $p \\mid f(y + (2xd - x))$. Then by the problem condition $p \\mid f(x) + f(2xd - x)$, $p \\mid f(y) + f(2xd - x)$, and hence $f(x) \\equiv -f(2xd - x) \\equiv f(y) (\\bmod p)$.\n\nOn the other hand, assume that $f(x) \\equiv f(y) (\\bmod p)$. Again we have $p \\mid f(x) + f(2xd - x)$ which by our assumption implies that $p \\mid f(x) + f(2xd - x) + (f(y) - f(x)) = f(y) + f(2xd - x)$. Hence by the problem condition $p \\mid f(y + (2xd - x))$. Using (1) we get $0 \\equiv y + (2xd - x) \\equiv y - x (\\bmod d)$.\n\nThus, we have proved that\n$$\n\\begin{equation*}\nx \\equiv y \\ (\\bmod d) \\Longleftrightarrow f(x) \\equiv f(y) \\ (\\bmod p). \\tag{2}\n\\end{equation*}\n$$\nWe are left to show that $p = d$: in this case (1) and (2) provide the desired statements.\n\nThe numbers $1, 2, \\ldots, d$ have distinct residues modulo $d$. By (2), numbers $f(1), f(2), \\ldots, f(d)$ have distinct residues modulo $p$; hence there are at least $d$ distinct residues, and $p \\geq d$. On the other hand, by the surjectivity of $f$, there exist $x_1, \\ldots, x_p \\in \\mathbb{N}$ such that $f(x_i) = i$ for any $i = 1, 2, \\ldots, p$. By (2), all these $x_i$'s have distinct residues modulo $d$. For the same reasons, $d \\geq p$. Hence, $d = p$.\n\nNow we prove that $f(n) = n$ by induction on $n$. If $n = 1$ then, by the Lemma, $p \\nmid f(1)$ for any prime $p$, so $f(1) = 1$, and the base is established. Suppose that $n > 1$ and denote $k = f(n)$. Note that there exists a prime $q \\mid n$, so by the Lemma $q \\mid k$ and $k > 1$.\n\nIf $k > n$ then $k - n + 1 > 1$, and there exists a prime $p \\mid k - n + 1$; we have $k \\equiv n - 1 (\\bmod p)$. By the induction hypothesis we have $f(n-1) = n-1 \\equiv k = f(n) (\\bmod p)$. Now, by the Lemma we obtain $n-1 \\equiv n (\\bmod p)$ which cannot be true.\n\nAnalogously, if $k < n$, then $f(k-1) = k-1$ by induction hypothesis. Moreover, $n - k + 1 > 1$, so there exists a prime $p \\mid n - k + 1$ and $n \\equiv k - 1 (\\bmod p)$. By the Lemma again, $k = f(n) \\equiv f(k-1) = k-1 (\\bmod p)$, which is also false. The only remaining case is $k = n$, so $f(n) = n$.\n\nFinally, the function $f(n) = n$ obviously satisfies the condition.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31877,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. Prove that the number $\\left(4 k^{2}-1\\right)^{2}$ has a positive divisor of the form $8 k n-1$ if and only if $k$ is even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The statement follows from the following fact.\n\nLemma. For arbitrary positive integers $x$ and $y$, the number $4 x y-1$ divides $\\left(4 x^{2}-1\\right)^{2}$ if and only if $x=y$.\n\nProof. If $x=y$ then $4 x y-1=4 x^{2}-1$ obviously divides $\\left(4 x^{2}-1\\right)^{2}$ so it is sufficient to consider the opposite direction.\n\nCall a pair $(x, y)$ of positive integers bad if $4 x y-1$ divides $\\left(4 x^{2}-1\\right)^{2}$ but $x \\neq y$. In order to prove that bad pairs do not exist, we present two properties of them which provide an infinite descent.\n\nProperty (i). If $(x, y)$ is a bad pair and $x<y$ then there exists a positive integer $z<x$ such that $(x, z)$ is also bad.\n\nLet $r=\\frac{\\left(4 x^{2}-1\\right)^{2}}{4 x y-1}$. Then\n$$\nr=-r \\cdot(-1) \\equiv -r(4 x y-1) = -\\left(4 x^{2}-1\\right)^{2} \\equiv -1 \\quad (\\bmod 4 x)\n$$\nand $r=4 x z-1$ with some positive integer $z$. From $x<y$ we obtain that\n$$\n4 x z-1=\\frac{\\left(4 x^{2}-1\\right)^{2}}{4 x y-1}<4 x^{2}-1\n$$\nand therefore $z<x$. By the construction, the number $4 x z-1$ is a divisor of $\\left(4 x^{2}-1\\right)^{2}$ so $(x, z)$ is a bad pair.\n\nProperty (ii). If $(x, y)$ is a bad pair then $(y, x)$ is also bad.\n\nSince $1=1^{2} \\equiv (4 x y)^{2} (\\bmod 4 x y-1)$, we have\n$$\n\\left(4 y^{2}-1\\right)^{2} \\equiv \\left(4 y^{2}-(4 x y)^{2}\\right)^{2}=16 y^{4}\\left(4 x^{2}-1\\right)^{2} \\equiv 0 \\quad (\\bmod 4 x y-1)\n$$\nHence, the number $4 x y-1$ divides $\\left(4 y^{2}-1\\right)^{2}$ as well.\n\nNow suppose that there exists at least one bad pair. Take a bad pair $(x, y)$ such that $2 x+y$ attains its smallest possible value. If $x<y$ then property (i) provides a bad pair $(x, z)$ with $z<y$ and thus $2 x+z<2 x+y$. Otherwise, if $y<x$, property (ii) yields that pair $(y, x)$ is also bad while $2 y+x<2 x+y$. Both cases contradict the assumption that $2 x+y$ is minimal; the Lemma is proved.\n\nTo prove the problem statement, apply the Lemma for $x=k$ and $y=2 n$; the number $8 k n-1$ divides $\\left(4 k^{2}-1\\right)^{2}$ if and only if $k=2 n$. Hence, there is no such $n$ if $k$ is odd and $n=k / 2$ is the only solution if $k$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31878,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a prime $p$ and a positive integer $n$, denote by $\\nu_{p}(n)$ the exponent of $p$ in the prime factorization of $n!$. Given a positive integer $d$ and a finite set $\\{p_{1}, \\ldots, p_{k}\\}$ of primes. Show that there are infinitely many positive integers $n$ such that $d \\mid \\nu_{p_{i}}(n)$ for all $1 \\leq i \\leq k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For arbitrary prime $p$ and positive integer $n$, denote by $\\operatorname{ord}_{p}(n)$ the exponent of $p$ in $n$. Thus,\n$$\n\\nu_{p}(n)=\\operatorname{ord}_{p}(n!)=\\sum_{i=1}^{n} \\operatorname{ord}_{p}(i)\n$$\nLemma. Let $p$ be a prime number, $q$ be a positive integer, $k$ and $r$ be positive integers such that $p^{k}>r$. Then $\\nu_{p}\\left(q p^{k}+r\\right)=\\nu_{p}\\left(q p^{k}\\right)+\\nu_{p}(r)$.\nProof. We claim that $\\operatorname{ord}_{p}\\left(q p^{k}+i\\right)=\\operatorname{ord}_{p}(i)$ for all $0<i<p^{k}$. Actually, if $d=\\operatorname{ord}_{p}(i)$ then $d<k$, so $q p^{k}+i$ is divisible by $p^{d}$, but only the first term is divisible by $p^{d+1}$; hence the sum is not.\nUsing this claim, we obtain\n$$\n\\nu_{p}\\left(q p^{k}+r\\right)=\\sum_{i=1}^{q p^{k}} \\operatorname{ord}_{p}(i)+\\sum_{i=q p^{k}+1}^{q p^{k}+r} \\operatorname{ord}_{p}(i)=\\sum_{i=1}^{q p^{k}} \\operatorname{ord}_{p}(i)+\\sum_{i=1}^{r} \\operatorname{ord}_{p}(i)=\\nu_{p}\\left(q p^{k}\\right)+\\nu_{p}(r)\n$$\nFor any integer $a$, denote by $\\bar{a}$ its residue modulo $d$. The addition of residues will also be performed modulo $d$, i.e. $\\bar{a}+\\bar{b}=\\overline{a+b}$. For any positive integer $n$, let $f(n)=(f_{1}(n), \\ldots, f_{k}(n))$, where $f_{i}(n)=\\overline{\\nu_{p_{i}}(n)}$.\nDefine the sequence $n_{1}=1, n_{\\ell+1}=(p_{1} p_{2} \\ldots p_{k})^{n_{\\ell}}$. We claim that\n$$\nf\\left(n_{\\ell_{1}}+n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f\\left(n_{\\ell_{1}}\\right)+f\\left(n_{\\ell_{2}}\\right)+\\ldots+f\\left(n_{\\ell_{m}}\\right)\n$$\nfor any $\\ell_{1}<\\ell_{2}<\\ldots<\\ell_{m}$. (The addition of $k$-tuples is componentwise.) The base case $m=1$ is trivial.\nSuppose that $m>1$. By the construction of the sequence, $p_{i}^{n_{\\ell_{1}}}$ divides $n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}$; clearly, $p_{i}^{n_{\\ell_{1}}}>n_{\\ell_{1}}$ for all $1 \\leq i \\leq k$. Therefore the Lemma can be applied for $p=p_{i}, k=r=n_{\\ell_{1}}$ and $q p^{k}=n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}$ to obtain\n$$\nf_{i}\\left(n_{\\ell_{1}}+n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f_{i}\\left(n_{\\ell_{1}}\\right)+f_{i}\\left(n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right) \\quad \\text{ for all } 1 \\leq i \\leq k\n$$\nand hence\n$$\nf\\left(n_{\\ell_{1}}+n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f\\left(n_{\\ell_{1}}\\right)+f\\left(n_{\\ell_{2}}+\\ldots+n_{\\ell_{m}}\\right)=f\\left(n_{\\ell_{1}}\\right)+f\\left(n_{\\ell_{2}}\\right)+\\ldots+f\\left(n_{\\ell_{m}}\\right)\n$$\nby the induction hypothesis.\nNow consider the values $f\\left(n_{1}\\right), f\\left(n_{2}\\right), \\ldots$ There exist finitely many possible values of $f$. Hence, there exists an infinite sequence of indices $\\ell_{1}<\\ell_{2}<\\ldots$ such that $f\\left(n_{\\ell_{1}}\\right)=f\\left(n_{\\ell_{2}}\\right)=\\ldots$. and thus\n$$\nf\\left(n_{\\ell_{m+1}}+n_{\\ell_{m+2}}+\\ldots+n_{\\ell_{m+d}}\\right)=f\\left(n_{\\ell_{m+1}}\\right)+\\ldots+f\\left(n_{\\ell_{m+d}}\\right)=d \\cdot f\\left(n_{\\ell_{1}}\\right)=(\\overline{0}, \\ldots, \\overline{0})\n$$\nfor all $m$. We have found infinitely many suitable numbers.\nWe use the same Lemma and definition of the function $f$.\nLet $S=\\{f(n): n \\in \\mathbb{N}\\}$. Obviously, set $S$ is finite. For every $s \\in S$ choose the minimal $n_{s}$ such that $f\\left(n_{s}\\right)=s$. Denote $N=\\max_{s \\in S} n_{s}$. Moreover, let $g$ be an integer such that $p_{i}^{g}>N$ for each $i=1,2, \\ldots, k$. Let $P=(p_{1} p_{2} \\ldots p_{k})^{g}$.\nWe claim that\n$$\n\\{f(n) \\mid n \\in[m P, m P+N]\\}=S \\tag{1}\n$$\nfor every positive integer $m$. In particular, since $(\\overline{0}, \\ldots, \\overline{0})=f(1) \\in S$, it follows that for an arbitrary $m$ there exists $n \\in[m P, m P+N]$ such that $f(n)=(\\overline{0}, \\ldots, \\overline{0})$. So there are infinitely many suitable numbers.\nTo prove (1), let $a_{i}=f_{i}(m P)$. Consider all numbers of the form $n_{m, s}=m P+n_{s}$ with $s=(s_{1}, \\ldots, s_{k}) \\in S$ (clearly, all $n_{m, s}$ belong to $[m P, m P+N]$ ). Since $n_{s} \\leq N<p_{i}^{g}$ and $p_{i}^{g} \\mid m P$, we can apply the Lemma for the values $p=p_{i}, r=n_{s}, k=g, q p^{k}=m P$ to obtain\n$$\nf_{i}\\left(n_{m, s}\\right)=f_{i}(m P)+f_{i}\\left(n_{s}\\right)=a_{i}+s_{i} ;\n$$\nhence for distinct $s, t \\in S$ we have $f\\left(n_{m, s}\\right) \\neq f\\left(n_{m, t}\\right)$.\nThus, the function $f$ attains at least $|S|$ distinct values in $[m P, m P+N]$. Since all these values belong to $S, f$ should attain all possible values in $[m P, m P+N]$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31879,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S \\subseteq \\mathbb{R}$ be a set of real numbers. We say that a pair $(f, g)$ of functions from $S$ into $S$ is a Spanish Couple on $S$, if they satisfy the following conditions:\n(i) Both functions are strictly increasing, i.e. $f(x) < f(y)$ and $g(x) < g(y)$ for all $x, y \\in S$ with $x < y$;\n(ii) The inequality $f(g(g(x))) < g(f(x))$ holds for all $x \\in S$.\nDecide whether there exists a Spanish Couple\n\na. on the set $S = \\mathbb{N}$ of positive integers;\nb. on the set $S = \\{ a - 1 / b : a, b \\in \\mathbb{N} \\}$.",
    "options": [],
    "answer": "a: no; b: yes",
    "solution": "We show that the answer is NO for part (a), and YES for part (b).\n\na.\nThroughout the solution, we will use the notation $g_{k}(x) = \\overbrace{g(g(\\ldots g}^{k}(x) \\ldots))$, including $g_{0}(x) = x$ as well.\nSuppose that there exists a Spanish Couple $(f, g)$ on the set $\\mathbb{N}$. From property (i) we have $f(x) \\geq x$ and $g(x) \\geq x$ for all $x \\in \\mathbb{N}$.\n\nWe claim that $g_{k}(x) \\leq f(x)$ for all $k \\geq 0$ and all positive integers $x$. The proof is done by induction on $k$. We already have the base case $k = 0$ since $x \\leq f(x)$. For the induction step from $k$ to $k+1$, apply the induction hypothesis on $g_{2}(x)$ instead of $x$, then apply (ii):\n$$\ng\\left(g_{k+1}(x)\\right) = g_{k}\\left(g_{2}(x)\\right) \\leq f\\left(g_{2}(x)\\right) < g(f(x))\n$$\nSince $g$ is increasing, it follows that $g_{k+1}(x) < f(x)$. The claim is proven.\n\nIf $g(x) = x$ for all $x \\in \\mathbb{N}$ then $f(g(g(x))) = f(x) = g(f(x))$, and we have a contradiction with (ii). Therefore one can choose an $x_{0} \\in S$ for which $x_{0} < g\\left(x_{0}\\right)$. Now consider the sequence $x_{0}, x_{1}, \\ldots$ where $x_{k} = g_{k}\\left(x_{0}\\right)$. The sequence is increasing. Indeed, we have $x_{0} < g\\left(x_{0}\\right) = x_{1}$, and $x_{k} < x_{k+1}$ implies $x_{k+1} = g\\left(x_{k}\\right) < g\\left(x_{k+1}\\right) = x_{k+2}$.\n\nHence, we obtain a strictly increasing sequence $x_{0} < x_{1} < \\ldots$ of positive integers which on the other hand has an upper bound, namely $f\\left(x_{0}\\right)$. This cannot happen in the set $\\mathbb{N}$ of positive integers, thus no Spanish Couple exists on $\\mathbb{N}$.\n\nb.\nWe present a Spanish Couple on the set $S = \\{ a - 1 / b : a, b \\in \\mathbb{N} \\}$.\nLet\n$$\n\\begin{aligned}\n& f(a - 1 / b) = a + 1 - 1 / b \\\\\n& g(a - 1 / b) = a - 1 / (b + 3^{a})\n\\end{aligned}\n$$\nThese functions are clearly increasing. Condition (ii) holds, since\n$$\nf(g(g(a - 1 / b))) = (a + 1) - 1 / (b + 2 \\cdot 3^{a}) < (a + 1) - 1 / (b + 3^{a+1}) = g(f(a - 1 / b))\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31880,
    "subject": "Mathematics (Multi-modal)",
    "question": "For an integer $m$, denote by $t(m)$ the unique number in $\\{1,2,3\\}$ such that $m+t(m)$ is a multiple of $3$. A function $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ satisfies $f(-1)=0$, $f(0)=1$, $f(1)=-1$ and\n$$\nf\\left(2^{n}+m\\right)=f\\left(2^{n}-t(m)\\right)-f(m) \\quad \\text{ for all integers } m, n \\geq 0 \\text{ with } 2^{n}>m.\n$$\n\nProve that $f(3p) \\geq 0$ holds for all integers $p \\geq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given conditions determine $f$ uniquely on the positive integers. The signs of $f(1), f(2), \\ldots$ seem to change quite erratically. However, values of the form $f\\left(2^{n}-t(m)\\right)$ are sufficient to compute directly any functional value. Indeed, let $n>0$ have base $2$ representation $n=2^{a_{0}}+2^{a_{1}}+\\cdots+2^{a_{k}}, a_{0}>a_{1}>\\cdots>a_{k} \\geq 0$, and let $n_{j}=2^{a_{j}}+2^{a_{j-1}}+\\cdots+2^{a_{k}}, j=0, \\ldots, k$. Repeated applications of the recurrence show that $f(n)$ is an alternating sum of the quantities $f\\left(2^{a_{j}}-t\\left(n_{j+1}\\right)\\right)$ plus $(-1)^{k+1}$. (The exact formula is not needed for our proof.)\n\nSo we focus attention on the values $f\\left(2^{n}-1\\right), f\\left(2^{n}-2\\right)$ and $f\\left(2^{n}-3\\right)$. Six cases arise; more specifically,\n$t\\left(2^{2k}-3\\right)=2$, $t\\left(2^{2k}-2\\right)=1$, $t\\left(2^{2k}-1\\right)=3$, $t\\left(2^{2k+1}-3\\right)=1$, $t\\left(2^{2k+1}-2\\right)=3$, $t\\left(2^{2k+1}-1\\right)=2$.\n\nClaim. For all integers $k \\geq 0$ the following equalities hold:\n$$\n\\begin{array}{lll}\nf\\left(2^{2k+1}-3\\right)=0, & f\\left(2^{2k+1}-2\\right)=3^{k}, & f\\left(2^{2k+1}-1\\right)=-3^{k}, \\\\\nf\\left(2^{2k+2}-3\\right)=-3^{k}, & f\\left(2^{2k+2}-2\\right)=-3^{k}, & f\\left(2^{2k+2}-1\\right)=2 \\cdot 3^{k}\n\\end{array}\n$$\n\nProof. By induction on $k$. The base $k=0$ comes down to checking that $f(2)=-1$ and $f(3)=2$; the given values $f(-1)=0, f(0)=1, f(1)=-1$ are also needed. Suppose the claim holds for $k-1$. For $f\\left(2^{2k+1}-t(m)\\right)$, the recurrence formula and the induction hypothesis yield\n$$\n\\begin{aligned}\n& f\\left(2^{2k+1}-3\\right)=f\\left(2^{2k}+(2^{2k}-3)\\right)=f\\left(2^{2k}-2\\right)-f\\left(2^{2k}-3\\right)=-3^{k-1}+3^{k-1}=0, \\\\\n& f\\left(2^{2k+1}-2\\right)=f\\left(2^{2k}+(2^{2k}-2)\\right)=f\\left(2^{2k}-1\\right)-f\\left(2^{2k}-2\\right)=2 \\cdot 3^{k-1}+3^{k-1}=3^{k}, \\\\\n& f\\left(2^{2k+1}-1\\right)=f\\left(2^{2k}+(2^{2k}-1)\\right)=f\\left(2^{2k}-3\\right)-f\\left(2^{2k}-1\\right)=-3^{k-1}-2 \\cdot 3^{k-1}=-3^{k} .\n\\end{aligned}\n$$\nFor $f\\left(2^{2k+2}-t(m)\\right)$ we use the three equalities just established:\n$$\n\\begin{aligned}\n& f\\left(2^{2k+2}-3\\right)=f\\left(2^{2k+1}+(2^{2k+1}-3)\\right)=f\\left(2^{2k+1}-1\\right)-f\\left(2^{2k+1}-3\\right)=-3^{k}-0=-3^{k}, \\\\\n& f\\left(2^{2k+2}-2\\right)=f\\left(2^{2k+1}+(2^{2k+1}-2)\\right)=f\\left(2^{2k+1}-3\\right)-f\\left(2^{2k}-2\\right)=0-3^{k}=-3^{k}, \\\\\n& f\\left(2^{2k+2}-1\\right)=f\\left(2^{2k+1}+(2^{2k+1}-1)\\right)=f\\left(2^{2k+1}-2\\right)-f\\left(2^{2k+1}-1\\right)=3^{k}+3^{k}=2 \\cdot 3^{k} .\n\\end{aligned}\n$$\nThe claim follows.\n\nA closer look at the six cases shows that $f\\left(2^{n}-t(m)\\right) \\geq 3^{(n-1)/2}$ if $2^{n}-t(m)$ is divisible by $3$, and $f\\left(2^{n}-t(m)\\right) \\leq 0$ otherwise. On the other hand, note that $2^{n}-t(m)$ is divisible by $3$ if and only if $2^{n}+m$ is. Therefore, for all nonnegative integers $m$ and $n$,\n(i) $f\\left(2^{n}-t(m)\\right) \\geq 3^{(n-1)/2}$ if $2^{n}+m$ is divisible by $3$ ;\n(ii) $f\\left(2^{n}-t(m)\\right) \\leq 0$ if $2^{n}+m$ is not divisible by $3$ .\n\nOne more (direct) consequence of the claim is that $\\left|f\\left(2^{n}-t(m)\\right)\\right| \\leq \\frac{2}{3} \\cdot 3^{n/2}$ for all $m, n \\geq 0$.\n\nThe last inequality enables us to find an upper bound for $|f(m)|$ for $m$ less than a given power of $2$. We prove by induction on $n$ that $|f(m)| \\leq 3^{n/2}$ holds true for all integers $m, n \\geq 0$ with $2^{n}>m$.\n\nThe base $n=0$ is clear as $f(0)=1$. For the inductive step from $n$ to $n+1$, let $m$ and $n$ satisfy $2^{n+1}>m$. If $m<2^{n}$, we are done by the inductive hypothesis. If $m \\geq 2^{n}$ then $m=2^{n}+k$ where $2^{n}>k \\geq 0$. Now, by $\\left|f\\left(2^{n}-t(k)\\right)\\right| \\leq \\frac{2}{3} \\cdot 3^{n/2}$ and the inductive assumption,\n$$\n|f(m)|=\\left|f\\left(2^{n}-t(k)\\right)-f(k)\\right| \\leq\\left|f\\left(2^{n}-t(k)\\right)\\right|+|f(k)| \\leq \\frac{2}{3} \\cdot 3^{n/2}+3^{n/2}<3^{(n+1)/2}\n$$\nThe induction is complete.\n\nWe proceed to prove that $f(3p) \\geq 0$ for all integers $p \\geq 0$. Since $3p$ is not a power of $2$, its binary expansion contains at least two summands. Hence one can write $3p=2^{a}+2^{b}+c$ where $a>b$ and $2^{b}>c \\geq 0$. Applying the recurrence formula twice yields\n$$\nf(3p)=f\\left(2^{a}+2^{b}+c\\right)=f\\left(2^{a}-t\\left(2^{b}+c\\right)\\right)-f\\left(2^{b}-t(c)\\right)+f(c)\n$$\nSince $2^{a}+2^{b}+c$ is divisible by $3$, we have $f\\left(2^{a}-t\\left(2^{b}+c\\right)\\right) \\geq 3^{(a-1)/2}$ by (i). Since $2^{b}+c$ is not divisible by $3$, we have $f\\left(2^{b}-t(c)\\right) \\leq 0$ by (ii). Finally $|f(c)| \\leq 3^{b/2}$ as $2^{b}>c \\geq 0$, so that $f(c) \\geq -3^{b/2}$. Therefore $f(3p) \\geq 3^{(a-1)/2}-3^{b/2}$ which is nonnegative because $a>b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31881,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be positive real numbers such that\n$$\na b c d = 1 \\quad \\text{ and } \\quad a + b + c + d > \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a}\n$$\nProve that\n$$\na + b + c + d < \\frac{b}{a} + \\frac{c}{b} + \\frac{d}{c} + \\frac{a}{d}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show that if $a b c d = 1$, the sum $a + b + c + d$ cannot exceed a certain weighted mean of the expressions $\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a}$ and $\\frac{b}{a} + \\frac{c}{b} + \\frac{d}{c} + \\frac{a}{d}$.\nBy applying the AM-GM inequality to the numbers $\\frac{a}{b}$, $\\frac{a}{b}$, $\\frac{b}{c}$ and $\\frac{a}{d}$, we obtain\n$$\na = \\sqrt[4]{\\frac{a^4}{a b c d}} = \\sqrt[4]{\\frac{a}{b} \\cdot \\frac{a}{b} \\cdot \\frac{b}{c} \\cdot \\frac{a}{d}} \\leq \\frac{1}{4}\\left(\\frac{a}{b} + \\frac{a}{b} + \\frac{b}{c} + \\frac{a}{d}\\right)\n$$\nAnalogously,\n$$\nb \\leq \\frac{1}{4}\\left(\\frac{b}{c} + \\frac{b}{c} + \\frac{c}{d} + \\frac{b}{a}\\right), \\quad c \\leq \\frac{1}{4}\\left(\\frac{c}{d} + \\frac{c}{d} + \\frac{d}{a} + \\frac{c}{b}\\right) \\quad \\text{ and } \\quad d \\leq \\frac{1}{4}\\left(\\frac{d}{a} + \\frac{d}{a} + \\frac{a}{b} + \\frac{d}{c}\\right).\n$$\nSumming up these estimates yields\n$$\na + b + c + d \\leq \\frac{3}{4}\\left(\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a}\\right) + \\frac{1}{4}\\left(\\frac{b}{a} + \\frac{c}{b} + \\frac{d}{c} + \\frac{a}{d}\\right).\n$$\nIn particular, if $a + b + c + d > \\frac{a}{b} + \\frac{b}{c} + \\frac{c}{d} + \\frac{d}{a}$ then $a + b + c + d < \\frac{b}{a} + \\frac{c}{b} + \\frac{d}{c} + \\frac{a}{d}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31882,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\rightarrow \\mathbb{N}$ be a function which satisfies\n$$\nf\\left(x+\\frac{1}{f(y)}\\right)=f\\left(y+\\frac{1}{f(x)}\\right) \\quad \\text{ for all } x, y \\in \\mathbb{R}\n$$\nProve that there is a positive integer which is not a value of $f$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that the statement is false and $f(\\mathbb{R})=\\mathbb{N}$. We prove several properties of the function $f$ in order to reach a contradiction.\n\nTo start with, observe that one can assume $f(0)=1$. Indeed, let $a \\in \\mathbb{R}$ be such that $f(a)=1$, and consider the function $g(x)=f(x+a)$. By substituting $x+a$ and $y+a$ for $x$ and $y$ in (1), we have\n$$\ng\\left(x+\\frac{1}{g(y)}\\right)=f\\left(x+a+\\frac{1}{f(y+a)}\\right)=f\\left(y+a+\\frac{1}{f(x+a)}\\right)=g\\left(y+\\frac{1}{g(x)}\\right) .\n$$\nSo $g$ satisfies the functional equation (1), with the additional property $g(0)=1$. Also, $g$ and $f$ have the same set of values: $g(\\mathbb{R})=f(\\mathbb{R})=\\mathbb{N}$. Henceforth we assume $f(0)=1$.\n\nClaim 1. For an arbitrary fixed $c \\in \\mathbb{R}$ we have $\\left\\{f\\left(c+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\mathbb{N}$.\n\nProof. Equation (1) and $f(\\mathbb{R})=\\mathbb{N}$ imply\n$f(\\mathbb{R})=\\left\\{f\\left(x+\\frac{1}{f(c)}\\right): x \\in \\mathbb{R}\\right\\}=\\left\\{f\\left(c+\\frac{1}{f(x)}\\right): x \\in \\mathbb{R}\\right\\} \\subset\\left\\{f\\left(c+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\} \\subset f(\\mathbb{R})$.\nThe claim follows.\n\nWe will use Claim 1 in the special cases $c=0$ and $c=1 / 3$ :\n$$\n\\left\\{f\\left(\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\left\\{f\\left(\\frac{1}{3}+\\frac{1}{n}\\right): n \\in \\mathbb{N}\\right\\}=\\mathbb{N} .\n$$\n\nClaim 2. If $f(u)=f(v)$ for some $u, v \\in \\mathbb{R}$ then $f(u+q)=f(v+q)$ for all nonnegative rational $q$. Furthermore, if $f(q)=1$ for some nonnegative rational $q$ then $f(k q)=1$ for all $k \\in \\mathbb{N}$.\n\nProof. For all $x \\in \\mathbb{R}$ we have by (1)\n$$\nf\\left(u+\\frac{1}{f(x)}\\right)=f\\left(x+\\frac{1}{f(u)}\\right)=f\\left(x+\\frac{1}{f(v)}\\right)=f\\left(v+\\frac{1}{f(x)}\\right)\n$$\nSince $f(x)$ attains all positive integer values, this yields $f(u+1 / n)=f(v+1 / n)$ for all $n \\in \\mathbb{N}$. Let $q=k / n$ be a positive rational number. Then $k$ repetitions of the last step yield\n$$\nf(u+q)=f\\left(u+\\frac{k}{n}\\right)=f\\left(v+\\frac{k}{n}\\right)=f(v+q)\n$$\nNow let $f(q)=1$ for some nonnegative rational $q$, and let $k \\in \\mathbb{N}$. As $f(0)=1$, the previous conclusion yields successively $f(q)=f(2 q), f(2 q)=f(3 q), \\ldots, f((k-1) q)=f(k q)$, as needed.\n\nClaim 3. The equality $f(q)=f(q+1)$ holds for all nonnegative rational $q$.\n\nProof. Let $m$ be a positive integer such that $f(1 / m)=1$. Such an $m$ exists by (2). Applying the second statement of Claim 2 with $q=1 / m$ and $k=m$ yields $f(1)=1$.\nGiven that $f(0)=f(1)=1$, the first statement of Claim 2 implies $f(q)=f(q+1)$ for all nonnegative rational $q$.\n\nClaim 4. The equality $f\\left(\\frac{1}{n}\\right)=n$ holds for every $n \\in \\mathbb{N}$.\n\nProof. For a nonnegative rational $q$ we set $x=q, y=0$ in (1) and use Claim 3 to obtain\n$$\nf\\left(\\frac{1}{f(q)}\\right)=f\\left(q+\\frac{1}{f(0)}\\right)=f(q+1)=f(q) .\n$$\nBy (2), for each $n \\in \\mathbb{N}$ there exists a $k \\in \\mathbb{N}$ such that $f(1 / k)=n$. Applying the last equation with $q=1 / k$, we have\n$$\nn=f\\left(\\frac{1}{k}\\right)=f\\left(\\frac{1}{f(1 / k)}\\right)=f\\left(\\frac{1}{n}\\right) .\n$$\nNow we are ready to obtain a contradiction. Let $n \\in \\mathbb{N}$ be such that $f(1 / 3+1 / n)=1$. Such an $n$ exists by (2). Let $1 / 3+1 / n=s / t$, where $s, t \\in \\mathbb{N}$ are coprime. Observe that $t>1$ as $1 / 3+1 / n$ is not an integer. Choose $k, l \\in \\mathbb{N}$ so that that $k s-l t=1$.\nBecause $f(0)=f(s / t)=1$, Claim 2 implies $f(k s / t)=1$. Now $f(k s / t)=f(1 / t+l)$; on the other hand $f(1 / t+l)=f(1 / t)$ by $l$ successive applications of Claim 3. Finally, $f(1 / t)=t$ by Claim 4, leading to the impossible $t=1$. The solution is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31883,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for any four positive real numbers $a$, $b$, $c$, $d$ the inequality\n$$\n\\frac{(a-b)(a-c)}{a+b+c}+\\frac{(b-c)(b-d)}{b+c+d}+\\frac{(c-d)(c-a)}{c+d+a}+\\frac{(d-a)(d-b)}{d+a+b} \\geq 0\n$$\nholds. Determine all cases of equality.",
    "options": [],
    "answer": "Equality holds if and only if a = c and b = d.",
    "solution": "Solution 1. Denote the four terms by\n$$\nA=\\frac{(a-b)(a-c)}{a+b+c}, \\quad B=\\frac{(b-c)(b-d)}{b+c+d}, \\quad C=\\frac{(c-d)(c-a)}{c+d+a}, \\quad D=\\frac{(d-a)(d-b)}{d+a+b} .\n$$\nThe expression $2A$ splits into two summands as follows,\n$$\n2A=A' + A'' \\quad \\text{ where } \\quad A' = \\frac{(a-c)^2}{a+b+c}, \\quad A'' = \\frac{(a-c)(a-2b+c)}{a+b+c}\n$$\nthis is easily verified. We analogously represent $2B = B' + B''$, $2C = C' + C''$, $2D = D' + D''$ and examine each of the sums $A'+B'+C'+D'$ and $A''+B''+C''+D''$ separately.\nWrite $s = a+b+c+d$; the denominators become $s-d$, $s-a$, $s-b$, $s-c$. By the Cauchy-Schwarz inequality,\n$$\n\\begin{aligned}\n& \\left(\\frac{|a-c|}{\\sqrt{s-d}} \\cdot \\sqrt{s-d} + \\frac{|b-d|}{\\sqrt{s-a}} \\cdot \\sqrt{s-a} + \\frac{|c-a|}{\\sqrt{s-b}} \\cdot \\sqrt{s-b} + \\frac{|d-b|}{\\sqrt{s-c}} \\cdot \\sqrt{s-c}\\right)^2 \\\\\n& \\quad \\leq \\left(\\frac{(a-c)^2}{s-d} + \\frac{(b-d)^2}{s-a} + \\frac{(c-a)^2}{s-b} + \\frac{(d-b)^2}{s-c}\\right)(4s-s) = 3s\\left(A'+B'+C'+D'\\right)\n\\end{aligned}\n$$\nHence\n$$\n\\begin{equation*}\nA'+B'+C'+D' \\geq \\frac{(2|a-c|+2|b-d|)^2}{3s} \\geq \\frac{16 \\cdot |a-c| \\cdot |b-d|}{3s} \\tag{1}\n\\end{equation*}\n$$\nNext we estimate the absolute value of the other sum. We couple $A''$ with $C''$ to obtain\n$$\n\\begin{aligned}\nA''+C'' & = \\frac{(a-c)(a+c-2b)}{s-d} + \\frac{(c-a)(c+a-2d)}{s-b} \\\\\n& = \\frac{(a-c)(a+c-2b)(s-b) + (c-a)(c+a-2d)(s-d)}{(s-d)(s-b)} \\\\\n& = \\frac{(a-c)(-2b(s-b)-b(a+c)+2d(s-d)+d(a+c))}{s(a+c)+bd} \\\\\n& = \\frac{3(a-c)(d-b)(a+c)}{M}, \\quad \\text{ with } \\quad M = s(a+c)+bd\n\\end{aligned}\n$$\nHence by cyclic shift\n$$\nB''+D'' = \\frac{3(b-d)(a-c)(b+d)}{N}, \\quad \\text{ with } \\quad N = s(b+d)+ca\n$$\nThus\n$$\n\\begin{equation*}\nA''+B''+C''+D'' = 3(a-c)(b-d)\\left(\\frac{b+d}{N} - \\frac{a+c}{M}\\right) = \\frac{3(a-c)(b-d)W}{MN} \\tag{2}\n\\end{equation*}\n$$\nwhere\n$$\n\\begin{equation*}\nW = (b+d)M - (a+c)N = bd(b+d) - ac(a+c) \\tag{3}\n\\end{equation*}\n$$\nNote that\n$$\n\\begin{equation*}\nMN > ac(a+c) + bd(b+d) s \\geq |W| \\cdot s . \\tag{4}\n\\end{equation*}\n$$\nNow (2) and (4) yield\n$$\n\\begin{equation*}\n\\left|A''+B''+C''+D''\\right| \\leq \\frac{3 \\cdot |a-c| \\cdot |b-d|}{s} . \\tag{5}\n\\end{equation*}\n$$\nCombined with (1) this results in\n$$\n\\begin{aligned}\n& 2(A+B+C+D) = \\left(A'+B'+C'+D'\\right) + \\left(A''+B''+C''+D''\\right) \\\\\n& \\quad \\geq \\frac{16 \\cdot |a-c| \\cdot |b-d|}{3s} - \\frac{3 \\cdot |a-c| \\cdot |b-d|}{s} = \\frac{7 \\cdot |a-c| \\cdot |b-d|}{3(a+b+c+d)} \\geq 0\n\\end{aligned}\n$$\nThis is the required inequality. From the last line we see that equality can be achieved only if either $a = c$ or $b = d$. Since we also need equality in (1), this implies that actually $a = c$ and $b = d$ must hold simultaneously, which is obviously also a sufficient condition.\n\n\nSolution 2. We keep the notations $A$, $B$, $C$, $D$, $s$, and also $M$, $N$, $W$ from the preceding solution; the definitions of $M$, $N$, $W$ and relations (3), (4) in that solution did not depend on the foregoing considerations. Starting from\n$$\n2A = \\frac{(a-c)^2 + 3(a+c)(a-c)}{a+b+c} - 2a + 2c,\n$$\nwe get\n$$\n\\begin{aligned}\n& 2(A+C) = (a-c)^2\\left(\\frac{1}{s-d} + \\frac{1}{s-b}\\right) + 3(a+c)(a-c)\\left(\\frac{1}{s-d} - \\frac{1}{s-b}\\right) \\\\\n& \\quad = (a-c)^2 \\frac{2s-b-d}{M} + 3(a+c)(a-c) \\cdot \\frac{d-b}{M} = \\frac{p(a-c)^2 - 3(a+c)(a-c)(b-d)}{M}\n\\end{aligned}\n$$\nwhere $p = 2s-b-d = s+a+c$. Similarly, writing $q = s+b+d$ we have\n$$\n2(B+D) = \\frac{q(b-d)^2 - 3(b+d)(b-d)(c-a)}{N} ;\n$$\nspecific grouping of terms in the numerators has its aim. Note that $pq > 2s^2$. By adding the fractions expressing $2(A+C)$ and $2(B+D)$,\n$$\n2(A+B+C+D) = \\frac{p(a-c)^2}{M} + \\frac{3(a-c)(b-d)W}{MN} + \\frac{q(b-d)^2}{N}\n$$\nwith $W$ defined by (3).\nSubstitution $x = (a-c)/M$, $y = (b-d)/N$ brings the required inequality to the form\n$$\n\\begin{equation*}\n2(A+B+C+D) = Mp x^2 + 3W x y + Nq y^2 \\geq 0 . \\tag{6}\n\\end{equation*}\n$$\nIt will be enough to verify that the discriminant $\\Delta = 9W^2 - 4MNpq$ of the quadratic trinomial $Mp t^2 + 3W t + Nq$ is negative; on setting $t = x/y$ one then gets (6). The first inequality in (4) together with $pq > 2s^2$ imply $4MNpq > 8s^3(ac(a+c)+bd(b+d))$. Since\n$$\n(a+c)s^3 > (a+c)^4 \\geq 4ac(a+c)^2 \\quad \\text{ and likewise } \\quad (b+d)s^3 > 4bd(b+d)^2\n$$\nthe estimate continues as follows,\n$$\n4MNpq > 8\\left(4(ac)^2(a+c)^2 + 4(bd)^2(b+d)^2\\right) > 32(bd(b+d)-ac(a+c))^2 = 32W^2 \\geq 9W^2 .\n$$\nThus indeed $\\Delta < 0$. The desired inequality (6) hence results. It becomes an equality if and only if $x = y = 0$; equivalently, if and only if $a = c$ and simultaneously $b = d$.\n\n\nSolution 3.\n$$\n\\begin{gathered}\n(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\\\\n=((a-b)(a+b+d))((a-c)(a+c+d))(b+c+d)= \\\\\n=\\left(a^2+ad-b^2-bd\\right)\\left(a^2+ad-c^2-cd\\right)(b+c+d)= \\\\\n=\\left(a^4+2a^3d-a^2b^2-a^2bd-a^2c^2-a^2cd+a^2d^2-ab^2d-abd^2-ac^2d-acd^2+b^2c^2+b^2cd+b c^2d+bcd^2\\right)(b+c+d)= \\\\\na^4b+a^4c+a^4d+\\left(b^3c^2+a^2d^3\\right)-a^2c^3+\\left(2a^3d^2-b^3a^2+c^3b^2\\right)+ \\\\\n+\\left(b^3cd-c^3da-d^3ab\\right)+\\left(2a^3bd+c^3db-d^3ac\\right)+\\left(2a^3cd-b^3da+d^3bc\\right) \\\\\n+\\left(-a^2b^2c+3b^2c^2d-2ac^2d^2\\right)+\\left(-2a^2b^2d+2bc^2d^2\\right)+\\left(-a^2bc^2-2a^2c^2d-2ab^2d^2+2b^2cd^2\\right)+ \\\\\n+\\left(-2a^2bcd-ab^2cd-abc^2d-2abcd^2\\right)\n\\end{gathered}\n$$\nIntroducing the notation $S_{xyzw} = \\sum_{cyc} a^x b^y c^z d^w$, one can write\n$$\n\\begin{gathered}\n\\sum_{cyc}(a-b)(a-c)(a+b+d)(a+c+d)(b+c+d)= \\\\\n=S_{4100}+S_{4010}+S_{4001}+2S_{3200}-S_{3020}+2S_{3002}-S_{3110}+2S_{3101}+2S_{3011}-3S_{2120}-6S_{2111}= \\\\\n+\\left(S_{4100}+S_{4001}+\\frac{1}{2}S_{3110}+\\frac{1}{2}S_{3011}-3S_{2120}\\right)+ \\\\\n+\\left(S_{4010}-S_{3020}-\\frac{3}{2}S_{3110}+\\frac{3}{2}S_{3011}+\\frac{9}{16}S_{2210}+\\frac{9}{16}S_{2201}-\\frac{9}{8}S_{2111}\\right)+ \\\\\n+\\frac{9}{16}\\left(S_{3200}-S_{2210}-S_{2201}+S_{3002}\\right)+\\frac{23}{16}\\left(S_{3200}-2S_{3101}+S_{3002}\\right)+\\frac{39}{8}\\left(S_{3101}-S_{2111}\\right),\n\\end{gathered}\n$$\nwhere the expressions\n$$\n\\begin{gathered}\nS_{4100}+S_{4001}+\\frac{1}{2}S_{3110}+\\frac{1}{2}S_{3011}-3S_{2120}=\\sum_{cyc}\\left(a^4b+bc^4+\\frac{1}{2}a^3bc+\\frac{1}{2}abc^3-3a^2bc^2\\right), \\\\\nS_{4010}-S_{3020}-\\frac{3}{2}S_{3110}+\\frac{3}{2}S_{3011}+\\frac{9}{16}S_{2210}+\\frac{9}{16}S_{2201}-\\frac{9}{8}S_{2111}=\\sum_{cyc} a^2c\\left(a-c-\\frac{3}{4}b+\\frac{3}{4}d\\right)^2, \\\\\nS_{3200}-S_{2210}-S_{2201}+S_{3002}=\\sum_{cyc} b^2\\left(a^3-a^2c-ac^2+c^3\\right)=\\sum_{cyc} b^2(a+c)(a-c)^2, \\\\\nS_{3200}-2S_{3101}+S_{3002}=\\sum_{cyc} a^3(b-d)^2 \\quad \\text{ and } \\quad S_{3101}-S_{2111}=\\frac{1}{3} \\sum_{cyc} bd\\left(2a^3+c^3-3a^2c\\right)\n\\end{gathered}\n$$\nare all nonnegative.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31884,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane we consider rectangles whose sides are parallel to the coordinate axes and have positive length. Such a rectangle will be called a box. Two boxes intersect if they have a common point in their interior or on their boundary.\nFind the largest $n$ for which there exist $n$ boxes $B_{1}, \\ldots, B_{n}$ such that $B_{i}$ and $B_{j}$ intersect if and only if $i \\not \\equiv j \\pm 1(\\bmod n)$.",
    "options": [],
    "answer": "6",
    "solution": "The maximum number of such boxes is $6$. One example is shown in the figure.\n\n![](attached_image_1.png)\n\nNow we show that $6$ is the maximum. Suppose that boxes $B_{1}, \\ldots, B_{n}$ satisfy the condition. Let the closed intervals $I_{k}$ and $J_{k}$ be the projections of $B_{k}$ onto the $x$- and $y$-axis, for $1 \\leq k \\leq n$.\n\nIf $B_{i}$ and $B_{j}$ intersect, with a common point $(x, y)$, then $x \\in I_{i} \\cap I_{j}$ and $y \\in J_{i} \\cap J_{j}$. So the intersections $I_{i} \\cap I_{j}$ and $J_{i} \\cap J_{j}$ are nonempty. Conversely, if $x \\in I_{i} \\cap I_{j}$ and $y \\in J_{i} \\cap J_{j}$ for some real numbers $x, y$, then $(x, y)$ is a common point of $B_{i}$ and $B_{j}$. Putting it around, $B_{i}$ and $B_{j}$ are disjoint if and only if their projections on at least one coordinate axis are disjoint.\n\nFor brevity we call two boxes or intervals adjacent if their indices differ by $1$ modulo $n$, and nonadjacent otherwise.\n\nThe adjacent boxes $B_{k}$ and $B_{k+1}$ do not intersect for each $k=1, \\ldots, n$. Hence $(I_{k}, I_{k+1})$ or $(J_{k}, J_{k+1})$ is a pair of disjoint intervals, $1 \\leq k \\leq n$. So there are at least $n$ pairs of disjoint intervals among $\\left(I_{1}, I_{2}\\right), \\ldots,\\left(I_{n-1}, I_{n}\\right),\\left(I_{n}, I_{1}\\right);\\left(J_{1}, J_{2}\\right), \\ldots,\\left(J_{n-1}, J_{n}\\right),\\left(J_{n}, J_{1}\\right)$.\n\nNext, every two nonadjacent boxes intersect, hence their projections on both axes intersect, too. Then the claim below shows that at most $3$ pairs among $\\left(I_{1}, I_{2}\\right), \\ldots,\\left(I_{n-1}, I_{n}\\right),\\left(I_{n}, I_{1}\\right)$ are disjoint, and the same holds for $\\left(J_{1}, J_{2}\\right), \\ldots,\\left(J_{n-1}, J_{n}\\right),\\left(J_{n}, J_{1}\\right)$. Consequently $n \\leq 3+3=6$, as stated. Thus we are left with the claim and its justification.\n\n**Claim.** Let $\\Delta_{1}, \\Delta_{2}, \\ldots, \\Delta_{n}$ be intervals on a straight line such that every two nonadjacent intervals intersect. Then $\\Delta_{k}$ and $\\Delta_{k+1}$ are disjoint for at most three values of $k=1, \\ldots, n$.\n\n**Proof.** Denote $\\Delta_{k}=[a_{k}, b_{k}], 1 \\leq k \\leq n$. Let $\\alpha=\\max \\left(a_{1}, \\ldots, a_{n}\\right)$ be the rightmost among the left endpoints of $\\Delta_{1}, \\ldots, \\Delta_{n}$, and let $\\beta=\\min \\left(b_{1}, \\ldots, b_{n}\\right)$ be the leftmost among their right endpoints. Assume that $\\alpha=a_{2}$ without loss of generality.\n\nIf $\\alpha \\leq \\beta$ then $a_{i} \\leq \\alpha \\leq \\beta \\leq b_{i}$ for all $i$. Every $\\Delta_{i}$ contains $\\alpha$, and thus no disjoint pair $(\\Delta_{i}, \\Delta_{i+1})$ exists.\n\nIf $\\beta<\\alpha$ then $\\beta=b_{i}$ for some $i$ such that $a_{i}<b_{i}=\\beta<\\alpha=a_{2}<b_{2}$, hence $\\Delta_{2}$ and $\\Delta_{i}$ are disjoint. Now $\\Delta_{2}$ intersects all remaining intervals except possibly $\\Delta_{1}$ and $\\Delta_{3}$, so $\\Delta_{2}$ and $\\Delta_{i}$ can be disjoint only if $i=1$ or $i=3$. Suppose by symmetry that $i=3$; then $\\beta=b_{3}$. Since each of the intervals $\\Delta_{4}, \\ldots, \\Delta_{n}$ intersects $\\Delta_{2}$, we have $a_{i} \\leq \\alpha \\leq b_{i}$ for $i=4, \\ldots, n$. Therefore $\\alpha \\in \\Delta_{4} \\cap \\ldots \\cap \\Delta_{n}$, in particular $\\Delta_{4} \\cap \\ldots \\cap \\Delta_{n} \\neq \\emptyset$. Similarly, $\\Delta_{5}, \\ldots, \\Delta_{n}, \\Delta_{1}$ all intersect $\\Delta_{3}$, so that $\\Delta_{5} \\cap \\ldots \\cap \\Delta_{n} \\cap \\Delta_{1} \\neq \\emptyset$ as $\\beta \\in \\Delta_{5} \\cap \\ldots \\cap \\Delta_{n} \\cap \\Delta_{1}$. This leaves $(\\Delta_{1}, \\Delta_{2})$, $(\\Delta_{2}, \\Delta_{3})$ and $\\left(\\Delta_{3}, \\Delta_{4}\\right)$ as the only candidates for disjoint interval pairs, as desired.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 31885,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$ determine the number of permutations $(a_{1}, a_{2}, \\ldots, a_{n})$ of the set $\\{1,2, \\ldots, n\\}$ with the following property:\n$$\n2(a_{1}+a_{2}+\\cdots+a_{k}) \\quad \\text{is divisible by} \\ k \\quad \\text{for} \\ k=1,2, \\ldots, n.\n$$",
    "options": [],
    "answer": "F_1 = 1, F_2 = 2, and for n ≥ 3, F_n = 3 · 2^(n−2)",
    "solution": "For each $n$ let $F_{n}$ be the number of permutations of $\\{1,2, \\ldots, n\\}$ with the required property; call them nice. For $n=1,2,3$ every permutation is nice, so $F_{1}=1$, $F_{2}=2$, $F_{3}=6$.\n\nTake an $n>3$ and consider any nice permutation $(a_{1}, a_{2}, \\ldots, a_{n})$ of $\\{1,2, \\ldots, n\\}$. Then $n-1$ must be a divisor of the number\n$$\n\\begin{aligned}\n& 2(a_{1}+a_{2}+\\cdots+a_{n-1})=2((1+2+\\cdots+n)-a_{n}) \\\\\n& \\quad=n(n+1)-2 a_{n}=(n+2)(n-1)+(2-2 a_{n}) .\n\\end{aligned}\n$$\nSo $2 a_{n}-2$ must be divisible by $n-1$, hence equal to $0$ or $n-1$ or $2 n-2$. This means that\n$$\na_{n}=1 \\quad \\text{or} \\quad a_{n}=\\frac{n+1}{2} \\quad \\text{or} \\quad a_{n}=n\n$$\nSuppose that $a_{n}=(n+1)/2$. Since the permutation is nice, taking $k=n-2$ we get that $n-2$ has to be a divisor of\n$$\n\\begin{aligned}\n2(a_{1}+a_{2}+\\cdots+a_{n-2}) & =2((1+2+\\cdots+n)-a_{n}-a_{n-1}) \\\\\n& =n(n+1)-(n+1)-2 a_{n-1}=(n+2)(n-2)+(3-2 a_{n-1})\n\\end{aligned}\n$$\nSo $2 a_{n-1}-3$ should be divisible by $n-2$, hence equal to $0$ or $n-2$ or $2 n-4$. Obviously $0$ and $2 n-4$ are excluded because $2 a_{n-1}-3$ is odd. The remaining possibility ($2 a_{n-1}-3=n-2$) leads to $a_{n-1}=(n+1)/2=a_{n}$, which also cannot hold. This eliminates $(n+1)/2$ as a possible value of $a_{n}$. Consequently $a_{n}=1$ or $a_{n}=n$.\n\nIf $a_{n}=n$ then $(a_{1}, a_{2}, \\ldots, a_{n-1})$ is a nice permutation of $\\{1,2, \\ldots, n-1\\}$. There are $F_{n-1}$ such permutations. Attaching $n$ to any one of them at the end creates a nice permutation of $\\{1,2, \\ldots, n\\}$.\n\nIf $a_{n}=1$ then $(a_{1}-1, a_{2}-1, \\ldots, a_{n-1}-1)$ is a permutation of $\\{1,2, \\ldots, n-1\\}$. It is also nice because the number\n$$\n2((a_{1}-1)+\\cdots+(a_{k}-1))=2(a_{1}+\\cdots+a_{k})-2k\n$$\nis divisible by $k$, for any $k \\leq n-1$. And again, any one of the $F_{n-1}$ nice permutations $(b_{1}, b_{2}, \\ldots, b_{n-1})$ of $\\{1,2, \\ldots, n-1\\}$ gives rise to a nice permutation of $\\{1,2, \\ldots, n\\}$ whose last term is $1$, namely $(b_{1}+1, b_{2}+1, \\ldots, b_{n-1}+1, 1)$.\n\nThe bijective correspondences established in both cases show that there are $F_{n-1}$ nice permutations of $\\{1,2, \\ldots, n\\}$ with the last term $1$ and also $F_{n-1}$ nice permutations of $\\{1,2, \\ldots, n\\}$ with the last term $n$. Hence follows the recurrence $F_{n}=2 F_{n-1}$. With the base value $F_{3}=6$ this gives the outcome formula $F_{n}=3 \\cdot 2^{n-2}$ for $n \\geq 3$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31886,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the coordinate plane consider the set $S$ of all points with integer coordinates. For a positive integer $k$, two distinct points $A, B \\in S$ will be called $k$-friends if there is a point $C \\in S$ such that the area of the triangle $A B C$ is equal to $k$. A set $T \\subset S$ will be called a $k$-clique if every two points in $T$ are $k$-friends. Find the least positive integer $k$ for which there exists a $k$-clique with more than 200 elements.",
    "options": [],
    "answer": "180180",
    "solution": "To begin, let us describe those points $B \\in S$ which are $k$-friends of the point $(0,0)$. By definition, $B=(u, v)$ satisfies this condition if and only if there is a point $C=(x, y) \\in S$ such that $\\frac{1}{2}|u y-v x|=k$. (This is a well-known formula expressing the area of triangle $A B C$ when $A$ is the origin.)\n\nTo say that there exist integers $x, y$ for which $|u y-v x|=2 k$, is equivalent to saying that the greatest common divisor of $u$ and $v$ is also a divisor of $2 k$. Summing up, a point $B=(u, v) \\in S$ is a $k$-friend of $(0,0)$ if and only if $\\operatorname{gcd}(u, v)$ divides $2 k$.\n\nTranslation by a vector with integer coordinates does not affect $k$-friendship; if two points are $k$-friends, so are their translates. It follows that two points $A, B \\in S$, $A=(s, t)$, $B=(u, v)$, are $k$-friends if and only if the point $(u-s, v-t)$ is a $k$-friend of $(0,0)$; i.e., if $\\operatorname{gcd}(u-s, v-t) \\mid 2 k$.\n\nLet $n$ be a positive integer which does not divide $2 k$. We claim that a $k$-clique cannot have more than $n^{2}$ elements.\n\nIndeed, all points $(x, y) \\in S$ can be divided into $n^{2}$ classes determined by the remainders that $x$ and $y$ leave in division by $n$. If a set $T$ has more than $n^{2}$ elements, some two points $A, B \\in T$, $A=(s, t)$, $B=(u, v)$, necessarily fall into the same class. This means that $n \\mid u-s$ and $n \\mid v-t$. Hence $n \\mid d$ where $d=\\operatorname{gcd}(u-s, v-t)$. And since $n$ does not divide $2 k$, also $d$ does not divide $2 k$. Thus $A$ and $B$ are not $k$-friends and the set $T$ is not a $k$-clique.\n\nNow let $M(k)$ be the least positive integer which does not divide $2 k$. Write $M(k)=m$ for the moment and consider the set $T$ of all points $(x, y)$ with $0 \\leq x, y < m$. There are $m^{2}$ of them. If $A=(s, t)$, $B=(u, v)$ are two distinct points in $T$ then both differences $|u-s|,|v-t|$ are integers less than $m$ and at least one of them is positive. By the definition of $m$, every positive integer less than $m$ divides $2 k$. Therefore $u-s$ (if nonzero) divides $2 k$, and the same is true of $v-t$. So $2 k$ is divisible by $\\operatorname{gcd}(u-s, v-t)$, meaning that $A, B$ are $k$-friends. Thus $T$ is a $k$-clique.\n\nIt follows that the maximum size of a $k$-clique is $M(k)^{2}$, with $M(k)$ defined as above. We are looking for the minimum $k$ such that $M(k)^{2}>200$.\n\nBy the definition of $M(k)$, $2 k$ is divisible by the numbers $1,2, \\ldots, M(k)-1$, but not by $M(k)$ itself. If $M(k)^{2}>200$ then $M(k) \\geq 15$. Trying to hit $M(k)=15$ we get a contradiction immediately ($2 k$ would have to be divisible by $3$ and $5$, but not by $15$).\n\nSo let us try $M(k)=16$. Then $2 k$ is divisible by the numbers $1,2, \\ldots, 15$, hence also by their least common multiple $L$, but not by $16$. And since $L$ is not a multiple of $16$, we infer that $k=L / 2$ is the least $k$ with $M(k)=16$.\n\nFinally, observe that if $M(k) \\geq 17$ then $2 k$ must be divisible by the least common multiple of $1,2, \\ldots, 16$, which is equal to $2 L$. Then $2 k \\geq 2 L$, yielding $k>L / 2$.\n\nIn conclusion, the least $k$ with the required property is equal to $L / 2=180180$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31887,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S=\\{x_{1}, x_{2}, \\ldots, x_{k+\\ell}\\}$ be a $(k+\\ell)$-element set of real numbers contained in the interval $[0,1]$; $k$ and $\\ell$ are positive integers. A $k$-element subset $A \\subset S$ is called nice if\n$$\n\\left|\\frac{1}{k} \\sum_{x_{i} \\in A} x_{i}-\\frac{1}{\\ell} \\sum_{x_{j} \\in S \\backslash A} x_{j}\\right| \\leq \\frac{k+\\ell}{2 k \\ell} .\n$$\nProve that the number of nice subsets is at least $\\frac{2}{k+\\ell}\\binom{k+\\ell}{k}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31888,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given trapezoid $A B C D$ with parallel sides $A B$ and $C D$, assume that there exist points $E$ on line $B C$ outside segment $B C$, and $F$ inside segment $A D$, such that $\\angle D A E=\\angle C B F$. Denote by $I$ the point of intersection of $C D$ and $E F$, and by $J$ the point of intersection of $A B$ and $E F$. Let $K$ be the midpoint of segment $E F$; assume it does not lie on line $A B$.\nProve that $I$ belongs to the circumcircle of $A B K$ if and only if $K$ belongs to the circumcircle of $C D J$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that the disposition of points is as in the diagram.\nSince $\\angle E B F=180^{\\circ}-\\angle C B F=180^{\\circ}-\\angle E A F$ by hypothesis, the quadrilateral $A E B F$ is cyclic. Hence $A J \\cdot J B=F J \\cdot J E$. In view of this equality, $I$ belongs to the circumcircle of $A B K$ if and only if $I J \\cdot J K=F J \\cdot J E$. Expressing $I J=I F+F J$, $J E=F E-F J$, and $J K=\\frac{1}{2} F E-F J$, we find that $I$ belongs to the circumcircle of $A B K$ if and only if\n$$\nF J=\\frac{I F \\cdot F E}{2 I F+F E}\n$$\nSince $A E B F$ is cyclic and $A B, C D$ are parallel, $\\angle F E C=\\angle F A B=180^{\\circ}-\\angle C D F$. Then $C D F E$ is also cyclic, yielding $I D \\cdot I C=I F \\cdot I E$. It follows that $K$ belongs to the circumcircle of $C D J$ if and only if $I J \\cdot I K=I F \\cdot I E$. Expressing $I J=I F+F J$, $I K=I F+\\frac{1}{2} F E$, and $I E=I F+F E$, we find that $K$ is on the circumcircle of $C D J$ if and only if\n$$\nF J=\\frac{I F \\cdot F E}{2 I F+F E} .\n$$\nThe conclusion follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31889,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral and let $P$ and $Q$ be points in $ABCD$ such that $PQDA$ and $QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $E$ on the line segment $PQ$ such that $\\angle PAE = \\angle QDE$ and $\\angle PBE = \\angle QCE$. Show that the quadrilateral $ABCD$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $F$ be the point on the line $AD$ such that $EF \\parallel PA$. By hypothesis, the quadrilateral $PQDA$ is cyclic. So if $F$ lies between $A$ and $D$ then $\\angle EFD = \\angle PAD = 180^{\\circ} - \\angle EQD$; the points $F$ and $Q$ are on distinct sides of the line $DE$ and we infer that $EFDQ$ is a cyclic quadrilateral. And if $D$ lies between $A$ and $F$ then a similar argument shows that $\\angle EFD = \\angle EQD$; but now the points $F$ and $Q$ lie on the same side of $DE$, so that $EDFQ$ is a cyclic quadrilateral.\n\nIn either case we obtain the equality $\\angle EFQ = \\angle EDQ = \\angle PAE$ which implies that $FQ \\parallel AE$. So the triangles $EFQ$ and $PAE$ are either homothetic or parallel-congruent. More specifically, triangle $EFQ$ is the image of $PAE$ under the mapping $f$ which carries the points $P, E$ respectively to $E, Q$ and is either a homothety or translation by a vector. Note that $f$ is uniquely determined by these conditions and the position of the points $P, E, Q$ alone.\n\nLet now $G$ be the point on the line $BC$ such that $EG \\parallel PB$. The same reasoning as above applies to points $B, C$ in place of $A, D$, implying that the triangle $EGQ$ is the image of $PBE$ under the same mapping $f$. So $f$ sends the four points $A, P, B, E$ respectively to $F, E, G, Q$.\n\nIf $PE \\neq QE$, so that $f$ is a homothety with a centre $X$, then the lines $AF, PE, BG$—i.e. the lines $AD, PQ, BC$—are concurrent at $X$. And since $PQDA$ and $QPBC$ are cyclic quadrilaterals, the equalities $XA \\cdot XD = XP \\cdot XQ = XB \\cdot XC$ hold, showing that the quadrilateral $ABCD$ is cyclic.\n\nFinally, if $PE = QE$, so that $f$ is a translation, then $AD \\parallel PQ \\parallel BC$. Thus $PQDA$ and $QPBC$ are isosceles trapezoids. Then also $ABCD$ is an isosceles trapezoid, hence a cyclic quadrilateral.\nHere is another way to reach the conclusion that the lines $AD$, $BC$ and $PQ$ are either concurrent or parallel. From the cyclic quadrilateral $PQDA$ we get\n$$\n\\angle PAD = 180^{\\circ} - \\angle PQD = \\angle QDE + \\angle QED = \\angle PAE + \\angle QED.\n$$\nHence $\\angle QED = \\angle PAD - \\angle PAE = \\angle EAD$. This in view of the tangent-chord theorem means that the circumcircle of triangle $EAD$ is tangent to the line $PQ$ at $E$. Analogously, the circumcircle of triangle $EBC$ is tangent to $PQ$ at $E$.\n\nSuppose that the line $AD$ intersects $PQ$ at $X$. Since $XE$ is tangent to the circle $(EAD)$, $XE^2 = XA \\cdot XD$. Also, $XA \\cdot XD = XP \\cdot XQ$ because $P, Q, D, A$ lie on a circle. Therefore $XE^2 = XP \\cdot XQ$.\n\nIt is not hard to see that this equation determines the position of the point $X$ on the line $PQ$ uniquely. Thus, if $BC$ also cuts $PQ$, say at $Y$, then the analogous equation for $Y$ yields $X = Y$, meaning that the three lines indeed concur. In this case, as well as in the case where $AD \\parallel PQ \\parallel BC$, the concluding argument is the same as in the first solution.\n\nIt remains to eliminate the possibility that e.g. $AD$ meets $PQ$ at $X$ while $BC \\parallel PQ$. Indeed, $QPBC$ would then be an isosceles trapezoid and the angle equality $\\angle PBE = \\angle QCE$ would force that $E$ is the midpoint of $PQ$. So the length of $XE$, which is the geometric mean of the lengths of $XP$ and $XQ$, should also be their arithmetic mean—impossible, as $XP \\neq XQ$. The proof is now complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31890,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ and $n$ be integers with $0 \\leq k \\leq n-2$. Consider a set $L$ of $n$ lines in the plane such that no two of them are parallel and no three have a common point. Denote by $I$ the set of intersection points of lines in $L$. Let $O$ be a point in the plane not lying on any line of $L$.\nA point $X \\in I$ is colored red if the open line segment $O X$ intersects at most $k$ lines in $L$. Prove that $I$ contains at least $\\frac{1}{2}(k+1)(k+2)$ red points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There are at least $\\frac{1}{2}(k+1)(k+2)$ points in the intersection set $I$ in view of the condition $n \\geq k+2$.\nFor each point $P \\in I$, define its order as the number of lines that intersect the open line segment $O P$. By definition, $P$ is red if its order is at most $k$. Note that there is always at least one point $X \\in I$ of order $0$. Indeed, the lines in $L$ divide the plane into regions, bounded or not, and $O$ belongs to one of them. Clearly any corner of this region is a point of $I$ with order $0$.\n\nClaim. Suppose that two points $P, Q \\in I$ lie on the same line of $L$, and no other line of $L$ intersects the open line segment $P Q$. Then the orders of $P$ and $Q$ differ by at most $1$.\n\nProof. Let $P$ and $Q$ have orders $p$ and $q$, respectively, with $p \\geq q$. Consider triangle $O P Q$. Now $p$ equals the number of lines in $L$ that intersect the interior of side $O P$. None of these lines intersects the interior of side $P Q$, and at most one can pass through $Q$. All remaining lines must intersect the interior of side $O Q$, implying that $q \\geq p-1$. The conclusion follows.\n\nWe prove the main result by induction on $k$. The base $k=0$ is clear since there is a point of order $0$ which is red. Assuming the statement true for $k-1$, we pass on to the inductive step. Select a point $P \\in I$ of order $0$, and consider one of the lines $\\ell \\in L$ that pass through $P$. There are $n-1$ intersection points on $\\ell$, one of which is $P$. Out of the remaining $n-2$ points, the $k$ closest to $P$ have orders not exceeding $k$ by the Claim. It follows that there are at least $k+1$ red points on $\\ell$.\n\nLet us now consider the situation with $\\ell$ removed (together with all intersection points it contains). By hypothesis of induction, there are at least $\\frac{1}{2}k(k+1)$ points of order not exceeding $k-1$ in the resulting configuration. Restoring $\\ell$ back produces at most one new intersection point on each line segment joining any of these points to $O$, so their order is at most $k$ in the original configuration. The total number of points with order not exceeding $k$ is therefore at least $(k+1)+\\frac{1}{2}k(k+1)=\\frac{1}{2}(k+1)(k+2)$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31891,
    "subject": "Mathematics (Multi-modal)",
    "question": "There is given a convex quadrilateral $A B C D$. Prove that there exists a point $P$ inside the quadrilateral such that\n$$\n\\angle P A B+\\angle P D C=\\angle P B C+\\angle P A D=\\angle P C D+\\angle P B A=\\angle P D A+\\angle P C B=90^{\\circ} \\tag{1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "if and only if the diagonals $A C$ and $B D$ are perpendicular.\n\nFor a point $P$ in $A B C D$ which satisfies (1), let $K, L, M, N$ be the feet of perpendiculars from $P$ to lines $A B, B C, C D, D A$, respectively. Note that $K, L, M, N$ are interior to the sides as all angles in (1) are acute. The cyclic quadrilaterals $A K P N$ and $D N P M$ give\n$$\n\\angle P A B+\\angle P D C=\\angle P N K+\\angle P N M=\\angle K N M .\n$$\nAnalogously, $\\angle P B C+\\angle P A D=\\angle L K N$ and $\\angle P C D+\\angle P B A=\\angle M L K$. Hence the equalities (1) imply $\\angle K N M=\\angle L K N=\\angle M L K=90^{\\circ}$, so that $K L M N$ is a rectangle. The converse also holds true, provided that $K, L, M, N$ are interior to sides $A B, B C, C D, D A$.\n\ni. Suppose that there exists a point $P$ in $A B C D$ such that $K L M N$ is a rectangle. We show that $A C$ and $B D$ are parallel to the respective sides of $K L M N$.\nLet $O_{A}$ and $O_{C}$ be the circumcentres of the cyclic quadrilaterals $A K P N$ and $C M P L$. Line $O_{A} O_{C}$ is the common perpendicular bisector of $L M$ and $K N$, therefore $O_{A} O_{C}$ is parallel to $K L$ and $M N$. On the other hand, $O_{A} O_{C}$ is the midline in the triangle $A C P$ that is parallel to $A C$. Therefore the diagonal $A C$ is parallel to the sides $K L$ and $M N$ of the rectangle. Likewise, $B D$ is parallel to $K N$ and $L M$. Hence $A C$ and $B D$ are perpendicular.\n\n![](attached_image_1.png)\n\nii. Suppose that $A C$ and $B D$ are perpendicular and meet at $R$. If $A B C D$ is a rhombus, $P$ can be chosen to be its centre. So assume that $A B C D$ is not a rhombus, and let $B R<D R$ without loss of generality.\nDenote by $U_{A}$ and $U_{C}$ the circumcentres of the triangles $A B D$ and $C D B$, respectively. Let $A V_{A}$ and $C V_{C}$ be the diameters through $A$ and $C$ of the two circumcircles. Since $A R$ is an altitude in triangle $A D B$, lines $A C$ and $A V_{A}$ are isogonal conjugates, i.e. $\\angle D A V_{A}=\\angle B A C$. Now $B R<D R$ implies that ray $A U_{A}$ lies in $\\angle D A C$. Similarly, ray $C U_{C}$ lies in $\\angle D C A$. Both diameters $A V_{A}$ and $C V_{C}$ intersect $B D$ as the angles at $B$ and $D$ of both triangles are acute. Also $U_{A} U_{C}$ is parallel to $A C$ as it is the perpendicular bisector of $B D$. Hence $V_{A} V_{C}$ is parallel to $A C$, too. We infer that $A V_{A}$ and $C V_{C}$ intersect at a point $P$ inside triangle $A C D$, hence inside $A B C D$.\nConstruct points $K, L, M, N, O_{A}$ and $O_{C}$ in the same way as in the introduction. It follows from the previous paragraph that $K, L, M, N$ are interior to the respective sides. Now $O_{A} O_{C}$ is a midline in triangle $A C P$ again. Therefore lines $A C, O_{A} O_{C}$ and $U_{A} U_{C}$ are parallel.\nThe cyclic quadrilateral $A K P N$ yields $\\angle N K P=\\angle N A P$. Since $\\angle N A P=\\angle D A U_{A}= \\angle B A C$, as specified above, we obtain $\\angle N K P=\\angle B A C$. Because $P K$ is perpendicular to $A B$, it follows that $N K$ is perpendicular to $A C$, hence parallel to $B D$. Likewise, $L M$ is parallel to $B D$.\nConsider the two homotheties with centres $A$ and $C$ which transform triangles $A B D$ and $C D B$ into triangles $A K N$ and $C M L$, respectively. The images of points $U_{A}$ and $U_{C}$ are $O_{A}$ and $O_{C}$, respectively. Since $U_{A} U_{C}$ and $O_{A} O_{C}$ are parallel to $A C$, the two ratios of homothety are the same, equal to $\\lambda=A N / A D=A K / A B=A O_{A} / A U_{A}=C O_{C} / C U_{C}=C M / C D=C L / C B$. It is now straightforward that $D N / D A=D M / D C=B K / B A=B L / B C=1-\\lambda$. Hence $K L$ and $M N$ are parallel to $A C$, implying that $K L M N$ is a rectangle and completing the proof.\n\n\nSolution 2:\n\nFor a point $P$ distinct from $A, B, C, D$, let circles ($A P D$) and ($B P C$) intersect again at $Q$ ($Q=P$ if the circles are tangent). Next, let circles ($A Q B$) and ($C Q D$) intersect again at $R$. We show that if $P$ lies in $A B C D$ and satisfies (1) then $A C$ and $B D$ intersect at $R$ and are perpendicular; the converse is also true. It is convenient to use directed angles. Let $\\measuredangle(U V, X Y)$ denote the angle of counterclockwise rotation that makes line $U V$ parallel to line $X Y$. Recall that four noncollinear points $U, V, X, Y$ are concyclic if and only if $\\measuredangle(U X, V X)=\\measuredangle(U Y, V Y)$.\nThe definitions of points $P, Q$ and $R$ imply\n$$\n\\begin{aligned}\n\\measuredangle(A R, B R) & =\\measuredangle(A Q, B Q)=\\measuredangle(A Q, P Q)+\\measuredangle(P Q, B Q)=\\measuredangle(A D, P D)+\\measuredangle(P C, B C), \\\\\n\\measuredangle(C R, D R) & =\\measuredangle(C Q, D Q)=\\measuredangle(C Q, P Q)+\\measuredangle(P Q, D Q)=\\measuredangle(C B, P B)+\\measuredangle(P A, D A), \\\\\n\\measuredangle(B R, C R) & =\\measuredangle(B R, R Q)+\\measuredangle(R Q, C R)=\\measuredangle(B A, A Q)+\\measuredangle(D Q, C D) \\\\\n& =\\measuredangle(B A, A P)+\\measuredangle(A P, A Q)+\\measuredangle(D Q, D P)+\\measuredangle(D P, C D) \\\\\n& =\\measuredangle(B A, A P)+\\measuredangle(D P, C D) .\n\\end{aligned}\n$$\nObserve that the whole construction is reversible. One may start with point $R$, define $Q$ as the second intersection of circles ($A R B$) and ($C R D$), and then define $P$ as the second intersection of circles ($A Q D$) and ($B Q C$). The equalities above will still hold true.\nAssume in addition that $P$ is interior to $A B C D$. Then\n$$\n\\begin{aligned}\n\\measuredangle(A D, P D)=\\angle P D A, \\measuredangle(P C, B C) & =\\angle P C B, \\measuredangle(C B, P B)=\\angle P B C, \\measuredangle(P A, D A)=\\angle P A D \\\\\n\\measuredangle(B A, A P) & =\\angle P A B, \\measuredangle(D P, C D)=\\angle P D C .\n\\end{aligned}\n$$\n\ni. Suppose that $P$ lies in $A B C D$ and satisfies (1). Then $\\measuredangle(A R, B R)=\\angle P D A+\\angle P C B=90^{\\circ}$ and similarly $\\measuredangle(B R, C R)=\\measuredangle(C R, D R)=90^{\\circ}$. It follows that $R$ is the common point of lines $A C$ and $B D$, and that these lines are perpendicular.\n\nii. Suppose that $A C$ and $B D$ are perpendicular and intersect at $R$. We show that the point $P$ defined by the reverse construction (starting with $R$ and ending with $P$) lies in $A B C D$. This is enough to finish the solution, because then the angle equalities above will imply (1).\nOne can assume that $Q$, the second common point of circles ($A B R$) and ($C D R$), lies in $\\angle A R D$. Then in fact $Q$ lies in triangle $A D R$ as angles $A Q R$ and $D Q R$ are obtuse. Hence $\\angle A Q D$ is obtuse, too, so that $B$ and $C$ are outside circle ($A D Q$) ($\\angle A B D$ and $\\angle A C D$ are acute).\nNow $\\angle C A B+\\angle C D B=\\angle B Q R+\\angle C Q R=\\angle C Q B$ implies $\\angle C A B<\\angle C Q B$ and $\\angle C D B< \\angle C Q B$. Hence $A$ and $D$ are outside circle ($B C Q$). In conclusion, the second common point $P$ of circles ($A D Q$) and ($B C Q$) lies on their $\\operatorname{arcs} A D Q$ and $B C Q$.\nWe can assume that $P$ lies in $\\angle C Q D$. Since\n$$\n\\begin{gathered}\n\\angle Q P C+\\angle Q P D=\\left(180^{\\circ}-\\angle Q B C\\right)+\\left(180^{\\circ}-\\angle Q A D\\right)= \\\\\n=360^{\\circ}-(\\angle R B C+\\angle Q B R)-(\\angle R A D-\\angle Q A R)=360^{\\circ}-\\angle R B C-\\angle R A D>180^{\\circ},\n\\end{gathered}\n$$\npoint $P$ lies in triangle $C D Q$, and hence in $A B C D$. The proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31892,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $p$ be a prime number. Prove that if $a, b, c$ are integers (not necessarily positive) satisfying the equations\n$$\na^{n}+p b=b^{n}+p c=c^{n}+p a,\n$$\nthen $a=b=c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If two of $a, b, c$ are equal, it is immediate that all the three are equal. So we may assume that $a \\neq b \\neq c \\neq a$. Subtracting the equations we get $a^{n}-b^{n}=-p(b-c)$ and two cyclic copies of this equation, which upon multiplication yield\n$$\n\\frac{a^{n}-b^{n}}{a-b} \\cdot \\frac{b^{n}-c^{n}}{b-c} \\cdot \\frac{c^{n}-a^{n}}{c-a}=-p^{3} .\n$$\nIf $n$ is odd then the differences $a^{n}-b^{n}$ and $a-b$ have the same sign and the product on the left is positive, while $-p^{3}$ is negative. So $n$ must be even.\nLet $d$ be the greatest common divisor of the three differences $a-b, b-c, c-a$, so that $a-b=d u, b-c=d v, c-a=d w ; \\operatorname{gcd}(u, v, w)=1, u+v+w=0$.\nFrom $a^{n}-b^{n}=-p(b-c)$ we see that $(a-b) \\mid p(b-c)$, i.e., $u \\mid p v$; and cyclically $v|p w, w| p u$. As $\\operatorname{gcd}(u, v, w)=1$ and $u+v+w=0$, at most one of $u, v, w$ can be divisible by $p$. Supposing that the prime $p$ does not divide any one of them, we get $u|v, v| w, w \\mid u$, whence $|u|=|v|=|w|=1$; but this quarrels with $u+v+w=0$.\nThus $p$ must divide exactly one of these numbers. Let e.g. $p \\mid u$ and write $u=p u_{1}$. Now we obtain, similarly as before, $u_{1}|v, v| w, w \\mid u_{1}$ so that $\\left|u_{1}\\right|=|v|=|w|=1$. The equation $p u_{1}+v+w=0$ forces that the prime $p$ must be even; i.e. $p=2$. Hence $v+w=-2 u_{1}= \\pm 2$, implying $v=w(= \\pm 1)$ and $u=-2 v$. Consequently $a-b=-2(b-c)$.\nKnowing that $n$ is even, say $n=2 k$, we rewrite the equation $a^{n}-b^{n}=-p(b-c)$ with $p=2$ in the form\n$$\n\\left(a^{k}+b^{k}\\right)\\left(a^{k}-b^{k}\\right)=-2(b-c)=a-b .\n$$\nThe second factor on the left is divisible by $a-b$, so the first factor ( $a^{k}+b^{k}$ ) must be $\\pm 1$. Then exactly one of $a$ and $b$ must be odd; yet $a-b=-2(b-c)$ is even. Contradiction ends the proof.\nThe beginning is as in the first solution. Assuming that $a, b, c$ are not all equal, hence are all distinct, we derive equation (1) with the conclusion that $n$ is even. Write $n=2 k$.\nSuppose that $p$ is odd. Then the integer\n$$\n\\frac{a^{n}-b^{n}}{a-b}=a^{n-1}+a^{n-2} b+\\cdots+b^{n-1},\n$$\nwhich is a factor in (1), must be odd as well. This sum of $n=2 k$ summands is odd only if $a$ and $b$ have different parities. The same conclusion holding for $b, c$ and for $c, a$, we get that $a, b, c, a$ alternate in their parities, which is clearly impossible.\nThus $p=2$. The original system shows that $a, b, c$ must be of the same parity. So we may divide (1) by $p^{3}$, i.e. $2^{3}$, to obtain the following product of six integer factors:\n$$\n\\frac{a^{k}+b^{k}}{2} \\cdot \\frac{a^{k}-b^{k}}{a-b} \\cdot \\frac{b^{k}+c^{k}}{2} \\cdot \\frac{b^{k}-c^{k}}{b-c} \\cdot \\frac{c^{k}+a^{k}}{2} \\cdot \\frac{c^{k}-a^{k}}{c-a}=-1 .\n$$\nEach one of the factors must be equal to $\\pm 1$. In particular, $a^{k}+b^{k}= \\pm 2$. If $k$ is even, this becomes $a^{k}+b^{k}=2$ and yields $|a|=|b|=1$, whence $a^{k}-b^{k}=0$, contradicting (2).\nLet now $k$ be odd. Then the sum $a^{k}+b^{k}$, with value $\\pm 2$, has $a+b$ as a factor. Since $a$ and $b$ are of the same parity, this means that $a+b= \\pm 2$; and cyclically, $b+c= \\pm 2, c+a= \\pm 2$. In some two of these equations the signs must coincide, hence some two of $a, b, c$ are equal. This is the desired contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31893,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, \\ldots, a_{n}$ be distinct positive integers, $n \\geq 3$. Prove that there exist distinct indices $i$ and $j$ such that $a_{i}+a_{j}$ does not divide any of the numbers $3 a_{1}, 3 a_{2}, \\ldots, 3 a_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, let $0<a_{1}<a_{2}<\\cdots<a_{n}$. One can also assume that $a_{1}, a_{2}, \\ldots, a_{n}$ are coprime. Otherwise division by their greatest common divisor reduces the question to the new sequence whose terms are coprime integers.\n\nSuppose that the claim is false. Then for each $i<n$ there exists a $j$ such that $a_{n}+a_{i}$ divides $3 a_{j}$. If $a_{n}+a_{i}$ is not divisible by $3$ then $a_{n}+a_{i}$ divides $a_{j}$ which is impossible as $0<a_{j} \\leq a_{n}<a_{n}+a_{i}$. Thus $a_{n}+a_{i}$ is a multiple of $3$ for $i=1, \\ldots, n-1$, so that $a_{1}, a_{2}, \\ldots, a_{n-1}$ are all congruent (to $-a_{n}$) modulo $3$.\n\nNow $a_{n}$ is not divisible by $3$ or else so would be all remaining $a_{i}$'s, meaning that $a_{1}, a_{2}, \\ldots, a_{n}$ are not coprime. Hence $a_{n} \\equiv r \\pmod{3}$ where $r \\in\\{1,2\\}$, and $a_{i} \\equiv 3-r \\pmod{3}$ for all $i=1, \\ldots, n-1$.\n\nConsider a sum $a_{n-1}+a_{i}$ where $1 \\leq i \\leq n-2$. There is at least one such sum as $n \\geq 3$. Let $j$ be an index such that $a_{n-1}+a_{i}$ divides $3 a_{j}$. Observe that $a_{n-1}+a_{i}$ is not divisible by $3$ since $a_{n-1}+a_{i} \\equiv 2 a_{i} \\not\\equiv 0 \\pmod{3}$. It follows that $a_{n-1}+a_{i}$ divides $a_{j}$, in particular $a_{n-1}+a_{i} \\leq a_{j}$. Hence $a_{n-1}<a_{j} \\leq a_{n}$, implying $j=n$. So $a_{n}$ is divisible by all sums $a_{n-1}+a_{i}$, $1 \\leq i \\leq n-2$. In particular $a_{n-1}+a_{i} \\leq a_{n}$ for $i=1, \\ldots, n-2$.\n\nLet $j$ be such that $a_{n}+a_{n-1}$ divides $3 a_{j}$. If $j \\leq n-2$ then $a_{n}+a_{n-1} \\leq 3 a_{j}<a_{j}+2 a_{n-1}$. This yields $a_{n}<a_{n-1}+a_{j}$; however $a_{n-1}+a_{j} \\leq a_{n}$ for $j \\leq n-2$. Therefore $j=n-1$ or $j=n$.\n\nFor $j=n-1$ we obtain $3 a_{n-1}=k\\left(a_{n}+a_{n-1}\\right)$ with $k$ an integer, and it is straightforward that $k=1$ ($k \\leq 0$ and $k \\geq 3$ contradict $0<a_{n-1}<a_{n}$; $k=2$ leads to $a_{n-1}=2 a_{n}>a_{n-1}$). Thus $3 a_{n-1}=a_{n}+a_{n-1}$, i.e. $a_{n}=2 a_{n-1}$.\n\nSimilarly, if $j=n$ then $3 a_{n}=k\\left(a_{n}+a_{n-1}\\right)$ for some integer $k$, and only $k=2$ is possible. Hence $a_{n}=2 a_{n-1}$ holds true in both cases remaining, $j=n-1$ and $j=n$.\n\nNow $a_{n}=2 a_{n-1}$ implies that the sum $a_{n-1}+a_{1}$ is strictly between $a_{n}/2$ and $a_{n}$. But $a_{n-1}$ and $a_{1}$ are distinct as $n \\geq 3$, so it follows from the above that $a_{n-1}+a_{1}$ divides $a_{n}$. This provides the desired contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31894,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{0}, a_{1}, a_{2}, \\ldots$ be a sequence of positive integers such that the greatest common divisor of any two consecutive terms is greater than the preceding term; in symbols, $\\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$. Prove that $a_{n} \\geq 2^{n}$ for all $n \\geq 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $a_{i} \\geq \\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$, the sequence is strictly increasing. In particular $a_{0} \\geq 1, a_{1} \\geq 2$. For each $i \\geq 1$ we also have $a_{i+1}-a_{i} \\geq \\operatorname{gcd}\\left(a_{i}, a_{i+1}\\right)>a_{i-1}$, and consequently $a_{i+1} \\geq a_{i}+a_{i-1}+1$. Hence $a_{2} \\geq 4$ and $a_{3} \\geq 7$. The equality $a_{3}=7$ would force equalities in the previous estimates, leading to $\\operatorname{gcd}\\left(a_{2}, a_{3}\\right)=\\operatorname{gcd}(4,7)>a_{1}=2$, which is false. Thus $a_{3} \\geq 8$; the result is valid for $n=0,1,2,3$. These are the base cases for a proof by induction.\n\nTake an $n \\geq 3$ and assume that $a_{i} \\geq 2^{i}$ for $i=0,1, \\ldots, n$. We must show that $a_{n+1} \\geq 2^{n+1}$. Let $\\operatorname{gcd}\\left(a_{n}, a_{n+1}\\right)=d$. We know that $d>a_{n-1}$. The induction claim is reached immediately in the following cases:\n$$\n\\begin{aligned}\n& \\text { if } a_{n+1} \\geq 4 d \\quad \\text { then } a_{n+1}>4 a_{n-1} \\geq 4 \\cdot 2^{n-1}=2^{n+1} ; \\\\\n& \\text { if } \\quad a_{n} \\geq 3 d \\quad \\text { then } \\quad a_{n+1} \\geq a_{n}+d \\geq 4 d>4 a_{n-1} \\geq 4 \\cdot 2^{n-1}=2^{n+1} ; \\\\\n& \\text { if } \\quad a_{n}=d \\quad \\text { then } \\quad a_{n+1} \\geq a_{n}+d=2 a_{n} \\geq 2 \\cdot 2^{n}=2^{n+1} .\n\\end{aligned}\n$$\nThe only remaining possibility is that $a_{n}=2 d$ and $a_{n+1}=3 d$, which we assume for the sequel. So $a_{n+1}=\\frac{3}{2} a_{n}$.\n\nLet now $\\operatorname{gcd}\\left(a_{n-1}, a_{n}\\right)=d^{\\prime}$; then $d^{\\prime}>a_{n-2}$. Write $a_{n}=m d^{\\prime}$ ( $m$ an integer). Keeping in mind that $d^{\\prime} \\leq a_{n-1}<d$ and $a_{n}=2 d$, we get that $m \\geq 3$. Also $a_{n-1}<d=\\frac{1}{2} m d^{\\prime}$, $a_{n+1}=\\frac{3}{2} m d^{\\prime}$. Again we single out the cases which imply the induction claim immediately:\n$$\n\\begin{aligned}\n& \\text { if } m \\geq 6 \\quad \\text { then } a_{n+1}=\\frac{3}{2} m d^{\\prime} \\geq 9 d^{\\prime}>9 a_{n-2} \\geq 9 \\cdot 2^{n-2}>2^{n+1} ; \\\\\n& \\text { if } 3 \\leq m \\leq 4 \\text { then } a_{n-1}<\\frac{1}{2} \\cdot 4 d^{\\prime}, \\text { and hence } a_{n-1}=d^{\\prime}, \\\\\n& \\quad a_{n+1}=\\frac{3}{2} m a_{n-1} \\geq \\frac{3}{2} \\cdot 3 a_{n-1} \\geq \\frac{9}{2} \\cdot 2^{n-1}>2^{n+1} .\n\\end{aligned}\n$$\nSo we are left with the case $m=5$, which means that $a_{n}=5 d^{\\prime}, a_{n+1}=\\frac{15}{2} d^{\\prime}, a_{n-1}<d=\\frac{5}{2} d^{\\prime}$. The last relation implies that $a_{n-1}$ is either $d^{\\prime}$ or $2 d^{\\prime}$. Anyway, $a_{n-1} \\mid 2 d^{\\prime}$.\n\nThe same pattern repeats once more. We denote $\\operatorname{gcd}\\left(a_{n-2}, a_{n-1}\\right)=d^{\\prime \\prime}$; then $d^{\\prime \\prime}>a_{n-3}$. Because $d^{\\prime \\prime}$ is a divisor of $a_{n-1}$, hence also of $2 d^{\\prime}$, we may write $2 d^{\\prime}=m^{\\prime} d^{\\prime \\prime}$ ( $m^{\\prime}$ an integer). Since $d^{\\prime \\prime} \\leq a_{n-2}<d^{\\prime}$, we get $m^{\\prime} \\geq 3$. Also, $a_{n-2}<d^{\\prime}=\\frac{1}{2} m^{\\prime} d^{\\prime \\prime}, a_{n+1}=\\frac{15}{2} d^{\\prime}=\\frac{15}{4} m^{\\prime} d^{\\prime \\prime}$. As before, we consider the cases:\n$$\n\\begin{aligned}\n& \\text { if } m^{\\prime} \\geq 5 \\quad \\text { then } a_{n+1}=\\frac{15}{4} m^{\\prime} d^{\\prime \\prime} \\geq \\frac{75}{4} d^{\\prime \\prime}>\\frac{75}{4} a_{n-3} \\geq \\frac{75}{4} \\cdot 2^{n-3}>2^{n+1} ; \\\\\n& \\text { if } 3 \\leq m^{\\prime} \\leq 4 \\text { then } a_{n-2}<\\frac{1}{2} \\cdot 4 d^{\\prime \\prime}, \\text { and hence } a_{n-2}=d^{\\prime \\prime}, \\\\\n& \\qquad a_{n+1}=\\frac{15}{4} m^{\\prime} a_{n-2} \\geq \\frac{15}{4} \\cdot 3 a_{n-2} \\geq \\frac{45}{4} \\cdot 2^{n-2}>2^{n+1} .\n\\end{aligned}\n$$\nBoth of them have produced the induction claim. But now there are no cases left. Induction is complete; the inequality $a_{n} \\geq 2^{n}$ holds for all $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31895,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Show that the numbers\n$$\n\\binom{2^{n}-1}{0}, \\quad\\binom{2^{n}-1}{1}, \\quad\\binom{2^{n}-1}{2}, \\quad \\ldots, \\quad\\binom{2^{n}-1}{2^{n-1}-1}\n$$\nare congruent modulo $2^{n}$ to $1,3,5, \\ldots, 2^{n}-1$ in some order.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is well-known that all these numbers are odd. So the assertion that their remainders $(\\bmod\\ 2^{n})$ make up a permutation of $\\{1,3, \\ldots, 2^{n}-1\\}$ is equivalent just to saying that these remainders are all distinct. We begin by showing that\n$$\n\\begin{equation*}\n\\binom{2^{n}-1}{2 k}+\\binom{2^{n}-1}{2 k+1} \\equiv 0\\left(\\bmod 2^{n}\\right) \\quad \\text{ and } \\quad\\binom{2^{n}-1}{2 k} \\equiv(-1)^{k}\\binom{2^{n-1}-1}{k}\\left(\\bmod 2^{n}\\right) . \\tag{1}\n\\end{equation*}\n$$\nThe first relation is immediate, as the sum on the left is equal to $\\binom{2^{n}}{2 k+1}=\\frac{2^{n}}{2 k+1}\\binom{2^{n}-1}{2 k}$, hence is divisible by $2^{n}$. The second relation:\n$$\n\\binom{2^{n}-1}{2 k}=\\prod_{j=1}^{2 k} \\frac{2^{n}-j}{j}=\\prod_{i=1}^{k} \\frac{2^{n}-(2 i-1)}{2 i-1} \\cdot \\prod_{i=1}^{k} \\frac{2^{n-1}-i}{i} \\equiv(-1)^{k}\\binom{2^{n-1}-1}{k} \\quad\\left(\\bmod 2^{n}\\right) .\n$$\nThis prepares ground for a proof of the required result by induction on $n$. The base case $n=1$ is obvious. Assume the assertion is true for $n-1$ and pass to $n$, denoting $a_{k}=\\binom{2^{n-1}-1}{k}$, $b_{m}=\\binom{2^{n}-1}{m}$. The induction hypothesis is that all the numbers $a_{k}\\left(0 \\leq k<2^{n-2}\\right)$ are distinct $\\left(\\bmod 2^{n-1}\\right)$; the claim is that all the numbers $b_{m}\\left(0 \\leq m<2^{n-1}\\right)$ are distinct $\\left(\\bmod 2^{n}\\right)$.\nThe congruence relations (1) are restated as\n$$\n\\begin{equation*}\nb_{2 k} \\equiv(-1)^{k} a_{k} \\equiv-b_{2 k+1} \\quad\\left(\\bmod 2^{n}\\right) \\tag{2}\n\\end{equation*}\n$$\nShifting the exponent in the first relation of (1) from $n$ to $n-1$ we also have the congruence $a_{2 i+1} \\equiv-a_{2 i}\\left(\\bmod 2^{n-1}\\right)$. We hence conclude:\nIf, for some $j, k<2^{n-2}, a_{k} \\equiv-a_{j}\\left(\\bmod 2^{n-1}\\right)$, then $\\{j, k\\}=\\{2 i, 2 i+1\\}$ for some $i$.\nThis is so because in the sequence $\\left(a_{k}: k<2^{n-2}\\right)$ each term $a_{j}$ is complemented to $0\\left(\\bmod 2^{n-1}\\right)$ by only one other term $a_{k}$, according to the induction hypothesis.\nFrom (2) we see that $b_{4 i} \\equiv a_{2 i}$ and $b_{4 i+3} \\equiv a_{2 i+1}\\left(\\bmod 2^{n}\\right)$. Let\n$$\nM=\\{m: 0 \\leq m<2^{n-1}, m \\equiv 0 \\text{ or } 3(\\bmod 4)\\}, \\quad L=\\{l: 0 \\leq l<2^{n-1}, l \\equiv 1 \\text{ or } 2(\\bmod 4)\\} .\n$$\nThe last two congruences take on the unified form\n$$\n\\begin{equation*}\nb_{m} \\equiv a_{\\lfloor m / 2\\rfloor} \\quad\\left(\\bmod 2^{n}\\right) \\quad \\text{ for all } \\quad m \\in M \\tag{4}\n\\end{equation*}\n$$\nThus all the numbers $b_{m}$ for $m \\in M$ are distinct $\\left(\\bmod 2^{n}\\right)$ because so are the numbers $a_{k}$ (they are distinct $\\left(\\bmod 2^{n-1}\\right)$, hence also $\\left.\\left(\\bmod 2^{n}\\right)\\right)$.\nEvery $l \\in L$ is paired with a unique $m \\in M$ into a pair of the form $\\{2 k, 2 k+1\\}$. So (2) implies that also all the $b_{l}$ for $l \\in L$ are distinct $\\left(\\bmod 2^{n}\\right)$. It remains to eliminate the possibility that $b_{m} \\equiv b_{l}\\left(\\bmod 2^{n}\\right)$ for some $m \\in M, l \\in L$.\nSuppose that such a situation occurs. Let $m^{\\prime} \\in M$ be such that $\\{m^{\\prime}, l\\}$ is a pair of the form $\\{2 k, 2 k+1\\}$, so that (see (2)) $b_{m^{\\prime}} \\equiv-b_{l}\\left(\\bmod 2^{n}\\right)$. Hence $b_{m^{\\prime}} \\equiv-b_{m}\\left(\\bmod 2^{n}\\right)$. Since both $m^{\\prime}$ and $m$ are in $M$, we have by (4) $b_{m^{\\prime}} \\equiv a_{j}, b_{m} \\equiv a_{k}\\left(\\bmod 2^{n}\\right)$ for $j=\\left\\lfloor m^{\\prime} / 2\\right\\rfloor, k=\\lfloor m / 2\\rfloor$.\nThen $a_{j} \\equiv-a_{k}\\left(\\bmod 2^{n}\\right)$. Thus, according to $(3), j=2 i, k=2 i+1$ for some $i$ (or vice versa). The equality $a_{2 i+1} \\equiv-a_{2 i}\\left(\\bmod 2^{n}\\right)$ now means that $\\binom{2^{n-1}-1}{2 i}+\\binom{2^{n-1}-1}{2 i+1} \\equiv 0\\left(\\bmod 2^{n}\\right)$. However, the sum on the left is equal to $\\binom{2^{n-1}}{2 i+1}$. A number of this form cannot be divisible by $2^{n}$. This is a contradiction which concludes the induction step and proves the result.\nWe again proceed by induction, writing for brevity $N=2^{n-1}$ and keeping notation $a_{k}=\\binom{N-1}{k}, b_{m}=\\binom{2 N-1}{m}$. Assume that the result holds for the sequence $\\left(a_{0}, a_{1}, a_{2}, \\ldots, a_{N / 2-1}\\right)$. In view of the symmetry $a_{N-1-k}=a_{k}$ this sequence is a permutation of $\\left(a_{0}, a_{2}, a_{4}, \\ldots, a_{N-2}\\right)$. So the induction hypothesis says that this latter sequence, taken $(\\bmod N)$, is a permutation of $(1,3,5, \\ldots, N-1)$. Similarly, the induction claim is that $\\left(b_{0}, b_{2}, b_{4}, \\ldots, b_{2 N-2}\\right)$, taken $(\\bmod 2 N)$, is a permutation of $(1,3,5, \\ldots, 2 N-1)$.\nIn place of the congruence relations (2) we now use the following ones,\n$$\n\\begin{equation*}\nb_{4 i} \\equiv a_{2 i}(\\bmod N) \\quad \\text{ and } \\quad b_{4 i+2} \\equiv b_{4 i}+N(\\bmod 2 N) \\tag{5}\n\\end{equation*}\n$$\nGiven this, the conclusion is immediate: the first formula of (5) together with the induction hypothesis tells us that $\\left(b_{0}, b_{4}, b_{8}, \\ldots, b_{2 N-4}\\right)(\\bmod N)$ is a permutation of $(1,3,5, \\ldots, N-1)$. Then the second formula of (5) shows that $\\left(b_{2}, b_{6}, b_{10}, \\ldots, b_{2 N-2}\\right)(\\bmod N)$ is exactly the same permutation; moreover, this formula distinguishes $(\\bmod 2 N)$ each $b_{4 i}$ from $b_{4 i+2}$.\nConsequently, these two sequences combined represent $(\\bmod 2 N)$ a permutation of the sequence $(1,3,5, \\ldots, N-1, N+1, N+3, N+5, \\ldots, N+N-1)$, and this is precisely the induction claim.\nNow we prove formulas (5); we begin with the second one. Since $b_{m+1}=b_{m} \\cdot \\frac{2 N-m-1}{m+1}$,\n$$\nb_{4 i+2}=b_{4 i} \\cdot \\frac{2 N-4 i-1}{4 i+1} \\cdot \\frac{2 N-4 i-2}{4 i+2}=b_{4 i} \\cdot \\frac{2 N-4 i-1}{4 i+1} \\cdot \\frac{N-2 i-1}{2 i+1} .\n$$\nThe desired congruence $b_{4 i+2} \\equiv b_{4 i}+N$ may be multiplied by the odd number ( $4 i+1$ ) ( $2 i+1$ ), giving rise to a chain of successively equivalent congruences:\n$$\n\\begin{aligned}\nb_{4 i}(2 N-4 i-1)(N-2 i-1) & \\equiv\\left(b_{4 i}+N\\right)(4 i+1)(2 i+1) & & (\\bmod 2 N), \\\\\nb_{4 i}(2 i+1-N) & \\equiv\\left(b_{4 i}+N\\right)(2 i+1) & & (\\bmod 2 N), \\\\\n\\left(b_{4 i}+2 i+1\\right) N & \\equiv 0 & & (\\bmod 2 N) ;\n\\end{aligned}\n$$\nand the last one is satisfied, as $b_{4 i}$ is odd. This settles the second relation in (5).\nThe first one is proved by induction on $i$. It holds for $i=0$. Assume $b_{4 i} \\equiv a_{2 i}(\\bmod 2 N)$ and consider $i+1$ :\n$$\nb_{4 i+4}=b_{4 i+2} \\cdot \\frac{2 N-4 i-3}{4 i+3} \\cdot \\frac{2 N-4 i-4}{4 i+4} ; \\quad a_{2 i+2}=a_{2 i} \\cdot \\frac{N-2 i-1}{2 i+1} \\cdot \\frac{N-2 i-2}{2 i+2} .\n$$\nBoth expressions have the fraction $\\frac{N-2 i-2}{2 i+2}$ as the last factor. Since $2 i+2<N=2^{n-1}$, this fraction reduces to $\\ell / m$ with $\\ell$ and $m$ odd. In showing that $b_{4 i+4} \\equiv a_{2 i+2}(\\bmod 2 N)$, we may ignore this common factor $\\ell / m$. Clearing other odd denominators reduces the claim to\n$$\nb_{4 i+2}(2 N-4 i-3)(2 i+1) \\equiv a_{2 i}(N-2 i-1)(4 i+3) \\quad(\\bmod 2 N)\n$$\nBy the inductive assumption (saying that $b_{4 i} \\equiv a_{2 i}(\\bmod 2 N)$ ) and by the second relation of (5), this is equivalent to\n$$\n\\left(b_{4 i}+N\\right)(2 i+1) \\equiv b_{4 i}(2 i+1-N) \\quad(\\bmod 2 N)\n$$\na congruence which we have already met in the preceding proof a few lines above. This completes induction (on $i$ ) and the proof of (5), hence also the whole solution.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 31896,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every $n \\in \\mathbb{N}$ let $d(n)$ denote the number of (positive) divisors of $n$. Find all functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ with the following properties:\n(i) $d(f(x))=x$ for all $x \\in \\mathbb{N}$;\n(ii) $f(x y)$ divides $(x-1) y^{x y-1} f(x)$ for all $x, y \\in \\mathbb{N}$.",
    "options": [],
    "answer": "The unique function is given by f(1)=1 and, for n>1 with prime factorization n=∏ p_i^{a_i}, f(n)=∏ p_i^{p_i^{a_i}−1}.",
    "solution": "There is a unique solution: the function $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ defined by $f(1)=1$ and\n$$\nf(n)=p_{1}^{p_{1}^{a_{1}}-1} p_{2}^{p_{2}^{a_{2}}-1} \\cdots p_{k}^{p_{k}^{a_{k}}-1} \\text{ where } n=p_{1}^{a_{1}} p_{2}^{a_{2}} \\cdots p_{k}^{a_{k}} \\text{ is the prime factorization of } n>1.\n$$\nDirect verification shows that this function meets the requirements.\n\nConversely, let $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ satisfy (i) and (ii). Applying (i) for $x=1$ gives $d(f(1))=1$, so $f(1)=1$. In the sequel we prove that (1) holds for all $n>1$. Notice that $f(m)=f(n)$ implies $m=n$ in view of (i). The formula $d\\left(p_{1}^{b_{1}} \\cdots p_{k}^{b_{k}}\\right)=\\left(b_{1}+1\\right) \\cdots\\left(b_{k}+1\\right)$ will be used throughout.\n\nLet $p$ be a prime. Since $d(f(p))=p$, the formula just mentioned yields $f(p)=q^{p-1}$ for some prime $q$; in particular $f(2)=q^{2-1}=q$ is a prime. We prove that $f(p)=p^{p-1}$ for all primes $p$.\n\nSuppose that $p$ is odd and $f(p)=q^{p-1}$ for a prime $q$. Applying (ii) first with $x=2$, $y=p$ and then with $x=p, y=2$ shows that $f(2 p)$ divides both $(2-1) p^{2 p-1} f(2)=p^{2 p-1} f(2)$ and $(p-1) 2^{2 p-1} f(p)=(p-1) 2^{2 p-1} q^{p-1}$. If $q \\neq p$ then the odd prime $p$ does not divide $(p-1) 2^{2 p-1} q^{p-1}$, hence the greatest common divisor of $p^{2 p-1} f(2)$ and $(p-1) 2^{2 p-1} q^{p-1}$ is a divisor of $f(2)$. Thus $f(2 p)$ divides $f(2)$ which is a prime. As $f(2 p)>1$, we obtain $f(2 p)=f(2)$ which is impossible. So $q=p$, i.e. $f(p)=p^{p-1}$.\n\nFor $p=2$ the same argument with $x=2, y=3$ and $x=3, y=2$ shows that $f(6)$ divides both $3^{5} f(2)$ and $2^{6} f(3)=2^{6} 3^{2}$. If the prime $f(2)$ is odd then $f(6)$ divides $3^{2}=9$, so $f(6) \\in\\{1,3,9\\}$. However then $6=d(f(6)) \\in\\{d(1), d(3), d(9)\\}=\\{1,2,3\\}$ which is false. In conclusion $f(2)=2$.\n\nNext, for each $n>1$ the prime divisors of $f(n)$ are among the ones of $n$. Indeed, let $p$ be the least prime divisor of $n$. Apply (ii) with $x=p$ and $y=n / p$ to obtain that $f(n)$ divides $(p-1) y^{n-1} f(p)=(p-1) y^{n-1} p^{p-1}$. Write $f(n)=\\ell P$ where $\\ell$ is coprime to $n$ and $P$ is a product of primes dividing $n$. Since $\\ell$ divides $(p-1) y^{n-1} p^{p-1}$ and is coprime to $y^{n-1} p^{p-1}$, it divides $p-1$; hence $d(\\ell) \\leq \\ell<p$. But (i) gives $n=d(f(n))=d(\\ell P)$, and $d(\\ell P)=d(\\ell) d(P)$ as $\\ell$ and $P$ are coprime. Therefore $d(\\ell)$ is a divisor of $n$ less than $p$, meaning that $\\ell=1$ and proving the claim.\n\nNow (1) is immediate for prime powers. If $p$ is a prime and $a \\geq 1$, by the above the only prime factor of $f\\left(p^{a}\\right)$ is $p$ (a prime factor does exist as $f\\left(p^{a}\\right)>1$). So $f\\left(p^{a}\\right)=p^{b}$ for some $b \\geq 1$, and (i) yields $p^{a}=d\\left(f\\left(p^{a}\\right)\\right)=d\\left(p^{b}\\right)=b+1$. Hence $f\\left(p^{a}\\right)=p^{p^{a}-1}$, as needed.\n\nLet us finally show that (1) is true for a general $n>1$ with prime factorization $n=p_{1}^{a_{1}} \\cdots p_{k}^{a_{k}}$. We saw that the prime factorization of $f(n)$ has the form $f(n)=p_{1}^{b_{1}} \\cdots p_{k}^{b_{k}}$. For $i=1, \\ldots, k$, set $x=p_{i}^{a_{i}}$ and $y=n / x$ in (ii) to infer that $f(n)$ divides $\\left(p_{i}^{a_{i}}-1\\right) y^{n-1} f\\left(p_{i}^{a_{i}}\\right)$. Hence $p_{i}^{b_{i}}$ divides $\\left(p_{i}^{a_{i}}-1\\right) y^{n-1} f\\left(p_{i}^{a_{i}}\\right)$, and because $p_{i}^{b_{i}}$ is coprime to $\\left(p_{i}^{a_{i}}-1\\right) y^{n-1}$, it follows that $p_{i}^{b_{i}}$ divides $f\\left(p_{i}^{a_{i}}\\right)=p_{i}^{p_{i}^{a_{i}}-1}$. So $b_{i} \\leq p_{i}^{a_{i}}-1$ for all $i=1, \\ldots, k$. Combined with (i), these conclusions imply\n$$\np_{1}^{a_{1}} \\cdots p_{k}^{a_{k}}=n=d(f(n))=d\\left(p_{1}^{b_{1}} \\cdots p_{k}^{b_{k}}\\right)=\\left(b_{1}+1\\right) \\cdots\\left(b_{k}+1\\right) \\leq p_{1}^{a_{1}} \\cdots p_{k}^{a_{k}}\n$$\nHence all inequalities $b_{i} \\leq p_{i}^{a_{i}}-1$ must be equalities, $i=1, \\ldots, k$, implying that (1) holds true. The proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31897,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = a + b + c$. Prove that\n$$\n\\frac{1}{(2a + b + c)^2} + \\frac{1}{(2b + c + a)^2} + \\frac{1}{(2c + a + b)^2} \\leq \\frac{3}{16}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For positive real numbers $x$, $y$, $z$, from the arithmetic-geometric-mean inequality,\n$$\n2x + y + z = (x + y) + (x + z) \\geq 2 \\sqrt{(x + y)(x + z)}\n$$\nwe obtain\n$$\n\\frac{1}{(2x + y + z)^2} \\leq \\frac{1}{4(x + y)(x + z)}\n$$\nApplying this to the left-hand side terms of the inequality to prove, we get\n$$\n\\begin{align*}\n\\frac{1}{(2a + b + c)^2} &+ \\frac{1}{(2b + c + a)^2} + \\frac{1}{(2c + a + b)^2} \\\\\n&\\leq \\frac{1}{4(a + b)(a + c)} + \\frac{1}{4(b + c)(b + a)} + \\frac{1}{4(c + a)(c + b)} \\\\\n&= \\frac{(b + c) + (c + a) + (a + b)}{4(a + b)(b + c)(c + a)} = \\frac{a + b + c}{2(a + b)(b + c)(c + a)} \\tag{1}\n\\end{align*}\n$$\nA second application of the inequality of the arithmetic-geometric mean yields\n$$\na^2 b + a^2 c + b^2 a + b^2 c + c^2 a + c^2 b \\geq 6abc\n$$\nor, equivalently,\n$$\n\\begin{equation*}\n9(a + b)(b + c)(c + a) \\geq 8(a + b + c)(ab + bc + ca) \\tag{2}\n\\end{equation*}\n$$\nThe supposition $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = a + b + c$ can be written as\n$$\n\\begin{equation*}\nab + bc + ca = abc(a + b + c) . \\tag{3}\n\\end{equation*}\n$$\nApplying the arithmetic-geometric-mean inequality $x^2 y^2 + x^2 z^2 \\geq 2x^2 y z$ thrice, we get\n$$\na^2 b^2 + b^2 c^2 + c^2 a^2 \\geq a^2 b c + a b^2 c + a b c^2\n$$\nwhich is equivalent to\n$$\n\\begin{equation*}\n(ab + bc + ca)^2 \\geq 3abc(a + b + c) \\tag{4}\n\\end{equation*}\n$$\nCombining (1), (2), (3), and (4), we will finish the proof:\n$$\n\\begin{aligned}\n\\frac{a + b + c}{2(a + b)(b + c)(c + a)} &= \\frac{(a + b + c)(ab + bc + ca)}{2(a + b)(b + c)(c + a)} \\cdot \\frac{ab + bc + ca}{abc(a + b + c)} \\cdot \\frac{abc(a + b + c)}{(ab + bc + ca)^2} \\\\\n&\\leq \\frac{9}{2 \\cdot 8} \\cdot 1 \\cdot \\frac{1}{3} = \\frac{3}{16}\n\\end{aligned}\n$$\nEquivalently, we prove the homogenized inequality\n$$\n\\frac{(a + b + c)^2}{(2a + b + c)^2} + \\frac{(a + b + c)^2}{(a + 2b + c)^2} + \\frac{(a + b + c)^2}{(a + b + 2c)^2} \\leq \\frac{3}{16}(a + b + c)\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right)\n$$\nfor all positive real numbers $a$, $b$, $c$. Without loss of generality we choose $a + b + c = 1$. Thus, the problem is equivalent to prove for all $a$, $b$, $c > 0$, fulfilling this condition, the inequality\n$$\n\\begin{equation*}\n\\frac{1}{(1 + a)^2} + \\frac{1}{(1 + b)^2} + \\frac{1}{(1 + c)^2} \\leq \\frac{3}{16}\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) \\tag{5}\n\\end{equation*}\n$$\nApplying Jensen's inequality to the function $f(x) = \\frac{x}{(1 + x)^2}$, which is concave for $0 \\leq x \\leq 2$ and increasing for $0 \\leq x \\leq 1$, we obtain\n$$\n\\alpha \\frac{a}{(1 + a)^2} + \\beta \\frac{b}{(1 + b)^2} + \\gamma \\frac{c}{(1 + c)^2} \\leq (\\alpha + \\beta + \\gamma) \\frac{A}{(1 + A)^2}, \\quad \\text{ where } \\quad A = \\frac{\\alpha a + \\beta b + \\gamma c}{\\alpha + \\beta + \\gamma} .\n$$\nChoosing $\\alpha = \\frac{1}{a}$, $\\beta = \\frac{1}{b}$, and $\\gamma = \\frac{1}{c}$, we can apply the harmonic-arithmetic-mean inequality\n$$\nA = \\frac{3}{\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}} \\leq \\frac{a + b + c}{3} = \\frac{1}{3} < 1\n$$\nFinally we prove (5):\n$$\n\\begin{aligned}\n\\frac{1}{(1 + a)^2} + \\frac{1}{(1 + b)^2} + \\frac{1}{(1 + c)^2} &\\leq \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) \\frac{A}{(1 + A)^2} \\\\\n&\\leq \\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right) \\frac{\\frac{1}{3}}{\\left(1 + \\frac{1}{3}\\right)^2} = \\frac{3}{16}\\left(\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}\\right)\n\\end{aligned}",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31898,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ be any function that maps the set of real numbers into the set of real numbers. Prove that there exist real numbers $x$ and $y$ such that\n$$\nf(x-f(y))>y f(x)+x .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that\n$$\nf(x-f(y)) \\leq y f(x)+x \\quad \\text{ for all real } x, y . \\tag{1}\n$$\nLet $a=f(0)$. Setting $y=0$ in (1) gives $f(x-a) \\leq x$ for all real $x$ and, equivalently,\n$$\nf(y) \\leq y+a \\quad \\text{ for all real } y . \\tag{2}\n$$\nSetting $x=f(y)$ in (1) yields in view of (2)\n$$\na=f(0) \\leq y f(f(y))+f(y) \\leq y f(f(y))+y+a .\n$$\nThis implies $0 \\leq y(f(f(y))+1)$ and thus\n$$\nf(f(y)) \\geq -1 \\quad \\text{ for all } y>0 . \\tag{3}\n$$\nFrom (2) and (3) we obtain $-1 \\leq f(f(y)) \\leq f(y)+a$ for all $y>0$, so\n$$\nf(y) \\geq -a-1 \\quad \\text{ for all } y>0 \\tag{4}\n$$\nNow we show that\n$$\nf(x) \\leq 0 \\quad \\text{ for all real } x . \\tag{5}\n$$\nAssume the contrary, i.e. there is some $x$ such that $f(x)>0$. Take any $y$ such that\n$$\ny<x-a \\quad \\text{ and } \\quad y<\\frac{-a-x-1}{f(x)} .\n$$\nThen in view of (2)\n$$\nx-f(y) \\geq x-(y+a)>0\n$$\nand with (1) and (4) we obtain\n$$\ny f(x)+x \\geq f(x-f(y)) \\geq -a-1,\n$$\nwhence\n$$\ny \\geq \\frac{-a-x-1}{f(x)}\n$$\ncontrary to our choice of $y$. Thereby, we have established (5).\nSetting $x=0$ in (5) leads to $a=f(0) \\leq 0$ and (2) then yields\n$$\nf(x) \\leq x \\quad \\text{ for all real } x . \\tag{6}\n$$\nNow choose $y$ such that $y>0$ and $y>-f(-1)-1$ and set $x=f(y)-1$. From (1), (5) and\n(6) we obtain\n$$\nf(-1)=f(x-f(y)) \\leq y f(x)+x=y f(f(y)-1)+f(y)-1 \\leq y(f(y)-1)-1 \\leq -y-1,\n$$\ni.e. $y \\leq -f(-1)-1$, a contradiction to the choice of $y$.\nAssume that\n$$\nf(x-f(y)) \\leq y f(x)+x \\quad \\text{ for all real } x, y . \\tag{7}\n$$\nLet $a=f(0)$. Setting $y=0$ in (7) gives $f(x-a) \\leq x$ for all real $x$ and, equivalently,\n$$\nf(y) \\leq y+a \\quad \\text{ for all real } y . \\tag{8}\n$$\nNow we show that\n$$\nf(z) \\geq 0 \\quad \\text{ for all } z \\geq 1 \\tag{9}\n$$\nLet $z \\geq 1$ be fixed, set $b=f(z)$ and assume that $b<0$. Setting $x=w+b$ and $y=z$ in (7) gives\n$$\nf(w)-z f(w+b) \\leq w+b \\text{ for all real } w . \\tag{10}\n$$\nApplying (10) to $w, w+b, \\ldots, w+(n-1) b$, where $n=1,2, \\ldots$, leads to\n$$\n\\begin{aligned}\nf(w)-z^{n} f(w+n b)= & (f(w)-z f(w+b))+z(f(w+b)-z f(w+2 b)) \\\\\n& +\\cdots+z^{n-1}(f(w+(n-1) b)-z f(w+n b)) \\\\\n\\leq & (w+b)+z(w+2 b)+\\cdots+z^{n-1}(w+n b)\n\\end{aligned}\n$$\nFrom (8) we obtain\n$$\nf(w+n b) \\leq w+n b+a\n$$\nand, thus, we have for all positive integers $n$\n$$\nf(w) \\leq\\left(1+z+\\cdots+z^{n-1}+z^{n}\\right) w+\\left(1+2 z+\\cdots+n z^{n-1}+n z^{n}\\right) b+z^{n} a . \\tag{11}\n$$\nWith $w=0$ we get\n$$\na \\leq\\left(1+2 z+\\cdots+n z^{n-1}+n z^{n}\\right) b+a z^{n} \\tag{12}\n$$\nIn view of the assumption $b<0$ we find some $n$ such that\n$$\na>(n b+a) z^{n} \\tag{13}\n$$\nbecause the right hand side tends to $-\\infty$ as $n \\rightarrow \\infty$. Now (12) and (13) give the desired contradiction and (9) is established. In addition, we have for $z=1$ the strict inequality\n$$\nf(1)>0 . \\tag{14}\n$$\nIndeed, assume that $f(1)=0$. Then setting $w=-1$ and $z=1$ in (11) leads to\n$$\nf(-1) \\leq-(n+1)+a\n$$\nwhich is false if $n$ is sufficiently large.\nTo complete the proof we set $t=\\min \\{-a,-2 / f(1)\\}$. Setting $x=1$ and $y=t$ in (7) gives\n$$\nf(1-f(t)) \\leq t f(1)+1 \\leq -2+1=-1 \\tag{15}\n$$\nOn the other hand, by (8) and the choice of $t$ we have $f(t) \\leq t+a \\leq 0$ and hence $1-f(t) \\geq 1$. The inequality (9) yields\n$$\nf(1-f(t)) \\geq 0\n$$\nwhich contradicts (15).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31899,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f$ from the set of real numbers into the set of real numbers which satisfy for all real $x, y$ the identity\n$$\nf(x f(x+y))=f(y f(x))+x^{2} .\n$$",
    "options": [],
    "answer": "f(x) = x and f(x) = -x",
    "solution": "It is no hard to see that the two functions given by $f(x)=x$ and $f(x)=-x$ for all real $x$ respectively solve the functional equation. In the sequel, we prove that there are no further solutions.\nLet $f$ be a function satisfying the given equation. It is clear that $f$ cannot be a constant. Let us first show that $f(0)=0$. Suppose that $f(0) \\neq 0$. For any real $t$, substituting $(x, y)=\\left(0, \\frac{t}{f(0)}\\right)$ into the given functional equation, we obtain\n$$\n\\begin{equation*}\nf(0)=f(t), \\tag{1}\n\\end{equation*}\n$$\ncontradicting the fact that $f$ is not a constant function. Therefore, $f(0)=0$. Next for any $t$, substituting $(x, y)=(t, 0)$ and $(x, y)=(t,-t)$ into the given equation, we get\n$$\nf(t f(t))=f(0)+t^{2}=t^{2}\n$$\nand\n$$\nf(t f(0))=f(-t f(t))+t^{2}\n$$\nrespectively. Therefore, we conclude that\n$$\n\\begin{equation*}\nf(t f(t))=t^{2}, \\quad f(-t f(t))=-t^{2}, \\quad \\text{ for every real } t . \\tag{2}\n\\end{equation*}\n$$\nConsequently, for every real $v$, there exists a real $u$, such that $f(u)=v$. We also see that if $f(t)=0$, then $0=f(t f(t))=t^{2}$ so that $t=0$, and thus 0 is the only real number satisfying $f(t)=0$.\nWe next show that for any real number $s$,\n$$\n\\begin{equation*}\nf(-s)=-f(s) \\tag{3}\n\\end{equation*}\n$$\nThis is clear if $f(s)=0$. Suppose now $f(s)<0$, then we can find a number $t$ for which $f(s)=-t^{2}$. As $t \\neq 0$ implies $f(t) \\neq 0$, we can also find number $a$ such that $a f(t)=s$. Substituting $(x, y)=(t, a)$ into the given equation, we get\n$$\nf(t f(t+a))=f(a f(t))+t^{2}=f(s)+t^{2}=0\n$$\nand therefore, $t f(t+a)=0$, which implies $t+a=0$, and hence $s=-t f(t)$. Consequently, $f(-s)=f(t f(t))=t^{2}=-\\left(-t^{2}\\right)=-f(s)$ holds in this case.\nFinally, suppose $f(s)>0$ holds. Then there exists a real number $t \\neq 0$ for which $f(s)=t^{2}$. Choose a number $a$ such that $t f(a)=s$. Substituting $(x, y)=(t, a-t)$ into the given equation, we get $f(s)=f(t f(a))=f((a-t) f(t))+t^{2}=f((a-t) f(t))+f(s)$. So we have $f((a-t) f(t))=0$, from which we conclude that $(a-t) f(t)=0$. Since $f(t) \\neq 0$, we get $a=t$ so that $s=t f(t)$ and thus we see $f(-s)=f(-t f(t))=-t^{2}=-f(s)$ holds in this case also. This observation finishes the proof of (3).\nBy substituting $(x, y)=(s, t),(x, y)=(t,-s-t)$ and $(x, y)=(-s-t, s)$ into the given equation,\nwe obtain\n$$\n\\begin{array}{r}\nf(s f(s+t)))=f(t f(s))+s^{2} \\\\\nf(t f(-s))=f((-s-t) f(t))+t^{2}\n\\end{array}\n$$\nand\n$$\nf((-s-t) f(-t))=f(s f(-s-t))+(s+t)^{2}\n$$\nrespectively. Using the fact that $f(-x)=-f(x)$ holds for all $x$ to rewrite the second and the third equation, and rearranging the terms, we obtain\n$$\n\\begin{aligned}\nf(t f(s))-f(s f(s+t)) & =-s^{2} \\\\\nf(t f(s))-f((s+t) f(t)) & =-t^{2} \\\\\nf((s+t) f(t))+f(s f(s+t)) & =(s+t)^{2}\n\\end{aligned}\n$$\nAdding up these three equations now yields $2 f(t f(s))=2 t s$, and therefore, we conclude that $f(t f(s))=t s$ holds for every pair of real numbers $s, t$. By fixing $s$ so that $f(s)=1$, we obtain $f(x)=s x$. In view of the given equation, we see that $s= \\pm 1$. It is easy to check that both functions $f(x)=x$ and $f(x)=-x$ satisfy the given functional equation, so these are the desired solutions.\nAs in Solution 1 we obtain (1), (2) and (3).\nNow we prove that $f$ is injective. For this purpose, let us assume that $f(r)=f(s)$ for some $r \\neq s$. Then, by (2)\n$$\nr^{2}=f(r f(r))=f(r f(s))=f((s-r) f(r))+r^{2}\n$$\nwhere the last statement follows from the given functional equation with $x=r$ and $y=s-r$. Hence, $h=(s-r) f(r)$ satisfies $f(h)=0$ which implies $h^{2}=f(h f(h))=f(0)=0$, i.e., $h=0$. Then, by $s \\neq r$ we have $f(r)=0$ which implies $r=0$, and finally $f(s)=f(r)=f(0)=0$. Analogously, it follows that $s=0$ which gives the contradiction $r=s$.\nTo prove $|f(1)|=1$ we apply (2) with $t=1$ and also with $t=f(1)$ and obtain $f(f(1))=1$ and $(f(1))^{2}=f(f(1) \\cdot f(f(1)))=f(f(1))=1$.\nNow we choose $\\eta \\in\\{-1,1\\}$ with $f(1)=\\eta$. Using that $f$ is odd and the given equation with $x=1, y=z$ (second equality) and with $x=-1, y=z+2$ (fourth equality) we obtain\n$$\n\\begin{align*}\n& f(z)+2 \\eta=\\eta(f(z \\eta)+2)=\\eta(f(f(z+1))+1)=\\eta(-f(-f(z+1))+1) \\\\\n& \\quad=-\\eta f((z+2) f(-1))=-\\eta f((z+2)(-\\eta))=\\eta f((z+2) \\eta)=f(z+2) \\tag{4}\n\\end{align*}\n$$\nHence,\n$$\nf(z+2 \\eta)=\\eta f(\\eta z+2)=\\eta(f(\\eta z)+2 \\eta)=f(z)+2 .\n$$\nUsing this argument twice we obtain\n$$\nf(z+4 \\eta)=f(z+2 \\eta)+2=f(z)+4\n$$\nSubstituting $z=2 f(x)$ we have\n$$\nf(2 f(x))+4=f(2 f(x)+4 \\eta)=f(2 f(x+2)),\n$$\nwhere the last equality follows from (4). Applying the given functional equation we proceed to\n$$\nf(2 f(x+2))=f(x f(2))+4=f(2 \\eta x)+4\n$$\nwhere the last equality follows again from (4) with $z=0$, i.e., $f(2)=2 \\eta$. Finally, $f(2 f(x))= f(2 \\eta x)$ and by injectivity of $f$ we get $2 f(x)=2 \\eta x$ and hence the two solutions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31900,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $2009$ cards, each having one gold side and one black side, lying in parallel on a long table. Initially all cards show their gold sides. Two players, standing by the same long side of the table, play a game with alternating moves. Each move consists of choosing a block of $50$ consecutive cards, the leftmost of which is showing gold, and turning them all over, so those which showed gold now show black and vice versa. The last player who can make a legal move wins.\n\na. Does the game necessarily end?\n\nb. Does there exist a winning strategy for the starting player?",
    "options": [],
    "answer": "a. Yes. b. No; the second player wins.",
    "solution": "a.\nWe interpret a card showing black as the digit $0$ and a card showing gold as the digit $1$. Thus each position of the $2009$ cards, read from left to right, corresponds bijectively to a nonnegative integer written in binary notation of $2009$ digits, where leading zeros are allowed. Each move decreases this integer, so the game must end.\n\nb.\nWe show that there is no winning strategy for the starting player. We label the cards from right to left by $1, \\ldots, 2009$ and consider the set $S$ of cards with labels $50i$, $i=1,2, \\ldots, 40$. Let $g_{n}$ be the number of cards from $S$ showing gold after $n$ moves. Obviously, $g_{0}=40$. Moreover, $|g_{n}-g_{n+1}|=1$ as long as the play goes on. Thus, after an odd number of moves, the nonstarting player finds a card from $S$ showing gold and hence can make a move. Consequently, this player always wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31901,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any integer $n \\geq 2$, let $N(n)$ be the maximal number of triples $(a_{i}, b_{i}, c_{i})$, $i=1, \\ldots, N(n)$, consisting of nonnegative integers $a_{i}, b_{i}$ and $c_{i}$ such that the following two conditions are satisfied:\n(1) $a_{i}+b_{i}+c_{i}=n$ for all $i=1, \\ldots, N(n)$,\n(2) If $i \\neq j$, then $a_{i} \\neq a_{j}, b_{i} \\neq b_{j}$ and $c_{i} \\neq c_{j}$.\nDetermine $N(n)$ for all $n \\geq 2$.",
    "options": [],
    "answer": "N(n) = floor(2n/3) + 1",
    "solution": "Let $n \\geq 2$ be an integer and let $\\{T_{1}, \\ldots, T_{N}\\}$ be any set of triples of nonnegative integers satisfying the conditions (1) and (2). Since the $a$-coordinates are pairwise distinct we have\n$$\n\\sum_{i=1}^{N} a_{i} \\geq \\sum_{i=1}^{N}(i-1)=\\frac{N(N-1)}{2}\n$$\nAnalogously,\n$$\n\\sum_{i=1}^{N} b_{i} \\geq \\frac{N(N-1)}{2} \\quad \\text{ and } \\quad \\sum_{i=1}^{N} c_{i} \\geq \\frac{N(N-1)}{2} .\n$$\nSumming these three inequalities and applying (1) yields\n$$\n3 \\frac{N(N-1)}{2} \\leq \\sum_{i=1}^{N} a_{i}+\\sum_{i=1}^{N} b_{i}+\\sum_{i=1}^{N} c_{i}=\\sum_{i=1}^{N}\\left(a_{i}+b_{i}+c_{i}\\right)=n N,\n$$\nhence $3 \\frac{N-1}{2} \\leq n$ and, consequently,\n$$\nN \\leq\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1\n$$\nBy constructing examples, we show that this upper bound can be attained, so $N(n)=\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1$.\nWe distinguish the cases $n=3 k-1, n=3 k$ and $n=3 k+1$ for $k \\geq 1$ and present the extremal examples in form of a table.\n\n| $n=3 k-1$ |   |   |\n| :---: | :---: | :---: |\n| $\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1=2 k$ |   |   |\n| $a_{i}$ | $b_{i}$ | $c_{i}$ |\n| 0 | $k+1$ | $2 k-2$ |\n| 1 | $k+2$ | $2 k-4$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $k-1$ | $2 k$ | 0 |\n| $k$ | 0 | $2 k-1$ |\n| $k+1$ | 1 | $2 k-3$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $2 k-1$ | $k-1$ | 1 |\n\n| $n=3 k$ |   |   |\n| :---: | :---: | :---: |\n| $\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1=2 k+1$ |   |   |\n| $a_{i}$ | $b_{i}$ | $c_{i}$ |\n| 0 | $k$ | $2 k$ |\n| 1 | $k+1$ | $2 k-2$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $k$ | $2 k$ | 0 |\n| $k+1$ | 0 | $2 k-1$ |\n| $k+2$ | 1 | $2 k-3$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $2 k$ | $k-1$ | 1 |\n\n| $n=3 k+1$ |   |   |\n| :---: | :---: | :---: |\n| $\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1=2 k+1$ |   |   |\n| $a_{i}$ | $b_{i}$ | $c_{i}$ |\n| 0 | $k$ | $2 k+1$ |\n| 1 | $k+1$ | $2 k-1$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $k$ | $2 k$ | 1 |\n| $k+1$ | 0 | $2 k$ |\n| $k+2$ | 1 | $2 k-2$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ |\n| $2 k$ | $k-1$ | 2 |\n\nIt can be easily seen that the conditions (1) and (2) are satisfied and that we indeed have $\\left\\lfloor\\frac{2 n}{3}\\right\\rfloor+1$ triples in each case.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31902,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Given a sequence $\\varepsilon_{1}, \\ldots, \\varepsilon_{n-1}$ with $\\varepsilon_{i}=0$ or $\\varepsilon_{i}=1$ for each $i=1, \\ldots, n-1$, the sequences $a_{0}, \\ldots, a_{n}$ and $b_{0}, \\ldots, b_{n}$ are constructed by the following rules:\n$$\n\\begin{gathered}\na_{0}=b_{0}=1, \\quad a_{1}=b_{1}=7 \\\\\na_{i+1}=\\left\\{\\begin{array}{ll}\n2 a_{i-1}+3 a_{i}, & \\text{ if } \\varepsilon_{i}=0, \\\\\n3 a_{i-1}+a_{i}, & \\text{ if } \\varepsilon_{i}=1,\n\\end{array}\\right. \\quad \\text{ for each } i=1, \\ldots, n-1, \\\\\nb_{i+1}=\\left\\{\\begin{array}{ll}\n2 b_{i-1}+3 b_{i}, & \\text{ if } \\varepsilon_{n-i}=0, \\\\\n3 b_{i-1}+b_{i}, & \\text{ if } \\varepsilon_{n-i}=1,\n\\end{array}\\right. \\text{ for each } i=1, \\ldots, n-1\n\\end{gathered}\n$$\nProve that $a_{n}=b_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For a binary word $w=\\sigma_{1} \\ldots \\sigma_{n}$ of length $n$ and a letter $\\sigma \\in\\{0,1\\}$ let $w \\sigma= \\sigma_{1} \\ldots \\sigma_{n} \\sigma$ and $\\sigma w=\\sigma \\sigma_{1} \\ldots \\sigma_{n}$. Moreover let $\\bar{w}=\\sigma_{n} \\ldots \\sigma_{1}$ and let $\\emptyset$ be the empty word (of length 0 and with $\\bar{\\emptyset}=\\emptyset$ ). Let $( u, v )$ be a pair of two real numbers. For binary words $w$ we define recursively the numbers $(u, v)^{w}$ as follows:\n$$\n\\begin{gathered}\n(u, v)^{\\emptyset}=v, \\quad(u, v)^{0}=2 u+3 v, \\quad(u, v)^{1}=3 u+v \\\\\n(u, v)^{w \\sigma \\varepsilon}= \\begin{cases}2(u, v)^{w}+3(u, v)^{w \\sigma}, & \\text{ if } \\varepsilon=0 \\\\\n3(u, v)^{w}+(u, v)^{w \\sigma}, & \\text{ if } \\varepsilon=1\\end{cases}\n\\end{gathered}\n$$\nIt easily follows by induction on the length of $w$ that for all real numbers $u_{1}, v_{1}, u_{2}, v_{2}, \\lambda_{1}$ and $\\lambda_{2}$\n$$\n\\begin{equation*}\n\\left(\\lambda_{1} u_{1}+\\lambda_{2} u_{2}, \\lambda_{1} v_{1}+\\lambda_{2} v_{2}\\right)^{w}=\\lambda_{1}\\left(u_{1}, v_{1}\\right)^{w}+\\lambda_{2}\\left(u_{2}, v_{2}\\right)^{w} \\tag{1}\n\\end{equation*}\n$$\nand that for $\\varepsilon \\in\\{0,1\\}$\n$$\n\\begin{equation*}\n(u, v)^{\\varepsilon w}=\\left(v,(u, v)^{\\varepsilon}\\right)^{w} \\tag{2}\n\\end{equation*}\n$$\nObviously, for $n \\geq 1$ and $w=\\varepsilon_{1} \\ldots \\varepsilon_{n-1}$, we have $a_{n}=(1,7)^{w}$ and $b_{n}=(1,7)^{\\bar{w}}$. Thus it is sufficient to prove that\n$$\n\\begin{equation*}\n(1,7)^{w}=(1,7)^{\\bar{w}} \\tag{3}\n\\end{equation*}\n$$\nfor each binary word $w$. We proceed by induction on the length of $w$. The assertion is obvious if $w$ has length 0 or 1. Now let $w \\sigma \\varepsilon$ be a binary word of length $n \\geq 2$ and suppose that the assertion is true for all binary words of length at most $n-1$.\nNote that $(2,1)^{\\sigma}=7=(1,7)^{\\emptyset}$ for $\\sigma \\in\\{0,1\\}, (1,7)^{0}=23$, and $(1,7)^{1}=10$.\n\nFirst let $\\varepsilon=0$. Then in view of the induction hypothesis and the equalities (1) and (2), we obtain\n$$\n\\begin{aligned}\n(1,7)^{w \\sigma 0}&=2(1,7)^{w}+3(1,7)^{w \\sigma}=2(1,7)^{\\bar{w}}+3(1,7)^{\\sigma \\bar{w}}=2(2,1)^{\\sigma \\bar{w}}+3(1,7)^{\\sigma \\bar{w}} \\\\\n&=(7,23)^{\\sigma \\bar{w}}=(1,7)^{0 \\sigma \\bar{w}}\n\\end{aligned}\n$$\nNow let $\\varepsilon=1$. Analogously, we obtain\n$$\n\\begin{aligned}\n(1,7)^{w \\sigma 1}&=3(1,7)^{w}+(1,7)^{w \\sigma}=3(1,7)^{\\bar{w}}+(1,7)^{\\sigma \\bar{w}}=3(2,1)^{\\sigma \\bar{w}}+(1,7)^{\\sigma \\bar{w}} \\\\\n&=(7,10)^{\\sigma \\bar{w}}=(1,7)^{1 \\sigma \\bar{w}}\n\\end{aligned}\n$$\nThus the induction step is complete, (3) and hence also $a_{n}=b_{n}$ are proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31903,
    "subject": "Mathematics (Multi-modal)",
    "question": "For an integer $m \\geq 1$, we consider partitions of a $2^{m} \\times 2^{m}$ chessboard into rectangles consisting of cells of the chessboard, in which each of the $2^{m}$ cells along one diagonal forms a separate rectangle of side length 1. Determine the smallest possible sum of rectangle perimeters in such a partition.",
    "options": [],
    "answer": "(m+1) 2^{m+2}",
    "solution": "Solution 1. For a $k \\times k$ chessboard, we introduce in a standard way coordinates of the vertices of the cells and assume that the cell $C_{ij}$ in row $i$ and column $j$ has vertices $(i-1, j-1), (i-1, j), (i, j-1), (i, j)$, where $i, j \\in \\{1, \\ldots, k\\}$. Without loss of generality assume that the cells $C_{ii}$, $i=1, \\ldots, k$, form a separate rectangle. Then we may consider the boards $B_{k} = \\bigcup_{1 \\leq i < j \\leq k} C_{ij}$ below that diagonal and the congruent board $B_{k}' = \\bigcup_{1 \\leq j < i \\leq k} C_{ij}$ above that diagonal separately because no rectangle can simultaneously cover cells from $B_{k}$ and $B_{k}'$. We will show that for $k = 2^{m}$ the smallest total perimeter of a rectangular partition of $B_{k}$ is $m 2^{m+1}$. Then the overall answer to the problem is $2 \\cdot m 2^{m+1} + 4 \\cdot 2^{m} = (m+1) 2^{m+2}$.\n\nFirst we inductively construct for $m \\geq 1$ a partition of $B_{2^{m}}$ with total perimeter $m 2^{m+1}$. If $m=0$, the board $B_{2^{m}}$ is empty and the total perimeter is 0. For $m \\geq 0$, the board $B_{2^{m+1}}$ consists of a $2^{m} \\times 2^{m}$ square in the lower right corner with vertices $(2^{m}, 2^{m}), (2^{m}, 2^{m+1}), (2^{m+1}, 2^{m}), (2^{m+1}, 2^{m+1})$ to which two boards congruent to $B_{2^{m}}$ are glued along the left and the upper margin. The square together with the inductive partitions of these two boards yield a partition with total perimeter $4 \\cdot 2^{m} + 2 \\cdot m 2^{m+1} = (m+1) 2^{m+2}$ and the induction step is complete.\n\nLet\n$$\nD_{k} = 2k \\log_{2} k.\n$$\nNote that $D_{k} = m 2^{m+1}$ if $k = 2^{m}$. Now we show by induction on $k$ that the total perimeter of a rectangular partition of $B_{k}$ is at least $D_{k}$. The case $k=1$ is trivial (see $m=0$ from above). Let the assertion be true for all positive integers less than $k$. We investigate a fixed rectangular partition of $B_{k}$ that attains the minimal total perimeter. Let $R$ be the rectangle that covers the cell $C_{1k}$ in the lower right corner. Let $(i, j)$ be the upper left corner of $R$. First we show that $i = j$. Assume that $i < j$. Then the line from $(i, j)$ to $(i+1, j)$ or from $(i, j)$ to $(i, j-1)$ must belong to the boundary of some rectangle in the partition. Without loss of generality assume that this is the case for the line from $(i, j)$ to $(i+1, j)$.\n\nCase 1. No line from $(i, l)$ to $(i+1, l)$ where $j < l < k$ belongs to the boundary of some rectangle of the partition.\nThen there is some rectangle $R'$ of the partition that has with $R$ the common side from $(i, j)$ to $(i, k)$. If we join these two rectangles to one rectangle we get a partition with smaller total perimeter, a contradiction.\n\nCase 2. There is some $l$ such that $j < l < k$ and the line from $(i, l)$ to $(i+1, l)$ belongs to the boundary of some rectangle of the partition.\nThen we replace the upper side of $R$ by the line $(i+1, j)$ to $(i+1, k)$ and for the rectangles whose lower side belongs to the line from $(i, j)$ to $(i, k)$ we shift the lower side upwards so that the new lower side belongs to the line from $(i+1, j)$ to $(i+1, k)$. In such a way we obtain a rectangular partition of $B_{k}$ with smaller total perimeter, a contradiction.\n\nNow the fact that the upper left corner of $R$ has the coordinates $(i, i)$ is established. Consequently, the partition consists of $R$, of rectangles of a partition of a board congruent to $B_{i}$ and of rectangles of a partition of a board congruent to $B_{k-i}$. By the induction hypothesis, its total perimeter is at least\n$$\n2(k-i) + 2i + D_{i} + D_{k-i} \\geq 2k + 2i \\log_{2} i + 2(k-i) \\log_{2}(k-i).\n$$\nSince the function $f(x) = 2x \\log_{2} x$ is convex for $x > 0$, Jensen's inequality immediately shows that the minimum of the right hand side is attained for $i = k/2$. Hence the total perimeter of the optimal partition of $B_{k}$ is at least $2k + 2k/2 \\log_{2} k/2 + 2(k/2) \\log_{2}(k/2) = D_{k}$.\n\n\nSolution 2. We start as in Solution 1 and present another proof that $m 2^{m+1}$ is a lower bound for the total perimeter of a partition of $B_{2^{m}}$ into $n$ rectangles. Let briefly $M = 2^{m}$. For $1 \\leq i \\leq M$, let $r_{i}$ denote the number of rectangles in the partition that cover some cell from row $i$ and let $c_{j}$ be the number of rectangles that cover some cell from column $j$. Note that the total perimeter $p$ of all rectangles in the partition is\n$$\np = 2\\left(\\sum_{i=1}^{M} r_{i} + \\sum_{i=1}^{M} c_{i}\\right).\n$$\nNo rectangle can simultaneously cover cells from row $i$ and from column $i$ since otherwise it would also cover the cell $C_{ii}$. We classify subsets $S$ of rectangles of the partition as follows. We say that $S$ is of type $i$, $1 \\leq i \\leq M$, if $S$ contains all $r_{i}$ rectangles that cover some cell from row $i$, but none of the $c_{i}$ rectangles that cover some cell from column $i$. Altogether there are $2^{n - r_{i} - c_{i}}$ subsets of type $i$. Now we show that no subset $S$ can be simultaneously of type $i$ and of type $j$ if $i \\neq j$. Assume the contrary and let without loss of generality $i < j$. The cell $C_{ij}$ must be covered by some rectangle $R$. The subset $S$ is of type $i$, hence $R$ is contained in $S$. $S$ is of type $j$, thus $R$ does not belong to $S$, a contradiction. Since there are $2^{n}$ subsets of rectangles of the partition, we infer\n$$\n2^{n} \\geq \\sum_{i=1}^{M} 2^{n - r_{i} - c_{i}} = 2^{n} \\sum_{i=1}^{M} 2^{-\\left(r_{i} + c_{i}\\right)}\n$$\nBy applying Jensen's inequality to the convex function $f(x) = 2^{-x}$ we derive\n$$\n\\frac{1}{M} \\sum_{i=1}^{M} 2^{-\\left(r_{i} + c_{i}\\right)} \\geq 2^{-\\frac{1}{M} \\sum_{i=1}^{M} \\left(r_{i} + c_{i}\\right)} = 2^{-\\frac{p}{2M}}\n$$\nFrom the previous inequalities we obtain\n$$\n1 \\geq M 2^{-\\frac{p}{2M}}\n$$\nand equivalently\n$$\np \\geq m 2^{m+1}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31904,
    "subject": "Mathematics (Multi-modal)",
    "question": "Five identical empty buckets of 2-liter capacity stand at the vertices of a regular pentagon. Cinderella and her wicked Stepmother go through a sequence of rounds: At the beginning of every round, the Stepmother takes one liter of water from the nearby river and distributes it arbitrarily over the five buckets. Then Cinderella chooses a pair of neighboring buckets, empties them into the river, and puts them back. Then the next round begins. The Stepmother's goal is to make one of these buckets overflow. Cinderella's goal is to prevent this. Can the wicked Stepmother enforce a bucket overflow?",
    "options": [],
    "answer": "No; the Stepmother cannot enforce an overflow, and Cinderella can prevent any bucket from overflowing indefinitely.",
    "solution": "No, the Stepmother cannot enforce a bucket overflow and Cinderella can keep playing forever. Throughout we denote the five buckets by $B_{0}, B_{1}, B_{2}, B_{3}$, and $B_{4}$, where $B_{k}$ is adjacent to bucket $B_{k-1}$ and $B_{k+1}$ ($k=0,1,2,3,4$) and all indices are taken modulo $5$. Cinderella enforces that the following three conditions are satisfied at the beginning of every round:\n\n(1) Two adjacent buckets (say $B_{1}$ and $B_{2}$) are empty.\n\n(2) The two buckets standing next to these adjacent buckets (here $B_{0}$ and $B_{3}$) have total contents at most $1$.\n\n(3) The remaining bucket (here $B_{4}$) has contents at most $1$.\n\nThese conditions clearly hold at the beginning of the first round, when all buckets are empty.\n\nAssume that Cinderella manages to maintain them until the beginning of the $r$-th round ($r \\geq 1$). Denote by $x_{k}$ ($k=0,1,2,3,4$) the contents of bucket $B_{k}$ at the beginning of this round and by $y_{k}$ the corresponding contents after the Stepmother has distributed her liter of water in this round.\n\nBy the conditions, we can assume $x_{1}=x_{2}=0$, $x_{0}+x_{3} \\leq 1$ and $x_{4} \\leq 1$. Then, since the Stepmother adds one liter, we conclude $y_{0}+y_{1}+y_{2}+y_{3} \\leq 2$. This inequality implies $y_{0}+y_{2} \\leq 1$ or $y_{1}+y_{3} \\leq 1$. For reasons of symmetry, we only consider the second case.\n\nThen Cinderella empties buckets $B_{0}$ and $B_{4}$.\n\nAt the beginning of the next round $B_{0}$ and $B_{4}$ are empty (condition (1) is fulfilled), due to $y_{1}+y_{3} \\leq 1$ condition (2) is fulfilled and finally since $x_{2}=0$ we also must have $y_{2} \\leq 1$ (condition (3) is fulfilled).\n\nTherefore, Cinderella can indeed manage to maintain the three conditions (1)-(3) also at the beginning of the $(r+1)$-th round. By induction, she thus manages to maintain them at the beginning of every round. In particular she manages to keep the contents of every single bucket at most $1$ liter. Therefore, the buckets of $2$-liter capacity will never overflow.\nWe prove that Cinderella can maintain the following two conditions and hence she can prevent the buckets from overflow:\n\n(1') Every two non-adjacent buckets contain a total of at most $1$.\n\n(2') The total contents of all five buckets is at most $\\frac{3}{2}$.\n\nWe use the same notations as in the first solution. The two conditions again clearly hold at the beginning. Assume that Cinderella maintained these two conditions until the beginning of the $r$-th round. A pair of non-neighboring buckets $(B_{i}, B_{i+2}), i=0,1,2,3,4$ is called critical if $y_{i}+y_{i+2}>1$. By condition (2), after the Stepmother has distributed her water we have $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \\leq \\frac{5}{2}$. Therefore,\n\n$$\n(y_{0}+y_{2})+(y_{1}+y_{3})+(y_{2}+y_{4})+(y_{3}+y_{0})+(y_{4}+y_{1})=2(y_{0}+y_{1}+y_{2}+y_{3}+y_{4}) \\leq 5\n$$\n\nand hence there is a pair of non-neighboring buckets which is not critical, say $(B_{0}, B_{2})$. Now, if both of the pairs $(B_{3}, B_{0})$ and $(B_{2}, B_{4})$ are critical, we must have $y_{1}<\\frac{1}{2}$ and Cinderella can empty the buckets $B_{3}$ and $B_{4}$. This clearly leaves no critical pair of buckets and the total contents of all the buckets is then $y_{1}+(y_{0}+y_{2}) \\leq \\frac{3}{2}$. Therefore, conditions (1') and (2') are fulfilled.\n\nNow suppose that without loss of generality the pair $(B_{3}, B_{0})$ is not critical. If in this case $y_{0} \\leq \\frac{1}{2}$, then one of the inequalities $y_{0}+y_{1}+y_{2} \\leq \\frac{3}{2}$ and $y_{0}+y_{3}+y_{4} \\leq \\frac{3}{2}$ must hold. But then Cinderella can empty $B_{3}$ and $B_{4}$ or $B_{1}$ and $B_{2}$, respectively and clearly fulfill the conditions.\n\nFinally consider the case $y_{0}>\\frac{1}{2}$. By $y_{0}+y_{1}+y_{2}+y_{3}+y_{4} \\leq \\frac{5}{2}$, at least one of the pairs $(B_{1}, B_{3})$ and $(B_{2}, B_{4})$ is not critical. Without loss of generality let this be the pair $(B_{1}, B_{3})$. Since the pair $(B_{3}, B_{0})$ is not critical and $y_{0}>\\frac{1}{2}$, we must have $y_{3} \\leq \\frac{1}{2}$. But then, as before, Cinderella can maintain the two conditions at the beginning of the next round by either emptying $B_{1}$ and $B_{2}$ or $B_{4}$ and $B_{0}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31905,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a $999 \\times 999$ board a limp rook can move in the following way: From any square it can move to any of its adjacent squares, i.e. a square having a common side with it, and every move must be a turn, i.e. the directions of any two consecutive moves must be perpendicular. A nonintersecting route of the limp rook consists of a sequence of pairwise different squares that the limp rook can visit in that order by an admissible sequence of moves. Such a non-intersecting route is called cyclic, if the limp rook can, after reaching the last square of the route, move directly to the first square of the route and start over.\nHow many squares does the longest possible cyclic, non-intersecting route of a limp rook visit?",
    "options": [],
    "answer": "996000",
    "solution": "The answer is $998^{2}-4=4 \\cdot\\left(499^{2}-1\\right)$ squares.\n\nFirst we show that this number is an upper bound for the number of cells a limp rook can visit. To do this we color the cells with four colors $A, B, C$ and $D$ in the following way: for $(i, j) \\equiv(0,0) \\bmod 2$ use $A$, for $(i, j) \\equiv(0,1) \\bmod 2$ use $B$, for $(i, j) \\equiv(1,0) \\bmod 2$ use $C$ and for $(i, j) \\equiv(1,1) \\bmod 2$ use $D$. From an $A$-cell the rook has to move to a $B$-cell or a $C$-cell. In the first case, the order of the colors of the cells visited is given by $A, B, D, C, A, B, D, C, A, \\ldots$, in the second case it is $A, C, D, B, A, C, D, B, A, \\ldots$. Since the route is closed it must contain the same number of cells of each color. There are only $499^{2}$ $A$-cells. In the following we will show that the rook cannot visit all the $A$-cells on its route and hence the maximum possible number of cells in a route is $4 \\cdot\\left(499^{2}-1\\right)$.\n\nAssume that the route passes through every single $A$-cell. Color the $A$-cells in black and white in a chessboard manner, i.e. color any two $A$-cells at distance 2 in different color. Since the number of $A$-cells is odd the rook cannot always alternate between visiting black and white $A$-cells along its route. Hence there are two $A$-cells of the same color which are four rook-steps apart that are visited directly one after the other. Let these two $A$-cells have row and column numbers $(a, b)$ and $(a+2, b+2)$ respectively.\n\n![](attached_image_1.png)\n\nThere is up to reflection only one way the rook can take from $(a, b)$ to $(a+2, b+2)$. Let this way be $(a, b) \\rightarrow(a, b+1) \\rightarrow(a+1, b+1) \\rightarrow(a+1, b+2) \\rightarrow(a+2, b+2)$. Also let without loss of generality the color of the cell $(a, b+1)$ be $B$ (otherwise change the roles of columns and rows).\n\nNow consider the $A$-cell $(a, b+2)$. The only way the rook can pass through it is via $(a-1, b+2) \\rightarrow(a, b+2) \\rightarrow(a, b+3)$ in this order, since according to our assumption after every $A$-cell the rook passes through a $B$-cell. Hence, to connect these two parts of the path, there must be a path connecting the cell $(a, b+3)$ and $(a, b)$ and also a path connecting $(a+2, b+2)$ and $(a-1, b+2)$.\n\nBut these four cells are opposite vertices of a convex quadrilateral and the paths are outside of that quadrilateral and hence they must intersect. This is due to the following fact:\nThe path from $(a, b)$ to $(a, b+3)$ together with the line segment joining these two cells form a closed loop that has one of the cells $(a-1, b+2)$ and $(a+2, b+2)$ in its inside and the other one on the outside. Thus the path between these two points must cross the previous path.\n\nBut an intersection is only possible if a cell is visited twice. This is a contradiction.\n\nHence the number of cells visited is at most $4 \\cdot\\left(499^{2}-1\\right)$.\n\nThe following picture indicates a recursive construction for all $n \\times n$-chessboards with $n \\equiv 3$ mod 4 which clearly yields a path that misses exactly one $A$-cell (marked with a dot, the center cell of the $15 \\times 15$-chessboard) and hence, in the case of $n=999$ crosses exactly $4 \\cdot\\left(499^{2}-1\\right)$ cells.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31906,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any integer $n \\geq 2$, we compute the integer $h(n)$ by applying the following procedure to its decimal representation. Let $r$ be the rightmost digit of $n$.\n(1) If $r=0$, then the decimal representation of $h(n)$ results from the decimal representation of $n$ by removing this rightmost digit $0$.\n(2) If $1 \\leq r \\leq 9$ we split the decimal representation of $n$ into a maximal right part $R$ that solely consists of digits not less than $r$ and into a left part $L$ that either is empty or ends with a digit strictly smaller than $r$. Then the decimal representation of $h(n)$ consists of the decimal representation of $L$, followed by two copies of the decimal representation of $R-1$. For instance, for the number $n=17,151,345,543$, we will have $L=17,151$, $R=345,543$ and $h(n)=17,151,345,542,345,542$.\nProve that, starting with an arbitrary integer $n \\geq 2$, iterated application of $h$ produces the integer $1$ after finitely many steps.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We identify integers $n \\geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from $0$ to $9$. We recursively define ten functions $f_{0}, \\ldots, f_{9}$ that map some strings into integers for $k=9,8, \\ldots, 1,0$. The function $f_{9}$ is only defined on strings $x$ (including the empty string $\\varepsilon$) that entirely consist of nines. If $x$ consists of $m$ nines, then $f_{9}(x)=m+1$, $m=0,1, \\ldots$. For $k \\leq 8$, the domain of $f_{k}(x)$ is the set of all strings consisting only of digits that are $\\geq k$. We write $x$ in the form $x_{0} k x_{1} k x_{2} k \\ldots x_{m-1} k x_{m}$ where the strings $x_{s}$ only consist of digits $\\geq k+1$. Note that some of these strings might equal the empty string $\\varepsilon$ and that $m=0$ is possible, i.e. the digit $k$ does not appear in $x$. Then we define\n$$\nf_{k}(x)=\\sum_{s=0}^{m} 4^{f_{k+1}\\left(x_{s}\\right)} .\n$$\nWe will use the following obvious fact:\n**Fact 1.** If $x$ does not contain digits smaller than $k$, then $f_{i}(x)=4^{f_{i+1}(x)}$ for all $i=0, \\ldots, k-1$. In particular, $f_{i}(\\varepsilon)=4^{9-i}$ for all $i=0,1, \\ldots, 9$.\nMoreover, by induction on $k=9,8, \\ldots, 0$ it follows easily:\n**Fact 2.** If the nonempty string $x$ does not contain digits smaller than $k$, then $f_{i}(x)>f_{i}(\\varepsilon)$ for all $i=0, \\ldots, k$.\nWe will show the essential fact:\n**Fact 3.** $f_{0}(n)>f_{0}(h(n))$.\nThen the empty string will necessarily be reached after a finite number of applications of $h$. But starting from a string without leading zeros, $\\varepsilon$ can only be reached via the strings $1 \\rightarrow 00 \\rightarrow 0 \\rightarrow \\varepsilon$. Hence also the number $1$ will appear after a finite number of applications of $h$.\n\n**Proof of Fact 3.** If the last digit $r$ of $n$ is $0$, then we write $n=x_{0} 0 \\ldots 0 x_{m-1} 0 \\varepsilon$ where the $x_{i}$ do not contain the digit $0$. Then $h(n)=x_{0} 0 \\ldots 0 x_{m-1}$ and $f_{0}(n)-f_{0}(h(n))=f_{0}(\\varepsilon)>0$.\nSo let the last digit $r$ of $n$ be at least $1$. Let $L=y k$ and $R=z r$ be the corresponding left and right parts where $y$ is some string, $k \\leq r-1$ and the string $z$ consists only of digits not less than $r$. Then $n=y k z r$ and $h(n)=y k z(r-1) z(r-1)$. Let $d(y)$ be the smallest digit of $y$. We consider two cases which do not exclude each other.\n\n**Case 1.** $d(y) \\geq k$.\nThen\n$$\nf_{k}(n)-f_{k}(h(n))=f_{k}(z r)-f_{k}(z(r-1) z(r-1)) .\n$$\nIn view of Fact 1 this difference is positive if and only if\n$$\nf_{r-1}(z r)-f_{r-1}(z(r-1) z(r-1))>0\n$$\nWe have, using Fact 2,\n$$\nf_{r-1}(z r)=4^{f_{r}(z r)}=4^{f_{r}(z)+4^{f_{r+1}(\\varepsilon)}} \\geq 4 \\cdot 4^{f_{r}(z)}>4^{f_{r}(z)}+4^{f_{r}(z)}+4^{f_{r}(\\varepsilon)}=f_{r-1}(z(r-1) z(r-1)) .\n$$\nHere we use the additional definition $f_{10}(\\varepsilon)=0$ if $r=9$. Consequently, $f_{k}(n)-f_{k}(h(n))>0$ and according to Fact 1, $f_{0}(n)-f_{0}(h(n))>0$.\n\n**Case 2.** $d(y) \\leq k$.\nWe prove by induction on $d(y)=k, k-1, \\ldots, 0$ that $f_{i}(n)-f_{i}(h(n))>0$ for all $i=0, \\ldots, d(y)$. By Fact 1, it suffices to do so for $i=d(y)$. The initialization $d(y)=k$ was already treated in Case 1. Let $t=d(y)<k$. Write $y$ in the form $u t v$ where $v$ does not contain digits $\\leq t$. Then, in view of the induction hypothesis,\n$$\nf_{t}(n)-f_{t}(h(n))=f_{t}(v k z r)-f_{t}(v k z(r-1) z(r-1))=4^{f_{t+1}(v k z r)}-4^{f_{t+1}(v k z(r-1) z(r-1))}>0 .\n$$\nThus the inequality $f_{d(y)}(n)-f_{d(y)}(h(n))>0$ is established and from Fact 1 it follows that $f_{0}(n)-f_{0}(h(n))>0$.\nWe identify integers $n \\geq 2$ with the digit-strings, briefly strings, of their decimal representation and extend the definition of $h$ to all non-empty strings with digits from $0$ to $9$. Moreover, let us define that the empty string, $\\varepsilon$, is being mapped to the empty string. In the following all functions map the set of strings into the set of strings. For two functions $f$ and $g$ let $g \\circ f$ be defined by $(g \\circ f)(x)=g(f(x))$ for all strings $x$ and let, for non-negative integers $n$, $f^{n}$ denote the $n$-fold application of $f$. For any string $x$ let $s(x)$ be the smallest digit of $x$, and for the empty string let $s(\\varepsilon)=\\infty$. We define nine functions $g_{1}, \\ldots, g_{9}$ as follows: Let $k \\in\\{1, \\ldots, 9\\}$ and let $x$ be a string. If $x=\\varepsilon$ then $g_{k}(x)=\\varepsilon$. Otherwise, write $x$ in the form $x=y z r$ where $y$ is either the empty string or ends with a digit smaller than $k$, $s(z) \\geq k$ and $r$ is the rightmost digit of $x$. Then $g_{k}(x)=z r$.\n\n**Lemma 1.** We have $g_{k} \\circ h=g_{k} \\circ h \\circ g_{k}$ for all $k=1, \\ldots, 9$.\n\n**Proof of Lemma 1.** Let $x=y z r$ be as in the definition of $g_{k}$. If $y=\\varepsilon$, then $g_{k}(x)=x$, whence\n$$\ng_{k}(h(x))=g_{k}\\left(h\\left(g_{k}(x)\\right)\\right.\n$$\nSo let $y \\neq \\varepsilon$.\n\nCase 1. $z$ contains a digit smaller than $r$.\nLet $z=u a v$ where $a<r$ and $s(v) \\geq r$. Then\n$$\nh(x)= \\begin{cases}yuav & \\text{if } r=0 \\\\ yuav(r-1) v(r-1) & \\text{if } r>0\\end{cases}\n$$\nand\n$$\nh\\left(g_{k}(x)\\right)=h(z r)=h(u a v r)= \\begin{cases}uav & \\text{if } r=0 \\\\ uav(r-1) v(r-1) & \\text{if } r>0\\end{cases}\n$$\nSince $y$ ends with a digit smaller than $k$, the equality is obviously true.\n\nCase 2. $z$ does not contain a digit smaller than $r$.\nLet $y=u v$ where $u$ is either the empty string or ends with a digit smaller than $r$ and $s(v) \\geq r$. We have\n$$\nh(x)= \\begin{cases}u v z & \\text{if } r=0 \\\\ u v z(r-1) v z(r-1) & \\text{if } r>0\\end{cases}\n$$\nand\n$$\nh\\left(g_{k}(x)\\right)=h(z r)= \\begin{cases}z & \\text{if } r=0 \\\\ z(r-1) z(r-1) & \\text{if } r>0\\end{cases}\n$$\nRecall that $y$ and hence $v$ ends with a digit smaller than $k$, but all digits of $v$ are at least $r$. Now if $r>k$, then $v=\\varepsilon$, whence the terminal digit of $u$ is smaller than $k$, which entails\n$$\ng_{k}(h(x))=z(r-1) z(r-1)=g_{k}\\left(h\\left(g_{k}(x)\\right)\\right)\n$$\nIf $r \\leq k$, then\n$$\ng_{k}(h(x))=z(r-1)=g_{k}\\left(h\\left(g_{k}(x)\\right)\\right)\n$$\nso that in both cases the equality is true. Thus Lemma 1 is proved.\n\n**Lemma 2.** Let $k \\in\\{1, \\ldots, 9\\}$, let $x$ be a non-empty string and let $n$ be a positive integer. If $h^{n}(x)=\\varepsilon$ then $\\left(g_{k} \\circ h\\right)^{n}(x)=\\varepsilon$.\n\n**Proof of Lemma 2.** We proceed by induction on $n$. If $n=1$ we have\n$$\n\\varepsilon=h(x)=g_{k}(h(x))=\\left(g_{k} \\circ h\\right)(x)\n$$\nNow consider the step from $n-1$ to $n$ where $n \\geq 2$. Let $h^{n}(x)=\\varepsilon$ and let $y=h(x)$. Then $h^{n-1}(y)=\\varepsilon$ and by the induction hypothesis $\\left(g_{k} \\circ h\\right)^{n-1}(y)=\\varepsilon$. In view of Lemma 1,\n$$\n\\begin{aligned}\n& \\varepsilon=\\left(g_{k} \\circ h\\right)^{n-2}\\left(\\left(g_{k} \\circ h\\right)(y)\\right)=\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}(h(y))\\right) \\\\\n& \\quad=\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}\\left(h\\left(g_{k}(y)\\right)\\right)=\\left(g_{k} \\circ h\\right)^{n-2}\\left(g_{k}\\left(h\\left(g_{k}(h(x))\\right)\\right)=\\left(g_{k} \\circ h\\right)^{n}(x) .\\right.\\right.\n\\end{aligned}\n$$\nThus the induction step is complete and Lemma 2 is proved.\n\nWe say that the non-empty string $x$ terminates if $h^{n}(x)=\\varepsilon$ for some non-negative integer $n$.\n\n**Lemma 3.** Let $x=y z r$ where $s(y) \\geq k, s(z) \\geq k, y$ ends with the digit $k$ and $z$ is possibly empty. If $y$ and $z r$ terminate then also $x$ terminates.\n\n**Proof of Lemma 3.** Suppose that $y$ and $z r$ terminate. We proceed by induction on $k$. Let $k=0$. Obviously, $h(y w)=y h(w)$ for any non-empty string $w$. Let $h^{n}(z r)=\\epsilon$. It follows easily by induction on $m$ that $h^{m}(y z r)=y h^{m}(z r)$ for $m=1, \\ldots, n$. Consequently, $h^{n}(y z r)=y$. Since $y$ terminates, also $x=y z r$ terminates.\n\nNow let the assertion be true for all nonnegative integers less than $k$ and let us prove it for $k$ where $k \\geq 1$. It turns out that it is sufficient to prove that $y g_{k}(h(z r))$ terminates. Indeed:\n\nCase 1. $r=0$.\nThen $h(y z r)=y z=y g_{k}(h(z r))$.\n\nCase 2. $0<r \\leq k$.\nWe have $h(z r)=z(r-1) z(r-1)$ and $g_{k}(h(z r))=z(r-1)$. Then $h(y z r)=y z(r-1) y z(r-1)=y g_{k}(h(z r)) y g_{k}(h(z r))$ and we may apply the induction hypothesis to see that if $y g_{k}(h(z r))$ terminates, then $h(y z r)$ terminates.\n\nCase 3. $r>k$.\nThen $h(y z r)=y h(z r)=y g_{k}(h(z r))$.\nNote that $y g_{k}(h(z r))$ has the form $y z^{\\prime} r^{\\prime}$ where $s\\left(z^{\\prime}\\right) \\geq k$. By the same arguments it is sufficient to prove that $y g_{k}\\left(h\\left(z^{\\prime} r^{\\prime}\\right)\\right)=y\\left(g_{k} \\circ h\\right)^{2}(z r)$ terminates and, by induction, that $y\\left(g_{k} \\circ h\\right)^{m}(z r)$ terminates for some positive integer $m$. In view of Lemma 2 there is some $m$ such that $(g_{k} \\circ h)^{m}(z r)=\\epsilon$, so $x=y z r$ terminates if $y$ terminates. Thus Lemma 3 is proved.\n\nNow assume that there is some string $x$ that does not terminate. We choose $x$ minimal. If $x \\geq 10$, we can write $x$ in the form $x=y z r$ of Lemma 3 and by this lemma $x$ terminates since $y$ and $z r$ are smaller than $x$. If $x \\leq 9$, then $h(x)=(x-1)(x-1)$ and $h(x)$ terminates again by Lemma 3 and the minimal choice of $x$.\nWe commence by introducing some terminology. Instead of integers, we will consider the set $S$ of all strings consisting of the digits $0,1, \\ldots, 9$, including the empty string $\\epsilon$. If $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ is a nonempty string, we let $\\rho(a)=a_{n}$ denote the terminal digit of $a$ and $\\lambda(a)$ be the string with the last digit removed. We also define $\\lambda(\\epsilon)=\\epsilon$ and denote the set of non-negative integers by $\\mathbb{N}_{0}$.\n\nNow let $k \\in\\{0,1,2, \\ldots, 9\\}$ denote any digit. We define a function $f_{k}: S \\longrightarrow S$ on the set of strings: First, if the terminal digit of $n$ belongs to $\\{0,1, \\ldots, k\\}$, then $f_{k}(n)$ is obtained from $n$ by deleting this terminal digit, i.e $f_{k}(n)=\\lambda(n)$. Secondly, if the terminal digit of $n$ belongs to $\\{k+1, \\ldots, 9\\}$, then $f_{k}(n)$ is obtained from $n$ by the process described in the problem. We also define $f_{k}(\\epsilon)=\\epsilon$. Note that up to the definition for integers $n \\leq 1$, the function $f_{0}$ coincides with the function $h$ in the problem, through interpreting integers as digit strings. The argument will be roughly as follows. We begin by introducing a straightforward generalization of our claim about $f_{0}$. Then it will be easy to see that $f_{9}$ has all these stronger properties, which means that is suffices to show for $k \\in\\{0,1, \\ldots, 8\\}$ that $f_{k}$ possesses these properties provided that $f_{k+1}$ does.\n\nWe continue to use $k$ to denote any digit. The operation $f_{k}$ is said to be separating, if the following holds: Whenever $a$ is an initial segment of $b$, there is some $N \\in \\mathbb{N}_{0}$ such that $f_{k}^{N}(b)=a$. The following two notions only apply to the case where $f_{k}$ is indeed separating, otherwise they remain undefined. For every $a \\in S$ we denote the least $N \\in \\mathbb{N}_{0}$ for which $f_{k}^{N}(a)=\\epsilon$ occurs by $g_{k}(a)$ (because $\\epsilon$ is an initial segment of $a$, such an $N$ exists if $f_{k}$ is separating). If for every two strings $a$ and $b$ such that $a$ is a terminal segment of $b$ one has $g_{k}(a) \\leq g_{k}(b)$, we say that $f_{k}$ is coherent. In case that $f_{k}$ is separating and coherent we call the digit $k$ seductive.\n\nAs $f_{9}(a)=\\lambda(a)$ for all $a$, it is obvious that $9$ is seductive. Hence in order to show that $0$ is seductive, which clearly implies the statement of the problem, it suffices to take any $k \\in\\{0,1, \\ldots, 8\\}$ such that $k+1$ is seductive and to prove that $k$ has to be seductive as well. Note that in doing so, we have the function $g_{k+1}$ at our disposal. We have to establish two things and we begin with\n\n**Step 1.** $f_{k}$ is separating.\n\nBefore embarking on the proof of this, we record a useful observation which is easily proved by induction on $M$.\n\n**Claim 1.** For any strings $A, B$ and any positive integer $M$ such that $f_{k}^{M-1}(B) \\neq \\epsilon$, we have\n$$\nf_{k}^{M}(A k B)=A k f_{k}^{M}(B) .\n$$\n\nNow we call a pair $(a, b)$ of strings wicked provided that $a$ is an initial segment of $b$, but there is no $N \\in \\mathbb{N}_{0}$ such that $f_{k}^{N}(b)=a$. We need to show that there are none, so assume that there were such pairs. Choose a wicked pair $(a, b)$ for which $g_{k+1}(b)$ attains its minimal possible value. Obviously $b \\neq \\epsilon$ for any wicked pair $(a, b)$. Let $z$ denote the terminal digit of $b$. Observe that $a \\neq b$, which means that $a$ is also an initial segment of $\\lambda(b)$. To facilitate the construction of the eventual contradiction, we prove\n\n**Claim 2.** There cannot be an $N \\in \\mathbb{N}_{0}$ such that\n$$\nf_{k}^{N}(b)=\\lambda(b)\n$$\n\n**Proof of Claim 2.** For suppose that such an $N$ existed. Because $g_{k+1}(\\lambda(b))<g_{k+1}(b)$ in view of the coherency of $f_{k+1}$, the pair $(a, \\lambda(b))$ is not wicked. But then there is some $N'$ for which $f_{k}^{N'}(\\lambda(b))=a$ which entails $f_{k}^{N+N'}(b)=a$, contradiction. Hence Claim 2 is proved.\n\nIt follows that $z \\leq k$ is impossible, for otherwise $N=1$ violated Claim 2.\nAlso $z>k+1$ is impossible: Set $B=f_{k}(b)$. Then also $f_{k+1}(b)=B$, but $g_{k+1}(B)<g_{k+1}(b)$ and $a$ is an initial segment of $B$. Thus the pair $(a, B)$ is not wicked. Hence there is some $N \\in \\mathbb{N}_{0}$ with $a=f_{k}^{N}(B)$, which, however, entails $a=f_{k}^{N+1}(b)$.\n\nWe are left with the case $z=k+1$. Let $L$ denote the left part and $R=R^{*}(k+1)$ the right part of $b$. Then we have symbolically\n$$\nf_{k}(b)=L R^{*} k R^{*} k, \\quad f_{k}^{2}(b)=L R^{*} k R^{*} \\quad \\text{and} \\quad f_{k+1}(b)=L R^{*} .\n$$\nUsing that $R^{*}$ is a terminal segment of $L R^{*}$ and the coherency of $f_{k+1}$, we infer\n$$\ng_{k+1}\\left(R^{*}\\right) \\leq g_{k+1}\\left(L R^{*}\\right)<g_{k+1}(b)\n$$\nHence the pair $(\\epsilon, R^{*})$ is not wicked, so there is some minimal $M \\in \\mathbb{N}_{0}$ with $f_{k}^{M}\\left(R^{*}\\right)=\\epsilon$ and by Claim 1 it follows that $f_{k}^{2+M}(b)=L R^{*} k$. Finally, we infer that $\\lambda(b)=L R^{*}=f_{k}\\left(L R^{*} k\\right)= f_{k}^{3+M}(b)$, which yields a contradiction to Claim 2.\n\nThis final contradiction establishes that $f_{k}$ is indeed separating.\n\n**Step 2.** $f_{k}$ is coherent.\n\nTo prepare the proof of this, we introduce some further pieces of terminology. A nonempty string $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ is called a hypostasis, if $a_{n}<a_{i}$ for all $i=1, \\ldots, n-1$. Reading an arbitrary string $a$ backwards, we easily find a, possibly empty, sequence $(A_{1}, A_{2}, \\ldots, A_{m})$ of hypostases such that $\\rho(A_{1}) \\leq \\rho(A_{2}) \\leq \\cdots \\leq \\rho(A_{m})$ and, symbolically, $a=A_{1} A_{2} \\ldots A_{m}$. The latter sequence is referred to as the decomposition of $a$. So, for instance, $(20,0,9)$ is the decomposition of $2009$ and the string $50$ is a hypostasis. Next we explain when we say about two strings $a$ and $b$ that $a$ is injectible into $b$. The definition is by induction on the length of $b$. Let $(B_{1}, B_{2}, \\ldots, B_{n})$ be the decomposition of $b$ into hypostases. Then $a$ is injectible into $b$ if for the decomposition $(A_{1}, A_{2}, \\ldots, A_{m})$ of $a$ there is a strictly increasing function $H:\\{1,2, \\ldots, m\\} \\longrightarrow\\{1,2, \\ldots, n\\}$ satisfying\n$$\n\\rho(A_{i})=\\rho(B_{H(i)}) \\text{ for all } i=1, \\ldots, m ;\n$$\n$\\lambda(A_{i})$ is injectible into $\\lambda(B_{H(i)})$ for all $i=1, \\ldots, m$.\nIf one can choose $H$ with $H(m)=n$, then we say that $a$ is strongly injectible into $b$. Obviously, if $a$ is a terminal segment of $b$, then $a$ is strongly injectible into $b$.\n\n**Claim 3.** If $a$ and $b$ are two nonempty strings such that $a$ is strongly injectible into $b$, then $\\lambda(a)$ is injectible into $\\lambda(b)$.\n\n**Proof of Claim 3.** Let $(B_{1}, B_{2}, \\ldots, B_{n})$ be the decomposition of $b$ and let $(A_{1}, A_{2}, \\ldots, A_{m})$ be the decomposition of $a$. Take a function $H$ exemplifying that $a$ is strongly injectible into $b$. Let $(C_{1}, C_{2}, \\ldots, C_{r})$ be the decomposition of $\\lambda(A_{m})$ and let $(D_{1}, D_{2}, \\ldots, D_{s})$ be the decomposition of $\\lambda(B_{n})$. Choose a strictly increasing $H':\\{1,2, \\ldots, r\\} \\longrightarrow\\{1,2, \\ldots s\\}$ witnessing that $\\lambda(A_{m})$ is injectible into $\\lambda(B_{n})$. Clearly, $(A_{1}, A_{2}, \\ldots, A_{m-1}, C_{1}, C_{2}, \\ldots, C_{r})$ is the decomposition of $\\lambda(a)$ and $(B_{1}, B_{2}, \\ldots, B_{n-1}, D_{1}, D_{2}, \\ldots, D_{s})$ is the decomposition of $\\lambda(b)$. Then the function $H'':\\{1,2, \\ldots, m+r-1\\} \\longrightarrow\\{1,2, \\ldots, n+s-1\\}$ given by $H''(i)=H(i)$ for $i=1,2, \\ldots, m-1$ and $H''(m-1+i)=n-1+H'(i)$ for $i=1,2, \\ldots, r$ exemplifies that $\\lambda(a)$ is injectible into $\\lambda(b)$, which finishes the proof of the claim.\n\nA pair $(a, b)$ of strings is called aggressive if $a$ is injectible into $b$ and nevertheless $g_{k}(a)>g_{k}(b)$. Observe that if $f_{k}$ was incoherent, which we shall assume from now on, then such pairs existed. Now among all aggressive pairs we choose one, say $(a, b)$, for which $g_{k}(b)$ attains its least possible value. Obviously $f_{k}(a)$ cannot be injectible into $f_{k}(b)$, for otherwise the pair $(f_{k}(a), f_{k}(b))$ was aggressive and contradicted our choice of $(a, b)$. Let $(A_{1}, A_{2}, \\ldots, A_{m})$ and $(B_{1}, B_{2}, \\ldots, B_{n})$ be the decompositions of $a$ and $b$ and take a function $H:\\{1,2, \\ldots, m\\} \\longrightarrow\\{1,2, \\ldots, n\\}$ exemplifying that $a$ is indeed injectible into $b$. If we had $H(m)<n$, then $a$ was also injectible into the number $b'$ whose decomposition is $(B_{1}, B_{2}, \\ldots, B_{n-1})$ and by separativity of $f_{k}$ we obtained $g_{k}(b')<g_{k}(b)$, whence the pair $(a, b')$ was also aggressive, contrary to the minimality condition imposed on $b$. Therefore $a$ is strongly injectible into $b$. In particular, $a$ and $b$ have a common terminal digit, say $z$. If we had $z \\leq k$, then $f_{k}(a)=\\lambda(a)$ and $f_{k}(b)=\\lambda(b)$, so that by Claim 3, $f_{k}(a)$ was injectible into $f_{k}(b)$, which is a contradiction. Hence, $z \\geq k+1$.\n\nNow let $r$ be the minimal element of $\\{1,2, \\ldots, m\\}$ for which $\\rho(A_{r})=z$. Then the maximal right part of $a$ consisting of digits $\\geq z$ is equal to $R_{a}$, the string whose decomposition is $(A_{r}, A_{r+1}, \\ldots, A_{m})$. Then $R_{a}-1$ is a hypostasis and $(A_{1}, \\ldots, A_{r-1}, R_{a}-1, R_{a}-1)$ is the decomposition of $f_{k}(a)$. Defining $s$ and $R_{b}$ in a similar fashion with respect to $b$, we see that $(B_{1}, \\ldots, B_{s-1}, R_{b}-1, R_{b}-1)$ is the decomposition of $f_{k}(b)$. The definition of injectibility then easily entails that $R_{a}$ is strongly injectible into $R_{b}$. It follows from Claim 3 that $\\lambda(R_{a})= \\lambda(R_{a}-1)$ is injectible into $\\lambda(R_{b})=\\lambda(R_{b}-1)$, whence the function $H':\\{1,2, \\ldots, r+1\\} \\longrightarrow \\{1,2, \\ldots, s+1\\}$, given by $H'(i)=H(i)$ for $i=1,2, \\ldots, r-1$, $H'(r)=s$ and $H'(r+1)=s+1$ exemplifies that $f_{k}(a)$ is injectible into $f_{k}(b)$, which yields a contradiction as before.\n\nThis shows that aggressive pairs cannot exist, whence $f_{k}$ is indeed coherent, which finishes the proof of the seductivity of $k$, whereby the problem is finally solved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31907,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. The incircle of $ABC$ touches the sides $AB$ and $AC$ at the points $Z$ and $Y$, respectively. Let $G$ be the point where the lines $BY$ and $CZ$ meet, and let $R$ and $S$ be points such that the two quadrilaterals $BCYR$ and $BCSZ$ are parallelograms.\nProve that $GR = GS$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $k$ the incircle and by $k_{a}$ the excircle opposite to $A$ of triangle $ABC$. Let $k$ and $k_{a}$ touch the side $BC$ at the points $X$ and $T$, respectively, let $k_{a}$ touch the lines $AB$ and $AC$ at the points $P$ and $Q$, respectively. We use several times the fact that opposing sides of a parallelogram are of equal length, that points of contact of the excircle and incircle to a side of a triangle lie symmetric with respect to the midpoint of this side and that segments on two tangents to a circle defined by the points of contact and their point of intersection have the same length. So we conclude\n$$\n\\begin{gathered}\nZP = ZB + BP = XB + BT = BX + CX = ZS \\text{ and } \\\\\nCQ = CT = BX = BZ = CS.\n\\end{gathered}\n$$\n![](attached_image_1.png)\nSo for each of the points $Z, C$, their distances to $S$ equal the length of a tangent segment from this point to $k_{a}$. It is well-known, that all points with this property lie on the line $ZC$, which is the radical axis of $S$ and $k_{a}$. Similar arguments yield that $BY$ is the radical axis of $R$ and $k_{a}$. So the point of intersection of $ZC$ and $BY$, which is $G$ by definition, is the radical center of $R, S$ and $k_{a}$, from which the claim $GR = GS$ follows immediately.\nDenote $x = AZ = AY$, $y = BZ = BX$, $z = CX = CY$, $p = ZG$, $q = GC$. Several lengthy calculations (Menelaos' theorem in triangle $AZC$, law of Cosines in triangles $ABC$ and $AZC$ and Stewart's theorem in triangle $ZCS$) give four equations for $p, q, \\cos \\alpha$ and $GS$ in terms of $x, y$, and $z$ that can be resolved for $GS$. The result is symmetric in $y$ and $z$, so $GR = GS$. More in detail this means:\nThe line $BY$ intersects the sides of triangle $AZC$, so Menelaos' theorem yields $\\frac{p}{q} \\cdot \\frac{z}{x} \\cdot \\frac{x+y}{y} = 1$, hence\n$$\n\\begin{equation*}\n\\frac{p}{q} = \\frac{x y}{y z + z x} . \\tag{1}\n\\end{equation*}\n$$\nSince we only want to show that the term for $GS$ is symmetric in $y$ and $z$, we abbreviate terms that are symmetric in $y$ and $z$ by capital letters, starting with $N = x y + y z + z x$. So (1) implies\n$$\n\\begin{equation*}\n\\frac{p}{p+q} = \\frac{x y}{x y + y z + z x} = \\frac{x y}{N} \\quad \\text{ and } \\quad \\frac{q}{p+q} = \\frac{y z + z x}{x y + y z + z x} = \\frac{y z + z x}{N} . \\tag{2}\n\\end{equation*}\n$$\nNow the law of Cosines in triangle $ABC$ yields\n$$\n\\cos \\alpha = \\frac{(x+y)^2 + (x+z)^2 - (y+z)^2}{2(x+y)(x+z)} = \\frac{2x^2 + 2x y + 2x z - 2y z}{2(x+y)(x+z)} = 1 - \\frac{2 y z}{(x+y)(x+z)} .\n$$\nWe use this result to apply the law of Cosines in triangle $AZC$:\n$$\n\\begin{align*}\n(p+q)^2 & = x^2 + (x+z)^2 - 2x(x+z) \\cos \\alpha \\\\\n& = x^2 + (x+z)^2 - 2x(x+z) \\cdot \\left(1 - \\frac{2 y z}{(x+y)(x+z)}\\right) \\\\\n& = z^2 + \\frac{4 x y z}{x+y} \\tag{3}\n\\end{align*}\n$$\nNow in triangle $ZCS$ the segment $GS$ is a cevian, so with Stewart's theorem we have $p y^2 + q(y+z)^2 = (p+q)\\left(GS^2 + p q\\right)$, hence\n$$\nGS^2 = \\frac{p}{p+q} \\cdot y^2 + \\frac{q}{p+q} \\cdot (y+z)^2 - \\frac{p}{p+q} \\cdot \\frac{q}{p+q} \\cdot (p+q)^2 .\n$$\nReplacing the $p$'s and $q$'s herein by (2) and (3) yields\n$$\n\\begin{aligned}\nGS^2 & = \\frac{x y}{N} y^2 + \\frac{y z + z x}{N}(y+z)^2 - \\frac{x y}{N} \\cdot \\frac{y z + z x}{N} \\cdot \\left(z^2 + \\frac{4 x y z}{x+y}\\right) \\\\\n& = \\frac{x y^3}{N} + \\underbrace{\\frac{y z (y+z)^2}{N}}_{M_1} + \\frac{z x (y+z)^2}{N} - \\frac{x y z^3 (x+y)}{N^2} - \\underbrace{\\frac{4 x^2 y^2 z^2}{N^2}}_{M_2} \\\\\n& = \\frac{x y^3 + z x (y+z)^2}{N} - \\frac{x y z^3 (x+y)}{N^2} + M_1 - M_2 \\\\\n& = \\underbrace{\\frac{x\\left(y^3 + y^2 z + y z^2 + z^3\\right)}{N}}_{M_3} + \\frac{x y z^2 N}{N^2} - \\frac{x y z^3 (x+y)}{N^2} + M_1 - M_2 \\\\\n& = \\frac{x^2 y^2 z^2 + x y^2 z^3 + x^2 y z^3 - x^2 y z^3 - x y^2 z^3}{N^2} + M_1 - M_2 + M_3 \\\\\n& = \\frac{x^2 y^2 z^2}{N^2} + M_1 - M_2 + M_3,\n\\end{aligned}\n$$\na term that is symmetric in $y$ and $z$, indeed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31908,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a polygon that is convex and symmetric to some point $O$. Prove that for some parallelogram $R$ satisfying $P \\subset R$ we have\n$$\n\\frac{|R|}{|P|} \\leq \\sqrt{2}\n$$\nwhere $|R|$ and $|P|$ denote the area of the sets $R$ and $P$, respectively.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will construct two parallelograms $R_{1}$ and $R_{3}$, each of them containing $P$, and prove that at least one of the inequalities $|R_{1}| \\leq \\sqrt{2}|P|$ and $|R_{3}| \\leq \\sqrt{2}|P|$ holds (see Figure 1).\nFirst we will construct a parallelogram $R_{1} \\supseteq P$ with the property that the midpoints of the sides of $R_{1}$ are points of the boundary of $P$.\nChoose two points $A$ and $B$ of $P$ such that the triangle $OAB$ has maximal area. Let $a$ be the line through $A$ parallel to $OB$ and $b$ the line through $B$ parallel to $OA$. Let $A'$, $B'$, $a'$ and $b'$ be the points or lines, that are symmetric to $A$, $B$, $a$ and $b$, respectively, with respect to $O$. Now let $R_{1}$ be the parallelogram defined by $a$, $b$, $a'$ and $b'$.\n![](attached_image_1.png)\nFigure 1\nObviously, $A$ and $B$ are located on the boundary of the polygon $P$, and $A$, $B$, $A'$ and $B'$ are midpoints of the sides of $R_{1}$. We note that $P \\subseteq R_{1}$. Otherwise, there would be a point $Z \\in P$ but $Z \\notin R_{1}$, i.e., one of the lines $a$, $b$, $a'$ or $b'$ were between $O$ and $Z$. If it is $a$, we have $|OZB| > |OAB|$, which is contradictory to the choice of $A$ and $B$. If it is one of the lines $b$, $a'$ or $b'$ almost identical arguments lead to a similar contradiction.\nLet $R_{2}$ be the parallelogram $ABA'B'$. Since $A$ and $B$ are points of $P$, segment $AB \\subset P$ and so $R_{2} \\subset R_{1}$. Since $A$, $B$, $A'$ and $B'$ are midpoints of the sides of $R_{1}$, an easy argument yields\n$$\n|R_{1}| = 2 \\cdot |R_{2}| . \\tag{1}\n$$\nLet $R_{3}$ be the smallest parallelogram enclosing $P$ defined by lines parallel to $AB$ and $BA'$. Obviously $R_{2} \\subset R_{3}$ and every side of $R_{3}$ contains at least one point of the boundary of $P$. Denote by $C$ the intersection point of $a$ and $b$, by $X$ the intersection point of $AB$ and $OC$, and by $X'$ the intersection point of $XC$ and the boundary of $R_{3}$. In a similar way denote by $D$ the intersection point of $b$ and $a'$, by $Y$ the intersection point of $A'B$ and $OD$, and by $Y'$ the intersection point of $YD$ and the boundary of $R_{3}$.\nNote that $OC = 2 \\cdot OX$ and $OD = 2 \\cdot OY$, so there exist real numbers $x$ and $y$ with $1 \\leq x, y \\leq 2$ and $OX' = x \\cdot OX$ and $OY' = y \\cdot OY$. Corresponding sides of $R_{3}$ and $R_{2}$ are parallel which yields\n$$\n|R_{3}| = x y \\cdot |R_{2}| . \\tag{2}\n$$\nThe side of $R_{3}$ containing $X'$ contains at least one point $X^*$ of $P$; due to the convexity of $P$ we have $AX^*B \\subset P$. Since this side of the parallelogram $R_{3}$ is parallel to $AB$ we have $|AX^*B| = |AX'B|$, so $|OAX'B|$ does not exceed the area of $P$ confined to the sector defined by the rays $OB$ and $OA$. In a similar way we conclude that $|OB'Y'A'|$ does not exceed the area of $P$ confined to the sector defined by the rays $OB$ and $OA'$. Putting things together we have $|OAX'B| = x \\cdot |OAB|$, $|OBDA'| = y \\cdot |OBA'|$. Since $|OAB| = |OBA'|$, we conclude that $|P| \\geq 2 \\cdot |AX'BY'A'| = 2 \\cdot (x \\cdot |OAB| + y \\cdot |OBA'|) = 4 \\cdot \\frac{x+y}{2} \\cdot |OAB| = \\frac{x+y}{2} \\cdot |R_{2}|$; this is in short\n$$\n\\frac{x+y}{2} \\cdot |R_{2}| \\leq |P| . \\tag{3}\n$$\nSince all numbers concerned are positive, we can combine (1)-(3). Using the arithmetic-geometric-mean inequality we obtain\n$$\n|R_{1}| \\cdot |R_{3}| = 2 \\cdot |R_{2}| \\cdot x y \\cdot |R_{2}| \\leq 2 \\cdot |R_{2}|^{2} \\left(\\frac{x+y}{2}\\right)^{2} \\leq 2 \\cdot |P|^{2} .\n$$\nThis implies immediately the desired result $|R_{1}| \\leq \\sqrt{2} \\cdot |P|$ or $|R_{3}| \\leq \\sqrt{2} \\cdot |P|$.\nWe construct the parallelograms $R_{1}$, $R_{2}$ and $R_{3}$ in the same way as in Solution 1 and will show that $\\frac{|R_{1}|}{|P|} \\leq \\sqrt{2}$ or $\\frac{|R_{3}|}{|P|} \\leq \\sqrt{2}$.\n![](attached_image_2.png)\nFigure 2\nRecall that affine one-to-one maps of the plane preserve the ratio of areas of subsets of the plane. On the other hand, every parallelogram can be transformed with an affine map onto a square. It follows that without loss of generality we may assume that $R_{1}$ is a square (see Figure 2).\nThen $R_{2}$, whose vertices are the midpoints of the sides of $R_{1}$, is a square too, and $R_{3}$, whose sides are parallel to the diagonals of $R_{1}$, is a rectangle.\nLet $a > 0$, $b \\geq 0$ and $c \\geq 0$ be the distances introduced in Figure 2. Then $|R_{1}| = 2a^{2}$ and $|R_{3}| = (a + 2b)(a + 2c)$.\nPoints $A$, $A'$, $B$ and $B'$ are in the convex polygon $P$. Hence the square $ABA'B'$ is a subset of $P$. Moreover, each of the sides of the rectangle $R_{3}$ contains a point of $P$, otherwise $R_{3}$ would not be minimal. It follows that\n$$\n|P| \\geq a^{2} + 2 \\cdot \\frac{ab}{2} + 2 \\cdot \\frac{ac}{2} = a(a + b + c)\n$$\nNow assume that both $\\frac{|R_{1}|}{|P|} > \\sqrt{2}$ and $\\frac{|R_{3}|}{|P|} > \\sqrt{2}$, then\n$$\n2a^{2} = |R_{1}| > \\sqrt{2} \\cdot |P| \\geq \\sqrt{2} \\cdot a(a + b + c)\n$$\nand\n$$\n(a + 2b)(a + 2c) = |R_{3}| > \\sqrt{2} \\cdot |P| \\geq \\sqrt{2} \\cdot a(a + b + c) .\n$$\nAll numbers concerned are positive, so after multiplying these inequalities we get\n$$\n2a^{2}(a + 2b)(a + 2c) > 2a^{2}(a + b + c)^{2}\n$$\nBut the arithmetic-geometric-mean inequality implies the contradictory result\n$$\n2a^{2}(a + 2b)(a + 2c) \\leq 2a^{2}\\left(\\frac{(a + 2b) + (a + 2c)}{2}\\right)^{2} = 2a^{2}(a + b + c)^{2} .\n$$\nHence $\\frac{|R_{1}|}{|P|} \\leq \\sqrt{2}$ or $\\frac{|R_{3}|}{|P|} \\leq \\sqrt{2}$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31909,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the sides $AD$ and $BC$ of the quadrilateral $ABCD$ (such that $AB$ is not parallel to $CD$) intersect at point $P$. Points $O_{1}$ and $O_{2}$ are the circumcenters and points $H_{1}$ and $H_{2}$ are the orthocenters of triangles $ABP$ and $DCP$, respectively. Denote the midpoints of segments $O_{1}H_{1}$ and $O_{2}H_{2}$ by $E_{1}$ and $E_{2}$, respectively. Prove that the perpendicular from $E_{1}$ on $CD$, the perpendicular from $E_{2}$ on $AB$ and the line $H_{1}H_{2}$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We keep triangle $ABP$ fixed and move the line $CD$ parallel to itself uniformly, i.e. linearly dependent on a single parameter $\\lambda$ (see Figure 1). Then the points $C$ and $D$ also move uniformly. Hence, the points $O_{2}, H_{2}$ and $E_{2}$ move uniformly, too. Therefore also the perpendicular from $E_{2}$ on $AB$ moves uniformly. Obviously, the points $O_{1}, H_{1}, E_{1}$ and the perpendicular from $E_{1}$ on $CD$ do not move at all. Hence, the intersection point $S$ of these two perpendiculars moves uniformly. Since $H_{1}$ does not move, while $H_{2}$ and $S$ move uniformly along parallel lines (both are perpendicular to $CD$), it is sufficient to prove their collinearity for two different positions of $CD$.\n\n![](attached_image_1.png)\nFigure 1\n\nLet $CD$ pass through either point $A$ or point $B$. Note that by hypothesis these two cases are different. We will consider the case $A \\in CD$, i.e. $A = D$. So we have to show that the perpendiculars from $E_{1}$ on $AC$ and from $E_{2}$ on $AB$ intersect on the altitude $AH$ of triangle $ABC$ (see Figure 2).\n\n![](attached_image_2.png)\nFigure 2\n\nTo this end, we consider the midpoints $A_{1}, B_{1}, C_{1}$ of $BC, CA, AB$, respectively. As $E_{1}$ is the center of FEUERBACH's circle (nine-point circle) of $\\triangle ABP$, we have $E_{1}C_{1} = E_{1}H$. Similarly, $E_{2}B_{1} = E_{2}H$. Note further that a point $X$ lies on the perpendicular from $E_{1}$ on $A_{1}C_{1}$ if and only if\n$$\nXC_{1}^{2} - XA_{1}^{2} = E_{1}C_{1}^{2} - E_{1}A_{1}^{2}.\n$$\nSimilarly, the perpendicular from $E_{2}$ on $A_{1}B_{1}$ is characterized by\n$$\nXA_{1}^{2} - XB_{1}^{2} = E_{2}A_{1}^{2} - E_{2}B_{1}^{2}\n$$\nThe line $H_{1}H_{2}$, which is perpendicular to $B_{1}C_{1}$ and contains $A$, is given by\n$$\nXB_{1}^{2} - XC_{1}^{2} = AB_{1}^{2} - AC_{1}^{2}.\n$$\nThe three lines are concurrent if and only if\n$$\n\\begin{aligned}\n0 & = XC_{1}^{2} - XA_{1}^{2} + XA_{1}^{2} - XB_{1}^{2} + XB_{1}^{2} - XC_{1}^{2} \\\\\n& = E_{1}C_{1}^{2} - E_{1}A_{1}^{2} + E_{2}A_{1}^{2} - E_{2}B_{1}^{2} + AB_{1}^{2} - AC_{1}^{2} \\\\\n& = -E_{1}A_{1}^{2} + E_{2}A_{1}^{2} + E_{1}H^{2} - E_{2}H^{2} + AB_{1}^{2} - AC_{1}^{2}\n\\end{aligned}\n$$\ni.e. it suffices to show that\n$$\nE_{1}A_{1}^{2} - E_{2}A_{1}^{2} - E_{1}H^{2} + E_{2}H^{2} = \\frac{AC^{2} - AB^{2}}{4}\n$$\nWe have\n$$\n\\frac{AC^{2} - AB^{2}}{4} = \\frac{HC^{2} - HB^{2}}{4} = \\frac{(HC + HB)(HC - HB)}{4} = \\frac{HA_{1} \\cdot BC}{2}.\n$$\nLet $F_{1}, F_{2}$ be the projections of $E_{1}, E_{2}$ on $BC$. Obviously, these are the midpoints of $HP_{1}$, $HP_{2}$, where $P_{1}, P_{2}$ are the midpoints of $PB$ and $PC$ respectively. Then\n$$\n\\begin{aligned}\n& E_{1}A_{1}^{2} - E_{2}A_{1}^{2} - E_{1}H^{2} + E_{2}H^{2} \\\\\n& = F_{1}A_{1}^{2} - F_{1}H^{2} - F_{2}A_{1}^{2} + F_{2}H^{2} \\\\\n& = \\left(F_{1}A_{1} - F_{1}H\\right)\\left(F_{1}A_{1} + F_{1}H\\right) - \\left(F_{2}A_{1} - F_{2}H\\right)\\left(F_{2}A_{1} + F_{2}H\\right) \\\\\n& = A_{1}H \\cdot \\left(A_{1}P_{1} - A_{1}P_{2}\\right) \\\\\n& = \\frac{A_{1}H \\cdot BC}{2} \\\\\n& = \\frac{AC^{2} - AB^{2}}{4},\n\\end{aligned}\n$$\nwhich proves the claim.\nLet the perpendicular from $E_{1}$ on $CD$ meet $PH_{1}$ at $X$, and the perpendicular from $E_{2}$ on $AB$ meet $PH_{2}$ at $Y$ (see Figure 3). Let $\\varphi$ be the intersection angle of $AB$ and $CD$. Denote by $M, N$ the midpoints of $PH_{1}, PH_{2}$ respectively.\n\n![](attached_image_3.png)\nFigure 3\n\nWe will prove now that triangles $E_{1}XM$ and $E_{2}YN$ have equal angles at $E_{1}, E_{2}$, and supplementary angles at $X, Y$.\nIn the following, angles are understood as oriented, and equalities of angles modulo $180^{\\circ}$.\nLet $\\alpha = \\angle H_{2}PD$, $\\psi = \\angle DPC$, $\\beta = \\angle CPH_{1}$. Then $\\alpha + \\psi + \\beta = \\varphi$, $\\angle E_{1}XH_{1} = \\angle H_{2}YE_{2} = \\varphi$, thus $\\angle MXE_{1} + \\angle NYE_{2} = 180^{\\circ}$.\nBy considering the Feuerbach circle of $\\triangle ABP$ whose center is $E_{1}$ and which goes through $M$, we have $\\angle E_{1}MH_{1} = \\psi + 2\\beta$. Analogous considerations with the Feuerbach circle of $\\triangle DCP$ yield $\\angle H_{2}NE_{2} = \\psi + 2\\alpha$. Hence indeed $\\angle XE_{1}M = \\varphi - (\\psi + 2\\beta) = (\\psi + 2\\alpha) - \\varphi = \\angle YE_{2}N$. It follows now that\n$$\n\\frac{XM}{ME_{1}} = \\frac{YN}{NE_{2}}.\n$$\nFurthermore, $ME_{1}$ is half the circumradius of $\\triangle ABP$, while $PH_{1}$ is the distance of $P$ to the orthocenter of that triangle, which is twice the circumradius times the cosine of $\\psi$. Together with analogous reasoning for $\\triangle DCP$ we have\n$$\n\\frac{ME_{1}}{PH_{1}} = \\frac{1}{4 \\cos \\psi} = \\frac{NE_{2}}{PH_{2}}.\n$$\nBy multiplication,\n$$\n\\frac{XM}{PH_{1}} = \\frac{YN}{PH_{2}},\n$$\nand therefore\n$$\n\\frac{PX}{XH_{1}} = \\frac{H_{2}Y}{YP}.\n$$\nLet $E_{1}X, E_{2}Y$ meet $H_{1}H_{2}$ in $R, S$ respectively.\nApplying the intercept theorem to the parallels $E_{1}X, PH_{2}$ and center $H_{1}$ gives\n$$\n\\frac{H_{2}R}{RH_{1}} = \\frac{PX}{XH_{1}},\n$$\nwhile with parallels $E_{2}Y, PH_{1}$ and center $H_{2}$ we obtain\n$$\n\\frac{H_{2}S}{SH_{1}} = \\frac{H_{2}Y}{YP}\n$$\nCombination of the last three equalities yields that $R$ and $S$ coincide.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31910,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with incenter $I$ and let $X$, $Y$ and $Z$ be the incenters of the triangles $BIC$, $CIA$ and $AIB$, respectively. Let the triangle $XYZ$ be equilateral. Prove that $ABC$ is equilateral too.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$AZ$, $AI$ and $AY$ divide $\\angle BAC$ into four equal angles; denote them by $\\alpha$. In the same way we have four equal angles $\\beta$ at $B$ and four equal angles $\\gamma$ at $C$. Obviously $\\alpha + \\beta + \\gamma = \\frac{180^\\circ}{4} = 45^\\circ$; and $0^\\circ < \\alpha, \\beta, \\gamma < 45^\\circ$.\n\n![](attached_image_1.png)\n\nEasy calculations in various triangles yield $\\angle BIC = 180^\\circ - 2\\beta - 2\\gamma = 180^\\circ - (90^\\circ - 2\\alpha) = 90^\\circ + 2\\alpha$, hence (for $X$ is the incenter of triangle $BCI$, so $IX$ bisects $\\angle BIC$) we have $\\angle XIC = \\angle BIX = \\frac{1}{2} \\angle BIC = 45^\\circ + \\alpha$ and with similar arguments $\\angle CIY = \\angle YIA = 45^\\circ + \\beta$ and $\\angle AIZ = \\angle ZIB = 45^\\circ + \\gamma$.\n\nFurthermore, we have $\\angle XIY = \\angle XIC + \\angle CIY = (45^\\circ + \\alpha) + (45^\\circ + \\beta) = 135^\\circ - \\gamma$, $\\angle YIZ = 135^\\circ - \\alpha$, and $\\angle ZIX = 135^\\circ - \\beta$.\n\nNow we calculate the lengths of $IX$, $IY$ and $IZ$ in terms of $\\alpha$, $\\beta$ and $\\gamma$. The perpendicular from $I$ on $CX$ has length $IX \\cdot \\sin \\angle CXI = IX \\cdot \\sin (90^\\circ + \\beta) = IX \\cdot \\cos \\beta$. But $CI$ bisects $\\angle YCX$, so the perpendicular from $I$ on $CY$ has the same length, and we conclude\n$$\nIX \\cdot \\cos \\beta = IY \\cdot \\cos \\alpha\n$$\nTo make calculations easier we choose a length unit that makes $IX = \\cos \\alpha$. Then $IY = \\cos \\beta$ and with similar arguments $IZ = \\cos \\gamma$.\n\nSince $XYZ$ is equilateral we have $ZX = ZY$. The law of cosines in triangles $XYI$, $YZI$ yields\n$$\n\\begin{aligned}\n& ZX^2 = ZY^2 \\\\\n\\Longrightarrow & IZ^2 + IX^2 - 2 \\cdot IZ \\cdot IX \\cdot \\cos \\angle ZIX = IZ^2 + IY^2 - 2 \\cdot IZ \\cdot IY \\cdot \\cos \\angle YIZ \\\\\n\\Longrightarrow & IX^2 - IY^2 = 2 \\cdot IZ \\cdot (IX \\cdot \\cos \\angle ZIX - IY \\cdot \\cos \\angle YIZ) \\\\\n\\Longrightarrow & \\underbrace{\\cos^2 \\alpha - \\cos^2 \\beta}_{\\text{L.H.S.}} = \\underbrace{2 \\cdot \\cos \\gamma \\cdot (\\cos \\alpha \\cdot \\cos (135^\\circ - \\beta) - \\cos \\beta \\cdot \\cos (135^\\circ - \\alpha))}_{\\text{R.H.S.}}.\n\\end{aligned}\n$$\nA transformation of the left-hand side (L.H.S.) yields\n$$\n\\begin{aligned}\n\\text{L.H.S.} & = \\cos^2 \\alpha \\cdot (\\sin^2 \\beta + \\cos^2 \\beta) - \\cos^2 \\beta \\cdot (\\sin^2 \\alpha + \\cos^2 \\alpha) \\\\\n& = \\cos^2 \\alpha \\cdot \\sin^2 \\beta - \\cos^2 \\beta \\cdot \\sin^2 \\alpha\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\n& = (\\cos \\alpha \\cdot \\sin \\beta + \\cos \\beta \\cdot \\sin \\alpha) \\cdot (\\cos \\alpha \\cdot \\sin \\beta - \\cos \\beta \\cdot \\sin \\alpha) \\\\\n& = \\sin (\\beta + \\alpha) \\cdot \\sin (\\beta - \\alpha) = \\sin (45^\\circ - \\gamma) \\cdot \\sin (\\beta - \\alpha)\n\\end{aligned}\n$$\nwhereas a transformation of the right-hand side (R.H.S.) leads to\n$$\n\\begin{aligned}\n\\text{R.H.S.} & = 2 \\cdot \\cos \\gamma \\cdot (\\cos \\alpha \\cdot (-\\cos (45^\\circ + \\beta)) - \\cos \\beta \\cdot (-\\cos (45^\\circ + \\alpha))) \\\\\n& = 2 \\cdot \\frac{\\sqrt{2}}{2} \\cdot \\cos \\gamma \\cdot (\\cos \\alpha \\cdot (\\sin \\beta - \\cos \\beta) + \\cos \\beta \\cdot (\\cos \\alpha - \\sin \\alpha)) \\\\\n& = \\sqrt{2} \\cdot \\cos \\gamma \\cdot (\\cos \\alpha \\cdot \\sin \\beta - \\cos \\beta \\cdot \\sin \\alpha) \\\\\n& = \\sqrt{2} \\cdot \\cos \\gamma \\cdot \\sin (\\beta - \\alpha)\n\\end{aligned}\n$$\nEquating L.H.S. and R.H.S. we obtain\n$$\n\\begin{aligned}\n& \\sin (45^\\circ - \\gamma) \\cdot \\sin (\\beta - \\alpha) = \\sqrt{2} \\cdot \\cos \\gamma \\cdot \\sin (\\beta - \\alpha) \\\\\n\\Longrightarrow & \\sin (\\beta - \\alpha) \\cdot (\\sqrt{2} \\cdot \\cos \\gamma - \\sin (45^\\circ - \\gamma)) = 0 \\\\\n\\Longrightarrow & \\alpha = \\beta \\text{ or } \\sqrt{2} \\cdot \\cos \\gamma = \\sin (45^\\circ - \\gamma).\n\\end{aligned}\n$$\nBut $\\gamma < 45^\\circ$; so $\\sqrt{2} \\cdot \\cos \\gamma > \\cos \\gamma > \\cos 45^\\circ = \\sin 45^\\circ > \\sin (45^\\circ - \\gamma)$. This leaves $\\alpha = \\beta$.\n\nWith similar reasoning we have $\\alpha = \\gamma$, which means triangle $ABC$ must be equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31911,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a circumscribed quadrilateral. Let $g$ be a line through $A$ which meets the segment $BC$ in $M$ and the line $CD$ in $N$. Denote by $I_{1}, I_{2}$, and $I_{3}$ the incenters of $\\triangle ABM$, $\\triangle MNC$, and $\\triangle NDA$, respectively. Show that the orthocenter of $\\triangle I_{1} I_{2} I_{3}$ lies on $g$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $k_{1}, k_{2}$ and $k_{3}$ be the incircles of triangles $ABM$, $MNC$, and $NDA$, respectively (see Figure 1). We shall show that the tangent $h$ from $C$ to $k_{1}$ which is different from $CB$ is also tangent to $k_{3}$.\n\n![](attached_image_1.png)\nFigure 1\n\nTo this end, let $X$ denote the point of intersection of $g$ and $h$. Then $ABCX$ and $ABCD$ are circumscribed quadrilaterals, whence\n$$\nCD - CX = (AB + CD) - (AB + CX) = (BC + AD) - (BC + AX) = AD - AX,\n$$\ni.e.\n$$\nAX + CD = CX + AD\n$$\nwhich in turn reveals that the quadrilateral $AXCD$ is also circumscribed. Thus $h$ touches indeed the circle $k_{3}$.\n\nMoreover, we find that $\\angle I_{3} C I_{1} = \\angle I_{3} CX + \\angle XCI_{1} = \\frac{1}{2}(\\angle DCX + \\angle XCB) = \\frac{1}{2} \\angle DCB = \\frac{1}{2}(180^{\\circ} - \\angle MCN) = 180^{\\circ} - \\angle MI_{2}N = \\angle I_{3} I_{2} I_{1}$, from which we conclude that $C, I_{1}, I_{2}, I_{3}$ are concyclic.\n\nLet now $L_{1}$ and $L_{3}$ be the reflection points of $C$ with respect to the lines $I_{2} I_{3}$ and $I_{1} I_{2}$ respectively. Since $I_{1} I_{2}$ is the angle bisector of $\\angle NMC$, it follows that $L_{3}$ lies on $g$. By analogous reasoning, $L_{1}$ lies on $g$.\n\nLet $H$ be the orthocenter of $\\triangle I_{1} I_{2} I_{3}$. We have $\\angle I_{2} L_{3} I_{1} = \\angle I_{1} C I_{2} = \\angle I_{1} I_{3} I_{2} = 180^{\\circ} - \\angle I_{1} H I_{2}$, which entails that the quadrilateral $I_{2} H I_{1} L_{3}$ is cyclic. Analogously, $I_{3} H L_{1} I_{2}$ is cyclic.\n\nThen, working with oriented angles modulo $180^{\\circ}$, we have\n$$\n\\angle L_{3} H I_{2} = \\angle L_{3} I_{1} I_{2} = \\angle I_{2} I_{1} C = \\angle I_{2} I_{3} C = \\angle L_{1} I_{3} I_{2} = \\angle L_{1} H I_{2},\n$$\nwhence $L_{1}, L_{3}$, and $H$ are collinear. By $L_{1} \\neq L_{3}$, the claim follows.\nWe start by proving that $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic.\n\n![](attached_image_2.png)\nFigure 2\n\nTo this end, notice first that $I_{2}, M, I_{1}$ are collinear, as are $N, I_{2}, I_{3}$ (see Figure 2). Denote by $\\alpha, \\beta, \\gamma, \\delta$ the internal angles of $ABCD$. By considerations in triangle $CMN$, it follows that $\\angle I_{3} I_{2} I_{1} = \\frac{\\gamma}{2}$. We will show that $\\angle I_{3} C I_{1} = \\frac{\\gamma}{2}$, too. Denote by $I$ the incenter of $ABCD$. Clearly, $I_{1} \\in BI$, $I_{3} \\in DI$, $\\angle I_{1} A I_{3} = \\frac{\\alpha}{2}$.\n\nUsing the abbreviation $[X, YZ]$ for the distance from point $X$ to the line $YZ$, we have because of $\\angle BAI_{1} = \\angle IAI_{3}$ and $\\angle I_{1} AI = \\angle I_{3} AD$ that\n$$\n\\frac{[I_{1}, AB]}{[I_{1}, AI]} = \\frac{[I_{3}, AI]}{[I_{3}, AD]}\n$$\nFurthermore, consideration of the angle sums in $AIB$, $BIC$, $CID$ and $DIA$ implies $\\angle AIB + \\angle CID = \\angle BIC + \\angle DIA = 180^{\\circ}$, from which we see\n$$\n\\frac{[I_{1}, AI]}{[I_{3}, CI]} = \\frac{I_{1}I}{I_{3}I} = \\frac{[I_{1}, CI]}{[I_{3}, AI]}\n$$\nBecause of $[I_{1}, AB] = [I_{1}, BC]$, $[I_{3}, AD] = [I_{3}, CD]$, multiplication yields\n$$\n\\frac{[I_{1}, BC]}{[I_{3}, CI]} = \\frac{[I_{1}, CI]}{[I_{3}, CD]}\n$$\nBy $\\angle DCI = \\angle ICB = \\gamma / 2$ it follows that $\\angle I_{1} CB = \\angle I_{3} CI$ which concludes the proof of the above statement.\n\nLet the perpendicular from $I_{1}$ on $I_{2} I_{3}$ intersect $g$ at $Z$. Then $\\angle MI_{1}Z = 90^{\\circ} - \\angle I_{3} I_{2} I_{1} = 90^{\\circ} - \\gamma / 2 = \\angle MCI_{2}$. Since we have also $\\angle ZMI_{1} = \\angle I_{2} MC$, triangles $MZI_{1}$ and $MI_{2}C$ are similar. From this one easily proves that also $MI_{2}Z$ and $MCI_{1}$ are similar. Because $C, I_{1}, I_{2}$, and $I_{3}$ are concyclic, $\\angle MZI_{2} = \\angle MI_{1}C = \\angle NI_{3}C$, thus $NI_{2}Z$ and $NCI_{3}$ are similar, hence $NCI_{2}$ and $NI_{3}Z$ are similar. We conclude $\\angle ZI_{3}I_{2} = \\angle I_{2}CN = 90^{\\circ} - \\gamma / 2$, hence $I_{1} I_{2} \\perp ZI_{3}$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31912,
    "subject": "Mathematics (Multi-modal)",
    "question": "A social club has $n$ members. They have the membership numbers $1, 2, \\ldots, n$, respectively. From time to time members send presents to other members, including items they have already received as presents from other members. In order to avoid the embarrassing situation that a member might receive a present that he or she has sent to other members, the club adds the following rule to its statutes at one of its annual general meetings:\n\n\"A member with membership number $a$ is permitted to send a present to a member with membership number $b$ if and only if $a(b-1)$ is a multiple of $n$.\"\n\nProve that, if each member follows this rule, none will receive a present from another member that he or she has already sent to other members.\n\nAlternative formulation: Let $G$ be a directed graph with $n$ vertices $v_{1}, v_{2}, \\ldots, v_{n}$, such that there is an edge going from $v_{a}$ to $v_{b}$ if and only if $a$ and $b$ are distinct and $a(b-1)$ is a multiple of $n$. Prove that this graph does not contain a directed cycle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. Suppose there is an edge from $v_{i}$ to $v_{j}$. Then $i(j-1) = ij - i = k n$ for some integer $k$, which implies $i = ij - k n$. If $\\operatorname{gcd}(i, n) = d$ and $\\operatorname{gcd}(j, n) = e$, then $e$ divides $ij - k n = i$ and thus $e$ also divides $d$. Hence, if there is an edge from $v_{i}$ to $v_{j}$, then $\\operatorname{gcd}(j, n) \\mid \\operatorname{gcd}(i, n)$.\nIf there is a cycle in $G$, say $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$, then we have\n$$\n\\operatorname{gcd}\\left(i_{1}, n\\right)\\left|\\operatorname{gcd}\\left(i_{r}, n\\right)\\right| \\operatorname{gcd}\\left(i_{r-1}, n\\right)|\\ldots| \\operatorname{gcd}\\left(i_{2}, n\\right) \\mid \\operatorname{gcd}\\left(i_{1}, n\\right)\n$$\nwhich implies that all these greatest common divisors must be equal, say be equal to $t$.\nNow we pick any of the $i_{k}$, without loss of generality let it be $i_{1}$. Then $i_{r}(i_{1}-1)$ is a multiple of $n$ and hence also (by dividing by $t$), $i_{1}-1$ is a multiple of $\\frac{n}{t}$. Since $i_{1}$ and $i_{1}-1$ are relatively prime, also $t$ and $\\frac{n}{t}$ are relatively prime. So, by the Chinese remainder theorem, the value of $i_{1}$ is uniquely determined modulo $n = t \\cdot \\frac{n}{t}$ by the value of $t$. But, as $i_{1}$ was chosen arbitrarily among the $i_{k}$, this implies that all the $i_{k}$ have to be equal, a contradiction.\n\n\nSolution 2. If $a, b, c$ are integers such that $ab - a$ and $bc - b$ are multiples of $n$, then also $ac - a = a(bc - b) + (ab - a) - (ab - a)c$ is a multiple of $n$. This implies that if there is an edge from $v_{a}$ to $v_{b}$ and an edge from $v_{b}$ to $v_{c}$, then there also must be an edge from $v_{a}$ to $v_{c}$. Therefore, if there are any cycles at all, the smallest cycle must have length 2. But suppose the vertices $v_{a}$ and $v_{b}$ form such a cycle, i.e., $ab - a$ and $ab - b$ are both multiples of $n$. Then $a - b$ is also a multiple of $n$, which can only happen if $a = b$, which is impossible.\n\n\nSolution 3. Suppose there was a cycle $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$. Then $i_{1}(i_{2}-1)$ is a multiple of $n$, i.e., $i_{1} \\equiv i_{1} i_{2} \\bmod n$. Continuing in this manner, we get $i_{1} \\equiv i_{1} i_{2} \\equiv i_{1} i_{2} i_{3} \\equiv i_{1} i_{2} i_{3} \\ldots i_{r} \\bmod n$. But the same holds for all $i_{k}$, i.e., $i_{k} \\equiv i_{1} i_{2} i_{3} \\ldots i_{r} \\bmod n$. Hence $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\bmod n$, which means $i_{1} = i_{2} = \\cdots = i_{r}$, a contradiction.\n\n\nSolution 4. Let $n = k$ be the smallest value of $n$ for which the corresponding graph has a cycle. We show that $k$ is a prime power.\nIf $k$ is not a prime power, it can be written as a product $k = d e$ of relatively prime integers greater than 1. Reducing all the numbers modulo $d$ yields a single vertex or a cycle in the corresponding graph on $d$ vertices, because if $a(b-1) \\equiv 0 \\bmod k$ then this equation also holds modulo $d$. But since the graph on $d$ vertices has no cycles, by the minimality of $k$, we must have that all the indices of the cycle are congruent modulo $d$. The same holds modulo $e$ and hence also modulo $k = d e$. But then all the indices are equal, which is a contradiction.\nThus $k$ must be a prime power $k = p^{m}$. There are no edges ending at $v_{k}$, so $v_{k}$ is not contained in any cycle. All edges not starting at $v_{k}$ end at a vertex belonging to a non-multiple of $p$, and all edges starting at a non-multiple of $p$ must end at $v_{1}$. But there is no edge starting at $v_{1}$. Hence there is no cycle.\n\n\nSolution 5. Suppose there was a cycle $v_{i_{1}} \\rightarrow v_{i_{2}} \\rightarrow \\cdots \\rightarrow v_{i_{r}} \\rightarrow v_{i_{1}}$. Let $q = p^{m}$ be a prime power dividing $n$. We claim that either $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\equiv 0 \\bmod q$ or $i_{1} \\equiv i_{2} \\equiv \\cdots \\equiv i_{r} \\equiv 1 \\bmod q$.\nSuppose that there is an $i_{s}$ not divisible by $q$. Then, as $i_{s}(i_{s+1}-1)$ is a multiple of $q$, $i_{s+1} \\equiv 1 \\bmod p$. Similarly, we conclude $i_{s+2} \\equiv 1 \\bmod p$ and so on. So none of the labels is divisible by $p$, but since $i_{s}(i_{s+1}-1)$ is a multiple of $q = p^{m}$ for all $s$, all $i_{s+1}$ are congruent to 1 modulo $q$. This proves the claim.\nNow, as all the labels are congruent modulo all the prime powers dividing $n$, they must all be equal by the Chinese remainder theorem. This is a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31913,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $N$ is called balanced, if $N=1$ or if $N$ can be written as a product of an even number of not necessarily distinct primes. Given positive integers $a$ and $b$, consider the polynomial $P$ defined by $P(x)=(x+a)(x+b)$.\n\na. Prove that there exist distinct positive integers $a$ and $b$ such that all the numbers $P(1), P(2), \\ldots, P(50)$ are balanced.\n\nb. Prove that if $P(n)$ is balanced for all positive integers $n$, then $a=b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define a function $f$ on the set of positive integers by $f(n)=0$ if $n$ is balanced and $f(n)=1$ otherwise. Clearly, $f(n m) \\equiv f(n)+f(m) \\bmod 2$ for all positive integers $n, m$.\n\na. Now for each positive integer $n$ consider the binary sequence $(f(n+1), f(n+2), \\ldots, f(n+50))$. As there are only $2^{50}$ different such sequences there are two different positive integers $a$ and $b$ such that\n$$\n(f(a+1), f(a+2), \\ldots, f(a+50))=(f(b+1), f(b+2), \\ldots, f(b+50)).\n$$\nBut this implies that for the polynomial $P(x)=(x+a)(x+b)$ all the numbers $P(1), P(2), \\ldots, P(50)$ are balanced, since for all $1 \\leq k \\leq 50$ we have $f(P(k)) \\equiv f(a+k)+f(b+k) \\equiv 2 f(a+k) \\equiv 0 \\bmod 2$.\n\nb. Now suppose $P(n)$ is balanced for all positive integers $n$ and $a<b$. Set $n=k(b-a)-a$ for sufficiently large $k$, such that $n$ is positive. Then $P(n)=k(k+1)(b-a)^2$, and this number can only be balanced, if $f(k)=f(k+1)$ holds. Thus, the sequence $f(k)$ must become constant for sufficiently large $k$. But this is not possible, as for every prime $p$ we have $f(p)=1$ and for every square $t^2$ we have $f\\left(t^2\\right)=0$.\nHence $a=b$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31914,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f$ be a non-constant function from the set of positive integers into the set of positive integers, such that $a-b$ divides $f(a)-f(b)$ for all distinct positive integers $a, b$. Prove that there exist infinitely many primes $p$ such that $p$ divides $f(c)$ for some positive integer $c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that there are only finitely many primes $p_{1}, p_{2}, \\ldots, p_{m}$ that divide some function value produced of $f$.\nThere are infinitely many positive integers $a$ such that $v_{p_{i}}(a)>v_{p_{i}}(f(1))$ for all $i=1,2, \\ldots, m$, e.g. $a=\\left(p_{1} p_{2} \\ldots p_{m}\\right)^{\\alpha}$ with $\\alpha$ sufficiently large. Pick any such $a$. The condition of the problem then yields $a \\mid (f(a+1)-f(1))$. Assume $f(a+1) \\neq f(1)$. Then we must have $v_{p_{i}}(f(a+1)) \\neq v_{p_{i}}(f(1))$ for at least one $i$. This yields $v_{p_{i}}(f(a+1)-f(1))=\\min \\left\\{v_{p_{i}}(f(a+1)), v_{p_{i}}(f(1))\\right\\} \\leq v_{p_{1}}(f(1))<v_{p_{i}}(a)$. But this contradicts the fact that $a \\mid (f(a+1)-f(1))$.\nHence we must have $f(a+1)=f(1)$ for all such $a$.\nNow, for any positive integer $b$ and all such $a$, we have $(a+1-b) \\mid (f(a+1)-f(b))$, i.e., $(a+1-b) \\mid (f(1)-f(b))$. Since this is true for infinitely many positive integers $a$ we must have $f(b)=f(1)$. Hence $f$ is a constant function, a contradiction. Therefore, our initial assumption was false and there are indeed infinitely many primes $p$ dividing $f(c)$ for some positive integer $c$.\nAssume that there are only finitely many primes $p_{1}, p_{2}, \\ldots, p_{m}$ that divide some function value of $f$. Since $f$ is not identically $1$, we must have $m \\geq 1$.\nThen there exist non-negative integers $\\alpha_{1}, \\ldots, \\alpha_{m}$ such that\n$$\nf(1)=p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{m}^{\\alpha_{m}}\n$$\nWe can pick a positive integer $r$ such that $f(r) \\neq f(1)$. Let\n$$\nM=1+p_{1}^{\\alpha_{1}+1} p_{2}^{\\alpha_{2}+1} \\ldots p_{m}^{\\alpha_{m}+1} \\cdot (f(r)+r)\n$$\nThen for all $i \\in \\{1, \\ldots, m\\}$ we have that $p_{i}^{\\alpha_{i}+1}$ divides $M-1$ and hence by the condition of the problem also $f(M)-f(1)$. This implies that $f(M)$ is divisible by $p_{i}^{\\alpha_{i}}$ but not by $p_{i}^{\\alpha_{i}+1}$ for all $i$ and therefore $f(M)=f(1)$.\nHence\n$$\n\\begin{aligned}\nM-r & >p_{1}^{\\alpha_{1}+1} p_{2}^{\\alpha_{2}+1} \\ldots p_{m}^{\\alpha_{m}+1} \\cdot (f(r)+r)-r \\\\\n& \\geq p_{1}^{\\alpha_{1}+1} p_{2}^{\\alpha_{2}+1} \\ldots p_{m}^{\\alpha_{m}+1}+(f(r)+r)-r \\\\\n& >p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{m}^{\\alpha_{m}}+f(r) \\\\\n& \\geq |f(M)-f(r)|\n\\end{aligned}\n$$\nBut since $M-r$ divides $f(M)-f(r)$ this can only be true if $f(r)=f(M)=f(1)$, which contradicts the choice of $r$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31915,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ such that there exists a sequence of positive integers $a_{1}, a_{2}, \\ldots, a_{n}$ satisfying\n$$\na_{k+1}=\\frac{a_{k}^{2}+1}{a_{k-1}+1}-1\n$$\nfor every $k$ with $2 \\leq k \\leq n-1$.",
    "options": [],
    "answer": "n = 1, 2, 3, 4",
    "solution": "Such a sequence exists for $n=1,2,3,4$ and no other $n$. Since the existence of such a sequence for some $n$ implies the existence of such a sequence for all smaller $n$, it suffices to prove that $n=5$ is not possible and $n=4$ is possible.\n\nAssume first that for $n=5$ there exists a sequence of positive integers $a_{1}, a_{2}, \\ldots, a_{5}$ satisfying the conditions\n$$\n\\begin{aligned}\n& a_{2}^{2}+1=\\left(a_{1}+1\\right)\\left(a_{3}+1\\right), \\\\\n& a_{3}^{2}+1=\\left(a_{2}+1\\right)\\left(a_{4}+1\\right), \\\\\n& a_{4}^{2}+1=\\left(a_{3}+1\\right)\\left(a_{5}+1\\right) .\n\\end{aligned}\n$$\nAssume $a_{1}$ is odd, then $a_{2}$ has to be odd as well and as then $a_{2}^{2}+1 \\equiv 2 \\bmod 4, a_{3}$ has to be even. But this is a contradiction, since then the even number $a_{2}+1$ cannot divide the odd number $a_{3}^{2}+1$.\n\nHence $a_{1}$ is even.\nIf $a_{2}$ is odd, $a_{3}^{2}+1$ is even (as a multiple of $a_{2}+1$ ) and hence $a_{3}$ is odd, too. Similarly we must have $a_{4}$ odd as well. But then $a_{3}^{2}+1$ is a product of two even numbers $\\left(a_{2}+1\\right)\\left(a_{4}+1\\right)$ and thus is divisible by 4 , which is a contradiction as for odd $a_{3}$ we have $a_{3}^{2}+1 \\equiv 2 \\bmod 4$.\n\nHence $a_{2}$ is even. Furthermore $a_{3}+1$ divides the odd number $a_{2}^{2}+1$ and so $a_{3}$ is even. Similarly, $a_{4}$ and $a_{5}$ are even as well.\n\nNow set $x=a_{2}$ and $y=a_{3}$. From the given condition we get $(x+1) \\mid\\left(y^{2}+1\\right)$ and $(y+1) \\mid\\left(x^{2}+1\\right)$. We will prove that there is no pair of positive even numbers $(x, y)$ satisfying these two conditions, thus yielding a contradiction to the assumption.\n\nAssume there exists a pair ( $x_{0}, y_{0}$ ) of positive even numbers satisfying the two conditions $\\left(x_{0}+1\\right) \\mid\\left(y_{0}^{2}+1\\right)$ and $\\left(y_{0}+1\\right) \\mid\\left(x_{0}^{2}+1\\right)$.\nThen one has $\\left(x_{0}+1\\right) \\mid\\left(y_{0}^{2}+1+x_{0}^{2}-1\\right)$, i.e., $\\left(x_{0}+1\\right) \\mid\\left(x_{0}^{2}+y_{0}^{2}\\right)$, and similarly $\\left(y_{0}+1\\right) \\mid\\left(x_{0}^{2}+y_{0}^{2}\\right)$. Any common divisor $d$ of $x_{0}+1$ and $y_{0}+1$ must hence also divide the number $\\left(x_{0}^{2}+1\\right)+\\left(y_{0}^{2}+1\\right)-\\left(x_{0}^{2}+y_{0}^{2}\\right)=2$. But as $x_{0}+1$ and $y_{0}+1$ are both odd, we must have $d=1$. Thus $x_{0}+1$ and $y_{0}+1$ are relatively prime and therefore there exists a positive integer $k$ such that\n$$\nk(x+1)(y+1)=x^{2}+y^{2}\n$$\nhas the solution $\\left(x_{0}, y_{0}\\right)$. We will show that the latter equation has no solution $(x, y)$ in positive even numbers.\n\nAssume there is a solution. Pick the solution $\\left(x_{1}, y_{1}\\right)$ with the smallest sum $x_{1}+y_{1}$ and assume $x_{1} \\geq y_{1}$. Then $x_{1}$ is a solution to the quadratic equation\n$$\nx^{2}-k\\left(y_{1}+1\\right) x+y_{1}^{2}-k\\left(y_{1}+1\\right)=0\n$$\nLet $x_{2}$ be the second solution, which by Vieta's theorem fulfills $x_{1}+x_{2}=k\\left(y_{1}+1\\right)$ and $x_{1} x_{2}=y_{1}^{2}-k\\left(y_{1}+1\\right)$. If $x_{2}=0$, the second equation implies $y_{1}^{2}=k\\left(y_{1}+1\\right)$, which is impossible, as $y_{1}+1>1$ cannot divide the relatively prime number $y_{1}^{2}$. Therefore $x_{2} \\neq 0$.\n\nAlso we get $\\left(x_{1}+1\\right)\\left(x_{2}+1\\right)=x_{1} x_{2}+x_{1}+x_{2}+1=y_{1}^{2}+1$ which is odd, and hence $x_{2}$ must be even and positive. Also we have $x_{2}+1=\\frac{y_{1}^{2}+1}{x_{1}+1} \\leq \\frac{y_{1}^{2}+1}{y_{1}+1} \\leq y_{1} \\leq x_{1}$. But this means that the pair ( $x^{\\prime}, y^{\\prime}$ ) with $x^{\\prime}=y_{1}$ and $y^{\\prime}=x_{2}$ is another solution of $k(x+1)(y+1)=x^{2}+y^{2}$ in even positive numbers with $x^{\\prime}+y^{\\prime}<x_{1}+y_{1}$, a contradiction.\n\nTherefore we must have $n \\leq 4$.\nWhen $n=4$, a possible example of a sequence is $a_{1}=4, a_{2}=33, a_{3}=217$ and $a_{4}=1384$.\nIt is easy to check that for $n=4$ the sequence $a_{1}=4, a_{2}=33, a_{3}=217$ and $a_{4}=1384$ is possible.\n\nNow assume there is a sequence with $n \\geq 5$. Then we have in particular\n$$\n\\begin{aligned}\n& a_{2}^{2}+1=\\left(a_{1}+1\\right)\\left(a_{3}+1\\right), \\\\\n& a_{3}^{2}+1=\\left(a_{2}+1\\right)\\left(a_{4}+1\\right), \\\\\n& a_{4}^{2}+1=\\left(a_{3}+1\\right)\\left(a_{5}+1\\right) .\n\\end{aligned}\n$$\nAlso assume without loss of generality that among all such quintuples $\\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}\\right)$ we have chosen one with minimal $a_{1}$.\n\nOne shows quickly the following fact:\n$$\n\\begin{align*}\n& \\text{ If three positive integers } x, y, z \\text{ fulfill } y^{2}+1=(x+1)(z+1) \\text{ and if } y \\text{ is even, then } \\\\\n& x \\text{ and } z \\text{ are even as well and either } x<y<z \\text{ or } z<y<x \\text{ holds. } \\tag{1}\n\\end{align*}\n$$\nIndeed, the first part is obvious and from $x<y$ we conclude\n$$\nz+1=\\frac{y^{2}+1}{x+1} \\geq \\frac{y^{2}+1}{y}>y\n$$\nand similarly in the other case.\n\nNow, if $a_{3}$ was odd, then $\\left(a_{2}+1\\right)\\left(a_{4}+1\\right)=a_{3}^{2}+1 \\equiv 2 \\bmod 4$ would imply that one of $a_{2}$ or $a_{4}$ is even, this contradicts (1). Thus $a_{3}$ and hence also $a_{1}, a_{2}, a_{4}$ and $a_{5}$ are even. According to (1), one has $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}$ or $a_{1}>a_{2}>a_{3}>a_{4}>a_{5}$ but due to the minimality of $a_{1}$ the first series of inequalities must hold.\n\nConsider the identity\n$$\n\\left(a_{3}+1\\right)\\left(a_{1}+a_{3}\\right)=a_{3}^{2}-1+\\left(a_{1}+1\\right)\\left(a_{3}+1\\right)=a_{2}^{2}+a_{3}^{2}=a_{2}^{2}-1+\\left(a_{2}+1\\right)\\left(a_{4}+1\\right)=\\left(a_{2}+1\\right)\\left(a_{2}+a_{4}\\right) .\n$$\nAny common divisor of the two odd numbers $a_{2}+1$ and $a_{3}+1$ must also divide ( $a_{2}+1$ )( $a_{4}+ 1)-\\left(a_{3}+1\\right)\\left(a_{3}-1\\right)=2$, so these numbers are relatively prime. Hence the last identity shows that $a_{1}+a_{3}$ must be a multiple of $a_{2}+1$, i.e. there is an integer $k$ such that\n$$\n\\begin{equation*}\na_{1}+a_{3}=k\\left(a_{2}+1\\right) . \\tag{2}\n\\end{equation*}\n$$\nNow set $a_{0}=k\\left(a_{1}+1\\right)-a_{2}$. This is an integer and we have\n$$\n\\begin{aligned}\n\\left(a_{0}+1\\right)\\left(a_{2}+1\\right) & =k\\left(a_{1}+1\\right)\\left(a_{2}+1\\right)-\\left(a_{2}-1\\right)\\left(a_{2}+1\\right) \\\\\n& =\\left(a_{1}+1\\right)\\left(a_{1}+a_{3}\\right)-\\left(a_{1}+1\\right)\\left(a_{3}+1\\right)+2 \\\\\n& =\\left(a_{1}+1\\right)\\left(a_{1}-1\\right)+2=a_{1}^{2}+1\n\\end{aligned}\n$$\nThus $a_{0} \\geq 0$. If $a_{0}>0$, then by (1) we would have $a_{0}<a_{1}<a_{2}$ and then the quintuple ( $a_{0}, a_{1}, a_{2}, a_{3}, a_{4}$ ) would contradict the minimality of $a_{1}$.\n\nHence $a_{0}=0$, implying $a_{2}=a_{1}^{2}$. But also $a_{2}=k\\left(a_{1}+1\\right)$, which finally contradicts the fact that $a_{1}+1>1$ is relatively prime to $a_{1}^{2}$ and thus cannot be a divisior of this number.\n\nHence $n \\geq 5$ is not possible.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31916,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be distinct integers greater than $1$. Prove that there exists a positive integer $n$ such that $(a^{n}-1)(b^{n}-1)$ is not a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "At first we notice that\n$$\n\\begin{align*}\n(1-\\alpha)^{\\frac{1}{2}}(1-\\beta)^{\\frac{1}{2}} & =\\left(1-\\frac{1}{2} \\cdot \\alpha-\\frac{1}{8} \\cdot \\alpha^{2}-\\cdots\\right)\\left(1-\\frac{1}{2} \\cdot \\beta-\\frac{1}{8} \\cdot \\beta^{2}-\\cdots\\right) \\\\\n& =\\sum_{k, \\ell \\geq 0} c_{k, \\ell} \\cdot \\alpha^{k} \\beta^{\\ell} \\quad \\text{ for all } \\alpha, \\beta \\in(0,1)\n\\end{align*}\n$$\nwhere $c_{0,0}=1$ and $c_{k, \\ell}$ are certain coefficients.\nFor an indirect proof, we suppose that $x_{n}=\\sqrt{(a^{n}-1)(b^{n}-1)} \\in \\mathbb{Z}$ for all positive integers $n$. Replacing $a$ by $a^{2}$ and $b$ by $b^{2}$ if necessary, we may assume that $a$ and $b$ are perfect squares, hence $\\sqrt{a b}$ is an integer.\nAt first we shall assume that $a^{\\mu} \\neq b^{\\nu}$ for all positive integers $\\mu, \\nu$. We have\n$$\n\\begin{equation*}\nx_{n}=(\\sqrt{a b})^{n}\\left(1-\\frac{1}{a^{n}}\\right)^{\\frac{1}{2}}\\left(1-\\frac{1}{b^{n}}\\right)^{\\frac{1}{2}}=\\sum_{k, \\ell \\geq 0} c_{k, \\ell}\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n} .\n\\end{equation*}\n$$\nChoosing $k_{0}$ and $\\ell_{0}$ such that $a^{k_{0}}>\\sqrt{a b}, b^{\\ell_{0}}>\\sqrt{a b}$, we define the polynomial\n$$\nP(x)=\\prod_{k=0, \\ell=0}^{k_{0}-1, \\ell_{0}-1}\\left(a^{k} b^{\\ell} x-\\sqrt{a b}\\right)=: \\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x^{i}\n$$\nwith integer coefficients $d_{i}$. By our assumption, the zeros\n$$\n\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}, \\quad k=0, \\ldots, k_{0}-1, \\quad \\ell=0, \\ldots, \\ell_{0}-1,\n$$\nof $P$ are pairwise distinct.\nFurthermore, we consider the integer sequence\n$$\n\\begin{equation*}\ny_{n}=\\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x_{n+i}, \\quad n=1,2, \\ldots\n\\end{equation*}\n$$\nBy the theory of linear recursions, we obtain\n$$\n\\begin{equation*}\ny_{n}=\\sum_{\\substack{k, \\ell \\geq 0 \\\\ k \\geq k_{0} \\text{ or } \\ell \\geq \\ell_{0}}} e_{k, \\ell}\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}, \\quad n=1,2, \\ldots,\n\\end{equation*}\n$$\nwith real numbers $e_{k, \\ell}$. We have\n$$\n|y_{n}| \\leq \\sum_{\\substack{k, \\ell \\geq 0 \\\\ k \\geq k_{0} \\text{ or } \\ell \\geq \\ell_{0}}}|e_{k, \\ell}|\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}=: M_{n} .\n$$\nBecause the series in (4) is obtained by a finite linear combination of the absolutely convergent series (1), we conclude that in particular $M_{1}<\\infty$. Since\n$$\n\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}} \\leq \\lambda:=\\max \\left\\{\\frac{\\sqrt{a b}}{a^{k_{0}}}, \\frac{\\sqrt{a b}}{b^{\\ell_{0}}}\\right\\} \\quad \\text{ for all } k, \\ell \\geq 0 \\text{ such that } k \\geq k_{0} \\text{ or } \\ell \\geq \\ell_{0},\n$$\nwe get the estimates $M_{n+1} \\leq \\lambda M_{n}, n=1,2, \\ldots$. Our choice of $k_{0}$ and $\\ell_{0}$ ensures $\\lambda<1$, which implies $M_{n} \\rightarrow 0$ and consequently $y_{n} \\rightarrow 0$ as $n \\rightarrow \\infty$. It follows that $y_{n}=0$ for all sufficiently large $n$.\nSo, equation (3) reduces to $\\sum_{i=0}^{k_{0} \\cdot \\ell_{0}} d_{i} x_{n+i}=0$.\nUsing the theory of linear recursions again, for sufficiently large $n$ we have\n$$\nx_{n}=\\sum_{k=0, \\ell=0}^{k_{0}-1, \\ell_{0}-1} f_{k, \\ell}\\left(\\frac{\\sqrt{a b}}{a^{k} b^{\\ell}}\\right)^{n}\n$$\nfor certain real numbers $f_{k, \\ell}$.\nComparing with (2), we see that $f_{k, \\ell}=c_{k, \\ell}$ for all $k, \\ell \\geq 0$ with $k<k_{0}, \\ell<\\ell_{0}$, and $c_{k, \\ell}=0$ if $k \\geq k_{0}$ or $\\ell \\geq \\ell_{0}$, since we assumed that $a^{\\mu} \\neq b^{\\nu}$ for all positive integers $\\mu, \\nu$.\nIn view of (1), this means\n$$\n\\begin{equation*}\n(1-\\alpha)^{\\frac{1}{2}}(1-\\beta)^{\\frac{1}{2}}=\\sum_{k=0, \\ell=0}^{k_{0}-1, \\ell_{0}-1} c_{k, \\ell} \\cdot \\alpha^{k} \\beta^{\\ell}\n\\end{equation*}\n$$\nfor all real numbers $\\alpha, \\beta \\in(0,1)$. We choose $k^{*}<k_{0}$ maximal such that there is some $i$ with $c_{k^{*}, i} \\neq 0$. Squaring (5) and comparing coefficients of $\\alpha^{2 k^{*}} \\beta^{2 i^{*}}$, where $i^{*}$ is maximal with $c_{k^{*}, i^{*}} \\neq 0$, we see that $k^{*}=0$. This means that the right hand side of (5) is independent of $\\alpha$, which is clearly impossible.\nWe are left with the case that $a^{\\mu}=b^{\\nu}$ for some positive integers $\\mu$ and $\\nu$. We may assume that $\\mu$ and $\\nu$ are relatively prime. Then there is some positive integer $c$ such that $a=c^{\\nu}$ and $b=c^{\\mu}$. Now starting with the expansion (2), i.e.,\n$$\nx_{n}=\\sum_{j \\geq 0} g_{j}\\left(\\frac{\\sqrt{c^{\\mu+\\nu}}}{c^{j}}\\right)^{n}\n$$\nfor certain coefficients $g_{j}$, and repeating the arguments above, we see that $g_{j}=0$ for sufficiently large $j$, say $j>j_{0}$. But this means that\n$$\n\\left(1-x^{\\mu}\\right)^{\\frac{1}{2}}\\left(1-x^{\\nu}\\right)^{\\frac{1}{2}}=\\sum_{j=0}^{j_{0}} g_{j} x^{j}\n$$\nfor all real numbers $x \\in(0,1)$. Squaring, we see that\n$$\n\\left(1-x^{\\mu}\\right)\\left(1-x^{\\nu}\\right)\n$$\nis the square of a polynomial in $x$. In particular, all its zeros are of order at least $2$, which implies $\\mu=\\nu$ by looking at roots of unity. So we obtain $\\mu=\\nu=1$, i.e., $a=b$, a contradiction.\nWe set $a^{2}=A, b^{2}=B$, and $z_{n}=\\sqrt{(A^{n}-1)(B^{n}-1)}$. Let us assume that $z_{n}$ is an integer for $n=1,2, \\ldots$ Without loss of generality, we may suppose that $b<a$. We determine an integer $k \\geq 2$ such that $b^{k-1} \\leq a<b^{k}$, and define a sequence $\\gamma_{1}, \\gamma_{2}, \\ldots$ of rational numbers such that\n$$\n2 \\gamma_{1}=1 \\quad \\text{ and } \\quad 2 \\gamma_{n+1}=\\sum_{i=1}^{n} \\gamma_{i} \\gamma_{n-i} \\text{ for } n=1,2, \\ldots\n$$\nIt could easily be shown that $\\gamma_{n}=\\frac{1 \\cdot 1 \\cdot 3 \\ldots(2 n-3)}{2 \\cdot 4 \\cdot 6 \\ldots 2 n}$, for instance by reading Vandermonde's convolution as an equation between polynomials, but we shall have no use for this fact.\nUsing Landau's $O$-Notation in the usual way, we have\n$$\n\\begin{aligned}\n& \\left\\{(a b)^{n}-\\gamma_{1}\\left(\\frac{a}{b}\\right)^{n}-\\gamma_{2}\\left(\\frac{a}{b^{3}}\\right)^{n}-\\cdots-\\gamma_{k}\\left(\\frac{a}{b^{2 k-1}}\\right)^{n}+O\\left(\\frac{b}{a}\\right)^{n}\\right\\}^{2} \\\\\n& =A^{n} B^{n}-2 \\gamma_{1} A^{n}-\\sum_{i=2}^{k}\\left(2 \\gamma_{i}-\\sum_{j=1}^{i-1} \\gamma_{j} \\gamma_{i-j}\\right)\\left(\\frac{A}{B^{i-1}}\\right)^{n}+O\\left(\\frac{A}{B^{k}}\\right)^{n}+O\\left(B^{n}\\right) \\\\\n& =A^{n} B^{n}-A^{n}+O\\left(B^{n}\\right)\n\\end{aligned}\n$$\nwhence\n$$\nz_{n}=(a b)^{n}-\\gamma_{1}\\left(\\frac{a}{b}\\right)^{n}-\\gamma_{2}\\left(\\frac{a}{b^{3}}\\right)^{n}-\\cdots-\\gamma_{k}\\left(\\frac{a}{b^{2 k-1}}\\right)^{n}+O\\left(\\frac{b}{a}\\right)^{n} .\n$$\nNow choose rational numbers $r_{1}, r_{2}, \\ldots, r_{k+1}$ such that\n$$\n(x-a b) \\cdot\\left(x-\\frac{a}{b}\\right) \\ldots\\left(x-\\frac{a}{b^{2 k-1}}\\right)=x^{k+1}-r_{1} x^{k}+\\cdots \\pm r_{k+1},\n$$\nand then a natural number $M$ for which $M r_{1}, M r_{2}, \\ldots M r_{k+1}$ are integers. For known reasons,\n$$\nM\\left(z_{n+k+1}-r_{1} z_{n+k}+\\cdots \\pm r_{k+1} z_{n}\\right)=O\\left(\\frac{b}{a}\\right)^{n}\n$$\nfor all $n \\in \\mathbb{N}$ and thus there is a natural number $N$ which is so large, that\n$$\nz_{n+k+1}=r_{1} z_{n+k}-r_{2} z_{n+k-1}+\\cdots \\mp r_{k+1} z_{n}\n$$\nholds for all $n \\geqslant N$. Now the theory of linear recursions reveals that there are some rational numbers $\\delta_{0}, \\delta_{1}, \\delta_{2}, \\ldots, \\delta_{k}$ such that\n$$\nz_{n}=\\delta_{0}(a b)^{n}-\\delta_{1}\\left(\\frac{a}{b}\\right)^{n}-\\delta_{2}\\left(\\frac{a}{b^{3}}\\right)^{n}-\\cdots-\\delta_{k}\\left(\\frac{a}{b^{2 k-1}}\\right)^{n}\n$$\nfor sufficiently large $n$, where $\\delta_{0}>0$ as $z_{n}>0$. As before, one obtains\n$$\n\\begin{aligned}\n& A^{n} B^{n}-A^{n}-B^{n}+1=z_{n}^{2} \\\\\n& =\\left\\{\\delta_{0}(a b)^{n}-\\delta_{1}\\left(\\frac{a}{b}\\right)^{n}-\\delta_{2}\\left(\\frac{a}{b^{3}}\\right)^{n}-\\cdots-\\delta_{k}\\left(\\frac{a}{b^{2 k-1}}\\right)^{n}\\right\\}^{2} \\\\\n& =\\delta_{0}^{2} A^{n} B^{n}-2 \\delta_{0} \\delta_{1} A^{n}-\\sum_{i=2}^{i=k}\\left(2 \\delta_{0} \\delta_{i}-\\sum_{j=1}^{j=i-1} \\delta_{j} \\delta_{i-j}\\right)\\left(\\frac{A}{B^{i-1}}\\right)^{n}+O\\left(\\frac{A}{B^{k}}\\right)^{n} .\n\\end{aligned}\n$$\nEasy asymptotic calculations yield $\\delta_{0}=1, \\delta_{1}=\\frac{1}{2}, \\delta_{i}=\\frac{1}{2} \\sum_{j=1}^{j=i-1} \\delta_{j} \\delta_{i-j}$ for $i=2,3, \\ldots, k-2$, and then $a=b^{k-1}$. It follows that $k>2$ and there is some $P \\in \\mathbb{Q}[X]$ for which $(X-1)(X^{k-1}-1)= P(X)^{2}$. But this cannot occur, for instance as $X^{k-1}-1$ has no double zeros. Thus our\nassumption that $z_{n}$ was an integer for $n=1,2, \\ldots$ turned out to be wrong, which solves the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31917,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let the real numbers $a, b, c, d$ satisfy the relations $a+b+c+d=6$ and $a^{2}+b^{2}+c^{2}+d^{2}=12$. Prove that\n$$\n36 \\leq 4\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right)-\\left(a^{4}+b^{4}+c^{4}+d^{4}\\right) \\leq 48\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that\n$$\n\\begin{gathered}\n4\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right)-\\left(a^{4}+b^{4}+c^{4}+d^{4}\\right)=-\\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\\right) \\\\\n+6\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-4(a+b+c+d)+4 \\\\\n=-\\left((a-1)^{4}+(b-1)^{4}+(c-1)^{4}+(d-1)^{4}\\right)+52\n\\end{gathered}\n$$\nNow, introducing $x=a-1, y=b-1, z=c-1, t=d-1$, we need to prove the inequalities\n$$\n16 \\geq x^{4}+y^{4}+z^{4}+t^{4} \\geq 4\n$$\nunder the constraint\n$$\n\\begin{equation*}\nx^{2}+y^{2}+z^{2}+t^{2}=\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)-2(a+b+c+d)+4=4 \\tag{1}\n\\end{equation*}\n$$\n(we will not use the value of $x+y+z+t$ though it can be found).\nNow the rightmost inequality in (1) follows from the power mean inequality:\n$$\nx^{4}+y^{4}+z^{4}+t^{4} \\geq \\frac{\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}}{4}=4\n$$\nFor the other one, expanding the brackets we note that\n$$\n\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}=\\left(x^{4}+y^{4}+z^{4}+t^{4}\\right)+q\n$$\nwhere $q$ is a nonnegative number, so\n$$\nx^{4}+y^{4}+z^{4}+t^{4} \\leq\\left(x^{2}+y^{2}+z^{2}+t^{2}\\right)^{2}=16\n$$\nand we are done.\nFirst, we claim that $0 \\leq a, b, c, d \\leq 3$. Actually, we have\n$$\na+b+c=6-d, \\quad a^{2}+b^{2}+c^{2}=12-d^{2}\n$$\nhence the power mean inequality\n$$\na^{2}+b^{2}+c^{2} \\geq \\frac{(a+b+c)^{2}}{3}\n$$\nrewrites as\n$$\n12-d^{2} \\geq \\frac{(6-d)^{2}}{3} \\quad \\Longleftrightarrow \\quad 2 d(d-3) \\leq 0\n$$\nwhich implies the desired inequalities for $d$; since the conditions are symmetric, we also have the same estimate for the other variables.\nNow, to prove the rightmost inequality, we use the obvious inequality $x^{2}(x-2)^{2} \\geq 0$ for each real $x$; this inequality rewrites as $4 x^{3}-x^{4} \\leq 4 x^{2}$. It follows that\n$$\n\\left(4 a^{3}-a^{4}\\right)+\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\leq 4\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=48\n$$\nas desired.\n\nNow we prove the leftmost inequality in an analogous way. For each $x \\in[0,3]$, we have $(x+1)(x-1)^{2}(x-3) \\leq 0$ which is equivalent to $4 x^{3}-x^{4} \\geq 2 x^{2}+4 x-3$. This implies that $\\left(4 a^{3}-a^{4}\\right)+\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\geq 2\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)+4(a+b+c+d)-12=36$, as desired.\nFirst, expanding $48=4\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)$ and applying the AM-GM inequality, we have\n$$\n\\begin{aligned}\na^{4}+b^{4}+c^{4}+d^{4}+48 & =\\left(a^{4}+4 a^{2}\\right)+\\left(b^{4}+4 b^{2}\\right)+\\left(c^{4}+4 c^{2}\\right)+\\left(d^{4}+4 d^{2}\\right) \\\\\n& \\geq 2\\left(\\sqrt{a^{4} \\cdot 4 a^{2}}+\\sqrt{b^{4} \\cdot 4 b^{2}}+\\sqrt{c^{4} \\cdot 4 c^{2}}+\\sqrt{d^{4} \\cdot 4 d^{2}}\\right) \\\\\n& =4\\left(\\left|a^{3}\\right|+\\left|b^{3}\\right|+\\left|c^{3}\\right|+\\left|d^{3}\\right|\\right) \\geq 4\\left(a^{3}+b^{3}+c^{3}+d^{3}\\right)\n\\end{aligned}\n$$\nwhich establishes the rightmost inequality.\nTo prove the leftmost inequality, we first show that $a, b, c, d \\in[0,3]$ as in the previous solution. Moreover, we can assume that $0 \\leq a \\leq b \\leq c \\leq d$. Then we have $a+b \\leq b+c \\leq \\frac{2}{3}(b+c+d) \\leq \\frac{2}{3} \\cdot 6=4$.\nNext, we show that $4 b-b^{2} \\leq 4 c-c^{2}$. Actually, this inequality rewrites as $(c-b)(b+c-4) \\leq 0$, which follows from the previous estimate. The inequality $4 a-a^{2} \\leq 4 b-b^{2}$ can be proved analogously.\nFurther, the inequalities $a \\leq b \\leq c$ together with $4 a-a^{2} \\leq 4 b-b^{2} \\leq 4 c-c^{2}$ allow us to apply the Chebyshev inequality obtaining\n$$\n\\begin{aligned}\na^{2}\\left(4 a-a^{2}\\right)+b^{2}\\left(4 b-b^{2}\\right)+c^{2}\\left(4 c-c^{2}\\right) & \\geq \\frac{1}{3}\\left(a^{2}+b^{2}+c^{2}\\right)\\left(4(a+b+c)-\\left(a^{2}+b^{2}+c^{2}\\right)\\right) \\\\\n& =\\frac{\\left(12-d^{2}\\right)\\left(4(6-d)-\\left(12-d^{2}\\right)\\right)}{3} .\n\\end{aligned}\n$$\nThis implies that\n$$\n\\begin{align*}\n\\left(4 a^{3}-a^{4}\\right) & +\\left(4 b^{3}-b^{4}\\right)+\\left(4 c^{3}-c^{4}\\right)+\\left(4 d^{3}-d^{4}\\right) \\geq \\frac{\\left(12-d^{2}\\right)\\left(d^{2}-4 d+12\\right)}{3}+4 d^{3}-d^{4} \\\\\n& =\\frac{144-48 d+16 d^{3}-4 d^{4}}{3}=36+\\frac{4}{3}(3-d)(d-1)\\left(d^{2}-3\\right) \\tag{2}\n\\end{align*}\n$$\nFinally, we have $d^{2} \\geq \\frac{1}{4}\\left(a^{2}+b^{2}+c^{2}+d^{2}\\right)=3$ (which implies $d>1$ ); so, the expression $\\frac{4}{3}(3-d)(d-1)\\left(d^{2}-3\\right)$ in the right-hand part of (2) is nonnegative, and the desired inequality is proved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31918,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence $x_{1}, x_{2}, \\ldots$ is defined by $x_{1}=1$ and $x_{2k}=-x_{k}$, $x_{2k-1}=(-1)^{k+1} x_{k}$ for all $k \\geq 1$. Prove that $x_{1}+x_{2}+\\cdots+x_{n} \\geq 0$ for all $n \\geq 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start with some observations. First, from the definition of $x_{i}$ it follows that for each positive integer $k$ we have\n$$\nx_{4k-3}=x_{2k-1}=-x_{4k-2} \\quad \\text{ and } \\quad x_{4k-1}=x_{4k}=-x_{2k}=x_{k} .\n$$\nHence, denoting $S_{n}=\\sum_{i=1}^{n} x_{i}$, we have\n$$\nS_{4k}=\\sum_{i=1}^{k}\\left(\\left(x_{4k-3}+x_{4k-2}\\right)+\\left(x_{4k-1}+x_{4k}\\right)\\right)=\\sum_{i=1}^{k}\\left(0+2x_{k}\\right)=2S_{k}\n$$\n$$\nS_{4k+2}=S_{4k}+\\left(x_{4k+1}+x_{4k+2}\\right)=S_{4k}\n$$\nObserve also that $S_{n}=\\sum_{i=1}^{n} x_{i} \\equiv \\sum_{i=1}^{n} 1=n \\pmod{2}$.\nNow we prove by induction on $k$ that $S_{i} \\geq 0$ for all $i \\leq 4k$. The base case is valid since $x_{1}=x_{3}=x_{4}=1$, $x_{2}=-1$. For the induction step, assume that $S_{i} \\geq 0$ for all $i \\leq 4k$. Using the relations above, we obtain\n$$\nS_{4k+4}=2S_{k+1} \\geq 0, \\quad S_{4k+2}=S_{4k} \\geq 0, \\quad S_{4k+3}=S_{4k+2}+x_{4k+3}=\\frac{S_{4k+2}+S_{4k+4}}{2} \\geq 0\n$$\nSo, we are left to prove that $S_{4k+1} \\geq 0$. If $k$ is odd, then $S_{4k}=2S_{k} \\geq 0$; since $k$ is odd, $S_{k}$ is odd as well, so we have $S_{4k} \\geq 2$ and hence $S_{4k+1}=S_{4k}+x_{4k+1} \\geq 1$.\nConversely, if $k$ is even, then we have $x_{4k+1}=x_{2k+1}=x_{k+1}$, hence $S_{4k+1}=S_{4k}+x_{4k+1}=2S_{k}+x_{k+1}=S_{k}+S_{k+1} \\geq 0$. The step is proved.\nWe will use the notation of $S_{n}$ and the relations above from the previous solution.\nAssume the contrary and consider the minimal $n$ such that $S_{n+1}<0$; surely $n \\geq 1$, and from $S_{n} \\geq 0$ we get $S_{n}=0$, $x_{n+1}=-1$. Hence, we are especially interested in the set $M=\\{n: S_{n}=0\\}$; our aim is to prove that $x_{n+1}=1$ whenever $n \\in M$ thus coming to a contradiction.\nFor this purpose, we first describe the set $M$ inductively. We claim that (i) $M$ consists only of even numbers, (ii) $2 \\in M$, and (iii) for every even $n \\geq 4$ we have $n \\in M \\Longleftrightarrow [n/4] \\in M$. Actually, (i) holds since $S_{n} \\equiv n \\pmod{2}$, (ii) is straightforward, while (iii) follows from the relations $S_{4k+2}=S_{4k}=2S_{k}$.\nNow, we are left to prove that $x_{n+1}=1$ if $n \\in M$. We use the induction on $n$. The base case is $n=2$, that is, the minimal element of $M$; here we have $x_{3}=1$, as desired.\nFor the induction step, consider some $4 \\leq n \\in M$ and let $m=[n/4] \\in M$; then $m$ is even, and $x_{m+1}=1$ by the induction hypothesis. We prove that $x_{n+1}=x_{m+1}=1$. If $n=4m$ then we have $x_{n+1}=x_{2m+1}=x_{m+1}$ since $m$ is even; otherwise, $n=4m+2$, and $x_{n+1}=-x_{2m+2}=x_{m+1}$, as desired. The proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31919,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $\\mathbb{Q}^{+}$ the set of all positive rational numbers. Determine all functions $f: \\mathbb{Q}^{+} \\rightarrow \\mathbb{Q}^{+}$ which satisfy the following equation for all $x, y \\in \\mathbb{Q}^{+}$:\n$$\nf\\left(f(x)^{2} y\\right)=x^{3} f(x y)\n$$",
    "options": [],
    "answer": "f(x) = 1/x for all positive rational x",
    "solution": "By substituting $y=1$, we get\n$$\nf\\left(f(x)^{2}\\right)=x^{3} f(x)\n$$\nThen, whenever $f(x)=f(y)$, we have\n$$\nx^{3}=\\frac{f\\left(f(x)^{2}\\right)}{f(x)}=\\frac{f\\left(f(y)^{2}\\right)}{f(y)}=y^{3}\n$$\nwhich implies $x=y$, so the function $f$ is injective.\n\nNow replace $x$ by $x y$ in the previous equation, and apply the original equation twice, the second time to $(y, f(x)^{2})$ instead of $(x, y)$:\n$$\nf\\left(f(x y)^{2}\\right)=(x y)^{3} f(x y)=y^{3} f\\left(f(x)^{2} y\\right)=f\\left(f(x)^{2} f(y)^{2}\\right)\n$$\nSince $f$ is injective, we get\n$$\n\\begin{aligned}\nf(x y)^{2} & =f(x)^{2} f(y)^{2} \\\\\nf(x y) & =f(x) f(y)\n\\end{aligned}\n$$\nTherefore, $f$ is multiplicative. This also implies $f(1)=1$ and $f\\left(x^{n}\\right)=f(x)^{n}$ for all integers $n$.\n\nThen the functional equation can be re-written as\n$$\n\\begin{align*}\nf(f(x))^{2} f(y) & =x^{3} f(x) f(y), \\\\\nf(f(x)) & =\\sqrt{x^{3} f(x)}\n\\end{align*}\n$$\nLet $g(x)=x f(x)$. Then, we have\n$$\n\\begin{aligned}\ng(g(x)) & =g(x f(x))=x f(x) \\cdot f(x f(x))=x f(x)^{2} f(f(x))= \\\\\n& =x f(x)^{2} \\sqrt{x^{3} f(x)}=(x f(x))^{5 / 2}=(g(x))^{5 / 2}\n\\end{aligned}\n$$\nand, by induction,\n$$\n\\underbrace{g(g(\\ldots g}_{n+1}(x) \\ldots))=(g(x))^{(5 / 2)^{n}}\n$$\nfor every positive integer $n$.\n\nConsider this for a fixed $x$. The left-hand side is always rational, so $(g(x))^{(5 / 2)^{n}}$ must be rational for every $n$. We show that this is possible only if $g(x)=1$. Suppose that $g(x) \\neq 1$, and let the prime factorization of $g(x)$ be $g(x)=p_{1}^{\\alpha_{1}} \\ldots p_{k}^{\\alpha_{k}}$ where $p_{1}, \\ldots, p_{k}$ are distinct primes and $\\alpha_{1}, \\ldots, \\alpha_{k}$ are nonzero integers. Then the unique prime factorization is\n$$\n\\underbrace{g(g(\\ldots g}_{n+1}(x) \\ldots))=(g(x))^{(5 / 2)^{n}}=p_{1}^{(5 / 2)^{n} \\alpha_{1}} \\ldots p_{k}^{(5 / 2)^{n} \\alpha_{k}}\n$$\nwhere the exponents should be integers. But this is not true for large values of $n$, for example $\\left(\\frac{5}{2}\\right)^{n} \\alpha_{1}$ cannot be an integer when $2^{n} \\nmid \\alpha_{1}$. Therefore, $g(x) \\neq 1$ is impossible.\n\nHence, $g(x)=1$ and thus $f(x)=\\frac{1}{x}$ for all $x$.\n\nThe function $f(x)=\\frac{1}{x}$ satisfies the equation:\n$$\nf\\left(f(x)^{2} y\\right)=\\frac{1}{f(x)^{2} y}=\\frac{1}{\\left(\\frac{1}{x}\\right)^{2} y}=\\frac{x^{3}}{x y}=x^{3} f(x y)\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31920,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $f$ and $g$ are two functions defined on the set of positive integers and taking positive integer values. Suppose also that the equations $f(g(n))=f(n)+1$ and $g(f(n))= g(n)+1$ hold for all positive integers. Prove that $f(n)=g(n)$ for all positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Throughout the solution, by $\\mathbb{N}$ we denote the set of all positive integers. For any function $h: \\mathbb{N} \\rightarrow \\mathbb{N}$ and for any positive integer $k$, define $h^{k}(x)=\\underbrace{h(h(\\ldots h(x) \\ldots))}_{k}$ (in particular, $h^{0}(x)=x$).\n\nObserve that $f\\left(g^{k}(x)\\right)=f\\left(g^{k-1}(x)\\right)+1=\\cdots=f(x)+k$ for any positive integer $k$, and similarly $g\\left(f^{k}(x)\\right)=g(x)+k$. Now let $a$ and $b$ be the minimal values attained by $f$ and $g$, respectively; say $f\\left(n_{f}\\right)=a, g\\left(n_{g}\\right)=b$. Then we have $f\\left(g^{k}\\left(n_{f}\\right)\\right)=a+k, g\\left(f^{k}\\left(n_{g}\\right)\\right)=b+k$, so the function $f$ attains all values from the set $N_{f}=\\{a, a+1, \\ldots\\}$, while $g$ attains all the values from the set $N_{g}=\\{b, b+1, \\ldots\\}$.\n\nNext, note that $f(x)=f(y)$ implies $g(x)=g(f(x))-1=g(f(y))-1=g(y)$; surely, the converse implication also holds. Now, we say that $x$ and $y$ are similar (and write $x \\sim y$) if $f(x)=f(y)$ (equivalently, $g(x)=g(y)$). For every $x \\in \\mathbb{N}$, we define $[x]=\\{y \\in \\mathbb{N}: x \\sim y\\}$; surely, $y_{1} \\sim y_{2}$ for all $y_{1}, y_{2} \\in[x]$, so $[x]=[y]$ whenever $y \\in[x]$.\n\nNow we investigate the structure of the sets $[x]$.\n\nClaim 1. Suppose that $f(x) \\sim f(y)$; then $x \\sim y$, that is, $f(x)=f(y)$. Consequently, each class $[x]$ contains at most one element from $N_{f}$, as well as at most one element from $N_{g}$.\n\nProof. If $f(x) \\sim f(y)$, then we have $g(x)=g(f(x))-1=g(f(y))-1=g(y)$, so $x \\sim y$. The second statement follows now from the sets of values of $f$ and $g$.\n\nNext, we clarify which classes do not contain large elements.\n\nClaim 2. For any $x \\in \\mathbb{N}$, we have $[x] \\subseteq\\{1,2, \\ldots, b-1\\}$ if and only if $f(x)=a$. Analogously, $[x] \\subseteq\\{1,2, \\ldots, a-1\\}$ if and only if $g(x)=b$.\n\nProof. We will prove that $[x] \\subseteq \\{1,2, \\ldots, b-1\\} \\Longleftrightarrow f(x)>a$; the proof of the second statement is similar.\n\nNote that $f(x)>a$ implies that there exists some $y$ satisfying $f(y)=f(x)-1$, so $f(g(y))= f(y)+1=f(x)$, and hence $x \\sim g(y) \\geq b$. Conversely, if $b \\leq c \\sim x$ then $c=g(y)$ for some $y \\in \\mathbb{N}$, which in turn follows $f(x)=f(g(y))=f(y)+1 \\geq a+1$, and hence $f(x)>a$.\n\nClaim 2 implies that there exists exactly one class contained in $\\{1, \\ldots, a-1\\}$ (that is, the class $[n_{g}]$), as well as exactly one class contained in $\\{1, \\ldots, b-1\\}$ (the class $[n_{f}]$). Assume for a moment that $a \\leq b$; then $[n_{g}]$ is contained in $\\{1, \\ldots, b-1\\}$ as well, hence it coincides with $[n_{g}]$. So, we get that\n$$\n\\begin{equation*}\nf(x)=a \\Longleftrightarrow g(x)=b \\Longleftrightarrow x \\sim n_{f} \\sim n_{g} . \\tag{1}\n\\end{equation*}\n$$\n\nClaim 3. $a=b$.\n\nProof. By Claim 2, we have $[a] \\neq[n_{f}]$, so $[a]$ should contain some element $a' \\geq b$ by Claim 2 again. If $a \\neq a'$, then $[a]$ contains two elements $\\geq a$ which is impossible by Claim 1. Therefore, $a=a' \\geq b$. Similarly, $b \\geq a$.\n\nNow we are ready to prove the problem statement. First, we establish the following\n\nClaim 4. For every integer $d \\geq 0, f^{d+1}\\left(n_{f}\\right)=g^{d+1}\\left(n_{f}\\right)=a+d$.\n\nProof. Induction on $d$. For $d=0$, the statement follows from (1) and Claim 3. Next, for $d>1$ from the induction hypothesis we have $f^{d+1}\\left(n_{f}\\right)=f\\left(f^{d}\\left(n_{f}\\right)\\right)=f\\left(g^{d}\\left(n_{f}\\right)\\right)=f\\left(n_{f}\\right)+d=a+d$. The equality $g^{d+1}\\left(n_{f}\\right)=a+d$ is analogous.\n\nFinally, for each $x \\in \\mathbb{N}$, we have $f(x)=a+d$ for some $d \\geq 0$, so $f(x)=f\\left(g^{d}\\left(n_{f}\\right)\\right)$ and hence $x \\sim g^{d}\\left(n_{f}\\right)$. It follows that $g(x)=g\\left(g^{d}\\left(n_{f}\\right)\\right)=g^{d+1}\\left(n_{f}\\right)=a+d=f(x)$ by Claim 4.\nWe start with the same observations, introducing the relation $\\sim$ and proving Claim 1 from the previous solution.\n\nNote that $f(a)>a$ since otherwise we have $f(a)=a$ and hence $g(a)=g(f(a))=g(a)+1$, which is false.\n\nClaim 2'. $\\quad a=b$.\n\nProof. We can assume that $a \\leq b$. Since $f(a) \\geq a+1$, there exists some $x \\in \\mathbb{N}$ such that $f(a)=f(x)+1$, which is equivalent to $f(a)=f(g(x))$ and $a \\sim g(x)$. Since $g(x) \\geq b \\geq a$, by Claim 1 we have $a=g(x) \\geq b$, which together with $a \\leq b$ proves the Claim.\n\nNow, almost the same method allows to find the values $f(a)$ and $g(a)$.\n\nClaim $3'$. $f(a)=g(a)=a+1$.\n\nProof. Assume the contrary; then $f(a) \\geq a+2$, hence there exist some $x, y \\in \\mathbb{N}$ such that $f(x)=f(a)-2$ and $f(y)=g(x)$ (as $g(x) \\geq a=b$). Now we get $f(a)=f(x)+2=f\\left(g^{2}(x)\\right)$, so $a \\sim g^{2}(x) \\geq a$, and by Claim 1 we get $a=g^{2}(x)=g(f(y))=1+g(y) \\geq 1+a$; this is impossible. The equality $g(a)=a+1$ is similar.\n\nNow, we are prepared for the proof of the problem statement. First, we prove it for $n \\geq a$.\n\nClaim 4'. For each integer $x \\geq a$, we have $f(x)=g(x)=x+1$.\n\nProof. Induction on $x$. The base case $x=a$ is provided by Claim $3'$, while the induction step follows from $f(x+1)=f(g(x))=f(x)+1=(x+1)+1$ and the similar computation for $g(x+1)$.\n\nFinally, for an arbitrary $n \\in \\mathbb{N}$ we have $g(n) \\geq a$, so by Claim $4'$ we have $f(n)+1= f(g(n))=g(n)+1$, hence $f(n)=g(n)$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31921,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, \\ldots, a_{r}$ be positive real numbers. For $n>r$, we inductively define\n$$\na_{n}=\\max _{1 \\leq k \\leq n-1}\\left(a_{k}+a_{n-k}\\right) \\tag{1}\n$$\nProve that there exist positive integers $\\ell \\leq r$ and $N$ such that $a_{n}=a_{n-\\ell}+a_{\\ell}$ for all $n \\geq N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, from the problem conditions we have that each $a_{n}(n>r)$ can be expressed as $a_{n}=a_{j_{1}}+a_{j_{2}}$ with $j_{1}, j_{2}<n, j_{1}+j_{2}=n$. If, say, $j_{1}>r$ then we can proceed in the same way with $a_{j_{1}}$, and so on. Finally, we represent $a_{n}$ in a form\n$$\na_{n}=a_{i_{1}}+\\cdots+a_{i_{k}}  \\tag{2}\\\\\n1 \\leq i_{j} \\leq r, \\quad i_{1}+\\cdots+i_{k}=n \\tag{3}\n$$\nMoreover, if $a_{i_{1}}$ and $a_{i_{2}}$ are the numbers in (2) obtained on the last step, then $i_{1}+i_{2}>r$. Hence we can adjust (3) as\n$$\n1 \\leq i_{j} \\leq r, \\quad i_{1}+\\cdots+i_{k}=n, \\quad i_{1}+i_{2}>r . \\tag{4}\n$$\nOn the other hand, suppose that the indices $i_{1}, \\ldots, i_{k}$ satisfy the conditions (4). Then, denoting $s_{j}=i_{1}+\\cdots+i_{j}$, from (1) we have\n$$\na_{n}=a_{s_{k}} \\geq a_{s_{k-1}}+a_{i_{k}} \\geq a_{s_{k-2}}+a_{i_{k-1}}+a_{i_{k}} \\geq \\cdots \\geq a_{i_{1}}+\\cdots+a_{i_{k}}\n$$\nSummarizing these observations we get the following\nClaim. For every $n>r$, we have\n$$\na_{n}=\\max \\left\\{a_{i_{1}}+\\cdots+a_{i_{k}}: \\text{ the collection }\\left(i_{1}, \\ldots, i_{k}\\right) \\text{ satisfies }(4)\\right\\} .\n$$\nNow we denote\n$$\ns=\\max _{1 \\leq i \\leq r} \\frac{a_{i}}{i}\n$$\nand fix some index $\\ell \\leq r$ such that $s=\\frac{a_{\\ell}}{\\ell}$.\nConsider some $n \\geq r^{2} \\ell+2 r$ and choose an expansion of $a_{n}$ in the form (2), (4). Then we have $n=i_{1}+\\cdots+i_{k} \\leq r k$, so $k \\geq n / r \\geq r \\ell+2$. Suppose that none of the numbers $i_{3}, \\ldots, i_{k}$ equals $\\ell$. Then by the pigeonhole principle there is an index $1 \\leq j \\leq r$ which appears among $i_{3}, \\ldots, i_{k}$ at least $\\ell$ times, and surely $j \\neq \\ell$. Let us delete these $\\ell$ occurrences of $j$ from $\\left(i_{1}, \\ldots, i_{k}\\right)$, and add $j$ occurrences of $\\ell$ instead, obtaining a sequence $\\left(i_{1}, i_{2}, i_{3}', \\ldots, i_{k'}'\\right)$ also satisfying (4). By Claim, we have\n$$\na_{i_{1}}+\\cdots+a_{i_{k}}=a_{n} \\geq a_{i_{1}}+a_{i_{2}}+a_{i_{3}'}+\\cdots+a_{i_{k'}'}\n$$\nor, after removing the coinciding terms, $\\ell a_{j} \\geq j a_{\\ell}$, so $\\frac{a_{\\ell}}{\\ell} \\leq \\frac{a_{j}}{j}$. By the definition of $\\ell$, this means that $\\ell a_{j}=j a_{\\ell}$, hence\n$$\na_{n}=a_{i_{1}}+a_{i_{2}}+a_{i_{3}'}+\\cdots+a_{i_{k'}'} .\n$$\nThus, for every $n \\geq r^{2} \\ell+2 r$ we have found a representation of the form (2), (4) with $i_{j}=\\ell$ for some $j \\geq 3$. Rearranging the indices we may assume that $i_{k}=\\ell$.\nFinally, observe that in this representation, the indices $\\left(i_{1}, \\ldots, i_{k-1}\\right)$ satisfy the conditions (4) with $n$ replaced by $n-\\ell$. Thus, from the Claim we get\n$$\na_{n-\\ell}+a_{\\ell} \\geq\\left(a_{i_{1}}+\\cdots+a_{i_{k-1}}\\right)+a_{\\ell}=a_{n}\n$$\nwhich by (1) implies\n$$\na_{n}=a_{n-\\ell}+a_{\\ell} \\quad \\text{ for each } n \\geq r^{2} \\ell+2 r\n$$\nas desired.\nAs in the previous solution, we involve the expansion (2), (3), and we fix some index $1 \\leq \\ell \\leq r$ such that\n$$\n\\frac{a_{\\ell}}{\\ell}=s=\\max _{1 \\leq i \\leq r} \\frac{a_{i}}{i}\n$$\nNow, we introduce the sequence $\\left(b_{n}\\right)$ as $b_{n}=a_{n}-s n$; then $b_{\\ell}=0$.\nWe prove by induction on $n$ that $b_{n} \\leq 0$, and $\\left(b_{n}\\right)$ satisfies the same recurrence relation as $\\left(a_{n}\\right)$. The base cases $n \\leq r$ follow from the definition of $s$. Now, for $n>r$ from the induction hypothesis we have\n$$\nb_{n}=\\max _{1 \\leq k \\leq n-1}\\left(a_{k}+a_{n-k}\\right)-n s=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}+n s\\right)-n s=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}\\right) \\leq 0,\n$$\nas required.\nNow, if $b_{k}=0$ for all $1 \\leq k \\leq r$, then $b_{n}=0$ for all $n$, hence $a_{n}=s n$, and the statement is trivial. Otherwise, define\n$$\nM=\\max _{1 \\leq i \\leq r}\\left|b_{i}\\right|, \\quad \\varepsilon=\\min \\left\\{\\left|b_{i}\\right|: 1 \\leq i \\leq r, b_{i}<0\\right\\} .\n$$\nThen for $n>r$ we obtain\n$$\nb_{n}=\\max _{1 \\leq k \\leq n-1}\\left(b_{k}+b_{n-k}\\right) \\geq b_{\\ell}+b_{n-\\ell}=b_{n-\\ell},\n$$\nso\n$$\n0 \\geq b_{n} \\geq b_{n-\\ell} \\geq b_{n-2 \\ell} \\geq \\cdots \\geq-M\n$$\nThus, in view of the expansion (2), (3) applied to the sequence $\\left(b_{n}\\right)$, we get that each $b_{n}$ is contained in a set\n$$\nT=\\left\\{b_{i_{1}}+b_{i_{2}}+\\cdots+b_{i_{k}}: i_{1}, \\ldots, i_{k} \\leq r\\right\\} \\cap[-M, 0]\n$$\nWe claim that this set is finite. Actually, for any $x \\in T$, let $x=b_{i_{1}}+\\cdots+b_{i_{k}}\\left(i_{1}, \\ldots, i_{k} \\leq r\\right)$. Then among $b_{i_{j}}$ 's there are at most $\\frac{M}{\\varepsilon}$ nonzero terms (otherwise $x<\\frac{M}{\\varepsilon} \\cdot(-\\varepsilon)<-M$ ). Thus $x$ can be expressed in the same way with $k \\leq \\frac{M}{\\varepsilon}$, and there is only a finite number of such sums.\nFinally, for every $t=1,2, \\ldots, \\ell$ we get that the sequence\n$$\nb_{r+t}, b_{r+t+\\ell}, b_{r+t+2 \\ell}, \\ldots\n$$\nis non-decreasing and attains the finite number of values; therefore it is constant from some index. Thus, the sequence $\\left(b_{n}\\right)$ is periodic with period $\\ell$ from some index $N$, which means that\n$$\nb_{n}=b_{n-\\ell}=b_{n-\\ell}+b_{\\ell} \\quad \\text{ for all } n>N+\\ell\n$$\nand hence\n$$\na_{n}=b_{n}+n s=\\left(b_{n-\\ell}+(n-\\ell) s\\right)+\\left(b_{\\ell}+\\ell s\\right)=a_{n-\\ell}+a_{\\ell} \\quad \\text{ for all } n>N+\\ell\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31922,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given six positive numbers $a, b, c, d, e, f$ such that $a < b < c < d < e < f$. Let $a + c + e = S$ and $b + d + f = T$. Prove that\n$$\n2 S T > \\sqrt{3(S + T)(S(b d + b f + d f) + T(a c + a e + c e))}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We define also $\\sigma = a c + c e + a e$, $\\tau = b d + b f + d f$. The idea of the solution is to interpret (1) as a natural inequality on the roots of an appropriate polynomial.\n\nActually, consider the polynomial\n$$\nP(x) = (b + d + f)(x - a)(x - c)(x - e) + (a + c + e)(x - b)(x - d)(x - f)\n$$\n$$\n= T\\left(x^3 - S x^2 + \\sigma x - a c e\\right) + S\\left(x^3 - T x^2 + \\tau x - b d f\\right)\n$$\nSurely, $P$ is cubic with leading coefficient $S + T > 0$. Moreover, we have\n$$\nP(a) = S(a - b)(a - d)(a - f) < 0, \\quad P(c) = S(c - b)(c - d)(c - f) > 0 \\\\\nP(e) = S(e - b)(e - d)(e - f) < 0, \\quad P(f) = T(f - a)(f - c)(f - e) > 0\n$$\nHence, each of the intervals $(a, c), (c, e), (e, f)$ contains at least one root of $P(x)$. Since there are at most three roots at all, we obtain that there is exactly one root in each interval (denote them by $\\alpha \\in (a, c), \\beta \\in (c, e), \\gamma \\in (e, f)$). Moreover, the polynomial $P$ can be factorized as\n$$\nP(x) = (T + S)(x - \\alpha)(x - \\beta)(x - \\gamma)\n$$\nEquating the coefficients in the two representations of $P(x)$ provides\n$$\n\\alpha + \\beta + \\gamma = \\frac{2 T S}{T + S}, \\quad \\alpha \\beta + \\alpha \\gamma + \\beta \\gamma = \\frac{S \\tau + T \\sigma}{T + S}\n$$\nNow, since the numbers $\\alpha, \\beta, \\gamma$ are distinct, we have\n$$\n0 < (\\alpha - \\beta)^2 + (\\alpha - \\gamma)^2 + (\\beta - \\gamma)^2 = 2(\\alpha + \\beta + \\gamma)^2 - 6(\\alpha \\beta + \\alpha \\gamma + \\beta \\gamma)\n$$\nwhich implies\n$$\n\\frac{4 S^2 T^2}{(T + S)^2} = (\\alpha + \\beta + \\gamma)^2 > 3(\\alpha \\beta + \\alpha \\gamma + \\beta \\gamma) = \\frac{3(S \\tau + T \\sigma)}{T + S}\n$$\nor\n$$\n4 S^2 T^2 > 3(T + S)(T \\sigma + S \\tau)\n$$\nwhich is exactly what we need.\n\n\n$U = \\frac{1}{2}\\left((e - a)^2 + (c - a)^2 + (e - c)^2\\right) = S^2 - 3(a c + a e + c e)$\nand\n$V = \\frac{1}{2}\\left((f - b)^2 + (f - d)^2 + (d - b)^2\\right) = T^2 - 3(b d + b f + d f)$\nThen\n$$\n(\\text{L.H.S.})^2 - (\\text{R.H.S.})^2 = (2 S T)^2 - (S + T)(S \\cdot 3(b d + b f + d f) + T \\cdot 3(a c + a e + c e)) = \\\\\n= 4 S^2 T^2 - (S + T)\\left(S(T^2 - V) + T(S^2 - U)\\right) = (S + T)(S V + T U) - S T(T - S)^2\n$$\nand the statement is equivalent with\n$$\n(S + T)(S V + T U) > S T(T - S)^2\n$$\nBy the Cauchy-Schwarz inequality,\n$$\n(S + T)(T U + S V) \\geq (\\sqrt{S \\cdot T U} + \\sqrt{T \\cdot S V})^2 = S T(\\sqrt{U} + \\sqrt{V})^2\n$$\nEstimate the quantities $\\sqrt{U}$ and $\\sqrt{V}$ by the QM-AM inequality with the positive terms $(e - c)^2$ and $(d - b)^2$ being omitted:\n$$\n\\sqrt{U} + \\sqrt{V} > \\sqrt{\\frac{(e - a)^2 + (c - a)^2}{2}} + \\sqrt{\\frac{(f - b)^2 + (f - d)^2}{2}} \\\\\n> \\frac{(e - a) + (c - a)}{2} + \\frac{(f - b) + (f - d)}{2} = \\left(f - \\frac{d}{2} - \\frac{b}{2}\\right) + \\left(\\frac{e}{2} + \\frac{c}{2} - a\\right) \\\\\n= (T - S) + \\frac{3}{2}(e - d) + \\frac{3}{2}(c - b) > T - S\n$$\nThe estimates prove the required inequality.\n\n\nWe keep using the notations $\\sigma$ and $\\tau$ from Solution 1. Moreover, let $s = c + e$. Note that\n$$\n(c - b)(c - d) + (e - f)(e - d) + (e - f)(c - b) < 0\n$$\nsince each summand is negative. This rewrites as\n$$\n(b d + b f + d f) - (a c + c e + a e) < (c + e)(b + d + f - a - c - e), \\text{ or } \\\\\n\\tau - \\sigma < s(T - S)\n$$\nThen we have\n$$\nS \\tau + T \\sigma = S(\\tau - \\sigma) + (S + T)\\sigma < S s(T - S) + (S + T)(c e + a s) \\\\\n\\leq S s(T - S) + (S + T)\\left(\\frac{s^2}{4} + (S - s)s\\right) = s\\left(2 S T - \\frac{3}{4}(S + T)s\\right)\n$$\nUsing this inequality together with the AM-GM inequality we get\n$$\n\\sqrt{\\frac{3}{4}(S + T)(S \\tau + T \\sigma)} < \\sqrt{\\frac{3}{4}(S + T)s\\left(2 S T - \\frac{3}{4}(S + T)s\\right)} \\\\\n\\leq \\frac{\\frac{3}{4}(S + T)s + 2 S T - \\frac{3}{4}(S + T)s}{2} = S T\n$$\n\n\nWe introduce the expressions $\\sigma$ and $\\tau$ as in the previous solutions. The idea of the solution is to change the values of variables $a, \\ldots, f$ keeping the left-hand side unchanged and increasing the right-hand side; it will lead to a simpler inequality which can be proved in a direct way.\nNamely, we change the variables (i) keeping the (non-strict) inequalities $a \\leq b \\leq c \\leq d \\leq e \\leq f$; (ii) keeping the values of sums $S$ and $T$ unchanged; and finally (iii) increasing the values of $\\sigma$ and $\\tau$. Then the left-hand side of (1) remains unchanged, while the right-hand side increases. Hence, the inequality (1) (and even a non-strict version of (1)) for the changed values would imply the same (strict) inequality for the original values.\nFirst, we find the sufficient conditions for (ii) and (iii) to be satisfied.\nLemma. Let $x, y, z > 0$; denote $U(x, y, z) = x + y + z$, $v(x, y, z) = x y + x z + y z$. Suppose that $x' + y' = x + y$ but $|x - y| \\geq |x' - y'|$; then we have $U(x', y', z) = U(x, y, z)$ and $v(x', y', z) \\geq v(x, y, z)$ with equality achieved only when $|x - y| = |x' - y'|$.\nProof. The first equality is obvious. For the second, we have\n$$\nv(x', y', z) = z(x' + y') + x' y' = z(x' + y') + \\frac{(x' + y')^2 - (x' - y')^2}{4} \\\\\n\\geq z(x + y) + \\frac{(x + y)^2 - (x - y)^2}{4} = v(x, y, z)\n$$\nwith the equality achieved only for $(x' - y')^2 = (x - y)^2 \\Longleftrightarrow |x' - y'| = |x - y|$, as desired.\nNow, we apply Lemma several times making the following changes. For each change, we denote the new values by the same letters to avoid cumbersome notations.\n1. Let $k = \\frac{d - c}{2}$. Replace $(b, c, d, e)$ by $(b + k, c + k, d - k, e - k)$. After the change we have $a < b < c = d < e < f$, the values of $S, T$ remain unchanged, but $\\sigma, \\tau$ strictly increase by Lemma.\n2. Let $\\ell = \\frac{e - d}{2}$. Replace $(c, d, e, f)$ by $(c + \\ell, d + \\ell, e - \\ell, f - \\ell)$. After the change we have $a < b < c = d = e < f$, the values of $S, T$ remain unchanged, but $\\sigma, \\tau$ strictly increase by the Lemma.\n3. Finally, let $m = \\frac{c - b}{3}$. Replace $(a, b, c, d, e, f)$ by $(a + 2 m, b + 2 m, c - m, d - m, e - m, f - m)$. After the change, we have $a < b = c = d = e < f$ and $S, T$ are unchanged. To check (iii), we observe that our change can be considered as a composition of two changes: $(a, b, c, d) \\rightarrow (a + m, b + m, c - m, d - m)$ and $(a, b, e, f) \\rightarrow (a + m, b + m, e - m, f - m)$. It is easy to see that each of these two consecutive changes satisfy the conditions of the Lemma, hence the values of $\\sigma$ and $\\tau$ increase.\nFinally, we come to the situation when $a < b = c = d = e < f$, and we need to prove the inequality\n$$\n2(a + 2 b)(2 b + f) \\geq \\sqrt{3(a + 4 b + f)\\left((a + 2 b)(b^2 + 2 b f) + (2 b + f)(2 a b + b^2)\\right)} \\\\\n= \\sqrt{3 b(a + 4 b + f) \\cdot ((a + 2 b)(b + 2 f) + (2 b + f)(2 a + b))}\n$$\nNow, observe that\n$$\n2 \\cdot 2(a + 2 b)(2 b + f) = 3 b(a + 4 b + f) + ((a + 2 b)(b + 2 f) + (2 a + b)(2 b + f))\n$$\nHence the inequality rewrites as\n$$\n3 b(a + 4 b + f) + ((a + 2 b)(b + 2 f) + (2 a + b)(2 b + f)) \\geq 2 \\sqrt{3 b(a + 4 b + f) \\cdot ((a + 2 b)(b + 2 f) + (2 b + f)(2 a + b))}\n$$\nwhich is simply the AM-GM inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31923,
    "subject": "Mathematics (Multi-modal)",
    "question": "On some planet, there are $2^{N}$ countries $(N \\geq 4)$. Each country has a flag $N$ units wide and one unit high composed of $N$ fields of size $1 \\times 1$, each field being either yellow or blue. No two countries have the same flag.\n\nWe say that a set of $N$ flags is diverse if these flags can be arranged into an $N \\times N$ square so that all $N$ fields on its main diagonal will have the same color. Determine the smallest positive integer $M$ such that among any $M$ distinct flags, there exist $N$ flags forming a diverse set.",
    "options": [],
    "answer": "2^{N-2}+1",
    "solution": "When speaking about the diagonal of a square, we will always mean the main diagonal.\n\nLet $M_{N}$ be the smallest positive integer satisfying the problem condition. First, we show that $M_{N}>2^{N-2}$. Consider the collection of all $2^{N-2}$ flags having yellow first squares and blue second ones. Obviously, both colors appear on the diagonal of each $N \\times N$ square formed by these flags.\n\nWe are left to show that $M_{N} \\leq 2^{N-2}+1$, thus obtaining the desired answer. We start with establishing this statement for $N=4$.\n\nSuppose that we have 5 flags of length 4. We decompose each flag into two parts of 2 squares each; thereby, we denote each flag as $L R$, where the $2 \\times 1$ flags $L, R \\in \\mathcal{S}=\\{\\mathrm{BB}, \\mathrm{BY}, \\mathrm{YB}, \\mathrm{YY}\\}$ are its left and right parts, respectively. First, we make two easy observations on the flags $2 \\times 1$ which can be checked manually.\n\n(i) For each $A \\in \\mathcal{S}$, there exists only one $2 \\times 1$ flag $C \\in \\mathcal{S}$ (possibly $C=A$) such that $A$ and $C$ cannot form a $2 \\times 2$ square with monochrome diagonal (for part BB, that is YY, and for BY that is YB).\n\n(ii) Let $A_{1}, A_{2}, A_{3} \\in \\mathcal{S}$ be three distinct elements; then two of them can form a $2 \\times 2$ square with yellow diagonal, and two of them can form a $2 \\times 2$ square with blue diagonal (for all parts but BB, a pair (BY, YB) fits for both statements, while for all parts but BY, these pairs are $(\\mathrm{YB}, \\mathrm{YY})$ and $(\\mathrm{BB}, \\mathrm{YB})$).\n\nNow, let $\\ell$ and $r$ be the numbers of distinct left and right parts of our 5 flags, respectively. The total number of flags is $5 \\leq r \\ell$, hence one of the factors (say, $r$) should be at least 3. On the other hand, $\\ell, r \\leq 4$, so there are two flags with coinciding right part; let them be $L_{1} R_{1}$ and $L_{2} R_{1}$ ($L_{1} \\neq L_{2}$).\n\nNext, since $r \\geq 3$, there exist some flags $L_{3} R_{3}$ and $L_{4} R_{4}$ such that $R_{1}, R_{3}, R_{4}$ are distinct. Let $L' R'$ be the remaining flag. By (i), one of the pairs ($L'$, $L_{1}$) and ($L'$, $L_{2}$) can form a $2 \\times 2$ square with monochrome diagonal; we can assume that $L'$, $L_{2}$ form a square with a blue diagonal. Finally, the right parts of two of the flags $L_{1} R_{1}, L_{3} R_{3}, L_{4} R_{4}$ can also form a $2 \\times 2$ square with a blue diagonal by (ii). Putting these $2 \\times 2$ squares on the diagonal of a $4 \\times 4$ square, we find a desired arrangement of four flags.\n\nWe are ready to prove the problem statement by induction on $N$; actually, above we have proved the base case $N=4$. For the induction step, assume that $N>4$, consider any $2^{N-2}+1$ flags of length $N$, and arrange them into a large flag of size $(2^{N-2}+1) \\times N$. This flag contains a non-monochrome column since the flags are distinct; we may assume that this column is the first one. By the pigeonhole principle, this column contains at least $\\left\\lceil\\frac{2^{N-2}+1}{2}\\right\\rceil=2^{N-3}+1$ squares of one color (say, blue). We call the flags with a blue first square good.\n\nConsider all the good flags and remove the first square from each of them. We obtain at least $2^{N-3}+1 \\geq M_{N-1}$ flags of length $N-1$; by the induction hypothesis, $N-1$ of them can form a square $Q$ with the monochrome diagonal. Now, returning the removed squares, we obtain a rectangle $(N-1) \\times N$, and our aim is to supplement it on the top by one more flag.\n\nIf $Q$ has a yellow diagonal, then we can take each flag with a yellow first square (it exists by a choice of the first column; moreover, it is not used in $Q$). Conversely, if the diagonal of $Q$ is blue then we can take any of the $\\geq 2^{N-3}+1-(N-1)>0$ remaining good flags. So, in both cases we get a desired $N \\times N$ square.\nWe present a different proof of the estimate $M_{N} \\leq 2^{N-2}+1$. We do not use the induction, involving Hall's lemma on matchings instead.\n\nConsider arbitrary $2^{N-2}+1$ distinct flags and arrange them into a large $(2^{N-2}+1) \\times N$ flag. Construct two bipartite graphs $G_{\\mathrm{y}}=(V \\cup V', E_{\\mathrm{y}})$ and $G_{\\mathrm{b}}=(V \\cup V', E_{\\mathrm{b}})$ with the common set of vertices as follows. Let $V$ and $V'$ be the set of columns and the set of flags under consideration, respectively. Next, let the edge $(c, f)$ appear in $E_{\\mathrm{y}}$ if the intersection of column $c$ and flag $f$ is yellow, and $(c, f) \\in E_{\\mathrm{b}}$ otherwise. Then we have to prove exactly that one of the graphs $G_{\\mathrm{y}}$ and $G_{\\mathrm{b}}$ contains a matching with all the vertices of $V$ involved.\n\nAssume that these matchings do not exist. By Hall's lemma, it means that there exist two sets of columns $S_{\\mathrm{y}}, S_{\\mathrm{b}} \\subset V$ such that $|E_{\\mathrm{y}}(S_{\\mathrm{y}})| \\leq |S_{\\mathrm{y}}|-1$ and $|E_{\\mathrm{b}}(S_{\\mathrm{b}})| \\leq |S_{\\mathrm{b}}|-1$ (in the left-hand sides, $E_{\\mathrm{y}}(S_{\\mathrm{y}})$ and $E_{\\mathrm{b}}(S_{\\mathrm{b}})$ denote respectively the sets of all vertices connected to $S_{\\mathrm{y}}$ and $S_{\\mathrm{b}}$ in the corresponding graphs). Our aim is to prove that this is impossible. Note that $S_{\\mathrm{y}}, S_{\\mathrm{b}} \\neq V$ since $N \\leq 2^{N-2}+1$.\n\nFirst, suppose that $S_{\\mathrm{y}} \\cap S_{\\mathrm{b}} \\neq \\varnothing$, so there exists some $c \\in S_{\\mathrm{y}} \\cap S_{\\mathrm{b}}$. Note that each flag is connected to $c$ either in $G_{\\mathrm{y}}$ or in $G_{\\mathrm{b}}$, hence $E_{\\mathrm{y}}(S_{\\mathrm{y}}) \\cup E_{\\mathrm{b}}(S_{\\mathrm{b}})=V'$. Hence we have $2^{N-2}+1=|V'| \\leq |E_{\\mathrm{y}}(S_{\\mathrm{y}})|+|E_{\\mathrm{b}}(S_{\\mathrm{b}})| \\leq |S_{\\mathrm{y}}|+|S_{\\mathrm{b}}|-2 \\leq 2N-4$; this is impossible for $N \\geq 4$.\n\nSo, we have $S_{\\mathrm{y}} \\cap S_{\\mathrm{b}}=\\varnothing$. Let $y=|S_{\\mathrm{y}}|, b=|S_{\\mathrm{b}}|$. From the construction of our graph, we have that all the flags in the set $V''=V' \\setminus (E_{\\mathrm{y}}(S_{\\mathrm{y}}) \\cup E_{\\mathrm{b}}(S_{\\mathrm{b}}))$ have blue squares in the columns of $S_{\\mathrm{y}}$ and yellow squares in the columns of $S_{\\mathrm{b}}$. Hence the only undetermined positions in these flags are the remaining $N-y-b$ ones, so $2^{N-y-b} \\geq |V''| \\geq |V'|-|E_{\\mathrm{y}}(S_{\\mathrm{y}})|-|E_{\\mathrm{b}}(S_{\\mathrm{b}})| \\geq 2^{N-2}+1-(y-1)-(b-1)$, or, denoting $c=y+b$, $2^{N-c}+c>2^{N-2}+2$. This is impossible since $N \\geq c \\geq 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31924,
    "subject": "Mathematics (Multi-modal)",
    "question": "Six stacks $S_{1}, \\ldots, S_{6}$ of coins are standing in a row. In the beginning every stack contains a single coin. There are two types of allowed moves:\n\nMove 1: If stack $S_{k}$ with $1 \\leq k \\leq 5$ contains at least one coin, you may remove one coin from $S_{k}$ and add two coins to $S_{k+1}$.\n\nMove 2: If stack $S_{k}$ with $1 \\leq k \\leq 4$ contains at least one coin, then you may remove one coin from $S_{k}$ and exchange stacks $S_{k+1}$ and $S_{k+2}$.\n\nDecide whether it is possible to achieve by a sequence of such moves that the first five stacks are empty, whereas the sixth stack $S_{6}$ contains exactly $2010^{2010^{2010}}$ coins.",
    "options": [],
    "answer": "Yes, it is possible.",
    "solution": "Denote by $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right) \\rightarrow\\left(a_{1}^{\\prime}, a_{2}^{\\prime}, \\ldots, a_{n}^{\\prime}\\right)$ the following: if some consecutive stacks contain $a_{1}, \\ldots, a_{n}$ coins, then it is possible to perform several allowed moves such that the stacks contain $a_{1}^{\\prime}, \\ldots, a_{n}^{\\prime}$ coins respectively, whereas the contents of the other stacks remain unchanged.\n\nLet $A=2010^{2010}$ or $A=2010^{2010^{2010}}$, respectively. Our goal is to show that\n$$\n(1,1,1,1,1,1) \\rightarrow(0,0,0,0,0, A)\n$$\nFirst we prove two auxiliary observations.\n\nLemma 1. $\\quad(a, 0,0) \\rightarrow\\left(0,2^{a}, 0\\right)$ for every $a \\geq 1$.\n\nProof. We prove by induction that $(a, 0,0) \\rightarrow\\left(a-k, 2^{k}, 0\\right)$ for every $1 \\leq k \\leq a$. For $k=1$, apply Move 1 to the first stack:\n$$\n(a, 0,0) \\rightarrow(a-1,2,0)=\\left(a-1,2^{1}, 0\\right)\n$$\nNow assume that $k<a$ and the statement holds for some $k<a$. Starting from $\\left(a-k, 2^{k}, 0\\right)$, apply Move 1 to the middle stack $2^{k}$ times, until it becomes empty. Then apply Move 2 to the first stack:\n$$\n\\left(a-k, 2^{k}, 0\\right) \\rightarrow\\left(a-k, 2^{k}-1,2\\right) \\rightarrow \\cdots \\rightarrow\\left(a-k, 0,2^{k+1}\\right) \\rightarrow\\left(a-k-1,2^{k+1}, 0\\right)\n$$\nHence,\n$$\n(a, 0,0) \\rightarrow\\left(a-k, 2^{k}, 0\\right) \\rightarrow\\left(a-k-1,2^{k+1}, 0\\right)\n$$\n\nLemma 2. For every positive integer $n$, let $P_{n}=\\underbrace{2^{2^{.2}}}_{n}$ (e.g. $P_{3}=2^{2^{2}}=16$ ). Then $(a, 0,0,0) \\rightarrow\\left(0, P_{a}, 0,0\\right)$ for every $a \\geq 1$.\n\nProof. Similarly to Lemma 1, we prove that $(a, 0,0,0) \\rightarrow\\left(a-k, P_{k}, 0,0\\right)$ for every $1 \\leq k \\leq a$.\nFor $k=1$, apply Move 1 to the first stack:\n$$\n(a, 0,0,0) \\rightarrow(a-1,2,0,0)=\\left(a-1, P_{1}, 0,0\\right)\n$$\nNow assume that the lemma holds for some $k<a$. Starting from ( $a-k, P_{k}, 0,0$ ), apply Lemma 1, then apply Move 1 to the first stack:\n$$\n\\left(a-k, P_{k}, 0,0\\right) \\rightarrow\\left(a-k, 0,2^{P_{k}}, 0\\right)=\\left(a-k, 0, P_{k+1}, 0\\right) \\rightarrow\\left(a-k-1, P_{k+1}, 0,0\\right)\n$$\nTherefore,\n$$\n(a, 0,0,0) \\rightarrow\\left(a-k, P_{k}, 0,0\\right) \\rightarrow\\left(a-k-1, P_{k+1}, 0,0\\right)\n$$\n\nNow we prove the statement of the problem.\n\nFirst apply Move 1 to stack 5, then apply Move 2 to stacks $S_{4}, S_{3}, S_{2}$ and $S_{1}$ in this order. Then apply Lemma 2 twice:\n$$\n\\begin{gathered}\n(1,1,1,1,1,1) \\rightarrow(1,1,1,1,0,3) \\rightarrow(1,1,1,0,3,0) \\rightarrow(1,1,0,3,0,0) \\rightarrow(1,0,3,0,0,0) \\rightarrow \\\\\n\\rightarrow(0,3,0,0,0,0) \\rightarrow\\left(0,0, P_{3}, 0,0,0\\right)=(0,0,16,0,0,0) \\rightarrow\\left(0,0,0, P_{16}, 0,0\\right)\n\\end{gathered}\n$$\nWe already have more than $A$ coins in stack $S_{4}$, since\n$$\nA \\leq 2010^{2010^{2010}}<\\left(2^{11}\\right)^{2010^{2010}}=2^{11 \\cdot 2010^{2010}}<2^{2010^{2011}}<2^{\\left(2^{11}\\right)^{2011}}=2^{2^{11 \\cdot 2011}}<2^{2^{2^{15}}}<P_{16} .\n$$\nTo decrease the number of coins in stack $S_{4}$, apply Move 2 to this stack repeatedly until its size decreases to $A / 4$. (In every step, we remove a coin from $S_{4}$ and exchange the empty stacks $S_{5}$ and $S_{6}$.)\n$$\n\\begin{aligned}\n\\left(0,0,0, P_{16}, 0,0\\right) \\rightarrow & \\left(0,0,0, P_{16}-1,0,0\\right) \\rightarrow\\left(0,0,0, P_{16}-2,0,0\\right) \\rightarrow \\\\\n& \\rightarrow \\cdots \\rightarrow(0,0,0, A / 4,0,0)\n\\end{aligned}\n$$\nFinally, apply Move 1 repeatedly to empty stacks $S_{4}$ and $S_{5}$ :\n$$\n(0,0,0, A / 4,0,0) \\rightarrow \\cdots \\rightarrow(0,0,0,0, A / 2,0) \\rightarrow \\cdots \\rightarrow(0,0,0,0,0, A)\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31925,
    "subject": "Mathematics (Multi-modal)",
    "question": "$n \\geq 4$ players participated in a tennis tournament. Any two players have played exactly one game, and there was no tie game. We call a company of four players bad if one player was defeated by the other three players, and each of these three players won a game and lost another game among themselves. Suppose that there is no bad company in this tournament. Let $w_{i}$ and $\\ell_{i}$ be respectively the number of wins and losses of the $i$th player. Prove that\n$$\n\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)^{3} \\geq 0\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any tournament $T$ satisfying the problem condition, denote by $S(T)$ sum under consideration, namely\n$$\nS(T)=\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)^{3}\n$$\nFirst, we show that the statement holds if a tournament $T$ has only 4 players. Actually, let $A=\\left(a_{1}, a_{2}, a_{3}, a_{4}\\right)$ be the number of wins of the players; we may assume that $a_{1} \\geq a_{2} \\geq a_{3} \\geq a_{4}$. We have $a_{1}+a_{2}+a_{3}+a_{4}=\\binom{4}{2}=6$, hence $a_{4} \\leq 1$. If $a_{4}=0$, then we cannot have $a_{1}=a_{2}=a_{3}=2$, otherwise the company of all players is bad. Hence we should have $A=(3,2,1,0)$, and $S(T)=3^{3}+1^{3}+(-1)^{3}+(-3)^{3}=0$. On the other hand, if $a_{4}=1$, then only two possibilities, $A=(3,1,1,1)$ and $A=(2,2,1,1)$ can take place. In the former case we have $S(T)=3^{3}+3 \\cdot(-2)^{3}>0$, while in the latter one $S(T)=1^{3}+1^{3}+(-1)^{3}+(-1)^{3}=0$, as desired.\n\nNow we turn to the general problem. Consider a tournament $T$ with no bad companies and enumerate the players by the numbers from 1 to $n$. For every 4 players $i_{1}, i_{2}, i_{3}, i_{4}$ consider a \"sub-tournament\" $T_{i_{1} i_{2} i_{3} i_{4}}$ consisting of only these players and the games which they performed with each other. By the abovementioned, we have $S\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right) \\geq 0$. Our aim is to prove that\n$$\nS(T)=\\sum_{i_{1}, i_{2}, i_{3}, i_{4}} S\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right)\n$$\nwhere the sum is taken over all 4-tuples of distinct numbers from the set $\\{1, \\ldots, n\\}$. This way the problem statement will be established.\n\nWe interpret the number $\\left(w_{i}-\\ell_{i}\\right)^{3}$ as following. For $i \\neq j$, let $\\varepsilon_{ij}=1$ if the $i$th player wins against the $j$th one, and $\\varepsilon_{ij}=-1$ otherwise. Then\n$$\n\\left(w_{i}-\\ell_{i}\\right)^{3}=\\left(\\sum_{j \\neq i} \\varepsilon_{ij}\\right)^{3}=\\sum_{j_{1}, j_{2}, j_{3} \\neq i} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}} .\n$$\nHence,\n$$\nS(T)=\\sum_{i \\notin\\{j_{1}, j_{2}, j_{3}\\}} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}\n$$\nTo simplify this expression, consider all the terms in this sum where two indices are equal. If, for instance, $j_{1}=j_{2}$, then the term contains $\\varepsilon_{i j_{1}}^{2}=1$, so we can replace this term by $\\varepsilon_{i j_{3}}$. Make such replacements for each such term; obviously, after this change each term of the form $\\varepsilon_{i j_{3}}$ will appear $P(T)$ times, hence\n$$\nS(T)=\\sum_{|\\{i, j_{1}, j_{2}, j_{3}\\}|=4} \\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}+P(T) \\sum_{i \\neq j} \\varepsilon_{ij}=S_{1}(T)+P(T) S_{2}(T)\n$$\nWe show that $S_{2}(T)=0$ and hence $S(T)=S_{1}(T)$ for each tournament. Actually, note that $\\varepsilon_{ij}=-\\varepsilon_{ji}$, and the whole sum can be split into such pairs. Since the sum in each pair is 0, so is $S_{2}(T)$.\n\nThus the desired equality rewrites as\n$$\nS_{1}(T)=\\sum_{i_{1}, i_{2}, i_{3}, i_{4}} S_{1}\\left(T_{i_{1} i_{2} i_{3} i_{4}}\\right)\n$$\nNow, if all the numbers $j_{1}, j_{2}, j_{3}$ are distinct, then the set $\\{i, j_{1}, j_{2}, j_{3}\\}$ is contained in exactly one 4-tuple, hence the term $\\varepsilon_{i j_{1}} \\varepsilon_{i j_{2}} \\varepsilon_{i j_{3}}$ appears in the right-hand part of this equality exactly once, as well as in the left-hand part. Clearly, there are no other terms in both parts, so the equality is established.\nSimilarly to the first solution, we call the subsets of players as companies, and the $k$-element subsets will be called as $k$-companies.\n\nIn any company of the players, call a player the local champion of the company if he defeated all other members of the company. Similarly, if a player lost all his games against the others in the company then call him the local loser of the company. Obviously every company has at most one local champion and at most one local loser. By the condition of the problem, whenever a 4-company has a local loser, then this company has a local champion as well.\n\nSuppose that $k$ is some positive integer, and let us count all cases when a player is the local champion of some $k$-company. The $i$th player won against $w_{i}$ other player. To be the local champion of a $k$-company, he must be a member of the company, and the other $k-1$ members must be chosen from those whom he defeated. Therefore, the $i$th player is the local champion of $\\binom{w_{i}}{k-1}$ $k$-companies. Hence, the total number of local champions of all $k$-companies is $\\sum_{i=1}^{n}\\binom{w_{i}}{k-1}$.\n\nSimilarly, the total number of local losers of the $k$-companies is $\\sum_{i=1}^{n}\\binom{\\ell_{i}}{k-1}$.\n\nNow apply this for $k=2,3$ and 4.\n\nSince every game has a winner and a loser, we have $\\sum_{i=1}^{n} w_{i}=\\sum_{i=1}^{n} \\ell_{i}=\\binom{n}{2}$, and hence\n$$\n\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)=0\n$$\nIn every 3-company, either the players defeated one another in a cycle or the company has both a local champion and a local loser. Therefore, the total number of local champions and local losers in the 3-companies is the same, $\\sum_{i=1}^{n}\\binom{w_{i}}{2}=\\sum_{i=1}^{n}\\binom{\\ell_{i}}{2}$. So we have\n$$\n\\sum_{i=1}^{n}\\left(\\binom{w_{i}}{2}-\\binom{\\ell_{i}}{2}\\right)=0\n$$\nIn every 4-company, by the problem's condition, the number of local losers is less than or equal to the number of local champions. Then the same holds for the total numbers of local champions and local losers in all 4-companies, so $\\sum_{i=1}^{n}\\binom{w_{i}}{3} \\geq \\sum_{i=1}^{n}\\binom{\\ell_{i}}{3}$. Hence,\n$$\n\\sum_{i=1}^{n}\\left(\\binom{w_{i}}{3}-\\binom{\\ell_{i}}{3}\\right) \\geq 0\n$$\nNow we establish the problem statement as a linear combination of the above. It is easy check that\n$$\n(x-y)^{3}=24\\left(\\binom{x}{3}-\\binom{y}{3}\\right)+24\\left(\\binom{x}{2}-\\binom{y}{2}\\right)-\\left(3(x+y)^{2}-4\\right)(x-y)\n$$\nApply this identity to $x=w_{i}$ and $y=\\ell_{i}$. Since every player played $n-1$ games, we have $w_{i}+\\ell_{i}=n-1$, and thus\n$$\n\\left(w_{i}-\\ell_{i}\\right)^{3}=24\\left(\\binom{w_{i}}{3}-\\binom{\\ell_{i}}{3}\\right)+24\\left(\\binom{w_{i}}{2}-\\binom{\\ell_{i}}{2}\\right)-\\left(3(n-1)^{2}-4\\right)\\left(w_{i}-\\ell_{i}\\right) .\n$$\nThen\n$$\n\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)^{3}=24 \\underbrace{\\sum_{i=1}^{n}\\left(\\binom{w_{i}}{3}-\\binom{\\ell_{i}}{3}\\right)}_{\\geq 0}+24 \\underbrace{\\sum_{i=1}^{n}\\left(\\binom{w_{i}}{2}-\\binom{\\ell_{i}}{2}\\right)}_{0}-\\left(3(n-1)^{2}-4\\right) \\underbrace{\\sum_{i=1}^{n}\\left(w_{i}-\\ell_{i}\\right)}_{0} \\geq 0 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31926,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P_{1}, \\ldots, P_{s}$ be arithmetic progressions of integers, the following conditions being satisfied:\n(i) each integer belongs to at least one of them;\n(ii) each progression contains a number which does not belong to other progressions.\nDenote by $n$ the least common multiple of steps of these progressions; let $n=p_{1}^{\\alpha_{1}} \\ldots p_{k}^{\\alpha_{k}}$ be its prime factorization. Prove that\n$$\ns \\geq 1+\\sum_{i=1}^{k} \\alpha_{i}\\left(p_{i}-1\\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we prove the key lemma, and then we show how to apply it to finish the solution.\nLet $n_{1}, \\ldots, n_{k}$ be positive integers. By an $n_{1} \\times n_{2} \\times \\cdots \\times n_{k}$ grid we mean the set $N= \\left\\{\\left(a_{1}, \\ldots, a_{k}\\right): a_{i} \\in \\mathbb{Z}, 0 \\leq a_{i} \\leq n_{i}-1\\right\\}$; the elements of $N$ will be referred to as points. In this grid, we define a subgrid as a subset of the form\n$$\n\\begin{equation*}\nL=\\left\\{\\left(b_{1}, \\ldots, b_{k}\\right) \\in N: b_{i_{1}}=x_{i_{1}}, \\ldots, b_{i_{t}}=x_{i_{t}}\\right\\} \\tag{1}\n\\end{equation*}\n$$\nwhere $I=\\left\\{i_{1}, \\ldots, i_{t}\\right\\}$ is an arbitrary nonempty set of indices, and $x_{i_{j}} \\in [0, n_{i_{j}}-1]$ $(1 \\leq j \\leq t)$ are fixed integer numbers. Further, we say that a subgrid (1) is orthogonal to the $i$th coordinate axis if $i \\in I$, and that it is parallel to the $i$th coordinate axis otherwise.\n\nLemma. Assume that the grid $N$ is covered by subgrids $L_{1}, L_{2}, \\ldots, L_{s}$ (this means $N=\\bigcup_{i=1}^{s} L_{i}$ ) so that\n(ii') each subgrid contains a point which is not covered by other subgrids;\n(iii) for each coordinate axis, there exists a subgrid $L_{i}$ orthogonal to this axis.\nThen\n$$\ns \\geq 1+\\sum_{i=1}^{k}\\left(n_{i}-1\\right)\n$$\nProof. Assume to the contrary that $s \\leq \\sum_{i}\\left(n_{i}-1\\right)=s'$. Our aim is to find a point that is not covered by $L_{1}, \\ldots, L_{s}$.\nThe idea of the proof is the following. Imagine that we expand each subgrid to some maximal subgrid so that for the $i$th axis, there will be at most $n_{i}-1$ maximal subgrids orthogonal to this axis. Then the desired point can be found easily: its $i$th coordinate should be that not covered by the maximal subgrids orthogonal to the $i$th axis. Surely, the conditions for existence of such expansion are provided by Hall's lemma on matchings. So, we will follow this direction, although we will apply Hall's lemma to some subgraph instead of the whole graph.\nConstruct a bipartite graph $G=\\left(V \\cup V', E\\right)$ as follows. Let $V=\\left\\{L_{1}, \\ldots, L_{s}\\right\\}$, and let $V'=\\left\\{v_{ij}: 1 \\leq i \\leq s, 1 \\leq j \\leq n_{i}-1\\right\\}$ be some set of $s'$ elements. Further, let the edge $\\left(L_{m}, v_{ij}\\right)$ appear iff $L_{m}$ is orthogonal to the $i$th axis.\nFor each subset $W \\subset V$, denote\n$$\nf(W)=\\left\\{v \\in V':(L, v) \\in E \\text{ for some } L \\in W\\right\\}\n$$\nNotice that $f(V)=V'$ by (iii).\nNow, consider the set $W \\subset V$ containing the maximal number of elements such that $|W|> |f(W)|$; if there is no such set then we set $W=\\varnothing$. Denote $W'=f(W), U=V \\backslash W, U'=V' \\backslash W'$.\nBy our assumption and the Lemma condition, $|f(V)|=|V'| \\geq|V|$, hence $W \\neq V$ and $U \\neq \\varnothing$. Permuting the coordinates, we can assume that $U'=\\left\\{v_{ij}: 1 \\leq i \\leq \\ell\\right\\}, W'=\\left\\{v_{ij}: \\ell+1 \\leq i \\leq k\\right\\}$.\nConsider the induced subgraph $G'$ of $G$ on the vertices $U \\cup U'$. We claim that for every $X \\subset U$, we get $|f(X) \\cap U'| \\geq|X|$ (so $G'$ satisfies the conditions of Hall's lemma). Actually, we have $|W| \\geq|f(W)|$, so if $|X|>|f(X) \\cap U'|$ for some $X \\subset U$, then we have\n$$\n|W \\cup X|=|W|+|X|>|f(W)|+|f(X) \\cap U'|=|f(W) \\cup (f(X) \\cap U')|=|f(W \\cup X)| .\n$$\nThis contradicts the maximality of $|W|$.\nThus, applying Hall's lemma, we can assign to each $L \\in U$ some vertex $v_{ij} \\in U'$ so that to distinct elements of $U$, distinct vertices of $U'$ are assigned. In this situation, we say that $L \\in U$ corresponds to the $i$th axis, and write $g(L)=i$. Since there are $n_{i}-1$ vertices of the form $v_{ij}$, we get that for each $1 \\leq i \\leq \\ell$, not more than $n_{i}-1$ subgrids correspond to the $i$th axis.\nFinally, we are ready to present the desired point. Since $W \\neq V$, there exists a point $b=\\left(b_{1}, b_{2}, \\ldots, b_{k}\\right) \\in N \\backslash\\left(\\cup_{L \\in W} L\\right)$. On the other hand, for every $1 \\leq i \\leq \\ell$, consider any subgrid $L \\in U$ with $g(L)=i$. This means exactly that $L$ is orthogonal to the $i$th axis, and hence all its elements have the same $i$th coordinate $c_{L}$. Since there are at most $n_{i}-1$ such subgrids, there exists a number $0 \\leq a_{i} \\leq n_{i}-1$ which is not contained in a set $\\left\\{c_{L}: g(L)=i\\right\\}$. Choose such number for every $1 \\leq i \\leq \\ell$. Now we claim that point $a=\\left(a_{1}, \\ldots, a_{\\ell}, b_{\\ell+1}, \\ldots, b_{k}\\right)$ is not covered, hence contradicting the Lemma condition.\nSurely, point $a$ cannot lie in some $L \\in U$, since all the points in $L$ have $g(L)$th coordinate $c_{L} \\neq a_{g(L)}$. On the other hand, suppose that $a \\in L$ for some $L \\in W$; recall that $b \\notin L$. But the points $a$ and $b$ differ only at first $\\ell$ coordinates, so $L$ should be orthogonal to at least one of the first $\\ell$ axes, and hence our graph contains some edge $\\left(L, v_{ij}\\right)$ for $i \\leq \\ell$. It contradicts the definition of $W'$. The Lemma is proved.\n\nNow we turn to the problem. Let $d_{j}$ be the step of the progression $P_{j}$. Note that since $n=$ l.c.m. $\\left(d_{1}, \\ldots, d_{s}\\right)$, for each $1 \\leq i \\leq k$ there exists an index $j(i)$ such that $p_{i}^{\\alpha_{i}} \\mid d_{j(i)}$. We assume that $n>1$; otherwise the problem statement is trivial.\nFor each $0 \\leq m \\leq n-1$ and $1 \\leq i \\leq k$, let $m_{i}$ be the residue of $m$ modulo $p_{i}^{\\alpha_{i}}$, and let $m_{i}=\\overline{r_{i\\alpha_{i}} \\ldots r_{i1}}$ be the base $p_{i}$ representation of $m_{i}$ (possibly, with some leading zeroes). Now, we put into correspondence to $m$ the sequence $r(m)=\\left(r_{11}, \\ldots, r_{1\\alpha_{1}}, r_{21}, \\ldots, r_{k\\alpha_{k}}\\right)$. Hence $r(m)$ lies in a $\\underbrace{p_{1} \\times \\cdots \\times p_{1}}_{\\alpha_{1} \\text{ times}} \\times \\cdots \\times \\underbrace{p_{k} \\times \\cdots \\times p_{k}}_{\\alpha_{k} \\text{ times}} \\operatorname{grid} N$.\nSurely, if $r(m)=r\\left(m'\\right)$ then $p_{i}^{\\alpha_{i}} \\mid m_{i}-m_{i}'$, which follows $p_{i}^{\\alpha_{i}} \\mid m-m'$ for all $1 \\leq i \\leq k$; consequently, $n \\mid m-m'$. So, when $m$ runs over the set $\\{0, \\ldots, n-1\\}$, the sequences $r(m)$ do not repeat; since $|N|=n$, this means that $r$ is a bijection between $\\{0, \\ldots, n-1\\}$ and $N$. Now we will show that for each $1 \\leq i \\leq s$, the set $L_{i}=\\left\\{r(m): m \\in P_{i}\\right\\}$ is a subgrid, and that for each axis there exists a subgrid orthogonal to this axis. Obviously, these subgrids cover $N$, and the condition (ii') follows directly from (ii). Hence the Lemma provides exactly the estimate we need.\nConsider some $1 \\leq j \\leq s$ and let $d_{j}=p_{1}^{\\gamma_{1}} \\ldots p_{k}^{\\gamma_{k}}$. Consider some $q \\in P_{j}$ and let $r(q)= \\left(r_{11}, \\ldots, r_{k\\alpha_{k}}\\right)$. Then for an arbitrary $q'$, setting $r\\left(q'\\right)=\\left(r_{11}', \\ldots, r_{k\\alpha_{k}}'\\right)$ we have\n$$\nq' \\in P_{j} \\quad \\Longleftrightarrow \\quad p_{i}^{\\gamma_{i}} \\mid q-q' \\text{ for each } 1 \\leq i \\leq k \\quad \\Longleftrightarrow \\quad r_{i, t}=r_{i, t}' \\text{ for all } t \\leq \\gamma_{i} .\n$$\nHence $L_{j}=\\left\\{\\left(r_{11}', \\ldots, r_{k\\alpha_{k}}'\\right) \\in N: r_{i, t}=r_{i, t}'\\right.$ for all $\\left.t \\leq \\gamma_{i}\\right\\}$ which means that $L_{j}$ is a subgrid containing $r(q)$. Moreover, in $L_{j(i)}$, all the coordinates corresponding to $p_{i}$ are fixed, so it is orthogonal to all of their axes, as desired.\n\n\nComment 1. The estimate in the problem is sharp for every $n$. One of the possible examples is the following one. For each $1 \\leq i \\leq k, 0 \\leq j \\leq \\alpha_{i}-1,1 \\leq k \\leq p-1$, let\n$$\nP_{i, j, k}=k p_{i}^{j}+p_{i}^{j+1} \\mathbb{Z}\n$$\nand add the progression $P_{0}=n \\mathbb{Z}$. One can easily check that this set satisfies all the problem conditions. There also exist other examples.\nOn the other hand, the estimate can be adjusted in the following sense. For every $1 \\leq i \\leq k$, let $0=\\alpha_{i0}, \\alpha_{i1}, \\ldots, \\alpha_{ih_{i}}$ be all the numbers of the form $\\operatorname{ord}_{p_{i}}\\left(d_{j}\\right)$ in an increasing order (we delete the repeating occurences of a number, and add a number $0=\\alpha_{i0}$ if it does not occur). Then, repeating the arguments from the solution one can obtain that\n$$\ns \\geq 1+\\sum_{i=1}^{k} \\sum_{j=1}^{h_{i}}\\left(p^{\\alpha_{j}-\\alpha_{j-1}}-1\\right)\n$$\nNote that $p^{\\alpha}-1 \\geq \\alpha(p-1)$, and the equality is achieved only for $\\alpha=1$. Hence, for reaching the minimal number of the progressions, one should have $\\alpha_{i, j}=j$ for all $i, j$. In other words, for each $1 \\leq j \\leq \\alpha_{i}$, there should be an index $t$ such that $\\operatorname{ord}_{p_{i}}\\left(d_{t}\\right)=j$.\nWe start with introducing some notation. For positive integer $r$, we denote $[r]=\\{1,2, \\ldots, r\\}$. Next, we say that a set of progressions $\\mathcal{P}=\\{P_{1}, \\ldots, P_{s}\\}$ cover $\\mathbb{Z}$ if each integer belongs to some of them; we say that this covering is minimal if no proper subset of $\\mathcal{P}$ covers $\\mathbb{Z}$. Obviously, each covering contains a minimal subcovering.\n\nNext, for a minimal covering $\\{P_{1}, \\ldots, P_{s}\\}$ and for every $1 \\leq i \\leq s$, let $d_{i}$ be the step of progression $P_{i}$, and $h_{i}$ be some number which is contained in $P_{i}$ but in none of the other progressions. We assume that $n>1$, otherwise the problem is trivial. This implies $d_{i}>1$, otherwise the progression $P_{i}$ covers all the numbers, and $n=1$.\n\nWe will prove a more general statement, namely the following\n\nClaim. Assume that the progressions $P_{1}, \\ldots, P_{s}$ and number $n=p_{1}^{\\alpha_{1}} \\ldots p_{k}^{\\alpha_{k}}>1$ are chosen as in the problem statement. Moreover, choose some nonempty set of indices $I=\\{i_{1}, \\ldots, i_{t}\\} \\subseteq[k]$ and some positive integer $\\beta_{i} \\leq \\alpha_{i}$ for every $i \\in I$. Consider the set of indices\n$$\nT=\\{j: 1 \\leq j \\leq s, \\text{ and } p_{i}^{\\alpha_{i}-\\beta_{i}+1} \\mid d_{j} \\text{ for some } i \\in I\\}\n$$\nThen\n$$\n\\begin{equation*}\n|T| \\geq 1+\\sum_{i \\in I} \\beta_{i}\\left(p_{i}-1\\right) \\tag{2}\n\\end{equation*}\n$$\nObserve that the Claim for $I=[k]$ and $\\beta_{i}=\\alpha_{i}$ implies the problem statement, since the left-hand side in (2) is not greater than $s$. Hence, it suffices to prove the Claim.\n\n1. First, we prove the Claim assuming that all $d_{j}$'s are prime numbers. If for some $1 \\leq i \\leq k$ we have at least $p_{i}$ progressions with the step $p_{i}$, then they do not intersect and hence cover all the integers; it means that there are no other progressions, and $n=p_{i}$; the Claim is trivial in this case.\nNow assume that for every $1 \\leq i \\leq k$, there are not more than $p_{i}-1$ progressions with step $p_{i}$; each such progression covers the numbers with a fixed residue modulo $p_{i}$, therefore there exists a residue $q_{i} \\bmod p_{i}$ which is not touched by these progressions. By the Chinese Remainder Theorem, there exists a number $q$ such that $q \\equiv q_{i}\\left(\\bmod p_{i}\\right)$ for all $1 \\leq i \\leq k$; this number cannot be covered by any progression with step $p_{i}$, hence it is not covered at all. A contradiction.\n\n2. Now, we assume that the general Claim is not valid, and hence we consider a counterexample $\\{P_{1}, \\ldots, P_{s}\\}$ for the Claim; we can choose it to be minimal in the following sense:\n- the number $n$ is minimal possible among all the counterexamples;\n- the sum $\\sum_{i} d_{i}$ is minimal possible among all the counterexamples having the chosen value of $n$.\nAs was mentioned above, not all numbers $d_{i}$ are primes; hence we can assume that $d_{1}$ is composite, say $p_{1} \\mid d_{1}$ and $d_{1}'=\\frac{d_{1}}{p_{1}}>1$. Consider a progression $P_{1}'$ having the step $d_{1}'$, and containing $P_{1}$. We will focus on two coverings constructed as follows.\n(i) Surely, the progressions $P_{1}', P_{2}, \\ldots, P_{s}$ cover $\\mathbb{Z}$, though this covering in not necessarily minimal. So, choose some minimal subcovering $\\mathcal{P}'$ in it; surely $P_{1}' \\in \\mathcal{P}'$ since $h_{1}$ is not covered by $P_{2}, \\ldots, P_{s}$, so we may assume that $\\mathcal{P}'=\\{P_{1}', P_{2}, \\ldots, P_{s'}\\}$ for some $s' \\leq s$. Furthermore, the period of the covering $\\mathcal{P}'$ can appear to be less than $n$; so we denote this period by\n$$\nn'=p_{1}^{\\alpha_{1}-\\sigma_{1}} \\ldots p_{k}^{\\alpha_{k}-\\sigma_{k}}=\\text{ l.c.m. }\\left(d_{1}', d_{2}, \\ldots, d_{s'}\\right)\n$$\nObserve that for each $P_{j} \\notin \\mathcal{P}'$, we have $h_{j} \\in P_{1}'$, otherwise $h_{j}$ would not be covered by $\\mathcal{P}$.\n(ii) On the other hand, each nonempty set of the form $R_{i}=P_{i} \\cap P_{1}'(1 \\leq i \\leq s)$ is also a progression with a step $r_{i}=$ l.c.m. $\\left(d_{i}, d_{1}'\\right)$, and such sets cover $P_{1}'$. Scaling these progressions with the ratio $1 / d_{1}'$, we obtain the progressions $Q_{i}$ with steps $q_{i}=r_{i} / d_{1}'$ which cover $\\mathbb{Z}$. Now we choose a minimal subcovering $\\mathcal{Q}$ of this covering; again we should have $Q_{1} \\in \\mathcal{Q}$ by the reasons of $h_{1}$. Now, denote the period of $\\mathcal{Q}$ by\n$$\nn''=\\text{ l.c.m. }\\{q_{i}: Q_{i} \\in \\mathcal{Q}\\}=\\frac{\\text{ l.c.m. }\\{r_{i}: Q_{i} \\in \\mathcal{Q}\\}}{d_{1}'}=\\frac{p_{1}^{\\gamma_{1}} \\ldots p_{k}^{\\gamma_{k}}}{d_{1}'}\n$$\nNote that if $h_{j} \\in P_{1}'$, then the image of $h_{j}$ under the scaling can be covered by $Q_{j}$ only; so, in this case we have $Q_{j} \\in \\mathcal{Q}$.\nOur aim is to find the desired number of progressions in coverings $\\mathcal{P}$ and $\\mathcal{Q}$. First, we have $n \\geq n'$, and the sum of the steps in $\\mathcal{P}'$ is less than that in $\\mathcal{P}$; hence the Claim is valid for $\\mathcal{P}'$. We apply it to the set of indices $I'=\\{i \\in I: \\beta_{i}>\\sigma_{i}\\}$ and the exponents $\\beta_{i}'=\\beta_{i}-\\sigma_{i}$; hence the set under consideration is\n$$\nT'=\\{j: 1 \\leq j \\leq s', \\text{ and } p_{i}^{(\\alpha_{i}-\\sigma_{i})-\\beta_{i}'+1}=p_{i}^{\\alpha_{i}-\\beta_{i}+1} \\mid d_{j} \\text{ for some } i \\in I'\\} \\subseteq T \\cap[s']\n$$\nand we obtain that\n$$\n|T \\cap[s']| \\geq|T'| \\geq 1+\\sum_{i \\in I'}(\\beta_{i}-\\sigma_{i})(p_{i}-1)=1+\\sum_{i \\in I}(\\beta_{i}-\\sigma_{i})_{+}(p_{i}-1),\n$$\nwhere $(x)_{+}=\\max \\{x, 0\\}$; the latter equality holds as for $i \\notin I'$ we have $\\beta_{i} \\leq \\sigma_{i}$.\nObserve that $x=(x-y)_{+}+\\min \\{x, y\\}$ for all $x, y$. So, if we find at least\n$$\nG=\\sum_{i \\in I} \\min \\{\\beta_{i}, \\sigma_{i}\\}(p_{i}-1)\n$$\nindices in $T \\cap\\{s'+1, \\ldots, s\\}$, then we would have\n$$\n|T|=|T \\cap[s']|+|T \\cap\\{s'+1, \\ldots, s\\}| \\geq 1+\\sum_{i \\in I}\\left((\\beta_{i}-\\sigma_{i})_{+}+\\min \\{\\beta_{i}, \\sigma_{i}\\}\\right)(p_{i}-1)=1+\\sum_{i \\in I} \\beta_{i}(p_{i}-1),\n$$\nthus leading to a contradiction with the choice of $\\mathcal{P}$. We will find those indices among the indices of progressions in $\\mathcal{Q}$.\n\n3. Now denote $I''=\\{i \\in I: \\sigma_{i}>0\\}$ and consider some $i \\in I''$; then $p_{i}^{\\alpha_{i}} \\nmid n'$. On the other hand, there exists an index $j(i)$ such that $p_{i}^{\\alpha_{i}} \\mid d_{j(i)}$; this means that $d_{j(i)} \\nmid n'$ and hence $P_{j(i)}$ cannot appear in $\\mathcal{P}'$, so $j(i)>s'$. Moreover, we have observed before that in this case $h_{j(i)} \\in P_{1}'$, hence $Q_{j(i)} \\in \\mathcal{Q}$. This means that $q_{j(i)} \\mid n''$, therefore $\\gamma_{i}=\\alpha_{i}$ for each $i \\in I''$ (recall here that $q_{i}=r_{i} / d_{1}'$ and hence $d_{j(i)}|r_{j(i)}| d_{1}' n''$).\nLet $d_{1}'=p_{1}^{\\tau_{1}} \\ldots p_{k}^{\\tau_{k}}$. Then $n''=p_{1}^{\\gamma_{1}-\\tau_{1}} \\ldots p_{k}^{\\gamma_{i}-\\tau_{i}}$. Now, if $i \\in I''$, then for every $\\beta$ the condition $p_{i}^{(\\gamma_{i}-\\tau_{i})-\\beta+1} \\mid q_{j}$ is equivalent to $p_{i}^{\\alpha_{i}-\\beta+1} \\mid r_{j}$.\nNote that $n'' \\leq n / d_{1}'<n$, hence we can apply the Claim to the covering $\\mathcal{Q}$. We perform this with the set of indices $I''$ and the exponents $\\beta_{i}''=\\min \\{\\beta_{i}, \\sigma_{i}\\}>0$. So, the set under consideration is\n$$\n\\begin{aligned}\nT'' & =\\{j: Q_{j} \\in \\mathcal{Q}, \\text{ and } p_{i}^{(\\gamma_{i}-\\tau_{i})-\\min \\{\\beta_{i}, \\sigma_{i}\\}+1} \\mid q_{j} \\text{ for some } i \\in I''\\} \\\\\n& =\\{j: Q_{j} \\in \\mathcal{Q}, \\text{ and } p_{i}^{\\alpha_{i}-\\min \\{\\beta_{i}, \\sigma_{i}\\}+1} \\mid r_{j} \\text{ for some } i \\in I''\\}\n\\end{aligned}\n$$\nand we obtain $|T''| \\geq 1+G$. Finally, we claim that $T'' \\subseteq T \\cap(\\{1\\} \\cup\\{s'+1, \\ldots, s\\})$; then we will obtain $|T \\cap\\{s'+1, \\ldots, s\\}| \\geq G$, which is exactly what we need.\nTo prove this, consider any $j \\in T''$. Observe first that $\\alpha_{i}-\\min \\{\\beta_{i}, \\sigma_{i}\\}+1>\\alpha_{i}-\\sigma_{i} \\geq \\tau_{i}$, hence from $p_{i}^{\\alpha_{i}-\\min \\{\\beta_{i}, \\sigma_{i}\\}+1} \\mid r_{j}=$ l.c.m. $\\left(d_{1}', d_{j}\\right)$ we have $p_{i}^{\\alpha_{i}-\\min \\{\\beta_{i}, \\sigma_{i}\\}+1} \\mid d_{j}$, which means that $j \\in T$. Next, the exponent of $p_{i}$ in $d_{j}$ is greater than that in $n'$, which means that $P_{j}$ \\notin $\\mathcal{P}'$. This may appear only if $j=1$ or $j>s'$, as desired. This completes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31927,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_{1} A_{2} \\ldots A_{n}$ be a convex polygon. Point $P$ inside this polygon is chosen so that its projections $P_{1}, \\ldots, P_{n}$ onto lines $A_{1} A_{2}, \\ldots, A_{n} A_{1}$ respectively lie on the sides of the polygon. Prove that for arbitrary points $X_{1}, \\ldots, X_{n}$ on sides $A_{1} A_{2}, \\ldots, A_{n} A_{1}$ respectively,\n$$\n\\max \\left\\{\\frac{X_{1} X_{2}}{P_{1} P_{2}}, \\ldots, \\frac{X_{n} X_{1}}{P_{n} P_{1}}\\right\\} \\geq 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote $P_{n+1}=P_{1}, X_{n+1}=X_{1}, A_{n+1}=A_{1}$.\n\nLemma. Let point $Q$ lies inside $A_{1} A_{2} \\ldots A_{n}$. Then it is contained in at least one of the circumcircles of triangles $X_{1} A_{2} X_{2}, \\ldots, X_{n} A_{1} X_{1}$.\n\nProof. If $Q$ lies in one of the triangles $X_{1} A_{2} X_{2}, \\ldots, X_{n} A_{1} X_{1}$, the claim is obvious. Otherwise $Q$ lies inside the polygon $X_{1} X_{2} \\ldots X_{n}$ (see Fig. 1). Then we have\n$$\n\\begin{aligned}\n& \\left(\\angle X_{1} A_{2} X_{2}+\\angle X_{1} Q X_{2}\\right)+\\cdots+\\left(\\angle X_{n} A_{1} X_{1}+\\angle X_{n} Q X_{1}\\right) \\\\\n& \\quad=\\left(\\angle X_{1} A_{1} X_{2}+\\cdots+\\angle X_{n} A_{1} X_{1}\\right)+\\left(\\angle X_{1} Q X_{2}+\\cdots+\\angle X_{n} Q X_{1}\\right)=(n-2) \\pi+2 \\pi=n \\pi\n\\end{aligned}\n$$\nhence there exists an index $i$ such that $\\angle X_{i} A_{i+1} X_{i+1}+\\angle X_{i} Q X_{i+1} \\geq \\frac{\\pi n}{n}=\\pi$. Since the quadrilateral $Q X_{i} A_{i+1} X_{i+1}$ is convex, this means exactly that $Q$ is contained the circumcircle of $\\triangle X_{i} A_{i+1} X_{i+1}$, as desired. $\\square$\n\nNow we turn to the solution. Applying lemma, we get that $P$ lies inside the circumcircle of triangle $X_{i} A_{i+1} X_{i+1}$ for some $i$. Consider the circumcircles $\\omega$ and $\\Omega$ of triangles $P_{i} A_{i+1} P_{i+1}$ and $X_{i} A_{i+1} X_{i+1}$ respectively (see Fig. 2); let $r$ and $R$ be their radii. Then we get $2 r=A_{i+1} P \\leq 2 R$ (since $P$ lies inside $\\Omega$ ), hence\n$$\nP_{i} P_{i+1}=2 r \\sin \\angle P_{i} A_{i+1} P_{i+1} \\leq 2 R \\sin \\angle X_{i} A_{i+1} X_{i+1}=X_{i} X_{i+1},\n$$\nQED.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\nAs in Solution 1, we assume that all indices of points are considered modulo $n$.\n\nWe will prove a bit stronger inequality, namely\n$$\n\\max \\left\\{\\frac{X_{1} X_{2}}{P_{1} P_{2}} \\cos \\alpha_{1}, \\ldots, \\frac{X_{n} X_{1}}{P_{n} P_{1}} \\cos \\alpha_{n}\\right\\} \\geq 1,\n$$\nwhere $\\alpha_{i}(1 \\leq i \\leq n)$ is the angle between lines $X_{i} X_{i+1}$ and $P_{i} P_{i+1}$. We denote $\\beta_{i}=\\angle A_{i} P_{i} P_{i-1}$ and $\\gamma_{i}=\\angle A_{i+1} P_{i} P_{i+1}$ for all $1 \\leq i \\leq n$.\n\nSuppose that for some $1 \\leq i \\leq n$, point $X_{i}$ lies on the segment $A_{i} P_{i}$, while point $X_{i+1}$ lies on the segment $P_{i+1} A_{i+2}$. Then the projection of the segment $X_{i} X_{i+1}$ onto the line $P_{i} P_{i+1}$ contains segment $P_{i} P_{i+1}$, since $\\gamma_{i}$ and $\\beta_{i+1}$ are acute angles (see Fig. 3). Therefore, $X_{i} X_{i+1} \\cos \\alpha_{i} \\geq P_{i} P_{i+1}$, and in this case the statement is proved.\n\nSo, the only case left is when point $X_{i}$ lies on segment $P_{i} A_{i+1}$ for all $1 \\leq i \\leq n$ (the case when each $X_{i}$ lies on segment $A_{i} P_{i}$ is completely analogous).\n\nNow, assume to the contrary that the inequality\n$$\n\\begin{equation*}\nX_{i} X_{i+1} \\cos \\alpha_{i}<P_{i} P_{i+1} \\tag{1}\n\\end{equation*}\n$$\nholds for every $1 \\leq i \\leq n$. Let $Y_{i}$ and $Y_{i+1}^{\\prime}$ be the projections of $X_{i}$ and $X_{i+1}$ onto $P_{i} P_{i+1}$. Then inequality (1) means exactly that $Y_{i} Y_{i+1}^{\\prime}<P_{i} P_{i+1}$, or $P_{i} Y_{i}>P_{i+1} Y_{i+1}^{\\prime}$ (again since $\\gamma_{i}$ and $\\beta_{i+1}$ are acute; see Fig. 4). Hence, we have\n$$\nX_{i} P_{i} \\cos \\gamma_{i}>X_{i+1} P_{i+1} \\cos \\beta_{i+1}, \\quad 1 \\leq i \\leq n .\n$$\nMultiplying these inequalities, we get\n$$\n\\begin{equation*}\n\\cos \\gamma_{1} \\cos \\gamma_{2} \\cdots \\cos \\gamma_{n}>\\cos \\beta_{1} \\cos \\beta_{2} \\cdots \\cos \\beta_{n} . \\tag{2}\n\\end{equation*}\n$$\nOn the other hand, the sines theorem applied to triangle $P P_{i} P_{i+1}$ provides\n$$\n\\frac{P P_{i}}{P P_{i+1}}=\\frac{\\sin \\left(\\frac{\\pi}{2}-\\beta_{i+1}\\right)}{\\sin \\left(\\frac{\\pi}{2}-\\gamma_{i}\\right)}=\\frac{\\cos \\beta_{i+1}}{\\cos \\gamma_{i}} .\n$$\nMultiplying these equalities we get\n$$\n1=\\frac{\\cos \\beta_{2}}{\\cos \\gamma_{1}} \\cdot \\frac{\\cos \\beta_{3}}{\\cos \\gamma_{2}} \\cdots \\frac{\\cos \\beta_{1}}{\\cos \\gamma_{n}}\n$$\nwhich contradicts (2).\n\n![](attached_image_3.png)\nFig. 3\n![](attached_image_4.png)\nFig. 4",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31928,
    "subject": "Mathematics (Multi-modal)",
    "question": "The vertices $X$, $Y$, $Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC$, $CA$, $AB$ of an acute-angled triangle $ABC$. Prove that the incenter of triangle $ABC$ lies inside triangle $XYZ$.\n\nThe vertices $X$, $Y$, $Z$ of an equilateral triangle $XYZ$ lie respectively on the sides $BC$, $CA$, $AB$ of a triangle $ABC$. Prove that if the incenter of triangle $ABC$ lies outside triangle $XYZ$, then one of the angles of triangle $ABC$ is greater than $120^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove a stronger fact; namely, we will show that the incenter $I$ of triangle $ABC$ lies inside the incircle of triangle $XYZ$ (and hence surely inside triangle $XYZ$ itself). We denote by $d(U, VW)$ the distance between point $U$ and line $VW$.\n\nDenote by $O$ the incenter of $\\triangle XYZ$ and by $r$, $r'$, and $R'$ the inradii of triangles $ABC$, $XYZ$ and the circumradius of $XYZ$, respectively. Then we have $R' = 2r'$, and the desired inequality is $OI \\leq r'$. We assume that $O \\neq I$; otherwise the claim is trivial.\n\nLet the incircle of $\\triangle ABC$ touch its sides $BC$, $AC$, $AB$ at points $A_1$, $B_1$, $C_1$ respectively. The lines $IA_1$, $IB_1$, $IC_1$ cut the plane into 6 acute angles, each one containing one of the points $A_1$, $B_1$, $C_1$ on its border. We may assume that $O$ lies in an angle defined by lines $IA_1$, $IC_1$ and containing point $C_1$ (see Fig. 1). Let $A'$ and $C'$ be the projections of $O$ onto lines $IA_1$ and $IC_1$, respectively.\n\nSince $OX = R'$, we have $d(O, BC) \\leq R'$. Since $OA' \\parallel BC$, it follows that $d(A', BC) = A'I + r \\leq R'$, or $A'I \\leq R' - r$. On the other hand, the incircle of $\\triangle XYZ$ lies inside $\\triangle ABC$, hence $d(O, AB) \\geq r'$, and analogously we get $d(O, AB) = C'C_1 = r - IC' \\geq r'$, or $IC' \\leq r - r'$.\n\n![](attached_image_1.png)\nFig. 1\n![](attached_image_2.png)\nFig. 2\n\nFinally, the quadrilateral $IA'O C'$ is circumscribed due to the right angles at $A'$ and $C'$ (see Fig. 2). On its circumcircle, we have $\\widehat{A'O C'} = 2 \\angle A'I C' < 180^{\\circ} = \\widehat{O C'I}$, hence $180^{\\circ} \\geq \\widetilde{IC'} > \\widetilde{A'O}$. This means that $IC' > A'O$. Finally, we have $OI \\leq IA' + A'O \\leq IA' + IC' \\leq (R' - r) + (r - r') = R' - r' = r'$, as desired.\nAssume the contrary. Then the incenter $I$ should lie in one of triangles $AYZ$, $BXZ$, $CXY$ - assume that it lies in $\\triangle AYZ$. Let the incircle $\\omega$ of $\\triangle ABC$ touch sides $BC$, $AC$ at point $A_1$, $B_1$ respectively. Without loss of generality, assume that point $A_1$ lies on segment $CX$. In this case we will show that $\\angle C > 90^{\\circ}$ thus leading to a contradiction.\n\nNote that $\\omega$ intersects each of the segments $XY$ and $YZ$ at two points; let $U$, $U'$, and $V$, $V'$ be the points of intersection of $\\omega$ with $XY$ and $YZ$, respectively ($UY > U'Y$, $VY > V'Y$; see Figs. 3 and 4). Note that $60^{\\circ} = \\angle XYZ = \\frac{1}{2}(\\widehat{UV} - \\widehat{U'V'}) \\leq \\frac{1}{2} \\widehat{UV}$, hence $\\widehat{UV} \\geq 120^{\\circ}$.\n\nOn the other hand, since $I$ lies in $\\triangle AYZ$, we get $\\widehat{UU'} < 180^{\\circ}$, hence $\\widehat{UA_1U'} \\leq \\widehat{UA_1V'} < 180^{\\circ} - \\widehat{UV} \\leq 60^{\\circ}$.\n\nNow, two cases are possible due to the order of points $Y$, $B_1$ on segment $AC$.\n\n![](attached_image_3.png)\nFig. 3\n![](attached_image_4.png)\nFig. 4\n\nCase 1. Let point $Y$ lie on the segment $AB_1$ (see Fig. 3). Then we have $\\angle YXC = \\frac{1}{2}(\\widehat{A_1U'} - \\widehat{A_1U}) \\leq \\frac{1}{2} \\widehat{A_1U'} < 30^{\\circ}$; analogously, we get $\\angle XYC \\leq \\frac{1}{2} \\widehat{A_1U'} < 30^{\\circ}$. Therefore, $\\angle YCX = 180^{\\circ} - \\angle YXC - \\angle XYC > 120^{\\circ}$, as desired.\n\nCase 2. Now let point $Y$ lie on the segment $CB_1$ (see Fig. 4). Analogously, we obtain $\\angle YXC < 30^{\\circ}$. Next, $\\angle IYX > \\angle ZYX = 60^{\\circ}$, but $\\angle IYX < \\angle IYB_1$, since $YB_1$ is a tangent and $YX$ is a secant line to circle $\\omega$ from point $Y$. Hence, we get $120^{\\circ} < \\angle IYB_1 + \\angle IYX = \\angle B_1YX = \\angle YXC + \\angle YCX < 30^{\\circ} + \\angle YCX$, hence $\\angle YCX > 120^{\\circ} - 30^{\\circ} = 90^{\\circ}$, as desired.\nAssume the contrary. As in Solution 2, we assume that the incenter $I$ of $\\triangle ABC$ lies in $\\triangle AYZ$, and the tangency point $A_1$ of $\\omega$ and $BC$ lies on segment $CX$. Surely, $\\angle YZA \\leq 180^{\\circ} - \\angle YZX = 120^{\\circ}$, hence points $I$ and $Y$ lie on one side of the perpendicular bisector to $XY$; therefore $IX > IY$. Moreover, $\\omega$ intersects segment $XY$ at two points, and therefore the projection $M$ of $I$ onto $XY$ lies on the segment $XY$. In this case, we will prove that $\\angle C > 120^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31929,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the smallest number $n$ such that there exist polynomials $f_{1}, f_{2}, \\ldots, f_{n}$ with rational coefficients satisfying\n$$\nx^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+\\cdots+f_{n}(x)^{2}\n$$",
    "options": [],
    "answer": "5",
    "solution": "The equality $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$ shows that $n \\leq 5$. It remains to show that $x^{2}+7$ is not a sum of four (or less) squares of polynomials with rational coefficients.\nSuppose by way of contradiction that $x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+f_{3}(x)^{2}+f_{4}(x)^{2}$, where the coefficients of polynomials $f_{1}, f_{2}, f_{3}$ and $f_{4}$ are rational (some of these polynomials may be zero).\nClearly, the degrees of $f_{1}, f_{2}, f_{3}$ and $f_{4}$ are at most $1$. Thus $f_{i}(x)=a_{i} x+b_{i}$ for $i=1,2,3,4$ and some rationals $a_{1}, b_{1}, a_{2}, b_{2}, a_{3}, b_{3}, a_{4}, b_{4}$. It follows that $x^{2}+7=\\sum_{i=1}^{4}\\left(a_{i} x+b_{i}\\right)^{2}$ and hence\n$$\n\\begin{equation*}\n\\sum_{i=1}^{4} a_{i}^{2}=1, \\quad \\sum_{i=1}^{4} a_{i} b_{i}=0, \\quad \\sum_{i=1}^{4} b_{i}^{2}=7 . \\tag{1}\n\\end{equation*}\n$$\nLet $p_{i}=a_{i}+b_{i}$ and $q_{i}=a_{i}-b_{i}$ for $i=1,2,3,4$. Then\n$$\n\\begin{aligned}\n\\sum_{i=1}^{4} p_{i}^{2} & =\\sum_{i=1}^{4} a_{i}^{2}+2 \\sum_{i=1}^{4} a_{i} b_{i}+\\sum_{i=1}^{4} b_{i}^{2}=8, \\\\\n\\sum_{i=1}^{4} q_{i}^{2} & =\\sum_{i=1}^{4} a_{i}^{2}-2 \\sum_{i=1}^{4} a_{i} b_{i}+\\sum_{i=1}^{4} b_{i}^{2}=8 \\\\\n\\text{ and } \\sum_{i=1}^{4} p_{i} q_{i} & =\\sum_{i=1}^{4} a_{i}^{2}-\\sum_{i=1}^{4} b_{i}^{2}=-6,\n\\end{aligned}\n$$\nwhich means that there exist a solution in integers $x_{1}, y_{1}, x_{2}, y_{2}, x_{3}, y_{3}, x_{4}, y_{4}$ and $m>0$ of the system of equations\n$$\n\\text{ (i) } \\sum_{i=1}^{4} x_{i}^{2}=8 m^{2}, \\quad \\text{ (ii) } \\sum_{i=1}^{4} y_{i}^{2}=8 m^{2}, \\quad \\text{ (iii) } \\sum_{i=1}^{4} x_{i} y_{i}=-6 m^{2}\n$$\nWe will show that such a solution does not exist.\nAssume the contrary and consider a solution with minimal $m$. Note that if an integer $x$ is odd then $x^{2} \\equiv 1(\\bmod 8)$. Otherwise (i.e., if $x$ is even) we have $x^{2} \\equiv 0(\\bmod 8)$ or $x^{2} \\equiv 4 (\\bmod 8)$. Hence, by (i), we get that $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are even. Similarly, by (ii), we get that $y_{1}, y_{2}, y_{3}$ and $y_{4}$ are even. Thus the LHS of (iii) is divisible by $4$ and $m$ is also even. It follows that $\\left(\\frac{x_{1}}{2}, \\frac{y_{1}}{2}, \\frac{x_{2}}{2}, \\frac{y_{2}}{2}, \\frac{x_{3}}{2}, \\frac{y_{3}}{2}, \\frac{x_{4}}{2}, \\frac{y_{4}}{2}, \\frac{m}{2}\\right)$ is a solution of the system of equations (i), (ii) and (iii), which contradicts the minimality of $m$.\n\n\nSolution 2:\n\nWe prove that $n \\leq 4$ is impossible. Define the numbers $a_{i}, b_{i}$ for $i=1,2,3,4$ as in the previous solution.\nBy Euler's identity we have\n$$\n\\begin{aligned}\n\\left(a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}\\right)\\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}+b_{4}^{2}\\right) & =\\left(a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}+a_{4} b_{4}\\right)^{2}+\\left(a_{1} b_{2}-a_{2} b_{1}+a_{3} b_{4}-a_{4} b_{3}\\right)^{2} \\\\\n& +\\left(a_{1} b_{3}-a_{3} b_{1}+a_{4} b_{2}-a_{2} b_{4}\\right)^{2}+\\left(a_{1} b_{4}-a_{4} b_{1}+a_{2} b_{3}-a_{3} b_{2}\\right)^{2}\n\\end{aligned}\n$$\nSo, using the relations (1) from the Solution 1 we get that\n$$\n\\begin{equation*}\n7=\\left(\\frac{m_{1}}{m}\\right)^{2}+\\left(\\frac{m_{2}}{m}\\right)^{2}+\\left(\\frac{m_{3}}{m}\\right)^{2} \\tag{2}\n\\end{equation*}\n$$\nwhere\n$$\n\\begin{aligned}\n& \\frac{m_{1}}{m}=a_{1} b_{2}-a_{2} b_{1}+a_{3} b_{4}-a_{4} b_{3} \\\\\n& \\frac{m_{2}}{m}=a_{1} b_{3}-a_{3} b_{1}+a_{4} b_{2}-a_{2} b_{4} \\\\\n& \\frac{m_{3}}{m}=a_{1} b_{4}-a_{4} b_{1}+a_{2} b_{3}-a_{3} b_{2}\n\\end{aligned}\n$$\nand $m_{1}, m_{2}, m_{3} \\in \\mathbb{Z}, m \\in \\mathbb{N}$.\nLet $m$ be a minimum positive integer number for which (2) holds. Then\n$$\n8 m^{2}=m_{1}^{2}+m_{2}^{2}+m_{3}^{2}+m^{2}\n$$\nAs in the previous solution, we get that $m_{1}, m_{2}, m_{3}, m$ are all even numbers. Then $\\left(\\frac{m_{1}}{2}, \\frac{m_{2}}{2}, \\frac{m_{3}}{2}, \\frac{m}{2}\\right)$ is also a solution of (2) which contradicts the minimality of $m$. So, we have $n \\geq 5$. The example with $n=5$ is already shown in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31930,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b$ be integers, and let $P(x) = a x^{3} + b x$. For any positive integer $n$ we say that the pair $(a, b)$ is $n$-good if $n \\mid P(m) - P(k)$ implies $n \\mid m - k$ for all integers $m, k$. We say that $(a, b)$ is very good if $(a, b)$ is $n$-good for infinitely many positive integers $n$.\n\na. Find a pair $(a, b)$ which is $51$-good, but not very good.\n\nb. Show that all $2010$-good pairs are very good.\n\n(Turkey)",
    "options": [],
    "answer": "a: (1, -51^2). b: Every 2010-good pair is very good (indeed it is 67^i-good for all positive integers i).",
    "solution": "a. We show that the pair $(1, -51^{2})$ is $51$-good but not very good. Let $P(x) = x^{3} - 51^{2} x$. Since $P(51) = P(0)$, the pair $(1, -51^{2})$ is not $n$-good for any positive integer that does not divide $51$. Therefore, $(1, -51^{2})$ is not very good.\n\nOn the other hand, if $P(m) \\equiv P(k) \\pmod{51}$, then $m^{3} \\equiv k^{3} \\pmod{51}$. By Fermat's theorem, from this we obtain\n$$\nm \\equiv m^{3} \\equiv k^{3} \\equiv k \\quad (\\bmod 3) \\quad \\text{and} \\quad m \\equiv m^{33} \\equiv k^{33} \\equiv k \\quad (\\bmod 17).\n$$\nHence we have $m \\equiv k \\pmod{51}$. Therefore $(1, -51^{2})$ is $51$-good.\n\nb. We will show that if a pair $(a, b)$ is $2010$-good then $(a, b)$ is $67^{i}$-good for all positive integer $i$.\n\nClaim 1. If $(a, b)$ is $2010$-good then $(a, b)$ is $67$-good.\n\nProof. Assume that $P(m) \\equiv P(k) \\pmod{67}$. Since $67$ and $30$ are coprime, there exist integers $m'$ and $k'$ such that $k' \\equiv k \\pmod{67}$, $k' \\equiv 0 \\pmod{30}$, and $m' \\equiv m \\pmod{67}$, $m' \\equiv 0 \\pmod{30}$. Then we have $P(m') \\equiv P(0) \\equiv P(k') \\pmod{30}$ and $P(m') \\equiv P(m) \\equiv P(k) \\equiv P(k') \\pmod{67}$, hence $P(m') \\equiv P(k') \\pmod{2010}$. This implies $m' \\equiv k' \\pmod{2010}$ as $(a, b)$ is $2010$-good. It follows that $m \\equiv m' \\equiv k' \\equiv k \\pmod{67}$. Therefore, $(a, b)$ is $67$-good.\n\nClaim 2. If $(a, b)$ is $67$-good then $67 \\mid a$.\n\nProof. Suppose that $67 \\nmid a$. Consider the sets $\\{a t^{2} \\pmod{67}: 0 \\leq t \\leq 33\\}$ and $\\{-3 a s^{2} - b \\pmod{67}: 0 \\leq s \\leq 33\\}$. Since $a \\not\\equiv 0 \\pmod{67}$, each of these sets has $34$ elements. Hence they have at least one element in common. If $a t^{2} \\equiv -3 a s^{2} - b \\pmod{67}$ then for $m = t \\pm s$, $k = \\mp 2s$ we have\n$$\n\\begin{aligned}\nP(m) - P(k) = a(m^{3} - k^{3}) + b(m - k) & = (m - k)\\left(a(m^{2} + m k + k^{2}) + b\\right) \\\\\n& = (t \\pm 3s)\\left(a t^{2} + 3 a s^{2} + b\\right) \\equiv 0 \\quad (\\bmod 67)\n\\end{aligned}\n$$\nSince $(a, b)$ is $67$-good, we must have $m \\equiv k \\pmod{67}$ in both cases, that is, $t \\equiv 3s \\pmod{67}$ and $t \\equiv -3s \\pmod{67}$. This means $t \\equiv s \\equiv 0 \\pmod{67}$ and $b \\equiv -3 a s^{2} - a t^{2} \\equiv 0 \\pmod{67}$. But then $67 \\mid P(7) - P(2) = 67 \\cdot 5 a + 5 b$ and $67 \\nmid 7 - 2$, contradicting that $(a, b)$ is $67$-good.\n\nClaim 3. If $(a, b)$ is $2010$-good then $(a, b)$ is $67^{i}$-good for all $i \\geq 1$.\n\nProof. By Claim 2, we have $67 \\mid a$. If $67 \\mid b$, then $P(x) \\equiv P(0) \\pmod{67}$ for all $x$, contradicting that $(a, b)$ is $67$-good. Hence, $67 \\nmid b$.\n\nSuppose that $67^{i} \\mid P(m) - P(k) = (m - k)\\left(a(m^{2} + m k + k^{2}) + b\\right)$. Since $67 \\mid a$ and $67 \\nmid b$, the second factor $a(m^{2} + m k + k^{2}) + b$ is coprime to $67$ and hence $67^{i} \\mid m - k$. Therefore, $(a, b)$ is $67^{i}$-good.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31931,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N}$ be the set of all positive integers. Find all functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that the number $(f(m)+n)(m+f(n))$ is a square for all $m, n \\in \\mathbb{N}$.\n(U.S.A.)",
    "options": [],
    "answer": "All functions of the form f(n) = n + c, where c is a fixed nonnegative integer.",
    "solution": "First, it is clear that all functions of the form $f(n)=n+c$ with a constant nonnegative integer $c$ satisfy the problem conditions since $(f(m)+n)(f(n)+m)=(n+m+c)^{2}$ is a square.\n\nWe are left to prove that there are no other functions. We start with the following Lemma. Suppose that $p \\mid f(k)-f(\\ell)$ for some prime $p$ and positive integers $k, \\ell$. Then $p \\mid k-\\ell$.\n\nProof. Suppose first that $p^{2} \\mid f(k)-f(\\ell)$, so $f(\\ell)=f(k)+p^{2} a$ for some integer $a$. Take some positive integer $D>\\max \\{f(k), f(\\ell)\\}$ which is not divisible by $p$ and set $n=p D-f(k)$. Then the positive numbers $n+f(k)=p D$ and $n+f(\\ell)=p D+(f(\\ell)-f(k))=p(D+p a)$ are both divisible by $p$ but not by $p^{2}$. Now, applying the problem conditions, we get that both the numbers $(f(k)+n)(f(n)+k)$ and $(f(\\ell)+n)(f(n)+\\ell)$ are squares divisible by $p$ (and thus by $p^{2}$ ); this means that the multipliers $f(n)+k$ and $f(n)+\\ell$ are also divisible by $p$, therefore $p \\mid(f(n)+k)-(f(n)+\\ell)=k-\\ell$ as well.\n\nOn the other hand, if $f(k)-f(\\ell)$ is divisible by $p$ but not by $p^{2}$, then choose the same number $D$ and set $n=p^{3} D-f(k)$. Then the positive numbers $f(k)+n=p^{3} D$ and $f(\\ell)+n= p^{3} D+(f(\\ell)-f(k))$ are respectively divisible by $p^{3}$ (but not by $p^{4}$ ) and by $p$ (but not by $p^{2}$ ). Hence in analogous way we obtain that the numbers $f(n)+k$ and $f(n)+\\ell$ are divisible by $p$, therefore $p \\mid(f(n)+k)-(f(n)+\\ell)=k-\\ell$.\n\nWe turn to the problem. First, suppose that $f(k)=f(\\ell)$ for some $k, \\ell \\in \\mathbb{N}$. Then by Lemma we have that $k-\\ell$ is divisible by every prime number, so $k-\\ell=0$, or $k=\\ell$. Therefore, the function $f$ is injective.\n\nNext, consider the numbers $f(k)$ and $f(k+1)$. Since the number $(k+1)-k=1$ has no prime divisors, by Lemma the same holds for $f(k+1)-f(k)$; thus $|f(k+1)-f(k)|=1$.\n\nNow, let $f(2)-f(1)=q,|q|=1$. Then we prove by induction that $f(n)=f(1)+q(n-1)$. The base for $n=1,2$ holds by the definition of $q$. For the step, if $n>1$ we have $f(n+1)= f(n) \\pm q=f(1)+q(n-1) \\pm q$. Since $f(n) \\neq f(n-2)=f(1)+q(n-2)$, we get $f(n)=f(1)+q n$, as desired.\n\nFinally, we have $f(n)=f(1)+q(n-1)$. Then $q$ cannot be -1 since otherwise for $n \\geq f(1)+1$ we have $f(n) \\leq 0$ which is impossible. Hence $q=1$ and $f(n)=(f(1)-1)+n$ for each $n \\in \\mathbb{N}$, and $f(1)-1 \\geq 0$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31932,
    "subject": "Mathematics (Multi-modal)",
    "question": "The rows and columns of a $2^{n} \\times 2^{n}$ table are numbered from $0$ to $2^{n}-1$. The cells of the table have been colored with the following property being satisfied: for each $0 \\leq i, j \\leq 2^{n}-1$, the $j$th cell in the $i$th row and the $(i+j)$th cell in the $j$th row have the same color. (The indices of the cells in a row are considered modulo $2^{n}$.)\n\nProve that the maximal possible number of colors is $2^{n}$.\n\n(Iran)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Throughout the solution we denote the cells of the table by coordinate pairs; $(i, j)$ refers to the $j$th cell in the $i$th row.\n\nConsider the directed graph, whose vertices are the cells of the board, and the edges are the arrows $(i, j) \\rightarrow (j, i+j)$ for all $0 \\leq i, j \\leq 2^{n}-1$. From each vertex $(i, j)$, exactly one edge passes (to $(j, i+j \\bmod 2^{n})$); conversely, to each cell $(j, k)$ exactly one edge is directed (from the cell $(k-j \\bmod 2^{n}, j)$). Hence, the graph splits into cycles.\n\nNow, in any coloring considered, the vertices of each cycle should have the same color by the problem condition. On the other hand, if each cycle has its own color, the obtained coloring obviously satisfies the problem conditions. Thus, the maximal possible number of colors is the same as the number of cycles, and we have to prove that this number is $2^{n}$.\n\nNext, consider any cycle $(i_{1}, j_{1}), (i_{2}, j_{2}), \\ldots$; we will describe it in other terms. Define a sequence $(a_{0}, a_{1}, \\ldots)$ by the relations $a_{0}=i_{1}, a_{1}=j_{1}, a_{n+1}=a_{n}+a_{n-1}$ for all $n \\geq 1$ (we say that such a sequence is a Fibonacci-type sequence). Then an obvious induction shows that $i_{k} \\equiv a_{k-1} \\pmod{2^{n}}, j_{k} \\equiv a_{k} \\pmod{2^{n}}$. Hence we need to investigate the behavior of Fibonacci-type sequences modulo $2^{n}$.\n\nDenote by $F_{0}, F_{1}, \\ldots$ the Fibonacci numbers defined by $F_{0}=0, F_{1}=1$, and $F_{n+2}=F_{n+1}+F_{n}$ for $n \\geq 0$. We also set $F_{-1}=1$ according to the recurrence relation.\n\nFor every positive integer $m$, denote by $\\nu(m)$ the exponent of $2$ in the prime factorization of $m$, i.e. for which $2^{\\nu(m)} \\mid m$ but $2^{\\nu(m)+1} \\nmid m$.\n\n**Lemma 1.** For every Fibonacci-type sequence $a_{0}, a_{1}, a_{2}, \\ldots$, and every $k \\geq 0$, we have $a_{k}=F_{k-1} a_{0}+F_{k} a_{1}$.\n\n*Proof.* Apply induction on $k$. The base cases $k=0,1$ are trivial. For the step, from the induction hypothesis we get\n$$\na_{k+1}=a_{k}+a_{k-1}=(F_{k-1} a_{0}+F_{k} a_{1})+(F_{k-2} a_{0}+F_{k-1} a_{1})=F_{k} a_{0}+F_{k+1} a_{1}.\n$$\n\n**Lemma 2.** For every $m \\geq 3$,\n(a) we have $\\nu(F_{3 \\cdot 2^{m-2}})=m$;\n(b) $d=3 \\cdot 2^{m-2}$ is the least positive index for which $2^{m} \\mid F_{d}$;\n(c) $F_{3 \\cdot 2^{m-2}+1} \\equiv 1+2^{m-1} \\pmod{2^{m}}$.\n\n*Proof.* Apply induction on $m$. In the base case $m=3$ we have $\\nu(F_{3 \\cdot 2^{m-2}})=F_{6}=8$, so $\\nu(F_{3 \\cdot 2^{m-2}})=\\nu(8)=3$, the preceding Fibonacci-numbers are not divisible by $8$, and indeed $F_{3 \\cdot 2^{m-2}+1}=F_{7}=13 \\equiv 1+4 \\pmod{8}$.\n\nNow suppose that $m>3$ and let $k=3 \\cdot 2^{m-3}$. By applying Lemma 1 to the Fibonacci-type sequence $F_{k}, F_{k+1}, \\ldots$ we get\n$$\n\\begin{gathered}\nF_{2k}=F_{k-1} F_{k}+F_{k} F_{k+1}=(F_{k+1}-F_{k}) F_{k}+F_{k+1} F_{k}=2 F_{k+1} F_{k}-F_{k}^{2} \\\\\nF_{2k+1}=F_{k}^2+F_{k+1}^2\n\\end{gathered}\n$$\nBy the induction hypothesis, $\\nu(F_{k})=m-1$, and $F_{k+1}$ is odd. Therefore we get $\\nu(F_{k}^{2})=2(m-1)>(m-1)+1=\\nu(2 F_{k} F_{k+1})$, which implies $\\nu(F_{2k})=m$, establishing statement (a).\n\nMoreover, since $F_{k+1}=1+2^{m-2}+a 2^{m-1}$ for some integer $a$, we get\n$$\nF_{2k+1}=F_{k}^{2}+F_{k+1}^{2} \\equiv 0+(1+2^{m-2}+a 2^{m-1})^{2} \\equiv 1+2^{m-1} \\pmod{2^{m}}\n$$\nas desired in statement (c).\n\nWe are left to prove that $2^{m} \\nmid F_{\\ell}$ for $\\ell<2k$. Assume the contrary. Since $2^{m-1} \\mid F_{\\ell}$, from the induction hypothesis it follows that $\\ell>k$. But then we have $F_{\\ell}=F_{k-1} F_{\\ell-k}+F_{k} F_{\\ell-k+1}$, where the second summand is divisible by $2^{m-1}$ but the first one is not (since $F_{k-1}$ is odd and $\\ell-k<k$). Hence the sum is not divisible even by $2^{m-1}$. A contradiction.\n\nNow, for every pair of integers $(a, b) \\neq (0,0)$, let $\\mu(a, b)=\\min\\{\\nu(a), \\nu(b)\\}$. By an obvious induction, for every Fibonacci-type sequence $A=(a_{0}, a_{1}, \\ldots)$ we have $\\mu(a_{0}, a_{1})=\\mu(a_{1}, a_{2})=\\ldots$; denote this common value by $\\mu(A)$. Also denote by $p_{n}(A)$ the period of this sequence modulo $2^{n}$, that is, the least $p>0$ such that $a_{k+p} \\equiv a_{k} \\pmod{2^{n}}$ for all $k \\geq 0$.\n\n**Lemma 3.** Let $A=(a_{0}, a_{1}, \\ldots)$ be a Fibonacci-type sequence such that $\\mu(A)=k<n$. Then $p_{n}(A)=3 \\cdot 2^{n-1-k}$.\n\n*Proof.* First, we note that the sequence $(a_{0}, a_{1}, \\ldots)$ has period $p$ modulo $2^{n}$ if and only if the sequence $(a_{0}/2^{k}, a_{1}/2^{k}, \\ldots)$ has period $p$ modulo $2^{n-k}$. Hence, passing to this sequence we can assume that $k=0$.\n\nWe prove the statement by induction on $n$. It is easy to see that for $n=1,2$ the claim is true; actually, each Fibonacci-type sequence $A$ with $\\mu(A)=0$ behaves as $0,1,1,0,1,1, \\ldots$ modulo $2$, and as $0,1,1,2,3,1,0,1,1,2,3,1, \\ldots$ modulo $4$ (all pairs of residues from which at least one is odd appear as a pair of consecutive terms in this sequence).\n\nNow suppose that $n \\geq 3$ and consider an arbitrary Fibonacci-type sequence $A=(a_{0}, a_{1}, \\ldots)$ with $\\mu(A)=0$. Obviously we should have $p_{n-1}(A) \\mid p_{n}(A)$, or, using the induction hypothesis, $s=3 \\cdot 2^{n-2} \\mid p_{n}(A)$. Next, we may suppose that $a_{0}$ is even; hence $a_{1}$ is odd, and $a_{0}=2 b_{0}$, $a_{1}=2 b_{1}+1$ for some integers $b_{0}, b_{1}$.\n\nConsider the Fibonacci-type sequence $B=(b_{0}, b_{1}, \\ldots)$ starting with $(b_{0}, b_{1})$. Since $a_{0}=2 b_{0}+F_{0}, a_{1}=2 b_{1}+F_{1}$, by an easy induction we get $a_{k}=2 b_{k}+F_{k}$ for all $k \\geq 0$. By the induction hypothesis, we have $p_{n-1}(B) \\mid s$, hence the sequence $(2 b_{0}, 2 b_{1}, \\ldots)$ is $s$-periodic modulo $2^{n}$. On the other hand, by Lemma 2 we have $F_{s+1} \\equiv 1+2^{n-1} \\pmod{2^{n}}, F_{2s} \\equiv 0 \\pmod{2^{n}}, F_{2s+1} \\equiv 1 \\pmod{2^{n}}$, hence\n$$\n\\begin{gathered}\na_{s+1}=2 b_{s+1}+F_{s+1} \\equiv 2 b_{1}+1+2^{n-1} \\not\\equiv 2 b_{1}+1=a_{1} \\pmod{2^{n}} \\\\\na_{2s}=2 b_{2s}+F_{2s} \\equiv 2 b_{0}+0=a_{0} \\pmod{2^{n}} \\\\\na_{2s+1}=2 b_{2s+1}+F_{2s+1} \\equiv 2 b_{1}+1=a_{1} \\pmod{2^{n}}\n\\end{gathered}\n$$\nThe first line means that $A$ is not $s$-periodic, while the other two provide that $a_{2s} \\equiv a_{0}$, $a_{2s+1} \\equiv a_{1}$ and hence $a_{2s+t} \\equiv a_{t}$ for all $t \\geq 0$. Hence $s \\mid p_{n}(A) \\mid 2s$ and $p_{n}(A) \\neq s$, which means that $p_{n}(A)=2s$, as desired.\n\nFinally, Lemma 3 provides a straightforward method of counting the number of cycles. Actually, take any number $0 \\leq k \\leq n-1$ and consider all the cells $(i, j)$ with $\\mu(i, j)=k$. The total number of such cells is $2^{2(n-k)}-2^{2(n-k-1)}=3 \\cdot 2^{2n-2k-2}$. On the other hand, they are split into cycles, and by Lemma 3 the length of each cycle is $3 \\cdot 2^{n-1-k}$. Hence the number of cycles consisting of these cells is exactly $\\frac{3 \\cdot 2^{2n-2k-2}}{3 \\cdot 2^{n-1-k}}=2^{n-k-1}$. Finally, there is only one cell $(0,0)$ which is not mentioned in the previous computation, and it forms a separate cycle. So the total number of cycles is\n$$\n1+\\sum_{k=0}^{n-1} 2^{n-1-k}=1+(1+2+4+\\cdots+2^{n-1})=2^{n}\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 31933,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all sequences $\\left(x_{1}, x_{2}, \\ldots, x_{2011}\\right)$ of positive integers such that for every positive integer $n$ there is an integer $a$ with\n$$\nx_{1}^{n}+2 x_{2}^{n}+\\cdots+2011 x_{2011}^{n}=a^{n+1}+1 .\n$$",
    "options": [],
    "answer": "The unique sequence is x1 = 1 and x2 = x3 = … = x2011 = 2023065.",
    "solution": "Throughout this solution, the set of positive integers will be denoted by $\\mathbb{Z}_{+}$.\nPut $k=2+3+\\cdots+2011=2023065$. We have\n$$\n1^{n}+2 k^{n}+\\cdots 2011 k^{n}=1+k \\cdot k^{n}=k^{n+1}+1\n$$\nfor all $n$, so $(1, k, \\ldots, k)$ is a valid sequence. We shall prove that it is the only one.\nLet a valid sequence $\\left(x_{1}, \\ldots, x_{2011}\\right)$ be given. For each $n \\in \\mathbb{Z}_{+}$ we have some $y_{n} \\in \\mathbb{Z}_{+}$ with\n$$\nx_{1}^{n}+2 x_{2}^{n}+\\cdots+2011 x_{2011}^{n}=y_{n}^{n+1}+1 .\n$$\nNote that $x_{1}^{n}+2 x_{2}^{n}+\\cdots+2011 x_{2011}^{n}<\\left(x_{1}+2 x_{2}+\\cdots+2011 x_{2011}\\right)^{n+1}$, which implies that the sequence $\\left(y_{n}\\right)$ is bounded. In particular, there is some $y \\in \\mathbb{Z}_{+}$ with $y_{n}=y$ for infinitely many $n$.\nLet $m$ be the maximum of all the $x_{i}$. Grouping terms with equal $x_{i}$ together, the sum $x_{1}^{n}+ 2 x_{2}^{n}+\\cdots+2011 x_{2011}^{n}$ can be written as\n$$\nx_{1}^{n}+2 x_{2}^{n}+\\cdots+x_{2011}^{n}=a_{m} m^{n}+a_{m-1}(m-1)^{n}+\\cdots+a_{1}\n$$\nwith $a_{i} \\geq 0$ for all $i$ and $a_{1}+\\cdots+a_{m}=1+2+\\cdots+2011$. So there exist arbitrarily large values of $n$, for which\n$$\n\\begin{equation*}\na_{m} m^{n}+\\cdots+a_{1}-1-y \\cdot y^{n}=0 . \\tag{1}\n\\end{equation*}\n$$\nThe following lemma will help us to determine the $a_{i}$ and $y$ :\nLemma. Let integers $b_{1}, \\ldots, b_{N}$ be given and assume that there are arbitrarily large positive integers $n$ with $b_{1}+b_{2} 2^{n}+\\cdots+b_{N} N^{n}=0$. Then $b_{i}=0$ for all $i$.\nProof. Suppose that not all $b_{i}$ are zero. We may assume without loss of generality that $b_{N} \\neq 0$.\nDividing through by $N^{n}$ gives\n$$\n\\left|b_{N}\\right|=\\left|b_{N-1}\\left(\\frac{N-1}{N}\\right)^{n}+\\cdots+b_{1}\\left(\\frac{1}{N}\\right)^{n}\\right| \\leq\\left(\\left|b_{N-1}\\right|+\\cdots+\\left|b_{1}\\right|\\right)\\left(\\frac{N-1}{N}\\right)^{n} .\n$$\nThe expression $\\left(\\frac{N-1}{N}\\right)^{n}$ can be made arbitrarily small for $n$ large enough, contradicting the assumption that $b_{N}$ be non-zero. $\\square$\nWe obviously have $y>1$. Applying the lemma to (1) we see that $a_{m}=y=m, a_{1}=1$, and all the other $a_{i}$ are zero. This implies $\\left(x_{1}, \\ldots, x_{2011}\\right)=(1, m, \\ldots, m)$. But we also have $1+m=a_{1}+\\cdots+a_{m}=1+\\cdots+2011=1+k$ so $m=k$, which is what we wanted to show.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31934,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all pairs $(f, g)$ of functions from the set of positive integers to itself that satisfy\n$$\nf^{g(n)+1}(n) + g^{f(n)}(n) = f(n+1) - g(n+1) + 1\n$$\nfor every positive integer $n$. Here, $f^{k}(n)$ means $\\underbrace{f(f(\\ldots f}_{k}(n) \\ldots))$.",
    "options": [],
    "answer": "f(n) = n for all positive integers n, and g(n) = 1 for all positive integers n",
    "solution": "The given relation implies\n$$\n\\begin{equation*}\nf\\left(f^{g(n)}(n)\\right) < f(n+1) \\quad \\text{ for all } n, \\tag{1}\n\\end{equation*}\n$$\nwhich will turn out to be sufficient to determine $f$.\nLet $y_1 < y_2 < \\ldots$ be all the values attained by $f$ (this sequence might be either finite or infinite). We will prove that for every positive $n$ the function $f$ attains at least $n$ values, and we have $(\\mathrm{i})_n$: $f(x) = y_n$ if and only if $x = n$, and $(\\mathrm{ii})_n$: $y_n = n$. The proof will follow the scheme\n$$\n\\begin{equation*}\n(\\mathrm{i})_1, (\\mathrm{ii})_1, (\\mathrm{i})_2, (\\mathrm{ii})_2, \\ldots, (\\mathrm{i})_n, (\\mathrm{ii})_n, \\ldots \\tag{2}\n\\end{equation*}\n$$\nTo start, consider any $x$ such that $f(x) = y_1$. If $x > 1$, then (1) reads $f\\left(f^{g(x-1)}(x-1)\\right) < y_1$, contradicting the minimality of $y_1$. So we have that $f(x) = y_1$ is equivalent to $x = 1$, establishing $(\\mathrm{i})_1$.\nNext, assume that for some $n$ statement $(\\mathrm{i})_n$ is established, as well as all the previous statements in (2). Note that these statements imply that for all $k \\geq 1$ and $a < n$ we have $f^{k}(x) = a$ if and only if $x = a$.\nNow, each value $y_i$ with $1 \\leq i \\leq n$ is attained at the unique integer $i$, so $y_{n+1}$ exists. Choose an arbitrary $x$ such that $f(x) = y_{n+1}$; we necessarily have $x > n$. Substituting $x-1$ into (1) we have $f\\left(f^{g(x-1)}(x-1)\\right) < y_{n+1}$, which implies\n$$\n\\begin{equation*}\nf^{g(x-1)}(x-1) \\in \\{1, \\ldots, n\\} \\tag{3}\n\\end{equation*}\n$$\nSet $b = f^{g(x-1)}(x-1)$. If $b < n$ then we would have $x-1 = b$ which contradicts $x > n$. So $b = n$, and hence $y_n = n$, which proves (ii)$_n$. Next, from (i)$_n$ we now get $f(k) = n \\Longleftrightarrow k = n$, so removing all the iterations of $f$ in (3) we obtain $x-1 = b = n$, which proves (i)$_{n+1}$.\nSo, all the statements in (2) are valid and hence $f(n) = n$ for all $n$. The given relation between $f$ and $g$ now reads $n + g^{n}(n) = n + 1 - g(n+1) + 1$ or $g^{n}(n) + g(n+1) = 2$, from which it\nimmediately follows that we have $g(n) = 1$ for all $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31935,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ be a positive integer and consider a checkerboard consisting of $m$ by $m$ unit squares. At the midpoints of some of these unit squares there is an ant. At time 0, each ant starts moving with speed 1 parallel to some edge of the checkerboard. When two ants moving in opposite directions meet, they both turn $90^{\\circ}$ clockwise and continue moving with speed 1. When more than two ants meet, or when two ants moving in perpendicular directions meet, the ants continue moving in the same direction as before they met. When an ant reaches one of the edges of the checkerboard, it falls off and will not re-appear.\n\nConsidering all possible starting positions, determine the latest possible moment at which the last ant falls off the checkerboard or prove that such a moment does not necessarily exist.",
    "options": [],
    "answer": "3m/2 - 1",
    "solution": "For $m=1$ the answer is clearly correct, so assume $m>1$. In the sequel, the word collision will be used to denote meeting of exactly two ants, moving in opposite directions.\n\nIf at the beginning we place an ant on the southwest corner square facing east and an ant on the southeast corner square facing west, then they will meet in the middle of the bottom row at time $\\frac{m-1}{2}$. After the collision, the ant that moves to the north will stay on the board for another $m-\\frac{1}{2}$ time units and thus we have established an example in which the last ant falls off at time $\\frac{m-1}{2}+m-\\frac{1}{2}=\\frac{3 m}{2}-1$. So, we are left to prove that this is the latest possible moment.\n\nConsider any collision of two ants $a$ and $a'$. Let us change the rule for this collision, and enforce these two ants to turn anticlockwise. Then the succeeding behavior of all the ants does not change; the only difference is that $a$ and $a'$ swap their positions. These arguments may be applied to any collision separately, so we may assume that at any collision, either both ants rotate clockwise or both of them rotate anticlockwise by our own choice.\n\nFor instance, we may assume that there are only two types of ants, depending on their initial direction: NE-ants, which move only north or east, and $SW$-ants, moving only south and west. Then we immediately obtain that all ants will have fallen off the board after $2 m-1$ time units. However, we can get a better bound by considering the last moment at which a given ant collides with another ant.\n\nChoose a coordinate system such that the corners of the checkerboard are $(0,0),(m, 0),(m, m)$ and $(0, m)$. At time $t$, there will be no NE-ants in the region $\\{(x, y): x+y<t+1\\}$ and no SW-ants in the region $\\{(x, y): x+y>2 m-t-1\\}$. So if two ants collide at $(x, y)$ at time $t$, we have\n$$\n\\begin{equation*}\nt+1 \\leq x+y \\leq 2 m-t-1 . \\tag{1}\n\\end{equation*}\n$$\n\nAnalogously, we may change the rules so that each ant would move either alternatingly north and west, or alternatingly south and east. By doing so, we find that apart from (1) we also have $|x-y| \\leq m-t-1$ for each collision at point $(x, y)$ and time $t$.\n\nTo visualize this, put\n$$\nB(t)=\\left\\{(x, y) \\in[0, m]^2: t+1 \\leq x+y \\leq 2 m-t-1 \\text{ and } |x-y| \\leq m-t-1\\right\\} .\n$$\n\nAn ant can thus only collide with another ant at time $t$ if it happens to be in the region $B(t)$. The following figure displays $B(t)$ for $t=\\frac{1}{2}$ and $t=\\frac{7}{2}$ in the case $m=6$ :\n\n![](attached_image_1.png)\n\nNow suppose that an NE-ant has its last collision at time $t$ and that it does so at the point $(x, y)$ (if the ant does not collide at all, it will fall off the board within $m-\\frac{1}{2}<\\frac{3 m}{2}-1$ time units, so this case can be ignored). Then we have $(x, y) \\in B(t)$ and thus $x+y \\geq t+1$ and $x-y \\geq-(m-t-1)$.\n\nSo we get\n$$\nx \\geq \\frac{(t+1)-(m-t-1)}{2}=t+1-\\frac{m}{2} .\n$$\n\nBy symmetry we also have $y \\geq t+1-\\frac{m}{2}$, and hence $\\min \\{x, y\\} \\geq t+1-\\frac{m}{2}$. After this collision, the ant will move directly to an edge, which will take at most $m-\\min \\{x, y\\}$ units of time. In sum, the total amount of time the ant stays on the board is at most\n$$\nt+(m-\\min \\{x, y\\}) \\leq t+m-\\left(t+1-\\frac{m}{2}\\right)=\\frac{3 m}{2}-1 .\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 31936,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $W = \\ldots x_{-1} x_{0} x_{1} x_{2} \\ldots$ be an infinite periodic word consisting of the letters $a$ and $b$. Suppose that the minimal period $N$ of $W$ is greater than $2^{n}$.\n\nA finite nonempty word $U$ is said to appear in $W$ if there exist indices $k \\leq \\ell$ such that $U = x_{k} x_{k+1} \\ldots x_{\\ell}$. A finite word $U$ is called ubiquitous if the four words $U a$, $U b$, $a U$, and $b U$ all appear in $W$. Prove that there are at least $n$ ubiquitous finite nonempty words.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Throughout the solution, all the words are nonempty. For any word $R$ of length $m$, we call the number of indices $i \\in \\{1,2, \\ldots, N\\}$ for which $R$ coincides with the subword $x_{i+1} x_{i+2} \\ldots x_{i+m}$ of $W$ the multiplicity of $R$ and denote it by $\\mu(R)$. Thus a word $R$ appears in $W$ if and only if $\\mu(R) > 0$. Since each occurrence of a word in $W$ is both succeeded by either the letter $a$ or the letter $b$ and similarly preceded by one of those two letters, we have\n$$\n\\begin{equation*}\n\\mu(R) = \\mu(R a) + \\mu(R b) = \\mu(a R) + \\mu(b R) \\tag{1}\n\\end{equation*}\n$$\nfor all words $R$.\n\nWe claim that the condition that $N$ is in fact the minimal period of $W$ guarantees that each word of length $N$ has multiplicity $1$ or $0$ depending on whether it appears or not. Indeed, if the words $x_{i+1} x_{i+2} \\ldots x_{i+N}$ and $x_{j+1} \\ldots x_{j+N}$ are equal for some $1 \\leq i < j \\leq N$, then we have $x_{i+a} = x_{j+a}$ for every integer $a$, and hence $j-i$ is also a period.\n\nMoreover, since $N > 2^{n}$, at least one of the two words $a$ and $b$ has a multiplicity that is strictly larger than $2^{n-1}$.\n\nFor each $k = 0, 1, \\ldots, n-1$, let $U_{k}$ be a subword of $W$ whose multiplicity is strictly larger than $2^{k}$ and whose length is maximal subject to this property. Note that such a word exists in view of the two observations made in the two previous paragraphs.\n\nFix some index $k \\in \\{0,1, \\ldots, n-1\\}$. Since the word $U_{k} b$ is longer than $U_{k}$, its multiplicity can be at most $2^{k}$, so in particular $\\mu\\left(U_{k} b\\right) < \\mu\\left(U_{k}\\right)$. Therefore, the word $U_{k} a$ has to appear by (1). For a similar reason, the words $U_{k} b$, $a U_{k}$, and $b U_{k}$ have to appear as well. Hence, the word $U_{k}$ is ubiquitous. Moreover, if the multiplicity of $U_{k}$ were strictly greater than $2^{k+1}$, then by (1) at least one of the two words $U_{k} a$ and $U_{k} b$ would have multiplicity greater than $2^{k}$ and would thus violate the maximality condition imposed on $U_{k}$.\n\nSo we have $\\mu\\left(U_{0}\\right) \\leq 2 < \\mu\\left(U_{1}\\right) \\leq 4 < \\ldots \\leq 2^{n-1} < \\mu\\left(U_{n-1}\\right)$, which implies in particular that the words $U_{0}, U_{1}, \\ldots, U_{n-1}$ have to be distinct. As they have been proved to be ubiquitous as well, the problem is solved.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31937,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a square table of $2011$ by $2011$ cells we place a finite number of napkins that each cover a square of $52$ by $52$ cells. In each cell we write the number of napkins covering it, and we record the maximal number $k$ of cells that all contain the same nonzero number. Considering all possible napkin configurations, what is the largest value of $k$?",
    "options": [],
    "answer": "3986729",
    "solution": "Let $m=39$, then $2011=52 m-17$. We begin with an example showing that there can exist $3986729$ cells carrying the same positive number.\n\n![](attached_image_1.png)\n\nTo describe it, we number the columns from the left to the right and the rows from the bottom to the top by $1,2, \\ldots, 2011$. We will denote each napkin by the coordinates of its lower left cell. There are four kinds of napkins: first, we take all napkins $(52 i+36,52 j+1)$ with $0 \\leq j \\leq i \\leq m-2$; second, we use all napkins $(52 i+1,52 j+36)$ with $0 \\leq i \\leq j \\leq m-2$; third, we use all napkins $(52 i+36,52 i+36)$ with $0 \\leq i \\leq m-2$; and finally the napkin $(1,1)$. Different groups of napkins are shown by different types of hatchings in the picture.\n\nNow except for those squares that carry two or more different hatchings, all squares have the number $1$ written into them. The number of these exceptional cells is easily computed to be $\\left(52^{2}-35^{2}\\right) m-17^{2}=57392$.\n\nWe are left to prove that $3986729$ is an upper bound for the number of cells containing the same number. Consider any configuration of napkins and any positive integer $M$. Suppose there are $g$ cells with a number different from $M$. Then it suffices to show $g \\geq 57392$. Throughout the solution, a line will mean either a row or a column.\n\nConsider any line $\\ell$. Let $a_{1}, \\ldots, a_{52 m-17}$ be the numbers written into its consecutive cells. For $i=1,2, \\ldots, 52$, let $s_{i}=\\sum_{t \\equiv i(\\bmod 52)} a_{t}$. Note that $s_{1}, \\ldots, s_{35}$ have $m$ terms each, while $s_{36}, \\ldots, s_{52}$ have $m-1$ terms each. Every napkin intersecting $\\ell$ contributes exactly $1$ to each $s_{i}$; hence the number $s$ of all those napkins satisfies $s_{1}=\\cdots=s_{52}=s$. Call the line $\\ell$ rich if $s>(m-1) M$ and poor otherwise.\n\nSuppose now that $\\ell$ is rich. Then in each of the sums $s_{36}, \\ldots, s_{52}$ there exists a term greater than $M$; consider all these terms and call the corresponding cells the rich bad cells for this line. So, each rich line contains at least $17$ cells that are bad for this line.\n\nIf, on the other hand, $\\ell$ is poor, then certainly $s<m M$ so in each of the sums $s_{1}, \\ldots, s_{35}$ there exists a term less than $M$; consider all these terms and call the corresponding cells the poor bad cells for this line. So, each poor line contains at least $35$ cells that are bad for this line.\n\nLet us call all indices congruent to $1,2, \\ldots$, or $35$ modulo $52$ small, and all other indices, i.e. those congruent to $36,37, \\ldots$, or $52$ modulo $52$, big. Recall that we have numbered the columns from the left to the right and the rows from the bottom to the top using the numbers $1,2, \\ldots, 52 m-17$; we say that a line is big or small depending on whether its index is big or small. By definition, all rich bad cells for the rows belong to the big columns, while the poor ones belong to the small columns, and vice versa.\n\nIn each line, we put a strawberry on each cell that is bad for this line. In addition, for each small rich line we put an extra strawberry on each of its (rich) bad cells. A cell gets the strawberries from its row and its column independently.\n\nNotice now that a cell with a strawberry on it contains a number different from $M$. If this cell gets a strawberry by the extra rule, then it contains a number greater than $M$. Moreover, it is either in a small row and in a big column, or vice versa. Suppose that it is in a small row, then it is not bad for its column. So it has not more than two strawberries in this case. On the other hand, if the extra rule is not applied to some cell, then it also has not more than two strawberries. So, the total number $N$ of strawberries is at most $2 g$.\n\nWe shall now estimate $N$ in a different way. For each of the $2 \\cdot 35 m$ small lines, we have introduced at least $34$ strawberries if it is rich and at least $35$ strawberries if it is poor, so at least $34$ strawberries in any case. Similarly, for each of the $2 \\cdot 17(m-1)$ big lines, we put at least $\\min (17,35)=17$ strawberries. Summing over all lines we obtain\n$$\n2 g \\geq N \\geq 2(35 m \\cdot 34+17(m-1) \\cdot 17)=2(1479 m-289)=2 \\cdot 57392\n$$\nas desired.\nWe present a different proof of the estimate which is the hard part of the problem. Let $S=35$, $H=17$, $m=39$; so the table size is $2011=S m+H(m-1)$, and the napkin size is $52=S+H$. Fix any positive integer $M$ and call a cell vicious if it contains a number distinct from $M$. We will prove that there are at least $H^{2}(m-1)+2 S H m$ vicious cells.\n\nFirstly, we introduce some terminology. As in the previous solution, we number rows and columns and we use the same notions of small and big indices and lines; so, an index is small if it is congruent to one of the numbers $1,2, \\ldots, S$ modulo $(S+H)$. The numbers $1,2, \\ldots, S+H$ will be known as residues. For two residues $i$ and $j$, we say that a cell is of type $(i, j)$ if the index of its row is congruent to $i$ and the index of its column to $j$ modulo $(S+H)$. The number of vicious cells of this type is denoted by $v_{ij}$.\n\nLet $s, s'$ be two variables ranging over small residues and let $h, h'$ be two variables ranging over big residues. A cell is said to be of class $A, B, C$, or $D$ if its type is of shape $(s, s')$, $(s, h)$, $(h, s)$, or $(h, h')$, respectively. The numbers of vicious cells belonging to these classes are denoted in this order by $a, b, c$, and $d$. Observe that each cell belongs to exactly one class.\n\n**Claim 1.** We have\n$$\nm \\leq \\frac{a}{S^{2}}+\\frac{b+c}{2 S H} . \\tag{1}\n$$\nProof. Consider an arbitrary small row $r$. Denote the numbers of vicious cells on $r$ belonging to the classes $A$ and $B$ by $\\alpha$ and $\\beta$, respectively. As in the previous solution, we obtain that $\\alpha \\geq S$ or $\\beta \\geq H$. So in each case we have $\\frac{\\alpha}{S}+\\frac{\\beta}{H} \\geq 1$.\n\nPerforming this argument separately for each small row and adding up all the obtained inequalities, we get $\\frac{a}{S}+\\frac{b}{H} \\geq m S$. Interchanging rows and columns we similarly get $\\frac{a}{S}+\\frac{c}{H} \\geq m S$. Summing these inequalities and dividing by $2 S$ we get what we have claimed.\n\n**Claim 2.** Fix two small residues $s, s'$ and two big residues $h, h'$. Then $2 m-1 \\leq v_{s s'}+v_{s h'}+v_{h h'}$.\n\nProof. Each napkin covers exactly one cell of type $(s, s')$. Removing all napkins covering a vicious cell of this type, we get another collection of napkins, which covers each cell of type $(s, s')$ either $0$ or $M$ times depending on whether the cell is vicious or not. Hence $(m^{2}-v_{s s'}) M$ napkins are left and throughout the proof of Claim 2 we will consider only these remaining napkins. Now, using a red pen, write in each cell the number of napkins covering it. Notice that a cell containing a red number greater than $M$ is surely vicious.\n\nWe call two cells neighbors if they can be simultaneously covered by some napkin. So, each cell of type $(h, h')$ has not more than four neighbors of type $(s, s')$, while each cell of type $(s, h')$ has not more than two neighbors of each of the types $(s, s')$ and $(h, h')$. Therefore, each red number at a cell of type $(h, h')$ does not exceed $4 M$, while each red number at a cell of type $(s, h')$ does not exceed $2 M$.\n\nLet $x, y$, and $z$ be the numbers of cells of type $(h, h')$ whose red number belongs to $(M, 2 M]$, $(2 M, 3 M]$, and $(3 M, 4 M]$, respectively. All these cells are vicious, hence $x+y+z \\leq v_{h h'}$. The red numbers appearing in cells of type $(h, h')$ clearly sum up to $(m^{2}-v_{s s'}) M$. Bounding each of these numbers by a multiple of $M$ we get\n$$\n(m^{2}-v_{s s'}) M \\leq ((m-1)^{2}-(x+y+z)) M+2 x M+3 y M+4 z M,\n$$\ni.e.\n$$\n2 m-1 \\leq v_{s s'}+x+2 y+3 z \\leq v_{s s'}+v_{h h'}+y+2 z .\n$$\nSo, to prove the claim it suffices to prove that $y+2 z \\leq v_{s h'}$.\n\nFor a cell $\\delta$ of type $(h, h')$ and a cell $\\beta$ of type $(s, h')$ we say that $\\delta$ forces $\\beta$ if there are more than $M$ napkins covering both of them. Since each red number in a cell of type $(s, h')$ does not exceed $2 M$, it cannot be forced by more than one cell.\n\nOn the other hand, if a red number in a $(h, h')$-cell belongs to $(2 M, 3 M]$, then it forces at least one of its neighbors of type $(s, h')$ (since the sum of red numbers in their cells is greater than $2 M$). Analogously, a $(h, h')$-cell with the red number in $(3 M, 4 M]$ forces both its neighbors of type $(s, h')$, since their red numbers do not exceed $2 M$. Therefore there are at least $y+2 z$ forced cells and clearly all of them are vicious, as desired.\n\n**Claim 3.** We have\n$$\n2 m-1 \\leq \\frac{a}{S^{2}}+\\frac{b+c}{2 S H}+\\frac{d}{H^{2}} \\tag{2}\n$$\nProof. Averaging the previous result over all $S^{2} H^{2}$ possibilities for the quadruple $(s, s', h, h')$, we get $2 m-1 \\leq \\frac{a}{S^{2}}+\\frac{b}{S H}+\\frac{d}{H^{2}}$. Due to the symmetry between rows and columns, the same estimate holds with $b$ replaced by $c$. Averaging these two inequalities we arrive at our claim.\n\nNow let us multiply (2) by $H^{2}$, multiply (1) by $(2 S H-H^{2})$ and add them; we get\n$H^{2}(2 m-1)+(2 S H-H^{2}) m \\leq a \\cdot \\frac{H^{2}+2 S H-H^{2}}{S^{2}}+(b+c) \\frac{H^{2}+2 S H-H^{2}}{2 S H}+d=a \\cdot \\frac{2 H}{S}+b+c+d$.\n\nThe left-hand side is exactly $H^{2}(m-1)+2 S H m$, while the right-hand side does not exceed $a+b+c+d$ since $2 H \\leq S$. Hence we come to the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31938,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with circumcircle $\\Omega$. Let $B_{0}$ be the midpoint of $AC$ and let $C_{0}$ be the midpoint of $AB$. Let $D$ be the foot of the altitude from $A$, and let $G$ be the centroid of the triangle $ABC$. Let $\\omega$ be a circle through $B_{0}$ and $C_{0}$ that is tangent to the circle $\\Omega$ at a point $X \\neq A$. Prove that the points $D$, $G$, and $X$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $AB = AC$, then the statement is trivial. So without loss of generality we may assume $AB < AC$. Denote the tangents to $\\Omega$ at points $A$ and $X$ by $a$ and $x$, respectively.\nLet $\\Omega_{1}$ be the circumcircle of triangle $AB_{0}C_{0}$. The circles $\\Omega$ and $\\Omega_{1}$ are homothetic with center $A$, so they are tangent at $A$, and $a$ is their radical axis. Now, the lines $a$, $x$, and $B_{0}C_{0}$ are the three radical axes of the circles $\\Omega$, $\\Omega_{1}$, and $\\omega$. Since $a \\nmid\\nmid B_{0}C_{0}$, these three lines are concurrent at some point $W$.\nThe points $A$ and $D$ are symmetric with respect to the line $B_{0}C_{0}$; hence $WX = WA = WD$. This means that $W$ is the center of the circumcircle $\\gamma$ of triangle $ADX$. Moreover, we have $\\angle WAO = \\angle WXO = 90^{\\circ}$, where $O$ denotes the center of $\\Omega$. Hence $\\angle AWX + \\angle AOX = 180^{\\circ}$.\n\n![](attached_image_1.png)\n\nDenote by $T$ the second intersection point of $\\Omega$ and the line $DX$. Note that $O$ belongs to $\\Omega_{1}$. Using the circles $\\gamma$ and $\\Omega$, we find\n$$\n\\angle DAT = \\angle ADX - \\angle ATD = \\frac{1}{2}\\left(360^{\\circ} - \\angle AWX\\right) - \\frac{1}{2} \\angle AOX = 180^{\\circ} - \\frac{1}{2}(\\angle AWX + \\angle AOX) = 90^{\\circ}.\n$$\nSo, $AD \\perp AT$, and hence $AT \\parallel BC$. Thus, $ATCB$ is an isosceles trapezoid inscribed in $\\Omega$.\nDenote by $A_{0}$ the midpoint of $BC$, and consider the image of $ATCB$ under the homothety $h$ with center $G$ and factor $-\\frac{1}{2}$. We have $h(A) = A_{0}$, $h(B) = B_{0}$, and $h(C) = C_{0}$. From the symmetry about $B_{0}C_{0}$, we have $\\angle TCB = \\angle CBA = \\angle B_{0}C_{0}A = \\angle DC_{0}B_{0}$. Using $AT \\parallel DA_{0}$, we conclude $h(T) = D$. Hence the points $D$, $G$, and $T$ are collinear, and $X$ lies on the same line.\nWe define the points $A_{0}$, $O$, and $W$ as in the previous solution and we concentrate on the case $AB < AC$. Let $Q$ be the perpendicular projection of $A_{0}$ on $B_{0}C_{0}$.\nSince $\\angle WAO = \\angle WQO = \\angle OXW = 90^{\\circ}$, the five points $A$, $W$, $X$, $O$, and $Q$ lie on a common circle. Furthermore, the reflections with respect to $B_{0}C_{0}$ and $OW$ map $A$ to $D$ and $X$, respectively. For these reasons, we have\n$$\n\\angle WQD = \\angle AQW = \\angle AXW = \\angle WAX = \\angle WQX.\n$$\nThus the three points $Q$, $D$, and $X$ lie on a common line, say $\\ell$.\n\n![](attached_image_2.png)\n\nTo complete the argument, we note that the homothety centered at $G$ sending the triangle $ABC$ to the triangle $A_{0}B_{0}C_{0}$ maps the altitude $AD$ to the altitude $A_{0}Q$. Therefore it maps $D$ to $Q$, so the points $D$, $G$, and $Q$ are collinear. Hence $G$ lies on $\\ell$ as well.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31939,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = AC$, and let $D$ be the midpoint of $AC$. The angle bisector of $\\angle BAC$ intersects the circle through $D$, $B$, and $C$ in a point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A$, $E$, and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet at a point $I$, and the lines $CI$ and $BD$ meet at a point $K$. Show that $I$ is the incenter of triangle $KAB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D'$ be the midpoint of the segment $AB$, and let $M$ be the midpoint of $BC$. By symmetry at line $AM$, the point $D'$ has to lie on the circle $BCD$. Since the $\\operatorname{arcs} D'E$ and $ED$ of that circle are equal, we have $\\angle ABI = \\angle D'BE = \\angle EBD = IBK$, so $I$ lies on the angle bisector of $\\angle ABK$. For this reason it suffices to prove in the sequel that the ray $AI$ bisects the angle $\\angle BAK$.\n\nFrom\n$$\n\\angle DFA = 180^\\circ - \\angle BFA = 180^\\circ - \\angle BEA = \\angle MEB = \\frac{1}{2} \\angle CEB = \\frac{1}{2} \\angle CDB\n$$\nwe derive $\\angle DFA = \\angle DAF$ so the triangle $AFD$ is isosceles with $AD = DF$.\n\n![](attached_image_1.png)\n\nApplying Menelaus's theorem to the triangle $ADF$ with respect to the line $CIK$, and applying the angle bisector theorem to the triangle $ABF$, we infer\n$$\n1 = \\frac{AC}{CD} \\cdot \\frac{DK}{KF} \\cdot \\frac{FI}{IA} = 2 \\cdot \\frac{DK}{KF} \\cdot \\frac{BF}{AB} = 2 \\cdot \\frac{DK}{KF} \\cdot \\frac{BF}{2 \\cdot AD} = \\frac{DK}{KF} \\cdot \\frac{BF}{AD}\n$$\nand therefore\n$$\n\\frac{BD}{AD} = \\frac{BF + FD}{AD} = \\frac{BF}{AD} + 1 = \\frac{KF}{DK} + 1 = \\frac{DF}{DK} = \\frac{AD}{DK} .\n$$\nIt follows that the triangles $ADK$ and $BDA$ are similar, hence $\\angle DAK = \\angle ABD$. Then\n$$\n\\angle IAB = \\angle AFD - \\angle ABD = \\angle DAF - \\angle DAK = \\angle KAI\n$$\nshows that the point $K$ is indeed lying on the angle bisector of $\\angle BAK$.\nIt can be shown in the same way as in the first solution that $I$ lies on the angle bisector of $\\angle ABK$. Here we restrict ourselves to proving that $KI$ bisects $\\angle AKB$.\n\n![](attached_image_2.png)\n\nDenote the circumcircle of triangle $BCD$ and its center by $\\omega_1$ and by $O_1$, respectively. Since the quadrilateral $ABFE$ is cyclic, we have $\\angle DFE = \\angle BAE = \\angle DAE$. By the same reason, we have $\\angle EAF = \\angle EBF = \\angle ABE = \\angle AFE$. Therefore $\\angle DAF = \\angle DFA$, and hence $DF = DA = DC$. So triangle $AFC$ is inscribed in a circle $\\omega_2$ with center $D$.\n\nDenote the circumcircle of triangle $ABD$ by $\\omega_3$, and let its center be $O_3$. Since the $\\operatorname{arcs} BE$ and $EC$ of circle $\\omega_1$ are equal, and the triangles $ADE$ and $FDE$ are congruent, we have $\\angle AO_1B = 2 \\angle BDE = \\angle BDA$, so $O_1$ lies on $\\omega_3$. Hence $\\angle O_3O_1D = \\angle O_3DO_1$.\n\nThe line $BD$ is the radical axis of $\\omega_1$ and $\\omega_3$. Point $C$ belongs to the radical axis of $\\omega_1$ and $\\omega_2$, and $I$ also belongs to it since $AI \\cdot IF = BI \\cdot IE$. Hence $K = BD \\cap CI$ is the radical center of $\\omega_1$, $\\omega_2$, and $\\omega_3$, and $AK$ is the radical axis of $\\omega_2$ and $\\omega_3$. Now, the radical axes $AK$, $BK$ and $IK$ are perpendicular to the central lines $O_3D$, $O_3O_1$ and $O_1D$, respectively. By $\\angle O_3O_1D = \\angle O_3DO_1$, we get that $KI$ is the angle bisector of $\\angle AKB$.\nAgain, let $M$ be the midpoint of $BC$. As in the previous solutions, we can deduce $\\angle ABI = \\angle IBK$. We show that the point $I$ lies on the angle bisector of $\\angle KAB$.\n\nLet $G$ be the intersection point of the circles $AFC$ and $BCD$, different from $C$. The lines $CG$, $AF$, and $BE$ are the radical axes of the three circles $AGFC$, $CDB$, and $ABFE$, so $I = AF \\cap BE$ is the radical center of the three circles and $CG$ also passes through $I$.\n\n![](attached_image_3.png)\n\nThe angle between line $DE$ and the tangent to the circle $BCD$ at $E$ is equal to $\\angle EBD = \\angle EAF = \\angle ABE = \\angle AFE$. As the tangent at $E$ is perpendicular to $AM$, the line $DE$ is perpendicular to $AF$. The triangle $AFE$ is isosceles, so $DE$ is the perpendicular bisector of $AF$ and thus $AD = DF$. Hence, the point $D$ is the center of the circle $AFC$, and this circle passes through $M$ as well since $\\angle AMC = 90^\\circ$.\n\nLet $B'$ be the reflection of $B$ in the point $D$, so $ABCB'$ is a parallelogram. Since $DC = DG$ we have $\\angle GCD = \\angle DBC = \\angle KB'A$. Hence, the quadrilateral $AKCB'$ is cyclic and thus $\\angle CAK = \\angle CB'K = \\angle ABD = 2 \\angle MAI$. Then\n$$\n\\angle IAB = \\angle MAB - \\angle MAI = \\frac{1}{2} \\angle CAB - \\frac{1}{2} \\angle CAK = \\frac{1}{2} \\angle KAB\n$$\nand therefore $AI$ is the angle bisector of $\\angle KAB$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31940,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C D E F$ be a convex hexagon all of whose sides are tangent to a circle $\\omega$ with center $O$. Suppose that the circumcircle of triangle $A C E$ is concentric with $\\omega$. Let $J$ be the foot of the perpendicular from $B$ to $C D$. Suppose that the perpendicular from $B$ to $D F$ intersects the line $E O$ at a point $K$. Let $L$ be the foot of the perpendicular from $K$ to $D E$. Prove that $D J=D L$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $\\omega$ and the circumcircle of triangle $A C E$ are concentric, the tangents from $A$, $C$, and $E$ to $\\omega$ have equal lengths; that means that $A B=B C, C D=D E$, and $E F=F A$. Moreover, we have $\\angle B C D=\\angle D E F=\\angle F A B$.\n\n![](attached_image_1.png)\n\nConsider the rotation around point $D$ mapping $C$ to $E$; let $B^{\\prime}$ and $L^{\\prime}$ be the images of the points $B$ and $J$, respectively, under this rotation. Then one has $D J=D L^{\\prime}$ and $B^{\\prime} L^{\\prime} \\perp D E$; moreover, the triangles $B^{\\prime} E D$ and $B C D$ are congruent. Since $\\angle D E O<90^{\\circ}$, the lines $E O$ and $B^{\\prime} L^{\\prime}$ intersect at some point $K^{\\prime}$. We intend to prove that $K^{\\prime} B \\perp D F$; this would imply $K=K^{\\prime}$, therefore $L=L^{\\prime}$, which proves the problem statement.\n\nAnalogously, consider the rotation around $F$ mapping $A$ to $E$; let $B^{\\prime \\prime}$ be the image of $B$ under this rotation. Then the triangles $F A B$ and $F E B^{\\prime \\prime}$ are congruent. We have $E B^{\\prime \\prime}=A B=B C= E B^{\\prime}$ and $\\angle F E B^{\\prime \\prime}=\\angle F A B=\\angle B C D=\\angle D E B^{\\prime}$, so the points $B^{\\prime}$ and $B^{\\prime \\prime}$ are symmetrical with respect to the angle bisector $E O$ of $\\angle D E F$. So, from $K^{\\prime} B^{\\prime} \\perp D E$ we get $K^{\\prime} B^{\\prime \\prime} \\perp E F$.\n\nFrom these two relations we obtain\n$$\nK^{\\prime} D^{2}-K^{\\prime} E^{2}=B^{\\prime} D^{2}-B^{\\prime} E^{2} \\quad \\text { and } \\quad K^{\\prime} E^{2}-K^{\\prime} F^{2}=B^{\\prime \\prime} E^{2}-B^{\\prime \\prime} F^{2} .\n$$\nAdding these equalities and taking into account that $B^{\\prime} E=B^{\\prime \\prime} E$ we obtain\n$$\nK^{\\prime} D^{2}-K^{\\prime} F^{2}=B^{\\prime} D^{2}-B^{\\prime \\prime} F^{2}=B D^{2}-B F^{2}\n$$\nwhich means exactly that $K^{\\prime} B \\perp D F$.\nLet us denote the points of tangency of $A B, B C, C D, D E, E F$, and $F A$ to $\\omega$ by $R, S, T, U, V$, and $W$, respectively. As in the previous solution, we mention that $A R= A W=C S=C T=E U=E V$.\n\nThe reflection in the line $B O$ maps $R$ to $S$, therefore $A$ to $C$ and thus also $W$ to $T$. Hence, both lines $R S$ and $W T$ are perpendicular to $O B$, therefore they are parallel. On the other hand, the lines $U V$ and $W T$ are not parallel, since otherwise the hexagon $A B C D E F$ is symmetric with respect to the line $B O$ and the lines defining the point $K$ coincide, which contradicts the conditions of the problem. Therefore we can consider the intersection point $Z$ of $U V$ and $W T$.\n\n![](attached_image_2.png)\n\nNext, we recall a well-known fact that the points $D, F, Z$ are collinear. Actually, $D$ is the pole of the line $U T, F$ is the pole of $V W$, and $Z=T W \\cap U V$; so all these points belong to the polar line of $T U \\cap V W$.\n\nNow, we put $O$ into the origin, and identify each point (say $X$ ) with the vector $\\overrightarrow{O X}$. So, from now on all the products of points refer to the scalar products of the corresponding vectors.\n\nSince $O K \\perp U Z$ and $O B \\perp T Z$, we have $K \\cdot(Z-U)=0=B \\cdot(Z-T)$. Next, the condition $B K \\perp D Z$ can be written as $K \\cdot(D-Z)=B \\cdot(D-Z)$. Adding these two equalities we get\n$$\nK \\cdot(D-U)=B \\cdot(D-T)\n$$\nBy symmetry, we have $D \\cdot(D-U)=D \\cdot(D-T)$. Subtracting this from the previous equation, we obtain $(K-D) \\cdot(D-U)=(B-D) \\cdot(D-T)$ and rewrite it in vector form as\n$$\n\\overrightarrow{D K} \\cdot \\overrightarrow{U D}=\\overrightarrow{D B} \\cdot \\overrightarrow{T D} .\n$$\nFinally, projecting the vectors $\\overrightarrow{D K}$ and $\\overrightarrow{D B}$ onto the lines $U D$ and $T D$ respectively, we can rewrite this equality in terms of segment lengths as $D L \\cdot U D=D J \\cdot T D$, thus $D L=D J$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31941,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with circumcircle $\\omega$. Let $t$ be a tangent line to $\\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $BC, CA$, and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is tangent to the circle $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To avoid a large case distinction, we will use the notion of oriented angles. Namely, for two lines $\\ell$ and $m$, we denote by $\\angle(\\ell, m)$ the angle by which one may rotate $\\ell$ anticlockwise to obtain a line parallel to $m$. Thus, all oriented angles are considered modulo $180^{\\circ}$.\n\n![](attached_image_1.png)\n\nDenote by $T$ the point of tangency of $t$ and $\\omega$. Let $A' = t_{b} \\cap t_{c}$, $B' = t_{a} \\cap t_{c}$, $C' = t_{a} \\cap t_{b}$. Introduce the point $A''$ on $\\omega$ such that $TA = AA''$ ($A'' \\neq T$ unless $TA$ is a diameter). Define the points $B''$ and $C''$ in a similar way.\n\nSince the points $C$ and $B$ are the midpoints of $\\operatorname{arcs} TC''$ and $TB''$, respectively, we have\n$$\n\\begin{aligned}\n\\angle\\left(t, B''C''\\right) &= \\angle\\left(t, TC''\\right) + \\angle\\left(TC'', B''C''\\right) = 2\\angle(t, TC) + 2\\angle\\left(TC'', BC''\\right) \\\\\n&= 2(\\angle(t, TC) + \\angle(TC, BC)) = 2\\angle(t, BC) = \\angle\\left(t, t_{a}\\right)\n\\end{aligned}\n$$\nIt follows that $t_{a}$ and $B''C''$ are parallel. Similarly, $t_{b} \\parallel A''C''$ and $t_{c} \\parallel A''B''$. Thus, either the triangles $A'B'C'$ and $A''B''C''$ are homothetic, or they are translates of each other. Now we will prove that they are in fact homothetic, and that the center $K$ of the homothety belongs to $\\omega$. It would then follow that their circumcircles are also homothetic with respect to $K$ and are therefore tangent at this point, as desired.\n\nWe need the two following claims.\n\nClaim 1. The point of intersection $X$ of the lines $B''C$ and $BC''$ lies on $t_{a}$.\n\nProof. Actually, the points $X$ and $T$ are symmetric about the line $BC$, since the lines $CT$ and $CB''$ are symmetric about this line, as are the lines $BT$ and $BC''$.\n\nClaim 2. The point of intersection $I$ of the lines $BB'$ and $CC'$ lies on the circle $\\omega$.\n\nProof. We consider the case that $t$ is not parallel to the sides of $ABC$; the other cases may be regarded as limit cases. Let $D = t \\cap BC$, $E = t \\cap AC$, and $F = t \\cap AB$.\n\nDue to symmetry, the line $DB$ is one of the angle bisectors of the lines $B'D$ and $FD$; analogously, the line $FB$ is one of the angle bisectors of the lines $B'F$ and $DF$. So $B$ is either the incenter or one of the excenters of the triangle $B'DF$. In any case we have $\\angle(BD, DF) + \\angle(DF, FB) + \\angle\\left(B'B, B'D\\right) = 90^{\\circ}$, so\n$$\n\\angle\\left(B'B, B'C'\\right) = \\angle\\left(B'B, B'D\\right) = 90^{\\circ} - \\angle(BC, DF) - \\angle(DF, BA) = 90^{\\circ} - \\angle(BC, AB).\n$$\nAnalogously, we get $\\angle\\left(C'C, B'C'\\right) = 90^{\\circ} - \\angle(BC, AC)$. Hence,\n$$\n\\angle(BI, CI) = \\angle\\left(B'B, B'C'\\right) + \\angle\\left(B'C', C'C\\right) = \\angle(BC, AC) - \\angle(BC, AB) = \\angle(AB, AC),\n$$\nwhich means exactly that the points $A, B, I, C$ are concyclic.\n\nNow we can complete the proof. Let $K$ be the second intersection point of $B'B''$ and $\\omega$. Applying Pascal's theorem to hexagon $KB''CIBC''$ we get that the points $B' = KB'' \\cap IB$ and $X = B''C \\cap BC''$ are collinear with the intersection point $S$ of $CI$ and $C''K$. So $S = CI \\cap B'X = C'$, and the points $C', C'', K$ are collinear. Thus $K$ is the intersection point of $B'B''$ and $C'C''$ which implies that $K$ is the center of the homothety mapping $A'B'C'$ to $A''B''C''$, and it belongs to $\\omega$.\nDefine the points $T, A', B'$, and $C'$ in the same way as in the previous solution. Let $X, Y$, and $Z$ be the symmetric images of $T$ about the lines $BC, CA$, and $AB$, respectively. Note that the projections of $T$ on these lines form a Simson line of $T$ with respect to $ABC$, therefore the points $X, Y, Z$ are also collinear. Moreover, we have $X \\in B'C'$, $Y \\in C'A'$, $Z \\in A'B'$.\n\nDenote $\\alpha = \\angle(t, TC) = \\angle(BT, BC)$. Using the symmetry in the lines $AC$ and $BC$, we get\n$$\n\\angle(BC, BX) = \\angle(BT, BC) = \\alpha \\quad \\text{and} \\quad \\angle\\left(XC, XC'\\right) = \\angle(t, TC) = \\angle\\left(YC, YC'\\right) = \\alpha.\n$$\nSince $\\angle\\left(XC, XC'\\right) = \\angle\\left(YC, YC'\\right)$, the points $X, Y, C, C'$ lie on some circle $\\omega_{c}$. Define the circles $\\omega_{a}$ and $\\omega_{b}$ analogously. Let $\\omega'$ be the circumcircle of triangle $A'B'C'$.\n\nNow, applying Miquel's theorem to the four lines $A'B', A'C', B'C'$, and $XY$, we obtain that the circles $\\omega', \\omega_{a}, \\omega_{b}, \\omega_{c}$ intersect at some point $K$. We will show that $K$ lies on $\\omega$, and that the tangent lines to $\\omega$ and $\\omega'$ at this point coincide; this implies the problem statement.\n\nDue to symmetry, we have $XB = TB = ZB$, so the point $B$ is the midpoint of one of the $\\operatorname{arcs} XZ$ of circle $\\omega_{b}$. Therefore $\\angle(KB, KX) = \\angle(XZ, XB)$. Analogously, $\\angle(KX, KC) = \\angle(XC, XY)$. Adding these equalities and using the symmetry in the line $BC$ we get\n$$\n\\angle(KB, KC) = \\angle(XZ, XB) + \\angle(XC, XZ) = \\angle(XC, XB) = \\angle(TB, TC).\n$$\nTherefore, $K$ lies on $\\omega$.\n\nNext, let $k$ be the tangent line to $\\omega$ at $K$. We have\n$$\n\\begin{aligned}\n\\angle\\left(k, KC'\\right) &= \\angle(k, KC) + \\angle\\left(KC, KC'\\right) = \\angle(KB, BC) + \\angle\\left(XC, XC'\\right) \\\\\n&= (\\angle(KB, BX) - \\angle(BC, BX)) + \\alpha = \\angle\\left(KB', B'X\\right) - \\alpha + \\alpha = \\angle\\left(KB', B'C'\\right)\n\\end{aligned}\n$$\nwhich means exactly that $k$ is tangent to $\\omega'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31942,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 1$ be an odd integer. Determine all functions $f$ from the set of integers to itself such that for all integers $x$ and $y$ the difference $f(x)-f(y)$ divides $x^{n}-y^{n}$.",
    "options": [],
    "answer": "All functions of the form f(x) = s · x^d + t, where s is either 1 or −1, t is any integer, and d is a positive integer dividing the given odd integer n.",
    "solution": "Obviously, all functions in the answer satisfy the condition of the problem. We will show that there are no other functions satisfying that condition.\nLet $f$ be a function satisfying the given condition. For each integer $n$, the function $g$ defined by $g(x)=f(x)+n$ also satisfies the same condition. Therefore, by subtracting $f(0)$ from $f(x)$ we may assume that $f(0)=0$.\nFor any prime $p$, the condition on $f$ with $(x, y)=(p, 0)$ states that $f(p)$ divides $p^{n}$. Since the set of primes is infinite, there exist integers $d$ and $\\varepsilon$ with $0 \\leq d \\leq n$ and $\\varepsilon \\in\\{1,-1\\}$ such that for infinitely many primes $p$ we have $f(p)=\\varepsilon p^{d}$. Denote the set of these primes by $P$. Since a function $g$ satisfies the given condition if and only if $-g$ satisfies the same condition, we may suppose $\\varepsilon=1$.\nThe case $d=0$ is easily ruled out, because 0 does not divide any nonzero integer. Suppose $d \\geq 1$ and write $n$ as $m d+r$, where $m$ and $r$ are integers such that $m \\geq 1$ and $0 \\leq r \\leq d-1$. Let $x$ be an arbitrary integer. For each prime $p$ in $P$, the difference $f(p)-f(x)$ divides $p^{n}-x^{n}$. Using the equality $f(p)=p^{d}$, we get\n$$\np^{n}-x^{n}=p^{r}\\left(p^{d}\\right)^{m}-x^{n} \\equiv p^{r} f(x)^{m}-x^{n} \\equiv 0 \\quad\\left(\\bmod p^{d}-f(x)\\right)\n$$\nSince we have $r<d$, for large enough primes $p \\in P$ we obtain\n$$\n\\left|p^{r} f(x)^{m}-x^{n}\\right|<p^{d}-f(x) .\n$$\nHence $p^{r} f(x)^{m}-x^{n}$ has to be zero. This implies $r=0$ and $x^{n}=\\left(x^{d}\\right)^{m}=f(x)^{m}$. Since $m$ is odd, we obtain $f(x)=x^{d}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31943,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x)$ and $Q(x)$ be two polynomials with integer coefficients such that no nonconstant polynomial with rational coefficients divides both $P(x)$ and $Q(x)$. Suppose that for every positive integer $n$ the integers $P(n)$ and $Q(n)$ are positive, and $2^{Q(n)}-1$ divides $3^{P(n)}-1$. Prove that $Q(x)$ is a constant polynomial.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we show that there exists an integer $d$ such that for all positive integers $n$ we have $\\operatorname{gcd}(P(n), Q(n)) \\leq d$.\nSince $P(x)$ and $Q(x)$ are coprime (over the polynomials with rational coefficients), Euclid's algorithm provides some polynomials $R_{0}(x), S_{0}(x)$ with rational coefficients such that $P(x) R_{0}(x)- Q(x) S_{0}(x)=1$. Multiplying by a suitable positive integer $d$, we obtain polynomials $R(x)= d \\cdot R_{0}(x)$ and $S(x)=d \\cdot S_{0}(x)$ with integer coefficients for which $P(x) R(x)-Q(x) S(x)=d$. Then we have $\\operatorname{gcd}(P(n), Q(n)) \\leq d$ for any integer $n$.\n\nTo prove the problem statement, suppose that $Q(x)$ is not constant. Then the sequence $Q(n)$ is not bounded and we can choose a positive integer $m$ for which\n$$\n\\begin{equation*}\nM=2^{Q(m)}-1 \\geq 3^{\\max \\{P(1), P(2), \\ldots, P(d)\\}} . \\tag{1}\n\\end{equation*}\n$$\nSince $M=2^{Q(n)}-1 \\mid 3^{P(n)}-1$, we have $2,3 \\nmid M$. Let $a$ and $b$ be the multiplicative orders of $2$ and $3$ modulo $M$, respectively. Obviously, $a=Q(m)$ since the lower powers of $2$ do not reach $M$. Since $M$ divides $3^{P(m)}-1$, we have $b \\mid P(m)$. Then $\\operatorname{gcd}(a, b) \\leq \\operatorname{gcd}(P(m), Q(m)) \\leq d$. Since the expression $a x-b y$ attains all integer values divisible by $\\operatorname{gcd}(a, b)$ when $x$ and $y$ run over all nonnegative integer values, there exist some nonnegative integers $x, y$ such that $1 \\leq m+a x-b y \\leq d$.\n\nBy $Q(m+a x) \\equiv Q(m)\\ (\\bmod\\ a)$ we have\n$$\n2^{Q(m+a x)} \\equiv 2^{Q(m)} \\equiv 1 \\quad(\\bmod\\ M)\n$$\nand therefore\n$$\nM\\left|2^{Q(m+a x)}-1\\right| 3^{P(m+a x)}-1 .\n$$\nThen, by $P(m+a x-b y) \\equiv P(m+a x)(\\bmod\\ b)$ we have\n$$\n3^{P(m+a x-b y)} \\equiv 3^{P(m+a x)} \\equiv 1 \\quad(\\bmod\\ M) .\n$$\nSince $P(m+a x-b y)>0$ this implies $M \\leq 3^{P(m+a x-b y)}-1$. But $P(m+a x-b y)$ is listed among $P(1), P(2), \\ldots, P(d)$, so\n$$\nM<3^{P(m+a x-b y)} \\leq 3^{\\max \\{P(1), P(2), \\ldots, P(d)\\}}\n$$\nwhich contradicts (1).",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31944,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime number. For every integer $a$, define the number\n$$\nS_{a} = \\frac{a}{1} + \\frac{a^{2}}{2} + \\cdots + \\frac{a^{p-1}}{p-1}.\n$$\nLet $m$ and $n$ be integers such that\n$$\nS_{3} + S_{4} - 3 S_{2} = \\frac{m}{n}\n$$\nProve that $p$ divides $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For rational numbers $p_{1} / q_{1}$ and $p_{2} / q_{2}$ with the denominators $q_{1}, q_{2}$ not divisible by $p$, we write $p_{1} / q_{1} \\equiv p_{2} / q_{2} (\\bmod p)$ if the numerator $p_{1} q_{2} - p_{2} q_{1}$ of their difference is divisible by $p$.\n\nWe start with finding an explicit formula for the residue of $S_{a}$ modulo $p$. Note first that for every $k = 1, \\ldots, p-1$ the number $\\binom{p}{k}$ is divisible by $p$, and\n$$\n\\frac{1}{p} \\binom{p}{k} = \\frac{(p-1)(p-2) \\cdots (p-k+1)}{k!} \\equiv \\frac{(-1) \\cdot (-2) \\cdots (-k+1)}{k!} = \\frac{(-1)^{k-1}}{k} (\\bmod p)\n$$\nTherefore, we have\n$$\nS_{a} = -\\sum_{k=1}^{p-1} \\frac{(-a)^{k} (-1)^{k-1}}{k} \\equiv -\\sum_{k=1}^{p-1} (-a)^{k} \\cdot \\frac{1}{p} \\binom{p}{k} \\quad (\\bmod p).\n$$\nThe number on the right-hand side is integer. Using the binomial formula we express it as\n$$\n-\\sum_{k=1}^{p-1} (-a)^{k} \\cdot \\frac{1}{p} \\binom{p}{k} = -\\frac{1}{p} \\left(-1 - (-a)^{p} + \\sum_{k=0}^{p} (-a)^{k} \\binom{p}{k} \\right) = \\frac{(a-1)^{p} - a^{p} + 1}{p}\n$$\nsince $p$ is odd. So, we have\n$$\nS_{a} \\equiv \\frac{(a-1)^{p} - a^{p} + 1}{p} (\\bmod p).\n$$\nFinally, using the obtained formula we get\n$$\n\\begin{aligned}\nS_{3} + S_{4} - 3 S_{2} & \\equiv \\frac{\\left(2^{p} - 3^{p} + 1\\right) + \\left(3^{p} - 4^{p} + 1\\right) - 3\\left(1^{p} - 2^{p} + 1\\right)}{p} \\\\\n& = \\frac{4 \\cdot 2^{p} - 4^{p} - 4}{p} = -\\frac{\\left(2^{p} - 2\\right)^{2}}{p} (\\bmod p)\n\\end{aligned}\n$$\nBy Fermat's theorem, $p \\mid 2^{p} - 2$, so $p^{2} \\mid \\left(2^{p} - 2\\right)^{2}$ and hence $S_{3} + S_{4} - 3 S_{2} \\equiv 0 (\\bmod p)$.\nOne may solve the problem without finding an explicit formula for $S_{a}$. It is enough to find the following property.\n\n**Lemma.** For every integer $a$, we have $S_{a+1} \\equiv S_{-a} (\\bmod p)$.\n\n*Proof.* We expand $S_{a+1}$ using the binomial formula as\n$$\nS_{a+1} = \\sum_{k=1}^{p-1} \\frac{1}{k} \\sum_{j=0}^{k} \\binom{k}{j} a^{j} = \\sum_{k=1}^{p-1} \\left( \\frac{1}{k} + \\sum_{j=1}^{k} a^{j} \\cdot \\frac{1}{k} \\binom{k}{j} \\right ) = \\sum_{k=1}^{p-1} \\frac{1}{k} + \\sum_{j=1}^{p-1} a^{j} \\sum_{k=j}^{p-1} \\frac{1}{k} \\binom{k}{j} a^{k}.\n$$\nNote that $\\frac{1}{k} + \\frac{1}{p-k} = \\frac{p}{k(p-k)} \\equiv 0 (\\bmod p)$ for all $1 \\leq k \\leq p-1$; hence the first sum vanishes modulo $p$. For the second sum, we use the relation $\\frac{1}{k} \\binom{k}{j} = \\frac{1}{j} \\binom{k-1}{j-1}$ to obtain\n$$\nS_{a+1} \\equiv \\sum_{j=1}^{p-1} \\frac{a^{j}}{j} \\sum_{k=1}^{p-1} \\binom{k-1}{j-1} \\quad (\\bmod p).\n$$\nFinally, from the relation\n$$\n\\sum_{k=1}^{p-1} \\binom{k-1}{j-1} = \\binom{p-1}{j} = \\frac{(p-1)(p-2) \\ldots (p-j)}{j!} \\equiv (-1)^{j} \\quad (\\bmod p)\n$$\nwe obtain\n$$\nS_{a+1} \\equiv \\sum_{j=1}^{p-1} \\frac{a^{j} (-1)^{j}}{j!} = S_{-a}.\n$$\nNow we turn to the problem. Using the lemma we get\n$$\n\\begin{equation*}\nS_{3} - 3 S_{2} \\equiv S_{-2} - 3 S_{2} = \\sum_{\\substack{1 \\leq k \\leq p-1 \\\\ k \\text{ is even}}} \\frac{-2 \\cdot 2^{k}}{k} + \\sum_{\\substack{1 \\leq k \\leq p-1 \\\\ k \\text{ is odd}}} \\frac{-4 \\cdot 2^{k}}{k} (\\bmod p) \\tag{1}\n\\end{equation*}\n$$\nThe first sum in (1) expands as\n$$\n\\sum_{\\ell=1}^{(p-1)/2} \\frac{-2 \\cdot 2^{2\\ell}}{2\\ell} = -\\sum_{\\ell=1}^{(p-1)/2} \\frac{4^{\\ell}}{\\ell}\n$$\nNext, using Fermat's theorem, we expand the second sum in (1) as\n$$\n-\\sum_{\\ell=1}^{(p-1)/2} \\frac{2^{2\\ell+1}}{2\\ell-1} \\equiv -\\sum_{\\ell=1}^{(p-1)/2} \\frac{2^{p+2\\ell}}{p+2\\ell-1} = -\\sum_{m=(p+1)/2}^{p-1} \\frac{2 \\cdot 4^{m}}{2m} = -\\sum_{m=(p+1)/2}^{p-1} \\frac{4^{m}}{m} (\\bmod p)\n$$\n(here we set $m = \\ell + \\frac{p-1}{2}$). Hence,\n$$\nS_{3} - 3 S_{2} \\equiv -\\sum_{\\ell=1}^{(p-1)/2} \\frac{4^{\\ell}}{\\ell} - \\sum_{m=(p+1)/2}^{p-1} \\frac{4^{m}}{m} = -S_{4} \\quad (\\bmod p)\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31945,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer and set $n=2^{k}+1$. Prove that $n$ is a prime number if and only if the following holds: there is a permutation $a_{1}, \\ldots, a_{n-1}$ of the numbers $1,2, \\ldots, n-1$ and a sequence of integers $g_{1}, g_{2}, \\ldots, g_{n-1}$ such that $n$ divides $g_{i}^{a_{i}}-a_{i+1}$ for every $i \\in\\{1,2, \\ldots, n-1\\}$, where we set $a_{n}=a_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N=\\{1,2, \\ldots, n-1\\}$. For $a, b \\in N$, we say that $b$ follows $a$ if there exists an integer $g$ such that $b \\equiv g^{a} (\\bmod n)$ and denote this property as $a \\rightarrow b$. This way we have a directed graph with $N$ as set of vertices. If $a_{1}, \\ldots, a_{n-1}$ is a permutation of $1,2, \\ldots, n-1$ such that $a_{1} \\rightarrow a_{2} \\rightarrow \\ldots \\rightarrow a_{n-1} \\rightarrow a_{1}$ then this is a Hamiltonian cycle in the graph.\n\nStep I. First consider the case when $n$ is composite. Let $n=p_{1}^{\\alpha_{1}} \\ldots p_{s}^{\\alpha_{s}}$ be its prime factorization. All primes $p_{i}$ are odd.\n\nSuppose that $\\alpha_{i}>1$ for some $i$. For all integers $a, g$ with $a \\geq 2$, we have $g^{a} \\not \\equiv p_{i} \\left(\\bmod p_{i}^{2}\\right)$, because $g^{a}$ is either divisible by $p_{i}^{2}$ or it is not divisible by $p_{i}$. It follows that in any Hamiltonian cycle $p_{i}$ comes immediately after $1$. The same argument shows that $2 p_{i}$ also should come immediately after $1$, which is impossible. Hence, there is no Hamiltonian cycle in the graph.\n\nNow suppose that $n$ is square-free. We have $n=p_{1} p_{2} \\ldots p_{s}>9$ and $s \\geq 2$. Assume that there exists a Hamiltonian cycle. There are $\\frac{n-1}{2}$ even numbers in this cycle, and each number which follows one of them should be a quadratic residue modulo $n$. So, there should be at least $\\frac{n-1}{2}$ nonzero quadratic residues modulo $n$. On the other hand, for each $p_{i}$ there exist exactly $\\frac{p_{i}+1}{2}$ quadratic residues modulo $p_{i}$; by the Chinese Remainder Theorem, the number of quadratic residues modulo $n$ is exactly $\\frac{p_{1}+1}{2} \\cdot \\frac{p_{2}+1}{2} \\cdot \\ldots \\cdot \\frac{p_{s}+1}{2}$, including $0$. Then we have a contradiction by\n$$\n\\frac{p_{1}+1}{2} \\cdot \\frac{p_{2}+1}{2} \\cdot \\ldots \\cdot \\frac{p_{s}+1}{2} \\leq \\frac{2 p_{1}}{3} \\cdot \\frac{2 p_{2}}{3} \\cdot \\ldots \\cdot \\frac{2 p_{s}}{3}=\\left(\\frac{2}{3}\\right)^{s} n \\leq \\frac{4 n}{9}<\\frac{n-1}{2} .\n$$\nThis proves the \"if\"-part of the problem.\n\nStep II. Now suppose that $n$ is prime. For any $a \\in N$, denote by $\\nu_{2}(a)$ the exponent of $2$ in the prime factorization of $a$, and let $\\mu(a)=\\max \\left\\{t \\in[0, k] \\mid 2^{t} \\rightarrow a\\right\\}$.\n\nLemma. For any $a, b \\in N$, we have $a \\rightarrow b$ if and only if $\\nu_{2}(a) \\leq \\mu(b)$.\n\nProof. Let $\\ell=\\nu_{2}(a)$ and $m=\\mu(b)$.\n\nSuppose $\\ell \\leq m$. Since $b$ follows $2^{m}$, there exists some $g_{0}$ such that $b \\equiv g_{0}^{2^{m}} (\\bmod n)$. By $\\operatorname{gcd}(a, n-1)=2^{\\ell}$ there exist some integers $p$ and $q$ such that $p a-q(n-1)=2^{\\ell}$. Choosing $g=g_{0}^{2^{m-\\ell} p}$ we have $g^{a}=g_{0}^{2^{m-\\ell} p a}=g_{0}^{2^{m}+2^{m-\\ell} q(n-1)} \\equiv g_{0}^{2^{m}} \\equiv b (\\bmod n)$ by Fermat's theorem. Hence, $a \\rightarrow b$.\n\nTo prove the reverse statement, suppose that $a \\rightarrow b$, so $b \\equiv g^{a} (\\bmod n)$ with some $g$. Then $b \\equiv\\left(g^{a / 2^{\\ell}}\\right)^{2^{\\ell}}$, and therefore $2^{\\ell} \\rightarrow b$. By the definition of $\\mu(b)$, we have $\\mu(b) \\geq \\ell$. The lemma is proved.\n\nNow for every $i$ with $0 \\leq i \\leq k$, let\n$$\n\\begin{aligned}\nA_{i} & =\\left\\{a \\in N \\mid \\nu_{2}(a)=i\\right\\}, \\\\\nB_{i} & =\\{a \\in N \\mid \\mu(a)=i\\}, \\\\\n\\text{and } C_{i} & =\\{a \\in N \\mid \\mu(a) \\geq i\\}=B_{i} \\cup B_{i+1} \\cup \\ldots \\cup B_{k} .\n\\end{aligned}\n$$\nWe claim that $\\left|A_{i}\\right|=\\left|B_{i}\\right|$ for all $0 \\leq i \\leq k$. Obviously we have $\\left|A_{i}\\right|=2^{k-i-1}$ for all $i= 0, \\ldots, k-1$, and $\\left|A_{k}\\right|=1$. Now we determine $\\left|C_{i}\\right|$. We have $\\left|C_{0}\\right|=n-1$ and by Fermat's theorem we also have $C_{k}=\\{1\\}$, so $\\left|C_{k}\\right|=1$. Next, notice that $C_{i+1}=\\left\\{x^{2} \\bmod n \\mid x \\in C_{i}\\right\\}$. For every $a \\in N$, the relation $x^{2} \\equiv a (\\bmod n)$ has at most two solutions in $N$. Therefore we have $2\\left|C_{i+1}\\right| \\leq\\left|C_{i}\\right|$, with the equality achieved only if for every $y \\in C_{i+1}$, there exist distinct elements $x, x' \\in C_{i}$ such that $x^{2} \\equiv x'^{2} \\equiv y (\\bmod n)$ (this implies $x+x'=n$). Now, since $2^{k}\\left|C_{k}\\right|=\\left|C_{0}\\right|$, we obtain that this equality should be achieved in each step. Hence $\\left|C_{i}\\right|=2^{k-i}$ for $0 \\leq i \\leq k$, and therefore $\\left|B_{i}\\right|=2^{k-i-1}$ for $0 \\leq i \\leq k-1$ and $\\left|B_{k}\\right|=1$.\n\nFrom the previous arguments we can see that for each $z \\in C_{i} (0 \\leq i<k)$ the equation $x^{2} \\equiv z^{2} (\\bmod n)$ has two solutions in $C_{i}$, so we have $n-z \\in C_{i}$. Hence, for each $i=0,1, \\ldots, k-1$, exactly half of the elements of $C_{i}$ are odd. The same statement is valid for $B_{i}=C_{i} \\backslash C_{i+1}$ for $0 \\leq i \\leq k-2$. In particular, each such $B_{i}$ contains an odd number. Note that $B_{k}=\\{1\\}$ also contains an odd number, and $B_{k-1}=\\left\\{2^{k}\\right\\}$ since $C_{k-1}$ consists of the two square roots of $1$ modulo $n$.\n\nStep III. Now we construct a Hamiltonian cycle in the graph. First, for each $i$ with $0 \\leq i \\leq k$, connect the elements of $A_{i}$ to the elements of $B_{i}$ by means of an arbitrary bijection. After performing this for every $i$, we obtain a subgraph with all vertices having in-degree $1$ and outdegree $1$, so the subgraph is a disjoint union of cycles. If there is a unique cycle, we are done. Otherwise, we modify the subgraph in such a way that the previous property is preserved and the number of cycles decreases; after a finite number of steps we arrive at a single cycle.\n\nFor every cycle $C$, let $\\lambda(C)=\\min _{c \\in C} \\nu_{2}(c)$. Consider a cycle $C$ for which $\\lambda(C)$ is maximal. If $\\lambda(C)=0$, then for any other cycle $C'$ we have $\\lambda\\left(C'\\right)=0$. Take two arbitrary vertices $a \\in C$ and $a' \\in C'$ such that $\\nu_{2}(a)=\\nu_{2}\\left(a'\\right)=0$; let their direct successors be $b$ and $b'$, respectively. Then we can unify $C$ and $C'$ to a single cycle by replacing the edges $a \\rightarrow b$ and $a' \\rightarrow b'$ by $a \\rightarrow b'$ and $a' \\rightarrow b$.\n\nNow suppose that $\\lambda=\\lambda(C) \\geq 1$; let $a \\in C \\cap A_{\\lambda}$. If there exists some $a' \\in A_{\\lambda} \\backslash C$, then $a'$ lies in another cycle $C'$ and we can merge the two cycles in exactly the same way as above. So, the only remaining case is $A_{\\lambda} \\subset C$. Since the edges from $A_{\\lambda}$ lead to $B_{\\lambda}$, we get also $B_{\\lambda} \\subset C$. If $\\lambda \\neq k-1$ then $B_{\\lambda}$ contains an odd number; this contradicts the assumption $\\lambda(C)>0$. Finally, if $\\lambda=k-1$, then $C$ contains $2^{k-1}$ which is the only element of $A_{k-1}$. Since $B_{k-1}=\\left\\{2^{k}\\right\\}=A_{k}$ and $B_{k}=\\{1\\}$, the cycle $C$ contains the path $2^{k-1} \\rightarrow 2^{k} \\rightarrow 1$ and it contains an odd number again. This completes the proof of the \"only if\"-part of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31946,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a function $f: \\mathbb{R}^k \\rightarrow \\mathbb{R}$ is a metapolynomial if, for some positive integers $m$ and $n$, it can be represented in the form\n$$\nf\\left(x_1, \\ldots, x_k\\right)=\\max_{i=1, \\ldots, m} \\min_{j=1, \\ldots, n} P_{i, j}\\left(x_1, \\ldots, x_k\\right)\n$$\nwhere $P_{i, j}$ are multivariate polynomials. Prove that the product of two metapolynomials is also a metapolynomial.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the notation $f(x)=f\\left(x_1, \\ldots, x_k\\right)$ for $x=\\left(x_1, \\ldots, x_k\\right)$ and $[m]=\\{1,2, \\ldots, m\\}$. Observe that if a metapolynomial $f(x)$ admits a representation like the one in the statement for certain positive integers $m$ and $n$, then they can be replaced by any $m' \\geq m$ and $n' \\geq n$. For instance, if we want to replace $m$ by $m+1$ then it is enough to define $P_{m+1, j}(x)=P_{m, j}(x)$ and note that repeating elements of a set do not change its maximum nor its minimum. So one can assume that any two metapolynomials are defined with the same $m$ and $n$. We reserve letters $P$ and $Q$ for polynomials, so every function called $P, P_{i, j}, Q, Q_{i, j}, \\ldots$ is a polynomial function.\n\nWe start with a lemma that is useful to change expressions of the form min max $f_{i, j}$ to ones of the form max min $g_{i, j}$.\n\nLemma. Let $\\{a_{i, j}\\}$ be real numbers, for all $i \\in[m]$ and $j \\in[n]$. Then\n$$\n\\min_{i \\in[m]} \\max_{j \\in[n]} a_{i, j}=\\max_{j_1, \\ldots, j_m \\in[n]} \\min_{i \\in[m]} a_{i, j_i},\n$$\nwhere the max in the right-hand side is over all vectors $\\left(j_1, \\ldots, j_m\\right)$ with $j_1, \\ldots, j_m \\in[n]$.\n\nProof. We can assume for all $i$ that $a_{i, n}=\\max \\{a_{i, 1}, \\ldots, a_{i, n}\\}$ and $a_{m, n}=\\min \\{a_{1, n}, \\ldots, a_{m, n}\\}$. The left-hand side is $=a_{m, n}$ and hence we need to prove the same for the right-hand side. If $\\left(j_1, j_2, \\ldots, j_m\\right)=(n, n, \\ldots, n)$ then $\\min \\{a_{1, j_1}, \\ldots, a_{m, j_m}\\}=\\min \\{a_{1, n}, \\ldots, a_{m, n}\\}=a_{m, n}$ which implies that the right-hand side is $\\geq a_{m, n}$. It remains to prove the opposite inequality and this is equivalent to $\\min \\{a_{1, j_1}, \\ldots, a_{m, j_m}\\} \\leq a_{m, n}$ for all possible $\\left(j_1, j_2, \\ldots, j_m\\right)$. This is true because $\\min \\{a_{1, j_1}, \\ldots, a_{m, j_m}\\} \\leq a_{m, j_m} \\leq a_{m, n}$.\n\nWe need to show that the family $\\mathcal{M}$ of metapolynomials is closed under multiplication, but it turns out easier to prove more: that it is also closed under addition, maxima and minima.\n\nFirst we prove the assertions about the maxima and the minima. If $f_1, \\ldots, f_r$ are metapolynomials, assume them defined with the same $m$ and $n$. Then\n$$\nf=\\max \\{f_1, \\ldots, f_r\\}=\\max \\{\\max_{i \\in[m]} \\min_{j \\in[n]} P_{i, j}^1, \\ldots, \\max_{i \\in[m]} \\min_{j \\in[n]} P_{i, j}^r\\}=\\max_{s \\in[r], i \\in[m]} \\min_{j \\in[n]} P_{i, j}^s\n$$\nIt follows that $f=\\max \\{f_1, \\ldots, f_r\\}$ is a metapolynomial. The same argument works for the minima, but first we have to replace min max by max min, and this is done via the lemma.\n\nAnother property we need is that if $f=\\max \\min P_{i, j}$ is a metapolynomial then so is $-f$. Indeed, $-f=\\min \\left(-\\min P_{i, j}\\right)=\\min \\max P_{i, j}$.\n\nTo prove $\\mathcal{M}$ is closed under addition let $f=\\max \\min P_{i, j}$ and $g=\\max \\min Q_{i, j}$. Then\n$$\n\\begin{gathered}\nf(x)+g(x)=\\max_{i \\in[m]} \\min_{j \\in[n]} P_{i, j}(x)+\\max_{i \\in[m]} \\min_{j \\in[n]} Q_{i, j}(x) \\\\\n=\\max_{i_1, i_2 \\in[m]}\\left(\\min_{j \\in[n]} P_{i_1, j}(x)+\\min_{j \\in[n]} Q_{i_2, j}(x)\\right)=\\max_{i_1, i_2 \\in[m]} \\min_{j_1, j_2 \\in[n]}\\left(P_{i_1, j_1}(x)+Q_{i_2, j_2}(x)\\right),\n\\end{gathered}\n$$\nand hence $f(x)+g(x)$ is a metapolynomial.\n\nWe proved that $\\mathcal{M}$ is closed under sums, maxima and minima, in particular any function that can be expressed by sums, max, min, polynomials or even metapolynomials is in $\\mathcal{M}$.\n\nWe would like to proceed with multiplication along the same lines like with addition, but there is an essential difference. In general the product of the maxima of two sets is not equal to the maximum of the product of the sets. We need to deal with the fact that $a<b$ and $c<d$ do not imply $a c<b d$. However this is true for $a, b, c, d \\geq 0$.\n\nIn view of this we decompose each function $f(x)$ into its positive part $f^{+}(x)=\\max \\{f(x), 0\\}$ and its negative part $f^{-}(x)=\\max \\{0,-f(x)\\}$. Note that $f=f^{+}-f^{-}$ and $f^{+}, f^{-} \\in \\mathcal{M}$ if $f \\in \\mathcal{M}$. The whole problem reduces to the claim that if $f$ and $g$ are metapolynomials with $f, g \\geq 0$ then $f g$ is also a metapolynomial.\n\nAssuming this claim, consider arbitrary $f, g \\in \\mathcal{M}$. We have\n$$\nf g=\\left(f^{+}-f^{-}\\right)\\left(g^{+}-g^{-}\\right)=f^{+} g^{+}-f^{+} g^{-}-f^{-} g^{+}+f^{-} g^{-}\n$$\nand hence $f g \\in \\mathcal{M}$. Indeed, $\\mathcal{M}$ is closed under addition, also $f^{+} g^{+}, f^{+} g^{-}, f^{-} g^{+}, f^{-} g^{-} \\in \\mathcal{M}$ because $f^{+}, f^{-}, g^{+}, g^{-} \\geq 0$.\n\nIt remains to prove the claim. In this case $f, g \\geq 0$, and one can try to repeat the argument for the sum. More precisely, let $f=\\max \\min P_{i j} \\geq 0$ and $g=\\max \\min Q_{i j} \\geq 0$. Then\n$$\nf g=\\max \\min P_{i, j} \\cdot \\max \\min Q_{i, j}=\\max \\min P_{i, j}^{+} \\cdot \\max \\min Q_{i, j}^{+}=\\max \\min P_{i_1, j_1}^{+} \\cdot Q_{i_2, j_2}^{+}\n$$\nHence it suffices to check that $P^{+} Q^{+} \\in \\mathcal{M}$ for any pair of polynomials $P$ and $Q$. This reduces to the identity\n$$\nu^{+} v^{+}=\\max \\left\\{0, \\min \\{u v, u, v\\}, \\min \\left\\{u v, u v^{2}, u^{2} v\\right\\}, \\min \\left\\{u v, u, u^{2} v\\right\\}, \\min \\left\\{u v, u v^{2}, v\\right\\}\\right\\},\n$$\nwith $u$ replaced by $P(x)$ and $v$ replaced by $Q(x)$. The formula is proved by a case-by-case analysis. If $u \\leq 0$ or $v \\leq 0$ then both sides equal 0. In case $u, v \\geq 0$, the right-hand side is clearly $\\leq u v$. To prove the opposite inequality we use that $u v$ equals\n$$\n\\begin{array}{ll}\n\\min \\{u v, u, v\\} & \\text{ if } 0 \\leq u, v \\leq 1, \\\\\n\\min \\left\\{u v, u v^{2}, u^{2} v\\right\\} & \\text{ if } 1 \\leq u, v, \\\\\n\\min \\left\\{u v, u, u^{2} v\\right\\} & \\text{ if } 0 \\leq v \\leq 1 \\leq u, \\\\\n\\min \\left\\{u v, u v^{2}, v\\right\\} & \\text{ if } 0 \\leq u \\leq 1 \\leq v .\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31947,
    "subject": "Mathematics (Multi-modal)",
    "question": "Several positive integers are written in a row. Iteratively, Alice chooses two adjacent numbers $x$ and $y$ such that $x > y$ and $x$ is to the left of $y$, and replaces the pair $(x, y)$ by either $(y+1, x)$ or $(x-1, x)$. Prove that she can perform only finitely many such iterations.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note first that the allowed operation does not change the maximum $M$ of the initial sequence. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be the numbers obtained at some point of the process. Consider the sum\n$$\nS = a_{1} + 2 a_{2} + \\cdots + n a_{n}\n$$\nWe claim that $S$ increases by a positive integer amount with every operation. Let the operation replace the pair $(a_{i}, a_{i+1})$ by a pair $(c, a_{i})$, where $a_{i} > a_{i+1}$ and $c = a_{i+1} + 1$ or $c = a_{i} - 1$. Then the new and the old value of $S$ differ by\n$$\nd = \\left(i c + (i+1) a_{i}\\right) - \\left(i a_{i} + (i+1) a_{i+1}\\right) = a_{i} - a_{i+1} + i (c - a_{i+1})\n$$\nThe integer $d$ is positive since $a_{i} - a_{i+1} \\geq 1$ and $c - a_{i+1} \\geq 0$.\n\nOn the other hand $S \\leq (1 + 2 + \\cdots + n) M$ as $a_{i} \\leq M$ for all $i = 1, \\ldots, n$. Since $S$ increases by at least 1 at each step and never exceeds the constant $(1 + 2 + \\cdots + n) M$, the process stops after a finite number of iterations.\nLike in the first solution, note that the operations do not change the maximum $M$ of the initial sequence. Now consider the reverse lexicographical order for $n$-tuples of integers. We say that $(x_{1}, \\ldots, x_{n}) < (y_{1}, \\ldots, y_{n})$ if $x_{n} < y_{n}$, or if $x_{n} = y_{n}$ and $x_{n-1} < y_{n-1}$, or if $x_{n} = y_{n}$, $x_{n-1} = y_{n-1}$ and $x_{n-2} < y_{n-2}$, etc. Each iteration creates a sequence that is greater than the previous one with respect to this order, and no sequence occurs twice during the process. On the other hand, there are finitely many possible sequences because their terms are always positive integers not exceeding $M$. Hence the process cannot continue forever.\nLet the current numbers be $a_{1}, a_{2}, \\ldots, a_{n}$. Define the score $s_{i}$ of $a_{i}$ as the number of $a_{j}$'s that are less than $a_{i}$. Call the sequence $s_{1}, s_{2}, \\ldots, s_{n}$ the score sequence of $a_{1}, a_{2}, \\ldots, a_{n}$.\n\nLet us say that a sequence $x_{1}, \\ldots, x_{n}$ dominates a sequence $y_{1}, \\ldots, y_{n}$ if the first index $i$ with $x_{i} \\neq y_{i}$ is such that $x_{i} < y_{i}$. We show that after each operation the new score sequence dominates the old one. Score sequences do not repeat, and there are finitely many possibilities for them, no more than $(n-1)^{n}$. Hence the process will terminate.\n\nConsider an operation that replaces $(x, y)$ by $(a, x)$, with $a = y + 1$ or $a = x - 1$. Suppose that $x$ was originally at position $i$. For each $j < i$ the score $s_{j}$ does not increase with the change because $y \\leq a$ and $x \\leq x$. If $s_{j}$ decreases for some $j < i$ then the new score sequence dominates the old one. Assume that $s_{j}$ stays the same for all $j < i$ and consider $s_{i}$. Since $x > y$ and $y \\leq a \\leq x$, we see that $s_{i}$ decreases by at least 1. This concludes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31948,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geq 1$ be an integer. What is the maximum number of disjoint pairs of elements of the set $\\{1,2, \\ldots, n\\}$ such that the sums of the different pairs are different integers not exceeding $n$?",
    "options": [],
    "answer": "floor((2n−1)/5)",
    "solution": "Consider $x$ such pairs in $\\{1,2, \\ldots, n\\}$. The sum $S$ of the $2x$ numbers in them is at least $1+2+\\cdots+2x$ since the pairs are disjoint. On the other hand $S \\leq n+(n-1)+\\cdots+(n-x+1)$ because the sums of the pairs are different and do not exceed $n$. This gives the inequality\n$$\n\\frac{2x(2x+1)}{2} \\leq nx-\\frac{x(x-1)}{2}\n$$\nwhich leads to $x \\leq \\frac{2n-1}{5}$. Hence there are at most $\\left\\lfloor\\frac{2n-1}{5}\\right\\rfloor$ pairs with the given properties.\n\nWe show a construction with exactly $\\left\\lfloor\\frac{2n-1}{5}\\right\\rfloor$ pairs. First consider the case $n=5k+3$ with $k \\geq 0$, where $\\left\\lfloor\\frac{2n-1}{5}\\right\\rfloor=2k+1$. The pairs are displayed in the following table.\n\n| Pairs | $3k+1$ | $3k$ | $\\cdots$ | $2k+2$ | $4k+2$ | $4k+1$ | $\\cdots$ | $3k+3$ | $3k+2$ |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n|       | 2     | 4     | $\\cdots$ | $2k$   | 1     | 3     | $\\cdots$ | $2k-1$ | $2k+1$ |\n| Sums  | $3k+3$ | $3k+4$ | $\\cdots$ | $4k+2$ | $4k+3$ | $4k+4$ | $\\cdots$ | $5k+2$ | $5k+3$ |\n\nThe $2k+1$ pairs involve all numbers from $1$ to $4k+2$; their sums are all numbers from $3k+3$ to $5k+3$. The same construction works for $n=5k+4$ and $n=5k+5$ with $k \\geq 0$. In these cases the required number $\\left\\lfloor\\frac{2n-1}{5}\\right\\rfloor$ of pairs equals $2k+1$ again, and the numbers in the table do not exceed $5k+3$. In the case $n=5k+2$ with $k \\geq 0$ one needs only $2k$ pairs. They can be obtained by ignoring the last column of the table (thus removing $5k+3$). Finally, $2k$ pairs are also needed for the case $n=5k+1$ with $k \\geq 0$. Now it suffices to ignore the last column of the table and then subtract $1$ from each number in the first row.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31949,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a $999 \\times 999$ square table some cells are white and the remaining ones are red. Let $T$ be the number of triples $(C_{1}, C_{2}, C_{3})$ of cells, the first two in the same row and the last two in the same column, with $C_{1}$ and $C_{3}$ white and $C_{2}$ red. Find the maximum value $T$ can attain.",
    "options": [],
    "answer": "4*999^4/27",
    "solution": "We prove that in an $n \\times n$ square table there are at most $\\frac{4 n^{4}}{27}$ such triples.\nLet row $i$ and column $j$ contain $a_{i}$ and $b_{j}$ white cells respectively, and let $R$ be the set of red cells. For every red cell $(i, j)$ there are $a_{i} b_{j}$ admissible triples $(C_{1}, C_{2}, C_{3})$ with $C_{2}=(i, j)$, therefore\n$$\nT=\\sum_{(i, j) \\in R} a_{i} b_{j}\n$$\nWe use the inequality $2 a b \\leq a^{2}+b^{2}$ to obtain\n$$\nT \\leq \\frac{1}{2} \\sum_{(i, j) \\in R}\\left(a_{i}^{2}+b_{j}^{2}\\right)=\\frac{1}{2} \\sum_{i=1}^{n}\\left(n-a_{i}\\right) a_{i}^{2}+\\frac{1}{2} \\sum_{j=1}^{n}\\left(n-b_{j}\\right) b_{j}^{2} .\n$$\nThis is because there are $n-a_{i}$ red cells in row $i$ and $n-b_{j}$ red cells in column $j$. Now we maximize the right-hand side.\nBy the AM-GM inequality we have\n$$\n(n-x) x^{2}=\\frac{1}{2}(2 n-2 x) \\cdot x \\cdot x \\leq \\frac{1}{2}\\left(\\frac{2 n}{3}\\right)^{3}=\\frac{4 n^{3}}{27}\n$$\nwith equality if and only if $x=\\frac{2 n}{3}$. By putting everything together, we get\n$$\nT \\leq \\frac{n}{2} \\frac{4 n^{3}}{27}+\\frac{n}{2} \\frac{4 n^{3}}{27}=\\frac{4 n^{4}}{27}\n$$\nIf $n=999$ then any coloring of the square table with $x=\\frac{2 n}{3}=666$ white cells in each row and column attains the maximum as all inequalities in the previous argument become equalities. For example color a cell $(i, j)$ white if $i-j \\equiv 1,2, \\ldots, 666(\\bmod 999)$, and red otherwise.\nTherefore the maximum value $T$ can attain is $T=\\frac{4 \\cdot 999^{4}}{27}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31950,
    "subject": "Mathematics (Multi-modal)",
    "question": "Players $A$ and $B$ play a game with $N \\geq 2012$ coins and $2012$ boxes arranged around a circle. Initially $A$ distributes the coins among the boxes so that there is at least $1$ coin in each box. Then the two of them make moves in the order $B, A, B, A, \\ldots$ by the following rules:\n- On every move of his $B$ passes $1$ coin from every box to an adjacent box.\n- On every move of hers $A$ chooses several coins that were not involved in $B$'s previous move and are in different boxes. She passes every chosen coin to an adjacent box.\n\nPlayer $A$'s goal is to ensure at least $1$ coin in each box after every move of hers, regardless of how $B$ plays and how many moves are made. Find the least $N$ that enables her to succeed.",
    "options": [],
    "answer": "4022",
    "solution": "We argue for a general $n \\geq 7$ instead of $2012$ and prove that the required minimum $N$ is $2n-2$. For $n=2012$ this gives $N_{\\text{min}}=4022$.\n\na. If $N=2n-2$ player $A$ can achieve her goal. Let her start the game with a regular distribution: $n-2$ boxes with $2$ coins and $2$ boxes with $1$ coin. Call the boxes of the two kinds red and white respectively. We claim that on her first move $A$ can achieve a regular distribution again, regardless of $B$'s first move $M$. She acts according as the following situation $S$ occurs after $M$ or not: The initial distribution contains a red box $R$ with $2$ white neighbors, and $R$ receives no coins from them on move $M$.\n\nSuppose that $S$ does not occur. Exactly one of the coins $c_1$ and $c_2$ in a given red box $X$ is involved in $M$, say $c_1$. If $M$ passes $c_1$ to the right neighbor of $X$, let $A$ pass $c_2$ to its left neighbor, and vice versa. By doing so with all red boxes $A$ performs a legal move $M'$. Thus $M$ and $M'$ combined move the $2$ coins of every red box in opposite directions. Hence after $M$ and $M'$ are complete each neighbor of a red box $X$ contains exactly $1$ coin that was initially in $X$. So each box with a red neighbor is non-empty after $M'$. If initially there is a box $X$ with $2$ white neighbors ($X$ is red and unique) then $X$ receives a coin from at least one of them on move $M$ since $S$ does not occur. Such a coin is not involved in $M'$, so $X$ is also non-empty after $M'$. Furthermore each box $Y$ has given away its initial content after $M$ and $M'$. A red neighbor of $Y$ adds $1$ coin to it; a white neighbor adds at most $1$ coin because it is not involved in $M'$. Hence each box contains $1$ or $2$ coins after $M'$. Because $N=2n-2$, such a distribution is regular.\n\nNow let $S$ occur after move $M$. Then $A$ leaves untouched the exceptional red box $R$. With all remaining red boxes she proceeds like in the previous case, thus making a legal move $M''$. Box $R$ receives no coins from its neighbors on either move, so there is $1$ coin in it after $M''$. Like above $M$ and $M''$ combined pass exactly $1$ coin from every red box different from $R$ to each of its neighbors. Every box except $R$ has a red neighbor different from $R$, hence all boxes are non-empty after $M''$. Next, each box $Y$ except $R$ loses its initial content after $M$ and $M''$. A red neighbor of $Y$ adds at most $1$ coin to it; a white neighbor also adds at most $1$ coin as it does not participate in $M''$. Thus each box has $1$ or $2$ coins after $M''$, and the obtained distribution is regular.\n\nPlayer $A$ can apply the described strategy indefinitely, so $N=2n-2$ enables her to succeed.\n\nb. For $N \\leq 2n-3$ player $B$ can achieve an empty box after some move of $A$. Let $\\alpha$ be a set of $\\ell$ consecutive boxes containing a total of $N(\\alpha)$ coins. We call $\\alpha$ an arc if $\\ell \\leq n-2$ and $N(\\alpha) \\leq 2\\ell-3$. Note that $\\ell \\geq 2$ by the last condition. Moreover if both extremes of $\\alpha$ are non-empty boxes then $N(\\alpha) \\geq 2$, so that $N(\\alpha) \\leq 2\\ell-3$ implies $\\ell \\geq 3$. Observe also that if an extreme $X$ of $\\alpha$ has more than $1$ coin then ignoring $X$ yields a shorter arc. It follows that every arc contains an arc whose extremes have at most $1$ coin each.\n\nGiven a clockwise labeling $1,2, \\ldots, n$ of the boxes, suppose that boxes $1,2, \\ldots, \\ell$ form an $\\operatorname{arc} \\alpha$, with $\\ell \\leq n-2$ and $N(\\alpha) \\leq 2\\ell-3$. Suppose also that all $n \\geq 7$ boxes are non-empty. Then $B$ can move so that an arc $\\alpha'$ with $N\\left(\\alpha'\\right)<N(\\alpha)$ will appear after any response of $A$.\n\nOne may assume exactly $1$ coin in boxes $1$ and $\\ell$ by a previous remark. Let $B$ pass $1$ coin in counterclockwise direction from box $1$ and box $n$, and in clockwise direction from each remaining box. This leaves $N(\\alpha)-2$ coins in the boxes of $\\alpha$. In addition, due to $3 \\leq \\ell \\leq n-2$, box $\\ell$ has exactly $1$ coin $c$, the one received from box $\\ell-1$.\n\nLet player $A$'s next move $M$ pass $k \\leq 2$ coins to boxes $1,2, \\ldots, \\ell$ from the remaining ones. Only boxes $1$ and $\\ell$ can receive such coins, at most $1$ each. If $k<2$ then after move $M$ boxes $1,2, \\ldots, \\ell$ form an arc $\\alpha'$ with $N\\left(\\alpha'\\right)<N(\\alpha)$. If $k=2$ then $M$ adds a coin to box $\\ell$. Also $M$ does not move coin $c$ from $\\ell$ because $c$ is involved in the previous move of $B$. In summary boxes $1,2, \\ldots, \\ell$ contain $N(\\alpha)$ coins like before, so they form an arc. However there are $2$ coins now in the extreme $\\ell$ of the arc. Ignore $\\ell$ to obtain a shorter $\\operatorname{arc} \\alpha'$ with $N\\left(\\alpha'\\right)<N(\\alpha)$.\n\nConsider any initial distribution without empty boxes. Since $N \\leq 2n-3$, there are at least $3$ boxes in it with exactly $1$ coin. It follows from $n \\geq 7$ that some $2$ of them are the extremes of an arc $\\alpha$. Hence $B$ can make the move described above, which leads to an arc $\\alpha'$ with $N\\left(\\alpha'\\right)<N(\\alpha)$ after $A$'s response. If all boxes in the new distribution are non-empty he can repeat the same, and so on. Because $N(\\alpha)$ cannot decrease indefinitely, an empty box will occur after some move of $A$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31951,
    "subject": "Mathematics (Multi-modal)",
    "question": "The columns and the rows of a $3n \\times 3n$ square board are numbered $1, 2, \\ldots, 3n$. Every square $(x, y)$ with $1 \\leq x, y \\leq 3n$ is colored asparagus, byzantium or citrine according as the modulo $3$ remainder of $x+y$ is $0$, $1$ or $2$ respectively. One token colored asparagus, byzantium or citrine is placed on each square, so that there are $3n^{2}$ tokens of each color.\n\nSuppose that one can permute the tokens so that each token is moved to a distance of at most $d$ from its original position, each asparagus token replaces a byzantium token, each byzantium token replaces a citrine token, and each citrine token replaces an asparagus token. Prove that it is possible to permute the tokens so that each token is moved to a distance of at most $d+2$ from its original position, and each square contains a token with the same color as the square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality it suffices to prove that the A-tokens can be moved to distinct A-squares in such a way that each A-token is moved to a distance at most $d+2$ from its original place. This means we need a perfect matching between the $3n^{2}$ A-squares and the $3n^{2}$ A-tokens such that the distance in each pair of the matching is at most $d+2$.\n\nTo find the matching, we construct a bipartite graph. The A-squares will be the vertices in one class of the graph; the vertices in the other class will be the A-tokens.\n\nSplit the board into $3 \\times 1$ horizontal triminos; then each trimino contains exactly one A-square. Take a permutation $\\pi$ of the tokens which moves A-tokens to B-tokens, B-tokens to C-tokens, and C-tokens to A-tokens, in each case to a distance at most $d$. For each A-square $S$, and for each A-token $T$, connect $S$ and $T$ by an edge if $T$, $\\pi(T)$ or $\\pi^{-1}(T)$ is on the trimino containing $S$. We allow multiple edges; it is even possible that the same square and the same token are connected with three edges. Obviously the lengths of the edges in the graph do not exceed $d+2$. By length of an edge we mean the distance between the A-square and the A-token it connects.\n\nEach A-token $T$ is connected with the three A-squares whose triminos contain $T$, $\\pi(T)$ and $\\pi^{-1}(T)$. Therefore in the graph all tokens are of degree $3$. We show that the same is true for the A-squares. Let $S$ be an arbitrary A-square, and let $T_{1}, T_{2}, T_{3}$ be the three tokens on the trimino containing $S$. For $i=1,2,3$, if $T_{i}$ is an A-token, then $S$ is connected with $T_{i}$; if $T_{i}$ is a B-token then $S$ is connected with $\\pi^{-1}\\left(T_{i}\\right)$; finally, if $T_{i}$ is a C-token then $S$ is connected with $\\pi\\left(T_{i}\\right)$. Hence in the graph the A-squares also are of degree $3$.\n\nSince the A-squares are of degree $3$, from every set $\\mathcal{S}$ of A-squares exactly $3|\\mathcal{S}|$ edges start. These edges end in at least $|\\mathcal{S}|$ tokens because the A-tokens also are of degree $3$. Hence every set $\\mathcal{S}$ of A-squares has at least $|\\mathcal{S}|$ neighbors among the A-tokens.\n\nTherefore, by Hall's marriage theorem, the graph contains a perfect matching between the two vertex classes. So there is a perfect matching between the A-squares and A-tokens with edges no longer than $d+2$. It follows that the tokens can be permuted as specified in the problem statement.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 31952,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the triangle $A B C$ the point $J$ is the center of the excircle opposite to $A$. This excircle is tangent to the side $B C$ at $M$, and to the lines $A B$ and $A C$ at $K$ and $L$ respectively. The lines $L M$ and $B J$ meet at $F$, and the lines $K M$ and $C J$ meet at $G$. Let $S$ be the point of intersection of the lines $A F$ and $B C$, and let $T$ be the point of intersection of the lines $A G$ and $B C$. Prove that $M$ is the midpoint of $S T$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha=\\angle C A B$, $\\beta=\\angle A B C$ and $\\gamma=\\angle B C A$. The line $A J$ is the bisector of $\\angle C A B$, so $\\angle J A K=\\angle J A L=\\frac{\\alpha}{2}$. By $\\angle A K J=\\angle A L J=90^{\\circ}$ the points $K$ and $L$ lie on the circle $\\omega$ with diameter $A J$.\n\nThe triangle $K B M$ is isosceles as $B K$ and $B M$ are tangents to the excircle. Since $B J$ is the bisector of $\\angle K B M$, we have $\\angle M B J=90^{\\circ}-\\frac{\\beta}{2}$ and $\\angle B M K=\\frac{\\beta}{2}$. Likewise $\\angle M C J=90^{\\circ}-\\frac{\\gamma}{2}$ and $\\angle C M L=\\frac{\\gamma}{2}$. Also $\\angle B M F=\\angle C M L$, therefore\n$$\n\\angle L F J=\\angle M B J-\\angle B M F=\\left(90^{\\circ}-\\frac{\\beta}{2}\\right)-\\frac{\\gamma}{2}=\\frac{\\alpha}{2}=\\angle L A J .\n$$\nHence $F$ lies on the circle $\\omega$. (By the angle computation, $F$ and $A$ are on the same side of $B C$.) Analogously, $G$ also lies on $\\omega$. Since $A J$ is a diameter of $\\omega$, we obtain $\\angle A F J=\\angle A G J=90^{\\circ}$.\n\n![](attached_image_1.png)\n\nThe lines $A B$ and $B C$ are symmetric with respect to the external bisector $B F$. Because $A F \\perp B F$ and $K M \\perp B F$, the segments $S M$ and $A K$ are symmetric with respect to $B F$, hence $S M=A K$. By symmetry $T M=A L$. Since $A K$ and $A L$ are equal as tangents to the excircle, it follows that $S M=T M$, and the proof is complete.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31953,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D$, $H$, $F$, $G$ are concyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We show first that the triangles $FDG$ and $FBE$ are similar. Since $ABCD$ is cyclic, the triangles $EAB$ and $EDC$ are similar, as well as $FAB$ and $FCD$. The parallelogram $ECGD$ yields $GD = EC$ and $\\angle CDG = \\angle DCE$; also $\\angle DCE = \\angle DCA = \\angle DBA$ by inscribed angles. Therefore\n$$\n\\begin{gathered}\n\\angle FDG = \\angle FDC + \\angle CDG = \\angle FBA + \\angle ABD = \\angle FBE, \\\\\n\\frac{GD}{EB} = \\frac{CE}{EB} = \\frac{CD}{AB} = \\frac{FD}{FB} .\n\\end{gathered}\n$$\nIt follows that $FDG$ and $FBE$ are similar, and so $\\angle FGD = \\angle FEB$.\n![](attached_image_1.png)\nSince $H$ is the reflection of $E$ with respect to $FD$, we conclude that\n$$\n\\angle FHD = \\angle FED = 180^{\\circ} - \\angle FEB = 180^{\\circ} - \\angle FGD .\n$$\n\nThis proves that $D$, $H$, $F$, $G$ are concyclic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31954,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $A B C$ the points $D$, $E$ and $F$ are the feet of the altitudes through $A$, $B$ and $C$ respectively. The incenters of the triangles $A E F$ and $B D F$ are $I_{1}$ and $I_{2}$ respectively; the circumcenters of the triangles $A C I_{1}$ and $B C I_{2}$ are $O_{1}$ and $O_{2}$ respectively. Prove that $I_{1} I_{2}$ and $O_{1} O_{2}$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle C A B=\\alpha$, $\\angle A B C=\\beta$, $\\angle B C A=\\gamma$. We start by showing that $A$, $B$, $I_{1}$ and $I_{2}$ are concyclic. Since $A I_{1}$ and $B I_{2}$ bisect $\\angle C A B$ and $\\angle A B C$, their extensions beyond $I_{1}$ and $I_{2}$ meet at the incenter $I$ of the triangle. The points $E$ and $F$ are on the circle with diameter $B C$, so $\\angle A E F=\\angle A B C$ and $\\angle A F E=\\angle A C B$. Hence the triangles $A E F$ and $A B C$ are similar with ratio of similitude $\\frac{A E}{A B}=\\cos \\alpha$. Because $I_{1}$ and $I$ are their incenters, we obtain $I_{1} A=I A \\cos \\alpha$ and $I I_{1}=I A-I_{1} A=2 I A \\sin ^{2} \\frac{\\alpha}{2}$. By symmetry $I I_{2}=2 I B \\sin ^{2} \\frac{\\beta}{2}$. The law of sines in the triangle $A B I$ gives $I A \\sin \\frac{\\alpha}{2}=I B \\sin \\frac{\\beta}{2}$. Hence\n$$\nI I_{1} \\cdot I A=2\\left(I A \\sin \\frac{\\alpha}{2}\\right)^{2}=2\\left(I B \\sin \\frac{\\beta}{2}\\right)^{2}=I I_{2} \\cdot I B .\n$$\nTherefore $A$, $B$, $I_{1}$ and $I_{2}$ are concyclic, as claimed.\n\n![](attached_image_1.png)\n\nIn addition $I I_{1} \\cdot I A=I I_{2} \\cdot I B$ implies that $I$ has the same power with respect to the circles $\\left(A C I_{1}\\right)$, $\\left(B C I_{2}\\right)$ and $\\left(A B I_{1} I_{2}\\right)$. Then $C I$ is the radical axis of $\\left(A C I_{1}\\right)$ and $\\left(B C I_{2}\\right)$; in particular $C I$ is perpendicular to the line of centers $O_{1} O_{2}$.\n\nNow it suffices to prove that $C I \\perp I_{1} I_{2}$. Let $C I$ meet $I_{1} I_{2}$ at $Q$, then it is enough to check that $\\angle I I_{1} Q+\\angle I_{1} I Q=90^{\\circ}$. Since $\\angle I_{1} I Q$ is external for the triangle $A C I$, we have\n$$\n\\angle I I_{1} Q+\\angle I_{1} I Q=\\angle I I_{1} Q+(\\angle A C I+\\angle C A I)=\\angle I I_{1} I_{2}+\\angle A C I+\\angle C A I .\n$$\nIt remains to note that $\\angle I I_{1} I_{2}=\\frac{\\beta}{2}$ from the cyclic quadrilateral $A B I_{1} I_{2}$, and $\\angle A C I=\\frac{\\gamma}{2}$, $\\angle C A I=\\frac{\\alpha}{2}$. Therefore $\\angle I I_{1} Q+\\angle I_{1} I Q=\\frac{\\alpha}{2}+\\frac{\\beta}{2}+\\frac{\\gamma}{2}=90^{\\circ}$, completing the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31955,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $AB C$ be a triangle with $AB \\neq AC$ and circumcenter $O$. The bisector of $\\angle BAC$ intersects $BC$ at $D$. Let $E$ be the reflection of $D$ with respect to the midpoint of $BC$. The lines through $D$ and $E$ perpendicular to $BC$ intersect the lines $AO$ and $AD$ at $X$ and $Y$ respectively. Prove that the quadrilateral $BX CY$ is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The bisector of $\\angle BAC$ and the perpendicular bisector of $BC$ meet at $P$, the midpoint of the minor arc $\\widehat{BC}$ (they are different lines as $AB \\neq AC$). In particular $OP$ is perpendicular to $BC$ and intersects it at $M$, the midpoint of $BC$.\n\nDenote by $Y'$ the reflection of $Y$ with respect to $OP$. Since $\\angle BYC = \\angle BY'C$, it suffices to prove that $BX CY'$ is cyclic.\n\n![](attached_image_1.png)\n\nWe have\n$$\n\\angle XAP = \\angle OPA = \\angle EYP.\n$$\nThe first equality holds because $OA = OP$, and the second one because $EY$ and $OP$ are both perpendicular to $BC$ and hence parallel. But $\\{Y, Y'\\}$ and $\\{E, D\\}$ are pairs of symmetric points with respect to $OP$, it follows that $\\angle EYP = \\angle DY'P$ and hence\n$$\n\\angle XAP = \\angle DY'P = \\angle XY'P.\n$$\nThe last equation implies that $XAY'P$ is cyclic. By the powers of $D$ with respect to the circles $(XAY'P)$ and $(ABPC)$ we obtain\n$$\nXD \\cdot DY' = AD \\cdot DP = BD \\cdot DC\n$$\nIt follows that $BX CY'$ is cyclic, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31956,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle BCA = 90^{\\circ}$, and let $C_0$ be the foot of the altitude from $C$. Choose a point $X$ in the interior of the segment $CC_0$, and let $K, L$ be the points on the segments $AX, BX$ for which $BK = BC$ and $AL = AC$ respectively. Denote by $M$ the intersection of $AL$ and $BK$. Show that $MK = ML$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $C'$ be the reflection of $C$ in the line $AB$, and let $\\omega_1$ and $\\omega_2$ be the circles with centers $A$ and $B$, passing through $L$ and $K$ respectively. Since $AC' = AC = AL$ and $BC' = BC = BK$, both $\\omega_1$ and $\\omega_2$ pass through $C$ and $C'$. By $\\angle BCA = 90^{\\circ}$, $AC$ is tangent to $\\omega_2$ at $C$, and $BC$ is tangent to $\\omega_1$ at $C$. Let $K_1 \\neq K$ be the second intersection of $AX$ and $\\omega_2$, and let $L_1 \\neq L$ be the second intersection of $BX$ and $\\omega_1$.\n![](attached_image_1.png)\nBy the powers of $X$ with respect to $\\omega_2$ and $\\omega_1$,\n$$\nXK \\cdot XK_1 = XC \\cdot XC' = XL \\cdot XL_1,\n$$\nso the points $K_1, L, K, L_1$ lie on a circle $\\omega_3$.\nThe power of $A$ with respect to $\\omega_2$ gives\n$$\nAL^2 = AC^2 = AK \\cdot AK_1,\n$$\nindicating that $AL$ is tangent to $\\omega_3$ at $L$. Analogously, $BK$ is tangent to $\\omega_3$ at $K$. Hence $MK$ and $ML$ are the two tangents from $M$ to $\\omega_3$ and therefore $MK = ML$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31957,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with circumcenter $O$ and incenter $I$. The points $D$, $E$ and $F$ on the sides $BC$, $CA$ and $AB$ respectively are such that $BD + BF = CA$ and $CD + CE = AB$. The circumcircles of the triangles $BFD$ and $CDE$ intersect at $P \\neq D$. Prove that $OP = OI$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\omega_{A}$, $\\omega_{B}$ and $\\omega_{C}$ meet the bisectors $AI$, $BI$ and $CI$ at $A \\neq A'$, $B \\neq B'$ and $C \\neq C'$ respectively. The key observation is that $A'$, $B'$ and $C'$ do not depend on the particular choice of $D$, $E$ and $F$, provided that $BD + BF = CA$, $CD + CE = AB$ and $AE + AF = BC$ hold true (the last equality follows from the other two). For a proof we need the following fact.\n\nLemma. Given is an angle with vertex $A$ and measure $\\alpha$. A circle $\\omega$ through $A$ intersects the angle bisector at $L$ and sides of the angle at $X$ and $Y$. Then $AX + AY = 2AL \\cos \\frac{\\alpha}{2}$.\n\nProof. Note that $L$ is the midpoint of $\\operatorname{arc} \\widehat{XLY}$ in $\\omega$ and set $XL = YL = u$, $XY = v$. By Ptolemy's theorem $AX \\cdot YL + AY \\cdot XL = AL \\cdot XY$, which rewrites as $(AX + AY)u = AL \\cdot v$. Since $\\angle LXY = \\frac{\\alpha}{2}$ and $\\angle XLY = 180^\\circ - \\alpha$, we have $v = 2 \\cos \\frac{\\alpha}{2} u$ by the law of sines, and the claim follows. $\\square$\n\n![](attached_image_1.png)\n\nApply the lemma to $\\angle BAC = \\alpha$ and the circle $\\omega = \\omega_{A}$, which intersects $AI$ at $A'$. This gives $2AA' \\cos \\frac{\\alpha}{2} = AE + AF = BC$; by symmetry analogous relations hold for $BB'$ and $CC'$. It follows that $A'$, $B'$ and $C'$ are independent of the choice of $D$, $E$ and $F$, as stated.\n\nWe use the lemma two more times with $\\angle BAC = \\alpha$. Let $\\omega$ be the circle with diameter $AI$. Then $X$ and $Y$ are the tangency points of the incircle of $ABC$ with $AB$ and $AC$, and hence $AX = AY = \\frac{1}{2}(AB + AC - BC)$. So the lemma yields $2AI \\cos \\frac{\\alpha}{2} = AB + AC - BC$. Next, if $\\omega$ is the circumcircle of $ABC$ and $AI$ intersects $\\omega$ at $M \\neq A$ then $\\{X, Y\\} = \\{B, C\\}$, and so $2AM \\cos \\frac{\\alpha}{2} = AB + AC$ by the lemma. To summarize,\n\n$$\n\\begin{equation*}\n2AA' \\cos \\frac{\\alpha}{2} = BC, \\quad 2AI \\cos \\frac{\\alpha}{2} = AB + AC - BC, \\quad 2AM \\cos \\frac{\\alpha}{2} = AB + AC. \\tag{*}\n\\end{equation*}\n$$\n\nThese equalities imply $AA' + AI = AM$, hence the segments $AM$ and $IA'$ have a common midpoint. It follows that $I$ and $A'$ are equidistant from the circumcenter $O$. By symmetry $OI = OA' = OB' = OC'$, so $I$, $A'$, $B'$, $C'$ are on a circle centered at $O$.\n\nTo prove $OP = OI$, now it suffices to show that $I$, $A'$, $B'$, $C'$ and $P$ are concyclic. Clearly one can assume $P \\neq I, A', B', C'$.\n\nWe use oriented angles to avoid heavy case distinction. The oriented angle between the lines $l$ and $m$ is denoted by $\\angle(l, m)$. We have $\\angle(l, m) = -\\angle(m, l)$ and $\\angle(l, m) + \\angle(m, n) = \\angle(l, n)$ for arbitrary lines $l$, $m$ and $n$. Four distinct non-collinear points $U$, $V$, $X$, $Y$ are concyclic if and only if $\\angle(UX, VX) = \\angle(UY, VY)$.\n\n![](attached_image_2.png)\n\nSuppose for the moment that $A'$, $B'$, $P$, $I$ are distinct and noncollinear; then it is enough to check the equality $\\angle(A'P, B'P) = \\angle(A'I, B'I)$. Because $A$, $F$, $P$, $A'$ are on the circle $\\omega_{A}$, we have $\\angle(A'P, FP) = \\angle(A'A, FA) = \\angle(A'I, AB)$. Likewise $\\angle(B'P, FP) = \\angle(B'I, AB)$. Therefore\n\n$$\n\\angle(A'P, B'P) = \\angle(A'P, FP) + \\angle(FP, B'P) = \\angle(A'I, AB) - \\angle(B'I, AB) = \\angle(A'I, B'I).\n$$\n\nBy Miquel's theorem the circles $(AEF) = \\omega_{A}$, $(BFD) = \\omega_{B}$ and $(CDE) = \\omega_{C}$ have a common point, for arbitrary points $D$, $E$ and $F$ on $BC$, $CA$ and $AB$. So $\\omega_{A}$ passes through the common point $P \\neq D$ of $\\omega_{B}$ and $\\omega_{C}$.\n\nHere we assumed that $P \\neq F$. If $P = F$ then $P \\neq D, E$ and the conclusion follows similarly (use $\\angle(A'F, B'F) = \\angle(A'F, EF) + \\angle(EF, DF) + \\angle(DF, B'F)$ and inscribed angles in $\\omega_{A}$, $\\omega_{B}$, $\\omega_{C}$).\n\nThere is no loss of generality in assuming $A'$, $B'$, $P$, $I$ distinct and noncollinear. If $ABC$ is an equilateral triangle then the equalities $(*)$ imply that $A'$, $B'$, $C'$, $I$, $O$ and $P$ coincide, so $OP = OI$. Otherwise at most one of $A'$, $B'$, $C'$ coincides with $I$. If say $C' = I$ then $OI \\perp CI$ by the previous reasoning. It follows that $A'$, $B' \\neq I$ and hence $A' \\neq B'$. Finally $A'$, $B'$ and $I$ are noncollinear because $I$, $A'$, $B'$, $C'$ are concyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31958,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral with non-parallel sides $BC$ and $AD$. Assume that there is a point $E$ on the side $BC$ such that the quadrilaterals $ABED$ and $AECD$ are circumscribed. Prove that there is a point $F$ on the side $AD$ such that the quadrilaterals $ABCF$ and $BCDF$ are circumscribed if and only if $AB$ is parallel to $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\omega_1$ and $\\omega_2$ be the incircles and $O_1$ and $O_2$ the incenters of the quadrilaterals $ABED$ and $AECD$ respectively. A point $F$ with the stated property exists only if $\\omega_1$ and $\\omega_2$ are also the incircles of the quadrilaterals $ABCF$ and $BCDF$.\n\n![](attached_image_1.png)\n\nLet the tangents from $B$ to $\\omega_2$ and from $C$ to $\\omega_1$ (other than $BC$) meet $AD$ at $F_1$ and $F_2$ respectively. We need to prove that $F_1 = F_2$ if and only if $AB \\parallel CD$.\n\nLemma. The circles $\\omega_1$ and $\\omega_2$ with centers $O_1$ and $O_2$ are inscribed in an angle with vertex $O$. The points $P, S$ on one side of the angle and $Q, R$ on the other side are such that $\\omega_1$ is the incircle of the triangle $PQO$, and $\\omega_2$ is the excircle of the triangle $RSO$ opposite to $O$. Denote $p = OO_1 \\cdot OO_2$. Then exactly one of the following relations holds:\n$$\nOP \\cdot OR < p < OQ \\cdot OS, \\quad OP \\cdot OR > p > OQ \\cdot OS, \\quad OP \\cdot OR = p = OQ \\cdot OS.\n$$\nProof. Denote $\\angle OPO_1 = u$, $\\angle OQO_1 = v$, $\\angle OO_2R = x$, $\\angle OO_2S = y$, $\\angle POQ = 2\\varphi$. Because $PO_1$, $QO_1$, $RO_2$, $SO_2$ are internal or external bisectors in the triangles $PQO$ and $RSO$, we have\n$$\n\\begin{equation*}\nu + v = x + y \\left(= 90^\\circ - \\varphi\\right). \\tag{1}\n\\end{equation*}\n$$\n![](attached_image_2.png)\n\nBy the law of sines\n$$\n\\frac{OP}{OO_1} = \\frac{\\sin(u+\\varphi)}{\\sin u} \\quad \\text{ and } \\quad \\frac{OO_2}{OR} = \\frac{\\sin(x+\\varphi)}{\\sin x}.\n$$\nTherefore, since $x, u$ and $\\varphi$ are acute,\n$OP \\cdot OR \\geq p \\Leftrightarrow \\frac{OP}{OO_1} \\geq \\frac{OO_2}{OR} \\Leftrightarrow \\sin x \\sin(u+\\varphi) \\geq \\sin u \\sin(x+\\varphi) \\Leftrightarrow \\sin(x-u) \\geq 0 \\Leftrightarrow x \\geq u$.\nThus $OP \\cdot OR \\geq p$ is equivalent to $x \\geq u$, with $OP \\cdot OR = p$ if and only if $x = u$.\nAnalogously, $p \\geq OQ \\cdot OS$ is equivalent to $v \\geq y$, with $p = OQ \\cdot OS$ if and only if $v = y$. On the other hand $x \\geq u$ and $v \\geq y$ are equivalent by (1), with $x = u$ if and only if $v = y$. The conclusion of the lemma follows from here.\n\nGoing back to the problem, apply the lemma to the quadruples $\\{B, E, D, F_1\\}, \\{A, B, C, D\\}$ and $\\{A, E, C, F_2\\}$. Assuming $OE \\cdot OF_1 > p$, we obtain\n$$\nOE \\cdot OF_1 > p \\Rightarrow OB \\cdot OD < p \\Rightarrow OA \\cdot OC > p \\Rightarrow OE \\cdot OF_2 < p.\n$$\nIn other words, $OE \\cdot OF_1 > p$ implies\n$$\nOB \\cdot OD < p < OA \\cdot OC \\text{ and } OE \\cdot OF_1 > p > OE \\cdot OF_2.\n$$\nSimilarly, $OE \\cdot OF_1 < p$ implies\n$$\nOB \\cdot OD > p > OA \\cdot OC \\text{ and } OE \\cdot OF_1 < p < OE \\cdot OF_2.\n$$\nIn these cases $F_1 \\neq F_2$ and $OB \\cdot OD \\neq OA \\cdot OC$, so the lines $AB$ and $CD$ are not parallel.\n\nThere remains the case $OE \\cdot OF_1 = p$. Here the lemma leads to $OB \\cdot OD = p = OA \\cdot OC$ and $OE \\cdot OF_1 = p = OE \\cdot OF_2$. Therefore $F_1 = F_2$ and $AB \\parallel CD$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31959,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call admissible a set $A$ of integers that has the following property:\nIf $x, y \\in A$ (possibly $x = y$) then $x^{2} + k x y + y^{2} \\in A$ for every integer $k$.\nDetermine all pairs $m, n$ of nonzero integers such that the only admissible set containing both $m$ and $n$ is the set of all integers.",
    "options": [],
    "answer": "gcd(m, n) = 1",
    "solution": "A pair of integers $m, n$ fulfills the condition if and only if $\\gcd(m, n) = 1$. Suppose that $\\gcd(m, n) = d > 1$. The set\n$$\nA = \\{ \\ldots, -2d, -d, 0, d, 2d, \\ldots \\}\n$$\nis admissible, because if $d$ divides $x$ and $y$ then it divides $x^{2} + k x y + y^{2}$ for every integer $k$. Also $m, n \\in A$ and $A \\neq \\mathbb{Z}$.\n\nNow let $\\gcd(m, n) = 1$, and let $A$ be an admissible set containing $m$ and $n$. We use the following observations to prove that $A = \\mathbb{Z}$:\n\n(i) $k x^{2} \\in A$ for every $x \\in A$ and every integer $k$.\n\n(ii) $(x + y)^{2} \\in A$ for all $x, y \\in A$.\n\nTo justify (i) let $y = x$ in the definition of an admissible set; to justify (ii) let $k = 2$.\n\nSince $\\gcd(m, n) = 1$, we also have $\\gcd(m^{2}, n^{2}) = 1$. Hence one can find integers $a, b$ such that $a m^{2} + b n^{2} = 1$. It follows from (i) that $a m^{2} \\in A$ and $b n^{2} \\in A$. Now we deduce from (ii) that $1 = (a m^{2} + b n^{2})^{2} \\in A$. But if $1 \\in A$ then (i) implies $k \\in A$ for every integer $k$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31960,
    "subject": "Mathematics (Multi-modal)",
    "question": "An integer $a$ is called friendly if the equation $(m^{2}+n)(n^{2}+m)=a(m-n)^{3}$ has a solution over the positive integers.\n\na. Prove that there are at least 500 friendly integers in the set $\\{1,2, \\ldots, 2012\\}$.\n\nb. Decide whether $a=2$ is friendly.",
    "options": [],
    "answer": "There are at least 502 friendly integers in {1,2,…,2012} (for example, all numbers 5, 9, …, 2009). The integer 2 is not friendly.",
    "solution": "a. Every $a$ of the form $a=4k-3$ with $k \\geq 2$ is friendly. Indeed the numbers $m=2k-1>0$ and $n=k-1>0$ satisfy the given equation with $a=4k-3$:\n$$\n((2k-1)^{2}+(k-1))((k-1)^{2}+(2k-1)) = (4k-3)k^{3} = a(m-n)^{3}.\n$$\nHence $5, 9, \\ldots, 2009$ are friendly and so $\\{1,2, \\ldots, 2012\\}$ contains at least 502 friendly numbers.\n\nb. We show that $a=2$ is not friendly. Consider the equation with $a=2$ and rewrite its left-hand side as a difference of squares:\n$$\n\\frac{1}{4}\\left((m^{2}+n+n^{2}+m)^{2}-(m^{2}+n-n^{2}-m)^{2}\\right) = 2(m-n)^{3}\n$$\nSince $m^{2}+n-n^{2}-m = (m-n)(m+n-1)$, we can further reformulate the equation as\n$$\n(m^{2}+n+n^{2}+m)^{2} = (m-n)^{2}\\left(8(m-n)+(m+n-1)^{2}\\right)\n$$\nIt follows that $8(m-n)+(m+n-1)^{2}$ is a perfect square. Clearly $m>n$, hence there is an integer $s \\geq 1$ such that\n$$\n(m+n-1+2s)^{2} = 8(m-n)+(m+n-1)^{2}\n$$\nSubtracting the squares gives $s(m+n-1+s) = 2(m-n)$. Since $m+n-1+s > m-n$, we conclude that $s<2$. Therefore the only possibility is $s=1$ and $m=3n$. However then the left-hand side of the given equation (with $a=2$) is greater than $m^{3}=27n^{3}$, whereas its right-hand side equals $16n^{3}$. The contradiction proves that $a=2$ is not friendly.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31961,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a nonnegative integer $n$ define $\\operatorname{rad}(n)=1$ if $n=0$ or $n=1$, and $\\operatorname{rad}(n)=p_{1} p_{2} \\cdots p_{k}$ where $p_{1}<p_{2}<\\cdots<p_{k}$ are all prime factors of $n$. Find all polynomials $f(x)$ with nonnegative integer coefficients such that $\\operatorname{rad}(f(n))$ divides $\\operatorname{rad}\\left(f\\left(n^{\\operatorname{rad}(n)}\\right)\\right)$ for every nonnegative integer $n$.",
    "options": [],
    "answer": "f(x) = a x^m with nonnegative integers a and m (including the zero polynomial).",
    "solution": "We are going to prove that $f(x)=a x^{m}$ for some nonnegative integers $a$ and $m$. If $f(x)$ is the zero polynomial we are done, so assume that $f(x)$ has at least one positive coefficient. In particular $f(1)>0$.\n\nLet $p$ be a prime number. The condition is that $f(n) \\equiv 0\\ (\\bmod\\ p)$ implies\n$$\nf\\left(n^{\\operatorname{rad}(n)}\\right) \\equiv 0 \\quad(\\bmod\\ p)\n$$\nSince $\\operatorname{rad}\\left(n^{\\operatorname{rad}(n)^{k}}\\right)=\\operatorname{rad}(n)$ for all $k$, repeated applications of the preceding implication show that if $p$ divides $f(n)$ then\n$$\nf\\left(n^{\\operatorname{rad}(n)^{k}}\\right) \\equiv 0 \\quad(\\bmod\\ p) \\quad \\text{ for all } k\n$$\nThe idea is to construct a prime $p$ and a positive integer $n$ such that $p-1$ divides $n$ and $p$ divides $f(n)$. In this case, for $k$ large enough $p-1$ divides $\\operatorname{rad}(n)^{k}$. Hence if $(p, n)=1$ then $n^{\\operatorname{rad}(n)^{k}} \\equiv 1\\ (\\bmod\\ p)$ by Fermat's little theorem, so that\n$$\nf(1) \\equiv f\\left(n^{\\operatorname{rad}(n)^{k}}\\right) \\equiv 0 \\quad(\\bmod\\ p)\n$$\nSuppose that $f(x)=g(x) x^{m}$ with $g(0) \\neq 0$. Let $t$ be a positive integer, $p$ any prime factor of $g(-t)$ and $n=(p-1) t$. So $p-1$ divides $n$ and $f(n)=f((p-1) t) \\equiv f(-t) \\equiv 0\\ (\\bmod\\ p)$, hence either $(p, n)>1$ or the previous congruence holds. If $(p,(p-1) t)>1$ then $p$ divides $t$ and $g(0) \\equiv g(-t) \\equiv 0\\ (\\bmod\\ p)$, meaning that $p$ divides $g(0)$.\n\nIn conclusion we proved that each prime factor of $g(-t)$ divides $g(0) f(1) \\neq 0$, and thus the set of prime factors of $g(-t)$ when $t$ ranges through the positive integers is finite. This is known to imply that $g(x)$ is a constant polynomial, and so $f(x)=a x^{m}$.\nLet $f(x)$ be a polynomial with integer coefficients (not necessarily nonnegative) such that $\\operatorname{rad}(f(n))$ divides $\\operatorname{rad}\\left(f\\left(n^{\\operatorname{rad}(n)}\\right)\\right)$ for any nonnegative integer $n$. We give a complete description of all polynomials with this property. More precisely, we claim that if $f(x)$ is such a polynomial and $\\xi$ is a root of $f(x)$ then so is $\\xi^{d}$ for every positive integer $d$.\n\nTherefore each root of $f(x)$ is zero or a root of unity. In particular, if a root of unity $\\xi$ is a root of $f(x)$ then $1=\\xi^{d}$ is a root too (for some positive integer $d$). In the original problem $f(x)$ has nonnegative coefficients. Then either $f(x)$ is the zero polynomial or $f(1)>0$ and $\\xi=0$ is the only possible root. In either case $f(x)=a x^{m}$ with $a$ and $m$ nonnegative integers.\n\nTo prove the claim let $\\xi$ be a root of $f(x)$, and let $g(x)$ be an irreducible factor of $f(x)$ such that $g(\\xi)=0$. If 0 or 1 are roots of $g(x)$ then either $\\xi=0$ or $\\xi=1$ (because $g(x)$ is irreducible) and we are done. So assume that $g(0), g(1) \\neq 0$. By decomposing $d$ as a product of prime numbers, it is enough to consider the case $d=p$ prime. We argue for $p=2$. Since $\\operatorname{rad}\\left(2^{k}\\right)=2$ for every $k$, we have\n$$\n\\operatorname{rad}\\left(f\\left(2^{k}\\right)\\right) \\mid \\operatorname{rad}\\left(f\\left(2^{2 k}\\right)\\right).\n$$\nNow we prove that $g(x)$ divides $f\\left(x^{2}\\right)$. Suppose that this is not the case. Then, since $g(x)$ is irreducible, there are integer-coefficient polynomials $a(x), b(x)$ and an integer $N$ such that\n$$\na(x) g(x)+b(x) f\\left(x^{2}\\right)=N\n$$\nEach prime factor $p$ of $g\\left(2^{k}\\right)$ divides $f\\left(2^{k}\\right)$, so by $\\operatorname{rad}\\left(f\\left(2^{k}\\right)\\right) \\mid \\operatorname{rad}\\left(f\\left(2^{2 k}\\right)\\right)$ it also divides $f\\left(2^{2 k}\\right)$. From the equation above with $x=2^{k}$ it follows that $p$ divides $N$.\n\nIn summary, each prime divisor of $g\\left(2^{k}\\right)$ divides $N$, for all $k \\geq 0$. Let $p_{1}, \\ldots, p_{n}$ be the odd primes dividing $N$, and suppose that\n$$\ng(1)=2^{\\alpha} p_{1}^{\\alpha_{1}} \\cdots p_{n}^{\\alpha_{n}}\n$$\nIf $k$ is divisible by $\\varphi\\left(p_{1}^{\\alpha_{1}+1} \\cdots p_{n}^{\\alpha_{n}+1}\\right)$ then\n$$\n2^{k} \\equiv 1 \\quad\\left(\\bmod p_{1}^{\\alpha_{1}+1} \\cdots p_{n}^{\\alpha_{n}+1}\\right)\n$$\nyielding\n$$\ng\\left(2^{k}\\right) \\equiv g(1) \\quad\\left(\\bmod p_{1}^{\\alpha_{1}+1} \\cdots p_{n}^{\\alpha_{n}+1}\\right).\n$$\nIt follows that for each $i$ the maximal power of $p_{i}$ dividing $g\\left(2^{k}\\right)$ and $g(1)$ is the same, namely $p_{i}^{\\alpha_{i}}$. On the other hand, for large enough $k$, the maximal power of 2 dividing $g\\left(2^{k}\\right)$ and $g(0) \\neq 0$ is the same. From the above, for $k$ divisible by $\\varphi\\left(p_{1}^{\\alpha_{1}+1} \\cdots p_{n}^{\\alpha_{n}+1}\\right)$ and large enough, we obtain that $g\\left(2^{k}\\right)$ divides $g(0) \\cdot g(1)$. This is impossible because $g(0), g(1) \\neq 0$ are fixed and $g\\left(2^{k}\\right)$ is arbitrarily large.\n\nIn conclusion, $g(x)$ divides $f\\left(x^{2}\\right)$. Recall that $\\xi$ is a root of $f(x)$ such that $g(\\xi)=0$; then $f\\left(\\xi^{2}\\right)=0$, i.e. $\\xi^{2}$ is a root of $f(x)$.\n\nLikewise if $\\xi$ is a root of $f(x)$ and $p$ an arbitrary prime then $\\xi^{p}$ is a root too. The argument is completely analogous, in the proof above just replace 2 by $p$ and \"odd prime\" by \"prime different from $p$.\"",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31962,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$ and $y$ be positive integers. If $x^{2^{n}}-1$ is divisible by $2^{n} y+1$ for every positive integer $n$, prove that $x=1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First we prove the following fact: For every positive integer $y$ there exist infinitely many primes $p \\equiv 3 \\pmod{4}$ such that $p$ divides some number of the form $2^{n} y+1$.\n\nClearly it is enough to consider the case $y$ odd. Let\n$$\n2y+1 = p_{1}^{e_{1}} \\cdots p_{r}^{e_{r}}\n$$\nbe the prime factorization of $2y+1$. Suppose on the contrary that there are finitely many primes $p_{r+1}, \\ldots, p_{r+s} \\equiv 3 \\pmod{4}$ that divide some number of the form $2^{n} y+1$ but do not divide $2y+1$.\n\nWe want to find an $n$ such that $p_{i}^{e_{i}} \\mid\\mid 2^{n} y+1$ for $1 \\leq i \\leq r$ and $p_{i} \\nmid 2^{n} y+1$ for $r+1 \\leq i \\leq r+s$. For this it suffices to take\n$$\nn = 1 + \\varphi\\left(p_{1}^{e_{1}+1} \\cdots p_{r}^{e_{r}+1} p_{r+1}^{1} \\cdots p_{r+s}^{1}\\right)\n$$\nbecause then\n$$\n2^{n} y+1 \\equiv 2y+1 \\pmod{p_{1}^{e_{1}+1} \\cdots p_{r}^{e_{r}+1} p_{r+1}^{1} \\cdots p_{r+s}^{1}}\n$$\nThe last congruence means that $p_{1}^{e_{1}}, \\ldots, p_{r}^{e_{r}}$ divide exactly $2^{n} y+1$ and no prime $p_{r+1}, \\ldots, p_{r+s}$ divides $2^{n} y+1$. It follows that the prime factorization of $2^{n} y+1$ consists of the prime powers $p_{1}^{e_{1}}, \\ldots, p_{r}^{e_{r}}$ and powers of primes $\\equiv 1 \\pmod{4}$. Because $y$ is odd, we obtain\n$$\n2^{n} y+1 \\equiv p_{1}^{e_{1}} \\cdots p_{r}^{e_{r}} \\equiv 2y+1 \\equiv 3 \\pmod{4}\n$$\nThis is a contradiction since $n>1$, and so $2^{n} y+1 \\equiv 1 \\pmod{4}$.\n\nNow we proceed to the problem. If $p$ is a prime divisor of $2^{n} y+1$ the problem statement implies that $x^{d} \\equiv 1 \\pmod{p}$ for $d=2^{n}$. By Fermat's little theorem the same congruence holds for $d=p-1$, so it must also hold for $d=\\gcd(2^{n}, p-1)$. For $p \\equiv 3 \\pmod{4}$ we have $\\gcd(2^{n}, p-1)=2$, therefore in this case $x^{2} \\equiv 1 \\pmod{p}$.\n\nIn summary, we proved that every prime $p \\equiv 3 \\pmod{4}$ that divides some number of the form $2^{n} y+1$ also divides $x^{2}-1$. This is possible only if $x=1$, otherwise by the above $x^{2}-1$ would be a positive integer with infinitely many prime factors.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31963,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that for every prime $p > 100$ and every integer $r$ there exist two integers $a$ and $b$ such that $p$ divides $a^{2} + b^{5} - r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Fix $p$, and let $\\mathcal{P} = \\{0, 1, \\ldots, p-1\\}$ be the set of residue classes modulo $p$. For every $r \\in \\mathcal{P}$, let $S_{r} = \\{(a, b) \\in \\mathcal{P} \\times \\mathcal{P} : a^{2} + b^{5} \\equiv r\\}$, and let $s_{r} = |S_{r}|$. Our aim is to prove $s_{r} > 0$ for all $r \\in \\mathcal{P}$.\nWe will use the well-known fact that for every residue class $r \\in \\mathcal{P}$ and every positive integer $k$, there are at most $k$ values $x \\in \\mathcal{P}$ such that $x^{k} \\equiv r$.\n\nLemma. Let $N$ be the number of quadruples $(a, b, c, d) \\in \\mathcal{P}^{4}$ for which $a^{2} + b^{5} \\equiv c^{2} + d^{5}$. Then\n$$\nN = \\sum_{r \\in \\mathcal{P}} s_{r}^{2}\n$$\nand\n$$\nN \\leq p(p^{2} + 4p - 4).\n$$\n\nProof. (a) For each residue class $r$ there exist exactly $s_{r}$ pairs $(a, b)$ with $a^{2} + b^{5} \\equiv r$ and $s_{r}$ pairs $(c, d)$ with $c^{2} + d^{5} \\equiv r$. So there are $s_{r}^{2}$ quadruples with $a^{2} + b^{5} \\equiv c^{2} + d^{5} \\equiv r$. Taking the sum over all $r \\in \\mathcal{P}$, the statement follows.\n\n(b) Choose an arbitrary pair $(b, d) \\in \\mathcal{P}$ and look for the possible values of $a, c$.\n1. Suppose that $b^{5} \\equiv d^{5}$, and let $k$ be the number of such pairs $(b, d)$. The value $b$ can be chosen in $p$ different ways. For $b \\equiv 0$ only $d = 0$ has this property; for the nonzero values of $b$ there are at most 5 possible values for $d$. So we have $k \\leq 1 + 5(p-1) = 5p - 4$.\nThe values $a$ and $c$ must satisfy $a^{2} \\equiv c^{2}$, so $a \\equiv \\pm c$, and there are exactly $2p - 1$ such pairs $(a, c)$.\n2. Now suppose $b^{5} \\not\\equiv d^{5}$. In this case $a$ and $c$ must be distinct. By $(a-c)(a+c) = d^{5} - b^{5}$, the value of $a-c$ uniquely determines $a+c$ and thus $a$ and $c$ as well. Hence, there are $p-1$ suitable pairs $(a, c)$.\nThus, for each of the $k$ pairs $(b, d)$ with $b^{5} \\equiv d^{5}$ there are $2p - 1$ pairs $(a, c)$, and for each of the other $p^{2} - k$ pairs $(b, d)$ there are $p-1$ pairs $(a, c)$. Hence,\n$$\nN = k(2p - 1) + (p^{2} - k)(p - 1) = p^{2}(p - 1) + kp \\leq p^{2}(p - 1) + (5p - 4)p = p(p^{2} + 4p - 4).\n$$\n\nTo prove the statement of the problem, suppose that $S_{r} = \\emptyset$ for some $r \\in \\mathcal{P}$; obviously $r \\not\\equiv 0$. Let $T = \\{x^{10} : x \\in \\mathcal{P} \\setminus \\{0\\}\\}$ be the set of nonzero 10th powers modulo $p$. Since each residue class is the 10th power of at most 10 elements in $\\mathcal{P}$, we have $|T| \\geq \\frac{p-1}{10} \\geq 4$ by $p > 100$.\nFor every $t \\in T$, we have $S_{tr} = \\emptyset$. Indeed, if $(x, y) \\in S_{tr}$ and $t \\equiv z^{10}$ then\n$$\n(z^{-5}x)^{2} + (z^{-2}y)^{5} \\equiv t^{-1}(x^{2} + y^{5}) \\equiv r\n$$\nso $(z^{-5}x, z^{-2}y) \\in S_{r}$.\nSo, there are at least $\\frac{p-1}{10} \\geq 4$ empty sets among $S_{1}, \\ldots, S_{p-1}$, and there are at most $p-4$ nonzero values among $s_{0}, s_{2}, \\ldots, s_{p-1}$. Then by the AM-QM inequality we obtain\n$$\nN = \\sum_{r \\in \\mathcal{P} \\setminus rT} s_{r}^{2} \\geq \\frac{1}{p-4}\\left(\\sum_{r \\in \\mathcal{P} \\setminus rT} s_{r}\\right)^{2} = \\frac{|\\mathcal{P} \\times \\mathcal{P}|^{2}}{p-4} = \\frac{p^{4}}{p-4} > p(p^{2} + 4p - 4),\n$$\nwhich is impossible by the lemma.\nIf $5 \\nmid p-1$, then all modulo $p$ residue classes are complete fifth powers and the statement is trivial. So assume that $p = 10k + 1$ where $k \\geq 10$. Let $g$ be a primitive root modulo $p$.\n\nWe will use the following facts:\n(F1) If some residue class $x$ is not quadratic then $x^{(p-1)/2} \\equiv -1 \\pmod{p}$.\n(F2) For every integer $d$, as a simple corollary of the summation formula for geometric progressions,\n$$\n\\sum_{i=0}^{2k-1} g^{5di} \\equiv \\begin{cases}\n2k & \\text{if } 2k \\mid d \\\\\n0 & \\text{if } 2k \\nmid d\n\\end{cases} \\pmod{p}\n$$\n\nSuppose that, contrary to the statement, some modulo $p$ residue class $r$ cannot be expressed as $a^{2} + b^{5}$. Of course $r \\not\\equiv 0 \\pmod{p}$. By (F1) we have $(r - b^{5})^{(p-1)/2} = (r - b^{5})^{5k} \\equiv -1 \\pmod{p}$ for all residue classes $b$.\nFor $t = 1, 2, \\ldots, k-1$ consider the sums\n$$\nS(t) = \\sum_{i=0}^{2k-1} (r - g^{5i})^{5k} g^{5ti}\n$$\nBy the indirect assumption and (F2),\n$$\nS(t) = \\sum_{i=0}^{2k-1} (r - (g^{i})^{5})^{5k} g^{5ti} \\equiv \\sum_{i=0}^{2k-1} (-1) g^{5ti} \\equiv -\\sum_{i=0}^{2k-1} g^{5ti} \\equiv 0 \\pmod{p}\n$$\nbecause $2k$ cannot divide $t$.\nOn the other hand, by the binomial theorem,\n$$\n\\begin{aligned}\nS(t) &= \\sum_{i=0}^{2k-1} \\left( \\sum_{j=0}^{5k} \\binom{5k}{j} r^{5k-j} (-g^{5i})^{j} \\right) g^{5ti} \\\\\n&= \\sum_{j=0}^{5k} (-1)^{j} \\binom{5k}{j} r^{5k-j} \\left( \\sum_{i=0}^{2k-1} g^{5(j+t)i} \\right) \\\\\n&\\equiv \\sum_{j=0}^{5k} (-1)^{j} \\binom{5k}{j} r^{5k-j} \\begin{cases}\n2k & \\text{if } 2k \\mid j+t \\\\\n0 & \\text{if } 2k \\nmid j+t\n\\end{cases} \\pmod{p}\n\\end{aligned}\n$$\nSince $1 \\leq j + t < 6k$, the number $2k$ divides $j + t$ only for $j = 2k - t$ and $j = 4k - t$. Hence,\n$$\n\\begin{gathered}\n0 \\equiv S(t) \\equiv (-1)^{t} \\left( \\binom{5k}{2k-t} r^{3k+t} + \\binom{5k}{4k-t} r^{k+t} \\right) \\cdot 2k \\pmod{p} \\\\\n\\binom{5k}{2k-t} r^{2k} + \\binom{5k}{4k-t} \\equiv 0 \\pmod{p}\n\\end{gathered}\n$$\nTaking this for $t = 1, 2$ and eliminating $r$, we get\n$$\n\\begin{aligned}\n0 &\\equiv \\binom{5k}{2k-2} \\left( \\binom{5k}{2k-1} r^{2k} + \\binom{5k}{4k-1} \\right ) - \\binom{5k}{2k-1} \\left( \\binom{5k}{2k-2} r^{2k} + \\binom{5k}{4k-2} \\right ) \\\\\n&= \\binom{5k}{2k-2} \\binom{5k}{4k-1} - \\binom{5k}{2k-1} \\binom{5k}{4k-2} \\\\\n&= \\frac{(5k)!^{2}}{(2k-1)! (3k+2)! (4k-1)! (k+2)!} \\left( (2k-1)(k+2) - (3k+2)(4k-1) \\right ) \\\\\n&= \\frac{-(5k)!^{2} \\cdot 2k(5k+1)}{(2k-1)! (3k+2)! (4k-1)! (k+2)!} \\pmod{p}\n\\end{aligned}\n$$\nBut in the last expression none of the numbers is divisible by $p = 10k + 1$, a contradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31964,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer and let $a_{1}, \\ldots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_{0}, \\ldots, u_{n}$ and $v_{0}, \\ldots, v_{n}$ inductively by $u_{0}=u_{1}=v_{0}=1$, and\n$$\nu_{k+1}=u_{k}+a_{k} u_{k-1}, \\quad v_{k+1}=v_{k}+a_{n-k} v_{k-1} \\quad \\text{ for } k=1, \\ldots, n-1 .$$\nProve that $u_{n}=v_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove by induction on $k$ that\n$$\n\\begin{equation*}\nu_{k}=\\sum_{\\substack{0<i_{1}<\\ldots<i_{t}<k \\\\ i_{j+1}-i_{j} \\geqslant 2}} a_{i_{1}} \\ldots a_{i_{t}} \\tag{1}\\end{equation*}\n$$\nNote that we have one trivial summand equal to 1 (which corresponds to $t=0$ and the empty sequence, whose product is 1).\nFor $k=0,1$ the sum on the right-hand side only contains the empty product, so (1) holds due to $u_{0}=u_{1}=1$. For $k \\geqslant 1$, assuming the result is true for $0,1, \\ldots, k$, we have\n$$\n\\begin{aligned}\nu_{k+1} & =\\sum_{\\substack{0<i_{1}<\\ldots<i_{t}<k, i_{j+1}-i_{j} \\geqslant 2}} a_{i_{1}} \\ldots a_{i_{t}}+\\sum_{\\substack{0<i_{1}<\\ldots<i_{t}<k-1, i_{j+1}-i_{j} \\geqslant 2}} a_{i_{1}} \\ldots a_{i_{t}} \\cdot a_{k} \\\\\n& =\\sum_{\\substack{0<i_{1}<\\ldots<i_{t}<k+1, i_{j+1}-i_{j} \\geqslant 2, k \\notin\\left\\{i_{1}, \\ldots, i_{t}\\right\\}}} a_{i_{1}} \\ldots a_{i_{t}}+\\sum_{\\substack{0<i_{1}<\\ldots<i_{t}<k+1, i_{j+1}-i_{j} \\geqslant 2, k \\in\\left\\{i_{1}, \\ldots, i_{t}\\right\\}}} a_{i_{1}} \\ldots a_{i_{t}} \\\\\n& =\\sum_{\\substack{0<i_{1}<\\ldots<i_{t}<k+1, i_{j+1}-i_{j} \\geqslant 2}} a_{i_{1}} \\ldots a_{i_{t}},\\end{aligned}\n$$\nas required.\nApplying (1) to the sequence $b_{1}, \\ldots, b_{n}$ given by $b_{k}=a_{n-k}$ for $1 \\leqslant k \\leqslant n$, we get\n$$\n\\begin{equation*}\nv_{k}=\\sum_{\\substack{0<i_{1}<\\ldots<i_{t}<k \\\\ i_{j+1}-i_{j} \\geqslant 2}} b_{i_{1}} \\ldots b_{i_{t}}=\\sum_{\\substack{n>i_{1}>\\ldots>i_{t}>n-k \\\\ i_{j}-i_{j+1} \\geqslant 2}} a_{i_{1}} \\ldots a_{i_{t}} \\tag{2}\\end{equation*}\n$$\nFor $k=n$ the expressions (1) and (2) coincide, so indeed $u_{n}=v_{n}$.\n\n\nSolution 2:\nDefine recursively a sequence of multivariate polynomials by\n$$\nP_{0}=P_{1}=1, \\quad P_{k+1}\\left(x_{1}, \\ldots, x_{k}\\right)=P_{k}\\left(x_{1}, \\ldots, x_{k-1}\\right)+x_{k} P_{k-1}\\left(x_{1}, \\ldots, x_{k-2}\\right),\n$$\nso $P_{n}$ is a polynomial in $n-1$ variables for each $n \\geqslant 1$. Two easy inductive arguments show that\n$$\nu_{n}=P_{n}\\left(a_{1}, \\ldots, a_{n-1}\\right), \\quad v_{n}=P_{n}\\left(a_{n-1}, \\ldots, a_{1}\\right)$$\nso we need to prove $P_{n}\\left(x_{1}, \\ldots, x_{n-1}\\right)=P_{n}\\left(x_{n-1}, \\ldots, x_{1}\\right)$ for every positive integer $n$. The cases $n=1,2$ are trivial, and the cases $n=3,4$ follow from $P_{3}(x, y)=1+x+y$ and $P_{4}(x, y, z)= 1+x+y+z+x z$.\nNow we proceed by induction, assuming that $n \\geqslant 5$ and the claim hold for all smaller cases. Using $F(a, b)$ as an abbreviation for $P_{|a-b|+1}\\left(x_{a}, \\ldots, x_{b}\\right)$ (where the indices $a, \\ldots, b$ can be either in increasing or decreasing order),\n$$\n\\begin{aligned}\nF(n, 1) & =F(n, 2)+x_{1} F(n, 3)=F(2, n)+x_{1} F(3, n) \\\\\n& =\\left(F(2, n-1)+x_{n} F(2, n-2)\\right)+x_{1}\\left(F(3, n-1)+x_{n} F(3, n-2)\\right) \\\\\n& =\\left(F(n-1,2)+x_{1} F(n-1,3)\\right)+x_{n}\\left(F(n-2,2)+x_{1} F(n-2,3)\\right) \\\\\n& =F(n-1,1)+x_{n} F(n-2,1)=F(1, n-1)+x_{n} F(1, n-2) \\\\\n& =F(1, n)\n\\end{aligned}\n$$\nas we wished to show.\n\n\nSolution 3:\nUsing matrix notation, we can rewrite the recurrence relation as\n$$\n\\binom{u_{k+1}}{u_{k+1}-u_{k}}=\\binom{u_{k}+a_{k} u_{k-1}}{a_{k} u_{k-1}}=\\left(\\begin{array}{cc}\n1+a_{k} & -a_{k} \\\\\na_{k} & -a_{k}\n\\end{array}\\right)\\binom{u_{k}}{u_{k}-u_{k-1}}\n$$\nfor $1 \\leqslant k \\leqslant n-1$, and similarly\n$$\n\\left(v_{k+1} ; v_{k}-v_{k+1}\\right)=\\left(v_{k}+a_{n-k} v_{k-1} ;-a_{n-k} v_{k-1}\\right)=\\left(v_{k} ; v_{k-1}-v_{k}\\right)\\left(\\begin{array}{cl}\n1+a_{n-k} & -a_{n-k} \\\\\na_{n-k} & -a_{n-k}\n\\end{array}\\right)\n$$\nfor $1 \\leqslant k \\leqslant n-1$. Hence, introducing the $2 \\times 2$ matrices $A_{k}=\\left(\\begin{array}{cl}1+a_{k} & -a_{k} \\\\ a_{k} & -a_{k}\\end{array}\\right)$ we have\n$$\n\\binom{u_{k+1}}{u_{k+1}-u_{k}}=A_{k}\\binom{u_{k}}{u_{k}-u_{k-1}} \\quad \\text{ and } \\quad\\left(v_{k+1} ; v_{k}-v_{k+1}\\right)=\\left(v_{k} ; v_{k-1}-v_{k}\\right) A_{n-k} .\n$$\nfor $1 \\leqslant k \\leqslant n-1$. Since $\\binom{u_{1}}{u_{1}-u_{0}}=\\binom{1}{0}$ and $\\left(v_{1} ; v_{0}-v_{1}\\right)=(1 ; 0)$, we get\n$$\n\\binom{u_{n}}{u_{n}-u_{n-1}}=A_{n-1} A_{n-2} \\cdots A_{1} \\cdot\\binom{1}{0} \\quad \\text{ and } \\quad\\left(v_{n} ; v_{n-1}-v_{n}\\right)=(1 ; 0) \\cdot A_{n-1} A_{n-2} \\cdots A_{1} .\n$$\nIt follows that\n$$\n\\left(u_{n}\\right)=(1 ; 0)\\binom{u_{n}}{u_{n}-u_{n-1}}=(1 ; 0) \\cdot A_{n-1} A_{n-2} \\cdots A_{1} \\cdot\\binom{1}{0}=\\left(v_{n} ; v_{n-1}-v_{n}\\right)\\binom{1}{0}=\\left(v_{n}\\right) .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31965,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that in any set of $2000$ distinct real numbers there exist two pairs $a > b$ and $c > d$ with $a \\neq c$ or $b \\neq d$, such that\n$$\n\\left|\\frac{a-b}{c-d}-1\\right|<\\frac{1}{100000}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any set $S$ of $n=2000$ distinct real numbers, let $D_{1} \\leqslant D_{2} \\leqslant \\cdots \\leqslant D_{m}$ be the distances between them, displayed with their multiplicities. Here $m = n(n-1)/2$. By rescaling the numbers, we may assume that the smallest distance $D_{1}$ between two elements of $S$ is $D_{1} = 1$. Let $D_{1} = 1 = y - x$ for $x, y \\in S$. Evidently $D_{m} = v - u$ is the difference between the largest element $v$ and the smallest element $u$ of $S$.\n\nIf $D_{i+1}/D_{i} < 1 + 10^{-5}$ for some $i = 1, 2, \\ldots, m-1$ then the required inequality holds, because $0 \\leqslant D_{i+1}/D_{i} - 1 < 10^{-5}$. Otherwise, the reverse inequality\n$$\n\\frac{D_{i+1}}{D_{i}} \\geqslant 1 + \\frac{1}{10^{5}}\n$$\nholds for each $i = 1, 2, \\ldots, m-1$, and therefore\n$$\nv-u = D_{m} = \\frac{D_{m}}{D_{1}} = \\frac{D_{m}}{D_{m-1}} \\cdots \\frac{D_{3}}{D_{2}} \\cdot \\frac{D_{2}}{D_{1}} \\geqslant \\left(1 + \\frac{1}{10^{5}}\\right)^{m-1}\n$$\nFrom $m-1 = n(n-1)/2 - 1 = 1000 \\cdot 1999 - 1 > 19 \\cdot 10^{5}$, together with the fact that for all $n \\geqslant 1$, $\\left(1 + \\frac{1}{n}\\right)^{n} \\geqslant 1 + \\binom{n}{1} \\cdot \\frac{1}{n} = 2$, we get\n$$\n\\left(1 + \\frac{1}{10^{5}}\\right)^{19 \\cdot 10^{5}} = \\left(\\left(1 + \\frac{1}{10^{5}}\\right)^{10^{5}}\\right)^{19} \\geqslant 2^{19} = 2^{9} \\cdot 2^{10} > 500 \\cdot 1000 > 2 \\cdot 10^{5}\n$$\nand so $v-u = D_{m} > 2 \\cdot 10^{5}$.\n\nSince the distance of $x$ to at least one of the numbers $u, v$ is at least $(u-v)/2 > 10^{5}$, we have\n$$\n|x-z| > 10^{5}\n$$\nfor some $z \\in \\{u, v\\}$. Since $y-x = 1$, we have either $z > y > x$ (if $z = v$) or $y > x > z$ (if $z = u$). If $z > y > x$, selecting $a = z$, $b = y$, $c = z$ and $d = x$ (so that $b \\neq d$), we obtain\n$$\n\\left|\\frac{a-b}{c-d} - 1\\right| = \\left|\\frac{z-y}{z-x} - 1\\right| = \\left|\\frac{x-y}{z-x}\\right| = \\frac{1}{z-x} < 10^{-5}.\n$$\nOtherwise, if $y > x > z$, we may choose $a = y$, $b = z$, $c = x$ and $d = z$ (so that $a \\neq c$), and obtain\n$$\n\\left|\\frac{a-b}{c-d} - 1\\right| = \\left|\\frac{y-z}{x-z} - 1\\right| = \\left|\\frac{y-x}{x-z}\\right| = \\frac{1}{x-z} < 10^{-5}.\n$$\nThe desired result follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31966,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and consider a sequence $a_{1}, a_{2}, \\ldots, a_{n}$ of positive integers. Extend it periodically to an infinite sequence $a_{1}, a_{2}, \\ldots$ by defining $a_{n+i}=a_{i}$ for all $i \\geqslant 1$. If\n$$\na_{1} \\leqslant a_{2} \\leqslant \\cdots \\leqslant a_{n} \\leqslant a_{1}+n\n$$\nand\n$$\na_{a_{i}} \\leqslant n+i-1 \\quad \\text{ for } i=1,2, \\ldots, n\n$$\nprove that\n$$\na_{1}+\\cdots+a_{n} \\leqslant n^{2} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, we claim that\n$$\n\\begin{equation*}\na_{i} \\leqslant n+i-1 \\quad \\text{ for } i=1,2, \\ldots, n . \\tag{3}\n\\end{equation*}\n$$\nAssume contrariwise that $i$ is the smallest counterexample. From $a_{n} \\geqslant a_{n-1} \\geqslant \\cdots \\geqslant a_{i} \\geqslant n+i$ and $a_{a_{i}} \\leqslant n+i-1$, taking into account the periodicity of our sequence, it follows that\n$$\n\\begin{equation*}\na_{i} \\text{ cannot be congruent to } i, i+1, \\ldots, n-1 \\text{, or } n(\\bmod n) \\text{. } \\tag{4}\n\\end{equation*}\n$$\nThus our assumption that $a_{i} \\geqslant n+i$ implies the stronger statement that $a_{i} \\geqslant 2 n+1$, which by $a_{1}+n \\geqslant a_{n} \\geqslant a_{i}$ gives $a_{1} \\geqslant n+1$. The minimality of $i$ then yields $i=1$, and (4) becomes contradictory. This establishes our first claim.\n\nIn particular we now know that $a_{1} \\leqslant n$. If $a_{n} \\leqslant n$, then $a_{1} \\leqslant \\cdots \\leqslant a_{n} \\leqslant n$ and the desired inequality holds trivially. Otherwise, consider the number $t$ with $1 \\leqslant t \\leqslant n-1$ such that\n$$\n\\begin{equation*}\na_{1} \\leqslant a_{2} \\leqslant \\ldots \\leqslant a_{t} \\leqslant n<a_{t+1} \\leqslant \\ldots \\leqslant a_{n} \\tag{5}\n\\end{equation*}\n$$\nSince $1 \\leqslant a_{1} \\leqslant n$ and $a_{a_{1}} \\leqslant n$ by (2), we have $a_{1} \\leqslant t$ and hence $a_{n} \\leqslant n+t$. Therefore if for each positive integer $i$ we let $b_{i}$ be the number of indices $j \\in\\{t+1, \\ldots, n\\}$ satisfying $a_{j} \\geqslant n+i$, we have\n$$\nb_{1} \\geqslant b_{2} \\geqslant \\ldots \\geqslant b_{t} \\geqslant b_{t+1}=0\n$$\nNext we claim that $a_{i}+b_{i} \\leqslant n$ for $1 \\leqslant i \\leqslant t$. Indeed, by $n+i-1 \\geqslant a_{a_{i}}$ and $a_{i} \\leqslant n$, each $j$ with $a_{j} \\geqslant n+i$ (thus $a_{j}>a_{a_{i}}$ ) belongs to $\\left\\{a_{i}+1, \\ldots, n\\right\\}$, and for this reason $b_{i} \\leqslant n-a_{i}$.\n\nIt follows from the definition of the $b_{i}$\\,s and (5) that\n$$\na_{t+1}+\\ldots+a_{n} \\leqslant n(n-t)+b_{1}+\\ldots+b_{t} .\n$$\nAdding $a_{1}+\\ldots+a_{t}$ to both sides and using that $a_{i}+b_{i} \\leqslant n$ for $1 \\leqslant i \\leqslant t$, we get\n$$\na_{1}+a_{2}+\\cdots+a_{n} \\leqslant n(n-t)+n t=n^{2}\n$$\nas we wished to prove.\nIn the first quadrant of an infinite grid, consider the increasing \"staircase\" obtained by shading in dark the bottom $a_{i}$ cells of the $i$th column for $1 \\leqslant i \\leqslant n$. We will prove that there are at most $n^{2}$ dark cells.\n\nTo do it, consider the $n \\times n$ square $S$ in the first quadrant with a vertex at the origin. Also consider the $n \\times n$ square directly to the left of $S$. Starting from its lower left corner, shade in light the leftmost $a_{j}$ cells of the $j$th row for $1 \\leqslant j \\leqslant n$. Equivalently, the light shading is obtained by reflecting the dark shading across the line $x=y$ and translating it $n$ units to the left. The figure below illustrates this construction for the sequence $6,6,6,7,7,7,8,12,12,14$.\n\n![](attached_image_1.png)\n\nWe claim that there is no cell in $S$ which is both dark and light. Assume, contrariwise, that there is such a cell in column $i$. Consider the highest dark cell in column $i$ which is inside $S$. Since it is above a light cell and inside $S$, it must be light as well. There are two cases:\n\nCase 1. $a_{i} \\leqslant n$\n\nIf $a_{i} \\leqslant n$ then this dark and light cell is $(i, a_{i})$, as highlighted in the figure. However, this is the $(n+i)$-th cell in row $a_{i}$, and we only shaded $a_{a_{i}}<n+i$ light cells in that row, a contradiction.\n\nCase 2. $a_{i} \\geqslant n+1$\n\nIf $a_{i} \\geqslant n+1$, this dark and light cell is $(i, n)$. This is the $(n+i)$-th cell in row $n$ and we shaded $a_{n} \\leqslant a_{1}+n$ light cells in this row, so we must have $i \\leqslant a_{1}$. But $a_{1} \\leqslant a_{a_{1}} \\leqslant n$ by (1) and (2), so $i \\leqslant a_{1}$ implies $a_{i} \\leqslant a_{a_{1}} \\leqslant n$, contradicting our assumption.\n\nWe conclude that there are no cells in $S$ which are both dark and light. It follows that the number of shaded cells in $S$ is at most $n^{2}$.\n\nFinally, observe that if we had a light cell to the right of $S$, then by symmetry we would have a dark cell above $S$, and then the cell $(n, n)$ would be dark and light. It follows that the number of light cells in $S$ equals the number of dark cells outside of $S$, and therefore the number of shaded cells in $S$ equals $a_{1}+\\cdots+a_{n}$. The desired result follows.\nAs in Solution 1, we first establish that $a_{i} \\leqslant n+i-1$ for $1 \\leqslant i \\leqslant n$. Now define $c_{i}=\\max \\left(a_{i}, i\\right)$ for $1 \\leqslant i \\leqslant n$ and extend the sequence $c_{1}, c_{2}, \\ldots$ periodically modulo $n$. We claim that this sequence also satisfies the conditions of the problem.\n\nFor $1 \\leqslant i<j \\leqslant n$ we have $a_{i} \\leqslant a_{j}$ and $i<j$, so $c_{i} \\leqslant c_{j}$. Also $a_{n} \\leqslant a_{1}+n$ and $n<1+n$ imply $c_{n} \\leqslant c_{1}+n$. Finally, the definitions imply that $c_{c_{i}} \\in\\left\\{a_{a_{i}}, a_{i}, a_{i}-n, i\\right\\}$ so $c_{c_{i}} \\leqslant n+i-1$ by (2) and (3). This establishes (1) and (2) for $c_{1}, c_{2}, \\ldots$.\n\nOur new sequence has the additional property that\n$$\n\\begin{equation*}\nc_{i} \\geqslant i \\quad \\text{ for } i=1,2, \\ldots, n \\tag{6}\n\\end{equation*}\n$$\nwhich allows us to construct the following visualization: Consider $n$ equally spaced points on a circle, sequentially labelled $1,2, \\ldots, n(\\bmod n)$, so point $k$ is also labelled $n+k$. We draw arrows from vertex $i$ to vertices $i+1, \\ldots, c_{i}$ for $1 \\leqslant i \\leqslant n$, keeping in mind that $c_{i} \\geqslant i$ by (6). Since $c_{i} \\leqslant n+i-1$ by (3), no arrow will be drawn twice, and there is no arrow from a vertex to itself. The total number of arrows is\n$$\n\\text{ number of arrows }=\\sum_{i=1}^{n}\\left(c_{i}-i\\right)=\\sum_{i=1}^{n} c_{i}-\\binom{n+1}{2}\n$$\nNow we show that we never draw both arrows $i \\rightarrow j$ and $j \\rightarrow i$ for $1 \\leqslant i<j \\leqslant n$. Assume contrariwise. This means, respectively, that\n$$\ni<j \\leqslant c_{i} \\quad \\text{ and } \\quad j<n+i \\leqslant c_{j} .\n$$\nWe have $n+i \\leqslant c_{j} \\leqslant c_{1}+n$ by (1), so $i \\leqslant c_{1}$. Since $c_{1} \\leqslant n$ by (3), this implies that $c_{i} \\leqslant c_{c_{1}} \\leqslant n$ using (1) and (3). But then, using (1) again, $j \\leqslant c_{i} \\leqslant n$ implies $c_{j} \\leqslant c_{c_{i}}$, which combined with $n+i \\leqslant c_{j}$ gives us that $n+i \\leqslant c_{c_{i}}$. This contradicts (2).\n\nThis means that the number of arrows is at most $\\binom{n}{2}$, which implies that\n$$\n\\sum_{i=1}^{n} c_{i} \\leqslant\\binom{ n}{2}+\\binom{n+1}{2}=n^{2} .\n$$\nRecalling that $a_{i} \\leqslant c_{i}$ for $1 \\leqslant i \\leqslant n$, the desired inequality follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31967,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}_{\\geqslant 0}$ be the set of all nonnegative integers. Find all the functions $f: \\mathbb{Z}_{\\geqslant 0} \\rightarrow \\mathbb{Z}_{\\geqslant 0}$ satisfying the relation\n$$\nf(f(f(n)))=f(n+1)+1\n$$\nfor all $n \\in \\mathbb{Z}_{\\geqslant 0}$.",
    "options": [],
    "answer": "Two functions: (1) f(n) = n + 1 for all nonnegative integers n; (2) the piecewise function by residue mod four: f(n) = n + 1 if n is congruent to 0 or 2 modulo 4, f(n) = n + 5 if n is congruent to 1 modulo 4, and f(n) = n − 3 if n is congruent to 3 modulo 4.",
    "solution": "Answer. There are two such functions: $f(n)=n+1$ for all $n \\in \\mathbb{Z}_{\\geqslant 0}$, and\n$$\nf(n)=\\left\\{\\begin{array}{ll}\nn+1, & n \\equiv 0(\\bmod 4) \\text{ or } n \\equiv 2(\\bmod 4),  \\\\\nn+5, & n \\equiv 1(\\bmod 4), \\\\\nn-3, & n \\equiv 3(\\bmod 4)\n\\end{array} \\quad \\text{ for all } n \\in \\mathbb{Z}_{\\geqslant 0}\\right.\n$$\nThroughout all the solutions, we write $h^{k}(x)$ to abbreviate the $k$th iteration of function $h$, so $h^{0}$ is the identity function, and $h^{k}(x)=\\underbrace{h(\\ldots h}_{k \\text{ times}}(x) \\ldots))$ for $k \\geqslant 1$.\n\nSolution 1. To start, we get from (*) that\n$$\nf^{4}(n)=f(f^{3}(n))=f(f(n+1)+1) \\quad \\text{ and } \\quad f^{4}(n+1)=f^{3}(f(n+1))=f(f(n+1)+1)+1\n$$\nthus\n$$\n\\begin{equation*}\nf^{4}(n)+1=f^{4}(n+1) \\tag{2}\n\\end{equation*}\n$$\nI. Let us denote by $R_{i}$ the range of $f^{i}$; note that $R_{0}=\\mathbb{Z}_{\\geqslant 0}$ since $f^{0}$ is the identity function. Obviously, $R_{0} \\supseteq R_{1} \\supseteq \\ldots$. Next, from (2) we get that if $a \\in R_{4}$ then also $a+1 \\in R_{4}$. This implies that $\\mathbb{Z}_{\\geqslant 0} \\backslash R_{4}$ - and hence $\\mathbb{Z}_{\\geqslant 0} \\backslash R_{1}$ - is finite. In particular, $R_{1}$ is unbounded.\nAssume that $f(m)=f(n)$ for some distinct $m$ and $n$. Then from $(*)$ we obtain $f(m+1)= f(n+1)$; by an easy induction we then get that $f(m+c)=f(n+c)$ for every $c \\geqslant 0$. So the function $f(k)$ is periodic with period $|m-n|$ for $k \\geqslant m$, and thus $R_{1}$ should be bounded, which is false. So, $f$ is injective.\nII. Denote now $S_{i}=R_{i-1} \\backslash R_{i}$; all these sets are finite for $i \\leqslant 4$. On the other hand, by the injectivity we have $n \\in S_{i} \\Longleftrightarrow f(n) \\in S_{i+1}$. By the injectivity again, $f$ implements a bijection between $S_{i}$ and $S_{i+1}$, thus $\\left|S_{1}\\right|=\\left|S_{2}\\right|=\\ldots$; denote this common cardinality by $k$. If $0 \\in R_{3}$ then $0=f(f(f(n)))$ for some $n$, thus from (*) we get $f(n+1)=-1$ which is impossible. Therefore $0 \\in R_{0} \\backslash R_{3}=S_{1} \\cup S_{2} \\cup S_{3}$, thus $k \\geqslant 1$.\nNext, let us describe the elements $b$ of $R_{0} \\backslash R_{3}=S_{1} \\cup S_{2} \\cup S_{3}$. We claim that each such element satisfies at least one of three conditions (i) $b=0$, (ii) $b=f(0)+1$, and (iii) $b-1 \\in S_{1}$. Otherwise $b-1 \\in \\mathbb{Z}_{\\geqslant 0}$, and there exists some $n>0$ such that $f(n)=b-1$; but then $f^{3}(n-1)=f(n)+1=b$, so $b \\in R_{3}$.\nThis yields\n$$\n3 k=\\left|S_{1} \\cup S_{2} \\cup S_{3}\\right| \\leqslant 1+1+\\left|S_{1}\\right|=k+2\n$$\nor $k \\leqslant 1$. Therefore $k=1$, and the inequality above comes to equality. So we have $S_{1}=\\{a\\}$, $S_{2}=\\{f(a)\\}$, and $S_{3}=\\left\\{f^{2}(a)\\right\\}$ for some $a \\in \\mathbb{Z}_{\\geqslant 0}$, and each one of the three options (i), (ii), and (iii) should be realized exactly once, which means that\n$$\n\\begin{equation*}\n\\left\\{a, f(a), f^{2}(a)\\right\\}=\\{0, a+1, f(0)+1\\} \\tag{3}\n\\end{equation*}\n$$\nIII. From (3), we get $a+1 \\in\\left\\{f(a), f^{2}(a)\\right\\}$ (the case $a+1=a$ is impossible). If $a+1=f^{2}(a)$ then we have $f(a+1)=f^{3}(a)=f(a+1)+1$ which is absurd. Therefore\n$$\n\\begin{equation*}\nf(a)=a+1 \\tag{4}\n\\end{equation*}\n$$\nNext, again from (3) we have $0 \\in\\left\\{a, f^{2}(a)\\right\\}$. Let us consider these two cases separately.\nCase 1. Assume that $a=0$, then $f(0)=f(a)=a+1=1$. Also from (3) we get $f(1)=f^{2}(a)= f(0)+1=2$. Now, let us show that $f(n)=n+1$ by induction on $n$; the base cases $n \\leqslant 1$ are established. Next, if $n \\geqslant 2$ then the induction hypothesis implies\n$$\nn+1=f(n-1)+1=f^{3}(n-2)=f^{2}(n-1)=f(n)\n$$\nestablishing the step. In this case we have obtained the first of two answers; checking that is satisfies (*) is straightforward.\nCase 2. Assume now that $f^{2}(a)=0$; then by (3) we get $a=f(0)+1$. By (4) we get $f(a+1)= f^{2}(a)=0$, then $f(0)=f^{3}(a)=f(a+1)+1=1$, hence $a=f(0)+1=2$ and $f(2)=3$ by (4). To summarize,\n$$\nf(0)=1, \\quad f(2)=3, \\quad f(3)=0\n$$\nNow let us prove by induction on $m$ that (1) holds for all $n=4 k, 4 k+2,4 k+3$ with $k \\leqslant m$ and for all $n=4 k+1$ with $k<m$. The base case $m=0$ is established above. For the step, assume that $m \\geqslant 1$. From ( $*$ ) we get $f^{3}(4 m-3)=f(4 m-2)+1=4 m$. Next, by (2) we have\n$$\nf(4 m)=f^{4}(4 m-3)=f^{4}(4 m-4)+1=f^{3}(4 m-3)+1=4 m+1\n$$\nThen by the induction hypothesis together with (*) we successively obtain\n$$\n\\begin{aligned}\n& f(4 m-3)=f^{3}(4 m-1)=f(4 m)+1=4 m+2 \\\\\n& f(4 m+2)=f^{3}(4 m-4)=f(4 m-3)+1=4 m+3 \\\\\n& f(4 m+3)=f^{3}(4 m-3)=f(4 m-2)+1=4 m\n\\end{aligned}\n$$\nthus finishing the induction step.\nFinally, it is straightforward to check that the constructed function works:\n$$\n\\begin{aligned}\nf^{3}(4 k) & =4 k+7=f(4 k+1)+1, & & f^{3}(4 k+1)=4 k+4=f(4 k+2)+1, \\\\\nf^{3}(4 k+2) & =4 k+1=f(4 k+3)+1, & & f^{3}(4 k+3)=4 k+6=f(4 k+4)+1 .\n\\end{aligned}\n$$\n\n\nSolution 2. I. For convenience, let us introduce the function $g(n)=f(n)+1$. Substituting $f(n)$ instead of $n$ into (*) we obtain\n$$\n\\begin{equation*}\nf^{4}(n)=f(f(n)+1)+1, \\quad \\text{ or } \\quad f^{4}(n)=g^{2}(n) \\tag{5}\n\\end{equation*}\n$$\nApplying $f$ to both parts of (*) and using (5) we get\n$$\n\\begin{equation*}\nf^{4}(n)+1=f(f(n+1)+1)+1=f^{4}(n+1) \\tag{6}\n\\end{equation*}\n$$\nThus, if $g^{2}(0)=f^{4}(0)=c$ then an easy induction on $n$ shows that\n$$\n\\begin{equation*}\ng^{2}(n)=f^{4}(n)=n+c, \\quad n \\in \\mathbb{Z}_{\\geqslant 0} \\tag{7}\n\\end{equation*}\n$$\nThis relation implies that both $f$ and $g$ are injective: if, say, $f(m)=f(n)$ then $m+c= f^{4}(m)=f^{4}(n)=n+c$. Next, since $g(n) \\geqslant 1$ for every $n$, we have $c=g^{2}(0) \\geqslant 1$. Thus from (7) again we obtain $f(n) \\neq n$ and $g(n) \\neq n$ for all $n \\in \\mathbb{Z}_{\\geqslant 0}$.\nII. Next, application of $f$ and $g$ to (7) yields\n$$\n\\begin{equation*}\nf(n+c)=f^{5}(n)=f^{4}(f(n))=f(n)+c \\quad \\text{ and } \\quad g(n+c)=g^{3}(n)=g(n)+c . \\tag{8}\n\\end{equation*}\n$$\nIn particular, this means that if $m \\equiv n(\\bmod c)$ then $f(m) \\equiv f(n)(\\bmod c)$. Conversely, if $f(m) \\equiv f(n)(\\bmod c)$ then we get $m+c=f^{4}(m) \\equiv f^{4}(n)=n+c(\\bmod c)$. Thus,\n$$\n\\begin{equation*}\nm \\equiv n(\\bmod c) \\Longleftrightarrow f(m) \\equiv f(n) \\quad(\\bmod c) \\Longleftrightarrow g(m) \\equiv g(n) \\quad(\\bmod c) . \\tag{9}\n\\end{equation*}\n$$\nNow, let us introduce the function $\\delta(n)=f(n)-n=g(n)-n-1$. Set\n$$\nS=\\sum_{n=0}^{c-1} \\delta(n)\n$$\nUsing (8), we get that for every complete residue system $n_{1}, \\ldots, n_{c}$ modulo $c$ we also have\n$$\nS=\\sum_{i=1}^{c} \\delta\\left(n_{i}\\right)\n$$\nBy (9), we get that $\\left\\{f^{k}(n): n=0, \\ldots, c-1\\right\\}$ and $\\left\\{g^{k}(n): n=0, \\ldots, c-1\\right\\}$ are complete residue systems modulo $c$ for all $k$. Thus we have\n$$\nc^{2}=\\sum_{n=0}^{c-1}\\left(f^{4}(n)-n\\right)=\\sum_{k=0}^{3} \\sum_{n=0}^{c-1}\\left(f^{k+1}(n)-f^{k}(n)\\right)=\\sum_{k=0}^{3} \\sum_{n=0}^{c-1} \\delta\\left(f^{k}(n)\\right)=4 S\n$$\nand similarly\n$$\nc^{2}=\\sum_{n=0}^{c-1}\\left(g^{2}(n)-n\\right)=\\sum_{k=0}^{1} \\sum_{n=0}^{c-1}\\left(g^{k+1}(n)-g^{k}(n)\\right)=\\sum_{k=0}^{1} \\sum_{n=0}^{c-1}\\left(\\delta\\left(g^{k}(n)\\right)+1\\right)=2 S+2 c\n$$\nTherefore $c^{2}=4 S=2 \\cdot 2 S=2\\left(c^{2}-2 c\\right)$, or $c^{2}=4 c$. Since $c \\neq 0$, we get $c=4$. Thus, in view of (8) it is sufficient to determine the values of $f$ on the numbers $0,1,2,3$.\nIII. Let $d=g(0) \\geqslant 1$. Then $g(d)=g^{2}(0)=0+c=4$. Now, if $d \\geqslant 4$, then we would have $g(d-4)=g(d)-4=0$ which is impossible. Thus $d \\in\\{1,2,3\\}$. If $d=1$ then we have $f(0)=g(0)-1=0$ which is impossible since $f(n) \\neq n$ for all $n$. If $d=3$ then $g(3)=g^{2}(0)=4$ and hence $f(3)=3$ which is also impossible. Thus $g(0)=2$ and hence $g(2)=g^{2}(0)=4$.\nNext, if $g(1)=1+4 k$ for some integer $k$, then $5=g^{2}(1)=g(1+4 k)=g(1)+4 k=1+8 k$ which is impossible. Thus, since $\\{g(n): n=0,1,2,3\\}$ is a complete residue system modulo 4 , we get $g(1)=3+4 k$ and hence $g(3)=g^{2}(1)-4 k=5-4 k$, leading to $k=0$ or $k=1$. So, we obtain iether\n$$\nf(0)=1, f(1)=2, f(2)=3, f(3)=4, \\quad \\text{ or } \\quad f(0)=1, f(1)=6, f(2)=3, f(3)=0\n$$\nthus arriving to the two functions listed in the answer.\nFinally, one can check that these two function work as in Solution 1. One may simplify the checking by noticing that (8) allows us to reduce it to $n=0,1,2,3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31968,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m \\neq 0$ be an integer. Find all polynomials $P(x)$ with real coefficients such that\n$$\n\\left(x^{3}-m x^{2}+1\\right) P(x+1)+\\left(x^{3}+m x^{2}+1\\right) P(x-1)=2\\left(x^{3}-m x+1\\right) P(x)\n$$\nfor all real numbers $x$.",
    "options": [],
    "answer": "P(x) = t x for any real t",
    "solution": "Let $P(x)=a_{n} x^{n}+\\cdots+a_{0} x^{0}$ with $a_{n} \\neq 0$. Comparing the coefficients of $x^{n+1}$ on both sides gives $a_{n}(n-2 m)(n-1)=0$, so $n=1$ or $n=2 m$.\nIf $n=1$, one easily verifies that $P(x)=x$ is a solution, while $P(x)=1$ is not. Since the given condition is linear in $P$, this means that the linear solutions are precisely $P(x)=t x$ for $t \\in \\mathbb{R}$.\nNow assume that $n=2 m$. The polynomial $x P(x+1)-(x+1) P(x)=(n-1) a_{n} x^{n}+\\cdots$ has degree $n$, and therefore it has at least one (possibly complex) root $r$. If $r \\notin\\{0,-1\\}$, define $k=P(r) / r=P(r+1) /(r+1)$. If $r=0$, let $k=P(1)$. If $r=-1$, let $k=-P(-1)$. We now consider the polynomial $S(x)=P(x)-k x$. It also satisfies (1) because $P(x)$ and $k x$ satisfy it. Additionally, it has the useful property that $r$ and $r+1$ are roots.\nLet $A(x)=x^{3}-m x^{2}+1$ and $B(x)=x^{3}+m x^{2}+1$. Plugging in $x=s$ into (1) implies that:\nIf $s-1$ and $s$ are roots of $S$ and $s$ is not a root of $A$, then $s+1$ is a root of $S$.\nIf $s$ and $s+1$ are roots of $S$ and $s$ is not a root of $B$, then $s-1$ is a root of $S$.\nLet $a \\geqslant 0$ and $b \\geqslant 1$ be such that $r-a, r-a+1, \\ldots, r, r+1, \\ldots, r+b-1, r+b$ are roots of $S$, while $r-a-1$ and $r+b+1$ are not. The two statements above imply that $r-a$ is a root of $B$ and $r+b$ is a root of $A$.\nSince $r-a$ is a root of $B(x)$ and of $A(x+a+b)$, it is also a root of their greatest common divisor $C(x)$ as integer polynomials. If $C(x)$ was a non-trivial divisor of $B(x)$, then $B$ would have a rational root $\\alpha$. Since the first and last coefficients of $B$ are $1, \\alpha$ can only be 1 or -1 ; but $B(-1)=m>0$ and $B(1)=m+2>0$ since $n=2 m$.\nTherefore $B(x)=A(x+a+b)$. Writing $c=a+b \\geqslant 1$ we compute\n$$\n0=A(x+c)-B(x)=(3 c-2 m) x^{2}+c(3 c-2 m) x+c^{2}(c-m) .\n$$\nThen we must have $3 c-2 m=c-m=0$, which gives $m=0$, a contradiction. We conclude that $f(x)=t x$ is the only solution.\nMultiplying (1) by $x$, we rewrite it as\n$$\nx\\left(x^{3}-m x^{2}+1\\right) P(x+1)+x\\left(x^{3}+m x^{2}+1\\right) P(x-1)=[(x+1)+(x-1)]\\left(x^{3}-m x+1\\right) P(x) .\n$$\nAfter regrouping, it becomes\n$$\n\\begin{equation*}\n\\left(x^{3}-m x^{2}+1\\right) Q(x)=\\left(x^{3}+m x^{2}+1\\right) Q(x-1), \\tag{2}\n\\end{equation*}\n$$\nwhere $Q(x)=x P(x+1)-(x+1) P(x)$. If $\\operatorname{deg} P \\geqslant 2$ then $\\operatorname{deg} Q=\\operatorname{deg} P$, so $Q(x)$ has a finite multiset of complex roots, which we denote $R_{Q}$. Each root is taken with its multiplicity. Then the multiset of complex roots of $Q(x-1)$ is $R_{Q}+1=\\{z+1: z \\in R_{Q}\\}$.\nLet $\\{x_{1}, x_{2}, x_{3}\\}$ and $\\{y_{1}, y_{2}, y_{3}\\}$ be the multisets of roots of the polynomials $A(x)=x^{3}-m x^{2}+1$ and $B(x)=x^{3}+m x^{2}+1$, respectively. From (2) we get the equality of multisets\n$$\n\\{x_{1}, x_{2}, x_{3}\\} \\cup R_{Q}=\\{y_{1}, y_{2}, y_{3}\\} \\cup(R_{Q}+1) .\n$$\nFor every $r \\in R_{Q}$, since $r+1$ is in the set of the right hand side, we must have $r+1 \\in R_{Q}$ or $r+1=x_{i}$ for some $i$. Similarly, since $r$ is in the set of the left hand side, either $r-1 \\in R_{Q}$ or $r=y_{i}$ for some $i$. This implies that, possibly after relabelling $y_{1}, y_{2}, y_{3}$, all the roots of (2) may be partitioned into three chains of the form $\\{y_{i}, y_{i}+1, \\ldots, y_{i}+k_{i}=x_{i}\\}$ for $i=1,2,3$ and some integers $k_{1}, k_{2}, k_{3} \\geqslant 0$.\nNow we analyze the roots of the polynomial $A_{a}(x)=x^{3}+a x^{2}+1$. Using calculus or elementary methods, we find that the local extrema of $A_{a}(x)$ occur at $x=0$ and $x=-2 a / 3$; their values are $A_{a}(0)=1>0$ and $A_{a}(-2 a / 3)=1+4 a^{3} / 27$, which is positive for integers $a \\geqslant-1$ and negative for integers $a \\leqslant-2$. So when $a \\in \\mathbb{Z}, A_{a}$ has three real roots if $a \\leqslant-2$ and one if $a \\geqslant-1$.\nNow, since $y_{i}-x_{i} \\in \\mathbb{Z}$ for $i=1,2,3$, the cubics $A_{m}$ and $A_{-m}$ must have the same number of real roots. The previous analysis then implies that $m=1$ or $m=-1$. Therefore the real root $\\alpha$ of $A_{1}(x)=x^{3}+x^{2}+1$ and the real root $\\beta$ of $A_{-1}(x)=x^{3}-x^{2}+1$ must differ by an integer. But this is impossible, because $A_{1}\\left(-\\frac{3}{2}\\right)=-\\frac{1}{8}$ and $A_{1}(-1)=1$ so $-1.5<\\alpha<-1$, while $A_{-1}(-1)=-1$ and $A_{-1}\\left(-\\frac{1}{2}\\right)=\\frac{5}{8}$, so $-1<\\beta<-0.5$.\nIt follows that $\\operatorname{deg} P \\leqslant 1$. Then, as shown in Solution 1, we conclude that the solutions are $P(x)=t x$ for all real numbers $t$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 31969,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find the smallest integer $k$ with the following property: Given any real numbers $a_{1}, \\ldots, a_{d}$ such that $a_{1}+a_{2}+\\cdots+a_{d}=n$ and $0 \\leqslant a_{i} \\leqslant 1$ for $i=1,2, \\ldots, d$, it is possible to partition these numbers into $k$ groups (some of which may be empty) such that the sum of the numbers in each group is at most $1$.",
    "options": [],
    "answer": "2n-1",
    "solution": "Answer. $k=2n-1$.\n\nSolution 1. If $d=2n-1$ and $a_{1}=\\cdots=a_{2n-1}=n/(2n-1)$, then each group in such a partition can contain at most one number, since $2n/(2n-1)>1$. Therefore $k \\geqslant 2n-1$. It remains to show that a suitable partition into $2n-1$ groups always exists.\nWe proceed by induction on $d$. For $d \\leqslant 2n-1$ the result is trivial. If $d \\geqslant 2n$, then since\n$$\n\\left(a_{1}+a_{2}\\right)+\\ldots+\\left(a_{2n-1}+a_{2n}\\right) \\leqslant n\n$$\nwe may find two numbers $a_{i}$, $a_{i+1}$ such that $a_{i}+a_{i+1} \\leqslant 1$. We \"merge\" these two numbers into one new number $a_{i}+a_{i+1}$. By the induction hypothesis, a suitable partition exists for the $d-1$ numbers $a_{1}, \\ldots, a_{i-1}, a_{i}+a_{i+1}, a_{i+2}, \\ldots, a_{d}$. This induces a suitable partition for $a_{1}, \\ldots, a_{d}$.\n\n\nSolution 2. We will show that it is even possible to split the sequence $a_{1}, \\ldots, a_{d}$ into $2n-1$ contiguous groups so that the sum of the numbers in each group does not exceed $1$. Consider a segment $S$ of length $n$, and partition it into segments $S_{1}, \\ldots, S_{d}$ of lengths $a_{1}, \\ldots, a_{d}$, respectively, as shown below. Consider a second partition of $S$ into $n$ equal parts by $n-1$ \"empty dots\".\n![](attached_image_1.png)\nAssume that the $n-1$ empty dots are in segments $S_{i_{1}}, \\ldots, S_{i_{n-1}}$. (If a dot is on the boundary of two segments, we choose the right segment). These $n-1$ segments are distinct because they have length at most $1$. Consider the partition:\n$$\n\\left\\{a_{1}, \\ldots, a_{i_{1}-1}\\right\\},\\left\\{a_{i_{1}}\\right\\},\\left\\{a_{i_{1}+1}, \\ldots, a_{i_{2}-1}\\right\\},\\left\\{a_{i_{2}}\\right\\}, \\ldots,\\left\\{a_{i_{n-1}}\\right\\},\\left\\{a_{i_{n-1}+1}, \\ldots, a_{d}\\right\\}.\n$$\nIn the example above, this partition is $\\left\\{a_{1}, a_{2}\\right\\},\\left\\{a_{3}\\right\\},\\left\\{a_{4}, a_{5}\\right\\},\\left\\{a_{6}\\right\\}, \\varnothing,\\left\\{a_{7}\\right\\},\\left\\{a_{8}, a_{9}, a_{10}\\right\\}$. We claim that in this partition, the sum of the numbers in each group is at most $1$.\nFor the sets $\\left\\{a_{i_{t}}\\right\\}$ this is obvious since $a_{i_{t}} \\leqslant 1$. For the sets $\\left\\{a_{i_{t}+1}, \\ldots, a_{i_{t+1}-1}\\right\\}$ this follows from the fact that the corresponding segments lie between two neighboring empty dots, or between an endpoint of $S$ and its nearest empty dot. Therefore the sum of their lengths cannot exceed $1$.\n\n\nSolution 3. First put all numbers greater than $\\frac{1}{2}$ in their own groups. Then, form the remaining groups as follows: For each group, add new $a_{i}$'s one at a time until their sum exceeds $\\frac{1}{2}$. Since the last summand is at most $\\frac{1}{2}$, this group has sum at most $1$. Continue this procedure until we have used all the $a_{i}$'s. Notice that the last group may have sum less than $\\frac{1}{2}$. If the sum of the numbers in the last two groups is less than or equal to $1$, we merge them into one group. In the end we are left with $m$ groups. If $m=1$ we are done. Otherwise the first $m-2$ have sums greater than $\\frac{1}{2}$ and the last two have total sum greater than $1$. Therefore $n>(m-2)/2+1$ so $m \\leqslant 2n-1$ as desired.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 31970,
    "subject": "Mathematics (Multi-modal)",
    "question": "In the plane, 2013 red points and 2014 blue points are marked so that no three of the marked points are collinear. One needs to draw $k$ lines not passing through the marked points and dividing the plane into several regions. The goal is to do it in such a way that no region contains points of both colors.\nFind the minimal value of $k$ such that the goal is attainable for every possible configuration of 4027 points.",
    "options": [],
    "answer": "2013",
    "solution": "Answer. $k=2013$.\n\nFirstly, let us present an example showing that $k \\geqslant 2013$. Mark 2013 red and 2013 blue points on some circle alternately, and mark one more blue point somewhere in the plane. The circle is thus split into 4026 arcs, each arc having endpoints of different colors. Thus, if the goal is reached, then each arc should intersect some of the drawn lines. Since any line contains at most two points of the circle, one needs at least $4026 / 2=2013$ lines.\n\nIt remains to prove that one can reach the goal using 2013 lines. First of all, let us mention that for every two points $A$ and $B$ having the same color, one can draw two lines separating these points from all other ones. Namely, it suffices to take two lines parallel to $A B$ and lying on different sides of $A B$ sufficiently close to it: the only two points between these lines will be $A$ and $B$.\n\nNow, let $P$ be the convex hull of all marked points. Two cases are possible.\n\nCase 1. Assume that $P$ has a red vertex $A$. Then one may draw a line separating $A$ from all the other points, pair up the other 2012 red points into 1006 pairs, and separate each pair from the other points by two lines. Thus, 2013 lines will be used.\n\nCase 2. Assume now that all the vertices of $P$ are blue. Consider any two consecutive vertices of $P$, say $A$ and $B$. One may separate these two points from the others by a line parallel to $A B$. Then, as in the previous case, one pairs up all the other 2012 blue points into 1006 pairs, and separates each pair from the other points by two lines. Again, 2013 lines will be used.\nIf $n$ points in the plane, no three of which are collinear, are colored in red and blue arbitrarily, then it suffices to draw $\\lfloor n / 2\\rfloor$ lines to reach the goal.\n\nWe proceed by induction on $n$. If $n \\leqslant 2$ then the statement is obvious. Now assume that $n \\geqslant 3$, and consider a line $\\ell$ containing two marked points $A$ and $B$ such that all the other marked points are on one side of $\\ell$; for instance, any line containing a side of the convex hull works.\n\nRemove for a moment the points $A$ and $B$. By the induction hypothesis, for the remaining configuration it suffices to draw $\\lfloor n / 2\\rfloor-1$ lines to reach the goal. Now return the points $A$ and $B$ back. Three cases are possible.\n\nCase 1. If $A$ and $B$ have the same color, then one may draw a line parallel to $\\ell$ and separating $A$ and $B$ from the other points. Obviously, the obtained configuration of $\\lfloor n / 2\\rfloor$ lines works.\n\nCase 2. If $A$ and $B$ have different colors, but they are separated by some drawn line, then again the same line parallel to $\\ell$ works.\n\nCase 3. Finally, assume that $A$ and $B$ have different colors and lie in one of the regions defined by the drawn lines. By the induction assumption, this region contains no other points of one of the colors - without loss of generality, the only blue point it contains is $A$. Then it suffices to draw a line separating $A$ from all other points.\n\nThus the step of the induction is proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31971,
    "subject": "Mathematics (Multi-modal)",
    "question": "A crazy physicist discovered a new kind of particle which he called an imon, after some of them mysteriously appeared in his lab. Some pairs of imons in the lab can be entangled, and each imon can participate in many entanglement relations. The physicist has found a way to perform the following two kinds of operations with these particles, one operation at a time.\n(i) If some imon is entangled with an odd number of other imons in the lab, then the physicist can destroy it.\n(ii) At any moment, he may double the whole family of imons in his lab by creating a copy $I^{\\prime}$ of each imon $I$. During this procedure, the two copies $I^{\\prime}$ and $J^{\\prime}$ become entangled if and only if the original imons $I$ and $J$ are entangled, and each copy $I^{\\prime}$ becomes entangled with its original imon $I$; no other entanglements occur or disappear at this moment.\nProve that the physicist may apply a sequence of such operations resulting in a family of imons, no two of which are entangled.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider a graph with the imons as vertices, and two imons being connected if and only if they are entangled. Recall that a proper coloring of a graph $G$ is a coloring of its vertices in several colors so that every two connected vertices have different colors.\n\nLemma. Assume that a graph $G$ admits a proper coloring in $n$ colors $(n>1)$. Then one may perform a sequence of operations resulting in a graph which admits a proper coloring in $n-1$ colors.\n\nProof. Let us apply repeatedly operation (i) to any appropriate vertices while it is possible. Since the number of vertices decreases, this process finally results in a graph where all the degrees are even. Surely this graph also admits a proper coloring in $n$ colors $1, \\ldots, n$; let us fix this coloring.\n\nNow apply the operation (ii) to this graph. A proper coloring of the resulting graph in $n$ colors still exists: one may preserve the colors of the original vertices and color the vertex $I^{\\prime}$ in a color $k+1(\\bmod n)$ if the vertex $I$ has color $k$. Then two connected original vertices still have different colors, and so do their two connected copies. On the other hand, the vertices $I$ and $I^{\\prime}$ have different colors since $n>1$.\n\nAll the degrees of the vertices in the resulting graph are odd, so one may apply operation (i) to delete consecutively all the vertices of color $n$ one by one; no two of them are connected by an edge, so their degrees do not change during the process. Thus, we obtain a graph admitting a proper coloring in $n-1$ colors, as required. The lemma is proved.\n\nNow, assume that a graph $G$ has $n$ vertices; then it admits a proper coloring in $n$ colors. Applying repeatedly the lemma we finally obtain a graph admitting a proper coloring in one color, that is - a graph with no edges, as required.\nAgain, we will use the graph language.\n\nI. We start with the following observation.\nLemma. Assume that a graph $G$ contains an isolated vertex $A$, and a graph $G^{\\circ}$ is obtained from $G$ by deleting this vertex. Then, if one can apply a sequence of operations which makes a graph with no edges from $G^{\\circ}$, then such a sequence also exists for $G$.\n\nProof. Consider any operation applicable to $G^{\\circ}$ resulting in a graph $G_{1}^{\\circ}$; then there exists a sequence of operations applicable to $G$ and resulting in a graph $G_{1}$ differing from $G_{1}^{\\circ}$ by an addition of an isolated vertex $A$. Indeed, if this operation is of type (i), then one may simply repeat it in $G$.\nOtherwise, the operation is of type (ii), and one may apply it to $G$ and then delete the vertex $A^{\\prime}$ (it will have degree 1).\nThus one may change the process for $G^{\\circ}$ into a corresponding process for $G$ step by step.\nIn view of this lemma, if at some moment a graph contains some isolated vertex, then we may simply delete it; let us call this operation (iii).\n\nII. Let $V=\\left\\{A_{1}^{0}, \\ldots, A_{n}^{0}\\right\\}$ be the vertices of the initial graph. Let us describe which graphs can appear during our operations. Assume that operation (ii) was applied $m$ times. If these were the only operations applied, then the resulting graph $G_{n}^{m}$ has the set of vertices which can be enumerated as\n$$\nV_{n}^{m}=\\left\\{A_{i}^{j}: 1 \\leqslant i \\leqslant n, 0 \\leqslant j \\leqslant 2^{m}-1\\right\\},\n$$\nwhere $A_{i}^{0}$ is the common \"ancestor\" of all the vertices $A_{i}^{j}$, and the binary expansion of $j$ (adjoined with some zeroes at the left to have $m$ digits) \"keeps the history\" of this vertex: the $d$th digit from the right is 0 if at the $d$th doubling the ancestor of $A_{i}^{j}$ was in the original part, and this digit is 1 if it was in the copy.\n\nNext, the two vertices $A_{i}^{j}$ and $A_{k}^{\\ell}$ in $G_{n}^{m}$ are connected with an edge exactly if either (1) $j=\\ell$ and there was an edge between $A_{i}^{0}$ and $A_{k}^{0}$ (so these vertices appeared at the same application of operation (ii)); or (2) $i=k$ and the binary expansions of $j$ and $\\ell$ differ in exactly one digit (so their ancestors became connected as a copy and the original vertex at some application of (ii)).\n\nNow, if some operations (i) were applied during the process, then simply some vertices in $G_{n}^{m}$ disappeared. So, in any case the resulting graph is some induced subgraph of $G_{n}^{m}$.\n\nIII. Finally, we will show that from each (not necessarily induced) subgraph of $G_{n}^{m}$ one can obtain a graph with no vertices by applying operations (i), (ii) and (iii). We proceed by induction on $n$; the base case $n=0$ is trivial.\n\nFor the induction step, let us show how to apply several operations so as to obtain a graph containing no vertices of the form $A_{n}^{j}$ for $j \\in \\mathbb{Z}$. We will do this in three steps.\n\nStep 1. We apply repeatedly operation (i) to any appropriate vertices while it is possible. In the resulting graph, all vertices have even degrees.\n\nStep 2. Apply operation (ii) obtaining a subgraph of $G_{n}^{m+1}$ with all degrees being odd. In this graph, we delete one by one all the vertices $A_{n}^{j}$ where the sum of the binary digits of $j$ is even; it is possible since there are no edges between such vertices, so all their degrees remain odd. After that, we delete all isolated vertices.\n\nStep 3. Finally, consider any remaining vertex $A_{n}^{j}$ (then the sum of digits of $j$ is odd). If its degree is odd, then we simply delete it. Otherwise, since $A_{n}^{j}$ is not isolated, we consider any vertex adjacent to it. It has the form $A_{k}^{j}$ for some $k<n$ (otherwise it would have the form $A_{n}^{\\ell}$, where $\\ell$ has an even digit sum; but any such vertex has already been deleted at Step 2). No neighbor of $A_{k}^{j}$ was deleted at Steps 2 and 3, so it has an odd degree. Then we successively delete $A_{k}^{j}$ and $A_{n}^{j}$.\n\nNotice that this deletion does not affect the applicability of this step to other vertices, since no two vertices $A_{i}^{j}$ and $A_{k}^{\\ell}$ for different $j, \\ell$ with odd digit sum are connected with an edge. Thus we will delete all the remaining vertices of the form $A_{n}^{j}$, obtaining a subgraph of $G_{n-1}^{m+1}$. The application of the induction hypothesis finishes the proof.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31972,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer, and let $A$ be a subset of $\\{1, \\ldots, n\\}$. An $A$-partition of $n$ into $k$ parts is a representation of $n$ as a sum $n=a_{1}+\\cdots+a_{k}$, where the parts $a_{1}, \\ldots, a_{k}$ belong to $A$ and are not necessarily distinct. The number of different parts in such a partition is the number of (distinct) elements in the set $\\left\\{a_{1}, a_{2}, \\ldots, a_{k}\\right\\}$.\nWe say that an $A$-partition of $n$ into $k$ parts is optimal if there is no $A$-partition of $n$ into $r$ parts with $r<k$. Prove that any optimal $A$-partition of $n$ contains at most $\\sqrt[3]{6 n}$ different parts.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If there are no $A$-partitions of $n$, the result is vacuously true. Otherwise, let $k_{\\text{min}}$ be the minimum number of parts in an $A$-partition of $n$, and let $n=a_{1}+\\cdots+a_{k_{\\min}}$ be an optimal partition. Denote by $s$ the number of different parts in this partition, so we can write $S=\\left\\{a_{1}, \\ldots, a_{k_{\\text{min}}}\\right\\}=\\left\\{b_{1}, \\ldots, b_{s}\\right\\}$ for some pairwise different numbers $b_{1}<\\cdots<b_{s}$ in $A$.\nIf $s>\\sqrt[3]{6 n}$, we will prove that there exist subsets $X$ and $Y$ of $S$ such that $|X|<|Y|$ and $\\sum_{x \\in X} x=\\sum_{y \\in Y} y$. Then, deleting the elements of $Y$ from our partition and adding the elements of $X$ to it, we obtain an $A$-partition of $n$ into less than $k_{\\text{min}}$ parts, which is the desired contradiction.\nFor each positive integer $k \\leqslant s$, we consider the $k$-element subset\n$$\nS_{1,0}^{k}:=\\left\\{b_{1}, \\ldots, b_{k}\\right\\}\n$$\nas well as the following $k$-element subsets $S_{i, j}^{k}$ of $S$ :\n$$\nS_{i, j}^{k}:=\\left\\{b_{1}, \\ldots, b_{k-i}, b_{k-i+j+1}, b_{s-i+2}, \\ldots, b_{s}\\right\\}, \\quad i=1, \\ldots, k, \\quad j=1, \\ldots, s-k\n$$\nPictorially, if we represent the elements of $S$ by a sequence of dots in increasing order, and represent a subset of $S$ by shading in the appropriate dots, we have:\n![](attached_image_1.png)\nDenote by $\\Sigma_{i, j}^{k}$ the sum of elements in $S_{i, j}^{k}$. Clearly, $\\Sigma_{1,0}^{k}$ is the minimum sum of a $k$-element subset of $S$. Next, for all appropriate indices $i$ and $j$ we have\n$$\n\\Sigma_{i, j}^{k}=\\Sigma_{i, j+1}^{k}+b_{k-i+j+1}-b_{k-i+j+2}<\\Sigma_{i, j+1}^{k} \\quad \\text{ and } \\quad \\Sigma_{i, s-k}^{k}=\\Sigma_{i+1,1}^{k}+b_{k-i}-b_{k-i+1}<\\Sigma_{i+1,1}^{k} .\n$$\nTherefore\n$$\n1 \\leqslant \\Sigma_{1,0}^{k}<\\Sigma_{1,1}^{k}<\\Sigma_{1,2}^{k}<\\cdots<\\Sigma_{1, s-k}^{k}<\\Sigma_{2,1}^{k}<\\cdots<\\Sigma_{2, s-k}^{k}<\\Sigma_{3,1}^{k}<\\cdots<\\Sigma_{k, s-k}^{k} \\leqslant n .\n$$\nTo see this in the picture, we start with the $k$ leftmost points marked. At each step, we look for the rightmost point which can move to the right, and move it one unit to the right. We continue until the $k$ rightmost points are marked. As we do this, the corresponding sums clearly increase.\nFor each $k$ we have found $k(s-k)+1$ different integers of the form $\\sum_{i, j}^{k}$ between 1 and $n$. As we vary $k$, the total number of integers we are considering is\n$$\n\\sum_{k=1}^{s}(k(s-k)+1)=s \\cdot \\frac{s(s+1)}{2}-\\frac{s(s+1)(2 s+1)}{6}+s=\\frac{s\\left(s^{2}+5\\right)}{6}>\\frac{s^{3}}{6}>n .\n$$\nSince they are between 1 and $n$, at least two of these integers are equal. Consequently, there exist $1 \\leqslant k<k^{\\prime} \\leqslant s$ and $X=S_{i, j}^{k}$ as well as $Y=S_{i^{\\prime}, j^{\\prime}}^{k^{\\prime}}$ such that\n$$\n\\sum_{x \\in X} x=\\sum_{y \\in Y} y, \\quad \\text{ but } \\quad|X|=k<k^{\\prime}=|Y|\n$$\nas required. The result follows.\nAssume, to the contrary, that the statement is false, and choose the minimum number $n$ for which it fails. So there exists a set $A \\subseteq\\{1, \\ldots, n\\}$ together with an optimal $A$ partition $n=a_{1}+\\cdots+a_{k_{\\text{min}}}$ of $n$ refuting our statement, where, of course, $k_{\\text{min}}$ is the minimum number of parts in an $A$-partition of $n$. Again, we define $S=\\left\\{a_{1}, \\ldots, a_{k_{\\text{min}}}\\right\\}=\\left\\{b_{1}, \\ldots, b_{s}\\right\\}$ with $b_{1}<\\cdots<b_{s}$; by our assumption we have $s>\\sqrt[3]{6 n}>1$. Without loss of generality we assume that $a_{k_{\\min}}=b_{s}$. Let us distinguish two cases.\n\nCase 1. $b_{s} \\geqslant \\frac{s(s-1)}{2}+1$.\nConsider the partition $n-b_{s}=a_{1}+\\cdots+a_{k_{\\text{min}}-1}$, which is clearly a minimum $A$-partition of $n-b_{s}$ with at least $s-1 \\geqslant 1$ different parts. Now, from $n<\\frac{s^{3}}{6}$ we obtain\n$$\nn-b_{s} \\leqslant n-\\frac{s(s-1)}{2}-1<\\frac{s^{3}}{6}-\\frac{s(s-1)}{2}-1<\\frac{(s-1)^{3}}{6}\n$$\nso $s-1>\\sqrt[3]{6\\left(n-b_{s}\\right)}$, which contradicts the choice of $n$.\n\nCase 2. $b_{s} \\leqslant \\frac{s(s-1)}{2}$.\nSet $b_{0}=0, \\Sigma_{0,0}=0$, and $\\Sigma_{i, j}=b_{1}+\\cdots+b_{i-1}+b_{j}$ for $1 \\leqslant i \\leqslant j<s$. There are $\\frac{s(s-1)}{2}+1>b_{s}$ such sums; so at least two of them, say $\\Sigma_{i, j}$ and $\\Sigma_{i^{\\prime}, j^{\\prime}}$, are congruent modulo $b_{s}$ (where $(i, j) \\neq\\left(i^{\\prime}, j^{\\prime}\\right)$ ). This means that $\\Sigma_{i, j}-\\Sigma_{i^{\\prime}, j^{\\prime}}=r b_{s}$ for some integer $r$. Notice that for $i \\leqslant j<k<s$ we have\n$$\n0<\\Sigma_{i, k}-\\Sigma_{i, j}=b_{k}-b_{j}<b_{s},\n$$\nso the indices $i$ and $i^{\\prime}$ are distinct, and we may assume that $i>i^{\\prime}$. Next, we observe that $\\Sigma_{i, j}-\\Sigma_{i^{\\prime}, j^{\\prime}}=\\left(b_{i^{\\prime}}-b_{j^{\\prime}}\\right)+b_{j}+b_{i^{\\prime}+1}+\\cdots+b_{i-1}$ and $b_{i^{\\prime}} \\leqslant b_{j^{\\prime}}$ imply\n$$\n-b_{s}<-b_{j^{\\prime}}<\\Sigma_{i, j}-\\Sigma_{i^{\\prime}, j^{\\prime}}<\\left(i-i^{\\prime}\\right) b_{s}\n$$\nso $0 \\leqslant r \\leqslant i-i^{\\prime}-1$.\nThus, we may remove the $i$ terms of $\\Sigma_{i, j}$ in our $A$-partition, and replace them by the $i^{\\prime}$ terms of $\\Sigma_{i^{\\prime}, j^{\\prime}}$ and $r$ terms equal to $b_{s}$, for a total of $r+i^{\\prime}<i$ terms. The result is an $A$-partition of $n$ into a smaller number of parts, a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31973,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $r$ be a positive integer, and let $a_{0}, a_{1}, \\ldots$ be an infinite sequence of real numbers. Assume that for all nonnegative integers $m$ and $s$ there exists a positive integer $n \\in [m+1, m+r]$ such that\n$$\na_{m}+a_{m+1}+\\cdots+a_{m+s}=a_{n}+a_{n+1}+\\cdots+a_{n+s}\n$$\nProve that the sequence is periodic, i.e. there exists some $p \\geqslant 1$ such that $a_{n+p}=a_{n}$ for all $n \\geqslant 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For every indices $m \\leqslant n$ we will denote $S(m, n)=a_{m}+a_{m+1}+\\cdots+a_{n-1}$; thus $S(n, n)=0$. Let us start with the following lemma.\n\nLemma. Let $b_{0}, b_{1}, \\ldots$ be an infinite sequence. Assume that for every nonnegative integer $m$ there exists a nonnegative integer $n \\in [m+1, m+r]$ such that $b_{m}=b_{n}$. Then for every indices $k \\leqslant \\ell$ there exists an index $t \\in [\\ell, \\ell+r-1]$ such that $b_{t}=b_{k}$. Moreover, there are at most $r$ distinct numbers among the terms of $(b_{i})$.\n\nProof. To prove the first claim, let us notice that there exists an infinite sequence of indices $k_{1}=k, k_{2}, k_{3}, \\ldots$ such that $b_{k_{1}}=b_{k_{2}}=\\cdots=b_{k}$ and $k_{i}<k_{i+1} \\leqslant k_{i}+r$ for all $i \\geqslant 1$. This sequence is unbounded from above, thus it hits each segment of the form $[\\ell, \\ell+r-1]$ with $\\ell \\geqslant k$, as required.\n\nTo prove the second claim, assume, to the contrary, that there exist $r+1$ distinct numbers $b_{i_{1}}, \\ldots, b_{i_{r+1}}$. Let us apply the first claim to $k=i_{1}, \\ldots, i_{r+1}$ and $\\ell=\\max \\{i_{1}, \\ldots, i_{r+1}\\}$; we obtain that for every $j \\in \\{1, \\ldots, r+1\\}$ there exists $t_{j} \\in [s, s+r-1]$ such that $b_{t_{j}}=b_{i_{j}}$. Thus the segment $[s, s+r-1]$ should contain $r+1$ distinct integers, which is absurd.\n\nSetting $s=0$ in the problem condition, we see that the sequence $(a_{i})$ satisfies the condition of the lemma, thus it attains at most $r$ distinct values. Denote by $A_{i}$ the ordered $r$-tuple $(a_{i}, \\ldots, a_{i+r-1})$; then among $A_{i}$'s there are at most $r^{r}$ distinct tuples, so for every $k \\geqslant 0$ two of the tuples $A_{k}, A_{k+1}, \\ldots, A_{k+r^{r}}$ are identical. This means that there exists a positive integer $p \\leqslant r^{r}$ such that the equality $A_{d}=A_{d+p}$ holds infinitely many times. Let $D$ be the set of indices $d$ satisfying this relation.\n\nNow we claim that $D$ coincides with the set of all nonnegative integers. Since $D$ is unbounded, it suffices to show that $d \\in D$ whenever $d+1 \\in D$. For that, denote $b_{k}=S(k, p+k)$. The sequence $b_{0}, b_{1}, \\ldots$ satisfies the lemma conditions, so there exists an index $t \\in [d+1, d+r]$ such that $S(t, t+p)=S(d, d+p)$. This last relation rewrites as $S(d, t)=S(d+p, t+p)$. Since $A_{d+1}=A_{d+p+1}$, we have $S(d+1, t)=S(d+p+1, t+p)$, therefore we obtain\n$$\na_{d}=S(d, t)-S(d+1, t)=S(d+p, t+p)-S(d+p+1, t+p)=a_{d+p}\n$$\nand thus $A_{d}=A_{d+p}$, as required.\n\nFinally, we get $A_{d}=A_{d+p}$ for all $d$, so in particular $a_{d}=a_{d+p}$ for all $d$, QED.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31974,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 2$ be an integer. Consider all circular arrangements of the numbers $0,1, \\ldots, n$; the $n+1$ rotations of an arrangement are considered to be equal. A circular arrangement is called beautiful if, for any four distinct numbers $0 \\leqslant a, b, c, d \\leqslant n$ with $a+c=b+d$, the chord joining numbers $a$ and $c$ does not intersect the chord joining numbers $b$ and $d$.\nLet $M$ be the number of beautiful arrangements of $0,1, \\ldots, n$. Let $N$ be the number of pairs ( $x, y$ ) of positive integers such that $x+y \\leqslant n$ and $\\operatorname{gcd}(x, y)=1$. Prove that\n$$\nM=N+1 .\n$$",
    "options": [],
    "answer": "M = N + 1",
    "solution": "Given a circular arrangement of $[0, n]=\\{0,1, \\ldots, n\\}$, we define a $k$-chord to be a (possibly degenerate) chord whose (possibly equal) endpoints add up to $k$. We say that three chords of a circle are aligned if one of them separates the other two. Say that $m \\geqslant 3$ chords are aligned if any three of them are aligned. For instance, in Figure 1, $A, B$, and $C$ are aligned, while $B, C$, and $D$ are not.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n\nClaim. In a beautiful arrangement, the $k$-chords are aligned for any integer $k$.\n\nProof. We proceed by induction. For $n \\leqslant 3$ the statement is trivial. Now let $n \\geqslant 4$, and proceed by contradiction. Consider a beautiful arrangement $S$ where the three $k$-chords $A, B, C$ are not aligned. If $n$ is not among the endpoints of $A, B$, and $C$, then by deleting $n$ from $S$ we obtain a beautiful arrangement $S \\backslash\\{n\\}$ of $[0, n-1]$, where $A, B$, and $C$ are aligned by the induction hypothesis. Similarly, if 0 is not among these endpoints, then deleting 0 and decreasing all the numbers by 1 gives a beautiful arrangement $S \\backslash\\{0\\}$ where $A, B$, and $C$ are aligned. Therefore both 0 and $n$ are among the endpoints of these segments. If $x$ and $y$ are their respective partners, we have $n \\geqslant 0+x=k=n+y \\geqslant n$. Thus 0 and $n$ are the endpoints of one of the chords; say it is $C$.\n\nLet $D$ be the chord formed by the numbers $u$ and $v$ which are adjacent to 0 and $n$ and on the same side of $C$ as $A$ and $B$, as shown in Figure 2. Set $t=u+v$. If we had $t=n$, the $n$-chords $A$, $B$, and $D$ would not be aligned in the beautiful arrangement $S \\backslash\\{0, n\\}$, contradicting the induction hypothesis. If $t<n$, then the $t$-chord from 0 to $t$ cannot intersect $D$, so the chord $C$ separates $t$ and $D$. The chord $E$ from $t$ to $n-t$ does not intersect $C$, so $t$ and $n-t$ are on the same side of $C$. But then the chords $A, B$, and $E$ are not aligned in $S \\backslash\\{0, n\\}$, a contradiction. Finally, the case $t>n$ is equivalent to the case $t<n$ via the beauty-preserving relabelling $x \\mapsto n-x$ for $0 \\leqslant x \\leqslant n$, which sends $t$-chords to $(2 n-t)$-chords. This proves the Claim.\n\nHaving established the Claim, we prove the desired result by induction. The case $n=2$ is trivial. Now assume that $n \\geqslant 3$. Let $S$ be a beautiful arrangement of $[0, n]$ and delete $n$ to obtain the beautiful arrangement $T$ of $[0, n-1]$. The $n$-chords of $T$ are aligned, and they contain every point except 0. Say $T$ is of Type 1 if 0 lies between two of these $n$-chords, and it is of Type 2 otherwise; i.e., if 0 is aligned with these $n$-chords. We will show that each Type 1 arrangement of $[0, n-1]$ arises from a unique arrangement of $[0, n]$, and each Type 2 arrangement of $[0, n-1]$ arises from exactly two beautiful arrangements of $[0, n]$.\n\nIf $T$ is of Type 1, let 0 lie between chords $A$ and $B$. Since the chord from 0 to $n$ must be aligned with $A$ and $B$ in $S, n$ must be on the other arc between $A$ and $B$. Therefore $S$ can be recovered uniquely from $T$. In the other direction, if $T$ is of Type 1 and we insert $n$ as above, then we claim the resulting arrangement $S$ is beautiful. For $0<k<n$, the $k$-chords of $S$ are also $k$-chords of $T$, so they are aligned. Finally, for $n<k<2 n$, notice that the $n$-chords of $S$ are parallel by construction, so there is an antisymmetry axis $\\ell$ such that $x$ is symmetric to $n-x$ with respect to $\\ell$ for all $x$. If we had two $k$-chords which intersect, then their reflections across $\\ell$ would be two $(2 n-k)$-chords which intersect, where $0<2 n-k<n$, a contradiction.\n\nIf $T$ is of Type 2, there are two possible positions for $n$ in $S$, on either side of 0. As above, we check that both positions lead to beautiful arrangements of $[0, n]$.\n\nHence if we let $M_{n}$ be the number of beautiful arrangements of $[0, n]$, and let $L_{n}$ be the number of beautiful arrangements of $[0, n-1]$ of Type 2, we have\n$$\nM_{n}=\\left(M_{n-1}-L_{n-1}\\right)+2 L_{n-1}=M_{n-1}+L_{n-1}\n$$\nIt then remains to show that $L_{n-1}$ is the number of pairs $(x, y)$ of positive integers with $x+y=n$ and $\\operatorname{gcd}(x, y)=1$. Since $n \\geqslant 3$, this number equals $\\varphi(n)=\\#\\{x: 1 \\leqslant x \\leqslant n, \\operatorname{gcd}(x, n)=1\\}$.\n\nTo prove this, consider a Type 2 beautiful arrangement of $[0, n-1]$. Label the positions $0, \\ldots, n-1(\\bmod n)$ clockwise around the circle, so that number 0 is in position 0. Let $f(i)$ be the number in position $i$; note that $f$ is a permutation of $[0, n-1]$. Let $a$ be the position such that $f(a)=n-1$.\n\nSince the $n$-chords are aligned with 0, and every point is in an $n$-chord, these chords are all parallel and\n$$\nf(i)+f(-i)=n \\quad \\text{ for all } i\n$$\nSimilarly, since the $(n-1)$-chords are aligned and every point is in an $(n-1)$-chord, these chords are also parallel and\n$$\nf(i)+f(a-i)=n-1 \\quad \\text{ for all } i\n$$\nTherefore $f(a-i)=f(-i)-1$ for all $i$; and since $f(0)=0$, we get\n$$\n\\begin{equation*}\nf(-a k)=k \\quad \\text{ for all } k \\tag{1}\n\\end{equation*}\n$$\nRecall that this is an equality modulo $n$. Since $f$ is a permutation, we must have $(a, n)=1$. Hence $L_{n-1} \\leqslant \\varphi(n)$.\n\nTo prove equality, it remains to observe that the labeling (1) is beautiful. To see this, consider four numbers $w, x, y, z$ on the circle with $w+y=x+z$. Their positions around the circle satisfy $(-a w)+(-a y)=(-a x)+(-a z)$, which means that the chord from $w$ to $y$ and the chord from $x$ to $z$ are parallel. Thus (1) is beautiful, and by construction it has Type 2. The desired result follows.\nNotice that there are exactly $N$ irreducible fractions $f_{1}<\\cdots<f_{N}$ in $(0,1)$ whose denominator is at most $n$, since the pair $(x, y)$ with $x+y \\leqslant n$ and $(x, y)=1$ corresponds to the fraction $x/(x+y)$. Write $f_{i}=\\frac{a_{i}}{b_{i}}$ for $1 \\leqslant i \\leqslant N$.\n\nWe begin by constructing $N+1$ beautiful arrangements. Take any $\\alpha \\in(0,1)$ which is not one of the above $N$ fractions. Consider a circle of perimeter 1. Successively mark points $0,1,2, \\ldots, n$ where 0 is arbitrary, and the clockwise distance from $i$ to $i+1$ is $\\alpha$. The point $k$ will be at clockwise distance $\\{k \\alpha\\}$ from 0, where $\\{r\\}$ denotes the fractional part of $r$. Call such a circular arrangement cyclic and denote it by $A(\\alpha)$. If the clockwise order of the points is the same in $A\\left(\\alpha_{1}\\right)$ and $A\\left(\\alpha_{2}\\right)$, we regard them as the same circular arrangement. Figure 3 shows the cyclic arrangement $A(3/5+\\epsilon)$ of $[0,13]$ where $\\epsilon>0$ is very small.\n\n![](attached_image_3.png)\nFigure 3\n\nIf $0 \\leqslant a, b, c, d \\leqslant n$ satisfy $a+c=b+d$, then $a \\alpha+c \\alpha=b \\alpha+d \\alpha$, so the chord from $a$ to $c$ is parallel to the chord from $b$ to $d$ in $A(\\alpha)$. Hence in a cyclic arrangement all $k$-chords are parallel. In particular every cyclic arrangement is beautiful.\n\nNext we show that there are exactly $N+1$ distinct cyclic arrangements. To see this, let us see how $A(\\alpha)$ changes as we increase $\\alpha$ from 0 to 1. The order of points $p$ and $q$ changes precisely when we cross a value $\\alpha=f$ such that $\\{p f\\}=\\{q f\\}$; this can only happen if $f$ is one of the $N$ fractions $f_{1}, \\ldots, f_{N}$. Therefore there are at most $N+1$ different cyclic arrangements.\n\nTo show they are all distinct, recall that $f_{i}=a_{i}/b_{i}$ and let $\\epsilon>0$ be a very small number. In the arrangement $A\\left(f_{i}+\\epsilon\\right)$, point $k$ lands at $\\frac{k a_{i}\\left(\\bmod b_{i}\\right)}{b_{i}}+k \\epsilon$. Therefore the points are grouped into $b_{i}$ clusters next to the points $0, \\frac{1}{b_{i}}, \\ldots, \\frac{b_{i}-1}{b_{i}}$ of the circle. The cluster following $\\frac{k}{b_{i}}$ contains the numbers congruent to $k a_{i}^{-1}$ modulo $b_{i}$, listed clockwise in increasing order. It follows that the first number after 0 in $A\\left(f_{i}+\\epsilon\\right)$ is $b_{i}$, and the first number after 0 which is less than $b_{i}$ is $a_{i}^{-1}\\left(\\bmod b_{i}\\right)$, which uniquely determines $a_{i}$. In this way we can recover $f_{i}$ from the cyclic arrangement. Note also that $A\\left(f_{i}+\\epsilon\\right)$ is not the trivial arrangement where we list $0,1, \\ldots, n$ in order clockwise. It follows that the $N+1$ cyclic arrangements $A(\\epsilon), A\\left(f_{1}+\\epsilon\\right), \\ldots, A\\left(f_{N}+\\epsilon\\right)$ are distinct.\n\nLet us record an observation which will be useful later:\n$$\n\\begin{equation*}\n\\text{ if } f_{i}<\\alpha<f_{i+1} \\text{ then } 0 \\text{ is immediately after } b_{i+1} \\text{ and before } b_{i} \\text{ in } A(\\alpha)\\text{. } \\tag{2}\n\\end{equation*}\n$$\nIndeed, we already observed that $b_{i}$ is the first number after 0 in $A\\left(f_{i}+\\epsilon\\right)=A(\\alpha)$. Similarly we see that $b_{i+1}$ is the last number before 0 in $A\\left(f_{i+1}-\\epsilon\\right)=A(\\alpha)$.\n\nFinally, we show that any beautiful arrangement of $[0, n]$ is cyclic by induction on $n$. For $n \\leqslant 2$ the result is clear. Now assume that all beautiful arrangements of $[0, n-1]$ are cyclic, and consider a beautiful arrangement $A$ of $[0, n]$. The subarrangement $A_{n-1}=A \\backslash\\{n\\}$ of $[0, n-1]$ obtained by deleting $n$ is cyclic; say $A_{n-1}=A_{n-1}(\\alpha)$.\n\nLet $\\alpha$ be between the consecutive fractions $\\frac{p_{1}}{q_{1}}<\\frac{p_{2}}{q_{2}}$ among the irreducible fractions of denominator at most $n-1$. There is at most one fraction $\\frac{i}{n}$ in $\\left(\\frac{p_{1}}{q_{1}}, \\frac{p_{2}}{q_{2}}\\right)$, since $\\frac{i}{n}<\\frac{i}{n-1} \\leqslant \\frac{i+1}{n}$ for $0<i \\leqslant n-1$.\n\nCase 1. There is no fraction with denominator $n$ between $\\frac{p_{1}}{q_{1}}$ and $\\frac{p_{2}}{q_{2}}$.\n\nIn this case the only cyclic arrangement extending $A_{n-1}(\\alpha)$ is $A_{n}(\\alpha)$. We know that $A$ and $A_{n}(\\alpha)$ can only differ in the position of $n$. Assume $n$ is immediately after $x$ and before $y$ in $A_{n}(\\alpha)$. Since the neighbors of 0 are $q_{1}$ and $q_{2}$ by (2), we have $x, y \\geqslant 1$.\n\n![](attached_image_4.png)\nFigure 4\n\nIn $A_{n}(\\alpha)$ the chord from $n-1$ to $x$ is parallel and adjacent to the chord from $n$ to $x-1$, so $n-1$ is between $x-1$ and $x$ in clockwise order, as shown in Figure 4. Similarly, $n-1$ is between $y$ and $y-1$. Therefore $x, y, x-1, n-1$, and $y-1$ occur in this order in $A_{n}(\\alpha)$ and hence in $A$ (possibly with $y=x-1$ or $x=y-1$).\n\nNow, $A$ may only differ from $A_{n}(\\alpha)$ in the location of $n$. In $A$, since the chord from $n-1$ to $x$ and the chord from $n$ to $x-1$ do not intersect, $n$ is between $x$ and $n-1$. Similarly, $n$ is between $n-1$ and $y$. Then $n$ must be between $x$ and $y$ and $A=A_{n}(\\alpha)$. Therefore $A$ is cyclic as desired.\n\nCase 2. There is exactly one $i$ with $\\frac{p_{1}}{q_{1}}<\\frac{i}{n}<\\frac{p_{2}}{q_{2}}$.\n\nIn this case there exist two cyclic arrangements $A_{n}\\left(\\alpha_{1}\\right)$ and $A_{n}\\left(\\alpha_{2}\\right)$ of the numbers $0, \\ldots, n$ extending $A_{n-1}(\\alpha)$, where $\\frac{p_{1}}{q_{1}}<\\alpha_{1}<\\frac{i}{n}$ and $\\frac{i}{n}<\\alpha_{2}<\\frac{p_{2}}{q_{2}}$. In $A_{n-1}(\\alpha), 0$ is the only number between $q_{2}$ and $q_{1}$ by (2). For the same reason, $n$ is between $q_{2}$ and 0 in $A_{n}\\left(\\alpha_{1}\\right)$, and between 0 and $q_{1}$ in $A_{n}\\left(\\alpha_{2}\\right)$.\n\nLetting $x=q_{2}$ and $y=q_{1}$, the argument of Case 1 tells us that $n$ must be between $x$ and $y$ in $A$. Therefore $A$ must equal $A_{n}\\left(\\alpha_{1}\\right)$ or $A_{n}\\left(\\alpha_{2}\\right)$, and therefore it is cyclic.\n\nThis concludes the proof that every beautiful arrangement is cyclic. It follows that there are exactly $N+1$ beautiful arrangements of $[0, n]$ as we wished to show.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31975,
    "subject": "Mathematics (Multi-modal)",
    "question": "Players $A$ and $B$ play a paintful game on the real line. Player $A$ has a pot of paint with four units of black ink. A quantity $p$ of this ink suffices to blacken a (closed) real interval of length $p$. In every round, player $A$ picks some positive integer $m$ and provides $1 / 2^{m}$ units of ink from the pot. Player $B$ then picks an integer $k$ and blackens the interval from $k / 2^{m}$ to $(k+1) / 2^{m}$ (some parts of this interval may have been blackened before). The goal of player $A$ is to reach a situation where the pot is empty and the interval $[0,1]$ is not completely blackened.\nDecide whether there exists a strategy for player $A$ to win in a finite number of moves.",
    "options": [],
    "answer": "No; player A has no such winning strategy.",
    "solution": "No. Such a strategy for player $A$ does not exist.\n\nWe will present a strategy for player $B$ that guarantees that the interval $[0,1]$ is completely blackened, once the paint pot has become empty.\n\nAt the beginning of round $r$, let $x_{r}$ denote the largest real number for which the interval between $0$ and $x_{r}$ has already been blackened; for completeness we define $x_{1}=0$. Let $m$ be the integer picked by player $A$ in this round; we define an integer $y_{r}$ by\n$$\n\\frac{y_{r}}{2^{m}} \\leqslant x_{r}<\\frac{y_{r}+1}{2^{m}}\n$$\nNote that $I_{0}^{r}=[y_{r} / 2^{m}, (y_{r}+1) / 2^{m}]$ is the leftmost interval that may be painted in round $r$ and that still contains some uncolored point.\n\nPlayer $B$ now looks at the next interval $I_{1}^{r}=[(y_{r}+1) / 2^{m}, (y_{r}+2) / 2^{m}]$. If $I_{1}^{r}$ still contains an uncolored point, then player $B$ blackens the interval $I_{1}^{r}$; otherwise he blackens the interval $I_{0}^{r}$. We make the convention that, at the beginning of the game, the interval $[1,2]$ is already blackened; thus, if $y_{r}+1=2^{m}$, then $B$ blackens $I_{0}^{r}$.\n\nOur aim is to estimate the amount of ink used after each round. Firstly, we will prove by induction that, if before $r$th round the segment $[0,1]$ is not completely colored, then, before this move,\n(i) the amount of ink used for the segment $[0, x_{r}]$ is at most $3 x_{r}$; and\n(ii) for every $m$, $B$ has blackened at most one interval of length $1 / 2^{m}$ to the right of $x_{r}$.\n\nObviously, these conditions are satisfied for $r=0$. Now assume that they were satisfied before the $r$th move, and consider the situation after this move; let $m$ be the number $A$ has picked at this move.\n\nIf $B$ has blackened the interval $I_{1}^{r}$ at this move, then $x_{r+1}=x_{r}$, and (i) holds by the induction hypothesis. Next, had $B$ blackened before the $r$th move any interval of length $1 / 2^{m}$ to the right of $x_{r}$, this interval would necessarily coincide with $I_{1}^{r}$. By our strategy, this cannot happen. So, condition (ii) also remains valid.\n\nAssume now that $B$ has blackened the interval $I_{0}^{r}$ at the $r$th move, but the interval $[0,1]$ still contains uncolored parts (which means that $I_{1}^{r}$ is contained in $[0,1]$). Then condition (ii) clearly remains true, and we need to check (i) only. In our case, the intervals $I_{0}^{r}$ and $I_{1}^{r}$ are completely colored after the $r$th move, so $x_{r+1}$ either reaches the right endpoint of $I_{1}$ or moves even further to the right. So, $x_{r+1}=x_{r}+\\alpha$ for some $\\alpha>1 / 2^{m}$.\n\nNext, any interval blackened by $B$ before the $r$th move which intersects $(x_{r}, x_{r+1})$ should be contained in $[x_{r}, x_{r+1}]$; by (ii), all such intervals have different lengths not exceeding $1 / 2^{m}$, so the total amount of ink used for them is less than $2 / 2^{m}$. Thus, the amount of ink used for the segment $[0, x_{r+1}]$ does not exceed the sum of $2 / 2^{m}$, $3 x_{r}$ (used for $[0, x_{r}]$), and $1 / 2^{m}$ used for the segment $I_{0}^{r}$. In total it gives at most $3(x_{r}+1 / 2^{m})<3(x_{r}+\\alpha)=3 x_{r+1}$. Thus condition (i) is also verified in this case. The claim is proved.\n\nFinally, we can perform the desired estimation. Consider any situation in the game, say after the $(r-1)$st move; assume that the segment $[0,1]$ is not completely black. By (ii), in the segment $[x_{r}, 1]$ player $B$ has colored several segments of different lengths; all these lengths are negative powers of $2$ not exceeding $1-x_{r}$; thus the total amount of ink used for this interval is at most $2(1-x_{r})$. Using (i), we obtain that the total amount of ink used is at most $3 x_{r}+2(1-x_{r})<3$. Thus the pot is not empty, and therefore $A$ never wins.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31976,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be the circumcircle of a triangle $A B C$. Denote by $M$ and $N$ the midpoints of the sides $A B$ and $A C$, respectively, and denote by $T$ the midpoint of the $\\operatorname{arc} B C$ of $\\omega$ not containing $A$. The circumcircles of the triangles $A M T$ and $A N T$ intersect the perpendicular bisectors of $A C$ and $A B$ at points $X$ and $Y$, respectively; assume that $X$ and $Y$ lie inside the triangle $A B C$. The lines $M N$ and $X Y$ intersect at $K$. Prove that $K A=K T$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $O$ be the center of $\\omega$, thus $O = M Y \\cap N X$. Let $\\ell$ be the perpendicular bisector of $A T$ (it also passes through $O$). Denote by $r$ the operation of reflection about $\\ell$. Since $A T$ is the angle bisector of $\\angle B A C$, the line $r(A B)$ is parallel to $A C$. Since $O M \\perp A B$ and $O N \\perp A C$, this means that the line $r(O M)$ is parallel to the line $O N$ and passes through $O$, so $r(O M) = O N$. Finally, the circumcircle $\\gamma$ of the triangle $A M T$ is symmetric about $\\ell$, so $r(\\gamma) = \\gamma$. Thus the point $M$ maps to the common point of $O N$ with the arc $A M T$ of $\\gamma$ - that is, $r(M) = X$.\n\nSimilarly, $r(N) = Y$. Thus, we get $r(M N) = X Y$, and the common point $K$ of $M N$ and $X Y$ lies on $\\ell$. This means exactly that $K A = K T$.\n\nLet $X^{\\prime}$ be the common point of the line $N X$ with the external bisector of $\\angle B A C$; notice that it lies outside the triangle $A B C$. Then we have $\\angle T A X^{\\prime} = 90^{\\circ}$ and $X^{\\prime} A = X^{\\prime} C$, so we get $\\angle X^{\\prime} A M = 90^{\\circ} + \\angle B A C / 2 = 180^{\\circ} - \\angle X^{\\prime} A C = 180^{\\circ} - \\angle X^{\\prime} C A = \\angle X^{\\prime} C L$. Thus the triangles $X^{\\prime} A M$ and $X^{\\prime} C L$ are congruent, and therefore\n$$\n\\angle M X^{\\prime} L = \\angle A X^{\\prime} C + (\\angle C X^{\\prime} L - \\angle A X^{\\prime} M) = \\angle A X^{\\prime} C = 180^{\\circ} - 2 \\angle X^{\\prime} A C = \\angle B A C = \\angle M A L .\n$$\nThis means that $X^{\\prime}$ lies on $\\gamma$.\nThus we have $\\angle T X N = \\angle T X X^{\\prime} = \\angle T A X^{\\prime} = 90^{\\circ}$, so $T X \\parallel A C$. Then $\\angle X T A = \\angle T A C = \\angle T A M$, so the cyclic quadrilateral $M A T X$ is an isosceles trapezoid. Similarly, $N A T Y$ is an isosceles trapezoid, so again the lines $M N$ and $X Y$ are the reflections of each other about the perpendicular bisector of $A T$. Thus $K$ belongs to this perpendicular bisector.\nLet $L$ be the second common point of the line $A C$ with the circumcircle $\\gamma$ of the triangle $A M T$. From the cyclic quadrilaterals $A B T C$ and $A M T L$ we get $\\angle B T C = 180^{\\circ} - \\angle B A C = \\angle M T L$, which implies $\\angle B T M = \\angle C T L$. Since $A T$ is an angle bisector in these quadrilaterals, we have $B T = T C$ and $M T = T L$. Thus the triangles $B T M$ and $C T L$ are congruent, so $C L = B M = A M$.\n\nLet $X^{\\prime}$ be the common point of the line $N X$ with the external bisector of $\\angle B A C$; notice that it lies outside the triangle $A B C$. Then we have $\\angle T A X^{\\prime} = 90^{\\circ}$ and $X^{\\prime} A = X^{\\prime} C$, so we get $\\angle X^{\\prime} A M = 90^{\\circ} + \\angle B A C / 2 = 180^{\\circ} - \\angle X^{\\prime} A C = 180^{\\circ} - \\angle X^{\\prime} C A = \\angle X^{\\prime} C L$. Thus the triangles $X^{\\prime} A M$ and $X^{\\prime} C L$ are congruent, and therefore\n\nSimilarly, $r(N) = Y$. Thus, we get $r(M N) = X Y$, and the common point $K$ of $M N$ and $X Y$ lies on $\\ell$. This means exactly that $K A = K T$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31977,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $A B C$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $A E D B$ (all vertices of the rhombus lie on different sides of $A E D B$). Let $\\varphi$ be the non-obtuse angle of the rhombus. Prove that $\\varphi \\leqslant \\max \\{\\angle B A C, \\angle A B C\\}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K, L, M$, and $N$ be the vertices of the rhombus lying on the sides $A E, E D, D B$, and $B A$, respectively. Denote by $d(X, Y Z)$ the distance from a point $X$ to a line $Y Z$. Since $D$ and $E$ are the feet of the bisectors, we have $d(D, A B)=d(D, A C), d(E, A B)=d(E, B C)$, and $d(D, B C)=d(E, A C)=0$, which implies\n$$\nd(D, A C)+d(D, B C)=d(D, A B) \\quad \\text{ and } \\quad d(E, A C)+d(E, B C)=d(E, A B) .\n$$\nSince $L$ lies on the segment $D E$ and the relation $d(X, A C)+d(X, B C)=d(X, A B)$ is linear in $X$ inside the triangle, these two relations imply\n$$\n\\begin{equation*}\nd(L, A C)+d(L, B C)=d(L, A B) \\tag{1}\n\\end{equation*}\n$$\nDenote the angles as in the figure below, and denote $a=K L$. Then we have $d(L, A C)=a \\sin \\mu$ and $d(L, B C)=a \\sin \\nu$. Since $K L M N$ is a parallelogram lying on one side of $A B$, we get\n$$\nd(L, A B)=d(L, A B)+d(N, A B)=d(K, A B)+d(M, A B)=a(\\sin \\delta+\\sin \\varepsilon)\n$$\nThus the condition (1) reads\n$$\n\\begin{equation*}\n\\sin \\mu+\\sin \\nu=\\sin \\delta+\\sin \\varepsilon . \\tag{2}\n\\end{equation*}\n$$\n![](attached_image_1.png)\nIf one of the angles $\\alpha$ and $\\beta$ is non-acute, then the desired inequality is trivial. So we assume that $\\alpha, \\beta<\\pi / 2$. It suffices to show then that $\\psi=\\angle N K L \\leqslant \\max \\{\\alpha, \\beta\\}$.\nAssume, to the contrary, that $\\psi>\\max \\{\\alpha, \\beta\\}$. Since $\\mu+\\psi=\\angle C K N=\\alpha+\\delta$, by our assumption we obtain $\\mu=(\\alpha-\\psi)+\\delta<\\delta$. Similarly, $\\nu<\\varepsilon$. Next, since $K N \\| M L$, we have $\\beta=\\delta+\\nu$, so $\\delta<\\beta<\\pi / 2$. Similarly, $\\varepsilon<\\pi / 2$. Finally, by $\\mu<\\delta<\\pi / 2$ and $\\nu<\\varepsilon<\\pi / 2$, we obtain\n$$\n\\sin \\mu<\\sin \\delta \\quad \\text{ and } \\quad \\sin \\nu<\\sin \\varepsilon\n$$\nThis contradicts (2).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31978,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle B > \\angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\\angle PBA = \\angle QBA = \\angle ACB$ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD = PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \\neq A$. Prove that $QB = QR$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $\\omega$ the circumcircle of the triangle $ABC$, and let $\\angle ACB = \\gamma$. Note that the condition $\\gamma < \\angle CBA$ implies $\\gamma < 90^\\circ$. Since $\\angle PBA = \\gamma$, the line $PB$ is tangent to $\\omega$, so $PA \\cdot PC = PB^2 = PD^2$. By $\\frac{PA}{PD} = \\frac{PD}{PC}$ the triangles $PAD$ and $PDC$ are similar, and $\\angle ADP = \\angle DCP$.\n\nNext, since $\\angle ABQ = \\angle ACB$, the triangles $ABC$ and $AQB$ are also similar. Then $\\angle AQB = \\angle ABC = \\angle ARC$, which means that the points $D, R, C$, and $Q$ are concyclic. Therefore $\\angle DRQ = \\angle DCQ = \\angle ADP$.\n\n![](attached_image_1.png)\nFigure 1\n\nNow from $\\angle ARB = \\angle ACB = \\gamma$ and $\\angle PDB = \\angle PBD = 2\\gamma$ we get\n$$\n\\angle QBR = \\angle ADB - \\angle ARB = \\angle ADP + \\angle PDB - \\angle ARB = \\angle DRQ + \\gamma = \\angle QRB,\n$$\nso the triangle $QRB$ is isosceles, which yields $QB = QR$.\nAgain, denote by $\\omega$ the circumcircle of the triangle $ABC$. Denote $\\angle ACB = \\gamma$. Since $\\angle PBA = \\gamma$, the line $PB$ is tangent to $\\omega$.\n\nLet $E$ be the second intersection point of $BQ$ with $\\omega$. If $V'$ is any point on the ray $CE$ beyond $E$, then $\\angle BEV' = 180^\\circ - \\angle BEC = 180^\\circ - \\angle BAC = \\angle PAB$; together with $\\angle ABQ = \\angle PBA$ this shows firstly, that the rays $BA$ and $CE$ intersect at some point $V$, and secondly that the triangle $VEB$ is similar to the triangle $PAB$. Thus we have $\\angle BVE = \\angle BPA$. Next, $\\angle AEV = \\angle BEV - \\gamma = \\angle PAB - \\angle ABQ = \\angle AQB$; so the triangles $PBQ$ and $VAE$ are also similar.\n\nLet $PH$ be an altitude in the isosceles triangle $PBD$; then $BH = HD$. Let $G$ be the intersection point of $PH$ and $AB$. By the symmetry with respect to $PH$, we have $\\angle BDG = \\angle DBG = \\gamma = \\angle BEA$; thus $DG \\parallel AE$ and hence $\\frac{BG}{GA} = \\frac{BD}{DE}$. Thus the points $G$ and $D$ correspond to each other in the similar triangles $PAB$ and $VEB$, so $\\angle DVB = \\angle GPB = 90^\\circ - \\angle PBQ = 90^\\circ - \\angle VAE$. Thus $VD \\perp AE$.\n\nLet $T$ be the common point of $VD$ and $AE$, and let $DS$ be an altitude in the triangle $BDR$. The points $S$ and $T$ are the feet of corresponding altitudes in the similar triangles $ADE$ and $BDR$, so $\\frac{BS}{SR} = \\frac{AT}{TE}$. On the other hand, the points $T$ and $H$ are feet of corresponding altitudes in the similar triangles $VAE$ and $PBQ$, so $\\frac{AT}{TE} = \\frac{BH}{HQ}$. Thus $\\frac{BS}{SR} = \\frac{AT}{TE} = \\frac{BH}{HQ}$, and the triangles $BHS$ and $BQR$ are similar.\n\nFinally, $SH$ is a median in the right-angled triangle $SBD$; so $BH = HS$, and hence $BQ = QR$.\n\n![](attached_image_2.png)\nFigure 2\nDenote by $\\omega$ and $O$ the circumcircle of the triangle $ABC$ and its center, respectively. From the condition $\\angle PBA = \\angle BCA$ we know that $BP$ is tangent to $\\omega$.\n\nLet $E$ be the second point of intersection of $\\omega$ and $BD$. Due to the isosceles triangle $BDP$, the tangent of $\\omega$ at $E$ is parallel to $DP$ and consequently it intersects $BP$ at some point $L$. Of course, $PD \\parallel LE$. Let $M$ be the midpoint of $BE$, and let $H$ be the midpoint of $BR$. Notice that $\\angle AEB = \\angle ACB = \\angle ABQ = \\angle ABE$, so $A$ lies on the perpendicular bisector of $BE$; thus the points $L, A, M$, and $O$ are collinear. Let $\\omega_1$ be the circle with diameter $BO$. Let $Q' = HO \\cap BE$; since $HO$ is the perpendicular bisector of $BR$, the statement of the problem is equivalent to $Q' = Q$.\n\nConsider the following sequence of projections (see Fig. 3).\n1. Project the line $BE$ to the line $LB$ through the center $A$. (This maps $Q$ to $P$.)\n2. Project the line $LB$ to $BE$ in parallel direction with $LE$. ($P \\mapsto D$.)\n3. Project the line $BE$ to the circle $\\omega$ through its point $A$. ($D \\mapsto R$.)\n4. Scale $\\omega$ by the ratio $\\frac{1}{2}$ from the point $B$ to the circle $\\omega_1$. ($R \\mapsto H$.)\n5. Project $\\omega_1$ to the line $BE$ through its point $O$. ($H \\mapsto Q'$.)\n\nWe prove that the composition of these transforms, which maps the line $BE$ to itself, is the identity. To achieve this, it suffices to show three fixed points. An obvious fixed point is $B$ which is fixed by all the transformations above. Another fixed point is $M$, its path being $M \\mapsto L \\mapsto E \\mapsto E \\mapsto M \\mapsto M$.\n\n![](attached_image_3.png)\nFigure 3\n![](attached_image_4.png)\nFigure 4\n\nIn order to show a third fixed point, draw a line parallel with $LE$ through $A$; let that line intersect $BE, LB$ and $\\omega$ at $X, Y$ and $Z \\neq A$, respectively (see Fig. 4). We show that $X$ is a fixed point. The images of $X$ at the first three transformations are $X \\mapsto Y \\mapsto X \\mapsto Z$. From $\\angle XBZ = \\angle EAZ = \\angle AEL = \\angle LBA = \\angle BZX$ we can see that the triangle $XBZ$ is isosceles. Let $U$ be the midpoint of $BZ$; then the last two transformations do $Z \\mapsto U \\mapsto X$, and the point $X$ is fixed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31979,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C D E F$ be a convex hexagon with $A B = D E$, $B C = E F$, $C D = F A$, and $\\angle A - \\angle D = \\angle C - \\angle F = \\angle E - \\angle B$. Prove that the diagonals $A D$, $B E$, and $C F$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In all three solutions, we denote $\\theta = \\angle A - \\angle D = \\angle C - \\angle F = \\angle E - \\angle B$ and assume without loss of generality that $\\theta \\geqslant 0$.\n\nSolution 1. Let $x = A B = D E$, $y = C D = F A$, $z = E F = B C$. Consider the points $P, Q$, and $R$ such that the quadrilaterals $C D E P$, $E F A Q$, and $A B C R$ are parallelograms. We compute\n$$\n\\begin{aligned}\n\\angle P E Q & = \\angle F E Q + \\angle D E P - \\angle E = (180^\\circ - \\angle F) + (180^\\circ - \\angle D) - \\angle E \\\\\n& = 360^\\circ - \\angle D - \\angle E - \\angle F = \\frac{1}{2}(\\angle A + \\angle B + \\angle C - \\angle D - \\angle E - \\angle F) = \\theta / 2\n\\end{aligned}\n$$\nSimilarly, $\\angle Q A R = \\angle R C P = \\theta / 2$.\n\n![](attached_image_1.png)\n\nIf $\\theta = 0$, since $\\triangle R C P$ is isosceles, $R = P$. Therefore $A B \\parallel R C = P C \\parallel E D$, so $A B D E$ is a parallelogram. Similarly, $B C E F$ and $C D F A$ are parallelograms. It follows that $A D$, $B E$ and $C F$ meet at their common midpoint.\n\nNow assume $\\theta > 0$. Since $\\triangle P E Q$, $\\triangle Q A R$, and $\\triangle R C P$ are isosceles and have the same angle at the apex, we have $\\triangle P E Q \\sim \\triangle Q A R \\sim \\triangle R C P$ with ratios of similarity $y : z : x$. Thus\n$$\n\\begin{equation*}\n\\triangle P Q R \\text{ is similar to the triangle with sidelengths } y, z, \\text{ and } x. \\tag{1}\n\\end{equation*}\n$$\nNext, notice that\n$$\n\\frac{R Q}{Q P} = \\frac{z}{y} = \\frac{R A}{A F}\n$$\nand, using directed angles between rays,\n$$\n\\begin{aligned}\n\\nsucceq(R Q, Q P) & = \\Varangle(R Q, Q E) + \\not(Q E, Q P) \\\\\n& = \\Varangle(R Q, Q E) + \\not(R A, R Q) = \\Varangle(R A, Q E) = \\Varangle(R A, A F) .\n\\end{aligned}\n$$\nThus $\\triangle P Q R \\sim \\triangle F A R$. Since $F A = y$ and $A R = z$, (1) then implies that $F R = x$. Similarly $F P = x$. Therefore $C R F P$ is a rhombus.\n\nWe conclude that $C F$ is the perpendicular bisector of $P R$. Similarly, $B E$ is the perpendicular bisector of $P Q$ and $A D$ is the perpendicular bisector of $Q R$. It follows that $A D$, $B E$, and $C F$ are concurrent at the circumcenter of $P Q R$.\n\n\nSolution 2. Let $X = C D \\cap E F$, $Y = E F \\cap A B$, $Z = A B \\cap C D$, $X' = F A \\cap B C$, $Y' = B C \\cap D E$, and $Z' = D E \\cap F A$. From $\\angle A + \\angle B + \\angle C = 360^\\circ + \\theta / 2$ we get $\\angle A + \\angle B > 180^\\circ$ and $\\angle B + \\angle C > 180^\\circ$, so $Z$ and $X'$ are respectively on the opposite sides of $B C$ and $A B$ from the hexagon. Similar conclusions hold for $X, Y, Y'$, and $Z'$. Then\n$$\n\\angle Y Z X = \\angle B + \\angle C - 180^\\circ = \\angle E + \\angle F - 180^\\circ = \\angle Y' Z' X'\n$$\nand similarly $\\angle Z X Y = \\angle Z' X' Y'$ and $\\angle X Y Z = \\angle X' Y' Z'$, so $\\triangle X Y Z \\sim \\triangle X' Y' Z'$. Thus there is a rotation $R$ which sends $\\triangle X Y Z$ to a triangle with sides parallel to $\\triangle X' Y' Z'$. Since $A B = D E$ we have $R(\\overrightarrow{A B}) = \\overrightarrow{D E}$. Similarly, $R(\\overrightarrow{C D}) = \\overrightarrow{F A}$ and $R(\\overrightarrow{E F}) = \\overrightarrow{B C}$. Therefore\n$$\n\\overrightarrow{0} = \\overrightarrow{A B} + \\overrightarrow{B C} + \\overrightarrow{C D} + \\overrightarrow{D E} + \\overrightarrow{E F} + \\overrightarrow{F A} = (\\overrightarrow{A B} + \\overrightarrow{C D} + \\overrightarrow{E F}) + R(\\overrightarrow{A B} + \\overrightarrow{C D} + \\overrightarrow{E F})\n$$\nIf $R$ is a rotation by $180^\\circ$, then any two opposite sides of our hexagon are equal and parallel, so the three diagonals meet at their common midpoint. Otherwise, we must have\n$$\n\\overrightarrow{A B} + \\overrightarrow{C D} + \\overrightarrow{E F} = \\overrightarrow{0}\n$$\nor else we would have two vectors with different directions whose sum is $\\overrightarrow{0}$.\n\n![](attached_image_2.png)\n\nThis allows us to consider a triangle $L M N$ with $\\overrightarrow{L M} = \\overrightarrow{E F}$, $\\overrightarrow{M N} = \\overrightarrow{A B}$, and $\\overrightarrow{N L} = \\overrightarrow{C D}$. Let $O$ be the circumcenter of $\\triangle L M N$ and consider the points $O_1, O_2, O_3$ such that $\\triangle A O_1 B$, $\\triangle C O_2 D$, and $\\triangle E O_3 F$ are translations of $\\triangle M O N$, $\\triangle N O L$, and $\\triangle L O M$, respectively. Since $F O_3$ and $A O_1$ are translations of $M O$, quadrilateral $A F O_3 O_1$ is a parallelogram and $O_3 O_1 = F A = C D = N L$. Similarly, $O_1 O_2 = L M$ and $O_2 O_3 = M N$. Therefore $\\triangle O_1 O_2 O_3 \\cong \\triangle L M N$. Moreover, by means of the rotation $R$ one may check that these triangles have the same orientation.\n\nLet $T$ be the circumcenter of $\\triangle O_1 O_2 O_3$. We claim that $A D$, $B E$, and $C F$ meet at $T$. Let us show that $C$, $T$, and $F$ are collinear. Notice that $C O_2 = O_2 T = T O_3 = O_3 F$ since they are all equal to the circumradius of $\\triangle L M N$. Therefore $\\triangle T O_3 F$ and $\\triangle C O_2 T$ are isosceles. Using directed angles between rays again, we get\n$$\n\\begin{equation*}\n\\nsucceq\\left(T F, T O_3\\right) = \\Varangle\\left(F O_3, F T\\right) \\quad \\text{ and } \\quad \\not \\quad\\left(T O_2, T C\\right) = \\Varangle\\left(C T, C O_2\\right) . \\tag{2}\n\\end{equation*}\n$$\nAlso, $T$ and $O$ are the circumcenters of the congruent triangles $\\triangle O_1 O_2 O_3$ and $\\triangle L M N$ so we have $\\Varangle\\left(T O_3, T O_2\\right) = \\Varangle(O N, O M)$. Since $C O_2$ and $F O_3$ are translations of $N O$ and $M O$ respectively, this implies\n$$\n\\begin{equation*}\n\\nsucceq\\left(T O_3, T O_2\\right) = \\Varangle\\left(C O_2, F O_3\\right) . \\tag{3}\n\\end{equation*}\n$$\nAdding the three equations in (2) and (3) gives\n$$\n\\nsucceq(T F, T C) = \\Varangle(C T, F T) = -\\Varangle(T F, T C)\n$$\nwhich implies that $T$ is on $C F$. Analogous arguments show that it is on $A D$ and $B E$ also. The desired result follows.\n\n\nSolution 3. Place the hexagon on the complex plane, with $A$ at the origin and vertices labelled clockwise. Now $A, B, C, D, E, F$ represent the corresponding complex numbers. Also consider the complex numbers $a, b, c, a', b', c'$ given by $B - A = a$, $D - C = b$, $F - E = c$, $E - D = a'$, $A - F = b'$, and $C - B = c'$. Let $k = |a| / |b|$. From $a / b' = -k e^{i \\angle A}$ and $a' / b = -k e^{i \\angle D}$ we get that $\\left(a' / a\\right)\\left(b' / b\\right) = e^{-i \\theta}$ and similarly $\\left(b' / b\\right)\\left(c' / c\\right) = e^{-i \\theta}$ and $\\left(c' / c\\right)\\left(a' / a\\right) = e^{-i \\theta}$. It follows that $a' = a r$, $b' = b r$, and $c' = c r$ for a complex number $r$ with $|r| = 1$, as shown below.\n\n![](attached_image_3.png)\n\nWe have\n$$\n0 = a + c r + b + a r + c + b r = (a + b + c)(1 + r) .\n$$\nIf $r = -1$, then the hexagon is centrally symmetric and its diagonals intersect at its center of symmetry. Otherwise\n$$\na + b + c = 0 .\n$$\nTherefore\n$$\nA = 0, \\quad B = a, \\quad C = a + c r, \\quad D = c(r - 1), \\quad E = -b r - c, \\quad F = -b r .\n$$\nNow consider a point $W$ on $A D$ given by the complex number $c(r - 1) \\lambda$, where $\\lambda$ is a real number with $0 < \\lambda < 1$. Since $D \\neq A$, we have $r \\neq 1$, so we can define $s = 1 / (r - 1)$. From $r \\bar{r} = |r|^2 = 1$ we get\n$$\n1 + s = \\frac{r}{r - 1} = \\frac{r}{r - r \\bar{r}} = \\frac{1}{1 - \\bar{r}} = -\\bar{s} .\n$$\nNow,\n$$\n\\begin{aligned}\nW \\text{ is on } B E & \\Longleftrightarrow c(r - 1) \\lambda - a \\| a - (-b r - c) = b(r - 1) \\Longleftrightarrow c \\lambda - a s \\| b \\\\\n& \\Longleftrightarrow -a \\lambda - b \\lambda - a s \\| b \\Longleftrightarrow a(\\lambda + s) \\| b .\n\\end{aligned}\n$$\nOne easily checks that $r \\neq \\pm 1$ implies that $\\lambda + s \\neq 0$ since $s$ is not real. On the other hand,\n$$\n\\begin{aligned}\nW \\text{ on } C F & \\Longleftrightarrow c(r - 1) \\lambda + b r \\| -b r - (a + c r) = a(r - 1) \\Longleftrightarrow c \\lambda + b(1 + s) \\| a \\\\\n& \\Longleftrightarrow -a \\lambda - b \\lambda - b \\bar{s} \\| a \\Longleftrightarrow b(\\lambda + \\bar{s}) \\| a \\Longleftrightarrow b \\| a(\\lambda + s),\n\\end{aligned}\n$$\nwhere in the last step we use that $(\\lambda + s)(\\lambda + \\bar{s}) = |\\lambda + s|^2 \\in \\mathbb{R}_{>0}$. We conclude that $A D \\cap B E = C F \\cap B E$, and the desired result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31980,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. Find all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that\n$$\nm^{2}+f(n) \\mid m f(m)+n\n$$\nfor all positive integers $m$ and $n$.",
    "options": [],
    "answer": "f(n) = n for all positive integers n",
    "solution": "Answer. $f(n)=n$.\n\nSolution 1. Setting $m=n=2$ tells us that $4+f(2) \\mid 2 f(2)+2$. Since $2 f(2)+2<2(4+f(2))$, we must have $2 f(2)+2=4+f(2)$, so $f(2)=2$. Plugging in $m=2$ then tells us that $4+f(n) \\mid 4+n$, which implies that $f(n) \\leqslant n$ for all $n$.\nSetting $m=n$ gives $n^{2}+f(n) \\mid n f(n)+n$, so $n f(n)+n \\geqslant n^{2}+f(n)$ which we rewrite as $(n-1)(f(n)-n) \\geqslant 0$. Therefore $f(n) \\geqslant n$ for all $n \\geqslant 2$. This is trivially true for $n=1$ also.\nIt follows that $f(n)=n$ for all $n$. This function obviously satisfies the desired property.\n\n\nSolution 2. Setting $m=f(n)$ we get $f(n)(f(n)+1) \\mid f(n) f(f(n))+n$. This implies that $f(n) \\mid n$ for all $n$.\nNow let $m$ be any positive integer, and let $p>2 m^{2}$ be a prime number. Note that $p>m f(m)$ also. Plugging in $n=p-m f(m)$ we learn that $m^{2}+f(n)$ divides $p$. Since $m^{2}+f(n)$ cannot equal 1, it must equal $p$. Therefore $p-m^{2}=f(n) \\mid n=p-m f(m)$. But $p-m f(m)<p<2\\left(p-m^{2}\\right)$, so we must have $p-m f(m)=p-m^{2}$, i.e., $f(m)=m$.\n\n\nSolution 3. Plugging $m=1$ we obtain $1+f(n) \\leqslant f(1)+n$, so $f(n) \\leqslant n+c$ for the constant $c= f(1)-1$. Assume that $f(n) \\neq n$ for some fixed $n$. When $m$ is large enough (e.g. $m \\geqslant \\max (n, c+1)$ ) we have\n$$\nm f(m)+n \\leqslant m(m+c)+n \\leqslant 2 m^{2}<2\\left(m^{2}+f(n)\\right),\n$$\nso we must have $m f(m)+n=m^{2}+f(n)$. This implies that\n$$\n0 \\neq f(n)-n=m(f(m)-m),\n$$\nwhich is impossible for $m>|f(n)-n|$. It follows that $f$ is the identity function.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31981,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine whether there exists an infinite sequence of nonzero digits $a_{1}, a_{2}, a_{3}, \\ldots$ and a positive integer $N$ such that for every integer $k > N$, the number $\\overline{a_{k} a_{k-1} \\ldots a_{1}}$ is a perfect square.",
    "options": [],
    "answer": "No",
    "solution": "Answer. No.\n\nAssume that $a_{1}, a_{2}, a_{3}, \\ldots$ is such a sequence. For each positive integer $k$, let $y_{k}= \\overline{a_{k} a_{k-1} \\ldots a_{1}}$. By the assumption, for each $k > N$ there exists a positive integer $x_{k}$ such that $y_{k}=x_{k}^{2}$.\n\nI. For every $n$, let $5^{\\gamma_{n}}$ be the greatest power of $5$ dividing $x_{n}$. Let us show first that $2 \\gamma_{n} \\geqslant n$ for every positive integer $n > N$.\n\nAssume, to the contrary, that there exists a positive integer $n > N$ such that $2 \\gamma_{n} < n$, which yields\n$$\ny_{n+1}=\\overline{a_{n+1} a_{n} \\ldots a_{1}}=10^{n} a_{n+1}+\\overline{a_{n} a_{n-1} \\ldots a_{1}}=10^{n} a_{n+1}+y_{n}=5^{2 \\gamma_{n}}\\left(2^{n} 5^{n-2 \\gamma_{n}} a_{n+1}+\\frac{y_{n}}{5^{2 \\gamma_{n}}}\\right) .\n$$\nSince $5 \\nmid y_{n} / 5^{2 \\gamma_{n}}$, we obtain $\\gamma_{n+1}=\\gamma_{n}<n<n+1$. By the same arguments we obtain that $\\gamma_{n}=\\gamma_{n+1}=\\gamma_{n+2}=\\ldots$. Denote this common value by $\\gamma$.\n\nNow, for each $k \\geqslant n$ we have\n$$\n\\left(x_{k+1}-x_{k}\\right)\\left(x_{k+1}+x_{k}\\right)=x_{k+1}^{2}-x_{k}^{2}=y_{k+1}-y_{k}=a_{k+1} \\cdot 10^{k}\n$$\nOne of the numbers $x_{k+1}-x_{k}$ and $x_{k+1}+x_{k}$ is not divisible by $5^{\\gamma+1}$ since otherwise one would have $5^{\\gamma+1} \\mid \\left(\\left(x_{k+1}-x_{k}\\right)+\\left(x_{k+1}+x_{k}\\right)\\right)=2 x_{k+1}$. On the other hand, we have $5^{k} \\mid \\left(x_{k+1}-x_{k}\\right)\\left(x_{k+1}+x_{k}\\right)$, so $5^{k-\\gamma}$ divides one of these two factors. Thus we get\n$$\n5^{k-\\gamma} \\leqslant \\max \\left\\{x_{k+1}-x_{k}, x_{k+1}+x_{k}\\right\\}<2 x_{k+1}=2 \\sqrt{y_{k+1}}<2 \\cdot 10^{(k+1) / 2}\n$$\nwhich implies $5^{2 k}<4 \\cdot 5^{2 \\gamma} \\cdot 10^{k+1}$, or $(5 / 2)^{k}<40 \\cdot 5^{2 \\gamma}$. The last inequality is clearly false for sufficiently large values of $k$. This contradiction shows that $2 \\gamma_{n} \\geqslant n$ for all $n > N$.\n\nII. Consider now any integer $k > \\max \\{N / 2, 2\\}$. Since $2 \\gamma_{2 k+1} \\geqslant 2 k+1$ and $2 \\gamma_{2 k+2} \\geqslant 2 k+2$, we have $\\gamma_{2 k+1} \\geqslant k+1$ and $\\gamma_{2 k+2} \\geqslant k+1$. So, from $y_{2 k+2}=a_{2 k+2} \\cdot 10^{2 k+1}+y_{2 k+1}$ we obtain $5^{2 k+2} \\mid y_{2 k+2}-y_{2 k+1}=a_{2 k+2} \\cdot 10^{2 k+1}$ and thus $5 \\mid a_{2 k+2}$, which implies $a_{2 k+2}=5$. Therefore,\n$$\n\\left(x_{2 k+2}-x_{2 k+1}\\right)\\left(x_{2 k+2}+x_{2 k+1}\\right)=x_{2 k+2}^{2}-x_{2 k+1}^{2}=y_{2 k+2}-y_{2 k+1}=5 \\cdot 10^{2 k+1}=2^{2 k+1} \\cdot 5^{2 k+2} .\n$$\nSetting $A_{k}=x_{2 k+2} / 5^{k+1}$ and $B_{k}=x_{2 k+1} / 5^{k+1}$, which are integers, we obtain\n$$\n\\begin{equation*}\n\\left(A_{k}-B_{k}\\right)\\left(A_{k}+B_{k}\\right)=2^{2 k+1} \\tag{1}\n\\end{equation*}\n$$\nBoth $A_{k}$ and $B_{k}$ are odd, since otherwise $y_{2 k+2}$ or $y_{2 k+1}$ would be a multiple of $10$ which is false by $a_{1} \\neq 0$; so one of the numbers $A_{k}-B_{k}$ and $A_{k}+B_{k}$ is not divisible by $4$. Therefore (1) yields $A_{k}-B_{k}=2$ and $A_{k}+B_{k}=2^{2 k}$, hence $A_{k}=2^{2 k-1}+1$ and thus\n$$\nx_{2 k+2}=5^{k+1} A_{k}=10^{k+1} \\cdot 2^{k-2}+5^{k+1}>10^{k+1}\n$$\nsince $k \\geqslant 2$. This implies that $y_{2 k+2}>10^{2 k+2}$ which contradicts the fact that $y_{2 k+2}$ contains $2 k+2$ digits. The desired result follows.\nAgain, we assume that a sequence $a_{1}, a_{2}, a_{3}, \\ldots$ satisfies the problem conditions, introduce the numbers $x_{k}$ and $y_{k}$ as in the previous solution, and notice that\n$$\n\\begin{equation*}\ny_{k+1}-y_{k}=\\left(x_{k+1}-x_{k}\\right)\\left(x_{k+1}+x_{k}\\right)=10^{k} a_{k+1} \\tag{2}\n\\end{equation*}\n$$\nfor all $k > N$. Consider any such $k$. Since $a_{1} \\neq 0$, the numbers $x_{k}$ and $x_{k+1}$ are not multiples of $10$, and therefore the numbers $p_{k}=x_{k+1}-x_{k}$ and $q_{k}=x_{k+1}+x_{k}$ cannot be simultaneously multiples of $20$, and hence one of them is not divisible either by $4$ or by $5$. In view of (2), this means that the other one is divisible by either $5^{k}$ or by $2^{k-1}$. Notice also that $p_{k}$ and $q_{k}$ have the same parity, so both are even.\n\nOn the other hand, we have $x_{k+1}^{2}=x_{k}^{2}+10^{k} a_{k+1} \\geqslant x_{k}^{2}+10^{k}>2 x_{k}^{2}$, so $x_{k+1} / x_{k}>\\sqrt{2}$, which implies that\n$$\n\\begin{equation*}\n1<\\frac{q_{k}}{p_{k}}=1+\\frac{2}{x_{k+1} / x_{k}-1}<1+\\frac{2}{\\sqrt{2}-1}<6 \\tag{3}\n\\end{equation*}\n$$\nThus, if one of the numbers $p_{k}$ and $q_{k}$ is divisible by $5^{k}$, then we have\n$$\n10^{k+1}>10^{k} a_{k+1}=p_{k} q_{k} \\geqslant \\frac{\\left(5^{k}\\right)^{2}}{6}\n$$\nand hence $(5 / 2)^{k}<60$ which is false for sufficiently large $k$. So, assuming that $k$ is large, we get that $2^{k-1}$ divides one of the numbers $p_{k}$ and $q_{k}$. Hence\n$$\n\\left\\{p_{k}, q_{k}\\right\\}=\\left\\{2^{k-1} \\cdot 5^{r_{k}} b_{k}, 2 \\cdot 5^{k-r_{k}} c_{k}\\right\\} \\quad \\text{ with nonnegative integers } b_{k}, c_{k}, r_{k} \\text{ such that } b_{k} c_{k}=a_{k+1} .\n$$\nMoreover, from (3) we get\n$$\n6>\\frac{2^{k-1} \\cdot 5^{r_{k}} b_{k}}{2 \\cdot 5^{k-r_{k}} c_{k}} \\geqslant \\frac{1}{36} \\cdot\\left(\\frac{2}{5}\\right)^{k} \\cdot 5^{2 r_{k}} \\quad \\text{ and } \\quad 6>\\frac{2 \\cdot 5^{k-r_{k}} c_{k}}{2^{k-1} \\cdot 5^{r_{k}} b_{k}} \\geqslant \\frac{4}{9} \\cdot\\left(\\frac{5}{2}\\right)^{k} \\cdot 5^{-2 r_{k}}\n$$\nso\n$$\n\\begin{equation*}\n\\alpha k+c_{1}<r_{k}<\\alpha k+c_{2} \\quad \\text{ for } \\alpha=\\frac{1}{2} \\log _{5}\\left(\\frac{5}{2}\\right)<1 \\text{ and some constants } c_{2}>c_{1} . \\tag{4}\n\\end{equation*}\n$$\nConsequently, for $C=c_{2}-c_{1}+1-\\alpha>0$ we have\n$$\n\\begin{equation*}\n(k+1)-r_{k+1} \\leqslant k-r_{k}+C \\tag{5}\n\\end{equation*}\n$$\nNext, we will use the following easy lemma.\n\nLemma. Let $s$ be a positive integer. Then $5^{s+2^{s}} \\equiv 5^{s}\\left(\\bmod 10^{s}\\right)$.\n\nProof. Euler's theorem gives $5^{2^{s}} \\equiv 1\\left(\\bmod 2^{s}\\right)$, so $5^{s+2^{s}}-5^{s}=5^{s}\\left(5^{2^{s}}-1\\right)$ is divisible by $2^{s}$ and $5^{s}$.\n\nNow, for every large $k$ we have\n$$\n\\begin{equation*}\nx_{k+1}=\\frac{p_{k}+q_{k}}{2}=5^{r_{k}} \\cdot 2^{k-2} b_{k}+5^{k-r_{k}} c_{k} \\equiv 5^{k-r_{k}} c_{k} \\quad\\left(\\bmod 10^{r_{k}}\\right) \\tag{6}\n\\end{equation*}\n$$\nsince $r_{k} \\leqslant k-2$ by (4); hence $y_{k+1} \\equiv 5^{2\\left(k-r_{k}\\right)} c_{k}^{2}\\left(\\bmod 10^{r_{k}}\\right)$. Let us consider some large integer $s$, and choose the minimal $k$ such that $2\\left(k-r_{k}\\right) \\geqslant s+2^{s}$; it exists by (4). Set $d=2\\left(k-r_{k}\\right)-\\left(s+2^{s}\\right)$. By (4) we have $2^{s}<2\\left(k-r_{k}\\right)<\\left(\\frac{2}{\\alpha}-2\\right) r_{k}-\\frac{2 c_{1}}{\\alpha}$; if $s$ is large this implies $r_{k}>s$, so (6) also holds modulo $10^{s}$. Then (6) and the lemma give\n$$\n\\begin{equation*}\ny_{k+1} \\equiv 5^{2\\left(k-r_{k}\\right)} c_{k}^{2}=5^{s+2^{s}} \\cdot 5^{d} c_{k}^{2} \\equiv 5^{s} \\cdot 5^{d} c_{k}^{2} \\quad\\left(\\bmod 10^{s}\\right) \\tag{7}\n\\end{equation*}\n$$\nBy (5) and the minimality of $k$ we have $d \\leqslant 2 C$, so $5^{d} c_{k}^{2} \\leqslant 5^{2 C} \\cdot 81=D$. Using $5^{4}<10^{3}$ we obtain\n$$\n5^{s} \\cdot 5^{d} c_{k}^{2}<10^{3 s / 4} D<10^{s-1}\n$$\nfor sufficiently large $s$. This, together with (7), shows that the $s$\\text{th} digit from the right in $y_{k+1}$, which is $a_{s}$, is zero. This contradicts the problem condition.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31982,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{Q} \\longrightarrow \\mathbb{Z}$ satisfying\n$$\nf\\left(\\frac{f(x)+a}{b}\\right)=f\\left(\\frac{x+a}{b}\\right)\n$$\nfor all $x \\in \\mathbb{Q}$, $a \\in \\mathbb{Z}$, and $b \\in \\mathbb{Z}_{>0}$. (Here, $\\mathbb{Z}_{>0}$ denotes the set of positive integers.)",
    "options": [],
    "answer": "All constant functions, the floor function, and the ceiling function.",
    "solution": "Answer. There are three kinds of such functions, which are: all constant functions, the floor function, and the ceiling function.\n\nSolution 1.\nI. We start by verifying that these functions do indeed satisfy (1). This is clear for all constant functions. Now consider any triple $(x, a, b) \\in \\mathbb{Q} \\times \\mathbb{Z} \\times \\mathbb{Z}_{>0}$ and set\n$$\nq=\\left\\lfloor\\frac{x+a}{b}\\right\\rfloor\n$$\nThis means that $q$ is an integer and $b q \\leqslant x+a<b(q+1)$. It follows that $b q \\leqslant\\lfloor x\\rfloor+a<b(q+1)$ holds as well, and thus we have\n$$\n\\left\\lfloor\\frac{\\lfloor x\\rfloor+a}{b}\\right\\rfloor=\\left\\lfloor\\frac{x+a}{b}\\right\\rfloor\n$$\nmeaning that the floor function does indeed satisfy (1). One can check similarly that the ceiling function has the same property.\n\nII. Let us now suppose conversely that the function $f: \\mathbb{Q} \\longrightarrow \\mathbb{Z}$ satisfies (1) for all $(x, a, b) \\in \\mathbb{Q} \\times \\mathbb{Z} \\times \\mathbb{Z}_{>0}$. According to the behaviour of the restriction of $f$ to the integers we distinguish two cases.\n\nCase 1: There is some $m \\in \\mathbb{Z}$ such that $f(m) \\neq m$.\nWrite $f(m)=C$ and let $\\eta \\in\\{-1,+1\\}$ and $b$ denote the sign and absolute value of $f(m)-m$, respectively. Given any integer $r$, we may plug the triple ( $m, r b-C, b$ ) into ( 1 ), thus getting $f(r)=f(r-\\eta)$. Starting with $m$ and using induction in both directions, we deduce from this that the equation $f(r)=C$ holds for all integers $r$. Now any rational number $y$ can be written in the form $y=\\frac{p}{q}$ with $(p, q) \\in \\mathbb{Z} \\times \\mathbb{Z}_{>0}$, and substituting $(C-p, p-C, q)$ into (1) we get $f(y)=f(0)=C$. Thus $f$ is the constant function whose value is always $C$.\n\nCase 2: One has $f(m)=m$ for all integers $m$.\nNote that now the special case $b=1$ of (1) takes a particularly simple form, namely\n$$\n\\begin{equation*}\nf(x)+a=f(x+a) \\quad \\text{ for all }(x, a) \\in \\mathbb{Q} \\times \\mathbb{Z} \\tag{2}\n\\end{equation*}\n$$\nDefining $f\\left(\\frac{1}{2}\\right)=\\omega$ we proceed in three steps.\n\nStep A. We show that $\\omega \\in\\{0,1\\}$.\nIf $\\omega \\leqslant 0$, we may plug $\\left(\\frac{1}{2},-\\omega, 1-2 \\omega\\right)$ into (1), obtaining $0=f(0)=f\\left(\\frac{1}{2}\\right)=\\omega$. In the contrary case $\\omega \\geqslant 1$ we argue similarly using the triple $\\left(\\frac{1}{2}, \\omega-1,2 \\omega-1\\right)$.\n\nStep B. We show that $f(x)=\\omega$ for all rational numbers $x$ with $0<x<1$.\nAssume that this fails and pick some rational number $\\frac{a}{b} \\in(0,1)$ with minimal $b$ such that $f\\left(\\frac{a}{b}\\right) \\neq \\omega$. Obviously, $\\operatorname{gcd}(a, b)=1$ and $b \\geqslant 2$. If $b$ is even, then $a$ has to be odd and we can substitute $\\left(\\frac{1}{2}, \\frac{a-1}{2}, \\frac{b}{2}\\right)$ into (1), which yields\n$$\n\\begin{equation*}\nf\\left(\\frac{\\omega+(a-1) / 2}{b / 2}\\right)=f\\left(\\frac{a}{b}\\right) \\neq \\omega \\tag{3}\n\\end{equation*}\n$$\nRecall that $0 \\leqslant(a-1) / 2<b / 2$. Thus, in both cases $\\omega=0$ and $\\omega=1$, the left-hand part of (3) equals $\\omega$ either by the minimality of $b$, or by $f(\\omega)=\\omega$. A contradiction.\n\nThus $b$ has to be odd, so $b=2 k+1$ for some $k \\geqslant 1$. Applying (1) to $\\left(\\frac{1}{2}, k, b\\right)$ we get\n$$\n\\begin{equation*}\nf\\left(\\frac{\\omega+k}{b}\\right)=f\\left(\\frac{1}{2}\\right)=\\omega \\tag{4}\n\\end{equation*}\n$$\nSince $a$ and $b$ are coprime, there exist integers $r \\in\\{1,2, \\ldots, b\\}$ and $m$ such that $r a-m b=k+\\omega$. Note that we actually have $1 \\leqslant r<b$, since the right hand side is not a multiple of $b$. If $m$ is negative, then we have $r a-m b>b \\geqslant k+\\omega$, which is absurd. Similarly, $m \\geqslant r$ leads to $r a-m b<b r-b r=0$, which is likewise impossible; so we must have $0 \\leqslant m \\leqslant r-1$.\n\nWe finally substitute $\\left(\\frac{k+\\omega}{b}, m, r\\right)$ into (1) and use (4) to learn\n$$\nf\\left(\\frac{\\omega+m}{r}\\right)=f\\left(\\frac{a}{b}\\right) \\neq \\omega\n$$\nBut as above one may see that the left hand side has to equal $\\omega$ due to the minimality of $b$. This contradiction concludes our step B.\n\nStep $C$. Now notice that if $\\omega=0$, then $f(x)=\\lfloor x\\rfloor$ holds for all rational $x$ with $0 \\leqslant x<1$ and hence by (2) this even holds for all rational numbers $x$. Similarly, if $\\omega=1$, then $f(x)=\\lceil x\\rceil$ holds for all $x \\in \\mathbb{Q}$. Thereby the problem is solved.\n\nComment 1. An alternative treatment of Steps B and C from the second case, due to the proposer, proceeds as follows. Let square brackets indicate the floor function in case $\\omega=0$ and the ceiling function if $\\omega=1$. We are to prove that $f(x)=[x]$ holds for all $x \\in \\mathbb{Q}$, and because of Step A and (2) we already know this in case $2 x \\in \\mathbb{Z}$. Applying (1) to ( $2 x, 0,2$ ) we get\n$$\nf(x)=f\\left(\\frac{f(2 x)}{2}\\right)\n$$\nand by the previous observation this yields\n$$\n\\begin{equation*}\nf(x)=\\left[\\frac{f(2 x)}{2}\\right] \\quad \\text{ for all } x \\in \\mathbb{Q} \\tag{5}\n\\end{equation*}\n$$\nAn easy induction now shows\n$$\n\\begin{equation*}\nf(x)=\\left[\\frac{f\\left(2^{n} x\\right)}{2^{n}}\\right] \\quad \\text{ for all }(x, n) \\in \\mathbb{Q} \\times \\mathbb{Z}_{>0} \\tag{6}\n\\end{equation*}\n$$\nNow suppose first that $x$ is not an integer but can be written in the form $\\frac{p}{q}$ with $p \\in \\mathbb{Z}$ and $q \\in \\mathbb{Z}_{>0}$ both being odd. Let $d$ denote the multiplicative order of 2 modulo $q$ and let $m$ be any large integer. Plugging $n=d m$ into (6) and using (2) we get\n$$\nf(x)=\\left[\\frac{f\\left(2^{d m} x\\right)}{2^{d m}}\\right]=\\left[\\frac{f(x)+\\left(2^{d m}-1\\right) x}{2^{d m}}\\right]=\\left[x+\\frac{f(x)-x}{2^{d m}}\\right]\n$$\nSince $x$ is not an integer, the square bracket function is continuous at $x$; hence as $m$ tends to infinity the above fomula gives $f(x)=[x]$. To complete the argument we just need to observe that if some $y \\in \\mathbb{Q}$ satisfies $f(y)=[y]$, then (5) yields $f\\left(\\frac{y}{2}\\right)=f\\left(\\frac{[y]}{2}\\right)=\\left[\\frac{[y]}{2}\\right]=\\left[\\frac{y}{2}\\right]$.\nHere we just give another argument for the second case of the above solution. Again we use equation (2). It follows that the set $S$ of all zeros of $f$ contains for each $x \\in \\mathbb{Q}$ exactly one term from the infinite sequence $\\ldots, x-2, x-1, x, x+1, x+2, \\ldots$.\n\nNext we claim that\n$$\n\\begin{equation*}\n\\text{ if }(p, q) \\in \\mathbb{Z} \\times \\mathbb{Z}_{>0} \\text{ and } \\frac{p}{q} \\in S, \\text{ then } \\frac{p}{q+1} \\in S \\text{ holds as well. } \\tag{7}\n\\end{equation*}\n$$\nTo see this we just plug $\\left(\\frac{p}{q}, p, q+1\\right)$ into (1), thus getting $f\\left(\\frac{p}{q+1}\\right)=f\\left(\\frac{p}{q}\\right)=0$.\n\nFrom this we get that\n$$\n\\begin{equation*}\n\\text{ if } x, y \\in \\mathbb{Q}, x>y>0, \\text{ and } x \\in S, \\text{ then } y \\in S . \\tag{8}\n\\end{equation*}\n$$\nIndeed, if we write $x=\\frac{p}{q}$ and $y=\\frac{r}{s}$ with $p, q, r, s \\in \\mathbb{Z}_{>0}$, then $p s>q r$ and (7) tells us\n$$\n0=f\\left(\\frac{p}{q}\\right)=f\\left(\\frac{p r}{q r}\\right)=f\\left(\\frac{p r}{q r+1}\\right)=\\ldots=f\\left(\\frac{p r}{p s}\\right)=f\\left(\\frac{r}{s}\\right) .\n$$\nEssentially the same argument also establishes that\n$$\n\\begin{equation*}\n\\text{ if } x, y \\in \\mathbb{Q}, x<y<0, \\text{ and } x \\in S, \\text{ then } y \\in S . \\tag{9}\n\\end{equation*}\n$$\nFrom (8) and (9) we get $0 \\in S \\subseteq(-1,+1)$ and hence the real number $\\alpha=\\sup (S)$ exists and satisfies $0 \\leqslant \\alpha \\leqslant 1$.\n\nLet us assume that we actually had $0<\\alpha<1$. Note that $f(x)=0$ if $x \\in(0, \\alpha) \\cap \\mathbb{Q}$ by (8), and $f(x)=1$ if $x \\in(\\alpha, 1) \\cap \\mathbb{Q}$ by (9) and (2). Let $K$ denote the unique positive integer satisfying $K \\alpha<1 \\leqslant(K+1) \\alpha$. The first of these two inequalities entails $\\alpha<\\frac{1+\\alpha}{K+1}$, and thus there is a rational number $x \\in\\left(\\alpha, \\frac{1+\\alpha}{K+1}\\right)$. Setting $y=(K+1) x-1$ and substituting $(y, 1, K+1)$ into ( 1 ) we learn\n$$\nf\\left(\\frac{f(y)+1}{K+1}\\right)=f\\left(\\frac{y+1}{K+1}\\right)=f(x) .\n$$\nSince $\\alpha<x<1$ and $0<y<\\alpha$, this simplifies to\n$$\nf\\left(\\frac{1}{K+1}\\right)=1\n$$\nBut, as $0<\\frac{1}{K+1} \\leqslant \\alpha$, this is only possible if $\\alpha=\\frac{1}{K+1}$ and $f(\\alpha)=1$. From this, however, we get the contradiction\n$$\n0=f\\left(\\frac{1}{(K+1)^{2}}\\right)=f\\left(\\frac{\\alpha+0}{K+1}\\right)=f\\left(\\frac{f(\\alpha)+0}{K+1}\\right)=f(\\alpha)=1 .\n$$\nThus our assumption $0<\\alpha<1$ has turned out to be wrong and it follows that $\\alpha \\in\\{0,1\\}$. If $\\alpha=0$, then we have $S \\subseteq(-1,0]$, whence $S=(-1,0] \\cap \\mathbb{Q}$, which in turn yields $f(x)=\\lceil x\\rceil$ for all $x \\in \\mathbb{Q}$ due to (2). Similarly, $\\alpha=1$ entails $S=[0,1) \\cap \\mathbb{Q}$ and $f(x)=\\lfloor x\\rfloor$ for all $x \\in \\mathbb{Q}$. Thereby the solution is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31983,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\nu$ be an irrational positive number, and let $m$ be a positive integer. A pair $(a, b)$ of positive integers is called good if\n$$\na\\lceil b \\nu\\rceil-b\\lfloor a \\nu\\rfloor=m\n$$\nA good pair $(a, b)$ is called excellent if neither of the pairs $(a-b, b)$ and $(a, b-a)$ is good. (As usual, by $\\lfloor x\\rfloor$ and $\\lceil x\\rceil$ we denote the integer numbers such that $x-1<\\lfloor x\\rfloor \\leqslant x$ and $x \\leqslant\\lceil x\\rceil<x+1$.)\nProve that the number of excellent pairs is equal to the sum of the positive divisors of $m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For positive integers $a$ and $b$, let us denote\n$$\nf(a, b)=a\\lceil b \\nu\\rceil-b\\lfloor a \\nu\\rfloor\n$$\nWe will deal with various values of $m$; thus it is convenient to say that a pair $(a, b)$ is $m$-good or $m$-excellent if the corresponding conditions are satisfied.\nTo start, let us investigate how the values $f(a+b, b)$ and $f(a, b+a)$ are related to $f(a, b)$. If $\\{a \\nu\\}+\\{b \\nu\\}<1$, then we have $\\lfloor(a+b) \\nu\\rfloor=\\lfloor a \\nu\\rfloor+\\lfloor b \\nu\\rfloor$ and $\\lceil(a+b) \\nu\\rceil=\\lceil a \\nu\\rceil+\\lceil b \\nu\\rceil-1$, so\n$$\nf(a+b, b)=(a+b)\\lceil b \\nu\\rceil-b(\\lfloor a \\nu\\rfloor+\\lfloor b \\nu\\rfloor)=f(a, b)+b(\\lceil b \\nu\\rceil-\\lfloor b \\nu\\rfloor)=f(a, b)+b\n$$\nand\n$$\nf(a, b+a)=a(\\lceil b \\nu\\rceil+\\lceil a \\nu\\rceil-1)-(b+a)\\lfloor a \\nu\\rfloor=f(a, b)+a(\\lceil a \\nu\\rceil-1-\\lfloor a \\nu\\rfloor)=f(a, b) .\n$$\nSimilarly, if $\\{a \\nu\\}+\\{b \\nu\\} \\geqslant 1$ then one obtains\n$$\nf(a+b, b)=f(a, b) \\quad \\text{ and } \\quad f(a, b+a)=f(a, b)+a\n$$\nSo, in both cases one of the numbers $f(a+b, a)$ and $f(a, b+a)$ is equal to $f(a, b)$ while the other is greater than $f(a, b)$ by one of $a$ and $b$. Thus, exactly one of the pairs $(a+b, b)$ and $(a, b+a)$ is excellent (for an appropriate value of $m$).\n\nNow let us say that the pairs $(a+b, b)$ and $(a, b+a)$ are the children of the pair $(a, b)$, while this pair is their parent. Next, if a pair $(c, d)$ can be obtained from $(a, b)$ by several passings from a parent to a child, we will say that $(c, d)$ is a descendant of $(a, b)$, while $(a, b)$ is an ancestor of $(c, d)$ (a pair is neither an ancestor nor a descendant of itself). Thus each pair of distinct positive integers has a unique ancestor of the form $(a, a)$; our aim is now to find how many $m$-excellent descendants each such pair has.\n\nNotice now that if a pair $(a, b)$ is $m$-excellent then $\\min \\{a, b\\} \\leqslant m$. Indeed, if $a=b$ then $f(a, a)=a=m$, so the statement is valid. Otherwise, the pair $(a, b)$ is a child of some pair $(a', b')$. If $b=b'$ and $a=a'+b'$, then we should have $m=f(a, b)=f(a', b')+b'$, so $b=b'=m-f(a', b')<m$. Similarly, if $a=a'$ and $b=b'+a'$ then $a<m$.\n\nLet us consider the set $S_{m}$ of all pairs $(a, b)$ such that $f(a, b) \\leqslant m$ and $\\min \\{a, b\\} \\leqslant m$. Then all the ancestors of the elements in $S_{m}$ are again in $S_{m}$, and each element in $S_{m}$ either is of the form $(a, a)$ with $a \\leqslant m$, or has a unique ancestor of this form. From the arguments above we see that all $m$-excellent pairs lie in $S_{m}$.\n\nWe claim now that the set $S_{m}$ is finite. Indeed, assume, for instance, that it contains infinitely many pairs $(c, d)$ with $d>2m$. Such a pair is necessarily a child of $(c, d-c)$, and thus a descendant of some pair $(c, d')$ with $m<d' \\leqslant 2m$. Therefore, one of the pairs $(a, b) \\in S_{m}$ with $m<b \\leqslant 2m$ has infinitely many descendants in $S_{m}$, and all these descendants have the form $(a, b+k a)$ with $k$ a positive integer. Since $f(a, b+k a)$ does not decrease as $k$ grows, it becomes constant for $k \\geqslant k_{0}$, where $k_{0}$ is some positive integer. This means that $\\{a \\nu\\}+\\{(b+k a) \\nu\\}<1$ for all $k \\geqslant k_{0}$. But this yields $1>\\{(b+k a) \\nu\\}=\\left\\{\\left(b+k_{0} a\\right) \\nu\\right\\}+\\left(k-k_{0}\\right)\\{a \\nu\\}$ for all $k>k_{0}$, which is absurd.\n\nSimilarly, one can prove that $S_{m}$ contains finitely many pairs $(c, d)$ with $c>2m$, thus finitely many elements at all.\n\nWe are now prepared for proving the following crucial lemma.\n\nLemma. Consider any pair $(a, b)$ with $f(a, b) \\neq m$. Then the number $g(a, b)$ of its $m$-excellent descendants is equal to the number $h(a, b)$ of ways to represent the number $t=m-f(a, b)$ as $t=k a+\\ell b$ with $k$ and $\\ell$ being some nonnegative integers.\n\nProof. We proceed by induction on the number $N$ of descendants of $(a, b)$ in $S_{m}$. If $N=0$ then clearly $g(a, b)=0$. Assume that $h(a, b)>0$; without loss of generality, we have $a \\leqslant b$. Then, clearly, $m-f(a, b) \\geqslant a$, so $f(a, b+a) \\leqslant f(a, b)+a \\leqslant m$ and $a \\leqslant m$, hence $(a, b+a) \\in S_{m}$ which is impossible. Thus in the base case we have $g(a, b)=h(a, b)=0$, as desired.\n\nNow let $N>0$. Assume that $f(a+b, b)=f(a, b)+b$ and $f(a, b+a)=f(a, b)$ (the other case is similar). If $f(a, b)+b \\neq m$, then by the induction hypothesis we have\n$$\ng(a, b)=g(a+b, b)+g(a, b+a)=h(a+b, b)+h(a, b+a) .\n$$\nNotice that both pairs $(a+b, b)$ and $(a, b+a)$ are descendants of $(a, b)$ and thus each of them has strictly less descendants in $S_{m}$ than $(a, b)$ does.\n\nNext, each one of the $h(a+b, b)$ representations of $m-f(a+b, b)=m-b-f(a, b)$ as the sum $k'(a+b)+\\ell' b$ provides the representation $m-f(a, b)=k a+\\ell b$ with $k=k'<k'+\\ell'+1=\\ell$. Similarly, each one of the $h(a, b+a)$ representations of $m-f(a, b+a)=m-f(a, b)$ as the sum $k' a+\\ell'(b+a)$ provides the representation $m-f(a, b)=k a+\\ell b$ with $k=k'+\\ell' \\geqslant \\ell'=\\ell$. This correspondence is obviously bijective, so\n$$\nh(a, b)=h(a+b, b)+h(a, b+a)=g(a, b)\n$$\nas required.\n\nFinally, if $f(a, b)+b=m$ then $(a+b, b)$ is $m$-excellent, so $g(a, b)=1+g(a, b+a)=1+h(a, b+a)$ by the induction hypothesis. On the other hand, the number $m-f(a, b)=b$ has a representation $0 \\cdot a+1 \\cdot b$ and sometimes one more representation as $k a+0 \\cdot b$; this last representation exists simultaneously with the representation $m-f(a, b+a)=k a+0 \\cdot(b+a)$, so $h(a, b)=1+h(a, b+a)$ as well. Thus in this case the step is also proved.\n\nNow it is easy to finish the solution. There exists a unique $m$-excellent pair of the form $(a, a)$, and each other $m$-excellent pair $(a, b)$ has a unique ancestor of the form $(x, x)$ with $x<m$. By the lemma, for every $x<m$ the number of its $m$-excellent descendants is $h(x, x)$, which is the number of ways to represent $m-f(x, x)=m-x$ as $k x+\\ell x$ (with nonnegative integer $k$ and $\\ell$). This number is 0 if $x \\nmid m$, and $m / x$ otherwise. So the total number of excellent pairs is\n$$\n1+\\sum_{x \\mid m, x<m} \\frac{m}{x}=1+\\sum_{d \\mid m, d>1} d=\\sum_{d \\mid m} d\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31984,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $z_{0} < z_{1} < z_{2} < \\cdots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n \\geqslant 1$ such that\n$$\nz_{n} < \\frac{z_{0} + z_{1} + \\cdots + z_{n}}{n} \\leqslant z_{n+1} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For $n = 1, 2, \\ldots$ define\n$$\nd_{n} = \\left(z_{0} + z_{1} + \\cdots + z_{n}\\right) - n z_{n}\n$$\nThe sign of $d_{n}$ indicates whether the first inequality in (1) holds; i.e., it is satisfied if and only if $d_{n} > 0$.\nNotice that\n$$\nn z_{n+1} - \\left(z_{0} + z_{1} + \\cdots + z_{n}\\right) = (n+1) z_{n+1} - \\left(z_{0} + z_{1} + \\cdots + z_{n} + z_{n+1}\\right) = -d_{n+1},\n$$\nso the second inequality in (1) is equivalent to $d_{n+1} \\leqslant 0$. Therefore, we have to prove that there is a unique index $n \\geqslant 1$ that satisfies $d_{n} > 0 \\geqslant d_{n+1}$.\nBy its definition the sequence $d_{1}, d_{2}, \\ldots$ consists of integers and we have\n$$\nd_{1} = \\left(z_{0} + z_{1}\\right) - 1 \\cdot z_{1} = z_{0} > 0 .\n$$\nFrom\n$d_{n+1} - d_{n} = \\left(\\left(z_{0} + \\cdots + z_{n} + z_{n+1}\\right) - (n+1) z_{n+1}\\right) - \\left(\\left(z_{0} + \\cdots + z_{n}\\right) - n z_{n}\\right) = n\\left(z_{n} - z_{n+1}\\right) < 0$\nwe can see that $d_{n+1} < d_{n}$ and thus the sequence strictly decreases.\nHence, we have a decreasing sequence $d_{1} > d_{2} > \\ldots$ of integers such that its first element $d_{1}$ is positive. The sequence must drop below 0 at some point, and thus there is a unique index $n$, that is the index of the last positive term, satisfying $d_{n} > 0 \\geqslant d_{n+1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31985,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a sequence $x_{1}, x_{2}, \\ldots, x_{n}$ of real numbers, we define its price as\n$$\n\\max_{1 \\leqslant i \\leqslant n}\\left|x_{1}+\\cdots+x_{i}\\right| .\n$$\nGiven $n$ real numbers, Dave and George want to arrange them into a sequence with a low price. Diligent Dave checks all possible ways and finds the minimum possible price $D$. Greedy George, on the other hand, chooses $x_{1}$ such that $\\left|x_{1}\\right|$ is as small as possible; among the remaining numbers, he chooses $x_{2}$ such that $\\left|x_{1}+x_{2}\\right|$ is as small as possible, and so on. Thus, in the $i^{\\text {th }}$ step he chooses $x_{i}$ among the remaining numbers so as to minimise the value of $\\left|x_{1}+x_{2}+\\cdots+x_{i}\\right|$. In each step, if several numbers provide the same value, George chooses one at random. Finally he gets a sequence with price $G$.\nFind the least possible constant $c$ such that for every positive integer $n$, for every collection of $n$ real numbers, and for every possible sequence that George might obtain, the resulting values satisfy the inequality $G \\leqslant c D$.",
    "options": [],
    "answer": "2",
    "solution": "If the initial numbers are $1,-1,2$, and $-2$, then Dave may arrange them as $1,-2,2,-1$, while George may get the sequence $1,-1,2,-2$, resulting in $D=1$ and $G=2$. So we obtain $c \\geqslant 2$.\n\nTherefore, it remains to prove that $G \\leqslant 2 D$. Let $x_{1}, x_{2}, \\ldots, x_{n}$ be the numbers Dave and George have at their disposal. Assume that Dave and George arrange them into sequences $d_{1}, d_{2}, \\ldots, d_{n}$ and $g_{1}, g_{2}, \\ldots, g_{n}$, respectively. Put\n$$\nM=\\max_{1 \\leqslant i \\leqslant n}\\left|x_{i}\\right|, \\quad S=\\left|x_{1}+\\cdots+x_{n}\\right|, \\quad \\text{ and } \\quad N=\\max \\{M, S\\} .\n$$\nWe claim that\n$$\n\\begin{align*}\n& D \\geqslant S,  \\tag{1}\\\\\n& D \\geqslant \\frac{M}{2}, \\quad \\text{ and }  \\tag{2}\\\\\n& G \\leqslant N=\\max \\{M, S\\} . \\tag{3}\n\\end{align*}\n$$\nThese inequalities yield the desired estimate, as $G \\leqslant \\max \\{M, S\\} \\leqslant \\max \\{M, 2 S\\} \\leqslant 2 D$.\n\nThe inequality (1) is a direct consequence of the definition of the price.\n\nTo prove (2), consider an index $i$ with $\\left|d_{i}\\right|=M$. Then we have\n$$\nM=\\left|d_{i}\\right|=\\left|\\left(d_{1}+\\cdots+d_{i}\\right)-\\left(d_{1}+\\cdots+d_{i-1}\\right)\\right| \\leqslant\\left|d_{1}+\\cdots+d_{i}\\right|+\\left|d_{1}+\\cdots+d_{i-1}\\right| \\leqslant 2 D,\n$$\nas required.\n\nIt remains to establish (3). Put $h_{i}=g_{1}+g_{2}+\\cdots+g_{i}$. We will prove by induction on $i$ that $\\left|h_{i}\\right| \\leqslant N$. The base case $i=1$ holds, since $\\left|h_{1}\\right|=\\left|g_{1}\\right| \\leqslant M \\leqslant N$. Notice also that $\\left|h_{n}\\right|=S \\leqslant N$.\n\nFor the induction step, assume that $\\left|h_{i-1}\\right| \\leqslant N$. We distinguish two cases.\n\nCase 1. Assume that no two of the numbers $g_{i}, g_{i+1}, \\ldots, g_{n}$ have opposite signs.\nWithout loss of generality, we may assume that they are all nonnegative. Then one has $h_{i-1} \\leqslant h_{i} \\leqslant \\cdots \\leqslant h_{n}$, thus\n$$\n\\left|h_{i}\\right| \\leqslant \\max \\{\\left|h_{i-1}\\right|,\\left|h_{n}\\right|\\} \\leqslant N .\n$$\n\nCase 2. Among the numbers $g_{i}, g_{i+1}, \\ldots, g_{n}$ there are positive and negative ones.\nThen there exists some index $j \\geqslant i$ such that $h_{i-1} g_{j} \\leqslant 0$. By the definition of George's sequence we have\n$$\n\\left|h_{i}\\right|=\\left|h_{i-1}+g_{i}\\right| \\leqslant\\left|h_{i-1}+g_{j}\\right| \\leqslant \\max \\{\\left|h_{i-1}\\right|,\\left|g_{j}\\right|\\} \\leqslant N .\n$$\nThus, the induction step is established.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 31986,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that\n$$\nn^{2}+4 f(n)=f(f(n))^{2}\n$$\nfor all $n \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "All functions f from the integers to the integers satisfying the equation are exactly the following:\n1) f(n) = n + 1 for all integers n.\n2) For any integer a ≥ 1, the piecewise function given by f(n) = 1 − n for all n ≤ −a and f(n) = n + 1 for all n ≥ −a + 1.\n3) The sign-symmetric piecewise function with f(n) = n + 1 for n > 0, f(0) = 0, and f(n) = 1 − n for n < 0.",
    "solution": "Part I. Let us first check that each of the functions above really satisfies the given functional equation. If $f(n)=n+1$ for all $n$, then we have\n$$\nn^{2}+4 f(n)=n^{2}+4 n+4=(n+2)^{2}=f(n+1)^{2}=f(f(n))^{2} .\n$$\nIf $f(n)=n+1$ for $n>-a$ and $f(n)=-n+1$ otherwise, then we have the same identity for $n>-a$ and\n$$\nn^{2}+4 f(n)=n^{2}-4 n+4=(2-n)^{2}=f(1-n)^{2}=f(f(n))^{2}\n$$\notherwise. The same applies to the third solution (with $a=0$ ), where in addition one has\n$$\n0^{2}+4 f(0)=0=f(f(0))^{2}\n$$\nPart II. It remains to prove that these are really the only functions that satisfy our functional equation. We do so in three steps:\nStep 1: We prove that $f(n)=n+1$ for $n>0$.\nConsider the sequence ($a_{k}$) given by $a_{k}=f^{k}(1)$ for $k \\geqslant 0$. Setting $n=a_{k}$ in (1), we get\n$$\na_{k}^{2}+4 a_{k+1}=a_{k+2}^{2}\n$$\nOf course, $a_{0}=1$ by definition. Since $a_{2}^{2}=1+4 a_{1}$ is odd, $a_{2}$ has to be odd as well, so we set $a_{2}=2 r+1$ for some $r \\in \\mathbb{Z}$. Then $a_{1}=r^{2}+r$ and consequently\n$$\na_{3}^{2}=a_{1}^{2}+4 a_{2}=(r^{2}+r)^{2}+8 r+4\n$$\nSince $8 r+4 \\neq 0, a_{3}^{2} \\neq(r^{2}+r)^{2}$, so the difference between $a_{3}^{2}$ and $(r^{2}+r)^{2}$ is at least the distance from $(r^{2}+r)^{2}$ to the nearest even square (since $8 r+4$ and $r^{2}+r$ are both even). This implies that\n$$\n|8 r+4|=|a_{3}^{2}-(r^{2}+r)^{2}| \\geqslant(r^{2}+r)^{2}-(r^{2}+r-2)^{2}=4(r^{2}+r-1)\n$$\n(for $r=0$ and $r=-1$, the estimate is trivial, but this does not matter). Therefore, we have\n$$\n4 r^{2} \\leqslant|8 r+4|-4 r+4\n$$\nIf $|r| \\geqslant 4$, then\n$$\n4 r^{2} \\geqslant 16|r| \\geqslant 12|r|+16>8|r|+4+4|r|+4 \\geqslant|8 r+4|-4 r+4\n$$\na contradiction. Thus $|r|<4$. Checking all possible remaining values of $r$, we find that $(r^{2}+r)^{2}+8 r+4$ is only a square in three cases: $r=-3, r=0$ and $r=1$. Let us now distinguish these three cases:\n- $r=-3$, thus $a_{1}=6$ and $a_{2}=-5$. For each $k \\geqslant 1$, we have\n$$\na_{k+2}= \\pm \\sqrt{a_{k}^{2}+4 a_{k+1}}\n$$\nand the sign needs to be chosen in such a way that $a_{k+1}^{2}+4 a_{k+2}$ is again a square. This yields $a_{3}=-4, a_{4}=-3, a_{5}=-2, a_{6}=-1, a_{7}=0, a_{8}=1, a_{9}=2$. At this point we have reached a contradiction, since $f(1)=f(a_{0})=a_{1}=6$ and at the same time $f(1)=f(a_{8})=a_{9}=2$.\n- $r=0$, thus $a_{1}=0$ and $a_{2}=1$. Then $a_{3}^{2}=a_{1}^{2}+4 a_{2}=4$, so $a_{3}= \\pm 2$. This, however, is a contradiction again, since it gives us $f(1)=f(a_{0})=a_{1}=0$ and at the same time $f(1)=f(a_{2})=a_{3}= \\pm 2$.\n- $r=1$, thus $a_{1}=2$ and $a_{2}=3$. We prove by induction that $a_{k}=k+1$ for all $k \\geqslant 0$ in this case, which we already know for $k \\leqslant 2$ now. For the induction step, assume that $a_{k-1}=k$ and $a_{k}=k+1$. Then\n$$\na_{k+1}^{2}=a_{k-1}^{2}+4 a_{k}=k^{2}+4 k+4=(k+2)^{2}\n$$\nso $a_{k+1}= \\pm(k+2)$. If $a_{k+1}=-(k+2)$, then\n$$\na_{k+2}^{2}=a_{k}^{2}+4 a_{k+1}=(k+1)^{2}-4 k-8=k^{2}-2 k-7=(k-1)^{2}-8\n$$\nThe latter can only be a square if $k=4$ (since 1 and 9 are the only two squares whose difference is 8 ). Then, however, $a_{4}=5, a_{5}=-6$ and $a_{6}= \\pm 1$, so\n$$\na_{7}^{2}=a_{5}^{2}+4 a_{6}=36 \\pm 4\n$$\nbut neither 32 nor 40 is a perfect square. Thus $a_{k+1}=k+2$, which completes our induction. This also means that $f(n)=f(a_{n-1})=a_{n}=n+1$ for all $n \\geqslant 1$.\n\nStep 2: We prove that either $f(0)=1$, or $f(0)=0$ and $f(n) \\neq 0$ for $n \\neq 0$.\nSet $n=0$ in (1) to get\n$$\n4 f(0)=f(f(0))^{2}\n$$\nThis means that $f(0) \\geqslant 0$. If $f(0)=0$, then $f(n) \\neq 0$ for all $n \\neq 0$, since we would otherwise have\n$$\nn^{2}=n^{2}+4 f(n)=f(f(n))^{2}=f(0)^{2}=0\n$$\nIf $f(0)>0$, then we know that $f(f(0))=f(0)+1$ from the first step, so\n$$\n4 f(0)=(f(0)+1)^{2}\n$$\nwhich yields $f(0)=1$.\n\nStep 3: We discuss the values of $f(n)$ for $n<0$.\nLemma. For every $n \\geqslant 1$, we have $f(-n)=-n+1$ or $f(-n)=n+1$. Moreover, if $f(-n)= -n+1$ for some $n \\geqslant 1$, then also $f(-n+1)=-n+2$.\nProof. We prove this statement by strong induction on $n$. For $n=1$, we get\n$$\n1+4 f(-1)=f(f(-1))^{2}\n$$\nThus $f(-1)$ needs to be nonnegative. If $f(-1)=0$, then $f(f(-1))=f(0)= \\pm 1$, so $f(0)=1$ (by our second step). Otherwise, we know that $f(f(-1))=f(-1)+1$, so\n$$\n1+4 f(-1)=(f(-1)+1)^{2}\n$$\nwhich yields $f(-1)=2$ and thus establishes the base case. For the induction step, we consider two cases:\n- If $f(-n) \\leqslant-n$, then\n$$\nf(f(-n))^{2}=(-n)^{2}+4 f(-n) \\leqslant n^{2}-4 n<(n-2)^{2}\n$$\nso $|f(f(-n))| \\leqslant n-3$ (for $n=2$, this case cannot even occur). If $f(f(-n)) \\geqslant 0$, then we already know from the first two steps that $f(f(f(-n)))=f(f(-n))+1$, unless perhaps if $f(0)=0$ and $f(f(-n))=0$. However, the latter would imply $f(-n)=0$ (as shown in Step 2) and thus $n=0$, which is impossible. If $f(f(-n))<0$, we can apply the induction hypothesis to $f(f(-n))$. In either case, $f(f(f(-n)))= \\pm f(f(-n))+1$. Therefore,\n$$\nf(-n)^{2}+4 f(f(-n))=f(f(f(-n)))^{2}=( \\pm f(f(-n))+1)^{2}\n$$\nwhich gives us\n$$\n\\begin{aligned}\nn^{2} & \\leqslant f(-n)^{2}=( \\pm f(f(-n))+1)^{2}-4 f(f(-n)) \\leqslant f(f(-n))^{2}+6|f(f(-n))|+1 \\\\\n& \\leqslant(n-3)^{2}+6(n-3)+1=n^{2}-8\n\\end{aligned}\n$$\na contradiction.\n- Thus, we are left with the case that $f(-n)>-n$. Now we argue as in the previous case: if $f(-n) \\geqslant 0$, then $f(f(-n))=f(-n)+1$ by the first two steps, since $f(0)=0$ and $f(-n)=0$ would imply $n=0$ (as seen in Step 2) and is thus impossible. If $f(-n)<0$, we can apply the induction hypothesis, so in any case we can infer that $f(f(-n))= \\pm f(-n)+1$. We obtain\n$$\n(-n)^{2}+4 f(-n)=( \\pm f(-n)+1)^{2}\n$$\nso either\n$$\nn^{2}=f(-n)^{2}-2 f(-n)+1=(f(-n)-1)^{2}\n$$\nwhich gives us $f(-n)= \\pm n+1$, or\n$$\nn^{2}=f(-n)^{2}-6 f(-n)+1=(f(-n)-3)^{2}-8 .\n$$\nSince 1 and 9 are the only perfect squares whose difference is 8 , we must have $n=1$, which we have already considered.\nFinally, suppose that $f(-n)=-n+1$ for some $n \\geqslant 2$. Then\n$$\nf(-n+1)^{2}=f(f(-n))^{2}=(-n)^{2}+4 f(-n)=(n-2)^{2}\n$$\nso $f(-n+1)= \\pm(n-2)$. However, we already know that $f(-n+1)=-n+2$ or $f(-n+1)=n$, so $f(-n+1)=-n+2$.\nCombining everything we know, we find the solutions as stated in the answer:\n- One solution is given by $f(n)=n+1$ for all $n$.\n- If $f(n)$ is not always equal to $n+1$, then there is a largest integer $m$ (which cannot be positive) for which this is not the case. In view of the lemma that we proved, we must then have $f(n)=-n+1$ for any integer $n<m$. If $m=-a<0$, we obtain $f(n)=-n+1$ for $n \\leqslant-a$ (and $f(n)=n+1$ otherwise). If $m=0$, we have the additional possibility that $f(0)=0, f(n)=-n+1$ for negative $n$ and $f(n)=n+1$ for positive $n$.\nLet us provide an alternative proof for Part II, which also proceeds in several steps.\nStep 1. Let $a$ be an arbitrary integer and $b=f(a)$. We first concentrate on the case where $|a|$ is sufficiently large.\n1. If $b=0$, then (1) applied to $a$ yields $a^{2}=f(f(a))^{2}$, thus\n$$\n\\begin{equation*}\nf(a)=0 \\quad \\Rightarrow \\quad a= \\pm f(0) \\tag{2}\n\\end{equation*}\n$$\nFrom now on, we set $D=|f(0)|$. Throughout Step 1, we will assume that $a \\notin\\{-D, 0, D\\}$, thus $b \\neq 0$.\n2. From (1), noticing that $f(f(a))$ and $a$ have the same parity, we get\n$$\n0 \\neq 4|b|=|f(f(a))^{2}-a^{2}| \\geqslant a^{2}-(|a|-2)^{2}=4|a|-4\n$$\nHence we have\n$$\n\\begin{equation*}\n|b|=|f(a)| \\geqslant|a|-1 \\quad \\text{ for } a \\notin\\{-D, 0, D\\} . \\tag{3}\n\\end{equation*}\n$$\nFor the rest of Step 1, we also assume that $|a| \\geqslant E=\\max \\{D+2,10\\}$. Then by (3) we have $|b| \\geqslant D+1$ and thus $|f(b)| \\geqslant D$.\n3. Set $c=f(b)$; by (3), we have $|c| \\geqslant|b|-1$. Thus (1) yields\n$$\na^{2}+4 b=c^{2} \\geqslant(|b|-1)^{2}\n$$\nwhich implies\n$$\na^{2} \\geqslant(|b|-1)^{2}-4|b|=(|b|-3)^{2}-8>(|b|-4)^{2}\n$$\nbecause $|b| \\geqslant|a|-1 \\geqslant 9$. Thus (3) can be refined to\n$$\n|a|+3 \\geqslant|f(a)| \\geqslant|a|-1 \\quad \\text{ for }|a| \\geqslant E .\n$$\nNow, from $c^{2}=a^{2}+4 b$ with $|b| \\in[|a|-1,|a|+3]$ we get $c^{2}=(a \\pm 2)^{2}+d$, where $d \\in\\{-16,-12,-8,-4,0,4,8\\}$. Since $|a \\pm 2| \\geqslant 8$, this can happen only if $c^{2}=(a \\pm 2)^{2}$, which in turn yields $b= \\pm a+1$. To summarise,\n$$\n\\begin{equation*}\nf(a)=1 \\pm a \\quad \\text{ for }|a| \\geqslant E \\tag{4}\n\\end{equation*}\n$$\nWe have shown that, with at most finitely many exceptions, $f(a)=1 \\pm a$. Thus it will be convenient for our second step to introduce the sets\n$$\nZ_{+}=\\{a \\in \\mathbb{Z}: f(a)=a+1\\}, \\quad Z_{-}=\\{a \\in \\mathbb{Z}: f(a)=1-a\\}, \\quad \\text{ and } \\quad Z_{0}=\\mathbb{Z} \\backslash\\left(Z_{+} \\cup Z_{-}\\right) .\n$$\nStep 2. Now we investigate the structure of the sets $Z_{+}, Z_{-}$, and $Z_{0}$.\n4. Note that $f(E+1)=1 \\pm(E+1)$. If $f(E+1)=E+2$, then $E+1 \\in Z_{+}$. Otherwise we have $f(1+E)=-E$; then the original equation (1) with $n=E+1$ gives us $(E-1)^{2}=f(-E)^{2}$, so $f(-E)= \\pm(E-1)$. By (4) this may happen only if $f(-E)=1-E$, so in this case $-E \\in Z_{+}$. In any case we find that $Z_{+} \\neq \\varnothing$.\n5. Now take any $a \\in Z_{+}$. We claim that every integer $x \\geqslant a$ also lies in $Z_{+}$. We proceed by induction on $x$, the base case $x=a$ being covered by our assumption. For the induction step, assume that $f(x-1)=x$ and plug $n=x-1$ into ( 1 ). We get $f(x)^{2}=(x+1)^{2}$, so either $f(x)=x+1$ or $f(x)=-(x+1)$.\nAssume that $f(x)=-(x+1)$ and $x \\neq-1$, since otherwise we already have $f(x)=x+1$. Plugging $n=x$ into (1), we obtain $f(-x-1)^{2}=(x-2)^{2}-8$, which may happen only if $x-2= \\pm 3$ and $f(-x-1)= \\pm 1$. Plugging $n=-x-1$ into (1), we get $f( \\pm 1)^{2}=(x+1)^{2} \\pm 4$, which in turn may happen only if $x+1 \\in\\{-2,0,2\\}$.\nThus $x \\in\\{-1,5\\}$ and at the same time $x \\in\\{-3,-1,1\\}$, which gives us $x=-1$. Since this has already been excluded, we must have $f(x)=x+1$, which completes our induction.\n6. Now we know that either $Z_{+}=\\mathbb{Z}$ (if $Z_{+}$is not bounded below), or $Z_{+}=\\left\\{a \\in \\mathbb{Z}: a \\geqslant a_{0}\\right\\}$, where $a_{0}$ is the smallest element of $Z_{+}$. In the former case, $f(n)=n+1$ for all $n \\in \\mathbb{Z}$, which is our first solution. So we assume in the following that $Z_{+}$is bounded below and has a smallest element $a_{0}$.\nIf $Z_{0}=\\varnothing$, then we have $f(x)=x+1$ for $x \\geqslant a_{0}$ and $f(x)=1-x$ for $x<a_{0}$. In particular, $f(0)=1$ in any case, so $0 \\in Z_{+}$and thus $a_{0} \\leqslant 0$. Thus we end up with the second solution listed in the answer. It remains to consider the case where $Z_{0} \\neq \\varnothing$.\n7. Assume that there exists some $a \\in Z_{0}$ with $b=f(a) \\notin Z_{0}$, so that $f(b)=1 \\pm b$. Then we have $a^{2}+4 b=(1 \\pm b)^{2}$, so either $a^{2}=(b-1)^{2}$ or $a^{2}=(b-3)^{2}-8$. In the former case we have $b=1 \\pm a$, which is impossible by our choice of $a$. So we get $a^{2}=(b-3)^{2}-8$, which implies $f(b)=1-b$ and $|a|=1,|b-3|=3$.\nIf $b=0$, then we have $f(b)=1$, so $b \\in Z_{+}$and therefore $a_{0} \\leqslant 0$; hence $a=-1$. But then $f(a)=0=a+1$, so $a \\in Z_{+}$, which is impossible.\nIf $b=6$, then we have $f(6)=-5$, so $f(-5)^{2}=16$ and $f(-5) \\in\\{-4,4\\}$. Then $f(f(-5))^{2}= 25+4 f(-5) \\in\\{9,41\\}$, so $f(-5)=-4$ and $-5 \\in Z_{+}$. This implies $a_{0} \\leqslant-5$, which contradicts our assumption that $\\pm 1=a \\notin Z_{+}$.\n8. Thus we have shown that $f\\left(Z_{0}\\right) \\subseteq Z_{0}$, and $Z_{0}$ is finite. Take any element $c \\in Z_{0}$, and consider the sequence defined by $c_{i}=f^{i}(c)$. All elements of the sequence $\\left(c_{i}\\right)$ lie in $Z_{0}$, hence it is bounded. Choose an index $k$ for which $\\left|c_{k}\\right|$ is maximal, so that in particular $\\left|c_{k+1}\\right| \\leqslant\\left|c_{k}\\right|$ and $\\left|c_{k+2}\\right| \\leqslant\\left|c_{k}\\right|$. Our functional equation (1) yields\n$$\n\\left(\\left|c_{k}\\right|-2\\right)^{2}-4=\\left|c_{k}\\right|^{2}-4\\left|c_{k}\\right| \\leqslant c_{k}^{2}+4 c_{k+1}=c_{k+2}^{2}\n$$\nSince $c_{k}$ and $c_{k+2}$ have the same parity and $\\left|c_{k+2}\\right| \\leqslant\\left|c_{k}\\right|$, this leaves us with three possibilities: $\\left|c_{k+2}\\right|=\\left|c_{k}\\right|,\\left|c_{k+2}\\right|=\\left|c_{k}\\right|-2$, and $\\left|c_{k}\\right|-2= \\pm 2, c_{k+2}=0$.\nIf $\\left|c_{k+2}\\right|=\\left|c_{k}\\right|-2$, then $f\\left(c_{k}\\right)=c_{k+1}=1-\\left|c_{k}\\right|$, which means that $c_{k} \\in Z_{-}$or $c_{k} \\in Z_{+}$, and we reach a contradiction.\nIf $\\left|c_{k+2}\\right|=\\left|c_{k}\\right|$, then $c_{k+1}=0$, thus $c_{k+3}^{2}=4 c_{k+2}$. So either $c_{k+3} \\neq 0$ or (by maximality of $\\left.\\left|c_{k+2}\\right|=\\left|c_{k}\\right|\\right) c_{i}=0$ for all $i$. In the former case, we can repeat the entire argument\nwith $c_{k+2}$ in the place of $c_{k}$. Now $\\left|c_{k+4}\\right|=\\left|c_{k+2}\\right|$ is not possible any more since $c_{k+3} \\neq 0$, leaving us with the only possibility $\\left|c_{k}\\right|-2=\\left|c_{k+2}\\right|-2= \\pm 2$.\nThus we know now that either all $c_{i}$ are equal to 0 , or $\\left|c_{k}\\right|=4$. If $c_{k}= \\pm 4$, then either $c_{k+1}=0$ and $\\left|c_{k+2}\\right|=\\left|c_{k}\\right|=4$, or $c_{k+2}=0$ and $c_{k+1}=-4$. From this point onwards, all elements of the sequence are either 0 or $\\pm 4$.\nLet $c_{r}$ be the last element of the sequence that is not equal to 0 or $\\pm 4$ (if such an element exists). Then $c_{r+1}, c_{r+2} \\in\\{-4,0,4\\}$, so\n$$\nc_{r}^{2}=c_{r+2}^{2}-4 c_{r+1} \\in\\{-16,0,16,32\\}\n$$\nwhich gives us a contradiction. Thus all elements of the sequence are equal to 0 or $\\pm 4$, and since the choice of $c_{0}=c$ was arbitrary, $Z_{0} \\subseteq\\{-4,0,4\\}$.\n9. Finally, we show that $4 \\notin Z_{0}$ and $-4 \\notin Z_{0}$. Suppose that $4 \\in Z_{0}$. Then in particular $a_{0}$ (the smallest element of $Z_{+}$) cannot be less than 4 , since this would imply $4 \\in Z_{+}$. So $-3 \\in Z_{-}$, which means that $f(-3)=4$. Then $25=(-3)^{2}+4 f(-3)=f(f(-3))^{2}=f(4)^{2}$, so $f(4)= \\pm 5 \\notin Z_{0}$, and we reach a contradiction.\nSuppose that $-4 \\in Z_{0}$. The only possible values for $f(-4)$ that are left are 0 and -4 . Note that $4 f(0)=f(f(0))^{2}$, so $f(0) \\geqslant 0$. If $f(-4)=0$, then we get $16=(-4)^{2}+0=f(0)^{2}$, thus $f(0)=4$. But then $f(f(-4)) \\notin Z_{0}$, which is impossible. Thus $f(-4)=-4$, which gives us $0=(-4)^{2}+4 f(-4)=f(f(-4))^{2}=16$, and this is clearly absurd.\nNow we are left with $Z_{0}=\\{0\\}$ and $f(0)=0$ as the only possibility. If $1 \\in Z_{-}$, then $f(1)=0$, so $1=1^{2}+4 f(1)=f(f(1))^{2}=f(0)^{2}=0$, which is another contradiction. Thus $1 \\in Z_{+}$, meaning that $a_{0} \\leqslant 1$. On the other hand, $a_{0} \\leqslant 0$ would imply $0 \\in Z_{+}$, so we can only have $a_{0}=1$. Thus $Z_{+}$comprises all positive integers, and $Z_{-}$comprises all negative integers. This gives us the third solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31987,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 2$ be an integer. Consider an $n \\times n$ chessboard divided into $n^{2}$ unit squares. We call a configuration of $n$ rooks on this board happy if every row and every column contains exactly one rook. Find the greatest positive integer $k$ such that for every happy configuration of rooks, we can find a $k \\times k$ square without a rook on any of its $k^{2}$ unit squares.",
    "options": [],
    "answer": "⌊√(n−1)⌋",
    "solution": "Answer. $\\lfloor\\sqrt{n-1}\\rfloor$.\n\nLet $\\ell$ be a positive integer. We will show that (i) if $n>\\ell^{2}$ then each happy configuration contains an empty $\\ell \\times \\ell$ square, but (ii) if $n \\leqslant \\ell^{2}$ then there exists a happy configuration not containing such a square. These two statements together yield the answer.\n\ni). Assume that $n>\\ell^{2}$. Consider any happy configuration. There exists a row $R$ containing a rook in its leftmost square. Take $\\ell$ consecutive rows with $R$ being one of them. Their union $U$ contains exactly $\\ell$ rooks. Now remove the $n-\\ell^{2} \\geqslant 1$ leftmost columns from $U$ (thus at least one rook is also removed). The remaining part is an $\\ell^{2} \\times \\ell$ rectangle, so it can be split into $\\ell$ squares of size $\\ell \\times \\ell$, and this part contains at most $\\ell-1$ rooks. Thus one of these squares is empty.\n\n(ii). Now we assume that $n \\leqslant \\ell^{2}$. Firstly, we will construct a happy configuration with no empty $\\ell \\times \\ell$ square for the case $n=\\ell^{2}$. After that we will modify it to work for smaller values of $n$.\n\nLet us enumerate the rows from bottom to top as well as the columns from left to right by the numbers $0,1, \\ldots, \\ell^{2}-1$. Every square will be denoted, as usual, by the pair ( $r, c$ ) of its row and column numbers. Now we put the rooks on all squares of the form ( $i \\ell+j, j \\ell+i$ ) with $i, j=0,1, \\ldots, \\ell-1$ (the picture below represents this arrangement for $\\ell=3$ ). Since each number from 0 to $\\ell^{2}-1$ has a unique representation of the form $i \\ell+j(0 \\leqslant i, j \\leqslant \\ell-1)$, each row and each column contains exactly one rook.\n\n![](attached_image_1.png)\n\nNext, we show that each $\\ell \\times \\ell$ square $A$ on the board contains a rook. Consider such a square $A$, and consider $\\ell$ consecutive rows the union of which contains $A$. Let the lowest of these rows have number $p \\ell+q$ with $0 \\leqslant p, q \\leqslant \\ell-1$ (notice that $p \\ell+q \\leqslant \\ell^{2}-\\ell$ ). Then the rooks in this union are placed in the columns with numbers $q \\ell+p,(q+1) \\ell+p, \\ldots,(\\ell-1) \\ell+p$, $p+1, \\ell+(p+1), \\ldots,(q-1) \\ell+p+1$, or, putting these numbers in increasing order,\n$$\np+1, \\ell+(p+1), \\ldots,(q-1) \\ell+(p+1), q \\ell+p,(q+1) \\ell+p, \\ldots,(\\ell-1) \\ell+p .\n$$\nOne readily checks that the first number in this list is at most $\\ell-1$ (if $p=\\ell-1$, then $q=0$, and the first listed number is $q \\ell+p=\\ell-1$ ), the last one is at least ( $\\ell-1$ ) $\\ell$, and the difference between any two consecutive numbers is at most $\\ell$. Thus, one of the $\\ell$ consecutive columns intersecting $A$ contains a number listed above, and the rook in this column is inside $A$, as required. The construction for $n=\\ell^{2}$ is established.\n\nIt remains to construct a happy configuration of rooks not containing an empty $\\ell \\times \\ell$ square for $n<\\ell^{2}$. In order to achieve this, take the construction for an $\\ell^{2} \\times \\ell^{2}$ square described above and remove the $\\ell^{2}-n$ bottom rows together with the $\\ell^{2}-n$ rightmost columns. We will have a rook arrangement with no empty $\\ell \\times \\ell$ square, but several rows and columns may happen to be empty. Clearly, the number of empty rows is equal to the number of empty columns, so one can find a bijection between them, and put a rook on any crossing of an empty row and an empty column corresponding to each other.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31988,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider $n \\geqslant 3$ lines in the plane such that no two lines are parallel and no three have a common point. These lines divide the plane into polygonal regions; let $\\mathcal{F}$ be the set of regions having finite area. Prove that it is possible to colour $\\lceil\\sqrt{n / 2}\\rceil$ of the lines blue in such a way that no region in $\\mathcal{F}$ has a completely blue boundary. (For a real number $x,\\lceil x\\rceil$ denotes the least integer which is not smaller than $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $L$ be the given set of lines. Choose a maximal (by inclusion) subset $B \\subseteq L$ such that when we colour the lines of $B$ blue, no region in $\\mathcal{F}$ has a completely blue boundary. Let $|B|=k$. We claim that $k \\geqslant\\lceil\\sqrt{n / 2}\\rceil$.\n\nLet us colour all the lines of $L \\backslash B$ red. Call a point blue if it is the intersection of two blue lines. Then there are $\\binom{k}{2}$ blue points.\n\nNow consider any red line $\\ell$. By the maximality of $B$, there exists at least one region $A \\in \\mathcal{F}$ whose only red side lies on $\\ell$. Since $A$ has at least three sides, it must have at least one blue vertex. Let us take one such vertex and associate it to $\\ell$.\n\nSince each blue point belongs to four regions (some of which may be unbounded), it is associated to at most four red lines. Thus the total number of red lines is at most $4\\binom{k}{2}$. On the other hand, this number is $n-k$, so\n$$\nn-k \\leqslant 2 k(k-1), \\quad \\text{ thus } \\quad n \\leqslant 2 k^{2}-k \\leqslant 2 k^{2}\n$$\nand finally $k \\geqslant\\lceil\\sqrt{n / 2}\\rceil$, which gives the desired result.\n\n**Comment 1.** The constant factor in the estimate can be improved in different ways; we sketch two of them below. On the other hand, the Problem Selection Committee is not aware of any results showing that it is sometimes impossible to colour $k$ lines satisfying the desired condition for $k \\gg \\sqrt{n}$. In this situation we find it more suitable to keep the original formulation of the problem.\n\n1. Firstly, we show that in the proof above one has in fact $k=|B| \\geqslant\\lceil\\sqrt{2 n / 3}\\rceil$.\nLet us make weighted associations as follows. Let a region $A$ whose only red side lies on $\\ell$ have $k$ vertices, so that $k-2$ of them are blue. We associate each of these blue vertices to $\\ell$, and put the weight $\\frac{1}{k-2}$ on each such association. So the sum of the weights of all the associations is exactly $n-k$.\nNow, one may check that among the four regions adjacent to a blue vertex $v$, at most two are triangles. This means that the sum of the weights of all associations involving $v$ is at most $1+1+\\frac{1}{2}+\\frac{1}{2}=3$. This leads to the estimate\n$$\nn-k \\leqslant 3\\binom{k}{2}\n$$\nor\n$$\n2 n \\leqslant 3 k^{2}-k<3 k^{2}\n$$\nwhich yields $k \\geqslant\\lceil\\sqrt{2 n / 3}\\rceil$.\n\n2. Next, we even show that $k=|B| \\geqslant[\\sqrt{n}]$. For this, we specify the process of associating points to red lines in one more different way.\nCall a point red if it lies on a red line as well as on a blue line. Consider any red line $\\ell$, and take an arbitrary region $A \\in \\mathcal{F}$ whose only red side lies on $\\ell$. Let $r^{\\prime}, r, b_{1}, \\ldots, b_{k}$ be its vertices in clockwise order with $r^{\\prime}, r \\in \\ell$; then the points $r^{\\prime}, r$ are red, while all the points $b_{1}, \\ldots, b_{k}$ are blue. Let us associate to $\\ell$ the red point $r$ and the blue point $b_{1}$. One may notice that to each pair of a red point $r$ and a blue point $b$, at most one red line can be associated, since there is at most one region $A$ having $r$ and $b$ as two clockwise consecutive vertices.\nWe claim now that at most two red lines are associated to each blue point $b$; this leads to the desired bound\n$$\nn-k \\leqslant 2\\binom{k}{2} \\quad \\Longleftrightarrow \\quad n \\leqslant k^{2} .\n$$\nAssume, to the contrary, that three red lines $\\ell_{1}, \\ell_{2}$, and $\\ell_{3}$ are associated to the same blue point $b$. Let $r_{1}, r_{2}$, and $r_{3}$ respectively be the red points associated to these lines; all these points are distinct. The point $b$ defines four blue rays, and each point $r_{i}$ is the red point closest to $b$ on one of these rays. So we may assume that the points $r_{2}$ and $r_{3}$ lie on one blue line passing through $b$, while $r_{1}$ lies on the other one.\n![](attached_image_1.png)\nNow consider the region $A$ used to associate $r_{1}$ and $b$ with $\\ell_{1}$. Three of its clockwise consecutive vertices are $r_{1}, b$, and either $r_{2}$ or $r_{3}$ (say, $r_{2}$ ). Since $A$ has only one red side, it can only be the triangle $r_{1} b r_{2}$; but then both $\\ell_{1}$ and $\\ell_{2}$ pass through $r_{2}$, as well as some blue line. This is impossible by the problem assumptions.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31989,
    "subject": "Mathematics (Multi-modal)",
    "question": "We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of $100$ cards each from this deck. We would like to define a rule that declares one of them a winner. This rule should satisfy the following conditions:\n1. The winner only depends on the relative order of the $200$ cards: if the cards are laid down in increasing order face down and we are told which card belongs to which player, but not what numbers are written on them, we can still decide the winner.\n2. If we write the elements of both sets in increasing order as $A=\\{a_{1}, a_{2}, \\ldots, a_{100}\\}$ and $B=\\{b_{1}, b_{2}, \\ldots, b_{100}\\}$, and $a_{i}>b_{i}$ for all $i$, then $A$ beats $B$.\n3. If three players draw three disjoint sets $A, B, C$ from the deck, $A$ beats $B$ and $B$ beats $C$, then $A$ also beats $C$.\nHow many ways are there to define such a rule? Here, we consider two rules as different if there exist two sets $A$ and $B$ such that $A$ beats $B$ according to one rule, but $B$ beats $A$ according to the other.",
    "options": [],
    "answer": "100",
    "solution": "Answer. $100$.\n\nSolution 1. We prove a more general statement for sets of cardinality $n$ (the problem being the special case $n=100$, then the answer is $n$). In the following, we write $A>B$ or $B<A$ for \"$A$ beats $B$\".\n\nPart I. Let us first define $n$ different rules that satisfy the conditions. To this end, fix an index $k \\in \\{1,2, \\ldots, n\\}$. We write both $A$ and $B$ in increasing order as $A=\\{a_{1}, a_{2}, \\ldots, a_{n}\\}$ and $B=\\{b_{1}, b_{2}, \\ldots, b_{n}\\}$ and say that $A$ beats $B$ if and only if $a_{k}>b_{k}$. This rule clearly satisfies all three conditions, and the rules corresponding to different $k$ are all different. Thus there are at least $n$ different rules.\n\nPart II. Now we have to prove that there is no other way to define such a rule. Suppose that our rule satisfies the conditions, and let $k \\in \\{1,2, \\ldots, n\\}$ be minimal with the property that\n$$\nA_{k}=\\{1,2, \\ldots, k, n+k+1, n+k+2, \\ldots, 2n\\} \\prec B_{k}=\\{k+1, k+2, \\ldots, n+k\\}.\n$$\nClearly, such a $k$ exists, since this holds for $k=n$ by assumption. Now consider two disjoint sets $X=\\{x_{1}, x_{2}, \\ldots, x_{n}\\}$ and $Y=\\{y_{1}, y_{2}, \\ldots, y_{n}\\}$, both in increasing order (i.e., $x_{1}<x_{2}<\\cdots<x_{n}$ and $y_{1}<y_{2}<\\cdots<y_{n}$). We claim that $X<Y$ if (and only if - this follows automatically) $x_{k}<y_{k}$.\n\nTo prove this statement, pick arbitrary real numbers $u_{i}, v_{i}, w_{i} \\notin X \\cup Y$ such that\n$$\nu_{1}<u_{2}<\\cdots<u_{k-1}<\\min(x_{1}, y_{1}), \\quad \\max(x_{n}, y_{n})<v_{k+1}<v_{k+2}<\\cdots<v_{n},$$\nand\n$$x_{k}<v_{1}<v_{2}<\\cdots<v_{k}<w_{1}<w_{2}<\\cdots<w_{n}<u_{k}<u_{k+1}<\\cdots<u_{n}<y_{k},$$\nand set\n$$U=\\{u_{1}, u_{2}, \\ldots, u_{n}\\}, V=\\{v_{1}, v_{2}, \\ldots, v_{n}\\}, W=\\{w_{1}, w_{2}, \\ldots, w_{n}\\}.$$ \nThen\n- $u_{i}<y_{i}$ and $x_{i}<v_{i}$ for all $i$, so $U<Y$ and $X<V$ by the second condition.\n- The elements of $U \\cup W$ are ordered in the same way as those of $A_{k-1} \\cup B_{k-1}$, and since $A_{k-1}>B_{k-1}$ by our choice of $k$, we also have $U>W$ (if $k=1$, this is trivial).\n- The elements of $V \\cup W$ are ordered in the same way as those of $A_{k} \\cup B_{k}$, and since $A_{k}<B_{k}$ by our choice of $k$, we also have $V<W$.\nIt follows that\n$$\nX<V<W<U<Y\n$$\nso $X<Y$ by the third condition, which is what we wanted to prove.\n\n\nSolution 2. Another possible approach to Part II of this problem is induction on $n$. For $n=1$, there is trivially only one rule in view of the second condition.\n\nIn the following, we assume that our claim (namely, that there are no possible rules other than those given in Part I) holds for $n-1$ in place of $n$. We start with the following observation:\n\nClaim. At least one of the two relations\n$$\n(\\{2\\} \\cup \\{2i-1 \\mid 2 \\leqslant i \\leqslant n\\}) \\prec (\\{1\\} \\cup \\{2i \\mid 2 \\leqslant i \\leqslant n\\})\n$$\nand\n$$\n(\\{2i-1 \\mid 1 \\leqslant i \\leqslant n-1\\} \\cup \\{2n\\}) \\prec (\\{2i \\mid 1 \\leqslant i \\leqslant n-1\\} \\cup \\{2n-1\\})\n$$\nholds.\n\nProof. Suppose that the first relation does not hold. Since our rule may only depend on the relative order, we must also have\n$$\n(\\{2\\} \\cup \\{3i-2 \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup \\{3n-2\\}) > (\\{1\\} \\cup \\{3i-1 \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup \\{3n\\}).\n$$\nLikewise, if the second relation does not hold, then we must also have\n$$\n(\\{1\\} \\cup \\{3i-1 \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup \\{3n\\}) > (\\{3\\} \\cup \\{3i \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup \\{3n-1\\}).\n$$\nNow condition 3 implies that\n$$\n(\\{2\\} \\cup \\{3i-2 \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup \\{3n-2\\}) > (\\{3\\} \\cup \\{3i \\mid 2 \\leqslant i \\leqslant n-1\\} \\cup \\{3n-1\\}),\n$$\nwhich contradicts the second condition.\n\nNow we distinguish two cases, depending on which of the two relations actually holds:\n\nFirst case: $(\\{2\\} \\cup \\{2i-1 \\mid 2 \\leqslant i \\leqslant n\\}) \\prec (\\{1\\} \\cup \\{2i \\mid 2 \\leqslant i \\leqslant n\\})$.\nLet $A=\\{a_{1}, a_{2}, \\ldots, a_{n}\\}$ and $B=\\{b_{1}, b_{2}, \\ldots, b_{n}\\}$ be two disjoint sets, both in increasing order. We claim that the winner can be decided only from the values of $a_{2}, \\ldots, a_{n}$ and $b_{2}, \\ldots, b_{n}$, while $a_{1}$ and $b_{1}$ are actually irrelevant. Suppose that this was not the case, and assume without loss of generality that $a_{2}<b_{2}$. Then the relative order of $a_{1}, a_{2}, \\ldots, a_{n}, b_{2}, \\ldots, b_{n}$ is fixed, and the position of $b_{1}$ has to decide the winner. Suppose that for some value $b_{1}=x$, $B$ wins, while for some other value $b_{1}=y$, $A$ wins.\n\nWrite $B_{x}=\\{x, b_{2}, \\ldots, b_{n}\\}$ and $B_{y}=\\{y, b_{2}, \\ldots, b_{n}\\}$, and let $\\varepsilon>0$ be smaller than half the distance between any two of the numbers in $B_{x} \\cup B_{y} \\cup A$. For any set $M$, let $M \\pm \\varepsilon$ be the set obtained by adding/subtracting $\\varepsilon$ to all elements of $M$. By our choice of $\\varepsilon$, the relative order of the elements of $(B_{y}+\\varepsilon) \\cup A$ is still the same as for $B_{y} \\cup A$, while the relative order of the elements of $(B_{x}-\\varepsilon) \\cup A$ is still the same as for $B_{x} \\cup A$. Thus $A \\prec B_{x}-\\varepsilon$, but $A \\succ B_{y}+\\varepsilon$. Moreover, if $y>x$, then $B_{x}-\\varepsilon < B_{y}+\\varepsilon$ by condition 2, while otherwise the relative order of the elements in $(B_{x}-\\varepsilon) \\cup (B_{y}+\\varepsilon)$ is the same as for the two sets $\\{2\\} \\cup \\{2i-1 \\mid 2 \\leqslant i \\leqslant n\\}$ and $\\{1\\} \\cup \\{2i \\mid 2 \\leqslant i \\leqslant n\\}$, so that $B_{x}-\\varepsilon < B_{y}+\\varepsilon$. In either case, we obtain\n$$\nA \\prec B_{x}-\\varepsilon \\prec B_{y}+\\varepsilon \\prec A\n$$\nwhich contradicts condition 3.\n\nSo we know now that the winner does not depend on $a_{1}, b_{1}$. Therefore, we can define a new rule $<^{*}$ on sets of cardinality $n-1$ by saying that $A<^{*} B$ if and only if $A \\cup \\{a\\} < B \\cup \\{b\\}$ for some $a, b$ (or equivalently, all $a, b$) such that $a<\\min A$, $b<\\min B$ and $A \\cup \\{a\\}$ and $B \\cup \\{b\\}$ are disjoint. The rule $<^{*}$ satisfies all conditions again, so by the induction hypothesis, there exists an index $i$ such that $A<^{*} B$ if and only if the $i^{\\text{th}}$ smallest element of $A$ is less than the $i^{\\text{th}}$ smallest element of $B$. This implies that $C<D$ if and only if the $(i+1)^{\\text{th}}$ smallest element of $C$ is less than the $(i+1)^{\\text{th}}$ smallest element of $D$, which completes our induction.\n\nSecond case: $(\\{2i-1 \\mid 1 \\leqslant i \\leqslant n-1\\} \\cup \\{2n\\}) \\prec (\\{2i \\mid 1 \\leqslant i \\leqslant n-1\\} \\cup \\{2n-1\\})$.\nSet $-A=\\{-a \\mid a \\in A\\}$ for any $A \\subseteq \\mathbb{R}$. For any two disjoint sets $A, B \\subseteq \\mathbb{R}$ of cardinality $n$, we write $A \\prec^{\\circ} B$ to mean $(-B)<(-A)$. It is easy to see that $\\prec^{\\circ}$ defines a rule to determine a winner that satisfies the three conditions of our problem as well as the relation of the first case. So it follows in the same way as in the first case that for some $i$, $A \\prec^{\\circ} B$ if and only if the $i^{\\text{th}}$ smallest element of $A$ is less than the $i^{\\text{th}}$ smallest element of $B$, which is equivalent to the condition that the $i^{\\text{th}}$ largest element of $-A$ is greater than the $i^{\\text{th}}$ largest element of $-B$. This proves that the original rule $\\prec$ also has the desired form.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31990,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $M$ be a set of $n \\geqslant 4$ points in the plane, no three of which are collinear. Initially these points are connected with $n$ segments so that each point in $M$ is the endpoint of exactly two segments. Then, at each step, one may choose two segments $AB$ and $CD$ sharing a common interior point and replace them by the segments $AC$ and $BD$ if none of them is present at this moment. Prove that it is impossible to perform $n^{3} / 4$ or more such moves.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "A line is said to be red if it contains two points of $M$. As no three points of $M$ are collinear, each red line determines a unique pair of points of $M$. Moreover, there are precisely $\\binom{n}{2} < \\frac{n^{2}}{2}$ red lines. By the value of a segment we mean the number of red lines intersecting it in its interior, and the value of a set of segments is defined to be the sum of the values of its elements. We will prove that (i) the value of the initial set of segments is smaller than $n^{3} / 2$ and that (ii) each step decreases the value of the set of segments present by at least 2. Since such a value can never be negative, these two assertions imply the statement of the problem.\n\nTo show (i) we just need to observe that each segment has a value that is smaller than $n^{2} / 2$. Thus the combined value of the $n$ initial segments is indeed below $n \\cdot n^{2} / 2 = n^{3} / 2$.\n\nIt remains to establish (ii). Suppose that at some moment we have two segments $AB$ and $CD$ sharing an interior point $S$, and that at the next moment we have the two segments $AC$ and $BD$ instead. Let $X_{AB}$ denote the set of red lines intersecting the segment $AB$ in its interior and let the sets $X_{AC}$, $X_{BD}$, and $X_{CD}$ be defined similarly. We are to prove that $|X_{AC}| + |X_{BD}| + 2 \\leqslant |X_{AB}| + |X_{CD}|$.\n\nAs a first step in this direction, we claim that\n$$\n|X_{AC} \\cup X_{BD}| + 2 \\leqslant |X_{AB} \\cup X_{CD}|. \\tag{1}\n$$\nIndeed, if $g$ is a red line intersecting, e.g. the segment $AC$ in its interior, then it has to intersect the triangle $ACS$ once again, either in the interior of its side $AS$, or in the interior of its side $CS$, or at $S$, meaning that it belongs to $X_{AB}$ or to $X_{CD}$ (see Figure 1). Moreover, the red lines $AB$ and $CD$ contribute to $X_{AB} \\cup X_{CD}$ but not to $X_{AC} \\cup X_{BD}$. Thereby (1) is proved.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n![](attached_image_3.png)\nFigure 3\n\nSimilarly but more easily one obtains\n$$\n|X_{AC} \\cap X_{BD}| \\leqslant |X_{AB} \\cap X_{CD}|. \\tag{2}\n$$\nIndeed, a red line $h$ appearing in $X_{AC} \\cap X_{BD}$ belongs, for similar reasons as above, also to $X_{AB} \\cap X_{CD}$. To make the argument precise, one may just distinguish the cases $S \\in h$ (see Figure 2) and $S \\notin h$ (see Figure 3). Thereby (2) is proved.\n\nAdding (1) and (2) we obtain the desired conclusion, thus completing the solution of this problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31991,
    "subject": "Mathematics (Multi-modal)",
    "question": "A card deck consists of 1024 cards. On each card, a set of distinct decimal digits is written in such a way that no two of these sets coincide (thus, one of the cards is empty). Two players alternately take cards from the deck, one card per turn. After the deck is empty, each player checks if he can throw out one of his cards so that each of the ten digits occurs on an even number of his remaining cards. If one player can do this but the other one cannot, the one who can is the winner; otherwise a draw is declared.\nDetermine all possible first moves of the first player after which he has a winning strategy.",
    "options": [],
    "answer": "All nonempty first moves; if the first player starts with the empty card, the second player can force a win.",
    "solution": "Let us identify each card with the set of digits written on it. For any collection of cards $C_{1}, C_{2}, \\ldots, C_{k}$ denote by their sum the set $C_{1} \\triangle C_{2} \\triangle \\cdots \\triangle C_{k}$ consisting of all elements belonging to an odd number of the $C_{i}$'s. Denote the first and the second player by $\\mathcal{F}$ and $\\mathcal{S}$, respectively.\n\nSince each digit is written on exactly 512 cards, the sum of all the cards is $\\varnothing$. Therefore, at the end of the game the sum of all the cards of $\\mathcal{F}$ will be the same as that of $\\mathcal{S}$; denote this sum by $C$. Then the player who took $C$ can throw it out and get the desired situation, while the other one cannot. Thus, the player getting card $C$ wins, and no draw is possible.\n\nNow, given a nonempty card $B$, one can easily see that all the cards can be split into 512 pairs of the form ($X$, $X \\triangle B$) because ($X \\triangle B$) $\\triangle B = X$. The following lemma shows a property of such a partition that is important for the solution.\n\n**Lemma.** Let $B \\neq \\varnothing$ be some card. Let us choose 512 cards so that exactly one card is chosen from every pair $(X, X \\triangle B)$. Then the sum of all chosen cards is either $\\varnothing$ or $B$.\n\nProof. Let $b$ be some element of $B$. Enumerate the pairs; let $X_{i}$ be the card not containing $b$ in the $i^{\\text{th}}$ pair, and let $Y_{i}$ be the other card in this pair. Then the sets $X_{i}$ are exactly all the sets not containing $b$, therefore each digit $a \\neq b$ is written on exactly 256 of these cards, so $X_{1} \\triangle X_{2} \\triangle \\cdots \\triangle X_{512} = \\varnothing$. Now, if we replace some summands in this sum by the other elements from their pairs, we will simply add $B$ several times to this sum, thus the sum will either remain unchanged or change by $B$, as required. $\\square$\n\nNow we consider two cases.\n\n**Case 1.** Assume that $\\mathcal{F}$ takes the card $\\varnothing$ on his first move. In this case, we present a winning strategy for $\\mathcal{S}$.\n\nLet $\\mathcal{S}$ take an arbitrary card $A$. Assume that $\\mathcal{F}$ takes card $B$ after that; then $\\mathcal{S}$ takes $A \\triangle B$. Split all 1024 cards into 512 pairs of the form $(X, X \\triangle B)$; we call two cards in one pair partners. Then the four cards taken so far form two pairs ($\\varnothing, B$) and ($A, A \\triangle B$) belonging to $\\mathcal{F}$ and $\\mathcal{S}$, respectively. On each of the subsequent moves, when $\\mathcal{F}$ takes some card, $\\mathcal{S}$ should take the partner of this card in response.\n\nConsider the situation at the end of the game. Let us for a moment replace card $A$ belonging to $\\mathcal{S}$ by $\\varnothing$. Then he would have one card from each pair; by our lemma, the sum of all these cards would be either $\\varnothing$ or $B$. Now, replacing $\\varnothing$ back by $A$ we get that the actual sum of the cards of $\\mathcal{S}$ is either $A$ or $A \\triangle B$, and he has both these cards. Thus $\\mathcal{S}$ wins.\n\n**Case 2.** Now assume that $\\mathcal{F}$ takes some card $A \\neq \\varnothing$ on his first move. Let us present a winning strategy for $\\mathcal{F}$ in this case.\n\nAssume that $\\mathcal{S}$ takes some card $B \\neq \\varnothing$ on his first move; then $\\mathcal{F}$ takes $A \\triangle B$. Again, let us split all the cards into pairs of the form $(X, X \\triangle B)$; then the cards which have not been taken yet form several complete pairs and one extra element (card $\\varnothing$ has not been taken while its partner $B$ has). Now, on each of the subsequent moves, if $\\mathcal{S}$ takes some element from a complete pair, then $\\mathcal{F}$ takes its partner. If $\\mathcal{S}$ takes the extra element, then $\\mathcal{F}$ takes an arbitrary card $Y$, and the partner of $Y$ becomes the new extra element.\n\nThus, on his last move $\\mathcal{S}$ is forced to take the extra element. After that player $\\mathcal{F}$ has cards $A$ and $A \\triangle B$, player $\\mathcal{S}$ has cards $B$ and $\\varnothing$, and $\\mathcal{F}$ has exactly one element from every other pair. Thus the situation is the same as in the previous case with roles reversed, and $\\mathcal{F}$ wins.\n\nFinally, if $\\mathcal{S}$ takes $\\varnothing$ on his first move then $\\mathcal{F}$ denotes any card which has not been taken yet by $B$ and takes $A \\triangle B$. After that, the same strategy as above is applicable.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 31992,
    "subject": "Mathematics (Multi-modal)",
    "question": "The points $P$ and $Q$ are chosen on the side $B C$ of an acute-angled triangle $A B C$ so that $\\angle P A B = \\angle A C B$ and $\\angle Q A C = \\angle C B A$. The points $M$ and $N$ are taken on the rays $A P$ and $A Q$, respectively, so that $A P = P M$ and $A Q = Q N$. Prove that the lines $B M$ and $C N$ intersect on the circumcircle of the triangle $A B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $S$ the intersection point of the lines $B M$ and $C N$. Let moreover $\\beta = \\angle Q A C = \\angle C B A$ and $\\gamma = \\angle P A B = \\angle A C B$. From these equalities it follows that the triangles $A B P$ and $C A Q$ are similar (see Figure 1). Therefore we obtain\n$$\n\\frac{B P}{P M} = \\frac{B P}{P A} = \\frac{A Q}{Q C} = \\frac{N Q}{Q C}.\n$$\nMoreover,\n$$\n\\angle B P M = \\beta + \\gamma = \\angle C Q N.\n$$\nHence the triangles $B P M$ and $N Q C$ are similar. This gives $\\angle B M P = \\angle N C Q$, so the triangles $B P M$ and $B S C$ are also similar. Thus we get\n$$\n\\angle C S B = \\angle B P M = \\beta + \\gamma = 180^\\circ - \\angle B A C,\n$$\nwhich completes the solution.\n\n![](attached_image_1.png)\nFigure 1\nAs in the previous solution, denote by $S$ the intersection point of the lines $B M$ and $N C$. Let moreover the circumcircle of the triangle $A B C$ intersect the lines $A P$ and $A Q$ again at $K$ and $L$, respectively (see Figure 2).\n\nNote that $\\angle L B C = \\angle L A C = \\angle C B A$ and similarly $\\angle K C B = \\angle K A B = \\angle B C A$. It implies that the lines $B L$ and $C K$ meet at a point $X$, being symmetric to the point $A$ with respect to the line $B C$. Since $A P = P M$ and $A Q = Q N$, it follows that $X$ lies on the line $M N$. Therefore, using Pascal's theorem for the hexagon $A L B S C K$, we infer that $S$ lies on the circumcircle of the triangle $A B C$, which finishes the proof.\n\n![](attached_image_2.png)\nFigure 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31993,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. The points $K$, $L$, and $M$ lie on the segments $BC$, $CA$, and $AB$, respectively, such that the lines $AK$, $BL$, and $CM$ intersect in a common point. Prove that it is possible to choose two of the triangles $ALM$, $BMK$, and $CKL$ whose inradii sum up to at least the inradius of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote\n$$\na = \\frac{BK}{KC}, \\quad b = \\frac{CL}{LA}, \\quad c = \\frac{AM}{MB}\n$$\nBy Ceva's theorem, $abc = 1$, so we may, without loss of generality, assume that $a \\geqslant 1$. Then at least one of the numbers $b$ or $c$ is not greater than $1$. Therefore at least one of the pairs $(a, b)$, $(b, c)$ has its first component not less than $1$ and the second one not greater than $1$. Without loss of generality, assume that $1 \\leqslant a$ and $b \\leqslant 1$.\n\nTherefore, we obtain $bc \\leqslant 1$ and $1 \\leqslant ca$, or equivalently\n$$\n\\frac{AM}{MB} \\leqslant \\frac{LA}{CL} \\quad \\text{and} \\quad \\frac{MB}{AM} \\leqslant \\frac{BK}{KC}\n$$\nThe first inequality implies that the line passing through $M$ and parallel to $BC$ intersects the segment $AL$ at a point $X$ (see Figure 1). Therefore the inradius of the triangle $ALM$ is not less than the inradius $r_1$ of triangle $AMX$.\n\nSimilarly, the line passing through $M$ and parallel to $AC$ intersects the segment $BK$ at a point $Y$, so the inradius of the triangle $BMK$ is not less than the inradius $r_2$ of the triangle $BMY$. Thus, to complete our solution, it is enough to show that $r_1 + r_2 \\geqslant r$, where $r$ is the inradius of the triangle $ABC$. We prove that in fact $r_1 + r_2 = r$.\n\n![](attached_image_1.png)\nFigure 1\n\nSince $MX \\parallel BC$, the dilation with centre $A$ that takes $M$ to $B$ takes the incircle of the triangle $AMX$ to the incircle of the triangle $ABC$. Therefore\n$$\n\\frac{r_1}{r} = \\frac{AM}{AB}, \\quad \\text{and similarly} \\quad \\frac{r_2}{r} = \\frac{MB}{AB}\n$$\nAdding these equalities gives $r_1 + r_2 = r$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31994,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $A B C$ with $A B > B C$. The angle bisector of $\\angle A B C$ intersects $\\Omega$ at $M \\neq B$. Let $\\Gamma$ be the circle with diameter $B M$. The angle bisectors of $\\angle A O B$ and $\\angle B O C$ intersect $\\Gamma$ at points $P$ and $Q$, respectively. The point $R$ is chosen on the line $P Q$ so that $B R = M R$. Prove that $B R \\parallel A C$. (Here we always assume that an angle bisector is a ray.)\n(Russia)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $K$ be the midpoint of $B M$, i.e., the centre of $\\Gamma$. Notice that $A B \\neq B C$ implies $K \\neq O$. Clearly, the lines $O M$ and $O K$ are the perpendicular bisectors of $A C$ and $B M$, respectively. Therefore, $R$ is the intersection point of $P Q$ and $O K$.\nLet $N$ be the second point of intersection of $\\Gamma$ with the line $O M$. Since $B M$ is a diameter of $\\Gamma$, the lines $B N$ and $A C$ are both perpendicular to $O M$. Hence $B N \\parallel A C$, and it suffices to prove that $B N$ passes through $R$. Our plan for doing this is to interpret the lines $B N, O K$, and $P Q$ as the radical axes of three appropriate circles.\nLet $\\omega$ be the circle with diameter $B O$. Since $\\angle B N O = \\angle B K O = 90^\\circ$, the points $N$ and $K$ lie on $\\omega$.\nNext we show that the points $O, K, P$, and $Q$ are concyclic. To this end, let $D$ and $E$ be the midpoints of $B C$ and $A B$, respectively. Clearly, $D$ and $E$ lie on the rays $O Q$ and $O P$, respectively. By our assumptions about the triangle $A B C$, the points $B, E, O, K$, and $D$ lie in this order on $\\omega$. It follows that $\\angle E O R = \\angle E B K = \\angle K B D = \\angle K O D$, so the line $K O$ externally bisects the angle $P O Q$. Since the point $K$ is the centre of $\\Gamma$, it also lies on the perpendicular bisector of $P Q$. So $K$ coincides with the midpoint of the $\\operatorname{arc} P O Q$ of the circumcircle $\\gamma$ of triangle $P O Q$.\nThus the lines $O K, B N$, and $P Q$ are pairwise radical axes of the circles $\\omega, \\gamma$, and $\\Gamma$. Hence they are concurrent at $R$, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31995,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a convex quadrilateral with $\\angle B = \\angle D = 90^{\\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. The points $S$ and $T$ are chosen on the sides $AB$ and $AD$, respectively, in such a way that $H$ lies inside triangle $SCT$ and\n$$\n\\angle SHC - \\angle BSC = 90^{\\circ}, \\quad \\angle THC - \\angle DTC = 90^{\\circ}.\n$$\nProve that the circumcircle of triangle $SHT$ is tangent to the line $BD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the line passing through $C$ and perpendicular to the line $SC$ intersect the line $AB$ at $Q$ (see Figure 1). Then\n$$\n\\angle SQC = 90^{\\circ} - \\angle BSC = 180^{\\circ} - \\angle SHC,\n$$\nwhich implies that the points $C$, $H$, $S$, and $Q$ lie on a common circle. Moreover, since $SQ$ is a diameter of this circle, we infer that the circumcentre $K$ of triangle $SHC$ lies on the line $AB$. Similarly, we prove that the circumcentre $L$ of triangle $CHT$ lies on the line $AD$.\n\n![](attached_image_1.png)\nFigure 1\n\nIn order to prove that the circumcircle of triangle $SHT$ is tangent to $BD$, it suffices to show that the perpendicular bisectors of $HS$ and $HT$ intersect on the line $AH$. However, these two perpendicular bisectors coincide with the angle bisectors of angles $AKH$ and $ALH$. Therefore, in order to complete the solution, it is enough (by the bisector theorem) to show that\n$$\n\\begin{equation*}\n\\frac{AK}{KH} = \\frac{AL}{LH} . \\tag{1}\n\\end{equation*}\n$$\nWe present two proofs of this equality.\n\nFirst proof. Let the lines $KL$ and $HC$ intersect at $M$ (see Figure 2). Since $KH = KC$ and $LH = LC$, the points $H$ and $C$ are symmetric to each other with respect to the line $KL$. Therefore $M$ is the midpoint of $HC$. Denote by $O$ the circumcentre of quadrilateral $ABCD$. Then $O$ is the midpoint of $AC$. Therefore we have $OM \\parallel AH$ and hence $OM \\perp BD$. This together with the equality $OB = OD$ implies that $OM$ is the perpendicular bisector of $BD$ and therefore $BM = DM$.\n\nSince $CM \\perp KL$, the points $B$, $C$, $M$, and $K$ lie on a common circle with diameter $KC$. Similarly, the points $L$, $C$, $M$, and $D$ lie on a circle with diameter $LC$. Thus, using the sine law, we obtain\n$$\n\\frac{AK}{AL} = \\frac{\\sin \\angle ALK}{\\sin \\angle AKL} = \\frac{DM}{CL} \\cdot \\frac{CK}{BM} = \\frac{CK}{CL} = \\frac{KH}{LH},\n$$\nwhich finishes the proof of (1).\n\n![](attached_image_2.png)\nFigure 2\n\nSecond proof. If the points $A$, $H$, and $C$ are collinear, then $AK = AL$ and $KH = LH$, so the equality (1) follows. Assume therefore that the points $A$, $H$, and $C$ do not lie in a line and consider the circle $\\omega$ passing through them (see Figure 3). Since the quadrilateral $ABCD$ is cyclic,\n$$\n\\angle BAC = \\angle BDC = 90^{\\circ} - \\angle ADH = \\angle HAD.\n$$\nLet $N \\neq A$ be the intersection point of the circle $\\omega$ and the angle bisector of $\\angle CAH$. Then $AN$ is also the angle bisector of $\\angle BAD$. Since $H$ and $C$ are symmetric to each other with respect to the line $KL$ and $HN = NC$, it follows that both $N$ and the centre of $\\omega$ lie on the line $KL$. This means that the circle $\\omega$ is an Apollonius circle of the points $K$ and $L$. This immediately yields (1).\n\n![](attached_image_3.png)\nFigure 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31996,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$. Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$, respectively. We call the pair $(E, F)$ interesting, if the quadrilateral $KSAT$ is cyclic.\nSuppose that the pairs $(E_1, F_1)$ and $(E_2, F_2)$ are interesting. Prove that\n$$\n\\frac{E_1E_2}{AB} = \\frac{F_1F_2}{AC}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. For any interesting pair $(E, F)$, we will say that the corresponding triangle $EFK$ is also interesting.\nLet $EFK$ be an interesting triangle. Firstly, we prove that $\\angle KEF = \\angle KFE = \\angle A$, which also means that the circumcircle $\\omega_1$ of the triangle $AEF$ is tangent to the lines $KE$ and $KF$.\nDenote by $\\omega$ the circle passing through the points $K, S, A$, and $T$. Let the line $AM$ intersect the line $ST$ and the circle $\\omega$ (for the second time) at $N$ and $L$, respectively (see Figure 1).\nSince $EF \\parallel TS$ and $M$ is the midpoint of $EF$, $N$ is the midpoint of $ST$. Moreover, since $K$ and $M$ are symmetric to each other with respect to the line $ST$, we have $\\angle KNS = \\angle MNS = \\angle LNT$. Thus the points $K$ and $L$ are symmetric to each other with respect to the perpendicular bisector of $ST$. Therefore $KL \\parallel ST$.\nLet $G$ be the point symmetric to $K$ with respect to $N$. Then $G$ lies on the line $EF$, and we may assume that it lies on the ray $MF$. One has\n$$\n\\angle KGE = \\angle KNS = \\angle SNM = \\angle KLA = 180^\\circ - \\angle KSA\n$$\n(if $K = L$, then the angle $KLA$ is understood to be the angle between $AL$ and the tangent to $\\omega$ at $L$). This means that the points $K, G, E$, and $S$ are concyclic. Now, since $KSGT$ is a parallelogram, we obtain $\\angle KEF = \\angle KSG = 180^\\circ - \\angle TKS = \\angle A$. Since $KE = KF$, we also have $\\angle KFE = \\angle KEF = \\angle A$.\n\nFirst method. We have just proved that all interesting triangles are similar to each other. This allows us to use the following lemma.\n**Lemma.** Let $ABC$ be an arbitrary triangle. Choose two points $E_1$ and $E_2$ on the side $AC$, two points $F_1$ and $F_2$ on the side $AB$, and two points $K_1$ and $K_2$ on the side $BC$, in a way that the triangles $E_1F_1K_1$ and $E_2F_2K_2$ are similar. Then the six circumcircles of the triangles $AE_iF_i$, $BF_iK_i$, and $CE_iK_i$ $(i=1,2)$ meet at a common point $Z$. Moreover, $Z$ is the centre of the spiral similarity that takes the triangle $E_1F_1K_1$ to the triangle $E_2F_2K_2$.\n**Proof.** Firstly, notice that for each $i=1,2$, the circumcircles of the triangles $AE_iF_i$, $BF_iK_i$, and $CK_iE_i$ have a common point $Z_i$ by Miquel's theorem. Moreover, we have\n$\\nsucceq(Z_iF_i, Z_iE_i) = \\Varangle(AB, CA), \\quad \\Varangle(Z_iK_i, Z_iF_i) = \\Varangle(BC, AB), \\quad \\nprec(Z_iE_i, Z_iK_i) = \\Varangle(CA, BC)$.\nThis yields that the points $Z_1$ and $Z_2$ correspond to each other in similar triangles $E_1F_1K_1$ and $E_2F_2K_2$. Thus, if they coincide, then this common point is indeed the desired centre of a spiral similarity.\nFinally, in order to show that $Z_1 = Z_2$, one may notice that $\\nsucceq(AB, AZ_1) = \\Varangle(E_1F_1, E_1Z_1) = \\Varangle(E_2F_2, E_2Z_2) = \\Varangle(AB, AZ_2)$ (see Figure 2). Similarly, one has $\\Varangle(BC, BZ_1) = \\Varangle(BC, BZ_2)$ and $\\nsucceq(CA, CZ_1) = \\Varangle(CA, CZ_2)$. This yields $Z_1 = Z_2$.\nNow, let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $AC$ and $AB$, respectively, and let $R$ be the midpoint of $BC$ (see Figure 3). Then $R$ is the circumcentre of the cyclic quadrilateral $BCPQ$. Thus we obtain $\\angle APQ = \\angle B$ and $\\angle RPC = \\angle C$, which yields $\\angle QPR = \\angle A$. Similarly, we show that $\\angle PQR = \\angle A$. Thus, all interesting triangles are similar to the triangle $PQR$.\n![](attached_image_1.png)\nFigure 3\n![](attached_image_2.png)\nFigure 4\nDenote now by $Z$ the common point of the circumcircles of $APQ$, $BQR$, and $CPR$. Let $E_1F_1K_1$ and $E_2F_2K_2$ be two interesting triangles. By the lemma, $Z$ is the centre of any spiral similarity taking one of the triangles $E_1F_1K_1$, $E_2F_2K_2$, and $PQR$ to some other of them. Therefore the triangles $ZE_1E_2$ and $ZF_1F_2$ are similar, as well as the triangles $ZE_1F_1$ and $ZPQ$. Hence\n$$\n\\frac{E_1E_2}{F_1F_2} = \\frac{ZE_1}{ZF_1} = \\frac{ZP}{ZQ}\n$$\nMoreover, the equalities $\\angle AZQ = \\angle APQ = \\angle ABC = 180^\\circ - \\angle QZR$ show that the point $Z$ lies on the line $AR$ (see Figure 4). Therefore the triangles $AZP$ and $ACR$ are similar, as well as the triangles $AZQ$ and $ABR$. This yields\n$$\n\\frac{ZP}{ZQ} = \\frac{ZP}{RC} \\cdot \\frac{RB}{ZQ} = \\frac{AZ}{AC} \\cdot \\frac{AB}{AZ} = \\frac{AB}{AC}\n$$\nwhich completes the solution.\n\nSecond method. Now we will start from the fact that $\\omega_1$ is tangent to the lines $KE$ and $KF$ (see Figure 5). We prove that if $(E, F)$ is an interesting pair, then\n$$\n\\begin{equation*}\n\\frac{AE}{AB} + \\frac{AF}{AC} = 2 \\cos \\angle A. \\tag{1}\n\\end{equation*}\n$$\nLet $Y$ be the intersection point of the segments $BE$ and $CF$. The points $B, K$, and $C$ are collinear, hence applying Pascal's theorem to the degenerated hexagon $AFFYEE$, we infer that $Y$ lies on the circle $\\omega_1$.\nDenote by $Z$ the second intersection point of the circumcircle of the triangle $BFY$ with the line $BC$ (see Figure 6). By Miquel's theorem, the points $C, Z, Y$, and $E$ are concyclic. Therefore we obtain\n$$\nBF \\cdot AB + CE \\cdot AC = BY \\cdot BE + CY \\cdot CF = BZ \\cdot BC + CZ \\cdot BC = BC^2.\n$$\nOn the other hand, $BC^2 = AB^2 + AC^2 - 2AB \\cdot AC \\cos \\angle A$, by the cosine law. Hence\n$$\n(AB - AF) \\cdot AB + (AC - AE) \\cdot AC = AB^2 + AC^2 - 2AB \\cdot AC \\cos \\angle A,\n$$\nwhich simplifies to the desired equality (1).\nLet now $(E_1, F_1)$ and $(E_2, F_2)$ be two interesting pairs of points. Then we get\n$$\n\\frac{AE_1}{AB} + \\frac{AF_1}{AC} = \\frac{AE_2}{AB} + \\frac{AF_2}{AC},\n$$\nwhich gives the desired result.\n\nThird method. Again, we make use of the fact that all interesting triangles are similar (and equi-oriented). Let us put the picture onto a complex plane such that $A$ is at the origin, and identify each point with the corresponding complex number.\nLet $EFK$ be any interesting triangle. The equalities $\\angle KEF = \\angle KFE = \\angle A$ yield that the ratio $\\nu = \\frac{K-E}{F-E}$ is the same for all interesting triangles. This in turn means that the numbers $E$, $F$, and $K$ satisfy the linear equation\n$$\n\\begin{equation*}\nK = \\mu E + \\nu F, \\quad \\text{ where } \\quad \\mu = 1 - \\nu. \\tag{2}\n\\end{equation*}\n$$\nNow let us choose the points $X$ and $Y$ on the rays $AB$ and $AC$, respectively, so that $\\angle CXA = \\angle AYB = \\angle A = \\angle KEF$ (see Figure 7). Then each of the triangles $AXC$ and $YAB$ is similar to any interesting triangle, which also means that\n$$\n\\begin{equation*}\nC = \\mu A + \\nu X = \\nu X \\quad \\text{ and } \\quad B = \\mu Y + \\nu A = \\mu Y. \\tag{3}\n\\end{equation*}\n$$\nMoreover, one has $X / Y = \\overline{C / B}$.\nSince the points $E, F$, and $K$ lie on $AC, AB$, and $BC$, respectively, one gets\n$$\nE = \\rho Y, \\quad F = \\sigma X, \\quad \\text{ and } \\quad K = \\lambda B + (1 - \\lambda) C\n$$\nfor some real $\\rho, \\sigma$, and $\\lambda$. In view of (3), the equation (2) now reads $\\lambda B + (1 - \\lambda) C = K = \\mu E + \\nu F = \\rho B + \\sigma C$, or\n$$\n(\\lambda - \\rho) B = (\\sigma + \\lambda - 1) C.\n$$\nSince the nonzero complex numbers $B$ and $C$ have different arguments, the coefficients in the brackets vanish, so $\\rho = \\lambda$ and $\\sigma = 1 - \\lambda$. Therefore,\n$$\n\\begin{equation*}\n\\frac{E}{Y} + \\frac{F}{X} = \\rho + \\sigma = 1. \\tag{4}\n\\end{equation*}\n$$\nNow, if $(E_1, F_1)$ and $(E_2, F_2)$ are two distinct interesting pairs, one may apply (4) to both pairs. Subtracting, we get\n$$\n\\frac{E_1 - E_2}{Y} = \\frac{F_2 - F_1}{X}, \\quad \\text{ so } \\quad \\frac{E_1 - E_2}{F_2 - F_1} = \\frac{Y}{X} = \\frac{\\bar{B}}{\\bar{C}}.\n$$\nTaking absolute values provides the required result.\nSolution 2. Let $(E, F)$ be an interesting pair. This time we prove that\n$$\n\\begin{equation*}\n\\frac{AM}{AK} = \\cos \\angle A. \\tag{5}\n\\end{equation*}\n$$\nAs in Solution 1, we introduce the circle $\\omega$ passing through the points $K, S, A$, and $T$, together with the points $N$ and $L$ at which the line $AM$ intersect the line $ST$ and the circle $\\omega$ for the second time, respectively. Let moreover $O$ be the centre of $\\omega$ (see Figures 8 and 9). As in Solution 1, we note that $N$ is the midpoint of $ST$ and show that $KL \\parallel ST$, which implies $\\angle FAM = \\angle EAK$.\n![](attached_image_3.png)\nFigure 8\n![](attached_image_4.png)\nFigure 9\nSuppose now that $K \\neq L$ (see Figure 8). Then $KL \\parallel ST$, and consequently the lines $KM$ and $KL$ are perpendicular. It implies that the lines $LO$ and $KM$ meet at a point $X$ lying on the circle $\\omega$. Since the lines $ON$ and $XM$ are both perpendicular to the line $ST$, they are parallel to each other, and hence $\\angle LON = \\angle LXK = \\angle MAK$. On the other hand, $\\angle OLN = \\angle MKA$, so we infer that triangles $NOL$ and $MAK$ are similar. This yields\n$$\n\\frac{AM}{AK} = \\frac{ON}{OL} = \\frac{ON}{OT} = \\cos \\angle TON = \\cos \\angle A.\n$$\nIf, on the other hand, $K = L$, then the points $A, M, N$, and $K$ lie on a common line, and this line is the perpendicular bisector of $ST$ (see Figure 9). This implies that $AK$ is a diameter of $\\omega$, which yields $AM = 2OK - 2NK = 2ON$. So also in this case we obtain\n$$\n\\frac{AM}{AK} = \\frac{2ON}{2OT} = \\cos \\angle TON = \\cos \\angle A.\n$$\n\nThus (5) is proved.\nLet $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $AC$ and $AB$, respectively (see Figure 10). We claim that the point $M$ lies on the line $PQ$. Consider now the composition of the dilatation with factor $\\cos \\angle A$ and centre $A$, and the reflection with respect to the angle bisector of $\\angle BAC$. This transformation is a similarity that takes $B, C$, and $K$ to $P, Q$, and $M$, respectively. Since $K$ lies on the line $BC$, the point $M$ lies on the line $PQ$.\n![](attached_image_5.png)\nFigure 10",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31997,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with circumcircle $\\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $CI$ intersect the segment $BC$ and the arc $BC$ (not containing $A$) of $\\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $AI$ intersect $AV$ at $X$, and let the line passing through $V$ and parallel to $AI$ intersect $AB$ at $Y$. Let $W$ and $Z$ be the midpoints of $AX$ and $BC$, respectively. Prove that if the points $I$, $X$, and $Y$ are collinear, then the points $I$, $W$, and $Z$ are also collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start with some general observations. Set $\\alpha = \\angle A / 2$, $\\beta = \\angle B / 2$, $\\gamma = \\angle C / 2$. Then obviously $\\alpha + \\beta + \\gamma = 90^{\\circ}$. Since $\\angle UIC = 90^{\\circ}$, we obtain $\\angle IUC = \\alpha + \\beta$. Therefore $\\angle BIV = \\angle IUC - \\angle IBC = \\alpha = \\angle BAI = \\angle BYV$, which implies that the points $B$, $Y$, $I$, and $V$ lie on a common circle (see Figure 1).\n\nAssume now that the points $I$, $X$ and $Y$ are collinear. We prove that $\\angle YIA = 90^{\\circ}$.\nLet the line $XU$ intersect $AB$ at $N$. Since the lines $AI$, $UX$, and $VY$ are parallel, we get\n$$\n\\frac{NX}{AI} = \\frac{YN}{YA} = \\frac{VU}{VI} = \\frac{XU}{AI},\n$$\nimplying $NX = XU$. Moreover, $\\angle BIU = \\alpha = \\angle BNU$. This implies that the quadrilateral $BUIN$ is cyclic, and since $BI$ is the angle bisector of $\\angle UBN$, we infer that $NI = UI$. Thus in the isosceles triangle $NIU$, the point $X$ is the midpoint of the base $NU$. This gives $\\angle IXN = 90^{\\circ}$, i.e., $\\angle YIA = 90^{\\circ}$.\n\n![](attached_image_1.png)\nFigure 1\n\nLet $S$ be the midpoint of the segment $VC$. Let moreover $T$ be the intersection point of the lines $AX$ and $SI$, and set $x = \\angle BAV = \\angle BCV$. Since $\\angle CIA = 90^{\\circ} + \\beta$ and $SI = SC$, we obtain\n$$\n\\angle TIA = 180^{\\circ} - \\angle AIS = 90^{\\circ} - \\beta - \\angle CIS = 90^{\\circ} - \\beta - \\gamma - x = \\alpha - x = \\angle TAI,\n$$\nwhich implies that $TI = TA$. Therefore, since $\\angle XIA = 90^{\\circ}$, the point $T$ is the midpoint of $AX$, i.e., $T = W$.\n\nTo complete our solution, it remains to show that the intersection point of the lines $IS$ and $BC$ coincide with the midpoint of the segment $BC$. But since $S$ is the midpoint of the segment $VC$, it suffices to show that the lines $BV$ and $IS$ are parallel.\n\nSince the quadrilateral $BYIV$ is cyclic, $\\angle VBI = \\angle VYI = \\angle YIA = 90^{\\circ}$. This implies that $BV$ is the external angle bisector of the angle $ABC$, which yields $\\angle VAC = \\angle VCA$. Therefore $2\\alpha - x = 2\\gamma + x$, which gives $\\alpha = \\gamma + x$. Hence $\\angle SCI = \\alpha$, so $\\angle VSI = 2\\alpha$.\n\nOn the other hand, $\\angle BVC = 180^{\\circ} - \\angle BAC = 180^{\\circ} - 2\\alpha$, which implies that the lines $BV$ and $IS$ are parallel. This completes the solution.\nAs in Solution 1, we first prove that the points $B$, $Y$, $I$, $V$ lie on a common circle and $\\angle YIA = 90^{\\circ}$. The remaining part of the solution is based on the following lemma, which holds true for any triangle $ABC$, not necessarily with the property that $I$, $X$, $Y$ are collinear.\n\nLemma. Let $ABC$ be the triangle inscribed in a circle $\\Gamma$ and let $I$ be its incentre. Assume that the line passing through $I$ and perpendicular to the line $AI$ intersects the side $AB$ at the point $Y$. Let the circumcircle of the triangle $BYI$ intersect the circle $\\Gamma$ for the second time at $V$, and let the excircle of the triangle $ABC$ opposite to the vertex $A$ be tangent to the side $BC$ at $E$. Then\n$$\n\\angle BAV = \\angle CAE.\n$$\nProof. Let $\\rho$ be the composition of the inversion with centre $A$ and radius $\\sqrt{AB \\cdot AC}$, and the symmetry with respect to $AI$. Clearly, $\\rho$ interchanges $B$ and $C$.\n\nLet $J$ be the excentre of the triangle $ABC$ opposite to $A$ (see Figure 2). Then we have $\\angle JAC = \\angle BAI$ and $\\angle JCA = 90^{\\circ} + \\gamma = \\angle BIA$, so the triangles $ACJ$ and $AIB$ are similar, and therefore $AB \\cdot AC = AI \\cdot AJ$. This means that $\\rho$ interchanges $I$ and $J$. Moreover, since $Y$ lies on $AB$ and $\\angle AIY = 90^{\\circ}$, the point $Y' = \\rho(Y)$ lies on $AC$, and $\\angle JY'A = 90^{\\circ}$. Thus $\\rho$ maps the circumcircle $\\gamma$ of the triangle $BYI$ to a circle $\\gamma'$ with diameter $JC$.\n\nFinally, since $V$ lies on both $\\Gamma$ and $\\gamma$, the point $V' = \\rho(V)$ lies on the line $\\rho(\\Gamma) = AB$ as well as on $\\gamma'$, which in turn means that $V' = E$. This implies the desired result.\n\n![](attached_image_2.png)\n\nNow we turn to the solution of the problem.\nAssume that the incircle $\\omega_1$ of the triangle $ABC$ is tangent to $BC$ at $D$, and let the excircle $\\omega_2$ of the triangle $ABC$ opposite to the vertex $A$ touch the side $BC$ at $E$ (see Figure 3). The homothety with centre $A$ that takes $\\omega_2$ to $\\omega_1$ takes the point $E$ to some point $F$, and the tangent to $\\omega_1$ at $F$ is parallel to $BC$. Therefore $DF$ is a diameter of $\\omega_1$. Moreover, $Z$ is the midpoint of $DE$. This implies that the lines $IZ$ and $FE$ are parallel.\n\nLet $K = YI \\cap AE$. Since $\\angle YIA = 90^{\\circ}$, the lemma yields that $I$ is the midpoint of $XK$. This implies that the segments $IW$ and $AK$ are parallel. Therefore, the points $W$, $I$ and $Z$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 31998,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1} < a_{2} < \\cdots < a_{n}$ be pairwise coprime positive integers with $a_{1}$ being prime and $a_{1} \\geqslant n+2$. On the segment $I = [0, a_{1} a_{2} \\cdots a_{n}]$ of the real line, mark all integers that are divisible by at least one of the numbers $a_{1}, \\ldots, a_{n}$. These points split $I$ into a number of smaller segments. Prove that the sum of the squares of the lengths of these segments is divisible by $a_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A = a_{1} \\cdots a_{n}$. Throughout the solution, all intervals will be nonempty and have integer end-points. For any interval $X$, the length of $X$ will be denoted by $|X|$.\n\nDefine the following two families of intervals:\n$$\n\\begin{aligned}\n\\mathcal{S} & = \\{[x, y]: x < y \\text{ are consecutive marked points}\\} \\\\\n\\mathcal{T} & = \\{[x, y]: x < y \\text{ are integers, } 0 \\leqslant x \\leqslant A-1, \\text{ and no point is marked in } (x, y)\\}\n\\end{aligned}\n$$\nWe are interested in computing $\\sum_{X \\in \\mathcal{S}} |X|^{2}$ modulo $a_{1}$.\n\nNote that the number $A$ is marked, so in the definition of $\\mathcal{T}$ the condition $y \\leqslant A$ is enforced without explicitly prescribing it.\n\nAssign weights to the intervals in $\\mathcal{T}$, depending only on their lengths. The weight of an arbitrary interval $Y \\in \\mathcal{T}$ will be $w(|Y|)$, where\n$$\nw(k) = \\begin{cases} 1 & \\text{ if } k = 1, \\\\ 2 & \\text{ if } k \\geqslant 2. \\end{cases}\n$$\n\nConsider an arbitrary interval $X \\in \\mathcal{S}$ and its sub-intervals $Y \\in \\mathcal{T}$. Clearly, $X$ has one sub-interval of length $|X|$, two sub-intervals of length $|X|-1$ and so on; in general $X$ has $|X|-d+1$ sub-intervals of length $d$ for every $d = 1, 2, \\ldots, |X|$. The sum of the weights of the sub-intervals of $X$ is\n$$\n\\sum_{Y \\in \\mathcal{T}, Y \\subseteq X} w(|Y|) = \\sum_{d=1}^{|X|} (|X|-d+1) \\cdot w(d) = |X| \\cdot 1 + ((|X|-1)+(|X|-2)+\\cdots+1) \\cdot 2 = |X|^{2}.\n$$\nSince the intervals in $\\mathcal{S}$ are non-overlapping, every interval $Y \\in \\mathcal{T}$ is a sub-interval of a single interval $X \\in \\mathcal{S}$. Therefore,\n$$\n\\begin{equation*}\n\\sum_{X \\in \\mathcal{S}} |X|^{2} = \\sum_{X \\in \\mathcal{S}} \\left( \\sum_{Y \\in \\mathcal{T}, Y \\subseteq X} w(|Y|) \\right) = \\sum_{Y \\in \\mathcal{T}} w(|Y|). \\tag{1}\n\\end{equation*}\n$$\n\nFor every $d = 1, 2, \\ldots, a_{1}$, we count how many intervals in $\\mathcal{T}$ are of length $d$. Notice that the multiples of $a_{1}$ are all marked, so the lengths of the intervals in $\\mathcal{S}$ and $\\mathcal{T}$ cannot exceed $a_{1}$. Let $x$ be an arbitrary integer with $0 \\leqslant x \\leqslant A-1$ and consider the interval $[x, x+d]$. Let $r_{1}, \\ldots, r_{n}$ be the remainders of $x$ modulo $a_{1}, \\ldots, a_{n}$, respectively. Since $a_{1}, \\ldots, a_{n}$ are pairwise coprime, the number $x$ is uniquely identified by the sequence $(r_{1}, \\ldots, r_{n})$, due to the Chinese remainder theorem.\n\nFor every $i = 1, \\ldots, n$, the property that the interval $(x, x+d)$ does not contain any multiple of $a_{i}$ is equivalent with $r_{i} + d \\leqslant a_{i}$, i.e. $r_{i} \\in \\{0, 1, \\ldots, a_{i} - d\\}$, so there are $a_{i} - d + 1$ choices for the number $r_{i}$ for each $i$. Therefore, the number of the remainder sequences $(r_{1}, \\ldots, r_{n})$ that satisfy $[x, x+d] \\in \\mathcal{T}$ is precisely $(a_{1} + 1 - d) \\cdots (a_{n} + 1 - d)$. Denote this product by $f(d)$.\n\nNow we can group the last sum in (1) by length of the intervals. As we have seen, for every $d = 1, \\ldots, a_{1}$ there are $f(d)$ intervals $Y \\in \\mathcal{T}$ with $|Y| = d$. Therefore, (1) can be continued as\n$$\n\\begin{equation*}\n\\sum_{X \\in \\mathcal{S}} |X|^{2} = \\sum_{Y \\in \\mathcal{T}} w(|Y|) = \\sum_{d=1}^{a_{1}} f(d) \\cdot w(d) = 2 \\sum_{d=1}^{a_{1}} f(d) - f(1). \\tag{2}\n\\end{equation*}\n$$\n\nHaving the formula (2), the solution can be finished using the following well-known fact:\n\n**Lemma.** If $p$ is a prime, $F(x)$ is a polynomial with integer coefficients, and $\\deg F \\leqslant p-2$, then $\\sum_{x=1}^{p} F(x)$ is divisible by $p$.\n\nProof. Obviously, it is sufficient to prove the lemma for monomials of the form $x^{k}$ with $k \\leqslant p-2$. Apply induction on $k$. If $k=0$ then $F=1$, and the statement is trivial.\n\nLet $1 \\leqslant k \\leqslant p-2$, and assume that the lemma is proved for all lower degrees. Then\n$$\n\\begin{aligned}\n0 & \\equiv p^{k+1} = \\sum_{x=1}^{p} \\left( x^{k+1} - (x-1)^{k+1} \\right) = \\sum_{x=1}^{p} \\left( \\sum_{\\ell=0}^{k} (-1)^{k-\\ell} \\binom{k+1}{\\ell} x^{\\ell} \\right) \\\\\n& = (k+1) \\sum_{x=1}^{p} x^{k} + \\sum_{\\ell=0}^{k-1} (-1)^{k-\\ell} \\binom{k+1}{\\ell} \\sum_{x=1}^{p} x^{\\ell} \\equiv (k+1) \\sum_{x=1}^{p} x^{k} \\pmod{p}\n\\end{aligned}\n$$\nSince $0 < k+1 < p$, this proves $\\sum_{x=1}^{p} x^{k} \\equiv 0 \\pmod{p}$.\n\nIn (2), by applying the lemma to the polynomial $f$ and the prime $a_{1}$, we obtain that $\\sum_{d=1}^{a_{1}} f(d)$ is divisible by $a_{1}$. The term $f(1) = a_{1} \\cdots a_{n}$ is also divisible by $a_{1}$; these two facts together prove that $\\sum_{X \\in \\mathcal{S}} |X|^{2}$ is divisible by $a_{1}$.\nThe conventions from the first paragraph of the first solution are still in force. We shall prove the following more general statement:\n\n(⊞) Let $p$ denote a prime number, let $p = a_{1} < a_{2} < \\cdots < a_{n}$ be $n$ pairwise coprime positive integers, and let $d$ be an integer with $1 \\leqslant d \\leqslant p-n$. Mark all integers that are divisible by at least one of the numbers $a_{1}, \\ldots, a_{n}$ on the interval $I = [0, a_{1} a_{2} \\cdots a_{n}]$ of the real line. These points split $I$ into a number of smaller segments, say of lengths $b_{1}, \\ldots, b_{k}$. Then the sum $\\sum_{i=1}^{k} \\binom{b_{i}}{d}$ is divisible by $p$.\n\nApplying $(\\boxplus)$ to $d=1$ and $d=2$ and using the equation $x^{2} = 2 \\binom{x}{2} + \\binom{x}{1}$, one easily gets the statement of the problem.\n\nTo prove $(\\boxplus)$ itself, we argue by induction on $n$. The base case $n=1$ follows from the known fact that the binomial coefficient $\\binom{p}{d}$ is divisible by $p$ whenever $1 \\leqslant d \\leqslant p-1$.\n\nLet us now assume that $n \\geqslant 2$, and that the statement is known whenever $n-1$ rather than $n$ coprime integers are given together with some integer $d \\in [1, p-n+1]$. Suppose that the numbers $p = a_{1} < a_{2} < \\cdots < a_{n}$ and $d$ are as above. Write $A' = \\prod_{i=1}^{n-1} a_{i}$ and $A = A' a_{n}$. Mark the points on the real axis divisible by one of the numbers $a_{1}, \\ldots, a_{n-1}$ green and those divisible by $a_{n}$ red. The green points divide $[0, A']$ into certain sub-intervals, say $J_{1}, J_{2}, \\ldots, J_{\\ell}$.\n\nTo translate intervals we use the notation $[a, b] + m = [a + m, b + m]$ whenever $a, b, m \\in \\mathbb{Z}$.\n\nFor each $i \\in \\{1, 2, \\ldots, \\ell\\}$ let $\\mathcal{F}_{i}$ be the family of intervals into which the red points partition the intervals $J_{i}, J_{i} + A', \\ldots, J_{i} + (a_{n} - 1) A'$. We are to prove that\n$$\n\\sum_{i=1}^{\\ell} \\sum_{X \\in \\mathcal{F}_{i}} \\binom{|X|}{d}\n$$\nis divisible by $p$.\n\nLet us fix any index $i$ with $1 \\leqslant i \\leqslant \\ell$ for a while. Since the numbers $A'$ and $a_{n}$ are coprime by hypothesis, the numbers $0, A', \\ldots, (a_{n} - 1) A'$ form a complete system of residues modulo $a_{n}$. Moreover, we have $|J_{i}| \\leqslant p < a_{n}$, as in particular all multiples of $p$ are green. So each of the intervals $J_{i}, J_{i} + A', \\ldots, J_{i} + (a_{n} - 1) A'$ contains at most one red point. More precisely, for each $j \\in \\{1, \\ldots, |J_{i}| - 1\\}$ there is exactly one amongst those intervals containing a red point splitting it into an interval of length $j$ followed by an interval of length $|J_{i}| - j$, while the remaining $a_{n} - |J_{i}| + 1$ such intervals have no red points in their interiors. For these reasons\n$$\n\\begin{aligned}\n\\sum_{X \\in \\mathcal{F}_{i}} \\binom{|X|}{d} & = 2 \\left( \\binom{1}{d} + \\cdots + \\binom{|J_{i}| - 1}{d} \\right) + (a_{n} - |J_{i}| + 1) \\binom{|J_{i}|}{d} \\\\\n& = 2 \\binom{|J_{i}|}{d+1} + (a_{n} - d + 1) \\binom{|J_{i}|}{d} - (d+1) \\binom{|J_{i}|}{d+1} \\\\\n& = (1 - d) \\binom{|J_{i}|}{d+1} + (a_{n} - d + 1) \\binom{|J_{i}|}{d}\n\\end{aligned}\n$$\nSo it remains to prove that\n$$\n(1 - d) \\sum_{i=1}^{\\ell} \\binom{|J_{i}|}{d+1} + (a_{n} - d + 1) \\sum_{i=1}^{\\ell} \\binom{|J_{i}|}{d}\n$$\nis divisible by $p$. By the induction hypothesis, however, it is even true that both summands are divisible by $p$, for $1 \\leqslant d < d+1 \\leqslant p - (n-1)$. This completes the proof of $(\\boxplus)$ and hence the solution of the problem.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 31999,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $c \\geqslant 1$ be an integer. Define a sequence of positive integers by $a_{1}=c$ and\n$$\na_{n+1}=a_{n}^{3}-4 c \\cdot a_{n}^{2}+5 c^{2} \\cdot a_{n}+c\n$$\nfor all $n \\geqslant 1$. Prove that for each integer $n \\geqslant 2$ there exists a prime number $p$ dividing $a_{n}$ but none of the numbers $a_{1}, \\ldots, a_{n-1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us define $x_{0}=0$ and $x_{n}=a_{n} / c$ for all integers $n \\geqslant 1$. It is easy to see that the sequence ($x_{n}$) thus obtained obeys the recursive law\n$$\n\\begin{equation*}\nx_{n+1}=c^{2}\\left(x_{n}^{3}-4 x_{n}^{2}+5 x_{n}\\right)+1 \\tag{1}\n\\end{equation*}\n$$\nfor all integers $n \\geqslant 0$. In particular, all of its terms are positive integers; notice that $x_{1}=1$ and $x_{2}=2 c^{2}+1$. Since\n$$\n\\begin{equation*}\nx_{n+1}=c^{2} x_{n}\\left(x_{n}-2\\right)^{2}+c^{2} x_{n}+1>x_{n} \\tag{2}\n\\end{equation*}\n$$\nholds for all integers $n \\geqslant 0$, it is also strictly increasing. Since $x_{n+1}$ is by (1) coprime to $c$ for any $n \\geqslant 0$, it suffices to prove that for each $n \\geqslant 2$ there exists a prime number $p$ dividing $x_{n}$ but none of the numbers $x_{1}, \\ldots, x_{n-1}$. Let us begin by establishing three preliminary claims.\n\n**Claim 1.** If $i \\equiv j(\\bmod m)$ holds for some integers $i, j \\geqslant 0$ and $m \\geqslant 1$, then $x_{i} \\equiv x_{j}\\left(\\bmod x_{m}\\right)$ holds as well.\n\n*Proof.* Evidently, it suffices to show $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}\\right)$ for all integers $i \\geqslant 0$ and $m \\geqslant 1$. For this purpose we may argue for fixed $m$ by induction on $i$ using $x_{0}=0$ in the base case $i=0$. Now, if we have $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}\\right)$ for some integer $i$, then the recursive equation (1) yields\n$$\nx_{i+m+1} \\equiv c^{2}\\left(x_{i+m}^{3}-4 x_{i+m}^{2}+5 x_{i+m}\\right)+1 \\equiv c^{2}\\left(x_{i}^{3}-4 x_{i}^{2}+5 x_{i}\\right)+1 \\equiv x_{i+1}\\left(\\bmod x_{m}\\right),\n$$\nwhich completes the induction. $\\square$\n\n**Claim 2.** If the integers $i, j \\geqslant 2$ and $m \\geqslant 1$ satisfy $i \\equiv j(\\bmod m)$, then $x_{i} \\equiv x_{j}\\left(\\bmod x_{m}^{2}\\right)$ holds as well.\n\n*Proof.* Again it suffices to prove $x_{i+m} \\equiv x_{i}\\left(\\bmod x_{m}^{2}\\right)$ for all integers $i \\geqslant 2$ and $m \\geqslant 1$. As above, we proceed for fixed $m$ by induction on $i$. The induction step is again easy using (1), but this time the base case $i=2$ requires some calculation. Set $L=5 c^{2}$. By (1) we have $x_{m+1} \\equiv L x_{m}+1\\left(\\bmod x_{m}^{2}\\right)$, and hence\n$$\n\\begin{aligned}\nx_{m+1}^{3}-4 x_{m+1}^{2}+5 x_{m+1} & \\equiv\\left(L x_{m}+1\\right)^{3}-4\\left(L x_{m}+1\\right)^{2}+5\\left(L x_{m}+1\\right) \\\\\n& \\equiv\\left(3 L x_{m}+1\\right)-4\\left(2 L x_{m}+1\\right)+5\\left(L x_{m}+1\\right) \\equiv 2 \\quad\\left(\\bmod x_{m}^{2}\\right)\n\\end{aligned}\n$$\nwhich in turn gives indeed $x_{m+2} \\equiv 2 c^{2}+1 \\equiv x_{2}\\left(\\bmod x_{m}^{2}\\right)$. $\\square$\n\n**Claim 3.** For each integer $n \\geqslant 2$, we have $x_{n}>x_{1} \\cdot x_{2} \\cdots x_{n-2}$.\n\n*Proof.* The cases $n=2$ and $n=3$ are clear. Arguing inductively, we assume now that the claim holds for some $n \\geqslant 3$. Recall that $x_{2} \\geqslant 3$, so by monotonicity and (2) we get $x_{n} \\geqslant x_{3} \\geqslant x_{2}\\left(x_{2}-2\\right)^{2}+x_{2}+1 \\geqslant 7$. It follows that\n$$\nx_{n+1}>x_{n}^{3}-4 x_{n}^{2}+5 x_{n}>7 x_{n}^{2}-4 x_{n}^{2}>x_{n}^{2}>x_{n} x_{n-1},\n$$\nwhich by the induction hypothesis yields $x_{n+1}>x_{1} \\cdot x_{2} \\cdots x_{n-1}$, as desired. $\\square$\n\nNow we direct our attention to the problem itself: let any integer $n \\geqslant 2$ be given. By Claim 3 there exists a prime number $p$ appearing with a higher exponent in the prime factorisation of $x_{n}$ than in the prime factorisation of $x_{1} \\cdots x_{n-2}$. In particular, $p \\mid x_{n}$, and it suffices to prove that $p$ divides none of $x_{1}, \\ldots, x_{n-1}$.\n\nOtherwise let $k \\in\\{1, \\ldots, n-1\\}$ be minimal such that $p$ divides $x_{k}$. Since $x_{n-1}$ and $x_{n}$ are coprime by (1) and $x_{1}=1$, we actually have $2 \\leqslant k \\leqslant n-2$. Write $n=q k+r$ with some integers $q \\geqslant 0$ and $0 \\leqslant r<k$. By Claim 1 we have $x_{n} \\equiv x_{r}\\left(\\bmod x_{k}\\right)$, whence $p \\mid x_{r}$. Due to the minimality of $k$ this entails $r=0$, i.e. $k \\mid n$.\n\nThus from Claim 2 we infer\n$$\nx_{n} \\equiv x_{k} \\quad\\left(\\bmod x_{k}^{2}\\right)\n$$\nNow let $\\alpha \\geqslant 1$ be maximal with the property $p^{\\alpha} \\mid x_{k}$. Then $x_{k}^{2}$ is divisible by $p^{\\alpha+1}$ and by our choice of $p$ so is $x_{n}$. So by the previous congruence $x_{k}$ is a multiple of $p^{\\alpha+1}$ as well, contrary to our choice of $\\alpha$. This is the final contradiction concluding the solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32000,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every real number $x$, let $\\|x\\|$ denote the distance between $x$ and the nearest integer. Prove that for every pair $(a, b)$ of positive integers there exist an odd prime $p$ and a positive integer $k$ satisfying\n$$\n\\left\\|\\frac{a}{p^{k}}\\right\\|+\\left\\|\\frac{b}{p^{k}}\\right\\|+\\left\\|\\frac{a+b}{p^{k}}\\right\\|=1 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice first that $\\left\\lfloor x+\\frac{1}{2}\\right\\rfloor$ is an integer nearest to $x$, so $\\|x\\|=\\left|\\left\\lfloor x+\\frac{1}{2}\\right\\rfloor-x\\right|$. Thus we have\n$$\n\\left\\lfloor x+\\frac{1}{2}\\right\\rfloor=x \\pm\\|x\\| .\n$$\nFor every rational number $r$ and every prime number $p$, denote by $v_{p}(r)$ the exponent of $p$ in the prime factorisation of $r$. Recall the notation $(2n-1)!!$ for the product of all odd positive integers not exceeding $2n-1$, i.e., $(2n-1)!!=1 \\cdot 3 \\cdots (2n-1)$.\n\nLemma. For every positive integer $n$ and every odd prime $p$, we have\n$$\nv_{p}((2n-1)!!)=\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{p^{k}}+\\frac{1}{2}\\right\\rfloor .\n$$\nProof. For every positive integer $k$, let us count the multiples of $p^{k}$ among the factors $1,3, \\ldots, 2n-1$. If $\\ell$ is an arbitrary integer, the number $(2\\ell-1)p^{k}$ is listed above if and only if\n$$\n0<(2\\ell-1)p^{k} \\leqslant 2n \\quad \\Longleftrightarrow \\quad \\frac{1}{2}<\\ell \\leqslant \\frac{n}{p^{k}}+\\frac{1}{2} \\quad \\Longleftrightarrow \\quad 1 \\leqslant \\ell \\leqslant \\left\\lfloor\\frac{n}{p^{k}}+\\frac{1}{2}\\right\\rfloor .\n$$\nHence, the number of multiples of $p^{k}$ among the factors is precisely $m_{k}=\\left\\lfloor\\frac{n}{p^{k}}+\\frac{1}{2}\\right\\rfloor$. Thus we obtain\n$$\nv_{p}((2n-1)!!)=\\sum_{i=1}^{n} v_{p}(2i-1)=\\sum_{i=1}^{n} \\sum_{k=1}^{v_{p}(2i-1)} 1=\\sum_{k=1}^{\\infty} \\sum_{\\ell=1}^{m_{k}} 1=\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{n}{p^{k}}+\\frac{1}{2}\\right\\rfloor .\n$$\n$\\square$\n\nIn order to prove the problem statement, consider the rational number\n$$\nN=\\frac{(2a+2b-1)!!}{(2a-1)!!\\cdot(2b-1)!!}=\\frac{(2a+1)(2a+3) \\cdots (2a+2b-1)}{1 \\cdot 3 \\cdots (2b-1)} .\n$$\nObviously, $N>1$, so there exists a prime $p$ with $v_{p}(N)>0$. Since $N$ is a fraction of two odd numbers, $p$ is odd.\n\nBy our lemma,\n$$\n0<v_{p}(N)=\\sum_{k=1}^{\\infty}\\left(\\left\\lfloor\\frac{a+b}{p^{k}}+\\frac{1}{2}\\right\\rfloor-\\left\\lfloor\\frac{a}{p^{k}}+\\frac{1}{2}\\right\\rfloor-\\left\\lfloor\\frac{b}{p^{k}}+\\frac{1}{2}\\right\\rfloor\\right) .\n$$\nTherefore, there exists some positive integer $k$ such that the integer number\n$$\nd_{k}=\\left\\lfloor\\frac{a+b}{p^{k}}+\\frac{1}{2}\\right\\rfloor-\\left\\lfloor\\frac{a}{p^{k}}+\\frac{1}{2}\\right\\rfloor-\\left\\lfloor\\frac{b}{p^{k}}+\\frac{1}{2}\\right\\rfloor\n$$\nis positive, so $d_{k} \\geqslant 1$. By the earlier formula we have\n$$\n1 \\leqslant d_{k}=\\frac{a+b}{p^{k}}-\\frac{a}{p^{k}}-\\frac{b}{p^{k}} \\pm \\left\\|\\frac{a+b}{p^{k}}\\right\\| \\pm \\left\\|\\frac{a}{p^{k}}\\right\\| \\pm \\left\\|\\frac{b}{p^{k}}\\right\\|= \\pm \\left\\|\\frac{a+b}{p^{k}}\\right\\| \\pm \\left\\|\\frac{a}{p^{k}}\\right\\| \\pm \\left\\|\\frac{b}{p^{k}}\\right\\| .\n$$\nSince $\\|x\\|<\\frac{1}{2}$ for every rational $x$ with odd denominator, the relation above can only be satisfied if all three signs on the right-hand side are positive and $d_{k}=1$. Thus we get\n$$\n\\left\\|\\frac{a}{p^{k}}\\right\\|+\\left\\|\\frac{b}{p^{k}}\\right\\|+\\left\\|\\frac{a+b}{p^{k}}\\right\\|=d_{k}=1\n$$\nas required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32001,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a fixed positive integer. Find the maximum possible value of\n$$\n\\sum_{1 \\leqslant r<s \\leqslant 2 n}(s-r-n) x_{r} x_{s}\n$$\nwhere $-1 \\leqslant x_{i} \\leqslant 1$ for all $i=1,2, \\ldots, 2 n$.",
    "options": [],
    "answer": "n(n-1)",
    "solution": "Let $Z$ be the expression to be maximized. Since this expression is linear in every variable $x_{i}$ and $-1 \\leqslant x_{i} \\leqslant 1$, the maximum of $Z$ will be achieved when $x_{i}=-1$ or $1$. Therefore, it suffices to consider only the case when $x_{i} \\in\\{-1,1\\}$ for all $i=1,2, \\ldots, 2 n$.\nFor $i=1,2, \\ldots, 2 n$, we introduce auxiliary variables\n$$\ny_{i}=\\sum_{r=1}^{i} x_{r}-\\sum_{r=i+1}^{2 n} x_{r} .\n$$\nTaking squares of both sides, we have\n$$\n\\begin{align*}\ny_{i}^{2} & =\\sum_{r=1}^{2 n} x_{r}^{2}+\\sum_{r<s \\leqslant i} 2 x_{r} x_{s}+\\sum_{i<r<s} 2 x_{r} x_{s}-\\sum_{r \\leqslant i<s} 2 x_{r} x_{s} \\\\\n& =2 n+\\sum_{r<s \\leqslant i} 2 x_{r} x_{s}+\\sum_{i<r<s} 2 x_{r} x_{s}-\\sum_{r \\leqslant i<s} 2 x_{r} x_{s} \\tag{1}\n\\end{align*}\n$$\nwhere the last equality follows from the fact that $x_{r} \\in\\{-1,1\\}$. Notice that for every $r<s$, the coefficient of $x_{r} x_{s}$ in (1) is $2$ for each $i=1, \\ldots, r-1, s, \\ldots, 2 n$, and this coefficient is $-2$ for each $i=r, \\ldots, s-1$. This implies that the coefficient of $x_{r} x_{s}$ in $\\sum_{i=1}^{2 n} y_{i}^{2}$ is $2(2 n-s+r)-2(s-r)= 4(n-s+r)$. Therefore, summing (1) for $i=1,2, \\ldots, 2 n$ yields\n$$\n\\begin{equation*}\n\\sum_{i=1}^{2 n} y_{i}^{2}=4 n^{2}+\\sum_{1 \\leqslant r<s \\leqslant 2 n} 4(n-s+r) x_{r} x_{s}=4 n^{2}-4 Z . \\tag{2}\n\\end{equation*}\n$$\nHence, it suffices to find the minimum of the left-hand side.\nSince $x_{r} \\in\\{-1,1\\}$, we see that $y_{i}$ is an even integer. In addition, $y_{i}-y_{i-1}=2 x_{i}= \\pm 2$, and so $y_{i-1}$ and $y_{i}$ are consecutive even integers for every $i=2,3, \\ldots, 2 n$. It follows that $y_{i-1}^{2}+y_{i}^{2} \\geqslant 4$, which implies\n$$\n\\begin{equation*}\n\\sum_{i=1}^{2 n} y_{i}^{2}=\\sum_{j=1}^{n}\\left(y_{2 j-1}^{2}+y_{2 j}^{2}\\right) \\geqslant 4 n . \\tag{3}\n\\end{equation*}\n$$\nCombining (2) and (3), we get\n$$\n\\begin{equation*}\n4 n \\leqslant \\sum_{i=1}^{2 n} y_{i}^{2}=4 n^{2}-4 Z \\tag{4}\n\\end{equation*}\n$$\nHence, $Z \\leqslant n(n-1)$.\nIf we set $x_{i}=1$ for odd indices $i$ and $x_{i}=-1$ for even indices $i$, then we obtain equality in (3) (and thus in (4)). Therefore, the maximum possible value of $Z$ is $n(n-1)$, as desired.\nWe present a different method of obtaining the bound $Z \\leqslant n(n-1)$. As in the previous solution, we reduce the problem to the case $x_{i} \\in\\{-1,1\\}$. For brevity, we use the notation $[2 n]=\\{1,2, \\ldots, 2 n\\}$.\nConsider any $x_{1}, x_{2}, \\ldots, x_{2 n} \\in\\{-1,1\\}$. Let\n$$\nA=\\left\\{i \\in[2 n]: x_{i}=1\\right\\} \\quad \\text{ and } \\quad B=\\left\\{i \\in[2 n]: x_{i}=-1\\right\\} .\n$$\nFor any subsets $X$ and $Y$ of $[2 n]$ we define\n$$\ne(X, Y)=\\sum_{r<s, r \\in X, s \\in Y}(s-r-n) .\n$$\nOne may observe that\n$e(A, A)+e(A, B)+e(B, A)+e(B, B)=e([2 n],[2 n])=\\sum_{1 \\leqslant r<s \\leqslant 2 n}(s-r-n)=-\\frac{(n-1) n(2 n-1)}{3}$.\nTherefore, we have\n$$\n\\begin{equation*}\nZ=e(A, A)-e(A, B)-e(B, A)+e(B, B)=2(e(A, A)+e(B, B))+\\frac{(n-1) n(2 n-1)}{3} . \\tag{5}\n\\end{equation*}\n$$\nThus, we need to maximize $e(A, A)+e(B, B)$, where $A$ and $B$ form a partition of $[2n]$.\nDue to the symmetry, we may assume that $|A|=n-p$ and $|B|=n+p$, where $0 \\leqslant p \\leqslant n$. From now on, we fix the value of $p$ and find an upper bound for $Z$ in terms of $n$ and $p$.\nLet $a_{1}<a_{2}<\\cdots<a_{n-p}$ and $b_{1}<b_{2}<\\cdots<b_{n+p}$ list all elements of $A$ and $B$, respectively. Then\n$$\n\\begin{equation*}\ne(A, A)=\\sum_{1 \\leqslant i<j \\leqslant n-p}\\left(a_{j}-a_{i}-n\\right)=\\sum_{i=1}^{n-p}(2 i-1-n+p) a_{i}-\\binom{n-p}{2} \\cdot n \\tag{6}\n\\end{equation*}\n$$\nand similarly\n$$\n\\begin{equation*}\ne(B, B)=\\sum_{i=1}^{n+p}(2 i-1-n-p) b_{i}-\\binom{n+p}{2} \\cdot n . \\tag{7}\n\\end{equation*}\n$$\nThus, now it suffices to maximize the value of\n$$\n\\begin{equation*}\nM=\\sum_{i=1}^{n-p}(2 i-1-n+p) a_{i}+\\sum_{i=1}^{n+p}(2 i-1-n-p) b_{i} . \\tag{8}\n\\end{equation*}\n$$\nIn order to get an upper bound, we will apply the rearrangement inequality to the sequence $a_{1}, a_{2}, \\ldots, a_{n-p}, b_{1}, b_{2}, \\ldots, b_{n+p}$ (which is a permutation of $1,2, \\ldots, 2 n$ ), together with the sequence of coefficients of these numbers in (8). The coefficients of $a_{i}$ form the sequence\n$$\nn-p-1, n-p-3, \\ldots, 1-n+p\n$$\nand those of $b_{i}$ form the sequence\n$$\nn+p-1, n+p-3, \\ldots, 1-n-p\n$$\nAltogether, these coefficients are, in descending order:\n- $n+p+1-2 i$, for $i=1,2, \\ldots, p$;\n- $n-p+1-2 i$, counted twice, for $i=1,2, \\ldots, n-p$; and\n$-(n+p+1-2 i)$, for $i=p, p-1, \\ldots, 1$.\nThus, the rearrangement inequality yields\n$$\n\\begin{align*}\nM \\leqslant \\sum_{i=1}^{p}(n+ & p+1-2 i)(2 n+1-i) \\\\\n& +\\sum_{i=1}^{n-p}(n-p+1-2 i)((2 n+2-p-2 i)+(2 n+1-p-2 i)) \\\\\n& \\quad-\\sum_{i=1}^{p}(n+p+1-2 i) i \\tag{9}\n\\end{align*}\n$$\nFinally, combining the information from (5), (6), (7), and (9), we obtain\n$$\n\\begin{aligned}\nZ \\leqslant & \\frac{(n-1) n(2 n-1)}{3}-2 n\\left(\\binom{n-p}{2}+\\binom{n+p}{2}\\right) \\\\\n& +2 \\sum_{i=1}^{p}(n+p+1-2 i)(2 n+1-2 i)+2 \\sum_{i=1}^{n-p}(n-p+1-2 i)(4 n-2 p+3-4 i),\n\\end{aligned}\n$$\nwhich can be simplified to\n$$\nZ \\leqslant n(n-1)-\\frac{2}{3} p(p-1)(p+1) .\n$$\nSince $p$ is a nonnegative integer, this yields $Z \\leqslant n(n-1)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32002,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $2 \\mathbb{Z} + 1$ denote the set of odd integers. Find all functions $f: \\mathbb{Z} \\rightarrow 2 \\mathbb{Z} + 1$ satisfying\n$$\nf(x + f(x) + y) + f(x - f(x) - y) = f(x + y) + f(x - y)\n$$\nfor every $x, y \\in \\mathbb{Z}$.",
    "options": [],
    "answer": "All solutions are as follows: choose an odd integer d, an integer k, and odd integers ℓ_0, ℓ_1, …, ℓ_{d−1}. For every integer x written uniquely as x = m d + i with m ∈ ℤ and i ∈ {0, 1, …, d−1}, define\n\nf(m d + i) = d(2 k m + ℓ_i).\n\nEquivalently, Δ_d f(x) = f(x + d) − f(x) = 2 d k is constant, and for each residue class i modulo d the values are f(m d + i) = ℓ_i d + 2 k m d. These and only these functions map ℤ to odd integers and satisfy the given equation.",
    "solution": "Throughout the solution, all functions are assumed to map integers to integers.\nFor any function $g$ and any nonzero integer $t$, define\n$$\n\\Delta_{t} g(x) = g(x + t) - g(x)\n$$\nFor any nonzero integers $a$ and $b$, notice that $\\Delta_{a} \\Delta_{b} g = \\Delta_{b} \\Delta_{a} g$. Moreover, if $\\Delta_{a} g = 0$ and $\\Delta_{b} g = 0$, then $\\Delta_{a + b} g = 0$ and $\\Delta_{a t} g = 0$ for all nonzero integers $t$. We say that $g$ is $t$-quasiperiodic if $\\Delta_{t} g$ is a constant function (in other words, if $\\Delta_{1} \\Delta_{t} g = 0$, or $\\Delta_{1} g$ is $t$-periodic). In this case, we call $t$ a quasi-period of $g$. We say that $g$ is quasi-periodic if it is $t$-quasi-periodic for some nonzero integer $t$.\nNotice that a quasi-period of $g$ is a period of $\\Delta_{1} g$. So if $g$ is quasi-periodic, then its minimal positive quasi-period $t$ divides all its quasi-periods.\nWe now assume that $f$ satisfies (1). First, by setting $a = x + y$, the problem condition can be rewritten as\n$$\n\\begin{equation*}\n\\Delta_{f(x)} f(a) = \\Delta_{f(x)} f(2x - a - f(x)) \\quad \\text{ for all } x, a \\in \\mathbb{Z} \\tag{2}\n\\end{equation*}\n$$\nLet $b$ be an arbitrary integer and let $k$ be an arbitrary positive integer. Applying (2) when $a$ is substituted by $b, b + f(x), \\ldots, b + (k - 1) f(x)$ and summing up all these equations, we get\n$$\n\\Delta_{k f(x)} f(b) = \\Delta_{k f(x)} f(2x - b - k f(x))\n$$\nNotice that a similar argument works when $k$ is negative, so that\n$$\n\\begin{equation*}\n\\Delta_{M} f(b) = \\Delta_{M} f(2x - b - M) \\quad \\text{ for any nonzero integer } M \\text{ such that } f(x) \\mid M . \\tag{3}\n\\end{equation*}\n$$\nWe now prove two lemmas.\n**Lemma 1.** For any distinct integers $x$ and $y$, the function $\\Delta_{\\operatorname{lcm}(f(x), f(y))} f$ is $2(y - x)$-periodic.\nProof. Denote $L = \\operatorname{lcm}(f(x), f(y))$. Applying (3) twice, we obtain\n$$\n\\Delta_{L} f(b) = \\Delta_{L} f(2x - b - L) = \\Delta_{L} f(2y - (b + 2(y - x)) - L) = \\Delta_{L} f(b + 2(y - x)) .\n$$\nThus, the function $\\Delta_{L} f$ is $2(y - x)$-periodic, as required.\n**Lemma 2.** Let $g$ be a function. If $t$ and $s$ are nonzero integers such that $\\Delta_{t s} g = 0$ and $\\Delta_{t} \\Delta_{t} g = 0$, then $\\Delta_{t} g = 0$.\nProof. Assume, without loss of generality, that $s$ is positive. Let $a$ be an arbitrary integer. Since $\\Delta_{t} \\Delta_{t} g = 0$, we have\n$$\n\\Delta_{t} g(a) = \\Delta_{t} g(a + t) = \\cdots = \\Delta_{t} g(a + (s - 1) t)\n$$\nThe sum of these $s$ equal numbers is $\\Delta_{t s} g(a) = 0$, so each of them is zero, as required.\nWe now return to the solution.\n**Step 1.** We prove that $f$ is quasi-periodic.\nLet $Q = \\operatorname{lcm}(f(0), f(1))$. Applying Lemma 1, we get that the function $g = \\Delta_{Q} f$ is 2-periodic. In other words, the values of $g$ are constant on even numbers and on odd numbers separately. Moreover, setting $M = Q$ and $x = b = 0$ in (3), we get $g(0) = g(-Q)$. Since 0 and $-Q$ have different parities, the value of $g$ at even numbers is the same as that at odd numbers. Thus, $g$ is constant, which means that $Q$ is a quasi-period of $f$.\n**Step 2.** Denote the minimal positive quasi-period of $f$ by $T$. We prove that $T \\mid f(x)$ for all integers $x$.\nSince an odd number $Q$ is a quasi-period of $f$, the number $T$ is also odd. Now suppose, to the contrary, that there exist an odd prime $p$, a positive integer $\\alpha$, and an integer $u$ such that $p^{\\alpha} \\mid T$ but $p^{\\alpha} \\nmid f(u)$. Setting $x = u$ and $y = 0$ in (1), we have $2 f(u) = f(u + f(u)) + f(u - f(u))$, so $p^{\\alpha}$ does not divide the value of $f$ at one of the points $u + f(u)$ or $u - f(u)$. Denote this point by $v$.\nLet $L = \\operatorname{lcm}(f(u), f(v))$. Since $|u - v| = f(u)$, from Lemma 1 we get $\\Delta_{2 f(u)} \\Delta_{L} f = 0$. Hence the function $\\Delta_{L} f$ is $2 f(u)$-periodic as well as $T$-periodic, so it is $\\operatorname{gcd}(T, 2 f(u))$-periodic, or $\\Delta_{\\operatorname{gcd}(T, 2 f(u))} \\Delta_{L} f = 0$. Similarly, observe that the function $\\Delta_{\\operatorname{gcd}(T, 2 f(u))} f$ is $L$-periodic as well as $T$-periodic, so we may conclude that $\\Delta_{\\operatorname{gcd}(T, L)} \\Delta_{\\operatorname{gcd}(T, 2 f(u))} f = 0$. Since $p^{\\alpha} \\nmid L$, both $\\operatorname{gcd}(T, 2 f(u))$ and $\\operatorname{gcd}(T, L)$ divide $T / p$. We thus obtain $\\Delta_{T / p} \\Delta_{T / p} f = 0$, which yields\n$$\n\\Delta_{T / p} \\Delta_{T / p} \\Delta_{1} f = 0\n$$\nSince $\\Delta_{T} \\Delta_{1} f = 0$, we can apply Lemma 2 to the function $\\Delta_{1} f$, obtaining $\\Delta_{T / p} \\Delta_{1} f = 0$. However, this means that $f$ is $(T / p)$-quasi-periodic, contradicting the minimality of $T$. Our claim is proved.\n**Step 3.** We describe all functions $f$.\nLet $d$ be the greatest common divisor of all values of $f$. Then $d$ is odd. By Step 2, $d$ is a quasi-period of $f$, so that $\\Delta_{d} f$ is constant. Since the value of $\\Delta_{d} f$ is even and divisible by $d$, we may denote this constant by $2 d k$, where $k$ is an integer. Next, for all $i = 0, 1, \\ldots, d - 1$, define $\\ell_{i} = f(i) / d$; notice that $\\ell_{i}$ is odd. Then\n$$\nf(m d + i) = \\Delta_{m d} f(i) + f(i) = 2 k m d + \\ell_{i} d \\quad \\text{ for all } m \\in \\mathbb{Z} \\quad \\text{ and } i = 0, 1, \\ldots, d - 1 .\n$$\nThis shows that all functions satisfying (1) are listed in the answer.\nIt remains to check that all such functions indeed satisfy (1). This is equivalent to checking (2), which is true because for every integer $x$, the value of $f(x)$ is divisible by $d$, so that $\\Delta_{f(x)} f$ is constant.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32003,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a fixed integer with $n \\geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \\in\\{1,2, \\ldots, n\\}$ the sequences\n$$\n\\begin{aligned}\n& P(2015 i), P(2015 i-1), \\ldots, P(2015 i-2014) \\quad \\text{ and } \\\\\n& Q(2015 i), Q(2015 i-1), \\ldots, Q(2015 i-2014)\n\\end{aligned}\n$$\nare permutations of each other.\n\na. Prove that there exist distinct block-similar polynomials of degree $n+1$.\n\nb. Prove that there do not exist distinct block-similar polynomials of degree $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For convenience, we set $k=2015=2 \\ell+1$.\n\n**a.**\nConsider the following polynomials of degree $n+1$ :\n$$\nP(x)=\\prod_{i=0}^{n}(x-i k) \\quad \\text{ and } \\quad Q(x)=\\prod_{i=0}^{n}(x-i k-1) .\n$$\nSince $Q(x)=P(x-1)$ and $P(0)=P(k)=P(2 k)=\\cdots=P(n k)$, these polynomials are block-similar (and distinct).\n\n**b.**\nFor every polynomial $F(x)$ and every nonnegative integer $m$, define $\\Sigma_{F}(m)= \\sum_{i=1}^{m} F(i)$; in particular, $\\Sigma_{F}(0)=0$. It is well-known that for every nonnegative integer $d$ the sum $\\sum_{i=1}^{m} i^{d}$ is a polynomial in $m$ of degree $d+1$. Thus $\\Sigma_{F}$ may also be regarded as a real polynomial of degree $\\operatorname{deg} F+1$ (with the exception that if $F=0$, then $\\Sigma_{F}=0$ as well). This allows us to consider the values of $\\Sigma_{F}$ at all real points (where the initial definition does not apply).\n\nAssume for the sake of contradiction that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Then both polynomials $\\Sigma_{P-Q}(x)$ and $\\Sigma_{P^{2}-Q^{2}}(x)$ have roots at the points $0, k, 2 k, \\ldots, n k$. This motivates the following lemma, where we use the special polynomial\n$$\nT(x)=\\prod_{i=0}^{n}(x-i k)\n$$\n\n**Lemma.** Assume that $F(x)$ is a nonzero polynomial such that $0, k, 2 k, \\ldots, n k$ are among the roots of the polynomial $\\Sigma_{F}(x)$. Then $\\operatorname{deg} F \\geqslant n$, and there exists a polynomial $G(x)$ such that $\\operatorname{deg} G=\\operatorname{deg} F-n$ and $F(x)=T(x) G(x)-T(x-1) G(x-1)$.\n\n**Proof.** If $\\operatorname{deg} F<n$, then $\\Sigma_{F}(x)$ has at least $n+1$ roots, while its degree is less than $n+1$. Therefore, $\\Sigma_{F}(x)=0$ and hence $F(x)=0$, which is impossible. Thus $\\operatorname{deg} F \\geqslant n$.\n\nThe lemma condition yields that $\\Sigma_{F}(x)=T(x) G(x)$ for some polynomial $G(x)$ such that $\\operatorname{deg} G=\\operatorname{deg} \\Sigma_{F}-(n+1)=\\operatorname{deg} F-n$.\n\nNow, let us define $F_{1}(x)=T(x) G(x)-T(x-1) G(x-1)$. Then for every positive integer $n$ we have\n$$\n\\Sigma_{F_{1}}(n)=\\sum_{i=1}^{n}(T(x) G(x)-T(x-1) G(x-1))=T(n) G(n)-T(0) G(0)=T(n) G(n)=\\Sigma_{F}(n)\n$$\nso the polynomial $\\Sigma_{F-F_{1}}(x)=\\Sigma_{F}(x)-\\Sigma_{F_{1}}(x)$ has infinitely many roots. This means that this polynomial is zero, which in turn yields $F(x)=F_{1}(x)$, as required.\n\nFirst, we apply the lemma to the nonzero polynomial $R_{1}(x)=P(x)-Q(x)$. Since the degree of $R_{1}(x)$ is at most $n$, we conclude that it is exactly $n$. Moreover, $R_{1}(x)=\\alpha \\cdot(T(x)-T(x-1))$ for some nonzero constant $\\alpha$.\n\nOur next aim is to prove that the polynomial $S(x)=P(x)+Q(x)$ is constant. Assume the contrary. Then, notice that the polynomial $R_{2}(x)=P(x)^{2}-Q(x)^{2}=R_{1}(x) S(x)$ is also nonzero and satisfies the lemma condition. Since $n<\\operatorname{deg} R_{1}+\\operatorname{deg} S=\\operatorname{deg} R_{2} \\leqslant 2 n$, the lemma yields\n$$\nR_{2}(x)=T(x) G(x)-T(x-1) G(x-1)\n$$\nwith some polynomial $G(x)$ with $0<\\operatorname{deg} G \\leqslant n$.\n\nSince the polynomial $R_{1}(x)=\\alpha(T(x)-T(x-1))$ divides the polynomial\n$$\nR_{2}(x)=T(x)(G(x)-G(x-1))+G(x-1)(T(x)-T(x-1)),\n$$\nwe get $R_{1}(x) \\mid T(x)(G(x)-G(x-1))$. On the other hand,\n$$\n\\operatorname{gcd}\\left(T(x), R_{1}(x)\\right)=\\operatorname{gcd}(T(x), T(x)-T(x-1))=\\operatorname{gcd}(T(x), T(x-1))=1\n$$\nsince both $T(x)$ and $T(x-1)$ are the products of linear polynomials, and their roots are distinct. Thus $R_{1}(x) \\mid G(x)-G(x-1)$. However, this is impossible since $G(x)-G(x-1)$ is a nonzero polynomial of degree less than $n=\\operatorname{deg} R_{1}$.\n\nThus, our assumption is wrong, and $S(x)$ is a constant polynomial, say $S(x)=\\beta$. Notice that the polynomials $(2 P(x)-\\beta) / \\alpha$ and $(2 Q(x)-\\beta) / \\alpha$ are also block-similar and distinct. So we may replace the initial polynomials by these ones, thus obtaining two block-similar polynomials $P(x)$ and $Q(x)$ with $P(x)=-Q(x)=T(x)-T(x-1)$. It remains to show that this is impossible.\n\nFor every $i=1,2 \\ldots, n$, the values $T(i k-k+1)$ and $T(i k-1)$ have the same sign. This means that the values $P(i k-k+1)=T(i k-k+1)$ and $P(i k)=-T(i k-1)$ have opposite signs, so $P(x)$ has a root in each of the $n$ segments $[i k-k+1, i k]$. Since $\\operatorname{deg} P=n$, it must have exactly one root in each of them.\n\nThus, the sequence $P(1), P(2), \\ldots, P(k)$ should change sign exactly once. On the other hand, since $P(x)$ and $-P(x)$ are block-similar, this sequence must have as many positive terms as negative ones. Since $k=2 \\ell+1$ is odd, this shows that the middle term of the sequence above must be zero, so $P(\\ell+1)=0$, or $T(\\ell+1)=T(\\ell)$. However, this is not true since\n$$\n|T(\\ell+1)|=|\\ell+1| \\cdot|\\ell| \\cdot \\prod_{i=2}^{n}|\\ell+1-i k|<|\\ell| \\cdot|\\ell+1| \\cdot \\prod_{i=2}^{n}|\\ell-i k|=|T(\\ell)|,\n$$\nwhere the strict inequality holds because $n \\geqslant 2$. We come to the final contradiction.\nWe provide an alternative argument for part (b).\n\nAssume again that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Let $R(x)=P(x)-Q(x)$ and $S(x)=P(x)+Q(x)$. For brevity, we also denote the segment $[(i-1) k+1, i k]$ by $I_{i}$, and the set $\\{(i-1) k+1,(i-1) k+2, \\ldots, i k\\}$ of all integer points in $I_{i}$ by $Z_{i}$.\n\n**Step 1.** We prove that $R(x)$ has exactly one root in each segment $I_{i}, i=1,2, \\ldots, n$, and all these roots are simple.\n\nIndeed, take any $i \\in\\{1,2, \\ldots, n\\}$ and choose some points $p^{-}, p^{+} \\in Z_{i}$ so that\n$$\nP\\left(p^{-}\\right)=\\min _{x \\in Z_{i}} P(x) \\quad \\text{ and } \\quad P\\left(p^{+}\\right)=\\max _{x \\in Z_{i}} P(x)\n$$\nSince the sequences of values of $P$ and $Q$ in $Z_{i}$ are permutations of each other, we have $R\\left(p^{-}\\right)=P\\left(p^{-}\\right)-Q\\left(p^{-}\\right) \\leqslant 0$ and $R\\left(p^{+}\\right)=P\\left(p^{+}\\right)-Q\\left(p^{+}\\right) \\geqslant 0$. Since $R(x)$ is continuous, there exists at least one root of $R(x)$ between $p^{-}$ and $p^{+}$ - thus in $I_{i}$.\n\nSo, $R(x)$ has at least one root in each of the $n$ disjoint segments $I_{i}$ with $i=1,2, \\ldots, n$. Since $R(x)$ is nonzero and its degree does not exceed $n$, it should have exactly one root in each of these segments, and all these roots are simple, as required.\n\n**Step 2.** We prove that $S(x)$ is constant.\n\nWe start with the following claim.\n\n**Claim.** For every $i=1,2, \\ldots, n$, the sequence of values $S((i-1) k+1), S((i-1) k+2), \\ldots$, $S(i k)$ cannot be strictly increasing.\n\n**Proof.** Fix any $i \\in\\{1,2, \\ldots, n\\}$. Due to the symmetry, we may assume that $P(i k) \\leqslant Q(i k)$. Choose now $p^{-}$ and $p^{+}$ as in Step 1. If we had $P\\left(p^{+}\\right)=P\\left(p^{-}\\right)$, then $P$ would be constant on $Z_{i}$, so all the elements of $Z_{i}$ would be the roots of $R(x)$, which is not the case. In particular, we have $p^{+} \\neq p^{-}$. If $p^{-}>p^{+}$, then $S\\left(p^{-}\\right)=P\\left(p^{-}\\right)+Q\\left(p^{-}\\right) \\leqslant Q\\left(p^{+}\\right)+P\\left(p^{+}\\right)=S\\left(p^{+}\\right)$, so our claim holds.\n\nWe now show that the remaining case $p^{-}<p^{+}$ is impossible. Assume first that $P\\left(p^{+}\\right)> Q\\left(p^{+}\\right)$. Then, like in Step 1, we have $R\\left(p^{-}\\right) \\leqslant 0, R\\left(p^{+}\\right)>0$, and $R(i k) \\leqslant 0$, so $R(x)$ has a root in each of the intervals $\\left[p^{-}, p^{+}\\right)$ and $\\left(p^{+}, i k\\right]$. This contradicts the result of Step 1.\n\nWe are left only with the case $p^{-}<p^{+}$ and $P\\left(p^{+}\\right)=Q\\left(p^{+}\\right)$ (thus $p^{+}$ is the unique root of $R(x)$ in $I_{i}$ ). If $p^{+}=i k$, then the values of $R(x)$ on $Z_{i} \\backslash\\{i k\\}$ are all of the same sign, which is absurd since their sum is zero. Finally, if $p^{-}<p^{+}<i k$, then $R\\left(p^{-}\\right)$ and $R(i k)$ are both negative. This means that $R(x)$ should have an even number of roots in $\\left[p^{-}, i k\\right]$, counted with multiplicity. This also contradicts the result of Step 1.\n\nIn a similar way, one may prove that for every $i=1,2, \\ldots, n$, the sequence $S((i-1) k+1)$, $S((i-1) k+2), \\ldots, S(i k)$ cannot be strictly decreasing. This means that the polynomial $\\Delta S(x)=S(x)-S(x-1)$ attains at least one nonnegative value, as well as at least one nonpositive value, on the set $Z_{i}$ (and even on $Z_{i} \\backslash\\{(i-1) k+1\\}$ ); so $\\Delta S$ has a root in $I_{i}$.\n\nThus $\\Delta S$ has at least $n$ roots; however, its degree is less than $n$, so $\\Delta S$ should be identically zero. This shows that $S(x)$ is a constant, say $S(x) \\equiv \\beta$.\n\n**Step 3.** Notice that the polynomials $P(x)-\\beta / 2$ and $Q(x)-\\beta / 2$ are also block-similar and distinct. So we may replace the initial polynomials by these ones, thus reaching $P(x)=-Q(x)$.\n\nThen $R(x)=2 P(x)$, so $P(x)$ has exactly one root in each of the segments $I_{i}, i=1,2, \\ldots, n$. On the other hand, $P(x)$ and $-P(x)$ should attain the same number of positive values on $Z_{i}$. Since $k$ is odd, this means that $Z_{i}$ contains exactly one root of $P(x)$; moreover, this root should be at the center of $Z_{i}$, because $P(x)$ has the same number of positive and negative values on $Z_{i}$.\n\nThus we have found all $n$ roots of $P(x)$, so\n$$\nP(x)=c \\prod_{i=1}^{n}(x-i k+\\ell) \\quad \\text{ for some } c \\in \\mathbb{R} \\backslash\\{0\\}\n$$\nwhere $\\ell=(k-1) / 2$. It remains to notice that for every $t \\in Z_{1} \\backslash\\{1\\}$ we have\n$$\n|P(t)|=|c| \\cdot|t-\\ell-1| \\cdot \\prod_{i=2}^{n}|t-i k+\\ell|<|c| \\cdot \\ell \\cdot \\prod_{i=2}^{n}|1-i k+\\ell|=|P(1)|\n$$\nso $P(1) \\neq-P(t)$ for all $t \\in Z_{1}$. This shows that $P(x)$ is not block-similar to $-P(x)$. The final contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32004,
    "subject": "Mathematics (Multi-modal)",
    "question": "In Lineland there are $n \\geqslant 1$ towns, arranged along a road running from left to right. Each town has a left bulldozer (put to the left of the town and facing left) and a right bulldozer (put to the right of the town and facing right). The sizes of the $2n$ bulldozers are distinct. Every time when a right and a left bulldozer confront each other, the larger bulldozer pushes the smaller one off the road. On the other hand, the bulldozers are quite unprotected at their rears; so, if a bulldozer reaches the rear-end of another one, the first one pushes the second one off the road, regardless of their sizes.\nLet $A$ and $B$ be two towns, with $B$ being to the right of $A$. We say that town $A$ can sweep town $B$ away if the right bulldozer of $A$ can move over to $B$ pushing off all bulldozers it meets. Similarly, $B$ can sweep $A$ away if the left bulldozer of $B$ can move to $A$ pushing off all bulldozers of all towns on its way.\nProve that there is exactly one town which cannot be swept away by any other one.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T_{1}, T_{2}, \\ldots, T_{n}$ be the towns enumerated from left to right. Observe first that, if town $T_{i}$ can sweep away town $T_{j}$, then $T_{i}$ also can sweep away every town located between $T_{i}$ and $T_{j}$.\nWe prove the problem statement by strong induction on $n$. The base case $n=1$ is trivial.\nFor the induction step, we first observe that the left bulldozer in $T_{1}$ and the right bulldozer in $T_{n}$ are completely useless, so we may forget them forever. Among the other $2n-2$ bulldozers, we choose the largest one. Without loss of generality, it is the right bulldozer of some town $T_{k}$ with $k<n$.\nSurely, with this large bulldozer $T_{k}$ can sweep away all the towns to the right of it. Moreover, none of these towns can sweep $T_{k}$ away; so they also cannot sweep away any town to the left of $T_{k}$. Thus, if we remove the towns $T_{k+1}, T_{k+2}, \\ldots, T_{n}$, none of the remaining towns would change its status of being (un)sweepable away by the others.\nApplying the induction hypothesis to the remaining towns, we find a unique town among $T_{1}, T_{2}, \\ldots, T_{k}$ which cannot be swept away. By the above reasons, it is also the unique such town in the initial situation. Thus the induction step is established.\nWe start with the same enumeration and the same observation as in Solution 1. We also denote by $\\ell_{i}$ and $r_{i}$ the sizes of the left and the right bulldozers belonging to $T_{i}$, respectively. One may easily see that no two towns $T_{i}$ and $T_{j}$ with $i<j$ can sweep each other away, for this would yield $r_{i}>\\ell_{j}>r_{i}$.\nClearly, there is no town which can sweep $T_{n}$ away from the right. Then we may choose the leftmost town $T_{k}$ which cannot be swept away from the right. One can observe now that no town $T_{i}$ with $i>k$ may sweep away some town $T_{j}$ with $j<k$, for otherwise $T_{i}$ would be able to sweep $T_{k}$ away as well.\nNow we prove two claims, showing together that $T_{k}$ is the unique town which cannot be swept away, and thus establishing the problem statement.\n\nClaim 1. $T_{k}$ also cannot be swept away from the left.\nProof. Let $T_{m}$ be some town to the left of $T_{k}$. By the choice of $T_{k}$, town $T_{m}$ can be swept away from the right by some town $T_{p}$ with $p>m$. As we have already observed, $p$ cannot be greater than $k$. On the other hand, $T_{m}$ cannot sweep $T_{p}$ away, so a fortiori it cannot sweep $T_{k}$ away.\n\nClaim 2. Any town $T_{m}$ with $m \\neq k$ can be swept away by some other town.\nProof. If $m<k$, then $T_{m}$ can be swept away from the right due to the choice of $T_{k}$. In the remaining case we have $m>k$.\nLet $T_{p}$ be a town among $T_{k}, T_{k+1}, \\ldots, T_{m-1}$ having the largest right bulldozer. We claim that $T_{p}$ can sweep $T_{m}$ away. If this is not the case, then $r_{p}<\\ell_{q}$ for some $q$ with $p<q \\leqslant m$. But this means that $\\ell_{q}$ is greater than all the numbers $r_{i}$ with $k \\leqslant i \\leqslant m-1$, so $T_{q}$ can sweep $T_{k}$ away. This contradicts the choice of $T_{k}$.\nWe separately prove that (i) there exists a town which cannot be swept away, and that (ii) there is at most one such town. We also make use of the two observations from the previous solutions.\nTo prove $(i)$, assume contrariwise that every town can be swept away. Let $t_{1}$ be the leftmost town; next, for every $k=1,2, \\ldots$ we inductively choose $t_{k+1}$ to be some town which can sweep $t_{k}$ away. Now we claim that for every $k=1,2, \\ldots$, the town $t_{k+1}$ is to the right of $t_{k}$; this leads to the contradiction, since the number of towns is finite.\nInduction on $k$. The base case $k=1$ is clear due to the choice of $t_{1}$. Assume now that for all $j$ with $1 \\leqslant j<k$, the town $t_{j+1}$ is to the right of $t_{j}$. Suppose that $t_{k+1}$ is situated to the left of $t_{k}$; then it lies between $t_{j}$ and $t_{j+1}$ (possibly coinciding with $t_{j}$ ) for some $j<k$. Therefore, $t_{k+1}$ can be swept away by $t_{j+1}$, which shows that it cannot sweep $t_{j+1}$ away - so $t_{k+1}$ also cannot sweep $t_{k}$ away. This contradiction proves the induction step.\nTo prove (ii), we also argue indirectly and choose two towns $A$ and $B$ neither of which can be swept away, with $A$ being to the left of $B$. Consider the largest bulldozer $b$ between them (taking into consideration the right bulldozer of $A$ and the left bulldozer of $B$ ). Without loss of generality, $b$ is a left bulldozer; then it is situated in some town to the right of $A$, and this town may sweep $A$ away since nothing prevents it from doing that. A contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32005,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{V}$ be a finite set of points in the plane. We say that $\\mathcal{V}$ is balanced if for any two distinct points $A, B \\in \\mathcal{V}$, there exists a point $C \\in \\mathcal{V}$ such that $AC = BC$. We say that $\\mathcal{V}$ is center-free if for any distinct points $A, B, C \\in \\mathcal{V}$, there does not exist a point $P \\in \\mathcal{V}$ such that $PA = PB = PC$.\n\na. Show that for all $n \\geqslant 3$, there exists a balanced set consisting of $n$ points.\n\nb. For which $n \\geqslant 3$ does there exist a balanced, center-free set consisting of $n$ points?",
    "options": [],
    "answer": "a) For every integer n at least three, there exists a balanced set of n points. b) Balanced, center-free sets exist exactly for odd n at least three; they do not exist for even n.",
    "solution": "Part (a).\nAssume that $n$ is odd. Consider a regular $n$-gon. Label the vertices of the $n$-gon as $A_{1}, A_{2}, \\ldots, A_{n}$ in counter-clockwise order, and set $\\mathcal{V} = \\{A_{1}, \\ldots, A_{n}\\}$. We check that $\\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{j}$, let $k \\in \\{1,2, \\ldots, n\\}$ be the solution of $2k \\equiv i + j \\pmod{n}$. Then, since $k - i \\equiv j - k \\pmod{n}$, we have $A_{i}A_{k} = A_{j}A_{k}$, as required.\n\nNow assume that $n$ is even. Consider a regular $(3n - 6)$-gon, and let $O$ be its circumcenter. Again, label its vertices as $A_{1}, \\ldots, A_{3n-6}$ in counter-clockwise order, and choose $\\mathcal{V} = \\{O, A_{1}, A_{2}, \\ldots, A_{n-1}\\}$. We check that $\\mathcal{V}$ is balanced. For any two distinct vertices $A_{i}$ and $A_{j}$, we always have $OA_{i} = OA_{j}$. We now consider the vertices $O$ and $A_{i}$. First note that the triangle $OA_{i}A_{n/2-1+i}$ is equilateral for all $i \\leqslant \\frac{n}{2}$. Hence, if $i \\leqslant \\frac{n}{2}$, then we have $OA_{n/2-1+i} = A_{i}A_{n/2-1+i}$; otherwise, if $i > \\frac{n}{2}$, then we have $OA_{i-n/2+1} = A_{i}A_{i-n/2+1}$. This completes the proof.\n\n\nPart (b).\nWe now show that there exists a balanced, center-free set containing $n$ points for all odd $n \\geqslant 3$, and that one does not exist for any even $n \\geqslant 3$.\nIf $n$ is odd, then let $\\mathcal{V}$ be the set of vertices of a regular $n$-gon. We have shown in part (a) that $\\mathcal{V}$ is balanced. We claim that $\\mathcal{V}$ is also center-free. Indeed, if $P$ is a point such that $PA = PB = PC$ for some three distinct vertices $A, B$ and $C$, then $P$ is the circumcenter of the $n$-gon, which is not contained in $\\mathcal{V}$.\n\nNow suppose that $\\mathcal{V}$ is a balanced, center-free set of even cardinality $n$. We will derive a contradiction. For a pair of distinct points $A, B \\in \\mathcal{V}$, we say that a point $C \\in \\mathcal{V}$ is associated with the pair $\\{A, B\\}$ if $AC = BC$. Since there are $\\frac{n(n-1)}{2}$ pairs of points, there exists a point $P \\in \\mathcal{V}$ which is associated with at least $\\left\\lceil \\frac{n(n-1)}{2} / n \\right\\rceil = \\frac{n}{2}$ pairs. Note that none of these $\\frac{n}{2}$ pairs can contain $P$, so that the union of these $\\frac{n}{2}$ pairs consists of at most $n-1$ points. Hence there exist two such pairs that share a point. Let these two pairs be $\\{A, B\\}$ and $\\{A, C\\}$. Then $PA = PB = PC$, which is a contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32006,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider an infinite sequence $a_{1}, a_{2}, \\ldots$ of positive integers with $a_{i} \\leqslant 2015$ for all $i \\geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \\neq j+a_{j}$.\nProve that there exist two positive integers $b$ and $N$ such that\n$$\n\\left|\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)\\right| \\leqslant 1007^{2}\n$$\nwhenever $n>m \\geqslant N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \\neq n+a_{n}$ for $m \\neq n$, each positive integer receives at most one arrow. There are some positive integers, such as 1, that receive no arrows; these will be referred to as starting points in the sequel. When one starts at any of the starting points and keeps following the arrows, one is led to an infinite path, called its ray, that visits a strictly increasing sequence of positive integers. Since the length of any arrow is at most 2015, such a ray, say with starting point $s$, meets every interval of the form $[n, n+2014]$ with $n \\geqslant s$ at least once.\n\nSuppose for the sake of contradiction that there would be at least 2016 starting points. Then we could take an integer $n$ that is larger than the first 2016 starting points. But now the interval $[n, n+2014]$ must be met by at least 2016 rays in distinct points, which is absurd. We have thereby shown that the number $b$ of starting points satisfies $1 \\leqslant b \\leqslant 2015$. Let $N$ denote any integer that is larger than all starting points. We contend that $b$ and $N$ are as required.\n\nTo see this, let any two integers $m$ and $n$ with $n>m \\geqslant N$ be given. The sum $\\sum_{i=m+1}^{n} a_{i}$ gives the total length of the arrows emerging from $m+1, \\ldots, n$. Taken together, these arrows form $b$ subpaths of our rays, some of which may be empty. Now on each ray we look at the first number that is larger than $m$; let $x_{1}, \\ldots, x_{b}$ denote these numbers, and let $y_{1}, \\ldots, y_{b}$ enumerate in corresponding order the numbers defined similarly with respect to $n$. Then the list of differences $y_{1}-x_{1}, \\ldots, y_{b}-x_{b}$ consists of the lengths of these paths and possibly some zeros corresponding to empty paths. Consequently, we obtain\n$$\n\\sum_{i=m+1}^{n} a_{i}=\\sum_{j=1}^{b}\\left(y_{j}-x_{j}\\right)\n$$\nwhence\n$$\n\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)=\\sum_{j=1}^{b}\\left(y_{j}-n\\right)-\\sum_{j=1}^{b}\\left(x_{j}-m\\right)\n$$\nNow each of the $b$ rays meets the interval $[m+1, m+2015]$ at some point and thus $x_{1}- m, \\ldots, x_{b}-m$ are $b$ distinct members of the set $\\{1,2, \\ldots, 2015\\}$. Moreover, since $m+1$ is not a starting point, it must belong to some ray; so 1 has to appear among these numbers, wherefore\n$$\n1+\\sum_{j=1}^{b-1}(j+1) \\leqslant \\sum_{j=1}^{b}\\left(x_{j}-m\\right) \\leqslant 1+\\sum_{j=1}^{b-1}(2016-b+j)\n$$\nThe same argument applied to $n$ and $y_{1}, \\ldots, y_{b}$ yields\n$$\n1+\\sum_{j=1}^{b-1}(j+1) \\leqslant \\sum_{j=1}^{b}\\left(y_{j}-n\\right) \\leqslant 1+\\sum_{j=1}^{b-1}(2016-b+j)\n$$\nSo altogether we get\n$$\n\\begin{gathered}\n\\left|\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)\\right| \\leqslant \\sum_{j=1}^{b-1}((2016-b+j)-(j+1))=(b-1)(2015-b) \\\\\n\\leqslant\\left(\\frac{(b-1)+(2015-b)}{2}\\right)^{2}=1007^{2}\n\\end{gathered}\n$$\nas desired.\n\n\nSolution 2:\n\nSet $s_{n}=n+a_{n}$ for all positive integers $n$. By our assumptions, we have\n$$\nn+1 \\leqslant s_{n} \\leqslant n+2015\n$$\nfor all $n \\in \\mathbb{Z}_{>0}$. The members of the sequence $s_{1}, s_{2}, \\ldots$ are distinct. We shall investigate the set\n$$\nM=\\mathbb{Z}_{>0} \\backslash\\left\\{s_{1}, s_{2}, \\ldots\\right\\}\n$$\nClaim. At most 2015 numbers belong to $M$.\n\nProof. Otherwise let $m_{1}<m_{2}<\\cdots<m_{2016}$ be any 2016 distinct elements from $M$. For $n=m_{2016}$ we have\n$$\n\\left\\{s_{1}, \\ldots, s_{n}\\right\\} \\cup\\left\\{m_{1}, \\ldots, m_{2016}\\right\\} \\subseteq\\{1,2, \\ldots, n+2015\\}\n$$\nwhere on the left-hand side we have a disjoint union containing altogether $n+2016$ elements. But the set on the right-hand side has only $n+2015$ elements. This contradiction proves our claim. $\\square$\n\nNow we work towards proving that the positive integers $b=|M|$ and $N=\\max (M)$ are as required. Recall that we have just shown $b \\leqslant 2015$.\n\nLet us consider any integer $r \\geqslant N$. As in the proof of the above claim, we see that\n$$\n\\begin{equation*}\nB_{r}=M \\cup\\left\\{s_{1}, \\ldots, s_{r}\\right\\} \\tag{1}\n\\end{equation*}\n$$\nis a subset of $[1, r+2015] \\cap \\mathbb{Z}$ with precisely $b+r$ elements. Due to the definitions of $M$ and $N$, we also know $[1, r+1] \\cap \\mathbb{Z} \\subseteq B_{r}$. It follows that there is a set $C_{r} \\subseteq\\{1,2, \\ldots, 2014\\}$ with $\\left|C_{r}\\right|=b-1$ and\n$$\n\\begin{equation*}\nB_{r}=([1, r+1] \\cap \\mathbb{Z}) \\cup\\left\\{r+1+x \\mid x \\in C_{r}\\right\\} . \\tag{2}\n\\end{equation*}\n$$\nFor any finite set of integers $J$ we denote the sum of its elements by $\\sum J$. Now the equations (1) and (2) give rise to two ways of computing $\\sum B_{r}$ and the comparison of both methods leads to\n$$\n\\sum M+\\sum_{i=1}^{r} s_{i}=\\sum_{i=1}^{r} i+b(r+1)+\\sum C_{r}\n$$\nor in other words to\n$$\n\\begin{equation*}\n\\sum M+\\sum_{i=1}^{r}\\left(a_{i}-b\\right)=b+\\sum C_{r} \\tag{3}\n\\end{equation*}\n$$\nAfter this preparation, we consider any two integers $m$ and $n$ with $n>m \\geqslant N$. Plugging $r=n$ and $r=m$ into (3) and subtracting the estimates that result, we deduce\n$$\n\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)=\\sum C_{n}-\\sum C_{m}\n$$\nSince $C_{n}$ and $C_{m}$ are subsets of $\\{1,2, \\ldots, 2014\\}$ with $\\left|C_{n}\\right|=\\left|C_{m}\\right|=b-1$, it is clear that the absolute value of the right-hand side of the above inequality attains its largest possible value if either $C_{m}=\\{1,2, \\ldots, b-1\\}$ and $C_{n}=\\{2016-b, \\ldots, 2014\\}$, or the other way around. In these two cases we have\n$$\n\\left|\\sum C_{n}-\\sum C_{m}\\right|=(b-1)(2015-b),\n$$\nso in the general case we find\n$$\n\\left|\\sum_{i=m+1}^{n}\\left(a_{i}-b\\right)\\right| \\leqslant(b-1)(2015-b) \\leqslant\\left(\\frac{(b-1)+(2015-b)}{2}\\right)^{2}=1007^{2}\n$$\nas desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32007,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a nonempty set of positive integers. We say that a positive integer $n$ is clean if it has a unique representation as a sum of an odd number of distinct elements from $S$. Prove that there exist infinitely many positive integers that are not clean.\n(U.S.A.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define an odd (respectively, even) representation of $n$ to be a representation of $n$ as a sum of an odd (respectively, even) number of distinct elements of $S$. Let $\\mathbb{Z}_{>0}$ denote the set of all positive integers.\nSuppose, to the contrary, that there exist only finitely many positive integers that are not clean. Therefore, there exists a positive integer $N$ such that every integer $n > N$ has exactly one odd representation.\nClearly, $S$ is infinite. We now claim the following properties of odd and even representations.\n\nProperty 1. Any positive integer $n$ has at most one odd and at most one even representation.\n\nProof. We first show that every integer $n$ has at most one even representation. Since $S$ is infinite, there exists $x \\in S$ such that $x > \\max \\{ n, N \\}$. Then, the number $n + x$ must be clean, and $x$ does not appear in any even representation of $n$. If $n$ has more than one even representation, then we obtain two distinct odd representations of $n + x$ by adding $x$ to the even representations of $n$, which is impossible. Therefore, $n$ can have at most one even representation.\n\nSimilarly, there exist two distinct elements $y, z \\in S$ such that $y, z > \\max \\{ n, N \\}$. If $n$ has more than one odd representation, then we obtain two distinct odd representations of $n + y + z$ by adding $y$ and $z$ to the odd representations of $n$. This is again a contradiction.\n\nProperty 2. Fix $s \\in S$. Suppose that a number $n > N$ has no even representation. Then $n + 2 a s$ has an even representation containing $s$ for all integers $a \\geqslant 1$.\n\nProof. It is sufficient to prove the following statement: If $n$ has no even representation without $s$, then $n + 2s$ has an even representation containing $s$ (and hence no even representation without $s$ by Property 1).\n\nNotice that the odd representation of $n + s$ does not contain $s$; otherwise, we have an even representation of $n$ without $s$. Then, adding $s$ to this odd representation of $n + s$, we get that $n + 2s$ has an even representation containing $s$, as desired.\n\nProperty 3. Every sufficiently large integer has an even representation.\n\nProof. Fix any $s \\in S$, and let $r$ be an arbitrary element in $\\{ 1, 2, \\ldots, 2s \\}$. Then, Property 2 implies that the set $Z_r = \\{ r + 2 a s : a \\geqslant 0 \\}$ contains at most one number exceeding $N$ with no even representation. Therefore, $Z_r$ contains finitely many positive integers with no even representation, and so does $\\mathbb{Z}_{>0} = \\bigcup_{r=1}^{2s} Z_r$.\n\nIn view of Properties 1 and 3, we may assume that $N$ is chosen such that every $n > N$ has exactly one odd and exactly one even representation. In particular, each element $s > N$ of $S$ has an even representation.\n\nProperty 4. For any $s, t \\in S$ with $N < s < t$, the even representation of $t$ contains $s$.\n\nProof. Suppose the contrary. Then, $s + t$ has at least two odd representations: one obtained by adding $s$ to the even representation of $t$ and one obtained by adding $t$ to the even representation of $s$. Since the latter does not contain $s$, these two odd representations of $s + t$ are distinct, a contradiction.\n\nLet $s_1 < s_2 < \\cdots$ be all the elements of $S$, and set $\\sigma_n = \\sum_{i=1}^n s_i$ for each nonnegative integer $n$. Fix an integer $k$ such that $s_k > N$. Then, Property 4 implies that for every $i > k$ the even representation of $s_i$ contains all the numbers $s_k, s_{k+1}, \\ldots, s_{i-1}$. Therefore,\n$$\n\\begin{equation*}\ns_i = s_k + s_{k+1} + \\cdots + s_{i-1} + R_i = \\sigma_{i-1} - \\sigma_{k-1} + R_i \\tag{1}\n\\end{equation*}\n$$\nwhere $R_i$ is a sum of some of $s_1, \\ldots, s_{k-1}$. In particular, $0 \\leqslant R_i \\leqslant s_1 + \\cdots + s_{k-1} = \\sigma_{k-1}$.\n\nLet $j_0$ be an integer satisfying $j_0 > k$ and $\\sigma_{j_0} > 2 \\sigma_{k-1}$. Then (1) shows that, for every $j > j_0$,\n$$\n\\begin{equation*}\ns_{j+1} \\geqslant \\sigma_j - \\sigma_{k-1} > \\sigma_j / 2 \\tag{2}\n\\end{equation*}\n$$\nNext, let $p > j_0$ be an index such that $R_p = \\min_{i > j_0} R_i$. Then,\n$$\ns_{p+1} = s_k + s_{k+1} + \\cdots + s_p + R_{p+1} = (s_p - R_p) + s_p + R_{p+1} \\geqslant 2 s_p .\n$$\nTherefore, there is no element of $S$ larger than $s_p$ but smaller than $2 s_p$. It follows that the even representation $\\tau$ of $2 s_p$ does not contain any element larger than $s_p$. On the other hand, inequality (2) yields $2 s_p > s_1 + \\cdots + s_{p-1}$, so $\\tau$ must contain a term larger than $s_{p-1}$. Thus, it must contain $s_p$. After removing $s_p$ from $\\tau$, we have that $s_p$ has an odd representation not containing $s_p$, which contradicts Property 1 since $s_p$ itself also forms an odd representation of $s_p$.\nWe will also use Property 1 from Solution 1.\n\nWe first define some terminology and notations used in this solution. Let $\\mathbb{Z}_{\\geqslant 0}$ denote the set of all nonnegative integers. All sums mentioned are regarded as sums of distinct elements of $S$. Moreover, a sum is called even or odd depending on the parity of the number of terms in it. All closed or open intervals refer to sets of all integers inside them, e.g., $[a, b] = \\{ x \\in \\mathbb{Z} : a \\leqslant x \\leqslant b \\}$.\n\nAgain, let $s_1 < s_2 < \\cdots$ be all elements of $S$, and denote $\\sigma_n = \\sum_{i=1}^n s_i$ for each positive integer $n$. Let $O_n$ (respectively, $E_n$) be the set of numbers representable as an odd (respectively, even) sum of elements of $\\{ s_1, \\ldots, s_n \\}$. Set $E = \\bigcup_{n=1}^{\\infty} E_n$ and $O = \\bigcup_{n=1}^{\\infty} O_n$. We assume that $0 \\in E_n$ since 0 is representable as a sum of 0 terms.\n\nWe now proceed to our proof. Assume, to the contrary, that there exist only finitely many positive integers that are not clean and denote the number of non-clean positive integers by $m-1$. Clearly, $S$ is infinite. By Property 1 from Solution 1, every positive integer $n$ has at most one odd and at most one even representation.\n\nStep 1. We estimate $s_{n+1}$ and $\\sigma_{n+1}$.\n\nUpper bounds: Property 1 yields $|O_n| = |E_n| = 2^{n-1}$, so $|[1, 2^{n-1} + m] \\setminus O_n| \\geqslant m$. Hence, there exists a clean integer $x_n \\in [1, 2^{n-1} + m] \\setminus O_n$. The definition of $O_n$ then yields that the odd representation of $x_n$ contains a term larger than $s_n$. Therefore, $s_{n+1} \\leqslant x_n \\leqslant 2^{n-1} + m$ for every positive integer $n$. Moreover, since $s_1$ is the smallest clean number, we get $\\sigma_1 = s_1 \\leqslant m$. Then,\n$$\n\\sigma_{n+1} = \\sum_{i=2}^{n+1} s_i + s_1 \\leqslant \\sum_{i=2}^{n+1} (2^{i-2} + m) + m = 2^n - 1 + (n+1)m\n$$\nfor every positive integer $n$. Notice that this estimate also holds for $n = 0$.\n\nLower bounds: Since $O_{n+1} \\subseteq [1, \\sigma_{n+1}]$, we have $\\sigma_{n+1} \\geqslant |O_{n+1}| = 2^n$ for all positive integers $n$. Then,\n$$\ns_{n+1} = \\sigma_{n+1} - \\sigma_n \\geqslant 2^n - (2^{n-1} - 1 + n m) = 2^{n-1} + 1 - n m\n$$\nfor every positive integer $n$.\n\nCombining the above inequalities, we have\n$$\n\\begin{equation*}\n2^{n-1} + 1 - n m \\leqslant s_{n+1} \\leqslant 2^{n-1} + m \\quad \\text{ and } \\quad 2^n \\leqslant \\sigma_{n+1} \\leqslant 2^n - 1 + (n+1)m, \\tag{3}\n\\end{equation*}\n$$\nfor every positive integer $n$.\n\nStep 2. We prove Property 3 from Solution 1.\n\nFor every integer $x$ and set of integers $Y$, define $x \\pm Y = \\{ x \\pm y : y \\in Y \\}$.\n\nIn view of Property 1, we get\n$$\nE_{n+1} = E_n \\sqcup (s_{n+1} + O_n) \\quad \\text{ and } \\quad O_{n+1} = O_n \\sqcup (s_{n+1} + E_n)\n$$\nwhere $\\sqcup$ denotes the disjoint union operator. Notice also that $s_{n+2} \\geqslant 2^n + 1 - (n+1)m > 2^{n-1} - 1 + n m \\geqslant \\sigma_n$ for every sufficiently large $n$. We now claim the following.\n\nClaim 1. $\\left( \\sigma_n - s_{n+1}, s_{n+2} - s_{n+1} \\right) \\subseteq E_n$ for every sufficiently large $n$.\n\nProof. For sufficiently large $n$, all elements of $\\left( \\sigma_n, s_{n+2} \\right)$ are clean. Clearly, the elements of $\\left( \\sigma_n, s_{n+2} \\right)$ can be in neither $O_n$ nor $O \\setminus O_{n+1}$. So, $\\left( \\sigma_n, s_{n+2} \\right) \\subseteq O_{n+1} \\setminus O_n = s_{n+1} + E_n$, which yields the claim.\n\nNow, Claim 1 together with inequalities (3) implies that, for all sufficiently large $n$,\n$$\nE \\supseteq E_n \\supseteq \\left( \\sigma_n - s_{n+1}, s_{n+2} - s_{n+1} \\right) \\supseteq \\left( 2 n m, 2^{n-1} - (n+2)m \\right)\n$$\nThis easily yields that $\\mathbb{Z}_{\\geqslant 0} \\setminus E$ is also finite. Since $\\mathbb{Z}_{\\geqslant 0} \\setminus O$ is also finite, by Property 1, there exists a positive integer $N$ such that every integer $n > N$ has exactly one even and one odd representation.\n\nStep 3. We investigate the structures of $E_n$ and $O_n$.\n\nSuppose that $z \\in E_{2n}$. Since $z$ can be represented as an even sum using $\\{ s_1, s_2, \\ldots, s_{2n} \\}$, so can its complement $\\sigma_{2n} - z$. Thus, we get $E_{2n} = \\sigma_{2n} - E_{2n}$. Similarly, we have\n$$\n\\begin{equation*}\nE_{2n} = \\sigma_{2n} - E_{2n}, \\quad O_{2n} = \\sigma_{2n} - O_{2n}, \\quad E_{2n+1} = \\sigma_{2n+1} - O_{2n+1}, \\quad O_{2n+1} = \\sigma_{2n+1} - E_{2n+1} \\tag{4}\n\\end{equation*}\n$$\n\nClaim 2. For every sufficiently large $n$, we have\n$$\n[0, \\sigma_n] \\supseteq O_n \\supseteq (N, \\sigma_n - N) \\quad \\text{ and } \\quad [0, \\sigma_n] \\supseteq E_n \\supseteq (N, \\sigma_n - N)\n$$\n\nProof. Clearly $O_n, E_n \\subseteq [0, \\sigma_n]$ for every positive integer $n$. We now prove $O_n, E_n \\supseteq (N, \\sigma_n - N)$. Taking $n$ sufficiently large, we may assume that $s_{n+1} \\geqslant 2^{n-1} + 1 - n m > \\frac{1}{2} (2^{n-1} - 1 + n m) \\geqslant \\sigma_n / 2$. Therefore, the odd representation of every element of $(N, \\sigma_n / 2]$ cannot contain a term larger than $s_n$. Thus, $(N, \\sigma_n / 2] \\subseteq O_n$. Similarly, since $s_{n+1} + s_1 > \\sigma_n / 2$, we also have $(N, \\sigma_n / 2] \\subseteq E_n$. Equations (4) then yield that, for sufficiently large $n$, the interval $(N, \\sigma_n - N)$ is a subset of both $O_n$ and $E_n$, as desired.\n\nStep 4. We obtain a final contradiction.\n\nNotice that $0 \\in \\mathbb{Z}_{\\geqslant 0} \\setminus O$ and $1 \\in \\mathbb{Z}_{\\geqslant 0} \\setminus E$. Therefore, the sets $\\mathbb{Z}_{\\geqslant 0} \\setminus O$ and $\\mathbb{Z}_{\\geqslant 0} \\setminus E$ are nonempty. Denote $o = \\max (\\mathbb{Z}_{\\geqslant 0} \\setminus O)$ and $e = \\max (\\mathbb{Z}_{\\geqslant 0} \\setminus E)$. Observe also that $e, o \\leqslant N$.\n\nTaking $k$ sufficiently large, we may assume that $\\sigma_{2k} > 2N$ and that Claim 2 holds for all $n \\geqslant 2k$. Due to (4) and Claim 2, we have that $\\sigma_{2k} - e$ is the minimal number greater than $N$ which is not in $E_{2k}$, i.e., $\\sigma_{2k} - e = s_{2k+1} + s_1$. Similarly,\n$$\n\\sigma_{2k} - o = s_{2k+1}, \\quad \\sigma_{2k+1} - e = s_{2k+2}, \\quad \\text{ and } \\quad \\sigma_{2k+1} - o = s_{2k+2} + s_1 .\n$$\nTherefore, we have\n$$\n\\begin{aligned}\ns_1 & = (s_{2k+1} + s_1) - s_{2k+1} = (\\sigma_{2k} - e) - (\\sigma_{2k} - o) = o - e \\\\\n& = (\\sigma_{2k+1} - e) - (\\sigma_{2k+1} - o) = s_{2k+2} - (s_{2k+2} + s_1) = -s_1\n\\end{aligned}\n$$\nwhich is impossible since $s_1 > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32008,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a company of people some pairs are enemies. A group of people is called unsociable if the number of members in the group is odd and at least $3$, and it is possible to arrange all its members around a round table so that every two neighbors are enemies. Given that there are at most $2015$ unsociable groups, prove that it is possible to partition the company into $11$ parts so that no two enemies are in the same part.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove the following more general statement.\n\n**Claim.** Let $G$ be a graph with chromatic number $k \\geqslant 3$. Then $G$ contains at least $2^{k-1}-k$ unsociable groups.\n\nRecall that the chromatic number of $G$ is the least $k$ such that a proper coloring\n$$\nV=V_{1} \\sqcup \\cdots \\sqcup V_{k}\n$$\nexists. In view of $2^{11}-12>2015$, the claim implies the problem statement.\n\nLet $G$ be a graph with chromatic number $k$. We say that a proper coloring (1) of $G$ is leximinimal, if the $k$-tuple $( |V_{1}|,|V_{2}|, \\ldots,|V_{k}| )$ is lexicographically minimal; in other words, the following conditions are satisfied: the number $n_{1}=|V_{1}|$ is minimal; the number $n_{2}=|V_{2}|$ is minimal, subject to the previously chosen value of $n_{1}$; $\\ldots$; the number $n_{k-1}=|V_{k-1}|$ is minimal, subject to the previously chosen values of $n_{1}, \\ldots, n_{k-2}$.\n\nThe following lemma is the core of the proof.\n\n**Lemma 1.** Suppose that $G=(V, E)$ is a graph with odd chromatic number $k \\geqslant 3$, and let (1) be one of its leximinimal colorings. Then $G$ contains an odd cycle which visits all color classes $V_{1}, V_{2}, \\ldots, V_{k}$.\n\n*Proof of Lemma 1.* Let us call a cycle colorful if it visits all color classes.\n\nDue to the definition of the chromatic number, $V_{1}$ is nonempty. Choose an arbitrary vertex $v \\in V_{1}$. We construct a colorful odd cycle that has only one vertex in $V_{1}$, and this vertex is $v$.\n\nWe draw a subgraph of $G$ as follows. Place $v$ in the center, and arrange the sets $V_{2}, V_{3}, \\ldots, V_{k}$ in counterclockwise circular order around it. For convenience, let $V_{k+1}=V_{2}$. We will draw arrows to add direction to some edges of $G$, and mark the vertices these arrows point to. First we draw arrows from $v$ to all its neighbors in $V_{2}$, and mark all those neighbors. If some vertex $u \\in V_{i}$ with $i \\in\\{2,3, \\ldots, k\\}$ is already marked, we draw arrows from $u$ to all its neighbors in $V_{i+1}$ which are not marked yet, and we mark all of them. We proceed doing this as long as it is possible. The process of marking is exemplified in Figure 1.\n\nNotice that by the rules of our process, in the final state, marked vertices in $V_{i}$ cannot have unmarked neighbors in $V_{i+1}$. Moreover, $v$ is connected to all marked vertices by directed paths.\n\nNow move each marked vertex to the next color class in circular order (see an example in Figure 3). In view of the arguments above, the obtained coloring $V_{1} \\sqcup W_{2} \\sqcup \\cdots \\sqcup W_{k}$ is proper. Notice that $v$ has a neighbor $w \\in W_{2}$, because otherwise\n$$\n\\left(V_{1} \\backslash\\{v\\}\\right) \\sqcup\\left(W_{2} \\cup\\{v\\}\\right) \\sqcup W_{3} \\sqcup \\cdots \\sqcup W_{k}\n$$\nwould be a proper coloring lexicographically smaller than (1). If $w$ was unmarked, i.e., $w$ was an element of $V_{2}$, then it would be marked at the beginning of the process and thus moved to $V_{3}$, which did not happen. Therefore, $w$ is marked and $w \\in V_{k}$.\n\n![](attached_image_1.png)\nFigure 1\nFigure 2\n\nSince $w$ is marked, there exists a directed path from $v$ to $w$. This path moves through the sets $V_{2}, \\ldots, V_{k}$ in circular order, so the number of edges in it is divisible by $k-1$ and thus even. Closing this path by the edge $w \\rightarrow v$, we get a colorful odd cycle, as required. $\\square$\n\n*Proof of the claim.* Let us choose a leximinimal coloring (1) of $G$. For every set $C \\subseteq\\{1,2, \\ldots, k\\}$ such that $|C|$ is odd and greater than $1$, we will provide an odd cycle visiting exactly those color classes whose indices are listed in the set $C$. This property ensures that we have different cycles for different choices of $C$, and it proves the claim because there are $2^{k-1}-k$ choices for the set $C$.\n\nLet $V_{C}=\\bigcup_{c \\in C} V_{c}$, and let $G_{C}$ be the induced subgraph of $G$ on the vertex set $V_{C}$. We also have the induced coloring of $V_{C}$ with $|C|$ colors; this coloring is of course proper. Notice further that the induced coloring is leximinimal: if we had a lexicographically smaller coloring $\\left(W_{c}\\right)_{c \\in C}$ of $G_{C}$, then these classes, together the original color classes $V_{i}$ for $i \\notin C$, would provide a proper coloring which is lexicographically smaller than (1). Hence Lemma 1, applied to the subgraph $G_{C}$ and its leximinimal coloring $\\left(V_{c}\\right)_{c \\in C}$, provides an odd cycle that visits exactly those color classes that are listed in the set $C$. $\\square$\nWe provide a different proof of the claim from the previous solution.\n\nWe say that a graph is critical if deleting any vertex from the graph decreases the graph's chromatic number. Obviously every graph contains a critical induced subgraph with the same chromatic number.\n\n**Lemma 2.** Suppose that $G=(V, E)$ is a critical graph with chromatic number $k \\geqslant 3$. Then every vertex $v$ of $G$ is contained in at least $2^{k-2}-1$ unsociable groups.\n\n*Proof.* For every set $X \\subseteq V$, denote by $n(X)$ the number of neighbors of $v$ in the set $X$.\n\nSince $G$ is critical, there exists a proper coloring of $G \\backslash\\{v\\}$ with $k-1$ colors, so there exists a proper coloring $V=V_{1} \\sqcup V_{2} \\sqcup \\cdots \\sqcup V_{k}$ of $G$ such that $V_{1}=\\{v\\}$. Among such colorings, take one for which the sequence $\\left(n\\left(V_{2}\\right), n\\left(V_{3}\\right), \\ldots, n\\left(V_{k}\\right)\\right)$ is lexicographically minimal. Clearly, $n\\left(V_{i}\\right)>0$ for every $i=2,3, \\ldots, k$; otherwise $V_{2} \\sqcup \\ldots \\sqcup V_{i-1} \\sqcup\\left(V_{i} \\cup V_{1}\\right) \\sqcup V_{i+1} \\sqcup \\ldots V_{k}$ would be a proper coloring of $G$ with $k-1$ colors.\n\nWe claim that for every $C \\subseteq\\{2,3, \\ldots, k\\}$ with $|C| \\geqslant 2$ being even, $G$ contains an unsociable group so that the set of its members' colors is precisely $C \\cup\\{1\\}$. Since the number of such sets $C$ is $2^{k-2}-1$, this proves the lemma. Denote the elements of $C$ by $c_{1}, \\ldots, c_{2 \\ell}$ in increasing order. For brevity, let $U_{i}=V_{c_{i}}$. Denote by $N_{i}$ the set of neighbors of $v$ in $U_{i}$.\n\nWe show that for every $i=1, \\ldots, 2 \\ell-1$ and $x \\in N_{i}$, the subgraph induced by $U_{i} \\cup U_{i+1}$ contains a path that connects $x$ with another point in $N_{i+1}$. For the sake of contradiction, suppose that no such path exists. Let $S$ be the set of vertices that lie in the connected component of $x$ in the subgraph induced by $U_{i} \\cup U_{i+1}$, and let $P=U_{i} \\cap S$, and $Q=U_{i+1} \\cap S$ (see Figure 3). Since $x$ is separated from $N_{i+1}$, the sets $Q$ and $N_{i+1}$ are disjoint. So, if we re-color $G$ by replacing $U_{i}$ and $U_{i+1}$ by $\\left(U_{i} \\cup Q\\right) \\backslash P$ and $\\left(U_{i+1} \\cup P\\right) \\backslash Q$, respectively, we obtain a proper coloring such that $n\\left(U_{i}\\right)=n\\left(V_{c_{i}}\\right)$ is decreased and only $n\\left(U_{i+1}\\right)=n\\left(V_{c_{i+1}}\\right)$ is increased. That contradicts the lexicographical minimality of $\\left(n\\left(V_{2}\\right), n\\left(V_{3}\\right), \\ldots, n\\left(V_{k}\\right)\\right)$.\n\n![](attached_image_2.png)\nFigure 3\n\nNext, we build a path through $U_{1}, U_{2}, \\ldots, U_{2 \\ell}$ as follows. Let the starting point of the path be an arbitrary vertex $v_{1}$ in the set $N_{1}$. For $i \\leqslant 2 \\ell-1$, if the vertex $v_{i} \\in N_{i}$ is already defined, connect $v_{i}$ to some vertex in $N_{i+1}$ in the subgraph induced by $U_{i} \\cup U_{i+1}$, and add these edges to the path. Denote the new endpoint of the path by $v_{i+1}$; by the construction we have $v_{i+1} \\in N_{i+1}$ again, so the process can be continued. At the end we have a path that starts at $v_{1} \\in N_{1}$ and ends at some $v_{2 \\ell} \\in N_{2 \\ell}$. Moreover, all edges in this path connect vertices in neighboring classes: if a vertex of the path lies in $U_{i}$, then the next vertex lies in $U_{i+1}$ or $U_{i-1}$. Notice that the path is not necessary simple, so take a minimal subpath of it. The minimal subpath is simple and connects the same endpoints $v_{1}$ and $v_{2 \\ell}$. The property that every edge steps to a neighboring color class (i.e., from $U_{i}$ to $U_{i+1}$ or $U_{i-1}$ ) is preserved. So the resulting path also visits all of $U_{1}, \\ldots, U_{2 \\ell}$, and its length must be odd. Closing the path with the edges $v v_{1}$ and $v_{2 \\ell} v$ we obtain the desired odd cycle (see Figure 4).\n\n![](attached_image_3.png)\nFigure 4\n\nNow we prove the claim by induction on $k \\geqslant 3$. The base case $k=3$ holds by applying Lemma 2 to a critical subgraph. For the induction step, let $G_{0}$ be a critical $k$-chromatic subgraph of $G$, and let $v$ be an arbitrary vertex of $G_{0}$. By Lemma 2, $G_{0}$ has at least $2^{k-2}-1$ unsociable groups containing $v$. On the other hand, the graph $G_{0} \\backslash\\{v\\}$ has chromatic number $k-1$, so it contains at least $2^{k-2}-(k-1)$ unsociable groups by the induction hypothesis. Altogether, this gives $2^{k-2}-1+2^{k-2}-(k-1)=2^{k-1}-k$ distinct unsociable groups in $G_{0}$ (and thus in $G$ ).",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32009,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with orthocenter $H$. Let $G$ be the point such that the quadrilateral $ABGH$ is a parallelogram. Let $I$ be the point on the line $GH$ such that $AC$ bisects $HI$. Suppose that the line $AC$ intersects the circumcircle of the triangle $GCI$ at $C$ and $J$. Prove that $IJ = AH$.\n(Australia)\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Since $HG \\parallel AB$ and $BG \\parallel AH$, we have $BG \\perp BC$ and $CH \\perp GH$. Therefore, the quadrilateral $BGCH$ is cyclic. Since $H$ is the orthocenter of the triangle $ABC$, we have $\\angle HAC = 90^\\circ - \\angle ACB = \\angle CBH$. Using that $BGCH$ and $CGJI$ are cyclic quadrilaterals, we get\n$$\n\\angle CJI = \\angle CGH = \\angle CBH = \\angle HAC.\n$$\nLet $M$ be the intersection of $AC$ and $GH$, and let $D \\neq A$ be the point on the line $AC$ such that $AH = HD$. Then $\\angle MJI = \\angle HAC = \\angle MDH$.\nSince $\\angle MJI = \\angle MDH$, $\\angle IMJ = \\angle HMD$, and $IM = MH$, the triangles $IMJ$ and $HMD$ are congruent, and thus $IJ = HD = AH$.\nObtain $\\angle CGH = \\angle HAC$ as in the previous solution. In the parallelogram $ABGH$ we have $\\angle BAH = \\angle HGB$. It follows that\n$$\n\\angle HMC = \\angle BAC = \\angle BAH + \\angle HAC = \\angle HGB + \\angle CGH = \\angle CGB.\n$$\nSo the right triangles $CMH$ and $CGB$ are similar. Also, in the circumcircle of triangle $GCI$ we have similar triangles $MIJ$ and $MCG$. Therefore,\n$$\n\\frac{IJ}{CG} = \\frac{MI}{MC} = \\frac{MH}{MC} = \\frac{GB}{GC} = \\frac{AH}{CG}.\n$$\nHence $IJ = AH$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32010,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle inscribed into a circle $\\Omega$ with center $O$. A circle $\\Gamma$ with center $A$ meets the side $BC$ at points $D$ and $E$ such that $D$ lies between $B$ and $E$. Moreover, let $F$ and $G$ be the common points of $\\Gamma$ and $\\Omega$. We assume that $F$ lies on the arc $AB$ of $\\Omega$ not containing $C$, and $G$ lies on the arc $AC$ of $\\Omega$ not containing $B$. The circumcircles of the triangles $BDF$ and $CEG$ meet the sides $AB$ and $AC$ again at $K$ and $L$, respectively. Suppose that the lines $FK$ and $GL$ are distinct and intersect at $X$. Prove that the points $A$, $X$, and $O$ are collinear.\n\n![](attached_image_1.png)\nFigure 1\n",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It suffices to prove that the lines $FK$ and $GL$ are symmetric about $AO$. Now the segments $AF$ and $AG$, being chords of $\\Omega$ with the same length, are clearly symmetric with respect to $AO$. Hence it is enough to show\n$$\n\\begin{equation*}\n\\angle KFA = \\angle AGL . \\tag{1}\n\\end{equation*}\n$$\nLet us denote the circumcircles of $BDF$ and $CEG$ by $\\omega_B$ and $\\omega_C$, respectively. To prove (1), we start from\n$$\n\\angle KFA = \\angle DFG + \\angle GFA - \\angle DFK .\n$$\nIn view of the circles $\\omega_B$, $\\Gamma$, and $\\Omega$, this may be rewritten as\n$$\n\\angle KFA = \\angle CEG + \\angle GBA - \\angle DBK = \\angle CEG - \\angle CBG .\n$$\nDue to the circles $\\omega_C$ and $\\Omega$, we obtain $\\angle KFA = \\angle CLG - \\angle CAG = \\angle AGL$. Thereby the problem is solved.\nAgain, we denote the circumcircle of $BDKF$ by $\\omega_B$. In addition, we set $\\alpha= \\angle BAC$, $\\varphi=\\angle ABF$, and $\\psi=\\angle EDA=\\angle AED$ (see Figure 2). Notice that $AF=AG$ entails $\\varphi=\\angle GCA$, so all three of $\\alpha, \\varphi$, and $\\psi$ respect the \"symmetry\" between $B$ and $C$ of our configuration. Again, we reduce our task to proving (1).\nThis time, we start from\n$$\n2 \\angle KFA = 2(\\angle DFA - \\angle DFK) .\n$$\nSince the triangle $AFD$ is isosceles, we have\n$$\n\\angle DFA = \\angle ADF = \\angle EDF - \\psi = \\angle BFD + \\angle EBF - \\psi .\n$$\nMoreover, because of the circle $\\omega_B$ we have $\\angle DFK = \\angle CBA$. Altogether, this yields\n$$\n2 \\angle KFA = \\angle DFA + (\\angle BFD + \\angle EBF - \\psi) - 2 \\angle CBA,\n$$\nwhich simplifies to\n$$\n2 \\angle KFA = \\angle BFA + \\varphi - \\psi - \\angle CBA .\n$$\nNow the quadrilateral $AFBC$ is cyclic, so this entails $2 \\angle KFA = \\alpha + \\varphi - \\psi$.\nDue to the \"symmetry\" between $B$ and $C$ alluded to above, this argument also shows that $2 \\angle AGL = \\alpha + \\varphi - \\psi$. This concludes the proof of (1).\n\n![](attached_image_2.png)\nFigure 2",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32011,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C$ be an acute triangle, and let $M$ be the midpoint of $A C$. A circle $\\omega$ passing through $B$ and $M$ meets the sides $A B$ and $B C$ again at $P$ and $Q$, respectively. Let $T$ be the point such that the quadrilateral $B P T Q$ is a parallelogram. Suppose that $T$ lies on the circumcircle of the triangle $A B C$. Determine all possible values of $B T / B M$.",
    "options": [],
    "answer": "sqrt(2)",
    "solution": "Let $S$ be the center of the parallelogram $B P T Q$, and let $B' \\neq B$ be the point on the ray $B M$ such that $B M = M B'$ (see Figure 1).\nIt follows that $A B C B'$ is a parallelogram. Then, $\\angle A B B' = \\angle P Q M$ and $\\angle B B' A = \\angle B' B C = \\angle M P Q$, and so the triangles $A B B'$ and $M Q P$ are similar. It follows that $A M$ and $M S$ are corresponding medians in these triangles. Hence,\n$$\n\\angle S M P = \\angle B' A M = \\angle B C A = \\angle B T A .\n$$\nSince $\\angle A C T = \\angle P B T$ and $\\angle T A C = \\angle T B C = \\angle B T P$, the triangles $T C A$ and $P B T$ are similar. Again, as $T M$ and $P S$ are corresponding medians in these triangles, we have\n$$\n\\angle M T A = \\angle T P S = \\angle B Q P = \\angle B M P .\n$$\nNow we deal separately with two cases.\n\nCase 1. $S$ does not lie on $B M$. Since the configuration is symmetric between $A$ and $C$, we may assume that $S$ and $A$ lie on the same side with respect to the line $B M$.\nApplying the previous results, we get\n$$\n\\angle B M S = \\angle B M P - \\angle S M P = \\angle M T A - \\angle B T A = \\angle M T B,\n$$\nand so the triangles $B S M$ and $B M T$ are similar. We now have $B M^2 = B S \\cdot B T = B T^2 / 2$, so $B T = \\sqrt{2} B M$.\n\nCase 2. $S$ lies on $B M$. It follows from the previous results that $\\angle B C A = \\angle M T A = \\angle B Q P = \\angle B M P$ (see Figure 2). Thus, $P Q \\parallel A C$ and $P M \\parallel A T$. Hence, $B S / B M = B P / B A = B M / B T$, so $B T^2 = 2 B M^2$ and $B T = \\sqrt{2} B M$.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\nAgain, we denote by $\\Omega$ the circumcircle of the triangle $A B C$.\nChoose the points $X$ and $Y$ on the rays $B A$ and $B C$ respectively, so that $\\angle M X B = \\angle M B C$ and $\\angle B Y M = \\angle A B M$ (see Figure 4). Then the triangles $B M X$ and $Y M B$ are similar. Since $\\angle X P M = \\angle B Q M$, the points $P$ and $Q$ correspond to each other in these triangles. So, if $\\overrightarrow{B P} = \\mu \\cdot \\overrightarrow{B X}$, then $\\overrightarrow{B Q} = (1-\\mu) \\cdot \\overrightarrow{B Y}$. Thus\n$$\n\\overrightarrow{B T} = \\overrightarrow{B P} + \\overrightarrow{B Q} = \\overrightarrow{B Y} + \\mu \\cdot (\\overrightarrow{B X} - \\overrightarrow{B Y}) = \\overrightarrow{B Y} + \\mu \\cdot \\overrightarrow{Y X},\n$$\nwhich means that $T$ lies on the line $X Y$.\nLet $B' \\neq B$ be the point on the ray $B M$ such that $B M = M B'$. Then $\\angle M B' A = \\angle M B C = \\angle M X B$ and $\\angle C B' M = \\angle A B M = \\angle B Y M$. This means that the triangles $B M X$, $B A B'$, $Y M B$, and $B' C B$ are all similar; hence $B A \\cdot B X = B M \\cdot B B' = B C \\cdot B Y$. Thus there exists an inversion centered at $B$ which swaps $A$ with $X$, $M$ with $B'$, and $C$ with $Y$. This inversion then swaps $\\Omega$ with the line $X Y$, and hence it preserves $T$. Therefore, we have $B T^2 = B M \\cdot B B' = 2 B M^2$, and $B T = \\sqrt{2} B M$.\n\n![](attached_image_3.png)\nFigure 4\nWe begin with the following lemma.\n\nLemma. Let $A B C T$ be a cyclic quadrilateral. Let $P$ and $Q$ be points on the sides $B A$ and $B C$ respectively, such that $B P T Q$ is a parallelogram. Then $B P \\cdot B A + B Q \\cdot B C = B T^2$.\n\nProof. Let the circumcircle of the triangle $Q T C$ meet the line $B T$ again at $J$ (see Figure 5). The power of $B$ with respect to this circle yields\n$$\nB Q \\cdot B C = B J \\cdot B T.\n$$\nWe also have $\\angle T J Q = 180^{\\circ} - \\angle Q C T = \\angle T A B$ and $\\angle Q T J = \\angle A B T$, and so the triangles $T J Q$ and $B A T$ are similar. We now have $T J / T Q = B A / B T$. Therefore,\n$$\nT J \\cdot B T = T Q \\cdot B A = B P \\cdot B A.\n$$\nCombining the above, we get the desired result. $\\square$\n\nLet $X$ and $Y$ be the midpoints of $B A$ and $B C$ respectively (see Figure 6). Applying the lemma to the cyclic quadrilaterals $P B Q M$ and $A B C T$, we obtain\n$$\nB X \\cdot B P + B Y \\cdot B Q = B M^2\n$$\nand\n$$\nB P \\cdot B A + B Q \\cdot B C = B T^2.\n$$\nSince $B A = 2 B X$ and $B C = 2 B Y$, we have $B T^2 = 2 B M^2$, and so $B T = \\sqrt{2} B M$.\n\n![](attached_image_4.png)\nFigure 5\n![](attached_image_5.png)\nFigure 6",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32012,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C$ be an acute triangle with $A B > A C$, and let $\\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\\Gamma$ that satisfy $\\angle A Q H = 90^{\\circ}$ and $\\angle Q K H = 90^{\\circ}$. Prove that the circumcircles of the triangles $K Q H$ and $K F M$ are tangent to each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider any point $T$ such that $T K$ is tangent to the circle $K Q H$ at $K$ with $Q$ and $T$ lying on different sides of $K H$ (see Figure 1). Then $\\angle H K T = \\angle H Q K$ and we are to prove that $\\angle M K T = \\angle C F K$. Thus it remains to show that $\\angle H Q K = \\angle C F K + \\angle H K M$. Due to $\\angle H Q K = 90^{\\circ} - \\angle Q' H A'$, and $\\angle C F K = 90^{\\circ} - \\angle K F A$, this means the same as $\\angle Q' H A' = \\angle K F A - \\angle H K M$. Now, since the triangles $K H E$ and $A H Q'$ are similar with $F$ and $J$ being the midpoints of corresponding sides, we have $\\angle K F A = \\angle H J A$, and analogously one may obtain $\\angle H K M = \\angle J Q H$. Thereby our task is reduced to verifying\n$$\n\\angle Q' H A' = \\angle H J A - \\angle J Q H .\n$$\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\nTo avoid confusion, let us draw a new picture at this moment (see Figure 2). Owing to $\\angle Q' H A' = \\angle J Q H + \\angle H J Q$ and $\\angle H J A = \\angle Q J A + \\angle H J Q$, we just have to show that $2 \\angle J Q H = \\angle Q J A$. To this end, it suffices to remark that $A Q A' Q'$ is a rectangle and that $J$, being defined to be the midpoint of $H Q'$, has to lie on the mid parallel of $Q A'$ and $Q' A$.\nWe define the points $A'$ and $E$ and prove that the ray $M H$ passes through $Q$ in the same way as in the first solution. Notice that the points $A'$ and $E$ can play analogous roles to the points $Q$ and $K$, respectively: point $A'$ is the second intersection of the line $M H$ with $\\Gamma$, and $E$ is the point on $\\Gamma$ with the property $\\angle H E A' = 90^{\\circ}$ (see Figure 3).\n\nIn the circles $K Q H$ and $E A' H$, the line segments $H Q$ and $H A'$ are diameters, respectively; so, these circles have a common tangent $t$ at $H$, perpendicular to $M H$. Let $R$ be the radical center of the circles $A B C, K Q H$ and $E A' H$. Their pairwise radical axes are the lines $Q K$, $A' E$ and the line $t$; they all pass through $R$. Let $S$ be the midpoint of $H R$; by $\\angle Q K H = \\angle H E A' = 90^{\\circ}$, the quadrilateral $H E R K$ is cyclic and its circumcenter is $S$; hence we have $S K = S E = S H$. The line $B C$, being the perpendicular bisector of $H E$, passes through $S$.\nThe circle $H M F$ also is tangent to $t$ at $H$; from the power of $S$ with respect to the circle $H M F$ we have\n$$\nS M \\cdot S F = S H^{2} = S K^{2} .\n$$\nSo, the power of $S$ with respect to the circles $K Q H$ and $K F M$ is $S K^{2}$. Therefore, the line segment $S K$ is tangent to both circles at $K$.\n\n![](attached_image_3.png)\nFigure 3",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32013,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all triples $(a, b, c)$ of positive integers for which $a b - c$, $b c - a$, and $c a - b$ are powers of $2$.\n\nExplanation: A power of $2$ is an integer of the form $2^{n}$, where $n$ denotes some nonnegative integer.",
    "options": [],
    "answer": "All permutations of (2,2,2), (2,2,3), (3,5,7), and (2,6,11).",
    "solution": "It can easily be verified that these sixteen triples are as required. Now let $(a, b, c)$ be any triple with the desired property. If we would have $a = 1$, then both $b - c$ and $c - b$ were powers of $2$, which is impossible since their sum is zero; because of symmetry, this argument shows $a, b, c \\geqslant 2$.\n\nCase 1. Among $a, b$, and $c$ there are at least two equal numbers.\n\nWithout loss of generality we may suppose that $a = b$. Then $a^{2} - c$ and $a(c - 1)$ are powers of $2$. The latter tells us that actually $a$ and $c - 1$ are powers of $2$. So there are nonnegative integers $\\alpha$ and $\\gamma$ with $a = 2^{\\alpha}$ and $c = 2^{\\gamma} + 1$. Since $a^{2} - c = 2^{2\\alpha} - 2^{\\gamma} - 1$ is a power of $2$ and thus incongruent to $-1$ modulo $4$, we must have $\\gamma \\leqslant 1$. Moreover, each of the terms $2^{2\\alpha} - 2$ and $2^{2\\alpha} - 3$ can only be a power of $2$ if $\\alpha = 1$. It follows that the triple $(a, b, c)$ is either $(2, 2, 2)$ or $(2, 2, 3)$.\n\nCase 2. The numbers $a, b$, and $c$ are distinct.\n\nDue to symmetry we may suppose that\n$$\n2 \\leqslant a < b < c\n$$\nWe are to prove that the triple $(a, b, c)$ is either $(2, 6, 11)$ or $(3, 5, 7)$. By our hypothesis, there exist three nonnegative integers $\\alpha, \\beta$, and $\\gamma$ such that\n$$\n\\begin{align*}\nbc - a & = 2^{\\alpha}, \\\\\nac - b & = 2^{\\beta}, \\\\\n\\text{and}\\quad ab - c & = 2^{\\gamma}.\n\\end{align*}\n$$\nEvidently we have\n$$\n\\alpha > \\beta > \\gamma.\n$$\nDepending on how large $a$ is, we divide the argument into two further cases.\n\nCase 2.1. $a = 2$.\n\nWe first prove that $\\gamma = 0$. Assume for the sake of contradiction that $\\gamma > 0$. Then $c$ is even by (4) and, similarly, $b$ is even by (5) and (3). So the left-hand side of (2) is congruent to $2$ modulo $4$, which is only possible if $bc = 4$. As this contradicts (1), we have thereby shown that $\\gamma = 0$, i.e., that $c = 2b - 1$.\n\nNow (3) yields $3b - 2 = 2^{\\beta}$. Due to $b > 2$ this is only possible if $\\beta \\geqslant 4$. If $\\beta = 4$, then we get $b = 6$ and $c = 2 \\cdot 6 - 1 = 11$, which is a solution. It remains to deal with the case $\\beta \\geqslant 5$. Now (2) implies\n$$\n9 \\cdot 2^{\\alpha} = 9b(2b - 1) - 18 = (3b - 2)(6b + 1) - 16 = 2^{\\beta}(2^{\\beta + 1} + 5) - 16\n$$\nand by $\\beta \\geqslant 5$ the right-hand side is not divisible by $32$. Thus $\\alpha \\leqslant 4$ and we get a contradiction to (5).\n\nCase 2.2. $a \\geqslant 3$.\n\nPick an integer $\\vartheta \\in \\{-1, +1\\}$ such that $c - \\vartheta$ is not divisible by $4$. Now\n$$\n2^{\\alpha} + \\vartheta \\cdot 2^{\\beta} = (bc - a\\vartheta^{2}) + \\vartheta(ca - b) = (b + a\\vartheta)(c - \\vartheta)\n$$\nis divisible by $2^{\\beta}$ and, consequently, $b + a\\vartheta$ is divisible by $2^{\\beta - 1}$. On the other hand, $2^{\\beta} = ac - b > (a - 1)c \\geqslant 2c$ implies in view of (1) that $a$ and $b$ are smaller than $2^{\\beta - 1}$. All this is only possible if $\\vartheta = 1$ and $a + b = 2^{\\beta - 1}$. Now (3) yields\n$$\nac - b = 2(a + b)\n$$\nwhence $4b > a + 3b = a(c - 1) \\geqslant ab$, which in turn yields $a = 3$.\n\nSo (6) simplifies to $c = b + 2$ and (2) tells us that $b(b + 2) - 3 = (b - 1)(b + 3)$ is a power of $2$. Consequently, the factors $b - 1$ and $b + 3$ are powers of $2$ themselves. Since their difference is $4$, this is only possible if $b = 5$ and thus $c = 7$. Thereby the solution is complete.\nAs in the beginning of the first solution, we observe that $a, b, c \\geqslant 2$. Depending on the parities of $a, b$, and $c$ we distinguish three cases.\n\nCase 1. The numbers $a, b$, and $c$ are even.\n\nLet $2^{A}, 2^{B}$, and $2^{C}$ be the largest powers of $2$ dividing $a, b$, and $c$ respectively. We may assume without loss of generality that $1 \\leqslant A \\leqslant B \\leqslant C$. Now $2^{B}$ is the highest power of $2$ dividing $ac - b$, whence $ac - b = 2^{B} \\leqslant b$. Similarly, we deduce $bc - a = 2^{A} \\leqslant a$. Adding both estimates we get $(a + b)c \\leqslant 2(a + b)$, whence $c \\leqslant 2$. So $c = 2$ and thus $A = B = C = 1$; moreover, we must have had equality throughout, i.e., $a = 2^{A} = 2$ and $b = 2^{B} = 2$. We have thereby found the solution $(a, b, c) = (2, 2, 2)$.\n\nCase 2. The numbers $a, b$, and $c$ are odd.\n\nIf any two of these numbers are equal, say $a = b$, then $ac - b = a(c - 1)$ has a nontrivial odd divisor and cannot be a power of $2$. Hence $a, b$, and $c$ are distinct. So we may assume without loss of generality that $a < b < c$.\n\nLet $\\alpha$ and $\\beta$ denote the nonnegative integers for which $bc - a = 2^{\\alpha}$ and $ac - b = 2^{\\beta}$ hold. Clearly, we have $\\alpha > \\beta$, and thus $2^{\\beta}$ divides\n$$\na \\cdot 2^{\\alpha} - b \\cdot 2^{\\beta} = a(bc - a) - b(ac - b) = b^{2} - a^{2} = (b + a)(b - a).\n$$\nSince $a$ is odd, it is not possible that both factors $b + a$ and $b - a$ are divisible by $4$. Consequently, one of them has to be a multiple of $2^{\\beta - 1}$. Hence one of the numbers $2(b + a)$ and $2(b - a)$ is divisible by $2^{\\beta}$ and in either case we have\n$$\nac - b = 2^{\\beta} \\leqslant 2(a + b).\n$$\nThis in turn yields $(a - 1)b < ac - b < 4b$ and thus $a = 3$ (recall that $a$ is odd and larger than $1$). Substituting this back into (7) we learn $c \\leqslant b + 2$. But due to the parity $b < c$ entails that $b + 2 \\leqslant c$ holds as well. So we get $c = b + 2$ and from $bc - a = (b - 1)(b + 3)$ being a power of $2$ it follows that $b = 5$ and $c = 7$.\n\nCase 3. Among $a, b$, and $c$ both parities occur.\n\nWithout loss of generality, we suppose that $c$ is odd and that $a \\leqslant b$. We are to show that $(a, b, c)$ is either $(2, 2, 3)$ or $(2, 6, 11)$. As at least one of $a$ and $b$ is even, the expression $ab - c$ is odd; since it is also a power of $2$, we obtain\n$$\nab - c = 1.\n$$\nIf $a = b$, then $c = a^{2} - 1$, and from $ac - b = a(a^{2} - 2)$ being a power of $2$ it follows that both $a$ and $a^{2} - 2$ are powers of $2$, whence $a = 2$. This gives rise to the solution $(2, 2, 3)$.\n\nWe may suppose $a < b$ from now on. As usual, we let $\\alpha > \\beta$ denote the integers satisfying\n$$\n2^{\\alpha} = bc - a \\quad \\text{and} \\quad 2^{\\beta} = ac - b\n$$\nIf $\\beta = 0$ it would follow that $ac - b = ab - c = 1$ and hence that $b = c = 1$, which is absurd. So $\\beta$ and $\\alpha$ are positive and consequently $a$ and $b$ are even. Substituting $c = ab - 1$ into (9) we obtain\n$$\n\\begin{align*}\n2^{\\alpha} & = ab^{2} - (a + b) \\\\\n2^{\\beta} & = a^{2}b - (a + b)\n\\end{align*}\n$$\nThe addition of both equation yields $2^{\\alpha} + 2^{\\beta} = (ab - 2)(a + b)$. Now $ab - 2$ is even but not divisible by $4$, so the highest power of $2$ dividing $a + b$ is $2^{\\beta - 1}$. For this reason, the equations (10) and (11) show that the highest powers of $2$ dividing either of the numbers $ab^{2}$ and $a^{2}b$ is likewise $2^{\\beta - 1}$. Thus there is an integer $\\tau \\geqslant 1$ together with odd integers $A, B$, and $C$ such that $a = 2^{\\tau}A, b = 2^{\\tau}B, a + b = 2^{3\\tau}C$, and $\\beta = 1 + 3\\tau$.\n\nNotice that $A + B = 2^{2\\tau}C \\geqslant 4C$. Moreover, (11) entails $A^{2}B - C = 2$. Thus $8 = 4A^{2}B - 4C \\geqslant 4A^{2}B - A - B \\geqslant A^{2}(3B - 1)$. Since $A$ and $B$ are odd with $A < B$, this is only possible if $A = 1$ and $B = 3$. Finally, one may conclude $C = 1, \\tau = 1, a = 2, b = 6$, and $c = 11$. We have thereby found the triple $(2, 6, 11)$. This completes the discussion of the third case, and hence the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32014,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}_{>0}$ denote the set of positive integers. Consider a function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$. For any $m, n \\in \\mathbb{Z}_{>0}$ we write $f^{n}(m)=\\underbrace{f(f(\\ldots f}_{n}(m) \\ldots))$. Suppose that $f$ has the following two properties:\n(i) If $m, n \\in \\mathbb{Z}_{>0}$, then $\\frac{f^{n}(m)-m}{n} \\in \\mathbb{Z}_{>0}$;\n(ii) The set $\\mathbb{Z}_{>0} \\backslash\\left\\{f(n) \\mid n \\in \\mathbb{Z}_{>0}\\right\\}$ is finite.\nProve that the sequence $f(1)-1, f(2)-2, f(3)-3, \\ldots$ is periodic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We split the solution into three steps. In the first of them, we show that the function $f$ is injective and explain how this leads to a useful visualization of $f$. Then comes the second step, in which most of the work happens: its goal is to show that for any $n \\in \\mathbb{Z}_{>0}$ the sequence $n, f(n), f^{2}(n), \\ldots$ is an arithmetic progression. Finally, in the third step we put everything together, thus solving the problem.\n\nStep 1. We commence by checking that $f$ is injective. For this purpose, we consider any $m, k \\in \\mathbb{Z}_{>0}$ with $f(m)=f(k)$. By (i), every positive integer $n$ has the property that\n$$\n\\frac{k-m}{n}=\\frac{f^{n}(m)-m}{n}-\\frac{f^{n}(k)-k}{n}\n$$\nis a difference of two integers and thus integral as well. But for $n=|k-m|+1$ this is only possible if $k=m$. Thereby, the injectivity of $f$ is established.\n\nNow recall that due to condition (ii) there are finitely many positive integers $a_{1}, \\ldots, a_{k}$ such that $\\mathbb{Z}_{>0}$ is the disjoint union of $\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$ and $\\left\\{f(n) \\mid n \\in \\mathbb{Z}_{>0}\\right\\}$. Notice that by plugging $n=1$ into condition (i) we get $f(m)>m$ for all $m \\in \\mathbb{Z}_{>0}$.\n\nWe contend that every positive integer $n$ may be expressed uniquely in the form $n=f^{j}\\left(a_{i}\\right)$ for some $j \\geqslant 0$ and $i \\in\\{1, \\ldots, k\\}$. The uniqueness follows from the injectivity of $f$. The existence can be proved by induction on $n$ in the following way. If $n \\in\\left\\{a_{1}, \\ldots, a_{k}\\right\\}$, then we may take $j=0$; otherwise there is some $n' < n$ with $f\\left(n'\\right)=n$ to which the induction hypothesis may be applied.\n\nThe result of the previous paragraph means that every positive integer appears exactly once in the following infinite picture, henceforth referred to as \"the Table\":\n\n| $a_{1}$ | $f\\left(a_{1}\\right)$ | $f^{2}\\left(a_{1}\\right)$ | $f^{3}\\left(a_{1}\\right)$ | $\\ldots$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $a_{2}$ | $f\\left(a_{2}\\right)$ | $f^{2}\\left(a_{2}\\right)$ | $f^{3}\\left(a_{2}\\right)$ | $\\ldots$ |\n| $\\vdots$ | $\\vdots$ | $\\vdots$ | $\\vdots$ |  |\n| $a_{k}$ | $f\\left(a_{k}\\right)$ | $f^{2}\\left(a_{k}\\right)$ | $f^{3}\\left(a_{k}\\right)$ | $\\ldots$ |\n\nThe Table\n\nStep 2. Our next goal is to prove that each row of the Table is an arithmetic progression. Assume contrariwise that the number $t$ of rows which are arithmetic progressions would satisfy $0 \\leqslant t < k$. By permuting the rows if necessary we may suppose that precisely the first $t$ rows are arithmetic progressions, say with steps $T_{1}, \\ldots, T_{t}$. Our plan is to find a further row that is \"not too sparse\" in an asymptotic sense, and then to prove that such a row has to be an arithmetic progression as well.\n\nLet us write $T=\\operatorname{lcm}\\left(T_{1}, T_{2}, \\ldots, T_{t}\\right)$ and $A=\\max \\left\\{a_{1}, a_{2}, \\ldots, a_{t}\\right\\}$ if $t>0$; and $T=1$ and $A=0$ if $t=0$. For every integer $n \\geqslant A$, the interval $\\Delta_{n}=[n+1, n+T]$ contains exactly $T / T_{i}$ elements of the $i^{\\text{th}}$ row $(1 \\leqslant i \\leqslant t)$. Therefore, the number of elements from the last $(k-t)$ rows of the Table contained in $\\Delta_{n}$ does not depend on $n \\geqslant A$. It is not possible that none of these intervals $\\Delta_{n}$ contains an element from the $k-t$ last rows, because infinitely many numbers appear in these rows. It follows that for each $n \\geqslant A$ the interval $\\Delta_{n}$ contains at least one member from these rows.\n\nThis yields that for every positive integer $d$, the interval $[A+1, A+(d+1)(k-t) T]$ contains at least $(d+1)(k-t)$ elements from the last $k-t$ rows; therefore, there exists an index $x$ with $t+1 \\leqslant x \\leqslant k$, possibly depending on $d$, such that our interval contains at least $d+1$ elements from the $x^{\\text{th}}$ row. In this situation we have\n$$\nf^{d}\\left(a_{x}\\right) \\leqslant A+(d+1)(k-t) T\n$$\nFinally, since there are finitely many possibilities for $x$, there exists an index $x \\geqslant t+1$ such that the set\n$$\nX=\\left\\{d \\in \\mathbb{Z}_{>0} \\mid f^{d}\\left(a_{x}\\right) \\leqslant A+(d+1)(k-t) T\\right\\}\n$$\nis infinite. Thereby we have found the \"dense row\" promised above.\n\nBy assumption (i), for every $d \\in X$ the number\n$$\n\\beta_{d}=\\frac{f^{d}\\left(a_{x}\\right)-a_{x}}{d}\n$$\nis a positive integer not exceeding\n$$\n\\frac{A+(d+1)(k-t) T}{d} \\leqslant \\frac{A d+2 d(k-t) T}{d}=A+2(k-t) T\n$$\nThis leaves us with finitely many choices for $\\beta_{d}$, which means that there exists a number $T_{x}$ such that the set\n$$\nY=\\left\\{d \\in X \\mid \\beta_{d}=T_{x}\\right\\}\n$$\nis infinite. Notice that we have $f^{d}\\left(a_{x}\\right)=a_{x}+d \\cdot T_{x}$ for all $d \\in Y$.\n\nNow we are prepared to prove that the numbers in the $x^{\\text{th}}$ row form an arithmetic progression, thus coming to a contradiction with our assumption. Let us fix any positive integer $j$. Since the set $Y$ is infinite, we can choose a number $y \\in Y$ such that $y-j>\\left|f^{j}\\left(a_{x}\\right)-\\left(a_{x}+j T_{x}\\right)\\right|$. Notice that both numbers\n$$\nf^{y}\\left(a_{x}\\right)-f^{j}\\left(a_{x}\\right)=f^{y-j}\\left(f^{j}\\left(a_{x}\\right)\\right)-f^{j}\\left(a_{x}\\right) \\quad \\text{ and } \\quad f^{y}\\left(a_{x}\\right)-\\left(a_{x}+j T_{x}\\right)=(y-j) T_{x}\n$$\nare divisible by $y-j$. Thus, the difference between these numbers is also divisible by $y-j$. Since the absolute value of this difference is less than $y-j$, it has to vanish, so we get $f^{j}\\left(a_{x}\\right)= a_{x}+j \\cdot T_{x}$.\n\nHence, it is indeed true that all rows of the Table are arithmetic progressions.\n\nStep 3. Keeping the above notation in force, we denote the step of the $i^{\\text{th}}$ row of the table by $T_{i}$. **Now we claim that we have $f(n)-n=f(n+T)-(n+T)$ for all $n \\in \\mathbb{Z}_{>0}$, where**\n$$\nT=\\operatorname{lcm}\\left(T_{1}, \\ldots, T_{k}\\right)\n$$\nTo see this, let any $n \\in \\mathbb{Z}_{>0}$ be given and denote the index of the row in which it appears in the Table by $i$. Then we have $f^{j}(n)=n+j \\cdot T_{i}$ for all $j \\in \\mathbb{Z}_{>0}$, and thus indeed\n$$\nf(n+T)-f(n)=f^{1+T / T_{i}}(n)-f(n)=\\left(n+T+T_{i}\\right)-\\left(n+T_{i}\\right)=T .\n$$\nThis concludes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32015,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}_{>0}$ denote the set of positive integers. For any positive integer $k$, a function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ is called $k$-good if $\\operatorname{gcd}(f(m)+n, f(n)+m) \\leqslant k$ for all $m \\neq n$. Find all $k$ such that there exists a $k$-good function.",
    "options": [],
    "answer": "All positive integers k greater than or equal to 2",
    "solution": "We first show that there is no $1$-good function. Suppose that there exists a function $f$ such that $G_{f}(m, n)=1$ for all $m \\neq n$. Now, if there are two distinct even numbers $m$ and $n$ such that $f(m)$ and $f(n)$ are both even, then $2 \\mid G_{f}(m, n)$, a contradiction. A similar argument holds if there are two distinct odd numbers $m$ and $n$ such that $f(m)$ and $f(n)$ are both odd. Hence we can choose an even $m$ and an odd $n$ such that $f(m)$ is odd and $f(n)$ is even. This also implies that $2 \\mid G_{f}(m, n)$, a contradiction.\n\nWe now construct a $2$-good function. Define $f(n)=2^{g(n)+1}-n-1$, where $g$ is defined recursively by $g(1)=1$ and $g(n+1)=\\left(2^{g(n)+1}\\right)!$.\n\nFor any positive integers $m>n$, set\n$$\nA=f(m)+n=2^{g(m)+1}-m+n-1, \\quad B=f(n)+m=2^{g(n)+1}-n+m-1 .\n$$\nWe need to show that $\\operatorname{gcd}(A, B) \\leqslant 2$. First, note that $A+B=2^{g(m)+1}+2^{g(n)+1}-2$ is not divisible by $4$, so that $4 \\nmid \\operatorname{gcd}(A, B)$. Now we suppose that there is an odd prime $p$ for which $p \\mid \\operatorname{gcd}(A, B)$ and derive a contradiction.\n\nWe first claim that $2^{g(m-1)+1} \\geqslant B$. This is a rather weak bound; one way to prove it is as follows. Observe that $g(k+1)>g(k)$ and hence $2^{g(k+1)+1} \\geqslant 2^{g(k)+1}+1$ for every positive integer $k$. By repeatedly applying this inequality, we obtain $2^{g(m-1)+1} \\geqslant 2^{g(n)+1}+(m-1)-n=B$.\n\nNow, since $p \\mid B$, we have $p-1<B \\leqslant 2^{g(m-1)+1}$, so that $p-1 \\mid\\left(2^{g(m-1)+1}\\right)!=g(m)$. Hence $2^{g(m)} \\equiv 1(\\bmod p)$, which yields $A+B \\equiv 2^{g(n)+1}(\\bmod p)$. However, since $p \\mid A+B$, this implies that $p=2$, a contradiction.\nWe provide an alternative construction of a $2$-good function $f$.\nLet $\\mathcal{P}$ be the set consisting of $4$ and all odd primes. For every $p \\in \\mathcal{P}$, we say that a number $a \\in\\{0,1, \\ldots, p-1\\}$ is $p$-useful if $a \\not \\equiv -a(\\bmod p)$. Note that a residue modulo $p$ which is neither $0$ nor $2$ is $p$-useful (the latter is needed only when $p=4$).\n\nWe will construct $f$ recursively; in some steps, we will also define a $p$-useful number $a_{p}$. After the $m^{\\text{th}}$ step, the construction will satisfy the following conditions:\n(i) The values of $f(n)$ have already been defined for all $n \\leqslant m$, and $p$-useful numbers $a_{p}$ have already been defined for all $p \\leqslant m+2$;\n(ii) If $n \\leqslant m$ and $p \\leqslant m+2$, then $f(n)+n \\not \\equiv a_{p}(\\bmod p)$;\n(iii) $\\operatorname{gcd}\\left(f\\left(n_{1}\\right)+n_{2}, f\\left(n_{2}\\right)+n_{1}\\right) \\leqslant 2$ for all $n_{1}<n_{2} \\leqslant m$.\nIf these conditions are satisfied, then $f$ will be a $2$-good function.\n\n**Step 1.** Set $f(1)=1$ and $a_{3}=1$. Clearly, all the conditions are satisfied.\n\n**Step $m$, for $m \\geqslant 2$.** We need to determine $f(m)$ and, if $m+2 \\in \\mathcal{P}$, the number $a_{m+2}$.\n\nDefining $f(m)$. Let $X_{m}=\\{p \\in \\mathcal{P}: p \\mid f(n)+m$ for some $n<m\\}$. We will determine $f(m) \\bmod p$ for all $p \\in X_{m}$ and then choose $f(m)$ using the Chinese Remainder Theorem.\nTake any $p \\in X_{m}$. If $p \\leqslant m+1$, then we define $f(m) \\equiv -a_{p}-m(\\bmod p)$. Otherwise, if $p \\geqslant m+2$, then we define $f(m) \\equiv 0(\\bmod p)$.\n\nDefining $a_{m+2}$. Now let $p=m+2$ and suppose that $p \\in \\mathcal{P}$. We choose $a_{p}$ to be a residue modulo $p$ that is not congruent to $0,2$, or $f(n)+n$ for any $n \\leqslant m$. Since $f(1)+1=2$, there are at most $m+1<p$ residues to avoid, so we can always choose a remaining residue.\n\nWe first check that (ii) is satisfied. We only need to check it if $p=m+2$ or $n=m$. In the former case, we have $f(n)+n \\not \\equiv a_{p}(\\bmod p)$ by construction. In the latter case, if $n=m$ and $p \\leqslant m+1$, then we have $f(m)+m \\equiv -a_{p} \\not \\equiv a_{p}(\\bmod p)$, where we make use of the fact that $a_{p}$ is $p$-useful.\n\nNow we check that (iii) holds. Suppose, to the contrary, that $p \\mid \\operatorname{gcd}(f(n)+m, f(m)+n)$ for some $n<m$. Then $p \\in X_{m}$ and $p \\mid f(m)+n$. If $p \\geqslant m+2$, then $0 \\equiv f(m)+n \\equiv n(\\bmod p)$, which is impossible since $n<m<p$.\nOtherwise, if $p \\leqslant m+1$, then\n$$\n0 \\equiv (f(m)+n)+(f(n)+m) \\equiv (f(n)+n)+(f(m)+m) \\equiv (f(n)+n)-a_{p}(\\bmod p) .\n$$\nThis implies that $f(n)+n \\equiv a_{p}(\\bmod p)$, a contradiction with (ii).",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32016,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every positive integer $n$ with prime factorization $n=\\prod_{i=1}^{k} p_{i}^{\\alpha_{i}}$, define\n$$\n\\mho(n)=\\sum_{i: p_{i}>10^{100}} \\alpha_{i}\n$$\nThat is, $\\mho(n)$ is the number of prime factors of $n$ greater than $10^{100}$, counted with multiplicity.\nFind all strictly increasing functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that\n$$\n\\mho(f(a)-f(b)) \\leqslant \\mho(a-b) \\quad \\text{ for all integers } a \\text{ and } b \\text{ with } a>b .\n$$",
    "options": [],
    "answer": "All functions of the form f(x) = a x + b, where b is any integer and a is a positive integer whose prime factors are all at most ten to the one hundred.",
    "solution": "Answer. $f(x)=a x+b$, where $b$ is an arbitrary integer, and $a$ is an arbitrary positive integer with $\\mho(a)=0$.\n\nA straightforward check shows that all the functions listed in the answer satisfy the problem condition. It remains to show the converse.\n\nAssume that $f$ is a function satisfying the problem condition. Notice that the function $g(x)=f(x)-f(0)$ also satisfies this condition. Replacing $f$ by $g$, we assume from now on that $f(0)=0$; then $f(n)>0$ for any positive integer $n$. Thus, we aim to prove that there exists a positive integer $a$ with $\\mho(a)=0$ such that $f(n)=a n$ for all $n \\in \\mathbb{Z}$.\n\nWe start by introducing some notation. Set $N=10^{100}$. We say that a prime $p$ is large if $p>N$, and $p$ is small otherwise; let $\\mathcal{S}$ be the set of all small primes. Next, we say that a positive integer is large or small if all its prime factors are such (thus, the number 1 is the unique number which is both large and small). For a positive integer $k$, we denote the greatest large divisor of $k$ and the greatest small divisor of $k$ by $L(k)$ and $S(k)$, respectively; thus, $k=L(k) S(k)$.\n\nWe split the proof into three steps.\n\nStep 1. We prove that for every large $k$, we have $k|f(a)-f(b) \\Longleftrightarrow k| a-b$. In other words, $L(f(a)-f(b))=L(a-b)$ for all integers $a$ and $b$ with $a>b$.\n\nWe use induction on $k$. The base case $k=1$ is trivial. For the induction step, assume that $k_{0}$ is a large number, and that the statement holds for all large numbers $k$ with $k<k_{0}$.\n\nClaim 1. For any integers $x$ and $y$ with $0<x-y<k_{0}$, the number $k_{0}$ does not divide $f(x)-f(y)$.\n\nProof. Assume, to the contrary, that $k_{0} \\mid f(x)-f(y)$. Let $\\ell=L(x-y)$; then $\\ell \\leqslant x-y<k_{0}$. By the induction hypothesis, $\\ell \\mid f(x)-f(y)$, and thus $\\operatorname{lcm}\\left(k_{0}, \\ell\\right) \\mid f(x)-f(y)$. Notice that $\\operatorname{lcm}\\left(k_{0}, \\ell\\right)$ is large, and $\\operatorname{lcm}\\left(k_{0}, \\ell\\right) \\geqslant k_{0}>\\ell$. But then\n$$\n\\mho(f(x)-f(y)) \\geqslant \\mho\\left(\\operatorname{lcm}\\left(k_{0}, \\ell\\right)\\right)>\\mho(\\ell)=\\mho(x-y)\n$$\nwhich is impossible.\n\nNow we complete the induction step. By Claim 1, for every integer $a$ each of the sequences\n$$\nf(a), f(a+1), \\ldots, f\\left(a+k_{0}-1\\right) \\quad \\text{ and } \\quad f(a+1), f(a+2), \\ldots, f\\left(a+k_{0}\\right)\n$$\nforms a complete residue system modulo $k_{0}$. This yields $f(a) \\equiv f\\left(a+k_{0}\\right)\\left(\\bmod k_{0}\\right)$. Thus, $f(a) \\equiv f(b)\\left(\\bmod k_{0}\\right)$ whenever $a \\equiv b\\left(\\bmod k_{0}\\right)$.\n\nFinally, if $a \\not \\equiv b\\left(\\bmod k_{0}\\right)$ then there exists an integer $b^{\\prime}$ such that $b^{\\prime} \\equiv b\\left(\\bmod k_{0}\\right)$ and $\\left|a-b^{\\prime}\\right|<k_{0}$. Then $f(b) \\equiv f\\left(b^{\\prime}\\right) \\not \\equiv f(a)\\left(\\bmod k_{0}\\right)$. The induction step is proved.\n\nStep 2. We prove that for some small integer a there exist infinitely many integers $n$ such that $f(n)=a n$. In other words, $f$ is linear on some infinite set.\n\nWe start with the following general statement.\n\nClaim 2. There exists a constant $c$ such that $f(t)<c t$ for every positive integer $t>N$.\n\nProof. Let $d$ be the product of all small primes, and let $\\alpha$ be a positive integer such that $2^{\\alpha}>f(N)$. Then, for every $p \\in \\mathcal{S}$ the numbers $f(0), f(1), \\ldots, f(N)$ are distinct modulo $p^{\\alpha}$. Set $P=d^{\\alpha}$ and $c=P+f(N)$.\n\nChoose any integer $t>N$. Due to the choice of $\\alpha$, for every $p \\in \\mathcal{S}$ there exists at most one nonnegative integer $i \\leqslant N$ with $p^{\\alpha} \\mid f(t)-f(i)$. Since $|\\mathcal{S}|<N$, we can choose a nonnegative integer $j \\leqslant N$ such that $p^{\\alpha} \\nmid f(t)-f(j)$ for all $p \\in \\mathcal{S}$. Therefore, $S(f(t)-f(j))<P$.\n\nOn the other hand, Step 1 shows that $L(f(t)-f(j))=L(t-j) \\leqslant t-j$. Since $0 \\leqslant j \\leqslant N$, this yields\n$$\nf(t)=f(j)+L(f(t)-f(j)) \\cdot S(f(t)-f(j))<f(N)+(t-j) P \\leqslant(P+f(N)) t=c t\n$$\n\nNow let $\\mathcal{T}$ be the set of large primes. For every $t \\in \\mathcal{T}$, Step 1 implies $L(f(t))=t$, so the ratio $f(t) / t$ is an integer. Now Claim 2 leaves us with only finitely many choices for this ratio, which means that there exists an infinite subset $\\mathcal{T}^{\\prime} \\subseteq \\mathcal{T}$ and a positive integer a such that $f(t)=a t$ for all $t \\in \\mathcal{T}^{\\prime}$, as required.\n\nSince $L(t)=L(f(t))=L(a) L(t)$ for all $t \\in \\mathcal{T}^{\\prime}$, we get $L(a)=1$, so the number $a$ is small.\n\nStep 3. We show that $f(x)=a x$ for all $x \\in \\mathbb{Z}$.\n\nLet $R_{i}=\\{x \\in \\mathbb{Z}: x \\equiv i(\\bmod N!)\\}$ denote the residue class of $i$ modulo $N!$.\n\nClaim 3. Assume that for some $r$, there are infinitely many $n \\in R_{r}$ such that $f(n)=a n$. Then $f(x)=a x$ for all $x \\in R_{r+1}$.\n\nProof. Choose any $x \\in R_{r+1}$. By our assumption, we can select $n \\in R_{r}$ such that $f(n)=a n$ and $|n-x|>|f(x)-a x|$. Since $n-x \\equiv r-(r+1)=-1(\\bmod N!)$, the number $|n-x|$ is large. Therefore, by Step 1 we have $f(x) \\equiv f(n)=a n \\equiv a x(\\bmod n-x)$, so $n-x \\mid f(x)-a x$. Due to the choice of $n$, this yields $f(x)=a x$.\n\nTo complete Step 3, notice that the set $\\mathcal{T}^{\\prime}$ found in Step 2 contains infinitely many elements of some residue class $R_{i}$. Applying Claim 3, we successively obtain that $f(x)=a x$ for all $x \\in R_{i+1}, R_{i+2}, \\ldots, R_{i+N!}=R_{i}$. This finishes the solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32017,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\geqslant 3$ with the following property: for all real numbers $a_{1}, a_{2}, \\ldots, a_{n}$ and $b_{1}, b_{2}, \\ldots, b_{n}$ satisfying $\\left|a_{k}\\right|+\\left|b_{k}\\right|=1$ for $1 \\leqslant k \\leqslant n$, there exist $x_{1}, x_{2}, \\ldots, x_{n}$, each of which is either $-1$ or $1$, such that\n$$\n\\left|\\sum_{k=1}^{n} x_{k} a_{k}\\right|+\\left|\\sum_{k=1}^{n} x_{k} b_{k}\\right| \\leqslant 1\n$$",
    "options": [],
    "answer": "All odd integers greater than or equal to 3",
    "solution": "Answer. $n$ can be any odd integer greater than or equal to $3$.\n\nFor any even integer $n \\geqslant 4$, we consider the case\n$$\na_{1}=a_{2}=\\cdots=a_{n-1}=b_{n}=0 \\quad \\text{and} \\quad b_{1}=b_{2}=\\cdots=b_{n-1}=a_{n}=1.\n$$\nThe condition $\\left|a_{k}\\right|+\\left|b_{k}\\right|=1$ is satisfied for each $1 \\leqslant k \\leqslant n$. No matter how we choose each $x_{k}$, both sums $\\sum_{k=1}^{n} x_{k} a_{k}$ and $\\sum_{k=1}^{n} x_{k} b_{k}$ are odd integers. This implies $\\left|\\sum_{k=1}^{n} x_{k} a_{k}\\right| \\geqslant 1$ and $\\left|\\sum_{k=1}^{n} x_{k} b_{k}\\right| \\geqslant 1$, which shows (1) cannot hold.\n\nFor any odd integer $n \\geqslant 3$, we may assume without loss of generality $b_{k} \\geqslant 0$ for $1 \\leqslant k \\leqslant n$ (this can be done by flipping the pair $\\left(a_{k}, b_{k}\\right)$ to $\\left(-a_{k},-b_{k}\\right)$ and $x_{k}$ to $-x_{k}$ if necessary) and $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{m} \\geqslant 0>a_{m+1} \\geqslant \\cdots \\geqslant a_{n}$. We claim that the choice $x_{k}=(-1)^{k+1}$ for $1 \\leqslant k \\leqslant n$ will work. Define\n$$\ns=\\sum_{k=1}^{m} x_{k} a_{k} \\quad \\text{and} \\quad t=-\\sum_{k=m+1}^{n} x_{k} a_{k}\n$$\nNote that\n$$\ns=\\left(a_{1}-a_{2}\\right)+\\left(a_{3}-a_{4}\\right)+\\cdots \\geqslant 0\n$$\nby the assumption $a_{1} \\geqslant a_{2} \\geqslant \\cdots \\geqslant a_{m}$ (when $m$ is odd, there is a single term $a_{m}$ at the end, which is also positive). Next, we have\n$$\ns=a_{1}-\\left(a_{2}-a_{3}\\right)-\\left(a_{4}-a_{5}\\right)-\\cdots \\leqslant a_{1} \\leqslant 1\n$$\nSimilarly,\n$$\nt=\\left(-a_{n}+a_{n-1}\\right)+\\left(-a_{n-2}+a_{n-3}\\right)+\\cdots \\geqslant 0\n$$\nand\n$$\nt=-a_{n}+\\left(a_{n-1}-a_{n-2}\\right)+\\left(a_{n-3}-a_{n-4}\\right)+\\cdots \\leqslant-a_{n} \\leqslant 1.\n$$\nFrom the condition, we have $a_{k}+b_{k}=1$ for $1 \\leqslant k \\leqslant m$ and $-a_{k}+b_{k}=1$ for $m+1 \\leqslant k \\leqslant n$. It follows that $\\sum_{k=1}^{n} x_{k} a_{k}=s-t$ and $\\sum_{k=1}^{n} x_{k} b_{k}=1-s-t$. Hence it remains to prove\n$$\n|s-t|+|1-s-t| \\leqslant 1\n$$\nunder the constraint $0 \\leqslant s, t \\leqslant 1$. By symmetry, we may assume $s \\geqslant t$. If $1-s-t \\geqslant 0$, then we have\n$$\n|s-t|+|1-s-t|=s-t+1-s-t=1-2 t \\leqslant 1\n$$\nIf $1-s-t \\leqslant 0$, then we have\n$$\n|s-t|+|1-s-t|=s-t-1+s+t=2 s-1 \\leqslant 1\n$$\nHence, the inequality is true in both cases.\n\nThese show $n$ can be any odd integer greater than or equal to $3$.\nThe even case can be handled in the same way as Solution 1. For the odd case, we prove by induction on $n$.\n\nFirstly, for $n=3$, we may assume without loss of generality $a_{1} \\geqslant a_{2} \\geqslant a_{3} \\geqslant 0$ and $b_{1}=a_{1}-1$ (if $b_{1}=1-a_{1}$, we may replace each $b_{k}$ by $-b_{k}$).\n\n- Case 1. $b_{2}=a_{2}-1$ and $b_{3}=a_{3}-1$, in which case we take $(x_{1}, x_{2}, x_{3})=(1,-1,1)$.\nLet $c=a_{1}-a_{2}+a_{3}$ so that $0 \\leqslant c \\leqslant 1$. Then $|b_{1}-b_{2}+b_{3}|=|a_{1}-a_{2}+a_{3}-1|=1-c$ and hence $|c|+|b_{1}-b_{2}+b_{3}|=1$.\n\n- Case 2. $b_{2}=1-a_{2}$ and $b_{3}=1-a_{3}$, in which case we take $(x_{1}, x_{2}, x_{3})=(1,-1,1)$.\nLet $c=a_{1}-a_{2}+a_{3}$ so that $0 \\leqslant c \\leqslant 1$. Since $a_{3} \\leqslant a_{2}$ and $a_{1} \\leqslant 1$, we have\n$$\nc-1 \\leqslant b_{1}-b_{2}+b_{3}=a_{1}+a_{2}-a_{3}-1 \\leqslant 1-c.\n$$\nThis gives $|b_{1}-b_{2}+b_{3}| \\leqslant 1-c$ and hence $|c|+|b_{1}-b_{2}+b_{3}| \\leqslant 1$.\n\n- Case 3. $b_{2}=a_{2}-1$ and $b_{3}=1-a_{3}$, in which case we take $(x_{1}, x_{2}, x_{3})=(-1,1,1)$.\nLet $c=-a_{1}+a_{2}+a_{3}$. If $c \\geqslant 0$, then $a_{3} \\leqslant 1$ and $a_{2} \\leqslant a_{1}$ imply\n$$\nc-1 \\leqslant-b_{1}+b_{2}+b_{3}=-a_{1}+a_{2}-a_{3}+1 \\leqslant 1-c\n$$\nIf $c<0$, then $a_{1} \\leqslant a_{2}+1$ and $a_{3} \\geqslant 0$ imply\n$$\n-c-1 \\leqslant-b_{1}+b_{2}+b_{3}=-a_{1}+a_{2}-a_{3}+1 \\leqslant 1+c.\n$$\nIn both cases, we get $|-b_{1}+b_{2}+b_{3}| \\leqslant 1-|c|$ and hence $|c|+|-b_{1}+b_{2}+b_{3}| \\leqslant 1$.\n\n- Case 4. $b_{2}=1-a_{2}$ and $b_{3}=a_{3}-1$, in which case we take $(x_{1}, x_{2}, x_{3})=(-1,1,1)$.\nLet $c=-a_{1}+a_{2}+a_{3}$. If $c \\geqslant 0$, then $a_{2} \\leqslant 1$ and $a_{3} \\leqslant a_{1}$ imply\n$$\nc-1 \\leqslant-b_{1}+b_{2}+b_{3}=-a_{1}-a_{2}+a_{3}+1 \\leqslant 1-c.\n$$\nIf $c<0$, then $a_{1} \\leqslant a_{3}+1$ and $a_{2} \\geqslant 0$ imply\n$$\n-c-1 \\leqslant-b_{1}+b_{2}+b_{3}=-a_{1}-a_{2}+a_{3}+1 \\leqslant 1+c.\n$$\nIn both cases, we get $|-b_{1}+b_{2}+b_{3}| \\leqslant 1-|c|$ and hence $|c|+|-b_{1}+b_{2}+b_{3}| \\leqslant 1$.\n\nWe have found $x_{1}, x_{2}, x_{3}$ satisfying (1) in each case for $n=3$.\n\nNow, let $n \\geqslant 5$ be odd and suppose the result holds for any smaller odd cases. Again we may assume $a_{k} \\geqslant 0$ for each $1 \\leqslant k \\leqslant n$. By the Pigeonhole Principle, there are at least three indices $k$ for which $b_{k}=a_{k}-1$ or $b_{k}=1-a_{k}$. Without loss of generality, suppose $b_{k}=a_{k}-1$ for $k=1,2,3$. Again by the Pigeonhole Principle, as $a_{1}, a_{2}, a_{3}$ lies between $0$ and $1$, the difference of two of them is at most $\\frac{1}{2}$. By changing indices if necessary, we may assume $0 \\leqslant d=a_{1}-a_{2} \\leqslant \\frac{1}{2}$.\n\nBy the inductive hypothesis, we can choose $x_{3}, x_{4}, \\ldots, x_{n}$ such that $a' = \\sum_{k=3}^{n} x_{k} a_{k}$ and $b' = \\sum_{k=3}^{n} x_{k} b_{k}$ satisfy $|a'|+|b'| \\leqslant 1$. We may further assume $a' \\geqslant 0$.\n\n- Case 1. $b' \\geqslant 0$, in which case we take $(x_{1}, x_{2})=(-1,1)$.\nWe have $|-a_{1}+a_{2}+a'|+|-(a_{1}-1)+(a_{2}-1)+b'|=|-d+a'|+|-d+b'| \\leqslant \\max \\{a'+b'-2d, a'-b', b'-a', 2d-a'-b'\\} \\leqslant 1$ since $0 \\leqslant a', b', a'+b' \\leqslant 1$ and $0 \\leqslant d \\leqslant \\frac{1}{2}$.\n\n- Case 2. $0 > b' \\geqslant -a'$, in which case we take $(x_{1}, x_{2})=(-1,1)$.\nWe have $|-a_{1}+a_{2}+a'|+|-(a_{1}-1)+(a_{2}-1)+b'|=|-d+a'|+|-d+b'|$. If $-d+a' \\geqslant 0$, this equals $a'-b'=|a'|+|b'| \\leqslant 1$. If $-d+a'<0$, this equals $2d-a'-b' \\leqslant 2d \\leqslant 1$.\n\n- Case 3. $b'<-a'$, in which case we take $(x_{1}, x_{2})=(1,-1)$.\nWe have $|a_{1}-a_{2}+a'|+|(a_{1}-1)-(a_{2}-1)+b'|=|d+a'|+|d+b'|$. If $d+b' \\geqslant 0$, this equals $2d+a'+b'<2d \\leqslant 1$. If $d+b'<0$, this equals $a'-b'=|a'|+|b'| \\leqslant 1$.\n\nTherefore, we have found $x_{1}, x_{2}, \\ldots, x_{n}$ satisfying (1) in each case. By induction, the property holds for all odd integers $n \\geqslant 3$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32018,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest real number $a$ such that for all $n \\geqslant 1$ and for all real numbers $x_{0}, x_{1}, \\ldots, x_{n}$ satisfying $0=x_{0}<x_{1}<x_{2}<\\cdots<x_{n}$, we have\n$$\n\\frac{1}{x_{1}-x_{0}}+\\frac{1}{x_{2}-x_{1}}+\\cdots+\\frac{1}{x_{n}-x_{n-1}} \\geqslant a\\left(\\frac{2}{x_{1}}+\\frac{3}{x_{2}}+\\cdots+\\frac{n+1}{x_{n}}\\right)\n$$",
    "options": [],
    "answer": "4/9",
    "solution": "We first show that $a=\\frac{4}{9}$ is admissible. For each $2 \\leqslant k \\leqslant n$, by the Cauchy-Schwarz Inequality, we have\n$$\n\\left(x_{k-1}+\\left(x_{k}-x_{k-1}\\right)\\right)\\left(\\frac{(k-1)^{2}}{x_{k-1}}+\\frac{3^{2}}{x_{k}-x_{k-1}}\\right) \\geqslant(k-1+3)^{2}\n$$\nwhich can be rewritten as\n$$\n\\frac{9}{x_{k}-x_{k-1}} \\geqslant \\frac{(k+2)^{2}}{x_{k}}-\\frac{(k-1)^{2}}{x_{k-1}} \\tag{2}\n$$\nSumming (2) over $k=2,3, \\ldots, n$ and adding $\\frac{9}{x_{1}}$ to both sides, we have\n$$\n9 \\sum_{k=1}^{n} \\frac{1}{x_{k}-x_{k-1}} \\geqslant 4 \\sum_{k=1}^{n} \\frac{k+1}{x_{k}}+\\frac{n^{2}}{x_{n}}>4 \\sum_{k=1}^{n} \\frac{k+1}{x_{k}}\n$$\nThis shows (1) holds for $a=\\frac{4}{9}$.\n\nNext, we show that $a=\\frac{4}{9}$ is the optimal choice. Consider the sequence defined by $x_{0}=0$ and $x_{k}=x_{k-1}+k(k+1)$ for $k \\geqslant 1$, that is, $x_{k}=\\frac{1}{3} k(k+1)(k+2)$. Then the left-hand side of (1) equals\n$$\n\\sum_{k=1}^{n} \\frac{1}{k(k+1)}=\\sum_{k=1}^{n}\\left(\\frac{1}{k}-\\frac{1}{k+1}\\right)=1-\\frac{1}{n+1}\n$$\nwhile the right-hand side equals\n$$\na \\sum_{k=1}^{n} \\frac{k+1}{x_{k}}=3 a \\sum_{k=1}^{n} \\frac{1}{k(k+2)}=\\frac{3}{2} a \\sum_{k=1}^{n}\\left(\\frac{1}{k}-\\frac{1}{k+2}\\right)=\\frac{3}{2}\\left(1+\\frac{1}{2}-\\frac{1}{n+1}-\\frac{1}{n+2}\\right) a .\n$$\nWhen $n$ tends to infinity, the left-hand side tends to 1 while the right-hand side tends to $\\frac{9}{4} a$. Therefore $a$ has to be at most $\\frac{4}{9}$.\n\nHence the largest value of $a$ is $\\frac{4}{9}$.\nWe shall give an alternative method to establish (1) with $a=\\frac{4}{9}$. We define $y_{k}=x_{k}-x_{k-1}>0$ for $1 \\leqslant k \\leqslant n$. By the Cauchy-Schwarz Inequality, for $1 \\leqslant k \\leqslant n$, we have\n$$\n\\left(y_{1}+y_{2}+\\cdots+y_{k}\\right)\\left(\\sum_{j=1}^{k} \\frac{1}{y_{j}}\\binom{j+1}{2}^{2}\\right) \\geqslant\\left(\\binom{2}{2}+\\binom{3}{2}+\\cdots+\\binom{k+1}{2}\\right)^{2}=\\binom{k+2}{3}^{2} .\n$$\nThis can be rewritten as\n$$\n\\frac{k+1}{y_{1}+y_{2}+\\cdots+y_{k}} \\leqslant \\frac{36}{k^{2}(k+1)(k+2)^{2}}\\left(\\sum_{j=1}^{k} \\frac{1}{y_{j}}\\binom{j+1}{2}^{2}\\right) \\tag{3}\n$$\nSumming (3) over $k=1,2, \\ldots, n$, we get\n$$\n\\frac{2}{y_{1}}+\\frac{3}{y_{1}+y_{2}}+\\cdots+\\frac{n+1}{y_{1}+y_{2}+\\cdots+y_{n}} \\leqslant \\frac{c_{1}}{y_{1}}+\\frac{c_{2}}{y_{2}}+\\cdots+\\frac{c_{n}}{y_{n}} \\tag{4}\n$$\nwhere for $1 \\leqslant m \\leqslant n$,\n$$\n\\begin{aligned}\nc_{m} & =36\\binom{m+1}{2}^{2} \\sum_{k=m}^{n} \\frac{1}{k^{2}(k+1)(k+2)^{2}} \\\\\n& =\\frac{9 m^{2}(m+1)^{2}}{4} \\sum_{k=m}^{n}\\left(\\frac{1}{k^{2}(k+1)^{2}}-\\frac{1}{(k+1)^{2}(k+2)^{2}}\\right) \\\\\n& =\\frac{9 m^{2}(m+1)^{2}}{4}\\left(\\frac{1}{m^{2}(m+1)^{2}}-\\frac{1}{(n+1)^{2}(n+2)^{2}}\\right)<\\frac{9}{4}\n\\end{aligned}\n$$\nFrom (4), the inequality (1) holds for $a=\\frac{4}{9}$. This is also the upper bound as can be verified in the same way as Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32019,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which all positive divisors of $n$ can be put into the cells of a rectangular table under the following constraints:\n- each cell contains a distinct divisor;\n- the sums of all rows are equal; and\n- the sums of all columns are equal.",
    "options": [],
    "answer": "1",
    "solution": "Solution 1. Suppose all positive divisors of $n$ can be arranged into a rectangular table of size $k \\times l$ where the number of rows $k$ does not exceed the number of columns $l$. Let the sum of numbers in each column be $s$. Since $n$ belongs to one of the columns, we have $s \\geqslant n$, where equality holds only when $n=1$.\nFor $j=1,2, \\ldots, l$, let $d_{j}$ be the largest number in the $j$-th column. Without loss of generality, assume $d_{1}>d_{2}>\\cdots>d_{l}$. Since these are divisors of $n$, we have\n$$\nd_{l} \\leqslant \\frac{n}{l} \\tag{1}\n$$\nAs $d_{l}$ is the maximum entry of the $l$-th column, we must have\n$$\nd_{l} \\geqslant \\frac{s}{k} \\geqslant \\frac{n}{k} \\tag{2}\n$$\nThe relations (1) and (2) combine to give $\\frac{n}{l} \\geqslant \\frac{n}{k}$, that is, $k \\geqslant l$. Together with $k \\leqslant l$, we conclude that $k=l$. Then all inequalities in (1) and (2) are equalities. In particular, $s=n$ and so $n=1$, in which case the conditions are clearly satisfied.\n\n\nSolution 2. Clearly $n=1$ works. Then we assume $n>1$ and let its prime factorization be $n=p_{1}^{r_{1}} p_{2}^{r_{2}} \\cdots p_{t}^{r_{t}}$. Suppose the table has $k$ rows and $l$ columns with $1<k \\leqslant l$. Note that $k l$ is the number of positive divisors of $n$ and the sum of all entries is the sum of positive divisors of $n$, which we denote by $\\sigma(n)$. Consider the column containing $n$. Since the column sum is $\\frac{\\sigma(n)}{l}$, we must have $\\frac{\\sigma(n)}{l}>n$. Therefore, we have\n$$\n\\begin{aligned}\n\\left(r_{1}+1\\right)\\left(r_{2}+1\\right) \\cdots\\left(r_{t}+1\\right) & =k l \\leqslant l^{2}<\\left(\\frac{\\sigma(n)}{n}\\right)^{2} \\\\\n& =\\left(1+\\frac{1}{p_{1}}+\\cdots+\\frac{1}{p_{1}^{r_{1}}}\\right)^{2} \\cdots\\left(1+\\frac{1}{p_{t}}+\\cdots+\\frac{1}{p_{t}^{r_{t}}}\\right)^{2} .\n\\end{aligned}\n$$\nThis can be rewritten as\n$$\n\\begin{equation*}\nf\\left(p_{1}, r_{1}\\right) f\\left(p_{2}, r_{2}\\right) \\cdots f\\left(p_{t}, r_{t}\\right)<1 \\tag{3}\n\\end{equation*}\n$$\nwhere\n$$\nf(p, r)=\\frac{r+1}{\\left(1+\\frac{1}{p}+\\cdots+\\frac{1}{p^{r}}\\right)^{2}}=\\frac{(r+1)\\left(1-\\frac{1}{p}\\right)^{2}}{\\left(1-\\frac{1}{p^{r+1}}\\right)^{2}}\n$$\nDirect computation yields\n$$\nf(2,1)=\\frac{8}{9}, \\quad f(2,2)=\\frac{48}{49}, \\quad f(3,1)=\\frac{9}{8}\n$$\nAlso, we find that\n$$\n\\begin{aligned}\n& f(2, r) \\geqslant\\left(1-\\frac{1}{2^{r+1}}\\right)^{-2}>1 \\quad \\text{ for } r \\geqslant 3 \\\\\n& f(3, r) \\geqslant \\frac{4}{3}\\left(1-\\frac{1}{3^{r+1}}\\right)^{-2}>\\frac{4}{3}>\\frac{9}{8} \\quad \\text{ for } r \\geqslant 2, \\text{ and } \\\\\n& f(p, r) \\geqslant \\frac{32}{25}\\left(1-\\frac{1}{p^{r+1}}\\right)^{-2}>\\frac{32}{25}>\\frac{9}{8} \\quad \\text{ for } p \\geqslant 5\n\\end{aligned}\n$$\nFrom these values and bounds, it is clear that (3) holds only when $n=2$ or 4 . In both cases, it is easy to see that the conditions are not satisfied. Hence, the only possible $n$ is 1 .",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32020,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ for which we can fill in the entries of an $n \\times n$ table with the following properties:\n- each entry can be one of $I$, $M$ and $O$;\n- in each row and each column, the letters $I$, $M$ and $O$ occur the same number of times; and\n- in any diagonal whose number of entries is a multiple of three, the letters $I$, $M$ and $O$ occur the same number of times.",
    "options": [],
    "answer": "All positive integers divisible by 9",
    "solution": "We first show that such a table exists when $n$ is a multiple of $9$. Consider the following $9 \\times 9$ table.\n$$\n\\left(\\begin{array}{ccccccccc}\nI & I & I & M & M & M & O & O & O  \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M \\\\\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M \\\\\nI & I & I & M & M & M & O & O & O \\\\\nM & M & M & O & O & O & I & I & I \\\\\nO & O & O & I & I & I & M & M & M\n\\end{array}\\right)\n$$\nIt is a direct checking that the table (1) satisfies the requirements. For $n=9k$ where $k$ is a positive integer, we form an $n \\times n$ table using $k \\times k$ copies of (1). For each row and each column of the table of size $n$, since there are three $I$'s, three $M$'s and three $O$'s for any nine consecutive entries, the numbers of $I$, $M$ and $O$ are equal. In addition, every diagonal of the large table whose number of entries is divisible by $3$ intersects each copy of (1) at a diagonal with number of entries divisible by $3$ (possibly zero). Therefore, every such diagonal also contains the same number of $I$, $M$ and $O$.\n\nNext, consider any $n \\times n$ table for which the requirements can be met. As the number of entries of each row should be a multiple of $3$, we let $n=3k$ where $k$ is a positive integer. We divide the whole table into $k \\times k$ copies of $3 \\times 3$ blocks. We call the entry at the centre of such a $3 \\times 3$ square a vital entry. We also call any row, column or diagonal that contains at least one vital entry a vital line. We compute the number of pairs ($l$, $c$) where $l$ is a vital line and $c$ is an entry belonging to $l$ that contains the letter $M$. We let this number be $N$.\n\nOn the one hand, since each vital line contains the same number of $I$, $M$ and $O$, it is obvious that each vital row and each vital column contain $k$ occurrences of $M$. For vital diagonals in either direction, we count there are exactly\n$$\n1+2+\\cdots+(k-1)+k+(k-1)+\\cdots+2+1=k^{2}\n$$\noccurrences of $M$. Therefore, we have $N=4k^{2}$.\n\nOn the other hand, there are $3k^{2}$ occurrences of $M$ in the whole table. Note that each entry belongs to exactly $1$ or $4$ vital lines. Therefore, $N$ must be congruent to $3k^{2} \\bmod 3$.\n\nFrom the double counting, we get $4k^{2} \\equiv 3k^{2} (\\bmod 3)$, which forces $k$ to be a multiple of $3$. Therefore, $n$ has to be a multiple of $9$ and the proof is complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32021,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n \\geqslant 3$ islands in a city. Initially, the ferry company offers some routes between some pairs of islands so that it is impossible to divide the islands into two groups such that no two islands in different groups are connected by a ferry route.\n\nAfter each year, the ferry company will close a ferry route between some two islands $X$ and $Y$. At the same time, in order to maintain its service, the company will open new routes according to the following rule: for any island which is connected by a ferry route to exactly one of $X$ and $Y$, a new route between this island and the other of $X$ and $Y$ is added.\n\nSuppose at any moment, if we partition all islands into two nonempty groups in any way, then it is known that the ferry company will close a certain route connecting two islands from the two groups after some years. Prove that after some years there will be an island which is connected to all other islands by ferry routes.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Initially, we pick any pair of islands $A$ and $B$ which are connected by a ferry route and put $A$ in set $\\mathcal{A}$ and $B$ in set $\\mathcal{B}$. From the condition, without loss of generality there must be another island which is connected to $A$. We put such an island $C$ in set $\\mathcal{B}$. We say that two sets of islands form a network if each island in one set is connected to each island in the other set.\n\nNext, we shall include all islands to $\\mathcal{A} \\cup \\mathcal{B}$ one by one. Suppose we have two sets $\\mathcal{A}$ and $\\mathcal{B}$ which form a network where $3 \\leqslant |\\mathcal{A} \\cup \\mathcal{B}| < n$. This relation no longer holds only when a ferry route between islands $A \\in \\mathcal{A}$ and $B \\in \\mathcal{B}$ is closed. In that case, we define $\\mathcal{A}' = \\{A, B\\}$, and $\\mathcal{B}' = (\\mathcal{A} \\cup \\mathcal{B}) - \\{A, B\\}$. Note that $\\mathcal{B}'$ is nonempty. Consider any island $C \\in \\mathcal{A} - \\{A\\}$. From the relation of $\\mathcal{A}$ and $\\mathcal{B}$, we know that $C$ is connected to $B$. If $C$ was not connected to $A$ before the route between $A$ and $B$ closes, then there will be a route added between $C$ and $A$ afterwards. Hence, $C$ must now be connected to both $A$ and $B$. The same holds true for any island in $\\mathcal{B} - \\{B\\}$. Therefore, $\\mathcal{A}'$ and $\\mathcal{B}'$ form a network, and $\\mathcal{A}' \\cup \\mathcal{B}' = \\mathcal{A} \\cup \\mathcal{B}$. Hence these islands can always be partitioned into sets $\\mathcal{A}$ and $\\mathcal{B}$ which form a network.\n\nAs $|\\mathcal{A} \\cup \\mathcal{B}| < n$, there are some islands which are not included in $\\mathcal{A} \\cup \\mathcal{B}$. From the condition, after some years there must be a ferry route between an island $A$ in $\\mathcal{A} \\cup \\mathcal{B}$ and an island $D$ outside $\\mathcal{A} \\cup \\mathcal{B}$ which closes. Without loss of generality assume $A \\in \\mathcal{A}$. Then each island in $\\mathcal{B}$ must then be connected to $D$, no matter it was or not before. Hence, we can put $D$ in set $\\mathcal{A}$ so that the new sets $\\mathcal{A}$ and $\\mathcal{B}$ still form a network and the size of $\\mathcal{A} \\cup \\mathcal{B}$ is increased by $1$. The same process can be done to increase the size of $\\mathcal{A} \\cup \\mathcal{B}$. Eventually, all islands are included in this way so we may now assume $|\\mathcal{A} \\cup \\mathcal{B}| = n$.\n\nSuppose a ferry route between $A \\in \\mathcal{A}$ and $B \\in \\mathcal{B}$ is closed after some years. We put $A$ and $B$ in set $\\mathcal{A}'$ and all remaining islands in set $\\mathcal{B}'$. Then $\\mathcal{A}'$ and $\\mathcal{B}'$ form a network. This relation no longer holds only when a route between $A$, without loss of generality, and $C \\in \\mathcal{B}'$ is closed. Since this must eventually occur, at that time island $B$ will be connected to all other islands and the result follows.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32022,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time when he claps his hands, all the snails move along their own segments and stay at the next intersection points until the next clap. Since there are $n-1$ intersection points on each segment, all snails will reach the furthest intersection points from their starting points after $n-1$ claps.\n\na) Prove that if $n$ is odd then Jeff can always place the snails so that no two of them ever occupy the same intersection point.\n\nb) Prove that if $n$ is even then there must be a moment when some two snails occupy the same intersection point no matter how Jeff places the snails.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) For odd $n$, we travel along the circumference of the disk and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' alternately. Since every pair of lines intersect in the disk, there are exactly $n-1$ points between $A_{i}$ and $B_{i}$ for any fixed $1 \\leqslant i \\leqslant n$. As $n$ is odd, this means one of $A_{i}$ and $B_{i}$ is marked 'in' and the other is marked 'out'. Then Jeff can put a snail on the endpoint of each segment which is closer to the 'in' side of the corresponding line. We claim that the snails on $l_{i}$ and $l_{j}$ do not meet for any pairs $i, j$, hence proving part (a).\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nWithout loss of generality, we may assume the snails start at $A_{i}$ and $A_{j}$ respectively. Let $l_{i}$ intersect $l_{j}$ at $P$. Note that there is an odd number of points between $\\operatorname{arc} A_{i} A_{j}$. Each of these points belongs to a line $l_{k}$. Such a line $l_{k}$ must intersect exactly one of the segments $A_{i} P$ and $A_{j} P$, making an odd number of intersections. For the other lines, they may intersect both segments $A_{i} P$ and $A_{j} P$, or meet none of them. Therefore, the total number of intersection points on segments $A_{i} P$ and $A_{j} P$ (not counting $P$) is odd. However, if the snails arrive at $P$ at the same time, then there should be the same number of intersections on $A_{i} P$ and $A_{j} P$, which gives an even number of intersections. This is a contradiction so the snails do not meet each other.\n\n(b) For even $n$, we consider any way that Jeff places the snails and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' according to the directions travelled by the snails. In this case there must be two neighbouring points $A_{i}$ and $A_{j}$ both of which are marked 'in'. Let $P$ be the intersection of the segments $A_{i} B_{i}$ and $A_{j} B_{j}$. Then any other segment meeting one of the segments $A_{i} P$ and $A_{j} P$ must also meet the other one, and so the number of intersections on $A_{i} P$ and $A_{j} P$ are the same. This shows the snails starting from $A_{i}$ and $A_{j}$ will meet at $P$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32023,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\\angle F B C = 90^{\\circ}$. Suppose triangles $A B F$, $A C D$ and $A D E$ are similar isosceles triangles with\n$$\n\\angle F A B = \\angle F B A = \\angle D A C = \\angle D C A = \\angle E A D = \\angle E D A.\n$$\nLet $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D$, $E M$ and $F X$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the common angle in (1) by $\\theta$. As $\\triangle A B F \\sim \\triangle A C D$, we have $\\frac{A B}{A C} = \\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. From $E A = E D$, we get\n$$\n\\angle A F D = \\angle A B C = 90^{\\circ} + \\theta = 180^{\\circ} - \\frac{1}{2} \\angle A E D.\n$$\nHence, $F$ lies on the circle with centre $E$ and radius $E A$. In particular, $E F = E A = E D$. As $\\angle E F A = \\angle E A F = 2 \\theta = \\angle B F C$, points $B, F, E$ are collinear.\nAs $\\angle E D A = \\angle M A D$, we have $E D \\parallel A M$ and hence $E, D, X$ are collinear. As $M$ is the midpoint of $C F$ and $\\angle C B F = 90^{\\circ}$, we get $M F = M B$. In the isosceles triangles $E F A$ and $M F B$, we have $\\angle E F A = \\angle M F B$ and $A F = B F$. Therefore, they are congruent to each other. Then we have $B M = A E = X M$ and $B E = B F + F E = A F + F M = A M = E X$. This shows $\\triangle E M B \\cong \\triangle E M X$. As $F$ and $D$ lie on $E B$ and $E X$ respectively and $E F = E D$, we know that lines $B D$ and $X F$ are symmetric with respect to $E M$. It follows that the three lines are concurrent.\nFrom $\\angle C A D = \\angle E D A$, we have $A C \\parallel E D$. Together with $A C \\parallel E X$, we know that $E, D, X$ are collinear. Denote the common angle in (1) by $\\theta$. From $\\triangle A B F \\sim \\triangle A C D$, we get $\\frac{A B}{A C} = \\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. This yields $\\angle A F D = \\angle A B C = 90^{\\circ} + \\theta$ and hence $\\angle F D C = 90^{\\circ}$, implying that $B C D F$ is cyclic. Let $\\Gamma_{1}$ be its circumcircle.\nNext, from $\\triangle A B F \\sim \\triangle A D E$, we have $\\frac{A B}{A D} = \\frac{A F}{A E}$ so that $\\triangle A B D \\sim \\triangle A F E$. Therefore,\n$$\n\\angle A F E = \\angle A B D = \\theta + \\angle F B D = \\theta + \\angle F C D = 2 \\theta = 180^{\\circ} - \\angle B F A.\n$$\nThis implies $B, F, E$ are collinear. Note that $F$ is the incentre of triangle $D A B$. Point $E$ lies on the internal angle bisector of $\\angle D B A$ and lies on the perpendicular bisector of $A D$. It follows that $E$ lies on the circumcircle $\\Gamma_{2}$ of triangle $A B D$, and $E A = E F = E D$.\nAlso, since $C F$ is a diameter of $\\Gamma_{1}$ and $M$ is the midpoint of $C F$, $M$ is the centre of $\\Gamma_{1}$ and hence $\\angle A M D = 2 \\theta = \\angle A B D$. This shows $M$ lies on $\\Gamma_{2}$. Next, $\\angle M D X = \\angle M A E = \\angle D X M$ since $A M X E$ is a parallelogram. Hence $M D = M X$ and $X$ lies on $\\Gamma_{1}$.\n![](attached_image_1.png)\nWe now have two ways to complete the solution.\n- Method 1. From $E F = E A = X M$ and $E X \\parallel F M$, $E F M X$ is an isosceles trapezoid and is cyclic. Denote its circumcircle by $\\Gamma_{3}$. Since $B D$, $E M$, $F X$ are the three radical axes of $\\Gamma_{1}$, $\\Gamma_{2}$, $\\Gamma_{3}$, they must be concurrent.\n- Method 2. As $\\angle D M F = 2 \\theta = \\angle B F M$, we have $D M \\parallel E B$. Also,\n$$\n\\angle B F D + \\angle X B F = \\angle B F C + \\angle C F D + 90^{\\circ} - \\angle C B X = 2 \\theta + (90^{\\circ} - \\theta) + 90^{\\circ} - \\theta = 180^{\\circ}\n$$\nimplies $D F \\parallel X B$. These show the corresponding sides of triangles $D M F$ and $B E X$ are parallel. By Desargues' Theorem, the two triangles are perspective and hence $D B$, $M E$, $F X$ meet at a point.\nLet the common angle in (1) be $\\theta$. From $\\triangle A B F \\sim \\triangle A C D$, we have $\\frac{A B}{A C} = \\frac{A F}{A D}$ so that $\\triangle A B C \\sim \\triangle A F D$. Then $\\angle A D F = \\angle A C B = 90^{\\circ} - 2 \\theta = 90^{\\circ} - \\angle B A D$ and hence $D F \\perp A B$. As $F A = F B$, this implies $\\triangle D A B$ is isosceles with $D A = D B$. Then $F$ is the incentre of $\\triangle D A B$.\nNext, from $\\angle A E D = 180^{\\circ} - 2 \\theta = 180^{\\circ} - \\angle D B A$, points $A, B, D, E$ are concyclic. Since we also have $E A = E D$, this shows $E, F, B$ are collinear and $E A = E F = E D$.\n![](attached_image_2.png)\nNote that $C$ lies on the internal angle bisector of $\\angle B A D$ and lies on the external angle bisector of $\\angle D B A$. It follows that it is the $A$-excentre of triangle $D A B$. As $M$ is the midpoint of $C F$, $M$ lies on the circumcircle of triangle $D A B$ and it is the centre of the circle passing through $D, F, B, C$. By symmetry, $D E F M$ is a rhombus. Then the midpoints of $A X$, $E M$ and $D F$ coincide, and it follows that $D A F X$ is a parallelogram.\nLet $P$ be the intersection of $B D$ and $E M$, and $Q$ be the intersection of $A D$ and $B E$. From $\\angle B A C = \\angle D C A$, we know that $D C$, $A B$, $E M$ are parallel. Thus we have $\\frac{D P}{P B} = \\frac{C M}{M A}$. This is further equal to $\\frac{A E}{B E}$ since $C M = D M = D E = A E$ and $M A = B E$. From $\\triangle A E Q \\sim \\triangle B E A$, we find that\n$$\n\\frac{D P}{P B} = \\frac{A E}{B E} = \\frac{A Q}{B A} = \\frac{Q F}{F B}\n$$\nby the Angle Bisector Theorem. This implies $Q D \\parallel F P$ and hence $F, P, X$ are collinear, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32024,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with circumcircle $\\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $BC$. Denote by $D$ the foot of perpendicular from $I$ to side $BC$. The line through $I$ perpendicular to $AI$ meets sides $AB$ and $AC$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $AEF$ intersects $\\Gamma$ at a point $X$ other than $A$. Prove that lines $XD$ and $AM$ meet on $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AM$ meet $\\Gamma$ again at $Y$ and $XY$ meet $BC$ at $D'$. It suffices to show $D' = D$. We shall apply the following fact.\n\n- Claim. For any cyclic quadrilateral $PQRS$ whose diagonals meet at $T$, we have\n$$\n\\frac{QT}{TS} = \\frac{PQ \\cdot QR}{PS \\cdot SR}\n$$\nProof. We use $[W_1 W_2 W_3]$ to denote the area of $W_1 W_2 W_3$. Then\n$$\n\\frac{QT}{TS} = \\frac{[PQR]}{[PSR]} = \\frac{\\frac{1}{2} PQ \\cdot QR \\sin \\angle PQR}{\\frac{1}{2} PS \\cdot SR \\sin \\angle PSR} = \\frac{PQ \\cdot QR}{PS \\cdot SR}\n$$\nApplying the Claim to $ABYC$ and $XBYC$ respectively, we have $1 = \\frac{BM}{MC} = \\frac{AB \\cdot BY}{AC \\cdot CY}$ and $\\frac{BD'}{D'C} = \\frac{XB \\cdot BY}{XC \\cdot CY}$. These combine to give\n$$\n\\begin{equation*}\n\\frac{BD'}{CD'} = \\frac{XB}{XC} \\cdot \\frac{BY}{CY} = \\frac{XB}{XC} \\cdot \\frac{AC}{AB} \\tag{1}\n\\end{equation*}\n$$\nNext, we use directed angles to find that $\\measuredangle XBF = \\measuredangle XBA = \\measuredangle XCA = \\measuredangle XCE$ and $\\measuredangle XFB = \\measuredangle XFA = \\measuredangle XEA = \\measuredangle XEC$. This shows triangles $XBF$ and $XCE$ are directly similar. In particular, we have\n$$\n\\begin{equation*}\n\\frac{XB}{XC} = \\frac{BF}{CE} \\tag{2}\n\\end{equation*}\n$$\nIn the following, we give two ways to continue the proof.\n\n- Method 1. Here is a geometrical method. As $\\angle FIB = \\angle AIB - 90^{\\circ} = \\frac{1}{2} \\angle ACB = \\angle ICB$ and $\\angle FBI = \\angle IBC$, the triangles $FBI$ and $IBC$ are similar. Analogously, triangles $EIC$ and $IBC$ are also similar. Hence, we get\n$$\n\\begin{equation*}\n\\frac{FB}{IB} = \\frac{BI}{BC} \\quad \\text{ and } \\quad \\frac{EC}{IC} = \\frac{IC}{BC} \\tag{3}\n\\end{equation*}\n$$\n![](attached_image_1.png)\nNext, construct a line parallel to $BC$ and tangent to the incircle. Suppose it meets sides $AB$ and $AC$ at $B_1$ and $C_1$ respectively. Let the incircle touch $AB$ and $AC$ at $B_2$ and $C_2$ respectively. By homothety, the line $B_1I$ is parallel to the external angle bisector of $\\angle ABC$, and hence $\\angle B_1IB = 90^{\\circ}$. Since $\\angle BB_2I = 90^{\\circ}$, we get $BB_2 \\cdot BB_1 = BI^2$, and similarly $CC_2 \\cdot CC_1 = CI^2$. Hence,\n$$\n\\begin{equation*}\n\\frac{BI^2}{CI^2} = \\frac{BB_2 \\cdot BB_1}{CC_2 \\cdot CC_1} = \\frac{BB_1}{CC_1} \\cdot \\frac{BD}{CD} = \\frac{AB}{AC} \\cdot \\frac{BD}{CD} . \\tag{4}\n\\end{equation*}\n$$\nCombining (1), (2), (3) and (4), we conclude\n$$\n\\frac{BD'}{CD'} = \\frac{XB}{XC} \\cdot \\frac{AC}{AB} = \\frac{BF}{CE} \\cdot \\frac{AC}{AB} = \\frac{BI^2}{CI^2} \\cdot \\frac{AC}{AB} = \\frac{BD}{CD}\n$$\nso that $D' = D$. The result then follows.\n\n\n- Method 2. We continue the proof of Solution 1 using trigonometry. Let $\\beta = \\frac{1}{2} \\angle ABC$ and $\\gamma = \\frac{1}{2} \\angle ACB$. Observe that $\\angle FIB = \\angle AIB - 90^{\\circ} = \\gamma$. Hence, $\\frac{BF}{FI} = \\frac{\\sin \\angle FIB}{\\sin \\angle IBF} = \\frac{\\sin \\gamma}{\\sin \\beta}$. Similarly, $\\frac{CE}{EI} = \\frac{\\sin \\beta}{\\sin \\gamma}$. As $FI = EI$, we get\n$$\n\\frac{BD'}{CD'} = \\frac{AC}{AB} \\cdot \\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^2 = \\frac{\\sin 2\\beta}{\\sin 2\\gamma} \\cdot \\left(\\frac{\\sin \\gamma}{\\sin \\beta}\\right)^2 = \\frac{\\tan \\gamma}{\\tan \\beta} = \\frac{ID / CD}{ID / BD} = \\frac{BD}{CD} .\n$$\nThis shows $D' = D$ and the result follows.\nLet $\\omega_A$ be the $A$-mixtilinear incircle of triangle $ABC$. From the properties of mixtilinear incircles, $\\omega_A$ touches sides $AB$ and $AC$ at $F$ and $E$ respectively. Suppose $\\omega_A$ is tangent to $\\Gamma$ at $T$. Let $AM$ meet $\\Gamma$ again at $Y$, and let $D_1, T_1$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $BC$ respectively. It is well-known that $\\angle BAT = \\angle D_1AC$ so that $A, D_1, T_1$ are collinear.\n![](attached_image_2.png)\nWe then show that $X, M, T_1$ are collinear. Let $R$ be the radical centre of $\\omega_A, \\Gamma$ and the circumcircle of triangle $AEF$. Then $R$ lies on $AX, EF$ and the tangent at $T$ to $\\Gamma$. Let $AT$ meet $\\omega_A$ again at $S$ and meet $EF$ at $P$. Obviously, $SFTE$ is a harmonic quadrilateral. Projecting from $T$, the pencil ($R, P ; F, E$) is harmonic. We further project the pencil onto $\\Gamma$ from $A$, so that $XBTC$ is a harmonic quadrilateral. As $TT_1 \\parallel BC$, the projection from $T_1$ onto $BC$ maps $T$ to a point at infinity, and hence maps $X$ to the midpoint of $BC$, which is $M$. This shows $X, M, T_1$ are collinear.\nWe have two ways to finish the proof.\n\n- Method 1. Note that both $AY$ and $XT_1$ are chords of $\\Gamma$ passing through the midpoint $M$ of the chord $BC$. By the Butterfly Theorem, $XY$ and $AT_1$ cut $BC$ at a pair of symmetric points with respect to $M$, and hence $X, D, Y$ are collinear. The proof is thus complete.\n\n- Method 2. Here, we finish the proof without using the Butterfly Theorem. As $DTT_1D_1$ is an isosceles trapezoid, we have\n$$\n\\measuredangle YTD = \\measuredangle YTT_1 + \\measuredangle T_1TD = \\measuredangle YAT_1 + \\measuredangle AD_1D = \\measuredangle YMD\n$$\nso that $D, T, Y, M$ are concyclic. As $X, M, T_1$ are collinear, we have\n$$\n\\measuredangle AYD = \\measuredangle MTD = \\measuredangle D_1T_1M = \\measuredangle AT_1X = \\measuredangle AYX\n$$\nThis shows $X, D, Y$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32025,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$. Prove that the circumcentre of triangle $X S Y$ is equidistant from $P$ and $M$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the perpendicular from $S$ to $X Y$ meet line $Q M$ at $S'$. Let $E$ be the foot of altitude from $B$ to side $A C$. Since $Q$ and $S$ lie on the perpendicular bisector of $A D$, using directed angles, we have\n$$\n\\begin{aligned}\n\\measuredangle S D Q & =\\measuredangle Q A S=\\measuredangle X A S-\\measuredangle X A Q=\\left(\\frac{\\pi}{2}-\\measuredangle A Y X\\right)-\\measuredangle B A P=\\measuredangle C B A-\\measuredangle A Y X \\\\\n& =(\\measuredangle C B A-\\measuredangle A C B)-\\measuredangle B C A-\\measuredangle A Y X=\\measuredangle P E M-(\\measuredangle B C A+\\measuredangle A Y X) \\\\\n& =\\measuredangle P Q M-\\measuredangle(B C, X Y)=\\frac{\\pi}{2}-\\measuredangle\\left(S' Q, B C\\right)-\\measuredangle(B C, X Y)=\\measuredangle S S' Q\n\\end{aligned}\n$$\nThis shows $D, S', S, Q$ are concyclic.\n\n![](attached_image_1.png)\n\nLet the perpendicular from $N$ to $B C$ intersect line $S S'$ at $O_1$. (Note that the two lines coincide when $S$ is the midpoint of $A O$, in which case the result is true since the circumcentre of triangle $X S Y$ must lie on this line.) It suffices to show that $O_1$ is the circumcentre of triangle $X S Y$ since $N$ lies on the perpendicular bisector of $P M$. From\n$$\n\\measuredangle D S' O_1=\\measuredangle D Q S=\\measuredangle S Q A=\\angle(S Q, Q A)=\\angle\\left(O D, O_1 N\\right)=\\measuredangle D N O_1\n$$\nsince $S Q \\parallel O D$ and $Q A \\parallel O_1 N$, we know that $D, O_1, S', N$ are concyclic. Therefore, we get\n$$\n\\measuredangle S D S'=\\measuredangle S Q S'=\\angle\\left(S Q, Q S'\\right)=\\angle\\left(N D, N S'\\right)=\\measuredangle D N S'\n$$\nso that $S D$ is a tangent to the circle through $D, O_1, S', N$. Then we have\n$$\n\\begin{equation*}\nS S' \\cdot S O_1=S D^2=S X^2 \\tag{1}\n\\end{equation*}\n$$\nNext, we show that $S$ and $S'$ are symmetric with respect to $X Y$. By the Sine Law, we have\n$$\n\\frac{S S'}{\\sin \\angle S Q S'}=\\frac{S Q}{\\sin \\angle S S' Q}=\\frac{S Q}{\\sin \\angle S D Q}=\\frac{S Q}{\\sin \\angle S A Q}=\\frac{S A}{\\sin \\angle S Q A} .\n$$\nIt follows that\n$$\nS S'=S A \\cdot \\frac{\\sin \\angle S Q S'}{\\sin \\angle S Q A}=S A \\cdot \\frac{\\sin \\angle H O A}{\\sin \\angle O H A}=S A \\cdot \\frac{A H}{A O}=S A \\cdot 2 \\cos A,\n$$\nwhich is twice the distance from $S$ to $X Y$. Note that $S$ and $C$ lie on the same side of the perpendicular bisector of $P M$ if and only if $\\angle S A C<\\angle O A C$ if and only if $\\angle Y X A>\\angle C B A$. This shows $S$ and $O_1$ lie on different sides of $X Y$. As $S'$ lies on ray $S O_1$, it follows that $S$ and $S'$ cannot lie on the same side of $X Y$. Therefore, $S$ and $S'$ are symmetric with respect to $X Y$.\nLet $d$ be the diameter of the circumcircle of triangle $X S Y$. As $S S'$ is twice the distance from $S$ to $X Y$ and $S X=S Y$, we have $S S'=2 \\frac{S X^2}{d}$. It follows from (1) that $d=2 S O_1$. As $S O_1$ is the perpendicular bisector of $X Y$, point $O_1$ is the circumcentre of triangle $X S Y$.\nDenote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $O_1$ be the circumcentre of triangle $X S Y$. Consider two other possible positions of $S$. We name them $S'$ and $S''$ and define the analogous points $X', Y', O_1', X'', Y'' O_1''$ accordingly. Note that $S, S', S''$ lie on the perpendicular bisector of $A D$.\nAs $X X'$ and $Y Y'$ meet at $A$ and the circumcircles of triangles $A X Y$ and $A X' Y'$ meet at $D$, there is a spiral similarity with centre $D$ mapping $X Y$ to $X' Y'$. We find that\n$$\n\\measuredangle S X Y=\\frac{\\pi}{2}-\\measuredangle Y A X=\\frac{\\pi}{2}-\\measuredangle Y' A X'=\\measuredangle S' X' Y'\n$$\nand similarly $\\measuredangle S Y X=\\measuredangle S' Y' X'$. This shows triangles $S X Y$ and $S' X' Y'$ are directly similar. Then the spiral similarity with centre $D$ takes points $S, X, Y, O_1$ to $S', X', Y', O_1'$. Similarly, there is a spiral similarity with centre $D$ mapping $S, X, Y, O_1$ to $S'', X'', Y'', O_1''$. From these, we see that there is a spiral similarity taking the corresponding points $S, S', S''$ to points $O_1, O_1', O_1''$. In particular, $O_1, O_1', O_1''$ are collinear.\n\n![](attached_image_2.png)\n\nIt now suffices to show that $O_1$ lies on the perpendicular bisector of $P M$ for two special cases.\nFirstly, we take $S$ to be the midpoint of $A H$. Then $X$ and $Y$ are the feet of altitudes from $C$ and $B$ respectively. It is well-known that the circumcircle of triangle $X S Y$ is the nine-point circle of triangle $A B C$. Then $O_1$ is the nine-point centre and $O_1 P=O_1 M$. Indeed, $P$ and $M$ also lies on the nine-point circle.\nSecondly, we take $S'$ to be the midpoint of $A O$. Then $X'$ and $Y'$ are the midpoints of $A B$ and $A C$ respectively. Then $X' Y' \\parallel B C$. Clearly, $S'$ lies on the perpendicular bisector of $P M$. This shows the perpendicular bisectors of $X' Y'$ and $P M$ coincide. Hence, we must have $O_1' P=O_1' M$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32026,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incentre of a non-equilateral triangle $ABC$, $I_{A}$ be the $A$-excentre, $I_{A}'$ be the reflection of $I_{A}$ in $BC$, and $l_{A}$ be the reflection of line $A I_{A}'$ in $A I$. Define points $I_{B}, I_{B}'$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l_{A}$ and $l_{B}$.\n\na. Prove that $P$ lies on line $OI$ where $O$ is the circumcentre of triangle $ABC$.\n\nb. Let one of the tangents from $P$ to the incircle of triangle $ABC$ meet the circumcircle at points $X$ and $Y$. Show that $\\angle X I Y = 120^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a. Let $A'$ be the reflection of $A$ in $BC$ and let $M$ be the second intersection of line $A I$ and the circumcircle $\\Gamma$ of triangle $ABC$. As triangles $ABA'$ and $AOC$ are isosceles with $\\angle ABA' = 2\\angle ABC = \\angle AOC$, they are similar to each other. Also, triangles $ABI_{A}$ and $AIC$ are similar. Therefore we have\n$$\n\\frac{AA'}{AI_{A}} = \\frac{AA'}{AB} \\cdot \\frac{AB}{AI_{A}} = \\frac{AC}{AO} \\cdot \\frac{AI}{AC} = \\frac{AI}{AO}.\n$$\nTogether with $\\angle A'AI_{A} = \\angle IAO$, we find that triangles $AA'I_{A}$ and $AIO$ are similar.\n![](attached_image_1.png)\nDenote by $P'$ the intersection of line $AP$ and line $OI$. Using directed angles, we have\n$$\n\\begin{aligned}\n\\measuredangle MAP' &= \\measuredangle I_{A}'AI_{A} = \\measuredangle I_{A}'AA' - \\measuredangle I_{A}AA' = \\measuredangle AA'I_{A} - \\measuredangle (AM, OM) \\\\\n&= \\measuredangle AIO - \\measuredangle AMO = \\measuredangle MOP'\n\\end{aligned}\n$$\nThis shows $M, O, A, P'$ are concyclic.\nDenote by $R$ and $r$ the circumradius and inradius of triangle $ABC$. Then\n$$\nIP' = \\frac{IA \\cdot IM}{IO} = \\frac{IO^{2} - R^{2}}{IO}\n$$\nis independent of $A$. Hence, $BP$ also meets line $OI$ at the same point $P'$ so that $P' = P$, and $P$ lies on $OI$.\n\nb. By Poncelet's Porism, the other tangents to the incircle of triangle $ABC$ from $X$ and $Y$ meet at a point $Z$ on $\\Gamma$. Let $T$ be the touching point of the incircle to $XY$, and let $D$ be the midpoint of $XY$. We have\n$$\n\\begin{aligned}\nOD &= IT \\cdot \\frac{OP}{IP} = r\\left(1 + \\frac{OI}{IP}\\right) = r\\left(1 + \\frac{OI^{2}}{OI \\cdot IP}\\right) = r\\left(1 + \\frac{R^{2} - 2Rr}{R^{2} - IO^{2}}\\right) \\\\\n&= r\\left(1 + \\frac{R^{2} - 2Rr}{2Rr}\\right) = \\frac{R}{2} = \\frac{OX}{2}\n\\end{aligned}\n$$\nThis shows $\\angle XZY = 60^{\\circ}$ and hence $\\angle XIY = 120^{\\circ}$.\na. Note that triangles $A I_{B} C$ and $I_{A} B C$ are similar since their corresponding interior angles are equal. Therefore, the four triangles $A I_{B}' C$, $A I_{B} C$, $I_{A} B C$ and $I_{A}' B C$ are all similar. From $\\triangle A I_{B}' C \\sim \\triangle I_{A}' B C$, we get $\\triangle A I_{A}' C \\sim \\triangle I_{B}' B C$. From $\\measuredangle ABP = \\measuredangle I_{B}' B C = \\measuredangle A I_{A}' C$ and $\\measuredangle BAP = \\measuredangle I_{A}' AC$, the triangles $ABP$ and $A I_{A}' C$ are directly similar.\n![](attached_image_2.png)\nConsider the inversion with centre $A$ and radius $\\sqrt{AB \\cdot AC}$ followed by the reflection in $AI$. Then $B$ and $C$ are mapped to each other, and $I$ and $I_{A}$ are mapped to each other.\nFrom the similar triangles obtained, we have $AP \\cdot AI_{A}' = AB \\cdot AC$ so that $P$ is mapped to $I_{A}'$ under the transformation. In addition, line $AO$ is mapped to the altitude from $A$, and hence $O$ is mapped to the reflection of $A$ in $BC$, which we call point $A'$. Note that $AA'I_{A}I_{A}'$ is an isosceles trapezoid, which shows it is inscribed in a circle. The preimage of this circle is a straight line, meaning that $O, I, P$ are collinear.\n\nb. Denote by $R$ and $r$ the circumradius and inradius of triangle $ABC$. Note that by the above transformation, we have $\\triangle APO \\sim \\triangle AA'I_{A}'$ and $\\triangle AA'I_{A} \\sim \\triangle AIO$. Therefore, we find that\n$$\nPO = A'I_{A}' \\cdot \\frac{AO}{AI_{A}'} = AI_{A} \\cdot \\frac{AO}{A'I_{A}} = \\frac{AI_{A}}{A'I_{A}} \\cdot AO = \\frac{AO}{IO} \\cdot AO\n$$\nThis shows $PO \\cdot IO = R^{2}$, and it follows that $P$ and $I$ are mapped to each other under the inversion with respect to the circumcircle $\\Gamma$ of triangle $ABC$. Then $PX \\cdot PY$, which is the power of $P$ with respect to $\\Gamma$, equals $PI \\cdot PO$. This yields $X, I, O, Y$ are concyclic.\nLet $T$ be the touching point of the incircle to $XY$, and let $D$ be the midpoint of $XY$. Then\n$$\nOD = IT \\cdot \\frac{PO}{PI} = r \\cdot \\frac{PO}{PO - IO} = r \\cdot \\frac{R^{2}}{R^{2} - IO^{2}} = r \\cdot \\frac{R^{2}}{2Rr} = \\frac{R}{2}.\n$$\nThis shows $\\angle DOX = 60^{\\circ}$ and hence $\\angle XIY = \\angle XOY = 120^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32027,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_{1}$, $B_{1}$ and $C_{1}$ be points on sides $BC$, $CA$ and $AB$ of an acute triangle $ABC$ respectively, such that $AA_{1}$, $BB_{1}$ and $CC_{1}$ are the internal angle bisectors of triangle $ABC$. Let $I$ be the incentre of triangle $ABC$, and $H$ be the orthocentre of triangle $A_{1}B_{1}C_{1}$. Show that\n$$\nAH + BH + CH \\geqslant AI + BI + CI.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality, assume $\\alpha = \\angle BAC \\leqslant \\beta = \\angle CBA \\leqslant \\gamma = \\angle ACB$. Denote by $a$, $b$, $c$ the lengths of $BC$, $CA$, $AB$ respectively. We first show that triangle $A_{1}B_{1}C_{1}$ is acute.\nChoose points $D$ and $E$ on side $BC$ such that $B_{1}D \\parallel AB$ and $B_{1}E$ is the internal angle bisector of $\\angle BB_{1}C$. As $\\angle B_{1}DB = 180^{\\circ} - \\beta$ is obtuse, we have $BB_{1} > B_{1}D$. Thus,\n$$\n\\frac{BE}{EC} = \\frac{BB_{1}}{B_{1}C} > \\frac{DB_{1}}{B_{1}C} = \\frac{BA}{AC} = \\frac{BA_{1}}{A_{1}C}\n$$\nTherefore, $BE > BA_{1}$ and $\\frac{1}{2} \\angle BB_{1}C = \\angle BB_{1}E > \\angle BB_{1}A_{1}$. Similarly, $\\frac{1}{2} \\angle BB_{1}A > \\angle BB_{1}C_{1}$. It follows that\n$$\n\\angle A_{1}B_{1}C_{1} = \\angle BB_{1}A_{1} + \\angle BB_{1}C_{1} < \\frac{1}{2}(\\angle BB_{1}C + \\angle BB_{1}A) = 90^{\\circ}\n$$\nis acute. By symmetry, triangle $A_{1}B_{1}C_{1}$ is acute.\nLet $BB_{1}$ meet $A_{1}C_{1}$ at $F$. From $\\alpha \\leqslant \\gamma$, we get $a \\leqslant c$, which implies\n$$\nBA_{1} = \\frac{ca}{b + c} \\leqslant \\frac{ac}{a + b} = BC_{1}\n$$\nand hence $\\angle BC_{1}A_{1} \\leqslant \\angle BA_{1}C_{1}$. As $BF$ is the internal angle bisector of $\\angle A_{1}BC_{1}$, this shows $\\angle B_{1}FC_{1} = \\angle BFA_{1} \\leqslant 90^{\\circ}$. Hence, $H$ lies on the same side of $BB_{1}$ as $C_{1}$. This shows $H$ lies inside triangle $BB_{1}C_{1}$. Similarly, from $\\alpha \\leqslant \\beta$ and $\\beta \\leqslant \\gamma$, we know that $H$ lies inside triangles $CC_{1}B_{1}$ and $AA_{1}C_{1}$.\n![](attached_image_1.png)\nAs $\\alpha \\leqslant \\beta \\leqslant \\gamma$, we have $\\alpha \\leqslant 60^{\\circ} \\leqslant \\gamma$. Then $\\angle BIC \\leqslant 120^{\\circ} \\leqslant \\angle AIB$. Firstly, suppose $\\angle AIC \\geqslant 120^{\\circ}$.\nRotate points $B$, $I$, $H$ through $60^{\\circ}$ about $A$ to $B'$, $I'$, $H'$ so that $B'$ and $C$ lie on different sides of $AB$. Since triangle $AI'I$ is equilateral, we have\n$$\n\\begin{equation*}\nAI + BI + CI = I'I + B'I' + IC = B'I' + I'I + IC \\tag{1}\n\\end{equation*}\n$$\nSimilarly,\n$$\n\\begin{equation*}\nAH + BH + CH = H'H + B'H' + HC = B'H' + H'H + HC \\tag{2}\n\\end{equation*}\n$$\nAs $\\angle AII' = \\angle AI'I = 60^{\\circ}$, $\\angle AI'B' = \\angle AIB \\geqslant 120^{\\circ}$ and $\\angle AIC \\geqslant 120^{\\circ}$, the quadrilateral $B'I'I C$ is convex and lies on the same side of $B'C$ as $A$.\nNext, since $H$ lies inside triangle $ACC_{1}$, $H$ lies outside $B'I'I C$. Also, $H$ lying inside triangle $ABI$ implies $H'$ lies inside triangle $AB'I'$. This shows $H'$ lies outside $B'I'I C$ and hence the convex quadrilateral $B'I'I C$ is contained inside the quadrilateral $B'H'H C$. It follows that the perimeter of $B'I'I C$ cannot exceed the perimeter of $B'H'H C$. From (1) and (2), we conclude that\n$$\nAH + BH + CH \\geqslant AI + BI + CI.\n$$\nFor the case $\\angle AIC < 120^{\\circ}$, we can rotate $B$, $I$, $H$ through $60^{\\circ}$ about $C$ to $B'$, $I'$, $H'$ so that $B'$ and $A$ lie on different sides of $BC$. The proof is analogous to the previous case and we still get the desired inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32028,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define $P(n) = n^{2} + n + 1$. For any positive integers $a$ and $b$, the set\n$$\n\\{P(a), P(a+1), P(a+2), \\ldots, P(a+b)\\}\n$$\nis said to be fragrant if none of its elements is relatively prime to the product of the other elements. Determine the smallest size of a fragrant set.",
    "options": [],
    "answer": "6",
    "solution": "We have the following observations.\n(i) $(P(n), P(n+1)) = 1$ for any $n$.\nWe have $(P(n), P(n+1)) = \\left(n^{2} + n + 1, n^{2} + 3n + 3\\right) = \\left(n^{2} + n + 1, 2n + 2\\right)$. Noting that $n^{2} + n + 1$ is odd and $\\left(n^{2} + n + 1, n + 1\\right) = (1, n + 1) = 1$, the claim follows.\n\n(ii) $(P(n), P(n+2)) = 1$ for $n \\not\\equiv 2 \\pmod{7}$ and $(P(n), P(n+2)) = 7$ for $n \\equiv 2 \\pmod{7}$.\nFrom $(2n + 7) P(n) - (2n - 1) P(n+2) = 14$ and the fact that $P(n)$ is odd, $(P(n), P(n+2))$ must be a divisor of $7$. The claim follows by checking $n \\equiv 0, 1, \\ldots, 6 \\pmod{7}$ directly.\n\n(iii) $(P(n), P(n+3)) = 1$ for $n \\not\\equiv 1 \\pmod{3}$ and $3 \\mid (P(n), P(n+3))$ for $n \\equiv 1 \\pmod{3}$.\nFrom $(n + 5) P(n) - (n - 1) P(n+3) = 18$ and the fact that $P(n)$ is odd, $(P(n), P(n+3))$ must be a divisor of $9$. The claim follows by checking $n \\equiv 0, 1, 2 \\pmod{3}$ directly.\n\nSuppose there exists a fragrant set with at most $5$ elements. We may assume it contains exactly $5$ elements $P(a), P(a+1), \\ldots, P(a+4)$ since the following argument also works with fewer elements. Consider $P(a+2)$. From (i), it is relatively prime to $P(a+1)$ and $P(a+3)$. Without loss of generality, assume $(P(a), P(a+2)) > 1$. From (ii), we have $a \\equiv 2 \\pmod{7}$. The same observation implies $(P(a+1), P(a+3)) = 1$. In order that the set is fragrant, $(P(a), P(a+3))$ and $(P(a+1), P(a+4))$ must both be greater than $1$. From (iii), this holds only when both $a$ and $a+1$ are congruent to $1 \\pmod{3}$, which is a contradiction.\n\nIt now suffices to construct a fragrant set of size $6$. By the Chinese Remainder Theorem, we can take a positive integer $a$ such that\n$$\na \\equiv 7 \\pmod{19}, \\quad a+1 \\equiv 2 \\pmod{7}, \\quad a+2 \\equiv 1 \\pmod{3} .\n$$\nFor example, we may take $a = 197$. From (ii), both $P(a+1)$ and $P(a+3)$ are divisible by $7$. From (iii), both $P(a+2)$ and $P(a+5)$ are divisible by $3$. One also checks from $19 \\mid P(7) = 57$ and $19 \\mid P(11) = 133$ that $P(a)$ and $P(a+4)$ are divisible by $19$. Therefore, the set $\\{P(a), P(a+1), \\ldots, P(a+5)\\}$ is fragrant.\n\nTherefore, the smallest size of a fragrant set is $6$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32029,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive integer which is not a square number. Denote by $A$ the set of all positive integers $k$ such that\n$$\nk = \\frac{x^{2} - a}{x^{2} - y^{2}}\n$$\nfor some integers $x$ and $y$ with $x > \\sqrt{a}$. Denote by $B$ the set of all positive integers $k$ such that (1) is satisfied for some integers $x$ and $y$ with $0 \\leqslant x < \\sqrt{a}$. Prove that $A = B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first prove the following preliminary result.\n- Claim. For fixed $k$, let $x, y$ be integers satisfying (1). Then the numbers $x_{1}, y_{1}$ defined by\n$$\nx_{1} = \\frac{1}{2}\\left(x - y + \\frac{(x - y)^{2} - 4a}{x + y}\\right), \\quad y_{1} = \\frac{1}{2}\\left(x - y - \\frac{(x - y)^{2} - 4a}{x + y}\\right)\n$$\nare integers and satisfy (1) (with $x, y$ replaced by $x_{1}, y_{1}$ respectively).\n\nProof. Since $x_{1} + y_{1} = x - y$ and\n$$\nx_{1} = \\frac{x^{2} - x y - 2a}{x + y} = -x + \\frac{2(x^{2} - a)}{x + y} = -x + 2k(x - y),\n$$\nboth $x_{1}$ and $y_{1}$ are integers. Let $u = x + y$ and $v = x - y$. The relation (1) can be rewritten as\n$$\nu^{2} - (4k - 2) u v + (v^{2} - 4a) = 0\n$$\nBy Vieta's Theorem, the number $z = \\frac{v^{2} - 4a}{u}$ satisfies\n$$\nv^{2} - (4k - 2) v z + (z^{2} - 4a) = 0\n$$\nSince $x_{1}$ and $y_{1}$ are defined so that $v = x_{1} + y_{1}$ and $z = x_{1} - y_{1}$, we can reverse the process and verify (1) for $x_{1}, y_{1}$.\n\nWe first show that $B \\subset A$. Take any $k \\in B$ so that (1) is satisfied for some integers $x, y$ with $0 \\leqslant x < \\sqrt{a}$. Clearly, $y \\neq 0$ and we may assume $y$ is positive. Since $a$ is not a square, we have $k > 1$. Hence, we get $0 \\leqslant x < y < \\sqrt{a}$. Define\n$$\nx_{1} = \\frac{1}{2}\\left|x - y + \\frac{(x - y)^{2} - 4a}{x + y}\\right|, \\quad y_{1} = \\frac{1}{2}\\left(x - y - \\frac{(x - y)^{2} - 4a}{x + y}\\right).\n$$\nBy the Claim, $x_{1}, y_{1}$ are integers satisfying (1). Also, we have\n$$\nx_{1} \\geqslant -\\frac{1}{2}\\left(x - y + \\frac{(x - y)^{2} - 4a}{x + y}\\right) = \\frac{2a + x(y - x)}{x + y} \\geqslant \\frac{2a}{x + y} > \\sqrt{a}.\n$$\nThis implies $k \\in A$ and hence $B \\subset A$.\n\nNext, we shall show that $A \\subset B$. Take any $k \\in A$ so that (1) is satisfied for some integers $x, y$ with $x > \\sqrt{a}$. Again, we may assume $y$ is positive. Among all such representations of $k$, we choose the one with smallest $x + y$. Define\n$$\nx_{1} = \\frac{1}{2}\\left|x - y + \\frac{(x - y)^{2} - 4a}{x + y}\\right|, \\quad y_{1} = \\frac{1}{2}\\left(x - y - \\frac{(x - y)^{2} - 4a}{x + y}\\right).\n$$\nBy the Claim, $x_{1}, y_{1}$ are integers satisfying (1). Since $k > 1$, we get $x > y > \\sqrt{a}$. Therefore, we have $y_{1} > \\frac{4a}{x + y} > 0$ and $\\frac{4a}{x + y} < x + y$. It follows that\n$$\nx_{1} + y_{1} \\leqslant \\max \\left\\{x - y, \\frac{4a - (x - y)^{2}}{x + y}\\right\\} < x + y\n$$\nIf $x_{1} > \\sqrt{a}$, we get a contradiction due to the minimality of $x + y$. Therefore, we must have $0 \\leqslant x_{1} < \\sqrt{a}$, which means $k \\in B$ so that $A \\subset B$.\n\nThe two subset relations combine to give $A = B$.\nThe relation (1) is equivalent to\n$$\nk y^{2} - (k - 1) x^{2} = a\n$$\nMotivated by Pell's Equation, we prove the following, which is essentially the same as the Claim in Solution 1.\n- Claim. If $(x_{0}, y_{0})$ is a solution to (2), then $((2k - 1)x_{0} \\pm 2k y_{0}, (2k - 1)y_{0} \\pm 2(k - 1)x_{0})$ is also a solution to (2).\n\nProof. We check directly that\n$$\n\\begin{aligned}\n& k\\left((2k - 1)y_{0} \\pm 2(k - 1)x_{0}\\right)^{2} - (k - 1)\\left((2k - 1)x_{0} \\pm 2k y_{0}\\right)^{2} \\\\\n= & \\left(k(2k - 1)^{2} - (k - 1)(2k)^{2}\\right) y_{0}^{2} + \\left(k(2(k - 1))^{2} - (k - 1)(2k - 1)^{2}\\right) x_{0}^{2} \\\\\n= & k y_{0}^{2} - (k - 1) x_{0}^{2} = a.\n\\end{aligned}\n$$\nIf (2) is satisfied for some $0 \\leqslant x < \\sqrt{a}$ and nonnegative integer $y$, then clearly (1) implies $y > x$. Also, we have $k > 1$ since $a$ is not a square number. By the Claim, consider another solution to (2) defined by\n$$\nx_{1} = (2k - 1)x + 2k y, \\quad y_{1} = (2k - 1)y + 2(k - 1)x.\n$$\nIt satisfies $x_{1} \\geqslant (2k - 1)x + 2k(x + 1) = (4k - 1)x + 2k > x$. Then we can replace the old solution by a new one which has a larger value in $x$. After a finite number of replacements, we must get a solution with $x > \\sqrt{a}$. This shows $B \\subset A$.\n\nIf (2) is satisfied for some $x > \\sqrt{a}$ and nonnegative integer $y$, by the Claim we consider another solution to (2) defined by\n$$\nx_{1} = |(2k - 1)x - 2k y|, \\quad y_{1} = (2k - 1)y - 2(k - 1)x.\n$$\nFrom (2), we get $\\sqrt{k} y > \\sqrt{k - 1} x$. This implies $k y > \\sqrt{k(k - 1)} x > (k - 1)x$ and hence $(2k - 1)x - 2k y < x$. On the other hand, the relation (1) implies $x > y$. Then it is clear that $(2k - 1)x - 2k y > -x$. These combine to give $x_{1} < x$, which means we have found a solution to (2) with $x$ having a smaller absolute value. After a finite number of steps, we shall obtain a solution with $0 \\leqslant x < \\sqrt{a}$. This shows $A \\subset B$.\n\nThe desired result follows from $B \\subset A$ and $A \\subset B$.\nIt suffices to show $A \\cup B$ is a subset of $A \\cap B$. We take any $k \\in A \\cup B$, which means there exist integers $x, y$ satisfying (1). Since $a$ is not a square, it follows that $k \\neq 1$. As in Solution 2, the result follows readily once we have proved the existence of a solution ($x_{1}, y_{1}$) to (1) with $|x_{1}| > |x|$, and, in case of $x > \\sqrt{a}$, another solution ($x_{2}, y_{2}$) with $|x_{2}| < |x|$.\n\nWithout loss of generality, assume $x, y \\geqslant 0$. Let $u = x + y$ and $v = x - y$. Then $u \\geqslant v$ and (1) becomes\n$$\nk = \\frac{(u + v)^{2} - 4a}{4u v}\n$$\nThis is the same as\n$$\nv^{2} + (2u - 4k u)v + u^{2} - 4a = 0\n$$\nLet $v_{1} = 4k u - 2u - v$. Then $u + v_{1} = 4k u - u - v \\geqslant 8u - u - v > u + v$. By Vieta's Theorem, $v_{1}$ satisfies\n$$\nv_{1}^{2} + (2u - 4k u)v_{1} + u^{2} - 4a = 0\n$$\nThis gives $k = \\frac{(u + v_{1})^{2} - 4a}{4u v_{1}}$. As $k$ is an integer, $u + v_{1}$ must be even. Therefore, $x_{1} = \\frac{u + v_{1}}{2}$ and $y_{1} = \\frac{v_{1} - u}{2}$ are integers. By reversing the process, we can see that $(x_{1}, y_{1})$ is a solution to (1), with $x_{1} = \\frac{u + v_{1}}{2} > \\frac{u + v}{2} = x \\geqslant 0$. This completes the first half of the proof.\n\nSuppose $x > \\sqrt{a}$. Then $u + v > 2\\sqrt{a}$ and (3) can be rewritten as\n$$\nu^{2} + (2v - 4k v)u + v^{2} - 4a = 0\n$$\nLet $u_{2} = 4k v - 2v - u$. By Vieta's Theorem, we have $u u_{2} = v^{2} - 4a$ and\n$$\nu_{2}^{2} + (2v - 4k v)u_{2} + v^{2} - 4a = 0\n$$\nBy $u > 0, u + v > 2\\sqrt{a}$ and (3), we have $v > 0$. If $u_{2} \\geqslant 0$, then $v u_{2} \\leqslant u u_{2} = v^{2} - 4a < v^{2}$. This shows $u_{2} < v \\leqslant u$ and $0 < u_{2} + v < u + v$. If $u_{2} < 0$, then $(u_{2} + v) + (u + v) = 4k v > 0$ and $u_{2} + v < u + v$ imply $|u_{2} + v| < u + v$. In any case, since $u_{2} + v$ is even from (4), we can define $x_{2} = \\frac{u_{2} + v}{2}$ and $y_{2} = \\frac{u_{2} - v}{2}$ so that (1) is satisfied with $|x_{2}| < x$, as desired. The proof is thus complete.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32030,
    "subject": "Mathematics (Multi-modal)",
    "question": "Denote by $\\mathbb{N}$ the set of all positive integers. Find all functions $f: \\mathbb{N} \\rightarrow \\mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-m n$ is nonzero and divides $m f(m)+n f(n)$.",
    "options": [],
    "answer": "f(n) = n^2 for all positive integers n",
    "solution": "It is given that\n$$\nf(m)+f(n)-m n \\mid m f(m)+n f(n) . \\tag{1}\n$$\nTaking $m=n=1$ in (1), we have $2 f(1)-1 \\mid 2 f(1)$. Then $2 f(1)-1 \\mid 2 f(1)-(2 f(1)-1)=1$ and hence $f(1)=1$.\nLet $p \\geqslant 7$ be a prime. Taking $m=p$ and $n=1$ in (1), we have $f(p)-p+1 \\mid p f(p)+1$ and hence\n$$\nf(p)-p+1 \\mid p f(p)+1-p(f(p)-p+1)=p^{2}-p+1\n$$\nIf $f(p)-p+1=p^{2}-p+1$, then $f(p)=p^{2}$. If $f(p)-p+1 \\neq p^{2}-p+1$, as $p^{2}-p+1$ is an odd positive integer, we have $p^{2}-p+1 \\geqslant 3(f(p)-p+1)$, that is,\n$$\n\\begin{equation*}\nf(p) \\leqslant \\frac{1}{3}\\left(p^{2}+2 p-2\\right) \\tag{2}\n\\end{equation*}\n$$\nTaking $m=n=p$ in (1), we have $2 f(p)-p^{2} \\mid 2 p f(p)$. This implies\n$$\n2 f(p)-p^{2} \\mid 2 p f(p)-p\\left(2 f(p)-p^{2}\\right)=p^{3} .\n$$\nBy (2) and $f(p) \\geqslant 1$, we get\n$$\n-p^{2}<2 f(p)-p^{2} \\leqslant \\frac{2}{3}\\left(p^{2}+2 p-2\\right)-p^{2}<-p\n$$\nsince $p \\geqslant 7$. This contradicts the fact that $2 f(p)-p^{2}$ is a factor of $p^{3}$. Thus we have proved that $f(p)=p^{2}$ for all primes $p \\geqslant 7$.\nLet $n$ be a fixed positive integer. Choose a sufficiently large prime $p$. Consider $m=p$ in (1). We obtain\n$$\nf(p)+f(n)-p n \\mid p f(p)+n f(n)-n(f(p)+f(n)-p n)=p f(p)-n f(p)+p n^{2} .\n$$\nAs $f(p)=p^{2}$, this implies $p^{2}-p n+f(n) \\mid p\\left(p^{2}-p n+n^{2}\\right)$. As $p$ is sufficiently large and $n$ is fixed, $p$ cannot divide $f(n)$, and so $\\left(p, p^{2}-p n+f(n)\\right)=1$. It follows that $p^{2}-p n+f(n) \\mid p^{2}-p n+n^{2}$ and hence\n$$\np^{2}-p n+f(n) \\mid\\left(p^{2}-p n+n^{2}\\right)-\\left(p^{2}-p n+f(n)\\right)=n^{2}-f(n) .\n$$\nNote that $n^{2}-f(n)$ is fixed while $p^{2}-p n+f(n)$ is chosen to be sufficiently large. Therefore, we must have $n^{2}-f(n)=0$ so that $f(n)=n^{2}$ for any positive integer $n$.\nFinally, we check that when $f(n)=n^{2}$ for any positive integer $n$, then\n$$\nf(m)+f(n)-m n=m^{2}+n^{2}-m n\n$$\nand\n$$\nm f(m)+n f(n)=m^{3}+n^{3}=(m+n)\\left(m^{2}+n^{2}-m n\\right) .\n$$\nThe latter expression is divisible by the former for any positive integers $m, n$. This shows $f(n)=n^{2}$ is the only solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32031,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an odd positive integer. In the Cartesian plane, a cyclic polygon $P$ with area $S$ is chosen. All its vertices have integral coordinates, and the squares of its side lengths are all divisible by $n$. Prove that $2 S$ is an integer divisible by $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P = A_{1} A_{2} \\ldots A_{k}$ and let $A_{k+i} = A_{i}$ for $i \\geqslant 1$. By the Shoelace Formula, the area of any convex polygon with integral coordinates is half an integer. Therefore, $2 S$ is an integer. We shall prove by induction on $k \\geqslant 3$ that $2 S$ is divisible by $n$. Clearly, it suffices to consider $n = p^{t}$ where $p$ is an odd prime and $t \\geqslant 1$.\n\nFor the base case $k = 3$, let the side lengths of $P$ be $\\sqrt{n a}$, $\\sqrt{n b}$, $\\sqrt{n c}$ where $a, b, c$ are positive integers. By Heron's Formula,\n$$\n16 S^{2} = n^{2} \\left(2 a b + 2 b c + 2 c a - a^{2} - b^{2} - c^{2}\\right).\n$$\nThis shows $16 S^{2}$ is divisible by $n^{2}$. Since $n$ is odd, $2 S$ is divisible by $n$.\n\nAssume $k \\geqslant 4$. If the square of length of one of the diagonals is divisible by $n$, then that diagonal divides $P$ into two smaller polygons, to which the induction hypothesis applies. Hence we may assume that none of the squares of diagonal lengths is divisible by $n$. As usual, we denote by $\\nu_{p}(r)$ the exponent of $p$ in the prime decomposition of $r$. We claim the following.\n\n- Claim. $\\nu_{p}\\left(A_{1} A_{m}^{2}\\right) > \\nu_{p}\\left(A_{1} A_{m+1}^{2}\\right)$ for $2 \\leqslant m \\leqslant k-1$.\n\nProof. The case $m = 2$ is obvious since $\\nu_{p}\\left(A_{1} A_{2}^{2}\\right) \\geqslant p^{t} > \\nu_{p}\\left(A_{1} A_{3}^{2}\\right)$ by the condition and the above assumption.\n\nSuppose $\\nu_{p}\\left(A_{1} A_{2}^{2}\\right) > \\nu_{p}\\left(A_{1} A_{3}^{2}\\right) > \\cdots > \\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$ where $3 \\leqslant m \\leqslant k-1$. For the induction step, we apply Ptolemy's Theorem to the cyclic quadrilateral $A_{1} A_{m-1} A_{m} A_{m+1}$ to get\n$$\nA_{1} A_{m+1} \\times A_{m-1} A_{m} + A_{1} A_{m-1} \\times A_{m} A_{m+1} = A_{1} A_{m} \\times A_{m-1} A_{m+1}\n$$\nwhich can be rewritten as\n$$\n\\begin{align*}\nA_{1} A_{m+1}^{2} \\times A_{m-1} A_{m}^{2} = & A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2} + A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2} \\\\\n& - 2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1} \\tag{1}\n\\end{align*}\n$$\nFrom this, $2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1}$ is an integer. We consider the component of $p$ of each term in (1). By the inductive hypothesis, we have $\\nu_{p}\\left(A_{1} A_{m-1}^{2}\\right) > \\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$. Also, we have $\\nu_{p}\\left(A_{m} A_{m+1}^{2}\\right) \\geqslant p^{t} > \\nu_{p}\\left(A_{m-1} A_{m+1}^{2}\\right)$. These give\n$$\n\\begin{equation*}\n\\nu_{p}\\left(A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2}\\right) > \\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right) \\tag{2}\n\\end{equation*}\n$$\nNext, we have $\\nu_{p}\\left(4 A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2} \\times A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right) = \\nu_{p}\\left(A_{1} A_{m-1}^{2} \\times A_{m} A_{m+1}^{2}\\right) + \\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right) > 2 \\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)$ from (2). This implies\n$$\n\\begin{equation*}\n\\nu_{p}\\left(2 A_{1} A_{m-1} \\times A_{m} A_{m+1} \\times A_{1} A_{m} \\times A_{m-1} A_{m+1}\\right) > \\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right) \\tag{3}\n\\end{equation*}\n$$\nCombining (1), (2) and (3), we conclude that\n$$\n\\nu_{p}\\left(A_{1} A_{m+1}^{2} \\times A_{m-1} A_{m}^{2}\\right) = \\nu_{p}\\left(A_{1} A_{m}^{2} \\times A_{m-1} A_{m+1}^{2}\\right)\n$$\nBy $\\nu_{p}\\left(A_{m-1} A_{m}^{2}\\right) \\geqslant p^{t} > \\nu_{p}\\left(A_{m-1} A_{m+1}^{2}\\right)$, we get $\\nu_{p}\\left(A_{1} A_{m+1}^{2}\\right) < \\nu_{p}\\left(A_{1} A_{m}^{2}\\right)$. The Claim follows by induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32032,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all polynomials $P(x)$ of odd degree $d$ and with integer coefficients satisfying the following property: for each positive integer $n$, there exist $n$ positive integers $x_{1}, x_{2}, \\ldots, x_{n}$ such that $\\frac{1}{2}<\\frac{P\\left(x_{i}\\right)}{P\\left(x_{j}\\right)}<2$ and $\\frac{P\\left(x_{i}\\right)}{P\\left(x_{j}\\right)}$ is the $d$-th power of a rational number for every pair of indices $i$ and $j$ with $1 \\leqslant i, j \\leqslant n$.",
    "options": [],
    "answer": "All such polynomials are P(x) = a (r x + s)^d with integers a, r, s and d odd.",
    "solution": "Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\\cdots+a_{0}$. Consider the substitution $y=d a_{d} x+a_{d-1}$. By defining $Q(y)=P(x)$, we find that $Q$ is a polynomial with rational coefficients without the term $y^{d-1}$. Let $Q(y)=b_{d} y^{d}+b_{d-2} y^{d-2}+b_{d-3} y^{d-3}+\\cdots+b_{0}$ and $B=\\max _{0 \\leqslant i \\leqslant d}\\left\\{\\left|b_{i}\\right|\\right\\}$ (where $b_{d-1}=0$ ).\nThe condition shows that for each $n \\geqslant 1$, there exist integers $y_{1}, y_{2}, \\ldots, y_{n}$ such that $\\frac{1}{2}<\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{j}\\right)}<2$ and $\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{j}\\right)}$ is the $d$-th power of a rational number for $1 \\leqslant i, j \\leqslant n$. Since $n$ can be arbitrarily large, we may assume all $x_{i}$'s and hence $y_{i}$'s are integers larger than some absolute constant in the following.\nBy Dirichlet's Theorem, since $d$ is odd, we can find a sufficiently large prime $p$ such that $p \\equiv 2(\\bmod d)$. In particular, we have $(p-1, d)=1$. For this fixed $p$, we choose $n$ to be sufficiently large. Then by the Pigeonhole Principle, there must be $d+1$ of $y_{1}, y_{2}, \\ldots, y_{n}$ which are congruent $\\bmod p$. Without loss of generality, assume $y_{i} \\equiv y_{j}(\\bmod p)$ for $1 \\leqslant i, j \\leqslant d+1$. We shall establish the following.\n- Claim. $\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{1}\\right)}=\\frac{y_{i}^{d}}{y_{1}^{d}}$ for $2 \\leqslant i \\leqslant d+1$.\nProof. Let $\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{1}\\right)}=\\frac{l^{d}}{m^{d}}$ where $(l, m)=1$ and $l, m>0$. This can be rewritten in the expanded form\n$$\n\\begin{equation*}\nb_{d}\\left(m^{d} y_{i}^{d}-l^{d} y_{1}^{d}\\right)=-\\sum_{j=0}^{d-2} b_{j}\\left(m^{d} y_{i}^{j}-l^{d} y_{1}^{j}\\right) \\tag{1}\n\\end{equation*}\n$$\nLet $c$ be the common denominator of $Q$, so that $c Q(k)$ is an integer for any integer $k$. Note that $c$ depends only on $P$ and so we may assume $(p, c)=1$. Then $y_{1} \\equiv y_{i}(\\bmod p)$ implies $c Q\\left(y_{1}\\right) \\equiv c Q\\left(y_{i}\\right)(\\bmod p)$.\n- Case 1. $p \\mid c Q\\left(y_{1}\\right)$.\nIn this case, there is a cancellation of $p$ in the numerator and denominator of $\\frac{c Q\\left(y_{i}\\right)}{c Q\\left(y_{1}\\right)}$, so that $m^{d} \\leqslant p^{-1}\\left|c Q\\left(y_{1}\\right)\\right|$. Noting $\\left|Q\\left(y_{1}\\right)\\right|<2 B y_{1}^{d}$ as $y_{1}$ is large, we get\n$$\n\\begin{equation*}\nm \\leqslant p^{-\\frac{1}{d}}(2 c B)^{\\frac{1}{d}} y_{1} . \\tag{2}\n\\end{equation*}\n$$\nFor large $y_{1}$ and $y_{i}$, the relation $\\frac{1}{2}<\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{1}\\right)}<2$ implies\n$$\n\\begin{equation*}\n\\frac{1}{3}<\\frac{y_{i}^{d}}{y_{1}^{d}}<3 \\tag{3}\n\\end{equation*}\n$$\nWe also have\n$$\n\\begin{equation*}\n\\frac{1}{2}<\\frac{l^{d}}{m^{d}}<2 \\tag{4}\n\\end{equation*}\n$$\nNow, the left-hand side of (1) is\n$$\nb_{d}\\left(m y_{i}-l y_{1}\\right)\\left(m^{d-1} y_{i}^{d-1}+m^{d-2} y_{i}^{d-2} l y_{1}+\\cdots+l^{d-1} y_{1}^{d-1}\\right) .\n$$\nSuppose on the contrary that $m y_{i}-l y_{1} \\neq 0$. Then the absolute value of the above expression is at least $\\left|b_{d}\\right| m^{d-1} y_{i}^{d-1}$. On the other hand, the absolute value of the right-hand side of (1) is at most\n$$\n\\begin{aligned}\n\\sum_{j=0}^{d-2} B\\left(m^{d} y_{i}^{j}+l^{d} y_{1}^{j}\\right) & \\leqslant(d-1) B\\left(m^{d} y_{i}^{d-2}+l^{d} y_{1}^{d-2}\\right) \\\\\n& \\leqslant(d-1) B\\left(7 m^{d} y_{i}^{d-2}\\right) \\\\\n& \\leqslant 7(d-1) B\\left(p^{-\\frac{1}{d}}(2 c B)^{\\frac{1}{d}} y_{1}\\right) m^{d-1} y_{i}^{d-2} \\\\\n& \\leqslant 21(d-1) B p^{-\\frac{1}{d}}(2 c B)^{\\frac{1}{d}} m^{d-1} y_{i}^{d-1}\n\\end{aligned}\n$$\nby using successively (3), (4), (2) and again (3). This shows\n$$\n\\left|b_{d}\\right| m^{d-1} y_{i}^{d-1} \\leqslant 21(d-1) B p^{-\\frac{1}{d}}(2 c B)^{\\frac{1}{d}} m^{d-1} y_{i}^{d-1}\n$$\nwhich is a contradiction for large $p$ as $b_{d}, B, c, d$ depend only on the polynomial $P$. Therefore, we have $m y_{i}-l y_{1}=0$ in this case.\n- Case 2. $\\left(p, c Q\\left(y_{1}\\right)\\right)=1$.\nFrom $c Q\\left(y_{1}\\right) \\equiv c Q\\left(y_{i}\\right)(\\bmod p)$, we have $l^{d} \\equiv m^{d}(\\bmod p)$. Since $(p-1, d)=1$, we use Fermat Little Theorem to conclude $l \\equiv m(\\bmod p)$. Then $p \\mid m y_{i}-l y_{1}$. Suppose on the contrary that $m y_{i}-l y_{1} \\neq 0$. Then the left-hand side of (1) has absolute value at least $\\left|b_{d}\\right| p m^{d-1} y_{i}^{d-1}$. Similar to Case 1, the right-hand side of (1) has absolute value at most\n$$\n21(d-1) B(2 c B)^{\\frac{1}{d}} m^{d-1} y_{i}^{d-1}\n$$\nwhich must be smaller than $\\left|b_{d}\\right| p m^{d-1} y_{i}^{d-1}$ for large $p$. Again this yields a contradiction and hence $m y_{i}-l y_{1}=0$.\nIn both cases, we find that $\\frac{Q\\left(y_{i}\\right)}{Q\\left(y_{1}\\right)}=\\frac{l^{d}}{m^{d}}=\\frac{y_{i}^{d}}{y_{1}^{d}}$.\nFrom the Claim, the polynomial $Q\\left(y_{1}\\right) y^{d}-y_{1}^{d} Q(y)$ has roots $y=y_{1}, y_{2}, \\ldots, y_{d+1}$. Since its degree is at most $d$, this must be the zero polynomial. Hence, $Q(y)=b_{d} y^{d}$. This implies $P(x)=a_{d}\\left(x+\\frac{a_{d-1}}{d a_{d}}\\right)^{d}$. Let $\\frac{a_{d-1}}{d a_{d}}=\\frac{s}{r}$ with integers $r, s$ where $r \\geqslant 1$ and $(r, s)=1$. Since $P$ has integer coefficients, we need $r^{d} \\mid a_{d}$. Let $a_{d}=r^{d} a$. Then $P(x)=a(r x+s)^{d}$. It is obvious that such a polynomial satisfies the conditions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32033,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, \\ldots, a_{n}, k$, and $M$ be positive integers such that\n$$\n\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\cdots+\\frac{1}{a_{n}}=k \\quad \\text{ and } \\quad a_{1} a_{2} \\ldots a_{n}=M .\n$$\nIf $M>1$, prove that the polynomial\n$$\nP(x)=M(x+1)^{k}-\\left(x+a_{1}\\right)\\left(x+a_{2}\\right) \\cdots\\left(x+a_{n}\\right)\n$$\nhas no positive roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first prove that, for $x>0$,\n$$\n\\begin{equation*}\na_{i}(x+1)^{1 / a_{i}} \\leqslant x+a_{i}, \\tag{1}\n\\end{equation*}\n$$\nwith equality if and only if $a_{i}=1$. It is clear that equality occurs if $a_{i}=1$.\nIf $a_{i}>1$, the AM-GM inequality applied to a single copy of $x+1$ and $a_{i}-1$ copies of 1 yields\n$$\n\\frac{(x+1)+\\overbrace{1+1+\\cdots+1}^{a_{i}-1 \\text{ ones }}}{a_{i}} \\geqslant \\sqrt[a_{i}]{(x+1) \\cdot 1^{a_{i}-1}} \\Longrightarrow a_{i}(x+1)^{1 / a_{i}} \\leqslant x+a_{i} .\n$$\nSince $x+1>1$, the inequality is strict for $a_{i}>1$.\nMultiplying the inequalities (1) for $i=1,2, \\ldots, n$ yields\n$$\n\\prod_{i=1}^{n} a_{i}(x+1)^{1 / a_{i}} \\leqslant \\prod_{i=1}^{n}\\left(x+a_{i}\\right) \\Longleftrightarrow M(x+1)^{\\sum_{i=1}^{n} 1 / a_{i}}-\\prod_{i=1}^{n}\\left(x+a_{i}\\right) \\leqslant 0 \\Longleftrightarrow P(x) \\leqslant 0\n$$\nwith equality iff $a_{i}=1$ for all $i \\in\\{1,2, \\ldots, n\\}$. But this implies $M=1$, which is not possible. Hence $P(x)<0$ for all $x \\in \\mathbb{R}^{+}$, and $P$ has no positive roots.\nWe will prove that, in fact, all coefficients of the polynomial $P(x)$ are non-positive, and at least one of them is negative, which implies that $P(x)<0$ for $x>0$.\nIndeed, since $a_{j} \\geqslant 1$ for all $j$ and $a_{j}>1$ for some $j$ (since $a_{1} a_{2} \\ldots a_{n}=M>1$ ), we have $k=\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\cdots+\\frac{1}{a_{n}}<n$, so the coefficient of $x^{n}$ in $P(x)$ is $-1<0$. Moreover, the coefficient of $x^{r}$ in $P(x)$ is negative for $k<r \\leqslant n=\\operatorname{deg}(P)$.\nFor $0 \\leqslant r \\leqslant k$, the coefficient of $x^{r}$ in $P(x)$ is\n$$\nM \\cdot\\binom{k}{r}-\\sum_{1 \\leqslant i_{1}<i_{2}<\\cdots<i_{n-r} \\leqslant n} a_{i_{1}} a_{i_{2}} \\cdots a_{i_{n-r}}=a_{1} a_{2} \\cdots a_{n} \\cdot\\binom{k}{r}-\\sum_{1 \\leqslant i_{1}<i_{2}<\\cdots<i_{n-r} \\leqslant n} a_{i_{1}} a_{i_{2}} \\cdots a_{i_{n-r}},\n$$\nwhich is non-positive iff\n$$\n\\begin{equation*}\n\\binom{k}{r} \\leqslant \\sum_{1 \\leqslant j_{1}<j_{2}<\\cdots<j_{r} \\leqslant n} \\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}}} . \\tag{2}\n\\end{equation*}\n$$\nWe will prove (2) by induction on $r$. For $r=0$ it is an equality because the constant term of $P(x)$ is $P(0)=0$, and if $r=1$, (2) becomes $k=\\sum_{i=1}^{n} \\frac{1}{a_{i}}$. For $r>1$, if (2) is true for a given $r<k$, we have\n$$\n\\binom{k}{r+1}=\\frac{k-r}{r+1} \\cdot\\binom{k}{r} \\leqslant \\frac{k-r}{r+1} . \\sum_{1 \\leqslant j_{1}<j_{2}<\\cdots<j_{r} \\leqslant n} \\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}}},\n$$\nand it suffices to prove that\n$$\n\\frac{k-r}{r+1} . \\sum_{1 \\leqslant j_{1}<j_{2}<\\cdots<j_{r} \\leqslant n} \\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}}} \\leqslant \\sum_{1 \\leqslant j_{1}<\\cdots<j_{r}<j_{r+1} \\leqslant n} \\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}} a_{j_{r+1}}},\n$$\nwhich is equivalent to\n$$\n\\left(\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\cdots+\\frac{1}{a_{n}}-r\\right) \\sum_{1 \\leqslant j_{1}<j_{2}<\\cdots<j_{r} \\leqslant n} \\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}}} \\leqslant(r+1) \\sum_{1 \\leqslant j_{1}<\\cdots<j_{r}<j_{r+1} \\leqslant n} \\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}} a_{j_{r+1}}} .\n$$\nSince there are $r+1$ ways to choose a fraction $\\frac{1}{a_{j_{i}}}$ from $\\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}} a_{j_{r+1}}}$ to factor out, every term $\\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}} a_{j_{r+1}}}$ in the right hand side appears exactly $r+1$ times in the product\n$$\n\\left(\\frac{1}{a_{1}}+\\frac{1}{a_{2}}+\\cdots+\\frac{1}{a_{n}}\\right) \\sum_{1 \\leqslant j_{1}<j_{2}<\\cdots<j_{r} \\leqslant n} \\frac{1}{a_{j_{1}} a_{j_{2}} \\cdots a_{j_{r}}} .\n$$\nHence all terms in the right hand side cancel out.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32034,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $q$ be a real number. Gugu has a napkin with ten distinct real numbers written on it, and he writes the following three lines of real numbers on the blackboard:\n- In the first line, Gugu writes down every number of the form $a-b$, where $a$ and $b$ are two (not necessarily distinct) numbers on his napkin.\n- In the second line, Gugu writes down every number of the form $q a b$, where $a$ and $b$ are two (not necessarily distinct) numbers from the first line.\n- In the third line, Gugu writes down every number of the form $a^{2}+b^{2}-c^{2}-d^{2}$, where $a, b, c, d$ are four (not necessarily distinct) numbers from the first line.\nDetermine all values of $q$ such that, regardless of the numbers on Gugu's napkin, every number in the second line is also a number in the third line.",
    "options": [],
    "answer": "-2, 0, 2",
    "solution": "Answer: $-2, 0, 2$.\n\nCall a number $q$ good if every number in the second line appears in the third line unconditionally. We first show that the numbers $0$ and $\\pm 2$ are good. The third line necessarily contains $0$, so $0$ is good. For any two numbers $a, b$ in the first line, write $a = x - y$ and $b = u - v$, where $x, y, u, v$ are (not necessarily distinct) numbers on the napkin. We may now write\n$$\n2 a b = 2(x - y)(u - v) = (x - v)^2 + (y - u)^2 - (x - u)^2 - (y - v)^2\n$$\nwhich shows that $2$ is good. By negating both sides of the above equation, we also see that $-2$ is good.\n\nWe now show that $-2, 0$, and $2$ are the only good numbers. Assume for sake of contradiction that $q$ is a good number, where $q \\notin \\{-2, 0, 2\\}$. We now consider some particular choices of numbers on Gugu's napkin to arrive at a contradiction.\n\nAssume that the napkin contains the integers $1, 2, \\ldots, 10$. Then, the first line contains the integers $-9, -8, \\ldots, 9$. The second line then contains $q$ and $81q$, so the third line must also contain both of them. But the third line only contains integers, so $q$ must be an integer. Furthermore, the third line contains no number greater than $162 = 9^2 + 9^2 - 0^2 - 0^2$ or less than $-162$, so we must have $-162 \\leqslant 81q \\leqslant 162$. This shows that the only possibilities for $q$ are $\\pm 1$.\n\nNow assume that $q = \\pm 1$. Let the napkin contain $0, 1, 4, 8, 12, 16, 20, 24, 28, 32$. The first line contains $\\pm 1$ and $\\pm 4$, so the second line contains $\\pm 4$. However, for every number $a$ in the first line, $a \\not\\equiv 2 \\pmod{4}$, so we may conclude that $a^2 \\equiv 0, 1 \\pmod{8}$. Consequently, every number in the third line must be congruent to $-2, -1, 0, 1, 2 \\pmod{8}$; in particular, $\\pm 4$ cannot be in the third line, which is a contradiction.\nLet $q$ be a good number, as defined in the first solution, and define the polynomial $P\\left(x_1, \\ldots, x_{10}\\right)$ as\n$$\n\\prod_{i<j} (x_i - x_j) \\prod_{a_i \\in S} \\left(q(x_1 - x_2)(x_3 - x_4) - (a_1 - a_2)^2 - (a_3 - a_4)^2 + (a_5 - a_6)^2 + (a_7 - a_8)^2\\right),\n$$\nwhere $S = \\{x_1, \\ldots, x_{10}\\}$.\n\nWe claim that $P\\left(x_1, \\ldots, x_{10}\\right) = 0$ for every choice of real numbers $\\left(x_1, \\ldots, x_{10}\\right)$. If any two of the $x_i$ are equal, then $P\\left(x_1, \\ldots, x_{10}\\right) = 0$ trivially. If no two are equal, assume that Gugu has those ten numbers $x_1, \\ldots, x_{10}$ on his napkin. Then, the number $q(x_1 - x_2)(x_3 - x_4)$ is in the second line, so we must have some $a_1, \\ldots, a_8$ so that\n$$\nq(x_1 - x_2)(x_3 - x_4) - (a_1 - a_2)^2 - (a_3 - a_4)^2 + (a_5 - a_6)^2 + (a_7 - a_8)^2 = 0\n$$\nand hence $P\\left(x_1, \\ldots, x_{10}\\right) = 0$.\n\nSince every polynomial that evaluates to zero everywhere is the zero polynomial, and the product of two nonzero polynomials is necessarily nonzero, we may define $F$ such that\n$$\n\\begin{equation*}\nF\\left(x_1, \\ldots, x_{10}\\right) \\equiv q(x_1 - x_2)(x_3 - x_4) - (a_1 - a_2)^2 - (a_3 - a_4)^2 + (a_5 - a_6)^2 + (a_7 - a_8)^2 \\equiv 0 \\tag{1}\n\\end{equation*}\n$$\nfor some particular choice $a_i \\in S$.\n\nEach of the sets $\\{a_1, a_2\\}, \\{a_3, a_4\\}, \\{a_5, a_6\\}$, and $\\{a_7, a_8\\}$ is equal to at most one of the four sets $\\{x_1, x_3\\}, \\{x_2, x_3\\}, \\{x_1, x_4\\}$, and $\\{x_2, x_4\\}$. Thus, without loss of generality, we may assume that at most one of the sets $\\{a_1, a_2\\}, \\{a_3, a_4\\}, \\{a_5, a_6\\}, \\{a_7, a_8\\}$ is equal to $\\{x_1, x_3\\}$. Let $u_1, u_3, u_5, u_7$ be the indicator functions for this equality of sets: that is, $u_i = 1$ if and only if $\\{a_i, a_{i+1}\\} = \\{x_1, x_3\\}$. By assumption, at least three of the $u_i$ are equal to $0$.\n\nWe now compute the coefficient of $x_1 x_3$ in $F$. It is equal to $q + 2(u_1 + u_3 - u_5 - u_7) = 0$, and since at least three of the $u_i$ are zero, we must have that $q \\in \\{-2, 0, 2\\}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32035,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of real numbers $a_{1}, a_{2}, \\ldots$ satisfies the relation\n$$\na_{n} = -\\max_{i+j=n} (a_{i} + a_{j}) \\quad \\text{ for all } n > 2017\n$$\nProve that this sequence is bounded, i.e., there is a constant $M$ such that $|a_{n}| \\leqslant M$ for all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Set $D = 2017$. Denote\n$$\nM_{n} = \\max_{k < n} a_{k} \\quad \\text{ and } \\quad m_{n} = -\\min_{k < n} a_{k} = \\max_{k < n} (-a_{k}) .\n$$\nClearly, the sequences $(m_{n})$ and $(M_{n})$ are nondecreasing. We need to prove that both are bounded.\n\nConsider an arbitrary $n > D$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$.\n\ni. There exist indices $p$ and $q$ such that $a_{n} = - (a_{p} + a_{q})$ and $p + q = n$. Since $a_{p}, a_{q} \\leqslant M_{n}$, we have $a_{n} \\geqslant -2 M_{n}$.\n\nii. On the other hand, choose an index $k < n$ such that $a_{k} = M_{n}$. Then, we have\n$$\na_{n} = -\\max_{\\ell < n} (a_{n-\\ell} + a_{\\ell}) \\leqslant - (a_{n-k} + a_{k}) = -a_{n-k} - M_{n} \\leqslant m_{n} - M_{n} .\n$$\nSummarizing (i) and (ii), we get\n$$\n-2 M_{n} \\leqslant a_{n} \\leqslant m_{n} - M_{n},\n$$\nwhence\n$$\n\\begin{equation*}\nm_{n} \\leqslant m_{n+1} \\leqslant \\max \\{ m_{n}, 2 M_{n} \\} \\quad \\text{ and } \\quad M_{n} \\leqslant M_{n+1} \\leqslant \\max \\{ M_{n}, m_{n} - M_{n} \\} . \\tag{1}\n\\end{equation*}\n$$\n\nNow, say that an index $n > D$ is lucky if $m_{n} \\leqslant 2 M_{n}$. Two cases are possible.\n\nCase 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \\leqslant 2 M_{n}$ and $M_{n} \\leqslant M_{n+1} \\leqslant M_{n}$. Therefore, $M_{n+1} = M_{n}$ and $m_{n+1} \\leqslant 2 M_{n} = 2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1} = M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k > n$ are lucky (i.e., $m_{k} \\leqslant 2 M_{k}$ for all these indices), and $M_{k} = M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$.\n\nCase 2. Assume now that there is no lucky index, i.e., $2 M_{n} < m_{n}$ for all $n > D$. Then (1) shows that for all $n > D$ we have $m_{n} \\leqslant m_{n+1} \\leqslant m_{n}$, so $m_{n} = m_{D+1}$ for all $n > D$. Since $M_{n} < m_{n} / 2$ for all such indices, all of the $m_{n}$ and $M_{n}$ are bounded by $m_{D+1}$.\n\nThus, in both cases the sequences $(m_{n})$ and $(M_{n})$ are bounded, as desired.\nAs in the previous solution, let $D = 2017$. If the sequence is bounded above, say, by $Q$, then we have that $a_{n} \\geqslant \\min \\{ a_{1}, \\ldots, a_{D}, -2 Q \\}$ for all $n$, so the sequence is bounded. Assume for sake of contradiction that the sequence is not bounded above. Let $\\ell = \\min \\{ a_{1}, \\ldots, a_{D} \\}$, and $L = \\max \\{ a_{1}, \\ldots, a_{D} \\}$. Call an index $n$ good if the following criteria hold:\n$$\n\\begin{equation*}\na_{n} > a_{i} \\text{ for each } i < n, \\quad a_{n} > -2 \\ell, \\quad \\text{ and } \\quad n > D \\tag{2}\n\\end{equation*}\n$$\nWe first show that there must be some good index $n$. By assumption, we may take an index $N$ such that $a_{N} > \\max \\{ L, -2 \\ell \\}$. Choose $n$ minimally such that $a_{n} = \\max \\{ a_{1}, a_{2}, \\ldots, a_{N} \\}$. Now, the first condition in (2) is satisfied because of the minimality of $n$, and the second and third conditions are satisfied because $a_{n} \\geqslant a_{N} > L, -2 \\ell$, and $L \\geqslant a_{i}$ for every $i$ such that $1 \\leqslant i \\leqslant D$.\n\nLet $n$ be a good index. We derive a contradiction. We have that\n$$\n\\begin{equation*}\na_{n} + a_{u} + a_{v} \\leqslant 0, \\tag{3}\n\\end{equation*}\n$$\nwhenever $u + v = n$.\n\nWe define the index $u$ to maximize $a_{u}$ over $1 \\leqslant u \\leqslant n-1$, and let $v = n - u$. Then, we note that $a_{u} \\geqslant a_{v}$ by the maximality of $a_{u}$.\n\nAssume first that $v \\leqslant D$. Then, we have that\n$$\na_{N} + 2 \\ell \\leqslant 0,\n$$\nbecause $a_{u} \\geqslant a_{v} \\geqslant \\ell$. But this contradicts our assumption that $a_{n} > -2 \\ell$ in the second criteria of (2).\n\nNow assume that $v > D$. Then, there exist some indices $w_{1}, w_{2}$ summing up to $v$ such that\n$$\na_{v} + a_{w_{1}} + a_{w_{2}} = 0\n$$\nBut combining this with (3), we have\n$$\na_{n} + a_{u} \\leqslant a_{w_{1}} + a_{w_{2}} .\n$$\nBecause $a_{n} > a_{u}$, we have that $\\max \\{ a_{w_{1}}, a_{w_{2}} \\} > a_{u}$. But since each of the $w_{i}$ is less than $v$, this contradicts the maximality of $a_{u}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32036,
    "subject": "Mathematics (Multi-modal)",
    "question": "An integer $n \\geqslant 3$ is given. We call an $n$-tuple of real numbers $(x_{1}, x_{2}, \\ldots, x_{n})$ Shiny if for each permutation $y_{1}, y_{2}, \\ldots, y_{n}$ of these numbers we have\n$$\n\\sum_{i=1}^{n-1} y_{i} y_{i+1} = y_{1} y_{2} + y_{2} y_{3} + y_{3} y_{4} + \\cdots + y_{n-1} y_{n} \\geqslant -1.\n$$\nFind the largest constant $K = K(n)$ such that\n$$\n\\sum_{1 \\leqslant i < j \\leqslant n} x_{i} x_{j} \\geqslant K\n$$\nholds for every Shiny $n$-tuple $(x_{1}, x_{2}, \\ldots, x_{n})$.",
    "options": [],
    "answer": "-(n-1)/2",
    "solution": "Answer: $K = - (n-1)/2$.\n\nSolution 1. First of all, we show that we may not take a larger constant $K$. Let $t$ be a positive number, and take $x_{2} = x_{3} = \\cdots = t$ and $x_{1} = -1/(2t)$. Then, every product $x_{i} x_{j}$ ($i \\neq j$) is equal to either $t^{2}$ or $-1/2$. Hence, for every permutation $y_{i}$ of the $x_{i}$, we have\n$$\ny_{1} y_{2} + \\cdots + y_{n-1} y_{n} \\geqslant (n-3) t^{2} - 1 \\geqslant -1.\n$$\nThis justifies that the $n$-tuple $(x_{1}, \\ldots, x_{n})$ is Shiny. Now, we have\n$$\n\\sum_{i<j} x_{i} x_{j} = -\\frac{n-1}{2} + \\frac{(n-1)(n-2)}{2} t^{2}.\n$$\nThus, as $t$ approaches $0$ from above, $\\sum_{i<j} x_{i} x_{j}$ gets arbitrarily close to $-(n-1)/2$. This shows that we may not take $K$ any larger than $-(n-1)/2$. It remains to show that $\\sum_{i<j} x_{i} x_{j} \\geqslant -(n-1)/2$ for any Shiny choice of the $x_{i}$.\n\nFrom now onward, assume that $(x_{1}, \\ldots, x_{n})$ is a Shiny $n$-tuple. Let the $z_{i}$ ($1 \\leqslant i \\leqslant n$) be some permutation of the $x_{i}$ to be chosen later. The indices for $z_{i}$ will always be taken modulo $n$. We will first split up the sum $\\sum_{i<j} x_{i} x_{j} = \\sum_{i<j} z_{i} z_{j}$ into $\\lfloor (n-1)/2 \\rfloor$ expressions, each of the form $y_{1} y_{2} + \\cdots + y_{n-1} y_{n}$ for some permutation $y_{i}$ of the $z_{i}$, and some leftover terms. More specifically, write\n$$\n\\begin{equation*}\n\\sum_{i<j} z_{i} z_{j} = \\sum_{q=0}^{n-1} \\sum_{\\substack{i+j \\equiv q \\ (\\bmod n) \\\\ i \\neq j \\ (\\bmod n)}} z_{i} z_{j} = \\sum_{p=1}^{\\left\\lfloor \\frac{n-1}{2} \\right\\rfloor} \\sum_{\\substack{i+j \\equiv 2p-1, 2p \\ (\\bmod n) \\\\ i \\neq j \\ (\\bmod n)}} z_{i} z_{j} + L, \\tag{1}\n\\end{equation*}\n$$\nwhere $L = z_{1} z_{-1} + z_{2} z_{-2} + \\cdots + z_{(n-1)/2} z_{-(n-1)/2}$ if $n$ is odd, and $L = z_{1} z_{-1} + z_{1} z_{-2} + z_{2} z_{-2} + \\cdots + z_{(n-2)/2} z_{-n/2}$ if $n$ is even. We note that for each $p = 1, 2, \\ldots, \\lfloor (n-1)/2 \\rfloor$, there is some permutation $y_{i}$ of the $z_{i}$ such that\n$$\n\\sum_{\\substack{i+j \\equiv 2p-1, 2p \\ (\\bmod n) \\\\ i \\neq j \\ (\\bmod n)}} z_{i} z_{j} = \\sum_{k=1}^{n-1} y_{k} y_{k+1},\n$$\nbecause we may choose $y_{2i-1} = z_{i+p-1}$ for $1 \\leqslant i \\leqslant (n+1)/2$ and $y_{2i} = z_{p-i}$ for $1 \\leqslant i \\leqslant n/2$.\n\nWe show (1) graphically for $n=6,7$ in the diagrams below. The edges of the graphs each represent a product $z_{i} z_{j}$, and the dashed and dotted series of lines represents the sum of the edges, which is of the form $y_{1} y_{2} + \\cdots + y_{n-1} y_{n}$ for some permutation $y_{i}$ of the $z_{i}$ precisely when the series of lines is a Hamiltonian path. The filled edges represent the summands of $L$.\n\n![](attached_image_1.png)\n\n![](attached_image_2.png)\n\nNow, because the $z_{i}$ are Shiny, we have that (1) yields the following bound:\n$$\n\\sum_{i<j} z_{i} z_{j} \\geqslant -\\left\\lfloor \\frac{n-1}{2} \\right\\rfloor + L.\n$$\nIt remains to show that, for each $n$, there exists some permutation $z_{i}$ of the $x_{i}$ such that $L \\geqslant 0$ when $n$ is odd, and $L \\geqslant -1/2$ when $n$ is even. We now split into cases based on the parity of $n$ and provide constructions of the permutations $z_{i}$.\n\nSince we have not made any assumptions yet about the $x_{i}$, we may now assume without loss of generality that\n$$\n\\begin{equation*}\nx_{1} \\leqslant x_{2} \\leqslant \\cdots \\leqslant x_{k} \\leqslant 0 \\leqslant x_{k+1} \\leqslant \\cdots \\leqslant x_{n}. \\tag{2}\n\\end{equation*}\n$$\n\n## Case 1: $n$ is odd.\n\nWithout loss of generality, assume that $k$ (from (2)) is even, because we may negate all the $x_{i}$ if $k$ is odd. We then have $x_{1} x_{2}, x_{3} x_{4}, \\ldots, x_{n-2} x_{n-1} \\geqslant 0$ because the factors are of the same sign. Let $L = x_{1} x_{2} + x_{3} x_{4} + \\cdots + x_{n-2} x_{n-1} \\geqslant 0$. We choose our $z_{i}$ so that this definition of $L$ agrees with the sum of the leftover terms in (1). Relabel the $x_{i}$ as $z_{i}$ such that\n$$\n\\{z_{1}, z_{n-1}\\}, \\{z_{2}, z_{n-2}\\}, \\ldots, \\{z_{(n-1)/2}, z_{(n+1)/2}\\}\n$$\nare some permutation of\n$$\n\\{x_{1}, x_{2}\\}, \\{x_{3}, x_{4}\\}, \\ldots, \\{x_{n-2}, x_{n-1}\\},\n$$\nand $z_{n} = x_{n}$. Then, we have $L = z_{1} z_{n-1} + \\cdots + z_{(n-1)/2} z_{(n+1)/2}$, as desired.\n\nCase 2: $n$ is even.\n\nLet $L = x_{1} x_{2} + x_{2} x_{3} + \\cdots + x_{n-1} x_{n}$. Assume without loss of generality $k \\neq 1$. Now, we have\n$$\n\\begin{gathered}\n2L = (x_{1} x_{2} + \\cdots + x_{n-1} x_{n}) + (x_{1} x_{2} + \\cdots + x_{n-1} x_{n}) \\geqslant (x_{2} x_{3} + \\cdots + x_{n-1} x_{n}) + x_{k} x_{k+1} \\\\\n\\geqslant x_{2} x_{3} + \\cdots + x_{n-1} x_{n} + x_{n} x_{1} \\geqslant -1\n\\end{gathered}\n$$\nwhere the first inequality holds because the only negative term in $L$ is $x_{k} x_{k+1}$, the second inequality holds because $x_{1} \\leqslant x_{k} \\leqslant 0 \\leqslant x_{k+1} \\leqslant x_{n}$, and the third inequality holds because the $x_{i}$ are assumed to be Shiny. We thus have that $L \\geqslant -1/2$. We now choose a suitable $z_{i}$ such that the definition of $L$ matches the leftover terms in (1).\n\nRelabel the $x_{i}$ with $z_{i}$ in the following manner: $x_{2i-1} = z_{-i}$, $x_{2i} = z_{i}$ (again taking indices modulo $n$). We have that\n$$\nL = \\sum_{\\substack{i+j \\equiv 0, -1 \\ (\\bmod n) \\\\ i \\neq j \\ (\\bmod n)}} z_{i} z_{j}\n$$\nas desired.\nWe present another proof that $\\sum_{i<j} x_{i} x_{j} \\geqslant - (n-1)/2$ for any Shiny $n$-tuple $(x_{1}, \\ldots, x_{n})$. Assume an ordering of the $x_{i}$ as in (2), and let $\\ell = n - k$. Assume without loss of generality that $k \\geqslant \\ell$. Also assume $k \\neq n$, (as otherwise, all of the $x_{i}$ are nonpositive, and so the inequality is trivial). Define the sets of indices $S = \\{1, 2, \\ldots, k\\}$ and $T = \\{k+1, \\ldots, n\\}$. Define the following sums:\n$$\nK = \\sum_{\\substack{i<j \\\\ i, j \\in S}} x_{i} x_{j}, \\quad M = \\sum_{\\substack{i \\in S \\\\ j \\in T}} x_{i} x_{j}, \\quad \\text{ and } \\quad L = \\sum_{\\substack{i<j \\\\ i, j \\in T}} x_{i} x_{j}\n$$\nBy definition, $K, L \\geqslant 0$ and $M \\leqslant 0$. We aim to show that $K + L + M \\geqslant - (n-1)/2$.\n\nWe split into cases based on whether $k = \\ell$ or $k > \\ell$.\n\n## Case 1: $k > \\ell$.\n\nConsider all permutations $\\phi: \\{1, 2, \\ldots, n\\} \\to \\{1, 2, \\ldots, n\\}$ such that $\\phi^{-1}(T) = \\{2, 4, \\ldots, 2\\ell\\}$. Note that there are $k! \\ell!$ such permutations $\\phi$. Define\n$$\nf(\\phi) = \\sum_{i=1}^{n-1} x_{\\phi(i)} x_{\\phi(i+1)}\n$$\nWe know that $f(\\phi) \\geqslant -1$ for every permutation $\\phi$ with the above property. Averaging $f(\\phi)$ over all $\\phi$ gives\n$$\n-1 \\leqslant \\frac{1}{k! \\ell!} \\sum_{\\phi} f(\\phi) = \\frac{2\\ell}{k\\ell} M + \\frac{2(k-\\ell-1)}{k(k-1)} K,\n$$\nwhere the equality holds because there are $k\\ell$ products in $M$, of which $2\\ell$ are selected for each $\\phi$, and there are $k(k-1)/2$ products in $K$, of which $k-\\ell-1$ are selected for each $\\phi$. We now have\n$$\nK + L + M \\geqslant K + L + \\left(-\\frac{k}{2} - \\frac{k-\\ell-1}{k-1} K\\right) = -\\frac{k}{2} + \\frac{\\ell}{k-1} K + L.\n$$\nSince $k \\leqslant n-1$ and $K, L \\geqslant 0$, we get the desired inequality.\n\nCase 2: $k = \\ell = n/2$.\n\nWe do a similar approach, considering all $\\phi: \\{1, 2, \\ldots, n\\} \\to \\{1, 2, \\ldots, n\\}$ such that $\\phi^{-1}(T) = \\{2, 4, \\ldots, 2\\ell\\}$, and defining $f$ the same way. Analogously to Case 1, we have\n$$\n-1 \\leqslant \\frac{1}{k! \\ell!} \\sum_{\\phi} f(\\phi) = \\frac{2\\ell-1}{k\\ell} M\n$$\nbecause there are $k\\ell$ products in $M$, of which $2\\ell-1$ are selected for each $\\phi$. Now, we have that\n$$\nK + L + M \\geqslant M \\geqslant -\\frac{n^{2}}{4(n-1)} \\geqslant -\\frac{n-1}{2},\n$$\nwhere the last inequality holds because $n \\geqslant 4$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32037,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{0}, a_{1}, a_{2}, \\ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \\ldots$ be a sequence of positive integers such that $a_{0}=0$, $a_{1}=1$, and\n$$\na_{n+1}=\\left\\{\\begin{array}{ll}\na_{n} b_{n}+a_{n-1}, & \\text{ if } b_{n-1}=1 \\\\\na_{n} b_{n}-a_{n-1}, & \\text{ if } b_{n-1}>1\n\\end{array} \\quad \\text{ for } n=1,2, \\ldots\\right.\n$$\nProve that at least one of the two numbers $a_{2017}$ and $a_{2018}$ must be greater than or equal to $2017$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The value of $b_{0}$ is irrelevant since $a_{0}=0$, so we may assume that $b_{0}=1$.\n\nLemma. We have $a_{n} \\geqslant 1$ for all $n \\geqslant 1$.\n\nProof. Let us suppose otherwise in order to obtain a contradiction. Let\n$$\n\\begin{equation*}\nn \\geqslant 1 \\text{ be the smallest integer with } a_{n} \\leqslant 0 \\text{. } \\tag{1}\n\\end{equation*}\n$$\nNote that $n \\geqslant 2$. It follows that $a_{n-1} \\geqslant 1$ and $a_{n-2} \\geqslant 0$. Thus we cannot have $a_{n}= a_{n-1} b_{n-1}+a_{n-2}$, so we must have $a_{n}=a_{n-1} b_{n-1}-a_{n-2}$. Since $a_{n} \\leqslant 0$, we have $a_{n-1} \\leqslant a_{n-2}$. Thus we have $a_{n-2} \\geqslant a_{n-1} \\geqslant a_{n}$.\n\nLet\n$$\n\\begin{equation*}\nr \\text{ be the smallest index with } a_{r} \\geqslant a_{r+1} \\geqslant a_{r+2} \\text{. } \\tag{2}\n\\end{equation*}\n$$\nThen $r \\leqslant n-2$ by the above, but also $r \\geqslant 2$ : if $b_{1}=1$, then $a_{2}=a_{1}=1$ and $a_{3}=a_{2} b_{2}+a_{1}>a_{2}$; if $b_{1}>1$, then $a_{2}=b_{1}>1=a_{1}$.\n\nBy the minimal choice (2) of $r$, it follows that $a_{r-1}<a_{r}$. And since $2 \\leqslant r \\leqslant n-2$, by the minimal choice (1) of $n$ we have $a_{r-1}, a_{r}, a_{r+1}>0$. In order to have $a_{r+1} \\geqslant a_{r+2}$, we must have $a_{r+2}=a_{r+1} b_{r+1}-a_{r}$ so that $b_{r} \\geqslant 2$. Putting everything together, we conclude that\n$$\na_{r+1}=a_{r} b_{r} \\pm a_{r-1} \\geqslant 2 a_{r}-a_{r-1}=a_{r}+\\left(a_{r}-a_{r-1}\\right)>a_{r},\n$$\nwhich contradicts (2). $\\square$\n\nTo complete the problem, we prove that $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$ by induction. The cases $n=0,1$ are given. Assume it is true for all non-negative integers strictly less than $n$, where $n \\geqslant 2$. There are two cases:\n\nCase 1: $b_{n-1}=1$.\n\nThen $a_{n+1}=a_{n} b_{n}+a_{n-1}$. By the inductive assumption one of $a_{n-1}, a_{n}$ is at least $n-1$ and the other, by the lemma, is at least 1. Hence\n$$\na_{n+1}=a_{n} b_{n}+a_{n-1} \\geqslant a_{n}+a_{n-1} \\geqslant(n-1)+1=n .\n$$\nThus $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$, as desired.\n\nCase 2: $b_{n-1}>1$.\n\nSince we defined $b_{0}=1$ there is an index $r$ with $1 \\leqslant r \\leqslant n-1$ such that\n$$\nb_{n-1}, b_{n-2}, \\ldots, b_{r} \\geqslant 2 \\quad \\text{ and } \\quad b_{r-1}=1 .\n$$\nWe have $a_{r+1}=a_{r} b_{r}+a_{r-1} \\geqslant 2 a_{r}+a_{r-1}$. Thus $a_{r+1}-a_{r} \\geqslant a_{r}+a_{r-1}$.\n\nNow we claim that $a_{r}+a_{r-1} \\geqslant r$. Indeed, this holds by inspection for $r=1$; for $r \\geqslant 2$, one of $a_{r}, a_{r-1}$ is at least $r-1$ by the inductive assumption, while the other, by the lemma, is at least 1. Hence $a_{r}+a_{r-1} \\geqslant r$, as claimed, and therefore $a_{r+1}-a_{r} \\geqslant r$ by the last inequality in the previous paragraph.\n\nSince $r \\geqslant 1$ and, by the lemma, $a_{r} \\geqslant 1$, from $a_{r+1}-a_{r} \\geqslant r$ we get the following two inequalities:\n$$\na_{r+1} \\geqslant r+1 \\quad \\text{ and } \\quad a_{r+1}>a_{r} .\n$$\nNow observe that\n$$\na_{m}>a_{m-1} \\Longrightarrow a_{m+1}>a_{m} \\text{ for } m=r+1, r+2, \\ldots, n-1 \\text{, }\n$$\nsince $a_{m+1}=a_{m} b_{m}-a_{m-1} \\geqslant 2 a_{m}-a_{m-1}=a_{m}+\\left(a_{m}-a_{m-1}\\right)>a_{m}$. Thus\n$$\na_{n}>a_{n-1}>\\cdots>a_{r+1} \\geqslant r+1 \\Longrightarrow a_{n} \\geqslant n\n$$\nSo $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$, as desired.\nWe say that an index $n>1$ is bad if $b_{n-1}=1$ and $b_{n-2}>1$; otherwise $n$ is good. The value of $b_{0}$ is irrelevant to the definition of $\\left(a_{n}\\right)$ since $a_{0}=0$; so we assume that $b_{0}>1$.\n\nLemma 1. (a) $a_{n} \\geqslant 1$ for all $n>0$.\n(b) If $n>1$ is good, then $a_{n}>a_{n-1}$.\n\nProof. Induction on $n$. In the base cases $n=1,2$ we have $a_{1}=1 \\geqslant 1, a_{2}=b_{1} a_{1} \\geqslant 1$, and finally $a_{2}>a_{1}$ if 2 is good, since in this case $b_{1}>1$.\n\nNow we assume that the lemma statement is proved for $n=1,2, \\ldots, k$ with $k \\geqslant 2$, and prove it for $n=k+1$. Recall that $a_{k}$ and $a_{k-1}$ are positive by the induction hypothesis.\n\nCase 1: $k$ is bad.\nWe have $b_{k-1}=1$, so $a_{k+1}=b_{k} a_{k}+a_{k-1} \\geqslant a_{k}+a_{k-1}>a_{k} \\geqslant 1$, as required.\n\nCase 2: $k$ is good.\nWe already have $a_{k}>a_{k-1} \\geqslant 1$ by the induction hypothesis. We consider three easy subcases.\n\nSubcase 2.1: $b_{k}>1$.\nThen $a_{k+1} \\geqslant b_{k} a_{k}-a_{k-1} \\geqslant a_{k}+\\left(a_{k}-a_{k-1}\\right)>a_{k} \\geqslant 1$.\n\nSubcase 2.2: $b_{k}=b_{k-1}=1$.\nThen $a_{k+1}=a_{k}+a_{k-1}>a_{k} \\geqslant 1$.\n\nSubcase 2.3: $b_{k}=1$ but $b_{k-1}>1$.\nThen $k+1$ is bad, and we need to prove only (a), which is trivial: $a_{k+1}=a_{k}-a_{k-1} \\geqslant 1$.\n\nSo, in all three subcases we have verified the required relations.\n\nLemma 2. Assume that $n>1$ is bad. Then there exists a $j \\in\\{1,2,3\\}$ such that $a_{n+j} \\geqslant a_{n-1}+j+1$, and $a_{n+i} \\geqslant a_{n-1}+i$ for all $1 \\leqslant i<j$.\n\nProof. Recall that $b_{n-1}=1$. Set\n$$\nm=\\inf \\left\\{i>0: b_{n+i-1}>1\\right\\}\n$$\n(possibly $m=+\\infty$ ). We claim that $j=\\min \\{m, 3\\}$ works. Again, we distinguish several cases, according to the value of $m$; in each of them we use Lemma 1 without reference.\n\nCase 1: $m=1$, so $b_{n}>1$.\nThen $a_{n+1} \\geqslant 2 a_{n}+a_{n-1} \\geqslant a_{n-1}+2$, as required.\n\nCase 2: $m=2$, so $b_{n}=1$ and $b_{n+1}>1$.\nThen we successively get\n$$\n\\begin{gathered}\na_{n+1}=a_{n}+a_{n-1} \\geqslant a_{n-1}+1, \\\\\na_{n+2} \\geqslant 2 a_{n+1}+a_{n}=2\\left(a_{n-1}+1\\right)+a_{n}=a_{n-1}+\\left(a_{n-1}+a_{n}+2\\right) \\geqslant a_{n-1}+4,\n\\end{gathered}\n$$\nwhich is even better than we need.\n\nCase 3: $m>2$, so $b_{n}=b_{n+1}=1$.\nThen we successively get\n$$\n\\begin{gathered}\na_{n+1}=a_{n}+a_{n-1} \\geqslant a_{n-1}+1, \\quad a_{n+2}=a_{n+1}+a_{n} \\geqslant a_{n-1}+1+a_{n} \\geqslant a_{n-1}+2, \\\\\na_{n+3} \\geqslant a_{n+2}+a_{n+1} \\geqslant\\left(a_{n-1}+1\\right)+\\left(a_{n-1}+2\\right) \\geqslant a_{n-1}+4,\n\\end{gathered}\n$$\nas required. $\\square$\n\nLemmas 1(b) and 2 provide enough information to prove that $\\max \\left\\{a_{n}, a_{n+1}\\right\\} \\geqslant n$ for all $n$ and, moreover, that $a_{n} \\geqslant n$ often enough. Indeed, assume that we have found some $n$ with $a_{n-1} \\geqslant n-1$. If $n$ is good, then by Lemma 1(b) we have $a_{n} \\geqslant n$ as well. If $n$ is bad, then Lemma 2 yields $\\max \\left\\{a_{n+i}, a_{n+i+1}\\right\\} \\geqslant a_{n-1}+i+1 \\geqslant n+i$ for all $0 \\leqslant i<j$ and $a_{n+j} \\geqslant a_{n-1}+j+1 \\geqslant n+j$; so $n+j$ is the next index to start with.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32038,
    "subject": "Mathematics (Multi-modal)",
    "question": "Assume that a function $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ satisfies the following condition:\nFor every $x, y \\in \\mathbb{R}$ such that $(f(x)+y)(f(y)+x)>0$, we have $f(x)+y=f(y)+x$.\nProve that $f(x)+y \\leqslant f(y)+x$ whenever $x>y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define $g(x)=x-f(x)$. The condition on $f$ then rewrites as follows:\nFor every $x, y \\in \\mathbb{R}$ such that $((x+y)-g(x))((x+y)-g(y))>0$, we have $g(x)=g(y)$.\nThis condition may in turn be rewritten in the following form:\nIf $g(x) \\neq g(y)$, then the number $x+y$ lies (non-strictly) between $g(x)$ and $g(y)$.\nNotice here that the function $g_{1}(x)=-g(-x)$ also satisfies $(*)$, since\n$$\n\\begin{gathered}\ng_{1}(x) \\neq g_{1}(y) \\Longrightarrow \\quad g(-x) \\neq g(-y) \\quad \\Longrightarrow \\quad-(x+y) \\text{ lies between } g(-x) \\text{ and } g(-y) \\\\\n\\Longrightarrow \\quad x+y \\text{ lies between } g_{1}(x) \\text{ and } g_{1}(y)\n\\end{gathered}\n$$\nOn the other hand, the relation we need to prove reads now as\n$$\n\\begin{equation*}\ng(x) \\leqslant g(y) \\quad \\text{ whenever } x<y \\tag{1}\n\\end{equation*}\n$$\nAgain, this condition is equivalent to the same one with $g$ replaced by $g_{1}$.\nIf $g(x)=2 x$ for all $x \\in \\mathbb{R}$, then (*) is obvious; so in what follows we consider the other case. We split the solution into a sequence of lemmas, strengthening one another. We always consider some value of $x$ with $g(x) \\neq 2 x$ and denote $X=g(x)$.\n\nLemma 1. Assume that $X<2 x$. Then on the interval $(X-x ; x]$ the function $g$ attains at most two values - namely, $X$ and, possibly, some $Y>X$. Similarly, if $X>2 x$, then $g$ attains at most two values on $[x ; X-x)$ - namely, $X$ and, possibly, some $Y<X$.\n\nProof. We start with the first claim of the lemma. Notice that $X-x<x$, so the considered interval is nonempty.\nTake any $a \\in(X-x ; x)$ with $g(a) \\neq X$ (if it exists). If $g(a)<X$, then $(*)$ yields $g(a) \\leqslant a+x \\leqslant g(x)=X$, so $a \\leqslant X-x$ which is impossible. Thus, $g(a)>X$ and hence by (*) we get $X \\leqslant a+x \\leqslant g(a)$.\nNow, for any $b \\in(X-x ; x)$ with $g(b) \\neq X$ we similarly get $b+x \\leqslant g(b)$. Therefore, the number $a+b$ (which is smaller than each of $a+x$ and $b+x$ ) cannot lie between $g(a)$ and $g(b)$, which by (*) implies that $g(a)=g(b)$. Hence $g$ may attain only two values on ( $X-x ; x$ ], namely $X$ and $g(a)>X$.\nTo prove the second claim, notice that $g_{1}(-x)=-X<2 \\cdot(-x)$, so $g_{1}$ attains at most two values on $(-X+x,-x]$, i.e., $-X$ and, possibly, some $-Y>-X$. Passing back to $g$, we get what we need.\n\nLemma 2. If $X<2 x$, then $g$ is constant on $(X-x ; x)$. Similarly, if $X>2 x$, then $g$ is constant on $(x ; X-x)$.\n\nProof. Again, it suffices to prove the first claim only. Assume, for the sake of contradiction, that there exist $a, b \\in(X-x ; x)$ with $g(a) \\neq g(b)$; by Lemma 1 , we may assume that $g(a)=X$ and $Y=g(b)>X$.\nNotice that $\\min \\{X-a, X-b\\}>X-x$, so there exists a $u \\in(X-x ; x)$ such that $u<\\min \\{X-a, X-b\\}$. By Lemma 1, we have either $g(u)=X$ or $g(u)=Y$. In the former case, by (*) we have $X \\leqslant u+b \\leqslant Y$ which contradicts $u<X-b$. In the second case, by (*) we have $X \\leqslant u+a \\leqslant Y$ which contradicts $u<X-a$. Thus the lemma is proved.\n\nLemma 3. If $X<2 x$, then $g(a)=X$ for all $a \\in(X-x ; x)$. Similarly, if $X>2 x$, then $g(a)=X$ for all $a \\in(x ; X-x)$.\n\nProof. Again, we only prove the first claim.\nBy Lemmas 1 and 2, this claim may be violated only if $g$ takes on a constant value $Y>X$ on ( $X-x, x$ ). Choose any $a, b \\in(X-x ; x)$ with $a<b$. By (*), we have\n$$\n\\begin{equation*}\nY \\geqslant b+x \\geqslant X \\tag{2}\n\\end{equation*}\n$$\nIn particular, we have $Y \\geqslant b+x>2 a$. Applying Lemma 2 to $a$ in place of $x$, we obtain that $g$ is constant on $(a, Y-a)$. By (2) again, we have $x \\leqslant Y-b<Y-a$; so $x, b \\in(a ; Y-a)$. But $X=g(x) \\neq g(b)=Y$, which is a contradiction.\n\nNow we are able to finish the solution. Assume that $g(x)>g(y)$ for some $x<y$. Denote $X=g(x)$ and $Y=g(y)$; by ( $*$ ), we have $X \\geqslant x+y \\geqslant Y$, so $Y-y \\leqslant x<y \\leqslant X-x$, and hence $(Y-y ; y) \\cap(x ; X-x)=(x, y) \\neq \\varnothing$. On the other hand, since $Y-y<y$ and $x<X-x$, Lemma 3 shows that $g$ should attain a constant value $X$ on ( $x ; X-x$ ) and a constant value $Y \\neq X$ on $(Y-y ; y)$. Since these intervals overlap, we get the final contradiction.\nAs in the previous solution, we pass to the function $g$ satisfying (*) and notice that we need to prove the condition (1). We will also make use of the function $g_{1}$.\nIf $g$ is constant, then (1) is clearly satisfied. So, in the sequel we assume that $g$ takes on at least two different values. Now we collect some information about the function $g$.\n\nClaim 1. For any $c \\in \\mathbb{R}$, all the solutions of $g(x)=c$ are bounded.\n\nProof. Fix any $y \\in \\mathbb{R}$ with $g(y) \\neq c$. Assume first that $g(y)>c$. Now, for any $x$ with $g(x)=c$, by (*) we have $c \\leqslant x+y \\leqslant g(y)$, or $c-y \\leqslant x \\leqslant g(y)-y$. Since $c$ and $y$ are constant, we get what we need.\nIf $g(y)<c$, we may switch to the function $g_{1}$ for which we have $g_{1}(-y)>-c$. By the above arguments, we obtain that all the solutions of $g_{1}(-x)=-c$ are bounded, which is equivalent to what we need.\nAs an immediate consequence, the function $g$ takes on infinitely many values, which shows that the next claim is indeed widely applicable.\n\nClaim 2. If $g(x)<g(y)<g(z)$, then $x<z$.\n\nProof. By (*), we have $g(x) \\leqslant x+y \\leqslant g(y) \\leqslant z+y \\leqslant g(z)$, so $x+y \\leqslant z+y$, as required.\n\nClaim 3. Assume that $g(x)>g(y)$ for some $x<y$. Then $g(a) \\in\\{g(x), g(y)\\}$ for all $a \\in[x ; y]$.\n\nProof. If $g(y)<g(a)<g(x)$, then the triple ( $y, a, x$ ) violates Claim 2. If $g(a)<g(y)<g(x)$, then the triple $(a, y, x)$ violates Claim 2. If $g(y)<g(x)<g(a)$, then the triple $(y, x, a)$ violates Claim 2. The only possible cases left are $g(a) \\in\\{g(x), g(y)\\}$.\n\nIn view of Claim 3, we say that an interval $I$ (which may be open, closed, or semi-open) is a Dirichlet interval* if the function $g$ takes on just two values on $I$.\nAssume now, for the sake of contradiction, that (1) is violated by some $x<y$. By Claim 3, $[x ; y]$ is a Dirichlet interval. Set\n$r=\\inf \\{a:(a ; y]$ is a Dirichlet interval $\\}$ and $s=\\sup \\{b:[x ; b)$ is a Dirichlet interval $\\}$.\nClearly, $r \\leqslant x<y \\leqslant s$. By Claim 1, $r$ and $s$ are finite. Denote $X=g(x), Y=g(y)$, and $\\Delta=(y-x) / 2$.\nSuppose first that there exists a $t \\in(r ; r+\\Delta)$ with $f(t)=Y$. By the definition of $r$, the interval ( $r-\\Delta ; y$ ] is not Dirichlet, so there exists an $r^{\\prime} \\in(r-\\Delta ; r]$ such that $g\\left(r^{\\prime}\\right) \\notin\\{X, Y\\}$.\nThe function $g$ attains at least three distinct values on $\\left[r^{\\prime} ; y\\right]$, namely $g\\left(r^{\\prime}\\right), g(x)$, and $g(y)$. Claim 3 now yields $g\\left(r^{\\prime}\\right) \\leqslant g(y)$; the equality is impossible by the choice of $r^{\\prime}$, so in fact $g\\left(r^{\\prime}\\right)<Y$. Applying (*) to the pairs ( $r^{\\prime}, y$ ) and ( $t, x$ ) we obtain $r^{\\prime}+y \\leqslant Y \\leqslant t+x$, whence $r-\\Delta+y<r^{\\prime}+y \\leqslant t+x<r+\\Delta+x$, or $y-x<2 \\Delta$. This is a contradiction.\n\nThus, $g(t)=X$ for all $t \\in(r ; r+\\Delta)$. Applying the same argument to $g_{1}$, we get $g(t)=Y$ for all $t \\in(s-\\Delta ; s)$.\nFinally, choose some $s_{1}, s_{2} \\in(s-\\Delta ; s)$ with $s_{1}<s_{2}$ and denote $\\delta=\\left(s_{2}-s_{1}\\right) / 2$. As before, we choose $r^{\\prime} \\in(r-\\delta ; r)$ with $g\\left(r^{\\prime}\\right) \\notin\\{X, Y\\}$ and obtain $g\\left(r^{\\prime}\\right)<Y$. Choose any $t \\in(r ; r+\\delta)$; by the above arguments, we have $g(t)=X$ and $g\\left(s_{1}\\right)=g\\left(s_{2}\\right)=Y$. As before, we apply (*) to the pairs $\\left(r^{\\prime}, s_{2}\\right)$ and $\\left(t, s_{1}\\right)$ obtaining $r-\\delta+s_{2}<r^{\\prime}+s_{2} \\leqslant Y \\leqslant t+s_{1}<r+\\delta+s_{1}$, or $s_{2}-s_{1}<2 \\delta$. This is a final contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32039,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Define a chameleon to be any sequence of $3n$ letters, with exactly $n$ occurrences of each of the letters $a$, $b$, and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$, there exists a chameleon $Y$ such that $X$ cannot be changed to $Y$ using fewer than $3n^{2}/2$ swaps.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "To start, notice that the swap of two identical letters does not change a chameleon, so we may assume there are no such swaps.\nFor any two chameleons $X$ and $Y$, define their distance $d(X, Y)$ to be the minimal number of swaps needed to transform $X$ into $Y$ (or vice versa). Clearly, $d(X, Y)+d(Y, Z) \\geqslant d(X, Z)$ for any three chameleons $X$, $Y$, and $Z$.\n\nLemma. Consider two chameleons\n$$\nP=\\underbrace{a a \\ldots a}_{n} \\underbrace{b b \\ldots b}_{n} \\underbrace{c c \\ldots c}_{n} \\text{ and } Q=\\underbrace{c c \\ldots c}_{n} \\underbrace{b b \\ldots b}_{n} \\underbrace{a a \\ldots a}_{n} \\text{.}\n$$\nThen $d(P, Q) \\geqslant 3n^{2}$.\n\nProof. For any chameleon $X$ and any pair of distinct letters $u, v \\in \\{a, b, c\\}$, we define $f_{u, v}(X)$ to be the number of pairs of positions in $X$ such that the left one is occupied by $u$, and the right one is occupied by $v$. Define $f(X)=f_{a, b}(X)+f_{a, c}(X)+f_{b, c}(X)$. Notice that $f_{a, b}(P)=f_{a, c}(P)=f_{b, c}(P)=n^{2}$ and $f_{a, b}(Q)=f_{a, c}(Q)=f_{b, c}(Q)=0$, so $f(P)=3n^{2}$ and $f(Q)=0$.\n\nNow consider some swap changing a chameleon $X$ to $X'$, say, the letters $a$ and $b$ are swapped. Then $f_{a, b}(X)$ and $f_{a, b}(X')$ differ by exactly 1, while $f_{a, c}(X)=f_{a, c}(X')$ and $f_{b, c}(X)=f_{b, c}(X')$. This yields $|f(X)-f(X')|=1$, i.e., on any swap the value of $f$ changes by 1. Hence $d(X, Y) \\geqslant |f(X)-f(Y)|$ for any two chameleons $X$ and $Y$. In particular, $d(P, Q) \\geqslant |f(P)-f(Q)|=3n^{2}$, as desired.\n\nBack to the problem, take any chameleon $X$ and notice that $d(X, P)+d(X, Q) \\geqslant d(P, Q) \\geqslant 3n^{2}$ by the lemma. Consequently, $\\max \\{d(X, P), d(X, Q)\\} \\geqslant \\frac{3n^{2}}{2}$, which establishes the problem statement.\nWe use the notion of distance from Solution 1, but provide a different lower bound for it.\nIn any chameleon $X$, we enumerate the positions in it from left to right by $1,2, \\ldots, 3n$. Define $s_{c}(X)$ as the sum of positions occupied by $c$. The value of $s_{c}$ changes by at most 1 on each swap, but this fact alone does not suffice to solve the problem; so we need an improvement.\n\nFor every chameleon $X$, denote by $X_{\\bar{c}}$ the sequence obtained from $X$ by removing all $n$ letters $c$. Enumerate the positions in $X_{\\bar{c}}$ from left to right by $1,2, \\ldots, 2n$, and define $s_{\\bar{c}, b}(X)$ as the sum of positions in $X_{\\bar{c}}$ occupied by $b$. (In other words, here we consider the positions of the $b$'s relatively to the $a$'s only.) Finally, denote\n$$\nd'(X, Y):=|s_{c}(X)-s_{c}(Y)|+|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}(Y)|.\n$$\nNow consider any swap changing a chameleon $X$ to $X'$. If no letter $c$ is involved into this swap, then $s_{c}(X)=s_{c}(X')$; on the other hand, exactly one letter $b$ changes its position in $X_{\\bar{c}}$, so $|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}(X')|=1$. If a letter $c$ is involved into a swap, then $X_{\\bar{c}}=X_{\\bar{c}}'$, so $s_{\\bar{c}, b}(X)=s_{\\bar{c}, b}(X')$ and $|s_{c}(X)-s_{c}(X')|=1$. Thus, in all cases we have $d'(X, X')=1$.\n\nAs in the previous solution, this means that $d(X, Y) \\geqslant d'(X, Y)$ for any two chameleons $X$ and $Y$. Now, for any chameleon $X$ we will indicate a chameleon $Y$ with $d'(X, Y) \\geqslant 3n^{2}/2$, thus finishing the solution.\n\nThe function $s_{c}$ attains all integer values from $1+\\cdots+n=\\frac{n(n+1)}{2}$ to $(2n+1)+\\cdots+3n=2n^{2}+\\frac{n(n+1)}{2}$. If $s_{c}(X) \\leqslant n^{2}+\\frac{n(n+1)}{2}$, then we put the letter $c$ into the last $n$ positions in $Y$; otherwise we put the letter $c$ into the first $n$ positions in $Y$. In either case we already have $|s_{c}(X)-s_{c}(Y)| \\geqslant n^{2}$.\n\nSimilarly, $s_{\\bar{c}, b}$ ranges from $\\frac{n(n+1)}{2}$ to $n^{2}+\\frac{n(n+1)}{2}$. So, if $s_{\\bar{c}, b}(X) \\leqslant \\frac{n^{2}}{2}+\\frac{n(n+1)}{2}$, then we put the letter $b$ into the last $n$ positions in $Y$ which are still free; otherwise, we put the letter $b$ into the first $n$ such positions. The remaining positions are occupied by $a$. In any case, we have $|s_{\\bar{c}, b}(X)-s_{\\bar{c}, b}(Y)| \\geqslant \\frac{n^{2}}{2}$, thus $d'(X, Y) \\geqslant n^{2}+\\frac{n^{2}}{2}=\\frac{3n^{2}}{2}$, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32040,
    "subject": "Mathematics (Multi-modal)",
    "question": "Sir Alex plays the following game on a row of 9 cells. Initially, all cells are empty. In each move, Sir Alex is allowed to perform exactly one of the following two operations:\n(1) Choose any number of the form $2^{j}$, where $j$ is a non-negative integer, and put it into an empty cell.\n(2) Choose two (not necessarily adjacent) cells with the same number in them; denote that number by $2^{j}$. Replace the number in one of the cells with $2^{j+1}$ and erase the number in the other cell.\nAt the end of the game, one cell contains the number $2^{n}$, where $n$ is a given positive integer, while the other cells are empty. Determine the maximum number of moves that Sir Alex could have made, in terms of $n$.",
    "options": [],
    "answer": "2 * sum_{j=0}^{8} C(n, j) - 1",
    "solution": "Solution 1. We will solve a more general problem, replacing the row of 9 cells with a row of $k$ cells, where $k$ is a positive integer. Denote by $m(n, k)$ the maximum possible number of moves Sir Alex can make starting with a row of $k$ empty cells, and ending with one cell containing the number $2^{n}$ and all the other $k-1$ cells empty. Call an operation of type (1) an insertion, and an operation of type (2) a merge.\nOnly one move is possible when $k=1$, so we have $m(n, 1)=1$. From now on we consider $k \\geqslant 2$, and we may assume Sir Alex's last move was a merge. Then, just before the last move, there were exactly two cells with the number $2^{n-1}$, and the other $k-2$ cells were empty.\nPaint one of those numbers $2^{n-1}$ blue, and the other one red. Now trace back Sir Alex's moves, always painting the numbers blue or red following this rule: if $a$ and $b$ merge into $c$, paint $a$ and $b$ with the same color as $c$. Notice that in this backward process new numbers are produced only by reversing merges, since reversing an insertion simply means deleting one of the numbers. Therefore, all numbers appearing in the whole process will receive one of the two colors.\nSir Alex's first move is an insertion. Without loss of generality, assume this first number inserted is blue. Then, from this point on, until the last move, there is always at least one cell with a blue number.\nBesides the last move, there is no move involving a blue and a red number, since all merges involve numbers with the same color, and insertions involve only one number. Call an insertion of a blue number or merge of two blue numbers a blue move, and define a red move analogously.\nThe whole sequence of blue moves could be repeated on another row of $k$ cells to produce one cell with the number $2^{n-1}$ and all the others empty, so there are at most $m(n-1, k)$ blue moves.\nNow we look at the red moves. Since every time we perform a red move there is at least one cell occupied with a blue number, the whole sequence of red moves could be repeated on a row of $k-1$ cells to produce one cell with the number $2^{n-1}$ and all the others empty, so there are at most $m(n-1, k-1)$ red moves. This proves that\n$$\nm(n, k) \\leqslant m(n-1, k)+m(n-1, k-1)+1\n$$\nOn the other hand, we can start with an empty row of $k$ cells and perform $m(n-1, k)$ moves to produce one cell with the number $2^{n-1}$ and all the others empty, and after that perform $m(n-1, k-1)$ moves on those $k-1$ empty cells to produce the number $2^{n-1}$ in one of them, leaving $k-2$ empty. With one more merge we get one cell with $2^{n}$ and the others empty, proving that\n$$\nm(n, k) \\geqslant m(n-1, k)+m(n-1, k-1)+1.\n$$\nIt follows that\n$$\n\\begin{equation*}\nm(n, k)=m(n-1, k)+m(n-1, k-1)+1, \\tag{1}\n\\end{equation*}\n$$\nfor $n \\geqslant 1$ and $k \\geqslant 2$.\nIf $k=1$ or $n=0$, we must insert $2^{n}$ on our first move and immediately get the final configuration, so $m(0, k)=1$ and $m(n, 1)=1$, for $n \\geqslant 0$ and $k \\geqslant 1$. These initial values, together with the recurrence relation (1), determine $m(n, k)$ uniquely.\nFinally, we show that\n$$\n\\begin{equation*}\nm(n, k)=2 \\sum_{j=0}^{k-1}\\binom{n}{j}-1, \\tag{2}\n\\end{equation*}\n$$\nfor all integers $n \\geqslant 0$ and $k \\geqslant 1$.\nWe use induction on $n$. Since $m(0, k)=1$ for $k \\geqslant 1$, (2) is true for the base case. We make the induction hypothesis that (2) is true for some fixed positive integer $n$ and all $k \\geqslant 1$. We have $m(n+1,1)=1=2\\binom{n+1}{0}-1$, and for $k \\geqslant 2$ the recurrence relation (1) and the induction hypothesis give us\n$$\n\\begin{aligned}\n& m(n+1, k)=m(n, k)+m(n, k-1)+1=2 \\sum_{j=0}^{k-1}\\binom{n}{j}-1+2 \\sum_{j=0}^{k-2}\\binom{n}{j}-1+1 \\\\\n& \\quad=2 \\sum_{j=0}^{k-1}\\binom{n}{j}+2 \\sum_{j=0}^{k-1}\\binom{n}{j-1}-1=2 \\sum_{j=0}^{k-1}\\left(\\binom{n}{j}+\\binom{n}{j-1}\\right)-1=2 \\sum_{j=0}^{k-1}\\binom{n+1}{j}-1,\n\\end{aligned}\n$$\nwhich completes the proof.\n\n\nSolution 2. Define merges and insertions as in Solution 1. After each move made by Sir Alex we compute the number $N$ of empty cells, and the sum $S$ of all the numbers written in the cells. Insertions always increase $S$ by some power of 2, and increase $N$ exactly by 1. Merges do not change $S$ and decrease $N$ exactly by 1. Since the initial value of $N$ is 0 and its final value is 1, the total number of insertions exceeds that of merges by exactly one. So, to maximize the number of moves, we need to maximize the number of insertions.\nWe will need the following lemma.\nLemma. If the binary representation of a positive integer $A$ has $d$ nonzero digits, then $A$ cannot be represented as a sum of fewer than $d$ powers of 2. Moreover, any representation of $A$ as a sum of $d$ powers of 2 must coincide with its binary representation.\nProof. Let $s$ be the minimum number of summands in all possible representations of $A$ as sum of powers of 2. Suppose there is such a representation with $s$ summands, where two of the summands are equal to each other. Then, replacing those two summands with the result of their sum, we obtain a representation with fewer than $s$ summands, which is a contradiction. We deduce that in any representation with $s$ summands, the summands are all distinct, so any such representation must coincide with the unique binary representation of $A$, and $s=d$.\nNow we split the solution into a sequence of claims.\nClaim 1. After every move, the number $S$ is the sum of at most $k-1$ distinct powers of 2.\nProof. If $S$ is the sum of $k$ (or more) distinct powers of 2, the Lemma implies that the $k$ cells are filled with these numbers. This is a contradiction since no more merges or insertions can be made.\nLet $A(n, k-1)$ denote the set of all positive integers not exceeding $2^{n}$ with at most $k-1$ nonzero digits in its base 2 representation. Since every insertion increases the value of $S$, by Claim 1, the total number of insertions is at most $|A(n, k-1)|$. We proceed to prove that it is possible to achieve this number of insertions.\nClaim 2. Let $A(n, k-1)=\\{a_{1}, a_{2}, \\ldots, a_{m}\\}$, with $a_{1}<a_{2}<\\cdots<a_{m}$. If after some of Sir Alex's moves the value of $S$ is $a_{j}$, with $j \\in\\{1,2, \\ldots, m-1\\}$, then there is a sequence of moves after which the value of $S$ is exactly $a_{j+1}$.\nProof. Suppose $S=a_{j}$. Performing all possible merges, we eventually get different powers of 2 in all nonempty cells. After that, by Claim 1 there will be at least one empty cell, in which we want to insert $a_{j+1}-a_{j}$. It remains to show that $a_{j+1}-a_{j}$ is a power of 2.\nFor this purpose, we notice that if $a_{j}$ has less than $k-1$ nonzero digits in base 2 then $a_{j+1}=a_{j}+1$. Otherwise, we have $a_{j}=2^{b_{k-1}}+\\cdots+2^{b_{2}}+2^{b_{1}}$ with $b_{1}<b_{2}<\\cdots<b_{k-1}$. Then, adding any number less than $2^{b_{1}}$ to $a_{j}$ will result in a number with more than $k-1$ nonzero binary digits. On the other hand, $a_{j}+2^{b_{1}}$ is a sum of $k$ powers of 2, not all distinct, so by the Lemma it will be a sum of less than $k$ distinct powers of 2. This means that $a_{j+1}-a_{j}=2^{b_{1}}$, completing the proof.\nClaims 1 and 2 prove that the maximum number of insertions is $|A(n, k-1)|$. We now compute this number.\nClaim 3. $|A(n, k-1)|=\\sum_{j=0}^{k-1}\\binom{n}{j}$.\nProof. The number $2^{n}$ is the only element of $A(n, k-1)$ with $n+1$ binary digits. Any other element has at most $n$ binary digits, at least one and at most $k-1$ of them are nonzero (so they are ones). For each $j \\in\\{1,2, \\ldots, k-1\\}$, there are $\\binom{n}{j}$ such elements with exactly $j$ binary digits equal to one. We conclude that $|A(n, k-1)|=1+\\sum_{j=1}^{k-1}\\binom{n}{j}=\\sum_{j=0}^{k-1}\\binom{n}{j}$.\nRecalling that the number of insertions exceeds that of merges by exactly 1, we deduce that the maximum number of moves is $2 \\sum_{j=0}^{k-1}\\binom{n}{j}-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32041,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\text{th}}$ smallest positive integer not in $X$, and let\n$$\nX * Y = X \\cup \\{ f_{X}(y) : y \\in Y \\} .\n$$\nLet $A$ be a set of $a > 0$ positive integers, and let $B$ be a set of $b > 0$ positive integers. Prove that if $A * B = B * A$, then\n$$\n\\begin{equation*}\n\\underbrace{A *(A * \\cdots *(A *(A * A)) \\ldots)}_{A \\text{ appears } b \\text{ times }} = \\underbrace{B *(B * \\cdots *(B *(B * B)) \\ldots)}_{B \\text{ appears } a \\text{ times }} . \\tag{U.S.A.}\n\\end{equation*}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For any function $g: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ and any subset $X \\subset \\mathbb{Z}_{>0}$, we define $g(X) = \\{ g(x) : x \\in X \\}$. We have that the image of $f_{X}$ is $f_{X}(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{>0} \\setminus X$. We now show a general lemma about the operation $*$, with the goal of showing that $*$ is associative.\n\n**Lemma 1.** Let $X$ and $Y$ be finite sets of positive integers. The functions $f_{X * Y}$ and $f_{X} \\circ f_{Y}$ are equal.\n\n*Proof.* We have\n$$\nf_{X * Y}(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{>0} \\setminus (X * Y) = (\\mathbb{Z}_{>0} \\setminus X) \\setminus f_{X}(Y) = f_{X}(\\mathbb{Z}_{>0}) \\setminus f_{X}(Y) = f_{X}(\\mathbb{Z}_{>0} \\setminus Y) = f_{X}(f_{Y}(\\mathbb{Z}_{>0}))\n$$\nThus, the functions $f_{X * Y}$ and $f_{X} \\circ f_{Y}$ are strictly increasing functions with the same range. Because a strictly increasing function is uniquely defined by its range, we have $f_{X * Y} = f_{X} \\circ f_{Y}$.\n\nLemma 1 implies that $*$ is associative, in the sense that $(A * B) * C = A * (B * C)$ for any finite sets $A, B$, and $C$ of positive integers. We prove the associativity by noting\n$$\n\\begin{gathered}\n\\mathbb{Z}_{>0} \\setminus ((A * B) * C) = f_{(A * B) * C}(\\mathbb{Z}_{>0}) = f_{A * B}(f_{C}(\\mathbb{Z}_{>0})) = f_{A}(f_{B}(f_{C}(\\mathbb{Z}_{>0}))) \\\\\n= f_{A}(f_{B * C}(\\mathbb{Z}_{>0})) = f_{A * (B * C)}(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{>0} \\setminus (A * (B * C))\n\\end{gathered}\n$$\nIn light of the associativity of $*$, we may drop the parentheses when we write expressions like $A * (B * C)$. We also introduce the notation\n$$\nX^{* k} = \\underbrace{X *(X * \\cdots *(X *(X * X)) \\ldots)}_{X \\text{ appears } k \\text{ times }}\n$$\nOur goal is then to show that $A * B = B * A$ implies $A^{* b} = B^{* a}$. We will do so via the following general lemma.\n\n**Lemma 2.** Suppose that $X$ and $Y$ are finite sets of positive integers satisfying $X * Y = Y * X$ and $|X| = |Y|$. Then, we must have $X = Y$.\n\n*Proof.* Assume that $X$ and $Y$ are not equal. Let $s$ be the largest number in exactly one of $X$ and $Y$. Without loss of generality, say that $s \\in X \\setminus Y$. The number $f_{X}(s)$ counts the $s^{th}$ number not in $X$, which implies that\n$$\n\\begin{equation*}\nf_{X}(s) = s + | X \\cap \\{ 1, 2, \\ldots, f_{X}(s) \\} | \\tag{1}\n\\end{equation*}\n$$\nSince $f_{X}(s) \\geqslant s$, we have that\n$$\n\\{ f_{X}(s) + 1, f_{X}(s) + 2, \\ldots \\} \\cap X = \\{ f_{X}(s) + 1, f_{X}(s) + 2, \\ldots \\} \\cap Y\n$$\nwhich, together with the assumption that $|X| = |Y|$, gives\n$$\n\\begin{equation*}\n| X \\cap \\{ 1, 2, \\ldots, f_{X}(s) \\} | = | Y \\cap \\{ 1, 2, \\ldots, f_{X}(s) \\} | \\tag{2}\n\\end{equation*}\n$$\nNow consider the equation\n$$\nt - | Y \\cap \\{ 1, 2, \\ldots, t \\} | = s\n$$\nThis equation is satisfied only when $t \\in [ f_{Y}(s), f_{Y}(s+1) )$, because the left hand side counts the number of elements up to $t$ that are not in $Y$. We have that the value $t = f_{X}(s)$ satisfies the above equation because of (1) and (2). Furthermore, since $f_{X}(s) \\notin X$ and $f_{X}(s) \\geqslant s$, we have that $f_{X}(s) \\notin Y$ due to the maximality of $s$. Thus, by the above discussion, we must have $f_{X}(s) = f_{Y}(s)$.\n\nFinally, we arrive at a contradiction. The value $f_{X}(s)$ is neither in $X$ nor in $f_{X}(Y)$, because $s$ is not in $Y$ by assumption. Thus, $f_{X}(s) \\notin X * Y$. However, since $s \\in X$, we have $f_{Y}(s) \\in Y * X$, a contradiction.\nWe will use Lemma 1 from Solution 1. Additionally, let $X^{* k}$ be defined as in Solution 1. If $X$ and $Y$ are finite sets, then\n$$\n\\begin{equation*}\nf_{X} = f_{Y} \\Longleftrightarrow f_{X}(\\mathbb{Z}_{>0}) = f_{Y}(\\mathbb{Z}_{>0}) \\Longleftrightarrow (\\mathbb{Z}_{>0} \\setminus X) = (\\mathbb{Z}_{>0} \\setminus Y) \\Longleftrightarrow X = Y, \\tag{3}\n\\end{equation*}\n$$\nwhere the first equivalence is because $f_{X}$ and $f_{Y}$ are strictly increasing functions, and the second equivalence is because $f_{X}(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{>0} \\setminus X$ and $f_{Y}(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{>0} \\setminus Y$.\n\nDenote $g = f_{A}$ and $h = f_{B}$. The given relation $A * B = B * A$ is equivalent to $f_{A * B} = f_{B * A}$ because of (3), and by Lemma 1 of the first solution, this is equivalent to $g \\circ h = h \\circ g$. Similarly, the required relation $A^{* b} = B^{* a}$ is equivalent to $g^{b} = h^{a}$. We will show that\n$$\n\\begin{equation*}\ng^{b}(n) = h^{a}(n) \\tag{4}\n\\end{equation*}\n$$\nfor all $n \\in \\mathbb{Z}_{>0}$, which suffices to solve the problem.\n\nTo start, we claim that (4) holds for all sufficiently large $n$. Indeed, let $p$ and $q$ be the maximal elements of $A$ and $B$, respectively; we may assume that $p \\geqslant q$. Then, for every $n \\geqslant p$ we have $g(n) = n + a$ and $h(n) = n + b$, whence $g^{b}(n) = n + a b = h^{a}(n)$, as was claimed.\n\nIn view of this claim, if (4) is not identically true, then there exists a maximal $s$ with $g^{b}(s) \\neq h^{a}(s)$. Without loss of generality, we may assume that $g(s) \\neq s$, for if we had $g(s) = h(s) = s$, then $s$ would satisfy (4). As $g$ is increasing, we then have $g(s) > s$, so (4) holds for $n = g(s)$. But then we have\n$$\ng(g^{b}(s)) = g^{b+1}(s) = g^{b}(n) = h^{a}(n) = h^{a}(g(s)) = g(h^{a}(s))\n$$\nwhere the last equality holds in view of $g \\circ h = h \\circ g$. By the injectivity of $g$, the above equality yields $g^{b}(s) = h^{a}(s)$, which contradicts the choice of $s$. Thus, we have proved that (4) is identically true on $\\mathbb{Z}_{>0}$, as desired.\n\nComment 2. We present another proof of Lemma 2 of the first solution.\n\nLet $x = |X| = |Y|$. Say that $u$ is the smallest number in $X$ and $v$ is the smallest number in $Y$; assume without loss of generality that $u \\leqslant v$.\n\nLet $T$ be any finite set of positive integers, and define $t = |T|$. Enumerate the elements of $X$ as $x_{1} < x_{2} < \\cdots < x_{n}$. Define $S_{m} = f_{T * X^{*(m-1)}}(X)$, and enumerate its elements $s_{m, 1} < s_{m, 2} < \\cdots < s_{m, n}$. Note that the $S_{m}$ are pairwise disjoint; indeed, if we have $m < m'$, then\n$$\nS_{m} \\subset T * X^{* m} \\subset T * X^{*(m' - 1)} \\quad \\text{and} \\quad S_{m'} = (T * X^{* m'}) \\setminus (T * X^{*(m' - 1)})\n$$\nWe claim the following statement, which essentially says that the $S_{m}$ are eventually linear translates of each other:\n\n**Claim.** For every $i$, there exists some $m_{i}$ and $c_{i}$ such that for all $m > m_{i}$, we have that $s_{m, i} = t + m n - c_{i}$. Furthermore, the $c_{i}$ do not depend on the choice of $T$.\n\nFirst, we show that this claim implies Lemma 2. We may choose $T = X$ and $T = Y$. Then, there is some $m'$ such that for all $m \\geqslant m'$, we have\n$$\n\\begin{equation*}\nf_{X^{* m}}(X) = f_{Y * X^{*(m-1)}}(X) \\tag{5}\n\\end{equation*}\n$$\nBecause $u$ is the minimum element of $X$, $v$ is the minimum element of $Y$, and $u \\leqslant v$, we have that\n$$\n\\left( \\bigcup_{m = m'}^{\\infty} f_{X * m}(X) \\right) \\cup X^{* m'} = \\left( \\bigcup_{m = m'}^{\\infty} f_{Y * X^{*(m-1)}}(X) \\right) \\cup (Y * X^{*(m' - 1)}) = \\{ u, u + 1, \\ldots \\}\n$$\nand in both the first and second expressions, the unions are of pairwise distinct sets. By (5), we obtain $X^{* m'} = Y * X^{*(m' - 1)}$. Now, because $X$ and $Y$ commute, we get $X^{* m'} = X^{*(m' - 1)} * Y$, and so $X = Y$.\n\nWe now prove the claim.\n\n*Proof of the claim.* We induct downwards on $i$, first proving the statement for $i = n$, and so on.\n\nAssume that $m$ is chosen so that all elements of $S_{m}$ are greater than all elements of $T$ (which is possible because $T$ is finite). For $i = n$, we have that $s_{m, n} > s_{k, n}$ for every $k < m$. Thus, all $(m - 1) n$ numbers of the form $s_{k, u}$ for $k < m$ and $1 \\leqslant u \\leqslant n$ are less than $s_{m, n}$. We then have that $s_{m, n}$ is the $((m - 1) n + x_{n})^{\\text{th}}$ number not in $T$, which is equal to $t + (m - 1) n + x_{n}$. So we may choose $c_{n} = x_{n} - n$, which does not depend on $T$, which proves the base case for the induction.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32042,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a given positive integer. In the Cartesian plane, each lattice point with nonnegative coordinates initially contains a butterfly, and there are no other butterflies. The neighborhood of a lattice point $c$ consists of all lattice points within the axis-aligned $(2 n+1) \\times (2 n+1)$ square centered at $c$, apart from $c$ itself. We call a butterfly lonely, crowded, or comfortable, depending on whether the number of butterflies in its neighborhood $N$ is respectively less than, greater than, or equal to half of the number of lattice points in $N$.\n\nEvery minute, all lonely butterflies fly away simultaneously. This process goes on for as long as there are any lonely butterflies. Assuming that the process eventually stops, determine the number of comfortable butterflies at the final state.\n\n(Bulgaria)",
    "options": [],
    "answer": "n^2 + 1",
    "solution": "We always identify a butterfly with the lattice point it is situated at. For two points $p$ and $q$, we write $p \\geqslant q$ if each coordinate of $p$ is at least the corresponding coordinate of $q$. Let $O$ be the origin, and let $\\mathcal{Q}$ be the set of initially occupied points, i.e., of all lattice points with nonnegative coordinates. Let $\\mathcal{R}_{\\mathrm{H}}=\\{(x, 0): x \\geqslant 0\\}$ and $\\mathcal{R}_{\\mathrm{V}}=\\{(0, y): y \\geqslant 0\\}$ be the sets of the lattice points lying on the horizontal and vertical boundary rays of $\\mathcal{Q}$. Denote by $N(a)$ the neighborhood of a lattice point $a$.\n\n1. Initial observations. We call a set of lattice points up-right closed if its points stay in the set after being shifted by any lattice vector $(i, j)$ with $i, j \\geqslant 0$. Whenever the butterflies form a up-right closed set $\\mathcal{S}$, we have $|N(p) \\cap \\mathcal{S}| \\geqslant |N(q) \\cap \\mathcal{S}|$ for any two points $p, q \\in \\mathcal{S}$ with $p \\geqslant q$. So, since $\\mathcal{Q}$ is up-right closed, the set of butterflies at any moment also preserves this property. We assume all forthcoming sets of lattice points to be up-right closed.\n\nWhen speaking of some set $\\mathcal{S}$ of lattice points, we call its points lonely, comfortable, or crowded with respect to this set (i.e., as if the butterflies were exactly at all points of $\\mathcal{S}$). We call a set $\\mathcal{S} \\subset \\mathcal{Q}$ stable if it contains no lonely points. In what follows, we are interested only in those stable sets whose complements in $\\mathcal{Q}$ are finite, because one can easily see that only a finite number of butterflies can fly away on each minute.\n\nIf the initial set $\\mathcal{Q}$ of butterflies contains some stable set $\\mathcal{S}$, then, clearly no butterfly of this set will fly away. On the other hand, the set $\\mathcal{F}$ of all butterflies in the end of the process is stable. This means that $\\mathcal{F}$ is the largest (with respect to inclusion) stable set within $\\mathcal{Q}$, and we are about to describe this set.\n\n2. A description of a final set. The following notion will be useful. Let $\\mathcal{U}=\\{\\vec{u}_1, \\vec{u}_2, \\ldots, \\vec{u}_d\\}$ be a set of $d$ pairwise non-parallel lattice vectors, each having a positive $x$- and a negative $y$-coordinate. Assume that they are numbered in increasing order according to slope. We now define a $\\mathcal{U}$-curve to be the broken line $p_0 p_1 \\ldots p_d$ such that $p_0 \\in \\mathcal{R}_{\\mathrm{V}}, p_d \\in \\mathcal{R}_{\\mathrm{H}}$, and $\\overrightarrow{p_{i-1} p_i}=\\vec{u}_i$ for all $i=1,2, \\ldots, m$ (see the Figure below to the left).\n\n![](attached_image_1.png)\n\nConstruction of $\\mathcal{U}$-curve\n\n![](attached_image_2.png)\n\nConstruction of $\\mathcal{D}$\n\nNow, let $\\mathcal{K}_n=\\{(i, j): 1 \\leqslant i \\leqslant n, -n \\leqslant j \\leqslant -1\\}$. Consider all the rays emerging at $O$ and passing through a point from $\\mathcal{K}_n$; number them as $r_1, \\ldots, r_m$ in increasing order according to slope. Let $A_i$ be the farthest from $O$ lattice point in $r_i \\cap \\mathcal{K}_n$, set $k_i=|r_i \\cap \\mathcal{K}_n|$, let $\\vec{v}_i=\\overrightarrow{O A_i}$, and finally denote $\\mathcal{V}=\\{\\vec{v}_i: 1 \\leqslant i \\leqslant m\\}$; see the Figure above to the right. We will concentrate on the $\\mathcal{V}$-curve $d_0 d_1 \\ldots d_m$; let $\\mathcal{D}$ be the set of all lattice points $p$ such that $p \\geqslant p'$ for some (not necessarily lattice) point $p'$ on the $\\mathcal{V}$-curve. In fact, we will show that $\\mathcal{D}=\\mathcal{F}$.\n\nClearly, the $\\mathcal{V}$-curve is symmetric in the line $y=x$. Denote by $D$ the convex hull of $\\mathcal{D}$.\n\n3. We prove that the set $\\mathcal{D}$ contains all stable sets. Let $\\mathcal{S} \\subset \\mathcal{Q}$ be a stable set (recall that it is assumed to be up-right closed and to have a finite complement in $\\mathcal{Q}$). Denote by $S$ its convex hull; clearly, the vertices of $S$ are lattice points. The boundary of $S$ consists of two rays (horizontal and vertical ones) along with some $\\mathcal{V}_*$-curve for some set of lattice vectors $\\mathcal{V}_*$.\n\nClaim 1. For every $\\vec{v}_i \\in \\mathcal{V}$, there is a $\\vec{v}_i^* \\in \\mathcal{V}_*$ co-directed with $\\vec{v}$ with $|\\vec{v}_i^*| \\geqslant |\\vec{v}|$.\n\nProof. Let $\\ell$ be the supporting line of $S$ parallel to $\\vec{v}_i$ (i.e., $\\ell$ contains some point of $S$, and the set $S$ lies on one side of $\\ell$). Take any point $b \\in \\ell \\cap \\mathcal{S}$ and consider $N(b)$. The line $\\ell$ splits the set $N(b) \\backslash \\ell$ into two congruent parts, one having an empty intersection with $\\mathcal{S}$. Hence, in order for $b$ not to be lonely, at least half of the set $\\ell \\cap N(b)$ (which contains $2 k_i$ points) should lie in $S$. Thus, the boundary of $S$ contains a segment $\\ell \\cap S$ with at least $k_i+1$ lattice points (including $b$) on it; this segment corresponds to the required vector $\\vec{v}_i^* \\in \\mathcal{V}_*$. $\\square$\n\n![](attached_image_3.png)\n\nClaim 2. Each stable set $\\mathcal{S} \\subseteq \\mathcal{Q}$ lies in $\\mathcal{D}$.\n\nProof. To show this, it suffices to prove that the $\\mathcal{V}_*$-curve lies in $D$, i.e., that all its vertices do so. Let $p'$ be an arbitrary vertex of the $\\mathcal{V}_*$-curve; $p'$ partitions this curve into two parts, $\\mathcal{X}$ (being down-right of $p$) and $\\mathcal{Y}$ (being up-left of $p$). The set $\\mathcal{V}$ is split now into two parts: $\\mathcal{V}_{\\mathcal{X}}$ consisting of those $\\vec{v}_i \\in \\mathcal{V}$ for which $\\vec{v}_i^*$ corresponds to segment in $\\mathcal{X}$, and a similar part $\\mathcal{V}_{\\mathcal{Y}}$. Notice that the $\\mathcal{V}$-curve consists of several segments corresponding to $\\mathcal{V}_{\\mathcal{X}}$, followed by those corresponding to $\\mathcal{V}_{\\mathcal{Y}}$. Hence there is a vertex $p$ of the $\\mathcal{V}$-curve separating $\\mathcal{V}_{\\mathcal{X}}$ from $\\mathcal{V}_{\\mathcal{Y}}$. Claim 1 now yields that $p' \\geqslant p$, so $p' \\in \\mathcal{D}$, as required. $\\square$\n\nClaim 2 implies that the final set $\\mathcal{F}$ is contained in $\\mathcal{D}$.\n\n4. $\\mathcal{D}$ is stable, and its comfortable points are known. Recall the definitions of $r_i$; let $r_i'$ be the ray complementary to $r_i$. By our definitions, the set $N(O)$ contains no points between the rays $r_i$ and $r_{i+1}$, as well as between $r_i'$ and $r_{i+1}'$.\n\nClaim 3. In the set $\\mathcal{D}$, all lattice points of the $\\mathcal{V}$-curve are comfortable.\n\nProof. Let $p$ be any lattice point of the $\\mathcal{V}$-curve, belonging to some segment $d_i d_{i+1}$. Draw the line $\\ell$ containing this segment. Then $\\ell \\cap \\mathcal{D}$ contains exactly $k_i+1$ lattice points, all of which lie in $N(p)$ except for $p$. Thus, exactly half of the points in $N(p) \\cap \\ell$ lie in $\\mathcal{D}$. It remains to show that all points of $N(p)$ above $\\ell$ lie in $\\mathcal{D}$ (recall that all the points below $\\ell$ lack this property).\n\nNotice that each vector in $\\mathcal{V}$ has one coordinate greater than $n / 2$; thus the neighborhood of $p$ contains parts of at most two segments of the $\\mathcal{V}$-curve succeeding $d_i d_{i+1}$, as well as at most two of those preceding it.\n\nThe angles formed by these consecutive segments are obtained from those formed by $r_j$ and $r_{j-1}'$ (with $i-1 \\leqslant j \\leqslant i+2$) by shifts; see the Figure below. All the points in $N(p)$ above $\\ell$ which could lie outside $\\mathcal{D}$ lie in shifted angles between $r_j, r_{j+1}$ or $r_j', r_{j-1}'$. But those angles, restricted to $N(p)$, have no lattice points due to the above remark. The claim is proved. $\\square$\n\n![](attached_image_4.png)\n\nProof of Claim 3\n\nClaim 4. All the points of $\\mathcal{D}$ which are not on the boundary of $D$ are crowded.\n\nProof. Let $p \\in \\mathcal{D}$ be such a point. If it is to the up-right of some point $p'$ on the curve, then the claim is easy: the shift of $N(p') \\cap \\mathcal{D}$ by $\\overrightarrow{p' p}$ is still in $\\mathcal{D}$, and $N(p)$ contains at least one more point of $\\mathcal{D}$ - either below or to the left of $p$. So, we may assume that $p$ lies in a right triangle constructed on some hypothenuse $d_i d_{i+1}$. Notice here that $d_i, d_{i+1} \\in N(p)$.\n\nDraw a line $\\ell \\parallel d_i d_{i+1}$ through $p$, and draw a vertical line $h$ through $d_i$; see Figure below. Let $\\mathcal{D}_{\\mathrm{L}}$ and $\\mathcal{D}_{\\mathrm{R}}$ be the parts of $\\mathcal{D}$ lying to the left and to the right of $h$, respectively (points of $\\mathcal{D} \\cap h$ lie in both parts).\n\n![](attached_image_5.png)\n\nProof of Claim 4\n\nNotice that the vectors $\\overrightarrow{d_i p}, \\overrightarrow{d_{i+1} d_{i+2}}, \\overrightarrow{d_i d_{i+1}}, \\overrightarrow{d_{i-1} d_i}$, and $\\overrightarrow{p d_{i+1}}$ are arranged in non-increasing order by slope. This means that $\\mathcal{D}_{\\mathrm{L}}$ shifted by $\\overrightarrow{d_i p}$ still lies in $\\mathcal{D}$, as well as $\\mathcal{D}_{\\mathrm{R}}$ shifted by $\\overrightarrow{d_{i+1} p}$. As we have seen in the proof of Claim 3, these two shifts cover all points of $N(p)$ above $\\ell$, along with those on $\\ell$ to the left of $p$. Since $N(p)$ contains also $d_i$ and $d_{i+1}$, the point $p$ is crowded. $\\square$\n\nThus, we have proved that $\\mathcal{D}=\\mathcal{F}$, and have shown that the lattice points on the $\\mathcal{V}$-curve are exactly the comfortable points of $\\mathcal{D}$. It remains to find their number.\n\nRecall the definition of $\\mathcal{K}_n$ (see Figure on the first page of the solution). Each segment $d_i d_{i+1}$ contains $k_i$ lattice points different from $d_i$. Taken over all $i$, these points exhaust all the lattice points in the $\\mathcal{V}$-curve, except for $d_1$, and thus the number of lattice points on the $\\mathcal{V}$-curve is $1+\\sum_{i=1}^m k_i$. On the other hand, $\\sum_{i=1}^m k_i$ is just the number of points in $\\mathcal{K}_n$, so it equals $n^2$. Hence the answer to the problem is $n^2+1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32043,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C D E$ be a convex pentagon such that $A B = B C = C D$, $\\angle E A B = \\angle B C D$, and $\\angle E D C = \\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Throughout the solution, we refer to $\\angle A, \\angle B, \\angle C, \\angle D$, and $\\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \\perp C I$ and, similarly, $A C \\perp B I$. Hence $A C$ and $B D$ meet at the orthocenter $H$ of the triangle $B I C$, and $I H \\perp B C$. It remains to prove that $E$ lies on the line $I H$ or, equivalently, $E I \\perp B C$.\n\nLines $I B$ and $I C$ bisect $\\angle B$ and $\\angle C$, respectively. Since $I A = I C, I B = I D$, and $A B = B C = C D$, the triangles $I A B, I C B$ and $I C D$ are congruent. Hence $\\angle I A B = \\angle I C B = \\angle C / 2 = \\angle A / 2$, so the line $I A$ bisects $\\angle A$. Similarly, the line $I D$ bisects $\\angle D$. Finally, the line $I E$ bisects $\\angle E$ because $I$ lies on all the other four internal bisectors of the angles of the pentagon.\n\nThe sum of the internal angles in a pentagon is $540^{\\circ}$, so\n$$\n\\angle E = 540^{\\circ} - 2 \\angle A + 2 \\angle B .\n$$\nIn quadrilateral $A B I E$,\n$$\n\\begin{aligned}\n\\angle B I E & = 360^{\\circ} - \\angle E A B - \\angle A B I - \\angle A E I = 360^{\\circ} - \\angle A - \\frac{1}{2} \\angle B - \\frac{1}{2} \\angle E \\\\\n& = 360^{\\circ} - \\angle A - \\frac{1}{2} \\angle B - \\left(270^{\\circ} - \\angle A - \\angle B\\right) \\\\\n& = 90^{\\circ} + \\frac{1}{2} \\angle B = 90^{\\circ} + \\angle I B C,\n\\end{aligned}\n$$\nwhich means that $E I \\perp B C$, completing the proof.\nWe present another proof of the fact that $E$ lies on line $I H$. Since all five internal bisectors of $A B C D E$ meet at $I$, this pentagon has an inscribed circle with center $I$. Let this circle touch side $B C$ at $T$.\n\nApplying Brianchon's theorem to the (degenerate) hexagon $A B T C D E$ we conclude that $A C, B D$ and $E T$ are concurrent, so point $E$ also lies on line $I H T$, completing the proof.\nWe present yet another proof that $E I \\perp B C$. In pentagon $A B C D E$, $\\angle E < 180^{\\circ} \\Longleftrightarrow \\angle A + \\angle B + \\angle C + \\angle D > 360^{\\circ}$. Then $\\angle A + \\angle B = \\angle C + \\angle D > 180^{\\circ}$, so rays $E A$ and $C B$ meet at a point $P$, and rays $B C$ and $E D$ meet at a point $Q$. Now,\n$$\n\\angle P B A = 180^{\\circ} - \\angle B = 180^{\\circ} - \\angle D = \\angle Q D C\n$$\nand, similarly, $\\angle P A B = \\angle Q C D$. Since $A B = C D$, the triangles $P A B$ and $Q C D$ are congruent with the same orientation. Moreover, $P Q E$ is isosceles with $E P = E Q$.\n\n![](attached_image_1.png)\n\nIn Solution 1 we have proved that triangles $I A B$ and $I C D$ are also congruent with the same orientation. Then we conclude that quadrilaterals $P B I A$ and $Q D I C$ are congruent, which implies $I P = I Q$. Then $E I$ is the perpendicular bisector of $P Q$ and, therefore, $E I \\perp P Q \\Longleftrightarrow E I \\perp B C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32044,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $R$ and $S$ be distinct points on circle $\\Omega$, and let $t$ denote the tangent line to $\\Omega$ at $R$. Point $R'$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $RS$ of $\\Omega$ so that the circumcircle $\\Gamma$ of triangle $ISR'$ intersects $t$ at two different points. Denote by $A$ the common point of $\\Gamma$ and $t$ that is closest to $R$. Line $AI$ meets $\\Omega$ again at $J$. Show that $JR'$ is tangent to $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In the circles $\\Omega$ and $\\Gamma$ we have $\\angle JRS = \\angle JIS = \\angle AR'S$. On the other hand, since $RA$ is tangent to $\\Omega$, we get $\\angle SJR = \\angle SRA$. So the triangles $ARR'$ and $SJR$ are similar, and\n$$\n\\frac{R'R}{RJ} = \\frac{AR'}{SR} = \\frac{AR'}{SR'}.\n$$\nThe last relation, together with $\\angle AR'S = \\angle JRR'$, yields $\\triangle ASR' \\sim \\triangle R'JR$, hence $\\angle SAR' = \\angle RR'J$. It follows that $JR'$ is tangent to $\\Gamma$ at $R'$.\nAs in Solution 1, we notice that $\\angle JRS = \\angle JIS = \\angle AR'S$, so we have $RJ \\parallel AR'$. Let $A'$ be the reflection of $A$ about $S$; then $ARA'R'$ is a parallelogram with center $S$, and hence the point $J$ lies on the line $RA'$. From $\\angle SR'A' = \\angle SRA = \\angle SJR$ we get that the points $S, J, A', R'$ are concyclic. This proves that $\\angle SR'J = \\angle SA'J = \\angle SA'R = \\angle SAR'$, so $JR'$ is tangent to $\\Gamma$ at $R'$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32045,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B = B C$, and suppose that the line segments $A A_{1}$, $B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\\omega$ the circle $A B C$. Let $\\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \\neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. If $A A_{1} = C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may assume $A A_{1} < C C_{1}$.\nLet $R$ be the radical center of the circles $A E B C$ and $A_{1} E B C_{1}$, and the circumcircle of the symmetric trapezoid $A C C_{1} A_{1}$; that is the common point of the pairwise radical axes $A C$, $A_{1} C_{1}$, and $B E$. By the symmetry of $A C$ and $A_{1} C_{1}$, the point $R$ lies on the common perpendicular bisector of $A A_{1}$ and $C C_{1}$, which is the external bisector of $\\angle A D C$.\nLet $F$ be the second intersection of the line $D R$ and the circle $A C D$. From the power of $R$ with respect to the circles $\\omega$ and $A C F D$ we have $R B \\cdot R E = R A \\cdot R C = R D \\cdot D F$, so the points $B, E, D$ and $F$ are concyclic.\nThe line $R D F$ is the external bisector of $\\angle A D C$, so the point $F$ bisects the $\\operatorname{arc} \\overline{C D A}$. By $A B = B C$, on circle $\\omega$, the point $B$ is the midpoint of $\\operatorname{arc} \\overline{A E C}$; let $M$ be the point diametrically opposite to $B$, that is the midpoint of the opposite $\\operatorname{arc} \\widetilde{C A}$ of $\\omega$. Notice that the points $B, F$ and $M$ lie on the perpendicular bisector of $A C$, so they are collinear.\n![](attached_image_1.png)\nFinally, let $X$ be the second intersection point of $\\omega$ and the line $D E$. Since $B M$ is a diameter in $\\omega$, we have $\\angle B X M = 90^{\\circ}$. Moreover,\n$$\n\\angle E X M = 180^{\\circ} - \\angle M B E = 180^{\\circ} - \\angle F B E = \\angle E D F,\n$$\nso $M X$ and $F D$ are parallel. Since $B X$ is perpendicular to $M X$ and $B B_{1}$ is perpendicular to $F D$, this shows that $X$ lies on line $B B_{1}$.\nSolution 2. Define point $M$ as the point opposite to $B$ on circle $\\omega$, and point $R$ as the intersection of lines $A C$, $A_{1} C_{1}$ and $B E$, and show that $R$ lies on the external bisector of $\\angle A D C$, like in the first solution.\nSince $B$ is the midpoint of the $\\operatorname{arc} \\overline{A E C}$, the line $B E R$ is the external bisector of $\\angle C E A$. Now we show that the internal angle bisectors of $\\angle A D C$ and $\\angle C E A$ meet on the segment $A C$. Let the angle bisector of $\\angle A D C$ meet $A C$ at $S$, and let the angle bisector of $\\angle C E A$, which is line $E M$, meet $A C$ at $S^{\\prime}$. By applying the angle bisector theorem to both internal and external bisectors of $\\angle A D C$ and $\\angle C E A$,\n$$\nA S : C S = A D : C D = A R : C R = A E : C E = A S^{\\prime} : C S^{\\prime},\n$$\nso indeed $S = S^{\\prime}$.\nBy $\\angle R D S = \\angle S E R = 90^{\\circ}$ the points $R, S, D$ and $E$ are concyclic.\n![](attached_image_2.png)\nNow let the lines $B B_{1}$ and $D E$ meet at point $X$. Notice that $\\angle E X B = \\angle E D S$ because both $B B_{1}$ and $D S$ are perpendicular to the line $D R$, we have that $\\angle E D S = \\angle E R S$ in circle $S R D E$, and $\\angle E R S = \\angle E M B$ because $S R \\perp B M$ and $E R \\perp M E$. Therefore, $\\angle E X B = \\angle E M B$, so indeed, the point $X$ lies on $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32046,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 3$ be an integer. Two regular $n$-gons $\\mathcal{A}$ and $\\mathcal{B}$ are given in the plane. Prove that the vertices of $\\mathcal{A}$ that lie inside $\\mathcal{B}$ or on its boundary are consecutive.\n(That is, prove that there exists a line separating those vertices of $\\mathcal{A}$ that lie inside $\\mathcal{B}$ or on its boundary from the other vertices of $\\mathcal{A}$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start with finding a regular $n$-gon $\\mathcal{C}$ which (i) is inscribed into $\\mathcal{B}$ (that is, all vertices of $\\mathcal{C}$ lie on the perimeter of $\\mathcal{B}$); and (ii) is either a translation of $\\mathcal{A}$, or a homothetic image of $\\mathcal{A}$ with a positive factor.\nSuch a polygon may be constructed as follows. Let $O_{A}$ and $O_{B}$ be the centers of $\\mathcal{A}$ and $\\mathcal{B}$, respectively, and let $A$ be an arbitrary vertex of $\\mathcal{A}$. Let $\\overrightarrow{O_{B} C}$ be the vector co-directional to $\\overrightarrow{O_{A} A}$, with $C$ lying on the perimeter of $\\mathcal{B}$. The rotations of $C$ around $O_{B}$ by multiples of $2\\pi/n$ form the required polygon. Indeed, it is regular, inscribed into $\\mathcal{B}$ (due to the rotational symmetry of $\\mathcal{B}$), and finally the translation/homothety mapping $\\overrightarrow{O_{A} A}$ to $\\overrightarrow{O_{B} C}$ maps $\\mathcal{A}$ to $\\mathcal{C}$.\n\nNow we separate two cases.\n\n![](attached_image_1.png)\nConstruction of $\\mathcal{C}$\n![](attached_image_2.png)\nCase 1: Translation\n\nCase 1: $\\mathcal{C}$ is a translation of $\\mathcal{A}$ by a vector $\\vec{v}$.\nDenote by $t$ the translation transform by vector $\\vec{v}$. We need to prove that the vertices of $\\mathcal{C}$ which stay in $\\mathcal{B}$ under $t$ are consecutive. To visualize the argument, we refer the plane to Cartesian coordinates so that the $x$-axis is co-directional with $\\vec{v}$. This way, the notions of right/left and top/bottom are also introduced, according to the $x$- and $y$-coordinates, respectively.\n\nLet $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ be the top and the bottom vertices of $\\mathcal{B}$ (if several vertices are extremal, we take the rightmost of them). They split the perimeter of $\\mathcal{B}$ into the right part $\\mathcal{B}_{\\mathrm{R}}$ and the left part $\\mathcal{B}_{\\mathrm{L}}$ (the vertices $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ are assumed to lie in both parts); each part forms a connected subset of the perimeter of $\\mathcal{B}$. So the vertices of $\\mathcal{C}$ are also split into two parts $\\mathcal{C}_{\\mathrm{L}} \\subset \\mathcal{B}_{\\mathrm{L}}$ and $\\mathcal{C}_{\\mathrm{R}} \\subset \\mathcal{B}_{\\mathrm{R}}$, each of which consists of consecutive vertices.\n\nNow, all the points in $\\mathcal{B}_{\\mathrm{R}}$ (and hence in $\\mathcal{C}_{\\mathrm{R}}$) move out from $\\mathcal{B}$ under $t$, since they are the rightmost points of $\\mathcal{B}$ on the corresponding horizontal lines. It remains to prove that the vertices of $\\mathcal{C}_{\\mathrm{L}}$ which stay in $\\mathcal{B}$ under $t$ are consecutive.\n\nFor this purpose, let $C_{1}, C_{2}$, and $C_{3}$ be three vertices in $\\mathcal{C}_{\\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $t\\left(C_{1}\\right)$ and $t\\left(C_{3}\\right)$ lie in $\\mathcal{B}$; we need to prove that $t\\left(C_{2}\\right) \\in \\mathcal{B}$ as well. Let $A_{i}=t\\left(C_{i}\\right)$. The line through $C_{2}$ parallel to $\\vec{v}$ crosses the segment $C_{1}C_{3}$ to the right of $C_{2}$; this means that this line crosses $A_{1}A_{3}$ to the right of $A_{2}$, so $A_{2}$ lies inside the triangle $A_{1}C_{2}A_{3}$ which is contained in $\\mathcal{B}$. This yields the desired result.\n\nCase 2: $\\mathcal{C}$ is a homothetic image of $\\mathcal{A}$ centered at $X$ with factor $k>0$.\nDenote by $h$ the homothety mapping $\\mathcal{C}$ to $\\mathcal{A}$. We need now to prove that the vertices of $\\mathcal{C}$ which stay in $\\mathcal{B}$ after applying $h$ are consecutive. If $X \\in \\mathcal{B}$, the claim is easy. Indeed, if $k<1$, then the vertices of $\\mathcal{A}$ lie on the segments of the form $XC$ ($C$ being a vertex of $\\mathcal{C}$) which lie in $\\mathcal{B}$. If $k>1$, then the vertices of $\\mathcal{A}$ lie on the extensions of such segments $XC$ beyond $C$, and almost all these extensions lie outside $\\mathcal{B}$. The exceptions may occur only in case when $X$ lies on the boundary of $\\mathcal{B}$, and they may cause one or two vertices of $\\mathcal{A}$ stay on the boundary of $\\mathcal{B}$. But even in this case those vertices are still consecutive.\n\nSo, from now on we assume that $X \\notin \\mathcal{B}$.\n\nNow, there are two vertices $B_{\\mathrm{T}}$ and $\\mathcal{B}_{\\mathrm{B}}$ of $\\mathcal{B}$ such that $\\mathcal{B}$ is contained in the angle $\\angle B_{\\mathrm{T}} X B_{\\mathrm{B}}$; if there are several options, say, for $B_{\\mathrm{T}}$, then we choose the farthest one from $X$ if $k>1$, and the nearest one if $k<1$. For the visualization purposes, we refer the plane to Cartesian coordinates so that the $y$-axis is co-directional with $\\overrightarrow{B_{\\mathrm{B}} B_{\\mathrm{T}}}$, and $X$ lies to the left of the line $B_{\\mathrm{T}} B_{\\mathrm{B}}$. Again, the perimeter of $\\mathcal{B}$ is split by $B_{\\mathrm{T}}$ and $B_{\\mathrm{B}}$ into the right part $\\mathcal{B}_{\\mathrm{R}}$ and the left part $\\mathcal{B}_{\\mathrm{L}}$, and the set of vertices of $\\mathcal{C}$ is split into two subsets $\\mathcal{C}_{\\mathrm{R}} \\subset \\mathcal{B}_{\\mathrm{R}}$ and $\\mathcal{C}_{\\mathrm{L}} \\subset \\mathcal{B}_{\\mathrm{L}}$.\n\n![](attached_image_3.png)\nCase 2, $X$ inside $\\mathcal{B}$\n![](attached_image_4.png)\nSubcase 2.1: $k>1$\n\n## Subcase 2.1: $k>1$.\n\nIn this subcase, all points from $\\mathcal{B}_{\\mathrm{R}}$ (and hence from $\\mathcal{C}_{\\mathrm{R}}$) move out from $\\mathcal{B}$ under $h$, because they are the farthest points of $\\mathcal{B}$ on the corresponding rays emanated from $X$. It remains to prove that the vertices of $\\mathcal{C}_{\\mathrm{L}}$ which stay in $\\mathcal{B}$ under $h$ are consecutive.\n\nAgain, let $C_{1}, C_{2}, C_{3}$ be three vertices in $\\mathcal{C}_{\\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $h\\left(C_{1}\\right)$ and $h\\left(C_{3}\\right)$ lie in $\\mathcal{B}$. Let $A_{i}=h\\left(C_{i}\\right)$. Then the ray $X C_{2}$ crosses the segment $C_{1}C_{3}$ beyond $C_{2}$, so this ray crosses $A_{1}A_{3}$ beyond $A_{2}$; this implies that $A_{2}$ lies in the triangle $A_{1}C_{2}A_{3}$, which is contained in $\\mathcal{B}$.\n\n![](attached_image_5.png)\nSubcase 2.2: $k<1$\n\n## Subcase 2.2: $k<1$.\n\nThis case is completely similar to the previous one. All points from $\\mathcal{B}_{\\mathrm{L}}$ (and hence from $\\mathcal{C}_{\\mathrm{L}}$) move out from $\\mathcal{B}$ under $h$, because they are the nearest points of $\\mathcal{B}$ on the corresponding rays emanated from $X$. Assume that $C_{1}, C_{2}$, and $C_{3}$ are three vertices in $\\mathcal{C}_{\\mathrm{R}}$ such that $C_{2}$ lies between $C_{1}$ and $C_{3}$, and $h\\left(C_{1}\\right)$ and $h\\left(C_{3}\\right)$ lie in $\\mathcal{B}$; let $A_{i}=h\\left(C_{i}\\right)$. Then $A_{2}$ lies on the segment $X C_{2}$, and the segments $X A_{2}$ and $A_{1}A_{3}$ cross each other. Thus $A_{2}$ lies in the triangle $A_{1}C_{2}A_{3}$, which is contained in $\\mathcal{B}$.\n\nNow, choose a vertex $A_{1}$ of $\\mathcal{A}$ such that the vector $\\overrightarrow{O_{A} A_{1}}$ points \"mostly outside $\\mathcal{H}_{1}$\"; strictly speaking, this means that the scalar product $\\left\\langle\\overrightarrow{O_{A} A_{1}}, \\overrightarrow{b_{1}}\\right\\rangle$ is minimal. Starting from $A_{1}$, enumerate the vertices of $\\mathcal{A}$ clockwise as $A_{1}, A_{2}, \\ldots, A_{n}$; by the rotational symmetry, the choice of $A_{1}$ yields that the vector $\\overrightarrow{O_{A} A_{i}}$ points \"mostly outside $\\mathcal{H}_{i}$\", i.e.,\n$$\n\\begin{equation*}\n\\left\\langle\\overrightarrow{O_{A} A_{i}}, \\overrightarrow{b_{i}}\\right\\rangle=\\min_{j \\in[n]}\\left\\langle\\overrightarrow{O_{A} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle . \\tag{1}\n\\end{equation*}\n$$\n![](attached_image_6.png)\n\nWe intend to reformulate the problem in more combinatorial terms, for which purpose we introduce the following notion. Say that a subset $I \\subseteq[n]$ is connected if the elements of this set are consecutive in the cyclic order (in other words, if we join each $i$ with $i+1 \\bmod n$ by an edge, this subset is connected in the usual graph sense). Clearly, the union of two connected subsets sharing at least one element is connected too. Next, for any half-plane $\\mathcal{H}$ the indices of vertices of, say, $\\mathcal{A}$ that lie in $\\mathcal{H}$ form a connected set.\n\nTo access the problem, we denote\n$$\nM=\\left\\{j \\in[n]: A_{j} \\notin \\mathcal{B}\\right\\}, \\quad M_{i}=\\left\\{j \\in[n]: A_{j} \\notin \\mathcal{H}_{i}\\right\\} \\quad \\text{ for } i \\in[n] .\n$$\nWe need to prove that $[n] \\backslash M$ is connected, which is equivalent to $M$ being connected. On the other hand, since $\\mathcal{B}=\\bigcap_{i \\in[n]} \\mathcal{H}_{i}$, we have $M=\\bigcup_{i \\in[n]} M_{i}$, where the sets $M_{i}$ are easier to investigate. We will utilize the following properties of these sets; the first one holds by the definition of $M_{i}$, along with the above remark.\n\n![](attached_image_7.png)\nThe sets $M_{i}$\n\nProperty 1: Each set $M_{i}$ is connected. $\\square$\n\nProperty 2: If $M_{i}$ is nonempty, then $i \\in M_{i}$.\n\nProof. Indeed, we have\n$$\n\\begin{equation*}\nj \\in M_{i} \\Longleftrightarrow A_{j} \\notin \\mathcal{H}_{i} \\Longleftrightarrow\\left\\langle\\overrightarrow{B_{i} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle<0 \\Longleftrightarrow\\left\\langle\\overrightarrow{O_{A} A_{j}}, \\overrightarrow{b_{i}}\\right\\rangle<\\left\\langle\\overrightarrow{O_{A} B_{i}}, \\overrightarrow{b_{i}}\\right\\rangle . \\tag{2}\n\\end{equation*}\n$$\nThe right-hand part of the last inequality does not depend on $j$. Therefore, if some $j$ lies in $M_{i}$, then by (1) so does $i$. $\\square$\n\nIn view of Property 2, it is useful to define the set\n$$\nM^{\\prime}=\\left\\{i \\in[n]: i \\in M_{i}\\right\\}=\\left\\{i \\in[n]: M_{i} \\neq \\varnothing\\right\\} .\n$$\n\nProperty 3: The set $M^{\\prime}$ is connected.\n\nProof. To prove this property, we proceed on with the investigation started in (2) to write\n$$\ni \\in M^{\\prime} \\Longleftrightarrow A_{i} \\in M_{i} \\Longleftrightarrow\\left\\langle\\overrightarrow{B_{i} A_{i}}, \\overrightarrow{b_{i}}\\right\\rangle<0 \\Longleftrightarrow\\left\\langle\\overrightarrow{O_{B} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle<\\left\\langle\\overrightarrow{O_{B} B_{i}}, \\overrightarrow{b_{i}}\\right\\rangle+\\left\\langle\\overrightarrow{A_{i} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle .\n$$\nThe right-hand part of the obtained inequality does not depend on $i$, due to the rotational symmetry; denote its constant value by $\\mu$. Thus, $i \\in M^{\\prime}$ if and only if $\\left\\langle\\overrightarrow{O_{B} O_{A}}, \\overrightarrow{b_{i}}\\right\\rangle<\\mu$. This condition is in turn equivalent to the fact that $B_{i}$ lies in a certain (open) half-plane whose boundary line is orthogonal to $O_{B} O_{A}$; thus, it defines a connected set. $\\square$\n\nNow we can finish the solution. Since $M^{\\prime} \\subseteq M$, we have\n$$\nM=\\bigcup_{i \\in[n]} M_{i}=M^{\\prime} \\cup \\bigcup_{i \\in[n]} M_{i},\n$$\nso $M$ can be obtained from $M^{\\prime}$ by adding all the sets $M_{i}$ one by one. All these sets are connected, and each nonempty $M_{i}$ contains an element of $M^{\\prime}$ (namely, $i$). Thus their union is also connected.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32047,
    "subject": "Mathematics (Multi-modal)",
    "question": "A convex quadrilateral $ABCD$ has an inscribed circle with center $I$. Let $I_{a}$, $I_{b}$, $I_{c}$, and $I_{d}$ be the incenters of the triangles $DAB$, $ABC$, $BCD$, and $CDA$, respectively. Suppose that the common external tangents of the circles $A I_{b} I_{d}$ and $C I_{b} I_{d}$ meet at $X$, and the common external tangents of the circles $B I_{a} I_{c}$ and $D I_{a} I_{c}$ meet at $Y$. Prove that $\\angle X I Y = 90^{\\circ}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $\\omega_{a}$, $\\omega_{b}$, $\\omega_{c}$ and $\\omega_{d}$ the circles $A I_{b} I_{d}$, $B I_{a} I_{c}$, $C I_{b} I_{d}$, and $D I_{a} I_{c}$, let their centers be $O_{a}$, $O_{b}$, $O_{c}$ and $O_{d}$, and let their radii be $r_{a}$, $r_{b}$, $r_{c}$ and $r_{d}$, respectively.\n\nClaim 1. $I_{b} I_{d} \\perp AC$ and $I_{a} I_{c} \\perp BD$.\n\nProof. Let the incircles of triangles $ABC$ and $ACD$ be tangent to the line $AC$ at $T$ and $T'$, respectively. (See the figure to the left.) We have $AT = \\frac{AB + AC - BC}{2}$ in triangle $ABC$, $AT' = \\frac{AD + AC - CD}{2}$ in triangle $ACD$, and $AB - BC = AD - CD$ in quadrilateral $ABCD$, so\n$$\nAT = \\frac{AC + AB - BC}{2} = \\frac{AC + AD - CD}{2} = AT'.\n$$\nThis shows $T = T'$. As an immediate consequence, $I_{b} I_{d} \\perp AC$.\n\nThe second statement can be shown analogously. $\\square$\n\n![](attached_image_1.png)\n![](attached_image_2.png)\n\nClaim 2. The points $O_{a}$, $O_{b}$, $O_{c}$ and $O_{d}$ lie on the lines $AI$, $BI$, $CI$ and $DI$, respectively.\n\nProof. By symmetry it suffices to prove the claim for $O_{a}$. (See the figure to the right above.)\n\nNotice first that the incircles of triangles $ABC$ and $ACD$ can be obtained from the incircle of the quadrilateral $ABCD$ with homothety centers $B$ and $D$, respectively, and homothety factors less than $1$, therefore the points $I_{b}$ and $I_{d}$ lie on the line segments $BI$ and $DI$, respectively.\n\nAs is well-known, in every triangle the altitude and the diameter of the circumcircle starting from the same vertex are symmetric about the angle bisector. By Claim 1, in triangle $A I_{d} I_{b}$, the segment $AT$ is the altitude starting from $A$. Since the foot $T$ lies inside the segment $I_{b} I_{d}$, the circumcenter $O_{a}$ of triangle $A I_{d} I_{b}$ lies in the angle domain $I_{b} A I_{d}$ in such a way that $\\angle I_{b} A T = \\angle O_{a} A I_{d}$. The points $I_{b}$ and $I_{d}$ are the incenters of triangles $ABC$ and $ACD$, so the lines $A I_{b}$ and $A I_{d}$ bisect the angles $\\angle BAC$ and $\\angle CAD$, respectively. Then\n$$\n\\angle O_{a} A D = \\angle O_{a} A I_{d} + \\angle I_{d} A D = \\angle I_{b} A T + \\angle I_{d} A D = \\frac{1}{2} \\angle BAC + \\frac{1}{2} \\angle CAD = \\frac{1}{2} \\angle BAD,\n$$\nso $O_{a}$ lies on the angle bisector of $\\angle BAD$, that is, on line $AI$. $\\square$\n\nThe point $X$ is the external similitude center of $\\omega_{a}$ and $\\omega_{c}$; let $U$ be their internal similitude center. The points $O_{a}$ and $O_{c}$ lie on the perpendicular bisector of the common chord $I_{b} I_{d}$ of $\\omega_{a}$ and $\\omega_{c}$, and the two similitude centers $X$ and $U$ lie on the same line; by Claim 2, that line is parallel to $AC$.\n\n![](attached_image_3.png)\n\nFrom the similarity of the circles $\\omega_{a}$ and $\\omega_{c}$, from $O_{a} I_{b} = O_{a} I_{d} = O_{a} A = r_{a}$ and $O_{c} I_{b} = O_{c} I_{d} = O_{c} C = r_{c}$, and from $AC \\parallel O_{a} O_{c}$ we can see that\n$$\n\\frac{O_{a} X}{O_{c} X} = \\frac{O_{a} U}{O_{c} U} = \\frac{r_{a}}{r_{c}} = \\frac{O_{a} I_{b}}{O_{c} I_{b}} = \\frac{O_{a} I_{d}}{O_{c} I_{d}} = \\frac{O_{a} A}{O_{c} C} = \\frac{O_{a} I}{O_{c} I}.\n$$\nSo the points $X, U, I_{b}, I_{d}, I$ lie on the Apollonius circle of the points $O_{a}, O_{c}$ with ratio $r_{a} : r_{c}$. In this Apollonius circle $XU$ is a diameter, and the lines $IU$ and $IX$ are respectively the internal and external bisectors of $\\angle O_{a} I O_{c} = \\angle A I C$, according to the angle bisector theorem. Moreover, in the Apollonius circle the diameter $UX$ is the perpendicular bisector of $I_{b} I_{d}$, so the lines $IX$ and $IU$ are the internal and external bisectors of $\\angle I_{b} I I_{d} = \\angle B I D$, respectively.\n\nRepeating the same argument for the points $B, D$ instead of $A, C$, we get that the line $IY$ is the internal bisector of $\\angle A I C$ and the external bisector of $\\angle B I D$. Therefore, the lines $IX$ and $IY$ respectively are the internal and external bisectors of $\\angle B I D$, so they are perpendicular.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32048,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $2017$ mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard, one at a time, so that no tangent segment intersects any other circles or previously drawn tangent segments. Luciano keeps drawing tangent segments until no more can be drawn. Find all possible numbers of tangent segments when he stops drawing.",
    "options": [],
    "answer": "6048",
    "solution": "First, consider a particular arrangement of circles $C_{1}, C_{2}, \\ldots, C_{n}$ where all the centers are aligned and each $C_{i}$ is eclipsed from the other circles by its neighbors - for example, taking $C_{i}$ with center $(i^{2}, 0)$ and radius $i / 2$ works. Then the only tangent segments that can be drawn are between adjacent circles $C_{i}$ and $C_{i+1}$, and exactly three segments can be drawn for each pair. So Luciano will draw exactly $3(n-1)$ segments in this case.\n\n![](attached_image_1.png)\n\nFor the general case, start from a final configuration (that is, an arrangement of circles and segments in which no further segments can be drawn). The idea of the solution is to continuously resize and move the circles around the plane, one by one (in particular, making sure we never have 4 circles with a common tangent line), and show that the number of segments drawn remains constant as the picture changes. This way, we can reduce any circle/segment configuration to the particular one mentioned above, and the final number of segments must remain at $3 n-3$.\n\nSome preliminary considerations: look at all possible tangent segments joining any two circles. A segment that is tangent to a circle $A$ can do so in two possible orientations - it may come out of $A$ in clockwise or counterclockwise orientation. Two segments touching the same circle with the same orientation will never intersect each other. Each pair $(A, B)$ of circles has 4 choices of tangent segments, which can be identified by their orientations - for example, $(A+, B-)$ would be the segment which comes out of $A$ in clockwise orientation and comes out of $B$ in counterclockwise orientation. In total, we have $2 n(n-1)$ possible segments, disregarding intersections.\n\nNow we pick a circle $C$ and start to continuously move and resize it, maintaining all existing tangent segments according to their identifications, including those involving $C$. We can keep our choice of tangent segments until the configuration reaches a transition. We lose nothing if we assume that $C$ is kept at least $\\varepsilon$ units away from any other circle, where $\\varepsilon$ is a positive, fixed constant; therefore at a transition either: (1) a currently drawn tangent segment $t$ suddenly becomes obstructed; or (2) a currently absent tangent segment $t$ suddenly becomes unobstructed and available.\n\nClaim. A transition can only occur when three circles $C_{1}, C_{2}, C_{3}$ are tangent to a common line $\\ell$ containing $t$, in a way such that the three tangent segments lying on $\\ell$ (joining the three circles pairwise) are not obstructed by any other circles or tangent segments (other than $C_{1}, C_{2}, C_{3}$).\n\nProof. Since (2) is effectively the reverse of (1), it suffices to prove the claim for (1). Suppose $t$ has suddenly become obstructed, and let us consider two cases.\n\nCase 1: $t$ becomes obstructed by a circle\n\n![](attached_image_2.png)\n\nThen the new circle becomes the third circle tangent to $\\ell$, and no other circles or tangent segments are obstructing $t$.\n\nCase 2: $t$ becomes obstructed by another tangent segment $t'$\n\nWhen two segments $t$ and $t'$ first intersect each other, they must do so at a vertex of one of them. But if a vertex of $t'$ first crossed an interior point of $t$, the circle associated to this vertex was already blocking $t$ (absurd), or is about to (we already took care of this in case 1). So we only have to analyze the possibility of $t$ and $t'$ suddenly having a common vertex. However, if that happens, this vertex must belong to a single circle (remember we are keeping different circles at least $\\varepsilon$ units apart from each other throughout the moving/resizing process), and therefore they must have different orientations with respect to that circle.\n\n![](attached_image_3.png)\n\nThus, at the transition moment, both $t$ and $t'$ are tangent to the same circle at a common point, that is, they must be on the same line $\\ell$ and hence we again have three circles simultaneously tangent to $\\ell$. Also no other circles or tangent segments are obstructing $t$ or $t'$ (otherwise, they would have disappeared before this transition). $\\square$\n\nNext, we focus on the maximality of a configuration immediately before and after a transition, where three circles share a common tangent line $\\ell$. Let the three circles be $C_{1}, C_{2}, C_{3}$, ordered by their tangent points. The only possibly affected segments are the ones lying on $\\ell$, namely $t_{12}, t_{23}$ and $t_{13}$. Since $C_{2}$ is in the middle, $t_{12}$ and $t_{23}$ must have different orientations with respect to $C_{2}$. For $C_{1}$, $t_{12}$ and $t_{13}$ must have the same orientation, while for $C_{3}$, $t_{13}$ and $t_{23}$ must have the same orientation. The figure below summarizes the situation, showing alternative positions for $C_{1}$ (namely, $C_{1}$ and $C_{1}'$ ) and for $C_{3}$ ($C_{3}$ and $C_{3}'$ ).\n\n![](attached_image_4.png)\n\nNow perturb the diagram slightly so the three circles no longer have a common tangent, while preserving the definition of $t_{12}, t_{23}$ and $t_{13}$ according to their identifications. First note that no other circles or tangent segments can obstruct any of these segments. Also recall that tangent segments joining the same circle at the same orientation will never obstruct each other.\n\nThe availability of the tangent segments can now be checked using simple diagrams.\n\nCase 1: $t_{13}$ passes through $C_{2}$\n\n![](attached_image_5.png)\n\nIn this case, $t_{13}$ is not available, but both $t_{12}$ and $t_{23}$ are.\n\nCase 2: $t_{13}$ does not pass through $C_{2}$\n\n![](attached_image_6.png)\n\nNow $t_{13}$ is available, but $t_{12}$ and $t_{23}$ obstruct each other, so only one can be drawn.\n\nIn any case, exactly 2 out of these 3 segments can be drawn. Thus the maximal number of segments remains constant as we move or resize the circles, and we are done.\n\nAnswer: If there were $n$ circles, there would always be exactly $3(n-1)$ segments; so the only possible answer is $3 \\cdot 2017-3=6048$.\nFirst note that all tangent segments lying on the boundary of the convex hull of the circles are always drawn since they do not intersect anything else. Now in the final picture, aside from the $n$ circles, the blackboard is divided into regions. We can consider the picture as a plane (multi-)graph $G$ in which the circles are the vertices and the tangent segments are the edges. The idea of this solution is to find a relation between the number of edges and the number of regions in $G$; then, once we prove that $G$ is connected, we can use Euler's formula to finish the problem.\n\nThe boundary of each region consists of 1 or more (for now) simple closed curves, each made of arcs and tangent segments. The segment and the arc might meet smoothly (as in $S_{i}$, $i=1,2, \\ldots, 6$ in the figure below) or not (as in $P_{1}, P_{2}, P_{3}, P_{4}$; call such points sharp corners of the boundary). In other words, if a person walks along the border, her direction would suddenly turn an angle of $\\pi$ at a sharp corner.\n\n![](attached_image_7.png)\n\nClaim 1. The outer boundary $B_{1}$ of any internal region has at least 3 sharp corners.\n\nProof. Let a person walk one lap along $B_{1}$ in the counterclockwise orientation. As she does so, she will turn clockwise as she moves along the circle arcs, and not turn at all when moving along the lines. On the other hand, her total rotation after one lap is $2 \\pi$ in the counterclockwise direction! Where could she be turning counterclockwise? She can only do so at sharp corners, and, even then, she turns only an angle of $\\pi$ there. But two sharp corners are not enough, since at least one arc must be present-so she must have gone through at least 3 sharp corners. $\\square$\n\nClaim 2. Each internal region is simply connected, that is, has only one boundary curve.\n\nProof. Suppose, by contradiction, that some region has an outer boundary $B_{1}$ and inner boundaries $B_{2}, B_{3}, \\ldots, B_{m}(m \\geqslant 2)$. Let $P_{1}$ be one of the sharp corners of $B_{1}$.\n\nNow consider a car starting at $P_{1}$ and traveling counterclockwise along $B_{1}$. It starts in reverse, i.e., it is initially facing the corner $P_{1}$. Due to the tangent conditions, the car may travel in a way so that its orientation only changes when it is moving along an arc. In particular, this means the car will sometimes travel forward. For example, if the car approaches a sharp corner when driving in reverse, it would continue travel forward after the corner, instead of making an immediate half-turn. This way, the orientation of the car only changes in a clockwise direction since the car always travels clockwise around each arc.\n\nNow imagine there is a laser pointer at the front of the car, pointing directly ahead. Initially, the laser endpoint hits $P_{1}$, but, as soon as the car hits an arc, the endpoint moves clockwise around $B_{1}$. In fact, the laser endpoint must move continuously along $B_{1}$ ! Indeed, if the endpoint ever jumped (within $B_{1}$, or from $B_{1}$ to one of the inner boundaries), at the moment of the jump the interrupted laser would be a drawable tangent segment that Luciano missed (see figure below for an example).\n\n![](attached_image_8.png)\n\nNow, let $P_{2}$ and $P_{3}$ be the next two sharp corners the car goes through, after $P_{1}$ (the previous lemma assures their existence). At $P_{2}$ the car starts moving forward, and at $P_{3}$ it will start to move in reverse again. So, at $P_{3}$, the laser endpoint is at $P_{3}$ itself. So while the car moved counterclockwise between $P_{1}$ and $P_{3}$, the laser endpoint moved clockwise between $P_{1}$ and $P_{3}$. That means the laser beam itself scanned the whole region within $B_{1}$, and it should have crossed some of the inner boundaries.\n\nClaim 3. Each region has exactly 3 sharp corners.\n\nProof. Consider again the car of the previous claim, with its laser still firmly attached to its front, traveling the same way as before and going through the same consecutive sharp corners $P_{1}, P_{2}$ and $P_{3}$. As we have seen, as the car goes counterclockwise from $P_{1}$ to $P_{3}$, the laser endpoint goes clockwise from $P_{1}$ to $P_{3}$, so together they cover the whole boundary. If there were a fourth sharp corner $P_{4}$, at some moment the laser endpoint would pass through it. But, since $P_{4}$ is a sharp corner, this means the car must be on the extension of a tangent segment going through $P_{4}$. Since the car is not on that segment itself (the car never goes through $P_{4}$ ), we would have 3 circles with a common tangent line, which is not allowed.\n\n![](attached_image_9.png)\n\nWe are now ready to finish the solution. Let $r$ be the number of internal regions, and $s$ be the number of tangent segments. Since each tangent segment contributes exactly 2 sharp corners to the diagram, and each region has exactly 3 sharp corners, we must have $2 s=3 r$. Since the graph corresponding to the diagram is connected, we can use Euler's formula $n-s+r=1$ and find $s=3 n-3$ and $r=2 n-2$.\n\nAnswer: If there were $n$ circles, there would always be exactly $3(n-1)$ segments; so the only possible answer is $3 \\cdot 2017-3=6048$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32049,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\geqslant 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\\{0,1, \\ldots, p-1\\}$ that was not chosen before by either of the two players and then chooses an element $a_{i}$ of the set $\\{0,1,2,3,4,5,6,7,8,9\\}$. Eduardo has the first move. The game ends after all the indices $i \\in\\{0,1, \\ldots, p-1\\}$ have been chosen. Then the following number is computed:\n$$\nM=a_{0}+10 \\cdot a_{1}+\\cdots+10^{p-1} \\cdot a_{p-1}=\\sum_{j=0}^{p-1} a_{j} \\cdot 10^{j}\n$$\nThe goal of Eduardo is to make the number $M$ divisible by $p$, and the goal of Fernando is to prevent this.\nProve that Eduardo has a winning strategy.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_{0}=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple of $10$.\n\nNow assume that the prime number $p$ does not belong to $\\{2,5\\}$. Eduardo chooses $i=p-1$ and $a_{p-1}=0$ in the first move. By Fermat's Little Theorem, $\\left(10^{(p-1) / 2}\\right)^{2}=10^{p-1} \\equiv 1\\pmod{p}$, so $p \\mid\\left(10^{(p-1) / 2}\\right)^{2}-1=\\left(10^{(p-1) / 2}+1\\right)\\left(10^{(p-1) / 2}-1\\right)$. Since $p$ is prime, either $p \\mid 10^{(p-1) / 2}+1$ or $p \\mid 10^{(p-1) / 2}-1$. Thus we have two cases:\n\nCase a: $10^{(p-1) / 2} \\equiv -1\\pmod{p}$\n\nIn this case, for each move $\\left(i, a_{i}\\right)$ of Fernando, Eduardo immediately makes the move $\\left(j, a_{j}\\right)= \\left(i+\\frac{p-1}{2}, a_{i}\\right)$, if $0 \\leqslant i \\leqslant \\frac{p-3}{2}$, or $\\left(j, a_{j}\\right)=\\left(i-\\frac{p-1}{2}, a_{i}\\right)$, if $\\frac{p-1}{2} \\leqslant i \\leqslant p-2$. We will have $10^{j} \\equiv -10^{i} \\pmod{p}$, and so $a_{j} \\cdot 10^{j}=a_{i} \\cdot 10^{j} \\equiv -a_{i} \\cdot 10^{i}\\pmod{p}$. Notice that this move by Eduardo is always possible. Indeed, immediately before a move by Fernando, for any set of the type $\\{r, r+(p-1) / 2\\}$ with $0 \\leqslant r \\leqslant(p-3) / 2$, either no element of this set was chosen as an index by the players in the previous moves or else both elements of this set were chosen as indices by the players in the previous moves. Therefore, after each of his moves, Eduardo always makes the sum of the numbers $a_{k} \\cdot 10^{k}$ corresponding to the already chosen pairs $\\left(k, a_{k}\\right)$ divisible by $p$, and thus wins the game.\n\nCase b: $10^{(p-1) / 2} \\equiv 1\\pmod{p}$\n\nIn this case, for each move $\\left(i, a_{i}\\right)$ of Fernando, Eduardo immediately makes the move $\\left(j, a_{j}\\right)= \\left(i+\\frac{p-1}{2}, 9-a_{i}\\right)$, if $0 \\leqslant i \\leqslant \\frac{p-3}{2}$, or $\\left(j, a_{j}\\right)=\\left(i-\\frac{p-1}{2}, 9-a_{i}\\right)$, if $\\frac{p-1}{2} \\leqslant i \\leqslant p-2$. The same argument as above shows that Eduardo can always make such move. We will have $10^{j} \\equiv 10^{i} \\pmod{p}$, and so $a_{j} \\cdot 10^{j}+a_{i} \\cdot 10^{i} \\equiv\\left(a_{i}+a_{j}\\right) \\cdot 10^{i}=9 \\cdot 10^{i}\\pmod{p}$. Therefore, at the end of the game, the sum of all terms $a_{k} \\cdot 10^{k}$ will be congruent to\n$$\n\\sum_{i=0}^{\\frac{p-3}{2}} 9 \\cdot 10^{i}=10^{(p-1) / 2}-1 \\equiv 0 \\quad\\pmod{p}\n$$\nand Eduardo wins the game.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32050,
    "subject": "Mathematics (Multi-modal)",
    "question": "Call a rational number short if it has finitely many digits in its decimal expansion. For a positive integer $m$, we say that a positive integer $t$ is $m$-tastic if there exists a number $c \\in \\{1,2,3, \\ldots, 2017\\}$ such that $\\frac{10^{t}-1}{c \\cdot m}$ is short, and such that $\\frac{10^{k}-1}{c \\cdot m}$ is not short for any $1 \\leqslant k<t$. Let $S(m)$ be the set of $m$-tastic numbers. Consider $S(m)$ for $m=1,2, \\ldots$. What is the maximum number of elements in $S(m)$ ?",
    "options": [],
    "answer": "807",
    "solution": "If $m=2^{a} \\cdot 5^{b} \\cdot s$, with $\\operatorname{gcd}(s, 10)=1$, then $\\frac{10^{t}-1}{m}$ is short if and only if $s$ divides $10^{t}-1$. So we may (and will) suppose without loss of generality that $\\operatorname{gcd}(m, 10)=1$. Define\n$$\nC=\\{1 \\leqslant c \\leqslant 2017: \\operatorname{gcd}(c, 10)=1\\} .\n$$\nThe $m$-tastic numbers are then precisely the smallest exponents $t>0$ such that $10^{t} \\equiv 1 \\pmod{c m}$ for some integer $c \\in C$, that is, the set of orders of $10$ modulo $c m$. In other words,\n$$\nS(m)=\\left\\{\\operatorname{ord}_{c m}(10): c \\in C\\right\\} .\n$$\nSince there are $4 \\cdot 201+3=807$ numbers $c$ with $1 \\leqslant c \\leqslant 2017$ and $\\operatorname{gcd}(c, 10)=1$, namely those such that $c \\equiv 1,3,7,9 \\pmod{10}$,\n$$\n|S(m)| \\leqslant |C|=807 .\n$$\nNow we find $m$ such that $|S(m)|=807$. Let\n$$\nP=\\{1<p \\leqslant 2017: p \\text{ is prime, } p \\neq 2,5\\}\n$$\nand choose a positive integer $\\alpha$ such that every $p \\in P$ divides $10^{\\alpha}-1$ (e.g. $\\alpha=\\varphi(T)$, $T$ being the product of all primes in $P$), and let $m=10^{\\alpha}-1$.\n\nClaim. For every $c \\in C$, we have\n$$\n\\operatorname{ord}_{c m}(10)=c \\alpha\n$$\nAs an immediate consequence, this implies $|S(m)|=|C|=807$, finishing the problem.\n\nProof. Obviously $\\operatorname{ord}_{m}(10)=\\alpha$. Let $t=\\operatorname{ord}_{c m}(10)$. Then\n$$\nc m \\mid 10^{t}-1 \\quad \\Longrightarrow \\quad m \\mid 10^{t}-1 \\quad \\Longrightarrow \\quad \\alpha \\mid t .\n$$\nHence $t=k \\alpha$ for some $k \\in \\mathbb{Z}_{>0}$. We will show that $k=c$.\n\nDenote by $\\nu_{p}(n)$ the number of prime factors $p$ in $n$, that is, the maximum exponent $\\beta$ for which $p^{\\beta} \\mid n$. For every $\\ell \\geqslant 1$ and $p \\in P$, the Lifting the Exponent Lemma provides\n$$\n\\nu_{p}\\left(10^{\\ell \\alpha}-1\\right)=\\nu_{p}\\left(\\left(10^{\\alpha}\\right)^{\\ell}-1\\right)=\\nu_{p}\\left(10^{\\alpha}-1\\right)+\\nu_{p}(\\ell)=\\nu_{p}(m)+\\nu_{p}(\\ell)\n$$\nSo\n$$\n\\begin{aligned}\nc m \\mid 10^{k \\alpha}-1 & \\Longleftrightarrow \\forall p \\in P ; \\quad \\nu_{p}(c m) \\leqslant \\nu_{p}\\left(10^{k \\alpha}-1\\right) \\\\\n& \\Longleftrightarrow \\forall p \\in P ; \\quad \\nu_{p}(m)+\\nu_{p}(c) \\leqslant \\nu_{p}(m)+\\nu_{p}(k) \\\\\n& \\Longleftrightarrow \\forall p \\in P ; \\quad \\nu_{p}(c) \\leqslant \\nu_{p}(k) \\\\\n& \\Longleftrightarrow c \\mid k .\n\\end{aligned}\n$$\nThe first such $k$ is $k=c$, so $\\operatorname{ord}_{c m}(10)=c \\alpha$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32051,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(p, q)$ of prime numbers with $p > q$ for which the number\n$$\n\\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}\n$$\nis an integer.",
    "options": [],
    "answer": "(3, 2)",
    "solution": "Let $M = (p+q)^{p-q}(p-q)^{p+q} - 1$, which is relatively prime with both $p+q$ and $p-q$. Denote by $(p-q)^{-1}$ the multiplicative inverse of $(p-q)$ modulo $M$.\nBy eliminating the term $-1$ in the numerator,\n$$\n\\begin{align*}\n(p+q)^{p+q}(p-q)^{p-q}-1 &\\equiv (p+q)^{p-q}(p-q)^{p+q}-1 \\quad (\\bmod M) \\\\\n(p+q)^{2q} &\\equiv (p-q)^{2q} \\quad (\\bmod M)  \\tag{1}\\\\\n\\left((p+q) \\cdot (p-q)^{-1}\\right)^{2q} &\\equiv 1 \\quad (\\bmod M) \\tag{2}\n\\end{align*}\n$$\n\nCase 1: $q \\geqslant 5$.\nConsider an arbitrary prime divisor $r$ of $M$. Notice that $M$ is odd, so $r \\geqslant 3$. By (2), the multiplicative order of $\\left((p+q) \\cdot (p-q)^{-1}\\right)$ modulo $r$ is a divisor of the exponent $2q$ in (2), so it can be $1, 2, q$ or $2q$.\nBy Fermat's theorem, the order divides $r-1$. So, if the order is $q$ or $2q$ then $r \\equiv 1 (\\bmod q)$. If the order is $1$ or $2$ then $r \\mid (p+q)^2 - (p-q)^2 = 4pq$, so $r = p$ or $r = q$. The case $r = p$ is not possible, because, by applying Fermat's theorem,\n$$\nM = (p+q)^{p-q}(p-q)^{p+q} - 1 \\equiv q^{p-q}(-q)^{p+q} - 1 = \\left(q^2\\right)^p - 1 \\equiv q^2 - 1 = (q+1)(q-1) \\pmod{p}\n$$\nand the last factors $q-1$ and $q+1$ are less than $p$ and thus $p \\nmid M$. Hence, all prime divisors of $M$ are either $q$ or of the form $kq+1$; it follows that all positive divisors of $M$ are congruent to $0$ or $1$ modulo $q$.\nNow notice that\n$$\nM = \\left((p+q)^{\\frac{p-q}{2}}(p-q)^{\\frac{p+q}{2}} - 1\\right)\\left((p+q)^{\\frac{p-q}{2}}(p-q)^{\\frac{p+q}{2}} + 1\\right)\n$$\nis the product of two consecutive positive odd numbers; both should be congruent to $0$ or $1$ modulo $q$. But this is impossible by the assumption $q \\geqslant 5$. So, there is no solution in Case 1.\n\nCase 2: $q = 2$.\nBy (1), we have $M \\mid (p+q)^{2q} - (p-q)^{2q} = (p+2)^4 - (p-2)^4$, so\n$$\n\\begin{gathered}\n(p+2)^{p-2}(p-2)^{p+2} - 1 = M \\leqslant (p+2)^4 - (p-2)^4 \\leqslant (p+2)^4 - 1, \\\\\n(p+2)^{p-6}(p-2)^{p+2} \\leqslant 1.\n\\end{gathered}\n$$\nIf $p \\geqslant 7$ then the left-hand side is obviously greater than $1$. For $p = 5$ we have $(p+2)^{p-6}(p-2)^{p+2} = 7^{-1} \\cdot 3^7$ which is also too large.\nThere remains only one candidate, $p = 3$, which provides a solution:\n$$\n\\frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1} = \\frac{5^5 \\cdot 1^1 - 1}{5^1 \\cdot 1^5 - 1} = \\frac{3124}{4} = 781.\n$$\nSo in Case 2 the only solution is $(p, q) = (3, 2)$.\n\nCase 3: $q = 3$.\nSimilarly to Case 2, we have\n$$\nM \\left\\lvert\\, (p+q)^{2q} - (p-q)^{2q} = 64 \\cdot \\left(\\left(\\frac{p+3}{2}\\right)^6 - \\left(\\frac{p-3}{2}\\right)^6\\right) .\\right.\n$$\nSince $M$ is odd, we conclude that\n$$\nM \\left\\lvert\\, \\left(\\frac{p+3}{2}\\right)^6 - \\left(\\frac{p-3}{2}\\right)^6\\right.\n$$\nand\n$$\n\\begin{gathered}\n(p+3)^{p-3}(p-3)^{p+3} - 1 = M \\leqslant \\left(\\frac{p+3}{2}\\right)^6 - \\left(\\frac{p-3}{2}\\right)^6 \\leqslant \\left(\\frac{p+3}{2}\\right)^6 - 1, \\\\\n64(p+3)^{p-9}(p-3)^{p+3} \\leqslant 1.\n\\end{gathered}\n$$\nIf $p \\geqslant 11$ then the left-hand side is obviously greater than $1$. If $p = 7$ then the left-hand side is $64 \\cdot 10^{-2} \\cdot 4^{10} > 1$. If $p = 5$ then the left-hand side is $64 \\cdot 8^{-4} \\cdot 2^8 = 2^2 > 1$. Therefore, there is no solution in Case 3.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32052,
    "subject": "Mathematics (Multi-modal)",
    "question": "Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1, such that $f(x, y)=1$ for each $(x, y)$ in the set $S$.\n\nNote: A homogenous polynomial of degree $n$ is any nonzero polynomial of the form\n$$\nf(x, y)=a_{0} x^{n}+a_{1} x^{n-1} y+a_{2} x^{n-2} y^{2}+\\cdots+a_{n-1} x y^{n-1}+a_{n} y^{n} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \\pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $(x_{1}, y_{1})$ through $(x_{n}, y_{n})$. If any two of these lattice points $(x_{i}, y_{i})$ and $(x_{j}, y_{j})$ lie on the same line through the origin, then $(x_{j}, y_{j})=(-x_{i},-y_{i})$ because both of the points are irreducible. We then have $f(x_{j}, y_{j})= \\pm f(x_{i}, y_{i})$ whenever $f$ is homogenous, so we can assume that no two of the lattice points are collinear with the origin by ignoring the extra lattice points.\n\nConsider the homogenous polynomials $\\ell_{i}(x, y)=y_{i} x-x_{i} y$ and define\n$$\ng_{i}(x, y)=\\prod_{j \\neq i} \\ell_{j}(x, y) .\n$$\nThen $\\ell_{i}(x_{j}, y_{j})=0$ if and only if $j=i$, because there is only one lattice point on each line through the origin. Thus, $g_{i}(x_{j}, y_{j})=0$ for all $j \\neq i$. Define $a_{i}=g_{i}(x_{i}, y_{i})$, and note that $a_{i} \\neq 0$.\n\nNote that $g_{i}(x, y)$ is a degree $n-1$ polynomial with the following two properties:\n1. $g_{i}(x_{j}, y_{j})=0$ if $j \\neq i$.\n2. $g_{i}(x_{i}, y_{i})=a_{i}$.\n\nFor any $N \\geqslant n-1$, there also exists a polynomial of degree $N$ with the same two properties. Specifically, let $I_{i}(x, y)$ be a degree 1 homogenous polynomial such that $I_{i}(x_{i}, y_{i})=1$, which exists since $(x_{i}, y_{i})$ is irreducible. Then $I_{i}(x, y)^{N-(n-1)} g_{i}(x, y)$ satisfies both of the above properties and has degree $N$.\n\nWe may now reduce the problem to the following claim:\n**Claim:** For each positive integer $a$, there is a homogenous polynomial $f_{a}(x, y)$, with integer coefficients, of degree at least 1, such that $f_{a}(x, y) \\equiv 1\\pmod{a}$ for all relatively prime $(x, y)$.\n\nTo see that this claim solves the problem, take $a$ to be the least common multiple of the numbers $a_{i}$ ($1 \\leqslant i \\leqslant n$). Take $f_{a}$ given by the claim, choose some power $f_{a}(x, y)^{k}$ that has degree at least $n-1$, and subtract appropriate multiples of the $g_{i}$ constructed above to obtain the desired polynomial.\n\nWe prove the claim by factoring $a$. First, if $a$ is a power of a prime ($a=p^{k}$), then we may choose either:\n- $f_{a}(x, y)=(x^{p-1}+y^{p-1})^{\\phi(a)}$ if $p$ is odd;\n- $f_{a}(x, y)=(x^{2}+x y+y^{2})^{\\phi(a)}$ if $p=2$.\n\nNow suppose $a$ is any positive integer, and let $a=q_{1} q_{2} \\cdots q_{k}$, where the $q_{i}$ are prime powers, pairwise relatively prime. Let $f_{q_{i}}$ be the polynomials just constructed, and let $F_{q_{i}}$ be powers of these that all have the same degree. Note that\n$$\n\\frac{a}{q_{i}} F_{q_{i}}(x, y) \\equiv \\frac{a}{q_{i}}\\pmod{a}\n$$\nfor any relatively prime $x, y$. By Bézout's lemma, there is an integer linear combination of the $\\frac{a}{q_{i}}$ that equals 1. Thus, there is a linear combination of the $F_{q_{i}}$ such that $F_{q_{i}}(x, y) \\equiv 1 \\pmod{a}$ for any relatively prime $(x, y)$; and this polynomial is homogenous because all the $F_{q_{i}}$ have the same degree.\nAs in the previous solution, label the irreducible lattice points $(x_{1}, y_{1}), \\ldots,(x_{n}, y_{n})$ and assume without loss of generality that no two of the points are collinear with the origin. We induct on $n$ to construct a homogenous polynomial $f(x, y)$ such that $f(x_{i}, y_{i})=1$ for all $1 \\leqslant i \\leqslant n$.\n\nIf $n=1$ : Since $x_{1}$ and $y_{1}$ are relatively prime, there exist some integers $c, d$ such that $c x_{1}+d y_{1}=1$. Then $f(x, y)=c x+d y$ is suitable.\n\nIf $n \\geqslant 2$ : By the induction hypothesis we already have a homogeneous polynomial $g(x, y)$ with $g(x_{1}, y_{1})=\\ldots=g(x_{n-1}, y_{n-1})=1$. Let $j=\\deg g$,\n$$\ng_{n}(x, y)=\\prod_{k=1}^{n-1}(y_{k} x-x_{k} y)\n$$\nand $a_{n}=g_{n}(x_{n}, y_{n})$. By assumption, $a_{n} \\neq 0$. Take some integers $c, d$ such that $c x_{n}+d y_{n}=1$. We will construct $f(x, y)$ in the form\n$$\nf(x, y)=g(x, y)^{K}-C \\cdot g_{n}(x, y) \\cdot(c x+d y)^{L}\n$$\nwhere $K$ and $L$ are some positive integers and $C$ is some integer. We assume that $L=K j-n+1$ so that $f$ is homogenous.\n\nDue to $g(x_{1}, y_{1})=\\ldots=g(x_{n-1}, y_{n-1})=1$ and $g_{n}(x_{1}, y_{1})=\\ldots=g_{n}(x_{n-1}, y_{n-1})=0$, the property $f(x_{1}, y_{1})=\\ldots=f(x_{n-1}, y_{n-1})=1$ is automatically satisfied with any choice of $K, L$, and $C$.\n\nFurthermore,\n$$\nf(x_{n}, y_{n})=g(x_{n}, y_{n})^{K}-C \\cdot g_{n}(x_{n}, y_{n}) \\cdot(c x_{n}+d y_{n})^{L}=g(x_{n}, y_{n})^{K}-C a_{n} .\n$$\nIf we have an exponent $K$ such that $g(x_{n}, y_{n})^{K} \\equiv 1\\pmod{a_{n}}$, then we may choose $C$ such that $f(x_{n}, y_{n})=1$. We now choose such a $K$.\n\nConsider an arbitrary prime divisor $p$ of $a_{n}$. By\n$$\np \\mid a_{n}=g_{n}(x_{n}, y_{n})=\\prod_{k=1}^{n-1}(y_{k} x_{n}-x_{k} y_{n})\n$$\nthere is some $1 \\leqslant k<n$ such that $x_{k} y_{n} \\equiv x_{n} y_{k}\\pmod{p}$. We first show that $x_{k} x_{n}$ or $y_{k} y_{n}$ is relatively prime with $p$. This is trivial in the case $x_{k} y_{n} \\equiv x_{n} y_{k} \\not \\equiv 0\\pmod{p}$. In the other case, we have $x_{k} y_{n} \\equiv x_{n} y_{k} \\equiv 0\\pmod{p}$. If, say $p \\mid x_{k}$, then $p \\nmid y_{k}$ because $(x_{k}, y_{k})$ is irreducible, so $p \\mid x_{n}$; then $p \\nmid y_{n}$ because $(x_{n}, y_{n})$ is irreducible. In summary, $p \\mid x_{k}$ implies $p \\nmid y_{k} y_{n}$. Similarly, $p \\mid y_{n}$ implies $p \\nmid x_{k} x_{n}$.\n\nBy the homogeneity of $g$ we have the congruences\n$$\nx_{k}^{d} \\cdot g(x_{n}, y_{n})=g(x_{k} x_{n}, x_{k} y_{n}) \\equiv g(x_{k} x_{n}, y_{k} x_{n})=x_{n}^{d} \\cdot g(x_{k}, y_{k})=x_{n}^{d}\\pmod{p}\n$$\nand\n$$\ny_{k}^{d} \\cdot g(x_{n}, y_{n})=g(y_{k} x_{n}, y_{k} y_{n}) \\equiv g(x_{k} y_{n}, y_{k} y_{n})=y_{n}^{d} \\cdot g(x_{k}, y_{k})=y_{n}^{d} \\pmod{p} .\n$$\nIf $p \\nmid x_{k} x_{n}$, then take the $(p-1)^{st}$ power of the first congruence; otherwise take the $(p-1)^{st}$ power of the second; by Fermat's theorem, in both cases we get\n$$\ng(x_{n}, y_{n})^{p-1} \\equiv 1 \\pmod{p}\n$$\nIf $p^{\\alpha} \\mid m$, then we have\n$$\ng(x_{n}, y_{n})^{p^{\\alpha-1}(p-1)} \\equiv 1 \\pmod{p^{\\alpha}}\n$$\nwhich implies that the exponent $K=n \\cdot \\varphi(a_{n})$, which is a multiple of all $p^{\\alpha-1}(p-1)$, is a suitable choice. (The factor $n$ is added only so that $K \\geqslant n$ and so $L>0$.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32053,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be an odd prime number and $\\mathbb{Z}_{>0}$ be the set of positive integers. Suppose that a function $f: \\mathbb{Z}_{>0} \\times \\mathbb{Z}_{>0} \\rightarrow \\{0,1\\}$ satisfies the following properties:\n- $f(1,1)=0$;\n- $f(a, b)+f(b, a)=1$ for any pair of relatively prime positive integers $(a, b)$ not both equal to $1$;\n- $f(a+b, b)=f(a, b)$ for any pair of relatively prime positive integers $(a, b)$.\nProve that\n$$\n\\sum_{n=1}^{p-1} f\\left(n^{2}, p\\right) \\geqslant \\sqrt{2 p}-2 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $\\mathbb{A}$ the set of all pairs of coprime positive integers. Notice that for every $(a, b) \\in \\mathbb{A}$ there exists a pair $(u, v) \\in \\mathbb{Z}^{2}$ with $u a+v b=1$. Moreover, if $(u_{0}, v_{0})$ is one such pair, then all such pairs are of the form $(u, v)=\\left(u_{0}+k b, v_{0}-k a\\right)$, where $k \\in \\mathbb{Z}$. So there exists a unique such pair $(u, v)$ with $-b / 2<u \\leqslant b / 2$; we denote this pair by $(u, v)=g(a, b)$.\n\nLemma. Let $(a, b) \\in \\mathbb{A}$ and $(u, v)=g(a, b)$. Then $f(a, b)=1 \\Longleftrightarrow u>0$.\n\nProof. We induct on $a+b$. The base case is $a+b=2$. In this case, we have that $a=b=1$, $g(a, b)=g(1,1)=(0,1)$ and $f(1,1)=0$, so the claim holds.\n\nAssume now that $a+b>2$, and so $a \\neq b$, since $a$ and $b$ are coprime. Two cases are possible.\n\nCase 1: $a>b$.\nNotice that $g(a-b, b)=(u, v+u)$, since $u(a-b)+(v+u) b=1$ and $u \\in(-b / 2, b / 2]$. Thus $f(a, b)=1 \\Longleftrightarrow f(a-b, b)=1 \\Longleftrightarrow u>0$ by the induction hypothesis.\n\nCase 2: $a<b$. (Then, clearly, $b \\geqslant 2$.)\nNow we estimate $v$. Since $v b=1-u a$, we have\n$$\n1+\\frac{a b}{2}>v b \\geqslant 1-\\frac{a b}{2}, \\quad \\text{ so } \\quad \\frac{1+a}{2} \\geqslant \\frac{1}{b}+\\frac{a}{2}>v \\geqslant \\frac{1}{b}-\\frac{a}{2}>-\\frac{a}{2} .\n$$\nThus $1+a>2 v>-a$, so $a \\geqslant 2 v>-a$, hence $a / 2 \\geqslant v>-a / 2$, and thus $g(b, a)=(v, u)$.\n\nObserve that $f(a, b)=1 \\Longleftrightarrow f(b, a)=0 \\Longleftrightarrow f(b-a, a)=0$. We know from Case 1 that $g(b-a, a)=(v, u+v)$. We have $f(b-a, a)=0 \\Longleftrightarrow v \\leqslant 0$ by the inductive hypothesis. Then, since $b>a \\geqslant 1$ and $u a+v b=1$, we have $v \\leqslant 0 \\Longleftrightarrow u>0$, and we are done.\n\nThe Lemma proves that, for all $(a, b) \\in \\mathbb{A}$, $f(a, b)=1$ if and only if the inverse of $a$ modulo $b$, taken in $\\{1,2, \\ldots, b-1\\}$, is at most $b / 2$. Then, for any odd prime $p$ and integer $n$ such that $n \\not \\equiv 0(\\bmod p)$, $f\\left(n^{2}, p\\right)=1$ iff the inverse of $n^{2} \\bmod p$ is less than $p / 2$. Since $\\left\\{n^{2} \\bmod p: 1 \\leqslant n \\leqslant p-1\\right\\}=\\left\\{n^{-2} \\bmod p: 1 \\leqslant n \\leqslant p-1\\right\\}$, including multiplicities (two for each quadratic residue in each set), we conclude that the desired sum is twice the number of quadratic residues that are less than $p / 2$, i.e.,\n$$\n\\left.\\sum_{n=1}^{p-1} f\\left(n^{2}, p\\right)=2 \\left\\lvert\\,\\left\\{k: 1 \\leqslant k \\leqslant \\frac{p-1}{2} \\text{ and } k^{2} \\bmod p<\\frac{p}{2}\\right\\}\\right. \\right\\rvert\\, .\n$$\nSince the number of perfect squares in the interval $[1, p / 2)$ is $\\lfloor\\sqrt{p / 2}\\rfloor>\\sqrt{p / 2}-1$, we conclude that\n$$\n\\sum_{n=1}^{p-1} f\\left(n^{2}, p\\right)>2\\left(\\sqrt{\\frac{p}{2}}-1\\right)=\\sqrt{2 p}-2 .\n$$\nWe provide a different proof for the Lemma. For this purpose, we use continued fractions to find $g(a, b)=(u, v)$ explicitly.\n\nThe function $f$ is completely determined on $\\mathbb{A}$ by the following\n\nClaim. Represent $a / b$ as a continued fraction; that is, let $a_{0}$ be an integer and $a_{1}, \\ldots, a_{k}$ be positive integers such that $a_{k} \\geqslant 2$ and\n$$\n\\frac{a}{b}=a_{0}+\\frac{1}{a_{1}+\\frac{1}{a_{2}+\\frac{1}{\\cdots+\\frac{1}{a_{k}}}}}=\\left[a_{0} ; a_{1}, a_{2}, \\ldots, a_{k}\\right] .\n$$\nThen $f(a, b)=0 \\Longleftrightarrow k$ is even.\n\nProof. We induct on $b$. If $b=1$, then $a / b=[a]$ and $k=0$. Then, for $a \\geqslant 1$, an easy induction shows that $f(a, 1)=f(1,1)=0$.\n\nNow consider the case $b>1$. Perform the Euclidean division $a=q b+r$, with $0 \\leqslant r<b$. We have $r \\neq 0$ because $\\operatorname{gcd}(a, b)=1$. Hence\n$$\nf(a, b)=f(r, b)=1-f(b, r), \\quad \\frac{a}{b}=\\left[q ; a_{1}, \\ldots, a_{k}\\right], \\quad \\text{ and } \\quad \\frac{b}{r}=\\left[a_{1} ; a_{2}, \\ldots, a_{k}\\right] .\n$$\nThen the number of terms in the continued fraction representations of $a / b$ and $b / r$ differ by one. Since $r<b$, the inductive hypothesis yields\n$$\nf(b, r)=0 \\Longleftrightarrow k-1 \\text{ is even, }\n$$\nand thus\n$$\nf(a, b)=0 \\Longleftrightarrow f(b, r)=1 \\Longleftrightarrow k-1 \\text{ is odd } \\Longleftrightarrow k \\text{ is even. }\n$$\n$\\square$\n\nNow we use the following well-known properties of continued fractions to prove the Lemma:\nLet $p_{i}$ and $q_{i}$ be coprime positive integers with $\\left[a_{0} ; a_{1}, a_{2}, \\ldots, a_{i}\\right]=p_{i} / q_{i}$, with the notation borrowed from the Claim. In particular, $a / b=\\left[a_{0} ; a_{1}, a_{2}, \\ldots, a_{k}\\right]=p_{k} / q_{k}$. Assume that $k>0$ and define $q_{-1}=0$ if necessary. Then\n- $q_{k}=a_{k} q_{k-1}+q_{k-2}$, \\quad and\n- $a q_{k-1}-b p_{k-1}=p_{k} q_{k-1}-q_{k} p_{k-1}=(-1)^{k-1}$.\n\nAssume that $k>0$. Then $a_{k} \\geqslant 2$, and\n$$\nb=q_{k}=a_{k} q_{k-1}+q_{k-2} \\geqslant a_{k} q_{k-1} \\geqslant 2 q_{k-1} \\Longrightarrow q_{k-1} \\leqslant \\frac{b}{2},\n$$\nwith strict inequality for $k>1$, and\n$$\n(-1)^{k-1} q_{k-1} a+(-1)^{k} p_{k-1} b=1 .\n$$\nNow we finish the proof of the Lemma. It is immediate for $k=0$. If $k=1$, then $(-1)^{k-1}=1$, so\n$$\n-b / 2<0 \\leqslant(-1)^{k-1} q_{k-1} \\leqslant b / 2 .\n$$\nIf $k>1$, we have $q_{k-1}<b / 2$, so\n$$\n-b / 2<(-1)^{k-1} q_{k-1}<b / 2 .\n$$\nThus, for any $k>0$, we find that $g(a, b)=\\left((-1)^{k-1} q_{k-1},(-1)^{k} p_{k-1}\\right)$, and so\n$$\nf(a, b)=1 \\Longleftrightarrow k \\text{ is odd } \\Longleftrightarrow u=(-1)^{k-1} q_{k-1}>0 .\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32054,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given any set $S$ of positive integers, show that at least one of the following two assertions holds:\n(1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\\sum_{x \\in F} 1 / x = \\sum_{x \\in G} 1 / x$;\n(2) There exists a positive rational number $r < 1$ such that $\\sum_{x \\in F} 1 / x \\neq r$ for all finite subsets $F$ of $S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. Argue indirectly. Agree, as usual, that the empty sum is $0$ to consider rationals in $[0,1)$; adjoining $0$ causes no harm, since $\\sum_{x \\in F} 1 / x = 0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\\sum_{x \\in F_{r}} 1 / x = r$. The argument hinges on the lemma below.\n\nLemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q - r = 1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$.\n\nProof. If $x$ is a member of $F_{q}$, then\n$$\n\\sum_{y \\in F_{q} \\setminus \\{x\\}} \\frac{1}{y} = \\sum_{y \\in F_{q}} \\frac{1}{y} - \\frac{1}{x} = q - \\frac{1}{x} = r = \\sum_{y \\in F_{r}} \\frac{1}{y}\n$$\nso $F_{r} = F_{q} \\setminus \\{x\\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then\n$$\n\\sum_{y \\in F_{r} \\cup \\{x\\}} \\frac{1}{y} = \\sum_{y \\in F_{r}} \\frac{1}{y} + \\frac{1}{x} = r + \\frac{1}{x} = q = \\sum_{y \\in F_{q}} \\frac{1}{y}\n$$\nso $F_{q} = F_{r} \\cup \\{x\\}$, and $x$ is a member of $F_{q}$.\n\nConsider now an element $x$ of $S$ and a positive rational $r < 1$. Let $n = \\lfloor r x \\rfloor$ and consider the sets $F_{r - k / x}, k = 0, \\ldots, n$. Since $0 \\leqslant r - n / x < 1 / x$, the set $F_{r - n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r - (n - 2k) / x}$ do not contain $x$, whereas the $F_{r - (n - 2k - 1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd.\nSolution 2. A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \\setminus \\{1\\}$. We may and will therefore assume that $S$ consists of integers greater than $1$. Label the elements of $S$ increasingly $x_{1} < x_{2} < \\cdots$, where $x_{1} \\geqslant 2$.\n\nWe first show that $S$ satisfies (2) if $x_{n+1} \\geqslant 2 x_{n}$ for all $n$. In this case, $x_{n} \\geqslant 2^{n-1} x_{1}$ for all $n$, so\n$$\ns = \\sum_{n \\geqslant 1} \\frac{1}{x_{n}} \\leqslant \\sum_{n \\geqslant 1} \\frac{1}{2^{n-1} x_{1}} = \\frac{2}{x_{1}}\n$$\nIf $x_{1} \\geqslant 3$, or $x_{1} = 2$ and $x_{n+1} > 2 x_{n}$ for some $n$, then $\\sum_{x \\in F} 1 / x < s < 1$ for every finite subset $F$ of $S$, so $S$ satisfies (2); and if $x_{1} = 2$ and $x_{n+1} = 2 x_{n}$ for all $n$, that is, $x_{n} = 2^{n}$ for all $n$, then every finite subset $F$ of $S$ consists of powers of $2$, so $\\sum_{x \\in F} 1 / x \\neq 1 / 3$ and again $S$ satisfies (2).\n\nFinally, we deal with the case where $x_{n+1} < 2 x_{n}$ for some $n$. Consider the positive rational $r = 1 / x_{n} - 1 / x_{n+1} < 1 / x_{n+1}$. If $r = \\sum_{x \\in F} 1 / x$ for no finite subset $F$ of $S$, then $S$ satisfies (2).\n\nWe now assume that $r = \\sum_{x \\in F_{0}} 1 / x$ for some finite subset $F_{0}$ of $S$, and show that $S$ satisfies (1). Since $\\sum_{x \\in F_{0}} 1 / x = r < 1 / x_{n+1}$, it follows that $x_{n+1}$ is not a member of $F_{0}$, so\n$$\n\\sum_{x \\in F_{0} \\cup \\{x_{n+1}\\}} \\frac{1}{x} = \\sum_{x \\in F_{0}} \\frac{1}{x} + \\frac{1}{x_{n+1}} = r + \\frac{1}{x_{n+1}} = \\frac{1}{x_{n}}.\n$$\nConsequently, $F = F_{0} \\cup \\{x_{n+1}\\}$ and $G = \\{x_{n}\\}$ are distinct finite subsets of $S$ such that $\\sum_{x \\in F} 1 / x = \\sum_{x \\in G} 1 / x$, and $S$ satisfies (1).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32055,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f:(0, \\infty) \\rightarrow \\mathbb{R}$ satisfying\n$$\n\\left(x+\\frac{1}{x}\\right) f(y)=f(x y)+f\\left(\\frac{y}{x}\\right)\n$$\nfor all $x, y>0$.",
    "options": [],
    "answer": "f(x) = C1*x + C2/x for arbitrary real constants C1 and C2",
    "solution": "Answer: $f(x)=C_{1} x+\\frac{C_{2}}{x}$ with arbitrary constants $C_{1}$ and $C_{2}$.\n\nSolution 1. Fix a real number $a>1$, and take a new variable $t$. For the values $f(t), f\\left(t^{2}\\right)$, $f(a t)$ and $f\\left(a^{2} t^{2}\\right)$, the relation (1) provides a system of linear equations:\n$$\n\\begin{array}{ll}\nx=y=t: & \\left(t+\\frac{1}{t}\\right) f(t) \\\\\nx=\\frac{t}{a}, y=a t: & \\left(\\frac{t}{a}+\\frac{a}{t}\\right) f(a t)=f\\left(t^{2}\\right)+f\\left(t^{2}\\right)+f\\left(a^{2}\\right) \\\\\nx=a^{2} t, y=t: & \\left(a^{2} t+\\frac{1}{a^{2} t}\\right) f(t)=f\\left(a^{2} t^{2}\\right)+f\\left(\\frac{1}{a^{2}}\\right) \\\\\nx=y=a t: & \\left(a t+\\frac{1}{a t}\\right) f(a t)=f\\left(a^{2} t^{2}\\right)+f(1)\n\\end{array}\n$$\nIn order to eliminate $f\\left(t^{2}\\right)$, take the difference of (2a) and (2b); from (2c) and (2d) eliminate $f\\left(a^{2} t^{2}\\right)$; then by taking a linear combination, eliminate $f(a t)$ as well:\n$$\n\\begin{gathered}\n\\left(t+\\frac{1}{t}\\right) f(t)-\\left(\\frac{t}{a}+\\frac{a}{t}\\right) f(a t)=f(1)-f\\left(a^{2}\\right) \\quad \\text{ and } \\\\\n\\left(a^{2} t+\\frac{1}{a^{2} t}\\right) f(t)-\\left(a t+\\frac{1}{a t}\\right) f(a t)=f\\left(1 / a^{2}\\right)-f(1), \\quad \\text{ so } \\\\\n\\left(\\left(a t+\\frac{1}{a t}\\right)\\left(t+\\frac{1}{t}\\right)-\\left(\\frac{t}{a}+\\frac{a}{t}\\right)\\left(a^{2} t+\\frac{1}{a^{2} t}\\right)\\right) f(t) \\\\\n=\\left(a t+\\frac{1}{a t}\\right)\\left(f(1)-f\\left(a^{2}\\right)\\right)-\\left(\\frac{t}{a}+\\frac{a}{t}\\right)\\left(f\\left(1 / a^{2}\\right)-f(1)\\right)\n\\end{gathered}\n$$\nNotice that on the left-hand side, the coefficient of $f(t)$ is nonzero and does not depend on $t$ :\n$$\n\\left(a t+\\frac{1}{a t}\\right)\\left(t+\\frac{1}{t}\\right)-\\left(\\frac{t}{a}+\\frac{a}{t}\\right)\\left(a^{2} t+\\frac{1}{a^{2} t}\\right)=a+\\frac{1}{a}-\\left(a^{3}+\\frac{1}{a^{3}}\\right)<0 .\n$$\nAfter dividing by this fixed number, we get\n$$\n\\begin{equation*}\nf(t)=C_{1} t+\\frac{C_{2}}{t}\n\\end{equation*}\n$$\nwhere the numbers $C_{1}$ and $C_{2}$ are expressed in terms of $a, f(1), f\\left(a^{2}\\right)$ and $f\\left(1 / a^{2}\\right)$, and they do not depend on $t$.\nThe functions of the form (3) satisfy the equation:\n$$\n\\left(x+\\frac{1}{x}\\right) f(y)=\\left(x+\\frac{1}{x}\\right)\\left(C_{1} y+\\frac{C_{2}}{y}\\right)=\\left(C_{1} x y+\\frac{C_{2}}{x y}\\right)+\\left(C_{1} \\frac{y}{x}+C_{2} \\frac{x}{y}\\right)=f(x y)+f\\left(\\frac{y}{x}\\right) .\n$$\n\n\nSolution 2. We start with an observation. If we substitute $x=a \\neq 1$ and $y=a^{n}$ in (1), we obtain\n$$\nf\\left(a^{n+1}\\right)-\\left(a+\\frac{1}{a}\\right) f\\left(a^{n}\\right)+f\\left(a^{n-1}\\right)=0 .\n$$\nFor the sequence $z_{n}=a^{n}$, this is a homogeneous linear recurrence of the second order, and its characteristic polynomial is $t^{2}-\\left(a+\\frac{1}{a}\\right) t+1=(t-a)\\left(t-\\frac{1}{a}\\right)$ with two distinct nonzero roots, namely $a$ and $1 / a$. As is well-known, the general solution is $z_{n}=C_{1} a^{n}+C_{2}(1 / a)^{n}$ where the index $n$ can be as well positive as negative. Of course, the numbers $C_{1}$ and $C_{2}$ may depend of the choice of $a$, so in fact we have two functions, $C_{1}$ and $C_{2}$, such that\n$$\n\\begin{equation*}\nf\\left(a^{n}\\right)=C_{1}(a) \\cdot a^{n}+\\frac{C_{2}(a)}{a^{n}} \\text{ for every } a \\neq 1 \\text{ and every integer } n \\text{. }\n\\end{equation*}\n$$\nThe relation (4) can be easily extended to rational values of $n$, so we may conjecture that $C_{1}$ and $C_{2}$ are constants, and whence $f(t)=C_{1} t+\\frac{C_{2}}{t}$. As it was seen in the previous solution, such functions indeed satisfy (1).\nThe equation (1) is linear in $f$; so if some functions $f_{1}$ and $f_{2}$ satisfy (1) and $c_{1}, c_{2}$ are real numbers, then $c_{1} f_{1}(x)+c_{2} f_{2}(x)$ is also a solution of (1). In order to make our formulas simpler, define\n$$\nf_{0}(x)=f(x)-f(1) \\cdot x .\n$$\nThis function is another one satisfying (1) and the extra constraint $f_{0}(1)=0$. Repeating the same argument on linear recurrences, we can write $f_{0}(a)=K(a) a^{n}+\\frac{L(a)}{a^{n}}$ with some functions $K$ and $L$. By substituting $n=0$, we can see that $K(a)+L(a)=f_{0}(1)=0$ for every $a$. Hence,\n$$\nf_{0}\\left(a^{n}\\right)=K(a)\\left(a^{n}-\\frac{1}{a^{n}}\\right) .\n$$\nNow take two numbers $a>b>1$ arbitrarily and substitute $x=(a / b)^{n}$ and $y=(a b)^{n}$ in (1):\n$$\n\\begin{align*}\n\\left(\\frac{a^{n}}{b^{n}}+\\frac{b^{n}}{a^{n}}\\right) f_{0}\\left((a b)^{n}\\right) & =f_{0}\\left(a^{2 n}\\right)+f_{0}\\left(b^{2 n}\\right), \\quad \\text{ so } \\\\\n\\left(\\frac{a^{n}}{b^{n}}+\\frac{b^{n}}{a^{n}}\\right) K(a b)\\left((a b)^{n}-\\frac{1}{(a b)^{n}}\\right) & =K(a)\\left(a^{2 n}-\\frac{1}{a^{2 n}}\\right)+K(b)\\left(b^{2 n}-\\frac{1}{b^{2 n}}\\right), \\quad \\text{ or equivalently } \\\\\nK(a b)\\left(a^{2 n}-\\frac{1}{a^{2 n}}+b^{2 n}-\\frac{1}{b^{2 n}}\\right) & =K(a)\\left(a^{2 n}-\\frac{1}{a^{2 n}}\\right)+K(b)\\left(b^{2 n}-\\frac{1}{b^{2 n}}\\right) .\n\\end{align*}\n$$\nBy dividing (5) by $a^{2 n}$ and then taking limit with $n \\rightarrow+\\infty$ we get $K(a b)=K(a)$. Then (5) reduces to $K(a)=K(b)$. Hence, $K(a)=K(b)$ for all $a>b>1$.\nFix $a>1$. For every $x>0$ there is some $b$ and an integer $n$ such that $1<b<a$ and $x=b^{n}$. Then\n$$\nf_{0}(x)=f_{0}\\left(b^{n}\\right)=K(b)\\left(b^{n}-\\frac{1}{b^{n}}\\right)=K(a)\\left(x-\\frac{1}{x}\\right) .\n$$\nHence, we have $f(x)=f_{0}(x)+f(1) x=C_{1} x+\\frac{C_{2}}{x}$ with $C_{1}=K(a)+f(1)$ and $C_{2}=-K(a)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32056,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m, n \\geqslant 2$ be integers. Let $f\\left(x_{1}, \\ldots, x_{n}\\right)$ be a polynomial with real coefficients such that\n$$\nf\\left(x_{1}, \\ldots, x_{n}\\right)=\\left\\lfloor\\frac{x_{1}+\\ldots+x_{n}}{m}\\right\\rfloor \\text{ for every } x_{1}, \\ldots, x_{n} \\in\\{0,1, \\ldots, m-1\\} .\n$$\nProve that the total degree of $f$ is at least $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We transform the problem to a single variable question by the following\n\nLemma. Let $a_{1}, \\ldots, a_{n}$ be nonnegative integers and let $G(x)$ be a nonzero polynomial with $\\operatorname{deg} G \\leqslant a_{1}+\\ldots+a_{n}$. Suppose that some polynomial $F\\left(x_{1}, \\ldots, x_{n}\\right)$ satisfies\n$$\nF\\left(x_{1}, \\ldots, x_{n}\\right)=G\\left(x_{1}+\\ldots+x_{n}\\right) \\quad \\text{ for }\\left(x_{1}, \\ldots, x_{n}\\right) \\in\\left\\{0,1, \\ldots, a_{1}\\right\\} \\times \\ldots \\times\\left\\{0,1, \\ldots, a_{n}\\right\\} \\text{. }\n$$\nThen $F$ cannot be the zero polynomial, and $\\operatorname{deg} F \\geqslant \\operatorname{deg} G$.\n\nFor proving the lemma, we will use forward differences of polynomials. If $p(x)$ is a polynomial with a single variable, then define $(\\Delta p)(x)=p(x+1)-p(x)$. It is well-known that if $p$ is a nonconstant polynomial then $\\operatorname{deg} \\Delta p=\\operatorname{deg} p-1$.\n\nIf $p\\left(x_{1}, \\ldots, x_{n}\\right)$ is a polynomial with $n$ variables and $1 \\leqslant k \\leqslant n$ then let\n$$\n\\left(\\Delta_{k} p\\right)\\left(x_{1}, \\ldots, x_{n}\\right)=p\\left(x_{1}, \\ldots, x_{k-1}, x_{k}+1, x_{k+1}, \\ldots, x_{n}\\right)-p\\left(x_{1}, \\ldots, x_{n}\\right)\n$$\nIt is also well-known that either $\\Delta_{k} p$ is the zero polynomial or $\\operatorname{deg}\\left(\\Delta_{k} p\\right) \\leqslant \\operatorname{deg} p-1$.\n\nProof of the lemma. We apply induction on the degree of $G$. If $G$ is a constant polynomial then we have $F(0, \\ldots, 0)=G(0) \\neq 0$, so $F$ cannot be the zero polynomial.\n\nSuppose that $\\operatorname{deg} G \\geqslant 1$ and the lemma holds true for lower degrees. Since $a_{1}+\\ldots+a_{n} \\geqslant \\operatorname{deg} G>0$, at least one of $a_{1}, \\ldots, a_{n}$ is positive; without loss of generality suppose $a_{1} \\geqslant 1$.\n\nConsider the polynomials $F_{1}=\\Delta_{1} F$ and $G_{1}=\\Delta G$. On the grid $\\left\\{0, \\ldots, a_{1}-1\\right\\} \\times\\left\\{0, \\ldots, a_{2}\\right\\} \\times \\ldots \\times\\left\\{0, \\ldots, a_{n}\\right\\}$ we have\n$$\n\\begin{aligned}\nF_{1}\\left(x_{1}, \\ldots, x_{n}\\right) & =F\\left(x_{1}+1, x_{2}, \\ldots, x_{n}\\right)-F\\left(x_{1}, x_{2}, \\ldots, x_{n}\\right)= \\\\\n& =G\\left(x_{1}+\\ldots+x_{n}+1\\right)-G\\left(x_{1}+\\ldots+x_{n}\\right)=G_{1}\\left(x_{1}+\\ldots+x_{n}\\right)\n\\end{aligned}\n$$\nSince $G$ is nonconstant, we have $\\operatorname{deg} G_{1}=\\operatorname{deg} G-1 \\leqslant\\left(a_{1}-1\\right)+a_{2}+\\ldots+a_{n}$. Therefore we can apply the induction hypothesis to $F_{1}$ and $G_{1}$ and conclude that $F_{1}$ is not the zero polynomial and $\\operatorname{deg} F_{1} \\geqslant \\operatorname{deg} G_{1}$. Hence, $\\operatorname{deg} F \\geqslant \\operatorname{deg} F_{1}+1 \\geqslant \\operatorname{deg} G_{1}+1=\\operatorname{deg} G$. That finishes the proof.\n\nTo prove the problem statement, take the unique polynomial $g(x)$ so that $g(x)=\\left\\lfloor\\frac{x}{m}\\right\\rfloor$ for $x \\in\\{0,1, \\ldots, n(m-1)\\}$ and $\\operatorname{deg} g \\leqslant n(m-1)$. Notice that precisely $n(m-1)+1$ values of $g$ are prescribed, so $g(x)$ indeed exists and is unique. Notice further that the constraints $g(0)=g(1)=0$ and $g(m)=1$ together enforce $\\operatorname{deg} g \\geqslant 2$.\n\nBy applying the lemma to $a_{1}=\\ldots=a_{n}=m-1$ and the polynomials $f$ and $g$, we achieve $\\operatorname{deg} f \\geqslant \\operatorname{deg} g$. Hence we just need a suitable lower bound on $\\operatorname{deg} g$.\n\nConsider the polynomial $h(x)=g(x+m)-g(x)-1$. The degree of $g(x+m)-g(x)$ is $\\operatorname{deg} g-1 \\geqslant 1$, so $\\operatorname{deg} h=\\operatorname{deg} g-1 \\geqslant 1$, and therefore $h$ cannot be the zero polynomial. On the other hand, $h$ vanishes at the points $0,1, \\ldots, n(m-1)-m$, so $h$ has at least $(n-1)(m-1)$ roots. Hence,\n$$\n\\operatorname{deg} f \\geqslant \\operatorname{deg} g=\\operatorname{deg} h+1 \\geqslant(n-1)(m-1)+1 \\geqslant n\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32057,
    "subject": "Mathematics (Multi-modal)",
    "question": "Queenie and Horst play a game on a $20 \\times 20$ chessboard. In the beginning the board is empty. In every turn, Horst places a black knight on an empty square in such a way that his new knight does not attack any previous knights. Then Queenie places a white queen on an empty square. The game gets finished when somebody cannot move.\nFind the maximal positive $K$ such that, regardless of the strategy of Queenie, Horst can put at least $K$ knights on the board.",
    "options": [],
    "answer": "100",
    "solution": "We show two strategies, one for Horst to place at least $100$ knights, and another strategy for Queenie that prevents Horst from putting more than $100$ knights on the board.\n\nA strategy for Horst: Put knights only on black squares, until all black squares get occupied.\n\nColour the squares of the board black and white in the usual way, such that the white and black squares alternate, and let Horst put his knights on black squares as long as it is possible. Two knights on squares of the same colour never attack each other. The number of black squares is $20^{2} / 2 = 200$. The two players occupy the squares in turn, so Horst will surely find empty black squares in his first $100$ steps.\n\nA strategy for Queenie: Group the squares into cycles of length $4$, and after each step of Horst, occupy the opposite square in the same cycle.\n\nConsider the squares of the board as vertices of a graph; let two squares be connected if two knights on those squares would attack each other. Notice that in a $4 \\times 4$ board the squares can be grouped into $4$ cycles of length $4$, as shown in Figure 1. Divide the board into parts of size $4 \\times 4$, and perform the same grouping in every part; this way we arrange the $400$ squares of the board into $100$ cycles (Figure 2).\n\n![](attached_image_1.png)\nFigure 1\n\n![](attached_image_2.png)\nFigure 2\n\n![](attached_image_3.png)\nFigure 3\n\nThe strategy of Queenie can be as follows: Whenever Horst puts a new knight to a certain square $A$, which is part of some cycle $A-B-C-D-A$, let Queenie put her queen on the opposite square $C$ in that cycle (Figure 3). From this point, Horst cannot put any knight on $A$ or $C$ because those squares are already occupied, neither on $B$ or $D$ because those squares are attacked by the knight standing on $A$. Hence, Horst can put at most one knight on each cycle, that is at most $100$ knights in total.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32058,
    "subject": "Mathematics (Multi-modal)",
    "question": "An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once:\n$$\n\\begin{aligned}\n& 4 \\\\\n& 2\\ 6 \\\\\n& 5\\ 7\\ 1 \\\\\n& 8\\ 3\\ 10\\ 9 .\n\\end{aligned}\n$$\nIs it possible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\\cdots+2018$ exactly once?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Answer: No, it is not possible.\n\nLet $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\\cdots+n$, it follows that they form a permutation of $1,2, \\ldots, n$.\n\n![](attached_image_1.png)\nFigure 1\n![](attached_image_2.png)\nFigure 2\n\nConsider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\\prime}$, has side length $\\ell \\geqslant\\lceil(n-2) / 2\\rceil$. Since $T^{\\prime}$ obeys the anti-Pascal rule, it contains $\\ell$ pairwise distinct positive integers $a_{1}^{\\prime}, a_{2}^{\\prime}, \\ldots, a_{\\ell}^{\\prime}$, where $a_{1}^{\\prime}$ is at the apex, and $a_{k}^{\\prime}$ and $b_{k}^{\\prime}=a_{1}^{\\prime}+a_{2}^{\\prime}+\\cdots+a_{k}^{\\prime}$ are the two neighbours below $b_{k-1}^{\\prime}$ for each $k=2,3 \\ldots, \\ell$. Since the $a_{k}$ all lie outside $T^{\\prime}$, and they form a permutation of $1,2, \\ldots, n$, the $a_{k}^{\\prime}$ are all greater than $n$. Consequently,\n$$\n\\begin{array}{r}\nb_{\\ell}^{\\prime} \\geqslant(n+1)+(n+2)+\\cdots+(n+\\ell)=\\frac{\\ell(2 n+\\ell+1)}{2} \\\\\n\\geqslant \\frac{1}{2} \\cdot \\frac{n-2}{2}\\left(2 n+\\frac{n-2}{2}+1\\right)=\\frac{5 n(n-2)}{8}\n\\end{array}\n$$\nwhich is greater than $1+2+\\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32059,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k$ be a positive integer. The organising committee of a tennis tournament is to schedule the matches for $2k$ players so that every two players play once, each day exactly one match is played, and each player arrives to the tournament site the day of his first match, and departs the day of his last match. For every day a player is present on the tournament, the committee has to pay 1 coin to the hotel. The organisers want to design the schedule so as to minimise the total cost of all players' stays. Determine this minimum cost.",
    "options": [],
    "answer": "k(4k^2 + k - 1) / 2",
    "solution": "Enumerate the days of the tournament $1,2, \\ldots, \\binom{2k}{2}$. Let $b_{1} \\leqslant b_{2} \\leqslant \\cdots \\leqslant b_{2k}$ be the days the players arrive to the tournament, arranged in nondecreasing order; similarly, let $e_{1} \\geqslant \\cdots \\geqslant e_{2k}$ be the days they depart arranged in nonincreasing order (it may happen that a player arrives on day $b_{i}$ and departs on day $e_{j}$, where $i \\neq j$). If a player arrives on day $b$ and departs on day $e$, then his stay cost is $e-b+1$. Therefore, the total stay cost is\n$$\n\\Sigma=\\sum_{i=1}^{2k} e_{i}-\\sum_{i=1}^{2k} b_{i}+n=\\sum_{i=1}^{2k}\\left(e_{i}-b_{i}+1\\right)\n$$\n\nBounding the total cost from below. To this end, estimate $e_{i+1}-b_{i+1}+1$. Before day $b_{i+1}$, only $i$ players were present, so at most $\\binom{i}{2}$ matches could be played. Therefore, $b_{i+1} \\leqslant \\binom{i}{2}+1$. Similarly, at most $\\binom{i}{2}$ matches could be played after day $e_{i+1}$, so $e_{i} \\geqslant \\binom{2k}{2}-\\binom{i}{2}$. Thus,\n$$\ne_{i+1}-b_{i+1}+1 \\geqslant \\binom{2k}{2}-2\\binom{i}{2}=k(2k-1)-i(i-1)\n$$\nThis lower bound can be improved for $i>k$: List the $i$ players who arrived first, and the $i$ players who departed last; at least $2i-2k$ players appear in both lists. The matches between these players were counted twice, though the players in each pair have played only once. Therefore, if $i>k$, then\n$$\ne_{i+1}-b_{i+1}+1 \\geqslant \\binom{2k}{2}-2\\binom{i}{2}+\\binom{2i-2k}{2}=(2k-i)^{2}.\n$$\n\nAn optimal tournament. We now describe a schedule in which the lower bounds above are all achieved simultaneously. Split players into two groups $X$ and $Y$, each of cardinality $k$. Next, partition the schedule into three parts. During the first part, the players from $X$ arrive one by one, and each newly arrived player immediately plays with everyone already present. During the third part (after all players from $X$ have already departed) the players from $Y$ depart one by one, each playing with everyone still present just before departing.\n\nIn the middle part, everyone from $X$ should play with everyone from $Y$. Let $S_{1}, S_{2}, \\ldots, S_{k}$ be the players in $X$, and let $T_{1}, T_{2}, \\ldots, T_{k}$ be the players in $Y$. Let $T_{1}, T_{2}, \\ldots, T_{k}$ arrive in this order; after $T_{j}$ arrives, he immediately plays with all the $S_{i}, i>j$. Afterwards, players $S_{k}$, $S_{k-1}, \\ldots, S_{1}$ depart in this order; each $S_{i}$ plays with all the $T_{j}, i \\leqslant j$, just before his departure, and $S_{k}$ departs the day $T_{k}$ arrives. For $0 \\leqslant s \\leqslant k-1$, the number of matches played between $T_{k-s}$'s arrival and $S_{k-s}$'s departure is\n$$\n\\sum_{j=k-s}^{k-1}(k-j)+1+\\sum_{j=k-s}^{k-1}(k-j+1)=\\frac{1}{2} s(s+1)+1+\\frac{1}{2} s(s+3)=(s+1)^{2}\n$$\nThus, if $i>k$, then the number of matches that have been played between $T_{i-k+1}$'s arrival, which is $b_{i+1}$, and $S_{i-k+1}$'s departure, which is $e_{i+1}$, is $(2k-i)^{2}$; that is, $e_{i+1}-b_{i+1}+1=(2k-i)^{2}$, showing the second lower bound achieved for all $i>k$.\n\nIf $i \\leqslant k$, then the matches between the $i$ players present before $b_{i+1}$ all fall in the first part of the schedule, so there are $\\binom{i}{2}$ such, and $b_{i+1}=\\binom{i}{2}+1$. Similarly, after $e_{i+1}$, there are $i$ players left, all $\\binom{i}{2}$ matches now fall in the third part of the schedule, and $e_{i+1}=\\binom{2k}{2}-\\binom{i}{2}$. The first lower bound is therefore also achieved for all $i \\leqslant k$.\n\nConsequently, all lower bounds are achieved simultaneously, and the schedule is indeed optimal.\n\nEvaluation. Finally, evaluate the total cost for the optimal schedule:\n$$\n\\begin{aligned}\n\\Sigma & =\\sum_{i=0}^{k}(k(2k-1)-i(i-1))+\\sum_{i=k+1}^{2k-1}(2k-i)^{2}=(k+1)k(2k-1)-\\sum_{i=0}^{k} i(i-1)+\\sum_{j=1}^{k-1} j^{2} \\\\\n& =k(k+1)(2k-1)-k^{2}+\\frac{1}{2}k(k+1)=\\frac{1}{2}k\\left(4k^{2}+k-1\\right)\n\\end{aligned}\nConsider any tournament schedule. Label players $P_{1}, P_{2}, \\ldots, P_{2k}$ in order of their arrival, and label them again $Q_{2k}, Q_{2k-1}, \\ldots, Q_{1}$ in order of their departure, to define a permutation $a_{1}, a_{2}, \\ldots, a_{2k}$ of $1,2, \\ldots, 2k$ by $P_{i}=Q_{a_{i}}$.\n\nWe first describe an optimal tournament for any given permutation $a_{1}, a_{2}, \\ldots, a_{2k}$ of the indices $1,2, \\ldots, 2k$. Next, we find an optimal permutation and an optimal tournament.\n\nOptimisation for a fixed $a_{1}, \\ldots, a_{2k}$. We say that the cost of the match between $P_{i}$ and $P_{j}$ is the number of players present at the tournament when this match is played. Clearly, the Committee pays for each day the cost of the match of that day. Hence, we are to minimise the total cost of all matches.\n\nNotice that $Q_{2k}$'s departure does not precede $P_{2k}$'s arrival. Hence, the number of players at the tournament monotonically increases (non-strictly) until it reaches $2k$, and then monotonically decreases (non-strictly). So, the best time to schedule the match between $P_{i}$ and $P_{j}$ is either when $P_{\\max (i, j)}$ arrives, or when $Q_{\\max \\left(a_{i}, a_{j}\\right)}$ departs, in which case the cost is $\\min \\left(\\max (i, j), \\max \\left(a_{i}, a_{j}\\right)\\right)$.\n\nConversely, assuming that $i>j$, if this match is scheduled between the arrivals of $P_{i}$ and $P_{i+1}$, then its cost will be exactly $i=\\max (i, j)$. Similarly, one can make it cost $\\max \\left(a_{i}, a_{j}\\right)$. Obviously, these conditions can all be simultaneously satisfied, so the minimal cost for a fixed sequence $a_{1}, a_{2}, \\ldots, a_{2k}$ is\n$$\n\\Sigma\\left(a_{1}, \\ldots, a_{2k}\\right)=\\sum_{1 \\leqslant i<j \\leqslant 2k} \\min \\left(\\max (i, j), \\max (a_{i}, a_{j})\\right) .\n$$\n\nOptimising the sequence $\\left(a_{i}\\right)$. Optimisation hinges on the lemma below.\n\nLemma. If $a \\leqslant b$ and $c \\leqslant d$, then\n$$\n\\begin{aligned}\n\\min (\\max (a, x), \\max (c, y))+\\min (\\max (b, x), \\max (d, y)) & \\\\\n& \\geqslant \\min (\\max (a, x), \\max (d, y))+\\min (\\max (b, x), \\max (c, y))\n\\end{aligned}\n$$\n\nProof. Write $a^{\\prime}=\\max (a, x) \\leqslant \\max (b, x)=b^{\\prime}$ and $c^{\\prime}=\\max (c, y) \\leqslant \\max (d, y)=d^{\\prime}$ and check that $\\min \\left(a^{\\prime}, c^{\\prime}\\right)+\\min \\left(b^{\\prime}, d^{\\prime}\\right) \\geqslant \\min \\left(a^{\\prime}, d^{\\prime}\\right)+\\min \\left(b^{\\prime}, c^{\\prime}\\right)$.\n\nConsider a permutation $a_{1}, a_{2}, \\ldots, a_{2k}$ such that $a_{i}<a_{j}$ for some $i<j$. Swapping $a_{i}$ and $a_{j}$ does not change the $(i, j)$th summand in (1), and for $\\ell \\notin\\{i, j\\}$ the sum of the $(i, \\ell)$th and the $(j, \\ell)$th summands does not increase by the Lemma. Hence the optimal value does not increase, but the number of disorders in the permutation increases. This process stops when $a_{i}=2k+1-i$ for all $i$, so the required minimum is\n$$\n\\begin{aligned}\nS(2k, 2k-1, \\ldots, 1) & =\\sum_{1 \\leqslant i<j \\leqslant 2k} \\min (\\max (i, j), \\max (2k+1-i, 2k+1-j)) \\\\\n& =\\sum_{1 \\leqslant i<j \\leqslant 2k} \\min (j, 2k+1-i)\n\\end{aligned}\n$$\nThe latter sum is fairly tractable and yields the stated result; we omit the details.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32060,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be distinct positive integers. The following infinite process takes place on an initially empty board.\n(i) If there is at least a pair of equal numbers on the board, we choose such a pair and increase one of its components by $a$ and the other by $b$.\n(ii) If no such pair exists, we write down two times the number $0$.\nProve that, no matter how we make the choices in (i), operation (ii) will be performed only finitely many times.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We may assume $\\gcd(a, b)=1$; otherwise we work in the same way with multiples of $d=\\gcd(a, b)$.\nSuppose that after $N$ moves of type (ii) and some moves of type (i) we have to add two new zeros. For each integer $k$, denote by $f(k)$ the number of times that the number $k$ appeared on the board up to this moment. Then $f(0)=2N$ and $f(k)=0$ for $k<0$. Since the board contains at most one $k-a$, every second occurrence of $k-a$ on the board produced, at some moment, an occurrence of $k$; the same stands for $k-b$. Therefore,\n$$\nf(k)=\\left\\lfloor\\frac{f(k-a)}{2}\\right\\rfloor+\\left\\lfloor\\frac{f(k-b)}{2}\\right\\rfloor, \\tag{1}\n$$\nyielding\n$$\nf(k) \\geqslant \\frac{f(k-a)+f(k-b)}{2}-1 . \\tag{2}\n$$\nSince $\\gcd(a, b)=1$, every integer $x>ab-a-b$ is expressible in the form $x=sa+tb$, with integer $s, t \\geqslant 0$.\nWe will prove by induction on $s+t$ that if $x=sa+bt$, with $s, t$ nonnegative integers, then\n$$\nf(x)>\\frac{f(0)}{2^{s+t}}-2 \\tag{3}\n$$\nThe base case $s+t=0$ is trivial. Assume now that (3) is true for $s+t=v$. Then, if $s+t=v+1$ and $x=sa+tb$, at least one of the numbers $s$ and $t$—say $s$—is positive, hence by (2),\n$$\nf(x)=f(sa+tb) \\geqslant \\frac{f((s-1)a+tb)}{2}-1>\\frac{1}{2}\\left(\\frac{f(0)}{2^{s+t-1}}-2\\right)-1=\\frac{f(0)}{2^{s+t}}-2 .\n$$\nAssume now that we must perform moves of type (ii) ad infinitum. Take $n=ab-a-b$ and suppose $b>a$. Since each of the numbers $n+1, n+2, \\ldots, n+b$ can be expressed in the form $sa+tb$, with $0 \\leqslant s \\leqslant b$ and $0 \\leqslant t \\leqslant a$, after moves of type (ii) have been performed $2^{a+b+1}$ times and we have to add a new pair of zeros, each $f(n+k), k=1,2, \\ldots, b$, is at least $2$. In this case (1) yields inductively $f(n+k) \\geqslant 2$ for all $k \\geqslant 1$. But this is absurd: after a finite number of moves, $f$ cannot attain nonzero values at infinitely many points.\nWe start by showing that the result of the process in the problem does not depend on the way the operations are performed. For that purpose, it is convenient to modify the process a bit.\n\nClaim 1. Suppose that the board initially contains a finite number of nonnegative integers, and one starts performing type (i) moves only. Assume that one had applied $k$ moves which led to a final arrangement where no more type (i) moves are possible. Then, if one starts from the same initial arrangement, performing type (i) moves in an arbitrary fashion, then the process will necessarily stop at the same final arrangement.\n\nProof. Throughout this proof, all moves are supposed to be of type (i).\nInduct on $k$; the base case $k=0$ is trivial, since no moves are possible. Assume now that $k \\geqslant 1$. Fix some canonical process, consisting of $k$ moves $M_{1}, M_{2}, \\ldots, M_{k}$, and reaching the final arrangement $A$. Consider any sample process $m_{1}, m_{2}, \\ldots$ starting with the same initial arrangement and proceeding as long as possible; clearly, it contains at least one move. We need to show that this process stops at $A$.\nLet move $m_{1}$ consist in replacing two copies of $x$ with $x+a$ and $x+b$. If move $M_{1}$ does the same, we may apply the induction hypothesis to the arrangement appearing after $m_{1}$. Otherwise, the canonical process should still contain at least one move consisting in replacing $(x, x) \\mapsto (x+a, x+b)$, because the initial arrangement contains at least two copies of $x$, while the final one contains at most one such.\nLet $M_{i}$ be the first such move. Since the copies of $x$ are indistinguishable and no other copy of $x$ disappeared before $M_{i}$ in the canonical process, the moves in this process can be permuted as $M_{i}, M_{1}, \\ldots, M_{i-1}, M_{i+1}, \\ldots, M_{k}$, without affecting the final arrangement. Now it suffices to perform the move $m_{1}=M_{i}$ and apply the induction hypothesis as above.\n\nClaim 2. Consider any process starting from the empty board, which involved exactly $n$ moves of type (ii) and led to a final arrangement where all the numbers are distinct. Assume that one starts with the board containing $2n$ zeroes (as if $n$ moves of type (ii) were made in the beginning), applying type (i) moves in an arbitrary way. Then this process will reach the same final arrangement.\n\nProof. Starting with the board with $2n$ zeros, one may indeed model the first process mentioned in the statement of the claim, omitting the type (ii) moves. This way, one reaches the same final arrangement. Now, Claim 1 yields that this final arrangement will be obtained when type (i) moves are applied arbitrarily.\n\nClaim 2 allows now to reformulate the problem statement as follows: There exists an integer $n$ such that, starting from $2n$ zeroes, one may apply type (i) moves indefinitely.\n\nIn order to prove this, we start with an obvious induction on $s+t=k \\geqslant 1$ to show that if we start with $2^{s+t}$ zeros, then we can get simultaneously on the board, at some point, each of the numbers $sa+tb$, with $s+t=k$.\n\nSuppose now that $a<b$. Then, an appropriate use of separate groups of zeros allows us to get two copies of each of the numbers $sa+tb$, with $1 \\leqslant s, t \\leqslant b$.\n\nDefine $N=ab-a-b$, and notice that after representing each of numbers $N+k, 1 \\leqslant k \\leqslant b$, in the form $sa+tb, 1 \\leqslant s, t \\leqslant b$ we can get, using enough zeros, the numbers $N+1, N+2, \\ldots, N+a$ and the numbers $N+1, N+2, \\ldots, N+b$.\n\nFrom now on we can perform only moves of type (i). Indeed, if $n \\geqslant N$, the occurrence of the numbers $n+1, n+2, \\ldots, n+a$ and $n+1, n+2, \\ldots, n+b$ and the replacement $(n+1, n+1) \\mapsto (n+b+1, n+a+1)$ leads to the occurrence of the numbers $n+2, n+3, \\ldots, n+a+1$ and $n+2, n+3, \\ldots, n+b+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32061,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider 2018 pairwise crossing circles no three of which are concurrent. These circles subdivide the plane into regions bounded by circular edges that meet at vertices. Notice that there are an even number of vertices on each circle. Given the circle, alternately colour the vertices on that circle red and blue. In doing so for each circle, every vertex is coloured twice once for each of the two circles that cross at that point. If the two colourings agree at a vertex, then it is assigned that colour; otherwise, it becomes yellow. Show that, if some circle contains at least 2061 yellow points, then the vertices of some region are all yellow.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Letting $n=2018$, we will show that, if every region has at least one non-yellow vertex, then every circle contains at most $n+\\lfloor\\sqrt{n-2}\\rfloor-2$ yellow points. In the case at hand, the latter equals $2018+44-2=2060$, contradicting the hypothesis.\n\nConsider the natural geometric graph $G$ associated with the configuration of $n$ circles. Fix any circle $C$ in the configuration, let $k$ be the number of yellow points on $C$, and find a suitable lower bound for the total number of yellow vertices of $G$ in terms of $k$ and $n$. It turns out that $k$ is even, and $G$ has at least\n$$\n\\begin{equation*}\nk+2\\binom{k / 2}{2}+2\\binom{n-k / 2-1}{2}=\\frac{k^{2}}{2}-(n-2) k+(n-2)(n-1) \\tag{*}\n\\end{equation*}\n$$\nyellow vertices. The proof hinges on the two lemmata below.\n\nLemma 1. Let two circles in the configuration cross at $x$ and $y$. Then $x$ and $y$ are either both yellow or both non-yellow.\n\nProof. This is because the numbers of interior vertices on the four arcs $x$ and $y$ determine on the two circles have like parities.\nIn particular, each circle in the configuration contains an even number of yellow vertices.\n\nLemma 2. If $\\overparen{x y}$, $\\overparen{y z}$, and $\\overparen{z x}$ are circular arcs of three pairwise distinct circles in the configuration, then the number of yellow vertices in the set $\\{x, y, z\\}$ is odd.\n\nProof. Let $C_{1}, C_{2}, C_{3}$ be the three circles under consideration. Assume, without loss of generality, that $C_{2}$ and $C_{3}$ cross at $x$, $C_{3}$ and $C_{1}$ cross at $y$, and $C_{1}$ and $C_{2}$ cross at $z$. Let $k_{1}$, $k_{2}, k_{3}$ be the numbers of interior vertices on the three circular arcs under consideration. Since each circle in the configuration, different from the $C_{i}$, crosses the cycle $\\overparen{x y} \\cup \\overparen{y z} \\cup \\overparen{z x}$ at an even number of points (recall that no three circles are concurrent), and self-crossings are counted twice, the sum $k_{1}+k_{2}+k_{3}$ is even.\n\nLet $Z_{1}$ be the colour $z$ gets from $C_{1}$ and define the other colours similarly. By the preceding, the number of bichromatic pairs in the list $\\left(Z_{1}, Y_{1}\\right),\\left(X_{2}, Z_{2}\\right),\\left(Y_{3}, X_{3}\\right)$ is odd. Since the total number of colour changes in a cycle $Z_{1}-Y_{1}-Y_{3}-X_{3}-X_{2}-Z_{2}-Z_{1}$ is even, the number of bichromatic pairs in the list $\\left(X_{2}, X_{3}\\right),\\left(Y_{1}, Y_{3}\\right),\\left(Z_{1}, Z_{2}\\right)$ is odd, and the lemma follows.\n\nWe are now in a position to prove that (*) bounds the total number of yellow vertices from below. Refer to Lemma 1 to infer that the $k$ yellow vertices on $C$ pair off to form the pairs of points where $C$ is crossed by $k / 2$ circles in the configuration. By Lemma 2, these circles cross pairwise to account for another $2\\binom{k / 2}{2}$ yellow vertices. Finally, the remaining $n-k / 2-1$ circles in the configuration cross $C$ at non-yellow vertices, by Lemma 1, and Lemma 2 applies again to show that these circles cross pairwise to account for yet another $2\\binom{n-k / 2-1}{2}$ yellow vertices. Consequently, there are at least (*) yellow vertices.\n\nNext, notice that $G$ is a plane graph on $n(n-1)$ degree 4 vertices, having exactly $2 n(n-1)$ edges and exactly $n(n-1)+2$ faces (regions), the outer face inclusive (by Euler's formula for planar graphs).\n\nLemma 3. Each face of $G$ has equally many red and blue vertices. In particular, each face has an even number of non-yellow vertices.\n\nProof. Trace the boundary of a face once in circular order, and consider the colours each vertex is assigned in the colouring of the two circles that cross at that vertex, to infer that colours of non-yellow vertices alternate.\n\nConsequently, if each region has at least one non-yellow vertex, then it has at least two such. Since each vertex of $G$ has degree 4, consideration of vertex-face incidences shows that $G$ has at least $n(n-1) / 2+1$ non-yellow vertices, and hence at most $n(n-1) / 2-1$ yellow vertices. (In fact, Lemma 3 shows that there are at least $n(n-1) / 4+1 / 2$ red, respectively blue, vertices.)\n\nFinally, recall the lower bound (*) for the total number of yellow vertices in $G$, to write $n(n-1) / 2-1 \\geqslant k^{2} / 2-(n-2) k+(n-2)(n-1)$, and conclude that $k \\leqslant n+\\lfloor\\sqrt{n-2}\\rfloor-2$, as claimed in the first paragraph.\nThe first two lemmata in Solution 1 show that the circles in the configuration split into two classes: Consider any circle $C$ along with all circles that cross $C$ at yellow points to form one class; the remaining circles then form the other class. Lemma 2 shows that any pair of circles in the same class cross at yellow points; otherwise, they cross at non-yellow points.\n\nCall the circles from the two classes white and black, respectively. Call a region yellow if its vertices are all yellow. Let $w$ and $b$ be the numbers of white and black circles, respectively; clearly, $w+b=n$. Assume that $w \\geqslant b$, and that there is no yellow region. Clearly, $b \\geqslant 1$, otherwise each region is yellow. The white circles subdivide the plane into $w(w-1)+2$ larger regions - call them white. The white regions (or rather their boundaries) subdivide each black circle into black arcs. Since there are no yellow regions, each white region contains at least one black arc.\n\nConsider any white region; let it contain $t \\geqslant 1$ black arcs. We claim that the number of points at which these $t$ \\operatorname{arcs}$ cross does not exceed $t-1$. To prove this, consider a multigraph whose vertices are these black arcs, two vertices being joined by an edge for each point at which the corresponding arcs cross. If this graph had more than $t-1$ edges, it would contain a cycle, since it has $t$ vertices; this cycle would correspond to a closed contour formed by black sub-arcs, lying inside the region under consideration. This contour would, in turn, define at least one yellow region, which is impossible.\n\nLet $t_{i}$ be the number of black arcs inside the $i^{\\text{th}}$ white region. The total number of black arcs is $\\sum_{i} t_{i}=2 w b$, and they cross at $2\\binom{b}{2}=b(b-1)$ points. By the preceding,\n$$\nb(b-1) \\leqslant \\sum_{i=1}^{w^{2}-w+2}\\left(t_{i}-1\\right)=\\sum_{i=1}^{w^{2}-w+2} t_{i}-\\left(w^{2}-w+2\\right)=2 w b-\\left(w^{2}-w+2\\right)\n$$\nor, equivalently, $(w-b)^{2} \\leqslant w+b-2=n-2$, which is the case if and only if $w-b \\leqslant\\lfloor\\sqrt{n-2}\\rfloor$. Consequently, $b \\leqslant w \\leqslant(n+\\lfloor\\sqrt{n-2}\\rfloor) / 2$, so there are at most $2(w-1) \\leqslant n+\\lfloor\\sqrt{n-2}\\rfloor-2$ yellow vertices on each circle - a contradiction.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32062,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $O$ be the circumcentre, and $\\Omega$ be the circumcircle of an acute-angled triangle $A B C$. Let $P$ be an arbitrary point on $\\Omega$, distinct from $A, B, C$, and their antipodes in $\\Omega$. Denote the circumcentres of the triangles $A O P, B O P$, and $C O P$ by $O_{A}, O_{B}$, and $O_{C}$, respectively. The lines $\\ell_{A}, \\ell_{B}$, and $\\ell_{C}$ perpendicular to $B C, C A$, and $A B$ pass through $O_{A}, O_{B}$, and $O_{C}$, respectively. Prove that the circumcircle of the triangle formed by $\\ell_{A}, \\ell_{B}$, and $\\ell_{C}$ is tangent to the line $O P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the lines $\\ell_{B}$ and $\\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B} L_{C}$; moreover, $O_{A}, O_{B}$, $O_{C}$ all lie on the perpendicular bisector of $O P$.\n\n![](attached_image_1.png)\n\nClaim 1. The points $L_{B}, P, O_{A}$, and $O_{C}$ are concyclic.\n\nProof. Since $O$ is symmetric to $P$ in $O_{A} O_{C}$, we have\n$$\n\\Varangle\\left(O_{A} P, O_{C} P\\right)=\\Varangle\\left(O_{C} O, O_{A} O\\right)=\\Varangle(C P, A P)=\\Varangle(C B, A B)=\\Varangle\\left(O_{A} L_{B}, O_{C} L_{B}\\right) .\n$$\nDenote the circle through $L_{B}, P, O_{A}$ and $O_{C}$ by $\\omega_{B}$. Define the circles $\\omega_{A}$ and $\\omega_{C}$ similarly.\n\nClaim 2. The circumcircle of the triangle $L_{A} L_{B} L_{C}$ passes through $P$.\n\nProof. From cyclic quadruples of points in the circles $\\omega_{B}$ and $\\omega_{C}$, we have\n$$\n\\begin{aligned}\n\\Varangle\\left(L_{C} L_{A}, L_{C} P\\right) & =\\Varangle\\left(L_{C} O_{B}, L_{C} P\\right)=\\Varangle\\left(O_{A} O_{B}, O_{A} P\\right) \\\\\n& =\\Varangle\\left(O_{A} O_{C}, O_{A} P\\right)=\\Varangle\\left(L_{B} O_{C}, L_{B} P\\right)=\\Varangle\\left(L_{B} L_{A}, L_{B} P\\right) .\n\\end{aligned}\n$$\n\nClaim 3. The points $P, L_{C}$, and $C$ are collinear.\n\nProof. We have $\\Varangle\\left(P L_{C}, L_{C} L_{A}\\right)=\\Varangle\\left(P L_{C}, L_{C} O_{B}\\right)=\\Varangle\\left(P O_{A}, O_{A} O_{B}\\right)$. Further, since $O_{A}$ is the centre of the circle $A O P$, $\\Varangle\\left(P O_{A}, O_{A} O_{B}\\right)=\\Varangle(P A, A O)$. As $O$ is the circumcentre of the triangle $P C A, \\Varangle(P A, A O)=\\pi / 2-\\Varangle(C A, C P)=\\Varangle\\left(C P, L_{C} L_{A}\\right)$. We obtain $\\Varangle\\left(P L_{C}, L_{C} L_{A}\\right)= \\Varangle\\left(C P, L_{C} L_{A}\\right)$, which shows that $P \\in C L_{C}$.\n\nSimilarly, the points $P, L_{A}, A$ are collinear, and the points $P, L_{B}, B$ are also collinear. Finally, the computation above also shows that\n$$\n\\Varangle\\left(O P, P L_{A}\\right)=\\Varangle(P A, A O)=\\Varangle\\left(P L_{C}, L_{C} L_{A}\\right),\n$$\nwhich means that $O P$ is tangent to the circle $P L_{A} L_{B} L_{C}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32063,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n > 1$ be a positive integer. Each cell of an $n \\times n$ table contains an integer. Suppose that the following conditions are satisfied:\n(i) Each number in the table is congruent to $1$ modulo $n$;\n(ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$.\nLet $R_{i}$ be the product of the numbers in the $i^{\\text{th}}$ row, and $C_{j}$ be the product of the numbers in the $j^{\\text{th}}$ column. Prove that the sums $R_{1} + \\cdots + R_{n}$ and $C_{1} + \\cdots + C_{n}$ are congruent modulo $n^{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_{i, j}$ be the entry in the $i^{\\text{th}}$ row and the $j^{\\text{th}}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j} = A_{i, j} - 1$ and $r_{i} = R_{i} - 1$. We show that\n$$\n\\sum_{i=1}^{n} R_{i} \\equiv (n-1) + P \\quad (\\bmod\\ n^{4}) \\tag{1}\n$$\nDue to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1) + P$ modulo $n^{4}$, whence the conclusion.\nBy condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence\n$R_{i} = \\prod_{j=1}^{n} (1 + a_{i, j}) = 1 + \\sum_{j=1}^{n} a_{i, j} + \\sum_{1 \\leqslant j_{1} < j_{2} \\leqslant n} a_{i, j_{1}} a_{i, j_{2}} + \\cdots \\equiv 1 + \\sum_{j=1}^{n} a_{i, j} \\equiv 1 - n + \\sum_{j=1}^{n} A_{i, j} \\ (\\bmod\\ n^{2})$\nfor every index $i$. Using condition (ii), we obtain $R_{i} \\equiv 1\\ (\\bmod\\ n^{2})$, and so $n^{2} \\mid r_{i}$.\nTherefore, every product of at least two of the $r_{i}$ is divisible by $n^{4}$. Repeating the same argument, we obtain\n$$\nP = \\prod_{i=1}^{n} R_{i} = \\prod_{i=1}^{n} (1 + r_{i}) \\equiv 1 + \\sum_{i=1}^{n} r_{i} \\quad (\\bmod\\ n^{4}),\n$$\nwhence\n$$\n\\sum_{i=1}^{n} R_{i} = n + \\sum_{i=1}^{n} r_{i} \\equiv n + (P - 1) \\quad (\\bmod\\ n^{4})\n$$\nas desired.\nWe present a more straightforward (though lengthier) way to establish (1). We also use the notation of $a_{i, j}$.\nBy condition (i), all the $a_{i, j}$ are divisible by $n$. Therefore, we have\n$$\n\\begin{aligned}\nP = \\prod_{i=1}^{n} \\prod_{j=1}^{n} (1 + a_{i, j}) &\\equiv 1 + \\sum_{(i, j)} a_{i, j} + \\sum_{(i_{1}, j_{1}), (i_{2}, j_{2})} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} \\\\\n&+ \\sum_{(i_{1}, j_{1}), (i_{2}, j_{2}), (i_{3}, j_{3})} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \\quad (\\bmod\\ n^{4})\n\\end{aligned}\n$$\nwhere the last two sums are taken over all unordered pairs/triples of pairwise different pairs $(i, j)$; such conventions are applied throughout the solution.\nSimilarly,\n$$\n\\sum_{i=1}^{n} R_{i} = \\sum_{i=1}^{n} \\prod_{j=1}^{n} (1 + a_{i, j}) \\equiv n + \\sum_{i} \\sum_{j} a_{i, j} + \\sum_{i} \\sum_{j_{1}, j_{2}} a_{i, j_{1}} a_{i, j_{2}} + \\sum_{i} \\sum_{j_{1}, j_{2}, j_{3}} a_{i, j_{1}} a_{i, j_{2}} a_{i, j_{3}} \\quad (\\bmod\\ n^{4}) .\n$$\nTherefore,\n$$\n\\begin{aligned}\nP + (n-1) - \\sum_{i} R_{i} &\\equiv \\sum_{\\substack{(i_{1}, j_{1}), (i_{2}, j_{2}) \\\\ i_{1} \\neq i_{2}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} \\\\\n&+ \\sum_{\\substack{(i_{1}, j_{1}), (i_{2}, j_{2}), (i_{3}, j_{3}) \\\\ i_{1} \\neq i_{2} \\neq i_{3} \\neq i_{1}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \\\\\n&+ \\sum_{\\substack{(i_{1}, j_{1}), (i_{2}, j_{2}), (i_{3}, j_{3}) \\\\ i_{1} \\neq i_{2} = i_{3}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \\quad (\\bmod\\ n^{4}) .\n\\end{aligned}\n$$\nWe show that in fact each of the three sums appearing in the right-hand part of this congruence is divisible by $n^{4}$; this yields (1). Denote those three sums by $\\Sigma_{1}, \\Sigma_{2}$, and $\\Sigma_{3}$ in order of appearance. Recall that by condition (ii) we have\n$$\n\\sum_{j} a_{i, j} \\equiv 0 \\quad (\\bmod\\ n^{2}) \\quad \\text{for all indices } i\n$$\nFor every two indices $i_{1} < i_{2}$ we have\n$$\n\\sum_{j_{1}} \\sum_{j_{2}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} = \\left(\\sum_{j_{1}} a_{i_{1}, j_{1}}\\right) \\cdot \\left(\\sum_{j_{2}} a_{i_{2}, j_{2}}\\right) \\equiv 0 \\quad (\\bmod\\ n^{4})\n$$\nsince each of the two factors is divisible by $n^{2}$. Summing over all pairs $(i_{1}, i_{2})$ we obtain $n^{4} \\mid \\Sigma_{1}$.\nSimilarly, for every three indices $i_{1} < i_{2} < i_{3}$ we have\n$$\n\\sum_{j_{1}} \\sum_{j_{2}} \\sum_{j_{3}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} = \\left(\\sum_{j_{1}} a_{i_{1}, j_{1}}\\right) \\cdot \\left(\\sum_{j_{2}} a_{i_{2}, j_{2}}\\right) \\cdot \\left(\\sum_{j_{3}} a_{i_{3}, j_{3}}\\right)\n$$\nwhich is divisible even by $n^{6}$. Hence $n^{4} \\mid \\Sigma_{2}$.\nFinally, for every indices $i_{1} \\neq i_{2} = i_{3}$ and $j_{2} < j_{3}$ we have\n$$\na_{i_{2}, j_{2}} \\cdot a_{i_{2}, j_{3}} \\cdot \\sum_{j_{1}} a_{i_{1}, j_{1}} \\equiv 0 \\quad (\\bmod\\ n^{4})\n$$\nsince the three factors are divisible by $n, n$, and $n^{2}$, respectively. Summing over all 4-tuples of indices $(i_{1}, i_{2}, j_{2}, j_{3})$ we get $n^{4} \\mid \\Sigma_{3}$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 32064,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f:\\{1,2,3, \\ldots\\} \\rightarrow\\{2,3, \\ldots\\}$ be a function such that $f(m+n) \\mid f(m)+f(n)$ for all pairs $m, n$ of positive integers. Prove that there exists a positive integer $c>1$ which divides all values of $f$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For every positive integer $m$, define $S_{m}=\\{n: m \\mid f(n)\\}$.\n\nLemma. If the set $S_{m}$ is infinite, then $S_{m}=\\{d, 2 d, 3 d, \\ldots\\}=d \\cdot \\mathbb{Z}_{>0}$ for some positive integer $d$.\n\nProof. Let $d=\\min S_{m}$; the definition of $S_{m}$ yields $m \\mid f(d)$.\nWhenever $n \\in S_{m}$ and $n>d$, we have $m|f(n)| f(n-d)+f(d)$, so $m \\mid f(n-d)$ and therefore $n-d \\in S_{m}$. Let $r \\leqslant d$ be the least positive integer with $n \\equiv r(\\bmod d)$; repeating the same step, we can see that $n-d, n-2 d, \\ldots, r \\in S_{m}$. By the minimality of $d$, this shows $r=d$ and therefore $d \\mid n$.\nStarting from an arbitrarily large element of $S_{m}$, the process above reaches all multiples of $d$; so they all are elements of $S_{m}$.\n\nThe solution for the problem will be split into two cases.\n\nCase 1: The function $f$ is bounded.\n\nCall a prime $p$ frequent if the set $S_{p}$ is infinite, i.e., if $p$ divides $f(n)$ for infinitely many positive integers $n$; otherwise call $p$ sporadic. Since the function $f$ is bounded, there are only a finite number of primes that divide at least one $f(n)$; so altogether there are finitely many numbers $n$ such that $f(n)$ has a sporadic prime divisor. Let $N$ be a positive integer, greater than all those numbers $n$.\nLet $p_{1}, \\ldots, p_{k}$ be the frequent primes. By the lemma we have $S_{p_{i}}=d_{i} \\cdot \\mathbb{Z}_{>0}$ for some $d_{i}$. Consider the number\n$$\nn=N d_{1} d_{2} \\cdots d_{k}+1\n$$\nDue to $n>N$, all prime divisors of $f(n)$ are frequent primes. Let $p_{i}$ be any frequent prime divisor of $f(n)$. Then $n \\in S_{p_{i}}$, and therefore $d_{i} \\mid n$. But $n \\equiv 1\\left(\\bmod d_{i}\\right)$, which means $d_{i}=1$. Hence $S_{p_{i}}=1 \\cdot \\mathbb{Z}_{>0}=\\mathbb{Z}_{>0}$ and therefore $p_{i}$ is a common divisor of all values $f(n)$.\n\nCase 2: $f$ is unbounded.\n\nWe prove that $f(1)$ divides all $f(n)$.\nLet $a=f(1)$. Since $1 \\in S_{a}$, by the lemma it suffices to prove that $S_{a}$ is an infinite set.\nCall a positive integer $p$ a peak if $f(p)>\\max (f(1), \\ldots, f(p-1))$. Since $f$ is not bounded, there are infinitely many peaks. Let $1=p_{1}<p_{2}<\\ldots$ be the sequence of all peaks, and let $h_{k}=f\\left(p_{k}\\right)$. Notice that for any peak $p_{i}$ and for any $k<p_{i}$, we have $f\\left(p_{i}\\right) \\mid f(k)+f\\left(p_{i}-k\\right)< 2 f\\left(p_{i}\\right)$, hence\n$$\n\\begin{equation*}\nf(k)+f\\left(p_{i}-k\\right)=f\\left(p_{i}\\right)=h_{i} \\tag{1}\n\\end{equation*}\n$$\nBy the pigeonhole principle, among the numbers $h_{1}, h_{2}, \\ldots$ there are infinitely many that are congruent modulo $a$. Let $k_{0}<k_{1}<k_{2}<\\ldots$ be an infinite sequence of positive integers such that $h_{k_{0}} \\equiv h_{k_{1}} \\equiv \\ldots(\\bmod a)$. Notice that\n$$\nf\\left(p_{k_{i}}-p_{k_{0}}\\right)=f\\left(p_{k_{i}}\\right)-f\\left(p_{k_{0}}\\right)=h_{k_{i}}-h_{k_{0}} \\equiv 0 \\quad(\\bmod a),\n$$\nso $p_{k_{i}}-p_{k_{0}} \\in S_{a}$ for all $i=1,2, \\ldots$. This provides infinitely many elements in $S_{a}$.\nHence, $S_{a}$ is an infinite set, and therefore $f(1)=a$ divides $f(n)$ for every $n$.\nLet $d_{n}=\\operatorname{gcd}(f(n), f(1))$. From $d_{n+1} \\mid f(1)$ and $d_{n+1}|f(n+1)| f(n)+f(1)$, we can see that $d_{n+1} \\mid f(n)$; then $d_{n+1} \\mid \\operatorname{gcd}(f(n), f(1))=d_{n}$. So the sequence $d_{1}, d_{2}, \\ldots$ is nonincreasing in the sense that every element is a divisor of the previous elements. Let $d=\\min \\left(d_{1}, d_{2}, \\ldots\\right)=\\operatorname{gcd}\\left(d_{1} . d_{2}, \\ldots\\right)=\\operatorname{gcd}(f(1), f(2), \\ldots) ;$ we have to prove $d \\geqslant 2$.\nFor the sake of contradiction, suppose that the statement is wrong, so $d=1$; that means there is some index $n_{0}$ such that $d_{n}=1$ for every $n \\geqslant n_{0}$, i.e., $f(n)$ is coprime with $f(1)$.\n\nClaim 1. If $2^{k} \\geqslant n_{0}$ then $f\\left(2^{k}\\right) \\leqslant 2^{k}$.\n\nProof. By the condition, $f(2 n) \\mid 2 f(n)$; a trivial induction yields $f\\left(2^{k}\\right) \\mid 2^{k} f(1)$. If $2^{k} \\geqslant n_{0}$ then $f\\left(2^{k}\\right)$ is coprime with $f(1)$, so $f\\left(2^{k}\\right)$ is a divisor of $2^{k}$.\n\nClaim 2. There is a constant $C$ such that $f(n)<n+C$ for every $n$.\n\nProof. Take the first power of 2 which is greater than or equal to $n_{0}$ : let $K=2^{k} \\geqslant n_{0}$. By Claim 1, we have $f(K) \\leqslant K$. Notice that $f(n+K) \\mid f(n)+f(K)$ implies $f(n+K) \\leqslant f(n)+f(K) \\leqslant f(n)+K$. If $n=t K+r$ for some $t \\geqslant 0$ and $1 \\leqslant r \\leqslant K$, then we conclude\n$f(n) \\leqslant K+f(n-K) \\leqslant 2 K+f(n-2 K) \\leqslant \\ldots \\leqslant t K+f(r)<n+\\max (f(1), f(2), \\ldots, f(K))$, so the claim is true with $C=\\max (f(1), \\ldots, f(K))$.\n\nClaim 3. If $a, b \\in \\mathbb{Z}_{>0}$ are coprime then $\\operatorname{gcd}(f(a), f(b)) \\mid f(1)$. In particular, if $a, b \\geqslant n_{0}$ are coprime then $f(a)$ and $f(b)$ are coprime.\n\nProof. Let $d=\\operatorname{gcd}(f(a), f(b))$. We can replicate Euclid's algorithm. Formally, apply induction on $a+b$. If $a=1$ or $b=1$ then we already have $d \\mid f(1)$.\nWithout loss of generality, suppose $1<a<b$. Then $d \\mid f(a)$ and $d|f(b)| f(a)+f(b-a)$, so $d \\mid f(b-a)$. Therefore $d$ divides $\\operatorname{gcd}(f(a), f(b-a))$ which is a divisor of $f(1)$ by the induction hypothesis.\n\nLet $p_{1}<p_{2}<\\ldots$ be the sequence of all prime numbers; for every $k$, let $q_{k}$ be the lowest power of $p_{k}$ with $q_{k} \\geqslant n_{0}$. (Notice that there are only finitely many positive integers with $q_{k} \\neq p_{k}$.)\nTake a positive integer $N$, and consider the numbers\n$$\nf(1), f\\left(q_{1}\\right), f\\left(q_{2}\\right), \\ldots, f\\left(q_{N}\\right)\n$$\nHere we have $N+1$ numbers, each being greater than 1 , and they are pairwise coprime by Claim 3. Therefore, they have at least $N+1$ different prime divisors in total, and their greatest prime divisor is at least $p_{N+1}$. Hence, $\\max \\left(f(1), f\\left(q_{1}\\right), \\ldots, f\\left(q_{N}\\right)\\right) \\geqslant p_{N+1}$.\nChoose $N$ such that $\\max \\left(q_{1}, \\ldots, q_{N}\\right)=p_{N}$ (this is achieved if $N$ is sufficiently large), and $p_{N+1}-p_{N}>C$ (that is possible, because there are arbitrarily long gaps between the primes). Then we establish a contradiction\n$$\np_{N+1} \\leqslant \\max \\left(f(1), f\\left(q_{1}\\right), \\ldots, f\\left(q_{N}\\right)\\right)<\\max \\left(1+C, q_{1}+C, \\ldots, q_{N}+C\\right)=p_{N}+C<p_{N+1}\n$$\nwhich proves the statement.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32065,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 2018$ be an integer, and let $a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n}$ be pairwise distinct positive integers not exceeding $5n$. Suppose that the sequence\n$$\n\\frac{a_{1}}{b_{1}}, \\frac{a_{2}}{b_{2}}, \\ldots, \\frac{a_{n}}{b_{n}}\n$$\nforms an arithmetic progression. Prove that the terms of the sequence are equal.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose that (1) is an arithmetic progression with nonzero difference. Let the difference be $\\Delta=\\frac{c}{d}$, where $d>0$ and $c, d$ are coprime.\nWe will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \\leqslant i \\leqslant n$ and any prime divisor $p$ of $d$, say that the index $i$ is $p$-wrong, if $v_{p}\\left(b_{i}\\right)<v_{p}(d)$. ($v_{p}(x)$ stands for the exponent of $p$ in the prime factorisation of $x$.)\n\nClaim 1. For any prime $p$, all $p$-wrong indices are congruent modulo $p$. In other words, the $p$-wrong indices (if they exist) are included in an arithmetic progression with difference $p$.\n\nProof. Let $\\alpha=v_{p}(d)$. For the sake of contradiction, suppose that $i$ and $j$ are $p$-wrong indices (i.e., none of $b_{i}$ and $b_{j}$ is divisible by $p^{\\alpha}$ ) such that $i \\not \\equiv j\\pmod{p}$. Then the least common denominator of $\\frac{a_{i}}{b_{i}}$ and $\\frac{a_{j}}{b_{j}}$ is not divisible by $p^{\\alpha}$. But this is impossible because in their difference, $(i-j) \\Delta=\\frac{(i-j) c}{d}$, the numerator is coprime to $p$, but $p^{\\alpha}$ divides the denominator $d$.\n\nClaim 2. $d$ has no prime divisors greater than $5$.\n\nProof. Suppose that $p \\geqslant 7$ is a prime divisor of $d$. Among the indices $1,2, \\ldots, n$, at most $\\left\\lceil\\frac{n}{p}\\right\\rceil<\\frac{n}{p}+1$ are $p$-wrong, so $p$ divides at least $\\frac{p-1}{p} n-1$ of $b_{1}, \\ldots, b_{n}$. Since these denominators are distinct,\n$$\n5 n \\geqslant \\max \\left\\{b_{i}: p \\mid b_{i}\\right\\} \\geqslant\\left(\\frac{p-1}{p} n-1\\right) p=(p-1)(n-1)-1 \\geqslant 6(n-1)-1>5 n\n$$\na contradiction.\n\nClaim 3. For every $0 \\leqslant k \\leqslant n-30$, among the denominators $b_{k+1}, b_{k+2}, \\ldots, b_{k+30}$, at least $\\varphi(30)=8$ are divisible by $d$.\n\nProof. By Claim 1, the $2$-wrong, $3$-wrong and $5$-wrong indices can be covered by three arithmetic progressions with differences $2,3$ and $5$. By a simple inclusion-exclusion, $(2-1) \\cdot(3-1) \\cdot(5-1)=8$ indices are not covered; by Claim 2, we have $d \\mid b_{i}$ for every uncovered index $i$.\n\nClaim 4. $|\\Delta|<\\frac{20}{n-2}$ and $d>\\frac{n-2}{20}$.\n\nProof. From the sequence (1), remove all fractions with $b_{n}<\\frac{n}{2}$. There remain at least $\\frac{n}{2}$ fractions, and they cannot exceed $\\frac{5 n}{n / 2}=10$. So we have at least $\\frac{n}{2}$ elements of the arithmetic progression (1) in the interval $(0,10]$, hence the difference must be below $\\frac{10}{n / 2-1}=\\frac{20}{n-2}$.\nThe second inequality follows from $\\frac{1}{d} \\leqslant \\frac{|c|}{d}=|\\Delta|$.\n\nNow we have everything to get the final contradiction. By Claim 3, we have $d \\mid b_{i}$ for at least $\\left\\lfloor\\frac{n}{30}\\right\\rfloor \\cdot 8$ indices $i$. By Claim 4, we have $d \\geqslant \\frac{n-2}{20}$. Therefore,\n$$\n5 n \\geqslant \\max \\left\\{b_{i}: d \\mid b_{i}\\right\\} \\geqslant\\left(\\left\\lfloor\\frac{n}{30}\\right\\rfloor \\cdot 8\\right) \\cdot d>\\left(\\frac{n}{30}-1\\right) \\cdot 8 \\cdot \\frac{n-2}{20}>5 n\n$$\nwhich is a contradiction. Therefore, the difference $\\Delta$ must be zero, i.e., all terms of the sequence are equal.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32066,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}$ be the set of integers. Determine all functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ such that, for all integers $a$ and $b$,\n$$\nf(2a) + 2f(b) = f(f(a+b)).\n$$",
    "options": [],
    "answer": "All functions are either f(n) = 0 for all integers n, or f(n) = 2n + K for any integer constant K.",
    "solution": "Substituting $a=0$, $b=n+1$ gives $f(f(n+1)) = f(0) + 2f(n+1)$. Substituting $a=1$, $b=n$ gives $f(f(n+1)) = f(2) + 2f(n)$.\nIn particular, $f(0) + 2f(n+1) = f(2) + 2f(n)$, and so $f(n+1) - f(n) = \\frac{1}{2}(f(2) - f(0))$. Thus $f(n+1) - f(n)$ must be constant. Since $f$ is defined only on $\\mathbb{Z}$, this tells us that $f$ must be a linear function; write $f(n) = M n + K$ for arbitrary constants $M$ and $K$, and we need only determine which choices of $M$ and $K$ work.\n\nNow, (1) becomes\n$$\n2M a + K + 2(M b + K) = M(M(a+b) + K) + K\n$$\nwhich we may rearrange to form\n$$\n(M-2)(M(a+b) + K) = 0\n$$\nThus, either $M = 2$, or $M(a+b) + K = 0$ for all values of $a+b$. In particular, the only possible solutions are $f(n) = 0$ and $f(n) = 2n + K$ for any constant $K \\in \\mathbb{Z}$, and these are easily seen to work.\nLet $K = f(0)$.\nFirst, put $a = 0$ in (1); this gives\n$$\n\\begin{equation*}\nf(f(b)) = 2f(b) + K \\tag{2}\n\\end{equation*}\n$$\nfor all $b \\in \\mathbb{Z}$.\nNow put $b = 0$ in (1); this gives\n$$\nf(2a) + 2K = f(f(a)) = 2f(a) + K,\n$$\nwhere the second equality follows from (2). Consequently,\n$$\n\\begin{equation*}\nf(2a) = 2f(a) - K \\tag{3}\n\\end{equation*}\n$$\nfor all $a \\in \\mathbb{Z}$.\nSubstituting (2) and (3) into (1), we obtain\n$$\n\\begin{aligned}\nf(2a) + 2f(b) & = f(f(a+b)) \\\\\n2f(a) - K + 2f(b) & = 2f(a+b) + K \\\\\nf(a) + f(b) & = f(a+b) + K .\n\\end{aligned}\n$$\nThus, if we set $g(n) = f(n) - K$ we see that $g$ satisfies the Cauchy equation $g(a+b) = g(a) + g(b)$. The solution to the Cauchy equation over $\\mathbb{Z}$ is well-known; indeed, it may be proven by an easy induction that $g(n) = M n$ for each $n \\in \\mathbb{Z}$, where $M = g(1)$ is a constant.\nTherefore, $f(n) = M n + K$, and we may proceed as in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32067,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $u_{1}, u_{2}, \\ldots, u_{2019}$ be real numbers satisfying\n$$\nu_{1}+u_{2}+\\cdots+u_{2019}=0 \\quad \\text{and} \\quad u_{1}^{2}+u_{2}^{2}+\\cdots+u_{2019}^{2}=1 .$$\nLet $a=\\min \\left(u_{1}, u_{2}, \\ldots, u_{2019}\\right)$ and $b=\\max \\left(u_{1}, u_{2}, \\ldots, u_{2019}\\right)$. Prove that\n$$\na b \\leqslant-\\frac{1}{2019}$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $P=\\left\\{i: u_{i}>0\\right\\}$ and $N=\\left\\{i: u_{i} \\leqslant 0\\right\\}$ be the indices of positive and nonpositive elements in the sequence, and let $p=|P|$ and $n=|N|$ be the sizes of these sets; then $p+n=2019$. By the condition $\\sum_{i=1}^{2019} u_{i}=0$ we have $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i \\in P} u_{i}-\\sum_{i \\in N}|u_{i}|$, so\n$$\n\\begin{equation*}\n\\sum_{i \\in P} u_{i}=\\sum_{i \\in N}|u_{i}| . \\tag{1}\n\\end{equation*}\n$$\nAfter this preparation, estimate the sum of squares of the positive and nonpositive elements as follows:\n$$\n\\begin{align*}\n& \\sum_{i \\in P} u_{i}^{2} \\leqslant \\sum_{i \\in P} b u_{i}=b \\sum_{i \\in P} u_{i}=b \\sum_{i \\in N}|u_{i}| \\leqslant b \\sum_{i \\in N}|a|=-n a b ;  \\tag{2}\\\\\n& \\sum_{i \\in N} u_{i}^{2} \\leqslant \\sum_{i \\in N}|a| \\cdot|u_{i}|=|a| \\sum_{i \\in N}|u_{i}|=|a| \\sum_{i \\in P} u_{i} \\leqslant|a| \\sum_{i \\in P} b=-p a b . \\tag{3}\n\\end{align*}\n$$\nThe sum of these estimates is\n$$\n1=\\sum_{i=1}^{2019} u_{i}^{2}=\\sum_{i \\in P} u_{i}^{2}+\\sum_{i \\in N} u_{i}^{2} \\leqslant-(p+n) a b=-2019 a b ;\n$$\nthat proves $a b \\leqslant \\frac{-1}{2019}$.\nAs in the previous solution we conclude that $a<0$ and $b>0$.\nFor every index $i$, the number $u_{i}$ is a convex combination of $a$ and $b$, so\n$$\nu_{i}=x_{i} a+y_{i} b \\quad \\text{with some weights} \\ 0 \\leqslant x_{i}, y_{i} \\leqslant 1, \\text{with} \\ x_{i}+y_{i}=1 .$$\nLet $X=\\sum_{i=1}^{2019} x_{i}$ and $Y=\\sum_{i=1}^{2019} y_{i}$. From $0=\\sum_{i=1}^{2019} u_{i}=\\sum_{i=1}^{2019}\\left(x_{i} a+y_{i} b\\right)=-|a| X+b Y$, we get\n$$\n\\begin{equation*}\n|a| X=b Y . \\tag{4}\n\\end{equation*}\n$$\nFrom $\\sum_{i=1}^{2019}\\left(x_{i}+y_{i}\\right)=2019$ we have\n$$\n\\begin{equation*}\nX+Y=2019 \\tag{5}\n\\end{equation*}\n$$\nThe system of linear equations $(4,5)$ has a unique solution:\n$$\nX=\\frac{2019 b}{|a|+b}, \\quad Y=\\frac{2019|a|}{|a|+b}\n$$\nNow apply the following estimate to every $u_{i}^{2}$ in their sum:\n$$\nu_{i}^{2}=x_{i}^{2} a^{2}+2 x_{i} y_{i} a b+y_{i}^{2} b^{2} \\leqslant x_{i} a^{2}+y_{i} b^{2}$$\nwe obtain that\n$$\n1=\\sum_{i=1}^{2019} u_{i}^{2} \\leqslant \\sum_{i=1}^{2019}\\left(x_{i} a^{2}+y_{i} b^{2}\\right)=X a^{2}+Y b^{2}=\\frac{2019 b}{|a|+b}|a|^{2}+\\frac{2019|a|}{|a|+b} b^{2}=2019|a| b=-2019 a b\n$$\nHence, $a b \\leqslant \\frac{-1}{2019}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32068,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 3$ be a positive integer and let $(a_{1}, a_{2}, \\ldots, a_{n})$ be a strictly increasing sequence of $n$ positive real numbers with sum equal to $2$. Let $X$ be a subset of $\\{1,2, \\ldots, n\\}$ such that the value of\n$$\n\\left|1-\\sum_{i \\in X} a_{i}\\right|\n$$\nis minimised. Prove that there exists a strictly increasing sequence of $n$ positive real numbers $(b_{1}, b_{2}, \\ldots, b_{n})$ with sum equal to $2$ such that\n$$\n\\sum_{i \\in X} b_{i}=1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. Without loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, and we may assume strict inequality as otherwise $b_{i}=a_{i}$ works. Also, $X$ clearly cannot be empty.\nIf $n \\in X$, add $\\Delta$ to $a_{n}$, producing a sequence of $c_{i}$ with $\\sum_{i \\in X} c_{i}=\\sum_{i \\in X^{c}} c_{i}$, and then scale as described above to make the sum equal to $2$. Otherwise, there is some $k$ with $k \\in X$ and $k+1 \\in X^{c}$. Let $\\delta=a_{k+1}-a_{k}$.\n- If $\\delta>\\Delta$, add $\\Delta$ to $a_{k}$ and then scale.\n- If $\\delta<\\Delta$, then considering $X \\cup\\{k+1\\} \\backslash\\{k\\}$ contradicts $X$ being $(a_{i})$-minimising.\n- If $\\delta=\\Delta$, choose any $j \\neq k, k+1$ (possible since $n \\geqslant 3$), and any $\\epsilon$ less than the least of $a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $j \\in X$ then add $\\Delta-\\epsilon$ to $a_{k}$ and $\\epsilon$ to $a_{j}$, then scale; otherwise, add $\\Delta$ to $a_{k}$ and $\\epsilon / 2$ to $a_{k+1}$, and subtract $\\epsilon / 2$ from $a_{j}$, then scale.\nSolution 2. This is similar to Solution 1, but without scaling. As in that solution, without loss of generality, assume $\\sum_{i \\in X} a_{i}<1$.\nSuppose there exists $1 \\leqslant j \\leqslant n-1$ such that $j \\in X$ but $j+1 \\in X^{c}$. Then $a_{j+1}-a_{j} \\geqslant \\Delta$, because otherwise considering $X \\cup\\{j+1\\} \\backslash\\{j\\}$ contradicts $X$ being $(a_{i})$-minimising.\nIf $a_{j+1}-a_{j}>\\Delta$, put\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text{ if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text{ if } i=j+1 \\\\ a_{i}, & \\text{ otherwise }\\end{cases}\n$$\nIf $a_{j+1}-a_{j}=\\Delta$, choose any $\\epsilon$ less than the least of $\\Delta / 2, a_{1}$ and all the differences $a_{i+1}-a_{i}$. If $|X| \\geqslant 2$, choose $k \\in X$ with $k \\neq j$, and put\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2-\\epsilon, & \\text{ if } i=j \\\\ a_{j+1}-\\Delta / 2, & \\text{ if } i=j+1 \\\\ a_{k}+\\epsilon, & \\text{ if } i=k \\\\ a_{i}, & \\text{ otherwise }\\end{cases}\n$$\nOtherwise, $|X^{c}| \\geqslant 2$, so choose $k \\in X^{c}$ with $k \\neq j+1$, and put\n$$\nb_{i}= \\begin{cases}a_{j}+\\Delta / 2, & \\text{ if } i=j ; \\\\ a_{j+1}-\\Delta / 2+\\epsilon, & \\text{ if } i=j+1 ; \\\\ a_{k}-\\epsilon, & \\text{ if } i=k ; \\\\ a_{i}, & \\text{ otherwise. }\\end{cases}\n$$\nIf there is no $1 \\leqslant j \\leqslant n$ such that $j \\in X$ but $j+1 \\in X^{c}$, there must be some $1<k \\leqslant n$ such that $X=[k, n]$ (certainly $X$ cannot be empty). We must have $a_{1}>\\Delta$, as otherwise considering $X \\cup\\{1\\}$ contradicts $X$ being $(a_{i})$-minimising. Now put\n$$\nb_{i}= \\begin{cases}a_{1}-\\Delta / 2, & \\text{ if } i=1 \\\\ a_{n}+\\Delta / 2, & \\text{ if } i=n \\\\ a_{i}, & \\text{ otherwise }\\end{cases}\n$$\nSolution 3. Without loss of generality, assume $\\sum_{i \\in X} a_{i} \\leqslant 1$, so $\\Delta \\geqslant 0$. If $\\Delta=0$ we can take $b_{i}=a_{i}$, so now assume that $\\Delta>0$.\nSuppose that there is some $k \\leqslant n$ such that $|X \\cap[k, n]|>|X^{c} \\cap[k, n]|$. If we choose the largest such $k$ then $|X \\cap[k, n]|-|X^{c} \\cap[k, n]|=1$. We can now find the required sequence $(b_{i})$ by starting with $c_{i}=a_{i}$ for $i<k$ and $c_{i}=a_{i}+\\Delta$ for $i \\geqslant k$, and then scaling as described above.\nIf no such $k$ exists, we will derive a contradiction. For each $i \\in X$ we can choose $i<j_{i} \\leqslant n$ in such a way that $j_{i} \\in X^{c}$ and all the $j_{i}$ are different. (For instance, note that necessarily $n \\in X^{c}$ and now just work downwards; each time an $i \\in X$ is considered, let $j_{i}$ be the least element of $X^{c}$ greater than $i$ and not yet used.) Let $Y$ be the (possibly empty) subset of $[1, n]$ consisting of those elements in $X^{c}$ that are also not one of the $j_{i}$. In any case\n$$\n\\Delta=\\sum_{i \\in X}\\left(a_{j_{i}}-a_{i}\\right)+\\sum_{j \\in Y} a_{j}\n$$\nwhere each term in the sums is positive. Since $n \\geqslant 3$ the total number of terms above is at least two. Take a least such term and its corresponding index $i$ and consider the set $Z$ which we form from $X$ by removing $i$ and adding $j_{i}$ (if it is a term of the first type) or just by adding $j$ if it is a term of the second type. The corresponding expression of $\\Delta$ for $Z$ has the sign of its least term changed, meaning that the sum is still nonnegative but strictly less than $\\Delta$, which contradicts $X$ being $(a_{i})$-minimising.\nSolution 4. This uses some similar ideas to Solution 3, but describes properties of the index sets $X$ that are sufficient to describe a corresponding sequence $(b_{i})$ that is not derived from $(a_{i})$.\nNote that, for two subsets $X, Y$ of $[1, n]$, the following are equivalent:\n- $|X \\cap[i, n]| \\leqslant|Y \\cap[i, n]|$ for all $1 \\leqslant i \\leqslant n$;\n- $Y$ is at least as large as $X$, and for all $1 \\leqslant j \\leqslant|Y|$, the $j^{\\text{th}}$ largest element of $Y$ is at least as big as the $j^{\\text{th}}$ largest element of $X$;\n- there is an injective function $f: X \\rightarrow Y$ such that $f(i) \\geqslant i$ for all $i \\in X$.\nIf these equivalent conditions are satisfied, we write $X \\leq Y$. We write $X<Y$ if $X \\leq Y$ and $X \\neq Y$.\nNote that if $X \\prec Y$, then $\\sum_{i \\in X} a_{i}<\\sum_{i \\in Y} a_{i}$ (the second description above makes this clear).\nWe claim first that, if $n \\geqslant 3$ and $X \\prec X^{c}$, then there exists $Y$ with $X \\prec Y \\prec X^{c}$. Indeed, as $|X| \\leqslant|X^{c}|$, we have $|X^{c}| \\geqslant 2$. Define $Y$ to consist of the largest element of $X^{c}$, together with all but the largest element of $X$; it is clear both that $Y$ is distinct from $X$ and $X^{c}$, and that $X \\leq Y \\leq X^{c}$, which is what we need.\nBut, in this situation, we have\n$$\n\\sum_{i \\in X} a_{i}<\\sum_{i \\in Y} a_{i}<\\sum_{i \\in X^{c}} a_{i} \\quad \\text{ and } \\quad 1-\\sum_{i \\in X} a_{i}=-\\left(1-\\sum_{i \\in X^{c}} a_{i}\\right),\n$$\nso $\\left|1-\\sum_{i \\in Y} a_{i}\\right|<\\left|1-\\sum_{i \\in X} a_{i}\\right|$.\nHence if $X$ is $(a_{i})$-minimising, we do not have $X<X^{c}$, and similarly we do not have $X^{c}<X$.\nConsidering the first description above, this immediately implies the following Claim.\nClaim. There exist $1 \\leqslant k, \\ell \\leqslant n$ such that $|X \\cap[k, n]|>\\frac{n-k+1}{2}$ and $|X \\cap[\\ell, n]|<\\frac{n-\\ell+1}{2}$.\nWe now construct our sequence $(b_{i})$ using this claim. Let $k$ and $\\ell$ be the greatest values satisfying the claim, and without loss of generality suppose $k=n$ and $\\ell<n$ (otherwise replace $X$ by its complement). As $\\ell$ is maximal, $n-\\ell$ is even and $|X \\cap[\\ell, n]|=\\frac{n-\\ell}{2}$. For sufficiently small positive $\\epsilon$, we take\n$$\nb_{i}=i \\epsilon+ \\begin{cases}0, & \\text{ if } i<\\ell \\\\ \\delta, & \\text{ if } \\ell \\leqslant i \\leqslant n-1 \\\\ \\gamma, & \\text{ if } i=n\\end{cases}\n$$\nLet $M=\\sum_{i \\in X} i$. So we require\n$$\nM \\epsilon+\\left(\\frac{n-\\ell}{2}-1\\right) \\delta+\\gamma=1\n$$\nand\n$$\n\\frac{n(n+1)}{2} \\epsilon+(n-\\ell) \\delta+\\gamma=2\n$$\nThese give\n$$\n\\gamma=2 \\delta+\\left(\\frac{n(n+1)}{2}-2 M\\right) \\epsilon\n$$\nand for sufficiently small positive $\\epsilon$, solving for $\\gamma$ and $\\delta$ gives $0<\\delta<\\gamma$ (since $\\epsilon=0$ gives $\\delta=1 /\\left(\\frac{n-\\ell}{2}+1\\right)$ and $\\gamma=2 \\delta$), so the sequence is strictly increasing and has positive values.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32069,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 2$ be a positive integer and $a_{1}, a_{2}, \\ldots, a_{n}$ be real numbers such that\n$$\na_{1}+a_{2}+\\cdots+a_{n}=0 .\n$$\nDefine the set $A$ by\n$$\nA=\\left\\{(i, j)\\left|1 \\leqslant i<j \\leqslant n,\\left|a_{i}-a_{j}\\right| \\geqslant 1\\right\\} .\\right.\n$$\nProve that, if $A$ is not empty, then\n$$\n\\sum_{(i, j) \\in A} a_{i} a_{j}<0 .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Define sets $B$ and $C$ by\n$$\n\\begin{aligned}\n& B=\\left\\{(i, j)\\left|1 \\leqslant i, j \\leqslant n,\\left|a_{i}-a_{j}\\right| \\geqslant 1\\right\\},\\right. \\\\\n& C=\\left\\{(i, j)\\left|1 \\leqslant i, j \\leqslant n,\\left|a_{i}-a_{j}\\right|<1\\right\\} .\\right.\n\\end{aligned}\n$$\nWe have\n$$\n\\begin{aligned}\n\\sum_{(i, j) \\in A} a_{i} a_{j} & =\\frac{1}{2} \\sum_{(i, j) \\in B} a_{i} a_{j} \\\\\n\\sum_{(i, j) \\in B} a_{i} a_{j} & =\\sum_{1 \\leqslant i, j \\leqslant n} a_{i} a_{j}-\\sum_{(i, j) \\notin B} a_{i} a_{j}=0-\\sum_{(i, j) \\in C} a_{i} a_{j} .\n\\end{aligned}\n$$\nSo it suffices to show that if $A$ (and hence $B$) are nonempty, then\n$$\n\\sum_{(i, j) \\in C} a_{i} a_{j}>0 .\n$$\nPartition the indices into sets $P, Q, R$, and $S$ such that\n$$\n\\begin{array}{ll}\nP=\\left\\{i \\mid a_{i} \\leqslant-1\\right\\} & R=\\left\\{i \\mid 0<a_{i}<1\\right\\} \\\\\nQ=\\left\\{i \\mid-1<a_{i} \\leqslant 0\\right\\} & S=\\left\\{i \\mid 1 \\leqslant a_{i}\\right\\} .\n\\end{array}\n$$\nThen\n$$\n\\sum_{(i, j) \\in C} a_{i} a_{j} \\geqslant \\sum_{i \\in P \\cup S} a_{i}^{2}+\\sum_{i, j \\in Q \\cup R} a_{i} a_{j}=\\sum_{i \\in P \\cup S} a_{i}^{2}+\\left(\\sum_{i \\in Q \\cup R} a_{i}\\right)^{2} \\geqslant 0 .\n$$\nThe first inequality holds because all of the positive terms in the RHS are also in the LHS, and all of the negative terms in the LHS are also in the RHS. The first inequality attains equality only if both sides have the same negative terms, which implies $\\left|a_{i}-a_{j}\\right|<1$ whenever $i, j \\in Q \\cup R$; the second inequality attains equality only if $P=S=\\varnothing$. But then we would have $A=\\varnothing$. So $A$ nonempty implies that the inequality holds strictly, as required.\nConsider $P, Q, R, S$ as in Solution 1, set\n$$\np=\\sum_{i \\in P} a_{i}, \\quad q=\\sum_{i \\in Q} a_{i}, \\quad r=\\sum_{i \\in R} a_{i}, \\quad s=\\sum_{i \\in S} a_{i},\n$$\nand let\n$$\nt_{+}=\\sum_{(i, j) \\in A, a_{i} a_{j} \\geqslant 0} a_{i} a_{j}, \\quad t_{-}=\\sum_{(i, j) \\in A, a_{i} a_{j} \\leqslant 0} a_{i} a_{j} .\n$$\nWe know that $p+q+r+s=0$, and we need to prove that $t_{+}+t_{-}<0$.\nNotice that $t_{+} \\leqslant p^{2} / 2+p q+r s+s^{2} / 2$ (with equality only if $p=s=0$), and $t_{-} \\leqslant p r+p s+q s$ (with equality only if there do not exist $i \\in Q$ and $j \\in R$ with $a_{j}-a_{i}>1$). Therefore,\n$$\nt_{+}+t_{-} \\leqslant \\frac{p^{2}+s^{2}}{2}+p q+r s+p r+p s+q s=\\frac{(p+q+r+s)^{2}}{2}-\\frac{(q+r)^{2}}{2}=-\\frac{(q+r)^{2}}{2} \\leqslant 0\n$$\nIf $A$ is not empty and $p=s=0$, then there must exist $i \\in Q, j \\in R$ with $\\left|a_{i}-a_{j}\\right|>1$, and hence the earlier equality conditions cannot both occur.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32070,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_{1}, x_{2}, \\ldots, x_{n}$ be different real numbers. Prove that\n$$\n\\sum_{1 \\leqslant i \\leqslant n} \\prod_{j \\neq i} \\frac{1-x_{i} x_{j}}{x_{i}-x_{j}}= \\begin{cases}0, & \\text{ if } n \\text{ is even } \\\\ 1, & \\text{ if } n \\text{ is odd }\\end{cases}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1 (Lagrange interpolation). Since both sides of the identity are rational functions, it suffices to prove it when all $x_{i} \\notin\\{ \\pm 1\\}$. Define\n$$\nf(t)=\\prod_{i=1}^{n}\\left(1-x_{i} t\\right)\n$$\nand note that\n$$\nf\\left(x_{i}\\right)=\\left(1-x_{i}^{2}\\right) \\prod_{j \\neq i} 1-x_{i} x_{j}\n$$\nUsing the nodes $+1,-1, x_{1}, \\ldots, x_{n}$, the Lagrange interpolation formula gives us the following expression for $f$ :\n$$\n\\sum_{i=1}^{n} f\\left(x_{i}\\right) \\frac{(x-1)(x+1)}{\\left(x_{i}-1\\right)\\left(x_{i}+1\\right)} \\prod_{j \\neq i} \\frac{x-x_{j}}{x_{i}-x_{j}}+f(1) \\frac{x+1}{1+1} \\prod_{1 \\leqslant i \\leqslant n} \\frac{x-x_{i}}{1-x_{i}}+f(-1) \\frac{x-1}{-1-1} \\prod_{1 \\leqslant i \\leqslant n} \\frac{x-x_{i}}{1-x_{i}} .\n$$\nThe coefficient of $t^{n+1}$ in $f(t)$ is 0 , since $f$ has degree $n$. The coefficient of $t^{n+1}$ in the above expression of $f$ is\n$$\n\\begin{aligned}\n0 & =\\sum_{1 \\leqslant i \\leqslant n} \\frac{f\\left(x_{i}\\right)}{\\prod_{j \\neq i}\\left(x_{i}-x_{j}\\right) \\cdot\\left(x_{i}-1\\right)\\left(x_{i}+1\\right)}+\\frac{f(1)}{\\prod_{1 \\leqslant j \\leqslant n}\\left(1-x_{j}\\right) \\cdot(1+1)}+\\frac{f(-1)}{\\prod_{1 \\leqslant j \\leqslant n}\\left(-1-x_{j}\\right) \\cdot(-1-1)} \\\\\n& =-G\\left(x_{1}, \\ldots, x_{n}\\right)+\\frac{1}{2}+\\frac{(-1)^{n+1}}{2}\n\\end{aligned}\n$$\n\n\nSolution 2 (using symmetries). Observe that $G$ is symmetric in the variables $x_{1}, \\ldots, x_{n}$. Define $V=\\prod_{i<j}\\left(x_{j}-x_{i}\\right)$ and let $F=G \\cdot V$, which is a polynomial in $x_{1}, \\ldots, x_{n}$. Since $V$ is alternating, $F$ is also alternating (meaning that, if we exchange any two variables, then $F$ changes sign). Every alternating polynomial in $n$ variables $x_{1}, \\ldots, x_{n}$ vanishes when any two variables $x_{i}, x_{j}(i \\neq j)$ are equal, and is therefore divisible by $x_{i}-x_{j}$ for each pair $i \\neq j$. Since these linear factors are pairwise coprime, $V$ divides $F$ exactly as a polynomial. Thus $G$ is in fact a symmetric polynomial in $x_{1}, \\ldots, x_{n}$.\nNow observe that if all $x_{i}$ are nonzero and we set $y_{i}=1 / x_{i}$ for $i=1, \\ldots, n$, then we have\n$$\n\\frac{1-y_{i} y_{j}}{y_{i}-y_{j}}=\\frac{1-x_{i} x_{j}}{x_{i}-x_{j}}\n$$\nso that\n$$\nG\\left(\\frac{1}{x_{1}}, \\ldots, \\frac{1}{x_{n}}\\right)=G\\left(x_{1}, \\ldots, x_{n}\\right)\n$$\nBy continuity this is an identity of rational functions. Since $G$ is a polynomial, it implies that $G$ is constant. (If $G$ were not constant, we could choose a point $\\left(c_{1}, \\ldots, c_{n}\\right)$ with all $c_{i} \\neq 0$, such that $G\\left(c_{1}, \\ldots, c_{n}\\right) \\neq G(0, \\ldots, 0)$; then $g(x):=G\\left(c_{1} x, \\ldots, c_{n} x\\right)$ would be a nonconstant polynomial in the variable $x$, so $|g(x)| \\rightarrow \\infty$ as $x \\rightarrow \\infty$, hence $\\left|G\\left(\\frac{y}{c_{1}}, \\ldots, \\frac{y}{c_{n}}\\right)\\right| \\rightarrow \\infty$ as $y \\rightarrow 0$, which is impossible since $G$ is a polynomial.)\nWe may identify the constant by substituting $x_{i}=\\zeta^{i}$, where $\\zeta$ is a primitive $n^{\\text{th}}$ root of unity in $\\mathbb{C}$. In the $i^{\\text{th}}$ term in the sum in the original expression we have a factor $1-\\zeta^{i} \\zeta^{n-i}=0$, unless $i=n$ or $2 i=n$. In the case where $n$ is odd, the only exceptional term is $i=n$, which gives the value $\\prod_{j \\neq n} \\frac{1-\\zeta^{j}}{1-\\zeta^{j}}=1$. When $n$ is even, we also have the term $\\prod_{j \\neq \\frac{n}{2}} \\frac{1+\\zeta^{j}}{-1-\\zeta^{j}}=(-1)^{n-1}=-1$, so the sum is 0 .\n\n\n$$\nG\\left(x_{1}, \\ldots, x_{n}\\right)=\\frac{P\\left(x_{n}\\right)}{\\prod_{j \\neq n}\\left(x_{n}-x_{j}\\right)}\n$$\nwhere $P\\left(x_{n}\\right)$ is a polynomial in $x_{n}$ whose coefficients are rational functions in the other variables. We then have\n$$\nP\\left(x_{n}\\right)=\\left(\\prod_{j \\neq n}\\left(1-x_{n} x_{j}\\right)\\right)+\\sum_{1 \\leqslant i \\leqslant n-1}\\left(x_{i} x_{n}-1\\right)\\left(\\prod_{j \\neq i, n}\\left(x_{n}-x_{j}\\right)\\right)\\left(\\prod_{j \\neq i, n} \\frac{1-x_{i} x_{j}}{x_{i}-x_{j}}\\right) .\n$$\nFor any $k \\neq n$, substituting $x_{n}=x_{k}$ (which is valid when manipulating the numerator $P\\left(x_{n}\\right)$\non its own), we have (noting that $x_{n}-x_{j}$ vanishes when $j=k$ )\n$$\n\\begin{aligned}\nP\\left(x_{k}\\right) & =\\left(\\prod_{j \\neq n}\\left(1-x_{k} x_{j}\\right)\\right)+\\sum_{1 \\leqslant i \\leqslant n-1}\\left(x_{i} x_{k}-1\\right)\\left(\\prod_{j \\neq i, n}\\left(x_{k}-x_{j}\\right)\\right)\\left(\\prod_{j \\neq i, n} \\frac{1-x_{i} x_{j}}{x_{i}-x_{j}}\\right) \\\\\n& =\\left(\\prod_{j \\neq n}\\left(1-x_{k} x_{j}\\right)\\right)+\\left(x_{k}^{2}-1\\right)\\left(\\prod_{j \\neq k, n}\\left(x_{k}-x_{j}\\right)\\right)\\left(\\prod_{j \\neq k, n} \\frac{1-x_{k} x_{j}}{x_{k}-x_{j}}\\right) \\\\\n& =\\left(\\prod_{j \\neq n}\\left(1-x_{k} x_{j}\\right)\\right)+\\left(x_{k}^{2}-1\\right)\\left(\\prod_{j \\neq k, n}\\left(1-x_{k} x_{j}\\right)\\right) \\\\\n& =0\n\\end{aligned}\n$$\nNote that $P$ is a polynomial in $x_{n}$ of degree $n-1$. For any choice of distinct real numbers $x_{1}, \\ldots, x_{n-1}, P$ has those real numbers as its roots, and the denominator has the same degree and the same roots. This shows that $G$ is constant in $x_{n}$, for any fixed choice of distinct $x_{1}, \\ldots, x_{n-1}$. Now, $G$ is symmetric in all $n$ variables, so it must be also be constant in each of the other variables. $G$ is therefore a constant that depends only on $n$. The constant may be identified as in the previous solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32071,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polynomial $P(x, y, z)$ in three variables with real coefficients satisfies the identities\n$$\nP(x, y, z) = P(x, y, x y - z) = P(x, z x - y, z) = P(y z - x, y, z)\n$$\nProve that there exists a polynomial $F(t)$ in one variable such that\n$$\nP(x, y, z) = F\\left(x^{2} + y^{2} + z^{2} - x y z\\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. In the first two steps, we deal with any polynomial $P(x, y, z)$ satisfying $P(x, y, z) = P(x, y, x y - z)$. Call such a polynomial weakly symmetric, and call a polynomial satisfying the full conditions in the problem symmetric.\n\nStep 1. We start with the description of weakly symmetric polynomials. We claim that they are exactly the polynomials in $x, y$, and $z(x y - z)$. Clearly, all such polynomials are weakly symmetric. For the converse statement, consider $P_{1}(x, y, z) := P\\left(x, y, z + \\frac{1}{2} x y\\right)$, which satisfies $P_{1}(x, y, z) = P_{1}(x, y, -z)$ and is therefore a polynomial in $x, y$, and $z^{2}$. This means that $P$ is a polynomial in $x, y$, and $\\left(z - \\frac{1}{2} x y\\right)^{2} = -z(x y - z) + \\frac{1}{4} x^{2} y^{2}$, and therefore a polynomial in $x, y$, and $z(x y - z)$.\n\nStep 2. Suppose that $P$ is weakly symmetric. Consider the monomials in $P(x, y, z)$ of highest total degree. Our aim is to show that in each such monomial $\\mu x^{a} y^{b} z^{c}$ we have $a, b \\geqslant c$. Consider the expansion\n$$\nP(x, y, z) = \\sum_{i, j, k} \\mu_{i j k} x^{i} y^{j} (z(x y - z))^{k} \\tag{1.1}\n$$\nThe maximal total degree of a summand in (1.1) is $m = \\max_{i, j, k: \\mu_{i j k} \\neq 0}(i + j + 3k)$. Now, for any $i, j, k$ satisfying $i + j + 3k = m$ the summand $\\mu_{i, j, k} x^{i} y^{j} (z(x y - z))^{k}$ has leading term of the form $\\mu x^{i + k} y^{j + k} z^{k}$. No other nonzero summand in (1.1) may have a term of this form in its expansion, hence this term does not cancel in the whole sum. Therefore, $\\operatorname{deg} P = m$, and the leading component of $P$ is exactly\n$$\n\\sum_{i + j + 3k = m} \\mu_{i, j, k} x^{i + k} y^{j + k} z^{k}\n$$\nand each summand in this sum satisfies the condition claimed above.\n\nStep 3. We now prove the problem statement by induction on $m = \\operatorname{deg} P$. For $m = 0$ the claim is trivial. Consider now a symmetric polynomial $P$ with $\\operatorname{deg} P > 0$. By Step 2, each of its monomials $\\mu x^{a} y^{b} z^{c}$ of the highest total degree satisfies $a, b \\geqslant c$. Applying other weak symmetries, we obtain $a, c \\geqslant b$ and $b, c \\geqslant a$; therefore, $P$ has a unique leading monomial of the form $\\mu(x y z)^{c}$. The polynomial $P_{0}(x, y, z) = P(x, y, z) - \\mu\\left(x y z - x^{2} - y^{2} - z^{2}\\right)^{c}$ has smaller total degree. Since $P_{0}$ is symmetric, it is representable as a polynomial function of $x y z - x^{2} - y^{2} - z^{2}$. Then $P$ is also of this form, completing the inductive step.\n\n\nSolution 2. We will rely on the well-known identity\n$$\n\\cos^{2} u + \\cos^{2} v + \\cos^{2} w - 2 \\cos u \\cos v \\cos w - 1 = 0 \\quad \\text{whenever} \\ u + v + w = 0. \\tag{2.1}\n$$\nClaim 1. The polynomial $P(x, y, z)$ is constant on the surface\n$$\n\\mathfrak{S} = \\{(2 \\cos u, 2 \\cos v, 2 \\cos w): u + v + w = 0\\}\n$$\nProof. Notice that for $x = 2 \\cos u, y = 2 \\cos v, z = 2 \\cos w$, the Vieta jumps $x \\mapsto y z - x$, $y \\mapsto z x - y, z \\mapsto x y - z$ in ( $*$ ) replace ( $u, v, w$ ) by ( $v - w, -v, w$ ), ( $u, w - u, -w$ ) and ( $-u, v, u - v$ ), respectively. For example, for the first type of jump we have\n$$\ny z - x = 4 \\cos v \\cos w - 2 \\cos u = 2 \\cos (v + w) + 2 \\cos (v - w) - 2 \\cos u = 2 \\cos (v - w).\n$$\nDefine $G(u, v, w) = P(2 \\cos u, 2 \\cos v, 2 \\cos w)$. For $u + v + w = 0$, the jumps give\n$$\n\\begin{aligned}\nG(u, v, w) & = G(v - w, -v, w) = G(w - v, -v, (v - w) - (-v)) = G(-u - 2v, -v, 2v - w) \\\\\n& = G(u + 2v, v, w - 2v)\n\\end{aligned}\n$$\nBy induction,\n$$\nG(u, v, w) = G(u + 2k v, v, w - 2k v) \\quad (k \\in \\mathbb{Z}). \\tag{2.2}\n$$\nSimilarly,\n$$\nG(u, v, w) = G(u, v - 2\\ell u, w + 2\\ell u) \\quad (\\ell \\in \\mathbb{Z}). \\tag{2.3}\n$$\nAnd, of course, we have\n$$\nG(u, v, w) = G(u + 2p \\pi, v + 2q \\pi, w - 2(p + q) \\pi) \\quad (p, q \\in \\mathbb{Z}) \\tag{2.4}\n$$\nTake two nonzero real numbers $u, v$ such that $u, v$ and $\\pi$ are linearly independent over $\\mathbb{Q}$. By combining (2.2-2.4), we can see that $G$ is constant on a dense subset of the plane $u + v + w = 0$. By continuity, $G$ is constant on the entire plane and therefore $P$ is constant on $\\mathfrak{S}$.\n\nClaim 2. The polynomial $T(x, y, z) = x^{2} + y^{2} + z^{2} - x y z - 4$ divides $P(x, y, z) - P(2,2,2)$.\nProof. By dividing $P$ by $T$ with remainders, there exist some polynomials $R(x, y, z), A(y, z)$ and $B(y, z)$ such that\n$$\nP(x, y, z) - P(2,2,2) = T(x, y, z) \\cdot R(x, y, z) + A(y, z) x + B(y, z). \\tag{2.5}\n$$\nOn the surface $\\mathfrak{S}$ the LHS of (2.5) is zero by Claim 1 (since $(2,2,2) \\in \\mathfrak{S}$ ) and $T = 0$ by (2.1). Hence, $A(y, z) x + B(y, z)$ vanishes on $\\mathfrak{S}$.\nNotice that for every $y = 2 \\cos v$ and $z = 2 \\cos w$ with $\\frac{\\pi}{3} < v, w < \\frac{2\\pi}{3}$, there are two distinct values of $x$ such that $(x, y, z) \\in \\mathfrak{S}$, namely $x_{1} = 2 \\cos (v + w)$ (which is negative), and $x_{2} = 2 \\cos (v - w)$ (which is positive). This can happen only if $A(y, z) = B(y, z) = 0$. Hence, $A(y, z) = B(y, z) = 0$ for $|y| < 1, |z| < 1$. The polynomials $A$ and $B$ vanish on an open set, so $A$ and $B$ are both the zero polynomial.\nThe quotient $(P(x, y, z) - P(2,2,2)) / T(x, y, z)$ is a polynomial of lower degree than $P$ and it also satisfies (*). The problem statement can now be proven by induction on the degree of $P$.\n\n\nSolution 3 (using algebraic geometry, just for interest). Let $Q = x^{2} + y^{2} + z^{2} - x y z$ and let $t \\in \\mathbb{C}$. Checking where $Q - t, \\frac{\\partial Q}{\\partial x}, \\frac{\\partial Q}{\\partial y}$ and $\\frac{\\partial Q}{\\partial z}$ vanish simultaneously, we find that the surface $Q = t$ is smooth except for the cases $t = 0$, when the only singular point is $(0,0,0)$, and $t = 4$, when the four points $(\\pm 2, \\pm 2, \\pm 2)$ that satisfy $x y z = 8$ are the only singular points. The singular points are the fixed points of the group $\\Gamma$ of polynomial automorphisms of $\\mathbb{C}^{3}$ generated by the three Vieta involutions\n$$\n\\iota_{1}: (x, y, z) \\mapsto (x, y, x y - z), \\quad \\iota_{2}: (x, y, z) \\mapsto (x, x z - y, z), \\quad \\iota_{3}: (x, y, z) \\mapsto (y z - x, y, z).\n$$\n$\\Gamma$ acts on each surface $\\mathcal{V}_{t}: Q - t = 0$. If $Q - t$ were reducible then the surface $Q = t$ would contain a curve of singular points. Therefore $Q - t$ is irreducible in $\\mathbb{C}[x, y, z]$. (One can also prove algebraically that $Q - t$ is irreducible, for example by checking that its discriminant as a quadratic polynomial in $x$ is not a square in $\\mathbb{C}[y, z]$, and likewise for the other two variables.) In the following solution we will only use the algebraic surface $\\mathcal{V}_{0}$.\nLet $U$ be the $\\Gamma$-orbit of $(3,3,3)$. Consider $\\iota_{3} \\circ \\iota_{2}$, which leaves $z$ invariant. For each fixed value of $z$, $\\iota_{3} \\circ \\iota_{2}$ acts linearly on $(x, y)$ by the matrix\n$$\nM_{z} := \\begin{pmatrix}\nz^{2} - 1 & -z \\\\\nz & -1\n\\end{pmatrix}\n$$\nThe reverse composition $\\iota_{2} \\circ \\iota_{3}$ acts by $M_{z}^{-1} = M_{z}^{\\text{adj}}$. Note $\\operatorname{det} M_{z} = 1$ and $\\operatorname{tr} M_{z} = z^{2} - 2$. When $z$ does not lie in the real interval $[-2, 2]$, the eigenvalues of $M_{z}$ do not have absolute value 1, so every orbit of the group generated by $M_{z}$ on $\\mathbb{C}^{2} \\setminus \\{(0,0)\\}$ is unbounded. For example, fixing $z = 3$ we find $(3 F_{2k+1}, 3 F_{2k-1}, 3) \\in U$ for every $k \\in \\mathbb{Z}$, where $(F_{n})_{n \\in \\mathbb{Z}}$ is the Fibonacci sequence with $F_{0} = 0, F_{1} = 1$.\nNow we may start at any point $(3 F_{2k+1}, 3 F_{2k-1}, 3)$ and iteratively apply $\\iota_{1} \\circ \\iota_{2}$ to generate another infinite sequence of distinct points of $U$, Zariski dense in the hyperbola cut out of $\\mathcal{V}_{0}$ by the plane $x - 3 F_{2k+1} = 0$. (The plane $x = a$ cuts out an irreducible conic when $a \\notin \\{-2, 0, 2\\}$.) Thus the Zariski closure $\\bar{U}$ of $U$ contains infinitely many distinct algebraic curves in $\\mathcal{V}_{0}$. Since $\\mathcal{V}_{0}$ is an irreducible surface this implies that $\\bar{U} = \\mathcal{V}_{0}$.\nFor any polynomial $P$ satisfying (*), we have $P - P(3,3,3) = 0$ at each point of $U$. Since $\\bar{U} = \\mathcal{V}_{0}$, $P - P(3,3,3)$ vanishes on $\\mathcal{V}_{0}$. Then Hilbert's Nullstellensatz and the irreducibility of $Q$ imply that $P - P(3,3,3)$ is divisible by $Q$. Now $(P - P(3,3,3)) / Q$ is a polynomial also satisfying (*), so we may complete the proof by an induction on the total degree, as in the other solutions.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32072,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}$ be the set of integers. We consider functions $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ satisfying\n$$\nf(f(x+y)+y)=f(f(x)+y)\n$$\nfor all integers $x$ and $y$. For such a function, we say that an integer $v$ is $f$-rare if the set\n$$\nX_{v}=\\{x \\in \\mathbb{Z}: f(x)=v\\}\n$$\nis finite and nonempty.\n\na. Prove that there exists such a function $f$ for which there is an $f$-rare integer.\n\nb. Prove that no such function $f$ can have more than one $f$-rare integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a) Let $f$ be the function where $f(0)=0$ and $f(x)$ is the largest power of $2$ dividing $2x$ for $x \\neq 0$. The integer $0$ is evidently $f$-rare, so it remains to verify the functional equation.\n\nSince $f(2x)=2f(x)$ for all $x$, it suffices to verify the functional equation when at least one of $x$ and $y$ is odd (the case $x=y=0$ being trivial). If $y$ is odd, then we have\n$$\nf(f(x+y)+y)=2=f(f(x)+y)\n$$\nsince all the values attained by $f$ are even. If, on the other hand, $x$ is odd and $y$ is even, then we already have\n$$\nf(x+y)=2=f(x)\n$$\nfrom which the functional equation follows immediately.\n\nb) An easy inductive argument (substituting $x+ky$ for $x$) shows that\n$$\nf(f(x+ky)+y)=f(f(x)+y)\n$$\nfor all integers $x$, $y$ and $k$. If $v$ is an $f$-rare integer and $a$ is the least element of $X_{v}$, then by substituting $y=a-f(x)$ in the above, we see that\n$$\nf(x+k \\cdot (a-f(x)))-f(x)+a \\in X_{v}\n$$\nfor all integers $x$ and $k$, so that in particular\n$$\nf(x+k \\cdot (a-f(x))) \\geqslant f(x)\n$$\nfor all integers $x$ and $k$, by assumption on $a$. This says that on the (possibly degenerate) arithmetic progression through $x$ with common difference $a-f(x)$, the function $f$ attains its minimal value at $x$.\n\nRepeating the same argument with $a$ replaced by the greatest element $b$ of $X_{v}$ shows that\n$$\nf(x+k \\cdot (b-f(x))) \\leqslant f(x)\n$$\nfor all integers $x$ and $k$. Combined with the above inequality, we therefore have\n$$\nf(x+k \\cdot (a-f(x)) \\cdot (b-f(x)))=f(x)\n$$\nfor all integers $x$ and $k$.\n\nThus if $f(x) \\neq a, b$, then the set $X_{f(x)}$ contains a nondegenerate arithmetic progression, so is infinite. So the only possible $f$-rare integers are $a$ and $b$.\n\nIn particular, the $f$-rare integer $v$ we started with must be one of $a$ or $b$, so that $f(v)=f(a)=f(b)=v$. This means that there cannot be any other $f$-rare integers $v'$, as they would on the one hand have to be either $a$ or $b$, and on the other would have to satisfy $f(v')=v'$. Thus $v$ is the unique $f$-rare integer.\nPart (b) only. Suppose $v$ is $f$-rare, and let $a$ and $b$ be the least and greatest elements of $X_{v}$, respectively. Substituting $x=v$ and $y=a-v$ into the equation shows that\n$$\nf(v)-v+a \\in X_{v}\n$$\nand in particular $f(v) \\geqslant v$. Repeating the same argument with $x=v$ and $y=b-v$ shows that $f(v) \\leqslant v$, and hence $f(v)=v$.\n\nSuppose now that $v'$ is a second $f$-rare integer. We may assume that $v=0$ (see Comment 1). We've seen that $f(v')=v'$; we claim that in fact $f(kv')=v'$ for all positive integers $k$. This gives a contradiction unless $v'=v=0$.\n\nThis claim is proved by induction on $k$. Supposing it to be true for $k$, we substitute $y=kv'$ and $x=0$ into the functional equation to yield\n$$\nf((k+1)v')=f(f(0)+kv')=f(kv')=v'\n$$\nusing that $f(0)=0$. This completes the induction, and hence the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32073,
    "subject": "Mathematics (Multi-modal)",
    "question": "The infinite sequence $a_{0}, a_{1}, a_{2}, \\ldots$ of (not necessarily different) integers has the following properties: $0 \\leqslant a_{i} \\leqslant i$ for all integers $i \\geqslant 0$, and\n$$\n\\binom{k}{a_{0}}+\\binom{k}{a_{1}}+\\cdots+\\binom{k}{a_{k}}=2^{k}\n$$\nfor all integers $k \\geqslant 0$.\nProve that all integers $N \\geqslant 0$ occur in the sequence (that is, for all $N \\geqslant 0$, there exists $i \\geqslant 0$ with $a_{i}=N$ ).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove by induction on $k$ that every initial segment of the sequence, $a_{0}, a_{1}, \\ldots, a_{k}$, consists of the following elements (counted with multiplicity, and not necessarily in order), for some $\\ell \\geqslant 0$ with $2 \\ell \\leqslant k+1$ :\n$$\n0,1, \\ldots, \\ell-1, \\quad 0,1, \\ldots, k-\\ell .\n$$\nFor $k=0$ we have $a_{0}=0$, which is of this form. Now suppose that for $k=m$ the elements $a_{0}, a_{1}, \\ldots, a_{m}$ are $0,0,1,1,2,2, \\ldots, \\ell-1, \\ell-1, \\ell, \\ell+1, \\ldots, m-\\ell-1, m-\\ell$ for some $\\ell$ with $0 \\leqslant 2 \\ell \\leqslant m+1$. It is given that\n$$\n\\binom{m+1}{a_{0}}+\\binom{m+1}{a_{1}}+\\cdots+\\binom{m+1}{a_{m}}+\\binom{m+1}{a_{m+1}}=2^{m+1},\n$$\nwhich becomes\n$$\n\\begin{aligned}\n\\left(\\binom{m+1}{0}+\\binom{m+1}{1}\\right. & \\left.+\\cdots+\\binom{m+1}{\\ell-1}\\right) \\\\\n& +\\left(\\binom{m+1}{0}+\\binom{m+1}{1}+\\cdots+\\binom{m+1}{m-\\ell}\\right)+\\binom{m+1}{a_{m+1}}=2^{m+1}\n\\end{aligned}\n$$\nor, using $\\binom{m+1}{i}=\\binom{m+1}{m+1-i}$, that\n$$\n\\begin{aligned}\n\\left(\\binom{m+1}{0}+\\binom{m+1}{1}\\right. & \\left.+\\cdots+\\binom{m+1}{\\ell-1}\\right) \\\\\n& +\\left(\\binom{m+1}{m+1}+\\binom{m+1}{m}+\\cdots+\\binom{m+1}{\\ell+1}\\right)+\\binom{m+1}{a_{m+1}}=2^{m+1}\n\\end{aligned}\n$$\nOn the other hand, it is well known that\n$$\n\\binom{m+1}{0}+\\binom{m+1}{1}+\\cdots+\\binom{m+1}{m+1}=2^{m+1},\n$$\nand so, by subtracting, we get\n$$\n\\binom{m+1}{a_{m+1}}=\\binom{m+1}{\\ell} .\n$$\nFrom this, using the fact that the binomial coefficients $\\binom{m+1}{i}$ are increasing for $i \\leqslant \\frac{m+1}{2}$ and decreasing for $i \\geqslant \\frac{m+1}{2}$, we conclude that either $a_{m+1}=\\ell$ or $a_{m+1}=m+1-\\ell$. In either case, $a_{0}, a_{1}, \\ldots, a_{m+1}$ is again of the claimed form, which concludes the induction.\nAs a result of this description, any integer $N \\geqslant 0$ appears as a term of the sequence $a_{i}$ for some $0 \\leqslant i \\leqslant 2 N$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32074,
    "subject": "Mathematics (Multi-modal)",
    "question": "You are given a set of $n$ blocks, each weighing at least $1$; their total weight is $2n$. Prove that for every real number $r$ with $0 \\leqslant r \\leqslant 2n-2$ you can choose a subset of the blocks whose total weight is at least $r$ but at most $r+2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Claim. Suppose that you have $n$ blocks, each of weight at least $1$, and of total weight $s \\leqslant 2n$. Then for every $r$ with $-2 \\leqslant r \\leqslant s$, you can choose some of the blocks whose total weight is at least $r$ but at most $r+2$.\n\nProof. The base case $n=1$ is trivial. To prove the inductive step, let $x$ be the largest block weight. Clearly, $x \\geqslant s/n$, so $s-x \\leqslant \\frac{n-1}{n} s \\leqslant 2(n-1)$. Hence, if we exclude a block of weight $x$, we can apply the inductive hypothesis to show the claim holds (for this smaller set) for any $-2 \\leqslant r \\leqslant s-x$. Adding the excluded block to each of those combinations, we see that the claim also holds when $x-2 \\leqslant r \\leqslant s$. So if $x-2 \\leqslant s-x$, then we have covered the whole interval $[-2, s]$. But each block weight is at least $1$, so we have $x-2 \\leqslant (s-(n-1))-2 = s-(2n-(n-1)) \\leqslant s-(s-(n-1)) \\leqslant s-x$, as desired.\nLet $x_1, \\ldots, x_n$ be the weights of the blocks in weakly increasing order. Consider the set $S$ of sums of the form $\\sum_{j \\in J} x_j$ for a subset $J \\subseteq \\{1,2, \\ldots, n\\}$. We want to prove that the mesh of $S$—i.e., the largest distance between two adjacent elements—is at most $2$.\n\nFor $0 \\leqslant k \\leqslant n$, let $S_k$ denote the set of sums of the form $\\sum_{i \\in J} x_i$ for a subset $J \\subseteq \\{1,2, \\ldots, k\\}$. We will show by induction on $k$ that the mesh of $S_k$ is at most $2$.\n\nThe base case $k=0$ is trivial (as $S_0 = \\{0\\}$). For $k > 0$ we have\n$$\nS_k = S_{k-1} \\cup \\left(x_k + S_{k-1}\\right)\n$$\n(where $\\left(x_k + S_{k-1}\\right)$ denotes $\\left\\{x_k + s : s \\in S_{k-1}\\right\\}$), so it suffices to prove that $x_k \\leqslant \\sum_{j<k} x_j + 2$. But if this were not the case, we would have $x_l > \\sum_{j<k} x_j + 2 \\geqslant k+1$ for all $l \\geqslant k$, and hence\n$$\n2n = \\sum_{j=1}^n x_j > (n+1-k)(k+1) + k-1\n$$\nThis rearranges to $n > k(n+1-k)$, which is false for $1 \\leqslant k \\leqslant n$, giving the desired contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32075,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Harry has $n$ coins lined up on his desk, each showing heads or tails. He repeatedly does the following operation: if there are $k$ coins showing heads and $k>0$, then he flips the $k^{\\text{th}}$ coin over; otherwise he stops the process. (For example, the process starting with $T H T$ would be $T H T \\rightarrow H H T \\rightarrow H T T \\rightarrow T T T$, which takes three steps.)\n\nLetting $C$ denote the initial configuration (a sequence of $n H$'s and $T$'s), write $\\ell(C)$ for the number of steps needed before all coins show $T$. Show that this number $\\ell(C)$ is finite, and determine its average value over all $2^{n}$ possible initial configurations $C$.",
    "options": [],
    "answer": "n(n+1)/4",
    "solution": "We represent the problem using a directed graph $G_{n}$ whose vertices are the length-$n$ strings of $H$'s and $T$'s. The graph features an edge from each string to its successor (except for $T T \\cdots T T$, which has no successor). We will also write $\\bar{H}=T$ and $\\bar{T}=H$.\n\nThe graph $G_{0}$ consists of a single vertex: the empty string. The main claim is that $G_{n}$ can be described explicitly in terms of $G_{n-1}$:\n- We take two copies, $X$ and $Y$, of $G_{n-1}$.\n- In $X$, we take each string of $n-1$ coins and just append a $T$ to it. In symbols, we replace $s_{1} \\cdots s_{n-1}$ with $s_{1} \\cdots s_{n-1} T$.\n- In $Y$, we take each string of $n-1$ coins, flip every coin, reverse the order, and append an $H$ to it. In symbols, we replace $s_{1} \\cdots s_{n-1}$ with $\\bar{s}_{n-1} \\bar{s}_{n-2} \\cdots \\bar{s}_{1} H$.\n- Finally, we add one new edge from $Y$ to $X$, namely $H H \\cdots H H H \\rightarrow H H \\cdots H H T$.\n\nWe depict $G_{4}$ below, in a way which indicates this recursive construction:\n\n![](attached_image_1.png)\n\nWe prove the claim inductively. Firstly, $X$ is correct as a subgraph of $G_{n}$, as the operation on coins is unchanged by an extra $T$ at the end: if $s_{1} \\cdots s_{n-1}$ is sent to $t_{1} \\cdots t_{n-1}$, then $s_{1} \\cdots s_{n-1} T$ is sent to $t_{1} \\cdots t_{n-1} T$.\n\nNext, $Y$ is also correct as a subgraph of $G_{n}$, as if $s_{1} \\cdots s_{n-1}$ has $k$ occurrences of $H$, then $\\bar{s}_{n-1} \\cdots \\bar{s}_{1} H$ has $(n-1-k)+1=n-k$ occurrences of $H$, and thus (provided that $k>0$), if $s_{1} \\cdots s_{n-1}$ is sent to $t_{1} \\cdots t_{n-1}$, then $\\bar{t}_{n-1} \\cdots \\bar{t}_{1} H$ is sent to $\\bar{t}_{n-1} \\cdots \\bar{t}_{1} H$.\n\nFinally, the one edge from $Y$ to $X$ is correct, as the operation does send $H H \\cdots H H H$ to $H H \\cdots H H T$.\n\nTo finish, note that the sequences in $X$ take an average of $E(n-1)$ steps to terminate, whereas the sequences in $Y$ take an average of $E(n-1)$ steps to reach $H H \\cdots H$ and then an additional $n$ steps to terminate. Therefore, we have\n$$\nE(n)=\\frac{1}{2}(E(n-1)+(E(n-1)+n))=E(n-1)+\\frac{n}{2}\n$$\nWe have $E(0)=0$ from our description of $G_{0}$. Thus, by induction, we have $E(n)=\\frac{1}{2}(1+\\cdots+ n)=\\frac{1}{4} n(n+1)$, which in particular is finite.\nWe consider what happens with configurations depending on the coins they start and end with.\n- If a configuration starts with $H$, the last $n-1$ coins follow the given rules, as if they were all the coins, until they are all $T$, then the first coin is turned over.\n- If a configuration ends with $T$, the last coin will never be turned over, and the first $n-1$ coins follow the given rules, as if they were all the coins.\n- If a configuration starts with $T$ and ends with $H$, the middle $n-2$ coins follow the given rules, as if they were all the coins, until they are all $T$. After that, there are $2 n-1$ more steps: first coins $1,2, \\ldots, n-1$ are turned over in that order, then coins $n, n-1, \\ldots, 1$ are turned over in that order.\n\nAs this covers all configurations, and the number of steps is clearly finite for 0 or 1 coins, it follows by induction on $n$ that the number of steps is always finite.\n\nWe define $E_{A B}(n)$, where $A$ and $B$ are each one of $H, T$ or $*$, to be the average number of steps over configurations of length $n$ restricted to those that start with $A$, if $A$ is not $*$, and that end with $B$, if $B$ is not $*$ (so $*$ represents \"either $H$ or $T$\"). The above observations tell us that, for $n \\geqslant 2$:\n- $E_{H *}(n)=E(n-1)+1$.\n- $E_{* T}(n)=E(n-1)$.\n- $E_{H T}(n)=E(n-2)+1$ (by using both the observations for $H *$ and for $* T$).\n- $E_{T H}(n)=E(n-2)+2 n-1$.\n\nNow $E_{H *}(n)=\\frac{1}{2}\\left(E_{H H}(n)+E_{H T}(n)\\right)$, so $E_{H H}(n)=2 E(n-1)-E(n-2)+1$. Similarly, $E_{T T}(n)=2 E(n-1)-E(n-2)-1$. So\n$$\nE(n)=\\frac{1}{4}\\left(E_{H T}(n)+E_{H H}(n)+E_{T T}(n)+E_{T H}(n)\\right)=E(n-1)+\\frac{n}{2} .\n$$\nWe have $E(0)=0$ and $E(1)=\\frac{1}{2}$, so by induction on $n$ we have $E(n)=\\frac{1}{4} n(n+1)$.\nLet $H_{i}$ be the number of heads in positions $1$ to $i$ inclusive (so $H_{n}$ is the total number of heads), and let $I_{i}$ be $1$ if the $i^{\\text{th}}$ coin is a head, $0$ otherwise. Consider the function\n$$\nt(i)=I_{i}+2\\left(\\min \\{i, H_{n}\\}-H_{i}\\right) .\n$$\nWe claim that $t(i)$ is the total number of times coin $i$ is turned over (which implies that the process terminates). Certainly $t(i)=0$ when all coins are tails, and $t(i)$ is always a nonnegative integer, so it suffices to show that when the $k^{\\text{th}}$ coin is turned over (where $k=H_{n}$), $t(k)$ goes down by $1$ and all the other $t(i)$ are unchanged. We show this by splitting into cases:\n- If $i<k$, $I_{i}$ and $H_{i}$ are unchanged, and $\\min \\{i, H_{n}\\}=i$ both before and after the coin flip, so $t(i)$ is unchanged.\n- If $i>k$, $\\min \\{i, H_{n}\\}=H_{n}$ both before and after the coin flip, and both $H_{n}$ and $H_{i}$ change by the same amount, so $t(i)$ is unchanged.\n- If $i=k$ and the coin is heads, $I_{i}$ goes down by $1$, as do both $\\min \\{i, H_{n}\\}=H_{n}$ and $H_{i}$; so $t(i)$ goes down by $1$.\n- If $i=k$ and the coin is tails, $I_{i}$ goes up by $1$, $\\min \\{i, H_{n}\\}=i$ is unchanged and $H_{i}$ goes up by $1$; so $t(i)$ goes down by $1$.\n\nWe now need to compute the average value of\n$$\n\\sum_{i=1}^{n} t(i)=\\sum_{i=1}^{n} I_{i}+2 \\sum_{i=1}^{n} \\min \\{i, H_{n}\\}-2 \\sum_{i=1}^{n} H_{i} .\n$$\nThe average value of the first term is $\\frac{1}{2} n$, and that of the third term is $-\\frac{1}{2} n(n+1)$. To compute the second term, we sum over choices for the total number of heads, and then over the possible values of $i$, getting\n$$\n2^{1-n} \\sum_{j=0}^{n}\\binom{n}{j} \\sum_{i=1}^{n} \\min \\{i, j\\}=2^{1-n} \\sum_{j=0}^{n}\\binom{n}{j}\\left(n j-\\binom{j}{2}\\right) .\n$$\nNow, in terms of trinomial coefficients,\n$$\n\\sum_{j=0}^{n} j\\binom{n}{j}=\\sum_{j=1}^{n}\\binom{n}{n-j, j-1,1\\}=n \\sum_{j=0}^{n-1}\\binom{n-1}{j}=2^{n-1} n\n$$\nand\n$$\n\\sum_{j=0}^{n}\\binom{j}{2}\\binom{n}{j}=\\sum_{j=2}^{n}\\binom{n}{n-j, j-2,2\\}=\\binom{n}{2} \\sum_{j=0}^{n-2}\\binom{n-2}{j}=2^{n-2}\\binom{n}{2} .\n$$\nSo the second term above is\n$$\n2^{1-n}\\left(2^{n-1} n^{2}-2^{n-2}\\binom{n}{2}\\right)=n^{2}-\\frac{n(n-1)}{4}\n$$\nand the required average is\n$$\nE(n)=\\frac{1}{2} n+n^{2}-\\frac{n(n-1)}{4}-\\frac{1}{2} n(n+1)=\\frac{n(n+1)}{4} .\n$$\nHarry has built a Turing machine to flip the coins for him. The machine is initially positioned at the $k^{\\text{th}}$ coin, where there are $k$ heads (and the position before the first coin is considered to be the $0^{\\text{th}}$ coin). The machine then moves according to the following rules, stopping when it reaches the position before the first coin: if the coin at its current position is $H$, it flips the coin and moves to the previous coin, while if the coin at its current position is $T$, it flips the coin and moves to the next position.\n\nConsider the maximal sequences of consecutive moves in the same direction. Suppose the machine has $a$ consecutive moves to the next coin, before a move to the previous coin. After those $a$ moves, the $a$ coins flipped in those moves are all heads, as is the coin the machine is now at, so at least the next $a+1$ moves will all be moves to the previous coin. Similarly, $a$ consecutive moves to the previous coin are followed by at least $a+1$ consecutive moves to the next coin. There cannot be more than $n$ consecutive moves in the same direction, so this proves that the process terminates (with a move from the first coin to the position before the first coin).\n\nThus we have a (possibly empty) sequence $a_{1}<\\cdots<a_{t} \\leqslant n$ giving the lengths of maximal sequences of consecutive moves in the same direction, where the final $a_{t}$ moves must be moves to the previous coin, ending before the first coin. We claim there is a bijection between initial configurations of the coins and such sequences. This gives\n$$\nE(n)=\\frac{1}{2}(1+2+\\cdots+n)=\\frac{n(n+1)}{4}\n$$\nas required, since each $i$ with $1 \\leqslant i \\leqslant n$ will appear in half of the sequences, and will contribute $i$ to the number of moves when it does.\n\nTo see the bijection, consider following the sequence of moves backwards, starting with the machine just before the first coin and all coins showing tails. This certainly determines a unique configuration of coins that could possibly correspond to the given sequence. Furthermore, every coin flipped as part of the $a_{j}$ consecutive moves is also flipped as part of all subsequent sequences of $a_{k}$ consecutive moves, for all $k>j$, meaning that, as we follow the moves backwards, each coin is always in the correct state when flipped to result in a move in the required direction. (Alternatively, since there are $2^{n}$ possible configurations of coins and $2^{n}$ possible such ascending sequences, the fact that the sequence of moves determines at most one configuration of coins, and thus that there is an injection from configurations of coins to such ascending sequences, is sufficient for it to be a bijection, without needing to show that coins are in the right state as we move backwards.)\nWe explicitly describe what happens with an arbitrary sequence $C$ of $n$ coins. Suppose that $C$ contain $k$ heads at positions $1 \\leqslant c_{1}<c_{2}<\\cdots<c_{k} \\leqslant n$.\n\nLet $i$ be the minimal index such that $c_{i} \\geqslant k$. Then the first few steps will consist of turning over the $k^{\\text{th}},(k+1)^{\\text{th}}, \\ldots, c_{i}^{\\text{th}},\\left(c_{i}-1\\right)^{\\text{th}},\\left(c_{i}-2\\right)^{\\text{th}}, \\ldots, k^{\\text{th}}$ coins in this order. After that we get a configuration with $k-1$ heads at the same positions as in the initial one, except for $c_{i}$. This part of the process takes $2\\left(c_{i}-k\\right)+1$ steps.\n\nAfter that, the process acts similarly; by induction on the number of heads we deduce that the process ends. Moreover, if the $c_{i}$ disappear in order $c_{i_{1}}, \\ldots, c_{i_{k}}$, the whole process takes\n$$\n\\ell(C)=\\sum_{j=1}^{k}\\left(2\\left(c_{i_{j}}-(k+1-j)\\right)+1\\right)=2 \\sum_{j=1}^{k} c_{j}-2 \\sum_{j=1}^{k}(k+1-j)+k=2 \\sum_{j=1}^{k} c_{j}-k^{2}\n$$\nsteps.\n\nNow let us find the total value $S_{k}$ of $\\ell(C)$ over all $\\binom{n}{k}$ configurations with exactly $k$ heads. To sum up the above expression over those, notice that each number $1 \\leqslant i \\leqslant n$ appears as $c_{j}$ exactly $\\binom{n-1}{k-1}$ times. Thus\n$$\n\\begin{array}{r}\nS_{k}=2\\binom{n-1}{k-1} \\sum_{i=1}^{n} i-\\binom{n}{k} k^{2}=2 \\frac{(n-1) \\cdots(n-k+1)}{(k-1)!} \\cdot \\frac{n(n+1)}{2}-\\frac{n \\cdots(n-k+1)}{k!} k^{2} \\\\\n=\\frac{n(n-1) \\cdots(n-k+1)}{(k-1)!}((n+1)-k)=n(n-1)\\binom{n-2}{k-1}+n\\binom{n-1}{k-1} .\n\\end{array}\n$$\nTherefore, the total value of $\\ell(C)$ over all configurations is\n$$\n\\sum_{k=1}^{n} S_{k}=n(n-1) \\sum_{k=1}^{n}\\binom{n-2}{k-1}+n \\sum_{k=1}^{n}\\binom{n-1}{k-1}=n(n-1) 2^{n-2}+n 2^{n-1}=2^{n} \\frac{n(n+1)}{4} .\n$$\nHence the required average is $E(n)=\\frac{n(n+1)}{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32076,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a flat plane in Camelot, King Arthur builds a labyrinth $\\mathfrak{L}$ consisting of $n$ walls, each of which is an infinite straight line. No two walls are parallel, and no three walls have a common point. Merlin then paints one side of each wall entirely red and the other side entirely blue.\nAt the intersection of two walls there are four corners: two diagonally opposite corners where a red side and a blue side meet, one corner where two red sides meet, and one corner where two blue sides meet. At each such intersection, there is a two-way door connecting the two diagonally opposite corners at which sides of different colours meet.\nAfter Merlin paints the walls, Morgana then places some knights in the labyrinth. The knights can walk through doors, but cannot walk through walls.\nLet $k(\\mathfrak{L})$ be the largest number $k$ such that, no matter how Merlin paints the labyrinth $\\mathfrak{L}$, Morgana can always place at least $k$ knights such that no two of them can ever meet. For each $n$, what are all possible values for $k(\\mathfrak{L})$, where $\\mathfrak{L}$ is a labyrinth with $n$ walls?",
    "options": [],
    "answer": "n+1",
    "solution": "First we show by induction that the $n$ walls divide the plane into $\\binom{n+1}{2}+1$ regions. The claim is true for $n=0$ as, when there are no walls, the plane forms a single region. When placing the $n^{\\text {th }}$ wall, it intersects each of the $n-1$ other walls exactly once and hence splits each of $n$ of the regions formed by those other walls into two regions. By the induction hypothesis, this yields $\\left(\\binom{n}{2}+1\\right)+n=\\binom{n+1}{2}+1$ regions, proving the claim.\n\nNow let $G$ be the graph with vertices given by the $\\binom{n+1}{2}+1$ regions, and with two regions connected by an edge if there is a door between them.\n\nWe now show that no matter how Merlin paints the $n$ walls, Morgana can place at least $n+1$ knights. No matter how the walls are painted, there are exactly $\\binom{n}{2}$ intersection points, each of which corresponds to a single edge in $G$. Consider adding the edges of $G$ sequentially and note that each edge reduces the number of connected components by at most one. Therefore the number of connected components of $G$ is at least $\\binom{n+1}{2}+1-\\binom{n}{2}=n+1$. If Morgana places a knight in regions corresponding to different connected components of $G$, then no two knights can ever meet.\n\nNow we give a construction showing that, no matter what shape the labyrinth is, Merlin can colour it such that there are exactly $n+1$ connected components, allowing Morgana to place at most $n+1$ knights.\n\nFirst, we choose a coordinate system on the labyrinth so that none of the walls run due north-south, or due east-west. We then have Merlin paint the west face of each wall red, and the east face of each wall blue. We label the regions according to how many walls the region is on the east side of: the labels are integers between 0 and $n$.\n\nWe claim that, for each $i$, the regions labelled $i$ are connected by doors. First, we note that for each $i$ with $0 \\leqslant i \\leqslant n$ there is a unique region labelled $i$ which is unbounded to the north.\n\nNow, consider a knight placed in some region with label $i$, and ask them to walk north (moving east or west by following the walls on the northern sides of regions, as needed). This knight will never get stuck: each region is convex, and so, if it is bounded to the north, it has a single northernmost vertex with a door northwards to another region with label $i$.\n\nEventually it will reach a region which is unbounded to the north, which will be the unique such region with label $i$. Hence every region with label $i$ is connected to this particular region, and so all regions with label $i$ are connected to each other.\n\nAs a result, there are exactly $n+1$ connected components, and Morgana can place at most $n+1$ knights.\nWe give another description of a strategy for Merlin to paint the walls so that Morgana can place no more than $n+1$ knights.\n\nMerlin starts by building a labyrinth of $n$ walls of his own design. He places walls in turn with increasing positive gradients, placing each so far to the right that all intersection points of previously-placed lines lie to the left of it. He paints each in such a way that blue is on the left and red is on the right.\n\nFor example, here is a possible sequence of four such lines $\\ell_{1}, \\ell_{2}, \\ell_{3}, \\ell_{4}$ :\n\n![](attached_image_1.png)\n\nWe say that a region is \"on the right\" if it has $x$-coordinate unbounded above (note that if we only have one wall, then both regions are on the right). We claim inductively that, after placing $n$ lines, there are $n+1$ connected components in the resulting labyrinth, each of which contains exactly one region on the right. This is certainly true after placing 0 lines, as then there is only one region (and hence one connected component) and it is on the right.\n\nWhen placing the $n^{\\text {th }}$ line, it then cuts every one of the $n-1$ previously placed lines, and since it is to the right of all intersection points, the regions it cuts are exactly the $n$ regions on the right.\n\n![](attached_image_2.png)\n\nThe addition of this line leaves all previous connected components with exactly one region on the right, and creates a new connected component containing exactly one region, and that region is also on the right. As a result, by induction, this particular labyrinth will have $n+1$ connected components.\n\nHaving built this labyrinth, Merlin then moves the walls one-by-one (by a sequence of continuous translations and rotations of lines) into the proper position of the given labyrinth, in such a way that no two lines ever become parallel.\n\nThe only time the configuration is changed is when one wall is moved through an intersection point of two others:\n\n![](attached_image_3.png)\n\n![](attached_image_4.png)\n\nNote that all moves really do switch between two configurations like this: all sets of three lines have this colour configuration initially, and the rules on rotations mean they are preserved (in particular, we cannot create three lines creating a triangle with three red edges inwards, or three blue edges inwards).\n\nHowever, as can be seen, such a move preserves the number of connected components, so in the painting this provides for Arthur's actual labyrinth, Morgana can still only place at most $n+1$ knights.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32077,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a certain social network, there are $2019$ users, some pairs of which are friends, where friendship is a symmetric relation. Initially, there are $1010$ people with $1009$ friends each and $1009$ people with $1010$ friends each. However, the friendships are rather unstable, so events of the following kind may happen repeatedly, one at a time:\n\nLet $A$, $B$, and $C$ be people such that $A$ is friends with both $B$ and $C$, but $B$ and $C$ are not friends; then $B$ and $C$ become friends, but $A$ is no longer friends with them.\n\nProve that, regardless of the initial friendships, there exists a sequence of such events after which each user is friends with at most one other user.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "All of the solutions presented below will use this reformulation.\n\nNote that the given graph is connected, since the total degree of any two vertices is at least $2018$ and hence they are either adjacent or have at least one neighbour in common. Hence the given graph satisfies the following condition:\n\n> Every connected component of $G$ with at least three vertices is not complete and has a vertex of odd degree.\n\nWe will show that if a graph $G$ satisfies condition (1) and has a vertex of degree at least $2$, then there is a refriending on $G$ that preserves condition (1). Since refriendings decrease the total number of edges of $G$, by using a sequence of such refriendings, we must reach a graph $G$ with maximal degree at most $1$, so we are done.\n\n![](attached_image_1.png)\n\nPick a vertex $A$ of degree at least $2$ in a connected component $G'$ of $G$. Since no component of $G$ with at least three vertices is complete we may assume that not all of the neighbours of $A$ are adjacent to one another. (For example, pick a maximal complete subgraph $K$ of $G'$. Some vertex $A$ of $K$ has a neighbour outside $K$, and this neighbour is not adjacent to every vertex of $K$ by maximality.) Removing $A$ from $G$ splits $G'$ into smaller connected components $G_1, \\ldots, G_k$ (possibly with $k=1$), to each of which $A$ is connected by at least one edge. We divide into several cases.\n\nCase 1: $k \\geqslant 2$ and $A$ is connected to some $G_i$ by at least two edges.\n\nChoose a vertex $B$ of $G_i$ adjacent to $A$, and a vertex $C$ in another component $G_j$ adjacent to $A$. The vertices $B$ and $C$ are not adjacent, and hence removing edges $AB$ and $AC$ and adding in edge $BC$ does not disconnect $G'$. It is easy to see that this preserves the condition, since the refriending does not change the parity of the degrees of vertices.\n\nCase 2: $k \\geqslant 2$ and $A$ is connected to each $G_i$ by exactly one edge.\n\nConsider the induced subgraph on any $G_i$ and the vertex $A$. The vertex $A$ has degree $1$ in this subgraph; since the number of odd-degree vertices of a graph is always even, we see that $G_i$ has a vertex of odd degree (in $G$). Thus if we let $B$ and $C$ be any distinct neighbours of $A$, then removing edges $AB$ and $AC$ and adding in edge $BC$ preserves the above condition: the refriending creates two new components, and if either of these components has at least three vertices, then it cannot be complete and must contain a vertex of odd degree (since each $G_i$ does).\n\nCase 3: $k=1$ and $A$ is connected to $G_1$ by at least three edges.\n\nBy assumption, $A$ has two neighbours $B$ and $C$ which are not adjacent to one another. Removing edges $AB$ and $AC$ and adding in edge $BC$ does not disconnect $G'$. We are then done as in Case 1.\n\nCase 4: $k=1$ and $A$ is connected to $G_1$ by exactly two edges.\n\nLet $B$ and $C$ be the two neighbours of $A$, which are not adjacent. Removing edges $AB$ and $AC$ and adding in edge $BC$ results in two new components: one consisting of a single vertex; and the other containing a vertex of odd degree. We are done unless this second component would be a complete graph on at least $3$ vertices. But in this case, $G_1$ would be a complete graph minus the single edge $BC$, and hence has at least $4$ vertices since $G'$ is not a $4$-cycle. If we let $D$ be a third vertex of $G_1$, then removing edges $BA$ and $BD$ and adding in edge $AD$ does not disconnect $G'$. We are then done as in Case 1.\n\n![](attached_image_2.png)\nAs in the previous solution, note that a refriending preserves the property that a graph has a vertex of odd degree and (trivially) the property that it is not complete; note also that our initial graph is connected. We describe an algorithm to reduce our initial graph to a graph of maximal degree at most $1$, proceeding in two steps.\n\nStep 1: There exists a sequence of refriendings reducing the graph to a tree.\n\nProof. Since the number of edges decreases with each refriending, it suffices to prove the following: as long as the graph contains a cycle, there exists a refriending such that the resulting graph is still connected. We will show that the graph in fact contains a cycle $Z$ and vertices $A, B, C$ such that $A$ and $B$ are adjacent in the cycle $Z$, $C$ is not in $Z$, and is adjacent to $A$ but not $B$. Removing edges $AB$ and $AC$ and adding in edge $BC$ keeps the graph connected, so we are done.\n\n![](attached_image_3.png)\n\nTo find this cycle $Z$ and vertices $A, B, C$, we pursue one of two strategies. If the graph contains a triangle, we consider a largest complete subgraph $K$, which thus contains at least three vertices. Since the graph itself is not complete, there is a vertex $C$ not in $K$ connected to a vertex $A$ of $K$. By maximality of $K$, there is a vertex $B$ of $K$ not connected to $C$, and hence we are done by choosing a cycle $Z$ in $K$ through the edge $AB$.\n\n![](attached_image_4.png)\n\nIf the graph is triangle-free, we consider instead a smallest cycle $Z$. This cycle cannot be Hamiltonian (i.e. it cannot pass through every vertex of the graph), since otherwise by minimality the graph would then have no other edges, and hence would have even degree at every vertex. We may thus choose a vertex $C$ not in $Z$ adjacent to a vertex $A$ of $Z$. Since the graph is triangle-free, it is not adjacent to any neighbour $B$ of $A$ in $Z$, and we are done.\n\nStep 2: Any tree may be reduced to a disjoint union of single edges and vertices by a sequence of refriendings.\n\nProof. The refriending preserves the property of being acyclic. Hence, after applying a sequence of refriendings, we arrive at an acyclic graph in which it is impossible to perform any further refriendings. The maximal degree of any such graph is $1$: if it had a vertex $A$ with two neighbours $B, C$, then $B$ and $C$ would necessarily be nonadjacent since the graph is cycle-free, and so a refriending would be possible. Thus we reach a graph with maximal degree at most $1$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32078,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n>1$ be an integer. Suppose we are given $2n$ points in a plane such that no three of them are collinear. The points are to be labelled $A_{1}, A_{2}, \\ldots, A_{2n}$ in some order. We then consider the $2n$ angles $\\angle A_{1}A_{2}A_{3}, \\angle A_{2}A_{3}A_{4}, \\ldots, \\angle A_{2n-2}A_{2n-1}A_{2n}, \\angle A_{2n-1}A_{2n}A_{1}, \\angle A_{2n}A_{1}A_{2}$. We measure each angle in the way that gives the smallest positive value (i.e. between $0^{\\circ}$ and $180^{\\circ}$). Prove that there exists an ordering of the given points such that the resulting $2n$ angles can be separated into two groups with the sum of one group of angles equal to the sum of the other group.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\ell$ be a line separating the points into two groups ($L$ and $R$) with $n$ points in each. Label the points $A_{1}, A_{2}, \\ldots, A_{2n}$ so that $L=\\{A_{1}, A_{3}, \\ldots, A_{2n-1}\\}$. We claim that this labelling works.\nTake a line $s = A_{2n}A_{1}$.\n(a) Rotate $s$ around $A_{1}$ until it passes through $A_{2}$; the rotation is performed in a direction such that $s$ is never parallel to $\\ell$.\n(b) Then rotate the new $s$ around $A_{2}$ until it passes through $A_{3}$ in a similar manner.\n(c) Perform $2n-2$ more such steps, after which $s$ returns to its initial position.\nThe total (directed) rotation angle $\\Theta$ of $s$ is clearly a multiple of $180^{\\circ}$. On the other hand, $s$ was never parallel to $\\ell$, which is possible only if $\\Theta=0$. Now it remains to partition all the $2n$ angles into those where $s$ is rotated anticlockwise, and the others.\nWhen tracing a cyclic path through the $A_{i}$ in order, with straight line segments between consecutive points, let $\\theta_{i}$ be the exterior angle at $A_{i}$, with a sign convention that it is positive if the path turns left and negative if the path turns right. Then $\\sum_{i=1}^{2n} \\theta_{i} = 360k^{\\circ}$ for some integer $k$. Let $\\phi_{i} = \\angle A_{i-1}A_{i}A_{i+1}$ (indices $\\bmod 2n$), defined as in the problem; thus $\\phi_{i} = 180^{\\circ} - |\\theta_{i}|$.\nLet $L$ be the set of $i$ for which the path turns left at $A_{i}$ and let $R$ be the set for which it turns right. Then $S = \\sum_{i \\in L} \\phi_{i} - \\sum_{i \\in R} \\phi_{i} = (180(|L| - |R|) - 360k)^{\\circ}$, which is a multiple of $360^{\\circ}$ since the number of points is even. We will show that the points can be labelled such that $S=0$, in which case $L$ and $R$ satisfy the required condition of the problem.\nNote that the value of $S$ is defined for a slightly larger class of configurations: it is OK for two points to coincide, as long as they are not consecutive, and OK for three points to be collinear, as long as $A_{i}, A_{i+1}$ and $A_{i+2}$ do not appear on a line in that order. In what follows it will be convenient, although not strictly necessary, to consider such configurations.\nConsider how $S$ changes if a single one of the $A_{i}$ is moved along some straight-line path (not passing through any $A_{j}$ and not lying on any line $A_{j}A_{k}$, but possibly crossing such lines). Because $S$ is a multiple of $360^{\\circ}$, and the angles change continuously, $S$ can only change when a point moves between $R$ and $L$. Furthermore, if $\\phi_{j}=0$ when $A_{j}$ moves between $R$ and $L$, $S$ is unchanged; it only changes if $\\phi_{j}=180^{\\circ}$ when $A_{j}$ moves between those sets.\nFor any starting choice of points, we will now construct a new configuration, with labels such that $S=0$, that can be perturbed into the original one without any $\\phi_{i}$ passing through $180^{\\circ}$, so that $S=0$ for the original configuration with those labels as well.\nTake some line such that there are $n$ points on each side of that line. The new configuration has $n$ copies of a single point on each side of the line, and a path that alternates between sides of the line; all angles are $0$, so this configuration has $S=0$. Perturbing the points into their original positions, while keeping each point on its side of the line, no angle $\\phi_{i}$ can pass through $180^{\\circ}$, because no straight line can go from one side of the line to the other and back. So the perturbation process leaves $S=0$.\nFirst, let $\\ell$ be a line in the plane such that there are $n$ points on one side and the other $n$ points on the other side. For convenience, assume $\\ell$ is horizontal (otherwise, we can rotate the plane). Then we can use the terms \"above\", \"below\", \"left\" and \"right\" in the usual way. We denote the $n$ points above the line in an arbitrary order as $P_{1}, P_{2}, \\ldots, P_{n}$, and the $n$ points below the line as $Q_{1}, Q_{2}, \\ldots, Q_{n}$.\nIf we connect $P_{i}$ and $Q_{j}$ with a line segment, the line segment will intersect with the line $\\ell$. Denote the intersection as $I_{ij}$. If $P_{i}$ is connected to $Q_{j}$ and $Q_{k}$, where $j<k$, then $I_{ij}$ and $I_{ik}$ are two different points, because $P_{i}, Q_{j}$ and $Q_{k}$ are not collinear.\nNow we define a \"sign\" for each angle $\\angle Q_{j}P_{i}Q_{k}$. Assume $j<k$. We specify that the sign is positive for the following two cases:\n- if $i$ is odd and $I_{ij}$ is to the left of $I_{ik}$,\n- if $i$ is even and $I_{ij}$ is to the right of $I_{ik}$.\nOtherwise the sign of the angle is negative. If $j>k$, then the sign of $\\angle Q_{j}P_{i}Q_{k}$ is taken to be the same as for $\\angle Q_{k}P_{i}Q_{j}$.\nSimilarly, we can define the sign of $\\angle P_{j}Q_{i}P_{k}$ with $j<k$ (or equivalently $\\angle P_{k}Q_{i}P_{j}$). For example, it is positive when $i$ is odd and $I_{ji}$ is to the left of $I_{ki}$.\nHenceforth, whenever we use the notation $\\angle Q_{j}P_{i}Q_{k}$ or $\\angle P_{j}Q_{i}P_{k}$ for a numerical quantity, it is understood to denote either the (geometric) measure of the angle or the negative of this measure, depending on the sign as specified above.\nWe now have the following important fact for signed angle measures:\n$$\n\\angle Q_{i_{1}}P_{k}Q_{i_{3}} = \\angle Q_{i_{1}}P_{k}Q_{i_{2}} + \\angle Q_{i_{2}}P_{k}Q_{i_{3}} \\tag{1}\n$$\nfor all points $P_{k}, Q_{i_{1}}, Q_{i_{2}}$ and $Q_{i_{3}}$ with $i_{1}<i_{2}<i_{3}$. The following figure shows a \"natural\" arrangement of the points. Equation (1) still holds for any other arrangement, as can be easily verified.\n\n![](attached_image_1.png)\n\nSimilarly, we have\n$$\n\\angle P_{i_{1}}Q_{k}P_{i_{3}} = \\angle P_{i_{1}}Q_{k}P_{i_{2}} + \\angle P_{i_{2}}Q_{k}P_{i_{3}} \\tag{2}\n$$\nfor all points $Q_{k}, P_{i_{1}}, P_{i_{2}}$ and $P_{i_{3}}$, with $i_{1}<i_{2}<i_{3}$.\nWe are now ready to specify the desired ordering $A_{1}, \\ldots, A_{2n}$ of the points:\n- if $i \\leqslant n$ is odd, put $A_{i} = P_{i}$ and $A_{2n+1-i} = Q_{i}$;\n- if $i \\leqslant n$ is even, put $A_{i} = Q_{i}$ and $A_{2n+1-i} = P_{i}$.\nFor example, for $n=3$ this ordering is $P_{1}, Q_{2}, P_{3}, Q_{3}, P_{2}, Q_{1}$. This sequence alternates between $P$'s and $Q$'s, so the above conventions specify a sign for each of the angles $A_{i-1}A_{i}A_{i+1}$. We claim that the sum of these $2n$ signed angles equals $0$. If we can show this, it would complete the proof.\nWe prove the claim by induction. For brevity, we use the notation $\\angle P_{i}$ to denote whichever of the $2n$ angles has its vertex at $P_{i}$, and $\\angle Q_{i}$ similarly.\nFirst let $n=2$. If the four points can be arranged to form a convex quadrilateral, then the four line segments $P_{1}Q_{1}, P_{1}Q_{2}, P_{2}Q_{1}$ and $P_{2}Q_{2}$ constitute a self-intersecting quadrilateral. We use several figures to illustrate the possible cases.\nThe following figure is one possible arrangement of the points.\n\n![](attached_image_2.png)\n\nThen $\\angle P_{1}$ and $\\angle Q_{1}$ are positive, $\\angle P_{2}$ and $\\angle Q_{2}$ are negative, and we have\n$$\n|\\angle P_{1}| + |\\angle Q_{1}| = |\\angle P_{2}| + |\\angle Q_{2}|.\n$$\nWith signed measures, we have\n$$\n\\angle P_{1} + \\angle Q_{1} + \\angle P_{2} + \\angle Q_{2} = 0. \\tag{3}\n$$\nIf we switch the labels of $P_{1}$ and $P_{2}$, we have the following picture:\n\n![](attached_image_3.png)\n\nSwitching labels $P_{1}$ and $P_{2}$ has the effect of flipping the sign of all four angles (as well as swapping the magnitudes on the relabelled points); that is, the new values of ($\\angle P_{1}, \\angle P_{2}, \\angle Q_{1}, \\angle Q_{2}$) equal the old values of ($-\\angle P_{2}, -\\angle P_{1}, -\\angle Q_{1}, -\\angle Q_{2}$). Consequently, equation (3) still holds. Similarly, when switching the labels of $Q_{1}$ and $Q_{2}$, or both the $P$'s and the $Q$'s, equation (3) still holds.\nThe remaining subcase of $n=2$ is that one point lies inside the triangle formed by the other three. We have the following picture.\n\n![](attached_image_4.png)\n\nWe have\n$$\n|\\angle P_{1}| + |\\angle Q_{1}| + |\\angle Q_{2}| = |\\angle P_{2}|.\n$$\nand equation (3) holds.\nAgain, switching the labels for $P$'s or the $Q$'s will not affect the validity of equation (3). Also, if the point lying inside the triangle of the other three is one of the $Q$'s rather than the $P$'s, the result still holds, since our sign convention is preserved when we relabel $Q$'s as $P$'s and vice-versa and reflect across $\\ell$.\nWe have completed the proof of the claim for $n=2$.\nAssume the claim holds for $n=k$, and we wish to prove it for $n=k+1$. Suppose we are given our $2(k+1)$ points. First ignore $P_{k+1}$ and $Q_{k+1}$, and form $2k$ angles from $P_{1}, \\ldots, P_{k}$, $Q_{1}, \\ldots, Q_{k}$ as in the $n=k$ case. By the induction hypothesis we have\n$$\n\\sum_{i=1}^{k} (\\angle P_{i} + \\angle Q_{i}) = 0.\n$$\nWhen we add in the two points $P_{k+1}$ and $Q_{k+1}$, this changes our angles as follows:\n- the angle at $P_{k}$ changes from $\\angle Q_{k-1}P_{k}Q_{k}$ to $\\angle Q_{k-1}P_{k}Q_{k+1}$;\n- the angle at $Q_{k}$ changes from $\\angle P_{k-1}Q_{k}P_{k}$ to $\\angle P_{k-1}Q_{k}P_{k+1}$;\n- two new angles $\\angle Q_{k}P_{k+1}Q_{k+1}$ and $\\angle P_{k}Q_{k+1}P_{k+1}$ are added.\nWe need to prove the changes have no impact on the total sum. In other words, we need to prove\n$$\n\\left(\\angle Q_{k-1}P_{k}Q_{k+1} - \\angle Q_{k-1}P_{k}Q_{k}\\right) + \\left(\\angle P_{k-1}Q_{k}P_{k+1} - \\angle P_{k-1}Q_{k}P_{k}\\right) + (\\angle P_{k+1} + \\angle Q_{k+1}) = 0. \\tag{4}\n$$\nIn fact, from equations (1) and (2), we have\n$$\n\\angle Q_{k-1}P_{k}Q_{k+1} - \\angle Q_{k-1}P_{k}Q_{k} = \\angle Q_{k}P_{k}Q_{k+1}\n$$\nand\n$$\n\\angle P_{k-1}Q_{k}P_{k+1} - \\angle P_{k-1}Q_{k}P_{k} = \\angle P_{k}Q_{k}P_{k+1}.\n$$\nTherefore, the left hand side of equation (4) becomes $\\angle Q_{k}P_{k}Q_{k+1} + \\angle P_{k}Q_{k}P_{k+1} + \\angle Q_{k}P_{k+1}Q_{k+1} + \\angle P_{k}Q_{k+1}P_{k+1}$, which equals $0$, simply by applying the $n=2$ case of the claim. This completes the induction.\nWe shall think instead of the problem as asking us to assign a weight $\\pm 1$ to each angle, such that the weighted sum of all the angles is zero.\nGiven an ordering $A_{1}, \\ldots, A_{2n}$ of the points, we shall assign weights according to the following recipe: walk in order from point to point, and assign the left turns $+1$ and the right turns $-1$. This is the same weighting as in Solution 3, and as in that solution, the weighted sum is a multiple of $360^{\\circ}$.\nWe now aim to show the following:\n\n**Lemma.** Transposing any two consecutive points in the ordering changes the weighted sum by $\\pm 360^{\\circ}$ or $0$.\n\nKnowing that, we can conclude quickly: if the ordering $A_{1}, \\ldots, A_{2n}$ has weighted angle sum $360k^{\\circ}$, then the ordering $A_{2n}, \\ldots, A_{1}$ has weighted angle sum $-360k^{\\circ}$ (since the angles are the same, but left turns and right turns are exchanged). We can reverse the ordering of $A_{1}, \\ldots, A_{2n}$ by a sequence of transpositions of consecutive points, and in doing so the weighted angle sum must become zero somewhere along the way.\n\nWe now prove that lemma:\n\nProof. Transposing two points amounts to taking a section $A_{k}A_{k+1}A_{k+2}A_{k+3}$ as depicted, reversing the central line segment $A_{k+1}A_{k+2}$, and replacing its two neighbours with the dotted lines.\n\n![](attached_image_5.png)\n\nFigure 1: Transposing two consecutive vertices: before (left) and afterwards (right)\n\nIn each triangle, we alter the sum by $\\pm 180^{\\circ}$. Indeed, using (anticlockwise) directed angles modulo $360^{\\circ}$, we either add or subtract all three angles of each triangle.\nHence both triangles together alter the sum by $\\pm 180 \\pm 180^{\\circ}$, which is $\\pm 360^{\\circ}$ or $0$. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32079,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are 60 empty boxes $B_{1}, \\ldots, B_{60}$ in a row on a table and an unlimited supply of pebbles. Given a positive integer $n$, Alice and Bob play the following game.\n\nIn the first round, Alice takes $n$ pebbles and distributes them into the 60 boxes as she wishes. Each subsequent round consists of two steps:\n\na. Bob chooses an integer $k$ with $1 \\leqslant k \\leqslant 59$ and splits the boxes into the two groups $B_{1}, \\ldots, B_{k}$ and $B_{k+1}, \\ldots, B_{60}$.\n\nb. Alice picks one of these two groups, adds one pebble to each box in that group, and removes one pebble from each box in the other group.\n\nBob wins if, at the end of any round, some box contains no pebbles. Find the smallest $n$ such that Alice can prevent Bob from winning.",
    "options": [],
    "answer": "960",
    "solution": "Solution 1 (Alice).\n\nAlice initially distributes pebbles according to $V_{\\left\\lceil\\frac{N}{2}\\right\\rceil}$. Suppose the current configuration of pebbles dominates $V_{i}$. If Bob makes a $k$-move with $k \\geqslant i$ then Alice picks the left group, which results in a configuration that dominates $V_{i+1}$. Likewise, if Bob makes a $k$-move with $k<i$ then Alice picks the right group, which results in a configuration that dominates $V_{i-1}$. Since none of $V_{1}, \\ldots, V_{N}$ contains an empty box, Alice can prevent Bob from ever winning.\n\n\nSolution 1 (Bob).\n\nThe key idea in this solution is the following claim.\n\n**Claim.** If there exist a positive integer $k$ such that there are at least $2k$ boxes that have at most $k$ pebbles each then Bob can force a win.\n\n**Proof.** We ignore the other boxes. First, Bob makes a $k$-move (splits the $2k$ boxes into two groups of $k$ boxes each). Without loss of generality, Alice picks the left group. Then Bob makes a $(k+1)$-move, $\\ldots$, a $(2k-1)$-move. By Observation A, we may suppose Alice always picks the left group. After Bob's $(2k-1)$-move, the rightmost box becomes empty and Bob wins.\n\nNow, we claim that if $n<M$ then either there already exists an empty box, or there exist a positive integer $k$ and $2k$ boxes with at most $k$ pebbles each (and thus Bob can force a win). Otherwise, assume each box contains at least 1 pebble, and for each $1 \\leqslant k \\leqslant \\left\\lfloor\\frac{N}{2}\\right\\rfloor$, at least $N-(2k-1)=N+1-2k$ boxes contain at least $k+1$ pebbles. Summing, there are at least as many pebbles in total as in $V_{\\left\\lceil\\frac{N}{2}\\right\\rceil}$; that is, at least $M$ pebbles, as desired.\n\n\nSolution 2 (Alice).\n\nLet $K=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor$. Alice starts with the boxes in the configuration $V_{K}$. For each of Bob's $N-1$ possible choices, consider the subset of rounds in which he makes that choice. In that subset of rounds, Alice alternates between picking the left group and picking the right group; the first time Bob makes that choice, Alice picks the group containing the $K^{\\text{th}}$ box. Thus, at any time during the game, the number of pebbles in each box depends only on which choices Bob has made an odd number of times. This means that the number of pebbles in a box could decrease by at most the number of choices for which Alice would have started by removing a pebble from the group containing that box. These numbers are, for each box,\n$$\n\\left\\lfloor\\frac{N}{2}\\right\\rfloor,\\left\\lfloor\\frac{N}{2}-1\\right\\rfloor, \\ldots, 1,0,1, \\ldots,\\left\\lceil\\frac{N}{2}-1\\right\\rceil\n$$\nThese are pointwise less than the numbers of pebbles the boxes started with, meaning that no box ever becomes empty with this strategy.\n\n\nSolution 2 (Bob).\n\nLet $K=\\left\\lfloor\\frac{N}{2}+1\\right\\rfloor$. For Bob's strategy, we consider a configuration $X$ with at most $M-1$ pebbles, and we make use of Observation A. Consider two configurations with $M$ pebbles: $V_{K}$ and $V_{N+1-K}$ (if $n$ is odd, they are the same configuration; if $n$ is even, one is the reverse of the other). The configuration $X$ has fewer pebbles than $V_{K}$ in at least one box, and fewer pebbles than $V_{N+1-K}$ in at least one box.\n\nSuppose first that, with respect to one of those configurations (without loss of generality $V_{K}$), $X$ has fewer pebbles in one of the boxes in the half where they have $1,2, \\ldots,\\left\\lceil\\frac{N}{2}\\right\\rceil$ pebbles (the right half in $V_{K}$ if $N$ is even; if $N$ is odd, we can take it to be the right half, without loss of generality, as the configuration is symmetric). Note that the number cannot be fewer in the box with 1 pebble in $V_{K}$, because then it would have 0 pebbles. Bob then does a $K$-move. If Alice picks the right group, the total number of pebbles goes down and we restart Bob's strategy with a smaller number of pebbles. If Alice picks the left group, Bob follows with a $(K+1)$-move, a $(K+2)$-move, and so on; by Observation A we may assume Alice always picks the left group. But whichever box in the right half had fewer pebbles in $X$ than in $V_{K}$ ends up with 0 pebbles at some point in this sequence of moves.\n\nOtherwise, $N$ is even, and for both of those configurations, there are fewer pebbles in $X$ only on the $2,3, \\ldots, \\frac{N}{2}+1$ side. That is, the numbers of pebbles in $X$ are at least\n$$\n\\frac{N}{2}, \\frac{N}{2}-1, \\ldots, 1,1, \\ldots, \\frac{N}{2}\n$$\nwith equality occurring at least once on each side. Bob does an $\\frac{N}{2}$-move. Whichever group Alice chooses, the total number of pebbles is unchanged, and the side from which pebbles are removed now has a box with fewer pebbles than in $(C)$, so the previous case of Bob's strategy can now be applied.\n\n\nSolution 3 (Bob).\n\nFor any configuration $C$, define $L(C)$ to be the greatest integer such that, for all $0 \\leqslant i \\leqslant N-1$, the box $B_{i+1}$ contains at least $L(C)-i$ pebbles. Similarly, define $R(C)$ to be greatest integer such that, for all $0 \\leqslant i \\leqslant N-1$, the box $B_{N-i}$ contains at least $R(C)-i$ pebbles. (Thus, $C$ dominates the 'left half' of $V_{L(C)}$ and the 'right half' of $V_{N+1-R(C)}$.) Then $C$ dominates a 'V$-$shaped' configuration if and only if $L(C)+R(C) \\geqslant N+1$. Note that if $C$ dominates a V-shaped configuration, it has at least $M$ pebbles.\n\nNow suppose that there are fewer than $M$ pebbles, so we have $L(C)+R(C) \\leqslant N$. Then Bob makes an $L(C)$-move (or more generally any move with at least $L(C)$ boxes on the left and $R(C)$ boxes on the right). Let $C'$ be the new configuration, and suppose that no box becomes empty (otherwise Bob has won). If Alice picks the left group, we have $L(C')=L(C)+1$ and $R(C')=R(C)-1$. Otherwise, we have $L(C')=L(C)-1$ and $R(C')=R(C)+1$. In either case, we have $L(C')+R(C') \\leqslant N$.\n\nBob then repeats this strategy, until one of the boxes becomes empty. Since the condition in Observation A holds, we may assume that Alice picks a group on the same side each time. Then one of $L$ and $R$ is strictly decreasing; without loss of generality assume that $L$ strictly decreases. At some point we reach $L=1$. If $B_{2}$ is still nonempty, then $B_{1}$ must contain a single pebble. Bob makes a 1-move, and by Observation A, Alice must (eventually) pick the right group, making this box empty.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32080,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice has a map of Wonderland, a country consisting of $n \\geqslant 2$ towns. For every pair of towns, there is a narrow road going from one town to the other. One day, all the roads are declared to be \"one way\" only. Alice has no information on the direction of the roads, but the King of Hearts has offered to help her. She is allowed to ask him a number of questions. For each question in turn, Alice chooses a pair of towns and the King of Hearts tells her the direction of the road connecting those two towns.\n\nAlice wants to know whether there is at least one town in Wonderland with at most one outgoing road. Prove that she can always find out by asking at most $4 n$ questions.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will show Alice needs to ask at most $4 n-7$ questions. Her strategy has the following phases. In what follows, $S$ is the set of towns that Alice, so far, does not know to have more than one outgoing road (so initially $|S|=n$ ).\n\nPhase 1. Alice chooses any two towns, say $A$ and $B$. Without loss of generality, suppose that the King of Hearts' answer is that the road goes from $A$ to $B$.\nAt the end of this phase, Alice has asked 1 question.\n\nPhase 2. During this phase there is a single (variable) town $T$ that is known to have at least one incoming road but not yet known to have any outgoing roads. Initially, $T$ is $B$. Alice does the following $n-2$ times: she picks a town $X$ she has not asked about before, and asks the direction of the road between $T$ and $X$. If it is from $X$ to $T, T$ is unchanged; if it is from $T$ to $X, X$ becomes the new choice of town $T$, as the previous $T$ is now known to have an outgoing road.\nAt the end of this phase, Alice has asked a total of $n-1$ questions. The final town $T$ is not yet known to have any outgoing roads, while every other town has exactly one outgoing road known. The undirected graph of roads whose directions are known is a tree.\n\nPhase 3. During this phase, Alice asks about the directions of all roads between $T$ and another town she has not previously asked about, stopping if she finds two outgoing roads from $T$. This phase involves at most $n-2$ questions. If she does not find two outgoing roads from $T$, she has answered her original question with at most $2 n-3 \\leqslant 4 n-7$ questions, so in what follows we suppose that she does find two outgoing roads, asking a total of $k$ questions in this phase, where $2 \\leqslant k \\leqslant n-2$ (and thus $n \\geqslant 4$ for what follows).\nFor every question where the road goes towards $T$, the town at the other end is removed from $S$ (as it already had one outgoing road known), while the last question resulted in $T$ being removed from $S$. So at the end of this phase, $|S|=n-k+1$, while a total of $n+k-1$ questions have been asked. Furthermore, the undirected graph of roads within $S$ whose directions are known contains no cycles (as $T$ is no longer a member of $S$, all questions asked in this phase involved $T$ and the graph was a tree before this phase started). Every town in $S$ has exactly one outgoing road known (not necessarily to another town in $S$ ).\n\nPhase 4. During this phase, Alice repeatedly picks any pair of towns in $S$ for which she does not know the direction of the road between them. Because every town in $S$ has exactly one outgoing road known, this always results in the removal of one of those two towns from $S$. Because there are no cycles in the graph of roads of known direction within $S$, this can continue until there are at most 2 towns left in $S$.\nIf it ends with $t$ towns left, $n-k+1-t$ questions were asked in this phase, so a total of $2 n-t$ questions have been asked.\n\nPhase 5. During this phase, Alice asks about all the roads from the remaining towns in $S$ that she has not previously asked about. She has definitely already asked about any road between those towns (if $t=2$ ). She must also have asked in one of the first two phases about\nat least one other road involving one of those towns (as those phases resulted in a tree with $n>2$ vertices). So she asks at most $t(n-t)-1$ questions in this phase.\nAt the end of this phase, Alice knows whether any town has at most one outgoing road. If $t=1$, at most $3 n-3 \\leqslant 4 n-7$ questions were needed in total, while if $t=2$, at most $4 n-7$ questions were needed in total.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32081,
    "subject": "Mathematics (Multi-modal)",
    "question": "For any two different real numbers $x$ and $y$, we define $D(x, y)$ to be the unique integer $d$ satisfying $2^{d} \\leqslant |x-y| < 2^{d+1}$. Given a set of reals $\\mathcal{F}$, and an element $x \\in \\mathcal{F}$, we say that the scales of $x$ in $\\mathcal{F}$ are the values of $D(x, y)$ for $y \\in \\mathcal{F}$ with $x \\neq y$.\nLet $k$ be a given positive integer. Suppose that each member $x$ of $\\mathcal{F}$ has at most $k$ different scales in $\\mathcal{F}$ (note that these scales may depend on $x$). What is the maximum possible size of $\\mathcal{F}$?",
    "options": [],
    "answer": "2^k",
    "solution": "We first construct a set $\\mathcal{F}$ with $2^{k}$ members, each member having at most $k$ different scales in $\\mathcal{F}$. Take $\\mathcal{F} = \\{0, 1, 2, \\ldots, 2^{k} - 1\\}$. The scale between any two members of $\\mathcal{F}$ is in the set $\\{0, 1, \\ldots, k-1\\}$.\n\nWe now show that $2^{k}$ is an upper bound on the size of $\\mathcal{F}$. For every finite set $\\mathcal{S}$ of real numbers, and every real $x$, let $r_{\\mathcal{S}}(x)$ denote the number of different scales of $x$ in $\\mathcal{S}$. That is, $r_{\\mathcal{S}}(x) = |\\{D(x, y): x \\neq y \\in \\mathcal{S}\\}|$. Thus, for every element $x$ of the set $\\mathcal{F}$ in the problem statement, we have $r_{\\mathcal{F}}(x) \\leqslant k$. The condition $|\\mathcal{F}| \\leqslant 2^{k}$ is an immediate consequence of the following lemma.\n\nLemma. Let $\\mathcal{S}$ be a finite set of real numbers, and define\n$$\nw(\\mathcal{S}) = \\sum_{x \\in \\mathcal{S}} 2^{-r_{\\mathcal{S}}(x)}\n$$\nThen $w(\\mathcal{S}) \\leqslant 1$.\n\nProof. Induction on $n = |\\mathcal{S}|$. If $\\mathcal{S} = \\{x\\}$, then $r_{\\mathcal{S}}(x) = 0$, so $w(\\mathcal{S}) = 1$.\n\nAssume now $n \\geqslant 2$, and let $x_{1} < \\cdots < x_{n}$ list the members of $\\mathcal{S}$. Let $d$ be the minimal scale between two distinct elements of $\\mathcal{S}$; then there exist neighbours $x_{t}$ and $x_{t+1}$ with $D(x_{t}, x_{t+1}) = d$. Notice that for any two indices $i$ and $j$ with $j - i > 1$ we have $D(x_{i}, x_{j}) > d$, since\n$$\n|x_{i} - x_{j}| = |x_{i+1} - x_{i}| + |x_{j} - x_{i+1}| \\geqslant 2^{d} + 2^{d} = 2^{d+1}\n$$\nNow choose the minimal $i \\leqslant t$ and the maximal $j \\geqslant t+1$ such that $D(x_{i}, x_{i+1}) = D(x_{i+1}, x_{i+2}) = \\cdots = D(x_{j-1}, x_{j}) = d$.\n\nLet $E$ be the set of all the $x_{s}$ with even indices $i \\leqslant s \\leqslant j$, $O$ be the set of those with odd indices $i \\leqslant s \\leqslant j$, and $R$ be the rest of the elements (so that $\\mathcal{S}$ is the disjoint union of $E$, $O$ and $R$). Set $\\mathcal{S}_{O} = R \\cup O$ and $\\mathcal{S}_{E} = R \\cup E$; we have $|\\mathcal{S}_{O}| < |\\mathcal{S}|$ and $|\\mathcal{S}_{E}| < |\\mathcal{S}|$, so $w(\\mathcal{S}_{O}), w(\\mathcal{S}_{E}) \\leqslant 1$ by the inductive hypothesis.\n\nClearly, $r_{\\mathcal{S}_{O}}(x) \\leqslant r_{\\mathcal{S}}(x)$ and $r_{\\mathcal{S}_{E}}(x) \\leqslant r_{\\mathcal{S}}(x)$ for any $x \\in R$, and thus\n$$\n\\begin{aligned}\n\\sum_{x \\in R} 2^{-r_{\\mathcal{S}}(x)} & = \\frac{1}{2} \\sum_{x \\in R} \\left(2^{-r_{\\mathcal{S}}(x)} + 2^{-r_{\\mathcal{S}}(x)}\\right) \\\\\n& \\leqslant \\frac{1}{2} \\sum_{x \\in R} \\left(2^{-r_{\\mathcal{S}_{O}}(x)} + 2^{-r_{\\mathcal{S}_{E}}(x)}\\right)\n\\end{aligned}\n$$\nOn the other hand, for every $x \\in O$, there is no $y \\in \\mathcal{S}_{O}$ such that $D_{\\mathcal{S}_{O}}(x, y) = d$ (as all candidates from $\\mathcal{S}$ were in $E$). Hence, we have $r_{\\mathcal{S}_{O}}(x) \\leqslant r_{\\mathcal{S}}(x) - 1$, and thus\n$$\n\\sum_{x \\in O} 2^{-r_{\\mathcal{S}}(x)} \\leqslant \\frac{1}{2} \\sum_{x \\in O} 2^{-r_{S_{O}}(x)}\n$$\nSimilarly, for every $x \\in E$, we have\n$$\n\\sum_{x \\in E} 2^{-r_{\\mathcal{S}}(x)} \\leqslant \\frac{1}{2} \\sum_{x \\in E} 2^{-r_{\\mathcal{S}_{E}}(x)}\n$$\nWe can then combine these to give\n$$\n\\begin{aligned}\nw(S) & = \\sum_{x \\in R} 2^{-r_{\\mathcal{S}}(x)} + \\sum_{x \\in O} 2^{-r_{\\mathcal{S}}(x)} + \\sum_{x \\in E} 2^{-r_{\\mathcal{S}}(x)} \\\\\n& \\leqslant \\frac{1}{2} \\sum_{x \\in R} \\left(2^{-r_{\\mathcal{S}_{O}}(x)} + 2^{-r_{\\mathcal{S}_{E}}(x)}\\right) + \\frac{1}{2} \\sum_{x \\in O} 2^{-r_{\\mathcal{S}_{O}}(x)} + \\frac{1}{2} \\sum_{x \\in E} 2^{-r_{\\mathcal{S}_{E}}(x)} \\\\\n& = \\frac{1}{2} \\left(\\sum_{x \\in \\mathcal{S}_{O}} 2^{-r_{\\mathcal{S}_{O}}(x)} + \\sum_{x \\in \\mathcal{S}_{E}} 2^{-r_{\\mathcal{S}_{E}}(x)}\\right) \\quad (\\text{since } \\mathcal{S}_{O} = O \\cup R \\text{ and } \\mathcal{S}_{E} = E \\cup R) \\\\\n& = \\frac{1}{2} \\left(w(\\mathcal{S}_{O}) + w(\\mathcal{S}_{E})\\right) \\quad \\text{(by definition of } w(\\cdot)) \\\\\n& \\leqslant 1 \\quad (\\text{by the inductive hypothesis})\n\\end{aligned}\n$$\nwhich completes the induction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32082,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle. Circle $\\Gamma$ passes through $A$, meets segments $AB$ and $AC$ again at points $D$ and $E$ respectively, and intersects segment $BC$ at $F$ and $G$ such that $F$ lies between $B$ and $G$. The tangent to circle $BDF$ at $F$ and the tangent to circle $CEG$ at $G$ meet at point $T$. Suppose that points $A$ and $T$ are distinct. Prove that line $AT$ is parallel to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that $\\angle TFB = \\angle FDA$ because $FT$ is tangent to circle $BDF$, and moreover $\\angle FDA = \\angle CGA$ because quadrilateral $ADFG$ is cyclic. Similarly, $\\angle TGB = \\angle GEC$ because $GT$ is tangent to circle $CEG$, and $\\angle GEC = \\angle CFA$. Hence,\n$$\n\\begin{equation*}\n\\angle TFB = \\angle CGA \\quad \\text{and} \\quad \\angle TGB = \\angle CFA. \\tag{1}\n\\end{equation*}\n$$\n![](attached_image_1.png)\nTriangles $FGA$ and $GFT$ have a common side $FG$, and by (1) their angles at $F, G$ are the same. So, these triangles are congruent. So, their altitudes starting from $A$ and $T$, respectively, are equal and hence $AT$ is parallel to line $BFGC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32083,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle and let $D$, $E$, and $F$ be the feet of altitudes from $A$, $B$, and $C$ to sides $BC$, $CA$, and $AB$, respectively. Denote by $\\omega_{B}$ and $\\omega_{C}$ the incircles of triangles $BDF$ and $CDE$, and let these circles be tangent to segments $DF$ and $DE$ at $M$ and $N$, respectively. Let line $MN$ meet circles $\\omega_{B}$ and $\\omega_{C}$ again at $P \\neq M$ and $Q \\neq N$, respectively. Prove that $MP = NQ$.\n(Vietnam)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote the centres of $\\omega_{B}$ and $\\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $BC$ be tangent to the two circles at $T$ and $U$, respectively.\n\n![](attached_image_1.png)\n\nFrom the cyclic quadrilaterals $AFDC$ and $ABDE$ we have\n$$\n\\angle MDO_{B} = \\frac{1}{2} \\angle FDB = \\frac{1}{2} \\angle BAC = \\frac{1}{2} \\angle CDE = \\angle O_{C}DN,\n$$\nso the right-angled triangles $DMO_{B}$ and $DNO_{C}$ are similar. The ratio of similarity between the two triangles is\n$$\n\\frac{DN}{DM} = \\frac{O_{C}N}{O_{B}M} = \\frac{r_{C}}{r_{B}}.\n$$\nLet $\\varphi = \\angle DMN$ and $\\psi = \\angle MND$. The lines $FM$ and $EN$ are tangent to $\\omega_{B}$ and $\\omega_{C}$, respectively, so\n$$\n\\angle MTP = \\angle FMP = \\angle DMN = \\varphi \\quad \\text{and} \\quad \\angle QUN = \\angle QNE = \\angle MND = \\psi.\n$$\n(It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $DMT$ or $DUN$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $MTP$ and $QUN$.)\n\nIn the circles $\\omega_{B}$ and $\\omega_{C}$ the lengths of chords $MP$ and $NQ$ are\n$$\nMP = 2r_{B} \\cdot \\sin \\angle MTP = 2r_{B} \\cdot \\sin \\varphi \\quad \\text{and} \\quad NQ = 2r_{C} \\cdot \\sin \\angle QUN = 2r_{C} \\cdot \\sin \\psi.\n$$\nBy applying the sine rule to triangle $DNM$ we get\n$$\n\\frac{DN}{DM} = \\frac{\\sin \\angle DMN}{\\sin \\angle MND} = \\frac{\\sin \\varphi}{\\sin \\psi}.\n$$\nFinally, putting the above observations together, we get\n$$\n\\frac{MP}{NQ} = \\frac{2r_{B} \\sin \\varphi}{2r_{C} \\sin \\psi} = \\frac{r_{B}}{r_{C}} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi} = \\frac{DM}{DN} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi} = \\frac{\\sin \\psi}{\\sin \\varphi} \\cdot \\frac{\\sin \\varphi}{\\sin \\psi} = 1,\n$$\nso $MP = NQ$ as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32084,
    "subject": "Mathematics (Multi-modal)",
    "question": "In triangle $A B C$, let $A_{1}$ and $B_{1}$ be two points on sides $B C$ and $A C$, and let $P$ and $Q$ be two points on segments $A A_{1}$ and $B B_{1}$, respectively, so that line $P Q$ is parallel to $A B$. On ray $P B_{1}$, beyond $B_{1}$, let $P_{1}$ be a point so that $\\angle P P_{1} C=\\angle B A C$. Similarly, on ray $Q A_{1}$, beyond $A_{1}$, let $Q_{1}$ be a point so that $\\angle C Q_{1} Q=\\angle C B A$. Show that points $P, Q, P_{1}$, and $Q_{1}$ are concyclic.\n(Ukraine)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. Throughout the solution we use oriented angles.\nLet rays $A A_{1}$ and $B B_{1}$ intersect the circumcircle of $\\triangle A C B$ at $A_{2}$ and $B_{2}$, respectively. By\n$$\n\\angle Q P A_{2}=\\angle B A A_{2}=\\angle B B_{2} A_{2}=\\angle Q B_{2} A_{2},\n$$\npoints $P, Q, A_{2}, B_{2}$ are concyclic; denote the circle passing through these points by $\\omega$. We shall prove that $P_{1}$ and $Q_{1}$ also lie on $\\omega$.\n![](attached_image_1.png)\nBy\n$$\n\\angle C A_{2} A_{1}=\\angle C A_{2} A=\\angle C B A=\\angle C Q_{1} Q=\\angle C Q_{1} A_{1},\n$$\npoints $C, Q_{1}, A_{2}, A_{1}$ are also concyclic. From that we get\n$$\n\\angle Q Q_{1} A_{2}=\\angle A_{1} Q_{1} A_{2}=\\angle A_{1} C A_{2}=\\angle B C A_{2}=\\angle B A A_{2}=\\angle Q P A_{2},\n$$\nso $Q_{1}$ lies on $\\omega$.\nIt follows similarly that $P_{1}$ lies on $\\omega$.\nSolution 2. First consider the case when lines $P P_{1}$ and $Q Q_{1}$ intersect each other at some point $R$.\nLet line $P Q$ meet the sides $A C$ and $B C$ at $E$ and $F$, respectively. Then\n$$\n\\angle P P_{1} C=\\angle B A C=\\angle P E C,\n$$\nso points $C, E, P, P_{1}$ lie on a circle; denote that circle by $\\omega_{P}$. It follows analogously that points $C, F, Q, Q_{1}$ lie on another circle; denote it by $\\omega_{Q}$.\nLet $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem to the lines $A A_{1} P$ and $B B_{1} Q$ provides that points $C=A B_{1} \\cap B A_{1}, R=A_{1} Q \\cap B_{1} P$ and $T=A Q \\cap B P$ are collinear.\nLet line $R C T$ meet $P Q$ and $A B$ at $S$ and $U$, respectively. From $A B \\| P Q$ we obtain\n$$\n\\frac{S P}{S Q}=\\frac{U B}{U A}=\\frac{S F}{S E},\n$$\nso\n$$\nS P \\cdot S E=S Q \\cdot S F .\n$$\n![](attached_image_2.png)\nSo, point $S$ has equal powers with respect to $\\omega_{P}$ and $\\omega_{Q}$, hence line $R C S$ is their radical axis; then $R$ also has equal powers to the circles, so $R P \\cdot R P_{1}=R Q \\cdot R Q_{1}$, proving that points $P, P_{1}, Q, Q_{1}$ are indeed concyclic.\nNow consider the case when $P P_{1}$ and $Q Q_{1}$ are parallel. Like in the previous case, let $A Q$ and $B P$ intersect at $T$. Applying Pappus' theorem again to the lines $A A_{1} P$ and $B B_{1} Q$, in this limit case it shows that line $C T$ is parallel to $P P_{1}$ and $Q Q_{1}$.\nLet line $C T$ meet $P Q$ and $A B$ at $S$ and $U$, as before. The same calculation as in the previous case shows that $S P \\cdot S E=S Q \\cdot S F$, so $S$ lies on the radical axis between $\\omega_{P}$ and $\\omega_{Q}$.\n![](attached_image_3.png)\nLine $C S T$, that is the radical axis between $\\omega_{P}$ and $\\omega_{Q}$, is perpendicular to the line $\\ell$ of centres of $\\omega_{P}$ and $\\omega_{Q}$. Hence, the chords $P P_{1}$ and $Q Q_{1}$ are perpendicular to $\\ell$. So the quadrilateral $P P_{1} Q_{1} Q$ is an isosceles trapezium with symmetry axis $\\ell$, and hence is cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32085,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a point inside triangle $ABC$. Let $AP$ meet $BC$ at $A_{1}$, let $BP$ meet $CA$ at $B_{1}$, and let $CP$ meet $AB$ at $C_{1}$. Let $A_{2}$ be the point such that $A_{1}$ is the midpoint of $PA_{2}$, let $B_{2}$ be the point such that $B_{1}$ is the midpoint of $PB_{2}$, and let $C_{2}$ be the point such that $C_{1}$ is the midpoint of $PC_{2}$. Prove that points $A_{2}, B_{2}$, and $C_{2}$ cannot all lie strictly inside the circumcircle of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. Since\n$$\n\\angle APB + \\angle BPC + \\angle CPA = 2\\pi = (\\pi - \\angle ACB) + (\\pi - \\angle BAC) + (\\pi - \\angle CBA),\n$$\nat least one of the following inequalities holds:\n$$\n\\angle APB \\geqslant \\pi - \\angle ACB, \\quad \\angle BPC \\geqslant \\pi - \\angle BAC, \\quad \\angle CPA \\geqslant \\pi - \\angle CBA.\n$$\nWithout loss of generality, we assume that $\\angle BPC \\geqslant \\pi - \\angle BAC$. We have $\\angle BPC > \\angle BAC$ because $P$ is inside $\\triangle ABC$. So $\\angle BPC \\geqslant \\max (\\angle BAC, \\pi - \\angle BAC)$ and hence\n$$\n\\begin{equation*}\n\\sin \\angle BPC \\leqslant \\sin \\angle BAC. \\tag{*}\n\\end{equation*}\n$$\nLet the rays $AP, BP$, and $CP$ cross the circumcircle $\\Omega$ again at $A_{3}, B_{3}$, and $C_{3}$, respectively. We will prove that at least one of the ratios $\\frac{PB_{1}}{B_{1}B_{3}}$ and $\\frac{PC_{1}}{C_{1}C_{3}}$ is at least 1, which yields that one of the points $B_{2}$ and $C_{2}$ does not lie strictly inside $\\Omega$.\nBecause $A, B, C, B_{3}$ lie on a circle, the triangles $CB_{1}B_{3}$ and $BB_{1}A$ are similar, so\n$$\n\\frac{CB_{1}}{B_{1}B_{3}} = \\frac{BB_{1}}{B_{1}A}.\n$$\nApplying the sine rule we obtain\n$$\n\\frac{PB_{1}}{B_{1}B_{3}} = \\frac{PB_{1}}{CB_{1}} \\cdot \\frac{CB_{1}}{B_{1}B_{3}} = \\frac{PB_{1}}{CB_{1}} \\cdot \\frac{BB_{1}}{B_{1}A} = \\frac{\\sin \\angle ACP}{\\sin \\angle BPC} \\cdot \\frac{\\sin \\angle BAC}{\\sin \\angle PBA}.\n$$\nSimilarly,\n$$\n\\frac{PC_{1}}{C_{1}C_{3}} = \\frac{\\sin \\angle PBA}{\\sin \\angle BPC} \\cdot \\frac{\\sin \\angle BAC}{\\sin \\angle ACP}.\n$$\nMultiplying these two equations we get\n$$\n\\frac{PB_{1}}{B_{1}B_{3}} \\cdot \\frac{PC_{1}}{C_{1}C_{3}} = \\frac{\\sin^{2} \\angle BAC}{\\sin^{2} \\angle BPC} \\geqslant 1\n$$\nusing (*), which yields the desired conclusion.\n\n\nSolution 2. Define points $A_{3}, B_{3}$, and $C_{3}$ as in Solution 1. Assume for the sake of contradiction that $A_{2}, B_{2}$, and $C_{2}$ all lie strictly inside circle $ABC$. It follows that $PA_{1} < A_{1}A_{3}$, $PB_{1} < B_{1}B_{3}$, and $PC_{1} < C_{1}C_{3}$.\nObserve that $\\triangle PBC_{3} \\sim \\triangle PCB_{3}$. Let $X$ be the point on side $PB_{3}$ that corresponds to point $C_{1}$ on side $PC_{3}$ under this similarity. In other words, $X$ lies on segment $PB_{3}$ and satisfies $PX : XB_{3} = PC_{1} : C_{1}C_{3}$. It follows that\n$$\n\\angle XCP = \\angle PBC_{1} = \\angle B_{3}BA = \\angle B_{3}CB_{1}\n$$\nHence lines $CX$ and $CB_{1}$ are isogonal conjugates in $\\triangle PCB_{3}$.\n![](attached_image_1.png)\nLet $Y$ be the foot of the bisector of $\\angle B_{3}CP$ in $\\triangle PCB_{3}$. Since $PC_{1} < C_{1}C_{3}$, we have $PX < XB_{3}$. Also, we have $PY < YB_{3}$ because $PB_{1} < B_{1}B_{3}$ and $Y$ lies between $X$ and $B_{1}$. By the angle bisector theorem in $\\triangle PCB_{3}$, we have $PY : YB_{3} = PC : CB_{3}$. So $PC < CB_{3}$ and it follows that $\\angle PB_{3}C < \\angle CPB_{3}$. Now since $\\angle PB_{3}C = \\angle BB_{3}C = \\angle BAC$, we have\n$$\n\\angle BAC < \\angle CPB_{3}.\n$$\nSimilarly, we have\n$$\n\\angle CBA < \\angle APC_{3} \\quad \\text{and} \\quad \\angle ACB < \\angle BPA_{3} = \\angle B_{3}PA.\n$$\nAdding these three inequalities yields $\\pi < \\pi$, and this contradiction concludes the proof.\n\n\nSolution 3. Choose coordinates such that the circumcentre of $\\triangle ABC$ is at the origin and the circumradius is 1. Then we may think of $A, B$, and $C$ as vectors in $\\mathbb{R}^{2}$ such that\n$$\n|A|^{2} = |B|^{2} = |C|^{2} = 1.\n$$\n$P$ may be represented as a convex combination $\\alpha A + \\beta B + \\gamma C$ where $\\alpha, \\beta, \\gamma > 0$ and $\\alpha + \\beta + \\gamma = 1$. Then\n$$\nA_{1} = \\frac{\\beta B + \\gamma C}{\\beta + \\gamma} = \\frac{1}{1-\\alpha} P - \\frac{\\alpha}{1-\\alpha} A,\n$$\nso\n$$\nA_{2} = 2A_{1} - P = \\frac{1+\\alpha}{1-\\alpha} P - \\frac{2\\alpha}{1-\\alpha} A\n$$\nHence\n$$\n|A_{2}|^{2} = \\left(\\frac{1+\\alpha}{1-\\alpha}\\right)^{2} |P|^{2} + \\left(\\frac{2\\alpha}{1-\\alpha}\\right)^{2} |A|^{2} - \\frac{4\\alpha(1+\\alpha)}{(1-\\alpha)^{2}} A \\cdot P.\n$$\nUsing $|A|^{2} = 1$ we obtain\n$$\n\\begin{equation*}\n\\frac{(1-\\alpha)^{2}}{2(1+\\alpha)} |A_{2}|^{2} = \\frac{1+\\alpha}{2} |P|^{2} + \\frac{2\\alpha^{2}}{1+\\alpha} - 2\\alpha A \\cdot P. \\tag{1}\n\\end{equation*}\n$$\nLikewise\n$$\n\\begin{equation*}\n\\frac{(1-\\beta)^{2}}{2(1+\\beta)} |B_{2}|^{2} = \\frac{1+\\beta}{2} |P|^{2} + \\frac{2\\beta^{2}}{1+\\beta} - 2\\beta B \\cdot P \\tag{2}\n\\end{equation*}\n$$\nand\n$$\n\\begin{equation*}\n\\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} |C_{2}|^{2} = \\frac{1+\\gamma}{2} |P|^{2} + \\frac{2\\gamma^{2}}{1+\\gamma} - 2\\gamma C \\cdot P. \\tag{3}\n\\end{equation*}\n$$\nSumming (1), (2) and (3) we obtain on the LHS the positive linear combination\n$$\n\\mathrm{LHS} = \\frac{(1-\\alpha)^{2}}{2(1+\\alpha)} |A_{2}|^{2} + \\frac{(1-\\beta)^{2}}{2(1+\\beta)} |B_{2}|^{2} + \\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} |C_{2}|^{2}\n$$\nand on the RHS the quantity\n$$\n\\left(\\frac{1+\\alpha}{2} + \\frac{1+\\beta}{2} + \\frac{1+\\gamma}{2}\\right) |P|^{2} + \\left(\\frac{2\\alpha^{2}}{1+\\alpha} + \\frac{2\\beta^{2}}{1+\\beta} + \\frac{2\\gamma^{2}}{1+\\gamma}\\right) - 2(\\alpha A \\cdot P + \\beta B \\cdot P + \\gamma C \\cdot P).\n$$\nThe first term is $2|P|^{2}$ and the last term is $-2P \\cdot P$, so\n$$\n\\begin{aligned}\n\\mathrm{RHS} & = \\left(\\frac{2\\alpha^{2}}{1+\\alpha} + \\frac{2\\beta^{2}}{1+\\beta} + \\frac{2\\gamma^{2}}{1+\\gamma}\\right) \\\\\n& = \\frac{3\\alpha-1}{2} + \\frac{(1-\\alpha)^{2}}{2(1+\\alpha)} + \\frac{3\\beta-1}{2} + \\frac{(1-\\beta)^{2}}{2(1+\\beta)} + \\frac{3\\gamma-1}{2} + \\frac{(1-\\gamma)^{2}}{2(1+\\gamma)} \\\\\n& = \\frac{(1-\\alpha)^{2}}{2(1+\\alpha)} + \\frac{(1-\\beta)^{2}}{2(1+\\beta)} + \\frac{(1-\\gamma)^{2}}{2(1+\\gamma)}\n\\end{aligned}\n$$\nHere we used the fact that\n$$\n\\frac{3\\alpha-1}{2} + \\frac{3\\beta-1}{2} + \\frac{3\\gamma-1}{2} = 0.\n$$\nWe have shown that a linear combination of $|A_{2}|^{2}, |B_{2}|^{2}$, and $|C_{2}|^{2}$ with positive coefficients is equal to the sum of the coefficients. Therefore at least one of $|A_{2}|^{2}, |B_{2}|^{2}$, and $|C_{2}|^{2}$ must be at least 1, as required.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32086,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C D E$ be a convex pentagon with $C D = D E$ and $\\angle E D C \\neq 2 \\cdot \\angle A D B$. Suppose that a point $P$ is located in the interior of the pentagon such that $A P = A E$ and $B P = B C$. Prove that $P$ lies on the diagonal $C E$ if and only if $\\operatorname{area}(B C D) + \\operatorname{area}(A D E) = \\operatorname{area}(A B D) + \\operatorname{area}(A B P)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nFor the equivalence with the collinearity condition, let $F$ denote the foot of the perpendicular from $P^{\\prime}$ to $A B$, so that $F$ is the midpoint of $P P^{\\prime}$. We have that $P$ lies on $C E$ if and only if $F$ lies on $M N$, which occurs if and only if we have the equality $\\angle A F M = \\angle B F N$ of signed angles modulo $\\pi$. By concyclicity of $A P^{\\prime} F M$ and $B F P^{\\prime} N$, this is equivalent to $\\angle A P^{\\prime} M = \\angle B P^{\\prime} N$, which occurs if and only if $A P^{\\prime} M$ and $B P^{\\prime} N$ are directly similar.\n![](attached_image_2.png)\nFor the other equivalence with the area condition, we have the equality of signed areas $\\operatorname{area}(A B D) + \\operatorname{area}(A B P) = \\operatorname{area}\\left(A P^{\\prime} B D\\right) = \\operatorname{area}\\left(A P^{\\prime} D\\right) + \\operatorname{area}\\left(B D P^{\\prime}\\right)$. Using the identity $\\operatorname{area}(A D E) - \\operatorname{area}\\left(A P^{\\prime} D\\right) = \\operatorname{area}(A D E) + \\operatorname{area}\\left(A D P^{\\prime}\\right) = 2 \\operatorname{area}(A D M)$, and similarly for $B$, we find that the area condition is equivalent to the equality\n$$\n\\operatorname{area}(D A M) = \\operatorname{area}(D B N)\n$$\nNow note that $A$ and $B$ lie on the perpendicular bisectors of $P^{\\prime} E$ and $P^{\\prime} C$, respectively. If we write $G$ and $H$ for the feet of the perpendiculars from $D$ to these perpendicular bisectors respectively, then this area condition can be rewritten as\n$$\nM A \\cdot G D = N B \\cdot H D\n$$\n(In this condition, we interpret all lengths as signed lengths according to suitable conventions: for instance, we orient $P^{\\prime} E$ from $P^{\\prime}$ to $E$, orient the parallel line $D H$ in the same direction, and orient the perpendicular bisector of $P^{\\prime} E$ at an angle $\\pi / 2$ clockwise from the oriented segment $P^{\\prime} E$ - we adopt the analogous conventions at $B$.)\n![](attached_image_3.png)\nTo relate the signed lengths $G D$ and $H D$ to the triangles $A P^{\\prime} M$ and $B P^{\\prime} N$, we use the following calculation.\nClaim. Let $\\Gamma$ denote the circle centred on $D$ with both $E$ and $C$ on the circumference, and $h$ the power of $P^{\\prime}$ with respect to $\\Gamma$. Then we have the equality\n$$\nG D \\cdot P^{\\prime} M = H D \\cdot P^{\\prime} N = \\frac{1}{4} h \\neq 0 .\n$$\nProof. Firstly, we have $h \\neq 0$, since otherwise $P^{\\prime}$ would lie on $\\Gamma$, and hence the internal angle bisectors of $\\angle E D P^{\\prime}$ and $\\angle P^{\\prime} D C$ would pass through $A$ and $B$ respectively. This would violate the angle inequality $\\angle E D C \\neq 2 \\cdot \\angle A D B$ given in the question.\nNext, let $E^{\\prime}$ denote the second point of intersection of $P^{\\prime} E$ with $\\Gamma$, and let $E^{\\prime \\prime}$ denote the point on $\\Gamma$ diametrically opposite $E^{\\prime}$, so that $E^{\\prime \\prime} E$ is perpendicular to $P^{\\prime} E$. The point $G$ lies on the perpendicular bisectors of the sides $P^{\\prime} E$ and $E E^{\\prime \\prime}$ of the right-angled triangle $P^{\\prime} E E^{\\prime \\prime}$; it follows that $G$ is the midpoint of $P^{\\prime} E^{\\prime \\prime}$. Since $D$ is the midpoint of $E^{\\prime} E^{\\prime \\prime}$, we have that $G D = \\frac{1}{2} P^{\\prime} E^{\\prime}$. Since $P^{\\prime} M = \\frac{1}{2} P^{\\prime} E$, we have $G D \\cdot P^{\\prime} M = \\frac{1}{4} P^{\\prime} E^{\\prime} \\cdot P^{\\prime} E = \\frac{1}{4} h$. The other equality $H D \\cdot P^{\\prime} N$ follows by exactly the same argument.\n![](attached_image_4.png) $\\square$\nFrom this claim, we see that the area condition is equivalent to the equality\n$$\n\\left(M A : P^{\\prime} M\\right) = \\left(N B : P^{\\prime} N\\right)\n$$\nof ratios of signed lengths, which is equivalent to direct similarity of $A P^{\\prime} M$ and $B P^{\\prime} N$, as desired.\nAlong the perpendicular bisector of $C E$, define the linear function\n$$\nf(X) = \\operatorname{area}(B C X) + \\operatorname{area}(A X E) - \\operatorname{area}(A B X) - \\operatorname{area}(A B P),\n$$\nwhere, from now on, we always use signed areas. Thus, we want to show that $C, P, E$ are collinear if and only if $f(D) = 0$.\n![](attached_image_5.png)\nLet $P^{\\prime}$ be the reflection of $P$ across line $A B$. The point $P^{\\prime}$ does not lie on the line $C E$. To see this, we let $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ be the points obtained from $A$ and $B$ by dilating with scale factor 2 about $P^{\\prime}$, so that $P$ is the orthogonal projection of $P^{\\prime}$ onto $A^{\\prime \\prime} B^{\\prime \\prime}$. Since $A$ lies on the perpendicular bisector of $P^{\\prime} E$, the triangle $A^{\\prime \\prime} E P^{\\prime}$ is right-angled at $E$ (and $B^{\\prime \\prime} C P^{\\prime}$ similarly). If $P^{\\prime}$ were to lie on $C E$, then the lines $A^{\\prime \\prime} E$ and $B^{\\prime \\prime} C$ would be perpendicular to $C E$ and $A^{\\prime \\prime}$ and $B^{\\prime \\prime}$ would lie on the opposite side of $C E$ to $D$. It follows that the line $A^{\\prime \\prime} B^{\\prime \\prime}$ does not meet triangle $C D E$, and hence point $P$ does not lie inside $C D E$. But then $P$ must lie inside $A B C E$, and it is clear that such a point cannot reflect to a point $P^{\\prime}$ on $C E$.\nWe thus let $O$ be the centre of the circle $C E P^{\\prime}$. The lines $A O$ and $B O$ are the perpendicular bisectors of $E P^{\\prime}$ and $C P^{\\prime}$, respectively, so\n$$\n\\begin{aligned}\n\\operatorname{area}(B C O) + \\operatorname{area}(A O E) & = \\operatorname{area}\\left(O P^{\\prime} B\\right) + \\operatorname{area}\\left(P^{\\prime} O A\\right) = \\operatorname{area}\\left(P^{\\prime} B O A\\right) \\\\\n& = \\operatorname{area}(A B O) + \\operatorname{area}\\left(B A P^{\\prime}\\right) = \\operatorname{area}(A B O) + \\operatorname{area}(A B P)\n\\end{aligned}\n$$\nand hence $f(O) = 0$.\nNotice that if point $O$ coincides with $D$ then points $A, B$ lie in angle domain $C D E$ and $\\angle E O C = 2 \\cdot \\angle A O B$, which is not allowed. So, $O$ and $D$ must be distinct. Since $f$ is linear and vanishes at $O$, it follows that $f(D) = 0$ if and only if $f$ is constant zero - we want to show this occurs if and only if $C, P, E$ are collinear.\n![](attached_image_6.png)\n![](attached_image_7.png)\nIn the one direction, suppose firstly that $C, P, E$ are not collinear, and let $T$ be the centre of the circle $C E P$. The same calculation as above provides\n$$\n\\operatorname{area}(B C T) + \\operatorname{area}(A T E) = \\operatorname{area}(P B T A) = \\operatorname{area}(A B T) - \\operatorname{area}(A B P)\n$$\nso\n$$\nf(T) = -2 \\operatorname{area}(A B P) \\neq 0 .\n$$\nHence, the linear function $f$ is nonconstant with its zero is at $O$, so that $f(D) \\neq 0$.\nIn the other direction, suppose that the points $C, P, E$ are collinear. We will show that $f$ is constant zero by finding a second point (other than $O$ ) at which it vanishes.\n![](attached_image_8.png)\nLet $Q$ be the reflection of $P$ across the midpoint of $A B$, so $P A Q B$ is a parallelogram. It is easy to see that $Q$ is on the perpendicular bisector of $C E$; for instance if $A^{\\prime}$ and $B^{\\prime}$ are the points produced from $A$ and $B$ by dilating about $P$ with scale factor 2, then the projection of $Q$ to $C E$ is the midpoint of the projections of $A^{\\prime}$ and $B^{\\prime}$, which are $E$ and $C$ respectively. The triangles $B C Q$ and $A Q E$ are indirectly congruent, so\n$$\nf(Q) = (\\operatorname{area}(B C Q) + \\operatorname{area}(A Q E)) - (\\operatorname{area}(A B Q) - \\operatorname{area}(B A P)) = 0 - 0 = 0 .\n$$\nThe points $O$ and $Q$ are distinct. To see this, consider the circle $\\omega$ centred on $Q$ with $P^{\\prime}$ on the circumference; since triangle $P P^{\\prime} Q$ is right-angled at $P^{\\prime}$, it follows that $P$ lies outside $\\omega$. On the other hand, $P$ lies between $C$ and $E$ on the line $C P E$. It follows that $C$ and $E$ cannot both lie on $\\omega$, so that $\\omega$ is not the circle $C E P^{\\prime}$ and $Q \\neq O$.\nSince $O$ and $Q$ are distinct zeroes of the linear function $f$, we have $f(D) = 0$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32087,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $I$ be the incentre of acute-angled triangle $A B C$. Let the incircle meet $B C$, $C A$, and $A B$ at $D$, $E$, and $F$, respectively. Let line $E F$ intersect the circumcircle of the triangle at $P$ and $Q$, such that $F$ lies between $E$ and $P$. Prove that $\\angle D P A + \\angle A Q D = \\angle Q I P$.\n\n(Slovakia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $N$ and $M$ be the midpoints of the $\\operatorname{arcs}\\ \\overparen{B C}$ of the circumcircle, containing and opposite vertex $A$, respectively. By $\\angle F A E = \\angle B A C = \\angle B N C$, the right-angled kites $A F I E$ and $N B M C$ are similar. Consider the spiral similarity $\\varphi$ (dilation in case of $A B = A C$) that moves $A F I E$ to $N B M C$. The directed angle in which $\\varphi$ changes directions is $\\angle(A F, N B)$, same as $\\angle(A P, N P)$ and $\\angle(A Q, N Q)$; so lines $A P$ and $A Q$ are mapped to lines $N P$ and $N Q$, respectively. Line $E F$ is mapped to $B C$; we can see that the intersection points $P = E F \\cap A P$ and $Q = E F \\cap A Q$ are mapped to points $B C \\cap N P$ and $B C \\cap N Q$, respectively. Denote these points by $P'$ and $Q'$, respectively.\n\n![](attached_image_1.png)\n\nLet $L$ be the midpoint of $B C$. We claim that points $P, Q, D, L$ are concyclic (if $D = L$ then line $B C$ is tangent to circle $P Q D$). Let $P Q$ and $B C$ meet at $Z$. By applying Menelaus' theorem to triangle $A B C$ and line $E F Z$, we have\n$$\n\\frac{B D}{D C} = \\frac{B F}{F A} \\cdot \\frac{A E}{E C} = -\\frac{B Z}{Z C},\n$$\nso the pairs $B, C$ and $D, Z$ are harmonic. It is well-known that this implies $Z B \\cdot Z C = Z D \\cdot Z L$. (The inversion with pole $Z$ that swaps $B$ and $C$ sends $Z$ to infinity and $D$ to the midpoint of $B C$, because the cross-ratio is preserved.) Hence, $Z D \\cdot Z L = Z B \\cdot Z C = Z P \\cdot Z Q$ by the power of $Z$ with respect to the circumcircle; this proves our claim.\n\nBy $\\angle M P P' = \\angle M Q Q' = \\angle M L P' = \\angle M L Q' = 90^{\\circ}$, the quadrilaterals $M L P P'$ and $M L Q Q'$ are cyclic. Then the problem statement follows by\n$$\n\\begin{aligned}\n\\angle D P A + \\angle A Q D & = 360^{\\circ} - \\angle P A Q - \\angle Q D P = 360^{\\circ} - \\angle P N Q - \\angle Q L P \\\\\n& = \\angle L P N + \\angle N Q L = \\angle P' M L + \\angle L M Q' = \\angle P' M Q' = \\angle P I Q\n\\end{aligned} .\n$$\nDefine the point $M$ and the same spiral similarity $\\varphi$ as in the previous solution. (The point $N$ is not necessary.) It is well-known that the centre of the spiral similarity that maps $F, E$ to $B, C$ is the Miquel point of the lines $F E, B C, B F$ and $C E$; that is, the second intersection of circles $A B C$ and $A E F$. Denote that point by $S$.\n\nBy $\\varphi(F) = B$ and $\\varphi(E) = C$ the triangles $S B F$ and $S C E$ are similar, so we have\n$$\n\\frac{S B}{S C} = \\frac{B F}{C E} = \\frac{B D}{C D} .\n$$\nBy the converse of the angle bisector theorem, that indicates that line $S D$ bisects $\\angle B S C$ and hence passes through $M$.\n\nLet $K$ be the intersection point of lines $E F$ and $S I$. Notice that $\\varphi$ sends points $S, F, E, I$ to $S, B, C, M$, so $\\varphi(K) = \\varphi(F E \\cap S I) = B C \\cap S M = D$. By $\\varphi(I) = M$, we have $K D \\parallel I M$.\n\n![](attached_image_2.png)\n\nWe claim that triangles $S P I$ and $S D Q$ are similar, and so are triangles $S P D$ and $S I Q$. Let ray $S I$ meet the circumcircle again at $L$. Note that the segment $E F$ is perpendicular to the angle bisector $A M$. Then by $\\angle A M L = \\angle A S L = \\angle A S I = 90^{\\circ}$, we have $M L \\parallel P Q$. Hence, $\\widehat{P L} = \\widehat{M Q}$ and therefore $\\angle P S L = \\angle M S Q = \\angle D S Q$. By $\\angle Q P S = \\angle Q M S$, the triangles $S P K$ and $S M Q$ are similar. Finally,\n$$\n\\frac{S P}{S I} = \\frac{S P}{S K} \\cdot \\frac{S K}{S I} = \\frac{S M}{S Q} \\cdot \\frac{S D}{S M} = \\frac{S D}{S Q}\n$$\nshows that triangles $S P I$ and $S D Q$ are similar. The second part of the claim can be proved analogously.\n\nNow the problem statement can be proved by\n$$\n\\angle D P A + \\angle A Q D = \\angle D P S + \\angle S Q D = \\angle Q I S + \\angle S I P = \\angle Q I P .\n$$\nDenote the circumcircle of triangle $A B C$ by $\\Gamma$, and let rays $P D$ and $Q D$ meet $\\Gamma$ again at $V$ and $U$, respectively. We will show that $A U \\perp I P$ and $A V \\perp I Q$. Then the problem statement will follow as\n$$\n\\angle D P A + \\angle A Q D = \\angle V U A + \\angle A V U = 180^{\\circ} - \\angle U A V = \\angle Q I P .\n$$\n\nLet $M$ be the midpoint of $\\operatorname{arc}\\ \\widehat{B U V C}$ and let $N$ be the midpoint of $\\operatorname{arc}\\ \\widehat{C A B}$; the lines $A I M$ and $A N$ being the internal and external bisectors of angle $B A C$, respectively, are perpendicular. Let the tangents drawn to $\\Gamma$ at $B$ and $C$ meet at $R$; let line $P Q$ meet $A U, A I, A V$ and $B C$ at $X, T, Y$ and $Z$, respectively.\n\nAs in Solution 1, we observe that the pairs $B, C$ and $D, Z$ are harmonic. Projecting these points from $Q$ onto the circumcircle, we can see that $B, C$ and $U, P$ are also harmonic. Analogously, the pair $V, Q$ is harmonic with $B, C$. Consider the inversion about the circle with centre $R$, passing through $B$ and $C$. Points $B$ and $C$ are fixed points, so this inversion exchanges every point of $\\Gamma$ by its harmonic pair with respect to $B, C$. In particular, the inversion maps points $B, C, N, U, V$ to points $B, C, M, P, Q$, respectively.\n\nCombine the inversion with projecting $\\Gamma$ from $A$ to line $P Q$; the points $B, C, M, P, Q$ are projected to $F, E, T, P, Q$, respectively.\n\n![](attached_image_3.png)\n\nThe combination of these two transformations is projective map from the lines $A B, A C$, $A N, A U, A V$ to $I F, I E, I T, I P, I Q$, respectively. On the other hand, we have $A B \\perp I F$, $A C \\perp I E$ and $A N \\perp A T$, so the corresponding lines in these two pencils are perpendicular. This proves $A U \\perp I P$ and $A V \\perp I Q$, and hence completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32088,
    "subject": "Mathematics (Multi-modal)",
    "question": "The incircle $\\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\\omega$ again at $R$. Line $A R$ meets $\\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet again at $Q \\neq P$. Prove that lines $D I$ and $P Q$ meet on the external bisector of angle $B A C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Step 1. The external bisector of $\\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\\omega$. Since $A N \\cdot A I = A E^{2} = A R \\cdot A P$, the points $R$, $N, I$, and $P$ are concyclic. As $I R = I P$, the line $N I$ is the external bisector of $\\angle P N R$, so $P N$ meets $\\omega$ again at the point symmetric to $R$ with respect to $A N$ - i.e. at $K$.\nLet $D N$ cross $\\omega$ again at $S$. Opposite sides of any quadrilateral inscribed in the circle $\\omega$ meet on the polar line of the intersection of the diagonals with respect to $\\omega$. Since $L$ lies on the polar line $A L$ of $N$ with respect to $\\omega$, the line $P S$ must pass through $L$. Thus it suffices to prove that the points $S, Q$, and $P$ are collinear.\n\n![](attached_image_1.png)\n\nStep 2. Let $\\Gamma$ be the circumcircle of $\\triangle B I C$. Notice that\n$$\n\\begin{aligned}\n\\angle(B Q, Q C) = \\angle & (B Q, Q P) + \\angle(P Q, Q C) = \\angle(B F, F P) + \\angle(P E, E C) \\\\\n& = \\angle(E F, E P) + \\angle(F P, F E) = \\angle(F P, E P) = \\angle(D F, D E) = \\angle(B I, I C)\n\\end{aligned}\n$$\nso $Q$ lies on $\\Gamma$. Let $Q P$ meet $\\Gamma$ again at $T$. It will now suffice to prove that $S, P$, and $T$ are collinear. Notice that $\\angle(B I, I T) = \\angle(B Q, Q T) = \\angle(B F, F P) = \\angle(F K, K P)$. Note $F D \\perp F K$ and $F D \\perp B I$ so $F K \\parallel B I$ and hence $I T$ is parallel to the line $K N P$. Since $D I = I K$, the line $I T$ crosses $D N$ at its midpoint $M$.\n\nStep 3. Let $F^{\\prime}$ and $E^{\\prime}$ be the midpoints of $D E$ and $D F$, respectively. Since $D E^{\\prime} \\cdot E^{\\prime} F = D E^{\\prime 2} = B E^{\\prime} \\cdot E^{\\prime} I$, the point $E^{\\prime}$ lies on the radical axis of $\\omega$ and $\\Gamma$; the same holds for $F^{\\prime}$. Therefore, this radical axis is $E^{\\prime} F^{\\prime}$, and it passes through $M$. Thus $I M \\cdot M T = D M \\cdot M S$, so $S, I, D$, and $T$ are concyclic. This shows $\\angle(D S, S T) = \\angle(D I, I T) = \\angle(D K, K P) = \\angle(D S, S P)$, whence the points $S, P$, and $T$ are collinear, as desired.\n\n![](attached_image_2.png)\nWe start as in Solution 1. Namely, we introduce the same points $K, L, N$, and $S$, and show that the triples $(P, N, K)$ and $(P, S, L)$ are collinear. We conclude that $K$ and $R$ are symmetric in $A I$, and reduce the problem statement to showing that $P, Q$, and $S$ are collinear.\n\nStep 1. Let $A R$ meet the circumcircle $\\Omega$ of $A B C$ again at $X$. The lines $A R$ and $A K$ are isogonal in the angle $B A C$; it is well known that in this case $X$ is the tangency point of $\\Omega$ with the $A$-mixtilinear circle. It is also well known that for this point $X$, the line $X I$ crosses $\\Omega$ again at the midpoint $M^{\\prime}$ of $\\operatorname{arc} B A C$.\n\nStep 2. Denote the circles $B F P$ and $C E P$ by $\\Omega_{B}$ and $\\Omega_{C}$, respectively. Let $\\Omega_{B}$ cross $A R$ and $E F$ again at $U$ and $Y$, respectively. We have\n$$\n\\angle(U B, B F) = \\angle(U P, P F) = \\angle(R P, P F) = \\angle(R F, F A),\n$$\nso $U B \\parallel R F$.\n\n![](attached_image_3.png)\n\nNext, we show that the points $B, I, U$, and $X$ are concyclic. Since\n$$\n\\angle(U B, U X) = \\angle(R F, R X) = \\angle(A F, A R) + \\angle(F R, F A) = \\angle\\left(M^{\\prime} B, M^{\\prime} X\\right) + \\angle(D R, D F),\n$$\nit suffices to prove $\\angle(I B, I X) = \\angle\\left(M^{\\prime} B, M^{\\prime} X\\right) + \\angle(D R, D F)$, or $\\angle\\left(I B, M^{\\prime} B\\right) = \\angle(D R, D F)$. But both angles equal $\\angle(C I, C B)$, as desired. (This is where we used the fact that $M^{\\prime}$ is the midpoint of arc $B A C$ of $\\Omega$.)\n\nIt follows now from circles BUIX and BPUFY that\n$$\n\\begin{aligned}\n\\angle(I U, U B) = \\angle(I X, B X) = \\angle\\left(M^{\\prime} X, B X\\right) = & \\frac{\\pi-\\angle A}{2} \\\\\n& = \\angle(E F, A F) = \\angle(Y F, B F) = \\angle(Y U, B U),\n\\end{aligned}\n$$\nso the points $Y, U$, and $I$ are collinear.\n\nLet $E F$ meet $B C$ at $W$. We have\n$$\n\\angle(I Y, Y W) = \\angle(U Y, F Y) = \\angle(U B, F B) = \\angle(R F, A F) = \\angle(C I, C W),\n$$\nso the points $W, Y, I$, and $C$ are concyclic.\n\nSimilarly, if $V$ and $Z$ are the second meeting points of $\\Omega_{C}$ with $A R$ and $E F$, we get that the 4-tuples ( $C, V, I, X$ ) and ( $B, I, Z, W$ ) are both concyclic.\n\nStep 3. Let $Q^{\\prime} = C Y \\cap B Z$. We will show that $Q^{\\prime} = Q$.\n\nFirst of all, we have\n$$\n\\begin{aligned}\n\\angle\\left(Q^{\\prime} Y, Q^{\\prime} B\\right) = \\angle(C Y, Z B) & = \\angle(C Y, Z Y) + \\angle(Z Y, B Z) \\\\\n& = \\angle(C I, I W) + \\angle(I W, I B) = \\angle(C I, I B) = \\frac{\\pi-\\angle A}{2} = \\angle(F Y, F B)\n\\end{aligned}\n$$\nso $Q^{\\prime} \\in \\Omega_{B}$. Similarly, $Q^{\\prime} \\in \\Omega_{C}$. Thus $Q^{\\prime} \\in \\Omega_{B} \\cap \\Omega_{C} = \\{P, Q\\}$ and it remains to prove that $Q^{\\prime} \\neq P$. If we had $Q^{\\prime} = P$, we would have $\\angle(P Y, P Z) = \\angle\\left(Q^{\\prime} Y, Q^{\\prime} Z\\right) = \\angle(I C, I B)$. This would imply\n$$\n\\angle(P Y, Y F) + \\angle(E Z, Z P) = \\angle(P Y, P Z) = \\angle(I C, I B) = \\angle(P E, P F),\n$$\nso circles $\\Omega_{B}$ and $\\Omega_{C}$ would be tangent at $P$. That is excluded in the problem conditions, so $Q^{\\prime} = Q$.\n\n![](attached_image_4.png)\n\nStep 4. Now we are ready to show that $P, Q$, and $S$ are collinear.\n\nNotice that $A$ and $D$ are the poles of $E W$ and $D W$ with respect to $\\omega$, so $W$ is the pole of $A D$. Hence, $W I \\perp A D$. Since $C I \\perp D E$, this yields $\\angle(I C, W I) = \\angle(D E, D A)$. On the other hand, $D A$ is a symmedian in $\\triangle D E F$, so $\\angle(D E, D A) = \\angle(D N, D F) = \\angle(D S, D F)$. Therefore,\n$$\n\\begin{aligned}\n\\angle(P S, P F) = \\angle(D S, D F) = \\angle(D E, D A) = & \\angle(I C, I W) \\\\\n& = \\angle(Y C, Y W) = \\angle(Y Q, Y F) = \\angle(P Q, P F)\n\\end{aligned}\n$$\nwhich yields the desired collinearity.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32089,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each line $\\ell \\in \\mathcal{L}$ a point $f(\\ell)$ on $\\ell$. Suppose that for any point $X$, and for any three lines $\\ell_{1}, \\ell_{2}, \\ell_{3}$ passing through $X$, the points $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right), f\\left(\\ell_{3}\\right)$ and $X$ lie on a circle.\nProve that there is a unique point $P$ such that $f(\\ell)=P$ for any line $\\ell$ passing through $P$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Write $\\angle\\left(\\ell_{1}, \\ell_{2}\\right)$ for the directed angle modulo $180^{\\circ}$ between the lines $\\ell_{1}$ and $\\ell_{2}$. Given a point $P$ and an angle $\\alpha \\in\\left(0,180^{\\circ}\\right)$, for each line $\\ell$, let $\\ell^{\\prime}$ be the line through $P$ satisfying $\\angle\\left(\\ell^{\\prime}, \\ell\\right)=\\alpha$, and let $h_{P, \\alpha}(\\ell)$ be the intersection point of $\\ell$ and $\\ell^{\\prime}$. We will prove that there is some pair ( $P, \\alpha$ ) such that $f$ and $h_{P, \\alpha}$ are the same function. Then $P$ is the unique point in the problem statement.\nGiven an angle $\\alpha$ and a point $P$, let a line $\\ell$ be called $(P, \\alpha)$-good if $f(\\ell)=h_{P, \\alpha}(\\ell)$. Let a point $X \\neq P$ be called ( $P, \\alpha$ )-good if the circle $g(X)$ passes through $P$ and some point $Y \\neq P, X$ on $g(X)$ satisfies $\\angle(P Y, Y X)=\\alpha$. It follows from this definition that if $X$ is $(P, \\alpha)$ good then every point $Y \\neq P, X$ of $g(X)$ satisfies this angle condition, so $h_{P, \\alpha}(X Y)=Y$ for every $Y \\in g(X)$. Equivalently, $f(\\ell) \\in\\left\\{X, h_{P, \\alpha}(\\ell)\\right\\}$ for each line $\\ell$ passing through $X$. This shows the following lemma.\nLemma 1. If $X$ is $(P, \\alpha)$-good and $\\ell$ is a line passing through $X$ then either $f(\\ell)=X$ or $\\ell$ is $(P, \\alpha)$-good.\nLemma 2. If $X$ and $Y$ are different ( $P, \\alpha$ )-good points, then line $X Y$ is $(P, \\alpha)$-good.\nProof. If $X Y$ is not $(P, \\alpha)$-good then by the previous Lemma, $f(X Y)=X$ and similarly $f(X Y)=Y$, but clearly this is impossible as $X \\neq Y$.\nLemma 3. If $\\ell_{1}$ and $\\ell_{2}$ are different ( $P, \\alpha$ )-good lines which intersect at $X \\neq P$, then either $f\\left(\\ell_{1}\\right)=X$ or $f\\left(\\ell_{2}\\right)=X$ or $X$ is $(P, \\alpha)$-good.\nProof. If $f\\left(\\ell_{1}\\right), f\\left(\\ell_{2}\\right) \\neq X$, then $g(X)$ is the circumcircle of $X, f\\left(\\ell_{1}\\right)$ and $f\\left(\\ell_{2}\\right)$. Since $\\ell_{1}$ and $\\ell_{2}$ are $(P, \\alpha)$-good lines, the angles\n$$\n\\angle\\left(P f\\left(\\ell_{1}\\right), f\\left(\\ell_{1}\\right) X\\right)=\\angle\\left(P f\\left(\\ell_{2}\\right), f\\left(\\ell_{2}\\right) X\\right)=\\alpha\n$$\nso $P$ lies on $g(X)$. Hence, $X$ is $(P, \\alpha)$-good.\nLemma 4. If $\\ell_{1}, \\ell_{2}$ and $\\ell_{3}$ are different $(P, \\alpha)$-good lines which intersect at $X \\neq P$, then $X$ is $(P, \\alpha)$-good.\nProof. This follows from the previous Lemma since at most one of the three lines $\\ell_{i}$ can satisfy $f\\left(\\ell_{i}\\right)=X$ as the three lines are all $(P, \\alpha)$-good.\nLemma 5. If $A B C$ is a triangle such that $A, B, C, f(A B), f(A C)$ and $f(B C)$ are all different points, then there is some point $P$ and some angle $\\alpha$ such that $A, B$ and $C$ are $(P, \\alpha)$-good points and $A B, B C$ and $C A$ are ( $P, \\alpha$ )-good lines.\n![](attached_image_1.png)\nProof. Let $D, E, F$ denote the points $f(B C), f(A C), f(A B)$, respectively. Then $g(A)$, $g(B)$ and $g(C)$ are the circumcircles of $A E F, B D F$ and $C D E$, respectively. Let $P \\neq F$ be the second intersection of circles $g(A)$ and $g(B)$ (or, if these circles are tangent at $F$, then $P=F$ ). By Miquel's theorem (or an easy angle chase), $g(C)$ also passes through $P$. Then by the cyclic quadrilaterals, the directed angles\n$$\n\\angle(P D, D C)=\\angle(P F, F B)=\\angle(P E, E A)=\\alpha,\n$$\nfor some angle $\\alpha$. Hence, lines $A B, B C$ and $C A$ are all ( $P, \\alpha$ )-good, so by Lemma 3, $A, B$ and $C$ are $(P, \\alpha)$-good. (In the case where $P=D$, the line $P D$ in the equation above denotes the line which is tangent to $g(B)$ at $P=D$. Similar definitions are used for $P E$ and $P F$ in the cases where $P=E$ or $P=F$.) $\\square$\nConsider the set $\\Omega$ of all points $(x, y)$ with integer coordinates $1 \\leqslant x, y \\leqslant 1000$, and consider the set $L_{\\Omega}$ of all horizontal, vertical and diagonal lines passing through at least one point in $\\Omega$. A simple counting argument shows that there are 5998 lines in $L_{\\Omega}$. For each line $\\ell$ in $L_{\\Omega}$ we colour the point $f(\\ell)$ red. Then there are at most 5998 red points. Now we partition the points in $\\Omega$ into 10000 ten by ten squares. Since there are at most 5998 red points, at least one of these squares $\\Omega_{10}$ contains no red points. Let $(m, n)$ be the bottom left point in $\\Omega_{10}$. Then the triangle with vertices $(m, n),(m+1, n)$ and $(m, n+1)$ satisfies the condition of Lemma 5, so these three vertices are all ( $P, \\alpha$ )-good for some point $P$ and angle $\\alpha$, as are the lines joining them. From this point on, we will simply call a point or line good if it is $(P, \\alpha)$-good for this particular pair $(P, \\alpha)$. Now by Lemma 1, the line $x=m+1$ is good, as is the line $y=n+1$. Then Lemma 3 implies that ( $m+1, n+1$ ) is good. By applying these two lemmas repeatedly, we can prove that the line $x+y=m+n+2$ is good, then the points ( $m, n+2$ ) and ( $m+2, n$ ) then the lines $x=m+2$ and $y=n+2$, then the points $(m+2, n+1),(m+1, n+2)$ and $(m+2, n+2)$ and so on until we have prove that all points in $\\Omega_{10}$ are good.\nNow we will use this to prove that every point $S \\neq P$ is good. Since $g(S)$ is a circle, it passes through at most two points of $\\Omega_{10}$ on any vertical line, so at most 20 points in total. Moreover, any line $\\ell$ through $S$ intersects at most 10 points in $\\Omega_{10}$. Hence, there are at least eight lines $\\ell$ through $S$ which contain a point $Q$ in $\\Omega_{10}$ which is not on $g(S)$. Since $Q$ is not on $g(S)$, the point $f(\\ell) \\neq Q$. Hence, by Lemma 1, the line $\\ell$ is good. Hence, at least eight good lines pass through $S$, so by Lemma 4, the point $S$ is good. Hence, every point $S \\neq P$ is good, so by Lemma 2, every line is good. In particular, every line $\\ell$ passing through $P$ is good, and therefore satisfies $f(\\ell)=P$, as required.\nNote that for any distinct points $X, Y$, the circles $g(X)$ and $g(Y)$ meet on $X Y$ at the point $f(X Y) \\in g(X) \\cap g(Y) \\cap(X Y)$. We write $s(X, Y)$ for the second intersection point of circles $g(X)$ and $g(Y)$.\nLemma 1. Suppose that $X, Y$ and $Z$ are not collinear, and that $f(X Y) \\notin\\{X, Y\\}$ and similarly for $Y Z$ and $Z X$. Then $s(X, Y)=s(Y, Z)=s(Z, X)$.\nProof. The circles $g(X), g(Y)$ and $g(Z)$ through the vertices of triangle $X Y Z$ meet pairwise on the corresponding edges (produced). By Miquel's theorem, the second points of intersection of any two of the circles coincide. (See the diagram for Lemma 5 of Solution 1.) $\\square$\nNow pick any line $\\ell$ and any six different points $Y_{1}, \\ldots, Y_{6}$ on $\\ell \\backslash\\{f(\\ell)\\}$. Pick a point $X$ not on $\\ell$ or any of the circles $g\\left(Y_{i}\\right)$. Reordering the indices if necessary, we may suppose that $Y_{1}, \\ldots, Y_{4}$ do not lie on $g(X)$, so that $f\\left(X Y_{i}\\right) \\notin\\left\\{X, Y_{i}\\right\\}$ for $1 \\leqslant i \\leqslant 4$. By applying the above lemma to triangles $X Y_{i} Y_{j}$ for $1 \\leqslant i<j \\leqslant 4$, we find that the points $s\\left(Y_{i}, Y_{j}\\right)$ and $s\\left(X, Y_{i}\\right)$ are all equal, to point $O$ say. Note that either $O$ does not lie on $\\ell$, or $O=f(\\ell)$, since $O \\in g\\left(Y_{i}\\right)$.\nNow consider an arbitrary point $X^{\\prime}$ not on $\\ell$ or any of the circles $g\\left(Y_{i}\\right)$ for $1 \\leqslant i \\leqslant 4$. As above, we see that there are two indices $1 \\leqslant i<j \\leqslant 4$ such that $Y_{i}$ and $Y_{j}$ do not lie on $g\\left(X^{\\prime}\\right)$. By applying the above lemma to triangle $X^{\\prime} Y_{i} Y_{j}$ we see that $s\\left(X^{\\prime}, Y_{i}\\right)=O$, and in particular $g\\left(X^{\\prime}\\right)$ passes through $O$.\nWe will now show that $f\\left(\\ell^{\\prime}\\right)=O$ for all lines $\\ell^{\\prime}$ through $O$. By the above note, we may assume that $\\ell^{\\prime} \\neq \\ell$. Consider a variable point $X^{\\prime} \\in \\ell^{\\prime} \\backslash\\{O\\}$ not on $\\ell$ or any of the circles $g\\left(Y_{i}\\right)$ for $1 \\leqslant i \\leqslant 4$. We know that $f\\left(\\ell^{\\prime}\\right) \\in g\\left(X^{\\prime}\\right) \\cap \\ell^{\\prime}=\\left\\{X^{\\prime}, O\\right\\}$. Since $X^{\\prime}$ was suitably arbitrary, we have $f\\left(\\ell^{\\prime}\\right)=O$ as desired.\nNotice that, for any two different points $X$ and $Y$, the point $f(X Y)$ lies on both $g(X)$ and $g(Y)$, so any two such circles meet in at least one point. We refer to two circles as cutting only in the case where they cross, and so meet at exactly two points, thus excluding the cases where they are tangent or are the same circle.\nLemma 1. Suppose there is a point $P$ such that all circles $g(X)$ pass through $P$. Then $P$ has the given property.\nProof. Consider some line $\\ell$ passing through $P$, and suppose that $f(\\ell) \\neq P$. Consider some $X \\in \\ell$ with $X \\neq P$ and $X \\neq f(\\ell)$. Then $g(X)$ passes through all of $P, f(\\ell)$ and $X$, but those three points are collinear, a contradiction.\nLemma 2. Suppose that, for all $\\epsilon>0$, there is a point $P_{\\epsilon}$ with $g\\left(P_{\\epsilon}\\right)$ of radius at most $\\epsilon$. Then there is a point $P$ with the given property.\nProof. Consider a sequence $\\epsilon_{i}=2^{-i}$ and corresponding points $P_{\\epsilon_{i}}$. Because the two circles $g\\left(P_{\\epsilon_{i}}\\right)$ and $g\\left(P_{\\epsilon_{j}}\\right)$ meet, the distance between $P_{\\epsilon_{i}}$ and $P_{\\epsilon_{j}}$ is at most $2^{1-i}+2^{1-j}$. As $\\sum_{i} \\epsilon_{i}$ converges, these points converge to some point $P$. For all $\\epsilon>0$, the point $P$ has distance at most $2 \\epsilon$ from $P_{\\epsilon}$, and all circles $g(X)$ pass through a point with distance at most $2 \\epsilon$ from $P_{\\epsilon}$, so distance at most $4 \\epsilon$ from $P$. A circle that passes distance at most $4 \\epsilon$ from $P$ for all $\\epsilon>0$ must pass through $P$, so by Lemma 1 the point $P$ has the given property.\nLemma 3. Suppose no two of the circles $g(X)$ cut. Then there is a point $P$ with the given property.\nProof. Consider a circle $g(X)$ with centre $Y$. The circle $g(Y)$ must meet $g(X)$ without cutting it, so has half the radius of $g(X)$. Repeating this argument, there are circles with arbitrarily small radius and the result follows by Lemma 2.\nLemma 4. Suppose there are six different points $A, B_{1}, B_{2}, B_{3}, B_{4}, B_{5}$ such that no three are collinear, no four are concyclic, and all the circles $g\\left(B_{i}\\right)$ cut pairwise at $A$. Then there is a point $P$ with the given property.\nProof. Consider some line $\\ell$ through $A$ that does not pass through any of the $B_{i}$ and is not tangent to any of the $g\\left(B_{i}\\right)$. Fix some direction along that line, and let $X_{\\epsilon}$ be the point on $\\ell$ that has distance $\\epsilon$ from $A$ in that direction. In what follows we consider only those $\\epsilon$ for which $X_{\\epsilon}$ does not lie on any $g\\left(B_{i}\\right)$ (this restriction excludes only finitely many possible values of $\\epsilon$ ).\nConsider the circle $g\\left(X_{\\epsilon}\\right)$. Because no four of the $B_{i}$ are concyclic, at most three of them lie on this circle, so at least two of them do not. There must be some sequence of $\\epsilon \\rightarrow 0$ such that it is the same two of the $B_{i}$ for all $\\epsilon$ in that sequence, so now restrict attention to that sequence, and suppose without loss of generality that $B_{1}$ and $B_{2}$ do not lie on $g\\left(X_{\\epsilon}\\right)$ for any $\\epsilon$ in that sequence.\nThen $f\\left(X_{\\epsilon} B_{1}\\right)$ is not $B_{1}$, so must be the other point of intersection of $X_{\\epsilon} B_{1}$ with $g\\left(B_{1}\\right)$, and the same applies with $B_{2}$. Now consider the three points $X_{\\epsilon}, f\\left(X_{\\epsilon} B_{1}\\right)$ and $f\\left(X_{\\epsilon} B_{2}\\right)$. As $\\epsilon \\rightarrow 0$, the angle at $X_{\\epsilon}$ tends to $\\angle B_{1} A B_{2}$ or $180^{\\circ}-\\angle B_{1} A B_{2}$, which is not 0 or $180^{\\circ}$ because no three of the points were collinear. All three distances between those points are bounded above by constant multiples of $\\epsilon$ (in fact, if the triangle is scaled by a factor of $1 / \\epsilon$, it tends to a fixed triangle). Thus the circumradius of those three points, which is the radius of $g\\left(X_{\\epsilon}\\right)$, is also bounded above by a constant multiple of $\\epsilon$, and so the result follows by Lemma 2.\nLemma 5. Suppose there are two points $A$ and $B$ such that $g(A)$ and $g(B)$ cut. Then there is a point $P$ with the given property.\nProof. Suppose that $g(A)$ and $g(B)$ cut at $C$ and $D$. One of those points, without loss of generality $C$, must be $f(A B)$, and so lie on the line $A B$. We now consider two cases, according to whether $D$ also lies on that line.\nCase 1: $D$ does not lie on that line.\nIn this case, consider a sequence of $X_{\\epsilon}$ at distance $\\epsilon$ from $D$, tending to $D$ along some line that is not a tangent to either circle, but perturbed slightly (by at most $\\epsilon^{2}$ ) to ensure that no three of the points $A, B$ and $X_{\\epsilon}$ are collinear and no four are concyclic.\nConsider the points $f\\left(X_{\\epsilon} A\\right)$ and $f\\left(X_{\\epsilon} B\\right)$, and the circles $g\\left(X_{\\epsilon}\\right)$ on which they lie. The point $f\\left(X_{\\epsilon} A\\right)$ might be either $A$ or the other intersection of $X_{\\epsilon} A$ with the circle $g(A)$, and the same applies for $B$. If, for some sequence of $\\epsilon \\rightarrow 0$, both those points are the other point of intersection, the same argument as in the proof of Lemma 4 applies to find arbitrarily small circles. Otherwise, we have either infinitely many of those circles passing through $A$, or infinitely many passing through $B$; without loss of generality, suppose infinitely many through $A$.\nWe now show we can find five points $B_{i}$ satisfying the conditions of Lemma 4 (together with $A$ ). Let $B_{1}$ be any of the $X_{\\epsilon}$ for which $g\\left(X_{\\epsilon}\\right)$ passes through $A$. Then repeat the following four times, for $2 \\leqslant i \\leqslant 5$.\nConsider some line $\\ell=X_{\\epsilon} A$ (different from those considered for previous $i$ ) that is not tangent to any of the $g\\left(B_{j}\\right)$ for $j<i$, and is such that $f(\\ell)=A$, so $g(Y)$ passes through $A$ for all $Y$ on that line. If there are arbitrarily small circles $g(Y)$ we are done by Lemma 2, so the radii of such circles must be bounded below. But as $Y \\rightarrow A$, along any line not tangent to $g\\left(B_{j}\\right)$, the radius of a circle through $Y$ and tangent to $g\\left(B_{j}\\right)$ at $A$ tends to 0 . So there must be some $Y$ such that $g(Y)$ cuts $g\\left(B_{j}\\right)$ at $A$ rather than being tangent to it there, for all of the previous $B_{j}$, and we may also pick it such that no three of the $B_{i}$ and $A$ are collinear and no four are concyclic. Let $B_{i}$ be this $Y$. Now the result follows by Lemma 4.\nCase 2: $D$ does lie on that line.\nIn this case, we follow a similar argument, but the sequence of $X_{\\epsilon}$ needs to be slightly different. $C$ and $D$ both lie on the line $A B$, so one must be $A$ and the other must be $B$. Consider a sequence of $X_{\\epsilon}$ tending to $B$. Rather than tending to $B$ along a straight line (with small perturbations), let the sequence be such that all the points are inside the two circles, with the angle between $X_{\\epsilon} B$ and the tangent to $g(B)$ at $B$ tending to 0 .\nAgain consider the points $f\\left(X_{\\epsilon} A\\right)$ and $f\\left(X_{\\epsilon} B\\right)$. If, for some sequence of $\\epsilon \\rightarrow 0$, both those points are the other point of intersection with the respective circles, we see that the angle at $X_{\\epsilon}$ tends to the angle between $A B$ and the tangent to $g(B)$ at $B$, which is not 0 or $180^{\\circ}$, while the distances tend to 0 (although possibly slower than any multiple of $\\epsilon$ ), so we have arbitrarily small circumradii and the result follows by Lemma 2. Otherwise, we have either infinitely many of the circles $g\\left(X_{\\epsilon}\\right)$ passing through $A$, or infinitely many passing through $B$, and the same argument as in the previous case enables us to reduce to Lemma 4.\nFor any point $X$, denote by $t(X)$ the line tangent to $g(X)$ at $X$; notice that $f(t(X))=X$, so $f$ is surjective.\nStep 1: We find a point $P$ for which there are at least two different lines $p_{1}$ and $p_{2}$ such that $f\\left(p_{i}\\right)=P$.\nChoose any point $X$. If $X$ does not have this property, take any $Y \\in g(X) \\backslash\\{X\\}$; then $f(X Y)=Y$. If $Y$ does not have the property, $t(Y)=X Y$, and the circles $g(X)$ and $g(Y)$ meet again at some point $Z$. Then $f(X Z)=Z=f(Y Z)$, so $Z$ has the required property.\nWe will show that $P$ is the desired point. From now on, we fix two different lines $p_{1}$ and $p_{2}$ with $f\\left(p_{1}\\right)=f\\left(p_{2}\\right)=P$. Assume for contradiction that $f(\\ell)=Q \\neq P$ for some line $\\ell$ through $P$. We fix $\\ell$, and note that $Q \\in g(P)$.\nStep 2: We prove that $P \\in g(Q)$.\nTake an arbitrary point $X \\in \\ell \\backslash\\{P, Q\\}$. Two cases are possible for the position of $t(X)$ in relation to the $p_{i}$; we will show that each case (and subcase) occurs for only finitely many positions of $X$, yielding a contradiction.\nCase 2.1: $t(X)$ is parallel to one of the $p_{i}$; say, to $p_{1}$.\nLet $t(X)$ cross $p_{2}$ at $R$. Then $g(R)$ is the circle ( $P R X$ ), as $f(R P)=P$ and $f(R X)=X$. Let $R Q$ cross $g(R)$ again at $S$. Then $f(R Q) \\in\\{R, S\\} \\cap g(Q)$, so $g(Q)$ contains one of the points $R$ and $S$.\nIf $R \\in g(Q)$, then $R$ is one of finitely many points in the intersection $g(Q) \\cap p_{2}$, and each of them corresponds to a unique position of $X$, since $R X$ is parallel to $p_{1}$.\nIf $S \\in g(Q)$, then $\\angle(Q S, S P)=\\angle(R S, S P)=\\angle(R X, X P)=\\angle\\left(p_{1}, \\ell\\right)$, so $\\angle(Q S, S P)$ is constant for all such points $X$, and all points $S$ obtained in such a way lie on one circle $\\gamma$ passing through $P$ and $Q$. Since $g(Q)$ does not contain $P$, it is different from $\\gamma$, so there are only finitely many points $S$. Each of them uniquely determines $R$ and thus $X$.\n![](attached_image_2.png)\nSo, Case 2.1 can occur for only finitely many points $X$.\nCase 2.2: $t(X)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively.\nClearly, $R_{1} \\neq R_{2}$, as $t(X)$ is the tangent to $g(X)$ at $X$, and $g(X)$ meets $\\ell$ only at $X$ and $Q$. Notice that $g\\left(R_{i}\\right)$ is the circle ( $P X R_{i}$ ). Let $R_{i} Q$ meet $g\\left(R_{i}\\right)$ again at $S_{i}$; then $S_{i} \\neq Q$, as $g\\left(R_{i}\\right)$ meets $\\ell$ only at $P$ and $X$. Then $f\\left(R_{i} Q\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases.\n![](attached_image_3.png)\nSubcase 2.2.1: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=S_{2}$; so $S_{1}, S_{2} \\in g(Q)$.\nIn this case we have $0=\\angle\\left(R_{1} X, X P\\right)+\\angle\\left(X P, R_{2} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} R_{2}\\right)= \\angle\\left(Q S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} Q\\right)$, which shows $P \\in g(Q)$.\nSubcase 2.2.2: $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=R_{2}$; so $R_{1}, R_{2} \\in g(Q)$.\nThis can happen for at most four positions of $X$ - namely, at the intersections of $\\ell$ with a line of the form $K_{1} K_{2}$, where $K_{i} \\in g(Q) \\cap p_{i}$.\nSubcase 2.2.3: $f\\left(R_{1} Q\\right)=S_{1}, f\\left(R_{2} Q\\right)=R_{2}$ (the case $f\\left(R_{1} Q\\right)=R_{1}, f\\left(R_{2} Q\\right)=S_{2}$ is similar).\nIn this case, there are at most two possible positions for $R_{2}$ - namely, the meeting points of $g(Q)$ with $p_{2}$. Consider one of them. Let $X$ vary on $\\ell$. Then $R_{1}$ is the projection of $X$ to $p_{1}$ via $R_{2}, S_{1}$ is the projection of $R_{1}$ to $g(Q)$ via $Q$. Finally, $\\angle\\left(Q S_{1}, S_{1} X\\right)=\\angle\\left(R_{1} S_{1}, S_{1} X\\right)= \\angle\\left(R_{1} P, P X\\right)=\\angle\\left(p_{1}, \\ell\\right) \\neq 0$, so $X$ is obtained by a fixed projective transform $g(Q) \\rightarrow \\ell$ from $S_{1}$. So, if there were three points $X$ satisfying the conditions of this subcase, the composition of the three projective transforms would be the identity. But, if we apply it to $X=Q$, we successively get some point $R_{1}^{\\prime}$, then $R_{2}$, and then some point different from $Q$, a contradiction.\nThus Case 2.2 also occurs for only finitely many points $X$, as desired.\nStep 3: We show that $f(P Q)=P$, as desired.\nThe argument is similar to that in Step 2, with the roles of $Q$ and $X$ swapped. Again, we show that there are only finitely many possible positions for a point $X \\in \\ell \\backslash\\{P, Q\\}$, which is absurd.\nCase 3.1: $t(Q)$ is parallel to one of the $p_{i}$; say, to $p_{1}$.\nLet $t(Q)$ cross $p_{2}$ at $R$; then $g(R)$ is the circle ( $P R Q$ ). Let $R X$ cross $g(R)$ again at $S$. Then $f(R X) \\in\\{R, S\\} \\cap g(X)$, so $g(X)$ contains one of the points $R$ and $S$.\n![](attached_image_4.png)\nSubcase 3.1.1: $S=f(R X) \\in g(X)$.\nWe have $\\angle(t(X), Q X)=\\angle(S X, S Q)=\\angle(S R, S Q)=\\angle(P R, P Q)=\\angle\\left(p_{2}, \\ell\\right)$. Hence $t(X) \\| p_{2}$. Now we recall Case 2.1: we let $t(X)$ cross $p_{1}$ at $R^{\\prime}$, so $g\\left(R^{\\prime}\\right)=\\left(P R^{\\prime} X\\right)$, and let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$; notice that $S^{\\prime} \\neq Q$. Excluding one position of $X$, we may assume that $R^{\\prime} \\notin g(Q)$, so $R^{\\prime} \\neq f\\left(R^{\\prime} Q\\right)$. Therefore, $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then, as in Case 2.1, we get $\\angle(t(Q), P Q)=\\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(p_{2}, \\ell\\right)$. This means that $t(Q)$ is parallel to $p_{2}$, which is impossible.\nSubcase 3.1.2: $R=f(R X) \\in g(X)$.\nIn this case, we have $\\angle(t(X), \\ell)=\\angle(R X, R Q)=\\angle\\left(R X, p_{1}\\right)$. Again, let $R^{\\prime}=t(X) \\cap p_{1}$; this point exists for all but at most one position of $X$. Then $g\\left(R^{\\prime}\\right)=\\left(R^{\\prime} X P\\right)$; let $R^{\\prime} Q$ meet $g\\left(R^{\\prime}\\right)$ again at $S^{\\prime}$. Due to $\\angle\\left(R^{\\prime} X, X R\\right)=\\angle(Q X, Q R)=\\angle\\left(\\ell, p_{1}\\right), R^{\\prime}$ determines $X$ in at most two ways, so for all but finitely many positions of $X$ we have $R^{\\prime} \\notin g(Q)$. Therefore, for those positions we have $S^{\\prime}=f\\left(R^{\\prime} Q\\right) \\in g(Q)$. But then $\\angle\\left(R X, p_{1}\\right)=\\angle\\left(R^{\\prime} X, X P\\right)=\\angle\\left(R^{\\prime} S^{\\prime}, S^{\\prime} P\\right)= \\angle\\left(Q S^{\\prime}, S^{\\prime} P\\right)=\\angle(t(Q), Q P)$ is fixed, so this case can hold only for one specific position of $X$ as well.\nThus, in Case 3.1, there are only finitely many possible positions of $X$, yielding a contradiction.\nCase 3.2: $t(Q)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively.\nBy Step 2, $R_{1} \\neq R_{2}$. Notice that $g\\left(R_{i}\\right)$ is the circle $\\left(P Q R_{i}\\right)$. Let $R_{i} X$ meet $g\\left(R_{i}\\right)$ at $S_{i}$; then $S_{i} \\neq X$. Then $f\\left(R_{i} X\\right) \\in\\left\\{R_{i}, S_{i}\\right\\}$, and we distinguish several subcases.\n![](attached_image_5.png)\nSubcase 3.2.1: $f\\left(R_{1} X\\right)=S_{1}$ and $f\\left(R_{2} X\\right)=S_{2}$, so $S_{1}, S_{2} \\in g(X)$.\nAs in Subcase 2.2.1, we have $0=\\angle\\left(R_{1} Q, Q P\\right)+\\angle\\left(Q P, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} P\\right)+\\angle\\left(S_{2} P, S_{2} X\\right)$, which shows $P \\in g(X)$. But $X, Q \\in g(X)$ as well, so $g(X)$ meets $\\ell$ at three distinct points, which is absurd.\nSubcase 3.2.2: $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=R_{2}$, so $R_{1}, R_{2} \\in g(X)$.\nNow three distinct collinear points $R_{1}, R_{2}$, and $Q$ belong to $g(X)$, which is impossible.\nSubcase 3.2.3: $f\\left(R_{1} X\\right)=S_{1}, f\\left(R_{2} X\\right)=R_{2}$ (the case $f\\left(R_{1} X\\right)=R_{1}, f\\left(R_{2} X\\right)=S_{2}$ is similar).\nWe have $\\angle\\left(X R_{2}, R_{2} Q\\right)=\\angle\\left(X S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} S_{1}, S_{1} Q\\right)=\\angle\\left(R_{1} P, P Q\\right)=\\angle\\left(p_{1}, \\ell\\right)$, so this case can occur for a unique position of $X$.\nThus, in Case 3.2, there is only a unique position of $X$, again yielding the required contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32090,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all pairs $(m, n)$ of positive integers satisfying the equation\n$$\n\\left(2^{n}-1\\right)\\left(2^{n}-2\\right)\\left(2^{n}-4\\right) \\cdots\\left(2^{n}-2^{n-1}\\right)=m!\n$$",
    "options": [],
    "answer": "(1,1), (3,2)",
    "solution": "We will get an upper bound on $n$ from the speed at which $v_{2}\\left(L_{n}\\right)$ grows.\nFrom\n$$\nL_{n}=\\left(2^{n}-1\\right)\\left(2^{n}-2\\right) \\cdots\\left(2^{n}-2^{n-1}\\right)=2^{1+2+\\cdots+(n-1)}\\left(2^{n}-1\\right)\\left(2^{n-1}-1\\right) \\cdots\\left(2^{1}-1\\right)\n$$\nwe read\n$$\nv_{2}\\left(L_{n}\\right)=1+2+\\cdots+(n-1)=\\frac{n(n-1)}{2} .\n$$\nOn the other hand, $v_{2}(m!)$ is expressed by the Legendre formula as\n$$\nv_{2}(m!)=\\sum_{i=1}^{\\infty}\\left\\lfloor\\frac{m}{2^{i}}\\right\\rfloor\n$$\nAs usual, by omitting the floor functions,\n$$\nv_{2}(m!)<\\sum_{i=1}^{\\infty} \\frac{m}{2^{i}}=m .\n$$\nThus, $L_{n}=m!$ implies the inequality\n$$\n\\begin{equation*}\n\\frac{n(n-1)}{2}<m . \\tag{2}\n\\end{equation*}\n$$\nIn order to obtain an opposite estimate, observe that\n$$\nL_{n}=\\left(2^{n}-1\\right)\\left(2^{n}-2\\right) \\cdots\\left(2^{n}-2^{n-1}\\right)<\\left(2^{n}\\right)^{n}=2^{n^{2}} .\n$$\nWe claim that\n$$\n\\begin{equation*}\n2^{n^{2}}<\\left(\\frac{n(n-1)}{2}\\right)!\\text{ for } n \\geqslant 6 . \\tag{3}\n\\end{equation*}\n$$\nFor $n=6$ the estimate (3) is true because $2^{6^{2}}<6.9 \\cdot 10^{10}$ and $\\left(\\frac{n(n-1)}{2}\\right)!=15!>1.3 \\cdot 10^{12}$.\nFor $n \\geqslant 7$ we prove (3) by the following inequalities:\n$$\n\\begin{aligned}\n\\left(\\frac{n(n-1)}{2}\\right)! & =15!\\cdot 16 \\cdot 17 \\cdots \\frac{n(n-1)}{2}>2^{36} \\cdot 16^{\\frac{n(n-1)}{2}-15} \\\\\n& =2^{2 n(n-1)-24}=2^{n^{2}} \\cdot 2^{n(n-2)-24}>2^{n^{2}} .\n\\end{aligned}\n$$\nPutting together (2) and (3), for $n \\geqslant 6$ we get a contradiction, since\n$$\nL_{n}<2^{n^{2}}<\\left(\\frac{n(n-1)}{2}\\right)!<m!=L_{n}\n$$\nHence $n \\geqslant 6$ is not possible.\nChecking manually the cases $n \\leqslant 5$ we find\n$$\n\\begin{gathered}\nL_{1}=1=1!, \\quad L_{2}=6=3!, \\quad 5!<L_{3}=168<6!, \\\\\n7!<L_{4}=20160<8!\\quad \\text{ and } \\quad 10!<L_{5}=9999360<11!.\n\\end{gathered}\n$$\nSo, there are two solutions:\n$$\n(m, n) \\in\\{(1,1),(3,2)\\} .\n$$\nLike in the previous solution, the cases $n=1,2,3,4$ are checked manually. We will exclude $n \\geqslant 5$ by considering the exponents of 3 and 31 in (1).\nFor odd primes $p$ and distinct integers $a, b$, coprime to $p$, with $p \\mid a-b$, the Lifting The Exponent lemma asserts that\n$$\nv_{p}\\left(a^{k}-b^{k}\\right)=v_{p}(a-b)+v_{p}(k) \\text{.}\n$$\nNotice that 3 divides $2^{k}-1$ if only if $k$ is even; moreover, by the Lifting The Exponent lemma we have\n$$\nv_{3}\\left(2^{2 k}-1\\right)=v_{3}\\left(4^{k}-1\\right)=1+v_{3}(k)=v_{3}(3 k)\n$$\nHence,\n$$\nv_{3}\\left(L_{n}\\right)=\\sum_{2 k \\leqslant n} v_{3}\\left(4^{k}-1\\right)=\\sum_{k \\leqslant\\left\\lfloor\\frac{n}{2}\\right\\rfloor} v_{3}(3 k) .\n$$\nNotice that the last expression is precisely the exponent of 3 in the prime factorisation of $\\left(3\\left\\lfloor\\frac{n}{2}\\right\\rfloor\\right)!$. Therefore\n$$\n\\begin{align*}\nv_{3}(m!)=v_{3}\\left(L_{n}\\right) & =v_{3}\\left(\\left(3\\left\\lfloor\\frac{n}{2}\\right\\rfloor\\right)!\\right) \\\\\n3\\left\\lfloor\\frac{n}{2}\\right\\rfloor & \\leqslant m \\leqslant 3\\left\\lfloor\\frac{n}{2}\\right\\rfloor+2 \\tag{4}\n\\end{align*}\n$$\nSuppose that $n \\geqslant 5$. Note that every fifth factor in $L_{n}$ is divisible by $31=2^{5}-1$, and hence we have $v_{31}\\left(L_{n}\\right) \\geqslant\\left\\lfloor\\frac{n}{5}\\right\\rfloor$. Then\n$$\n\\begin{equation*}\n\\frac{n}{10} \\leqslant\\left\\lfloor\\frac{n}{5}\\right\\rfloor \\leqslant v_{31}\\left(L_{n}\\right)=v_{31}(m!)=\\sum_{k=1}^{\\infty}\\left\\lfloor\\frac{m}{31^{k}}\\right\\rfloor<\\sum_{k=1}^{\\infty} \\frac{m}{31^{k}}=\\frac{m}{30} \\tag{5}\n\\end{equation*}\n$$\nBy combining (4) and (5),\n$$\n3 n<m \\leqslant \\frac{3 n}{2}+2\n$$\nso $n<\\frac{4}{3}$ which is inconsistent with the inequality $n \\geqslant 5$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32091,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a, b, c)$ of positive integers such that $a^{3} + b^{3} + c^{3} = (a b c)^{2}$.",
    "options": [],
    "answer": "All permutations of (1, 2, 3)",
    "solution": "Solution 1. We will start by proving that $c=1$. Note that\n$$\n3 a^{3} \\geqslant a^{3} + b^{3} + c^{3} > a^{3}\n$$\nSo $3 a^{3} \\geqslant (a b c)^{2} > a^{3}$ and hence $3 a \\geqslant b^{2} c^{2} > a$. Now $b^{3} + c^{3} = a^{2} (b^{2} c^{2} - a) \\geqslant a^{2}$, and so\n$$\n18 b^{3} \\geqslant 9 (b^{3} + c^{3}) \\geqslant 9 a^{2} \\geqslant b^{4} c^{4} \\geqslant b^{3} c^{5}\n$$\nso $18 \\geqslant c^{5}$ which yields $c=1$.\n\nNow, note that we must have $a > b$, as otherwise we would have $2 b^{3} + 1 = b^{4}$ which has no positive integer solutions. So\n$$\na^{3} - b^{3} \\geqslant (b + 1)^{3} - b^{3} > 1\n$$\nand\n$$\n2 a^{3} > 1 + a^{3} + b^{3} > a^{3},\n$$\nwhich implies $2 a^{3} > a^{2} b^{2} > a^{3}$ and so $2 a > b^{2} > a$. Therefore\n$$\n4 (1 + b^{3}) = 4 a^{2} (b^{2} - a) \\geqslant 4 a^{2} > b^{4}\n$$\nso $4 > b^{3} (b - 4)$; that is, $b \\leqslant 4$.\n\nNow, for each possible value of $b$ with $2 \\leqslant b \\leqslant 4$ we obtain a cubic equation for $a$ with constant coefficients. These are as follows:\n$$\n\\begin{array}{ll}\nb=2: & a^{3} - 4 a^{2} + 9 = 0 \\\\\nb=3: & a^{3} - 9 a^{2} + 28 = 0 \\\\\nb=4: & a^{3} - 16 a^{2} + 65 = 0 .\n\\end{array}\n$$\nThe only case with an integer solution for $a$ with $b \\leqslant a$ is $b=2$, leading to $(a, b, c) = (3, 2, 1)$.\n\n\nSolution 2. Again, we will start by proving that $c=1$. Suppose otherwise that $c \\geqslant 2$. We have $a^{3} + b^{3} + c^{3} \\leqslant 3 a^{3}$, so $b^{2} c^{2} \\leqslant 3 a$. Since $c \\geqslant 2$, this tells us that $b \\leqslant \\sqrt{3 a} / 4$. As the right-hand side of the original equation is a multiple of $a^{2}$, we have $a^{2} \\leqslant 2 b^{3} \\leqslant 2 (3 a / 4)^{3 / 2}$. In other words, $a \\leqslant \\frac{27}{16} < 2$, which contradicts the assertion that $a \\geqslant c \\geqslant 2$. So there are no solutions in this case, and so we must have $c=1$.\n\nNow, the original equation becomes $a^{3} + b^{3} + 1 = a^{2} b^{2}$. Observe that $a \\geqslant 2$, since otherwise $a = b = 1$ as $a \\geqslant b$.\nThe right-hand side is a multiple of $a^{2}$, so the left-hand side must be as well. Thus, $b^{3} + 1 \\geqslant a^{2}$. Since $a \\geqslant b$, we also have\n$$\nb^{2} = a + \\frac{b^{3} + 1}{a^{2}} \\leqslant 2 a + \\frac{1}{a^{2}}\n$$\nand so $b^{2} \\leqslant 2 a$ since $b^{2}$ is an integer. Thus $(2 a)^{3 / 2} + 1 \\geqslant b^{3} + 1 \\geqslant a^{2}$, from which we deduce $a \\leqslant 8$.\n\nNow, for each possible value of $a$ with $2 \\leqslant a \\leqslant 8$ we obtain a cubic equation for $b$ with constant coefficients. These are as follows:\n$$\n\\begin{array}{ll}\na=2: & b^{3} - 4 b^{2} + 9 = 0 \\\\\na=3: & b^{3} - 9 b^{2} + 28 = 0 \\\\\na=4: & b^{3} - 16 b^{2} + 65 = 0 \\\\\na=5: & b^{3} - 25 b^{2} + 126 = 0 \\\\\na=6: & b^{3} - 36 b^{2} + 217 = 0 \\\\\na=7: & b^{3} - 49 b^{2} + 344 = 0 \\\\\na=8: & b^{3} - 64 b^{2} + 513 = 0 .\n\\end{array}\n$$\nThe only case with an integer solution for $b$ with $a \\geqslant b$ is $a=3$, leading to $(a, b, c) = (3, 2, 1)$.\n\n\nSolution 3. Set $k = (b^{3} + c^{3}) / a^{2} \\leqslant 2 a$, and rewrite the original equation as $a + k = (b c)^{2}$. Since $b^{3}$ and $c^{3}$ are positive integers, we have $(b c)^{3} \\geqslant b^{3} + c^{3} - 1 = k a^{2} - 1$, so\n$$\na + k \\geqslant (k a^{2} - 1)^{2 / 3}\n$$\nAs in Comment 1.2, $k$ is a positive integer; for each value of $k \\geqslant 1$, this gives us a polynomial inequality satisfied by $a$:\n$$\nk^{2} a^{4} - a^{3} - 5 k a^{2} - 3 k^{2} a - (k^{3} - 1) \\leqslant 0 .\n$$\nWe now prove that $a \\leqslant 3$. Indeed,\n$$\n0 \\geqslant \\frac{k^{2} a^{4} - a^{3} - 5 k a^{2} - 3 k^{2} a - (k^{3} - 1)}{k^{2}} \\geqslant a^{4} - a^{3} - 5 a^{2} - 3 a - k \\geqslant a^{4} - a^{3} - 5 a^{2} - 5 a,\n$$\nwhich fails when $a \\geqslant 4$.\nThis leaves ten triples with $3 \\geqslant a \\geqslant b \\geqslant c \\geqslant 1$, which may be checked manually to give $(a, b, c) = (3, 2, 1)$.\n\n\nSolution 4. Again, observe that $b^{3} + c^{3} = a^{2} (b^{2} c^{2} - a)$, so $b \\leqslant a \\leqslant b^{2} c^{2} - 1$.\nWe consider the function $f(x) = x^{2} (b^{2} c^{2} - x)$. It can be seen that that on the interval $[0, b^{2} c^{2} - 1]$ the function $f$ is increasing if $x < \\frac{2}{3} b^{2} c^{2}$ and decreasing if $x > \\frac{2}{3} b^{2} c^{2}$. Consequently, it must be the case that\n$$\nb^{3} + c^{3} = f(a) \\geqslant \\min \\left(f(b), f(b^{2} c^{2} - 1)\\right)\n$$\nFirst, suppose that $b^{3} + c^{3} \\geqslant f(b^{2} c^{2} - 1)$. This may be written $b^{3} + c^{3} \\geqslant (b^{2} c^{2} - 1)^{2}$, and so\n$$\n2 b^{3} \\geqslant b^{3} + c^{3} \\geqslant (b^{2} c^{2} - 1)^{2} > b^{4} c^{4} - 2 b^{2} c^{2} \\geqslant b^{4} c^{4} - 2 b^{3} c^{4}\n$$\nThus, $(b - 2) c^{4} < 2$, and the only solutions to this inequality have $(b, c) = (2, 2)$ or $b \\leqslant 3$ and $c = 1$. It is easy to verify that the only case giving a solution for $a \\geqslant b$ is $(a, b, c) = (3, 2, 1)$.\n\nOtherwise, suppose that $b^{3} + c^{3} = f(a) \\geqslant f(b)$. Then, we have\n$$\n2 b^{3} \\geqslant b^{3} + c^{3} = a^{2} (b^{2} c^{2} - a) \\geqslant b^{2} (b^{2} c^{2} - b)\n$$\nConsequently $b c^{2} \\leqslant 3$, with strict inequality in the case that $b \\neq c$. Hence $c = 1$ and $b \\leqslant 2$. Both of these cases have been considered already, so we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32092,
    "subject": "Mathematics (Multi-modal)",
    "question": "We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_{0}, a_{1}, \\ldots, a_{n} \\in S$, all integer roots of the polynomial $a_{0}+a_{1} x+\\cdots+a_{n} x^{n}$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^{a}-2^{b}$ for positive integers $a$ and $b$.",
    "options": [],
    "answer": "all integers",
    "solution": "First, note that $0=2^{1}-2^{1} \\in S$ and $2=2^{2}-2^{1} \\in S$. Now, $-1 \\in S$, since it is a root of $2 x+2$, and $1 \\in S$, since it is a root of $2 x^{2}-x-1$. Also, if $n \\in S$ then $-n$ is a root of $x+n$, so it suffices to prove that all positive integers must be in $S$.\n\nNow, we claim that any positive integer $n$ has a multiple in $S$. Indeed, suppose that $n=2^{\\alpha} \\cdot t$ for $\\alpha \\in \\mathbb{Z}_{\\geqslant 0}$ and $t$ odd. Then $t \\mid 2^{\\phi(t)}-1$, so $n \\mid 2^{\\alpha+\\phi(t)+1}-2^{\\alpha+1}$. Moreover, $2^{\\alpha+\\phi(t)+1}-2^{\\alpha+1} \\in S$, and so $S$ contains a multiple of every positive integer $n$.\n\nWe will now prove by induction that all positive integers are in $S$. Suppose that $0,1, \\ldots, n-1 \\in S$; furthermore, let $N$ be a multiple of $n$ in $S$. Consider the base-$n$ expansion of $N$, say $N=a_{k} n^{k}+a_{k-1} n^{k-1}+\\cdots+a_{1} n+a_{0}$. Since $0 \\leqslant a_{i}<n$ for each $a_{i}$, we have that all the $a_{i}$ are in $S$. Furthermore, $a_{0}=0$ since $N$ is a multiple of $n$. Therefore, $a_{k} n^{k}+a_{k-1} n^{k-1}+\\cdots+a_{1} n-N=0$, so $n$ is a root of a polynomial with coefficients in $S$. This tells us that $n \\in S$, completing the induction.\nAs in the previous solution, we can prove that $0,1$ and $-1$ must all be in any rootiful set $S$ containing all numbers of the form $2^{a}-2^{b}$ for $a, b \\in \\mathbb{Z}_{>0}$.\n\nWe show that, in fact, every integer $k$ with $|k|>2$ can be expressed as a root of a polynomial whose coefficients are of the form $2^{a}-2^{b}$. Observe that it suffices to consider the case where $k$ is positive, as if $k$ is a root of $a_{n} x^{n}+\\cdots+a_{1} x+a_{0}=0$, then $-k$ is a root of $(-1)^{n} a_{n} x^{n}+\\cdots- a_{1} x+a_{0}=0$.\n\nNote that\n$$\n\\left(2^{a_{n}}-2^{b_{n}}\\right) k^{n}+\\cdots+\\left(2^{a_{0}}-2^{b_{0}}\\right)=0\n$$\nis equivalent to\n$$\n2^{a_{n}} k^{n}+\\cdots+2^{a_{0}}=2^{b_{n}} k^{n}+\\cdots+2^{b_{0}} .\n$$\nHence our aim is to show that two numbers of the form $2^{a_{n}} k^{n}+\\cdots+2^{a_{0}}$ are equal, for a fixed value of $n$. We consider such polynomials where every term $2^{a_{i}} k^{i}$ is at most $2 k^{n}$; in other words, where $2 \\leqslant 2^{a_{i}} \\leqslant 2 k^{n-i}$, or, equivalently, $1 \\leqslant a_{i} \\leqslant 1+(n-i) \\log_{2} k$. Therefore, there must be $1+\\left\\lfloor(n-i) \\log_{2} k\\right\\rfloor$ possible choices for $a_{i}$ satisfying these constraints.\n\nThe number of possible polynomials is then\n$$\n\\prod_{i=0}^{n}\\left(1+\\left\\lfloor(n-i) \\log_{2} k\\right\\rfloor\\right) \\geqslant \\prod_{i=0}^{n-1}(n-i) \\log_{2} k=n!\\left(\\log_{2} k\\right)^{n}\n$$\nwhere the inequality holds as $1+\\lfloor x\\rfloor \\geqslant x$.\n\nAs there are $(n+1)$ such terms in the polynomial, each at most $2 k^{n}$, such a polynomial must have value at most $2 k^{n}(n+1)$. However, for large $n$, we have $n!\\left(\\log_{2} k\\right)^{n}>2 k^{n}(n+1)$. Therefore there are more polynomials than possible values, so some two must be equal, as required.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32093,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\\binom{a n}{b}-1$ is divisible by $a n+1$ for all positive integers $n$ with $a n \\geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first show that $b$ is $a$-good if and only if $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$.\nTo start with, the condition that $a n+1 \\left\\lvert\\,\\binom{ a n}{b}-1\\right.$ can be rewritten as saying that\n$$\n\\frac{a n(a n-1) \\cdots(a n-b+1)}{b!} \\equiv 1 \\quad(\\bmod a n+1) .\n$$\nSuppose, on the one hand, there is a prime $p \\leqslant b$ with $p \\nmid a$. Take $t=v_{p}(b!)$. Then there exist positive integers $c$ such that $a c \\equiv 1\\left(\\bmod p^{t+1}\\right)$. If we take $c$ big enough, and then take $n=(p-1) c$, then $a n=a(p-1) c \\equiv p-1\\left(\\bmod p^{t+1}\\right)$ and $a n \\geqslant b$. Since $p \\leqslant b$, one of the terms of the numerator $a n(a n-1) \\cdots(a n-b+1)$ is $a n-p+1$, which is divisible by $p^{t+1}$. Hence the $p$-adic valuation of the numerator is at least $t+1$, but that of the denominator is exactly $t$. This means that $p \\left\\lvert\\,\\binom{ a n}{b}\\right.$, so $p \\nmid \\binom{a n}{b}-1$. As $p \\mid a n+1$, we get that $a n+1 \\nmid \\binom{a n}{b}-1$, so $b$ is not $a$-good.\nOn the other hand, if for all primes $p \\leqslant b$ we have $p \\mid a$, then every factor of $b!$ is coprime to $a n+1$, and hence invertible modulo $a n+1$ : hence $b!$ is also invertible modulo $a n+1$. Then equation above reduces to:\n$$\na n(a n-1) \\cdots(a n-b+1) \\equiv b!\\quad(\\bmod a n+1) .\n$$\nHowever, we can rewrite the left-hand side as follows:\n$$\na n(a n-1) \\cdots(a n-b+1) \\equiv(-1)(-2) \\cdots(-b) \\equiv(-1)^{b} b!\\quad(\\bmod a n+1) .\n$$\nProvided that $a n>1$, if $b$ is even we deduce $(-1)^{b} b!\\equiv b!$ as needed. On the other hand, if $b$ is odd, and we take $a n+1>2(b!)$, then we will not have $(-1)^{b} b!\\equiv b!$, so $b$ is not $a$-good. This completes the claim.\nTo conclude from here, suppose that $b$ is $a$-good, but $b+2$ is not. Then $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$, but there is a prime $q \\leqslant b+2$ for which $q \\nmid a$ : so $q=b+1$ or $q=b+2$. We cannot have $q=b+2$, as that is even too, so we have $q=b+1$ : in other words, $b+1$ is prime.\nWe show only half of the claim of the previous solution: we show that if $b$ is $a$-good, then $p \\mid a$ for all primes $p \\leqslant b$. We do this with Lucas' theorem.\nSuppose that we have $p \\leqslant b$ with $p \\nmid a$. Then consider the expansion of $b$ in base $p$; there will be some digit (not the final digit) which is nonzero, because $p \\leqslant b$. Suppose it is the $p^{t}$ digit for $t \\geqslant 1$.\nNow, as $n$ varies over the integers, $a n+1$ runs over all residue classes modulo $p^{t+1}$; in particular, there is a choice of $n$ (with $a n>b$ ) such that the $p^{0}$ digit of $a n$ is $p-1$ (so $p \\mid a n+1)$ and the $p^{t}$ digit of $a n$ is 0. Consequently, $p \\mid a n+1$ but $p \\mid \\binom{ a n}{b}$ (by Lucas' theorem) so $p \\nmid \\binom{a n}{b}-1$. Thus $b$ is not $a$-good.\nNow we show directly that if $b$ is $a$-good but $b+2$ fails to be so, then there must be a prime dividing $a n+1$ for some $n$, which also divides $(b+1)(b+2)$. Indeed, the ratio between $\\binom{a n}{b+2}$ and $\\binom{a n}{b}$ is $(b+1)(b+2) /(a n-b)(a n-b-1)$. We know that there must be a choice of $a n+1$ such that the former binomial coefficient is 1 modulo $a n+1$ but the latter is not, which means that the given ratio must not be $1 \\bmod a n+1$. If $b+1$ and $b+2$ are both coprime to $a n+1$ then the ratio is 1, so that must not be the case. In particular, as any prime less than $b$ divides $a$, it must be the case that either $b+1$ or $b+2$ is prime.\nHowever, we can observe that $b$ must be even by insisting that $a n+1$ is prime (which is possible by Dirichlet's theorem) and hence $\\binom{a n}{b} \\equiv(-1)^{b}=1$. Thus $b+2$ cannot be prime, so $b+1$ must be prime.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32094,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $H=\\{\\lfloor i \\sqrt{2}\\rfloor: i \\in \\mathbb{Z}_{>0}\\}=\\{1,2,4,5,7, \\ldots\\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \\subset\\{1,2, \\ldots, n\\}$ satisfies $|A| \\geqslant C \\sqrt{n}$, then there exist $a, b \\in A$ such that $a-b \\in H$. (Here $\\mathbb{Z}_{>0}$ is the set of positive integers, and $\\lfloor z\\rfloor$ denotes the greatest integer less than or equal to $z$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First, observe that if $n$ is a positive integer, then $n \\in H$ exactly when\n$$\n\\left\\{\\frac{n}{\\sqrt{2}}\\right\\}>1-\\frac{1}{\\sqrt{2}} \\tag{1}\n$$\nTo see why, observe that $n \\in H$ if and only if $0<i \\sqrt{2}-n<1$ for some $i \\in \\mathbb{Z}_{>0}$. In other words, $0<i-n / \\sqrt{2}<1 / \\sqrt{2}$, which is equivalent to (1).\n\nNow, write $A=\\{a_{1}<a_{2}<\\cdots<a_{k}\\}$, where $k=|A|$. Observe that the set of differences is not altered by shifting $A$, so we may assume that $A \\subseteq\\{0,1, \\ldots, n-1\\}$ with $a_{1}=0$.\n\nFrom (1), we learn that $\\left\\{a_{i} / \\sqrt{2}\\right\\}<1-1 / \\sqrt{2}$ for each $i>1$ since $a_{i}-a_{1} \\notin H$. Furthermore, we must have $\\left\\{a_{i} / \\sqrt{2}\\right\\}<\\left\\{a_{j} / \\sqrt{2}\\right\\}$ whenever $i<j$; otherwise, we would have\n$$\n-\\left(1-\\frac{1}{\\sqrt{2}}\\right)<\\left\\{\\frac{a_{j}}{\\sqrt{2}}\\right\\}-\\left\\{\\frac{a_{i}}{\\sqrt{2}}\\right\\}<0\n$$\nSince $\\left\\{\\left(a_{j}-a_{i}\\right) / \\sqrt{2}\\right\\}=\\left\\{a_{j} / \\sqrt{2}\\right\\}-\\left\\{a_{i} / \\sqrt{2}\\right\\}+1$, this implies that $\\left\\{\\left(a_{j}-a_{i}\\right) / \\sqrt{2}\\right\\}>1 / \\sqrt{2}> 1-1 / \\sqrt{2}$, contradicting (1).\n\nNow, we have a sequence $0=a_{1}<a_{2}<\\cdots<a_{k}<n$, with\n$$\n0=\\left\\{\\frac{a_{1}}{\\sqrt{2}}\\right\\}<\\left\\{\\frac{a_{2}}{\\sqrt{2}}\\right\\}<\\cdots<\\left\\{\\frac{a_{k}}{\\sqrt{2}}\\right\\}<1-\\frac{1}{\\sqrt{2}}\n$$\nWe use the following fact: for any $d \\in \\mathbb{Z}$, we have\n$$\n\\left\\{\\frac{d}{\\sqrt{2}}\\right\\}>\\frac{1}{2 d \\sqrt{2}} \\tag{2}\n$$\nTo see why this is the case, let $h=\\lfloor d / \\sqrt{2}\\rfloor$, so $\\{d / \\sqrt{2}\\}=d / \\sqrt{2}-h$. Then\n$$\n\\left\\{\\frac{d}{\\sqrt{2}}\\right\\}\\left(\\frac{d}{\\sqrt{2}}+h\\right)=\\frac{d^{2}-2 h^{2}}{2} \\geqslant \\frac{1}{2}\n$$\nsince the numerator is a positive integer. Because $d / \\sqrt{2}+h<2 d / \\sqrt{2}$, inequality (2) follows.\n\nLet $d_{i}=a_{i+1}-a_{i}$, for $1 \\leqslant i<k$. Then $\\left\\{a_{i+1} / \\sqrt{2}\\right\\}-\\left\\{a_{i} / \\sqrt{2}\\right\\}=\\left\\{d_{i} / \\sqrt{2}\\right\\}$, and we have\n$$\n1-\\frac{1}{\\sqrt{2}}>\\sum_{i}\\left\\{\\frac{d_{i}}{\\sqrt{2}}\\right\\}>\\frac{1}{2 \\sqrt{2}} \\sum_{i} \\frac{1}{d_{i}} \\geqslant \\frac{(k-1)^{2}}{2 \\sqrt{2}} \\frac{1}{\\sum_{i} d_{i}}>\\frac{(k-1)^{2}}{2 \\sqrt{2}} \\cdot \\frac{1}{n} . \\tag{3}\n$$\nHere, the first inequality holds because $\\left\\{a_{k} / \\sqrt{2}\\right\\}<1-1 / \\sqrt{2}$, the second follows from (2), the third follows from an easy application of the AM-HM inequality (or Cauchy-Schwarz), and the fourth follows from the fact that $\\sum_{i} d_{i}=a_{k}<n$.\n\nRearranging this, we obtain\n$$\n\\sqrt{2 \\sqrt{2}-2} \\cdot \\sqrt{n}>k-1\n$$\nwhich provides the required bound on $k$.\nLet $\\alpha=2+\\sqrt{2}$, so $(1 / \\alpha)+(1 / \\sqrt{2})=1$. Thus, $J=\\{\\lfloor i \\alpha\\rfloor: i \\in \\mathbb{Z}_{>0}\\}$ is the complementary Beatty sequence to $H$ (in other words, $H$ and $J$ are disjoint with $H \\cup J=\\mathbb{Z}_{>0}$ ). Write $A=\\{a_{1}<a_{2}<\\cdots<a_{k}\\}$. Suppose that $A$ has no differences in $H$, so all its differences are in $J$ and we can set $a_{i}-a_{1}=\\lfloor\\alpha b_{i}\\rfloor$ for $b_{i} \\in \\mathbb{Z}_{>0}$.\n\nFor any $j>i$, we have $a_{j}-a_{i}=\\lfloor\\alpha b_{j}\\rfloor-\\lfloor\\alpha b_{i}\\rfloor$. Because $a_{j}-a_{i} \\in J$, we also have $a_{j}-a_{i}=\\lfloor\\alpha t\\rfloor$ for some positive integer $t$. Thus, $\\lfloor\\alpha t\\rfloor=\\lfloor\\alpha b_{j}\\rfloor-\\lfloor\\alpha b_{i}\\rfloor$. The right hand side must equal either $\\lfloor\\alpha(b_{j}-b_{i})\\rfloor$ or $\\lfloor\\alpha(b_{j}-b_{i})\\rfloor-1$, the latter of which is not a member of $J$ as $\\alpha>2$. Therefore, $t=b_{j}-b_{i}$ and so we have $\\lfloor\\alpha b_{j}\\rfloor-\\lfloor\\alpha b_{i}\\rfloor=\\lfloor\\alpha(b_{j}-b_{i})\\rfloor$.\n\nFor $1 \\leqslant i<k$ we now put $d_{i}=b_{i+1}-b_{i}$, and we have\n$$\n\\lfloor\\alpha \\sum_{i} d_{i}\\rfloor=\\lfloor\\alpha b_{k}\\rfloor=\\sum_{i}\\lfloor\\alpha d_{i}\\rfloor\n$$\nthat is, $\\sum_{i}\\{\\alpha d_{i}\\}<1$. We also have\n$$\n1+\\lfloor\\alpha \\sum_{i} d_{i}\\rfloor=1+a_{k}-a_{1} \\leqslant a_{k} \\leqslant n\n$$\nso $\\sum_{i} d_{i} \\leqslant n / \\alpha$.\n\nWith the above inequalities, an argument similar to (3) (which uses the fact that $\\{\\alpha d\\}= \\{d \\sqrt{2}\\}>1 /(2 d \\sqrt{2})$ for positive integers $d$ ) proves that $1>\\left((k-1)^{2} /(2 \\sqrt{2})\\right)(\\alpha / n)$, which again rearranges to give\n$$\n\\sqrt{2 \\sqrt{2}-2} \\cdot \\sqrt{n}>k-1\n$$\nAgain, define $J=\\mathbb{Z}_{>0} \\backslash H$, so all differences between elements of $A$ are in $J$. We start by making the following observation. Suppose we have a set $B \\subseteq\\{1,2, \\ldots, n\\}$ such that all of the differences between elements of $B$ are in $H$. Then $|A| \\cdot|B| \\leqslant 2 n$.\n\nTo see why, observe that any two sums of the form $a+b$ with $a \\in A, b \\in B$ are different; otherwise, we would have $a_{1}+b_{1}=a_{2}+b_{2}$, and so $|a_{1}-a_{2}|=|b_{2}-b_{1}|$. However, then the left hand side is in $J$ whereas the right hand side is in $H$. Thus, $\\{a+b: a \\in A, b \\in B\\}$ is a set of size $|A| \\cdot|B|$ all of whose elements are no greater than $2 n$, yielding the claimed inequality.\n\nWith this in mind, it suffices to construct a set $B$, all of whose differences are in $H$ and whose size is at least $C' \\sqrt{n}$ for some constant $C'>0$.\n\nTo do so, we will use well-known facts about the negative Pell equation $X^{2}-2 Y^{2}=-1$; in particular, that there are infinitely many solutions and the values of $X$ are given by the recurrence $X_{1}=1, X_{2}=7$ and $X_{m}=6 X_{m-1}-X_{m-2}$. Therefore, we may choose $X$ to be a solution with $\\sqrt{n} / 6<X \\leqslant \\sqrt{n}$.\n\nNow, we claim that we may choose $B=\\{X, 2 X, \\ldots,\\lfloor(1 / 3) \\sqrt{n}\\rfloor X\\}$. Indeed, we have\n$$\n\\left(\\frac{X}{\\sqrt{2}}-Y\\right)\\left(\\frac{X}{\\sqrt{2}}+Y\\right)=\\frac{-1}{2}\n$$\nand so\n$$\n0>\\left(\\frac{X}{\\sqrt{2}}-Y\\right) \\geqslant \\frac{-3}{\\sqrt{2 n}},\n$$\nfrom which it follows that $\\{X / \\sqrt{2}\\}>1-(3 / \\sqrt{2 n})$. Combined with (1), this shows that all differences between elements of $B$ are in $H$.\nAs in Solution 3, we will provide a construction of a large set $B \\subseteq\\{1,2, \\ldots, n\\}$, all of whose differences are in $H$.\n\nChoose $Y$ to be a solution to the Pell-like equation $X^{2}-2 Y^{2}= \\pm 1$; such solutions are given by the recurrence $Y_{1}=1, Y_{2}=2$ and $Y_{m}=2 Y_{m-1}+Y_{m-2}$, and so we can choose $Y$ such that $n /(3 \\sqrt{2})<Y \\leqslant n / \\sqrt{2}$. Furthermore, it is known that for such a $Y$ and for $1 \\leqslant x<Y$,\n$$\n\\{x \\sqrt{2}\\}+\\{(Y-x) \\sqrt{2}\\}=\\{Y / \\sqrt{2}\\} \\tag{4}\n$$\nif $X^{2}-2 Y^{2}=1$, and\n$$\n\\{x \\sqrt{2}\\}+\\{(Y-x) \\sqrt{2}\\}=1+\\{Y / \\sqrt{2}\\} \\tag{5}\n$$\nif $X^{2}-2 Y^{2}=-1$. Indeed, this is a statement of the fact that $X / Y$ is a best rational approximation to $\\sqrt{2}$, from below in the first case and from above in the second.\n\nNow, consider the sequence $\\{\\sqrt{2}\\},\\{2 \\sqrt{2}\\}, \\ldots,\\{(Y-1) \\sqrt{2}\\}$. The Erdős-Szekeres theorem tells us that this sequence has a monotone subsequence with at least $\\sqrt{Y-2}+1>\\sqrt{Y}$ elements; if that subsequence is decreasing, we may reflect (using (4) or (5)) to ensure that it is increasing. Call the subsequence $\\{y_{1} \\sqrt{2}\\},\\{y_{2} \\sqrt{2}\\}, \\ldots,\\{y_{t} \\sqrt{2}\\}$ for $t>\\sqrt{Y}$.\n\nNow, set $B=\\{\\lfloor y_{i} \\sqrt{2}\\rfloor: 1 \\leqslant i \\leqslant t\\}$. We have $\\lfloor y_{j} \\sqrt{2}\\rfloor-\\lfloor y_{i} \\sqrt{2}\\rfloor=\\lfloor(y_{j}-y_{i}) \\sqrt{2}\\rfloor$ for $i<j$ (because the corresponding inequality for the fractional parts holds by the ordering assumption on the $\\{y_{i} \\sqrt{2}\\}$ ), which means that all differences between elements of $B$ are indeed in $H$. Since $|B|>\\sqrt{Y}>\\sqrt{n} / \\sqrt{3 \\sqrt{2}}$, this is the required set.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32095,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there is a constant $c>0$ and infinitely many positive integers $n$ with the following property: there are infinitely many positive integers that cannot be expressed as the sum of fewer than $c n \\log (n)$ pairwise coprime $n^{\\text{th}}$ powers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "$n$ is divisible by $\\varphi\\left(p^{e}\\right)$ for every prime power $p^{e}$ exactly dividing $N$.\nThis property ensures that all $n^{\\text{th}}$ powers are congruent to $0$ or $1$ modulo each such prime power $p^{e}$, and hence that any sum of $m$ pairwise coprime $n^{\\text{th}}$ powers is congruent to $m$ or $m-1$ modulo $p^{e}$, since at most one of the $n^{\\text{th}}$ powers is divisible by $p$. Thus, if $k$ denotes the number of distinct prime factors of $N$, we find by the Chinese Remainder Theorem at most $2^{k} m$ residue classes modulo $N$ which are sums of at most $m$ pairwise coprime $n^{\\text{th}}$ powers. In particular, if $N>2^{k} m$ then there are infinitely many positive integers not expressible as a sum of at most $m$ pairwise coprime $n^{\\text{th}}$ powers.\nIt thus suffices to prove that there are arbitrarily large pairs $(n, N)$ of integers satisfying $(\\dagger)$ such that\n$$\nN>c \\cdot 2^{k} n \\log (n)\n$$\nfor some positive constant $c$.\nWe construct such pairs as follows. Fix a positive integer $t$ and choose (distinct) prime numbers $p \\mid 2^{2^{t-1}}+1$ and $q \\mid 2^{2^{t}}+1$; we set $N=p q$. It is well-known that $2^{t} \\mid p-1$ and $2^{t+1} \\mid q-1$, hence\n$$\nn=\\frac{(p-1)(q-1)}{2^{t}}\n$$\nis an integer and the pair ( $n, N$ ) satisfies ( $\\dagger$ ).\nEstimating the size of $N$ and $n$ is now straightforward. We have\n$$\n\\log _{2}(n) \\leqslant 2^{t-1}+2^{t}-t<2^{t+1}<2 \\cdot \\frac{N}{n}\n$$\nwhich rearranges to\n$$\nN>\\frac{1}{8} \\cdot 2^{2} n \\log _{2}(n)\n$$\n\n\nSolution 2, obtaining better bounds. As in the preceding solution, we seek arbitrarily large pairs of integers $n$ and $N$ satisfying $(\\dagger)$ such that $N>c 2^{k} n \\log (n)$.\nThis time, to construct such pairs, we fix an integer $x \\geqslant 4$, set $N$ to be the lowest common multiple of $1,2, \\ldots, 2 x$, and set $n$ to be twice the lowest common multiple of $1,2, \\ldots, x$. The pair $(n, N)$ does indeed satisfy the condition, since if $p^{e}$ is a prime power divisor of $N$ then $\\frac{\\varphi\\left(p^{e}\\right)}{2} \\leqslant x$ is a factor of $\\frac{n}{2}=\\operatorname{lcm}_{r \\leqslant x}(r)$.\nNow $2 N / n$ is the product of all primes having a power lying in the interval ( $x, 2 x$ ], and hence $2 N / n>x^{\\pi(2 x)-\\pi(x)}$. Thus for sufficiently large $x$ we have\n$$\n\\log \\left(\\frac{2 N}{2^{\\pi(2 x)} n}\\right)>(\\pi(2 x)-\\pi(x)) \\log (x)-\\log (2) \\pi(2 x) \\sim x\n$$\nusing the Prime Number Theorem $\\pi(t) \\sim t / \\log (t)$.\nOn the other hand, $n$ is a product of at most $\\pi(x)$ prime powers less than or equal to $x$, and so we have the upper bound\n$$\n\\log (n) \\leqslant \\pi(x) \\log (x) \\sim x\n$$\nagain by the Prime Number Theorem. Combined with the above inequality, we find that for any $\\epsilon>0$, the inequality\n$$\n\\log \\left(\\frac{N}{2^{\\pi(2 x)} n}\\right)>(1-\\epsilon) \\log (n)\n$$\nholds for sufficiently large $x$. Rearranging this shows that\n$$\nN>2^{\\pi(2 x)} n^{2-\\epsilon}>2^{\\pi(2 x)} n \\log (n)\n$$\nfor all sufficiently large $x$ and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32096,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$ and $b$ be two positive integers. Prove that the integer\n$$\na^{2} + \\left\\lceil \\frac{4 a^{2}}{b} \\right\\rceil\n$$\nis not a square. (Here $\\lceil z\\rceil$ denotes the least integer greater than or equal to $z$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume, for contradiction, that\n$$\na^{2} + \\left\\lceil \\frac{4 a^{2}}{b} \\right\\rceil = (a + k)^{2}, \\quad \\text{or} \\quad \\left\\lceil \\frac{(2a)^{2}}{b} \\right\\rceil = (2a + k)k.\n$$\nClearly, $k \\geqslant 1$. In other words, the equation\n$$\n\\left\\lceil \\frac{c^{2}}{b} \\right\\rceil = (c + k)k \\tag{1}\n$$\nhas a positive integer solution $(c, k)$, with an even value of $c$.\n\nChoose a positive integer solution of (1) with minimal possible value of $k$, without regard to the parity of $c$. From\n$$\n\\frac{c^{2}}{b} > \\left\\lceil \\frac{c^{2}}{b} \\right\\rceil - 1 = c k + k^{2} - 1 \\geqslant c k\n$$\nand\n$$\n\\frac{(c - k)(c + k)}{b} < \\frac{c^{2}}{b} \\leqslant \\left\\lceil \\frac{c^{2}}{b} \\right\\rceil = (c + k)k\n$$\nit can be seen that $c > b k > c - k$, so\n$$\nc = k b + r \\quad \\text{with some} \\ 0 < r < k.\n$$\nBy substituting this in (1) we get\n$$\n\\left\\lceil \\frac{c^{2}}{b} \\right\\rceil = \\left\\lceil \\frac{(b k + r)^{2}}{b} \\right\\rceil = k^{2} b + 2 k r + \\left\\lceil \\frac{r^{2}}{b} \\right\\rceil\n$$\nand\n$$\n(c + k)k = (k b + r + k)k = k^{2} b + 2 k r + k(k - r),\n$$\nso\n$$\n\\left\\lceil \\frac{r^{2}}{b} \\right\\rceil = k(k - r) \\tag{2}\n$$\nNotice that relation (2) provides another positive integer solution of (1), namely $c' = r$ and $k' = k - r$, with $c' > 0$ and $0 < k' < k$. That contradicts the minimality of $k$, and hence finishes the solution.\nSuppose that\n$$\na^{2} + \\left\\lceil \\frac{4 a^{2}}{b} \\right\\rceil = c^{2}\n$$\nwith some positive integer $c > a$, so\n$$\n\\begin{align*}\n& c^{2} - 1 < a^{2} + \\frac{4 a^{2}}{b} \\leqslant c^{2} \\\\\n& 0 \\leqslant c^{2} b - a^{2}(b + 4) < b \\tag{3}\n\\end{align*}\n$$\nLet $d = c^{2} b - a^{2}(b + 4)$, $x = c + a$ and $y = c - a$; then we have $c = \\frac{x + y}{2}$ and $a = \\frac{x - y}{2}$, and (3) can be re-written as follows:\n$$\n\\begin{align*}\n\\left( \\frac{x + y}{2} \\right)^{2} b - \\left( \\frac{x - y}{2} \\right)^{2} (b + 4) & = d, \\\\\nx^{2} - (b + 2) x y + y^{2} + d = 0 ; \\quad 0 & \\leqslant d < b. \\tag{4}\n\\end{align*}\n$$\nSo, by the indirect assumption, the equation (4) has some positive integer solution $(x, y)$.\n\nFix $b$ and $d$, and take a pair $(x, y)$ of positive integers, satisfying (4), such that $x + y$ is minimal. By the symmetry in (4) we may assume that $x \\geqslant y \\geqslant 1$.\n\nNow we perform a usual \"Vieta jump\". Consider (4) as a quadratic equation in variable $x$, and let $z$ be its second root. By the Vieta formulas,\n$$\nx + z = (b + 2) y, \\quad \\text{and} \\quad z x = y^{2} + d\n$$\nso\n$$\nz = (b + 2) y - x = \\frac{y^{2} + d}{x}\n$$\nThe first formula shows that $z$ is an integer, and by the second formula $z$ is positive. Hence $(z, y)$ is another positive integer solution of (4). From\n$$\n\\begin{aligned}\n(x - 1)(z - 1) & = x z - (x + z) + 1 = (y^{2} + d) - (b + 2) y + 1 \\\\\n& < (y^{2} + b) - (b + 2) y + 1 = (y - 1)^{2} - b(y - 1) \\leqslant (y - 1)^{2} \\leqslant (x - 1)^{2}\n\\end{aligned}\n$$\nwe can see that $z < x$ and therefore $z + y < x + y$. But this contradicts the minimality of $x + y$ among the positive integer solutions of (4).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32097,
    "subject": "Mathematics (Multi-modal)",
    "question": "Version 1. Let $n$ be a positive integer, and set $N=2^{n}$. Determine the smallest real number $a_{n}$ such that, for all real $x$,\n$$\n\\sqrt[N]{\\frac{x^{2 N}+1}{2}} \\leqslant a_{n}(x-1)^{2}+x .\n$$\n\nVersion 2. For every positive integer $N$, determine the smallest real number $b_{N}$ such that, for all real $x$,\n$$\n\\sqrt[N]{\\frac{x^{2 N}+1}{2}} \\leqslant b_{N}(x-1)^{2}+x\n$$\n(Ireland)",
    "options": [],
    "answer": "Version 1: a_n = 2^{n-1}. Version 2: b_N = N/2.",
    "solution": "Solution 1 (for Version 1). First of all, assume that $a_{n}<N / 2$ satisfies the condition. Take $x=1+t$ for $t>0$, we should have\n$$\n\\frac{(1+t)^{2 N}+1}{2} \\leqslant\\left(1+t+a_{n} t^{2}\\right)^{N}\n$$\nExpanding the brackets we get\n$$\n\\begin{equation*}\n\\left(1+t+a_{n} t^{2}\\right)^{N}-\\frac{(1+t)^{2 N}+1}{2}=\\left(N a_{n}-\\frac{N^{2}}{2}\\right) t^{2}+c_{3} t^{3}+\\ldots+c_{2 N} t^{2 N} \\tag{1}\n\\end{equation*}\n$$\nwith some coefficients $c_{3}, \\ldots, c_{2 N}$. Since $a_{n}<N / 2$, the right hand side of (1) is negative for sufficiently small $t$. A contradiction.\nIt remains to prove the following inequality\n$$\n\\sqrt[N]{\\frac{1+x^{2 N}}{2}} \\leqslant x+\\frac{N}{2}(x-1)^{2}, \\quad \\mathcal{I}(N, x)\n$$\nwhere $N=2^{n}$.\nUse induction in $n$. The base case $n=0$ is trivial: $N=1$ and both sides of $\\mathcal{I}(N, x)$ are equal to $\\left(1+x^{2}\\right) / 2$. For completing the induction we prove $\\mathcal{I}(2 N, x)$ assuming that $\\mathcal{I}(N, y)$ is established for all real $y$. We have\n$$\n\\begin{aligned}\n\\left(x+N(x-1)^{2}\\right)^{2} & =x^{2}+N^{2}(x-1)^{4}+N(x-1)^{2} \\frac{(x+1)^{2}-(x-1)^{2}}{2} \\\\\n& =x^{2}+\\frac{N}{2}\\left(x^{2}-1\\right)^{2}+\\left(N^{2}-\\frac{N}{2}\\right)(x-1)^{4} \\geqslant x^{2}+\\frac{N}{2}\\left(x^{2}-1\\right)^{2} \\geqslant \\sqrt[N]{\\frac{1+x^{4 N}}{2}}\n\\end{aligned}\n$$\nwhere the last inequality is $\\mathcal{I}\\left(N, x^{2}\\right)$. Since\n$$\nx+N(x-1)^{2} \\geqslant x+\\frac{(x-1)^{2}}{2}=\\frac{x^{2}+1}{2} \\geqslant 0\n$$\ntaking square root we get $\\mathcal{I}(2 N, x)$. The inductive step is complete.\n\n\nSolution 2.1 (for Version 2). Like in Solution 1 of Version 1, we conclude that $b_{N} \\geqslant N / 2$. It remains to prove the inequality $\\mathcal{I}(N, x)$ for an arbitrary positive integer $N$.\nFirst of all, $\\mathcal{I}(N, 0)$ is obvious. Further, if $x>0$, then the left hand sides of $\\mathcal{I}(N,-x)$ and $\\mathcal{I}(N, x)$ coincide, while the right hand side of $\\mathcal{I}(N,-x)$ is larger than that of $\\mathcal{I}(N,-x)$ (their difference equals $2(N-1) x \\geqslant 0$ ). Therefore, $\\mathcal{I}(N,-x)$ follows from $\\mathcal{I}(N, x)$. So, hereafter we suppose that $x>0$.\nDivide $\\mathcal{I}(N, x)$ by $x$ and let $t=(x-1)^{2} / x=x-2+1 / x$; then $\\mathcal{I}(n, x)$ reads as\n$$\n\\begin{equation*}\nf_{N}:=\\frac{x^{N}+x^{-N}}{2} \\leqslant\\left(1+\\frac{N}{2} t\\right)^{N} . \\tag{2}\n\\end{equation*}\n$$\nThe key identity is the expansion of $f_{N}$ as a polynomial in $t$ :\nLemma.\n$$\n\\begin{equation*}\nf_{N}=N \\sum_{k=0}^{N} \\frac{1}{N+k}\\binom{N+k}{2 k} t^{k} \\tag{3}\n\\end{equation*}\n$$\nProof. Apply induction on $N$. We will make use of the straightforward recurrence relation\n$$\n\\begin{equation*}\nf_{N+1}+f_{N-1}=(x+1 / x) f_{N}=(2+t) f_{N} \\tag{4}\n\\end{equation*}\n$$\nThe base cases $N=1,2$ are straightforward:\n$$\nf_{1}=1+\\frac{t}{2}, \\quad f_{2}=\\frac{1}{2} t^{2}+2 t+1\n$$\nFor the induction step from $N-1$ and $N$ to $N+1$, we compute the coefficient of $t^{k}$ in $f_{N+1}$ using the formula $f_{N+1}=(2+t) f_{N}-f_{N-1}$. For $k=0$ that coefficient equals 1 , for $k>0$ it equals\n$$\n\\begin{aligned}\n& 2 \\frac{N}{N+k}\\binom{N+k}{2 k}+\\frac{N}{N+k-1}\\binom{N+k-1}{2 k-2}-\\frac{N-1}{N+k-1}\\binom{N+k-1}{2 k} \\\\\n& =\\frac{(N+k-1)!}{(2 k)!(N-k)!}\\left(2 N+\\frac{2 k(2 k-1) N}{(N+k-1)(N-k+1)}-\\frac{(N-1)(N-k)}{N+k-1}\\right) \\\\\n& =\\frac{(N+k-1)!}{(2 k)!(N-k+1)!}\\left(2 N(N-k+1)+3 k N+k-N^{2}-N\\right)=\\frac{\\binom{N+k+1}{2 k}}{(N+k+1)}(N+1),\n\\end{aligned}\n$$\nthat completes the induction.\nTurning back to the problem, in order to prove (2) we write\n$$\n\\left(1+\\frac{N}{2} t\\right)^{N}-f_{N}=\\left(1+\\frac{N}{2} t\\right)^{N}-N \\sum_{k=0}^{N} \\frac{1}{N+k}\\binom{N+k}{2 k} t^{k}=\\sum_{k=0}^{N} \\alpha_{k} t^{k}\n$$\nwhere\n$$\n\\begin{aligned}\n\\alpha_{k} & =\\left(\\frac{N}{2}\\right)^{k}\\binom{N}{k}-\\frac{N}{N+k}\\binom{N+k}{2 k} \\\\\n& =\\left(\\frac{N}{2}\\right)^{k}\\binom{N}{k}\\left(1-2^{k} \\frac{(1+1 / N)(1+2 / N) \\cdot \\ldots \\cdot(1+(k-1) / N)}{(k+1) \\cdot \\ldots \\cdot(2 k)}\\right) \\\\\n& \\geqslant\\left(\\frac{N}{2}\\right)^{k}\\binom{N}{k}\\left(1-2^{k} \\frac{2 \\cdot 3 \\cdot \\ldots \\cdot k}{(k+1) \\cdot \\ldots \\cdot(2 k)}\\right)=\\left(\\frac{N}{2}\\right)^{k}\\binom{N}{k}\\left(1-\\prod_{j=1}^{k} \\frac{2 j}{k+j}\\right) \\geqslant 0,\n\\end{aligned}\n$$\nand (2) follows.\n\n\nSolution 2.2 (for Version 2). Here we present another proof of the inequality (2) for $x>0$, or, equivalently, for $t=(x-1)^{2} / x \\geqslant 0$. Instead of finding the coefficients of the polynomial $f_{N}=f_{N}(t)$ we may find its roots, which is in a sense more straightforward. Note that the recurrence (4) and the initial conditions $f_{0}=1, f_{1}=1+t / 2$ imply that $f_{N}$ is a polynomial in $t$ of degree $N$. It also follows by induction that $f_{N}(0)=1, f_{N}^{\\prime}(0)=N^{2} / 2$ : the recurrence relations read as $f_{N+1}(0)+f_{N-1}(0)=2 f_{N}(0)$ and $f_{N+1}^{\\prime}(0)+f_{N-1}^{\\prime}(0)=2 f_{N}^{\\prime}(0)+f_{N}(0)$, respectively.\nNext, if $x_{k}=\\exp \\left(\\frac{i \\pi(2 k-1)}{2 N}\\right)$ for $k \\in\\{1,2, \\ldots, N\\}$, then\n$$\n-t_{k}:=2-x_{k}-\\frac{1}{x_{k}}=2-2 \\cos \\frac{\\pi(2 k-1)}{2 N}=4 \\sin ^{2} \\frac{\\pi(2 k-1)}{4 N}>0\n$$\nand\n$$\nf_{N}\\left(t_{k}\\right)=\\frac{x_{k}^{N}+x_{k}^{-N}}{2}=\\frac{\\exp \\left(\\frac{i \\pi(2 k-1)}{2}\\right)+\\exp \\left(-\\frac{i \\pi(2 k-1)}{2}\\right)}{2}=0 .\n$$\nSo the roots of $f_{N}$ are $t_{1}, \\ldots, t_{N}$ and by the AM-GM inequality we have\n$$\n\\begin{aligned}\nf_{N}(t)=\\left(1-\\frac{t}{t_{1}}\\right)\\left(1-\\frac{t}{t_{2}}\\right) \\ldots\\left(1-\\frac{t}{t_{N}}\\right) & \\leqslant\\left(1-\\frac{t}{N}\\left(\\frac{1}{t_{1}}+\\ldots+\\frac{1}{t_{n}}\\right)\\right)^{N}= \\\\\n\\left(1+\\frac{t f_{N}^{\\prime}(0)}{N}\\right)^{N} & =\\left(1+\\frac{N}{2} t\\right)^{N}\n\\end{aligned}\n$$\n\n\nSolution 2.3 (for Version 2). Here we solve the problem when $N \\geqslant 1$ is an arbitrary real number. For a real number $a$ let\n$$\nf(x)=\\left(\\frac{x^{2 N}+1}{2}\\right)^{\\frac{1}{N}}-a(x-1)^{2}-x\n$$\nThen $f(1)=0$,\n$$\n\\begin{gathered}\nf^{\\prime}(x)=\\left(\\frac{x^{2 N}+1}{2}\\right)^{\\frac{1}{N}-1} x^{2 N-1}-2 a(x-1)-1 \\quad \\text { and } \\quad f^{\\prime}(1)=0 \\\\\nf^{\\prime \\prime}(x)=(1-N)\\left(\\frac{x^{2 N}+1}{2}\\right)^{\\frac{1}{N}-2} x^{4 N-2}+(2 N-1)\\left(\\frac{x^{2 N}+1}{2}\\right)^{\\frac{1}{N}-1} x^{2 N-2}-2 a \\quad \\text { and } \\quad f^{\\prime \\prime}(1)=N-2 a\n\\end{gathered}\n$$\nSo if $a<\\frac{N}{2}$, the function $f$ has a strict local minimum at point 1 , and the inequality $f(x) \\leqslant 0=f(1)$ does not hold. This proves $b_{N} \\geqslant N / 2$.\nFor $a=\\frac{N}{2}$ we have $f^{\\prime \\prime}(1)=0$ and\n$$\nf^{\\prime \\prime \\prime}(x)=\\frac{1}{2}(1-N)(1-2 N)\\left(\\frac{x^{2 N}+1}{2}\\right)^{\\frac{1}{N}-3} x^{2 N-3}\\left(1-x^{2 N}\\right) \\quad \\begin{cases}>0 & \\text { if } 0<x<1 \\text { and } \\\\ <0 & \\text { if } x>1\\end{cases}\n$$\nHence, $f^{\\prime \\prime}(x)<0$ for $x \\neq 1$; $f^{\\prime}(x)>0$ for $x<1$ and $f^{\\prime}(x)<0$ for $x>1$, finally $f(x)<0$ for $x \\neq 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32098,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{A}$ denote the set of all polynomials in three variables $x, y, z$ with integer coefficients. Let $\\mathcal{B}$ denote the subset of $\\mathcal{A}$ formed by all polynomials which can be expressed as\n$$\n(x+y+z) P(x, y, z)+(x y+y z+z x) Q(x, y, z)+x y z R(x, y, z)\n$$\nwith $P, Q, R \\in \\mathcal{A}$. Find the smallest non-negative integer $n$ such that $x^{i} y^{j} z^{k} \\in \\mathcal{B}$ for all nonnegative integers $i, j, k$ satisfying $i+j+k \\geqslant n$.\n(Venezuela)",
    "options": [],
    "answer": "4",
    "solution": "We start by showing that $n \\leqslant 4$, i.e., any monomial $f=x^{i} y^{j} z^{k}$ with $i+j+k \\geqslant 4$ belongs to $\\mathcal{B}$. Assume that $i \\geqslant j \\geqslant k$, the other cases are analogous.\nLet $x+y+z=p$, $x y+y z+z x=q$ and $x y z=r$. Then\n$$\n0=(x-x)(x-y)(x-z)=x^{3}-p x^{2}+q x-r,\n$$\ntherefore $x^{3} \\in \\mathcal{B}$. Next, $x^{2} y^{2}=x y q-(x+y) r \\in \\mathcal{B}$.\nIf $k \\geqslant 1$, then $r$ divides $f$, thus $f \\in \\mathcal{B}$. If $k=0$ and $j \\geqslant 2$, then $x^{2} y^{2}$ divides $f$, thus we have $f \\in \\mathcal{B}$ again. Finally, if $k=0, j \\leqslant 1$, then $x^{3}$ divides $f$ and $f \\in \\mathcal{B}$ in this case also.\n\nIn order to prove that $n \\geqslant 4$, we show that the monomial $x^{2} y$ does not belong to $\\mathcal{B}$. Assume the contrary:\n$$\n\\begin{equation*}\nx^{2} y=p P+q Q+r R \\tag{1}\n\\end{equation*}\n$$\nfor some polynomials $P, Q, R$. If polynomial $P$ contains the monomial $x^{2}$ (with nonzero coefficient), then $p P+q Q+r R$ contains the monomial $x^{3}$ with the same nonzero coefficient. So $P$ does not contain $x^{2}, y^{2}, z^{2}$ and we may write\n$$\nx^{2} y=(x+y+z)(a x y+b y z+c z x)+(x y+y z+z x)(d x+e y+f z)+g x y z\n$$\nwhere $a, b, c ; d, e, f ; g$ are the coefficients of $x y, y z, z x ; x, y, z ; x y z$ in the polynomials $P$; $Q ; R$, respectively (the remaining coefficients do not affect the monomials of degree 3 in $p P+q Q+r R$ ). By considering the coefficients of $x y^{2}$ we get $e=-a$, analogously $e=-b$, $f=-b, f=-c, d=-c$, thus $a=b=c$ and $f=e=d=-a$, but then the coefficient of $x^{2} y$ in the right hand side equals $a+d=0 \\neq 1$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32099,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose that $a, b, c, d$ are positive real numbers satisfying $(a+c)(b+d)=a c+b d$. Find the smallest possible value of\n$$\nS=\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{d}+\\frac{d}{a}\n$$\n(Israel)",
    "options": [],
    "answer": "8",
    "solution": "To show that $S \\geqslant 8$, apply the AM-GM inequality twice as follows:\n$$\n\\left(\\frac{a}{b}+\\frac{c}{d}\\right)+\\left(\\frac{b}{c}+\\frac{d}{a}\\right) \\geqslant 2 \\sqrt{\\frac{a c}{b d}}+2 \\sqrt{\\frac{b d}{a c}}=\\frac{2(a c+b d)}{\\sqrt{a b c d}}=\\frac{2(a+c)(b+d)}{\\sqrt{a b c d}} \\geqslant 2 \\cdot \\frac{2 \\sqrt{a c} \\cdot 2 \\sqrt{b d}}{\\sqrt{a b c d}}=8.\n$$\nThe above inequalities turn into equalities when $a=c$ and $b=d$. Then the condition $(a+c)(b+d)=a c+b d$ can be rewritten as $4 a b=a^{2}+b^{2}$. So it is satisfied when $a / b=2 \\pm \\sqrt{3}$. Hence, $S$ attains value 8, e.g., when $a=c=1$ and $b=d=2+\\sqrt{3}$.\nBy homogeneity we may suppose that $a b c d=1$. Let $a b=C$, $b c=A$ and $c a=B$. Then $a, b, c$ can be reconstructed from $A, B$ and $C$ as $a=\\sqrt{B C / A}$, $b=\\sqrt{A C / B}$ and $c=\\sqrt{A B / C}$. Moreover, the condition $(a+c)(b+d)=a c+b d$ can be written in terms of $A, B, C$ as\n$$\nA+\\frac{1}{A}+C+\\frac{1}{C}=b c+a d+a b+c d=(a+c)(b+d)=a c+b d=B+\\frac{1}{B}\n$$\nWe then need to minimize the expression\n$$\n\\begin{aligned}\nS & :=\\frac{a d+b c}{b d}+\\frac{a b+c d}{a c}=\\left(A+\\frac{1}{A}\\right) B+\\left(C+\\frac{1}{C}\\right) \\frac{1}{B} \\\\\n& =\\left(A+\\frac{1}{A}\\right)\\left(B-\\frac{1}{B}\\right)+\\left(A+\\frac{1}{A}+C+\\frac{1}{C}\\right) \\frac{1}{B} \\\\\n& =\\left(A+\\frac{1}{A}\\right)\\left(B-\\frac{1}{B}\\right)+\\left(B+\\frac{1}{B}\\right) \\frac{1}{B}\n\\end{aligned}\n$$\nWithout loss of generality assume that $B \\geqslant 1$ (otherwise, we may replace $B$ by $1 / B$ and swap $A$ and $C$, this changes neither the relation nor the function to be maximized). Therefore, we can write\n$$\nS \\geqslant 2\\left(B-\\frac{1}{B}\\right)+\\left(B+\\frac{1}{B}\\right) \\frac{1}{B}=2 B+\\left(1-\\frac{1}{B}\\right)^{2}=: f(B) .\n$$\nClearly, $f$ increases on $[1, \\infty)$. Since\n$$\nB+\\frac{1}{B}=A+\\frac{1}{A}+C+\\frac{1}{C} \\geqslant 4,\n$$\nwe have $B \\geqslant B^{\\prime}$, where $B^{\\prime}=2+\\sqrt{3}$ is the unique root greater than 1 of the equation $B^{\\prime}+1 / B^{\\prime}=4$. Hence,\n$$\nS \\geqslant f(B) \\geqslant f\\left(B^{\\prime}\\right)=2\\left(B^{\\prime}-\\frac{1}{B^{\\prime}}\\right)+\\left(B^{\\prime}+\\frac{1}{B^{\\prime}}\\right) \\frac{1}{B^{\\prime}}=2 B^{\\prime}-\\frac{2}{B^{\\prime}}+\\frac{4}{B^{\\prime}}=8\n$$\nIt remains to note that when $A=C=1$ and $B=B^{\\prime}$ we have the equality $S=8$.\nWe present another proof of the inequality $S \\geqslant 8$. We start with the estimate\n$$\n\\left(\\frac{a}{b}+\\frac{c}{d}\\right)+\\left(\\frac{b}{c}+\\frac{d}{a}\\right) \\geqslant 2 \\sqrt{\\frac{a c}{b d}}+2 \\sqrt{\\frac{b d}{a c}}\n$$\nLet $y=\\sqrt{a c}$ and $z=\\sqrt{b d}$, and assume, without loss of generality, that $a c \\geqslant b d$. By the AM-GM inequality, we have\n$$\ny^{2}+z^{2}=a c+b d=(a+c)(b+d) \\geqslant 2 \\sqrt{a c} \\cdot 2 \\sqrt{b d}=4 y z .\n$$\nSubstituting $x=y / z$, we get $4 x \\leqslant x^{2}+1$. For $x \\geqslant 1$, this holds if and only if $x \\geqslant 2+\\sqrt{3}$.\nNow we have\n$$\n2 \\sqrt{\\frac{a c}{b d}}+2 \\sqrt{\\frac{b d}{a c}}=2\\left(x+\\frac{1}{x}\\right) .\n$$\nClearly, this is minimized by setting $x(\\geqslant 1)$ as close to 1 as possible, i.e., by taking $x=2+\\sqrt{3}$. Then $2(x+1 / x)=2((2+\\sqrt{3})+(2-\\sqrt{3}))=8$, as required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32100,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$, $d$ be four real numbers such that $a \\geqslant b \\geqslant c \\geqslant d > 0$ and $a + b + c + d = 1$. Prove that\n$$\n(a + 2b + 3c + 4d) a^{a} b^{b} c^{c} d^{d} < 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The weighted AM-GM inequality with weights $a$, $b$, $c$, $d$ gives\n$$\na^{a} b^{b} c^{c} d^{d} \\leqslant a \\cdot a + b \\cdot b + c \\cdot c + d \\cdot d = a^{2} + b^{2} + c^{2} + d^{2}\n$$\nso it suffices to prove that $(a + 2b + 3c + 4d)(a^{2} + b^{2} + c^{2} + d^{2}) < 1 = (a + b + c + d)^{3}$. This can be done in various ways, for example:\n$$\n\\begin{aligned}\n(a + b + c + d)^{3} > & a^{2}(a + 3b + 3c + 3d) + b^{2}(3a + b + 3c + 3d) \\\\\n& + c^{2}(3a + 3b + c + 3d) + d^{2}(3a + 3b + 3c + d) \\\\\n\\geqslant & (a^{2} + b^{2} + c^{2} + d^{2}) \\cdot (a + 2b + 3c + 4d)\n\\end{aligned}\n$$\nFrom $b \\geqslant d$ we get\n$$\na + 2b + 3c + 4d \\leqslant a + 3b + 3c + 3d = 3 - 2a .\n$$\nIf $a < \\frac{1}{2}$, then the statement can be proved by\n$$\n(a + 2b + 3c + 4d) a^{a} b^{b} c^{c} d^{d} \\leqslant (3 - 2a) a^{a} a^{b} a^{c} a^{d} = (3 - 2a) a = 1 - (1 - a)(1 - 2a) < 1\n$$\nFrom now on we assume $\\frac{1}{2} \\leqslant a < 1$.\nBy $b, c, d < 1 - a$ we have\n$$\nb^{b} c^{c} d^{d} < (1 - a)^{b} \\cdot (1 - a)^{c} \\cdot (1 - a)^{d} = (1 - a)^{1 - a}\n$$\nTherefore,\n$$\n(a + 2b + 3c + 4d) a^{a} b^{b} c^{c} d^{d} < (3 - 2a) a^{a} (1 - a)^{1 - a}\n$$\nFor $0 < x < 1$, consider the functions\n$f(x) = (3 - 2x) x^{x} (1 - x)^{1 - x}$ and $g(x) = \\log f(x) = \\log (3 - 2x) + x \\log x + (1 - x) \\log (1 - x)$; hereafter, log denotes the natural logarithm. It is easy to verify that\n$$\ng''(x) = -\\frac{4}{(3 - 2x)^{2}} + \\frac{1}{x} + \\frac{1}{1 - x} = \\frac{1 + 8(1 - x)^{2}}{x(1 - x)(3 - 2x)^{2}} > 0\n$$\nso $g$ is strictly convex on $(0, 1)$.\nBy $g\\left(\\frac{1}{2}\\right) = \\log 2 + 2 \\cdot \\frac{1}{2} \\log \\frac{1}{2} = 0$ and $\\lim_{x \\rightarrow 1-} g(x) = 0$, we have $g(x) \\leqslant 0$ (and hence $f(x) \\leqslant 1$) for all $x \\in \\left[\\frac{1}{2}, 1\\right)$, and therefore\n$$\n(a + 2b + 3c + 4d) a^{a} b^{b} c^{c} d^{d} < f(a) \\leqslant 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32101,
    "subject": "Mathematics (Multi-modal)",
    "question": "A magician intends to perform the following trick. She announces a positive integer $n$, along with $2n$ real numbers $x_{1} < \\ldots < x_{2n}$, to the audience. A member of the audience then secretly chooses a polynomial $P(x)$ of degree $n$ with real coefficients, computes the $2n$ values $P(x_{1}), \\ldots, P(x_{2n})$, and writes down these $2n$ values on the blackboard in non-decreasing order. After that the magician announces the secret polynomial to the audience.\nCan the magician find a strategy to perform such a trick?\n(Luxembourg)",
    "options": [],
    "answer": "No, she cannot.",
    "solution": "Answer: No, she cannot.\n\nLet $x_{1} < x_{2} < \\ldots < x_{2n}$ be real numbers chosen by the magician. We will construct two distinct polynomials $P(x)$ and $Q(x)$, each of degree $n$, such that the member of audience will write down the same sequence for both polynomials. This will mean that the magician cannot distinguish $P$ from $Q$.\n\n**Claim.** There exists a polynomial $P(x)$ of degree $n$ such that $P(x_{2i-1}) + P(x_{2i}) = 0$ for $i = 1, 2, \\ldots, n$.\n\n**Proof.** We want to find a polynomial $a_{n} x^{n} + \\ldots + a_{1} x + a_{0}$ satisfying the following system of equations:\n$$\n\\left\\{\n\\begin{array}{l}\n(x_{1}^{n} + x_{2}^{n}) a_{n} + (x_{1}^{n-1} + x_{2}^{n-1}) a_{n-1} + \\ldots + 2 a_{0} = 0 \\\\\n(x_{3}^{n} + x_{4}^{n}) a_{n} + (x_{3}^{n-1} + x_{4}^{n-1}) a_{n-1} + \\ldots + 2 a_{0} = 0 \\\\\n\\cdots \\\\\n(x_{2n-1}^{n} + x_{2n}^{n}) a_{n} + (x_{2n-1}^{n-1} + x_{2n}^{n-1}) a_{n-1} + \\ldots + 2 a_{0} = 0\n\\end{array}\n\\right.\n$$\nWe use the well known fact that a homogeneous system of $n$ linear equations in $n+1$ variables has a nonzero solution. (This fact can be proved using induction on $n$, via elimination of variables.) Applying this fact to the above system, we find a nonzero polynomial $P(x)$ of degree not exceeding $n$ such that its coefficients $a_{0}, \\ldots, a_{n}$ satisfy this system. Therefore $P(x_{2i-1}) + P(x_{2i}) = 0$ for all $i = 1, 2, \\ldots, n$. Notice that $P$ has a root on each segment $[x_{2i-1}, x_{2i}]$ by the Intermediate Value theorem, so $n$ roots in total. Since $P$ is nonzero, we get $\\operatorname{deg} P = n$.\n\nNow consider a polynomial $P(x)$ provided by the Claim, and take $Q(x) = -P(x)$. The properties of $P(x)$ yield that $P(x_{2i-1}) = Q(x_{2i})$ and $Q(x_{2i-1}) = P(x_{2i})$ for all $i = 1, 2, \\ldots, n$. It is also clear that $P \\neq -P = Q$ and $\\operatorname{deg} Q = \\operatorname{deg} P = n$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32102,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $k$ be positive integers. Prove that for $a_{1}, \\ldots, a_{n} \\in [1,2^{k}]$ one has\n$$\n\\sum_{i=1}^{n} \\frac{a_{i}}{\\sqrt{a_{1}^{2}+\\ldots+a_{i}^{2}}} \\leqslant 4 \\sqrt{k n}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Partition the set of indices $\\{1,2, \\ldots, n\\}$ into disjoint subsets $M_{1}, M_{2}, \\ldots, M_{k}$ so that $a_{\\ell} \\in [2^{j-1}, 2^{j}]$ for $\\ell \\in M_{j}$. Then, if $|M_{j}|=: p_{j}$, we have\n$$\n\\sum_{\\ell \\in M_{j}} \\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}} \\leqslant \\sum_{i=1}^{p_{j}} \\frac{2^{j}}{2^{j-1} \\sqrt{i}} = 2 \\sum_{i=1}^{p_{j}} \\frac{1}{\\sqrt{i}}\n$$\nwhere we used that $a_{\\ell} \\leqslant 2^{j}$ and in the denominator every index from $M_{j}$ contributes at least $(2^{j-1})^{2}$. Now, using $\\sqrt{i}-\\sqrt{i-1}=\\frac{1}{\\sqrt{i}+\\sqrt{i-1}} \\geqslant \\frac{1}{2 \\sqrt{i}}$, we deduce that\n$$\n\\sum_{\\ell \\in M_{j}} \\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}} \\leqslant 2 \\sum_{i=1}^{p_{j}} \\frac{1}{\\sqrt{i}} \\leqslant 2 \\sum_{i=1}^{p_{j}} 2(\\sqrt{i}-\\sqrt{i-1}) = 4 \\sqrt{p_{j}}.\n$$\nTherefore, summing over $j=1, \\ldots, k$ and using the QM-AM inequality, we obtain\n$$\n\\sum_{\\ell=1}^{n} \\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}} \\leqslant 4 \\sum_{j=1}^{k} \\sqrt{|M_{j}|} \\leqslant 4 \\sqrt{k \\sum_{j=1}^{k}|M_{j}|} = 4 \\sqrt{k n}\n$$\nApply induction on $n$. The base $n \\leqslant 16$ is clear: our sum does not exceed $n \\leqslant 4 \\sqrt{n k}$. For the inductive step from $1, \\ldots, n-1$ to $n \\geqslant 17$ consider two similar cases.\n\nCase 1: $n=2 t$.\nLet $x_{\\ell}=\\frac{a_{\\ell}}{\\sqrt{a_{1}^{2}+\\ldots+a_{\\ell}^{2}}}$. We have\n$$\n\\exp \\left(-x_{t+1}^{2}-\\ldots-x_{2 t}^{2}\\right) \\geqslant (1-x_{t+1}^{2}) \\ldots (1-x_{2 t}^{2}) = \\frac{a_{1}^{2}+\\ldots+a_{t}^{2}}{a_{1}^{2}+\\ldots+a_{2 t}^{2}} \\geqslant \\frac{1}{1+4^{k}},\n$$\nwhere we used that the product is telescopic and then an estimate $a_{t+i} \\leqslant 2^{k} a_{i}$ for $i=1, \\ldots, t$. Therefore, $x_{t+1}^{2}+\\ldots+x_{2 t}^{2} \\leqslant \\log (4^{k}+1) \\leqslant 2 k$, where $\\log$ denotes the natural logarithm. This implies $x_{t+1}+\\ldots+x_{2 t} \\leqslant \\sqrt{2 k t}$. Hence, using the inductive hypothesis for $n=t$ we get\n$$\n\\sum_{\\ell=1}^{2 t} x_{\\ell} \\leqslant 4 \\sqrt{k t}+\\sqrt{2 k t} \\leqslant 4 \\sqrt{2 k t}\n$$\n\nCase 2: $n=2 t+1$.\nAnalogously, we get $x_{t+2}^{2}+\\ldots+x_{2 t+1}^{2} \\leqslant \\log (4^{k}+1) \\leqslant 2 k$ and\n$$\n\\sum_{\\ell=1}^{2 t+1} x_{\\ell} \\leqslant 4 \\sqrt{k(t+1)}+\\sqrt{2 k t} \\leqslant 4 \\sqrt{k(2 t+1)}\n$$\nThe last inequality is true for all $t \\geqslant 8$ since\n$$\n4 \\sqrt{2 t+1}-\\sqrt{2 t} \\geqslant 3 \\sqrt{2 t}=\\sqrt{18 t} \\geqslant \\sqrt{16 t+16}=4 \\sqrt{t+1}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32103,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}^{+}$ be the set of positive real numbers. Determine all functions $f: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}^{+}$ such that, for all positive real numbers $x$ and $y$,\n$$\nf(x+f(xy))+y=f(x)f(y)+1\n$$",
    "options": [],
    "answer": "f(x) = x + 1",
    "solution": "A straightforward check shows that $f(x)=x+1$ satisfies (*). We divide the proof of the converse statement into a sequence of steps.\n\nStep 1: $f$ is injective.\nPut $x=1$ in (*) and rearrange the terms to get\n$$\ny=f(1)f(y)+1-f(1+f(y))\n$$\nTherefore, if $f(y_1)=f(y_2)$, then $y_1=y_2$.\n\nStep 2: $f$ is (strictly) monotone increasing.\nFor any fixed $y \\in \\mathbb{R}^{+}$, the function\n$$\ng(x):=f(x+f(xy))=f(x)f(y)+1-y\n$$\nis injective by Step 1. Therefore, $x_1+f(x_1y) \\neq x_2+f(x_2y)$ for all $y, x_1, x_2 \\in \\mathbb{R}^{+}$ with $x_1 \\neq x_2$. Plugging in $z_i=x_iy$, we arrive at\n$$\n\\frac{z_1-z_2}{y} \\neq f(z_2)-f(z_1), \\quad \\text{or} \\quad \\frac{1}{y} \\neq \\frac{f(z_2)-f(z_1)}{z_1-z_2}\n$$\nfor all $y, z_1, z_2 \\in \\mathbb{R}^{+}$ with $z_1 \\neq z_2$. This means that the right-hand side of the rightmost relation is always non-positive, i.e., $f$ is monotone non-decreasing. Since $f$ is injective, it is strictly monotone.\n\nStep 3: There exist constants $a$ and $b$ such that $f(y)=a y+b$ for all $y \\in \\mathbb{R}^{+}$.\nSince $f$ is monotone and bounded from below by $0$, for each $x_0 \\geqslant 0$, there exists a right $\\operatorname{limit} \\lim_{x \\searrow x_0} f(x) \\geqslant 0$. Put $p=\\lim_{x \\searrow 0} f(x)$ and $q=\\lim_{x \\searrow p} f(x)$.\nFix an arbitrary $y$ and take the limit of $(*)$ as $x \\searrow 0$. We have $f(xy) \\searrow p$ and hence $f(x+f(xy)) \\searrow q$; therefore, we obtain\n$$\nq+y=p f(y)+1, \\quad \\text{or} \\quad f(y)=\\frac{q+y-1}{p} .\n$$\n(Notice that $p \\neq 0$, otherwise $q+y=1$ for all $y$, which is absurd.) The claim is proved.\n\nStep 4: $f(x)=x+1$ for all $x \\in \\mathbb{R}^{+}$.\nBased on the previous step, write $f(x)=a x+b$. Putting this relation into (*) we get\n$$\na(x+a x y+b)+b+y=(a x+b)(a y+b)+1\n$$\nwhich can be rewritten as\n$$\n(a-a b)x+(1-a b)y+a b+b-b^2-1=0 \\quad \\text{for all} \\ x, y \\in \\mathbb{R}^{+} .\n$$\nThis identity may hold only if all the coefficients are $0$, i.e.,\n$$\na-a b=1-a b=a b+b-b^2-1=0 .\n$$\nHence, $a=b=1$.\nWe provide another proof that $f(x)=x+1$ is the only function satisfying (*).\nPut $a=f(1)$. Define the function $\\phi: \\mathbb{R}^{+} \\rightarrow \\mathbb{R}$ by\n$$\n\\phi(x)=f(x)-x-1\n$$\nThen equation (*) reads as\n$$\n\\phi(x+f(xy))=f(x)f(y)-f(xy)-x-y .\n$$\nSince the right-hand side is symmetric under swapping $x$ and $y$, we obtain\n$$\n\\phi(x+f(xy))=\\phi(y+f(xy)) .\n$$\nIn particular, substituting $(x, y)=(t, 1/t)$ we get\n$$\n\\phi(a+t)=\\phi\\left(a+\\frac{1}{t}\\right), \\quad t \\in \\mathbb{R}^{+} .\n$$\nNotice that the function $f$ is bounded from below by a positive constant. Indeed, for each $y \\in \\mathbb{R}^{+}$, the relation (*) yields $f(x)f(y)>y-1$, hence\n$$\nf(x)>\\frac{y-1}{f(y)} \\quad \\text{for all} \\ x \\in \\mathbb{R}^{+}\n$$\nIf $y>1$, this provides a desired positive lower bound for $f(x)$.\nNow, let $M=\\inf_{x \\in \\mathbb{R}^{+}} f(x)>0$. Then, for all $y \\in \\mathbb{R}^{+}$,\n$$\nM \\geqslant \\frac{y-1}{f(y)}, \\quad \\text{or} \\quad f(y) \\geqslant \\frac{y-1}{M}\n$$\nLemma 1. The function $f(x)$ (and hence $\\phi(x)$) is bounded on any segment $[p, q]$, where $0<p<q<+\\infty$.\nProof. $f$ is bounded from below by $M$. It remains to show that $f$ is bounded from above on $[p, q]$. Substituting $y=1$ into $(*)$, we get\n$$\nf(x+f(x))=a f(x)\n$$\nTake $z \\in [p, q]$ and put $s=f(z)$. By the above, we have\n$$\nf(z+s)=a s \\quad \\text{and} \\quad f(z+s+a s)=f(z+s+f(z+s))=a^2 s .\n$$\nPlugging in $(x, y)=\\left(z, 1+\\frac{s}{z}\\right)$ to $(*)$ and using the previous estimate, we obtain\n$$\nf(z+a s)=f(z+f(z+s))=s f\\left(1+\\frac{s}{z}\\right)-\\frac{s}{z} \\geqslant \\frac{s^2}{M z}-\\frac{s}{z}\n$$\nNow, substituting $(x, y)=\\left(z+a s, \\frac{z}{z+a s}\\right)$ to $(*)$ and applying the above estimate and the estimate $f(y) \\geqslant M$, we obtain\n$$\na^2 s=f(z+s+a s)=f(z+a s+f(z))=f(z+a s)f\\left(\\frac{z}{z+a s}\\right)+1-\\frac{z}{z+a s} \\\\\n\\geqslant M f(z+a s) \\geqslant \\frac{s^2}{z}-\\frac{M s}{z} \\geqslant \\frac{s^2}{q}-\\frac{M s}{p}\n$$\nThis yields $s \\leqslant q\\left(\\frac{M}{p}+a^2\\right)=: L$, and $f$ is bounded from above by $L$ on $[p, q]$.\nApplying Lemma 1 to the segment $[a, a+1]$, we see that $\\phi$ is bounded on it. By the previous symmetry, we get that $\\phi$ is also bounded on $[a+1,+\\infty)$, and hence on $[a,+\\infty)$. Put $C=\\max\\{a, 3\\}$.\n\nLemma 2. For all $x \\geqslant C$, we have $\\phi(x)=0$ (and hence $f(x)=x+1$).\nProof. Substituting $y=x$ to the earlier equation, we obtain\n$$\n\\phi\\left(x+f\\left(x^2\\right)\\right)=f(x)^2-f\\left(x^2\\right)-2x\n$$\nhence,\n$$\n\\phi\\left(x+f\\left(x^2\\right)\\right)+\\phi\\left(x^2\\right)=f(x)^2-(x+1)^2=\\phi(x)(f(x)+x+1)\n$$\nSince $f(x)+x+1 \\geqslant C+1 \\geqslant 4$, we obtain that\n$$\n|\\phi(x)| \\leqslant \\frac{1}{4}\\left(|\\phi\\left(x+f\\left(x^2\\right)\\right)|+|\\phi\\left(x^2\\right)|\\right)\n$$\nSince $C \\geqslant a$, there exists a finite supremum $S=\\sup_{x \\geqslant C}|\\phi(x)|$. For each $x \\in [C,+\\infty)$, both $x+f\\left(x^2\\right)$ and $x^2$ are greater than $x$; hence they also lie in $[C,+\\infty)$. Therefore, taking the supremum of the left-hand side over $x \\in [C,+\\infty)$, we obtain $S \\leqslant S/2$ and hence $S=0$. Thus, $\\phi(x)=0$ for all $x \\geqslant C$.\nIt remains to show that $f(y)=y+1$ when $0<y<C$. For each $y$, choose $x>\\max\\{C, \\frac{C}{y}\\}$. Then all three numbers $x, x y$, and $x+f(x y)$ are greater than $C$, so (*) reads as\n$$\n(x+x y+1)+1+y=(x+1)f(y)+1, \\quad \\text{hence} \\quad f(y)=y+1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32104,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find the number of permutations $a_{1}, a_{2}, \\ldots, a_{n}$ of the sequence $1,2, \\ldots, n$ satisfying\n$$\na_{1} \\leqslant 2 a_{2} \\leqslant 3 a_{3} \\leqslant \\ldots \\leqslant n a_{n}\n$$",
    "options": [],
    "answer": "F_{n+1}",
    "solution": "Denote by $P_{n}$ the number of permutations that satisfy (*). It is easy to see that $P_{1}=1$ and $P_{2}=2$.\n\nLemma 1. Let $n \\geqslant 3$. If a permutation $a_{1}, \\ldots, a_{n}$ satisfies (*) then either $a_{n}=n$, or $a_{n-1}=n$ and $a_{n}=n-1$.\n\nProof. Let $k$ be the index for which $a_{k}=n$. If $k=n$ then we are done.\nIf $k=n-1$ then, by $(*)$, we have $n(n-1)=(n-1) a_{n-1} \\leqslant n a_{n}$, so $a_{n} \\geqslant n-1$. Since $a_{n} \\neq a_{n-1}=n$, the only choice for $a_{n}$ is $a_{n}=n-1$.\nNow suppose that $k \\leqslant n-2$. For every $k<i<n$ we have $k n=k a_{k} \\leqslant i a_{i}<n a_{i}$, so $a_{i} \\geqslant k+1$. Moreover, $n a_{n} \\geqslant(n-1) a_{n-1} \\geqslant(n-1)(k+1)=n k+(n-1-k)>n k$, so $a_{n} \\geqslant k+1$. Now the $n-k+1$ numbers $a_{k}, a_{k+1}, \\ldots, a_{n}$ are all greater than $k$; but there are only $n-k$ such values; this is not possible.\nIf $a_{n}=n$ then $a_{1}, a_{2}, \\ldots, a_{n-1}$ must be a permutation of the numbers $1, \\ldots, n-1$ satisfying $a_{1} \\leqslant 2 a_{2} \\leqslant \\ldots \\leqslant(n-1) a_{n-1}$; there are $P_{n-1}$ such permutations. The last inequality in (*), $(n-1) a_{n-1} \\leqslant n a_{n}=n^{2}$, holds true automatically.\nIf $(a_{n-1}, a_{n})=(n, n-1)$, then $a_{1}, \\ldots, a_{n-2}$ must be a permutation of $1, \\ldots, n-2$ satisfying $a_{1} \\leqslant \\ldots \\leqslant(n-2) a_{n-2}$; there are $P_{n-2}$ such permutations. The last two inequalities in (*) hold true automatically by $(n-2) a_{n-2} \\leqslant(n-2)^{2}<n(n-1)=(n-1) a_{n-1}=n a_{n}$.\nHence, the sequence ( $P_{1}, P_{2}, \\ldots$ ) satisfies the recurrence relation $P_{n}=P_{n-1}+P_{n-2}$ for $n \\geqslant 3$. The first two elements are $P_{1}=F_{2}$ and $P_{2}=F_{3}$, so by a trivial induction we have $P_{n}=F_{n+1}$.\nWe claim that all sought permutations are of the following kind. Split $\\{1,2, \\ldots, n\\}$ into singletons and pairs of adjacent numbers. In each pair, swap the two numbers and keep the singletons unchanged.\nSuch permutations correspond to tilings of a $1 \\times n$ chessboard using dominoes and unit squares; it is well-known that the number of such tilings is the Fibonacci number $F_{n+1}$.\nThe claim follows by induction from\n\nLemma 2. Assume that $a_{1}, \\ldots, a_{n}$ is a permutation satisfying (*), and $k$ is an integer such that $1 \\leqslant k \\leqslant n$ and $\\{a_{1}, a_{2}, \\ldots, a_{k-1}\\}=\\{1,2, \\ldots, k-1\\}$. (If $k=1$, the condition is empty.) Then either $a_{k}=k$, or $a_{k}=k+1$ and $a_{k+1}=k$.\n\nProof. Choose $t$ with $a_{t}=k$. Since $k \\notin\\{a_{1}, \\ldots, a_{k-1}\\}$, we have either $t=k$ or $t>k$. If $t=k$ then we are done, so assume $t>k$.\nNotice that one of the numbers among the $t-k$ numbers $a_{k}, a_{k+1}, \\ldots, a_{t-1}$ is at least $t$, because there are only $t-k-1$ values between $k$ and $t$. Let $i$ be an index with $k \\leqslant i<t$ and $a_{i} \\geqslant t$; then $k t=t a_{t} \\geqslant i a_{i} \\geqslant i t \\geqslant k t$, so that all the inequalities turn into equalities, hence $i=k$ and $a_{k}=t$. If $t=k+1$, we are done.\nSuppose that $t>k+1$. Then the chain of inequalities $k t=k a_{k} \\leqslant \\ldots \\leqslant t a_{t}=k t$ should also turn into a chain of equalities. From this point we can find contradictions in several ways; for example by pointing to $a_{t-1}=\\frac{k t}{t-1}=k+\\frac{k}{t-1}$ which cannot be an integer, or considering\nthe product of the numbers $(k+1) a_{k+1}, \\ldots,(t-1) a_{t-1}$; the numbers $a_{k+1}, \\ldots, a_{t-1}$ are distinct and greater than $k$, so\n$$\n(k t)^{t-k-1}=(k+1) a_{k+1} \\cdot(k+2) a_{k+2} \\cdot \\ldots \\cdot(t-1) a_{t-1} \\geqslant((k+1)(k+2) \\cdot \\ldots \\cdot(t-1))^{2} .\n$$\nNotice that $(k+i)(t-i)=k t+i(t-k-i)>k t$ for $1 \\leqslant i<t-k$. This leads to the contradiction\n$$\n(k t)^{t-k-1} \\geqslant((k+1)(k+2) \\cdot \\ldots \\cdot(t-1))^{2}=\\prod_{i=1}^{t-k-1}(k+i)(t-i)>(k t)^{t-k-1}\n$$\nTherefore, the case $t>k+1$ is not possible.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32105,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a regular 100-gon, 41 vertices are colored black and the remaining 59 vertices are colored white. Prove that there exist 24 convex quadrilaterals $Q_{1}, \\ldots, Q_{24}$ whose corners are vertices of the 100-gon, so that\n- the quadrilaterals $Q_{1}, \\ldots, Q_{24}$ are pairwise disjoint, and\n- every quadrilateral $Q_{i}$ has three corners of one color and one corner of the other color.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Call a quadrilateral skew-colored, if it has three corners of one color and one corner of the other color. We will prove the following\n\nClaim. If the vertices of a convex $(4k+1)$-gon $P$ are colored black and white such that each color is used at least $k$ times, then there exist $k$ pairwise disjoint skew-colored quadrilaterals whose vertices are vertices of $P$. (One vertex of $P$ remains unused.)\n\nThe problem statement follows by removing 3 arbitrary vertices of the 100-gon and applying the Claim to the remaining 97 vertices with $k=24$.\n\nProof of the Claim. We prove by induction. For $k=1$ we have a pentagon with at least one black and at least one white vertex. If the number of black vertices is even then remove a black vertex; otherwise remove a white vertex. In the remaining quadrilateral, there are an odd number of black and an odd number of white vertices, so the quadrilateral is skew-colored.\n\nFor the induction step, assume $k \\geqslant 2$. Let $b$ and $w$ be the numbers of black and white vertices, respectively; then $b, w \\geqslant k$ and $b+w=4k+1$. Without loss of generality we may assume $w \\geqslant b$, so $k \\leqslant b \\leqslant 2k$ and $2k+1 \\leqslant w \\leqslant 3k+1$.\n\nWe want to find four consecutive vertices such that three of them are white, the fourth one is black. Denote the vertices by $V_{1}, V_{2}, \\ldots, V_{4k+1}$ in counterclockwise order, such that $V_{4k+1}$ is black, and consider the following $k$ groups of vertices:\n\n$$\n\\left(V_{1}, V_{2}, V_{3}, V_{4}\\right),\\left(V_{5}, V_{6}, V_{7}, V_{8}\\right), \\ldots,\\left(V_{4k-3}, V_{4k-2}, V_{4k-1}, V_{4k}\\right)\n$$\n\nIn these groups there are $w$ white and $b-1$ black vertices. Since $w > b-1$, there is a group, $\\left(V_{i}, V_{i+1}, V_{i+2}, V_{i+3}\\right)$ that contains more white than black vertices. If three are white and one is black in that group, we are done. Otherwise, if $V_{i}, V_{i+1}, V_{i+2}, V_{i+3}$ are all white then let $V_{j}$ be the first black vertex among $V_{i+4}, \\ldots, V_{4k+1}$ (recall that $V_{4k+1}$ is black); then $V_{j-3}, V_{j-2}$ and $V_{j-1}$ are white and $V_{j}$ is black.\n\nNow we have four consecutive vertices $V_{i}, V_{i+1}, V_{i+2}, V_{i+3}$ that form a skew-colored quadrilateral. The remaining vertices form a convex $(4k-3)$-gon; $w-3$ of them are white and $b-1$ are black. Since $b-1 \\geqslant k-1$ and $w-3 \\geqslant (2k+1)-3 > k-1$, we can apply the Claim with $k-1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32106,
    "subject": "Mathematics (Multi-modal)",
    "question": "$4 n$ coins of weights $1, 2, 3, \\ldots, 4 n$ are given. Each coin is colored in one of $n$ colors and there are four coins of each color. Show that all these coins can be partitioned into two sets with the same total weight, such that each set contains two coins of each color.\n(Hungary)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us pair the coins with weights summing up to $4 n + 1$, resulting in the set $S$ of $2 n$ pairs: $\\{1, 4 n\\}, \\{2, 4 n - 1\\}, \\ldots, \\{2 n, 2 n + 1\\}$. It suffices to partition $S$ into two sets, each consisting of $n$ pairs, such that each set contains two coins of each color.\n\nIntroduce a multi-graph $G$ (i.e., a graph with loops and multiple edges allowed) on $n$ vertices, so that each vertex corresponds to a color. For each pair of coins from $S$, we add an edge between the vertices corresponding to the colors of those coins. Note that each vertex has degree $4$. Also, a desired partition of the coins corresponds to a coloring of the edges of $G$ in two colors, say red and blue, so that each vertex has degree $2$ with respect to each color (i.e., each vertex has equal red and blue degrees).\n\nTo complete the solution, it suffices to provide such a coloring for each component $G'$ of $G$. Since all degrees of the vertices are even, in $G'$ there exists an Euler circuit $C$ (i.e., a circuit passing through each edge of $G'$ exactly once). Note that the number of edges in $C$ is even (it equals twice the number of vertices in $G'$). Hence all the edges can be colored red and blue so that any two edges adjacent in $C$ have different colors (one may move along $C$ and color the edges one by one alternating red and blue colors). Thus in $G'$ each vertex has equal red and blue degrees, as desired.\nAs in Solution 1, we will show that it is possible to partition $2 n$ pairs $\\{1, 4 n\\}$, $\\{2, 4 n - 1\\}, \\ldots, \\{2 n, 2 n + 1\\}$ into two sets, each consisting of $n$ pairs, such that each set contains two coins of each color.\n\nIntroduce a multi-graph (i.e., a graph with multiple edges allowed) $\\Gamma$ whose vertices correspond to coins; thus we have $4 n$ vertices of $n$ colors so that there are four vertices of each color. Connect pairs of vertices $\\{1, 4 n\\}, \\{2, 4 n - 1\\}, \\ldots, \\{2 n, 2 n + 1\\}$ by $2 n$ black edges.\n\nFurther, for each monochromatic quadruple of vertices $i, j, k, \\ell$ we add a pair of grey edges forming a matching, e.g., $(i, j)$ and $(k, \\ell)$. In each of $n$ colors of coins we can choose one of three possible matchings; this results in $3^n$ ways of constructing grey edges. Let us call each of $3^n$ possible graphs $\\Gamma$ a cyclic graph. Note that in a cyclic graph $\\Gamma$ each vertex has both black and grey degrees equal to $1$. Hence $\\Gamma$ is a union of disjoint cycles, and in each cycle black and grey edges alternate (in particular, all cycles have even lengths).\n\nIt suffices to find a cyclic graph with all its cycle lengths divisible by $4$. Indeed, in this case, for each cycle we start from some vertex, move along the cycle and recolor the black edges either to red or to blue, alternating red and blue colors. Now blue and red edges define the required partition, since for each monochromatic quadruple of vertices the grey edges provide a bijection between the endpoints of red and blue edges.\n\nAmong all possible cyclic graphs, let us choose graph $\\Gamma_0$ having the minimal number of components (i.e., cycles). The following claim completes the solution.\n\n**Claim.** In $\\Gamma_0$, all cycle lengths are divisible by $4$.\n\n**Proof.** Assuming the contrary, choose a cycle $C_1$ with an odd number of grey edges. For some color $c$ the cycle $C_1$ contains exactly one grey edge joining two vertices $i, j$ of color $c$, while the other edge joining two vertices $k, \\ell$ of color $c$ lies in another cycle $C_2$. Now delete edges $(i, j)$ and $(k, \\ell)$ and add edges $(i, k)$ and $(j, \\ell)$. By this switch we again obtain a cyclic graph $\\Gamma_0'$ and decrease the number of cycles by $1$. This contradicts the choice of $\\Gamma_0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32107,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider any rectangular table having finitely many rows and columns, with a real number $a(r, c)$ in the cell in row $r$ and column $c$. A pair ($R, C$), where $R$ is a set of rows and $C$ a set of columns, is called a saddle pair if the following two conditions are satisfied:\n(i) For each row $r'$, there is $r \\in R$ such that $a(r, c) \\geqslant a(r', c)$ for all $c \\in C$;\n(ii) For each column $c'$, there is $c \\in C$ such that $a(r, c) \\leqslant a(r, c')$ for all $r \\in R$.\nA saddle pair ($R, C$) is called a minimal pair if for each saddle pair ($R', C'$) with $R' \\subseteq R$ and $C' \\subseteq C$, we have $R' = R$ and $C' = C$.\nProve that any two minimal pairs contain the same number of rows.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. We say that a pair ($R', C'$) of nonempty sets is a subpair of a pair ($R, C$) if $R' \\subseteq R$ and $C' \\subseteq C$. The subpair is proper if at least one of the inclusions is strict.\nLet ($R_1, C_1$) and ($R_2, C_2$) be two saddle pairs with $|R_1| > |R_2|$. We will find a saddle subpair ($R', C'$) of ($R_1, C_1$) with $|R'| \\leqslant |R_2|$; clearly, this implies the desired statement.\n\nStep 1: We construct maps $\\rho: R_1 \\rightarrow R_1$ and $\\sigma: C_1 \\rightarrow C_1$ such that $|\\rho(R_1)| \\leqslant |R_2|$, and $a(\\rho(r_1), c_1) \\geqslant a(r_1, \\sigma(c_1))$ for all $r_1 \\in R_1$ and $c_1 \\in C_1$.\nSince $(R_1, C_1)$ is a saddle pair, for each $r_2 \\in R_2$ there is $r_1 \\in R_1$ such that $a(r_1, c_1) \\geqslant a(r_2, c_1)$ for all $c_1 \\in C_1$; denote one such $r_1$ by $\\rho_1(r_2)$. Similarly, we define four functions\n$$\n\\begin{align*}\n& \\rho_1: R_2 \\rightarrow R_1 \\quad \\text{such that} \\quad a(\\rho_1(r_2), c_1) \\geqslant a(r_2, c_1) \\quad \\text{for all} \\quad r_2 \\in R_2, \\quad c_1 \\in C_1 ; \\\\\n& \\rho_2: R_1 \\rightarrow R_2 \\quad \\text{such that} \\quad a(\\rho_2(r_1), c_2) \\geqslant a(r_1, c_2) \\quad \\text{for all} \\quad r_1 \\in R_1, \\quad c_2 \\in C_2 ; \\\\\n& \\sigma_1: C_2 \\rightarrow C_1 \\quad \\text{such that} \\quad a(r_1, \\sigma_1(c_2)) \\leqslant a(r_1, c_2) \\quad \\text{for all} \\quad r_1 \\in R_1, \\quad c_2 \\in C_2 ;  \\\\\n& \\sigma_2: C_1 \\rightarrow C_2 \\quad \\text{such that} \\quad a(r_2, \\sigma_2(c_1)) \\leqslant a(r_2, c_1) \\quad \\text{for all} \\quad r_2 \\in R_2, \\quad c_1 \\in C_1 .\n\\end{align*}\n$$\nSet now $\\rho = \\rho_1 \\circ \\rho_2: R_1 \\rightarrow R_1$ and $\\sigma = \\sigma_1 \\circ \\sigma_2: C_1 \\rightarrow C_1$. We have\n$$\n|\\rho(R_1)| = |\\rho_1(\\rho_2(R_1))| \\leqslant |\\rho_1(R_2)| \\leqslant |R_2| .\n$$\nMoreover, for all $r_1 \\in R_1$ and $c_1 \\in C_1$, we get\n$$\n\\begin{align*}\na(\\rho(r_1), c_1) = a(\\rho_1(\\rho_2(r_1)), c_1) \\geqslant a(\\rho_2(r_1), c_1) & \\geqslant a(\\rho_2(r_1), \\sigma_2(c_1)) \\\\\n& \\geqslant a(r_1, \\sigma_2(c_1)) \\geqslant a(r_1, \\sigma_1(\\sigma_2(c_1))) = a(r_1, \\sigma(c_1))\n\\end{align*}\n$$\nas desired.\n\nStep 2: Given maps $\\rho$ and $\\sigma$, we construct a proper saddle subpair $(R', C')$ of $(R_1, C_1)$.\nThe properties of $\\rho$ and $\\sigma$ yield that\n$$\na(\\rho^i(r_1), c_1) \\geqslant a(\\rho^{i-1}(r_1), \\sigma(c_1)) \\geqslant \\ldots \\geqslant a(r_1, \\sigma^i(c_1)),\n$$\nfor each positive integer $i$ and all $r_1 \\in R_1, c_1 \\in C_1$.\nConsider the images $R^i = \\rho^i(R_1)$ and $C^i = \\sigma^i(C_1)$. Clearly, $R_1 = R^0 \\supseteq R^1 \\supseteq R^2 \\supseteq \\ldots$ and $C_1 = C^0 \\supseteq C^1 \\supseteq C^2 \\supseteq \\ldots$. Since both chains consist of finite sets, there is an index $n$ such that $R^n = R^{n+1} = \\ldots$ and $C^n = C^{n+1} = \\ldots$. Then $\\rho^n(R^n) = R^{2n} = R^n$, so $\\rho^n$ restricted to $R^n$ is a bijection. Similarly, $\\sigma^n$ restricted to $C^n$ is a bijection from $C^n$ to itself. Therefore, there exists a positive integer $k$ such that $\\rho^{nk}$ acts identically on $R^n$, and $\\sigma^{nk}$ acts identically on $C^n$.\nWe claim now that ($R^n, C^n$) is a saddle subpair of ($R_1, C_1$), with $|R^n| \\leqslant |R^1| = |\\rho(R_1)| \\leqslant |R_2|$, which is what we needed. To check that this is a saddle pair, take any row $r'$, since ($R_1, C_1$) is a saddle pair, there exists $r_1 \\in R_1$ such that $a(r_1, c_1) \\geqslant a(r', c_1)$ for all $c_1 \\in C_1$. Set now $r_* = \\rho^{nk}(r_1) \\in R^n$. Then, for each $c \\in C^n$ we have $c = \\sigma^{nk}(c)$ and hence\n$$\na(r_*, c) = a(\\rho^{nk}(r_1), c) \\geqslant a(r_1, \\sigma^{nk}(c)) = a(r_1, c) \\geqslant a(r', c),\n$$\nwhich establishes condition (i). Condition (ii) is checked similarly.\n\n\nSolution 2. Denote by $\\mathcal{R}$ and $\\mathcal{C}$ the set of all rows and the set of all columns of the table, respectively. Let $\\mathcal{T}$ denote the given table; for a set $R$ of rows and a set $C$ of columns, let $\\mathcal{T}[R, C]$ denote the subtable obtained by intersecting rows from $R$ and columns from $C$.\nWe say that row $r_1$ exceeds row $r_2$ in range of columns $C$ (where $C \\subseteq \\mathcal{C}$) and write $r_1 \\succeq_C r_2$ or $r_2 \\leq_C r_1$, if $a(r_1, c) \\geqslant a(r_2, c)$ for all $c \\in C$. We say that a row $r_1$ is equal to a row $r_2$ in range of columns $C$ and write $r_1 \\equiv_C r_2$, if $a(r_1, c) = a(r_2, c)$ for all $c \\in C$. We introduce similar notions, and use the same notation, for columns. Then conditions (i) and (ii) in the definition of a saddle pair can be written as (i) for each $r' \\in \\mathcal{R}$ there exists $r \\in R$ such that $r \\geq_C r'$; and (ii) for each $c' \\in \\mathcal{C}$ there exists $c \\in C$ such that $c \\leq_R c'$.\n\nLemma. Suppose that $(R, C)$ is a minimal pair. Remove from the table several rows outside of $R$ and/or several columns outside of $C$. Then ($R, C$) remains a minimal pair in the new table.\n\nProof. Obviously, $(R, C)$ remains a saddle pair. Suppose $(R', C')$ is a proper subpair of $(R, C)$. Since ($R, C$) is a saddle pair, for each row $r^*$ of the initial table, there is a row $r \\in R$ such that $r \\geq_C r^*$. If ($R', C'$) became saddle after deleting rows not in $R$ and/or columns not in $C$, there would be a row $r' \\in R'$ satisfying $r' \\succeq_{C'} r$. Therefore, we would obtain that $r' \\succeq_{C'} r^*$, which is exactly condition (i) for the pair ($R', C'$) in the initial table; condition (ii) is checked similarly. Thus, $(R', C')$ was saddle in the initial table, which contradicts the hypothesis that $(R, C)$ was minimal. Hence, ($R, C$) remains minimal after deleting rows and/or columns.\n\nBy the Lemma, it suffices to prove the statement of the problem in the case $\\mathcal{R} = R_1 \\cup R_2$ and $\\mathcal{C} = C_1 \\cup C_2$. Further, suppose that there exist rows that belong both to $R_1$ and $R_2$. Duplicate every such row, and refer one copy of it to the set $R_1$, and the other copy to the set $R_2$. Then ($R_1, C_1$) and ($R_2, C_2$) will remain minimal pairs in the new table, with the same numbers of rows and columns, but the sets $R_1$ and $R_2$ will become disjoint. Similarly duplicating columns in $C_1 \\cap C_2$, we make $C_1$ and $C_2$ disjoint. Thus it is sufficient to prove the required statement in the case $R_1 \\cap R_2 = \\varnothing$ and $C_1 \\cap C_2 = \\varnothing$.\n\nThe rest of the solution is devoted to the proof of the following claim including the statement of the problem.\n\nClaim. Suppose that ($R_1, C_1$) and ($R_2, C_2$) are minimal pairs in table $\\mathcal{T}$ such that $R_2 = \\mathcal{R} \\setminus R_1$ and $C_2 = \\mathcal{C} \\setminus C_1$. Then $|R_1| = |R_2|$, $|C_1| = |C_2|$; moreover, there are four bijections\n$$\n\\begin{align*}\n& \\rho_1: R_2 \\rightarrow R_1 \\quad \\text{such that} \\quad \\rho_1(r_2) \\equiv_{C_1} r_2 \\quad \\text{for all} \\quad r_2 \\in R_2 ; \\\\\n& \\rho_2: R_1 \\rightarrow R_2 \\quad \\text{such that} \\quad \\rho_2(r_1) \\equiv_{C_2} r_1 \\quad \\text{for all} \\quad r_1 \\in R_1 ; \\\\\n& \\sigma_1: C_2 \\rightarrow C_1 \\quad \\text{such that} \\quad \\sigma_1(c_2) \\equiv_{R_1} c_2 \\quad \\text{for all} \\quad c_2 \\in C_2 ;  \\\\\n& \\sigma_2: C_1 \\rightarrow C_2 \\quad \\text{such that} \\quad \\sigma_2(c_1) \\equiv_{R_2} c_1 \\quad \\text{for all} \\quad c_1 \\in C_1 .\n\\end{align*}\n$$\n\nWe prove the Claim by induction on $|\\mathcal{R}| + |\\mathcal{C}|$. In the base case we have $|R_1| = |R_2| = |C_1| = |C_2| = 1$; let $R_i = \\{r_i\\}$ and $C_i = \\{c_i\\}$. Since ($R_1, C_1$) and ($R_2, C_2$) are saddle pairs, we have $a(r_1, c_1) \\geqslant a(r_2, c_1) \\geqslant a(r_2, c_2) \\geqslant a(r_1, c_2) \\geqslant a(r_1, c_1)$, hence, the table consists of four equal numbers, and the statement follows.\n\nTo prove the inductive step, introduce the maps $\\rho_1, \\rho_2, \\sigma_1$, and $\\sigma_2$ as in Solution 1, see above. Suppose first that all four maps are surjective. Then, in fact, we have $|R_1| = |R_2|$, $|C_1| = |C_2|$, and all maps are bijective. Moreover, for all $r_2 \\in R_2$ and $c_2 \\in C_2$ we have\n$$\n\\begin{align*}\na(r_2, c_2) \\leqslant a(r_2, \\sigma_2^{-1}(c_2)) \\leqslant a(\\rho_1(r_2), \\sigma_2^{-1}(c_2)) \\leqslant a(\\rho_1(r_2), \\sigma_1^{-1} \\circ \\sigma_2^{-1}(c_2)) \\\\\n\\leqslant a(\\rho_2 \\circ \\rho_1(r_2), \\sigma_1^{-1} \\circ \\sigma_2^{-1}(c_2))\n\\end{align*}\n$$\nSumming up, we get\n$$\n\\sum_{\\substack{r_2 \\in R_2 \\\\ c_2 \\in C_2}} a(r_2, c_2) \\leqslant \\sum_{\\substack{r_2 \\in R_2 \\\\ c_2 \\in C_2}} a(\\rho_2 \\circ \\rho_1(r_2), \\sigma_1^{-1} \\circ \\sigma_2^{-1}(c_2)) .\n$$\nSince $\\rho_1 \\circ \\rho_2$ and $\\sigma_1^{-1} \\circ \\sigma_2^{-1}$ are permutations of $R_2$ and $C_2$, respectively, this inequality is in fact equality. Therefore, all inequalities above turn into equalities, which establishes the inductive step in this case.\n\nIt remains to show that all four maps are surjective. For the sake of contradiction, we assume that $\\rho_1$ is not surjective. Now let $R_1' = \\rho_1(R_2)$ and $C_1' = \\sigma_1(C_2)$, and set $R^* = R_1 \\setminus R_1'$ and $C^* = C_1 \\setminus C_1'$. By our assumption, $R^* \\neq \\varnothing$.\nLet $\\mathcal{Q}$ be the table obtained from $\\mathcal{T}$ by removing the rows in $R^*$ and the columns in $C^*$; in other words, $\\mathcal{Q} = \\mathcal{T}[R_1' \\cup R_2, C_1' \\cup C_2]$. By the definition of $\\rho_1$, for each $r_2 \\in R_2$ we have $\\rho_1(r_2) \\geq_{C_1} r_2$, so a fortiori $\\rho_1(r_2) \\geq_{C_1'} r_2$; moreover, $\\rho_1(r_2) \\in R_1'$. Similarly, $C_1' \\ni \\sigma_1(c_2) \\leq_{R_1'} c_2$ for each $c_2 \\in C_2$. This means that ($R_1', C_1'$) is a saddle pair in $\\mathcal{Q}$. Recall that ($R_2, C_2$) remains a minimal pair in $\\mathcal{Q}$, due to the Lemma.\n\nTherefore, $\\mathcal{Q}$ admits a minimal pair $(\\bar{R}_1, \\bar{C}_1)$ such that $\\bar{R}_1 \\subseteq R_1'$ and $\\bar{C}_1 \\subseteq C_1'$. For a minute, confine ourselves to the subtable $\\overline{\\mathcal{Q}} = \\mathcal{Q}[\\bar{R}_1 \\cup R_2, \\bar{C}_1 \\cup C_2]$. By the Lemma, the pairs $(\\bar{R}_1, \\bar{C}_1)$ and ($R_2, C_2$) are also minimal in $\\overline{\\mathcal{Q}}$. By the inductive hypothesis, we have $|R_2| = |\\bar{R}_1| \\leqslant |R_1'| = |\\rho_1(R_2)| \\leqslant |R_2|$, so all these inequalities are in fact equalities. This implies that $\\bar{R}_2 = R_2'$ and that $\\rho_1$ is a bijection $R_2 \\rightarrow R_1'$. Similarly, $\\bar{C}_1 = C_1'$, and $\\sigma_1$ is a bijection $C_2 \\rightarrow C_1'$. In particular, ($R_1', C_1'$) is a minimal pair in $\\mathcal{Q}$.\n\nNow, by inductive hypothesis again, we have $|R_1'| = |R_2|$, $|C_1'| = |C_2|$, and there exist four bijections\n$$\n\\begin{array}{lllll}\n\\rho_1': R_2 \\rightarrow R_1' & \\text{such that} & \\rho_1'(r_2) \\equiv_{C_1'} r_2 & \\text{for all} & r_2 \\in R_2 ; \\\\\n\\rho_2': R_1' \\rightarrow R_2 & \\text{such that} & \\rho_2'(r_1) \\equiv_{C_2} r_1 & \\text{for all} & r_1 \\in R_1' ; \\\\\n\\sigma_1': C_2 \\rightarrow C_1' & \\text{such that} & \\sigma_1'(c_2) \\equiv_{R_1'} c_2 & \\text{for all} & c_2 \\in C_2 ; \\\\\n\\sigma_2': C_1' \\rightarrow C_2 & \\text{such that} & \\sigma_2'(c_1) \\equiv_{R_2} c_1 & \\text{for all} & c_1 \\in C_1' .\n\\end{array}\n$$\nNotice here that $\\sigma_1$ and $\\sigma_1'$ are two bijections $C_2 \\rightarrow C_1'$ satisfying $\\sigma_1'(c_2) \\equiv_{R_1'} c_2 \\geq_{R_1} \\sigma_1(c_2)$ for all $c_2 \\in C_2$. Now, if $\\sigma_1'(c_2) \\neq \\sigma_1(c_2)$ for some $c_2 \\in C_2$, then we could remove column $\\sigma_1'(c_2)$ from $C_1'$ obtaining another saddle pair $(R_1', C_1' \\setminus \\{\\sigma_1'(c_2)\\})$ in $\\mathcal{Q}$. This is impossible for a minimal pair ($R_1', C_1'$); hence the maps $\\sigma_1$ and $\\sigma_1'$ coincide.\n\nNow we are prepared to show that ($R_1', C_1'$) is a saddle pair in $\\mathcal{T}$, which yields a desired contradiction (since ($R_1, C_1$) is not minimal). By symmetry, it suffices to find, for each $r' \\in \\mathcal{R}$, a row $r_1 \\in R_1'$ such that $r_1 \\geq_{C_1'} r'$. If $r' \\in R_2$, then we may put $r_1 = \\rho_1(r')$; so, in the sequel we assume $r' \\in R_1$.\nThere exists $r_2 \\in R_2$ such that $r' \\leq_{C_2} r_2$; set $r_1 = (\\rho_2')^{-1}(r_2) \\in R_1'$ and recall that $r_1 \\equiv_{C_2} r_2 \\geq_{C_2} r'$. Therefore, implementing the bijection $\\sigma_1 = \\sigma_1'$, for each $c_1 \\in C_1'$ we get\n$$\na(r', c_1) \\leqslant a(r', \\sigma_1^{-1}(c_1)) \\leqslant a(r_1, \\sigma_1^{-1}(c_1)) = a(r_1, \\sigma_1' \\circ \\sigma_1^{-1}(c_1)) = a(r_1, c_1),\n$$\nwhich shows $r' \\leq_{C_1'} r_1$, as desired. The inductive step is completed.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32108,
    "subject": "Mathematics (Multi-modal)",
    "question": "For every integer $n \\geqslant 1$ consider the $n \\times n$ table with entry $\\left\\lfloor\\frac{ij}{n+1}\\right\\rfloor$ at the intersection of row $i$ and column $j$, for every $i=1, \\ldots, n$ and $j=1, \\ldots, n$. Determine all integers $n \\geqslant 1$ for which the sum of the $n^{2}$ entries in the table is equal to $\\frac{1}{4} n^{2}(n-1)$.",
    "options": [],
    "answer": "All integers n such that n+1 is prime.",
    "solution": "Answer: All integers $n$ for which $n+1$ is a prime.\n\nFirst, observe that every pair $x, y$ of real numbers for which the sum $x+y$ is integer satisfies\n$$\n\\lfloor x\\rfloor+\\lfloor y\\rfloor \\geqslant x+y-1 \\tag{1}\n$$\nThe inequality is strict if $x$ and $y$ are integers, and it holds with equality otherwise.\nWe estimate the sum $S$ as follows.\n$$\n\\begin{aligned}\n2 S=\\sum_{1 \\leqslant i, j \\leqslant n}\\left(\\left\\lfloor\\frac{ij}{n+1}\\right\\rfloor+\\left\\lfloor\\frac{ij}{n+1}\\right\\rfloor\\right)= & \\sum_{1 \\leqslant i, j \\leqslant n}\\left(\\left\\lfloor\\frac{ij}{n+1}\\right\\rfloor+\\left\\lfloor\\frac{(n+1-i)j}{n+1}\\right\\rfloor\\right) \\\\\n& \\geqslant \\sum_{1 \\leqslant i, j \\leqslant n}(j-1)=\\frac{(n-1)n^{2}}{2} .\n\\end{aligned}\n$$\nThe inequality in the last line follows from (1) by setting $x=ij/(n+1)$ and $y=(n+1-i)j/(n+1)$, so that $x+y=j$ is integral.\nNow $S=\\frac{1}{4} n^{2}(n-1)$ if and only if the inequality in the last line holds with equality, which means that none of the values $ij/(n+1)$ with $1 \\leqslant i, j \\leqslant n$ may be integral.\nHence, if $n+1$ is composite with factorisation $n+1=ab$ for $2 \\leqslant a, b \\leqslant n$, one gets a strict inequality for $i=a$ and $j=b$. If $n+1$ is a prime, then $ij/(n+1)$ is never integral and $S=\\frac{1}{4} n^{2}(n-1)$.\nTo simplify the calculation with indices, extend the table by adding a phantom column of index 0 with zero entries (which will not change the sum of the table). Fix a row $i$ with $1 \\leqslant i \\leqslant n$, and let $d:=\\operatorname{gcd}(i, n+1)$ and $k:=(n+1) / d$. For columns $j=0, \\ldots, n$, define the remainder $r_{j}:=ij \\bmod (n+1)$. We first prove the following\n\nClaim. For every integer $g$ with $1 \\leqslant g \\leqslant d$, the remainders $r_{j}$ with indices $j$ in the range\n$$\n(g-1)k \\leqslant j \\leqslant gk-1 \\tag{2}\n$$\nform a permutation of the $k$ numbers $0 \\cdot d, 1 \\cdot d, 2 \\cdot d, \\ldots, (k-1) \\cdot d$.\n\nProof. If $r_{j'}=r_{j}$ holds for two indices $j'$ and $j$ in (2), then $i(j'-j) \\equiv 0 \\bmod (n+1)$, so that $j'-j$ is a multiple of $k$; since $|j'-j| \\leqslant k-1$, this implies $j'=j$. Hence, the $k$ remainders are pairwise distinct. Moreover, each remainder $r_{j}=ij \\bmod (n+1)$ is a multiple of $d=\\operatorname{gcd}(i, n+1)$. This proves the claim.\n\nWe then have\n$$\n\\sum_{j=0}^{n} r_{j}=\\sum_{g=1}^{d} \\sum_{\\ell=0}^{(n+1)/d-1} \\ell d=d^{2} \\cdot \\frac{1}{2}\\left(\\frac{n+1}{d}-1\\right) \\frac{n+1}{d}=\\frac{(n+1-d)(n+1)}{2} . \\tag{3}\n$$\nBy using (3), compute the sum $S_{i}$ of row $i$ as follows:\n$$\n\\begin{align*}\nS_{i}=\\sum_{j=0}^{n}\\left\\lfloor\\frac{ij}{n+1}\\right\\rfloor & =\\sum_{j=0}^{n} \\frac{ij-r_{j}}{n+1}=\\frac{i}{n+1} \\sum_{j=0}^{n} j-\\frac{1}{n+1} \\sum_{j=0}^{n} r_{j} \\\\\n& =\\frac{i}{n+1} \\cdot \\frac{n(n+1)}{2}-\\frac{1}{n+1} \\cdot \\frac{(n+1-d)(n+1)}{2}=\\frac{(in-n-1+d)}{2} \\tag{4}\n\\end{align*}\n$$\nEquation (4) yields the following lower bound on the row sum $S_{i}$, which holds with equality if and only if $d=\\operatorname{gcd}(i, n+1)=1$ :\n$$\nS_{i} \\geqslant \\frac{(in-n-1+1)}{2}=\\frac{n(i-1)}{2} . \\tag{5}\n$$\nBy summing up the bounds (5) for the rows $i=1, \\ldots, n$, we get the following lower bound for the sum of all entries in the table\n$$\n\\sum_{i=1}^{n} S_{i} \\geqslant \\sum_{i=1}^{n} \\frac{n}{2}(i-1)=\\frac{n^{2}(n-1)}{4} . \\tag{6}\n$$\nIn (6) we have equality if and only if equality holds in (5) for each $i=1, \\ldots, n$, which happens if and only if $\\operatorname{gcd}(i, n+1)=1$ for each $i=1, \\ldots, n$, which is equivalent to the fact that $n+1$ is a prime. Thus the sum of the table entries is $\\frac{1}{4} n^{2}(n-1)$ if and only if $n+1$ is a prime.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32109,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a positive integer $n$, find the smallest value of\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor\n$$\nover all permutations $(a_{1}, a_{2}, \\ldots, a_{n})$ of $(1,2, \\ldots, n)$.",
    "options": [],
    "answer": "floor(log_2 n) + 1",
    "solution": "Answer: The minimum of such sums is $\\left\\lfloor\\log _{2} n\\right\\rfloor+1$; so if $2^{k} \\leqslant n<2^{k+1}$, the minimum is $k+1$.\n\nSolution 1. Suppose that $2^{k} \\leqslant n<2^{k+1}$ with some nonnegative integer $k$. First we show a permutation $(a_{1}, a_{2}, \\ldots, a_{n})$ such that\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor=k+1\n$$\nthen we will prove that\n$$\n\\left\\lfloor\\frac{a_{1}}{1}\\right\\rfloor+\\left\\lfloor\\frac{a_{2}}{2}\\right\\rfloor+\\cdots+\\left\\lfloor\\frac{a_{n}}{n}\\right\\rfloor \\geqslant k+1\n$$\nfor every permutation. Hence, the minimal possible value will be $k+1$.\n\nI. Consider the permutation\n$$\n\\begin{gathered}\n(a_{1})=(1), \\quad(a_{2}, a_{3})=(3,2), \\quad(a_{4}, a_{5}, a_{6}, a_{7})=(7,4,5,6), \\quad \\ldots \\\\\n(a_{2^{k-1}}, \\ldots, a_{2^{k}-1})=(2^{k}-1,2^{k-1}, 2^{k-1}+1, \\ldots, 2^{k}-2), \\\\\n(a_{2^{k}}, \\ldots, a_{n})=(n, 2^{k}, 2^{k}+1, \\ldots, n-1),\n\\end{gathered}\n$$\nThis permutation consists of $k+1$ cycles. In every cycle $(a_{p}, \\ldots, a_{q})=(q, p, p+1, \\ldots, q-1)$ we have $q<2p$, so\n$$\n\\sum_{i=p}^{q}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor=\\left\\lfloor\\frac{q}{p}\\right\\rfloor+\\sum_{i=p+1}^{q}\\left\\lfloor\\frac{i-1}{i}\\right\\rfloor=1\n$$\nThe total sum over all cycles is precisely $k+1$.\n\nII. In order to establish the lower bound, we prove a more general statement.\n\nClaim. If $b_{1}, \\ldots, b_{2^{k}}$ are distinct positive integers then\n$$\n\\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor \\geqslant k+1\n$$\nFrom the Claim it follows immediately that $\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant k+1$.\n\nProof of the Claim. Apply induction on $k$. For $k=1$ the claim is trivial, $\\left\\lfloor\\frac{b_{1}}{1}\\right\\rfloor \\geqslant 1$. Suppose the Claim holds true for some positive integer $k$, and consider $k+1$.\nIf there exists an index $j$ such that $2^{k}<j \\leqslant 2^{k+1}$ and $b_{j} \\geqslant j$ then\n$$\n\\sum_{i=1}^{2^{k+1}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor+\\left\\lfloor\\frac{b_{j}}{j}\\right\\rfloor \\geqslant(k+1)+1\n$$\nby the induction hypothesis, so the Claim is satisfied.\nOtherwise we have $b_{j}<j \\leqslant 2^{k+1}$ for every $2^{k}<j \\leqslant 2^{k+1}$. Among the $2^{k+1}$ distinct numbers $b_{1}, \\ldots, b_{2^{k+1}}$ there is some $b_{m}$ which is at least $2^{k+1}$; that number must be among $b_{1} \\ldots, b_{2^{k}}$. Hence, $1 \\leqslant m \\leqslant 2^{k}$ and $b_{m} \\geqslant 2^{k+1}$.\nWe will apply the induction hypothesis to the numbers\n$$\nc_{1}=b_{1}, \\ldots, c_{m-1}=b_{m-1}, \\quad c_{m}=b_{2^{k}+1}, \\quad c_{m+1}=b_{m+1}, \\ldots, c_{2^{k}}=b_{2^{k}}\n$$\nso take the first $2^{k}$ numbers but replace $b_{m}$ with $b_{2^{k}+1}$. Notice that\n$$\n\\left\\lfloor\\frac{b_{m}}{m}\\right\\rfloor \\geqslant\\left\\lfloor\\frac{2^{k+1}}{m}\\right\\rfloor=\\left\\lfloor\\frac{2^{k}+2^{k}}{m}\\right\\rfloor \\geqslant\\left\\lfloor\\frac{b_{2^{k}+1}+m}{m}\\right\\rfloor=\\left\\lfloor\\frac{c_{m}}{m}\\right\\rfloor+1\n$$\nFor the other indices $i$ with $1 \\leqslant i \\leqslant 2^{k}, i \\neq m$ we have $\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor=\\left\\lfloor\\frac{c_{i}}{i}\\right\\rfloor$, so\n$$\n\\sum_{i=1}^{2^{k+1}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor=\\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{b_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{2^{k}}\\left\\lfloor\\frac{c_{i}}{i}\\right\\rfloor+1 \\geqslant(k+1)+1\n$$\nThat proves the Claim and hence completes the solution.\nSolution 2. We present a different proof for the lower bound.\nAssume again $2^{k} \\leqslant n<2^{k+1}$, and let $P=\\{2^{0}, 2^{1}, \\ldots, 2^{k}\\}$ be the set of powers of 2 among $1,2, \\ldots, n$. Call an integer $i \\in\\{1,2, \\ldots, n\\}$ and the interval $[i, a_{i}]$ good if $a_{i} \\geqslant i$.\n\nLemma 1. The good intervals cover the integers $1,2, \\ldots, n$.\nProof. Consider an arbitrary $x \\in\\{1,2 \\ldots, n\\}$; we want to find a good interval $[i, a_{i}]$ that covers $x$; i.e., $i \\leqslant x \\leqslant a_{i}$. Take the cycle of the permutation that contains $x$, that is $(x, a_{x}, a_{a_{x}}, \\ldots)$. In this cycle, let $i$ be the first element with $a_{i} \\geqslant x$; then $i \\leqslant x \\leqslant a_{i}$.\n\nLemma 2. If a good interval $[i, a_{i}]$ covers $p$ distinct powers of 2 then $\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant p$; more formally, $\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant |[i, a_{i}] \\cap P|$.\nProof. The ratio of the smallest and largest powers of 2 in the interval is at least $2^{p-1}$. By Bernoulli's inequality, $\\frac{a_{i}}{i} \\geqslant 2^{p-1} \\geqslant p$; that proves the lemma.\n\nNow, by Lemma 1, the good intervals cover $P$. By applying Lemma 2 as well, we obtain that\n$$\n\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor=\\sum_{i \\text{ is good }}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i \\text{ is good }}^{n}|[i, a_{i}] \\cap P| \\geqslant |P|=k+1.\n$$\nSolution 3. We show yet another proof for the lower bound, based on the following inequality.\n\nLemma 3.\n$$\n\\left\\lfloor\\frac{a}{b}\\right\\rfloor \\geqslant \\log _{2} \\frac{a+1}{b}\n$$\nfor every pair $a, b$ of positive integers.\n\nProof. Let $t=\\left\\lfloor\\frac{a}{b}\\right\\rfloor$, so $t \\leqslant \\frac{a}{b}$ and $\\frac{a+1}{b} \\leqslant t+1$. By applying the inequality $2^{t} \\geqslant t+1$, we obtain\n$$\n\\left\\lfloor\\frac{a}{b}\\right\\rfloor=t \\geqslant \\log _{2}(t+1) \\geqslant \\log _{2} \\frac{a+1}{b}\n$$\nBy applying the lemma to each term, we get\n$$\n\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\sum_{i=1}^{n} \\log _{2} \\frac{a_{i}+1}{i}=\\sum_{i=1}^{n} \\log _{2}(a_{i}+1)-\\sum_{i=1}^{n} \\log _{2} i\n$$\nNotice that the numbers $a_{1}+1, a_{2}+1, \\ldots, a_{n}+1$ form a permutation of $2,3, \\ldots, n+1$. Hence, in the last two sums all terms cancel out, except for $\\log _{2}(n+1)$ in the first sum and $\\log _{2} 1=0$ in the second sum. Therefore,\n$$\n\\sum_{i=1}^{n}\\left\\lfloor\\frac{a_{i}}{i}\\right\\rfloor \\geqslant \\log _{2}(n+1)>k\n$$\nAs the left-hand side is an integer, it must be at least $k+1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32110,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 2$ be an integer, and let $a_{1}, a_{2}, \\ldots, a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\\cdots+a_{n}=1$. Prove that\n$$\n\\sum_{k=1}^{n} \\frac{a_{k}}{1-a_{k}}\\left(a_{1}+a_{2}+\\cdots+a_{k-1}\\right)^{2}<\\frac{1}{3} .\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For all $k \\leqslant n$, let\n$$\ns_{k}=a_{1}+a_{2}+\\cdots+a_{k} \\quad \\text{ and } \\quad b_{k}=\\frac{a_{k} s_{k-1}^{2}}{1-a_{k}},\n$$\nwith the convention that $s_{0}=0$. Note that $b_{k}$ is exactly a summand in the sum we need to estimate. We shall prove the inequality\n$$\n\\begin{equation*}\nb_{k}<\\frac{s_{k}^{3}-s_{k-1}^{3}}{3} \\tag{1}\n\\end{equation*}\n$$\nIndeed, it suffices to check that\n$$\n\\begin{aligned}\n(1) & \\Longleftrightarrow 0<\\left(1-a_{k}\\right)\\left(\\left(s_{k-1}+a_{k}\\right)^{3}-s_{k-1}^{3}\\right)-3 a_{k} s_{k-1}^{2} \\\\\n& \\Longleftrightarrow 0<\\left(1-a_{k}\\right)\\left(3 s_{k-1}^{2}+3 s_{k-1} a_{k}+a_{k}^{2}\\right)-3 s_{k-1}^{2} \\\\\n& \\Longleftrightarrow 0<-3 a_{k} s_{k-1}^{2}+3\\left(1-a_{k}\\right) s_{k-1} a_{k}+\\left(1-a_{k}\\right) a_{k}^{2} \\\\\n& \\Longleftrightarrow 0<3\\left(1-a_{k}-s_{k-1}\\right) s_{k-1} a_{k}+\\left(1-a_{k}\\right) a_{k}^{2}\n\\end{aligned}\n$$\nwhich holds since $a_{k}+s_{k-1}=s_{k} \\leqslant 1$ and $a_{k} \\in(0,1)$.\nThus, adding inequalities (1) for $k=1, \\ldots, n$, we conclude that\n$$\nb_{1}+b_{2}+\\cdots+b_{n}<\\frac{s_{n}^{3}-s_{0}^{3}}{3}=\\frac{1}{3}\n$$\nFirst, let us define\n$$\nS\\left(a_{1}, \\ldots, a_{n}\\right):=\\sum_{k=1}^{n} \\frac{a_{k}}{1-a_{k}}\\left(a_{1}+a_{2}+\\cdots+a_{k-1}\\right)^{2} .\n$$\nFor some index $i$, denote $a_{1}+\\cdots+a_{i-1}$ by $s$. If we replace $a_{i}$ with two numbers $a_{i} / 2$ and $a_{i} / 2$, i.e. replace the tuple ( $a_{1}, \\ldots, a_{n}$ ) with ( $a_{1}, \\ldots, a_{i-1}, a_{i} / 2, a_{i} / 2, a_{i+1}, \\ldots, a_{n}$ ), the sum will increase by\n$$\n\\begin{aligned}\nS\\left(a_{1}, \\ldots, a_{i} / 2, a_{i} / 2, \\ldots, a_{n}\\right)-S\\left(a_{1}, \\ldots, a_{n}\\right) & =\\frac{a_{i} / 2}{1-a_{i} / 2}\\left(s^{2}+\\left(s+a_{i} / 2\\right)^{2}\\right)-\\frac{a_{i}}{1-a_{i}} s^{2} \\\\\n& =a_{i} \\frac{\\left(1-a_{i}\\right)\\left(2 s^{2}+s a_{i}+a_{i}^{2} / 4\\right)-\\left(2-a_{i}\\right) s^{2}}{\\left(2-a_{i}\\right)\\left(1-a_{i}\\right)} \\\\\n& =a_{i} \\frac{\\left(1-a_{i}-s\\right) s a_{i}+\\left(1-a_{i}\\right) a_{i}^{2} / 4}{\\left(2-a_{i}\\right)\\left(1-a_{i}\\right)},\n\\end{aligned}\n$$\nwhich is strictly positive. So every such replacement strictly increases the sum. By repeating this process and making maximal number in the tuple tend to zero, we keep increasing the sum which will converge to\n$$\n\\int_{0}^{1} x^{2} d x=\\frac{1}{3}\n$$\nThis completes the proof.\nWe sketch a probabilistic version of the first solution. Let $x_{1}, x_{2}, x_{3}$ be drawn uniformly and independently at random from the segment [ 0,1 ]. Let $I_{1} \\cup I_{2} \\cup \\cdots \\cup I_{n}$ be a partition of $[0,1]$ into segments of length $a_{1}, a_{2}, \\ldots, a_{n}$ in this order. Let $J_{k}:=I_{1} \\cup \\cdots \\cup I_{k-1}$ for $k \\geqslant 2$ and $J_{1}:=\\varnothing$. Then\n$$\n\\begin{aligned}\n\\frac{1}{3}= & \\sum_{k=1}^{n} \\mathbb{P}\\left\\{x_{1} \\geqslant x_{2}, x_{3} ; x_{1} \\in I_{k}\\right\\} \\\\\n= & \\sum_{k=1}^{n}\\left(\\mathbb{P}\\left\\{x_{1} \\in I_{k} ; x_{2}, x_{3} \\in J_{k}\\right\\}+2 \\cdot \\mathbb{P}\\left\\{x_{1} \\geqslant x_{2} ; x_{1}, x_{2} \\in I_{k} ; x_{3} \\in J_{k}\\right\\}\\right. \\\\\n& \\left.+\\mathbb{P}\\left\\{x_{1} \\geqslant x_{2}, x_{3} ; x_{1}, x_{2}, x_{3} \\in I_{k}\\right\\}\\right) \\\\\n= & \\sum_{k=1}^{n}\\left(a_{k}\\left(a_{1}+\\cdots+a_{k-1}\\right)^{2}+2 \\cdot \\frac{a_{k}^{2}}{2} \\cdot\\left(a_{1}+\\cdots+a_{k-1}\\right)+\\frac{a_{k}^{3}}{3}\\right) \\\\\n> & \\sum_{k=1}^{n}\\left(a_{k}\\left(a_{1}+\\cdots+a_{k-1}\\right)^{2}+a_{k}^{2}\\left(a_{1}+\\cdots+a_{k-1}\\right) \\cdot \\frac{a_{1}+\\cdots+a_{k-1}}{1-a_{k}}\\right)\n\\end{aligned}\n$$\nwhere for the last inequality we used that $1-a_{k} \\geqslant a_{1}+\\cdots+a_{k-1}$. This completes the proof since\n$$\na_{k}+\\frac{a_{k}^{2}}{1-a_{k}}=\\frac{a_{k}}{1-a_{k}}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32111,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 1$ be an integer, and let $x_{0}, x_{1}, \\ldots, x_{n+1}$ be $n+2$ non-negative real numbers that satisfy $x_{i} x_{i+1}-x_{i-1}^{2} \\geqslant 1$ for all $i=1,2, \\ldots, n$. Show that\n$$\nx_{0}+x_{1}+\\cdots+x_{n}+x_{n+1}>\\left(\\frac{2 n}{3}\\right)^{3 / 2}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1.\nLemma 1.1. If $a, b, c$ are non-negative numbers such that $a b-c^{2} \\geqslant 1$, then\n$$\n(a+2 b)^{2} \\geqslant(b+2 c)^{2}+6\n$$\nProof. $(a+2 b)^{2}-(b+2 c)^{2}=(a-b)^{2}+2(b-c)^{2}+6\\left(a b-c^{2}\\right) \\geqslant 6$.\n\nLemma 1.2. $\\sqrt{1}+\\cdots+\\sqrt{n}>\\frac{2}{3} n^{3 / 2}$.\nProof. Bernoulli's inequality $(1+t)^{3 / 2}>1+\\frac{3}{2} t$ for $0>t \\geqslant-1$ (or, alternatively, a straightforward check) gives\n$$\n\\begin{equation*}\n(k-1)^{3 / 2}=k^{3 / 2}\\left(1-\\frac{1}{k}\\right)^{3 / 2}>k^{3 / 2}\\left(1-\\frac{3}{2 k}\\right)=k^{3 / 2}-\\frac{3}{2} \\sqrt{k} \\tag{*}\n\\end{equation*}\n$$\nSumming up ( $*$ ) over $k=1,2, \\ldots, n$ yields\n$$\n0>n^{3 / 2}-\\frac{3}{2}(\\sqrt{1}+\\cdots+\\sqrt{n}) .\n$$\nNow put $y_{i}:=2 x_{i}+x_{i+1}$ for $i=0,1, \\ldots, n$. We get $y_{0} \\geqslant 0$ and $y_{i}^{2} \\geqslant y_{i-1}^{2}+6$ for $i=1,2, \\ldots, n$ by Lemma 1.1. Thus, an easy induction on $i$ gives $y_{i} \\geqslant \\sqrt{6 i}$. Using this estimate and Lemma 1.2 we get\n$$\n3\\left(x_{0}+\\ldots+x_{n+1}\\right) \\geqslant y_{1}+\\ldots+y_{n} \\geqslant \\sqrt{6}(\\sqrt{1}+\\sqrt{2}+\\ldots+\\sqrt{n})>\\sqrt{6} \\cdot \\frac{2}{3} n^{3 / 2}=3\\left(\\frac{2 n}{3}\\right)^{3 / 2}\n$$\n\n\nSolution 2.\nSay that an index $i \\in\\{0,1, \\ldots, n+1\\}$ is $\\operatorname{good}$, if $x_{i} \\geqslant \\sqrt{\\frac{2}{3}} i$, otherwise call the index $i$ bad.\n\nLemma 2.1. There are no two consecutive bad indices.\nProof. Assume the contrary and consider two bad indices $j, j+1$ with minimal possible $j$. Since 0 is good, we get $j>0$, thus by minimality $j-1$ is a good index and we have\n$$\n\\frac{2}{3} \\sqrt{j(j+1)}>x_{j} x_{j+1} \\geqslant x_{j-1}^{2}+1 \\geqslant \\frac{2}{3}(j-1)+1=\\frac{2}{3} \\cdot \\frac{j+(j+1)}{2}\n$$\nthat contradicts the AM-GM inequality for numbers $j$ and $j+1$.\n\nLemma 2.2. If an index $j \\leqslant n-1$ is good, then\n$$\nx_{j+1}+x_{j+2} \\geqslant \\sqrt{\\frac{2}{3}}(\\sqrt{j+1}+\\sqrt{j+2})\n$$\nProof. We have\n$$\nx_{j+1}+x_{j+2} \\geqslant 2 \\sqrt{x_{j+1} x_{j+2}} \\geqslant 2 \\sqrt{x_{j}^{2}+1} \\geqslant 2 \\sqrt{\\frac{2}{3} j+1} \\geqslant \\sqrt{\\frac{2}{3} j+\\frac{2}{3}}+\\sqrt{\\frac{2}{3} j+\\frac{4}{3}},\n$$\nthe last inequality follows from concavity of the square root function, or, alternatively, from the AM-QM inequality for the numbers $\\sqrt{\\frac{2}{3} j+\\frac{2}{3}}$ and $\\sqrt{\\frac{2}{3} j+\\frac{4}{3}}$.\n\nLet $S_{i}=x_{1}+\\ldots+x_{i}$ and $T_{i}=\\sqrt{\\frac{2}{3}}(\\sqrt{1}+\\ldots+\\sqrt{i})$.\n\nLemma 2.3. If an index $i$ is good, then $S_{i} \\geqslant T_{i}$.\nProof. Induction on $i$. The base case $i=0$ is clear. Assume that the claim holds for good indices less than $i$ and prove it for a good index $i>0$.\nIf $i-1$ is good, then by the inductive hypothesis we get $S_{i}=S_{i-1}+x_{i} \\geqslant T_{i-1}+\\sqrt{\\frac{2}{3}} i=T_{i}$.\nIf $i-1$ is bad, then $i>1$, and $i-2$ is good by Lemma 2.1. Then using Lemma 2.2 and the inductive hypothesis we get\n$$\nS_{i}=S_{i-2}+x_{i-1}+x_{i} \\geqslant T_{i-2}+\\sqrt{\\frac{2}{3}}(\\sqrt{i-1}+\\sqrt{i})=T_{i}\n$$\nSince either $n$ or $n+1$ is good by Lemma 2.1, Lemma 2.3 yields in both cases $S_{n+1} \\geqslant T_{n}$, and it remains to apply Lemma 1.2 from Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32112,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all functions $f: \\mathbb{R} \\rightarrow \\mathbb{R}$ that satisfy\n$$\n(f(a)-f(b))(f(b)-f(c))(f(c)-f(a))=f\\left(a b^{2}+b c^{2}+c a^{2}\\right)-f\\left(a^{2} b+b^{2} c+c^{2} a\\right)\n$$\nfor all real numbers $a, b, c$.",
    "options": [],
    "answer": "f(x)=αx+β or f(x)=αx^3+β where α∈{−1,0,1} and β∈ℝ",
    "solution": "Answer: $f(x)=\\alpha x+\\beta$ or $f(x)=\\alpha x^{3}+\\beta$ where $\\alpha \\in\\{-1,0,1\\}$ and $\\beta \\in \\mathbb{R}$.\n\nIt is straightforward to check that above functions satisfy the equation. Now let $f(x)$ satisfy the equation, which we denote $E(a, b, c)$. Then clearly $f(x)+C$ also does; therefore, we may suppose without loss of generality that $f(0)=0$. We start with proving\n\nLemma. Either $f(x) \\equiv 0$ or $f$ is injective.\n\nProof. Denote by $\\Theta \\subseteq \\mathbb{R}^{2}$ the set of points $(a, b)$ for which $f(a)=f(b)$. Let $\\Theta^{*}=\\{(x, y) \\in \\Theta$ : $x \\neq y\\}$. The idea is that if $(a, b) \\in \\Theta$, then by $E(a, b, x)$ we get\n$$\nH_{a, b}(x):=\\left(a b^{2}+b x^{2}+x a^{2}, a^{2} b+b^{2} x+x^{2} a\\right) \\in \\Theta\n$$\nfor all real $x$. Reproducing this argument starting with $(a, b) \\in \\Theta^{*}$, we get more and more points in $\\Theta$. There are many ways to fill in the details, we give below only one of them.\n\nAssume that $(a, b) \\in \\Theta^{*}$. Note that\n$$\ng_{-}(x):=\\left(a b^{2}+b x^{2}+x a^{2}\\right)-\\left(a^{2} b+b^{2} x+x^{2} a\\right)=(a-b)(b-x)(x-a)\n$$\nand\n$$\ng_{+}(x):=\\left(a b^{2}+b x^{2}+x a^{2}\\right)+\\left(a^{2} b+b^{2} x+x^{2} a\\right)=\\left(x^{2}+a b\\right)(a+b)+x\\left(a^{2}+b^{2}\\right) .\n$$\nHence, there exists $x$ for which both $g_{-}(x) \\neq 0$ and $g_{+}(x) \\neq 0$. This gives a point $(\\alpha, \\beta)= H_{a, b}(x) \\in \\Theta^{*}$ for which $\\alpha \\neq-\\beta$. Now compare $E(\\alpha, 1,0)$ and $E(\\beta, 1,0)$. The left-hand side expressions coincide, on right-hand side we get $f(\\alpha)-f\\left(\\alpha^{2}\\right)=f(\\beta)-f\\left(\\beta^{2}\\right)$, respectively. Hence, $f\\left(\\alpha^{2}\\right)=f\\left(\\beta^{2}\\right)$ and we get a point $\\left(\\alpha_{1}, \\beta_{1}\\right):=\\left(\\alpha^{2}, \\beta^{2}\\right) \\in \\Theta^{*}$ with both coordinates $\\alpha_{1}, \\beta_{1}$ non-negative. Continuing squaring the coordinates, we get a point $(\\gamma, \\delta) \\in \\Theta^{*}$ for which $\\delta>5 \\gamma \\geqslant 0$. Our nearest goal is to get a point $(0, r) \\in \\Theta^{*}$. If $\\gamma=0$, this is already done. If $\\gamma>0$, denote by $x$ a real root of the quadratic equation $\\delta \\gamma^{2}+\\gamma x^{2}+x \\delta^{2}=0$, which exists since the discriminant $\\delta^{4}-4 \\delta \\gamma^{3}$ is positive. Also $x<0$ since this equation cannot have non-negative root. For the point $H_{\\delta, \\gamma}(x)=:(0, r) \\in \\Theta$ the first coordinate is 0 . The difference of coordinates equals $-r=(\\delta-\\gamma)(\\gamma-x)(x-\\delta)<0$, so $r \\neq 0$ as desired.\n\nNow, let $(0, r) \\in \\Theta^{*}$. We get $H_{0, r}(x)=\\left(r x^{2}, r^{2} x\\right) \\in \\Theta$. Thus $f\\left(r x^{2}\\right)=f\\left(r^{2} x\\right)$ for all $x \\in \\mathbb{R}$. Replacing $x$ to $-x$ we get $f\\left(r x^{2}\\right)=f\\left(r^{2} x\\right)=f\\left(-r^{2} x\\right)$, so $f$ is even: $(a,-a) \\in \\Theta$ for all $a$. Then $H_{a,-a}(x)=\\left(a^{3}-a x^{2}+x a^{2},-a^{3}+a^{2} x+x^{2} a\\right) \\in \\Theta$ for all real $a, x$. Putting $x=\\frac{1+\\sqrt{5}}{2} a$ we obtain $\\left(0,(1+\\sqrt{5}) a^{3}\\right) \\in \\Theta$ which means that $f(y)=f(0)=0$ for every real $y$.\n\nHereafter we assume that $f$ is injective and $f(0)=0$. By $E(a, b, 0)$ we get\n$$\n\\begin{equation*}\nf(a) f(b)(f(a)-f(b))=f\\left(a^{2} b\\right)-f\\left(a b^{2}\\right) \\tag{$\\triangleright$}\n\\end{equation*}\n$$\nLet $\\kappa:=f(1)$ and note that $\\kappa=f(1) \\neq f(0)=0$ by injectivity. Putting $b=1$ in ( $\\triangleright$ ) we get\n$$\n\\begin{equation*}\n\\kappa f(a)(f(a)-\\kappa)=f\\left(a^{2}\\right)-f(a) . \\tag{\\%}\n\\end{equation*}\n$$\nSubtracting the same equality for $-a$ we get\n$$\n\\kappa(f(a)-f(-a))(f(a)+f(-a)-\\kappa)=f(-a)-f(a) .\n$$\nNow, if $a \\neq 0$, by injectivity we get $f(a)-f(-a) \\neq 0$ and thus\n$$\n\\begin{equation*}\nf(a)+f(-a)=\\kappa-\\kappa^{-1}=: \\lambda \\tag{A}\n\\end{equation*}\n$$\nIt follows that\n$$\nf(a)-f(b)=f(-b)-f(-a)\n$$\nfor all non-zero $a, b$. Replace non-zero numbers $a, b$ in ( $\\circlearrowleft$ ) with $-a,-b$, respectively, and add the two equalities. Due to ( $\\hat{\\text { ) }}$ we get\n$$\n(f(a)-f(b))(f(a) f(b)-f(-a) f(-b))=0\n$$\nthus $f(a) f(b)=f(-a) f(-b)=(\\lambda-f(a))(\\lambda-f(b))$ for all non-zero $a \\neq b$. If $\\lambda \\neq 0$, this implies $f(a)+f(b)=\\lambda$ that contradicts injectivity when we vary $b$ with fixed $a$. Therefore, $\\lambda=0$ and $\\kappa= \\pm 1$. Thus $f$ is odd. Replacing $f$ with $-f$ if necessary (this preserves the original equation) we may suppose that $f(1)=1$.\n\nNow, (\\%) yields $f\\left(a^{2}\\right)=f^{2}(a)$. Summing relations ( $\\varnothing$ ) for pairs ( $a, b$ ) and ( $a,-b$ ), we get $-2 f(a) f^{2}(b)=-2 f\\left(a b^{2}\\right)$, i.e. $f(a) f\\left(b^{2}\\right)=f\\left(a b^{2}\\right)$. Putting $b=\\sqrt{x}$ for each non-negative $x$ we get $f(a x)=f(a) f(x)$ for all real $a$ and non-negative $x$. Since $f$ is odd, this multiplicativity relation is true for all $a, x$. Also, from $f\\left(a^{2}\\right)=f^{2}(a)$ we see that $f(x) \\geqslant 0$ for $x \\geqslant 0$. Next, $f(x)>0$ for $x>0$ by injectivity.\n\nAssume that $f(x)$ for $x>0$ does not have the form $f(x)=x^{\\tau}$ for a constant $\\tau$. The known property of multiplicative functions yields that the graph of $f$ is dense on $(0, \\infty)^{2}$. In particular, we may find positive $b<1 / 10$ for which $f(b)>1$. Also, such $b$ can be found if $f(x)=x^{\\tau}$ for some $\\tau<0$. Then for all $x$ we have $x^{2}+x b^{2}+b \\geqslant 0$ and so $E(1, b, x)$ implies that\n$$\nf\\left(b^{2}+b x^{2}+x\\right)=f\\left(x^{2}+x b^{2}+b\\right)+(f(b)-1)(f(x)-f(b))(f(x)-1) \\geqslant 0-\\left((f(b)-1)^{3} / 4\\right.\n$$\nis bounded from below (the quadratic trinomial bound $(t-f(1))(t-f(b)) \\geqslant-(f(b)-1)^{2} / 4$ for $t=f(x)$ is used). Hence, $f$ is bounded from below on ( $b^{2}-\\frac{1}{4 b},+\\infty$ ), and since $f$ is odd it is bounded from above on $\\left(0, \\frac{1}{4 b}-b^{2}\\right)$. This is absurd if $f(x)=x^{\\tau}$ for $\\tau<0$, and contradicts to the above dense graph condition otherwise.\n\nTherefore, $f(x)=x^{\\tau}$ for $x>0$ and some constant $\\tau>0$. Dividing $E(a, b, c)$ by $(a-b)(b- c)(c-a)=\\left(a b^{2}+b c^{2}+c a^{2}\\right)-\\left(a^{2} b+b^{2} c+c^{2} a\\right)$ and taking a limit when $a, b, c$ all go to 1 (the divided ratios tend to the corresponding derivatives, say, $\\frac{a^{\\tau}-b^{\\tau}}{a-b} \\rightarrow\\left(x^{\\tau}\\right)_{x=1}^{\\prime}=\\tau$ ), we get $\\tau^{3}=\\tau \\cdot 3^{\\tau-1}, \\tau^{2}=3^{\\tau-1}, F(\\tau):=3^{\\tau / 2-1 / 2}-\\tau=0$. Since function $F$ is strictly convex, it has at most two roots, and we get $\\tau \\in\\{1,3\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32113,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be an infinite set of positive integers, such that there exist four pairwise distinct $a, b, c, d \\in S$ with $\\operatorname{gcd}(a, b) \\neq \\operatorname{gcd}(c, d)$. Prove that there exist three pairwise distinct $x, y, z \\in S$ such that $\\operatorname{gcd}(x, y)=\\operatorname{gcd}(y, z) \\neq \\operatorname{gcd}(z, x)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "There exists $\\alpha \\in S$ so that $\\{\\operatorname{gcd}(\\alpha, s) \\mid s \\in S, s \\neq \\alpha\\}$ contains at least two elements. Since $\\alpha$ has only finitely many divisors, there is a $d \\mid \\alpha$ such that the set $B=\\{\\beta \\in S \\mid \\operatorname{gcd}(\\alpha, \\beta)=d\\}$ is infinite. Pick $\\gamma \\in S$ so that $\\operatorname{gcd}(\\alpha, \\gamma) \\neq d$. Pick $\\beta_{1}, \\beta_{2} \\in B$ so that $\\operatorname{gcd}\\left(\\beta_{1}, \\gamma\\right)=\\operatorname{gcd}\\left(\\beta_{2}, \\gamma\\right)=: d'$. If $d=d'$, then $\\operatorname{gcd}\\left(\\alpha, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\gamma, \\beta_{1}\\right) \\neq \\operatorname{gcd}(\\alpha, \\gamma)$. If $d \\neq d'$, then either $\\operatorname{gcd}\\left(\\alpha, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\alpha, \\beta_{2}\\right)=d$ and $\\operatorname{gcd}\\left(\\beta_{1}, \\beta_{2}\\right) \\neq d$ or $\\operatorname{gcd}\\left(\\gamma, \\beta_{1}\\right)=\\operatorname{gcd}\\left(\\gamma, \\beta_{2}\\right)=d'$ and $\\operatorname{gcd}\\left(\\beta_{1}, \\beta_{2}\\right) \\neq d'$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32114,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 3$ be an integer. An integer $m \\geqslant n+1$ is called $n$-colourful if, given infinitely many marbles in each of $n$ colours $C_{1}, C_{2}, \\ldots, C_{n}$, it is possible to place $m$ of them around a circle so that in any group of $n+1$ consecutive marbles there is at least one marble of colour $C_{i}$ for each $i=1, \\ldots, n$.\nProve that there are only finitely many positive integers which are not $n$-colourful. Find the largest among them.",
    "options": [],
    "answer": "n^2 - n - 1",
    "solution": "Answer: $m_{\\text{max}} = n^{2} - n - 1$.\n\nFirst suppose that there are $n(n-1)-1$ marbles. Then for one of the colours, say blue, there are at most $n-2$ marbles, which partition the non-blue marbles into at most $n-2$ groups with at least $(n-1)^{2} > n(n-2)$ marbles in total. Thus one of these groups contains at least $n+1$ marbles and this group does not contain any blue marble.\n\nNow suppose that the total number of marbles is at least $n(n-1)$. Then we may write this total number as $n k + j$ with some $k \\geqslant n-1$ and with $0 \\leqslant j \\leqslant n-1$. We place around a circle $k-j$ copies of the colour sequence $[1,2,3, \\ldots, n]$ followed by $j$ copies of the colour sequence $[1,1,2,3, \\ldots, n]$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32115,
    "subject": "Mathematics (Multi-modal)",
    "question": "A thimblerigger has 2021 thimbles numbered from 1 through 2021. The thimbles are arranged in a circle in arbitrary order. The thimblerigger performs a sequence of 2021 moves; in the $k^{\\text{th}}$ move, he swaps the positions of the two thimbles adjacent to thimble $k$.\nProve that there exists a value of $k$ such that, in the $k^{\\text{th}}$ move, the thimblerigger swaps some thimbles $a$ and $b$ such that $a < k < b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume the contrary. Say that the $k^{\\text{th}}$ thimble is the central thimble of the $k^{\\text{th}}$ move, and its position on that move is the central position of the move.\n\nStep 1: Black and white colouring.\nBefore the moves start, let us paint all thimbles in white. Then, after each move, we repaint its central thimble in black. This way, at the end of the process all thimbles have become black.\nBy our assumption, in every move $k$, the two swapped thimbles have the same colour (as their numbers are either both larger or both smaller than $k$). At every moment, assign the colours of the thimbles to their current positions; then the only position which changes its colour in a move is its central position. In particular, each position is central for exactly one move (when it is being repainted to black).\n\nStep 2: Red and green colouring.\nNow we introduce a colouring of the positions. If in the $k^{\\text{th}}$ move, the numbers of the two swapped thimbles are both less than $k$, then we paint the central position of the move in red; otherwise we paint that position in green. This way, each position has been painted in red or green exactly once. We claim that among any two adjacent positions, one becomes green and the other one becomes red; this will provide the desired contradiction since 2021 is odd.\n\nConsider two adjacent positions $A$ and $B$, which are central in the $a^{\\text{th}}$ and in the $b^{\\text{th}}$ moves, respectively, with $a < b$. Then in the $a^{\\text{th}}$ move the thimble at position $B$ is white, and therefore has a number greater than $a$. After the $a^{\\text{th}}$ move, position $A$ is green and the thimble at position $A$ is black. By the arguments from Step 1, position $A$ contains only black thimbles after the $a^{\\text{th}}$ step. Therefore, on the $b^{\\text{th}}$ move, position $A$ contains a black thimble whose number is therefore less than $b$, while thimble $b$ is at position $B$. So position $B$ becomes red, and hence $A$ and $B$ have different colours.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32116,
    "subject": "Mathematics (Multi-modal)",
    "question": "The kingdom of Anisotropy consists of $n$ cities. For every two cities there exists exactly one direct one-way road between them. We say that a path from $X$ to $Y$ is a sequence of roads such that one can move from $X$ to $Y$ along this sequence without returning to an already visited city. A collection of paths is called diverse if no road belongs to two or more paths in the collection.\nLet $A$ and $B$ be two distinct cities in Anisotropy. Let $N_{A B}$ denote the maximal number of paths in a diverse collection of paths from $A$ to $B$. Similarly, let $N_{B A}$ denote the maximal number of paths in a diverse collection of paths from $B$ to $A$. Prove that the equality $N_{A B}=N_{B A}$ holds if and only if the number of roads going out from $A$ is the same as the number of roads going out from $B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We write $X \\rightarrow Y$ (or $Y \\leftarrow X$) if the road between $X$ and $Y$ goes from $X$ to $Y$. Notice that, if there is any route moving from $X$ to $Y$ (possibly passing through some cities more than once), then there is a path from $X$ to $Y$ consisting of some roads in the route. Indeed, any cycle in the route may be removed harmlessly; after some removals one obtains a path.\n\nSay that a path is short if it consists of one or two roads.\n\nPartition all cities different from $A$ and $B$ into four groups, $\\mathcal{I}$, $\\mathcal{O}$, $\\mathcal{A}$, and $\\mathcal{B}$ according to the following rules: for each city $C$,\n$$\n\\begin{array}{ll}\nC \\in \\mathcal{I} \\Longleftrightarrow A \\rightarrow C \\leftarrow B ; & C \\in \\mathcal{O} \\Longleftrightarrow A \\leftarrow C \\rightarrow B ; \\\\\nC \\in \\mathcal{A} \\Longleftrightarrow A \\rightarrow C \\rightarrow B ; & C \\in \\mathcal{B} \\Longleftrightarrow A \\leftarrow C \\leftarrow B .\n\\end{array}\n$$\n\nLemma. Let $\\mathcal{P}$ be a diverse collection consisting of $p$ paths from $A$ to $B$. Then there exists a diverse collection consisting of at least $p$ paths from $A$ to $B$ and containing all short paths from $A$ to $B$.\n\nProof. In order to obtain the desired collection, modify $\\mathcal{P}$ as follows.\nIf there is a direct road $A \\rightarrow B$ and the path consisting of this single road is not in $\\mathcal{P}$, merely add it to $\\mathcal{P}$.\nNow consider any city $C \\in \\mathcal{A}$ such that the path $A \\rightarrow C \\rightarrow B$ is not in $\\mathcal{P}$. If $\\mathcal{P}$ contains at most one path containing a road $A \\rightarrow C$ or $C \\rightarrow B$, remove that path (if it exists), and add the path $A \\rightarrow C \\rightarrow B$ to $\\mathcal{P}$ instead. Otherwise, $\\mathcal{P}$ contains two paths of the forms $A \\rightarrow C \\rightarrow B$ and $A \\rightarrow C \\rightarrow B$, where $C \\rightarrow B$ and $A \\rightarrow C$ are some paths. In this case, we recombine the edges to form two new paths $A \\rightarrow C \\rightarrow B$ and $A \\rightarrow C \\rightarrow B$ (removing cycles from the latter if needed). Now we replace the old two paths in $\\mathcal{P}$ with the two new ones.\nAfter any operation described above, the number of paths in the collection does not decrease, and the collection remains diverse. Applying such operation to each $C \\in \\mathcal{A}$, we obtain the desired collection. $\\square$\n\nBack to the problem, assume, without loss of generality, that there is a road $A \\rightarrow B$, and let $a$ and $b$ denote the numbers of roads going out from $A$ and $B$, respectively. Choose a diverse collection $\\mathcal{P}$ consisting of $N_{A B}$ paths from $A$ to $B$. We will transform it into a diverse collection $\\mathcal{Q}$ consisting of at least $N_{A B}+(b-a)$ paths from $B$ to $A$. This construction yields\n$$\nN_{B A} \\geqslant N_{A B}+(b-a) ; \\quad \\text{similarly, we get} \\quad N_{A B} \\geqslant N_{B A}+(a-b),\n$$\nwhence $N_{B A}-N_{A B}=b-a$. This yields the desired equivalence.\n\nApply the lemma to get a diverse collection $\\mathcal{P}^{\\prime}$ of at least $N_{A B}$ paths containing all $|\\mathcal{A}|+1$ short paths from $A$ to $B$. Notice that the paths in $\\mathcal{P}^{\\prime}$ contain no edge of a short path from $B$ to $A$. Each non-short path in $\\mathcal{P}^{\\prime}$ has the form $A \\rightarrow C \\rightarrow D \\rightarrow B$, where $C \\rightarrow D$ is a path from some city $C \\in \\mathcal{I}$ to some city $D \\in \\mathcal{O}$. For each such path, put into $\\mathcal{Q}$ the path $B \\rightarrow C \\rightarrow D \\rightarrow A$; also put into $\\mathcal{Q}$ all short paths from $B$ to $A$. Clearly, the collection $\\mathcal{Q}$ is diverse.\n\nNow, all roads going out from $A$ end in the cities from $\\mathcal{I} \\cup \\mathcal{A} \\cup \\{B\\}$, while all roads going out from $B$ end in the cities from $\\mathcal{I} \\cup \\mathcal{B}$. Therefore,\n$$\na=|\\mathcal{I}|+|\\mathcal{A}|+1, \\quad b=|\\mathcal{I}|+|\\mathcal{B}|, \\quad \\text{and hence} \\quad a-b=|\\mathcal{A}|-|\\mathcal{B}|+1 .\n$$\nOn the other hand, since there are $|\\mathcal{A}|+1$ short paths from $A$ to $B$ (including $A \\rightarrow B$) and $|\\mathcal{B}|$ short paths from $B$ to $A$, we infer\n$$\n|\\mathcal{Q}|=\\left|\\mathcal{P}^{\\prime}\\right|-(|\\mathcal{A}|+1)+|\\mathcal{B}| \\geqslant N_{A B}+(b-a),\n$$\nas desired.\nWe recall some graph-theoretical notions. Let $G$ be a finite graph, and let $V$ be the set of its vertices; fix two distinct vertices $s, t \\in V$. An $(s, t)$-cut is a partition of $V$ into two parts $V=S \\sqcup T$ such that $s \\in S$ and $t \\in T$. The cut-edges in the cut $(S, T)$ are the edges going from $S$ to $T$, and the size $e(S, T)$ of the cut is the number of cut-edges.\n\nWe will make use of the following theorem (which is a partial case of the Ford-Fulkerson \"min-cut max-flow\" theorem).\n\nTheorem (Menger). Let $G$ be a directed graph, and let $s$ and $t$ be its distinct vertices. Then the maximal number of edge-disjoint paths from $s$ to $t$ is equal to the minimal size of an $(s, t)$-cut.\n\nBack to the problem. Consider a directed graph $G$ whose vertices are the cities, and edges correspond to the roads. Then $N_{A B}$ is the maximal number of edge-disjoint paths from $A$ to $B$ in this graph; the number $N_{B A}$ is interpreted similarly.\n\nAs in the previous solution, denote by $a$ and $b$ the out-degrees of vertices $A$ and $B$, respectively. To solve the problem, we show that for any $(A, B)$-cut $(S_{A}, T_{A})$ in our graph there exists a $(B, A)$-cut $(S_{B}, T_{B})$ satisfying\n$$\ne\\left(S_{B}, T_{B}\\right)=e\\left(S_{A}, T_{A}\\right)+(b-a)\n$$\nThis yields\n$$\nN_{B A} \\leqslant N_{A B}+(b-a) ; \\quad \\text{similarly, we get} \\quad N_{A B} \\leqslant N_{B A}+(a-b),\n$$\nwhence again $N_{B A}-N_{A B}=b-a$.\n\nThe construction is simple: we put $S_{B}=S_{A} \\cup \\{B\\} \\backslash \\{A\\}$ and hence $T_{B}=T_{A} \\cup \\{A\\} \\backslash \\{B\\}$. To show that it works, let A and B denote the sets of cut-edges in $(S_{A}, T_{A})$ and $(S_{B}, T_{B})$, respectively. Let $a_{s}$ and $a_{t}=a-a_{s}$ denote the numbers of edges going from $A$ to $S_{A}$ and $T_{A}$, respectively. Similarly, denote by $b_{s}$ and $b_{t}=b-b_{s}$ the numbers of edges going from $B$ to $S_{B}$ and $T_{B}$, respectively.\n\nNotice that any edge incident to none of $A$ and $B$ either belongs to both A and B, or belongs to none of them. Denote the number of such edges belonging to A by $c$. The edges in A which are not yet accounted for split into two categories: those going out from $A$ to $T_{A}$ (including $A \\rightarrow B$ if it exists), and those going from $S_{A} \\backslash \\{A\\}$ to $B$ - in other words, going from $S_{B}$ to $B$. The numbers of edges in the two categories are $a_{t}$ and $|S_{B}|-1-b_{s}$, respectively. Therefore,\n$$\n|\\mathrm{A}|=c+a_{t}+\\left(|S_{B}|-b_{s}-1\\right) . \\quad \\text{Similarly, we get} \\quad|\\mathrm{B}|=c+b_{t}+\\left(|S_{A}|-a_{s}-1\\right),\n$$\nand hence\n$$\n|\\mathrm{B}|-|\\mathrm{A}|=\\left(b_{t}+b_{s}\\right)-\\left(a_{t}+a_{s}\\right)=b-a,\n$$\nsince $|S_{A}|=|S_{B}|$. This finishes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32117,
    "subject": "Mathematics (Multi-modal)",
    "question": "A hunter and an invisible rabbit play a game on an infinite square grid. First the hunter fixes a colouring of the cells with finitely many colours. The rabbit then secretly chooses a cell to start in. Every minute, the rabbit reports the colour of its current cell to the hunter, and then secretly moves to an adjacent cell that it has not visited before (two cells are adjacent if they share a side). The hunter wins if after some finite time either\n- the rabbit cannot move; or\n- the hunter can determine the cell in which the rabbit started.\nDecide whether there exists a winning strategy for the hunter.",
    "options": [],
    "answer": "Yes, there exists a colouring that yields a winning strategy for the hunter.",
    "solution": "Answer: Yes, there exists a colouring that yields a winning strategy for the hunter.\n\nA central idea is that several colourings $C_{1}, C_{2}, \\ldots, C_{k}$ can be merged together into a single product colouring $C_{1} \\times C_{2} \\times \\cdots \\times C_{k}$ as follows: the colours in the product colouring are ordered tuples $(c_{1}, \\ldots, c_{n})$ of colours, where $c_{i}$ is a colour used in $C_{i}$, so that each cell gets a tuple consisting of its colours in the individual colourings $C_{i}$. This way, any information which can be determined from one of the individual colourings can also be determined from the product colouring.\n\nNow let the hunter merge the following colourings:\n- The first two colourings $C_{1}$ and $C_{2}$ allow the tracking of the horizontal and vertical movements of the rabbit.\nThe colouring $C_{1}$ colours the cells according to the residue of their $x$-coordinates modulo $3$, which allows to determine whether the rabbit moves left, moves right, or moves vertically. Similarly, the colouring $C_{2}$ uses the residues of the $y$-coordinates modulo $3$, which allows to determine whether the rabbit moves up, moves down, or moves horizontally.\n- Under the condition that the rabbit's $x$-coordinate is unbounded, colouring $C_{3}$ allows to determine the exact value of the $x$-coordinate:\nIn $C_{3}$, the columns are coloured white and black so that the gaps between neighboring black columns are pairwise distinct. As the rabbit's $x$-coordinate is unbounded, it will eventually visit two black cells in distinct columns. With the help of colouring $C_{1}$ the hunter can catch that moment, and determine the difference of $x$-coordinates of those two black cells, hence deducing the precise column.\nSymmetrically, under the condition that the rabbit's $y$-coordinate is unbounded, there is a colouring $C_{4}$ that allows the hunter to determine the exact value of the $y$-coordinate.\n- Finally, under the condition that the sum $x+y$ of the rabbit's coordinates is unbounded, colouring $C_{5}$ allows to determine the exact value of this sum: The diagonal lines $x+y=\\text{const}$ are coloured black and white, so that the gaps between neighboring black diagonals are pairwise distinct.\nUnless the rabbit gets stuck, at least two of the three values $x, y$ and $x+y$ must be unbounded as the rabbit keeps moving. Hence the hunter can eventually determine two of these three values; thus he does know all three. Finally the hunter works backwards with help of the colourings $C_{1}$ and $C_{2}$ and computes the starting cell of the rabbit.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32118,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a checkered $3m \\times 3m$ square, where $m$ is an integer greater than $1$. A frog sits on the lower left corner cell $S$ and wants to get to the upper right corner cell $F$. The frog can hop from any cell to either the next cell to the right or the next cell upwards.\n\nSome cells can be sticky, and the frog gets trapped once it hops on such a cell. A set $X$ of cells is called blocking if the frog cannot reach $F$ from $S$ when all the cells of $X$ are sticky. A blocking set is minimal if it does not contain a smaller blocking set.\n\na. Prove that there exists a minimal blocking set containing at least $3m^{2} - 3m$ cells.\n\nb. Prove that every minimal blocking set contains at most $3m^{2}$ cells.\n\nNote. An example of a minimal blocking set for $m=2$ is shown below. Cells of the set $X$ are marked by letters $x$.\n\n|   |   |   |   |   | $F$ |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| $x$ | $x$ |   |   |   |   |\n|   |   | $x$ |   |   |   |\n|   |   |   | $x$ |   |   |\n|   |   |   |   | $x$ |   |\n| $S$ |   | $x$ |   |   |   |",
    "options": [],
    "answer": "Detailed solution",
    "solution": "a.\nIn the following example the square is divided into $m$ stripes of size $3 \\times 3m$. It is easy to see that $X$ is a minimal blocking set. The first and the last stripe each contains $3m-1$ cells from the set $X$; every other stripe contains $3m-2$ cells, see Figure 1. The total number of cells in the set $X$ is $3m^{2} - 2m + 2$.\n\nb.\nSolution 1.\nFor a given blocking set $X$, say that a non-sticky cell is red if the frog can reach it from $S$ via some hops without entering set $X$. We call a non-sticky cell blue if the frog can reach $F$ from that cell via hops without entering set $X$. One can regard the blue cells as those reachable from $F$ by anti-hops, i.e. moves downwards and to the left. We also colour all cells in $X$ green. It follows from the definition of the blocking set that no cell will be coloured twice. In Figure 2 we show a sample of a blocking set and the corresponding colouring.\n\nNow assume that $X$ is a minimal blocking set. We denote by $R$ (resp., $B$ and $G$) the total number of red (resp., blue and green) cells.\n\nWe claim that $G \\leqslant R+1$ and $G \\leqslant B+1$. Indeed, there are at most $2R$ possible frog hops from red cells. Every green or red cell (except for $S$) is accessible by such hops. Hence $2R \\geqslant G + (R-1)$, or equivalently $G \\leqslant R+1$. In order to prove the inequality $G \\leqslant B+1$, we turn over the board and apply the similar arguments.\n\nTherefore we get $9m^{2} \\geqslant B+R+G \\geqslant 3G-2$, so $G \\leqslant 3m^{2}$.\n\n\nSolution 2.\nWe shall use the same colouring as in the above solution. Again, assume that $X$ is a minimal blocking set.\n\nNote that any $2 \\times 2$ square cannot contain more than 2 green cells. Indeed, on Figure 3(a) the cell marked with \"?\" does not block any path, while on Figure 3(b) the cell marked with \"?\" should be coloured red and blue simultaneously. So we can split all green cells into chains consisting of three types of links shown on Figure 4 (diagonal link in the other direction is not allowed, corresponding green cells must belong to different chains). For example, there are 3 chains in Figure 2(b).\n\n![](attached_image_1.png)\n(a)\n![](attached_image_2.png)\n(b)\n![](attached_image_3.png)\nFigure 3\n![](attached_image_4.png)\nFigure 4\n![](attached_image_5.png)\nFigure 5\n\nWe will inscribe green chains in disjoint axis-aligned rectangles so that the number of green cells in each rectangle will not exceed $1/3$ of the area of the rectangle. This will give us the bound $G \\leqslant 3m^{2}$. Sometimes the rectangle will be the minimal bounding rectangle of the chain, sometimes minimal bounding rectangles will be expanded in one or two directions in order to have sufficiently large area.\n\nNote that for any two consecutive cells in the chain the colouring of some neighbouring cells is uniquely defined (see Figure 5). In particular, this observation gives a corresponding rectangle for the chains of height (or width) 1 (see Figure 6(a)). A separate green cell can be inscribed in $1 \\times 3$ or $3 \\times 1$ rectangle with one red and one blue cell, see Figure 6(b)-(c), otherwise we get one of impossible configurations shown in Figure 3.\n\n![](attached_image_6.png)\nFigure 6\nFigure 7\n\nAny diagonal chain of length 2 is always inscribed in a $2 \\times 3$ or $3 \\times 2$ rectangle without another green cells. Indeed, one of the squares marked with \"?\" in Figure 7(a) must be red. If it is the bottom question mark, then the remaining cell in the corresponding $2 \\times 3$ rectangle must have the same colour, see Figure 7(b).\n\nA longer chain of height (or width) 2 always has a horizontal (resp., vertical) link and can be inscribed into a $3 \\times a$ rectangle. In this case we expand the minimal bounding rectangle across the long side which touches the mentioned link. On Figure 8(a) the corresponding expansion of the minimal bounding rectangle is coloured in light blue. The upper right corner cell must be also blue. Indeed it cannot be red or green. If it is not coloured in blue, see Figure 8(b), then all anti-hop paths from $F$ to \"?\" are blocked with green cells. And these green cells are surrounded by blue ones, what is impossible. In this case the green chain contains $a$ cells, which is exactly $1/3$ of the area of the rectangle.\n\n![](attached_image_7.png)\n\nIn the remaining case the minimal bounding rectangle of the chain is of size $a \\times b$ where $a, b \\geqslant 3$. Denote by $\\ell$ the length of the chain (i.e. the number of cells in the chain).\n\nIf the chain has at least two diagonal links (see Figure 9), then $\\ell \\leqslant a+b-3 \\leqslant ab/3$.\n\nIf the chain has only one diagonal link then $\\ell = a+b-2$. In this case the chain has horizontal and vertical end-links, and we expand the minimal bounding rectangle in two directions to get an $(a+1) \\times (b+1)$ rectangle. On Figure 10 a corresponding expansion of the minimal bounding rectangle is coloured in light red. Again the length of the chain does not exceed $1/3$ of the rectangle's area: $\\ell \\leqslant a+b-2 \\leqslant (a+1)(b+1)/3$.\n\nOn the next step we will use the following statement: all cells in constructed rectangles are coloured red, green or blue (the cells upwards and to the right of green cells are blue; the cells downwards and to the left of green cells are red). The proof repeats the same arguments as before (see Figure 8(b).)\n\n![](attached_image_8.png)\nFigure 9\n![](attached_image_9.png)\nFigure 10\n![](attached_image_10.png)\nFigure 11\n\nNote that all constructed rectangles are disjoint. Indeed, assume that two rectangles have a common cell. Using the above statement, one can see that the only such cell can be a common corner cell, as shown in Figure 11. Moreover, in this case both rectangles should be expanded, otherwise they would share a green corner cell.\n\nIf they were expanded along the same axis (see Figure 11(a)), then again the common corner cannot be coloured correctly. If they were expanded along different axes (see Figure 11(b)) then the two chains have a common point and must be connected in one chain. (These arguments work for $2 \\times 3$ and $1 \\times 3$ rectangles in a similar manner.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32119,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine the largest $N$ for which there exists a table $T$ of integers with $N$ rows and 100 columns that has the following properties:\n(i) Every row contains the numbers $1,2, \\ldots, 100$ in some order.\n(ii) For any two distinct rows $r$ and $s$, there is a column $c$ such that $|T(r, c)-T(s, c)| \\geqslant 2$.\nHere $T(r, c)$ means the number at the intersection of the row $r$ and the column $c$.",
    "options": [],
    "answer": "100!/2^{50}",
    "solution": "Answer: The largest such integer is $N=100!/2^{50}$.\n\n## Solution 1.\n\nNon-existence of a larger table. Let us consider some fixed row in the table, and let us replace (for $k=1,2, \\ldots, 50$) each of two numbers $2k-1$ and $2k$ respectively by the symbol $x_{k}$. The resulting pattern is an arrangement of 50 symbols $x_{1}, x_{2}, \\ldots, x_{50}$, where every symbol occurs exactly twice. Note that there are $N=100!/2^{50}$ distinct patterns $P_{1}, \\ldots, P_{N}$.\n\nIf two rows $r \\neq s$ in the table have the same pattern $P_{i}$, then $|T(r, c)-T(s, c)| \\leqslant 1$ holds for all columns $c$. As this violates property (ii) in the problem statement, different rows have different patterns. Hence there are at most $N=100!/2^{50}$ rows.\n\nExistence of a table with $N$ rows. We construct the table by translating every pattern $P_{i}$ into a corresponding row with the numbers $1,2, \\ldots, 100$. We present a procedure that inductively replaces the symbols by numbers. The translation goes through steps $k=1,2, \\ldots, 50$ in increasing order and at step $k$ replaces the two occurrences of symbol $x_{k}$ by $2k-1$ and $2k$.\n\n- The left occurrence of $x_{1}$ is replaced by 1, and its right occurrence is replaced by 2.\n- For $k \\geqslant 2$, we already have the number $2k-2$ somewhere in the row, and now we are looking for the places for $2k-1$ and $2k$. We make the three numbers $2k-2, 2k-1, 2k$ show up (ordered from left to right) either in the order $2k-2, 2k-1, 2k$, or as $2k, 2k-2, 2k-1$, or as $2k-1, 2k, 2k-2$. This is possible, since the number $2k-2$ has been placed in the preceding step, and shows up before / between / after the two occurrences of the symbol $x_{k}$.\n\nWe claim that the $N$ rows that result from the $N$ patterns yield a table with the desired property (ii). Indeed, consider the $r$-th and the $s$-th row ($r \\neq s$), which by construction result from patterns $P_{r}$ and $P_{s}$. Call a symbol $x_{i}$ aligned, if it occurs in the same two columns in $P_{r}$ and in $P_{s}$. Let $k$ be the largest index, for which symbol $x_{k}$ is not aligned. Note that $k \\geqslant 2$. Consider the column $c'$ with $T\\left(r, c'\\right)=2k$ and the column $c''$ with $T\\left(s, c''\\right)=2k$. Then $T\\left(r, c''\\right) \\leqslant 2k$ and $T\\left(s, c'\\right) \\leqslant 2k$, as all symbols $x_{i}$ with $i \\geqslant k+1$ are aligned.\n\n- If $T\\left(r, c''\\right) \\leqslant 2k-2$, then $\\left|T\\left(r, c''\\right)-T\\left(s, c''\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(s, c'\\right) \\leqslant 2k-2$, then $\\left|T\\left(r, c'\\right)-T\\left(s, c'\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(r, c''\\right)=2k-1$ and $T\\left(s, c'\\right)=2k-1$, then the symbol $x_{k}$ is aligned; contradiction.\n\nIn the only remaining case we have $c'=c''$, so that $T\\left(r, c'\\right)=T\\left(s, c'\\right)=2k$ holds. Now let us consider the columns $d'$ and $d''$ with $T\\left(r, d'\\right)=2k-1$ and $T\\left(s, d''\\right)=2k-1$. Then $d' \\neq d''$ (as the symbol $x_{k}$ is not aligned), and $T\\left(r, d''\\right) \\leqslant 2k-2$ and $T\\left(s, d'\\right) \\leqslant 2k-2$ (as all symbols $x_{i}$ with $i \\geqslant k+1$ are aligned).\n\n- If $T\\left(r, d''\\right) \\leqslant 2k-3$, then $\\left|T\\left(r, d''\\right)-T\\left(s, d''\\right)\\right| \\geqslant 2$ as desired.\n- If $T\\left(s, d'\\right) \\leqslant 2k-3$, then $\\left|T\\left(r, d'\\right)-T\\left(s, d'\\right)\\right| \\geqslant 2$ as desired.\n\nIn the only remaining case we have $T\\left(r, d''\\right)=2k-2$ and $T\\left(s, d'\\right)=2k-2$. Now the row $r$ has the numbers $2k-2, 2k-1, 2k$ in the three columns $d', d'', c'$. As one of these triples violates the ordering property of $2k-2, 2k-1, 2k$, we have the final contradiction.\nWe provide a bit different proof why the example from Solution 1 (see also Comment 1) works.\n\nLemma. Let $\\pi_{1}$ and $\\pi_{2}$ be two permutations of the set $\\{1,2, \\ldots, n\\}$ such that $\\left|\\pi_{1}(i)-\\pi_{2}(i)\\right| \\leqslant 1$ for every $i$. Then there exists a set of disjoint pairs ($i, i+1$) such that $\\pi_{2}$ is obtained from $\\pi_{1}$ by swapping elements in each pair from the set.\n\nProof. We may assume that $\\pi_{1}(i)=i$ for every $i$ and proceed by induction on $n$. The case $n=1$ is trivial. If $\\pi_{2}(n)=n$, we simply apply the induction hypothesis. If $\\pi_{2}(n)=n-1$, then $\\pi_{2}(i)=n$ for some $i<n$. It is clear that $i=n-1$, and we can also use the induction hypothesis.\n\nNow let $\\pi_{1}$ and $\\pi_{2}$ be two rows (which we identify with permutations of $\\{1,2, \\ldots, 100\\}$) of the table constructed in Solution 1. Assume that $\\left|\\pi_{1}(i)-\\pi_{2}(i)\\right| \\leqslant 1$ for any $i$. From the Lemma it follows that there exists a set $S \\subset \\{1, \\ldots, 99\\}$ such that any two numbers from $S$ differ by at least 2 and $\\pi_{2}$ is obtained from $\\pi_{1}$ by applying the permutations $(j, j+1)$, $j \\in S$. Let $r=\\min (S)$. If $r=2k-1$ is odd, then $\\pi_{1}$ and $\\pi_{2}$ induce two subpermutations of $\\{2k-2, 2k-1, 2k\\}$ (or of $\\{1,2\\}$ for $k=1$) of opposite parities. Thus $r=2k$ is even. Since $\\pi_{1}$ and $\\pi_{2}$ induce subpermutations of the same (even) parity of $\\{2k, 2k+1, 2k+2\\}$, we must have $2k+2 \\in S$. Next, $2k+4 \\in S$ and so on, we get $98 \\in S$, but then the parities of the subpermutations of $\\{98,99,100\\}$ in $\\pi_{1}, \\pi_{2}$ are opposite. A contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32120,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A B C D$ be a parallelogram such that $A C = B C$. A point $P$ is chosen on the extension of the segment $A B$ beyond $B$. The circumcircle of the triangle $A C D$ meets the segment $P D$ again at $Q$, and the circumcircle of the triangle $A P Q$ meets the segment $P C$ again at $R$. Prove that the lines $C D$, $A Q$, and $B R$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Common remarks. The introductory steps presented here are used in all solutions below.\nSince $A C = B C = A D$, we have $\\angle A B C = \\angle B A C = \\angle A C D = \\angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain\n$$\n\\angle C R A = 180^{\\circ} - \\angle A R P = 180^{\\circ} - \\angle A Q P = \\angle D Q A = \\angle D C A = \\angle C B A,\n$$\nso the points $A, B, C$, and $R$ lie on some circle $\\gamma$.\n\n\nSolution 1. Introduce the point $X = A Q \\cap C D$; we need to prove that $B, R$ and $X$ are collinear.\nBy means of the circle $(A P R Q)$ we have\n$$\n\\angle R Q X = 180^{\\circ} - \\angle A Q R = \\angle R P A = \\angle R C X\n$$\n(the last equality holds in view of $A B \\parallel C D$), which means that the points $C, Q, R$, and $X$ also lie on some circle $\\delta$.\nUsing the circles $\\delta$ and $\\gamma$ we finally obtain\n$$\n\\angle X R C = \\angle X Q C = 180^{\\circ} - \\angle C Q A = \\angle A D C = \\angle B A C = 180^{\\circ} - \\angle C R B,\n$$\nthat proves the desired collinearity.\n\n\nSolution 2. Let $\\alpha$ denote the circle $(A P R Q)$. Since\n$$\n\\angle C A P = \\angle A C D = \\angle A Q D = 180^{\\circ} - \\angle A Q P,\n$$\nthe line $A C$ is tangent to $\\alpha$.\nNow, let $A D$ meet $\\alpha$ again at a point $Y$ (which necessarily lies on the extension of $D A$ beyond $A$). Using the circle $\\gamma$, along with the fact that $A C$ is tangent to $\\alpha$, we have\n$$\n\\angle A R Y = \\angle C A D = \\angle A C B = \\angle A R B,\n$$\nso the points $Y, B$, and $R$ are collinear.\nApplying Pascal's theorem to the hexagon $A A Y R P Q$ (where $A A$ is regarded as the tangent to $\\alpha$ at $A$), we see that the points $A A \\cap R P = C$, $A Y \\cap P Q = D$, and $Y R \\cap Q A$ are collinear. Hence the lines $C D$, $A Q$, and $B R$ are concurrent.\n\n\nSolution 3. As in Solution 1, we introduce the point $X = A Q \\cap C D$ and aim at proving that the points $B, R$, and $X$ are collinear. As in Solution 2, we denote $\\alpha = (A P Q R)$; but now we define $Y$ to be the second meeting point of $R B$ with $\\alpha$.\nUsing the circle $\\alpha$ and noticing that $C D$ is tangent to $\\gamma$, we obtain\n$$\n\\begin{equation*}\n\\angle R Y A = \\angle R P A = \\angle R C X = \\angle R B C . \\tag{1}\n\\end{equation*}\n$$\nSo $A Y \\parallel B C$, and hence $Y$ lies on $D A$.\nNow the chain of equalities (1) shows also that $\\angle R Y D = \\angle R C X$, which implies that the points $C, D, Y$, and $R$ lie on some circle $\\beta$. Hence, the lines $C D$, $A Q$, and $Y B R$ are the pairwise radical axes of the circles $(A Q C D)$, $\\alpha$, and $\\beta$, so those lines are concurrent.\n\n![](attached_image_1.png)\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32121,
    "subject": "Mathematics (Multi-modal)",
    "question": "Version 1. Let $n$ be a fixed positive integer, and let $S$ be the set of points $(x, y)$ on the Cartesian plane such that both coordinates $x$ and $y$ are nonnegative integers smaller than $2n$ (thus $|S| = 4n^2$). Assume that $\\mathcal{F}$ is a set consisting of $n^2$ quadrilaterals such that all their vertices lie in $S$, and each point in $S$ is a vertex of exactly one of the quadrilaterals in $\\mathcal{F}$.\nDetermine the largest possible sum of areas of all $n^2$ quadrilaterals in $\\mathcal{F}$.\n\nVersion 2. Let $n$ be a fixed positive integer, and let $S$ be the set of points $(x, y)$ on the Cartesian plane such that both coordinates $x$ and $y$ are nonnegative integers smaller than $2n$ (thus $|S| = 4n^2$). Assume that $\\mathcal{F}$ is a set of polygons such that all vertices of polygons in $\\mathcal{F}$ lie in $S$, and each point in $S$ is a vertex of exactly one of the polygons in $\\mathcal{F}$.\nDetermine the largest possible sum of areas of all polygons in $\\mathcal{F}$.",
    "options": [],
    "answer": "n^2(2n+1)(2n-1)/3",
    "solution": "Answer for both Versions: The largest possible sum of areas is $\\Sigma(n) := \\frac{1}{3} n^2 (2n+1)(2n-1)$.\n\nCommon remarks. Throughout all solutions, the area of a polygon $P$ will be denoted by $[P]$.\nWe say that a polygon is legal if all its vertices belong to $S$. Let $O = \\left(n - \\frac{1}{2}, n - \\frac{1}{2}\\right)$ be the centre of $S$. We say that a legal square is central if its centre is situated at $O$. Finally, say that a set $\\mathcal{F}$ of polygons is acceptable if it satisfies the problem requirements, i.e. if all polygons in $\\mathcal{F}$ are legal, and each point in $S$ is a vertex of exactly one polygon in $\\mathcal{F}$. For an acceptable set $\\mathcal{F}$, we denote by $\\Sigma(\\mathcal{F})$ the sum of areas of polygons in $\\mathcal{F}$.\n\n\nSolution 1, for both Versions. Each point in $S$ is a vertex of a unique central square. Thus the set $\\mathcal{G}$ of central squares is acceptable. We will show that\n$$\n\\begin{equation*}\n\\Sigma(\\mathcal{F}) \\leqslant \\Sigma(\\mathcal{G}) = \\Sigma(n) \\tag{1}\n\\end{equation*}\n$$\nthus establishing the answer.\nWe will use the following key lemma.\n\nLemma 1. Let $P = A_1 A_2 \\ldots A_m$ be a polygon, and let $O$ be an arbitrary point in the plane. Then\n$$\n\\begin{equation*}\n[P] \\leqslant \\frac{1}{2} \\sum_{i=1}^m OA_i^2 \\tag{2}\n\\end{equation*}\n$$\nmoreover, if $P$ is a square centred at $O$, then the inequality (2) turns into an equality.\n\nProof. Put $A_{n+1} = A_1$. For each $i = 1, 2, \\ldots, m$, we have\n$$\n\\left[OA_i A_{i+1}\\right] \\leqslant \\frac{OA_i \\cdot OA_{i+1}}{2} \\leqslant \\frac{OA_i^2 + OA_{i+1}^2}{4}.\n$$\nTherefore,\n$$\n[P] \\leqslant \\sum_{i=1}^m \\left[OA_i A_{i+1}\\right] \\leqslant \\frac{1}{4} \\sum_{i=1}^m \\left(OA_i^2 + OA_{i+1}^2\\right) = \\frac{1}{2} \\sum_{i=1}^m OA_i^2\n$$\nwhich proves (2). Finally, all the above inequalities turn into equalities when $P$ is a square centred at $O$.\n\nBack to the problem, consider an arbitrary acceptable set $\\mathcal{F}$. Applying Lemma 1 to each element in $\\mathcal{F}$ and to each element in $\\mathcal{G}$ (achieving equality in the latter case), we obtain\n$$\n\\Sigma(\\mathcal{F}) \\leqslant \\frac{1}{2} \\sum_{A \\in S} OA^2 = \\Sigma(\\mathcal{G})\n$$\nwhich establishes the left inequality in (1).\n\nIt remains to compute $\\Sigma(\\mathcal{G})$. We have\n$$\n\\begin{aligned}\n\\Sigma(\\mathcal{G}) = & \\frac{1}{2} \\sum_{A \\in S} OA^2 = \\frac{1}{2} \\sum_{i=0}^{2n-1} \\sum_{j=0}^{2n-1} \\left(\\left(n - \\frac{1}{2} - i\\right)^2 + \\left(n - \\frac{1}{2} - j\\right)^2\\right) \\\\\n& = \\frac{1}{8} \\cdot 4 \\cdot 2n \\sum_{i=0}^{n-1} (2n - 2i - 1)^2 = n \\sum_{j=0}^{n-1} (2j + 1)^2 = n \\left(\\sum_{j=1}^{2n} j^2 - \\sum_{j=1}^n (2j)^2\\right) \\\\\n& = n \\left(\\frac{2n(2n+1)(4n+1)}{6} - 4 \\cdot \\frac{n(n+1)(2n+1)}{6}\\right) = \\frac{n^2 (2n+1)(2n-1)}{3} = \\Sigma(n)\n\\end{aligned}\n$$\n\n\nSolution 2, for Version 1. Let $\\mathcal{F}$ be an accessible set of quadrilaterals. For every quadrilateral $ABCD$ in $\\mathcal{F}$ write\n$$\n\\begin{equation*}\n[ABCD] = \\frac{AC \\cdot BD}{2} \\sin \\phi \\leqslant \\frac{AC^2 + BD^2}{4} \\tag{3}\n\\end{equation*}\n$$\nwhere $\\phi$ is the angle between $AC$ and $BD$. Applying this estimate to all members in $\\mathcal{F}$ we obtain\n$$\n\\Sigma(\\mathcal{F}) \\leqslant \\frac{1}{4} \\sum_{i=1}^{2n^2} A_i B_i^2\n$$\nwhere $A_1, A_2, \\ldots, A_{2n^2}, B_1, B_2, \\ldots, B_{2n^2}$ is some permutation of $S$. For brevity, denote\n$$\nf\\left(\\left(A_i\\right), \\left(B_i\\right)\\right) := \\sum_{i=1}^{2n^2} A_i B_i^2\n$$\nThe rest of the solution is based on the following lemma.\n\nLemma 2. The maximal value of $f\\left(\\left(A_i\\right), \\left(B_i\\right)\\right)$ over all permutations of $S$ equals $\\frac{4}{3} n^2 (4n^2 - 1)$ and is achieved when $A_i$ is symmetric to $B_i$ with respect to $O$, for every $i = 1, 2, \\ldots, 2n^2$.\n\nProof. Let $A_i = (p_i, q_i)$ and $B_i = (r_i, s_i)$, for $i = 1, 2, \\ldots, 2n^2$. We have\n$$\nf\\left(\\left(A_i\\right), \\left(B_i\\right)\\right) = \\sum_{i=1}^{2n^2} (p_i - r_i)^2 + \\sum_{i=1}^{2n^2} (q_i - s_i)^2\n$$\nit suffices to bound the first sum, the second is bounded similarly. This can be done, e.g., by means of the QM-AM inequality as follows:\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2n^2} (p_i - r_i)^2 & = \\sum_{i=1}^{2n^2} (2p_i^2 + 2r_i^2 - (p_i + r_i)^2) = 4n \\sum_{j=0}^{2n-1} j^2 - \\sum_{i=1}^{2n^2} (p_i + r_i)^2 \\\\\n\\leqslant 4n \\sum_{j=0}^{2n-1} j^2 & - \\frac{1}{2n^2} \\left(\\sum_{i=1}^{2n^2} (p_i + r_i)\\right)^2 = 4n \\sum_{j=0}^{2n-1} j^2 - \\frac{1}{2n^2} \\left(2n \\cdot \\sum_{j=0}^{2n-1} j\\right)^2 \\\\\n& = 4n \\cdot \\frac{2n(2n-1)(4n-1)}{6} - 2n^2 (2n-1)^2 = \\frac{2n^2 (2n-1)(2n+1)}{3}\n\\end{aligned}\n$$\nAll the estimates are sharp if $p_i + r_i = 2n-1$ for all $i$. Thus,\n$$\nf\\left(\\left(A_i\\right), \\left(B_i\\right)\\right) \\leqslant \\frac{4n^2 (4n^2 - 1)}{3}\n$$\nand the estimate is sharp when $p_i + r_i = q_i + s_i = 2n-1$ for all $i$, i.e. when $A_i$ and $B_i$ are symmetric with respect to $O$.\n\nLemma 2 yields\n$$\n\\Sigma(\\mathcal{F}) \\leqslant \\frac{1}{4} \\cdot \\frac{4n^2 (4n^2 - 1)}{3} = \\frac{n^2 (2n-1)(2n+1)}{3}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32122,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral whose sides have pairwise different lengths. Let $O$ be the circumcentre of $ABCD$. The internal angle bisectors of $\\angle ABC$ and $\\angle ADC$ meet $AC$ at $B_{1}$ and $D_{1}$, respectively. Let $O_{B}$ be the centre of the circle which passes through $B$ and is tangent to $AC$ at $D_{1}$. Similarly, let $O_{D}$ be the centre of the circle which passes through $D$ and is tangent to $AC$ at $B_{1}$.\nAssume that $B D_{1} \\parallel D B_{1}$. Prove that $O$ lies on the line $O_{B} O_{D}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Common remarks. We introduce some objects and establish some preliminary facts common for all solutions below.\nLet $\\Omega$ denote the circle $(ABCD)$, and let $\\gamma_{B}$ and $\\gamma_{D}$ denote the two circles from the problem statement (their centres are $O_{B}$ and $O_{D}$, respectively). Clearly, all three centres $O$, $O_{B}$, and $O_{D}$ are distinct.\nAssume, without loss of generality, that $AB > BC$. Suppose that $AD > DC$, and let $H = AC \\cap BD$. Then the rays $BB_{1}$ and $DD_{1}$ lie on one side of $BD$, as they contain the midpoints of the $\\operatorname{arcs} ADC$ and $ABC$, respectively. However, if $BD_{1} \\parallel DB_{1}$, then $B_{1}$ and $D_{1}$ should be separated by $H$. This contradiction shows that $AD < CD$.\nLet $\\gamma_{B}$ and $\\gamma_{D}$ meet $\\Omega$ again at $T_{B}$ and $T_{D}$, respectively. The common chord $BT_{B}$ of $\\Omega$ and $\\gamma_{B}$ is perpendicular to their line of\n![](attached_image_1.png)\ncentres $O_{B}O$; likewise, $DT_{D} \\perp O_{D}O$. Therefore, $O \\in O_{B}O_{D} \\Longleftrightarrow O_{B}O \\parallel O_{D}O \\Longleftrightarrow BT_{B} \\parallel DT_{D}$, and the problem reduces to showing that\n$$\n\\begin{equation*}\nBT_{B} \\parallel DT_{D}. \\tag{1}\n\\end{equation*}\n$$\n\n\nSolution 1. Let the diagonals $AC$ and $BD$ cross at $H$. Consider the homothety $h$ centred at $H$ and mapping $B$ to $D$. Since $BD_{1} \\parallel DB_{1}$, we have $h(D_{1}) = B_{1}$.\nLet the tangents to $\\Omega$ at $B$ and $D$ meet $AC$ at $L_{B}$ and $L_{D}$, respectively. We have\n$$\n\\angle L_{B}BB_{1} = \\angle L_{B}BC + \\angle CBB_{1} = \\angle BAL_{B} + \\angle B_{1}BA = \\angle BB_{1}L_{B},\n$$\nwhich means that the triangle $L_{B}BB_{1}$ is isosceles, $L_{B}B = L_{B}B_{1}$. The powers of $L_{B}$ with respect to $\\Omega$ and $\\gamma_{D}$ are $L_{B}B^{2}$ and $L_{B}B_{1}^{2}$, respectively; so they are equal, whence $L_{B}$ lies on the radical axis $T_{D}D$ of those two circles. Similarly, $L_{D}$ lies on the radical axis $T_{B}B$ of $\\Omega$ and $\\gamma_{B}$.\nBy the sine rule in the triangle $BHL_{B}$, we obtain\n$$\n\\begin{equation*}\n\\frac{HL_{B}}{\\sin \\angle HBL_{B}} = \\frac{BL_{B}}{\\sin \\angle BHL_{B}} = \\frac{B_{1}L_{B}}{\\sin \\angle BHL_{B}}; \\tag{2}\n\\end{equation*}\n$$\nsimilarly,\n$$\n\\begin{equation*}\n\\frac{HL_{D}}{\\sin \\angle HDL_{D}} = \\frac{DL_{D}}{\\sin \\angle DHL_{D}} = \\frac{D_{1}L_{D}}{\\sin \\angle DHL_{D}}. \\tag{3}\n\\end{equation*}\n$$\nClearly, $\\angle BHL_{B} = \\angle DHL_{D}$. In the circle $\\Omega$, tangent lines $BL_{B}$ and $DL_{D}$ form equal angles with the chord $BD$, so $\\sin \\angle HBL_{B} = \\sin \\angle HDL_{D}$ (this equality does not depend on the picture). Thus, dividing (2) by (3) we get\n$$\n\\frac{HL_{B}}{HL_{D}} = \\frac{B_{1}L_{B}}{D_{1}L_{D}}, \\quad \\text{and hence} \\quad \\frac{HL_{B}}{HL_{D}} = \\frac{HL_{B} - B_{1}L_{B}}{HL_{D} - D_{1}L_{D}} = \\frac{HB_{1}}{HD_{1}}.\n$$\nSince $h(D_{1}) = B_{1}$, the obtained relation yields $h(L_{D}) = L_{B}$, so $h$ maps the line $L_{D}B$ to $L_{B}D$, and these lines are parallel, as desired.\n\n\nSolution 2. Let $BD_{1}$ and $T_{B}D_{1}$ meet $\\Omega$ again at $X_{B}$ and $Y_{B}$, respectively. Then\n$$\n\\angle BD_{1}C = \\angle BT_{B}D_{1} = \\angle BT_{B}Y_{B} = \\angle BX_{B}Y_{B},\n$$\nwhich shows that $X_{B}Y_{B} \\parallel AC$. Similarly, let $DB_{1}$ and $T_{D}B_{1}$ meet $\\Omega$ again at $X_{D}$ and $Y_{D}$, respectively; then $X_{D}Y_{D} \\parallel AC$.\nLet $M_{D}$ and $M_{B}$ be the midpoints of the $\\operatorname{arcs} ABC$ and $ADC$, respectively; then the points $D_{1}$ and $B_{1}$ lie on $DM_{D}$ and $BM_{B}$, respectively. Let $K$ be the midpoint of $AC$ (which lies on $M_{B}M_{D}$). Applying Pascal's theorem to $M_{D}DX_{D}X_{B}BM_{B}$, we obtain that the points $D_{1} = M_{D}D \\cap X_{B}B$, $B_{1} = DX_{D} \\cap BM_{B}$, and $X_{D}X_{B} \\cap M_{B}M_{D}$ are collinear, which means that $X_{B}X_{D}$ passes through $K$. Due to symmetry, the diagonals of an isosceles trapezoid $X_{B}Y_{B}X_{D}Y_{D}$ cross at $K$.\n![](attached_image_2.png)\nLet $b$ and $d$ denote the distances from the lines $X_{B}Y_{B}$ and $X_{D}Y_{D}$, respectively, to $AC$. Then we get\n$$\n\\frac{X_{B}Y_{B}}{X_{D}Y_{D}} = \\frac{b}{d} = \\frac{D_{1}X_{B}}{B_{1}X_{D}},\n$$\nwhere the second equation holds in view of $D_{1}X_{B} \\parallel B_{1}X_{D}$. Therefore, the triangles $D_{1}X_{B}Y_{B}$ and $B_{1}X_{D}Y_{D}$ are similar. The triangles $D_{1}T_{B}B$ and $B_{1}T_{D}D$ are similar to them and hence to each other. Since $BD_{1} \\parallel DB_{1}$, these triangles are also homothetical. This yields $BT_{B} \\parallel DT_{D}$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32123,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $n \\geqslant 3$ satisfying the following property: every convex $n$-gon whose sides all have length $1$ contains an equilateral triangle of side length $1$.\n(Every polygon is assumed to contain its boundary.)",
    "options": [],
    "answer": "All odd n ≥ 3",
    "solution": "Answer: All odd $n \\geqslant 3$.\n\nFirst we show that for every even $n \\geqslant 4$ there exists a polygon violating the required statement. Consider a regular $k$-gon $A_{0} A_{1}, \\ldots A_{k-1}$ with side length $1$. Let $B_{1}, B_{2}, \\ldots, B_{n / 2-1}$ be the points symmetric to $A_{1}, A_{2}, \\ldots, A_{n / 2-1}$ with respect to the line $A_{0} A_{n / 2}$. Then $P= A_{0} A_{1} A_{2} \\ldots A_{n / 2-1} A_{n / 2} B_{n / 2-1} B_{n / 2-2} \\ldots B_{2} B_{1}$ is a convex $n$-gon whose sides all have length $1$. If $k$ is big enough, $P$ is contained in a strip of width $1 / 2$, which clearly does not contain any equilateral triangle of side length $1$.\n\n![](attached_image_1.png)\n\nAssume now that $n=2k+1$. As the case $k=1$ is trivially true, we assume $k \\geqslant 2$ henceforth. Consider a convex $(2k+1)$-gon $P$ whose sides all have length $1$. Let $d$ be its longest diagonal. The endpoints of $d$ split the perimeter of $P$ into two polylines, one of which has length at least $k+1$. Hence we can label the vertices of $P$ so that $P=A_{0} A_{1} \\ldots A_{2k}$ and $d=A_{0} A_{\\ell}$ with $\\ell \\geqslant k+1$. We will show that, in fact, the polygon $A_{0} A_{1} \\ldots A_{\\ell}$ contains an equilateral triangle of side length $1$.\n\nSuppose that $\\angle A_{\\ell} A_{0} A_{1} \\geqslant 60^{\\circ}$. Since $d$ is the longest diagonal, we have $A_{1} A_{\\ell} \\leqslant A_{0} A_{\\ell}$, so $\\angle A_{0} A_{1} A_{\\ell} \\geqslant \\angle A_{\\ell} A_{0} A_{1} \\geqslant 60^{\\circ}$. It follows that there exists a point $X$ inside the triangle $A_{0} A_{1} A_{\\ell}$ such that the triangle $A_{0} A_{1} X$ is equilateral, and this triangle is contained in $P$. Similar arguments apply if $\\angle A_{\\ell-1} A_{\\ell} A_{0} \\geqslant 60^{\\circ}$.\n\n![](attached_image_2.png)\n\nFrom now on, assume $\\angle A_{\\ell} A_{0} A_{1}<60^{\\circ}$ and $A_{\\ell-1} A_{\\ell} A_{0}<60^{\\circ}$.\nConsider an isosceles trapezoid $A_{0} Y Z A_{\\ell}$ such that $A_{0} A_{\\ell} \\parallel Y Z$, $A_{0} Y=Z A_{\\ell}=1$, and $\\angle A_{\\ell} A_{0} Y=\\angle Z A_{\\ell} A_{0}=60^{\\circ}$. Suppose that $A_{0} A_{1} \\ldots A_{\\ell}$ is contained in $A_{0} Y Z A_{\\ell}$. Note that the perimeter of $A_{0} A_{1} \\ldots A_{\\ell}$ equals $\\ell+A_{0} A_{\\ell}$ and the perimeter of $A_{0} Y Z A_{\\ell}$ equals $2 A_{0} A_{\\ell}+1$.\n\n![](attached_image_3.png)\n\nRecall a well-known fact stating that if a convex polygon $P_{1}$ is contained in a convex polygon $P_{2}$, then the perimeter of $P_{1}$ is at most the perimeter of $P_{2}$. Hence we obtain\n$$\n\\ell+A_{0} A_{\\ell} \\leqslant 2 A_{0} A_{\\ell}+1, \\quad \\text{ i.e. } \\quad \\ell-1 \\leqslant A_{0} A_{\\ell} .\n$$\nOn the other hand, the triangle inequality yields\n$$\nA_{0} A_{\\ell}<A_{\\ell} A_{\\ell+1}+A_{\\ell+1} A_{\\ell+2}+\\ldots+A_{2k} A_{0}=2k+1-\\ell \\leqslant \\ell-1,\n$$\nwhich gives a contradiction.\n\nTherefore, there exists a vertex $A_{m}$ of $A_{0} A_{1} \\ldots A_{\\ell}$ which lies outside $A_{0} Y Z A_{\\ell}$. Since\n$$\n\\begin{equation*}\n\\angle A_{\\ell} A_{0} A_{1}<60^{\\circ}=\\angle A_{\\ell} A_{0} Y \\quad \\text{ and } \\quad A_{\\ell-1} A_{\\ell} A_{0}<60^{\\circ}=\\angle Z A_{\\ell} A_{0}, \\tag{1}\n\\end{equation*}\n$$\nthe distance between $A_{m}$ and $A_{0} A_{\\ell}$ is at least $\\sqrt{3} / 2$.\nLet $P$ be the projection of $A_{m}$ to $A_{0} A_{\\ell}$. Then $P A_{m} \\geqslant \\sqrt{3} / 2$, and by (1) we have $A_{0} P>1 / 2$ and $P A_{\\ell}>1 / 2$. Choose points $Q \\in A_{0} P$, $R \\in P A_{\\ell}$, and $S \\in P A_{m}$ such that $P Q=P R=1 / 2$ and $P S=\\sqrt{3} / 2$. Then $Q R S$ is an equilateral triangle of side length $1$ contained in $A_{0} A_{1} \\ldots A_{\\ell}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32124,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\omega$ be the circumcircle of a triangle $A B C$, and let $\\Omega_{A}$ be its excircle which is tangent to the segment $B C$. Let $X$ and $Y$ be the intersection points of $\\omega$ and $\\Omega_{A}$. Let $P$ and $Q$ be the projections of $A$ onto the tangent lines to $\\Omega_{A}$ at $X$ and $Y$, respectively. The tangent line at $P$ to the circumcircle of the triangle $A P X$ intersects the tangent line at $Q$ to the circumcircle of the triangle $A Q Y$ at a point $R$. Prove that $A R \\perp B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $D$ be the point of tangency of $B C$ and $\\Omega_{A}$. Let $D'$ be the point such that $D D'$ is a diameter of $\\Omega_{A}$. Let $R'$ be (the unique) point such that $A R' \\perp B C$ and $R' D' \\parallel B C$. We shall prove that $R'$ coincides with $R$.\n\nLet $P X$ intersect $A B$ and $D' R'$ at $S$ and $T$, respectively. Let $U$ be the ideal common point of the parallel lines $B C$ and $D' R'$. Note that the (degenerate) hexagon $A S X T U C$ is circumscribed around $\\Omega_{A}$, hence by the Brianchon theorem $A T$, $S U$, and $X C$ concur at a point which we denote by $V$. Then $V S \\parallel B C$. It follows that $\\Varangle(S V, V X)=\\Varangle(B C, C X)= \\Varangle(B A, A X)$, hence $A X S V$ is cyclic. Therefore, $\\Varangle(P X, X A)=\\Varangle(S V, V A)=\\Varangle\\left(R' T, T A\\right)$. Since $\\angle A P T=\\angle A R' T=90^{\\circ}$, the quadrilateral $A P R' T$ is cyclic. Hence,\n$$\n\\Varangle(X A, A P)=90^{\\circ}-\\Varangle(P X, X A)=90^{\\circ}-\\Varangle\\left(R' T, T A\\right)=\\Varangle\\left(T A, A R'\\right)=\\Varangle\\left(T P, P R'\\right) .\n$$\nIt follows that $P R'$ is tangent to the circle $(A P X)$.\n\nAnalogous argument shows that $Q R'$ is tangent to the circle $(A Q Y)$. Therefore, $R=R'$ and $A R \\perp B C$.\n\n![](attached_image_1.png)\n\nLet $J X$ intersect $\\omega$ again at $L$. Then $J L=d$. Let $L K$ be a diameter of $\\omega$ and let $M$ be the midpoint of $J K$. Since $J L=L K$, we have $\\angle L M K=90^{\\circ}$, so $M$ lies on $\\omega$. Let $R'$ be the point such that $R' P$ is tangent to the circle $(A P X)$ and $A R' \\perp B C$. Note that the line $A R'$ is symmetric to the line $A O$ with respect to $A J$.\n\n![](attached_image_2.png)\n![](attached_image_3.png)\n\nLemma. Let $M$ be the midpoint of the side $J K$ in a triangle $A J K$. Let $X$ be a point on the circle $(A M K)$ such that $\\angle J X K=90^{\\circ}$. Then there exists a point $T$ on the line $K X$ such that the triangles $A K J$ and $A J T$ are similar and equioriented.\n\nProof. Note that $M X=M K$. We construct a parallelogram $A J N K$. Let $T$ be a point on $K X$ such that $\\Varangle(N J, J A)=\\Varangle(K J, J T)$. Then\n$$\n\\Varangle(J N, N A)=\\Varangle(K A, A M)=\\Varangle(K X, X M)=\\Varangle(M K, K X)=\\Varangle(J K, K T) .\n$$\nSo there exists a spiral similarity with center $J$ mapping the triangle $A J N$ to the triangle $T J K$. Therefore, the triangles $N J K$ and $A J T$ are similar and equioriented. It follows that the triangles $A K J$ and $A J T$ are similar and equioriented. $\\square$\n\n![](attached_image_4.png)\n\nBack to the problem, we construct a point $T$ as in the lemma. We perform the composition $\\phi$ of inversion with centre $A$ and radius $A J$ and reflection in $A J$. It is known that every triangle $A E F$ is similar and equioriented to $A \\phi(F) \\phi(E)$.\n\nSo $\\phi(K)=T$ and $\\phi(T)=K$. Let $P^{*}=\\phi(P)$ and $R^{*}=\\phi\\left(R'\\right)$. Observe that $\\phi(T K)$ is a circle with diameter $A P^{*}$. Let $A A'$ be a diameter of $\\omega$. Then $P^{*} K \\perp A K \\perp A' K$, so $A'$ lies on $P^{*} K$. The triangles $A R' P$ and $A P^{*} R^{*}$ are similar and equioriented, hence\n$$\n\\Varangle\\left(A A', A' P^{*}\\right)=\\Varangle\\left(A A', A' K\\right)=\\Varangle(A X, X P)=\\Varangle(A X, X P)=\\Varangle\\left(A P, P R'\\right)=\\Varangle\\left(A R^{*}, R^{*} P^{*}\\right),\n$$\nso $A, A', R^{*}$, and $P^{*}$ are concyclic. Since $A'$ and $R^{*}$ lie on $A O$, we obtain $R^{*}=A'$. So $R'=\\phi\\left(A'\\right)$, and $\\phi\\left(A'\\right) P$ is tangent to the circle $(A P X)$.\n\nAn identical argument shows that $\\phi\\left(A'\\right) Q$ is tangent to the circle $(A Q Y)$. Therefore, $R= \\phi\\left(A'\\right)$ and $A R \\perp B C$.\nLet $J$ and $r$ be the center and the radius of $\\Omega_{A}$. Denote the diameter of $\\omega$ by $d$ and its center by $O$. By Euler's formula, $O J^{2}=(d / 2)^{2}+d r$, so the power of $J$ with respect to $\\omega$ equals $d r$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32125,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice is given a rational number $r>1$ and a line with two points $B \\neq R$, where point $R$ contains a red bead and point $B$ contains a blue bead. Alice plays a solitaire game by performing a sequence of moves. In every move, she chooses a (not necessarily positive) integer $k$, and a bead to move. If that bead is placed at point $X$, and the other bead is placed at $Y$, then Alice moves the chosen bead to point $X^{\\prime}$ with $\\overrightarrow{Y X^{\\prime}}=r^{k} \\overrightarrow{Y X}$.\n\nAlice's goal is to move the red bead to the point $B$. Find all rational numbers $r>1$ such that Alice can reach her goal in at most 2021 moves.",
    "options": [],
    "answer": "All r = (b+1)/b for integers b from 1 to 1010",
    "solution": "Answer: All $r=(b+1) / b$ with $b=1, \\ldots, 1010$.\n\nDenote the red and blue beads by $\\mathcal{R}$ and $\\mathcal{B}$, respectively. Introduce coordinates on the line and identify the points with their coordinates so that $R=0$ and $B=1$. Then, during the game, the coordinate of $\\mathcal{R}$ is always smaller than the coordinate of $\\mathcal{B}$. Moreover, the distance between the beads always has the form $r^{\\ell}$ with $\\ell \\in \\mathbb{Z}$, since it only multiplies by numbers of this form. Denote the value of the distance after the $m^{\\text{th}}$ move by $d_{m}=r^{\\alpha_{m}}$, $m=0,1,2, \\ldots$ (after the $0^{\\text{th}}$ move we have just the initial position, so $\\alpha_{0}=0$ ).\n\nIf some bead is moved in two consecutive moves, then Alice could instead perform a single move (and change the distance from $d_{i}$ directly to $d_{i+2}$ ) which has the same effect as these two moves. So, if Alice can achieve her goal, then she may as well achieve it in fewer (or the same) number of moves by alternating the moves of $\\mathcal{B}$ and $\\mathcal{R}$. In the sequel, we assume that Alice alternates the moves, and that $\\mathcal{R}$ is shifted altogether $t$ times.\n\nIf $\\mathcal{R}$ is shifted in the $m^{\\text{th}}$ move, then its coordinate increases by $d_{m}-d_{m+1}$. Therefore, the total increment of $\\mathcal{R}$ 's coordinate, which should be 1 , equals\neither\n$$\n\\begin{aligned}\n\\left(d_{0}-d_{1}\\right)+\\left(d_{2}-d_{3}\\right)+\\cdots+\\left(d_{2 t-2}-d_{2 t-1}\\right) & =1+\\sum_{i=1}^{t-1} r^{\\alpha_{2 i}}-\\sum_{i=1}^{t} r^{\\alpha_{2 i-1}} \\\\\n\\left(d_{1}-d_{2}\\right)+\\left(d_{3}-d_{4}\\right)+\\cdots+\\left(d_{2 t-1}-d_{2 t}\\right) & =\\sum_{i=1}^{t} r^{\\alpha_{2 i-1}}-\\sum_{i=1}^{t} r^{\\alpha_{2 i}}\n\\end{aligned}\n$$\ndepending on whether $\\mathcal{R}$ or $\\mathcal{B}$ is shifted in the first move. Moreover, in the former case we should have $t \\leqslant 1011$, while in the latter one we need $t \\leqslant 1010$. So both cases reduce to an equation\n$$\n\\begin{equation*}\n\\sum_{i=1}^{n} r^{\\beta_{i}}=\\sum_{i=1}^{n-1} r^{\\gamma_{i}}, \\quad \\beta_{i}, \\gamma_{i} \\in \\mathbb{Z} \\tag{1}\n\\end{equation*}\n$$\nfor some $n \\leqslant 1011$. Thus, if Alice can reach her goal, then this equation has a solution for $n=1011$ (we can add equal terms to both sums in order to increase $n$ ).\n\nConversely, if (1) has a solution for $n=1011$, then Alice can compose a corresponding sequence of distances $d_{0}, d_{1}, d_{2}, \\ldots, d_{2021}$ and then realise it by a sequence of moves. So the problem reduces to the solvability of (1) for $n=1011$.\n\nAssume that, for some rational $r$, there is a solution of (1). Write $r$ in lowest terms as $r=a / b$. Substitute this into (1), multiply by the common denominator, and collect all terms on the left hand side to get\n$$\n\\begin{equation*}\n\\sum_{i=1}^{2 n-1}(-1)^{i} a^{\\mu_{i}} b^{N-\\mu_{i}}=0, \\quad \\mu_{i} \\in\\{0,1, \\ldots, N\\} \\tag{2}\n\\end{equation*}\n$$\nfor some $N \\geqslant 0$. We assume that there exist indices $j_{-}$ and $j_{+}$ such that $\\mu_{j_{-}}=0$ and $\\mu_{j_{+}}=N$.\n\nReducing (2) modulo $a-b$ (so that $a \\equiv b$ ), we get\n$$\n0=\\sum_{i=1}^{2 n-1}(-1)^{i} a^{\\mu_{i}} b^{N-\\mu_{i}} \\equiv \\sum_{i=1}^{2 n-1}(-1)^{i} b^{\\mu_{i}} b^{N-\\mu_{i}}=-b^{N} \\quad \\bmod (a-b) .\n$$\nSince $\\operatorname{gcd}(a-b, b)=1$, this is possible only if $a-b=1$.\n\nReducing (2) modulo $a+b$ (so that $a \\equiv-b$ ), we get\n$$\n0=\\sum_{i=1}^{2 n-1}(-1)^{i} a^{\\mu_{i}} b^{N-\\mu_{i}} \\equiv \\sum_{i=1}^{2 n-1}(-1)^{i}(-1)^{\\mu_{i}} b^{\\mu_{i}} b^{N-\\mu_{i}}=S b^{N} \\quad \\bmod (a+b)\n$$\nfor some odd (thus nonzero) $S$ with $|S| \\leqslant 2 n-1$. Since $\\operatorname{gcd}(a+b, b)=1$, this is possible only if $a+b \\mid S$. So $a+b \\leqslant 2 n-1$, and hence $b=a-1 \\leqslant n-1=1010$.\n\nThus we have shown that any sought $r$ has the form indicated in the answer. It remains to show that for any $b=1,2, \\ldots, 1010$ and $a=b+1$, Alice can reach the goal. For this purpose, in (1) we put $n=a, \\beta_{1}=\\beta_{2}=\\cdots=\\beta_{a}=0$, and $\\gamma_{1}=\\gamma_{2}=\\cdots=\\gamma_{b}=1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32126,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are only finitely many quadruples $(a, b, c, n)$ of positive integers such that\n$$\nn! = a^{n-1} + b^{n-1} + c^{n-1}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For fixed $n$ there are clearly finitely many solutions; we will show that there is no solution with $n > 100$. So, assume $n > 100$. By the AM-GM inequality,\n$$\n\\begin{aligned}\nn! & = 2 n(n-1)(n-2)(n-3) \\cdot (3 \\cdot 4 \\cdots (n-4)) \\\\\n& \\leqslant 2(n-1)^{4}\\left(\\frac{3+\\cdots+(n-4)}{n-6}\\right)^{n-6} = 2(n-1)^{4}\\left(\\frac{n-1}{2}\\right)^{n-6} < \\left(\\frac{n-1}{2}\\right)^{n-1}\n\\end{aligned}\n$$\nthus $a, b, c < (n-1)/2$.\n\nFor every prime $p$ and integer $m \\neq 0$, let $\\nu_{p}(m)$ denote the $p$-adic valuation of $m$; that is, the greatest non-negative integer $k$ for which $p^{k}$ divides $m$. Legendre's formula states that\n$$\n\\nu_{p}(n!) = \\sum_{s=1}^{\\infty} \\left\\lfloor \\frac{n}{p^{s}} \\right\\rfloor\n$$\nand a well-known corollary of this formula is that\n$$\n\\begin{equation*}\n\\nu_{p}(n!) < \\sum_{s=1}^{\\infty} \\frac{n}{p^{s}} = \\frac{n}{p-1} . \\tag{$\\ominus$}\n\\end{equation*}\n$$\nIf $n$ is odd then $a^{n-1}, b^{n-1}, c^{n-1}$ are squares, and by considering them modulo $4$ we conclude that $a, b$ and $c$ must be even. Hence, $2^{n-1} \\mid n!$ but that is impossible for odd $n$ because $\\nu_{2}(n!) = \\nu_{2}((n-1)!) < n-1$ by $(\\ominus)$.\n\nFrom now on we assume that $n$ is even. If all three numbers $a+b, b+c, c+a$ are powers of $2$ then $a, b, c$ have the same parity. If they all are odd, then $n! = a^{n-1} + b^{n-1} + c^{n-1}$ is also odd which is absurd. If all $a, b, c$ are divisible by $4$, this contradicts $\\nu_{2}(n!) \\leqslant n-1$. If, say, $a$ is not divisible by $4$, then $2a = (a+b) + (a+c) - (b+c)$ is not divisible by $8$, and since all $a+b, b+c, c+a$ are powers of $2$, we get that one of these sums equals $4$, so two of the numbers of $a, b, c$ are equal to $2$. Say, $a = b = 2$, then $c = 2^{r} - 2$ and, since $c \\mid n!$, we must have $c \\mid a^{n-1} + b^{n-1} = 2^{n}$ implying $r = 2$, and so $c = 2$, which is impossible because $n! \\equiv 0 \\not\\equiv 3 \\cdot 2^{n-1} (\\bmod 5)$.\n\nSo now we assume that the sum of two numbers among $a, b, c$, say $a+b$, is not a power of $2$, so it is divisible by some odd prime $p$. Then $p \\leqslant a+b < n$ and so $c^{n-1} = n! - (a^{n-1} + b^{n-1})$ is divisible by $p$. If $p$ divides $a$ and $b$, we get $p^{n-1} \\mid n!$, contradicting $(\\ominus)$. Next, using $(\\ominus)$ and the Lifting the Exponent Lemma we get\n$$\n\\begin{equation*}\n\\nu_{p}(1) + \\nu_{p}(2) + \\cdots + \\nu_{p}(n) = \\nu_{p}(n!) = \\nu_{p}(n! - c^{n-1}) = \\nu_{p}(a^{n-1} + b^{n-1}) = \\nu_{p}(a+b) + \\nu_{p}(n-1) . \\tag{$\\diamond$}\n\\end{equation*}\n$$\nIn view of $(\\diamond)$, no number of $1, 2, \\ldots, n$ can be divisible by $p$, except $a+b$ and $n-1 > a+b$. On the other hand, $p \\mid c$ implies that $p < n/2$ and so there must be at least two such numbers. Hence, there are two multiples of $p$ among $1, 2, \\ldots, n$, namely $a+b = p$ and $n-1 = 2p$. But this is another contradiction because $n-1$ is odd. This final contradiction shows that there is no solution of the equation for $n > 100$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32127,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all integers $n \\geqslant 2$ with the following property: every $n$ pairwise distinct integers whose sum is not divisible by $n$ can be arranged in some order $a_{1}, a_{2}, \\ldots, a_{n}$ so that $n$ divides $1 \\cdot a_{1}+2 \\cdot a_{2}+\\cdots+n \\cdot a_{n}$.",
    "options": [],
    "answer": "All odd integers and all powers of 2",
    "solution": "Answer: All odd integers and all powers of $2$.\n\nIf $n=2^{k} a$, where $a \\geqslant 3$ is odd and $k$ is a positive integer, we can consider a set containing the number $2^{k}+1$ and $n-1$ numbers congruent to $1$ modulo $n$. The sum of these numbers is congruent to $2^{k}$ modulo $n$ and therefore is not divisible by $n$; for any permutation $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$ of these numbers\n$$\n1 \\cdot a_{1}+2 \\cdot a_{2}+\\cdots+n \\cdot a_{n} \\equiv 1+\\cdots+n \\equiv 2^{k-1} a\\left(2^{k} a+1\\right) \\not \\equiv 0 \\quad\\left(\\bmod 2^{k}\\right)\n$$\nand a fortiori $1 \\cdot a_{1}+2 \\cdot a_{2}+\\cdots+n \\cdot a_{n}$ is not divisible by $n$.\n\nFrom now on, we suppose that $n$ is either odd or a power of $2$. Let $S$ be the given set of integers, and $s$ be the sum of elements of $S$.\n\n**Lemma 1.** If there is a permutation $\\left(a_{i}\\right)$ of $S$ such that $(n, s)$ divides $\\sum_{i=1}^{n} i a_{i}$, then there is a permutation $\\left(b_{i}\\right)$ of $S$ such that $n$ divides $\\sum_{i=1}^{n} i b_{i}$.\n\nProof. Let $r=\\sum_{i=1}^{n} i a_{i}$. Consider the permutation $\\left(b_{i}\\right)$ defined by $b_{i}=a_{i+x}$, where $a_{j+n}=a_{j}$. For this permutation, we have\n$$\n\\sum_{i=1}^{n} i b_{i}=\\sum_{i=1}^{n} i a_{i+x} \\equiv \\sum_{i=1}^{n}(i-x) a_{i} \\equiv r-s x \\quad(\\bmod n)\n$$\nSince $(n, s)$ divides $r$, the congruence $r-s x \\equiv 0(\\bmod n)$ admits a solution.\n\n**Lemma 2.** Every set $T$ of $k m$ integers, $m>1$, can be partitioned into $m$ sets of $k$ integers so that in every set either the sum of elements is not divisible by $k$ or all the elements leave the same remainder upon division by $k$.\n\nProof. The base case, $m=2$. If $T$ contains $k$ elements leaving the same remainder upon division by $k$, we form one subset $A$ of these elements; the remaining elements form a subset $B$. If $k$ does not divide the sum of all elements of $B$, we are done. Otherwise it is enough to exchange any element of $A$ with any element of $B$ not congruent to it modulo $k$, thus making sums of both $A$ and $B$ not divisible by $k$. This cannot be done only when all the elements of $T$ are congruent modulo $k$; in this case any partition will do.\n\nIf no $k$ elements of $T$ have the same residue modulo $k$, there are three elements $a, b, c \\in T$ leaving pairwise distinct remainders upon division by $k$. Let $t$ be the sum of elements of $T$. It suffices to find $A \\subset T$ such that $|A|=k$ and $\\sum_{x \\in A} x \\not \\equiv 0, t(\\bmod k)$: then neither the sum of elements of $A$ nor the sum of elements of $B=T \\backslash A$ is divisible by $k$. Consider $U^{\\prime} \\subset T \\backslash\\{a, b, c\\}$ with $\\left|U^{\\prime}\\right|=k-1$. The sums of elements of three sets $U^{\\prime} \\cup\\{a\\}, U^{\\prime} \\cup\\{b\\}, U^{\\prime} \\cup\\{c\\}$ leave three different remainders upon division by $k$, and at least one of them is not congruent either to $0$ or to $t$.\n\nNow let $m>2$. If $T$ contains $k$ elements leaving the same remainder upon division by $k$, we form one subset $A$ of these elements and apply the inductive hypothesis to the remaining $k(m-1)$ elements. Otherwise, we choose any $U \\subset T,|U|=k-1$. Since all the remaining elements cannot be congruent modulo $k$, there is $a \\in T \\backslash U$ such that $a \\not \\equiv-\\sum_{x \\in U} x(\\bmod k)$. Now we can take $A=U \\cup\\{a\\}$ and apply the inductive hypothesis to $T \\backslash A$.\n\nNow we are ready to prove the statement of the problem for all odd $n$ and $n=2^{k}$. The proof is by induction.\n\nIf $n$ is prime, the statement follows immediately from Lemma 1, since in this case $(n, s)=1$. Turning to the general case, we can find prime $p$ and an integer $t$ such that $p^{t} \\mid n$ and $p^{t} \\nmid s$. By Lemma 2, we can partition $S$ into $p$ sets of $\\frac{n}{p}=k$ elements so that in every set either the sum of numbers is not divisible by $k$ or all numbers have the same residue modulo $k$.\n\nFor sets in the first category, by the inductive hypothesis there is a permutation $\\left(a_{i}\\right)$ such that $k \\mid \\sum_{i=1}^{k} i a_{i}$.\n\nIf $n$ (and therefore $k$) is odd, then for each permutation $\\left(b_{i}\\right)$ of a set in the second category we have\n$$\n\\sum_{i=1}^{k} i b_{i} \\equiv b_{1} \\frac{k(k+1)}{2} \\equiv 0 \\quad(\\bmod k)\n$$\nBy combining such permutation for all sets of the partition, we get a permutation $\\left(c_{i}\\right)$ of $S$ such that $k \\mid \\sum_{i=1}^{n} i c_{i}$. Since this sum is divisible by $k$, and $k$ is divisible by $(n, s)$, we are done by Lemma 1.\n\nIf $n=2^{s}$, we have $p=2$ and $k=2^{s-1}$. Then for each of the subsets there is a permutation $\\left(a_{1}, \\ldots, a_{k}\\right)$ such that $\\sum_{i=1}^{k} i a_{i}$ is divisible by $2^{s-2}=\\frac{k}{2}$: if the subset belongs to the first category, the expression is divisible even by $k$, and if it belongs to the second one,\n$$\n\\sum_{i=1}^{k} i a_{i} \\equiv a_{1} \\frac{k(k+1)}{2} \\equiv 0\\left(\\bmod \\frac{k}{2}\\right)\n$$\nNow the numbers of each permutation should be multiplied by all the odd or all the even numbers not exceeding $n$ in increasing order so that the resulting sums are divisible by $k$:\n$$\n\\sum_{i=1}^{k}(2 i-1) a_{i} \\equiv \\sum_{i=1}^{k} 2 i a_{i} \\equiv 2 \\sum_{i=1}^{k} i a_{i} \\equiv 0 \\quad(\\bmod k)\n$$\nCombining these two sums, we again get a permutation $\\left(c_{i}\\right)$ of $S$ such that $k \\mid \\sum_{i=1}^{n} i c_{i}$, and finish the case by applying Lemma 1.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32128,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, a_{3}, \\ldots$ be an infinite sequence of positive integers such that $a_{n+2m}$ divides $a_{n} + a_{n+m}$ for all positive integers $n$ and $m$. Prove that this sequence is eventually periodic, i.e. there exist positive integers $N$ and $d$ such that $a_{n} = a_{n+d}$ for all $n > N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will make repeated use of the following simple observation:\n\nLemma 1. If a positive integer $d$ divides $a_{n}$ and $a_{n-m}$ for some $m$ and $n > 2m$, it also divides $a_{n-2m}$. If $d$ divides $a_{n}$ and $a_{n-2m}$, it also divides $a_{n-m}$.\n\nProof. Both parts are obvious since $a_{n}$ divides $a_{n-2m} + a_{n-m}$.\n\nClaim. The sequence $(a_{n})$ is bounded.\n\nProof. Suppose the contrary. Then there exist infinitely many indices $n$ such that $a_{n}$ is greater than each of the previous terms $a_{1}, a_{2}, \\ldots, a_{n-1}$. Let $a_{n} = k$ be such a term, $n > 10$. For each $s < \\frac{n}{2}$ the number $a_{n} = k$ divides $a_{n-s} + a_{n-2s} < 2k$, therefore\n$$\na_{n-s} + a_{n-2s} = k\n$$\nIn particular,\n$$\na_{n} = a_{n-1} + a_{n-2} = a_{n-2} + a_{n-4} = a_{n-4} + a_{n-8}\n$$\nthat is, $a_{n-1} = a_{n-4}$ and $a_{n-2} = a_{n-8}$. It follows from Lemma 1 that $a_{n-1}$ divides $a_{n-1-3s}$ for $3s < n-1$ and $a_{n-2}$ divides $a_{n-2-6s}$ for $6s < n-2$. Since at least one of the numbers $a_{n-1}$ and $a_{n-2}$ is at least $a_{n}/2$, so is some $a_{i}$ with $i \\leq 6$. However, $a_{n}$ can be arbitrarily large, a contradiction.\n\nSince $(a_{n})$ is bounded, there exist only finitely many $i$ for which $a_{i}$ appears in the sequence finitely many times. In other words, there exists $N$ such that if $a_{i} = t$ and $i > N$, then $a_{j} = t$ for infinitely many $j$.\n\nClearly the sequence $(a_{n+N})_{n > 0}$ satisfies the divisibility condition, and it is enough to prove that this sequence is eventually periodic. Thus truncating the sequence if necessary, we can assume that each number appears infinitely many times in the sequence. Let $k$ be the maximum number appearing in the sequence.\n\nLemma 2. If a positive integer $d$ divides $a_{n}$ for some $n$, then the numbers $i$ such that $d$ divides $a_{i}$ form an arithmetical progression with an odd difference.\n\nProof. Let $i_{1} < i_{2} < i_{3} < \\ldots$ be all the indices $i$ such that $d$ divides $a_{i}$. If $i_{s} + i_{s+1}$ is even, it follows from Lemma 1 that $d$ also divides $a_{\\frac{i_{s} + i_{s+1}}{2}}$, impossible since $i_{s} < \\frac{i_{s} + i_{s+1}}{2} < i_{s+1}$. Thus $i_{s}$ and $i_{s+1}$ are always of different parity, and therefore $i_{s} + i_{s+2}$ is even. Applying Lemma 1 again, we see that $d$ divides $a_{\\frac{i_{s} + i_{s+2}}{2}}$, hence $\\frac{i_{s} + i_{s+2}}{2} = i_{s+1}$.\n\nWe are ready now to solve the problem.\n\nThe number of positive divisors of all terms of the progression is finite. Let $d_{s}$ be the difference of the progression corresponding to $s$, that is, $s$ divides $a_{n}$ if and only if it divides $a_{n + t d_{s}}$ for any positive integer $t$. Let $D$ be the product of all $d_{s}$. Then each $s$ dividing a term of the progression divides $a_{n}$ if and only if it divides $a_{n+D}$. This means that the sets of divisors of $a_{n}$ and $a_{n+D}$ coincide, and $a_{n+D} = a_{n}$. Thus $D$ is a period of the sequence.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32129,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a polynomial $P(x)$ with integer coefficients let $P^{1}(x)=P(x)$ and $P^{k+1}(x)= P\\left(P^{k}(x)\\right)$ for $k \\geqslant 1$. Find all positive integers $n$ for which there exists a polynomial $P(x)$ with integer coefficients such that for every integer $m \\geqslant 1$, the numbers $P^{m}(1), \\ldots, P^{m}(n)$ leave exactly $\\left\\lceil n / 2^{m}\\right\\rceil$ distinct remainders when divided by $n$.",
    "options": [],
    "answer": "All powers of 2 and all primes",
    "solution": "Answer: All powers of 2 and all primes.\n\nDenote the set of residues modulo $\\ell$ by $\\mathbb{Z}_{\\ell}$. Observe that $P$ can be regarded as a function $\\mathbb{Z}_{\\ell} \\rightarrow \\mathbb{Z}_{\\ell}$ for any positive integer $\\ell$. Denote the cardinality of the set $P^{m}\\left(\\mathbb{Z}_{\\ell}\\right)$ by $f_{m, \\ell}$. Note that $f_{m, n}=\\left\\lceil n / 2^{m}\\right\\rceil$ for all $m \\geqslant 1$ if and only if $f_{m+1, n}=\\left\\lceil f_{m, n} / 2\\right\\rceil$ for all $m \\geqslant 0$.\n\nPart 1. The required polynomial exists when $n$ is a power of 2 or a prime.\nIf $n$ is a power of 2, set $P(x)=2x$.\nIf $n=p$ is an odd prime, every function $f: \\mathbb{Z}_{p} \\rightarrow \\mathbb{Z}_{p}$ coincides with some polynomial with integer coefficients. So we can pick the function that sends $x \\in \\{0,1, \\ldots, p-1\\}$ to $\\lfloor x / 2\\rfloor$.\n\nPart 2. The required polynomial does not exist when $n$ is not a prime power.\nLet $n=ab$ where $a, b>1$ and $\\operatorname{gcd}(a, b)=1$. Note that, since $\\operatorname{gcd}(a, b)=1$,\n$$\nf_{m, ab}=f_{m, a} f_{m, b}\n$$\nby the Chinese remainder theorem. Also, note that, if $f_{m, \\ell}=f_{m+1, \\ell}$, then $P$ permutes the image of $P^{m}$ on $\\mathbb{Z}_{\\ell}$, and therefore $f_{s, \\ell}=f_{m, \\ell}$ for all $s>m$. So, as $f_{m, ab}=1$ for sufficiently large $m$, we have for each $m$\n$$\nf_{m, a}>f_{m+1, a} \\quad \\text{or} \\quad f_{m, a}=1, \\quad f_{m, b}>f_{m+1, b} \\quad \\text{or} \\quad f_{m, b}=1.\n$$\nChoose the smallest $m$ such that $f_{m+1, a}=1$ or $f_{m+1, b}=1$. Without loss of generality assume that $f_{m+1, a}=1$. Then $f_{m+1, ab}=f_{m+1, b}<f_{m, b} \\leqslant f_{m, ab} / 2 \\leqslant f_{m+1, ab}$, a contradiction.\n\nPart 3. The required polynomial does not exist when $n$ is an odd prime power that is not a prime.\nLet $n=p^{k}$, where $p \\geqslant 3$ is prime and $k \\geqslant 2$. For $r \\in \\mathbb{Z}_{p}$ let $S_{r}$ denote the subset of $\\mathbb{Z}_{p^{k}}$ consisting of numbers congruent to $r$ modulo $p$. We denote the cardinality of a set $S$ by $|S|$.\n\nClaim. For any residue $r$ modulo $p$, either $|P(S_{r})|=p^{k-1}$ or $|P(S_{r})| \\leqslant p^{k-2}$.\n\nProof. Recall that $P(r+h)=P(r)+h P^{\\prime}(r)+h^{2} Q(r, h)$, where $Q$ is an integer polynomial.\nIf $p \\mid P^{\\prime}(r)$, then $P(r+ps) \\equiv P(r)\\pmod{p^{2}}$, hence all elements of $P(S_{r})$ are congruent modulo $p^{2}$. So in this case $|P(S_{r})| \\leqslant p^{k-2}$.\nNow we show that $p \\nmid P^{\\prime}(r)$ implies $|P(S_{r})|=p^{k-1}$ for all $k$.\nSuppose the contrary: $|P(S_{r})|<p^{k-1}$ for some $k>1$. Let us choose the smallest $k$ for which this is so. To each residue in $P(S_{r})$ we assign its residue modulo $p^{k-1}$; denote the resulting set by $\\bar{P}(S, r)$. We have $|\\bar{P}(S, r)|=p^{k-2}$ by virtue of minimality of $k$. Then $|P(S_{r})|<p^{k-1}=p \\cdot|\\bar{P}(S, r)|$, that is, there is $u=P(x) \\in P(S_{r})(x \\equiv r\\pmod{p})$ and $t \\not\\equiv 0 \\pmod{p}$ such that $u+p^{k-1} t \\notin P(S_{r})$.\nNote that $P(x+p^{k-1} s) \\equiv u+p^{k-1} s P^{\\prime}(x)\\pmod{p^{k}}$. Since $P(x+p^{k-1} s) \\not\\equiv u+p^{k-1} t \\pmod{p^{k}}$, the congruence $p^{k-1} s P^{\\prime}(x) \\equiv p^{k-1} t\\pmod{p^{k}}$ has no solutions. So the congruence $s P^{\\prime}(x) \\equiv t\\pmod{p}$ has no solutions, which contradicts $p \\nmid P^{\\prime}(r)$. $\\square$\n\nSince the image of $P^{m}$ consists of one element for sufficiently large $m$, we can take the smallest $m$ such that $|P^{m-1}(S_{r})|=p^{k-1}$ for some $r \\in \\mathbb{Z}_{p}$, but $|P^{m}(S_{q})| \\leqslant p^{k-2}$ for all $q \\in \\mathbb{Z}_{p}$.\nFrom now on, we fix $m$ and $r$.\nSince the image of $P^{m-1}(\\mathbb{Z}_{p^{k}}) \\backslash P^{m-1}(S_{r})$ under $P$ contains $P^{m}(\\mathbb{Z}_{p^{k}}) \\backslash P^{m}(S_{r})$, we have\n$$\na:=|P^{m}(\\mathbb{Z}_{p^{k}}) \\backslash P^{m}(S_{r})| \\leqslant|P^{m-1}(\\mathbb{Z}_{p^{k}}) \\backslash P^{m-1}(S_{r})|\n$$\nthus\n$$\na+p^{k-1} \\leqslant f_{m-1, p^{k}} \\leqslant 2 f_{m, p^{k}} \\leqslant 2 p^{k-2}+2 a\n$$\nso\n$$\n(p-2) p^{k-2} \\leqslant a\n$$\nSince $f_{i, p}=1$ for sufficiently large $i$, there is exactly one $t \\in \\mathbb{Z}_{p}$ such that $P(t) \\equiv t\\pmod{p}$. Moreover, as $i$ increases, the cardinality of the set $\\{s \\in \\mathbb{Z}_{p} \\mid P^{i}(s) \\equiv t\\pmod{p}\\}$ increases (strictly), until it reaches the value $p$. So either\n$$\n|\\{s \\in \\mathbb{Z}_{p} \\mid P^{m-1}(s) \\equiv t \\quad\\pmod{p}\\}|=p \\quad \\text{or} \\quad|\\{s \\in \\mathbb{Z}_{p} \\mid P^{m-1}(s) \\equiv t \\quad\\pmod{p}\\}| \\geqslant m\n$$\nTherefore, either $f_{m-1, p}=1$ or there exists a subset $X \\subset \\mathbb{Z}_{p}$ of cardinality at least $m$ such that $P^{m-1}(x) \\equiv t\\pmod{p}$ for all $x \\in X$.\nIn the first case $|P^{m-1}(\\mathbb{Z}_{p^{k}})| \\leqslant p^{k-1}=|P^{m-1}(S_{r})|$, so $a=0$, a contradiction.\nIn the second case let $Y$ be the set of all elements of $\\mathbb{Z}_{p^{k}}$ congruent to some element of $X$ modulo $p$. Let $Z=\\mathbb{Z}_{p^{k}} \\backslash Y$. Then $P^{m-1}(Y) \\subset S_{t}, P(S_{t}) \\subsetneq S_{t}$, and $Z=\\bigcup_{i \\in \\mathbb{Z}_{p} \\backslash X} S_{i}$, so\n$$\n|P^{m}(Y)| \\leqslant|P(S_{t})| \\leqslant p^{k-2} \\quad \\text{and} \\quad|P^{m}(Z)| \\leqslant|\\mathbb{Z}_{p} \\backslash X| \\cdot p^{k-2} \\leqslant(p-m) p^{k-2}\n$$\nHence,\n$$\n(p-2) p^{k-2} \\leqslant a<|P^{m}(\\mathbb{Z}_{p^{k}})| \\leqslant|P^{m}(Y)|+|P^{m}(Z)| \\leqslant(p-m+1) p^{k-2}\n$$\nand $m<3$. Then $|P^{2}(S_{q})| \\leqslant p^{k-2}$ for all $q \\in \\mathbb{Z}_{p}$, so\n$$\np^{k} / 4 \\leqslant|P^{2}(\\mathbb{Z}_{p^{k}})| \\leqslant p^{k-1}\n$$\nwhich is impossible for $p \\geqslant 5$. It remains to consider the case $p=3$.\nAs before, let $t$ be the only residue modulo 3 such that $P(t) \\equiv t\\pmod{3}$.\nIf $3 \\nmid P^{\\prime}(t)$, then $P(S_{t})=S_{t}$ by the proof of the Claim above, which is impossible.\nSo $3 \\mid P^{\\prime}(t)$. By substituting $h=3^{i} s$ into the formula $P(t+h)=P(t)+h P^{\\prime}(t)+h^{2} Q(t, h)$, we obtain $P\\left(t+3^{i} s\\right) \\equiv P(t)\\pmod{3^{i+1}}$. Using induction on $i$ we see that all elements of $P^{i}(S_{t})$ are congruent modulo $3^{i+1}$. Thus, $|P^{k-1}(S_{t})|=1$.\nNote that $f_{1,3} \\leqslant 2$ and $f_{2,3} \\leqslant 1$, so $P^{2}(\\mathbb{Z}_{3^{k}}) \\subset S_{t}$. Therefore, $|P^{k+1}(\\mathbb{Z}_{3^{k}})| \\leqslant|P^{k-1}(S_{t})|=1$. It follows that $3^{k} \\leqslant 2^{k+1}$, which is impossible for $k \\geqslant 2$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32130,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\geqslant 2$ be an integer. Find the smallest integer $n \\geqslant k+1$ with the property that there exists a set of $n$ distinct real numbers such that each of its elements can be written as a sum of $k$ other distinct elements of the set.",
    "options": [],
    "answer": "k+4",
    "solution": "First we show that $n \\geqslant k+4$. Suppose that there exists such a set with $n$ numbers and denote them by $a_{1} < a_{2} < \\cdots < a_{n}$.\nNote that in order to express $a_{1}$ as a sum of $k$ distinct elements of the set, we must have $a_{1} \\geqslant a_{2} + \\cdots + a_{k+1}$ and, similarly for $a_{n}$, we must have $a_{n-k} + \\cdots + a_{n-1} \\geqslant a_{n}$. We also know that $n \\geqslant k+1$.\nIf $n = k+1$ then we have $a_{1} \\geqslant a_{2} + \\cdots + a_{k+1} > a_{1} + \\cdots + a_{k} \\geqslant a_{k+1}$, which gives a contradiction.\nIf $n = k+2$ then we have $a_{1} \\geqslant a_{2} + \\cdots + a_{k+1} \\geqslant a_{k+2}$, that again gives a contradiction.\nIf $n = k+3$ then we have $a_{1} \\geqslant a_{2} + \\cdots + a_{k+1}$ and $a_{3} + \\cdots + a_{k+2} \\geqslant a_{k+3}$. Adding the two inequalities we get $a_{1} + a_{k+2} \\geqslant a_{2} + a_{k+3}$, again a contradiction.\nIt remains to give an example of a set with $k+4$ elements satisfying the condition of the problem. We start with the case when $k = 2l$ and $l \\geqslant 1$. In that case, denote by $A_{i} = \\{-i, i\\}$ and take the set $A_{1} \\cup \\cdots \\cup A_{l+2}$, which has exactly $k+4 = 2l+4$ elements. We are left to show that this set satisfies the required condition.\nNote that if a number $i$ can be expressed in the desired way, then so can $-i$ by negating the expression. Therefore, we consider only $1 \\leqslant i \\leqslant l+2$.\nIf $i < l+2$, we sum the numbers from some $l-1$ sets $A_{j}$ with $j \\neq 1, i+1$, and the numbers $i+1$ and $-1$.\nFor $i = l+2$, we sum the numbers from some $l-1$ sets $A_{j}$ with $j \\neq 1, l+1$, and the numbers $l+1$ and $1$.\nIt remains to give a construction for odd $k = 2l+1$ with $l \\geqslant 1$ (since $k \\geqslant 2$). To that end, we modify the construction for $k = 2l$ by adding $0$ to the previous set.\nThis is a valid set as $0$ can be added to each constructed expression, and $0$ can be expressed as follows: take the numbers $1, 2, -3$ and all the numbers from the remaining $l-1$ sets $A_{4}, A_{5}, \\cdots, A_{l+2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32131,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}_{>0}$ be the set of positive real numbers. Find all functions $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ such that, for every $x \\in \\mathbb{R}_{>0}$, there exists a unique $y \\in \\mathbb{R}_{>0}$ satisfying\n$$\nx f(y)+y f(x) \\leqslant 2 .\n$$",
    "options": [],
    "answer": "f(x) = 1/x for all x > 0",
    "solution": "First we prove that the function $f(x)=1 / x$ satisfies the condition of the problem statement. The AM-GM inequality gives\n$$\n\\frac{x}{y}+\\frac{y}{x} \\geqslant 2\n$$\nfor every $x, y>0$, with equality if and only if $x=y$. This means that, for every $x>0$, there exists a unique $y>0$ such that\n$$\n\\frac{x}{y}+\\frac{y}{x} \\leqslant 2\n$$\nnamely $y=x$.\n\nLet now $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ be a function that satisfies the condition of the problem statement. We say that a pair of positive real numbers $(x, y)$ is good if $x f(y)+y f(x) \\leqslant 2$. Observe that if $(x, y)$ is good, then so is $(y, x)$.\n\nLemma 1.0. If $(x, y)$ is good, then $x=y$.\n\nProof. Assume that there exist positive real numbers $x \\neq y$ such that $(x, y)$ is good. The uniqueness assumption says that $y$ is the unique positive real number such that $(x, y)$ is good. In particular, $(x, x)$ is not a good pair. This means that\n$$\nx f(x)+x f(x)>2\n$$\nand thus $x f(x)>1$. Similarly, $(y, x)$ is a good pair, so $(y, y)$ is not a good pair, which implies $y f(y)>1$. We apply the AM-GM inequality to obtain\n$$\nx f(y)+y f(x) \\geqslant 2 \\sqrt{x f(y) \\cdot y f(x)}=2 \\sqrt{x f(x) \\cdot y f(y)}>2 .\n$$\nThis is a contradiction, since $(x, y)$ is a good pair.\n\nBy assumption, for any $x>0$, there always exists a good pair containing $x$, however Lemma 1 implies that the only good pair that can contain $x$ is $(x, x)$, so\n$$\nx f(x) \\leqslant 1 \\quad \\Longleftrightarrow \\quad f(x) \\leqslant \\frac{1}{x},\n$$\nfor every $x>0$.\n\n\nSolution 1.1. We give an alternative way to prove that $f(x)=1 / x$ assuming $f(x) \\leqslant 1 / x$ for every $x>0$.\nIndeed, if $f(x)<1 / x$ then for every $a>0$ with $f(x)<1 / a<1 / x$ (and there are at least two of them), we have\n$$\na f(x)+x f(a)<1+\\frac{x}{a}<2 .\n$$\nHence $(x, a)$ is a good pair for every such $a$, a contradiction. We conclude that $f(x)=1 / x$.\n\n\nSolution 1.2. We can also conclude from Lemma 1 and $f(x) \\leqslant 1 / x$ as follows.\nLemma 2. The function $f$ is decreasing.\nProof. Let $y>x>0$. Lemma 1 says that $(x, y)$ is not a good pair, but $(y, y)$ is. Hence\n$$\nx f(y)+y f(x)>2 \\geqslant 2 y f(y)>y f(y)+x f(y),\n$$\nwhere we used $y>x$ (and $f(y)>0$ ) in the last inequality. This implies that $f(x)>f(y)$, showing that $f$ is decreasing.\n\nWe now prove that $f(x)=1 / x$ for all $x$. Fix a value of $x$ and note that for $y>x$ we must have $x f(x)+y f(x)>x f(y)+y f(x)>2$ (using that $f$ is decreasing for the first step), hence $f(x)>\\frac{2}{x+y}$. The last inequality is true for every $y>x>0$. If we fix $x$ and look for the supremum of the expression $\\frac{2}{x+y}$ over all $y>x$, we get\n$$\nf(x) \\geqslant \\frac{2}{x+x}=\\frac{1}{x} .\n$$\nSince we already know that $f(x) \\leqslant 1 / x$, we conclude that $f(x)=1 / x$.\nAs in the first solution, we note that $f(x)=1 / x$ is a solution, and we set out to prove that it is the only one. We write $g(x)$ for the unique positive real number such that $(x, g(x))$ is a good pair. In this solution, we prove Lemma 2 without assuming Lemma 1.\n\nLemma 2. The function $f$ is decreasing.\nProof. Consider $x<y$. It holds that $y f(g(y))+g(y) f(y) \\leqslant 2$. Moreover, because $y$ is the only positive real number such that $(g(y), y)$ is a good pair and $x \\neq y$, we have $x f(g(y))+g(y) f(x)>2$. Combining these two inequalities yields\n$$\nx f(g(y))+g(y) f(x)>2 \\geqslant y f(g(y))+g(y) f(y),\n$$\nor $f(g(y))(x-y)>g(y)(f(y)-f(x))$. Because $g(y)$ and $f(g(y))$ are both positive while $x-y$ is negative, it follows that $f(y)<f(x)$, showing that $f$ is decreasing.\n\nWe now prove Lemma 1 using Lemma 2. Suppose that $x \\neq y$ but $x f(y)+y f(x) \\leqslant 2$. As in the first solution, we get $x f(x)+x f(x)>2$ and $y f(y)+y f(y)>2$, which implies $x f(x)+y f(y)>2$. Now\n$$\nx f(x)+y f(y)>2 \\geqslant x f(y)+y f(x)\n$$\nimplies $(x-y)(f(x)-f(y))>0$, which contradicts the fact that $f$ is decreasing. So $y=x$ is the unique $y$ such that $(x, y)$ is a good pair, and in particular we have $f(x) \\leqslant 1 / x$.\n\nWe can now conclude the proof as in any of the Solutions 1.x.\nAs in the other solutions we verify that the function $f(x)=1 / x$ is a solution. We first want to prove the following lemma:\n\nLemma 3. For all $x \\in \\mathbb{R}_{>0}$ we actually have $x f(g(x))+g(x) f(x)=2$ (that is: the inequality is actually an equality).\n\nProof. We proceed by contradiction: Assume there exists some number $x>0$ such that for $y=g(x)$ we have $x f(y)+y f(x)<2$. Then for any $0<\\epsilon<\\frac{2-x f(y)-y f(x)}{2 f(x)}$ we have, by uniqueness of $y$, that $x f(y+\\epsilon)+(y+\\epsilon) f(x)>2$. Therefore\n$$\n\\begin{align*}\nf(y+\\epsilon) & >\\frac{2-(y+\\epsilon) f(x)}{x}=\\frac{2-y f(x)-\\epsilon f(x)}{x} \\\\\n& >\\frac{2-y f(x)-\\frac{2-x f(y)-y f(x)}{2}}{x} \\\\\n& =\\frac{2-x f(y)-y f(x)}{2 x}+f(y)>f(y) \\tag{1}\n\\end{align*}\n$$\nFurthermore, for every such $\\epsilon$ we have $g(y+\\epsilon) f(y+\\epsilon)+(y+\\epsilon) f(g(y+\\epsilon)) \\leqslant 2$ and $g(y+\\epsilon) f(y)+y f(g(y+\\epsilon))>2$ (since $y \\neq y+\\epsilon=g(g(y+\\epsilon)))$. This gives us the two inequalities\n$$\nf(g(y+\\epsilon)) \\leqslant \\frac{2-g(y+\\epsilon) f(y+\\epsilon)}{y+\\epsilon} \\quad \\text{ and } \\quad f(g(y+\\epsilon))>\\frac{2-g(y+\\epsilon) f(y)}{y} .\n$$\nCombining these two inequalities and rearranging the terms leads to the inequality\n$$\n2 \\epsilon<g(y+\\epsilon)[(y+\\epsilon) f(y)-y f(y+\\epsilon)] .\n$$\nMoreover combining with the inequality (1) we obtain\n$2 \\epsilon<g(y+\\epsilon)\\left[(y+\\epsilon) f(y)-y \\frac{2-x f(y)-y f(x)}{2 x}+f(y)\\right]=g(y+\\epsilon)\\left[\\epsilon f(y)-y \\frac{2-x f(y)-y f(x)}{2 x}\\right]$.\nWe now reach the desired contradiction, since for $\\epsilon$ sufficiently small we have that the left hand side is positive while the right hand side is negative.\n\nWith this lemma it then follows that for all $x, y \\in \\mathbb{R}_{>0}$ we have\n$$\nx f(y)+y f(x) \\geqslant 2\n$$\nsince for $y=g(x)$ we have equality and by uniqueness for $y \\neq g(x)$ the inequality is strict.\n\nIn particular for every $x \\in \\mathbb{R}_{>0}$ and for $y=x$ we have $2 x f(x) \\geqslant 2$, or equivalently $f(x) \\geqslant 1 / x$ for all $x \\in \\mathbb{R}_{>0}$. With this inequality we obtain for all $x \\in \\mathbb{R}_{>0}$\n$$\n2 \\geqslant x f(g(x))+g(x) f(x) \\geqslant \\frac{x}{g(x)}+\\frac{g(x)}{x} \\geqslant 2\n$$\nwhere the first inequality comes from the problem statement. Consequently each of these inequalities must actually be an equality, and in particular we obtain $f(x)=1 / x$ for all $x \\in \\mathbb{R}_{>0}$.\nAgain, let us prove that $f(x)=1 / x$ is the only solution. Let again $g(x)$ be the unique positive real number such that $(x, g(x))$ is a good pair.\n\nLemma 4. The function $f$ is strictly convex.\nProof. Consider the function $q_{s}(x)=f(x)+s x$ for some real number $s$. If $f$ is not strictly convex, then there exist $u<v$ and $t \\in(0,1)$ such that\n$$\nf(t u+(1-t) v) \\geqslant t f(u)+(1-t) f(v) .\n$$\nHence\n$$\n\\begin{aligned}\nq_{s}(t u+(1-t) v) & \\geqslant t f(u)+(1-t) f(v)+s(t u+(1-t) v) \\\\\n& =t q_{s}(u)+(1-t) q_{s}(v)\n\\end{aligned}\n$$\nLet $w=t u+(1-t) v$ and consider the case $s=f(g(w)) / g(w)$. For that particular choice of $s$, the function $q_{s}(x)$ has a unique minimum at $x=w$. However, since $q_{s}(w) \\geqslant t q_{s}(u)+(1-t) q_{s}(v)$, it must hold $q_{s}(u) \\leqslant q_{s}(w)$ or $q_{s}(v) \\leqslant q_{s}(w)$, a contradiction.\n\nLemma 5. The function $f$ is continuous.\nProof. Since $f$ is strictly convex and defined on an open interval, it is also continuous.\n\nAs in Solution 1, we can prove that $f(x) \\leqslant 1 / x$. If $f(x)<1 / x$, then we consider the function $h(y)=x f(y)+y f(x)$ which is continuous. Since $h(x)<2$, there exist at least two distinct $z \\neq x$ such that $h(z)<2$ giving that $(x, z)$ is good pair for both values of $z$, a contradiction. We conclude that $f(x)=1 / x$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32132,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$ we denote by $s(n)$ the sum of the digits of $n$. Let $P(x)= x^{n}+a_{n-1} x^{n-1}+\\cdots+a_{1} x+a_{0}$ be a polynomial, where $n \\geqslant 2$ and $a_{i}$ is a positive integer for all $0 \\leqslant i \\leqslant n-1$. Could it be the case that, for all positive integers $k$, $s(k)$ and $s(P(k))$ have the same parity?\n(Belarus)",
    "options": [],
    "answer": "No",
    "solution": "With the notation above, we begin by choosing a positive integer $t$ such that\n$$\n10^{t} > \\max \\left\\{ \\frac{100^{n-1} a_{n-1}}{\\left(10^{\\frac{1}{n-1}} - 9^{\\frac{1}{n-1}}\\right)^{n-1}}, \\frac{a_{n-1}}{9} 10^{n-1}, \\frac{a_{n-1}}{9} (10 a_{n-1})^{n-1}, \\ldots, \\frac{a_{n-1}}{9} (10 a_{0})^{n-1} \\right\\}\n$$\nAs a direct consequence of $10^{t}$ being bigger than the first quantity listed in the above set, we get that the interval\n$$\nI = \\left[ \\left( \\frac{9}{a_{n-1}} 10^{t} \\right)^{\\frac{1}{n-1}}, \\left( \\frac{1}{a_{n-1}} 10^{t+1} \\right)^{\\frac{1}{n-1}} \\right)\n$$\ncontains at least 100 consecutive positive integers.\nLet $X$ be a positive integer in $I$ such that $X$ is congruent to $1 \\bmod 100$. Since $X \\in I$ we have\n$$\n9 \\cdot 10^{t} \\leqslant a_{n-1} X^{n-1} < 10^{t+1}\n$$\nthus the first digit (from the left) of $a_{n-1} X^{n-1}$ must be 9.\nNext, we observe that $a_{n-1} (10 a_{i})^{n-1} < 9 \\cdot 10^{t} \\leqslant a_{n-1} X^{n-1}$, thus $10 a_{i} < X$ for all $i$, which immediately implies that $a_{0} < a_{1} X < \\cdots < a_{n} X^{n}$, and the number of digits of this strictly increasing sequence forms a strictly increasing sequence too. In other words, if $i < j$, the number of digits of $a_{i} X^{i}$ is less than the number of digits of $a_{j} X^{j}$.\nLet $\\alpha$ be the number of digits of $a_{n-1} X^{n-1}$, thus $10^{\\alpha-1} \\leqslant a_{n-1} X^{n-1} < 10^{\\alpha}$. We are now going to look at $P(10^{\\alpha} X)$ and $P(10^{\\alpha-1} X)$ and prove that the sum of their digits has different parities. This will finish the proof since $s(10^{\\alpha} X) = s(10^{\\alpha-1} X) = s(X)$.\nWe have $P(10^{\\alpha} X) = 10^{\\alpha n} X^{n} + a_{n-1} 10^{\\alpha(n-1)} X^{n-1} + \\cdots + a_{0}$, and since $10^{\\alpha(i+1)} > 10^{\\alpha i} a_{n-1} X^{n-1} > 10^{\\alpha i} a_{i} X^{i}$, the terms $a_{i} 10^{\\alpha i} X^{i}$ do not interact when added; in particular, there is no carryover caused by addition. Thus we have $s(P(10^{\\alpha} X)) = s(X^{n}) + s(a_{n-1} X^{n-1}) + \\cdots + s(a_{0})$.\nWe now look at $P(10^{\\alpha-1} X) = 10^{(\\alpha-1) n} X^{n} + a_{n-1} 10^{(\\alpha-1)(n-1)} X^{n-1} + \\cdots + a_{0}$. Firstly, if $i < n-1$, then $a_{n-1} X^{n-1}$ has more digits than $a_{i} X^{i}$ and $a_{n-1} X^{n-1} \\geqslant 10 a_{i} X^{i}$. It now follows that $10^{(\\alpha-1)(i+1)+1} > 10^{(\\alpha-1) i} a_{n-1} X^{n-1} \\geqslant 10^{(\\alpha-1) i+1} a_{i} X^{i}$, thus all terms $10^{(\\alpha-1) i} a_{i} X^{i}$ for $0 \\leqslant i \\leqslant n-1$ come in 'blocks', exactly as in the previous case.\nFinally, $10^{(\\alpha-1) n+1} > 10^{(\\alpha-1)(n-1)} a_{n-1} X^{n-1} \\geqslant 10^{(\\alpha-1) n}$, thus $10^{(\\alpha-1)(n-1)} a_{n-1} X^{n-1}$ has exactly $(\\alpha-1) n+1$ digits, and its first digit is 9, as established above. On the other hand, $10^{(\\alpha-1) n} X^{n}$ has exactly $(\\alpha-1) n$ zeros, followed by 01 (as $X$ is $1 \\bmod 100$). Therefore, when we add the terms, the 9 and 1 turn into 0, the 0 turns into 1, and nothing else is affected.\nPutting everything together, we obtain\n$$\ns(P(10^{\\alpha-1} X)) = s(X^{n}) + s(a_{n-1} X^{n-1}) + \\cdots + s(a_{0}) - 9 = s(P(10^{\\alpha} X)) - 9\n$$\nthus $s(P(10^{\\alpha} X))$ and $s(P(10^{\\alpha-1} X))$ have different parities, as claimed.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32133,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, an $n$-sequence is a sequence $(a_{0}, \\ldots, a_{n})$ of non-negative integers satisfying the following condition: if $i$ and $j$ are non-negative integers with $i+j \\leqslant n$, then $a_{i}+a_{j} \\leqslant n$ and $a_{a_{i}+a_{j}}=a_{i+j}$.\nLet $f(n)$ be the number of $n$-sequences. Prove that there exist positive real numbers $c_{1}, c_{2}$ and $\\lambda$ such that\n$$\nc_{1} \\lambda^{n}<f(n)<c_{2} \\lambda^{n}\n$$\nfor all positive integers $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In order to solve this, we will give a complete classification of $n$-sequences.\nLet $k=\\lfloor n / 2\\rfloor$. We will say that an $n$-sequence is large if $a_{i}>k$ for some $i$, and small if no such $i$ exists. For now we will assume that $(a_{i})$ is not the identity sequence (in other words, that $a_{i} \\neq i$ for some $i$).\n\nLemma 1. If $a_{r}=a_{s}$ and $r, s<n$, then $a_{r+1}=a_{s+1}$.\nProof. We have $a_{r+1}=a_{a_{r}+a_{1}}=a_{a_{s}+a_{1}}=a_{s+1}$.\n\nLemma 2. If $i \\leqslant k$, then $a_{i} \\leqslant k$.\nProof. We have $i+i \\leqslant n$, so $a_{i}+a_{i} \\leqslant n$ whence $a_{i} \\leqslant k$.\n\nLemma 3. There exist $r, s$ such that $a_{r}=a_{s}$ and $r \\neq s$.\nProof. If $a_{0} \\neq 0$ then $a_{2 a_{0}}=a_{0}$. Otherwise, $a_{a_{i}}=a_{i}$ for all $i$, so take $i$ such that $a_{i} \\neq i$ (which we can do by our earlier assumption).\n\nLemma 4. Let $r$ be the smallest index such that $a_{s}=a_{r}$ for some $s>r$, and let $d$ be the minimum positive integer such that $a_{r+d}=a_{r}$. Then\n1. The subsequence $(a_{r}, a_{r+1}, \\ldots, a_{n})$ is periodic with minimal period $d$. That is, for $u<v$, we have $a_{u}=a_{v}$ if and only if $u, v \\geqslant r$ and $d \\mid v-u$.\n2. $a_{i}=i$ for $i<r$ and $a_{i} \\geqslant r$ for $i \\geqslant r$.\nIn this case we say $(a_{i})$ has period $d$ and offset $r$.\nProof. We prove each in turn:\n1. The \"if\" implication is clear from Lemma 1. For the reverse direction, suppose $a_{u}=a_{v}$. Then there are integers $r \\leqslant u_{0}, v_{0}<r+d$ such that $d \\mid u-u_{0}, v-v_{0}$, so $a_{u_{0}}=a_{u}=a_{v}= a_{v_{0}}$. If $u_{0}<v_{0}$ then $a_{r+d+u_{0}-v_{0}}=a_{r+d}=a_{r}$, contradicting the minimality of $d$. There is a similar contradiction when $u_{0}>v_{0}$. Thus $u_{0}=v_{0}$, so $d \\mid u-v$.\n2. If $r=0$ there is nothing to prove. Otherwise $a_{0}=a_{2 a_{0}}$ so $2 a_{0}=0$. Then we have $a_{a_{i}}=a_{i}$ for all $i$, so $a_{i}=i$ for $i<r$.\n\nLemma 5. Either\n1. $d \\mid a_{i}-i$ for all $i$, or\n2. $r=0$ and $d \\mid a_{i}-i-d / 2$ for all $i$.\nProof. Note that Lemma 4 tells us that if $a_{u}=a_{v}$ then $d \\mid u-v$. Since $a_{a_{i}+a_{0}}=a_{i}$ for all $i$, we have $d \\mid a_{i}-i+a_{0}$. For $i=0$, this means that $d \\mid 2 a_{0}$. If $d \\mid a_{0}$ then that means that $d \\mid a_{i}-i$ for all $i$. Otherwise, if $d \\nmid a_{0}$, then $d \\mid a_{0}-d / 2$ and thus $d \\mid a_{i}-i-d / 2$ for all $i$. In addition, part 2 of Lemma 4 says that we must have $r=0$ in this case.\n\nLemma 6. If $d$ is even and $d \\mid a_{0}-d / 2$, then $(a_{i})$ is small. (Note that we must have $r=0$ in this case.)\nProof. Note that if $d \\leqslant k+1$, then by Lemma 2, $(a_{0}, \\ldots, a_{d-1})$ is a period for the sequence consisting of elements at most $k$, so $(a_{i})$ must be small. Now suppose $d>k+1$. We show that $a_{i} \\leqslant k$ for all $i$ by induction. Note that Lemma 2 already establishes this for $i \\leqslant k$. We must have $d \\mid a_{d / 2}$ and $a_{d / 2} \\leqslant k<d$ so $a_{d / 2}=0$. Thus, for $i>k$, if $a_{j} \\leqslant k$ for $j<i$, then $a_{i-d / 2} \\leqslant k$, so $a_{i}=a_{(i-d / 2)+d / 2}=a_{a_{i-d / 2}} \\leqslant k$.\n\nLemma 7. If $(a_{i})$ is small, then $r+d \\leqslant k+1$.\nProof. Since $(a_{i})$ is small, there exists $u, v \\leqslant k+1$ such that $u<v$ and $a_{u}=a_{v}$. Thus $u \\leqslant r$ and $d \\mid v-u$, so $r+d \\leqslant v \\leqslant k+1$.\n\nLemma 8. If $(a_{i})$ is large, then $r+d>k+1$ and $a_{i}=i$ for all $0 \\leqslant i<r+d$.\nProof. Since $(a_{i})$ is large and has period $d$ and offset $r$, the period $(a_{r}, \\ldots, a_{r+d-1})$ must have an element that is larger than $k$, so by Lemma 2 we must have $r+d-1>k$.\nWe already have $a_{i}=i$ for $i<r$. Now we show that $a_{i}=i$ for $r \\leqslant i \\leqslant k$. By Lemma 6 we have $d \\mid a_{i}-i$ but $r \\leqslant i \\leqslant k$. Since $k-r+1>d$, this means that $a_{i}=i$ for $i \\leqslant k$.\nFinally, one can show inductively that $a_{i}=i$ for $k<i<r+d$. Indeed, if $a_{j}=j$ for all $j<i$, then $a_{i} \\geqslant i$ (otherwise $a_{i}=a_{j}$ for some $j<i$, but then $r \\leqslant j$ and $i<r+d$ means that $d \\nmid j-i$.) However, $a_{i}+(n-i)=a_{i}+a_{n-i} \\leqslant n$, so $a_{i}=i$.\n\nThus large sequences are determined by $r$ and $d$. It is not hard to check that all sequences of the form $a_{i}=i$ for $i<r+d$ and with period $d$ and offset $r$ are $n$-sequences. There are $(n-k-1)(n+k+2) / 2$ possible choices of $(r, d)$ where $0 \\leqslant r<n, d \\geqslant 1$, and $k+1<r+d \\leqslant n$.\n\nFor small sequences, for a given period $d$ and offset $r$, we need to choose the period $(a_{r}, \\ldots, a_{r+d-1})$ satisfying $r \\leqslant a_{j} \\leqslant k$ and $d \\mid a_{j}-j$ for $r \\leqslant j<r+d$. There are $g(k+1-r, d)$ such choices, where we define $g(x, d)$ to be $(p+1)^{q} p^{d-q}$ with $p=\\lfloor x / d\\rfloor$ and $q=x-d p$.\n\nFurthermore, if $d$ is even then there are $g(k+1, d)$ choices for the period $(a_{0}, \\ldots, a_{d-1})$ satisfying $d \\mid a_{j}-j-d / 2$ for $j<d$. Again it is not hard to check that, once these choices are made, then the resulting sequence is an $n$-sequence.\n\nThus the total number of $n$-sequences is\n$$\n\\begin{equation*}\nf(n)=1+\\frac{(n-k-1)(n+k+2)}{2}+\\sum_{d'=1}^{\\lfloor(k+1) / 2\\rfloor} g\\left(k+1,2 d'\\right)+\\sum_{r=0}^{k} \\sum_{d=1}^{k+1-r} g(k+1-r, d) \\tag{1}\n\\end{equation*}\n$$\n\nNow, to show that $f(n)>c_{1} \\lambda^{n}$ for some $c_{1}$, we note that\n$$\nf(n)>g\\left(k+1,\\left\\lfloor\\frac{k+1}{3}\\right\\rfloor\\right) \\geqslant 3^{\\lfloor(k+1) / 3\\rfloor}>3^{n / 6}-1 .\n$$\n\nTo show that $f(n)<c_{2} \\lambda^{n}$ for some $c_{2}$, it actually suffices to show that there is a positive real number $c_{3}$ such that for all positive integers $x$,\n$$\n\\sum_{d=1}^{x} g(x, d) \\leqslant c_{3} 3^{x / 3}\n$$\nIn fact, the following lemma suffices, as it bounds the left hand side of the above inequality by a pair of geometric series with initial term $3^{x / 3}$ :\n\nLemma. For positive $d, x$, we have:\n$$\ng(x, d) \\leqslant \\begin{cases}3^{x / 3}\\left(\\frac{64}{81}\\right)^{x / 3-d}, & \\text{ if } d \\leqslant x / 3 ; \\\\ 3^{x / 3}\\left(\\frac{8}{9}\\right)^{d-x / 3}, & \\text{ if } d \\geqslant x / 3 .\\end{cases}\n$$\n\nProof. There are a few key observations needed, all of which are immediate from the definition:\n- $(x, d)$ is the maximum product of a sequence of $d$ integers that sums to $x$.\n- For any positive integer $k$, we have $g(k x, k d)=g(x, d)^{k}$.\n- If $2 d \\leqslant x \\leqslant 3 d$, then $g(x, d)=2^{3 d-x} 3^{x-2 d}$. Likewise, if $3 d \\leqslant x \\leqslant 4 d$ then $g(x, d)= 3^{4 d-x} 4^{x-3 d}$.\n\nWith these observations, if $d \\leqslant x / 3$, then\n$$\n3^{4(x-3 d)} g(3 x, 3 d) \\leqslant g(3 x+12(x-3 d), 3 d+4(x-3 d))=g(15 x-36 d, 4 x-9 d)\n$$\nTo calculate $g(15 x-36 d, 4 x-9 d)$, note that\n$$\n\\begin{aligned}\n& 3(4 x-9 d)=12 x-27 d \\leqslant 15 x-36 d, \\\\\n& 4(4 x-9 d)=16 x-36 d \\geqslant 15 x-36 d,\n\\end{aligned}\n$$\nso\n$$\ng(15 x-36 d, 4 x-9 d)=3^{4(4 x-9 d)-(15 x-36 d)} 4^{(15 x-36 d)-3(4 x-9 d)}=3^{x} 4^{3(x-3 d)}\n$$\nThus\n$$\ng(x, d)^{3}=g(3 x, 3 d)=\\frac{3^{4(x-3 d)} g(3 x, 3 d)}{3^{4(x-3 d)}} \\leqslant \\frac{3^{x} 4^{3(x-3 d)}}{3^{4(x-3 d)}}=3^{x}\\left(\\frac{64}{81}\\right)^{x-3 d},\n$$\nwhich completes the proof of the first claim. Likewise, if $d \\geqslant x / 3$,\n$$\n3^{2(3 d-x)} g(3 x, 3 d) 3 \\leqslant g(3 x+6(3 d-x), 3 d+2(3 d-x))=g(18 d-3 x, 9 d-2 x) .\n$$\nAgain we have\n$$\n2(9 d-2 x) \\leqslant 18-3 x \\leqslant 18-3 x+3(3 d-x)=3(9 d-2 x),\n$$\nso\n$$\ng(18 d-3 x, 9 d-2 x)=2^{3(9 d-2 x)-(18 d-3 x)} 3^{(18 d-3 x)-2(9 d-2 x)}=2^{3(3 d-x)} 3^{x} .\n$$\nThus\n$$\ng(x, d)^{3}=g(3 x, 3 d)=\\frac{2^{3(3 d-x)} g(3 x, 3 d)}{3^{2(3 d-x)}} \\leqslant \\frac{2^{3(3 d-x)} 3^{x}}{3^{2(3 d-x)}}=3^{x}\\left(\\frac{8}{9}\\right)^{3 d-x}\n$$\nfrom which the second claim follows.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32134,
    "subject": "Mathematics (Multi-modal)",
    "question": "A $\\pm 1$-sequence is a sequence of $2022$ numbers $a_{1}, \\ldots, a_{2022}$, each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\\pm 1$-sequence, there exists an integer $k$ and indices $1 \\leqslant t_{1}<\\ldots<t_{k} \\leqslant 2022$ so that $t_{i+1}-t_{i} \\leqslant 2$ for all $i$, and\n$$\n\\left|\\sum_{i=1}^{k} a_{t_{i}}\\right| \\geqslant C.\n$$",
    "options": [],
    "answer": "506",
    "solution": "First, we prove that this can always be achieved. Without loss of generality, suppose at least $\\frac{2022}{2}=1011$ terms of the $\\pm 1$-sequence are $+1$. Define a subsequence as follows: starting at $t=0$, if $a_{t}=+1$ we always include $a_{t}$ in the subsequence. Otherwise, we skip $a_{t}$ if we can (i.e. if we included $a_{t-1}$ in the subsequence), otherwise we include it out of necessity, and go to the next $t$. Clearly, this subsequence will include all $+1$s. Also, for each $-1$ included in the sequence, a $-1$ must have been skipped, so at most $\\left\\lfloor\\frac{1011}{2}\\right\\rfloor=505$ can be included. Hence the sum is at least $1011-505=506$, as desired.\n\nNext, we prove that, for the $\\pm 1$-sequence\n$$\n(\\{-1\\},\\{+1,+1\\},\\{-1,-1\\},\\{+1,+1\\}, \\ldots,\\{+1,+1\\},\\{-1,-1\\},\\{+1\\})\n$$\neach admissible subsequence $a_{t_{i}}$ has $-506 \\leqslant \\sum_{i} a_{t_{i}} \\leqslant 506$. We say that the terms inside each curly bracket is a block. In total, there are $1012$ blocks - $506$ of them hold $+1$s, and $506$ of them hold $-1$s. (The two blocks at each end hold $1$ number each, each other block holds $2$.)\n\nSuppose an admissible subsequence includes terms from $k$ blocks holding $+1$s. Then, in each $-1$-pair in between the $+1$-pairs, the subsequence must also include at least one $-1$. There can be at most two $+1$s included from each $+1$-block, and at least one $-1$ must be included from each $-1$-block, so the sum is at most $2k-(k-1)=k+1$.\n\nFor $k<506$, this is at most $506$. If $k=506$, one of the $+1$-blocks must be the one at the end, meaning it can only include one $+1$, so that the maximum in this case is only $k$, not $k+1$, so in this case the sum is also at most $506$.\n\nHence we have shown that for any admissible subsequence, $\\sum_{i} a_{t_{i}} \\leqslant 506$. Analogously we can show that $-506 \\leqslant \\sum_{i} a_{t_{i}}$, meaning that $C \\leqslant 506$ as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32135,
    "subject": "Mathematics (Multi-modal)",
    "question": "The Bank of Oslo issues coins made out of two types of metal: aluminium (denoted $A$) and copper (denoted $C$). Morgane has $n$ aluminium coins, and $n$ copper coins, and arranges her $2n$ coins in a row in some arbitrary initial order. Given a fixed positive integer $k \\leqslant 2n$, she repeatedly performs the following operation: identify the largest subsequence containing the $k$-th coin from the left which consists of consecutive coins made of the same metal, and move all coins in that subsequence to the left end of the row. For example, if $n=4$ and $k=4$, the process starting from the configuration $A A C C C A C A$ would be\n$$\nA A C C C A C A \\rightarrow C C C A A A C A \\rightarrow A A A C C C C A \\rightarrow C C C C A A A A \\rightarrow \\cdots .\n$$\nFind all pairs $(n, k)$ with $1 \\leqslant k \\leqslant 2n$ such that for every initial configuration, at some point of the process there will be at most one aluminium coin adjacent to a copper coin.",
    "options": [],
    "answer": "All pairs (n, k) with n ≤ k ≤ (3n+1)/2",
    "solution": "Define a block to be a maximal subsequence of consecutive coins made out of the same metal, and let $M^{b}$ denote a block of $b$ coins of metal $M$. The property that there is at most one aluminium coin adjacent to a copper coin is clearly equivalent to the configuration having two blocks, one consisting of all $A$-s and one consisting of all $C$-s.\n\nFirst, notice that if $k < n$, the sequence $A^{n-1} C^{n-1} A C$ remains fixed under the operation, and will therefore always have 4 blocks. Next, if $k > \\frac{3n+1}{2}$, let $a = k - n - 1$, $b = 2n - k + 1$. Then $k > 2a + b$, $k > 2b + a$, so the configuration $A^{a} C^{b} A^{b} C^{a}$ will always have four blocks:\n$$\nA^{a} C^{b} A^{b} C^{a} \\rightarrow C^{a} A^{a} C^{b} A^{b} \\rightarrow A^{b} C^{a} A^{a} C^{b} \\rightarrow C^{b} A^{b} C^{a} A^{a} \\rightarrow A^{a} C^{b} A^{b} C^{a} \\rightarrow \\ldots\n$$\nTherefore a pair $(n, k)$ can have the desired property only if $n \\leqslant k \\leqslant \\frac{3n+1}{2}$. We claim that all such pairs in fact do have the desired property. Clearly, the number of blocks in a configuration cannot increase, so whenever the operation is applied, it either decreases or remains constant. We show that unless there are only two blocks, after a finite amount of steps the number of blocks will decrease.\n\nConsider an arbitrary configuration with $c \\geqslant 3$ blocks. We note that as $k \\geqslant n$, the leftmost block cannot be moved, because in this case all $n$ coins of one type are in the leftmost block, meaning there are only two blocks. If a block which is not the leftmost or rightmost block is moved, its neighbor blocks will be merged, causing the number of blocks to decrease.\n\nHence the only case in which the number of blocks does not decrease in the next step is if the rightmost block is moved. If $c$ is odd, the leftmost and the rightmost blocks are made of the same metal, so this would merge two blocks. Hence $c \\geqslant 4$ must be even. Suppose there is a configuration of $c$ blocks with the $i$-th block having size $a_{i}$ so that the operation always moves the rightmost block:\n$$\nA^{a_{1}} \\ldots A^{a_{c-1}} C^{a_{c}} \\rightarrow C^{a_{c}} A^{a_{1}} \\ldots A^{a_{c-1}} \\rightarrow A^{a_{c-1}} C^{a_{c}} A^{a_{1}} \\ldots C^{a_{c-2}} \\rightarrow \\ldots\n$$\nBecause the rightmost block is always moved, $k \\geqslant 2n+1-a_{i}$ for all $i$. Because $\\sum a_{i} = 2n$, summing this over all $i$ we get $c k \\geqslant 2c n + c - \\sum a_{i} = 2c n + c - 2n$, so $k \\geqslant 2n + 1 - \\frac{2n}{c} \\geqslant \\frac{3n}{2} + 1$. But this contradicts $k \\leqslant \\frac{3n+1}{2}$. Hence at some point the operation will not move the rightmost block, meaning that the number of blocks will decrease, as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32136,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. We start with $n$ piles of pebbles, each initially containing a single pebble. One can perform moves of the following form: choose two piles, take an equal number of pebbles from each pile and form a new pile out of these pebbles. For each positive integer $n$, find the smallest number of non-empty piles that one can obtain by performing a finite sequence of moves of this form.",
    "options": [],
    "answer": "1 if n is a power of 2; otherwise 2",
    "solution": "We can combine two piles of $2^{k-1}$ pebbles to make one pile of $2^{k}$ pebbles. In particular, given $2^{k}$ piles of one pebble, we may combine them as follows:\n$$\n\\begin{array}{lll}\n2^{k} \\text{ piles of } 1 \\text{ pebble } & \\rightarrow & 2^{k-1} \\text{ piles of } 2 \\text{ pebbles } \\\\\n2^{k-1} \\text{ piles of } 2 \\text{ pebbles } & \\rightarrow & 2^{k-2} \\text{ piles of } 4 \\text{ pebbles } \\\\\n2^{k-2} \\text{ piles of } 4 \\text{ pebbles } & \\rightarrow & 2^{k-3} \\text{ piles of } 8 \\text{ pebbles } \\\\\n& \\vdots & \\\\\n2 \\text{ piles of } 2^{k-1} \\text{ pebbles } & \\rightarrow & 1 \\text{ pile of } 2^{k} \\text{ pebbles }\n\\end{array}\n$$\nThis proves the desired result in the case when $n$ is a power of $2$.\n\nIf $n$ is not a power of $2$, choose $N$ such that $2^{N}<n<2^{N+1}$. Let $m=n-2^{N}$. Then $0<m<2^{N}$. Make a pile of $2^{N}$ pebbles and call it the large pile. (Alternatively, one can be more efficient and make a pile of $2^{M}$ pebbles where $m \\leqslant 2^{M}$.) Since $n$ is not a power of two, there is at least one other pile with pebbles. All other piles have a single pebble (initial condition).\n\nChoose one single pebble pile and remove the pebble and one pebble from the large pile and form a pile of $2$ pebbles. If $m<2^{N}-1$, remove another pebble from the large pile and one pebble from the $2$-pile and form a new pile of $2$ pebbles. Repeat until the large pile contains exactly $m$ pebbles.\n\nAt this point we have one pile of $m$ pebbles, one pile of $2$ pebbles, and the rest are single pebble piles. There must be $n-m-2=2^{N}-2$ single piles. Combine two and two into piles of two pebbles. Then there are $2^{N-1}$ piles of two pebbles, which we can make into one pile of $2^{N}$ pebbles. We are left with exactly two piles of pebbles.\n\nLastly we will prove that it is not possible to form a single pile with all pebbles when $n$ is not a power of two. A move consists of choosing two piles of say $a$ and $b$ pebbles, then removing $c \\leqslant \\min (a, b)$ pebbles from both piles, and forming a new pile with $2c$ pebbles. If we include piles of zero pebbles, then this move changes the number of pebbles in three piles as follows (and leaves all other piles unchanged):\n$$\n\\begin{aligned}\n& a \\rightarrow a-c \\\\\n& b \\rightarrow b-c \\\\\n& 0 \\rightarrow 2c\n\\end{aligned}\n$$\nAssume that after the move the number of pebbles in each pile is divisible by an odd integer $m$. In particular, $m|2c, m| a-c$ and $m \\mid b-c$. Since $m$ is odd, it follows that $m \\mid c$, and then that $m \\mid a$ and $m \\mid b$. Hence also before the move the number of pebbles in each pile is divisible by the integer $m$.\n\nIf $n$ is not a power of $2$, then $n$ is divisible by an odd integer $m>1$. In order to make a single pile of $n$ pebbles, we would have to start with a distribution in which the number of pebbles in each pile is divisible by the integer $m$. This is impossible when we start with all piles containing a single pebble.\nWe show an alternative strategy if $n$ is not a power of $2$. Write $n$ in binary form: $n=2^{i_{1}}+2^{i_{2}}+\\cdots+2^{i_{k}}$, where $i_{1}>i_{2}>\\cdots>i_{k}$. Now we make piles of sizes $2^{i_{1}}, 2^{i_{2}}, \\ldots, 2^{i_{k}}$. We call the pile with $2^{i_{1}}$ the large pile, all the others are the small piles.\n\nThe strategy is the following: take the two smallest small piles. If they have the same size of $2^{a}$, we make a pile of size $2^{a+1}$. If they have different sizes, we double the smaller pile using the large pile (we allow the large pile to have a negative number of pebbles: later we prove that it is not possible). We call all the new piles small. When we have only one small pile, we terminate the process: we have at most $2$ piles.\n\nAfter each move we have one less number of piles, and all the piles have cardinality power of $2$. The number of pebbles is decreasing, and at the end of the process, it has a size of $n-2^{i_{2}+1} \\geqslant n-2^{i_{1}}>0$, thus we can manage to have two piles.\nThroughout the solution, we will consider the moves in reverse order. Namely, imagine we have some piles of pebbles, and we are allowed to perform moves as follows: take a pile with an even number of pebbles, split it into two equal halves and add the pebbles from each half to a different pile, possibly forming new piles (we may assume for convenience that there are infinitely many empty piles at any given moment). Given a configuration of piles $\\mathcal{C}$, we will use $|\\mathcal{C}|$ to denote the number of non-empty piles in $\\mathcal{C}$. Given two configurations $\\mathcal{C}_{1}, \\mathcal{C}_{2}$, we will say that $\\mathcal{C}_{2}$ is reachable from $\\mathcal{C}_{1}$ if $\\mathcal{C}_{2}$ can be obtained by performing a finite sequence of moves starting from $\\mathcal{C}_{1}$.\n\nCall a configuration of piles $\\mathcal{C}$\n- simple if each (non-empty) pile in $\\mathcal{C}$ consists of a single pebble;\n- good if at least one (non-empty) pile in $\\mathcal{C}$ has an even number of pebbles and the numbers of pebbles on the piles in $\\mathcal{C}$ have no non-trivial odd common divisor ($\\mathcal{C}$ has the odd divisor property);\n- solvable if there exists a simple configuration which is reachable from $\\mathcal{C}$.\n\nThe problem asks to find the smallest number of non-empty piles in a solvable configuration consisting of $n$ pebbles. We begin the process of answering this question by making the following observation:\n\n**Lemma 1.** Let $\\mathcal{C}$ be a configuration of piles. Let $\\mathcal{C}'$ be a configuration obtained by applying a single move to $\\mathcal{C}$. Then\n(i) if $\\mathcal{C}'$ has the odd divisor property, then so does $\\mathcal{C}$;\n(ii) the converse to (i) holds if $|\\mathcal{C}'| \\geqslant |\\mathcal{C}|$.\n\nProof. Suppose that the move consists of splitting a pile of size $2a$ and adding $a$ pebbles to each of two piles of sizes $b$ and $c$. Here, $a$ is a positive integer and $b, c$ are non-negative integers. Thus, $\\mathcal{C}'$ can be obtained from $\\mathcal{C}$ by replacing the piles of sizes $2a, b, c$ by two piles of sizes $a+b$ and $a+c$. Note that the extra assumption $|\\mathcal{C}'| \\geqslant |\\mathcal{C}|$ of part (ii) holds if and only if at least one of $b, c$ is zero.\n\n(i) Suppose $\\mathcal{C}$ doesn't have the odd divisor property, i.e. there exists an odd integer $d>1$ such that the size of each pile in $\\mathcal{C}$ is divisible by $d$. In particular, $2a, b, c$ are multiples of $d$, so since $d$ is odd, it follows that $a, b, c$ are all divisible by $d$. Thus, $a+b$ and $a+c$ are also divisible by $d$, so $d$ divides the size of each pile in $\\mathcal{C}'$. In conclusion, $\\mathcal{C}'$ doesn't have the odd divisor property.\n\n(ii) If $\\mathcal{C}'$ doesn't have the odd divisor property and at least one of $b, c$ is zero, then there exists an odd integer $d>1$ such that the size of each pile in $\\mathcal{C}'$ is divisible by $d$. In particular, $d$ divides $a+b$ and $a+c$, so since at least one of these numbers is equal to $a$, it follows that $d$ divides $a$. But then $d$ must divide all three of $a, b$ and $c$, and hence it certainly divides $2a, b$ and $c$. Thus, $\\mathcal{C}$ doesn't have the odd divisor property, as desired.\n\n**Lemma 2.** If $\\mathcal{C}_2$ is reachable from $\\mathcal{C}_1$ and $\\mathcal{C}_2$ has the odd divisor property, then so does $\\mathcal{C}_1$. In particular, any solvable configuration has the odd divisor property.\n\nProof. The first statement follows by inductively applying part (i) of Lemma 1. The second statement follows from the first because every simple configuration has the odd divisor property.\n\nThe main claim is the following:\n\n**Lemma 3.** Let $\\mathcal{C}$ be a good configuration. Then there exists a configuration $\\mathcal{C}'$ with the following properties:\n- $\\mathcal{C}'$ is reachable from $\\mathcal{C}$ and $|\\mathcal{C}'| > |\\mathcal{C}|$;\n- $\\mathcal{C}'$ is either simple or good.\n\nProof. Call a configuration terminal if it is a counterexample to the claim. The following claim is at the heart of the proof:\n\n**Claim.** Let $a_1, \\ldots, a_k$ be the numbers of pebbles on the non-empty piles of a terminal configuration $\\mathcal{C}$. Then there exists a unique $i \\in [k]$ such that $a_i$ is even. Moreover, for all $t \\geqslant 1$ we have $a_j \\equiv \\frac{a_i}{2} \\pmod{2^t}$ for all $j \\neq i$.\n\nProof of Claim. Since the configuration is good, there must exist $i \\in [k]$ such that $a_i$ is even. Moreover, by assumption, if we split the pile with $a_i$ pebbles into two equal halves, the resulting configuration will not be good. By part (ii) of Lemma 2, the only way this can happen is that $\\frac{a_i}{2}$ and $a_j$ for all $j \\neq i$ are odd. To prove the second assertion, we proceed by induction on $t$, with the case $t=1$ already being established. If $t \\geqslant 2$, then split the pile with $a_i$ pebbles into two equal halves and move one half to the pile with $a_j$ pebbles. Since $\\frac{a_i}{2}$ and $a_j$ are both odd, $a_j + \\frac{a_i}{2}$ is even, so by part (ii) of Lemma 2, the resulting configuration $\\mathcal{C}'$ is good. Thus, $\\mathcal{C}'$ is terminal, so by the induction hypothesis, we have $\\frac{a_i}{2} \\equiv \\frac{1}{2}(a_j + \\frac{a_i}{2}) \\pmod{2^{t-1}}$, whence $a_j \\equiv \\frac{a_i}{2} \\pmod{2^t}$, as desired.\n\nSuppose for contradiction that there exists a configuration as in the Claim. It follows that there exists $i \\in [k]$ and an odd integer $x$ such that $a_i = 2x$ and $a_j = x$ for all $j \\neq i$. Thus, $x$ is an odd common divisor of $a_1, \\ldots, a_k$, so by the odd divisor property, we must have $x=1$. But then we can obtain a simple configuration by splitting the only pile with two pebbles into two piles each consisting of a single pebble, which is a contradiction.\n\nWith the aid of Lemmas 2 and 3, it is not hard to characterise all solvable configurations:\n\n**Lemma 4.** A configuration of piles $\\mathcal{C}$ is solvable if and only if it is simple or good.\n\nProof. ($\\Longrightarrow$) Suppose $\\mathcal{C}$ is not simple. Then since we have to be able to perform at least one move, there must be at least one non-empty pile in $\\mathcal{C}$ with an even number of pebbles. Moreover, by Lemma 2, $\\mathcal{C}$ has the odd divisor property, so it must be good.\n\n($\\Longleftarrow$) This follows by repeatedly applying Lemma 3 until we reach a simple configuration. Note that the process must stop eventually since the number of non-empty piles is bounded from above.\n\nFinally, the answer to the problem is implied by the following corollary of Lemma 4:\n\n**Lemma 5.** Let $n$ be a positive integer. Then\n(i) a configuration consisting of a single pile of $n$ pebbles is solvable if and only if $n$ is a power of two;\n(ii) if $n \\geqslant 2$, then the configuration consisting of piles of sizes $2$ and $n-2$ is good.\n\nProof. (i) By Lemma 4, this configuration is solvable if and only if either $n=1$ or $n$ is even and has no non-trivial odd divisor, so the conclusion follows.\n\n(ii) Since $2$ is even and has no non-trivial odd divisor, this configuration is certainly good, so the conclusion follows by Lemma 4.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32137,
    "subject": "Mathematics (Multi-modal)",
    "question": "Lucy starts by writing $s$ integer-valued 2022-tuples on a blackboard. After doing that, she can take any two (not necessarily distinct) tuples $\\mathbf{v} = (v_1, \\ldots, v_{2022})$ and $\\mathbf{w} = (w_1, \\ldots, w_{2022})$ that she has already written, and apply one of the following operations to obtain a new tuple:\n$$\n\\begin{aligned}\n& \\mathbf{v} + \\mathbf{w} = (v_1 + w_1, \\ldots, v_{2022} + w_{2022}) \\\\\n& \\mathbf{v} \\vee \\mathbf{w} = (\\max(v_1, w_1), \\ldots, \\max(v_{2022}, w_{2022}))\n\\end{aligned}\n$$\nand then write this tuple on the blackboard.\nIt turns out that, in this way, Lucy can write any integer-valued 2022-tuple on the blackboard after finitely many steps. What is the smallest possible number $s$ of tuples that she initially wrote?",
    "options": [],
    "answer": "3",
    "solution": "We solve the problem for $n$-tuples for any $n \\geqslant 3$: we will show that the answer is $s = 3$, regardless of the value of $n$.\n\nFirst, let us briefly introduce some notation. For an $n$-tuple $\\mathbf{v}$, we will write $\\mathbf{v}_i$ for its $i$-th coordinate (where $1 \\leqslant i \\leqslant n$). For a positive integer $n$ and a tuple $\\mathbf{v}$ we will denote by $n \\cdot \\mathbf{v}$ the tuple obtained by applying addition on $\\mathbf{v}$ with itself $n$ times. Furthermore, we denote by $\\mathbf{e}(i)$ the tuple which has $i$-th coordinate equal to one and all the other coordinates equal to zero. We say that a tuple is positive if all its coordinates are positive, and negative if all its coordinates are negative.\n\nWe will show that three tuples suffice, and then that two tuples do not suffice.\n\nThree tuples suffice. Write $\\mathbf{c}$ for the constant-valued tuple $\\mathbf{c} = (-1, \\ldots, -1)$.\nWe note that it is enough for Lucy to be able to make the tuples $\\mathbf{e}(1), \\ldots, \\mathbf{e}(n), \\mathbf{c}$; from those any other tuple $\\mathbf{v}$ can be made as follows. First we choose some positive integer $k$ such that $k + \\mathbf{v}_i > 0$ for all $i$. Then, by adding a positive number of copies of $\\mathbf{c}, \\mathbf{e}(1), \\ldots, \\mathbf{e}(n)$, she can make\n$$\nk \\mathbf{c} + (k + \\mathbf{v}_1) \\cdot \\mathbf{e}(1) + \\cdots + (k + \\mathbf{v}_n) \\cdot \\mathbf{e}(n)\n$$\nwhich we claim is equal to $\\mathbf{v}$. Indeed, this can be checked by comparing coordinates: the $i$-th coordinate of the right-hand side is $-k + (k + \\mathbf{v}_i) = \\mathbf{v}_i$ as needed.\n\nLucy can take her three starting tuples to be $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$, such that $\\mathbf{a}_i = -i^2, \\mathbf{b}_i = i$ and $\\mathbf{c} = -1$ (as above).\n\nFor any $1 \\leqslant j \\leqslant n$, write $\\mathbf{d}(j)$ for the tuple $2 \\cdot \\mathbf{a} + 4j \\cdot \\mathbf{b} + (2j^2 - 1) \\cdot \\mathbf{c}$, which Lucy can make by adding together $\\mathbf{a}, \\mathbf{b}$ and $\\mathbf{c}$ repeatedly. This has $i$-th term\n$$\n\\begin{aligned}\n\\mathbf{d}(j)_i &= 2 \\mathbf{a}_i + 4j \\mathbf{b}_i + (2j^2 - 1) \\mathbf{c}_i \\\\\n&= -2i^2 + 4ij - (2j^2 - 1) \\\\\n&= 1 - 2(i - j)^2\n\\end{aligned}\n$$\nThis is 1 if $j = i$, and at most -1 otherwise. Hence Lucy can produce the tuple $\\mathbf{1} = (1, \\ldots, 1)$ as $\\mathbf{d}(1) \\vee \\cdots \\vee \\mathbf{d}(n)$.\n\nShe can then produce the constant tuple $\\mathbf{0} = (0, \\ldots, 0)$ as $\\mathbf{1} + \\mathbf{c}$, and for any $1 \\leqslant j \\leqslant n$ she can then produce the tuple $\\mathbf{e}(j)$ as $\\mathbf{d}(j) \\vee \\mathbf{0}$. Since she can now produce $\\mathbf{e}(1), \\ldots, \\mathbf{e}(n)$ and already had $\\mathbf{c}$, she can (as we argued earlier) produce any integer-valued tuple.\n\nTwo tuples do not suffice. We start with an observation: Let $a$ be a non-negative real number and suppose that two tuples $\\mathbf{v}$ and $\\mathbf{w}$ satisfy $\\mathbf{v}_j \\geqslant a \\mathbf{v}_k$ and $\\mathbf{w}_j \\geqslant a \\mathbf{w}_k$ for some $1 \\leqslant j, k \\leqslant n$. Then we claim that the same inequality holds for $\\mathbf{v} + \\mathbf{w}$ and $\\mathbf{v} \\vee \\mathbf{w}$: Indeed, the property for the sum is verified by an easy computation:\n$$\n(\\mathbf{v} + \\mathbf{w})_j = \\mathbf{v}_j + \\mathbf{w}_j \\geqslant a \\mathbf{v}_k + a \\mathbf{w}_k = a(\\mathbf{v} + \\mathbf{w})_k.\n$$\nFor the second operation, we denote by $\\mathbf{m}$ the tuple $\\mathbf{v} \\vee \\mathbf{w}$. Then $\\mathbf{m}_j \\geqslant \\mathbf{v}_j \\geqslant a \\mathbf{v}_k$ and $\\mathbf{m}_j \\geqslant \\mathbf{w}_j \\geqslant a \\mathbf{w}_k$. Since $\\mathbf{m}_k = \\mathbf{v}_k$ or $\\mathbf{m}_k = \\mathbf{w}_k$, the observation follows.\n\nAs a consequence of this observation we have that if all starting tuples satisfy such an inequality, then all generated tuples will also satisfy it, and so we would not be able to obtain every integer-valued tuple.\n\nLet us now prove that Lucy needs at least three starting tuples. For contradiction, let us suppose that Lucy started with only two tuples $\\mathbf{v}$ and $\\mathbf{w}$. We are going to distinguish two cases. In the first case, suppose we can find a coordinate $i$ such that $\\mathbf{v}_i, \\mathbf{w}_i \\geqslant 0$. Both operations preserve the sign, thus we can not generate any tuple that has negative $i$-th coordinate. Similarly for $\\mathbf{v}_i, \\mathbf{w}_i \\leqslant 0$.\n\nSuppose the opposite, i.e., for every $i$ we have either $\\mathbf{v}_i > 0 > \\mathbf{w}_i$, or $\\mathbf{v}_i < 0 < \\mathbf{w}_i$. Since we assumed that our tuples have at least three coordinates, by pigeonhole principle there exist two coordinates $j \\neq k$ such that $\\mathbf{v}_j$ has the same sign as $\\mathbf{v}_k$ and $\\mathbf{w}_j$ has the same sign as $\\mathbf{w}_k$ (because there are only two possible combinations of signs).\n\nWithout loss of generality assume that $\\mathbf{v}_j, \\mathbf{v}_k > 0$ and $\\mathbf{w}_j, \\mathbf{w}_k < 0$. Let us denote the positive real number $\\mathbf{v}_j / \\mathbf{v}_k$ by $a$. If $\\mathbf{w}_j / \\mathbf{w}_k \\leqslant a$, then both inequalities $\\mathbf{v}_j \\geqslant a \\mathbf{v}_k$ and $\\mathbf{w}_j \\geqslant a \\mathbf{w}_k$ are satisfied. On the other hand, if $\\mathbf{w}_j / \\mathbf{w}_k \\leqslant a$, then both $\\mathbf{v}_k \\geqslant (1 / a) \\mathbf{v}_j$ and $\\mathbf{w}_k \\geqslant (1 / a) \\mathbf{w}_j$ are satisfied. In either case, we have found the desired inequality satisfied by both starting tuples, a contradiction with the observation above.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32138,
    "subject": "Mathematics (Multi-modal)",
    "question": "Alice fills the fields of an $n \\times n$ board with numbers from $1$ to $n^{2}$, each number being used exactly once. She then counts the total number of good paths on the board. A good path is a sequence of fields of arbitrary length (including $1$) such that:\n(i) The first field in the sequence is one that is only adjacent to fields with larger numbers,\n(ii) Each subsequent field in the sequence is adjacent to the previous field,\n(iii) The numbers written on the fields in the sequence are in increasing order.\n\nTwo fields are considered adjacent if they share a common side. Find the smallest possible number of good paths Alice can obtain, as a function of $n$.",
    "options": [],
    "answer": "2n^2 - 2n + 1",
    "solution": "We will call any field that is only adjacent to fields with larger numbers a well. Other fields will be called non-wells. Let us make a second $n \\times n$ board $B$ where in each field we will write the number of good sequences which end on the corresponding field in the original board $A$. We will thus look for the minimal possible value of the sum of all entries in $B$.\n\nWe note that any well has just one good path ending in it, consisting of just the well, and that any other field has the number of good paths ending in it equal to the sum of this quantity for all the adjacent fields with smaller values, since a good path can only come into the field from a field of lower value. Therefore, if we fill in the fields in $B$ in increasing order with respect to their values in $A$, it follows that each field not adjacent to any already filled field will receive a $1$, while each field adjacent to already filled fields will receive the sum of the numbers already written on these adjacent fields.\n\nWe note that there is at least one well in $A$, that corresponding with the field with the entry $1$ in $A$. Hence, the sum of values of fields in $B$ corresponding to wells in $A$ is at least $1$. We will now try to minimize the sum of the non-well entries, i.e. of the entries in $B$ corresponding to the non-wells in $A$. We note that we can ascribe to each pair of adjacent fields the value of the lower assigned number and that the sum of non-well entries will then equal to the sum of the ascribed numbers. Since the lower number is still at least $1$, the sum of non-well entries will at least equal the number of pairs of adjacent fields, which is $2 n(n-1)$. Hence, the total minimum sum of entries in $B$ is at least $2 n^{2}-2 n+1$. The necessary conditions for the minimum to be achieved is for there to be only one well and for no two entries in $B$ larger than $1$ to be adjacent to each other.\n\nWe will now prove that the lower limit of $2 n^{2}-2 n+1$ entries can be achieved. This amounts to finding a way of marking a certain set of squares, those that have a value of $1$ in $B$, such that no two unmarked squares are adjacent and that the marked squares form a connected tree with respect to adjacency.\n\nFor $n=1$ and $n=2$ the markings are respectively the lone field and the L-trimino. Now, for $n>2$, let $s=2$ for $n \\equiv 0,2 \\bmod 3$ and $s=1$ for $n \\equiv 1 \\bmod 3$. We will take indices $k$ and $l$ to be arbitrary non-negative integers. For $n \\geqslant 3$ we will construct a path of marked squares in the first two columns consisting of all squares of the form $(1, i)$ where $i$ is not of the form $6k+s$ and $(2, j)$ where $j$ is of the form $6k+s-1, 6k+s$ or $6+s+1$. Obviously, this path is connected. Now, let us consider the fields $(2, 6k+s)$ and $(1, 6k+s+3)$. For each considered field $(i, j)$ we will mark all squares of the form $(l, j)$ for $l>i$ and $(i+2k, j \\pm 1)$. One can easily see that no set of marked fields will produce a cycle, that the only fields of the unmarked form $(1, 6k+s), (2+2l+1, 6k+s \\pm 1)$ and $(2+2l, 6k+s+3 \\pm 1)$ and that no two are adjacent, since the consecutive considered fields are in columns of opposite parity. Examples of markings are given for $n=3,4,5,6,7$, and the corresponding constructions for $A$ and $B$ are given for $n=5$.\n\n$n=3$\n![](attached_image_1.png)\n\n| $n=4$ |   |   |   |\n| :--- | :--- | :--- | :--- |\n| $\\bullet$ | $\\bullet$ | $\\bullet$ | $\\bullet$ |\n| $\\bullet$ |   | $\\bullet$ |   |\n| $\\bullet$ | $\\bullet$ |   | $\\bullet$ |\n|   | $\\bullet$ | $\\bullet$ | $\\bullet$ |\n\n| $\\bullet$ | $\\bullet$ | $\\bullet$ | $\\bullet$ | $\\bullet$ |\n| :--- | :--- | :--- | :--- | :--- |\n| $\\bullet$ |   | $\\bullet$ |   | $\\bullet$ |\n| $\\bullet$ | $\\bullet$ |   | $\\bullet$ |   |\n|   | $\\bullet$ | $\\bullet$ | $\\bullet$ | $\\bullet$ |\n| $\\bullet$ | $\\bullet$ |   | $\\bullet$ |   |\n\n$n=5$\n![](attached_image_2.png)\n\n$n=7$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32139,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}_{\\geqslant 0}$ be the set of non-negative integers, and let $f: \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\rightarrow \\mathbb{Z}_{\\geqslant 0}$ be a bijection such that whenever $f\\left(x_{1}, y_{1}\\right)>f\\left(x_{2}, y_{2}\\right)$, we have $f\\left(x_{1}+1, y_{1}\\right)>f\\left(x_{2}+1, y_{2}\\right)$ and $f\\left(x_{1}, y_{1}+1\\right)>f\\left(x_{2}, y_{2}+1\\right)$.\nLet $N$ be the number of pairs of integers $(x, y)$, with $0 \\leqslant x, y<100$, such that $f(x, y)$ is odd. Find the smallest and largest possible value of $N$.",
    "options": [],
    "answer": "smallest 2500, largest 7500",
    "solution": "We defer the constructions to the end of the solution. Instead, we begin by characterizing all such functions $f$, prove a formula and key property for such functions, and then solve the problem, providing constructions.\n\nCharacterization Suppose $f$ satisfies the given relation. The condition can be written more strongly as\n$$\n\\begin{aligned}\nf\\left(x_{1}, y_{1}\\right)>f\\left(x_{2}, y_{2}\\right) & \\Longleftrightarrow f\\left(x_{1}+1, y_{1}\\right)>f\\left(x_{2}+1, y_{2}\\right) \\\\\n& \\Longleftrightarrow f\\left(x_{1}, y_{1}+1\\right)>f\\left(x_{2}, y_{2}+1\\right)\n\\end{aligned}\n$$\nIn particular, this means for any $(k, l) \\in \\mathbb{Z}^{2}, f(x+k, y+l)-f(x, y)$ has the same sign for all $x$ and $y$.\n\nCall a non-zero vector $(k, l) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0}$ a needle if $f(x+k, y)-f(x, y+l)>0$ for all $x$ and $y$. It is not hard to see that needles and non-needles are both closed under addition, and thus under scalar division (whenever the quotient lives in $\\mathbb{Z}^{2}$ ).\n\nIn addition, call a positive rational number $\\frac{k}{l}$ a grade if the vector $(k, l)$ is a needle. (Since needles are closed under scalar multiples and quotients, this is well-defined.)\n\nClaim. Grades are closed upwards.\nProof. Consider positive rationals $k_{1} / l_{1}<k_{2} / l_{2}$ with $k_{1} / l_{1}$ a grade. Then:\n- $\\left(k_{1}, l_{1}\\right)$ is a needle\n- so $\\left(k_{1} l_{2}, l_{1} l_{2}\\right)$ is a needle,\n- so $\\left(k_{2} l_{1}, l_{1} l_{2}\\right)$ is a needle (as $k_{2} l_{1}-k_{1} l_{2}>0$ and $(1,0)$ is a needle).\nThus ( $k_{2}, l_{2}$ ) is a needle, as wanted.\n\nClaim. A grade exists.\nProof. If no positive integer $n$ is a grade, then $f(1,0)>f(0, n)$ for all $n$ which is impossible.\nSimilarly, there is an $n$ such that $f(0,1)<f(n, 0)$, thus $1 / n$ is not a grade for some large $n$. That means that small positive rational values are not grades, then there is a switch, and after that all values are grades. Call the place of that switch $\\alpha$. Here $\\alpha$ is the infimum of the grades.\n\nClaim (Key property). If $x_{1}+y_{1} \\alpha>x_{2}+y_{2} \\alpha$ then $f\\left(x_{1}, y_{1}\\right)>f\\left(x_{2}, y_{2}\\right)$.\nProof. If both $x_{1} \\geqslant x_{2}$ and $y_{1} \\geqslant y_{2}$ this is clear.\nSuppose $x_{1} \\geqslant x_{2}$ and $y_{1}<y_{2}$. Then $\\frac{x_{1}-x_{2}}{y_{2}-y_{1}}>\\alpha$ is a grade. This gives $f\\left(x_{1}, y_{1}\\right)>f\\left(x_{2}, y_{2}\\right)$. Suppose $x_{1}<x_{2}$ and $y_{1} \\geqslant y_{2}$. Then $\\frac{x_{2}-x_{1}}{u_{1}-u_{2}}<\\alpha$ is not a grade. This gives $f\\left(x_{2}, y_{2}\\right)<f\\left(x_{1}, y_{1}\\right)$.\n\nFrom those observations we get the following claim.\nClaim. The function $f$ orders pairs $(x, y)$ based on the value of $x+y \\alpha$. If $\\alpha$ is rational, tiebreaking is done by larger $x$ - or $y$-coordinate (depending on whether $\\alpha$ is a grade).\n\nWe can imagine this the following way: take a line with slope $-\\frac{1}{\\alpha}$ under the first quadrant of the plane. And we start to move this line upward (but it stays parallel to the original line). First it hits $(0,0)$, so $f(0,0)=0$. And each time the line hits a point $p, f(p)$ is the number of points hit before. If $\\alpha \\in \\mathbb{Q}$, it is possible that the line hits multiple points. Then those points are ordered the same way as their $x$ or $y$ coordinates, depending on whether $\\alpha$ is a grade.\n\nWe understood the behaviour of $f$, now we need to focus on the region of $A=\\{(x, y) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\mid x<100, y<100\\}$. First, we can assume that $\\alpha$ is irrational. If we change it a little bit in the right direction, the behaviour and values of the $f$ function does not change in $A$.\n\nClaim.\n$$\nf(x, y)+f(x+1, y+1)=f(x+1, y)+f(x, y+1)+1\n$$\nProof.\n$$\n\\begin{gathered}\nf(x+1, y+1)-f(x, y+1)= \\\\\n\\#\\{(a, b) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\mid x+(y+1) \\alpha \\leqslant a+b \\alpha<(x+1)+(y+1) \\alpha\\}= \\\\\n\\#\\{(a, b) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{>0} \\mid x+(y+1) \\alpha \\leqslant a+b \\alpha<(x+1)+(y+1) \\alpha\\}+ \\\\\n\\#\\{(a, 0) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\mid(x+1)+y \\alpha \\leqslant a<(x+1)+(y+1) \\alpha\\}= \\\\\n\\#\\{(a, b) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\mid x+y \\alpha \\leqslant a+b \\alpha<(x+1)+y \\alpha\\}+1=f(x+1, y)-f(x, y) .\n\\end{gathered}\n$$\n\nFrom this claim we immediately get that $2500 \\leqslant N \\leqslant 7500$; now we show that those bounds are indeed sharp.\n\nRemember that if $\\alpha$ is irrational then\n$$\nf(a, b)=\\#\\{(x, y) \\in \\mathbb{Z}_{\\geqslant 0} \\times \\mathbb{Z}_{\\geqslant 0} \\mid x+y \\alpha<a+b \\alpha\\}\n$$\n\nConstruction for 7500 Select $\\alpha \\approx 199.999$.\nClaim.\n1. $f(n, 0)=n$ for $0 \\leqslant n \\leqslant 100$.\n2. $f(0, k) \\equiv k \\bmod 2$ for $0 \\leqslant k \\leqslant 100$.\nProof.\n1. $f(n, 0)=\\#\\{(x, y) \\mid x+y \\alpha<n\\}=\\#\\{(x, y) \\mid x+199 y<n\\}=n$.\n2. \n$$\n\\begin{aligned}\n& f(0, k)=\\#\\{(x, y) \\mid x+y \\alpha<k \\alpha\\}=\\sum_{l=0}^{k-1} \\#\\{(x, l) \\mid x+l \\alpha<k \\alpha\\} \\\\\n& \\quad=\\sum_{l=0}^{k-1} \\#\\{x \\mid x<(k-l) \\alpha\\}=\\sum_{l=0}^{k-1} 200(k-l)-1=200 A-k\n\\end{aligned}\n$$\nfor some integer $A$.\n\nFrom this claim, using the equality $f(x, y)+f(x+1, y+1)=f(x+1, y)+f(x, y+1)+1$, we can prove that $\\bmod 2$ the region $A$ looks like the following: in the rows $(-, 2 y)$ the remainders modulo 2 alternate, while the rows $(-, 2 y+1)$ contain only odd numbers.\n\nConstruction for 2500 Select $\\alpha \\approx 200.001$.\nClaim.\n1. $f(n, 0)=n$ for $0 \\leqslant n \\leqslant 100$.\n2. $f(0, k) \\equiv 0 \\bmod 2$ for $0 \\leqslant k \\leqslant 100$.\nProof.\n1. As above.\n2. Similarly to the above:\n$$\n\\begin{aligned}\nf(0, k) & =\\#\\{(x, y) \\mid x+y \\alpha<k \\alpha\\}=\\sum_{l=0}^{k-1} \\#\\{(x, l) \\mid x+l \\alpha<k \\alpha\\} \\\\\n& =\\sum_{l=0}^{k-1} \\#\\{x \\mid x<(k-l) \\alpha\\}=\\sum_{l=0}^{k-1} 200(k-l)=200 A\n\\end{aligned}\n$$\nfor some integer $A$.\n\nSimilarly to the above, we can prove that mod 2 the region $A$ looks like the following: in the rows $(-, 2 y)$ the remainder modulo 2 alternate, while the rows $(-, 2 y+1)$ contain only even numbers.\n\nThus, the optimal bounds are $2500 \\leqslant N \\leqslant 7500$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32140,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD E$ be a convex pentagon such that $BC = DE$. Assume there is a point $T$ inside $ABCDE$ with $TB = TD$, $TC = TE$ and $\\angle TBA = \\angle AET$. Let lines $CD$ and $CT$ intersect line $AB$ at points $P$ and $Q$, respectively, and let lines $CD$ and $DT$ intersect line $AE$ at points $R$ and $S$, respectively. Assume that points $P, B, A, Q$ and $R, E, A, S$ respectively, are collinear and occur on their lines in this order. Prove that the points $P, S, Q, R$ are concyclic.\n\n(Slovakia)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "By the conditions we have $BC = DE$, $CT = ET$ and $TB = TD$, so the triangles $TBC$ and $TDE$ are congruent, in particular $\\angle BTC = \\angle DTE$.\n\nIn triangles $TBQ$ and $TES$ we have $\\angle TBQ = \\angle SET$ and $\\angle QTB = 180^\\circ - \\angle BTC = 180^\\circ - \\angle DTE = \\angle ETS$, so these triangles are similar to each other. It follows that $\\angle TSE = \\angle BQT$ and\n$$\n\\frac{TD}{TQ} = \\frac{TB}{TQ} = \\frac{TE}{TS} = \\frac{TC}{TS}.\n$$\nBy rearranging this relation we get $TD \\cdot TS = TC \\cdot TQ$, so $C, D, Q$ and $S$ are concyclic. (Alternatively, we can get $\\angle CQD = \\angle CSD$ from the similar triangles $TCS$ and $TDQ$.) Hence, $\\angle DCQ = \\angle DSQ$.\n\nFinally, from the angles of triangle $CQP$ we get\n$$\n\\angle RPQ = \\angle RCQ - \\angle PQC = \\angle DSQ - \\angle DSR = \\angle RSQ,\n$$\nwhich proves that $P, Q, R$ and $S$ are concyclic.\nAs in the previous solution, we note that triangles $TBC$ and $TDE$ are congruent. Denote the intersection point of $DT$ and $BA$ by $V$, and the intersection point of $CT$ and $EA$ by $W$. From triangles $BCQ$ and $DES$ we then have\n$$\n\\begin{aligned}\n\\angle VSW & = \\angle DSE = 180^\\circ - \\angle SET - \\angle TED - \\angle EDT \\\\\n& = 180^\\circ - \\angle TBA - \\angle TCB - \\angle CBT = 180^\\circ - \\angle QCB - \\angle CBQ = \\angle BQC = \\angle VQW\n\\end{aligned}\n$$\nmeaning that $VSQW$ is cyclic, and in particular $\\angle WVQ = \\angle WSQ$. Since\n$$\n\\angle VTB = 180^\\circ - \\angle BTC - \\angle CTD = 180^\\circ - \\angle CTD - \\angle DTE = \\angle ETW,\n$$\nand $\\angle TBV = \\angle WET$ by assumption, we have that the triangles $VTB$ and $WTE$ are similar, hence\n$$\n\\frac{VT}{WT} = \\frac{BT}{ET} = \\frac{DT}{CT}.\n$$\nThus $CD \\parallel VW$, and angle chasing yields\n$$\n\\angle RPQ = \\angle WVQ = \\angle WSQ = \\angle RSQ\n$$\nconcluding the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32141,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral. Assume that the points $Q$, $A$, $B$, $P$ are collinear in this order, in such a way that the line $AC$ is tangent to the circle $ADQ$, and the line $BD$ is tangent to the circle $BCP$. Let $M$ and $N$ be the midpoints of $BC$ and $AD$, respectively. Prove that the following three lines are concurrent: line $CD$, the tangent of circle $ANQ$ at point $A$, and the tangent to circle $BMP$ at point $B$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first prove that triangles $ADQ$ and $CDB$ are similar. Since $ABCD$ is cyclic, we have $\\angle DAQ = \\angle DCB$. By the tangency of $AC$ to the circle $AQD$ we also have $\\angle CBD = \\angle CAD = \\angle AQD$. The claimed similarity is proven.\n\nLet $R$ be the midpoint of $CD$. Points $N$ and $R$ correspond in the proven similarity, and so $\\angle QNA = \\angle BRC$.\n\n![](attached_image_1.png)\n\nLet $K$ be the second common point of line $CD$ with circle $ABR$ (i.e., if $CD$ intersects circle $ABR$, then $K \\neq R$ is the other intersection; otherwise, if $CD$ is tangent to $CD$, then $K = R$). In both cases, we have $\\angle BAK = \\angle BRC = \\angle QNA$; that indicates that $AK$ is tangent to circle $ANQ$. It can be showed analogously that $BK$ is tangent to circle $BMP$.\nWe present a second solution, without using the condition that $ABCD$ is cyclic. Again, $M$ and $N$ can be any points on lines $BC$ and $AD$ such that $BM : MC = DN : NA$.\n\nLet $AB$ and $CD$ meet at $T$ (if $AB \\parallel CD$ then $T$ is their common ideal point). Let $CD$ meet the tangent to the circle $ANQ$ at $A$, and the tangent to the circle $BMP$ at $B$ at points $K_1$ and $K_2$, respectively.\n\n![](attached_image_2.png)\n\nLet $I$ and $J$ be the ideal points of $AD$ and $BC$, respectively. Notice that the pencils ($AD$, $AC$, $AT$, $AK_1$) and ($QA$, $QD$, $QI$, $QN$) of lines are congruent, because $\\angle K_1AD = \\angle AQ N$, $\\angle CAD = \\angle AQD$ and $\\angle IAT = \\angle IQT$. Hence,\n\n$$\n(D, C ; T, K_1) = (AD, AC ; AT, AK_1) = (QA, QD ; QI, QN) = (A, D ; I, N) = \\frac{DN}{NA}.\n$$\n\nIt can be obtained analogously that\n\n$$\n(D, C ; T, K_2) = (BD, BC ; BT, BK_2) = (PC, PB ; PJ, PM) = (C, B ; I, N) = \\frac{BM}{MC}.\n$$\n\nFrom $BM : MC = DN : DA$ we get $(D, C ; T, K_1) = (D, C ; T, K_2)$ and hence $K_1 = K_2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32142,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with $AC > AB$, let $O$ be its circumcentre, and let $D$ be a point on the segment $BC$. The line through $D$ perpendicular to $BC$ intersects the lines $AO$, $AC$ and $AB$ at $W$, $X$ and $Y$, respectively. The circumcircles of triangles $AXY$ and $ABC$ intersect again at $Z \\neq A$.\nProve that if $OW = OD$, then $DZ$ is tangent to the circle $AXY$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AO$ intersect $BC$ at $E$. As $EDW$ is a right-angled triangle and $O$ is on $WE$, the condition $OW = OD$ means $O$ is the circumcentre of this triangle. So $OD = OE$ which establishes that $D, E$ are reflections in the perpendicular bisector of $BC$.\n\nNow observe:\n$$\n180^{\\circ} - \\angle DXZ = \\angle ZXY = \\angle ZAY = \\angle ZCD,\n$$\nwhich shows $CDXZ$ is cyclic.\n\n![](attached_image_1.png)\n\nWe next show that $AZ \\parallel BC$. To do this, introduce point $Z'$ on circle $ABC$ such that $AZ' \\parallel BC$. By the previous result, it suffices to prove that $CDXZ'$ is cyclic. Notice that triangles $BAE$ and $CZ'D$ are reflections in the perpendicular bisector of $BC$. Using this and that $A, O, E$ are collinear:\n$$\n\\angle DZ'C = \\angle BAE = \\angle BAO = 90^{\\circ} - \\frac{1}{2} \\angle AOB = 90^{\\circ} - \\angle C = \\angle DXC,\n$$\nso $DXZ'C$ is cyclic, giving $Z \\equiv Z'$ as desired.\n\nUsing $AZ \\parallel BC$ and $CDXZ$ cyclic we get:\n$$\n\\angle AZD = \\angle CDZ = \\angle CXZ = \\angle AYZ,\n$$\nwhich by the converse of alternate segment theorem shows $DZ$ is tangent to circle $AXY$.\nNotice that point $Z$ is the Miquel-point of lines $AC$, $BC$, $BA$ and $DY$; then $B, D, Z, Y$ and $C, D, X, Y$ are concyclic. Moreover, $Z$ is the centre of the spiral similarity that maps $BC$ to $YX$.\n\nBy $BC \\perp YX$, the angle of that similarity is $90^{\\circ}$; hence the circles $ABCZ$ and $AXYZ$ are perpendicular, therefore the radius $OZ$ in circle $ABCZ$ is tangent to circle $AXYZ$.\n\n![](attached_image_2.png)\n\nBy $OW = OD$, the triangle $OWD$ is isosceles, and\n$$\n\\angle ZOA = 2 \\angle ZBA = 2 \\angle ZBY = 2 \\angle ZDY = \\angle ODW + \\angle DWO,\n$$\nso $D$ lies on line $ZO$ that is tangent to circle $AXY$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32143,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle, and let $\\ell_{1}$ and $\\ell_{2}$ be two parallel lines. For $i=1,2$, let $\\ell_{i}$ meet the lines $BC$, $CA$, and $AB$ at $X_{i}$, $Y_{i}$, and $Z_{i}$, respectively. Suppose that the line through $X_{i}$ perpendicular to $BC$, the line through $Y_{i}$ perpendicular to $CA$, and finally the line through $Z_{i}$ perpendicular to $AB$, determine a non-degenerate triangle $\\Delta_{i}$.\nShow that the circumcircles of $\\Delta_{1}$ and $\\Delta_{2}$ are tangent to each other.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Throughout the solutions, $\\Varangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^{\\circ}$.\nLet the vertices of $\\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that lines $E_{i}F_{i}$, $F_{i}D_{i}$ and $D_{i}E_{i}$ are the perpendiculars through $X, Y$ and $Z$, respectively, and denote the circumcircle of $\\Delta_{i}$ by $\\omega_{i}$.\nIn triangles $D_{1}Y_{1}Z_{1}$ and $D_{2}Y_{2}Z_{2}$ we have $Y_{1}Z_{1} \\parallel Y_{2}Z_{2}$ because they are parts of $\\ell_{1}$ and $\\ell_{2}$. Moreover, $D_{1}Y_{1} \\parallel D_{2}Y_{2}$ are perpendicular to $AC$ and $D_{1}Z_{1} \\parallel D_{2}Z_{2}$ are perpendicular to $AB$, so the two triangles are homothetic and their homothetic centre is $Y_{1}Y_{2} \\cap Z_{1}Z_{2} = A$. Hence, line $D_{1}D_{2}$ passes through $A$. Analogously, line $E_{1}E_{2}$ passes through $B$ and $F_{1}F_{2}$ passes through $C$.\n![](attached_image_1.png)\nThe corresponding sides of $\\Delta_{1}$ and $\\Delta_{2}$ are parallel, because they are perpendicular to the respective sides of triangle $ABC$. Hence, $\\Delta_{1}$ and $\\Delta_{2}$ are either homothetic, or they can be translated to each other. Using that $B, X_{2}, Z_{2}$ and $E_{2}$ are concyclic, $C, X_{2}, Y_{2}$ and $F_{2}$ are concyclic, $Z_{2}E_{2} \\perp AB$ and $Y_{2}, F_{2} \\perp AC$ we can calculate\n$$\n\\begin{align*}\n\\Varangle\\left(E_{1}E_{2}, F_{1}F_{2}\\right) &= \\Varangle\\left(E_{1}E_{2}, X_{1}X_{2}\\right) + \\Varangle\\left(X_{1}X_{2}, F_{1}F_{2}\\right) = \\Varangle\\left(BE_{2}, BX_{2}\\right) + \\Varangle\\left(CX_{2}, CF_{2}\\right) \\\\\n&= \\Varangle\\left(Z_{2}E_{2}, Z_{2}X_{2}\\right) + \\Varangle\\left(Y_{2}X_{2}, Y_{2}F_{2}\\right) = \\Varangle\\left(Z_{2}E_{2}, \\ell_{2}\\right) + \\Varangle\\left(\\ell_{2}, Y_{2}F_{2}\\right) \\\\\n&= \\Varangle\\left(Z_{2}E_{2}, Y_{2}F_{2}\\right) = \\Varangle(AB, AC) \\not\\equiv 0, \\tag{1}\n\\end{align*}\n$$\nand conclude that lines $E_{1}E_{2}$ and $F_{1}F_{2}$ are not parallel. Hence, $\\Delta_{1}$ and $\\Delta_{2}$ are homothetic; the lines $D_{1}D_{2}$, $E_{1}E_{2}$, and $F_{1}F_{2}$ are concurrent at the homothetic centre of the two triangles. Denote this homothetic centre by $H$.\nFor $i=1,2$, using (1), and that $A, Y_{i}, Z_{i}$ and $D_{i}$ are concyclic,\n$$\n\\begin{aligned}\n\\Varangle\\left(HE_{i}, HF_{i}\\right) &= \\Varangle\\left(E_{1}E_{2}, F_{1}F_{2}\\right) = \\Varangle(AB, AC) \\\\\n&= \\Varangle\\left(AZ_{i}, AY_{i}\\right) = \\Varangle\\left(D_{i}Z_{i}, D_{i}Y_{i}\\right) = \\Varangle\\left(D_{i}E_{i}, D_{i}F_{i}\\right),\n\\end{aligned}\n$$\nso $H$ lies on circle $\\omega_{i}$.\nThe same homothety that maps $\\Delta_{1}$ to $\\Delta_{2}$, sends $\\omega_{1}$ to $\\omega_{2}$ as well. Point $H$, that is the centre of the homothety, is a common point of the two circles, That finishes proving that $\\omega_{1}$ and $\\omega_{2}$ are tangent to each other.\nAs in the first solution, let the vertices of $\\Delta_{i}$ be $D_{i}, E_{i}, F_{i}$, such that $E_{i}F_{i}$, $F_{i}D_{i}$ and $D_{i}E_{i}$ are the perpendiculars through $X_{i}, Y_{i}$ and $Z_{i}$, respectively. In the same way we conclude that $\\left(A, D_{1}, D_{2}\\right), \\left(B, E_{1}, E_{2}\\right)$ and $\\left(C, F_{1}, F_{2}\\right)$ are collinear.\nThe corresponding sides of triangles $ABC$ and $D_{i}E_{i}F_{i}$ are perpendicular to each other. Hence, there is a spiral similarity with rotation $\\pm 90^{\\circ}$ that maps $ABC$ to $D_{i}E_{i}F_{i}$; let $M_{i}$ be the centre of that similarity. Hence, $\\Varangle\\left(M_{i}A, M_{i}D_{i}\\right) = \\Varangle\\left(M_{i}B, M_{i}E_{i}\\right) = \\Varangle\\left(M_{i}C, M_{i}F_{i}\\right) = 90^{\\circ}$. The circle with diameter $AD_{i}$ passes through $M_{i}, Y_{i}, Z_{i}$, so $M_{i}, A, Y_{i}, Z_{i}, D_{i}$ are concyclic; analogously ($M_{i}, B, X_{i}, Z_{i}, E_{i}$) and ($M_{i}, C, X_{i}, Y_{i}, F_{i}$) are concyclic.\nBy applying Desargues' theorem to triangles $ABC$ and $D_{i}E_{i}F_{i}$ we conclude that the lines $AD_{i}, BE_{i}$ and $BF_{i}$ are concurrent; let their intersection be $H$. Since $\\left(A, D_{1}, D_{2}\\right), \\left(B, E_{1}, E_{2}\\right)$ and $\\left(C, F_{1}, F_{2}\\right)$ are collinear, we obtain the same point $H$ for $i=1$ and $i=2$.\n![](attached_image_2.png)\nBy $\\Varangle(CB, CH) = \\Varangle\\left(CX_{i}, CF_{i}\\right) = \\Varangle\\left(Y_{i}X_{i}, Y_{i}F_{i}\\right) = \\Varangle\\left(Y_{i}Z_{i}, Y_{i}D_{i}\\right) = \\Varangle\\left(AZ_{i}, AD_{i}\\right) = \\Varangle(AB, AH)$, point $H$ lies on circle $ABC$.\nAnalogously, from $\\Varangle\\left(F_{i}D_{i}, F_{i}H\\right) = \\Varangle\\left(F_{i}Y_{i}, F_{i}C\\right) = \\Varangle\\left(X_{i}Y_{i}, X_{i}C\\right) = \\Varangle\\left(X_{i}Z_{i}, X_{i}B\\right) = \\Varangle\\left(E_{i}Z_{i}, E_{i}B\\right) = \\Varangle\\left(E_{i}D_{i}, E_{i}H\\right)$, we can see that point $H$ lies on circle $D_{i}E_{i}F_{i}$ as well. Therefore, circles $ABC$ and $D_{i}E_{i}F_{i}$ intersect at point $H$.\nThe spiral similarity moves the circle $ABC$ to circle $D_{i}E_{i}F_{i}$, so the two circles are perpendicular. Hence, both circles $D_{1}E_{1}F_{1}$ and $D_{2}E_{2}F_{2}$ are tangent to the radius of circle $ABC$ at $H$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32144,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute-angled triangle $A B C$, point $H$ is the foot of the altitude from $A$. Let $P$ be a moving point such that the bisectors $k$ and $\\ell$ of angles $P B C$ and $P C B$, respectively, intersect each other on the line segment $A H$. Let $k$ and $A C$ meet at $E$, let $\\ell$ and $A B$ meet at $F$, and let $E F$ and $A H$ meet at $Q$. Prove that, as $P$ varies, the line $P Q$ passes through a fixed point.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the reflections of the line $B C$ with respect to the lines $A B$ and $A C$ intersect at point $K$. We will prove that $P, Q$ and $K$ are collinear, so $K$ is the common point of the varying line $P Q$.\nLet lines $B E$ and $C F$ intersect at $I$. For every point $O$ and $d>0$, denote by ($O, d$) the circle centred at $O$ with radius $d$, and define $\\omega_{I}=(I, I H)$ and $\\omega_{A}=(A, A H)$. Let $\\omega_{K}$ and $\\omega_{P}$ be the incircle of triangle $K B C$ and the $P$-excircle of triangle $P B C$, respectively.\nSince $I H \\perp B C$ and $A H \\perp B C$, the circles $\\omega_{A}$ and $\\omega_{I}$ are tangent to each other at $H$. So, $H$ is the external homothetic centre of $\\omega_{A}$ and $\\omega_{I}$. From the complete quadrangle $B C E F$ we have $(A, I ; Q, H)=-1$, therefore $Q$ is the internal homothetic centre of $\\omega_{A}$ and $\\omega_{I}$. Since $B A$ and $C A$ are the external bisectors of angles $\\angle K B C$ and $\\angle K C B$, circle $\\omega_{A}$ is the $K$-excircle in triangle $B K C$. Hence, $K$ is the external homothetic centre of $\\omega_{A}$ and $\\omega_{K}$. Also it is clear that $P$ is the external homothetic centre of $\\omega_{I}$ and $\\omega_{P}$.\nLet point $T$ be the tangency point of $\\omega_{P}$ and $B C$, and let $T'$ be the tangency point of $\\omega_{K}$ and $B C$. Since $\\omega_{I}$ is the incircle and $\\omega_{P}$ is the $P$-excircle of $P B C$, $T C=B H$ and since $\\omega_{K}$ is the incircle and $\\omega_{A}$ is the $K$-excircle of $K B C$, $T' C=B H$. Therefore $T C=T' C$ and $T \\equiv T'$. It yields that $\\omega_{K}$ and $\\omega_{P}$ are tangent to each other at $T$.\n![](attached_image_1.png)\nLet point $S$ be the internal homothetic centre of $\\omega_{A}$ and $\\omega_{P}$, and let $S'$ be the internal homothetic centre of $\\omega_{I}$ and $\\omega_{K}$. It's obvious that $S$ and $S'$ lie on $B C$. We claim that $S \\equiv S'$. To prove our claim, let $r_{A}, r_{I}, r_{P}$, and $r_{K}$ be the radii of $\\omega_{A}, \\omega_{I}, \\omega_{P}$ and $\\omega_{K}$, respectively.\nIt is well known that if the sides of a triangle are $a, b, c$, its semiperimeter is $s=(a+b+c)/2$, and the radii of the incircle and the $a$-excircle are $r$ and $r_{a}$, respectively, then $r \\cdot r_{a}=(s-b)(s-c)$. Applying this fact to triangle $P B C$ we get $r_{I} \\cdot r_{P}=B H \\cdot C H$. The same fact in triangle $K C B$ yields $r_{K} \\cdot r_{A}=C T \\cdot B T$. Since $B H=C T$ and $B T=C H$, from these two we get\n$$\n\\frac{H S}{S T}=\\frac{r_{A}}{r_{P}}=\\frac{r_{I}}{r_{K}}=\\frac{H S'}{S' T},\n$$\nso $S=S'$ indeed.\nFinally, by applying the generalised Monge's theorem to the circles $\\omega_{A}, \\omega_{I}$, and $\\omega_{K}$ (with two pairs of internal and one pair of external common tangents), we can see that points $Q$, $S$, and $K$ are collinear. Similarly one can show that $Q, S$ and $P$ are collinear, and the result follows.\nAgain, let $B E$ and $C F$ meet at $I$, that is the incentre in triangle $B C P$; then $P I$ is the third angle bisector. From the tangent segments of the incircle we have $B P-C P= B H-C H$; hence, the possible points $P$ lie on a branch of a hyperbola $\\mathcal{H}$ with foci $B, C$, and $H$ is a vertex of $\\mathcal{H}$. Since $P I$ bisects the angle between the radii $B P$ and $C P$ of the hyperbola, line $P I$ is tangent to $\\mathcal{H}$.\n![](attached_image_2.png)\nLet $K$ be the second intersection of $P Q$ and $\\mathcal{H}$, we will show that $A K$ is tangent to $\\mathcal{H}$ at $K$; this property determines $K$.\nLet $G=K I \\cap A P$ and $M=P I \\cap A K$. From the complete quadrangle $B C E F$ we can see that ($H, Q ; I, A$) is harmonic, so in the complete quadrangle $A P I K$, point $H$ lies on line $G M$.\nConsider triangle $A I M$. Its side $A I$ is tangent to $\\mathcal{H}$ at $H$, the side $I M$ is tangent to $\\mathcal{H}$ at $P$, and $K$ is a common point of the third side $A M$ and the hyperbola such that the lines $A P$, $I K$ and $M H$ are concurrent at the generalised Gergonne-point $G$. It follows that the third side, $A M$ is also tangent to $\\mathcal{H}$ at $K$.\n(Alternatively, in the last step we can apply the converse of Brianchon's theorem to the degenerate hexagon $A H I P M K$. By the theorem there is a conic section $\\mathcal{H}'$ such that lines $A I, I M$ and $M A$ are tangent to $\\mathcal{H}'$ at $H, P$ and $K$, respectively. But the three points $H, K$ and $P$, together with the tangents at $H$ and $P$ uniquely determine $\\mathcal{H}'$, so indeed $\\mathcal{H}'=\\mathcal{H}$.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32145,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ and $A'B'C'$ be two triangles having the same circumcircle $\\omega$, and the same orthocentre $H$. Let $\\Omega$ be the circumcircle of the triangle determined by the lines $AA'$, $BB'$, and $CC'$. Prove that $H$, the centre of $\\omega$, and the centre of $\\Omega$ are collinear.\n(Denmark)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "In what follows, $\\Varangle(p, q)$ will denote the directed angle between lines $p$ and $q$, taken modulo $180^\\circ$. Denote by $O$ the centre of $\\omega$. In any triangle, the homothety with ratio $-\\frac{1}{2}$ centred at the centroid of the triangle takes the vertices to the midpoints of the opposite sides and it takes the orthocentre to the circumcentre. Therefore the triangles $ABC$ and $A'B'C'$ share the same centroid $G$ and the midpoints of their sides lie on a circle $\\rho$ with centre on $OH$. We will prove that $\\omega$, $\\Omega$, and $\\rho$ are coaxial, so in particular it follows that their centres are collinear on $OH$.\n\nLet $D = BB' \\cap CC'$, $E = CC' \\cap AA'$, $F = AA' \\cap BB'$, $S = BC' \\cap B'C$, and $T = BC \\cap B'C'$. Since $D$, $S$, and $T$ are the intersections of opposite sides and of the diagonals in the quadrilateral $BB'CC'$ inscribed in $\\omega$, by Brocard's theorem triangle $DST$ is self-polar with respect to $\\omega$, i.e. each vertex is the pole of the opposite side. We apply this in two ways.\n\n![](attached_image_1.png)\n\nFirst, from $D$ being the pole of $ST$ it follows that the inverse $D^*$ of $D$ with respect to $\\omega$ is the projection of $D$ onto $ST$. In particular, $D^*$ lies on the circle with diameter $SD$. If $N$ denotes the midpoint of $SD$ and $R$ the radius of $\\omega$, then the power of $O$ with respect to this circle is $ON^2 - ND^2 = OD \\cdot OD^* = R^2$. By rearranging, we see that $ND^2$ is the power of $N$ with respect to $\\omega$.\n\nSecond, from $T$ being the pole of $SD$ it follows that $OT$ is perpendicular to $SD$. Let $M$ and $M'$ denote the midpoints of $BC$ and $B'C'$. Then since $OM \\perp BC$ and $OM' \\perp B'C'$, it follows that $OMM'T$ is cyclic and\n$$\n\\Varangle(SD, BC) = \\Varangle(OT, OM) = \\Varangle(B'C', MM').\n$$\nFrom $BB'CC'$ being cyclic we also have $\\Varangle(BC, BB') = \\Varangle(CC', B'C')$, hence we obtain\n$$\n\\begin{aligned}\n\\Varangle(SD, BB') &= \\Varangle(SD, BC) + \\Varangle(BC, BB') \\\\\n&= \\Varangle(B'C', MM') + \\Varangle(CC', B'C') = \\Varangle(CC', MM').\n\\end{aligned}\n$$\nNow from the homothety mentioned in the beginning, we know that $MM'$ is parallel to $AA'$, hence the above implies that $\\Varangle(SD, BB') = \\Varangle(CC', AA')$, which shows that $\\Omega$ is tangent to $SD$ at $D$. In particular, $ND^2$ is also the power of $N$ with respect to $\\Omega$.\n\nAdditionally, from $BB'CC'$ being cyclic it follows that triangles $DBC$ and $DC'B'$ are inversely similar, so $\\Varangle(BB', DM') = \\Varangle(DM, CC')$. This yields\n$$\n\\begin{aligned}\n\\Varangle(SD, DM') &= \\Varangle(SD, BB') + \\Varangle(BB', DM') \\\\\n&= \\Varangle(CC', MM') + \\Varangle(DM, CC') = \\Varangle(DM, MM'),\n\\end{aligned}\n$$\nwhich shows that the circle $DMM'$ is also tangent to $SD$. Since $N$, $M$, and $M'$ are collinear on the Newton-Gauss line of the complete quadrilateral determined by the lines $BB'$, $CC'$, $BC'$, and $B'C$, it follows that $ND^2 = NM \\cdot NM'$. Hence $N$ has the same power with respect to $\\omega$, $\\Omega$, and $\\rho$.\n\nBy the same arguments there exist points on the tangents to $\\Omega$ at $E$ and $F$ which have the same power with respect to $\\omega$, $\\Omega$, and $\\rho$. The tangents to a given circle at three distinct points cannot be concurrent, hence we obtain at least two distinct points with the same power with respect to $\\omega$, $\\Omega$, and $\\rho$. Hence the three circles are coaxial, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32146,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A A^{\\prime} B C C^{\\prime} B^{\\prime}$ be a convex cyclic hexagon such that $A C$ is tangent to the incircle of the triangle $A^{\\prime} B^{\\prime} C^{\\prime}$, and $A^{\\prime} C^{\\prime}$ is tangent to the incircle of the triangle $A B C$. Let the lines $A B$ and $A^{\\prime} B^{\\prime}$ meet at $X$ and let the lines $B C$ and $B^{\\prime} C^{\\prime}$ meet at $Y$.\nProve that if $X B Y B^{\\prime}$ is a convex quadrilateral, then it has an incircle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Denote by $\\omega$ and $\\omega^{\\prime}$ the incircles of $\\triangle A B C$ and $\\triangle A^{\\prime} B^{\\prime} C^{\\prime}$ and let $I$ and $I^{\\prime}$ be the centres of these circles. Let $N$ and $N^{\\prime}$ be the second intersections of $B I$ and $B^{\\prime} I^{\\prime}$ with $\\Omega$, the circumcircle of $A^{\\prime} B C C^{\\prime} B^{\\prime} A$, and let $O$ be the centre of $\\Omega$. Note that $O N \\perp A C$, $O N^{\\prime} \\perp A^{\\prime} C^{\\prime}$ and $O N=O N^{\\prime}$ so $N N^{\\prime}$ is parallel to the angle bisector $I I^{\\prime}$ of $A C$ and $A^{\\prime} C^{\\prime}$. Thus $I I^{\\prime} \\| N N^{\\prime}$ which is antiparallel to $B B^{\\prime}$ with respect to $B I$ and $B^{\\prime} I^{\\prime}$. Therefore $B, I, I^{\\prime}, B^{\\prime}$ are concyclic.\n![](attached_image_1.png)\n\nFurther define $P$ as the intersection of $A C$ and $A^{\\prime} C^{\\prime}$ and $M$ as the antipode of $N^{\\prime}$ in $\\Omega$. Consider the circle $\\Gamma_{1}$ with centre $N$ and radius $N A=N C$ and the circle $\\Gamma_{2}$ with centre $M$ and radius $M A^{\\prime}=M C^{\\prime}$. Their radical axis passes through $P$ and is perpendicular to $M N \\perp N N^{\\prime} \\| I P$, so $I$ lies on their radical axis. Therefore, since $I$ lies on $\\Gamma_{1}$, it must also lie on $\\Gamma_{2}$. Thus, if we define $Z$ as the second intersection of $M I$ with $\\Omega$, we have that $I$ is the incentre of triangle $Z A^{\\prime} C^{\\prime}$. (Note that the point $Z$ can also be constructed directly via Poncelet's porism.)\n\nConsider the incircle $\\omega_{c}$ with centre $I_{c}$ of triangle $C^{\\prime} B^{\\prime} Z$. Note that $\\angle Z I C^{\\prime}=90^{\\circ}+ \\frac{1}{2} \\angle Z A^{\\prime} C^{\\prime}=90^{\\circ}+\\frac{1}{2} \\angle Z B^{\\prime} C^{\\prime}=\\angle Z I_{c} C^{\\prime}$, so $Z, I, I_{c}, C^{\\prime}$ are concyclic. Similarly $B^{\\prime}, I^{\\prime}, I_{c}, C^{\\prime}$ are concyclic.\n\nThe external centre of dilation from $\\omega$ to $\\omega_{c}$ is the intersection of $I_{c}$ and $C^{\\prime} Z$ ( $D$ in the picture), that is the radical centre of circles $\\Omega, C^{\\prime} I_{c} I Z$ and $I I^{\\prime} I_{c}$. Similarly, the external centre of dilation from $\\omega^{\\prime}$ to $\\omega_{c}$ is the intersection of $I^{\\prime} I_{c}$ and $B^{\\prime} C^{\\prime}$ ( $D^{\\prime}$ in the picture), that is the radical centre of circles $\\Omega, B^{\\prime} I^{\\prime} I_{c} C^{\\prime}$ and $I I^{\\prime} I_{c}$. Therefore the Monge line of $\\omega, \\omega^{\\prime}$ and $\\omega_{c}$ is line $D D^{\\prime}$, and the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$ coincide. Hence the external centre $T$ of dilation from $\\omega$ to $\\omega^{\\prime}$ is also on the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$.\n![](attached_image_2.png)\n\nNow since $B, I, I^{\\prime}, B^{\\prime}$ are concyclic, the intersection $T^{\\prime}$ of $B B^{\\prime}$ and $I I^{\\prime}$ is on the radical axis of $\\Omega$ and circle $I I^{\\prime} I_{c}$. Thus $T^{\\prime}=T$ and $T$ lies on line $B B^{\\prime}$. Finally, construct a circle $\\Omega_{0}$ tangent to $A^{\\prime} B^{\\prime}, B^{\\prime} C^{\\prime}, A B$ on the same side of these lines as $\\omega^{\\prime}$. The centre of dilation from $\\omega^{\\prime}$ to $\\Omega_{0}$ is $B^{\\prime}$, so by Monge's theorem the external centre of dilation from $\\Omega_{0}$ to $\\omega$ must be on the line $T B B^{\\prime}$. However, it is on line $A B$, so it must be $B$ and $B C$ must be tangent to $\\Omega_{0}$ as desired.\n![](attached_image_3.png)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32147,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples of positive integers $(a, b, p)$ with $p$ prime and\n$$\na^{p} = b! + p.\n$$",
    "options": [],
    "answer": "[(2, 2, 2), (3, 4, 3)]",
    "solution": "Solution 1. Clearly, $a > 1$. We consider three cases.\n\nCase 1: We have $a < p$. Then we either have $a \\leqslant b$ which implies $a \\mid a^{p} - b! = p$ leading to a contradiction, or $a > b$ which is also impossible since in this case we have $b! \\leqslant a! < a^{p} - p$, where the last inequality is true for any $p > a > 1$.\n\nCase 2: We have $a > p$. In this case $b! = a^{p} - p > p^{p} - p \\geqslant p!$ so $b > p$ which means that $a^{p} = b! + p$ is divisible by $p$. Hence, $a$ is divisible by $p$ and $b! = a^{p} - p$ is not divisible by $p^{2}$. This means that $b < 2p$. If $a < p^{2}$ then $a / p < p$ divides both $a^{p}$ and $b!$ and hence it also divides $p = a^{p} - b!$ which is impossible. On the other hand, the case $a \\geqslant p^{2}$ is also impossible since then $a^{p} \\geqslant (p^{2})^{p} > (2p-1)! + p \\geqslant b! + p$.\n\nComment. The inequality $p^{2p} > (2p-1)! + p$ can be shown e.g. by using\n$$\n(2p-1)! = [1 \\cdot (2p-1)] \\cdot [2 \\cdot (2p-2)] \\cdots [(p-1)(p+1)] \\cdot p < \\left(\\left(\\frac{2p}{2}\\right)^{2}\\right)^{p-1} \\cdot p = p^{2p-1}\n$$\nwhere the inequality comes from applying AM-GM to each of the terms in square brackets.\n\nCase 3: We have $a = p$. In this case $b! = p^{p} - p$. One can check that the values $p = 2, 3$ lead to the claimed solutions and $p = 5$ does not lead to a solution. So we now assume that $p \\geqslant 7$. We have $b! = p^{p} - p > p!$ and so $b \\geqslant p + 1$ which implies that\n$$\nv_{2}((p+1)!) \\leqslant v_{2}(b!) = v_{2}\\left(p^{p-1} - 1\\right) \\stackrel{LTE}{=} 2 v_{2}(p-1) + v_{2}(p+1) - 1 = v_{2}\\left(\\frac{p-1}{2} \\cdot (p-1) \\cdot (p+1)\\right),\n$$\nwhere in the middle we used lifting-the-exponent lemma. On the RHS we have three factors of $(p+1)!$. But, due to $p+1 \\geqslant 8$, there are at least 4 even numbers among $1, 2, \\ldots, p+1$, so this case is not possible.\n\n\nSolution 2. The cases $a \\neq p$ are covered as in solution 1, as are $p = 2, 3$. For $p \\geqslant 5$ we have $b! = p(p^{p-1} - 1)$. By Zsigmondy's Theorem there exists some prime $q$ that divides $p^{p-1} - 1$ but does not divide $p^{k} - 1$ for $k < p-1$. It follows that $\\operatorname{ord}_{q}(p) = p-1$, and hence $q \\equiv 1 \\bmod (p-1)$. Note that $p \\neq q$. But then we must have $q \\geqslant 2p-1$, giving\n$$\nb! \\geqslant (2p-1)! = [1 \\cdot (2p-1)] \\cdot [2 \\cdot (2p-2)] \\cdots [(p-1) \\cdot (p+1)] \\cdot p > (2p-1)^{p-1} p > p^{p} > p^{p} - p,\n$$\na contradiction.\n\n\nSolution 3. The cases $a \\neq p$ are covered as in solution 1, as are $p = 2, 3$. Also $b > p$, as $p^{p} > p! + p$ for $p > 2$. The cases $p = 5, 7, 11$ are also checked manually, so assume $p \\geqslant 13$. Let $q \\mid p+1$ be an odd prime. By LTE\n$$\nv_{q}\\left(p^{p} - p\\right) = v_{q}\\left(\\left(p^{2}\\right)^{\\frac{p-1}{2}} - 1\\right) = v_{q}\\left(p^{2} - 1\\right) + v_{q}\\left(\\frac{p-1}{2}\\right) = v_{q}(p+1)\n$$\nBut $b \\geqslant p+1$, so then $v_{q}(b!) > v_{q}(p+1)$, since $q < p+1$, a contradiction. This means that $p+1$ has no odd prime divisor, i.e. $p+1 = 2^{k}$ for some $k$.\n\nNow let $q \\mid p-1$ be an odd prime. By LTE\n$$\nv_{q}\\left(p^{p} - p\\right) = 2 v_{q}(p-1)\n$$\nLet $d = v_{q}(p-1)$. Then $p \\geqslant 1 + q^{d}$, so\n$$\nv_{q}(b!) \\geqslant v_{q}(p!) \\geqslant v_{q}\\left(q^{d}!\\right) > q^{d-1} \\geqslant 2d\n$$\nprovided $d \\geqslant 2$ and $q > 3$, or $d \\geqslant 3$.\nIf $q = 3, d = 2$ and $p \\geqslant 13$ then $v_{q}(b!) \\geqslant v_{q}(p!) \\geqslant v_{q}(13!) = 5 > 2d$. Either way, $d \\leqslant 1$.\nIf $p > 2q + 1$ (so $p > 3q$, as $q \\mid p-1$) then\n$$\nv_{q}(b!) \\geqslant v_{q}((3q)!) = 3\n$$\nso we must have $q \\geqslant \\frac{p}{2}$, in other words, $p-1 = 2q$. This implies that $p = 2^{k} - 1$ and $q = 2^{k-1} - 1$ are both prime, but it is not possible to have two consecutive Mersenne primes.\n\n\nSolution 4. Let $a = p, b > p$ and $p \\geqslant 5$ (the remaining cases are dealt with as in solution 3). Modulo $(p+1)^{2}$ it holds that\n$$\np^{p} - p = (p+1-1)^{p} - p \\equiv \\binom{p}{1}(p+1)(-1)^{p-1} + (-1)^{p} - p = p(p+1) - 1 - p = p^{2} - 1 \\not\\equiv 0 \\pmod{(p+1)^{2}}\n$$\nSince $p \\geqslant 5$, the numbers $2$ and $\\frac{p+1}{2}$ are distinct and less than or equal to $p$. Therefore, $p+1 \\mid p!$, and so $(p+1)^{2} \\mid (p+1)!$.\nBut $b \\geqslant p+1$, so $b! \\equiv 0 \\not\\equiv p^{p} - p \\pmod{(p+1)^{2}}$, a contradiction.\n\nTherefore, the only solutions are $(2, 2, 2)$ and $(3, 4, 3)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32148,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each $1 \\leqslant i \\leqslant 9$ and $T \\in \\mathbb{N}$, define $d_{i}(T)$ to be the total number of times the digit $i$ appears when all the multiples of $1829$ between $1$ and $T$ inclusive are written out in base $10$.\nShow that there are infinitely many $T \\in \\mathbb{N}$ such that there are precisely two distinct values among $d_{1}(T), d_{2}(T), \\ldots, d_{9}(T)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $n := 1829$. First, we choose some $k$ such that $n \\mid 10^{k} - 1$. For instance, any multiple of $\\varphi(n)$ would work since $n$ is coprime to $10$. We will show that either $T = 10^{k} - 1$ or $T = 10^{k} - 2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrarily large.\n\nFor this it suffices to show that $\\#\\left\\{ d_{i}\\left(10^{k} - 1\\right) : 1 \\leqslant i \\leqslant 9 \\right\\} \\leqslant 2$. Indeed, if\n$$\n\\#\\left\\{ d_{i}\\left(10^{k} - 1\\right) : 1 \\leqslant i \\leqslant 9 \\right\\} = 1\n$$\nthen, since $10^{k} - 1$ which consists of all nines is a multiple of $n$, we have\n$$\nd_{i}\\left(10^{k} - 2\\right) = d_{i}\\left(10^{k} - 1\\right) \\text{ for } i \\in \\{1, \\ldots, 8\\}, \\text{ and } d_{9}\\left(10^{k} - 2\\right) < d_{9}\\left(10^{k} - 1\\right)\n$$\nThis means that $\\#\\left\\{ d_{i}\\left(10^{k} - 2\\right) : 1 \\leqslant i \\leqslant 9 \\right\\} = 2$.\n\nTo prove that $\\#\\left\\{ d_{i}\\left(10^{k} - 1\\right) \\right\\} \\leqslant 2$ we need an observation. Let $\\overline{a_{k-1} a_{k-2} \\ldots a_{0}} \\in \\{1, \\ldots, 10^{k} - 1\\}$ be the decimal expansion of an arbitrary number, possibly with leading zeroes. Then $\\overline{a_{k-1} a_{k-2} \\ldots a_{0}}$ is divisible by $n$ if and only if $\\overline{a_{k-2} \\ldots a_{0} a_{k-1}}$ is divisible by $n$. Indeed, this follows from the fact that\n$$\n10 \\cdot \\overline{a_{k-1} a_{k-2} \\ldots a_{0}} - \\overline{a_{k-2} \\ldots a_{0} a_{k-1}} = (10^{k} - 1) \\cdot a_{k-1}\n$$\nis divisible by $n$.\n\nThis observation shows that the set of multiples of $n$ between $1$ and $10^{k} - 1$ is invariant under simultaneous cyclic permutation of digits when numbers are written with leading zeroes. Hence, for each $i \\in \\{1, \\ldots, 9\\}$ the number $d_{i}\\left(10^{k} - 1\\right)$ is $k$ times larger than the number of $k$ digit numbers which start from the digit $i$ and are divisible by $n$. Since the latter number is either $\\left\\lfloor 10^{k-1} / n \\right\\rfloor$ or $1 + \\left\\lfloor 10^{k-1} / n \\right\\rfloor$, we conclude that $\\#\\left\\{ d_{i}\\left(10^{k} - 1\\right) \\right\\} \\leqslant 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32149,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that $5^{n}-3^{n}$ is not divisible by $2^{n}+65$ for any positive integer $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Notice that if $n$ is even, then $3 \\mid m$, but $3 \\nmid 5^{n}-3^{n}$, contradiction. So, from now on we assume that $n$ is odd, $n=2k+1$. Obviously $n=1$ is not possible, so $n \\geqslant 3$. Notice that $m$ is coprime to $2$, $3$ and $5$.\nLet $m_1$ be the smallest positive multiple of $m$ that can be written in the form of either $|5a^{2}-3b^{2}|$ or $|a^{2}-15b^{2}|$ with some integers $a$ and $b$.\nNote that $5^{n}-3^{n}=5(5^{k})^{2}-3(3^{k})^{2}$ is a multiple of $m$, so the set of such multiples is non-empty, and therefore $m_1$ is well-defined.\n\nI. First we show that $m_1 \\leqslant 5m$. Consider the numbers\n$$\n5^{k+1}x+3^{k+1}y, \\quad 0 \\leqslant x, y \\leqslant \\sqrt{m}\n$$\nThere are $\\lfloor\\sqrt{m}\\rfloor+1>\\sqrt{m}$ choices for $x$ and $y$, so there are more than $m$ possible pairs $(x, y)$. Hence, two of these sums are congruent modulo $m$: $5^{k+1}x_1+3^{k+1}y_1 \\equiv 5^{k+1}x_2+3^{k+1}y_2 \\pmod{m}$.\nNow choose $a=x_1-x_2$ and $b=y_1-y_2$; at least one of $a, b$ is nonzero, and\n$$\n5^{k+1}a+3^{k+1}b \\equiv 0 \\quad (\\bmod m), \\quad |a|,|b| \\leqslant \\sqrt{m}\n$$\nFrom\n$$\n0 \\equiv (5^{k+1}a)^{2}-(3^{k+1}b)^{2}=5^{n+1}a^{2}-3^{n+1}b^{2} \\equiv 5 \\cdot 3^{n}a^{2}-3^{n+1}b^{2}=3^{n}(5a^{2}-3b^{2}) \\pmod{m}\n$$\nwe can see that $|5a^{2}-3b^{2}|$ is a multiple of $m$. Since at least one of $a$ and $b$ is nonzero, $5a^{2} \\neq 3b^{2}$. Hence, by the choice of $a, b$, we have $0<|5a^{2}-3b^{2}| \\leqslant \\max(5a^{2}, 3b^{2}) \\leqslant 5m$. That shows that $m_1 \\leqslant 5m$.\n\nII. Next, we show that $m_1$ cannot be divisible by $2$, $3$ and $5$. Since $m_1$ equals either $|5a^{2}-3b^{2}|$ or $|a^{2}-15b^{2}|$ with some integers $a, b$, we have six cases to check. In all six cases, we will get a contradiction by presenting another multiple of $m$, smaller than $m_1$.\n- If $5 \\mid m_1$ and $m_1=|5a^{2}-3b^{2}|$, then $5 \\mid b$ and $|a^{2}-15(\\frac{b}{5})^{2}|=\\frac{m_1}{5}<m_1$.\n- If $5 \\mid m_1$ and $m_1=|a^{2}-15b^{2}|$, then $5 \\mid a$ and $|5(\\frac{a}{5})^{2}-3b^{2}|=\\frac{m_1}{5}<m_1$.\n- If $3 \\mid m_1$ and $m_1=|5a^{2}-3b^{2}|$, then $3 \\mid a$ and $|b^{2}-15(\\frac{a}{3})^{2}|=\\frac{m_1}{3}<m_1$.\n- If $3 \\mid m_1$ and $m_1=|a^{2}-15b^{2}|$, then $3 \\mid a$ and $|5b^{2}-3(\\frac{a}{3})^{2}|=\\frac{m_1}{3}<m_1$.\n- If $2 \\mid m_1$ and $m_1=|5a^{2}-3b^{2}|$, then $|\\left(\\frac{5a-3b}{2}\\right)^{2}-15\\left(\\frac{a-b}{2}\\right)^{2}|=\\frac{m_1}{2}<m_1$.\n- If $2 \\mid m_1$ and $m_1=|a^{2}-15b^{2}|$, then $|5\\left(\\frac{a-3b}{2}\\right)^{2}-3\\left(\\frac{a-5b}{2}\\right)^{2}|=\\frac{m_1}{2}<m_1$.\n(The last two expressions can be obtained from $(\\sqrt{5}a+\\sqrt{3}b)(\\sqrt{5}-\\sqrt{3})=(5a-3b)+\\sqrt{15}(b-a)$ and $(a+\\sqrt{15}b)(\\sqrt{5}-\\sqrt{3})=\\sqrt{5}(a-3b)+\\sqrt{3}(5b-a)$.)\nIn all six cases, we found that either $\\frac{m_1}{2}$, $\\frac{m_1}{3}$ or $\\frac{m_1}{5}$ is of the form $|5x^{2}-3y^{2}|$ or $|x^{2}-15y^{2}|$. Since $m$ is coprime to $2$, $3$ and $5$, the presented number is a multiple of $m$, but this contradicts the minimality of $m_1$.\n\nIII. The last remaining case is $m_1=m$, so either $m=|5a^{2}-3b^{2}|$ or $m=|a^{2}-15b^{2}|$. We will get a contradiction by considering the two sides modulo $3$, $4$ and $5$.\n- $2^{n}+65=5a^{2}-3b^{2}$ is not possible, because $2^{n}+65 \\equiv 1 \\pmod{3}$, but $5a^{2}-3b^{2} \\not\\equiv 1 \\pmod{3}$.\n- $2^{n}+65=3b^{2}-5a^{2}$ is not possible, because $2^{n}+65 \\equiv 1 \\pmod{4}$, but $3b^{2}-5a^{2} \\not\\equiv 1 \\pmod{4}$.\n- $2^{n}+65=a^{2}-15b^{2}$ is not possible, because $2^{n}+65 \\equiv \\pm 2 \\pmod{5}$, but $a^{2}-15b^{2} \\not\\equiv \\pm 2 \\pmod{5}$.\n- $2^{n}+65=15b^{2}-a^{2}$ is not possible, because $2^{n}+65 \\equiv 1 \\pmod{4}$, but $15b^{2}-a^{2} \\not\\equiv 1 \\pmod{4}$.\nWe found a contradiction in all cases, that completes the solution.\nSuppose again that $5^{n} \\equiv 3^{n} \\pmod{m=2^{n}+65}$. Like in the first solution, we conclude that $n$ must be odd, and $n \\geqslant 3$, so $8 \\mid 2^{n}$.\nUsing Jacobi symbols,\n$$\n-1=\\left(\\frac{2^{n}+65}{5}\\right)=\\left(\\frac{5}{2^{n}+65}\\right)=\\left(\\frac{5^{n}}{2^{n}+65}\\right)=\\left(\\frac{3^{n}}{2^{n}+65}\\right)=\\left(\\frac{3}{2^{n}+65}\\right)=\\left(\\frac{2^{n}+65}{3}\\right)=1,\n$$\ncontradiction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32150,
    "subject": "Mathematics (Multi-modal)",
    "question": "Professor Oak is feeding his 100 Pokémon. Each Pokémon has a bowl whose capacity is a positive real number of kilograms. These capacities are known to Professor Oak. The total capacity of all the bowls is 100 kilograms. Professor Oak distributes 100 kilograms of food in such a way that each Pokémon receives a non-negative integer number of kilograms of food (which may be larger than the capacity of their bowl). The dissatisfaction level of a Pokémon who received $N$ kilograms of food and whose bowl has a capacity of $C$ kilograms is equal to $|N-C|$.\nFind the smallest real number $D$ such that, regardless of the capacities of the bowls, Professor Oak can distribute the food in a way that the sum of the dissatisfaction levels over all the 100 Pokémon is at most $D$.",
    "options": [],
    "answer": "50",
    "solution": "Answer: The answer is $D=50$.\n\nSolution 1. First, consider the situation where 99 bowls have a capacity of 0.5 kilograms and the last bowl has a capacity of 50.5 kilograms. No matter how Professor Oak distributes the food, the dissatisfaction level of every Pokémon will be at least 0.5. This amounts to a total dissatisfaction level of at least 50, proving that $D \\geqslant 50$.\n\nNow we prove that no matter what the capacities of the bowls are, Professor Oak can always distribute food in a way that the total dissatisfaction level is at most 50. We start by fixing some notation. We number the Pokémon from 1 to 100. Let $C_{i}>0$ be the capacity of the bowl of the $i^{\\text{th}}$ Pokémon. By assumption, we have $C_{1}+C_{2}+\\cdots+C_{100}=100$. We write $F_{i}:=C_{i}-\\left\\lfloor C_{i}\\right\\rfloor$ for the fractional part of $C_{i}$. Without loss of generality, we may assume that $F_{1} \\leqslant F_{2} \\leqslant \\cdots \\leqslant F_{100}$.\n\nHere is a strategy: Professor Oak starts by giving $\\left\\lfloor C_{i}\\right\\rfloor$ kilograms of food to the $i^{\\text{th}}$ Pokémon. Let\n$$\nR:=100-\\left\\lfloor C_{1}\\right\\rfloor-\\left\\lfloor C_{2}\\right\\rfloor-\\cdots-\\left\\lfloor C_{100}\\right\\rfloor=F_{1}+F_{2}+\\cdots+F_{100} \\geqslant 0\n$$\nbe the amount of food left. He continues by giving an extra kilogram of food to the $R$ Pokémon numbered $100-R+1,100-R+2, \\ldots, 100$, i.e. the Pokémon with the $R$ largest values of $F_{i}$. By doing so, Professor Oak distributed 100 kilograms of food. The total dissatisfaction level with this strategy is\n$$\nd:=F_{1}+\\cdots+F_{100-R}+\\left(1-F_{100-R+1}\\right)+\\cdots+\\left(1-F_{100}\\right).\n$$\nWe can rewrite\n$$\n\\begin{aligned}\nd & =2\\left(F_{1}+\\cdots+F_{100-R}\\right)+R-\\left(F_{1}+\\cdots+F_{100}\\right) \\\\\n& =2\\left(F_{1}+\\cdots+F_{100-R}\\right).\n\\end{aligned}\n$$\nNow, observe that the arithmetic mean of $F_{1}, F_{2}, \\ldots, F_{100-R}$ is not greater than the arithmetic mean of $F_{1}, F_{2}, \\ldots, F_{100}$, because we assumed $F_{1} \\leqslant F_{2} \\leqslant \\cdots \\leqslant F_{100}$. Therefore\n$$\nd \\leqslant 2(100-R) \\cdot \\frac{F_{1}+\\cdots+F_{100}}{100}=2 \\cdot \\frac{R(100-R)}{100}.\n$$\nFinally, we use the AM-GM inequality to see that $R(100-R) \\leqslant \\frac{100^{2}}{2^{2}}$ which implies $d \\leqslant 50$. We conclude that there is always a distribution for which the total dissatisfaction level is at most 50, proving that $D \\leqslant 50$.\nWe adopt the same notation as in Solution 1. Let $C_{i}>0$ be the capacity of the bowl of the $i^{\\text{th}}$ Pokémon. By assumption, we have $C_{1}+C_{2}+\\cdots+C_{100}=100$. We write $F_{i}:=C_{i}-\\left\\lfloor C_{i}\\right\\rfloor$ for the fractional part of $C_{i}$, and $R=F_{1}+F_{2}+\\cdots+F_{100}$. Note that $R=100-\\left\\lfloor C_{1}\\right\\rfloor-\\cdots-\\left\\lfloor C_{100}\\right\\rfloor$ is an integer.\n\nThis solution uses the probabilistic method. We consider all distributions in which each Pokémon receives $\\left\\lfloor C_{i}\\right\\rfloor+\\varepsilon_{i}$ kilograms of food, where $\\varepsilon_{i} \\in\\{0,1\\}$ and $\\varepsilon_{1}+\\varepsilon_{2}+\\cdots+\\varepsilon_{100}=R$. There are $\\binom{100}{R}$ such distributions. Suppose each of them occurs in an equal probability. In other words,\n$$\n\\varepsilon_{i}= \\begin{cases}0 & \\text{ with probability } \\frac{100-R}{100} \\\\ 1 & \\text{ with probability } \\frac{R}{100}\\end{cases}\n$$\nThe expected value of the dissatisfaction level of the $i^{\\text{th}}$ Pokémon is\n$$\n\\frac{100-R}{100}\\left(C_{i}-\\left\\lfloor C_{i}\\right\\rfloor\\right)+\\frac{R}{100}\\left(\\left\\lfloor C_{i}\\right\\rfloor+1-C_{i}\\right)=\\frac{100-R}{100} F_{i}+\\frac{R}{100}\\left(1-F_{i}\\right).\n$$\nHence, the expected value of the total dissatisfaction level is\n$$\n\\begin{aligned}\n\\sum_{i=1}^{100}\\left(\\frac{100-R}{100} F_{i}+\\frac{R}{100}\\left(1-F_{i}\\right)\\right) & =\\frac{100-R}{100} \\sum_{i=1}^{100} F_{i}+\\frac{R}{100} \\sum_{i=1}^{100}\\left(1-F_{i}\\right) \\\\\n& =\\frac{100-R}{100} \\cdot R+\\frac{R}{100} \\cdot(100-R) \\\\\n& =2 \\cdot \\frac{R(100-R)}{100}.\n\\end{aligned}\n$$\nAs in Solution 1, this is at most 50. We conclude that there is at least one distribution for which the total dissatisfaction level is at most 50.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32151,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x_{1}, x_{2}, \\ldots, x_{2023}$ be distinct real positive numbers such that\n$$\na_{n}=\\sqrt{\\left(x_{1}+x_{2}+\\cdots+x_{n}\\right)\\left(\\frac{1}{x_{1}}+\\frac{1}{x_{2}}+\\cdots+\\frac{1}{x_{n}}\\right)}\n$$\nis an integer for every $n=1,2, \\ldots, 2023$. Prove that $a_{2023} \\geqslant 3034$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start with some basic observations. First note that the sequence $a_{1}, a_{2}, \\ldots, a_{2023}$ is increasing and thus, since all elements are integers, $a_{n+1}-a_{n} \\geqslant 1$. We also observe that $a_{1}=1$ and\n$$\na_{2}=\\sqrt{\\left(x_{1}+x_{2}\\right)\\left(\\frac{1}{x_{1}}+\\frac{1}{x_{2}}\\right)}>2\n$$\nby Cauchy-Schwarz inequality and using $x_{1} \\neq x_{2}$. So, $a_{2} \\geqslant 3$.\n\nNow, we proceed to the main part of the argument. We observe that 3034 is about three halves of 2023. Motivated by this observation, we will prove the following.\n\n**Claim.** If $a_{n+1}-a_{n}=1$, then $a_{n+2}-a_{n+1} \\geqslant 2$.\n\nIn other words, the sequence has to increase by at least 2 at least half of the times. Assuming the claim is true, since $a_{1}=1$, we would be done since\n$$\n\\begin{aligned}\na_{2023} & =\\left(a_{2023}-a_{2022}\\right)+\\left(a_{2022}-a_{2021}\\right)+\\cdots+\\left(a_{2}-a_{1}\\right)+a_{1} \\\\\n& \\geqslant(2+1) \\cdot 1011+1 \\\\\n& =3034\n\\end{aligned}\n$$\n\nWe now prove the claim. We start by observing that\n$$\n\\begin{aligned}\na_{n+1}^{2}= & \\left(x_{1}+\\cdots+x_{n+1}\\right)\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n+1}}\\right) \\\\\n= & \\left(x_{1}+\\cdots+x_{n}\\right)\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n}}\\right)+1 \\\\\n& +\\frac{1}{x_{n+1}}\\left(x_{1}+\\cdots+x_{n}\\right)+x_{n+1}\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n}}\\right) \\\\\n\\geqslant & a_{n}^{2}+1+2 \\sqrt{\\frac{1}{x_{n+1}}\\left(x_{1}+\\cdots+x_{n}\\right) \\cdot x_{n+1}\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n}}\\right)} \\\\\n= & a_{n}^{2}+1+2 a_{n} \\\\\n= & \\left(a_{n}+1\\right)^{2},\n\\end{aligned}\n$$\nwhere we used AM-GM to obtain the inequality. In particular, if $a_{n+1}=a_{n}+1$, then\n$$\n\\begin{equation*}\n\\frac{1}{x_{n+1}}\\left(x_{1}+\\cdots+x_{n}\\right)=x_{n+1}\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n}}\\right) . \\tag{1}\n\\end{equation*}\n$$\nNow, assume for the sake of contradiction that both $a_{n+1}=a_{n}+1$ and $a_{n+2}=a_{n+1}+1$ hold. In this case, (1) gives\n$$\n\\frac{1}{x_{n+2}}\\left(x_{1}+\\cdots+x_{n+1}\\right)=x_{n+2}\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n+1}}\\right) .\n$$\nWe can rewrite this relation as\n$$\n\\frac{x_{n+1}}{x_{n+2}}\\left(\\frac{1}{x_{n+1}}\\left(x_{1}+\\cdots+x_{n}\\right)+1\\right)=\\frac{x_{n+2}}{x_{n+1}}\\left(x_{n+1}\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n}}\\right)+1\\right) .\n$$\nFrom (1) again, we conclude that $x_{n+1}=x_{n+2}$ which is a contradiction.\nThe trick is to compare $a_{n+2}$ and $a_{n}$. Observe that\n$$\n\\begin{aligned}\na_{n+2}^{2}= & \\left(x_{1}+\\cdots+x_{n+2}\\right)\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n+2}}\\right) \\\\\n= & \\left(x_{1}+\\cdots+x_{n}\\right)\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n}}\\right)+\\left(x_{n+1}+x_{n+2}\\right)\\left(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}\\right) \\\\\n& \\quad+\\left(x_{1}+\\cdots+x_{n}\\right)\\left(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}\\right)+\\left(x_{n+1}+x_{n+2}\\right)\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n}}\\right) \\\\\n\\geqslant & a_{n}^{2}+\\left(x_{n+1}+x_{n+2}\\right)\\left(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}\\right) \\\\\n& \\quad+2 \\sqrt{\\left(x_{n+1}+x_{n+2}\\right)\\left(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}\\right)\\left(x_{1}+\\cdots+x_{n}\\right)\\left(\\frac{1}{x_{1}}+\\cdots+\\frac{1}{x_{n}}\\right)} \\\\\n= & a_{n}^{2}+\\left(x_{n+1}+x_{n+2}\\right)\\left(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}\\right)+2 a_{n} \\sqrt{\\left(x_{n+1}+x_{n+2}\\right)\\left(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}\\right)},\n\\end{aligned}\n$$\nwhere we used AM-GM to obtain the inequality. Furthermore, we have\n$$\n\\left(x_{n+1}+x_{n+2}\\right)\\left(\\frac{1}{x_{n+1}}+\\frac{1}{x_{n+2}}\\right)>4\n$$\nbecause $x_{n+1} \\neq x_{n+2}$ by assumption. Therefore, it follows that\n$$\na_{n+2}^{2}>a_{n}^{2}+4+4 a_{n}=\\left(a_{n}+2\\right)^{2}\n$$\nBecause $a_{n+2}$ and $a_{n}$ are both positive integers, we conclude that\n$$\na_{n+2} \\geqslant a_{n}+3\n$$\nA simple induction gives $a_{2 k+1} \\geqslant 3 k+a_{1}$ for every $k \\geqslant 0$. Since $a_{1}=1$, it follows that $a_{2 k+1} \\geqslant 3 k+1$. We get the desired conclusion for $k=1011$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32152,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}_{>0}$ be the set of positive real numbers. Determine all functions $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ such that\n$$\nx(f(x)+f(y)) \\geqslant (f(f(x))+y) f(y)\n$$\nfor every $x, y \\in \\mathbb{R}_{>0}$.",
    "options": [],
    "answer": "f(x) = c/x for some c > 0",
    "solution": "Answer: All functions $f(x)=\\frac{c}{x}$ for some $c>0$.\n\nSolution 1. Let $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. We will write $f^{k}(x)=f(f(\\cdots f(x) \\cdots))$ for the composition of $f$ with itself $k$ times, with the convention that $f^{0}(x)=x$. Substituting $y=x$ gives\n$$\nx \\geqslant f^{2}(x)\n$$\nSubstituting $x=f(y)$ instead leads to $f(y)+f^{2}(y) \\geqslant y+f^{3}(y)$, or equivalently\n$$\nf(y)-f^{3}(y) \\geqslant y-f^{2}(y)\n$$\nWe can generalise this inequality. If we replace $y$ by $f^{n-1}(y)$ in the above inequality, we get\n$$\nf^{n}(y)-f^{n+2}(y) \\geqslant f^{n-1}(y)-f^{n+1}(y)\n$$\nfor every $y \\in \\mathbb{R}_{>0}$ and for every integer $n \\geqslant 1$. In particular, $f^{n}(y)-f^{n+2}(y) \\geqslant y-f^{2}(y) \\geqslant 0$ for every $n \\geqslant 1$. Hereafter consider even integers $n=2 m$. Observe that\n$$\ny-f^{2 m}(y)=\\sum_{i=0}^{m-1}\\left(f^{2 i}(y)-f^{2 i+2}(y)\\right) \\geqslant m\\left(y-f^{2}(y)\\right) .\n$$\nSince $f$ takes positive values, it holds that $y-f^{2 m}(y)<y$ for every $m \\geqslant 1$. So, we have proved that $y>m\\left(y-f^{2}(y)\\right)$ for every $y \\in \\mathbb{R}_{>0}$ and every $m \\geqslant 1$. Since $y-f^{2}(y) \\geqslant 0$, this holds if only if\n$$\nf^{2}(y)=y\n$$\nfor every $y \\in \\mathbb{R}_{>0}$. The original inequality becomes\n$$\nx f(x) \\geqslant y f(y)\n$$\nfor every $x, y \\in \\mathbb{R}_{>0}$. Hence, $x f(x)$ is constant. We conclude that $f(x)=c / x$ for some $c>0$.\nWe now check that all the functions of the form $f(x)=c / x$ are indeed solutions of the original problem. First, note that all these functions satisfy $f(f(x))=c /(c / x)=x$. So it's sufficient to check that $x f(x) \\geqslant y f(y)$, which is true since $c \\geqslant c$.\nLet $f: \\mathbb{R}_{>0} \\rightarrow \\mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. As in Solution 1, we prove that\n$$\nf^{n}(y) \\geqslant f^{n+2}(y)\n$$\nfor every $y \\in \\mathbb{R}_{>0}$ and every $n \\geqslant 0$. Since $f$ takes positive values, this implies that\n$$\ny f(y) \\geqslant f(y) f^{2}(y) \\geqslant f^{2}(y) f^{3}(y) \\geqslant \\cdots\n$$\nIn other words, $y f(y) \\geqslant f^{n}(y) f^{n+1}(y)$ for every $y \\in \\mathbb{R}_{>0}$ and every $n \\geqslant 1$.\nWe replace $x$ by $f^{n}(x)$ in the original inequality and get\n$$\nf^{n}(x)-f^{n+2}(x) \\geqslant \\frac{y f(y)-f^{n}(x) f^{n+1}(x)}{f(y)}\n$$\nUsing that $x f(x) \\geqslant f^{n}(x) f^{n+1}(x)$, we obtain\n$$\nf^{n}(x)-f^{n+2}(x) \\geqslant \\frac{y f(y)-x f(x)}{f(y)}\n$$\nfor every $n \\geqslant 0$. The same trick as in Solution 1 gives\n$$\nx>x-f^{2 m}(x)=\\sum_{i=0}^{m-1}\\left(f^{2 i}(x)-f^{2 i+2}(x)\\right) \\geqslant m \\cdot \\frac{y f(y)-x f(x)}{f(y)}\n$$\nfor every $x, y \\in \\mathbb{R}_{>0}$ and every $m \\geqslant 1$. Possibly permuting $x$ and $y$, we may assume that $y f(y)-x f(x) \\geqslant 0$ then the above inequality implies $x f(x)=y f(y)$. We conclude as in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32153,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, \\ldots, a_{2023}$ be positive integers such that\n- $a_{1}, a_{2}, \\ldots, a_{2023}$ is a permutation of $1,2, \\ldots, 2023$, and\n- $\\left|a_{1}-a_{2}\\right|,\\left|a_{2}-a_{3}\\right|, \\ldots,\\left|a_{2022}-a_{2023}\\right|$ is a permutation of $1,2, \\ldots, 2022$.\nProve that $\\max \\left(a_{1}, a_{2023}\\right) \\geqslant 507$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For the sake of clarity, we consider and prove the following generalisation of the original problem (which is the case $N=1012$):\nLet $N$ be a positive integer and $a_{1}, a_{2}, \\ldots, a_{2N-1}$ be positive integers such that\n- $a_{1}, a_{2}, \\ldots, a_{2N-1}$ is a permutation of $1,2, \\ldots, 2N-1$, and\n- $\\left|a_{1}-a_{2}\\right|,\\left|a_{2}-a_{3}\\right|, \\ldots,\\left|a_{2N-2}-a_{2N-1}\\right|$ is a permutation of $1,2, \\ldots, 2N-2$.\nThen $a_{1}+a_{2N-1} \\geqslant N+1$ and hence $\\max \\left(a_{1}, a_{2N-1}\\right) \\geqslant\\left\\lceil\\frac{N+1}{2}\\right\\rceil$.\n\nNow we proceed to the proof of the generalised statement. We introduce the notion of score of a number $a \\in\\{1,2, \\ldots, 2N-1\\}$. The score of $a$ is defined to be\n$$\ns(a):=|a-N| .\n$$\nNote that, by the triangle inequality,\n$$\n|a-b| \\leqslant|a-N|+|N-b|=s(a)+s(b)\n$$\nConsidering the sum $\\left|a_{1}-a_{2}\\right|+\\left|a_{2}-a_{3}\\right|+\\cdots+\\left|a_{2N-2}-a_{2N-1}\\right|$, we find that\n$$\n\\begin{aligned}\n(N-1)(2N-1) & =\\left|a_{1}-a_{2}\\right|+\\left|a_{2}-a_{3}\\right|+\\cdots+\\left|a_{2N-2}-a_{2N-1}\\right| \\\\\n& \\leqslant 2\\left(s\\left(a_{1}\\right)+s\\left(a_{2}\\right)+\\cdots+s\\left(a_{2N-1}\\right)\\right)-\\left(s\\left(a_{1}\\right)+s\\left(a_{2N-1}\\right)\\right) \\\\\n& =2N(N-1)-\\left(s\\left(a_{1}\\right)+s\\left(a_{2N-1}\\right)\\right)\n\\end{aligned}\n$$\nFor the last equality we used that the numbers $s\\left(a_{1}\\right), s\\left(a_{2}\\right), \\ldots, s\\left(a_{2N-1}\\right)$ are a permutation of $0,1,1,2,2, \\ldots, N-1, N-1$.\nHence, $s\\left(a_{1}\\right)+s\\left(a_{2N-1}\\right) \\leqslant 2N(N-1)-(N-1)(2N-1)=N-1$. We conclude that\n$$\n\\left(N-a_{1}\\right)+\\left(N-a_{2N-1}\\right) \\leqslant s\\left(a_{1}\\right)+s\\left(a_{2N-1}\\right) \\leqslant N-1,\n$$\nwhich implies $a_{1}+a_{2N-1} \\geqslant N+1$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32154,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $k \\geqslant 2$ be an integer. Determine all sequences of positive integers $a_{1}, a_{2}, \\ldots$ for which there exists a monic polynomial $P$ of degree $k$ with non-negative integer coefficients such that\n$$\nP\\left(a_{n}\\right)=a_{n+1} a_{n+2} \\cdots a_{n+k}\n$$\nfor every integer $n \\geqslant 1$.",
    "options": [],
    "answer": "Exactly the arithmetic progressions of positive integers with nonnegative common difference. If the common difference is zero, the sequence is constant and the polynomial is the pure power of degree k. If the common difference is a positive integer d, then the polynomial equals the product of k consecutive linear terms with increments d, namely the product of x plus d, x plus two d, up to x plus k d.",
    "solution": "Answer: The sequence $\\left(a_{n}\\right)$ must be an arithmetic progression consisting of positive integers with common difference $d \\geqslant 0$, and $P(x)=(x+d) \\cdots(x+k d)$.\n\nCommon remarks. The following arguments and observations are implicit in the solutions given below.\nSuppose the sequence $\\left(a_{n}\\right)$ is an arithmetic progression with common difference $d \\geqslant 0$. Then it satisfies the condition with\n$$\nP(x)=(x+d) \\cdots(x+k d)\\text{.}\n$$\nThis settles one direction. Now suppose ($a_{n}$) is a sequence satisfying the condition. We will show that it is a non-decreasing arithmetic progression.\nSince $P(x)$ has non-negative integer coefficients, it is strictly increasing on the positive real line. In particular, it holds that, for any positive integer $x, y$,\n$$\nP(x)<P(y) \\quad \\Longleftrightarrow \\quad x<y .\n$$\nFurthermore, if the sequence ($a_{n}$) is eventually constant, then $P(x)=x^{k}$ and the sequence $\\left(a_{n}\\right)$ is actually constant. Indeed, if $P(x)$ were not the polynomial $x^{k}$, then $P\\left(a_{n}\\right)=a_{n+1} \\cdots a_{n+k}$ cannot be satisfied for $n$ such that $a_{n}=\\cdots=a_{n+k}$. By a descending induction, we conclude that $\\left(a_{n}\\right)$ is constant. Thus we can restrict to the case $\\left(a_{n}\\right)$ is not eventually constant.\n\nSolution 1. We assume that ($a_{n}$) is not eventually constant.\n\nStep 1. The first goal is to show that the sequence must be increasing, i.e. $a_{n}<a_{n+1}$ for all $n \\geqslant 1$.\nFirst, by comparing the two equalities\n$$\n\\begin{aligned}\nP\\left(a_{n}\\right) & =a_{n+1} a_{n+2} \\cdots a_{n+k}, \\\\\nP\\left(a_{n+1}\\right) & =a_{n+2} \\cdots a_{n+k} a_{n+k+1},\n\\end{aligned}\n$$\nwe observe that\n$$\n\\begin{align*}\n& a_{n}<a_{n+1} \\Longleftrightarrow P\\left(a_{n}\\right)<P\\left(a_{n+1}\\right) \\quad \\Longleftrightarrow \\quad a_{n+1}<a_{n+k+1},  \\\\\n& a_{n}>a_{n+1} \\quad \\Longleftrightarrow P\\left(a_{n}\\right)>P\\left(a_{n+1}\\right) \\quad \\Longleftrightarrow \\quad a_{n+1}>a_{n+k+1},  \\\\\n& a_{n}=a_{n+1} \\quad \\Longleftrightarrow P\\left(a_{n}\\right)=P\\left(a_{n+1}\\right) \\quad \\Longleftrightarrow \\quad a_{n+1}=a_{n+k+1} .\n\\end{align*}\n$$\n\nClaim 1. $\\quad a_{n} \\leqslant a_{n+1}$ for all $n \\geqslant 1$.\n\nProof. Suppose, to the contrary, that $a_{n(0)-1}>a_{n(0)}$ for some $n(0) \\geqslant 2$. We will give an infinite sequence of positive integers $n(0)<n(1)<\\cdots$ satisfying\n$$\na_{n(i)-1}>a_{n(i)} \\text{ and } a_{n(i)}>a_{n(i+1)}\n$$\nThen $a_{n(0)}, a_{n(1)}, a_{n(2)}, \\ldots$ is an infinite decreasing sequence of positive integers, which is absurd.\nWe construct such a sequence inductively. If we have chosen $n(i)$, then we let $n(i+1)$ be the smallest index larger than $n(i)$ such that $a_{n(i)}>a_{n(i+1)}$. Note that such an index always exists and satisfies $n(i)+1 \\leqslant n(i+1) \\leqslant n(i)+k$ because $a_{n(i)}>a_{n(i)+k}$ by (2). We need to check that $a_{n(i+1)-1}>a_{n(i+1)}$. This is immediate if $n(i+1)=n(i)+1$ by construction. If $n(i+1) \\geqslant n(i)+2$, then $a_{n(i+1)-1} \\geqslant a_{n(i)}$ by minimality of $n(i+1)$, and so $a_{n(i+1)-1} \\geqslant a_{n(i)}>a_{n(i+1)}$.\n\nWe are now ready to prove that the sequence $a_{n}$ is increasing. Suppose $a_{n}=a_{n+1}$ for some $n \\geqslant 1$. Then we also have $a_{n+1}=a_{n+k+1}$ by (3), and since the sequence is non-decreasing we have $a_{n}=a_{n+1}=a_{n+2}=\\cdots=a_{n+k+1}$. We repeat the argument for $a_{n+k}=a_{n+k+1}$ and get that the sequence is eventually constant, which contradicts our assumption. Hence\n$$\na_{n}<a_{n+1} \\text{ for all } n \\geqslant 1 .\n$$\n\nStep 2. The next and final goal is to prove that the sequence $a_{n}$ is an arithmetic progression. Observe that we can make differences of terms appear as follows\n$$\n\\begin{aligned}\nP\\left(a_{n}\\right) & =a_{n+1} a_{n+2} \\cdots a_{n+k} \\\\\n& =\\left(a_{n}+\\left(a_{n+1}-a_{n}\\right)\\right)\\left(a_{n}+\\left(a_{n+2}-a_{n}\\right)\\right) \\cdots\\left(a_{n}+\\left(a_{n+k}-a_{n}\\right)\\right) .\n\\end{aligned}\n$$\nWe will prove that, for $n$ large enough, the sum\n$$\n\\left(a_{n+1}-a_{n}\\right)+\\left(a_{n+2}-a_{n}\\right)+\\cdots+\\left(a_{n+k}-a_{n}\\right)\n$$\nis equal to the coefficient $b$ of the term $x^{k-1}$ in $P$. The argument is based on the following claim.\n\nClaim 2. There exists a bound $A$ with the following properties:\n1. If $\\left(c_{1}, \\ldots, c_{k}\\right)$ is a $k$-tuple of positive integers with $c_{1}+\\cdots+c_{k}>b$, then for every $x \\geqslant A$ we have $P(x)<\\left(x+c_{1}\\right)\\left(x+c_{2}\\right) \\cdots\\left(x+c_{k}\\right)$.\n2. If $\\left(c_{1}, \\ldots, c_{k}\\right)$ is a $k$-tuple of positive integers with $c_{1}+\\cdots+c_{k}<b$, then for every $x \\geqslant A$ we have $P(x)>\\left(x+c_{1}\\right)\\left(x+c_{2}\\right) \\cdots\\left(x+c_{k}\\right)$.\n\nProof. It suffices to show parts 1 and 2 separately, because then we can take the maximum of two bounds.\nWe first show part 1. For each single $\\left(c_{1}, \\ldots, c_{k}\\right)$ such a bound $A$ exists since\n$$\nP(x)-\\left(x+c_{1}\\right)\\left(x+c_{2}\\right) \\cdots\\left(x+c_{k}\\right)=\\left(b-\\left(c_{1}+\\cdots+c_{k}\\right)\\right) x^{k-1}+(\\text{ terms of degree } \\leqslant k-2)\n$$\nhas negative leading coefficient and hence takes negative values for $x$ large enough.\nSuppose $A$ is a common bound for all tuples $c=\\left(c_{1}, \\ldots, c_{k}\\right)$ satisfying $c_{1}+\\cdots+c_{k}=b+1$ (note that there are only finitely many such tuples). Then, for any tuple $c^{\\prime}=\\left(c_{1}^{\\prime}, \\ldots, c_{k}^{\\prime}\\right)$ with $c_{1}^{\\prime}+\\cdots+c_{k}^{\\prime}>b$, there exists a tuple $c=\\left(c_{1}, \\ldots, c_{k}\\right)$ with $c_{1}+\\cdots+c_{k}=b+1$ and $c_{i}^{\\prime} \\geqslant c_{i}$, and then the inequality for $c^{\\prime}$ follows from the inequality for $c$.\nWe can show part 2 either in a similar way, or by using that there are only finitely many such tuples.\n\nTake $A$ satisfying the assertion of Claim 2, and take $N$ such that $n \\geqslant N$ implies $a_{n} \\geqslant A$. Then for each $n \\geqslant N$, we have\n$$\n\\left(a_{n+1}-a_{n}\\right)+\\cdots+\\left(a_{n+k}-a_{n}\\right)=b\n$$\nBy taking the difference of this equality and the equality for $n+1$, we obtain\n$$\na_{n+k+1}-a_{n+1}=k\\left(a_{n+1}-a_{n}\\right)\n$$\nfor every $n \\geqslant N$.\nWe conclude using an extremal principle. Let $d=\\min \\left\\{a_{n+1}-a_{n} \\mid n \\geqslant N\\right\\}$, and suppose it is attained at some index $n \\geqslant N$. Since\n$$\nk d=k\\left(a_{n+1}-a_{n}\\right)=a_{n+k+1}-a_{n+1}=\\sum_{i=1}^{k}\\left(a_{n+i+1}-a_{n+i}\\right)\n$$\nand each summand is at least $d$, we conclude that $d$ is also attained at $n+1, \\ldots, n+k$, and inductively at all $n^{\\prime} \\geqslant n$. We see that the equation $P(x)=(x+d)(x+2 d) \\cdots(x+k d)$ is true for infinitely many values of $x$ (all $a_{n^{\\prime}}$ for $n^{\\prime} \\geqslant n$ ), hence this is an equality of polynomials. Finally we use (backward) induction to show that $a_{n+1}-a_{n}=d$ for every $n \\geqslant 1$.\n\n\nSolution 2. We assume that ($a_{n}$) is not eventually constant. In this solution, we first prove an alternative version of Claim 1.\n\nClaim 3. There exist infinitely many $n \\geqslant 1$ with\n$$\na_{n} \\leqslant \\min \\left\\{a_{n+1}, \\ldots, a_{n+k}\\right\\} .\n$$\n\nProof. Suppose not, then for all but finitely many $n \\geqslant 1$, it holds that $a_{n}>\\min \\left\\{a_{n+1}, \\ldots, a_{n+k}\\right\\}$. Hence for all large enough $n$, there always exist some $1 \\leqslant l \\leqslant k$ such that $a_{n}>a_{n+l}$. This induces an infinite decreasing sequence $a_{n}>a_{n+l_{1}}>a_{n+l_{2}}>\\cdots$ of positive integers, which is absurd.\n\nWe use Claim 3 to quickly settle the case $P(x)=x^{k}$. In that case, for every $n$ with $a_{n} \\leqslant \\min \\left\\{a_{n+1}, \\ldots, a_{n+k}\\right\\}$, since $a_{n+1} \\cdots a_{n+k}=a_{n}^{k}$, it implies $a_{n}=a_{n+1}=\\cdots=a_{n+k}$. This shows that the sequence is eventually constant, which contradicts our assumption.\n\nFrom now on, assume\n$$\nP(x)>x^{k} \\text{ for all } x>0 \\text{. }\n$$\n\nClaim 4. For every $M>0$, there exists some $N>0$ such that $a_{n}>M$ for all $n>N$.\n\nProof. Suppose there exists some $M>0$, such that $a_{n} \\leqslant M$ for infinitely many $n$. For each $i$ with $a_{i} \\leqslant M$, we consider the $k$-tuple ($a_{i+1}, \\ldots, a_{i+k}$). Then each of the terms in the $k$-tuple is bounded from above by $P\\left(a_{i}\\right)$, and hence by $P(M)$ too. Since the number of such $k$-tuples is bounded by $P(M)^{k}$, we deduce by the Pigeonhole Principle that there exist some indices $i<j$ such that $\\left(a_{i+1}, \\ldots, a_{i+k}\\right)=\\left(a_{j+1}, \\ldots, a_{j+k}\\right)$. Since $a_{n}$ is uniquely determined by the $k$ terms before it, we conclude that $a_{i+k+1}=a_{j+k+1}$ must hold, and similarly $a_{i+l}=a_{j+l}$ for all $l \\geqslant 0$, so the sequence is eventually periodic, for some period $p=j-i$.\nTake $K$ such that $a_{n}=a_{n+p}$ for every $n \\geqslant K$. Then, by taking the products of the inequalities\n$$\na_{n}^{k}<P\\left(a_{n}\\right)=a_{n+1} \\cdots a_{n+k}\n$$\nfor $K \\leqslant n \\leqslant K+p-1$, we obtain\n$$\n\\begin{aligned}\n\\prod_{n=K}^{K+p-1} a_{n}^{k} & <\\prod_{n=K}^{K+p-1} a_{n+1} \\cdots a_{n+k} \\\\\n& =a_{K+1} a_{K+2}^{2} \\cdots a_{K+k-1}^{k-1}\\left(\\prod_{n=K+k}^{K+p} a_{n}\\right)^{k} a_{K+p+1}^{k-1} \\cdots a_{K+p+k-2}^{2} a_{K+p+k-1} \\\\\n& =\\left(\\prod_{n=K}^{K+p-1} a_{n}\\right)^{k} \\quad \\text{ (by periodicity), }\n\\end{aligned}\n$$\nwhich is a contradiction.\n\nWrite $P(x)=x^{k}+b x^{k-1}+Q(x)$, where $Q(x)$ is of degree at most $k-2$. Take $M$ such that $x>M$ implies $x^{k-1}>Q(x)$.\n\nClaim 5. There exist non-negative integers $b_{1}, \\cdots, b_{k}$ such that $P(x)=\\left(x+b_{1}\\right) \\cdots\\left(x+b_{k}\\right)$, and such that, for infinitely many $n \\geqslant 1$, we have $a_{n+i}=a_{n}+b_{i}$ for every $1 \\leqslant i \\leqslant k$.\n\nProof. By Claims 3 and 4, there are infinitely many $n$ such that\n$$\na_{n}>M \\text{ and } a_{n} \\leqslant \\min \\left\\{a_{n+1}, \\ldots, a_{n+k}\\right\\} .\n$$\nCall such indices $n$ to be good. We claim that if $n$ is a good index then\n$$\n\\max \\left\\{a_{n+1}, \\ldots, a_{n+k}\\right\\} \\leqslant a_{n}+b\n$$\nIndeed, if $a_{n+i} \\geqslant a_{n}+b+1$, then together with $a_{n} \\leqslant \\min \\left\\{a_{n+1}, \\ldots, a_{n+k}\\right\\}$ and $a_{n}^{k-1}>Q\\left(a_{n}\\right)$, we have\n$$\na_{n}^{k}+(b+1) a_{n}^{k-1}>a_{n}^{k}+b a_{n}^{k-1}+Q\\left(a_{n}\\right)=P\\left(a_{n}\\right) \\geqslant\\left(a_{n}+b+1\\right) a_{n}^{k-1}\n$$\na contradiction.\nHence for each good index $n$, we may write $a_{n+i}=a_{n}+b_{i}$ for all $1 \\leqslant i \\leqslant k$ for some choices of $\\left(b_{1}, \\ldots, b_{k}\\right)$ (which may depend on $n$ ) and $0 \\leqslant b_{i} \\leqslant b$. Again by Pigeonhole Principle, some $k$-tuple $\\left(b_{1}, \\ldots, b_{k}\\right)$ must be chosen for infinitely such good indices $n$. This means that the equation $P\\left(a_{n}\\right)=\\left(a_{n}+b_{1}\\right) \\cdots\\left(a_{n}+b_{k}\\right)$ is satisfied by infinitely many good indices $n$. By Claim 4, $a_{n}$ is unbounded among these $a_{n}$ 's, hence $P(x)=\\left(x+b_{1}\\right) \\cdots\\left(x+b_{k}\\right)$ must hold identically.\n\nClaim 6. We have $b_{i}=i b_{1}$ for all $1 \\leqslant i \\leqslant k$.\n\nProof. Call an index $n$ excellent if $a_{n+i}=a_{n}+b_{i}$ for every $1 \\leqslant i \\leqslant k$. From Claim 5 we know there are infinitely many excellent $n$.\nWe first show that for any pair $1 \\leqslant i<j \\leqslant k$ there is $1 \\leqslant l \\leqslant k$ such that $b_{j}=b_{i}+b_{l}$. Indeed, for such $i$ and $j$ and for excellent $n, a_{n}+b_{j}$ (which is equal to $a_{n+j}$ ) divides $P\\left(a_{n+i}\\right)= \\prod_{l=1}^{k}\\left(a_{n}+b_{i}+b_{l}\\right)$, and hence divides $\\prod_{l=1}^{k}\\left(b_{i}+b_{l}-b_{j}\\right)$. Since $a_{n}+b_{j}$ is unbounded among excellent $n$, we have $\\prod_{l=1}^{k}\\left(b_{i}+b_{l}-b_{j}\\right)=0$, hence there is $l$ such that $b_{j}=b_{i}+b_{l}$.\nIn particular, $b_{j}=b_{i}+b_{l} \\geqslant b_{i}$, i.e. $\\left(b_{1}, \\ldots, b_{k}\\right)$ is non-decreasing.\nSuppose $b_{1}=0$ and $n$ is an excellent number. In particular, it holds that $a_{n}=a_{n+1}$. Moreover, since\n$$\na_{n+k+1} P\\left(a_{n}\\right)=a_{n+1} \\cdots a_{n+k+1}=a_{n+1} P\\left(a_{n+1}\\right)\n$$\nwe have $a_{n}=a_{n+1}=a_{n+k+1}$, which divides $P\\left(a_{n+i}\\right)=\\prod_{l=1}^{k}\\left(a_{n}+b_{i}+b_{l}\\right)$ for each $1 \\leqslant i \\leqslant k$. Hence $a_{n}$ divides $\\prod_{l=1}^{k}\\left(b_{i}+b_{l}\\right)$. By the same reasoning, we have $b_{i}+b_{l}=0$ for some $l$, but since $b_{i}, b_{l} \\geqslant 0$ we obtain $b_{i}=0$ for each $1 \\leqslant i \\leqslant k$.\nNow suppose $b_{1} \\geqslant 1$. Then, for each $1 \\leqslant i<j \\leqslant k$, we have $b_{j}-b_{i}=b_{l} \\geqslant b_{1} \\geqslant 1$, hence $\\left(b_{1}, \\ldots, b_{k}\\right)$ is strictly increasing. Therefore, the $k-1$ elements $b_{2}<b_{3}<\\cdots<b_{k}$ are exactly equal to $b_{1}+b_{1}<\\cdots<b_{1}+b_{k-1}$, since they cannot be equal to $b_{1}+b_{k}$. This gives $b_{i}=i b_{1}$ for all $1 \\leqslant i \\leqslant k$ as desired.\n\nClaim 6 implies $P(x)=(x+d)(x+2 d) \\cdots(x+k d)$ for some $d \\geqslant 1$, and there are infinitely many indices $n$ with $a_{n+i}=a_{n}+i d$ for $1 \\leqslant i \\leqslant k$. By backwards induction, $P\\left(a_{n-1}\\right)= a_{n} \\cdots a_{n+k-1}$ implies $a_{n-1}=a_{n}-d$, and so on. Thus $a_{1}, \\ldots, a_{n}$ forms an arithmetic progression with common difference $d$. Since $n$ can be arbitrarily large, the whole sequence is an arithmetic progression too, as desired.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32155,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. Prove that there exist three permutations $a_{1}, a_{2}, \\ldots, a_{N}$; $b_{1}, b_{2}, \\ldots, b_{N}$; and $c_{1}, c_{2}, \\ldots, c_{N}$ of $1,2, \\ldots, N$ such that\n$$\n\\left|\\sqrt{a_{k}}+\\sqrt{b_{k}}+\\sqrt{c_{k}}-2 \\sqrt{N}\\right|<2023\n$$\nfor every $k=1,2, \\ldots, N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The idea is to approximate the numbers $\\sqrt{1}, \\sqrt{2}, \\ldots, \\sqrt{N}$ by the nearest integer with errors $<0.5$. This gives the following sequence\n$$\n1,1,2,2,2,2,3,3,3,3,3,3,4, \\ldots .\n$$\nMore precisely, for each $k \\geqslant 1$, we round $\\sqrt{k^{2}-k+1}, \\ldots, \\sqrt{k^{2}+k}$ to $k$, so that there are $2 k$ copies of $k$.\n\nStep 1. We first consider the easier case when $N$ has the form\n$$\nN=m(m+1) .\n$$\nIn this case, the numbers $\\sqrt{1}, \\sqrt{2}, \\ldots, \\sqrt{N}$ are approximated by the elements of the multiset $\\left\\{1_{\\times 2}, 2_{\\times 4}, 3_{\\times 6}, \\ldots, m_{\\times 2 m}\\right\\}$. Let $T_{m}$ denote \"half of\" the multiset, i.e.\n$$\nT_{m}:=\\left\\{1_{\\times 1}, 2_{\\times 2}, 3_{\\times 3}, \\ldots, m_{\\times m}\\right\\} .\n$$\nWe will prove by induction that there exists three permutations $\\left(u_{k}\\right),\\left(v_{k}\\right)$, and $\\left(w_{k}\\right)$ of the elements in the multiset $T_{m}$ such that $u_{k}+v_{k}+w_{k}=2 m+1$ is constant for $k=1,2, \\ldots, \\frac{m(m+1)}{2}$.\n\nWhen $m=1$, take $1+1+1=3$. When $m=2$, take $(1,2,2)+(2,1,2)+(2,2,1)=(5,5,5)$. Suppose that we have constructed three permutations $\\left(u_{k}\\right),\\left(v_{k}\\right)$, and $\\left(w_{k}\\right)$ of $T_{m-1}$ satisfying $u_{k}+v_{k}+w_{k}=2 m-1$ for every $k=1,2, \\ldots, \\frac{m(m-1)}{2}$. For $T_{m}$, we note that\n$$\nT_{m}=T_{m-1} \\sqcup\\left\\{m_{\\times m}\\right\\}\n$$\nand also\n$$\n\\begin{equation*}\nT_{m}=\\left(T_{m-1}+1\\right) \\sqcup\\{1,2, \\ldots, m\\} \\tag{1}\n\\end{equation*}\n$$\nHere $T_{m-1}+1$ means to add 1 to all elements in $T_{m-1}$. We construct the permutations $\\left(u_{k}^{\\prime}\\right)$, ( $v_{k}^{\\prime}$ ), and ( $w_{k}^{\\prime}$ ) of $T_{m}$ as follows:\n- For $k=1,2, \\ldots, \\frac{m(m-1)}{2}$, we set $u_{k}^{\\prime}=u_{k}, v_{k}^{\\prime}=v_{k}+1, w_{k}^{\\prime}=w_{k}+1$.\n- For $k=\\frac{m(m-1)}{2}+r$ with $r=1,2, \\ldots, m$, we set $u_{k}^{\\prime}=m, v_{k}^{\\prime}=r, w_{k}^{\\prime}=m+1-r$.\nIt is clear from (1) that $\\left(u_{k}^{\\prime}\\right),\\left(v_{k}^{\\prime}\\right)$, and $\\left(w_{k}^{\\prime}\\right)$ give three permutations of $T_{m}$, and that they satisfy $u_{k}^{\\prime}+v_{k}^{\\prime}+w_{k}^{\\prime}=2 m+1$ for every $k=1,2, \\ldots, \\frac{m(m+1)}{2}$.\n\nThe inductive construction can be visualised by the $3 \\times \\frac{m(m+1)}{2}$ matrix\n$$\n\\left[\\begin{array}{cccccc}\nu_{1} & \\ldots & u_{m(m-1) / 2} & m & \\ldots & m \\\nv_{1}+1 & \\ldots & v_{m(m-1) / 2}+1 & 1 & \\ldots & m \\\nw_{1}+1 & \\ldots & w_{m(m-1) / 2}+1 & m & \\ldots & 1\n\\end{array}\\right],\n$$\nin which the three rows represent the permutations $\\left(u_{k}^{\\prime}\\right),\\left(v_{k}^{\\prime}\\right),\\left(w_{k}^{\\prime}\\right)$, and the sum of the three entries of each column is $2 m+1$.\n\nThus, when $N=m^{2}+m$, we can construct permutations $\\left(a_{k}\\right),\\left(b_{k}\\right)$, and $\\left(c_{k}\\right)$ of $1,2, \\ldots, N$ such that\n$$\n\\begin{equation*}\n2 m+1-1.5<\\sqrt{a_{k}}+\\sqrt{b_{k}}+\\sqrt{c_{k}}<2 m+1+1.5 . \\tag{2}\n\\end{equation*}\n$$\nThis gives\n$$\n\\left|\\sqrt{a_{k}}+\\sqrt{b_{k}}+\\sqrt{c_{k}}-2 \\sqrt{N}\\right|<2.5<2023\n$$\nwhere we used that $-1<2 m-2 \\sqrt{m^{2}+m}<0$ for positive $m$.\n\nStep 2. We now proceed to the general case. Let $m$ be such that\n$$\nm(m+1) \\leqslant N<(m+1)(m+2)\n$$\nWrite $N=m(m+1)+t$ for some $t \\in\\{0,1, \\ldots, 2 m+1\\}$ and let\n$$\nL:=\\left\\lfloor\\frac{4}{9} N\\right\\rfloor\n$$\nWe will make use of the following inequalities below:\n$$\nN>m^{2}, \\quad N<(m+2)^{2}, \\quad t \\leqslant 2 m+1, \\quad L+1>4 N / 9, \\quad L \\leqslant 4 N / 9\n$$\nAs above, we construct three permutations $\\left(a_{k}\\right),\\left(b_{k}\\right)$, and $\\left(c_{k}\\right)$ of $1,2, \\ldots, m(m+1)$ satisfying (2). Now we construct the three required permutations $\\left(A_{k}\\right),\\left(B_{k}\\right)$, and $\\left(C_{k}\\right)$ of $1,2, \\ldots, N$ as follows:\nFor $k=1,2, \\ldots, m(m+1)$, if $a_{k} \\leqslant L$, take $A_{k}=a_{k}$, and if $a_{k}>L$, take $A_{k}=a_{k}+t$. For $k=m(m+1)+r$ with $r=1,2, \\ldots, t$, set $A_{k}=L+r$. Define the permutations ( $B_{k}$ ) and ( $C_{k}$ ) similarly. Now for $k=1,2, \\ldots, m(m+1)$, we show $0 \\leqslant \\sqrt{A_{k}}-\\sqrt{a_{k}} \\leqslant 2$. The lower bound is obvious. If $m \\leqslant 1$, then $N \\leqslant 5$ and hence $\\sqrt{A_{k}}-\\sqrt{a_{k}} \\leqslant \\sqrt{5}-\\sqrt{1} \\leqslant 2$. If $m \\geqslant 2$, then\n$$\n\\sqrt{A_{k}}-\\sqrt{a_{k}}=\\frac{A_{k}-a_{k}}{\\sqrt{A_{k}}+\\sqrt{a_{k}}} \\leqslant \\frac{t}{2 \\sqrt{L+1}} \\leqslant \\frac{2 m+1}{\\frac{4}{3} m} \\leqslant 2\n$$\nWe have similar inequalities for ( $B_{k}$ ) and ( $C_{k}$ ). Thus\n$$\n2 \\sqrt{N}-4.5<2 m+1-1.5 \\leqslant \\sqrt{A_{k}}+\\sqrt{B_{k}}+\\sqrt{C_{k}} \\leqslant 2 m+1+1.5+6<2 \\sqrt{N}+8.5\n$$\nFor $k=m^{2}+m+1, \\ldots, m^{2}+m+t$, we have\n$$\n2 \\sqrt{N}<3 \\sqrt{L+1} \\leqslant \\sqrt{A_{k}}+\\sqrt{B_{k}}+\\sqrt{C_{k}} \\leqslant 3 \\sqrt{L+t} \\leqslant \\sqrt{4 N+9 t}<2 \\sqrt{N}+8.5\n$$\nTo sum up, we have defined three permutations $\\left(A_{k}\\right),\\left(B_{k}\\right)$, and $\\left(C_{k}\\right)$ of $1,2, \\ldots, N$, such that\n$$\n\\left|\\sqrt{A_{k}}+\\sqrt{B_{k}}+\\sqrt{C_{k}}-2 \\sqrt{N}\\right|<8.5<2023\n$$\nholds for every $k=1,2, \\ldots, N$.\nThis is a variation of Solution 1 that uses induction for Step 2.\nLet $n$ be an integer satisfying $0 \\leqslant n \\leqslant m+1$ and define the multiset $T_{m, n}$ by\n$$\nT_{m, n}:=\\left\\{1_{\\times 1}, 2_{\\times 2}, 3_{\\times 3}, \\ldots, m_{\\times m},(m+1)_{\\times n}\\right\\}\n$$\nIn other words, $T_{m, 0}=T_{m}, T_{m, n}=T_{m} \\sqcup\\left\\{(m+1)_{\\times n}\\right\\}$ and $T_{m, m+1}=T_{m+1}$, where $T_{m}$ is the set defined in Solution 1.\n\nClaim. There exist three permutations $\\left(u_{k}\\right),\\left(v_{k}\\right),\\left(w_{k}\\right)$ of $T_{m, n}$ such that\n$$\n\\begin{cases}u_{k}+v_{k}+w_{k}=2 m+1 & (n=0), \\\\ u_{k}+v_{k}+w_{k} \\in\\{2 m+1,2 m+2,2 m+3\\} & (1 \\leqslant n \\leqslant m), \\\\ u_{k}+v_{k}+w_{k}=2 m+3 & (n=m+1) .\\end{cases}\n$$\nProof. We proceed by induction on $m$. If $n=0$ or $n=m+1$, the assertion can be proved as in Solution 1. If $1 \\leqslant n \\leqslant m$, we note that\n$$\nT_{m, n}=T_{m-1, n} \\sqcup\\left\\{m_{\\times(m-n)},(m+1)_{\\times n}\\right\\}=\\left(T_{m-1, n}+1\\right) \\sqcup\\{1,2, \\ldots, m\\}\n$$\nFrom the hypothesis of induction, it follows that we have three permutations $\\left(u_{k}\\right),\\left(v_{k}\\right),\\left(w_{k}\\right)$ of $T_{m-1, n}$ satisfying $u_{k}+v_{k}+w_{k} \\in\\{2 m-1,2 m, 2 m+1\\}$ for every $k$. We construct the permutations $\\left(u_{k}^{\\prime}\\right),\\left(v_{k}^{\\prime}\\right)$, and $\\left(w_{k}^{\\prime}\\right)$ of $T_{m, n}$ as follows:\n- For $k=1,2, \\ldots, \\frac{m(m-1)}{2}+n$, we set $u_{k}^{\\prime}=u_{k}, v_{k}^{\\prime}=v_{k}+1$, and $w_{k}^{\\prime}=w_{k}+1$.\n- For $k=\\frac{m(m-1)}{2}+n+r$ with $r=1,2, \\ldots, m$, we set $u_{k}^{\\prime}=m$ if $1 \\leqslant r \\leqslant m-n$ while $u_{k}^{\\prime}=m+1$ if $m-n+1 \\leqslant r \\leqslant m, v_{k}^{\\prime}=r$, and $w_{k}^{\\prime}=m+1-r$.\nIt is clear from the construction that $\\left(u_{k}^{\\prime}\\right),\\left(v_{k}^{\\prime}\\right)$, and $\\left(w_{k}^{\\prime}\\right)$ give three permutations of $T_{m, n}$, and they satisfy $u_{k}^{\\prime}+v_{k}^{\\prime}+w_{k}^{\\prime} \\in\\{2 m+1,2 m+2,2 m+3\\}$ for every $k=1,2, \\ldots, \\frac{m(m+1)}{2}+n$.\n\nAgain, we can visualise the construction using the matrix\n$$\n\\left[\\begin{array}{ccccccccc}\nu_{1} & \\ldots & u_{m(m-1) / 2+n} & m & \\ldots & m & m+1 & \\ldots & m+1 \\\\v_{1}+1 & \\ldots & v_{m(m-1) / 2+n}+1 & 1 & \\ldots & \\ldots & \\ldots & \\ldots & m \\\\w_{1}+1 & \\ldots & w_{m(m-1) / 2+n}+1 & m & \\ldots & \\ldots & \\ldots & \\ldots & 1\n\\end{array}\\right] .\n$$\n\nIn general, we have $m(m+1) \\leqslant N<(m+1)(m+2)$ for some $m \\geqslant 0$. Set $N=m(m+1)+t$ for some $t \\in\\{0,1, \\ldots, 2 m+1\\}$. Then the approximation of $\\{\\sqrt{1}, \\sqrt{2}, \\ldots, \\sqrt{N}\\}$ by the nearest integer with errors $<0.5$ is a multiset\n$$\n\\left\\{1_{\\times 2}, 2_{\\times 4}, \\ldots, m_{\\times 2 m},(m+1)_{\\times t}\\right\\}=T_{m, n_{1}} \\sqcup T_{m, n_{2}}\n$$\nwith $n_{1}=\\lfloor t / 2\\rfloor$ and $n_{2}=\\lceil t / 2\\rceil$.\nSince $0 \\leqslant n_{1} \\leqslant n_{2} \\leqslant m+1$, by using the Claim we can construct permutations ( $a_{k}$ ), ( $b_{k}$ ), and $\\left(c_{k}\\right)$ to satisfy the following inequality:\n$$\n2 m+1-1.5<\\sqrt{a_{k}}+\\sqrt{b_{k}}+\\sqrt{c_{k}}<2 m+3+1.5 .\n$$\nSince $m<\\sqrt{N}<m+2$, it follows that\n$$\n2 \\sqrt{N}-4.5<2 m+1-1.5<\\sqrt{a_{k}}+\\sqrt{b_{k}}+\\sqrt{c_{k}}<2 m+3+1.5<2 \\sqrt{N}+4.5\n$$\nand so\n$$\n\\left|\\sqrt{A_{k}}+\\sqrt{B_{k}}+\\sqrt{C_{k}}-2 \\sqrt{N}\\right|<4.5<2023\n$$\nThis solution is based on the geometrical insight of equilateral triangles.\n\nStep 1. We first consider the easier case of triangle numbers\n$$\nN=\\frac{m(m+1)}{2} .\n$$\nAs shown in the following picture, consider the triangular shaped lattice points inside an equilateral triangle $A B C$ with a total of $N$ points. The lattice is built in a way that the $\\ell^{\\text {th }}$ row has exactly $\\ell$ points for each $\\ell=1,2, \\ldots, m$. Rows are numbered in three different ways, one for each vertex.\n\nEach point $P_{k}$ in the triangular lattice is labelled with a triple of integers $\\left(a_{k}, b_{k}, c_{k}\\right)$ as follows. The first coordinate is called the $A$-coordinate, and so on for $B, C$. To define the $A$-coordinate, denoted $W_{a}(\\bullet)$, first label the lattice points by $1,2,3, \\ldots$ starting with the point closest to $A$ and then going down the rows with the rule that within a row, the labelling is from left to right (see right picture). The $B$-coordinate, denoted $W_{b}(\\bullet)$, is defined by rotating the $A$-coordinate counterclockwise by $120^{\\circ}$. The $C$-coordinate, denoted $W_{c}(\\bullet)$, similarly, by rotating the $A$-coordinate counterclockwise by $240^{\\circ}$.\n\nAssume that a point $P$ lies in the $\\ell_{a}{ }^{\\text {th }}$ row from the vertex $A$, in the $\\ell_{b}{ }^{\\text {th }}$ row from the vertex $B$, and in the $\\ell_{c}{ }^{\\text {th }}$ row from the vertex $C$. Note that $\\ell_{a}$ is proportional to the height of $A$ in the triangle, minus the height of $P$. Since inside an equilateral triangle, the sum of the lengths of the heights from a point to the three sides is independent of the point, we must have\n$$\n\\ell_{a}+\\ell_{b}+\\ell_{c}=2 m+1=\\sqrt{8 N+1} .\n$$\nSince there are exactly $1+2+\\cdots+\\ell=\\frac{\\ell(\\ell+1)}{2}$ points in the first $\\ell$ rows, the $A$-labeling $W_{a}(P)$ of the point $P$ satisfies\n$$\n\\frac{\\ell_{a}(\\ell_{a}-1)}{2}+1 \\leqslant W_{a}(P) \\leqslant \\frac{\\ell_{a}(\\ell_{a}+1)}{2} .\n$$\nIn paticular,\n$$\n\\left(\\ell_{a}-\\frac{1}{2}\\right)^{2}<2 W_{a}(P)<\\left(\\ell_{a}+\\frac{1}{2}\\right)^{2} .\n$$\nTaking the cyclic sum gives\n$$\n\\left|\\sqrt{2 W_{a}(P)}+\\sqrt{2 W_{b}(P)}+\\sqrt{2 W_{c}(P)}-\\left(\\ell_{a}+\\ell_{b}+\\ell_{c}\\right)\\right|<\\frac{3}{2}\n$$\nand thus\n$$\n\\left|\\sqrt{W_{a}(P)}+\\sqrt{W_{b}(P)}+\\sqrt{W_{c}(P)}-2 \\sqrt{N+\\frac{1}{8}}\\right|<\\frac{3}{2} \\cdot \\frac{1}{\\sqrt{2}}=\\frac{3 \\sqrt{2}}{4} .\n$$\n\nStep 2. Now, for a general positive integer $N$, there exists a positive integer $m$ such that\n$$\n\\frac{m(m-1)}{2}+1 \\leqslant N \\leqslant \\frac{m(m+1)}{2} .\n$$\nWrite $N=\\frac{m(m+1)}{2}-t$ with $t \\in\\{0,1, \\ldots, m-1\\}$. We modify the above construction for $\\frac{m(m+1)}{2}$ points into a construction for $N$ points as follows. We remove $t$ arbitrary points from the $m^{\\text {th }}$ row (namely the bottom row) of the triangular lattice. The remaining triangular lattice has $\\frac{m(m+1)}{2}-t=N$ points, and we assign their $A-, B$-, and $C$-coordinates as before (in the same order, yet skipping over the points that are removed so that the coordinates exactly form permutations of $1,2, \\ldots, N$ ).\n\nFor each point $P$ in the triangular lattice (that was not removed earlier), suppose that it is in the $\\ell_{a}{ }^{\\text {th }}, \\ell_{b}{ }^{\\text {th }}$, and $\\ell_{c}{ }^{\\text {th }}$ row when viewed from $A, B$, and $C$, respectively. Now the $A$-coordinates $W_{a}(P)$ still satisfies\n$$\n\\frac{\\ell_{a}(\\ell_{a}-1)}{2}+1 \\leqslant W_{a}(P) \\leqslant \\frac{\\ell_{a}(\\ell_{a}+1)}{2} .\n$$\nThe $B$-coordinate $W_{b}(P)$ satisfies\n$$\n\\frac{(\\ell_{b}-1)(\\ell_{b}-2)}{2}+1 \\leqslant W_{b}(P) \\leqslant \\frac{\\ell_{b}(\\ell_{b}+1)}{2}\n$$\nbecause, viewing from point $B$, we have removed either 0 or 1 point from each row, and the first $\\ell_{b}-1$ rows have at least $0+1+\\cdots+(\\ell_{b}-2)=\\frac{(\\ell_{b}-1)(\\ell_{b}-2)}{2}$ points left. For the same reason, the $C$-labeling $W_{c}(P)$ satisfies\n$$\n\\frac{(\\ell_{c}-1)(\\ell_{c}-2)}{2}+1 \\leqslant W_{c}(P) \\leqslant \\frac{\\ell_{c}(\\ell_{c}+1)}{2} .\n$$\nFrom this, we deduce that\n$$\n\\begin{aligned}\n\\ell_{a}-\\frac{1}{2} & <\\sqrt{2 W_{a}(P)}<\\ell_{a}+\\frac{1}{2}, \\\\\n\\ell_{b}-\\frac{3}{2} & <\\sqrt{2 W_{b}(P)}<\\ell_{b}+\\frac{1}{2}, \\\\\n\\ell_{c}-\\frac{3}{2} & <\\sqrt{2 W_{c}(P)}<\\ell_{c}+\\frac{1}{2} .\n\\end{aligned}\n$$\nCombining all above with the inequalities $2 m-1<2 \\sqrt{2 N}<2 m+1$ and $\\ell_{a}+\\ell_{b}+\\ell_{c}=2 m+1$, we deduce that\n$$\n\\begin{aligned}\n2 \\sqrt{2 N}-\\frac{7}{2}<(2 m+1)-\\frac{7}{2}<\\sqrt{2 W_{a}(P)} & +\\sqrt{2 W_{b}(P)}+\\sqrt{2 W_{c}(P)} \\\\\n& <(2 m+1)+\\frac{3}{2}<2 \\sqrt{2 N}+\\frac{7}{2}\n\\end{aligned}\n$$\nTherefore, for each point $P$, we have\n$$\n\\left|\\sqrt{W_{a}(P)}+\\sqrt{W_{b}(P)}+\\sqrt{W_{c}(P)}-2 \\sqrt{N}\\right|<\\frac{7}{2} \\cdot \\frac{1}{\\sqrt{2}}<2.5<2013 .\n$$\nWe may finally order of the $N$ points in an arbitrary way. Then the $A$-labelings $W_{a}(\\bullet)$ give the permutation $a_{1}, \\ldots, a_{N}$, the $B$-labelings $W_{b}(\\bullet)$ give $b_{1}, \\ldots, b_{N}$, and the $C$-labelings $W_{c}(\\bullet)$ give $c_{1}, \\ldots, c_{N}$.\n\nFor each $k=1,2, \\ldots, N$, we have\n$$\n\\left|\\sqrt{a_{k}}+\\sqrt{b_{k}}+\\sqrt{c_{k}}-2 \\sqrt{N}\\right|<2.5<2023\n$$\n\n![](attached_image_1.png)",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32156,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. We arrange $1+2+\\cdots+n$ circles in a triangle with $n$ rows, such that the $i^{\\text{th}}$ row contains exactly $i$ circles. The following figure shows the case $n=6$.\n![](attached_image_1.png)\nIn this triangle, a ninja-path is a sequence of circles obtained by repeatedly going from a circle to one of the two circles directly below it. In terms of $n$, find the largest value of $k$ such that if one circle from every row is coloured red, we can always find a ninja-path in which at least $k$ of the circles are red.",
    "options": [],
    "answer": "1 + floor(log_2 n)",
    "solution": "Answer: The maximum value is $k=1+\\left\\lfloor\\log_{2} n\\right\\rfloor$.\n\nSolution 1. Write $N=\\left\\lfloor\\log_{2} n\\right\\rfloor$ so that we have $2^{N} \\leqslant n \\leqslant 2^{N+1}-1$.\nWe first provide a construction where every ninja-path passes through at most $N+1$ red circles. For the row $i=2^{a}+b$ for $0 \\leqslant a \\leqslant N$ and $0 \\leqslant b<2^{a}$, we colour the $(2b+1)^{\\text{th}}$ circle.\n![](attached_image_2.png)\nThen every ninja-path passes through at most one red circle in each of the rows $2^{a}, 2^{a}+ 1, \\ldots, 2^{a+1}-1$ for each $0 \\leqslant a \\leqslant N$. It follows that every ninja-path passes through at most $N+1$ red circles.\n\nWe now prove that for every colouring, there exists a ninja-path going through at least $N+1$ red circles. For each circle $C$, we assign the maximum number of red circles in a ninja-path that starts at the top of the triangle and ends at $C$.\n![](attached_image_3.png)\nNote that\n- if $C$ is not red, then the number assigned to $C$ is the maximum of the number assigned to the one or two circles above $C$, and\n- if $C$ is red, then the number assigned to $C$ is one plus the above maximum.\n\nWrite $v_{1}, \\ldots, v_{i}$ for the numbers in row $i$, and let $v_{m}$ be the maximum among these numbers. Then the numbers in row $i+1$ will be at least\n$$\nv_{1}, \\ldots, v_{m-1}, v_{m}, v_{m}, v_{m+1}, \\ldots, v_{i},\n$$\nnot taking into account the fact that one of the circles in row $i+1$ is red. On the other hand, for the red circle in row $i+1$, the lower bound on the assigned number can be increased by 1. Therefore the sum of the numbers in row $i+1$ is at least\n$$\n\\left(v_{1}+\\cdots+v_{i}\\right)+v_{m}+1.\n$$\nUsing this observation, we prove the following claim.\n\nClaim 1. Let $\\sigma_{k}$ be the sum of the numbers assigned to circles in row $k$. Then for $0 \\leqslant j \\leqslant N$, we have $\\sigma_{2^{j}} \\geqslant j \\cdot 2^{j}+1$.\n\nProof. We use induction on $j$. This is clear for $j=0$, since the number in the first row is always 1. For the induction step, suppose that $\\sigma_{2^{j}} \\geqslant j \\cdot 2^{j}+1$. Then the maximum value assigned to a circle in row $2^{j}$ is at least $j+1$. As a consequence, for every $k \\geqslant 2^{j}$, there is a circle on row $k$ with number at least $j+1$. Then by our observation above, we have\n$$\n\\sigma_{k+1} \\geqslant \\sigma_{k}+(j+1)+1=\\sigma_{k}+(j+2).\n$$\nThen we get\n$$\n\\sigma_{2^{j+1}} \\geqslant \\sigma_{2^{j}}+2^{j}(j+2) \\geqslant j \\cdot 2^{j}+1+2^{j}(j+2)=(j+j+2) 2^{j}+1=(j+1) 2^{j+1}+1.\n$$\nThis completes the inductive step. $\\square$\n\nFor $j=N$, this immediately implies that some circle in row $2^{N}$ has number at least $N+1$. This shows that there is a ninja-path passing through at least $N+1$ red circles.\nSolution 2. We give an alternative proof that there exists a ninja-path passing through at least $N+1$ red circles. Assign numbers to circles as in the previous solution, but we only focus on the numbers assigned to red circles.\n\nFor each positive integer $i$, denote by $e_{i}$ the number of red circles with number $i$.\n\nClaim 2. If the red circle on row $l$ has number $i$, then $e_{i} \\leqslant l$.\n\nProof. Note that if two circles $C$ and $C^{\\prime}$ are both assigned the same number $i$, then there cannot be a ninja-path joining the two circles. We partition the triangle into a smaller triangle with the red circle in row $l$ at its top along with $l-1$ lines that together cover all other circles.\n![](attached_image_4.png)\nIn each set, there can be at most one red circle with number $i$, and therefore $e_{i} \\leqslant l$. $\\square$\n\nWe observe that if there exists a red circle $C$ with number $i \\geqslant 2$, then there also exists a red circle with number $i-1$ in some row that is above the row containing $C$. This is because the second last red circle in the ninja-path ending at $C$ has number $i-1$.\n\nClaim 3. We have $e_{i} \\leqslant 2^{i-1}$ for every positive integer $i$.\n\nProof. We prove by induction on $i$. The base case $i=1$ is clear, since the only red circle with number 1 is the one at the top of the triangle. We now assume that the statement is true for $1 \\leqslant i \\leqslant j-1$ and prove the statement for $i=j$. If $e_{j}=0$, there is nothing to prove. Otherwise, let $l$ be minimal such that the red circle on row $l$ has number $j$. Then all the red circles on row $1, \\ldots, l-1$ must have number less than $j$. This shows that\n$$\nl-1 \\leqslant e_{1}+e_{2}+\\cdots+e_{j-1} \\leqslant 1+2+\\cdots+2^{j-2}=2^{j-1}-1.\n$$\nThis proves that $l \\leqslant 2^{j-1}$, and by Claim 2, we also have $e_{j} \\leqslant l$. Therefore $e_{j} \\leqslant 2^{j-1}$. $\\square$\n\nWe now see that\n$$\ne_{1}+e_{2}+\\cdots+e_{N} \\leqslant 1+\\cdots+2^{N-1}=2^{N}-1<n.\n$$\nTherefore there exists a red circle with number at least $N+1$, which means that there exists a ninja-path passing through at least $N+1$ red circles.\nSolution 3. We provide yet another proof that there exists a ninja-path passing through at least $N+1$ red circles. In this solution, we assign to a circle $C$ the maximum number of red circles on a ninja-path starting at $C$ (including $C$ itself).\n![](attached_image_5.png)\nDenote by $f_{i}$ the number of red circles with number $i$. Note that if a red circle $C$ has number $i$, and there is a ninja-path from $C$ to another red circle $C^{\\prime}$, then the number assigned to $C^{\\prime}$ must be less than $i$.\n\nClaim 4. If the red circle on row $l$ has number less than or equal to $i$, then $f_{i} \\leqslant l$.\n\nProof. This proof is same as the proof of Claim 2. The additional input is that if the red circle on row $l$ has number strictly less than $i$, then the smaller triangle cannot have a red circle with number $i$. $\\square$\n\nClaim 5. We have\n$$\nf_{1}+f_{2}+\\cdots+f_{i} \\leqslant n-\\left\\lfloor\\frac{n}{2^{i}}\\right\\rfloor\n$$\nfor all $0 \\leqslant i \\leqslant N$.\n\nProof. We use induction on $i$. The base case $i=0$ is clear as the left hand side is the empty sum and the right hand side is zero. For the induction step, we assume that $i \\geqslant 1$ and that the statement is true for $i-1$. Let $l$ be minimal such that the red circle on row $l$ has number less than or equal to $i$. Then all the red circles with number less than or equal to $i$ lie on rows $l, l+1, \\ldots, n$, and therefore\n$$\nf_{1}+f_{2}+\\cdots+f_{i} \\leqslant n-l+1.\n$$\nOn the other hand, the induction hypothesis together with the fact that $f_{i} \\leqslant l$ shows that\n$$\nf_{1}+\\cdots+f_{i-1}+f_{i} \\leqslant n-\\left\\lfloor\\frac{n}{2^{i-1}}\\right\\rfloor+l.\n$$\nAveraging the two inequalities gives\n$$\nf_{1}+\\cdots+f_{i} \\leqslant n-\\frac{1}{2}\\left\\lfloor\\frac{n}{2^{i-1}}\\right\\rfloor+\\frac{1}{2}.\n$$\nSince the left hand side is an integer, we conclude that\n$$\nf_{1}+\\cdots+f_{i} \\leqslant n-\\left\\lfloor\\frac{1}{2}\\left\\lfloor\\frac{n}{2^{i-1}}\\right\\rfloor\\right\\rfloor=n-\\left\\lfloor\\frac{n}{2^{i}}\\right\\rfloor\n$$\nThis completes the induction step.\n\nTaking $i=N$, we obtain\n$$\nf_{1}+f_{2}+\\cdots+f_{N} \\leqslant n-\\left\\lfloor\\frac{n}{2^{N}}\\right\\rfloor<n\n$$\nThis implies that there exists a ninja-path passing through at least $N+1$ red circles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32157,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\geqslant 2$ be a positive integer. Paul has a $1 \\times n^{2}$ rectangular strip consisting of $n^{2}$ unit squares, where the $i^{\\text{th}}$ square is labelled with $i$ for all $1 \\leqslant i \\leqslant n^{2}$. He wishes to cut the strip into several pieces, where each piece consists of a number of consecutive unit squares, and then translate (without rotating or flipping) the pieces to obtain an $n \\times n$ square satisfying the following property: if the unit square in the $i^{\\text{th}}$ row and $j^{\\text{th}}$ column is labelled with $a_{ij}$, then $a_{ij}-(i+j-1)$ is divisible by $n$.\n\nDetermine the smallest number of pieces Paul needs to make in order to accomplish this.",
    "options": [],
    "answer": "2n-1",
    "solution": "Answer: The minimum number of pieces is $2n-1$.\n\nSolution 1. For the entirety of the solution, we shall view the labels as taking values in $\\mathbb{Z} / n \\mathbb{Z}$, as only their values modulo $n$ play a role.\nHere are two possible constructions consisting of $2n-1$ pieces.\n1. Cut into pieces of sizes $n, 1, n, 1, \\ldots, n, 1, 1$, and glue the pieces of size 1 to obtain the last row.\n2. Cut into pieces of sizes $n, 1, n-1, 2, n-2, \\ldots, n-1, 1$, and switch the pairs of consecutive strips that add up to size $n$.\n\nWe now prove that using $2n-1$ pieces is optimal. It will be more helpful to think of the reverse process: start with $n$ pieces of size $1 \\times n$, where the $k^{\\text{th}}$ piece has squares labelled $k, k+1, \\ldots, k+n-1$. The goal is to restore the original $1 \\times n^{2}$ strip.\n\nNote that each piece, after cutting at appropriate places, is of the form $a, a+1, \\ldots, b-1$. Construct an (undirected but not necessarily simple) graph $\\Gamma$ with vertices labelled by $1, \\ldots, n$, where a piece of the form $a, a+1, \\ldots, b-1$ corresponds to an edge from $a$ to $b$. We make the following observations.\n- The cut pieces came from the $k^{\\text{th}}$ initial piece $k, k+1, \\ldots, k+n-1$ corresponds to a cycle $\\gamma_k$ (possibly of length 1) containing the vertex $k$.\n- Since it is possible to rearrange the pieces into one single $1 \\times n^{2}$ strip, the graph $\\Gamma$ has an Eulerian cycle.\n- The number of edges of $\\Gamma$ is equal to the total number of cut pieces.\n\nThe goal is to prove that $\\Gamma$ has at least $2n-1$ edges. Since $\\Gamma$ has an Eulerian cycle, it is connected. For every $1 \\leqslant k \\leqslant n$, pick one edge from $\\gamma_k$, delete it from $\\Gamma$ to obtain a new graph $\\Gamma'$. Since no two cycles $\\gamma_i$ and $\\gamma_j$ share a common edge, removing one edge from each cycle does not affect the connectivity of the graph. This shows that the new graph $\\Gamma'$ must also be connected. Therefore $\\Gamma'$ has at least $n-1$ edges, which means that $\\Gamma$ has at least $2n-1$ edges.\nWe provide an alternative proof that at least $2n-1$ pieces are needed. Instead of having a linear strip, we work with a number of circular strips, each having length a multiple of $n$ and labelled as\n$$\n1,2, \\ldots, n, 1,2, \\ldots, n, \\ldots, 1,2, \\ldots, n\n$$\nwhere there are $n^{2}$ cells in total across all circular strips. The goal is still to create the $n \\times n$ square by cutting and translating. Here, when we say \"translating\" the strips, we imagine that each cell has a number written on it and the final $n \\times n$ square is required to have every number written in the same upright, non-mirrored orientation.\n\nNote that the number of cuts will be equal to the number of pieces, because performing $l \\geqslant 1$ cuts on a single circular strip results in $l$ pieces.\n\nConsider any \"seam\" in the interior of the final square, between two squares $S$ and $T$, so that $S$ and $T$ belongs to two separate pieces. We are interested in the positions of these two squares in the original circular strips, with the aim of removing the seam.\n- If the two squares $S$ and $T$ come from the same circular strip and are adjacent, then the cut was unnecessary and we can simply remove the seam and reduce the number of required cuts by 1. The circular strips are not affected.\n- If these two squares $S$ and $T$ were not adjacent, then they are next to two different cuts (either from the same circular strip or two different circular strips). Denote the two cuts by $(S \\mid Y)$ and $(X \\mid T)$. We perform these two cuts and then glue the pieces back according to $(S \\mid T)$ and $(X \\mid Y)$. Performing this move would either split one circular strip into two or merge two circular strips into one, changing the number of circular strips by at most one. Afterwards, we may eliminate cut $(S \\mid T)$ since it is no longer needed, which also removes the corresponding seam from the final square.\n\nBy iterating this process, eventually we reach a state where there are some number of circular strips, but the final $n \\times n$ square no longer has any interior seams.\n\nSince no two rows of the square can be glued together while maintaining the consecutive numbering, the only possibility is to have exactly $n$ circular strips, each with length $n$. In this state at least $n$ cuts are required to reassemble the square. Recall that each seam removal operation changed the number of circular strips by at most one. So if we started with only one initial circular strip, then at least $n-1$ seams were removed. Hence in total, at least $n+(n-1)=2n-1$ cuts are required to transform one initial circular strip into the final square. Hence at least $2n-1$ pieces are required to achieve the desired outcome.\nAs with the previous solution, we again work with circular strips. In particular, we start out with $k$ circular strips, each having length a multiple of $n$ and labelled as\n$$\n1,2, \\ldots, n, 1,2, \\ldots, n, \\ldots, 1,2, \\ldots, n,\n$$\nwhere there are $n^{2}$ cells in total across all $k$ circular strips. The goal is still to create the $n \\times n$ square by cutting and translating the circular strips.\n\nClaim. Constructing the $n \\times n$ square requires at least $2n-k$ cuts (or alternatively, $2n-k$ pieces).\n\nProof. We prove by induction on $n$. The base case $n=1$ is clear, because we can only have $k=1$ and the only way of producing a $1 \\times 1$ square from a $1 \\times 1$ circular strip is by making a single cut. We now assume that $n \\geqslant 2$ and the statement is true for $n-1$.\n\nEach cut is a cut between a cell of label $i$ on the left and a cell of label $i+1$ on the right side, for a unique $1 \\leqslant i \\leqslant n$. Let $a_i$ be the number of such cuts, so that $a_1+a_2+\\cdots+a_n$ is the total number of cuts. Since all the left and right edges of the $n \\times n$ square at the end must be cut, we have $a_i \\geqslant 1$ for all $1 \\leqslant i \\leqslant n$.\n\nIf $a_i \\geqslant 2$ for all $i$, then\n$$\na_1+a_2+\\cdots+a_n \\geqslant 2n > 2n-k\n$$\nand hence there is nothing to prove. We therefore assume that there exist some $1 \\leqslant m \\leqslant n$ for which $a_m=1$. This unique cut must form the two ends of the linear strip\n$$\nm+1, m+2, \\ldots, m-1+n, m+n\n$$\nfrom the final product. There are two cases.\n\nCase 1: The strip is a single connected piece.\nIn this case, the strip must have come from a single circular strip of length exactly $n$. We now remove this circular strip from of the cutting and pasting process. By definition of $m$, none of the edges between $m$ and $m+1$ are cut. Therefore we may pretend that all the adjacent pairs of cells labelled $m$ and $m+1$ are single cells. The induction hypothesis then implies that\n$$\na_1+\\cdots+a_{m-1}+a_{m+1}+\\cdots+a_n \\geqslant 2(n-1)-(k-1).\n$$\nAdding back in $a_m$, we obtain\n$$\na_1+\\cdots+a_n \\geqslant 2(n-1)-(k-1)+1=2n-k.\n$$\n\nCase 2: The strip is not a single connected piece.\nSay the linear strip $m+1, \\ldots, m+n$ is composed of $l \\geqslant 2$ pieces $C_1, \\ldots, C_l$. We claim that if we cut the initial circular strips along both the left and right end points of the pieces $C_1, \\ldots, C_l$, and then remove them, the remaining part consists of at most $k+l-2$ connected pieces (where some of them may be circular and some of them may be linear). This is because $C_l, C_1$ form a consecutive block of cells on the circular strip, and removing $l-1$ consecutive blocks from $k$ circular strips results in at most $k+(l-1)-1$ connected pieces.\n\nOnce we have the connected pieces that form the complement of $C_1, \\ldots, C_l$, we may glue them back at appropriate endpoints to form circular strips. Say we get $k'$ circular strips after this procedure. As we are gluing back from at most $k+l-2$ connected pieces, we see that\n$$\nk' \\leqslant k+l-2.\n$$\nWe again observe that to get from the new circular strips to the $n-1$ strips of size $1 \\times n$, we never have to cut along the cell boundary between labels $m$ and $m+1$. Therefore the induction hypothesis applies, and we conclude that the total number of pieces is bounded below by\n$$\nl+\\left(2(n-1)-k'\\right) \\geqslant l+2(n-1)-(k+l-2)=2n-k.\n$$\nThis finishes the induction step, and therefore the statement holds for all $n$.\n\nTaking $k=1$ in the claim, we see that to obtain a $n \\times n$ square from a circular $1 \\times n^{2}$ strip, we need at least $2n-1$ connected pieces. This shows that constructing the $n \\times n$ square out of a linear $1 \\times n^{2}$ strip also requires at least $2n-1$ pieces.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32158,
    "subject": "Mathematics (Multi-modal)",
    "question": "Elisa has 2023 treasure chests, all of which are unlocked and empty at first. Each day, Elisa adds a new gem to one of the unlocked chests of her choice, and afterwards, a fairy acts according to the following rules:\n- if more than one chests are unlocked, it locks one of them, or\n- if there is only one unlocked chest, it unlocks all the chests.\nGiven that this process goes on forever, prove that there is a constant $C$ with the following property: Elisa can ensure that the difference between the numbers of gems in any two chests never exceeds $C$, regardless of how the fairy chooses the chests to lock.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We will prove that such a constant $C$ exists when there are $n$ chests for $n$ an odd positive integer. In fact we can take $C = n-1$. Elisa's strategy is simple: place a gem in the chest with the fewest gems (in case there are more than one such chests, pick one arbitrarily).\n\nFor each integer $t \\geqslant 0$, let $a_{1}^{t} \\leqslant a_{2}^{t} \\leqslant \\cdots \\leqslant a_{n}^{t}$ be the numbers of gems in the $n$ chests at the end of the $t^{\\text{th}}$ day. In particular, $a_{1}^{0} = \\cdots = a_{n}^{0} = 0$ and\n$$\na_{1}^{t} + a_{2}^{t} + \\cdots + a_{n}^{t} = t\n$$\nFor each $t \\geqslant 0$, there is a unique index $m = m(t)$ for which $a_{m}^{t+1} = a_{m}^{t} + 1$. We know that $a_{j}^{t} > a_{m(t)}^{t}$ for all $j > m(t)$, since $a_{m(t)}^{t} < a_{m(t)}^{t+1} \\leqslant a_{j}^{t+1} = a_{j}^{t}$. Elisa's strategy also guarantees that if an index $j$ is greater than the remainder of $t$ when divided by $n$ (i.e. the number of locked chests at the end of the $t^{\\text{th}}$ day), then $a_{j}^{t} \\geqslant a_{m(t)}^{t}$, because some chest with at most $a_{j}^{t}$ gems must still be unlocked at the end of the $t^{\\text{th}}$ day.\n\nRecall that a sequence $x_{1} \\leqslant x_{2} \\leqslant \\cdots \\leqslant x_{n}$ of real numbers is said to majorise another sequence $y_{1} \\leqslant y_{2} \\leqslant \\cdots \\leqslant y_{n}$ of real numbers when for all $1 \\leqslant k \\leqslant n$ we have\n$$\nx_{1} + x_{2} + \\cdots + x_{k} \\leqslant y_{1} + y_{2} + \\cdots + y_{k}\n$$\nand\n$$\nx_{1} + x_{2} + \\cdots + x_{n} = y_{1} + y_{2} + \\cdots + y_{n} .\n$$\nOur strategy for proving $a_{n}^{t} - a_{1}^{t} \\leqslant n-1$ is to inductively show that the sequence $(a_{i}^{t})$ is majorised by some other sequence $\\left(b_{i}^{t}\\right)$.\n\nWe define this other sequence $\\left(b_{i}^{t}\\right)$ as follows. Let $b_{k}^{0} = k - \\frac{n+1}{2}$ for $1 \\leqslant k \\leqslant n$. As $n$ is odd, this is a strictly increasing sequence of integers, and the sum of its terms is 0. Now define $b_{i}^{t} = b_{i}^{0} + \\left\\lfloor \\frac{t-i}{n} \\right\\rfloor + 1$ for $t \\geqslant 1$ and $1 \\leqslant i \\leqslant n$. Thus for $t \\geqslant 0$,\n$$\nb_{i}^{t+1} = \\begin{cases} b_{i}^{t} & \\text{ if } t+1 \\not\\equiv i (\\bmod n), \\\\ b_{i}^{t} + 1 & \\text{ if } t+1 \\equiv i (\\bmod n) . \\end{cases}\n$$\nFrom these properties it is easy to see that\n- $b_{1}^{t} + b_{2}^{t} + \\cdots + b_{n}^{t} = t$ for all $t \\geqslant 0$, and\n- $b_{i}^{t} \\leqslant b_{i+1}^{t}$ for all $t \\geqslant 0$ and $1 \\leqslant i \\leqslant n-1$, with the inequality being strict if $t \\not\\equiv i (\\bmod n)$.\n\n**Claim 1.** For each $t \\geqslant 0$, the sequence of integers $b_{1}^{t}, b_{2}^{t}, \\ldots, b_{n}^{t}$ majorises the sequence of integers $a_{1}^{t}, a_{2}^{t}, \\ldots, a_{n}^{t}$.\n\n*Proof.* We use induction on $t$. The base case $t=0$ is trivial. Assume $t \\geqslant 0$ and that $(b_{i}^{t})$ majorises $\\left(a_{i}^{t}\\right)$. We want to prove the same holds for $t+1$.\n\nFirst note that the two sequences $\\left(b_{i}^{t+1}\\right)$ and $\\left(a_{i}^{t+1}\\right)$ both sum up to $t+1$. Next, we wish to show that for $1 \\leqslant k < n$, we have\n$$\nb_{1}^{t+1} + b_{2}^{t+1} + \\cdots + b_{k}^{t+1} \\leqslant a_{1}^{t+1} + a_{2}^{t+1} + \\cdots + a_{k}^{t+1} .\n$$\nWhen $t+1$ is replaced by $t$, the above inequality holds by the induction hypothesis. For the sake of contradiction, suppose $k$ is the smallest index such that the inequality for $t+1$ fails. Since the left hand side increases by at most 1 during the transition from $t$ to $t+1$, the inequality for $t+1$ can fail only if all of the following occur:\n- $b_{1}^{t} + b_{2}^{t} + \\cdots + b_{k}^{t} = a_{1}^{t} + a_{2}^{t} + \\cdots + a_{k}^{t}$,\n- $t+1 \\equiv j (\\bmod n)$ for some $1 \\leqslant j \\leqslant k$ (so that $b_{j}^{t+1} = b_{j}^{t} + 1$),\n- $m(t) > k$ (so that $a_{i}^{t+1} = a_{i}^{t}$ for $1 \\leqslant i \\leqslant k$).\n\nThe first point and the minimality of $k$ tell us that $b_{1}^{t}, \\ldots, b_{k}^{t}$ majorises $a_{1}^{t}, \\ldots, a_{k}^{t}$ as well (again using the induction hypothesis), and in particular $b_{k}^{t} \\geqslant a_{k}^{t}$.\n\nThe second point tells us that the remainder of $t$ when divided by $n$ is at most $k-1$, so $a_{k}^{t} \\geqslant a_{m(t)}^{t}$ (by Elisa's strategy). But by the third point $(m(t) \\geqslant k+1)$ and the nondecreasing property of $a_{i}^{t}$, we must have the equalities $a_{k}^{t} = a_{k+1}^{t} = a_{m(t)}^{t}$. On the other hand, $a_{k}^{t} \\leqslant b_{k}^{t} < b_{k+1}^{t}$, with the second inequality being strict because $t \\not\\equiv k (\\bmod n)$. We conclude that\n$$\nb_{1}^{t} + b_{2}^{t} + \\cdots + b_{k+1}^{t} > a_{1}^{t} + a_{2}^{t} + \\cdots + a_{k+1}^{t},\n$$\na contradiction to the induction hypothesis. $\\square$\n\nThis completes the proof as it implies\n$$\na_{n}^{t} - a_{1}^{t} \\leqslant b_{n}^{t} - b_{1}^{t} \\leqslant b_{n}^{0} - b_{1}^{0} = n-1 .\n$$\nWe solve the problem when 2023 is replaced with an arbitrary integer $n$. We assume that Elisa uses the following strategy:\n\nAt the beginning of the $(n t + 1)^{\\text{th}}$ day, Elisa first labels her chests as $C_{1}^{t}, \\ldots, C_{n}^{t}$ so that before she adds in the gem, the number of gems in $C_{i}^{t}$ is less than or equal $C_{j}^{t}$ for all $1 \\leqslant i < j \\leqslant n$. Then for days $n t + 1, n t + 2, \\ldots, n t + n$, she adds a gem to chest $C_{i}^{t}$, where $i$ is chosen to be minimal such that $C_{i}^{t}$ is unlocked.\n\nDenote by $c_{i}^{t}$ the number of gems in chest $C_{i}^{t}$ at the beginning of the $(n t + 1)^{\\text{th}}$ day, so that\n$$\nc_{1}^{t} \\leqslant c_{2}^{t} \\leqslant \\cdots \\leqslant c_{n}^{t}\n$$\nby construction. Also, denote by $\\delta_{i}^{t}$ the total number of gems added to chest $C_{i}^{t}$ during days $n t + 1, \\ldots, n t + n$. We make the following observations.\n- We have $c_{1}^{0} = c_{2}^{0} = \\cdots = c_{n}^{0} = 0$.\n- We have $c_{1}^{t} + \\cdots + c_{n}^{t} = n t$, since $n$ gems are added every $n$ days.\n- The sequence $(c_{i}^{t+1})$ is a permutation of the sequence $(c_{i}^{t} + \\delta_{i}^{t})$ for all $t \\geqslant 0$.\n- We have $\\delta_{1}^{t} + \\cdots + \\delta_{n}^{t} = n$ for all $t \\geqslant 0$.\n- Since Elisa adds a gem to an unlocked chest $C_{i}^{t}$ with $i$ minimal, we have\n$$\n\\delta_{1}^{t} + \\delta_{2}^{t} + \\cdots + \\delta_{k}^{t} \\geqslant k\n$$\nfor every $1 \\leqslant k \\leqslant n$ and $t \\geqslant 0$.\n\nWe now define another sequence of sequences of integers as follows.\n$$\nd_{i}^{0} = 3 n \\left(i - \\frac{n+1}{2}\\right), \\quad d_{i}^{t} = d_{i}^{0} + t\n$$\nWe observe that\n$$\nd_{1}^{t} + \\cdots + d_{n}^{t} = c_{1}^{t} + \\cdots + c_{n}^{t} = n t\n$$\n**Claim 3.** For each $t \\geqslant 0$, the sequence $(d_{i}^{t})$ majorises the sequence $(c_{i}^{t})$.\n\n*Proof.* We induct on $t$. For $t=0$, this is clear as all the terms in the sequence $\\left(c_{i}^{t}\\right)$ are equal. For the induction step, we assume that $(d_{i}^{t})$ majorises $(c_{i}^{t})$. Given $1 \\leqslant k \\leqslant n-1$, we wish to show that\n$$\nd_{1}^{t+1} + \\cdots + d_{k}^{t+1} \\leqslant c_{1}^{t+1} + \\cdots + c_{k}^{t+1}\n$$\nCase 1: $c_{1}^{t+1}, \\ldots, c_{k}^{t+1}$ is a permutation of $c_{1}^{t} + \\delta_{1}^{t}, \\ldots, c_{k}^{t} + \\delta_{k}^{t}$.\n\nSince $d_{1}^{t} + \\cdots + d_{k}^{t} \\leqslant c_{1}^{t} + \\cdots + c_{k}^{t}$ by the induction hypothesis, we have\n$$\n\\sum_{i=1}^{k} d_{i}^{t+1} = k + \\sum_{i=1}^{k} d_{i}^{t} \\leqslant k + \\sum_{i=1}^{k} c_{i}^{t} \\leqslant \\sum_{i=1}^{k} (c_{i}^{t} + \\delta_{i}^{t}) = \\sum_{i=1}^{k} c_{i}^{t+1}\n$$\nCase 2: $c_{1}^{t+1}, \\ldots, c_{k}^{t+1}$ is not a permutation of $c_{1}^{t} + \\delta_{1}^{t}, \\ldots, c_{k}^{t} + \\delta_{k}^{t}$.\n\nIn this case, we have $c_{i}^{t} + \\delta_{i}^{t} > c_{j}^{t} + \\delta_{j}^{t}$ for some $i \\leqslant k < j$. It follows that\n$$\nc_{k}^{t} + n \\geqslant c_{i}^{t} + n \\geqslant c_{i}^{t} + \\delta_{i}^{t} > c_{j}^{t} + \\delta_{j}^{t} \\geqslant c_{j}^{t} \\geqslant c_{k+1}^{t} .\n$$\nUsing $d_{k}^{t} + 3 n = d_{k+1}^{t}$ and the induction hypothesis, we obtain\n$$\n\\begin{aligned}\n\\sum_{i=1}^{k} c_{i}^{t+1} & \\geqslant \\sum_{i=1}^{k} c_{i}^{t} > c_{1}^{t} + \\cdots + c_{k-1}^{t} + \\frac{1}{2} c_{k}^{t} + \\frac{1}{2} c_{k+1}^{t} - \\frac{n}{2} \\\\\n& = \\frac{1}{2} \\sum_{i=1}^{k-1} c_{i}^{t} + \\frac{1}{2} \\sum_{i=1}^{k+1} c_{i}^{t} - \\frac{n}{2} \\\\\n& \\geqslant \\frac{1}{2} \\sum_{i=1}^{k-1} d_{i}^{t} + \\frac{1}{2} \\sum_{i=1}^{k+1} d_{i}^{t} - \\frac{n}{2} = n + \\sum_{i=1}^{k} d_{i}^{t} \\geqslant k + \\sum_{i=1}^{k} d_{i}^{t} = \\sum_{i=1}^{k} d_{i}^{t+1}\n\\end{aligned}\n$$\nThis finishes the induction step.\n\nIt follows that\n$$\nc_{n}^{t} - c_{1}^{t} \\leqslant d_{n}^{t} - d_{1}^{t} = 3 n (n-1)\n$$\nFrom day $n t + 1$ to day $n(t+1) + 1$, Elisa adds $n$ gems, and therefore the difference may increase by at most $n$. This shows that the difference of the number of gems in two chests never exceeds $C = 3 n (n-1) + n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32159,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer, and consider an $N \\times N$ grid. A right-down path is a sequence of grid cells such that each cell is either one cell to the right of or one cell below the previous cell in the sequence. A right-up path is a sequence of grid cells such that each cell is either one cell to the right of or one cell above the previous cell in the sequence.\nProve that the cells of the $N \\times N$ grid cannot be partitioned into less than $N$ right-down or right-up paths. For example, the following partition of the $5 \\times 5$ grid uses 5 paths.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We define a good parallelogram to be a parallelogram composed of two isosceles right-angled triangles glued together as shown below.\n![](attached_image_2.png)\nGiven any partition into $k$ right-down or right-up paths, we can find a corresponding packing of good parallelograms that leaves an area of $k$ empty. Thus, it suffices to prove that we must leave an area of at least $N$ empty when we pack good parallelograms into an $N \\times N$ grid. This is actually equivalent to the original problem since we can uniquely recover the partition into right-down or right-up paths from the corresponding packing of good parallelograms.\n![](attached_image_3.png)\n![](attached_image_4.png)\nWe draw one of the diagonals in each cell so that it does not intersect any of the good parallelograms. Now, view these segments as mirrors, and consider a laser entering each of the $4 N$ boundary edges (with starting direction being perpendicular to the edge), bouncing along these mirrors until it exits at some other edge. When a laser passes through a good parallelogram, its direction goes back to the original one after bouncing two times. Thus, if the final direction of a laser is perpendicular to its initial direction, it must pass through at least one empty triangle. Similarly, if the final direction of a laser is opposite to its initial direction, it must pass though at least two empty triangles. Using this, we will estimate the number of empty triangles in the $N \\times N$ grid.\nWe associate the starting edge of a laser with the edge it exits at. Then, the boundary edges are divided into $2 N$ pairs. These pairs can be classified into three types:\n(1) a pair of a vertical and a horizontal boundary edge,\n(2) a pair of boundary edges from the same side, and\n(3) a pair of boundary edges from opposite sides.\nSince the beams do not intersect, we cannot have one type (3) pair from vertical boundary edges and another type (3) pair from horizontal boundary edges. Without loss of generality, we may assume that we have $t$ pairs of type (3) and they are all from vertical boundary edges. Then, out of the remaining boundary edges, there are $2 N$ horizontal boundary edges and $2 N-2 t$ vertical boundary edges. It follows that there must be at least $t$ pairs of type (2) from horizontal boundary edges. We know that a laser corresponding to a pair of type (1) passes through at least one empty triangle, and a laser corresponding to a pair of type (2) passes through at least two empty triangles. Thus, as the beams do not intersect, we have at least $(2 N-2 t)+2 \\cdot t=2 N$ empty triangles in the grid, leaving an area of at least $N$ empty as required.\nWe apply an induction on $N$. The base case $N=1$ is trivial. Suppose that the claim holds for $N-1$ and prove it for $N \\geqslant 2$.\nLet us denote the path containing the upper left corner by $P$. If $P$ is right-up, then every cell in $P$ is in the top row or in the leftmost column. By the induction hypothesis, there are at least $N-1$ paths passing through the lower right $(N-1) \\times(N-1)$ subgrid. Since $P$ is not amongst them, we have at least $N$ paths.\nNext, assume that $P$ is right-down. If $P$ contains the lower right corner, then we get an $(N-1) \\times(N-1)$ grid by removing $P$ and glueing the remaining two parts together. The main idea is to extend $P$ so that it contains the lower right corner and the above procedure gives a valid partition of an $(N-1) \\times(N-1)$ grid.\n![](attached_image_5.png)\nWe inductively construct $Q$, which denotes an extension of $P$ as a right-down path. Initially, $Q=P$. Let $A$ be the last cell of $Q, B$ be the cell below $A$, and $C$ be the cell to the right of $A$ (if they exist). Suppose that $A$ is not the lower right corner, and that (*) both $B$ and $C$ do not belong to the same path as $A$. Then, we can extend $Q$ as follows (in case we have two or more options, we can choose any one of them to extend $Q$ ).\n1. If $B$ belongs to a right-down path $R$, then we add the part of $R$, from $B$ to its end, to $Q$.\n2. If $C$ belongs to a right-down path $R$, then we add the part of $R$, from $C$ to its end, to $Q$.\n3. If $B$ belongs to a right-up path $R$ which ends at $B$, then we add the part of $R$ in the same column as $B$ to $Q$.\n4. If $C$ belongs to a right-up path $R$ which starts at $C$, then we add the part of $R$ in the same row as $C$ to $Q$.\n5. Otherwise, $B$ and $C$ must belong to the same right-up path $R$. In this case, we add $B$ and the cell to the right of $B$ to $Q$.\nNote that if $B$ does not exist, then case (4) must hold. If $C$ does not exist, then case (3) must hold.\nIt is easily seen that such an extension also satisfies the hypothesis (*), so we can repeat this construction to get an extension of $P$ containing the lower right corner, denoted by $Q$. We show that this is a desired extension, i.e. the partition of an $(N-1) \\times(N-1)$ grid obtained by removing $Q$ and glueing the remaining two parts together consists of right-down or right-up paths.\nTake a path $R$ in the partition of the $N \\times N$ grid intersecting $Q$. If the intersection of $Q$ and $R$ occurs in case (1) or case (2), then there exists a cell $D$ in $R$ such that the intersection of $Q$ and $R$ is the part of $R$ from $D$ to its end, so $R$ remains a right-down path after removal of $Q$. Similarly, if the intersection of $Q$ and $R$ occurs in case (3) or case (4), then $R$ remains a right-up path after removal of $Q$. If the intersection of $Q$ and $R$ occurs in case (5), then this intersection has exactly two adjacent cells. After the removal of these two cells (as we remove $Q), R$ is divided into two parts that are glued into a right-up path.\nThus, we may apply the induction hypothesis to the resulting partition of an $(N-1) \\times(N-1)$ grid, to find that it must contain at least $N-1$ paths. Since $P$ is contained in $Q$ and is not amongst these paths, the original partition must contain at least $N$ paths.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32160,
    "subject": "Mathematics (Multi-modal)",
    "question": "The Imomi archipelago consists of $n \\geqslant 2$ islands. Between each pair of distinct islands is a unique ferry line that runs in both directions, and each ferry line is operated by one of $k$ companies. It is known that if any one of the $k$ companies closes all its ferry lines, then it becomes impossible for a traveller, no matter where the traveller starts at, to visit all the islands exactly once (in particular, not returning to the island the traveller started at).\nDetermine the maximal possible value of $k$ in terms of $n$.",
    "options": [],
    "answer": "floor(log_2 n)",
    "solution": "Answer: The largest $k$ is $k=\\left\\lfloor\\log_{2} n\\right\\rfloor$.\n\nWe reformulate the problem using graph theory. We have a complete graph $K_{n}$ on $n$ nodes (corresponding to islands), and we want to colour the edges (corresponding to ferry lines) with $k$ colours (corresponding to companies), so that every Hamiltonian path contains all $k$ different colours. For a fixed set of $k$ colours, we say that an edge colouring of $K_{n}$ is good if every Hamiltonian path contains an edge of each one of these $k$ colours.\n\nWe first construct a good colouring of $K_{n}$ using $k=\\left\\lfloor\\log_{2} n\\right\\rfloor$ colours.\n\n**Claim 1.** Take $k=\\left\\lfloor\\log_{2} n\\right\\rfloor$. Consider the complete graph $K_{n}$ in which the nodes are labelled by $1,2, \\ldots, n$. Colour node $i$ with colour $\\min \\left(\\left\\lfloor\\log_{2} i\\right\\rfloor+1, k\\right)$ (so the colours of the first nodes are $1,2,2,3,3,3,3,4, \\ldots$ and the last $n-2^{k-1}+1$ nodes have colour $k$ ), and for $1 \\leqslant i<j \\leqslant n$, colour the edge $ij$ with the colour of the node $i$. Then the resulting edge colouring of $K_{n}$ is good.\n\n**Proof.** We need to check that every Hamiltonian path contains edges of every single colour. We first observe that the number of nodes assigned colour $k$ is $n-2^{k-1}+1$. Since $n \\geqslant 2^{k}$, we have\n$$\nn-2^{k-1}+1 \\geqslant \\frac{n}{2}+1\n$$\nThis implies that in any Hamiltonian path, there exists an edge between two nodes with colour $k$. Then that edge must have colour $k$.\n\nWe next show that for each $1 \\leqslant i<k$, every Hamiltonian path contains an edge of colour $i$. Suppose the contrary, that some Hamiltonian path does not contain an edge of colour $i$. Then nodes with colour $i$ can only be adjacent to nodes with colour less than $i$ inside the Hamiltonian path. Since there are $2^{i-1}$ nodes with colour $i$ and $2^{i-1}-1$ nodes with colour less than $i$, the Hamiltonian path must take the form\n$$\n(i) \\leftrightarrow(<i) \\leftrightarrow(i) \\leftrightarrow(<i) \\leftrightarrow \\cdots \\leftrightarrow(<i) \\leftrightarrow(i),\n$$\nwhere $(i)$ denotes a node with colour $i$, $(<i)$ denotes a node with colour less than $i$, and $\\leftrightarrow$ denotes an edge. But this is impossible, as the Hamiltonian path would not have any nodes with colours greater than $i$.\n\nFix a set of $k$ colours, we now prove that if there exists a good colouring of $K_{n}$, then $k \\leqslant\\left\\lfloor\\log_{2} n\\right\\rfloor$. For $n=2$, this is trivial, so we assume $n \\geqslant 3$. For any node $v$ of $K_{n}$ and $1 \\leqslant i \\leqslant k$, we denote by $d_{i}(v)$ the number of edges with colour $i$ incident with the node $v$.\n\n**Lemma 1.** Consider a good colouring of $K_{n}$, and let $AB$ be an arbitrary edge with colour $i$. If $d_{i}(A)+d_{i}(B) \\leqslant n-1$, then the colouring will remain good after recolouring edge $AB$ with any other colour.\n\n**Proof.** Suppose there exists a good colouring together with an edge $AB$ of colour $i$, such that if $AB$ is recoloured with another colour, the colouring will no longer be good. The failure of the new colouring being good will come from colour $i$, and thus there exists a Hamiltonian path containing edge $AB$ such that initially (i.e. before recolouring) $AB$ is the only edge of colour $i$ in this path. Writing $A=A_{0}$ and $B=B_{0}$, denote this Hamiltonian path by\n$$\nA_{s} \\leftrightarrow A_{s-1} \\leftrightarrow \\cdots \\leftrightarrow A_{1} \\leftrightarrow A_{0} \\leftrightarrow B_{0} \\leftrightarrow B_{1} \\leftrightarrow \\cdots \\leftrightarrow B_{t-1} \\leftrightarrow B_{t},\n$$\nwhere $s, t \\geqslant 0$ and $s+t+2=n$.\n\nIn the initial colouring, we observe the following.\n- The edge $B_{0}A_{s}$ must have colour $i$, since otherwise the path\n$$\nA_{0} \\leftrightarrow A_{1} \\leftrightarrow \\cdots \\leftrightarrow A_{s-1} \\leftrightarrow A_{s} \\leftrightarrow B_{0} \\leftrightarrow B_{1} \\leftrightarrow \\cdots \\leftrightarrow B_{t-1} \\leftrightarrow B_{t}\n$$\nhas no edges of colour $i$.\n- Similarly, the edge $A_{0}B_{t}$ must have colour $i$.\n- For each $0 \\leqslant p<s$, at least one of the edges $B_{0}A_{p}$ and $A_{0}A_{p+1}$ must have colour $i$, since otherwise the path\n$$\nA_{s} \\leftrightarrow \\cdots \\leftrightarrow A_{p+2} \\leftrightarrow A_{p+1} \\leftrightarrow A_{0} \\leftrightarrow A_{1} \\leftrightarrow \\cdots \\leftrightarrow A_{p-1} \\leftrightarrow A_{p} \\leftrightarrow B_{0} \\leftrightarrow B_{1} \\leftrightarrow \\cdots \\leftrightarrow B_{t}\n$$\nhas no edges of colour $i$.\n- Similarly, for each $0 \\leqslant q<t$, at least one of the edges $A_{0}B_{q}$ and $B_{0}B_{q+1}$ must have colour $i$.\n\nIn the above list, each edge $A_{0}X$ appears exactly once and also each edge $B_{0}X$ appears exactly once (where $A_{0}B_{0}$ and $B_{0}A_{0}$ are counted separately). Adding up the contributions to $d_{i}(A)+ d_{i}(B)$, we obtain\n$$\nd_{i}(A)+d_{i}(B) \\geqslant(s+1)+(t+1)=n .\n$$\nThis contradicts our assumption that $d_{i}(A)+d_{i}(B) \\leqslant n-1$.\n\nOur strategy now is to repeatedly recolour the edges using Lemma 1 until the colouring has a simple structure. For a node $v$, we define $m(v)$ to be the largest value of $d_{i}(v)$ over all colours $i$.\n\n**Lemma 2.** Assume we have a good colouring of $K_{n}$. Let $A, B$ be two distinct nodes, and let $j$ be the colour of edge $AB$ where $1 \\leqslant j \\leqslant k$. If\n- $m(A) \\geqslant m(B)$ and\n- $m(A)=d_{i}(A)$ for some $i \\neq j$,\nthen after recolouring edge $AB$ with colour $i$, the colouring remains good.\n\n**Proof.** Note that\n$$\nd_{j}(A)+d_{j}(B) \\leqslant(n-1-m(A))+m(B) \\leqslant n-1,\n$$\nand so we may apply Lemma 1.\n\n**Lemma 3.** Assume we have a good colouring of $K_{n}$. Let $S$ be a nonempty set of nodes. Let $A \\in S$ be a node such that $m(A) \\geqslant m(B)$ for all $B \\in S$, and choose $1 \\leqslant i \\leqslant k$ for which $d_{i}(A)=m(A)$. Then after recolouring the edge $AB$ with colour $i$ for all $B \\in S$ distinct from $A$, the colouring remains good.\n\n**Proof.** We repeatedly perform the following operation until all edges $AB$ with $B \\in S$ have colour $i$ :\n> choose an edge $AB$ with $B \\in S$ that does not have colour $i$, and recolour it with colour $i$.\nBy Lemma 2, the colouring remains good after one operation. Moreover, $m(A)$ increase by 1 during an operation, and all other $m(B)$ may increase by at most 1 . This shows that $m(A)$ will remain maximal amongst $m(B)$ for $B \\in S$. We will also have $d_{i}(A)=m(A)$ after the operation, since both sides increase by 1 . Therefore the operation can be performed repeatedly, and the colouring remains good.\n\nWe first apply Lemma 3 to the set of all $n$ nodes in $K_{n}$. After recolouring, there exists a node $A_{1}$ such that every edge incident with $A_{1}$ has colour $c_{1}$. We then apply Lemma 3 to the set of nodes excluding $A_{1}$, and we obtain a colouring where\n- every edge incident with $A_{1}$ has colour $c_{1}$,\n- every edge incident with $A_{2}$ except for $A_{1}A_{2}$ has colour $c_{2}$.\nRepeating this process, we arrive at the following configuration:\n- the $n$ nodes of $K_{n}$ are labelled $A_{1}, A_{2}, \\ldots, A_{n}$,\n- the node $A_{i}$ has a corresponding colour $c_{i}$ (as a convention, we also colour $A_{i}$ with $c_{i}$ ),\n- for all $1 \\leqslant u<v \\leqslant n$, the edge between $A_{u}$ and $A_{v}$ has colour $c_{u}$,\n- this colouring is good.\n\n**Claim 2.** For every colour $i$, there exists a $1 \\leqslant p \\leqslant n$ such that the number of nodes of colour $i$ amongst $A_{1}, \\ldots, A_{p}$ is greater than $p / 2$.\n\n**Proof.** Suppose the contrary, that for every $1 \\leqslant p \\leqslant n$, there are at most $\\lfloor p / 2\\rfloor$ nodes of colour $i$. We then construct a Hamiltonian path not containing any edge of colour $i$. Let $A_{x_{1}}, \\ldots, A_{x_{t}}$ be the nodes with colour $i$, where $x_{1}<x_{2}<\\cdots<x_{t}$, and let $A_{y_{1}}, A_{y_{2}}, \\ldots, A_{y_{s}}$ be the nodes with colour different from $i$, where $y_{1}<y_{2}<\\cdots<y_{s}$. We have $s+t=n$ and $t \\leqslant\\lfloor n / 2\\rfloor$, so $t \\leqslant s$. We also see that $y_{j}<x_{j}$ for all $1 \\leqslant j \\leqslant t$, because otherwise, $A_{1}, A_{2}, \\ldots, A_{x_{j}}$ will have $j$ nodes of colour $i$ and less than $j$ nodes of colour different from $i$. Then we can construct a Hamiltonian path\n$$\nA_{x_{1}} \\leftrightarrow A_{y_{1}} \\leftrightarrow A_{x_{2}} \\leftrightarrow A_{y_{2}} \\leftrightarrow A_{x_{3}} \\leftrightarrow \\cdots \\leftrightarrow A_{x_{t}} \\leftrightarrow A_{y_{t}} \\leftrightarrow A_{y_{t+1}} \\leftrightarrow \\cdots \\leftrightarrow A_{y_{s}}\n$$\nthat does not contain an edge with colour $i$. This contradicts that the colouring is good.\n\nSo for every colour $i$, there has to be an integer $p_{i}$ with $1 \\leqslant p_{i} \\leqslant n$ such that there are more than $p_{i} / 2$ nodes assigned colour $i$ amongst $A_{1}, \\ldots, A_{p_{i}}$. Choose the smallest such $p_{i}$ for every $i$, and without loss of generality assume\n$$\np_{1}<p_{2}<\\cdots<p_{k}\n$$\nNote that the inequalities are strict by the definition of $p_{i}$.\n\nThen amongst the nodes $A_{1}, \\ldots, A_{p_{i}}$, there are at least $\\left\\lceil\\left(p_{j}+1\\right) / 2\\right\\rceil$ nodes of colour $j$ for all $1 \\leqslant j \\leqslant i$. Then\n$$\np_{i} \\geqslant\\left\\lceil\\frac{p_{1}+1}{2}\\right\\rceil+\\left\\lceil\\frac{p_{2}+1}{2}\\right\\rceil+\\cdots+\\left\\lceil\\frac{p_{i}+1}{2}\\right\\rceil .\n$$\nThis inductively shows that\n$$\np_{i} \\geqslant 2^{i}-1\n$$\nfor all $1 \\leqslant i \\leqslant k$, and this already proves $n \\geqslant 2^{k}-1$.\n\nIt remains to show that $n=2^{k}-1$ is not possible. If $n=2^{k}-1$, then all inequalities have to be equalities, so $p_{i}=2^{i}-1$ and there must be exactly $2^{i-1}$ nodes of colour $i$. Moreover, there cannot be a node of colour $i$ amongst $A_{1}, A_{2}, \\ldots, A_{p_{i-1}}$, and so the set of nodes of colour $i$ must precisely be\n$$\nA_{2^{i-1}}, A_{2^{i-1}+1}, \\ldots, A_{2^{i}-1}\n$$\nThen we can form a Hamiltonian path\n$$\nA_{2^{k-1}} \\leftrightarrow A_{1} \\leftrightarrow A_{2^{k-1}+1} \\leftrightarrow A_{2} \\leftrightarrow A_{2^{k-1}+2} \\leftrightarrow A_{3} \\leftrightarrow \\ldots \\leftrightarrow A_{n},\n$$\nwhich does not contain an edge of colour $k$. This is a contradiction, and therefore $n \\geqslant 2^{k}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32161,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AC > BC$. Let $\\omega$ be the circumcircle of triangle $ABC$ and let $r$ be the radius of $\\omega$. Point $P$ lies on segment $AC$ such that $BC = CP$ and point $S$ is the foot of the perpendicular from $P$ to line $AB$. Let ray $BP$ intersect $\\omega$ again at $D$ and let $Q$ lie on line $SP$ such that $PQ = r$ and $S, P, Q$ lie on the line in that order. Finally, let the line perpendicular to $CQ$ from $A$ intersect the line perpendicular to $DQ$ from $B$ at $E$.\nProve that $E$ lies on $\\omega$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1 (Similar Triangles).\n![](attached_image_1.png)\nFirst observe that\n$$\n\\angle DPA = \\angle BPC \\stackrel{CP=CB}{=} \\angle CBP = \\angle CBD = \\angle CAD = \\angle PAD\n$$\nso $DP = DA$. Thus there is a symmetry in the problem statement swapping $(A, D) \\leftrightarrow (B, C)$.\nLet $O$ be the centre of $\\omega$ and let $E$ be the reflection of $P$ in $CD$ which, by\n$$\n\\angle CED = \\angle DPC = 180^{\\circ} - \\angle CPB \\stackrel{CP=CB}{=} 180^{\\circ} - \\angle PBC = 180^{\\circ} - \\angle DBC\n$$\nlies on $\\omega$. We claim the two lines concur at $E$. By the symmetry noted above, it suffices to prove that $BE \\perp DQ$ and then $AE \\perp CQ$ will follow by symmetry.\nWe have $AO = PQ$, $AD = DP$ and\n$$\n\\angle DAO = 90^{\\circ} - \\angle ABD \\stackrel{PQ \\perp AB}{=} \\angle DPQ.\n$$\nHence $\\triangle AOD \\cong \\triangle PQD$. Thus\n$\\angle QDB + \\angle DBE = \\angle ODA + \\angle DAE \\stackrel{DE=DA}{=} \\angle ODA + \\angle AED = (90^{\\circ} - \\angle AED) + \\angle AED = 90^{\\circ}$ giving $BE \\perp DQ$ as required.\nSolution 2 (Second Circle).\n![](attached_image_2.png)\nAs in Solution 1, we prove that $DA = DP$ and note the symmetry in the problem statement swapping $(A, D) \\leftrightarrow (B, C)$.\nLet $\\Gamma$ be the circumcircle of $\\triangle PCD$. Since $DP = DA$ and $\\angle ACD = \\angle PCD$, the radius of $\\Gamma$ is equal to that of $\\omega$. We have that\n$$\n\\angle DPQ = \\angle BPS = 90^{\\circ} - \\angle ABD = 90^{\\circ} - \\angle PCD.\n$$\nThis, combined with $PQ$ being equal to the common circumradius of $\\Gamma$ and $\\omega$, means that $Q$ is the circumcentre of $\\Gamma$.\nLet the perpendiculars to $CQ, DQ$ from $A, B$ intersect at $E$ then we have\n$$\n\\begin{aligned}\n& \\angle EAC = 90^{\\circ} - \\angle ACQ \\stackrel{QC=QP}{=} 90^{\\circ} - \\angle QPC = 90^{\\circ} - \\angle SPA = \\angle CAB \\Longrightarrow \\angle EAB = 2\\angle PAB \\\\\n& \\angle DBE = 90^{\\circ} - \\angle QDP \\stackrel{QD=QP}{=} 90^{\\circ} - \\angle DPQ = 90^{\\circ} - \\angle BPS = \\angle ABD \\Longrightarrow \\angle ABE = 2\\angle ABP .\n\\end{aligned}\n$$\nCombining these\n$$\n\\angle BEA = 180^{\\circ} - 2(\\angle PAB + \\angle ABP) = 180^{\\circ} - 2\\angle APD \\stackrel{DA=DP}{=} \\angle BDA\n$$\nwhich gives that $E$ lies on $\\omega$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32162,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral with $\\angle BAD < \\angle ADC$. Let $M$ be the midpoint of the arc $CD$ not containing $A$. Suppose there is a point $P$ inside $ABCD$ such that $\\angle ADB = \\angle CPD$ and $\\angle ADP = \\angle PCB$.\nProve that lines $AD$, $PM$, $BC$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X$ and $Y$ be the intersection points of $AM$ and $BM$ with $PD$ and $PC$ respectively. Since $ABCM D$ is cyclic and $CM = MD$, we have\n$$\n\\angle XAD = \\angle MAD = \\angle CBM = \\angle CBY.\n$$\nCombining this with $\\angle ADX = \\angle YCB$, we get $\\angle DXA = \\angle BYC$, and so $\\angle PXM = \\angle MYP$. Moreover, $\\angle YPX = \\angle CPD = \\angle ADB = \\angle AMB$. The quadrilateral $MXPY$ therefore has equal opposite angles and so is a parallelogram.\n![](attached_image_1.png)\nLet $R$ and $S$ be the intersection points of $AM$ and $BM$ with $BC$ and $AD$ respectively. Due to $AM \\parallel PC$ and $BM \\parallel PD$, we have $\\angle ASB = \\angle ADP = \\angle PCB = \\angle ARB$ and so the quadrilateral $ABRS$ is cyclic. We then have $\\angle SRB = 180^{\\circ} - \\angle BAS = \\angle DCB$ and so $SR \\parallel CD$. In triangles $PCD$ and $MRS$, the corresponding sides are parallel so they are homothetic meaning lines $DS$, $PM$, $CR$ concur at the centre of this homothety.\nLet $AD$ and $BC$ meet at $T$. Denote by $p_{a}, p_{b}, m_{a}$ and $m_{b}$ the distances between line $TA$ and $P$, $TB$ and $P$, $TA$ and $M$ and between $TB$ and $M$ respectively. Our goal is to prove $p_{a} : p_{b} = m_{a} : m_{b}$ which is equivalent to the collinearity of $T$, $P$ and $M$.\n![](attached_image_2.png)\nLet $\\angle BAC = \\angle BDC = \\alpha$, $\\angle DBA = \\angle DCA = \\beta$, $\\angle ADB = \\angle AMB = \\angle ACB = \\angle CPD = \\mu$, $\\angle ADP = \\angle PCB = \\nu$ and $\\angle MAD = \\angle CAM = \\angle MBD = \\angle CBM = \\chi$.\nFrom $\\angle ADP = \\angle PCB = \\nu$ and $\\angle MAD = \\angle CBM = \\chi$ we get\n$$\n\\frac{p_{a}}{p_{b}} = \\frac{PD \\cdot \\sin \\nu}{PC \\cdot \\sin \\nu} = \\frac{PD}{PC} \\quad \\text{and} \\quad \\frac{m_{a}}{m_{b}} = \\frac{MA \\cdot \\sin \\chi}{MB \\cdot \\sin \\chi} = \\frac{MA}{MB}.\n$$\nHence $p_{a} : p_{b} = m_{a} : m_{b}$ is equivalent to $PD : PC = MA : MB$, and since $\\angle CPD = \\angle AMB = \\mu$, this means we have to show that triangles $PDC$ and $MAB$ are similar.\nIn triangle $PDC$ we have\n$$\n\\begin{aligned}\n& \\angle PDC + \\angle DCP = 180^{\\circ} - \\angle CPD = 180^{\\circ} - \\mu, \\\\\n& \\angle PDC - \\angle DCP = (\\alpha + \\mu - \\nu) - (\\beta + \\mu - \\nu) = \\alpha - \\beta.\n\\end{aligned}\n$$\nSimilarly, in triangle $MAB$ we have\n$$\n\\begin{aligned}\n& \\angle BAM + \\angle MBA = 180^{\\circ} - \\angle AMB = 180^{\\circ} - \\mu, \\\\\n& \\angle BAM - \\angle MBA = (\\alpha + \\chi) - (\\beta + \\chi) = \\alpha - \\beta.\n\\end{aligned}\n$$\nTherefore, ($\\angle BAM$, $\\angle MBA$) and ($\\angle PDC$, $\\angle DCP$) satisfy the same system of linear equations. The common solution is\n$$\n\\angle BAM = \\angle PDC = \\frac{180^{\\circ} - \\mu + \\alpha - \\beta}{2} \\text{ and } \\angle MBA = \\angle DCP = \\frac{180^{\\circ} - \\mu - \\alpha + \\beta}{2}.\n$$\nHence triangles $PDC$ and $MAB$ have equal angles and so are similar. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32163,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with $AB < AC$. Denote its circumcircle by $\\Omega$ and denote the midpoint of $\\operatorname{arc} CAB$ by $S$. Let the perpendicular from $A$ to $BC$ meet $BS$ and $\\Omega$ at $D$ and $E \\neq A$ respectively. Let the line through $D$ parallel to $BC$ meet line $BE$ at $L$ and denote the circumcircle of triangle $BDL$ by $\\omega$. Let $\\omega$ meet $\\Omega$ again at $P \\neq B$.\nProve that the line tangent to $\\omega$ at $P$, and line $BS$ intersect on the internal bisector of $\\angle BAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $S'$ be the midpoint of $\\operatorname{arc} BC$ of $\\Omega$, diametrically opposite to $S$ so $SS'$ is a diameter in $\\Omega$ and $AS'$ is the angle bisector of $\\angle BAC$. Let the tangent of $\\omega$ at $P$ meet $\\Omega$ again at $Q \\neq P$, then we have $\\angle SQS' = 90^\\circ$.\nWe will show that triangles $APD$ and $S'QS$ are similar and their corresponding sides are parallel. Then it will follow that the lines connecting the corresponding vertices, namely line $AS'$, that is the angle bisector of $\\angle BAC$, line $PQ$, that is the tangent to $\\omega$ at $P$, and $DS$ are concurrent. Note that the sides $AD$ and $S'S$ have opposite directions, so the three lines cannot be parallel.\n\n![](attached_image_1.png)\n\nFirst we show that $AP \\perp DP$. Indeed, from cyclic quadrilaterals $APBE$ and $DPLB$ we can see that\n$$\n\\angle PAD = \\angle PAE = 180^\\circ - \\angle EBP = \\angle PBL = \\angle PDL = 90^\\circ - \\angle ADP.\n$$\nThen, in triangle $APD$ we have $\\angle DPA = 180^\\circ - \\angle PAD - \\angle ADP = 90^\\circ$.\n\nNow we can see that:\n- Both lines $ADE$ and $SS'$ are perpendicular to $BC$, so $AD \\parallel S'S$.\n- Line $PQ$ is tangent to circle $\\omega$ at $P$ so $\\angle DPQ = \\angle DBP = \\angle SBP = \\angle SQP$; it follows that $PD \\parallel QS$.\n- Finally, since $AP \\perp PD \\parallel QS \\perp S'Q$, we have $AP \\parallel S'Q$ as well.\n\nHence the corresponding sides of triangles $APD$ and $S'QS$ are parallel completing the solution.\nAgain, let $S'$ be the midpoint of arc $BC$, diametrically opposite to $S$, so $AES'S$ is an isosceles trapezoid, and $\\angle S'BS = \\angle S'PS = 90^\\circ$. Let lines $AE$ and $PS'$ meet at $T$ and let $AP$ and $S'B$ meet at point $M$.\nWe will need that points $L, P, S$ are collinear, and points $T$ and $M$ lie on circle $\\omega$.\n- From $\\angle LPB = \\angle LDB = 90^\\circ - \\angle BDE = 90^\\circ - \\angle BSS' = \\angle SS'B = 180^\\circ - \\angle BPS$ we get $\\angle LPB + \\angle BPS = 180^\\circ$, so $L, P$ and $S$ are indeed collinear.\n- Since $SS'$ is a diameter in $\\Omega$, lines $LPS$ and $PTS'$ are perpendicular. We also have $LD \\parallel BC \\perp AE$ hence $\\angle LDT = \\angle LPT = 90^\\circ$ and therefore $T \\in \\omega$.\n- By $\\angle LPM = \\angle SPA = \\angle SEA = \\angle EAS' = \\angle EBS' = \\angle LBM$, point $M$ is concyclic with $B, P, L$ so $M \\in \\omega$.\n\n![](attached_image_2.png)\n\nNow let $X$ be the intersection of line $BDS$ with the tangent of $\\omega$ at $P$ and apply Pascal's theorem to the degenerate cyclic hexagon $PPMBDT$. This gives points $PP \\cap BD = X$, $PM \\cap DT = A$ and $MB \\cap TP = S'$ are collinear so $X$ lies on line $AS'$, that is the bisector of $\\angle BAC$.\nLet $A'$ and $S'$ be the points of $\\Omega$ diametrically opposite to $A$ and $S$ respectively. It is well-known that $E$ and $A'$ are reflections with respect to $SS'$ so $AS'$ is the angle bisector of $\\angle EAA'$. Define point $T$ to be the intersection of $AE$ and $PS'$. As in the previous two solutions, we have: $\\angle DPA = 90^\\circ$ so $PD$ passes through $A'$; points $L, P, S$ are collinear; and $T \\in \\omega$.\nLet lines $AS'$ and $PDA'$ meet at $R$. From the angles of triangles $PRS'$ and $PTE$ we get\n$$\n\\angle ARP = \\angle AS'P + \\angle S'PA' = \\angle AEP + \\angle EPS' = \\angle ATP\n$$\nso points $A, P, T, R$ are concyclic. Denote their circle by $\\gamma$. Due to $\\angle RPA = \\angle DPA = 90^\\circ$, segment $AR$ is a diameter in $\\gamma$.\n\n![](attached_image_3.png)\n\nWe claim that circles $\\omega$ and $\\gamma$ are perpendicular. Let line $LPS$ meet $\\gamma$ again at $U \\neq P$, and consider triangles $PLT$ and $PTU$. By $\\angle LPT = \\angle TPU = 90^\\circ$ and\n$$\n\\angle PTL = \\angle PBL = 180^\\circ - \\angle EBP = \\angle PAE = \\angle PAT = \\angle PUT,\n$$\ntriangles $PLT$ and $PTU$ are similar. It follows that the spiral similarity that takes $PLT$ to $PTU$, maps $\\omega$ to $\\gamma$ and the angle of this similarity is $90^\\circ$, so circles $\\omega$ and $\\gamma$ are indeed perpendicular.\n\nFinally, let lines $BDS$ and $ARS'$ meet at $X$. We claim that $X$ bisects $AR$, so point $X$ is the centre of $\\gamma$ and, as $\\omega$ and $\\gamma$ are perpendicular, $PX$ is tangent to $\\omega$.\nLet $t$ be the tangent of $\\omega$ at $D$. From $\\angle(DT, t) = \\angle TPD = \\angle S'PA' = \\angle EAS'$ it can be seen that $t \\parallel AS'$. Let $I$ be the common point at infinity of $t$ and $AS'$. Moreover, let lines $LPS$ and $ADTE$ meet at $V$. By projecting line $AS'$ to circle $\\omega$ through $D$, then projecting $\\omega$ to line $AE$ through $L$, finally projecting $AE$ to $\\Omega$ through $P$, we find\n$$\n\\frac{AX}{RX} = (A, R; X, I) \\stackrel{D}{=} (T, P; B, D) \\stackrel{L}{=} (T, V; E, D) \\stackrel{P}{=} (S', S; E, A') = -1,\n$$\nso $X$ is the midpoint of $AR$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32164,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an equilateral triangle. Points $A_1$, $B_1$, $C_1$ lie inside triangle $ABC$ such that triangle $A_1B_1C_1$ is scalene, $BA_1 = A_1C$, $CB_1 = B_1A$, $AC_1 = C_1B$ and\n$$\n\\angle BA_1C + \\angle CB_1A + \\angle AC_1B = 480^\\circ.\n$$\nLines $BC_1$ and $CB_1$ intersect at $A_2$; lines $CA_1$ and $AC_1$ intersect at $B_2$; and lines $AB_1$ and $BA_1$ intersect at $C_2$.\nProve that the circumcircles of triangles $AA_1A_2$, $BB_1B_2$, $CC_1C_2$ have two common points.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\delta_A$, $\\delta_B$, $\\delta_C$ be the circumcircles of $\\triangle AA_1A_2$, $\\triangle BB_1B_2$, $\\triangle CC_1C_2$. The general strategy of the solution is to find two different points having equal power with respect to $\\delta_A$, $\\delta_B$, $\\delta_C$.\n\nClaim. $A_1$ is the circumcentre of $A_2BC$ and cyclic variations.\n\nProof. Since $A_1$ lies on the perpendicular bisector of $BC$ and inside $\\triangle BA_2C$, it suffices to prove $\\angle BA_1C = 2 \\angle BA_2C$. This follows from\n$$\n\\begin{aligned}\n\\angle BA_2C &= \\angle A_2BA + \\angle BAC + \\angle ACA_2 \\\\\n&= \\frac{1}{2}\\left((180^\\circ - \\angle AC_1B) + (180^\\circ - \\angle CB_1A)\\right) + 60^\\circ \\\\\n&= 240^\\circ - \\frac{1}{2}(480^\\circ - \\angle BA_1C) \\\\\n&= \\frac{1}{2} \\angle BA_1C\n\\end{aligned}\n$$\n$\\square$\n\n![](attached_image_1.png)\n\nThe circumcentres above give\n$$\n\\angle B_1B_2C_1 = \\angle B_1B_2A = \\angle B_2AB_1 = \\angle C_1AC_2 = \\angle AC_2C_1 = \\angle B_1C_2C_1\n$$\nand so $B_1C_1B_2C_2$ is cyclic. Likewise $C_1A_1C_2A_2$ and $A_1B_1A_2B_2$ are cyclic. Note that hexagon $A_1B_2C_1A_2B_1C_2$ is not cyclic since\n$$\n\\angle C_2A_1B_2 + \\angle B_2C_1A_2 + \\angle A_2B_1C_2 = 480^\\circ \\neq 360^\\circ.\n$$\nThus we can apply radical axis theorem to the three circles to show that $A_1A_2$, $B_1B_2$, $C_1C_2$ concur at a point $X$ and this point has equal power with respect to $\\delta_A$, $\\delta_B$, $\\delta_C$.\n\nLet the circumcircle of $\\triangle A_2BC$ meet $\\delta_A$ at $A_3 \\neq A_2$. Define $B_3$ and $C_3$ similarly.\n\nClaim. $BCB_3C_3$ cyclic.\n\nProof. Using directed angles\n$$\n\\begin{align*}\n\\npreceq BC_3C &= \\measuredangle BC_3C_2 + \\not C_2C_3C \\\\\n&= \\measuredangle BAC_2 + \\not C_2C_1C \\\\\n&= 90^\\circ + \\nmid(C_1C, AC_2) + \\not C_2C_1C \\tag{1} \\\\\n&= 90^\\circ + \\nmid C_1C_2B_1.\n\\end{align*}\n$$\nSimilarly $\\nsucceq CB_3B = 90^\\circ + \\measuredangle B_1B_2C_1$. Hence, using $B_1C_1B_2C_2$ cyclic\n$$\n\\Varangle BB_3C = 90^\\circ + \\not \\measuredangle C_1B_2B_1 = 90^\\circ + \\Varangle C_1C_2B_1 = \\Varangle BC_3C\n$$\nas required. $\\square$\n\n![](attached_image_2.png)\n\nSimilarly $CAC_3A_3$ and $ABA_3B_3$ are cyclic. $AC_3BA_3CB_3$ is not cyclic because then $AB_2CB_3$ cyclic would mean $B_2$ lies on $\\odot ABC$ which is impossible since $B_2$ lies inside $\\triangle ABC$. Thus we can apply radical axis theorem to the three circles to get $AA_3$, $BB_3$, $CC_3$ concur at a point $Y$ which has equal power with respect to $\\delta_A$, $\\delta_B$, $\\delta_C$.\n\nWe now make some technical observations before finishing.\n- Let $O$ be the centre of $\\triangle ABC$. We have that\n$$\n\\angle BA_1C = 480^\\circ - \\angle CB_1A - \\angle AC_1B > 480^\\circ - 180^\\circ - 180^\\circ = 120^\\circ.\n$$\nso $A_1$ lies inside $\\triangle BOC$. We have similar results for $B_1$, $C_1$ and thus $\\triangle BA_1C$, $\\triangle CB_1A$, $\\triangle AC_1B$ have disjoint interiors. It follows that $A_1B_2C_1A_2B_1C_2$ is a convex hexagon thus $X$ lies on segment $A_1A_2$ and therefore is inside $\\delta_A$.\n- Since $A_1$ is the centre of $A_2BC$ we have that $A_1A_2 = A_1A_3$ so, from cyclic quadrilateral $AA_2A_1A_3$ we get that lines $AA_2$ and $AA_3 \\equiv AY$ are reflections in line $AA_1$. As $X$ lies on segment $A_1A_2$, the only way $X \\equiv Y$ is if $A_1$ and $A_2$ both lie on the perpendicular bisector of $BC$. But this forces $B_1$ and $C_1$ to also be reflections in this line meaning $A_1B_1 = A_1C_1$ contradicting the scalene condition.\n\nSummarising, we have distinct points $X, Y$ with equal power with respect to $\\delta_A$, $\\delta_B$, $\\delta_C$ thus these circles have a common radical axis. As $X$ lies inside $\\delta_A$ (and similarly $\\delta_B$, $\\delta_C$), this radical axis intersects the circles at two points and so $\\delta_A$, $\\delta_B$, $\\delta_C$ have two points in common.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32165,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive, composite integers $n$ that satisfy the following property: if the positive divisors of $n$ are $1=d_{1}<d_{2}<\\cdots<d_{k}=n$, then $d_{i}$ divides $d_{i+1}+d_{i+2}$ for every $1 \\leqslant i \\leqslant k-2$.",
    "options": [],
    "answer": "n = p^r for a prime p and integer r ≥ 2",
    "solution": "Answer: $n=p^{r}$ is a prime power for some $r \\geqslant 2$.\n\nSolution 1. It is easy to see that such an $n=p^{r}$ with $r \\geqslant 2$ satisfies the condition as $d_{i}=p^{i-1}$ with $1 \\geqslant i \\geqslant k=r+1$ and clearly\n$$\np^{i-1} \\mid p^{i}+p^{i+1}\n$$\nNow, let us suppose that there is a positive integer $n$ that satisfies the divisibility condition of the problem and that has two different prime divisors $p$ and $q$. Without loss of generality, we assume $p<q$ and that they are the two smallest prime divisors of $n$. Then there is a positive integer $j$ such that\n$$\nd_{1}=1,\\ d_{2}=p,\\ \\ldots,\\ d_{j}=p^{j-1},\\ d_{j+1}=p^{j},\\ d_{j+2}=q\n$$\nand it follows that\n$$\nd_{k-j-1}=\\frac{n}{q},\\ d_{k-j}=\\frac{n}{p^{j}},\\ d_{k-j+1}=\\frac{n}{p^{j-1}},\\ \\ldots,\\ d_{k-1}=\\frac{n}{p},\\ d_{k}=n\n$$\nThus\n$$\n\\begin{equation*}\nd_{k-j-1}=\\frac{n}{q} \\left\\lvert\\, d_{k-j}+d_{k-j+1}=\\frac{n}{p^{j}}+\\frac{n}{p^{j-1}}=\\frac{n}{p^{j}}(p+1)\\right.\n\\tag{1}\n\\end{equation*}\n$$\nThis gives $p^{j} \\mid q(p+1)$, which is a contradiction since $\\operatorname{gcd}(p, p+1)=1$ and $p \\neq q$.\n\n\nSolution 2. Since $d_{i} d_{k+1-i}=n$, we have the equivalence:\n$$\nd_{k-i-1}\\left|d_{k-i}+d_{k-i+1} \\Longleftrightarrow \\frac{n}{d_{i+2}}\\right| \\frac{n}{d_{i+1}}+\\frac{n}{d_{i}}\n$$\nWe multiply both sides by $d_{i} d_{i+1} d_{i+2}$ and cancel the $n$'s to get\n$$\nd_{i} d_{i+1} \\mid d_{i} d_{i+2}+d_{i+1} d_{i+2}\n$$\nHence,\n$$\n\\begin{equation*}\nd_{i} \\mid d_{i+1} d_{i+2}\n\\tag{2}\n\\end{equation*}\n$$\nMoreover, by the condition of the problem,\n$$\nd_{i} \\mid d_{i+1}\\left(d_{i+1}+d_{i+2}\\right)=d_{i+1}^{2}+d_{i+1} d_{i+2}\n$$\nCombining this with (2) we get that $d_{i} \\mid d_{i+1}^{2}$ for all $1 \\leqslant i \\leqslant k-2$.\nLet $d_{2}=p$ be the smallest prime divisor of $n$. By induction on $i$ we prove that $p \\mid d_{i}$ for all $2 \\leqslant i \\leqslant k-1$. The base case $d_{2}=p$ is obvious. Let us suppose that $p \\mid d_{j}$ for some $2 \\leqslant j \\leqslant k-2$. Then we have that\n$$\np\\left|d_{j}\\right| d_{j+1}^{2} \\Longrightarrow p \\mid d_{j+1}\n$$\nas $p$ is prime, which completes the induction. This implies that $n$ has to be a prime power, as otherwise there would be another prime $q$ that divides $n$ and we would get that $p \\mid q$ which is obviously false.\nWe finally check that the powers of $p$ satisfy the condition in the statement of the problem as in Solution 1.\n\n\nSolution 3. We start by proving the following claim:\nClaim. $d_{i} \\mid d_{i+1}$ for every $1 \\leqslant i \\leqslant k-1$.\nProof. We prove the Claim by induction on $i$; it is trivial for $i=1$ because $d_{1}=1$. Suppose that $2 \\leqslant i \\leqslant k-1$ and the Claim is true for $i-1$, i.e. $d_{i-1} \\mid d_{i}$. By the induction hypothesis and the problem condition, $d_{i-1} \\mid d_{i}$ and $d_{i-1} \\mid d_{i}+d_{i+1}$, so $d_{i-1} \\mid d_{i+1}$.\nNow consider the divisors $d_{k-i}=\\frac{n}{d_{i+1}}, d_{k-i+1}=\\frac{n}{d_{i}}, d_{k-i+2}=\\frac{n}{d_{i-1}}$. By the problem condition,\n$$\n\\frac{d_{k-i+1}+d_{k-i+2}}{d_{k-i}}=\\frac{\\frac{n}{d_{i}}+\\frac{n}{d_{i-1}}}{\\frac{n}{d_{i+1}}}=\\frac{d_{i+1}}{d_{i}}+\\frac{d_{i+1}}{d_{i-1}}\n$$\nis an integer. We conclude that $\\frac{d_{i+1}}{d_{i}}$ is an integer, so $d_{i} \\mid d_{i+1}$.\nBy the Claim, $n$ cannot have two different prime divisors because the smallest one would divide the other one. Hence, $n$ must be a power of a prime, and powers of primes satisfy the condition of the problem as we saw in Solution 1.\n\n\nSolution 4. We present here a more technical way of finishing Solution 1 after obtaining (1). We let $v_{p}(m)$ denote the $p$-adic valuation of $m$. Notice that $v_{p}(n / q)=v_{p}(n)$ as $\\operatorname{gcd}(p, q)=1$ and that\n$$\nv_{p}\\left(\\frac{n}{p^{j}}(p+1)\\right)=v_{p}(n)-j\n$$\nas $\\operatorname{gcd}(p, p+1)=1$. But (1) implies\n$$\nv_{p}(n)=v_{p}(n / q) \\leqslant v_{p}\\left(\\frac{n}{p^{j}}(p+1)\\right)=v_{p}(n)-j\n$$\nwhich is a contradiction. Thus $n$ has only one prime divisor as desired.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32166,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n}$ be $2n$ positive integers such that the $n+1$ products\n$$\n\\begin{gathered}\na_{1} a_{2} a_{3} \\cdots a_{n}, \\\\\nb_{1} a_{2} a_{3} \\cdots a_{n}, \\\\\nb_{1} b_{2} a_{3} \\cdots a_{n}, \\\\\n\\vdots \\\\\nb_{1} b_{2} b_{3} \\cdots b_{n}\n\\end{gathered}\n$$\nform a strictly increasing arithmetic progression in that order. Determine the smallest positive integer that could be the common difference of such an arithmetic progression.",
    "options": [],
    "answer": "n!",
    "solution": "Answer: The smallest common difference is $n!$.\n\nSolution 1. The condition in the problem is equivalent to\n$$\nD=\\left(b_{1}-a_{1}\\right) a_{2} a_{3} \\cdots a_{n}=b_{1}\\left(b_{2}-a_{2}\\right) a_{3} a_{4} \\cdots a_{n}=\\cdots=b_{1} b_{2} \\cdots b_{n-1}\\left(b_{n}-a_{n}\\right),\n$$\nwhere $D$ is the common difference. Since the progression is strictly increasing, $D>0$, hence $b_{i}>a_{i}$ for every $1 \\leqslant i \\leqslant n$. Individually, these equalities simplify to\n$$\n\\begin{equation*}\n\\left(b_{i}-a_{i}\\right) a_{i+1}=b_{i}\\left(b_{i+1}-a_{i+1}\\right) \\text{ for every } 1 \\leqslant i \\leqslant n-1 \\tag{1}\n\\end{equation*}\n$$\nIf $g_{i}:=\\operatorname{gcd}\\left(a_{i}, b_{i}\\right)>1$ for some $1 \\leqslant i \\leqslant n$, then we can replace $a_{i}$ with $\\frac{a_{i}}{g_{i}}$ and $b_{i}$ with $\\frac{b_{i}}{g_{i}}$ to get a smaller common difference. Hence we may assume $\\operatorname{gcd}\\left(a_{i}, b_{i}\\right)=1$ for every $1 \\leqslant i \\leqslant n$.\nThen, we have $\\operatorname{gcd}\\left(b_{i}-a_{i}, b_{i}\\right)=\\operatorname{gcd}\\left(a_{i}, b_{i}\\right)=1$ and $\\operatorname{gcd}\\left(a_{i+1}, b_{i+1}-a_{i+1}\\right)=\\operatorname{gcd}\\left(a_{i+1}, b_{i+1}\\right)=1$ for every $1 \\leqslant i \\leqslant n-1$. The equality (1) implies $a_{i+1}=b_{i}$ and $b_{i}-a_{i}=b_{i+1}-a_{i+1}$. Thus,\n$$\na_{1}, \\quad b_{1}=a_{2}, \\quad b_{2}=a_{3}, \\quad \\ldots, \\quad b_{n-1}=a_{n}, \\quad b_{n}\n$$\nis an arithmetic progression with positive common difference. Since $a_{1} \\geqslant 1$, we have $a_{i} \\geqslant i$ for every $1 \\leqslant i \\leqslant n$, so\n$$\nD=\\left(b_{1}-a_{1}\\right) a_{2} a_{3} \\cdots a_{n} \\geqslant 1 \\cdot 2 \\cdot 3 \\cdots n=n!\n$$\nEquality is achieved when $b_{i}-a_{i}=1$ for $1 \\leqslant i \\leqslant n$ and $a_{1}=1$, i.e. $a_{i}=i$ and $b_{i}=i+1$ for every $1 \\leqslant i \\leqslant n$. Indeed, it is straightforward to check that these integers produce an arithmetic progression with common difference $n!$.\n\n\nSolution 2 (Variant of Solution 1). Similarly to Solution 1, we may assume $\\operatorname{gcd}\\left(a_{i}, b_{i}\\right)=1$ for every $1 \\leqslant i \\leqslant n$.\nDenote by $p_{1}, p_{2}, \\ldots, p_{n+1}$ the sequence obtained as the product in the problem statement. Then we have $\\frac{p_{i+1}}{p_{i}}=\\frac{b_{i}}{a_{i}}>1$, so $b_{i}>a_{i}$. Since $p_{1}, p_{2}, \\ldots, p_{n+1}$ is an arithmetic progression, we have $p_{i+2}=2 p_{i+1}-p_{i}$ hence\n$$\n2-\\frac{a_{i}}{b_{i}}=\\frac{2 b_{i}-a_{i}}{b_{i}}=\\frac{2 p_{i+1}-p_{i}}{p_{i+1}}=\\frac{p_{i+2}}{p_{i+1}}=\\frac{b_{i+1}}{a_{i+1}} .\n$$\nBut since the fractions on the left-hand side and the right-hand side are both irreducible, we conclude that $b_{i}=a_{i+1}$, so $2-\\frac{a_{i}}{a_{i+1}}=\\frac{a_{i+2}}{a_{i+1}}$. Then we have $a_{i}+a_{i+2}=2 a_{i+1}$, which means that $a_{1}, a_{2}, \\ldots, a_{n}$ is an arithmetic progression with positive common difference.\nWe conclude as in Solution 1.\n\n\nSolution 3. (The following solution is purely algebraic: it does not involve considerations on greatest common divisors.)\nWe retake Solution 1 from (1). Then we have\n$$\n\\frac{a_{i+1}}{b_{i+1}-a_{i+1}}=\\frac{b_{i}}{b_{i}-a_{i}}=1+\\frac{a_{i}}{b_{i}-a_{i}} .\n$$\nSo, for $1 \\leqslant i \\leqslant n$,\n$$\n\\frac{a_{i}}{b_{i}-a_{i}}=\\frac{a_{1}}{b_{1}-a_{1}}+(i-1)\n$$\nThen\n$$\na_{i} \\geqslant \\frac{a_{i}}{b_{i}-a_{i}}=\\frac{a_{1}}{b_{1}-a_{1}}+(i-1)>i-1\n$$\nsince $b_{i}-a_{i} \\geqslant 1$ and $b_{1}-a_{1}>0$. As $a_{i}$ is an integer, we have $a_{i} \\geqslant i$.\nWe again conclude as in Solution 1.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32167,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{1} < a_{2} < a_{3} < \\cdots$ be positive integers such that $a_{k+1}$ divides $2(a_{1} + a_{2} + \\cdots + a_{k})$ for every $k \\geqslant 1$. Suppose that for infinitely many primes $p$, there exists $k$ such that $p$ divides $a_{k}$. Prove that for every positive integer $n$, there exists $k$ such that $n$ divides $a_{k}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "For every $k \\geqslant 2$ define the quotient $b_{k} = 2(a_{1} + \\cdots + a_{k-1}) / a_{k}$, which must be a positive integer. We first prove the following properties of the sequence $(b_{k})$:\n\nClaim 1. We have $b_{k+1} \\leqslant b_{k} + 1$ for all $k \\geqslant 2$.\n\nProof. By subtracting $b_{k} a_{k} = 2(a_{1} + \\cdots + a_{k-1})$ from $b_{k+1} a_{k+1} = 2(a_{1} + \\cdots + a_{k})$, we find that $b_{k+1} a_{k+1} = b_{k} a_{k} + 2 a_{k} = (b_{k} + 2) a_{k}$. From $a_{k} < a_{k+1}$ it follows that $b_{k} + 2 > b_{k+1}$.\n\nClaim 2. The sequence $(b_{k})$ is unbounded.\n\nProof. We start by rewriting $b_{k+1} a_{k+1} = (b_{k} + 2) a_{k}$ as\n$$\na_{k+1} = a_{k} \\cdot \\frac{b_{k} + 2}{b_{k+1}} \\Longrightarrow a_{k+1} \\mid a_{k}(b_{k} + 2)\n$$\nIf the sequence $(b_{k})$ were bounded, say by some positive integer $B$, then the prime factors of the terms of the sequence $(a_{k})$ could only be primes less than or equal to $B + 2$ or those dividing $a_{1}$ or $a_{2}$, which contradicts the property in the statement of the problem.\n\nConsider now an arbitrary positive integer $n$. We assume $n > b_{2}$, otherwise we replace $n$ by an arbitrary multiple of $n$ that is bigger than $b_{2}$. By Claim 2, there exists $k$ such that $b_{k+1} \\geqslant n$. Consider the smallest such $k$. From Claim 1, it follows that we must have $b_{k} = n - 1$ and $b_{k+1} = n$ (we assumed $n > b_{2}$ to ensure that $k \\geqslant 2$). We now find that\n$$\na_{k+1} = a_{k} \\cdot \\frac{b_{k} + 2}{b_{k+1}} = a_{k} \\cdot \\frac{n + 1}{n}.\n$$\nBecause $n$ and $n + 1$ are coprime, this immediately implies that $a_{k}$ is divisible by $n$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32168,
    "subject": "Mathematics (Multi-modal)",
    "question": "A sequence of integers $a_{0}, a_{1}, a_{2}, \\ldots$ is called kawaii, if $a_{0}=0, a_{1}=1$, and, for any positive integer $n$, we have\n$$\n\\left(a_{n+1}-3 a_{n}+2 a_{n-1}\\right)\\left(a_{n+1}-4 a_{n}+3 a_{n-1}\\right)=0 .\n$$\nAn integer is called kawaii if it belongs to a kawaii sequence.\nSuppose that two consecutive positive integers $m$ and $m+1$ are both kawaii (not necessarily belonging to the same kawaii sequence). Prove that 3 divides $m$, and that $m / 3$ is kawaii.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We start by rewriting the condition in the problem as:\n$$\na_{n+1}=3 a_{n}-2 a_{n-1}, \\text{ or } a_{n+1}=4 a_{n}-3 a_{n-1}\n$$\nWe have $a_{n+1} \\equiv a_{n}$ or $a_{n-1}(\\bmod 2)$ and $a_{n+1} \\equiv a_{n-1}$ or $a_{n}(\\bmod 3)$ for all $n \\geqslant 1$. Now, since $a_{0}=0$ and $a_{1}=1$, we have that $a_{n} \\equiv 0,1 \\bmod 3$ for all $n \\geqslant 0$. Since $m$ and $m+1$ are kawaii integers, then necessarily $m \\equiv 0 \\bmod 3$.\nWe also observe that $a_{2}=3$ or $a_{2}=4$. Moreover,\n(1) If $a_{2}=3$, then $a_{n} \\equiv 1(\\bmod 2)$ for all $n \\geqslant 1$ since $a_{1} \\equiv a_{2} \\equiv 1(\\bmod 2)$.\n(2) If $a_{2}=4$, then $a_{n} \\equiv 1(\\bmod 3)$ for all $n \\geqslant 1$ since $a_{1} \\equiv a_{2} \\equiv 1(\\bmod 3)$.\nSince $m \\equiv 0(\\bmod 3)$, any kawaii sequence containing $m$ does not satisfy (2), so it must satisfy (1). Hence, $m$ is odd and $m+1$ is even.\nTake a kawaii sequence $\\left(a_{n}\\right)$ containing $m+1$. Let $t \\geqslant 2$ be such that $a_{t}=m+1$. As $\\left(a_{n}\\right)$ does not satisfy (1), it must satisfy (2). Then $a_{n} \\equiv 1(\\bmod 3)$ for all $n \\geqslant 1$. We define the sequence $a_{n}^{\\prime}=\\left(a_{n+1}-1\\right) / 3$. This is a kawaii sequence: $a_{0}^{\\prime}=0, a_{1}^{\\prime}=1$ and for all $n \\geqslant 1$, $\\left(a_{n+1}^{\\prime}-3 a_{n}^{\\prime}+2 a_{n-1}^{\\prime}\\right)\\left(a_{n+1}^{\\prime}-4 a_{n}^{\\prime}+3 a_{n-1}^{\\prime}\\right)=\\left(a_{n+2}-3 a_{n+1}+2 a_{n}\\right)\\left(a_{n+2}-4 a_{n+1}+3 a_{n}\\right) / 9=0$.\nFinally, we notice that the term $a_{t-1}^{\\prime}=m / 3$ which implies that $m / 3$ is kawaii.\nWe start by proving the following:\nClaim 1. We have $a_{n} \\equiv 0,1 \\bmod 3$ for all $n \\geqslant 0$.\nProof. We have $a_{n+1}=3 a_{n}-2 a_{n-1}=3\\left(a_{n}-a_{n-1}\\right)+a_{n-1}$ or $a_{n+1}=4 a_{n}-3 a_{n-1}=3\\left(a_{n}-a_{n-1}\\right)+a_{n}$, so $a_{n+1} \\equiv a_{n}$ or $a_{n-1} \\bmod 3$, and since $a_{0}=0$ and $a_{1}=1$ the result follows.\nHence if $m$ and $m+1$ are kawaii, then necessarily $m \\equiv 0 \\bmod 3$.\nClaim 2. An integer $\\geqslant 2$ is kawaii if and only if it can be written as $1+b_{2}+\\cdots+b_{n}$ for some $n \\geqslant 2$ with $b_{i}=2^{r_{i}} 3^{s_{i}}$ satisfying $r_{i}+s_{i}=i-1$ for $i=2, \\ldots, n$ and $b_{i} \\mid b_{i+1}$ for all $i=2, \\ldots, n-1$.\nProof. For a kawaii sequence ( $a_{n}$ ), we can write $a_{n+1}=3 a_{n}-2 a_{n-1}=a_{n}+2\\left(a_{n}-a_{n-1}\\right)$ or $a_{n+1}=4 a_{n}-3 a_{n-1}=a_{n}+3\\left(a_{n}-a_{n-1}\\right)$, so $a_{n+1}-a_{n}=2\\left(a_{n}-a_{n-1}\\right)$ or $3\\left(a_{n}-a_{n-1}\\right)$. Hence, $a_{n}=1+b_{2}+\\cdots+b_{n}$ where $b_{2}=2$ or 3 and $b_{i+1}=2 b_{i}$ or $3 b_{i}$.\nConversely, given a number that can be written in that way, we consider any sequence given by $a_{0}=0, a_{1}=1$ and $a_{i}=1+b_{2}+\\cdots+b_{i}$ for $2 \\leqslant i \\leqslant n$ and $a_{i}$ given by the kawaii condition for $i \\geqslant n+1$. This defines a kawaii sequence containing the given number as $a_{n}$.\nLet us suppose that $m$ and $m+1$ are kawaii, then they belong to some kawaii sequences and we can write them as in Claim 2 as $m=1+2+\\cdots+2^{\\ell}+2^{\\ell} \\cdot 3 \\cdot A$ and $m+1=1+2+\\cdots+2^{\\ell^{\\prime}}+2^{\\ell^{\\prime}} \\cdot 3 \\cdot A^{\\prime}$ where $\\ell$ is odd and $\\ell^{\\prime}$ is even because of modulo 3 reasons. Since $m+1 \\equiv m\\left(\\bmod 2^{\\min \\left(\\ell, \\ell^{\\prime}\\right)}\\right)$, we have $\\min \\left(\\ell, \\ell^{\\prime}\\right)=0$, so $\\ell^{\\prime}=0$.\nThen $m+1=1+b_{2}+\\cdots+b_{j}$ for some $b_{i}$ 's as in Claim 2 with $b_{2}=3$ and $b_{i} \\mid b_{i+1}$ : so with $3 \\mid b_{i}$ for all $i=2, \\ldots, j$. Then $\\frac{m}{3}=1+b_{1}^{\\prime}+\\cdots+b_{j-1}^{\\prime}$ with $b_{i}^{\\prime}=\\frac{b_{i+1}}{3}$ as in Claim 2 and $\\frac{m}{3}$ is a kawaii integer.\n(This solution is just a different combination of the ideas in Solutions 1 and 2) We first prove that, in a kawaii sequence $a_{0}, a_{1}, a_{2}, \\ldots$, every term $a_{t}$ with $t \\geqslant 0$ is congruent to 0 or 1 modulo 3.\nFor $n \\geqslant 1$, put $b_{n}=a_{n}-a_{n-1}$. We have\n$$\n\\begin{equation*}\na_{t}=a_{0}+\\sum_{k=1}^{t}\\left(a_{k}-a_{k-1}\\right)=\\sum_{k=1}^{t} b_{k} \\tag{*}\n\\end{equation*}\n$$\nNote that\n$$\na_{n+1}-3 a_{n}+2 a_{n-1}=b_{n+1}-2 b_{n} \\quad \\text{ and } \\quad a_{n+1}-4 a_{n}+3 a_{n-1}=b_{n+1}-3 b_{n} .\n$$\nThe conditions on the $b_{i}$ 's for defining a kawaii sequence are\n$$\nb_{1}=a_{1}-a_{0}=1, \\quad \\text{ and } \\quad \\frac{b_{n+1}}{b_{n}} \\in\\{2,3\\} \\quad \\text{ for } \\quad n \\geqslant 1 .\n$$\n1. If we have $\\frac{b_{n+1}}{b_{n}}=2$ for any $n$ with $1 \\leqslant n \\leqslant t-1$, then (*) implies that\n$$\na_{t}=\\sum_{k=1}^{t} 2^{k-1}=2^{t}-1 \\equiv 0,1(\\bmod 3)\n$$\n2. If there exists some integer $s$ with $2 \\leqslant s \\leqslant t-1$ such that\n$$\n\\frac{b_{2}}{b_{1}}=\\frac{b_{3}}{b_{2}}=\\cdots=\\frac{b_{s}}{b_{s-1}}=2, \\quad \\frac{b_{s+1}}{b_{s}}=3\n$$\nit implies that $3 \\mid b_{n}$ for any $n \\geqslant s+1$. Similarly to the argument in (1), we obtain $a_{t} \\equiv \\sum_{k=1}^{s} b_{k} \\equiv 0,1(\\bmod 3)$.\n3. If $\\frac{b_{2}}{b_{1}}=b_{2}=3$, we have $3 \\mid b_{n}$ for any $n \\geqslant 2$, and hence $a_{t} \\equiv 1(\\bmod 3)$.\nCombining these, we have proved that $a_{t} \\equiv 0,1(\\bmod 3)$.\nWe next prove that no positive kawaii integer is divisible by both 2 and 3 . If $b_{2}=2$ for some kawaii sequence, then $2 \\mid b_{n}$ and $a_{n} \\equiv 1(\\bmod 2)$ for all $n \\geqslant 2$ in it. If $b_{2}=3$ in some kawaii sequence, then $3 \\mid b_{n}$ and $a_{n} \\equiv 1(\\bmod 3)$ for all $n \\geqslant 2$ in it.\nNow, consider the original problem. Since $m$ and $m+1$ are both kawaii integer, it means\n$$\nm \\equiv 0,1(\\bmod 3), \\quad \\text{ and } \\quad m+1 \\equiv 0,1(\\bmod 3)\n$$\nand hence we easily obtain $3 \\mid m$. Since a kawaii integer $m$ is divisible by $3, m$ must be odd, and hence $m+1$ is even. Take a kawaii sequence $a_{0}, a_{1}, a_{2}, \\ldots$ containing $m+1$ as $a_{t}$. The fact that $m+1$ is even implies that $b_{2}=3$ and so $3 \\mid b_{n}$ for all $n \\geqslant 2$ in this sequence. Set $b_{n}^{\\prime}=\\frac{b_{n+1}}{3}$ for $n \\geqslant 1$. Thus $b_{1}^{\\prime}=\\frac{b_{2}}{3}=1$, and $\\frac{b_{n+1}^{\\prime}}{b_{n}^{\\prime}}=\\frac{b_{n+2}}{b_{n+1}} \\in\\{2,3\\}$ for all $n \\geqslant 1$. Define $a_{0}^{\\prime}=0$ and $a_{n}^{\\prime}=\\sum_{k=1}^{n} b_{k}^{\\prime}$ for $n \\geqslant 1$, then $a_{0}^{\\prime}, a_{1}^{\\prime}, a_{2}^{\\prime}, \\ldots$ is a kawaii sequence. Now,\n$$\na_{t-1}^{\\prime}=\\sum_{k=1}^{t-1} b_{k}^{\\prime}=\\frac{1}{3} \\sum_{k=2}^{t} b_{k}=\\frac{1}{3}\\left(-b_{1}+\\sum_{k=1}^{t} b_{k}\\right)=\\frac{1}{3}\\left(-1+a_{t}\\right)=\\frac{m}{3} .\n$$\nThis means that $\\frac{m}{3}$ is a kawaii integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32169,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c, d$ be positive integers satisfying\n$$\n\\frac{a b}{a+b}+\\frac{c d}{c+d}=\\frac{(a+b)(c+d)}{a+b+c+d}\n$$\nDetermine all possible values of $a+b+c+d$.",
    "options": [],
    "answer": "All positive integers that are not square-free.",
    "solution": "Answer: The possible values are the positive integers that are not square-free.\n\nFirst, note that if we take $a=\\ell$, $b=k \\ell$, $c=k \\ell$, $d=k^{2} \\ell$ for some positive integers $k$ and $\\ell$, then we have\n$$\n\\frac{a b}{a+b}+\\frac{c d}{c+d}=\\frac{k \\ell^{2}}{\\ell+k \\ell}+\\frac{k^{3} \\ell^{2}}{k \\ell+k^{2} \\ell}=\\frac{k \\ell}{k+1}+\\frac{k^{2} \\ell}{k+1}=k \\ell\n$$\nand\n$$\n\\frac{(a+b)(c+d)}{a+b+c+d}=\\frac{(\\ell+k \\ell)\\left(k \\ell+k^{2} \\ell\\right)}{\\ell+k \\ell+k \\ell+k^{2} \\ell}=\\frac{k(k+1)^{2} \\ell^{2}}{\\ell(k+1)^{2}}=k \\ell,\n$$\nso that\n$$\n\\frac{a b}{a+b}+\\frac{c d}{c+d}=k \\ell=\\frac{(a+b)(c+d)}{a+b+c+d}\n$$\nThis means that $a+b+c+d=\\ell\\left(1+2 k+k^{2}\\right)=\\ell(k+1)^{2}$ can be attained. We conclude that all non-square-free positive integers can be attained.\n\nNow, we will show that if\n$$\n\\frac{a b}{a+b}+\\frac{c d}{c+d}=\\frac{(a+b)(c+d)}{a+b+c+d}\n$$\nthen $a+b+c+d$ is not square-free. We argue by contradiction. Suppose that $a+b+c+d$ is square-free, and note that after multiplying by $(a+b)(c+d)(a+b+c+d)$, we obtain\n$$\n\\begin{equation*}\n(a b(c+d)+c d(a+b))(a+b+c+d)=(a+b)^{2}(c+d)^{2} . \\tag{1}\n\\end{equation*}\n$$\nA prime factor of $a+b+c+d$ must divide $a+b$ or $c+d$, and therefore divides both $a+b$ and $c+d$. Because $a+b+c+d$ is square-free, the fact that every prime factor of $a+b+c+d$ divides $a+b$ implies that $a+b+c+d$ itself divides $a+b$. Because $a+b<a+b+c+d$, this is impossible. So $a+b+c+d$ cannot be square-free.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32170,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that\n$$\nf^{b f(a)}(a+1) = (a+1) f(b)\n$$\nholds for all $a, b \\in \\mathbb{Z}_{>0}$, where $f^{k}(n) = f(f(\\cdots f(n) \\cdots))$ denotes the composition of $f$ with itself $k$ times.",
    "options": [],
    "answer": "f(n) = n + 1 for all positive integers n",
    "solution": "Answer: The only function satisfying the condition is $f(n) = n+1$ for all $n \\in \\mathbb{Z}_{>0}$.\nLet $P(a, b)$ be the equality in the statement.\n\nSolution 1. We divide the solution into 5 steps.\n\nStep 1. ($f$ is injective)\nClaim 1. For any $a \\geqslant 2$, the set $\\{f^{n}(a) \\mid n \\in \\mathbb{Z}_{>0}\\}$ is infinite.\nProof. First, we have $f^{f(a)}(a+1) \\stackrel{P(a, 1)}{=}(a+1) f(1)$. Varying $a$, we see that $f(\\mathbb{Z}_{>0})$ is infinite. Next, we have $f^{b f(a-1)}(a) \\stackrel{P(a-1, b)}{=} a f(b)$. So, varying $b$, $f^{b f(a-1)}(a)$ takes infinitely many values.\n\nClaim 2. For any $a \\geqslant 2$ and $n \\in \\mathbb{Z}_{>0}$, we have $f^{n}(a) \\neq a$.\nProof. Otherwise we would get a contradiction with Claim 1.\n\nAssume $f(b) = f(c)$ for some $b < c$. Then we have\n$$\n\\begin{aligned}\n(a+1) f(c) & \\stackrel{P(a, c)}{=} f^{c f(a)}(a+1) \\\\\n& = f^{(c-b) f(a)}(f^{b f(a)}(a+1)) \\\\\n& \\stackrel{P(a, b)}{=} f^{(c-b) f(a)}((a+1) f(b)) \\\\\n& = f^{(c-b) f(a)}((a+1) f(c))\n\\end{aligned}\n$$\nwhich contradicts Claim 2. So, $f$ is injective.\n\nStep 2. $(f(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{\\geqslant 2})$\nClaim 3. 1 is not in the range of $f$.\nProof. If $f(b) = 1$, then $f^{f(a)}(a+1) = a+1$ by $P(a, 1)$, which contradicts Claim 2.\n\nWe say that $a$ is a descendant of $b$ if $f^{n}(b) = a$ for some $n \\in \\mathbb{Z}_{>0}$.\nClaim 4. For any $a, b \\geqslant 1$, both of the following cannot happen at the same time:\n- $a$ is a descendant of $b$;\n- $b$ is a descendant of $a$.\nProof. If both of the above hold, then $a = f^{m}(b)$ and $b = f^{n}(a)$ for some $m, n \\in \\mathbb{Z}_{>0}$. Then $a = f^{m+n}(a)$, which contradicts Claim 2.\n\nClaim 5. For any $a, b \\geqslant 2$, exactly one of the following holds:\n- $a$ is a descendant of $b$;\n- $b$ is a descendant of $a$;\n- $a = b$.\nProof. For any $c \\geqslant 2$, taking $m = f^{c f(a-1)-1}(a)$ and $n = f^{c f(b-1)-1}(b)$, we have\n$$\nf(m) = f^{c f(a-1)}(a) \\stackrel{P(a-1, c)}{=} a f(c) \\text{ and } f(n) = f^{c f(b-1)}(b) \\stackrel{P(b-1, c)}{=} b f(c).\n$$\nHence\n$$\nf^{n f(a-1)}(a) \\stackrel{P(a-1, n)}{=} a f(n) = a b f(c) = b f(m) \\stackrel{P(a-1, m)}{=} f^{m f(b-1)}(b).\n$$\nThe assertion then follows from the injectivity of $f$ and Claim 2.\n\nNow, we show that any $a \\geqslant 2$ is in the range of $f$. Let $b = f(1)$. If $a = b$, then $a$ is in the range of $f$. If $a \\neq b$, either $a$ is a descendant of $b$, or $b$ is a descendant of $a$ by Claim 5. If $b$ is a descendant of $a$, then $b = f^{n}(a)$ for some $n \\in \\mathbb{Z}_{>0}$, so $1 = f^{n-1}(a)$. Then, by Claim 3, we have $n = 1$, so $1 = a$, which is absurd. So, $a$ is a descendant of $b$. In particular, $a$ is in the range of $f$. Thus, $f(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{\\geqslant 2}$.\n\nStep 3. $(f(1) = 2)$\nClaim 6. Let $a, n \\geqslant 2$, then $n a$ is a descendant of $a$.\nProof. We write $n = f(m)$ by Step 2. We have $n a = f(m) a \\stackrel{P(a-1, m)}{=} f^{m f(a-1)}(a)$, which shows $n a$ is a descendant of $a$.\n\nBy Claim 6, all even integers $\\geqslant 4$ are descendants of 2. Hence $2 = f(2k+1)$ for some $k \\geqslant 0$.\nNext, we show $f(2k+1) \\geqslant f(1)$, which implies $f(1) = 2$. It trivially holds if $k = 0$. If $k \\geqslant 1$, let $n$ be the integer such that $f^{n}(2) = 2k+2$. For any $b > n / f(1)$, we have\n$$\nf^{b f(1)-n}(2k+2) = f^{b f(1)}(2) \\stackrel{P(1, b)}{=} 2 f(b) \\text{ and } f^{b f(2k+1)}(2k+2) \\stackrel{P(2k+1, b)}{=} (2k+2) f(b)\n$$\nBy Claim 6, $(2k+2) f(b)$ is a descendant of $2 f(b)$. By Claim 2, we have $b f(2k+1) > b f(1) - n$. By taking $b$ large enough, we conclude $f(2k+1) \\geqslant f(1)$.\n\nStep 4. $(f(2) = 3$ and $f(3) = 4)$ From $f(1) = 2$ and $P(1, b)$, we have $f^{2b}(2) = 2 f(b)$. So taking $b = 1$, we obtain $f^{2}(2) = 2 f(1) = 4$; and taking $b = f(2)$, we have $f^{2 f(2)}(2) = 2 f^{2}(2) = 8$. Hence, $f^{2 f(2)-2}(4) = f^{2 f(2)}(2) = 8$ and $f^{f(3)}(4) \\stackrel{P(3,1)}{=} 8$ give $f(3) = 2 f(2) - 2$.\n\nClaim 7. For any $m, n \\in \\mathbb{Z}_{>0}$, if $f(m)$ divides $f(n)$, then $m \\leqslant n$.\nProof. If $f(m) = f(n)$, the assertion follows from the injectivity of $f$. If $f(m) < f(n)$, by $P(a, m), P(a, n)$ and Claim 6, we have that $f^{n f(a)}(a+1)$ is a descendant of $f^{m f(a)}(a+1)$ for any $a \\in \\mathbb{Z}_{>0}$. So $m f(a) < n f(a)$, and $m < n$.\n\nBy Claim 7, every possible divisor of $f(2)$ is in $\\{1, f(1) = 2, f(2)\\}$. Thus $f(2)$ is an odd prime or $f(2) = 4$. Since $f^{2}(2) = 4$, we have $f(2) \\neq 4$, and hence $f(2)$ is an odd prime. We set $p = f(2)$.\nNow, $f(3) = 2 f(2) - 2 = 2(p-1)$. Since $p-1$ divides $f(3)$, we have $p-1 \\in \\{1, f(1) = 2, f(2) = p\\}$ by Claim 7, so $p-1 = 2$. Thus, $f(2) = p = 3$ and $f(3) = 2(p-1) = 4$.\n\nStep 5. $(f(n) = n+1)$\nClaim 8. For any $b \\geqslant 1, f(2 f(b) - 1) = 2b + 2$.\nProof. Since $f^{2}(2) = 4$, we have $f^{2b-2}(4) = f^{2b}(2) = 2 f(b)$, so\n$$\n\\left.f^{f(2 f(b) - 1) + 2b - 2}(4) = f^{f(2 f(b) - 1)}(2 f(b))\\right)^{P(2 f(b) - 1, 1)} 4 f(b) \\stackrel{P(3, b)}{=} f^{4b}(4)\n$$\nwhich gives us $f(2 f(b) - 1) = 2b + 2$.\n\nFinally, we prove $f(n) = n+1$ by induction on $n$. Suppose $f(n) = n+1$ for all $1 \\leqslant n \\leqslant 2b+1$. Replace $b$ by $b+1$ in $f(2 f(b) - 1) = 2b + 2$ to get\n$$\nf(2b+3) = f(2 f(b+1) - 1) = 2(b+1) + 2 = 2b + 4.\n$$\nBy induction hypothesis, we have $f^{b}(b+2) = 2b + 2$. Hence\n$$\nf(f(2b+2)) = f^{b+2}(b+2) = f^{f(b+1)}(b+2) \\stackrel{P(b+1,1)}{=} 2(b+2) = f(2b+3).\n$$\nBy injectivity, $f(2b+2) = 2b+3$. Then $f(n) = n+1$ for all $n \\in \\mathbb{Z}_{>0}$, which is indeed a solution.\n\n\nSolution 2. In the same way as Steps 1-2 of Solution 1, we have that $f$ is injective and $f(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{\\geqslant 2}$.\nWe first note that Claim 2 in Solution 1 is also true for $a = 1$.\nClaim 2'. For any $a, n \\in \\mathbb{Z}_{>0}$, we have $f^{n}(a) \\neq a$.\nProof. If $a \\geqslant 2$, the assertion was proved in Claim 2 in Solution 1. If $a = 1$, we have that 1 is not in the range of $f$ by Claim 3 in Solution 1. So, $f^{n}(1) \\neq 1$ for every $n \\in \\mathbb{Z}_{>0}$.\n\nFor any $a, b \\in \\mathbb{Z}_{>0}$, we have\n$$\nf^{b f(f(a)-1) + 1}(a) = f^{b f(f(a)-1)}(f(a)) \\stackrel{P(f(a)-1, b)}{=} f(a) f(b)\n$$\nSince the right-hand side is symmetric in $a, b$, we have\n$$\nf^{b f(f(a)-1) + 1}(a) = f(a) f(b) = f^{a f(f(b)-1) + 1}(b).\n$$\nSince $f$ is injective, we have $f^{b f(f(a)-1)}(a) = f^{a f(f(b)-1)}(b)$. We set $g(n) = f(f(n)-1)$. Then we have $f^{b g(a)}(a) = f^{a g(b)}(b)$ for any $a, b \\in \\mathbb{Z}_{>0}$. We set $n_{a, b} = b g(a) - a g(b)$. Then, for sufficiently large $n$, we have $f^{n + n_{a, b}}(a) = f^{n}(b)$. For any $a, b, c \\in \\mathbb{Z}_{>0}$ and sufficiently large $n$, we have\n$$\nf^{n + n_{a, b} + n_{b, c} + n_{c, a}}(a) = f^{n}(a).\n$$\nBy Claim 2' above, we have $n_{a, b} + n_{b, c} + n_{c, a} = 0$, so\n$$\n(a-b) g(c) + (b-c) g(a) + (c-a) g(b) = 0.\n$$\nTaking $(a, b, c) = (n, n+1, n+2)$, we have $g(n+1) - g(n) = g(n+2) - g(n+1)$. So, $\\{g(n)\\}_{n \\geqslant 1}$ is an arithmetic progression.\n\nThere exist $C, D \\in \\mathbb{Z}$ such that $g(n) = f(f(n)-1) = C n + D$ for all $n \\in \\mathbb{Z}_{>0}$. By Step 2 of Solution 1, we have $f(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{\\geqslant 2}$, so $C = 1$. Since $2 = \\min_{n \\in \\mathbb{Z}_{>0}}\\{f(f(n)-1)\\}$, we have $D = 1$.\nThus, $g(n) = f(f(n)-1) = n+1$ for all $n \\geqslant 1$. For any $a, b \\in \\mathbb{Z}_{>0}$, we have $f^{b(a+1)}(a) = f^{a(b+1)}(b)$. By the injectivity of $f$, we have $f^{b}(a) = f^{a}(b)$. For any $n \\in \\mathbb{Z}_{>0}$, taking $(a, b) = (1, n)$, we have $f^{n}(1) = f(n)$, so $f^{n-1}(1) = n$ again by the injectivity of $f$. For any $n \\geqslant 1$, we have $f(n) = f(f^{n-1}(1)) = f^{n}(1) = n+1$.\n\n\nSolution 3. The following is another way of finishing Solution 2 after Claim 2' and having introduced $g(n) = f(f(n)-1)$. For $a, b \\in \\mathbb{Z}_{>0}$ satisfying $b = f^{k}(a)$, we have\n$$\n\\begin{aligned}\nf^{b g(a)}(f(a)) & = f^{b f(f(a)-1)}(f(a)) \\stackrel{P(f(a)-1, b)}{=} f(a) f(b) \\stackrel{P(f(b)-1, a)}{=} f^{a f(f(b)-1)}(f(b)) \\\\\n& = f^{a g(b) + k}(f(a)).\n\\end{aligned}\n$$\nBy Claim 2' in Solution 2, we have $b g(a) = a g(b) + k$, so $f^{k}(a) \\cdot g(a) = a \\cdot g(f^{k}(a)) + k$. Therefore, for any $n \\geqslant 0$, we put $a_{n} = f^{n}(1)$. We have\n$$\n\\left\\{\\begin{array}{l}\na_{n+1} \\cdot g(a_{n}) = a_{n} \\cdot g(a_{n+1}) + 1 \\\\\na_{n+2} \\cdot g(a_{n+1}) = a_{n+1} \\cdot g(a_{n+2}) + 1 \\\\\na_{n+2} \\cdot g(a_{n}) = a_{n} \\cdot g(a_{n+2}) + 2\n\\end{array}\\right.\n$$\nThen we have\n$$\n\\begin{aligned}\na_{n} \\cdot a_{n+1} \\cdot g(a_{n+2}) + 2 a_{n+1} & = a_{n+1} \\cdot a_{n+2} \\cdot g(a_{n}) \\\\\n& = a_{n+2}(a_{n} \\cdot g(a_{n+1}) + 1) \\\\\n& = a_{n} \\cdot a_{n+2} \\cdot g(a_{n+1}) + a_{n+2} \\\\\n& = a_{n}(a_{n+1} \\cdot g(a_{n+2}) + 1) + a_{n+2} \\\\\n& = a_{n} \\cdot a_{n+1} \\cdot g(a_{n+2}) + a_{n} + a_{n+2}\n\\end{aligned}\n$$\nFrom these, we have $2 a_{n+1} = a_{n} + a_{n+2}$. Thus $(a_{n})$ is an arithmetic progression, and we have $a_{n} = f^{n}(1) = C n + D$ for some $C, D \\in \\mathbb{Z}$.\nBy Step 2 in Solution 1, any $a \\in \\mathbb{Z}_{>0}$ is a descendant of 1, and $f(\\mathbb{Z}_{>0}) = \\mathbb{Z}_{\\geqslant 2}$. Hence $D = 1$ and $C = 1$, and so $f^{n}(1) = n+1$. For any $n \\geqslant 1$, we have $f^{n-1}(1) = n$, so $f(n) = f(f^{n-1}(1)) = f^{n}(1) = n+1$.\n\n\nSolution 4. We provide yet another (more technical) solution assuming Step 1 and Step 2 of Solution 1.\nBy Claim 5 in Solution 1, every $a \\geqslant 2$ is a descendant of 1. Let $g$ and $h$ be the functions on $\\mathbb{Z}_{\\geqslant 2}$ such that $f^{g(a)}(1) = a$ and $h(a) = f(a-1)$. Then, $g: \\mathbb{Z}_{\\geqslant 2} \\rightarrow \\mathbb{Z}_{\\geqslant 1}$ and $h: \\mathbb{Z}_{\\geqslant 2} \\rightarrow \\mathbb{Z}_{\\geqslant 2}$ are bijections. The equation $P(a, b)$ can be rewritten as\n$$\ng(a h(b)) = g(a) + (b-1) h(a)\n$$\nConsider the set $S_{a} = g(a \\cdot \\mathbb{Z}_{>0})$. Since $h$ is a bijection onto $\\mathbb{Z}_{\\geqslant 2}$, we have\n$$\nS_{a} = g(a) + h(a) \\cdot \\mathbb{Z}_{\\geqslant 0}\n$$\nConsider the intersection $S_{a} \\cap S_{b} = S_{\\operatorname{lcm}(a, b)}$. If we put $c = \\operatorname{lcm}(a, b)$, this gives\n$$\n\\left(g(a) + h(a) \\cdot \\mathbb{Z}_{\\geqslant 0}\\right) \\cap \\left(g(b) + h(b) \\cdot \\mathbb{Z}_{\\geqslant 0}\\right) = g(c) + h(c) \\cdot \\mathbb{Z}_{\\geqslant 0}.\n$$\nThen we have $h(c) = \\operatorname{lcm}(h(a), h(b))$ since the left hand side must be of the form $m + \\operatorname{lcm}(h(a), h(b)) \\cdot \\mathbb{Z}_{\\geqslant 0}$ for some $m$.\nIf $b$ is a multiple of $a$, then $\\operatorname{lcm}(a, b) = b$, so $h(b) = \\operatorname{lcm}(h(a), h(b))$, and hence $h(b)$ is a multiple of $h(a)$. Conversely, if $h(b)$ is a multiple of $h(a)$, then $h(b) = \\operatorname{lcm}(h(a), h(b))$. On the other hand, we have $h(c) = \\operatorname{lcm}(h(a), h(b))$. Since $h$ is injective, we have $c = b$, so $b$ is a multiple of $a$.\n\nWe apply the following claim for $H = h$.\nClaim. Suppose $H: \\mathbb{Z}_{\\geqslant 2} \\rightarrow \\mathbb{Z}_{\\geqslant 2}$ is a bijection such that $a$ divides $b$ if and only if $H(a)$ divides $H(b)$. Then:\n1. $H(p)$ is prime if and only if $p$ is prime;\n2. $H\\left(\\prod_{i=1}^{m} p_{i}^{e_{i}}\\right) = \\prod_{i=1}^{m} H(p_{i})^{e_{i}}$ i.e. $H$ is completely multiplicative;\n3. $H$ preserves $\\operatorname{gcd}$ and $\\operatorname{lcm}$.\nProof. We define $H(1) = 1$, and consider the bijection $H: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$. By the conditions on $H$, for any $n \\in \\mathbb{Z}_{\\geqslant 2}$, $n$ and $H(n)$ have the same number of divisors. Hence $H(p)$ is prime if and only if $p$ is prime.\nSince the only prime dividing $H(p^{r})$ is $H(p)$, we have $H(p^{r}) = H(p)^{s}$ for some $s \\geqslant 1$. Counting the number of divisors, we have $s = r$, so $H(p^{r}) = H(p)^{r}$ for any prime $p$ and $r \\geqslant 1$.\nFor $a, b \\in \\mathbb{Z}_{>0}$, recall that $\\operatorname{gcd}(a, b)$ is a unique positive integer satisfying the following condition: for any $c \\in \\mathbb{Z}_{>0}$, $c$ divides $\\operatorname{gcd}(a, b)$ if and only if $c$ divides both $a$ and $b$. By the condition on $H$, for any $c \\in \\mathbb{Z}_{>0}$, $H(c)$ divides $H(\\operatorname{gcd}(a, b))$ if and only if $H(c)$ divides both $H(a)$ and $H(b)$. Hence we have $H(\\operatorname{gcd}(a, b)) = \\operatorname{gcd}(H(a), H(b))$.\nSimilarly, we have $H(\\operatorname{lcm}(a, b)) = \\operatorname{lcm}(H(a), H(b))$. Hence we have\n$$\n\\begin{aligned}\nH\\left(\\prod_{i=1}^{m} p_{i}^{e_{i}}\\right) & = H\\left(\\operatorname{lcm}\\left(p_{1}^{e_{1}}, \\ldots, p_{r}^{e_{r}}\\right)\\right) = \\operatorname{lcm}\\left(H\\left(p_{1}^{e_{1}}\\right), \\ldots, H\\left(p_{r}^{e_{r}}\\right)\\right) = \\operatorname{lcm}\\left(H(p_{1})^{e_{1}}, \\ldots, H(p_{r})^{e_{r}}\\right) \\\\\n& = \\prod_{i=1}^{m} H(p_{i})^{e_{i}}\n\\end{aligned}\n$$\nsince $H(p_{i})$ and $H(p_{j})$ are different primes for $i \\neq j$.\n\nTake two primes $p \\neq q$, and let $x, y$ be positive integers such that\n$$\ng(p) + (x-1) h(p) = g(q) + (y-1) h(q).\n$$\nThis is possible as $h(p)$ and $h(q)$ are two distinct primes. For every $k \\geqslant 0$, by $P(p, x + k h(q))$ and $P(q, y + k h(p))$, we have\n$$\n\\left\\{\\begin{array}{l}\ng(p \\cdot h(x + k h(q))) = g(p) + (x + k h(q) - 1) h(p), \\\\\ng(q \\cdot h(y + k h(p))) = g(q) + (y + k h(p) - 1) h(q),\n\\end{array}\\right.\n$$\nwhere the right hand sides are equal. By the injectivity of $g$, we have\n$$\np \\cdot h(x + k h(q)) = q \\cdot h(y + k h(p)).\n$$\nSo, $h(y + k h(p))$ is divisible by $p$ for all $k \\geqslant 0$. By the above Claim, $h$ preserves gcd, so\n$$\nh(\\operatorname{gcd}(y, h(p))) = \\operatorname{gcd}(h(y), h(y + h(p)))\n$$\nis divisible by $p$. Since $h(p)$ is a prime, $y$ must be divisible by $h(p)$. Moreover, $h(h(p))$ is also a prime, so we have $h(h(p)) = p$. The function $h \\circ h$ is completely multiplicative, so we have $h(h(n)) = n$ for every $n \\geqslant 2$.\n\nBy $P(a, h(b))$ and $P(b, h(a))$, we have\n$$\n\\left\\{\\begin{array}{l}\ng(a b) = g(a \\cdot h(h(b))) \\stackrel{P(a, h(b))}{=} g(a) + (h(b) - 1) h(a), \\\\\ng(b a) = g(b \\cdot h(h(a))) \\stackrel{P(b, h(a))}{=} g(b) + (h(a) - 1) h(b),\n\\end{array}\\right.\n$$\nso\n$$\ng(a) + h(a)(h(b) - 1) = g(a b) = g(b) + h(b)(h(a) - 1).\n$$\nHence $g(a) - h(a) = g(b) - h(b)$ for any $a, b \\geqslant 2$, so $g - h$ is a constant function. By comparing the images of $g$ and $h$, the difference is $-1$, i.e. $g(a) - h(a) = 1$ for any $a \\geqslant 2$.\nSo, we have $g(h(a)) = h(h(a)) - 1 = a - 1$. By definition,\n$$\nf(a-1) = h(a) = f^{g(h(a))}(1) = f^{a-1}(1).\n$$\nBy the injectivity of $f$, we have $f^{a-2}(1) = a-1$ for every $a \\geqslant 2$. From this, we can deduce inductively that $f(a) = a+1$ for every $a \\geqslant 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32171,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all real numbers $\\alpha$ such that the number\n$$\n\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor\n$$\nis a multiple of $n$ for every positive integer $n$. (Here $\\lfloor z\\rfloor$ denotes the greatest integer less than or equal to $z$.)",
    "options": [],
    "answer": "All even integers",
    "solution": "Answer: All even integers satisfy the condition of the problem and no other real number $\\alpha$ does so.\n\nSolution 1. First we will show that even integers satisfy the condition. If $\\alpha=2 m$ where $m$ is an integer then\n$$\n\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor=2 m+4 m+\\cdots+2 m n=m n(n+1)\n$$\nwhich is a multiple of $n$.\n\nNow we will show that they are the only real numbers satisfying the conditions of the problem. Let $\\alpha=k+\\epsilon$ where $k$ is an integer and $0 \\leqslant \\epsilon<1$. Then the number\n$$\n\\begin{aligned}\n\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor & =k+\\lfloor\\epsilon\\rfloor+2 k+\\lfloor 2 \\epsilon\\rfloor+\\cdots+n k+\\lfloor n \\epsilon\\rfloor \\\\\n& =\\frac{k n(n+1)}{2}+\\lfloor\\epsilon\\rfloor+\\lfloor 2 \\epsilon\\rfloor+\\cdots+\\lfloor n \\epsilon\\rfloor\n\\end{aligned}\n$$\nhas to be a multiple of $n$. We consider two cases based on the parity of $k$.\n\nCase 1: $k$ is even.\nThen $\\frac{k n(n+1)}{2}$ is always a multiple of $n$. Thus\n$$\n\\lfloor\\epsilon\\rfloor+\\lfloor 2 \\epsilon\\rfloor+\\cdots+\\lfloor n \\epsilon\\rfloor\n$$\nalso has to be a multiple of $n$.\n\nWe will prove that $\\lfloor n \\epsilon\\rfloor=0$ for every positive integer $n$ by strong induction. The base case $n=1$ follows from the fact that $0 \\leqslant \\epsilon<1$. Let us suppose that $\\lfloor m \\epsilon\\rfloor=0$ for every $1 \\leqslant m<n$. Then the number\n$$\n\\lfloor\\epsilon\\rfloor+\\lfloor 2 \\epsilon\\rfloor+\\cdots+\\lfloor n \\epsilon\\rfloor=\\lfloor n \\epsilon\\rfloor\n$$\nhas to be a multiple of $n$. As $0 \\leqslant \\epsilon<1$ then $0 \\leqslant n \\epsilon<n$, which means that the number $\\lfloor n \\epsilon\\rfloor$ has to be equal to 0.\n\nThe equality $\\lfloor n \\epsilon\\rfloor=0$ implies $0 \\leqslant \\epsilon<1 / n$. Since this has to happen for all $n$, we conclude that $\\epsilon=0$ and then $\\alpha$ is an even integer.\n\nCase 2: $k$ is odd.\nWe will prove that $\\lfloor n \\epsilon\\rfloor=n-1$ for every natural number $n$ by strong induction. The base case $n=1$ again follows from the fact that $0 \\leqslant \\epsilon<1$. Let us suppose that $\\lfloor m \\epsilon\\rfloor=m-1$ for every $1 \\leqslant m<n$. We need the number\n$$\n1+\\lfloor n \\epsilon\\rfloor\n$$\nto be a multiple of $n$. As $k$ is odd, we need $1+\\lfloor n \\epsilon\\rfloor$ to be a multiple of $n$. Again, as $0 \\leqslant \\epsilon<1$ then $0 \\leqslant n \\epsilon<n$, so $\\lfloor n \\epsilon\\rfloor=n-1$ as we wanted.\n\nThis implies that $1-\\frac{1}{n} \\leqslant \\epsilon<1$ for all $n$ which is absurd. So there are no other solutions in this case.\n\nComment. An alternative ending to the previous solution is as follows.\nBy definition we have $S_{n} \\leqslant \\alpha \\frac{n(n+1)}{2}$, on the other hand (5) implies $S_{n} \\geqslant \\alpha n^{2}-n$ for all $n$ large enough, so $\\alpha=0$.\nAs in Solution 1 we check that for even integers the condition is satisfied. Then, without loss of generality we can assume $0 \\leqslant \\alpha<2$. We set $S_{n}=\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor$.\n\nNotice that\n$$\n\\begin{array}{ll}\nS_{n} \\equiv 0 & (\\bmod n) \\\\\nS_{n} \\equiv S_{n}-S_{n-1}=\\lfloor n \\alpha\\rfloor & (\\bmod n-1) \\tag{2}\n\\end{array}\n$$\nSince $\\operatorname{gcd}(n, n-1)=1$, (1) and (2) imply that\n$$\n\\begin{equation*}\nS_{n} \\equiv n\\lfloor n \\alpha\\rfloor \\quad(\\bmod n(n-1)) . \\tag{3}\n\\end{equation*}\n$$\nIn addition,\n$$\n\\begin{equation*}\n0 \\leqslant n\\lfloor n \\alpha\\rfloor-S_{n}=\\sum_{k=1}^{n}(\\lfloor n \\alpha\\rfloor-\\lfloor k \\alpha\\rfloor)<\\sum_{k=1}^{n}(n \\alpha-k \\alpha+1)=\\frac{n(n-1)}{2} \\alpha+n . \\tag{4}\n\\end{equation*}\n$$\nFor $n$ large enough, the RHS of (4) is less than $n(n-1)$. Then (3) forces\n$$\n\\begin{equation*}\n0=S_{n}-n\\lfloor n \\alpha\\rfloor=\\sum_{k=1}^{n}(\\lfloor n \\alpha\\rfloor-\\lfloor k \\alpha\\rfloor) \\tag{5}\n\\end{equation*}\n$$\nfor $n$ large enough.\n\nSince $\\lfloor n \\alpha\\rfloor-\\lfloor k \\alpha\\rfloor \\geqslant 0$ for $1 \\leqslant k \\leqslant n$, we get from (5) that, for all $n$ large enough, all these inequalities are equalities. In particular $\\lfloor\\alpha\\rfloor=\\lfloor n \\alpha\\rfloor$ for all $n$ large enough, which is absurd unless $\\alpha=0$.\nAs in other solutions, without loss of generality we may assume that $0 \\leqslant \\alpha<2$. Even integers satisfy the condition, so we assume $0<\\alpha<2$ and we will derive a contradiction.\n\nBy induction on $n$, we will simultaneously show that\n$$\n\\begin{gather*}\n\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor=n^{2}  \\tag{6}\\\\\n\\text{ and } \\quad \\frac{2 n-1}{n} \\leqslant \\alpha<2 . \\tag{7}\n\\end{gather*}\n$$\nThe base case is $n=1$ : If $\\alpha<1$, consider $m=\\left\\lceil\\frac{1}{\\alpha}\\right\\rceil>1$, then\n$$\n\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor m \\alpha\\rfloor=1\n$$\nis not a multiple of $m$, so we deduce (7). Hence, $\\lfloor\\alpha\\rfloor=1$ and (6) follows.\n\nFor the induction step: assume the induction hypothesis to be true for $n$, then by (7)\n$$\n2 n+1-\\frac{1}{n} \\leqslant(n+1) \\alpha<2 n+2\n$$\nHence,\n$$\nn^{2}+2 n \\leqslant\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor+\\lfloor(n+1) \\alpha\\rfloor=n^{2}+\\lfloor(n+1) \\alpha\\rfloor<n^{2}+2 n+2 .\n$$\nSo, necessarily $\\lfloor(n+1) \\alpha\\rfloor=2 n+1$ and\n$$\n\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor+\\lfloor(n+1) \\alpha\\rfloor=(n+1)^{2}\n$$\nin order to obtain a multiple of $n+1$. These two equalities give (6) and (7) respectively.\n\nFinally, we notice that condition (7) being true for all $n$ gives a contradiction.\nAs in other solutions without loss of generality we will assume that $0<\\alpha<2$ and derive a contradiction. For each $n$, we define\n$$\nb_{n}=\\frac{\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor}{n},\n$$\nwhich is a nonnegative integer by the problem condition and our assumption. Note that\n$$\n\\lfloor(n+1) \\alpha\\rfloor \\geqslant\\lfloor\\alpha\\rfloor,\\lfloor 2 \\alpha\\rfloor, \\ldots,\\lfloor n \\alpha\\rfloor \\quad \\text{ and } \\quad\\lfloor(n+1) \\alpha\\rfloor>\\lfloor\\alpha\\rfloor\n$$\nfor all $n>\\frac{1}{\\alpha}$. It follows that $b_{n+1}>b_{n} \\Longrightarrow b_{n+1} \\geqslant b_{n}+1$ for $n>\\frac{1}{\\alpha}$. Thus, for all such $n$,\n$$\nb_{n} \\geqslant n+C\n$$\nwhere $C$ is a fixed integer. On the other hand, the definition of $b_{n}$ gives\n$$\nb_{n}=\\frac{\\lfloor\\alpha\\rfloor+\\lfloor 2 \\alpha\\rfloor+\\cdots+\\lfloor n \\alpha\\rfloor}{n} \\leqslant \\frac{\\alpha+2 \\alpha+\\cdots+n \\alpha}{n}=\\frac{\\alpha}{2}(n+1),\n$$\nwhich is a contradiction for sufficiently large $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32172,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Find the minimum possible value of\n$$\nS = 2^{0} x_{0}^{2} + 2^{1} x_{1}^{2} + \\cdots + 2^{n} x_{n}^{2}\n$$\nwhere $x_{0}, x_{1}, \\ldots, x_{n}$ are nonnegative integers such that $x_{0} + x_{1} + \\cdots + x_{n} = n$.",
    "options": [],
    "answer": "n(n+1)/2",
    "solution": "Answer: The minimum value is $\\frac{n(n+1)}{2}$.\n\nFor a fixed $n$, let $f(n)$ denote the minimum possible value of $S$. Consider the following variant: among all infinite sequences of nonnegative integers $x_{0}, x_{1}, \\ldots$, only finitely many of which are nonzero, satisfying $x_{0} + x_{1} + \\cdots = n$, let $g(n)$ denote the minimum possible value of\n$$\nT = 2^{0} x_{0}^{2} + 2^{1} x_{1}^{2} + 2^{2} x_{2}^{2} + \\cdots\n$$\nIt is clear that $g(n) \\leqslant f(n)$. Conversely, it is easy to see that if a sequence $x_{0}, x_{1}, \\ldots$ achieves the minimum of $g(n)$, then $x_{0} \\geqslant x_{1} \\geqslant \\cdots$ and thus $x_{n+1} = x_{n+2} = \\cdots = 0$. In particular, $f(n) = g(n)$.\n\nNow, we hope to get an inductive formula for $g(n)$.\nNote that, in order to minimise $T$ for $n \\geqslant 1$, we must have $x_{0} \\geqslant 1$ since the sequence ($x_{i}$) is nonincreasing. Note that the minimal value of\n$$\n2^{1} x_{1}^{2} + 2^{2} x_{2}^{2} + \\cdots = 2\\left(2^{0} x_{1}^{2} + 2^{1} x_{2}^{2} + \\cdots\\right)\n$$\nover all infinite sequences of nonnegative integers with $x_{1} + x_{2} + \\cdots = m$ is exactly $2 g(m)$. As a result, for $n \\geqslant 1$ we have\n$$\ng(n) = \\min_{x_{0} \\in \\{1,2, \\ldots, n\\}} \\left(x_{0}^{2} + 2 g\\left(n - x_{0}\\right)\\right).\n$$\nWe now prove $g(n) = \\frac{n(n+1)}{2}$ by induction. It is clear that $g(0) = 0$. Assume that this has been proved for $n = 0, 1, \\ldots, N-1$. Then,\n$$\n\\begin{align*}\nx_{0}^{2} + 2 g\\left(N - x_{0}\\right) & = x_{0}^{2} + \\left(N - x_{0}\\right)\\left(N - x_{0} + 1\\right)  \\\n& = 2 x_{0}^{2} - (2N + 1)x_{0} + N(N+1) \\\\\n& = \\frac{1}{2}\\left[\\left(2x_{0} - N\\right)\\left(2x_{0} - N - 1\\right) + N^{2} + N\\right]\n\\end{align*}\n$$\nThe product of two consecutive integers $\\left(2x_{0} - N\\right)\\left(2x_{0} - N - 1\\right)$ is always nonnegative, and it is zero precisely when $2x_{0}$ is the even number in $\\{N, N+1\\}$. Thus the minimum of the final expression in equation (1) is $\\frac{1}{2}\\left(N^{2} + N\\right)$, so $g(N) = \\frac{N(N+1)}{2}$, completing the inductive proof.\nConsider the following table of numbers, where the row and column indices start from $0$, and $a_{i, j} = 2^{i}(2j+1)$ for $i, j \\geqslant 0$.\n\n|   | $j=0$ | 1 | 2 | 3 | 4 | 5 | $\\cdots$ |\n|---:|---:|---:|---:|---:|---:|---:|---:|\n| $i=0$ | 1 | 3 | 5 | 7 | 9 | 11 |  |\n| 1 | 2 | 6 | 10 | 14 | 18 | 22 |  |\n| 2 | 4 | 12 | 20 | 28 | 36 | 44 |  |\n| 3 | 8 | 24 | 40 | 56 | 72 | 88 |  |\n| 4 | 16 | 48 | 80 | 112 | 144 | 176 |  |\n| $\\vdots$ |  |  |  |  |  |  |  |\n\nEvery number can be written uniquely as a product of a power of $2$ and an odd number so every positive integer appears exactly once in the table above. It is easy to see that numbers in each row and each column are strictly increasing. Since the sum of the first $x$ odd positive integers is $x^{2}$, the sum of the first $x_{k}$ numbers in the $k^{\\text{th}}$ row is $2^{k} x_{k}^{2}$, the $k^{\\text{th}}$ term appearing in $S$.\n\nThus, the sum $S$ can be interpreted as the result of taking a total of $n$ numbers from the first $n$ rows of the table such that we take the leftmost $x_{k}$ numbers from row $k$ (where $\\sum_{k=1}^{n} x_{k} = n$), and then computing the sum of these $n$ numbers. In particular, the minimum possible value of $S$ is the same as the sum of the smallest $n$ numbers in this table, since every row and every column of the table is strictly increasing.\n\nMoreover, the smallest $n$ numbers, namely $1, 2, \\ldots, n$, appear in the first $n$ rows, so the minimum of $S$ is\n$$\n1 + 2 + \\cdots + n = \\frac{n(n+1)}{2}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32173,
    "subject": "Mathematics (Multi-modal)",
    "question": "Decide whether for every sequence $\\left(a_{n}\\right)$ of positive real numbers,\n$$\n\\frac{3^{a_{1}}+3^{a_{2}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{n}}\\right)^{2}}<\\frac{1}{2024}\n$$\nis true for at least one positive integer $n$.",
    "options": [],
    "answer": "Yes",
    "solution": "Solution 1. For every positive integer $n$, let $M_{n}=\\max \\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$. We first prove that\n$$\n\\frac{3^{a_{1}}+3^{a_{2}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{n}}\\right)^{2}} \\leqslant\\left(\\frac{3}{4}\\right)^{M_{n}}\n$$\nFor $i=1,2, \\ldots, n$, from $\\left(\\frac{3}{2}\\right)^{a_{i}} \\leqslant\\left(\\frac{3}{2}\\right)^{M_{n}}$ we can obtain $3^{a_{i}} \\leqslant\\left(\\frac{3}{4}\\right)^{M_{n}} \\cdot 2^{M_{n}} \\cdot 2^{a_{i}}$. By summing up over all $i$,\n$$\n\\sum_{i=1}^{n} 3^{a_{i}} \\leqslant\\left(\\frac{3}{4}\\right)^{M_{n}} \\cdot 2^{M_{n}} \\cdot \\sum_{i=1}^{n} 2^{a_{i}} \\leqslant\\left(\\frac{3}{4}\\right)^{M_{n}} \\cdot\\left(\\sum_{i=1}^{n} 2^{a_{i}}\\right)^{2}\n$$\nwhich is equivalent to the previous inequality.\nNow let $\\mu=\\log _{4 / 3} \\frac{1}{\\varepsilon}$, so that $\\mu$ is the positive real number with $\\left(\\frac{3}{4}\\right)^{\\mu}=\\varepsilon$. If there is an index $n$ such that $a_{n}>\\mu$, then $M_{n} \\geqslant a_{n}>\\mu$, and hence\n$$\n\\frac{3^{a_{1}}+3^{a_{2}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{n}}\\right)^{2}} \\leqslant\\left(\\frac{3}{4}\\right)^{M_{n}}<\\left(\\frac{3}{4}\\right)^{\\mu}=\\varepsilon .\n$$\nOtherwise we have $0<a_{i} \\leqslant \\mu$ for all positive integers $i$, so\n$$\n\\frac{3^{a_{1}}+3^{a_{2}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{n}}\\right)^{2}} \\leqslant \\frac{n \\cdot 3^{\\mu}}{(n \\cdot 1)^{2}}=\\frac{3^{\\mu}}{n}\n$$\nIf $n>\\left\\lfloor\\frac{3^{\\mu}}{\\varepsilon}\\right\\rfloor$, this is less than $\\varepsilon$.\n\n\nSolution 2. We will combine two upper bounds.\nFirst, start with the trivial estimate\n$$\n\\frac{3^{a_{1}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+\\cdots+2^{a_{n}}\\right)^{2}} \\leqslant \\frac{3^{a_{1}}+\\cdots+3^{a_{n}}}{4^{a_{1}}+\\cdots+4^{a_{n}}}\n$$\nBy applying Jensen's inequality to the convex function $x^{\\log _{3} 4}$ we get\n$$\n\\frac{4^{a_{1}}+\\cdots+4^{a_{n}}}{n}=\\frac{\\left(3^{a_{1}}\\right)^{\\log _{3} 4}+\\cdots+\\left(3^{a_{n}}\\right)^{\\log _{3} 4}}{n} \\geqslant\\left(\\frac{3^{a_{1}}+\\cdots+3^{a_{n}}}{n}\\right)^{\\log _{3} 4}\n$$\nso\n$$\n\\frac{3^{a_{1}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+\\cdots+2^{a_{n}}\\right)^{2}} \\leqslant \\frac{3^{a_{1}}+\\cdots+3^{a_{n}}}{4^{a_{1}}+\\cdots+4^{a_{n}}} \\leqslant\\left(\\frac{n}{3^{a_{1}}+\\cdots+3^{a_{n}}}\\right)^{\\log _{3} 4-1}\n$$\nHence, the original inequality holds true whenever\n$$\n3^{a_{1}}+\\cdots+3^{a_{n}}>\\left(\\frac{1}{\\varepsilon}\\right)^{\\frac{1}{\\log _{3} 4-1}} \\cdot n\n$$\nSecond, trivially\n$$\n\\frac{3^{a_{1}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+\\cdots+2^{a_{n}}\\right)^{2}} \\leqslant \\frac{3^{a_{1}}+\\cdots+3^{a_{n}}}{n^{2}}\n$$\nso the original inequality is satisfied if\n$$\n3^{a_{1}}+\\cdots+3^{a_{n}}<\\varepsilon \\cdot n^{2}\n$$\nIf $n>\\left(\\frac{1}{\\varepsilon}\\right)^{1+\\frac{1}{\\log _{3} 4-1}}$ then $\\left(\\frac{1}{\\varepsilon}\\right)^{\\frac{1}{\\log _{3} 4-1}} \\cdot n<\\varepsilon \\cdot n^{2}$, and therefore at least one of the two conditions is satisfied.\n\n\nSolution 3. Define $C=\\log _{4 / 3} \\frac{2}{\\varepsilon}$, so that if $a_{i}>C$ then $3^{a_{i}}<\\frac{\\varepsilon}{2} \\cdot 4^{a_{i}}$. We divide the sequence into \"small\" and \"large\" terms by how they compare to $C$ : let\n$$\n\\mathcal{S}_{n}=\\left\\{i \\leqslant n \\mid a_{i} \\leqslant C\\right\\} \\quad \\text{ and } \\quad \\mathcal{L}_{n}=\\left\\{i \\leqslant n \\mid a_{i}>C\\right\\}\n$$\nThen the original inequality is equivalent to\n$$\n\\frac{\\sum_{i \\in \\mathcal{S}_{n}} 3^{a_{i}}}{\\left(\\sum_{i \\in \\mathcal{S}_{n}} 2^{a_{i}}+\\sum_{i \\in \\mathcal{L}_{n}} 2^{a_{i}}\\right)^{2}}+\\frac{\\sum_{i \\in \\mathcal{L}_{n}} 3^{a_{i}}}{\\left(\\sum_{i \\in \\mathcal{S}_{n}} 2^{a_{i}}+\\sum_{i \\in \\mathcal{L}_{n}} 2^{a_{i}}\\right)^{2}}<\\frac{\\varepsilon}{2}+\\frac{\\varepsilon}{2}\n$$\nIf $\\mathcal{L}_{n}$ is nonempty, we have\n$$\n\\frac{\\sum_{i \\in \\mathcal{L}_{n}} 3^{a_{i}}}{\\left(\\sum_{i \\in \\mathcal{S}_{n}} 2^{a_{i}}+\\sum_{i \\in \\mathcal{L}_{n}} 2^{a_{i}}\\right)^{2}}<\\frac{\\varepsilon}{2} \\cdot \\frac{\\sum_{i \\in \\mathcal{L}_{n}} 4^{a_{i}}}{\\left(\\sum_{i \\in \\mathcal{L}_{n}} 2^{a_{i}}\\right)^{2}} \\leqslant \\frac{\\varepsilon}{2}\n$$\nand this also trivially holds when $\\mathcal{L}_{n}$ is empty (in which case the LHS is zero).\nNow suppose that $n \\geqslant \\frac{2}{\\varepsilon}\\left(\\frac{3}{2}\\right)^{C}$. Note that $3^{a_{i}} \\leqslant\\left(\\frac{3}{2}\\right)^{C} 2^{a_{i}}$ for $i \\in \\mathcal{S}_{n}$, so we have\n$$\n\\frac{\\sum_{i \\in \\mathcal{S}_{n}} 3^{a_{i}}}{\\left(\\sum_{i \\in \\mathcal{S}_{n}} 2^{a_{i}}+\\sum_{i \\in \\mathcal{L}_{n}} 2^{a_{i}}\\right)^{2}} \\leqslant \\frac{\\left(\\frac{3}{2}\\right)^{C} \\sum_{i \\in \\mathcal{S}_{n}} 2^{a_{i}}}{\\left(\\sum_{i \\in \\mathcal{S}_{n}} 2^{a_{i}}+\\sum_{i \\in \\mathcal{L}_{n}} 2^{a_{i}}\\right)^{2}} \\leqslant \\frac{\\left(\\frac{3}{2}\\right)^{C}}{\\sum_{i \\in \\mathcal{S}_{n}} 2^{a_{i}}+\\sum_{i \\in \\mathcal{L}_{n}} 2^{a_{i}}}<\\frac{\\left(\\frac{3}{2}\\right)^{C}}{n} \\leqslant \\frac{\\varepsilon}{2},\n$$\nso we have the original inequality.\n\n\nSolution 4. For every index $i=1,2, \\ldots, n$, apply the weighted AM-GM inequality to numbers $2^{a_{i}}$ and $(n-1)$ with weights $\\log _{2} \\frac{3}{2} \\approx 0.585$ and $\\log _{2} \\frac{4}{3} \\approx 0.415$ as\n$$\n\\begin{gathered}\n2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{n}} \\geqslant 2^{a_{i}}+(n-1) \\\\\n>\\log _{2} \\frac{3}{2} \\cdot 2^{a_{i}}+\\log _{2} \\frac{4}{3} \\cdot(n-1) \\geqslant\\left(2^{a_{i}}\\right)^{\\log _{2} \\frac{3}{2}} \\cdot(n-1)^{\\log _{2} \\frac{4}{3}} \\\\\n=\\left(\\frac{3}{2}\\right)^{a_{i}} \\cdot(n-1)^{\\log _{2} \\frac{4}{3}}>\\left(\\frac{3}{2}\\right)^{a_{i}} \\cdot(n-1)^{2 / 5} .\n\\end{gathered}\n$$\nBy summing up for $i=1,2, \\ldots, n$,\n$$\n\\left(2^{a_{1}}+\\cdots+2^{a_{n}}\\right)^{2}=\\sum_{i=1}^{n} 2^{a_{i}}\\left(2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{n}}\\right)>(n-1)^{2 / 5} \\sum_{i=1}^{n} 3^{a_{i}}\n$$\nso\n$$\n\\frac{3^{a_{1}}+3^{a_{2}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{n}}\\right)^{2}}<\\frac{1}{(n-1)^{2 / 5}}\n$$\nIf $n \\geqslant\\left(\\frac{1}{\\varepsilon}\\right)^{5 / 2}+1$ then $\\frac{1}{(n-1)^{2 / 5}}<\\varepsilon$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32174,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}_{>0}$ be the set of all positive integers. Determine all subsets $\\mathcal{S}$ of $\\left\\{2^{0}, 2^{1}, 2^{2}, \\ldots\\right\\}$ for which there exists a function $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ such that\n$$\n\\mathcal{S}=\\left\\{f(a+b)-f(a)-f(b) \\mid a, b \\in \\mathbb{Z}_{>0}\\right\\} .\n$$",
    "options": [],
    "answer": "All and only the nonempty subsets of {2^0, 2^1, 2^2, …} with at most two elements. Equivalently, S = {2^k} or S = {2^k, 2^ℓ} for some integers k, ℓ ≥ 0 with k ≠ ℓ.",
    "solution": "$$\nf(a+b)=2^{e(a, b)}+f(a)+f(b) .\n$$\n\nSolution 1. Clearly $\\mathcal{S}$ must be nonempty. We start with constructions when $1 \\leqslant|\\mathcal{S}| \\leqslant 2$.\n- If $\\mathcal{S}=\\left\\{2^{k}\\right\\}$, then take $f(x)=c x-2^{k}$ for any integer $c>2^{k}$.\n- If $\\mathcal{S}=\\left\\{2^{k}, 2^{\\ell}\\right\\}$ where $k>\\ell$, then take $f(x)=\\left(2^{k}-2^{\\ell}\\right)\\lfloor\\alpha x\\rfloor-2^{\\ell}$, where $\\alpha>2$ is not an integer. This works because $\\lfloor\\alpha(x+y)\\rfloor-(\\lfloor\\alpha x\\rfloor+\\lfloor\\alpha y\\rfloor) \\in\\{0,1\\}$ for all $x$ and $y$, and takes both values; the lower bound on $\\alpha$ ensures the values of $f$ are positive.\n\nObserve that, inductively,\n$$\nf(n)=2^{e(1,1)}+2^{e(2,1)}+\\cdots+2^{e(n-1,1)}+n f(1) .\n$$\n\nLemma 1. For any positive integers $n$ and $k$,\n$$\n\\{e(1,1), e(2,1), \\ldots, e(k-1,1)\\} \\subset\\{e(n, 1), e(n+1,1), \\ldots, e(n+k-1,1)\\} .\n$$\n\nProof. We work by induction on $k$; in the case $k=1$, the first multiset is empty, which provides our base case.\nFor the induction step, suppose $k \\geqslant 2$ and we know that\n$$\n\\{e(1,1), e(2,1), \\ldots, e(k-2,1)\\} \\subset\\{e(n, 1), e(n+1,1), \\ldots, e(n+k-2,1)\\} .\n$$\nBy definition, $f(n+k)-f(n)-f(k)=2^{e(n, k)}$, and using the first observation we see that\n$f(n+k)-f(n)-f(k)=\\left(2^{e(n, 1)}+2^{e(n+1,1)}+\\cdots 2^{e(n+k-1,1)}\\right)-\\left(2^{e(1,1)}+2^{e(2,1)}+\\cdots+2^{e(k-1,1)}\\right)$.\nFrom the induction hypothesis, we may write\n$$\n\\{e(n, 1), e(n+1,1), \\ldots, e(n+k-2,1)\\}=\\{e(1,1), e(2,1), \\ldots, e(k-2,1)\\} \\cup\\{a\\}\n$$\nfor some $a$. Thus\n$$\n2^{e(n, k)}=2^{a}+2^{e(n+k-1,1)}-2^{e(k-1,1)} .\n$$\nSo $\\{e(n, k), e(k-1,1)\\}=\\{a, e(n+k-1,1)\\}$. Thus $e(k-1,1)=a$ or $e(k-1,1)=e(n+k-1,1)$, and in either case we have our result. $\\square$\n\nLemma 2. The sequence $e(1,1), e(2,1), e(3,1), \\ldots$ takes at most two different values.\n\nProof. Suppose for a contradiction that $k \\geqslant 2$ is the least index with $e(k, 1) \\neq e(1,1)$, and that some $\\ell>k$ has $e(\\ell, 1) \\notin\\{e(k, 1), e(1,1)\\}$. By Lemma 1, any block of $k$ consecutive values of the sequence has at least $k-1$ values equal to $e(1,1)$. This forces\n$$\ne(\\ell-1,1)=e(\\ell-2,1)=\\cdots=e(\\ell-(k-1), 1)=e(1,1)\n$$\nand\n$$\ne(\\ell+1,1)=e(\\ell+2,1)=\\cdots=e(\\ell+(k-1), 1)=e(1,1)\n$$\nBut then the block $e(\\ell-1,1), e(\\ell, 1), e(\\ell+1,1), e(\\ell+2,1), \\ldots, e(\\ell+(k-1), 1)$ has length $k+1$ and does not contain $e(k, 1)$, a contradiction.\n\nFinally, for any $a$ and $b$ we have\n$$\n\\begin{aligned}\nf(a+b)-f(a)-f(b) & =\\left(2^{e(a, 1)}+2^{e(a+1,1)}+\\cdots 2^{e(a+b-1,1)}\\right)-\\left(2^{e(1,1)}+2^{e(2,1)}+\\cdots+2^{e(b-1,1)}\\right) \\\\\n& =2^{e(i, 1)}\n\\end{aligned}\n$$\nfor some $a \\leqslant i \\leqslant a+b-1$. So $|\\mathcal{S}| \\leqslant 2$.\n\n\nSolution 2. Subsets of size 1 or 2 can be achieved as in Solution 1, and $\\mathcal{S}$ must be nonempty. We consider such a set $\\mathcal{S}$ with $|\\mathcal{S}| \\geqslant 3$ and a corresponding function $f$ in order to achieve a contradiction. We will relate the $e(a, b)$ to values of $e(c, 1)$ with $c+1<a+b$, leading to a proof of Lemma 2 from Solution 1 that does not depend on Lemma 1 from that solution.\n\nSuppose $a>1$. We have $f(a+b)-f(a)-f(b)=2^{e(a, b)}$ and also $f(a)-f(a-1)-f(1)= 2^{e(a-1,1)}$, so\n$$\nf(a+b)-f(a-1)-f(1)-f(b)=2^{e(a, b)}+2^{e(a-1,1)}\n$$\nSimilarly, $f(a+b)-f(a-1)-f(b+1)=2^{e(a-1, b+1)}$ and $f(b+1)-f(1)-f(b)=2^{e(b, 1)}$, so\n$$\nf(a+b)-f(a-1)-f(1)-f(b)=2^{e(a-1, b+1)}+2^{e(b, 1)}\n$$\nThus either\n$$\ne(a, b)=e(a-1, b+1) \\quad \\text{ and } \\quad e(a-1,1)=e(b, 1)\n$$\nor\n$$\ne(a, b)=e(b, 1) \\quad \\text{ and } \\quad e(a-1, b+1)=e(a-1,1) .\n$$\nFor $n \\geqslant 4$, we consider these possibilities as $(a, b)$ ranges over all pairs with $a+b=n$. If the first case holds for every such pair (that is, if $e(c, 1)=e(d, 1)$ for all $c+d=n-1$ ), then all the $e(a, b)$ for $a+b=n$ are equal (and the above equations do not constrain whether or not the value is the same as any $e(c, 1)$ with $c+1<n)$. Otherwise, the values of $e(a, b)$ with $a+b=n$ are fully determined by the values of $e(c, 1)$ for which $e(c, 1) \\neq e(n-1-c, 1)$, and are not all equal.\n\nSpecifically, if $e(c, 1)=j$ and $e(n-1-c, 1)=k$ with $j \\neq k$, we have $e(c, n-c)=j=e(n-c, c)$ and $e(c+1, n-c-1)=k=e(n-c-1, c+1)$. Every other value of $e(a, b)$ with $a+b=n$ is then determined by the rule that $e(a, b)=e(a-1, b+1)$ if $e(a-1,1)=e(b, 1)$ : if we have $e(c, 1) \\neq e(n-1-c, 1)$, and $e\\left(c', 1\\right) \\neq e\\left(n-1-c', 1\\right)$, but $e(d, 1)=e(n-1-d, 1)$ for all $c<d<c'$, then if $c<c'-1$ we have $e\\left(c'-1, n-\\left(c'-1\\right)\\right)=e\\left(c', n-c'\\right)$, then if $c<c'-2$ we have $e\\left(c'-2, n-\\left(c'-2\\right)\\right)=e\\left(c'-1, n-\\left(c'-1\\right)\\right)=e\\left(c', n-c'\\right)$, and so on until $e(c+1, n-c-1)=e\\left(c', n-c'\\right)$ (yielding a contradiction if $e(n-c-1,1) \\neq e\\left(c', 1\\right)$; such a contradiction also arises trivially if $c+1=c'$ and $e(n-c-1,1) \\neq e\\left(c', 1\\right)$ ). If $c$ is the least integer such that $e(c, 1) \\neq e(n-1-c, 1)$, the values of $e(a, b)$ with $a<c$ are similarly determined to be equal to $e(c, n-c)$ (and likewise for $a>n-c$ ).\n\nIf $|\\mathcal{S}| \\geqslant 3$, that means that $e(c, 1)$ takes at least three different values. Let $m$ be such that $e(m, 1)$ does not equal any $e(c, 1)$ for $c<m$, and there are exactly two different values of $e(c, 1)$ for $c<m$ (and thus $m \\geqslant 3$ ).\nBecause $e(m, 1)$ does not equal any $e(c, 1)$ for $c<m$, we have that all $e(a, b)$ for $a+b=m+1$ are equal, and $e(c, 1)=e(d, 1)$ for all $c+d=m$. We now consider the values of $e(a, b)$ for $a+b=m+2$ determined by the above rules. Since $e(m, 1) \\neq e(1,1)$, we have $e(1, m+1)=e(1,1)$ and $e(2, m)=e(m, 1)$. If there were any other $e(d, 1) \\neq e(m+1-d, 1)$, consider the one with minimal $d>1$; because $e(m, 1) \\neq e(d, 1)$, we arrive at a contradiction. So every $e(c, 1)=e(d, 1)$ for $c+d=m+1$ except for $e(m, 1) \\neq e(1,1)$. But these equalities form a path connecting all $e(c, 1)$ for $c<m$ :\n$$\ne(1,1)=e(m-1,1)=e(2,1)=e(m-2,1)=e(3,1)=\\cdots\n$$\nwhich contradicts the assumption we made that there were exactly two different values of $e(c, 1)$ for $c<m$.\n\nSolution 3. Constructions for $1 \\leqslant|\\mathcal{S}| \\leqslant 2$ are shown in Solution 1, and $\\mathcal{S}$ must be nonempty. We suppose $|\\mathcal{S}| \\geqslant 3$ to derive a contradiction.\n\nClaim 1. $e(a, b), e(b, c)$, and $e(a, c)$ can take at most two different values.\n\nProof. By expanding $f(a+b+c)$ in three different ways, we get\n$$\n2^{e(a, b)}+2^{e(c, a+b)}=2^{e(b, c)}+2^{e(a, b+c)}=2^{e(a, c)}+2^{e(b, a+c)}\n$$\nThe result follows from the equality of the three multisets of exponents.\n\nFor Claims 2 to 4, we fix $k$ and let $N$ be the smallest integer such that $e(a, N-a+1)=k$ for some $a \\leqslant N$.\n\nClaim 2. For any $b$ with $b \\leqslant N$, we must have $e(b, N-b+1)=k$.\n\nProof. Suppose that $e(a, N-a+1)=k$ and $a<b$. Expanding $f(a+(b-a)+N-b+1)$ in two different ways, we see that\n$$\n2^{e(a, b-a)}+2^{e(b, N-b+1)}=2^{e(N-b+1, b-a)}+2^{e(N-a+1, a)}\n$$\nBy the minimality of $N$, we must have $e(b, N-b+1)=e(N-a+1, a)$. The case of $a>b$ follows by replacing $a$ and $b$ with $N-a+1$ and $N-b+1$.\n\nClaim 3. $e(a, 1)=e(N-a+1,1)$ for any a satisfying $1<a<N$.\n\nProof. By Claim 2, $e(a, N-a+1)=k$. Then by Claim 1, $e(a, N-a+1), e(a, 1)$, and $e(N-a+1,1)$ can take at most two different values. But by the minimality of $N$, we must have $e(a, N-a+1) \\neq e(a, 1)=e(N-a+1,1)$.\n\nClaim 4. $e(a, 1)=e(N-a, 1)$ for any a satisfying $1 \\leqslant a<N$.\n\nProof. By Claim 2, $e(a, N-a+1)=e(a+1, N-a)=k$. Expanding $f(1+a+(N-a))$ in two different ways, we see that\n$$\n2^{e(1, a)}+2^{e(a+1, N-a)}=2^{e(1, N-a)}+2^{e(N-a+1, a)}\n$$\nTherefore $e(1, a)=e(1, N-a)$, as required.\n\nIf $|\\mathcal{S}| \\geqslant 3$, then there exist $1<N_{k}<N_{\\ell}$ where $N_{k}$ and $N_{\\ell}$ are the minimal values corresponding to $k$ and $\\ell$. But Claims 3 and 4 imply that $e(a, 1)$ is constant for all $1 \\leqslant a<N_{\\ell}$, which is a contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32175,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_{0}, a_{1}, a_{2}, \\ldots$ be an infinite strictly increasing sequence of positive integers such that for each $n \\geqslant 1$ we have\n$$\na_{n} \\in\\left\\{\\frac{a_{n-1}+a_{n+1}}{2}, \\sqrt{a_{n-1} \\cdot a_{n+1}}\\right\\}\n$$\nLet $b_{1}, b_{2}, \\ldots$ be an infinite sequence of letters defined as\n$$\nb_{n}= \\begin{cases}A, & \\text{ if } a_{n}=\\frac{1}{2}\\left(a_{n-1}+a_{n+1}\\right) \\\\ G, & \\text{ otherwise }\\end{cases}\n$$\nProve that there exist positive integers $n_{0}$ and $d$ such that for all $n \\geqslant n_{0}$ we have $b_{n+d}=b_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. We will show that the eventual period of sequence $\\left(b_{n}\\right)$ consists of any fixed number of occurrences of $G$ (possibly zero) followed by a single $A$.\nWe look at the ratios of consecutive terms of the sequence $\\left(a_{n}\\right)$. Let $C$ and $D$ be coprime positive integers such that $a_{1} / a_{0}=(C+D) / C$. If $b_{n}=G$ then $a_{n} / a_{n-1}=a_{n+1} / a_{n}$. If $b_{n}=A$ and $a_{n} / a_{n-1}=(C+k D) /(C+(k-1) D)$ for some positive integer $k$ then\n$$\n\\frac{a_{n+1}}{a_{n}}=\\frac{2 a_{n}-a_{n-1}}{a_{n}}=\\frac{C+(k+1) D}{C+k D}\n$$\nThus, by induction, there is a sequence of positive integers $\\left(k_{n}\\right)$ for $n \\geqslant 1$ which satisfies $a_{n} / a_{n-1}=\\left(C+k_{n} D\\right) /\\left(C+\\left(k_{n}-1\\right) D\\right)$ for all positive integers $n$. Moreover, we have $k_{1}=1$ and\n$$\nk_{n+1}= \\begin{cases}k_{n}, & \\text{ if } b_{n}=G \\\\ k_{n}+1, & \\text{ if } b_{n}=A\\end{cases}\n$$\nIf there are only finitely many values of $n$ such that $b_{n}=A$ then the problem statement obviously holds (we can choose $d=1$ ). Thus, we may assume that $b_{n}=A$ for infinitely many $n$. This means that the sequence ( $k_{n}$ ) attains all positive integer values. Given a value $q \\geqslant 1$, denote by $m_{q}$ the last index where value $q$ occurs, that is, the index such that $k_{m_{q}}=q$ and $k_{m_{q}+1}=q+1$.\nOur aim is to prove that the sequence of differences $\\left(m_{q+1}-m_{q}\\right)$ is eventually constant. We first show that it is bounded above. To that end, fix $t \\geqslant 1$ (we will choose a suitably large $t$ later on) and consider a sequence $s(t)_{0}, s(t)_{1}, \\ldots$ defined for $q \\geqslant 1$ by $s(t)_{q}=a_{m_{q}} /(C+q D)^{t}$.\nWe note two properties of $s(t)_{q}$. First, simple algebra gives\n$$\n\\begin{gathered}\ns(t)_{q+1}=\\frac{a_{m_{q+1}}}{(C+(q+1) D)^{t}}=\\frac{a_{m_{q}}}{(C+(q+1) D)^{t}}\\left(\\frac{C+(q+1) D}{C+q D}\\right)^{m_{q+1}-m_{q}} \\\\\n=\\frac{a_{m_{q}}}{(C+q D)^{t}}\\left(\\frac{C+(q+1) D}{C+q D}\\right)^{m_{q+1}-m_{q}-t}=s(t)_{q}\\left(\\frac{C+(q+1) D}{C+q D}\\right)^{m_{q+1}-m_{q}-t}\n\\end{gathered}\n$$\nIt follows that\n$$\n\\left.\\begin{array}{l}\ns(t)_{q}>s(t)_{q+1} \\\\\ns(t)_{q}=s(t)_{q+1} \\\\\ns(t)_{q}<s(t)_{q+1}\n\\end{array}\\right\\} \\quad \\text{ if and only if } \\quad\\left\\{\\begin{array}{l}\nm_{q+1}-m_{q}<t \\\\\nm_{q+1}-m_{q}=t \\\\\nm_{q+1}-m_{q}>t\n\\end{array}\\right.\n$$\nSecond, suppose that $m_{q+1}-m_{q} \\geqslant t$ for some positive integer $q$. We claim that in that case $s(t)_{q}$ is a positive integer. Indeed, we have\n$$\na_{m_{q}+t}=a_{m_{q}}\\left(\\frac{C+(q+1) D}{C+q D}\\right)^{t},\n$$\nbecause $k_{m_{q}+1}=k_{m_{q}+2}=\\cdots=k_{m_{q}+t}=q+1$. Since $C+(q+1) D$ and $C+q D$ are coprime we have that\n$$\ns(t)_{q}=\\frac{a_{m_{q}}}{(C+q D)^{t}}\n$$\nis an integer.\nWe choose $T \\geqslant 1$ such that $s(T)_{1}<1$ (which exists since $C+D>1$ ). Then, by induction we can show that $s(T)_{q}<1$ for all $q$. Indeed, since $s(T)_{q}<1$, it is not a positive integer; this means that $m_{q+1}-m_{q}<T$ by the second property above. Hence by the first property above we have $s(T)_{q+1}<s(T)_{q}<1$, as needed.\nThis means that $m_{q+1}-m_{q}<T$ for all $q$. Thus there is a largest integer $T^{\\prime} \\leqslant T$ with the property that an equality $m_{q+1}-m_{q}=T^{\\prime}$ holds for infinitely many values of $q$.\nTherefore, for all sufficiently large values of $q$ we have the inequality $m_{q+1}-m_{q} \\leqslant T^{\\prime}$, which by the first property implies that the sequence $s\\left(T^{\\prime}\\right)$ is decreasing from some point on. Moreover, we know that the sequence attains infinitely many integer values since there are infinitely many values of $q$ for which we have the equality $m_{q+1}-m_{q}=T^{\\prime}$. As a consequence, the sequence $s\\left(T^{\\prime}\\right)$ is constant from some sufficiently large index $Q$ onwards.\nThis in turn means that the equality $m_{q+1}-m_{q}=T^{\\prime}$ holds for all $q \\geqslant Q$. Note that $b_{n}=A$ is equivalent to the fact that $n=m_{q}$ for some integer $q$. Thus, the sequence ( $b_{n}$ ) is periodic for $n \\geqslant Q$ with period $T^{\\prime}$, and the proof is complete.\n\n\nSolution 2. First, observe that the statement holds immediately if $b_{n}=G$ for all $n$; otherwise, there must be some $n$ for which $b_{n}=A$. Without loss of generality, we may assume that $n=1$, as we can translate the sequence without affecting the statement.\nWe define an arithmetic sequence $\\left(p_{n}\\right)$ by taking $p_{0}=a_{0} / \\operatorname{gcd}\\left(a_{0}, a_{1}\\right)$ and $p_{1}=a_{1} / \\operatorname{gcd}\\left(a_{0}, a_{1}\\right)$. Note that $p_{0}<p_{1}$, and hence that $\\left(p_{n}\\right)$ is an increasing sequence of positive integers, and also that $p_{2}=a_{2} / \\operatorname{gcd}\\left(a_{0}, a_{1}\\right)$.\nWe also define a sequence of positive integers $d_{n}=a_{n}-a_{n-1}$ and a sequence of positive rational numbers $q_{n}=a_{n} / a_{n-1}$.\nThen the following facts are immediate consequences of the definitions:\n- if $b_{n}=G$, then $q_{n+1}=q_{n}$ and $d_{n+1}=d_{n} q_{n}$;\n- if $b_{n}=A$, then $d_{n+1}=d_{n}$;\n- $q_{1}=p_{1} / p_{0}$;\n- if $b_{n}=A$ and $q_{n}=p_{i} / p_{i-1}$, then $q_{n+1}=p_{i+1} / p_{i}$.\nNow, let $k_{i}$ be the number of integers $n$ for which $b_{n}=G$ and $q_{n}=p_{i} / p_{i-1}$. If some $k_{i}$ is infinite then $b_{n}$ is eventually always $G$; otherwise, all values of $k_{i}$ are nonnegative integers.\nThe sequence of values for $d_{n}$ can be written as\n$$\nd_{0}, d_{0} \\frac{p_{1}}{p_{0}}, \\ldots, d_{0}\\left(\\frac{p_{1}}{p_{0}}\\right)^{k_{1}}, d_{0}\\left(\\frac{p_{1}}{p_{0}}\\right)^{k_{1}} \\frac{p_{2}}{p_{1}}, \\ldots, d_{0}\\left(\\frac{p_{1}}{p_{0}}\\right)^{k_{1}}\\left(\\frac{p_{2}}{p_{1}}\\right)^{k_{2}}, \\ldots\n$$\nand in particular all terms in this sequence are positive integers. Furthermore, $p_{i}$ and $p_{i+1}$ are coprime for all $i$, so the following sequence consists entirely of positive integers:\n$$\n\\begin{aligned}\nu_{0} & =d_{0} p_{0}^{-k_{1}} \\\\\nu_{1} & =d_{0} p_{0}^{-k_{1}} p_{1}^{k_{1}-k_{2}} \\\\\nu_{2} & =d_{0} p_{0}^{-k_{1}} p_{1}^{k_{1}-k_{2}} p_{2}^{k_{2}-k_{3}} \\\\\n& \\vdots\n\\end{aligned}\n$$\nWe will prove that $k_{i}$ is eventually constant, which implies that the sequence of $b_{n}$ is eventually periodic with period consisting of $k$ copies of $G$ followed by an $A$ (where $k$ is that constant value).\nObserve that either $k_{i}$ is unbounded, or is bounded with eventual maximum $k$ for some constant $k$. In the second case, let $r_{0}$ be minimal such that $k_{r_{0}}=k$; in the first case let $r_{0}=0$. We will construct an infinite sequence of integers as follows:\n- If $k_{r_{i}+1} \\geqslant k_{r_{i}}$, then $r_{i+1}=r_{i}+1$\n- If $k_{r_{i}+1}<k_{r_{i}}$, then $r_{i+1}$ is the minimal positive integer greater than $r_{i}$ such that $k_{r_{i+1}} \\geqslant k_{r_{i}}$. Observe that in the second case, such an $r_{i+1}$ must exist by our construction of $r_{0}$.\nWe claim that $u_{r_{i+1}} \\leqslant u_{r_{i}}$ with equality only if $k_{r_{i}+1}=k_{r_{i}}$ (so $r_{i+1}=r_{i}+1$ ). Indeed, if $k_{r_{i}+1} \\geqslant k_{r_{i}}$ then\n$$\nu_{r_{i+1}}=u_{r_{i}+1}=u_{r_{i}} p_{r_{i}}^{k_{r_{i}}-k_{r_{i}+1}} \\leqslant u_{r_{i}}\n$$\nwith equality if and only if $k_{r_{i}}=k_{r_{i}+1}$.\nOtherwise, we have\n$$\n\\frac{u_{r_{i+1}}}{u_{r_{i}}}=p_{r_{i}}^{k_{r_{i}}-k_{r_{i}+1}} p_{r_{i}+1}^{k_{r_{i}+1}-k_{r_{i}+2}} \\cdots p_{r_{i+1}-1}^{k_{r_{i+1}-1}-k_{r_{i+1}}}\n$$\nso we just need to show that the right hand side is strictly less than 1 . But this follows because\n$$\n\\begin{aligned}\np_{r_{i}}^{k_{r_{i}}-k_{r_{i}+1}} p_{r_{i}+1}^{k_{r_{i}+1}-k_{r_{i}+2}} \\cdots p_{r_{i+1}-1}^{k_{r_{i+1}-1}-k_{r_{i+1}}} & <p_{r_{i}+1}^{k_{r_{i}}-k_{r_{i}+2}} p_{r_{i}+2}^{k_{r_{i}+2}-k_{r_{i}+3}} \\cdots p_{r_{i+1}-1}^{k_{r_{i+1}-1}-k_{r_{i+1}}} \\\\\n& <p_{r_{i}+2}^{k_{r_{i}}-k_{r_{i}+3}} p_{r_{i}+3}^{k_{r_{i}+3}-k_{r_{i}+4}} \\cdots p_{r_{i+1}-1}^{k_{r_{i+1}-1}-k_{r_{i+1}}} \\\\\n& \\vdots \\\\\n& <p_{r_{i+1}-1}^{k_{r_{i}-k_{r_{i+1}}}} \\\\\n& \\leqslant 1\n\\end{aligned}\n$$\nwhere each inequality besides the last follows from the fact that $p_{j}<p_{j+1}$ and $k_{r_{i}}>k_{j}$ for $j<r_{i+1}$, and the last follows from the fact that $k_{r_{i}} \\leqslant k_{r_{i+1}}$.\nFinally, the sequence $u_{r_{i}}$ is an infinite nonincreasing sequence of positive integers so must eventually be constant, yielding the claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32176,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Q}$ be the set of rational numbers. Let $f: \\mathbb{Q} \\rightarrow \\mathbb{Q}$ be a function such that the following property holds: for all $x, y \\in \\mathbb{Q}$,\n$$\nf(x+f(y))=f(x)+y \\quad \\text{or} \\quad f(f(x)+y)=x+f(y).\n$$\n\nDetermine the maximum possible number of elements of $\\{f(x)+f(-x) \\mid x \\in \\mathbb{Q}\\}$.",
    "options": [],
    "answer": "2",
    "solution": "Solution 1. We begin by providing an example of a function $f$ for which there are two values of $g(x)$. We take the function $f(x)=\\lfloor x\\rfloor-\\{x\\}$, where $\\lfloor x\\rfloor$ denotes the floor of $x$ (that is, the largest integer less than or equal to $x$) and $\\{x\\}=x-\\lfloor x\\rfloor$ denotes the fractional part of $x$.\nFirst, we show that $f$ satisfies $P(x, y)$. Given $x, y \\in \\mathbb{Q}$, we have\n$$\n\\begin{aligned}\n& f(x)+y=\\lfloor x\\rfloor-\\{x\\}+\\lfloor y\\rfloor+\\{y\\}=(\\lfloor x\\rfloor+\\lfloor y\\rfloor)+(\\{y\\}-\\{x\\}) \\\\\n& x+f(y)=\\lfloor x\\rfloor+\\{x\\}+\\lfloor y\\rfloor-\\{y\\}=(\\lfloor x\\rfloor+\\lfloor y\\rfloor)+(\\{x\\}-\\{y\\})\n\\end{aligned}\n$$\nIf $\\{x\\}<\\{y\\}$, then we have that the fractional part of $f(x)+y$ is $\\{y\\}-\\{x\\}$ and the floor is $\\lfloor x\\rfloor+\\lfloor y\\rfloor$, so $f(x)+y \\rightarrow x+f(y)$. Likewise, if $\\{x\\}>\\{y\\}$, then $x+f(y) \\rightarrow f(x)+y$. Finally, if $\\{x\\}=\\{y\\}$, then $f(x)+y=x+f(y)=\\lfloor x\\rfloor+\\lfloor y\\rfloor$ is an integer. In all cases, the relation $P$ is satisfied.\nFinally, we observe that if $x$ is an integer then $g(x)=0$, and if $x$ is not an integer then $g(x)=-2$, so there are two values for $g(x)$ as required.\nNow, we prove that there cannot be more than two values of $g(x)$. $P(x, x)$ tells us that $x+f(x) \\sim x+f(x)$, or in other words, for all $x$,\n$$\n\\begin{equation*}\nf(x+f(x))=x+f(x) \\tag{1}\n\\end{equation*}\n$$\nWe begin with the following lemma.\nLemma 1. $f$ is a bijection, and satisfies\n$$\n\\begin{equation*}\nf(-f(-x))=x \\tag{2}\n\\end{equation*}\n$$\nProof. We first prove that $f$ is injective. Suppose that $f\\left(x_{1}\\right)=f\\left(x_{2}\\right)$; then $P\\left(x_{1}, x_{2}\\right)$ tells us that $f\\left(x_{1}\\right)+x_{2} \\sim f\\left(x_{2}\\right)+x_{1}$. Without loss of generality, suppose that $f\\left(x_{1}\\right)+x_{2} \\rightarrow f\\left(x_{2}\\right)+x_{1}$.\nBut $f\\left(x_{1}\\right)=f\\left(x_{2}\\right)$, so $f\\left(f\\left(x_{1}\\right)+x_{2}\\right)=f\\left(f\\left(x_{2}\\right)+x_{2}\\right)=f\\left(x_{2}\\right)+x_{2}$ by (1). Therefore, $f\\left(x_{2}\\right)+x_{1}=f\\left(x_{2}\\right)+x_{2}$, as required.\nNow, (1) with $x=0$ tells us that $f(f(0))=f(0)$ and so by injectivity $f(0)=0$.\nApplying $P(x,-f(x))$ tells us that $0 \\sim x+f(-f(x))$, so either $0=f(0)=x+f(-f(x))$ or $f(x+f(-f(x)))=0$ which implies that $x+f(-f(x))=0$ by injectivity. Either way, we deduce that $x=-f(-f(x))$, or $x=f(-f(-x))$ by replacing $x$ with $-x$.\nFinally, note that bijectivity follows immediately from (2).\nSince $f$ is bijective, it has an inverse, which we denote $f^{-1}$. Rearranging (2) (after replacing $x$ with $-x)$ gives that $f(-x)=-f^{-1}(x)$. We have $g(x)=f(x)+f(-x)=f(x)-f^{-1}(x)$.\nSuppose $g(x)=u$ and $g(y)=v$, where $u \\neq v$ are both nonzero. Define $x'=f^{-1}(x)$ and $y'=f^{-1}(y)$; by definition, we have\n$$\n\\begin{aligned}\n& x' \\rightarrow x \\rightarrow x'+u \\\\\n& y' \\rightarrow y \\rightarrow y'+v .\n\\end{aligned}\n$$\nPutting in $P\\left(x', y\\right)$ gives $x+y \\sim x'+y'+v$, and putting in $P\\left(x, y'\\right)$ gives $x+y \\sim x'+y'+u$. These are not equal since $u \\neq v$, and $x+y$ may have only one incoming and outgoing arrow because $f$ is a bijection, so we must have either $x'+y'+u \\rightarrow x+y \\rightarrow x'+y'+v$ or the same with the arrows reversed. Swapping $(x, u)$ and $(y, v)$ if necessary, we may assume without loss of generality that this is the correct direction for the arrows.\nAlso, we have $-x'-u \\rightarrow-x \\rightarrow-x'$ by Lemma 1. Putting in $P\\left(x+y,-x'-u\\right)$ gives $y \\sim y'+v-u$, and so $y'+v-u$ must be either $y'+v$ or $y'$. This means $u$ must be either 0 or $v$, and this contradicts our assumption about $u$ and $v$.\n\n\nSolution 2. We again start with Lemma 1, and note $f(0)=0$ as in the proof of that lemma.\n$P(x,-f(y))$ gives $x+f(-f(y)) \\sim f(x)-f(y)$, and using (2) this becomes $x-y \\sim f(x)-f(y)$. In other words, either $f(x-y)=f(x)-f(y)$ or $x-y=f(f(x)-f(y))$. In the latter case, we deduce that\n$$\n\\begin{aligned}\nf(-(x-y)) & =f(-f(f(x)-f(y))) \\\\\nf(y-x) & =f(-f(f(x)-f(y))) \\\\\n& =f(y)-f(x)\n\\end{aligned}\n$$\nThus, $f(y)-f(x)$ is equal to either $f(y-x)$ or $-f(x-y)$. Replacing $y$ with $x+d$, we deduce that $f(x+d)-f(x) \\in\\{f(d),-f(-d)\\}$.\nNow, we prove the following claim.\nClaim. For any $n \\in \\mathbb{Z}_{>0}$ and $d \\in \\mathbb{Q}$, we have that either $g(d)=0$ or $g(d)= \\pm g(d / n)$.\nIn particular, if $g(d / n)=0$ then $g(d)=0$.\nProof. We first prove that if $g(d / n)=0$ then $g(d)=0$. Suppose that $g(d / n)=0$. Then $f(d / n)=-f(-d / n)$ and so $f(x+d / n)-f(x)=f(d / n)$ for any $x$. Applying this repeatedly, we deduce that $f(x+d)-f(x)=n f(d / n)$ for any $x$. Applying this with $x=0$ and $x=-d$ and adding gives $f(d)+f(-d)=0$, so $g(d)=0$, and in particular the claim is true whenever $g(d)=0$.\nNow, select $n \\in \\mathbb{Z}_{>0}$ and $d \\in \\mathbb{Q}$ such that $g(d) \\neq 0$, and observe that we must have $g(d / n) \\neq 0$. Observe that for any $k \\in \\mathbb{Z}$ we have that $f(k d / n)-f((k-1) d / n) \\in\\{f(d / n),-f(-d / n)\\}$. Let $A_{i}$ be the number of $k \\in \\mathbb{Z}$ with $i-n<k \\leqslant i$ such that this difference equals $f(d / n)$.\nWe deduce that for any $i \\in \\mathbb{Z}$,\n$$\n\\begin{aligned}\nf(i d / n)-f(i d / n-d) & =\\sum_{i-n<k \\leqslant i} f(k d / n)-f((k-1) d / n) \\\\\n& =A_{i} f(d / n)-\\left(n-A_{i}\\right) f(-d / n) \\\\\n& =-n f(-d / n)+A_{i} g(d / n)\n\\end{aligned}\n$$\nSince $g(d / n)$ is nonzero, this is a nonconstant linear function of $A_{i}$. However, there are only two possible values for $f(i d / n)-f(i d / n-d)$, so there must be at most two possible values for $A_{i}$ as $i$ varies. And since $A_{i+1}-A_{i} \\in\\{-1,0,1\\}$, those two values must differ by 1 (if there are two values).\nNow, we have\n$$\n\\begin{aligned}\nf(d)-f(0) & =-n f(-d / n)+A_{n} g(d / n), \\quad \\text{and} \\\\\nf(0)-f(-d) & =-n f(-d / n)+A_{0} g(d / n)\n\\end{aligned}\n$$\nSubtracting these (using the fact that $f(0)=0$ ) we obtain\n$$\n\\begin{aligned}\nf(d)+f(-d) & =\\left(A_{n}-A_{0}\\right) g(d / n) \\\\\n& = \\pm g(d / n)\n\\end{aligned}\n$$\nwhere the last line follows from the fact that $g(d)$ is nonzero.\nIt immediately follows that there can only be one nonzero number of the form $g(x)$ up to sign; to see why, if $g(d)$ and $g\\left(d'\\right)$ are both nonzero, then for some $n, n' \\in \\mathbb{Z}_{>0}$ we have $d / n=d'/ n'$. But\n$$\ng(d)= \\pm g(d / n)= \\pm g\\left(d'\\right)\n$$\nFinally, suppose that for some $d, d'$ we have $g(d)=c$ and $g\\left(d'\\right)=-c$ for some nonzero $c$. So we have\n$$\nf(d)+f(-d)-f\\left(d'\\right)-f\\left(-d'\\right)=2 c\n$$\nwhich rearranges to become $\\left(f(d)-f\\left(d'\\right)\\right)-\\left(f\\left(-d'\\right)-f(-d)\\right)=2 c$.\nEach of the bracketed terms must be equal to either $f\\left(d-d'\\right)$ or $-f\\left(d'-d\\right)$. However, they cannot be equal since $c$ is nonzero, so $g\\left(d-d'\\right)=f\\left(d-d'\\right)+f\\left(d'-d\\right)= \\pm 2 c$. This contradicts the assertion that $g(-x)= \\pm c$ for all $x$.\n\n\nSolution 3. As in Solution 1, we start by establishing Lemma 1 as above, and write $f^{-1}(x)= -f(-x)$ for the inverse of $f$, and $g(x)=f(x)-f^{-1}(x)$.\nWe now prove the following.\nLemma 2. If $g(x) \\neq g(y)$, then $g(x+y)= \\pm(g(x)-g(y))$.\nProof. Assume $x$ and $y$ are such that $g(x) \\neq g(y)$. Applying $P\\left(x, f^{-1}(y)\\right)$ gives $x+y \\sim f(x)+f^{-1}(y)$, and applying $P\\left(f^{-1}(x), y\\right)$ gives $x+y \\sim f^{-1}(x)+f(y)$.\nObserve that\n$$\n\\begin{aligned}\n\\left(f(x)+f^{-1}(y)\\right)-\\left(f^{-1}(x)+f(y)\\right) & =\\left(f(x)-f^{-1}(x)\\right)-\\left(f(y)-f^{-1}(y)\\right) \\\\\n& =g(x)-g(y)\n\\end{aligned}\n$$\nBy assumption, $g(x) \\neq g(y)$, and so $f(x)+f^{-1}(y) \\neq f^{-1}(x)+f(y)$. Since $f$ is bijective, this means that these two values must be $f(x+y)$ and $f^{-1}(x+y)$ in some order, and so $g(x+y)=f(x+y)-f^{-1}(x+y)$ must be their difference up to sign, which is either $g(x)-g(y)$ or $g(y)-g(x)$.\nClaim. If $x$ and $q$ are rational numbers such that $g(q)=0$ and $n$ is an integer, then $g(x+n q)= g(x)$.\nProof. If $g(b)=0$ and $g(a) \\neq g(a+b)$, then the lemma tells us that $g(b)= \\pm(g(a+b)-g(a))$, which contradicts our assumptions. Therefore, $g(a)=g(a+b)$ whenever $g(b)=0$.\nA simple induction then gives that $g(n b)=0$ for any positive integer $n$, and $g(n b)=0$ for negative $n$ as $g(x)=g(-x)$. The claim follows immediately.\nLemma 3. There cannot be both positive and negative elements in the range of $g$.\nProof. Suppose that $g(x)>0$ and $g(y)<0$. Let $\\mathcal{S}$ be the set of numbers of the form $m x+n y$ for integers $m, n$. We first show that $g(\\mathcal{S})$ has infinitely many elements. Indeed, suppose $g(\\mathcal{S})$ is finite, and let $a \\in \\mathcal{S}$ maximise $g$ and $b \\in \\mathcal{S}$ maximise $-g$. Then $a+b \\in \\mathcal{S}$, and $g(a+b)=g(a)-g(b)$ or $g(b)-g(a)$. In the first case $g(a+b)>g(a)$ and in the second case $g(a+b)<g(b)$; in either case we get a contradiction.\nNow, we show that there must exist some nonzero rational number $q$ with $g(q)=0$. Indeed, suppose first that $a+f(a)=0$ for all $a$. Then $g(a)=f(a)+f(-a)=0$ for all $a$, and so $g$ takes no nonzero value. Otherwise, there is some $a$ with $a+f(a) \\neq 0$, and so (1) yields that $f(q)=0$ for $q=a+f(a) \\neq 0$. Noting that $f(-q)=0$ from Lemma 1 tells us that $g(q)=0$, as required.\nNow, there must exist integers $s$ and $s'$ such that $x s=q s'$ and integers $t$ and $t'$ such that $y t=q t'$. The claim above gives that the value of $g(m x+n y)$ depends only on the values of $m \\bmod s$ and $n \\bmod t$, so $g(m x+n y)$ can only take finitely many values.\nFinally, suppose that $g(x)=u$ and $g(y)=v$ where $u \\neq v$ have the same sign. Assume $u, v>0$ (the other case is similar) and assume $u>v$ without loss of generality.\n$P\\left(f^{-1}(x), f^{-1}(y)\\right)$ gives $x-y \\sim f^{-1}(x)-f^{-1}(y)=f(x)-f(y)-(u-v)$, and $P(x, y)$ gives $x-y \\sim f(x)-f(y)$. $u-v$ is nonzero, so $f(x-y)$ and $f^{-1}(x-y)$ must be $f(x)-f(y)-(u-v)$ and $f(x)-f(y)$ in some order, and since $g(x-y)$ must be nonnegative, we have\n$$\nf(x)-f(y)-(u-v) \\rightarrow x-y \\rightarrow f(x)-f(y) .\n$$\nThen, $P\\left(x-y, f^{-1}(y)\\right)$ tells us that $(x-y)+y \\sim(f(x)-f(y))+(f(y)-v)$, so $x \\sim f(x)-v$, contradicting either $v \\neq u$ or $v>0$.\n\n\nAnswer: 2 is the maximum number of elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32177,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p \\neq q$ be coprime positive integers. Determine all infinite sequences $a_{1}, a_{2}, \\ldots$ of positive integers such that the following conditions hold for all $n \\geqslant 1$:\n$$\n\\begin{aligned}\n& \\max \\left(a_{n}, a_{n+1}, \\ldots, a_{n+p}\\right)-\\min \\left(a_{n}, a_{n+1}, \\ldots, a_{n+p}\\right)=p \\quad \\text{ and } \\\\\n& \\max \\left(a_{n}, a_{n+1}, \\ldots, a_{n+q}\\right)-\\min \\left(a_{n}, a_{n+1}, \\ldots, a_{n+q}\\right)=q .\n\\end{aligned}\n$$",
    "options": [],
    "answer": "a_n = n + C for some nonnegative integer C",
    "solution": "Answer: The only such sequences are $a_{n}=n+C$, where $C$ is a nonnegative integer.\n\n## Common remarks.\n- Denote by $a_{[i, j]}$ the subsequence $a_{i}, a_{i+1}, \\ldots, a_{j}$.\n- Without loss of generality, in each solution we suppose $p<q$. It can be convenient to treat the case where $p=1$ separately.\n- The problem can also be posed for sequences of arbitrary integers (rather than positive). Refer to the comment after Solution 1 for a proof.\n\nSolution 1. Let $k=\\left\\lceil\\frac{q}{p}\\right\\rceil$. Note that $k \\geqslant 2$.\n\nLemma 1. If $i, j$ and $m$ are positive integers such that $|i-j| \\leqslant m p$ then $\\left|a_{i}-a_{j}\\right| \\leqslant m p$.\n\nProof. By the given condition, if $|i-j| \\leqslant p$ then $\\left|a_{i}-a_{j}\\right| \\leqslant p$. So the lemma follows from induction on $m$ and the triangle inequality.\n\nLemma 2. For a fixed $n$, suppose that $a_{i}$ is minimal over $i \\geqslant n$. Then $i \\leqslant n+p-1$.\n\nProof. Suppose for contradiction that $i \\geqslant n+p$. Then $\\min \\left(a_{[i-p, i+q-p]}\\right)=a_{i}$. Since $q-p \\leqslant (k-1) p$, it follows from Lemma 1 that $\\max \\left(a_{[i-p, i+q-p]}\\right) \\leqslant a_{i}+(k-1) p<a_{i}+q$, which is a contradiction.\n\nLemma 3. For a fixed $n>q$, suppose that $a_{i}$ is maximal over $i \\leqslant n$. Then $i \\geqslant n-p+1$.\n\nProof. Suppose $a_{j}$ is minimal over $j \\geqslant n-q$. Then by Lemma $2, j \\leqslant n-q+p-1$. So $\\min \\left(a_{[n-q, n]}\\right)=a_{j}$ and $a_{i} \\geqslant \\max \\left(a_{[n-q, n]}\\right)$, which implies that $a_{i} \\geqslant a_{j}+q$.\n\nLemma 2 also implies that if $j \\geqslant n$ then $a_{j} \\geqslant \\min \\left(a_{[n, n+p]}\\right)$. So if $i<j$, then we have $a_{j} \\geqslant a_{i}-p$, which contradicts $a_{i} \\geqslant a_{j}+q$. Hence we must have $i>j$.\n\nThe above inequality also gives $\\left|a_{i}-a_{j}\\right| \\geqslant q>(k-1) p$, so by Lemma 1 it follows that $|i-j|>(k-1) p$. Therefore $i>j+(k-1) p \\geqslant n-q+(k-1) p \\geqslant n-p+1$.\n\nLet $b_{n}$ be the minimal value of $a_{i}$ for $i \\geqslant n$. By Lemma 2, $b_{n+p}>b_{n}$ for all $n$. Hence $b_{n}=\\min \\left(a_{[n, n+p]}\\right)=\\min \\left(a_{[n, n+q]}\\right)$. Let $c_{n}$ be the maximal value of $a_{i}$ for $i \\leqslant n$. By Lemma 3, $c_{n-p}>c_{n}$ for all $n>q$. Hence $c_{n}=\\max \\left(a_{[n-p, n]}\\right)=\\max \\left(a_{[n-q, n]}\\right)$ for $n>q$.\n\nSo if $n>q$ then $b_{n}=c_{n+p}-p=c_{n+q}-q$. So for $n>q$ we get $b_{n+q-p}+p=c_{n+q}=b_{n}+q$, and hence $b_{n+q-p}=b_{n}+q-p$.\n\nNext note that $b_{n+p} \\leqslant a_{n+p} \\leqslant b_{n}+p$. So $b_{n+p}-b_{n} \\leqslant p$ for all $n>q$, and iterating this ($q-p$) times gives $b_{n+p(q-p)}-b_{n} \\leqslant p(q-p)$. But using $b_{n+q-p}=b_{n}+q-p$ gives $b_{n+p(q-p)}-b_{n}=p(q-p)$. Since equality occurs, we must have $b_{n+p}=b_{n}+p$.\n\nSo for $n>q, b_{n+p}=b_{n}+p$ and $b_{n+q-p}=b_{n}+q-p$. Since $p$ and $q-p$ are coprime, $b_{n+1}=b_{n}+1$ for all $n>q$. The only way for $b_{n}$ and $b_{n+1}$ to be different is if $b_{n}=a_{n}$, so we deduce that $a_{n+1}=a_{n}+1$ and there is a constant $C$ such that $a_{n}=n+C$ for all $n>q$.\n\nFinally, suppose $a_{n}=n+C$ for all $n \\geqslant N$. Then $p=\\max \\left(a_{N-1}, N+C+p-1\\right)- \\min \\left(a_{N-1}, N+C\\right)$. So $a_{N-1}=N+C+p$ or $N+C-1$. Similarly, $a_{N-1}=N+C+q$ or $N+C-1$. Hence $a_{N-1}=N+C-1$. So, by induction, we have $a_{n}=n+C$ for all positive integers $n$. Since $a_{1} \\geqslant 1, C$ is a nonnegative integer.\n\nIt is trivial to check that $a_{n}=n+C$ satisfies the given condition.\nFor $n, x \\geqslant 1$, let the $x$-width of $n$ be $\\max \\left(a_{[n, n+x]}\\right)-\\min \\left(a_{[n, n+x]}\\right)$. We call a positive integer $x$ good if the $x$-width of $n$ is less than or equal to $x$ for all sufficiently large $n$, and we call $x$ very good if the $x$-width of $n$ is equal to $x$ for sufficiently large $n$.\n\nLemma 1. If $p'$ is good and $q'$ is very good with $p'<q'<2p'$, then $2p'-q'$ is also good.\n\nProof. Note that $0<q'-p'<p'<q'$. Let $n$ be a sufficiently large positive integer. Then for $k \\in [n+q'-p', n+p']$, we have $a_{k} \\geqslant \\max \\left(a_{[n, n+p']}\\right)-p'$ and $a_{k} \\geqslant \\max \\left(a_{[n+q'-p', n+q']}\\right)-p'$ since $p'$ is good, which shows $a_{k} \\geqslant \\max \\left(a_{[n, n+q']}\\right)-p'$. Similarly we get $a_{k} \\leqslant \\min \\left(a_{[n, n+q']}\\right)+p'$.\n\nTherefore, for all $k \\in [n+q'-p', n+p']$ we have $a_{k} \\in [\\max \\left(a_{[n, n+q']}\\right)-p', \\min \\left(a_{[n, n+q']}\\right)+p']$. Thus, the $(2p'-q')$-width of $n+q'-p'$ is at most $(\\min \\left(a_{[n, n+q']}\\right)+p')-(\\max \\left(a_{[n, n+q']}\\right)-p')=2p'-q'$. The lemma follows.\n\nLemma 2. Let $p'$ be a good number and $q'$ a very good number with $p'<q'$. For sufficiently large $n$, take $s, t \\in [n, n+q']$ such that $\\min \\left(a_{[n, n+q']}\\right)=a_{s}$ and $\\max \\left(a_{[n, n+q']}\\right)=a_{t}$. Then $s \\in [n, n+p']$ and $t \\in [n+q'-p', n+q']$.\n\nProof. Lemma 2 and Lemma 3 from Solution 1 hold with $p$ and $q$ replaced by $p'$ and $q'$ by similar arguments. We can deduce the statement about $s$ from Lemma 2 of Solution 1. We can deduce the statement about $t$ from Lemma 3 of Solution 1.\n\nLemma 3. If $p'$ is good and $q'$ is very good with $2p'<q'$, then there exists a positive integer $r$ such that for all sufficiently large $n$, we have $a_{n+r}-a_{n} \\geqslant r$.\n\nProof. Let $r=q'-2p'$, and let $s$ and $t$ be as defined in Lemma 2. Then consider the identity\n$$\n\\left(a_{t}-a_{n+q'-p'}\\right)+\\left(a_{n+p'+r}-a_{n+p'}\\right)+\\left(a_{n+p'}-a_{s}\\right)=a_{t}-a_{s}=q'\n$$\nBy Lemma 2, we have $s \\in [n, n+p']$ and $t \\in [n+q'-p', n+q']$, so $a_{n+p'}-s \\leqslant p'$ and $a_{t}-a_{n+q'-p'} \\leqslant p'$. Combining these, we get $a_{n+p'+r}-a_{n+p'} \\geqslant q'-2p'=r$. This proves that $a_{n+r}-a_{n} \\geqslant r$ for sufficiently large $n$.\n\nLemma 4. Suppose $(p, q) \\neq (1,2)$. Then there exists a good number $p'$ such that $2p'<q$.\n\nProof. Let $p'$ be the smallest good positive integer. Note that $p$ is good, so $p'$ exists and is less than $q$.\n\nSuppose for contradiction that $2p' \\geqslant q$. If $2p'>q$, then by Lemma 1, $2p'-q$ is a good number strictly less than $p'$, which contradicts minimality of $p'$. If $2p'=q$, then $p'<p<2p'$. So we can apply Lemma 1 with $q_{0}=p$ to get that $2p'-p$ is a good number that is strictly less than $p'$, which again contradicts minimality.\n\nIf $(p, q)=(1,2)$ then the problem is easily solved. Otherwise, Lemmas 3 and 4 combined give us some $r>0$ such that $a_{n+r}-a_{n} \\geqslant r$ for $n$ sufficiently large.\n\nBy iterating, we get $a_{n+pr}-a_{n} \\geqslant pr$ for all sufficiently large $n$, and hence it follows that $a_{n+p}-a_{n}=p$. Similarly we get $a_{n+q}-a_{n}=q$. As $p$ and $q$ are coprime, we deduce that $a_{n+1}-a_{n}=1$ for sufficiently large $n$. Thus we get $a_{n}=n+C$ for sufficiently large $n$, and we can conclude by the same argument as Solution 1.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32178,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. A class of $n$ students run $n$ races, in each of which they are ranked with no draws. A student is eligible for a rating $(a, b)$ for positive integers $a$ and $b$ if they come in the top $b$ places in at least $a$ of the races. Their final score is the maximum possible value of $a-b$ across all ratings for which they are eligible.\nFind the maximum possible sum of all the scores of the $n$ students.",
    "options": [],
    "answer": "n(n-1)/2",
    "solution": "The answer can be achieved by the students finishing in the same order in every race. To show that this is the maximum, we will apply a series of modifications to the results of the races, each of which does not decrease the total score, such that after $k$ such modifications the first $k$ positions are the same in every race. Say that a student is scored on the $b^{\\text{th}}$ place if their score is $a-b$ because they came in the top $b$ places in $a$ of the races and $b$ is minimal with this property for that student.\n\nSupposing that the first $k-1$ positions are the same in every race, look at the students scored on the $k^{\\text{th}}$ place. If there are no such students, let $\\ell>k$ be minimal such that some student $S$ is scored on the $\\ell^{\\text{th}}$ place. Then, in every race where $S$ appears in any place from the $k^{\\text{th}}$ through the $\\ell^{\\text{th}}$ inclusive (of which there must be at least $\\ell$, otherwise $S$ would achieve a higher rating of 0 based on the $n^{\\text{th}}$ place), reorder the students in places $k$ through $\\ell$ so that $S$ finishes in the $k^{\\text{th}}$ place instead (and otherwise the ordering of those students is arbitrary). Now $S$ is scored on the $k^{\\text{th}}$ place, their score has gone up by $\\ell-k$ and no other scores have gone down (some might have gone up as well).\n\nNow we know that the first $k-1$ positions are the same in every race and at least one student is scored on the $k^{\\text{th}}$ place. Pick one such student $S$. In each race where $S$ finishes behind the $k^{\\text{th}}$ place, swap them with the student $T$ who finishes in the $k^{\\text{th}}$ place, leaving the positions of all other students unchanged. Each such swap increases the score of $S$ by 1 and decreases the score of $T$ by at most 1, so such swaps do not decrease the total score. At the end of this process, the first $k$ positions are the same in every race and the total score has not decreased.\n\nRepeating this $n$ times yields the required result.\n\nNote that taking $a=b=n$ shows each student has a nonnegative score. Consider a student who has race ranks $r_{1}, r_{2}, \\ldots, r_{n}$ and a final score of $s$. We first prove that\n$$\n\\sum_{i} r_{i} \\leqslant n(n-s) .\n$$\nWithout loss of generality, suppose that $r_{1} \\leqslant r_{2} \\leqslant \\cdots \\leqslant r_{n}$. There must exist some $k$ with $s+1 \\leqslant k \\leqslant n$ and $k-r_{k}=s$. In order to maximise $\\sum_{i} r_{i}$ while retaining the score of $s$, we can replace each of $r_{1}, \\ldots, r_{k-1}$ by $r_{k}$, and replace each of $r_{k+1}, \\ldots, r_{n}$ by $n$. Then the sum is\n$$\n\\sum_{i} r_{i} \\leqslant k r_{k}+(n-k) n=n^{2}-k\\left(n-r_{k}\\right)=n^{2}-k(n+s-k) \\leqslant n^{2}-s n \\tag{1}\n$$\nThe final inequality follows from the fact that given $s+1 \\leqslant k \\leqslant n$, the quantity $k(n+s-k)$ is minimised when $k=n$.\n\nThe sum of ranks of all students across all races is $\\frac{n^{2}(n+1)}{2}$. If the total of all student scores is $t$, then (1) implies\n$$\n\\frac{n^{2}(n+1)}{2} \\leqslant n^{3}-t n\n$$\nThis rearranges to $t \\leqslant \\frac{n(n-1)}{2}$, as required.\nThe answer can be achieved by having the same ranking for all $n$ races.\n\nNote that taking $a=b=n$ shows each student has a nonnegative score. Consider a student who has race ranks $r_{1}, r_{2}, \\ldots, r_{n}$ and a final score of $s$. We first prove that\n$$\n\\sum_{i} r_{i} \\leqslant n(n-s) .\n$$\nWithout loss of generality, suppose that $r_{1} \\leqslant r_{2} \\leqslant \\cdots \\leqslant r_{n}$. There must exist some $k$ with $s+1 \\leqslant k \\leqslant n$ and $k-r_{k}=s$. In order to maximise $\\sum_{i} r_{i}$ while retaining the score of $s$, we can replace each of $r_{1}, \\ldots, r_{k-1}$ by $r_{k}$, and replace each of $r_{k+1}, \\ldots, r_{n}$ by $n$. Then the sum is\n$$\n\\sum_{i} r_{i} \\leqslant k r_{k}+(n-k) n=n^{2}-k\\left(n-r_{k}\\right)=n^{2}-k(n+s-k) \\leqslant n^{2}-s n \\tag{1}\n$$\nThe final inequality follows from the fact that given $s+1 \\leqslant k \\leqslant n$, the quantity $k(n+s-k)$ is minimised when $k=n$.\n\nThe sum of ranks of all students across all races is $\\frac{n^{2}(n+1)}{2}$. If the total of all student scores is $t$, then (1) implies\n$$\n\\frac{n^{2}(n+1)}{2} \\leqslant n^{3}-t n\n$$\nThis rearranges to $t \\leqslant \\frac{n(n-1)}{2}$, as required.\nIn each race, assign the student in the $k^{\\text{th}}$ place a weight of $1-\\frac{k}{n}$. If a student finishes in the top $b$ places in at least $a$ of the races, the total of their weights is at least $a\\left(1-\\frac{b}{n}\\right)=a-b\\left(\\frac{a}{n}\\right) \\geqslant a-b$. Thus the sum of a student's weights across all races is at least their score, and so the sum of all weights for all students across all races is at least the sum of all the scores of all students. The sum of weights in each race is $\\frac{n-1}{2}$, so the sum of all weights across all races is $\\frac{n(n-1)}{2}$. Equality is achieved if and only if, for each student, the values of $b$ and $a$ determining that student's score have $a=n$ and they finish in exactly the $b^{\\text{th}}$ place in all $n$ races; that is, if the students are ranked the same in every race.\nGiven a positive integer $b(S)$ for each student $S$, define $a_{b}(S)$ to be the number of races in which $S$ finished in the top $b(S)$ places, and define $\\operatorname{score}_{b}(S)=a_{b}(S)-b(S)$; for a race $r$, let $I_{b}(S, r)$ be 1 if $S$ finished in the top $b(S)$ places in race $r$ and 0 otherwise, so\n$$\na_{b}(S)=\\sum_{r} I_{b}(S, r)\n$$\nThen the problem asks for the maximum across all possible results of the races of\n$$\n\\max_{b} \\sum_{S} \\operatorname{score}_{b}(S)=\\max_{b}\\left(\\sum_{r} \\sum_{S} I_{b}(S, r)-\\sum_{S} b(S)\\right) .\n$$\nGiven $b$, the sum $\\sum_{S} I_{b}(S, r)$ is maximised (not necessarily uniquely) for some choice of the rankings in race $r$, which is the same choice for every race. So the maximum possible sum of the scores of all the students occurs when all students are ranked the same in all races, which yields the given answer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32179,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. The integers $1, 2, 3, \\ldots, n^{2}$ are to be written in the cells of an $n \\times n$ board such that each integer is written in exactly one cell and each cell contains exactly one integer. For every integer $d$ with $d \\mid n$, the $d$-division of the board is the division of the board into $(n / d)^{2}$ nonoverlapping sub-boards, each of size $d \\times d$, such that each cell is contained in exactly one $d \\times d$ sub-board.\n\nWe say that $n$ is a cool number if the integers can be written on the $n \\times n$ board such that, for each integer $d$ with $d \\mid n$ and $1 < d < n$, in the $d$-division of the board, the sum of the integers written in each $d \\times d$ sub-board is not a multiple of $d$.\n\nDetermine all even cool numbers.",
    "options": [],
    "answer": "All even cool numbers are powers of two.",
    "solution": "We first show by induction that $n = 2^{k}$ is a cool number. The base case of $n = 2$ is trivial as there is no such $d$.\n\nFor induction, assume that $2^{k}$ is a cool number. We construct a numbering of a $2^{k+1} \\times 2^{k+1}$ board that satisfies the conditions.\n\nTake the $2^{k+1} \\times 2^{k+1}$ board and divide it into four $2^{k} \\times 2^{k}$ sub-boards. By assumption, there is some numbering $P$ of a $2^{k} \\times 2^{k}$ board that satisfies the required condition; we write down the numbering $P$ in each sub-board. Next, add $2^{2k}$ to every number in the second sub-board, add $2 \\times 2^{2k}$ to every number in the third sub-board, and add $3 \\times 2^{2k}$ to every number in the fourth sub-board. Then the numbers in the cells of the $2^{k+1} \\times 2^{k+1}$ board are the numbers $1$ to $2^{2(k+1)}$.\n\nNow locate $2^{2k}$ from the first sub-board, and swap it with $2^{2k} + 2^{k-1}$ from the second sub-board. Locate $3 \\times 2^{2k}$ from the third sub-board, and swap it with $3 \\times 2^{2k} + 2^{k-1}$ from the fourth sub-board.\n\nWe claim that this numbering of the $2^{k+1} \\times 2^{k+1}$ board satisfies the required conditions. For any $d = 2^{i}$ where $i < k$, consider any $2^{i} \\times 2^{i}$ sub-board. The sum of its cells modulo $2^{i}$ is not changed in the addition step or the swapping step, so the sum is congruent modulo $2^{i}$ to the sum of the corresponding $2^{i} \\times 2^{i}$ sub-board in $P$, which is nonzero, as required.\n\nIn the case of $d = 2^{k}$, we can directly evaluate the sum of the $(b+1)^{\\text{th}}$ sub-board for $b \\in \\{0, 1, 2, 3\\}$. The sum is given by\n$$\n2^{2k-1}(1 + 2^{2k}) + b 2^{4k} + (-1)^{b} 2^{k-1} \\equiv 2^{k-1} \\quad (\\bmod 2^{k})\n$$\nTherefore all sub-boards satisfy the required conditions and so $2^{k+1}$ is a cool number, completing the induction.\n\nIt remains to show that no other even number is a cool number. Let $n = 2^{s} m$ where $s$ is a positive integer and $m$ is an odd integer greater than $1$. For the sake of contradiction, suppose that there is a numbering of the $n \\times n$ board satisfying the required conditions.\n\nClaim. In the $2^{i}$-division of the board, where $1 \\leqslant i \\leqslant s$, the sum of numbers in each $2^{i} \\times 2^{i}$ sub-board is congruent to $2^{i-1}$ modulo $2^{i}$.\n\nProof. We prove the claim by induction on $i$. The base case of $i = 1$ holds as the sum of numbers in each $2 \\times 2$ sub-board must be odd. Next, suppose the claim is true for $2^{i}$. In the $2^{i+1}$-division, each $2^{i+1} \\times 2^{i+1}$ sub-board is made up of four $2^{i} \\times 2^{i}$ sub-boards, each with a sum congruent to $2^{i-1}$ modulo $2^{i}$. Hence the sum of each $2^{i+1} \\times 2^{i+1}$ sub-board is a multiple of $2^{i}$. It cannot be a multiple of $2^{i+1}$ because of the conditions, which means it must be congruent to $2^{i}$ modulo $2^{i+1}$. This proves the claim.\n\nBack to the problem, since $m$ is odd, summing up the $m^{2}$ sums of $2^{s} \\times 2^{s}$ sub-boards gives\n$$\n2^{s-1} m^{2} \\equiv 2^{s-1} \\quad (\\bmod 2^{s})\n$$\nHowever, the sum of the numbers from $1$ to $n^{2}$ is\n$$\n\\frac{n^{2}(n^{2} + 1)}{2} = 2^{2s-1} m^{2} (2^{2s} m^{2} + 1) \\equiv 0 \\quad (\\bmod 2^{s})\n$$\nThis is a contradiction. Therefore $n$ is not a cool number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32180,
    "subject": "Mathematics (Multi-modal)",
    "question": "On a board with $2024$ rows and $2023$ columns, Turbo the snail tries to move from the first row to the last row. On each attempt, he chooses to start on any cell in the first row, then moves one step at a time to an adjacent cell sharing a common side. He wins if he reaches any cell in the last row. However, there are $2022$ predetermined, hidden monsters in $2022$ of the cells, one in each row except the first and last rows, such that no two monsters share the same column. If Turbo unfortunately reaches a cell with a monster, his attempt ends and he is transported back to the first row to start a new attempt. The monsters do not move.\nSuppose Turbo is allowed to take $n$ attempts. Determine the minimum value of $n$ for which he has a strategy that guarantees reaching the last row, regardless of the locations of the monsters.\n(Hong Kong)\n\nComment. One of the main difficulties of solving this question is in determining the correct expression for $n$. Students may spend a long time attempting to prove bounds for the wrong value for $n$ before finding better strategies.\nStudents may incorrectly assume that Turbo is not allowed to backtrack to squares he has already visited within a single attempt. Fortunately, making this assumption does not change the answer to the problem, though it may make it slightly harder to find a winning strategy.",
    "options": [],
    "answer": "3",
    "solution": "First we demonstrate that there is no winning strategy if Turbo has $2$ attempts.\nSuppose that $(2, i)$ is the first cell in the second row that Turbo reaches on his first attempt. There can be a monster in this cell, in which case Turbo must return to the first row immediately, and he cannot have reached any other cells past the first row.\n\nNext, suppose that $(3, j)$ is the first cell in the third row that Turbo reaches on his second attempt. Turbo must have moved to this cell from $(2, j)$, so we know $j \\neq i$. So it is possible that there is a monster on $(3, j)$, in which case Turbo also fails on his second attempt. Therefore Turbo cannot guarantee to reach the last row in $2$ attempts.\n\nNext, we exhibit a strategy for $n=3$. On the first attempt, Turbo travels along the path\n$$\n(1,1) \\rightarrow (2,1) \\rightarrow (2,2) \\rightarrow \\cdots \\rightarrow (2,2023).\n$$\nThis path meets every cell in the second row, so Turbo will find the monster in row $2$ and his attempt will end.\n\nIf the monster in the second row is not on the edge of the board (that is, it is in cell $(2, i)$ with $2 \\leqslant i \\leqslant 2022$), then Turbo takes the following two paths in his second and third attempts:\n$$\n\\begin{aligned}\n& (1, i-1) \\rightarrow (2, i-1) \\rightarrow (3, i-1) \\rightarrow (3, i) \\rightarrow (4, i) \\rightarrow \\cdots \\rightarrow (2024, i). \\\\\n& (1, i+1) \\rightarrow (2, i+1) \\rightarrow (3, i+1) \\rightarrow (3, i) \\rightarrow (4, i) \\rightarrow \\cdots \\rightarrow (2024, i).\n\\end{aligned}\n$$\nThe only cells that may contain monsters in either of these paths are $(3, i-1)$ and $(3, i+1)$. At most one of these can contain a monster, so at least one of the two paths will be successful.\n\n![](attached_image_1.png)\nFigure 1: Turbo's first attempt, and his second and third attempts in the case where the monster on the second row is not on the edge. The cross indicates the location of a monster, and the shaded cells are cells guaranteed to not contain a monster.\n\nIf the monster in the second row is on the edge of the board, without loss of generality we may assume it is in $(2,1)$. Then, on the second attempt, Turbo takes the following path:\n$$\n(1,2) \\rightarrow (2,2) \\rightarrow (2,3) \\rightarrow (3,3) \\rightarrow \\cdots \\rightarrow (2022,2023) \\rightarrow (2023,2023) \\rightarrow (2024,2023).\n$$\n\n![](attached_image_2.png)\nFigure 2: Turbo's second and third attempts in the case where the monster on the second row is on the edge. The light gray cells on the right diagram indicate cells that were visited on the previous attempt. Note that not all safe cells have been shaded.\n\nIf there are no monsters on this path, then Turbo wins. Otherwise, let $(i, j)$ be the first cell on which Turbo encounters a monster. We have that $j=i$ or $j=i+1$. Then, on the third attempt, Turbo takes the following path:\n$$\n\\begin{aligned}\n(1,2) & \\rightarrow (2,2) \\rightarrow (2,3) \\rightarrow (3,3) \\rightarrow \\cdots \\rightarrow (i-2, i-1) \\rightarrow (i-1, i-1) \\\\\n& \\rightarrow (i, i-1) \\rightarrow (i, i-2) \\rightarrow \\cdots \\rightarrow (i, 2) \\rightarrow (i, 1) \\\\\n& \\rightarrow (i+1,1) \\rightarrow \\cdots \\rightarrow (2023,1) \\rightarrow (2024,1).\n\\end{aligned}\n$$\nNow note that\n- The cells from $(1,2)$ to $(i-1, i-1)$ do not contain monsters because they were reached earlier than $(i, j)$ on the previous attempt.\n- The cells $(i, k)$ for $1 \\leqslant k \\leqslant i-1$ do not contain monsters because there is only one monster in row $i$, and it lies in $(i, i)$ or $(i, i+1)$.\n- The cells $(k, 1)$ for $i \\leqslant k \\leqslant 2024$ do not contain monsters because there is at most one monster in column $1$, and it lies in $(2,1)$.\nTherefore Turbo will win on the third attempt.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32181,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer. Geoff and Ceri play a game in which they start by writing the numbers $1, 2, \\ldots, N$ on a board. They then take turns to make a move, starting with Geoff. Each move consists of choosing a pair of integers $(k, n)$, where $k \\geqslant 0$ and $n$ is one of the integers on the board, and then erasing every integer $s$ on the board such that $2^{k} \\mid n-s$. The game continues until the board is empty. The player who erases the last integer on the board loses.\n\nDetermine all values of $N$ for which Geoff can ensure that he wins, no matter how Ceri plays.\n\n(Indonesia)",
    "options": [],
    "answer": "Write N = t · 2^n with t odd. If t = 1 (i.e., N is a power of two), then Geoff wins if and only if n is odd. If t > 1, then Geoff wins if and only if n is even.",
    "solution": "Lemma 1. For any set $\\mathcal{S}$, $\\mathcal{S}$ wins if and only if $J(\\mathcal{S}, \\varnothing)$ wins. Similarly, $\\mathcal{S}$ wins if and only if $J(\\varnothing, \\mathcal{S})$ wins.\n\nProof. Let $(k, m)$ be a move on $\\mathcal{S}$, and let $\\mathcal{T}$ be the result of applying the move. Then we can reduce $J(\\mathcal{S}, \\varnothing)$ to $J(\\mathcal{T}, \\varnothing)$ by applying the move $(k+1, 2m-1)$.\n\nConversely, let $(k, m)$ be a move on $J(\\mathcal{S}, \\varnothing)$. We can express the result of this move as $J(\\mathcal{T}, \\varnothing)$ for some $\\mathcal{T}$. Then we can reduce $\\mathcal{S}$ to $\\mathcal{T}$ by applying the move $(\\max(k-1, 0), (k+1)/2)$.\n\nThis gives us a natural bijection between games starting with $\\mathcal{S}$ and games starting with $J(\\mathcal{S}, \\varnothing)$ and thus proves the first part of the lemma. The second part follows by a similar argument. $\\square$\n\nLemma 2. If $\\mathcal{S}$ and $\\mathcal{T}$ are nonempty and at least one of them loses, then $J(\\mathcal{S}, \\mathcal{T})$ wins.\n\nProof. If $\\mathcal{S}$ is losing, then we can delete $J(\\varnothing, \\mathcal{T})$ using the move $(1, t)$ for some $t \\in J(\\varnothing, \\mathcal{T})$, which leaves the losing set $J(\\mathcal{S}, \\varnothing)$. Similarly, if $\\mathcal{T}$ is losing, then we can delete $J(\\mathcal{S}, \\varnothing)$ using the move $(1, s)$ for some $s \\in J(\\mathcal{S}, \\varnothing)$, leaving the losing set $J(\\varnothing, \\mathcal{T})$. $\\square$\n\nLemma 3. If $\\mathcal{S}$ is nonempty and wins, then $J(\\mathcal{S}, \\mathcal{S})$ loses.\n\nProof. From this position, we can convert any sequence of moves into another valid sequence of moves by replacing $(k, 2n-1)$ with $(k, 2n)$, and vice versa. Thus we may assume that the initial move $(k, m)$ has $m$ odd. We want to show that any such move results in a winning position for the other player.\n\nThe move $(0, m)$ loses immediately. Otherwise, the move results in the set $J(\\mathcal{T}, \\mathcal{S})$ for some set $\\mathcal{T}$. There are three cases.\n\nIf $\\mathcal{T}$ is empty then the other player gets the winning set $J(\\varnothing, \\mathcal{S})$.\n\nIf $\\mathcal{T}$ is losing then the other player can choose the move $(1, s)$ for some $s \\in J(\\varnothing, \\mathcal{S})$, which leaves the losing set $J(\\mathcal{T}, \\varnothing)$.\n\nIf $\\mathcal{T}$ is nonempty winning then the other player can choose the move $(k, m+1)$, which results in the position $J(\\mathcal{T}, \\mathcal{T})$. We can then proceed by induction on $|\\mathcal{S}|$ to show that this is a losing set. $\\square$\n\nLemma 4. $[2n]$ wins if and only if $[n]$ loses.\n\nProof. Note that $[2n]=J([n],[n])$. The result then follows directly from the previous two lemmas. $\\square$\n\nLemma 5. For any integer $n \\geqslant 1$, $[2n+1]$ wins.\n\nProof. By Lemma 4, either $[n]$ or $[2n]$ loses. If $[n]$ loses, then by Lemma 2 we have that $[2n+1]=J([n+1],[n])$ wins. Otherwise, $[2n]$ loses, and therefore $[2n+1]$ wins by choosing the move $(k, 2n+1)$ for sufficiently large $k$ so that only $2n+1$ is eliminated. $\\square$\n\nIt remains to verify the original answer. We have two cases to consider:\n\n- Suppose $N=2^{n}$ for some $n$. For $N=1$, every move is an instant loss for Geoff. Then by Lemma 4, Geoff wins for $N=2^{n}$ if and only if Geoff loses for $N=2^{n-1}$, and thus by induction we have that Geoff wins for $N=2^{n}$ if and only if $n$ is odd.\n\n- Otherwise, $N=t 2^{n}$, for some $n$ and some $t>1$ with $t$ odd. By Lemma 5, Geoff wins when $n=0$. Then by Lemma 4, Geoff wins for $N=t 2^{n}$ if and only if Geoff loses for $N=t 2^{n-1}$, and thus by induction on $n$ we have that Geoff wins for $N=t 2^{n}$ if and only if $n$ is even.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32182,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ and $T$ be positive integers. James has $4 n$ marbles with weights $1,2, \\ldots, 4 n$. He places them on a balance scale, so that both sides have equal weight. Andrew may move a marble from one side of the scale to the other, so that the absolute difference in weights of the two sides remains at most $T$.\nFind, in terms of $n$, the minimum positive integer $T$ such that Andrew may make a sequence of moves such that each marble ends up on the opposite side of the scale, regardless of how James initially placed the marbles.",
    "options": [],
    "answer": "4n",
    "solution": "Consider partitioning the weights into pairs $(t, 4 n+1-t)$. Suppose that each side of the balance contains $n$ of those pairs. If one side of the balance contains the pair $(t, 4 n+1-t)$ for $1 \\leqslant t < 2 n$ and the other side contains $(2 n, 2 n+1)$, then the following sequence of moves swaps those pairs between the sides without ever increasing the absolute value of the difference above $4 n$.\n\n$$\n\\begin{align*}\nt, 4 n+1-t & \\mid 2 n, 2 n+1  \\tag{1}\\\\\nt, 2 n, 4 n+1-t & \\mid 2 n+1  \\tag{2}\\\\\nt, 2 n & \\mid 2 n+1, 4 n+1-t  \\tag{3}\\\\\nt, 2 n, 2 n+1 & \\mid 4 n+1-t  \\tag{4}\\\\\n2 n, 2 n+1 & \\mid t, 4 n+1-t \\tag{5}\n\\end{align*}\n$$\n\nApplying this sequence twice swaps any two pairs $(t, 4 n+1-t)$ and $(t', 4 n+1-t')$ between the sides. So we can achieve an arbitrary exchange of pairs between the sides, and $C$ can be any configuration where each side of the balance contains $n$ of those pairs.\n\nWe now show that any initial configuration can reach one where each side has $n$ of those pairs. Consider a configuration where one side has total weight $A-s$ and the other has total weight $A+s$, for some $0 \\leqslant s \\leqslant 2 n$, and where some pair is split between the two sides. (If no pair is split between the two sides, they must have equal weights and we are done.) Valid moves include moving any weight $w$ with $1 \\leqslant w \\leqslant 2 n+s$ from the $A+s$ side to the $A-s$ side, and moving any weight $w$ with $1 \\leqslant w \\leqslant 2 n-s$ from the $A-s$ side to the $A+s$ side. Suppose the pair $(t, 4 n+1-t)$, with $t \\leqslant 2 n$, is split between the sides. If $t$ is on the $A+s$ side, or on the $A-s$ side and $t \\leqslant 2 n-s$, it can be moved to the other side. Otherwise, $t$ is on the $A-s$ side and $t \\geqslant 2 n-s+1$, so $4 n+1-t \\leqslant 2 n+s$ is on the $A+s$ side and can be moved to the other side. So we can unite the two weights from that pair without splitting any other pair, and repeating this we reach a configuration where no pair is split between the sides.\n\n\nWe treat the $n=1$ case separately. The initial configuration has marbles $1,4$ on one side and $2,3$ on the other. So moving marbles $2, 4, 3, 1$ in that order is legal and every marble ends on the opposite side. Now assume $n \\geqslant 2$.\n\nMarbles of weight at most $2 n$ are called small. We will make use of the following lemmas:\n\n**Lemma 1.** If a pair of legal configurations differ only in the locations of small marbles then there is a sequence of legal moves to get from one to the other.\n\nProof. At first we only move marbles in the wrong position if they are not on the lighter side. (In the case of a tie, neither side is lighter.) Such a move is always legal. Since this reduces the number of marbles in the wrong position, eventually it will no longer be possible to perform such a move.\n\nThen the only marbles in the wrong position are on the lighter side. So moving one marble in the wrong position at a time will always increase $|\\delta|$, and $|\\delta| \\leqslant 4 n$ at the end. Hence every move is legal.\n\n**Lemma 2.** Let $k \\in \\mathbb{N}$. A positive integer can be expressed as a sum of distinct positive integers up to $k$ if and only if it is at most $k(k+1) / 2$.\n\nProof. The maximum possible sum of distinct positive integers up to $k$ is $k(k+1) / 2$. For the other direction we use induction on $k$. The case $k=1$ is trivial. Assume the statement is true for $k-1$. For positive integers up to $k$ we only need a single term. For larger integers, including $k$ in the expression means we are done by the inductive hypothesis.\n\nAlso note that $n(2 n+1) \\geqslant 4 n$ for $n \\geqslant 2$.\n\nLet $2 n < m \\leqslant 4 n$. Marbles of weight greater than $m$ are called big and marbles from $2 n+1$ to $m$ are called medium.\n\nSuppose all big marbles are on the correct side (that is, opposite where they started), $m$ is on the incorrect side and the configuration is legal. Then the following steps give a sequence of legal moves after which $m$ is on the correct side and the big marbles were never moved.\n\nAssume $m$ is on the left. In Step 2, we rearrange the small marbles so we can move $m$. But this is only possible if the weight of big and medium marbles on the right is not too large. So we may need to move some medium marbles from the right first, which we do in Step 1.\n\n**Step 1** Skip to Step 2 if the total weight of medium and big marbles on the right side is at most $n(4 n+1)+2 n-m$. Since the big marbles are in the correct position and $m$ is in the incorrect position, the big marbles on the right can weigh at most $n(4 n+1)-m$. So there must be a medium marble $m' < m$ on the right.\n\nFrom the first assumption, it is legal to move all small marbles to the left. Then by Lemma 2 we can move some of the small marbles to the right so the right side has weight exactly $n(4 n+1)+2 n$. Then moving $m'$ is legal. Repeat this step. Since the total weight of medium marbles on the right decreases, this step will occur a bounded number of times.\n\n**Step 2** Let the total weight of the right side be $n(4 n+1)+2 n-m+x$ and the weight of small marbles on the right side be $y$. Note that $y \\geqslant x$. If $x \\leqslant 0$ then moving $m$ is legal.\n\nOtherwise, by Lemma 2 there is a set of small marbles of weight $y-x$. By Lemma 1, there is a sequence of legal moves of small marbles such that the right side has weight exactly $n(4 n+1)+2 n-m$. Now moving $m$ is legal.\n\nApplying the process above for $m=4 n, 4 n-1, \\ldots, 2 n+1$ will move all nonsmall marbles to the opposite side. Then Lemma 1 completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32183,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $N$ be a positive integer and let $a_{1}, a_{2}, \\ldots$ be an infinite sequence of positive integers. Suppose that, for each $n > N$, $a_{n}$ is equal to the number of times $a_{n-1}$ appears in the list $a_{1}, a_{2}, \\ldots, a_{n-1}$.\nProve that at least one of the sequences $a_{1}, a_{3}, a_{5}, \\ldots$ and $a_{2}, a_{4}, a_{6}, \\ldots$ is eventually periodic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M > \\max \\left(a_{1}, \\ldots, a_{N}\\right)$. We first prove that some integer appears infinitely many times. If not, then the sequence contains arbitrarily large integers. The first time each integer larger than $M$ appears, it is followed by a $1$. So $1$ appears infinitely many times, which is a contradiction.\n\nNow we prove that every integer $x \\geqslant M$ appears at most $M-1$ times. If not, consider the first time that any $x \\geqslant M$ appears for the $M^{\\text{th}}$ time. Up to this point, each appearance of $x$ is preceded by an integer which has appeared $x \\geqslant M$ times. So there must have been at least $M$ numbers that have already appeared at least $M$ times before $x$ does, which is a contradiction.\n\nThus there are only finitely many numbers that appear infinitely many times. Let the largest of these be $k$. Since $k$ appears infinitely many times there must be infinitely many integers greater than $M$ which appear at least $k$ times in the sequence, so each integer $1,2, \\ldots, k-1$ also appears infinitely many times. Since $k+1$ doesn't appear infinitely often there must only be finitely many numbers which appear more than $k$ times. Let the largest such number be $l \\geqslant k$. From here on we call an integer $x$ big if $x > l$, medium if $l \\geqslant x > k$ and small if $x \\leqslant k$. To summarise, each small number appears infinitely many times in the sequence, while each big number appears at most $k$ times in the sequence.\n\nChoose a large enough $N'$ $>$ $N$ such that $a_{N'}$ is small, and in $a_{1}, \\ldots, a_{N'}$:\n- every medium number has already made all of its appearances;\n- every small number has made more than $\\max (k, N)$ appearances.\n\nSince every small number has appeared more than $k$ times, past this point each small number must be followed by a big number. Also, by definition each big number appears at most $k$ times, so it must be followed by a small number. Hence the sequence alternates between big and small numbers after $a_{N'}$.\n\nLemma 1. Let $g$ be a big number that appears after $a_{N'}$. If $g$ is followed by the small number $h$, then $h$ equals the amount of small numbers which have appeared at least $g$ times before that point.\n\nProof. By the definition of $N'$, the small number immediately preceding $g$ has appeared more than $\\max (k, N)$ times, so $g > \\max (k, N)$. And since $g > N$, the $g^{\\text{th}}$ appearance of every small number must occur after $a_{N}$ and hence is followed by $g$. Since there are $k$ small numbers and $g$ appears at most $k$ times, $g$ must appear exactly $k$ times, always following a small number after $a_{N}$. Hence on the $h^{\\text{th}}$ appearance of $g$, exactly $h$ small numbers have appeared at least $g$ times before that point.\n\nDenote by $a_{[i, j]}$ the subsequence $a_{i}, a_{i+1}, \\ldots, a_{j}$.\n\nLemma 2. Suppose that $i$ and $j$ satisfy the following conditions:\n(a) $j > i > N' + 2$,\n(b) $a_{i}$ is small and $a_{i} = a_{j}$,\n(c) no small value appears more than once in $a_{[i, j-1]}$.\n\nThen $a_{i-2}$ is equal to some small number in $a_{[i, j-1]}$.\n\nProof. Let $\\mathcal{I}$ be the set of small numbers that appear at least $a_{i-1}$ times in $a_{[1, i-1]}$. By Lemma 1, $a_{i} = |\\mathcal{I}|$. Similarly, let $\\mathcal{J}$ be the set of small numbers that appear at least $a_{j-1}$ times in $a_{[1, j-1]}$. Then by Lemma 1, $a_{j} = |\\mathcal{J}|$ and hence by (b), $|\\mathcal{I}| = |\\mathcal{J}|$. Also by definition, $a_{i-2} \\in \\mathcal{I}$ and $a_{j-2} \\in \\mathcal{J}$.\n\nSuppose the small number $a_{j-2}$ is not in $\\mathcal{I}$. This means $a_{j-2}$ has appeared less than $a_{i-1}$ times in $a_{[1, i-1]}$. By (c), $a_{j-2}$ has appeared at most $a_{i-1}$ times in $a_{[1, j-1]}$, hence $a_{j-1} \\leqslant a_{i-1}$. Combining with $a_{[1, i-1]} \\subset a_{[1, j-1]}$, this implies $\\mathcal{I} \\subseteq \\mathcal{J}$. But since $a_{j-2} \\in \\mathcal{J} \\backslash \\mathcal{I}$, this contradicts $|\\mathcal{I}| = |\\mathcal{J}|$. So $a_{j-2} \\in \\mathcal{I}$, which means it has appeared at least $a_{i-1}$ times in $a_{[1, i-1]}$ and one more time in $a_{[i, j-1]}$. Therefore $a_{j-1} > a_{i-1}$.\n\nBy (c), any small number appearing at least $a_{j-1}$ times in $a_{[1, j-1]}$ has also appeared $a_{j-1} - 1 \\geqslant a_{i-1}$ times in $a_{[1, i-1]}$. So $\\mathcal{J} \\subseteq \\mathcal{I}$ and hence $\\mathcal{I} = \\mathcal{J}$. Therefore, $a_{i-2} \\in \\mathcal{J}$, so it must appear at least $a_{j-1} - a_{i-1} = 1$ more time in $a_{[i, j-1]}$.\n\nFor each small number $a_{n}$ with $n > N' + 2$, let $p_{n}$ be the smallest number such that $a_{n + p_{n}} = a_{i}$ is also small for some $i$ with $n \\leqslant i < n + p_{n}$. In other words, $a_{n + p_{n}} = a_{i}$ is the first small number to occur twice after $a_{n-1}$. If $i > n$, Lemma 2 (with $j = n + p_{n}$) implies that $a_{i-2}$ appears again before $a_{n + p_{n}}$, contradicting the minimality of $p_{n}$. So $i = n$. Lemma 2 also implies that $p_{n} \\geqslant p_{n-2}$. So $p_{n}, p_{n+2}, p_{n+4}, \\ldots$ is a nondecreasing sequence bounded above by $2k$ (as there are only $k$ small numbers). Therefore, $p_{n}, p_{n+2}, p_{n+4}, \\ldots$ is eventually constant and the subsequence of small numbers is eventually periodic with period at most $k$.\nWe follow Solution 1 until after Lemma 1. For each $n > N'$, we keep track of how many times each of $1,2, \\ldots, k$ has appeared in $a_{1}, \\ldots, a_{n}$. We will record this information in an updating $(k+1)$-tuple\n$$\n\\left(b_{1}, b_{2}, \\ldots, b_{k} ; j\\right)\n$$\nwhere each $b_{i}$ records the number of times $i$ has appeared. The final element $j$ of the $(k+1)$ tuple, also called the active element, represents the latest small number that has appeared in $a_{1}, \\ldots, a_{n}$.\n\nAs $n$ increases, the value of $\\left(b_{1}, b_{2}, \\ldots, b_{k} ; j\\right)$ is updated whenever $a_{n}$ is small. The $(k+1)$ tuple updates deterministically based on its previous value. In particular, when $a_{n} = j$ is small, the active element is updated to $j$ and we increment $b_{j}$ by $1$. The next big number is $a_{n+1} = b_{j}$. By Lemma 1, the next value of the active element, or the next small number $a_{n+2}$, is given by the number of $b$ terms greater than or equal to the newly updated $b_{j}$, or\n$$\n\\left|\\left\\{i \\mid 1 \\leqslant i \\leqslant k, b_{i} \\geqslant b_{j}\\right\\}\\right|.\n$$\n\nEach sufficiently large integer which appears $i+1$ times must also appear $i$ times, with both of these appearances occurring after the initial block of $N$. So there exists a global constant $C$ such that $b_{i+1} - b_{i} \\leqslant C$. Suppose that for some $r$, $b_{r+1} - b_{r}$ is unbounded from below. Since the value of $b_{r+1} - b_{r}$ changes by at most $1$ when it is updated, there must be some update where $b_{r+1} - b_{r}$ decreases and $b_{r+1} - b_{r} < -(k-1)C$. Combining with the fact that $b_{i} - b_{i-1} \\leqslant C$ for all $i$, we see that at this particular point, by the triangle inequality\n$$\n\\min \\left(b_{1}, \\ldots, b_{r}\\right) > \\max \\left(b_{r+1}, \\ldots, b_{k}\\right)\n$$\nSince $b_{r+1} - b_{r}$ just decreased, the new active element is $r$. From this point on, if the new active element is at most $r$, by the previous formula, the next element to increase is once again from $b_{1}, \\ldots, b_{r}$. Thus only $b_{1}, \\ldots, b_{r}$ will increase from this point onwards, and $b_{k}$ will no longer increase, contradicting the fact that $k$ must appear infinitely often in the sequence. Therefore $|b_{r+1} - b_{r}|$ is bounded.\n\nSince $|b_{r+1} - b_{r}|$ is bounded, it follows that each of $|b_{i} - b_{1}|$ is bounded for $i = 1, \\ldots, k$. This means that there are only finitely many different states for $\\left(b_{1} - b_{1}, b_{2} - b_{1}, \\ldots, b_{k} - b_{1} ; j\\right)$. Since the next active element is completely determined by the relative sizes of $b_{1}, b_{2}, \\ldots, b_{k}$ to each other, and the update of $b$ terms depends on the active element, the active element must be eventually periodic. Therefore the small numbers subsequence, which is either $a_{1}, a_{3}, a_{5}, \\ldots$ or $a_{2}, a_{4}, a_{6}, \\ldots$, must be eventually periodic.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32184,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. Given an $n \\times n$ board, the unit cell in the top left corner is initially coloured black, and the other cells are coloured white. We then apply a series of colouring operations to the board. In each operation, we choose a $2 \\times 2$ square with exactly one cell coloured black and we colour the remaining three cells of that $2 \\times 2$ square black.\n\nDetermine all values of $n$ such that we can colour the whole board black.\n(Peru)",
    "options": [],
    "answer": "n is a power of 2",
    "solution": "Now we prove that if such a colouring is possible for $n$ then $n$ must be a power of 2. Suppose it is possible to colour an $n \\times n$ board where $n>1$. Identify the top left corner of the board by $(0,0)$ and the bottom right corner by $(n, n)$. Whenever an operation takes place in a $2 \\times 2$ square centred on $(i, j)$, we immediately draw an \"X\", joining the four cells' centres (see Figure 4). Also, identify this $\\mathbf{X}$ by $(i, j)$. The first operation implies there's an $\\mathbf{X}$ at $(1,1)$. Since the whole board is eventually coloured, every cell centre must be connected to at least one X. The collection of all $\\mathrm{X}_{\\mathrm{s}}$ forms a graph $G$.\n\n![](attached_image_1.png)\nFigure 4: L-trominoes placements corresponding to colouring operations (left) and the corresponding $\\mathbf{X}$ diagram (right).\n\n## Claim 1. The graph $G$ is a tree.\nProof. Since every operation requires a pre-existing black cell, each newly drawn $\\mathbf{X}$ apart from the first must connect to an existing $\\mathbf{X}$. So all $\\mathrm{X}_{\\mathrm{s}}$ are connected to the first X and $G$ must be connected. Now, suppose $G$ has a cycle. Consider the newest X involved in the cycle, it must connect to previous $\\mathrm{X}_{\\mathrm{s}}$ at at least two points. But this implies the corresponding operation will colour at most two cells, which is a contradiction. $\\square$\n\nNote that in the following arguments, Claims 2 to 4 only require the condition that $G$ is a tree and every cell is connected to $G$.\n\nClaim 2. If there's an X at $(i, j)$, then $1 \\leqslant i, j \\leqslant n-1$ and $i \\equiv j(\\bmod 2)$.\nProof. The inequalities $1 \\leqslant i, j \\leqslant n-1$ are clear. Call an X at $(i, j)$ good if $i \\equiv j(\\bmod 2)$, or bad if $i \\not \\equiv j(\\bmod 2)$. The first X at $(i, j)=(1,1)$ is good. Suppose some Xs are bad. Since $G$ is connected, there must exist a good $\\mathbf{X}$ connecting to a bad $\\mathbf{X}$. But this can only occur if they connect at two points, creating a cycle. This is a contradiction, thus all $\\mathrm{X}_{\\mathrm{s}}$ are good. $\\square$\n\nCall an X at $(i, j)$ odd if $i \\equiv j \\equiv 1(\\bmod 2)$, even if $i \\equiv j \\equiv 0(\\bmod 2)$.\n\nClaim 3. The integer $n$ must be even. Furthermore, there must be $4(n / 2-1)$ odd $\\mathrm{X}_{\\mathrm{s}}$ connecting the cells on the perimeter of the board as shown in Figure 5.\nProof. If $n$ is odd, the four corners of the bottom left cell are $(n, 0),(n-1,0),(n-1,1)$ and $(n, 1)$, none of which satisfies the conditions of Claim 2. So the bottom left cell cannot connect to any X. If $n$ is even, each cell on the edge of the board has exactly one corner satisfying the conditions of Claim 2, so the $\\mathbf{X}$ connecting it is uniquely determined. Therefore the cells on the perimeter of the board are connected to Xs according to Figure 5. $\\square$\n\n![](attached_image_2.png)\nFigure 5: Highlighting the permitted points for $\\mathrm{X}_{\\mathrm{s}}$ (left) and $\\mathrm{X}_{\\mathrm{s}}$ on the perimeter (right).\n\nDivide the $n \\times n$ board into $n^{2} / 4$ blocks of $2 \\times 2$ squares. Call each of these blocks a big-cell. We say a big-cell is filled if it contains an odd $\\mathbf{X}$ on its interior, empty otherwise. By Claim 3, each big-cell on the perimeter must be filled.\n\nClaim 4. Every big-cell is filled.\nProof. Recall that $\\mathrm{X}_{\\mathrm{s}}$ can only be at $(i, j)$ with $i \\equiv j(\\bmod 2)$. Suppose a big-cell centred at $(i, j)$ is empty. Then in order for its four cells to be coloured, there must be four even Xs on $(i-1, j-1),(i+1, j-1),(i-1, j+1)$ and $(i+1, j+1)$, \"surrounding\" the big-cell (see Figure 6).\n\nBy Claim 3, no empty big-cell can be on the perimeter. So if there exist some empty big-cells, the boundary between empty and filled big-cells must consist of a number of closed loops. Each closed loop is made up of several line segments of length 2, each of which separates a filled big-cell from an empty big-cell.\n\nSince every empty big-cell is surrounded by even $\\mathrm{X}_{\\mathrm{s}}$ and every filled big-cell contains an odd $\\mathbf{X}$, the two end points of each such line segment must be connected by $\\mathrm{X}_{\\mathrm{s}}$. Since these line segments form at least one closed loop, it implies the existence of a cycle made up of $\\mathrm{X}_{\\mathrm{s}}$ (see Figure 6). This is a contradiction, thus no big-cell can be empty. $\\square$\n\n![](attached_image_3.png)\nFigure 6: An empty big-cell surrounded by even $\\mathrm{X}_{\\mathrm{s}}$ (left) and the boundary between empty and filled $\\mathrm{X}_{\\mathrm{s}}$ creating a cycle (right).\n\nTherefore every big-cell is filled by an odd $\\mathbf{X}$, and the connections between them are provided by even Xs. We can now reduce the $n \\times n$ problem to an $n / 2 \\times n / 2$ problem in the following way. Perform a dilation of the board by a factor of $1 / 2$ with respect to $(0,0)$. Each big-cell is shrunk to a regular cell. For the Xs, replace each odd X at $(i, j)$ by the point $(i / 2, j / 2)$, and replace each even $\\mathbf{X}$ at $(i, j)$ by an $\\mathbf{X}$ at $(i / 2, j / 2)$.\n\nWe claim the new resulting graph of $\\mathrm{X}_{\\mathrm{s}}$ is a tree that connects all cells of an $n / 2 \\times n / 2$ board. First, two connected $\\mathrm{X}_{\\mathrm{s}}$ in the original $n \\times n$ board are still connected after their replacements (noting that some $\\mathrm{X}_{\\mathrm{s}}$ have been replaced by single points). For each cell in the $n / 2 \\times n / 2$ board, its centre corresponds to an odd $\\mathbf{X}$ from a filled big-cell in the original $n \\times n$ board, so it must be connected to the graph. Finally, suppose there exists a cycle in the new graph. The cycle consists of $\\mathrm{X}_{\\mathrm{s}}$ that correspond to even $\\mathrm{X}_{\\mathrm{s}}$ in the original graph connecting big-cells, forming a cycle of big-cells. Since in every big-cell, the four unit squares were connected by an odd $\\mathbf{X}$, this implies the existence of a cycle in the original graph, which is a contradiction.\n\nThus the new graph of $\\mathrm{X}_{\\mathrm{s}}$ must be a tree that connects all cells of an $n / 2 \\times n / 2$ board, which are the required conditions for Claims 2 to 4. Hence we can repeat our argument, halving the dimensions of the board each time, until we reach the base case of a $1 \\times 1$ board (where the tree is a single point). Therefore $n$ must be a power of 2, completing the solution.\nAs in Solution 1, it is possible the colour the whole board black for $n=2^{k}$.\nThe colouring operation is equivalent to the placement of $\\mathbf{L}$-trominoes. For each $\\mathbf{L}$-tromino we place on the board, we draw an arrow and a node as shown in Figure 7. We also draw a node in the top left corner of the board.\n![](attached_image_4.png)\nFigure 7: Tromino with corresponding arrow and node drawn.\n\nClaim 1. The arrows and nodes form a directed tree rooted at the top left corner.\nProof. The proof is similar to the proof of Claim 1 in Solution 1, with the additional note that the directions of the arrows inherit the order of the colouring operations, so they must be pointing away from the top left node. $\\square$\n\nNote that since all edges of the tree are diagonal, the nodes can only lie on points $(i, j)$ with $i+j \\equiv 0(\\bmod 2)$. This implies that we can only place down L-trominoes of one particular parity: that is, with the centre of the $\\mathbf{L}$-tromino on a point with $i+j \\equiv 0(\\bmod 2)$. In the remainder of the proof, we will implicitly use this parity property when determining possible positions of L-trominoes.\n\nNext, we show that certain configurations of edges of the tree are impossible.\n\nClaim 2. There cannot be two edges in a \"parallel\" configuration (see Figure 8).\nProof. In such a configuration, the two edges can either be directed in the same direction or opposite directions. If they point in the same direction (see Figure 8), then the $\\mathbf{L}$-trominoes corresponding to the two edges overlap.\n![](attached_image_5.png)\nFigure 8: Parallel configuration (left) and two parallel edges, case 1 (right).\n\nIf they point in opposite directions, then we get the diagram in Figure 9. The cells marked ($\\star$) must lie inside the $n \\times n$ board, so they must be covered by L-trominoes. There is only one possible way to cover these with a L-tromino of the right parity. But this makes the arrows form a cycle, which cannot happen. So we have a contradiction. $\\square$\n![](attached_image_6.png)\nFigure 9: Two parallel edges, case 2.\n\nClaim 3. There cannot be three edges in a \"zigzag\" configuration, shown in Figure 10.\n![](attached_image_7.png)\nFigure 10: Zigzag configuration.\n\nProof. Assume for contradiction that there is a zigzag. Then take the zigzag with maximal distance from the root of the tree (measured by distance along the graph from the root to the middle edge of the zigzag).\n\nWe may assume without loss of generality that the middle edge is directed down-right. Then the right edge must be directed up-right, since no two arrows can point to the same node. Next, we draw in the corresponding L-trominoes, and consider the cell marked ($\\star$). There are two possible ways to cover it with an L-tromino, because of the parity of L-tromino centres.\n\nWe could choose the centre of the L-tromino to be the top right corner of the cell (see Figure 11). This immediately gives another zigzag.\n![](attached_image_8.png)\nFigure 11: Zigzag configuration, case 1.\n\nThe other possibility is if we choose the centre of the L-tromino to be the bottom left corner of the cell (see Figure 12). Then we need to cover the cell marked ($\\star\\star$) with an L-tromino. If\n![](attached_image_9.png)\nFigure 12: Zigzag configuration, case 2.\n\nwe placed the centre of the $\\mathbf{L}$-tromino on the top left corner of the cell, this would give two parallel edges, contradicting Claim 2. So we must place the centre of the L-tromino on the bottom right corner of the cell, which gives a zigzag.\n\nIn each case, we get another zigzag further away from the root of the tree, which contradicts our assumption of maximality. So there cannot be any zigzags. $\\square$\n\nWe now colour the nodes of the tree. Colour the root node yellow. For all other nodes, we colour it white if it has an arrow coming out of it in a different direction to the arrow going in, and black otherwise.\n\nClaim 4. Any child of a black node is white.\n![](attached_image_10.png)\nFigure 13: Black node configuration.\n\nProof. Suppose we have a black node with a child. Then the arrow exiting the black node must be in the same direction as the arrow entering it by the definition of our colouring, giving the left diagram of Figure 13.\n\nThe cell marked ($\\star$) must be covered by an L-tromino. If the centre of this L-tromino is the bottom left corner, then this would give an arrow leaving the black node in a different direction, which cannot happen. So the centre of the $\\mathbf{L}$-tromino must instead be the top right corner, which gives an arrow leaving the upper node in a different direction. Thus the upper node must be white. $\\square$\n\nClaim 5. Every white node has three children, all of which are black.\n![](attached_image_11.png)\nFigure 14: White node configuration.\n\nProof. Refer to Figure 14. Suppose we have a white node, as in the leftmost diagram. The cell marked ($\\star$) must be covered by an L-tromino. If the centre of this L-tromino is the bottom right corner of the cell, then this would form a zigzag, which by Claim 3 is not allowed. So the centre must be the top left corner.\n\nNext, the cell marked ($\\star\\star$) must be covered by an L-tromino. If the centre of this L-tromino is the top right corner, this would form a zigzag, so the centre must be the bottom left corner instead. Thus we have shown that any white node has three children.\n\nFinally, note that if any of the child nodes had three children of their own, then this would give parallel edges in the diagram, which contradicts Claim 2. Therefore the child nodes of the white node must all be black. $\\square$\n\nWe now know that the node colours alternate between black and white as you go down the tree, so all white nodes lie on points with coordinates $(2i, 2j)$, and all black nodes lie on points with coordinates $(2i+1,2j+1)$.\n\nNow (assuming $n>1$) we will construct a new board whose cells are $2 \\times 2$ squares of our current board. We replace the root node and its child with a single big cell and a big root node,\n![](attached_image_12.png)\nFigure 15: Replacing with larger cells and $\\mathbf{L}$-trominoes.\n\nand we replace each white node and its three children with a big L-tromino, big arrow and big node as shown in Figure 15.\n\nEvery black node is the child of the root node or a white node, so every $\\mathbf{L}$-tromino is involved in exactly one replacement. Also, the parent of any white node is a black node, whose parent, in turn, is a white node or the root. So the starting point of every big arrow will be on a big node. Therefore we obtain an $\\mathbf{L}$-tromino tiling forming a tree.\n\nThis shows for $n>1$ that if an $n \\times n$ board can be tiled by L-trominoes forming a tree, then $n$ is even, and an $n / 2 \\times n / 2$ board can also be tiled by L-trominoes forming a tree. Since a $1 \\times 1$ board can trivially be tiled, we conclude that the only values of $n$ for which an $n \\times n$ board can be tiled are $n=2^{k}$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32185,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral such that $AC < BD < AD$ and $\\angle DBA < 90^{\\circ}$. Point $E$ lies on the line through $D$ parallel to $AB$ such that $E$ and $C$ lie on opposite sides of line $AD$, and $AC = DE$. Point $F$ lies on the line through $A$ parallel to $CD$ such that $F$ and $C$ lie on opposite sides of line $AD$, and $BD = AF$.\nProve that the perpendicular bisectors of segments $BC$ and $EF$ intersect on the circumcircle of $ABCD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $T$ be the midpoint of $\\operatorname{arc} \\overparen{BAC}$ and let lines $BA$ and $CD$ intersect $EF$ at $K$ and $L$, respectively. Note that $T$ lies on the perpendicular bisector of segment $BC$.\n\n![](attached_image_1.png)\n\nSince $ABCD$ is cyclic, $\\frac{BD}{\\sin \\angle BAD} = \\frac{AC}{\\sin \\angle ADC}$. From parallel lines we have $\\angle DAF = \\angle ADC$ and $\\angle BAD = \\angle EDA$. Hence,\n$$\nAF \\cdot \\sin \\angle DAF = BD \\cdot \\sin \\angle ADC = AC \\cdot \\sin \\angle BAD = DE \\cdot \\sin \\angle EDA.\n$$\nSo $F$ and $E$ are equidistant from the line $AD$, meaning that $EF$ is parallel to $AD$.\n\nWe have that $KADE$ and $FADL$ are parallelograms, hence we get $KA = DE = AC$ and $DL = AF = BD$. Also, $KE = AD = FL$ so it suffices to prove the perpendicular bisector of $KL$ passes through $T$.\n\nTriangle $AKC$ is isosceles so $\\angle BTC = \\angle BAC = 2 \\angle BKC$. Likewise, $\\angle BTC = 2 \\angle BLC$. Since $T, K$, and $L$ all lie on the same side of $BC$ and $T$ lies on the perpendicular bisector of $BC$, $T$ is the centre of circle $BKLC$. The result follows.\nLet $AF$ and $DE$ meet $\\omega$ at $X$ and $Y$, respectively, and let $T$ be as in Solution 1.\n\nAs $BD < AD$, $DY \\parallel AB$ and $\\angle BAY = \\angle DBA < 90^{\\circ}$, we have $DY < AB$ and $Y$ lies on the opposite side of line $AD$ to $C$. Also from $BD < AD$, we have $B, C$, and $D$ all lie on the same side of the perpendicular bisector of $AB$ which shows $AC > AB$. Combining these, we get $DY < AB < AC = DE$ and, as $Y$ and $E$ both lie on the same side of line $AD$, $Y$ lies in the interior of segment $DE$. Similarly, $X$ lies in the interior of segment $DF$.\n\nSince $AB$ is parallel to $DY$, we have $YA = BD = FA$. Likewise $XD = AC = ED$.\n\n![](attached_image_2.png)\n\nClaim 1. $T$ is the midpoint of $\\operatorname{arc} \\overparen{XY}$.\n\nProof. From $AX \\parallel CD$ and $AB \\parallel DY$ we have\n$$\n\\angle CAX = \\angle AXD = \\angle AYD = \\angle YDB.\n$$\nSince $T$ is the midpoint of $\\operatorname{arc} \\overparen{BAC}$, we have $\\angle BAT = \\angle TDC$, so\n$$\n\\angle TAX = \\angle CAX + \\angle BAC - \\angle BAT = \\angle YDB + \\angle BDC - \\angle TDC = \\angle YDT.\n$$\nRecall from above we have $AB < AC$ and analogously, $DC < DB$, which shows that $X, Y$ and $T$ all lie on the same side of line $AD$. In particular, $T$ and $A$ lie on opposite sides of $XY$ so $T$ lies on the internal angle bisector of $\\angle XAY$. Since $AF = AY$, we have $\\triangle ATF \\cong \\triangle ATY$, giving $TF = TY$.\n\nLikewise, $TE = TX$, so $TE = TF$, meaning that $T$ lies on the perpendicular bisector of segment $EF$ as required.\nFrom $AF = DB$, $AC = DE$ and\n$$\n\\angle(AC, AF) = \\angle(AC, CD) = \\angle(AB, BD) = \\angle(DE, DB),\n$$\ntriangles $ACF$ and $DEB$ are congruent, so $CF = BE$.\n\nLet $P = BE \\cap CF$. Since\n$$\n\\angle(CP, BP) = \\angle(CF, BE) = \\angle(AF, DB) = \\angle(DC, DB),\n$$\nwe have that $P$ lies on circle $ABCD$.\n\n![](attached_image_3.png)\n\nFinally, let $T$ be the Miquel point of the quadrilateral $BCFE$ so $T$ lies on circles $EFP$ and $ABCD$. Note that $T$ is the centre of spiral similarity taking segments $BE$ to $CF$ and since $BE = CF$, this is in fact just a rotation, so $TB = TC$ and $TE = TF$; that is, the perpendicular bisectors of $BC$ and $EF$ meet at $T$, on circle $ABCD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32186,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC < BC$, incentre $I$ and incircle $\\omega$. Let $X$ be the point in the interior of side $BC$ such that the line through $X$ parallel to $AC$ is tangent to $\\omega$. Similarly, let $Y$ be the point in the interior of side $BC$ such that the line through $Y$ parallel to $AB$ is tangent to $\\omega$. Let $AI$ intersect the circumcircle of triangle $ABC$ again at $P \\neq A$. Let $K$ and $L$ be the midpoints of $AB$ and $AC$, respectively.\nProve that $\\angle KIL + \\angle YPX = 180^{\\circ}$.\n(Poland)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nWe have $\\angle APB = \\angle ACB$ in the circumcircle and $\\angle ACB = \\angle A'XC$ because $A'X \\parallel AC$. Hence, $\\angle APB = \\angle A'XC$, and so quadrilateral $BPA'X$ is cyclic. Similarly, it follows that $CYA'P$ is cyclic.\nNow we are ready to transform $\\angle KIL + \\angle YPX$ to the sum of angles in triangle $A'CB$. By a homothety of factor 2 at $A$ we have $\\angle KIL = \\angle CA'B$. In circles $BPA'X$ and $CYA'P$ we have $\\angle APX = \\angle A'BC$ and $\\angle YPA = \\angle BCA'$, therefore\n$$\n\\angle KIL + \\angle YPX = \\angle CA'B + (\\angle YPA + \\angle APX) = \\angle CA'B + \\angle BCA' + \\angle A'BC = 180^{\\circ}.\n$$\nLet $BC = a$, $AC = b$, $AB = c$ and $s = \\frac{a+b+c}{2}$, and let the radii of the incircle, $B$-excircle and $C$-excircle be $r, r_b$ and $r_c$, respectively. Let the incircle be tangent to $AC$ and $AB$ at $B_0$ and $C_0$, respectively; let the $B$-excircle be tangent to $AC$ at $B_1$, and let the $C$-excircle be tangent to $AB$ at $C_1$. As is well-known, $AB_1 = s-c$ and $\\text{area}(\\triangle ABC) = rs = r_c(s-c)$.\nLet the line through $X$, parallel to $AC$ be tangent to the incircle at $E$, and the line through $Y$, parallel to $AB$ be tangent to the incircle at $D$. Finally, let $AP$ meet $BB_1$ at $F$.\n![](attached_image_2.png)\nIt is well-known that points $B, E$, and $B_1$ are collinear by the homothety between the incircle and the $B$-excircle, and $BE \\parallel IK$ because $IK$ is a midline in triangle $B_0EB_1$. Similarly, it follows that $C, D$, and $C_1$ are collinear and $CD \\parallel IL$. Hence, the problem reduces to proving $\\angle YPA = \\angle CBE$ (and its symmetric counterpart $\\angle APX = \\angle DCB$ with respect to the vertex $C$), so it suffices to prove that $FYPB$ is cyclic. Since $ACPB$ is cyclic, that is equivalent to $FY \\parallel B_1C$ and $\\frac{BF}{FB_1} = \\frac{BY}{YC}$.\nBy the angle bisector theorem we have\n$$\n\\frac{BF}{FB_1} = \\frac{AB}{AB_1} = \\frac{c}{s-c}.\n$$\nThe homothety at $C$ that maps the incircle to the $C$-excircle sends $Y$ to $B$, so\n$$\n\\frac{BC}{YC} = \\frac{r_c}{r} = \\frac{s}{s-c}.\n$$\nSo,\n$$\n\\frac{BY}{YC} = \\frac{BC}{YC} - 1 = \\frac{s}{s-c} - 1 = \\frac{c}{s-c} = \\frac{BF}{FB_1},\n$$\nwhich completes the solution.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32187,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral with $AB$ parallel to $CD$ and $AB < CD$. Lines $AD$ and $BC$ intersect at a point $P$. Point $X \\neq C$ on the circumcircle of triangle $ABC$ is such that $PC = PX$. Point $Y \\neq D$ on the circumcircle of triangle $ABD$ is such that $PD = PY$. Lines $AX$ and $BY$ intersect at $Q$.\n\nProve that $PQ$ is parallel to $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $M$ and $N$ be the midpoints of $AD$ and $BC$, respectively and let the perpendicular bisector of $AB$ intersect the line through $P$ parallel to $AB$ at $R$.\n\nLemma. Triangles $QAB$ and $RNM$ are similar.\n\nProof. Let $O$ be the circumcentre of triangle $ABC$, and let $S$ be the midpoint of $CX$. Since $N, S$, and $R$ are the respective perpendicular feet from $O$ to $BC, CX$, and $PR$, we have that quadrilaterals $PRNO$ and $CNSO$ are cyclic. Furthermore, $P, S$, and $O$ are collinear as $PC = PX$. Since $ABCX$ is also cyclic, we have that\n$$\n\\angle QAB = \\angle XCB = \\angle PON = 180^\\circ - \\angle NRP = \\angle MNR.\n$$\nAnalogously, we have that $\\angle ABQ = \\angle RMN$, so triangles $QAB$ and $RNM$ are similar. $\\square$\n\n![](attached_image_1.png)\n\nLet $d(Z, \\ell)$ denote the perpendicular distance from the point $Z$ to the line $\\ell$. Using that $PR \\parallel AB$ along with the similarities $QAB \\sim RNM$ and $PAB \\sim PMN$, we have that\n$$\n\\frac{d(Q, AB)}{AB} = \\frac{d(R, MN)}{MN} = \\frac{d(P, MN)}{MN} = \\frac{d(P, AB)}{AB},\n$$\nwhich implies that $PQ \\parallel AB$.\nLet $BD$ and $AC$ intersect at $T$ and let the line through $P$ parallel to $AB$ intersect $BD$ at $V$. Next, let $Q'$ be the foot of the perpendicular from $T$ to $PV$. Finally, let $Q'A$ intersect circle $ABC$ again at $X'$ and $Q'B$ intersect circle $ABD$ again at $Y'$.\n\n![](attached_image_2.png)\n\nClaim. $PQ'$ bisects $\\angle BQ'D$ externally.\n\nProof. Let $PT$ intersect $CD$ at $L$. Let $\\infty_{CD}$ be the point at infinity on line $CD$. From the standard Ceva-Menelaus configuration we have $(D, C ; L, \\infty_{CD})$ is harmonic. Hence projecting through $P$ we have\n$$\n-1 = (D, C ; L, \\infty_{CD}) = (D, B ; T, V).\n$$\nAs $(D, B ; T, V)$ is harmonic, and also $\\angle VQ'T = 90^\\circ$ (by construction), the claim follows. $\\square$\n\nNow as\n$$\n\\angle Q'PD = \\angle BAD = 180^\\circ - \\angle DY'B = 180^\\circ - \\angle DY'Q'\n$$\nwe have $Q'PDY'$ cyclic. By the claim, we have that $P$ is the midpoint of $\\operatorname{arc} \\widetilde{DQ'Y'}$, so $PD = PY'$.\n\nSince $Y$ is the unique point not equal to $D$ on circle $ABD$ satisfying $PD = PY$, we have $Y' = Y$.\nLikewise $X' = X$ so $Q' = Q$ and we are done.\nLet $AX$ intersect circle $PCX$ for the second time at $Q'$. Then\n$$\n\\angle AQ'P = \\angle XQ'P = \\angle XCP = \\angle XCB = 180^\\circ - \\angle BAX = \\angle Q'AB\n$$\nso $PQ'$ is parallel to $AB$. Hence, it suffices to show that $Q'$ is equal to $Q$. To do so, we aim to show the common chord of circles $PCX$ and $PDY$ is parallel to $AB$, since then by symmetry $Q'$ is also the second intersection of $BY$ and circle $PDY$.\n\n![](attached_image_3.png)\n\nLet the centres of circles $PCX$ and $PDY$ be $O_X$ and $O_Y$, respectively. Let the centres of circles $ABC$ and $ABD$ be $O_C$ and $O_D$, respectively.\n\nNote $P, O_X$, and $O_C$ are collinear since they all lie on the perpendicular bisector of $CX$. Likewise $P, O_Y$, and $O_D$ are collinear on the perpendicular bisector of $DY$. By considering the projections of $O_X$ and $O_C$ onto $BC$, and $O_Y$ and $O_D$ onto $AD$, we have\n$$\n\\frac{PO_X}{PO_C} = \\frac{\\frac{PC}{2}}{\\frac{PB + PC}{2}} = \\frac{\\frac{PD}{2}}{\\frac{PA + PD}{2}} = \\frac{PO_Y}{PO_D}.\n$$\nHence $O_XO_Y$ is parallel to $O_CO_D$, which is perpendicular to $AB$ as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32188,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with incentre $I$, and let $\\Omega$ be the circumcircle of triangle $BIC$. Let $K$ be a point in the interior of segment $BC$ such that $\\angle BAK < \\angle KAC$. The angle bisector of $\\angle BKA$ intersects $\\Omega$ at points $W$ and $X$ such that $A$ and $W$ lie on the same side of $BC$, and the angle bisector of $\\angle CKA$ intersects $\\Omega$ at points $Y$ and $Z$ such that $A$ and $Y$ lie on the same side of $BC$.\nProve that $\\angle WAY = \\angle ZAX$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution 1. Let $\\Gamma$ be circle $ABC$ and $\\omega$ be circle $AYZ$. Let $O, M$, and $S$ be the centres of $\\Gamma, \\Omega$, and $\\omega$, respectively. Let $AK$ intersect $\\Gamma$ again at $P$, and let the angle bisector of $\\angle ZAY$ intersect $\\omega$ again at $N$.\n\n![](attached_image_1.png)\n\nBy power of a point from $K$ to $\\Gamma$ and $\\Omega$, we have that $KA \\cdot KP = KB \\cdot KC = KY \\cdot KZ$, so $P$ also lies on $\\omega$. The pairwise common chords of $\\Gamma, \\Omega$, and $\\omega$ are then $AP \\perp OS, BC \\perp OM$, and $YZ \\perp MS$, so we have that $\\angle OMS = \\angle CKY = \\angle YKA = \\angle MSO$. As $M$ lies on $\\Gamma$ and $OM = OS$, $S$ also lies on $\\Gamma$. Note that $N$ lies on $MS$ as $NY = NZ$, so\n$$\n\\angle PAN = \\frac{1}{2} \\angle PSN = \\frac{1}{2} \\angle PSM = \\frac{1}{2} \\angle PAM.\n$$\nThus, $AN$ bisects $\\angle PAM$ in addition to $\\angle ZAY$, which means that $\\angle ZAK = \\angle IAY$ as $K$ lies on $AP$ and $I$ lies on $AM$.\n\n\nSolution 2. Define $M$ and $P$ as in Solution 1, and recall that $AYPZ$ is cyclic. Let $Q$ be the second intersection of the line parallel to $BC$ through $P$ with circle $ABC$ and let $J$ be the incentre of triangle $APQ$.\n\n![](attached_image_2.png)\n\nSince $PQ$ is parallel to $BC$ and $\\angle BAP < \\angle PAC$, the angle bisector of $\\angle APQ$ is parallel to the angle bisector of $\\angle AKC$. Hence, $PJ$ is parallel to $YZ$. As $M$ is the midpoint of $\\overparen{PQ}$ on circle $APQ$, we have that $MP = MJ$. Then since segments $YZ$ and $PJ$ are parallel and have a common point $M$ on their perpendicular bisectors, $PJYZ$ is cyclic with $JY = PZ$. It follows that $J$ also lies on circle $AYPZ$ and that $\\angle ZAP = \\angle JAY = \\angle IAY$.\n\n\nSolution 3. As in the previous solutions, let $M$ be the centre of $\\Omega$. Let $L$ be the intersection of $AM$ and $BC$, and let $L'$ be the reflection of $L$ over $YZ$. Let the circle $MYZ$ intersect $AM$ again at $T$.\n\n![](attached_image_3.png)\n\nNote that as $M$ is the midpoint of $\\overparen{BC}$ on circle $ABC$ and $L$ is the foot of the bisector of $\\angle BAC$, we have that $MA \\cdot ML = MI^2 = MY^2$. It follows by power of a point that $MY$ is tangent to circle $ALY$, so $\\angle LAY = \\angle LYM$. Using directed angles, we then have that\n$$\n\\Varangle AYT = \\Varangle MTY - \\Varangle MAY = \\Varangle MZY - \\Varangle LYM = \\Varangle ZYM - \\Varangle LYM = \\Varangle ZYL = \\Varangle L'YZ,\n$$\nwhere we use the fact that $MY = MZ$ and that $L$ and $L'$ are symmetric about $YZ$. Thus, $YT$ and $YL'$ are isogonal in $\\angle AYZ$. Analogously, we have that $ZT$ and $ZL'$ are isogonal in $\\angle YZA$. This means that $T$ and $L'$ are isogonal conjugates in triangle $AYZ$, which allows us to conclude that $\\angle ZAK = \\angle IAY$ since $L'$ lies on $AK$ and $T$ lies on $AI$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32189,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with $AB < AC$, and let $\\Gamma$ be the circumcircle of $ABC$. Points $X$ and $Y$ lie on $\\Gamma$ so that $XY$ and $BC$ intersect on the external angle bisector of $\\angle BAC$. Suppose that the tangents to $\\Gamma$ at $X$ and $Y$ intersect at a point $T$ on the same side of $BC$ as $A$, and that $TX$ and $TY$ intersect $BC$ at $U$ and $V$, respectively. Let $J$ be the centre of the excircle of triangle $TUV$ opposite the vertex $T$.\nProve that $AJ$ bisects $\\angle BAC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Claim. Quadrilateral $WXYZ$ is cyclic, and its circumcentre is $J$.\nProof. As $N$ is the midpoint of $\\widehat{BAC}$, $W$ and $Z$ lie on $BC$, and $X$ and $Y$ are the second intersections of $NW$ and $NZ$ with $\\Gamma$, we have that $WXYZ$ is cyclic.\nLet the parallel to $BC$ through $N$ intersect $TU$ and $TV$ at $U'$ and $V'$, respectively. Then $U'$ is the intersection of the tangents to $\\Gamma$ at $N$ and $X$, so $U'N = U'X$. As $NU' \\parallel BC$, $U'NX$ is similar to $UWX$, so $UW = UX$ as well. Hence, the perpendicular bisector of $WX$ is the internal bisector of $\\angle XUW$, which is the external bisector of $\\angle VUT$. Analogously, the perpendicular bisector of $YZ$ is the external bisector of $\\angle TVU$. This means that the circumcentre of $WXYZ$ is the intersection of the external bisectors of $\\angle VUT$ and $\\angle TVU$, which is $J$. $\\square$\n\n![](attached_image_1.png)\n\nLet $AN$ intersect $BC$ at $L$, so $XY$ passes through $L$ as well. By power of a point from $L$ to $\\Gamma$ and circle $WXYZ$, we have that $LA \\cdot LN = LX \\cdot LY = LW \\cdot LZ$, so $WANZ$ is also cyclic. Thus, $A$ is the Miquel point of quadrilateral $WXYZ$. As $WXYZ$ is cyclic with circumcentre $J$ and its opposite sides $WX$ and $YZ$ intersect at $N$, we have that $AN \\perp AJ$. Since $AN$ is the external bisector of $\\angle BAC$, this implies that $AJ$ is the internal bisector of $\\angle BAC$.\n\n\nSolution 2:\n\nLet the internal and external angle bisectors of $\\angle BAC$ intersect $BC$ at $K$ and $L$, respectively. Let $AK$ intersect circle $ABC$ again at $M$, and let $D$ be the intersection of the tangents to $\\Gamma$ at $B$ and $C$. Let $\\Omega$ be the $T$-excircle of $TUV$, and let $\\omega$ be the incircle of $DBC$.\n\nClaim. The points $T$, $K$, and $D$ are collinear.\nProof. Note that $BC$ and $XY$ are the polars of $T$ and $D$ with respect to $\\Gamma$. By La Hire's Theorem, $TD$ is the polar of $L$ with respect to $\\Gamma$. As $(B, C ; K, L) = -1$, $K$ also lies on the polar of $L$, thus proving the collinearity. $\\square$\n\nClaim. The incentre of $DBC$ is $M$.\nProof. We have that $\\angle MBC = \\angle MAC = \\frac{1}{2} \\angle BAC = \\frac{1}{2} \\angle DBC$, so $BM$ bisects $\\angle DBC$. Similarly, $CM$ bisects $\\angle BCD$, so $M$ is the incentre of $DBC$. $\\square$\n\n![](attached_image_2.png)\n\nClaim. The intersection of the common external tangents of $\\Omega$ and $\\omega$ is $K$.\nProof. Let $K'$ be the intersection of the common external tangents of $\\Omega$ and $\\omega$. As $\\Omega$ and $\\omega$ are both tangent to $BC$ and lie on the same side of $BC$ opposite to $A$, $K'$ lies on $BC$. As $T$ is the intersection of the common external tangents of $\\Gamma$ and $\\Omega$ and $D$ is the intersection of the common external tangents of $\\Gamma$ and $\\omega$, by Monge's theorem $K'$ lies on $TD$. As $K'$ lies on both $BC$ and $TD$, it is the same point as $K$. $\\square$\n\nHence, $K$ is collinear with the centres of $\\Omega$ and $\\omega$, which are $M$ and $J$, respectively. As $K$ and $M$ both lie on the bisector of $\\angle BAC$, so does $J$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32190,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB < AC < BC$, and let $D$ be a point in the interior of segment $BC$. Let $E$ be a point on the circumcircle of triangle $ABC$ such that $A$ and $E$ lie on opposite sides of line $BC$ and $\\angle BAD = \\angle EAC$. Let $I, I_{B}, I_{C}, J_{B}$, and $J_{C}$ be the incentres of triangles $ABC, ABD, ADC, ABE$, and $AEC$, respectively.\nProve that $I_{B}, I_{C}, J_{B}$, and $J_{C}$ are concyclic if and only if $AI, I_{B}J_{C}$, and $J_{B}I_{C}$ concur.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $X$ be the intersection of $I_{B}J_{C}$ and $J_{B}I_{C}$. We will prove that, provided that $AB < AC < BC$, the following two conditions are equivalent:\n(1) $AX$ bisects $\\angle BAC$;\n(2) $I_{B}, I_{C}, J_{B}$, and $J_{C}$ are concyclic.\nLet circles $AIB$ and $AIC$ meet $BC$ again at $P$ and $Q$, respectively. Note that $AB = BQ$ and $AC = CP$ because the centres of circles $AIB$ and $AIC$ lie on $CI$ and $BI$, respectively. Thus, $B, P, Q$, and $C$ are collinear in this order as $BQ + PC = AB + AC > BC$ by the triangle inequality.\n\n**Claim 1.** Points $P, J_{B}$, and $I_{C}$ are collinear, and points $Q, I_{B}$, and $J_{C}$ are collinear.\n\nProof. We have that\n$$\n\\angle AJ_{B}B = 90^{\\circ} + \\frac{1}{2} \\angle AEB = 90^{\\circ} + \\frac{1}{2} \\angle ACB = \\angle AIB = \\angle APB,\n$$\nso $ABJ_{B}P$ is cyclic. As $A$ is the centre of spiral similarity between $ABE$ and $ADC$, it is also the centre of spiral similarity between $ABJ_{B}$ and $ADI_{C}$. Hence, $A$ is the Miquel point of self-intersecting quadrilateral $BDI_{C}J_{B}$, so $P$ lies on $J_{B}I_{C}$. Analogously, we have that $Q$ lies on $I_{B}J_{C}$. $\\square$\n\n![](attached_image_1.png)\n\nThroughout the rest of the solution, we will use directed angles.\n\n**Proof of (1) $\\implies$ (2).** We assume that (1) holds.\nClaim 1 and the similarities $ABDI_{B} \\sim AECJ_{C}$ and $ABEJ_{B} \\sim ADCI_{C}$ tell us that\n$$\n\\nsucceq I_{B}XI_{C} = \\Varangle J_{C}QC + \\Varangle BPJ_{B} = \\Varangle J_{C}AC + \\Varangle BAJ_{B} = \\Varangle I_{B}AD + \\Varangle DAI_{C} = \\Varangle I_{B}AI_{C},\n$$\nso $AI_{B}XI_{C}$ is cyclic. Also, as $X \\in AI$, we have that\n$$\n\\nsucceq I_{B}AX = \\Varangle BAI - \\Varangle BAI_{B} = \\Varangle I_{B}AI_{C} - \\Varangle IB_{A}D = \\Varangle DAI_{C}.\n$$\nUsing these, we have that\n$$\n\\nsucceq I_{B}I_{C}P = \\Varangle I_{B}AX = \\Varangle DAI_{C} = \\Varangle BAJ_{B} = \\Varangle BPJ_{B},\n$$\nso $I_{B}I_{C} \\parallel BC$. Hence,\n$$\n\\nsucceq I_{B}I_{C}J_{B} = \\Varangle BPJ_{B} = \\Varangle BIJ_{B} = \\Varangle I_{B}IJ_{B},\n$$\nso $II_{B}J_{B}I_{C}$ is cyclic. Analogously, we have that $II_{C}J_{C}I_{B}$ is cyclic, so $I_{B}J_{B}J_{C}I_{C}$ is cyclic, thus proving (2).\n\n**Proof of (2) $\\implies$ (1).** We assume that (2) holds.\n\n**Claim 2.** Circles $IBC, IJ_{B}I_{C}$, and $II_{B}J_{C}$ are tangent at $I$.\n\nProof. Using the cyclic quadrilateral $BIJ_{B}P$, we have that\n$$\n\\nsucceq IBC = \\Varangle IBP = \\Varangle IJ_{B}P = \\Varangle IJ_{B}I_{C}.\n$$\nAs $C, I_{C}$, and $I$ are collinear, the tangents to circles $I_{B}I_{C}$ and $IBC$ at $I$ coincide, so circles $IJ_{B}I_{C}$ and $IBC$ are tangent at $I$. Analogously, circles $II_{B}J_{C}$ and $IBC$ are tangent at $I$, so all three circles are tangent at $I$.\n\n**Claim 3.** Point $I$ lies on circle $I_{B}J_{B}J_{C}I_{C}$.\n\nProof. Suppose that $I$ does not lie on circle $I_{B}J_{B}J_{C}I_{C}$. Then the circles $II_{B}J_{C}, IJ_{B}I_{C}$, and $I_{B}J_{B}J_{C}I_{C}$ are distinct. We apply the radical axis theorem to these three circles. By Claim 2, the radical axis of circles $II_{B}J_{C}$ and $IJ_{B}I_{C}$ is the tangent to circle $IBC$ at $I$. As $I_{B}J_{C}$ and $J_{B}I_{C}$ intersect at $X$, $IX$ must be tangent to circle $IBC$.\n\nHowever, by Claim 1 we have that $X$ is the intersection of $PI_{C}$ and $QI_{B}$. As $D$ lies on the interior of segment $BC$, $I_{B}$ lies on the interior of segment $BI$ and $I_{C}$ lies on the interior of segment $CI$. Hence, $I_{B}, P, Q$, and $I_{C}$ all lie on the perimeter of triangle $IBC$ in this order, so $X$ must be in the interior of triangle $IBC$. This means that $IX$ cannot be tangent to circle $BIC$, so $I$ must lie on circle $I_{B}J_{B}J_{C}I_{C}$.\n\nBy Claims 2 and 3, circles $II_{B}I_{C}$ and $IBC$ are tangent, so $I_{B}I_{C} \\parallel BC$. Since $I_{B}J_{B}J_{C}I_{C}$ is cyclic, we have that\n$$\n\\nsucceq PJ_{B}J_{C} = \\Varangle I_{C}J_{B}J_{C} = \\Varangle I_{C}I_{B}J_{C} = \\Varangle PQI_{B} = \\Varangle PQJ_{C},\n$$\nso $PJ_{B}J_{C}Q$ is cyclic. By the radical axis theorem on circles $AIPJ_{B}, AIQJ_{C}$, and $PJ_{B}J_{C}Q$, we have that $AI, I_{B}J_{C}$, and $J_{B}I_{C}$ concur at $X$, thus proving (1).\nLet $X$ be the intersection of $I_{B}J_{C}$ and $J_{B}I_{C}$. As in Solution 1, we will prove that conditions (1) and (2) are equivalent. To do so, we introduce the new condition:\n(3) $I_{B}I_{C} \\parallel BC$\nand show that (3) is equivalent to both (1) and (2), provided that $AB < AC < BC$.\nNote that $ABD \\stackrel{+}{\\sim} AEC$ and $ABE \\stackrel{+}{\\sim} ADC$, where $\\stackrel{+}{\\sim}$ denotes positive similarity. We will make use of the following fact.\n\n**Fact.** For points $P, P_{1}, P_{2}, P_{3}$, and $P_{4}$, the positive similarities\n$$\nPP_{1}P_{2} \\stackrel{\\ddagger}{\\sim} PP_{3}P_{4} \\quad \\text{ and } \\quad PP_{1}P_{3} \\stackrel{\\ddagger}{\\sim} PP_{2}P_{4}\n$$\nare equivalent.\n\n**Proof of (1) $\\iff$ (3).** Let $AI_{B}$ and $AI_{C}$ meet $BC$ at $S$ and $T$, respectively. Let $AJ_{B}$ meet $BE$ at $K$, $AJ_{C}$ meet $CE$ at $L$, and $KT$ and $SL$ meet at $Y$.\n\n![](attached_image_2.png)\n\n**Claim 1.** Line $AY$ bisects $\\angle BAC$.\n\nProof. Let $Y'$ be the intersection of $KT$ and the bisector of $\\angle BAC$. As\n$$\n\\angle BAK = \\frac{1}{2} \\angle BAE = \\frac{1}{2} \\angle DAC = \\angle TAC,\n$$\n$AY'$ also bisects $\\angle KAT$. Hence, $Y'$ is the foot of the bisector of $\\angle KAT$ in triangle $AKT$. Using the Fact, we have that\n$$\n\\begin{aligned}\nABE \\sim ADC &\\Longrightarrow ABEK \\sim ADCT \\\\\n&\\Longrightarrow ABD \\sim AKT \\sim AEC \\\\\n&\\Longrightarrow ABDS \\sim AKTY' \\sim AECL \\\\\n&\\Longrightarrow ABEK \\sim ASLY' \\sim ADCT.\n\\end{aligned}\n$$\nAs $K$ lies on $BE$, we have that $Y'$ lies on $SL$, so $Y = Y'$ and $AY$ bisects $\\angle BAC$. $\\square$\n\nWe show that $X$ lies on $AY$ if and only if $I_{B}I_{C} \\parallel BC$, which implies the equivalence of (1) and (3) by Claim 1. Let $AY$ meet $I_{B}J_{C}$ and $J_{B}I_{C}$ at $X_{1}$ and $X_{2}$, respectively. As $ABD$ and $AEC$ are similar, we have that $\\frac{AI_{B}}{AS} = \\frac{AJ_{C}}{AL}$, so $I_{B}J_{C} \\parallel SL$. Analogously, we have that $J_{B}I_{C} \\parallel KT$. Hence, $X_{1}$ and $X_{2}$ coincide with $X$ if and only if\n$$\n\\frac{AI_{B}}{AS} = \\frac{AX_{1}}{AY} = \\frac{AX_{2}}{AY} = \\frac{AI_{C}}{AT},\n$$\nwhich is equivalent to $I_{B}I_{C} \\parallel BC$. $\\square$\n\n**Proof of (2) $\\iff$ (3).** Let $AJ_{B}$ and $AJ_{C}$ meet circle $ABC$ at $M$ and $N$, respectively, and let $I_{B}'$ and $I_{C}'$ be the $A$-excentres of $ABD$ and $ADC$, respectively.\n\n![](attached_image_3.png)\n\n**Claim 2.** Lines $I_{B}I_{C}, I_{B}'I_{C}'$, and $BC$ are concurrent or pairwise parallel.\n\nProof. We work in the projective plane. Let $I_{B}I_{C}$ and $I_{B}'I_{C}'$ meet $BC$ at $Z$ and $Z'$, respectively. Note that $Z$ is the intersection of the external common tangents of the incircles of $ABD$ and $ADC$ and $AD$ is a common internal tangent of the incircles of $ABD$ and $ADC$, so $\\left(AD, AZ ; AI_{B}, AI_{C}\\right) = -1$. Applying the same argument to the $A$-excircles of $ABD$ and $ADC$ gives $\\left(AD, AZ' ; AI_{B}', AI_{C}'\\right) = -1$, which means that $Z = Z'$. Thus, $I_{B}I_{C}, I_{B}'I_{C}'$, and $BC$ concur, possibly at infinity. $\\square$\n\n![](attached_image_4.png)\n\n**Claim 3.** Lines $J_{B}I_{C}$ and $CM$ are parallel, and lines $I_{B}J_{C}$ and $BN$ are parallel.\n\nProof. Using the Fact, we have that\n$$\nABE \\stackrel{+}{\\sim} ADC \\Longrightarrow ABEJ_{B} \\stackrel{+}{\\sim} ADCI_{C} \\Longrightarrow AJ_{B}I_{C} \\stackrel{+}{\\sim} ABD.\n$$\nThus, $\\angle(BD, J_{B}I_{C}) = \\angle BAJ_{B} = \\angle BCM$, so $J_{B}I_{C} \\parallel CM$. Similarly, we have that $I_{B}J_{C} \\parallel BN$. $\\square$\n\n**Claim 4.** The centre of spiral similarity between $J_{B}J_{C}$ and $I_{B}'I_{C}'$ is $A$.\n\nProof. As $I_{B}$ and $I_{B}'$ are respectively the incentre and $A$-excentre of triangle $ABD$, we have that $ABI_{B}' \\stackrel{+}{\\sim} AI_{B}D$. Using the similarity $ABD \\stackrel{+}{\\sim} AEC$, this means that $ABI_{B}' \\stackrel{+}{\\sim} AJ_{C}C$, so $AB \\cdot AC = AI_{B}' \\cdot AJ_{C}$ and $\\angle BAI_{B}' = \\angle J_{C}AC$. Similarly, we have that $AB \\cdot AC = AJ_{B} \\cdot AI_{C}'$ and $\\angle BAJ_{B} = \\angle I_{C}'AC$. Together, these imply that $AI_{B}' \\cdot AJ_{C} = AJ_{B} \\cdot AI_{C}'$ and $\\angle J_{B}AJ_{C} = \\angle I_{B}'AI_{C}'$, so $AJ_{B}J_{C} \\stackrel{+}{\\sim} AI_{B}'I_{C}'$. $\\square$\n\nWe proceed using directed angles. By Claim 3, we have that $I_{B}J_{B}J_{C}I_{C}$ is cyclic if and only if\n$$\n\\begin{aligned}\n\\npreceq I_{B}I_{C}J_{B} = \\Varangle I_{B}J_{C}J_{B} &\\Longleftrightarrow \\Varangle I_{B}I_{C}J_{B} + \\nrightarrow MCB = \\Varangle I_{B}J_{C}J_{B} + \\Varangle MNB \\\\\n&\\Longleftrightarrow \\Varangle(I_{B}I_{C}, BC) = \\Varangle(MN, J_{B}J_{C}).\n\\end{aligned}\n$$\nBy Claim 4, we have that\n$$\n\\begin{aligned}\n\\nsucceq(J_{B}J_{C}, I_{B}'I_{C}') &= \\Varangle J_{B}AI_{B}' \\\\\n&= \\Varangle BAI_{B} + \\Varangle MAB \\\\\n&= \\Varangle EAJ_{C} + \\Varangle MAB \\\\\n&= \\Varangle NAC + \\Varangle MAB \\\\\n&= \\Varangle(MN, BC),\n\\end{aligned}\n$$\nwhich is equivalent to $\\nsucceq(BC, I_{B}'I_{C}') = \\Varangle(MN, J_{B}J_{C})$. Thus, $I_{B}J_{B}J_{C}I_{C}$ is cyclic if and only if\n$$\n\\begin{equation*}\n\\nsucceq(I_{B}I_{C}, BC) = \\Varangle(BC, I_{B}'I_{C}'). \\tag{*}\n\\end{equation*}\n$$\nSuppose that $I_{B}I_{C}$ is parallel to $BC$. By Claim 2, $I_{B}'I_{C}'$ is also parallel to $BC$, so we have that $\\Varangle(I_{B}I_{C}, BC) = \\Varangle(BC, I_{B}'I_{C}') = 0^{\\circ}$. Thus, (*) is satisfied, so $I_{B}J_{B}J_{C}I_{C}$ is cyclic.\n\n![](attached_image_5.png)\n\nSuppose now that $I_{B}I_{C}$ is not parallel to $BC$ while $I_{B}J_{B}J_{C}I_{C}$ is cyclic. By Claim 2, $I_{B}I_{C}$, $I_{B}'I_{C}'$, and $BC$ concur at a point $Z$. As $I_{B}$ and $I_{C}$ lie on segments $BI$ and $CI$, $Z$ must lie outside segment $BC$. Since $A$ is the intersection of the common external tangents of the incircle and $A$-excircle of $ABD$ and $ZD$ is a common internal tangent of the incircle and $A$-excircle of $ABD$, we have that $\\left(ZA, ZD ; ZI_{B}, ZI_{B}'\\right) = -1$. By (*), $ZD$ bisects $\\angle I_{B}ZI_{B}'$, so $\\angle AZD = 90^{\\circ}$: that is, $Z$ is the foot from $A$ to $BC$. But this implies that $\\angle ABC$ or $\\angle BCA$ is obtuse, contradicting the fact that $AB < AC < BC$. $\\square$\nLet $\\omega_{B}$ and $\\omega_{C}$ denote circles $AIB$ and $AIC$, respectively. Introduce $P, Q$ and $X$ as in Solution 1 and recall from Claim 1 in Solution 1 that $P, J_{B}$ and $I_{C}$ are collinear with $J_{B}$ lying on $\\omega_{B}$. From this, we can define $J_{B}$ and $I_{C}$ in terms of $X$ by $I_{C} = XP \\cap CI$ and $J_{B} \\neq P$ as the second intersection of line $XP$ with $\\omega_{B}$. Similarly, we can define $I_{B} = XQ \\cap BI$ and $J_{C} \\neq Q$ as the second intersection of line $XQ$ with $\\omega_{C}$. Note that this now detaches the definitions of points $I_{B}, I_{C}, J_{B}$, and $J_{C}$ from points $D$ and $E$.\n\nLet $\\ell$ be a line passing through $I$. We now allow $X$ to vary along $\\ell$ while fixing $\\triangle ABC$ and points $I, P$, and $Q$. We use the definitions from above to construct $I_{B}, I_{C}, J_{B}$, and $J_{C}$. We will classify all cases where these four points are concyclic. Throughout the rest of the solution we use directed angles and directed lengths.\n\nFor nondegeneracy reasons, we exclude cases where $X = I$ and $X$ lies on line $BC$, which means that $I_{B}, J_{B} \\neq B$ and $I_{C}, J_{C} \\neq C$. We also exclude the cases where $\\ell$ is tangent to either $\\omega_{B}$ or $\\omega_{C}$. Similar results hold in these cases and they can be treated as limit cases.\n\n![](attached_image_6.png)\n\n**Claim 1.** Line $I_{B}J_{B}$ passes through a fixed point on $\\omega_{B}$, and line $I_{C}J_{C}$ passes through a fixed point on $\\omega_{C}$ as $X$ varies on $\\ell$.\n\nProof. Let $U \\neq J_{B}$ be the second intersection of $I_{B}J_{B}$ with $\\omega_{B}$. We have by the law of sines that\n$$\n\\frac{\\sin \\measuredangle IJ_{B}U}{\\sin \\measuredangle UJ_{B}B} = \\frac{\\sin \\measuredangle IJ_{B}I_{B}}{\\sin \\measuredangle I_{B}J_{B}B} = \\frac{\\sin \\measuredangle J_{B}II_{B}}{\\sin \\measuredangle J_{B}BI_{B}} \\cdot \\frac{II_{B}}{I_{B}B} = \\frac{\\sin \\measuredangle J_{B}IB}{\\sin \\measuredangle J_{B}BI} \\cdot \\frac{II_{B}}{I_{B}B} = \\frac{\\sin \\measuredangle XPQ}{\\sin \\measuredangle XPI} \\cdot \\frac{II_{B}}{I_{B}B}.\n$$\nWe also have\n$$\n\\frac{II_{B}}{I_{B}B} = \\frac{\\sin \\measuredangle IQI_{B}}{\\sin \\measuredangle I_{B}QB} \\cdot \\frac{|IQ|}{|BQ|} = \\frac{\\sin \\measuredangle IQX}{\\sin \\measuredangle XQP} \\cdot \\frac{|IQ|}{|BQ|}.\n$$\nCombining these and applying Ceva's Theorem in $\\triangle PIQ$ with point $X$, we get\n$$\n\\frac{\\sin \\measuredangle IJ_{B}U}{\\sin \\measuredangle UJ_{B}B} = \\frac{\\sin \\measuredangle XPQ}{\\sin \\measuredangle XPI} \\cdot \\frac{\\sin \\measuredangle IQX}{\\sin \\measuredangle XQP} \\cdot \\frac{|IQ|}{|BQ|} = \\frac{\\sin \\measuredangle XIQ}{\\sin \\measuredangle XIP} \\cdot \\frac{|IQ|}{|BQ|} = \\frac{\\sin \\measuredangle (\\ell, IQ)}{\\sin \\measuredangle (\\ell, IP)} \\cdot \\frac{|IQ|}{|BQ|},\n$$\nwhich is independent of the choice of $X$ on $\\ell$. As $\\nsucceq IJ_{B}U + \\measuredangle UJ_{B}B = \\measuredangle IJ_{B}B = \\measuredangle IAB$ is fixed, this is enough to show point $U$ is fixed on $\\omega_{B}$.\n\nSimilarly, if we define $V \\neq J_{C}$ to be the second intersection of $I_{C}J_{C}$ with $\\omega_{C}$, we get that $V$ is fixed on $\\omega_{C}$.\n\nLet $G \\neq X$ and $H \\neq X$ be the second intersections of $\\ell$ with $\\omega_{B}$ and $\\omega_{C}$, respectively which exist as we are assuming $\\ell$ is not tangent to either $\\omega_{B}$ or $\\omega_{C}$.\n\n**Claim 2.** Points $U, G$ and $Q$ are collinear and points $V, H$ and $P$ are collinear.\n\nProof. Taking $X = G$, we have $J_{B} = G$ and $I_{B} = XQ \\cap BI$. Both of these points lie on line $QG$ which, by Claim 1, shows that $U, G, Q$ are collinear. Similarly, $V, H, P$ are collinear.\n\n**Claim 3.** Points $I_{B}, I_{C}, J_{B}, J_{C}$ are concyclic if and only if points $P, Q, G, H$ are concyclic. In particular, this depends only on $\\ell$, not on the choice of $X$ on $\\ell$.\n\nProof. We have that\n$$\n\\begin{aligned}\n&\\Varangle I_{C}J_{B}I_{B} = \\Varangle PJ_{B}U = \\Varangle PGU = \\Varangle PGQ \\\\\n&\\Varangle I_{C}J_{C}I_{B} = \\Varangle VJ_{C}Q = \\Varangle VHQ = \\Varangle PHQ.\n\\end{aligned}\n$$\nThus $\\nsucceq I_{C}J_{B}I_{B} = \\Varangle I_{C}J_{C}I_{B} \\Longleftrightarrow \\Varangle PGQ = \\measuredangle PHQ$ which proves the claim.\n\n**Claim 4.** $P, Q, G, H$ are concyclic if and only if $\\ell \\in \\{IA, IP, IQ, t\\}$ where $t$ is the tangent to circle $BIC$ at $I$.\n\nProof. When $\\ell = IA$, we have $G = H = A$ so the cyclic condition from Claim 3 holds. Similarly, when $\\ell = IP$ or $\\ell = IQ$, $G = P$ or $H = Q$, respectively, so again the cyclic condition holds.\n\nNow, consider the case where $\\ell \\notin \\{IP, IQ\\}$. In this case it is straightforward to see that the four points $P, Q, G$, and $H$ are distinct. We then have that $\\measuredangle QPG = \\measuredangle BPG = \\measuredangle BIG$, so\n$$\nPQGH \\text{ concyclic } \\Longleftrightarrow \\Varangle QHG = \\Varangle QPG \\Longleftrightarrow \\Varangle QHG = \\Varangle BIG \\Longleftrightarrow QH \\parallel BI.\n$$\nWe also have that $\\measuredangle CQH = \\measuredangle CIH$, so\n$$\n\\ell \\text{ tangent to circle } BIC \\Longleftrightarrow \\Varangle CIH = \\Varangle CBI \\Longleftrightarrow \\Varangle CQH = \\Varangle CBI \\Longleftrightarrow QH \\parallel BI.\n$$\nHence, in this case $P, Q, G, H$ are concyclic if and only if $\\ell$ is tangent to circle $BIC$, as claimed.\n\nWe now revert to using points $D$ and $E$ to define points $I_{B}, I_{C}, J_{B}, J_{C}$, and $X$, returning to the original set-up.\n\n**Claim 5.** Let $\\Gamma$ be the circle passing through $P$ and $Q$ that is tangent to $IP$ and $IQ$, which exists as $IP = IQ = IA$. Then $X$ lies on $\\Gamma$. Furthermore, $X$ lies on the same side of $BC$ as $A$ and does not lie on line $BC$.\n\nProof. We have that\n$$\n\\begin{aligned}\n\\nsucceq XPI &= \\Varangle J_{B}PI = \\Varangle J_{B}AI = \\Varangle BAI - \\Varangle BAJ_{B} = \\Varangle J_{B}AJ_{C} - \\Varangle J_{B}AE \\\\\n&= \\Varangle EAJ_{C} = \\Varangle J_{C}AC = \\Varangle J_{C}QC = \\Varangle XQP,\n\\end{aligned}\n$$\nso circle $XPQ$ is tangent to $IP$. Similarly, circle $XPQ$ is tangent to $IQ$, so $X$ lies on $\\Gamma$.\n\nAs $D$ lies in the interior of segment $BC$, $I_{C}$ lies in the interior of segment $CI$. Since $X$ is the second intersection of $PI_{C}$ with $\\Gamma$ and $IP$ is tangent to $\\Gamma$, $X$ lies in the interior of $\\overparen{PQ}$ on $\\Gamma$ on the same side of $BC$ as $A$. This implies the second part of the claim. $\\square$\n\nBy Claim 5, we cannot have $\\ell \\in \\{IP, IQ\\}$ in the original problem. Furthermore, as shown in Claim 2 of Solution 1, we have that $X$ lies inside triangle $IBC$, which means that $\\ell \\neq t$. Thus, the only remaining possibility in Claim 4 is $\\ell = AI$. We then have\n$$\nI_{B}I_{C}J_{B}J_{C} \\text{ concyclic } \\stackrel{\\text{ Claim 3}}{\\Longleftrightarrow} PQGH \\text{ concyclic } \\stackrel{\\text{ Claim 4}}{\\Longleftrightarrow} X \\text{ lies on } AI,\n$$\nfinishing the problem.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32191,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $n$ with the following property: for all positive divisors $d$ of $n$, we have that $d+1 \\mid n$ or $d+1$ is prime.",
    "options": [],
    "answer": "1, 2, 4, 12",
    "solution": "It is easy to verify that $n=1,2,4,12$ all work. We must show they are the only possibilities. We write $n=2^{k} m$, where $k$ is a nonnegative integer and $m$ is odd. Since $m \\mid n$, either $m+1$ is prime or $m+1 \\mid n$.\n\nIn the former case, since $m+1$ is even it must be $2$, so $n=2^{k}$. If $k \\geqslant 3$, we get a contradiction, since $8 \\mid n$ but $9 \\nmid n$. Hence $k \\leqslant 2$, so $n \\in\\{1,2,4\\}$.\n\nIn the latter case, we have $m+1 \\mid 2^{k} m$ and $m+1$ coprime to $m$, and hence $m+1 \\mid 2^{k}$. This means that $m+1=2^{j}$ with $2 \\leqslant j \\leqslant k$ (since $j=1$ gives $m=1$, which was considered earlier).\n\nThen we have $2^{k}+1 \\nmid n$: since $2^{k}+1$ is odd, it would have to divide $m$ but is larger than $m$. Hence, by the condition of the problem, $2^{k}+1$ is prime. If $k=2$, $j$ must be $2$ as well, and this gives the solution $n=12$. Also, $2^{k-1}+1 \\nmid n$ for $k>2$: since it is odd, it would have to divide $m$. However, we have no solutions to $2^{k-1}+1 \\mid 2^{j}-1$ with $j \\leqslant k$: the left-hand side is greater than the right unless $j=k$, when the left-hand side is just over half the right-hand side.\n\nSince we have $2^{k} \\mid n$ and $2^{k}+1 \\nmid n$, and $2^{k-1} \\mid n$ and $2^{k-1}+1 \\nmid n$, we must have $2^{k}+1$ and $2^{k-1}+1$ both prime. However, $2^{a}+1$ is a multiple of three if $a$ is odd, so we must have $2^{k}+1=3$ (impossible as this gives $k=1$) or $2^{k-1}+1=3$, which gives $j=k=2$, whence $n=12$.\nWe proceed as in Solution 1 as far as determining that $n=2^{k}\\left(2^{j}-1\\right)$ with $j \\leqslant k$.\n\nNow, we have $2^{j} \\mid n$ but $2^{j}+1 \\nmid n$, as it is odd and does not divide $2^{j}-1$. Thus $2^{j}+1$ is prime. The theory of Fermat primes tells us we must have $j=2^{h}$ with $h>0$.\n\nThen $2^{2^{h}}-1$ is congruent to $3$ or $6$ (modulo $9$) depending on whether $h$ is odd or even, respectively. In particular it is not divisible by $9$, so $n=2^{k}\\left(2^{2^{h}}-1\\right)$ is not divisible by $9$; so we must have $k \\leqslant 2$, since if $k \\geqslant 3$ then $8 \\mid n$ but $9 \\nmid n$ with $9$ not prime.\nLet $p$ be the smallest integer not dividing $n$. Since $p-1$ is a divisor of $n$, $p$ must be a prime. Let $1 \\leqslant r \\leqslant p-1$ be the remainder of $n$ modulo $p$. Since $p-r<p$, we have $p-r \\mid n$, so we may consider the divisor $d=\\frac{n}{p-r}$.\n\nSince $p \\mid n-r$, we have $p \\mid n+p-r$, whence $p \\mid d+1$. Thus $d+1 \\nmid n$; so it must be prime. On the other hand, this prime is divisible by $p$, so we conclude $d+1=p$, which means that $n=(p-1)(p-r)$.\n\nThen from $p-2, p-3 \\mid n$ we get $(p-2)(p-3) \\mid 2(p-r)$, from which we find\n$$(p-2)(p-3) \\leqslant 2(p-r) \\leqslant 2(p-1) .$$\nSolving this quadratic inequality gives $p \\leqslant 5$, which means that $n \\in\\{1,2,4,8,12,16\\}$. Of this set, $n=8$ and $n=16$ are not solutions.\nWe suppose that $n$ is not $1$ or $2$.\nSince $n \\mid n$ and $n+1 \\nmid n$, we know that $n+1$ is prime. Thus it is odd, so $2 \\mid n$; as $n>2$, we have $\\left.\\frac{n}{2} \\right\\rvert\\, n$ and $\\frac{n}{2}+1 \\nmid n$, so $\\frac{n}{2}+1$ is prime. Thus it is also odd, so $4 \\mid n$.\n\nWe must then have $\\left.\\frac{n}{4}+1 \\right\\rvert\\, n$ or $\\frac{n}{4}+1$ prime.\n\nIn the former case, $\\frac{n}{4} \\left\\lvert\\, 4\\left(\\frac{n}{4}+1\\right)-n\\right.$, so $\\left.\\frac{n}{4}+1 \\right\\rvert\\, 4$. This means that $n=4$ or $n=12$.\n\nIn the latter case, $\\frac{n}{4}+1$ must be odd if $n \\neq 4$. Thus we have $n=8 m$ where $2 m+1,4 m+1, 8 m+1$ are all prime; $n=8$ does not work, so $3 \\mid m$ (otherwise one of those numbers would be divisible by $3$). Thus $24 \\mid n$, so $25 \\mid n$ as $25$ is not prime.\n\nNow suppose that $p$ is the least positive integer not dividing $n$: as in Solution 3 we know that $p$ is prime, and what we have done so far shows that $p \\geqslant 7$. If $p^{2}-1=(p-1)(p+1)$ is the product of coprime integers less than $p$, it divides $n$, and $p^{2}$ is not prime so also divides $n$ (a contradiction); $p-1$ and $p+1$ are even and have no common factor higher than $2$, so all odd prime power divisors of their product are less than $p$ and the only case where $p^{2}-1$ is not a product of coprime integers less than $p$ is when one of $p-1$ and $p+1$ is a power of $2$, say $2^{m}$ (with $m \\geqslant 3$). If $p=2^{m}-1$, then $3 p-1=4\\left(3 \\times 2^{m-2}-1\\right)$ and $3 \\times 2^{m-2}-1$ is an odd integer less than $p$, so $3 p-1 \\mid n$ and so $3 p \\mid n$. Finally, if $p=2^{m}+1$, then $m$ is even and $2 p-1=2^{m+1}+1$ is a multiple of $3$; the only case where it is a power of $3$ is when $m=2$, but we have $m \\geqslant 3$, so $2 p-1$ is a product of coprime integers less than $p$ and again we have a contradiction.\nAs in Solution 4, we deduce that if $n>2$ then $n$ must be even. We write $n=2 \\cdot 3^{k} \\cdot r$, where $k$ is a nonnegative integer and $3 \\nmid r$.\n\nSince $r$ and $2 r$ are both different and nonzero modulo $3$, one of them must be congruent to $2$ modulo $3$. We'll say that it is $a r$, where $a \\in\\{1,2\\}$.\n\nSince $a r \\mid n$, we must have that $a r+1$ is either prime or a factor of $n$. In the first case, $a r+1=3$ because $3 \\mid a r+1$, and so $n=2 \\cdot 3^{k} \\cdot r$, where $r=2 / a$ is $1$ or $2$. Noting that we must have $k \\leqslant 1$ (else $9 \\mid n$ but $10 \\nmid n$), we can examine cases to deduce that $n \\in\\{2,4,12\\}$ are the only possibilities.\n\nOtherwise, $a r+1 \\mid n$. Since $a r+1$ is coprime to $r$, we must in fact have that $a r+1 \\mid 2 \\cdot 3^{k}$, and since $3 \\mid a r+1$ by assumption we deduce that $k \\geqslant 1$. In particular, $3^{k}+1$ is an even number that is at least $4$, so is not prime and must divide $n$. As it is coprime to $3$, we must in fact have $3^{k}+1 \\mid 2 r$.\n\nLet $q_{1}$ and $q_{2}$ be such that $q_{1}(a r+1)=2 \\cdot 3^{k}$ and $q_{2}\\left(3^{k}+1\\right)=2 r$. We have that $q_{1} a r<2 \\cdot 3^{k}$ and $q_{2} 3^{k}<2 r$, and multiplying these together gives $q_{1} q_{2} a<4$.\n\nIf $a=2$ then $q_{1}=q_{2}=1$, so $2 r+1=2 \\cdot 3^{k}$, which is not possible (considering both sides modulo $2$).\n\nIf $a=1$ then $r$ must be equivalent to $2$ modulo $3$, so $q_{2}\\left(3^{k}+1\\right)=2 r$ gives that $q_{2}$ is equivalent to $1$ modulo $3$, whence $q_{2}=1$. So we deduce that $2 r=3^{k}+1$. Thus, we deduce that $q_{1}\\left(3^{k}+3\\right)=4 \\cdot 3^{k}$, which rearranges to give $3^{k-1}\\left(4-q_{1}\\right)=q_{1}$, whence $3^{k-1} \\leqslant q_{1}<4$ and so $k \\leqslant 2$. We can examine cases to deduce that $n=12$ is the only possibility.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32192,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all sequences $a_{1}, a_{2}, \\ldots$ of positive integers such that, for any pair of positive integers $m \\leqslant n$, the arithmetic and geometric means\n$$\n\\frac{a_{m}+a_{m+1}+\\cdots+a_{n}}{n-m+1} \\quad \\text{ and } \\quad \\left(a_{m} a_{m+1} \\cdots a_{n}\\right)^{\\frac{1}{n-m+1}}\n$$\nare both integers.",
    "options": [],
    "answer": "All constant sequences of positive integers: a_m = c for all m, for any fixed positive integer c.",
    "solution": "Solution 1. Without loss of generality, we may divide all elements of $\\mathcal{S}$ by any common factor, after which they cannot all be even. As $a \\nmid b+2 c$ for $a$ even and $b$ odd, the elements of $\\mathcal{S}$ are all odd.\nWe now divide into three cases:\nCase 1: $|\\mathcal{S}|=1$.\nThe set $\\mathcal{S}=\\{t\\}$ clearly works.\nCase 2: $|\\mathcal{S}|=2$.\nSay $\\mathcal{S}=\\{r, s\\}$ with $r<s$, so either $s \\mid r+2 r$ or $s \\mid r+2 s$, and in either case $s \\mid 3 r$. We cannot have $s=3 r / 2$ as we assumed that $r$ is odd, so $s=3 r$ and $\\mathcal{S}=\\{r, 3 r\\}$, which clearly works by examining cases for $a$ and $b$.\nCase 3: $|\\mathcal{S}| \\geqslant 3$.\nIf all elements of $\\mathcal{S}$ are odd then for any $b, c \\in \\mathcal{S}, b+2 c \\not \\equiv b(\\bmod 4)$. If $a \\mid b+2 c$ with $a \\equiv b(\\bmod 4)$, this means there exists $k$ with $b+2 c=k a$ and $k \\equiv 3(\\bmod 4)$, so $k \\geqslant 3$. If $a$ is the greatest element of $\\mathcal{S}$ and $b<a$, we have $b+2 c<3 a$, a contradiction. Thus when $a$ is the greatest element, no $b<a$ has $b \\equiv a(\\bmod 4)$ (and thus all elements other than the greatest are congruent modulo 4).\nLet $d$ and $e$ be the largest and second largest element of $\\mathcal{S}$ respectively. Let $f \\neq d, e$ be any other element of $\\mathcal{S}$. There is some $c \\in \\mathcal{S}$ with $e \\mid f+2 c$, and $e \\not \\equiv f+2 c(\\bmod 4)$, so $f+2 c \\geqslant 3 e$, so $c>e$. Since $e$ is the second largest element of $\\mathcal{S}, c=d$, so $e \\mid f+2 d$, and this holds for all $f \\in \\mathcal{S}$ with $f<e$, but can only hold for at most one such $f$. So $|\\mathcal{S}| \\leqslant 3$.\nHence the elements of $\\mathcal{S}$ are $d>e>f$, and by the discussion above without loss of generality we may suppose these elements are all odd, $e \\equiv f(\\bmod 4)$ and $d \\not \\equiv e(\\bmod 4)$. We have above that $e \\mid f+2 d$. Furthermore, there exists some $c \\in \\mathcal{S}$ with $d \\mid f+2 c$, and $c \\neq d$ as $d>f$ so $d \\nmid f$, so $c \\leqslant e$; as $f+2 e<3 e$, we have $e>d / 3$. Since $f+2 c$ is odd and $f+2 c<3 d$, we have $f+2 c=d$.\nSubcase 3.1: $c=f$.\nHere $d=3 f$ and $e \\mid f+2 d=7 f$. As $e>f$ and $e \\equiv f(\\bmod 4)$, we have $e=7 f / 3$ and the elements are some multiples of $\\{3,7,9\\}$. But $a=7$ and $b=9$ have no corresponding value of $c$.\nSubcase 3.2: $c=e$.\nHere $d=f+2 e$ and $e \\mid f+2 d=3 f+4 e$ so $e \\mid 3 f$. But this is not possible with $e>f$ and $e \\equiv f(\\bmod 4)$.\nLet $d$ be the largest element. For any $e \\in \\mathcal{S}$ with $e \\neq d$ there must be a $f \\in \\mathcal{S}$ such that $d \\mid e+2 f$. This implies $2 f \\equiv-e(\\bmod d)$, hence $2 f \\equiv d-e(\\bmod d)$. Now $d-e$ is even (because all elements in $\\mathcal{S}$ are odd) and $d$ is odd, so $\\frac{d-e}{2}$ is an integer and we have $f \\equiv \\frac{d-e}{2} (\\bmod d)$. Further, $0<\\frac{d-e}{2}<d$, while we must also have $0<f \\leqslant d$, so $f=\\frac{d-e}{2}$. We conclude that for any $e \\in \\mathcal{S}$ with $e \\neq d$ the integer $\\frac{d-e}{2}$ is also in $\\mathcal{S}$ and not equal to $d$.\nDenote by $e_{1}<e_{2}<\\cdots<e_{k}<d$ the elements of $\\mathcal{S}$, where $k \\geqslant 1$. Then $\\frac{d-e_{1}}{2}>\\frac{d-e_{2}}{2}> \\cdots>\\frac{d-e_{k}}{2}$ are also elements of $\\mathcal{S}$, none of them equal to $d$. Hence we must have $e_{1}=\\frac{d-e_{k}}{2}$ and\n$e_{k}=\\frac{d-e_{1}}{2}$, so $2 e_{1}+e_{k}=d=2 e_{k}+e_{1}$. We conclude $e_{1}=e_{k}$, so $k=1$, and also $d=2 e_{k}+e_{1}=3 e_{1}$. Hence $\\mathcal{S}=\\left\\{e_{1}, 3 e_{1}\\right\\}$ for some positive integer $e_{1}$.\nLet $d$ be the largest element, and let $e \\in \\mathcal{S}$ be any other element. We will say that $x \\in \\mathcal{S} (\\bmod d)$ if the unique element $y$ in $\\{1, \\ldots, d\\}$ such that $x \\equiv y(\\bmod d)$ is an element of $\\mathcal{S}$. Note that by the choice of $d$ being the largest element, if $x \\neq d$, then $x \\not \\equiv 0(\\bmod d)$. The given condition implies that if $b \\in \\mathcal{S}$, then $-\\frac{b}{2} \\in \\mathcal{S}(\\bmod d)$. Repeating this gives $-\\frac{b}{2} \\in \\mathcal{S} \\Rightarrow \\frac{b}{4} \\in \\mathcal{S} (\\bmod d)$, and by iterating, we have $b \\in \\mathcal{S} \\Rightarrow \\frac{b}{(-2)^{k}} \\in \\mathcal{S}(\\bmod d)$ for all $k$. Since $d$ is odd, there is some $g$ such that $(-2)^{g} \\equiv 1(\\bmod d)$, so by setting $k=g-1$, we get that\n$$\n\\text { for all } d \\neq e \\in \\mathcal{S},-2 e \\in \\mathcal{S}(\\bmod d) \\text {. }\n$$\nNow, if $e>\\frac{d}{2}$, then $-2 e \\in \\mathcal{S}(\\bmod d)$ and $d-2 e<0$, so $2 d-2 e \\in \\mathcal{S}$, contradicting the lack of even elements. Then $e<\\frac{d}{2}$ for any $e \\in \\mathcal{S} \\backslash\\{d\\}$, so we have $e \\in \\mathcal{S} \\Rightarrow d-2 e \\in \\mathcal{S}$. Since $d-2 e \\neq d$, we must have $d-2 e<\\frac{d}{2}$, which rearranges to $e>\\frac{d}{4}$.\nLet $\\lambda \\in(0,1)$ be a positive real number and suppose we have proved that $e>\\lambda d$ for any $e \\in \\mathcal{S} \\backslash\\{d\\}$. Then $d-2 e>\\lambda d$, which rearranges to $e<\\frac{(1-\\lambda) d}{2}$. Then $d-2 e<\\frac{(1-\\lambda) d}{2}$, which rearranges to $e>\\frac{(1+\\lambda) d}{4}$. Defining $\\lambda_{0}=\\frac{1}{4}$ and $\\lambda_{i}=\\frac{1+\\lambda_{i-1}}{4}$ for $i \\geqslant 1$, we have shown that for all $e \\in \\mathcal{S} \\backslash\\{d\\}$ and all $\\lambda_{i}, e>\\lambda_{i} d$. Now note that the sequence $\\lambda_{i}$ is increasing and bounded above by $\\frac{1}{3}$, so it converges to some limit $\\ell$, which satisfies $\\ell=\\frac{1+\\ell}{4}$, so $\\ell=\\frac{1}{3}$. Hence $e \\geqslant \\frac{d}{3}$, but then $d-2 e \\geqslant \\frac{d}{3}$ implies $e \\leqslant \\frac{d}{3}$, so $e$ must be $\\frac{d}{3}$, and we are done.\nComment. We can finish Solution 3 alternatively as follows: after showing that if $e \\in \\mathcal{S} \\backslash\\{d\\}$ then $d-2 e \\in \\mathcal{S} \\backslash\\{d\\}$, note that\n$$\n(d-2 e)-\\frac{d}{3}=\\frac{2 d}{3}-2 e=-2\\left(e-\\frac{d}{3}\\right)\n$$\nSo consider $e \\in \\mathcal{S} \\backslash\\{d\\}$ maximising $\\left|e-\\frac{d}{3}\\right|$. If $e \\neq \\frac{d}{3}$, them the above shows that $\\left|(d-2 e)-\\frac{d}{3}\\right|>\\left|e-\\frac{d}{3}\\right|$, which is a contradiction. Thus $\\mathcal{S} \\backslash\\{d\\}$ is empty or equal to $\\left\\{\\frac{d}{3}\\right\\}$, which completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32193,
    "subject": "Mathematics (Multi-modal)",
    "question": "Determine all positive integers $a$ and $b$ such that there exists a positive integer $g$ such that $\\operatorname{gcd}\\left(a^{n}+b, b^{n}+a\\right)=g$ for all sufficiently large $n$.\n\n(Indonesia)",
    "options": [],
    "answer": "(1,1)",
    "solution": "It is clear that we may take $g=2$ for $(a, b)=(1,1)$. Supposing that $(a, b)$ satisfies the conditions in the problem, let $N$ be a positive integer such that $\\operatorname{gcd}\\left(a^{n}+b, b^{n}+a\\right)=g$ for all $n \\geqslant N$.\n\nLemma. We have that $g=\\operatorname{gcd}(a, b)$ or $g=2 \\operatorname{gcd}(a, b)$.\n\nProof. Note that both $a^{N}+b$ and $a^{N+1}+b$ are divisible by $g$. Hence\n$$\na\\left(a^{N}+b\\right)-\\left(a^{N+1}+b\\right)=a b-b=a(b-1)\n$$\nis divisible by $g$. Analogously, $b(a-1)$ is divisible by $g$. Their difference $a-b$ is then divisible by $g$, so $g$ also divides $a(b-1)+a(a-b)=a^{2}-a$. All powers of $a$ are then congruent modulo $g$, so $a+b \\equiv a^{N}+b \\equiv 0(\\bmod g)$. Then $2 a=(a+b)+(a-b)$ and $2 b=(a+b)-(a-b)$ are both divisible by $g$, so $g \\mid 2 \\operatorname{gcd}(a, b)$. On the other hand, it is clear that $\\operatorname{gcd}(a, b) \\mid g$, thus proving the Lemma.\n\nLet $d=\\operatorname{gcd}(a, b)$, and write $a=d x$ and $b=d y$ for coprime positive integers $x$ and $y$. We have that\n$$\n\\operatorname{gcd}\\left((d x)^{n}+d y,(d y)^{n}+d x\\right)=d \\operatorname{gcd}\\left(d^{n-1} x^{n}+y, d^{n-1} y^{n}+x\\right),\n$$\nso the Lemma tells us that\n$$\n\\operatorname{gcd}\\left(d^{n-1} x^{n}+y, d^{n-1} y^{n}+x\\right) \\leqslant 2\n$$\nfor all $n \\geqslant N$. Defining $K=d^{2} x y+1$, note that $K$ is coprime to each of $d, x$, and $y$. By Euler's theorem, for $n \\equiv-1(\\bmod \\varphi(K))$ we have that\n$$\nd^{n-1} x^{n}+y \\equiv d^{-2} x^{-1}+y \\equiv d^{-2} x^{-1}\\left(1+d^{2} x y\\right) \\equiv 0 \\quad(\\bmod K),\n$$\nso $K \\mid d^{n-1} x^{n}+y$. Analogously, we have that $K \\mid d^{n-1} y^{n}+x$. Taking such an $n$ which also satisfies $n \\geqslant N$ gives us that\n$$\nK \\mid \\operatorname{gcd}\\left(d^{n-1} x^{n}+y, d^{n-1} y^{n}+x\\right) \\leqslant 2\n$$\nThis is only possible when $d=x=y=1$, which yields the only solution $(a, b)=(1,1)$.\n\nFor any prime factor $p$ of $a b+1, p$ is coprime to $a$ and $b$. Take an $n \\geqslant N$ such that $n \\equiv-1 (\\bmod p-1)$. By Fermat's little theorem, we have that\n$$\n\\begin{aligned}\n& a^{n}+b \\equiv a^{-1}+b=a^{-1}(1+a b) \\equiv 0 \\quad(\\bmod p), \\\\\n& b^{n}+a \\equiv b^{-1}+a=b^{-1}(1+a b) \\equiv 0 \\quad(\\bmod p),\n\\end{aligned}\n$$\nthen $p$ divides $g$. By the Lemma, we have that $p \\mid 2 \\operatorname{gcd}(a, b)$, and thus $p=2$. Therefore, $a b+1$ is a power of 2, and $a$ and $b$ are both odd numbers.\n\nIf $(a, b) \\neq(1,1)$, then $a b+1$ is divisible by 4, hence $\\{a, b\\}=\\{-1,1\\}(\\bmod 4)$. For odd $n \\geqslant N$, we have that\n$$\na^{n}+b \\equiv b^{n}+a \\equiv(-1)+1=0 \\quad(\\bmod 4)\n$$\nthen $4 \\mid g$. But by the Lemma, we have that $\\nu_{2}(g) \\leqslant \\nu_{2}(2 \\operatorname{gcd}(a, b))=1$, which is a contradiction. So the only solution to the problem is $(a, b)=(1,1)$.\nAfter proving the Lemma, one can finish the solution as follows.\n\nFor any prime factor $p$ of $a b+1, p$ is coprime to $a$ and $b$. Take an $n \\geqslant N$ such that $n \\equiv-1 (\\bmod p-1)$. By Fermat's little theorem, we have that\n$$\n\\begin{aligned}\n& a^{n}+b \\equiv a^{-1}+b=a^{-1}(1+a b) \\equiv 0 \\quad(\\bmod p), \\\\\n& b^{n}+a \\equiv b^{-1}+a=b^{-1}(1+a b) \\equiv 0 \\quad(\\bmod p),\n\\end{aligned}\n$$\nthen $p$ divides $g$. By the Lemma, we have that $p \\mid 2 \\operatorname{gcd}(a, b)$, and thus $p=2$. Therefore, $a b+1$ is a power of 2, and $a$ and $b$ are both odd numbers.\n\nIf $(a, b) \\neq(1,1)$, then $a b+1$ is divisible by 4, hence $\\{a, b\\}=\\{-1,1\\}(\\bmod 4)$. For odd $n \\geqslant N$, we have that\n$$\na^{n}+b \\equiv b^{n}+a \\equiv(-1)+1=0 \\quad(\\bmod 4)\n$$\nthen $4 \\mid g$. But by the Lemma, we have that $\\nu_{2}(g) \\leqslant \\nu_{2}(2 \\operatorname{gcd}(a, b))=1$, which is a contradiction. So the only solution to the problem is $(a, b)=(1,1)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32194,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathcal{S}$ be a finite nonempty set of prime numbers. Let $1 = b_{1} < b_{2} < \\cdots$ be the sequence of all positive integers whose prime divisors all belong to $\\mathcal{S}$. Prove that, for all but finitely many positive integers $n$, there exist positive integers $a_{1}, a_{2}, \\ldots, a_{n}$ such that\n$$\n\\frac{a_{1}}{b_{1}} + \\frac{a_{2}}{b_{2}} + \\cdots + \\frac{a_{n}}{b_{n}} = \\left\\lceil \\frac{1}{b_{1}} + \\frac{1}{b_{2}} + \\cdots + \\frac{1}{b_{n}} \\right\\rceil.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $\\mathcal{S}$ has only one element $p$, then $b_{i} = p^{i-1}$ and we can easily find $a_{1}, \\ldots, a_{n}$ with $2 = \\left\\lceil \\sum_{i=0}^{n-1} \\frac{1}{p^{i}} \\right\\rceil = \\sum_{i=0}^{n-1} \\frac{a_{i}}{p^{i-1}}$ by taking $a_{1} = a_{2} = \\cdots = a_{n-1} = 1$ and choosing $a_{n} = p^{n-1} - (p + p^{2} + \\ldots + p^{n-2})$.\n\nMore generally, observe that the sum of $\\frac{1}{b_{i}}$ over all $i$ is\n$$\n\\begin{aligned}\n\\sum_{i} \\frac{1}{b_{i}} &= \\prod_{i} \\left(1 + \\frac{1}{p_{i}} + \\frac{1}{p_{i}^{2}} + \\ldots \\right) \\\\\n&= \\prod_{p \\in \\mathcal{S}} \\frac{p}{p-1}\n\\end{aligned}\n$$\nIn particular, if $n$ is large enough, then\n$$\n\\left\\lceil \\sum_{j=1}^{n} \\frac{1}{b_{j}} \\right\\rceil = \\left\\lceil \\prod_{p \\in \\mathcal{S}} \\frac{p}{p-1} \\right\\rceil\n$$\nFor the remainder of the proof, we will only consider $n$ large enough that this equality holds.\n\nNext, we handle the special case $\\mathcal{S} = \\{2,3\\}$, for which this product is $3$. Start by setting\n$$\na_{i} = \\begin{cases}1, & \\text{ if } 2 b_{i} \\leqslant b_{n} \\\\ 2, & \\text{ if } 2 b_{i} > b_{n}\\end{cases}\n$$\nThen,\n$$\n\\sum_{\\substack{i \\leqslant n \\\\ \\nu_{3}(b_{i}) = t}} \\frac{a_{i}}{b_{i}} = \\begin{cases} \\frac{2}{3^{t}}, & \\text{ if } b_{n} \\geqslant 3^{t} \\\\ 0, & \\text{ otherwise } \\end{cases}\n$$\nAs a result,\n$$\n\\begin{aligned}\n\\sum_{i \\leqslant n} \\frac{a_{i}}{b_{i}} &= \\sum_{\\substack{t \\geqslant 0 \\\\ 3^{t} \\leqslant b_{n}}} \\frac{2}{3^{t}} \\\\\n&= 3 - \\frac{1}{3^{T}}\n\\end{aligned}\n$$\nwhere $T$ is the largest $t \\geqslant 0$ with $3^{t} \\leqslant b_{n}$. Thus, increasing $a_{j}$ by one (where $b_{j} = 3^{T}$) gives a sequence of $a_{i}$ that works.\n\nOtherwise, we may assume that $|\\mathcal{S}| > 1$ and $\\mathcal{S} \\neq \\{2,3\\}$, which means that the product $\\prod_{p \\in \\mathcal{S}} \\frac{p}{p-1}$ is not an integer. Indeed,\n- if $|\\mathcal{S}| > 2$ then $2$ divides the denominator at least twice and so divides the denominator of the overall fraction;\n- if $|\\mathcal{S}| = 2$ and $2 \\notin \\mathcal{S}$ then $2$ divides the denominator and not the numerator;\n- if $\\mathcal{S} = \\{2, p\\}$ then the product is $2p/(p-1)$ which is not an integer for $p > 3$.\n\nIt follows that for some fixed $\\alpha > 0$, we have that\n$$\n\\left\\lceil \\prod_{p \\in \\mathcal{S}} \\frac{p}{p-1} \\right\\rceil = \\prod_{p \\in \\mathcal{S}} \\frac{p}{p-1} + \\alpha\n$$\nfrom which it follows that\n$$\n\\left\\lceil \\sum_{i=1}^{n} \\frac{1}{b_{i}} \\right\\rceil - \\sum_{i=1}^{n} \\frac{1}{b_{i}} > \\alpha\n$$\nIt will now suffice to prove the following claim.\n\n**Claim.** Suppose that $n$ is large enough, and let $e_{p}$ be the largest nonnegative integer such that $p^{e_{p}} \\leqslant b_{n}$. Let $M = \\prod_{p \\in \\mathcal{S}} p^{e_{p}}$. If $u$ is a positive integer such that $u / M > \\alpha$, then there exist nonnegative integers $a_{i}$ such that\n$$\n\\sum_{i} \\frac{a_{i}}{b_{i}} = \\frac{u}{M}.\n$$\nThe problem statement follows after replacing $a_{i}$ with $a_{i} + 1$ for each $i$.\n\nTo prove this, choose some constant $c$ such that $\\sum_{p \\in \\mathcal{S}} p^{-c} < \\alpha$, and suppose $n$ is large enough that $p^{c} < b_{n}$ for each $p \\in \\mathcal{S}$; in particular, $p^{c} \\mid M$ with $M$ defined as above.\n\nFor each $p \\in \\mathcal{S}$, let $i_{p}$ be such that $b_{i_{p}} = p^{e_{p}}$ and choose the smallest nonnegative integer $a_{i_{p}}$ satisfying\n$$\np^{e_{p}-c} \\left\\lvert\\, a_{i_{p}} \\left( \\frac{M}{p^{e_{p}}} \\right) - u \\right.\n$$\nSuch an $a_{i_{p}}$ must exist and be at most $p^{e_{p}-c}$; indeed, $\\frac{M}{p^{e_{p}}}$ is an integer coprime to $p$, so we can take $a_{i_{p}}$ to be equal to $u$ times its multiplicative inverse modulo $p^{e_{p}-c}$.\n\nThe sum of the contributions to the sum from the $a_{i_{p}}$ is at most\n$$\n\\sum_{p \\in \\mathcal{S}} \\frac{p^{e_{p}-c}}{p^{e_{p}}} = \\sum_{p \\in \\mathcal{S}} p^{-c} < \\alpha\n$$\nSo, we have\n$$\n\\frac{u}{M} = \\sum_{p \\in \\mathcal{S}} \\frac{a_{i_{p}}}{p^{e_{p}}} + \\frac{r}{\\prod_{p \\in \\mathcal{S}} p^{c}},\n$$\nwhere $r$ is an integer because of our choice of $a_{i_{p}}$ and $r$ is nonnegative because of the bound on $u$. Simply choose $a_{i} = r$ where $b_{i} = \\prod_{p \\in \\mathcal{S}} p^{c}$ to complete the proof.\nWe reduce to the claim as in Solution 1, and provide an alternative approach for constructing the $a_{i}$.\n\nLet $p_{0} \\in \\mathcal{S}$ be the smallest prime in $\\mathcal{S}$. Let $z_{0} = u / M$. We construct a sequence $z_{0}, z_{1}, z_{2}, \\ldots$ and values of $a_{i}$ by the following iterative process: to construct $z_{j+1}$,\n- select the largest prime $p \\in \\mathcal{S}$ dividing the denominator of $z_{j}$, and let $\\mu$ be the number of times $p$ divides the denominator of $z_{j}$;\n- choose the largest $\\nu$ such that $p_{0}^{\\nu} p^{\\mu} \\leqslant b_{n}$, and let $i \\leqslant n$ be such that $b_{i} = p_{0}^{\\nu} p^{\\mu}$;\n- choose $0 \\leqslant a_{i} < p$ such that the denominator of $z_{k} - a_{i} / b_{i}$ has at most $\\mu - 1$ factors of $p$, and let $z_{k+1} = z_{k} - a_{i} / b_{i}$;\n- continue until $p_{0}$ is the only prime dividing the denominator of $z_{k}$.\n\nNote that we can always choose $a_{i}$ in step 3; by construction, $z_{k} b_{i}$ has no factors of $p$ in its denominator, so must be realised as an element of $\\mathbb{Z}_{p}$.\n\nEach time we do this, $b_{i} > M / p_{0}$ by construction, so\n$$\n\\frac{a_{i}}{b_{i}} < \\frac{p p_{0}}{M} \\leqslant \\frac{p_{0} p_{1}}{M},\n$$\nwhere $p_{1}$ is the largest prime in $\\mathcal{S}$.\nAnd the number of times we do this operation is at most\n$$\n\\sum_{\\substack{p \\in \\mathcal{S} \\\\ p > p_{0}}} e_{p} \\leqslant |\\mathcal{S}| \\log_{2}(M)\n$$\nso the sum of the $a_{i} / b_{i}$ we have assigned is at most $|\\mathcal{S}| p_{0} p_{1} \\log_{2}(M) / M$.\n\nChoose $n$ large enough that $\\log_{2}(M) / M < \\alpha$; after subtracting the above choices of $a_{i} / b_{i}$ from $u / M$, we have a quantity of the form $r / p_{0}^{e_{p_{0}}}$, where $r$ is an integer by construction and $r$ is positive by the above bounds. Simply set $a_{i} = r$ where $b_{i} = p_{0}^{e_{P_{0}}}$ to complete the proof.\nAs in Solution 1, we may handle $|\\mathcal{S}| = 1$ and $\\mathcal{S} = \\{2,3\\}$ separately; otherwise, we can define $\\alpha$ as we did in that solution. Also define $e_{p}$ to be the largest nonnegative integer such that $p^{e_{p}} \\leqslant b_{n}$ as we did in Solution 1.\n\nWe will show that, for $n$ sufficiently large, we may choose some $j \\leqslant n$, and positive integers $a_{i}$, such that\n$$\n\\sum_{i \\neq j} \\frac{a_{i}}{b_{i}} - \\sum_{i \\neq j} \\frac{1}{b_{i}} < \\alpha\n$$\nand all $\\frac{a_{i}}{b_{i}}$ are integer multiples of $\\frac{1}{b_{j}}$. We then set $a_{j}$ to be the least positive integer such that the sum on the left is an integer, which will obviously have the required value.\n\nConcretely, choose $j$ such that $b_{j} = \\prod_{p \\in \\mathcal{S}} p^{\\left.e_{p} / |\\mathcal{S}| \\right]}$, which is less than $b_{n}$ by construction. For $i \\neq j$, set $a_{i} = b_{i} / \\operatorname{gcd}(b_{i}, b_{j})$. We have\n$$\n\\sum_{i \\neq j} \\frac{a_{i}}{b_{i}} - \\sum_{i \\neq j} \\frac{1}{b_{i}} < \\sum_{\\substack{i \\neq j \\\\ a_{i} > 1}} \\frac{a_{i}}{b_{i}}\n$$\nIf $a_{i} > 1$, then there must be some $p \\in \\mathcal{S}$ for which $p^{\\left\\lfloor e_{p} / |\\mathcal{S}| \\right] + 1} \\mid b_{i}$, and so\n$$\n\\frac{a_{i}}{b_{i}} = \\frac{1}{\\operatorname{gcd}(b_{i}, b_{j})} \\leqslant \\frac{1}{p^{\\left[e_{p} / |\\mathcal{S}| \\right]}} < \\frac{p}{b_{n}^{1 / |\\mathcal{S}|}}\n$$\nwhere the last inequality follows from the fact that $p^{e_{p} + 1} > b_{n}$.\n\nNow $n \\leqslant \\prod_{p \\in \\mathcal{S}} (\\log_{p}(b_{n}) + 1) \\leqslant (2 \\log b_{n})^{|\\mathcal{S}|}$, so\n$$\n\\sum_{\\substack{i \\neq j \\\\ a_{i} > 1}} \\frac{a_{i}}{b_{i}} \\leqslant \\frac{(2 \\log b_{n})^{|\\mathcal{S}|}}{b_{n}^{1 / |\\mathcal{S}|}}\n$$\nand so we can choose $n$ large enough that this quantity is less than $\\alpha$, as required.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32195,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer. We say that a polynomial $P$ with integer coefficients is $n$-good if there exists a polynomial $Q$ of degree 2 with integer coefficients such that $Q(k)(P(k)+Q(k))$ is never divisible by $n$ for any integer $k$.\nDetermine all integers $n$ such that every polynomial with integer coefficients is an $n$-good polynomial.",
    "options": [],
    "answer": "all positive integers n ≥ 3",
    "solution": "First, observe that no polynomial is $1$-good (because $Q(X)(P(X)+Q(X))$ always has roots modulo $1$) and the polynomial $P(X)=1$ is not $2$-good (because $Q(X)(Q(X)+1)$ is always divisible by $2$).\n\nNow, if $P$ is $d$-good with some $Q$, then $Q \\cdot (P+Q)$ has no roots $\\bmod d$. Therefore, it certainly has no roots $\\bmod n$ for $d \\mid n$, so $P$ must be $n$-good. Consequently, it suffices to show that all polynomials are $n$-good whenever $n$ is an odd prime, or $n=4$.\n\nWe start by handling the case $n=4$. We will construct a $Q$ such that $Q(X)$ is never divisible by $4$ and $Q(X)+P(X)$ is always odd; this will clearly show that $P$ is $4$-good. Note that any function modulo $2$ must be either constant or linear - in other words, there are $a, b \\in \\{0,1\\}$ such that $P(X)=a X+b \\bmod 2$ for all $X$. If $a=0$ then set $Q(X)=4 X^{2}+b+1$, and if $a=1$ then set $Q(X)=X^{2}+b+1$; in all cases, $Q$ will satisfy the required properties.\n\nIt remains to prove that any polynomial is $p$-good, where $p$ is an odd prime. We will prove that for any function $f$ defined $\\bmod p$, there is a quadratic $Q$ with no roots $\\bmod p$ such that $Q(x) \\neq f(x) \\bmod p$ for all $x$; the statement about $P$ then follows with $f$ replaced by $-P$. For the remainder of the proof, we will consider all equalities modulo $p$.\n\nSuppose that a function $f$ not satisfying the above exists; in other words, $f$ has the property that for any quadratic $Q$ with no roots $\\bmod p$, there is some $x$ such that $Q(x)=f(x)$. Without loss of generality, we may assume that $f$ has no roots $\\bmod p$. To see why, suppose that $f(u)=0$ for some $u$, and let $g$ be the function such that $g(x)=f(x)$ for $x \\neq u$ and $g(u)=1$. For any $Q$ with no roots, we know that there is some $x \\neq u$ such that $P(x)=f(x)$, and so $P(x)=g(x)$ for that choice of $x$. In particular, $g$ is also not $p$-good.\n\nNow, suppose first that there is some nonzero $t$ such that $t$ is not in the image of $f$. Then we may take $Q(X)=p X^{2}+t$; this quadratic is never equal to $f$ and is never zero. Thus, $f$ must be surjective onto the nonzero residues $\\bmod p$. There are $p$ choices for $X$ and $p-1$ nonzero residues $\\bmod p$, so there must be some $x_{1} \\neq x_{2} \\bmod p$ such that $f\\left(x_{1}\\right)=f\\left(x_{2}\\right)$, and $f$ is a bijection from the set of residues $\\bmod p$ not equal to $x_{2}$ to the set of nonzero residues $\\bmod p$.\n\nNow, note that we may choose any $b$ and $c$ with $b$ nonzero and replace $f(X)$ with $g(X)= f(b X+c)$; if there were some $Q$ with no roots such that $Q(x) \\neq g(x)$ for all $x$, then $Q(X / b-c / b)$ would work for $f$. Choose $b$ and $c$ such that $b x_{1}+c=1$ and $b x_{2}+c=-1$; such $b$ and $c$ must exist (we may take $b=2 /(x_{1}-x_{2})$ and $c=(x_{1}+x_{2}) /(x_{2}-x_{1})$ ). Renaming $g$ to $f$, we see that we may assume $f(1)=f(-1)$.\n\nLet $r'$ be a quadratic nonresidue $\\bmod p$. Choose $y \\neq 0$ such that $f(y)=(1-r') f(0)$, which must exist as the right hand side is nonzero and $1-r'$ is not equal to $1$. Choose $r=y^{2} / r'$, which is a quadratic nonresidue.\n\nConsider $\\phi(X)=f(X) /(X^{2}-r)$. By definition, $\\phi(1)=\\phi(-1)$ and $\\phi(0)=\\phi(y)$, so there are no more than $p-2$ values in the image of $\\phi$. Choose some nonzero $a$ not in the image of $\\phi$, so $f(X) /(X^{2}-r)$ is never equal to $a$. The quadratic $Q(X)=a(X^{2}-r)$ is never zero and also never equal to $f(X)$, which completes the proof.\nGiven $f$ a function $\\bmod p$ such that $f$ is surjective onto the nonzero elements of $\\mathbb{Z} / p \\mathbb{Z}$ and $f(1)=f(-1)$, we provide an alternative approach to construct a nonzero quadratic $Q(X)$ such that $Q(X) \\neq f(X)$. Let $r$ be the smallest quadratic nonresidue $\\bmod p$ (so $r-1$ is a square) and let $a$ vary over the nonzero elements $\\bmod p$; we will show that it is possible to choose $Q_{a}(X)=a(X^{2}-r)$ for some choice of $a$. Note that any quadratic of this form will be nowhere zero.\n\nSuppose that no such $Q_{a}$ works. Then, for each $a$, there exists $x$ such that $a(x^{2}-r)=f(x)$. We may assume that $x \\neq -1$, as if the equality holds for $x=-1$ then it also holds for $x=1$. However, $a(x^{2}-r)=f(x)$ implies $a=f(x)/(x^{2}-r)$, so $f(x)/(x^{2}-r)$ must be a surjection from $\\{x \\neq -1\\}$ to the set of nonzero $a$, and so this is a bijection. In particular, for each $a$, there exists a unique $x_{a}$ such that $f(x_{a})=a(x_{a}^{2}-r)$.\n\nWe now have\n$$\n\\begin{aligned}\n\\prod_{t \\neq 0} t & =\\prod_{a \\neq 0} f(x_{a}) \\\\\n& =\\prod_{a \\neq 0} a \\prod_{a \\neq 0}(x_{a}^{2}-r) \\\\\n& =\\prod_{a \\neq 0} a \\prod_{x \\neq -1}(x^{2}-r)\n\\end{aligned}\n$$\nwhere the first equality follows because $f$ is surjective onto the nonzero residues $\\bmod p$, and the second equality follows from the definition of $x_{a}$. The two products cancel, which means that $\\prod_{x \\neq -1}(x^{2}-r)=1$.\n\nHowever, we also get\n$$\n\\prod_{x \\neq -1}(x^{2}-r)=(-r)(1-r)\\left(\\prod_{x=2}^{(p-1)/2}(x^{2}-r)\\right)^{2} .\n$$\nHowever, this is a contradiction as $-r(1-r)=r(r-1)$, which is not a quadratic residue (by our choice of $r$ ).\nAs in Solution 1, we will reduce to the case of $p$ being an odd prime and $f$ being a function $\\bmod p$ with no roots which is surjective onto the set of nonzero residues $\\bmod p$, although we make no assumption about the values of $x_{1}$ and $x_{2}$ with $f(x_{1})=f(x_{2})$.\n\nWe will again consider quadratics of the form $Q_{a, b, c}(X)=a R(b X+c)$, where $R(X)=X^{2}-r$ for an arbitrary fixed quadratic nonresidue $r$, $a$ and $b$ are nonzero $\\bmod p$, and $c$ is any residue $\\bmod p$.\n\nFor each fixed $b$ and $c$, there must be $n$ pairs $(a, x)$ such that $a R(b x+c)=f(x)$, because there must be exactly one value of $a$ for each $x$. If any $a$ appears in no such pair then we are done, so assume otherwise. In other words, there must be exactly one $a$ such that there are two such $x$, and for all other $a$ there is only one such $x$.\n\nThus, for each $(b, c)$, there is exactly one unordered pair $\\{x_{1}, x_{2}\\}$ such that for some $a$ we have $f(x_{i})=a R(b x_{i}+c)$; in other words, there is exactly one unordered pair $\\{x_{1}, x_{2}\\}$ such that $f(x_{1}) / R(b x_{1}+c)=f(x_{2}) / R(b x_{2}+c)$.\n\nNow, we show that for each unordered pair $\\{x_{1}, x_{2}\\}$ there must be at least one pair $(b, c)$ such that $f(x_{1}) / R(b x_{1}+c)=f(x_{2}) / R(b x_{2}+c)$. Indeed, let $t=f(x_{1}) / f(x_{2})$. There must be some $x_{1}', x_{2}'$ such that $R(x_{1}') / R(x_{2}')=t$; this is because $R(X)$ and $t R(X)$ both take $\\frac{p+1}{2}$ nonzero values $\\bmod p$, so the intersection must be nonempty by the pigeonhole principle. Choosing $b$ and $c$ such that $b x_{1}+c=x_{1}'$ and $b x_{2}+c=x_{2}'$ gives the claim.\n\nNote further that if $(b, c)$ and $\\{x_{1}, x_{2}\\}$ satisfy the relation, then the same is true for $(-b,-c)$ and $\\{x_{1}, x_{2}\\}$ because $R(b x+c)=R(-b x-c)$. Since $b$ is nonzero, this means that each pair $\\{x_{1}, x_{2}\\}$ corresponds to at least two pairs ( $b, c$ ). However, since there are $p(p-1)$ pairs ( $b, c$ ) with $b$ nonzero and $p(p-1) / 2$ unordered pairs $\\{x_{1}, x_{2}\\}$, each $\\{x_{1}, x_{2}\\}$ must correspond to exactly two pairs $(b, c)$ and $(-b,-c)$ for some $(b, c)$.\n\nNow, since the image of $f$ has only $p-1$ elements, there must be some $x_{1}, x_{2}$ such that $f(x_{1})=f(x_{2})$. Choose any $b, c$ such that $b x_{1}+c=-\\left(b x_{2}+c\\right)$, so $R(b x_{1}+c)=R(b x_{2}+c)$ and so $f(x_{1}) / R(b x_{1}+c)=f(x_{2}) / R(b x_{2}+c)$. There is such a pair $b, c$ for any nonzero $b$, so there are at least $p-1$ such pairs, and this quantity is greater than $2$ for $p \\geqslant 5$.\n\nFinally, for the special case that $p=3$, we observe that there must be at least one allowed value for $Q(x)$ for each $x$, so there must exist such a quadratic $Q$ by Lagrange interpolation.\n\nComment. We may also handle the case $p=3$ as follows. Recall that we may assume $f$ is nonzero and surjective onto $\\{1,2\\} \\bmod 3$, so the image of $f$ must be $(1,1,2)$ or $(1,2,2)$ in some order. Without loss of generality $f(1)=f(2)$, so we either have $(f(0), f(1), f(2))=(1,2,2)$ or $(2,1,1)$. In the first case, take $Q(X)=2 X^{2}+2$, and in the second case take $Q(X)=X^{2}+1$.\nAgain, we reduce to the case of $p$ being an odd prime and $f$ being a function $\\bmod p$; we will show that there is a quadratic which is nowhere zero such that $Q(x)=f(x)$ has no root. We can handle the case of $p=3$ separately as in Solution 3, so assume that $p \\geqslant 5$.\n\nWe will prove the following more general statement: let $p \\geqslant 5$ be a prime and let $\\mathcal{A}_1, \\mathcal{A}_2, \\ldots, \\mathcal{A}_p$ be subsets of $\\mathbb{Z} / p \\mathbb{Z}$ with $|\\mathcal{A}_i|=2$ for all $i$. Then there exists a polynomial $Q \\in \\mathbb{Z} / p \\mathbb{Z}[X]$ of degree at most $2$ such that $Q(i) \\notin \\mathcal{A}_i$ for all $i$. Indeed, applying this statement to the sets $\\mathcal{A}_i=\\{0, f(i)\\}$ (and adding $p X^{2}$ if necessary) produces a quadratic $Q$ satisfying the desired property.\n\nChoose the coefficients of $Q$ uniformly at random from $\\mathbb{Z} / p \\mathbb{Z}$, and let $T$ be the random variable denoting the number of $i$ for which $Q(i) \\in \\mathcal{A}_i$. Observe that for $k \\leqslant 3$, we have\n$$\n\\mathbb{E}\\left[\\binom{T}{k}\\right]=2^{k}\\binom{p}{k} p^{-k} .\n$$\nTo see why, let $k \\leqslant 3$. If $\\mathcal{S} \\subseteq \\mathbb{Z} / p \\mathbb{Z}$ has size $k$ and $(a_i)_{i \\in \\mathcal{S}}$ is a $k$-tuple, the probability that $Q(i)=a_i$ on $\\mathcal{S}$ is equal to $p^{-k}$; for $k=3$ this follows by Lagrange interpolation, and for $k<3$ it follows from the $k=3$ case by summing. The expectation is therefore equal to the number of $\\mathcal{S} \\subseteq \\mathbb{Z} / p \\mathbb{Z}$ of size $k$ times the probability that $Q(i) \\in \\mathcal{A}_i$ for each $i \\in \\mathcal{S}$, which is equal to the right hand side as each $\\mathcal{A}_i$ has size $2$.\n\nNow, observe that we have the identity $(t-1)(t-3)(t-4)=-12+12\\binom{t}{1}-10\\binom{t}{2}+6\\binom{t}{3}$, so\n$$\n\\begin{aligned}\n\\mathbb{E}[(T-1)(T-3)(T-4)] & =-12+12 \\mathbb{E}\\left[\\binom{T}{1}\\right]-10 \\mathbb{E}\\left[\\binom{T}{2}\\right]+6 \\mathbb{E}\\left[\\binom{T}{3}\\right] \\\\\n& =-12+12 \\cdot 2-10 \\cdot 2\\left(1-\\frac{1}{p}\\right)+6 \\cdot \\frac{4}{3}\\left(1-\\frac{1}{p}\\right)\\left(1-\\frac{2}{p}\\right) \\\\\n& =-\\frac{4}{p}+\\frac{16}{p^{2}}\n\\end{aligned}\n$$\nThis is negative for $p \\geqslant 5$. Because $(t-1)(t-3)(t-4) \\geqslant 0$ for all integers $t>0$, it then follows that $T=0$ with positive probability, which implies that there must exist some $Q$ with $Q(i) \\notin \\mathcal{A}_i$ for all $i$, as desired.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32196,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{Z}_{>0}$ denote the set of positive integers. Let $f: \\mathbb{Z}_{>0} \\rightarrow \\mathbb{Z}_{>0}$ be a function satisfying the following property: for $m, n \\in \\mathbb{Z}_{>0}$, the equation\n$$\nf(m n)^2 = f\\left(m^2\\right) f(f(n)) f(m f(n))\n$$\nholds if and only if $m$ and $n$ are coprime.\nFor each positive integer $n$, determine all the possible values of $f(n)$.",
    "options": [],
    "answer": "For each positive integer n, the possible values of f(n) are exactly the positive integers whose set of prime divisors is the same as that of n; equivalently, f(n) can be any product of the primes dividing n with arbitrary positive exponents.",
    "solution": "- From $P(1,1)$, we have $f(1)^2 = f(1) f(f(1))^2$, so $f(1) = f(f(1))^2$.\n- From $P(1, f(1))$, we have $f(f(1))^2 = f(1) f(f(f(1))) f(f(f(1)))$, so $f(f(f(1))) = 1$.\n- From $P(1, f(f(1)))$, we have $f(f(f(1)))^2 = f(1) f(f(f(f(1)))) f(f(f(f(1))))$, which simplifies to $1 = f(1)^3$, so $f(1) = 1$.\n- From $P(1, n)$ we deduce $f(n) = f(f(n))$ for all $n$.\n- From $P(m, 1)$ we deduce $f(m) = f\\left(m^2\\right)$ for all $m$.\n- Simplifying $P(m, n)$, we have that\n$$\nf(m n)^2 = f(m) f(n) f(m f(n))\n$$\nif and only if $m$ and $n$ are coprime; refer to this as $Q(m, n)$.\n- From $Q(m, f(n))$, we have that $f(m f(n)) = f(m) f(n)$ if and only if $m$ and $f(n)$ are coprime; refer to this as $R(m, n)$.\n\nClaim. If $f(a) = 1$, then $a = 1$.\nProof. If $a \\neq 1$, then $Q(a, a)$ gives $f(a)^2 \\neq f(a)^2 f(a f(a))$. If $f(a) = 1$, then both sides simplify to 1, a contradiction.\n\nClaim. If $n \\neq 1$ then $\\operatorname{gcd}(n, f(n)) \\neq 1$.\nProof. If $\\operatorname{gcd}(n, f(n)) = 1$, then $Q(f(n), n)$ gives $f(n f(n))^2 = f(n)^3$, and $Q(n, f(n))$ gives $f(n f(n))^2 = f(n)^2 f(n f(n))$, which together yield $f(n) = 1$ for a contradiction.\n\nClaim. For all $n$ we have $\\operatorname{rad}(n) \\mid f(n)$.\nProof. For any prime $p \\mid n$, write $n = p^v n'$, with $p \\nmid n'$. From $Q\\left(p^v, n'\\right)$ we have $f(n)^2 = f\\left(p^v\\right) f\\left(n'\\right) f\\left(p^v f\\left(n'\\right)\\right)$. Since $\\operatorname{gcd}\\left(p^v, f\\left(p^v\\right)\\right) \\neq 1$, it follows that $p \\mid f\\left(p^v\\right)$, so $p \\mid f(n)$, and thus $\\operatorname{rad}(n) \\mid f(n)$.\n\nClaim. If $n$ is coprime to $f(k)$, then $f(n)$ is coprime to $f(k)$.\nProof. From $Q(f(k), n)$ we have $f(n f(k))^2 = f(k) f(n) f(f(k) f(n))$; applying $R(n, k)$ to the LHS, we conclude that $f(k) f(n) = f(f(k) f(n))$. Applying $R(f(n), k)$ we deduce that $f(n)$ is coprime to $f(k)$, as required.\n\nClaim. If $p$ is prime then $f(p)$ is a power of $p$.\nProof. Suppose otherwise. We know that $p \\mid f(p)$; let $q \\neq p$ be another prime with $q \\mid f(p)$.\nIf, for some positive integer $N$, we have $p \\nmid f(N)$, then $f(p)$ is coprime to $f(N)$, so $q \\nmid f(N)$, so $q \\nmid N$; thus, if $q \\mid N$, then $p \\mid f(N)$ (and in particular, $p \\mid f(q)$, by taking $N = q$).\nSimilarly, if $q \\nmid f(N)$ then $f(q)$ is coprime to $f(N)$; as $p \\mid f(q)$, this means $p \\nmid f(N)$, so $p \\nmid N$. So if $p \\mid N$, then $q \\mid f(N)$.\nTogether with $\\operatorname{rad}(n) \\mid f(n)$, this means that for any $n$ not coprime to $p q$, we have $p q \\mid f(n)$.\nLet $m = \\min \\{ \\nu_p(f(x)) \\mid x$ is not coprime to $p q \\}$, and let $X$ be a positive integer not coprime to $p q$ such that $\\nu_p(f(X)) = m$. The argument above shows $m \\geqslant 1$. We can write $f(X) = p^m q^y X'$, where $y \\geqslant 1, p \\nmid X'$ and $q \\nmid X'$. Since $f(f(X)) = f(X)$ we have $f\\left(p^m q^y X'\\right) = p^m q^y X'$. Applying $Q\\left(p^m, q^y X'\\right)$ gives $\\left(p^m q^y X'\\right)^2 = f\\left(p^m\\right) f\\left(q^y X'\\right) f\\left(p^m f\\left(q^y X'\\right)\\right)$. The RHS is divisible by $p^{3m}$ but the LHS is only divisible by $p^{2m}$, yielding a contradiction.\n\nClaim. For any integer $n, \\operatorname{rad}(f(n)) = \\operatorname{rad}(n)$.\nProof. We already have that $\\operatorname{rad}(n) \\mid f(n)$, so it remains only to show that no other primes divide $f(n)$. If $p$ is prime and $p \\nmid n$, the previous Claim shows that $n$ is coprime to $f(p)$, and thus $f(n)$ is coprime to $f(p)$; that is, $p \\nmid f(n)$. So exactly the same primes divide $f(n)$ as divide $n$.\n\nIt remains only to exhibit functions that show all values of $f(n)$ with $\\operatorname{rad}(f(n)) = \\operatorname{rad}(n)$ are possible. Given $e(p) \\geqslant 1$ for each prime $p$, take\n$$\nf(n) = \\prod_{p \\mid n} p^{e(p)}\n$$\nand we verify by examining exponents of each prime that this satisfies the conditions of the problem.\nAs in Solution 1, we see that there are indeed functions $f$ satisfying the given condition and producing all the given values of $f(n)$, and we follow Solution 1 to show the following facts:\n- $f(1) = 1$.\n- $f(m) = f\\left(m^2\\right)$ for all $m$.\n- $f(n) = f(f(n))$ for all $n$.\n- $f(m n)^2 = f(m) f(n) f(m f(n))$ if and only if $m$ and $n$ are coprime; refer to this as $Q(m, n)$.\n\nTaking $Q(m, n)$ together with $Q(n, m)$ gives that $f(m f(n)) = f(n f(m))$ if $m$ and $n$ are coprime.\nSuppose now that $m$ is coprime to both $n$ and $f(n)$. We have $f(m n)^2 = f(m) f(n) f(m f(n))$ and squaring both sides gives\n$$\n\\begin{aligned}\nf(m n)^4 & = f(m)^2 f(n)^2 f(m f(n))^2 \\\\\n& = f(m)^2 f(n)^2 f(m) f(f(n)) f(m f(f(n))) \\\\\n& = f(m)^3 f(n)^3 f(m f(n)) .\n\\end{aligned}\n$$\nThus $f(m f(n)) = f(m) f(n)$, so $f(m n)^2 = f(m)^2 f(n)^2$, so $f(m n) = f(m) f(n) = f(m f(n)) = f(n f(m))$.\nIf $m$ is coprime to both $n$ and $f(n)$ but however $n$ is not coprime to $f(m)$, we have\n$$\n\\begin{aligned}\nf(n f(m))^2 & \\neq f(n) f(f(m)) f(n f(f(m))) \\\\\n& = f(n) f(m) f(n f(m)) \\\\\n& = f(n f(m))^2,\n\\end{aligned}\n$$\na contradiction. Thus, given that $m$ and $n$ are coprime, we know that $m$ is coprime to $f(n)$ if and only if $n$ is coprime to $f(m)$. In particular, if $p$ and $q$ are different primes, then $p \\mid f(q)$ if and only if $q \\mid f(p)$, and likewise, for any positive integer $k, p \\mid f\\left(q^k\\right)$ if and only if $q \\mid f(p)$. More generally, if $p \\nmid n$, then $p \\mid f(n)$ if and only if $n$ is not coprime to $f(p)$.\n\nNow form a graph whose vertices are the primes, and where there is an edge between primes $p \\neq q$ if and only if $p \\mid f(q)$ (and so $q \\mid f(p)$ ); every vertex has finite degree. For any integer $n$, the primes dividing $f(n)$ are all the primes that are neighbours of any prime $q \\mid n$, together possibly with some further primes $p \\mid n$.\nIf $p$ and $q$ are different primes, we have $f(p f(q)) = f(q f(p))$. The LHS is divisible by all primes that (in the graph) are neighbours of $p$ or neighbours of neighbours of $q$, and possibly also by $p$ and by some primes that are neighbours of $q$, and a corresponding statement with $p$ and $q$ swapped applies to the RHS. Thus any prime that is a neighbour of a neighbour of $q$ must be one of: $p, q$, distance 1 from $q$, or distance 1 or 2 from $p$. For any prime $r$ that is distance 2 from $q$, there are only finitely many primes $p$ that it is distance 2 or less from, so by choosing a suitable prime $p$ (depending on $q$ ) we conclude that every prime that is a neighbour of a neighbour of $q$ is actually $q$ itself or a neighbour of $q$.\nSo the connected components of the graph are (finite) complete graphs. If $m$ is divisible only by primes in one component, and $n$ is divisible only by primes in another component, then $f(m n) = f(m) f(n)$. If $n$ is divisible by more than one prime from a component, considering the expression for $f(m n)^2$ as applied with successive prime power divisors of $n$ shows that $f(n)$ is divisible by all the primes in that component. However, while $f\\left(p^k\\right)$ is divisible by all the primes in the component of $p$ except possibly for $p$ itself, we do not yet know that $p \\mid f\\left(p^k\\right)$. We now consider cases for the order of a component.\nFor any prime $p$, we cannot have $f\\left(p^k\\right) = 1$, because $Q\\left(p^k, p^k\\right)$ gives\n$$\nf\\left(p^{2k}\\right)^2 \\neq f\\left(p^k\\right) f\\left(p^k\\right) f\\left(p^k f\\left(p^k\\right)\\right),\n$$\nand simplifying using $f\\left(m^2\\right) = f(m)$ results in $1 \\neq 1$. So for a component of order 1, $f\\left(p^k\\right)$ is a positive power of $p$, so has the same set of prime factors as $p$, as required.\nNow consider a component of order at least 2. Since $f(f(n)) = f(n)$, if the component has order at least 3, then for any $n \\neq 1$ whose prime divisors are in that component, $f(n)$ is divisible by all the primes in that component. If the component has order 2, we saw above that this is true except possibly for $n = p^k$. However, if the primes in the component are $p$ and $q$, and $f\\left(p^k\\right) = q^\\ell$, then $f\\left(q^\\ell\\right) = f\\left(f\\left(p^k\\right)\\right) = f\\left(p^k\\right) = q^\\ell$, which contradicts $p \\mid f\\left(q^\\ell\\right)$. So for any component of order at least 2, and any $n \\neq 1$ whose prime divisors are in that component, $f(n)$ is divisible by all the primes in that component.\nIn a component of order at least 2, let $m$ be the product of all the primes in that component, and let $t$ be maximal such that $m^t \\mid f(n)$ for all $n \\neq 1$ whose prime divisors are in that component; we have seen that $t \\geqslant 1$. If $m$ and $n$ are coprime numbers greater than 1, all of whose prime divisors are in that component, then $Q(m, n)$ tells us that $m^{3t/2} \\mid f(m n)$. For any $n' \\neq 1$, all of whose prime divisors are in that component, $f\\left(n'\\right)$ is divisible by all the primes in that component, so can be expressed as such a product, so $m^{3t/2} \\mid f\\left(f\\left(n'\\right)\\right) = f\\left(n'\\right)$. But this means $t \\geqslant 3t/2$, a contradiction, so all components have order 1, and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32197,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all integer solutions to: $a+b+c=24$, $a^{2}+b^{2}+c^{2}=210$, $a b c=440$.",
    "options": [],
    "answer": "All permutations of (5, 8, 11).",
    "solution": "Solution:\n\n$ab+bc+ca = \\left((a+b+c)^2 - (a^2+b^2+c^2)\\right)/2 = 183$, so $a$, $b$, $c$ are roots of the cubic $x^3 - 24x^2 + 183x - 440 = 0$. But it easily factorises as $(x-5)(x-8)(x-11)=0$, so the only solutions are permutations of $(5,8,11)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32198,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind the roots $r_{1}, r_{2}, r_{3}, r_{4}$ of the equation $4x^{4} - a x^{3} + b x^{2} - c x + 5 = 0$, given that they are positive reals satisfying $r_{1}/2 + r_{2}/4 + r_{3}/5 + r_{4}/8 = 1$.",
    "options": [],
    "answer": "r1 = 1/2, r2 = 1, r3 = 5/4, r4 = 2",
    "solution": "Solution:\n\nWe have $r_{1} r_{2} r_{3} r_{4} = 5/4$ and hence $(r_{1}/2)(r_{2}/4)(r_{3}/5)(r_{4}/8) = 1/4^{4}$. But AM/GM gives that $(r_{1}/2)(r_{2}/4)(r_{3}/5)(r_{4}/8) \\leq \\left( (r_{1}/2 + r_{2}/4 + r_{3}/5 + r_{4}/8)/4 \\right)^{4} = 1/4^{4}$ with equality iff $r_{1}/2 = r_{2}/4 = r_{3}/5 = r_{4}/8$. Hence we must have $r_{1} = 1/2$, $r_{2} = 1$, $r_{3} = 5/4$, $r_{4} = 2$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32199,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe reals $x, y, z$ satisfy $x \\neq 1$, $y \\neq 1$, $x \\neq y$, and $\\dfrac{y z - x^{2}}{1 - x} = \\dfrac{x z - y^{2}}{1 - y}$. Show that $\\dfrac{y z - x^{2}}{1 - x} = x + y + z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe have $y z - x^{2} - y^{2} z + y x^{2} = x z - y^{2} - x^{2} z + x y^{2}$. Hence $z (y - x - y^{2} + x^{2}) = -y^{2} + x y^{2} - x^{2} y + x^{2}$. Hence $z = \\dfrac{x + y - x y}{x + y - 1}$.\n\nSo $y z = x + y + z - x y - x z$, so $y z - x^{2} = x + y + z - x^{2} - x y - x z = (x + y + z)(1 - x)$, so $\\dfrac{y z - x^{2}}{1 - x} = x + y + z$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32200,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe function $f(n)$ is defined on the positive integers and takes non-negative integer values. It satisfies\n\n(1) $f(m n) = f(m) + f(n)$,\n\n(2) $f(n) = 0$ if the last digit of $n$ is $3$,\n\n(3) $f(10) = 0$.\n\nFind $f(1985)$.",
    "options": [],
    "answer": "0",
    "solution": "Solution:\n\nIf $f(m n) = 0$, then $f(m) + f(n) = 0$ (by (1)). But $f(m)$ and $f(n)$ are non-negative, so $f(m) = f(n) = 0$. Thus $f(10) = 0$ implies $f(5) = 0$. Similarly $f(3573) = 0$ by (2), so $f(397) = 0$. Hence $f(1985) = f(5) + f(397) = 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32201,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$O$ is the circumcenter of the triangle $ABC$. The lines $AO$, $BO$, $CO$ meet the opposite sides at $D$, $E$, $F$ respectively. Show that $1 / AD + 1 / BE + 1 / CF = 2 / AO$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nProjecting onto the altitude from $A$, we have $AD \\cos (C-B) = AC \\sin C = 2R \\sin B \\sin C$, so $2R / AD = \\cos (C-B) / (\\sin B \\sin C)$.\n\nHence $2R / AD + 2R / BE + 2R / CF = \\cos (C-B) / (\\sin B \\sin C) + \\cos (A-C) / (\\sin C \\sin A) + \\cos (B-A) / (\\sin A \\sin B)$.\n\nSo $2R \\sin A \\sin B \\sin C (1 / AD + 1 / BE + 1 / CF) = \\sin A \\cos (B-C) + \\sin B \\cos (C-A) + \\sin C \\cos (A-B)$\n\n$= 3 \\sin A \\sin B \\sin C + \\sin A \\cos B \\cos C + \\sin B \\cos A \\cos C + \\sin C \\cos A \\cos B$\n\n$= 3 \\sin A \\sin B \\sin C + \\sin (A+B) \\cos C + \\sin C \\cos A \\cos B$\n\n$= 3 \\sin A \\sin B \\sin C + \\sin C (\\cos C + \\cos A \\cos B)$\n\n$= 3 \\sin A \\sin B \\sin C + \\sin C (-\\cos (A+B) + \\cos A \\cos B)$\n\n$= 4 \\sin A \\sin B \\sin C$.\n\nHence $1 / AD + 1 / BE + 1 / CF = 2 / R$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32202,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind $f(x)$ such that $f(x)^{2} f\\left(\\frac{1-x}{1+x}\\right) = 64x$ for $x$ not $0, \\pm 1$.",
    "options": [],
    "answer": "f(x) = 4 ((x^2(1+x))/(1-x))^{1/3}",
    "solution": "Solution:\n\nPut $x = \\frac{1-y}{1+y}$, then $\\frac{1-x}{1+x} = y$, so $f\\left(\\frac{1-y}{1+y}\\right)^{2} f(y) = 64 \\frac{1-y}{1+y}$. Hence $f\\left(\\frac{1-x}{1+x}\\right)^{2} f(x) = 64 \\frac{1-x}{1+x}$.\n\nBut $f(x)^{4} f\\left(\\frac{1-x}{1+x}\\right)^{2} = 64^{2} x^{2}$, so $f(x)^{3} = 64 x^{2} \\frac{1+x}{1-x}$. Hence $f(x) = 4 \\left( \\frac{x^{2}(1+x)}{1-x} \\right)^{1/3}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32203,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the triangle $ABC$, the midpoints of $AC$ and $AB$ are $M$ and $N$ respectively. $BM$ and $CN$ meet at $P$. Show that if it is possible to inscribe a circle in the quadrilateral $AMPN$ (touching every side), then $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nIf the quadrilateral has an inscribed circle then $AM + PN = AN + PM$ (consider the tangents to the circle from $A, M, P, N$). But if $AB > AC$, then $BM > CN$ (see below). We have $AN = AB / 2$, $PM = BM / 3$, $AM = AC / 2$, $PN = CN / 3$, so it follows that $AM + PN < AN + PM$. Similarly, $AB < AC$ implies $AM + PN > AN + PM$, so the triangle must be isosceles.\n\nTo prove the result about the medians, note that $BM^2 = BC^2 + CM^2 - 2 BC \\cdot CM \\cos C = (BC - CM \\cos C)^2 + (CM \\sin C)^2$. Similarly, $CN^2 = (BC - BN \\cos B)^2 + (BN \\sin B)^2$. But $MN$ is parallel to $BC$, so $CM \\sin C = BN \\sin B$. But $AB > AC$, so $BN > CM$ and $B < C$, so $\\cos B > \\cos C$, hence $BN \\cos B > CM \\cos C$ and $BC - CM \\cos C > BC - BN \\cos B$. So $BM > CN$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32204,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that if $(2+\\sqrt{3})^{k}=1+m+n \\sqrt{3}$, for positive integers $m, n, k$ with $k$ odd, then $m$ is a perfect square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe have $(2+\\sqrt{3})^{4}=97+56 \\sqrt{3}=14(7+4 \\sqrt{3})-1=14(2+\\sqrt{3})^{2}-1$. Hence $(2+\\sqrt{3})^{k+2}=14 (2+\\sqrt{3})^{k}-(2+\\sqrt{3})^{k-2}$. Thus if $(2+\\sqrt{3})^{k}=a_{k}+b_{k} \\sqrt{3}$, then $a_{k+2}=14 a_{k}-a_{k-2}$.\n\nNow suppose the sequence $c_{k}$ satisfies $c_{1}=1, c_{2}=5, c_{k+1}=4 c_{k}-c_{k-1}$. We claim that $c_{k}^{2}-c_{k-1} c_{k+1}=6$. Induction on $k$. We have $c_{3}=19$, so $c_{2}^{2}-c_{1} c_{3}=25-19=6$. Thus the result is true for $k=2$. Suppose it is true for $k$. Then $c_{k+1}=4 c_{k}-c_{k-1}$, so $c_{k+1}^{2}=4 c_{k} c_{k+1}-c_{k-1} c_{k+1}=4 c_{k} c_{k+1}-c_{k}^{2}+6=c_{k}(4 c_{k+1}-c_{k})+6=c_{k} c_{k+2}+6$, so the result is true for $k+1$.\n\nNow put $d_{k}=c_{k}^{2}+1$. We show that $d_{k+2}=14 d_{k+1}-d_{k}$. Induction on $k$. We have $d_{1}=2, d_{2}=26, d_{3}=362=14 d_{2}-d_{1}$, so the result is true for $k=1$. Suppose it is true for $k$. We have $c_{k+3}-4 c_{k+2}+c_{k+1}=0$. Hence $12+2 c_{k+3} c_{k+1}-8 c_{k+2} c_{k+1}+2 c_{k+1}^{2}=12$. Hence $2 c_{k+2}^{2}-8 c_{k+2} c_{k+1}+2 c_{k+1}^{2}=12$. Hence $16 c_{k+2}^{2}-8 c_{k+2} c_{k+1}+c_{k+1}^{2}+1=14 c_{k+2}^{2}+14-c_{k+1}^{2}-1$, or $(4 c_{k+2}-c_{k+1})^{2}+1=14(c_{k+2}^{2}+1)-(c_{k+1}^{2}+1)$, or $c_{k+3}^{2}+1=14(c_{k+2}^{2}+1)-(c_{k+1}^{2}+1)$, or $d_{k+3}=14 d_{k+2}-d_{k+1}$. So the result is true for all $k$.\n\nBut $a_{1}=2, a_{3}=26$ and $a_{2k+3}=14 a_{2k+1}-a_{2k-1}$, and $d_{1}=2, d_{2}=26$ and $d_{k+1}=14 d_{k}-d_{k-1}$. Hence $a_{2k-1}=d_{k}=c_{k}^{2}+1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32205,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine the sequence $\\{p\\}_1, \\{p\\}_2, \\{p\\}_3, \\ldots$ as follows. $\\{p\\}_1 = 2$, and $\\{p\\}_n$ is the largest prime divisor of $\\{p\\}_1 \\{p\\}_2 \\ldots \\{p\\}_{n-1} + 1$. Prove that $5$ does not occur in the sequence.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32206,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow that the roots $r, s, t$ of the equation $x(x-2)(3x-7)=2$ are real and positive. Find $\\tan^{-1} r + \\tan^{-1} s + \\tan^{-1} t$.",
    "options": [],
    "answer": "3π/4",
    "solution": "Solution:\n\nPut $f(x) = x(x-2)(3x-7) - 2 = 3x^{3} - 13x^{2} + 14x - 2$. Then $f(0) = -2$, $f(1) = 2$, so there is a root between $0$ and $1$. $f(2) = -2$, so there is another root between $1$ and $2$. $f(3) = 4$, so the third root is between $2$ and $3$. $f(x) = 0$ has three roots, so they are all real and positive.\n\nWe have $\\tan(a + b + c) = \\dfrac{\\tan a + \\tan b + \\tan c - \\tan a \\tan b \\tan c}{1 - (\\tan a \\tan b + \\tan b \\tan c + \\tan c \\tan a)}$. So putting $a = \\tan^{-1} r$, $b = \\tan^{-1} s$, $c = \\tan^{-1} t$, we have\n\n$$\n\\tan(a + b + c) = \\frac{(r + s + t) - rst}{1 - (rs + st + tr)} = \\frac{13/3 - 2/3}{1 - 14/3} = \\frac{11/3}{-11/3} = -1.\n$$\n\nSo $a + b + c = -\\pi/4 + k\\pi$. But we know that each of $r, s, t$ is real and positive, so $a + b + c$ lies in the range $0$ to $3\\pi/2$. Hence $a + b + c = 3\\pi/4$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32207,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is a convex quadrilateral. $P$, $Q$ are points on the sides $AD$, $BC$ respectively such that $AP / PD = BQ / QC = AB / CD$. Show that the angle between the lines $PQ$ and $AB$ equals the angle between the lines $PQ$ and $CD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nIf $AB$ is parallel to $CD$, then it is obvious that $PQ$ is parallel to both. So assume $AB$ and $CD$ meet at $O$. Take $O$ as the origin for vectors. Let $\\mathbf{e}$ be a unit vector in the direction $OA$ and $\\mathbf{f}$ a unit vector in the direction $OC$. Take the vector $OA$ to be $a\\mathbf{e}$, $OB$ to be $b\\mathbf{e}$, $OC$ to be $c\\mathbf{f}$, and $OD$ to be $d\\mathbf{f}$. Then $OP$ is $((d-c)a\\mathbf{e} + (a-b)d\\mathbf{f})/(d-c+a-b)$ and $OQ$ is $((d-c)b\\mathbf{e} + (a-b)c\\mathbf{f})/(d-c+a-b)$. Hence $PQ$ is $(c-d)(a-b)(\\mathbf{e}+\\mathbf{f})/(d-c+a-b)$. But $\\mathbf{e}$ and $\\mathbf{f}$ are unit vectors, so $\\mathbf{e}+\\mathbf{f}$ makes the same angle with each of them and hence $PQ$ makes the same angle with $AB$ and $CD$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32208,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic progression. Show that the triangle must be equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the sides be $a$, $a + d$, $a + 2d$ with $d \\geq 0$. Then the altitudes are $k / a$, $k / (a + d)$, $k / (a + 2d)$, where $k$ is twice the area. We claim that $k / a + k / (a + 2d) > 2k / (a + d)$ unless $d = 0$. This is equivalent to $(a + d)(a + 2d) + a(a + d) > 2a(a + 2d)$ or $2d^2 > 0$, which is obviously true. So the altitudes can only form an arithmetic progression if $d = 0$ and hence the triangle is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32209,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe positive integers $a$, $b$, $c$, $d$, $p$, $q$ satisfy $ad - bc = 1$ and $a / b > p / q > c / d$. Show that $q \\geq b + d$ and that if $q = b + d$, then $p = a + c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$p / q > c / d$ implies $pd > cq$ and hence $pd \\geq cq + 1$, so $p / q \\geq c / d + 1 / (qd)$. Similarly, $a / b > p / q$ implies $a / b \\geq p / q + 1 / (bq)$. So $a / b - c / d \\geq 1 / (qd) + 1 / (qb) = (b + d) / (qbd)$. But $a / b - c / d = 1 / bd$. Hence $q \\geq b + d$.\n\nNow assume $q = b + d$. We have $ad - bc = 1 \\leq d$, so $ad + cd - d \\leq bc + cd$ and hence $(a + c - 1) / (b + d) \\leq c / d$. So $p \\geq a + c$. Similarly $ad - bc \\leq b$, so $bc + b + ab \\geq ad + ab$, so $(a + c + 1) / (b + d) \\geq a / b$. So $p \\leq a + c$. Hence $p = a + c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32210,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$P$ is a fixed point in the plane. Show that amongst triangles $ABC$ such that $PA = 3$, $PB = 5$, $PC = 7$, those with the largest perimeter have $P$ as incenter.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nGiven points $P$, $B$, $C$ and a fixed circle center $P$, we show that the point $A$ on the circle which maximises $AB + AC$ is such that $PA$ bisects angle $BAC$. Consider a point $A'$ close to $A$. Then the change in $AB + AC$ as we move $A$ to $A'$ is $AA'(\\sin PAC - \\sin PAB) + O(AA'^2)$. So for a maximal configuration we must have $\\sin PAC = \\sin PAB$, otherwise we could get a larger sum by taking $A'$ on one side or the other. This applies to each vertex of the triangle, so $P$ must be the incenter.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32211,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPoints $A_1, A_2, \\ldots, A_n$ are equally spaced on the side $BC$ of the triangle $ABC$ (so that $BA_1 = A_1A_2 = \\ldots = A_{n-1}A_n = A_nC$). Similarly, points $B_1, B_2, \\ldots, B_n$ are equally spaced on the side $CA$, and points $C_1, C_2, \\ldots, C_n$ are equally spaced on the side $AB$. Show that\n$$(AA_1^2 + AA_2^2 + \\ldots + AA_n^2 + BB_1^2 + BB_2^2 + \\ldots + BB_n^2 + CC_1^2 + \\ldots + CC_n^2)$$\nis a rational multiple of $$(AB^2 + BC^2 + CA^2).$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nUsing the cosine formula, $AA_k^2 = AB^2 + k^2 BC^2/(n+1)^2 - 2k AB \\cdot BC/(n+1) \\cos B$. So\n$$\n\\sum AA_k^2 = n AB^2 + BC^2/(n+1)^2 (1^2 + 2^2 + \\ldots + n^2) - 2 AB \\cdot BC \\cos B (1 + 2 + \\ldots + n)/(n+1).\n$$\nSimilarly for the other two sides.\n\nThus the total sum is\n$$\nn(AB^2 + BC^2 + CA^2) + n(2n+1)/(6(n+1))(AB^2 + BC^2 + CA^2) - n(AB \\cdot BC \\cos B + BC \\cdot CA \\cos C + CA \\cdot AB \\cos A).\n$$\nBut $AB \\cdot BC \\cos B = (AB^2 + BC^2 - CA^2)/2$, so\n$$\nAB \\cdot BC \\cos B + BC \\cdot CA \\cos C + CA \\cdot AB \\cos A = (AB^2 + BC^2 + CA^2)/2.\n$$\nThus the sum is a rational multiple of $(AB^2 + BC^2 + CA^2)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32212,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $k^{3} = 2$ and let $x$, $y$, $z$ be any rational numbers such that $x + y k + z k^{2}$ is non-zero. Show that there are rational numbers $u$, $v$, $w$ such that $(x + y k + z k^{2})(u + v k + w k^{2}) = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe need $x u + 2 z v + 2 y w = 1$, $y u + x v + 2 z w = 0$, $z u + y v + x w = 0$. This is just a straightforward set of linear equations. Solving, we get:\n\n$$\n\\begin{align*}\nu &= \\frac{x^{2} - 2 y z}{d}, \\\\\nv &= \\frac{2 z^{2} - x y}{d}, \\\\\nw &= \\frac{y^{2} - x z}{d},\n\\end{align*}\n$$\n\nwhere $d = x^{3} + 2 y^{3} + 4 z^{3} - 6 x y z$.\n\nThis would fail if $d = 0$. But if $d = 0$, then multiplying through by a suitable integer we have $6 m n r = m^{3} + 2 n^{3} + 4 r^{3}$ for some integers $m$, $n$, $r$. But we can divide by any common factor of $m$, $n$, $r$ to get them without any common factor. But $6 m n r$, $2 n^{3}$, $4 r^{3}$ are all even, so $m$ must be even. Put $m = 2 M$. Then $12 M n r = 8 M^{3} + 2 n^{3} + 4 r^{3}$, so $6 M n r = 4 M^{3} + n^{3} + 2 r^{3}$. But $6 M n r$, $4 M^{3}$ and $2 r^{3}$ are all even, so $n$ must be even. Put $n = 2 N$. Then $12 M N r = 4 M^{3} + 8 N^{3} + 2 r^{3}$, so $6 M N r = 2 M^{3} + 4 N^{3} + r^{3}$, so $r$ must be even. So $m$, $n$, $r$ had a common factor $2$. Contradiction. So $d$ cannot be zero.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32213,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $S$ be the collection of all sets of $n$ distinct positive integers, with no three in arithmetic progression. Show that there is a member of $S$ which has the largest sum of the inverses of its elements (you do not have to find it or to show that it is unique).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nInduction on $n$.\n\nFor $n = 1$, $\\{1\\}$ is obviously maximal.\n\nNow suppose $a_1 < a_2 < \\ldots < a_n$ is a maximal set for $n$.\n\nTake $a_{n+1}$ to be the smallest integer $> a_n$ such that $\\{a_1, a_2, \\ldots, a_{n+1}\\}$ has no three members in arithmetic progression.\n\nNow consider the sequences $b_1 < b_2 < \\ldots < b_n$ which have no three in arithmetic progression and $b_{n+1} \\leq a_{n+1}$. There are only finitely many such sequences. So we can find one which is maximal. Suppose it is $c_1 < c_2 < \\ldots < c_{n+1}$.\n\nNow take whichever of $a_i$, $c_i$ has the larger sum of inverses. It is clearly maximal with respect to sequences whose largest member is $\\leq a_{n+1}$.\n\nSuppose we have a sequence $x_1 < x_2 < \\ldots < x_{n+1}$ with no three in arithmetic progression and $x_{n+1} > a_{n+1}$. Then we have $1 / x_{n+1} < 1 / a_{n+1}$ and, by induction, $1 / x_1 + \\ldots + 1 / x_n \\leq 1 / a_1 + \\ldots + 1 / a_n$, so $1 / x_1 + \\ldots + 1 / x_{n+1} < 1 / a_1 + \\ldots + 1 / a_{n+1}$, so it is worse than the sequence we have chosen.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32214,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven positive real numbers $x$, $y$, $z$ each less than $\\pi / 2$, show that $\\pi / 2 + 2 \\sin x \\cos y + 2 \\sin y \\cos z > \\sin 2x + \\sin 2y + \\sin 2z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe have $\\sin 2x + \\sin 2y + \\sin 2z - 2 \\sin x \\cos y - 2 \\sin y \\cos z = 2 \\sin x (\\cos x - \\cos y) + 2 \\sin y (\\cos y - \\cos z) + 2 \\sin z \\cos z$, so we wish to show that\n\n$\\sin x (\\cos x - \\cos y) + \\sin y (\\cos y - \\cos z) + \\sin z \\cos z < \\pi / 2$ (*).\n\n![](attached_image_1.png)\n\nWe have to consider six cases:\n\n(1) $x \\leq y \\leq z$;\n\n(2) $x \\leq z \\leq y$;\n\n(3) $y \\leq x \\leq z$;\n\n(4) $y \\leq z \\leq x$;\n\n(5) $z \\leq x \\leq y$;\n\n(6) $z \\leq y \\leq x$.\n\nThe first case is obvious from the diagram, because the lhs represents the shaded area, and the rhs represents the whole quarter circle.\n\nIn cases (2) and (5) the second term is negative, and $-\\sin y < -\\sin x$, so the sum of the first two terms is less than $\\sin x (\\cos x - \\cos y) + \\sin x (\\cos y - \\cos z) = \\sin x (\\cos x - \\cos z)$. But by the same argument as the first case the two rectangles represented by $\\sin x (\\cos x - \\cos z)$ and $\\sin z \\cos z$ are disjoint and fit inside the quarter circle. So we have proved (2) and (5).\n\nIn cases (3) and (4), the first term is negative. The remaining two terms represent disjoint rectangles lying inside the quarter circle, so again the inequality holds.\n\nIn case (6) the first two terms are negative. The last term is $1/2 \\sin 2z \\leq 1/2 < \\pi / 2$, so the inequality certainly holds.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32215,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $a$, $b$, $c$, are the sides of a triangle, show that\n$$\n\\frac{a-b}{a+b} + \\frac{b-c}{b+c} + \\frac{c-a}{a+c} < \\frac{1}{16}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPut $f(a, b, c) = \\frac{a-b}{a+b} + \\frac{b-c}{b+c} + \\frac{c-a}{a+c}$. Let $A, B, C$ be a permutation of $a, b, c$, with $A \\leq B \\leq C$. If $(A, B, C) = (b, a, c), (a, c, b)$ or $(c, b, a)$, then $f(a, b, c) = X$, where\n$$\nX = \\frac{B-A}{B+A} + \\frac{C-B}{C+B} - \\frac{C-A}{A+C}.\n$$\nIf $(A, B, C) = (a, b, c), (b, c, a)$ or $(c, a, b)$, then $f(a, b, c) = -X$.\n\nPut $B = A + h$, $C = B + k = A + h + k$, where $h, k \\geq 0$. Since $A, B, C$ are the sides of a triangle, we also have $A + B > C$ or $A > k$. So\n$$\nX = \\frac{h}{2A + h} + \\frac{k}{2A + 2h + k} - \\frac{h + k}{2A + h + k} = \\frac{hk(h + k)}{(2A + h)(2A + h + k)(2A + 2h + k)}.\n$$\nThis is obviously non-negative. We claim also that it is $< \\frac{1}{20}$. That is equivalent to:\n$$\n20h^2k + 20hk^2 < (2A + h)(2A + h + k)(2A + 2h + k).\n$$\nSince $k < A$ it is sufficient to show that\n$$\n20h^2k + 20hk^2 \\leq (2k + h)(2k + h + k)(2k + 2h + k)\n$$\n$$\n= 18k^3 + 27hk^2 + 13h^2k + 2h^3\n$$\nor\n$$\n18k^3 + 7hk^2 - 7h^2k + 2h^3 \\geq 0.\n$$\nBut $7k^2 - 7hk + 2h^2 = 7(k - h/2)^2 + h^2/4 \\geq 0$ and $h$ and $k$ are non-negative, so\n$$\n18k^3 + h\\left(7k^2 - 7hk + 2h^2\\right) \\geq 0.\n$$\nThus we have established that $0 \\leq X < \\frac{1}{20}$, which shows that $f(a, b, c) < \\frac{1}{20}$, which is slightly stronger than the required result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32216,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe incircle of the triangle $ABC$ touches $AC$ at $M$ and $BC$ at $N$ and has center $O$. $AO$ meets $MN$ at $P$ and $BO$ meets $MN$ at $Q$. Show that $MP \\cdot OA = BC \\cdot OQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nThe key to getting started is to notice that angle $AQB = 90^\\circ$.\nAngle $BAQ = 90^\\circ - B/2$, so angle $OAQ = 90^\\circ - B/2 - A/2 = C/2$. So $OQ = AO \\sin (C/2)$. Thus we have to show that $MP = BC \\sin (C/2)$.\n\nLet the incircle touch $AB$ at $L$ and let $Y$ be the midpoint of $ML$ (also the intersection of $ML$ with $AO$). Angle $NMC = 90^\\circ - C/2$. It is also $A/2 +$ angle $MPY$, so angle $MPY = 90^\\circ - C/2 - A/2 = B/2$. Hence $MP = MY / \\sin (B/2)$. We have $MY = MO \\sin (MOA) = r \\cos (A/2)$ (where $r$ is the inradius, as usual). So $MP = (r \\cos (A/2)) / \\sin (B/2)$.\n\nWe have $BC = BN + NC = r (\\cot (B/2) + \\cot (C/2))$, so $MP / BC = (\\cos (A/2)) / (\\sin (B/2)(\\cot (B/2) + \\cot (C/2)))$. Hence $MP / (BC \\sin (C/2)) = (\\cos (A/2)) / (\\cos (B/2) \\sin (C/2) + \\sin (B/2) \\cos (C/2)) = \\cos (A/2) / \\sin (B/2 + C/2) = 1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32217,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe function $f$ on the positive integers satisfies $f(1)=1$, $f(2n+1)=f(2n)+1$ and $f(2n)=3 f(n)$. Find the set of all $m$ such that $m=f(n)$ for some $n$.",
    "options": [],
    "answer": "All positive integers whose base-three representation contains only the digits 0 and 1 (i.e., no digit 2).",
    "solution": "Solution:\n\nWe show that to obtain $f(n)$, one writes $n$ in base 2 and then reads it in base 3. For example, $12 = 1100_2$, so $f(12) = 1100_3 = 36$. Let $g(n)$ be defined in this way. Then certainly $g(1) = 1$. Now $2n+1$ has the same binary expansion as $2n$ except for a final 1, so $g(2n+1) = g(2n) + 1$. Similarly, $2n$ has the same binary expansion as $n$ with the addition of a final zero. Hence $g(2n) = 3g(n)$. So $g$ is the same as $f$. Hence the set of all $m$ such that $m = f(n)$ for some $n$ is the set of all $m$ which can be written in base 3 without a digit 2.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32218,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that there are infinitely many solutions in positive integers to $2 a^{2} - 3 a + 1 = 3 b^{2} + b$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nPut $A = a - 1$ and the equation becomes $A (2A + 1) = b (3b + 1)$. Let $d$ be the greatest common divisor of $A$ and $b$. Put $A = d x$, $b = d y$. Then $x (2 d x + 1) = y (3 d y + 1)$. Since $x$ and $y$ are coprime, $x$ must divide $3 d y + 1$. So put $3 d y + 1 = n x$. Then $2 d x + 1 = n y$. Solving for $x$ and $y$ in terms of $n$ and $d$ we get $x = \\frac{n + 3 d}{n^{2} - 6 d^{2}}$, $y = \\frac{n + 2 d}{n^{2} - 6 d^{2}}$.\n\nSo we would certainly be home if we could show that there were infinitely many solutions to $n^{2} - 6 d^{2} = 1$. It is not hard to find the first few: $1^{2} - 6 \\cdot 0^{2} = 1$, $5^{2} - 6 \\cdot 2^{2} = 1$, $49^{2} - 6 \\cdot 20^{2} = 1$. We notice that $49^{2} = 2 \\cdot 5^{2} - 1$, so we wonder whether $n = 2 \\cdot 49^{2} - 1$ might be another solution and indeed we find it gives $d = 1960 = 2 \\cdot 49 \\cdot 20$. This suggests we try $(2 n^{2} - 1)^{2} - 6 (2 n d)^{2} = 4 n^{4} - 4 n^{2} + 1 - 24 n^{2} d^{2} = 4 n^{2} (n^{2} - 6 d^{2} - 1) + 1 = 1$. So there are indeed infinitely many solutions to $n^{2} - 6 d^{2} = 1$ and we are done.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32219,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe function $f$ is defined on the non-negative integers. $f\\left(2^{n}-1\\right)=0$ for $n=0,1,2, \\ldots$. If $m$ is not of the form $2^{n}-1$, then $f(m)=f(m+1)+1$. Show that $f(n)+n=2^{k}-1$ for some $k$, and find $f\\left(2^{1990}\\right)$",
    "options": [],
    "answer": "f(2^{1990}) = 2^{1990} - 1",
    "solution": "Solution:\n\nWe claim that if $2^{m} \\leq n < 2^{m+1}$, then $f(n) = 2^{m+1} - n - 1$.\n\nPut $r = 2^{m+1} - n$. Then the claim follows by induction on $r$.\n\nHence $f\\left(2^{1990}\\right) = 2^{1990} - 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32220,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$I$ is the incenter of the triangle $ABC$ and the incircle touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $AD$ meets the incircle again at $P$. $M$ is the midpoint of $EF$. Show that $P$, $M$, $I$, $D$ are cyclic (or the points are collinear).",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\n$\\angle AEI = \\angle AME = 90^\\circ$, so $AEI$ and $AME$ are similar. Hence $AM / AE = AE / AI$ or $AM \\cdot AI = AE^2$. $AE$ is tangent to the incircle, so $AE^2 = AP \\cdot AD$. Hence $AM \\cdot AI = AP \\cdot AD$, so if $P$, $M$, $I$, $D$ are not collinear, then they are cyclic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32221,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$f(x) = (x + b)^2 + c$, where $b$ and $c$ are integers. If the prime $p$ divides $c$, but $p^2$ does not divide $c$, show that $f(n)$ is not divisible by $p^2$ for any integer $n$. If an odd prime $q$ does not divide $c$, but divides $f(n)$ for some $n$, show that for any $r$, we can find $N$ such that $q^r$ divides $f(N)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nThe first part is trivial. If $p$ does not divide $(x + b)$, then it does not divide $(x + b)^2$, so it does not divide $(x + b)^2 + c$. On the other hand, if $p$ does divide $x + b$, then $p^2$ divides $(x + b)^2$, so $p^2$ does not divide $(x + b)^2 + c$.\n\nFor the second part, we use induction on $r$. For $r = 1$, we are given that $q$ divides $f(n)$. Now suppose that $q^r$ divides $f(N)$ for some $N$. If $q^{r+1}$ divides $f(N)$, then we are done. So suppose $q^{r+1}$ does not divide $f(N)$, so $f(N) = q^r h$ where $q$ does not divide $h$. We have\n$$\nf\\left(N + kq^r\\right) = f(N) + q^r (2N + 2b) k = q^r h + q^r (2N + 2b) k.\n$$\nNow $q$ divides $(N + b)^2 + c$, and does not divide $c$, so it does not divide $(N + b)^2$ and hence does not divide $N + b$. It is odd, so it does not divide $2N + 2b$. Hence we can find $k$ such that $k(2N + 2b) \\equiv -h \\pmod{q}$. Then we have $q^{r+1}$ divides $f\\left(N + kq^r\\right)$, which completes the induction.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32222,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe circle $C$ has diameter $AB$. The tangent at $B$ is $T$. For each point $M$ (not equal to $A$) on $C$ there is a circle $C'$ which touches $T$ and touches $C$ at $M$. Find the point at which $C'$ touches $T$ and find the locus of the center of $C'$ as $M$ varies. Show that there is a circle orthogonal to all the circles $C'$.",
    "options": [],
    "answer": "The tangency point on T is N = AM ∩ T. The locus of centers O′ is the parabola y^2 = 4ax with vertex at B and axis along AB (with coordinates B = (0,0), A = (2a,0), O = (a,0)). A fixed circle orthogonal to every C′ is the circle centered at A with radius AB.",
    "solution": "Solution:\n\n![](attached_image_1.png)\nLet $O$ be the center of $C$. Let the line $AM$ meet $T$ at $N$. Let the perpendicular to $T$ at $N$ meet the line $OM$ at $O'$. Then $\\angle O'NM = \\angle MAB$ ($O'N$ parallel to $AB$, because both perpendicular to $T$) $= \\angle OMA$ ($OM = OA$) $= \\angle O'MN$. So $O'M = O'N$. Hence $O'$ is the center of $C'$.\n\nTake $B$ to be the origin and $A$ to be the point $(2a, 0)$, so $O$ is $(a, 0)$ and $C$ has radius $a$. If $O'$ is $(x, y)$, then we require that $O'O = x + a$ or $(x - a)^2 + y^2 = (x + a)^2$, or $y^2 = 4a x$, which is a parabola with vertex $B$ and axis the $x$-axis.\n\nTriangles $AMB$, $ABN$ are similar ($\\angle AMB = \\angle ABN = 90^\\circ$), so $AM / AB = AB / AN$ and hence $AM \\cdot AN = AB^2$. Now consider the circle center $A$ radius $AB$. It must meet the circle $C'$, because it contains the point $M$. Suppose it meets at $X$. Then $AX^2 = AB^2 = AM \\cdot AN$, so $AX$ is tangent to $C'$ and hence the circles are orthogonal.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32223,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$A$ and $B$ are opposite corners of an $n \\times n$ board, divided into $n^{2}$ squares by lines parallel to the sides. In each square the diagonal parallel to $AB$ is drawn, so that the board is divided into $2 n^{2}$ small triangles. The board has $(n+1)^{2}$ nodes and a large number of line segments, each of length $1$ or $\\sqrt{2}$. A piece moves from $A$ to $B$ along the line segments. It never moves along the same segment twice and its path includes exactly two sides of every small triangle on the board. For which $n$ is this possible?",
    "options": [],
    "answer": "n = 2",
    "solution": "Solution:\n\nThe diagram above shows that $n=2$ is possible (the path is $AHEFGHCDIHB$). Now suppose $n > 2$.\n\nNote that if $X$ is any vertex except $A$ or $B$, then an even number of segments with endpoint $X$ must be in the path.\n\nLet $F$ be the bottom left-hand vertex. Two sides of the triangle $EFG$ are in the path, so at least one of $EF$ and $FG$ is. But $EF$ and $EG$ are the only segments with endpoint $F$, so an even number of them must be in the path, so both are in the path. Hence, again considering $EFG$, $EG$ is not in the path. Hence, considering $EHG$, $EH$ and $HG$ are in the path.\n\n$E$ has an even number of segments on the path, so $CE$ is not on the path. Hence (considering $CEH$) $CH$ is on the path. Similarly, $GJ$ is not on the path and $HJ$ is on the path. An even number of segments at $H$ are on the path, so $DH$ and $HI$ are either both on the path or neither is on the path. But (considering $DHI$) at least one must be, so they both are. Hence $DI$ is not, and $CD$ is not.\n\nSince $n>2$, $C$ is not the top left vertex. Considering $MCD$, $MC$ and $MD$ are both on the path. Considering $DLI$, $DL$ is on the path. There must be an even number of segments at $D$, so $DP$ is on the path. Hence $MP$ is not. Now $M$ cannot be the top left vertex (with $n=3$) because then it should have an odd number of segments, whereas it would have two ($MC$ and $MD$). So there must be a vertex $N$ above $M$. Considering $NMP$, $MN$ must be in the path. But now $M$ has an odd number of segments. Contradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32224,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe number $1$ or the number $-1$ is assigned to each vertex of a cube. Then each face is given the product of its four vertices. What are the possible totals for the resulting $14$ numbers?",
    "options": [],
    "answer": "-10, -6, -2, 2, 6, 14",
    "solution": "Solution:\n\nIf every vertex is $1$, we get $14$ and that is clearly the highest possible total. The lowest possible total cannot be lower than $-14$, but we cannot even achieve that because if all the vertices are $-1$, then all the faces are $1$.\n\nIf we change a vertex, then we also change three faces. If the vertex and the three faces are all initially the same, then we make a change of $\\pm 8$. If three are of one kind and one the opposite, then we make a change of $\\pm 4$. If two are of one kind and two the opposite, then we make no change. Thus any sequence of changes must take us to $14 + 4n$ for some integer $n$. But we have already shown that the total is greater than $-14$ and at most $14$, so the only possibilities are $-10, -6, -2, 2, 6, 10$ and $14$.\n\nWe show that $10$ is not possible. If more than $2$ vertices are $-1$, then the vertex total is at most $2$, there are only $6$ faces, so the total is less than $10$. If all vertices are $1$, then the total is $14$. If all but one vertex is $1$, then the total is $6$. So the only possibility for $10$ is just two vertices $-1$. But however we choose any two vertices, there is always a face containing only one of them, so at least one face is $-1$, so the face total is at most $4$ and the vertex total is $4$, so the total is less than $10$. The other totals are possible, for example:\n\n$14$: all vertices $1$\n\n$6$: one vertex $-1$, rest $1$\n\n$2$: three vertices of one face $-1$, rest $1$\n\n$-2$: all vertices $-1$\n\n$-6$: all vertices but one $-1$\n\n$-10$: two opposite corners $1$, rest $-1$",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32225,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo perpendicular lines divide a square into four parts, three of which have area $1$. Show that the fourth part also has area $1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32226,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$f$ is a function defined on all reals in the interval $[0,1]$ and satisfies $f(0) = 0$, $f(x/3) = f(x)/2$, $f(1-x) = 1 - f(x)$. Find $f(18/1991)$.",
    "options": [],
    "answer": "5/128",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32227,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind a number $N$ with five digits, all different and none zero, which equals the sum of all distinct three digit numbers whose digits are all different and are all digits of $N$.",
    "options": [],
    "answer": "35964",
    "solution": "Solution:\nThere are $4 \\times 3 = 12$ numbers with a given digit of $n$ in the units place. Similarly, there are $12$ with it in the tens place and $12$ with it in the hundreds place. So the sum of the 3 digit numbers is $12 \\times 111(a+b+c+d+e)$, where $n=abcde$. So $8668a=332b+1232c+1322d+1331e$. We can easily see that $a=1$ is too small and $a=4$ is too big, so $a=2$ or $3$. Obviously $e$ must be even. $0$ is too small, so $e=2,4,6$ or $8$. Working mod $11$, we see that $0=2b+2d$, so $b+d=11$. Working $\\bmod 7$, we see that $2a=3b+6d+e$. Using the $\\bmod 11$ result, $b=2, d=9$ or $b=3, d=8$ or $b=4, d=7$ or $b=5, d=6$ or $b=6, d=5$ or $b=7, d=4$ or $b=8, d=3$ or $b=9, d=2$. Putting each of these into the $\\bmod 7$ result gives $2a-e=4,1,5,2,6,3,0,4$ mod $7$. So putting $a=2$ and remembering that $e$ must be $2,4,6,8$ and that all digits must be different gives $a, b, d, e = 2,4,7,6$ or $2,7,4,8$ or $2,8,3,4$ as the only possibilities. It is then straightforward but tiresome to check that none of these give a solution for $c$. Similarly putting $a=4$, gives $a, b, d, e=3,4,7,8$ or $3,5,6,4$ as the only possibilities. Checking, we find the solution above and no others.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32228,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p(m, n)$ be the polynomial $2 m^{2} - 6 m n + 5 n^{2}$. The range of $p$ is the set of all integers $k$ such that $k = p(m, n)$ for some integers $m$, $n$. Find which members of $\\{1,2, \\ldots, 100\\}$ are in the range of $p$. Show that if $h$ and $k$ are in the range of $p$, then so is $h k$.",
    "options": [],
    "answer": "The integers in 1 to 100 that are in the range are: 1, 2, 4, 5, 8, 9, 10, 13, 16, 17, 18, 20, 25, 26, 29, 32, 34, 36, 37, 40, 41, 45, 49, 50, 52, 53, 58, 61, 64, 65, 68, 72, 73, 74, 80, 81, 82, 85, 89, 90, 97, 98, 100.",
    "solution": "Solution:\n\nWe have $p(m, n) = (m - 2 n)^{2} + (m - n)^{2}$, so $p(2 a - b, a - b) = a^{2} + b^{2}$. Hence the range of $p$ is just the sums of two squares.\n\n$\\left(a^{2} + b^{2}\\right)\\left(c^{2} + d^{2}\\right) = (a c - b d)^{2} + (a d + b c)^{2}$, which establishes that if $h$ and $k$ are in the range, then so is $h k$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32229,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven three non-collinear points $M$, $N$, $H$ show how to construct a triangle which has $H$ as orthocenter and $M$ and $N$ as the midpoints of two sides.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nTake $H'$ so that $M$ is the midpoint of $HH'$. The circle diameter $NH'$ meets the line through $H$ perpendicular to $MN$ in two points (in general), either of which we may take as $A$. Then $B$ is the reflection of $A$ in $M$, and $C$ is the reflection of $A$ in $N$.\n\nTo see that this works, note that $M$ is the midpoint of $HH'$ and $AB$, so $AHBH'$ is a parallelogram. Hence $AH'$ is parallel to $BH$ and hence perpendicular to $AC$. In other words $\\angle NAH' = 90^{\\circ}$, so $A$ lies on the circle diameter $NH'$. $MN$ is parallel to $BC$, so $A$ lies on the perpendicular to $MN$ through $H$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32230,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$a_{n}$ is the last digit of $1+2+\\ldots+n$. Find $a_{1}+a_{2}+\\ldots+a_{1992}$.",
    "options": [],
    "answer": "6984",
    "solution": "Solution:\n\nIt is easy to compile the following table, from which we see that $a_{n}$ is periodic with period $20$, and indeed the sum for each decade (from $0$ to $9$) is $35$. Thus the sum for $1992$ is $199 \\cdot 35 + 5 + 6 + 8 = 6984$.\n\n| $n$ |   | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |\n| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |\n| 17 | 18 | 19 | 20 |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |\n| $a_{n}$ |   | 0 | 1 | 3 | 6 | 0 | 5 | 1 | 8 | 6 | 5 | 5 | 6 | 8 | 1 | 5 | 0 | 6 |\n| 3 | 1 | 0 | 0 |   |   |   |   |   |   |   |   |   | 6 |   | 1 | 5 |   |   |\n| sum |   | 0 | 1 | 4 | 10 | 10 | 15 | 16 | 24 | 30 | 35 | 40 | 46 | 54 | 55 | 60 | 60 | 66 |\n| 69 | 70 | 70 | 70 |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32231,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $f(x) = \\dfrac{a_{1}}{x + a_{1}} + \\dfrac{a_{2}}{x + a_{2}} + \\ldots + \\dfrac{a_{n}}{x + a_{n}}$, where $a_{i}$ are unequal positive reals. Find the sum of the lengths of the intervals in which $f(x) \\geq 1$.",
    "options": [],
    "answer": "a_1 + a_2 + \\cdots + a_n",
    "solution": "Solution:\n\nWLOG $a_{1} > a_{2} > \\ldots > a_{n}$. The graph of each $\\dfrac{a_{i}}{x + a_{i}}$ is a rectangular hyperbola with asymptotes $x = -a_{i}$ and $y = 0$. So it is not hard to see that the graph of $f(x)$ is made up of $n + 1$ strictly decreasing parts. For $x < -a_{1}$, $f(x)$ is negative. For $x \\in (-a_{i}, -a_{i+1})$, $f(x)$ decreases from $\\infty$ to $-\\infty$. Finally, for $x > -a_{n}$, $f(x)$ decreases from $\\infty$ to $0$. Thus $f(x) = 1$ at $n$ values $b_{1} < b_{2} < \\ldots < b_{n}$, and $f(x) \\geq 1$ on the $n$ intervals $(-a_{1}, b_{1}),\\ (-a_{2}, b_{2}),\\ \\ldots,\\ (-a_{n}, b_{n})$. So the sum of the lengths of these intervals is $\\sum (a_{i} + b_{i})$. We show that $\\sum b_{i} = 0$.\n\nMultiplying $f(x) = 1$ by $\\prod (x + a_{j})$ we get a polynomial of degree $n$:\n$$\n\\prod (x + a_{j}) - \\sum_{i} \\left(a_{i} \\prod_{j \\neq i} (x + a_{j})\\right) = 0\n$$\nThe coefficient of $x^{n}$ is $1$ and the coefficient of $x^{n-1}$ is $\\sum a_{j} - \\sum a_{i} = 0$. Hence the sum of the roots, which is $\\sum b_{i}$, is zero.\n\nTherefore, the sum of the lengths of the intervals is $\\sum a_{i}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32232,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABC$ is an equilateral triangle with side $2$. Show that any point $P$ on the incircle satisfies $PA^{2} + PB^{2} + PC^{2} = 5$. Show also that the triangle with side lengths $PA$, $PB$, $PC$ has area $\\sqrt{3}/4$.",
    "options": [],
    "answer": "PA^2 + PB^2 + PC^2 = 5; area = sqrt(3)/4",
    "solution": "Solution:\n\nTake vectors centered at the center $O$ of the triangle. Write the vector $OA$ as $\\mathbf{A}$ etc. Then\n$$\nPA^{2} + PB^{2} + PC^{2} = (\\mathbf{P} - \\mathbf{A})^{2} + (\\mathbf{P} - \\mathbf{B})^{2} + (\\mathbf{P} - \\mathbf{C})^{2} = 3 P^{2} + (A^{2} + B^{2} + C^{2}) - 2 \\mathbf{P} \\cdot (\\mathbf{A} + \\mathbf{B} + \\mathbf{C}) = 15 P^{2},\n$$\nsince $A^{2} = B^{2} = C^{2} = 4 P^{2}$ and $\\mathbf{A} + \\mathbf{B} + \\mathbf{C} = 0$. Finally, the side is $2$, so an altitude is $\\sqrt{3}$ and the inradius is $\\sqrt{3}/3 = 1/\\sqrt{3}$, so $PA^{2} + PB^{2} + PC^{2} = 15/3 = 5$.\n\nTake $Q$ outside the triangle so that $BQ = BP$ and $CQ = AP$. Then $BQC$ and $BPA$ are congruent, so $\\angle ABP = \\angle CBQ$ and hence $\\angle PBQ = 60^{\\circ}$, so $PBQ$ is equilateral. Hence $PQ$ is $PB$ and $PQC$ has sides equal to $PA$, $PB$, $PC$. If we construct two similar points outside the other two sides then we get a figure with total area equal to $2$ area $ABC$ and to $3$ area $PQC$ plus area of three equilateral triangles sides $PA$, $PB$, $PC$. Hence $3$ area $PQC = 2$ area $ABC$ $-$ area $ABC (PA^{2} + PB^{2} + PC^{2}) / PA^{2} = (3/4)$ area $ABC = (3\\sqrt{3})/4$. So area $PQC = (\\sqrt{3})/4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32233,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a_{n}, b_{n}$ be two sequences of integers such that:\n(1) $a_{0}=0$, $b_{0}=8$;\n(2) $a_{n+2}=2 a_{n+1}-a_{n}+2$, $b_{n+2}=2 b_{n+1}-b_{n}$,\n(3) $a_{n}^{2}+b_{n}^{2}$ is a square for $n>0$.\nFind at least two possible values for $\\left(a_{1992}, b_{1992}\\right)$.",
    "options": [],
    "answer": "(3976032, 7976) and (3960096, -7960)",
    "solution": "Solution:\n$a_{n}$ satisfies a standard linear recurrence relation with general solution $a_{n}=n^{2}+A n+k$. But $a_{0}=0$, so $k=0$. Hence $a_{n}=n^{2}+A n$. If you are not familiar with the general solution, then you can guess this solution and prove it by induction.\n\nSimilarly, $b_{n}=B n+8$. Hence $a_{n}^{2}+b_{n}^{2}=n^{4}+2 A n^{3}+(A^{2}+B^{2}) n^{2}+16 B n+64$. If this is a square, then looking at the constant and $n^{3}$ terms, it must be $\\left(n^{2}+A n+8\\right)^{2}$. Comparing the other terms, $A=B= \\pm 4$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32234,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConstruct a cyclic trapezium $ABCD$ with $AB$ parallel to $CD$, perpendicular distance $h$ between $AB$ and $CD$, and $AB + CD = m$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32235,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a triangle $ABC$, take $A'$ on the ray $BA$ (on the opposite side of $A$ to $B$) so that $AA' = BC$, and take $A''$ on the ray $CA$ (on the opposite side of $A$ to $C$) so that $AA'' = BC$. Similarly take $B'$, $B''$ on the rays $CB$, $AB$ respectively with $BB' = BB'' = CA$, and $C'$, $C''$ on the rays $AB$, $CB$. Show that the area of the hexagon $A''A'B''B'C'C''$ is at least 13 times the area of the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32236,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA palindrome is a positive integer which is unchanged if you reverse the order of its digits. For example, $23432$. If all palindromes are written in increasing order, what possible prime values can the difference between successive palindromes take?",
    "options": [],
    "answer": "2 and 11",
    "solution": "Solution:\nLet $x$ be a palindrome and $x'$ the next highest palindrome. If $x < 101$, then it is easy to see by inspection that $x' - x = 1, 2$ or $11$, so the only prime differences are $2$ and $11$.\n\nSo assume $x > 100$. If $x$ and $x'$ have the same final digit, then their difference is divisible by $10$ and hence not prime. So they must have different digits. Thus either $x = d 9 \\ldots 9 d$ and $x' = d' 0 \\ldots 0 d'$, where $d < 9$ and $d' = d + 1$, or $x'$ has one more digit than $x$ and $d = 9$, $d' = 1$. In the first case $x' - x = 11$. In the second case $x' - x = 2$. So again the only prime differences are $2$ and $11$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32237,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that any convex polygon of area $1$ is contained in some parallelogram of area $2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet the vertices $X$, $Y$ of the polygon be the two which are furthest apart. The polygon must lie between the lines through $X$ and $Y$ perpendicular to $XY$ (for if a vertex $Z$ lay outside the line through $Y$, then $ZY > XY$). Take two sides of a rectangle along these lines and the other two sides as close together as possible. There must be vertices $U$ and $V$ on each of the other two sides. But now the area of the rectangle is twice the area of $XUYV$, which is at most the area of the polygon. [In the case of a triangle one side of the rectangle will be $XY$.]",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32238,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all functions $f$ on the positive integers with positive integer values such that\n\n(1) if $x < y$, then $f(x) < f(y)$, and\n\n(2) $f(y f(x)) = x^{2} f(x y)$.",
    "options": [],
    "answer": "f(n) = n^2",
    "solution": "Solution:\n\nPut $y = 1$. Then $f(f(x)) = x^{2} f(x)$.\n\nPut $y = f(z)$, then $f(f(z) f(x)) = x^{2} f(x f(z)) = x^{2} z^{2} f(x z) = f(f(x z))$.\n\nBut $f$ is $(1,1)$ so $f(x z) = f(x) f(z)$.\n\nNow suppose $f(m) > m^{2}$ for some $m$. Then by (1), $f(f(m)) > f(m^{2}) = f(m \\cdot m) = f(m)^{2}$.\n\nBut $f(f(m)) = m^{2} f(m)$, so $m^{2} > f(m)$. Contradiction.\n\nSimilarly, suppose $f(m) < m^{2}$. Then $m^{2} f(m) = f(f(m)) < f(m^{2}) = f(m)^{2}$, so $m^{2} < f(m)$.\n\nContradiction. So we must have $f(m) = m^{2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32239,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABC$ is an equilateral triangle. $D$ is on the side $AB$ and $E$ is on the side $AC$ such that $DE$ touches the incircle. Show that $AD/DB + AE/EC = 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nPut $BD = x$, $CE = y$, $BC = a$. Then since the two tangents from $B$ to the incircle are of equal length, and similarly the two tangents from $D$ and $E$, we have $ED + BC = BD + CE$, or $ED = x + y - a$.\n\nBy the cosine law, $ED^2 = AE^2 + AD^2 - AE \\cdot AD$. Substituting and simplifying, we get $a = \\dfrac{3 x y}{x + y}$.\n\nHence $AD/DB = \\dfrac{2y - x}{x + y}$ and $AE/EC = \\dfrac{2x - y}{x + y}$ with sum $1$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32240,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $P$ and $Q$ are two points in the plane, let $m(PQ)$ be the perpendicular bisector of $PQ$. $S$ is a finite set of $n > 1$ points such that:\n\n(1) if $P$ and $Q$ belong to $S$, then some point of $m(PQ)$ belongs to $S$,\n\n(2) if $PQ$, $P'Q'$, $P''Q''$ are three distinct segments, whose endpoints are all in $S$, then if there is a point in all of $m(PQ)$, $m(P'Q')$, $m(P''Q'')$ it does not belong to $S$.\n\nWhat are the possible values of $n$?",
    "options": [],
    "answer": "3 and 5",
    "solution": "Solution:\n\nThere are $n(n-1)/2$ pairs of points. Each has a point of $S$ on its bisector. But each point of $S$ is on at most two bisectors, so $2n \\geq n(n-1)/2$. Hence $n \\leq 5$.\n\nThe equilateral triangle and regular pentagon show that $n = 3, 5$ are possible.\n\nConsider $n = 4$. There are 6 pairs of points, so at least one point of $S$ must be on two bisectors. WLOG $A$ is on the bisectors of $BC$ and $BD$. But then it is also on the bisector of $CD$.\n\nContradiction.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32241,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWe say that two non-negative integers are related if their sum uses only the digits $0$ and $1$. For example, $22$ and $79$ are related. Let $A$ and $B$ be two infinite sets of non-negative integers such that:\n\n(1) if $a \\in A$ and $b \\in B$, then $a$ and $b$ are related,\n\n(2) if $c$ is related to every member of $A$, then it belongs to $B$,\n\n(3) if $c$ is related to every member of $B$, then it belongs to $A$.\n\nShow that in one of the sets $A$, $B$ we can find an infinite number of pairs of consecutive numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose there is a member of $A$ with last digit $d$. Then every member of $B$ must have one of two possible last digits. Suppose there are members of $B$ with both possibilities. Then every member of $A$ must have last digit $d$. So either every member of $A$ has the same last digit or every member of $B$ has the same last digit (or both). Suppose every member of $A$ has the same last digit $d$.\n\nBut now if $n$ belongs to $B$ and $n+d$ has last digit $0$, then $n+1+d$ has last digit $1$. Moreover, if $m$ is any member of $A$, then $m+n$ has last digit $0$ and other digits all $0$ or $1$. Hence $m+n+1$ has last digit $1$ and other digits all $0$ or $1$, so $n+1$ must also belong to $B$. Similarly, if $n$ is in $B$ and $n+d$ has last digit $1$, then $n-1$ must also belong to $B$. So in either case there are infinitely many pairs of consecutive numbers in $B$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32242,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow that there is a number $1 < b < 1993$ such that if $1994$ is written in base $b$ then all its digits are the same. Show that there is no number $1 < b < 1992$ such that if $1993$ is written in base $b$ then all its digits are the same.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAny even number $2n$ can be written as $22$ in base $n - 1$. In particular, $1994 = 22_{996}$.\n\nWe have to show that we cannot write $1993 = aaa\\ldots a_b$. If the number has $n$ digits, then\n$$\n1993 = a(1 + b + \\ldots + b^{n-1}) = a\\frac{b^n - 1}{b - 1}.\n$$\nBut $1993$ is prime, so $a$ must be $1$. Hence\n$$\nb^{n-1} + \\ldots + b - 1992 = 0.\n$$\nSo $b$ must divide $1992 = 2^3 \\cdot 3 \\cdot 83$. We cannot have $n = 2$, for then $b = 1992$ and we require $b < 1992$. So $n > 2$. But $83^2 = 6889 > 1993$, so $b$ must divide $24$. Hence $b = 2, 3, 4, 6, 8, 12$, or $24$. But we can easily check that none of these work:\n\n$1 + 2 + 2^2 + \\ldots + 2^9 = 1023$, $1 + \\ldots + 2^{10} = 2047$.\n\n$1 + 3 + \\ldots + 3^6 = 1093$, $1 + \\ldots + 3^7 = 3280$\n\n$1 + 4 + \\ldots + 4^5 = 1365$, $1 + \\ldots + 4^6 = 5461$\n\n$1 + 6 + \\ldots + 6^4 = 1555$, $1 + \\ldots + 6^5 = 9331$\n\n$1 + 8 + 8^2 + 8^3 = 585$, $1 + \\ldots + 8^4 = 4681$\n\n$1 + 12 + 12^2 + 12^3 = 1885$, $1 + \\ldots + 12^4 = 22621$\n\n$1 + 24 + 24^2 = 601$, $1 + \\ldots + 24^3 = 14425$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32243,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is a cyclic quadrilateral. A circle whose center is on the side $AB$ touches the other three sides. Show that $AB = AD + BC$. What is the maximum possible area of $ABCD$ in terms of $|AB|$ and $|CD|$?",
    "options": [],
    "answer": "AB = AD + BC; maximum area = (AB + CD)/4 * sqrt(2*AB*CD - CD^2).",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nLet the circle have center $O$ on $AB$ and radius $r$. Let $\\angle OAD = \\theta$, $\\angle OBC = \\varphi$. Since $ABCD$ is cyclic, $\\angle ADC = 180^\\circ - \\varphi$, so $\\angle ODA = 90^\\circ - \\varphi / 2$.\n\nIf $AD$ touches the circle at $X$, then\n$$\nAD = AX + XD = r \\cot \\theta + r \\tan (\\varphi / 2).\n$$\nSimilarly,\n$$\nBC = r \\cot \\varphi + r \\tan (\\theta / 2).\n$$\nPut $t = \\tan (\\theta / 2)$. Then $\\cot \\theta = (1 - t^2) / 2t$, so\n$$\n\\cot \\theta + \\tan (\\theta / 2) = \\frac{1 + t^2}{2t} = \\frac{1}{\\sin \\theta}.\n$$\nSimilarly for $\\varphi$, so\n$$\nAD + BC = \\frac{r}{\\sin \\theta} + \\frac{r}{\\sin \\varphi} = AO + OB = AB.\n$$\n\nSuppose $AD$ and $BC$ meet at $H$ (we deal below with the case where they are parallel). Then $HCD$ and $HAB$ are similar, so\n$$\n\\text{area } HCD = \\left(\\frac{CD^2}{AB^2}\\right) \\text{area } HAB\n$$\nand\n$$\n\\text{area } ABCD = \\left(1 - \\frac{CD^2}{AB^2}\\right) \\text{area } HAB.\n$$\nAlso\n$$\n\\frac{AB}{CD} = \\frac{HA}{HC} = \\frac{HB}{HD} = \\frac{HA + HB}{HC + HD} = \\frac{HA + HB}{HB - BC + HA - DA} = \\frac{HA + HB}{HA + HB - AB}.\n$$\nHence\n$$\nHA + HB = \\frac{AB^2}{AB - CD},\n$$\nwhich is fixed. Now for fixed $HA + HB$ we maximise the area of $HAB$ by taking $HA = HB$ and hence $AD = BC$.\n\nPut $h = CD$, $k = AB$. So $k \\cos \\theta + h = k$. Hence $\\cos \\theta = 1 - h / k$. Hence\n$$\n\\sin \\theta = \\sqrt{2h / k - h^2 / k^2}.\n$$\nSo\n$$\n\\text{area } ABCD = \\frac{1}{2}(h + k) \\cdot \\frac{1}{2} k \\sin \\theta = (h / 2 + k / 2) \\sqrt{hk / 2 - h^2 / 4} \\quad (*).\n$$\n\nIf $AD$ and $BC$ are parallel then $A$ and $B$ must lie on the circle, so that $\\angle DAB = \\angle ABC = 90^\\circ$. But $ABCD$ is cyclic, so it must be a rectangle. Hence $AB = CD$ and\n$$\n\\text{area } ABCD = k^2 / 2.\n$$\nIn this case $(*)$ still gives the correct answer.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32244,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere is a bulb in each cell of an $n \\times n$ board. Initially all the bulbs are off. If a bulb is touched, that bulb and all the bulbs in the same row and column change state (those that are on, turn off, and those that are off, turn on). Show that it is possible by touching $m$ bulbs to turn all the bulbs on. What is the minimum possible value of $m$?",
    "options": [],
    "answer": "If n is odd: m = n. If n is even: m = n^2.",
    "solution": "Solution:\n\nIf $n$ is odd, touch each bulb in the first column. Then bulbs in the first column are each switched $n$ times, which is odd and so end up on. All other bulbs are switched just once, and so end up on. $n$ is obviously minimal, because if $m < n$, then there is a bulb which is not switched at all (there must be a column with no bulb touched and a row with no bulb touched, so the bulb in that column and row is not switched).\n\nIf $n$ is even, touch each bulb. Then each bulb is switched $2n-1$ times, so ends up on. We show that it is not possible to do better.\n\nNote first that there is no benefit in touching a bulb more than once, so each must be touched zero or one times. Thus we can represent the scheme as an array of $0$s and $1$s, where $0$ means that the corresponding bulb is not touched, and $1$ means that it is touched.\n\nLet $A, B, C, D$ be four values at the corners of a rectangle. We claim that $A+B$ has the same parity as $C+D$. Let $L_{AB}$ be the number of $1$s in the row $AB$ are touched, similarly $L_{BC}$ (the number of $1$s in the column $BC$), $L_{CD}$, $L_{DA}$. Since bulb $A$ is switched we must have $L_{AB} + L_{DA} + A$ odd (note that $L_{AB} + L_{DA}$ double-counts the number of touches of $A$). Similarly, $L_{BC} + L_{CD} + C$ is odd, so $A + C + (L_{AB} + L_{BC} + L_{CD} + L_{DA})$ is even. Similarly, considering $B$ and $D$, we find that $B + D + (L_{AB} + L_{BC} + L_{CD} + L_{DA})$ is even, so $A + C$ and $B + D$ have the same parity. Adding $B + C$ to both, we get that $A + B$ and $C + D$ have the same parity. It follows that either $A = D$ and $B = C$, or $A \\neq D$ and $B \\neq D$.\n\nKeeping $A$ and $B$ fixed, we can now vary $C$ (and hence $D$). It follows that either the row through $B$ matches that through $A$, or it has every cell different (to the corresponding cell in row $A$). Similarly for the other rows. So we have $k$ rows of one type and $n - k$ rows which are equal to its \"complement\". Suppose first that $k = n$, so that all rows are the same. If we have all $1$s, then we have a solution. If we have all $0$s, we obviously do not have a solution. So suppose there is a $0$ and a $1$ in each row. Then the total count at a $1$ is $n-1$ higher than at a $0$ (because of the extra $n-1$ $1$s in the same column). So they cannot both be odd (because $n$ is even). Contradiction.\n\nFinally suppose there is a row and a complement row. So position $A$ in one is $1$, then position $B$ in the same column in the other has $0$. If a row has $h$ $1$s, then a complement row has $n-h$ $1$s. The column has $z$ $1$s, so $A$ has $z + h - 1$ or $z + n - h - 1$ $1$s, and $B$ has $z + h$ or $z + n - h$ $1$s. But since $n$ is even, $z + h$ and $z + n - h$ have the same parity, so $A$ and $B$ have opposite parity. Contradiction. So the only solution for $n$ even is all $1$s.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32245,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABC$ is an acute-angled triangle. $P$ is a point inside its circumcircle. The rays $AP$, $BP$, $CP$ intersect the circle again at $D$, $E$, $F$. Find $P$ so that $DEF$ is equilateral.",
    "options": [],
    "answer": "P is the common intersection of the three Apollonius circles for triangle ABC, equivalently the point satisfying PB/PC = AB/AC (and the analogous conditions for the other pairs); take the intersection lying inside the triangle.",
    "solution": "![](attached_image_1.png)\n\n$PAB$ and $PED$ are similar, so $DE / AB = PD / PB$. Similarly, $DF / AC = PD / PC$, so $DE / DF = (AB / AC)(PC / PB)$. Thus we need $PB / PC = AB / AC$. So $P$ must lie on the circle of Apollonius, which is the circle we constructed with center $X$. Similarly, it must lie on the circle of Apollonius with center $Y$ and hence be one of their points of intersection. It also lies on the third circle and hence we choose the point of intersection inside the triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32246,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$n$ and $r$ are positive integers. Find the smallest $k$ for which we can construct $r$ subsets $A_1, A_2, \\ldots, A_r$ of $\\{0,1,2, \\ldots, n-1\\}$ each with $k$ elements such that each integer $0 \\leq m < n$ can be written as a sum of one element from each of the $r$ subsets.",
    "options": [],
    "answer": "ceil(n^(1/r))",
    "solution": "Solution:\n\nWe can form at most $k^r$ distinct sums, so $k^r$ must be $\\geq n$.\n\nNow consider $A_1 = \\{0, 1, 2, \\ldots, k-1\\}$, $A_2 = \\{0, k, 2k, \\ldots, (k-1)k\\}$, $A_3 = \\{0, k^2, 2k^2, \\ldots, (k-1)k^2\\}, \\ldots, A_r = \\{0, k^{r-1}, 2k^{r-1}, \\ldots, (k-1)k^{r-1}\\}$.\n\nThen for any non-negative integer $m < k^r$, we can write $m$ with $r$ digits in base $k$ (using leading zeros as necessary) and hence as a sum of one element from each $A_i$.\n\nThis subset works for $(k-1)k^{r-1} < n \\leq k^r$. For smaller $n$ above $(k-1)^r$ we cannot use all the elements given above, but we do not need them, so we just replace the elements which are too large by arbitrary elements under $n$.\n\nFor example, suppose $n = 17$, $r = 4$. We need $k = 3$. So we form $A_1 = \\{0, 1, 2\\}$, $A_2 = \\{0, 3, 6\\}$, $A_3 = \\{0, 9, 18\\}$, $A_4 = \\{0, 27, 54\\}$. Now $18, 27, 54$ are unnecessary, so we pad out $A_3$ and $A_4$ with other elements. We could take $A_3 = \\{0, 1, 9\\}$, $A_4 = \\{0, 1, 2\\}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32247,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow that given any integer $0 < n \\leq 2^{1000000}$ we can find a set $S$ of at most $1100000$ positive integers such that $S$ includes $1$ and $n$ and every element of $S$ except $1$ is a sum of two (possibly equal) smaller elements of $S$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32248,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all possible values for the sum of the digits of a square.",
    "options": [],
    "answer": "Exactly the nonnegative integers whose remainder upon division by nine is 0, 1, 4, or 7.",
    "solution": "Solution:\n$0^2 = 0$, $(\\pm 1)^2 = 1$, $(\\pm 2)^2 = 4$, $(\\pm 3)^2 = 0$, $(\\pm 4)^2 = 7 \\bmod 9$, so the condition is necessary.\n\nWe exhibit squares which give these values.\n\n$0 \\bmod 9$. Obviously $0^2 = 0$. We have $9^2 = 81$, $99^2 = 9801$ and in general $99\\ldots9^2 = (10^n - 1)^2 = 10^{2n} - 2 \\cdot 10^n + 1 = 99\\ldots980\\ldots01$, with digit sum $9n$.\n\n$1 \\bmod 9$. Obviously $1^2 = 1$ with digit sum $1$, and $8^2 = 64$ with digit sum $10$. We also have $98^2 = 9604$, $998^2 = 996004$, and in general $99\\ldots98^2 = (10^n - 2)^2 = 10^{2n} - 4 \\cdot 10^n + 4 = 99\\ldots960\\ldots04$, with digit sum $9n + 1$.\n\n$4 \\bmod 9$. Obviously $2^2 = 4$ with digit sum $4$, and $7^2 = 49$ with digit sum $13$. Also $97^2 = 9409$ with digit sum $22$, $997^2 = 994009$ with digit sum $31$, and in general $99\\ldots97^2 = (10^n - 3)^2 = 10^{2n} - 6 \\cdot 10^n + 9 = 99\\ldots940\\ldots09$, with digit sum $9n + 4$.\n\n$7 \\bmod 9$. Obviously $4^2 = 16$, with digit sum $7$. Also $95^2 = 9025$, digit sum $16$, $995^2 = 990025$ with digit sum $25$, and in general $99\\ldots95^2 = (10^n - 5)^2 = 10^{2n} - 10^{n+1} + 25 = 99\\ldots90\\ldots025$, with digit sum $9n - 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32249,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all solutions in real numbers $x_{1}, x_{2}, \\ldots, x_{n+1}$ all at least $1$ such that:\n\n(1) $x_{1}^{1/2} + x_{2}^{1/3} + x_{3}^{1/4} + \\ldots + x_{n}^{1/(n+1)} = n x_{n+1}^{1/2}$;\n\nand\n\n(2) $\\dfrac{x_{1} + x_{2} + \\ldots + x_{n}}{n} = x_{n+1}$.",
    "options": [],
    "answer": "x1 = x2 = ... = x_{n+1} = 1",
    "solution": "Solution:\n\nBy Cauchy-Schwarz, $\\left(\\sum x_{i}^{1/2}\\right)^{2} \\leq \\left(\\sum 1\\right)\\left(\\sum x_{i}\\right)$, with equality if and only if all $x_{i}$ are equal. In other words, if we put $x_{n+1} = \\dfrac{x_{1} + x_{2} + \\ldots + x_{n}}{n}$, then $\\sum x_{i}^{1/2} \\leq n x_{n+1}^{1/2}$.\n\nBut since all $x_{i} \\geq 1$, we have\n\n$x_{1}^{1/2} + x_{2}^{1/3} + x_{3}^{1/4} + \\ldots + x_{n}^{1/(n+1)} \\leq \\sum x_{i}^{1/2}$\n\nwith equality if and only if $x_{2} = x_{3} = \\ldots = x_{n} = 1$.\n\nHence\n\n$x_{1}^{1/2} + x_{2}^{1/3} + x_{3}^{1/4} + \\ldots + x_{n}^{1/(n+1)} \\leq x_{n+1}^{1/2}$\n\nwith equality if and only if all $x_{i} = 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32250,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$L$ and $L'$ are two perpendicular lines not in the same plane. $AA'$ is perpendicular to both lines, where $A$ belongs to $L$ and $A'$ belongs to $L'$. $S$ is the sphere with diameter $AA'$. For which points $P$ on $S$ can we find points $X$ on $L$ and $X'$ on $L'$ such that $XX'$ touches $S$ at $P$?",
    "options": [],
    "answer": "Exactly the points on the two great circles obtained by intersecting the sphere with the two planes through AA′ that bisect the angle between the directions of L and L′; equivalently, the intersection of S with the planes making equal angles with L and L′.",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Solid Geometry"
  },
  {
    "id": 32251,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABCD$ is an $n \\times n$ board. We call a diagonal row of cells a positive diagonal if it is parallel to $AC$. How many coins must be placed on an $n \\times n$ board such that every cell either has a coin or is in the same row, column or positive diagonal as a coin?",
    "options": [],
    "answer": "ceil((2n - 1)/3)",
    "solution": "Solution:\n\nThere must be at least $n - k$ rows without a coin and at least $n - k$ columns without a coin. Let $r_1, r_2, \\ldots, r_{n-k}$ be cells in the top row without a coin which are also in a column without a coin. Let $r_1, c_2, c_3, \\ldots, c_{n-k}$ be cells in the first column without a coin which are also in a row without a coin. Each of the $2n - 2k - 1$ $r_i$ and $c_j$ are on a different positive diagonal, so we must have $k \\geq 2n - 2k - 1$ and hence $k \\geq (2n - 1)/3$.\n\nLet $(i, j)$ denote the cell in row $i$, column $j$. For $n = 3m - 1$, put coins in $(m, 1), (m-1, 2), (m-2, 3), \\ldots, (1, m)$ and in $(2m-1, m+1), (2m-2, m+2), \\ldots, (m+1, 2m-1)$. It is easy to check that this works. For $n = 3m$, put an additional coin in $(2m, 2m)$, it is easy to check that works. For $n = 3m + 1$ we can use the same arrangement as for $3m + 2$.\n\n![](attached_image_1.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32252,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThe incircle of the triangle $ABC$ touches the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $AD$ meets the circle again at $X$ and $AX = XD$. $BX$ meets the circle again at $Y$ and $CX$ meets the circle again at $Z$. Show that $EY = FZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32253,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$\\mathrm{M}$ is the midpoint of the median $\\mathrm{AD}$ of the triangle $\\mathrm{ABC}$. The ray $\\mathrm{BM}$ meets $\\mathrm{AC}$ at $N$. Show that $\\mathrm{AB}$ is tangent to the circumcircle of $\\mathrm{NBC}$ iff $\\mathrm{BM} / \\mathrm{BN}=(\\mathrm{BC} / \\mathrm{BN})^{2}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nApplying Menelaus to the triangle $\\mathrm{ADC}$, we have ($\\mathrm{AM} / \\mathrm{MD})(\\mathrm{BD} / \\mathrm{DC})(\\mathrm{CN} / \\mathrm{NA})=1$, so $(\\mathrm{CN} / \\mathrm{NA})=2$. Hence $\\mathrm{AN} / \\mathrm{AC}=1 / 3$. Applying Menelaus to the triangle $\\mathrm{BNC}$, we have $(\\mathrm{BM} / \\mathrm{MN})(\\mathrm{AN} / \\mathrm{AC})(\\mathrm{CD} / \\mathrm{DB})=1$, so $\\mathrm{BM} / \\mathrm{MN}=3$. That is true irrespective of whether $\\mathrm{AB}$ is tangent to the circle NBC.\n\nIf $\\mathrm{AB}$ is tangent, then $\\mathrm{AB}^{2}=\\mathrm{AN} \\cdot \\mathrm{AC}=1 / 3 \\mathrm{AC}$. Also angle $\\mathrm{ABN}=$ angle $\\mathrm{BCN}$, so triangles $\\mathrm{ANB}$ and $\\mathrm{ABC}$ are similar. Hence $\\mathrm{BC} / \\mathrm{BN}=\\mathrm{AC} / \\mathrm{AB}$. Hence $(\\mathrm{BC} / \\mathrm{BN})^{2}=3=\\mathrm{BM} / \\mathrm{BN}$.\n\nConversely, if $(\\mathrm{BC} / \\mathrm{BN})^{2}=\\mathrm{BM} / \\mathrm{BN}$, then $(\\mathrm{BC} / \\mathrm{BN})^{2}=3$.\n\nNow applying the cosine formula to $\\mathrm{AMN}$ and $\\mathrm{AMB}$ and using $\\cos \\mathrm{AMN}+\\cos \\mathrm{AMB}=0$, we have $\\left(3 \\mathrm{AN}^{2}-3 \\mathrm{AM}^{2}-3 \\mathrm{MN}^{2}\\right)+\\left(\\mathrm{AB}^{2}-\\mathrm{AM}^{2}-\\mathrm{BM}^{2}\\right)=0$, so $A \\mathrm{~B}^{2}+\\mathrm{AC}^{2} / 3=\\mathrm{AD}^{2}+3 / 4 \\mathrm{BN}^{2}$. Similarly from triangles $\\mathrm{ADC}$ and $\\mathrm{ADB}$ we get $\\mathrm{AB}^{2}+\\mathrm{AC}^{2}=2 \\mathrm{AD}^{2}+\\mathrm{BC}^{2} / 2$. So using $\\mathrm{BN}^{2}=$ $\\mathrm{BC}^{2} / 3$ we get $2 \\mathrm{AB}^{2}+2 / 3 \\mathrm{AC}^{2}=A B^{2}+\\mathrm{AC}^{2}$ and hence $(\\mathrm{AC} / \\mathrm{AB})^{2}=3=(\\mathrm{BC} / \\mathrm{BN})^{2}$. So $\\mathrm{AC} / \\mathrm{AB}$ $=\\mathrm{BC} / \\mathrm{BN}$. Note that is not enough to conclude that triangles $\\mathrm{ABC}$ and $\\mathrm{BNC}$ are similar, because the common angle $\\mathrm{C}$ is not between $\\mathrm{AC}$ and $\\mathrm{AB}$. However, we have $\\mathrm{AN} / \\mathrm{AB}=(1 / 3)$ $\\mathrm{AC} / \\mathrm{AB}=\\mathrm{AB} / \\mathrm{AC}$, so $\\mathrm{AB}^{2}=\\mathrm{AN} \\cdot \\mathrm{AC}$, so $\\mathrm{AB}$ is tangent to the circle NBC.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32254,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind the smallest positive integer $n$ so that a cube with side $n$ can be divided into 1996 cubes each with side a positive integer.",
    "options": [],
    "answer": "13",
    "solution": "Solution:\nDivide all the cubes into unit cubes. Then the 1996 cubes must each contain at least one unit cube, so the large cube contains at least 1996 unit cubes. But $12^3 = 1728 < 1996 < 2197 = 13^3$, so it is certainly not possible for $n < 13$.\n\nIt can be achieved with 13 by $1 \\cdot 5^3 + 11 \\cdot 2^3 + 1984 \\cdot 1^3 = 13^3$ (actually packing the cubes together to form a $13 \\times 13 \\times 13$ cube is trivial since there are so many unit cubes).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32255,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$n = k^2 - k + 1$, where $k$ is a prime plus one. Show that we can color some squares of an $n \\times n$ board black so that each row and column has exactly $k$ black squares, but there is no rectangle with sides parallel to the sides of the board which has its four corner squares black.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe can regard the rows as lines and the columns as points. Black squares denote incidence. So line 3 contains point 4 iff square $(3,4)$ is black. The condition about rectangles then means that there is at most one line through two distinct points.\n\nSuppose we take the points to be $(a, b, c)$, where $a, b, c$ are residues $\\bmod p$, not all zero, and the coordinates are homogeneous, so that we regard $(a, b, c)$, $(2a, 2b, 2c), \\ldots, ((p-1)a, (p-1)b, (p-1)c)$ as the same point. That gives $\\frac{p^3-1}{p-1} = p^2 + p + 1$ points, which is the correct number.\n\nWe can take lines to be $l x + m y + n z = 0$, where the point is $(x, y, z)$. In other words, the lines are also triples $(l, m, n)$, with $l, m, n$ residues mod $p$, not all zero and $(l, m, n), (2l, 2m, 2n), \\ldots, ((p-1)l, (p-1)m, (p-1)n)$ representing the same line.\n\nOne way of writing the points is $p^2$ of the form $(a, b, 1)$, $p$ of the form $(a, 1, 0)$ and lastly $(1, 0, 0)$. Similarly for the lines. We must show that (1) each point is on $p+1$ lines (so each column has $p+1$ black squares), (2) each line has $p+1$ points (so each row has $p+1$ black squares), (3) two lines meet in just one point (so no rectangles).\n\n(1): Consider the point $P(a, b, 1)$ with $a$ non-zero. Then for any $m$, there is a unique $l$ such that $l a + m b + 1 \\cdot 1 = 0$, so there are $p$ lines of the form $(l, m, 1)$ which contain $P$. Similarly, there is a unique $l$ such that $l a + 1 b + 0 \\cdot 1 = 0$, so one line of the form $(l, 1, 0)$ contains $P$. The line $(1, 0, 0)$ does not contain $P$. So $P$ lies on just $p+1$ lines. Similarly for $(a, b, 1)$ with $b$ nonzero. The point $(0, 0, 1)$ does not lie on any lines $(l, m, 1)$, but lies on $(l, 1, 0)$ and $(1, 0, 0)$, so again it lies on $p+1$ lines.\n\nConsider the point $Q(a, 1, 0)$ with $a$ non-zero. For any $m$, there is a unique $l$ such that $Q$ lies on $(l, m, 0)$. There is also a unique $l$ such that $Q$ lies on $(l, 1, 0)$. $Q$ does not lie on $(1, 0, 0)$, so it lies on just $p+1$ lines. Similarly, the point $(0, 1, 0)$ lies on the $p$ lines $(l, 0, 0)$ and on $(1, 0, 0)$, but no others.\n\nFinally, the point $(1, 0, 0)$ lies on the $p$ lines $(0, m, 1)$, the line $(0, 1, 0)$ and no others. Thus in all cases a point lies on just $p+1$ lines. The proof of (2) is identical.\n\n(3). Suppose the lines are $(l, m, n)$ and $(L, M, N)$. If $l$ and $L$ are non-zero, then we can take the lines as $(1, m', n')$ and $(1, M', N')$. So any point $(x, y, z)$ on both satisfies $x + m' y + n' z = 0$ $(*)$ and $x + M' y + N' z = 0$. Subtracting, $(m' - M') y + (n' - N') z = 0$. The coefficients cannot both be zero, since the lines are distinct. So the ratio $y : z$ is fixed. Then $(*)$ gives the ratio $x : y$. So the point is uniquely determined. If just one of $l, L$ is non-zero, then we can take the lines as $(0, m', n')$, $(1, M', N')$. We cannot have both $m'$ and $n'$ zero, so the ratio $y : z$ is determined, then the other line determines the ratio $x : y$. So again the point is uniquely determined. Finally, suppose $l$ and $L$ are both zero. Then since the lines are distinct $y$ and $z$ must both be zero. So the unique point on both lines is $(1, 0, 0)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32256,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$n > 2$ is an integer. Consider the pairs $(a, b)$ of relatively prime positive integers, such that $a < b \\leq n$ and $a + b > n$. Show that the sum of $1 / ab$ taken over all such pairs is $1 / 2$.",
    "options": [],
    "answer": "1/2",
    "solution": "Solution:\n\nInduction on $n$. It is obvious for $n = 3$, because the only pairs are $(1, 3)$ and $(2, 3)$, and $1 / 3 + 1 / 6 = 1 / 2$.\n\nNow suppose it is true for $n$. As we move to $n + 1$, we introduce the new pairs $(a, n + 1)$ with $a$ relatively prime to $n + 1$ and we lose the pairs $(a, n + 1 - a)$ with $a$ relatively prime to $n + 1 - a$ and hence to $n + 1$.\n\nSo for each $a$ relatively prime to $n + 1$ and $a < (n + 1) / 2$ we gain $(a, n + 1)$ and $(n + 1 - a, n + 1)$ and lose $(a, n + 1 - a)$. But\n\n$$\n\\frac{1}{a(n + 1)} + \\frac{1}{(n + 1 - a)(n + 1)} = \\frac{n + 1 - a + a}{a(n + 1 - a)(n + 1)} = \\frac{1}{a(n + 1 - a)}.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32257,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAn equilateral triangle of side $n$ is divided into $n^{2}$ equilateral triangles of side 1 by lines parallel to the sides. Initially, all the sides of all the small triangles are painted blue. Three coins $A$, $B$, $C$ are placed at vertices of the small triangles. Each coin in turn is moved a distance 1 along a blue side to an adjacent vertex. The side it moves along is painted red, so once a coin has moved along a side, the side cannot be used again. More than one coin is allowed to occupy the same vertex. The coins are moved repeatedly in the order $A$, $B$, $C$, $A$, $B$, $C$, ... Show that it is possible to paint all the sides red in this way.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nWe use induction. It is obvious for $n=1$ and $2$ - see diagram above. Note that $A$, $B$, $C$ start and end at vertices of the large triangle.\n\n![](attached_image_2.png)\nNow assume that for $n$ we can find a solution with $A$, $B$, $C$ starting and ending at the vertices of the large triangle. Take $n+1$. We start with the paths shown which bring $A$, $B$, $C$ to $A'$, $B'$, $C'$ at the vertices of a triangle side $n-1$. Now by induction we can continue the paths so that we bring $A$, $B$, $C$, back to the vertices of that triangle after tracing out all its edges. Finally, note that for each of the points $A'$, $B'$, $C'$ there is a path length 2 over untraced segments to a vertex of the large triangle. So we get a solution for $n+1$ and hence for all $n$.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32258,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$A_{1}, A_{2}, \\ldots, A_{n}$ are points in the plane. A non-zero real number $k_{i}$ is assigned to each point, so that the square of the distance between $A_{i}$ and $A_{j}$ (for $i \\neq j$) is $k_{i} + k_{j}$. Show that $n$ is at most $4$ and that if $n = 4$, then $1 / k_{1} + 1 / k_{2} + 1 / k_{3} + 1 / k_{4} = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose we have four points $A$, $B$, $C$, $D$ with associated numbers $a$, $b$, $c$, $d$. Then $AB^{2} = a + b$, $AC^{2} = a + c$, so $AB^{2} - AC^{2} = b - c$. Similarly, $DB^{2} - DC^{2} = b - c$, so $AB^{2} - AC^{2} = DB^{2} - DC^{2}$.\n\nLet $X$ be the foot of the perpendicular from $A$ to $BC$, and $Y$ the foot of the perpendicular from $D$ to $BC$. Then\n$$\nAB^{2} - AC^{2} = (AX^{2} + XB^{2}) - (AX^{2} + XC^{2}) = XB^{2} - XC^{2}.\n$$\nSimilarly for $D$, so $XB^{2} - XC^{2} = YB^{2} - YC^{2}$. Hence $X = Y$, so $AD$ is perpendicular to $BC$. Similarly, $BD$ is perpendicular to $AC$, and $CD$ is perpendicular to $AB$. Hence $D$ is the (unique) orthocenter of $ABC$. So $n \\leq 4$.\n\nSuppose $n = 4$, so we have four points $A$, $B$, $C$, $D$ with associated numbers $a$, $b$, $c$, $d$. We have\n$$\nAB^{2} + AC^{2} - BC^{2} = (a + b) + (a + c) - (b + c) = 2a.\n$$\nBut by the cosine formula it is also $2 AB \\cdot AC \\cos BAC$. Hence $a = AB \\cdot AC \\cos BAC$. Similarly for $A$, $B$, $D$ etc.\n\nHence\n$$\na b / c d = (AB \\cdot AC \\cos BAC)(BA \\cdot BD \\cos ABD) / ((CA \\cdot CD \\cos ACD)(DB \\cdot DC \\cos BDC)) = (AB^{2} / CD^{2})(\\cos BAC / \\cos BDC)(\\cos ABD / \\cos ACD).\n$$\nTake $ABC$ to be acute with $D$ inside. Then angle $ABD = ACD (= 90^{\\circ} - BAC)$, and angle $BDC = 90^{\\circ} + ACD = 180^{\\circ} - BAC$. So $\\cos BAC / \\cos BDC = -1$. Thus $ab / cd = -AB^{2} / CD^{2} = -(a + b) / (c + d)$. Hence $ab(c + d) + cd(a + b) = 0$, so $1 / a + 1 / b + 1 / c + 1 / d = 0$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32259,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$I$ is the incenter of the triangle $ABC$. A circle with center $I$ meets the side $BC$ at $D$ and $P$, with $D$ nearer to $B$. Similarly, it meets the side $CA$ at $E$ and $Q$, with $E$ nearer to $C$, and it meets $AB$ at $F$ and $R$, with $F$ nearer to $A$. The lines $EF$ and $QR$ meet at $S$, the lines $FD$ and $RP$ meet at $T$, and the lines $DE$ and $PQ$ meet at $U$. Show that the circumcircles of $DUP$, $ESQ$ and $FTR$ have a single point in common.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\n$D$ and $P$ are the reflections of $Q$ and $E$ respectively in the line $CI$. Hence $PQ$ and $DE$ meet at a point on $CI$. So $U$ lies on $CI$. So $\\angle PIU = 1/2 \\angle PIE = \\angle PDE$ ($I$ is center of circle through $D$, $P$, $E$) $= \\angle PDU$ (same angle). Hence $PDIU$ is cyclic. In other words, $I$ lies on the circumcircle of $DUP$. Similarly, it lies on the circumcircles of $ESQ$ and $FTR$.\n\nBut the same argument shows that $\\angle DPT = \\angle DIT$, so $DPIT$ is cyclic. So $T$ lies on the circle through $D$, $P$ and $I$ and hence on the circumcircle of $DUP$. Similarly, for the other circles. So the circumcircles of $CUP$ and $FTR$ meet at $T$ and $I$. Similarly, the circumcircles of $FTR$ and $ESQ$ meet at $S$ and $I$, and the circumcircles of $ESQ$ and $DUP$ meet at $U$ and $I$. So the three circumcircles have just one point in common, namely $I$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32260,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$k \\geq 1$ is a real number such that if $m$ is a multiple of $n$, then $[mk]$ is a multiple of $[nk]$. Show that $k$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose $k$ is not an integer. Take an integer $n$ such that $nk > 1$, but $nk$ is not an integer. Now take a positive integer $c$ such that $\\frac{1}{c+1} \\leq nk - [nk] < \\frac{1}{c}$. Then $1 \\leq (c+1)nk - (c+1)[nk] < 1 + \\frac{1}{c}$. Hence $[(c+1) n k] = (c+1)[n k] + 1$. Put $m = (c+1) n$. Then $m$ is a multiple of $n$. But if $[mk]$ is a multiple of $[nk]$, then $[mk] - (c+1)[nk] = 1$ is a multiple of $[nk]$, which is impossible since $nk > 1$. So we have a contradiction. So $k$ must be an integer.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32261,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$n > 1$ is an integer. $D_n$ is the set of lattice points $(x, y)$ with $|x|, |y| \\leq n$. If the points of $D_n$ are colored with three colors (one for each point), show that there are always two points with the same color such that the line containing them does not contain any other points of $D_n$. Show that it is possible to color the points of $D_n$ with four colors (one for each point) so that if any line contains just two points of $D_n$ then those two points have different colors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nConsider the 4 points shown in the diagram. In each case the segment joining them is the diagonal of an $m \\times 1$ parallelogram or rectangle, so it cannot contain any other lattice points. The next points along each line are obviously outside set $D_n$. That proves the first part.\n\nThe second part is the standard parity argument. Color $(x, y)$ with color 1 if $x$ and $y$ are both even, 2 if $x$ is even and $y$ is odd, 3 if $x$ is odd and $y$ is even, and 4 if $x$ and $y$ are both odd. Then if two points are the same color, that means the first coordinates are the same parity and their second coordinates are the same parity. Hence the midpoint of the segment joining them is also a lattice point and they are not the only two points of $D_n$ on the line.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32262,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $o(n)$ be the number of $2n$-tuples $(a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n})$ such that each $a_{i}, b_{j} = 0$ or $1$ and $a_{1} b_{1} + a_{2} b_{2} + \\ldots + a_{n} b_{n}$ is odd. Similarly, let $e(n)$ be the number for which the sum is even. Show that $\\dfrac{o(n)}{e(n)} = \\dfrac{2^{n} - 1}{2^{n} + 1}$.",
    "options": [],
    "answer": "(2^n - 1)/(2^n + 1)",
    "solution": "Solution:\nWe prove by induction that $o(n) = 2^{2n-1} - 2^{n-1}$. For $n = 1$, this reads $o(1) = 2^{1} - 2^{0} = 1$, which is obviously true—the only such $2$-tuple is $(1, 1)$. Suppose it is true for $n$.\n\nIf $(a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n})$ gives an odd sum, then we can take $(a_{n+1}, b_{n+1})$ to be any of $(0, 0), (0, 1), (1, 0)$ and still get an odd sum for $(a_{1}, a_{2}, \\ldots, a_{n+1}, b_{1}, b_{2}, \\ldots, b_{n+1})$. On the other hand, if $(a_{1}, a_{2}, \\ldots, a_{n}, b_{1}, b_{2}, \\ldots, b_{n})$ is even, then we must have $a_{n+1} = b_{n+1} = 1$ to get an odd sum. Thus $o(n+1) = 3 o(n) + e(n)$.\n\nBut $o(n) = 2^{2n-1} - 2^{n-1}$ and $e(n) = (o(n) + e(n)) - o(n) = 2^{2n} - (2^{2n-1} - 2^{n-1}) = 2^{2n-1} + 2^{n-1}$.\n\nSo $o(n+1) = 3 o(n) + e(n) = 3(2^{2n-1} - 2^{n-1}) + (2^{2n-1} + 2^{n-1}) = 4 \\cdot 2^{2n-1} - 2 \\cdot 2^{n-1} = 2^{2(n+1)-1} - 2^{(n+1)-1}$, which establishes the result for $n+1$ and hence for all $n$.\n\nHence $e(n) = 2^{2n} - o(n) = 2^{2n-1} + 2^{n-1}$ and $\\dfrac{o(n)}{e(n)} = \\dfrac{2^{n} - 1}{2^{n} + 1}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32263,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$\\mathrm{ABC}$ is an acute-angled triangle with orthocenter $\\mathrm{H}$. $\\mathrm{AE}$ and $\\mathrm{BF}$ are altitudes. $\\mathrm{AE}$ is reflected in the angle bisector of angle $\\mathrm{A}$ and $\\mathrm{BF}$ is reflected in the angle bisector of angle $\\mathrm{B}$. The two reflections intersect at $\\mathrm{O}$. The rays $\\mathrm{AE}$ and $\\mathrm{AO}$ meet the circumcircle of $\\mathrm{ABC}$ at $\\mathrm{M}$ and $\\mathrm{N}$ respectively. $\\mathrm{P}$ is the intersection of $\\mathrm{BC}$ and $\\mathrm{HN}$, $\\mathrm{R}$ is the intersection of $\\mathrm{BC}$ and $\\mathrm{OM}$, and $\\mathrm{S}$ is the intersection of $\\mathrm{HR}$ and $\\mathrm{OP}$. Show that AHSO is a parallelogram.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\nWe show first that $\\mathrm{O}$ is the circumcenter of $\\mathrm{ABC}$. $\\angle \\mathrm{ABF} = 90^\\circ - \\mathrm{A}$. The line $\\mathrm{BC}$ is the reflection in $BD$ of the line $BA$ and the line $\\mathrm{BF}'$ is the reflection of $\\mathrm{BF}$, so angle $\\mathrm{CBF}' = 90^\\circ - \\mathrm{A}$. But if $\\mathrm{O}'$ is the circumcenter, then $\\angle \\mathrm{BO}'\\mathrm{C} = 2 \\angle \\mathrm{BAC} = 2\\mathrm{A}$, so $\\angle \\mathrm{O}'\\mathrm{BC} = 90^\\circ - \\mathrm{A}$. Hence $\\mathrm{O}'$ lies on $\\mathrm{BF}'$. Similarly, it lies on $\\mathrm{AE}'$ (the reflection of $\\mathrm{AE}$ in the angle bisector of $\\mathrm{A}$). Hence $\\mathrm{O} = \\mathrm{O}'$.\n\n![](attached_image_2.png)\n$\\angle \\mathrm{MBC} = \\angle \\mathrm{MAC} = 90^\\circ - \\mathrm{C}$ (since $\\mathrm{AH}$ is an altitude) $= \\angle \\mathrm{FBC}$ (since $\\mathrm{BF}$ is an altitude) $= \\angle \\mathrm{HBC}$ (same angle). So triangles HBE and MBE are congruent and $\\mathrm{HE} = \\mathrm{EM}$. [Note: this should be a familiar result.]\n$\\mathrm{AN}$ is a diameter, so angle $\\mathrm{AMN} = 90^\\circ = \\angle \\mathrm{AEC}$, so $\\mathrm{BC}$ and $\\mathrm{MN}$ are parallel. Hence $\\mathrm{P}$ is the midpoint of $\\mathrm{HN}$ and of $\\mathrm{BC}$. So $\\mathrm{OP}$ is perpendicular to $\\mathrm{BC}$. So $\\mathrm{AH}$ and $\\mathrm{OS}$ are parallel.\nSince $\\mathrm{R}$ lies on $\\mathrm{BC}$, triangles HER and MER are congruent, so $\\angle \\mathrm{EHR} = \\angle \\mathrm{EMR} = \\angle \\mathrm{AMO}$ (same angle) $= \\angle \\mathrm{MAO}$. Hence HS and AO are parallel. So AHSO is a parallelogram.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32264,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven $1997$ points inside a circle of radius $1$, one of them the center of the circle. For each point take the distance to the closest (distinct) point. Show that the sum of the squares of the resulting distances is at most $9$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet the points be $P_i$ for $i = 1, 2, \\ldots, 1997$. Take $P_1$ to be the center of the given unit circle. Let $x_i$ be the distance from $P_i$ to the closest of the other $1996$ points. Let $C_i$ be the circle centered at $P_i$ with radius $x_i / 2$. Then $C_i$ and $C_j$ cannot overlap by more than one point because $x_i$ and $x_j \\leq P_i P_j$. Also $x_i \\leq 1$, since $P_1 P_i \\leq 1$. Thus $C_i$ is entirely contained in the circle centered at $P_1$ with radius $3/2$. Since the circles $C_i$ do not overlap, their total area cannot exceed the area of the circle of radius $3/2$. Hence\n$$\n\\frac{x_1^2 + x_2^2 + \\ldots + x_{1997}^2}{4} \\leq \\frac{9}{4}\n$$\nwhich gives\n$$\nx_1^2 + x_2^2 + \\ldots + x_{1997}^2 \\leq 9.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32265,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThere are $98$ points on a circle. Two players play alternately as follows. Each player joins two points which are not already joined. The game ends when every point has been joined to at least one other. The winner is the last player to play. Does the first or second player have a winning strategy?",
    "options": [],
    "answer": "First player",
    "solution": "Solution:\n\nAssume there are $n$ points. The first to play so that $n-2$ points each have at least one segment loses, because the other player simply joins the last two points and the game ends. But there are $N = (n-3)(n-4)/2$ possible plays amongst the first $n-3$ points to get a segment. For $n = 1$ or $2 \\bmod 4$, $N$ is odd and for $n = 0$ or $3$ it is even. So the first player wins for $n = 1$ or $2 \\bmod 4$ (and in particular for $n = 98$) and the second player for $n = 0$ or $3 \\bmod 4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32266,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe incircle of the triangle $ABC$ touches $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. $AD$ meets the circle again at $Q$. Show that the line $EQ$ passes through the midpoint of $AF$ iff $AC = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\n$\\angle AQ M = \\angle EQ D$ (opposite angle) $= \\angle EDC$ ($CD$ tangent to circle $EQD$) $= \\left(180^\\circ - \\angle C\\right) / 2 = \\angle A / 2 + \\angle B / 2$ $(*)$.\n\n$MF^2 = MQ \\cdot ME$ ($MF$ tangent to circle $FQE$). So $AM = AF$ is equivalent to $AM^2 = MQ \\cdot ME$ or $AM / MQ = ME / AM$. But since triangles $AMQ$ and $EMA$ have a common angle $M$, $AM / MQ = ME / AM$ iff they are similar, and hence iff $\\angle AQ M = \\angle A$. Using $(*)$, $AM = AF$ iff $\\angle A = \\angle B$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32267,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nRepresentatives from $n > 1$ different countries sit around a table. If two people are from the same country then their respective right hand neighbors are from different countries. Find the maximum number of people who can sit at the table for each $n$.",
    "options": [],
    "answer": "n^2",
    "solution": "Solution:\n\nAnswer: $n^{2}$.\n\nObviously there cannot be more than $n^{2}$ people. For if there were, then at least one country would have more than $n$ representatives. But there are only $n$ different countries to choose their right-hand neighbours from. Contradiction.\n\nRepresent someone from country $i$ by $i$. Then for $n = 2$, the arrangement $1122$ works. [It wraps round, so that the second $2$ is adjacent to the first $1$.] Suppose we have an arrangement for $n$. Then each of $11, 22, \\ldots, nn$ must occur just once in the arrangement. Replace $11$ by $1(n+1)11$, $22$ by $2(n+1)22$, $\\ldots$, and $(n-1)(n-1)$ by $(n-1)(n+1)(n-1)(n-1)$. Finally replace $nn$ by $n(n+1)(n+1)nn$. It is easy to check that we now have an arrangement for $n+1$. We have added one additional representative for each of the countries $1$ to $n$ and $n+1$ representatives for country $n+1$, so we have indeed got $(n+1)^{2}$ people in all. We have also given a representative of each country $1$ to $n$ a neighbour from country $n+1$ on his right and we have given the ($n+1$) representatives from country $n+1$ neighbours (on their right) from each of the other countries. Otherwise we have left the seating unchanged.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32268,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$P_{1}, P_{2}, \\ldots, P_{n}$ are points in the plane and $r_{1}, r_{2}, \\ldots, r_{n}$ are real numbers such that the distance between $P_{i}$ and $P_{j}$ is $r_{i} + r_{j}$ (for $i$ not equal to $j$). Find the largest $n$ for which this is possible.",
    "options": [],
    "answer": "4",
    "solution": "Solution:\n\nDraw a circle radius $r_{i}$ at $P_{i}$. Then each pair of circles must touch. But that is possible iff $n \\leq 4$.\n\n![](attached_image_1.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32269,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$k$ is the positive root of the equation $x^{2} - 1998x - 1 = 0$. Define the sequence $x_{0}, x_{1}, x_{2}, \\ldots$ by $x_{0} = 1$, $x_{n+1} = \\left[ k x_{n} \\right]$. Find the remainder when $x_{1998}$ is divided by $1998$.",
    "options": [],
    "answer": "1000",
    "solution": "Solution:\n\nPut $p(x) = x^{2} - 1998x - 1$. Then $p(1998) = -1$, $p(1999) = 1998$, so $1998 < k < 1999$. Also $k$ is irrational (using the formula for the root of a quadratic).\n\nWe have $x_{n} = \\left[ k x_{n-1} \\right]$, so $x_{n} < k x_{n-1}$ and $> k x_{n-1} - 1$. Hence $x_{n} / k < x_{n-1} < x_{n} / k + 1 / k$, so $\\left[ x_{n} / k \\right] = x_{n-1} - 1$.\n\n$k = (1998 k + 1) / k = 1998 + 1 / k$. Hence $k x_{n} = 1998 x_{n} + x_{n} / k$. Hence $x_{n+1} = \\left[ k x_{n} \\right] = 1998 x_{n} + \\left[ x_{n} / k \\right] = 1998 x_{n} + x_{n-1} - 1$.\n\nHence $x_{n+1} = x_{n-1} - 1 \\bmod 1998$.\n\nSo $x_{1998} = 1 - 999 = 1000 \\bmod 1998$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32270,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all positive integers $n < 1000$ such that the cube of the sum of the digits of $n$ equals $n^{2}$.",
    "options": [],
    "answer": "1, 27",
    "solution": "Solution:\n\n$n < 1000$, so the sum of the digits is at most $27$, so $n^{2}$ is a cube not exceeding $27^{3}$. So we are looking for $m^{3}$ which is also a square. That implies $m$ is a square. So the only possibilities are $m = 1, 4, 9, 16, 25$. Giving $n = 1, 8, 27, 64, 125$. The corresponding cubes of the digit sums are $1, 512, 729, 1000, 512$, whereas the corresponding squares are $1, 64, 729, 4096, 15625$. Thus the only solutions are $n = 1, 27$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32271,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven two circles $C$ and $C'$, we say that $C$ bisects $C'$ if their common chord is a diameter of $C'$. Show that for any two circles which are not concentric, there are infinitely many circles which bisect them both. Find the locus of the centers of the bisecting circles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $C$, $C'$ have centers $O$, $O'$ respectively and radii $r$, $r'$ respectively. Let a circle center $P$ bisect $C$. Suppose it meets $C$ at $A$ and $B$. Then $AB$ is perpendicular to $OP$ and is a diameter of $C$. Hence $PA^{2} = OP^{2} + r^{2}$. Conversely, the circle center $P$, radius $\\sqrt{OP^{2} + r^{2}}$ bisects $C$. So $P$ will bisect $C$ and $C'$ iff $OP^{2} + r^{2} = OP'^{2} + r'^{2}$.\n\nIt is well-known that the locus of points $P'$ with equal tangents to $C$ and $C'$ is the radical axis. Call the radical axis $R$. For a point $P'$ on the radical axis we have $P'O^{2} - r^{2} = P'O'^{2} - r'^{2}$. If we reflect $P'$ in the perpendicular bisector of $OO'$ to get $P$, then $PO = P'O'$, $PO' = P'O$, so $PO'^{2} - r^{2} = PO^{2} - r^{2}$ and hence $PO^{2} + r^{2}$. Call the reflection of $R$ in the perpendicular bisector of $OO'$ the line $R'$. We have established that points on $R'$ form part of the locus. Conversely, if $P'$ is such that there is a circle center $P'$ bisecting both circles, then $OP'^{2} + r^{2} = O'P'^{2} + r'^{2}$, so if $P$ is the reflection of $P'$, then $OP^{2} - r^{2} = O'P^{2} - r'^{2}$ and hence $P$ lies on the radical axis $R$. Hence $P'$ must lie on $R'$.\n\n![](attached_image_1.png)\n\nWe have $PT^{2} = PO^{2} - r^{2} = PX^{2} + OX^{2} - r^{2}$, and similarly $PT'^{2} = PX^{2} + O'X^{2} - r'^{2}$. So $PT = PT'$ iff $OX^{2} - r^{2} = O'X^{2} - r'^{2}$. There is evidently a unique point $X$ for which that is true, so the locus of such $P$ is the line through $X$ perpendicular to $OO'$.\n\n![](attached_image_2.png)\n\nIf the circles intersect, then the point $X$ evidently lies on the line joining the two common points, because $OX^{2} - r^{2} = -XY^{2} = O'X^{2} - r'^{2}$. In any case the midpoint of each common tangent evidently lies on the line, so that provides a way of constructing it.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32272,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven points $P_1, P_2, \\ldots, P_n$ on a line we construct a circle on diameter $P_i P_j$ for each pair $i, j$ and we color the circle with one of $k$ colors. For each $k$, find all $n$ for which we can always find two circles of the same color with a common external tangent.",
    "options": [],
    "answer": "All n such that n ≥ k + 2",
    "solution": "Solution:\n\nThere are $n-1$ circles with diameter $P_i P_{i+1}$. Obviously, each pair has a common tangent. If $n-1 > k$, then two of them must have the same color.\n\nIf $n-1 \\leq k$, then color all circles with diameter $P_i P_j$ and $i < j$ with color $i$. Then if two circles have the same color, then both have a tangent at one of the points. Hence one lies inside the other and they do not have a common external tangent.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32273,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that any integer greater than $10$ whose digits are all members of $\\{1,3,7,9\\}$ has a prime factor $\\geq 11$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuch a number cannot be divisible by $2$ (or its last digit would be even) or by $5$ (or its last digit would be $0$ or $5$). So if the result is false then the number must be of the form $3^{m} 7^{n}$ for non-negative integers $m, n$. But we claim that a number of this form must have even $10$s digit.\n\nIt is easy to prove the claim by induction. It is true for $3$ and $7$ (the digit is $0$ in both cases). But if we multiply such a number by $3$ or $7$, then the new $10$s digit has the same parity as the carry from the units digit. But multiplying $1,3,7,9$ by $3$ gives a carry of $0,0,2,6$ respectively, which is always even, and multiplying by $7$ gives a carry of $0,2,4,6$, which is also always even. So the new number also has an even $10$s digit.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32274,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$O$ is the circumcenter of the acute-angled triangle $ABC$. The altitudes are $AD$, $BE$ and $CF$. The line $EF$ cuts the circumcircle at $P$ and $Q$. Show that $OA$ is perpendicular to $PQ$. If $M$ is the midpoint of $BC$, show that $AP^{2}=2 AD \\cdot OM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nLet $OA$ and $PQ$ meet at $T$. $\\angle AEH=\\angle AFH=90^{\\circ}$, so $AEHF$ is cyclic, so $\\angle AFT=\\angle AFE$ (same angle) $=\\angle AHE=90^{\\circ}-\\angle HAE=90^{\\circ}-\\angle DAC$ (same angle) $=\\angle C$. But $\\angle TAF=\\angle OAF$ (same angle) $=90^{\\circ}-(1 / 2) \\angle AOB=90^{\\circ}-\\angle C$. Hence $\\angle AFT=90^{\\circ}$, which establishes that $OA$ and $PQ$ are perpendicular.\n\nLet the circumradius be $R$ and let $AA'$ be a diameter. We have $AF=AC \\cos A=2R \\sin B \\cos A$. Hence $AT=AF \\cos OAB=AF \\sin C=2R \\cos A \\sin B \\sin C$. Now $PT^{2}=PT \\cdot TQ=AT \\cdot A'T=AT(2R-AT)$. Hence $AP^{2}=2R \\cdot AT=4R^{2} \\cos A \\sin B \\sin C$.\n\nWe have $AD=AC \\sin C=2R \\sin B \\sin C$, and $OM=OC \\cos COM=R \\cos A$. Hence $2 AD \\cdot OM=AP^{2}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32275,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven two points $A$ and $B$, take $C$ on the perpendicular bisector of $AB$. Define the sequence $C_{1}, C_{2}, C_{3}, \\ldots$ as follows. $C_{1}=C$. If $C_{n}$ is not on $AB$, then $C_{n+1}$ is the circumcenter of the triangle $ABC_{n}$. If $C_{n}$ lies on $AB$, then $C_{n+1}$ is not defined and the sequence terminates. Find all points $C$ such that the sequence is periodic from some point on.",
    "options": [],
    "answer": "All points C on the perpendicular bisector such that ∠ACB = 180°·r/s with r and s coprime integers and s not a power of 2.",
    "solution": "Solution:\n\nAnswer: any $C$ such that $\\angle ACB=180^{\\circ} r / s$, with $r$ and $s$ relatively prime integers and $s$ not a power of $2$.\n\nLet $\\angle AC_{n}B = x_{n}$, where the angle is measured clockwise, so that $x_{n}$ is positive on one side of $AB$ and negative on the other side. Then $x_{n}$ uniquely identifies $C_{n}$ on the perpendicular bisector.\n\nWe have $x_{n+1} = 2 x_{n}$. To make this work in all cases we have to take it mod $180^{\\circ}$ (so that if $AC_{n}B$ is obtuse, then $C_{n+1}$ lies on the other side of $AB$). If $x_{n}$ is eventually periodic then $x_{m+1} = x_{n+1}$, for some $n > m$, so $(2^{n} - 2^{m}) x_{1} = 0 \\bmod 180$. Hence $x_{1} = 180 r / s$ for some relatively prime integers $r, s$. Also $s$ cannot be a power of $2$ for then we would have $x_{k} = 180 r$ for some $k$, in which case the sequence would terminate rather than be periodic.\n\nConversely, suppose $x_{1} = 180 r / s$, with $r$ and $s$ relatively prime and $s$ not a power of $2$. Then $x_{n+1} = 180 \\cdot 2^{n} r / s$ cannot be $0 \\bmod 180$, so the sequence does not terminate. Put $s = 2^{b} c$, with $c$ odd. Let $d = \\varphi(c)$, where $\\varphi(m)$ is Euler's phi function, so that $2^{d} = 1 \\mod c$. Then $x_{b+1} = 180 r / c \\bmod 180$ and $2^{b+d} = 2^{b} \\bmod c$, so $x_{b+d+1} = 180 r / c \\mod 180$. Hence the sequence is periodic.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32276,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLabel the vertices of a regular $n$-gon from $1$ to $n > 3$. Draw all the diagonals. Show that if $n$ is odd then we can label each side and diagonal with a number from $1$ to $n$ different from the labels of its endpoints so that at each vertex the sides and diagonals all have different labels.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLabel the diagonal/side between $i$ and $j$ as $i + j$ (reduced if necessary mod $n$) almost works. The labels for all the lines at a given vertex will be different. But the line between $i$ and $n$ will have label $i$, the same as one endpoint. However, we are not using the label $2i$ for the lines from vertex $i$. So for the line between $i$ and $n$ we use $2i$ instead of $i + n$. The only points that need checking are (1) whether a line from $i$ to $n$ has a label different from $n$, and (2) whether all the lines at $n$ have different labels. Both points are ok because $n$ is odd.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32277,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all solutions to $ (m+1)^a = m^b + 1 $ in integers greater than $1$.",
    "options": [],
    "answer": "(m,a,b) = (2, 2, 3)",
    "solution": "Solution:\n\nTaking the equation modulo $m+1$ we get $(-1)^b = -1$, so $b$ is odd. Hence we can divide the right-hand side by $m+1$ to get $m^{b-1} - m^{b-2} + \\ldots - m + 1$. This has an odd number of terms. If $m$ is odd, then each term is odd and so the total is odd, but $(m+1)^{a-1}$ is even (note that $a > 1$). Contradiction, so $m$ is even.\n\nWe have $m^b = (m+1)^a - 1$. Expanding the right-hand side by the binomial theorem, and using $b > 1$, we see that $m$ must divide $a$. So $a$ is even also. Put $a = 2A$, $m = 2M$. We can factorise $(m+1)^a - 1$ as $((m+1)^A + 1)((m+1)^A - 1)$. The two factors have difference $2$, so their gcd divides $2$, but both factors are even, so their gcd is exactly $2$.\n\nIf $M = 1$ or a power of $2$, then the smaller factor $3^A - 1$ must be $2$, so $A = 1$ and we have $3^A + 1 = 4$, so $(2M)^b = 8$. Hence $M = 1$ and $b = 3$ and we have the solution $(m, a, b) = (2, 2, 3)$.\n\nIf $M$ is not a power of $2$, then $M^b > 2^b$, so we must have the larger factor $2 \\cdot M^b$ and the smaller factor $2^{b-1}$. But the larger factor is now $> 2^{b+1}$, so the difference between the factors is at least $3$. $2^{b-1} > 2$. Contradiction.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32278,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSome terms are deleted from an infinite arithmetic progression $1, x, y, \\ldots$ of real numbers to leave an infinite geometric progression $1, a, b, \\ldots$. Find all possible values of $a$.",
    "options": [],
    "answer": "a is any positive integer",
    "solution": "Solution:\n\nIf $a$ is negative, then the terms in the GP are alternately positive and negative, whereas either all terms in the AP from a certain point on are positive or all terms from a certain point on are negative. So $a$ cannot be negative. If $a$ is zero, then all terms in the GP except the first are zero, but at most one term of the AP is zero, so $a$ cannot be zero. Thus $a$ must be positive, so the AP must have infinitely many positive terms and hence $x \\geq 1$.\n\nLet $d = x - 1$, so all terms of the AP have the form $1 + n d$ for some positive integer $n$. Suppose $a = 1 + m d$, $a^2 = 1 + n d$, then $(1 + m d)^2 = 1 + n d$, so $d = (n - 2m)/m^2$, which is rational. Hence $a$ is rational. Suppose $a = b / c$, where $b$ and $c$ are relatively prime positive integers and $c > 1$. Then the denominator of the $n$th term of the GP is $c^n$, which becomes arbitrarily large as $n$ increases. But if $d = h / k$, then all terms of the AP have denominator at most $k$. So we cannot have $c > 1$. So $a$ must be a positive integer.\n\nOn the other hand, it is easy to see that any positive integer works. Take $x = 2$, then the AP includes all positive integers and hence includes any GP with positive integer terms.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32279,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a pile of 2000 stones, two players take turns in taking stones from the pile. Each player must remove $1, 2, 3, 4$, or $5$ stones from the pile at each turn, but may not take the same number as his opponent took on his last move. The player who takes the last stone wins. Does the first or second player have a winning strategy?",
    "options": [],
    "answer": "First player",
    "solution": "Solution:\n\nThe first player has a winning strategy. He takes $4$ on his first move leaving $7 \\bmod 13$ ($2000 = 153 \\times 13 + 7 + 4$). Now we claim that the first player can always leave: (1) $0 \\bmod 13$, (2) $3 \\bmod 13$ by taking away $3$, (3) $5 \\bmod 13$ by taking away $5$, or (4) $7 \\bmod 13$, and that the second player can never leave $0 \\bmod 13$.\n\nLet us look at each of these in turn. If the first player leaves $0 \\bmod 13$, then the second player can take $3$ and leave $10$. In that case the first player takes $5$ (a type (3) move). If the second player takes $1, 2, 4$ or $5$, leaving $12, 11, 9$ or $8 \\bmod 13$, then the first player takes $5, 4, 2, 1$ (respectively) and leaves $7 \\bmod 13$ (a type (4) move).\n\nIf the first player leaves $3 \\bmod 13$ by taking away $3$, then the second player cannot leave $0 \\bmod 13$, because he cannot take $3$ stones. If he takes $1, 2$ leaving $2, 1 \\bmod 13$ respectively, then the first player takes $2, 1$ leaving $0 \\bmod 13$ (a type (1) move). If the second player takes $4, 5$ leaving $12, 11 \\bmod 13$, then the first player takes $5, 4$ leaving $7 \\bmod 13$ (a type (4) move).\n\nIf the first player leaves $5 \\bmod 13$ by taking $5$, then the second player cannot leave $0 \\bmod 13$, because he cannot take $5$ stones. If he takes $1, 2, 3, 4$ stones, leaving $4, 3, 2, 1 \\bmod 13$, then the first player takes $4, 3, 2, 1$ stones leaving $0 \\bmod 13$ (a type (1) move).\n\nFinally, if the first player leaves $7 \\bmod 13$, and the second player takes $1$ stone, then the first player takes $3$ stones leaving $3 \\bmod 13$ (a type (2) move). If the second player takes $2, 3, 4$, or $5$ stones leaving $5, 4, 3, 2 \\bmod 13$, then the first player takes $5, 4, 3, 2$ stones leaving $0 \\bmod 13$ (a type (1) move).\n\nSo the second player can never leave $0 \\bmod 13$ and hence, in particular, can never take the last stone. But we have shown that the first player can always make a move of one of the four types, so can always move and hence must win (since after less than $2000$ moves there will be no stones left).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32280,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nA convex hexagon is called a unit if it has four diagonals of length $1$, whose endpoints include all the vertices of the hexagon. Show that there is a unit of area $k$ for any $0 < k \\leq 1$. What is the largest possible area for a unit?",
    "options": [],
    "answer": "3*sqrt(3)/4",
    "solution": "Solution:\nAnswer: We can get arbitrarily close to (but not achieve) $\\dfrac{3 \\sqrt{3}}{4}$ (approx $1.3$) by:\n\n![](attached_image_1.png)\n\nTo prove the first part, consider the diagram below. Take $AB = AC = 1$ and $\\angle BAC = 2\\theta$. Take $DE = DF = 1$ and take the points of intersection $X$ and $Y$ such that $AX = DX = AY = DY = 2/3$. It is easy to check that the area of the hexagon is $\\sin 2\\theta$. So by taking $\\theta$ in the interval $(0, \\pi/4]$ we can get any area $0 < k \\leq 1$.\n\n![](attached_image_2.png)\n\nIt is easy to check that there are six possible configurations for the unit diagonals, as shown in the diagram below.\n\n![](attached_image_3.png)\n\nConsider case 1.\n\n![](attached_image_4.png)\n\nThe area of the hexagon is area $AEDC$ + area $AFE$ + area $BAC$. The part of the segment $BF$ that lies inside $AEDC$ is wasted. The rest goes to provide height for the triangles on bases $AE$ and $AC$. So area $AFE$ + area $BAC$ can be maximised by taking $F$ close to $A$ and $\\angle BAC$ as close to a right angle as possible, so that the height of the triangle $BAC$ (on the base $AC$) is as large as possible. We can then get arbitrarily close to the area of:\n\n![](attached_image_5.png)\n\nWe obviously make $AEB$ a straight line. Now area $ADE$ + area $ADC$ = area $ACE$ + area $CDE$. So if we regard every point except $D$ as fixed, then we maximise the area by taking $\\angle EAD = \\angle CAD$, so that $D$ is the maximum distance from $CE$. Thus a maximal configuration must have $\\angle AED = \\angle CAD$. Similarly, it must have $\\angle CAD = \\angle CAB$, so all three angles must be equal. That disposes of case 1.\n\nIn cases 2 and 6 we find by a similar (but more tedious argument) the same maximum, although in one case we have to use the argument at the end for the final optimisation. In the other cases the maximum is smaller.\n\n![](attached_image_6.png)\n\n4\n5\n![](attached_image_7.png)\n6\n![](attached_image_8.png)\n\nHowever, all these details would take an already long solution way over length. Does anyone have a better approach?\n\nNo. 6 (second case) can be made arbitrarily close to the figure below (with $AB = AC = BD = 1$). To optimise it, suppose $\\angle ACB = \\theta$. Area $ABDC = $ area $ABC$ + area $BCD$. If we fix $\\theta$, then $BC$ is fixed, so to maximise area $BCD$ we must take $\\angle CBD = 90^\\circ$. But $\\theta$ cannot be optimal unless also $\\angle CAD = 90^\\circ$. We have $BA = BD$ and hence $\\angle BAD = \\angle BDA = 45^\\circ - \\theta/2$. Hence\n\n$$\n90^\\circ = \\angle CAD = \\angle BAC - \\angle BAD = (180^\\circ - 2\\theta) - (45^\\circ - \\theta/2)\n$$\n\nHence $\\theta = 30^\\circ$. So $\\angle ACD = \\angle BDC = 60^\\circ$ and $\\angle CAB = \\angle ABD = 120^\\circ$. It is easy to check that this has area $\\dfrac{3\\sqrt{3}}{4}$.\n\n![](attached_image_9.png)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32281,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that there are arbitrarily large numbers $n$ such that: (1) all its digits are 2 or more; and (2) the product of any four of its digits divides $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n$3232 = 16 \\times 202$ and $10000 = 16 \\times 625$. So any number with $3232$ as its last $4$ digits is divisible by $16$. So consider $N = 22223232$. Its sum of digits is $18$, so it is divisible by $9$. Hence it is divisible by $9 \\cdot 16 = 144$. But any four digits have at most four $2$'s and at most two $3$'s, so the product of any four digits divides $144$ and hence $N$. But now we can extend $N$ by inserting an additional $9m$ $2$'s at the front. Its digit sum is increased by $18m$, so it remains divisible by $144$ and it is still divisible by the product of any four digits.\nSolution:\nThe number $111111111$ with nine $1$'s is divisible by $9$. Hence the number with twenty-seven $1$'s, which equals $111111111 \\times 1000000001000000001$, is divisible by $27$. So $N$, the number with twenty-seven $3$'s, is divisible by $3^{4}$. Now the number with $27n$ $3$'s is divisible by $N$ and hence by $3^{4}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32282,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABC$ is a triangle. The incircle has center $I$ and touches the sides $BC$, $CA$, $AB$ at $D$, $E$, $F$ respectively. The rays $BI$ and $CI$ meet the line $EF$ at $P$ and $Q$ respectively. Show that if $DPQ$ is isosceles, then $ABC$ is isosceles.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\n$AF = AE$, so $\\angle AFE = 90^{\\circ} - A / 2$. Hence $\\angle BFP = 90^{\\circ} + A / 2$. But $\\angle FBP = B / 2$, so $\\angle FPB = C / 2$. But $BFP$ and $BDP$ are congruent ($BF = BD$, $BP$ common, $\\angle FBP = \\angle FDP$), so $\\angle DPB = C / 2$ and $\\angle DPQ = C$. Similarly, $\\angle DQP = B$. Hence $\\angle PDQ = A$. So $DQP$ and $ABC$ are similar. So if one is isosceles, so is the other.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32283,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $X$ be a set with $n$ elements. Given $k > 2$ subsets of $X$, each with at least $r$ elements, show that we can always find two of them whose intersection has at least $r - \\dfrac{n k}{4k - 4}$ elements.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32284,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nCall a set of 3 distinct elements which are in arithmetic progression a trio. What is the largest number of trios that can be subsets of a set of $n$ distinct real numbers?",
    "options": [],
    "answer": "If the set size is 2m, the maximum is m(m−1); if the set size is 2m+1, the maximum is m^2.",
    "solution": "Solution:\n\nLet $X$ be one of the elements. What is the largest number of trios that can have $X$ as middle element? Obviously, at most $\\max\\{b, a\\}$, where $b$ is the number of elements smaller than $X$ and $a$ is the number larger. Thus if $n=2m$, the number of trios is at most $0+1+2+\\ldots+m-1+m-1+m-2+\\ldots+1+0 = (m-1)m$. If $n=2m+1$, then the number is at most $0+1+2+\\ldots+m-1+m+m-1+\\ldots+1+0 = m^2$.\n\nThese maxima can be achieved by taking the numbers $1,2,3,\\ldots,n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32285,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nTwo players play a game on a $2000 \\times 2001$ board. Each has one piece and the players move their pieces alternately. A short move is one square in any direction (including diagonally) or no move at all. On his first turn each player makes a short move. On subsequent turns a player must make the same move as on his previous turn followed by a short move. This is treated as a single move. The board is assumed to wrap in both directions so a player on the edge of the board can move to the opposite edge. The first player wins if he can move his piece onto the same square as his opponent's piece. For example, suppose we label the squares from $(0,0)$ to $(1999,2000)$, and the first player's piece is initially at $(0,0)$ and the second player's at $(1996, 3)$. The first player could move to $(1999,2000)$, then the second player to $(1996,2)$. Then the first player could move to $(1998,1998)$, then the second player to $(1995,1)$. Can the first player always win irrespective of the initial positions of the two pieces?",
    "options": [],
    "answer": "Yes, the first player can always win.",
    "solution": "",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32286,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nShow that a square with side $1$ cannot be covered by five squares with side less than $1/2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose, for contradiction, that five squares, each with side less than $1/2$, can cover a unit square.\n\nThe diagonal of each small square is less than $1/2 \\sqrt{2} < 0.71$.\n\nConsider the four corners of the unit square. Each small square can cover at most one corner, since the distance between any two corners is $1$ (which is greater than the diagonal of a small square). Thus, at least four small squares are needed to cover the four corners.\n\nThis leaves at most one small square to cover the rest of the unit square. But the region not covered by the four corners is the interior of the unit square with small neighborhoods around the corners removed. This region contains a square of side close to $1 - 2 \\times (1/2) = 0$ (since the small squares are less than $1/2$ in side), but in fact, the remaining region is too large to be covered by a single small square of side less than $1/2$.\n\nTherefore, it is impossible to cover the unit square with five squares of side less than $1/2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32287,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe numbers $1, 2, \\ldots, 2002$ are written in order on a blackboard. Then the 1st, 4th, 7th, $\\ldots$, $3k+1$th, $\\ldots$ numbers in the list are erased. Then the 1st, 4th, 7th, $\\ldots$, $3k+1$th numbers in the remaining list are erased (leaving $3, 5, 8, 9, 12, \\ldots$). This process is carried out repeatedly until there are no numbers left. What is the last number to be erased?",
    "options": [],
    "answer": "1598",
    "solution": "Solution:\n\nLet $a_n$ be the first number remaining after $n$ iterations, so $a_0 = 1$, $a_1 = 2$, $a_2 = 3$, $a_3 = 5$, etc. We claim that:\n\n$\\displaystyle a_{n+1} = \\frac{3}{2} a_n$ if $a_n$ is even, and\n$\\displaystyle a_{n+1} = \\frac{3}{2}(a_n + 1) - 1$ if $a_n$ is odd.\n\nWe use induction on $n$.\n\nSuppose $a_n = 2N$. Consider the number $3N$. There are initially $N$ smaller numbers $\\equiv 1 \\pmod{3}$. So after the first iteration, it will lie in the $2N$th place. Hence, it will lie in first place after $n+1$ iterations.\n\nSimilarly, suppose $a_n = 2N + 1$. Consider $3N + 2$. There are initially $N + 1$ smaller numbers $\\equiv 1 \\pmod{3}$. So after the first iteration, it will lie in the $2N + 1$st place. Hence, it will lie in first place after $n+1$ iterations. That completes the induction.\n\nWe may now calculate successively the members of the sequence: $1, 2, 3, 5, 8, 12, 18, 27, 41, 62, 93, 140, 210, 315, 473, 710, 1065, 1598, 2397$. Hence $1598$ is the last surviving number from $1, 2, \\ldots, 2002$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32288,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nGiven a set of 9 points in the plane, no three collinear, show that for each point $P$ in the set, the number of triangles containing $P$ formed from the other 8 points in the set must be even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nJoin each pair of points, thus dividing the plane into polygonal regions. If a point $P$ moves around within one of the regions then the number of triangles it belongs to does not change. But if it crosses one of the lines then it leaves some triangles and enters others. Suppose the line is part of the segment joining the points $Q$ and $R$ of the set. Then it can only enter or leave a triangle $QRX$ for some $X$ in the set. Suppose $x$ points in the set lie on the same side of the line $QR$ as $P$. Then there are $6 - x$ points on the other side of the line $QR$. So $P$ leaves $x$ triangles and enters $6-x$. Thus the net change is even. Thus if we move $P$ until it is in the outer infinite region (outside the convex hull of the other 8 points), then we change the number of triangles by an even number. But in the outside region it belongs to no triangles.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32289,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABC$ is an equilateral triangle. $P$ is a variable interior point such that $\\angle APC = 120^\\circ$. The ray $CP$ meets $AB$ at $M$, and the ray $AP$ meets $BC$ at $N$. What is the locus of the circumcenter of the triangle $MBN$ as $P$ varies?",
    "options": [],
    "answer": "The locus is the segment of the perpendicular bisector of BG between the feet of the perpendiculars from A and from B to this line, where G denotes the center of the equilateral triangle ABC.",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\n$\\angle MPN = \\angle APC = 120^\\circ$ and $\\angle MBN = 60^\\circ$, so $MBNP$ is cyclic, in other words, $P$ lies on the circumcircle of $BMN$.\n\n$P$ also lies on the circle $AGC$, so $\\angle CPG = \\angle CAG$ (if $P$ is on the same side of $AG$ as $A$) $= 30^\\circ = \\angle MBG$. So $PMBG$ is cyclic. In other words, $G$ also lies on the circumcircle of $BMN$. If $P$ lies on the other side, the same conclusion follows from considering $\\angle APG$.\n\nSince $B$ and $G$ lie on the circumcircle, the center $O$ must lie on the perpendicular bisector of $BG$. But it is clear that the extreme positions of $O$ occur when $P$ is at $A$ and $B$ and that these are the feet of the perpendiculars from $A$ and $B$ to the perpendicular bisector.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32290,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$ABC$ is a triangle. $BD$ is an angle bisector. $E$, $F$ are the feet of the perpendiculars from $A$, $C$ respectively to the line $BD$. $M$ is the foot of the perpendicular from $D$ to the line $BC$. Show that $\\angle DME = \\angle DMF$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nLet $H$ be the foot of the perpendicular from $D$ to $AB$. $\\angle AHD = \\angle AED = 90^\\circ$, so $AHED$ is cyclic. Hence $\\angle DAE = \\angle DHE$. But $M$ is the reflection of $H$ in the line $BD$, so $\\angle DME = \\angle DAE$.\n\n$AE$ is parallel to $CD$, so $\\angle DAE = \\angle DCF$. $\\angle DFC = \\angle DMC$, so $DMCF$ is cyclic. Hence $\\angle DCF = \\angle DMF$. Hence $\\angle DME = \\angle DMF$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32291,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequence $a_{n}$ is defined as follows: $a_{1}=56$, $a_{n+1}=a_{n}-1/a_{n}$. Show that $a_{n}<0$ for some $n$ such that $0<n<2002$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nNote that whilst $a_{n}$ remains positive we have $a_{1}>a_{2}>a_{3}>\\ldots>a_{n}$. Hence if $a_{m}$ and $a_{m+n}$ are in this part of the sequence, then $a_{m+1}=a_{m}-1/a_{m}$, $a_{m+2}=a_{m+1}-1/a_{m+1}<a_{m+1}-1/a_{m}=a_{m}-2/a_{m}$. By a trivial induction $a_{m+n}<a_{m}-n/a_{m}$.\n\nIf we use one step then we need $56^{2}=3136$ terms to get $a_{1+3136}<56-56^{2}/56=0$, which is not good enough. So we try several steps.\n\nThus suppose that $a_{n}>0$ for all $n\\leq 2002$. Then we get successively:\n\n$a_{337}<56-336/56=50$\n\n$a_{837}<50-500/50=40$\n\n$a_{1237}<40-400/40=30$\n\n$a_{1537}<30-300/30=20$\n\n$a_{1737}<20-200/20=10$\n\n$a_{1837}<10-100/10=0$.\n\nContradiction. So we must have $a_{n}<0$ for some $n<2002$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32292,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA game is played on a $2001 \\times 2001$ board as follows. The first player's piece is the policeman, the second player's piece is the robber. Each piece can move one square south, one square east or one square northwest. In addition, the policeman (but not the robber) can move from the bottom right to the top left square in a single move. The policeman starts in the central square, and the robber starts one square diagonally northeast of the policeman. If the policeman moves onto the same square as the robber, then the robber is captured and the first player wins. However, the robber may move onto the same square as the policeman without being captured (and play continues). Show that the robber can avoid capture for at least $10000$ moves, but that the policeman can ultimately capture the robber.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nColor the squares with three colors as follows:\n\n```\n0 1 2 0 1 2 0 \\ldots 2\n1 2 0 1 2 0 1 \\cdots 0\n2 0 1 2 0 1 2 \\cdots 1\n```\n\nThe middle square is color $2$ (moving $999+1$ squares $E$ from the top left increases the color by $1$, then moving $999+1~S$ increases it by another $1$) and the square immediately $NE$ of it is also $2$. So both $P$ and $R$ start on color $2$. Note that any move increases the color by $1$ mod $3$, except for $P$'s special move which changes the color from $1$ to $0$.\n\nUntil $P$ has made this move, after each move of $P$, $P$'s color is always $1$ more than $R$'s color $(\\bmod 3)$, so $P$ cannot win (irrespective of the moves made by either player). Immediately after he makes the special move for the first time, $P$ is on color $0$ and $R$ is on color $1$, so immediately after his move $P$'s color is now $1$ less than $R$'s color mod $3$. Again $P$ cannot win. But after $P$ has made the special move for the second time, $P$'s color is the same as $R$'s (mod $3$) immediately after $P$'s move.\n\nNote that it takes $P$ at least $2001$ moves to complete his special move for the first time and at least $6002$ moves (in total) to complete his special move for the second time. This solves the first part of the question. Suppose $R$ just moves down to the bottom right and then moves in small circles (one move $NW$, one move $S$, one move $E$) waiting for $P$. It takes $P$ at least $6002 + 3999$ (moving from top left to the capture square, one square short of the bottom right) $= 10001$ to capture him, so $R$ makes at least $10000$ moves before being captured.\n\nWe claim that $P$ wins if he can get into any of the positions shown below relative to $R$, with $R$ to move $(*)$ :\n\n```\nX P X X X\nP \\quad x \\quad X \\quad P \\quad x\nx \\quad x \\quad R \\quad x \\quad x\nX \\quad P \\quad x \\quad X \\quad P\nx x x \\quad x\n```\n\nIt follows that $P$ can also win from the four positions below $(**)$ :\n\n```\nX X X P X X X\nx \\quad x \\quad x \\quad x \\quad x \\quad x \\quad x\nX x x x x x x\nP X X \\quad R \\quad x \\quad X \\quad P\nx \\quad x \\quad x \\quad x \\quad x \\quad x \\quad x\nx \\quad x \\quad x \\quad x \\quad x \\quad x \\quad x\nx \\quad x \\quad x \\quad P \\quad x \\quad x \\quad x\n```\n\nFor in each case at least one of $R$'s possible moves allow $P$ to move immediately into one of the winning positions at $(*)$. But $R$ can only make the other moves a limited number of times before running into the border. [That is obvious if the other two moves are $E$ and $S$. If they are $NW$ and $E$, then every $NW$ move takes $R$ closer to the top border, but his total number of $E$ moves can never exceed his total number of $NW$ moves by more than $2000$ because of the right border. Similarly, for $NW$ and $S$.]\n\nNow let $d$ be the number of rows plus the number of columns that $R$ and $P$ are apart. It is easy to check that the positions in $(*)$ and $(**)$ represent the only possibilities for $d=2$ and $3$. We show that $P$ can always get to $d=2$ or $3$. For $P$ can always copy $R$'s move, so he can certainly move so that $d$ never increases. But one of $R$'s moves will always allow $P$ to decrease $d$ by $1$ or $2$. There are three cases to consider:\n\nCase 1. If $P$ is east of $R$ and $R$ moves $E$, then $P$ moving $NW$ will decrease $d$ by $1$ or $2$. That is not possible if $P$ is in the top row, but then moving $S$ will decrease $d$ by $2$ unless $R$ is also in the top row. If both are in the top row, then $P$ moves $S$. Now after $R$'s next move, $P$ moves $NW$ which reduces $d$ by $2$.\n\nCase 2. If $P$ is south of $R$ and $R$ moves $S$, then a similar argument shows that $P$ can always decrease $d$ by $1$ or $2$ in one or two moves.\n\nCase 3. If $P$ is not south or east of $R$, and $R$ moves $NW$, then $P$ can always decrease $d$ by $1$ or $2$ by moving $S$ or $E$.\n\nBut repeated decreases by $1$ or $2$ must bring $d$ ultimately to $2$ or $3$ and hence to one of $(*)$ or $(**)$. So $P$ can always win.\n\nIt remains to prove the claim that $(*)$ are winning positions. The reason is that in each case $R$ has one move blocked off, so must make one of the other two. $P$ then copies $R$'s move, so next turn $R$ has the same move blocked off. Repeated use of the other two moves will bring him ultimately to one of the sides.\n\nWe start with the easiest case: in the two following positions. $R$ cannot move to $z$, so he must move east or south on each move. Hence he will (after at most $4000$ moves) reach the bottom right corner. He then loses moving out of it.\n\n```\nX P X\nP z x\nx X R\n```\n\nThe other cases of $(*)$ are slightly more complicated. Starting from either of the two positions below, we show that $R$ must eventually reach the extreme left column.\n\n```\nw \\quad x \\quad P \\quad x\nx \\quad R \\quad z \\quad x\nx y x P\n```\n\n$R$ cannot move to $z$, so he can only make $NW$ and $S$ moves. But his total number of $S$ moves can never exceed his total number of $NW$ moves by more than $2000$ because he cannot move off the bottom of the board, so he must eventually reach the extreme left column. [If he reaches the bottom row at $y$, then $P$ can always move to $z$ to preserve the configuration. If $R$ reaches the top row by moving to $w$, then $P$ can always move to $z$ to preserve the configuration.]\n\nHaving reached the extreme left column he is forced to move south. Eventually moving to $y$ will take him to the corner. $P$ then moves to $z$ and $R$ is captured on his next move.\n\nThe final case to consider is the two positions below. $R$ cannot move to $z$, so must move $E$ or $NW$. A similar argument to the previous case shows that he must eventually reach the top row. Having reached it at $w$, $P$ moves to $z$. So $R$ is forced to move right along the top row. When he reaches the corner at $y$, $P$ moves to $z$ and $R$ is captured when he moves out of the corner.\n\n```\nW x x\nx R y\nP z x\nX X P\n```",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32293,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A$, $B$ be two sets of $N$ consecutive integers. If $N=2003$, can we form $N$ pairs $(a,b)$ with $a \\in A$, $b \\in B$ such that the sums of the pairs are $N$ consecutive integers? What about $N=2004$?",
    "options": [],
    "answer": "Yes for N equals 2003; No for N equals 2004.",
    "solution": "Solution:\nWithout loss of generality, let $A = B = \\{1, 2, \\ldots, N\\}$—if we have a solution for $A = \\{a+1, a+2, \\ldots, a+N\\}$ and $B = \\{b+1, b+2, \\ldots, b+N\\}$, then subtracting $a$ from every element of $A$ and $b$ from every element of $B$ gives a solution for $A = B = \\{1, 2, \\ldots, N\\}$.\n\nSuppose the sum set is $(m+1), (m+2), \\ldots, (m+N)$. It has sum $N(2m + N + 1)/2$ and $A$ and $B$ each have sum $N(N+1)/2$, so we must have $2m = N + 1$, hence $N$ must be odd. So we cannot do it for $N = 2004$.\n\nSuppose $N = 2M + 1$, take the pairs $(1, M+1), (3, M), (5, M-1), \\ldots, (2M+1, 1), (2, 2M+1), (4, 2M), \\ldots, (2M, M+2)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32294,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$C$ is a point on the semicircle with diameter $AB$. $D$ is a point on the arc $BC$. $M$, $P$, $N$ are the midpoints of $AC$, $CD$ and $BD$. The circumcenters of $ACP$ and $BDP$ are $O$, $O'$. Show that $MN$ and $OO'$ are parallel.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nLet the center of the circle be $X$ and the radius $r$. Let $\\angle AXM = \\theta$, $\\angle BXN = \\varphi$. Note that $O$ is the intersection of $XM$ and the perpendicular to $CD$ at $Q$, the midpoint of $CP$. We have $XM = r \\cos \\theta$. Let $CD$ and $XM$ meet at $Y$. Then $\\angle PYX = 90^\\circ - \\angle PXY = 90^\\circ - \\angle PXC - \\angle CXM = \\theta + \\varphi - \\varphi = \\theta$. Hence $OX = PQ \\sec \\varphi$, so $OX / XM = PQ / (r \\cos \\theta \\cos \\varphi)$. Similarly, $O'X / ON = PQ / (r \\cos \\theta \\cos \\varphi)$, so $OO'$ and $MN$ are parallel.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32295,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nPablo was trying to solve the following problem: find the sequence $x_0, x_1, x_2, \\ldots, x_{2003}$ which satisfies $x_0 = 1$, $0 \\leq x_i \\leq 2 x_{i-1}$ for $1 \\leq i \\leq 2003$ and which maximises $S$. Unfortunately he could not remember the expression for $S$, but he knew that it had the form $S = \\pm x_1 \\pm x_2 \\pm \\ldots \\pm x_{2002} + x_{2003}$. Show that he can still solve the problem.",
    "options": [],
    "answer": "x_i = 2^i for i = 0, 1, ..., 2003",
    "solution": "Solution:\n\nFor any combination of signs the maximum is obtained by taking all $x_i$ as large as possible. Suppose we have a different set of $x_i$. Then for some $k$ we must have $x_k < 2 x_{k-1}$ and $x_i = 2 x_{i-1}$ for all $i > k$. Suppose $2 x^{k-1} - x^k = h > 0$. Then we can increase $x_k$ by $h$, $x_{k+1}$ by $2h$, $x_{k+2}$ by $4h$, $\\ldots$ So the sum will be increased by $h (\\pm 1 \\pm 2 \\pm \\ldots \\pm 2^{m-1} + 2^m)$ for some $m \\geq 0$. But $\\pm 1 \\pm 2 \\pm \\ldots \\pm 2^{m-1} \\geq - (1 + 2 + \\ldots + 2^{m-1}) = -2^m + 1$, so the overall sum will be increased by at least $1$. So the set of $x_i$ was not maximal.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32296,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nA $A \\subset \\{1,2,3, \\ldots, 49\\}$ does not contain six consecutive integers. Find the largest possible value of $|A|$. How many such subsets are there (of the maximum size)?",
    "options": [],
    "answer": "Maximum size = 41; Number of maximum subsets = 495",
    "solution": "Solution:\n\nWe must exclude at least one element of each of the 8 sets $\\{1,2, \\ldots, 6\\}, \\{7, \\ldots, 12\\}, \\{13, \\ldots, 18\\}, \\ldots, \\{43, \\ldots, 48\\}$. So $|A| \\leq 41$. But a value of 41 is certainly possible, for example, exclude $2,8,14, \\ldots, 44$.\n\nThe largest excluded element must be at least 44 (or we have the 6 consecutive elements 44, 45, 46, 47, 48, 49). The smallest excluded element must be at most 6. If we exclude 2 and 44, then the difference between them is $7 \\cdot 6$ and so the other 6 excluded elements are fixed. But if we exclude 3 and 44, for example, then there are several possible choices for the other elements.\n\nThere are 5 ways of choosing the smallest and largest excluded element to get a difference of $7 \\cdot 6$ between them (2 and 44, 3 and 45, 4 and 46, 5 and 47, 6 and 48). There are 4 ways to get a difference of $7 \\cdot 6 - 1$ (3 and 44, 4 and 45, 5 and 46, 6 and 47). There are 3 ways to get a difference of $7 \\cdot 6 - 2$ (4 and 44, 5 and 45, 6 and 46), 2 ways to get a difference of $7 \\cdot 6 - 3$ (5 and 44, 6 and 45), and 1 way to get a difference of $7 \\cdot 6 - 4$ (6 and 44).\n\nIf the difference is $7 \\cdot 6 - 1$, then we can shorten any of the 7 gaps, so there are 7 possibilities. For example, with 3 and 44, we could shorten the first gap, so excluding 3, 8, 14, 20, 26, 32, 38 and 44, or the second gap, so excluding $3,9,14,20,26,32,38$ and 44, and so on.\n\nIf the difference is $7 \\cdot 6 - 2$, then we can shorten one gap by two (7 possibilities) or two gaps by one (21 possibilities), total 28.\n\nIf the difference is $7 \\cdot 6 - 3$, then we can shorten one gap by three (7), one by two and one by one (42), or three by one (35), total 84.\n\nFinally, if the difference is $7 \\cdot 6 - 4$, we can shorten one by four (7), one by three and one by one (42), two by two (21), one by two and two by one (105), or four by one (35), total 210.\n\nSo the total number of possibilities is $5 \\cdot 1 + 4 \\cdot 7 + 3 \\cdot 28 + 2 \\cdot 84 + 1 \\cdot 210 = 495$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32297,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n$\\mathrm{ABCD}$ is a square. $\\mathrm{P}, \\mathrm{Q}$ are points on the sides $\\mathrm{BC}, \\mathrm{CD}$ respectively, distinct from the endpoints such that $\\mathrm{BP}=\\mathrm{CQ}$. $\\mathrm{X}, \\mathrm{Y}$ are points on $\\mathrm{AP}, \\mathrm{AQ}$ respectively. Show that there is a triangle with side lengths $BX, XY, YD$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n![](attached_image_1.png)\n\nWe have $DY < BY \\leq BX + XY$ (this is almost obvious, but to prove formally use the cosine formula for $BAY$ and $DAY$ and notice that $\\angle BAY > \\angle DAY$). Similarly, $BX < DX \\leq DY + YX$. So it remains to show that $XY < BX + DY$.\n\nTake $Q'$ on the extension of $BC$ so that $BQ' = DQ$, as shown in the diagram. Take $Y'$ on $AQ'$ so that $AY' = AY$. Then $XY' \\leq BX + BY' = BX + DY$. Now we claim that $\\angle PAQ' > \\angle PAQ$, so it follows by the same observation as above that $XY' > XY$. But the claim is almost obvious. Note that $PQ' = AB$.\n\n![](attached_image_2.png)\n\nSo take $P'$ on $AD$ with $\\angle P'PQ' = 90^\\circ$. Then $A$ lies inside the circle $P'PQ'$, so extend $PA$ to meet it again at $A'$. Then $\\angle PA'Q' = \\angle PP'Q' = 45^\\circ$, so $\\angle PAQ' = \\angle PA'Q' + \\angle AQ'Q' > 45^\\circ$. But $\\angle PAQ' + \\angle PAQ = 90^\\circ$, so $\\angle PAQ' > \\angle PAQ$ as claimed.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32298,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe sequences $a_0, a_1, a_2, \\ldots$ and $b_0, b_1, b_2, \\ldots$ are defined by $a_0 = 1$, $b_0 = 4$, $a_{n+1} = a_n^{2001} + b_n$, $b_{n+1} = b_n^{2001} + a_n$. Show that no member of either sequence is divisible by $2003$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n$2003$ is prime, so $a^{2002} = 1 \\bmod 2003$ for any $a$ not divisible by $2003$. Thus $a_{n+1} = a_n^{-1} + b_n \\bmod 2003$, $b_{n+1} = b_n^{-1} + a_n \\bmod 2003$. Put $c_n = a_n b_n$. Then $c_{n+1} = c_n + 1 / c_n + 2 = (c_n + 1)^2 / c_n \\bmod 2003$. So if $c_n \\neq 0 \\bmod 2003$, then $c_{n+1} \\neq 0 \\bmod 2003$ unless $c_n = -1 \\bmod 2003$. Then if $c_{n+1} = -1 \\bmod 2003$, we must have $(c_n^2 + 3c_n + 1) / c_n = 0 \\bmod 2003$, so $c_n^2 + 3c_n + 1 = 0 \\bmod 2003$. Note that $c_0 = 4$. So it is sufficient to show that there are no solutions to $x^2 + 3x + 1 = 0 \\bmod 2003$, or equivalently to $(x - 1000)^2 = 1000^2 - 1 = 502 \\bmod 2003$. In other words, we have to show that $502$ is a quadratic non-residue mod $2003$.\n\nThe easiest way to do that is to use the law of quadratic reciprocity, but that is almost certainly outside the syllabus. We note that $502 = 5 \\bmod 2003$, so $502$ is a square iff $5$ is a square. It is sufficient to show that $5^{1001} = -1 \\bmod 2003$, for then if we had $x^2 = 5$, we would have $x^{2002} = -1 \\bmod 2003$, whereas we know that $x^{2002} = 1 \\bmod 2003$. We note that $1001 = 7 \\cdot 11 \\cdot 13$. We start by showing that $5^7 = 8 \\bmod 2003$. We have $5^5 = 3125 = 1122 \\bmod 2003$, so $5^6 = 5610 = 1604 \\bmod 2003$, so $5^7 = 8020 = 8 \\bmod 2003$.\n\nWe calculate successively $2^{11} = 2048 = 45 \\bmod 2003$, so $2^{22} = 2025 = 22 \\bmod 2003$. Multiplying by $22$ is relatively easy, so $2^{44} = 484$, $2^{66} = 10648 = 633$, $2^{88} = 13926 = -95$, $2^{110} = -2090 = -87$, $2^{132} = -1914 = 89$, $2^{143} = 4005 = -1$ all $\\bmod 2003$. Hence $8^{11 \\cdot 13} = -1 \\bmod 2003$, so $5^{1001} = -1 \\bmod 2003$, as required, and we are done.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32299,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nThe in-circle of triangle $ABC$ touches the sides $BC$, $CA$ and $AB$ in $K$, $L$ and $M$ respectively. The line through $A$ and parallel to $LK$ meets $MK$ in $P$ and the line through $A$ and parallel to $MK$ meets $LK$ in $Q$. Show that the line $PQ$ bisects the sides $AB$ and $AC$ of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $AP$, $AQ$ produced meet $BC$ in $D$, $E$ respectively.\n\n![](attached_image_1.png)\n\nSince $MK$ is parallel to $AE$, we have $\\angle AEK = \\angle MKB$. Since $BK = BM$, both being tangents to the circle from $B$, $\\angle MKB = \\angle BMK$. This with the fact that $MK$ is parallel to $AE$ gives us $\\angle AEK = \\angle MAE$. This shows that $MAEK$ is an isosceles trapezoid. We conclude that $MA = KE$. Similarly, we can prove that $AL = DK$. But $AM = AL$. We get that $DK = KE$. Since $KP$ is parallel to $AE$, we get $DP = PA$ and similarly $EQ = QA$. This implies that $PQ$ is parallel to $DE$ and hence bisects $AB$, $AC$ when produced.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32300,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSolve for integers $x, y, z$ :\n$$\nx+y=1-z, \\quad x^{3}+y^{3}=1-z^{2}\n$$",
    "options": [],
    "answer": "All integer solutions are the infinite family (m, -m, 1) for any integer m, together with (0, 1, 0), (1, 0, 0), (0, -2, 3), (-2, 0, 3), (-2, -3, 6), (-3, -2, 6).",
    "solution": "Solution:\nEliminating $z$ from the given set of equations, we get\n$$\nx^{3}+y^{3}+(1-(x+y))^{2}=1\n$$\nThis factors to\n$$\n(x+y)\\left(x^{2}-x y+y^{2}+x+y-2\\right)=0\n$$\nCase 1. Suppose $x+y=0$. Then $z=1$ and $(x, y, z)=(m,-m, 1)$, where $m$ is an integer give one family of solutions.\n\nCase 2. Suppose $x+y \\neq 0$. Then we must have\n$$\nx^{2}-x y+y^{2}+x+y-2=0 .\n$$\nThis can be written in the form\n$$\n(2 x-y+1)^{2}+3(y+1)^{2}=12\n$$\nHere there are two possibilities:\n$$\n2 x-y+1=0,\\ y+1= \\pm 2 ; \\quad 2 x-y+1= \\pm 3,\\ y+1= \\pm 1\n$$\nAnalysing all these cases we get\n$$\n(x, y, z)=(0,1,0),(-2,-3,6),(1,0,0),(0,-2,3),(-2,0,3),(-3,-2,6) .\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32301,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a$, $b$, $c$, $x$ are real numbers such that $a b c \\neq 0$ and\n$$\n\\frac{x b + (1-x) c}{a} = \\frac{x c + (1-x) a}{b} = \\frac{x a + (1-x) b}{c}\n$$\nthen prove that either $a + b + c = 0$ or $a = b = c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose $a + b + c \\neq 0$ and let the common value be $\\lambda$. Then\n$$\n\\lambda = \\frac{x b + (1-x) c + x c + (1-x) a + x a + (1-x) b}{a + b + c} = 1\n$$\nWe get two equations:\n$$\n-a + x b + (1-x) c = 0, \\quad (1-x) a - b + x c = 0\n$$\n(The other equation is a linear combination of these two.) Using these two equations, we get the relations\n$$\n\\frac{a}{1-x + x^{2}} = \\frac{b}{x^{2} - x + 1} = \\frac{c}{(1-x)^{2} + x}\n$$\nSince $1-x + x^{2} \\neq 0$, we get $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32302,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a convex quadrilateral $P Q R S$, $P Q = R S$, $(\\sqrt{3}+1) Q R = S P$ and $\\angle R S P - \\angle S P Q = 30^{\\circ}$. Prove that\n$$\n\\angle P Q R - \\angle Q R S = 90^{\\circ}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $[$Fig$]$ denote the area of Fig. We have\n$$\n[P Q R S] = [P Q R] + [R S P] = [Q R S] + [S P Q]\n$$\nLet us write $P Q = p$, $Q R = q$, $R S = r$, $S P = s$. The above relations reduce to\n$$\np q \\sin \\angle P Q R + r s \\sin \\angle R S P = q r \\sin \\angle Q R S + s p \\sin \\angle S P Q\n$$\nUsing $p = r$ and $(\\sqrt{3}+1) q = s$ and dividing by $p q$, we get\n$$\n\\sin \\angle P Q R + (\\sqrt{3}+1) \\sin \\angle R S P = \\sin \\angle Q R S + (\\sqrt{3}+1) \\sin \\angle S P Q\n$$\nTherefore, $\\sin \\angle P Q R - \\sin \\angle Q R S = (\\sqrt{3}+1)(\\sin \\angle S P Q - \\sin \\angle R S P)$.\n\n![](attached_image_1.png)\n\nFig. 2.\n\nThis can be written in the form\n$$\n\\begin{aligned}\n2 \\sin & \\frac{\\angle P Q R - \\angle Q R S}{2} \\cos \\frac{\\angle P Q R + \\angle Q R S}{2} \\\\\n& = (\\sqrt{3}+1) 2 \\sin \\frac{\\angle S P Q - \\angle R S P}{2} \\cos \\frac{\\angle S P Q + \\angle R S P}{2}\n\\end{aligned}\n$$\nUsing the relations\n$$\n\\cos \\frac{\\angle P Q R + \\angle Q R S}{2} = -\\cos \\frac{\\angle S P Q + \\angle R S P}{2}\n$$\nand\n$$\n\\sin \\frac{\\angle S P Q - \\angle R S P}{2} = -\\sin 15^{\\circ} = -\\frac{(\\sqrt{3}-1)}{2 \\sqrt{2}}\n$$\nwe obtain\n$$\n\\sin \\frac{\\angle P Q R - \\angle Q R S}{2} = (\\sqrt{3}+1)\\left[-\\frac{(\\sqrt{3}-1)}{2 \\sqrt{2}}\\right] = \\frac{1}{\\sqrt{2}}\n$$\nThis shows that\n$$\n\\frac{\\angle P Q R - \\angle Q R S}{2} = \\frac{\\pi}{4} \\quad \\text{or} \\quad \\frac{3\\pi}{4}\n$$\nUsing the convexity of $P Q R S$, we can rule out the latter alternative. We obtain\n$$\n\\angle P Q R - \\angle Q R S = \\frac{\\pi}{2}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32303,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c$ be three real numbers such that $1 \\geq a \\geq b \\geq c \\geq 0$. Prove that if $\\lambda$ is a root of the cubic equation $x^{3}+a x^{2}+b x+c=0$ (real or complex), then $|\\lambda| \\leq 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSince $\\lambda$ is a root of the equation $x^{3}+a x^{2}+b x+c=0$, we have\n$$\n\\lambda^{3} = -a \\lambda^{2} - b \\lambda - c\n$$\nThis implies that\n$$\n\\begin{aligned}\n\\lambda^{4} & = -a \\lambda^{3} - b \\lambda^{2} - c \\lambda \\\\\n& = (1-a) \\lambda^{3} + (a-b) \\lambda^{2} + (b-c) \\lambda + c\n\\end{aligned}\n$$\nwhere we have used again\n$$\n-\\lambda^{3} - a \\lambda^{2} - b \\lambda - c = 0\n$$\nSuppose $|\\lambda| \\geq 1$. Then we obtain\n$$\n\\begin{aligned}\n|\\lambda|^{4} & \\leq (1-a)|\\lambda|^{3} + (a-b)|\\lambda|^{2} + (b-c)|\\lambda| + c \\\\\n& \\leq (1-a)|\\lambda|^{3} + (a-b)|\\lambda|^{3} + (b-c)|\\lambda|^{3} + c|\\lambda|^{3} \\\\\n& \\leq |\\lambda|^{3}\n\\end{aligned}\n$$\nThis shows that $|\\lambda| \\leq 1$. Hence the only possibility in this case is $|\\lambda|=1$. We conclude that $|\\lambda| \\leq 1$ is always true.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32304,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor any natural number $n$ ($n \\geq 3$), let $f(n)$ denote the number of non-congruent integer-sided triangles with perimeter $n$ (e.g., $f(3)=1$, $f(4)=0$, $f(7)=2$). Show that\n\na. $f(1999) > f(1996)$\n\nb. $f(2000) = f(1997)$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na. Let $a, b, c$ be the sides of a triangle with $a + b + c = 1996$, and each being a positive integer. Then $a + 1, b + 1, c + 1$ are also sides of a triangle with perimeter $1999$ because\n$$\na < b + c \\quad \\Longrightarrow \\quad a + 1 < (b + 1) + (c + 1)\n$$\nand so on. Moreover, $(999, 999, 1)$ form the sides of a triangle with perimeter $1999$, which is not obtainable in the form $(a + 1, b + 1, c + 1)$ where $a, b, c$ are the integers and the sides of a triangle with $a + b + c = 1996$. We conclude that $f(1999) > f(1996)$.\n\nb. As in the case (a) we conclude that $f(2000) \\geq f(1997)$. On the other hand, if $x, y, z$ are the integer sides of a triangle with $x + y + z = 2000$, and say $x \\geq y \\geq z \\geq 1$, then we cannot have $z = 1$; for otherwise we would get $x + y = 1999$ forcing $x, y$ to have opposite parity so that $x - y \\geq 1 = z$ violating triangle inequality for $x, y, z$. Hence $x \\geq y \\geq z > 1$. This implies that $x - 1 \\geq y - 1 \\geq z - 1 > 0$. We already have $x < y + z$. If $x \\geq y + z - 1$, then we see that $y + z - 1 \\leq x < y + z$, showing that $y + z - 1 = x$. Hence we obtain $2000 = x + y + z = 2x + 1$ which is impossible. We conclude that $x < y + z - 1$. This shows that $x - 1 < (y - 1) + (z - 1)$ and hence $x - 1, y - 1, z - 1$ are the sides of a triangle with perimeter $1997$. This gives $f(2000) \\leq f(1997)$. Thus we obtain the desired result.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32305,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle in which no angle is $90^{\\circ}$. For any point $P$ in the plane of the triangle, let $A_{1}, B_{1}, C_{1}$ denote the reflections of $P$ in the sides $BC, CA, AB$ respectively. Prove the following statements:\n\na. If $P$ is the incentre or an excentre of $ABC$, then $P$ is the circumcentre of $A_{1}B_{1}C_{1}$;\n\nb. If $P$ is the circumcentre of $ABC$, then $P$ is the orthocentre of $A_{1}B_{1}C_{1}$;\n\nc. If $P$ is the orthocentre of $ABC$, then $P$ is either the incentre or an excentre of $A_{1}B_{1}C_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\n![](attached_image_1.png)\nIf $P=I$ is the incentre of triangle $ABC$, and $r$ its inradius, then it is clear that $A_{1}I = B_{1}I = C_{1}I = 2r$. It follows that $I$ is the circumcentre of $A_{1}B_{1}C_{1}$. On the other hand if $P=I_{1}$ is the excentre of $ABC$ opposite $A$ and $r_{1}$ the corresponding exradius, then again we see that $A_{1}I_{1} = B_{1}I_{1} = C_{1}I_{1} = 2r_{1}$. Thus $I_{1}$ is the circumcentre of $A_{1}B_{1}C_{1}$.\n\n![](attached_image_2.png)\n\nb.\nLet $P=O$ be the circumcentre of $ABC$. By definition, it follows that $OA_{1}$ bisects and is bisected by $BC$ and so on. Let $D, E, F$ be the mid-points of $BC, CA, AB$ respectively. Then $FE$ is parallel to $BC$. But $E, F$ are also mid-points of $OB_{1}, OC_{1}$ and hence $FE$ is parallel to $B_{1}C_{1}$ as well. We conclude that $BC$ is parallel to $B_{1}C_{1}$. Since $OA_{1}$ is perpendicular to $BC$, it follows that $OA_{1}$ is perpendicular to $B_{1}C_{1}$. Similarly $OB_{1}$ is perpendicular to $C_{1}A_{1}$ and $OC_{1}$ is perpendicular to $A_{1}B_{1}$. These imply that $O$ is the orthocentre of $A_{1}B_{1}C_{1}$. (This applies whether $O$ is inside or outside $ABC$.)\n\nc.\nLet $P=H$, the orthocentre of $ABC$. We consider two possibilities; $H$ falls inside $ABC$ and $H$ falls outside $ABC$.\n\nSuppose $H$ is inside $ABC$; this happens if $ABC$ is an acute triangle. It is known that $A_{1}, B_{1}, C_{1}$ lie on the circumcircle of $ABC$. Thus $\\angle C_{1}A_{1}A = \\angle C_{1}CA = 90^{\\circ} - A$. Similarly $\\angle B_{1}A_{1}A = \\angle B_{1}BA = 90^{\\circ} - A$. These show that $\\angle C_{1}A_{1}A = \\angle B_{1}A_{1}A$. Thus $A_{1}A$ is an internal bisector of $\\angle C_{1}A_{1}B_{1}$. Similarly we can show that $B_{1}B$ bisects $\\angle A_{1}B_{1}C_{1}$ and $C_{1}C$ bisects $\\angle B_{1}C_{1}A_{1}$. Since $A_{1}A, B_{1}B, C_{1}C$ concur at $H$, we conclude that $H$ is the incentre of $A_{1}B_{1}C_{1}$.\n\nOR If $D, E, F$ are the feet of perpendiculars of $A, B, C$ to the sides $BC, CA, AB$ respectively, then we see that $EF, FD, DE$ are respectively parallel to $B_{1}C_{1}, C_{1}A_{1}, A_{1}B_{1}$. This implies that $\\angle C_{1}A_{1}H = \\angle FDH = \\angle ABE = 90^{\\circ} - A$, as $BDHF$ is a cyclic quadrilateral. Similarly, we can show that $\\angle B_{1}A_{1}H = 90^{\\circ} - A$. It follows that $A_{1}H$ is the internal bisector of $\\angle C_{1}A_{1}B_{1}$. We can proceed as in the earlier case.\n\nIf $H$ is outside $ABC$, the same proofs go through again, except that two of $A_{1}H, B_{1}H, C_{1}H$ are external angle bisectors and one of these is an internal angle bisector. Thus $H$ becomes an excentre of triangle $A_{1}B_{1}C_{1}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32306,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that the equation\n$$\nx^{2}+y^{2}+z^{2}=(x-y)(y-z)(z-x)\n$$\nhas infinitely many solutions in integers $x, y, z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe seek solutions $(x, y, z)$ which are in arithmetic progression. Let us put $y-x=z-y=d>0$ so that the equation reduces to the form\n$$\n3 y^{2}+2 d^{2}=2 d^{3}\n$$\nThus we get $3 y^{2}=2(d-1) d^{2}$. We conclude that $2(d-1)$ is 3 times a square. This is satisfied if $d-1=6 n^{2}$ for some $n$. Thus $d=6 n^{2}+1$ and $3 y^{2}=d^{2} \\cdot 2\\left(6 n^{2}\\right)$ giving us $y^{2}=4 d^{2} n^{2}$. Thus we can take $y=2 d n=2 n\\left(6 n^{2}+1\\right)$. From this we obtain $x=y-d=(2 n-1)\\left(6 n^{2}+1\\right), z=y+d=(2 n+1)\\left(6 n^{2}+1\\right)$. It is easily verified that\n$$\n(x, y, z)=\\left((2 n-1)\\left(6 n^{2}+1\\right), 2 n\\left(6 n^{2}+1\\right),(2 n+1)\\left(6 n^{2}+1\\right)\\right)\n$$\nis indeed a solution for a fixed $n$ and this gives an infinite set of solutions as $n$ varies over natural numbers.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32307,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $a, b, c$ are positive real numbers such that $a b c = 1$, prove that\n$$\na^{b+c} b^{c+a} c^{a+b} \\leq 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nNote that the inequality is symmetric in $a, b, c$ so that we may assume that $a \\geq b \\geq c$. Since $a b c = 1$, it follows that $a \\geq 1$ and $c \\leq 1$. Using $b = 1 / (a c)$, we get\n$$\na^{b+c} b^{c+a} c^{a+b} = \\frac{a^{b+c} c^{a+b}}{a^{c+a} c^{c+a}} = \\frac{c^{b-c}}{a^{a-b}} \\leq 1\n$$\nbecause $c \\leq 1$, $b \\geq c$, $a \\geq 1$ and $a \\geq b$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32308,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nGiven any nine integers show that it is possible to choose, from among them, four integers $a, b, c, d$ such that $a+b-c-d$ is divisible by $20$. Further show that such a selection is not possible if we start with eight integers instead of nine.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose there are four numbers $a, b, c, d$ among the given nine numbers which leave the same remainder modulo $20$. Then $a+b \\equiv c+d \\pmod{20}$ and we are done.\n\nIf not, there are two possibilities:\n\n(1) We may have two disjoint pairs $\\{a, c\\}$ and $\\{b, d\\}$ obtained from the given nine numbers such that $a \\equiv c \\pmod{20}$ and $b \\equiv d \\pmod{20}$. In this case we get $a+b \\equiv c+d \\pmod{20}$.\n\n(2) Or else there are at most three numbers having the same remainder modulo $20$ and the remaining six numbers leave distinct remainders which are also different from the first remainder (i.e., the remainder of the three numbers). Thus there are at least $7$ distinct remainders modulo $20$ that can be obtained from the given set of nine numbers. These $7$ remainders give rise to $\\binom{7}{2} = 21$ pairs of numbers. By pigeonhole principle, there must be two pairs $\\left(r_1, r_2\\right), \\left(r_3, r_4\\right)$ such that $r_1 + r_2 \\equiv r_3 + r_4 \\pmod{20}$. Going back we get four numbers $a, b, c, d$ such that $a+b \\equiv c+d \\pmod{20}$.\n\nIf we take the numbers $0, 0, 0, 1, 2, 4, 7, 12$, we check that the result is not true for these eight numbers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32309,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle and $D$ be the mid-point of side $BC$. Suppose $\\angle DAB = \\angle BCA$ and $\\angle DAC = 15^{\\circ}$. Show that $\\angle ADC$ is obtuse. Further, if $O$ is the circumcentre of $ADC$, prove that triangle $AOD$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nLet $\\alpha$ denote the equal angles $\\angle BAD = \\angle DCA$. Using sine rule in triangles $DAB$ and $DAC$, we get\n$$\n\\frac{AD}{\\sin B} = \\frac{BD}{\\sin \\alpha}, \\quad \\frac{CD}{\\sin 15^{\\circ}} = \\frac{AD}{\\sin \\alpha}\n$$\nEliminating $\\alpha$ (using $BD = DC$ and $2\\alpha + B + 15^{\\circ} = \\pi$), we obtain $1 + \\cos(B + 15^{\\circ}) = 2 \\sin B \\sin 15^{\\circ}$. But we know that $2 \\sin B \\sin 15^{\\circ} = \\cos(B - 15^{\\circ}) - \\cos(B + 15^{\\circ})$. Putting $\\beta = B - 15^{\\circ}$, we get a relation $1 + 2\\cos(\\beta + 30) = \\cos \\beta$. We write this in the form\n$$\n(1 - \\sqrt{3}) \\cos \\beta + \\sin \\beta = 1\n$$\nSince $\\sin \\beta \\leq 1$, it follows that $(1 - \\sqrt{3}) \\cos \\beta \\geq 0$. We conclude that $\\cos \\beta \\leq 0$ and hence that $\\beta$ is obtuse. So is angle $B$ and hence $\\angle ADC$.\n\nWe have the relation $(1 - \\sqrt{3}) \\cos \\beta + \\sin \\beta = 1$. If we set $x = \\tan(\\beta/2)$, then we get, using $\\cos \\beta = (1 - x^2)/(1 + x^2)$, $\\sin \\beta = 2x/(1 + x^2)$,\n$$\n(\\sqrt{3} - 2)x^2 + 2x - \\sqrt{3} = 0\n$$\nSolving for $x$, we obtain $x = 1$ or $x = \\sqrt{3}(2 + \\sqrt{3})$. If $x = \\sqrt{3}(2 + \\sqrt{3})$, then $\\tan(\\beta/2) > 2 + \\sqrt{3} = \\tan 75^{\\circ}$ giving us $\\beta > 150^{\\circ}$. This forces that $B > 165^{\\circ}$ and hence $B + A > 165^{\\circ} + 15^{\\circ} = 180^{\\circ}$, a contradiction. Thus $x = 1$ giving us $\\beta = \\pi/2$. This gives $B = 105^{\\circ}$ and hence $\\alpha = 30^{\\circ}$. Thus $\\angle DAO = 60^{\\circ}$. Since $OA = OD$, the result follows.\n\n\nSolution 2:\nLet $m_a$ denote the median $AD$. Then we can compute\n$$\n\\cos \\alpha = \\frac{c^2 + m_a^2 - (a^2/4)}{2cm_a}, \\quad \\sin \\alpha = \\frac{2\\Delta}{cm_a}\n$$\nwhere $\\Delta$ denotes the area of triangle $ABC$. These two expressions give\n$$\n\\cot \\alpha = \\frac{c^2 + m_a^2 - (a^2/4)}{4\\Delta}\n$$\nSimilarly, we obtain\n$$\n\\cot \\angle CAD = \\frac{b^2 + m_a^2 - (a^2/4)}{4\\Delta}\n$$\nThus we get\n$$\n\\cot \\alpha - \\cot 15^{\\circ} = \\frac{c^2 - a^2}{4\\Delta}\n$$\nSimilarly we can also obtain\n$$\n\\cot B - \\cot \\alpha = \\frac{c^2 - a^2}{4\\Delta}\n$$\ngiving us the relation\n$$\n\\cot B = 2\\cot \\alpha - \\cot 15^{\\circ}\n$$\nIf $B$ is acute then $2\\cot \\alpha > \\cot 15^{\\circ} = 2 + \\sqrt{3} > 2\\sqrt{3}$. It follows that $\\cot \\alpha > \\sqrt{3}$. This implies that $\\alpha < 30^{\\circ}$ and hence\n$$\nB = 180^{\\circ} - 2\\alpha - 15^{\\circ} > 105^{\\circ}\n$$\nThis contradiction forces that angle $B$ is obtuse and consequently $\\angle ADC$ is obtuse.\n\nSince $\\angle BAD = \\alpha = \\angle ACD$, the line $AB$ is tangent to the circumcircle $\\Gamma$ of $ADC$ at $A$. Hence $OA$ is perpendicular to $AB$. Draw $DE$ and $BF$ perpendicular to $AC$, and join $OD$. Since $\\angle DAC = 15^{\\circ}$, we see that $\\angle DOC = 30^{\\circ}$ and hence $DE = OD/2$. But $DE$ is parallel to $BF$ and $BD = DC$ shows that $BF = 2DE$. We conclude that $BF = DO$. But $DO = AO$, both being radii of $\\Gamma$. Thus $BF = AO$. Using right triangles $BFO$ and $BAO$, we infer that $AB = OF$. We conclude that $ABFO$ is a rectangle. In particular $\\angle AOF = 90^{\\circ}$. It follows that\n$$\n\\angle AOD = 90^{\\circ} - \\angle DOC = 90^{\\circ} - 30^{\\circ} = 60^{\\circ}\n$$\nSince $OA = OD$, we conclude that $AOD$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32310,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\mathbf{R}$ denote the set of all real numbers. Find all functions $f: \\mathbf{R} \\rightarrow \\mathbf{R}$ satisfying the condition\n$$\nf(x+y)=f(x) f(y) f(x y)\n$$\nfor all $x, y$ in $\\mathbf{R}$.",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = 1 for all real x; f(x) = -1 for all real x",
    "solution": "Solution:\nPutting $x=0, y=0$, we get $f(0)=f(0)^3$ so that $f(0)=0, 1$ or $-1$.\nIf $f(0)=0$, then taking $y=0$ in the given equation, we obtain $f(x)=f(x) f(0)^2=0$ for all $x$.\n\nSuppose $f(0)=1$. Taking $y=-x$, we obtain\n$$\n1=f(0)=f(x-x)=f(x) f(-x) f\\left(-x^2\\right)\n$$\nThis shows that $f(x) \\neq 0$ for any $x \\in \\mathbf{R}$.\n\nTaking $x=1, y=x-1$, we obtain\n$$\nf(x)=f(1) f(x-1)^2=f(1)[f(x) f(-x) f(-x)]^2\n$$\nUsing $f(x) \\neq 0$, we conclude that $1=k f(x)(f(-x))^2$, where $k=f(1)(f(-1))^2$.\nChanging $x$ to $-x$ here, we also infer that $1=k f(-x)(f(x))^2$.\nComparing these expressions we see that $f(-x)=f(x)$.\nIt follows that $1=k f(x)^3$.\nThus $f(x)$ is constant for all $x$.\nSince $f(0)=1$, we conclude that $f(x)=1$ for all real $x$.\n\nIf $f(0)=-1$, a similar analysis shows that $f(x)=-1$ for all $x \\in \\mathbf{R}$.\n\nWe can verify that each of these functions satisfies the given functional equation. Thus there are three solutions, all of them being constant functions.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32311,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. For a convex hexagon $A B C D E F$ in which each pair of opposite sides is unequal, consider the following six statements:\n$$\n\\begin{array}{ll}\n\\text{(a$_1$) } A B \\text{ is parallel to } D E ; & (a_2)\\ A E = B D \\\\\n(b_1)\\ B C \\text{ is parallel to } E F ; & (b_2)\\ B F = C E \\\\\n\\text{(c$_1$) } C D \\text{ is parallel to } F A ; & (c_2)\\ C A = D F\n\\end{array}\n$$\n(a) Show that if all the six statements are true, then the hexagon is cyclic (i.e., it can be inscribed in a circle).\n\n(b) Prove that, in fact, any five of these six statements also imply that the hexagon is cyclic.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\n(a) Suppose all the six statements are true. Then $A B D E$, $B C E F$, $C D F A$ are isosceles trapeziums; if $K, L, M, P, Q, R$ are the mid-points of $A B$, $B C$, $C D$, $D E$, $E F$, $F A$ respectively, then we see that $K P \\perp A B, E D$; $L Q \\perp B C, E F$ and $M R \\perp C D, F A$.\n\n![](attached_image_1.png)\n\nIf $A D$, $B E$, $C F$ themselves concur at a point $O$, then $O A = O B = O C = O D = O E = O F$. ($O$ is on the perpendicular bisector of each of the sides.) Hence $A, B, C, D, E, F$ are concyclic and lie on a circle with centre $O$.\n\nOtherwise these lines $A D$, $B E$, $C F$ form a triangle, say $X Y Z$. (See Fig.) Then $K X$, $M Y$, $Q Z$, when extended, become the internal angle bisectors of the triangle $X Y Z$ and hence concur at the incentre $O'$\nof $X Y Z$. As earlier $O'$ lies on the perpendicular bisector of each of the sides. Hence $O' A = O' B = O' C = O' D = O' E = O' F$, giving the concyclicity of $A, B, C, D, E, F$.\n\n(b) Suppose $(a_1)$, $(a_2)$, $(b_1)$, $(b_2)$ are true. Then we see that $A D = B E = C F$. Assume that $(c_1)$ is true. Then $C D$ is parallel to $A F$. It follows that triangles $Y C D$ and $Y F A$ are similar. This gives\n$$\n\\frac{F Y}{A Y} = \\frac{Y C}{Y D} = \\frac{F Y + Y C}{A Y + Y D} = \\frac{F C}{A D} = 1\n$$\nWe obtain $F Y = A Y$ and $Y C = Y D$. This forces that triangles $C Y A$ and $D Y F$ are congruent. In particular $A C = D F$ so that $(c_2)$ is true. The conclusion follows from (a).\n\nNow assume that $(c_2)$ is true; i.e., $A C = F D$. We have seen that $A D = B E = C F$. It follows that triangles $F D C$ and $A C D$ are congruent. In particular $\\angle A D C = \\angle F C D$. Similarly, we can show that $\\angle C F A = \\angle D A F$. We conclude that $C D$ is parallel to $A F$ giving $(c_1)$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32312,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDetermine the least positive value taken by the expression $a^{3}+b^{3}+c^{3}-3 a b c$ as $a, b, c$ vary over all positive integers. Find also all triples $(a, b, c)$ for which this least value is attained.",
    "options": [],
    "answer": "Least value is 4; attained exactly by the triples that are permutations of (1, 1, 2).",
    "solution": "Solution:\n\nWe observe that\n$$\nQ = a^{3} + b^{3} + c^{3} - 3 a b c = \\frac{1}{2}(a + b + c)\\left((a - b)^{2} + (b - c)^{2} + (c - a)^{2}\\right)\n$$\nSince we are looking for the least positive value taken by $Q$, it follows that $a, b, c$ are not all equal. Thus $a + b + c \\geq 1 + 1 + 2 = 4$ and $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} \\geq 1 + 1 + 0 = 2$. Thus we see that $Q \\geq 4$. Taking $a = 1$, $b = 1$ and $c = 2$, we get $Q = 4$. Therefore the least value of $Q$ is $4$ and this is achieved only by $a + b + c = 4$ and $(a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 2$. The triples for which $Q = 4$ are therefore given by\n$$\n(a, b, c) = (1, 1, 2),\\ (1, 2, 1),\\ (2, 1, 1)\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32313,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x, y$ be positive reals such that $x + y = 2$. Prove that\n$$\nx^{3} y^{3} (x^{3} + y^{3}) \\leq 2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe have from the AM-GM inequality, that\n$$\nx y \\leq \\left(\\frac{x + y}{2}\\right)^{2} = 1\n$$\nThus we obtain $0 < x y \\leq 1$. We write\n$$\n\\begin{aligned}\nx^{3} y^{3} (x^{3} + y^{3}) &= (x y)^{3} (x + y) (x^{2} - x y + y^{2}) \\\\\n&= 2 (x y)^{3} ((x + y)^{2} - 3 x y) \\\\\n&= 2 (x y)^{3} (4 - 3 x y)\n\\end{aligned}\n$$\nThus we need to prove that\n$$\n(x y)^{3} (4 - 3 x y) \\leq 1\n$$\nPutting $z = x y$, this inequality reduces to\n$$\nz^{3} (4 - 3 z) \\leq 1\n$$\nfor $0 < z \\leq 1$. We can prove this in different ways. We can put the inequality in the form\n$$\n3 z^{4} - 4 z^{3} + 1 \\geq 0\n$$\nHere the expression in the LHS factors to $(z - 1)^{2} (3 z^{2} + 2 z + 1)$ and $(3 z^{2} + 2 z + 1)$ is positive since its discriminant $D = -8 < 0$. Or applying the AM-GM inequality to the positive reals $4 - 3 z, z, z, z$, we obtain\n$$\nz^{3} (4 - 3 z) \\leq \\left(\\frac{4 - 3 z + 3 z}{4}\\right)^{4} \\leq 1\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32314,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDo there exist 100 lines in the plane, no three of them concurrent, such that they intersect exactly in 2002 points?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nAny set of 100 lines in the plane can be partitioned into a finite number of disjoint sets, say $A_{1}, A_{2}, A_{3}, \\ldots, A_{k}$, such that\n\n(i) Any two lines in each $A_{j}$ are parallel to each other, for $1 \\leq j \\leq k$ (provided, of course, $|A_{j}| \\geq 2$);\n\n(ii) for $j \\neq l$, the lines in $A_{j}$ and $A_{l}$ are not parallel.\n\nIf $|A_{j}| = m_{j}$, $1 \\leq j \\leq k$, then the total number of points of intersection is given by $\\sum_{1 \\leq j < l \\leq k} m_{j} m_{l}$, as no three lines are concurrent. Thus we have to find positive integers $m_{1}, m_{2}, \\ldots, m_{k}$ such that\n\n$$\n\\sum_{j=1}^{k} m_{j} = 100, \\quad \\sum_{1 \\leq j < l \\leq k} m_{j} m_{l} = 2002\n$$\n\nfor an affirmative answer to the given question.\n\nWe observe that\n\n$$\n\\begin{aligned}\n\\sum_{j=1}^{k} m_{j}^{2} &= \\left(\\sum_{j=1}^{k} m_{j}\\right)^{2} - 2\\left(\\sum_{1 \\leq j < l \\leq k} m_{j} m_{l}\\right) \\\\\n&= 100^{2} - 2(2002) = 5996\n\\end{aligned}\n$$\n\nThus we have to choose $m_{1}, m_{2}, \\ldots, m_{k}$ such that\n\n$$\n\\sum_{j=1}^{k} m_{j} = 100, \\quad \\sum_{j=1}^{k} m_{j}^{2} = 5996\n$$\n\nWe observe that $[\\sqrt{5996}] = 77$. So we may take $m_{1} = 77$, so that\n\n$$\n\\sum_{j=2}^{k} m_{j} = 23, \\quad \\sum_{j=2}^{k} m_{j}^{2} = 67\n$$\n\nNow we may choose $m_{2} = 5$, $m_{3} = m_{4} = 4$, $m_{5} = m_{6} = \\cdots = m_{14} = 1$. Finally, we can take\n\n$$\nk = 14, \\quad (m_{1}, m_{2}, \\ldots, m_{14}) = (77, 5, 4, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)\n$$\n\nproving the existence of 100 lines with exactly 2002 points of intersection.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32315,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDo there exist three distinct positive real numbers $a, b, c$ such that the numbers $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order?",
    "options": [],
    "answer": "No",
    "solution": "Solution:\n\nWe show that the answer is NO. Suppose, if possible, let $a, b, c$ be three distinct positive real numbers such that $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order. We may assume that $a < b < c$. Then there are only two cases we need to check:\n\n(I) $a+b-c < a < c+a-b < b < c < b+c-a < a+b+c$\n\nand\n\n(II) $a+b-c < a < b < c+a-b < c < b+c-a < a+b+c$.\n\nCase I. Suppose the chain of inequalities $a+b-c < a < c+a-b < b < c < b+c-a < a+b+c$ holds. Let $d$ be the common difference. Thus we see that\n$$\nc = a + b + c - 2d, \\quad b = a + b + c - 3d, \\quad a = a + b + c - 5d\n$$\nAdding these, we see that $a + b + c = 5d$. But then $a = 0$, contradicting the positivity of $a$.\n\nCase II. Suppose the inequalities $a+b-c < a < b < c+a-b < c < b+c-a < a+b+c$ are true. Again we see that\n$$\nc = a + b + c - 2d, \\quad b = a + b + c - 4d, \\quad a = a + b + c - 5d\n$$\nWe thus obtain $a + b + c = \\frac{11}{2} d$. This gives\n$$\na = \\frac{1}{2} d, \\quad b = \\frac{3}{2} d, \\quad c = \\frac{7}{2} d\n$$\nNote that $a + b - c = a + b + c - 6d = -\\frac{1}{2} d$. However, we also get $a + b - c = \\left(\\frac{1}{2} + \\frac{3}{2} - \\frac{7}{2}\\right) d = -\\frac{3}{2} d$. It follows that $3e = e$ giving $d = 0$. But this is impossible.\n\nThus there are no three distinct positive real numbers $a, b, c$ such that $a, b, c, b+c-a, c+a-b, a+b-c$ and $a+b+c$ form a 7-term arithmetic progression in some order.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32316,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nSuppose the $n^{2}$ numbers $1,2,3, \\ldots, n^{2}$ are arranged to form an $n$ by $n$ array consisting of $n$ rows and $n$ columns such that the numbers in each row (from left to right) and each column (from top to bottom) are in increasing order. Denote by $a_{j k}$ the number in $j$-th row and $k$-th column. Suppose $b_{j}$ is the maximum possible number of entries that can occur as $a_{j j}, 1 \\leq j \\leq n$. Prove that\n$$\nb_{1}+b_{2}+b_{3}+\\cdots b_{n} \\leq \\frac{n}{3}\\left(n^{2}-3 n+5\\right)\n$$\n(Example: In the case $n=3$, the only numbers which can occur as $a_{22}$ are 4, 5 or 6 so that $b_{2}=3$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $a_{j j}$ has to exceed all the numbers in the top left $j \\times j$ submatrix (excluding itself), and since there are $j^{2}-1$ entries, we must have $a_{j j} \\geq j^{2}$. Similarly, $a_{j j}$ must not exceed each of the numbers in the bottom right $(n-j+1) \\times (n-j+1)$ submatrix (other than itself) and there are $(n-j+1)^{2}-1$ such entries giving $a_{j j} \\leq n^{2}-(n-j+1)^{2}+1$. Thus we see that\n$$\na_{j j} \\in\\left\\{j^{2}, j^{2}+1, j^{2}+2, \\ldots, n^{2}-(n-j+1)^{2}+1\\right\\}\n$$\nThe number of elements in this set is $n^{2}-(n-j+1)^{2}-j^{2}+2$. This implies that\n$$\nb_{j} \\leq n^{2}-(n-j+1)^{2}-j^{2}+2=(2 n+2) j-2 j^{2}-(2 n-1)\n$$\nIt follows that\n$$\n\\begin{aligned}\n\\sum_{j=1}^{n} b_{j} & \\leq (2 n+2) \\sum_{j=1}^{n} j - 2 \\sum_{j=1}^{n} j^{2} - n(2 n-1) \\\\\n& = (2 n+2)\\left(\\frac{n(n+1)}{2}\\right) - 2\\left(\\frac{n(n+1)(2 n+1)}{6}\\right) - n(2 n-1) \\\\\n& = \\frac{n}{3}\\left(n^{2}-3 n+5\\right)\n\\end{aligned}\n$$\nwhich is the required bound.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32317,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nConsider an acute triangle $A B C$ and let $P$ be an interior point of $A B C$. Suppose the lines $B P$ and $C P$, when produced, meet $A C$ and $A B$ in $E$ and $F$ respectively. Let $D$ be the point where $A P$ intersects the line segment $E F$ and $K$ be the foot of perpendicular from $D$ on to $B C$. Show that $D K$ bisects $\\angle E K F$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nProduce $A P$ to meet $B C$ in $Q$. Join $K E$ and $K F$. Draw perpendiculars from $F$ and $E$ on to $B C$ to meet it in $M$ and $L$ respectively. Let us denote $\\angle B K F$ by $\\alpha$ and $\\angle C K E$ by $\\beta$. We show that $\\alpha=\\beta$ by proving $\\tan \\alpha=\\tan \\beta$. This implies that $\\angle D K F=\\angle D K E$. (See Figure below.)\n\n![](attached_image_1.png)\n\nSince the cevians $A Q$, $B E$ and $C F$ concur, we may write\n$$\n\\frac{B Q}{Q C}=\\frac{z}{y}, \\quad \\frac{C E}{E A}=\\frac{x}{z}, \\quad \\frac{A F}{F B}=\\frac{y}{x}\n$$\nWe observe that\n$$\n\\frac{F D}{D E}=\\frac{[A F D]}{[A E D]}=\\frac{[P F D]}{[P E D]}=\\frac{[A F P]}{[A E P]}\n$$\nHowever, standard computations involving bases give\n$$\n[A F P]=\\frac{y}{y+x}[A B P], \\quad [A E P]=\\frac{z}{z+x}[A C P]\n$$\nand\n$$\n[A B P]=\\frac{z}{x+y+z}[A B C], \\quad [A C P]=\\frac{y}{x+y+z}[A B C]\n$$\nThus we obtain\n$$\n\\frac{F D}{D E}=\\frac{x+z}{x+y}\n$$\nOn the other hand\n$$\n\\tan \\alpha=\\frac{F M}{K M}=\\frac{F B \\sin B}{K M}, \\quad \\tan \\beta=\\frac{E L}{K L}=\\frac{E C \\sin C}{K L}\n$$\nUsing $F B=\\left(\\frac{x}{x+y}\\right) A B$, $E C=\\left(\\frac{x}{x+z}\\right) A C$ and $A B \\sin B=A C \\sin C$, we obtain\n$$\n\\begin{aligned}\n\\frac{\\tan \\alpha}{\\tan \\beta} & =\\left(\\frac{x+z}{x+y}\\right)\\left(\\frac{K L}{K M}\\right) \\\\\n& =\\left(\\frac{x+z}{x+y}\\right)\\left(\\frac{D E}{F D}\\right) \\\\\n& =\\left(\\frac{x+z}{x+y}\\right)\\left(\\frac{x+y}{x+z}\\right)=1\n\\end{aligned}\n$$\nWe conclude that $\\alpha=\\beta$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32318,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all primes $p$ and $q$, and even numbers $n>2$, satisfying the equation\n$$\np^{n}+p^{n-1}+\\cdots+p+1=q^{2}+q+1\n$$",
    "options": [],
    "answer": "p=2, q=5, n=4",
    "solution": "Solution:\nObviously $p \\neq q$. We write this in the form\n$$\np\\left(p^{n-1}+p^{n-2}+\\cdots+1\\right)=q(q+1)\n$$\nIf $q \\leq p^{n / 2}-1$, then $q<p^{n / 2}$ and hence we see that $q^{2}<p^{n}$. Thus we obtain\n$$\nq^{2}+q<p^{n}+p^{n / 2}<p^{n}+p^{n-1}+\\cdots+p\n$$\nsince $n>2$. It follows that $q \\geq p^{n / 2}$. Since $n>2$ and is an even number, $n / 2$ is a natural number larger than 1. This implies that $q \\neq p^{n / 2}$ by the given condition that $q$ is a prime. We conclude that $q \\geq p^{n / 2}+1$. We may also write the above relation in the form\n$$\np\\left(p^{n / 2}-1\\right)\\left(p^{n / 2}+1\\right)=(p-1) q(q+1)\n$$\nThis shows that $q$ divides $\\left(p^{n / 2}-1\\right)\\left(p^{n / 2}+1\\right)$. But $q \\geq p^{n / 2}+1$ and $q$ is a prime. Hence the only possibility is $q=p^{n / 2}+1$. This gives\n$$\np\\left(p^{n / 2}-1\\right)=(p-1)(q+1)=(p-1)\\left(p^{n / 2}+2\\right)\n$$\nSimplification leads to $3 p=p^{n / 2}+2$. This shows that $p$ divides $2$. Thus $p=2$ and hence $q=5, n=4$. It is easy to verify that these indeed satisfy the given equation.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32319,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nShow that for every real number $a$ the equation\n$$\n8 x^{4}-16 x^{3}+16 x^{2}-8 x+a=0\n$$\nhas at least one non-real root and find the sum of all the non-real roots of the equation.",
    "options": [],
    "answer": "Sum of non-real roots = 1 if a ≤ 3/2; and = 2 if a > 3/2.",
    "solution": "Solution:\nSubstituting $x = y + \\frac{1}{2}$ in the equation, we obtain the equation in $y$:\n$$\n8 y^{4} + 4 y^{2} + a - \\frac{3}{2} = 0\n$$\nUsing the transformation $z = y^{2}$, we get a quadratic equation in $z$:\n$$\n8 z^{2} + 4 z + a - \\frac{3}{2} = 0\n$$\nThe discriminant of this equation is $32(2 - a)$ which is nonnegative if and only if $a \\leq 2$. For $a \\leq 2$, we obtain the roots\n$$\nz_{1} = \\frac{-1 + \\sqrt{2(2 - a)}}{4}, \\quad z_{2} = \\frac{-1 - \\sqrt{2(2 - a)}}{4}\n$$\nFor getting real $y$ we need $z \\geq 0$. Obviously $z_{2} < 0$ and hence it gives only non-real values of $y$. But $z_{1} \\geq 0$ if and only if $a \\leq \\frac{3}{2}$. In this case we obtain two real values for $y$ and hence two real roots for the original equation. Thus we conclude that there are two real roots and two non-real roots for $a \\leq \\frac{3}{2}$ and four non-real roots for $a > \\frac{3}{2}$. Obviously the sum of all the roots of the equation is $2$. For $a \\leq \\frac{3}{2}$, two real roots are given by $y_{1} = +\\sqrt{z_{1}}$ and $y_{2} = -\\sqrt{z_{1}}$. Hence the sum of real roots is $y_{1} + \\frac{1}{2} + y_{2} + \\frac{1}{2}$ which reduces to $1$. It follows the sum of the non-real roots is also $1$. Thus\n$$\n\\text{The sum of nonreal roots} = \\begin{cases} 1 & \\text{for } a \\leq \\frac{3}{2} \\\\ 2 & \\text{for } a > \\frac{3}{2} \\end{cases}\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32320,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFind all 7-digit numbers formed by using only the digits $5$ and $7$, and divisible by both $5$ and $7$.",
    "options": [],
    "answer": "7775775, 7757575, 5577775, 7575575, 5777555, 7755755, 5755575, 5557755, 7555555",
    "solution": "Solution:\n\nClearly, the last digit must be $5$ and we have to determine the remaining $6$ digits. For divisibility by $7$, it is sufficient to consider the number obtained by replacing $7$ by $0$; for example $5775755$ is divisible by $7$ if and only if $5005055$ is divisible by $7$. Each such number is obtained by adding some of the numbers from the set $\\{50, 500, 5000, 50000, 500000, 5000000\\}$ along with $5$. We look at the remainders of these when divided by $7$; they are $\\{1, 3, 2, 6, 4, 5\\}$. Thus it is sufficient to check for those combinations of remainders which add up to a number of the form $2 + 7k$, since the last digit is already $5$. These are $\\{2\\}, \\{3, 6\\}, \\{4, 5\\}, \\{2, 3, 4\\}, \\{1, 3, 5\\}, \\{1, 2, 6\\}, \\{2, 3, 5, 6\\}, \\{1, 4, 5, 6\\}$ and $\\{1, 2, 3, 4, 6\\}$. These correspond to the numbers $7775775, 7757575, 5577775, 7575575, 5777555, 7755755, 5755575, 5557755, 755555$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32321,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with sides $a, b, c$. Consider a triangle $A_1 B_1 C_1$ with sides equal to $a+\\frac{b}{2}$, $b+\\frac{c}{2}$, $c+\\frac{a}{2}$. Show that\n$$\n\\left[A_1 B_1 C_1\\right] \\geq \\frac{9}{4}[ABC]\n$$\nwhere $[XYZ]$ denotes the area of the triangle $XYZ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nIt is easy to observe that there is a triangle with sides $a+\\frac{b}{2}$, $b+\\frac{c}{2}$, $c+\\frac{a}{2}$. Using Heron's formula, we get\n$$\n16[ABC]^2 = (a+b+c)(a+b-c)(b+c-a)(c+a-b)\n$$\nand\n$$\n16\\left[A_1 B_1 C_1\\right]^2 = \\frac{3}{16}(a+b+c)(-a+b+3c)(-b+c+3a)(-c+a+3b)\n$$\nSince $a, b, c$ are the sides of a triangle, there are positive real numbers $p, q, r$ such that $a = q + r$, $b = r + p$, $c = p + q$. Using these relations we obtain\n$$\n\\frac{[ABC]^2}{\\left[A_1 B_1 C_1\\right]^2} = \\frac{16pqr}{3(2p+q)(2q+r)(2r+p)}\n$$\nThus it is sufficient to prove that\n$$\n(2p+q)(2q+r)(2r+p) \\geq 27pqr\n$$\nfor positive real numbers $p, q, r$. Using AM-GM inequality, we get\n$$\n2p+q \\geq 3(p^2 q)^{1/3}, \\quad 2q+r \\geq 3(q^2 r)^{1/3}, \\quad 2r+p \\geq 3(r^2 p)^{1/3}\n$$\nMultiplying these, we obtain the desired result. We also observe that equality holds if and only if $p = q = r$. This is equivalent to the statement that $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32322,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nConsider a convex quadrilateral $ABCD$, in which $K, L, M, N$ are the midpoints of the sides $AB$, $BC$, $CD$, $DA$ respectively. Suppose\n(a) $BD$ bisects $KM$ at $Q$;\n(b) $QA = QB = QC = QD$; and\n(c) $LK / LM = CD / CB$.\nProve that $ABCD$ is a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n![](attached_image_1.png)\nFig. 1.\nObserve that $KLMN$ is a parallelogram, $Q$ is the midpoint of $MK$ and hence $NL$ also passes through $Q$. Let $T$ be the point of intersection of $AC$ and $BD$; and let $S$ be the point of intersection of $BD$ and $MN$.\n\nConsider the triangle $MNK$. Note that $SQ$ is parallel to $NK$ and $Q$ is the midpoint of $MK$. Hence $S$ is the midpoint of $MN$. Since $MN$ is parallel to $AC$, it follows that $T$ is the midpoint of $AC$. Now $Q$ is the circumcentre of $\\triangle ABC$ and the median $BT$ passes through $Q$. Here there are two possibilities:\n(i) $ABC$ is a right triangle with $\\angle ABC = 90^\\circ$ and $T = Q$; and\n(ii) $T \\neq Q$ in which case $BT$ is perpendicular to $AC$.\n\nSuppose $\\angle ABC = 90^\\circ$ and $T = Q$. Observe that $Q$ is the circumcentre of the triangle $DCB$ and hence $\\angle DCB = 90^\\circ$. Similarly $\\angle DAB = 90^\\circ$. It follows that $\\angle ADC = 90^\\circ$ and $ABCD$ is a rectangle. This implies that $KLMN$ is a rhombus. Hence $LK / LM = 1$ and this gives $CD = CB$. Thus $ABCD$ is a square.\n\nIn the second case, observe that $BD$ is perpendicular to $AC$, $KL$ is parallel to $AC$ and $LM$ is parallel to $BD$. Hence it follows that $ML$ is perpendicular to $LK$. Similar reasoning shows that $KLMN$ is a rectangle.\n\nUsing $LK / LM = CD / CB$, we get that $CBD$ is similar to $LMK$. In particular, $\\angle LMK = \\angle CBD = \\alpha$ say. Since $LM$ is parallel to $DB$, we also get $\\angle BQK = \\alpha$. Since $KLMN$ is a cyclic quadrilateral we also get $\\angle LNK = \\angle LMK = \\alpha$. Using the fact that $BD$ is parallel to $NK$, we get $\\angle LQB = \\angle LNK = \\alpha$. Since $BD$ bisects $\\angle CBA$, we also have $\\angle KBQ = \\alpha$. Thus\n$$\nQK = KB = BL = LQ\n$$\nand $BL$ is parallel to $QK$. This gives $QM$ is parallel to $LC$ and\n$$\nQM = QL = BL = LC\n$$\nIt follows that $QLCM$ is a parallelogram. But $\\angle LCM = 90^\\circ$. Hence $\\angle MQL = 90^\\circ$. This implies that $KLMN$ is a square. Also observe that $\\angle LQK = 90^\\circ$ and hence $\\angle CBA = \\angle LQK = 90^\\circ$. This gives $\\angle CDA = 90^\\circ$ and hence $ABCD$ is a rectangle. Since $BA = BC$, it follows that $ABCD$ is a square.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32323,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose $p$ is a prime greater than $3$. Find all pairs of integers $(a, b)$ satisfying the equation\n$$\na^{2} + 3 a b + 2 p(a + b) + p^{2} = 0\n$$",
    "options": [],
    "answer": "Let p > 3 be prime.\n- If p = 3k + 1, with k = (p − 1)/3, the solutions are (a, b) = (−2k − 1, k^2), (−p, 0), (−(k + 1)p, k^2).\n- If p = 3k − 1, with k = (p + 1)/3, the solutions are (a, b) = (1 − 2k, −k^2), (−p, 0), ((k − 1)p, −k^2).",
    "solution": "Solution:\nWe write the equation in the form\n$$\na^{2} + 2 a p + p^{2} + b(3 a + 2 p) = 0\n$$\nHence\n$$\nb = \\frac{-(a + p)^{2}}{3 a + 2 p}\n$$\nis an integer. This shows that $3 a + 2 p$ divides $(a + p)^{2}$ and hence also divides $(3 a + 3 p)^{2}$. But, we have\n$$\n(3 a + 3 p)^{2} = (3 a + 2 p + p)^{2} = (3 a + 2 p)^{2} + 2 p(3 a + 2 p) + p^{2}\n$$\nIt follows that $3 a + 2 p$ divides $p^{2}$. Since $p$ is a prime, the only divisors of $p^{2}$ are $\\pm 1, \\pm p$ and $\\pm p^{2}$. Since $p > 3$, we also have $p = 3 k + 1$ or $3 k + 2$.\n\nCase 1: Suppose $p = 3 k + 1$. Obviously $3 a + 2 p = 1$ is not possible. In fact, we get $1 = 3 a + 2 p = 3 a + 2(3 k + 1) \\Rightarrow 3 a + 6 k = -1$ which is impossible. On the other hand $3 a + 2 p = -1$ gives $3 a = -2 p - 1 = -6 k - 3 \\Rightarrow a = -2 k - 1$ and $a + p = -2 k - 1 + 3 k + 1 = k$.\nThus $b = \\frac{-(a + p)^{2}}{3 a + 2 p} = k^{2}$. Thus $(a, b) = \\left(-2 k - 1, k^{2}\\right)$ when $p = 3 k + 1$. Similarly, $3 a + 2 p = p \\Rightarrow 3 a = -p$ which is not possible. Considering $3 a + 2 p = -p$, we get $3 a = -3 p$ or $a = -p \\Rightarrow b = 0$. Hence $(a, b) = (-3 k - 1, 0)$ where $p = 3 k + 1$.\nLet us consider $3 a + 2 p = p^{2}$. Hence $3 a = p^{2} - 2 p = p(p - 2)$ and neither $p$ nor $p - 2$ is divisible by $3$. If $3 a + 2 p = -p^{2}$, then $3 a = -p(p + 2) \\Rightarrow a = -(3 k + 1)(k + 1)$.\nHence $a + p = (3 k + 1)(-k - 1 + 1) = -(3 k + 1) k$. This gives $b = k^{2}$. Again $(a, b) = (-(k + 1)(3 k + 1), k^{2})$ when $p = 3 k + 1$.\n\nCase 2: Suppose $p = 3 k - 1$. If $3 a + 2 p = 1$, then $3 a = -6 k + 3$ or $a = -2 k + 1$. We also get\n$$\nb = \\frac{-(a + p)^{2}}{1} = \\frac{-(-2 k + 1 + 3 k - 1)^{2}}{1} = -k^{2}\n$$\nand we get the solution $(a, b) = \\left(-2 k + 1, k^{2}\\right)$. On the other hand $3 a + 2 p = -1$ does not have any integral solution for $a$. Similarly, there is no solution in the case $3 a + 2 p = p$. Taking $3 a + 2 p = -p$, we get $a = -p$ and hence $b = 0$. We get the solution $(a, b) = (-3 k + 1, 0)$. If $3 a + 2 p = p^{2}$, then $3 a = p(p - 2) = (3 k - 1)(3 k - 3)$ giving $a = (3 k - 1)(k - 1)$ and hence $a + p = (3 k - 1)(1 + k - 1) = k(3 k - 1)$. This gives $b = -k^{2}$ and hence $(a, b) = \\left(3 k - 1, -k^{2}\\right)$. Finally $3 a + 2 p = -p^{2}$ does not have any solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32324,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIf $\\alpha$ is a real root of the equation $x^{5}-x^{3}+x-2=0$, prove that $\\left[\\alpha^{6}\\right]=3$. (For any real number $a$, we denote by $[a]$ the greatest integer not exceeding $a$.)",
    "options": [],
    "answer": "3",
    "solution": "Solution:\n\nSuppose $\\alpha$ is a real root of the given equation. Then\n$$\n\\alpha^{5}-\\alpha^{3}+\\alpha-2=0\n$$\nThis gives $\\alpha^{5}-\\alpha^{3}+\\alpha-1=1$ and hence $(\\alpha-1)\\left(\\alpha^{4}+\\alpha^{3}+1\\right)=1$.\n\nObserve that $\\alpha^{4}+\\alpha^{3}+1 \\geq 2 \\alpha^{2}+\\alpha^{3}=\\alpha^{2}(\\alpha+2)$. If $-1 \\leq \\alpha<0$, then $\\alpha+2>0$, giving $\\alpha^{2}(\\alpha+2)>0$ and hence $(\\alpha-1)\\left(\\alpha^{4}+\\alpha^{3}+1\\right)<0$. If $\\alpha<-1$, then $\\alpha^{4}+\\alpha^{3}=\\alpha^{3}(\\alpha+1)>0$ and hence $\\alpha^{4}+\\alpha^{3}+1>0$. This again gives $(\\alpha-1)\\left(\\alpha^{4}+\\alpha^{3}+1\\right)<0$.\n\nThe above reasoning shows that for $\\alpha<0$, we have $\\alpha^{5}-\\alpha^{3}+\\alpha-1<0$ and hence cannot be equal to $1$. We conclude that a real root $\\alpha$ of $x^{5}-x^{3}+x-2=0$ is positive (obviously $\\alpha \\neq 0$).\n\nNow using $\\alpha^{5}-\\alpha^{3}+\\alpha-2=0$, we get\n$$\n\\alpha^{6}=\\alpha^{4}-\\alpha^{2}+2 \\alpha\n$$\nThe statement $\\left[\\alpha^{6}\\right]=3$ is equivalent to $3 \\leq \\alpha^{6}<4$.\n\nConsider $\\alpha^{4}-\\alpha^{2}+2 \\alpha<4$. Since $\\alpha>0$, this is equivalent to $\\alpha^{5}-\\alpha^{3}+2 \\alpha^{2}<4 \\alpha$. Using the relation (1), we can write $2 \\alpha^{2}-\\alpha+2<4 \\alpha$ or $2 \\alpha^{2}-5 \\alpha+2<0$. Treating this as a quadratic, we get this is equivalent to $\\frac{1}{2}<\\alpha<2$.\n\nNow observe that if $\\alpha \\geq 2$ then $1=(\\alpha-1)\\left(\\alpha^{4}+\\alpha^{3}+1\\right) \\geq 25$ which is impossible. If $0<\\alpha \\leq \\frac{1}{2}$, then $1=(\\alpha-1)\\left(\\alpha^{4}+\\alpha^{3}+1\\right)<0$ which again is impossible. We conclude that $\\frac{1}{2}<\\alpha<2$.\n\nSimilarly $\\alpha^{4}-\\alpha^{2}+2 \\alpha \\geq 3$ is equivalent to $\\alpha^{5}-\\alpha^{3}+2 \\alpha^{2}-3 \\alpha \\geq 0$ which is equivalent to $2 \\alpha^{2}-4 \\alpha+2 \\geq 0$. But this is $2(\\alpha-1)^{2} \\geq 0$ which is valid. Hence $3 \\leq \\alpha^{6}<4$ and we get $\\left[\\alpha^{6}\\right]=3$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32325,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $R$ denote the circumradius of a triangle $ABC$; $a, b, c$ its sides $BC, CA, AB$; and $r_{a}, r_{b}, r_{c}$ its exradii opposite $A, B, C$. If $2R \\leq r_{a}$, prove that\n(i) $a > b$ and $a > c$;\n(ii) $2R > r_{b}$ and $2R > r_{c}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe know that $2R = \\frac{abc}{2\\triangle}$ and $r_{a} = \\frac{\\triangle}{s-a}$, where $a, b, c$ are the sides of the triangle $ABC$, $s = \\frac{a+b+c}{2}$ and $\\triangle$ is the area of $ABC$. Thus the given condition $2R \\leq r_{a}$ translates to\n$$\nabc \\leq \\frac{2\\triangle^{2}}{s-a}\n$$\nPutting $s-a = p$, $s-b = q$, $s-c = r$, we get $a = q + r$, $b = r + p$, $c = p + q$ and the condition now is\n$$\np(p+q)(q+r)(r+p) \\leq 2\\triangle^{2}\n$$\nBut Heron's formula gives, $\\triangle^{2} = s(s-a)(s-b)(s-c) = pqr(p+q+r)$. We obtain $(p+q)(q+r)(r+p) \\leq 2qr(p+q+r)$. Expanding and effecting some cancellations, we get\n$$\np^{2}(q+r) + p(q^{2} + r^{2}) \\leq qr(q+r)\n$$\nSuppose $a \\leq b$. This implies that $q + r \\leq r + p$ and hence $q \\leq p$. This implies that $q^{2} r \\leq p^{2} r$ and $q r^{2} \\leq p r^{2}$ giving $qr(q+r) \\leq p^{2} r + p r^{2} < p^{2} r + p r^{2} + p^{2} q + p q^{2} = p^{2}(q+r) + p(q^{2} + r^{2})$ which contradicts $(\\star)$. Similarly, $a \\leq c$ is also not possible. This proves (i).\n\nSuppose $2R \\leq r_{b}$. As above this takes the form\n$$\nq^{2}(r+p) + q(r^{2} + p^{2}) \\leq pr(p+r)\n$$\nSince $a > b$ and $a > c$, we have $q > p$, $r > p$. Thus $q^{2} r > p^{2} r$ and $q r^{2} > p r^{2}$. Hence\n$$\nq^{2}(r+p) + q(r^{2} + p^{2}) > q^{2} r + q r^{2} > p^{2} r + p r^{2} = pr(p+r)\n$$\nwhich contradicts $(\\star\\star)$. Hence $2R > r_{b}$. Similarly, we can prove that $2R > r_{c}$. This proves (ii).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32326,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $S$ denote the set of all 6-tuples $(a, b, c, d, e, f)$ of positive integers such that $a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=f^{2}$. Consider the set\n$$\nT=\\{a b c d e f:(a, b, c, d, e, f) \\in S\\}\n$$\nFind the greatest common divisor of all the members of $T$.",
    "options": [],
    "answer": "24",
    "solution": "Solution:\nWe show that the required gcd is 24. Consider an element $(a, b, c, d, e, f) \\in S$. We have\n$$\na^{2}+b^{2}+c^{2}+d^{2}+e^{2}=f^{2}\n$$\nWe first observe that not all $a, b, c, d, e$ can be odd. Otherwise, we have $a^{2} \\equiv b^{2} \\equiv c^{2} \\equiv d^{2} \\equiv e^{2} \\equiv 1$ $(\\bmod 8)$ and hence $f^{2} \\equiv 5(\\bmod 8)$, which is impossible because no square can be congruent to 5 modulo 8. Thus at least one of $a, b, c, d, e$ is even.\n\nSimilarly if none of $a, b, c, d, e$ is divisible by 3, then $a^{2} \\equiv b^{2} \\equiv c^{2} \\equiv d^{2} \\equiv e^{2} \\equiv 1(\\bmod 3)$ and hence $f^{2} \\equiv 2(\\bmod 3)$ which again is impossible because no square is congruent to 2 modulo 3. Thus 3 divides $a b c d e f$.\n\nThere are several possibilities for $a, b, c, d, e$.\n\nCase 1: Suppose one of them is even and the other four are odd; say $a$ is even, $b, c, d, e$ are odd. Then $b^{2}+c^{2}+d^{2}+e^{2} \\equiv 4(\\bmod 8)$. If $a^{2} \\equiv 4(\\bmod 8)$, then $f^{2} \\equiv 0(\\bmod 8)$ and hence $2|a, 4| f$ giving $8 \\mid a f$. If $a^{2} \\equiv 0(\\bmod 8)$, then $f^{2} \\equiv 4(\\bmod 8)$ which again gives that $4 \\mid a$ and $2 \\mid f$ so that $8 \\mid a f$. It follows that $8 \\mid a b c d e f$ and hence $24 \\mid a b c d e f$.\n\nCase 2: Suppose $a, b$ are even and $c, d, e$ are odd. Then $c^{2}+d^{2}+e^{2} \\equiv 3(\\bmod 8)$. Since $a^{2}+b^{2} \\equiv 0$ or $4$ modulo 8, it follows that $f^{2} \\equiv 3$ or $7(\\bmod 8)$ which is impossible. Hence this case does not arise.\n\nCase 3: If three of $a, b, c, d, e$ are even and two odd, then $8 \\mid a b c d e f$ and hence $24|a b c d e f$.\n\nCase 4: If four of $a, b, c, d, e$ are even, then again $8 \\mid a b c d e f$ and $24|a b c d e f$.\n\nHere again for any six tuple $(a, b, c, d, e, f)$ in $S$, we observe that $24 \\mid a b c d e f$. Since\n$$\n1^{2}+1^{2}+1^{2}+2^{2}+3^{2}=4^{2}\n$$\nWe see that $(1,1,1,2,3,4) \\in S$ and hence $24 \\in T$. Thus 24 is the gcd of $T$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32327,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that the number of 5-tuples of positive integers $(a, b, c, d, e)$ satisfying the equation\n$$\na b c d e = 5(b c d e + a c d e + a b d e + a b c e + a b c d)\n$$\nis an odd integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe write the equation in the form:\n$$\n\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} + \\frac{1}{d} + \\frac{1}{e} = \\frac{1}{5}\n$$\nThe number of five tuples $(a, b, c, d, e)$ which satisfy the given relation and for which $a \\neq b$ is even, because if $(a, b, c, d, e)$ is a solution, then so is $(b, a, c, d, e)$ which is distinct from $(a, b, c, d, e)$. Similarly, the number of five tuples which satisfy the equation and for which $c \\neq d$ is also even. Hence it suffices to count only those five tuples $(a, b, c, d, e)$ for which $a = b$, $c = d$. Thus the equation reduces to\n$$\n\\frac{2}{a} + \\frac{2}{c} + \\frac{1}{e} = \\frac{1}{5}\n$$\nHere again, the tuple $(a, a, c, c, e)$ for which $a \\neq c$ is even because we can associate a different solution $(c, c, a, a, e)$ to this five tuple. Thus it suffices to consider the equation\n$$\n\\frac{4}{a} + \\frac{1}{e} = \\frac{1}{5}\n$$\nand show that the number of pairs $(a, e)$ satisfying this equation is odd.\nThis reduces to\n$$\na e = 20 e + 5 a\n$$\nor\n$$\n(a - 20)(e - 5) = 100\n$$\nBut observe that\n$$\n\\begin{aligned}\n& 100 = 1 \\times 100 = 2 \\times 50 = 4 \\times 25 = 5 \\times 20 \\\\\n& \\quad = 10 \\times 10 = 20 \\times 5 = 25 \\times 4 = 50 \\times 2 = 100 \\times 1\n\\end{aligned}\n$$\nNote that no factorisation of $100$ as product of two negative numbers yields a positive tuple $(a, e)$. Hence we get these $9$ solutions. This proves that the total number of five tuples $(a, b, c, d, e)$ satisfying the given equation is odd.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32328,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $M$ be the midpoint of side $BC$ of a triangle $ABC$. Let the median $AM$ intersect the incircle of $ABC$ at $K$ and $L$, $K$ being nearer to $A$ than $L$. If $AK = KL = LM$, prove that the sides of triangle $ABC$ are in the ratio $5 : 10 : 13$ in some order.",
    "options": [],
    "answer": "5:10:13",
    "solution": "Solution:\n\nLet $I$ be the incentre of triangle $ABC$ and $D$ be its projection on $BC$. Observe that $AB \\neq AC$ as $AB = AC$ implies that $D = L = M$. So assume that $AC > AB$. Let $N$ be the projection of $I$ on $KL$. Then the perpendicular $IN$ from $I$ to $KL$ is a bisector of $KL$ and as $AK = LM$, it is a bisector of $AM$ also. Hence $AI = IM$.\n\n![](attached_image_1.png)\n\nFig. 1.\n\nBut $AI = \\frac{r}{\\sin(A/2)} = r \\operatorname{cosec}(A/2)$ and\n$$\n\\begin{aligned}\nIM^2 & = ID^2 + DM^2 = r^2 + (BM - BD)^2 \\\\\n& = r^2 + \\left(\\frac{a}{2} - (s-b)\\right)^2\n\\end{aligned}\n$$\nHence $r^2 \\operatorname{cosec}^2(A/2) = r^2 + \\left((a/2) - (s-b)\\right)^2$ giving $r^2 \\cot^2(A/2) = ((b-c)/2)^2$. Since $b > c$, we obtain $r \\cot(A/2) = ((b-c)/2)$. So $s-a = ((b-c)/2)$. This gives $a = 2c$.\n\nAs $KN = NL$ and $AK = KL = LM$, we have $NL = AM/6$. We also have $AN = NM$. Now\n$$\n\\begin{aligned}\nr^2 = IL^2 = IN^2 + NL^2 & = AI^2 - AN^2 + NL^2 \\\\\n& = AI^2 - \\frac{1}{4} m_a^2 + \\frac{1}{36} m_a^2 \\\\\n& = r^2 \\operatorname{cosec}^2(A/2) - \\frac{2}{9} m_a^2\n\\end{aligned}\n$$\nHence $r^2 \\cot^2(A/2) = \\frac{2}{9} m_a^2$. From the above, we get\n$$\n\\left(\\frac{b-c}{2}\\right)^2 = \\frac{2}{9} \\cdot \\frac{1}{4} (2b^2 + 2c^2 - a^2)\n$$\nSimplification gives $5b^2 + 13c^2 - 18bc = 0$. This can be written as $(b-c)(5b-13c) = 0$. As $b \\neq c$, we get $5b - 13c = 0$. To conclude, $a = 2c$, $5b = 13c$ yield\n$$\n\\frac{a}{10} = \\frac{b}{13} = \\frac{c}{5}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32329,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\alpha$ and $\\beta$ be positive integers such that\n$$\n\\frac{43}{197}<\\frac{\\alpha}{\\beta}<\\frac{17}{77}\n$$\n\nFind the minimum possible value of $\\beta$.",
    "options": [],
    "answer": "32",
    "solution": "Solution:\nWe have\n$$\n\\frac{77}{17}<\\frac{\\beta}{\\alpha}<\\frac{197}{43}\n$$\nThat is,\n$$\n4+\\frac{9}{17}<\\frac{\\beta}{\\alpha}<4+\\frac{25}{43}\n$$\nThus $4<\\frac{\\beta}{\\alpha}<5$. Since $\\alpha$ and $\\beta$ are positive integers, we may write $\\beta=4 \\alpha+x$, where $0<x<\\alpha$. Now we get\n$$\n4+\\frac{9}{17}<4+\\frac{x}{\\alpha}<4+\\frac{25}{43}\n$$\nSo $\\frac{9}{17}<\\frac{x}{\\alpha}<\\frac{25}{43}$; that is, $\\frac{43 x}{25}<\\alpha<\\frac{17 x}{9}$.\nWe find the smallest value of $x$ for which $\\alpha$ becomes a well-defined integer. For $x=1,2,3$ the bounds of $\\alpha$ are respectively $\\left(1 \\frac{18}{25}, 1 \\frac{8}{9}\\right),\\left(3 \\frac{11}{25}, 3 \\frac{7}{9}\\right),\\left(5 \\frac{4}{9}, 5 \\frac{2}{3}\\right)$. None of these pairs contain an integer between them.\nFor $x=4$, we have $\\frac{43 x}{25}=6 \\frac{12}{25}$ and $\\frac{17 x}{9}=7 \\frac{5}{9}$. Hence, in this case $\\alpha=7$, and $\\beta=4 \\alpha+x=28+4=32$.\nThis is also the least possible value, because, if $x \\geq 5$, then $\\alpha>\\frac{43 x}{25} \\geq \\frac{43}{5}>8$, and so $\\beta>37$.\nHence the minimum possible value of $\\beta$ is 32.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32330,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $p, q, r$ be positive real numbers, not all equal, such that some two of the equations\n$$\np x^{2}+2 q x+r=0, \\quad q x^{2}+2 r x+p=0, \\quad r x^{2}+2 p x+q=0\n$$\nhave a common root, say $\\alpha$. Prove that\n(a) $\\alpha$ is real and negative; and\n(b) the third equation has non-real roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nConsider the discriminants of the three equations\n$$\n\\begin{array}{r}\np x^{2}+2 q x+r=0 \\\\\nq x^{2}+2 r x+p=0 \\\\\nr x^{2}+2 p x+q=0\n\\end{array}\n$$\nLet us denote them by $D_{1}, D_{2}, D_{3}$ respectively. Then we have\n$$\nD_{1}=4\\left(q^{2}-r p\\right), \\quad D_{2}=4\\left(r^{2}-p q\\right), \\quad D_{3}=4\\left(p^{2}-q r\\right)\n$$\nWe observe that\n$$\n\\begin{aligned}\nD_{1}+D_{2}+D_{3} & =4\\left(p^{2}+q^{2}+r^{2}-p q-q r-r p\\right) \\\\\n& =2\\left\\{(p-q)^{2}+(q-r)^{2}+(r-p)^{2}\\right\\}>0\n\\end{aligned}\n$$\nsince $p, q, r$ are not all equal. Hence at least one of $D_{1}, D_{2}, D_{3}$ must be positive. We may assume $D_{1}>0$.\nSuppose $D_{2}<0$ and $D_{3}<0$. In this case both the equations (2) and (3) have only non-real roots and equation (1) has only real roots. Hence the common root $\\alpha$ must be between (2) and (3). But then $\\bar{\\alpha}$ is the other root of both (2) and (3). Hence it follows that (2) and (3) have same set of roots. This implies that\n$$\n\\frac{q}{r}=\\frac{r}{p}=\\frac{p}{q}\n$$\nThus $p=q=r$ contradicting the given condition. Hence both $D_{2}$ and $D_{3}$ cannot be negative. We may assume $D_{2} \\geq 0$. Thus we have\n$$\nq^{2}-r p>0, \\quad r^{2}-p q \\geq 0\n$$\nThese two give\n$$\nq^{2} r^{2}>p^{2} q r\n$$\nsince $p, q, r$ are all positive. Hence we obtain $q r>p^{2}$ or $D_{3}<0$. We conclude that the common root must be between equations (1) and (2).\nThus\n$$\n\\begin{aligned}\n& p \\alpha^{2}+2 q \\alpha+r=0 \\\\\n& q \\alpha^{2}+2 r \\alpha+p=0\n\\end{aligned}\n$$\nEliminating $\\alpha^{2}$, we obtain\n$$\n2\\left(q^{2}-p r\\right) \\alpha=p^{2}-q r\n$$\nSince $q^{2}-p r>0$ and $p^{2}-q r<0$, we conclude that $\\alpha<0$.\nThe condition $p^{2}-q r<0$ implies that the equation (3) has only non-real roots.\n\nAlternately one can argue as follows. Suppose $\\alpha$ is a common root of two equations, say, (1) and (2). If $\\alpha$ is non-real, then $\\bar{\\alpha}$ is also a root of both (1) and (2). Hence the coefficients of (1) and (2) are proportional. This forces $p=q=r$, a contradiction. Hence the common root between any two equations cannot be non-real. Looking at the coefficients, we conclude that the common root $\\alpha$ must be negative. If (1) and (2) have common root $\\alpha$, then $q^{2} \\geq r p$ and $r^{2} \\geq p q$. Here at least one inequality is strict for $q^{2}=p r$ and $r^{2}=p q$ forces $p=q=r$. Hence $q^{2} r^{2}>p^{2} q r$. This gives $p^{2}<q r$ and hence (3) has non-real roots.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32331,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAll possible 6-digit numbers, in each of which the digits occur in non-increasing order (from left to right, e.g., $877550$) are written as a sequence in increasing order. Find the $2005$-th number in this sequence.",
    "options": [],
    "answer": "864110",
    "solution": "Solution:\n\nConsider a 6-digit number whose digits from left to right are in non-increasing order. If $1$ is the first digit of such a number, then the subsequent digits cannot exceed $1$. The set of all such numbers with initial digit equal to $1$ is\n$$\n\\{100000, 110000, 111000, 111100, 111110, 111111\\}.\n$$\nThere are $6$ elements in this set.\n\nLet us consider 6-digit numbers with initial digit $2$. Starting from $200000$, we can go up to $222222$. We count these numbers as follows:\n\n| $200000$ | - | $211111$ | $:$ | $6$ |\n| :--- | :--- | :--- | :--- | :--- |\n| $220000$ | - | $221111$ | $:$ | $5$ |\n| $222000$ | - | $222111$ | $:$ | $4$ |\n| $222200$ | - | $222211$ | $:$ | $3$ |\n| $222220$ | - | $222221$ | $:$ | $2$ |\n| $222222$ | - | $222222$ | $:$ | $1$ |\n\nThe number of such numbers is $21$.\n\nSimilarly we count numbers with initial digit $3$; the sequence starts from $300000$ and ends with $333333$. We have\n\n| $300000$ | - | $322222$ | $:$ | $21$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $330000$ | - | $332222$ | $:$ | $15$ |\n| $333000$ | - | $333222$ | $:$ | $10$ |\n| $333300$ | - | $333322$ | $:$ | $6$ |\n| $333330$ | - | $333332$ | $:$ | $3$ |\n| $333333$ | - | $333333$ | $:$ | $1$ |\n\nWe obtain the total number of numbers starting from $3$ equal to $56$.\n\nSimilarly,\n\n| $400000$ | $433333$ | $56$ |\n| :---: | :---: | :---: |\n| $440000$ | $443333$ | $35$ |\n| $444000$ | $444333$ | $20$ |\n| $444400$ | $444433$ | $10$ |\n| $444440$ | $444443$ | $4$ |\n| $444444$ | $444444$ | $1$ |\n|  |  | $126$ |\n\n| $500000$ | $544444$ | $126$ |\n| $550000$ | $554444$ | $70$ |\n| $555000$ | $555444$ | $35$ |\n| $555500$ | $555544$ | $15$ |\n| $555550$ | $555554$ | $5$ |\n| $555555$ | $555555$ | $1$ |\n|  |  | $252$ |\n\n| $600000$ | $655555$ | $252$ |\n| $660000$ | $665555$ | $126$ |\n| $666000$ | $666555$ | $56$ |\n| $666600$ | $666655$ | $21$ |\n| $666660$ | $666665$ | $6$ |\n| $666666$ | $666666$ | $1$ |\n|  |  | $462$ |\n\n| $700000$ | $766666$ | $462$ |\n| $770000$ | $776666$ | $210$ |\n| $777000$ | $777666$ | $84$ |\n| $777700$ | $777766$ | $28$ |\n| $777770$ | $777776$ | $7$ |\n| $777777$ | $777777$ | $1$ |\n|  |  | $792$ |\n\nThus the number of 6-digit numbers where digits are non-increasing starting from $100000$ and ending with $777777$ is\n$$\n792 + 462 + 252 + 126 + 56 + 21 + 6 = 1715\n$$\nSince $2005 - 1715 = 290$, we have to consider only $290$ numbers in the sequence with initial digit $8$. We have\n$$\n\\begin{aligned}\n& 800000 - 855555 \\quad: \\quad 252 \\\\\n& 860000 - 863333: 35 \\\\\n& 864000 - 864110 \\quad: \\quad 3\n\\end{aligned}\n$$\nThus the required number is $\\underline{864110}$.\n\n\nIt is known that the number of ways of choosing $r$ objects from $n$ different types of objects (with repetitions allowed) is $\\binom{n+r-1}{r}$. In particular, if we want to write $r$-digit numbers using $n$ digits allowing for repetitions with the additional condition that the digits appear in non-increasing order, we see that this can be done in $\\binom{n+r-1}{r}$ ways.\n\nNow we group the given numbers into different classes and write the number of ways in which each class can be obtained. To keep track we also write the cumulative sums of the number of numbers so obtained. Observe that the numbers themselves are written in ascending order. So we exhaust numbers beginning with $1$, then beginning with $2$ and so on.\n\n| Numbers | Digits used other than the fixed part | $n$ | $r$ | $\\overline{\\binom{n+r-1}{r}}$ | Cumulative sum |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| beginning with $1$ | $1,0$ | $2$ | $5$ | $\\binom{6}{5} = 6$ | $6$ |\n| $2$ | $2,1,0$ | $3$ | $5$ | $\\binom{7}{5} = 21$ | $27$ |\n| $3$ | $3,2,1,0$ | $4$ | $5$ | $\\binom{8}{5} = 56$ | $83$ |\n| $4$ | $4,3,2,1,0$ | $5$ | $5$ | $\\binom{9}{5} = 126$ | $209$ |\n| $5$ | $5,4,3,2,1,0$ | $6$ | $5$ | $\\binom{10}{5} = 252$ | $461$ |\n| $6$ | $6,5,4,3,2,1,0$ | $7$ | $5$ | $\\binom{11}{5} = 462$ | $923$ |\n| $7$ | $7,6,5,4,3,2,1,0$ | $8$ | $5$ | $\\binom{12}{5} = 792$ | $1715$ |\n| from $800000$ to $855555$ | $5,4,3,2,1,0$ | $6$ | $5$ | $\\binom{10}{5} = 252$ | $1967$ |\n| from $860000$ to $863333$ | $3,2,1,0$ | $4$ | $4$ | $\\binom{7}{4} = 35$ | $2002$ |\n\nThe next three 6-digit numbers are $864000, 864100, 864110$.\n\nHence the $2005$-th number in the sequence is $864110$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32332,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $x_{1}$ be a given positive integer. A sequence $\\langle x_{n}\\rangle_{n=1}^{\\infty} = \\langle x_{1}, x_{2}, x_{3}, \\cdots \\rangle$ of positive integers is such that $x_{n}$, for $n \\geq 2$, is obtained from $x_{n-1}$ by adding some nonzero digit of $x_{n-1}$. Prove that\n(a) the sequence has an even number;\n(b) the sequence has infinitely many even numbers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n(a) Let us assume that there are no even numbers in the sequence. This means that $x_{n+1}$ is obtained from $x_{n}$, by adding a nonzero even digit of $x_{n}$ to $x_{n}$, for each $n \\geq 1$.\nLet $E$ be the left most even digit in $x_{1}$ which may be taken in the form\n$$\nx_{1} = O_{1} O_{2} \\cdots O_{k} E D_{1} D_{2} \\cdots D_{l}\n$$\nwhere $O_{1}, O_{2}, \\ldots, O_{k}$ are odd digits $(k \\geq 0)$; $D_{1}, D_{2}, \\ldots, D_{l-1}$ are even or odd; and $D_{l}$ odd, $l \\geq 1$.\nSince each time we are adding at least 2 to a term of the sequence to get the next term, at some stage, we will have a term of the form\n$$\nx_{r} = O_{1} O_{2} \\cdots O_{k} E 999 \\cdots 9 F\n$$\nwhere $F = 3, 5, 7$ or $9$. Now we are forced to add $E$ to $x_{r}$ to get $x_{r+1}$, as it is the only even digit available. After at most four steps of addition, we see that some next term is of the form\n$$\nx_{s} = O_{1} O_{2} \\cdots O_{k} G 000 \\cdots M\n$$\nwhere $G$ replaces $E$ of $x_{r}$, $G = E+1$, $M = 1, 3, 5$, or $7$. But $x_{s}$ has no nonzero even digit contradicting our assumption. Hence the sequence has some even number as its term.\n\n(b) If there are only finitely many even terms and $x_{t}$ is the last term, then the sequence $\\langle x_{n}\\rangle_{n=t+1}^{\\infty} = \\langle x_{t+1}, x_{t+2}, \\ldots \\rangle$ is obtained in a similar manner and hence must have an even term by (a), a contradiction. Thus $\\langle x_{n}\\rangle_{n=1}^{\\infty}$ has infinitely many even terms.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32333,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all functions $f: \\mathbf{R} \\rightarrow \\mathbf{R}$ such that\n$$\nf\\left(x^{2}+y f(z)\\right)=x f(x)+z f(y)\n$$\nfor all $x, y, z$ in $\\mathbf{R}$. (Here $\\mathbf{R}$ denotes the set of all real numbers.)",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x for all real x",
    "solution": "Solution:\nTaking $x=y=0$ in (1), we get $z f(0)=f(0)$ for all $z \\in \\mathbf{R}$. Hence we obtain $f(0)=0$.\n\nTaking $y=0$ in (1), we get\n$$\nf\\left(x^{2}\\right)=x f(x)\n$$\nSimilarly $x=0$ in (1) gives\n$$\nf(y f(z))=z f(y)\n$$\nPutting $y=1$ in (3), we get\n$$\nf(f(z))=z f(1) \\quad \\forall z \\in \\mathbf{R}\n$$\nNow using (2) and (4), we obtain\n$$\nf(x f(x))=f\\left(f\\left(x^{2}\\right)\\right)=x^{2} f(1)\n$$\nPut $y=z=x$ in (3) also given\n$$\nf(x f(x))=x f(x)\n$$\nComparing (5) and (6), it follows that $x^{2} f(1)=x f(x)$. If $x \\neq 0$, then $f(x)=c x$, for some constant $c$. Since $f(0)=0$, we have $f(x)=c x$ for $x=0$ as well. Substituting this in (1), we see that\n$$\nc\\left(x^{2}+c y z\\right)=c x^{2}+c y z\n$$\nor\n$$\nc^{2} y z=c y z \\quad \\forall y, z \\in \\mathbf{R}\n$$\nThis implies that $c^{2}=c$. Hence $c=0$ or $1$. We obtain $f(x)=0$ for all $x$ or $f(x)=x$ for all $x$. It is easy to verify that these two are solutions of the given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32334,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a positive integer divisible by $4$. Find the number of permutations $\\sigma$ of $(1, 2, 3, \\ldots, n)$ which satisfy the condition $\\sigma(j) + \\sigma^{-1}(j) = n + 1$, for all $j \\in \\{1, 2, 3, \\ldots, n\\}$.",
    "options": [],
    "answer": "(n/2)! / (n/4)!",
    "solution": "Let us take $n = 4k$. First we observe that such $\\sigma$ cannot have a fixed point; for if $\\sigma(j) = j$ for some $j$, then $\\sigma^{-1}(j) = j$ and hence $n + 1 = \\sigma(j) + \\sigma^{-1}(j) = 2j$, which is impossible because $n$ is divisible by $4$.\n\nSuppose $\\sigma(a) = b$, where $a \\neq b$. Then $\\sigma^{-1}(a) = n + 1 - b$; $\\sigma(b) = n + 1 - a$ and $\\sigma(n + 1 - a) = n + 1 - b$. Thus we obtain a $4$-cycle $(a, b, n+1-a, n+1-b)$. It is easy to check that all these are distinct. Thus any such permutation $\\sigma$ is a product of $n/4$ disjoint $4$-cycles of the above type. Note that $b$ and $n + 1 - b$ may be interchanged to get another admissible cycle $(a, n+1-b, n+1-a, b)$.\n\nThus we need to split the set $\\{1, 2, 3, \\ldots, n\\}$ into $n/2$ pairs $\\{j, n+1-j\\}$, $1 \\leq j \\leq n/2$ and partition these $n/2$ pairs into $n/4$ pairs of pairs. Each such pair of pairs gives $2$ admissible $4$-cycles. The first splitting can be done in $\\frac{1}{k!} \\binom{2k}{2} \\binom{2k-2}{2} \\cdots \\binom{2}{2} = \\frac{(2k)!}{2^k k!}$ ways and there are $2^k$ further ways of getting $4$-cycles. Hence the desired number is\n$$\n\\frac{(2k)!}{2^k k!} = \\frac{(2k)!}{k!} = 2 \\cdot 6 \\cdot 10 \\cdots (n-2).\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32335,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n1. In a non-equilateral triangle $A B C$, the sides $a, b, c$ form an arithmetic progression. Let $I$ and $O$ denote the incentre and circumcentre of the triangle respectively.\n\na. Prove that $I O$ is perpendicular to $B I$.\n\nb. Suppose $B I$ extended meets $A C$ in $K$, and $D, E$ are the midpoints of $B C, B A$ respectively. Prove that $I$ is the circumcentre of triangle $D K E$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nExtend $B I$ to meet the circumcircle in $F$. Then we know that $F A = F I = F C$. (See Figure)\n\n![](attached_image_1.png)\n\nLet $B I : I F = \\lambda : \\mu$. Applying Stewart's theorem to triangle $B A F$, we get\n$$\n\\lambda A F^{2} + \\mu A B^{2} = (\\lambda + \\mu)\\left(A I^{2} + B I \\cdot I F\\right)\n$$\nSimilarly, Stewart's theorem to triangle $B C F$ gives\n$$\n\\lambda C F^{2} + \\mu B C^{2} = (\\lambda + \\mu)\\left(C I^{2} + B I \\cdot I F\\right)\n$$\nSince $C F = A F$, subtraction gives\n$$\n\\mu\\left(A B^{2} - B C^{2}\\right) = (\\lambda + \\mu)\\left(A I^{2} - C I^{2}\\right)\n$$\nUsing the standard notations $A B = c$, $B C = a$, $C A = b$ and $s = (a + b + c) / 2$, we get $A I^{2} = r^{2} + (s - a)^{2}$ and $C I^{2} = r^{2} + (s - c)^{2}$ where $r$ is the in-radius of $A B C$. Thus\n$$\n\\mu\\left(c^{2} - a^{2}\\right) = (\\lambda + \\mu)\\left((s - a)^{2} - (s - c)^{2}\\right) = (\\lambda + \\mu)(c - a) b\n$$\nIt follows that either $c = a$ or $\\mu(c + a) = (\\lambda + \\mu) b$. But $c = a$ implies that $a = b = c$ since $a, b, c$ are in arithmetic progression. However, we have taken a non-equilateral triangle $A B C$. Thus $c \\neq a$ and we have $\\mu(c + a) = (\\lambda + \\mu) b$. But $c + a = 2 b$ and we obtain\n$$\n2 b \\mu = (\\lambda + \\mu) b.\n$$\nWe conclude that $\\lambda = \\mu$. This in turn tells that $I$ is the mid-point of $B F$. Since $O F = O B$, we conclude that $O I$ is perpendicular to $B F$.\n\n\nAlternatively\n\nApplying Ptolemy's theorem to the cyclic quadrilateral $A B C F$, we get\n$$\nA B \\cdot C F + A F \\cdot B C = B F \\cdot C A\n$$\nSince $C F = A F$, we get $C F(c + a) = B F \\cdot b = B F(c + a) / 2$. This gives $B F = 2 C F = 2 I F$. Hence $I$ is the mid-point of $B F$ and as earlier we conclude that $O I$ is perpendicular to $B F$.\n\n\nAlternatively\n\nJoin $B O$. We have to prove that $\\angle B I O = 90^{\\circ}$, which is equivalent to $B I^{2} + I O^{2} = B O^{2}$. Draw $I L$ perpendicular to $A B$. Let $R$ denote the circumradius of $A B C$ and let $\\triangle$ denote its area. Observe that $B O = R$, $I O^{2} = R^{2} - 2 R r$,\n$$\nB I = \\frac{B L}{\\cos (B / 2)} = (s - b) \\sqrt{\\frac{c a}{s(s - b)}}\n$$\nThus we obtain\n$$\nB I^{2} = a c (s - b) / s = \\frac{a c}{3}\n$$\nsince $a, b, c$ are in arithmetic progression. Thus we need to prove that\n$$\n\\frac{a c}{3} + R^{2} - 2 R r = R^{2}\n$$\nThis reduces to proving $2 R r = a c / 3$. But\n$$\n2 R r = 2 \\cdot \\frac{a b c}{4 \\Delta} \\cdot \\frac{\\Delta}{s} = \\frac{a b c}{2 s} = \\frac{a b c}{a + b + c} = \\frac{a c}{3}\n$$\nusing $a + c = 2 b$. This proves the claim.\n\n\nb.\nJoin $I D$. Note that $\\angle B I O = \\angle B D O = 90^{\\circ}$. Hence $B, D, I, O$ are concyclic and hence $\\angle B I D = \\angle B O D = A$. Since $\\angle D B I = \\angle K B A = B / 2$, it follows that triangles $B A K$ and $B I D$ are similar. This gives\n$$\n\\frac{B A}{B I} = \\frac{B K}{B D} = \\frac{A K}{I D}\n$$\nHowever, we have seen earlier that $B I = a c / 3$. Moreover $A K = b c / (a + c)$. Thus we obtain\n$$\nB K = \\frac{B A \\cdot B D}{B I} = \\frac{1}{2} \\sqrt{3 a c}, \\quad I D = \\frac{A K \\cdot B I}{B A} = \\frac{1}{2} \\sqrt{\\frac{a c}{3}}.\n$$\nBy symmetry, we must have $I E = \\frac{1}{2} \\sqrt{\\frac{a c}{3}}$. Finally\n$$\nI K = \\frac{b}{a + b + c} \\cdot B K = \\frac{1}{3} B K = \\frac{1}{2} \\sqrt{\\frac{a c}{3}}\n$$\nThus $I D = I E = I K$ and $I$ is the circumcentre of $D K E$.\n\n\nAlternatively\n\nObserve that $A K = b c / (a + c) = c / 2 = A E$. Since $A I$ bisects angle $A$, we see that $A I E$ is congruent to $A I K$. This gives $I E = I K$. Similarly $C I D$ is congruent to $C I K$ giving $I D = I K$. We conclude that $I D = I K = I E$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32336,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all triples $(a, b, c)$ integers such that $a^2 + b^2 = c^2$, $\\text{gcd}(a, b, c) = 1$ and $2000 \\le a, b, c \\le 3000$.",
    "options": [],
    "answer": "(2100, 2059, 2941)",
    "solution": "We use the known characterisation of primitive Pythagorean triples $(a, b, c)$: $a = 2pq$, $b = p^2 - q^2$, $c = p^2 + q^2$, where $p$ and $q$ are relatively prime integers, one even and $q < p$. The given condition entails $c^2 = a^2 + b^2 > 2 \\times 2000^2$. Hence $2000\\sqrt{2} < c = p^2 + q^2 < 3000$. Using the approximation $\\sqrt{2} = 1.4142$, we get $2828 < p^2 + q^2 < 3000$. We also have $2000 < b = p^2 - q^2$. Adding these two, we obtain $4828 < 2p^2$, whence $50 \\le p$. Again, we have $2000 < a = 2pq$. Subtracting this from $p^2+q^2 < 3000$, we get $(p-q)^2 < 1000$ giving $p-q \\le 31$. Using $50 \\le p$ and $p-q \\le 31$, we obtain $19 \\le q$ and hence $3000 > p^2+q^2 \\ge p^2+19^2$. It follows that $p^2 < 2639$ and thus $p \\le 51$. We obtain $2000/51 < 2pq/p = 2q$.\n\nWe thus obtain a better bound for $q$: $20 \\le q$ and hence a better bound for $p$: $p^2 < 2639$ and hence $p \\le 50$. We have seen earlier that $p \\ge 50$, thus getting $p = 50$. Using $p^2 + q^2 < 3000$, we now get $q^2 < 500$. This implies that $q \\le 22$. Thus $20 \\le q \\le 22$. However $q$ has to be odd since $p$ is even. We conclude that $q = 21$. We see that $a = 2100$, $b = 2059$ and $c = 2941$ give only one triple in the given range.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32337,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(u_{jk})$, $1 \\le j \\le 3$, $1 \\le k \\le 2$ be real numbers and let $N$ be an integer such that\n$$\n\\max_{1 \\le k \\le 2} \\sum_{j=1}^{3} |u_{jk}| \\le N.\n$$\nSuppose $M$ and $l$ are positive integers such that $l^2 < (M+1)^3$. Prove that there exist integers $\\xi_1, \\xi_2, \\xi_3$, not all zero, such that\n$$\n\\max_{1 \\le j \\le 3} |\\xi_j| \\le M \\quad \\text{and} \\quad \\left| \\sum_{j=1}^{3} u_{jk} \\xi_j \\right| \\le \\frac{MN}{l}, \\quad \\text{for } k = 1, 2.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $a_1, a_2, a_3$ be integers such that $0 \\le a_1, a_2, a_3 \\le M$. Let\n$$\nS_k = \\left| \\sum_{j=1}^{3} u_{jk} a_j \\right|, \\quad k = 1, 2.\n$$\nThen Cauchy-Schwartz inequality gives\n$$\nS_k \\le \\sqrt{\\sum_{j=1}^{3} a_j^2} \\sqrt{\\sum_{j=1}^{3} u_{jk}^2} \\le \\sqrt{3M^2} \\sqrt{N^2} = \\sqrt{3MN^2}.\n$$\nWe have used $a_j^2 \\le M^2$ and $\\sum_{j=1}^3 u_{jk}^2 \\le (\\sum_{j=1}^3 |u_{jk}|)^2 \\le N^2$.\nSince $M$ is a positive integer, $M \\ge 1$, so that $\\sqrt{(M+1)^3 - 1} \\ge \\sqrt{3}$. Thus we have\n$$\n0 \\le S_k \\le MN \\sqrt{(M+1)^3 - 1},\n$$\nfor $k = 1, 2$. Divide the interval $[0, MN\\sqrt{(M+1)^3 - 1}]$ into $(M + N)^3 - 1$ equal parts, each of length $MN/\\sqrt{(M+1)^3 - 1}$. If we consider ordered triples $(a_1, a_2, a_3)$ of integers such that $0 \\le a_1, a_2, a_3 \\le M$, there are $(M+1)^3$ such triples and these give $(M+1)^3$ values for $S_k$. Hence by pigeon-hole principle, we can find two triples $(A_1, A_2, A_3)$ and $(B_1, B_2, B_3)$ such that $0 \\le A_1, A_2, A_3, B_1, B_2, B_3 \\le M$ and the corresponding $S$-values lie in the same subinterval. Thus\n$$\n|S_k(A_1, A_2, A_3) - S_k(B_1, B_2, B_3)| \\le \\frac{MN}{\\sqrt{(M+1)^3 - 1}} \\le \\frac{MN}{l},\n$$\nbecause $l^2 < (M+1)^3$ implies that $l^2 \\le (M+1)^3 - 1$. Taking $\\xi_j = A_j - B_j$, $j = 1, 2, 3$, we see that $|\\xi_j| \\le M$ and\n$$\n\\left| \\sum_{j=1}^{3} u_{jk} \\xi_j \\right| \\le |S_k(A_1, A_2, A_3) - S_k(B_1, B_2, B_3)| \\le \\frac{MN}{l}.\n$$\nNote that estimate does not depend on $k$ and hence valid for $k = 1, 2$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32338,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1, A_2, A_3, \\dots, A_n$ be $n$ subsets of a finite set $S$ such that $|A_j| = 8$ for each $j$, $1 \\le j \\le n$. For a subset $B$ of $S$, let $F(B) = \\{j : 1 \\le j \\le n \\text{ and } A_j \\subset B\\}$. Suppose for each subset $B$ of $S$, at least one of the following conditions holds:\n(i) $|B| > 25$;\n(ii) $F(B) = \\emptyset$;\n(iii) $\\cap_{j \\in F(B)} A_j \\ne \\emptyset$.\n\nProve that $A_1 \\cap A_2 \\cap A_3 \\cap \\dots \\cap A_n \\ne \\emptyset$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use induction on $n$. The case $n = 1$ is immediate. Assume the result for $n$ and consider the case of $n+1$. Thus $A_1, A_2, \\dots, A_{n+1}$ are subsets of $S$ with $|A_j| = 8$.\nNow if we consider the sets $A_1, A_2, \\dots, A_{j-1}, A_{j+1}, \\dots, A_{n+1}$, these satisfy induction hypothesis for each $j = 1, 2, 3, \\dots, n+1$. Hence there exists\n$$\nx_j \\in A_1 \\cap A_2 \\cap \\dots \\cap A_{j-1} \\cap A_{j+1} \\cap \\dots \\cap A_{n+1},\n$$\nfor each $j = 1, 2, \\dots, n+1$. If $x_1, x_2, \\dots, x_{n+1}$ are not all distinct, then we are through. Suppose they are all distinct. Consider $B = A_1 \\cup A_2 \\cup \\dots \\cup A_{n+1}$. We show that $|B| \\le 25$. Note that each $x_j$ is over counted $n-1$ times in the union. Thus\n$$\n|B| \\le |A_1| + |A_2| + \\dots + |A_{n+1}| - (n-1)(n+1) = 8(n+1) - (n^2-1) \\le 25,\n$$\nsince $n^2 - 8n + 16 = (n-4)^2 \\ge 0$. Thus the condition (i) fails. Clearly, the condition (2) also fails. Hence (iii) must hold.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32339,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a parallelogram. A variable line $l$ through the point $A$ intersects the rays $BC$ and $DC$ at $X$ and $Y$ respectively. Let $K$ and $L$ be the centres of the excircles of triangles $ABX$ and $ADY$, touching the sides $BX$ and $DY$ respectively. Prove that the size of $\\angle KCL$ does not depend on the choice of the line $l$. (Short-list, IMO-2005)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\angle DAX = 2\\gamma$ and $\\angle YAP = 2\\alpha$. Now $K$ being the ex-centre of triangle $ABX$, it lies on the bisectors of $\\angle XAB$ and $\\angle XBP$. Note that $\\angle ABX = 180^\\circ - 2\\alpha - 2\\gamma$ and hence $\\angle XBK = (2\\alpha + 2\\gamma)/2 = \\alpha + \\gamma$. Thus the angles of triangle $ABK$ are $\\alpha$, $\\gamma$, $180^\\circ - \\alpha - \\gamma$.\n\nSimilarly, the angles of $ADL$ are also $x$, $y$, $180^\\circ - x - y$. Thus $ABK$ is similar to $LDA$ and hence $AB/LD = BK/AD$. However $AB = DC$ and $AD = BC$. Thus we obtain $DC/LD = BK/BC$. Note also that $\\angle CBK = \\angle LDC$. We hence get the similarity of triangles $CBK$ and $LDC$. If we take $\\angle ALC = \\theta$, then $\\angle DCL = 180^\\circ - 2x - \\theta$. But by similarity of $CBK$ and $LDC$, we have $\\angle DCL = \\angle BKC$, giving $\\angle BKC = 180^\\circ - 2x - \\theta$. This gives $\\angle KCX = x + \\theta$.\n\n![](attached_image_1.png)\n\nThis gives\n$$\n\\angle KCL = 360^\\circ - (x + \\theta) - (2x + 2y) - (180^\\circ - 2x - y - \\theta) \\\\\n= 180^\\circ + x + y = 180^\\circ + (\\angle A)/2\n$$\nwhich depends only on the parallelogram but not on the line $l$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32340,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nProve that for every positive integer $n$ there exists a unique ordered pair $(a, b)$ of positive integers such that\n$$\nn = \\frac{1}{2}(a + b - 1)(a + b - 2) + a\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe have to prove that $f: \\mathbb{N} \\times \\mathbb{N} \\rightarrow \\mathbb{N}$ defined by\n$$\nf(a, b) = \\frac{1}{2}(a + b - 1)(a + b - 2) + a, \\quad \\forall a, b \\in \\mathbb{N}\n$$\nis a bijection. (Note that the right side is a natural number.) To this end define\n$$\nT(n) = \\frac{n(n+1)}{2}, \\quad n \\in \\mathbb{N} \\cup \\{0\\}\n$$\nAn idea of the proof can be obtained by looking at the following table of values of $f(a, b)$ for some small values of $a, b$.\n![](attached_image_1.png)\nWe observe that the $n$-th diagonal runs from $(1, n)$-th position to $(n, 1)$-th position and the entries are $n$ consecutive integers; the first entry in the $n$-th diagonal is one more than the last entry of the $(n-1)$-th diagonal. For example the first entry in 5-th diagonal is 11 which is one more than the last entry of 4-th diagonal which is 10. Observe that 5-th diagonal starts from 11 and ends with 15 which accounts for 5 consecutive natural numbers. Thus we see that $f(n-1, 1) + 1 = f(1, n)$. We also observe that the first $n$ diagonals exhaust all the natural numbers from 1 to $T(n)$. (Thus a kind of visual bijection is already there. We formally prove the property.)\n\nWe first observe that\n$$\nf(a, b) - T(a + b - 2) = a > 0\n$$\nand\n$$\nT(a + b - 1) - f(a, b) = \\frac{(a + b - 1)(a + b)}{2} - \\frac{(a + b - 1)(a + b - 2)}{2} - a = b - 1 \\geq 0\n$$\nThus we have\n$$\nT(a + b - 2) < f(a, b) = \\frac{(a + b - 1)(a + b - 2)}{2} + a \\leq T(a + b - 1)\n$$\nSuppose $f(a_1, b_1) = f(a_2, b_2)$. Then the previous observation shows that\n$$\n\\begin{aligned}\n& T(a_1 + b_1 - 2) < f(a_1, b_1) \\leq T(a_1 + b_1 - 1) \\\\\n& T(a_2 + b_2 - 2) < f(a_2, b_2) \\leq T(a_2 + b_2 - 1)\n\\end{aligned}\n$$\nSince the sequence $\\langle T(n) \\rangle_{n=0}^{\\infty}$ is strictly increasing, it follows that $a_1 + b_1 = a_2 + b_2$. But then the relation $f(a_1, b_1) = f(a_2, b_2)$ implies that $a_1 = a_2$ and $b_1 = b_2$. Hence $f$ is one-one.\n\nLet $n$ be any natural number. Since the sequence $\\langle T(n) \\rangle_{n=0}^{\\infty}$ is strictly increasing, we can find a natural number $k$ such that\n$$\nT(k-1) < n \\leq T(k)\n$$\nEquivalently,\n$$\n\\frac{(k-1)k}{2} < n \\leq \\frac{k(k+1)}{2}\n$$\nNow set $a = n - \\frac{k(k-1)}{2}$ and $b = k - a + 1$. Observe that $a > 0$. Now (1) shows that\n$$\na = n - \\frac{k(k-1)}{2} \\leq \\frac{k(k+1)}{2} - \\frac{k(k-1)}{2} = k\n$$\nHence $b = k - a + 1 \\geq 1$. Thus $a$ and $b$ are both positive integers and\n$$\nf(a, b) = \\frac{1}{2}(a + b - 1)(a + b - 2) + a = \\frac{k(k-1)}{2} + a = n\n$$\nThis shows that every natural number is in the range of $f$. Thus $f$ is also onto. We conclude that $f$ is a bijection.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32341,
    "subject": "Mathematics (Multi-modal)",
    "question": "There are $n$ markers, each with one side white and the other side black, aligned in a row so that their white sides are all up in the beginning. In each step, if possible, we choose a marker with the white side up (but not the one of outermost markers), remove it, and reverse the closest marker to the left and the closest marker to the right of it. Prove that one can achieve the state with only two markers remaining if and only if $(n-1)$ is not divisible by $3$. (Short-list, IMO-2005)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to handle the cases $n = 3, 5, 6$:\n\na. $n=3$: $WWW \\rightarrow BB$, by removing middle $W$ and reversing the other two;\n\nb. $n=5$: $(WWW)WW \\rightarrow (BB)WW = B(BWW) \\rightarrow BWB \\rightarrow WW$;\n\nc. $n = 6$: $(WWW)WWW \\rightarrow BB(WWW) = B(BWW)W \\rightarrow B(WB)W = (BWB)W \\rightarrow WWW$.\n\nThe last reduction shows that for $n > 6$, we can pick any six $W$'s and reduce it three $W$'s there by reducing the case to that of $n - 3$. Since we have ended with 2 markers for $n = 3$ and $5$, it follows by induction that the 'two marker state' is achievable for $n = 3k$ or $3k + 2$.\n\nWe have to show that this is not possible for $n = 3k + 1$. Observe that the number of $B$'s is invariant modulo $2$ in each move: $WWW \\rightarrow BB$; $WWB \\rightarrow BW$; $BWW \\rightarrow WB$; $BWB \\rightarrow WW$. Since all are $W$ in the beginning, the final two markers must be either $BB$ or $WW$. We assign weights to $W$ and $B$.\n\nTake the group of symmetries of an equilateral triangle with centre at origin and $y$-axis passing through a vertex; let $\\phi$ denote rotation by $120^\\circ$ about the centre in clock-wise direction; and let $\\psi$ denote the reflection in $y$ axis. Then it is easy to see that\n$$\n\\phi^3 = \\psi^2 = 1, \\phi^2\\psi = \\psi\\phi, \\psi\\phi^2 = \\phi\\psi, \\psi\\phi\\psi = \\phi^2.\n$$\nThus we may assign $\\omega(W) = \\phi$, $\\omega(B) = \\psi$. The properties of $\\phi$ and $\\psi$ show that they obey the reductions $WWW \\rightarrow BB$, $WWB \\rightarrow BW$; $BWW \\rightarrow WB$; $BWB \\rightarrow WW$ and the total weights remain unchanged after each move. Hence if the 'two marker state is achievable, then the end weights must be either $\\phi^2$ or $\\psi^2$ (corresponding to $WW$ and $BB$ respectively). If $n = 3k + 1$, the total weight at the beginning is $\\phi^{(3k+1)} = \\phi$. But $\\phi \\neq \\phi^2$ or $\\psi^2$ as is evident from the definitions of these functions. Hence 'two marker state' is not achievable for $n = 3k + 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32342,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $X$ denote the set of all triples $(a, b, c)$ of integers. Define a function $f: X \\rightarrow X$ by\n$$\nf(a, b, c) = (a + b + c, ab + bc + ca, abc)\n$$\nFind all triples $(a, b, c)$ in $X$ such that $f(f(a, b, c)) = (a, b, c)$.",
    "options": [],
    "answer": "All solutions are the triples (t, 0, 0) for any integer t, and the triple (-1, -1, 1).",
    "solution": "Solution:\nWe show that the solution set consists of $\\{(t, 0, 0) ; t \\in \\mathbb{Z}\\} \\cup \\{(-1, -1, 1)\\}$. Let us put $a + b + c = d$, $ab + bc + ca = e$ and $abc = f$. The given condition $f(f(a, b, c)) = (a, b, c)$ implies that\n$$\nd + e + f = a, \\quad de + ef + fd = b, \\quad def = c\n$$\nThus $abcdef = fc$ and hence either $cf = 0$ or $abde = 1$.\n\nCase I: Suppose $cf = 0$. Then either $c = 0$ or $f = 0$. However $c = 0$ implies $f = 0$ and vice-versa. Thus we obtain $a + b = d$, $d + e = a$, $ab = e$ and $de = b$. The first two relations give $b = -e$. Thus $e = ab = -ae$ and $de = b = -e$. We get either $e = 0$ or $a = d = -1$.\nIf $e = 0$, then $b = 0$ and $a = d = t$, say. We get the triple $(a, b, c) = (t, 0, 0)$, where $t \\in \\mathbb{Z}$. If $e \\neq 0$, then $a = d = -1$. But then $d + e + f = a$ implies that $-1 + e + 0 = -1$ forcing $e = 0$. Thus we get the solution family $(a, b, c) = (t, 0, 0)$, where $t \\in \\mathbb{Z}$.\n\nCase II: Suppose $cf \\neq 0$. In this case $abde = 1$. Hence either all are equal to $1$; or two equal to $1$ and the other two equal to $-1$; or all equal to $-1$.\nSuppose $a = b = d = e = 1$. Then $a + b + c = d$ shows that $c = -1$. Similarly $f = -1$. Hence $e = ab + bc + ca = 1 - 1 - 1 = -1$ contradicting $e = 1$.\nSuppose $a = b = 1$ and $d = e = -1$. Then $a + b + c = d$ gives $c = -3$ and $d + e + f = a$ gives $f = 3$. But then $f = abc = 1 \\cdot 1 \\cdot (-3) = -3$, a contradiction. Similarly $a = b = -1$ and $d = e = 1$ is not possible.\nIf $a = 1, b = -1, d = 1, e = -1$, then $a + b + c = d$ gives $c = 1$. Similarly $f = 1$. But then $f = abc = 1 \\cdot 1 \\cdot (-1) = -1$ a contradiction. If $a = 1, b = -1, d = -1, e = 1$, then $c = -1$ and $e = ab + bc + ca = -1 + 1 - 1 = -1$ and a contradiction to $e = 1$. The symmetry between $(a, b, c)$ and $(d, e, f)$ shows that $a = -1, b = 1, d = 1, e = -1$ is not possible. Finally if $a = -1, b = 1, d = -1$ and $e = 1$, then $c = -1$ and $f = -1$. But then $f = abc$ is not satisfied.\nThe only case left is that of $a, b, d, e$ being all equal to $-1$. Then $c = 1$ and $f = 1$. It is easy to check that $(-1, -1, 1)$ is indeed a solution.\n\nAlternatively\n$cf \\neq 0$ implies that $|c| \\geq 1$ and $|f| \\geq 1$. Observe that\n$$\nd^{2} - 2e = a^{2} + b^{2} + c^{2}, \\quad a^{2} - 2b = d^{2} + e^{2} + f^{2}\n$$\nAdding these two, we get $-2(b + e) = b^{2} + c^{2} + e^{2} + f^{2}$. This may be written in the form\n$$\n(b + 1)^{2} + (e + 1)^{2} + c^{2} + f^{2} - 2 = 0\n$$\nWe conclude that $c^{2} + f^{2} \\leq 2$. Using $|c| \\geq 1$ and $|f| \\geq 1$, we obtain $|c| = 1$ and $|f| = 1$, $b + 1 = 0$ and $e + 1 = 0$. Thus $b = e = -1$. Now $a + d = d + e + f + a + b + c$ and this gives $b + c + e + f = 0$. It follows that $c = f = 1$ and finally $a = d = -1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32343,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $P$ be a point in its plane inside the region of the angle $BAC$, but lying outside triangle $ABC$.\n\na. Prove that any two of the following statements imply the third:\n(i) the circumcentre of triangle $BPC$ lies on the ray $\\vec{PA}$;\n(ii) the circumcentre of triangle $CPA$ lies on the ray $\\vec{PB}$;\n(iii) the circumcentre of triangle $APB$ lies on the ray $\\vec{PC}$.\n\nb. Prove that when the conditions in (a) hold, the circumcentres of triangles $BPC$, $CPA$, $APB$ lie on the circumcircle of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The problem may be rephrased as follows: line segments $PA$, $PB$, $PC$ are given. Let $l_1$, $l_2$, $l_3$ denote the perpendicular bisectors of $PA$, $PB$, $PC$ respectively. Given that $l_1$, $l_3$ meet on $PB$ and $l_2$, $l_3$ meet on $PA$, we have to show that $l_1$, $l_2$ intersect on $PC$.\n\nLet $l_1$, $l_3$ intersect in $R$ on $PB$; and let $l_2$, $l_3$ intersect in $H$ on $PA$. If we consider $l_1$ and $l_2$, we observe that $l_1$ is perpendicular to $PH$, $l_2$ is perpendicular to $PR$, and $PC$ is perpendicular to $l_3$, a part of which is $HR$. It follows that $l_1$, $l_2$, $PC$ are concurrent. The other propositions may be proved similarly.\n\nLet $A'$, $B'$, $C'$ denote the mid-points of the line segments $PA$, $PB$, $PC$ respectively. We have to show that the circumcentre of $ABC$ passes through the circumcentres of $PAB$, $PBC$ and $PCA$. Note that $PBC$ is a homothetic expansion of $PB'C'$. If $O$, $O'$ denote the circumcentres of $PBC$ and $PB'C'$, then we get $PO' = PO/2$. However, $O'$ is the mid-point of $HP$. (This can be seen easily from right-triangle $PHB'$.) It follows that $O = H$. Thus $H$ is the circumcentre of $PBC$.\n\n![](attached_image_1.png)\n\nWe also observe that $PRA'C'$ is a cyclic quadrilateral. Hence the circumcentre of $PC'A'$ is the same as that of $PC'R$ and this is the mid-point of $PR$ since $PC'R$ is a right-triangle. But $PCA$ is the homothetic expansion of $PC'A'$ with centre $P$ and ratio $2$. Hence the circumcentre of $PCA$ is along $PR$ and moved by a factor of $2$ from the circumcentre of $PC'A'$; we get the point $R$. Hence $R$ is the circumcentre of $PCA$. Similarly, we may prove that the circumcentre of $PBA$ is $Q$.\n\nNow $\\angle CHB = 2\\angle P$ since $H$ is the circumcentre of $PBC$ and\n$$\n\\angle BAC = \\angle BAP + \\angle PAC = \\angle B'A'P + \\angle PA'C.\n$$\nBut the cyclic quadrilateral $B'HA'R$ shows that $\\angle B'A'P = \\angle PRH = 90^\\circ - \\angle P$. Similarly, $\\angle PA'C' = 90^\\circ - \\angle P$. Thus $\\angle BAC = 180^\\circ - 2\\angle P$. Thus $A$, $B$, $H$, $C$ are concyclic. It follows that $H$ lies on the circumcircle of $ABC$.\n\nWe also observe that $\\angle BQC = 2\\angle B'QC = 2(90^\\circ - \\angle P)$ and similarly $\\angle BRC = 2(90^\\circ - \\angle P)$. Thus $BC$ subtends the same angles at $A$, $Q$, $R$.\n\nIt follows that $A$, $B$, $C$, $Q$, $R$ all lie on the same circle. Since $H$ is already on the circumcircle of $ABC$, we conclude that $Q$, $R$, $H$ lie on the circumcircle of $ABC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32344,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $p$ be a prime and $X$ be a finite set containing at least $p$ elements. A collection of mutually disjoint $p$-element subsets of $X$ is called a $p$-family. The empty collection itself is regarded as a $p$-family. Let $A$ (respectively, $B$) denote the number of $p$-families having an even (respectively, odd) number of $p$-element subsets of $X$. Prove that $A$ and $B$ differ by a multiple of $p$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We observe that\n$$\nA = \\sum_{r \\ge 0,\\ r\\ \\text{even}} \\binom{n}{rp} \\frac{(rp)!}{(p!)^r r!}\n$$\nand\n$$\nB = \\sum_{r \\ge 0,\\ r\\ \\text{odd}} \\binom{n}{rp} \\frac{(rp)!}{(p!)^r r!}\n$$\nThe empty collection corresponds to $r=0$ in the summation for $A$. If $u_r = \\frac{(rp)!}{(p!)^r r!}$, $r \\ge 0$, then we have\n$$\nu_{r+1} = \\frac{(rp+1)(rp+2)\\cdots}{ }\n$$\nHence we obtain\n$$\n(p-1)!\\, u_{r+1} = (rp+1)(rp+2)\\cdots (rp+p-1) u_r.\n$$\nReading this modulo $p$, we get $u_{r+1} \\equiv u_r \\pmod{p}$, $r \\ge 0$. Using $u_0 = 1$, we conclude that $u_r \\equiv 1 \\pmod{p}$, for all $r \\ge 1$. This implies that\n$$\nA - B \\equiv \\sum_{r \\ge 0} (-1)^r \\binom{n}{rp} \\pmod{p}.\n$$\nThus we need to show that $p$ divides $M = \\sum (-1)^r \\binom{n}{rp}$. Suppose $p$ is odd. Taking $f(x) = (1-x)^n$, and $\\omega$ to be a primitive $p$-th root of unity, we see that\n$$\n\\begin{align*}\n\\sum_{k=0}^{p-1} f(\\omega^k) &= \\sum_{k=0}^{p-1} (1 - \\omega^k)^n \\\\\n&= \\sum_{k=0}^{p-1} \\sum_{j=0}^{n} \\binom{n}{j} (-1)^j \\omega^{kj} \\\\\n&= \\sum_{j=0}^{n} \\binom{n}{j} (-1)^j \\sum_{k=0}^{p-1} \\omega^{kj}.\n\\end{align*}\n$$\nWe observe that\n$$\n\\sum_{k=0}^{p-1} \\omega^{kj} = \\begin{cases} \\frac{1-\\omega^{pj}}{1-\\omega^j} = 0 & \\text{if } p \\text{ does not divide } j \\\\ p & \\text{if } p \\text{ divides } j. \\end{cases}\n$$\nThus the above sum reduces to\n$$\np \\left\\{ \\binom{n}{0} - \\binom{n}{p} + \\binom{n}{2p} + \\dots \\right\\} = pM.\n$$\nHowever $1, \\omega, \\omega^2, \\dots, \\omega^{p-1}$ are the roots of $x^p - 1 = 0$. If we set $\\alpha_k = 1 - \\omega^k$, $0 \\le k \\le p-1$, then $\\alpha_0, \\alpha_1, \\dots, \\alpha_{p-1}$ are the roots of $(x-1)^p+1 = 0$. If we set $c_j = (-1)^j \\binom{p}{j}$ for $j = 1, 2, \\dots, p-1$, then we have\n$$\n(x-\\alpha_0)(x-\\alpha_1)(x-\\alpha_2)\\cdots(x-\\alpha_{p-1}) = x^p+c_1x^{p-1}+c_2x^{p-2}+\\cdots+c_{p-1}x+c_p\n$$\nwhere we take $c_p = 0$. We observe that $p$ divides $c_j$ for $0 \\le j \\le p-1$. Taking $s_k = \\sum_{j=0}^{p-1} \\alpha_j^k$, use of Newton's identities show that $s_k$ is divisible by $p$ for $k \\ge 0$. Moreover for $n \\ge p$, we also have\n$$\ns_n + c_1 s_{n-1} + c_2 s_{n-2} + \\dots + c_{p-1} s_{n-(p-1)} + c_p s_{n-p} = 0.\n$$\nThus $p^2$ divides $s_n$. But\n$$\ns_n = \\sum_{j=0}^{p-1} \\alpha_j^n = \\sum_{j=0}^{p-1} (1 - \\omega^j)^n = pM.\n$$\nIt follows that $p$ divides $M$.\nIf $p$ is even, then $M \\equiv \\sum_{j \\ge 0} \\binom{n}{2j} \\equiv 2^{n-1} \\pmod{2}$. Since $n \\ge p=2$, we conclude that $2$ divides $M$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32345,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a cyclic quadrilateral $ABCD$, $AB = a$, $BC = b$, $CD = c$, $\\angle ABC = 120^\\circ$, and $\\angle ABD = 30^\\circ$. Prove that\n\ni) $c \\geq a + b$;\n\nii) $|\\sqrt{c+a} - \\sqrt{c+b}| = \\sqrt{c-a-b}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nApplying cosine rule to triangle $ABC$, we get\n$$\nAC^2 = a^2 + b^2 - 2ab \\cos 120^\\circ = a^2 + b^2 + ab\n$$\nObserve that $\\angle DAC = \\angle DBC = 120^\\circ - 30^\\circ = 90^\\circ$. Thus we get\n$$\nc^2 = \\frac{AC^2}{\\cos^2 30^\\circ} = \\frac{4}{3}(a^2 + b^2 + ab)\n$$\nSo\n$$\nc^2 - (a+b)^2 = \\frac{4}{3}(a^2 + b^2 + ab) - (a^2 + b^2 + 2ab) = \\frac{(a-b)^2}{3} \\geq 0\n$$\nThis proves $c \\geq a + b$ and thus (i) is true.\n\n![](attached_image_1.png)\n\nFor proving (ii), consider the product\n$$\nQ = (\\alpha + \\beta + \\gamma)(\\alpha - \\beta - \\gamma)(\\alpha + \\beta - \\gamma)(\\alpha - \\beta + \\gamma)\n$$\nwhere $\\alpha = \\sqrt{c+a}$, $\\beta = \\sqrt{c+b}$ and $\\gamma = \\sqrt{c-a-b}$. Expanding the product, we get\n$$\n\\begin{aligned}\nQ &= (c+a)^2 + (c+b)^2 + (c-a-b)^2 - 2(c+a)(c+b) - 2(c+a)(c-a-b) - 2(c+b)(c-a-b) \\\\\n&= -3c^2 + 4a^2 + 4b^2 + 4ab \\\\\n&= 0\n\\end{aligned}\n$$\nThus at least one of the factors must be equal to $0$. Since $\\alpha + \\beta + \\gamma > 0$ and $\\alpha + \\beta - \\gamma > 0$, it follows that the product of the remaining two factors is $0$. This gives\n$$\n\\sqrt{c+a} - \\sqrt{c+b} = \\sqrt{c-a-b} \\quad \\text{or} \\quad \\sqrt{c+a} - \\sqrt{c+b} = -\\sqrt{c-a-b}\n$$\nWe conclude that\n$$\n|\\sqrt{c+a} - \\sqrt{c+b}| = \\sqrt{c-a-b}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32346,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an equilateral triangle, and $D, E, F$ be points on $BC, CA, AB$ respectively. Let $\\angle BAD = \\alpha$, $\\angle CBE = \\beta$, $\\angle ACF = \\gamma$. Suppose $\\alpha + \\beta + \\gamma \\ge 120^\\circ$. Prove that the union $\\Sigma$ of triangular regions $BAD, CBE, ACF$ covers the triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose some point $P$ does not lie in any of the triangles $BAD, CBE, ACF$. Then $\\angle BAD < \\angle BAP$, $\\angle CBE < \\angle CBP$ and $\\angle ACF < \\angle ACP$. Thus\n$$\n\\alpha + \\beta + \\gamma < \\angle BAP + \\angle CBP + \\angle ACP.\n$$\nIf $P$ is the circumcentre of $ABC$, then $\\angle BAP = \\angle CBP = \\angle ACP = 30^\\circ$ and thus $\\alpha + \\beta + \\gamma < 90^\\circ$ contradicting $\\alpha + \\beta + \\gamma \\ge 120^\\circ$. Thus $P$ cannot be the circumcentre of $ABC$. Hence $PA, PB, PC$ are not all equal. We may assume $PA > PB$.\n\n![](attached_image_1.png)\n\nIn the above figure, triangles $PAB$ and $PLK$ are similar. If $PA > PB$, then $PL > PK$ and hence $\\angle PKL > \\angle PLK = \\angle PAB$. Thus\n$$\n\\begin{align*}\n\\angle BAP + \\angle CBP + \\angle ACP &= \\angle PLK + \\angle LKC + \\angle AKM \\\\\n&< \\angle PKL + \\angle LKC + \\angle AKM \\\\\n&< \\angle BKC = 120^\\circ.\n\\end{align*}\n$$\nIt follows that if $\\alpha + \\beta + \\gamma \\ge 120^\\circ$, then no such point $P$ which lies outside all the triangles $BAD, CBE, ACF$ can exist.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32347,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\na. Prove that if $n$ is a positive integer such that $n \\geq 4011^{2}$, then there exists an integer $l$ such that $n < l^{2} < \\left(1 + \\frac{1}{2005}\\right) n$.\n\nb. Find the smallest positive integer $M$ for which whenever an integer $n$ is such that $n \\geq M$, there exists an integer $l$, such that $n < l^{2} < \\left(1 + \\frac{1}{2005}\\right) n$.",
    "options": [],
    "answer": "4010^2 + 1",
    "solution": "Solution:\n\na. Let $n \\geq 4011^{2}$ and $m \\in \\mathbb{N}$ be such that $m^{2} \\leq n < (m+1)^{2}$. Then\n$$\n\\begin{aligned}\n\\left(1+\\frac{1}{2005}\\right) n - (m+1)^{2} & \\geq \\left(1+\\frac{1}{2005}\\right) m^{2} - (m+1)^{2} \\\\\n& = \\frac{m^{2}}{2005} - 2m - 1 \\\\\n& = \\frac{1}{2005}\\left(m^{2} - 4010m - 2005\\right) \\\\\n& = \\frac{1}{2005}\\left((m-2005)^{2} - 2005^{2} - 2005\\right) \\\\\n& \\geq \\frac{1}{2005}\\left((4011-2005)^{2} - 2005^{2} - 2005\\right) \\\\\n& = \\frac{1}{2005}\\left(2006^{2} - 2005^{2} - 2005\\right) \\\\\n& = \\frac{1}{2005}(4011 - 2005) = \\frac{2006}{2005} > 0\n\\end{aligned}\n$$\nThus we get\n$$\nn < (m+1)^{2} < \\left(1 + \\frac{1}{2005}\\right) n\n$$\nand $l^{2} = (m+1)^{2}$ is the desired square.\n\nb. We show that $M = 4010^{2} + 1$ is the required least number. Suppose $n \\geq M$. Write $n = 4010^{2} + k$, where $k$ is a positive integer. Note that we may assume $n < 4011^{2}$ by part (a). Now\n$$\n\\begin{aligned}\n\\left(1 + \\frac{1}{2005}\\right) n - 4011^{2} & = \\left(1 + \\frac{1}{2005}\\right)\\left(4010^{2} + k\\right) - 4011^{2} \\\\\n& = 4010^{2} + 2 \\cdot 4010 + k + \\frac{k}{2005} - 4011^{2} \\\\\n& = (4010 + 1)^{2} + (k - 1) + \\frac{k}{2005} - 4011^{2} \\\\\n& = (k - 1) + \\frac{k}{2005} > 0\n\\end{aligned}\n$$\nThus we obtain\n$$\n4010^{2} < n < 4011^{2} < \\left(1 + \\frac{1}{2005}\\right) n\n$$\nWe check that $M = 4010^{2}$ will not work. For suppose $n = 4010^{2}$. Then\n$$\n\\left(1 + \\frac{1}{2005}\\right) 4010^{2} = 4010^{2} + 2 \\cdot 4010 = 4011^{2} - 1 < 4011^{2}\n$$\nThus there is no square integer between $n$ and $\\left(1 + \\frac{1}{2005}\\right) n$.\nThis proves (b).",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32348,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with sides $a$, $b$, $c$, circumradius $R$ and inradius $r$.\nProve that\n$$\n\\frac{R}{2r} \\ge \\left( \\frac{64a^2 b^2 c^2}{(4a^2 - (b-c)^2)(4b^2 - (c-a)^2)(4c^2 - (a-b)^2)} \\right)^2\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(by Riddhipratim Basu) We introduce $s - a = x$, $s - b = y$, $s - c = z$, where $s = (a + b + c)/2$. Then $x$, $y$, $z$ are positive, and $a = y + z$, $b = z + x$, $c = x + y$. We may express\n$$\n\\frac{R}{2r} = \\frac{abcs}{4\\Delta^2} = \\frac{(x+y)(y+z)(z+x)}{8xyz}.\n$$\nSimilarly\n$$\n\\begin{align*}\n4a^2 - (b-c)^2 &= (3y+z)(3z+y), & 4b^2 - (c-a)^2 &= (3x+z)(3z+x), \\\\\n4c^2 - (a-b)^2 &= (3x+y)(3y+x).\n\\end{align*}\n$$\nThus we need to prove\n$$\n\\frac{(x+y)(y+z)(z+x)}{8xyz} \\ge \\frac{64^2(x+y)^4(y+z)^4(z+x)^4}{(3x+y)^2(3y+x)^2(3y+z)^2(3z+y)^2(3x+z)^2(3z+x)^2}\n$$\nThis may be written in the form\n$$\n(3x+y)^2(3y+x)^2(3y+z)^2(3z+y)^2(3x+z)^2(3z+x)^2 \\geq 32^3 xyz (x+y)^3 (y+z)^4 (z+x)^3\n$$\nUsing AM-GM inequality, we have\n$$\n\\begin{align*}\n(3x + y)(3y + x) &= 3x^2 + 3y^2 + 10xy \\\\\n&= (x + y)^2 + (x + y)^2 + (x + y)^2 + 4xy \\\\\n&\\geq 4\\{(x + y)^6 \\cdot 4xy}^{1/4}.\n\\end{align*}\n$$\nIt follows that $(3x+y)^2(3y+x)^2 \\geq 32\\sqrt{xy}(x+y)^3$ and similar expressions\nfor the other products. Thus\n$$\n(3x+y)^2(3y+x)^2(3y+z)^2(3z+y)^2(3x+z)^2(3z+x)^2 \\geq 32^3 xyz (x+y)^3 (y+z)^4 (z+x)^3,\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32349,
    "subject": "Mathematics (Multi-modal)",
    "question": "The positive divisors $d_1, d_2, \\dots, d_l$ of a natural number $n$ are arranged in the form\n$$1 = d_1 < d_2 < \\dots < d_l = n.$$ \nSuppose it is known that $d_1^2 + d_{15}^2 = d_{16}^2$. Find all possible values of $d_{17}$.",
    "options": [],
    "answer": "28",
    "solution": "We use the well known fact that given a triple $(a, b, c)$ of integers satisfying $a^2 + b^2 = c^2$, then one of $a, b$ is divisible by $4$; one of $a, b$ is divisible by $3$; and one of $a, b, c$ is divisible by $5$. Thus $4, 3$ and $5$ are divisors of $n$. Let us write\n$$n = 2^a 3^b 5^c t$$\nwhere $\\alpha \\ge 2$ and $t$ is not divisible by $2, 3, 5$. Note that $\\beta \\ge 1, \\gamma \\ge 1$.\nHence\n$$d_1 = 1,\\ d_2 = 2,\\ d_3 = 3,\\ d_4 = 4,\\ d_5 = 5,\\ d_6 = 6.$$ \nMoreover $10 = 2 \\cdot 5$ is also a divisor of $n$. Hence $d_7 \\le 10$. Thus $d_7 \\in \\{7, 8, 9, 10\\}$. We explore each of them.\n\n**Case 1:** Suppose $d_7 = 7$. In this case\n$$\n(d_{16} - d_{15}) (d_{16} + d_{15}) = d_{16}^2 - d_{15}^2 = d_7^2 = 49.\n$$\nIt follows that $d_{16} - d_{15} = 1$ and $d_{16} + d_{15} = 49$. We thus get $d_{16} = 25$, $d_{15} = 24$. Since $d_{15}|n$, we see that $8$ divides $n$. Similarly $25 = 5^2$ also divides $n$. It follows that $\\alpha \\ge 3, \\gamma \\ge 2$. Since $d_7 = 7$ is also a divisor of $n$, we may now write\n$$n = 2^a 3^b 5^c 7^d t_1$$\nwhere $t_i$ is not divisible by $2, 3, 5, 7$; $\\alpha \\ge 3, \\beta \\ge 1, \\gamma \\ge 2, \\delta \\ge 1$.\n\nIf $\\beta \\ge 2$, then $9$ divides $n$ and hence $d_1 = 1, d_2 = 2, d_3 = 3, d_4 = 4, d_5 = 5, d_6 = 6, d_7 = 7, d_8 = 8, d_9 = 9, d_{10} = 10, d_{11} \\le 12, d_{12} \\le 14, d_{13} \\le 15, d_{14} \\le 18, d_{15} \\le 20$. This contradicts $d_{15} = 24$. We conclude that $\\beta = 1$ and hence $n = 2^a \\cdot 3 \\cdot 5^\\gamma \\cdot 7^\\delta t_1$.\n\nIf $\\alpha \\ge 4$, then we have\n$d_1 = 1, d_2 = 2, d_3 = 3, d_4 = 4, d_5 = 5, d_6 = 6, d_7 = 7, d_8 = 8, d_9 = 10, d_{10} \\le 12, d_{11} \\le 14, d_{12} \\le 15, d_{13} \\le 16, d_{14} \\le 20, d_{15} \\le 21$, which again contradicts $d_{15} = 24$. This shows that $\\alpha = 3$ and hence $n = 2^3 3 \\cdot 5^7 7^\\delta t_1$ where $\\text{gcd}(t_1, 210) = 1$.\n\nThus we get\n$d_1 = 1, d_2 = 2, d_3 = 3, d_4 = 4, d_5 = 5, d_6 = 6, d_7 = 7, d_8 = 8, d_9 = 10, d_{10} \\le 12, d_{11} \\le 14, d_{12} \\le 15, d_{13} \\le 20, d_{14} \\le 21, d_{15} \\le 24$.\nSince $d_{15} = 24$ and $d_{16} = 25$, $t_1$ is not divisible by $11, 13, 17, 19$ or $23$. Otherwise $d_{15} < 24$. This shows that $\\gamma \\ge 2$ and $d_{17}$ cannot be equal to $26 = 2 \\cdot 13$ nor be equal to $27 = 3^3$. Since $4$ and $7$ are divisors of $n$, it follows that $28 = 4 \\cdot 7$ also divides $n$. Hence $d_{17} = 28$.\n\n**Case 2:** Suppose $d_7 = 8$. Then\n$$\n(d_{16} - d_{15}) (d_{16} + d_{15}) = 64 = 2 \\times 32 = 4 \\times 16\n$$\nso that $d_{16} = 17, d_{15} = 15$ or $d_{16} = 10, d_{15} = 6$. Since $d_6 = 6$, we can immediately rule out $d_{15} = 6$. If $d_7 = 8$ and $d_{15} = 15$, then\n$$\n8 = d_7 < d_8 < d_9 < d_{10} < d_{11} < d_{12} < d_{13} < d_{14} < d_{15} = 15,\n$$\nwhich is impossible.\n\n**Case 3:** If $d_7 = 9$, then\n$$\n(d_{16} - d_{15}) (d_{16} + d_{15}) = 81 = 1 \\times 81 = 3 \\times 27.\n$$\nWe get $d_{16} = 41, d_{15} = 40$ or $d_{16} = 15, d_{15} = 12$.\nSince\n$$\n9 = d_7 < d_8 < \\dots < d_{15},\n$$\nwe may rule out $d_{15} = 12$. If $d_{15} = 40$, then $40$ divides $n$ and hence $8$ also divides $n$. But $d_6 = 6$ and $d_7 = 9$ and $8$ cannot be a divisor of $n$.\n\n**Case 4:** Suppose $d_7 = 10$. In this case\n$$\n(d_{16} - d_{15}) (d_{16} + d_{15}) = 100 = 2 \\times 50\n$$\nand hence $d_{16} = 26, d_{15} = 24$. This shows $8|n$. But then $d_6 = 6, d_{10} = 10$ is impossible. We conclude that the only possible value is $d_{17} = 28$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32350,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A_1, A_2, \\dots, A_n$ be arithmetic progressions of integers, each of $k$ terms, such that any two of these arithmetic progressions have at least two common elements. Suppose $b$ of these arithmetic progressions have common difference $d_1$ and the remaining arithmetic progressions have common difference $d_2$, where $0 < b < n$. Prove that\n$$\nb \\le 2 \\left( k - \\frac{d_2}{\\text{gcd}(d_1, d_2)} \\right) - 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $[d_1, d_2]$ denote the least common multiple of $d_1$ and $d_2$. Let $P_j$ denote the union of all arithmetic progressions with common difference $d_j$, $j = 1, 2$, and let $S = P_1 \\cap P_2$. Then $S$ is an arithmetic progression with common difference $[d_1, d_2]$. Let $y$ be the least element of $P_1$ and $x$ the least element of $S$.\nSince any two arithmetic progressions with common difference $d_1$ have at least two common elements $b \\le k-1$. If $\\frac{d_2}{\\text{gcd}(d_1, d_2)} \\le \\frac{k}{2}$, then the result follows from $b \\le k-1$. Suppose $\\frac{d_2}{\\text{gcd}(d_1, d_2)} > \\frac{k}{2}$. Let\n$$\nm_0 = 2 \\frac{d_2}{\\text{gcd}(d_1, d_2)} - k.\n$$\nEach arithmetic progression with common difference $d_1$ contains at least 2 elements of $S$ and starts at one of the points $y+md_1$, $0 \\le m \\le k-2$. However an arithmetic progression with common difference $d_1$ starting at one of the points $x+d_1, x+2d_1, \\dots, x+m_0d_1$ contains only one point of $S$ namely $x+[d_1, d_2]$. Hence\n$$\nb \\le k-1 - m_0 = 2 \\left( k - \\frac{d_2}{\\text{gcd}(d_1, d_2)} \\right) - 1.\n$$",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32351,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $AB = AC$, and let $\\Gamma$ be its circumcircle. Suppose the incircle $\\gamma$ of $ABC$ moves (slides) on $BC$ in the direction of $B$. Prove that when $\\gamma$ touches $\\Gamma$ internally, it also touches the altitude through $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\gamma'$ be the position of $\\gamma$, when it touches $\\Gamma$ internally, and let $K$ be its centre. Let $O$ be the circumcentre and $I$ be the in-centre of $ABC$. Since $AB = AC$, both of these lie on the altitude $AD$. If $T$ is the point of contact of $\\Gamma$ and $\\gamma'$, then $T, K, O$ are collinear. Hence $OK = OT - KT = R - r$, where $R$ and $r$ are respectively the circumradius and inradius of $ABC$. Note that $K$ and $I$ are at same distance $r$ from $BC$. Thus $KI$ is perpendicular to $AD$ at $I$. Using the right-angled triangle $OKI$, we have $OK^2 = OI^2 + IK^2$. But $OI^2 = R^2 - 2Rr$. Hence we obtain\n\n![](attached_image_1.png)\n\n$$\nIK^2 = OK^2 - OI^2 = (R-r)^2 - (R^2 - 2Rr) = r^2.\n$$\n\nThus $IK = r$ showing that $\\gamma'$ touches $AD$ at $I$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32352,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a triangle $ABC$ right-angled at $C$, the median through $B$ bisects the angle between $BA$ and the bisector of $\\angle B$. Prove that\n$$\n\\frac{5}{2}<\\frac{AB}{BC}<3\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $E$ is the mid-point of $AC$, we have $AE = EC = b/2$. Since $BD$ bisects $\\angle ABC$, we also know that $CD = ab/(a+c)$. Since $BE$ bisects $\\angle ABD$, we also have\n$$\n\\frac{BD^{2}}{BA^{2}} = \\frac{DE^{2}}{EA^{2}}\n$$\nHowever,\n$$\n\\begin{aligned}\nBD^{2} & = BC^{2} + CD^{2} = a^{2} + \\frac{a^{2}b^{2}}{(a+c)^{2}} \\\\\nDE^{2} & = \\left(\\frac{b}{2} - \\frac{ab}{a+c}\\right)^{2}\n\\end{aligned}\n$$\n![](attached_image_1.png)\nUsing these in the above expression and simplifying, we get\n$$\na^{2}\\left\\{(a+c)^{2}+b^{2}\\right\\}=c^{2}(c-a)^{2}\n$$\nUsing $c^{2}=a^{2}+b^{2}$ and eliminating $b$, we obtain\n$$\nc^{3}-2ac^{2}-a^{2}c-2a^{3}=0\n$$\nIntroducing $t=c/a$, this reduces to a cubic equation;\n$$\nt^{3}-2t^{2}-t-2=0\n$$\nConsider the function $f(t)=t^{3}-2t^{2}-t-2$ for $t>0$ (as $c/a$ is positive). For $0<t \\leq 2$, we see that $f(t)=t^{2}(t-2)-t-2<0$. We also observe that $f(t)=(t-2)(t^{2}-1)-4$ is strictly increasing on $(2, \\infty)$. It is easy to compute\n$$\nf(5/2)=-\\frac{11}{8}<0, \\quad \\text{and} \\quad f(3)=4>0\n$$\nHence there is a unique value of $t$ in the interval $(5/2,3)$ such that $f(t)=0$. We conclude that\n$$\n\\frac{5}{2}<\\frac{c}{a}<3\n$$\nSolution:\n\nLet us take $\\angle B/4=\\theta$. Then $\\angle EBC=\\angle DBE=\\theta$ and $\\angle CBD=2\\theta$. Using sine rule in triangles $BEA$ and $BEC$, we get\n$$\n\\begin{aligned}\n\\frac{BE}{\\sin A} & = \\frac{AE}{\\sin \\theta} \\\\\n\\frac{BE}{\\sin 90^{\\circ}} & = \\frac{CE}{\\sin 3\\theta}\n\\end{aligned}\n$$\nSince $AE=CE$, we obtain $\\sin 3\\theta \\sin A=\\sin \\theta$. However $A=90^{\\circ}-4\\theta$. Thus we get $\\sin 3\\theta \\cos 4\\theta=\\sin \\theta$. Note that\n$$\n\\frac{c}{a}=\\frac{1}{\\cos 4\\theta}=\\frac{\\sin 3\\theta}{\\sin \\theta}=3-4\\sin^{2}\\theta\n$$\nThis shows that $c/a<3$. Using $c/a=3-4\\sin^{2}\\theta$, it is easy to compute $\\cos 2\\theta=((c/a)-1)/2$. Hence\n$$\n\\frac{a}{c}=\\cos 4\\theta=\\frac{1}{2}\\left(\\frac{c}{a}-1\\right)^{2}-1\n$$\nSuppose $c/a \\leq 5/2$. Then $((c/a)-1)^{2} \\leq 9/4$ and $a/c \\geq 2/5$. Thus\n$$\n\\frac{2}{5} \\leq \\frac{a}{c}=\\frac{1}{2}\\left(\\frac{c}{a}-1\\right)^{2}-1 \\leq \\frac{9}{8}-1=\\frac{1}{8}\n$$\nwhich is absurd. We conclude that $c/a>5/2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32353,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integer solutions $(x, y)$ of the equation $y^2 = x^3 - p^2x$, where $p$ is a prime such that $p \\equiv 3 \\pmod{4}$.",
    "options": [],
    "answer": "(x, y) = (0, 0), (p, 0), (-p, 0)",
    "solution": "$$\ny^2 = (x-p)(x)(x+p).\n$$\nWe split the analysis into the following cases.\n\nCase $p \\nmid y$. In this case, we have $(x-p,x) = (x,x+p) = 1$. If $x$ is even, then $(x-p,x+p) = 1$, so it follows that $x-p,x$ and $x+p$ are all squares. This is not possible since $x+p \\equiv 3 \\pmod{8}$. Thus, $x$ has to be odd. In this case we have $(x-p,x+p) = 2$. So we get\n$$\n\\begin{aligned}\nx &= r^2, \\\\\n(x-p) &= 2s^2, \\\\\n(x+p) &= 2t^2.\n\\end{aligned}\n$$\nThe last two equations imply $(t^2 - s^2) = p$ solving which we get $t = (p+1)/2$ and $s = (p-1)/2$. Now,\n$$\n\\begin{aligned}\nr^2 &= x = (x-p)+p \\\\\n&= (p-1)^2/2 + p \\\\\n&= (p^2+1)/2 \\\\\n&= ((8k+3)^2 + 1)/2 \\quad (\\text{here } p = 8k+3) \\\\\n&= (64k^2 + 48k + 10)/2 \\equiv 5 \\pmod{8}\n\\end{aligned}\n$$\nwhich is not possible. Thus there are no solutions in case $p \\nmid y$.\n\nCase $p \\mid y$. We first note that if $y=0$ then $x=0, p$ or $-p$, and this gives us three solutions. Now, assume that $y \\neq 0$. Since $p \\mid y$ it follows that $(x-p,x,x+p) = p$. Thus, $p^2 \\mid y$. Canceling $p^3$ from both the sides, we have\n$$\npb^2 = (a-1)(a)(a+1),\n$$\nwhere $a = x/p$ and $b = y/p^2$. Note that $a-1, a$ and $a+1$ are all positive since $y \\neq 0$. Now, $p$ divides exactly one of $(a-1), a$ and $(a+1)$. If $p$ divides $(a-1)$ then, $a$ and $(a+1)$ have to be squares, which is not possible since $a \\neq 0$. So, $p \\nmid (a-1)$. Similarly, $p \\nmid (a+1)$. Thus, it should be the case that $p \\mid a$.\nIf $a \\neq 0$ is even, then $(a-1), a/p$ and $(a+1)$ are relatively prime to each other, so they all have to be squares. But it is not possible that $(a-1)$ and $(a+1)$ are both squares. Thus, $a$ should be odd.\nNow, $(a-1, a+1) = 2$, so $(a-1) = 2r^2$ and $(a+1) = 2s^2$. This gives $s^2 - r^2 = 1$ which implies $s=1$ and $r=0$. This gives $a=1$ which is not divisible by $p$. Thus there are no solutions in this case.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32354,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ satisfying the equation\n$$\nf(x+y) + f(x)f(y) = (1+y)f(x) + (1+x)f(y) + f(xy), \\quad (1)\n$$\nfor all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "f(x) ≡ 0; f(x) = x^2 + x; f(x) = 3x",
    "solution": "Taking $x = y = 0$ in (1), we obtain $f(0)^2 = 2f(0)$ and hence $f(0) = 0$ or $2$. If $f(0) = 2$, then $y = 0$ in (1) gives that $f(x) = 2 + x$ for all $x$. Putting this in (1), we get $2xy = 0$ for all $x, y$. This contradiction shows that there are no solutions with $f(0) = 2$. Thus $f(0) = 0$.\nPutting $x = y = 1$ in (1), we have $f(2) + f(1)^2 = 5f(1)$. Similarly, $x = y = 2$ gives $f(2)^2 = 6f(2)$. If we take $f(2) = 0$, then $f(1) = 0$ or $5$. If $f(2) = 6$, then $f(1) = 2$ or $3$. Thus we have to consider 4 cases: $(f(2), f(1)) = (0, 0), (0, 5), (6, 2)$ and $(6, 3)$.\n\n**Case 1:** $(f(2), f(1)) = (0, 0)$.\nTaking $y = 1$ in (1), we obtain\n$$\nf(x+1) = 3f(x). \\quad (2)\n$$\nReplacing $x$ by $x+1$, $y$ by $2$ in (1) and using $f(2) = 0$, we also get\n$$\nf(x+3) = 3f(x+2) + f(2x+2),\n$$\nfor all $x$. This simplifies, in view of (2), to $f(2x) = 0$ for all $x$. Thus $f(x) \\equiv 0$ is a solution.\n\nAgain $y = 1$ in (1) gives\n$$\nf(x+1)+2f(x)=5(x+1), \\quad (3)\n$$\nfor all $x$. Taking $y = 2$ in (1) and using $f(2) = 0$, we also obtain $f(x+2) = 3f(x)+f(2x)$, for all $x$. Writing $x+2 = (x+1)+1$ and using (3) repeatedly, we get\n$$\nf(2x) = f(x) - 5x. \\quad (4)\n$$\nPutting $y = x$ in (1), we also get $f(2x) + f(x)^2 = 2(x+1)f(x) + f(x^2)$. Substituting for $f(2x)$ from (4), we get\n$$\nf(x)^2 - f(x^2) = 2xf(x) + f(x) + 5x,\n$$\nfor all $x$. Here we change $x$ to $2x$ to get\n$$\nf(2x)^2 - f(4x^2) = 4xf(2x) + f(2x) + 10x.\n$$\nUsing (4) repeatedly, this simplifies to $12xf(2x) = 0$ for all $x$. This implies that $f(2x) = 0$ for all $x \\neq 0$. Taking $x = 1/2$, we obtain $f(1) = 0$, a contradiction. Thus there is no solution in this case.\n\n**Case 2:** $(f(2), f(1)) = (0, 5)$.\nAgain $y = 1$ in (1) gives\n$$\nf(x+1) + 5f(x) = 5(x+1) + f(x),\n$$\nwhich simplifies to $f(x+1) = 5(x+1)$. This implies $f(x) = 5x$ for all $x$. Substituting into (1), we get $5(x+y) + 25xy = (1+y)5x + (1+x)5y + 5xy$, which simplifies to $5(x+y) + 25xy = 5x + 5xy + 5y + 5xy + 5xy$, or $5(x+y) + 25xy = 5(x+y) + 15xy$, which is $25xy = 15xy$, so $10xy = 0$ for all $x, y$, which is not possible. Thus, there is no solution in this case.\n\n**Case 3:** $(f(2), f(1)) = (6, 2)$.\nAgain $y = 1$ in (1) gives\n$$\nf(x+1) = f(x) + 2(x+1).\n$$\nReplacing $y$ by $2$ in (1), we also obtain\n$$\nf(x+2)+6f(x) = 3f(x) + 6(x+1) + f(2x).\n$$\nUsing the expression for $f(x+1)$, this reduces to\n$$\n4f(x) = 2x + f(2x), \\quad (5)\n$$\nfor all $x \\in \\mathbb{R}$. Taking $y = x$ in (1), we also get $f(2x)+f(x)^2 = 2(x+1)f(x)+f(x^2)$. This reduces to\n$$\nf(x)^2 - f(x^2) = 2xf(x) - 2f(x) + 2x.\n$$\nReplacing $x$ by $2x$ and repeatedly using (5), we obtain $f(x) = x^2+x$ for all $x$. It is easy to check that this is a solution.\n\n**Case 4:** $(f(2), f(1)) = (6, 3)$ and $f(2) = 6$.\nAs in the earlier cases, we put $y = 1$ in (1) to get\n$$\nf(x+1) = 3(x+1)\n$$\nfor all real $x$. This implies that $f(x) = 3x$, which is in fact a solution.\nThus there are three solutions: $f(x) \\equiv 0$; $f(x) = x^2+x$; and $f(x) = 3x$.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32355,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider the quadratic polynomial $p(x) = x^2 + ax + b$, where $a, b$ are in the interval $[-2, 2]$. Find the range of the real roots of $p(x) = 0$, as $a$ and $b$ vary over $[-2, 2]$.",
    "options": [],
    "answer": "[-1 - sqrt(3), 1 + sqrt(3)]",
    "solution": "Suppose $a, b \\in [-2, 2]$ and $\\alpha$ is a real root of $x^2 + ax + b = 0$. Then\n$$\n\\alpha = \\frac{-a \\pm \\sqrt{a^2 - 4b}}{2},\n$$\nwhich shows that\n$$\n\\alpha \\leq \\frac{2 + \\sqrt{4 + 8}}{2} = 1 + \\sqrt{3}.\n$$\n(Take $a = -2$, $b = -2$.) Similarly, we see that $\\alpha \\ge -1 - \\sqrt{3}$, by taking $a = 2$ and $b = -2$. Take any real number $\\lambda$ such that $|\\lambda| \\le 1$. We have\n$$\n(\\lambda\\alpha)^2 + (\\lambda a)(\\lambda\\alpha) + \\lambda^2 b = \\lambda^2 (\\alpha^2 + a\\alpha + b) = 0.\n$$\nThus $\\lambda\\alpha$ is also a root of the equation $x^2 + cx + d$, where\n$c = \\lambda a, \\ d = \\lambda^2 b.$\nNote that $|c| \\le |a|$ and $|d| \\le |b|$. Thus for each real root $\\alpha$ of $x^2 + ax + b = 0$ and each real $\\lambda$ with $|\\lambda| \\le 1$, we see that $\\lambda\\alpha$ is also a root of $x^2 + cx + d$, where $c, d \\in [-2, 2]$. Since $1 + \\sqrt{3}$ is a root of $x^2 - 2x - 2 = 0$, it follows that $\\lambda(1 + \\sqrt{3})$ is in the range for any $\\lambda$ with $|\\lambda| \\le 1$. Thus the range is $[-1 - \\sqrt{3}, 1 + \\sqrt{3}]$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32356,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a natural number such that $n = a^{2} + b^{2} + c^{2}$, for some natural numbers $a, b, c$. Prove that\n$$\n9 n = (p_{1} a + q_{1} b + r_{1} c)^{2} + (p_{2} a + q_{2} b + r_{2} c)^{2} + (p_{3} a + q_{3} b + r_{3} c)^{2}\n$$\nwhere $p_{j}$'s, $q_{j}$'s, $r_{j}$'s are all nonzero integers. Further, if $3$ does not divide at least one of $a, b, c$, prove that $9 n$ can be expressed in the form $x^{2} + y^{2} + z^{2}$, where $x, y, z$ are natural numbers none of which is divisible by $3$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIt can be easily seen that\n$$\n9 n = (2 b + 2 c - a)^{2} + (2 c + 2 a - b)^{2} + (2 a + 2 b - c)^{2}\n$$\nThus we can take $p_{1} = p_{2} = p_{3} = 2$, $q_{1} = q_{2} = q_{3} = 2$ and $r_{1} = r_{2} = r_{3} = -1$. Suppose $3$ does not divide $\\gcd(a, b, c)$. Then $3$ does divide at least one of $a, b, c$; say $3$ does not divide $a$. Note that each of $2 b + 2 c - a$, $2 c + 2 a - b$ and $2 a + 2 b - c$ is either divisible by $3$ or none of them is divisible by $3$, as the difference of any two sums is always divisible by $3$. If $3$ does not divide $2 b + 2 c - a$, then we have the required representation. If $3$ divides $2 b + 2 c - a$, then $3$ does not divide $2 b + 2 c + a$. On the other hand, we also note that\n$$\n9 n = (2 b + 2 c + a)^{2} + (2 c - 2 a - b)^{2} + (-2 a + 2 b - c)^{2} = x^{2} + y^{2} + z^{2}\n$$\nwhere $x = 2 b + 2 c + a$, $y = 2 c - 2 a - b$ and $z = -2 a + 2 b - c$. Since $x - y = 3(b + a)$ and $3$ does not divide $x$, it follows that $3$ does not divide $y$ as well. Similarly, we conclude that $3$ does not divide $z$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32357,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $w_a, w_b, w_c$ be the lengths of the internal angle bisectors of a triangle $ABC$ with sides $a, b, c$.\nLet $R$ be its circum-radius. Prove that\n$$\n\\frac{b^2 + c^2}{w_a} + \\frac{c^2 + a^2}{w_b} + \\frac{a^2 + b^2}{w_c} > 4R.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We use the standard\n$$\nw_a = \\frac{2bc \\cos(A/2)}{(b+c)}, \\text{ etc.}\n$$\n\nThe inequality takes the form\n$$\n\\sum_{\\text{cyclic}} \\frac{(b^2 + c^2)(b+c)}{4Rbc \\cos(A/2)} > 2.\n$$\nThis may be put in the form\n$$\n\\sum_{\\text{cyclic}} \\frac{(b^2 + c^2)(b+c) \\sin(A/2)}{2abc} > 1.\n$$\nBut note that $b^2+c^2 \\ge 2bc$ and $b+c > a$. Hence it is sufficient to prove that $\\sum_{\\text{cyclic}} \\sin(A/2) > 1$. We start with the identity\n$$\n\\sum_{\\text{cyclic}} \\cos A = 1 + 4 \\prod_{\\text{cyclic}} \\sin(A/2),\n$$\nwhich shows that $\\sum_{\\text{cyclic}} \\cos A > 1$, in any triangle $ABC$. But, whenever $A, B, C$ are the angles of a triangle, we know that $(\\pi - A)/2, (\\pi - B)/2, (\\pi - C)/2$ are also the angles of some other triangle. For this triangle, we get\n$$\n\\sum_{\\text{cyclic}} \\cos \\left( \\frac{\\pi - A}{2} \\right) > 1.\n$$\nIt follows that\n$$\n\\sum_{\\text{cyclic}} \\sin \\frac{A}{2} > 1,\n$$\nwhich is to be proved.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32358,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $m$ and $n$ be positive integers such that the equation $x^{2}-m x+n=0$ has real roots $\\alpha$ and $\\beta$. Prove that $\\alpha$ and $\\beta$ are integers if and only if $[m \\alpha]+[m \\beta]$ is the square of an integer. (Here $[x]$ denotes the largest integer not exceeding $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nIf $\\alpha$ and $\\beta$ are both integers, then\n$$\n[m \\alpha]+[m \\beta]=m \\alpha+m \\beta=m(\\alpha+\\beta)=m^{2}\n$$\nThis proves one implication.\n\nObserve that $\\alpha+\\beta=m$ and $\\alpha \\beta=n$. We use the property of integer function: $x-1<[x] \\leq x$ for any real number $x$. Thus\n$$\nm^{2}-2=m(\\alpha+\\beta)-2=m \\alpha-1+m \\beta-1<[m \\alpha]+[m \\beta] \\leq m(\\alpha+\\beta)=m^{2}.\n$$\nSince $m$ and $n$ are positive integers, both $\\alpha$ and $\\beta$ must be positive. If $m \\geq 2$, we observe that there is no square between $m^{2}-2$ and $m^{2}$. Hence, either $m=1$ or $[m \\alpha]+[m \\beta]=m^{2}$. If $m=1$, then $\\alpha+\\beta=1$ implies that both $\\alpha$ and $\\beta$ are positive reals smaller than 1. Hence $n=\\alpha \\beta$ cannot be a positive integer. We conclude that $[m \\alpha]+[m \\beta]=m^{2}$.\n\nPutting $m=\\alpha+\\beta$ in this relation, we get\n$$\n\\left[\\alpha^{2}+n\\right]+\\left[\\beta^{2}+n\\right]= (\\alpha+\\beta)^{2}\n$$\nUsing $[x+k]=[x]+k$ for any real number $x$ and integer $k$, this reduces to\n$$\n\\left[\\alpha^{2}\\right]+\\left[\\beta^{2}\\right]=\\alpha^{2}+\\beta^{2}\n$$\nThis shows that $\\alpha^{2}$ and $\\beta^{2}$ are both integers. On the other hand,\n$$\n\\alpha^{2}-\\beta^{2}=(\\alpha+\\beta)(\\alpha-\\beta)=m(\\alpha-\\beta)\n$$\nThus\n$$\n(\\alpha-\\beta)=\\frac{\\alpha^{2}-\\beta^{2}}{m}\n$$\nis a rational number. Since $\\alpha+\\beta=m$ is a rational number, it follows that both $\\alpha$ and $\\beta$ are rational numbers. However, both $\\alpha^{2}$ and $\\beta^{2}$ are integers. Hence each of $\\alpha$ and $\\beta$ is an integer.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32359,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a_1, a_2, \\dots, a_n$ be an arbitrary arrangement of numbers $1, 2, \\dots, n$ on a circle. Find\n$$\n\\min \\sum_{j=1}^{n} |a_j - a_{j+1}| \\quad \\text{and} \\quad \\max \\sum_{j=1}^{n} |a_j - a_{j+1}|,\n$$\nwhere $a_{n+1} = a_1$ and the extrema are taken over all possible arrangements of $1, 2, \\dots, n$.",
    "options": [],
    "answer": "Minimum: 2(n − 1). Maximum: n^2/2 if n is even; (n^2 − 1)/2 if n is odd.",
    "solution": "**Minimum:** Consider $1$ and $n$ on the circle. They divide the circle into two arcs. The sum of the numbers on either of the arc is at least $n-1$. Suppose for example the numbers $1 = b_1, b_2, \\dots, b_k = n$ appear on one of the arcs between $1$ and $n$, in that order. Then the sum of absolute differences of adjacent numbers on this arc is\n$$\n|1 - b_2| + |b_2 - b_3| + \\dots + |b_{k-1} - n| \\geq |1 - n| = n - 1.\n$$\nSimilarly, the least sum of absolute differences on the other arc is also $n-1$. Hence we see that\n$$\n\\sum_{j=1}^{n} |a_{j+1} - a_{j}| \\geq 2(n-1).\n$$\nThis is achieved by the permutation $(a_1, a_2, \\dots, a_n)$, where $a_j = j$ for $1 \\le j \\le n$.\n\n**Maximum:** We have\n$$\n\\sum_{j=1}^{n} |a_{j+1} - a_{j}| = \\sum_{j=1}^{n} \\pm (a_{j+1} - a_{j}).\n$$\nEach of the numbers $1, 2, \\dots, n$ appear in the right side sum twice. Hence to get a maximum sum we should have positive sign to larger numbers in both occurrences and smaller number should go with negative sign. Thus $n, (n-1), (n-2), \\dots, [n/2]$ should get positive signs and $1, 2, 3, \\dots, [n/2] - 1$ should get negative signs. This happens when:\nfor even $n$, the arrangement is $1, n, 2, (n-1), 3, (n-2), \\dots, n/2, (n/2) + 1$;\nfor odd $n$, the arrangement is $1, n, 2, (n-1), 3, (n-2), \\dots, [n/2] + 2, [n/2] + 1$.\nThe corresponding sums are:\n$$\n\\frac{n^2}{2} \\text{ when } n \\text{ is even, } \\quad \\frac{n^2-1}{2} \\text{ when } n \\text{ is odd.}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32360,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\sigma=\\left(a_{1}, a_{2}, a_{3}, \\ldots, a_{n}\\right)$ be a permutation of $(1,2,3, \\ldots, n)$. A pair $\\left(a_{i}, a_{j}\\right)$ is said to correspond to an inversion of $\\sigma$, if $i<j$ but $a_{i}>a_{j}$. (Example: In the permutation $(2,4,5,3,1)$, there are 6 inversions corresponding to the pairs $(2,1)$, $(4,3),(4,1),(5,3),(5,1),(3,1)$.) How many permutations of $(1,2,3, \\ldots n)$, $(n \\geq 3)$, have exactly two inversions?",
    "options": [],
    "answer": "((n+1)(n-2))/2",
    "solution": "Solution:\nIn a permutation of $(1,2,3, \\ldots, n)$, two inversions can occur in only one of the following two ways:\n(A) Two disjoint consecutive pairs are interchanged:\n$$\n\\begin{aligned}\n& (1,2,3, \\ldots, j-1, j, j+1, \\ldots, k-1, k, k+1, k+2, \\ldots, n) \\\\\n& \\quad \\longrightarrow (1,2, \\ldots, j-1, j+1, j, j+2, \\ldots, k-1, k+1, k, k+2, \\ldots, n)\n\\end{aligned}\n$$\n(B) Each block of three consecutive integers can be permuted in any of the following 2 ways:\n$$\n\\begin{aligned}\n& (1,2,3, \\ldots, k, k+1, k+2, \\ldots, n) \\longrightarrow (1,2, \\ldots, k+2, k, k+1, \\ldots, n) \\\\\n& (1,2,3, \\ldots, k, k+1, k+2, \\ldots, n) \\longrightarrow (1,2, \\ldots, k+1, k+2, k, \\ldots, n)\n\\end{aligned}\n$$\nConsider case (A). For $j=1$, there are $n-3$ possible values of $k$; for $j=2$, there are $n-4$ possibilities for $k$ and so on. Thus the number of permutations with two inversions of this type is\n$$\n1+2+\\cdots+(n-3)=\\frac{(n-3)(n-2)}{2}\n$$\nIn case (B), we see that there are $n-2$ permutations of each type, since $k$ can take values from $1$ to $n-2$. Hence we get $2(n-2)$ permutations of this type.\nFinally, the number of permutations with two inversions is\n$$\n\\frac{(n-3)(n-2)}{2}+2(n-2)=\\frac{(n+1)(n-2)}{2}\n$$",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32361,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that in a non-equilateral triangle, the following are equivalent:\n(a) the angles of the triangle are in arithmetic progression;\n(b) the common tangent to the nine-point circle and the in-circle is parallel to the Euler line.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A$, $B$, $C$ be the vertices, $I$ the in-center, $H$ the ortho-centre, $O$ the circum-center and $N$ the nine-point center of the triangle.\n\n**Claim 1:** (b) $\\Leftrightarrow IH = IO$ (for a non-equilateral triangle).\n\nThe line $NI$ joins the centre of nine-point circle and the centre of in-circle, so $NI$ is perpendicular to the common tangent to in-circle and nine-point circle (we use $N \\neq I$, as triangle is not equilateral). Thus (b) is equivalent to the statement that $NI$ is perpendicular to the Euler-line.\n\nSince $O$, $H$, $N$ are on the Euler-line and $N$ is the midpoint of $OH$, we have:\n(b) $\\Leftrightarrow NI$ is the perpendicular bisector of $OH$.\n\nBut the perpendicular bisector of $O$, $H$ is simply the locus of all points $P$ such that $OP = HP$. Thus (b) $\\Leftrightarrow IH = IO$.\n\n**Claim 2:** $IH = IO \\Leftrightarrow AH = AO$ or $A$, $H$, $O$, $I$ are conicyclic.\n\nBecause $H$ and $O$ are isogonal conjugates $\\angle HAI = \\angle OAH$. Also $AI$ is common. Thus the condition $AH = AO$ gives that $\\triangle AHI$ and $\\triangle OHI$ are congruent. This gives $IH = IO$.\n\nIf on the other hand $A$, $H$, $O$, $I$ are con-cyclic, then $IH$ and $IO$ subtend the same angle at $A$. This implies that $IH = IO$.\n\nThus $AH = AO$ or $A$, $H$, $O$, $I$ con-cyclic implies that $IH = IO$.\n\nConversely, suppose $IH = IO$. Then $\\frac{IH}{AI} = \\frac{IO}{AI}$. This implies that\n$$\n\\frac{\\sin \\angle HAI}{\\sin \\angle AHI} = \\frac{\\sin \\angle OAI}{\\sin \\angle AOI}\n$$\nIt follows that $\\sin \\angle AHI = \\sin \\angle AOI$ since $AI$ bisects $\\angle HAO$. This tells that $\\angle AHI + \\angle AOI = \\pi$ or $\\angle AHI = \\angle AOI$. In the first case $A$, $H$, $O$, $I$ are conicyclic. In the second case $AH = AO$.\n\nThus we have shown that $IH = IO$ if and only if $A$, $H$, $O$, $I$ are con-cyclic or $AH = AO$.\n\n**Claim 3:** $IH = IO$ if and only if $AH = AO$, or $BH = BO$, or $CH = CO$\n\nSuppose $IH = IO$. Clearly the circum-circle of $\\triangle HOI$ does not pass through all of $A$, $B$, $C$, because then $O$ would lie on the circum-circle of $\\triangle A$, $B$, $C$. Without loss of generality, assume that it does not pass through $A$. Then $AH = AO$, because $IH = IO$ and $A$, $H$, $O$, $I$ are not conicyclic. (In the other two cases, we have $BH = BO$ or $CH = CO$ depending on whether the circum-circle does not pass through $B$ or $C$ respectively.) The other part follows directly from claim 2.\n\n**Claim 4:** $AH = AO$, $BH = BO$ or $CH = CO$ if and only if angles of $ABC$ are in arithmetic progression.\n\nLet $R$ be the circum-radius so that $AO = BO = CO = R$. Then $AH = 2R\\cos A$, $BH = 2R\\cos B$, and $CH = 2R\\cos C$. Thus $AH = AO \\Leftrightarrow A = \\frac{\\pi}{3}$; $BH = BO \\Leftrightarrow B = \\frac{\\pi}{3}$; $CH = CO \\Leftrightarrow C = \\frac{\\pi}{3}$.\n\nNow $AH = AO$, $BH = BO$ or $CH = CO$ implies that one of the angles of triangle $ABC$ is $\\pi/3$. But clearly, one angle is $\\pi/3$ if and only if angles are in arithmetic progression.\n\nThus (b) $\\Leftrightarrow IH = IO$ (by claim 1) $\\Leftrightarrow AH = AO$, $BH = BO$, or $CH = CO$ (by claim 3) $\\Leftrightarrow$ angles are in arithmetic progression which is (a).",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32362,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle in which $AB = AC$. Let $D$ be the mid-point of $BC$ and $P$ be a point on $AD$. Suppose $E$ is the foot of perpendicular from $P$ on $AC$. If $\\frac{AP}{PD} = \\frac{BP}{PE} = \\lambda$, $\\frac{BD}{AD} = m$ and $z = m^2(1+\\lambda)$, prove that\n$$\nz^2 - (\\lambda^3 - \\lambda^2 - 2)z + 1 = 0\n$$\nHence show that $\\lambda \\geq 2$ and $\\lambda = 2$ if and only if $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $AD = h$, $PD = y$ and $BD = DC = a$. We\n![](attached_image_1.png)\nobserve that $BP^2 = a^2 + y^2$. Moreover,\n$PE = PA \\sin \\angle DAC = (h - y) \\frac{DC}{AC} = \\frac{a(h - y)}{b}$,\nwhere $b = AC = AB$. Using $AP / PD = (h - y) / y$, we obtain $y = h / (1 + \\lambda)$. Thus\n$$\n\\lambda^2 = \\frac{BP^2}{PE^2} = \\frac{(a^2 + y^2) b^2}{(h - y)^2 a^2}\n$$\nBut $(h - y) = \\lambda y = \\lambda h / (1 + \\lambda)$ and $b^2 = a^2 + h^2$. Thus we obtain\n$$\n\\lambda^4 = \\frac{(a^2(1 + \\lambda)^2 + h^2)(a^2 + h^2)}{a^2 h^2}\n$$\nUsing $m = a / h$ and $z = m^2(1 + \\lambda)$, this simplifies to\n$$\nz^2 - z(\\lambda^3 - \\lambda^2 - 2) + 1 = 0\n$$\nDividing by $z$, this gives\n$$\nz + \\frac{1}{z} = \\lambda^3 - \\lambda^2 - 2\n$$\nHowever $z + (1 / z) \\geq 2$ for any positive real number $z$. Thus $\\lambda^3 - \\lambda^2 - 4 \\geq 0$. This may be written in the form $(\\lambda - 2)(\\lambda^2 + \\lambda + 2) \\geq 0$. But $\\lambda^2 + \\lambda + 2 > 0$. (For example, one may check that its discriminant is negative.) Hence $\\lambda \\geq 2$. If $\\lambda = 2$, then $z + (1 / z) = 2$ and hence $z = 1$. This gives $m^2 = 1 / 3$ or $\\tan (A / 2) = m = 1 / \\sqrt{3}$. Thus $A = 60^\\circ$ and hence $ABC$ is equilateral.\n\nConversely, if triangle $ABC$ is equilateral, then $m = \\tan (A / 2) = 1 / \\sqrt{3}$ and hence $z = (1 + \\lambda) / 3$. Substituting this in the equation satisfied by $z$, we obtain\n$$\n(1 + \\lambda)^2 - 3(1 + \\lambda)(\\lambda^3 - \\lambda^2 - 2) + 9 = 0\n$$\nThis may be written in the form $(\\lambda - 2)(3\\lambda^3 + 6\\lambda^2 + 8\\lambda + 8) = 0$. Here the second factor is positive because $\\lambda > 0$. We conclude that $\\lambda = 2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32363,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $X$ be the set of all bijective functions from the set $S = \\{1, 2, 3, ..., n\\}$ to itself. For each $f \\in X$, define\n$$\nT_f(j) = \\begin{cases} 1, & \\text{if } f^{(12)}(j) = j, \\\\ 0, & \\text{otherwise.} \\end{cases}\n$$\n\nDetermine\n$$ \\sum_{f \\in X} \\sum_{j=1}^{n} T_f(j). $$\n(Here $f^{(k)}(x) = f(f^{(k-1)}(x))$ for $k \\ge 2$.)",
    "options": [],
    "answer": "t(n) · n!, where t(n) is the number of positive divisors of 12 that are ≤ n (so t(n)=6 for n≥12; for n=1..11, t(n)=1,2,3,4,4,5,5,5,5,5,5 respectively).",
    "solution": "Suppose $n \\ge 12$. The elements of $X$ are permutations of $\\{1, 2, 3, ..., n\\}$. If $j$ belongs to a $k$-cycle of $f$, where $k$ is a divisor of $12$, then $f^{(k)}(j) = j$. The number of divisors of $12$ is $6$. Let $m(j,k)$ be the number of elements $f$ of $T$ in which $j$ belongs to a $k$-cycle of $f$, $1 \\le j \\le n$. Then\n$$\nm(j,k) = (n-1)(n-2)\\cdots(n-k+1)(n-k)! = (n-1)!\n$$\nThe required sum is, therefore, $n(n-1)! \\times 8 = 6(n!)$ If $n < 12$, the answer is $t(n)(n!)$, where $t(n)$ is the number of positive divisors of $12$ which are less than or equal to $n$. Thus, they are $1, 2(2!), 3(3!), 4(4!), 4(5!), 5(6!), 5(7!), 5(8!), 5(9!), 5(10!), 5(11!)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32364,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nIf $x, y, z$ are positive real numbers, prove that\n$$\n(x+y+z)^{2}(y z+z x+x y)^{2} \\leq 3\\left(y^{2}+y z+z^{2}\\right)\\left(z^{2}+z x+x^{2}\\right)\\left(x^{2}+x y+y^{2}\\right)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nWe begin with the observation that\n$$\nx^{2}+x y+y^{2}=\\frac{3}{4}(x+y)^{2}+\\frac{1}{4}(x-y)^{2} \\geq \\frac{3}{4}(x+y)^{2}\n$$\nand similar bounds for $y^{2}+y z+z^{2}, z^{2}+z x+x^{2}$. Thus\n$$\n3\\left(x^{2}+x y+y^{2}\\right)\\left(y^{2}+y z+z^{2}\\right)\\left(z^{2}+z x+x^{2}\\right) \\geq \\frac{81}{64}(x+y)^{2}(y+z)^{2}(z+x)^{2}\n$$\nThus it is sufficient to prove that\n$$\n(x+y+z)(x y+y z+z x) \\leq \\frac{9}{8}(x+y)(y+z)(z+x)\n$$\nEquivalently, we need to prove that\n$$\n8(x+y+z)(x y+y z+z x) \\leq 9(x+y)(y+z)(z+x)\n$$\nHowever, we note that\n$$\n(x+y)(y+z)(z+x)=(x+y+z)(y z+z x+x y)-x y z\n$$\nThus the required inequality takes the form\n$$\n(x+y)(y+z)(z+x) \\geq 8 x y z\n$$\nThis follows from AM-GM inequalities;\n$$\nx+y \\geq 2 \\sqrt{x y}, \\quad y+z \\geq 2 \\sqrt{y z}, \\quad z+x \\geq 2 \\sqrt{z x}\n$$\nSolution:\nLet us introduce $x+y=c, y+z=a$ and $z+x=b$. Then $a, b, c$ are the sides of a triangle. If $s=(a+b+c) / 2$, then it is easy to calculate $x=s-a, y=s-b, z=s-c$ and $x+y+z=s$. We also observe that\n$x^{2}+x y+y^{2}=(x+y)^{2}-x y=c^{2}-\\frac{1}{4}(c+a-b)(c+b-a)=\\frac{3}{4} c^{2}+\\frac{1}{4}(a-b)^{2} \\geq \\frac{3}{4} c^{2}$.\nMoreover, $x y+y z+z x=(s-a)(s-b)+(s-b)(s-c)+(s-c)(s-a)$. Thus it is sufficient to prove that\n$$\ns \\sum(s-a)(s-b) \\leq \\frac{9}{8} a b c\n$$\nBut, $\\sum(s-a)(s-b)=r(4 R+r)$, where $r, R$ are respectively the in-radius, the circum-radius of the triangle whose sides are $a, b, c$, and $a b c=4 R r s$. Thus the inequality reduces to\n$$\nr(4 R+r) \\leq \\frac{9}{2} R r\n$$\nThis is simply $2 r \\leq R$. This follows from $I O^{2}=R(R-2 r)$, where $I$ is the incentre and $O$ the circumcentre.\nSolution:\nIf we set $x=\\lambda a, y=\\lambda b, z=\\lambda c$, then the inequality changes to\n$$\n(a+b+c)^{2}(a b+b c+c a)^{2} \\leq 3\\left(a^{2}+a b+b^{2}\\right)\\left(b^{2}+b c+c^{2}\\right)\\left(c^{2}+c a+a^{2}\\right)\n$$\nThis shows that we may assume $x+y+z=1$. Let $\\alpha=x y+y z+z x$. We see that\n$$\n\\begin{aligned}\nx^{2}+x y+y^{2} & =(x+y)^{2}-x y \\\\\n& =(x+y)(1-z)-x y \\\\\n& =x+y-\\alpha=1-z-\\alpha\n\\end{aligned}\n$$\nThus\n$$\n\\begin{aligned}\n\\prod\\left(x^{2}+x y+y^{2}\\right) & =(1-\\alpha-z)(1-\\alpha-x)(1-\\alpha-y) \\\\\n& =(1-\\alpha)^{3}-(1-\\alpha)^{2}+(1-\\alpha) \\alpha-x y z \\\\\n& =\\alpha^{2}-\\alpha^{3}-x y z\n\\end{aligned}\n$$\nThus we need to prove that $\\alpha^{2} \\leq 3\\left(\\alpha^{2}-\\alpha^{3}-x y z\\right)$. This reduces to\n$$\n3 x y z \\leq \\alpha^{2}(2-3 \\alpha)\n$$\nHowever\n$$\n3 \\alpha=3(x y+y z+z x) \\leq(x+y+z)^{2}=1\n$$\nso that $2-3 \\alpha \\geq 1$. Thus it suffices to prove that $3 x y z \\leq \\alpha^{2}$. But\n$$\n\\begin{aligned}\n\\alpha^{2}-3 x y z & =(x y+y z+z x)^{2}-3 x y z(x+y+z) \\\\\n& =\\sum_{\\text {cyclic }} x^{2} y^{2}-x y z(x+y+z) \\\\\n& =\\frac{1}{2} \\sum_{\\text {cyclic }}(x y-y z)^{2} \\geq 0\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32365,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a, b, c$ be non-negative real numbers such that $a+b \\le c+1$, $b+c \\le a+1$ and $c+a \\le b+1$. Prove that\n$$\na^2 + b^2 + c^2 \\le 2abc + 1.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Adding the first two, we get $2b \\le 2$ so that $b \\le 1$. Similarly, we get $c \\le 1$ and $a \\le 1$. Put $\\alpha = 1 - a$, $\\beta = 1 - b$ and $\\gamma = 1 - c$. Then $0 \\le \\alpha, \\beta, \\gamma \\le 1$ and\n$$\n\\alpha + \\beta\\gamma, \\quad \\beta + \\gamma \\ge \\alpha, \\quad \\gamma + \\alpha \\ge \\alpha.\n$$\nThe inequality to be proved reduces to\n$$\n\\alpha^2 + \\beta^2 + \\gamma^2 \\le 2(\\alpha\\beta + \\beta\\gamma + \\gamma\\alpha) - 2\\alpha\\beta\\gamma.\n$$\nWe may assume that $\\gamma$ is the largest among the three, so that $\\alpha \\le \\gamma$ and $\\beta \\le \\gamma$. Using $\\gamma \\le \\alpha + \\beta$, we get $\\gamma^2 \\le \\gamma(\\alpha + \\beta)$. Also $\\alpha^2 \\le \\alpha\\gamma$ and $\\beta^2 \\le \\beta\\gamma$. Thus $\\alpha^2 + \\beta^2 + \\gamma^2 \\le 2(\\alpha\\gamma + \\beta\\gamma)$. Hence it suffices to prove that $2\\alpha\\beta\\gamma \\le 2\\alpha\\beta$. This follows from $\\gamma \\le 1$.\n(Amar Arpit Goel, Utkarsh Tripati). As in the first solution, we conclude that $0 \\le a, b, c \\le 1$. The symmetry shows that we may assume $a \\ge b \\ge c$. We may write the inequality in the form\n$$\n(a-b)^2 \\le (1-c)(1+c-2ab).\n$$\nObserve that $1-c \\ge a-b \\ge 0$, by $b+c \\le a+1$. Using $1+c \\ge a+b$, we get $1+c \\ge a+b \\ge a+ab \\ge a(1+b)+b(a-1)$, since $b \\ge 0$ and $a \\le 1$. Thus $1+c-2b \\ge a-b \\ge 0$. It follows that\n$$\n(a-b)^2 \\le (1-c)(1+c-2ab),\n$$\nas required.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32366,
    "subject": "Mathematics (Multi-modal)",
    "question": "Given a finite string $S$ of symbols $X$ and $O$, we write $\\Delta(S)$ for the number $X$'s in $S$ minus the number of $O$'s. (For example, $\\Delta(XOOXOOX) = -1$.) We call a string $S$ balanced if every sub-string $T$ of (consecutive symbols) $S$ has the property $-1 \\le \\Delta(T) \\le 2$. (Thus $XOOXOOX$ is not balanced, since it contains the sub-string $OOXOO$ whose $\\Delta$-value is $-3$.) Find, with proof, the number of balanced strings of length $n$.",
    "options": [],
    "answer": "For length 2n−1: 2^(n+1) − 2. For length 2n: 3·2^n − 2.",
    "solution": "The balanced strings consists of $X$'s and $O$'s arranged alternately, or with as many as two consecutive letters of the same kind. Such occurrences of double letters must happen alternately with an even number (possibly zero) of single letters between any two double occurrence. If $b_n$ is the number of balanced strings of length $n$, we show that\n$$\nb_{n+2} = 2b_n + 2.\n$$\nFor $n=1$, we have the strings $X$ and $O$, so that $b_1 = 2$. For $n=2$, the balanced strings are $XX$, $XO$, $OX$, $OO$ so that $b_2 = 4$. Similarly, we have for $n=3$ the balanced strings $XXO$, $XOX$, $XOO$, $OXX$, $OXO$, $OOX$, giving $b_3 = 6$. However, for $n=4$, we see that the strings are $XXOX$, $XXOO$, $XOXX$, $XOXO$, $XOOX$, $OXXO$, $OXOX$, $OXOO$, $OOXX$, $OOXO$ and hence $b_4 = 10$.\nLet $x_n, y_n, z_n$ respectively denote the number of balanced strings of length $n$ that end with $XX$; that end with $XO$ and whose last occurrence of a double letter was $XX$; and that end with $XO$, but whose last occurrence of a double letter was $OO$. Note that $X$ and $O$ can be interchanged\n\nin any balanced string. Hence $x_n, y_n, z_n$ also denote the number of balanced strings of length $n$ that end with $OO$; that end with $OX$ and whose last occurrence of a double letter was $OO$; and that end with $OX$, but whose last occurrence of a double letter was $XX$. This shows that $b_n = 2x_n + 2y_n + 2z_n - 2$ (we count purely alternating strings $XOXO... XO$ and $OXOX... OX$ twice; once among $y_n$ and once among $z_n$). Thus $b_n + 2 = 2(x_n + y_n + z_n)$. We can form the strings of length $n+2$ from such string of length $n$ in exactly the following ways;\n(i) if already ending with $XX$, we may add either $OO$ or $OX$, but nothing else;\n(ii) if ending with $XO$, with last double occurrence $XX$, we can add $OX$ or $XO$, but nothing else;\n(iii) if ending with $OX$, with last double occurrence $OO$, we can add $XO$ or $OX$, but nothing else.\nThis gives $2(x_n + y_n + z_n)$ strings. Similarly, by interchanging $X$ and $O$, we get $2(x_n + y_n + z_n)$ more strings. Thus\n$$\nb_{n+2} + 2 = 4(x_n + y_n + z_n) = 2(b_n + 2).\n$$\nUsing the initial conditions, we get\n$$\nb_{2n} = 2 \\cdot 2^n - 2, \\quad b_{2n-1} = 2^{n+1} - 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32367,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define $f, g, h$ on $\\mathbb{Z} \\times \\mathbb{Z} \\times \\mathbb{Z}$ as follows:\n$$\nf(x, y, z) = (3x + 2y + 2z, 2x + 2y + z, 2x + y + 2z),\n$$\n$$\ng(x, y, z) = (3x + 2y - 2z, 2x + 2y - z, 2x + y - 2z),\n$$\n$$\nh(x, y, z) = (3x - 2y + 2z, 2x - y + 2z, 2x - 2y + z).\n$$\nGiven a primitive Pythagorean triplet $(x, y, z)$, with $x > y > z$, prove that starting from $(5, 4, 3)$, the triplet $(x, y, z)$ can be obtained, in a unique way, by repeated application of $f, g, h$ in some order. (Example: $(697, 528, 455) = f \\circ h \\circ g \\circ h(5, 4, 3)$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let\n$$\nu = u(x, y, z) = 3x - 2y - 2z,$$\n$$v = v(x, y, z) = -2x + 2y + z,$$\n$$w = w(x, y, z) = -2x + y + 2z.$$\nFirst of all, if $(x, y, z) \\neq (5, 4, 3)$ is a primitive Pythagorean triplet with $x > y > z > 0$, then, it is easy to check that $u^2 = v^2 + w^2$. Further, $x^2 + (2y - 2z)^2 > 0 \\Rightarrow 9x^2 > 4y^2 + 4z^2 + 8yz \\Rightarrow 3x > 2y + 2z \\Rightarrow u(x, y, z) > 0$.\nSimilarly, we have $4y > 3z \\Rightarrow 4yz - 3z^2 > 0 \\Rightarrow 4y^2 + z^2 + 4yz > 4x^2 \\Rightarrow 2y + z > 2x \\Rightarrow v(x, y, z) > 0$.\nNext, $(3y - 4z)^2 > 0 \\Rightarrow 25x^2 + (3y - 4z)^2 > 25x^2 \\Rightarrow 25x^2 > 16y^2 + 9z^2 + 24yz \\Rightarrow 25x^2 > (4y + 3x)^2 \\Rightarrow 5x > 4y + 3z \\Rightarrow u(x, y, z) > v(x, y, z)$.\n\nSimilarly, we have $u(x, y, z) > w(x, y, z)$. Also, since $x > y$, it follows that $u(x, y, z) > -w(x, y, z)$. And, $y > z$ implies $v(x, y, z) > w(x, y, z)$. Finally, we notice that if $y + z > x$ then it follows that $u(x, y, z) < x$. Now, let\n$$\nf_1(x, y, z) = (u, v, w),\n$$\n$$\ng_1(x, y, z) = (u, v, -w),\n$$\n$$\nh_1(x, y, z) = (u, -w, v).\n$$\nThen, it is easy to see that $f_1(f(x, y, z)) = g_1(g(x, y, z)) = h_1(h(x, y, z)) = (x, y, z)$. Further, it is easy to check that $\\text{gcd}(u, v, w) = \\text{gcd}(x, y, z)$.\n\nNow, we shall prove the result by contradiction. If there are primitive Pythagorean triplets that are not obtained from $(5, 4, 3)$ by repeated application of $f, g$ and $h$ in some order, then let $(a, b, c)$ be the triplet that cannot be obtained from $(5, 4, 3)$ with $a > b > c > 0$ and with least possible value for $a$. So, all the primitive Pythagorean triplets $(x, y, z)$ with $a > x > y > z > 0$ are obtained from $(5, 4, 3)$ (uniquely). Clearly, we have $a > 5$.\nNow, let $f_1(a, b, c) = (k, l, m)$. Then, $g_1(a, b, c) = (k, l, -m)$ and $h_1(a, b, c) = (k, -m, l)$. As noted before, $k$ and $l$ are positive, and $k > l, k > m$ and $k > -m$. Thus, there is a unique $\\Delta \\in \\{f_1, g_1, h_1\\}$ such that $\\Delta(a, b, c) = (r, s, t)$ with $r > s > t > 0$. Also, from the previous calculations, we have\n\nthat $a > r$. Thus, by the minimality of $a$, it follows that $(r, s, t)$ is obtained from $(5, 4, 3)$, in a unique way, by repeated applications of $f, g, h$ in some order. But, now $\\Delta(a, b, c) = (r, s, t)$ implies that by applying one of $f, g$ and $h$ to $(r, s, t)$ we would get $(a, b, c)$. Therefore, we can obtain $(a, b, c)$ from $(5, 4, 3)$ by repeated applications of $f, g, h$ in some order. This is a contradiction. Uniqueness follows from the uniqueness of $\\Delta$. This completes the solution.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32368,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many pairs $(m, n)$ of positive integers such that $m < n$ and $\\frac{(m+n)(m+n+1)}{mn}$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If we take $m = 1$, $n = 2$, we see that $\\frac{(m+n)(m+n+1)}{mn} = 6$. (Or we can start with $m = 2$, $n = 3$ as well.)\n\nSuppose we have some pair $(m, n)$ of positive integers such that $m < n$ and\n$$\n\\frac{(m+n)(m+n+1)}{mn} = k\n$$\nis an integer. This may be written in the form\n$$\nnk = m + 2n + 1 + \\frac{n(n+1)}{m}\n$$\nThus $\\frac{n(n+1)}{m}$ is also an integer, say equal to $l$. Observe that\n$$\n\\begin{aligned}\n(n+l)(n+l+1) &= \\left(n + \\frac{n(n+1)}{m}\\right) \\left(n + 1 + \\frac{n(n+1)}{m}\\right) \\\\\n&= \\frac{l(m+n+1)(m+n)}{m^2}.\n\\end{aligned}\n$$\nThus\n$$\n\\frac{(n+l)(n+l+1)}{nl} = \\frac{(m+n)(m+n+1)}{mn} = k.\n$$\nMoreover $lm = n(n+1) > n^2 > nm$ showing that $l > n$. Hence $(m, n) \\neq (n, l)$. Thus starting with the pair $(1, 2)$, we can generate new pair $(2, 6)$ and the process may be continued indefinitely.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32369,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be a triangle, $I$ its in-centre; $A_{1}, B_{1}, C_{1}$ be the reflections of $I$ in $BC, CA, AB$ respectively. Suppose the circum-circle of triangle $A_{1}B_{1}C_{1}$ passes through $A$. Prove that $B_{1}, C_{1}, I, I_{1}$ are concyclic, where $I_{1}$ is the in-centre of triangle $A_{1}B_{1}C_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nNote that $IA_{1} = IB_{1} = IC_{1} = 2r$, where $r$ is the in-radius of the triangle $ABC$. Hence $I$ is the circum-centre of the triangle $A_{1}B_{1}C_{1}$.\n\nLet $K$ be the point of intersection of $IB_{1}$ and $AC$. Then $IK = r$, $IA = 2r$ and $\\angle IKA = 90^{\\circ}$. It follows that $\\angle IAK = 30^{\\circ}$ and hence $\\angle IAB_{1} = 60^{\\circ}$. Thus $AIB_{1}$ is an equilateral triangle. Similarly triangle $AIC_{1}$ is also equilateral. We hence obtain $AB_{1} = AC_{1} = AI = IB_{1} = IC_{1} = 2r$.\n\nWe also observe that $\\angle B_{1}IC_{1} = 120^{\\circ}$ and $IB_{1}AC_{1}$ is a rhombus. Thus $\\angle B_{1}AC_{1} = 120^{\\circ}$ and by concyclicity $\\angle A_{1} = 60^{\\circ}$. Since $AB_{1} = AC_{1}$, $A$ is the midpoint of the arc $B_{1}AC_{1}$. It follows that $A_{1}A$ bisects $\\angle A_{1}$ and $I_{1}$ lies on the line $A_{1}A$. This implies that\n$$\n\\angle B_{1}I_{1}C_{1} = 90^{\\circ} + \\angle A_{1}/2 = 90^{\\circ} + 30^{\\circ} = 120^{\\circ}\n$$\nSince $\\angle B_{1}IC_{1} = 120^{\\circ}$, we conclude that $B_{1}, I, I_{1}, C_{1}$ are concyclic. (Further $A$ is the centre.)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32370,
    "subject": "Mathematics (Multi-modal)",
    "question": "The in-circle $\\Gamma$ of a triangle $ABC$ touches the side $BC$ at $D$. Let $D'$ be the point which is diametrically opposite to $D$ on the circle $\\Gamma$. The tangent through $D'$ to $\\Gamma$ meets $AD$ in $X$. The tangent to $\\Gamma$ through $X$, other than $XD'$, touches $\\Gamma$ at $N$. Prove that the circum-circle of triangle $BCN$ touches $\\Gamma$ at $N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that $NX$ is the polar of $Y$, $EF$ is the polar of $A$ and $BC$ is the polar of $D$. Since $A$, $Y$, $D$ are collinear, it follows that $NX$, $EF$, $BC$ are concurrent. Let the point of concurrency be $D'$. Let $S$ be the point of intersection of $EF$ and $AD$. Since \\{$E, S, F, D'$\\} form a harmonic range, \\{AE, AD, AB, AD'\\} is a harmonic pencil. It follows that \\{D', B, D, C\\} is a harmonic range. We also observe that $\\angle DND' = 90^\\circ$. Therefore $ND$ bisects $\\angle BNC$. This implies that $D$ is the mid-point of the minor arc $PQ$ of $\\Gamma$, where $P$, $Q$ are the points of intersection of $NB$, $NC$ with $\\Gamma$. Hence $PQ$ is parallel to $BC$. Now\n$$\n\\angle YNQ = \\angle NPQ = \\angle NBC.\n$$\nIt follows that $YN$ is tangent to the circum-circle of the triangle $BNC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32371,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $n$ is a composite number such that $\\varphi(n)$ divides $n-1$, where $\\varphi$ denotes Euler's totient function. Show that $n$ has at least four distinct prime factors.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose $n$ has only one prime factor. Then $n = p^\\alpha$ for some $\\alpha > 1$ and prime $p$. Then $\\varphi(n) = p^{\\alpha-1}(p-1)$ divides $p^\\alpha - 1$, which is impossible since $\\alpha \\ge 2$. Thus $n$ has at least two prime factors.\n\nSuppose $n = p^\\alpha q^\\beta$, where $p \\ne q$ are primes, $\\alpha \\ge 1$ and $\\beta \\ge 1$. Here $\\varphi(n) = p^{\\alpha-1}q^{\\beta-1}(p-1)(q-1)$ and $n-1 = p^\\alpha q^\\beta - 1$. Thus $\\varphi(n)|((n-1)(p-1))$ implies that $\\alpha = 1 = \\beta$ and $(p-1)(q-1)$ divides $pq-1$. This implies that $(p-1)|(q-1)$ and $(q-1)|(p-1)$, forcing $p=q$.\n\nIf $n$ has three distinct prime factors, say $p, q, r$, we see as in the earlier case $n = pqr$ and $(p-1)(q-1)(r-1)$ divides $pqr-1$. Taking $p-1=x, q-1=y$ and $r-1=z$, we see that\n$$\nt = \\frac{pqr - 1}{(p-1)(q-1)(r-1)} = 1 + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx},\n$$\nis an integer. We also observe that no prime can be even. (Otherwise $pqr-1$ is odd whereas one of $p-1, q-1, r-1$ is even.) Thus we may assume $3 \\le p < q < r$, so that $x \\ge 2, y \\ge 4$ and $z \\ge 6$. Thus\n$$\n1 < t \\le 1 + \\frac{1}{2} + \\frac{1}{4} + \\frac{1}{6} + \\frac{1}{8} + \\frac{1}{12} + \\frac{1}{24} = 1 + \\frac{28}{24} < 3.\n$$\nThus $t=2$ and hence\n$$\n\\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} = 1.\n$$\nIf $p \\ge 5$, we have $x \\ge 4, y \\ge 6$ and $z \\ge 10$. In this case,\n$$\n1 = \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} + \\frac{1}{xy} + \\frac{1}{yz} + \\frac{1}{zx} \\le \\frac{1}{4} + \\frac{1}{6} + \\frac{1}{10} + \\frac{1}{24} + \\frac{1}{60} + \\frac{1}{40} = \\frac{72}{120} < 1.\n$$\nThus $p=3$ and $x=2$. We obtain\n$$\n\\frac{1}{y} + \\frac{1}{z} + \\frac{1}{2} \\left( \\frac{1}{y} + \\frac{1}{z} \\right) + \\frac{1}{yz} = \\frac{1}{2}.\n$$\nThis may be written in the form $(y-3)(z-3) = 11$. We thus get $y=4, z=14$. In turn, we have $q=5$ and $r=15$. But then $r$ is not a prime. Thus the number of distinct prime factors of $n$ is more than 3.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32372,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(x) = x^2 + a x + b$ be a real polynomial with $a < 2$. Suppose $P(P(x)) = 0$ has four distinct real roots and the sum of some two of them is $\\le -1$. Prove that $P(x+y) \\ge P(x) + P(y)$ for all non-negative real numbers $x, y$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have to consider two possibilities: $x_1 + x_2 \\le -1$ or $x_1 + x_3 \\le -1$. (Both the roots are from the same equation or each root coming from a different equation.) Suppose $x_1 + x_2 \\le -1$. Then $-a = x_1 + x_2$ shows that $a > 0$. Since the equations $P(x) = \\alpha$ and $P(x) = \\beta$ have distinct real roots, we have\n$$\n4(b - \\alpha) < a^2, \\quad 4(b - \\beta) < a^2.\n$$\nThus $4b < a^2 + 2(\\alpha + \\beta) = a^2 - 2a = a(a - 2) < 0$. Hence $b < 0$ in this case.\nSuppose $x_1 + x_3 \\le -1$. We have $P(x_1) = \\alpha$, $P(x_3) = \\beta$, so that\n$$\nx_1^2 + x_3^2 + a(x_1 + x_3) + 2b = \\alpha + \\beta = -a.\n$$\nIf $a > 0$, we see that\n$$\n(x_1 + \\frac{a}{2})^2 + (x_3 + \\frac{a}{2})^2 + 2b = -a + \\frac{a^2}{2} = \\frac{a(a-2)}{2},\n$$\nso that $2b \\le a(a - 2)/2 < 0$. If $a \\le 0$, we have\n$$\nx_1^2 + x_3^2 + 2b = -a(x_1 + x_3 + 1) \\le 0,\n$$\nso that $b \\le 0$. We conclude that $b \\le 0$ in all cases.\nNow it is easy to see that\n$$\nP(x+y) = P(x) + P(y) + 2xy - b \\ge P(x) + P(y),\n$$\nfor all non-negative $x, y$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32373,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a non-isosceles triangle, and let $\\Gamma$ be its in-circle. Let $D$, $E$, $F$ be the points of contact of $\\Gamma$ with the sides $BC$, $CA$, $AB$ respectively. Suppose $FD$, $DE$, $EF$ intersect $CA$, $AB$, $BC$ in $U$, $V$, $W$ respectively. If $L$, $M$, $N$ are respectively the mid-points of $DW$, $EU$, $FV$ prove that $L$, $M$, $N$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nDraw a line through $N$ which is parallel to $ED$. This passes through the midpoints $P$ of $DF$ and $Q$ of $FE$. Similarly the line through $M$ parallel to $FD$ passes through the mid-point $Q$ of $FE$ and $R$ of $ED$; the line through $L$ parallel to $FE$ passes through the mid-point $P$ of $FD$ and $R$ of $ED$. We thus obtain the medial triangle $PRQ$ of $ABC$. Since $AF = AE$, the line $AQ$ is also the bisector of $\\angle A$. Similarly $BP$ bisects $\\angle B$ and $CR$ bisects $\\angle C$.\n\nThus $AQ$, $BP$, $CR$ concur at $I$, the in-centre of $ABC$. Now Desargues' theorem is applicable to the triangles $ABC$ and $QPR$. It follows that $PR \\cap BC$, $RQ \\cap CA$ and $QP \\cap AB$ are collinear. Thus $L$, $M$ and $N$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32374,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all triples $(p, x, y)$ such that $p^{x} = y^{4} + 4$, where $p$ is a prime and $x, y$ are natural numbers.",
    "options": [],
    "answer": "(5, 1, 1)",
    "solution": "Solution:\nWe begin with the standard factorisation\n$$\ny^{4}+4 = \\left(y^{2}-2y+2\\right)\\left(y^{2}+2y+2\\right)\n$$\nThus we have $y^{2}-2y+2 = p^{m}$ and $y^{2}+2y+2 = p^{n}$ for some positive integers $m$ and $n$ such that $m+n = x$. Since $y^{2}-2y+2 < y^{2}+2y+2$, we have $m < n$ so that $p^{m}$ divides $p^{n}$. Thus $y^{2}-2y+2$ divides $y^{2}+2y+2$. Writing $y^{2}+2y+2 = y^{2}-2y+2 + 4y$, we infer that $y^{2}-2y+2$ divides $4y$ and hence $y^{2}-2y+2$ divides $4y^{2}$. But\n$$\n4y^{2} = 4\\left(y^{2}-2y+2\\right) + 8(y-1)\n$$\nThus $y^{2}-2y+2$ divides $8(y-1)$. Since $y^{2}-2y+2$ divides both $4y$ and $8(y-1)$, we conclude that it also divides $8$. This gives $y^{2}-2y+2 = 1, 2, 4$ or $8$.\nIf $y^{2}-2y+2 = 1$, then $y = 1$ and $y^{4}+4 = 5$, giving $p = 5$ and $x = 1$. If $y^{2}-2y+2 = 2$, then $y^{2}-2y = 0$ giving $y = 2$. But then $y^{4}+4 = 20$ is not the power of a prime. The equations $y^{2}-2y+2 = 4$ and $y^{2}-2y+2 = 8$ have no integer solutions. We conclude that $(p, x, y) = (5, 1, 1)$ is the only solution.\n\nAlternatively, using $y^{2}-2y+2 = p^{m}$ and $y^{2}+2y+2 = p^{n}$, we may get\n$$\n4y = p^{m}\\left(p^{n-m}-1\\right)\n$$\nIf $m > 0$, then $p$ divides $4$ or $y$. If $p$ divides $4$, then $p = 2$. If $p$ divides $y$, then $y^{2}-2y+2 = p^{m}$ shows that $p$ divides $2$ and hence $p = 2$. But then $2^{x} = y^{4}+4$, which shows that $y$ is even. Taking $y = 2z$, we get $2^{x-2} = 4z^{4}+1$. This implies that $z = 0$ and hence $y = 0$, which is a contradiction. Thus $m = 0$ and $y^{2}-2y+2 = 1$. This gives $y = 1$ and hence $p = 5, x = 1$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32375,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $d$, $u$, $v$, $w$ are positive integers such that $u$, $v$, $w$ are distinct and\n$$\nd^3 - d(uv + vw + wu) - 2uvw = 0.\n$$\nProve that $d$ cannot be a prime. Find also the least possible value of $d$.",
    "options": [],
    "answer": "d is not prime; the least possible value of d is 8.",
    "solution": "Note that $u = 1$, $v = 2$, $w = 3$ gives $d^3 - 11d - 12 = 0$ which has no integer solutions. Thus $uvw > 8$ and $uv + vw + wu \\ge 3(uvw)^{2/3} > 12$. This shows that $d^3 > 12d + 16$ and we infer that $d \\ge 5$.\n\nIf $d$ is a prime, $d$ has to be an odd prime, say $d = p$. Then $p^3 = p(uv + vw + wu) + 2uvw$, so that $p$ divides $uvw$. Assume $p|u$, and write $u = pu_1$. If $p$ divides $vw$, it must divide either $v$ or $w$. But then\n$$\np(uv + vw + wu) > pu(v + w) > p^2(p + 1) > p^3.\n$$\nThus $p$ cannot divide $vw$. The relation now reduces to\n$$\np^2 = pu_1(v + w) + vw(1 + 2u_1).\n$$\nIt follows that $p$ divides $1 + 2u_1$. Thus $1 + 2u_1 \\ge p$ or $u_1 \\ge (p-1)/2$. Since $v$, $w$ are distinct, we have $v + w \\ge 3$. Thus\n$$\np(uv + vw + wu) > pu(v + w) \\ge 3p^2 u_1 \\ge \\frac{3p^2(p-1)}{2} > p^3,\n$$\nif $p \\ge 3$. We conclude that $d$ cannot be an odd prime.\n\nSuppose $d = 6$. Substituting in the given relation, we see that $3|uvw$. Assume $3|u$, so that $u = 3u_1$. We obtain\n$$\n36 = 3u_1(v + w) + vw(1 + u_1).\n$$\nThus $3|vw(1 + u_1)$. Observe that $3u_1(v + w) < 36$ and $v + w \\ge 3$. Hence $u_1 < 4$. Thus either $3|vw$ or $u_1 = 2$. But $u_1 = 2$ shows that $12 = 2(v + w) + vw$ forcing $v = w = 2$. Since $v \\ne w$, the other possibility is $3|vw$. Taking $3|v$, we have $v = 3v_1$ and\n$$\n12 = 3u_1v_1 + w(u_1 + v_1) + u_1v_1w.\n$$\nThus $u_1v_1(3 + w) < 12$. This forces $u_1v_1 < 3$ giving $(u_1, v_1) = (1, 2)$ or $(2, 1)$. But then $u_1 + v_1 = 3$ and $u_1v_1 = 2$, giving $12 = 6 + 5w$. This is impossible for an integer $w$.\n\nWe conclude that $d \\ge 8$. For $d = 8$, we may take $(u, v, w) = (2, 4, 7)$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32376,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $A$ be a set of real numbers such that $A$ has at least four elements. Suppose $A$ has the property that $a^{2} + b c$ is a rational number for all distinct numbers $a, b, c$ in $A$. Prove that there exists a positive integer $M$ such that $a \\sqrt{M}$ is a rational number for every $a$ in $A$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose $0 \\in A$. Then $a^{2} = a^{2} + 0 \\times b$ is rational and $a b = 0^{2} + a b$ is also rational for all $a, b$ in $A$, $a \\neq 0$, $b \\neq 0$, $a \\neq b$. Hence $a = a_{1} \\sqrt{M}$ for some rational $a_{1}$ and natural number $M$. For any $b \\neq 0$, we have\n$$\nb \\sqrt{M} = \\frac{a b}{a_{1}}\n$$\nwhich is a rational number.\nHence we may assume $0$ is not in $A$. If there is a number $a$ in $A$ such that $-a$ is also in $A$, then again we can get the conclusion as follows. Consider two other elements $c, d$ in $A$. Then $c^{2} + d a$ is rational and $c^{2} - d a$ is also rational. It follows that $c^{2}$ is rational and $d a$ is rational. Similarly, $d^{2}$ and $c a$ are also rationals. Thus $d / c = (d a) / (c a)$ is rational. Note that we can vary $d$ over $A$ with $d \\neq c$ and $d \\neq a$. Again $c^{2}$ is rational implies that $c = c_{1} \\sqrt{M}$ for some rational $c_{1}$ and natural number $M$. We observe that $c \\sqrt{M} = c_{1} M$ is rational, and\n$$\na \\sqrt{M} = \\frac{c a}{c_{1}}\n$$\nso that $a \\sqrt{M}$ is a rational number. Similarly is the case with $-a \\sqrt{M}$. For any other element $d$,\n$$\nb \\sqrt{M} = M c_{1} \\frac{d}{c}\n$$\nis a rational number.\nThus we may now assume that $0$ is not in $A$ and $a + b \\neq 0$ for any $a, b$ in $A$. Let $a, b, c, d$ be four distinct elements of $A$. We may assume $|a| > |b|$. Then $d^{2} + a b$ and $d^{2} + b c$ are rational numbers and so is their difference $a b - b c$. Writing $a^{2} + a b = a^{2} + b c + (a b - b c)$, and using the facts $a^{2} + b c$, $a b - b c$ are rationals, we conclude that $a^{2} + a b$ is also a rational number. Similarly, $b^{2} + a b$ is also a rational number.\nConsider\n$$\nq = \\frac{a}{b} = \\frac{a^{2} + a b}{b^{2} + a b}\n$$\nNote that $a^{2} + a b > 0$. Thus $q$ is a rational number and $a = b q$. This gives $a^{2} + a b = b^{2}(q^{2} + q)$. Let us take $b^{2}(q^{2} + q) = l$. Then\n$$\n|b| = \\sqrt{\\frac{l}{q^{2} + q}} = \\sqrt{\\frac{x}{y}}\n$$\nwhere $x$ and $y$ are natural numbers. Take $M = x y$. Then $|b| \\sqrt{M} = x$ is a rational number. Finally, for any $c$ in $A$, we have\n$$\nc \\sqrt{M} = b \\sqrt{M} \\frac{c}{b}\n$$\nis also a rational number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32377,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\alpha$ be a real number larger than $1$. Define a sequence $(s_n)_{n \\ge 1}$ as follows: $s_1 = 1$, $s_2 = \\alpha$; if $s_1, s_2, s_3, \\dots, s_{2^n}$ are defined for some $n \\ge 1$, then $s_{2^n+1}, s_{2^n+2}, \\dots, s_{2^{n+1}}$ are defined by $s_j = \\alpha s_{j-2^n}$, for $2^n+1 \\le j \\le 2^{n+1}$. (Thus the first few terms are $1, \\alpha, \\alpha^2, \\alpha, \\alpha^2, \\alpha^3, \\dots$) Let $c_n = s_1+s_2+s_3+\\dots+s_n$. If $n = 2^{e_0} + 2^{e_1} + \\dots + 2^{e_k}$, where $e_0 > e_1 > \\dots > e_k \\ge 0$ is the binary representation of a positive integer $n$, prove that\n$$\nc_n = (1 + \\alpha)^{e_0} + \\alpha(1 + \\alpha)^{e_1} + \\alpha^2(1 + \\alpha)^{e_2} + \\dots + \\alpha^k(1 + \\alpha)^{e_k}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove this in four steps. We use $s_n = s(n)$ and $c_n = c(n)$,\n\n(A) If $b(n)$ denotes the number of $1$'s in the binary representation of a non-negative integer $n$, we have $s(n) = \\alpha^{b(n-1)}$, $n \\ge 1$.\n**Proof:** We use induction. Note that $s(1) = 1 = \\alpha^{b(0)}$. Suppose $s(r) = \\alpha^{b(r-1)}$, for $1 \\le r \\le 2^{n-1}$. Take any $r$ such that $2^{n-1} + 1 \\le r \\le 2^n$. We have\n$$\n\\begin{aligned}\ns(r) &= \\alpha s(r - 2^{n-1}) \\\\\n&= \\alpha \\alpha^{b(r-2^{n-1}-1)} \\\\\n&= \\alpha^{b(r-2^{n-1}-1)+1} = \\alpha^{b(r-1)},\n\\end{aligned}\n$$\nas $r-1$ has exactly one more $1$ in its binary representation than $r-2^{n-1}-1$.\n\n(B) If $m \\ge 1$, we have $s(2m+1) = s(m+1)$ and $s(2m) = \\alpha s(m)$.\n**Proof:** Since $m$ and $2m$ have same number of $1$'s, we have\n$$\ns(2m+1) = \\alpha^{b(2m)} = \\alpha^{b(m)} = s(m+1).\n$$\nFor even numbers, first we show that $s(2m) = \\alpha s(2m-1)$. This is easy to establish by induction. Observe $s(2) = \\alpha = \\alpha s(1)$. If $2^{n-1} + 1 \\le 2m-1 < 2n \\le 2^n$, we get by induction hypothesis\n$$\ns(2m) = \\alpha s(2m - 2^{n-1}) = \\alpha^2 s(2m - 1 - 2^{n-1}) = \\alpha s(2m - 1).\n$$\nThus we get\n$$\ns(2m) = \\alpha s(2m - 1) = \\alpha s(2(m - 1) + 1) = \\alpha s(m - 1 + 1) = \\alpha s(m),\n$$\nwhere we have used $s(2k+1) = s(k+1)$.\n\n(C) For $m \\ge 1$, we have $c(2m) = (1+\\alpha)c(m)$ and $c(2m+1) = \\alpha c(m) + c(m+1)$.\n**Proof:** We have\n$$\n\\begin{aligned}\nc(2m) &= (s(1) + s(3) + \\cdots + s(2m-1)) + (s(2) + s(4) + \\cdots s(2m)) \\\\\n&= (s(1) + s(2) + \\cdots s(m)) + \\alpha(s(1) + s(2) + \\cdots s(m)) \\\\\n&= (1 + \\alpha)(s(1) + s(2) + \\cdots s(m)) = (1 + \\alpha)c(m).\n\\end{aligned}\n$$\nSimilarly,\n$$\n\\begin{aligned}\nc(2m+1) &= (s(1) + s(3) + \\cdots + s(2m-1) + s(2m+1)) + (s(2) + s(4) + \\cdots s(2m)) \\\\\n&= (s(1) + s(2) + \\cdots s(m) + s(m+1)) + \\alpha(s(1) + s(2) + \\cdots s(m)) \\\\\n&= \\alpha c(m) + c(m+1).\n\\end{aligned}\n$$\n\n(D) We complete the deduction. Let $n \\ge 1$ and $n = 2^{e_0} + 2^{e_1} + \\dots + 2^{e_k}$, where $e_0 > e_1 > \\dots > e_k \\ge 0$ is the binary representation of a positive integer $n$. We prove\n$$\nc(n) = (1 + \\alpha)^{e_0} + \\alpha(1 + \\alpha)^{e_1} + \\alpha^2(1 + \\alpha)^{e_2} + \\dots + \\alpha^k(1 + \\alpha)^{e_k}.\n$$\n\nFor $n = 1$, we have $1 = 2^0$, and $C(1) = s(1) = 1 = (1 + \\alpha)^0$. Suppose the statement is true for $c(1), c(2), \\dots, c(n-1)$, $n \\ge 2$. If $n = 2m$ and $m = 2^{e_0} + 2^{e_1} + \\dots + 2^{e_k}$, where $e_0 > e_1 > \\dots > e_k$, we have by (C),\n$$\nc(n) = (1 + \\alpha)c(m) = (1 + \\alpha)((1 + \\alpha)^{e_0} + \\dots + \\alpha^k(1 + \\alpha)^{e_k}) = (1 + \\alpha)^{e_0+1} + \\dots + \\alpha^k(1 + \\alpha)^{e_k+1}.\n$$\nAs $n = 2m = 2^{e_0+1} + \\dots + 2^{e_k+1}$, the desired result follows for $n = 2m$. Suppose $n = 2m + 1$, where $m = 2^{e_0} + 2^{e_1} + \\dots + 2^{e_k}$, where $e_0 > e_1 > \\dots > e_k$. In this case $n = 2^{e_0+1} + \\dots + 2^{e_k+1} + 2^0$ and $m+1 = 2^{e_0} + \\dots + 2^{e_k} + 2^0$. Thus\n$$\nc(2m+1) = ac(m) + c(m+1) = \\alpha((1+\\alpha)^{e_0} + \\dots + \\alpha^k(1+\\alpha)^{e_k}) + ((1+\\alpha)^{e_0} + \\dots + \\alpha^k(1+\\alpha)^{e_k} + \\alpha^{k+1}) = (1+\\alpha)^{e_0+1} + \\dots + \\alpha^k(1+\\alpha)^{e_k+1} + \\alpha^{k+1}(1+\\alpha)^0.\n$$\nThus the relation is true for $n = 2m + 1$ as well. This completes the proof by induction on $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32378,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAll the points with integer coordinates in the $x y$-plane are coloured using three colours, red, blue and green, each colour being used at least once. It is known that the point $(0,0)$ is coloured red and the point $(0,1)$ is coloured blue. Prove that there exist three points with integer coordinates of distinct colours which form the vertices of a right-angled triangle.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider the lattice points (points with integer coordinates) on the lines $y=0$ and $y=1$, other than $(0,0)$ and $(0,1)$. If one of them, say $A=(p, 1)$, is coloured green, then we have a right-angled triangle with $(0,0)$, $(0,1)$ and $A$ as vertices, all having different colours. (See Figures 1 and 2.)\n\n![](attached_image_1.png)\n\nIf not, the lattice points on $y=0$ and $y=1$ are all red or blue. We consider three different cases.\n\nCase 1. Suppose a point $B=(c, 0)$ is blue. Consider a green point $D=(p, q)$ in the plane. Suppose $p \\neq 0$. If its projection $(p, 0)$ on the $x$-axis is red, then $(p, q)$, $(p, 0)$ and $(c, 0)$ are the vertices of a required type of right-angled triangle. If $(p, 0)$ is blue, then we can consider the triangle whose vertices are $(0,0)$, $(p, 0)$ and $(p, q)$. If $p=0$, then the points $D$, $(0,0)$ and $(c, 0)$ will work. (Figure 3.)\n\nCase 2. A point $D=(c, 1)$, on the line $y=1$, is red. A similar argument works in this case.\n\n![](attached_image_2.png)\n\nFig-4\n\nCase 3. Suppose all the lattice points on the line $y=0$ are red and all on the line $y=1$ are blue points. Consider a green point $E=(p, q)$, where $q \\neq 0$ and $q \\neq 1$. (See Figure 4.) Consider an isosceles right-angled triangle $E K M$ with $\\angle E=90^{\\circ}$ such that the hypotenuse $K M$ is a part of the $x$-axis. Let $E M$ intersect $y=$ in $L$. Then $K$ is a red point and $L$ is a blue point. Hence $E K L$ is a desired triangle.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32379,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle and $D$, $E$, $F$ be points on the sides $BC$, $CA$, $AB$ respectively such that $AD = BE = CF$. Suppose the line segments $AD$, $BE$, $CF$ are not concurrent and enclose an equilateral triangle. Is $ABC$ necessarily equilateral?",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Yes. Draw $AK$ parallel to $BC$ and $AK = BD$. Join $K$ to $E$ and $F$. Choose $L$ on $BC$ with $L$ between $B$ and $C$, such that $BD = LC$. Note that $KB$ is parallel to $AD$ and $KB = AD$. Hence $\\angle KBE = \\angle APE = 60^\\circ$. On the other hand, $KB = AD = BE$. Thus $KBE$ is an equilateral triangle. This gives $\\angle BEK = 60^\\circ = \\angle BQR$. Thus $CF$ is parallel to $EK$ but $CF = BE = EK$. We conclude that $FCEK$ is a parallelogram. Since $CL$ is parallel to $AK$ and $CF$ is parallel to $KE$ we have $\\angle LCF = \\angle AKE$.\n\nConsider the triangles $LCF$ and $AKE$. We have $LC = AK$, $CF = KE$ and the included angles are also the same. Hence $LCF$ is congruent to $AKE$. Moreover $LF$ is parallel to $CA$. Thus\n$$\n\\frac{BF}{FA} = \\frac{BL}{LC} = \\frac{CD}{BD} = \\frac{1}{\\lambda}, \\text{ say.}\n$$\nConsider the transversal $FQC$ of the triangle $ABD$. By Menelaus' theorem\n$$\n\\frac{BF}{FA} \\cdot \\frac{AP}{PD} \\cdot \\frac{CD}{CB} = 1.\n$$\nThus\n$$\n\\frac{AP}{PD} = \\frac{FA}{BF} \\cdot \\frac{CB}{CD} = \\lambda \\left( \\frac{CD + DB}{CD} \\right) = \\lambda(1+\\lambda).\n$$\nSimilarly, we get\n$$\n\\frac{BQ}{QE} = \\frac{CR}{RF} = \\lambda(1+\\lambda).\n$$\nWe thus have\n$$\n\\frac{AP}{PD} = \\frac{BQ}{QE} = \\frac{CR}{RF}.\n$$\nThis equality implies\n$$\n\\frac{AD}{PD} = \\frac{BE}{QE} = \\frac{CF}{RF}.\n$$\nUsing $AD = BE = CF$, we get $PD = QE = RF$. But then $AP = BQ = CR$ and hence $AR = BP = CQ$. Consider the triangles $ABP$ and $BCQ$. Observe that $\\angle BPA = 120^\\circ = \\angle CQB$, $BP = CQ$ and $AP = BQ$. Thus $ABP$ and $BCQ$ are congruent triangles. This gives $AB = BC$. Similarly, we obtain $BC = CA$.\nLet us write $PQ = QR = RP = x$, $BE = CF = AD = y$, $AR = u$, $BP = v$ and $CQ = w$. Using the collinearity of the points $A$, $F$, $B$ with respect to the triangle $PQR$, Menelaus' theorem gives\n$$\n\\frac{RA}{AP} \\cdot \\frac{PB}{BQ} \\cdot \\frac{QF}{FR} = 1.\n$$\n(we are using only the lengths). This may be written in the form\n$$\n\\frac{u}{u+x} \\cdot \\frac{v}{v+x} \\cdot \\frac{y-w}{y-(w+x)} = 1.\n$$\nSimplification gives\n$$\nu+v = \\frac{(x+u)(x+v)(x+w) - uvw}{yx} - x.\n$$\nNote that the expression on the right side is symmetric in $u, v, w$. It follows that\n$$\nu + v = v + w = w + u.\n$$\nThus $u = v = w$ and hence $AP = BQ = CR$. As in the earlier solution, it follows that the triangles $ABP, BCQ, CAR$ are congruent, and hence $AB = BC = CA$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32380,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be positive real numbers such that $a^2 + b^2 + c^2 < 2(a + b + c)$. Prove that\n$$\n3abc < 4(a + b + c).\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The Cauchy-Schwarz inequality gives\n$$\n(a + b + c)^2 \\le 3(a^2 + b^2 + c^2).\n$$\nThus $(a + b + c) < 6$ and hence $\\frac{(a + b + c)^3}{9} < 4(a + b + c)$. Now the AM-GM inequality gives\n$$\n(a + b + c)^3 \\ge 27abc.\n$$\nThus\n$$\n3abc \\le \\frac{(a + b + c)^3}{9} < 4(a + b + c).\n$$\n\nAlternately, we have $a + b + c < 6$ so that $a^2 + b^2 + c^2 < 2(a + b + c) < 12$. Thus\n$$\n(a^2 + b^2 + c^2)(a + b + c) < 12(a + b + c).\n$$\nUsing the AM-GM inequality, we have\n$$\na^2 + b^2 + c^2 \\ge 3(abc)^{2/3}, \\quad a + b + c \\ge 3(abc)^{1/3}.\n$$\nThus\n$$\n9abc \\le (a^2 + b^2 + c^2)(a + b + c) < 12(a + b + c),\n$$\ngiving the required inequality. Thus",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32381,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $P(x)$ be a given polynomial with integer coefficients. Prove that there exist two polynomials $Q(x)$ and $R(x)$, again with integer coefficients, such that (i) $P(x) Q(x)$ is a polynomial in $x^{2}$; and (ii) $P(x) R(x)$ is a polynomial in $x^{3}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\\cdots+a_{n} x^{n}$ be a polynomial with integer coefficients.\n\nPart (i) We may write\n$$\nP(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\\cdots+x\\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\\cdots\\right)\n$$\nDefine\n$$\nQ(x)=a_{0}+a_{2} x^{2}+a_{4} x^{4}+\\cdots-x\\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\\cdots\\right)\n$$\nThen $Q(x)$ is also a polynomial with integer coefficients and\n$$\nP(x) Q(x)=\\left(a_{0}+a_{2} x^{2}+a_{4} x^{4}+\\cdots\\right)^{2}-x^{2}\\left(a_{1}+a_{3} x^{2}+a_{5} x^{5}+\\cdots\\right)^{2}\n$$\nis a polynomial in $x^{2}$.\n\nPart (ii) We write again\n$$\nP(x)=A(x)+x B(x)+x^{2} C(x)\n$$\nwhere\n$$\n\\begin{aligned}\n& A(x)=a_{0}+a_{3} x^{3}+a_{6} x^{6}+\\cdots \\\\\n& B(x)=a_{1}+a_{4} x^{3}+a_{7} x^{6}+\\cdots \\\\\n& C(x)=a_{2}+a_{5} x^{3}+a_{8} x^{6}+\\cdots\n\\end{aligned}\n$$\nNote that $A(x), B(x)$ and $C(x)$ are polynomials with integer coefficients and each of these is a polynomial in $x^{3}$. We may introduce\n$$\n\\begin{aligned}\n& S(x)=A(x)+\\omega x B(x)+\\omega^{2} x^{2} C(x) \\\\\n& T(x)=A(x)+\\omega^{2} x B(x)+\\omega x^{2} C(x)\n\\end{aligned}\n$$\nwhere $\\omega$ is an imaginary cube-root of unity. Then\n$$\n\\begin{aligned}\nS(x) T(x)=(A(x))^{2}+x^{2}(B(x))^{2}+x^{4}(C(x))^{2} & \\\\\n& -x A(x) B(x)-x^{3} B(x) C(x)-x^{2} C(x) A(x)\n\\end{aligned}\n$$\nsince $\\omega^{3}=1$ and $\\omega+\\omega^{2}=-1$. Taking $R(x)=S(x) T(x)$, we obtain\n$$\nP(x) R(x)=(A(x))^{3}+x^{3}(B(x))^{3}+x^{6}(C(x))^{3}-3 x^{3} A(x) B(x) C(x)\n$$\nwhich is a polynomial in $x^{3}$. This follows from the identity\n$$\n(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)=a^{3}+b^{3}+c^{3}-3 a b c\n$$\nAlternately, $R(x)$ may be directly defined by\n$$\n\\begin{aligned}\n& R(x)=(A(x))^{2}+x^{2}(B(x))^{2}+x^{4}(C(x))^{2} \\\\\n&-x A(x) B(x)-x^{3} B(x) C(x)-x^{2} C(x) A(x)\n\\end{aligned}\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32382,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $r$, $t$ and $a$ be three positive integers such that $r \\le t$ and let\n$$\nA = \\{1, 2, 3, \\dots, r\\}, \\quad B = \\{a+r+1, a+r+2, a+r+3, \\dots, a+r+t\\}.\n$$\nFor any set $P$ of integers and any integer $x$, we define $P + x = \\{p + x : p \\in P\\}$, called the translate of $P$ by $x$.\n\na. If $r = t$, prove that $Z$ (the set of all integers) is a union of a certain class of mutually disjoint translates of $A \\cup B$ if and only if $a = kr$ for some positive integer $k$.\n\nb. If $r < t$, prove that $Z$ is a union of a certain class of mutually disjoint translates of $A \\cup B$ if and only if $a = k(r + t)$ for some positive integer $k$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "First observe that $A$ and $B$ are separated by $a$ numbers and as such $A \\cup B$ is not a set of consecutive integers.\n\na. ($\\Rightarrow$) If $r = t$ and $a = kr$ for some $k \\in \\mathbb{N}$, then it is clear that $Z$ is covered by the mutually disjoint sets $(A \\cup B) + s$, where $s$ ranges over the set\n$$\n\\bigcup_{m \\in \\mathbb{Z}} \\{2(k+1)rm, 2(k+1)rm + r, \\dots, 2(k+1)rm + kr\\} = 2(k+1)rZ \\cup \\{2(k+1)rZ + r\\} \\cup \\dots \\cup \\{2(k+1)rZ + kr\\}.\n$$\n(Here $xP$ denotes the set $\\{x : p \\in P\\}$.)\n\nb. ($\\Rightarrow$) If $r < t$ and $a = k(r + t)$ for some $k \\in \\mathbb{N}$, then $Z$ is covered by the mutually disjoint sets $(A \\cup B) + s'$, where $s'$ ranges the set $\\{(r+t)s : s \\in Z\\} = (r+t)Z$.\n\na. ($\\Leftarrow$) If $r = t$, and $Z$ is covered by mutually disjoint sets of the form $(A \\cup B) + s$, for $s$ in a fixed subset of $Z$, then any translate of $A \\cup B$ has a gap of size $a+r+1-(r+1)=a$, which should be covered by mutually disjoint sets of size $r$. This proves that $a = kr$, for some $k \\in \\mathbb{N}$.\n\nb. ($\\Leftarrow$) If $r < t$, and $Z$ is covered by a certain class of mutually disjoint translates of $A \\cup B$, we will prove that the translates of $A$ and $B$ must alternate in any covering of $Z$ by translates of $A \\cup B$. This would imply that $a = k(r + t)$ for some $k \\in \\mathbb{N}$. If a covering contained two consecutive translates $A + x$ and $A + x + r$ of $A$, it would contain the corresponding matching translates $B + x$ and $B + x + r$ of $B$. But\n\n$B+x = \\{a+r+1+x, a+r+2+x, ..., a+r+t+x\\}$ and $B+x+r = \\{a+2r+1+x, a+2r+2+x, ..., a+2r+t+x\\}$. As $a+2r+1+x \\le a+r+t+x$, we see that $B+x$ and $B+x+r$ overlap. Hence this is impossible.\n\nSuppose now that a covering of $Z$ contained two consecutive translates $B+x$ and $B+x+t$ of $B$. Then the covering must contain the corresponding matching translates $A+x$ and $A+x+t$ of $A$.\n\nNow\n$$\nA+x = \\{1+x, 2+x, ..., r+x\\} \\text{ and } A+x+t = \\{1+x+t, 2+x+t, ..., r+x+t\\}.\n$$\nThe gap between $A+x$ and $A+x+t$ corresponds to the set $\\{r+x+1, r+x+2, ..., t+x\\}$ which is of size $t-r$, which is less than $t$. So $A+x$ must be followed immediately by another translate of $A$ itself. But this is exactly what we proved to be impossible just above. Thus two consecutive translates of neither $A$ nor $B$ can occur in any covering of $Z$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32383,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that three distinct non-zero integers $a$, $b$, $c$ satisfy the equation\n$$\n\\frac{a}{b} + \\frac{b}{c} + \\frac{c}{a} = 3\n$$\nif and only if $a$, $b$, $c$ are given by\n$$\na = kuv^2, \\quad b = -ku^2(u+v), \\quad c = kv(u+v)^2,\n$$\n(up to cyclic permutations) for some integers $u$, $v$, $k$.",
    "options": [],
    "answer": "All solutions are, up to cyclic permutation, a = k u v^2, b = −k u^2(u+v), c = k v (u+v)^2 for some integers u, v, k.",
    "solution": "Let $a$, $b$, $c$ be a solution of the given equation. Putting\n$$\n\\frac{a}{b} = X, \\quad \\frac{b}{c} = Y,\n$$\nwe have $c/a = (1/XY)$. Note that $X$ and $Y$ are rational numbers. The equation takes the form\n$$\nX^2Y + XY^2 + 1 = 3XY. \\quad (1)\n$$\nThis is a cubic curve. We observe that $(1, 1)$ is a point on the curve (1). If we know the rational points on (1), then we can get integer solutions of the given equation.\nSuppose $(x, y)$ is a rational point on the curve (1). Then the line joining $(x, y)$ to $(1, 1)$ has rational slope. Conversely, any line through $(1, 1)$ having rational slope intersects the curve (1) in a point having rational coordinates.\nConsider the line\n$$\n\\frac{y-1}{x-1} = m,\n$$\nwhere $m \\ne -1$ is a non-zero rational number. Let its intersection with (1) be $(X_0, Y_0) \\ne (1, 1)$. Then we have $Y_0 = m(X_0 - 1) + 1$. Putting this in (1), we get\n$$\nX_0^2(m(X_0 - 1) + 1) + X_0(m(X_0 - 1) + 1)^2 + 1 = 3X_0(m(X_0 - 1) + 1).\n$$\nThis reduces to\n$$\n(X_0 - 1)^2(m(1 + m)X_0 + 1) = 0.\n$$\nSince $X_0 \\ne 1$, we have\n$$\nX_0 = -\\frac{1}{m(1+m)}, \\quad Y_0 = m(X_0 - 1) + 1 = -\\frac{m^2}{(1+m)}.\n$$\nThus\n$$\n\\frac{a}{b} = -\\frac{1}{m(1+m)}, \\quad \\frac{b}{c} = -\\frac{m^2}{(1+m)}, \\quad \\frac{a}{c} = \\frac{m}{(1+m)^2}.\n$$\nSuppose $m = v/u$, where $v$, $u$ are integers and $\\gcd(v, u) = 1$. We get\n$$\n\\frac{a}{c} = \\frac{uv^2}{v(v+u)^2}.\n$$\nThis gives\n$$\n\\frac{a}{uv^2} = \\frac{c}{v(v+u)^2}.\n$$\nSimilarly, the expression for $a/b$ gives\n$$\n\\frac{a}{v^2} = -\\frac{b}{u(u+v)}.\n$$\nWe thus obtain\n$$\n\\frac{a}{uv^2} = -\\frac{b}{u^2(u+v)} = \\frac{c}{v(v+u)^2}.\n$$\nLet each of these ratios be $p/q$ for some integers $p$, $q$. We obtain\n$$\nqa = puv^2, \\quad qb = -pu^2(u+v), \\quad qc = pv(u+v)^2.\n$$\nThis shows that $\\gcd(qb, qc) = \\pm p(u + v)$. Thus\n$$\n\\pm q(b, c) = \\pm p(u + v).\n$$\nBut $qa = puv^2$. It follows that $\\pm q \\gcd(a, b, c) = p$. Hence\n$$\n\\frac{p}{q} = \\pm \\gcd(a, b, c) = k,\n$$\nsay. Finally, we have\n$$\na = kuv^2, \\quad b = -ku^2(u+v), \\quad c = kv(u+v)^2.\n$$\nIt is easy to check that $\\{a, b, c\\}$ satisfies the given relation.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32384,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : (0, \\infty) \\to (0, \\infty)$ such that\n$$\nf(f(x) + y) = x f(1 + x y),\n$$\nfor all $x, y$ in $(0, \\infty)$.",
    "options": [],
    "answer": "f(x) = 1/x",
    "solution": "We show that $f(x) = 1/x$ is the only solution. We do this in several steps.\n\nStep 1 We show that $f(x)$ is a non-increasing function. Suppose the contrary; say $0 < a < b$ implies that $f(a) < f(b)$, for some $a, b$. Then\n$$\nw = \\frac{b f(b) - a f(a)}{b - a}\n$$\nis in $(0, \\infty)$. It is easy to check that $w > f(b)$ using $f(b) > f(a)$. Thus $w > f(b) > f(a)$. Taking $x = a$ and $y = w - f(a)$, we get\n$$\nf(w) = a f \\left( 1 + a (w - f(a)) \\right).\n$$\nSimilarly, $x = b$ and $y = w - f(b)$ gives\n$$\nf(w) = b f \\left( 1 + b (w - f(b)) \\right).\n$$\nThus $a = b$ contradicting $a < b$. We conclude that $a < b$ implies that $f(b) \\le f(a)$.\n\nStep 2 Taking $x = y = 1$, we get $f(f(1) + 1) = f(2)$. Similarly, putting $x = 1, y = 2$, we obtain $f(f(1) + 2) = f(3)$; $x = 2, y = 1$ implies $f(f(2) + 1) = 2 f(3)$. Thus\n$$\n\\begin{aligned}\n2 f(3) &= f(f(2) + 1) = f(f(f(1) + 1) + 1) \\\\\n&= (f(1) + 1) f(1 + (f(1) + 1)) = (f(1) + 1) f(f(1) + 2) = (f(1) + 1) f(3).\n\\end{aligned}\n$$\nThis gives $f(1) + 1 = 2$ and hence $f(1) = 1$.\n\nStep 3 Suppose $x > 1$, and put $y = 1 - (1/x)$. Then\n$$\nf\\left(f(x) - \\frac{1}{x} + 1\\right) = x f(1 + x - 1) = x f(x).\n$$\nIf $f(x) > (1/x)$, we see that $f(x) - (1/x) + 1 > 1$ and the monotonicity of $f(x)$ gives\n$$\nf\\left(f(x) - \\frac{1}{x} + 1\\right) \\le f(1) = 1.\n$$\nThus $x f(x) \\le 1$, contradicting $f(x) > (1/x)$.\nIf $f(x) < (1/x)$, we see that $f(x) - (1/x) + 1 < 1$, and\n$$\nf\\left(f(x) - \\frac{1}{x} + 1\\right) \\ge f(1) = 1,\n$$\nand hence\n$$\n1 \\le f\\left(f(x) - \\frac{1}{x} + 1\\right) = x f(x) < 1,\n$$\nwhich again is impossible. We conclude that $f(x) = \\frac{1}{x}$, for all $x > 1$.\n\nNow, take any $x > 0$. Then $f(x) + 1 > 1$, so that\n$$\nf(f(x) + 1) = \\frac{1}{f(x) + 1}.\n$$\nPutting $y = 1$ in the given relation, we get\n$$\nf(f(x) + 1) = x f(1 + x) = \\frac{x}{1 + x};\n$$\nwe have used $1 + x > 1$ in the last equality. Thus we obtain\n$$\n\\frac{1}{f(x) + 1} = \\frac{x}{1 + x},\n$$\nfor all $x > 0$. This gives $f(x) = 1/x$ for all $x > 0$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32385,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all positive integers $a, b$ such that $(a-b)^{a+b} = a^a$.",
    "options": [],
    "answer": "(a, b) = (8, 4)",
    "solution": "Let $d = \\gcd(a, b)$. Write $a = dx$ and $b = dy$. Then the equation reduces to $d^b(x - y)^{a+b} = x^a$. Since $\\gcd(x, x-y) = 1$, it follows that $x - y = \\pm 1$ and $d^b = x^a$. Using $b = dx$ and $a = dy$, this further reduces to $dy = x^x$ or\n$$\nd^{x \\pm 1} = x^x\n$$\nNote that $d = 1 = x$ does not lead to any solution. Let $p$ be a prime dividing $d$ and hence $x$. Let $p^\\alpha \\mid d$, $p^\\beta \\mid x$. Considering the powers of $p$, we see that\n$$\n\\alpha(x \\pm 1) = \\beta x.\n$$\nThis implies that $x \\nmid \\alpha$. Putting $\\alpha = x\\gamma$, we obtain $\\gamma(x \\pm 1) = \\beta$. If $\\gamma(x+1) = \\beta$, then we have $x+1 = k\\gamma^\\beta + 1 > \\beta$ and the relation is not tenable. If $\\gamma(x-1) = \\beta$ and $\\beta > 1$, then again we see that\n$$\nx - 1 = k\\gamma^\\beta - 1 \\ge 2^\\beta - 1 > \\beta,\n$$\nso that the relation is again not valid. The only possibility is $\\beta = 1$. This gives $\\gamma(kp - 1) = 1$, which leads to $p = 2, k = 1, \\gamma = 1$. In turn, we get $x = 2, d = 4, y = 1, a = 8, b = 4$. Observe $(8 - 4)^{8+4} = 4^{12} = 8^8$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32386,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle and let $P$ be an interior point such that $\\angle BPC = 90^{\\circ}$, $\\angle BAP = \\angle BCP$. Let $M, N$ be the mid-points of $AC, BC$ respectively. Suppose $BP = 2PM$. Prove that $A, P, N$ are collinear.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nExtend $CP$ to $D$ such that $CP = PD$. Let $\\angle BCP = \\alpha = \\angle BAP$. Observe that $BP$ is the perpendicular bisector of $CD$. Hence $BC = BD$ and $BCD$ is an isosceles triangle. Thus $\\angle BDP = \\alpha$. But then $\\angle BDP = \\alpha = \\angle BAP$. This implies that $B, P, A, D$ all lie on a circle. In turn, we conclude that $\\angle DAB = \\angle DPB = 90^{\\circ}$. Since $P$ is the midpoint of $CP$ (by construction) and $M$ is the mid-point of $CA$ (given), it follows that $PM$ is parallel to $DA$ and $DA = 2PM = BP$. Thus $DBPA$ is an isosceles trapezium and $DB$ is parallel to $PA$.\n\n![](attached_image_1.png)\n\nWe hence get\n\n$$\n\\angle DPA = \\angle BAP = \\angle BCP = \\angle NPC\n$$\n\nthe last equality follows from the fact that $\\angle BPC = 90^{\\circ}$, and $N$ is the mid-point of $CB$ so that $NP = NC = NB$ for the right-angled triangle $BPC$. It follows that $A, P, N$ are collinear.\nSolution:\n\nWe use coordinate geometry. Let us take $P = (0, 0)$, and the coordinate axes along $PC$ and $PB$; we take $C = (c, 0)$ and $B = (0, b)$. Let $A = (u, v)$. We see that $N = (c/2, b/2)$ and $M = ((u + c)/2, v/2)$. The condition $PB = 2PM$ translates to\n\n$$\n(u + c)^2 + v^2 = b^2\n$$\n\nWe observe that the slope of $CP = 0$; that of $CB$ is $-b/c$; that of $PA$ is $v/u$; and that of $BA$ is $(v - b)/u$. Taking proper signs, we can convert $\\angle PCB = \\angle PAB$, via tan function, to the following relation:\n\n$$\nu^2 + v^2 - vb = -cu$$\n\nThus we obtain\n\n$$\nu(u + c) = v(b - v), \\quad c(c + u) = b(b - v)\n$$\n\nIt follows that $v/u = b/c$. But then we get that the slope of $AP$ and $PN$ are the same. We conclude that $A, P, N$ are collinear.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32387,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be an interior point of a triangle $ABC$. Show, with usual notations, that\n$$\n\\frac{PA}{a} + \\frac{PB}{b} + \\frac{PC}{c} \\ge \\sqrt{3}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $G$ denote the centroid of $ABC$. We have\n$$\n\\sum \\frac{PA}{a} = \\sum \\frac{PAGA}{aGA} \\ge \\frac{\\sum PAGA}{\\max\\{aGA, bGB, cGC\\}}.\n$$\nWe observe that $\\vec{PA} \\cdot \\vec{GA} \\le PAGA$. Thus\n$$\n\\begin{align*}\n\\sum PAGA & \\ge \\sum \\vec{PA} \\cdot \\vec{GA} \\\\\n&= \\sum (\\vec{PG} + \\vec{GA}) \\cdot \\vec{GA} \\\\\n&= \\vec{PG} \\cdot (\\vec{GA} + \\vec{GB} + \\vec{GC}) + \\sum GA^2 \\\\\n&= GA^2 + GB^2 + GC^2.\n\\end{align*}\n$$\n\n$$\n\\begin{align*}\n\\sqrt{3}am_a &= \\frac{\\sqrt{3}}{2}a(2m_a) \\\\\n&= \\frac{\\sqrt{3}}{2}a\\sqrt{2(b^2+c^2)-a^2} \\\\\n&\\le \\frac{1}{4}(3a^2+2b^2+2c^2-a^2) \\\\\n&= \\frac{1}{2}(a^2+b^2+c^2).\n\\end{align*}\n$$\nThis implies that $3\\sqrt{3}aGA \\le a^2 + b^2 + c^2$. Hence we obtain\n$$\n\\sqrt{3}aGA \\le \\frac{1}{3}(a^2 + b^2 + c^2) = \\frac{4}{9} \\sum m_a^2 = \\sum GA^2.\n$$\nSimilarly, we obtain\n$$\n\\sqrt{3}bGB \\le \\sum GA^2, \\quad \\sqrt{3}cGC \\le \\sum GA^2.\n$$\n\nIt follows that\n$$\n\\max\\{aGA, bGB, cGC\\} \\leq \\sum GA^2.\n$$\nTherefore\n$$\n\\sum \\frac{PA}{a} \\geq \\frac{\\sum GA^2}{\\sum GA^2 / \\sqrt{3}} = \\sqrt{3}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32388,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 2$ be an integer. If $n$ divides $3^n + 4^n$, prove that $7$ divides $n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that $n$ is odd and $3$ does not divide $n$. Let $p$ be the least prime dividing $n$. Let $c$ be an integer such that $0 < c < p$ and $4c \\equiv 3 \\pmod{p}$. Such a number $c$ exists since $\\gcd(3, 4) = 1$. Thus we have\n$$\n\\begin{align*}\n4^n(c^{2n} - 1) &\\equiv (c^n - 1)((4c)^n + 4^n) \\\\\n&\\equiv (c^n - 1)(3^n + 4^n) \\pmod{p} \\\\\n&\\equiv 0 \\pmod{p}.\n\\end{align*}\n$$\nThus $p$ divides $c^{2n} - 1$. Let $d$ be the order of $c$ modulo $p$. Then $d$ divides $2n$. By Fermat's little theorem $d$ also divides $p-1$. Thus $d|\\gcd(2n, p-1)$. Let $q$ be a prime divisor of $d$. Then $q$ divides $2n$ and $p-1$. If $q$ divides $n$, the choice of $p$ shows that $q \\ge p$. Thus $q$ cannot divide $p-1$. We conclude that $q|2$ hence $q=2$. Since $2|(p-1)$ and $2$ does not divide $n$, it follows that $d=2$. Thus $c^2 \\equiv 1 \\pmod{p}$.\nIf $p$ divides $c-1$, then $4 \\equiv 4c \\equiv 3 \\pmod{p}$, which is impossible. Hence $p$ divides $c+1$. This gives\n$$\n0 \\equiv 4c + 4 \\equiv 3 + 4 \\equiv 7 \\pmod{p}.\n$$\nHence $p$ divides $7$. This forces $p=7$. Thus $7$ divides $n$.\n(Ananth Shankar) As in the first solution, let $p$ be the least prime dividing $n$. Then $p$ cannot be $2$ or $3$. Hence $\\gcd(p, 2) = \\gcd(p, 3) = 1$. Moreover $\\gcd(n, p-1) = 1$, as $p$ is the least prime dividing $n$. Hence we can find a natural number $a$ such that $an \\equiv 1 \\pmod{p-1}$. Since $p-1$ is even, $a$ must be odd. We observe that $3^n + 4^n$ divides $3^{an} + 4^{an}$, since $a$ is odd. Thus $p$ divides $3^{an} + 4^{an}$. But Fermat's little theorem gives $3^{p-1} \\equiv 1 \\pmod{p}$. Since $an-1$ is divisible by $p-1$, it follows that $3^{an} \\equiv 3 \\pmod{p}$. Similarly, we get $4^{an} \\equiv 4 \\pmod{p}$. It follows that\n$$\n3^{an} + 4^{an} \\equiv 3 + 4 = 7 \\pmod{p}.\n$$\nWe conclude that $p$ divides $7$. Hence $p=7$. This implies that $7$ divides $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32389,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x) = \\sum_{k=1}^{n} a_k x^k$ and $g(x) = \\sum_{k=1}^{n} \\frac{a_k}{2^k - 1} x^k$ be two polynomials with real coefficients, where $n \\ge 3$. Suppose $1, 2^{n+1}$ are roots of $g(x) = 0$. Prove that $f(x) = 0$ has a positive root smaller than $2^n$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "It is easy to see that $g(2x) - g(x) = f(x)$. Thus\n$$\n\\sum_{k=0}^{n} f(2^k) = g(2^{n+1}) - g(1) = 0.\n$$\nConsider the relation $f(1) + f(2) + \\dots + f(2^n) = 0$. If $f(2^n) \\neq 0$, then $f(1)$ and $f(2^n)$ have opposite signs. Hence there exists $\\alpha$ in $(1, 2^n)$ such that $f(\\alpha) = 0$. If $f(2^n) = 0$, omit this and the argument applies to $f(1) + f(2) + \\dots + f(2^{n-1}) = 0$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32390,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose the altitudes of a triangle are all integers whose sum does not exceed $20$ and its in-radius is also an integer. Find all possible values for the area of the triangle.",
    "options": [],
    "answer": "3√3, 12√3, 100/√15",
    "solution": "We begin with the standard expression:\n$$\n\\frac{1}{r} = \\frac{1}{h_a} + \\frac{1}{h_b} + \\frac{1}{h_c},\n$$\nwhere $r$ is the inradius and $h_a, h_b, h_c$ are the altitudes of the given triangle, in standard notation.\n\nThe arithmetic mean - harmonic mean inequality gives\n$$\n\\frac{1}{r} \\ge \\frac{9}{h_a + h_b + h_c} \\ge \\frac{9}{20}.\n$$\nThus $r \\le 20/9$. It follows that $r = 1$ or $2$.\n\n**Case 1.** Suppose $r = 1$. We may assume that $h_a \\le h_b \\le h_c$. This implies that $c \\le b \\le a$. Using $a < b + c$, we must have\n$$\n\\frac{1}{h_a} < \\frac{1}{h_b} + \\frac{1}{h_c}.\n$$\nThus $3h_a \\le h_a + h_b + h_c \\le 20$ and we get $h_a \\le 6$. We may easily rule out $h_a = 1$. If $h_a = 2$, then we see that $1/h_a = 1/h_b + 1/h_c$, which is impossible.\nIf $h_a = 3$, then we get\n$$\n\\frac{1}{h_b} + \\frac{1}{h_c} = 1 - \\frac{1}{3} = \\frac{2}{3}.\n$$\nThis leads to\n$$\n(2h_b - 3)(2h_c - 3) = 9.\n$$\nThis gives $(h_b, h_c) = (2, 6)$ or $(3, 3)$. The first pair is not possible, since $h_b \\ge h_a = 3$. In the second case we have $(h_a, h_b, h_c) = (3, 3, 3)$.\nA similar analysis shows that there are no possible triples in the cases $h_a = 4, 5, 6$.\n\n**Case 2.** Suppose $r = 2$. We see that $3 \\le h_a \\le 6$, in this case. Again a case by case analysis shows that there are no possible triangles when $h_a = 3$ or $4$. If $h_a = 5$, we have one possibility: $(h_a, h_b, h_c) = (5, 5, 10)$. If $h_a = 6$, we have the triple $(6, 6, 6)$.\n\nIn the cases $(h_a, h_b, h_c) = (3, 3, 3)$ and $(6, 6, 6)$, we have equilateral triangles. It is easy to get area in terms of the altitude: $\\Delta = h^2/\\sqrt{3}$. Thus we get areas equal to $3\\sqrt{3}$ and $12\\sqrt{3}$. In the other case $(h_a, h_b, h_c) = (5, 5, 10)$, we see that $a = b = 2c$. Heron's formula can be used to compute $\\Delta$. We get $\\Delta = 100/\\sqrt{15}$.\n\nThus the possible values for $\\Delta$ are $3\\sqrt{3}$, $12\\sqrt{3}$ and $100/\\sqrt{15}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32391,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all integers $n \\ge 2$ and all primes $p, q, r$ with the property: whenever distinct positive integers $a_1, a_2, \\dots, a_n$ are such that the difference $a_j - a_k$, $1 \\le j < k \\le n$ is divisible by at least one of $p, q, r$ implies that one of $p, q, r$ divides all the differences $a_j - a_k$, $1 \\le j < k \\le n$.",
    "options": [],
    "answer": "Only n = 2 works, for any primes p, q, r. For every n ≥ 3, the property fails for all choices of primes.",
    "solution": "If $n = 2$, the statement holds for any primes $p, q, r$. We show that the result fails for $n \\ge 3$. Suppose $n = 3$ and $p < q < r$ are some primes. Let $a_1 = 1$ and $a_2 = r+1$. Consider the numbers $r+jq$, $0 \\le j < p$. One of these numbers is divisible by $p$, say $r+1$ or $r+2$. Take $a_3 = r+1$ or $a_3 = r+2$. Then $a_3 - a_1$ is divisible by $p$; $a_3 - a_2$ is divisible by $q$; and $a_2 - a_1$ is divisible by $r$. None of $p, q, r$ divide all the differences. For $n > 3$, consider $a_1, a_2, a_3$ as above; and $a_j = pqr(j-3) + 1$ for $j \\ge 4$. We may easily check that each of the differences $a_j - a_k$ is divisible by at least one of $p, q, r$, but there is no single prime among $p, q, r$ which divides all the differences.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32392,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P$ be a path on the vertex set $V = \\{1, 2, \\dots, n\\}$, where $j$ is joined to $j+1$, $1 \\le j \\le n-1$. For each subset $A \\subset V$ and the induced subgraph $G(A)$ of $P$, define $\\mu(A) = |A| + O(G(A))$, where $O(G(A))$ is the number of components of $G(A)$, each with an odd number of vertices; $\\mu(\\emptyset) = 0$.\nLet\n$$\nT(p, r) = \\{ A \\subset V \\mid |A| = p, \\mu(A) = 2r \\},\n$$\nfor $r \\le p \\le 2r$. Prove that\n$$\n|T(p, r)| = \\binom{n-r}{p-r} \\binom{n-p+1}{2r-p}.\n$$\n(Example: Let $V = \\{1, 2, \\dots, 9\\}$ and $A = \\{\\{1, 2\\}, \\{4, 5, 6\\}, \\{9\\}\\}$. In this case $\\mu(A) = 6+2=8$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "If $p = r = 0$, we have $A = \\emptyset$ so that $\\mu(A) = 0$. In this case $T(0,0) = \\{\\emptyset\\}$ and $|T(0,0)| = 1$, which agrees with the formula.\n\nAssume $p \\ge r \\ge 1$. Add two two dummy vertices $0$, $n+1$ to $P$ such that $0$ is joined to $1$ and $n$ is joined to $n+1$. Let\n$$\n\\Pi = (l_1, m_1, l_2, m_2, \\dots, l_k, m_k, l_{k+1})\n$$\nbe an ordered $(2k+1)$-tuple of positive integers such that\n$$\nl_1 + m_1 + l_2 + m_2 + \\cdots + l_k + m_k + l_{k+1} = n + 2. \\quad (*)\n$$\n(Thus $\\Pi$ is an ordered partition of $n+2$ into $2k+1$ parts.) We obtain a subset $A = A(\\Pi)$ of $V$ from this as follows: starting from the left end of the sequence $\\langle 0, 1, 2, \\dots, n+1 \\rangle$, we omit the first $l_1$ numbers; choose the next $m_1$ numbers; omit the next $l_2$ numbers; choose the next $m_2$ numbers and so on alternately. Finally, we omit the last $l_{k+1}$ numbers. (Observe, that $l_1 \\ge 1$ and $l_{k+1} \\ge 1$ implies that the dummy vertices $0$ and $n+1$ are not used at all.) The union of $k$ chosen sets of numbers consisting of $m_1, m_2, \\dots, m_k$ elements respectively is defined as $A$. We see that $A \\subseteq \\{1, 2, 3, \\dots, n\\}$. Conversely, any $A \\subseteq \\{1, 2, 3, \\dots, n\\}$ gives rise to an ordered $2k+1$ tuple of positive integers in a unique way; since $0 \\notin A$ and $n+1 \\notin A$, we have $l_1 \\ge 1$ and $l_{k+1} \\ge 1$. Since the $l$'s are positive, we see that $G(A)$ has $k$ components of vertex sizes $m_1, m_2, \\dots, m_k$. If $u$ of these $k$ numbers, say, $m_{i_1}, m_{i_2}, \\dots, m_{i_u}$ are odd and the remaining $v$ numbers, say, $m_{j_1}, m_{j_2}, \\dots, m_{j_v}$ are even, $0 \\le u, v \\le k$, then $u+v=k$, and $O(G(A)) = u$. Thus $\\mu(A) = m_1 + m_2 + \\cdots + m_k + u$. We count $A$ for which $\\mu(A) = 2r$, $1 \\le r \\le n$. Let $m_{i_1}, m_{i_2}, \\dots, m_{i_u}$ be respectively equal to $2m'_{i_1} - 1, 2m'_{i_2} - 1, \\dots, 2m'_{i_u} - 1$ and $m_{j_1}, m_{j_2}, \\dots, m_{j_v}$ be equal to $2m'_{j_1}, 2m'_{j_2}, \\dots, 2m'_{j_v}$. Then\n$$\n\\mu(A) = 2m'_{i_1} + 2m'_{i_2} + \\cdots + 2m'_{i_u} + 2m'_{j_1} + 2m'_{j_2} + \\cdots 2m'_{j_v} = 2r.\n$$\nThus we get\n$$\nm'_{i_1} + m'_{i_2} + \\cdots + m'_{i_u} + m'_{j_1} + m'_{j_2} + \\cdots m'_{j_v} = r.\n$$\nThe number of positive solutions of this is $\\binom{r-1}{k-1}$. Also we have\n$$\nl_1 + l_2 + \\cdots + l_{k+1} = n + 2 - (2r - u) = n - 2r + u + 2.\n$$\nThe number of positive solutions to this is $\\binom{n-2r+u+1}{k}$. Thus the number of positive solutions of $(*)$ with $\\mu(A) = 2r$ is\n$$\n\\sum_{k \\ge 1} \\binom{r-1}{k-1} \\binom{n-2r+u+1}{k}.\n$$\nSince $p+u=2r$, we have\n$$\n|T(p, r)| = \\sum_{k \\ge 1} \\binom{r-1}{k-1} \\binom{n-2r+u+1}{k} \\binom{k}{u}.\n$$\nas any $u$ of the $k$ $m$'s may be chosen to be odd and the rest even, giving rise to the factor $\\binom{k}{u}$. Using\n$$\n\\binom{n}{k} \\binom{k}{m} = \\binom{n}{m} \\binom{n-m}{n-k},\n$$\nwe get\n$$\n\\begin{align*}\n|T(p, r)| &= \\binom{n-p+1}{u} \\sum_{k \\ge 1} \\binom{r-1}{k-1} \\binom{n-p+1-u}{n-p+1-k} \\\\\n&= \\binom{n-p+1}{u} \\binom{n-p-u+r}{n-p} \\\\\n&= \\binom{n-p+1}{2r-p} \\binom{n-r}{n-p} \\\\\n&= \\binom{n-r}{p-r} \\binom{n-p+1}{2r-p};\n\\end{align*}\n$$\nwhere we have used Vander Monde identity.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32393,
    "subject": "Mathematics (Multi-modal)",
    "question": "If $a_1, a_2, \\dots, a_n$ are $n$ non-zero complex numbers, not necessarily distinct, and $k, l$ are distinct positive integers such that $a_1^k, a_2^k, \\dots, a_n^k$ and $a_1^l, a_2^l, \\dots, a_n^l$ are two identical collections of numbers. Prove that each $a_j$, $1 \\le j \\le n$, is a root of unity.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "The given hypothesis implies that there is a bijection $f : \\{1, 2, 3, \\dots, n\\} \\to \\{1, 2, 3, \\dots, n\\}$ such that $f(j) = m$ implies $a_j^k = a_m^l$. Consider the sequence\n$$\n1, f(1), f^{(2)}(1), \\dots,\n$$\nSince $f$ is a bijection on a finite set, there are positive integers $r, s$ such that $f^{(r)}(1) = f^{(s)}(1)$, where $0 \\le r < s \\le n-1$. (Here $f^{(0)}(1) = 1$). Since $f$ is a bijection, we get $f^{(t)}(1) = 1$, where $t = s - r$. We have\n$$\na_1^k = a_{f(1)}^l, \\quad a_{f(1)}^k = a_{f^{(2)}(1)}^l.\n$$\nThus $a_1^{k^2} = a_{f^{(2)}(1)}^{l^2}$. Using induction, we get\n$$\na_1^{k^t} = a_{f^{(t)}(1)}^{l^t} = a_1^{l^t}.\n$$\nSince $l \\neq k$, it follows that $a_1$ is a root of unity. This holds for any $a_j$ in place of $a_1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32394,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine a sequence $\\left\\langle a_{n}\\right\\rangle_{n=1}^{\\infty}$ as follows:\n\n$$\na_{n}= \\begin{cases}0, & \\text{ if the number of positive divisors of } n \\text{ is odd } \\\\ 1, & \\text{ if the number of positive divisors of } n \\text{ is even }\\end{cases}\n$$\n\n(The positive divisors of $n$ include 1 as well as $n$.) Let $x=0 . a_{1} a_{2} a_{3} \\ldots$ be the real number whose decimal expansion contains $a_{n}$ in the $n$-th place, $n \\geq 1$. Determine, with proof, whether $x$ is rational or irrational.",
    "options": [],
    "answer": "x is irrational",
    "solution": "Solution:\n\nWe show that $x$ is irrational. Suppose that $x$ is rational. Then the sequence $\\left\\langle a_{n}\\right\\rangle_{n=1}^{\\infty}$ is periodic after some stage; there exist natural numbers $k, l$ such that $a_{n}=a_{n+l}$ for all $n \\geq k$. Choose $m$ such that $m l \\geq k$ and $m l$ is a perfect square. Let\n\n$$\nm=p_{1}^{\\alpha_{1}} p_{2}^{\\alpha_{2}} \\ldots p_{r}^{\\alpha_{r}}, \\quad l=p_{1}^{\\beta_{1}} p_{2}^{\\beta_{2}} \\ldots p_{r}^{\\beta_{r}}\n$$\n\nbe the prime decompositions of $m, l$ so that $\\alpha_{j}+\\beta_{j}$ is even for $1 \\leq j \\leq r$. Now take a prime $p$ different from $p_{1}, p_{2}, \\ldots, p_{r}$. Consider $m l$ and $p m l$. Since $p m l-m l$ is divisible by $l$, we have $a_{p m l}=a_{m l}$. Hence $d(p m l)$ and $d(m l)$ have same parity. But $d(p m l)=2 d(m l)$, since $\\operatorname{gcd}(p, m l)=1$ and $p$ is a prime. Since $m l$ is a square, $d(m l)$ is odd. It follows that $d(p m l)$ is even and hence $a_{p m l} \\neq a_{m l}$. This contradiction implies that $x$ is irrational.\n\n\nAlternative Solution:\n\nAs earlier, assume that $x$ is rational and choose natural numbers $k, l$ such that $a_{n}=a_{n+l}$ for all $n \\geq k$. Consider the numbers $a_{m+1}, a_{m+2}, \\ldots, a_{m+l}$, where $m \\geq k$ is any number. This must contain at least one $0$. Otherwise $a_{n}=1$ for all $n \\geq k$. But $a_{r}=0$ if and only if $r$ is a square. Hence it follows that there are no squares for $n>k$, which is absurd. Thus every $l$ consecutive terms of the sequence $\\left\\langle a_{n}\\right\\rangle$ must contain a $0$ after certain stage. Let $t=\\max \\{k, l\\}$, and consider $t^{2}$ and $(t+1)^{2}$. Since there are no squares between $t^{2}$ and $(t+1)^{2}$, we conclude that $a_{t^{2}+j}=1$ for $1 \\leq j \\leq 2 t$. But then, we have $2 t(>l)$ consecutive terms of the sequence $\\left\\langle a_{n}\\right\\rangle$ which miss $0$, contradicting our earlier observation.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32395,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle in which $AB > AC$; $AM$ be the median and $AK$ be the angle bisector with $M, K$ on $BC$. Let $L$ be a point on $AM$ such that $KL$ is parallel to $AC$. Prove that $CL$ is perpendicular to $AK$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nExtend $AK$ to meet the circum-circle of $ABC$ in $D$, and join $MD$. Let $P$ be the point of intersection of $AK$ and $CL$. Observe that $\\angle DMK = 90^\\circ$ and $D, M, O$ are collinear. We show that $DMK$ is similar to $CPK$, which proves that $CL$ is perpendicular to $AK$. It is sufficient to prove that $KD/KC = KM/KP$. But $AK \\cdot KD = BK \\cdot KC$, which gives $KD/KC = BK/AK$. Thus we need to prove that\n$$\n\\frac{KM}{KP} = \\frac{BK}{AK}.\n$$\n\nSince $CL$ is a transversal in the triangle $AMK$. Menelaus' theorem gives\n$$\n\\frac{PA}{KP} = \\frac{AL \\cdot MC}{LM \\cdot CK}.\n$$\nBut $KL$ is parallel to $CA$, so that $AL/LM = CK/KM$. This implies that\n$$\n\\frac{PA}{KP} = \\frac{CM}{KM}.\n$$\nThus\n$$\n\\frac{AK}{KP} = \\frac{CM + MK}{KM}.\n$$\nAll we need to show is $BK = CM + MK$. Since $BM = MC$, the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32396,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real numbers $x$ such that\n\n$$\n\\left[x^{2}+2x\\right] = [x]^{2} + 2[x]\n$$\n\n(Here $[x]$ denotes the largest integer not exceeding $x$.)",
    "options": [],
    "answer": "{-1, -2, -3, ...} ∪ ⋃_{n=-1}^{∞} [n, sqrt(1 + (n+1)^2) - 1)",
    "solution": "Solution:\nAdding $1$ to both sides, the equation reduces to\n\n$$\n\\left[(x+1)^{2}\\right] = ([x+1])^{2}\n$$\n\nWe have used $[x] + m = [x + m]$ for every integer $m$.\n\nSuppose $x+1 \\leq 0$. Then $[x+1] \\leq x+1 \\leq 0$. Thus\n\n$$\n([x+1])^{2} \\geq (x+1)^{2} \\geq \\left[(x+1)^{2}\\right] = ([x+1])^{2}\n$$\n\nThus equality holds everywhere. This gives $[x+1] = x+1$ and thus $x+1$ is an integer. Using $x+1 \\leq 0$, we conclude that\n\n$$\nx \\in \\{-1, -2, -3, \\ldots\\}\n$$\n\nSuppose $x+1 > 0$. We have\n\n$$\n(x+1)^{2} \\geq \\left[(x+1)^{2}\\right] = ([x+1])^{2}\n$$\n\nMoreover, we also have\n\n$$\n(x+1)^{2} \\leq 1 + \\left[(x+1)^{2}\\right] = 1 + ([x+1])^{2}\n$$\n\nThus we obtain\n\n$$\n[x] + 1 = [x+1] \\leq (x+1) < \\sqrt{1 + ([x+1])^{2}} = \\sqrt{1 + ([x] + 1)^{2}}\n$$\n\nThis shows that\n\n$$\nx \\in \\left[n, \\sqrt{1 + (n+1)^{2}} - 1\\right)\n$$\n\nwhere $n \\geq -1$ is an integer. Thus the solution set is\n\n$$\n\\{-1, -2, -3, \\ldots\\} \\cup \\left\\{\\bigcup_{n=-1}^{\\infty} \\left[n, \\sqrt{1 + (n+1)^{2}} - 1\\right)\\right\\}\n$$\n\nIt is easy to verify that all the real numbers in this set indeed satisfy the given equation.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32397,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $x$, $y$, $z$ and $a$, $b$, $c$ be positive real numbers such that $x + y + z = a + b + c$ and $xyz = abc$. Suppose $\\max\\{x, y, z\\} \\ge \\max\\{a, b, c\\}$. Prove that\n$$\nab + bc + ca \\ge xy + yz + zx.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We may assume $x \\ge y \\ge z$ and $a \\ge b \\ge c$. We are given $x \\ge a$. Consider the cubics\n$$\nP(t) = (t - x)(t - y)(t - z), \\quad Q(t) = (t - a)(t - b)(t - c).\n$$\nSince $a$ is the largest root of $Q(t) = 0$, it follows that $Q(s) \\ge 0$ for $s \\ge a$. In particular, $Q(x) \\ge 0$. Observe that\n$$\nP(t) - Q(t) = (\\alpha - \\beta)t,\n$$\nwhere $\\alpha = xy + yz + zx$ and $\\beta = ab + bc + ca$. We thus get\n$$\n(\\alpha - \\beta)x = P(x) - Q(x) = -Q(x) \\le 0.\n$$\nSince $x$ is positive, we conclude that $\\alpha \\le \\beta$. This gives the desired inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32398,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nAll the points in the plane are coloured using three colours. Prove that there exists a triangle with vertices having the same colour such that either it is isosceles or its angles are in geometric progression.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nConsider a circle of positive radius in the plane and inscribe a regular heptagon $A B C D E F G$ in it. Since the seven vertices of this heptagon are coloured by three colours, some three vertices have the same colour, by pigeon-hole principle. Consider the triangle formed by these three vertices. Let us call the part of the circumference separated by any two consecutive vertices of the heptagon an arc. The three vertices of the same colour are separated by arcs of length $l, m, n$ as we move, say counter-clockwise, along the circle, starting from a fixed vertex among these three, where $l+m+n=7$. Since, the order of $l, m, n$ does not matter for a triangle, there are four possibilities: $1+1+5=7 ; 1+2+4=7 ; 1+3+3=7 ; 2+2+3=7$. In the first, third and fourth cases, we have isosceles triangles. In the second case, we have a triangle whose angles are in geometric progression. The four corresponding figures are shown below.\n\n![](attached_image_1.png)\n\n(i)\n\n![](attached_image_2.png)\n\n(ii)\n\n![](attached_image_3.png)\n\n(iii)\n\n![](attached_image_4.png)\n\n(iv)\n\nIn (i), $A B=B C$; in (iii), $A E=B E$; in (iv), $A C=C E$; and in (ii) we see that $\\angle D=\\pi / 7$, $\\angle A=2 \\pi / 7$ and $\\angle B=4 \\pi / 7$ which are in geometric progression.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32399,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $v_1, v_2, \\dots, v_n$ be $n (\\ge 2)$ unit vectors in the plane. Prove that there exist $\\lambda_1, \\lambda_2, \\dots, \\lambda_n$, each equal to $+1$ or $-1$, such that\n$$\n|\\lambda_1 v_1 + \\lambda_2 v_2 + \\dots + \\lambda_n v_n| \\le \\sqrt{2}.\n$$\n(Here $|v|$ denotes the length of the vector $v$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We prove the result for a more general class of vectors having magnitude not exceeding $1$. We use induction on $n$. If $n = 2$, parallelogram law gives\n$$\n|v_1 + v_2|^2 + |v_1 - v_2|^2 = 2(|v_1|^2 + |v_2|^2) \\le 4.\n$$\nHence either $|v_1 + v_2| \\le \\sqrt{2}$ or $|v_1 - v_2| \\le \\sqrt{2}$, giving the result. Suppose $n \\ge 3$ and the result is true for $n-1$ vectors. Consider $n$ vectors $v_1, v_2, \\dots, v_n$ such that $|v_j| \\le 1$, for $1 \\le j \\le n$. Among the six vectors $\\pm v_1, \\pm v_2, \\pm v_3$, there are two vectors, the angle $\\theta$ between which is $\\le 60^\\circ$. If we denote these two by $u_1$ and $u_2$, we have\n$$\n\\begin{aligned}\n|u_1 - u_2|^2 &= |u_1|^2 + |u_2|^2 - 2|u_1||u_2| \\cos \\theta \\\\\n&\\le 2 - 2 \\cos \\theta \\\\\n&\\le 2 - 2 \\cos 60^\\circ = 1.\n\\end{aligned}\n$$\nThus $|u_1 - u_2| \\le 1$. After renaming if necessary, we can take $u_1 - u_2 = \\lambda_1' v_1 + \\lambda_2' v_2$. Now consider $n-1$ vectors $u_1 - u_2, v_3, \\dots, v_n$. Induction hypothesis applies to this set of vectors. We therefore get\n$$\n|\\lambda'(\\lambda_1' v_1 + \\lambda_2' v_2) + \\lambda_3 v_3 + \\dots + \\lambda_n v_n| \\le \\sqrt{2}.\n$$\nTaking $\\lambda_j = \\lambda'\\lambda'_j$, for $j = 1, 2$, we get\n$$\n|\\lambda_1 v_1 + \\lambda_2 v_2 + \\lambda_3 v_3 + \\dots + \\lambda_n v_n| \\le \\sqrt{2}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32400,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be an acute-angled triangle and let $H$ be its orthocentre. Let $h_{\\max}$ denote the largest altitude of the triangle $ABC$. Prove that\n\n$$\nAH + BH + CH \\leq 2 h_{\\max}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $\\angle C$ be the smallest angle, so that $CA \\geq AB$ and $CB \\geq AB$. In this case the altitude through $C$ is the longest one. Let the altitude through $C$ meet $AB$ in $D$ and let $H$ be the orthocentre of $ABC$. Let $CD$ extended meet the circumcircle of $ABC$ in $K$. We have $CD = h_{\\max}$ so that the inequality to be proved is\n\n$$\nAH + BH + CH \\leq 2 CD\n$$\n\nUsing $CD = CH + HD$, this reduces to $AH + BH \\leq CD + HD$. However, we observe that $AH = AK$, $BH = BK$ and $HD = DK$. (For example, $BH = BK$ and $DH = DK$ follow from the congruency of the right-angled triangles $DBK$ and $DBH$.)\n\nThus we need to prove that $AK + BK \\leq CK$. Applying Ptolemy's theorem to the cyclic quadrilateral $BCAK$, we get\n\n$$\nAB \\cdot CK = AC \\cdot BK + BC \\cdot AK \\geq AB \\cdot BK + AB \\cdot AK\n$$\n\nThis implies that $CK \\geq AK + BK$, which is precisely what we are looking for.\n\nThere were other beautiful solutions given by students who participated in INMO-2009. We record them here.\n\n1. Let $AD, BE, CF$ be the altitudes and $H$ be the orthocentre. Observe that\n\n$$\n\\frac{AH}{AD} = \\frac{[AHB]}{[ADB]} = \\frac{[AHC]}{[ADC]}\n$$\n\nThis gives\n\n$$\n\\frac{AH}{AD} = \\frac{[AHB] + [AHC]}{[ADB] + [ADC]} = 1 - \\frac{[BHC]}{[ABC]}\n$$\n\nSimilar expressions for the ratios $BH/BE$ and $CH/CF$ may be obtained. Adding, we get\n\n$$\n\\frac{AH}{AD} + \\frac{BH}{BE} + \\frac{CH}{CF} = 2\n$$\n\nSuppose $AD$ is the largest altitude. We get\n\n$$\n\\frac{AH}{AD} + \\frac{BH}{AD} + \\frac{CH}{AD} \\leq \\frac{AH}{AD} + \\frac{BH}{BE} + \\frac{CH}{CF} = 2\n$$\n\nThis gives the result.\n\n2. Let $O$ be the circumcentre and let $L, M, N$ be the midpoints of $BC, CA, AB$ respectively. Then we know that $AH = 2OL$, $BH = 2OM$ and $CH = 2ON$. As earlier, assume $AD$ is the largest altitude. Then $BC$ is the least side. We have\n\n$$\n\\begin{aligned}\n4[ABC] &= 4[BOC] + 4[COA] + 4[AOB] \\\\\n&= BC \\times 2OL + CA \\times 2OM + AB \\times 2ON \\\\\n&= BC \\times AH + CA \\times BH + AB \\times CH \\\\\n&\\geq AB(AH + BH + CH)\n\\end{aligned}\n$$\n\nThus\n\n$$\nAH + BH + CH \\leq \\frac{4[ABC]}{AB} = 2AD\n$$\n\n3. We make use of the fact that $AH = 2R \\cos \\angle A$, $BH = 2R \\cos \\angle B$, $CH = 2R \\cos \\angle C$ and $AD = 2R \\sin \\angle B \\sin \\angle C$, where $R$ is the circumradius of $ABC$. We are assuming that $AD$ is the largest altitude so that $\\angle A$ is the least angle. Thus we have to prove that\n\n$$\n\\cos \\angle A + \\cos \\angle B + \\cos \\angle C \\leq 2 \\sin \\angle B \\sin \\angle C\n$$\n\nunder the assumption $\\angle A \\leq \\angle B$ and $\\angle A \\leq \\angle C$. On multiplying this by $2 \\sin \\angle A$, this is equivalent to\n\n$$\n\\begin{aligned}\n&2(\\sin \\angle A \\cos \\angle A + \\sin \\angle A \\cos \\angle B + \\sin \\angle A \\cos \\angle C) \\\\\n&\\leq 4 \\sin \\angle A \\sin \\angle B \\sin \\angle C = \\sin 2A + \\sin 2B + \\sin 2C\n\\end{aligned}\n$$\n\nThis is equivalent to\n\n$$\n\\cos \\angle B (\\sin \\angle A - \\sin \\angle B) + \\cos \\angle C (\\sin \\angle A - \\sin \\angle C) \\leq 0\n$$\n\nSince $ABC$ is acute-angled and $A$ is the least angle, the result follows.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32401,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle in which $\\angle A = 60^\\circ$. Let $T$ be the point where the in-circle and the nine-point circle of $ABC$ touch each other. If $r$ is the in-radius of $ABC$, prove that $AT = r$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\nWe first observe that $AI = r \\cos\\frac{A}{2} = 2r$. Let $T'$ be point on $AI$ such that $AT' = T'I = r$. Obviously $T'$ lies on the in-circle of $ABC$. We show that $T'$ also lies on the nine-point circle of $ABC$.\nLet $D, E, F$ be the midpoints of $BC, CA, AB$ respectively. Note that $T'F$ is parallel to $IB$. Hence $\\angle AFT' = \\angle ABI = \\frac{B}{2}$. We also observe that $DF$ is parallel to $AC$, so that $\\angle BFD = \\angle BAC = A$. It follows that\n$$\n\\angle T'FD = 180^\\circ - (A + \\frac{B}{2}).\n$$\nSimilarly, we obtain\n$$\n\\angle T'ED = 180^\\circ - (A + \\frac{C}{2}).\n$$\nIt is easy to check that\n$$\n\\angle T'FD + \\angle T'ED = 180^\\circ.\n$$\nHence $T', F, D, E$ are concyclic. Since $F, D, E$ are on the nine-point circle, it follows that $T'$ is also on the same circle.\nHowever, there is only one point at which the in-circle and the nine-point circle touch each other (Feuerbach's theorem). It follows that $T = T'$. Hence: $AT = r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32402,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c$ be positive real numbers such that $a^{3}+b^{3}=c^{3}$. Prove that\n\n$$\na^{2}+b^{2}-c^{2}>6(c-a)(c-b)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe given inequality may be written in the form\n\n$$\n7 c^{2}-6(a+b) c-\\left(a^{2}+b^{2}-6 a b\\right)<0\n$$\n\nPutting $x=7 c^{2}$, $y=-6(a+b) c$, $z=-\\left(a^{2}+b^{2}-6 a b\\right)$, we have to prove that $x+y+z<0$. Observe that $x, y, z$ are not all equal ($x>0, y<0$). Using the identity\n\n$$\nx^{3}+y^{3}+z^{3}-3 x y z=\\frac{1}{2}(x+y+z)\\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\\right]\n$$\n\nwe infer that it is sufficient to prove $x^{3}+y^{3}+z^{3}-3 x y z<0$. Substituting the values of $x, y, z$, we see that this is equivalent to\n\n$$\n343 c^{6}-216(a+b)^{3} c^{3}-\\left(a^{2}+b^{2}-6 a b\\right)^{3}-126 c^{3}(a+b)\\left(a^{2}+b^{2}-6 a b\\right)<0\n$$\n\nUsing $c^{3}=a^{3}+b^{3}$, this reduces to\n\n$$\n343\\left(a^{3}+b^{3}\\right)^{2}-216(a+b)^{3}\\left(a^{3}+b^{3}\\right)-\\left(a^{2}+b^{2}-6 a b\\right)^{3}-126\\left(a^{3}+b^{3}\\right)(a+b)\\left(a^{2}+b^{2}-6 a b\\right)<0\n$$\n\nThis may be simplified (after some tedious calculations) to,\n\n$$\n-a^{2} b^{2}\\left(129 a^{2}-254 a b+129 b^{2}\\right)<0\n$$\n\nBut $129 a^{2}-254 a b+129 b^{2}=129(a-b)^{2}+4 a b>0$. Hence the result follows.\n\n\n1. We have\n\n$$\na^{3}=c^{3}-b^{3}=(c-b)\\left(c^{2}+c b+b^{2}\\right)\n$$\n\nwhich is same as\n\n$$\n\\frac{a^{2}}{c-b}=\\frac{c^{2}+c b+b^{2}}{a}\n$$\n\nSimilarly, we get\n\n$$\n\\frac{b^{2}}{c-a}=\\frac{c^{2}+c a+a^{2}}{b}\n$$\n\nWe observe that\n\n$$\n\\frac{a^{2}}{c-b}+\\frac{b^{2}}{c-a}=\\frac{c\\left(a^{2}+b^{2}\\right)-a^{3}-b^{3}}{(c-a)(c-b)}=\\frac{c\\left(a^{2}+b^{2}-c^{2}\\right)}{(c-a)(c-b)}\n$$\n\nThis shows that\n\n$$\n\\frac{a^{2}+b^{2}-c^{2}}{(c-a)(c-b)}=\\frac{c^{2}+c b+b^{2}}{c a}+\\frac{c^{2}+c a+a^{2}}{c b}\n$$\n\nThus it is sufficient to prove that\n\n$$\n\\frac{c^{2}+c b+b^{2}}{c a}+\\frac{c^{2}+c a+a^{2}}{c b} \\geq 6\n$$\n\nHowever, we have $c^{2}+b^{2} \\geq 2 c b$ and $c^{2}+a^{2} \\geq 2 c a$. Hence\n\n$$\n\\frac{c^{2}+c b+b^{2}}{c a}+\\frac{c^{2}+c a+a^{2}}{c b} \\geq 3\\left(\\frac{b}{a}+\\frac{a}{b}\\right) \\geq 3 \\times 2=6\n$$\n\nWe have used AM-GM inequality.\n\n\n2. Let us set $x=a / c$ and $y=b / c$. Then $x^{3}+y^{3}=1$ and the inequality to be proved is $x^{2}+y^{2}-1>6(1-x)(1-y)$. This reduces to\n\n$$\n(x+y)^{2}+6(x+y)-8 x y-7>0\n$$\n\nBut\n\n$$\n1=x^{3}+y^{3}=(x+y)\\left(x^{2}-x y+y^{2}\\right)\n$$\n\nwhich gives $x y=\\left((x+y)^{3}-1\\right) / 3(x+y)$. Substituting this in (1) and introducing $x+y=t$, the inequality takes the form\n\n$$\nt^{2}+6 t-\\frac{8}{3} \\frac{\\left(t^{3}-1\\right)}{t}-7>0\n$$\n\nThis may be simplified to $-5 t^{3}+18 t^{2}-2 t+8>0$. Equivalently\n\n$$\n-(5 t-8)(t-1)^{2}>0\n$$\n\nThus we need to prove that $5 t<8$. Observe that $(x+y)^{3}>x^{3}+y^{3}=1$, so that $t>1$. We also have\n\n$$\n\\left(\\frac{x+y}{2}\\right) \\leq \\frac{x^{3}+y^{3}}{2}=\\frac{1}{2}\n$$\n\nThis shows that $t^{3} \\leq 4$. Thus\n\n$$\n\\left(\\frac{5 t}{8}\\right)^{3} \\leq \\frac{125 \\times 4}{512}=\\frac{500}{512}<1\n$$\n\nHence $5 t<8$, which proves the given inequality.\n\n\n3. We write $b^{3}=c^{3}-a^{3}$ and $a^{3}=c^{3}-b^{3}$ so that\n\n$$\nc-a=\\frac{b^{3}}{c^{2}-c a+a^{2}}, \\quad c-b=\\frac{a^{3}}{c^{2}-c b+b^{2}}\n$$\n\nThus the inequality reduces to\n\n$$\na^{2}+b^{2}-c^{2}>6 \\frac{a^{3} b^{3}}{\\left(c^{2}-c a+a^{2}\\right)\\left(c^{2}-c b+b^{2}\\right)}\n$$\n\nThis simplifies (after some lengthy calculations) to\n\n$$\n\\begin{aligned}\n& -c^{6}-(a+b) c^{5}-a b c^{4}+\\left(a^{3}+b^{3}\\right) c^{3}+\\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\\right) c^{2} \\\\\n& \\quad+\\left(a^{2} b+a b^{2}+a^{3}+b^{3}\\right) a b c+\\left(a^{4} b^{2}-6 a^{3} b^{3}+a^{2} b^{4}\\right)>0\n\\end{aligned}\n$$\n\nSubstituting\n$$\nc^{3}=a^{3}+b^{3}, \\quad c^{4}=c\\left(a^{3}+b^{3}\\right), \\quad c^{5}=c^{2}\\left(a^{3}+b^{3}\\right), \\quad c^{6}=\\left(a^{3}+b^{3}\\right)^{2}\n$$\nthe inequality further reduces to\n$$\na^{2} b^{2}\\left(a^{2}+b^{2}+c^{2}+a c+b c-6 a b\\right)>0\n$$\n\nThus we need to prove that $a^{2}+b^{2}+c^{2}+a c+b c-6 a b>0$. Since $a^{2}+b^{2} \\geq 2 a b$, it is enough to prove that $c^{2}+c(a+b)-4 a b>0$. Multiplying this by $c$ and using $a^{3}+b^{3}=c^{3}$, we need to prove that\n$$\na^{3}+b^{3}+c^{2} a+c^{2} b>4 a b c\n$$\n\nUsing AM-GM inequality to these 4 terms and using $c>a, c>b$ we get\n$$\na^{3}+b^{3}+c^{2} a+c^{2} b>4\\left(a^{3} b^{3} c^{2} a c^{2} b\\right)^{1 / 4}=4 a b c\n$$\nwhich proves the inequality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32403,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$ be two distinct odd positive integers. Define a sequence $(a_n)_{n \\ge 1}$ by\n$$\na_1 = a,\\quad a_2 = b,\\quad a_n \\text{ is the largest odd divisor of } a_{n-1} + a_{n-2}, \\text{ for all } n \\ge 3.\n$$\nShow that there exists a natural number $N$ such that $a_n = \\text{gcd}(a, b)$ for all $n \\ge N$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Each $a_n$ is odd for $n \\ge 1$. Hence $a_{n-1} + a_{n-2}$ is even for $n \\ge 3$. This implies $a_n \\le (a_{n-1} + a_{n-2})/2$, for all $n \\ge 3$. An easy induction shows that $a_n \\ge \\max\\{a, b\\}$, for all $n$. Thus the number of pairs of the form $\\{a_{n-1}, a_{n-2}\\}$ is finite. It follows that $a_n$ is eventually periodic.\n\nLet $M$ be the largest number in the cycle into which the sequence settles. Let $u$, $v$ be the numbers preceding it. Then we see that $u \\le M$, $v \\le M$ and $M \\le (u+v)/2$. Thus $u = v = M$. This implies that the sequence is eventually constant.\n\nLet $d = \\text{gcd}(a, b)$. Then $d$ is odd. An easy induction proves that $d|a_n$ for all $n$. Hence $d$ divides $M$. Since $a \\ne b$, it cannot be the case that $a_n = M$ for all $n$. Consider the position $a_k$, $M$, $M$ in the sequence, where $a_k \\ne M$. By definition of the sequence, we have\n$$\nM = \\frac{M + a_k}{2^r},\n$$\nwhere $r \\ge 1$. This gives $a_k = (2^r - 1)M$. Hence $M$ divides $a_k$. Now consider the position just one earlier: $a_{k-1}$, $a_k$, $M$. Again we have\n$$\nM = \\frac{a_{k-1} + a_k}{2^s},\n$$\nfor some $s \\ge 1$. This gives $a_{k-1} = 2^s M - a_k$. It follows that $M$ divides $a_{k-1}$. Since $M$ is odd, induction shows that $M$ divides $a_n$ for all $n$. Hence $M$ divides $d$. We conclude that $M = d$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32404,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f(x)$ and $g(y)$ be two monic polynomials with complex coefficients, and of the same degree $n$, such that\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} (a_j x + b_j y + c_j),\n$$\nwhere $a_j, b_j, c_j$ are complex numbers, $1 \\le j \\le n$. Prove that there exist complex numbers $a, b, c$ such that\n$$\nf(x) = (x+a)^n + c, \\quad g(y) = (y+b)^n + c.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that $\\prod_{j=1}^{n} a_j = 1$. Hence by dividing by this product, we may write\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} x - \\alpha_j y + \\beta_j.\n$$\nWe observe that\n$$\n\\prod_{j=1}^{n} (x - \\alpha_j y) = x^n - y^n = \\prod_{j=1}^{n} (x - w^j y),\n$$\nwhere $w$ is the primitive $d$-th root of unity. Hence, after renumbering if necessary, we may write $\\alpha_j = w^j$, $1 \\le j \\le n$. Thus we have\n$$\nf(x) - g(y) = \\prod_{j=1}^{n} (x - w^j y + \\beta_j).\n$$\nDefine\n$$\na = \\frac{\\beta_1 - w\\beta_n}{w - 1}, \\quad b = \\frac{\\beta_1 - \\beta_n}{w - 1}.\n$$\nConsider $F(x) = f(x + a)$, $G(y) = g(y + b)$. It is easy to check that\n$$\nF(x) - G(y) = (x - y)(x - wy) \\prod_{j=2}^{n-1} (x - w^j y + \\gamma_j),\n$$\nwhere $\\gamma_j = \\beta_j + a - w^j b$, $2 \\le j \\le n-1$. Putting $y = x$, we see that $F(x) = G(x)$. Putting $x = wy$, we also see that $F(wy) = G(y) = F(y)$. Thus whenever $\\alpha$ is a root of $F(x) = 0$, so is $w\\alpha$. Thus $F(y) = 0$ has roots $\\alpha, w\\alpha, \\dots, w^{n-1}\\alpha$. We thus obtain\n$$\nF(y) = G(y) = (y - \\alpha)(y - w\\alpha) \\cdots (y - w^{n-1}\\alpha) = y^n - \\alpha^n.\n$$\nThis gives\n$$\nf(x) = F(x - a) = (x - a)^n - \\alpha^n, \\quad g(y) = (y - b)^n - \\alpha^n.\n$$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32405,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove the identity:\n$$\n\\sum_{j=0}^{n} \\left\\{ \\binom{3n+2-j}{j} 2^j - \\binom{3n+1-j}{j-1} 2^{j-1} \\right\\} = 2^{3n},\n$$\nwhere the second term on the left side is to be interpreted as $0$ for $j = 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider $j$-th term in the above sum. It is equal to\n$(\\text{the coefficient of } x^j \\text{ in } (1+2x)^{3n+2-j}) - (\\text{the coefficient of } x^{j-1} \\text{ in } (1+2x)^{3n+1-j})$.\nThis in turn is equal to\n$$\n\\text{the coefficient of } x^{3n+2} \\text{ in } x^{3n+2-j} \\left( (1+2x)^{3n+2-j} - x(1+2x)^{3n+1-j} \\right).\n$$\nA little simplification shows that this is equal to\nthe coefficient of $x^{3n+2}$ in $(x+x^2)(x+2x^2)^{3n+1-j}$.\n\nThus the sum above is equal to the coefficient of $x^{3n+2}$ in\n$$\nx(1+x)((x+2x^2)^{3n+1} + (x+2x^2)^{3n} + \\dots + (x+2x^2)^{2n+1}).\n$$\nThe last expression reduces to\n$$\nx^{2n+2}(1+2x)^{2n+1} \\frac{1-(x+2x^2)^{n+1}}{1-2x}.\n$$\nNow the second term in the above product does not contribute to the coefficient of $x^{3n+2}$. Hence this coefficient is precisely equal to the coefficient of $x^n$ in\n$$\n\\frac{(1+2x)^{2n+1}}{1-2x}.\n$$\nUsing binomial expansion for $1/(1-2x)$, this may be seen to be equal to\n$$\n2^n \\left( 1 + \\binom{2n+1}{1} + \\binom{2n+1}{2} + \\dots + \\binom{2n+1}{n} \\right).\n$$\nThis is precisely\n$$\n2^n \\frac{1}{2} \\left( \\sum_{j=0}^{2n+1} \\binom{2n+1}{j} \\right).\n$$\nwhich reduces to $2^{3n}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32406,
    "subject": "Mathematics (Multi-modal)",
    "question": "Is there a positive integer $n$, which is a multiple of $103$, such that $2^{2n+1} \\equiv 2 \\pmod n$?",
    "options": [],
    "answer": "No",
    "solution": "We show that there is no such positive integer $n$. Suppose the contrary; assume that a positive integer $n$ exists such that $2^{2n+1} \\equiv 2 \\pmod n$ and $103 \\nmid n$. Then $2^{2n+1} \\equiv 2 \\pmod{103}$ as well; and as such $2^{2n} \\equiv 1 \\pmod{103}$. Since $103$ is prime, Fermat's little theorem gives $2^{102} \\equiv 1 \\pmod{103}$. If $d_1 = \\gcd(102, 2n)$, it follows that $2^{d_1} \\equiv 1 \\pmod{103}$. But $102 = 2 \\times 3 \\times 17$. It is easy to rule out $d_1 = 2, 3, 6$. Hence $17 \\mid d_1$. In turn $17 \\nmid n$.\n\nAgain, using that $17$ is a factor of $n$, we get $2^{2n+1} \\equiv 2 \\pmod{17}$ or $2^{2n} \\equiv 1 \\pmod{17}$. Now $17$ being a prime, Fermat's little theorem implies that $2^{16} \\equiv 1 \\pmod{17}$. If $d_2 = \\gcd(16, 2n)$, we see that $d_2$ is a power of $2$ and $2^{d_2} \\equiv 1 \\pmod{17}$. We see that $d_2 = 2, 4$ do not fit in. Hence $d_2 = 8$ or $16$. But then $d_2 \\nmid 2n$ shows that $4 \\nmid n$. Hence $2^{2n+1} \\equiv 2 \\pmod 4$, which may be seen to be impossible. Hence no such $n$ exists.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 32407,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDEF$ be a convex hexagon in which the diagonals $AD$, $BE$, $CF$ are concurrent at $O$. Suppose the area of triangle $OAF$ is the geometric mean of those of $OAB$ and $OEF$; and the area of triangle $OBC$ is the geometric mean of those of $OAB$ and $OCD$. Prove that the area of triangle $OED$ is the geometric mean of those of $OCD$ and $OEF$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $OA = a$, $OB = b$, $OC = c$, $OD = d$, $OE = e$, $OF = f$, $[OAB] = x$, $[OCD] = y$, $[OEF] = z$, $[ODE] = u$, $[OFA] = v$ and $[OBC] = w$. We are given that $v^2 = zx$, $w^2 = xy$ and we have to prove that $u^2 = yz$. Since $\\angle AOB = \\angle DOE$, we have\n\n$$\n\\frac{u}{x} = \\frac{\\frac{1}{2} de \\sin \\angle DOE}{\\frac{1}{2} ab \\sin \\angle AOB} = \\frac{de}{ab}.\n$$\n\n$$\n\\frac{v}{y} = \\frac{fa}{cd}, \\quad \\frac{w}{z} = \\frac{bc}{ef}.\n$$\n\nMultiplying these three equalities, we get $uvw = xyz$. Hence\n\n$$\nx^2 y^2 z^2 = u^2 v^2 w^2 = u^2 (zx)(xy).\n$$\n\nThis gives $u^2 = yz$, as desired.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32408,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with circumcircle $\\Gamma$. Let $M$ be a point in the interior of triangle $ABC$ which is also on the bisector of $\\angle A$. Let $AM$, $BM$, $CM$ meet $\\Gamma$ in $A_1$, $B_1$, $C_1$ respectively. Suppose $P$ is the point of intersection of $A_1C_1$ with $AB$; and $Q$ is the point of intersection of $A_1B_1$ with $AC$. Prove that $PQ$ is parallel to $BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $A = 2\\alpha$. Then $\\angle A_1AC = \\angle BAA_1 = \\alpha$. Thus\n$$\n\\angle A_1B_1C = \\alpha = \\angle BB_1A_1 = \\angle A_1C_1C = \\angle BC_1A_1\n$$\nWe also have $\\angle B_1CQ = \\angle AA_1B_1 = \\beta$, say. It follows that triangles $MA_1B_1$ and $QCB_1$ are similar and hence\n$$\n\\frac{QC}{MA_1} = \\frac{B_1C}{B_1A_1}\n$$\n![](attached_image_1.png)\nSimilarly, triangles $ACM$ and $C_1A_1M$ are similar and we get\n$$\n\\frac{AC}{AM} = \\frac{C_1A_1}{C_1M}\n$$\nUsing the point $P$, we get similar ratios:\n$$\n\\frac{PB}{MA_1} = \\frac{C_1B}{A_1C_1}, \\quad \\frac{AB}{AM} = \\frac{A_1B_1}{MB_1}\n$$\nThus,\n$$\n\\frac{QC}{PB} = \\frac{A_1C_1 \\cdot B_1C}{C_1B \\cdot B_1A_1}\n$$\nand\n$$\n\\begin{aligned}\n\\frac{AC}{AB} & = \\frac{MB_1 \\cdot C_1A_1}{A_1B_1 \\cdot C_1M} \\\\\n& = \\frac{MB_1}{C_1M} \\frac{C_1A_1}{A_1B_1} = \\frac{MB_1}{C_1M} \\frac{C_1B \\cdot QC}{PB \\cdot B_1C}\n\\end{aligned}\n$$\nHowever, triangles $C_1BM$ and $B_1CM$ are similar, which gives\n$$\n\\frac{B_1C}{C_1B} = \\frac{MB_1}{MC_1}\n$$\nPutting this in the last expression, we get\n$$\n\\frac{AC}{AB} = \\frac{QC}{PB}\n$$\nWe conclude that $PQ$ is parallel to $BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32409,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be integers such that $b$ is even. Suppose the equation $x^3 + a x^2 + b x + c = 0$ has roots $\\alpha$, $\\beta$, $\\gamma$ such that $\\alpha^2 = \\beta + \\gamma$. Prove that $\\alpha$ is an integer and $\\beta \\neq \\gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $\\alpha$, $\\beta$, $\\gamma$ be the roots of the cubic. Then we have\n$$\n\\alpha + \\beta + \\gamma = -a, \\quad \\alpha\\beta + \\beta\\gamma + \\gamma\\alpha = b, \\quad \\alpha\\beta\\gamma = -c.\n$$\nThus $\\alpha^2 = \\beta + \\gamma = -a - \\alpha$. We also observe that $\\alpha \\neq 0$; otherwise $c = 0$ contradicting that $c$ is odd. Hence $\\beta\\gamma = -c/\\alpha$ and\n$$\nb = \\alpha(\\beta + \\gamma) + \\beta\\gamma = \\alpha^3 - \\frac{c}{\\alpha},\n$$\nor $\\alpha^4 - b\\alpha - c = 0$. This gives $(\\alpha + \\alpha)^2 - b\\alpha - c = 0$, which simplifies to $a(2\\alpha - b - 1) = c + a(a - 1)$. Note that the right side is odd and hence is not equal to 0. We thus get\n$$\n\\alpha = \\frac{c + a(a - 1)}{2a - b - 1}.\n$$\nThis shows that $\\alpha$ is a rational number. Now $\\alpha$ is a rational root of a monic polynomial with integer coefficients. Hence it must be an integer. Since $a = -\\alpha(\\alpha + 1)$, $a$ is even.\nSuppose $\\beta = \\gamma$. Then $2\\beta = \\beta + \\gamma = -a - \\alpha$, so that $\\beta$ is a rational number. Hence $\\beta(= \\gamma)$ is an integer. Again $2\\beta = -a - \\alpha$ shows that $\\alpha$ is even. But then $c = -\\alpha\\beta\\gamma$ shows that $c$ is even. Hence $\\beta = \\gamma$ is not possible.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32410,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P_1(x) = a x^2 - b x - c$, $P_2(x) = b x^2 - c x - a$, $P_3(x) = c x^2 - a x - b$ be three quadratic polynomials where $a$, $b$, $c$ are non-zero real numbers. Suppose there exists a real number $\\alpha$ such that $P_1(\\alpha) = P_2(\\alpha) = P_3(\\alpha)$. Prove that $a = b = c$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We have three relations:\n$$\n\\begin{aligned}\na \\alpha^2 - b \\alpha - c &= \\lambda, \\\\\nb \\alpha^2 - c \\alpha - a &= \\lambda, \\\\\nc \\alpha^2 - a \\alpha - b &= \\lambda,\n\\end{aligned}\n$$\nwhere $\\lambda$ is the common value. Eliminating $\\alpha^2$ from these, taking these equations pairwise, we get three relations:\n$$\n\\begin{aligned}\n(c a - b^2) \\alpha - (b c - a^2) &= \\lambda (b - a), \\\\\n(a b - c^2) \\alpha - (c a - b^2) &= \\lambda (c - b), \\\\\n(b c - a^2) - (a b - c^2) &= \\lambda (a - c).\n\\end{aligned}\n$$\nAdding these three, we get\n$$\n(a b + b c + c a - a^2 - b^2 - c^2)(\\alpha - 1) = 0.\n$$\n(Alternatively, multiplying above relations respectively by $b-c$, $c-a$ and $a-b$, and adding also leads to this.) Thus either $a b + b c + c a - a^2 - b^2 - c^2 = 0$ or $\\alpha = 1$. In the first case\n$$\n0 = a b + b c + c a - a^2 - b^2 - c^2 = \\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2)\n$$\nshows that $a = b = c$. If $\\alpha = 1$, then we obtain\n$$\na - b - c = b - c - a = c - a - b,\n$$\nand once again we obtain $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32411,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all natural numbers $n > 1$ such that $n^{2}$ does not divide $(n-2)!$.",
    "options": [],
    "answer": "All primes; all numbers equal to twice a prime; and the numbers 8 and 9.",
    "solution": "Solution:\nSuppose $n = p q r$, where $p < q$ are primes and $r > 1$. Then $p \\geq 2$, $q \\geq 3$ and $r \\geq 2$, not necessarily a prime. Thus we have\n$$\n\\begin{aligned}\n& n-2 \\geq n-p = p q r - p \\geq 5p > p \\\\\n& n-2 \\geq n-q = q(p r - 1) \\geq 3q > q \\\\\n& n-2 \\geq n-p r = p r(q-1) \\geq 2 p r > p r \\\\\n& n-2 \\geq n-q r = q r(p-1) \\geq q r\n\\end{aligned}\n$$\nObserve that $p, q, p r, q r$ are all distinct. Hence their product divides $(n-2)!$. Thus $n^{2} = p^{2} q^{2} r^{2}$ divides $(n-2)!$ in this case. We conclude that either $n = p q$ where $p, q$ are distinct primes or $n = p^{k}$ for some prime $p$.\n\nCase 1. Suppose $n = p q$ for some primes $p, q$, where $2 < p < q$. Then $p \\geq 3$ and $q \\geq 5$. In this case\n$$\n\\begin{aligned}\n& n-2 > n-p = p(q-1) \\geq 4p \\\\\n& n-2 > n-q = q(p-1) \\geq 2q\n\\end{aligned}\n$$\nThus $p, q, 2p, 2q$ are all distinct numbers in the set $\\{1, 2, 3, \\ldots, n-2\\}$. We see that $n^{2} = p^{2} q^{2}$ divides $(n-2)!$. We conclude that $n = 2q$ for some prime $q \\geq 3$. Note that $n-2 = 2q-2 < 2q$ in this case so that $n^{2}$ does not divide $(n-2)!$.\n\nCase 2. Suppose $n = p^{k}$ for some prime $p$. We observe that $p, 2p, 3p, \\ldots, (p^{k-1}-1)p$ all lie in the set $\\{1, 2, 3, \\ldots, n-2\\}$. If $p^{k-1}-1 \\geq 2k$, then there are at least $2k$ multiples of $p$ in the set $\\{1, 2, 3, \\ldots, n-2\\}$. Hence $n^{2} = p^{2k}$ divides $(n-2)!$. Thus $p^{k-1}-1 < 2k$.\n\nIf $k \\geq 5$, then $p^{k-1}-1 \\geq 2^{k-1}-1 \\geq 2k$, which may be proved by an easy induction. Hence $k \\leq 4$. If $k=1$, we get $n=p$, a prime. If $k=2$, then $p-1<4$ so that $p=2$ or $3$; we get $n=2^{2}=4$ or $n=3^{2}=9$. For $k=3$, we have $p^{2}-1<6$ giving $p=2$; $n=2^{3}=8$ in this case. Finally, $k=4$ gives $p^{3}-1<8$. Again $p=2$ and $n=2^{4}=16$. However $n^{2}=2^{8}$ divides $14!$ and hence is not a solution.\n\nThus $n = p, 2p$ for some prime $p$ or $n = 8, 9$. It is easy to verify that these satisfy the conditions of the problem.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32412,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find the number of 4-digit numbers (in base 10) having non-zero digits and which are divisible by 4 but not by 8.",
    "options": [],
    "answer": "729",
    "solution": "If we take any four consecutive even numbers and divide them by 8, we get remainders 0, 2, 4, 6 in some order. Thus there is only one number of the form $8k + 4$ among them which is divisible by 4 but not by 8. Hence if we take four even consecutive numbers\n\n$$\n1000a + 100b + 10c + 2, \\quad 1000a + 100b + 10c + 4,\n$$\n$$\n1000a + 100b + 10c + 6, \\quad 1000a + 100b + 10c + 8,\n$$\n\nthere is exactly one among these four which is divisible by 4 but not by 8. Now we can divide the set of all 4-digit even numbers with non-zero digits into groups of 4 such consecutive even numbers with $a, b, c$ nonzero. And in each group, there is exactly one number which is divisible by 4 but not by 8. The number of such groups is precisely equal to $9 \\times 9 \\times 9 = 729$, since we can vary $a, b, c$ in the set $\\{1, 2, 3, 4, 5, 6, 7, 8, 9\\}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32413,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle in which $BC < AC$. Let $M$ be the mid-point of $AB$; $AP$ be the altitude from $A$ on to $BC$; and $BQ$ be the altitude from $B$ on to $AC$. Suppose $QP$ produced meets $AB$ (extended) in $T$. If $H$ is the ortho-centre of $ABC$, prove that $TH$ is perpendicular to $CM$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Complete the parallelogram $ADBC$. Join $CD$, $CH$ and $HD$. Let $S$ and $L$ be the midpoints of $CH$ and $HD$ respectively. Observe that $CPHQ$ is a cyclic quadrilateral and $CH$ is a diameter of the circumscribing circle. Thus $S$ is the centre of a circle $\\Gamma_1$ passing through $H, P, Q, C$. Similarly, $L$ is the centre of a circle $\\Gamma_2$ passing through $A, D, B, H$. Hence the radical axis of these circles, which passes through $H$ is perpendicular to $SL$. However $SL$ is parallel to $CD$ (which passes through $M$) and hence the radical axis of $\\Gamma_1$ and $\\Gamma_2$ is perpendicular to $CM$. But $BPQA$ is also cyclic so that $TP \\cdot TQ = TB \\cdot TA$. This shows that $T$ has same power with respect to $\\Gamma_1$ and $\\Gamma_2$. Thus $T$ is on the radical axis of these circles. It follows that $TH$ is perpendicular to $CM$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32414,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all non-zero real numbers $x, y, z$ which satisfy the system of equations:\n$$\n\\begin{aligned}\n\\left(x^{2}+x y+y^{2}\\right)\\left(y^{2}+y z+z^{2}\\right)\\left(z^{2}+z x+x^{2}\\right) & =x y z \\\\\n\\left(x^{4}+x^{2} y^{2}+y^{4}\\right)\\left(y^{4}+y^{2} z^{2}+z^{4}\\right)\\left(z^{4}+z^{2} x^{2}+x^{4}\\right) & =x^{3} y^{3} z^{3}\n\\end{aligned}\n$$",
    "options": [],
    "answer": "x = y = z = 1/3",
    "solution": "Solution:\nSince $x y z \\neq 0$, we can divide the second relation by the first. Observe that\n$$\nx^{4}+x^{2} y^{2}+y^{4}=\\left(x^{2}+x y+y^{2}\\right)\\left(x^{2}-x y+y^{2}\\right)\n$$\nholds for any $x, y$. Thus we get\n$$\n\\left(x^{2}-x y+y^{2}\\right)\\left(y^{2}-y z+z^{2}\\right)\\left(z^{2}-z x+x^{2}\\right)=x^{2} y^{2} z^{2}\n$$\nHowever, for any real numbers $x, y$, we have\n$$\nx^{2}-x y+y^{2} \\geq |x y|\n$$\nSince $x^{2} y^{2} z^{2}=|x y||y z||z x|$, we get\n$$\n|x y||y z||z x|=\\left(x^{2}-x y+y^{2}\\right)\\left(y^{2}-y z+z^{2}\\right)\\left(z^{2}-z x+x^{2}\\right) \\geq |x y||y z||z x|\n$$\nThis is possible only if\n$$\nx^{2}-x y+y^{2}=|x y|, \\quad y^{2}-y z+z^{2}=|y z|, \\quad z^{2}-z x+x^{2}=|z x|\n$$\nhold simultaneously. However $|x y|= \\pm x y$. If $x^{2}-x y+y^{2}=-x y$, then $x^{2}+y^{2}=0$ giving $x=y=0$. Since we are looking for nonzero $x, y, z$, we conclude that $x^{2}-x y+y^{2}=x y$ which is same as $x=y$. Using the other two relations, we also get $y=z$ and $z=x$. The first equation now gives $27 x^{6}=x^{3}$. This gives $x^{3}=1 / 27$ (since $x \\neq 0$ ), or $x=1 / 3$. We thus have $x=y=z=1 / 3$. These also satisfy the second relation, as may be verified.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32415,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a cyclic quadrilateral and let $E$ be the point of intersection of its diagonals $AC$ and $BD$. Suppose $AD$ and $BC$ meet in $F$. Let the midpoints of $AB$ and $CD$ be $G$ and $H$ respectively. If $\\Gamma$ is the circumcircle of triangle $EGH$, prove that $FE$ is tangent to $\\Gamma$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $AB$ and $DC$ meet in $M$; let $FE$ meet $HG$ in $T$, $AB$ in $P$ and $DC$ in $Q$ respectively. We use the following fact: in a triangle $ABC$ with circumcircle $\\gamma$, given a point $D$ outside $\\gamma$, $AD$ is tangent to $\\gamma$ if and only if\n\n![](attached_image_1.png)\n$$\n\\frac{DB}{DC} = \\frac{AB^2}{AC^2}.\n$$\nThus it suffices to prove that\n$$\n\\frac{TG}{TH} = \\frac{GE^2}{EH^2}.\n$$\nLet us put $AB = a$, $DC = b$, $GM = \\lambda$, $HM = \\mu$. Observe that $DB$, $CA$ and $FQ$ are concurrent cevians in the triangle $FDC$. Ceva's theorem gives\n$$\n\\frac{FA}{AD} \\cdot \\frac{DQ}{QC} \\cdot \\frac{CB}{BF} = 1.\n$$\nLooking at the transversal $A-B-M$ in the triangle $FDC$, we can apply Menelaus' theorem to get\n$$\n\\frac{FA}{AD} \\cdot \\frac{DM}{MC} \\cdot \\frac{CB}{BF} = 1.\n$$\nIt follows that $\\frac{DQ}{QC} = \\frac{DM}{MC}$.\nSimilarly, we get $\\frac{AP}{PB} = \\frac{AM}{MB}$. Observe\n$$\n\\frac{DH+HQ}{CH-HQ} = \\frac{DQ}{QC} = \\frac{DM}{MC} = \\frac{DH+HM}{HM-HC}.\n$$\nSimplification gives (using $DH = CH$) $HC^2 = HQ \\cdot HM$. In similar way, we get $GP \\cdot GM = GB^2$. Thus\n$$\n\\begin{aligned} GP &= \\frac{GB^2}{GM} = \\frac{a^2}{4\\lambda}, \\\\ PM &= \\lambda - \\frac{a^2}{4\\lambda} = \\frac{4\\lambda^2 - a^2}{4\\lambda}. \\end{aligned}\n$$\nThus $\\frac{PM}{GP} = \\frac{4\\lambda^2 - a^2}{a^2}$. Similarly, $\\frac{QM}{HQ} = \\frac{4\\mu^2 - b^2}{b^2}$. Using the transversal $T-P-Q$ in the triangle $MGH$, we have $\\frac{MP}{PG} \\cdot \\frac{GT}{TH} \\cdot \\frac{HQ}{QM} = 1$. Thus\n$$\n\\frac{GT}{TH} = \\frac{PG}{MP} \\cdot \\frac{QM}{HQ} = \\frac{a^2}{4\\lambda^2 - a^2} \\cdot \\frac{4\\mu^2 - b^2}{b^2}.\n$$\nSuppose we show that $4\\lambda^2 - a^2 = 4\\mu^2 - b^2$. We get $GT/TH = a^2/b^2$. We observe that triangles $EBA$ and $ECD$ are similar; and they have respective medians $EG$ and $EH$. Hence $EG^2/EH^2 = AB^2/CD^2 = a^2/b^2$. It follows that $GT/TH = GE^2/EH^2$, giving what we required. We use coordinate geometry to prove $4\\lambda^2 - a^2 = 4\\mu^2 - b^2$.\nLet us fix $M = (0,0)$. The circumcircle of $ABCD$ has equation $x^2 + y^2 + 2gx + 2fy + c = 0$. Let the equation of $MA$ be $y = mx$. Let $A = (x_1, y_1)$, $B = (x_2, y_2)$. Then $x_1, x_2$ are the solutions of $(1+m^2)x^2 + (2g+2fm)x + c = 0$, so that\n$$\n x_1 + x_2 = -\\frac{2g + 2fm}{1 + m^2}, \\quad x_1x_2 = \\frac{c}{1 + m^2}.\n$$\nThe coordinates of $G$ are $\\left(\\frac{x_1+x_2}{2}, m\\left(\\frac{x_1+x_2}{2}\\right)\\right)$. Thus\n$$\n\\lambda^2 = GM^2 = \\left(\\frac{x_1+x_2}{2}\\right)^2 + m^2 \\left(\\frac{x_1+x_2}{2}\\right)^2 = \\frac{(g+fm)^2}{1+m^2}.\n$$\nWe also have\n$$\n a^2 = AB^2 = (1+m^2)(x_1-x_2)^2 = (1+m^2)((x_1+x_2)^2 - 4x_1x_2) = 4\\frac{(g+fm)^2}{1+m^2} - 4c.\n$$\n\nThus\n$$\n4\\lambda^2 - a^2 = 4c\n$$\nSimilarly, we get $4\\mu^2 - b^2 = 4c$. It follows that $4\\lambda^2 - a^2 = 4\\mu^2 - b^2$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32416,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find three distinct positive integers with the least possible sum such that the sum of the reciprocals of any two integers among them is an integral multiple of the reciprocal of the third integer.",
    "options": [],
    "answer": "(2, 3, 6)",
    "solution": "We first observe that $(1, a, b)$ is not a solution whenever $1 < a < b$. Otherwise we should have $\\frac{1}{a} + \\frac{1}{b} = l \\cdot \\frac{1}{1} = l$ for some integer $l$. Hence we obtain $\\frac{a+b}{ab} = l$ showing that $a|b$ and $b|a$. But then $a=b$ contradicting $a \\neq b$. Thus the least number should be $2$. It is easy to verify that $(2, 3, 4)$ and $(2, 3, 5)$ are not solutions and $(2, 3, 6)$ satisfies all the conditions. (We may observe $(2, 4, 5)$ is also not a solution.) Since $3+4+5=12 > 11 = 2+3+6$, it follows that $(2, 3, 6)$ has the required minimality.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32417,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nHow many 6-tuples $\\left(a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\\right)$ are there such that each of $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$ is from the set $\\{1,2,3,4\\}$ and the six expressions\n$$\na_{j}^{2}-a_{j} a_{j+1}+a_{j+1}^{2}\n$$\nfor $j=1,2,3,4,5,6$ (where $a_{7}$ is to be taken as $a_{1}$) are all equal to one another?",
    "options": [],
    "answer": "40",
    "solution": "Solution:\n\nWithout loss of generality, we may assume that $a_{1}$ is the largest among $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. Consider the relation\n$$\na_{1}^{2}-a_{1} a_{2}+a_{2}^{2}=a_{2}^{2}-a_{2} a_{3}+a_{3}^{2}\n$$\nThis leads to\n$$\n\\left(a_{1}-a_{3}\\right)\\left(a_{1}+a_{3}-a_{2}\\right)=0\n$$\nObserve that $a_{1} \\geq a_{2}$ and $a_{3}>0$ together imply that the second factor on the left side is positive. Thus $a_{1}=a_{3}=\\max \\left\\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\\right\\}$. Using this and the relation\n$$\na_{3}^{2}-a_{3} a_{4}+a_{4}^{2}=a_{4}^{2}-a_{4} a_{5}+a_{5}^{2}\n$$\nwe conclude that $a_{3}=a_{5}$ as above. Thus we have\n$$\na_{1}=a_{3}=a_{5}=\\max \\left\\{a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}\\right\\}\n$$\nLet us consider the other relations. Using\n$$\na_{2}^{2}-a_{2} a_{3}+a_{3}^{2}=a_{3}^{2}-a_{3} a_{4}+a_{4}^{2}\n$$\nwe get $a_{2}=a_{4}$ or $a_{2}+a_{4}=a_{3}=a_{1}$. Similarly, two more relations give either $a_{4}=a_{6}$ or $a_{4}+a_{6}=a_{5}=a_{1}$; and either $a_{6}=a_{2}$ or $a_{6}+a_{2}=a_{1}$. Let us give values to $a_{1}$ and count the number of six-tuples in each case.\n\n(A) Suppose $a_{1}=1$. In this case all $a_{j}$'s are equal and we get only one six-tuple $(1,1,1,1,1,1)$.\n\n(B) If $a_{1}=2$, we have $a_{3}=a_{5}=2$. We observe that $a_{2}=a_{4}=a_{6}=1$ or $a_{2}=a_{4}=a_{6}=2$. We get two more six-tuples: $(2,1,2,1,2,1), (2,2,2,2,2,2)$.\n\n(C) Taking $a_{1}=3$, we see that $a_{3}=a_{5}=3$. In this case we get nine possibilities for $\\left(a_{2}, a_{4}, a_{6}\\right)$\n$$\n(1,1,1), (2,2,2), (3,3,3), (1,1,2), (1,2,1), (2,1,1), (1,2,2), (2,1,2), (2,2,1)\n$$\n\n(D) In the case $a_{1}=4$, we have $a_{3}=a_{5}=4$ and\n$$\n\\begin{aligned}\n\\left(a_{2}, a_{4}, a_{6}\\right)=(2,2,2), (4,4,4), (1,1,1), (3,3,3), \\\\\n(1,1,3), (1,3,1), (3,1,1), (1,3,3), (3,1,3), (3,3,1)\n\\end{aligned}\n$$\nThus we get $1+2+9+10=22$ solutions. Since $\\left(a_{1}, a_{3}, a_{5}\\right)$ and $\\left(a_{2}, a_{4}, a_{6}\\right)$ may be interchanged, we get 22 more six-tuples. However there are 4 common among these, namely, $(1,1,1,1,1,1), (2,2,2,2,2,2), (3,3,3,3,3,3)$ and $(4,4,4,4,4,4)$. Hence the total number of six-tuples is $22+22-4=40$.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32418,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $A = (a_{jk})$ be a $10 \\times 10$ array of positive real numbers such that the sum of numbers in each row as well as in each column is $1$. Show that there exist $j < k$ and $l < m$ such that\n$$\na_{jl}a_{km} + a_{jm}a_{kl} \\geq \\frac{1}{50}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the first column. Let $M$ be the maximum value of the entries in this column. Then $M \\ge 1/10$, by pigeonhole principle. Writing the first column $a_{1,1}, a_{2,1}, \\dots, a_{10,1}$, let $M = a_{j,1}$, for some $j$. Consider the sum\n$$\n\\sum_{l=2}^{10} (a_{j,1}a_{k,l} + a_{j,l}a_{k,1}),\n$$\nwhere $k \\neq j$. This is equal to\n$$\na_{j,1} \\left( \\sum_{l=2}^{10} a_{k,l} \\right) + a_{k,1} \\left( \\sum_{l=2}^{10} a_{j,l} \\right) = a_{j,1} (1 - a_{k,1}) + a_{k,1} (1 - a_{j,1}).\n$$\nVarying $k$ from $1$ to $10$, $k \\neq j$, we get\n$$\n\\begin{align*}\n\\sum_{k \\neq j} \\sum_{l=2}^{10} (a_{j,1}a_{k,l} + a_{j,l}a_{k,1}) &= \\sum_{k \\neq j} a_{j,1}(1-a_{k,1}) + a_{k,1}(1-a_{j,1}) \\\\\n&= a_{j,1} \\sum_{k \\neq j} (1-a_{k,1}) + (1-a_{j,1}) \\sum_{k \\neq j} a_{k,1} \\\\\n&= a_{j,1}(8+a_{j,1}) + (1-a_{j,1})^2 \\\\\n&= M^2 + 8M + (1-M)^2 \\\\\n&= 2M^2 + 6M + 1 \\ge 2 \\left( \\frac{1}{100} + \\frac{3}{10} \\right) + 1 = \\frac{81}{50}.\n\\end{align*}\n$$\nThere are $9 \\times 9 = 81$ summands in the above sum. Hence we can find $k,l$ such that\n$$\na_{j,1}a_{k,l} + a_{j,l}a_{k,1} \\ge \\frac{1}{50}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32419,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each integer $n \\ge 1$, define $a_n = \\lfloor \\frac{n}{\\sqrt{n}} \\rfloor$, where $\\lfloor x \\rfloor$ denotes the largest integer not exceeding $x$, for any real number $x$. Find the number of all $n$ in the set $\\{1, 2, 3, \\dots, 2010\\}$ for which $a_n > a_{n+1}$.",
    "options": [],
    "answer": "43",
    "solution": "Let us examine the first few natural numbers: $1, 2, 3, 4, 5, 6, 7, 8, 9$. Here we see that $a_n = 1, 2, 3, 2, 2, 3, 3, 4, 3$. We observe that $a_n \\le a_{n+1}$ for all $n$ except when $n+1$ is a square in which case $a_n > a_{n+1}$. We prove that this observation is valid in general. Consider the range\n\n$$\nm^2,\\ m^2+1,\\ m^2+2,\\ \\dots,\\ m^2+m,\\ m^2+m+1,\\ \\dots,\\ m^2+2m.\n$$\n\nLet $n$ take values in this range so that $n = m^2 + r$, where $0 \\le r \\le 2m$. Then we see that $\\lfloor \\sqrt{n} \\rfloor = m$ and hence\n$$\n\\lfloor \\frac{n}{\\sqrt{n}} \\rfloor = \\lfloor \\frac{m^2 + r}{m} \\rfloor = m + \\lfloor \\frac{r}{m} \\rfloor.\n$$\n\nThus $a_n$ takes the values $\\underbrace{m, m, m, \\dots, m}_{m \\text{ times}}, \\underbrace{m+1, m+1, m+1, \\dots, m+1}_{m \\text{ times}}, m+2$, in this range.\n\nBut when $n = (m+1)^2$, we see that $a_n = m+1$. This shows that $a_{n-1} > a_n$ whenever $n = (m+1)^2$. When we take $n$ in the set $\\{1, 2, 3, \\dots, 2010\\}$, we see that the only squares are $1^2, 2^2, \\dots, 44^2$ (since $44^2 = 1936$ and $45^2 = 2025$) and $n = (m+1)^2$ is possible for only 43 values of $m$. Thus $a_n > a_{n+1}$ for 43 values of $n$. (These are $2^2-1, 3^2-1, \\dots, 44^2-1$.)",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32420,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABC$ be an acute-angled triangle with altitude $AK$. Let $H$ be its orthocentre and $O$ be its circumcentre. Suppose $KOH$ is an acute-angled triangle and $P$ its circumcentre. Let $Q$ be the reflection of $P$ in the line $HO$. Show that $Q$ lies on the line joining the mid-points of $AB$ and $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $D$ be the mid-point of $BC$; $M$ that of $HK$; and $T$ that of $OH$. Then $PM$ is perpendicular to $HK$ and $PT$ is perpendicular to $OH$. Since $Q$ is the reflection of $P$ in $HO$, we observe that $P, T, Q$ are collinear, and $PT = TQ$. Let $QL$, $TN$ and $OS$ be the perpendiculars drawn respectively from $Q$, $T$ and $O$ onto the altitude $AK$. (See the figure.)\n\n![](attached_image_1.png)\n\nWe have $LN = NM$, since $T$ is the mid-point of $QP$; $HN = NS$, since $T$ is the mid-point of $OH$; and $HM = MK$, as $P$ is the circumcentre of $KHO$. We obtain\n$$\nLH + HN = LN = NM = NS + SM\n$$\nwhich gives $LH = SM$. We know that $AH = 2OD$. Thus\n$$\n\\begin{aligned}\nAL = AH & - LH = 2OD - LH = 2SK - SM = SK + (SK - SM) = SK + MK \\\\\n& = SK + HM = SK + HS + SM = SK + HS + LH = SK + LS = LK\n\\end{aligned}\n$$\nThis shows that $L$ is the mid-point of $AK$ and hence lies on the line joining the mid-points of $AB$ and $AC$. We observe that the line joining the mid-points of $AB$ and $AC$ is also perpendicular to $AK$. Since $QL$ is perpendicular to $AK$, we conclude that $Q$ also lies on the line joining the mid-points of $AB$ and $AC$.\n\nRemark: It may happen that $H$ is above $L$ as in the adjoining figure, but the result remains true here as well. We have $HN = NS$, $LN = NM$, and $HM = MK$ as earlier. Thus $HN = HL + LN$ and $NS = SM + NM$ give $HL = SM$. Now $AL = AH + HL = 2OD + SM = 2SK + SM = SK + (SK + SM) = SK + MK = SK + HM = SK + HL + LM = SK + SM + LM = LK$. The conclusion that $Q$ lies on the line joining the mid-points of $AB$ and $AC$ follows as earlier.\n\n![](attached_image_2.png)",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32421,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a triangle. Let $AD$, $BE$, $CF$ be cevians such that $\\angle BAD = \\angle CBE = \\angle ACF$. Suppose these cevians concur at a point $\\Omega$. (Such a point exists for each triangle and it is called a Brocard point.) Prove that\n$$\n\\frac{A\\Omega^2}{BC^2} + \\frac{B\\Omega^2}{CA^2} + \\frac{C\\Omega^2}{AB^2} \\ge 1\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that $\\angle B\\Omega D = \\angle AB\\Omega + BA\\Omega = B - \\alpha + \\alpha = B$, where $\\alpha$ is the Brocard angle. Thus triangles $B\\Omega D$ and $ABD$ are similar. It follows that $BD^2 = AD \\cdot \\Omega D$. Let $BD : DC = x : y$, $CE : EA = z : x$, $AF : FB = y : z$. Then $BD = a x/(x+y)$. We also have\n$$\n\\frac{\\Omega D}{AD} = \\frac{z}{x+y+z}, \\quad \\frac{A\\Omega}{AD} = \\frac{x+y}{x+y+z}.\n$$\nUsing this we get $x^2 a^2/(x+y)^2 = z AD^2/(x+y+z)$. This implies that\n$$\nA\\Omega^2 = \\frac{(x+y)^2}{(x+y+z)^2} AD^2 = \\frac{1}{x+y+z} \\cdot \\frac{x^2}{z} \\cdot a^2.\n$$\nThus\n$$\n\\frac{A\\Omega^2}{BC^2} = \\frac{1}{x+y+z} \\cdot \\frac{x^2}{z}.\n$$\n\nWe obtain\n$$\n\\sum_{\\text{cyclic}} \\frac{A\\Omega^2}{BC^2} = \\frac{1}{x+y+z} \\sum_{\\text{cyclic}} \\frac{x^2}{z} \\geq 1,\n$$\nas\n$$\n(x + y + z)^2 \\leq (x + y + z) \\sum_{\\text{cyclic}} \\frac{x^2}{z},\n$$\nby Cauchy-Schwarz inequality.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32422,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nDefine a sequence $\\langle a_{n}\\rangle_{n \\geq 0}$ by $a_{0}=0$, $a_{1}=1$ and\n$$\na_{n}=2 a_{n-1}+a_{n-2}\n$$\nfor $n \\geq 2$.\n\na. For every $m>0$ and $0 \\leq j \\leq m$, prove that $2 a_{m}$ divides $a_{m+j}+(-1)^{j} a_{m-j}$.\n\nb. Suppose $2^{k}$ divides $n$ for some natural numbers $n$ and $k$. Prove that $2^{k}$ divides $a_{n}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\na.\nConsider $f(j)=a_{m+j}+(-1)^{j} a_{m-j}$, $0 \\leq j \\leq m$, where $m$ is a natural number. We observe that $f(0)=2 a_{m}$ is divisible by $2 a_{m}$. Similarly,\n$$\nf(1)=a_{m+1}-a_{m-1}=2 a_{m}\n$$\nis also divisible by $2 a_{m}$. Assume that $2 a_{m}$ divides $f(j)$ for all $0 \\leq j<l$, where $l \\leq m$. We prove that $2 a_{m}$ divides $f(l)$. Observe\n$$\n\\begin{aligned}\n& f(l-1)=a_{m+l-1}+(-1)^{l-1} a_{m-l+1} \\\\\n& f(l-2)=a_{m+l-2}+(-1)^{l-2} a_{m-l+2}\n\\end{aligned}\n$$\nThus we have\n$$\n\\begin{aligned}\na_{m+l} & =2 a_{m+l-1}+a_{m+l-2} \\\\\n& =2 f(l-1)-2(-1)^{l-1} a_{m-l+1}+f(l-2)-(-1)^{l-2} a_{m-l+2} \\\\\n& =2 f(l-1)+f(l-2)+(-1)^{l-1}\\left(a_{m-l+2}-2 a_{m-l+1}\\right) \\\\\n& =2 f(l-1)+f(l-2)+(-1)^{l-1} a_{m-l}\n\\end{aligned}\n$$\nThis gives\n$$\nf(l)=2 f(l-1)+f(l-2)\n$$\nBy induction hypothesis $2 a_{m}$ divides $f(l-1)$ and $f(l-2)$. Hence $2 a_{m}$ divides $f(l)$. We conclude that $2 a_{m}$ divides $f(j)$ for $0 \\leq j \\leq m$.\n\nb.\nWe see that $f(m)=a_{2 m}$. Hence $2 a_{m}$ divides $a_{2 m}$ for all natural numbers $m$. Let $n=2^{k} l$ for some $l \\geq 1$. Taking $m=2^{k-1} l$, we see that $2 a_{m}$ divides $a_{n}$. Using an easy induction, we conclude that $2^{k} a_{l}$ divides $a_{n}$. In particular $2^{k}$ divides $a_{n}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32423,
    "subject": "Mathematics (Multi-modal)",
    "question": "Prove that there are infinitely many positive integers $m$ for which there exist consecutive odd positive integers $p_m, q_m$ ($\\equiv p_m + 2$) such that the pairs $(p_m, q_m)$ are all distinct and\n$$\np_m^2 + p_m q_m + q_m^2, \\quad p_m^2 + m p_m q_m + q_m^2\n$$\nare both perfect squares.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider the relations $p^2 + p q + q^2 = u^2$ and $p^2 + m p q + q^2 = v^2$. Thus $(m - 1) p q = (v - u)(v + u)$. Suppose we choose $m - 1 = r^2$ where $r$ is a positive integer and $v - u = r p$ and $v + u = r q$. Then $u = (r q - r p)/2$ and $4(p^2 + p q + q^2) = (r q - r p)^2$. This leads to\n$$\n(r^2 - 4) p^2 - (2 r^2 + 4) p q + (r^2 - 4) q^2 = 0.\n$$\nSolving for $p/q$, we get\n$$\n\\frac{p}{q} = \\frac{2 r^2 + 4 \\pm \\sqrt{4(r^2 + 2)^2 - 4(r^2 - 4)^2}}{2(r^2 - 4)}.\n$$\nWe want the discriminant to be a perfect square. This forces $3(r^2 - 1)$ to be a perfect square, which leads to the Pell's equation $r^2 - 3 t^2 = 1$. The equation has infinitely many solutions $(r_n, t_n)$ given by\n$$\nr_n + t_n \\sqrt{3} = (2 + \\sqrt{3})^n.\n$$\nWe also have recurrence relations:\n$$\nr_{n+1} = 2 r_n + 3 t_n, \\quad t_{n+1} = r_n + 2 t_n,\n$$\nwhere $r_1 = 2$ and $t_1 = 1$. Induction shows that $t_{2l}$ is even for all $l \\ge 1$. We can express $p/q$ in terms of $t$:\n$$\n\\frac{p}{q} = \\frac{(t \\pm 1)^2}{t^2 - 1} = \\frac{t + 1}{t - 1} \\text{ or } \\frac{t - 1}{t + 1}.\n$$\nTake $m = r_{2l}^2 + 1$, $p_m = t_{2l} - 1$, $q_m = t_{2l} + 1$; we see that $p_l, q_l$ are consecutive odd integers. Moreover,\n$$\n\\begin{align*}\np_m^2 + p_m q_m + q_m^2 &= r_{2l}^2, \\\\\np_m^2 + m p_m q_m + q_m^2 &= (t_{2l} r_{2l})^2.\n\\end{align*}\n$$\nThis proves our claim.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32424,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $D$, $E$, $F$ be points on the sides $BC$, $CA$, $AB$ respectively of a triangle $ABC$ such that $BD = CE = AF$ and $\\angle BDF = \\angle CED = \\angle AFE$. Prove that $ABC$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nConsider the triangle $BDF$, $CED$ and $AFE$ with $BD$, $CE$ and $AF$ as bases. The sides $DF$, $ED$ and $FE$ make equal angles $\\theta$ with the bases of respective triangles. If $B \\geq C \\geq A$, then it is easy to see that $FD \\geq DE \\geq EF$. Now using the triangle $FDE$, we see that $B \\geq C \\geq A$ gives $DE \\geq EF \\geq FD$. Combining, you get $FD = DE = EF$ and hence $A = B = C = 60^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32425,
    "subject": "Mathematics (Multi-modal)",
    "question": "In an acute triangle $ABC$, $AD$ is the median, $BE$ is the internal angle bisector and $CF$ is the altitude, with $D$, $E$, $F$ respectively on sides $BC$, $CA$, $AB$. If $DEF$ is equilateral, prove that $ABC$ is also equilateral.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that in the right triangle $BFC$, $BD = DC = a/2$. Hence $DF = a/2$ and this gives $DE = EF = a/2$. Now in triangle $BEC$, $D$ is the midpoint of $BC$ and $DE = DB = DC$. Hence $\\angle BEC = 90^\\circ$. Since $BE$ bisects $\\angle ABC$, we conclude that $BA = BC$ and $CE = EA$.\n\nIn the right triangle $CFA$, $E$ is the midpoint of $AC$. Hence $EC = EA = EF = a/2$. We conclude that $CA = a = CB$.\n\nWe obtain $AB = BC = CA$. Hence $ABC$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32426,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider two polynomials $P(x) = a_n x^n + a_{n-1} x^{n-1} + \\dots + a_1 x + a_0$ and $Q(x) = b_n x^n + b_{n-1} x^{n-1} + \\dots + b_1 x + b_0$ with integer coefficients such that $a_n - b_n$ is a prime, $a_{n-1} = b_{n-1}$ and $a_n b_0 - a_{n-1} b_n \\neq 0$. Suppose there exists a rational number $r$ such that $P(r) = Q(r) = 0$. Prove that $r$ is an integer.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $r = u/v$ where $\\gcd(u, v) = 1$. Then we get\n\n$$\n\\begin{align*}\na_n u^n + a_{n-1} u^{n-1} v + \\dots + a_1 u v^{n-1} + a_0 v^n &= 0, \\\\\nb_n u^n + b_{n-1} u^{n-1} v + \\dots + b_1 u v^{n-1} + b_0 v^n &= 0.\n\\end{align*}\n$$\n\nSubtraction gives\n$$\n(a_n - b_n)u^n + (a_{n-2} - b_{n-2})u^{n-2}v^2 + \\dots + (a_1 - b_1)u v^{n-1} + (a_0 - b_0)v^n = 0,\n$$\n\nsince $a_{n-1} = b_{n-1}$. This shows that $v$ divides $(a_n - b_n)u^n$ and hence it divides $a_n - b_n$. Since $a_n - b_n$ is a prime, either $v=1$ or $v = a_n - b_n$. Suppose the latter holds. The relation takes the form\n$$\nu^n + (a_{n-2} - b_{n-2})u^{n-2}v + \\dots + (a_1 - b_1)u v^{n-2} + (a_0 - b_0)v^{n-1} = 0.$$\n\n(Here we have divided throughout by $v$.) If $n > 1$, this forces $v \\mid u$, which is impossible since $\\gcd(v, u) = 1$ ($v > 1$ since it is equal to the prime $a_n - b_n$). If $n = 1$, then we get two equations:\n\n$$\n\\begin{align*}\na_1 u + a_0 v &= 0, \\\\\nb_1 u + b_0 v &= 0.\n\\end{align*}\n$$\n\nThis forces $a_1 b_0 - a_0 b_1 = 0$ contradicting $a_n b_0 - a_0 b_n \\neq 0$. (Note: The condition $a_n b_0 - a_0 b_n \\neq 0$ is extraneous. The condition $a_{n-1} = b_{n-1}$ forces that for $n=1$, we have $a_0 = b_0$. Thus we obtain, after subtraction\n$(a_1 - b_1)u = 0.$\nThis implies that $u = 0$ and hence $r = 0$ is an integer.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32427,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle each of whose angles is greater than $30^\\circ$. Suppose a circle, with centre $T$, cuts the segments $BC$ in $P, Q$; $CA$ in $K, L$; and $AB$ in $M, N$, such that $P, Q, K, L, M, N$ are on the circle in counter-clockwise direction in that order. Suppose further that the triangles $TQK$, $TLM$ and $TNP$ are all equilateral. Prove that:\n(i) the radius of the circle is $\\frac{2abc}{a^2 + b^2 + c^2 + 4\\sqrt{3}\\Delta}$;\n(ii) $a \\cdot AT = b \\cdot BT = c \\cdot CT$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) Write $BC = a$, $CA = b$ and $AB = c$. Let $\\angle PTQ = 2x$, $\\angle KTL = 2y$ and $\\angle MTN = 2z$. Since $TP = TQ$, we get $\\angle TQP = \\angle TPQ = 90^\\circ - x$. Thus $\\angle 30^\\circ + x$. Similarly, $\\angle BNP = 30^\\circ + z$. It follows that $z + x = 120^\\circ - B$. Likewise $x + y = 120^\\circ - C$ and $y + z = 120^\\circ - A$. Solving these, we get $x = A - 30^\\circ$, $y = B - 30^\\circ$ and $z = C - 30^\\circ$. As each angle $A, B, C$ is greater than $30^\\circ$, we see that $x, y, z$ are all positive. Further, $\\angle BPN = 30^\\circ + x = A$ and $\\angle BNP = 30^\\circ + z = C$. So the triangle $PBN$ is similar to $ABC$. So are $QKC$ and $ALM$. If we take $BN = ka$, $NP = kb$ and $PB = kc$, then $QC = kb^2/c$ (Note $PN = KQ$). Thus\n$$\n\\begin{align*}\na = BP + PQ + QC = kc + 2kb \\sin x + \\frac{kb^2}{c} &= \\frac{k}{c}(c^2 + b^2 + 2bc \\sin(A - 30^\\circ)) \\\\\n&= \\frac{k}{c}(b^2 + c^2 - 2bc \\cos(A + 60^\\circ)) \\\\\n&= \\frac{k}{c}f^2,\n\\end{align*}\n$$\nwhere $f^2 = (a^2 + b^2 + c^2 + 4\\sqrt{3}F)/2$. This shows that $k = ac/f^2$; $PN = kb = abc/f^2$, the radius of the circle as desired.\n\n(b) Using cosine rule,\n$$\n\\begin{align*}\nBT^2 &= BP^2 + PT^2 - 2BP \\cdot PT \\cdot \\cos \\angle BPT \\\\\n&= k^2(c^2 + b^2 - 2bc \\cos(90^\\circ + x)) \\\\\n&= k^2(b^2 + c^2 - 2bc \\cos(A + 60^\\circ)) \\\\\n&= k^2 f^2.\n\\end{align*}\n$$\nSo $BT = kf = ac/f$. This shows that $BT \\cdot AC = abc/f$. The symmetry of right side shows that\n$$\nBT \\cdot AC = AT \\cdot BC = CT \\cdot AB.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32428,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\n2. Call a natural number $n$ faithful, if there exist natural numbers $a < b < c$ such that $a$ divides $b$, $b$ divides $c$ and $n = a + b + c$.\n\n(i) Show that all but a finite number of natural numbers are faithful.\n\n(ii) Find the sum of all natural numbers which are not faithful.",
    "options": [],
    "answer": "65",
    "solution": "Solution:\n\nSuppose $n \\in \\mathbb{N}$ is faithful. Let $k \\in \\mathbb{N}$ and consider $k n$. Since $n = a + b + c$, with $a < b < c$, $a \\mid b$ and $b \\mid c$, we see that $k n = k a + k b + k c$ which shows that $k n$ is faithful.\n\nLet $p > 5$ be a prime. Then $p$ is odd and $p = (p - 3) + 2 + 1$ shows that $p$ is faithful. If $n \\in \\mathbb{N}$ contains a prime factor $p > 5$, then the above observation shows that $n$ is faithful. This shows that a number which is not faithful must be of the form $2^{\\alpha} 3^{\\beta} 5^{\\gamma}$. We also observe that $2^{4} = 16 = 12 + 3 + 1$, $3^{2} = 9 = 6 + 2 + 1$ and $5^{2} = 25 = 22 + 2 + 1$, so that $2^{4}$, $3^{2}$ and $5^{2}$ are faithful. Hence $n \\in \\mathbb{N}$ is also faithful if it contains a factor of the form $2^{\\alpha}$ where $\\alpha \\geq 4$; a factor of the form $3^{\\beta}$ where $\\beta \\geq 2$; or a factor of the form $5^{\\gamma}$ where $\\gamma \\geq 2$. Thus the numbers which are not faithful are of the form $2^{\\alpha} 3^{\\beta} 5^{\\gamma}$, where $\\alpha \\leq 3$, $\\beta \\leq 1$ and $\\gamma \\leq 1$. We may enumerate all such numbers:\n$$\n1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\n$$\nAmong these $120 = 112 + 7 + 1$, $60 = 48 + 8 + 4$, $40 = 36 + 3 + 1$, $30 = 18 + 9 + 3$, $20 = 12 + 6 + 2$, $15 = 12 + 2 + 1$, and $10 = 6 + 3 + 1$. It is easy to check that the other numbers cannot be written in the required form. Hence the only numbers which are not faithful are\n$$\n1, 2, 3, 4, 5, 6, 8, 12, 24\n$$\nTheir sum is $65$.\nSolution:\n\nIf $n = a + b + c$ with $a < b < c$ is faithful, we see that $a \\geq 1$, $b \\geq 2$ and $c \\geq 4$. Hence $n \\geq 7$. Thus $1, 2, 3, 4, 5, 6$ are not faithful. As observed earlier, $k n$ is faithful whenever $n$ is. We also notice that for odd $n \\geq 7$, we can write $n = 1 + 2 + (n - 3)$ so that all odd $n \\geq 7$ are faithful. Consider $2 n, 4 n, 8 n$, where $n \\geq 7$ is odd. By observation, they are all faithful. Let us list a few of them:\n$$\n\\begin{aligned}\n2 n & : \\quad 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, \\ldots \\\\\n4 n & : \\quad 28, 36, 44, 52, 60, 68, \\ldots \\\\\n8 n & : \\quad 56, 72, \\ldots\n\\end{aligned}\n$$\nWe observe that $16 = 12 + 3 + 1$ and hence it is faithful. Thus all multiples of $16$ are also faithful. Thus we see that $16, 32, 48, 64, \\ldots$ are faithful. Any even number which is not a multiple of $16$ must be either an odd multiple of $2$, or that of $4$, or that of $8$. Hence, the only numbers not covered by this process are $8, 10, 12, 20, 24, 40$. Of these, we see that\n$$\n10 = 1 + 3 + 6, \\quad 20 = 2 \\times 10, \\quad 40 = 4 \\times 10\n$$\nso that $10, 20, 40$ are faithful. Thus the only numbers which are not faithful are\n$$\n1, 2, 3, 4, 5, 6, 8, 12, 24\n$$\nTheir sum is $65$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32429,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose five of the nine vertices of a regular nine-sided polygon are arbitrarily chosen. Show that one can select four among these five such that they are the vertices of a trapezium.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Join the vertices of the nine-sided regular polygon. We get $\\binom{9}{2} = 36$ line segments. All these fall into $9$ sets of parallel lines. Now using any $5$ points, we get $\\binom{5}{2} = 10$ line segments. By the pigeon-hole principle, two of these must be parallel. But, these parallel lines determine a trapezium.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32430,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a convex pentagon in which $BC \\parallel AE$, $AB = BC + AE$ and $\\angle ABC = \\angle CDE$. Let $M$ be the mid-point of $CE$ and let $O$ be the circumcentre of triangle $BCD$. Suppose $\\angle DMO = 90^\\circ$. Prove that $2\\angle BDA = \\angle CDE$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let the circum-circle of $BCD$ be $\\Gamma$ and let the circle with diameter $OD$ be $\\Gamma'$. Let the mid-point of $OD$ be $O'$ (which is also the centre of $\\Gamma'$). Now $OO'$ passes through $D$ and hence $\\Gamma'$ and $\\Gamma$ are tangent to each other at $D$. Hence there is a homothety with centre $D$ taking $\\Gamma'$ to $\\Gamma$, with ratio $2$. Since $\\angle OMD = 90^\\circ$, the point $M$ is on $\\Gamma'$. Let the image of $M$ under the homothety be $D'$; it lies on $\\Gamma$ and $M$ is the midpoint of $DD'$.\n\nConsider the half-turn centred at $M$ taking $D$ to $D'$. This also takes $C$ to $E$ as $M$ is the mid-point of $CE$. Let $B'$ be the image of $B$ under this transformation. Since the half turn takes a line to another line parallel to it, the image of $BC$ is $B'E$; and $BC \\parallel B'E$. Thus $AE \\parallel BC \\parallel B'E$. It follows that $A$, $B'$, $E$ are collinear. Observe that $A$, $B$ lie on the same side of $CE$. Since we have performed a rigid transformation preserving $C$, $E$, the images of $A$, $B$ must also lie on the same side of $CE$. This implies that $A$ and $B'$ lie on different sides of $CE$. Thus $E$ lies between $A$ and $B'$. Because of this, we have\n$$\n\\begin{aligned}\nAB' = AE + EB' &= AE + BC \\quad (\\text{half-turn takes } BC \\text{ to } B'E) \\\\\n&= AB.\n\\end{aligned}\n$$\nWe also observe that the half-turn takes triangle $CD'B$ to $EDB'$. Therefore $\\angle CD'B = \\angle EDB'$. Thus we get\n$$\n\\begin{aligned}\n\\angle BDB' = \\angle BDE + \\angle EDB' &= \\angle BDE + \\angle CD'B = \\angle BDE + \\angle CDB \\\\\n&= \\angle CDE = \\angle CBA = 180^\\circ - \\angle BAB',\n\\end{aligned}\n$$\nsince $BC \\parallel AE$. This shows that $B$, $D$, $B'$, $A$ are concyclic. Since $AB = AB'$, they subtend equal angle at $D$. Thus $\\angle ADB' = \\angle BDA'$. Thus\n$$\n2\\angle BDA = \\angle BDA + \\angle ADB' = \\angle BDA + \\angle ADE + \\angle EDB'.\n$$\nBut observe $\\angle EDB' = \\angle CD'B = \\angle CDB$. Using this we obtain\n$$\n2\\angle BDA = \\angle BDA + \\angle ADE + \\angle CDB = \\angle CDE.\n$$\nThis completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32431,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $(a_0, a_1, a_2, ...)$ and $(b_0, b_1, b_2, ...)$ be two infinite sequences of integers such that\n$$\n(a_n - a_{n-1})(a_n - a_{n-2}) + (b_n - b_{n-1})(b_n - b_{n-2}) = 0,\n$$\nfor all integers $n \\ge 2$. Prove that there exists a positive integer $K$ such that\n$$\na_{K+2011} = a_{K+(2011)^{2011}}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Consider points $P_j = (a_j, b_j)$ in the plane. The slope of the lines $P_nP_{n-1}$ and $P_nP_{n-2}$ are\n$$\nr = \\frac{b_n - b_{n-1}}{a_n - a_{n-1}}, \\quad s = \\frac{b_n - b_{n-2}}{a_n - a_{n-2}},\n$$\nrespectively. The given condition implies that $rs = -1$. Hence it follows that the lines $P_nP_{n-1}$ and $P_nP_{n-2}$ are perpendicular to each other. This implies that $P_n$ lies on the circle $C_n$ with diameter $P_{n-1}P_{n-2}$. Let $f(n) = |P_{n-1} - P_{n-2}|^2$. Then $f(n)$ is an integer. The observation that $P_n$ lies on the circle $C_n$ shows that $f(n) \\le f(n-1)$. Thus we get a non-increasing sequence of positive integers. This must be constant after certain stage. Thus $f(n)$ is constant for $n \\ge N$, for some positive integer $N$. This implies that the diameter of $C_n$ is constant for all $n \\ge N$. Hence $C_n$ are all equal circles for $n \\ge N$. This implies that $P_n = P_{n+2}$ for all $n \\ge N$. But then $a_n = a_{n+2}$ for all $n \\ge N$. Hence\n$$\na_{N+2011} = a_{N+(2011)^{2011}}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32432,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a quadrilateral inscribed in a circle $\\Gamma$. Let $E$, $F$, $G$, $H$ be the midpoints of the arcs $AB$, $BC$, $CD$, $DA$ of the circle $\\Gamma$. Suppose $AC \\cdot BD = EG \\cdot FH$. Prove that $AC$, $BD$, $EG$, $FH$ are concurrent.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "![](attached_image_1.png)\n\nLet $R$ be the radius of the circle $\\Gamma$. Observe that $\\angle EDF = \\frac{1}{2}\\angle D$. Hence $EF = 2R \\sin \\frac{D}{2}$. Similarly, $HG = 2R \\sin \\frac{B}{2}$. But $\\angle B = 180^\\circ - \\angle D$.\nThus $HG = 2R \\cos \\frac{D}{2}$. We hence get\n$$\nEF \\cdot GH = 4R^2 \\sin \\frac{D}{2} \\cos \\frac{D}{2} = 2R^2 \\sin D = R \\cdot AC.\n$$\n\nSimilarly, we obtain $EH \\cdot FG = R \\cdot BD$.\n\nTherefore\n$$\nR(AC + BD) = EF \\cdot GH + EH \\cdot FG = EG \\cdot FH,\n$$\nby Ptolemy's theorem. By the given hypothesis, this gives $R(AC + BD) = AC \\cdot BD$. Thus\n$$\nAC \\cdot BD = R(AC + BD) \\geq 2R\\sqrt{AC \\cdot BD},\n$$\nusing AM-GM inequality. This implies that $AC \\cdot BD \\geq 4R^2$. But $AC$ and $BD$ are the chords of $\\Gamma$, so that $AC \\leq 2R$ and $BD \\leq 2R$. We obtain $AC \\cdot BD \\leq 4R^2$. It follows that $AC \\cdot BD = 4R^2$, implying that $AC = BD = 2R$. Thus $AC$ and $BD$ are two diameters of $\\Gamma$. Using $EG \\cdot FH = AC \\cdot BD$, we conclude that $EG$ and $FH$ are also two diameters of $\\Gamma$. Hence $AC$, $BD$, $EG$ and $FH$ all pass through the centre of $\\Gamma$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32433,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nSuppose five of the nine vertices of a regular nine-sided polygon are arbitrarily chosen. Show that one can select four among these five such that they are the vertices of a trapezium.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose four distinct points $P, Q, R, S$ (in that order on the circle) among these five are such that $\\widehat{P Q}=\\widehat{R S}$. Then $P Q R S$ is an isosceles trapezium, with $P S \\parallel Q R$. We use this in our argument.\n\n- If four of the five points chosen are adjacent, then we are through as observed earlier. (In this case four points $A, B, C, D$ are such that $\\widehat{A B}=\\widehat{B C}=\\widehat{C D}$.) See Fig 1.\n\n![](attached_image_1.png)\nFig 1.\n![](attached_image_2.png)\nFig 2.\n![](attached_image_3.png)\nFig 3.\n\n- Suppose only three of the vertices are adjacent, say $A, B, C$ (see Fig 2.) Then the remaining two must be among $E, F, G, H$. If these two are adjacent vertices, we can pair them with $A, B$ or $B, C$ to get equal arcs. If they are not adjacent, then they must be either $E, G$ or $F, H$ or $E, H$. In the first two cases, we can pair them with $A, C$ to get equal arcs. In the last case, we observe that $\\widehat{H A}=\\widehat{C E}$ and $A H E C$ is an isosceles trapezium.\n\n- Suppose only two among the five are adjacent, say $A, B$. Then the remaining three are among $D, E, F, G, H$. (See Fig 3.) If any two of these are adjacent, we can combine them with $A, B$ to get equal arcs. If no two among these three vertices are adjacent, then they must be $D, F, H$. In this case $\\widehat{H A}=\\widehat{B D}$ and $A H D B$ is an isosceles trapezium.\n\nFinally, if we choose 5 among the 9 vertices of a regular nine-sided polygon, then some two must be adjacent. Thus any choice of 5 among 9 must fall into one of the above three possibilities.\nSolution:\nHere is another solution used by many students. Suppose you join the vertices of the nine-sided regular polygon. You get $\\binom{9}{2}=36$ line segments. All these fall into 9 sets of parallel lines. Now using any 5 points, you get $\\binom{5}{2}=10$ line segments. By pigeon-hole principle, two of these must be parallel. But, these parallel lines determine a trapezium.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32434,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\rightarrow \\mathbb{R}$ such that\n\n$$\nf(x+y)f(x-y) = (f(x)+f(y))^2 - 4x^2f(y), \\quad (1)\n$$\n\nfor all $x, y \\in \\mathbb{R}$, where $\\mathbb{R}$ denotes the set of all real numbers.",
    "options": [],
    "answer": "f(x) = 0 for all real x; f(x) = x^2 for all real x",
    "solution": "Put $x = y = 0$; we get $f(0)^2 = 4f(0)^2$ and hence $f(0) = 0$.\n\n$$\nf(y)(f(y) - f(-y)) = 0.\n$$\n\nWe may conclude that either $f(y) = 0$ or $f(y) = f(-y)$ for each $y \\in \\mathbb{R}$. Replacing $y$ by $-y$, we may also conclude that $f(-y)(f(-y) - f(y)) = 0$. If $f(y) = 0$ and $f(-y) \\neq 0$ for some $y$, then we must have $f(-y) = f(y) = 0$, a contradiction. Hence either $f(y) = f(-y) = 0$ or $f(y) = f(-y)$ for each $y$. This forces $f$ is an even function.\n\nTaking $y = 1$ in (1), we get\n$$\nf(x+1)f(x-1) = (f(x)+f(1))^2 - 4x^2f(1).\n$$\n\nReplacing $y$ by $x$ and $x$ by $1$, you also get\n$$\nf(1+x)f(1-x) = (f(1)+f(x))^2 - 4f(x).\n$$\n\nComparing these two using the even nature of $f$, we get $f(x) = cx^2$, where $c = f(1)$. Putting $x = y = 1$ in (1), you get $4c^2 - 4c = 0$. Hence $c = 0$ or $1$. We get $f(x) = 0$ for all $x$ or $f(x) = x^2$ for all $x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32435,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an isosceles triangle with $AB = AC$. Let $D$ be a point on the segment $BC$ such that $BD = 2DC$. Let $P$ be a point on the segment $AD$ such that $\\angle BAC = \\angle BPD$. Prove that $\\angle BAC = 2\\angle DPC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Extend $AD$ to $E$ such that $PE = PB$. Join $EB$ and $EC$.\n\n![](attached_image_1.png)\n\n$$\n\\angle BPE = \\angle BAC \\text{ and } \\frac{PB}{PE} = \\frac{AB}{AC} = 1.\n$$\n\nHence it follows that $CAB$ is similar to $EFB$. Thus $\\angle PEB$. This shows that $A$, $B$, $E$, $C$ are concyclic. In turn we obtain\n$$\n\\angle AEC = \\angle ABC = \\angle ACB = \\angle AEB.\n$$\n\nWe conclude that $AE$ bisects $\\angle BEC$.\n\nLet $M$ be the mid-point of $BE$. Join $PM$ and $PC$. Since $EA$ bisects $\\angle BEC$, we have\n$$\n\\frac{CE}{EB} = \\frac{CD}{DB} = \\frac{CD}{2CD} = \\frac{1}{2}.\n$$\n\nThus $CE = EB/2 = EM$. We also observe that $\\angle MEP = \\angle CEP$. This shows that $CEP$ is congruent to $MEP$. This implies that\n$$\n\\angle BAC = \\angle BPE = 2\\angle MPE = 2\\angle DPC.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32436,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a quadrilateral inscribed in a circle. Suppose $AB = \\sqrt{2+\\sqrt{2}}$ and $AB$ subtends $135^{\\circ}$ at the centre of the circle. Find the maximum possible area of $ABCD$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "(5+3√3)/(4√2)",
    "solution": "Solution:\nLet $O$ be the centre of the circle in which $ABCD$ is inscribed and let $R$ be its radius. Using cosine rule in triangle $AOB$, we have\n$$\n2+\\sqrt{2}=2R^{2}\\left(1-\\cos 135^{\\circ}\\right)=R^{2}(2+\\sqrt{2})\n$$\nHence $R=1$.\n\nConsider quadrilateral $ABCD$ as in the second figure above. Join $AC$. For $[ADC]$ to be maximum, it is clear that $D$ should be the mid-point of the arc $AC$ so that its distance from the segment $AC$ is maximum. Hence $AD = DC$ for $[ABCD]$ to be maximum. Similarly, we conclude that $BC = CD$. Thus $BC = CD = DA$ which fixes the quadrilateral $ABCD$. Therefore each of the sides $BC, CD, DA$ subtends equal angles at the centre $O$.\n\nLet $\\angle BOC = \\alpha$, $\\angle COD = \\beta$ and $\\angle DOA = \\gamma$. Observe that\n$$\n[ABCD] = [AOB] + [BOC] + [COD] + [DOA] = \\frac{1}{2} \\sin 135^{\\circ} + \\frac{1}{2}(\\sin \\alpha + \\sin \\beta + \\sin \\gamma)\n$$\nNow $[ABCD]$ has maximum area if and only if $\\alpha = \\beta = \\gamma = \\left(360^{\\circ} - 135^{\\circ}\\right)/3 = 75^{\\circ}$. Thus\n$$\n[ABCD] = \\frac{1}{2} \\sin 135^{\\circ} + \\frac{3}{2} \\sin 75^{\\circ} = \\frac{1}{2}\\left(\\frac{1}{\\sqrt{2}} + 3 \\frac{\\sqrt{3}+1}{2\\sqrt{2}}\\right) = \\frac{5+3\\sqrt{3}}{4\\sqrt{2}}\n$$\n\nAlternatively, we can use Jensen's inequality. Observe that $\\alpha, \\beta, \\gamma$ are all less than $180^{\\circ}$. Since $\\sin x$ is concave on $(0, \\pi)$, Jensen's inequality gives\n$$\n\\frac{\\sin \\alpha + \\sin \\beta + \\sin \\gamma}{3} \\leq \\sin\\left(\\frac{\\alpha + \\beta + \\gamma}{3}\\right) = \\sin 75^{\\circ}\n$$\nHence\n$$\n[ABCD] \\leq \\frac{1}{2\\sqrt{2}} + \\frac{3}{2} \\sin 75^{\\circ} = \\frac{5+3\\sqrt{3}}{4\\sqrt{2}}\n$$\nwith equality if and only if $\\alpha = \\beta = \\gamma = 75^{\\circ}$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32437,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{R}^+$ denote the set of all positive real numbers. Find all functions $f: \\mathbb{R}^+ \\to \\mathbb{R}$ satisfying\n$$\nf(x) + f(y) \\le \\frac{f(x+y)}{2}, \\quad \\frac{f(x)}{x} + \\frac{f(y)}{y} \\ge \\frac{f(x+y)}{x+y},\n$$\nfor all $x, y \\in \\mathbb{R}^+$.",
    "options": [],
    "answer": "f(x) = a x^2 for all x > 0, where a ≤ 0",
    "solution": "Put $x = y = t$ ($t > 0$). We get\n$$\n4f(t) \\le f(2t), \\quad f(2t) \\ge 4f(t),\n$$\nfor all $t > 0$. Hence $f(2t) = 4f(t)$, for all $t > 0$. By induction\n$$\nf(2^m t) = 2^{2m} f(t), \\text{ for all } t > 0.\n$$\nLet $g(x) = f(x)/x$, $x > 0$, $g(0) = 0$. We show that $g(nt) = ng(t)$ for all $n \\in \\mathbb{N}$ and $t \\ge 0$. We have proved this for $n = 2^m$. We observe that $g(x+y) \\le g(x) + g(y)$ for all $x \\ge 0$, $y \\ge 0$. Hence by induction $g(nt) \\le ng(t)$ for all $n \\in \\mathbb{N}$ and $t \\ge 0$. Choose $m$ such that $2^{m-1} \\le n < 2^m$. Then\n$$\n2^m g(t) = g(2^m t) \\le g(nt) + g((2^m - n)t) \\le ng(t) + (2^m - n)g(t) = 2^m g(t).\n$$\nHence equality holds and $g(nt) = ng(t)$.\n\nNext we show that $g$ is a non-increasing function. We have\n$$\nf(t) + f(2t) \\le \\frac{f(t+2t)}{2} = \\frac{f(3t)}{2}.\n$$\nHence\n$$\n\\operatorname{tg}(t) + 2\\operatorname{tg}(2t) \\le (3t/2)\\operatorname{g}(3t).\n$$\nThis gives\n$$\ng(t) + 4g(t) \\le (9/2)g(t).\n$$\nHence $g(t) \\le 0$, for all $t \\ge 0$. Thus for any $0 < x \\le y$, we have\n$$\ng(x) \\ge g(y) + g(x - y) \\ge g(y).\n$$\nHence $g$ is a non-increasing function.\n\nLet $g(1) = a \\le 0$. We show that $g(t) = at$ for all $t > 0$. Suppose $g(t) < at$ for some $t > 0$. Choose positive rational $p/q$ such that $(p/q) > \\text{and } g(t) < a(p/q)$. But $g(nt) = ng(t)$ for all $n \\in \\mathbb{N}$. Hence\n$$\ng\\left(\\frac{p}{q}\\right) = \\frac{p}{q}g(1) = \\frac{pa}{q}.\n$$\nHence\n$$\ng(t) \\ge g\\left(\\frac{p}{q}\\right) = \\frac{pa}{q},\n$$\na contradiction to $g(t) < a(p/q)$. Similarly we can show that $g(t) > at$ is also not possible. We conclude that $g(t) = at$ for all $t > 0$. This gives $f(x) = ax^2$, where $a \\ge 0$. It is easy to verify that this gives indeed a solution to our equations.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32438,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a \\ge b$ and $c \\ge d$ be real numbers. Prove that the equation\n$$\n(x + a)(x + d) + (x + b)(x + c) = 0\n$$\nhas real roots.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $f(x) = (x + a)(x + d) + (x + b)(x + c)$. The leading coefficient of $f(x)$ is $2$. Hence $f(x)$ is positive for large values of $x$. We have\n$$\n\\begin{aligned}\nf(-a) + f(-b) &= (b-a)(c-a) + (a-b)(d-b) = (a-b)(-c+a+d-b), \\\\\nf(-c) + f(-d) &= (a-c)(d-c) + (b-d)(c-d) = (c-d)(-a+c+b-d).\n\\end{aligned}\n$$\nWe know that $a \\ge b$ and $c \\ge d$. If $f(-a) + f(-b) \\le 0$, then we are done; then either $f(-a) \\le 0$ or $f(-b) \\le 0$ so that $f(x)$ crosses the $x$-axis at some point. Otherwise $-c+a+d-b > 0$. This implies that $-a+c+b-d < 0$. But then\n$$\nf(-c) + f(-d) = (c - d)(-a + c + b - d) \\le 0.\n$$\nWe conclude that either $f(-c) \\le 0$ or $f(-d) \\le 0$. Again $f(x)$ crosses the $x$-axis somewhere.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32439,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $p_{1} < p_{2} < p_{3} < p_{4}$ and $q_{1} < q_{2} < q_{3} < q_{4}$ be two sets of prime numbers such that $p_{4} - p_{1} = 8$ and $q_{4} - q_{1} = 8$. Suppose $p_{1} > 5$ and $q_{1} > 5$. Prove that $30$ divides $p_{1} - q_{1}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSince $p_{4} - p_{1} = 8$, and no prime is even, we observe that $\\{p_{1}, p_{2}, p_{3}, p_{4}\\}$ is a subset of $\\{p_{1}, p_{1} + 2, p_{1} + 4, p_{1} + 6, p_{1} + 8\\}$. Moreover $p_{1}$ is larger than $3$. If $p_{1} \\equiv 1 \\pmod{3}$, then $p_{1} + 2$ and $p_{1} + 8$ are divisible by $3$. Hence we do not get $4$ primes in the set $\\{p_{1}, p_{1} + 2, p_{1} + 4, p_{1} + 6, p_{1} + 8\\}$. Thus $p_{1} \\equiv 2 \\pmod{3}$ and $p_{1} + 4$ is not a prime. We get $p_{2} = p_{1} + 2$, $p_{3} = p_{1} + 6$, $p_{4} = p_{1} + 8$.\n\nConsider the remainders of $p_{1}, p_{1} + 2, p_{1} + 6, p_{1} + 8$ when divided by $5$. If $p_{1} \\equiv 2 \\pmod{5}$, then $p_{1} + 8$ is divisible by $5$ and hence is not a prime. If $p_{1} \\equiv 3 \\pmod{5}$, then $p_{1} + 2$ is divisible by $5$. If $p_{1} \\equiv 4 \\pmod{5}$, then $p_{1} + 6$ is divisible by $5$. Hence the only possibility is $p_{1} \\equiv 1 \\pmod{5}$.\n\nThus we see that $p_{1} \\equiv 1 \\pmod{2}$, $p_{1} \\equiv 2 \\pmod{3}$ and $p_{1} \\equiv 1 \\pmod{5}$. We conclude that $p_{1} \\equiv 11 \\pmod{30}$.\n\nSimilarly $q_{1} \\equiv 11 \\pmod{30}$. It follows that $30$ divides $p_{1} - q_{1}$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32440,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCD$ be a trapezium with $AB \\parallel CD$. Let $P$ be a point on $AC$ such that $C$ is between $A$ and $P$; and let $X$, $Y$ be the mid-points of $AB$, $CD$ respectively. Let $PX$ intersect $BC$ in $N$ and $PY$ intersect $AD$ in $M$. Prove that $MN \\parallel AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe that\n$$\n\\frac{BN}{NC} = \\frac{[PNB]}{[PNC]}\n$$\nHowever $[PNB] + [XNC] = [PXB] = [PXA] = [PCN] + [ACN] + [AXN]$. Since $[XNB] = [AXN]$, we obtain $[PNB] = [PNC] + [ACN]$. Thus\n$$\n\\frac{BN}{NC} = \\frac{[PNC] + [ACN]}{[PNC]} = 1 + \\frac{[ACN]}{[PNC]} = 1 + \\frac{AC}{PC}.\n$$\nSimilarly, we can prove that\n$$\n\\frac{AM}{MD} = 1 + \\frac{AC}{CP}.\n$$\nComparison shows that $AM/MD = BN/NC$. We conclude that $MN \\parallel AB$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32441,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nDefine a sequence $\\langle f_{0}(x), f_{1}(x), f_{2}(x), \\ldots \\rangle$ of functions by\n$$\nf_{0}(x)=1, \\quad f_{1}(x)=x, \\quad (f_{n}(x))^{2}-1=f_{n+1}(x) f_{n-1}(x), \\text{ for } n \\geq 1\n$$\nProve that each $f_{n}(x)$ is a polynomial with integer coefficients.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nObserve that\n$$\nf_{n}^{2}(x)-f_{n-1}(x) f_{n+1}(x)=1=f_{n-1}^{2}(x)-f_{n-2}(x) f_{n}(x)\n$$\nThis gives\n$$\nf_{n}(x)\\left(f_{n}(x)+f_{n-2}(x)\\right)=f_{n-1}\\left(f_{n-1}(x)+f_{n+1}(x)\\right)\n$$\nWe write this as\n$$\n\\frac{f_{n-1}(x)+f_{n+1}(x)}{f_{n}(x)}=\\frac{f_{n-2}(x)+f_{n}(x)}{f_{n-1}(x)}\n$$\nUsing induction, we get\n$$\n\\frac{f_{n-1}(x)+f_{n+1}(x)}{f_{n}(x)}=\\frac{f_{0}(x)+f_{2}(x)}{f_{1}(x)}\n$$\nObserve that\n$$\nf_{2}(x)=\\frac{f_{1}^{2}(x)-1}{f_{0}(x)}=x^{2}-1\n$$\nHence\n$$\n\\frac{f_{n-1}(x)+f_{n+1}(x)}{f_{n}(x)}=\\frac{1+\\left(x^{2}-1\\right)}{x}=x\n$$\nThus we obtain\n$$\nf_{n+1}(x)=x f_{n}(x)-f_{n-1}(x)\n$$\nSince $f_{0}(x), f_{1}(x)$ and $f_{2}(x)$ are polynomials with integer coefficients, induction again shows that $f_{n}(x)$ is a polynomial with integer coefficients.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32442,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function such that $f(x + y + xy) = f(x) + f(y) + f(xy)$ for all $x, y \\in \\mathbb{R}$. Prove that $f$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Put $x = y = 0$; we get $f(0) = 3f(0)$ so that $f(0) = 0$. Taking $y = -1$, we get $f(-1) = f(x) + f(-1) + f(-x)$ so that $f(-x) = -f(x)$ for all $x \\in \\mathbb{R}$. Taking $y = 1$, we get $f(2x + 1) = 2f(x) + 1$, for all $x \\in \\mathbb{R}$. Replace $x$ by $u + v + uv$ in this, we get\n$$\nf(2(u + v + uv) + 1) = 2f(u + v + uv) + f(1) = 2f(u) + 2f(v) + 2f(uv) + f(1), \\quad (1)\n$$\n\nfor all $u, v \\in \\mathbb{R}$. Taking $x = u$, $y = 2v + 1$ in the original equation, we get\n$$\nf(u + 2v + 1 + 2uv + u) = f(u) + f(2v + 1) + f(2uv + u),\n$$\nand this reduces to\n$$\nf(2(u + v + uv) + 1) = f(u) + f(2v + 1) + f(2uv + u),\n$$\n(2)\nfor all $u, v \\in \\mathbb{R}$. Comparing (1) and (2) and using $f(2x + 1) = 2f(x) + 1$, we get\n$$\nf(2uv + u) = f(u) + 2f(uv),\n$$\n(3)\nfor all $u, v \\in \\mathbb{R}$. Put $v = -1/2$ in (3) to get\n$$\n0 = f(0) = f(u) + 2f(-u/2) = f(u) - 2f(u/2).\n$$\nThis shows that $f(u/2) = f(u)/2$ or $f(2u) = 2f(u)$ for all $u \\in \\mathbb{R}$. Thus\n$$\nf(2uv + u) = f(2uv) + f(u),\n$$\nfor all $u, v \\in \\mathbb{R}$. Given any $x \\neq 0$ and $y$, we can find $u, v$ such that $2uv = y$ and $u = x$. Thus $f(x+y) = f(x) + f(y)$ for all $x \\neq 0$ and $y$. However this also valid for $x = 0$. Thus $f(x+y) = f(x) + f(y)$ for all $x, y \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32443,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle. An interior point $P$ of $ABC$ is said to be good if we can find exactly $27$ rays emanating from $P$ intersecting the sides of the triangle $ABC$ such that the triangle is divided by these rays into $27$ smaller triangles of equal area. Determine the number of good points for a given triangle $ABC$.",
    "options": [],
    "answer": "325",
    "solution": "Solution:\n\nLet $P$ be a good point. Let $l, m, n$ be respectively the number of parts the sides $BC, CA, AB$ are divided by the rays starting from $P$. Note that a ray must pass through each of the vertices of the triangle $ABC$; otherwise we get some quadrilaterals.\n\nLet $h_1$ be the distance of $P$ from $BC$. Then $h_1$ is the height for all the triangles with their bases on $BC$. Equality of areas implies that all these bases have equal length. If we denote this by $x$, we get $l x = a$. Similarly, taking $y$ and $z$ as the lengths of the bases of triangles on $CA$ and $AB$ respectively, we get $m y = b$ and $n z = c$. Let $h_2$ and $h_3$ be the distances of $P$ from $CA$ and $AB$ respectively. Then\n\n$$\nh_1 x = h_2 y = h_3 z = \\frac{2 \\Delta}{27}\n$$\n\nwhere $\\Delta$ denotes the area of the triangle $ABC$. These lead to\n\n$$\nh_1 = \\frac{2 \\Delta}{27} \\frac{l}{a}, \\quad h_2 = \\frac{2 \\Delta}{27} \\frac{m}{b}, \\quad h_3 = \\frac{2 \\Delta}{27} \\frac{n}{c}\n$$\n\nBut\n\n$$\n\\frac{2 \\Delta}{a} = h_a, \\quad \\frac{2 \\Delta}{b} = h_b, \\quad \\frac{2 \\Delta}{c} = h_c\n$$\n\nThus we get\n\n$$\n\\frac{h_1}{h_a} = \\frac{l}{27}, \\quad \\frac{h_2}{h_b} = \\frac{m}{27}, \\quad \\frac{h_3}{h_c} = \\frac{n}{27}\n$$\n\nHowever, we also have\n\n$$\n\\frac{h_1}{h_a} = \\frac{[PBC]}{\\Delta}, \\quad \\frac{h_2}{h_b} = \\frac{[PCA]}{\\Delta}, \\quad \\frac{h_3}{h_c} = \\frac{[PAB]}{\\Delta}\n$$\n\nAdding these three relations,\n\n$$\n\\frac{h_1}{h_a} + \\frac{h_2}{h_b} + \\frac{h_3}{h_c} = 1\n$$\n\nThus\n\n$$\n\\frac{l}{27} + \\frac{m}{27} + \\frac{n}{27} = \\frac{h_1}{h_a} + \\frac{h_2}{h_b} + \\frac{h_3}{h_c} = 1\n$$\n\nWe conclude that $l + m + n = 27$. Thus every good point $P$ determines a partition $(l, m, n)$ of $27$ such that there are $l, m, n$ equal segments respectively on $BC, CA, AB$.\n\nConversely, take any partition $(l, m, n)$ of $27$. Divide $BC, CA, AB$ respectively into $l, m, n$ equal parts. Define\n\n$$\nh_1 = \\frac{2 l \\Delta}{27 a}, \\quad h_2 = \\frac{2 m \\Delta}{27 b}\n$$\n\nDraw a line parallel to $BC$ at a distance $h_1$ from $BC$; draw another line parallel to $CA$ at a distance $h_2$ from $CA$. Both lines are drawn such that they intersect at a point $P$ inside the triangle $ABC$. Then\n\n$$\n[PBC] = \\frac{1}{2} a h_1 = \\frac{l \\Delta}{27}, \\quad [PCA] = \\frac{m \\Delta}{27}\n$$\n\nHence\n\n$$\n[PAB] = \\frac{n \\Delta}{27}\n$$\n\nThis shows that the distance of $P$ from $AB$ is\n\n$$\nh_3 = \\frac{2 n \\Delta}{27 c}\n$$\n\nTherefore each triangle with base on $CA$ has area $\\frac{\\Delta}{27}$. We conclude that all the triangles which partition $ABC$ have equal areas. Hence $P$ is a good point.\n\nThus the number of good points is equal to the number of positive integral solutions of the equation $l + m + n = 27$. This is equal to\n\n$$\n\\binom{26}{2} = 325\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32444,
    "subject": "Mathematics (Multi-modal)",
    "question": "The circumcenter of the cyclic quadrilateral $ABCD$ is $O$. The second intersection point of the circles $ABO$ and $CDO$, other than $O$, is $P$, which lies in the interior of the triangle $DAO$. Choose a point $Q$ on the extension of $OP$ beyond $P$, and a point $R$ on the extension of $OP$ beyond $O$. Prove that $\\angle QAP = \\angle OBR$ holds if and only if $\\angle PDQ = \\angle RCO$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $H$ be the radical center of the circles $ABCD$, $ABOP$ and $CDPO$. Then the radical axes of any two of these circles, i.e. the lines $AB$, $CD$ and $OP$, pass through $H$. Since $P$ lies on the shorter arcs $AO$ and $DO$, it follows that $H$ lies on the extension of $OP$ beyond $P$. The radical center satisfies\n$$\nHA \\cdot HB = HC \\cdot HD = HO \\cdot HP.\n$$\n(1)\nSince the quadrilateral $ABOP$ is cyclic,\n![](attached_image_1.png)\n$$\n\\begin{aligned}\n\\angle QAB + \\angle BRQ &= (\\angle PAB + \\angle QAP) + (\\angle BOP - \\angle OBR) \\\\\n&= (\\angle PAB + \\angle BOP) + (\\angle QAP - \\angle OBR) \\\\\n&= 180^\\circ + (\\angle QAP - \\angle OBR).\n\\end{aligned}\n$$\nTherefore, $\\angle QAP = \\angle OBR$ holds if and only if the quadrilateral $ABRQ$ is cyclic, which is equivalent to $HQ \\cdot HR = HA \\cdot HB$.\nSimilarly, $\\angle QAP = \\angle OBR$ holds if and only if $HQ \\cdot HR = HC \\cdot HD$.\nCombining with (1),\n$$\n\\angle QAP = \\angle OBR \\Leftrightarrow HQ \\cdot HR = HA \\cdot HB \\Leftrightarrow HQ \\cdot HR = HC \\cdot HD \\Leftrightarrow \\angle PDQ = \\angle RCO.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32445,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $P(z) = a_n z^n + a_{n-1} z^{n-1} + \\dots + a_m z^m$ be polynomial with complex coefficients such that $a_m \\neq 0$, $a_n \\neq 0$ and $n > m$. Prove that\n$$\n\\max_{|z|=1}\\{|P(z)|\\} \\ge \\sqrt{\\sum_{k=m}^{n} |a_k|^2 + 2|a_m a_n|}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Note that we may assume $m = 0$, since $P(z) = z^m Q(z)$ for some polynomial $Q(z)$ where $|P(z)| = |Q(z)|$ on the unit circle. Thus we may write\n$$\nP(z) = \\sum_{k=0}^{n} a_k z^k, \\quad a_n \\neq 0, a_0 \\neq 0,\n$$\nand we have to prove that\n$$\n\\max_{|z|=1}\\{|P(z)|\\} \\ge \\sqrt{\\sum_{k=0}^{n} |a_k|^2 + 2|a_0 a_n|}.\n$$\nObserve that $|P(z)|^2 = \\sum_{k=0}^{n} \\sum_{j=0}^{n} a_k \\bar{a}_j z^{k-j}$. Let $\\omega$ be a primitive $n$-th root of unity. Then\n$$\n\\sum_{l=0}^{n-1} |P(\\omega^l z)|^2 = \\sum_{l=0}^{n-1} \\left( \\sum_{k=0}^{n} \\sum_{j=0}^{n} a_k \\bar{a}_j \\omega^{l(k-j)} z^{k-j} \\right) = \\sum_{k=0}^{n} \\sum_{j=0}^{n} a_k \\bar{a}_j z^{k-j} \\sum_{l=0}^{n-1} \\omega^{l(k-j)}.\n$$\nThe last sum is $n$ if $k-j \\equiv 0 \\pmod{n}$ and zero otherwise. Hence\n$$\n\\frac{1}{n} \\sum_{l=0}^{n-1} |P(\\omega^l z)|^2 = \\sum_{k=0}^{n} |a_k|^2 + a_n \\bar{a}_0 z^n + \\bar{a}_n a_0 z^{-n}.\n$$\nChoosing $z_0$ such that $z_0^n = \\frac{a_0|a_n|}{|a_0|a_n}$, we obtain\n$$\n\\frac{1}{n} \\sum_{l=0}^{n-1} |P(\\omega^l z_0)|^2 = \\sum_{k=0}^{n} |a_k|^2 + 2|a_0 a_n|.\n$$\nHence we can find $l$ such that\n$$\n|P(\\omega^l z_0)|^2 \\geq \\sum_{k=0}^{n} |a_k|^2 + 2|a_0 a_n|.\n$$\nThe result follows.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32446,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $f: \\mathbb{Z} \\rightarrow \\mathbb{Z}$ be a function satisfying $f(0) \\neq 0$, $f(1)=0$ and\n\n(i) $f(xy)+f(x)f(y)=f(x)+f(y)$;\n\n(ii) $(f(x-y)-f(0)) f(x) f(y)=0$,\n\nfor all $x, y \\in \\mathbb{Z}$, simultaneously.\n\na. Find the set of all possible values of the function $f$.\n\nb. If $f(10) \\neq 0$ and $f(2)=0$, find the set of all integers $n$ such that $f(n) \\neq 0$.",
    "options": [],
    "answer": "a) The function values are exactly {0, 1}. b) Under f(10) nonzero and f(2) zero, the integers with nonzero function value are precisely the multiples of five.",
    "solution": "Solution:\n\nSetting $y=0$ in the condition (ii), we get\n$$\n(f(x)-f(0)) f(x)=0\n$$\nfor all $x$ (since $f(0) \\neq 0$). Thus either $f(x)=0$ or $f(x)=f(0)$, for all $x \\in \\mathbb{Z}$. Now taking $x=y=0$ in (i), we see that $f(0)+f(0)^2=2 f(0)$. This shows that $f(0)=0$ or $f(0)=1$. Since $f(0) \\neq 0$, we must have $f(0)=1$. We conclude that\n$$\n\\text{either } f(x)=0 \\text{ or } f(x)=1 \\text{ for each } x \\in \\mathbb{Z}\n$$\nThis shows that the set of all possible values of $f(x)$ is $\\{0,1\\}$. This completes (a).\n\nLet $S=\\{n \\in \\mathbb{Z} \\mid f(n) \\neq 0\\}$. Hence we must have $S=\\{n \\in \\mathbb{Z} \\mid f(n)=1\\}$ by (a). Since $f(1)=0$, $1$ is not in $S$. And $f(0)=1$ implies that $0 \\in S$. Take any $x \\in \\mathbb{Z}$ and $y \\in S$. Using (ii), we get\n$$\nf(xy)+f(x)=f(x)+1\n$$\nThis shows that $xy \\in S$. If $x \\in \\mathbb{Z}$ and $y \\in \\mathbb{Z}$ are such that $xy \\in S$, then (ii) gives\n$$\n1+f(x)f(y)=f(x)+f(y)\n$$\nThus $(f(x)-1)(f(y)-1)=0$. It follows that $f(x)=1$ or $f(y)=1$; i.e., either $x \\in S$ or $y \\in S$. We also observe from (ii) that $x \\in S$ and $y \\in S$ implies that $f(x-y)=1$ so that $x-y \\in S$. Thus $S$ has the properties:\n\n(A) $x \\in \\mathbb{Z}$ and $y \\in S$ implies $xy \\in S$;\n\n(B) $x, y \\in \\mathbb{Z}$ and $xy \\in S$ implies $x \\in S$ or $y \\in S$;\n\n(C) $x, y \\in S$ implies $x-y \\in S$.\n\nNow we know that $f(10) \\neq 0$ and $f(2)=0$. Hence $f(10)=1$ and $10 \\in S$; and $2 \\notin S$. Writing $10=2 \\times 5$ and using (B), we conclude that $5 \\in S$ and $f(5)=1$. Hence $f(5k)=1$ for all $k \\in \\mathbb{Z}$ by (A).\n\nSuppose $f(5k+l)=1$ for some $l$, $1 \\leq l \\leq 4$. Then $5k+l \\in S$. Choose $u \\in \\mathbb{Z}$ such that $lu \\equiv 1 \\pmod{5}$. We have $(5k+l)u \\in S$ by (A). Moreover, $lu=1+5m$ for some $m \\in \\mathbb{Z}$ and\n$$\n(5k+l)u=5ku+lu=5ku+5m+1=5(ku+m)+1\n$$\nThis shows that $5(ku+m)+1 \\in S$. However, we know that $5(ku+m) \\in S$. By (C), $1 \\in S$ which is a contradiction. We conclude that $5k+l \\notin S$ for any $l$, $1 \\leq l \\leq 4$. Thus\n$$\nS=\\{5k \\mid k \\in \\mathbb{Z}\\}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32447,
    "subject": "Mathematics (Multi-modal)",
    "question": "Show that there exist infinitely many pairs $(a, b)$ of positive integers with the property that $a + b$ divides $ab + 1$, $a - b$ divides $ab - 1$, $b > 1$ and $a > b\\sqrt{3} - 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Observe $b^2 - 1 = b(a+b) - (ab+1)$ and $b^2 - 1 = b(a-b) - (ab-1)$. Hence $a+b$ and $a-b$ both divide $b^2 - 1$. Thus $\\text{lcm}(a+b, a-b)$ divides $b^2 - 1$. Since both are positive, $\\text{lcm}(a+b, a-b) \\le b^2 - 1$.\nLet $d = \\text{gcd}(a, b)$. Then $d|ab$ and $d|a+b|ab+1$. Hence $d|1$ showing $d=1$. If $e = \\text{gcd}(a+b, a-b)$, then $e|2a$ and $e|2b$ so that $e|\\text{gcd}(2a, 2b)$. But $\\text{gcd}(2a, 2b) = 2\\ \\text{gcd}(a, b) = 2$. Thus $e|2$ and hence $e \\le 2$. Hence\n$$\n\\text{lcm}(a+b, a-b) = \\frac{(a+b)(a-b)}{\\text{gcd}(a+b, a-b)} \\ge \\frac{a^2-b^2}{2}.\n$$\nIt follows that $a^2 - b^2 \\le 2(b^2 - 1)$ or $a^2 - 3b^2 \\le -2$.\nSuppose $a$ and $b$ are positive integers such that $a^2 - 3b^2 = -2$. Then $a$ and $b$ have same parity. For such a pair $(a, b)$, we have\n$$\nab + 1 = ab + \\frac{3b^2 - a^2}{2} = (a+b)\\frac{3b-a}{2},\n$$\n$$\nab - 1 = ab - \\frac{3b^2 - a^2}{2} = (a-b)\\frac{a+3b}{2}.\n$$\nHence $a+b$ divides $ab+1$ and $a-b$ divides $ab-1$. We also observe that\n$$\n\\sqrt{3}b = \\sqrt{a^2 + 2} < \\sqrt{a^2 + 2a + 1} = a + 1,\n$$\nso that $a > \\sqrt{3}b - 1$.\nThus we look for solutions of the equation $x^2 - 3y^2 = -2$ in positive integers. This equation has infinitely many solutions which may be described as follows:\nThe equation $x^2 - 3y^2 = -2$ has a particular solution $(1, 1)$. Consider the equation $x^2 - 3y^2 = 1$. This has infinitely many solutions $(u_n, v_n)$ given by\n$$\nu_n + \\sqrt{3}v_n = (2 + \\sqrt{3})^n.\n$$\nLet $a_n = u_n + 3v_n$ and $b_n = u_n + v_n$. Then\n$$\na_n^2 - 3v_n^2 = -2(u_n^2 - 3v_n^2) = -2.\n$$\nFor $n \\ge 1$ we have $b_n > 1$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32448,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $S$ be a nonempty set of primes satisfying the property that for each proper subset $P$ of $S$, all the prime factors of the number $\\left(\\prod_{p \\in P} p\\right) - 1$ are also in $S$. Determine all possible such sets $S$.",
    "options": [],
    "answer": "Exactly the following sets: any singleton containing one prime; any two element set consisting of two and a Fermat prime; and the set of all prime numbers.",
    "solution": "If $S$ is singleton, then the statement is vacuously true. Hence $S = \\{p\\}$, $p$ a prime.\n\nSuppose $S$ contains two elements. One of these must be odd primes and hence $2$ is in $S$. Moreover $p-1$ must be a power of $2$. Hence $p$ is Fermat prime.\n\nSuppose $|S| \\ge 3$. Here we consider two cases: $|S| < \\infty$ and $S$ is an infinite set.\n\nIf $|S| < \\infty$, let $S = \\{p_1(=2), p_2, \\dots, p_r\\}$, with $r \\ge 3$. Since $|S| \\ge 3$, there is a prime of the form $3k+1$ or $3k+2$ in $S$. Observe $(3k+1)-1=3k$ and $2(3k+2)-1=6k+3$ both have $3$ as a prime factor. Hence $3 \\in S$. Now\n$$\n(2 \\cdot p_3 \\cdots p_r) - 1 = 3^k \\text{ and } (3p_3 \\cdots p_r) - 1 = 2^l,\n$$\nfor some $k > 0$ and $l > 0$. Hence we get\n$$\n3(3^k + 1) = 2(2^l + 1).\n$$\nThis gives $2^{l+1} - 3^{k+1} = 1$. The only possibility is $l+1=2$ and $k+1=1$. This forces $k=0$, a contradiction. Hence $3 \\le |S| < \\infty$ is not possible.\n\nFinally assume $S$ is an infinite set. Suppose there exists a prime $p$ not in $S$. Take $S = \\{p_1(=2), p_2, \\dots\\}$\n\nConsider $p_1 - 1, p_1p_2 - 1, \\dots, p_1p_2 \\dots p_p - 1$ modulo $p$. Since $p$ is not in $S$, none of them is equal to $0$. By pigeonhole principle, we can find $j < k$ such that\n$$\np_1p_2 \\cdots p_j - 1 \\equiv p_1p_2 \\cdots p_j p_{j+1} \\cdots p_k - 1 \\pmod{p}.\n$$\nHence $p$ divides $p_1p_2 \\cdots p_j(p_{j+1} \\cdots p_k - 1)$. But this contradicts the choice of $p$ that it is not in $S$. Hence $S$ is the set of all primes now.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32449,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $2n$ real numbers are placed in the cells of a $2 \\times n$ grid such that the sum of the numbers in each of the $n$ columns is $1$. Prove that one can erase one of the two numbers in each column such that the sum of the remaining numbers in each of the rows does not exceed $\\frac{n+1}{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Assume that the numbers in the first row are $a_1 \\le a_2 \\le \\dots \\le a_n$ in some order. The numbers in the second row are $b_j = 1 - a_j$, $1 \\le j \\le n$. Hence $b_1 \\ge b_2 \\ge \\dots \\ge b_n$.\n\nIf $a_1 + a_2 + \\dots + a_n \\le \\frac{n+1}{4}$, we are done (we can erase all the numbers in the second row).\n\nLet $k$ be the least positive integer such that $a_1 + a_2 + \\dots + a_k > \\frac{n+1}{4}$. Then\n$$\na_1 + a_2 + \\dots + a_{k-1} \\le \\frac{n+1}{4}.\n$$\nWe show that\n$$\nb_k + b_{k+1} + \\dots + b_n \\le \\frac{n+1}{4}.\n$$\nObserve\n$$\na_k \\ge \\frac{a_1 + a_2 + \\dots + a_k}{k} > \\frac{n+1}{4k}.\n$$\nHence\n$$\nb_k + b_{k+1} + \\dots + b_n \\le (n+1-k)b_k = (n+1-k)(1-a_k) \\\\\n< (n+1-k) \\left(1 - \\frac{n+1}{4k}\\right) = \\frac{5}{4}(n+1) - \\left(\\frac{(n+1)^2 + 4k^2}{4k}\\right) \\\\\n\\le (\\text{AM-GM}) \\frac{5}{4}(n+1) - \\frac{2(n+1)(2k)}{4k} = \\frac{n+1}{4}.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32450,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f$ from the set of real numbers to itself satisfying\n$$\nf(x(1+y)) = f(x)(1+f(y))\n$$\nfor all real numbers $x, y$.",
    "options": [],
    "answer": "f(x) = 0 for all real x, and f(x) = x for all real x",
    "solution": "If $f$ is not identically zero, then by standard substitutions we get that $f(x) = x$ for $x = 0, \\pm 1$. Using these it is easy to see that $f$ is additive and multiplicative on the set of real numbers. It then follows by induction and continuity that $f(x) = x$ for all real $x$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32451,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, a cubic polynomial $p(x)$ is said to be $n$-good if there exist $n$ distinct integers $a_1, a_2, \\dots, a_n$ such that all the roots of the polynomial $p(x) + a_i = 0$ are integers for $1 \\le i \\le n$. Given a positive integer $n$ prove that there exists an $n$-good cubic polynomial.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $f(x) = x^3 - mx^2 + nx$, $k$ an integer, and $a, b, c$ the roots of $f(x) + k = 0$. Then\n$$\n4m^2 - 12n = 3(a-c)^2 + (a - 2b + c)^2.\n$$\nSince the equation $x^2 + 3y^2 = 1$ has a rational solution, it has infinitely many rational solutions. Therefore one can find $D$ such that $4D$ can be expressed as $3s^2 + t^2$ in at least $6n$ ways. By choosing $m$ and $n$ such that $4m^2 - 12n = 4D$ we get an $n$-good cubic polynomial. $\\square$",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32452,
    "subject": "Mathematics (Multi-modal)",
    "question": "For a positive integer $n$, a sum-friendly odd partition of $n$ is a sequence $(a_1, a_2, \\dots, a_k)$ of odd positive integers with $a_1 \\le a_2 \\le \\dots \\le a_k$ and $a_1 + a_2 + \\dots + a_k = n$ such that for all positive integers $m \\le n$, $m$ can be uniquely written as a subsum $m = a_{i_1} + a_{i_2} + \\dots + a_{i_r}$. (Two subsums $a_{i_1} + a_{i_2} + \\dots + a_{i_r}$, and $a_{j_1} + a_{j_2} + \\dots + a_{j_s}$ with $i_1 < i_2 < \\dots < i_r$ and $j_1 < j_2 < \\dots < j_s$ are considered the same if $r = s$ and $a_{i_l} = a_{j_l}$ for $1 \\le l \\le r$.) For example, $(1, 1, 3, 3)$ is a sum-friendly odd partition of $8$. Find the number of sum-friendly odd partitions of $9999$.",
    "options": [],
    "answer": "16",
    "solution": "We consider the sum-friendly odd partitions of a positive integer $n$. Clearly $(1, 1, \\dots, 1)$ is a sum-friendly odd partition. On the other hand, if $a_i > 1$ for some $i$, then let $r$ be the smallest such that $a_r > 1$. It follows that $a_r = r$ and that $r$ divides $a_i$ for all $i \\ge r$. Therefore $r$ is odd and it divides $n+1$. Moreover, $(\\frac{a_r}{r}, \\frac{a_{r+1}}{r}, \\dots, \\frac{a_k}{r})$ is a sum-friendly odd partition of $(n+1)/r - 1$. Thus by induction it follows that the number of sum-friendly odd partitions of $n$ equals the number of factorisations $(n+1) = d_1d_2 \\dots d_l$ in which $d_1, d_2, \\dots, d_{l-1}$ are odd. Hence there are $16$ sum-friendly odd partitions of $9999$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32453,
    "subject": "Mathematics (Multi-modal)",
    "question": "A marker is placed at the origin of an integer lattice. Calvin and Hobbes play the following game. Calvin starts the game and each of them takes turns alternatively. At each turn, one can choose two (not necessarily distinct) integers $a$, $b$, neither of which was chosen earlier by any player and move the marker by $a$ units in the horizontal direction and $b$ units in the vertical direction. Hobbes wins if the marker is back at the origin any time after the first move. Prove that Calvin can prevent Hobbes from winning.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_n$ denote the set of chosen integers after $n$ turns. We claim (by induction) that after Calvin's move he can ensure that if $a \\in A_n$ then $-a \\in A_n$, and that the marker is at $(r, -r)$ for some non-zero integer $r$ in $A_n$.\n\nLet Calvin move the marker to $(-1, 1)$ in his first turn. Suppose that, after $n$ turns, the marker is at $(r, -r)$ and that Hobbes then moves it to $(r+a, -r+b)$. If $a \\ne b$ then Calvin can move it back to $(r, -r)$. If $a = -b$ then Calvin can move the marker to $(c, -c)$ or $(-c, c)$ where $c$ is the largest element of $A_{n+1}$, so the claim follows. Hence Calvin can prevent Hobbes from winning. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32454,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$ with $B = 90^\\circ$, $D$ is a point on the segment $BC$ such that the inradii of triangles $ABD$ and $ADC$ are equal. If $\\angle ADB = \\varphi$ then prove that $\\tan^2(\\varphi/2) = \\tan(C/2)$. $\\square$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "This follows easily from formulae for inradius of a triangle and simple trigonometric identities. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32455,
    "subject": "Mathematics (Multi-modal)",
    "question": "In a triangle $ABC$, with $AB \\neq BC$, $E$ is a point on the line $AC$ such that $BE$ is perpendicular to $AC$. A circle passing through $A$ and touching the line $BE$ at a point $P \\neq B$ intersects the line $AB$ for the second time at $X$. Let $Q$ be a point on the line $PB$ different from $P$ such that $BQ = BP$. Let $Y$ be the point of intersection of the lines $CP$ and $AQ$. Prove that the points $C, X, Y, A$ are concyclic if and only if $CX$ is perpendicular to $AB$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We need the following well-known result.\n**Lemma.** In a triangle $KLM$ with $KL \\neq KM$, $R$ is a point on the $LM$ such that $KR$ is perpendicular to $LM$. Let $U$ be a point on the line $KR$. Let the lines $LU$ and $MU$ intersect $KM$ and $KL$, respectively, at $S$ and $T$ respectively. Then $L, T, S, M$ are concyclic if and only if $U$ is the orthocenter of triangle $KLM$.\n\nSuppose that $C, X, Y, A$ are concyclic. Let the lines $AP$ and $CQ$ intersect at $Z$. Since $BP = BQ$ we have $BQ^2 = BX \\cdot BA$, so $BQ$ is tangent to the circumcircle of triangle $AXQ$. Hence $\\angle BQX = \\angle XAQ = \\angle XCY$. Therefore the points $Q, X, P, C$ are concyclic. Further, $\\angle XAP = \\angle XPQ = \\angle XCQ$. Adding the two we get $\\angle YCZ = \\angle YAZ$. This proves that the points $A, Y, Z, C$ are concyclic. Applying the lemma to triangle $QAC$ we get that $P$ is the orthocenter of triangle $QAC$. Hence $\\angle CXA = \\angle CYA = 90^\\circ$.\n\nFor the converse, suppose that $CX$ is perpendicular to $AB$. Let the line $CP$ intersect the circumcircle of triangle $AXC$ at $Y'$, and let the lines $AY'$ and $BP$ intersect at $Q'$. Note that $P$ is the orthocenter of triangle $Q'AC$. Hence if $AP$ intersects $Q'C$ at $Z$, then $Z$ lies on the circle $\\gamma$. Note that $\\angle Q'CX = \\angle Q'CY' - \\angle XCY' = \\angle ZAY' - \\angle XAY' = \\angle XAP = \\angle Q'PX$. This shows that the points $Q', X, P, C$ are concyclic. Hence $\\angle XQ'B = \\angle XCP = \\angle XAY'$, so $BQ'$ is tangent to the circumcircle of triangle $AXQ'$. Therefore $BQ'^2 = BX \\cdot BA = BP^2$ and hence $Q = Q'$. This completes the solution. $\\square$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32456,
    "subject": "Mathematics (Multi-modal)",
    "question": "A positive integer $a$ is called a *double number* if it has an even number of digits (in base 10) and its base 10 representation has the form $a = a_1a_2\\cdots a_k a_1a_2\\cdots a_k$ with $0 \\le a_i \\le 9$ for $1 \\le i \\le k$, and $a_1 \\ne 0$. For example, 283283 is a double number. Determine whether or not there are infinitely many double numbers $a$ such that $a + 1$ is a square and $a + 1$ is not a power of 10.",
    "options": [],
    "answer": "Yes, there are infinitely many.",
    "solution": "The answer is affirmative. Let $k \\ge 0$ be such that $k \\equiv 15 \\pmod{42}$ and $b = 5(10^k + 1)/7 + 1$ (which is an integer). Then $c = \\frac{5}{7}(b+1)$ is an integer and $b^2 - 1 = (10^k + 1)c$ is a double number.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32457,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $h \\ge 3$ be an integer and $X$ the set of all positive integers that are greater than or equal to $2h$. Let $S$ be a nonempty subset of $X$ such that the following two conditions hold:\n* if $a + b \\in S$ with $a \\ge h, b \\ge h$, then $ab \\in S$;\n* if $ab \\in S$ with $a \\ge h, b \\ge h$, then $a+b \\in S$.\n\nProve that $S = X$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $f: X \\to \\{0, 1\\}$ be such that $f(x) = 1$ if and only if $x \\in S$. Then $f(a+b) = f(ab)$ whenever $a \\ge h, b \\ge h$. If $a \\ge h+2$ then\n$$\nf(2a - 1) = f(a^2 - a) = f(a^3 - 2a^2) = f(a^2 + a - 2) = f(2a + 1).\n$$\nGiven $n \\ge 2h$, it is easy to see that there exists an $a \\ge h+2$ such that $f(n) = f(2a-1)$. This proves that $f$ is constant and hence $S = X$. $\\square$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32458,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $\\Gamma_{1}$ and $\\Gamma_{2}$ be two circles touching each other externally at $R$. Let $l_{1}$ be a line which is tangent to $\\Gamma_{2}$ at $P$ and passing through the center $O_{1}$ of $\\Gamma_{1}$. Similarly, let $l_{2}$ be a line which is tangent to $\\Gamma_{2}$ at $Q$ and passing through the center $O_{2}$ of $\\Gamma_{2}$. Suppose $l_{1}$ and $l_{2}$ are not parallel and intersect at $K$. If $K P = K Q$, prove that the triangle $P Q R$ is equilateral.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nSuppose that $P$ and $Q$ lie on the opposite sides of line joining $O_{1}$ and $O_{2}$. By symmetry we may assume that the configuration is as shown in the figure below. Then we have $K P > K O_{1} > K Q$ since $K O_{1}$ is the hypotenuse of triangle $K Q O_{1}$. This is a contradiction to the given assumption, and therefore $P$ and $Q$ lie on the same side of the line joining $O_{1}$ and $O_{2}$.\n![](attached_image_1.png)\nSince $K P = K Q$ it follows that $K$ lies on the radical axis of the given circles, which is the common tangent at $R$. Therefore $K P = K Q = K R$ and hence $K$ is the circumcenter of $\\triangle P Q R$.\n![](attached_image_2.png)\nOn the other hand, $\\triangle K Q O_{1}$ and $\\triangle K R O_{1}$ are both right-angled triangles with $K Q = K R$ and $Q O_{1} = R O_{1}$, and hence the two triangles are congruent. Therefore $\\widehat{Q K O_{1}} = \\widehat{R K O_{1}}$, so $K O_{1}$, and hence $P K$ is perpendicular to $Q R$. Similarly, $Q K$ is perpendicular to $P R$, so it follows that $K$ is the orthocenter of $\\triangle P Q R$. Hence we have that $\\triangle P Q R$ is equilateral.\nSolution:\nWe again rule out the possibility that $P$ and $Q$ are on the opposite side of the line joining $O_{1} O_{2}$, and assume that they are on the same side.\nObserve that $\\triangle K P O_{2}$ is congruent to $\\triangle K Q O_{1}$ (since $K P = K Q$). Therefore $O_{1} P = O_{2} Q = r$ (say). In $\\triangle O_{1} O_{2} Q$, we have $\\widehat{O_{1} Q O_{2}} = \\pi / 2$ and $R$ is the midpoint of the hypotenuse, so $R Q = R O_{1} = r$. Therefore $\\triangle O_{1} R Q$ is equilateral, so $\\widehat{Q R O_{1}} = \\pi / 3$. Similarly, $P R = r$ and $\\widehat{P R O_{2}} = \\pi / 3$, hence $\\widehat{P R Q} = \\pi / 3$. Since $P R = Q R$ it follows that $\\triangle P Q R$ is equilateral.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32459,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all positive integers $m$, $n$, and primes $p \\geq 5$ such that\n$$\nm\\left(4 m^{2}+m+12\\right)=3\\left(p^{n}-1\\right)\n$$",
    "options": [],
    "answer": "(m, n, p) = (12, 4, 7)",
    "solution": "Solution:\nRewriting the given equation we have\n$$\n4 m^{3}+m^{2}+12 m+3=3 p^{n}\n$$\nThe left hand side equals $(4 m+1)\\left(m^{2}+3\\right)$.\nSuppose that $\\left(4 m+1, m^{2}+3\\right)=1$. Then $\\left(4 m+1, m^{2}+3\\right)=\\left(3 p^{n}, 1\\right),\\left(3, p^{n}\\right),\\left(p^{n}, 3\\right)$ or $\\left(1,3 p^{n}\\right)$, a contradiction since $4 m+1, m^{2}+3 \\geq 4$. Therefore $\\left(4 m+1, m^{2}+3\\right)>1$.\nSince $4 m+1$ is odd we have $\\left(4 m+1, m^{2}+3\\right)=\\left(4 m+1,16 m^{2}+48\\right)=(4 m+1,49)=7$ or $49$. This proves that $p=7$, and $4 m+1=3 \\cdot 7^{k}$ or $7^{k}$ for some natural number $k$. If $(4 m+1,49)=7$ then we have $k=1$ and $4 m+1=21$ which does not lead to a solution. Therefore $\\left(4 m+1, m^{2}+3\\right)=49$. If $7^{3}$ divides $4 m+1$ then it does not divide $m^{2}+3$, so we get $m^{2}+3 \\leq 3 \\cdot 7^{2}<7^{3} \\leq 4 m+1$. This implies $(m-2)^{2}<2$, so $m \\leq 3$, which does not lead to a solution. Therefore we have $4 m+1=49$ which implies $m=12$ and $n=4$. Thus $(m, n, p)=(12,4,7)$ is the only solution.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32460,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a, b, c, d$ be positive integers such that $a \\geq b \\geq c \\geq d$. Prove that the equation $x^{4}-a x^{3}-b x^{2}-c x-d=0$ has no integer solution.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nSuppose that $m$ is an integer root of $x^{4}-a x^{3}-b x^{2}-c x-d=0$. As $d \\neq 0$, we have $m \\neq 0$.\n\nSuppose now that $m>0$. Then $m^{4}-a m^{3}=b m^{2}+c m+d>0$ and hence $m>a \\geq d$. On the other hand $d=m\\left(m^{3}-a m^{2}-b m-c\\right)$ and hence $m$ divides $d$, so $m \\leq d$, a contradiction.\n\nIf $m<0$, then writing $n=-m>0$ we have $n^{4}+a n^{3}-b n^{2}+c n-d=n^{4}+n^{2}(a n-b)+(c n-d)>0$, a contradiction.\n\nThis proves that the given polynomial has no integer roots.",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32461,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $n$ be a positive integer. Call a nonempty subset $S$ of $\\{1,2, \\ldots, n\\}$ good if the arithmetic mean of the elements of $S$ is also an integer. Further let $t_{n}$ denote the number of good subsets of $\\{1,2, \\ldots, n\\}$. Prove that $t_{n}$ and $n$ are both odd or both even.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe show that $T_{n}-n$ is even. Note that the subsets $\\{1\\},\\{2\\}, \\cdots,\\{n\\}$ are good. Among the other good subsets, let $A$ be the collection of subsets with an integer average which belongs to the subset, and let $B$ be the collection of subsets with an integer average which is not a member of the subset. Then there is a bijection between $A$ and $B$, because removing the average takes a member of $A$ to a member of $B$; and including the average in a member of $B$ takes it to its inverse. So $T_{n}-n=|A|+|B|$ is even.\nSolution:\n\nLet $S=\\{1,2, \\ldots, n\\}$. For a subset $A$ of $S$, let $\\bar{A}=\\{n+1-a \\mid a \\in A\\}$. We call a subset $A$ symmetric if $\\bar{A}=A$. Note that the arithmetic mean of a symmetric subset is $(n+1)/2$. Therefore, if $n$ is even, then there are no symmetric good subsets, while if $n$ is odd then every symmetric subset is good.\n\nIf $A$ is a proper good subset of $S$, then so is $\\bar{A}$. Therefore, all the good subsets that are not symmetric can be paired. If $n$ is even then this proves that $t_{n}$ is even. If $n$ is odd, we have to show that there are odd number of symmetric subsets. For this, we note that a symmetric subset contains the element $(n+1)/2$ if and only if it has odd number of elements. Therefore, for any natural number $k$, the number of symmetric subsets of size $2k$ equals the number of symmetric subsets of size $2k+1$. The result now follows since there is exactly one symmetric subset with only one element.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32462,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn an acute triangle $ABC$, $O$ is the circumcenter, $H$ is the orthocenter and $G$ is the centroid. Let $OD$ be perpendicular to $BC$ and $HE$ be perpendicular to $CA$, with $D$ on $BC$ and $E$ on $CA$. Let $F$ be the midpoint of $AB$. Suppose the areas of triangles $ODC$, $HEA$ and $GFB$ are equal. Find all the possible values of $\\widehat{C}$.",
    "options": [],
    "answer": "π/3 or π/4",
    "solution": "Solution:\n\nLet $R$ be the circumradius of $\\triangle ABC$ and $\\Delta$ its area. We have $OD = R \\cos A$ and $DC = \\frac{a}{2}$, so\n$$\n[ODC] = \\frac{1}{2} \\cdot OD \\cdot DC = \\frac{1}{2} \\cdot R \\cos A \\cdot R \\sin A = \\frac{1}{2} R^2 \\sin A \\cos A\n$$\nAgain $HE = 2R \\cos C \\cos A$ and $EA = c \\cos A$. Hence\n$$\n[HEA] = \\frac{1}{2} \\cdot HE \\cdot EA = \\frac{1}{2} \\cdot 2R \\cos C \\cos A \\cdot c \\cos A = 2R^2 \\sin C \\cos C \\cos^2 A\n$$\nFurther\n$$\n[GFB] = \\frac{\\Delta}{6} = \\frac{1}{6} \\cdot 2R^2 \\sin A \\sin B \\sin C = \\frac{1}{3} R^2 \\sin A \\sin B \\sin C\n$$\nEquating (1) and (2) we get $\\tan A = 4 \\sin C \\cos C$. And equating (1) and (3), and using this relation we get\n$$\n\\begin{aligned}\n3 \\cos A & = 2 \\sin B \\sin C = 2 \\sin (C + A) \\sin C \\\\\n& = 2(\\sin C + \\cos C \\tan A) \\sin C \\cos A \\\\\n& = 2 \\sin^2 C (1 + 4 \\cos^2 C) \\cos A\n\\end{aligned}\n$$\nSince $\\cos A \\neq 0$ we get $3 = 2t(-4t + 5)$ where $t = \\sin^2 C$. This implies $(4t - 3)(2t - 1) = 0$ and therefore, since $\\sin C > 0$, we get $\\sin C = \\sqrt{3}/2$ or $\\sin C = 1/\\sqrt{2}$. Because $\\triangle ABC$ is acute, it follows that $\\widehat{C} = \\pi/3$ or $\\pi/4$.\n\nWe observe that the given conditions are satisfied in an equilateral triangle, so $\\widehat{C} = \\pi/3$ is a possibility. Also, the conditions are satisfied in a triangle where $\\widehat{C} = \\pi/4$, $\\widehat{A} = \\tan^{-1} 2$ and $\\widehat{B} = \\tan^{-1} 3$. Therefore $\\widehat{C} = \\pi/4$ is also a possibility.\n\nThus the two possible values of $\\widehat{C}$ are $\\pi/3$ and $\\pi/4$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32463,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $a, b, c, x, y, z$ be positive real numbers such that $a + b + c = x + y + z$ and $a b c = x y z$. Further, suppose that $a \\leq x < y < z \\leq c$ and $a < b < c$. Prove that $a = x$, $b = y$ and $c = z$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet\n$$\nf(t) = (t - x)(t - y)(t - z) - (t - a)(t - b)(t - c)\n$$\nThen $f(t) = k t$ for some constant $k$. Note that $k a = f(a) = (a - x)(a - y)(a - z) \\leq 0$ and hence $k \\leq 0$. Similarly, $k c = f(c) = (c - x)(c - y)(c - z) \\geq 0$ and hence $k \\geq 0$. Combining the two, it follows that $k = 0$ and that $f(a) = f(c) = 0$. These equalities imply that $a = x$ and $c = z$, and then it also follows that $b = y$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32464,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn a triangle $A B C$, let $D$ be a point on the segment $B C$ such that $A B + B D = A C + C D$. Suppose that the points $B$, $C$ and the centroids of triangles $A B D$ and $A C D$ lie on a circle. Prove that $A B = A C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nLet $G_{1}, G_{2}$ denote the centroids of triangles $A B D$ and $A C D$. Then $G_{1}, G_{2}$ lie on the line parallel to $B C$ that passes through the centroid of triangle $A B C$. Therefore $B G_{1} G_{2} C$ is an isosceles trapezoid. Therefore it follows that $B G_{1} = C G_{2}$. This proves that $A B^{2} + B D^{2} = A C^{2} + C D^{2}$. Hence it follows that $A B \\cdot B D = A C \\cdot C D$. Therefore the sets $\\{A B, B D\\}$ and $\\{A C, C D\\}$ are the same (since they are both equal to the set of roots of the same polynomial). Note that if $A B = C D$ then $A C = B D$ and then $A B + A C = B C$, a contradiction. Therefore it follows that $A B = A C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32465,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a natural number. Prove that\n$$\n\\left[\\frac{n}{1}\\right]+\\left[\\frac{n}{2}\\right]+\\left[\\frac{n}{3}\\right]+\\cdots+\\left[\\frac{n}{n}\\right]+[\\sqrt{n}]\n$$\nis even. (Here $[x]$ denotes the largest integer smaller than or equal to $x$.)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $f(n)$ denote the given expression. Then $f(1)=2$ which is even. Now suppose that $f(n)$ is even for some $n \\geq 1$. Then\n$$\n\\begin{aligned}\nf(n+1) & =\\left[\\frac{n+1}{1}\\right]+\\left[\\frac{n+1}{2}\\right]+\\left[\\frac{n+1}{3}\\right]+\\cdots+\\left[\\frac{n+1}{n+1}\\right]+[\\sqrt{n+1}] \\\\\n& =\\left[\\frac{n}{1}\\right]+\\left[\\frac{n}{2}\\right]+\\left[\\frac{n}{3}\\right]+\\cdots+\\left[\\frac{n}{n}\\right]+[\\sqrt{n+1}]+\\sigma(n+1)\n\\end{aligned}\n$$\nwhere $\\sigma(n+1)$ denotes the number of positive divisors of $n+1$. This follows from $\\left[\\frac{n+1}{k}\\right]=\\left[\\frac{n}{k}\\right]+1$ if $k$ divides $n+1$, and $\\left[\\frac{n+1}{k}\\right]=\\left[\\frac{n}{k}\\right]$ otherwise. Note that $[\\sqrt{n+1}]=[\\sqrt{n}]$ unless $n+1$ is a square, in which case $[\\sqrt{n+1}]=[\\sqrt{n}]+1$. On the other hand $\\sigma(n+1)$ is odd if and only if $n+1$ is a square. Therefore it follows that $f(n+1)=f(n)+2l$ for some integer $l$. This proves that $f(n+1)$ is even.\n\nThus it follows by induction that $f(n)$ is even for all natural numbers $n$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32466,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $a$, $b$ be natural numbers with $a b > 2$. Suppose that the sum of their greatest common divisor and least common multiple is divisible by $a + b$. Prove that the quotient is at most $(a + b) / 4$. When is this quotient exactly equal to $(a + b) / 4$?",
    "options": [],
    "answer": "The quotient is at most (a + b)/4. Equality holds if and only if either a = b = 2 or a and b are consecutive odd integers.",
    "solution": "Solution:\n\nLet $g$ and $l$ denote the greatest common divisor and the least common multiple, respectively, of $a$ and $b$. Then $g l = a b$. Therefore $g + l \\leq a b + 1$.\n\nSuppose that $(g + l) / (a + b) > (a + b) / 4$. Then we have $a b + 1 > (a + b)^2 / 4$, so we get $(a - b)^2 < 4$.\n\nAssuming $a \\geq b$ we either have $a = b$ or $a = b + 1$.\n\nIn the former case, $g = l = a$ and the quotient is $(g + l) / (a + b) = 1 \\leq (a + b) / 4$.\n\nIn the latter case, $g = 1$ and $l = b(b + 1)$ so we get that $2b + 1$ divides $b^2 + b + 1$. Therefore $2b + 1$ divides $4(b^2 + b + 1) - (2b + 1)^2 = 3$ which implies that $b = 1$ and $a = 2$, a contradiction to the given assumption that $a b > 2$.\n\nThis shows that $(g + l) / (a + b) \\leq (a + b) / 4$.\n\nNote that for the equality to hold, we need that either $a = b = 2$ or, $(a - b)^2 = 4$ and $g = 1$, $l = a b$. The latter case happens if and only if $a$ and $b$ are two consecutive odd numbers. (If $a = 2k + 1$ and $b = 2k - 1$ then $a + b = 4k$ divides $a b + 1 = 4k^2$ and the quotient is precisely $(a + b) / 4$.)",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32467,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nWritten on a blackboard is the polynomial $x^{2} + x + 2014$. Calvin and Hobbes take turns alternatively (starting with Calvin) in the following game. During his turn, Calvin should either increase or decrease the coefficient of $x$ by $1$. And during his turn, Hobbes should either increase or decrease the constant coefficient by $1$. Calvin wins if at any point of time the polynomial on the blackboard at that instant has integer roots. Prove that Calvin has a winning strategy.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nFor $i \\geq 0$, let $f_{i}(x)$ denote the polynomial on the blackboard after Hobbes' $i$-th turn. We let Calvin decrease the coefficient of $x$ by $1$. Therefore $f_{i+1}(2) = f_{i}(2) - 1$ or $f_{i+1}(2) = f_{i}(2) - 3$ (depending on whether Hobbes increases or decreases the constant term). So for some $i$, we have $0 \\leq f_{i}(2) \\leq 2$. If $f_{i}(2) = 0$ then Calvin has won the game. If $f_{i}(2) = 2$ then Calvin wins the game by reducing the coefficient of $x$ by $1$. If $f_{i}(2) = 1$ then $f_{i+1}(2) = 0$ or $f_{i+1}(2) = -2$. In the former case, Calvin has won the game and in the latter case Calvin wins the game by increasing the coefficient of $x$ by $1$.",
    "topic": "Algebra",
    "subtopic": "Abstract Algebra"
  },
  {
    "id": 32468,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn an acute-angled triangle $A B C$, a point $D$ lies on the segment $B C$. Let $O_{1}, O_{2}$ denote the circumcentres of triangles $A B D$ and $A C D$, respectively. Prove that the line joining the circumcentre of triangle $A B C$ and the orthocentre of triangle $O_{1} O_{2} D$ is parallel to $B C$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWithout loss of generality assume that $\\angle A D C \\geq 90^{\\circ}$. Let $O$ denote the circumcenter of triangle $A B C$ and $K$ the orthocentre of triangle $O_{1} O_{2} D$. We shall first show that the points $O$ and $K$ lie on the circumcircle of triangle $A O_{1} O_{2}$. Note that circumcircles of triangles $A B D$ and $A C D$ pass through the points $A$ and $D$, so $A D$ is perpendicular to $O_{1} O_{2}$ and, triangle $A O_{1} O_{2}$ is congruent to triangle $D O_{1} O_{2}$. In particular, $\\angle A O_{1} O_{2}=\\angle O_{2} O_{1} D=\\angle B$ since $O_{2} O_{1}$ is the perpendicular bisector of $A D$. On the other hand since $O O_{2}$ is the perpendicular bisector of $A C$ it follows that $\\angle A O O_{2}=\\angle B$. This shows that $O$ lies on the circumcircle of triangle $A O_{1} O_{2}$. Note also that, since $A D$ is perpendicular to $O_{1} O_{2}$, we have $\\angle O_{2} K A=90^{\\circ}-\\angle O_{1} O_{2} K=\\angle O_{2} O_{1} D=\\angle B$. This proves that $K$ also lies on the circumcircle of triangle $A O_{1} O_{2}$.\n\nTherefore $\\angle A K O=180^{\\circ}-\\angle A O_{2} O=\\angle A D C$ and hence $O K$ is parallel to $B C$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32469,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $n$ be a natural number and $X = \\{1, 2, \\ldots, n\\}$. For subsets $A$ and $B$ of $X$ we define $A \\Delta B$ to be the set of all those elements of $X$ which belong to exactly one of $A$ and $B$. Let $\\mathcal{F}$ be a collection of subsets of $X$ such that for any two distinct elements $A$ and $B$ in $\\mathcal{F}$ the set $A \\Delta B$ has at least two elements. Show that $\\mathcal{F}$ has at most $2^{n-1}$ elements. Find all such collections $\\mathcal{F}$ with $2^{n-1}$ elements.",
    "options": [],
    "answer": "Maximum size is 2^{n-1}. The extremal families are exactly the collections of all even-cardinality subsets of X or all odd-cardinality subsets of X.",
    "solution": "Solution:\nFor each subset $A$ of $\\{1, 2, \\ldots, n-1\\}$, we pair it with $A \\cup \\{n\\}$. Note that for any such pair $(A, B)$ not both $A$ and $B$ can be in $\\mathcal{F}$. Since there are $2^{n-1}$ such pairs it follows that $\\mathcal{F}$ can have at most $2^{n-1}$ elements.\n\nWe shall show by induction on $n$ that if $|\\mathcal{F}| = 2^{n-1}$ then $\\mathcal{F}$ contains either all the subsets with odd number of elements or all the subsets with even number of elements. The result is easy to see for $n = 1$. Suppose that the result is true for $n = m-1$. We now consider the case $n = m$. Let $\\mathcal{F}_1$ be the set of those elements in $\\mathcal{F}$ which contain $m$ and $\\mathcal{F}_2$ be the set of those elements which do not contain $m$. By induction, $\\mathcal{F}_2$ can have at most $2^{m-2}$ elements. Further, for each element $A$ of $\\mathcal{F}_1$ we consider $A \\setminus \\{m\\}$. This new collection also satisfies the required property, so it follows that $\\mathcal{F}_1$ has at most $2^{m-2}$ elements. Thus, if $|\\mathcal{F}| = 2^{m-1}$ then it follows that $|\\mathcal{F}_1| = |\\mathcal{F}_2| = 2^{m-2}$. Further, by induction hypothesis, $\\mathcal{F}_2$ contains all those subsets of $\\{1, 2, \\ldots, m-1\\}$ with (say) even number of elements. It then follows that $\\mathcal{F}_1$ contains all those subsets of $\\{1, 2, \\ldots, m\\}$ which contain the element $m$ and which contain an even number of elements. This proves that $\\mathcal{F}$ contains either all the subsets with odd number of elements or all the subsets with even number of elements.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32470,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a right-angled triangle with $\\angle B = 90^{\\circ}$. Let $BD$ be the altitude from $B$ onto $AC$. Let $P$, $Q$ and $I$ be the incentres of triangles $ABD$, $CBD$ and $ABC$ respectively. Show that the circumcentre of the triangle $PIQ$ lies on the hypotenuse $AC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nWe begin with the following lemma:\n\nLemma: Let $XYZ$ be a triangle with $\\angle XYZ = 90 + \\alpha$. Construct an isosceles triangle $XEZ$, externally on the side $XZ$, with base angle $\\alpha$. Then $E$ is the circumcentre of $\\triangle XYZ$.\n\nProof of the Lemma: Draw $ED \\perp XZ$. Then $DE$ is the perpendicular bisector of $XZ$. We also observe that $\\angle XED = \\angle ZED = 90 - \\alpha$. Observe that $E$ is on the perpendicular bisector of $XZ$. Construct the circumcircle of $XYZ$. Draw the perpendicular bisector of $XY$ and let it meet $DE$ in $F$. Then $F$ is the circumcentre of $\\triangle XYZ$. Join $XF$. Then $\\angle XFD = 90 - \\alpha$. But we know that $\\angle XED = 90 - \\alpha$. Hence $E = F$.\n\nLet $r_1$, $r_2$ and $r$ be the inradii of the triangles $ABD$, $CBD$ and $ABC$ respectively. Join $PD$ and $DQ$. Observe that $\\angle PDQ = 90^{\\circ}$. Hence\n\n$$\nPQ^2 = PD^2 + DQ^2 = 2r_1^2 + 2r_2^2\n$$\n\nLet $s_1 = (AB + BD + DA)/2$. Observe that $BD = ca/b$ and $AD = \\sqrt{AB^2 - BD^2} = \\sqrt{c^2 - \\left(\\frac{ca}{b}\\right)^2} = c^2/b$. This gives $s_1 = cs/b$. But $r_1 = s_1 - c = (c/b)(s-b) = cr/b$. Similarly, $r_2 = ar/b$. Hence\n\n$$\nPQ^2 = 2r^2 \\left(\\frac{c^2 + a^2}{b^2}\\right) = 2r^2\n$$\n\nConsider $\\triangle PIQ$. Observe that $\\angle PIQ = 90 + (B/2) = 135^{\\circ}$. Hence $PQ$ subtends $90^{\\circ}$ on the circumference of the circumcircle of $\\triangle PIQ$. But we have seen that $\\angle PDQ = 90^{\\circ}$. Now construct a circle with $PQ$ as diameter. Let it cut $AC$ again in $K$. It follows that $\\angle PKQ = 90^{\\circ}$ and the points $P, D, K, Q$ are concyclic. We also notice $\\angle KPQ = \\angle KDQ = 45^{\\circ}$ and $\\angle PQK = \\angle PDA = 45^{\\circ}$.\n\n![](attached_image_1.png)\n\nThus $PKQ$ is an isosceles right-angled triangle with $KP = KQ$. Therefore $KP^2 + KQ^2 = PQ^2 = 2r^2$ and hence $KP = KQ = r$.\n\nNow $\\angle PIQ = 90 + 45$ and $\\angle PKQ = 2 \\times 45^{\\circ} = 90^{\\circ}$ with $KP = KQ = r$.\n\nHence $K$ is the circumcentre of $\\triangle PIQ$.\nSolution:\n\nHere we use computation to prove that the point of contact $K$ of the incircle with $AC$ is the circumcentre of $\\triangle PIQ$. We show that $KP = KQ = r$.\n\nLet $r_1$ and $r_2$ be the inradii of triangles $ABD$ and $CBD$ respectively. Draw $PL \\perp AC$ and $QM \\perp AC$. If $s_1$ is the semiperimeter of $\\triangle ABD$, then $AL = s_1 - BD$.\n\n![](attached_image_2.png)\n\nBut\n$$\ns_1 = \\frac{AB + BD + DA}{2}, \\quad BD = \\frac{ca}{b}, \\quad AD = \\frac{c^2}{b}\n$$\n\nHence $s_1 = cs/b$. This gives $r_1 = s_1 - c = cr/b$, $AL = s_1 - BD = c(s-a)/b$. Hence $KL = AK - AL = (s-a) - \\frac{c(s-a)}{b} = \\frac{(b-c)(s-a)}{b}$.\n\nWe observe that $2r^2 = \\frac{(c+a-b)^2}{2} = \\frac{c^2 + a^2 + b^2 - 2bc - 2ab + 2ca}{2} = (b^2 - ba - bc + ac) = (b-c)(b-a)$.\n\n$$\n\\begin{aligned}\n(s-a)(b-c) &= (s-b + b-a)(b-c) \\\\\n&= r(b-c) + (b-a)(b-c) \\\\\n&= r(b-c) + 2r^2 = r(b-c + c + a - b) = ra\n\\end{aligned}\n$$\n\nThus $KL = ra/b$. Finally,\n$$\nKP^2 = KL^2 + LP^2 = \\frac{r^2 a^2}{b^2} + \\frac{r^2 + c^2}{b^2} = r^2\n$$\n\nThus $KP = r$. Similarly, $KQ = r$. This gives $KP = KI = KQ = r$ and therefore $K$ is the circumcentre of $\\triangle PIQ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32471,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nFor any natural number $n > 1$, write the infinite decimal expansion of $1 / n$ (for example, we write $1 / 2 = 0.4\\overline{9}$ as its infinite decimal expansion, not $0.5$). Determine the length of the non-periodic part of the (infinite) decimal expansion of $1 / n$.",
    "options": [],
    "answer": "r = max(ν_2(n), ν_5(n))",
    "solution": "Solution:\n\nFor any prime $p$, let $\\nu_{p}(n)$ be the maximum power of $p$ dividing $n$; i.e., $p^{\\nu_{p}(n)}$ divides $n$ but not a higher power. Let $r$ be the length of the non-periodic part of the infinite decimal expansion of $1 / n$.\n\nWrite\n$$\n\\frac{1}{n} = 0 . a_{1} a_{2} \\cdots a_{r} \\overline{b_{1} b_{2} \\cdots b_{s}}\n$$\n\nWe show that $r = \\max \\left(\\nu_{2}(n), \\nu_{5}(n)\\right)$.\n\nLet $a$ and $b$ be the numbers $a_{1} a_{2} \\cdots a_{r}$ and $b = b_{1} b_{2} \\cdots b_{s}$ respectively. (Here $a_{1}$ and $b_{1}$ can both be $0$.) Then\n$$\n\\frac{1}{n} = \\frac{1}{10^{r}}\\left(a + \\sum_{k \\geq 1} \\frac{b}{\\left(10^{s}\\right)^{k}}\\right) = \\frac{1}{10^{r}}\\left(a + \\frac{b}{10^{s} - 1}\\right)\n$$\n\nThus we get $10^{r}\\left(10^{s} - 1\\right) = n\\left(\\left(10^{s} - 1\\right)a + b\\right)$. It shows that $r \\geq \\max \\left(\\nu_{2}(n), \\nu_{5}(n)\\right)$. Suppose $r > \\max \\left(\\nu_{2}(n), \\nu_{5}(n)\\right)$. Then $10$ divides $b - a$. Hence the last digits of $a$ and $b$ are equal: $a_{r} = b_{s}$. This means\n$$\n\\frac{1}{n} = 0 . a_{1} a_{2} \\cdots a_{r-1} \\overline{b_{s} b_{1} b_{2} \\cdots b_{s-1}}\n$$\n\nThis contradicts the definition of $r$. Therefore $r = \\max \\left(\\nu_{2}(n), \\nu_{5}(n)\\right)$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32472,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFind all real functions $f$ from $\\mathbb{R} \\rightarrow \\mathbb{R}$ satisfying the relation\n$$\nf\\left(x^{2}+y f(x)\\right)=x f(x+y)\n$$",
    "options": [],
    "answer": "f(x) = 0 for all real x, or f(x) = x for all real x",
    "solution": "Solution:\nPut $x=0$ and we get $f(y f(0))=0$. If $f(0) \\neq 0$, then $y f(0)$ takes all real values when $y$ varies over real line. We get $f(x) \\equiv 0$. Suppose $f(0)=0$. Taking $y=-x$, we get $f\\left(x^{2}-x f(x)\\right)=0$ for all real $x$.\n\nSuppose there exists $x_{0} \\neq 0$ in $\\mathbb{R}$ such that $f\\left(x_{0}\\right)=0$. Putting $x=x_{0}$ in the given relation we get\n$$\nf\\left(x_{0}^{2}\\right)=x_{0} f\\left(x_{0}+y\\right)\n$$\nfor all $y \\in \\mathbb{R}$. Now the left side is a constant and hence it follows that $f$ is a constant function. But the only constant function which satisfies the equation is identically zero function, which is already obtained. Hence we may consider the case where $f(x) \\neq 0$ for all $x \\neq 0$.\n\nSince $f\\left(x^{2}-x f(x)\\right)=0$, we conclude that $x^{2}-x f(x)=0$ for all $x \\neq 0$. This implies that $f(x)=x$ for all $x \\neq 0$. Since $f(0)=0$, we conclude that $f(x)=x$ for all $x \\in \\mathbb{R}$.\nThus we have two functions: $f(x) \\equiv 0$ and $f(x)=x$ for all $x \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32473,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nThere are four basket-ball players $A$, $B$, $C$, $D$. Initially, the ball is with $A$. The ball is always passed from one person to a different person. In how many ways can the ball come back to $A$ after seven passes? (For example $A \\rightarrow C \\rightarrow B \\rightarrow D \\rightarrow A \\rightarrow B \\rightarrow C \\rightarrow A$ and $A \\rightarrow D \\rightarrow A \\rightarrow D \\rightarrow C \\rightarrow A \\rightarrow B \\rightarrow A$ are two ways in which the ball can come back to $A$ after seven passes.)",
    "options": [],
    "answer": "546",
    "solution": "Solution:\nLet $x_{n}$ be the number of ways in which $A$ can get back the ball after $n$ passes. Let $y_{n}$ be the number of ways in which the ball goes back to a fixed person other than $A$ after $n$ passes. Then\n$$\nx_{n}=3 y_{n-1},\n$$\nand\n$$\ny_{n}=x_{n-1}+2 y_{n-1}\n$$\nWe also have $x_{1}=0$, $x_{2}=3$, $y_{1}=1$ and $y_{2}=2$.\n\nEliminating $y_{n}$ and $y_{n-1}$, we get $x_{n+1}=3 x_{n-1}+2 x_{n}$. Thus\n$$\n\\begin{aligned}\n& x_{3}=3 x_{1}+2 x_{2}=2 \\times 3=6 \\\\\n& x_{4}=3 x_{2}+2 x_{3}=(3 \\times 3)+(2 \\times 6)=9+12=21 \\\\\n& x_{5}=3 x_{3}+2 x_{4}=(3 \\times 6)+(2 \\times 21)=18+42=60 \\\\\n& x_{6}=3 x_{4}+2 x_{5}=(3 \\times 21)+(2 \\times 60)=63+120=183 \\\\\n& x_{7}=3 x_{5}+2 x_{6}=(3 \\times 60)+(2 \\times 183)=180+366=546\n\\end{aligned}\n$$\n\n\nAlternate solution:\nSince the ball goes back to one of the other 3 persons, we have\n$$\nx_{n}+3 y_{n}=3^{n}\n$$\nsince there are $3^{n}$ ways of passing the ball in $n$ passes. Using $x_{n}=3 y_{n-1}$, we obtain\n$$\nx_{n-1}+x_{n}=3^{n-1}\n$$\nwith $x_{1}=0$. Thus\n$$\n\\begin{array}{r}\nx_{7}=3^{6}-x_{6}=3^{6}-3^{5}+x_{5}=3^{6}-3^{5}+3^{4}-x_{4}=3^{6}-3^{5}+3^{4}-3^{3}+x_{3} \\\\\n=3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-x_{2}=3^{6}-3^{5}+3^{4}-3^{3}+3^{2}-3 \\\\\n=\\left(2 \\times 3^{5}\\right)+\\left(2 \\times 3^{3}\\right)+(2 \\times 3)=486+54+6=546\n\\end{array}\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32474,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nLet $ABCD$ be a convex quadrilateral. Let the diagonals $AC$ and $BD$ intersect in $P$. Let $PE$, $PF$, $PG$ and $PH$ be the altitudes from $P$ onto the sides $AB$, $BC$, $CD$ and $DA$ respectively. Show that $ABCD$ has an incircle if and only if\n$$\n\\frac{1}{PE} + \\frac{1}{PG} = \\frac{1}{PF} + \\frac{1}{PH}\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nLet $AP = p$, $BP = q$, $CP = r$, $DP = s$; $AB = a$, $BC = b$, $CD = c$ and $DA = d$. Let $\\angle APB = \\angle CPD = \\theta$. Then $\\angle BPC = \\angle DPA = \\pi - \\theta$. Let us also write $PE = h_1$, $PF = h_2$, $PG = h_3$ and $PH = h_4$.\n\n![](attached_image_1.png)\n\nObserve that\n$$\nh_1 a = p q \\sin \\theta, \\quad h_2 b = q r \\sin \\theta, \\quad h_3 c = r s \\sin \\theta, \\quad h_4 d = s p \\sin \\theta\n$$\n\nHence\n$$\n\\frac{1}{h_1} + \\frac{1}{h_3} = \\frac{1}{h_2} + \\frac{1}{h_4}\n$$\nis equivalent to\n$$\n\\frac{a}{p q} + \\frac{c}{r s} = \\frac{b}{q r} + \\frac{d}{s p}\n$$\n\nThis is the same as\n$$\na r s + c p q = b s p + d q r\n$$\n\nThus we have to prove that $a + c = b + d$ if and only if $a r s + c p q = b s p + d q r$. Now we can write $a + c = b + d$ as\n$$\na^2 + c^2 + 2 a c = b^2 + d^2 + 2 b d\n$$\n\nBut we know that\n$$\n\\begin{aligned}\n& a^2 = p^2 + q^2 - 2 p q \\cos \\theta, \\quad c^2 = r^2 + s^2 - 2 r s \\cos \\theta \\\\\n& b^2 = q^2 + r^2 + 2 q r \\cos \\theta, \\quad d^2 = p^2 + s^2 + 2 p s \\cos \\theta\n\\end{aligned}\n$$\n\nHence $a + c = b + d$ is equivalent to\n$$\n- p q \\cos \\theta - r s \\cos \\theta + a c = p s \\cos \\theta + q r \\cos \\theta + b d\n$$\n\nSimilarly, by squaring $a r s + c p q = b s p + d q r$ we can show that it is equivalent to\n$$\n- p q \\cos \\theta - r s \\cos \\theta + a c = p s \\cos \\theta + q r \\cos \\theta + b d\n$$\n\nWe conclude that $a + c = b + d$ is equivalent to $c p q + a r s = b p s + d q r$. Hence $ABCD$ has an incircle if and only if\n$$\n\\frac{1}{h_1} + \\frac{1}{h_3} = \\frac{1}{h_2} + \\frac{1}{h_4}\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32475,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\nFrom a set of 11 square integers, show that one can choose 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that\n$$\na^{2}+b^{2}+c^{2} \\equiv d^{2}+e^{2}+f^{2} \\quad(\\bmod 12)\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\nThe first observation is that we can find 5 pairs of squares such that the two numbers in a pair have the same parity. We can see this as follows:\n\n| Odd numbers | Even numbers | Odd pairs | Even pairs | Total pairs |\n| :---: | :---: | :---: | :---: | :---: |\n| 0 | 11 | 0 | 5 | 5 |\n| 1 | 10 | 0 | 5 | 5 |\n| 2 | 9 | 1 | 4 | 5 |\n| 3 | 8 | 1 | 4 | 5 |\n| 4 | 7 | 2 | 3 | 5 |\n| 5 | 6 | 2 | 3 | 5 |\n| 6 | 5 | 3 | 2 | 5 |\n| 7 | 4 | 3 | 2 | 5 |\n| 8 | 3 | 4 | 1 | 5 |\n| 9 | 2 | 4 | 1 | 5 |\n| 10 | 1 | 5 | 0 | 5 |\n| 11 | 0 | 5 | 0 | 5 |\n\nLet us take such 5 pairs: say $\\left(x_{1}^{2}, y_{1}^{2}\\right),\\left(x_{2}^{2}, y_{2}^{2}\\right), \\ldots,\\left(x_{5}^{2}, y_{5}^{2}\\right)$. Then $x_{j}^{2}-y_{j}^{2}$ is divisible by 4 for $1 \\leq j \\leq 5$. Let $r_{j}$ be the remainder when $x_{j}^{2}-y_{j}^{2}$ is divisible by $3, 1 \\leq j \\leq 5$. We have 5 remainders $r_{1}, r_{2}, r_{3}, r_{4}, r_{5}$. But these can be 0, 1 or 2. Hence either one of the remainders occurs 3 times or each of the remainders occurs once. If, for example, $r_{1}=r_{2}=r_{3}$, then 3 divides $r_{1}+r_{2}+r_{3}$; if $r_{1}=0, r_{2}=1$ and $r_{3}=2$, then again 3 divides $r_{1}+r_{2}+r_{3}$. Thus we can always find three remainders whose sum is divisible by 3. This means we can find 3 pairs, say, $\\left(x_{1}^{2}, y_{1}^{2}\\right),\\left(x_{2}^{2}, y_{2}^{2}\\right),\\left(x_{3}^{2}, y_{3}^{2}\\right)$ such that 3 divides $\\left(x_{1}^{2}-y_{1}^{2}\\right)+\\left(x_{2}^{2}-y_{2}^{2}\\right)+\\left(x_{3}^{2}-y_{3}^{2}\\right)$. Since each difference is divisible by 4, we conclude that we can find 6 numbers $a^{2}, b^{2}, c^{2}, d^{2}, e^{2}, f^{2}$ such that\n$$\na^{2}+b^{2}+c^{2} \\equiv d^{2}+e^{2}+f^{2} \\quad(\\bmod 12)\n$$",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 32476,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be triangle in which $AB = AC$. Suppose the orthocentre of the triangle lies on the in-circle. Find the ratio $AB/BC$.",
    "options": [],
    "answer": "3/4",
    "solution": "Since the triangle is isosceles, the orthocentre lies on the perpendicular $AD$ from $A$ on to $BC$. Let it cut the in-circle at $H$. Now we are given that $H$ is the orthocentre of the triangle. Let $AB = AC = b$ and $BC = 2a$. Then $BD = a$. Observe that $b > a$ since $b$ is the hypotenuse and $a$ is a leg of a right-angled triangle. Let $BH$ meet $AC$ in $E$ and $CH$ meet $AB$ in $F$. By Pythagoras theorem applied to $\\triangle BDH$, we get\n\n![](attached_image_1.png)\n\n$$\nBH^2 = HD^2 + BD^2 = 4r^2 + a^2,\n$$\nwhere $r$ is the in-radius of $ABC$. We want to compute $BH$ in another way. Since $A, F, H, E$ are con-cyclic, we have\n$$\nBH \\cdot BE = BF \\cdot BA.\n$$\nBut $BF \\cdot BA = BD \\cdot BC = 2a^2$, since $A, F, D, C$ are con-cyclic. Hence $BH^2 = 4a^4/BE^2$. But\n$$\nBE^2 = 4a^2 - CE^2 = 4a^2 - BF^2 = 4a^2 - \\left(\\frac{2a^2}{b}\\right)^2 = \\frac{4a^2(b^2 - a^2)}{b^2}.\n$$\nThis leads to\n$$\nBH^2 = \\frac{a^2b^2}{b^2 - a^2}.\n$$\nThus we get\n$$\n\\frac{a^2b^2}{b^2 - a^2} = a^2 + 4r^2.\n$$\nThis simplifies to $(a^4/(b^2 - a^2)) = 4r^2$. Now we relate $a, b, r$ in another way using area. We know that $[ABC] = rs$, where $s$ is the semi-perimeter of $ABC$. We have $s = (b + b + 2a)/2 = b + a$. On the other hand area can be calculated using Heron's formula:\n$$\n[ABC]^2 = s(s - 2a)(s - b)(s - b) = (b + a)(b - a)a^2 = a^2(b^2 - a^2).\n$$\nHence\n$$\nr^2 = \\frac{[ABC]^2}{s^2} = \\frac{a^2(b^2 - a^2)}{(b + a)^2}.\n$$\nUsing this we get\n$$\n\\frac{a^4}{b^2 - a^2} = 4 \\left( \\frac{a^2(b^2 - a^2)}{(b + a)^2} \\right).\n$$\nTherefore $a^2 = 4(b - a)^2$, which gives $a = 2(b - a)$ or $2b = 3a$. Finally,\n$$\n\\frac{AB}{BC} = \\frac{b}{2a} = \\frac{3}{4}.\n$$\nWe use the known facts $BH = 2R \\cos B$ and $r = 4R \\sin(A/2) \\sin(B/2) \\sin(C/2)$, where $R$ is the circumradius of $\\triangle ABC$ and $r$ its in-radius. Therefore\n$$\nHD = BH \\sin \\angle HBD = 2R \\cos B \\sin \\left(\\frac{\\pi}{2} - C\\right) = 2R \\cos^2 B,\n$$\nsince $\\angle C = \\angle B$. But $\\angle B = (\\pi - \\angle A)/2$, since $ABC$ is isosceles. Thus we obtain\n$$\nHD = \\cos^2 \\left( \\frac{\\pi}{2} - \\frac{A}{2} \\right).\n$$\nHowever $HD$ is also the diameter of the in circle. Therefore $HD = 2r$. Thus we get\n$$\n2R \\cos^2 \\left( \\frac{\\pi}{2} - \\frac{A}{2} \\right) = 2r = 8R \\sin(A/2) \\sin^2\\left(\\frac{(\\pi - A)}{4}\\right).\n$$\nThis reduces to\n$$\n\\sin(A/2) = 2(1 - \\sin(A/2)).\n$$\nTherefore $\\sin(A/2) = 2/3$. We also observe that $\\sin(A/2) = BD/AB$. Finally\n$$\n\\frac{AB}{BC} = \\frac{AB}{2BD} = \\frac{1}{2\\sin(A/2)} = \\frac{3}{4}.\n$$\nLet $D$ be the mid-point of $BC$. Extend $AD$ to meet the circumcircle in $L$. Then we know that $HD = DL$. But $HD = 2r$. Thus $DL = 2r$. Therefore $IL = ID + DL = r + 2r = 3r$. We also know that $LB = LI$. Therefore $LB = 3r$. This gives\n$$\n\\frac{BL}{LD} = \\frac{3r}{2r} = \\frac{3}{2}.\n$$\nBut $\\triangle BLD$ is similar to $\\triangle ABD$. So\n$$\n\\frac{AB}{BD} = \\frac{BL}{LD} = \\frac{3}{2}.\n$$\nFinally,\n$$\n\\frac{AB}{BC} = \\frac{AB}{2BD} = \\frac{3}{4}.\n$$\nLet $D$ be the mid-point of $BC$ and $E$ be the mid-point of $DC$. Since $DI = IH(= r)$ and $DE = EC$, the mid-point theorem implies that $IE \\parallel CH$. But $CH \\perp AB$. Therefore $EI \\perp AB$. Let $EI$ meet $AB$ in $F$. Then $F$ is the point of tangency of the incircle of $\\triangle ABC$ with $AB$. Since the incircle is also tangent to $BC$ at $D$, we have $BF = BD$. Observe that $\\triangle BFE$ is similar to $\\triangle BDA$. Hence\n$$\n\\frac{AB}{BD} = \\frac{BE}{BF} = \\frac{BE}{BD} = \\frac{BD + DE}{BD} = 1 + \\frac{DE}{BD} = \\frac{3}{2}.\n$$\nThis gives\n$$\n\\frac{AB}{BC} = \\frac{3}{4}.\n$$",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32477,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n$$\nf(x^2 + x f(y)) = x f(x + y)\n$$\nfor all reals $x, y$.",
    "options": [],
    "answer": "f(x) = 0 for all real x; and f(x) = x for all real x",
    "solution": "It is easy to see that $f(0) = 0$ and $f(x^2) = x f(x)$ for all $x \\in \\mathbb{R}$. If $f(\\alpha) = 0$ for some $\\alpha$, then\n$$\nf(x^2 + x f(\\alpha)) = x f(x + \\alpha),\n$$\nfor all $x \\in \\mathbb{R}$. Therefore\n$$\nx f(x + \\alpha) = f(x^2) = x f(x),\n$$\nfor all $x \\in \\mathbb{R}$. For $x \\neq 0$, we get $f(x) = f(x + \\alpha)$. Note that this is also valid for $x = 0$. Suppose $f(1) = 0$. Then\n$$\nf(1 + f(x)) = f(x + 1),\n$$\nso that $f(f(x)) = f(x)$ for all $x$. If there exists a $\\lambda \\in \\mathbb{R}$ such that $f(\\lambda) \\neq 0$, then for any $t \\in \\mathbb{R}$, taking $s = t / f(\\lambda)$, we have\n$$\nf(s^2 + s f(\\lambda - s)) = s f(s + \\lambda - s) = s f(\\lambda) = t\n$$\nwhich shows that $f$ is onto. Hence there exists $x_0$ such that $f(x_0) = 1$. This gives\n$$\n1 = f(x_0) = f(f(x_0)) = f(1) = 0,\n$$\nwhich is absurd. Thus $f(1) = 0$ forces $f(x) = 0$ for all $x \\in \\mathbb{R}$.\n\nSuppose $f(1) \\neq 0$. Let $\\alpha \\in \\mathbb{R}$ be such that $f(\\alpha) = 0$. As we have seen earlier, $f(x + \\alpha) = f(x) = 0$ for all $x$. Therefore\n$$\n\\begin{aligned}\nf(\\alpha^2 + 1) &= f((1 + \\alpha)^2 + (1 + \\alpha) f(\\alpha)) = (\\alpha + 1) f(1 + 2\\alpha) = (\\alpha + 1) f(1), \\\\\nf(\\alpha^2 + 1) &= f((1 - \\alpha)^2 + (1 - \\alpha) f(\\alpha)) = (1 - \\alpha) f(1).\n\\end{aligned}\n$$\nThese show that $1 + \\alpha = 1 - \\alpha$. Therefore $\\alpha = 0$. Thus $f(\\alpha) = 0$ implies that $\\alpha = 0$.\n\nTaking $x = -y$ in the equation, we get $f(y^2 - y f(y)) = y f(y - y) = 0$. Hence $y^2 - y f(y) = 0$ for all $y \\in \\mathbb{R}$. This gives $f(y) = y$ for all $y \\neq 0$. Since $f(0) = 0$, we conclude that $f(y) = y$ for all $y \\in \\mathbb{R}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32478,
    "subject": "Mathematics (Multi-modal)",
    "question": "For positive real numbers $a$, $b$, $c$, which of the following statements necessarily implies $a = b = c$:\n(I) $a(b^3 + c^3) = b(c^3 + a^3) = c(a^3 + b^3)$,\n(II) $a(a^3 + b^3) = b(b^3 + c^3) = c(c^3 + a^3)$?\nJustify your answer.",
    "options": [],
    "answer": "Only (II) necessarily implies a = b = c; (I) does not (e.g., a = b = ((1 + √5)/2) c gives (I) without a = b = c).",
    "solution": "We show that (I) need not imply that $a = b = c$ whereas (II) always implies $a = b = c$.\n\nObserve that $a(b^3 + c^3) = b(c^3 + a^3)$ gives $c^3(a - b) = ab(a^2 - b^2)$. This gives either $a = b$ or $ab(a + b) = c^3$. Similarly, $b = c$ or $bc(b + c) = a^3$. If $a \\neq b$ and $b \\neq c$, we obtain\n$$\nab(a + b) = c^3, \\quad bc(b + c) = a^3.\n$$\nTherefore\n$$\nb(a^2 - c^2) + b^2(a - c) = c^3 - a^3.\n$$\nThis gives $(a-c)(a^2 + b^2 + c^2 + ab + bc + ca) = 0$. Since $a$, $b$, $c$ are positive, the only possibility is $a = c$. We have therefore 4 possibilities: $a = b = c$; $a \\neq b$, $b \\neq c$ and $c = a$; $b \\neq c$, $c \\neq a$ and $a = b$; $c \\neq a$, $a \\neq b$ and $b = c$.\n\nSuppose $a = b$ and $b$, $a \\neq c$. Then $b(c^3 + a^3) = c(a^3 + b^3)$ gives $ac^3 + a^4 = 2ca^3$. This implies that $a(a-c)(a^2 - ac - c^2) = 0$. Therefore $a^2 - ac - c^2 = 0$. Putting $a/c = x$, we get the quadratic equation $x^2 - x - 1 = 0$. Hence $x = (1 + \\sqrt{5})/2$. Thus we get\n$$\na = b = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) c, \\quad c \\text{ arbitrary positive real number.}\n$$\nSimilarly, we get other two cases:\n$$\nb = c = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) a, \\quad a \\text{ arbitrary positive real number;}\n$$\n$$\nc = a = \\left( \\frac{1 + \\sqrt{5}}{2} \\right) b, \\quad b \\text{ arbitrary positive real number.}\n$$\nAnd $a = b = c$ is the fourth possibility.\n\nConsider (II): $a(a^3 + b^3) = b(b^3 + c^3) = c(c^3 + a^3)$. Suppose $a$, $b$, $c$ are mutually distinct. We may assume $a = \\max\\{a, b, c\\}$. Hence $a > b$ and $a > c$. Using $a > b$, we get from the first relation that $a^3 + b^3 < b^3 + c^3$. Therefore $a^3 < c^3$ forcing $a < c$. This contradicts $a > c$. We conclude that $a$, $b$, $c$ cannot be mutually distinct. This means some two must be equal. If $a = b$, the equality of the first two expressions give $a^3 + b^3 = b^3 + c^3$ so that $a = c$. Similarly, we can show that $b = c$ implies $b = a$ and $c = a$ gives $c = b$.\n\n\nAlternate for (II) by a contestant:\n\nWe can write\n$$\n\\begin{aligned}\n\\frac{a^3}{c} + \\frac{b^3}{c} &= \\frac{c^3}{a} + a^2, \\\\\n\\frac{b^3}{a} + \\frac{c^3}{a} &= \\frac{a^3}{b} + b^2, \\\\\n\\frac{c^3}{b} + \\frac{a^3}{b} &= \\frac{b^3}{c} + c^2.\n\\end{aligned}\n$$\nAdding, we get\n$$\n\\frac{a^3}{c} + \\frac{b^3}{a} + \\frac{c^3}{b} = a^2 + b^2 + c^2.\n$$\n\nUsing C-S inequality, we have\n$$\n(a^2 + b^2 + c^2)^2 = \\left( \\frac{\\sqrt{a^3}}{\\sqrt{c}} \\cdot \\sqrt{ac} + \\frac{\\sqrt{b^3}}{\\sqrt{a}} \\cdot \\sqrt{ba} + \\frac{\\sqrt{c^3}}{\\sqrt{b}} \\cdot \\sqrt{cb} \\right)^2 \\\\\n\\le \\left( \\frac{a^3}{c} + \\frac{b^3}{a} + \\frac{c^3}{b} \\right) (ac + ba + cb) \\\\\n= (a^2 + b^2 + c^2)(ab + bc + ca).\n$$\nThus we obtain\n$$\na^2 + b^2 + c^2 \\le ab + bc + ca.\n$$\nHowever this implies $(a-b)^2 + (b-c)^2 + (c-a)^2 \\le 0$ and hence $a = b = c$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32479,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number. A sequence $x_1, x_2, \\dots, x_{n^2}$ is called *n-*good if each $x_i$ is an element of $\\{1, 2, \\dots, n\\}$ and the ordered pairs $(x_i, x_{i+1})$ are all different for $i = 1, 2, \\dots, n^2$ (here we consider the subscripts modulo $n^2$). Two *n-*good sequences $x_1, x_2, \\dots, x_{n^2}$ and $y_1, y_2, \\dots, y_{n^2}$ are called *similar* if there exists an integer $k$ such that $y_i = x_{i+k}$ for all $i = 1, 2, \\dots, n^2$ (again taking the subscripts modulo $n^2$). Suppose that there exists a non-trivial permutation $\\sigma$ of $\\{1, 2, \\dots, n\\}$ and an *n-*good sequence $x_1, x_2, \\dots, x_{n^2}$ which is similar to $\\sigma(x_1), \\sigma(x_2), \\dots, \\sigma(x_{n^2})$. Show that $n \\equiv 2 \\pmod{4}$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Without loss of generality we assume that $\\sigma(1) \\neq 1$. Also assume that $x_1 = x_2 = 1$. Let $k$ be the smallest natural number such that $\\sigma^k(1) = 1$. And let $r$ be the smallest natural number such that $\\sigma(x_i) = x_{i+r}$ for all $i = 1, 2, \\dots, n$. Therefore $\\sigma^k(x_i) = x_{i+kr}$. Since $x_1 = x_2 = 1$, it follows that $1+kr \\equiv 1 \\pmod{n^2}$, so $n^2$ divides $kr$. Further, it follows that $\\sigma^k$ is the identity permutation. In fact, by looking at the pairs $(a, a)$ appearing in the sequence, it follows that for any $i$ and $1 \\le j < k$ we have $\\sigma^j(i) \\neq i$. So $\\sigma$ is made up of $n/k$ $k$-cycles. Let $l = n/k$. Without loss of generality, we can assume that $\\sigma(i) = i+l$.\n\nConsider $\\{r, 2r, 3r, \\dots, (k-1)r\\} \\pmod{n^2}$. Let $s$ be the smallest natural number such that $s \\equiv jr \\pmod{n^2}$. Replacing $\\sigma$ by $\\sigma^j$, we may assume that $s = r$. It then follows that $kr = n^2$, so $r = nl$.\n\nFor each $i = 1, 2, \\dots, n^2$, let $a_i$ be an integer such that $0 \\le a_i \\le n-1$ and $a_i \\equiv x_{i+1} - x_i \\pmod{n^2}$. Note that for any $j = 0, 1, 2, \\dots, n-1$, there exists exactly $n$ values of $i$ for which $a_i = j$. Since $\\sigma(i) = i+l$, it follows that $a_{i+r} = a_i$. Therefore for any $j = 0, 1, 2, \\dots, n-1$, there exists exactly $l$ values of $i$ with $1 \\le i \\le r$ and $a_i = j$. Hence we get $l = x_{r+1} - x_1 \\equiv \\sum_{i=1}^r a_i \\equiv \\ln(n-1)/2 \\pmod{n}$. If $n$ is odd then it follows that $l = n$. If $n$ is even, then we have $n/2$ divides $l$. In this case, if $l$ is even then again $l = n$. Since $k > 1$, it follows that $l < n$ and hence $n \\equiv 2 \\pmod{4}$.",
    "topic": "Number Theory",
    "subtopic": "Modular Arithmetic"
  },
  {
    "id": 32480,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\mathbb{N}$ denote the set of all natural numbers. Define a function $T: \\mathbb{N} \\to \\mathbb{N}$ by $T(2k) = k$ and $T(2k + 1) = 2k + 2$. We write $T^2(n) = T(T(n))$ and in general $T^k(n) = T^{k-1}(T(n))$ for any $k > 1$.\n(i) Show that for each $n \\in \\mathbb{N}$, there exists $k$ such that $T^k(n) = 1$.\n(ii) For $k \\in \\mathbb{N}$, let $c_k$ denote the number of elements in the set $\\{n : T^k(n) = 1\\}$. Prove that $c_{k+2} = c_{k+1} + c_k$, for $k \\ge 1$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(i) For $n = 1$, we have $T(1) = 2$ and $T^2(1) = T(2) = 1$. Hence we may assume that $n > 1$.\nSuppose $n > 1$ is even. Then $T(n) = n/2$. We observe that $(n/2) \\le n - 1$ for $n > 1$.\nSuppose $n > 1$ is odd so that $n \\ge 3$. Then $T(n) = n + 1$ and $T^2(n) = (n + 1)/2$. Again we see that $(n + 1)/2 \\le (n - 1)$ for $n \\ge 3$.\nThus we see that in at most $2(n-1)$ steps $T$ sends $n$ to 1. Hence $k \\le 2(n-1)$. (Here $2(n-1)$ is only a bound. In reality, less number of steps will do.)\n\n(ii) We show that $c_n = f_{n+1}$, where $f_n$ is the $n$-th Fibonacci number.\nLet $n \\in \\mathbb{N}$ and let $k \\in \\mathbb{N}$ be such that $T^k(n) = 1$. Here $n$ can be odd or even. If $n$ is even, it can be either of the form $4d + 2$ or of the form $4d$.\nIf $n$ is odd, then $1 = T^k(n) = T^{k-1}(n+1)$. (Observe that $k > 1$; otherwise we get $n+1 = 1$ which is impossible since $n \\in \\mathbb{N}$.) Here $n+1$ is even.\nIf $n = 4d + 2$, then again $1 = T^k(4d + 2) = T^{k-1}(2d + 1)$. Here $2d + 1 = n/2$ is odd.\nThus each solution of $T^{k-1}(m) = 1$ produces exactly one solution of $T^k(n) = 1$ and $n$ is either odd or of the form $4d + 2$.\nIf $n = 4d$, we see that $1 = T^k(4d) = T^{k-1}(2d) = T^{k-2}(d)$. This shows that each solution of $T^{k-2}(m) = 1$ produces exactly one solution of $T^k(n) = 1$ of the form $4d$.\nThus the number of solutions of $T^k(n) = 1$ is equal to the number of solutions of $T^{k-1}(m) = 1$ and the number of solutions of $T^{k-2}(l) = 1$ for $k > 2$. This shows that $c_k = c_{k-1} + c_{k-2}$ for $k > 2$. We also observe that 2 is the only number which goes to 1 in one step and 4 is the only number which goes to 1 in two steps. Hence $c_1 = 1$ and $c_2 = 2$. This proves that $c_n = f_{n+1}$ for all $n \\in \\mathbb{N}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32481,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $\\alpha, \\beta$ are two positive rational numbers. Assume for some positive integers $m, n$, it is known that $\\alpha^{1/n} + \\beta^{1/m}$ is a rational number. Prove that each of $\\alpha^{1/n}$ and $\\beta^{1/m}$ is a rational number.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $l = \\text{lcm}(m, n)$ and write $n = l/u, m = l/v$. Then\n$$\n\\alpha^{1/n} = (\\alpha^u)^{1/l}, \\quad \\beta^{1/m} = (\\beta^v)^{1/l}.\n$$\nNote that $\\alpha^u$ and $\\beta^v$ are also rationals. Thus it is sufficient to consider the case $n = m$. Let us write $a = \\alpha^{1/n}$, $b = \\beta^{1/m}$ and $c = a + b$. We are given that $c$ is rational.\n\n**Lemma 1:** Let $a, b$ be real numbers such that $a + b = c$ is a rational number. Let $f(x)$ and $g(x)$ be the irreducible polynomials of $a$ and $b$ over $\\mathbb{Q}$. Then $\\deg(f(x)) = \\deg(g(x))$.\n\n**Proof of Lemma 1:** Let $p = \\deg(f(x))$ and $q = \\deg(g(x))$. Then\n$$\nf(x) = x^p + a_{p-1}x^{p-1} + \\dots + a_0, \\quad g(x) = x^q + b_{q-1}x^{q-1} + \\dots + b_0\n$$\nand these are polynomials with rational coefficients. Since $f(a) = 0$, we have $f(c-b) = 0$. Thus $b$ satisfies the polynomial $f(c-x)$ over $\\mathbb{Q}$. Since $g(x)$ is the irreducible polynomial of $b$ over $\\mathbb{Q}$, it follows that $g(x)$ divides $f(c-x)$. Hence $q \\le p$. Similarly we show that $p \\le q$ and we obtain $p = q$.\n\nWe prove the result by induction on $n$. If $n = 1$, then $\\alpha, \\beta$ are rational by the given condition. If $n = 2$, then\n$$\n\\alpha^{1/2} - \\beta^{1/2} = \\frac{\\alpha - \\beta}{\\alpha^{1/2} - \\beta^{1/2}}\n$$\nwhich shows that $\\alpha^{1/2} - \\beta^{1/2}$ is also rational. Combined with the given condition that $\\alpha^{1/2} + \\beta^{1/2}$ is rational, it follows that each of $\\alpha^{1/2}, \\beta^{1/2}$ is rational.\n\nSuppose the result is true for $k = 0, 1, 2, \\dots, n-1$. Let $f(x)$ be the irreducible polynomial of $a = \\alpha^{1/n}$ over $\\mathbb{Q}$ and $g(x)$ be that of $b = \\beta^{1/n}$ over $\\mathbb{Q}$. Then the lemma 1 shows that $\\deg(f(x)) = \\deg(g(x)) = m$, say. But we know that $a$ is a root of $x^n - \\alpha = 0$, $b$ is a root of $x^n - \\beta = 0$ and $b = c - a$. Thus $(c-a)^n = b^n = \\beta$. Hence $(a-c)^n = (-1)^n \\beta$. This shows that $a$ is a root of $(x-c)^n - (-1)^n \\beta = 0$. This is a polynomial with rational coefficients. Thus $a$ is a root of\n$$\nh(x) = (x - c)^n - (-1)^n \\beta - (x^n - \\alpha).\n$$\nNow $\\deg(h(x)) = n-1$, it follows that $\\deg(f(x)) \\le n-1$. Thus we have $m \\le n-1 < n$. Let $\\omega$ be a primitive $n$-th root of unity. Then\n$$\nx^n - \\alpha = \\prod_{j=0}^{n-1} (x - a\\omega^j).\n$$\nSince $f(x)$ is the irreducible polynomial of $a$ over $\\mathbb{Q}$, and $a$ is a root of $x^n - \\alpha$, we see that $f(x)$ divides $x^n - \\alpha$. Hence $x^n - \\alpha = f(x)q(x)$ for some rational polynomial $q(x)$. Now the factors of $f(x)$ are all of the form $(x - a\\omega^j)$. Hence the constant coefficient of $f(x)$ which is a rational number must be of the form $\\pm a^m$, where $m = \\deg(f(x))$. Since $m < n$, we see that $a^m \\in \\mathbb{Q}$ for some $m < n$. Similarly, we see that $b^m$ is also in $\\mathbb{Q}$. Since $m < n$, induction hypothesis shows that $a, b$ are also in $\\mathbb{Q}$. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32482,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $2016$ points of the circumference of a circle are coloured red and the remaining points are coloured blue. Given any natural number $n \\ge 3$, prove that there is a regular $n$-sided polygon all of whose vertices are blue.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $A_1, A_2, \\dots, A_{2016}$ be $2016$ points on the circle which are coloured *red* and the remaining blue. Let $n \\ge 3$ and let $B_1, B_2, \\dots, B_n$ be a regular $n$-sided polygon inscribed in this circle with the vertices chosen in anti-clock-wise direction. We place $B_1$ at $A_1$. (It is possible, in this position, some other $B$'s also coincide with some other $A$'s.) Rotate the polygon in anti-clock-wise direction gradually till some $B$'s coincide with (an equal number of) $A$'s second time. We again rotate the polygon in the same direction till some $B$'s coincide with an equal number of $A$'s third time, and so on until we return to the original position, i.e., $B_1$ at $A_1$. We see that the number of rotations will not be more than $2016 \\times n$, that is, at most these many times some $B$'s would have coincided with an equal number of $A$'s. Since the interval $(0, 360^\\circ)$ has infinitely many points, we can find a value $\\alpha^\\circ \\in (0, 360^\\circ)$ through which the polygon can be rotated from its initial position such that no $B$ coincides with any $A$. This gives a $n$-sided regular polygon having only blue vertices.\n\n\nAlternate Solution:\n\nConsider a regular $2017 \\times n$-gon on the circle; say, $A_1A_2A_3\\cdots A_{2017n}$. For each $j$, $1 \\le j \\le 2017$, consider the points $\\{A_k : k \\equiv j \\pmod{2017}\\}$. These are the vertices of a regular $n$-gon, say $S_j$. We get $2017$ regular $n$-gons; $S_1, S_2, \\dots, S_{2017}$. Since there are only $2016$ red points, by pigeon-hole principle there must be some $n$-gon among these $2017$ which does not contain any red point. But then it is a blue $n$-gon.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32483,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be a natural number. We define sequences $a_i$ and $b_i$ as follows. We let $a_0 = 1$ and $b_0 = n$. For $i > 0$, we let\n$$\n(a_i, b_i) = \\begin{cases} (2a_{i-1} + 1, b_{i-1} - a_{i-1} - 1) & \\text{if } a_{i-1} < b_{i-1}, \\\\ (a_{i-1} - b_{i-1} - 1, 2b_{i-1} + 1) & \\text{if } a_{i-1} > b_{i-1}, \\\\ (a_{i-1}, b_{i-1}) & \\text{if } a_{i-1} = b_{i-1}. \\end{cases}\n$$\nGiven that $a_k = b_k$ for some natural number $k$, prove that $n + 3$ is a power of 2.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We first note that $a_i + b_i = n + 1$ for all natural numbers $i$. We define three sequences $\\{r_i\\}$, $\\{s_i\\}$, $\\{t_i\\}$ as follows. Let $r_0 = 1$, $s_0 = t_0 = 0$. For $i > 0$, we let\n$$\n(r_i, s_i, t_i) = \\begin{cases} (2r_{i-1}, 2s_{i-1}, 2t_{i-1} + 1) & \\text{if } a_{i-1} < b_{i-1}, \\\\ (2r_{i-1} - 1, 2s_{i-1} - 1, 2t_{i-1} - 1) & \\text{if } a_{i-1} > b_{i-1}, \\\\ (r_{i-1}, s_{i-1}, t_{i-1}) & \\text{if } a_{i-1} = b_{i-1}. \\end{cases}\n$$\nIt is then easy to verify that $a_i = r_i a_0 + s_i b_0 + t_i$ and $b_i = (1 - r_i)a_0 + (1 - s_i)b_0 - t_i$ for all $i \\ge 0$.\n\nNow let $k$ be the smallest non-negative integer such that $a_k = b_k$. For any integer $l$ with $0 \\le l \\le k$ we claim that $r_l - s_l = t_l + 1 - 2s_l = 2^l$. We prove this claim by induction on $l$. The claim is straightforward to check for $l = 0$. Now, suppose that the claim is true for some non-negative integer $l < k$. Since $l + 1 \\le k$ it follows that $(r_{l+1}, s_{l+1}, t_{l+1})$ equals either $(2r_l, 2s_l, 2t_l + 1)$ or $(2r_l - 1, 2s_l - 1, 2t_l - 1)$. In either case, $r_{l+1} - s_{l+1} = 2(r_l - s_l)$ and $t_{l+1} + 1 - 2s_{l+1} = 2(t_l + 1 - 2s_l)$. Hence the claim.\n\nNow if $a_k = b_k$ then $r_k a_0 + s_k b_0 + t_k = (1 - r_k)a_0 + (1 - s_k)b_0 - t_k$. This implies $(2r_k - 1) + 2t_k = (1 - 2s_k)n$. Therefore $(n+3)(1 - 2s_k) = 2(r_k - s_k) + 2t_k - 4s_k + 2 = 2^{k+2}$. Hence $n+3$ is a power of 2. In fact, since $1 - 2s_k$ is also a power of 2, it follows that $s_k = 0$ and hence $n + 3 = 2^{k+2}$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32484,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $\\triangle ABC$ be a right-angled triangle with $\\angle B = 90^\\circ$. Let $D$ be a point on $AC$ such that the in-radii of the triangles $ABD$ and $CBD$ are equal. If this common value is $r'$ and if $r$ is the in-radius of triangle $ABC$, prove that\n$$\n\\frac{1}{r'} = \\frac{1}{r} + \\frac{1}{BD}.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $E$ and $F$ be the incentres of triangles $ABD$ and $CBD$ respectively. Let the incircles of triangles $ABD$ and $CBD$ touch $AC$ in $P$ and $Q$ respectively. If $\\angle BDA = \\theta$, we see that\n![](attached_image_1.png)\n$$\nr' = PD \\tan(\\theta/2) = QD \\cot(\\theta/2).\n$$\nHence\n$$\nPQ = PD + QD = r' \\left( \\cot \\frac{\\theta}{2} + \\tan \\frac{\\theta}{2} \\right) = \\frac{2r'}{\\sin \\theta}.\n$$\nBut we observe that\n$$\nDP = \\frac{BD + DA - AB}{2}, \\quad DQ = \\frac{BD + DC - BC}{2}.\n$$\nThus $PQ = (b - c - a + 2BD)/2$. We also have\n$$\n\\begin{aligned}\n\\frac{ac}{2} &= [ABC] = [ABD] + [CBD] = r' \\frac{(AB + BD + DA)}{2} + r' \\frac{(CB + BD + DC)}{2} \\\\\n&= r' \\frac{(c+a+b+2BD)}{2} = r'(s+BD).\n\\end{aligned}\n$$\nBut\n$$\nr' = \\frac{PQ \\sin \\theta}{2} = \\frac{PQ \\cdot h}{2BD},\n$$\nwhere $h$ is the altitude from $B$ on to $AC$. But we know that $h = ac/b$. Thus we get\n$$\nac = 2 \\times r'(s + BD) = 2 \\times \\frac{PQ \\cdot h}{2 \\times BD} (s + BD) = \\frac{(b - c - a + 2BD)ca(s + BD)}{2 \\times BD \\times b}.\n$$\nThus we get\n$$\n2 \\times BD \\times b = 2 \\times (BD - (s - b))(s + BD).\n$$\nThis gives $BD^2 = s(s - b)$. Since $ABC$ is a right-angled triangle $r = s - b$. Thus we get $BD^2 = rs$. On the other hand, we also have $[ABC] = r'(s + BD)$. Thus we get\n$$\nrs = [ABC] = r'(s + BD).\n$$\nHence\n$$\n\\frac{1}{r'} = \\frac{1}{r} + \\frac{BD}{rs} = \\frac{1}{r} + \\frac{1}{BD}.\n$$\n\n\nAlternate Solution 1:\n\nObserve that\n$$\n\\frac{r'}{r} = \\frac{AP}{AX} = \\frac{CQ}{CX} = \\frac{AP + CQ}{AC},\n$$\nwhere $X$ is the point at which the incircle of $ABC$ touches the side $AC$. If $s_1$ and $s_2$ are respectively the semi-perimeters of triangles $ABD$ and $CBD$, we know $AP = s_1 - BD$ and $CQ = s_2 - BD$. Therefore\n$$\n\\frac{r'}{r} = \\frac{(s_1 - BD) + (s_2 - BD)}{b} = \\frac{s_1 + s_2 - 2BD}{b}.\n$$\nBut\n$$\ns_1 + s_2 = \\frac{AD + BD + c}{2} + \\frac{CD + BD + a}{2} = \\frac{(a + b + c) + 2BD}{2} = \\frac{s + BD}{2}.\n$$\nThis gives\n$$\n\\frac{r'}{r} = \\frac{s + BD - 2BD}{b} = \\frac{s - BD}{b}.\n$$\nWe also have\n$$\nr' = \\frac{[ABD]}{s_1} = \\frac{[CBD]}{s_2} = \\frac{[ABD] + [CBD]}{s_1 + s_2} = \\frac{[ABC]}{s + BD} = \\frac{rs}{s + BD}.\n$$\nThis implies that\n$$\n\\frac{r'}{r} = \\frac{s}{s + BD}.\n$$\nComparing the two expressions for $r'/r$, we see that\n$$\n\\frac{s - BD}{b} = \\frac{s}{s + BD}.\n$$\nTherefore $s^2 - BD^2 = bs$, or $BD^2 = s(s-b)$. Thus we get $BD = \\sqrt{s(s-b)}$.\nWe know now that\n$$\n\\frac{r'}{r} = \\frac{s}{s + BD} = \\frac{s - BD}{b} = \\frac{BD}{(s-b) + BD} = \\frac{\\sqrt{s(s-b)}}{(s-b) + \\sqrt{s(s-b)}} = \\frac{\\sqrt{s}}{\\sqrt{s-b} + \\sqrt{s}}.\n$$\nTherefore\n$$\n\\frac{r}{r'} = 1 + \\sqrt{\\frac{s-b}{s}}.\n$$\nThis gives\n$$\n\\frac{1}{r'} = \\frac{1}{r} + \\left( \\sqrt{\\frac{s-b}{s}} \\right) \\frac{1}{r}.\n$$\nBut\n$$\n\\left( \\sqrt{\\frac{s-b}{s}} \\right) \\frac{1}{r} = \\left( \\frac{s-b}{\\sqrt{s(s-b)}} \\right) \\frac{1}{r} = \\left( \\frac{s-b}{BD} \\right) \\frac{1}{r}.\n$$\nIf $\\angle B = 90^\\circ$, we know that $r = s - b$. Therefore we get\n$$\n\\frac{1}{r'} = \\frac{1}{r} + \\left( \\frac{s-b}{BD} \\right) \\frac{1}{r} = \\frac{1}{r} + \\frac{1}{BD}.\n\n\nAlternate Solution 2:\n\nObserve that $\\angle EDF = 90^\\circ$. Hence $\\triangle EDP$ is similar to $\\triangle DFQ$. Therefore $DP \\cdot DQ = EP \\cdot FQ$. Taking $DP = x_2$ and $DQ = x_1$, we get $x_1x_2 = (r')^2$. We also observe that $BD = x_2 + y_2 = x_1 + y_1$. Since $\\angle EBF = 45^\\circ$, we get\n![](attached_image_2.png)\n$$\n1 = \\tan 45^\\circ = \\tan(\\beta_1 + \\beta_2) = \\frac{\\tan \\beta_1 + \\tan \\beta_2}{1 - \\tan \\beta_1 \\tan \\beta_2}.\n$$\nBut $\\tan \\beta_1 = r'/y_2$ and $\\tan \\beta_2 = r'/x_2$. Hence we obtain\n$$\n1 = \\frac{(r'/y_2) + (r'/x_2)}{1 - (r')^2/x_2y_2}.\n$$\nSolving for $r'$, we get\n$$\nr' = \\frac{x_2 y_2 - x_1 y_1}{x_2 + y_2}.\n$$\nWe also know\n$$\nr = \\frac{AB + BC - AC}{2} = \\frac{x_2 + y_2 - (x_1 + y_1)}{2} = \\frac{(x_2 - x_1) + (y_2 - y_1)}{2}.\n$$\nFinally,\n$$\n\\begin{aligned}\n\\frac{1}{r} + \\frac{1}{BD} &= \\frac{2}{(x_2 - x_1) + (y_2 - y_1)} + \\frac{1}{x_1 + x_2} \\\\\n&= \\frac{2x_1 + 2x_2 + (x_2 - x_1) + (y_2 - y_1)}{(x_1 + x_2)((x_2 - x_1) + (y_2 - y_1))}.\n\\end{aligned}\n$$\nBut we can write\n$$\n2x_1 + 2x_2 + (x_2 - x_1) + (y_2 - y_1) = (x_1 + x_2 + x_2 - x_1) + (y_1 + y_2 + y_2 - y_1) = 2(x_2 + y_2),\n$$\nand\n$$\n(x_1 + x_2)((x_2 - x_1) + (y_2 - y_1)) = 2(x_1 + x_2)(x_2 - y_1) \\\\\n= 2(x_2(x_2 + x_1 - y_1) - x_1y_1) = 2(x_2y_2 - x_1y_1).\n$$\nTherefore\n$$\n\\frac{1}{r} + \\frac{1}{BD} = \\frac{2(x_2 + y_2)}{2(x_2y_2 - x_1y_1)} = \\frac{1}{r'}.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32485,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n$ be an odd natural number. We consider an $n$ by $n$ grid which is made up of $n^2$ unit squares and $2n(n+1)$ edges. We colour each of these edges either red or blue. If there are at most $n^2$ red edges, then show that there exists a unit square at least three of whose edges are blue.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Suppose on the contrary that each unit square has at least two red edges. Each red edge is part of at most two unit squares. Therefore\n$$\n2n^2 \\le \\sum_{\\text{unit squares}} \\text{(red edges of the square)} = \\sum_{\\text{red edges}} \\text{(unit squares containing the edge)} \\le 2n^2.\n$$\nThis implies that each unit square has exactly two red edges and that there are a total of $n^2$ red edges.\nWe colour each of the unit squares black and white such that no two unit squares which share a common edge have the same colour (like in a chess board). Note that each red edge is part of exactly one black square and one white square. Since every unit square has exactly two red edges, the number of white squares is therefore $n^2/2$, a contradiction since this is not an integer.\nThis shows that there is a unit square with at most one red edge.",
    "topic": "Discrete Mathematics",
    "subtopic": "Graph Theory"
  },
  {
    "id": 32486,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute triangle with circumcircle $\\Gamma$. Let $A_1$, $B_1$ and $C_1$ be respectively the midpoints of the arcs $BAC$, $CBA$ and $ACB$ of $\\Gamma$. Show that the inradius of triangle $A_1B_1C_1$ is not less than the inradius of triangle $ABC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Rotate the triangle around the centre of $\\Gamma$ by $180^\\circ$. Then $A_1$, $B_1$, $C_1$ respectively go to $A_2$, $B_2$, $C_2$. These are respectively the midpoints of the minor arc $BC$, $CA$ and $AB$. We see that $A_1B_1C_1$ and $A_2B_2C_2$ are congruent triangles and hence have the same in-radii. But we know that $A_2$ is the intersection of the angle bisector of $\\angle BAC$ with $\\Gamma$ and similar descriptions hold for the other two points.\n![](attached_image_1.png)\n---\n\nObserve that $\\angle C_2A_2A = \\angle C_2CA = \\angle C/2$ and $\\angle B_2A_2A = \\angle B_2BA = \\angle B/2$. Hence $\\angle A_2 = (\\angle B/2) + (\\angle C/2) = (\\pi - \\angle A)/2$. Similarly we get other angles:\n$$\n\\angle B_2 = (\\pi - \\angle B)/2, \\quad \\angle C_2 = (\\pi - \\angle C)/2.\n$$\nWe know that for any triangle $ABC$ its inradius is given by\n$$\nr = 4R \\sin(A/2) \\sin(B/2) \\sin(C/2),\n$$\nwhere $R$ is its circumradius. If $r$ and $r_1$ are respectively the inradii of $\\triangle ABC$ and $\\triangle A_2B_2C_2$, then we have\n$$\n\\frac{r}{r_1} = \\frac{\\sin(A/2) \\sin(B/2) \\sin(C/2)}{\\sin((\\pi - A)/4) \\sin((\\pi - B)/4) \\sin((\\pi - C)/4)}.\n$$\nNow we use another known identity in a triangle:\n$$\n1 + 4 \\prod \\sin \\left( \\frac{\\pi - A}{4} \\right) = \\sum \\sin \\frac{A}{2}.\n$$\nThus we obtain\n$$\n\\frac{r}{r_1} = \\frac{4 \\prod \\sin(A/2)}{(\\sum \\sin(A/2)) - 1}.\n$$\nWe have to show that $r \\le r_1$. Equivalently we have to prove that\n$$\n1 + 4 \\prod \\sin(A/2) \\le \\sum \\sin(A/2).\n$$\nBut we know that\n$$\n\\sum \\sin(A/2) \\ge 3 \\left( \\prod \\sin(A/2) \\right)^{1/3}.\n$$\nThus it is sufficient to prove that\n$$\n1 + 4 \\prod \\sin(A/2) \\le 3 \\left( \\prod \\sin(A/2) \\right)^{1/3}.\n$$\nIntroducing $x = \\left( \\prod \\sin(A/2) \\right)^{1/3}$, this inequality is equivalent to\n$$\n4x^3 - 3x + 1 \\ge 0.\n$$\nHowever it is easy to see that $4x^3 - 3x + 1 = (2x - 1)^2(x + 1)$. Hence $4x^3 - 3x + 1 \\ge 0$ for all positive $x$. This completes the proof.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32487,
    "subject": "Mathematics (Multi-modal)",
    "question": "Consider a non-constant arithmetic progression $a_1, a_2, \\dots, a_n, \\dots$. Suppose there exist relatively prime positive integers $p > 1$ and $q > 1$ such that $a_1^2, a_{p+1}^2$ and $a_{q+1}^2$ are also the terms of the same arithmetic progression. Prove that the terms of the arithmetic progression are all integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us take $a_1 = a$. We have\n$$\na^2 = a + kd, \\quad (a + pd)^2 = a + ld, \\quad (a + qd)^2 = a + md.\n$$\nThus we have\n$$\na + ld = (a + pd)^2 = a^2 + 2pad + p^2d^2 = a + kd + 2pad + p^2d^2.\n$$\nSince we have non-constant AP, we see that $d \\neq 0$. Hence we obtain $2pa + p^2d = l - k$. Similarly, we get $2qa + q^2d = m - k$. Observe that $p^2q - pq^2 \\neq 0$. Otherwise $p = q$ and $\\text{gcd}(p, q) = p > 1$ which is a contradiction to the given hypothesis that $\\text{gcd}(p, q) = 1$. Hence we can solve the two equations for $a, d$:\n$$\na = \\frac{p^2(m-k) - q^2(l-k)}{2(p^2q - pq^2)}, \\quad d = \\frac{q(l-k) - p(m-k)}{p^2q - pq^2}.\n$$\nIt follows that $a, d$ are rational numbers. We also have\n$$\np^2a^2 = p^2a + kp^2d.\n$$\nBut $p^2d = l - k - 2pa$. Thus we get\n$$\np^2a^2 = p^2a + k(l - k - 2pa) = (p - 2k)pa + k(l - k).\n$$\nThis shows that $pa$ satisfies the equation\n$$\nx^2 - (p - 2k)x - k(l - k) = 0.\n$$\nSince $a$ is rational, we see that $pa$ is rational. Write $pa = w/z$, where $w$ is an integer and $z$ is a natural number such that $\\text{gcd}(w, z) = 1$. Substituting in the equation, we obtain\n$$\nw^2 - (p - 2k)wz - k(l - k)z^2 = 0.\n$$\nThis shows $z$ divides $w$. Since $\\text{gcd}(w, z) = 1$, it follows that $z = 1$ and $pa = w$ an integer. (In fact any rational solution of a monic polynomial with integer coefficients is necessarily an integer.) Similarly, we can prove that $qa$ is an integer. Since $\\text{gcd}(p, q) = 1$, there are integers $u$ and $v$ such that $pu + qv = 1$. Therefore $a = (pa)u + (qa)v$. It follows that $a$ is an integer. But $p^2d = l - k - 2pa$. Hence $p^2d$ is an integer. Similarly, $q^2d$ is also an integer. Since $\\text{gcd}(p^2, q^2) = 1$, it follows that $d$ is an integer. Combining these two, we see that all the terms of the AP are integers.\n\n**Alternatively**, we can prove that $a$ and $d$ are integers in another way. We have seen that $a$ and $d$ are rationals; and we have three relations:\n$$\na^2 = a + kd, \\quad p^2d + 2pa = n_1, \\quad q^2d + 2qa = n_2,\n$$\nwhere $n_1 = l - k$ and $n_2 = m - k$. Let $a = u/v$ and $d = x/y$ where $u, x$ are integers and $v, y$ are natural numbers, and $\\text{gcd}(u, v) = 1$, $\\text{gcd}(x, y) = 1$. Putting this in these relations, we obtain\n$$\nu^2 y = uvy + kxv^2, \\qquad (1)\n$$\n$$\n2puy + p^2vx = vyn_1, \\qquad (2)\n$$\n$$\n2quy + q^2vx = vyn_2. \\qquad (3)\n$$\nNow (1) shows that $v|u^2y$. Since $\\text{gcd}(u, v) = 1$, it follows that $v|y$. Similarly (2) shows that $y|p^2vx$. Using $\\text{gcd}(y, x) = 1$, we get that $y|p^2v$. Similarly, (3) shows that $y|q^2v$. Therefore $y$ divides $\\text{gcd}(p^2v, q^2v) = v$. The two results $v|y$ and $y|v$ imply $v = y$, since both $v, y$ are positive. Substitute this in (1) to get\n$$\nu^2 = uv + kxv.\n$$\nThis shows that $v|u^2$. Since $\\text{gcd}(u, v) = 1$, it follows that $v = 1$. This gives $v = y = 1$. Finally $a = u$ and $d = x$ which are integers.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32488,
    "subject": "Mathematics (Multi-modal)",
    "question": "Find all functions $f : \\mathbf{R} \\to \\mathbf{R}$ such that\n$$\nf(x^3 + f(y)) = x^2 f(x) + y,\n$$\nfor all $x, y \\in \\mathbf{R}$.",
    "options": [],
    "answer": "f(x) = x or f(x) = -x",
    "solution": "Let $f(0) = \\lambda$. Put $x = y = 0$ in (1) to get $f(\\lambda) = 0$. Taking $y = \\lambda$, we get $f(x^3) = x^2 f(x) + \\lambda$. Put $x = \\lambda, y = 0$ in (1) and we get\n$$\nf(\\lambda^3 + \\lambda) = \\lambda^2 f(\\lambda). \\qquad (2)\n$$\nBut $x = \\lambda, y = \\lambda$ gives $f(\\lambda^3) = \\lambda$. Now put $x = \\lambda, y = \\lambda^3$ and we get\n$$\nf(\\lambda^3 + \\lambda) = \\lambda^2 f(\\lambda) + \\lambda^3. \\qquad (3)\n$$\nComparing (2) and (3), we get $\\lambda = 0$. Therefore\n$$\nf(x^3) = x^2 f(x), \\quad f(f(x)) = x,\n$$\nfor all $x$. Now changing $y$ to $f(y)$, we get\n$$\nf(x^3 + y) = x^2 f(x) + f(y) = f(x^3) + f(y)\n$$\nfor all $x, y$. Since $x \\to x^3$ is a bijection of $\\mathbf{R}$ onto $\\mathbf{R}$, we obtain\n$$\nf(x + y) = f(x) + f(y), \\quad f(f(x)) = x\n$$\nfor all $x, y$.\nConsider $f(x^3) = x^2f(x)$. Replacing $x$ by $x+1$, we obtain\n$$\nf(x^3 + 3x^2 + 3x + 1) = (x^2 + 2x + 1)f(x + 1).\n$$\nUsing the additivity of $f$, we get\n$$\n\\begin{aligned}\nf(x^3) + 3f(x^2) + 3f(x) + f(1) &= (x^2 + 2x + 1)(f(x) + f(1)) \\\\\n&= x^2 f(x) + 2x f(x) + f(x) + x^2 f(1) + 2x f(1) + f(1).\n\\end{aligned}\n$$\nThis reduces to\n$$\n3f(x^2) + 2f(x) = cx^2 + 2cx + 2xf(x) \\quad (4)\n$$\nfor all $x \\in \\mathbb{R}$. Replace $x$ by $(x+1)$ and expand using additivity, we get\n$$\n3f(x^2) + 6f(x) = cx^2 + 6cx + 2xf(x) \\quad (5).\n$$\nComparing (4) and (5), we see that $f(x) = cx$. Replacing $x$ by $f(x)$, we get $x = cf(x)$. Together we get $c^2x = x$. Hence $c^2 = \\pm 1$. Therefore $f(x) = \\pm x$.",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32489,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $a$, $b$, $c$ be distinct positive real numbers such that $abc = 1$. Prove that\n$$\n\\sum_{\\text{cyclic}} \\frac{a^6}{(a-b)(a-c)} > 15.\n$$",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let us consider a cubic polynomial whose roots are $a$, $b$, $c$. We get $P(x) = x^3 - px^2 + qx - r$, where $p = a + b + c$, $q = ab + bc + ca$ and $r = abc$. We observe that\n$$\n\\frac{a^n}{(a-b)(a-c)} + \\frac{b^n}{(b-c)(b-a)} + \\frac{c^n}{(c-a)(c-b)} = \\sum_{\\text{cyclic}} \\frac{-a^n(b-c)}{(a-b)(b-c)(c-a)}\n$$\nLet us write\n$$\nS_n = \\sum_{\\text{cyclic}} \\frac{a^n (b-c)}{-(a-b)(b-c)(c-a)}\n$$\nIt is easy to see that $S_1 = 0$ and $S_2 = 1$. Since $a$, $b$, $c$ are the roots of $P(x) = 0$, we have $a^3 - pa^2 + qa - r = 0$, $b^3 - pb^2 + qb - r = 0$, $c^3 - pc^2 + qc - r = 0$. Multiply the first by $b-c$, the second by $c-a$ and the third by $a-b$, and adding all these and dividing the sum by $-(a-b)(b-c)(c-a)$, we obtain $S_3 - pS_2 + qS_1 = 0$. Hence $S_3 = p$. Now multiply the first by $a$, the second by $b$ and the third by $c$ and divide throughout by $-(a-b)(b-c)(c-a)$ to get $S_4 - pS_3 + qS_2 - rS_1 = 0$. Hence $S_4 = p^2 - q$. Similarly, $S_5 = p(p^2 - q) - qp + r = p^3 - 2pq + r$. We also get $S_6 - pS_5 + qS_4 - rS_3 = 0$. This gives\n$$\nS_6 = p(p^3 - 2pq + r) - q(p^2 - q) + rp = p^4 - 3p^2q + 2pr + q^2.\n$$\nWe can write it as $S_6 = p^2(p^2 - 3q) + 2pr + q^2$. But $p^2 - 3q = (a + b + c)^2 - 3(ab + bc + ca) = a^2 + b^2 + c^2 - ab - bc - ca > 0$. Hence\n$$\nS_6 > 2pr + q^2 = 2abc(a + b + c) + (ab + bc + ca)^2 \\geq 6 + 9 = 15.\n$$",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32490,
    "subject": "Mathematics (Multi-modal)",
    "question": "Suppose $n \\ge 0$ is an integer and all the roots of $x^3 + \\alpha x + 4 - (2 \\times 2016^n) = 0$ are integers. Find all possible values of $\\alpha$.",
    "options": [],
    "answer": "-3",
    "solution": "",
    "topic": "Algebra",
    "subtopic": "Intermediate Algebra"
  },
  {
    "id": 32491,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nIn the given figure, $ABCD$ is a square paper. It is folded along $EF$ such that $A$ goes to a point $A'$ ($A' \\neq C, B$) on the side $BC$ and $D$ goes to $D'$. The line $A'D'$ cuts $CD$ in $G$. Show that the inradius of the triangle $GCA'$ is the sum of the inradii of the triangles $GD'F$ and $A'BE$.\n\n![](attached_image_1.png)",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Solution:\n\nObserve that the triangles $GCA'$ and $A'BE$ are similar to the triangle $GD'F$. If $GF = u$, $GD' = v$ and $D'F = w$, then we have\n$$\nA'G = p u, \\quad CG = p v, \\quad A'C = p w, \\quad A'E = q u, \\quad BE = q w, \\quad A'B = q v\n$$\nIf $r$ is the inradius of $\\triangle GD'F$, then $p r$ and $q r$ are respectively the inradii of triangles $GCA'$ and $A'BE$. We have to show that $p r = r + q r$. We also observe that\n$$\nAE = EA', \\quad DF = FD'\n$$\nTherefore\n$$\np w + q v = q w + q u = w + u + p v = v + p u\n$$\nThe last two equalities give $(p-1)(u-v) = w$. The first two equalities give $(p-q)w = q(u-v)$. Hence\n$$\n\\frac{p-q}{q} = \\frac{u-v}{w} = \\frac{1}{p-1}\n$$\nThis simplifies to $p(p-q-1) = 0$. Since $p \\neq 0$, we get $p = q + 1$. This implies that $p r = q r + r$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32492,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $m$ and $n$ be two integers and define $a_0 = m$, $a_1 = n$ and $a_{k+1} = 4a_k - 5a_{k-1}$ for $k \\ge 1$. If $p > 5$ is a prime such that $p-1$ is divisible by $4$, then show that there are integers $m$ and $n$ such that $p$ does not divide $a_k$ for any $k \\ge 0$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Let $t$ be an integer such that $p$ divides $t^2 + 1$. There exists such an integer since $p-1$ is divisible by $4$. Let $m = 1$ and $n = t + 2$. Then\n$$\nn^2 = t^2 + 4t + 4 \\equiv 4t + 3 \\equiv 4n - 5m \\pmod{p}.\n$$\nTherefore, if $a_0 = 1$ and $a_1 = n$ then $a_2 \\equiv n^2 \\pmod{p}$. By induction, it is easy to see that $a_k \\equiv n_k \\pmod{p}$. Since $p > 5$, it follows that $p$ does not divide $n$. Therefore $p$ does not divide $a_k$ for any $k \\ge 0$.",
    "topic": "Number Theory",
    "subtopic": "Divisibility / Factorization"
  },
  {
    "id": 32493,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABCDE$ be a convex pentagon in which $\\angle A = \\angle B = \\angle C = \\angle D = 120^\\circ$ and side lengths are five consecutive integers in some order. Find all possible values of $AB + BC + CD$.",
    "options": [],
    "answer": "6, 7, 9",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32494,
    "subject": "Mathematics (Multi-modal)",
    "question": "Define an $n$-magic square to mean an $n \\times n$ square matrix of non-negative integers such that the sum of all the entries in each row and each column is $m$ for some $m \\in \\mathbb{N}$. Also define an $n$-permutation matrix to mean an $n \\times n$ square matrix of $n(n-1)$ zeroes and $n$ ones such that every row and every column of the array contains exactly one 1. Show that every $n$-magic square can be written as a sum of finitely many $n$-permutation matrices.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Fix an $n \\in \\mathbb{N}$. We will prove the following statement by Induction:\n$P(m)$ : Every $n$-magic square with common sum $m$ can be written as a sum of $m$ permutation matrices.\nFor the base case, we verify that $P(1)$ is true. Since all the entries are non-negative integers, the only way possible $n$-magic squares with common sum 1 are exactly $n$-permutation matrices. Hence, every such matrix can be written as a sum of $1$ $n$-permutation matrix, i.e., itself.\nFor the Induction Hypothesis, assume that the statement $P(k)$ is true for some $k \\in \\mathbb{N}$. Hence, every $n$-magic square with common sum $k$ can be written as a sum of $k$ $n$-permutation matrices. Consider any $n$-magic square $A_0$ with common sum $k+1$. We will prove that $A_0$ can be written as a sum of an $n$-permutation matrix and an $n$-magic square $A_1$ with common sum $k$.\nConsider the graph $\\mathcal{G} = (\\mathcal{W}, \\mathcal{E})$ defined as follows. Let $R_1, R_2, \\dots, R_n$ be the rows of $A_0$ and let $C_1, C_2, \\dots, C_n$ be the columns of $A_0$. Consider $\\mathcal{W} = \\{R_1, R_2, \\dots, R_n\\} \\cup \\{C_1, C_2, \\dots, C_n\\}$. Each of the sets $R_i = \\{R_1, R_2, \\dots, R_n\\}$ and $C_i = \\{C_1, C_2, \\dots, C_n\\}$ are independent sets. For every $i, j \\in \\mathbb{N}_n$, $R_i, C_j$ are connected by an edge if and only if $a_{ij} > 0$, where $a_{ij}$ is the entry of $A_0$ in the $i$-th row and $j$-th column. Observe that from our definition, $\\mathcal{G}$ is a bipartite graph.\nConsider some $S \\subseteq \\mathbb{R}$. By $N(S)$, we will denote the set of vertices which are adjacent to some vertex in $S$. We will show that $|N(S)| \\ge |S|$.\nFor every edge having an endpoint in $S$, label that edge with $a_{ij}$. Consider the 'sum' of all such edges. Clearly, for each $R_i \\in S$, the edges incident with $R_i$ contribute $m$ to this total. Hence, the total sum is given by $m|S|$. Now these are edges are a subset of all edges with an endpoint in $N(S)$. Hence, if we consider a similar such sum for the set $N(S)$, we get the sum to be $m|N(S)|$. As observed before, we see that $m|S| \\le m|N(S)|$. Hence, we get that $|N(S)| \\ge |S|$. Since $S$ was any arbitrary subset of $\\mathbb{R}$, we that this is true for all subsets of $\\mathbb{R}$.\nSince the hypothesis for Hall's Matching Theorem is satisfied, an application of it implies that there exists a matching in the above bipartite graph $\\mathcal{G}$. Since $|\\mathbb{R}| = |\\mathbb{C}| = n$, we see that the matching is nothing but a bijective function $\\sigma: \\mathbb{N}_n \\to \\mathbb{N}_n$ where $\\sigma(i)$ is defined to be the index of the element in $\\mathbb{C}$ which is matched with $R_i$. Note that $\\sigma$ can also be thought of as a permutation of $\\mathbb{N}_n$. We also know that $R_i \\cap C_j = \\{a_{ij}\\}, \\forall i, j \\in \\mathbb{N}_n$. Since $\\sigma$ is a matching, we know that $a_{i\\sigma(i)} > 0$ for all $i \\in \\mathbb{N}_n$. Consider the permutation matrix $P_\\sigma$ where each entry $p_{ij}$ is given by\n$$\np_{ij} = \\begin{cases} 1 & \\text{if } j = \\sigma(i) \\\\ 0 & \\text{if otherwise.} \\end{cases} \\quad \\forall i, j \\in \\mathbb{N}_n\n$$\nAlso consider the matrix $A_1 = A_0 - P_\\sigma$. Note that since $a_{i\\sigma(i)} > 0, \\forall i \\in \\mathbb{N}_n$, every entry of $A_1$ is non-negative. Also observe that since $P_\\sigma$ is a permutation matrix, the sum of entries in each row and each column reduces by 1. In effect, $A_1$ is still an $n$-magic square with common sum $k-1$. This completes the inductive step. Hence by Induction, we see that the statement $P(m)$ is true for all $m \\in \\mathbb{N}$. The proof is now complete.",
    "topic": "Discrete Mathematics",
    "subtopic": "Combinatorics"
  },
  {
    "id": 32495,
    "subject": "Mathematics (Multi-modal)",
    "question": "A polynomial $p(x)$ with real coefficients is called a square if and only if it is not a constant and there exists a polynomial $q(x)$ with real coefficients such that $p(x) = q(x)^2$. Suppose that $f(x)$ and $g(x)$ are non-constant polynomials with real coefficients such that neither of them is a square, but $f(g(x))$ is. Show that $g(f(x))$ is not a square.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "We can easily extend the definition of a square polynomial to polynomials with complex coefficients. In all the arguments below we consider polynomials with complex coefficients.\n\n**Lemma:** If $p(x)$ is a square and $\\alpha$ is a non-zero complex number then $p(x) - \\alpha$ is not a square.\n\n**Proof of Lemma:** Suppose $p(x) = q(x)^2$ and $p(x) - \\alpha = r(x)^2$, with both $q(x)$ and $r(x)$ being non-constant polynomials. Then $\\alpha = (q(x) - r(x))(q(x) + r(x))$. Clearly, either $q(x) - r(x)$ or $q(x) + r(x)$ is not a constant polynomial, and hence a contradiction. This proves the lemma.\n\n**Continuation of the solution:** We can write $f(x)$ as $f_1(x)^2(x - \\alpha_1)(x - \\alpha_2) \\cdots (x - \\alpha_k)$, where $f_1(x)$ is a polynomial and $\\alpha_1, \\alpha_2, \\cdots, \\alpha_k$ are distinct complex numbers. Then $f(g(\\tau)) = f_1(g(\\tau))^2(g(\\tau) - \\alpha_1) \\cdots (g(\\tau) - \\alpha_k)$ is a square. It follows that $(g(x) - \\alpha_1)(g(x) - \\alpha_2) \\cdots (g(x) - \\alpha_k) = h(x)^2$ for some polynomial $h(x)$. Let $\\beta$ be such that $g(\\beta) = \\alpha_1$. Then $h(\\beta) = 0$ and hence $h(x) = (x - \\beta)h_1(x)$. Note that $g(\\beta) - \\alpha_i = 0$ for any $i = 1$. Therefore it follows that $(x - \\beta)^2$ divides $g(x) - \\alpha_1$. We can then conclude that $g(x) - \\alpha_1$ is a square. Similarly, $g(x) - \\alpha_i$ is a square for $i = 1, 2, \\cdots, k$. By the above lemma, it follows that $k = 1$, so $f(x) = f_1(x)^2(x - \\alpha)$ and $g(x) = g_1(x)^2 + \\alpha$ for some non-zero complex number $\\alpha$. Therefore $g(f(x)) = g_1(f(x))^2 + \\alpha$ and hence by the above lemma it follows that $g(f(x))$ is not a square. This completes the proof.",
    "topic": "Algebra",
    "subtopic": "Algebraic Expressions"
  },
  {
    "id": 32496,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be a triangle with $\\angle A = 90^{\\circ}$ and $AB < AC$. Let $AD$ be the altitude from $A$ to $BC$. Let $P$, $Q$ and $I$ denote respectively the incentres of triangles $ABD$, $ACD$ and $ABC$. Prove that $AI$ is perpendicular to $PQ$ and $AI = PQ$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32497,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $n \\ge 1$ be an integer and consider the sum\n$$\nx = \\sum_{k \\ge 0} \\binom{n}{2k} 2^{n-2k} 3^k = \\binom{n}{0} 2^n + \\binom{n}{2} 2^{n-2} \\cdot 3 + \\binom{n}{4} 2^{n-4} \\cdot 3^2 + \\dots\n$$\nShow that $2x - 1$, $2x$, $2x + 1$ form the sides of a triangle whose area and inradius are also integers.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32498,
    "subject": "Mathematics (Multi-modal)",
    "question": "Let $ABC$ be an acute-angled triangle with incentre $I$. Draw a line perpendicular to $BI$ at $I$ and let it intersect $BC$ and $BA$ at $D$ and $E$ respectively. Let $P$ and $Q$ be respectively the incentres of the triangles $BIA$ and $BIC$. Suppose the four points $D, E, P, Q$ are concyclic. Prove that $BA = BC$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "Join $DE$ and $PQ$. Let $S$ be the centre of the circle $\\Gamma$ passing through $E, P, Q, D$. Let $P'$ denote the reflection of $P$ in $BI$. Since $\\angle PBI = \\angle QBI$, it follows that $P'$ lies on $BQ$. Since $E$ and $D$ are symmetric about the line $EI$, and $P$ and $P'$ are also symmetric about $BI$, it follows that $P'$ lies on $\\Gamma$. Since $Q$ lies on $\\Gamma$ and $P'$ lies on the line $BQ$, we conclude that $Q = P'$. This means $BI$ is the perpendicular bisector of $PQ$. Since $BI$ is already the perpendicular bisector of $ED$, it follows that $ED$ is parallel to $PQ$.\nConsider the circumcircle of the triangle $AIC$. We have\n![](attached_image_1.png)\n$$\n\\angle BIC = 180^\\{\\circ\\} - (B/2 + C/2) = 90^\\{\\circ\\} + (A/2).\n$$\nSince $\\angle BID = 90^\\{\\circ\\}$, we see that $\\angle DIC = A/2 = \\angle IAC$. Hence $ED$ is tangent to the circumcircle of $\\triangle AIC$ at $I$. Hence the circumcentre of the triangle $AIC$ lies on the line $BS$. This implies that $BS$ is the perpendicular bisector of $AC$ as well. It follows that $ED \\parallel AC$. Thus $PQ \\parallel ED \\parallel AB$.\nSince $BS$ is the perpendicular bisector of $PQ$ and that of $AC$, we see that $PQCA$ is an isosceles trapezium. Hence $P, Q, C, A$ are concyclic. This gives $\\angle PAC = \\angle QCA$. Therefore $\\angle A = \\angle C$ and hence $BA = BC$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  },
  {
    "id": 32499,
    "subject": "Mathematics (Multi-modal)",
    "question": "For each positive integer $n \\ge 2$, define the polynomial $f_n(x)$ by $f_n(x) = x^n - x^{n-1} - x^{n-2} - \\cdots - x - 1$. Prove that\n(a) for each positive integer $n \\ge 2$, the equation $f_n(x) = 0$ has a unique real positive root, say, $\\alpha_n$;\n(b) $(\\alpha_n)_{n \\ge 2}$ is a strictly increasing sequence; and\n(c) $\\lim_{n \\to \\infty} \\alpha_n = 2$.",
    "options": [],
    "answer": "Detailed solution",
    "solution": "(a) Since $f_n(x)$ has one change of sign, it follows that $f_n(x) = 0$ has at most one positive real root by Descartes' Rule. As $f_n(0) = -1 < 0$ and $f_n(2) = 2^n - 2^{n-1} - 2^{n-2} - \\cdots - 2 - 1 = 1 > 0$, we see that $f_n(x) = 0$ has at least one root between $0$ and $2$. Thus $f_n(x) = 0$ has exactly one root $\\alpha_n$ between $0$ and $2$, $n \\ge 2$.\n\n(b) We show that $\\alpha_n < \\alpha_{n+1}$, for $n \\ge 2$. Because $f_n(\\alpha_n) = 0$, we have $\\alpha_n^n - \\alpha_n^{n-1} - \\alpha_n^{n-2} - \\cdots - \\alpha_n - 1 = 0$. Now $f_{n+1}(\\alpha_n) = \\alpha_n(\\alpha_n^n - \\alpha_n^{n-1} - \\alpha_n^{n-2} - \\cdots - \\alpha_n - 1) - 1 = \\alpha_n \\cdot 0 - 1 = -1 < 0$. As $f_{n+1}(2) = 1 > 0$, as above, we infer that there is a root of $f_{n+1}(x) = 0$ between $\\alpha_n$ and $2$. Since $\\alpha_n > 0$, we conclude that $\\alpha_n < \\alpha_{n+1} < 2$, as desired.\n\n(c) Again\n$$\n(x-1)f_n(x) = (x-1)(x^n - x^{n-1} - x^{n-2} - \\cdots - x - 1) = x^{n+1} - 2x^n + 1.\n$$\nSo\n$$\n\\alpha_n^{n+1} - 2\\alpha_n^n + 1 = 0,\n$$\ngiving $\\alpha_n = 2 - \\frac{1}{\\alpha_n^n} \\ge 2 - \\frac{1}{\\alpha_2^n} = \\frac{1+\\sqrt{5}}{2}$, where $\\alpha_2 = \\frac{1+\\sqrt{5}}{2}$, being the positive root of $x^2 - x - 1 = 0$. But $\\alpha_2 > 1$. So $\\frac{1}{\\alpha_2^n} \\to 0$, as $n \\to \\infty$. Thus, in $2 - \\frac{1}{\\alpha_2^n} \\le \\alpha_n < 2$, we let $n \\to \\infty$, to get\n$$\n\\lim_{n \\to \\infty} \\alpha_n = 2.\n$$",
    "topic": "Algebra",
    "subtopic": "Equations and Inequalities"
  },
  {
    "id": 32500,
    "subject": "Mathematics (Multi-modal)",
    "question": "Problem:\n\nLet $ABC$ be a triangle with $\\angle BAC > 90^\\circ$. Let $D$ be a point on the segment $BC$ and $E$ be a point on the line $AD$ such that $AB$ is tangent to the circumcircle of triangle $ACD$ at $A$ and $BE$ is perpendicular to $AD$. Given that $CA = CD$ and $AE = CE$, determine $\\angle BCA$ in degrees.",
    "options": [],
    "answer": "45",
    "solution": "Solution:\n\nLet $\\angle C = 2\\alpha$. Then $\\angle CAD = \\angle CDA = 90^\\circ - \\alpha$. Moreover, $\\angle BAD = 2\\alpha$ as $AB$ is tangent to the circumcircle of $\\triangle CAD$. Since $AE = AD$, it gives $\\angle AEC = 2\\alpha$. Thus $\\triangle AEC$ is similar to $\\triangle ACD$. Hence\n$$\n\\frac{AE}{AC} = \\frac{AC}{AD}\n$$\nBut the condition that $BE \\perp AD$ gives $AE = AB \\cos 2\\alpha = c \\cos 2\\alpha$. It is easy to see that $\\angle B = 90^\\circ - 3\\alpha$. Using sine rule in triangle $ADC$, we get\n$$\n\\frac{AD}{\\sin 2\\alpha} = \\frac{AC}{\\sin (90^\\circ - \\alpha)}\n$$\nThis gives $AD = 2b \\sin \\alpha$. Thus we get\n$$\nb^2 = AC^2 = AE \\cdot AD = (c \\cos 2\\alpha) \\cdot 2b \\sin \\alpha\n$$\nUsing $b = 2R \\sin B$ and $c = 2R \\sin C$, this leads to\n$$\n\\cos 3\\alpha = 2 \\sin 2\\alpha \\cos 2\\alpha \\sin \\alpha = \\sin 4\\alpha \\sin \\alpha\n$$\nWriting $\\cos 3\\alpha = \\cos (4\\alpha - \\alpha)$ and expanding, we get $\\cos 4\\alpha \\cos \\alpha = 0$. Therefore $\\alpha = 90^\\circ$ or $4\\alpha = 90^\\circ$. But $\\alpha = 90^\\circ$ is not possible as $\\angle C = 2\\alpha$. Therefore $4\\alpha = 90^\\circ$ which gives $\\angle C = 2\\alpha = 45^\\circ$.",
    "topic": "Geometry",
    "subtopic": "Plane Geometry"
  }
]